Student Workbook with Solutions Applied Statistics and Probability for Engineers, 3rd Edition

Student Workbook with Solutions Heecheon You Pohang University of Science and Technology Applied Statistics and Probabi...

Author: Douglas C. Montgomery

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Student Workbook with Solutions Heecheon You Pohang University of Science and Technology

Applied Statistics and Probability lor Engineers Third Edition

Douglas C. Montgomery Arizona State University

George C. Runger Arizona State University

JOHN WILEY & SONS, INC.

Cover Image: Norm Christiansen

T o order books or for customer service call 1-800-CALL-WILEY (225-5945).

Copyright O 2003 by John Wiley & Sons, Inc. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-601 1, fax (201) 748-6008, E-Mail: [email protected]. ISBN 0-47 1-42682-2 Printed in the United States of America.

Printed and bound by Courier Kendallville, Inc.

This book is dedicated to: Jiyeon, Sophia, Jaechang, Sookhyun, Young, and Isik

Preface Objectives of the Book As a supplement of Applied Statistics and Probability for Engineers (3rdedition) authored by Drs. Douglas C. Montgomery and George C. Runger (abbreviated by MR hereinafter), this book is developed to serve as an effective, learner-friendly guide for students who wish to establish a solid understanding of probability and statistics. First, this study guide outlines learning goals and presents detailed information accordingly; this goal-oriented presentation is intended to help the reader see the destination before learning and check histher achievement after learning. Second, this book helps the reader build both scientific knowledge and computer application skill in probability and statistics. Third, this book provides concise explanations, step-by-step algorithms, comparison tables, graphic charts, daily-life examples, and real-world engineering problems, which are designed to help the reader to efficiently learn concepts and applications of probability and statistics. Lastly, this book covers essential concepts and techniques of probability and statistics; thus, the reader who has interest in the advanced topics indicated as CD only in MR should refer to the MR CD.

Features of the Book The reader will enjoy the following learner-friendly features of the book: Learning Goals: Each section starts with a list of learning goals which states topics covered and competency levels targeted. Organized, Concise Explanation: According to the learning goals defined, topics are presented with concise, clear explanations. Learning Aids: Tables, charts, and algorithms are provided wherever applicable to help the reader form integrated mental pictures by contrasting or visualizing concepts and techniques. Easy-to Understand Examples: For topics requiring problem-solving practices, daily-life examples (which are readily understandable) are used to help the reader better see the application process of statistical techniques. Engineering-Context Exercises: Together with daily-life examples, engineering problems (selected from exercises in the MR textbook; the original exercise numbers are indicated) are used to have the reader explore the application of statistical techniques to the real-world engineering domain. Complete Solutions of Exercises: At the end of each chapter, complete solutions of exercises are provided so that the reader checks hidher own answers with ease. Software Demonstration: To help students learn use of Excel and Minitab (student version) in solving probability and statistics problems, respectively, step-by-step directions and screen snapshots are provided for examples as applicable.

Organization of the Book This book presents various topics of probability and statistics with the structure displayed in the figure on next page. Chapters 2 through 5 describe probability and probability distributions. After explaining fundamental concepts, rules, and theorems of

vi

Preface probability, this book presents various discrete and continuous probability distributions of single and multiple random variables. Then, chapters 6 through 15 explain descriptive and inferential statistics. Descriptive statistics deals with measures and charts to describe collected data, whereas inferential statistics covers categorical data analysis, statistical inference on singleltwo samples, regression analysis, design of experiment, and nonparametric analysis.

Multiple Discrete Probability Distributions

1

Multiple Continuous J Probability Distributions

Ch.5

Ch. 11 Ch. 12 Ch. 13 Ch. 14

Acknowledgements I would like to express my considerable gratitude to Jenny Welter for her sincere support, trust, and encouragement in developing this book. I also wish to thank Drs. Douglas C. Montgomery and George C. Runger at Arizona State University for their pleasant partnership and helpful advice in producing this book. Finally, I greatly appreciate the insightful feedback and useful suggestions from the professors and students who reviewed the various parts of the manuscript: Dr. Yasser M. Dessouky at San Jose State University; Major Brian M. Layton at U.S. Military Academy; Dr. Arunkumar Pennathur at University of Texas at El Paso; Dr. Manuel D. Rossetti at University of Arkansas; Dr. Bruce Schmeiser at Purdue University; Dr. Lora Zimmer at Arizona State University; Christi Barb, Shea Kupper, Maria Fernanda Funez-Mejia, Tram Huynh, Usman Javaid, Thuan Mai, Prabaharan Veluswamy, Cheewei Yip, and Ronda Young at Wichita State University. Without their invaluable contributions, this book may not have been completed. Peace be with them all. Heecheon You

Contents Chapter 1

The Role of Statistics in Engineering 1- 1 1-2 1-3 1-4

Chapter 2

The Engineering Method and Statistical Thinking Collecting Engineering Data 1-2 Mechanistic and Empirical Models 1-3 Probability and Probability Models 1-3

Probability

1- 1

2-1

2-1 Sample Spaces and Events 2-1 2-2 Interpretations of Probability 2-4 2-3 Addition Rules 2-5 2-4 Conditional Probability 2-6 2-5 Multiplication and Total Probability Rules 2-6 Independence 2-8 2-7 Bayes' Theorem 2-9 2-8 Random Variables 2- 10 Answers to Exercises 2-1 1

Chapter 3

1-1

2-6

Discrete Random Variables and Probability Distributions 3-1 3- 1 Discrete Random Variables 3- 1 3-2 Probability Distributions and Probability Mass Functions 3-1 3-3 Cumulative Distribution Functions 3-3 3-4 Mean and Variance of a Discrete Random Variable 3-4 3-5 Discrete Uniform Distribution 3-5 3-6 Binomial Distribution 3-6 3-7 Geometric and Negative Binomial Distributions 3-8 3-8 Hypergeometric Distribution 3-13 3-9 Poisson Distribution 3- 16 Summary of Discrete Probability Distributions 3-20 EXCELApplications 3-2 1 Answers to Exercises 3-23

vii

viii

Contents

Chapter 4

Continuous Random Variables and Probability Distributions 4-1 4- 1 4-2 4-3 4-4 4-5 4-6 4-7

Continuous Random Variables 4- 1 Probability Distributions and Probability Density Functions 4-1 Cumulative Distribution Functions 4-3 Mean and Variance of a Continuous Random Variable 4-4 Continuous Uniform Distribution 4-5 Normal Distribution 4-6 Normal Approximation to the Binomial and Poisson Distributions 4-8 4-9 Exponential Distribution 4-1 1 4- 10 Erlang and Gamma Distributions 4- 14 4-1 1 Weibull Distribution 4-1 8 4- 12 Lognormal Distribution 4-20 Summary of Continuous Probability Distributions 4-21 EXCELApplications 4-22 Answers to Exercises 4-25

Chapter 5

Joint Probability Distributions

5-1

5-1 Two Discrete Random Variables 5-1 5-2 Multiple Discrete Random Variables 5-5 5-3 Two Continuous Random Variables 5-7 5-4 Multiple Continuous Random Variables 5- 12 5-5 Covariance and Correlation 5- 13 5-6 Bivariate Normal Distribution 5-15 5-7 Linear Combinations of Random Variables 5- 16 5-9 Moment Generating Functions 5-18 5- 10 Chebyshev's Inequality 5-20 Answers to Exercises 5-2 1

Chapter 6

Random Sampling and Data Description 6- 1 Data Summary and Display 6- 1 6-2 Random Sampling 6-3 6-3 Stem-and-Leaf Diagrams 6-4 6-4 Frequency Distributions and Histograms 6-5 BOXPlots 6-12 6-6 Time Sequence Plots 6-13 6-7 Probability Plots 6-14 MINITABApplications 6- 15 Answers to Exercises 6-20

6-8

6-1

Contents

Chapter 7

Point Estimation of Parameters

ix

7-1

7- 1 Introduction 7- 1 7-2 General Concepts of Point Estimation 7-2 7-3 Methods of Point Estimation 7-6 7-4 Sampling Distributions 7-7 7-5 Sampling Distributions of Means 7-7 Answers to Exercises 7-10

Chapter 8

Statistical Intervals for a Single Sample

8-1

8- 1 8-2

Introduction 8- 1 Confidence Interval on the Mean of a Normal Distribution, Variance Known 8-2 8-3 Confidence Interval on the Mean of a Normal Distribution, Variance Unknown 8-5 8-4 Confidence Interval on the Variance and Standard Deviation of a Normal Population 8-7 8-5 A Large-Sample Confidence Interval for a Population Proportion 8-9 8-6 A Prediction Interval for a Future Observation 8-1 1 8-7 Tolerance Intervals for a Normal Distribution 8- 13 Summary of Statistical Intervals for a Single Sample 8-14 MINITABApplications 8- 15 Answers to Exercises 8- 18

Chapter 9

Tests of Hypotheses for a Single Sample 9-1 9-2 9-3

9-1

Hypothesis Testing 9-1 Tests on the Mean of a Normal Distribution, Variance Known 9-6 Tests on the Mean of a Normal Distribution, Variance Unknown 9- 10 9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal Population 9- 13 9-5 Tests on a Population Proportion 9-15 9-7 Testing for Goodness of Fit 9- 17 9-8 Contingency Table Tests 9-22 MINITABApplications 9-28 Answers to Exercises 9-37

X

Contents

Chapter 10

Statistical Inference for Two Samples

10-1

""

10-2 Inference for a Difference in Means of Two Normal Distributions, Variances Known 10-1 10-3 Inference for a Difference in Means of Two Normal Distributions, Variances Unknown 10-6 10-4 Paired t-Test 10-11 10-5 Inference on the Variances of Two Normal Populations 10-14 10-6 Inference on Two Population Proportions 10-19 MINITABApplications 10-24 Answers to Exercises 10-30

Chapter 11

Simple Linear Regression and Correlation

11-1

11- 1 Empirical Models 11- 1 11-2 Simple Linear Regression 11-3 11-3 Properties of the Least Squares Estimators 11-8 11-4 Some Comments on Uses of Regression 11- 10 11-5 Hypothesis Tests in Simple Linear Regression 11- 11 11-6 Confidence Intervals 11- 17 11-7 Prediction of New Observations 11-21 11-8 Adequacy of the Regression Model 11-23 11-9 Transformations to a Straight Line 11-29 11-11 Correlation 11-30 MINITABApplications 11-34 Answers to Exercises 11-37

Chapter 12

Multiple Linear Regression

12-1

12-1 Multiple Linear Regression Model 12- 1 12-2 Hypothesis Tests in Multiple Linear Regression 12-5 12-3 Confidence Intervals in Multiple Linear Regression 12-13 12-4 Prediction of New Observations 12-15 12-5 Model Adequacy Checking 12-16 12-6 Aspects of Multiple Regression Modeling 12-18 MINITABApplications 12-28 Answers to Exercises 12-35

Contents

Chapter 13

Design and Analysis of Single-Factor Experiments: The Analysis of Variance 13-1 13- 1 Designing Engineering Experiments 13- 1 13-2 The Completely Randomized Single-Factor Experiment 13-3 The Random Effects Model 13- 13 13-4 Randomized Complete Block Design 13- 15 MINITABApplications 13-2 1 Answers to Exercises 13-26

Chapter 14

Design of Experiment with Several Factors

13-2

14-1

14-1 Introduction 14- 1 14-3 Factorial Experiments 14- 1 14-4 Two-Factor Factorial Experiments 14-4 14-5 General Factorial Experiments 14-1 1 14-7 2k Factorial Designs 14-1 1 14-8 Blocking and Confounding in the 2k Design 14-18 14-9 Fractional Replication of the 2k Design 14-2 1 MINITABApplications 14-29 Answers to Exercises 14-32

Chapter 15

Nonparametric Statistics

i 5-1

15- 1 Introduction 15- 1 15-2 Sign Test 15-2 15-3 Wilcoxon Signed-Rank Test 15-7 15-4 Wilcoxon Rank-Sum Test 15-12 15-5 Nonparametric Methods in the Analysis of Variance MINITABApplications 15- 18 Answers to Exercises 15-24

15-14

xi

The Role of Statistics in Engineering

1-1 The Engineering Method and Statistical Thinking 1-2 Collecting Engineering Data

1-1

1-3 Mechanistic and Empirical Models 1-4 Probability and Probability Models

The Engineering Method and Statistical Thinking

O Explain engineering problem solving process. 0 Describe the application of statistics in the engineering problem solving process. 0 Distinguish between enumerative and analytic studies. "

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Engineers develop practical solutions and techniques for engineering problems by applying scientific principles and methodologies. Existing systems are improved and/or new systems are introduced by the engineering approach for better harmony between humans, systems, and environments. In general, the engineering problem solving process includes the following steps: 1 . Problem definition: Describe the problem to solve. 2. Factor identification: Identify primary factors which cause the problem. 3. Model (hypothesis) suggestion: Propose a model (hypothesis) that explains the relationship between the problem and factors. 4. Experiment: Design and run an experiment to test the tentative model. 5. Analysis: Analyze data collected in the experiment. 6. Model modification: Refine the tentative model. 7. Model validation: Validate the engineering model by a follow-up experiment. 8. Conclusion (recommendation): Draw conclusions or make recommendations based on the analysis results. (Note) Some of these steps may be iterated as necessary.

1-2

Chapter 1

Application of Statistics

The Role of Statistics in Engineering Statistical methods are applied to interpret data with variability. Throughout the engineering problem solving process, engineers often encounter data showing variability. Statistics provides essential tools to deal with the observed variability. Examples of the application of statistics in the engineering problem solving process include, but are not limited to, the following: 1. Summarizing and presenting data: numerical summary and visualization in descriptive statistics (Chapter 6). 2. Inferring the characteristics (mean, median, proportion, and variance) of single/two populations: z, t, and F tests in parametric statistics (Chapters 7-10); signed, signed-rank, and rank-sum tests in nonparametric statistics (Chapter 15). 3. Testing the relationship between variables: correlation analysis (Chapter 11); categorical data analysis (Chapter 8). 4. Modeling the causal relationship between the response and independent variables: regression analysis (Chapters 1 1-12); analysis of variance (Chapters 13-14). 5. Identifying the sources of variability in response: analysis of variance (Chapters 13-14). 6. Evaluating the relative importance of factors for the response variable: regression analysis (Chapter 12); analysis of variance (Chapter 14). 7. Designing an efficient, effective experiment: design of experiment (Chapters 13-14). These applications would lead to development of general laws and principles such as Ohm's law and design guidelines.

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Enumerative vs. Analytic Studies

1-2

Two types of studies are defined depending on the use of a sample in statistical inference: 1. Enumerative study: Makes an inference to the well-defined population from which the sample is selected. (e.g.) defective rate of products in a lot 2. Analytic study: Makes an inference to a future (conceptual) population. (e.g.) defective rate of products at a production line

Collecting Engineering Data

0 Explain three methods of data collection: retrospective study, observational study, and designed experiment. Data Collection Methods

Three methods are available for data collection: 1. Retrospective study: Use existing records of the population. Some crucial information may be unavailable and the validity of data be questioned. 2. Observational study: Collect data by observing the population with as minimal interference as possible. Information of the population for some conditions of interest may be unavailable and some observations be contaminated by extraneous variables. 3. Designed experiment: Collect data by observing the population while controlling conditions on the experiment plan. The findings would obtain scientific rigorousness through deliberate control of extraneous variables.

1-3 Mechanistic and Empirical Models

1-3

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Models (explaining the relationship between variables) can be divided into two categories: 1. Mechanistic model: Established based on the underlying theory, principle, or law of a physical mechanism. E (e.g.) I = - + E (Ohm's law) R where: I = current, E = voltage, R= resistance, and E = random error 2. Empirical model: Established based on the experience, observation, or experiment of a system (population) under study. (e.g.1 Y = P o + P I X + E where: y = sitting height, x = stature, and E = random error

Mechanistic vs. Empirical Models

1-4

Probability and Probability Models

O Describe the application of probability in the engineering problem solving process. X.__

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Along with statistics, the concepts and models of probability are applied in the engineering problem solving process for the following: 1. Modeling the stochastic behavior of the system: discrete and continuous probability distributions. 2. Quantifying the risks involved in statistical inference: error probabilities in hypothesis testing. 3. Determining the sample size of an experiment for a designated test condition: sample size selection

Probability

2-1 2-2 2-3 2-4 2-5

Sample Spaces and Events Interpretations of Probability Addition Rules Conditional Probability Multiplication and Total Probability Rules

2-1

2-6 Independence 2-7 Bayes' Theorem 2-8 Random Variables Answers to Exercises

Sample Spaces and Events

0 Explain the terms random experiment, sample space, and event. Define the sample space and events of a random experiment. 0 Define a new joint event from existing events by using set operations. 0 Assess if events are mutually exclusive andlor exhaustive. Use a Venn diagram and/or tree diagram to display a sample space and events.

Random Experiment

When different outcomes are obtained in repeated trials, the experiment is called a random experiment. Some sources of the variation in outcomes are controllable and some are not in the random experiment. (e.g.) Sources of variability in testing the life length of an INFINITY light bulb (I) Material (2) Manufacturing process (3) Production environment (temperature, humidity, etc.) (4) Measurement instrument (5) Drift of current (6) Observer

Sample Space

A sample space is the set of possible outcomes of a random experiment (denoted as S). Two types of sample space are defined: 1. Discrete Sample Space: Consists of finite (or countably infinite) outcomes. (e.g.) Tossing a coin: S = {head, tail) 2. Continuous Sample Space: Consists of infinite and innumerable outcomes. (e.g.) The life length of the INFINITY light bulb (X):S = {x; x 2 0)

(9

2-2

Chapter 2 Event

Probability An event is a subset of the sample space of a random experiment (denoted by E).

0 Set Operations

Three set operations can be applied to define a new joint event from existing events: 1. Union ( E l u E 2 ): Combines all outcomes of E l and E2. 2. Intersection ( E l n E2 ): Includes outcomes that are common in El and E2.

3. Complement ( E' or (Ef)' = E . Laws for Set Operations

Mutually Exclusive1 Exhaustive Events

E ): Contains outcomes that are

in E. Note that

The following laws are applicable to the set operations: 1. Commutative Law E,nE2=E2nEl EluE2=E2uEl 2. Distributive Law (ElnE2)uE3=(EluE,)n(E,uE,) (El u E,) n E, = (El n E,) u (E, n E,) 3. DeMorgan's Law (E, ~ E , ) ' = E , ' u E ; (E, u E2)'= E; n E; A collection of events El, E2,. .. ,E k is said to be mutuallv exclusive (see Figure 2-I), if the events do not have any outcomes in common, i.e., E, n E, = 0 for all pairs (i,j), i # j

Next, the collection of events El, E2,. .. ,Ek is said to be exhaustive (see Figure 2I), if the union of the events is equal to S, i.e., El u E , u . . . u E , = S

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Figure 2-1 Mutually exclusive and exhaustive events. Example 2.1

Two dice are cast at the same time in an experiment. 1. (Sample Space) Define the sample space of the experiment. h The sample space is S = {(x, y); x = 1,... , 6, y = 1,. .. , 6 ) , where x and y represent the outcome of each die. Since there are 6 possible outcomes with the first and second dice each, the sample space consists of a total of 6 x 6 = 36 outcomes. 2. (Event) Of the sample space, find the pairs whose sum is 5 (El) and the pairs whose first die is odd (E2).

2-1 Sample Spaces and Events

2-3

Example 2.1 (cont.)

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3. (Set Operation) Find El

u E , , E, n E , , and E;

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5. (Exhaustive Events) Are El and E2 exhaustive? h No,

Exercise 2.1 (MR 2-33)

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The rise times (unit: min.) of a reactor for two batches are measured in an experiment. 1. Define the sample space of the experiment. 2. Define El where the reactor rise time of the first batch is less than 55 min. and E2 where the reactor rise time of the second batch is greater than 70 min. 3. Find E,

u E , , El n E, ,and E ; .

4. Are El and E2 mutually exclusive? 5. Are El and E2 exhaustive?

Diagrams

Diagrams are often used to display a sample space and events in an experiment: 1. Venn Diagram: A rectangle represents the sample space and circles indicate individual events, as illustrated in Figure 2-2.

Figure 2-2 Venn diagrams: (a) El u E , ; (b) E, n E , ; (c) E; . The areas shaded indicate the results of union, intersection, and complement, respectively.

724 Chapter 2 Probability Diagrams (cont.)

2. Tree Diagram: Branches represent possible outcomes (see Figure 2-3). The tree diagram method is useful when the sample space is established through a series of steps or stages.

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2-2

Interpretations of Probability

0 Explain the termprobability.

0 Determine the probability of an event. Probability

The probability of an event means the likelihood of the event occurring in a random experiment. If S denotes the sample space and E denotes an event, the following conditions should be met: (1) P(S)=I (2) 0 5 P(E)I1

( 3 ) P(E, u E, u E, u = P(E, ) + P(E,) are mutually exclusive. . a = )

Probability of Event

+ P(E,) + -.., where the events

If a sample space includes n outcomes that are equally likely, the probability of each outcome is lln. Thus, the probability of an event E consisting of k equally likely outcomes is 1 k P(E)=kx-=n n where: n = number of possible outcomes in S k = number of equally likely elements in E (Note) For any event E, P ( E f )= 1- P ( E )

2-5

2-3 Addition Rules

2-3

Addition Rules

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Probability o f Joint Event

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(Probability o f Joint Event) An instructor of a statistics class tells students that the probabilities of earning an A, B, C, and D or below are 3 , &, and respectively. Find the probabilities of ( I ) earning an A or B and ( 2 ) earning a B or below. Ohr Let E l , E2, E3, and E4 denote the events of earning an A, B, C, and D or below, respectively. These individual events are mutually exclusive and exhaustive because P(El u E, u E, u E , ) = P ( E l )+ P ( E , ) + P ( E , ) + P ( E 4 )

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2. When a disk is selected at random, find the probability that the scratch or shock resistance of the disk is high. 3. Consider the event that a disk has high scratch resistance and the event that a disk has high shock resistance. Are these two events mutually exclusive?

Exercise 2.2 (cont.)

2-4

Conditional Probability

O Calculate the conditional probability of events.

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Conditional Probability

The conditional probability P(B 1 A) is the probability of an event B, given an event A. The following formula is used to calculate the conditional probability: P(A n B ) ,where P(A) > 0 P(B ( A) = P(A)

Example 2.3

(Conditional Probability) A novel method to screen carpal tunnel syndrome (CTS) at the workplace is tested with two groups of people: 50 workers having CTS and 50 healthy workers without CTS. Let A and A' denote the event that a worker has CTS and the event that a worker does not have CTS, respectively. Let B and B ' denote the event that a CTS test is positive and the event that a CTS 1 test is negative. resnectivelv. The summarv of CTS test results is as follows: " Test Result Positive ( 8 ) Negative ( B ') CTS (A) 10 Group 45 Healthy ( A ' )

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Exercise 2.3

2-5

In Exercise 2.2, find the probability that a disk has high scratch resistance (A) when it has high shock resistance (B).

Multiplication and Total Probability Rules

R Apply a total probability rule to find the probability of an event when the event is partitioned into several mutually exclusive and exhaustive subsets.

2-5 Multiplication and Total Probability Rules

Multiplication Rule Total Probability Rule

2-7

From the definition of conditional probability, P(B n A) = P(B I A)P(A) = P(A I B)P(B) When event B is partitioned into two mutually exclusive subsets B n A and B n A' (see Figure 2-4), P(B) = P(B n A) + P(B n A') = P(B I A)P(A) + P(B I A')P(A')

Figure 2-4 Partitioning event B into two mutually exclusive subsets: B n A and ~n A'. As an extension, when event B is partitioned by a collection of k mutually exclusive and exhaustive events E l , E2,... ,Ek, P ( B ) = P ( B n E l ) +P ( B n E 2 ) + . . . + P ( B n E,)

= P ( B I E l ) P ( E l )+ P(B I E 2 ) P ( E 2 )

+ - . a +

P(B I E,)P(E,)

Figure 2-5 Partitioning event B into k mutually exclusive subsets: B n E l ,

r

Example 2.4

(Total Probability Rule) In Example 2.3, the CTS screening method experiment indicates that the probability of screening a worker having CTS ( A ) as positive (B) is 0.8 and the probability of screening a worker without CTS ( A ') as positive (B) is 0.1, i.e., P(B I A) = 0.8 and P(B I A') = 0.1 Suppose that the industry-wide incidence rate of CTS is P(A)= 0.0017. Find the probability that a worker shows a positive test (B) for CTS at the workplace. P(A) = 0.0017 and P(Ar)= 1 - P(A) = 0.9983 By applying a total probability rule, P(B) = P(B I A)P( A) + P(B I A')P(Ar) =0.8~0.0017+0.1~0.9983=0.101

2-8

d

Chapter 2

Probability

Exercise 2.4 ( M R 2-97)

2-6

Customers are used to evaluate preliminary product designs. In the past, 95% of highly successful products, 60% of moderately successful products, and 10% of poor products received good reviews. In addition, 40% of product designs have been highly successful, 35% have been moderately successful, and 25% have been poor products. Find the probability that a product receives a good review.

Independence

0 Explain the term independence between events. Assess the independence of two events. Two events A and B are stochastically independent if the occurrence of A does affect the probability of B and vice versa. In other words, two events A and B. are independent if and only if (1) P(A I B) = P(A) (2) P(B I A) = P(B) ( 3 ) P ( A n B) = P(A)P(B)

Independence of Events

r

(Derivation) P(A nB ) = P(A)P(B) P ( A n B) P(B I A) = = P ( B ) if A and B are independent. P(A)

Example 2.5

I

(Independence) For the CTS test results in Example 2.3, the following probabilities have been calculated: 4 45 P(B) = - and P(B I A) = 100 5 Check if events A and B are independent, 4 45 Ikr Since P(B I A) = - # P(B) = -, A and B are not independent. This 5 100 indicates that information from the CTS test is useful to screen workers having CTS at the workplace.

I

I

Eb

I Exercise 2.5

I

I

I

For the resistance test results in Exercise 2.2, the following probabilities have been calculated: 89 40 P ( A ) = - and P ( A I B ) = 100 43 Assess if events A and B are independent.

2-7 Bayes' Theorem

2-7

2-9

Bayes' Theorem p

m

m -7 --a -

Apply Bayes' theorem to find the conditional probability of an event when the event is partitioned into several mutually exclusive and exhaustive subsets.

Bayes' Theorem

From the definition of conditional probability, P(A n B ) - P(B I A)P(A) P(B I A)P(A) P(A I B ) = P(B) P(B) P(B n A) + P(B n A')

The multiplication rule for a collection of k mutually exclusive and exhaustive events E l , E2,... ,Ek and any event B is P(B) = P(B I E l ) P ( E , )+ P(B I E 2 ) P ( E 2 + ) ... + P(B I Ek ) P ( E k ) From the two expressions above, the following general result (known as Bayes' theorem) is derived:

Example 2.6

Exercise 2.6

(Bayes' Theorem) In Examples 2.3 and 2.4, the following probabilities have been identified: P(B I A) = 0.8, P(B I A') =0.1, P(A) = 0.0017, and P(B) = 0.101 Find the probability that a worker has CTS ( A )when the test is positive (B). h By applying Bayes' theorem, P(B I A)P(A) - P(B I A)P(A) P(AI B ) = P(B I A)P(A) + P(B I A')P( A') P(B) - 0.8 x 0.0017 = 0.013 0.101 Since the incidence rate of CTS in industry is low, the probability that a worker has CTS is quite small even if the test is positive.

I

In Exercise 2.4, what is the probability that a new design will be highly successful if it receives a good review?

2-10

Chapter 2

2-8

Random Variables

Probability

O Describe the terms random variable X and range o f X . R Distinguish between discrete and continuous random variables. Random Variable

A random variable, denoted by an uppercase such as X, associates real numbers with individual outcomes of a random experiment. Note that a measured value of Xis denoted by a lowercase such as x = 10. The set of possible numbers of X is referred to as the range of X. Depending on the type of range, two categories of random variables are defined: 1. Discrete Random Variable: Has a finite (or countably infinite) range. (e.g.) Tossing a coin: X = 0 for head and X = 1 for tail 2. Continuous Random Variable: Has an interval of real numbers for its infinite range. (e.g.) The life length of an INFINITY light bulb: X2 0

Answers to Exercises

2-1 1

Answers to Exercises Exercise 2.1

1. (Sample Space) S = {x; x > 0), where x represents the rise time of the reactor for a certain batch.

2. (Event) E l = {x;O<x 70) 3. (Set Operation) E, u E , = { x ; O < x < 5 5 o r x > 7 0 ) E, n E, = 0 E; = (x; x 2 55) 4. (Mutually Exclusive Events)

Yes, because E, n E ,

= 0.

5. (Exhaustive Events) No, because E , u E , Exercise 2.2

I (Probability of Joint Event)

I Exercise 2.3

Exercise 2.4

+S .

80 3. Since P(A nB) = -# 0 , A and B are not mutually exclusive. 100

I (Conditional Probability)

(Total Probability Rule) Let E l , E2, and E3 represent the event of being a highly successful product, the event of being a moderately successful product, and the event of being a poor product, respectively. Also, let G denote the event of receiving good reviews from customers.

P

2-12

Chapter 2

Probability

Then, P ( G I E l ) = 0 . 9 5 , P ( G I E 2 ) = 0 . 6 0 ,P(GIE,)=O.lO,


P ( E l )= 0.40, P ( E 2 )= 0.35, and P ( E 3 )= 0.25 The events E l , E2, and E3 are mutually exclusive and exhaustive because P(El u E , u E,) = P ( E l )+ P ( E 2 )+ P ( E 3 )= 1 = P ( S )

By applying a total probability rule, P(G) = P(G I El )P(El)+ P(G I E2 )P(E2 + P(G I E, )P(E3 = 0.95 x 0.40 + 0.60 x 0.35 + 0.10 x 0.25 = 0.62 Exercise 2.5

( (Independence) 40 89 Since P(A I B) = -# P(A) = -,A and B are not independent. This

I I

Exercise 2.6

43 100 indicates that the scratch resistance and shock resistance of disk are stochastically related.

(Bayes' Theorem) By applying Bayes' theorem, P(G I El )P(E,) P(G I E, )P(E,) + P(G I E2 )P(E, + P(G I E, )P(E3 0.95 x 0.40 = 0.62 0.95 x 0.40 + 0.60 x 0.35 + 0.10 x 0.25

P(E1 I G ) =

Discrete Random Variables and Probability Distributions 3- 1 Discrete Random Variables 3-2 Probability Distributions and Probability Mass Functions 3-3 Cumulative Distribution Functions 3-4 Mean and Variance of a Discrete Random Variable 3-5 Discrete Uniform Distribution 3-6 Binomial Distribution

3-1

3-7 Geometric and Negative Binomial Distributions 3-7.1 Geometric Distribution 3-7.2 Negative Binomial Distribution 3-8 Hypergeometric Distribution 3-9 Poisson Distribution Summary of Discrete Probability Distributions EXCELApplications Answers to Exercises

Discrete Random Variables

See Section 2-8. Random Variables to review the notion of discrete random variable.

3-2

Probability Distributions and Probability Mass Functions

Distinguish between probability mass function and cumulative distribution function.

0 Determine the probability mass function of a discrete random variable. Probability Distribution

A probability distribution indicates how probabilities are distributed over possible values of X.

Two types of functions are used to express the probability distribution of a discrete random variable X 1. Probability Mass Function (p.m.f.): Describes the probability of a value of X, i.e., P(X= xi).

3-2

r

Chapter 3

Discrete Random Variables and Probability Distributions

Probability Distribution (cont.)

2. Cumulative Distribution Function (c.d.f.): Describes the sum of the probabilities of values ofXthat are less than or equal to a specified value, i.e., P(X I xi).

Probability Mass Function (p.m.f.)

The probability mass function of a discrete random variable X, denoted asf(xi), is f ( x i ) = P ( X = x i ) ,xi=x1,x2,... , x n and satisfies the following properties (see Figure 3- 1): (1) f (xi) 2 0 for all xi

Figure 3-1 Probability mass function of casting a die. Example 3.1

(Probability Mass Function) The grades of n = 50 students in a statistics class are summarized as follows:

Grade (X)

B

C

(x = 2) 20

(x = 3)

A (X =

No. of students

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P

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I

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~

X

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I

X

T

X

E

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,

-

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,

~

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~

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m .il".a (i ~

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15 ~

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= 4)

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-lil

Let X denote a grade in statistics. Let the values 1,2, 3, and 4 represent an A, B, C, and D or below, respectively. Determine the probability mass function of X and plot f (xi) .

Probability mass function of X

3-3

3-3 Cumulative Distribution Functions


I

I

The orders from n = 100 customers for wooden panels of various thickness (X) are summarized as follows: Wooden Panel Thickness (X; unit: inch)

+

i 20

No. of customer orders *

e_l*_m_Xm

*-r_U-l_j___**jle-s-*~~x

**-rx;xr

-38

70 n ir

mmmx#d^naea.dm

rlaixrxi

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Determine the probability mass function of X and plot f (x, ) .

3-3

Cumulative Distribution Functions

Determine the cumulative distribution function of a discrete random variable.

Cumulative Distribution Function (c.d.f.)

The cumulative distribution function of a discrete random variable X, denoted as F(x), is F(x)= P(X l x ) = f(xi) X,

SX

and satisfies the following properties (see Figure 3-2): (1) 0 I F(x) 51 (2) F(x) 5 F ( y ) ifx l y

Figure 3-2 Cumulative distribution function of casting a die. Example 3.2

(Cumulative Distribution Function) In Example 3-1, the following probabilities have been calculated: 1 2 3 1 P ( X = l ) = - , P ( X = 2 ) = - , P ( X = 3 ) = - , and P ( X = 4 ) = 5 5 10 10 Determine the cumulative distribution function of X and plot F(x) h By

using the probability mass function of X,

3-4

Chapter 3


Example 3.2 (cont.)

Cumulative distribution function of X

Exercise 3.2

In Exercise 3-1, the following probabilities have been calculated: P(X=$)=0.2, P(X=$)=0.7,and ~ ( ~ = $ ) = 0 . l

I Determine the cumulative distribution function of X and plot F(x) . 3-4

Mean and Variance of a Discrete Random Variable ---=-

amMm #ww-mmmwm-*8-m---m-

-mmm-*-

Calculate the mean, variance, and standard deviation of a discrete random variable.

Mean

Variance of X (oZ)

The mean of X, denoted as p or E(X) , means the expected value o f X

The variance ofX, denoted as o2or V(X) , indicates the dispersion ofXabout p: 02 = V ( X ) = C ( x

- p ) l f (x) = E x 2f (X) -pi

X

X

The standard deviation of X is o =

,/V(X)

(Derivation) V(X) = Z x 2f (x) - p2 (see page 67 of MR) X

Example 3.3

(Mean, Variance, and Standard Deviation) Suppose that, in an experiment of casting a single die, six outcomes (X)are equally likely. Determine the mean, variance, and standard deviation of X.

BBm

3-5 Discrete Uniform Distribution

Example 3.3

( chr

The probability mass function of X is

I

Therefore,

3-5

I

Exercise 3.3

In Exercise 3-1, the following probabilities have been calculated: P(X=$)=0.2, P(X=$)=0.7,and P(X=;)=O.~ Determine the mean, variance, and standard deviation of X.

3-5

Discrete Uniform Distribution

Describe the probability distribution of a discrete uniform random variable.

0 Determine the probability mass function, mean, and variance of a discrete uniform random variable.

Discrete Uniform Random Variable

A discrete uniform random variable X has an equal probability for each value in the range o f X = [a, b], a < b (see Figure 3-3). Thus, the probability mass function of X is 1 f(x) = b - a + 1 , w h e r e x = a , a + l , ..., b

Figure 3-3 A discrete uniform distribution. The mean and variance of X are p=- b + a and o = (b-a+112 -1 2 12

3-6

Chapter 3

Example 3.4


Suppose that six outcomes are equally likely in the experiment of casting a single ldie. 1. (Probability Mass Function; Discrete Uniform Distribution) Determine the probability mass function of the number (X)of the die. h Since x = 1,2,. . . , 6, a = 1 and b = 6. Therefore, the probability mass function of X is

I

2. (Probability) Find the probability that the number ( X ) of the die in the experiment is greater than three.

1 3. (Mean and Variance) Calculate the mean and variance of X (hSince a =

I Exercise 3.4 (MR 3-50)

(

3-6

1 and b = 6, the mean and variance of X are

(Note) Check that the values of p and o2above are the same with the answers in Example 3.2. Suppose that product codes of 2, 3, or 4 letters are equally likely. 1. Determine the probability mass function of the number of letters (X)in a product code. 2. Calculate the mean and variance o f X 3. Calculate the mean and variance of the number of letters in 100 product codes (Y = 1002'). Note that E(cX) = cE(X) and V(cX) = c2v(X) (refer to Section 5-7. Linear Combinations of Random Variables for further detail).

Binomial Distribution

Describe the terms Bernoulli trial and binomial experiment.

O Describe the probability distribution of a binomial random variable. Determine the probability mass function, mean, and variance of a binomial random variable. Binomial Experiment

A binomial experiment refers to a random experiment consisting of n repeated trials which satisfy the following conditions: (1) The trials are independent, i.e., the outcome of a trial does not affect the outcomes of other trials, (2) Each trial has only two outcomes, labeled as 'success' and 'failure,' and (3) The probability of a success (p) in each trial is constant.

3-6 Binomial Distribution

Binomial Experiment (cont.) Bernoulli Trial

3-7

In other words, a binomial experiment consists of a series of n independent Bernoulli trials (see the definition of Bernoulli trial below) with a constant probability of success (p) in each trial. A Bernoulli refers to a trial that has only two possible outcomes. (e.g.) Bernoulli trials (1) Flipping a coin: S = {head, tail) (2) Truth of an answer: S = {right, wrong) (3) Status of a machine: S = {working, broken) (4) Quality of a product: S = {good, defective) (5) Accomplishment of a task: S = {success, failure) The probability mass function of a Bernoulli random variable X is

The mean and variance of a Bernoulli random variable are p = p and o 2 = p ( l - p )

(Derivation) p and 02of a Bernoulli random variable p=~xf(x)=ox(l-p)+lxp=p

Binomial Random Variable

A binomial random variable Xrepresents the number of trials whose outcome is a success out of n trials in a binomial experiment with a probability of success p (see Table 3-1).

* If an item selected from a population is replaced before the next trial, the size of the population is considered infinite even if it may be finite. If the probability of success p is constant, the trials are considered independent; otherwise, the trials are dependent. The probability mass function of X, denoted as B(n, p ) , is

(Note) The number of combinations of x from n: C," = The mean and variance of X are p = np and o 2 = np(1- p )

n!

3-8

r

Chapter 3

Example 3.5


A test has n = 50 multiple-choice questions; each question has four choices but only one answer is right. Suppose that a student gives hisfher answers by simple guess. 1. (Probability Mass Function; Binomial Distribution) Determine the probability mass function of the number of right answers (X)that the student gives in the test. ~ . r rSince a 'right answer' is a success, the probability of a success for each question is p = $ . Thus, the probability mass function of X is

I

2. (Probability) Find the probability that the student answers at least 30

questions correctly.

3. (Mean and Variance) Calculate the mean and variance of X. = np = 50 x 0.25 = 12.5 right answers

~ . r rp

o2 = np(1- p) = 50x 0.25 x 0.75 = 9.4 = 3.12 Exercise 3.5 (MR 3-67)

3-7 3-7.1

Because all airline passengers do not show up for their reserved seat, an airline sells n = 125 tickets for a flight that holds only 120 passengers. The probability that a passenger does not show up is 0.10. Assume that the passengers behave independently. 1. Determine the probability mass function of the number of passengers (X) who show up for the flight. 2. Find the probability that the flight departs with empty seats. 3. Calculate the mean and variance ofX.

Geometric and Negative Binomial Distributions Geometric Distribution

0 Describe the probability distribution of a geometric random variable. O Compare the geometric distribution with the binomial distribution. 0 Explain the lack of memory property of the geometric distribution. Determine the probability mass function, mean, and variance of a geometric random variable. 111/e--*w-*wa

_iawaw-

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3-9

3-7 Geometric and Negative Binomial Distributions

Geometric Random Variable

A geometric random variable Xrepresents the number of trials conducted until the first success in a binomial experiment with a probability of success p: Fail Fail Fail Success

u

P(X=x)

Fail (x - 1) times ( ~ - ~ ) ~ - l

=

t

Succeed at the'x time X

P

The probability mass function of X is f ( ~ ) = ( l - ~ ) " - ' ~x,= 1,2, ... The mean and variance of X are 2 1-P 1 p = - and o =P p2 The probability of X decreases geometricallv as illustrated in Figure 3-4.

0

2

4

8

6

10

12

Figure 3-4 Geometric distributions with selected values ofp. Geometric vs. Binomial Distributions

In the geometric distribution the number of trials is a random variable and the number of successes is one, whereas in the binomial distribution the number of trials is constant and the number of successes is a random variable (see Table 32).

Table 3-2 The Characteristics of Binomial and Geometric Distributions Distribution

Probability of No.of Succe~sfp)~i

Population*

i

No.of

Successes f

Geometric Infinite Constant i.. Variable -....... .......(..X..)...... ......... 1...... .. .+ * If an item selected from a population is replaced before the next trial, the size of the wB-

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-

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m

w

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~

~

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-

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~

population is considered infinite even if itmay be finite. If the probability of successp is constant, the trials are considered independent; otherwise, the trials are dependent.

~

~

~

-

-

~

-

3-10

Chapter 3

Lack of Memory Property

Discrete Random Variables and Probability Distributions The lack of memory property means that the system does not remember the history of previous outcomes so that the timer of the system can be reset at any time, i.e., P(t<X t) The lack of memory property of the geometric distribution implies that the value o f p does not change with trial. Due to the memoryless property of the geometric distribution, the number of trials (X) until the next success can be counted from any trial. The geometric distribution is the & discrete distribution having the lack of memory property.

(Proof) The lack of memory property The conditional probability P ( X < t + At ( X > t) is

Note that the sum of the first n terms of the geometric sequence a , ar, ar 2 , ...,ar is n-1

S,

=Car k=O

k

=-

a - urn , where r # 1 l-r

Accordingly,

t -1

(t+At-1)-1

C ~ ( 1 - p )- -~C P ( ~ - P ) ~ -

k=O

k=O

, where k = x - 1

t-1

1 - C P ( ~- P)' k=O

P-~(1-P) -

l-(l-P)

t+At-l

- P -~

( 1 P)' -

l-(l-p) P - ~ ( -PIt 1 1l-(l-P)

3-7 Geometric and Negative Binomial Distributions

3-11

Lack of Memory Property (cont.)

The probability P(X < At) is At-I

P(X < At) = P ( X 2 At - 1) =

(1 - p)"-' p

(At-I)-]

= ~ ( I - ~ ) " where - ~ ~k =, x - l

Therefore, P(Xt)=P(X O

(2)

1- f ( x W

=1

The integral off (x) from points a to b represents the probability that X belongs to the interval [a, b] (i.e., the area under the density function between a and b as shown in Figure 4-1). (4) P(X = x) = 0 The probability of every single point is zero because the width of any point is zero. Thus, P(a<XIb)=P(a<X O = 0, elsewhere Determine the cumulative distribution function of X. h By using the probability density function of X,

6;

F(x) = P ( X I x) = f (u)du =

1e-"du

= - e-"

I,

= -e-"

+ e-O = 1- e-"

Therefore, F(x) = Exercise 4.2

1"

1-e-",

x 1.57) = 1- P(Z 11.57) = 1 - 0.942 = 0.058 (4) P(-1.57 1Z 11.57) = P(Z 11.57) - P(Z 1-1.57) = Q(1.57) - @(-I .57) = 0.942 - 0.058 = 0.884

4-8

Chapter 4

Standardization

Continuous Random Variables and Probability Distributions An arbitrary normal random variable X with p and o2can be transformed into Z (with p = 0 and o2= l), as illustrated in Figure 4-5, by using the algebraic relationship between X and Z:

x-c1 z=(3

Figure 4-5 Standardizing a normal random variable. After standardizing the normal random variable X, probabilities of X are determined by using the z table. (Standardization) Suppose that the life length of an INFINITYlight bulb follows a normal distribution with p = 750 hours o2= 402. Calculate P(740 1X 5 770).


4-7

The sick-leave time of employees in a firm in a month is normally distributed with a mean of 100 hours and a standard deviation of 20 hours. Find the probability that the sick-leave time of an employee in a month exceeds 130 hours.

Normal Approximation to the Binomial and Poisson Distributions

R Apply the normal approximation to the binomial, hypergeometric, and Poisson random variables. Normal Approximation to Binomial Distribution

The binomial distribution B(n, p ) approximates to the normal distribution with p = np and o2= np(1 - p) if n p > 5 and n(l - p ) > 5 (see Figure 4-19 on page 119 of MR). However, whenp is so close to 0 or 1 that np < 5 or n(l - p ) < 5, corresponding binomial distribution is significantly skewed (see Figure 4-20 on page 120 of MR); thus, the normal approximation is inappropriate.

4-7 Normal Approximation to the Binomial and Poisson Distributions

Normal Approximation to Binomial Distribution

4-9

When normal approximation is applicable, the probability of a binomial random variable X with p = np and o2= np(1 - p ) can be determined by using the standard normal random variable - X-u X-no L=--

-

(cant.)

Normal Approximation to Hypergeometric Distribution

Recall that the binomial approximation is applicable to a hypergeometric distribution if the sample size n is relatively small to the population size N, i.e., nlN < 0.1 (see Section 3-8. Hypergeometric Distribution). Consequently, the normal approximation can be applied to the hypergeometric distribution withp = KIN (K: number of successes in N ) if nlN < 0.1, np > 5. and n(1 - p) > 5. Thus, when normal approximation is applicable, the probability of a hypergeometric random variable Xwith p = np and o2= np(1 - p)(N - n)/(N - 1) can be determined by using the standard normal random variable X - np K x - P -, where p = o ,/np(l - P ) ( N - n) /(N - I) N

z=-

Normal Approximation to Poisson Distribution

Recall that a Poisson distribution with k = h and o2= h is the limiting form of a binomial distribution with p = np and o2= np(1 - p ) (see Section 3-9. Poisson Distribution). Therefore, the normal approximation is applicable to a Poisson distribution if

m.

Accordingly, when normal approximation is applicable, the probability of a Poisson random variable X with p = h and oZ= h can be determined by using the standard normal random variable X-p --X - h

z=-

o

ii

Example 4.6

J

x

1. (Normal Approximation; Binomial Distribution) Suppose that X is a binomial random variable with n = 100 andp = 0.1. Find the probability P(X I 15) based on the corresponding binomial distribution and approximate normal distribution. Is the normal approximation reasonable? ~ . r rThe probability mass function, mean, and variance of the binomial random variable X are

I

Thus,

4-10

Chapter 4

Example 4.6 (cont.)

Continuous Random Variables and Probability Distributions By applying the normal approximation to the binomial random variable X,

---)

3

= P(Z I 1.6'7) = 0.953

Since np = 10 > 5 and n(l - p ) = 90 > 5, the normal approximation to the binomial random variable X i s satisfactory. 2. (Normal Approximation; Hypergeometric Distribution) Suppose that X has a hypergeometric distribution with N = 1,000, K = 100, and n = 100. Find the probability P ( X I 15) based on the corresponding hypergeometric distribution and approximate normal distribution. Is the normal approximation reasonable? FF The probability mass function, mean, and variance of the hypergeometric random variable X are

(Note) max (0, n - ( N - K ) ) = max (0, 100 - (2 000 - 100)) = max{O, -800) = 0 min{K,n) = min(100, 100) = 100

Thus,

By applying the normal approximation to the hypergeometric random variable X, P(X615)=P

-

----)

2.85

= P(Z 6 1.75) = 0.960

Since nlN = 0.1, np = 10 > 5 and n(l - p ) = 90 > 5, the normal approximation to the hypergeometric random variable Xis satisfactory.

. (Normal Approximation; Poisson Distribution) Suppose that X has a Poisson distribution with A. = 10. Find the probability P(X I 15) based on the corresponding Poisson distribution and approximate normal distribution. Is the normal approximation reasonable?

4-11

4-9 Exponential Distribution

Example 4.6 (cont.)

R R R R

The probability mass function, mean, and variance of the Poisson random variable X are

I

Thus,

I

By applying the normal approximation to the Poisson random variable X,

I

Since h =10 > 5, the normal approximation to the Poisson random variable X is satisfactory. 1. Suppose that X is a binomial random variable with n = 200 andp = 0.4. Find the probability P ( X I 70) based on the corresponding binomial distribution and approximate normal distribution. Is the normal approximation reasonable? 2. Suppose that X has a hypergeometric distribution with N = 2,000, K = 800, and n = 200. Find the probability P ( X I 70) based on the corresponding hypergeometric distribution and approximate normal distribution. Is the normal approximation reasonable? 3. Suppose that X has a Poisson distribution with h = 80. Find the probability P(X 1 70) based on the corresponding Poisson distribution and approximate normal distribution. Is the normal approximation reasonable?


4-9

I

Exponential Distribution

Describe the probability distribution of an exponential random variable. Explain the relationship between exponential and Poisson random variables. Explain the lack of memory property of an exponential random variable. Determine the probability density function, mean, and variance of an exponential random variable.

-ms*.?"+,w"M

%*

* s.B-d

Exponential Random Variable

*--

-- -.

*- "

d--*

-

" *

a

*

..

-

*w

An exponential random variable represents the length of an interval (in time, space, etc.) from a certain point until the next success in a Poisson process. The exponential distribution parallels the geometric distribution by being concerned with the probability of the next success.

* a s

4-12

Chapter 4

Exponential Random Variable (cont.)

Continuous Random Variables and Probability Distributions The probability density function and cumulative distribution function of exponential random variable X with parameter h (average number of successes per unit interval) are f ( x ) = ~ e - ~ and ~(x)=l-e-k, x20, h > 0 The mean and variance of X are 1 2 1 p = - and o =h h2

(Caution) Like the Poisson distribution, use consistent units to define an exponential random variable X and the corresponding parameter h. An exponential curve is skewed to the right, decreasing exponentially (see Figure 4-6; as h increases, the skewness of the curve increases.

Figure 4-6 Exponential curves with selected values of h. Relationship between Exponential and Poisson Distributions

The exponential and Poisson distributions are interchangeable. Let an exponential random variable X with parameter 3L (average number of successes during the unit length of time) denote the time until the next success. Let N denote the number of successes during a specific time interval x, i.e., N is a Poisson random variable with parameter h . Then, the following relationship exists between X and N: e-k(h)O -h =e P(X > x) = P ( N = 0) = 0! Therefore, F(X)=P(X<X)=~-P(X>X)=~-~-~ By differentiating F(x), the probability density function of the exponential random variable X i s

4-9 Exponential Distribution

Exponential vs. Poisson Distributions

4-13

In the exponential distribution the number of successes is fixed as one and the length of a time interval between successes is a random variable, whereas in the Poisson distribution the number of successes is a random variable and the length of a time interval is fixed (see Table 4-1). Table 4-1 The Characteristics of Poisson and Ex~onentialDistributions

Distribution

Population

Poisson

Infinite

Constant (h)

Infinite

Constant (A)

Exponential %-,~~".m~",z~,~~."

Lack of Memory Property -

- . .-- .-

-
O

x d G e

The mean and variance of X are k = e e+w2 1 2 and o2= e 2e+m2 Example 4.10

- 1)

Assume that the time (unit: hours) to failure of a newly developed electronic device follows a lognormal distribution with parameters 8 = - 1 and m2 = 22. 1. (Probability Density Function; Lognormal Distribution) Determine the probability density function of the time (X)until failure of the device. M

1

f(x)=-e

(lnx-e)* -

2@2

x&

I

I

=----e 1 2 G x

(ln x+l)* -

,x>O

2. (Probability) Find the probability that the device lasts longer than 100 hours without failure.

-\F,

3. (Mean and Variance) ind the mean and variance of X. e+w2 I 2 2 1 h p=e = e-1+2 = 2.7 hour /

2

o =e Exercise 4.10 (MR 4-120)

T

2e+w2

-4 =

e 2 (-1)+2*

(e 2' - 1 ) = 1 9 . 9 ~

Suppose that the length of time (in seconds) that a user views a page on a Web site before moving to another page is a lognormal random variable with parameters 8 = 0.5 and m2 = 1. 1. Determine the probability density function of the time (X)until a user moves from the page. 2. Find the probability that a page is viewed for more than 10 seconds. 3. Find the mean and variance of X.

Summary of Continuous Probability Distributions

4-21

Summary of Continuous Probability Distributions A. Probability Distributions Distribution

'I

Probability Density Function

Mean

e -hx , x > O , k > O , r = 1 , 2,... ( r - I)!

h

/'

\

Variance

Section

Uniform

Normal

yXr-l

Erlang

r

B. Comparison Parameters Distribution

Population

aate of Success

Len*h of Time Intervd

NO.of ~uccesses

Poisson

Infinite

Constant (A)

Constant

Variable (X)

Exponential

Infinite

Constant (A)

Variable (X)

Constant (1 )

Erlang

Infinite

Constant (A)

Variable (X)

Constant (positive integer r )

Gamma

Infinite

Constant (A)

Variable (X)

Constant (positive real number r)

Weibull

Infinite

Variable (X)

Constant ( I )

Changeable

(P, 6)

4-22

Chapter 4

Continuous Random Variables and Probability Distributions

EXCELApplications (Reading the z Table)

(1) Choose Insert > Function. (2) Select the category Statistical and the function NORMSDIST. (3) Enter the value of Z (Z). (4) Find the result circled with a dotted line below.

---- - -

WRJ&DIST

i?

I

---

_

--

I

= -1.57

-

= 0.058207562 Returns the standard normalcumdat~vedinbution (has a mean of zero and a standard deviation of one).

1 is the value For whlch you want the distr~butcon.

-

"-

"

Formularesult =

-

-- --

__*-----_

::---___--Q,USSM~S@:;

a

-

Helo on th~sf u n c t ~ n

Example 4.5

(Standardization) (1) Choose Insert % Function. (2) Select the category Statistical and the function NORMDIST. (3) Enter the value of X (X), mean (Mean; p), standard deviation (Standard-dev; o), and logical value (Cumulative; true for the cumulative distribution function and false for the probability density function). (4) Find the result circled with a dotted line below.

I

= 0.691462467 Returns the normal cumulat~vedatr~butcon for the speclhed mean and standard deviatm.

---

Cumulative IS a bgical value: f o r t h cumulative dlstribut~on funct~on,use TRUE; for the probainy mass funct~on,use FALSE.

--

.

Formda result = Hekl on ttus functlorl

-"

--------

Function. (2) Select the category Statistical and the function LOGNORMDIST. (3) Enter the values of X (X) and two parameters (Mean, 8;Standard-dev, a) of the lognormal distribution. (4) Find the results circled with a dotted line below.

=.

Mean

1 i

-

X

I

Standard-dev

I

100

1

-1

=2

I--_

,--

1

j

':4-[;.99746519$ Returns the cumufakive lognmal distnbubon of x, where In(x) IS n&%llydis&ib'ded with parametersMean and Standard-dev.

Standard-dew

IS

the standard d e W m of In(x), a poskive numbsc.

-Fornula result = Help on this funct~on

__-----.._

;:! WES~_?~:: I-,

---



(Probability Density Function) 3

(I) P ( X < t ) = ly(x)dx= f i 3 3 x 2 d ~ = x 3 1=(f) ~3

1

- o = - 27

19

Exercise 4.2

(Cumulative Distribution Function) By using the probability density function of X, F ( x ) = P ( X I x ) = [f(u)du=

13u2du=u31x= x 3 - O = x

Therefore, x = J0.038=0.19

0

3

4-25

4-26

Chapter 4

Exercise 4.4

Continuous Probability Distributions

1. (Probability Density Function; Continuous Uniform Distribution) a = 0.95, b = 1.05 1 I - 1 - 10, 0.95 1x 11.05 f(x)=-= b-a 1.05-0.95 0.1

2. (Probability) P ( X >1.02)=

f

.05

.02

f(x)dx=

3. (Mean and Variance) p=- b + a - 1.05+0.95 =1.0 2 2 (b - a12 (1.05 - 0.95)~= 8.3 x (3 = 12 12 Exercise 4.5

Exercise 4.6

1.05

=10(1.05-1.02)=0.3

= 0.029~

1 (Standardization)

1. (Normal Approximation; Binomial Distribution) The probability mass function, mean, and variance of the binomial random variable X are

Thus,

By applying the normal approximation to the binomial random variable X,

Since np = 80 > 5 and n(1 - p) = 120 > 5, the normal approximation to the binomial random variable X i s satisfactory.



4-27

2. (Normal Approximation; Hypergeometric Distribution) The probability mass function, mean, and variance of Xare K N-K 800 2000-800 800 1200

f(XI= ( x ) ( n - x ) -

( )( x

20-x

)( -

.

)(200-I)

,x = 0 to 200

(Note) max(0, n - (N - K)) = max{0,200 - (2000 - 800)) = max (0, - 1000) = 0 min{K,n) = min{800,200) = 200

Thus,

By applying the normal approximation to X

Since n/N = 0.1, np = 80 > 5 and n(l - p) = 120 > 5, the normal approximation to the hypergeometric random variable Xis satisfactory. 3. (Normal Approximation; Poisson Distribution) The probability mass function, mean, and variance of X are

Thus,

By applying the normal approximation to X

Since h = 80 > 5, the normal approximation to Xis satisfactory.

4-28

Chapter 4

Exercise 4.7

Continuous Probability Distributions

1. (Probability Density Function; Exponential Distribution) For a consistent unit for X and h, h = 1 failure1400 hrs = 0.0025 failurelhr Therefore, the probability density function of X is f (x) = he-h" = 0 . 0 0 2 5 e - ~ . ~ ~x~2~0"(unit: , hr) 2. (Probability) By using the exponential distribution, P ( X < loo)= F(100)=1 -e-O.:'= 0.221 The parameter of the corresponding Poisson distribution is x = 100 hrs, ?a= 0.0025 failurelhr x 100 hrs = 0.25 Therefore, P(NZ1)=1-P(N=0)=1-

e-)" (hx)' = l - e - aJ - 1- e-0.25 O!

= 0.22 1

As shown above, the probability from the exponential distribution is equal to the probability from the Poisson distribution. 3. (Mean and Variance) 1 1 p = - = -= 400 hrs until failure h 11400

Exercise 4.8

1. (Probability Density Function; Erlang Distribution) For a consistent unit for X and ?L, , h = 1 problem130 days= 1/30 problemlday Therefore, the probability density function of X with r = 4 problems is

where x 2 0 (unit: day) 2. (Probability) By using the Erlang distribution, P ( X > 120) =

2.06 x lo-' x3e-0.033xdr = 0.433 (this integral is solved

20

by integration by parts, which is not shown here) The parameter of the corresponding Poisson distribution is hx = 1/30 problemfday x 120 days = 4



1

Therefore,

4-29

t$=

e-hx(~)k P(NO

The mean and variance of X are

Conditional Mass Function

Recall that P(B ( A) = P(A nB) I P(A) (see Section 2-4. Conditional Probability). In parallel, the conditional p.rn.f. of X given Y = y, denoted as fxl, (x) ,with the joint p.m.f. f, (x, y ) is P ( X = x , Y = y ) - f,(x,y) fXly(x)=P(X=xIY=y)= P(Y = Y) f r (Y) The conditional p.m.f. fxl, (x) satisfies the following:

The conditional mean and conditional variance of X given Y = y are E ( X I Y ) = PA, =

Independence

Cxfxly(XI

Two random variables X and Yare independent if knowledge of the values of X does not affect any probabilities of the values of Y and vice versa. In other words, X and Yare not stochastically related. Recall that P(A I B) = P(A) , P(B I A ) = P(B) , and P(A nB) = P(A)P(B) for two independent events A and B (see Section 2-4. Conditional Probability). Likewise, two independent discrete random variables X and Y satisfy any of the following: (1 f Xiy (XI= f x (XI

r

Independence (cont.) Example 5.1

5-4

Chapter 5

Example 5.1 (cont.)

Joint Probability Distributions The mean and variance of X are 4

E ( X ) = p x =~xfx(r)=lfx(1)+2.fx(2)+3.fx(3)+4fx(4) x=l

=

1 x 0.21 + 2 x 0.445 + 3 x 0.26 + 4 x 0.085 = 2.22

3 . (Discrete Conditional p.m.f.) Determine the conditional probability distribution of statistics grade (Y) given that ergonomics grade is an A ( X = 1 ) . Find also the conditional mean and conditional variance of Y given X = 1 . h The conditional marginal probabilities of Y given X = 1 are

The conditional mean and conditional variance of Y given X = 1 are

4 . (Independence) Check if ergonomics grade (X)and statistics grade (Y) are independent. mr Check if f , (1,1> = f x (1) . fY ( 1 )

By using the probabilities f , (1,l) = 0.12, f x ( 1 ) = 0.21, and f y (1) = 0.245, f , (1,l)= 0.12 + f,(l) f y (1) = 0.21 x 0.245 = 0.052 Thus, it is concluded that ergonomics grade (3 and statistics grade (Y) are not independent; in other words, ergonomics grade and statistics grade are related.

1

5-2 Multiple Discrete Random Variables


5-2

5-5

The number of defects on the front side ( X ) of a wooden panel and the number of defects on the rear side (Y) of the panel are under study. 1. Suppose that the joint p.m.f of X and Y is modeled as f x r ( x , y ) = c ( x + y ) , x = 1 , 2 , 3 and y = 1 , 2 , 3 Determine the value of c. 2. Determine the marginal p.m.f. of X. Find also the mean and variance ofX. 3. Determine the conditional p.m.f. of Y given X = 2. Find also the conditional mean and conditional variance of Y given X = 2. 4. Check if the number of defects on the front side (X) of a wooden panel and the number of defects on the rear side (Y) of the panel are independent.

Multiple Discrete Random Variables Joint Probability Distributions

5-2.1

Explain the joint, marginal, and conditional probability distributions of multiple discrete random variables. Explain the independence of multiple discrete random variables. _

_

j

~

m

_

_

_

~

-

Joint Probability Mass Function

P

l

_

a

_

l

(

n

~

-

w

-

w

~

~

-

~

~

~

.

~

~m

~

h

n

~

~

~

~

m

-

(

r

~

w*ua--*l~i-*-a~-"s ~ M B I ~

A joint p.m.f. of discrete random variables X I ,X 2 , ... ,X,, which describes the probabilities of simultaneous outcomes ofXl, X2,... ,X,, is f R ( ~ I , ~ 2 , . . . , x p ) = =P ( x 1 ~ = X ,2 , . ~. . , X ~p =x,) Next, the joint p.m.f. of K = { X I ,X2,... ,Xk), a subset of R = { X I ,X2, ... ,Xp) with the joint p.m.f. f R ( x ,,x, ,...,x, ) , is

f

x

x

~fR(x12x2,...9xp) Rx,x2...Xk

where RX,X2...Xk denotes the set of all points in the range of R for which X , = x, ,

Marginal Probability Mass Function

A marginal p.m.f. of& which belongs to R = { X I ,X2,... ,Xp) with the joint p.m.f. f R ( x l , x 2,..., x , ) , is

f x L ( x i ) = P ( X i= ~ i ) = C f R ( x l , x ~ , . . . , X p ) where R , denotes the set of all points in the range of R for which X i = xi .

Conditional Probability Mass Function

A conditional p.m.f. of K = {x,,X 2 , ... ,Xk) given Kr={Xk+l,Xk+2,... ,Xp) describes the probabilities of outcomes of K at the given condition of K' . Note that K and K' are mutually exclusive. As an extension of the bivariate conditional p.m.f., the conditional p.m.f. of K

w

B

a

5-6

Chapter 5

Conditional Probability Mass Function (cont.)

Independence

5-2.2

Joint Probability Distributions given K' with the joint p.m.f. f , ( x ,,x, ,...,x , ) is P ( X , = x l , X 2= x 2 ; - , X p = x P ) fKIK' ( X I X 2 , .

.-,X k ) =

P(Xk+l= X k + ~X k + 2 = Xk+2 ..., X p = X p )

As an extension of the bivariate independence conditions, multiple independent discrete random variables R = { X I ,X2,. .. ,X p ) , consisting of two mutually exclusive subsets K = { X I ,X2,. .. ,Xk) and Kf={Xk+,,Xk+2,... ,X p ) , satisfy the following: ( 1 ) fKIKr(x1,x2 ,... r X k Ixk+l , X ~ + ~ , . . . ~ X ~ ) = ~ K ( X ~ , X ~ , . . . , X ~ )

Multinomial Probability Distributions

O Determine the marginal probability distribution of multinomial random variables. Multinomial Random Variables

Suppose that ( 1 ) A random experiment consists of n trials. ( 2 ) The outcome of each trial can be categorized into one of k classes. ( 3 ) The multinomial random variables XI,X2,... ,Xk represent the number of outcomes which belong to class 1 , class 2,. .. , class k, respectively. (4) pl, p2, ... ,pk denote the probability of an outcome belonging to class 1, class 2,. .. , and class k, respectively. ( 5 ) pl,p2,. .. ,pk are constant over n repeated trials. Then, X l , X2,. .. ,Xk have a multinomial distribution with the joint p.m.f.

where x, + x , + . . . + x k = n and p, + p , + . . . + p k = 1 Note that the marginal p.m.f. of &,,i = 1 to k, is a binomial distribution with mean and variance E ( X i )=px, = npi and v ( x , ) =o& =npi(l - p i )

Example 5.2

The maximum grip strength of an adult is categorized into one of three groups: 'low' for < 50 Ibs., 'moderate' for 50 to 100 lbs., and 'high' for > 100 lbs. Suppose the probabilities that an adult belongs to the low, moderate, and high groups are 0.1, 0.7, and 0.2, respectively. A group of n = 50 people is measured for grip strength.

5-3 Two Continuous Random Variables

5-7

1. (Joint p.m.f.; Multinomial Distribution) Find the probability that 6 people have low grip strength, 36 have moderate grip strength, and 8 people have high grip strength. h Let XI,X2, X3 denote the number of people that have low, moderate, and high grip strength, respectively. Since n = 50,pl = 0 . 1 , = ~ 0.7, ~ andp3 = 0.2, the joint p.m.f. ofXl, XZ,X3 is

Example 5.2 (cont.)

where x, + x 2 + - . . + x , = n

I

Thus,

2. (Marginal p.m.f.; Multinomial Distribution) Find the probability that 36 people in the group have moderate grip strength. ~ . r rThe marginal p.m.f. of X2 follows a binomial distribution with n = 50 andp2 = 0.7, i.e.,

I

i


L

1

1

5-3

Thus,

In the transmission of digital information, the probabilities that a bit has high, moderate, and low distortion are 0.01, 0.04, and 0.95, respectively. Suppose that n = 5 bits are transmitted and the amount of distortion of each bit is independent. 1. Find the probability that one bit has high distortion, one has moderate distortion, and three bits have low distortion. 2. Find the probability that four bits have low distortion.

Two Continuous Random Variables w..we"~.a.'w*e,~"," s ~ m * B a A . " " A " ~ ~ ~ * ~ " " ~ e w a .~ B" ~Bv * ~ , ' ~ . ~ " m ~ ~ a . " . ~ . - v e

inuous random variables X and Y by ;sing corresponding probability distributions. Calculate the mean and variance of X (or Y) a continuous random variable by using the corresponding marginal probability distribution. C1 Calculate the conditional mean and conditional variance of X given Y = y (or Y given X = x) by using the corresponding conditional probability distribution. C1 Assess the independence of X and Y.

5-8

Chapter 5

Joint Probability Density Function

Joint Probability Distributions The joint probability density function (p.d.f.) of two continuous random variables X and Y satisfies the following: (1) f m ( x 7 Y ) 2 0

(2)

Marginal Probability Density Function

[f,(~~~)mQ=l

The marginal p.d.f.'s of X and Y with the joint p.d.f. f , ( x ,y ) are

fx(x)= L f x r ( x , Y ) ~

The marginal p.d.f. of X satisfies the following: (1) fx(x) 2 0


E ( X ) = P x = [xfx ( x )A

Conditional Probability Distribution

The conditional p.d.f. of X given Y = y7 denoted as f x , ( x ) ,with joint p.d.f.

f m ( x 7 Y >is fXiY

(XI=

fxr (x7 Y ) fY ( Y )

The conditional p.d.f. f,,, ( x ) satisfies the following: (1)

fXlY

( x )2 0

The conditional mean and conditional variance of X given Y = y are

E ( X I Y ) = Pxly = [ x f x l y( x )A

Independence

Two continuous random variables X and Yare independent if any of the following is true: ( 1 ) f X l Y ( X I = f x (XI

( 2 ) fYiX(Y)= f y ( Y ) (3)

f , ( 4 Y )= f x ( x ) f y ( Y )

5-3 Two Continuous Random Variables

ff

5-9

The sizes of college books are under study in terms of width (X)and height (Y). 1. (Continuous Joint p.d.f.) Suppose that the joint p.d.f of Xand Y is modeled as fxr(x,y)=cxy, 5 < x < 9 and x + 1 0 and y > 0. Next, for the second condition,

Therefore, the joint p.d.f. of X and Y is fxr(x,y)=&xy, 5 < x < 9 and x + l < y < x + 3 2. (Marginal Mean and Variance; Continuous Marginal p.d.f.) Determine the marginal probability distribution of X. Find also the mean and variance of X. h The marginal p.d.f. of X is

= L x . 2l [(X + 3)2 - (X+ 112] 515

=&(x+2)x,

I


5<x - k x y --f x (x) &x(x + 2) 2(x + 2) - f y ( y )

The marginal p.d.f. of Y is 515

By the way, the range of Xvaries along y (see the graph on next page) as follows: 5<x 0, it can be concluded that ergonomics grade ( X ) and statistics grade (Y) have a positive linear relationship. In Exercise 5-1, the joint p.m.f. ofX(the number of defects on the front side of a wooden panel) and Y (the number of defects on the rear side of the panel) is f x r ( x , y ) = $ ( x + y ) , x = 1 , 2 , 3 and y = 1 , 2 , 3 The means and variances of X and Yare

I

E(X) = E(Y) = 2.17 and V(X) = V(Y) = 0.80' Determine the covariance and correlation of X and Y.

5-6 Bivariate Normal Distribution

5-6

5-15

Bivariate Normal Distribution

0 Explain the joint probability distribution of bivariate normal random variables. Determine the joint, marginal, and conditional probabilities of bivariate normal random variables. -jimn*l(*^.^.wm

xmsw

ma

Bivariate Normal Random Variables

rs-w#

--*

M*mrut

a

.n-

-

mrxsariarxarva

MM+B

-

a\xnna - #

.a urw*awm in*a uw -

---

-nannnenana

xlxr n r a m aa x

;."*ware

xa*

The joint probability density function of two normal random variables Xand Y 2 2 with means p~and py, variances ox and o y , and correlation p, (-1 < pxu < 1) is

f,

(x7 Y)=

1 2noxoy~l-p,

Bivariate normal distributions with different values of pxy are shown in Figure 52. The contour plots of the binomial curves are elliptical. The center of each ellipse is located at the point (px, py) in the xy plane. If p, > 0 (pxy< O), the major axis of each ellipse is positive (negative); if pxr = 0, the ellipses become circles.

Probability density function

(a)

(b)

(c)

Figure 5-2 Bivariate normal distributions with different values of p ~(a): 0 < pxr< 1; (b) pxr=O; (c) -1 < p,yy< 0. The marginal probability distributions of X and Yare normal with means px and p y and variances ox2 and o y2 ,respectively. The conditional probability distribution of Y given X = x is normal with mean and variance E(Yx)=p, + p , ~ ( x - p x ) ox

2

and V(Ylx)=o,(l-pH)

m

5-16

r

Chapter 5

Example 5.5


(Bivariate Normal Distribution) The heights (X; unit: in.) and weights (Y; unit: lbs.) of adult males in the US are normally distributed with means = 68.3 and p y = 162.8, variances = 2.3' and or2 = 23.02, and correlation pm = 0.49. Determine the probability that the weight (Y) of an adult male is between 160 and 180 Ibs given that his height (3 is 70 in. D-rr The conditional probability density function of Y given X = 70 is normal with mean and variance

0x2

I


Therefore,

Let Xand Yrepresent two dimensions of an injection molded part. Suppose that X and Y have a binomial distribution with means k = 3.00 and py = 7.70, and = 0.04' and or2 = 0.08~.Assume that Xand Yare independent, i.e., variances p ~ 0.=Determine P(2.95 < X < 3.05, 7.60 < Y < 7.80).

02

5-7

Linear Combinations of Random Variables

0 Explain the term linear combination of random variables. 0 Determine the mean and variance of a linear combination of random variables. Linear Combination

A random variable Y is sometimes defined by a linear combination of several random variables X I ,X2,. ..,Xp: Y = c,XI + c2X 2 + + cpX p , where ti's are constants - a .

5-7 Linear Combinations of Random Variables

Rules for Linear Combination

5-17

The following rules are useful to determine the mean and variance of a linear combination of X and Y: 1. Rules for Means (1) If b is a constant, then E(b) = 0 . (2) If a is a constant, then E(aX) = aE(X) . (3) E(aX fby) = aE(X) f bE(Y) 2. Rules for Variances (1) V(b) = 0 . (2) V(aX) = a 2 v ( x ) . = a V(X) + b2V(Y), if X and Y are independent.

(Note) cov(X, Y) = o,

= p,oxoy

The rules 1.3 and 2.3 can be extended for Y = c, X1 + c2X2+ .-.+ c p X p as follows: E(Y) = c l E ( X l ) + c2E(X2)+

r

= ~ ? V ( X ,+ ) c;v(x~)

Example 5.6

cpE(Xp)

+ + c i V(Xp) , i f z s are independent

(Linear Combination) For adult males in the US, the length of the shoulderelbow (X; unit: in.) and the length of the elbow-fingertip (Y;unit: in.) have a = bivariate normal distribution with means px= 14.4 and y y = 18.9, variances or2 = 0.g2, and correlation ,p = 0.737. Determine the mean and variance of the length of the shoulder-fingertip (X+ Y). D-rr The mean and variance of the length of the shoulder-fingertip Z = X + Y is E(X + Y) = E(X) + E(Y) = 14.4 + 18.9 = 33.3

0x2

Note that cov(X, Y) = p,oxoy

= 0.737 x 0.8 x 0.8 = 0.472 . Thus,

V(X + Y) = 0 . 8 ~+ 0 . 8 ~+ 2 x 0.472 = 1 . 4 9 ~ I v

i

1


The width of a casing (X; unit: in.) and the width of a door (Y; unit: in.) are normally distributed with means px = 24 and y, = 23 +,and standard

I deviations ox =

and o, = &, respectively. Assume that the width of the casing (X) and the width of the door (X)are independent. Determine the mean and standard deviation of the difference between the width of the casing (X) and the width of the door (Y).

5-18

Chapter 5

5-9

Moment Generating Functions


R Explain the term moment generatingfunction. Determine the moment generating function of a random variable X and the mth moments of X. R Find the mean and variance of X by using the first and second moments of X. Moment Generating Function

The moment generating function of a random variable X (denoted as M(t)) is the expected value of elx, i.e., ~ ( t=)E(etX) =

Ir

elxf (x),

if X is a discrete random variable

"

etxf (x)A, if X is a continuous random variable

The moment generating function of X is unique if it exists and completely determines the probability distribution of X. Thus, if two random variables have the same moment generating function, they have the same probability distribution. If ~ ( ~ ) (denotes t ) the mth derivative of M(t), the rnth moment of X about the origin (zero) (denoted by E(X")) is xmf (x), if X is a discrete random variable E(x")

= M(")(o) =

(r "

xmf(x)dx, if X is a continuous random variable

(Derivation) E ( r ) = M(") (0) The mth derivative of M(t) is xmetxf (x), ~ ( ~ )= ( t ) ' dt"

If

[[


xmetxf (x)dr, if X is a continuous random variable

By setting t = 0,

M (") (0) =

Application of Moments

xmf (x),


xrn (x)A, if X is a continuous random variable

The mean and variance ofXcan be determined by using the first and second moments of X: p = E(X) = M'(0)

o2= E(x*)

- [E(x)]~ = M"(0) - [M'(0)12

5-9 Moments Generating Functions

Example 5.7

1

5-19

Suppose that X has an exponential distribution f(x)=~e-*.x>o 1. (Moment Generating Function) Find the moment generating function of X. h From the definition of moment generating function,

--

'

fort-h w . " ~ ~ " ~ % % - - - * ~ w w * - - - . ~ , ~ ~ ~ v ~ d - ~

Population vs. Sample

-

~

w

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-

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m

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~ . B ~ w " ~ ~ - . ~ - ' ~ " v a ~ - w ",~-B,,wuB"w-""w*, * * -~ . ~ ~ . ~ " #~,--wm""*-,-'*-" mm e. w~~ ,-~ ~w e a

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1. Population: The collection of all the elements of a universe of interest. The size of a population may be either (1) small, (2) large but finite, or (3) infinite. (e.g.) Population size (1) Students in a statistics class: Small (2) College students in the United States: Large but finite (3) College students in the world: Infinite 2. Sample: A set of elements taken from the population under study, i.e., a subset of the uouulation (see Figure 6-1). Since it is impossible and/or impractical to examine the entire population in most cases, a sample is often used for statistical inference about the population. Population

Sample

Figure 6-1 Population versus sample.

6-4

Chapter 6

Random Sample

Random Sampling and Data Description

A random sample refers to a sample selected from the population under study by

certain chance mechanism to avoid bias (over- or underestimation). Thus, selecting a random sample that properly represents the population is important for valid statistical inference about the population. A random sample of size n is denoted by random variables XI, X2,... ,Xn where X, represents the i'hobservation of the sample. The random sample XI, X2,... ,X, has the following characteristics: (I) The X ' s are independent of each other, and (2) Every has the same probability distribution f (x) . In other words, XI,X2,... ,Xn are independently and identically distributed (Lid.). Thus, the joint probability density function of XI, X2,. .. ,Xn is f x , ,,..., , ~ X, ( x I , x z , . . . , x ~ )(xl)f(x2)...f(xn) =~

Example 6.2

(Random Sample) A public opinion poll on contraception is conducted by calling people's homes, which are randomly selected, between 8 AM and 5 PM on weekdays. Discuss this study design. h Even if the telephone numbers are randomly chosen, the survey sample is biased because most working class people are not at home during the survey time period. Therefore, the survey result is most likely an over- or underestimate.

Exercise 6.2

Defects on sheet metal panels in three categories of sizes (small, medium, and large) are under study. The metal panels are made of the same material. The analyst randomly selects 20 large-size panels (not including small- and mediumsize panels) and draws a conclusion regarding the average number of defects/ft2. Discuss this study design.

Statistic

A statistic is a function of random variables X I ,X2,... ,Xn; therefore, a statistic is also a random variable. (e.g.) Statistic (I) Sample mean

X=

n

X, I n i=I

(2) Sample variance

s2=

n

(Xi - 2 ) ' /(n - 1) i=l

6-3

Stem-and-Leaf Diagrams

0 Explain the use of a stem-and-leaf diagram. 0 Construct a stem-and-leaf diagram to visualize a set of data. 0 Determine the 100kthpercentiles, quartiles, interquartile range, median, and mode of a set of data. 0 Compare the variability measures variance, range, and interquartile range with each other.

1 i iI 8

6-3 Stem-and-Leaf Diagrams

Stem-and-Leaf Diagram

6-5

A stem-and-leaf diagram is a graphical display that presents both the individual values and corresponding freauencv distribution simultaneously by using 'stems' and 'leaves.' The following procedures are applied to construct a stem-and-leaf diagram (see Figure 6-2): Step 1: Determine stems and leaves. The digits of each data is divided into two parts: (I) stem for the 'significant' digits (one or two digits in most cases) and (2) leaf for the 'less significant' digit (last digit usually). The analyst should exercise histher own discretion to determine which digits are most significant with consideration of the range of data. The number of stems between 5 and 20 is commonly used. A stem can be further subdivided as necessary. As an example, the stem 5 can be divided into two stems 5L and 5U for the leaves 0 to 4 and for the leaves 5 to 9, respectively. Step 2: Arrange the stems and leaves. The stems are arranged in ascending (or descending) order. Then, beside each stem, corresponding leaves are listed next to each other in a row. Step 3: Summarize the frequency of leaves for each stem.

Figure 6-2 A stem-and-leaf diagram. Example 6.3

(Stem-and-Leaf Diagram) In Example 6-1, construct a stem-and-leaf diagram of the test scores. Step 1: Determine stems and leaves. The test scores range from 65 to 98. Each test score consists of two digits and the first digit has higher significance. Thus, the first and last digits of a test score are defined as the stem and leaf of the data, respectively: 6 to 9 for stems and 0 to 9 for leaves. To better represent the distribution of the test scores, the stems are further split into 6U, 7L, 7U, 8L, 8U, 9L, and 9U where L is for the lower half leaves (0 to 4) and U is for the upper half leaves (5 to 9).

6-6


Chapter 6

Example 6.3 (cont.)

I

Exercise 6.3 Percentile @k)

Step 3: 6U 7L 7u 8L 8U 9L 9U

58 02234 559 1344 56788888 0224 6678

2 5 3 4 8 4 4

1 In Exercise 6-1, construct a stem-and-leaf diagram of the test results. The 1OOkthpercentile (0 < k I 1; denoted as pk) of a set of data means the 1OOkth largest value of the data set. The procedures to findpk from n observations, x, ,x,, ...,x, , are as follows: Step 1: Arrange data in ascending order, say x(,),x(,) ,...,x(,, .

.Ink3

Step 2: Calculate the position number r by using n and k

r=

if nisodd (n + l)k, if n is even Step 3: Determine p, based on r

Pk

=I

x(,, , if r is an integer x(lr) + (x(l,.l)- xIlr))(r- Lrb, if r is not an integer

(Note) The symbols [rl (called ceiling) and Lv] (called floor) denote 'roundup' and 'round-down' r to the nearest integer, respectively. Quartile (qk)

The following three quartiles divide a set of data into four equal parts in ascending order: (1) First (lower) quartile: q, = (2) Second quartile (median): 9, = p,,,, (3) Third (upper) quartile: 9, = p,,,,

Interquartile Range (IQR)

The interquartile range (denoted as IQR) of a set of data is the difference between the upper and lower quartiles of the dataset, i.e., *QR = 93 - ql = p . 7 5 - ~ 0 . 2 5

i

I

6-3 Stem-and-Leaf Diagrams

Comparison of Variability Measures

6-7

The variance, range, and interquartile range of a data set have different sensitivity (stability): variance is most sensitive (least stable) and interquartile is least sensitive (most stable). For example, Figure 6-3(a) shows that two different data sets with the same range can have different variances; and Figure 6-3(b) displays that two different data sets with the same interquartile range can have different ranges and variances. These illustrations imply that variance, range, and interquartile range have high (low), moderate, and low (high) sensitivities (stabilities), respectively.

(a)

(b)

Figure 6-3 Sensitivities of variance, range, and interquartile range: (a) R1 = R2, while 012# 0 z 2 ; (b) IQRl = IQR2, while R1 # R2 and ol2# 0 2 ~ . Of the three variability measures, the appropriate variability measure is selected depending on the analyst's judgment on which measure best represents the variability of the data.

Median

Mode

The median of a set of data is the value that divides its ranked data set into two equal halves, i.e., p,,,, = 9 , . The mode of a set of data is the value that is observed most frequently. A data set can have no mode, one mode (unimodal), or multiple modes (bimodal, trimodal, etc.), as illustrated in Figure 6-4.

Figure 6-4 Modes of data distributions: (a) no mode; (b) unimodal; (c) bimodal. Example 6.4

(Percentile, Quartile, Interquartile Range, Median, and Mode) In Example 61, determine the (1) 8othpercentile, (2) first quartile, (3) interquartile range, (4) median, and (5) mode of the test scores. Note that n = 30 and 9 , = 90.5.

6-8

Chapter 6


Example 6.4 (cont.)

h

( 1 ) Since n = 30 is even, r = ( n + l)k = 3 1 x 0.8 = 24.8, which is not an integer. Therefore,

PO8 = X(lrj)+ ( ~ ( r r ) x ( [ r)('~ -

1j

1'9

= X(124.8j) + ( ~ ( l 2 8) 4 - X(j24 81) )(24.8 - 124.8B

ii

= x ( ~+~( x) ( ~-~~ )( ~ ~ ) ) ( 2 24) 4 . 8= 92 + (92 - 92) X 0.8 = 92

(2) Since r = (n+ l)k = 3 1x0.25 = 7.75 is not an integer,

I

-Lrb

91 = PO25 = X([rl) + ( ~ d r )-'([r])

- ~ ( 1 77 5 l(7.75 ~ - L7.75b = x ( ~+, ( x ( ~-)~ ( ~ ) ) ( 7 . 775)= 74 + (75 - 74)X 0.75 = 74.75 = ~ ( 1 77

5 + ~ ( ~ ( r 751) 7

(3) IQR = q3 - q1 = pR7, - ~ 0 . = 2 90.5 ~ - 74.75 = 15.75 ( 4 ) Since r = (n+ l)k = 3 1x0.5 = 15.5 is not an integer, ~ 0 . 5=

+ (413-xirr1))(r-rr)

'(1.1)

- ~ ( [ l ~ I(l5.5 . ~ l ) - 115.5)

= X ( i l 5 . 5 n + ('(l15.5l)

= x ( , ~+) ( x ( ~-~~)( , ~ ) ) ( 1 5-.15) 5 = 85 + (86 - 8 5 ) 0.5 ~ = 85.5

( 5 )Mode = 88 (unimodal) In Exercise 6-1, determine the ( 1 ) 9othpercentile, (2) third quartile, (3) interquartile range, ( 4 )median, and ( 5 ) mode of the test results. Note that n = 70 and q, = 1,097.75.

Exercise 6.4

6-4

Frequency Distributions and Histograms

0 Explain the termsfrequency, cumulativefrequency, relativefrequency, and relative cumulative frequency.

0 Construct a frequency table of a set of data. 0 Construct a bar graph and histogram to visualize a set of data.

O Explain the term skewness and describe the relationships between the central tendency measures mean, median, and mode depending on the skewness of data. wi--*-~Y".~Ln

-a*

--n_i

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A frequency indicates the number of observations that belong to a bin (class interval, cell, or category). The frequency of the fh bin is denoted by .

The sum of the frequencies of bins is called cumulative frequency (denoted by

6. ):

=z i

4

f,,

j=1

i = 1,2, ..., k

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u

n

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r

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I

6-4 Frequency Distributions and Histograms

Relative Frequency

6-9

A relative frequency is the ratio of a frequency to the total number of observations (n). The relative frequency of the i~ bin (denoted by pi) is p . =-A , n = total number of observations n

The sum of the relative frequencies of bins is called cumulative relative frequency (denoted by P, ):

r

Example 6.5

(Frequency Measures) In Example 6-1, prepare a frequency table (including frequency, relative frequency, and cumulative relative frequency) of the test scores with five class intervals: below 60,60 to 69, 70 to 79, 80 to 89, and 90 or above. [kr

Class Interval ( X )

d

Frequency

Relative Frequency

Cumulative Relative Frequency

Exercise 6.5

In Exercise 6-1, prepare a frequency table (including frequency, relative frequency, and cumulative relative frequency) of the failure test results with five class intervals: below 500, 500 to 999, 1,000 to 1,499, 1,500 to 1,999, and 2,000 or above.

Bar Graph

A bar graph is a graphical display with bars, the length of each bar being proportional to the frequency andlor relative frequency of the corresponding category. A bar graph is used to provide a visual comparison of the frequencies and/or relative frequencies of multiple categories.

Histogram

A histogram is a bar graph in which the length and width of each bar are proportional to the frequency (or relative frequency) and size of the corresponding bin, respectively. The shape of a histogram of a small set of data may vary significantly as the number of bins and corresponding bin width change. As the size of a data set becomes large (say, 75 or above), the shape of the histogram becomes stable. Different histograms can be constructed for the same data set depending on the analyst's judgment. The following procedures may be of use to construct a histogram with bins of an equal width for a data set with size n: Step 1: Determine the number of bins (k). The following formula can be used for k: kG& I (Note) The number of bins between 5 and 20 is used in most cases.

6-10

Chapter 6


Histogram (cont.)

Step 2: Determine the width of a bin (w). The bin width is selected based on the range of the data set and the number of bins: R max- min W>-= k k Step 3: Determine the limits of the histogram. The lower limit (I) of the first bin and the upper limit (u) of the last bin are slightly lower than the minimum and larger than the maximum, respectively, i.e.,

Step 4: Prepare a frequency table based on the bins determined. Count the number of observations that belong to each bin and calculate relative frequencies accordingly. Step 5: Develop a histogram. Construct a bar graph by using the frequencies (frequency histogram) and/or relative frequencies (relative frequency histogram) of the bins.

(Bar GraphEIistogram) In Example 6-1, construct a histogram of the test scores by using the following summary quantities: n = 3 0. minimum = 65. and maximum = 98 h Step 1: Determine the number of bins (k). = 5.5, k = 6. Since & =

I

Step 2: Determine the width of a bin (w). R max- min - 98 - 65 - 5.5 , w = 6. Since - = k k 6

I

Step 3: Determine the limits of the histogram.

6-4 Frequency Distributions and Histograms Step 5: Develop a histogram.

Example 6.6 (cont.)

I Exercise 6.6

Skewness

6-11

Test Scores

In Exercise 6-1, construct a histogram of the test results by using the following summary quantities: n = 70, minimum = 375, and maximum = 2,265 The skewness (denoted by y) of a set of data indicates the degree of asymmetry of the data set around the mean. As illustrated in Figure 6-5, the distribution with a long tail to the left and distribution with a long tail to the right have negative and positive values of skewness, respectively; the symmetric curve has a zero skewness.

Figure 6-5 Distributions with different values of skewness (y): (a) y < 0 (leftskewed); (b) y =O (symmetric); (c) y > 0 (right-skewed). Comparison of Central Tendency Measures

The mean, median, and mode of a data set can be different depending on the skewness of the data distribution, as illustrated in Figure 6-6. Both the mean and mode shift toward the direction of skewness with high sensitivity, whereas the median moves with low sensitivity. Of the three central tendency measures, the appropriate measure is selected depending on the analyst's judgment on which measure best suits the purpose of analysis: mean for arithmetic average, median for relative standing, and mode for frequency.

6-12

Chapter 6


Comparison of Central Tendency Measures

f(x)

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(x)

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iA iA -

(cont.)

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n

--

+.

x"

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=x

X

---u

z

X

x

mean < median < mode mean = mode = median mode < median < mean (a) (b) (c) Figure 6-6 Comparisons of means, medians, and modes of various distributions: (a) left-skewed; (b) symmetric; (c) right-skewed.

6-5

BOXPlots

Explain the use of a box plot.

O Construct a box plot to visualize a set of data. _"-

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IIII*mYeV

*-.XX

Box Plot

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A box plot is a graphical display that presents the center, dispersion, symmetry, and outliers of a set of data simultaneously. Note that outliers are observations that are unusually low or high compared to most of the data. A box plot consists of three components (see Figure 6-7): (1) Box: The left edge, middle line, and right edge of the box indicate the first (q,), second (q2),and third (q3) quartiles, respectively; therefore, the length of the box indicates the interquartile range (lQR) of the data set. (2) Whiskers: The left whisker extends from the left edge of the box to the smallest data point within 1.5 1QR from ql; and the right whisker extends from the right edge of the box to the largest data point within 1.5 IQR from 93. (3) Outliers: Points beyond the whiskers are outliers. outliers

4.l

1.5IQR

92

IQR

q3

outlier

& 1.SIQR

Figure 6-7 A box plot. Example 6.7

(Box Plot) In Example 6-1, construct a box plot of the test scores by using the following summary quantities: minimum = 65, maximum = 98, q, = 74.75, q2 = 85.5, q3 = 90.5, and IQR = 15.75

'i

1 I

i

-

-

6-13

6-6 Time Sequence Plots

uv

Since the minimum = 65 is greater than ql - 1.5 x IQR = 5 1.1 and the maximum = 98 is less than q 3 + 1.5 x IQR = 1 14.1, the left and right whiskers extend to 65 and 98, respectively. Thus, there are no outliers in the test scores which exceed 1.5 IQR from the edges of the box.

IIII

IIIIIIIII

60

70

111111111l111l

IIII

IIII

90

80

100

Test Scores Exercise 6.7

6-6

In Exercise 6-1, construct a box plot of the test results by using the following summary quantities: minimum = 375, maximum = 2,265, q l = 1,097.75, q2 = 1,436.5, q3 = 1,735, and lQR = 637.25

Time Sequence Plots

O Explain the use of a time sequence plot. I

-,-=

i

Time Sequence Plot

A time sequence plot is a graphical display that presents a time sequence (time series; data recorded in the order of time) to show the trend of the data over time. As an example, Figure 6-8 shows two time sequence plots of a company's sales of three years by year and by quarter. Each horizontal axis denotes time (years or quarters; t) and each vertical axis denotes sales (in year or quarter; x). Figure 68(a) displays the annual sales are increasing over the years and Figure 6-8(b) indicates that the annual sales have a cyclic variation by quarter-the first- and second-quarter sales are higher than the sales of the other quarters.

sales

sales

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(a) (b) Figure 6-8 Time sequence plots of a company' sales: (a) annual sales; (b) quarterly sales.

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6-14

Chapter 6

6-7

Probability Plots


0 Explain the use of a probability plot. -"9".wa*w,.--a-~eeM

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A probability plot is a graphical display that presents the cumulative relative frequencies of ordered data x(l),xcz), ... ,xc,, on the probability paper of a hypothetical distribution (such as normal, Weibull, and gamma distributions) to examine if the hypothesized distribution fits the data set. For example, Figure 6-9 shows a normal probability plot of 20 ordered observations xu) against corresponding cumulative relative frequencies (j- 0.5)ln. Statistical software is usually used to produce a probability plot. As the plotted points closely lie along a straight line (chosen subjectively), the adequacy of the hypothesized distribution is supported. The probability plot method is useful to examine if the data conforms to a hypothetical distribution, but lacks objectivity by relying on subjective judgment. A formal method to test the goodness-of-fit of a probability distribution is presented in Section 9-7.

60

70

80

90

100

Observation (xu))

Figure 6-9 A normal probability plot.

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MINITABApplications

MINITABApplications

I

Examples 6.1 and 6.4

1

(Descriptive Statistics)

(1) Choose File % New, click Minitab Project, and click OK. :he worksheet.

(3) Choose Stat % Basic Statistics % Display Descriptive Statistics. In

6-16

Chapter 6

Examples 6.2 and 6.4 (cont.)


I

I

I

Descriptive Statistics

80

82

84

86

88

95% Confdence Interval tor Slgma 74316

I

125444

95% Confidence Interval for Medlen

79 4574

Example 6.3

88 0000

(Stem-and-Leaf Diagram) (1) Choose Graph > Stem-and-Leaf. In Variables, select Score. In Increment, enter the difference between the smallest values of adiacent stems. Then click Ok.

Character Stem-and-Leaf Display

I

Stem-and-leaf of Score Leaf Unlt = 1.0

N

= 30

I

-

MINITAB

Example 6.5

I

Applications

6-17

(Frequency Measures) (1) Choose Graph > Histogram. In Graph variables, select Score. Then

(2) Under Type of Histogram, check Frequency or Cumulative Frequency. Under Type of Intervals, check Cutpoint. Under Definition of Intervals, check Midpointlcutpoint positions and enter the set of cut

Score

6-18

Chapter 6

Example 6.6


(Bar Graph/Histogram) ( 1) Choose Graph k Histogram. In Graph variables, select Score. Then

Jnder Type of Histogram, check Frequency or Cumulative Frequency. Under Type of Intervals, check Cutpoint. Under Definition of Intervals, check Midpointlcutpoint positions and enter the set of cut

64

70

76

82

Score

88

94

100

MINITAB Applications

Example 6.7

I (Box Plot)

Score

6-19

6-20

Chapter 6



I (Mean, Variance, and Range)

= 161,913.9= 402.4~

(3) Minimum = 375 (4) Maximum = 2,265 (5) R = 2,265 - 375 = 1,890 Exercise 6.2

1 (Random Sample) Since the same material is used for the sheet metal panels in various sizes, the number of defects per unit panel area will not be significantly different depending on the size of the panel. Thus, the random sample selected only from the large-size panels will properly represent the entire roup of the sheet metal panels to study the average number of defectslft .

B

Exercise 6.3

(Stem-and-Leaf Diagram) Step 1: Determine stems and leaves. The test results range from 375 to 2,265. For simplicity, round each data at the 10s place and use only the 1,000s and 100s digits for plot. Since the 1,000s digit is most meaningful, the 1,000s and 100s digits of a test result are defined as the stem and leaf of the data, respectively: 0 to 2 for stems and 0 to 9 for leaves. However, the number of the stems is only three, therefore, the stems are further divided into Of,Os,Oe, lz, It, If, Is, le,2z,and2t where 'z' is for the leaves 0 and 1, 't' is for 2 and 3, 'f' is for 4 and 5, 's' is for 6 and 7, and 'e' is for 8 and 9. Step 2:

0s Oe lz It 1f Is 1e 22 *B*vse.e*.,*",.a.-*b,+B.***,*

I

7 8888999 0000001111111 22233333333 44555555555 666666677 888888899999 011 23 '*

1 7 13 11 11 9 12 3

%--

Step 3: Summarize the frequency of leaves for each stem as shown above.


6-21

(Percentile, Quartile, Interquartile Range, Median, and Mode) (1) Since n = 70 is even, r = (n + l)k = 71 x 0.9 = 63.9, which is not an integer. Therefore, PO.^ = X(Lr]) + (~([rl) -~ ( l r i -

Exercise 6.4

L'D

- x(163,9) + (x([a.sl)- ~ ( 1 6 3 . I(63.9 9~ - l63.9D

+

= x ( ~ ~( )x

( ~-~~)( ~ ~ ) ) ( 6-363) . 9 = 1,890 + (1,910 - 1,890)X 0.9

= 1,908 (2) Since r = (n + l)k = 7 1x 0.75 = 53.25 is not an integer,

43 = P0.75 = x(1.h + ( ~ ( r r h-xILr1,

- LrD - ~ ( ~ 5 3 . 2+5(~(153.25) ~ - ~ ( 1 5 3 . 2 5l(53.25 ) - 153.258

= x ( ~+~( )x ( ~-~~)( ~ ~ ) ) ( 5 3-.53) 2 5 = 1,730 + (1,750 - 1,730)x 0.25 = 1,735 = 1,735 - 1,097.75 = 637.25 (3) IQR = q3 - 4, = (4) Since r = (n+l)k = 71x0.5 = 35.5 is not an integer, ~0.= 5 ~(1.j) + ( ~ ( l r-i ~ ( j r j)(' )

-

1'4

9

- X ( ~ a . 5 ) + (~([3i.il) - ~([3i,i~)(35-5 - 135.5D

+

= x ( ~ ~( )x

( ~-~~)( ~ ~ ) ) ( 3- 535) . 5= 1,421 + (1,452 - 1,421)x 0.5

= 1,436.5

(5) Modes = 1,102, 1,315, and 1,750 (trimodal) Exercise 6.5

1 (Frequency Measures)

Exercise 6.6

1 (Bar GraphIHistogram) Step 1: Determine the number of bins (k). Since & = f i = 8.4 , k = 9.

I

Step 2: Determine the width of a bin (w). R max- min 2,265 -375 =210,usew=220. Since - = k k 9

I

Step 3: Determine the limits of the histogram. . kw-R 9 x 220 - 1890 1G min- --- = 375 = 330 2 2 u=l+kw=330+9~220=2,310

6-22

Chapter 6

i



1

Step 4: Prepare a frequency table based on the bins determined. Cumulative Relative Class IntervaI ( X ) Frequency Relative Frequency Frequency

Step 5: Develop a histogram. -

T

I

02g

Test Results

Exercise 6.7

(Box Plot) Since the minimum = 375 is greater than ql - 1.5 x IQR = 141.9 and the maximum = 2,265 is less than 93 + 1.5 x IQR = 2,690.9, the left and right whiskers extend to the minimum and maximum, respectively. Thus, there are no outliers in the test results which exceed 1.5 IQR from the edges of the box.

I

Test Results (unit: hr)

7-4 Sampling Distributions 7-5 Sampling Distributions of Means Answers to Exercises

7- 1 Introduction 7-2 General Concepts of Point Estimation 7-3 Methods of Point Estimation

7-1

Introduction

8

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sm

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a

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a"

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O Describe the terms parameter, point estimator, and point estimate. O Identify two major areas of statistical inference. O Distinguish between point estimation and interval estimation. O Determine the point estimate of a parameter.

Parameter (0)

A parameter (denoted by 0) represents a characteristic of the vovulation under study. It is constant but unknown in most cases. (e.g.) Parameters Mean (p), variance (02),proportion (p),correlation coefficient (p), and regression coefficient (p)

Statistical Inference

Statistical inference refers to making decisions or drawing conclusions about a population by analyzing a sample from the population. Two major areas of statistical inference can be defined: 1. Parameter estimation: Estimates the value of 0. (e.g.) p. = 150 2. Hypothesis testing: Tests an assertion of 0. (e.g.) Ho: p = 150

Parameter Estimation

Parameter estimation is further divided into two areas: 1. Point estimation: Estimates the exact location of 0. (e.g.) p = 150 2. Interval estimation: Establishes an interval that includes the true value of 0 with a designated probability. (e.g.) 145 < p < 155

7-2

Chapter 7


A point estimator (denoted by 6 ) is a statistic used to estimate 0. It is a random variable because a statistic is a random variable.

Point Estima?r (0)

(e.g.) Point estimator of p: sample mean

X=

n

Xi l n

A single numerical value of 6 determined by a particular random sample is called a point estimate of 8 (denoted by 0).

m

Example 7.1

(Point Estimate) Suppose that the life length of an INFINITY light bulb (3 has a normal distribution with mean p and variance 02.A random sample of size n = 25 light bulbs was examined and the sum of the life lengths was 14,900 hours. light bulb. Estimate the mean life length (F) of an INFINITY

Exercise 7.1

The line width (X) of a tool used for semiconductor manufacturing is assumed to A random sample of size n be normally distributed with mean p and variance 02. = 20 tools was measured and the sum of the line width measurements was 13.2 micrometers (pm). Estimate the mean line width (p) of the tool.

General Concepts of Point Estimation

7-2 w

---

'Bw*

0 Explain the terms unbiased estimator, minimum variance unbiased estimator (MVUE), standard error, and mean square error ( M S E ) ofan estimator.

0 Select the appropriate point estimator of 8 in terms of unbiasedness, minimum variance, and

-

-_

minimum mean square error each. w*m"

I*

JXnXr

a_nlsje*(

Unbiased Estimator

w_(_aV*._^_--Pmw_'*&*~--B~Pm----e-

---

An unbiased estimator is a point estimator ( 6 ) whose expected value is equal to the true value of 8, i.e., ~ ( 6=)8 . Note that several unbiased estimators can be defined for a single parameter 8. As shown in Figure 7-1, the bias of 6 from the true value of 8: bias = ~ ( 6-)8

6 is the difference of the expected value of

7-2 General Concepts of Point Estimation

7-3

Unbiased Estimator (cont.)

Figure 7-1 Bias of a point estimator

6.

(Unbiased Estimator) Let X I , X2,... ,Xn denote a random sample of size n from a probability distribution with E(X) = p and V(X)= 02.Show if the sample mean

1 is an unbiased estimator of the population mean p. I Exercise 7.2 (MR 7-21

I

Since E(X) = p , X is an unbiased estimator of p. Let X I ,X2,. .. ,X7 denote a random sample from a population having mean y and variance d.Show if the following estimator

I is an unbiased estimator of p. Minimum Variance Unbiased Estimator (MVUE)

A minimum variance unbiased estimator (MVUE) of 0 is the unbiased 6 with the smallest variance (see Figure 7-2). By using the MVUE of 0, the unknown parameter 0 can be estimated accuratelv and precisely.

7-2 Variances of unbiased estimators of 0. Figure -

Example 7.3

(Minimum Variance Unbiased Estimator) Let XI,X2,. .. ,Xn denote a random sample of size n (>I) from a population with E(X) = p and V(X)= 0'. Both X , and X are unbiased estimators of p because their expected values are equal to p. Of the two estimators, which is preferred to estimate p and why?

7-4

1 1 Chapter 7


I

I i

Example 7.3 (cont.)

o-rr The variances of X,and

X

are

V(X, ) = o2

Since V(X,) = o2> ~ ( 2=)o21n for n > 1, with higher accuracy and precision.

c$

X

is preferred to estimate p

Exercise 7.3

Let X I ,X2,. .. ,Xn denote a random sample of size n (>3) from a population with E(X) = p and V(X) = 02.Both (X, + X 2 + X,) / 3 and X are unbiased estimators of p because their expected values are equal to p. Of the two estimators, which is preferred to estimate p and why?

Standard Error (061

The standard error of a point estimator (denoted by o6 ) is the standard deviation of a point estimator 6 . It can be used as a measure to indicate the precision of parameter estimation. If o6 includes unknown parameters that can be estimated, use of the estimates of the parameters in calculating o6 produces an estimated standard error (denoted by s6 or se(6)).

t'

Example 7.4

In Example 7-1, X has a normal distribution with mean p and variance o2and a random sample of size n = 25 was examined. 1. (Standard Error) Assuming o2= 402, determine the standard error of the n

sample mean ( X ). Note that

X=

XiI n - N(p, o21n) .

i=l

o 40 o- =-=-=8.0

0-rr

hrs.

"&,I5

2. (Estimated Standard Error) Suppose that o2is unknown and the sample variance sX2= 352. Calculate the estimated standard error of the sample mean

(XI. h

6

S-=-=-=--

B

s,

35 - 7.0 hrs.

X & & J 2 5

Exercise 7.4

In Exercise 7-1, X i s normally distributed with mean p and variance o2and a random sample of size n = 20 was measured. 1. Assuming o2= 0.05~,determine the standard error of the sample mean ( 2). 2. Suppose that o2is unknown and the sample variance s$ = 0.04~.Calculate the estimated standard error of the sample mean ( 2 ).

7-2 General Concepts of Point Estimation

Mean Square Error (MSE) o f Estimator

The mean square error (MSE)of a point estimator difference between

6

6

7-5

is the expected squared

and 0:

M S E ( ~=) E ( B - 0)' = E[G - E ( B ) ] + ~ [e- E ( B ) ~ =

v (6)+ (bias)'

=

~ ( 6 for ) an unbiased 6

because bias = 0.

Note that a biased estimator of 0 with the smallest MSE is sometimes used for precise estimation of 0 while sacrificing the accuracy of estimation on 0 (see Figure 7-3).

Figure 7-3 A biased

r

unbiased

Example 7.5

6 , with a smaller mean square error than that of the

6 ,.

(Mean Square Error o f Estimator) Suppose that the means and variances of 6 , and 6 , are ~ ( 6=,0), ~ ( 6= 0.90, ~ )~ ( 6=, 5), and ~ ( 6 =' 4) , respectively. Which estimator is preferred to estimate 0 and why? F O ~6 , ,

biasgl = E ( O , ) - 0 = 0 - 0 = 0 M S E ( ~ ,=) ~ ( 6+, (biasgl ) )' = 5 - 0 = 5 For

62, biasg, = E ( 0 2 )- 0 = 0.90 - 0 = -0.10 M S E ( ~ ,= ) ~

( 6+ (bias6, ~ ) )2

=4

By subtracting M S E ( ~), from MS.(@

+ 0.010' ),

M S E ( ~ ,-) M S E ( ~ , = ) 5 - ( 4 + 0 . 0 1 0 ~=) 1 - 0.010' The preferred estimator of 0 for precise estimation depends on the range of 0 as follows: if 0 2 10 because M S E ( ~),I M S E ( ~ , )and 6 , is unbiased

62,

if 0 < 10 because M S E ( ~ ,>) M S E ( ~ , )

7-6

Chapter 7

Exercise 7.5


I

Let XI, X2,... ,X5 denote a random sample of size n = 5 from a population with

E(X) = 70 and V(X) = 52. Two estimators of p are proposed:

I Of thc two estimators, which is preferred to estimate p and why? 7-3

Methods of Point Estimation

0 Explain the utility of the maximum likelihood method.

0 Find a point estimator of 0 by using the maximum likelihood method. Maximum Likelihood Method

The method of maximum likelihood is used to derive a point estimator of 0. This method finds a maximum likelihood estimator of 9 which maximizes the likelihood function of a random sample Xl, X2, ... ,Xn L(9)= f(xl;0)f(x2;9)... f ( x n ; 9 ) , whereXl,Xz,..., Xn-i.i.d. f(x;0) (Maximum Likelihood Estimator) Let X I ,X2,... ,X,,denote a random sample of size n from an exponential distribution with the parameter h. Find the maximum likelihood estimator of h. U-W The probability density function of an exponential distribution is f (x; h) = he-h" Thus, the likelihood function of X I ,X2, ... ,Xn is

Then, the log likelihood function is n

l n ~ ( h ) = n i n h -h C x i

I

Now, the derivative of in L(h) is

By equating this derivative of In L(h) to zero, the point estimator of h which maximizes L(h) is

7-7

7-4 Sampling Distribution


7-4

I

Let XI, X2,... ,Xnrepresent a random sample of size n from a geometric distribution with parameterp. Find the maximum likelihood estimators ofp.

Sampling Distribution

Explain the term sampling distribution.

Sampling Distribution

A sampling distribution is the probabilitv distribution of a statistic (a function of random variables such as sample mean and sample variance). The sampling distribution of a statistic depends on the following: (1) The distribution of the population (2) The size of the sample (3) The method of sample selection (e.g.) Sampling distribution of a sample mean

7-5

Sampling Distributions of Means

Explain the central limit theorem (CLT). ~ e t e r m i n ethe distribution of a sample mean by applying the central limit theorem. IX-e_^el

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Sampling Distribution of

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Suppose that a random sample of size n is taken from a normal distribution with mean p and variance 02. Then, the sampling distribution of the sample mean is

n

(Derivation)

X= i=l

0

X i1 n - N(p,-)

n

,where X

- N(p,

02)

Since XI, X2,. .. ,X, are independent and normally distributed with the same E(X) = p and V ( X )= 0 2 , the distribution of X is normal with mean and variance

P

7-8

Chapter 7

Sampling Distribution of


(x, + x 2 n + . . . + X n v(X) = V

=-[v(x,)+

V ( X , ) + . - . + v(x,)]

Tr

A

(cont.)

Central Limit (CLT)

Let XI,X2,... ,Xn denote a random sample of size n taken from a population ( X ) with mean y and variance 02.Then, the limiting- form of the distribution of the sample mean X is n

o2

x=zXi/n-~(p,-) n i=l

x-p a Z=--

o/&

as n approaches M.This normal approximation of theorem (CLT).

i'J(0,l)

X

is called the central limit

As displayed in Figure 7-4, the distributions of the sample means from uniform, binomial, and exponential distributions become normal distributions as their sample sizes n become large. In most cases, if n 2 30, normal approximation of X will be satisfactory regardless of the distribution of X.In case that n < 30 but the distribution of X is close to the normal, normal approximation of X will be still held.

(a)

(b)

(c)

-

Figure 7-4 Sampling distribution of X : (a) X uniform; (b) X - binomial with p = 0.2; (c) X exponential with h = 1.

-

Based on the central limit theorem, the sampling distribution of normal or non-normal is as follows: 1. Case 1: Normal population, X N(p, o2)

-

2

where X is

7-5 Sampling Distributions of Means

7-9

2. Case 2: Non-normal population with p and o2 (1) Case 2.1: Normal approximation applicable

Central Limit Theorem (CLT)

n

2= i=l

(cant.)

o

XiI n - N(p,-)

n

, if n 2 30 or the distribution of X is close to

the normal (2) Case 2.2: Normal approximation inapplicable It would be difficult to find the distribution of 2 if n < 30 and the distribution of X is significantly deviated from the normal. In this case, use non-parametric statistics for statistical inference (see Chapter 15). Example 7.7

(Central Limit Theorem; Sampling Distribution of Sample Mean) Suppose that the waiting time (X; unit: min.) of a customer to pick up hislher prescription at a drug store follows an exponential distribution with E(X) = 20 min. and V(X) = 52. A random sample of size n = 40 customers is observed. What is the distribution of the sample mean?

I

o-

Exercise 7.7

Since n 2 30, normal approximation is applicable to X even i f X is exponentially distributed. Thus, the sampling distribution of X is

Suppose that the number of messages (X) arriving to a computer server follows a Poisson distribution with E(X) = 10 messages/hour and V(X) = 10. The number of messages per hour was recorded for n = 48 hours. What is the distribution of the sample mean?

1

7-10

Chapter 7



1 1. (Point Estimate)

Exercise 7.2

I (Unbiased Estimator)

1 Exercise 7.3

Since ~ ( 6=)p,

6 is an unbiased estimator of p.

(Minimum Variance Unbiased Estimator) The variances of ( X I + X 2 + X , ) I3 and

1 1

1 Exercise 7.4

are

o > v(X) = - for n > 3, X is preferred as n the point estimator of p for more accurate and precise estimation on p. Since

-

v

)-3-

, 1 . (Standard Error) 2. (Estimated Standard Error)

Exercise 7.5

X

(Mean Square Error of Estimator) For

6,,

bias. = E ( @ 1 ) - p = 7 0 - 7 0 = 0 01


Since M S E ( ~ , )= 5.0 < M S E ( ~ , )= 9.4, and precise estimation on p.

7-11

6, is selected for more accurate

(Maximum Likelihood Estimator) The probability mass function of a geometric distribution is

Exercise 7.6

f (x; P) = (1 - PIX-'P Thus, the likelihood fhnction of XI, X2,... ,Xn is

Then, the log likelihood function is

Now, the derivative of In L(p) is

By equating the derivative of In L(p) to zero, the point estimator o f p is

The maximum likelihood estimator o f p is the reciprocal of

X

(Central Limit Theorem; Sampling Distribution of Sample Mean)

Exercise 7.7

I

Since n = 48 2 30, normal approximation is applicable to X even if X follows a Poison distribution. Therefore, the approximate distribution of is

X

8


8- 1 Introduction 8-2 Confidence Interval on the Mean of a Normal Distribution, Variance Known 8-3 Confidence Interval on the Mean of a Normal Distribution, Variance Unknown 8-4 Confidence Interval on the Variance and Standard Deviation of a Normal Population

8-1

8-5 A Large-Sample Confidence Interval for a Population Proportion 8-6 A Prediction Interval for a Future Observation 8-7 Tolerance Intervals for a Normal Distribution Summary of Statistical Intervals for a Single Sample MINITABApplications Answers to Exercises

Introduction

0 Distinguish between confidence, prediction, and tolerance intervals. __&

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Statistical Intervals

While point estimation estimates the exact location of a parameter (8), interval estimation establishes bounds of plausible values for 8. Three types of statistical intervals are defined: 1. Confidence Interval (CI): Bounds a parameter of the population distribution. (e.g.) A 90% CI on p where X- N(p, 02)indicates that the CI contains the value of p with 90% confidence (see Section 8.2). 2. Prediction Interval (PI): Bounds a fbture observation. (e.g.) A 90% PI on a new observation where X- N(y, 02)indicates that the PI contains a new observation with 90% confidence (see Section 8.6). 3. Tolerance Interval (TI): Bounds a selected proportion of the population distribution. (e.g.) A 95% TI on X with 90% confidence indicates that the TI contains 95% ofxvalues with 90% confidence (see Section 8.7).

8-2

Chapter 8

8-2

Confidence Interval on the Mean of a Normal Distribution, Variance Known


0 Interpret a 100(1 - a)% confidence interval (CI).

Cl Explain the relationship between the length of a CI and the precision of estimation. Ll Determine a sample size to satisfy a predetermined level of error (E) in estimating p. 0 Identify the point estimator of y when o2is known and the sampling distribution of the point estimator. Ll Establish a lOO(1 - a)% CI on p when o2is known. _fVIwa_U_#m-w-w-.w*-

Confidence Interval

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A 100(1 - a)% confidence interval on a parameter ( 8 )has lower and/or upper bounds ( I I 0 I u), which are computed by using a sample from the population. Since different samples will produce different values of I and u, the lower- and upper-confidence limits are considered values of random variables L and U which satisfy the following: P(LlOIU)=l-a, O l a l l Since L and U are random variables, a CI is a random interval. A 100(1 - a)% CI indicates that, if CIS are established from an infinite number of random samples, 100(1 - a)% of the CIS will contain the true value of 8. The types of confidence intervals include: 1 . Two-Sided CI: Specifies both the lower- and upper-confidence limits of 8, such as 1I 0 I u. (e.g.) 730 I k 5 750 hrs, where X = life length of an INFINITY light bulb 2. One-Sided CI: Defines either the lower- or upper-confidence limit of 8. (e.g.) 730 I p~ or pxI 750 hrs, whereX= life length of an INFINITY light bulb

Length of C I and Precision of Estimation

The length of a C I refers to the distance between the upper and lower limits (u I). The wider the CI, the more confident we are that the interval actually contains the true value of 8 (see Figure 8-I), but less informed we are about the true value of 0.

X

710

I

730

(f95%CI--+I

750

770

I

(life length; hrs.)

Figure 8-1 Confidence intervals on the mean life length (8 = k)of an INFINITY light bulb with selected levels of confidence. The length of a CI on 8 is inversely related to the precision of estimation on 8: the wider the CI, the less precise the estimation of 8.

8-2 Confidence Interval on the Mean of a Normal Distribution, Variance Known

Sample Size Selection (cont.)

8-3

The distance of a point estimate of 0 from the true value of 0 represents an error (denoted by E) in parameter estimation: ~=10-01

=I

T - p 1 in estimating p

To establish a 100(1 - a)% CI on p which does not exceed a predefined level of E, the sample size is determined by the formula 2

(in case n is not an integer, round up the value)

Inference Context

Parameter of interest: y (3

X - N(p,-)

Point estimator of y:

n

x-p z z=--

, o2known N(071)

"I& Confidence Interval Formula

A 100(1 - a)% CI on p when 0,is known is -

-

(J

X-Z,,~- x ) = 0.1, 2

=XE.1,14

Z21.06

(2) P ( X ~< x 2 ) = 0 . 1 2 ~ - P ( x : ~ > x 2 ) = 0 . 9 ,

Inference Context

Parameter of interest:

G~

Point estimator of 02:

s2=

n -1

=Xi,9,14 =7.79

,X,,X2,..., Xn-i.i.d. ~ ( p , o ~ )

8-8

Chapter 8 Confidence Interval Formula

Statistical Intervals for a Single Sample (, - 1)s2

lo l

(n - 1)s2

,

X 1-a i 2

Xai2,v

(n - 1)s

, ~

I o2

2 Xai2,v

for two-sided CI for lower-confidence bound

(n-1)s2

o 1 2 X I-a/ 2,v

for upper-confidence bound

(Derivation) Two-sided confidence interval on o2 (n - 1)s2 By using the test statistic x2= X 2 (v) ,

-

Therefore,

L= Xa12,v

r

Note that, for a one-sided CI, use limit. Example 8.3

instead of

xil2 to derive the corresponding

(Confidence Interval on oZ;Two-Sided CI) Suppose that the life length of an INFINITY light bulb (X)follows a normal distribution with mean p and variance 02. A random sample of n = 16 bulbs is tested, and the sample variance is found to be s2= 44'. Construct a 95% two-sided confidence interval on the variance of the life length of an INFINITYlight bulb (0'). OT. 1 - a = 0 . 9 5 : a = 0 . 0 5 ; v = n - 1 = 16-1 = 1 5 CI on 02: 95% two-sided (n-1)s' (n-1)s' lo I 2 Xai2,v

XI-a12,v

8-5 A Large-Sample Confidence Interval for a Population Proportion


8-5

8-9

A rivet is to be inserted into a hole. A random sample of n = 15 holes is selected, resulting in the sample standard deviation of the hold diameter = 0.008 mm. Assume that the hole diameter is normally distributed. Construct a 99% upperconfidence bound on the variance of the hole diameter (02).

A Large-Sample Confidence Interval for a Population Proportion

Identify the point estimator o f p and the sampling distribution of the point estimator Establish a 100(1 - a)%CI onp.

Inference Context

Parameter of interest: p Point estimator ofp:

X n

p = -,where

X - B(n, p )

- N(0,l) Sampling Distribution of

if np and n(1- p ) > 5

The mean and variance of a binomial random variable X E(X) = np and V(X) = np(1- p )

- B(n, p ) are

i, Thus,

Recall that in Section 4-7 a binomial distribution B(n,p) approximates to a normal distribution i f p is not close to either zero or one and n is relatively large so that both np and n(1-p) are greater than five. Therefore, if np and n(1- p ) > 5, the approximate distribution of P is

8-10

Chapter 8

Confidence Interval Formula


A

P - z,,

,

for two-sided CI

Ip 750 hrs

Test Statistic

A test statistic refers to a statistic used for statistical inference about 0. (e.g.) Test statistic for inference on k w h e r e X ~ ( p , a and ~ ) a2is known:

Test Regions

Two regions of a test statistic (see Figure 9-1) are established for testing Ho against H I : 1. Acceptance region: The region of a test statistic that will lead to failure to E & C J Ho. 2. Rejection (critical) region: The region of a test statistic that will lead to reiection of Ho. Note that the boundaries between the acceptance and rejection regions are called critical values.

-

For example, in Figure 9-1, in testing Ho: px= 750 vs. HI: px# 750 where X denotes the life length of an INFINITY light bulb, the test statistic of p is the sample mean and the critical values of the sample mean are 730 and 770 hours. Therefore, we will conclude the following: (1) If 730 I T 5 770 (acceptance region), fail to reject Ho. (2) if Z < 730 or 2 > 770 (rejection region), reject Ho. Rejection region

b 730 770 Mean life length of light bulb ( X ) Figure 9-1 Acceptance and rejection rejections for Ho: px = 750 vs. HI: px + 750, where X (unit: hours) is the life length of an INFINITY light bulb. Test Errors

The truth or falsity of a hypothesis can never be known with certainty unless the entire population is examined accurately and thoroughly. Therefore, a hypothesis test based on a random sample may lead to one of the two types of wrong conclusions (see Table 9-1): (1) Type I error: Reject Ho when Ho is true. (2) Type I1 error: Fail to reject Howhen Hois false.

......_............... - ................__.............

m...

(~vrrr.)

~~

~-

U ~ ~ j J g I v n i

Hois true (HI is false)

Truth

Fail to reject HO Correct decision (Correct acceptance)

Reject Ho Type I error (False rejection)

The probabilib of type Ierror (denotedas a)and the probabi/ity of type 11 % \ \ % \ \ L % \ % * ~ \ \ \ \ % ~ \ < < \ % ~ L \ \ L \ L L < \ ~ \ \ \ < % \ s .

a = P(type1 enor) = P(xe1ect Ho\ HQ-IS tme)

p = P(type I1 error) = P(fai1 to reject HOI HOis false)

Note that the type I error probability (a) is also called the level of significance of a test. Small values of a such as 0.05 and 0.01 are typically used to make it difficult to reject Ho if it is true. Power of Test

The power of a statistical test indicates the probability of rejecting Ho when Ho is false, indicating the ability (sensitivity) of the test to detect evidence against Ho: power = P(reject HoI HOis false) = 1 - P(fai1 to reject Ho I Ho is false) = 1 - P

Test Regions,

The test regions, hypotheses, a , P, and power of a test are related to each other. The acceptance and reiection regions of a test statistic 6 are determined based on a and the hvuothesized value of 0 (denoted as 00) in Ho (note that a and 00 are G i f i e d by the analyst). If L and U denote the lower and upper limits of an acceptance region, respectively, and we are testing Ho: 0 = 80, the acceptance region of 6 is determined as follows: 1 - a = P(fai1 to reject Ho I Ho is true) = P(fai1 to reject Ho 1 0 = 80) [P(L S 6 I U 1 0 = go), for two- sided H ,

CG P' and

Power

= { ~ ( ~ ~ 6 ~ 0 = 0 , )for , lower- sided H , k ( 6 5 U I 8 = go),

for upper- sided H ,

On the other hand, the P and power of the test are determined based on the acceptance region of the test and the true value of 0 as follows: f3 = P(fai1 to reject Ho I Hois false) = P(fai1 to reject Ho( 0 # 80) [P(L 5 6 I u 1 0 f go), for two- sided H , = / ~ ( ~ < 6 ~ 0 # 0 , ) , for lower- sided H , [ ~ ( S6U I 0 + go), and power

=

1- p

for upper- sided H ,

9-3

9-1 Hypothesis Testing

Test Errors (cont.)

is truck

-Mw-

( ~ a l s eacceptancei mm-w-eM

w -e-

-m%ms

-

-*

(Correct * *rejection) - -d

a

s

m

The probability of type I error (denoted as a ) and the probability of type I1 error (denoted as P) are conditional probabilities as follows: a = P(type I error) = P(reject HoI Hois true) p = P(type I1 error) = P(fai1 to reject Ho I Ho is false) Note that the type I error probability ( a ) is also called the level of significance of a test. Small values of a such as 0.05 and 0.01 are typically used to make it difficult to reject Hoif it is true. Power of Test

The power of a statistical test indicates the probability of rejecting Howhen Hois false, indicating the ability (sensitivity) of the test to detect evidence against No: power = P(reject HoI Ho is false) = 1 - P(fai1 to reject HoI Hois false) = 1 - P

Test Regions,

The test regions, hypotheses, a , P, and power of a test are related to each other. The acceptance and reiection regions of a test statistic 6 are determined based on a and the hvpothesized value of 8 (denoted as 4)in Ho(note that a and Bo are specified by the analyst). If L and U denote the lower and upper limits of an acceptance region, respectively, and we are testing Ho:8 = €lo,the acceptance region of 6 is determined as follows: 1 - a = P(fai1 to reject HoI Hois true) = P(fail to reject HO1 8 = 80) [P(L i 6 5 u I 8 = go), for two- sided HI

P9 and Power

={p(~i6~8=e,),

for lower- sided HI

b(65 U I 8 = 8,),

for upper- sided H,

On the other hand, the P and vower of the test are determined based on the acceptance repion of the test and the true value of 8 as follows: p = P(fail to reject Ho1 Hois false) = P(fai1 to reject Ho 1 8 # 00)

I

P ( L i 6 5 ~ l 8 # 8 , ) , fortwo-sidedH,

= ~(~ 764.3

I

2. (p and Power of Test) Assume that the true value of px= 730 hrs. Find the and power of the test if the acceptance region is 735.7 I 2 I 764.3.

I

6 = P(fai1 to reject Ho 1 Hois false)

I

power = 1 - p = 0.782

h

=

P(735.7 I

X 5764.3 ( p = 730)

P

9-1 Hypothesis Testing

9-5


The heat evolved from a cement mixture (X;, unit: callg) is approximately normally distributed with o2= 22. We wish to test Ho:b = 100 versus HI: b # 100 with a sample of n = 9 specimens. 1. Find the acceptance region of the test at a = 0.01. 2. Suppose that the acceptance region is defined as 98.3 I T 5 101.7. Find the P and power of the test if the true mean heat evolved is 103.

Relationship between a and p

The relationships between a and P are explained with illustrations in Figure 9-2. under H ,

under H ,

under H ,

under H,

Hypothetical distributions of X N ( p , o21n) for Ho: 8 = 80andH1:8 # 80

-

1. As the acceptance region widens, a decreases but p increases, provided that the sample size n does not change.

-- -

under H,

under H ,

2. As n increases, both a and p decrease, provided that the critical values are held constant.

under H , under H,

3. When Ho: 0 = 8 0 is false, p increases as the true value of 8 approaches to 8 0 and vice versa.

Figure 9-2 Relationships between a and P.

9-6

Chapter 9

Tests of Hypotheses for a Single Sample

Strong vs. Weak Conclusion

The analyst can directly control both a (probability of rejecting Ho when HOis true) and (hypothetical value of €9, but not P (probability of accepting Ho when HOis false; p depends on the true value of 8). Thus, it is considered that "rejection of Ho"is a strong conclusion and "acceptance of Ho" is a weak conclusion. Consequently, rather than saying we "accept HO,"we prefer saying "fail to reject Ho,"which indicates that we have not found sufficient evidence to reject HO.

Hypothesis Testing Procedure

The formal procedure to make a statistical decision about hypotheses consists of the following four steps throughout the rest of the chapters: Step 1: State Ho and HI. Step 2: Determine a test statistic and its value. Step 3: Determine a critical value(s) for a. Step 4: Make a conclusion, such as we reject orfail to reject HOat a. (Note) The four-step procedure of hypothesis testing is basically same with the eight-step procedure of hypothesis testing in MR.

9-2

Tests on the Mean of a Normal Distribution, Variance Known

R Test a hypothesis on p when o2is known (z-test). R Calculate the P-value of a z-test. O Compare the a-value approach with the P-value approach in reporting hypothesis test results. R Explain the relationship between confidence interval estimation and hypothesis testing. O Determine the sample size of a z-test for statistical inference on y by applying an appropriate sample size formula and operating characteristic (OC) curve. O Explain the effect of the sample size n on the statistical significance and power of the test.

Inference Context

Parameter of interest: y

o

Point estimator of p:

X - N(y,-)

n

, 02known

-

- p N(0,l) Test statistic of p: Z = -

o/,L

Test Procedure (2-test)

Step 1: State Ho and H I .

Ho: ~ = p l j H I : y f po for two-sided test y > plj or p < po for one-sided test Step 2: Determine a test statistic and its value.

2, =-

?-PO

of&

i!

9-2 Tests on the Mean of a Normal Distribution, Variance Known

z-test (cont.)

9-7

Step 3: Determine a critical value(s) for a. z,,, for two-sided test; z, for one-sided test Step 4: Make a conclusion. Reject HOif for two-sided test Izo > za,

I

zo > z, zo < -z,

Significance Probability (P-value)

for upper-sided test for lower-sided test

The significabce probability (denoted as P) of a test statistic refers to the smallest level of significance (a) that would lead to rejection of Ho. Thus, the smaller the P-value, the higher the statistical significance (an inverse relationship between Pvalue and statistical significance). For example, the P-value of a test statistic zo can be computed by using the following formulas: 11 for two-sided test for upper-sided test for lower-sided test (Note) @(z) = P(Z 5 z) : cumulative distribution function of Z

a-value vs. P-value Approach

Two approaches are available to use in reporting the result of a hypothesis test: 1. a-value approach: States the test result at the value of a preselected. This reporting method lacks providing the weight of evidence against Ho. 2. P-value approach: Specifies how far the test statistic is from the critical value(s). Once the P-value is known, the decision maker can draw a conclusion at any specified level of significance (a) as follows: ifPIa Reject Ho at a , otherwise Fail to reject Hoat a , Note that the P-value approach is more flexible and informative than the a-value approach.

Interval Estimation and Hypothesis Test

A close relationship exists between a two-sided 100(1 - a)% confidence interval (CI) on 0 (1 5 0 I u) and the test of a null hypothesis about 0 (Ho: 8 = 80) at a as follows: if the two- sided CI does not include 8, Reject Ho at a , Fail to reject Ho at a ,

otherwise

For example, suppose that a 95% CI on p is 75 1 I p I 779 and we are testing Ho: p = 750 vs. HI: p # 750 at a = 0.05. Since the CI of p does not include the hypothesized value p = 750, we will fail to reject Hoat a = 0.05.

Sample Size Formula

Formulas are used to determine the sample size of a particular test for particular levels of p (or power = 1 - p), a , and 6 (= 10 - 001, difference between the true value of 0 and its hypothesized value). For a z- test on a single sample, the following formulas are applied:

9-8

Chapter 9


Sample Size Formula

n=

( ~ a i 2+ z p ) 2 0 2

,where 6 = p - p o for two-sided test

6

(cant.)

Note that the sample size requirement increases as a , P, and 6 decrease and o increases.

Operating Characteristic (OC) Curve

Operating characteristic (OC) curves in Appendix A of MR plot P against d (for and t-tests) or 3L (for X2- and F-tests) for various sample sizes n at a = 0.01 and 0.05, i.e., P =An, d or h, a ) By using an appropriate OC chart, an adequate sample size can be determined for a particular test to satisfy a predefined test condition.

Z-

Table 9-2 displays a list of OC charts in MR and a formula of the OC parameter d for a z-test on p. By using the table, the appropriate OC chart for a particular ztest is chosen (e.g., for a one-sided z-test at a = 0.05, chart VIc is selected).

Table 9-2 Operating Characteristic Charts for z-test - Single Sample Chart VI a Test OC parameter (Appendix A in MR)

Two-sided z-test One-sided

Effect of Sample Size

0.05 0.01 0.05

(c)

-. (7

(7

As the sample size n increases, both the statistical significance (inverse to Pvalue) and power (= 1 - P) of the test increase. For example, Table 9-3 presents P-values and powers of testing on p for the following conditions:

I:[

X - N p,-

and x = 5 0 . 5

Ho: px= 50 VS. H I :px+ 50 a = 0.05 and the truevalue o f p = 50.5 The P-value column indicates that, for the same value of i? = 50.5, Ho is rejected a , while fro i s ' h t rejected at a = 0.05 at a = 0.05 when n = 100 because P I when n 2 50 because P $ a.

9-2 Tests on the Mean of a Normal Distribution, Variance Known

tr

9-9

Statistical vs. Practical Significance

The statistical significance of a test does not necessarily indicate its practical significance. Sincie the power of a test increases as the sample size increases, any small departure of 9 from the hypothesized value O0 will be detected (in other words, Ho: 0 = O0 will be rejected) for a large sample, even when the departure is of little practical significance. Therefore, the analyst should check if the statistical test result has also practical significance.

Example 9.2

For the light bulb life length data in Example 8.1, the following results have been obtained: n=30, T=780,

0'

=402

95% two-sided CI on p: 765.7 I p I 794.3 1. (Hypothesis Test on k, o2Known; Two-Sided Test) Test Ho: p H I : p # 765 hrs at a = 0.05. Step 1: State Ho and H I . Ho: p = 765 HI: ~ $ 7 6 5

= 765

hrs vs.

Step 2: Determine a test statistic and its value. X - p o - 780 - 765 = 2.05 zo =-0/& 40/@ Step 3: Determine a critical value(s) for a. za12

= z0.0512 = '0.025 =

Step 4: Make a conclusion. Since lzol = 2.05 > z,,,,

= 1.96, reject Hoat a = 0.05.

2. (P-value Approach) Find the P-value for this two-sided z-test. h P = 2[1- @(I z0 I)] = 2[1- @(I 2.05 I)] = 2[1- 0.9801 = 0.04 Since P = 0.04 I a = 0.05, reject Hoat a = 0.05. 3. (Relationship Between CI and Hypothesis Test) Test Ho: p = 765 hrs vs. HI: y # 765 hrs at a = 0.05 based on the 95% two-sided CI on p. h Since the 95% two-sided CI on y, 765.7 I p 2794.3, does not include the hypothesized value 765 hrs, reject Hoat a = 0.05. 4. (Sample Size Determination) Determine the sample size n required for this two-sided z-test to detect the true mean as high as 785 hours with power of 0.9. Apply an appropriate sample size formula and OC curve. h (1) Sample Size Formula P = 0.1 power = P(reject HOI Hois false) = 1 - p = 0.9 6=y-yo ~ 7 8 5 - 7 6 5 ~ 2 0

9-10

Chapter 9


Example 9.2 (cont.)

(2) OC Curve To design a two-sided z-test at a = 0.05 for a single sample, OC Chart VIa is applicable with the parameter

By using d = 0.5 and P = 0.1, the required sample size is determined as n as displayed below. Note that this sample size is similar with the sample size n = 42 determined by using a sample size formula.

= 44

(a)

O.C. curves fur different values of n for the two-sided nomu4 test for a level of significance a = 0.05

I


I

For the hole diameter data in Exercise 8.1, the following results have been obtained: n = 10, ?=1.5045,and o2=0.012

1. Test the hypothesis that the true mean hole diameter is greater than 1.50 in. at a = 0.01. 2. Find the P-value for this one-sided z-test. 3. Determine the sample size n required for this one-sided z-test to detect the true mean as high as 1.505 inches with power of 0.9. Apply an appropriate sample size formula and OC curve.

9-3

Tests on the Mean of a Normal Distribution, Variance / Unknown

Test a hypothesis on y when o2is unknown (t-test). O Determine the sample size of a t-test for statistical inference on p by using an appropriate operating characteristic (OC) curve.

9-11

9-3 Tests on the Mean of a Normal Distribution, Variance Unknown

Inference Context

Parameter of interest: y Point estimator of p:

X - N(p,-)0 ,o2unknown n

,

X-p Test statistic of p: T = ----- - t ( v ) , v = n - 1

s/&

Test Procedure (t-test)

Step 1: State Ho and H I . Ho: y = p o HI: p # pg for two-sided test, p > po or y < yo for one-sided test. Step 2: Determine a test statistic and its value.

2-PO To = ------

s/&

- t (n - 1)

Step 3: Determine a critical value(s) for a. t,,,,,-, for two-sided test; t u n ,

for one-sided test

Step 4: Make a conclusion. Reject Ho if t o > t n 1 for two-sided test

1


to > t,,,-,

for upper-sided test

to < -t,,,-,

for lower-sided test

Table 9-4 displays a list of OC charts in MR and a formula of the OC parameter d for a t-test on p. By using the table, the appropriate OC chart for a particular t-test is selected (e.g., for a two-sided t-test at a = 0.01, chart VIfis selected). Table 9-4 Operating Characteristic Charts for 1-test - Single Sample

Test

Chart VI

a

OC parmeter

(Appendix A in b4R)

- ,

..

Two-sided t-test One-sided

r

*

--

-"-"----

-" * -------*

*-*---*-w

-

-

w w

- -- - ----

(Note) For 6 , use a sample standard deviation or subjective estimate. Example 9.3

For the life length data in Example 8.2, the following results have been obtained: n = 3 0 , Z=780,and s 2 =402 1. (Hypothesis Test on p, 0' Unknown; Two-Sided Test) Test Ho: p = 765 hrs vs. HI: 1-1 # 765 hrs at a = 0.05.

9-12

Chapter 9

Example 9.3 (cont.)


* Step 1: State HOand H I . Ho: P = 765 HI: ~ $ 7 6 5 Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. f a /2,n-1 - '0.05 /2,30-1 = t0.025.29= 2.045 Step 4: Make a conclusion. Since ltol = 2.05 >

= 2.045 ,reject Ho at

a = 0.05.

2. (Sample Size Determination) Determine the sample size n required for this two-sided t-test to detect the true mean as high as 785 hours with power of 0.9. Apply an appropriate OC curve. hr power = P(reject Ho / H o is false) = 1 - p = 0.9 3 p = 0.1 6=p-j.10 =785-765=20 To design a two-sided t-test at a = 0.05 for a single sample, OC Chart VIe is applicable with the parameter

By using d = 0.5 and P = 0.1, the required sample size is determined as n = 45 as displayed below.

d

(e)

O.C. curves for different values of n for the two-sided t-test for a level of siBnif~cancea = 0.05.

9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal Population 7

m

9-13

/


I

For the Izod impact test data of a PVC pipe product in Exercise 8.2, the following results have been obtained:

II

The ASTM standard requires that Izod impact strength must be greater than 1.0 ft-lblin. 1. Test if the Izod impact test results satisfy the ASTM standard at a = 0.05. 2. Determine the sample size n required for this one-sided t-test to detect the true mean as high as 1.10 with power of 0.8. Apply an appropriate OC curve.

I 9-4

Hypothesis Tests on the Variance and Standard Deviation of a Normal Population

Test a hypothesis on o2(x2-test). O Determine the sample size of a %'-test for statistical inference on o2by using an appropriate operating characteristic (OC) curve.

Inference Context

Test Procedure a2-test)

Parameter of interest: o2


s2= i=l


x

=

n-1 (n-1)s'

o

,Xl,X2,..., Xnwi.i.d. N(~,O~)

- X (v),v=n-1

Step 1: State Ho and H I . Ho: 02 = o2o HI: o2# 0; for two-sided test,

o2> 0; for upper-sided test o2< (3; for lower-sided test. Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. 2 2 2,n-1 and x ~ -2,n-l ~ , for two-sided test

xa; 2


x 2~ - ~ for , ~lower-sided -~ test

9-14

Chapter 9

Tests of Hypotheses for a Single Sample Step 4: Make a conclusion. Reject Hoif 2 > 2 or 2 < x 2 ~ - ~

X2-test (cont.)

xo x ~ ~ ~ xo, ~ - ~ ~ for ~ ,two-sided ~ - ~ test 2 for upper-sided test xo2 > x,,~-, 2 2 xo < x ~ - ~ ,for~ -lower-sided ~ test


Table 9-5 displays a list of OC charts in MR and a formula of the OC parameter h for a X2-teston 2.By using the table, the appropriate OC chart for a particular x2-test is selected (e.g., for an upper-sided x2-test at a = 0.05, chart VIk is selected).

Table 9-5 Operating " Characteristic Charts for r2-test ,* Chart VI Test a OC parameter (Appendix A in MR) 0.05 Two-sided 0.01

r

Example 9.4

For the light bulb life length data in Example 8.3, the following results have been obtained: n = 16, s2 = 442 95% two-sided CI on 02: 32S2 l o2 5 68. l 2

1 . (Hypothesis Test on 02; Two-Sided Test) Test Ho: o2= 402 vs. HI: o2f 402 at a = 0.05. ~l-rrStep 1: State Ho and H I . Ho: o2=402 HI: o2+402 Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. 2 X0.0512,16-1

2

= X0.025,15 = 28.65 and

Step 4: Make a conclusion. Since xi = 18.15 P = 18.15

I

rr xi,,,

F/I

2 XI-0.05/2,16-1

2

= X0.975,15 = 6.91

~ i =,28.65 ~ and ~ ~ , ~ ~ = 6.91, fail to reject Ho at a = 0.05.

2. (Relationship Between and Hypothesis Test) Test Ho: o2= 402 vs. HI: o2 # 402 at a = 0.05 based pn the 95% two-sided CI on 02.

9-5 Tests on a Population Proportion Example 9.4 (cont.)

n-

9-15

Since the 95% two-sided CI on 02,%2S2I o' 5 68. 1 2 ,includes the hypothesized value 402, fail to reject Ho dt a = 0.05.

3. (Sample Size Determination) Determine the sample size n required for this two-sided x2-test to detect the true standard deviation as high as 50 with power of 0.8. Apply an appropriate OC curve. h power = P(reject Ho I Ho is false) = 1 - B = 0.8 s /3 = 0.2

To design a two-sided X2-testat a = 0.05, OC Chart VIi is applicable with the parameter

By using h = 1.25 and /3 = 0.2, the required sample size is determined as n = 75 as displayed below.

1.25

A

(I) O.C.curves for d i i k m l values ofn f a the tvm-dddd chi-squaretest fa a lml of significancea = 0.05


9-5

For the hole diameter data in Exercise 8.3, the following results have been obtained: n = 15 and s2= 0.008~ I . Test if there is strong evidence to indicate that the standard deviation of the hole diameter exceeds 0.01 mm. Use a = 0.01. 2. Determine the sample size n required for this upper-sided x2-test to detect the true standard deviation mean which exceeds the hypothesized value by 50% with power of 0.9. Apply an appropriate OC curve.

Tests on a Population Proportion

0 Test a hypothesis o n p (2-test) for a large sample. 0 Determine the sample size for statistical inference o n p by using an appropriate sample size formula.

9-16

Chapter 9

Inference Context


Parameter of interest: p

-

X Point estimator ofp: P = -, where X - B(n, p ) n Test statistic ofp:

Test Procedure (2-test)

-

- N(0,l)

if np and n(1- p) > 5

Step 1 : State Ho and H I . Ho: p 'PO HI: p stpo for two-sided test, p > po or p < p~ for one-sided test. Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value@) for a. z,,, for two-sided test; z , for one-sided test Step 4: Make a conclusion. Reject HOif )zO> za,, for two-sided test

1

zo > z, z, < -z,

Sample Size Formula

for upper-sided test for lower-sided test

For a hypothesis test onp, the following formulas are applied to determine the sample size: L

for two-sided test L

for one-sided test

Y

Example 9.5

For the corroded bridge data in Example 8.4, the following results have been obtained: x 28 n = 4 0 and j = - = - = 0 . 7 n 40 1. (Hypothesis Test onp; Two-Sided Test) Test Ho:p = 0.5 vs. H I :p # 0.5 at a = 0.05. h Step 1: State Ho and H I . Ho: p = 0.5 HI: ~ $ 0 . 5

9-7 Testing for Goodness of Fit

Example 9.5 (cont.)

I

9-17

Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. za12

= z0.05/2 = '0.025 =

Step 4: Make a conclusion. Since lz,l= 2.53 > z,,,~, = 1.96, reject Hoat a = 0.05.

2. (Sample Size Determination) Determine the sample size n required for this two-sided z-test to detect the true proportion as high as 70% with power of 0.9. Apply an appropriate sample size formula. b power = P(reject Ho I Hois false) = 1 - p = 0.9 p = 0.1


For the circuit test data in Exercise 8.4, the following results have been obtained: x 13 n = 3 0 0 and p=-=-=0.043 n 300 A

1. Test if the fraction of defective units produced is than 0.05. Use a = 0.05. 2. Determine the sample size n required for this one-sided z-test to detect the true proportion as low as 2% with power of 0.8. Apply an appropriate sample size formula.

9-7

Testing for Goodness of Fit

0 Distinguish between nominal and ordinal variables. Ll Explain why the expected frequency of each class interval should be at least three in the goodnessof-fit test. bution. X I ( * * i i ~ _ n ~ q X I M & * l m ~ ~

XUT

Ch.mlii--e

9-18

Chapter 9

Categorical Variable

Inference Context

Tests of Hypotheses for a Single Sample A categorical variable is used to represent a set of categories. Two types of categorical variables are defined depending on the significance of the order of the category listing: meaningful. 1. Nominal variable: The order of listing of categories is (e.g.) gender (male and female) hand dominance (left-handed, right-handed, and ambidextrous) 2. Ordinal variable: The order of listing of categories is meaningful. (e.g.) education (< 9 years, 9 - 12 years, and > 12 years) symptom severity (none, mild, moderate, and severe) The underlying probabilitv distribution of the population is unknown. Thus, we wish to test if a particular distribution fits the population. (e.g.) Ho: X - Poisson distribution with 3L (discrete distribution) Ho: X ~ ( p02 , ) (continuous distribution)

-

Test Statistic

X2 =

c k

(oi

i=l

where: k = 0, = E, = p =

-'i12

- ~ ~ ( k - ~ - l )

Ei Number of class intervals (bins) Observed frequency of class interval i Expected frequency of class interval i Number of parameters of the hypothesized distribution that are estimated by sample statistics

As the observed frequencies are close to the corresponding expected frequencies, the statistic x2becomes small. Thus, x2is used to test the null hypothesis Ho: X follows a articular distribution. Table 9-6 can be used to calculate the test statistic

d

Table 9-6 Goodness-of-Fit Test Table

class 0be;erved btewals Frequency i

Qi

Probability

Expact4 Frequency

Pi

Bi(=npi)

Oi-Ei

(Caution) Minimum expected frequenc If an expected frequency is too small, X can be improperly large by a small departure of the corresponding observed frequency from the expected frequency. Although there is no consensus on the minimum value of an expected frequency, the value 3,4, or 5 is widely used as the minimum. To avoid this undesirable case, when an expected frequency is very small (say, < 3), the corresponding class interval should be combined with an adjacent class interval and the number of class intervals k is reduced by one.

Y


Test Procedure &'-test)

9-19

Step 1: State Ho and H I . Ho: X A particular distribution. HI: X -/- The hypothesized distribution.

-

Step 2: Determine a test statistic and its value. 1. Estimate the parameteds) of the hypothesized distribution if their values are not provided. 2. Define class intervals and summarize observed frequencies (Oi's) accordingly. 3. Estimate the probabilities (pi's) of the class intervals.

4. Calculate the expected frequencies (Ei= npi)of the class intervals. If an expected frequency of a class interval is too small (< 3), combine it to an adjacent class interval. Then, repeat steps 2.2 to 2.4. k

5. Calculate the test statistic X; = i=l

(Oi - E , ) ~

Ei

Step 3: Determine a critical value for a. 2 Xa,k-p-1

Step 4: Make a conclusion. Reject Ho if

(Caution) Upper-sided critical region becomes smaller as the hypothesized distribution fits Since the test statistic xo2 better, no lower limit is set as a critical value in the goodness-of-fit test. Example 9.6

(Goodness-of-Fit Test; Discrete Distribution) The number of e-mails per hour (3 coming to Ms. Young's e-mail account is assumed to follow a Poisson distribution. The following hourly e-mail arrival data are obtained during n = 100 hours:

**--

s-e-D

2 3

e*--w*w*ws-

---me

7 5

~ ~ ~ w B M ~ ~ - - = m M m w - ~ - ~ - - # a m

--~wm"wa-,"-mm~em"

Conduct a goodness-of-fit test at a = 0.05. Step 1: State Ho and H I . Ho: X Poisson distribution with h (Note) E(X) = V(X)= 3L Hl: X -/- Poisson distribution

-

9-20

Chapter 9

Example 9.6 (cont.)

Tests of Hypotheses for a Single Sample Step 2: Determine a test statistic and its value. 1. Estimate the parameter of the hypothesized distribution. Ox60+lx28+2x7+3x5 = E(X) = = 0.57 100 The number of parameters estimated is p = 1. 2. Define class intervals and summarize observed frequencies (0;'s)

3. Estimate the probabilities (pj's) of the class intervals.

4. Calculate the expected frequencies (El = np;) of the class intervals. If an expected frequency is too small (< 3), adjust the class intervals. Since the expected frequency of the last class interval in the above table is less than three, combine the last two cells as follows: Observed Expected Frequency Probability Frequency oi- Ei (Oi -E;) # e-mails per hour Oi Ei (= n j i )

Fi

3

5. Calculate the test statistic:

(0; - E ; ) ~

= 3.14

= i=l

Ei

Step 3: Determine a critical value for a. 2 Xa,k-pl

2 - 2 - X0.05,3-1-1 = X0.05,l

= 3.84

Step 4: Make a conclusion. = 3.14 2 Xi,05,, = 3.84, fail to reject Ho at a = 0.05. In Since other words, the number of e-mail arrivals per hour follows a Poisson distribution at a = 0.05.

9-21


4

Exercise 9.6 (MR9-59)

1 Consider the following frequency table (n = 100) of a random variable X

1

0

,y

1

2

3

4

I

(Goodness-of-Fit Test; Continuous Distribution) The final scores (X)of n = 40 students in a statistics class are summarized as follows:

Example 9.7

I

Bins in X

-

Frequency

The mean and variance of the scores are 82.2 and 13.3~, respectively. Test if a normal distribution fits the test scores at a = 0.05. Step 1: State Ho and H I .

Ho: X HI: X

I I

+-

~(82.2J3.3~) ~(82.2J3.3~)


1. Estimate the parameter of the hypothesized distribution. Since y and o2are known, skip this step and the number of parameters estimated is p = 0. 2. Define class intervals and summarize observed frequencies (Oi's) accordingly. Bins in Z Observed Expected Xi - p Frequency Probability Frequency Bins in X d Oi Pi Ei (= nPi

3. Estimate the probabilities (pi's) of the class intervals. pl = P ( X < 60) = P(Z < -1.67) = 0.05 p, =P(6OIX" ,I,I,",I,",I,I,I"',I,I,I,I.,I",I,I,I,I,I,I,I,I,I,I,I,I,I , I ",I,I,I,I,I,I,I"..,I,I,I,I,I,I,I",I, I

"11"/11

n sw"

,m,mwmvm*A.#

"l".

Step 3: Determine a critical value for a. 2 Xa,(r-l)(c-1)

2

2

= X0.05,(3-1)(3-1) = X0.05,4 = 9'49

Step 4: Make a conclusion. = 19.5> = 9.49 , reject HO at a = 0.05. ~t is Since concluded that grades in ergonomics (X) and grades in statistics (Y) are not independent at a = 0.05.

Xi,05,4

9-26

m

Chapter 9

4


4

: 1


II I

The failures of an electronic component are under study. There are four types of failures (Y) and two mounting positions (X) at the device. A summary of n = 134 failures are as follows:

Mounting Positionlq

a e * " *

ese

~

~

~

Front Back

w

~

m

~

~

~

~

a

~

~

A

Failure Type (Y) 3 C

D

22 4

46 17

9 12

-

s

.

~

w

~

v

~

m

8

~

.

~

~

A

~

18 6 ~

~

~

~

~

~

~

~

B

.

~

1

Test if the type of failure (Y) is independent of the mounting position (X)at a = 0.05.

I

(Contingency Table Test; Homogeneity) A random sample of n = 300 adults with different hand sizes (2')evaluates two mouse designs (Y). The evaluation results are summarized as follows: -

I

~

w

~

Mouse Designs (Y) Conventional New Hand Size Groups (X) a

eu

--

w-s--a-*-

Small Medium La%e mmm.w

a-wa

flmvam

35 20 30

----- -----

%**

--

m"-mw--asmmwmmm-..

65 80 70 -"-

--s--*-aw--"m-

Test if users in different hand-size groups (X) have homogeneous opinions on the mouse designs at a = 0.05. mStep 1: State Ho and HI. Ho: Users in different hand-size groups are homogeneous in terms of opinions on the mouse designs. HI: Users in different hand-size groups are not homogeneous in terms of opinions on the mouse designs. Step 2: Determine a test statistic and its value.

Mouse Designs (Y) Conventional New Hand Size Groups (X)

",.

Totals ,a-,.,-,B

Totals

Medium

~ e # w w ~ ~ - ~ ~ m - * eb*m-e - - ~ "

ewmm,e-mm--a

1

30 70 B 28.3 71.7 100 ,, Iggggggggggggggg ,,*-,,,,-,,," ., ., .., i 85 ".~"2 15 .,,.," ,. . 1 300 ,... .,, ,a1 ,-,-,,

Large .w.vm...-a~w-~ws..~.*a-~

j

mm-

w,smsm.-.*a*wB, gggg.gggggggggg.".gg"

w. 8A8*wfl

~

~ w*d,a*s*%m--,7

B*B.s,w,

,%*,m-.,**

*w .,- a.".,,


2

2

= X0.05,(3-1)(2-1) = X0.05,2 = 5.99

Step 4: Make a conclusion. Since = 5.7 > ~ i , = ~5.99 ~ , fail , ~to reject Hoat a = 0.05. It is concluded that users in different hand-size groups do not have significantly different opinions on the mouse designs at a = 0.05.

Xi

~

~

~

w

9-27

9-8 Contingency Table Tests


A random sample of n = 630 students in different class standings (X)is asked their opinions (Y) on a proposed change in core curriculum. The survey results are as follows:

Class Standing ( X ) e--s-"ms---

-mm-.-m'a-----~m'm

Sophomore Junior Senior

70 60 40

wwmswm----me--=-m

*

- -----

sw

-w*e

*

130 70 60

a mw

-**+

a"B

B a wa

Test if students in different class standings (X) have homogeneous opinions on the curriculum change at a = 0.05.

9-28

Chapter 9



-

Example 9.1

(Power of Test) (1) Choose Stat ) Power and Sample Size > 1-Sample Z. Click Calculate power for each sample size. Enter the sample size selected in Sample sizes, the difference between the true mean and hypothesized mean to be detected in Difference, and the population standard deviation in Sigma. Then click O~tions.

Jnder Alternative Hypothesis select the type of alternative hypothesis, and in Significance level enter the probability of type I error (a).Then click OK twice.

I

I

Testing mean = null (versus n o t = null) Calculating power f o r mean = null + 20 Alpha = 0.05 Sigma = 40

..................

Sample ,

....................i

3 0 ; 0.7819

9-29

MINITABApplications

Example 9.2

I

(Inference on p, c2Known) (1) Choose File % New, click Minitab Project, and click OK. ITY light bulbs on the worksheet.

(2) r---A--------m

r

INFINITY

1

-

(3) Choose Stat % Basic Statistics % 1-Sample Z. Click Test mean, enter the hypothesized mean life length, and select not equal (type of the alternative hypothesis) in Alternative. Enter the assumed population

2-Test Test of mu = 765.00 vs mu not = 765.00 The assumed sigma = 40.0 Varlable INFINITY

Mean 780.00

N 30

StDev 40.02

SE Mean 7.30

................................... i Z 2.05 ,..................

P /0.040 i -5.

...........

l . I I

"y"

,5,

.ii..

3:.:

9.2.1,2

-30

Chapter 9


(5) Choose Stat > Power and Sample Size % 1-Sample Z. Click Calculate sample size for each power value. Enter the power of the test predefined in Power values, the difference between the true mean and hypothesized mean to be detected in Difference, and the population standard deviation

Example 9.2 (cont.)

(6) Under Alternative Hypothesis select the type of alternative hypothesis, and in Significance level enter the probability of type I error (a).Then click OK twice.

1 I

(7) Obtain the s a m ~ l esize reauired to satisfv the medefined test condition.

I I

Power and Sample Size 1-Sample Z Test

?I

I

Testing mean = null (versus not = null) Calculating power for mean = null + 20 Alpha = 0.05 Sigma = 40

..................

isample i Target i Slze -z i 43 i 0.9000

..................

Actual

? -- p

0.9064

Y

9-31

MINITABApplications

Example 9.3

(Inference on p, cr2 Unknown) (I) Choose File New, click Minitab Project, and click OK.

*

(2) Enter the life length data of INFINITYlight bulbs on the worksheet.

*

*

2hoose Stat Basic Statistics 1-Sample t. Click Test mean, ente:r the hypothesized mean life length, and select not equal in Alternative. Then click OK.

T-Test of the Mean

I

T e s t of mu = 765.00 vs mu n o t = 765.00 Variable INFINITY

Qs

Mean 780.00

N

30

*

~dl

StDev 40.02

I

.................................

SE Mean

7.31

i T I2.05 0.049 ,................................

9.3.1

a

lu:

(5) Choose Stat Power and Sample Size k 1-Sample t. Click Calculate sample size for each power value. Enter the power of the test predefined in Power values, the difference between the true mean and hypothesized mean to be detected in Difference, and the sample standard deviation in

9-32

Chapter 9

Example 9.3 (cont.)


(6) Under Alternative Hypothesis select the type of alternative hypothesis, and in Significance level enter the probability of type I error (a).Then

Cancel

( 7 ) Ohtain the sample size reauired to satisfv the redefined test condition.

I I

Power and Sample Size l-Sample t T e s t

I.

*I

T e s t i n g mean = n u l l ( v e r s u s n o t = n u l l ) C a l c u l a t i n g power f o r mean = n u l l + 20 A l p h a = 0.05 Sigma = 40

I

.................. !Sample

i

i

i

Slze 44

i

.................

:

Target rower

Actual

0.9000

0.9000

ruwm

9.3.2 Y

Example 9.5

(Inference on p)

*

(1) Choose Stat Basic Statistics ) 1 Proportion. Click Summarized data and enter the number of bridges surveyed in Number of trials and the number of bridges corroded in Number of successes. Then click

9-33

MINITABApplications

Example 9.5 (cont.)

(2) Enter the level of confidence in Confidence level, the hypothesized proportion in Test proportion, and not equal in Alternative. Check Use twice.

I

I

I

Test and Confidence Interval for One Proportion Tesc o f p Sample

= 0.5 vs p

X 28

N 40

not

= 0.5

Sample p 0.700000

:.. ............................ 95.0 % C I j (0.557987, 0.842013);

2-Value

P-Value

2.53 0.011 ................................

9.5.1 --

(4) Choose Stat > Power and Sample Size > 1 Proportion. Click Calculate sample size for each power value. Enter the power of the test predefined in Power values, the true proportion to be detected in Alternative p, and

9-34

Chapter 9

Example 9.5 (cont.)



(61 Ohtain the samnle size reauired to satisfv the medefined test condition.

I I

1

Power and Sample Size

I

T e s t f o r One P r o p o r t i o n T e s t i n g proportion = 0 . 5 ( v e r s u s n o t = 0 . 5 ) C a l c u l a t i n g power f o r p r o p o r t i o n = 0 . 7 Alpha = 0 . 0 5 Difference = 0.2

................. Sample Size 62

Example 9.8

i i

Target

Actual

ruwrr

ruwrr

0.9000

0.9029

-

(Goodness-of-Fit Test; Independence) (1) Choose File

+ New, click Minitab Project, and click OK. CI-T - j --J-

I

Others

~2 A

r--T--d----

Others

21

3

MI

9.5.2

-

MINITABApplications

9-35

, ,

(3) Choose Stat Tables Chi-Square Test. In Columns containing the table, select A, B, and Others which contain the frequencies. Then click

Example 9.8 (cont.)

OK.

1

Chi-square Test

I

Expected counts are p n n t e d b e l o r observed counts

I Total

A

B

Others

12 5.46

5 6.72

4 8.82

26

32

42

Total 21

100

................................................................................................... !Chi-Sq = 7 . 8 3 4 + 0.440 + 2 . 6 3 4 + 0 . 3 2 1 + 1.244 + 0 . 2 7 9 + 2 . 4 4 5 + 0 . 6 2 1 + 3 . 6 7 8 = 19.496 ~ D F= 4 , P-Value = 0.0011 .................................................................................................. i

Example 9.9

'

I *!

(Goodness-of-Fit Test; Homogeneity) (1) Choose File > New, click Minitab Project, and click OK. (2) 1

on the mouse designs.

9-36

Chapter 9


*

Example 9.9 (cont.)

(3) Choose Stat % Tables Chi-Square Test. In Columns containing the table, select Conventional and New which contain the frequencies. Then click OK.

I 1

Chi-Square Test

I

I Expected 1

1

c o u n t s a r e p r i n t e d below observed c o u n t s

Conventi 35 28.33

New 65 71.67

Total 100

Total 85 2 15 300 ....................................................... !

1.569 + 0.620 2.451 + 0.969 0.098 0.039 ~ D F= 2 , P-Value = 0.057 I C h l - ~ q=

+

+

+ =

5.746

-

......................................................... $t

IB~,~

v: $9

31


9-37


1. (AcceptanceICritical Regions) The test statistic of p is the sample mean with the following sampling distribution:

The acceptance region I I 2 I u for Ho: px = 100 vs. HI: px # 100 satisfies the following: 1 - a = 1 - 0.01 = 0.99 = P(fai1 to reject Ho I p = 100) =~(11X1u1p=100)

Accordingly, the critical values are

Thus, acceptance region: 98.3 1F 1 101.7 rejection region: F < 98.3 and F > 101.7 2. (p and Power of Test) = P(fai1 to reject HO1 HOis false) = P(98.3 I2 1101.7 1 p = 103)

a

power = 1 -

= 0.973

9-38

Chapter 9

Exercise 9.2


1. (Hypothesis Test on y, o2Known; Upper-Sided Test) Step 1: State Ho and H I . Ho: p = 1.50 HI: p > 1.50 Step 2: Determine a test statistic and its value. X-po - 1.5045-1.50 = 1.42 zo =-CJ/& o.o1/4G Step 3: Determine a critical value(s) for a. z, = z,,,, = 2.33 Step 4: Make a conclusion. Since z, = 1.42 2 z,,,, = 2.33, fail to reject Ho at a = 0.01.

2. (P-value Approach)

P = 1- Q(z,) = 1 - Q(1.42) = 1 - 0.922 = 0.078 Since P = 0.078 5 a = 0.01, fail to reject Ho at a = 0.01. 3. (Sample Size Determination) (1) Sample Size Formula power = P(reject HoI Ho is false) = 1 - p = 0.9 6 = p - po= 1.50 - 1.505 = 0.005 n=

(z, +

~

P = 0.1

~() z ~~+. o z~~~~, ~ ) ~o(2.33 . o ~+ ~1.28)~~ 0 . 0 E 1 53 ~ 0.005~ 0.005~

6 (2) OC Curve To design a one-sided z-test at a = 0.01 for a single sample, OC Chart VId is applicable with the parameter 161 10.0051 = 0.5 -- -----d = IP-PO I CJ 0 0.01 By using d = 0.5 and P = 0.1, the required sample size is determined as n = 55 as displayed below. Note that this sample size is similar with the sample size n = 53 determined by using a sample size formula.

d

(4 0 C. c

w for &fiereat values ofn for the one-sided n

d test for a lml of s ~ l p a u= 0 01


Exercise 9.3

9-39

1. (Hypothesis Test on p, o2Unknown; Upper-Sided Test) Step 1: State Ho and H I . Ho: y = 1.0 H,: p > 1 . 0 Step 2: Determine a test statistic and its value. to =--x - P o - 1.25 -1.0 = 4.47 s/& 0.25/fi Step 3: Determine a critical value(s) for a. tct,n-l - t0.05,20-~ = t0.05,19 = .73 Step 4: Make a conclusion. Since to = 4.47 > to.,

,,,,= 1.73, reject Hoat a = 0.05. We conclude

that the PVC pipe product meets the ASTM standard at a = 0.05. 2. (Sample Size Determination) power = P(reject HoI Hois false) = 1 - p = 0.8 a P = 0.2 6 ~ p - =1.10-l.O=O.lO p ~

I

To design a one-sided t-test at a = 0.05 for a single sample, OC Chart VIg is applicable with the parameter

I


9-40

Chapter 9

Exercise 9.4


1. (Hypothesis Test on 02; Upper-Sided Test) Step 1: State Ho and H I . Ho: o2=0.012 HI: 0 2 > 0 . 0 1 2 Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value@) for a. 2 Xa,n-I

2

2

= X0.0l,l5-1 = X0.01,14 = 29.14

Step 4: Make a conclusion. Since = 8.96 2 0,,14 = 29.14 ,fail to reject Ho at a = 0.0 1. There is no significant evidence which indicates that the standard deviation of the hole diameter is greater than 0.01 mm at a = 0.01.

2. (Sample Size Determination) power = P(reject Ho 1 HOis false) = 1 - j 3 = 0.9

p = 0.1

To design an upper-sided X2-testat a = 0.01, OC Chart VII is applicable with the parameter 0

h=-=1.5

By using h = 1.5 and p = 0.1, the required sample size is determined as n = 50 as displayed below.

0.C curves for different values of n for the one-sided (upper tail) chi-square test for a level of significance a = 0.01.

(1)


Exercise 9.5

1. (Hypothesis Test onp; Lower-Sided Test) Step 1 : State Ho and H I . Ha: p = 0.05 H I : p < 0.05 Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. z , = zo,05= 1.64 Step 4: Make a conclusion. Since z0 = -0.53 4: -z,,,

= -1.64, fail to reject Ho at

a = 0.05.

2. (Sample Size Determination) power = P(reject Ho I Ho is false) = 1 - p = 0.8 3 P = 0.2

Exercise 9.6

(Goodness-of-Fit Test; Discrete Distribution) Step 1: State Ho and H I . Ho: X Poisson distribution with h = 1.2 H I: X -/- Poisson distribution

-

Step 2: Determine a test statistic and its value. 1. Estimate the parameter of the hypothesized distribution. Since the value of h is given, skip this step. The number of parameters estimated is p = 0. 2. Define class intervals and summarize observed frequencies (Oi's) accordingly.

9-41

9-42

Chapter 9



-

4 or more -*----

4 B-M-*mm

------

0.03

-e*-m-----

3

1

0.3

* . a a * . %

-m,

"**-=-

3. Estimate the probabilities (pi's) of the class intervals.

4. Calculate the expected frequencies (Ei = npi) of the class intervals. If an expected frequency is too small (< 3), adjust the class intervals. All the expected frequencies are three or above. 5

(Oi - E , ) ~

5. Calculate the test statistic: X: =

= 6.6 i=l

I

Ei

Step 3: Determine a critical value for a.

Step 4: Make a conclusion. = 6.6 2 Since

Xi.05,4 = 9.49 ,fail to reject Ho at a = 0.05.

Exercise 9.7

(Goodness-of-Fit Test; Continuous Distribution) Step 1: State HOand H I . Ho: X Uniform distribution f (x) = 1 , 0 I x I 1 HI: X Uniform distribution

-

+

I I

Step 2: Determine a test statistic and its value. 1. Estimate the parameter of the hypothesized distribution. Since the uniform distribution does not have any parameter, skip this step. The number of parameters estimated is p = 0. 2. Define class intervals and summarize observed frequencies (Oi's) accordingly.

9-43

Answers to Exercises Exercise 9.7 (cont.)

3. Estimate the probabilities (pi's) of the class intervals. .2

p , = P(X < 0.2) = fidx =:IX

= 0.2

Since the sizes of the bins are the same, the corresponding probabilities are equal to each other as 0.2. 4. Calculate the expected frequencies (Ei = npi) of the class intervals. If an expected frequency is too small (< 3), adjust the class intervals. None of the expected frequencies are less than three. (0, -E , ) ~ =1.70 Ei

Xt =

5. Calculate the test statistic:

i=l

Step 3: Determine a critical value for a. 2 - 2 Xc~,k-~-l - X0.05,5-0-1

2

= X0.05,4 = 9.49

Step 4: Make a conclusion. Since = 1.49 P = 7.81 , fail to reject Ho at a = 0.05. The data does not indicate any abnormal functioning of the random number generator at a = 0.05. (Contingency Table Test; Independence) Step 1: State Ho and H I . Ho: The type of failure is independent of the mounting position. H I : The type of failure is not independent of the mounting position.

Exercise 9.8

Step 2: Determine a test statistic and its value:

A

I

---*

Totals

--m-mx-w-

1 ~

P

Failure Type (Y) B C

i

-

~

26

m

~

~

-

~

-

-

-

-

63 m

-

~

~

-

-

~

-

m

24 -

a

.

w

~

m

-

-

~

21 m

s

~

"

~

L 134 -,---

9-44

Chapter 9

Tests of Hypotheses for a Single Sample Step 3: Determine a critical value for a.


2 Xa,(r-l)(c-1)

2

= XO.O5,(2-1)(4-1)

2

= X0.05,3 = 7.8

Step 4: Make a conclusion. Since = 1 0 . 8 > ~ i , , ~= ,7.81, ~ reject Ho at a = 0.05.

Exercise 9.9

(Contingency Table Test; Homogeneity) Step 1: State Ho and H I . Ho: Students in different class standings are homogeneous in terms of opinions on the curriculum change. H I : Students in different class standings are not homogeneous in terms of opinions on the curriculum change.

I

I

S t e 2: ~ Determine a test statistic and its value.

Opinion (Y) Favoring opposing

Totals

Totals


- 2 2 - X0.05,(4-1)(2-1) = X0.05,4 = 9.49

Step 4: Make a conclusion. Since = 27.0 > X i = 9.49, reject Ho at a = 0.05. It is concluded that students in different class standings have significantly different opinions on the curriculum change at a = 0.05.

.,,,

10


10-2 Inference for a Difference in Means of Two Normal Distributions, Variances Known 10-3 Inference for a Difference in Means of Two Normal Distributions, Variances Unknown

10-4 Paired t-Test 10-5 Inference on the Variances of Two Normal Populations 10-6 Inference on Two Population Proportions MINITABApplications Answers to Exercises

I

c

10-2

Inference for a Difference in Means of Two Normal Distributions, Variances Known

Cl Test a hypothesis on p~ - p2 when o12and 022 are known (z-test). Cl Determine the sample size of a z-test for statistical inference on pl - pz by using an appropriate

sample size formula and operating characteristic (OC) curve.

O Establish a 100(1 - a)%confidence interval (CI) on p1 - p2 when o12and 022are known. 0 Determine the sample size of a z-test to satisfy a preselected level of error (I$) in estimating pl - p2. Inference Context

Parameter of interest: p~ - p2 Point estimator of pl - p2:

(Note)

Xl - N(pl)' , 0

0:

o;

"'1

n2

X1 - X2 - N(pl - p 2 ,-+ -) and

X2 - N(p2,-)0;

n1 XI and X2 are independent

1'( Test statistic of pl - p2: Z =

-2'

2

n2

) - (PI - 112)

I

2

, o 1 and 0 2 known;

-

_ N(O,l)

Isal p a p y s - l a ~ o l ~ o j "z- > Oz 1sal paprs-laddn ~ o j " z < OZ 1sao pap!s-OMIlroj " " 2 < lozJ J! OH13afax .uo!snpuo3 e ayeK :p days

L

Og-('x- '_XI

=

0 2

'anIan s)! pua q+s!+a+s)sa+ e au!uualaa :Z dais

'Lia~!l~adsal ' 1

ED = ( Z

zTi- 'ri=(Zx)y-('x)g=(Zx- Ix)y a ~ u e ! . ~pue e ~ ueaur q y ! ~~euuouS! z~- ~

IX

-

pue) ' [U ~1 ;D = ( 'x)/i Z ~ )' [rl g = (Ix)g - '' d =( -

sa3uepeA put? sueaur y o ! leuuou ~ PUB ouapuadapu! ale

Z~

-

.ouapuadapu! ale z-x pup I X

pue

IX

-

a3u!S

10-2 Inference for a Difference in Means of Two Normal Distributions, Variances Known

Sample Size Formula (cont.)


n=

10-3

( z a + ~ ~ ) ~ +( ooi :)

for one-sided test (A - A d L where: A = p, - p2 (true mean difference), A + A, (hypothesized mean difference), and

Table 10-1 displays a list of OC charts in MR and a formula of the OC parameter d for a z-test on p1 - p2.By using the table, the appropriate OC chart for a particular z-test is chosen (e.g., for a one-sided z-test at a = 0.05, chart VIc is selected).

Table 10-1 Operating Characteristic Charts for z-test - Two Samples Chart VI OC p m e t e r Test a G ' e A in %)


A 100(1 - a)% CI on p, - p2when o12and ~2~ are known is -

5 p, -)I,

-

I X, - X,

+ z,,,

for two-sided CI

I p l - p2 for lower-confidence bound

pl - p2 I Xl - X2 + Z,

1:

-+

: A


(Derivation) Two-sided confidence interval* on p1 - p2, o12and 022known By using the test statistic Z =

P(LI~IU)=I-~ 3

P(- z,,, I z I z,,,)= 1- a

Therefore, -

-

L=X,-X,-z,,,

-

-

and U = X l - X 2 + z a 1 2

10-4

Chapter 10


Sample Size Formula for Predefined Error

As an extension of the sample size formula for estimation on p with a preselected level of error (described in Section 8-I), the following formula is used for estimation on pl - p2:

Example 10.1

The life lengths of INFINITY (XI; unit: hour) and FOREVER(X2; unit: hour) light bulbs are under study. Suppose that XI and X2 are normally distributed with oI2= 402 and 022= 302,respectively. A random sample of INFINITY light bulbs is presented in Example 8.1 and a random sample of FOREVERlight bulbs is shown below: Life Length No Life Length No Life Length No II 755 21 837 7 89

(a

(y)+ 2

n=

(0;

7 8

769 836 847

9

I

0 ; ) . where a, = n2 = n

17 18 19

787 788 794

The two random samples are summarized as follows: Brand of Light Bulb

Sample Size nl

= 30

Sample Mean -

x,

= 780

hrs

Variance

o12= 402

mean life length of an INFINITY light bulb is different from that of a FOREVER light bulb at a = 0.05. h Step 1: State Ho and H I . Ho: -p2 = 0 HI: p1-p2+0 Step 2: Determine a test statistic and its value. (780 - 800) - 0 z,, = (X, - X2) - 6O-

-+-

I

+ --

Step 3: Determine a critical value(s) for a. za12 = z0.0512

= z0.025 =

Step 4: Make a conclusion. Since lz,, = 2.12 > zo,02j= 1.96, reject Hoat a = 0.05.

1

10-2 Inference for a Difference in Means of Two Normal Distributions, Variances Known

Example 10.1 (cont.)

10-5

2. (Sample Size Determination for Predefined Power of Test) Determine the sample size n (= nl = n2) required for this two-sided z-test to detect the true difference in mean life length as high as 20 hours with 0.8 of power. Apply an appropriate sample size formula and OC curve. w (1) Sample Size Formula power = P(reject HOI Ho is false) = 1 - p = 0.8 3 P = 0.2

(2) OC Curve To design a two-sided z-test at a = 0.05 for two samples, OC Chart Vla is applicable with the parameter A 120-0) d=dT=dy=0.4

o +o

402 + 30


3. (Contidence Interval; Two-Sided CI) Construct a 95% two-sided confidence interval on the mean difference in life length (pl - p2). Based on this 95% twosided CI on p~ - p2, test Ho:p1- p2 = 0 VS.HI:p1- p2 # 0 at a = 0.05. h

1 -a=0.95 3 a=0.05 95% two-sided CI on pl - p2:

10-6

Chapter 10


Since this 95% two-sided CI on y l - y2 does not include the hypothesized value zero, reject Ho at a = 0.05.


4. (Sample Size Determination for Predefined Error) Find the sample size n (= nl = n2) to construct a two-sided confidence interval on yl - y2 within 20 hours of error at a = 0.05.


Two types of plastic (say, plastic 1 and plastic 2) are used for an electronics component. It is known that the breaking strengths (unit: psi) of plastic 1 (XI) and plastic 2 (X2)are normal with 0,' = 0 2 ~= 1.O. From two random samples of n, = = 155.0. 10 and n2 = 12, we obtain = 162.5 and 1. The company will not adopt plastic 1 unless its breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample test results, should they use plastic l? Use a = 0.05 in reaching a decision. 2. Determine the sample size n (= nl = n2) required for this one-sided z-test to detect the true difference in mean breaking strength as high as 11.5 psi with 0.9 of power. Apply an appropriate sample size formula and OC curve. 3. Construct a 95% upper-sided confidence interval on the mean difference in breaking strength (y1- y2).

10-3

Inference for a Difference in Means of Two Normal Distributions, Variances Unknown

O Test a hypothesis on yl - y2 when o12and 0 2 ~are unknown (t-test). 0 Determine the sample size of a z-test for statistical inference on yl - y2 by using an appropriate operating characteristic (OC) curve. R Establish a 100(1 - a)% confidence interval (CI) on y l - y2 when 0 1 2 and 0; are unknown. *---*-.#.aw--m

-*-----

Inference Context

w

M

m

w

m

-

-

-

m

-

u

m

m

~

-

~

s

m

-

m

~

~

m

-

-

~

~

-

~

~

*

~

-

wa**m

m-ms-s

.s-s-dm

-"*

B-%

Parameter of interest: P I - y2 Point estimator of y1 - y2: (Note)

XI - N(pl ,-)0:

0:

0;

"'1

n2

X1 - X2 - N(yl - y 2,- + -) and

0 X2 - N(y2,A), o12and oZ2 unknown;

n1 and X2 are independent

n2

----

10-3 Inference for a Difference in Means of Two Normal Distributions, Variances Unknown

Inference Context (cont.)

10-7

Test statistic of pl - p2: Different test statistics of p1 - ~2 are used depending on the equality of o12and 022 as follows: (1) Case 1: Equal variances (oI2= 022 = 02)

,

where: S, = (n, - 1)s: + (n2 - 1)s; n, + n 2 - 2

(pooled estimator of 02)

(2) Case 2: Unequal variances (oI2# 022)

(Note) The equality of two variances (o12= 022)can be checked by using an F test described in Section 10-5.

Test Procedure (t-test)

Step 1: State Ho and H I . Ho: PI - P2 = 60 HI: PI - p2 # 60 for two-sided test PI - p2 > 60 or pl - p2 < 60 for one-sided test Step 2: Determine a test statistic and its value. (1) Case 1: Equal Variances (oI2= 022

where: S, =

n,

= 02)

+ n2 - 2

(2) Case 2: Unequal Variances (oI2# 022)

Step 3: Determine a critical value(s) for a. t,,,, for two-sided test; t , , for one-sided test Step 4: Make a conclusion. Reject Hoif It,, > t 2 for two-sided test

1

to > t,,


to < - t , ,


10-8

Chapter 10

Operating Characteristic (OC) fhrve


Table 10-2 displays a list of OC charts in MR and a formula of the OC parameter d for a t-test on p1 - p2 where ol2 = o22 = o2 and nl = n2 = n. Note that OC curves are unavailable for a t-test when oI2# 0 z 2 because the corresponding t distribution is unknown. By using the table, the appropriate OC chart for a particular t-test is chosen (e.g., for a two-sided t-test at a = 0.01, chart VIfis selected). The sample size n* obtained from an OC curve is used to determine the sample size n (= nl = n2) as follows: n*+l n=, where n* from an OC curve 2

Table 10-2 Operating Characteristic Charts for t-test Chart VI Test a

-

Two Samples

(Appendix A in MR)

OC parameter

Two-sided t-test One-sided

subjective estimate.

Confidence Interval l70rmula

A 100(1 - a ) % CI on p1 - p2 when oI2and 022are unknown depends on the equality of o12 and 022as follows: (1) Case 1: Equal variances (oI2 = 022 = 02)

-

-

I

I p1 - p2 I X I - X2 + ta, 2,vSp

3,- 1,- t 1 2 , vS

1 +5 pl - p2

for two-sided CI

for lower-confidence bound

n2

'"1

- pl-& I Xl - X 2 +ta,2,vSp


(2) Case 2: Unequal variances (o12# 0 2 ~ ) -

-

I p, -p2 I X I - X , +ta12,,

for two-sided CI

5 p, - p2 for lower-confidence bound -

-

pl -& I Xl -X2 +ta,2,v CI and Hypothesis Test for Large Sample


If the sample sizes are large (nl and n2 L 30), the z-based C1 formulas and test procedure in Section 10-2 can be applied to inference on pl - p2 regardless of whether the underlying populations are normal or non-normal according to the central limit theorem (described in Section 7-5).

10-3 Inference for a Difference in Means of Two Normal Distributions, Variances Unknown

I

II I

10-9

For the light bulb life length data in Example 10.1, the following results have been obtained: Brand of Light Bulb

Sample Size

nl

INFINITY (1)

FOREVER(2)

Sample Mean -

= 30

x,

= 780

Sample Variance

hrs

S :

= 40'

-

n2 = 25

x2 = 800 hrs

Sided Test) Assuming 01' # 0 z 2 , test if the mean life length of an INFINITY light bulb is different from that of a FOREVERlight bulb at a = 0.05. Step 1: State Ho and H I . Ho: ~1 - ~2 = 0 H I : PI - P2+ 0 Step 2: Determine a test statistic and its value. (780 - 800) - 0 to = (XI - X2)- 60 -

-+v=

(st I n , (s: I n ,

+

+si In,), ( ~l ni d 2

--

(40, I 30 + 302 12512

-2 =

-2~54

(40, 1 3 0 ) ~L (302 1251,

Step 3: Determine a critical value(s) for a. t a ~ =~t0.025,54 , ~ = 2.00 Step 4: Make a conclusion. Since Ito/= 2.12 > to,o,5,5,= 2.00, reject Hoat a = 0.05.

I

2. (Sample Size Determination) Assuming 0 1 , = 0 z 2 , determine the sample size n (= nl = n2) required for this two-sided t-test to detect the true difference in mean life length as high as 20 hours with 0.8 of power. Apply an appropriate OC curve. h To design a two-sided t-test at a = 0.05 for two samples, OC Chart VIe is applicable with the parameter

I

(Note) s: =

(n, - 1)s; + (n, - 1)s; n, + n 2 - 2

10-10 Chapter 10 Example 10.2 (cont.)

Statistical Inference for Two Samples For d = 0.28 and P = 0.2, n*

=

100 as displayed below.

1.o j - - - - - - - r - - - - - - 7 - - ~ - - - - - 3

(e)

O.C. curves for different v d w of n for the two-sided t-test for a level of significance a = 0.05.

Thus, the required sample size n is

3. (Confidence Interval on p1- p2, o12and o; Unknown but Equal; TwoSided CI) Assuming 0 1 2 = 0z2, construct a 95% two-sided confidence interval on the difference in mean life length (pl - p2). 95% two-sided CI on p1- p2:

(Note) This t-based CI is wider than the corresponding z-based CI -38.5 I p l - p , 5 - 1 5 in Example 10-1. 4. (Confidence Interval on pl - p2, o12 and o; Unknown and Unequal; TwoSided CI) Assuming 0 1 2 # 0 2 2 , construct a 95% two-sided confidence interval on the difference in mean life length (pl - p2). Based on this 95% two-sided CIonp~-p2,te~tH~:p~-p~=O~~.H~:p~-p~#Oata=0.05

10-11

10-4 Paired t-Test


I I

Since this 95% two-sided CI on p~ - p2 when 0 1 2 and 02' are unknown and unequal does not include the hypothesized value zero, reject Ho at a = 0.05. Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears measured in foot-pounds is an important characteristic. A random sample of n, = 10 gears from supplier 1 results in Fl = 290 and sl = 12, while another random sample of n2 = 16 gears from supplier 2 results in Z2 = 32 1 and s2 = 22. Assume that XI and X2 are normally distributed. 1. Assuming o12= 022,test if supplier 2 provides gears with higher mean strength at a = 0.05. 2. Assuming oI2= oZ2,determine the sample size n (= nl = n2) required for this one-sided t-test to detect the true mean difference as high as 25 foot-pounds with 0.9 of power. Apply an appropriate OC curve. 3. Assuming oI2= 022, construct a 95% upper-confidence bound on pl - p2. 4. Assuming oI2# 022,construct a 95% upper-confidence bound on

10-4

- p2.

Paired t-Test

O Explain a paired experiment and its purpose. O Test a hypothesis on AD for paired observations when 0 D 2 is unknown (paired t-test). O Establish a 100(1 - a ) % confidence interval (CI) on p~ for paired observations when 0~~is unknown.

Paired Experiment

A paired experiment collects a pair of observations (XI and X2) for each specimen (experimental unit) and analyzes their differences (instead of the original data). This paired experiment is used when heterogeneity exists between specimens and this heterogeneity can significantly affect XI and X2; in other words, XI and X2 are not independent. For instance, in Table 10-3, the effect of a diet program on the change in weight is under study with n participants, who are heterogeneous. This heterogeneity is likely to confound the effect of the diet program on the weight change; in other words, the weight before (XI) and the weight after (X2) the diet program are not independent. To block the effect of the heterogeneity on the weight change, it is proper to analyze the differences (D) of the paired observations.

10-12 Chapter 10


Paired Experiment (cont.)

Inference Context

Parameter of interest: ~ 1 0 -

Point estimator of pD: D = X I - X ,

0; - N ( p D ,T),

X I and X , are -

oD2unknown;

independent.

D-PD -t(v),v=n- 1 Test statistic of PD: T = -----s, I&

Test Procedure (paired t-test)

Step 1: State Ho and H I . Hot pD = 60 H I : p~;t GO for two-sided test, p~> 60 or p~ < 60 for one-sided test. Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. t,,,,,-, for two-sided test; t ,

for one-sided test

Step 4: Make a conclusion. Reject Ho if t o > t , 2 , n 1 for two-sided test

1


to > t,,,-,


to < -t,,,-,


D - t,, , ,

SD

-

-< p D 5 D

&-

+ t,,

,,

&

for two-sided CI for lower-confidence bound

SD PD 5 D + t,, -

&

CI and Hypothesis Test for Large Sample


If the sample size is large (n 2 30), the z-based CI formulas and test procedure in Section 9-2 can be applied to inference on pD according to the central limit theorem.

r

10-4 Paired t-Test

Example 10.3

10-13

The weights (unit: lbs) before and after a diet program for 30 participants are measured below.

w

7 8 9 10

-

w w

A

198 165 180 172 m

-

9

-

182 160 178 171 a

w

w

17 18 19 20

184 173 179 168

w

-

w

176 156 167 152 e

A

The summary of the weight data is as follows: Sample Size Sample Mean (no. participants) (weight loss)

d

n = 30

=

10 lbs

27 28 29 30

-

215 206 165 170

206 20 1 156 154

Sample Variance s i = 52

1. (Hypothesis Test on PD, OD' Unknown; Two-Sided CI) Test if there is a significant effect of the diet program on weight loss. Use a = 0.05. h Step 1: State Ho and H I . Ho: pD=o HI: b # O Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. ta/2,n-l

= t0.05/2,30-1 = t0.025,29 = 2.045

Step 4: Make a conclusion. Since It, = 10.95 > t0,025,24 = 2.045 , reject Hoat a = 0.05.

1

2. (Confidence Interval on b,OD' Unknown; Two-Sided CI) Construct a 95% two-sided confidence interval on the mean weight loss (pD)due to the diet program. Based on this 95% two-sided CI on PD, test Ho: p . =~ 0 VS.HI:PD z 0 at a = 0.05. rn 1 - a = 0 . 9 5 3 a=0.05; v = n - 1 = 3 0 - 1 = 2 9 95% two-sided CI on

10-14 Chapter 1


Example 10.3 (cont.) Since this 95% two-sided CI on zero, reject Ha at a = 0.05.


does not include the hypothesized value

A computer scientist is investigating the usefulness of two different design languages in improving programming tasks. Twelve expert programmers, familiar with both languages, are asked to code a standard function in both languages, and their coding times in minutes are recorded as follows:

*-

d d

-

7 8 9

16 14 21

10 13 19

11 12

13 18

15 20

w - w w . -

--m--w---

smm-s----

*

-

ee--ms

m-ma-mm

6 1 2

--

-2 *-=---

=---

-2

-ms-

--

w-mm"

The summary quantities of the coding time data are n = 12, d = 0.67, and s i = 2.962 1. Does the data suggest that there is no difference in mean coding time for the two languages? Use a = 0.05 in drawing a conclusion. 2. Construct a 95% two-sided confidence interval on the difference between the two languages in coding time (pD). Based on this 95% CI on b,test Ha: = 0 vs. H I :pD# 0 at a = 0.05.

10-5

Inference on the Variances of Two Normal Populations

Explain the characteristics of an F distribution. R Read the F table. Test a hypothesis on the ratio of two variances (012/022)(F-test). O Determine the sample size of an F-test for statistical inference on 0 1 ~ 1 0by 2 ~using an appropriate operating characteristic (OC) curve. Establish a 100(1 - a)% confidence interval (CI) on 012/022.

1

10-5 Inference on the Variances of Two Normal Populations

F Distribution

10-15

The probability density function of an F distribution with u and v degrees of freedom is

The mean and variance of the F distribution are E ( X ) = v l(v - 2 ) for v > 2

V ( X )=

2v2(u + v - 2) for v > 4 u(v - 2),(v- 4)

An F distribution is unimodal and skewed to the right (see Figure 10-4 in MR) like a X 2 distribution. However, an F distribution is more flexible in shape by having an additional parameter of degrees of freedom. F Table

The F table (see Appendix Table V in MR) provides 1 OOa upper-tail percentage points (fa,,,, ) of F distributions with various values of u and v, i.e.,

P(Fu,v > fa,,,, The lOOa lower-tail percentage points ( fl-a,,,v) can be found as follows: fi-a,,,,

1

=-

fa,,,,

(e.g.) Reading the F table P(F7,15 > f ) = 0.05 : f = f0,05,7,15 = 2.71 (2) P(F7,,, > f)=0.95 : f = fO,,,,,,,, = 1 / f0,,,15,7 =l/3.51 = 0.28

Inference Context

0: Parameter of interest: 0:

Point estimator of

o2 + 0 2

s: s;

: -

(Note) S:

=C(xli

=C(X,, -x,)' i n , - 1 ; "2

"1

In, - 1 and S:

-

X 1 N ( p l , o : ) and X , - N ( F ~ , ~ ; ) ; X l and X , are independent

Test statistic of

o2 + 0 2

s; l o 2 s; 10;

: F = -----L F(nl - 1, n, - 1)

10-16 Chapter Test Procedure (F-test)

Statistical Inference for Two Samples Step I : State Ho and H I . 2 0 1

2

--0 1 , o

Ho:

7

HI:

, 2

0:

t

0 2

g >-

2 01,o

0;

for two-sided test,

4 , o

02,o

2

or

5< 0, for one-sided test. 0;

2 02,o


Step 3: Determine a critical value(s) for a. fa/2,n,-l,n2-l

and

)

1 fi-a/2,nl-l,n2-1

for two-sided test

fa/2,n2-~,n1-1


fa,n, - I , ~ ~ - I /

\

1


Step 4: Make a conclusion. Reject Ho if f~ > f a / 2 , n l - l , n 2 - 1 0' f~ < fi-a/2,n,-l,n2-1 for upper-sided test fo > fa,nI-l,n2-l fo < fi-a,n,-l,n,-l


for two-sided test


Table 10-4 displays a list of OC charts in MR and a formula of the OC parameter = nz = n. By using the table, the appropriate OC chart for a particular F-test is chosen.

h for an F-test on 012/022 where n l

Table 10-4 O~eratinp. " Characteristic Charts for F-test Chart VI Test a OC parameter ( ~ ~ p e n dA i xin MR)

10-17

10-5 Inference on the Variances of Two Normal Populations Confidence Interval Formula

o2 s 2 < - L < 2I f a/2,n2-l,n,-I

s; fl-a/ 2,n2 - ~ , n ,-I

0;

'2

:

s -

2 fi-a,n2-l,n,-l S2

for two-sided CI

S2

0 < - 1

for lower-confidence bound

0; 0:

s;

0 2

S2

-75


fa,n2-~,n,-I

(Derivation) Two-sided confidence interval* on 012/022 s; / 0 ; By using the test statistic F = ----- F(n2 - 1, n, - 1) ,

s: 1 0 ;

-

Therefore,

* For a one-sided CI, use Example 10.4

f a instead of f a , , to derive the corresponding limit.

For the light bulb life length data in Example 10.1, the following results have been obtained:

Brand of Light Bulb INFINITY (1) FOREVER ( 2 )

Sample Size nl

= 30

n2 = 25

Sample Mean Z, -

x,

Sample Variance

= 780 hrs

S:

= 800

s,2 = 302

hrs

1. (Hypothesis Test on o12/022; Two-Sided Test) Test Ho: 0 012/022 # 1 at a = 0.05. h Step 1: State Ho and H I . Ho: o : / ( T = ; l HI: of 10; $ 1 Step 2: Determine a test statistic and its value.

1 ~ 1 0= 2 ~1

= 402

VS.H I :


Statistical Inference for Two Samples Step 3: Determine a critical value(s) for a. fa1 2,nl -l,n2 -1

= f0.05

12,30-1,25-1

fl-a12,nl-l,n2-1 = f0.975,29,24

= f0.025,29,24 = 2'22 and

= f0.025,24,29

2. I

Step 4: Make a conclusion. Since, f0 = 1.78 2 f0025,29,2, = 2.22 and fo = 1.78 g f0,975,29,24 = 0.46, fail to reject HOat a = 0.05.

2. (Sample Size Determination) Determine the sample size n (= nl = n2) required for this two-sided F-test to detect the ratio of '31 to '32 as high as 1.5 with 0.8 of power. Apply an appropriate OC curve. ~hrTOdesign a two-sided F-test at a = 0.05, OC Chart Vlo is applicable with the parameter

~=0'=1.'j '32

By using h = 1.5 and P = 0.2 (because power = 1 - P = 0.8), the sample size required is determined as n (= nl = n2) = 50 as displayed below.

1.50 (0)

A

O.C. curves for diffnent values of n for the two-sided F-test for a level of significance a = 0.05,

3. (Confidence Interval on oI2/oz2; Two-Sided CI) Construct a 95% two-sided ~. on this 95% CI on '31~/'322, test Ho: confidence interval on 0 1 ~ 1 ' 3 2Based ~ ~ 1 ~ 1 '= 32 12VS.HI:'312/'322# 1 at a = 0.05. ~hr1 - a = 0.95 a = 0.05 95% two-sided CI on '31'102~:

=

10-19

10-6 Inference on Two Population Proportions


Since this 95% two-sided CI on '312/'322includes the hypothesized value unity, fail to reject Hoat a = 0.05.


In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etching rate is an important characteristic in this process. Two different etching solutions have been compared, using two random samples of 10 wafers for each solution. The observed etching rates (unit: milslmin) are as follows:

9.4 9.3 9.6

10.3 10.0 10.3

10.6 10.7 10.4

10.2 10.7 10.4

that the etching rates are normally distributed.

1. Test Ho: o12/'322= 1 VS. HI: o12/'322# 1 at a = 0.05. 2. Determine the sample size n (= nl = n2) required for this two-sided F-test to detect the ratio of '3, to o2as high as 2 with 0.9 of power. Apply an appropriate OC curve. 3. Construct a 95% two-sided confidence interval on '312/'322.Based on this 95% = 1 VS.HI: 0121'322f 1 at a = 0.05. two-sided CI on 012/'322,test Ho:

10-6

Inference on Two Population Proportions

0 Determine the sample size of a z-test for statistical inference onpl -p2 by using an appropriate

Inference


pl -p2

10-20 Chapter 10 Inference Context (cont.)


-

-

(Note) X1 B(n1, P I ) , X, B(n,, p,) ; n l p l , n,(l- p , ) , n2p,, and n2(1 - p,) are greater than 5; X, and X, are independent;

-

and P2 = n2

n1

Test statistic ofpl -p2: The test statistic ofpl -p2depends on the equality ofpl and p2 as follows: (1) Case 1: Unequal proportions (pl #p2)

(2) Case 2: Equal proportions @I

Test Procedure (z-test)

= p2= p )

Step 1: State Ho and H I . Ho: PI -p2'60 HI: PI -p2 f 60 for two-sided test, pl - p2 > 60 or p1 - p 2 < 6o for one-sided test. Step 2: Determine a test statistic and its value. (1) Case 1: Unequal proportions (pl # p2) 8 -P2 - 6 , 4-6-6,

zO=~ A

A

( 2 ) Case 2: Equal proportions (pl = pz = p )

Step 3: Determine a critical value(s) for a. z,,, for two-sided test; z , for one-sided test Step 4: Make a conclusion. Reject Hoif Izol > zai2 for two-sided test z, > z,


< -2,


2,


Sample Size Formula


10-21

For a hypothesis test o n p l - p2, the following formulas are applied to determine the sample size:

Like the test statistic o n p , -p2, a 100(1 - a)%C I onpl equality o f p l and p2as follows: 1. Case 1: Unequal proportions (pl #p2)

a

PI

A

-/

A

-p2

on the

d v

,:- p2 A

-p2 depends

i p, - p2 for lower-confidence bound

-za/2 A

PI-PZ SP,

A

-P2


+ z a ~

n2

2. Case 2: Equal proportions (pl =p2 = p )

A

A

C pl - p i C P, - P2 +

for two-sided CI

I pl - p2 for lower-confidence bound for upper-confidence bound

Example 10.5

Random samples of bridges are tested for metal corrosion in the HAPPYand


Statistical Inference for Two Samples 1. (Hypothesis Test o n p l -pz; Unequal Proportions; Upper-Sided Test) Assuming pl #p2, test if the proportion of corroded bridges of HAPPYCounty exceeds that of GREATCounty by at least 0.1. Use a = 0.05. h

Since n , j l = 4 0 x 0 . 7 = 2 8 , n l ( l - j l ) = 4 0 x 0 . 3 = 1 2 , n 2 j 2 =30x0.5=15, and n2(1 - j2) = 30 x 0.5 = 15 are greater than five, the sampling distributions of

4 and p2 are approximately normal.

Step 1: State Ho and HI. Ho: pl -p2= 0.1 Hi: pl -p2 > 0.1 Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value@) for a. za = zo,05= 1.64 Step 4: Make a conclusion. Since )zo = 0.86 2 z,,,, = 1.64 , fail to reject Ho at a = 0.05.

1

2. (Sample Size Determination) Suppose that p, = 0.7 and p, = 0.5. Determine the sample size n (= nl = n2) required for this two-sided z-test to detect the difference of the two proportions with power of 0.9. k power = P(reject Ho I Ho is false) = 1 - p = 0.9 p = 0.1

=

3. (Confidence Interval onpl -pz;Unequal Proportions; Upper-Confidence Bound) Assumingpl +p2, construct a 95% upper-confidence bound on the difference of the two corroded bridge proportions (pl - p2). 95% two-sided CI onpl - p2: PI-P* Sjl-i)2

+Za


10-23

Example 10.5 (cont.) Exercise 10.5 (MR 10-65)

A random sample of nl = 500 adult residents in MARICOPA County found that xl = 385 were in favor of increasing the highway speed limit to 70 mph, while another random sample of n2 = 400 adult residents in PIMACounty found that x2 = 267 were in favor of the increased speed limit. 1. Does the survey data indicate that there is a difference between the residents of the two counties in support of increasing the speed limit? Assume pl - p2 = 0 and use a = 0.05. 2. Suppose that p1 = 0.75 and p2= 0.65. Determine the sample size n (= nl = n2) required for this two-sided z-test to detect the difference of the two proportions with power of 0.8. 3. Assuming pl = p2, construct a 95% two-sided confidence interval on the difference between the favor proportions of the two counties for the speed limit increase (pl - p2). Based on this 95% two-sided CI onpl - p2, test Ho:pl , p2=Ovs. H l : p l - p 2 + O a t a = 0 . 0 5 . I

10-24 Chapter 10


MINITAB Applications Example 10.2

I

(Inference on 1 1 - pz, o12and oz2Unknown) (1) Choose File New, click Minitab Project, and click OK.

*

(2) Enter the life length data of the INFINITYand FOREVERlight bulbs on the worksheet cl-1 &

g

INFINITY /FOREVER,

*

a 3;

*

(3) Choose Stat Basic Statistics 2-Sample t. Click Samples in different columns and select INFINITYand FOREVERin First and Second, respectively. Select not equal in Alternative and type the level of = confidence in Confidence level. Check Assume equal variances if o12

MINITABApplications

10-25


II

Two Sample T-Test and Confidence Interval

I

TWO 3 m p T~ f o r INFINITY

I INFINITY FOREVER

N 30 25

Mean 780.0 800.0

vs FOREVER StDev 40.0 30.0

I

SE Mean 7.3 6.0

............+t ...........,....................

95t C I f o r m u I N F I N I T Y - m u FOREVER: i,(.:3?.;~-.:~-.ai T - T e s t m u I N F I N I T Y = m u FOREVER ( v s n o t = ) :;T..:.z?.:>?.*P.,y.j.;p.329i

I

1 10.2.41

DF = 52

Two Sample T-Test and Confidence Interval

(

TWO

s m p l e T f o r INFINITY

INFINITY FOREVER

N 30 25

Mean 780.0 800.0

vs FOREVER StDev 40.0 30.0

SE Mean 7.3 6.0

1

95% C I f o r m u I N F I N I T Y - m u FOREVER:;.!.;::, T - T e s t m u I N F I N I T Y = m u FOREVER ( v s Both u s e P o o l e d S t D e v = 35.8

(5) Choose Stat N Power and Sample Size N 2-Sample Z. Click Calculate sample size for each power value. Enter the power of the test predefined in Power values, the difference between the true mean and hypothesized mean to be detected in Difference, and the pooled estimate of standard


10-26 Chapter 10


Statistical Inference for Two Samples fined test condition.

I 1

Power and Sample Size 2-Sample t T e s t T e s t i n g mean 1 = mean 2 ( v e r s u s n o t = ) C a l c u l a t i n g power f o r mean 1 = mean 2 + 20 Alpha = 0.05 Sigma = 35.8

Example 10.3

I

(Inference on po, oDZ Unknown) (1) Choose File New, click Minitab Project, and click OK.

*

(3) Choose Stat

* Basic Statistics N Paired t. Select BEFOREand AFTERin

MINITABApplications


I

10-27

(4) Enter the level of confidence in Confidence level, the hypothesized difference in mean in Test mean, and not equal in Alternative. Then click OK twice.

( 5 ) Intemret the analvsis results.

I

Paired T-Test and Confidence Interval P a l r e d T f o r Before - A f t e r N 30 30 30

Before After Difference

Example 10.5

(Inference on p l - p2)

+

Mean 181.00 171.00 10.000

StDev 22.76 23.63 Sf

17

+

SE Mean 4.15 4.31 n

*II 1 10.3.2 1

(1) Choose Stat Basic Statistics 2 Proportions. Click Summarized data and enter the number of bridges surveyed in Trials and the number of Ins.


(2) Enter the level of confidence in Confidence level, the hypothesized difference in proportion in Test difference, and greater than in Alternative. Check Use pooled estimate of p for test if the proportions

/

I

Test and Confidence Interval for Two Proportions

10.5.3

.: Test for ~ ( 1 ) - ~ ( 2 =) 0.1 [vs > 0.1) :i~,~..~.,~:~~...~:~~.1.~~..:...9:,?~~ 5 5 5 5 5 5 ;

+

+

-

(4) Choose Stat Power and Sample Size 2 Proportions. Click Calculate sample size for each power value. Enter the power of the test predefined in Power values and the true proportions of two populations

10-29

MINITABApplications


(5) Under Alternative Hypothesis select the type of alternative hypothesis, and in Significance level enter the probability of type I error (a).Then

I Power and Sample Size

I

*I

T e s t f o r Two P r o p o r t i o n s T e s t i n g p r o p o r t l o n 1 = p r o p o r t l o n 2 ( v e r s u s >) C a l c u l a t i n g power f o r p r o p o r t l o n 1 = 0 . 7 a n d p r o p o r t l o n 2 = 0 . 5 Alpha = 0 . 0 5 D i f f e r e n c e = 0 . 2

1

I

10-30 Chapter 10



1. (Hypothesis Test on - p ~tsZ , Known; Upper-Sided Test) Step 1: State HOand H I . Ho: PI - k 2 = 10 HI: ~ 1 - ~ 2 > 1 0 Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. za = z ~ , =~1.65 , Step 4: Make a conclusion. Since z, = -5.84 ;6 z,,, = 1.65, fail to reject Ho at a = 0.05. We do not have significant evidence to support use of plastic 1 at a = 0.05. 2. (Sample Size Determination) (1) Sample Size Formula

n=

i z , + ~ , 3 ) ~ ( 0 : + 0 ; )- (z,.05+zo.,)2(12+12) (A - A,)2

(11.5-10)~

(2) OC Curve To design a one-sided z-test at a = 0.05 for two samples, OC Chart VIc is applicable with the parameter 11 1.5 - 101 A -A = 1.06 By using d = 1.06 and P = 0.1, the required sample size is determined as n (= nl = n2) = 8 as displayed below.

1.06

(c) 0.C NMS for di&rent vahtea of n for ttv ~ r t ~ s i &nd

d tcs for a lnrd of s@Sicanw a = 0.05.

.


10-31


3. (Confidence Interval; Upper-Confidence Bound) 1-a=0.95 3 a=0.05 95% upper-confidence bound on p1- p2:

Exercise 10.2

1. (Hypothesis Test on pl - pz, o12and o; Unknown but Equal; Lower-Sided Test) Step 1: State Ho and HI. Ho: p1- p 2 = 0 H1: p1 - p 2 < 0 Step 2: Determine a test statistic and its value.

(Note) si =

(n,

- 1)s;

n,

+ (n2 - l)s22

+ n2 - 2

Step 3: Determine a critical value@)for a. ta,v - t0.05,24 = Step 4: Make a conclusion. Since to =-4.07 f0,,, ,,,,, = 4.03 and f o = 3.33 < f0,9,5,9,9 = 0.25 , fail to reject Ho at a = 0.05.

1 2. (Sample Size Determination) To design a two-sided F-test at a = 0.05, OC Chart VIo is applicable with the parameter

When h = 2.0 and P = 0.1, the required sample size is determined as n (= nl n2)= 25 as displayed below.

I (0)

O.C.curves for d i f f m t values of n for the two-sided F-test for a level of significance a = 0.05.

3. (Confidence Interval on O ~ ~ I OTwo-Sided Z~; CI) 1 - a = 0 . 9 5 3 a=0.05 95% two-sided CI on 0 1 ~ 1 0 2 ~ :

Since this 95% two-sided CI on 0 1 unity, fail to reject Ho at a = 0.05.

~ 1 0 includes 2 ~

the hypothesized value

=

7


Exercise 10.5

10-35

1. (Hypothesis Test onpl -pz, Equal Proportions; Two-Sided Test) The survey data is summarized as follows:

X

Sample Size

County

Sample Proportion

Since n, j, = 500 x 0.77 = 385 , n, (1 - j , ) = 500 x 0.23 = 115, n 2 j 2 = 400x 0.67 = 267, and n2(1 - j 2 ) = 400x 0.33 = 133 are greater than five, the sampling distributions of

4 and

are approximately normal.

Step 1: State Ho and H I . Ho: pi - p 2 = 0 Hi: pi - p 2 + 0 Step 2: Determine a test statistic and its value.

x, + x, - 385 + 267 = 0.72 (Note) j = -n, + n2 500 + 400 Step 3: Determine a critical value(s) for a. 'a12 = z0.05!2 = '0.025 = Step 4: Make a conclusion. Since lz, = 3.32 > z,,,,,

1

= 1.96 ,reject Hoat a = 0.05.

2. (Sample Size Determination) power = P(reject Ho I Ho is false) = 1 - p = 0.8 a P = 0.2

10-36 Chapter 10

Exercise 10.5


3. (Confidence Interval onpl -pz; Equal Proportions; Two-Sided CI)

(cont.)

95% two-sided CI on pl - p2:

Since this 95% two-sided CI o n p l -p2 does not include the hypothesized value zero, reject Ho at a = 0.05.


11- 1 Empirical Models 11-2 Simple Linear Regression 11-3 Properties of the Least Squares Estimators 1 1-4 Some Comments on Uses of Regression 11-5 Hypothesis Tests in Simple Linear Regression 11-5.1 Use of t-Tests 11-5.2 Analysis of Variance Approach to Test Significance of Regression

11-1

11-6 Confidence Intervals 11-6.1 Confidence Intervals on the Slope and Intercept 11-6.2 Confidence Interval on the Mean Response 11-7 Prediction of New Observations 11-8 Adequacy of the Regression Model 11-8.1 Residual Analysis 11-8.2 Coefficient of Determination 11-8.3 Lack-of-Fit Test 11-9 Transformations to a Straight Line 11- 11 Correlation MINITABApplications Answers to Exercises

Empirical Models

independent variable, dependent variable, and random error. le linear model. Regression Analysis

Regression analysis is a statistical technique to model the relationship between two or more variables so that we can predict the response of a variable at a given condition of the other variable(s).

11-2

Chapter 11

Regression Analysis (cont.)

Simple Linear Regression and Correlation For example, in Figure 11-1, the scatter diagram of price versus sales for a' plywood product indicates that the volume of sales linearly decreases as the price of the product increases. By modeling this linear relationship, we may want to know how many pieces of plywood would be sold if the price is set at $8.50. It may not be easy to answer the question because there is no simple, linear curve which passes through all the points in the diagram-the data randomly scatter along a straight line.

1

0

2

I

I

I

I

4

6

8

10

*

Price per Piece (x)

Figure 11-1 Scatter diagram of plywood price versus sales. Simple Linear Model

A simple linear model which explains the linear relationship between two variables (x and Y) is Y=po +Plx+& where:

(1) Regression coefficients (Po and PI): The coefficients Po and PI denote the intercept and slope of the regression line, respectively. The slope PI measures the expected change in Y for a unit change in

z. (2) Independent variable (x; regressor, predictor): Since the independent variable x is under control with negligible error in an experiment, x represents controlled constants (not random outcomes). (3) Random error (E): The variable E represents the random variation of Y around the straight regression line Po + Pix. It is assumed that E is normally distributed with mean 0 and constant variance 02, i.e., E - N(o,o~) (4) Dependent (response) variable (Y): Since outcomes in Y at the same value of x can vary randomly, Y is a random variable with the following distribution: y-N(P0 + P , x , ~ ~ ) In summary, four assumptions are made for the simple linear regression model: 1. Linear relationship between x and Y: The variables x and Yare linearly related. 2. Randomness of error: The errors (spreads of Y along the regression line) are random (not following any particular pattern).

11-2 Simple Linear Regression

Simple Linear Model (cont.)

11-3

3. Constant oZ:The variance of the random errors is constant as 02.In other words, the spreads of Y along the regression line are the same for different values of x (see Figure 11-2). 4. Normality of error: The random errors are normally distributed.

0 0

XI

x2

X

Figure 11-2 The variability of Y over x. The analyst should check if the assumptions of the linear model are met for proper regression analysis, which is called assessment of model adequacy (presented in Section 11-8).

11-2

Simple Linear Regression

O Explain the method of least squares. O Estimate regression coefficients by using the least squares method. 0 Calculate mean responses and residuals by using a fitted regression line. Cl Estimate error variance 02. Least Squares Method

In the 1800s Karl Gauss proposed 'the method of least squares' to estimate the regression coefficients of a linear model. Suppose that we have n pairs of observations (xi, yi), i = 1 , 2 , . .. , n, and the n observations are modeled by a simple linear model

The least squares method finds the estimates of Do and (3, which minimize the sum of thesquares of the errors (denoted by SSE;called error sum of squares; sum of the squares of the vertical deviations of data from the regression line in Figure 11-3)

Chapter

Least Squares Method (cont.)

po

Dl

(Derivation) Least square estimators and Recall that, in calculus, when we want to find the value of a variable which minimizes (or maximizes) a certain function, the following procedures are used: (1) Take the derivative of the function with respect to the variable designated, (2) Set the derivative as equal to zero, and (3) Solve the equation. Since the function SSE (or L) has two unknowns (Do and PI), we need to take a partial derivative with respect to each regression coefficient as follows:

I

!

1 1-2 Simple Linear Regression

Least Squares Method (cont.)

11-5

n

By multiplying the first equation above by

xi and the last equation by n each, i=l

By subtracting the last equation above from the first equation above,

Thus,

Lastly, from Equation (1 1-2), 2 Y i

'=' So = Fitted Regression Model

- S l t x i i=l

n

=y-plx

By using the least squares estimates of Do and Dl, the fitted (estimated) regression line is determined as follows:

+So +Six

By applying the fitted regression line, the mean response at xi is estimated. Note that an actual response yi can be different from the corresponding predicted response ji . The difference between the actual and predicted responses is called the residual (denoted as e;),which is an estimate of the random error E~ : e i = y i - j j i , i = 1 , 2 ,..., n

1-6

Chapter

Estimation of Error Variance


Since the error variance o2is unknown in most cases, estimation on o2is needed. The point estimator of o2is

(02

n

y: - ny2 - bls,

(Derivation) SS, = i=l

SS, =f:ei2 i=l

=

f:[vi

=f:(yi

-ji)2 =

*

-(Bo + D , X ~ ) ]because ~ ji =Po + Pixi A

i=l

i=l

-(y -

f:Li

pi? + p 1 x i ) r

-

A

because

* -

Do = y -Dl

x

i=l

i=l

i=l

n

(Notes) SS, = S, =

(yi - j?)2 (called total sum of squares)

n

n

=SS, -2fils,

+P:S,

because S, = C y i ( ? - x i ) and C ( ? - x i ) = O i=l

>xy + PI -SY

= SS, - 2 B 1 ~ v

A

Sxx

=cy,?-B1s, n

-nj2

because

i=l

Axy P1 = A

Sxx

n

n

because SS, = C ( ~ , - J ) =~C y ' - n ~ '

r

1 1-2 Simple Linear Regression Example 11.1

I

11-7

The sales' volumes (Y) of a plywood product at various prices ( x ) are surveyed as follows (sorted by price):

z

z

The summary quantities of the plywood price-sales data are n =30, y, = 1,930, y,? = 138,656, z x i = 225,

z

z

xi = 1,775 , and

xi y, = 13,360

Assume that x and Yare linearly related. 1 . (Estimation of Regression Coefficients) Calculate the least squares estimates of the intercept and slope of the linear model for x and Y.

The fitted regression line of the plywood price-sales data is j = p, + p,x = 159.9 - 1 2 . 7 ~ A

I

n

2. (Calculation of Residual) Calculate the residual ofy = 84 at x = $6. h The predicted sales at x = $6 is .. jX=,= P o + / 3 , ~ = 1 5 9 . 9 - 1 2 . 7 ~ 6 = 8 3 . 4 Thus, - yX,, - jx=, = 84 - 83.4 = 0.6 (underestimate) ex=, A

11-8

Chapter 1 1


-


3. (Estimation of 02) Estimate the error variance 0'.

SS,

= t y i-ny2 -&s, = 1 3 8 , 6 5 6 - 3 O x ( z )

2

-(-12.7)x(-1,115)

i=l

An article in the Journal of Sound and Vibration (Vol. 151, 1991, pp. 383-394) described a study investigating the relationship between noise exposure (x) and hypertension (Y). The following data (sorted by sound pressure level) are representative of those reported in the article:


I The summary quantities of the noise-hypertension data are

I Assume that x and Yare linearly related.

I1 1 11-3

1. Calculate the least squares estimates of the intercept and slope of the linear model for x and Y. 2. Calculate the residual of y = 5 mmHg at x = 85 dB. 3. Estimate the error variance 4'

Properties of the Least Squares Estimators

"*

Identify the sampling distributions of the least squares slope and intercept estimators. 0 Check if the least square estimators fi, and are unbiased estimators.

Dl

11-3 Properties of the Least Squares Estimators

Distribution of Slope Estimator

The sampling distribution of the least squares slope estimator

(Dl) (Note) Estimated standard error of

11-9

8, is

8, = 8,

(Derivation) Sampling distribution of Recall that the least squares estimator of P 1 is n

Also recall that Y =Po + p 1 x + & - N(Po + P l x , o 2 ) ,where E - N ( o , o ~ ) ,andx represents constants. Since variables y i's,

0, is a linear combination of the normal random

Dl is normally distributed with the following mean and variance:

I

=-plCxi(xi-;) c

(Note) Since E(&) = pl ,

-

because x ( x i - x ) = O

8, is an unbiased estimator of P I .

11-10 Chapter 1 1

Distribution of


The sampling distribution of the least squares intercept estimator

po is

Estimator

Do

(Derivation) Sampling distribution of Recall that the least squares estimator of Po is

Also recall that Y constants. Since and

- N(Po + P,x, 0 2 ) , B, - N(Pl , 0 2 1S , ) ,

and x denotes

Do is a linear combination of the normal random variables y, 's

p, , So is normally distributed with the following mean and variance: E(B,) = E ( y - p , ~ =) E(y) - ~ ~ ( f = i ,E(Po ) + Dl?)

-PI?

=Po + p 1 x - p 1 x = p 0 v(po)=v(y-plx)=v(y)+x2v(pl)-2xcov(y,pl)=v(y)+x2v(pl) (Note) cov(j7,pl) = 0 because

(Note) Since E(p0) = P o ,

11-4

and

p, are independent.

Do is an unbiased estimator of Po .

Some Comments on Uses of Regression -*wI*IXImw~~wa

-

-d -Wa--in

*--Ixn-*a1Xzmwm-~---

m-

--I.

1)",XU*XIcrr.-.FW(-=

E1W

--- *-*-

L l Explain common misuses of regression. Common Misuses of Regression

Care should be taken on the following to avoid technical misuse of regression: 1. Practical validity of regressor: A statistical significance can be found on the relationship between independent (x) and dependent (Y) variables, although they are not related from a practical sense. Therefore, the practical significance of the relationship among the variables should be checked separately.

i

11-1 7

11-5 Hypothesis Tests in Simple Linear Regression

i

2. Empiricavtheoretical validity of regressor: The direction and/or magnitude of a regression coefficient in the model can contradict common understanding of the relationship between the variables. Therefore, it should be checked if the sign and magnitude of each regression coefficient are reasonable as compared with previous findings.

Common Misuses of Regression (cont.)

3. Association of regressor with predictor: A significant relationship between the regressor and predictor in the model does not necessarily indicate a causeeffect relationship between the variables. Designed experiments are the only way to determine cause-effect relationships between variables. 4. Limited generalizability of the model: The regression relationship established between the variables may be valid only over the range(s) of the regressor(s) examined in the study. To apply the regression model beyond the original range(s) of the regressor(s), a new study should be designed for the extrapolation purpose.

11-5

Hypothesis Tests in Simple Linear Regression

11-5.1 Use of f-Tests O Conduct hypothesis tests on PI and Po each. ~

U

X

*

l

w

~

X

(

_

*

J

U

w

Inference Context on Slope

_

I

~

~

-

~

m

~

M

-

~

- *Lw*M~~ *- % ~ -w ~ w ~ Mm ~ ~ * I

PI

Parameter of interest: Point estimator of

CP,) Test statistic of

-

"^*

#

[

PI = -- N P, ,-

PI :

PI :

% " .

'x'

s,

T

-

1'

- P ~ , o -t(n-2)

O - J i S ,

Hypothesis Test on P1

(t-test)

Step 1: State HOand H I . Ho: PI = PI,O H I : PI # P1,o Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(~)for a. t a t 2.n-2


I

It0 > tat2.n-2

lil*

XIX(*FX

%"&

*"

o2is unknown.

I .

iiX4aa

X

*

e.

XXlll',

t_X

I

11-12 Chapter 1 1

The Significance of Regression


It is typical to test if the regressor x is significant to explain the variability in Y by using Ho: = 0, which is called testing the significance of regression. Note that failure to reject Ho: [3, = 0 may imply that (1) The regressor x is of little value in explaining the variability in Y, or (2) The true relationship between x and Y is not linear. On the other hand, rejection of Ho: = 0 may imply that (1) The regressor x is of value in explaining the variability in Y, (2) The simple linear model is adequate, or (3) A better model could be obtained by adding a higher order term(s) of x while a linear effect of x on Yremains significant.

Inference Context on Intercept (Po)


Test statistic of

Hypothesis Test on Po (t-test)

Po :

Po

To =

Step 1: State Ho and H I . Ho: Po = Po,o H1: Po + P0,o

1


I

Step 3: Determine a critical value(s) for a.


r

10 '

Example 11.2

( > tai 2,n-2

For the plywood price-sales data in Example 11-1, the following summary quantities have been calculated: n = 30, b2 =3.2', E x i = 2 2 5 , Sxx=87.5, =-12.7, and =159.9

8,

1. (Hypothesis Test on PI) Test Ho: P1 = 0 at a = 0.05.

8,

11-13

11-5 Hypothesis Tests in Simple Linear Regression Step 1: State Ho and H I . Ho: p1 = 0


I

HI:

pl + O


Step 3: Determine a critical value(s) for a. 'u12,n-2 - t0.025,28 = 2.05

I

-

I

Step 4: Make a conclusion. Since It0\= 37.4 > t,12,n-, = 2.05, reject Ho at a = 0.05.

2. (Hypothesis Test on Po) Test Ho: Step 1: State Ho and HI. H0: Po = 0 HI: pozo

=0

at a = 0.05.


Step 3: Determine a critical value@)for a. 'a/2,n-2 = t0.025,28 2.05


Step 4: Make a conclusion. Since (t,(=61.0>taI,,,-, =2.05,rejectHoata=0.05. For the noise-hypertension data in Exercise 11- 1, the following summary quantities have been calculated: n = 20, ir2 = 1.4', = 1,646, S, = 3,010.2, = 0.17, and = -9.81

a,

Ex,

1. Test HO:pl = 0 at a = 0.05. 2. Test HO:Po = 0 at a = 0.05.

II-5.2 Analysis of Variance Approach to Test Significance of Regression Explain the analysis of variance (ANOVA) technique. Describe the segmentation of the total variability of a response variable (SST)into the regression sum of squares (SSR)and error sum of squares (SSE). Establish an ANOVA table and conduct a hypothesis test on P1. ---l_nP-X--lj---M

-

- ~ ~ - - - - _ j ~ l l ~ w w = ~ - a ~

_

_

*

-

_

I

*

~

~

=

P

_

L

-

a

a

a

X

w

-

~

r t * /,m -s -

s-^

11-14 Chapter 11

Analysis of Variance (ANOVA)


The analysis of variance (ANOVA) technique provides a statistical procedure to evaluate the contribution of a variable (X,) to the variability in the response variable (Y). As illustrated in Figure 11-4, the ANOVA method divides the total variance of Y ( S T ) into the segments ( SS,, 's ) explained by variables Xi's and the segment unexplained (SSE), implying that the larger the size of the segment SS,, ,the higher the contribution of XI to the total variability in Y.

Figure 11-4 Segmentation of the total variability in Y ( S T ) . ANOVA for Test on

The ANOVA method can be used to test the significance of regression (Ho:PI = 0). For a simple linear model, as shown in Table 11- 1, the total variability in Y (SSTor Syy,total sum of squares; degrees of freedom = n - 1) is partitioned into two components: (1) Variability explained by the regression model (SSR, regression sum of squares; degrees of freedom =I) (2) Variability due to random error (SSE, error sum of squares; degrees of freedom = n - 2 )

Table 11-1 Partitioning of the Total Variability in Y (SST) SST SSR ss. (Total SS) (Regression SS) (Emr SS) r=l

r=l

DF n-l 1 (Note) SS: Sum of Squares; DF:Degrees of Freedom m*&"w-am**m-mema

-*B

waa*s~-~-*--essB--b-~a*

+

w ~ - m ~ ~ a - # - - - - ~ e ~ m . - - ~ w " * m . m ~

r=l

n-2

*sea%a--"--mm--e

A graphical interpretation of the ANOVA quantities is provided in Figure 11-5. The deviation of an observation yi from the corresponding mean y is divided into the yi-deviation explained by the regression line and the yi-deviation due to error. These deviation are squared and then added to those of the other observations to calculate SST,SSR,and SSEas follows:

11 i

1 1-5 Hypothesis Tests in Simple Linear Regression

11-15

ANOVA for

Test on

PI

(cont.)

(variability ofy, due to regress~on)

Figure 11-5 Partitioning of the total variability ofy. The ratio of SSR/l to S S E / (-~2) follows an F distribution: S S ~l --M S ~ F(l, n - 2) F= SS, l ( n - 2 ) MS, as P1 = 0 is This statistic F is used to test Ho: P1 = 0, because F becomes true. The quantities MSR (regression mean square) and MSE (error mean square) are adjusted SSR and SSEby their degrees of freedom, respectively. Note that

ANOVA

Table

The variation quantities from the ANOVA analysis are summarized in Table 11-2 to test the significance of regression.

Table 11-2 ANOVA Table for Testing the Significance of Regression Source of Sum of Degrees of Mean Squares Freedom Square Fo Variation 1 MS, MS,IMS, SSR = fils, Regression Error

SS, = SS, - SS,

Total

SS,

n -2

n

Hypothesis Test on

PI

(F-test)

=zyi -ny2

n-1

Step 1: State Ho and H I . Ho: Pl = 0 H I : Pl f 0 Step 2: Determine a test statistic and its value. - MSR F(l, - 2) --F, = SSR I SS, l(n - 2) MS,

'

MSE

I

11-16 Chapter 11


1


Hypothesis Test on

i

fa,l,n-2

P1

(F-test) (cont.)

Step 4: Make a conclusion. Reject HOif f~ > fa,~,n-2 In testing the significance of regression, the ANOVA test procedure is equivalent to the t-test procedure in Section 11-5.1, because the square of the t-test statistic becomes the F-test statistic when pl = 0. Therefore, use of either test procedure will lead to the same conclusion on 01.

Relationship between t- and F-tests

--f i l

S~Y

because Sv,,

--M S ~ because -

fils, = SS,

= MSR and

e2= MSE

MSE Note that the t-test is more flexible than the F-test in testing a hypothesis on PI, because the t-test can have either a one- or two-sided alternative hypothesis whereas the F-test can have only a two-sided alternative hypothesis. However, the F-test is used to evaluate the relative contributions of multiple regressors to the variability in Y (see Section 12-2).

Example 11.3

I

(ANOVA for Hypothesis Test on PI) For the plywood price-sales data in Example 11- 1, the following summary quantities have been calculated: y? =138,656, S , =-1,115, and =-12.7 n = 30, =1930,

zyi

p1

Establish an ANOVA table and conduct a hypothesis test on PI at a = 0.05.

II I1 I

I

SS, =

Ay,?

2

- njJ2= 138,656 - 3Ox(%)

= 14,492.7

i=l

SS, = filsxy = -12.7 x (-1,115) = 14,208.3 SS," = SS, - S S P = 14,492.7 - 14,208.3 = 284.4

Mean

Source of Variation

Sum of Squares

Degrees of Freedom

Square

fo

Regression

14,208.3

1

14,208.3

1398.9

Error

284.4

28

10.2

Total

14,492.7

29

(Note) t: = (-37.41~ = 1398.9 = fo

1 1-6 Confidence Intervals


Step 1: State Ho and H I . Ho: p1 = 0 H1: PI # 0 Step 2: Determine a test statistic and its value. SSR11 - MSR - 14,208.3 = 1,398.9 fo = SS,l(n-2) MS, 10.2 Step 3: Determine a critical value(s) for a. for,l,n-2 = f0.05,1,30-2 = 4.20 Step 4: Make a conclusion. Since f , = 1,398.9 >fa,,,,-, = 4.20, reject Ho at a = 0.05.

d


For the noise-hypertension data in Exercise 11- 1, the following summary quantities have been calculated: Establish an ANOVA table and conduct a hypothesis test on P1 at a = 0.05.

I1 6

Confidence Intervals

11-6.1 Confidence Intervals on the Slope and Intercept Establish confidence intervals on P1 and PO each.

Inference Context on Slope

p1


(PI)

P1 :

P, =

(Note) Estimated standard error of

A

Confidence Interval on

A 100(1 - a ) % confidence interval on I

P1

Inference Context on Intercept (Po1

-2

-

2

4

I

p,

-

2

i-' 1sn 0

P,


is

+2

1

-

:x; /

:

D,

-

=y -

-

Pl x - N fi

(Note) Estimated standard error of

p, =

ij2 - + -

11-17

1 1


11-18 Chapter 1 1

A 100(1 - a)% confidence interval on Po is

Confidence Interval on

Po For the plywood price-sales data in Example 11-1, the following summary quantities have been calculated: n = 3 0 , x x i =225, 6' =3.22, S , =87.5, =-12.7,and =159.9

Example 11.4

a,

Po

1. (Confidence Interval on PI) Establish a 95% two-sided confidence interval on the slope. h- 1 - a = 0 . 9 5 z a=0.05 95% two-sided CI on PI:

BI - t a / 2 , n - 2 J K 5 P l

5fi1+ t a / 2 , n - 2 J E

d

I

1

I I


=. - 12.7 - t o , 0 5 , 2 , 3 , - 2 ~5-12.7 ~ ~ ~+to,05,2,30-2 I 3.22 187.5 3 -12.7 - 2.05 x 0.34 I PI 5 -12.7 + 2.05 x 0.34 a -13.4 I P , 5-12.0

2. (Confidence Interval on Po) Establish a 95% two-sided confidence interval on the intercept. 1-a=0.95 a a=0.05 95% two-sided CI on Po:

For the noise-hypertension data in Exercise 11- 1, the following summary quantities have been calculated: n = 20, z x i = 1,646, B2 =1.4', S , =3,010.2, S , =3,010.2, = 0.17,

0,

and

Po =-9.81

1. Establish a 95% two-sided confidence interval on the slope. 2. Establish a 95% two-sided confidence interval on the intercept.

!

11-19

11-6 Confidence Intervals

11-6.2 Confidence Interval on the Mean Response

0 Identify the sampling distribution of the mean response estimator at x = xo. Establish a confidence interval on the mean response at x = xo.

Sampling Distribution of Mean Response Estimator

The mean response at x = x o is Pyls = E ( Y XO)=E[(PO + P ~ ~ + E ) I ~ O ~+ =P PP oO The point estimator of pylxo is

PYlxo

The sampling distribution of

is

(Derivation) Sampling distribution of

fi Ylxo

Recall that

Do - N Since

fi ylxo

(

p0,o

-+-

and

B1 - N

2 [ : ] )

is a linear combination of

1:) PI,-

Po and p1, fi ylxo has a normal distribution

with the following mean and variance: E ( P y l x o ) = ~ ( B o + b l ~ o ) = ~ ( B o ) + x o ~ ( B i ) = ~ o+ x o h =Prix0

(Note) Since E(P

Inference Context on Mean Response atx=x, (PYlXo

) = pylxo , fi ylxo is an unbiased estimator of

Parameter of interest: p yl,o Point estimator of pyIxo :

Pylxo = b o + B ~ X -O N o2is unknown.

~YI,

.

11-8 Adequacy of the Regression Model


h

1

11-25

The normal probability plot displays that the residuals lie closely along a straight line. In addition, the standardized residual plot shows more than 95% of the standardized residuals (29130 = 96.7%) fall in the interval (-2,2). Therefore, it is concluded that the residuals are normally distributed. (Notice that there is no indication of an outlier in the plywood price-sales data because all the standardized residuals are within the interval (-3,3).) Next, the plots ei vs. j j and ei vs. xi indicate that the residuals are random and have a constant variance. The variability of the residuals is slightly increased in the mid ranges of y and x; however, this deviation from the constant variance is not serious enough to reject the adequacy of the regression model.

d

In the noise-hypertension data in Exercise 11-1, the following residual plots are obtained for the fitted regression model j = -9.8 1 + 0 . 1 7 ~:


Observation Number

-3

-2

-1

0

I

2

3

Residuals (e,)

(a) normal probability plot

E

5 * 6 ...............

............

0.

(b) standardized residual plot

Sound Pressure Level O (x: dB)

.

a".- - . . . . . . - - -

(c) e, vs. ji

(d) e, vs. xi

1 Discuss the adequacy of the regression model by using the residual plots. 11-8.2 Coefficient of Determination (I?)

O Explain the term coeflcient of determination. O calculate the coefficient of determination of a regression model. -,-.- ..... . ...... B2ws,BMsB

-s*m*mw"w,d

* - ~ s m , w , m ....ass.*W

"m%#*B.m.w.b ~%esem

w*m8aM-ea-

.m.sm*a

8acv,B*

. . ~ . d ~ ~ a . d . . m ~ . m . ~ v ~ ~ - - ~ - ~ ~ . . ~ m . ~ ~ ~ w A ~ * ~ * ~ ~ . ~ ~ ~ ~ . ~ ~ ~ m . . w m # ~ a * ~ . ~ 2 d . ~ . e ~ . ~ ' ~ . M m . a . # a ~ we#.mw.Bw#w~~d~..a. a..~*.~a~..~

The coefficient of determination (denoted as R2) indicates the amount of Coefficient of Determination variability of the data explained bv the regression model. In other words, R2 is the proportion of the total variability in Y which is accounted for by the regression (R~) model:

11-26 Chapter 11


I

1

Coefficient of Determination

5s ' -I-%, R 2 =RSST SST

(RZ) (cont.)

(Caution) Use of R~ A large value of R~ does not necessarily indicate the adequacy of the regression model. By adding new regressors or high-order terms of the regressors to an existing model, R~ will always increase even if the new model is less desirable than the existing model in terms of simplicity, stability, and statistical significance. For instance, n data points of x and y can be 'perfectly' fit by a polynomial of degree n - 1 (e.g., four data points can be perfectly fit by a polynomial with a degree of three). Although the value of the perfect-fit model is one, the high-order polynomial is complex to use and often unstable when a different sample of data is used. Lastly, unless the error sum of the squares (SSE) of the new model is reduced by the error mean square of the existing model, the new model will have a larger error mean square (MSE), resulting in a decreased level of statistical significance.

Example 11.8

(Coefficient of Determination) For the plywood price-sales data in Example 111, the following summary quantities have been calculated: SS, = 14,492.7, SS, = 14,208.3, and SS, = 284.4 ) the fitted regression model Calculate the coefficient of determination ( R ~of 9 = 159.9 - 1 2 . 7 ~ .

I I Exercise 11.8 (MR 11-47)

0 f a , , -,,,-, = 2.78, fail to reject HOat a = 0.05. It is concluded that the order of the model is adequate.

Exercise 11.9

I

I

For the noise-hypertension data in Exercise 11-1, the following summary quantities have been calculated: m

= G ( j i -ii)'=4.8.and

n=2O, m=10, SS, =35.7, SS,,

i=l

I Conduct a lack-of-fit test on the regression model j. I1 9

= -9.8 1+ 0 . 1 7 ~at a = 0.05.

Transformations to a Straight Line

O Explain the relationship between linear and non-linear models. -_(j.

-ls^---Y_--s----w=

Transformation of Nonlinear Model

-------

-

-

~

-

-

*

~

w

a

-

~

~

~

~

~w

" a ~ #-4 w *witm

(CUYP-

^(i

XIWIIXI

PI

-

i

-*-

A non-linear relationship between independent (x) and dependent (Y) variables can be transformed into a linear model by mathematical operations. Examples are presented in Table 11-5.

P

11-30 Chapter 11 Transformation of Nonlinear Model (cont.)


Table 11-5 Transformation of Nonlinear Models (illustrated)

Linear Model

Nonlinear Model Y=P

~

~

~

~

I~n Y =El n p o + P , x + E

11-11 Correlation

Calculate the sample correlation coefficient (R) between random variables X and Y. LI Construct a confidence interval on the population correlation coefficient (p) between X and Y. Conduct a hypothesis test on p.

Correlation vs. Regression

A close relationship exists between the correlation of random variables Xand Y and the regression model of independent and dependent variables x and Y. Recall that in the regression model the regressor x is assumed to be a mathematical variable, not a random variable (see Section 11-1). Suppose that X and Yare random variables and the observations (X., Yi), i = 1,2, ... ,n, are obtained from a bivariate normal distribution. The population correlation coefficient p, (see Section 5-5) is

The conditional probability distribution of Y given X = x is a normal distribution with mean and variance (see Section 5-6)

= P o + P 1 x , where p o = p y - p x p m -

CJ Y

CJx

and

PI = p H - -CJ Y '3,

Therefore, it is identified that the conditional probability distribution of Y given X = x is equivalent to the probability distribution of a simple linear regression model with mean DO + plx and variance 02.

11- 11 Correlation

Estimation of Correlation Coefficient (PXY)

The estimator of , ,p

11-31

called sample correlation coefficient (denoted as R), is n

z ( x i - X)(? R - 66x ,& y -

-

Y)

-1IR t,12,,-2 IzOI> zaI2

for testing H o : p = po (# 0), n 2 25

The relationship between ergonomics grade (X)and statistics grade (Y) is under study. A random sample of n = 50 students is selected who have taken both ergonomics and statistics courses. Their final scores in ergonomics and statistics are summarized as follows: n = 50, S , = 6,652.5, S , = 3,999.0, and S , = 3,303.1

Example 11.10

I I

Assume that ergonomics grade and statistics grade have a bivariate normal distribution. 1. (Sample Correlation Coefficient) Calculate the sample correlation coefficient (R) between ergonomics grade (X)and statistics grade (Y).

II"

2. (Confidence Interval on pxy) Construct a 95% confidence interval on the population correlation coefficient (p) between X and Y. arctanh r + -

tanh(arctanh r -

)

Jn-3

z0.05 12 arctanh 0.64 - zo,0512 < p 1tanh(arctanh 0.64 + -

JG' -

3. (Hypothesis Test on pm) Test if p m # 0 at a = 0.05. Step 1: State Ho and H I . Ho: p = O HI: p # O

I


I

Step 3: Determine a critical value(s) for a. ta/2,n-2

= t0.0512,50-2 = t0.025,48 = 2.01

, I s

I 14

-

1 1-1 1 Correlation

I1-33


Step 4: Make a conclusion. Since Itol = 5.78 > t,/2,n-2 = 2.01, reject Ho. It is concluded that ergonomics grade (3 and statistics grade (Y) are significantly related at a = 0.05.


The measurements of weight (X) and systolic blood pressure (Y) for n = 26 randomly selected males in the age group 25 to 30 are collected, resulting in the following summary quantities: n = 26, S , =4,502.2, S , =15,312.3, and S , =6,422.2 Assume that weight and systolic blood pressure are jointly normally distributed. 1. Calculate the sample correlation coefficient (R) between X and Y. 2. Construct a 95% confidence interval on the population correlation coefficient (p) between X and Y. 3. Test if p~ # 0 at a = 0.05.

11-34 Chapter 1 1


MINITABApplications Examples 11.1-3,5,6,8

I (Simple Linear Regression)

I

( I ) Choose File

(2)

eet.

&

1

1

* New, click Minitab Project, and click OK.

I

Pr~ce

I

*

1 Sales Volume / 95 /

II

I

*

(3) Choose Stat Regression Regression. In Res~0nSeselect Sales Volume and in predictors select Price.

(4) Click Graphs. For Residuals for Plots, select Regular or Standardized. Under Residual Plots, check Normal plot of residuals, Residuals versus fits, andlor Residuals versus order. In Residuals versus the

MINITABApplications

11-35

(5) Click Options. Check Fit intercept. In Prediction Intervals for new observations, enter the value of price of interest for the confidence interval on the mean response andor prediction interval of a future observation. In Confidence level, type the level of confidence. Then click

Examples 11.1-3,5,6,8 (cont.)

OK

1

(6) Click Results. Under Control the Display of Results, click the level of

,

,

II

Regression Analysis The r e g r e s s ~ o ne q u a t i o n 13 S a l e s Volume = 160 - 12.7 P r l c e

i

Source

MS

DF

1 R e s i d u a l E r r o r ; 28 Total : 29 Regression

A.,$;.

14208 10

5.,,..?,!4,!

13g8.9:

P 0,000

1

j

14493 ............................................................................. *.

:

Obs 1 2 3 4

11.1.2

Prlce 5.0 5.0 5.0 5.0 5.0

S a l e s Vo 95.000 98.000 96.000 92.000 99.000

. ,......8.8.... ..

Fit 96.190 96.190 96.190 96.190 96.190

. . .

b'...b'....Hj. /

StDev Flt 1.032 1.032 1.032 1.032 1.032

Resldual -1.190 1.810 -0.190 -4.190 2.810

........h.7.7. ........

-

...........BB.0.774 y4

0.552 : 1.3y?'*

44............

_

R d e n o t e 8 e n o b s e r v a t l o n w l t h a l a r g e standardized r e s l d u a l

( Ptedlcted

Values

S t Resld -0.39 0.60 -0.06 -1.39 0.93 -1.12 0.18 0.50

--

P

11.3

11-36 Chapter 11

Example 11.7


;rl

(Residual Analysis)

(1) Normal probability plot

(2) Standardized residual plot

Normal Probablll~Plot of the Reslduals @*on..

;E B

. ..

*I"

8.

Reslduals Versus the Order of the Data D .'m. ts ..,.s W)

la)

'*

I

;

0

2 , 2

6

.

.

16

(4) e, vs. x,

I

I

Reslduals Versus Me Fnted Values mipas.

8.

S'.I

W

Flned value

m

Ob~ewstlonOrder

Residual

I

.............

........................

I

30



(

Exercise 11.1

1. (Estimation of Regression Coefficients)

The fitted regression line of the noise-hypertension data is j = p o +PIX=-9.81 +0.17x

2. (Calculation of Residual)

jx=*5=Po A

+ P1x= -9.81 + 0.17 x 85 = 4.76 A

ex,85 = YxES5- jx=85= 5 - 4.76 = 0.24

1

Exercise 11.2

I

3. (Estimation of d )

1 . (Hypothesis Test on j31) Step 1 : State Ho and H I . Ho: p1 = 0

HI:

P l f O



-

tc(/z,n-2 - t0.025,18= 2.10 Step 4: Make a conclusion. Since Ito(=6.68 > t,,2,n-2 = 2.10, reject Ho at a = 0.05.

11-37

11-38 Chapter 1 1



2. (Hypothesis Test on Po) Step 1 : State Ho and H I . Ho: po = 0 H I : po#o Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. ta/2,n-2 = t0.025,18 = 2.10 Step 4: Make a conclusion. Since Ito = 4.6 > t,,, ,,-, = 2.10, reject Ho at a = 0.05.

1

Exercise 11.3

I (ANOVA for Hypothesis Test on PI)

Regression

88.5

1

88.5

Error

35.7

18

2 .O

Total n

_

%

~

l

x

b

124.2 _

_

e

m

~

~

~

-

~

-

m

*

B

~

-

~

W

19

-m~ u~ r ~r ama r- r - * x l ~ m x w ~

x ~ r r r r a m M ~ r a l - a . r - ~ ~ - ~ - - m

(Note) t i = 6.7' = 44.7 = f o Step 1 : State Ho and H I . Ho: P I = 0

HI:

PI # 0

Step 2: Determine a test statistic and its value. SSR 11 - MSR - 88.5 - 44.7 ----f0 = S S i(n ~ - 2) M S ~ 2.0 Step 3: Determine a critical value(s) for a. f a , l , n - ~ = f0.05,1,20-2 = 4.41 Step 4: Make a conclusion. Since fo = 44.7 > f ,,,,,-,

44.7

= 4.41, reject HOat a = 0.05.

. Answers to Exercises

Exercise 11.4

1. (Confidence Interval on PI) 1-a=0.95 a a=0.05 95% two-sided CI on P1: 81 =)

3

3

- t a / i 7 n - 2 J G ~ '81 ~ i +ta/i.n-2JlSlr

0.17 - to,05,2,20-2 /,I

0.17-2.10XO.O3IP, I 0 . 1 7 + 2 . 1 0 ~ 0 . 0 3 0.12 I P, 10.23

2. (Confidence Interval on Po) 1 - a = 0.95 a a = 0.05 95% two-sided CI on Po:

Exercise 11.5

(Confidence Interval on y ylx, )

0 yls5 = Po + 8,xo = -9.81 + 0.17 x 85 = 4.7 1-a=0.95 a a=0.05 95% two-sided CI on y ,Ig5 :

d-

I P1 I 0.17 + to.05,2,20-2

11-39

11-40 Chapter


Exercise 11.6

(Prediction Interval on yo)

jo

=so

+ s l x o=-9,81+0.17~85=4.7 1 - a = 0.95 3 a = 0.05 The 95% two-sided prediction interval on yo at x

= 85

dB is

(Note) The prediction interval 1.7 1 yo 17.8 is wider than the corresponding confidence interval on the mean response 4.1 5 p,,,, 5 5.4 in Exercise 11-5.

11 I

1 1

Exercise 11.7

(Residual Analysis) The normal probability plot displays that the residuals lie closely along a straight line. In addition, the standardized residual plot shows that all the standardized residuals fall in the interval (-2,2). Therefore, it is concluded that the residuals are normally distributed. Next, the plots ei vs. 5iand ei vs. xiindicate that the residuals are random and have a constant variance. The variability of the residuals is slightly reduced in the low ranges of y and x; however, this deviation from the constant variance is not serious enough to reject the adequacy of the regression model.

Exercise 11.8

1 (Coefficient of Determination)

I I

2 SSR - 88.5 R =--- - 71.3% SS, 124.2 The fitted regression model explains 71.3% of the variability in blood pressure rise.

11-41


Exercise 11.9

(Lack-of-Fit Test) An ANOVA table to test the lack-of-fit of the regression model is as tbllows:

Source of Variation

Sum of Squares

Degrees of Freedom

Lack-of-Fit

4.8

8

Pure Error

30.8

10

Error

35.7

18

Mean Square 0.6

Fo 0.20

3.1

Step 1: State Ho and H I . Ho: The order of the model is correct. HI: The order of the model is not correct. Step 2: Determine a test statistic and its value. MS,, 0.6 fo = ------- 0.20 MS, 3.1 Step 3: Determine a critical value(s) for a. fa,m-2,n-m = f0.05,8,10 = 3.07 Step 4: Make a conclusion. Since fo = 0.20 P = 3.07, fail to reject Hoat a = 0.05. It is concluded that the order of the model is adequate.

Exercise 11.10

1 . (Sample Correlation Coefficient)

2. (Confidence Interval on pxy)

arctanh r + -

arctanh r - arctanh 0.77 --

JZT

>

)

z0.051 2 < p < tanh(arctanh 0.77 + -

-

0.55 < p 1 0 . 8 9

3. (Hypothesis Test on pxy) Step 1: State Ho and H I . Ho: p=O HI: p $ 0 Step 2: Determine a test statistic and its value.

J26-3

12


12-1 Multiple Linear Regression Model 12-2 Hypothesis Tests in Multiple Linear Regression 12-2.1 Test on Significance of Regression 12-2.2 Test on Individual Regression Coefficients and Subsets of Coefficients 12-3 Confidence Intervals in Multiple Linear Regression 12-3.1 Confidence Intervals on Individual Regression Coefficients 12-3.2 Confidence Interval on the Mean Response

12-1

12-4 Prediction of New Observations 12-5 Model Adequacy Checking 12-5.1 Residual Analysis 12-5.2 Influential Observations 12-6 Aspects of Multiple Regression Modeling 12-6.1 Polynomial Regression Models 12-6.2 Categorical Regressors and Indicator Variables 12-6.3 Selection of Variables and Model Building 12-6.4 Multicollinearity MINITAB Applications Answers to Exercises

Multiple Linear Regression Model

Explain the assumptions of a multiple linear regression model. R Transform a non-linear model into a linear model. O Calculate residuals. O Estimate the error variance 02. Multiple Linear Model

As an extension of a simple linear model, a multiple linear model explains the linear relationship between the response variable (Y)and k (> 1) multiple regressors (xi's): Y = P O +Plxl + " ' + P k ~+k& where: (1) Regression coefficients (Po, PI,. .. ,Pk): The coefficient Po denotes

12-2

Chapter 12


Multiple Linear Model (cont.)

the intercept of the model and the coefficients Pj's, j = 1 to k, denote the slopes of the regressors. Each slope coefficient Pj measures the expected change in Y per unit change in xi when the other regressors are held constant (this is why Pj's are sometimes called partial regression coefficients). (2) Independent variables (XI,xz, ... ,xk;regressors, predictors): Since the regressors xi's will be controlled with negligible error, xj's represent controlled constants (not random outcomes). (3) Random error (E): The variable E represents the random variation of Y around the linear line Po + P,x, + ... + P xk . It is assumed that E is normally distributed with mean 0 and constant variance 02, i.e., E- N ( o , ~ ~ ) (4) Dependent (response) variable (Y): Since outcomes in Y at the same values of xl, x2,... ,xk can vary randomly, Y is a random variable with the normal distribution Y N(Po +P,x, +...+ P k x k , 0 2 )

1

-

Like the assumptions of a simple linear model (see Section 1 1-I), the multiple linear model has four assumptions: (1) Linear relationship between x,'s and Y (2) Randomness of error (3) Constant o2 (4) Normality of error The analyst should check if these four assumptions are met for proper regression analysis (see the assessment of model adequacy in Section 12-5). Note that computers are mostly used in multiple regression analysis so that the detailed calculation process is not presented in this chapter. Transformation of Nonlinear Model

Like the transformation of a non-linear model in simple linear regression analysis (see Section 11-9), a non-linear relationship between xj7sand Y can be transformed into a linear model by applying appropriate mathematical operations. Therefore, any regression models with regressors linearly combined can be treated as a linear model. (e.g.) Transformation of a non-linear model 2 P I Z X ~+XE ~ (1) Y=Po + P I X I + P 2 ~ + Y = P O + Plxl + P2x2 + P3x3 + E , where j3, = PI2 and x, = xlx2

3

Y = PO + Plxl + P,x,

+ P3x3+ P4x4 + E ,where x,

1

3

= - and x, = x, XI

Matrix Notation

Matrix notations and operations are useful in multiple regression. Suppose that n observations consisting of k regressors and the response variable, (xi,,xi2,. . . ,xik, yi), i = 1,2,. .. , n, are modeled by a multiple linear model yl =Po +Plxil + . . . + P k ~ i k+ E ~ ,i = 1 , 2,..., n

4

i

12-1 Multiple Linear Regression Model

Matrix Notation (cant.)

12-3

The system of n equations is expressed in matrix notation as follows: y=X$+& where:

I!'] =I] and

P= .

&

Pk

Model Building

En

The least squares method finds the estimates of regression coefficients $ which minimize the sum of the squares of the errors

The least squares estimators of $ are

fi = (xfx)-' x'y Thus, the fitted regression model is

i = xfi Residual and Error Variance

The residuals of the fitted model are e=y-i The estimate of the error variance G~is (J A

=- S S ~-

n-P

"' - "x" n-P

, where p = number of regression coefficients =k

Standard Error of Regression Coefficients

B

j

The covariances of

+ 1 ( k = number of regressors)

6 are

COV($)= 02(x'x))-l = a2c, where C = (x'x)-' This covariance matrix is a 07 x p ) symmetric matrix whosejjthcomponent is the variance of and whose ij" component is the covariance of and ,where i

B

and j

= 0 to

B,

k, i # j.

Then, the estimated standard error of fi se(B, ) = &&

is

fi

12-4

Chapter 12

Standard Error of

P,


The standard error of a regression coefficient is a useful measure to represent the precision of estimation for the regression coefficient: the smaller the standard error, the better the estimation precision.

(cont.) Example 12.1

The electric power (Y) consumed each month at a chemical plant is under study. It

By using a statistical software program, a fitted regression model of the power consumption data including all the four regressors (xl, x2, x3, and x4) and the corresponding error sum of squares (SSE)are obtained as follows: i = -102.71 + 0 . 6 1 ~+ ~8 . 9 2 ~+ ~1 . 4 4 ~+ ~0 . 0 1 ~and ~ SSE= 1,699.0 1. (Calculation of Residual) Calculate the residual ofy = 301 when XI = 65"F, x2 = 25 days, x3 = 91%, and x4 = 94 tons. U-F The estimate of the power consumption given that xl = 65, x;, = 25, x3 = 91, and xq = 94 is j = -102.71 + 0 . 6 1 ~ 6 5+ 8 . 9 2 ~ 2 + 5 1.44~91+ 0 . 0 1 ~ 9= 4 291.8 Thus, e = y - 9 = 301 - 291.8 =9.2 (underestimate)

I

2. (Estimation of 02) Estimate the error variance 02. n = 12 and p (number of regression coefficients) = 5 6 2 = - s S ~- -- 242.7 = 15.6' n-p 12-5

U-F

I

12-2 Hypothesis Tests in Multiple Linear Regression


I I

12-5

The pull strength of a wire bond is under study. Information of pull strength ('y), die height (XI),post height (xz), loop height (xg), wire length (x4), bond width on the die (xs), and bond width on the post (xg) is collected as follows:

By using a statistical software program, a fitted regression model of the pull strength data including four regressors ( ~ 2x3, , x4, and x ~ and ) the corresponding error sum of squares (SSE) are obtained as follows: 5= 7.46 - 0 . 0 3 ~+ ~0 . 5 2 ~- ~0. lox4 - 2 . 1 6 ~ and ~ SSE= 10.91 I. Calculate the residual of y = 9.0 when x2 = 18.6, x3 = 28.6, x4 = 86.5, and x5 = 2.0.

2. Estimate the error variance a2.

!

12-2

Hypothesis Tests in Multiple Linear Regression

12-2.1 Test on Significance of Regression

CI Test the significance of multiple regression.

12-6

Chapter

Significance of Regression

Multiple Linear Regression The test for significance of regression determines if there is at least one xj, out of k regressors, that has a significant linear relationship with the response variable Y, i.e., Ho: P I = P 2 = = P k = O (3 P = O )

HI:

Pi # 0

for at least onej

=

1 to k

Like simple regression, the ANOVA method is used to test the significance of regression. For the multiple linear model, in Table 12-1, the total variability in Y (SSTor S, total sum of squares; degrees of freedom = n - 1) is partitioned into two components: (1) Variability explained by the regression model (SSR,regression sum of squares; degrees of freedom = k) (2) Variability due to random error (SSE, error sum of squares; degrees of freedom=n-k-1 = n - ( k + l ) = n - p ) Table 12-1 Partitioning. of the Total Variabilitv in Y (SST) ,, SST SSR SSE (Total SS) (Regression SS) (Error SS) Q

-

\

DF n (Note) SS: Sum of w-waw*-#a*.A8~"*m~

~ w w ~ - * v ~ . -

The ratio of S&/k to SSE/(n - p ) follows an F distribution: SSRl k --M S ~ ~ ( k ,n p) F= SS, l(n - p ) MS, as Ho: Pl = P2 = ... = Pk = 0 is true, it is used to test the Since F becomes significance of regression. Note that MSR (regression mean square) and MSE (error mean square) are adjusted SSRand SSEby their corresponding degrees of freedom, respectively. Also note that MSE is an estimate of the error variance:

Test Procedure (F-test)

The variation quantities from the ANOVA analysis are summarized in Table 12-2 to test the significance of regression. Table 12-2 ANOVA Table for Testing the Significance of Regression Source of Sum of Degrees of ~ e & Fo P-value Variation Squares Freedom Square Regression SSR k MSR MSRIMSE v

Error Total

SSE

n-P n -1

MSE


Test Procedure (F-test) (cont.)

12-7

Step 1: State HOand H I . ff0: p1 = p 2 = ... = pk-0 HI: p j + O foratleastonej,j= 1 t o k Step 2: Determine a test statistic and its value. SS, 1k MSR F ( k , n - p ) F, = SS, l(n - p ) MSE

---

Step 3: Determine a critical value@) for a. fa,k,n-p

Step 4: Make a conclusion. Reject Ho if f 0

R' vs. Adjusted R~

'

fa,k,n-p

The coefficient of determination (denoted as R2) indicates the amount of variability in Y explained by the regressors X I x2, , ... ,xk: R 2 =-S S ~=I--, SSE 0 5 ~ ~ 1 1 ~ S T ~ S T As discussed in Section 11-8.2, a large value of R2 does necessarily imply that the regression model is adequate. As a new regressor is added to the existing model, R2 will always increase regardless of whether the additional regressor is statistically significant or not; thus, a regression model having a large value of R2 may unsatisfactorily estimate mean responses or predict new observations. In contrast, the measure adjusted R2 considers the number of regressors (p) included in the model in the computation: adjusted R2 = 1- SSE l(n -PI =I-- n-1 (1-R2), 0 fa,, ,,-,

=4.12,rejectHoat a = 0.01. This

indicates that at least one of the four regressors has a significant linear relationship with the monthly power consumption at a = 0.05. 2. ( R and ~ Adjusted R') Calculate the coefficient of determination (R2)and adjusted RZof the regression model. h R' = S S R--4z957 . 7 42 . 5 % SS, 6,656.3 The four regressors (xl, x2, x3, and x4) explain 74.5% of the variability in the power consumption Y. n -1 12-1 Adjusted R 2 = 1- -(1-R2)=1-(1 - 0.745) = 0.599 n-P 12-5


I1

For the pull strength data in Exercise 12-1, the following ANOVA table is obtained by multiple regression including the four regressors (x2, x,, x4, and x5): Source of Sum of BegrePsof Mean fo P-value Variation Squares Freedom Square Regression 22.31 4 5.58 7.16 cO.01 10.91 14 0.78 Error

-

Total -x_l

*a/*--_*1P_

18

33.22 *,

-w.w--Y,%

m

~

~

~

~

*

e

_

-

~

~

n

_

~

~

X

~

-

~

~

m

a

1. Test the significance of regression at a = 0.05. 2. Calculate the coefficient of determination (R2) and adjusted R2 of the regression model.

12-2.2 Tests on Individual Regression Coefficients and Subsets of Coefficients

O Test the significance of an individual regressor. Cl Test the significance of a subset of regressors.

-

~

e

e

-

11


Significance of Individual Regressor Inference Context on Pi

Like the testing on the significance of a regression coefficient in simple regression (see Section 1 1-5.l), the significance of any individual regressor x, is evaluated by t test (called partial or marginal test).

p


p, - ~ ( p ,0,2 c J 7 ) ,$ is unknown. p . p.-p. Test statistic of P : T = se(P, > (Note) Estimated standard error of p, : se(p, ) = ,where 8' = MSE Point estimator of

p,

:

J

J - J

J

T

Hypothesis Test on Pi

(t-test)

12-9

,/=

Step 1: State Ho and H I . Ho: pj = pj,0 HI Pj # P,,o Step 2: Determine a test statistic and its value.


Step 4: Make a conclusion. Reject Hoif

1

It0 > tat 2,n-p

Example 12.3

(Hypothesis Test on Pi) Consider the power consumption data in Example 12-1. The following quantities are obtained for production volume ( ~ 4in ) the regression analysis: = 0.014, se(b4) = 0.734, n = 12, andp = k + 1 = 5

p4

Test the significance of x4 at a = 0.05. * Step 1: State Ho and H I . Ho: p 4 = 0 H,: P 4 f O Step 2: Determine a test statistic and its value. to =

0 4 - p4.0

se(b4

>

0.014 - 0 = 0.02 0.734

Step 3: Determine a critical value(s) for a. ta/2,n-p - t0.0512,12-5 = '0.025,7 = 2.365 Step 4: Make a conclusion. Since )to]= 0.02 2

= 2.365 , fail to reject HOat a = 0.05.

12-10 Chapter 12


I 'I

Exercise 12.3

I

Consider the wire bond pull strength data in Exercise 12-1. The following quantities are obtained for bond width on the die (x5)in the regression analysis: =-2.16, se(b,) =2.39, n = 1 9 , a n d p = k + 1 = 5

b5

Test the significance ofxs at a = 0.05. Significance of the Subset of Regressors

The significance of a subset of regressors, xl, x2, ... ,x, (r < k), is tested by the extra sum of squares method (partial F-test; general regression significance test), which is an extension of the ANOVA method explained in Section 12-2.1. Like the test for significance of regression, this test determines if there is at least one xi, out of r regressors, that has a significant linear relationship with Y, i.e., Ho: p 1 = p 2 = ... = p r = 0 H I : P j # O foratleastonej,j= 1 t o r To test the significance of the subset of regressors, let the vector of regression coefficients ( p ) be partitioned into $ for the regressors xl,x2, ... ,xr and fl for the other regressors x,l, x ,+z, ... ,xk:

,

,

Also let X, and X2 denote the columns of the regressors X related to those related to P, ,respectively. Then, the multiple linear model is J'=X$+E=Xlfll + X 2 p 2+E

P,

and

In Section 12-2.1, the regression sum of squares (SSR)and mean error square (MSE)are

where (j = (x'x)-' X'Y . Note that SS, (p) indicates the contribution of the regressors X to the variability in Y. Now, to identify the contribution of the regressors X, ,fit a reduced model for X 2 , i.e., Y =X2P2

+E

The least squares estimate of regression sum of squares is

P, is $,

[k i=l

SS,(P,>= $;x',y -

yi

n

1

2

= (x;

X, )-' X i y and the corresponding

12-11


Significance of the Subset of Regressors (cont.)

Then, the regression sum of squares due to model is

ssR($l

IP~)=~~R(P)-~~R(P~)=~'x'Y

given that

P2

is already in the

-~;x;Y

In other words, SS, (PI I P 2 ) indicates an increase in regression sum of squares by adding XI to the reduced model, which is called extra sum of squares due to

. The ratio of SS, ( P I I P , ) lr to SSE/(n- p ) follows an F distribution:

Like the full model, the statistic F becomes true.

Test Procedure (F-test)

as Ho: PI = P2 =

= Pr = 0

is

The variation quantities of the ANOVA analysis for testing the significance of the subset of regressors P1 are summarized in Table 12-3.

Table 12-3 ANOVA Table for Testing the Significance of a Subset of Regressors PI Sum of Degrees of Mean Source of Fo Freedom Square Variation Squares Regression ( P ) SSR k MSR MS, I MSE $1

SS,

Pz

Error Total

(PI l Pz )

SSR ( P z ) SSE

r

k -r n-P n-1

MSR(P, I P z ) MSR(PZ) MSE

Step 1: State Ho and H I . Ho: P 1 = O (PI= P 2 = ... = P r = O ) H I : P1 $ 0 ( P , + O for at leastonej,j= 1 t o r ) Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. fa,r,n-p

Step 4: Make a conclusion. Reject Hoif

f o > fa,r,n-p

MSR(P1 I P z ) MSE

I

12-12 Chapter 12

I

Example 12.4

I


I

I (

I

I

(Test on the Subset of Regressors) Consider the power consumption data in Example 12-1. Suppose that the four regressors (P) are grouped into two categories: (1) Non-production regressor (PI): temperature (x,) (2) Production regressors (Pz): number of days (xz), product purity (xg) and production volume (x4) Multiple regression is performed separately with the four regressors (P) and the non-production regressor (PI), yielding the ANOVA results below. Source of Sum of Degrees of Mean Variation Squares Freedom Square fo Regression (P) 4,957.2 4 1,239.3 5.1 Error 1,699.0 7 242.7

Source of Variation Regression (PI) Error Total

Sum of Squares 3,758.9

Degrees of Freedom 1

2,897.3 6,656.3

10 11

Mean Square 3,758.9

fo 13.0

289.7

Test the significance of the production regressors (P2) at a = 0.05. ~ - r rStep I : State Ho and H I . Ho: p 2 = 0 (p2 = p 3 = p 4 = 0) HI: P 2 f 0 (pj#Oforat least o n e j , j = 2 to4) Step 2: Determine a test statistic and its value. SSR(P2 I PI) = SS,(P) - SSR(PI) = 4,957.2 - 3,758.9 = 1,198.3

Step 3: Determine a critical value(s) for a. fa,r,n-p = f0.05,3,7 = 4.35

I b

I Exercise 12.4

Step 4: Make a conclusion. Since fo = 1.65 > fa,,,,-, = 4.35, fail to reject HOat a = 0.05. This indicates that none of the production regressors have a significant linear relationship with the power consumption at a = 0.05. Consider the wire bond pull strength data in Exercise 12- 1. Suppose that the six regressors (p) are grouped into two categories: (1) Process regressors (PI): die height (XI),post height (xz), and loop height (~3) (2) Wire-bond regressors (P2): wire length (x4), bond width on the die (x5), and bond width on the post (x6)

i

1 3 I

,

12-13

12-3 Confidence Intervals in Multiple Regression

Exercise 12.4 (cont.1

I

I I I I

Multiple regression is performed separately with the six regressors (P) and the wire-bond regressors ((Jz), yielding - the ANOVA results below. . Source of Variation Regression (P)

Sum of Squares 23.63

Error

9.59

Total

33.22

Degrees of Freedom 6

Mean Square 3.94

12

0.80

II 1

Total w

an e-B.-MC_wM*I#

U m % # % w m m -mm

18

Sum of Squares 10.8 1 22.41

--

-m-m

wau#va-%"w

-A%w .

.,l*iP.

Degrees of Freedom 3 15

33.22 -m-

4.93

~ - ' ~ ~ ~ ~ ~ ~ " w ~ w a v ~ ~ # ~ w m " ~ ~ w*wal . ~ .in^.w ~ ~ ~ . *I_i-Xve.,i ~ ~ ~ ~ ~ ~ *"* w > , a~ ~ . a ~ . ~ . ~ , a

fl*ljY"UlfX U ~ j l X U . w m . m w ~ . me-sm.e-. '~-,~"s~ ~ ,~a . ~ m ~ " , w ~ ~ ~ " ~ A 8 m-mm~.*--*.~vaM.B~d. w ~ m ~ w ~ / ~ ' m

Source of Variation Regression (p2) Error

fo

-- - 18i

*

*msn

ne

aau-mxn-aar

Mean Square 3.61 1.49

fo 2.41

-

rn n n a nnnw*x-a

w r a a ~ ' , r n ~ W ~ ~e

rxraxx

Test the significance of the process regressors (PI) at a = 0.05.

12-3

Confidence Intervals in Multiple Regression

12-3.1 Confidence Intervals on Individual Regression Coefficients ,"*

*-.-

" . , m ~ . ~ , " . , ~ < ~ ~ ~ r ~ ~ " ~ ~ ~ " ~ ~ . - ~ . d ~ ~ B % ~ , ~ " ~ a ~ ~ ~ -.~~a%ms".m",m~~m~m ~ , w - " ~ . " , , ~ , w amwm.'* ~ , " * ~ ~ .","ea,.~~.-*.~m.~ ' ~ a ~ ~ ~ ~ ~ . b

"v"vA

,."*"""."v*'"w>~~*~"~

The regression models presented in previous sections are based on continuous (quantitative) variables. However, categorical (qualitative) variables such as gender and education level are sometimes considered in regression analysis. To incorporate categorical variables in regression, indicator (dummy) variables are used. A categorical variable with c levels can be modeled by c - 1 indicator variables. (e.g.) Indicator variables (1) gender = (male, female} 0 for male 1 for female (2) education = {high school, college, and postgraduate} for high school for college for postgraduate

r

Another method of analyzing data including categorical information is to fit a separate regression model to the data of each category. Compared to this approach, the indicator variable approach has the following advantages: 1. One regression model: One regression model integrates several regression models for different categories. 2. Increased statistical power: By pooling the data of different categories, the error term will have a smaller mean square error due to a larger number of degrees of freedom (n - p ) , which results in increased power in statistical testinn.

Example 12.10

(Indicator Variable) A relationship between liver weight @; unit: g) and body surface area (xl; unit: m2) is under study. The following measurements are obtained from 20 cadavers (10 males and 10 females; gender (x2) = 0 for male and 1 for female):

1

12-21

12-6 Aspects of Multiple Regression Modeling

By assuming that the interaction between xl and x2 is negligible, a fitted regression model is obtained for the liver weight data:


F =Po + Plxl + P2x2 = -108.7 + 9 7 7 . 0 ~-~1 . 3 ~ ~

I

Predictor Constant BSA (XI)

b


Standard error 829.2 502.7

Coefficient -108.7 977.0

to

P-value

-0.13 1.94

0.897 0.069

Since the P-value for P2 is greater than 0.05, gender (xl) does not have a significant effect on liver weight (y) at a = 0.05.

A mechanical engineer is investigating the surface finish 0/;unit: pm) of metal parts produced on a lathe and its relationship to the speed (xl; unit: RPM) of the lathe. The following measurements are obtained by using two different types of cutting tools (x2 = 0 for tool type 302 and 1 for tool type 416): Type of Type of Surface Lathe cutting Surface Lathe cutting finish speed tool finish speed tool No. O (~1) (~2) NO. (0 (XI) (~2)

7 8 9

52.26 50.52 45.58

265 259 22 1

0 0 0

17 18 19

32.13 35.47 33.49

224 25 1 232

1 1 1 1

,..,

".",*%"A'*""

I

I

,w

The following regression model is fit to the surface finish data by using software: 5 = + Plxl + P2x2+ Ij3X1X2 = 11.50 + 0 . 1 5 ~- ~6 . 0 9 ~- 0~ . 0 3 ~ ~ ~ ~

Po

Predictor Constant Lathe speed (xl) Tool type (x2) X1 X X2

Coefficient 11.50 0.15 -6.09 -0.03

Standard error 2.50 0.01 4.02 0.02

required to adequately model the tool life data.

to

4.59 14.43 -1.51 -1.79

P-value cO.01 m

Regressors in the model X4

---

X I , X wv"",-'*,a.w ~ , X ~ , .wsm.Me, X ~ *.,a*-"X ~*--. ,~---" X ~71.1 sm-m

4

56.7 * " "*

w,*~~*%.m~",sa

Discuss which model is preferable for the pull strength data.

'%.,,,*'"d"'.~

0.89 *,*

*.mm,ms

CP

7.0

~,,-*?w~,m *.- -,b.#,,m"


Forced Selection1 Elimination Forward Selection

Backward Elimination

12-25

The forced selection/elimination technique includes/excludes regressors designated by the analyst inlfrom the model regardless of their statistical significance. The forward selection technique adds to the existing model the regressor xi out of candidate regressors (not in the model) that has the largest partial F-value V;) iff; exceeds the entry criterionJn (corresponding probability pi, = 0.10 or 0.15; preselected by the analyst): s s (Pi ~ I B model ) f i = M s (fl ~ rnodeI,Pi In other words, regressors are added to the model one at a time as long as their partial F-values are greater thanj,,. The backward elimination technique includes all candidate regressors in the model and then eliminates from the model the regressor xj that has the smallest partial F-value G)if& is less than the removal criterion&,, (corresponding probability pout= 0.15 or 0.20; predefined by the analyst):

In other words, regressors are removed from the model one at a time as long as their partial F-values are than&,,.

Stepwise Regression

The stepwise regression technique is the mix of the forward selection and backward elimination techniques: forward selection followed by backward elimination. This mixed variable selection technique is widely used. The procedure of stepwise regression is as follows. Step 1: Out of candidate regressors, include xiin the model having the largest F-value iff; >J;, (corresponding pin= 0.10 or 0.15). Step 2: Among the regressors in the model, remove x, from the model having the smallest F-value if& > h u t(corresponding pout= 0.15 or 0.20; pin 5 pout). Step 3: Repeat steps 1 and 2 until no other regressors are added to or removed from the model.

Example 12.12

(Stepwise Regression) Consider the power consumption data in Example 12-1. Out of the four regressors (XI,x2, xg, and xq), determine an appropriate set of regressors by the stepwise regression technique to build a regression model. Use software withpi, =pout= 0.15 (equivalent toJ;, = h u t=A, 12-, = 2.5). h Two regressors, xl and x2, are selected for model building.


Consider the wire bond pull strength data in Exercise 12-1. Out of the six regressors (XI,x2, x3, x4, xg, and x ~ )determine , an appropriate set of regressors by the stepwise regression technique to build a regression model. Use software with Pin =pout= 0.15 (equivalent to& = h u t=A, 19_, = 2.3).

12-26 Chapter 12


12-6.4 Multicollinearity

Explain the terms multicollinearity and variance injlation factor (VIF).

O Describe undesirable effects due to multicollinearity in multiple regression analysis. Identify regressors having multicollinearity based on VIF.

O Describe remedial measures to resolve the problem of multicollinearity. Multicollinearity

When there are strong dependencies among the regressors xj7s,it is said that multicollinearity exists. The following undesirable effects can be observed due to multicollinearity: 1. Unstable estimation of regression coefficients: The estimates of regression coefficients, 's, are sensitive to a small change in data.

b,

2. High standard errors of regression coefficient estimates: The standard errors of regression coefficient estimates. se(b j) 's, are unreasonably large. 3. Contradicting test results: Although the full regression model is significant by F-test, no individual regressors are found significant by ttest. A regression model with the problem of multicollinearity will produce unsatisfactory estimations of mean responses or predictions of new observations beyond the region of the x-space surveyed (where extrapolation is needed). However, this regression model may produce satisfactory estimations on the response within the region of the x-space surveyed (where interpolation is performed). Variance Inflation Factor (VIF)

The measure variance inflation factor (VIF) indicates the extent of multicollinearity: 1 VIF(P,)=2 , j = 1 , 2 ,..., n I-Ri where R; is the coefficient of determination between xj and the other k - 1 regressors. The stronger the multicollinearity, the larger the value of R;. If a VIF exceeds 10 (some researchers suggest 4 or 5), the problem of multicollinearity exists.

Remedial Measures

Remedial measures against multicollinearity include 1. Inclusion of additional data: Collect additional data to explore the possibility to break the existing linear dependencies between the regressors. 2. Removal of regressors with large values of VIF: Exclude regressors having large values of VIF from the model.

'r

Example 12.13


12-27

(Multicollinearity) Consider the power consumption data in Example 12-1. The variance inflation factors of the four regressors (x,,x2,x3,and x4) are obtained by using software as follows:

-I

Number of days in the month (x2) Product purity (x3)

2.2 1.3 1.o

Interpret the VIF values. Since all the VIF values are less than 10, the regression model does not have the problem of multicollinearity.

&

Exercise 12.13

Consider the wire bond pull strength data in Exercise 12- 1. The variance inflation factors of the four regressors (XZ,x3, x4, and xg) are obtained by using software as follows:

Loop height (x3) Wire length (x4) Bond width .--,on t $"."*"..- ->*~"'.* %,*

s".m

**~a,r"

Interpret the VIF values.

1.07 1.41

12-28 Chapter 12


MINITABApplications Examples 12.1-3,5-8,13

(Multiple Linear Regression)

(1) Choose File

* New, click Minitab Project, and click OK. sheet.

(3) Choose Stat

* Regression * Regression. In Response select Y and in

(4) Click Graphs. For Residuals for Plots, select Regular or Standardized. Under Residual Plots, check Normal plot of residuals, Residuals versus fits, andlor Residuals versus order. In Residuals versus the

MINITABApplications

12-29

(5) Click Options. Check Fit intercept. Under Display, check Variance inflation factors. In Prediction intervals for new observations, enter the values of the regressors under consideration for the confidence interval on the mean response andlor prediction interval of a future observation. In Confidence level, type the level of confidence. Then click OK. -

Examples 12.1-3,543, 13 (cont.)

1

(6) Click Results. Under Control the Display of Results, click the level of

1

(7) Click Storage. Check Fits, Residuals, and Cook's distance. Then click

f

0 @QQ 1 ~ 1z!P1 ,$6- - -406 ~ ~ 0 € t 0 ~ ~ 8 5 ~ 8 1 - j 8 S L001 8 5 Z ,16 I1>1003 LlS3tl 1SlI4 Px ,f .--.-* m I i n t 9 D n

......

I

'

-

PZ

R

-9EZ OPZ

_ __

_z __ 1 T

TnnpTsaI P ~ Z T P I B P U W ~ a6lwl S w qirn uoraBnxasqo ue saiouap 1

I

MINITABApplications

Example 12.4

12-31

(Test on the Subset of Regressors) The

I1

Regression Analysis The r e g r e s s i o n e q u a t i o n is Y = 224 + 0.953 xl

I

Predictor Constant

Coef 224.38 0.9525

StDev 15.88 0.2645

T 14.13 3.60

P 0.000 0.005

...........................................................................................................

I ~ n a l y s ~ 0. s v a r i a n c e

:s o u r c e

i Regression

DF 1

:T o t a l

10 11

i ~ e s i d u a lE r r o r

ss 3758.9 2897.3 6656.31

MS 3758.9 289.7

F

12.97

p i 0.005 i

,..........................................................................................................

Example 12.9

w:

. .

.~"", ,

.$":

8 .,

I#,j

(Polynomial Model)

*

( I ) Choose Stat Regression P Fitted Line Plot. In Response select Y, in Predictor select the regressor under consideration, and under Type of Regression Model click the type of the polynomial to be developed. Then click OK.

I

Polynomial Regression

I

A n a l y s i s of V a r i a n c e

I

SOURCE Regression Error Total

SOURCE

I

.Lanear

MS 2181.38 254.83

F 8.56002

Seq SS

F

P

603.83

2.36953

0.158109

DF 2 9 11

DF

SS 4362.75 2293.50 6656.25

..............L.....3.7.56..92 ,.....12..2?.3.7 ....L R 3 & R 3 ...:

Quadratic

1

I

P 8.27E-03

I

12-32 Chapter 12

Example 12.10


II

, ,

(Indicator Variables) (1) Choose File New, click Minitab Project, and click OK.

I

(2) Enter the liver weight data on the worksheet.

,

(3) Choose Stat Regression Regression. In Response select Y and in Predictors select regressors under consideration.

I

RegressionAnalysis The r e g r e s s i o n e q u a t i o n is

y = - 109 + 9 7 7 x l - 1 x2 ...................................................................................................... -108.7 977.0 -1.3

x l x2

,

R-Sq = 18.45

Source Regression Resldual Error Total

Example 12.11

I

502.7 126.7

T

P

-0.13 1.94 -0.01

0.897 0.069 0.992

i 1

i

.....................................................................................................i S = 281.2

/

StDev 829.2

Coef

Predictor Constant

source

DF

DF 2 17 19

R-Sq(ad3) = 8.85

SS 303477 1344478 1647955

Seq SS

US 151738 79087

F 1.92

P 0.177

I

(All Possible Regression Evaluation) (1) Choose Stat Regression Best Subsets. In Response select Y and in

,

,

12-33

MINITABApplications

Example 12.1 1 (cont.)

(2) Click Options. In Minimum and Maximum enter the minimum and maximum numbers of regressors that will be included in regression models, respectively. In Models of each size to print, specify the number (1 to 5) of best regression models to be printed for each number of

Best Subsets Regression Response is y

64.5 60.9 1.7 X 15.381 56.5 52.1 3.9 X 17.022 1 0.2 0.0 19.4 25.769 X 1 0.0 0.0 19.4 X 25.799 67.2 1.4 2 73.1 X X 14.095 56.8 3.7 2 64.6 X X 16.173 56.6 3.7 2 64.5 X X 16.210 2 64.1 56.2 3.8 X X 16.289 . . .2. . . .5.6. ..5. . . .4. 6. ..8. . . . . 5. ..9. . . . . .1 .7.. 9. 4. .1 . . . X. . . . . . .X. 64.9 3.0 3 74.5 14.573 X X X 63.1 3.4 3 73.2 14.944 X X X 3 64.7 51.4 5.7 X X X 17.149 1

1

..............................................

I

-

12-34 Chapter

Example 12.12


(Stepwise Regression) (1) Choose Stat > Regression

> Stepwise. In Response select Y and in

respectively.

(3) Intemret the analvsis results.

Stepwise Regression

I

R e s p o n s e is Step Constant

,.............,..... I x2 iT-Value!

on

y

4 predictors, with N =

-30.1607

2 0.5287

15.2 4.26

10.3 2.36

1

12


12-35


(Calculation of Residual) The estimate of the pull strength when x2 = 18.6, x3 = 28.6, x4 = 86.5, and x5 = 2.0 is $=7.46 - 0 . 0 3 ~ 1 8 . 6 +0 . 5 2 ~ 2 8 . 6- 0 . 1 0 ~ 8 6 . 5- 2 . 1 6 ~ 2 . 0 = 8 . 6 6 Thus, e = y - j = 9.0 - 8.66 = 0.34

(underestimate)

2. (Estimation of c2) n = 19 and p (number of regression coefficients) = 5 -2 SSE --10.91 (J =-= 0.78 = 0 . 8 8 ~ n-p 19-5 Exercise 12.2

' 1. (Test on the Significance of Regression) Step 1: State Ho and H I . Ho: p 2 = p3 = p 4 = p5 = 0 H I : p J # O foratleastonej,j= 1 to4 Step 2: Determine a test statistic and its value. SSRl k --MSR - 5.58 - 7.16 f o = SS, l(n - p ) MS, 0.78 Step 3: Determine a critical value@) for a. fa,k,n-p

= f0.05,4,14 = 3.1

Step 4: Make a conclusion. Since fo =7.16> fa,,,n-,

=3.11,reject Hoat a = 0.05. This

indicates that at least one of the four regressors in the model has a significant linear relationship with the pull strength of a wire bond at a = 0.05. 2. (R2 and Adjusted R2) R 2 - ss R-22.31 - -= 67.2% SS, 33.22 The four regressors (x2,x3, x4, and x5) explain 67.2% of the variability in the pull strength Y. n -1 Adjusted R~ = 1- -(1-R 2 )=I-- 19-1 (1 - 0.672) = 0.578 n-P 19-5

12-36 Chapter 12 Exercise 12.3


(Hypothesis Test on Pi) Step 1: State Ho and H I . H,,: p5 = 0 H1: p5 + 0 Step 2: Determine a test statistic and its value.

to =

bs

-P5,0

- -2.16-0 -

se f a , ,

,-,

= 5.12, reject Ho at a = 0.05. This indicates

that the quadratic term significantly contributes to the model.

12-38 Chapter 12

Exercise 12.10


I (Indicator Variable)

For tool type 302 (x2 = O), the fitted model is j = + = 11.50 + 0 . 1 5 ~ ~

B, B,x,

I 1

I

I

1

I Exercise 12.11

In contrast, for tool type 416 (x2 = I), the fitted model is

j =

B,

+ B1.1

+ B 2 + 133x1 =(Do + 82) + (ljl + B3)4

= (1 1.50 - 6.09)

+ (0.15 - 0 . 0 3 ) ~=5.41+ ~

0.12~~

However, since the P-values for Pi and P3 are greater than 0.05, it is concluded that two regression equations are not needed to model the tool life data.

1

(All-Possible Regression Evaluation) Either a two-regressor model including x3 and x4 or three-regressor model including xl, x3 and x4 is recommended. The two-regressor model has the I minimum value of C, and the three-regressor model has the maximum value 1 of adjusted R; and minimum value of MS&). I I

Exercise 12.12

1 Exercise 12.13

(Stepwise Regression) Two regressors, x3 and x4, are selected for model building. (Multicollinearity) Since all the VIF values are less than 10, the regression model does not have the problem of multicollinearity.

13-2

Chapter 13

13-2

Design and Analysis of Single-Factor Experiments: The Analysis of Variance

The Completely Randomized Single-Factor Experiment

13-2.1 An Example

0 Identify dependent, independent, and extraneous variables in an experiment. 0 Explain the terms treatment and replicate. Compare the following pairs of concepts each: controllable vs. uncontrollable variables; fixedeffects vs. random-effects factors; balanced vs. unbalanced designs; complete vs. restricted randomizations. Explain the purpose of randomization. Tensile Strength Experiment (example)

A tensile strength experiment is introduced to explain fundamental terminologies of experimental design. Suppose that a manufacturer of paper is interested in improving the tensile strength of the product. The manufacturer wishes to identify if the tensile strength is affected by the hardwood concentration in the pulp used for the paper. The range of hardwood concentration of interest is between 5 and 20%; and, within the range, four levels (a = 4) of hardwood concentration are examined: 5, 10, 15, and 20%. At each concentration level, three specimens (ni = 3, i = 1 to a) are tested in a laboratory. The measurements by, j = 1 to ni,i = 1 to a) of this experiment can be summarized in a table format like Table 13-1. Table 13-1 Data Structure of a Single Factor Experiment

F'wcrar3hWle

I

Replicate

1

2

...

I

...

j

Ylj

Y2j

... ...

Yy

...

Y aj

... ... ...

Yin,

... ... ...

Y a,,

...

Ya.

**-

n1 **s

n_(__"

aiamla

-

-*

-*

Totals Averages :a m-invn*_?_f*

X*j.

_wwf.i._ea

Yln,

Y2n2

paa.-aarr----s----~---wM*w-

rr--arw

YI.

Y2.

-

-

y,. a

8*_ X _

n

l

*

~

~

8

x

Y2. /

~

l

_

e

_

n

~

~

Yl. -

.

~

~

~

Yi,

~

~

"

~

~

'

.

a

1

Ya.

11 ,

Y..

-

~ ax*-carulxi.in-nr-m---*->--d~w*u"%.r % ~ . ~ ~ ~ ~ ~

i

m-

a.-n-xnrtrmaran-n-M

-

%

~

~

~

~

Y.. w

~

~

id%-

Types of Variables

Three categories of variables exist in an experiment: 1. Response (dependent) variable: The outcome (response) of interest (e.g., tensile strength) 2. Independent variable (factor): The variable whose effect on the response variable is under study (e.g., hardwood concentration) 3. Extraneous variable (nuisance factor): The variable whose effect on the response is not under consideration but may affect the response variable (e.g., difference between specimens, temperature, etc.)

Treatment (Factor Level)

The term treatment (factor level) of a factor refers to an individual condition of the factor. For example, the tensile strength experiment has four factor levels: 5, 10, 15, and 20%.

~

~

'

~

~

are in the interval (-2,2). Residuals whose standardized values are beyond +3 may be outliers. Next, the constant variance and randomness of error can be checked by examining the following residual plots: (1) eij vs. factor level i, i = 1,2,. .. , a (2) eij vs.

7,

(3) eij vs. t (if time sequence is known) An undesirable residual pattern (see Section 11-8.1) in a residual plot may be corrected by transformation of the response variable ('y) into another form such as

r

J; or logy. Example 13.3

(Calculation of Residual) Consider the tensile strength data in Example 13-1. The following summary quantities are obtained from the analysis of variance: J1, =10.0, J,, =15.8, &, =16.8, and =21.4 Calculate the residual of yl2 = 8. (overestimate) ~h' el, = yI2 - j12 = y12- J1, = 8 - 10 = -2

L,

13-12 Chapter 13


Consider the tube conductivity data in Exercise 13-1. The following summary quantities are obtained from the analysis of variance: y,, =145.0, j2,=145.3, y,, =131.5, 7,. =129.3, and 7,. =145.3 Calculate the residual of y24 = 143.


13-2.6 Determining Sample Size

O Determine the sample size of a single-factor experiment by using an appropriate operating characteristic (OC) curve.


Operating characteristic (OC) curves in Figure 13-6 of MR plot required sample sizes n for a fixed-effects factor experiment at different levels of the parameters a,p,@,vl,andv2,i.e., n =f( a , S, a,vl, and ~ 2 )

1% n2.:

where:

=

v 2 = a(n - 1) (Notes) 1. If the error variance o2is unknown, an estimate of o2(such as MSE, conventional value, and subjective estimate) can be used. I

2. The analyst can define the ratio

1=1

r' lo2 (or 0: / 02) in sample size

determination.

r

3. The sample size required increases as a, P, and

XI=,

r: (or o: ) decrease and I

3

o increases. Example 13.4

(Sample Size Determination; Fixed-Effects Factor) Consider the tensile strength experiment in Example 13- 1. Four hardwood concentrations (a = 4) are selected and their tensile strength means tested at a = 0.05. Suppose that the four normal populations have common variance o2= 32 and means pl = 12, y2 = 15, p3 =

I

18, and p4 = 20 (i.e.,

xa r=l

I

T: / o2= 4.1 ). Determine the number of replicates

within each treatment to reject the hypothesis of equality of means with power of at least 0.90 (= 1 - B). Use the operating characteristic curve of Figure 13-6(a) in MR for a fixedeffects factor with vl = 3 (= a - 1 = 4 - I). From the OC curve, the power of the test according to n is found as follows: n Q, v2 = a(n - 1) fi Power (1 p)

-

13-13

13-3 The Random Effects Factor



Thus, at least n = 5 replicates are needed to achieve the designated power of the test. Suppose that five normal populations (a = 5) have common variance o2= 1o2 and means p l = 175, p2 = 190, p3 = 160, p4 = 200, and p5 = 215 (i.e.,

za r=l

T: lo2 =

18.3). How many observations per population must be taken so that the probability of rejecting the hypothesis of equality of means is at least 0.95? Use a = 0.01.

13-3

The Random Effects Factor

Test the significance of a random-effects factor in a completely randomized design.

0 Estimate the variances of error (02), treatment effect (03, and response variable (V(YV)). --- ---- *#

ms*

ws

*--*

*

-+ms

Context

Model

***

*---

%

dMFm

*

*

*

*

*

a

a a

*

a

*

1. Factor: Random-effects factor with a treatments 2. Blocking: No blocks defined 3. Repetition: n for each factor level (balanced design) (Note) N (total number of observations) = n x a 4. Randomization: Complete randomization

Y'J = p + z i where: p

zi pi E,

+&..

rJ

= p i +&..

= overall

'

mean

= ithtreatment

{

i = 1 , 2 , ..., a j = 1 , 2 , ..., n

effect

th

- i.i.d. N (0, 0):

treatment mean = random error i.i.d. N (0, 02)

=i

-

ww

A

a>

e

a

13-14 Chapter 13 Model (cont.)

Design and Analysis of Single-Factor Experiments: The Analysis of Variance (Notes) 1. The random variables zi and E,. are assumed independent of each other. 2. Y, ~ ( y , , o ;+ a 2 )

-

3. In the random-effects model, testing the individual treatment effects (zi) are meaningless because the treatments are selected at random. ANOVA

The analysis of variance procedure of the fixed-effects model is valid for the random-effects model. The computation procedure for an ANOVA table in the fixed-effects model (see Section 13-2.2) is identical to that of the random-effects model. However, the conclusion of ANOVA for the random-effects model can be extended to the entire population of treatments. This generalization of conclusion is due to random selection of treatments in the experiment. The variance components (0: and 02) of the random-effects model can be and MSE: estimated by using MSTreatnebts

Therefore, the estimate of the variance of YV is

B(y,)=8: Hypothesis Test (F-test)

+ir2

Step 1: State Ho and H I . Ho: o,2 = 0 HI: 0 : f O (Note) If all treatment effects are identical, between treatment effects, a:

#

o:

= 0 ; but if there is variability

0.


Fo= "Treatments '(a SS, l(N - a)

- MSTraatments MSE

- F(u-1,N-a)

Step 3: Determine a critical value(s) for a. fa,a-l,(\i-a

Step 4: Make a conclusion. Reject Hoif f 0

ft

Example 13.5

'fa,a-1.~-a

(Estimation of Variance Components) Consider the tensile strength data in Example 13-1. Assume that the four levels of hardwood concentration are randomly selected. The following summary quantities are obtained from the analysis of variance: n = 5 , MSTreatments = 109.7 , and MS, = 7.9 Estimate the (1) variability in tensile strength due to random error (02), (2) variability due to hardwood concentration (0,2) and , (3) total variability in tensile strength (V(Yg)).Also interpret these variability estimation results.

13-4 Randomized Complete Block Design

13-15


These variance estimates indicate that about 72% (= 20.4128.3) of the total variability in tensile strength is attributable to hardwood concentration.

bf;

Exercise 13.5

13-4

Consider the tube conductivity data in Exercise 13-1. Assume that the five coating types are chosen at random. The following summary quantities are obtained from the analysis of variance: n = 4 , MS,,e,,,ent, =265.1, and MS, =16.2 Estimate the (1) variability in tube conductivity due to random error (02), (2) variability due to coating type (o:), and (3) total variability in tube conductivity (V(Yij)).Also interpret these variability estimation results.

Randomized Complete Block Design

0 Distinguish between complete block design and incomplete block design. Explain the effects of blocking on the error term. Test the significance of a fixed-effects factor in a randomized complete block design. Perform multiple comparisons by Fisher's least significant difference (LSD) method.

Complete vs. Incomplete Block Designs

A complete block design contains all the treatments in each block (see Table 135), while an incomplete block design may omit one or more treatments in each block. Recall that blocking is used to balance out the effect of a nuisance factor (heterogeneity between experimental units) on the response variable.

Table 13-5 Data Structure of a Complete Block Design

Treatments

Blocks

1

2

...

i

...

a

Totals

Means

13-16 Chapter 13

Context

Model

Design and Analysis of Single-Factor Experiments: The Analysis of Variance 1 . Factor: Fixed-effects factor with a treatments 2. Blocking: Complete block design with b fixed-effect blocks 3. Repetition: One observation for each combination of factor levels and blocks (Note) N (total number observations) = a x b x 1 4. Randomization: Restricted randomization (randomization within each block) i=l,2,...,a

Yv = = + T i + P j +Eii = y i + P i

+E..

j = 1 , 2 , ..., b

where: y = overall mean

ri = fh treatment effect,

ri = 0 i=l

b

Pi

=jth block

P

effect,

=0

;=I

k i = ithtreatment mean cii (Notes) 1.

= random

error

- i.i.d. N (0, 02)

Y, - N(yi, 0 2 )

2. The treatments and blocks are assumed to be independent of (not interacting with) each other. 3. Since the treatments and blocks are assumed to be fixed-effects factors in the model, the treatment effects (ri) and block effects (Pj) are defined as deviations from the overall mean; thus, the sums of ri's and Pj's are zero. ANOVA

The analysis of variance method is used to test the significance of a factor in a randomized complete block design. The total variability in Y (SST, total sum of squares; degrees of freedom = N - 1) is partitioned into three components: (1) Variability explained by the treatments (SSTreatments, treatment sum of squares; degrees of freedom = a - 1) (2) Variability explained by the blocks (SSsloCh,block sum of squares; degrees of freedom =b - 1) (3) Variability due to random error (SSE, error sum of squares; degrees of freedom = (a - l)(b - I)) In other words,

a

b

a

b

2

where: S S , = z z ( Y , - ~ . ) 2 = ~ , ~2 Y--Y.. , i=l ~ T I i=1 ,=I ab


13-17

ANOVA (cont.)

The ratio of S S T ~ , , ~ ~-, ~1)~to, ~SSEI(a ( ~ - l)(b - 1) follows an F distribution:

This statistic F is used to test Ho:TI = TZ = ... = = 0, because F becomes as the effect of the factor is insignificant. The quantities MSTreatments(treatment ~ ~ mean ~ ~ square), ~ and MSE (error mean square) mean square), M S B (block ~ SSE ~ ~by ~ their , corresponding degrees of are adjusted SSTreatments, S S B ~and freedom, respectively. Note that

MS, =

ss~

(a - l)(b - 1)

=62

The variation quantities from the ANOVA analysis are summarized in Table 13-6 to test the significance of a factor under consideration. Table 13-6 ANOVA Table for Testing the Significance of a Factor: Randomized Com~lete ,~ .Block -Design -0.Source of Sum of Mean Degrees of Variation Squares Freedom Square Fo Treatments SSTreatmentS a-1 MS~reatments MSTreatmentS / MSE -

Blocks Error Total Effectiveness of Blocking

-

~

~-

S S ~ ~ o ~ k ~

SSE

b-1 (a - l)(b - 1)

M S ~ l ~ ~ k ~

MSE

N-1

Compared to a completely randomized design (presented in Section 13.-2.2), a randomized complete block design reduces the sum of squares and degrees of freedom for error. In the block design the error sum of squares (see Figure 13-1) and corresponding degrees of freedom are decreased by SSslocks and b - 1 (= (N a) - (a - l)(b - 1); N = ab), respectively.

Figure 13-1 The effect of blocking on error sum of squares (SSE):(a) completely randomized design; (b) randomized complete block design.

13-18 Chapter 13


Effectiveness of Blocking (cont.)

The effectiveness of blocking depends on the significance of block effects. If the effect of blocks is significant, a complete block design will result in a smaller error mean square (MSE) than a completely randomized design, which makes the test more sensitive to detect the effect of treatments. In general, when the significance of block effects is uncertain, apply blocking and check if the block effects exist. There would be a slight loss in the degrees of freedom for error, which may be negligible in the test unless the number of trials is very small.

Hypothesis Test (F-test)

Step 1: State Ho and H I . Ho: T I = T ~ ... = =z,=O HI: T ~ # O foratleastonei,i= 1 , 2,..., a Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. fa,u-l,(u-l)(b-1)

Step 4: Make a conclusion. Reject Hoif f 0

'

fa,a-l,(u-l)(b-1)

Multiple Comparison Methods

The multiple comparison methods for the completely randomized design (see Section 13-2.3) are applicable to the complete block design. As an example, the value of Fisher's LSD is

Residual Analysis

The complete block design model has three assumptions on the error term: (1) normality, (2) constant variance, and (3) randomness. These assumptions can be checked by analyzing residuals e, = Y , -5, = Y , -(Y, +Y, - F These residuals are analyzed by the model adequacy assessment methods ex~lainedin Section 13-2.5.

Example 13.6

The effect of grip angle on maximum grip strength (unit: Ib.) is under study. Three grip angles (a = 3; 20°, 50°, and 80" from the horizontal) are selected by the experimenter. Five participants are recruited in the experiment; and each participant exerts hisfher maximum strength at the three different grip angles presented at random. The grip strength measurements are as follows:

I

13-19



5 *--. Totals Averages

? X ~ . X ' X / " ~ - ~ n

62 297

"iW"r*I.w-"UIU

".-7 1

-~m-a~,mas--

65 317

338

198 952

66.0

~ ~ . B ~ B . . B ~ " . ~ - m " m m a ~,-mv..*m.m,.mmeasae

The following sum a

N = 15,

b

yxy y 2

a

= 62,132,

i=1 j = l

y i 2 = 302,942, and

b

Y,j2

= 185,794

j=l

i=l

1. (Test on the Significance of a Fixed-Effects Factor; Complete Block Design) Perform the analysis of variance to test if there is any significant difference in maximum grip strength due to grip angle at a = 0.05. D T ~The analysis of variance of the grip strength data are as follows:

Source of Variation

Sum of Squares

Grip Angle

168.1

2

84.1

Participant

1,511.1

4

377.8

Error

32.5

8

4.1

Total

1,711.7

14

Mean Square

Degrees of Freedom

Fo 20.7

Step 1: State HOand H I . Ho: z1 = z 2 = ... =I-,=() HI: zi#O foratleastonei,i= 1 t o 3 Step 2: Determine a test statistic and its value.

f o = "~reatments S ( a -1

'1

- - MS~reatments = - 1) MS,

841= 20.7 4.1

Step 3: Determine a critical value@) for a. fa,a-l,(a-l)(b-1)

= f0.05,3-1,(3-1)(5-1)

Step 4: Make a conclusion. Since fo = 20.7 > fa,a-l,(a-l)(b-l)

= f0.05,2,8 = 4.46

= 4.46, reject HOat a = 0.05.

I/ 11I

13-20 Chapter 13


I


2. (Least Significant Difference Method) Compare these three treatment means by using Fisher's LSD method at a = 0.05. ~hStep . 1: Determine the LSD for a.

I

Step 2: Arrange the means of a treatments in descending order. 7,. = 67.6, j7,, = 63.4, and 7,.= 59.4 Step 3: Compare the difference between the largest and smallest means with the LSD. Continue this comparison with the next smallest mean as long as the mean difference is greater than the LSD. 2 VS.1 = 67.6 - 59.4 = 8.2 > 3.0 2 VS.3 = 67.6 - 63.4 = 4.2 > 3.0 Step 4: Continue Step 3 for the next largest mean until this iteration reaches the second smallest mean. 3 vs. 1 = 63.4 - 59.4 = 4.0 > 3.0 Step 5: Summarize the painvise comparison results. 2 3 1 (significantly different from each other)


3. (Calculation of Residual) Calculate the residual of y34= 79. e34=y34- j 3 4= y 3 4-(Y3, +J.4 -j7,,)=79-(63.4+77.0-63.5)=2.1

An experiment is conducted to investigate leaking current in a near-micron SOS MOSFETS device. This experiment investigates how leakage current varies as the channel length changes. Four channel lengths (a = 4) are selected by the experimenter; and for each channel length five different channel widths (b = 5) are used. Suppose that channel width is considered a nuisance factor. The data are as follows:

i = ~ j=l

i=l

-

.j=1

1. Perform the analysis of variance to test if there is any significant difference in leaking current due to the difference in channel length at a = 0.05. 2. Compare these four treatment means by using Fisher's LSD method at a = 0.05. 3. Calculate the residual = 0.8.

I1

q

MINITABApplications

13-21

MINITABApplications Examples 13.1 - 3

I

(Fixed-Effects Factor; Completely Randomized Design)

(1) Choose File

* New, click Minitab Project, and click OK. orksheet.

Click Comparisons. Check Fisher's, individual er'ror rate and type '5'

E'EI -

,,.-Ji =----[

..

"

.-

.-..a

23

1

.--.-..

8

L3

SLL'Z SLL'b-

SZO'ESLS'OI-

ST

S U O S T l W d U O S aSTl&.lIWd S , I a q S I 4

...................................................................................... O'SZ

O'OZ

O'SI

0.01

8

=

n a a a s parood

+---------+---------+---------+-----(-___ *__-__)

I88'Z

OOb'IZ

S

0Z

snsJaa slanp!saa u~ ..xapJo snsJaa slanppaa pue ' s ~ gsnsJaa slanp!saa 'slanp!sa~jo $old l a u r ~ y3ay3 o ~ 's)old lanp!saa lapun 'sqda~f)y3q3 (s)

a3ue!leA jo s!sdleuv aqL :sluaurpadxg.roped-alZu!sjo s!sd1euv pue uZ!saa

I

sa1durex3 -

E 1 laldeq3

zz-c c

MINITABApplications

Example 13.4

13-23

I (Sample Size Determination; Fixed-Effects Factor) (1) Choose Stat % Power and Sample Size % One-way ANOVA. Enter the number of treatments (a)in Number of levels. Click Calculate power for each sample size and enter the range of sample size to be examined in Sample sizes and true means of the treatments in Means. Enter also

(2) In Significance level enter the probability of type I error (a).Then click OK twice.

edefined test condition.

I

I

Power and Sample Size

Sigma = 3 Alpha = 0.05 Nunber of Levels = 4 Corrected Sum of Squares of Means = 36.75 Means = 12, 15, 18, 20

......................... Sample Size 3

4 5

Power

i

0.6253 ; 0.8308 I 0.9322 j 0.9752

6 1 .........................

4

w

13-24 Chapter 13

Example 13.6

I


(Randomized Complete Block Design) (1) Choose File % New, click Minitab Project, and click OK.

Click Graphs. Under Residual Plots, check Normal plot of residuals. Residuals versus fits, and Residuals versus order. in Remsiduals versus - 2

7 MINITAB


Applications

13-25

Click Results. In Display means corresponding to the terms, select

I

Analysis of Variance (Balanced Designs)

Factor Angle Partlcip

Type Levels Values fixed 3 1 fixed 5 1

2 2

3 3

4

5

Analysis of Variance for Strength :"""'..."""" ....................................................................................... :Source :Angle ~Particip :Error :Total

DF

SS

BS

2 4 8 14

168.13 1511.07 32.53 1711.73

84.07 377.77 4.07

20.67

0.001

.........................................................................................................

:

I Means Angle

N

Strength

1 2 3

5 5 5

59.400 67.600 63.400

13.6.1


13-26 Chapter 13

Answers to Exercises 1. (Test on the Significance of a Fixed-Effects Factor; Completely Randomized Design) The sums of squares for the analysis of variance are as follows:

Exercise 13.1

This analysis of variance is summarized in the following table:

Source of Variation

Sum of Squares

Degrees of Freedom

Mean Square

fo

I

Coating type

1,060.5

4

265.1

16.3

Error

243.3

15

16.2

I

Total

1,303.8

19

I

I

Step 1: State HOand H I . Ho: z l = z2= ... = z a = O H I : zi#O f o r a t l e a s t o n e i , i = 1 t o 5 Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. fa,a-1,~-a

-

- f0.05,5-1,20-5

= f0.05,4,15 = 3'06

Step 4: Make a conclusion. Since f , = 16.3 >fa,a-l,N-a= 3.06, reject Ho at a = 0.05.

1 2. (Confidence Interval on pi)

1

1-a=0.95 3 a=0.05 95% two-sided CI on LI I :



Exercise 13.2

13-27

( 3. (Confidence Interval on pi - &)

1

95% two-sided CI on p 2 - p 3:

(Least Significant Difference Method) Step 1: Determine the LSD for a. LSD = ta,2,N-a--

- 2.131 x 2.846 = 6.07

Step 2: Arrange the means of a treatments in descending order. y2. = J 5 ,=145.3, J,, =145.0, J3. =131.5, and 7,. = 129.3 Step 3: Compare the difference between the largest and smallest means with the LSD. Continue this comparison with the next smallest mean as long as the mean difference is greater than the LSD. 2, 5 vs. 4 = 145.3 - 129.3 = 16.0 > 6.07 2, 5 vs. 3 = 145.3 - 131.5 = 13.8 > 6.07 2, 5 vs. 1 = 145.3 - 145.0 = 0.3 < 6.07 Step 4: Continue Step 3 for the next largest mean until this iteration reaches the second smallest mean. 1 vs. 4 = 145.0 - 129.3 = 15.7 > 6.07 1 vs. 3 = 145.0 - 131.5 = 13.5 > 6.07 3 vs. 4 = 131.5 - 129.3 = 2.2 < 6.07 Step 5: Summarize the painvise comparison results by arranging the a treatments in order of mean and then underlining treatments whose means are not significantly different. 2 5 1 3 4

Exercise 13.3

Exercise 13.4 I

E

(Calculation of Residual) e2, = y2, - y2, = y2, - J 2 , = 143 - 145.3= -2.3

(overestimate)

(Sample Size Determination; Fixed-Effects Factor) Use the operating characteristic curve of Figure 13-6(b) in MR for a fixedeffects factor with vl = 4 (= a - 1 = 5 - I). From the OC curve, the power of the test according to n is found as follows:

13-28 Chapter 13



Thus, at least n = 3 replicates are needed to achieve the designated power of the test.

Exercise 13.5

I (Estimation of Variance Components; Random-Effects Factor) I ( I ) B2 MS, = 16.2 =

I Exercise 13.6

These variance estimates indicate that about 79% (= 62.2178.4)of the total variability in tube conductivity is attributable to coating type. 1. (Test on the Significance of a Fixed-Effects Factor; Complete Block Design) The analysis of variance of the leaking current data is as follows:

1 SS~reatments

=-

" i=1

2 Y.. ~ i .-

-

219.4 27.72 T - -= 5.5 20

13-29



I

Source of Variation

Sum of Squares

Degrees of Freedom

Mean Square

fo

Channel Length

5.5

3

1.8

4.4

Channel Width

3.6

4

0.9

Error

5.0

12

0.4

Total

14.1

19

Step 1: State Ho and H I . Ho: z, = z 2 = ... = ~ , = 0 H I : z i + 0 for at least one i, i = 1 to 4 Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. fa,a-l,(a-l)(b-1)

= f0.05,4-1,(4-1)(5-1)

= f0.05,3,12 = 3'49

Step 4: Make a conclusion. Since f o =4.4> fa,a-l,(a-l)(b-l, =3.49, reject Ho at a = 0 . 0 5 .

2. (Least Significant Difference Method) Step 1: Determine the LSD for a .

Step 2: Arrange the means of a treatments in descending order. Y3.=1.9, Y4.=1.9, y2, =0.9, and y,, =0.8 Step 3: Compare the difference between the largest and smallest means with the LSD. Continue this comparison with the next smallest mean as long as the mean difference is greater than the LSD. 3 , 4 VS.1 = 1.9 - 0.8 = 1.1 > 0.87 3 , 4 vs. 2 = 1.9 - 0.9 = 1.0 > 0.87 Step 4: Continue Step 3 for the next largest mean until this iteration reaches the second smallest mean. 2 VS. 1 = 0.9 - 0.8 = 0.1 < 0.87 Step 5: Summarize the painvise comparison results by arranging the a treatments in order of mean and then underlining treatments whose means are not significantly different. 3 4 2 1

3. (Calculation of Residual) ezl = yZ1- j Z 1= y21- ( j 2 ,+Y,l -j,.)=0.8-(0.9+0.9-1.4)=0.4

I

1

14 14-1 14-3 14-4 14-5 14-7

14-1

Introduction Factorial Experiments Two-Factor Factorial Experiments General Factorial Experiments 2k Factorial Designs 14-7.1 22 Design 14-7.2 2k Design for k 2 3 Factors 14-7.3 Single Replicate of the 2k Design


14-8 Blocking and Confounding in the 2k Design 14-9 Fractional Replication of the 2k Design 14-9.1 One-Half Fraction of the 2k Design 14-9.2 Smaller Fractions: The 2k-p Fractional Factorial MINITABApplications Answers to Exercises

Introduction

This chapter presents statistical concepts and methods to design an experiment with multiple factors (2 2) and analyze the data obtained from the experiment. Most of the techniques introduced in Chapter

13 for a single-factor experiment are extended to this multiple-factor experiment.

14-3

Factorial Design

CI Explain the term factorial design. Distinguish between main and interaction effects. Explain why information of interactions between factors is more meaningful than that of the main effects when the interactions are significant. O Explain the potential pitfall of a one-factor-at-a-time design when interactions exist.

14-2

Chapter 14

Factorial Design

Main vs. Interaction Effects

Design of Experiment with Several Factors A factorial design is an experimental design where all combinations of factor levels under consideration are tested at each replicate. For example, if factors A and B have two and three levels, respectively, a factorial design of A and B runs six (= 2 x 3) combinations of the levels at each replicate. The effect of a factor indicates the change in response by a change in the level of the factor. Two types of factor effects are defined: 1. Main effect: The effect of a factor alone. 2. Interaction effect: The effect of an interaction between factors. An interaction exists between factors if the effect of one factor depends on the condition of the other factor(s). Figure 14-1 shows that the main effect of A is independent of the level of B and vice versa, indicating the absence of an interaction between A and B. In other words, the main effects of A at the low and high levels of B are the same as 20 (i.e., 30 - 10 = 20 at the low level of B and 40 - 20 = 20 at the high level of B). Similarly, the main effects of B at the low and high levels of A are the same as 10. This insignificant AB interaction can be identified by calculating the difference of the diagonal averages:

Notice that the Bhighand Blowlines in the figure are parallel to each other when the interaction of A and B is absent. Response 50

0

A low

A high

Factor A

Factor A A~OW Ahigh Factor B

Blow

Bhizh Totals Averages

~vm-~~*-wea~ B%v---

eBa v-s

-aBa"--a.e

a

10 20 30 - *15&V

?

mw

v

sv

ve

30 40 70 35 *

d b BA

*
fa,@-l)(b-l),N-ab

for AB

It is recommended that the test for interaction be conducted first followed by the tests for main effects. If interaction is insignificant, the tests for main effects are meaningful; otherwise, the tests for main effects are meaningless because the effect of a factor is subject to the condition of the other factor(s).

Multiple Comparison

For significant fixed factors, the means of their individual treatments are compared by a multiple comparison method such as Fisher's least significant difference (LSD) method (see Section 13-2.3). When interaction AB is insignificant, the treatment means of each significant main effect are compared. When AB is significant, the treatment means of one factor (say A) at a particular level of the other factor (say B) are compared. For a fixed two-factor factorial design, the LSDs are determined as follows: LSD = ta,2,N-ub

JT

2MSE

for A

.,/%

-ta/2,~-ab

for B for AB

Residual Analysis

Like the single-factor model (shown in Section 13-2.2), the two-factor factorial model has three assumptions on the error term: (1) normality, (2) constant variance, and (3) randomness. These assumptions can be checked by analyzing residuals (eyk= yijk- jijk = yqk- Yy, ) as explained in Section 13-2.5.

14-8

Chapter 14

Factorial Design with a Single Replicate

Design of Experiment with Several Factors Sometimes a two-factor factorial experiment has only one observation (n = 1) for each combination of the treatments. In this case, the error degrees of freedom become zero (= N - ab = nab - ab for n = 1) and thus the error mean square is unavailable. In this unreplicated design, the interaction mean square can be used as the error mean square, assuming the interaction effect is negligible. However, this nointeraction assumption can yield an improper conclusion if the interaction is significant; thus, data and residuals should be carefully examined to check the significance of the interaction. An engineer suspects that the surface finish (Y) of metal parts is influenced by drying time ( A ) and type of paint (B). Three drying times (a = 3; 20,25, and 30 min.) and two types of paint (b = 2) are specifically selected by the engineer and three parts (n = 3) are tested for each combination of the drying times and paint types, yielding the following: Drying T i m (A; unit: min.) 20 (1) 25 (2) 30 (3) Totals Means 74 73 78 1 64 61 85 62 1 69.0 Paint 50 44 92 Type (B) 2 86 73 45 70 1 77.9

Example 14.1 (MR 14-2)

.~~a*"d~8w-.**w'a.8""~w~~.B.~a"~.m

1

mmmwel--___-

*"-

Totals Means

72.3

,m,wa.a"sasw-b

'*,a"

72.8 -._,.. ' . " ~ . ~75. " .."^ ~.~,"~"~"

."m*d"e_"

The following summary quantities are also provided for the surface finish data:

zzx a

N = 18, and

b

n

yVk2= 101,598,

z b

a

y i 2 = 582,726,

y , j , 2= 877,042,

7'2

y,,' = 298,066

1 . (Test on the Significance of Main and Interaction Effects; Two-Factor Factorial Design) Test if the effects of A, B, and AB are significant each at a = 0.05 by the analysis of variance. The analysis of variance of the surface finish data is as follows:

1

--

-

14-9

14-4 Two-Factor Factorial Experiments


The ANOVA results are summarized in the following table: Saurce of Sumof Degrees of Mean VWw Squaw Freedom Square 27.4 2 13.7 Drying Time (A)

0.07

355.6

1

355.6

1.90

AB

1,878.8

2

939.4

5.03

Error

2,242.7

12

186.9

Total

4,504.4

17

Paint Type (B)

Step 1: State Ho and H I . Ho:zi = 0 for i = 1 to 3 HI:zi # 0 for at least one i

) for A

Ho:P j = O f o r j = l and2 HI:Pi # 0 for at least onej

) for B

Ho:( T P )=~ 0 for all combinations of i's and j's HI:( T P )#~ 0 for at least one combination of i andj

h

) for AB


MSB - 355.6 - 1 for B MS, 186.9 MSAB - 939.4 fo= -- -= 5.03 for AB MS, 186.9 f0 -----

Step 3: Determine a critical value(s) for a. fa,a-1,N-ab = f0.05,3-1,18-6 fa,b-1,N-ab

= f0.05,2,12 = 3'89 for A

= f0.05,2-1,18-6 = f0.05,1,12 = 4'75

fa,(a-l)(b-l),N-ab

for

- f0.05,(3-1)(2-1),18-6 = f0.05,2,12 = 3'89 for AB

Step 4: Make a conclusion. Since fo = 5.03 > fa,(a-l)(b-l),N-ab = 3.89, reject Hoat a = 0.05 for AB. Due to the significant interaction, tests on the main effects of A and B are meaningless.

14-10 Chapter 14

Design of Experiment with Several Factors 2. (Multiple Comparison) Compare the surface finish means of the three drying times for paint type 1. The corresponding surface means are =85.0 YII, =62.7, y2,.=59.3, and


rm

Step 1: Determine the LSD for a.

Step 2: Arrange the means of a treatments in descending order. y3,,=85.0, TI,,=62.7, and Y21, =59.3 Step 3: Compare the difference between the largest and smallest means with the LSD. Continue this comparison with the next smallest mean as long as the mean difference is greater than the LSD. 3 VS. 2 = 85.0 - 59.3 = 25.7 > 24.32 3 VS. 1 = 85.0 - 62.7 = 22.3 < 24.32 Step 4: Continue Step 3 for the next largest mean until this iteration reaches the second smallest mean. 1 vs. 2 = 62.7 - 59.3 = 3.40 < 24.32

I

Step 5: Summarize the painvise comparison results by arranging the a treatments in order of mean and then underlining treatments whose means are not significantly different. 3 1 2

I

3. (Calculation of Residual) Calculate the residual 0fy313= 92. h


e313= y,,, - j313 = Y , , ~ - j731, = 92 - 85 = 7 (underestimate)

An experiment is conducted to determine whether either firing temperature ( A ) or furnace position (B) affects the baked density (Y) of a carbon anode. Three firing temperatures (a = 3) and two furnace positions (b = 2) are specifically selected by the experimenter and three measurements (n = 3) are collected for each combination of the treatments. The measurements are as follows:

Temperature (A; unit: O C ) 1

SOO(1) 5.70 5.65

825 (2) 10.63 10.80

850 (3) 5.65 5.10

TOMS

Means

65.69

7.30

5.47 5.21 33.14 5.52

10.26 10.04 62.04 10.34

5.38 5.32 32.61 5.44

62.10

6.90

Position (B) 2 Totals Means

127.79

7.10 I1 The following summary quantities are also provided for the baked density data:

(

a

N =18, ~

b

n

~ 2 =1,003.1, ~

gy,'

b

=6,010.6, Y Z CY,~ =8,171.6, ,'

-

-

14-5 General Factorial Experiments


and

zC

14-11

yb,2 = 3,007.7

1 1. Test if the effects of A, B, and AB are significant each at a = 0.05 by the analysis of variance. 2. Test the differences between the baked density means of the three firing temperatures (A): TI,,= 5.52 , y,,, = 10.34, and y,,, = 5.44 3. Calculate the residual ofy213= 10.43.

14-5

General Factorial Experiments

Factorial Design with More Than Two Factors

14-7

An experiment can include more than two factors. The analysis of variance procedure for this general factorial experiment is just an extension of that of a two-factor factorial design but requires tedious computation; thus, a computer software program is mostly used for the analysis.

2k ~actorialDesigns

14-7.1 2* Design

O Estimate the main and interaction effects of a 22 factorial desi n.

0 Test the significance of the main and interaction effects of a 29 factorial design. 2k Design

A 2k factorial design is a factorial design with k factors, each having two levels. The low and high levels of each factor are denoted by - (or '-1 ') and + (or ' 1'), respectively. A complete replicate of this design requires 2k observations, which is the smallest number of observations for a complete factorial design with k factors. Due to the minimum number of runs, the 2k design is particularly useful for an experiment with many factors for screening purposes. Two treatments are chosen for each factor by assuming a linearity of the response over the range of the factor levels.

ANOVA for 22 Design

The totals of the treatment combinations in a 22 design (2k design where k = 2 factors) are denoted by a series of lowercase letters: (I), a , b, and ab (see Figure 14-3). The presence (absence) of a letter indicates that the corresponding factor is at the high (low) level in the treatment combination. For example, the letter a indicates that factor A is at the high level and factor B is at the low level.

14-12 Chapter 14

Design of Experiment with Several Factors high (+I; +)

ANOVA for 2' Design (cont.)

low (-1; -)

low (-1;

A

high (+I; +)

-1

Figure 14-3 The 22 factorial design. The main effects A and B and interaction AB with n replicates are estimated as follows:

The linear combinations in the numerators of the main and interaction equations are called orthogonal contrasts (denoted as C) as they have the following two properties: (1) Each linear combination has coefficients {ci) whose sum is zero i

x

(2) Any two contrasts have coefficients {ci) and {dif whose sum of the products of the coefficients is zero ( cidi = 0 ). i

By using the orthogonal contrasts, the sums of squares for A, B, and AB are calculated: C: - [a+ab-b-(1)12 SS, = 74n

nC

ci

i=l

[ab + (1) - a - b12 4n

-,

Note that the total and error sums of squares are calculated by 2

2

ssT = c c R Y : i=1 j=l k=l

2

Y...

14-13

14-7 2k Factorial Designs

ANOVA for 22 Design (cont.)

Since each contrast sum of squares has a single degree of freedom, the mean squares of A, B, and AB are equal to the corresponding sums of squares each. The mean squares of error is M s E =--- SS, - ss, 4(n-1) N - 4 The ratios MSA/MSE,MSBIMS~, and MSAB/MSEhave F distributions:

These F values become small as the effects of A, B, and AB are insignificant (i.e., Ho: zi = 0 , Ho: P , = 0 , and Ho:( T P ) ~= 0 are true). The variation quantities from ANOVA are summarized in Table 14-3.

*"" "

Hypothesis Test (F-test)

B AB

SSB SSAB

Error

SSE SST

Total -

*

_

X

m

*

~

-

-

~

~

*

-

.

^

I__

-

^"""*

",-

1 1 N-4 N-1

Step 1: State HOand H I . Ho: zi = 0 for i = 1 and 2 H I : zi # 0 for at least one i Ho: HI:

Pi = 0

Ho: HI:

( T P ) =~ 0

for j = 1 and 2 & # 0 for at least onej

#

MSE X1 L

X

I----

''A~

=--

-

XIWII-"-

'I

SS, I 4(n - 1 )

---

- MSAB

MSE

r

X1

awemIIX^X V

) for A ) for B

Step 2: Determine a test statistic and its value. ss, 11 - MSA F(1,4(n - I)) F,, = SS, I4(n - 1 ) MSE SS, I1 --MSB F(1,4(n - 1 ) ) F,, = SS, I 4(n - 1) MS,

F,, =

X1

for all combinations of i's and j's 0 for at least one combination of i and j

---

MSB I MSE MSABI MS,

MSB M~AB

for A for B

F(1,4(n - 1)) for AB

) for AB

14-14 Chapter 14 Hypothesis Test (cont.)

r

Design of Experiment wit11 Several Factors Step 3: Determine a critical value(s) for a. fa,l,4(n-l) for A, B, and AB Step 4: Make a conclusion. Reject HOif for A, B, and AB fo > f a , l , 4 ( n - l )

Example 14.2

Exercise 14.2

(Calculation of Effects and Sums of Squares; 22 Design) Consider the surface finish data in Example 14-1. Assume that only two drying times (20 and 30 min.; a = 2) and two types of paint (b = 2) were tested. The following summary quantities are provided for the surface finish data: n = 3 , (1)=188, a = 2 5 5 , b = 2 4 6 , a n d ab=196 Estimate the effects A, B, and AB and calculate the corresponding sums of squares. nThe orthogonal contrasts of the 22 design for the surface finish data are C , =a+ab-b-(1)=255+196-246-188=17 C B =b+ab-a-(1)=246+196-255-188=-1 CAB=ab+(l)-a-b=196+188-255-246=-117

I

Thus, the estimates of the main and interaction effects are

I

Then, the corresponding sums of squares are

Consider the carbon anode baked density data in Exercise 14- 1. Assume that only two firing temperatures (800 and 850 "C; a = 2) and two furnace positions (b = 2) were tested. The following summary quantities are provided: n = 3 , (1)=17.18, a =16.65, b = 15.96, and ab=15.96 Estimate the effects A, B, and AB and calculate the corresponding sums of squares.


14-15

14-7.2 2* Design for k 2 3 Factors

O Identify the orthogonal contrasts of a 2k design. Estimate main and interaction effects in a 2k factorial design. -mmwa

"mmeb*a-m--

Orthogonal Contrasts of 2k Design

sms

-

-

m e

*--

----

m-

aswsd-

---*bms

m h w

* *

--

--%.

As shown in the anal sis of variance for a 22 design, orthogonal contrasts are useful to analyze a 2Ydesign. The effects of factors and their interactions and corresponding sums of squares can be calculated by using orthogonal contrasts: Contrast Effet = n2k-'

To determine the orthogonal contrasts of a 2k design in a systematic manner, a table of plus and minus signs is used (see Table 14-4 for a 23 design). This table includes columns of treatment combinations, identity (I), main effects, and interactions. In a main effect column, a plus (minus) sign indicates the high (low) level of the corresponding factor for the treatment condition (e.g., in the main effect column A, a plus is assigned to the treatment combination ac because the level of A is high. Once the signs of the main effects are established, the signs of each interaction column are determined by multiplying associated columns row by row (e.g., AB =A x B). Except the identity column, each column includes an equal number of plus and minus signs. By using the sign table, main and interaction effects and their sums of squares are calculated. For example, in Table 14-4, the main effect A and the corresponding sum of squares when n = 2 are determined as follows:

Table 14-4 Signs for Orthogonal Contrasts in the Z 3 Design

Treatment Combinations

C

ac bc abc

f

+ + + +

Effi,cts AB C

A

3

-

-

+

+

-

-

-

+

+ +

-

+

+ + + +

AC

BC

-

-

+ -

+

+ +

ABC

+ -

+

14-16 Chapter 14

Example 14.3


(Calculation of Effects and Sums of Squares; 2k Design) Power grip strength is measured on a handle to examine the effects of three handle design factors: diameter (A), angle (B), and weight (C). Two treatments are chosen for each factor and a single replicate (n = 1) is run. The grip strength measurements (unit: kg) are as follows:

Estimate the main and interaction effects of this 23 design and calculate the corresponding sums of squares. ~ - r rFrom Table 14-4, the orthogonal contrasts of the 23 design for the grip strength data are C A=a+ab+ac+abc-(1)-b-c-bc =40+33+42+34-35-31-36-30=17 CB = b+ab+bc+abc-(1)-a-c-ac =31+33+30+34-35-40-36-42~-25 CAB= ( l ) + a b + c + a b c - a - b - a c - b c =35+33+36+34-40-31-42-30~-5 C , =c+ac+bc+abc-(1)-a-b-ab =36+42+30+34-35-40-31-33=3 C,, = ( l ) + b + a c + a b c - a - a b - c - b c =35+31+42+34-40-33-36-30=3 CB, =(l)+a+bc+abc-b-ab-c-ac =35+40+30+34-31-33-36-42~-3 CAB,=a+b+c+abc-(1)-ab-ac-bc =40+31+36+34-35-33-42-30~1

I


I

The corresponding sums of squares are

14-17



d

An engineer is interested in the effects of cutting speed (A),metal hardness (B), and cutting angle (C) on the life of a cutting tool. Two levels are chosen for each factor and two replicates (n = 2) of this 23 factorial design are run. The tool life data (unit: hours) are as follows:


a

b ab c ac bc

43 5 348 472 453 377 500

325 354 552 440 406 605

760 702 1024 893 783 1105

380.0 351.0 512.0 446.5 391.5 552.5

Estimate the main and interaction effects of the factors and calculate the corresponding sums of squares.

14-7.3 Single Replicate of the 2* Design

Explain a method to analyze a two-factor factorial design with a single replicate. O Identify significant high-order interactions in a 2k design by a normal probability plot.

- -

'

-

2k Design with a Single Replicate

--

I X("

--^

ni

le-w.#i-n^-w-t

--wv*rraixi-lea

a,"

%

w^wB

?A

ar ra..

*,"-*ax-

i . i

a a

-"

i

xi

rx

-

-.*

llxa

Like the factorial design with a single replicate (presented in Section 14-4), an assumption is necessary for an unreplicated 2k design to have the error mean square. Interactions higher than a second-order interaction are often assumed negligible and then used as the error mean square. Caution should be exercised to check if the interaction(s) used for the error term is significant. Significant high-order interactions can be identified by constructing a normal probability plot of main and interaction effects. Negligible effects will be normally distributed with mean zero and variance 02and thus lie along a straight line in the normal probability plot, whereas significant effects deviate from the straight line (see Figure 14-21 of MR).

14-18 Chapter 14

lr

Example 14.4


(Normal Probability Plot of Effects) Consider the grip strength experiment with a single replicate in Example 14-3. The estimates of the main and interaction effects are as follows: A = 4.25 AC= 0.75 B = -6.25 BC= -0.75 C = 0.75 ABC= 0.25 AB= -1.25 Construct a normal probability of these effects and check if the third-order interaction ABC is significantly important. w The normal probability plot of the main and interaction effects below indicates that the second- and third-order interactions (including ABC) are not important. Thus, ABC can be used as the error term to test the significance of the main and other second-order interaction effects.

-5

0

5

Effect

d


14-8

An experiment has run a single replicate of a 24 design and calculated the following factor effects: A = 80.25 AB = 53.25 ABC = -2.95 B = -65.50 AC = 11.00 ABD = -8.00 C = -9.25 AD = 9.75 ACD = 10.25 D = -20.50 BC = 18.36 BCD = -7.95 BD = 15.10 ABCD = -6.25 CD =-1.25 Construct a normal probability of these effects and check if the third- and fourthorder interactions are significantly important.

Blocking and Confounding in the 2* Design

CI Construct 2P blocks for a 2k design. 0 Estimate main and interaction effects of a 2k design with 2P blocks.

14-8 Blocking and Confounding in the 2k ~ e s i ~ n 14-19

2k Block Design and Confounding

Sometimes all treatment combinations cannot be run under homogeneous conditions for various reasons. For example, suppose that a 2, design (with 4 treatment combinations) requires a total of 16 hours (4 hours for each combination) to run a single replicate; thus, this experiment is conducted in two days (blocks) and randomization is restricted within each block. This kind of a factorial design is called 2k design with 2* blocks, p < k. Some effects are confounded with blocks in a 2k block design. As an example, Figure 14-4 displays a 2' block design with interaction AB is confounded with the blocks. In this block design, the contrasts (C) to estimate the effects of A, B, and AB are C, =ab+a-b-(1) C, = a b + b - a - ( 1 ) C,, = a b + ( I ) - a - b However, the AB contrast is identical to that of the block effect and thus AB is confounded with the blocks. Block 1

Block 2

Figure 14-4 A 22 block design. Construction of Blocks

To construct a 2k design with 2' blocks, first,p effects should be selected for confounding. In general high-order interactions are chosen to confound with blocks; main and low-order interaction effects are less likely selected due to their importance. In addition to thep effects, 2P - p - 1 effects (interactions of thep effects; called generalized interactions) are confounded with blocks. For example, in a 23 design with 2' blocks, if AB and BC are selected for confounding, their generalized interaction AC (= AB x BC = AB'C because B~ = I) is also confounded. Care should be taken not to confound effects of interest. Second, based on the selectedp effects, defining contrasts (L) are formulated: L = a , x , +a,,x, +...+ a,xk where: a; =

0,

if the ith factor is not in the counfounded effect

1,

if the ith factor is in the counfounded effect

0,

if the ithfactor is not in the treatment combination

if the ithfactor is in the treatment combinaiton 1, For example, in case that AB and BC are selected for confounding in a 23 design with 2, blocks, two defining contrasts are established from AB and BC as follows: L, = X , + x2 and L, = x, + x, where: xi =

i

0,

if the ithfactor is not in the treatment combination

1,

if the ithfactor is in the treatment combinaiton

14-20 Chapter 14


Construction of Blocks (cont.)

r

n

I

*

m

Lastly, treatment combinations having the same values of L (modulus 2) will be placed in the same block. For example, for the 23 block design with L1 = xl + x2 and L2 = x2 + xg, the four blocks of the treatment combinations are determined according to their values of Ll (mod 2) and L2 (mod 2) as follows:

m

a

ANOVA

Example 14.5

a b

(0 + 0) (mod 2) = 0 (0 + 0) (mod 2) = 0 (1 + 0) (mod 2) = 1 (0 + 0) (mod 2) = 0 (0 + 1) (mod 2) = 1 (1 + 0) (mod 2) = 1

1 2 3

ac bc abc

(1 + 0) (mod 2) = 1 (0 + 1) (mod 2) = 1 ( 0 + l ) ( m o d 2 ) = 1 (1 + l ) ( m o d 2 ) = 0 + l x m o d 2J = 0 Q*-+A(mod 2) = 0

3 2 1

rannaxrrx-

-- --

smw

--*-

mwr

-----

-*-

The analysis of variance procedure by using orthogonal contrasts presented in Section 14-7.2 is applicable to the 2k block design except treating the confounded effect(s). The confounded effect(s) is redefined as the block effect and the significance of the block effect can be tested by comparing with the error mean square. Recall that in case of a single replicate experiment, a high-order interaction(s) can be used as the error mean square. The effects of backrest angle (A) and location of lumbar support (B) on the disc pressure (unit: N ) at the low back are under study. Two backrest angles (90" and 120") and two lumbar support locations (1 and 5 cm) are selected in the experiment. 1. (Construction of Blocks; 2k Block Design) Suppose that the four treatment combinations cannot be examined under the same condition. Establish two blocks with AB confounded for the 22 design. h The defining contrast is formulated from AB: L=xl +X2

I

where: xi =

I

i

0,

if the ith factor is not in the treatment combination

1,

if the ith factor is in the treatment combinaiton

The four treatment combinations are assigned to two blocks according to the

a b ab

(0 + 0) (mod 2) = 0 (1 + 0) (mod 2) = 1 (0 + 1) (mod 2) = 1 l L (-m od2 = 0 1 -+ ,, ,-,

*-

1 2 2 1

----.

a

I

2. (Calculation of Effects and Sums of Squares) The following disc pressure data are collected:

I

Calculate the sums of squares for this 2' block design with AB confounded.

-


14-21

14-9 Fractional Replication of the 2kDesign The orthogonal contrasts of the 22 design for the disc pressure data are C, =a+ab-b-(1)=320+155-380-620=-525 CB =b+ab-a-(1)=380+155-320-620=-405 CAB=ab+(l)-a-b=155+620-320-380=75

I

Thus, the estimates of the effects are

75 5= = 75

lock effect = AB = n2k-1

I


Consider the tool life experiment in Exercise 14.3. Suppose that the eight treatment combinations cannot be examined under the same condition. 1. Establish four blocks with AB and AC confounded for the 23 design. 2. Calculate the sums of squares for this 23 design with four blocks. The orthogonal contrasts of the tool life length data have been calculated as follows: C , = 146, C B = 674, Cc = 574, CAB= -90, CAc = -954, CBc = -194, and CAB, = -278


14-9

22-1

Fractional Replication of the 2k Factorial Design

14-9.1 One Half Fraction of the 2" Design

0 Describe the use of a fractional factorial design. Explain the terms generator, defining relation, principal fraction, alternate fraction, aliasing, and dealiasing. D Construct a 2k-' design. 0 Estimate the main and interaction effects of a 2k-' design. Identify the alias structure of a 2k-1 design. am-

* -*

*-

#u.mwm

s

"+

a m

a*

#

-"

sms

ww*

me-*-*

a

>--

- w w w w

"*

-wwA-wm

meawawm A " m w e - *

d w w w m a

-

w

*a

14-22 Chapter 14 2k-p Fractional Factorial Design

Of 2k

-I

Design

Design of Experiment with Several Factors A 1/2P(p < k) fraction of a 2k design (denoted as 2k-p ) can be run when highorder interactions are negligible. As the number of factors (= k) increases, the number of runs required increases geometrically but a large portion of the runs are used to estimate high-order interactions, which are often negligible. For example, a 25 design (k = five factors) which requires 32 runs uses, out of the 3 1 degrees of freedom, 5, 10, and 16 degrees of freedom to estimate the main effects, two-factor interactions, and three-factor and higher order interactions, respectively. When the three-factor and higher order interactions can be assumed negligible, a 25-' fractional factorial (which requires 16 runs) can be used to analyze the main effects and low-order interactions of the five factors. A 2k-1 design is constructed by selecting treatment combinations having the same sign (say, plus) on an effect column selected (called generator) in the table of signs for the 2k design. For example, a 23-1 design can be formed by selecting four treatment combinations (a, b, c, and abc) that have the plus sign on the ABC column in the sign table of the 23 design (Table 14-5). The fraction with the plus sign is called principal fraction and the other fraction is called alternate fraction. Since the identity column has only the plus sign, a 2k-1 design has the following relations (called defining relation) between the identity and generator columns: the signs of the generator column are equal to those of the identity column for the principal fraction and the opposite becomes true for the alternate fraction. For example, a 23-1 design with generator ABC has the defining relation I = ABC for the principal fraction and I = -ABC for the alternate fraction.

C

ac bc abc

+ + + +

-

-

+

+

-

-

-

+

+ +

-

+

+ + + +

-

+ -

+

-

+

-

-

+ +

-

+

Another method to construct a 2k-1 design is available as follows: Step 1: Prepare the sign table of a factorial design with k - 1 factors (called basic design). (e.g.) For a 23-1 design in Table 14-6, the basic design starts with two factors ( A and B). Step 2: Add the kthfactor and identify the corresponding interaction by using the defining relation of the fractional design. (e.g.) I = ABC in Table 14-6 3 C.I = A B C ~ =C=AB (forc2=~) Step 3: Determine the signs of the interaction for the kthfactor column. Step 4: Identify the treatment combinations of the 2k-1 design row by row.

14-9 Fractional Replication of the 2k Design Construction of 2k

14-23

Table 14-6 Construct~onof the 23-' Design with I = ABC

-I

Design

.&MI , ', ,A ,

(cont.)

1 2 3

""r

(1) a b

-

+ -

-

+

+ -

-

C

a b abc

a.r*,aq-~.e*."'Aa-"

Effects

In a 23-1 design with I = ABC (principal fraction in Table 14-5), the linear combinations that are used to estimate the main effects and interactions have the following relationships: I A -1 -2[a-b-c+abc]=l,c

I , =+[-a+b-c+abc]=IAc I , =+[-a-b+c+abc]=I,, These relationships indicate that each liner combination estimates both a main effect and an interaction: l,=A+BC

I,=B+AC I, =C+AB In contrast, a 23-' design with I = -ABC (alternate fraction in Table 14-5) produces the following estimates of the main effects and interactions: 1; = + [ - ( l ) + a b + a c - b c ] = A - B C

It is not possible to differentiate between A and BC, B and AC, C and AB. Thus, assuming the second-order interactions are negligible, the linear combinations lA (or I; ), 1, (or I; ), and Ic (or I: ) are used to estimate the main effects. A normal probability plot can be used to identify significant interactions (see Section 14-7.3). Negligible interactions will lie on a straight line in the normal probability plot, whereas significant ones deviate far from the straight line.

Aliasing

In a fractional factorial design, some effects cannot be estimated separately and these effects are called aliases. For example, in a 23-1 design with I = ABC, A and BC, B and AC, and C and AB are aliased with each other. An alias of an effect is found by multiplying the defining relation to the effect. For example, a 23-' design with I = ABC has the following alias structure: A = A.I = A.ABC = A2Bc = BC (for A2 = I) B = B.1 = B.ABC = A B ~ C= AC (for B2 = I) C = C.I = C.ABC = A B =~AB (for C2= I)

14-24 Chapter 14


Aliasing (cont.)

It is a common practice to form a fractional factorial design which aliases the main effects and low-order interactions with high-order interactions (which are mostly negligible).

Dealiasing

A sequence of fractional factorial designs can be performed to identify more detailed information of interactions. For example, a 23-' design with I = ABC aliases the main effects with the two-factor interactions, assuming that the twofactor interactions are negligible. However, if this assumption is uncertain, the alternate fraction can be run after running the principal fraction. By using the data from the principal and alternate fractions, the main effects and interactions can be estimated separately (called dealiasing) as follows: ~1 ( 1 +, I > ) = $ ( A + BC+ A - B C ) = A +(IA -I>)=BC

Projection of 2k-1

Design

fif

Example 14.6

When one or more factors are excluded from a 2k-' design due to their lack of significance, the fractional design will project into a full factorial design to examine the active factors in detail. For example, a z3-' design will project into a 22 design as one factor is dropped.

A study examines the effects of four factors on maximum acceptable weight (MAW; unit: kg) in lifting: object width (A),lifting level (B),lifting distance (C), and lifting frequency (D).A 24-1 design with I = ABCD is employed, yielding the following data:

'1 1. (Alias Structure; ABCD.

2'-' Design) Identify the alias structure of this 2'-' design

with I =

[)"

The 24-1 design with I = ABCD has the following alias relationships: A = A.ABCD = A ~ B C D= BCD B = B.ABCD = A B ~ C D= ACD C = C.ABCD = A B ~ = DABD D = DSABCD= A B C D ~= ABC AB = AB.ABCD = A ~ B ~ C=DCD AC = AC.ABCD = A 2 B 2 D = BD AD = AD.ABCD = A ~ B C D= ~BC

I

14-25

14-9 Fractional Replication of the 2k Design


2. (Estimation of Effects) Estimate the factor effects. Which effects appear large? mThe main effects and interactions are estimated as follows: I , = A + BCD=$(-24+32-24+16-37+15-17+18)=-5.25

I

I D = D + ABC=~(-24+32+24-16+37-15-17+18)=9.75 I,,

=AB+cD=~(~~-32-24+16+37-15-17+18)=1.75

I,,

=AD+BC=+(24+32-24-16-37-15+17+18)=-0.25 (Note) The interactions are estimated by forming columns of AB, AC, and AD in the sign table.


The effects A, B, D, and AC are relatively large.

A 24-' design with I = ABCD is used to study four factors in a chemical process. The factors are A = temperature, B = pressure, C = concentration, and D = stirring rate, and the response is filtration rate. The design and data are as follows:

---**

6 7 8

--w

+ +

-a -va-*-*

-

+ +

r n W * ~ ~ ~ ~ e -ww - *

+ + +

sw *a-

-a

wv m-

d

+

m e M # s a*-

*w

a

--

ac bc abed

ms-B&*mm

*e

* w#m

60 80 96 *

swaw

1. Estimate the factor effects. Which factor effects appear large? 2. Project this design into a full factorial including only important factors and provide a practical interpretation of the results.

14-9.2 Smaller Fractions: The 2 k - pFractional Factorial

0 Explain the conditions of design resolutions 111, IV, and V in a fractional factorial design. Construct a 2k-p fractional factorial with a designated design resolution.

0 Identify the alias structure of a fractional factorial design. Design Resolution

The term design resolution indicates the pattern of aliases in a fractional factorial design. The alias patterns of resolution 111, IV, and V designs are compared with each other in Table 14-7. A resolution I11 design (such as 23-' design with I = ABC) does not alias any main effects with other main effects but aliases main

14-26 Chapter 14

Design (cant.)


effects with two-factor interactions and some two-factor interactions with each other. Next, a resolution IV design (such as 24-' design with I = ABCD) does not alias any main effects with other main effects or two-factor interactions but aliases two-factor interactions with other two-factor interactions. Lastly, a resolution V design (such as 25-' design with I = ABCDE) does not alias any main effects and two-factor interactions with other main effects and two-factor interactions. The resolution of a fractional factorial design is indicated with a subscript (e.g., 2;;' denotes a 23-' design with resolution 111). As the design resolution becomes higher, more information about interactions is produced. For example, a resolution 111 design provides information about main effects only, while a resolution IV design provides information about two-factor interactions as well as main effects. Resolution 111 and 1V designs are often used in experiments to screen factors.

Table 14-7 Alias Patterns of Resolution 111, IV, and V Designs

Main Effects Two-Factor

Construction of 2k-p Design

X

(Iv)

0 (111)

X

(Iv)

0 (111)

To achieve a designated resolution for a fractional design with k factors, a set o f p generators should be selected properly. Table 14-29 of MR presents recommended generators for various 2k-p designs (k I 15), design resolutions (111, IV, and V), and runs (n I 128). For example, the table recommends two design generators (D = AB and E = AC, which are equivalent to the design relation I = ABD =ACE) for a 2;;' design with n = 8 runs. With the recommended generators of a fractional design, a complete defining relation is determined. Incorporating the generalized interactions of the generators (identified by multiplying the generators with each other; 2*-p - 1 generalized interactions are produced by p generators), the complete defining relation is formulated. For example, the two ( p = 2) recommended generators, 2 ABD and ACE, of the 2;;' design produces one (=2P-p - 1 = 2 - 2 - 1) generalized interaction: ABD.ACE = A'BCDE = BCDE Thus, the complete defining relation of the 2:i2 design becomes I = ABD =ACE = BCDE

14-27

14-9 Fractional Replication of the 2kDesign

2 k -P

Design (cont.)

Like the method to construct a 2k-' design, a 2k-p design is constructed as follows (see Table 14-8 for an example): Step 1: Prepare the sign table of a factorial design with k -p factors (basic design). Step 2: Addp columns for recommended design generators. Step 3: Determine the signs of thep columns with the design generators. Step 4: Identify the treatment combinations of the 2k-p design row by row. Table 14-8 Construction of the 2&* Design with I =ABD =ACE = BCDE

*Iias Structure of ?k -P

A

Design

4

+

+

6

+

-

7 8

-

+

+ +

-

+ + +

+

-

-

+

-

-

+

+

abd ace bc abcde

The alias structure of a 2k-p design is found by multiplying each effect to the complete defining relation. For example, the 2;i2 design with I = ABD =ACE = BCDE has the following alias structure: A = BD = CE = ABCDE BC=ACD=ABE=DE B = A D = ABCE = CDE CD=ABC=ADE=BE C=ABCD=AE=BDE D = AB = ACDE = BCE E=ABDE=AC=BCD When a number of factors under consideration is large (say, k 2 5), high-order (say, third- or higher-order) interactions are assumed to be negligible, which greatly simplifies the alias structure. For example, the alias structure above of the 2&* design with I = ABD = ACE = BCDE can be simplified as follows by assuming that interactions higher than three factors are assumed negligible: A=BD=CE BC=ACD=ABE=DE B=AD=CDE CD = ABC = ADE = BE C=AE=BDE D=AB=BCE E=AC=BCD

14s7

1. (Construction of a 2 k - pDesign) Construct a 2fG2 fractional factorial design. Table 14-29 of MR recommends two (p = 2) design generators, E = ABC and F = BCD, for the 2fG2 design, which yield the defining relation I = ABCE = BCDF. The two generators produce one generalized interaction: ABCE.BCDF = AB*C?DEF = ADEF

~.rr

14-28 Chapter 14



Thus, the complete defining relation of the 2;' I = ABCE = BCDF = ADEF

I

~

m

design with I = ABCE = BCDF

By using the two design generators, the 2';:

1 2 3

-

-

-

-

+

-

-

-

-

-

+

-

-

+ +

+

-

+ +

+ + + +

+ + + +

+

-

-

-

10 11

+

-

-

~

-

-

+

-

,

"

+

~

~

,

+ +

~

-

~

+ +

+

-

~

~

~ ~ ~

+

~~

w

~~

~ -

*

+ + ~~

,

acf bc

adef bde

-

-

-

(1) ae bef

-

-

+ +

13 14 15 16

+ + -

-

+

-

-

+

-

-

-

6 7

-

design is

-

-#

m

x ~

cde acd bcdf abcdef

e~

-

~~

~

~~

~ ~ ~

2. (Alias Structure) Identify the aliases of the fractional design, assuming that interactions higher than three factors are negligible. h Assuming that four-factor or higher-order interactions are negligible, the following aliases are found by multiplying each effect to the complete ~ defining relation of the 2 : ~design: A = BCE = DEF AB = CE B =ACE = CDF AC=BE C = ABE = BDF AD = EF D=BCF=AEF AE=BC=DF E = ABC = ADF AF = DE F=BCD=ADE BD = CF BF= CD


1. Construct a 2:i3 fractional factorial design. 2. Identify the aliases of the fractional design, assuming that only the main effects and two-factor interactions are of interest.

.~

-

s~

m

MINITABApplications

14-29

MINITABApplications Example 14.1

I

(Two-Factor Factorial Design)

(1) Choose File ) New, click Minitab Project, and click OK. (2) Enter the surface finish data on the worksheet.

1

r

---.. L--Ll--".+ ---- ---- ' -l ~ r y l n gT~mePaint Type / Surface ~ l n l s h l g l

(3) Choose Stat > ANOVA ) Two-way. In Response select Surface Finish and in Row factor and Column factor select Drying Time and Paint Type, respectively. Check Display means and Store residuals. Then click OK.

I

(4) Interpret the analysis results.

FI

I

Two-way Analysis of Variance

jAnaly91= of Variance for Surface :$ource DT 55 ns iDrylnp- T 2 27 14 :Palnc Ty 1 356 356 Ilnteractxon 2 1879 939 !Error 12 2243 187 :ToCaI 17 4504

F

P

0.07 1.90 5.03

0.930 0.193 0.026

i i i

i

14-30 Chapter 14


Example 14.2

(2' Factorial Design)

(1) Choose File


*

(3) Choose Stat ANOVA ) Two-way. In Response select Surface Finish and in Row factor and Column factor select Drying Time and Paint

I J

I

I

Two-way Analysis of Variance

.............................................................................................................

: A n a l y s i s of Variance f o r S u r f a c e ; Source DF SS MS j Drying T 1 24 24 : P a i n t Ty 1 0 0 i Interaction 1 1141 1141 i Error 8 1501 188 Total 11 2666

F 0.13 0.00 6.08

P 0.729 0.984 0.039

.............................................................................................................

i

i i

: :

MINITABApplications

Example 14.4

I

14-31

(Normal Probability Plot of Effects) (1) Choose File

* New, click Minitab Project, and click OK. .+--. ..-.-..j,-, Effects E-Estimates

J.

1-

--I---

A

------.-ma-----.+

B

AB 4 C -- ------__^_l_.----.-._l_ 5 AC --

3

"""

^

Choose Graph

"

1

4.25J 6.2q -----.-I--1.25 ! "II"

4

l("-.l__X--._--_-

* Probability Plot. In Variables select E-Estimc

99 -

95

ML Erf rnstes

-

90 -

80 70 60 50 40 -

30

-

20 10 -

5 1 -

Data

Mean

0321428

StOev

2 92072

and

14-32 Chapter 14



II

1 . (Test on the Significance of Main and Interaction Effects; Two-Factor Factorial Design) The analysis of variance of the carbon anode density data is as follows:

I

a

b

n

2

i=1 j = 1 k=l

Source of Variation

I

127.79~ 1,003.1 ------- = 95.87 18

Sum of Squares

Mean

Degrees of Freedom

Square

94.53

2

47.27

0.54

12

0.04

95.87

17

Temperature (A) Furnace Position (B) AB Error Total t_wjn~,-~ew.Mwm~-*#*.~vm~B~8~a

-an*"~"~.ea;nwAw*~

Ho: Hl:

Pj = 0 Pj # 0

1,056.1

.e.nriw.-ien.a.na.

Step 1 : State HOand H I . Ho: zi = 0 for i = 1 to 3 H I : Ti f 0 for at least one i f o r j = 1 and 2 for at least one j

) for A ) for B

No: ( ~ ( 3=) 0~ for all combinations of i's and j's H I : ( T P )#~ 0 for at least one combination of i and j

I

h

-Firino -----D


) for AB

-

-

-

---

7


14-33

Example 14.1 (cont.) Step 3: Determine a critical value(s) for a. fa,,-1,N-ab

= f0.05,3-1,18-6

= f0.05,2,12 = 3'89 for A

fa,b-1,N-ab

= f0.05,2-1,18-6

= f0.05,1,12 = 4'75

fa,(,-l)(b-l),N-ab

- f0.05,(3-1)(2-1),18-6

Step 4: Make a conclusion. Since fo = 0.9 > f,,~,-l,(,-l,,N-ub

for

= f0.05,2,12 = 3'89

for AB

= 3.89, fail to reject Ho at a = 0.05

for AB, which indicates further tests on the main effects A and B are = 3.89 for A and meaningful. Next, since fo = 1,056.1 > f -,, ,,-, fo = 16.0 > fa,b-l,N-,b = 4.75 for B, we can conclude that both A (firing temperature) and B (furnace position) significantly affect the baked density of a carbon anode.

2. (Multiple Comparison) Step 1: Determine the LSD for a.

Step 2: Arrange the means of a treatments in descending order y2,,= 10.34, 2,. = 5.52, and &.,= 5.44 Step 3: Compare the difference between the largest and smallest means with the LSD. Continue this comparison with the next smallest mean as long as the mean difference is greater than the LSD. 2 vs. 3 = 10.34 - 5.44 = 4.90 > 0.25 2 VS. 1 = 10.34 - 5.52 = 4.82 > 0.25 Step 4: Continue Step 3 for the next largest mean until this iteration reaches the second smallest mean. 1 vs. 3 = 5.52 - 5.44 = 0.08 < 0.25 Step 5: Summarize the painvise comparison results by arranging the a treatments in order of mean and then underlining treatments whose means are not significantly different. 2 1 3 3. (Calculation of Residual)

J,,, = 10.62 e,,, = y 2 1 , - j 2 1 ,= y 2 1 3 - y 2=10.43-10.62=-0.19 ,~

(overestimate)

-

---

14-34 Chapter 14


(Calculation of Effects and Sums of Squares; 2' Design) The orthogonal contrasts of the 22 design for the baked density data are C, =a+ab-b-(1)=16.65+15.96-15.96-17.18=-0.53

Exercise 14.2

C, =b+ab-a-(1)=15.96+15.96-16.65-17.18=-1.91 CAB=ab+(l)-a-b=15.96+17.18-16.65-15.96=0.53

I


1

The corresponding sums of squares are (-1.91)~ C: - (-0.53)~ 2 SS, =1;= 0.02 SS, =--C~ = 0.30 n2 3x2' n2k 3 ~ 2 ~

I

(Calculation of Effects and Sums of Squares; 2k Design) The orthogonal contrasts of the 23 design for the tool life length data are CA =a+ab+ac+abc-(1)-b-c-bc = 760 + 1,024 + 783 + 81 1- 532 - 702 - 893 - 1,105 = 146 CB =b+ab+bc+abc-(1)-a-c-ac =702+1,024+1,105+811-532-760-893-783=674 CAB= ( l ) + a b + c + a b c - a - b - a c - b c =532+1,024+893+811-760-702-783-1,105=-90 C, = c + a c + b c + a b c - ( 1 ) - a - b - a b = 893 + 783 + 1,105+ 81 1- 532 - 760 - 702 - 1,024 = 574 CAC= ( l ) + b + a c + a b c - a - a b - c - b c =532+702+783+811-760-1,024-893-1,105=-954 CBc =(l)+a+bc+abc-b-ab-c-ac =532+760+1,105+811-702-1,024-893-783 = -194 CAB, =a+b+c+abc-(1)-ab-ac-bc =760+702+893+811-532-1,024-783-1,105 = -278

Exercise 14.3

I


14-35

Answers to Exercises Exercise 14.3 (cont.)

CABC- -278 ABC=---=-34.8 n2k-1 2 x 23-1

I

Exercise 14.4


1 (Normal Probability Plot of Effects) The normal probability plot of the effects below indicates that all the thirdand fourth-order interactions are not significant so that we can use these insignificant interactions to form the error mean square.

Effect

Exercise 14.5

(

I

1. (Construction of Blocks; 2' Block Design) The defining contrasts are formulated from AB and AC as follows: L1 = x1 + x2 and L2 = X, + x3 The eight treatment combinations are assigned to four blocks according to

--

a*-

a b

(1 + 0) (mod 2) = 1 (1 + 0) (mod 2) = 1 (0 + 1) (mod 2) = 1 (0 + 0) (mod 2) = 0

2 3

ac bc

(1 + 0 ) (mod2)- 1 (1 + 1) (mod2)=O (0 + 1) (mod 2) = 1 (0 + 1) (mod 2) = 1 Lmod22 = 0 (1 + 1)" (mod 2) = 0

3 2 1 --

l a

i-w

XX(IIIX

ali-m

*X.

iXlNIXWVX

X

14-36 Chapter 14


2. (Calculation of Effects and Sums of Squares) The generalized interaction of AB and AC is ABXAC=A~BC=BC


I I

Thus, the effects AB, AC, and BC are confounded with the blocks. The estimates of the effects are

574 C = % - = = 7 1 . 8 n2k-' 2 x 23-1

ABC = ' A B C - - 278 - 34.8 dk-' 2 x 23-1

Block effect = AB + AC + BC = CAB+ CAC+ CBC n2 k-' - -90 - 954 - 194 = 154.8 2 x 23-1 The corresponding sums of squares are 2

2

ssA ---C", =--14" nzk

Exercise 14.6

2

~

- 1,332.3

2

~

CB SS, = 7 -6742 = 28,392.3 n2 2 ~ 2 ~

1. (Estimation of Effects) The main effects and interactions are estimated as follows: I , =A+BCD=t(-45+100-45+65-75+60-80+96)=19.0

I , =B+ACD=$(-45-100+45+65-75-60+80+96)=1.5 1, = C + ABD=$(-45-100-45-65+75+60+80+96)=14.0 I , =D+ABC=$(-45+100+45-65+75-60-80+96)=16.5 I,,

=AB+CD=$(45-100-45+65+75-60-80+96)=-1.0

I,,

=AC+BD=$(45-100+45-65-75+60-80+96)=-18.5

I,,

=AD+BC=$(45+100-45-65-75-60+80+96)=19.0

Three factors A, C, and D show large effects on filtration rate, but factor B displays a relatively very small effect. 2. (Projection into a Factorial Design) The 24-1 design projects into a 23 design with a single replicate including factors A, C, and D. The main effects and interactions of this unreplicated 23 design can be easily determined by using the effect estimates of the design as follows:

-



A = 19.0 C = 14.0 D = 16.5

' Exercise 14.7

14-37

AC =-18.5 AD = 19.0 CD = -1.0 ACD = 1.5

The effects A, C, D, AC, and AD have relatively large influence on filtration rate. 1 . (Construction o f a 2k-PDesign) Table 14-29 of MR recommends three @ = 3) design generators, D = AB, E = AC, and F = BC, for the 2;i3 design, which yield the defining relation I = ABD = ACE = BCF. These three generators produce four (= 2P- p - 1 = 233 - 1) generalized interactions: ABD.ACE = A~BCDE= BCDE ABD.BCF = AB~CDF= ACDF ACE.BCF = A B ~ E F = ABEF ABD.ACE BCF = A~B~CL?DEF = DEF Thus, the complete defining relation of the 2fi3 design is I=ABD=ACE=BCF=BCDE=ACDF=ABEF=DEF By using the three design generators, the 2 f i 3 design with I = ABD =ACE =

6 7

+ -

-

+

+ +

-

+ -

-

+

abd cd ace bcf

2. (Alias Structure) Assuming that three-factor or higher-order interactions are negligible, the following aliases are found by multiplying each effect to the complete defining relation of the 2;i3 design: A=BD=CE B=AD=CF C=AE=BF D=AB=EF E=AC=DF F=BC=DE AF=CD=BE

-


15-1 Introduction 15-2 Sign Test 15-3 Wilcoxon Signed-Rank Test

15-5 Nonparametric Methods in the Analysis of Variance MINITABApplications

15-4 Wilcoxon Rank-Sum Test


15-1

Introduction

0 Explain nonparametric statistics in comparison with parametric statistics. sm.,m,sm.~m,ae

~ ~ > ~ * s s . , a ~ - . ~ ~ , m M . Buw, . aw.m&"mM-


.m.m,mm ,mm,~.-a*,~'%~*,-~d*e~~.d.~**a,e~#whw#

~ . ~ . m ~ ~ . ~ s ~ m . ~ , ~ ~ ~ ~ ~ ~ > ~ . ~ ~ ~ ~ , . w . ~ ~ " ~ ' ~ ~ ~ a ~ ~ ~ v , ~".BMBw"v"s" w m ' s .a.wm, ~ ~ ~ ms.aea.s,"a.*a % * ~ - ~ , ~ a ~ ~ ~ d ~ b ~ ~ . -

Nonparametric methods do not use information of the distribution of the underlying population, whereas parametric methods (such as t and F tests) are based on the sampling distribution of a particular statistic for the parameter of interest. The nonparametric methods make no assumptions except the underlying distribution is continuous, using only signs andlor &-the original observations may need to be transformed. Therefore, the nonparametric statistics is considered a distribution-free method and does not include the procedure of parameter estimation (point and interval estimations). Parametric and nonparametric statistics have their own advantages and disadvantages in terms of ease of use, time efficiency, applicability, and sensitivity (statistical power), as summarized in Table 15-1. Nonparametric tests are easier and quicker to perform and their applications are extended to the following contexts where parametric tests are inappropriate: (1) Underlying distribution unknown: Information of the underlying distribution is unavailable. (2) Normal approximation inapplicable: Normal approximation is not applicable when the underlying distribution is significantly deviated from a normal distribution and the sample size is small (say, n < 30) (see Section 7-5). (3) Categorical data: Categorical data include nominal and ordinal data (see Section 9-7).

15-2

Chapter 15


Nonparametric Statistics However, nonparametric tests are less sensitive (statically powerful) in detecting a small departure from the hypothesized value because they do not use all the information of the sample. Therefore, the nonparametric methods usually require a larger sample size to achieve the same statistical power.

Table 15-1 Comparison of Parametric and Nonparametric Statistics Ctitepia Parametric Nonpammtric Ease of use 0 Time efficiency 0 Applicability 0 Sensitivity 0 * * - %*" - * (Note) 0 : preferred --w

15-2

-mss wmm-*-mw*

-*w*ma

wa a

a#vaa

-#

#

es

-

w

-#

& 6, (2) R +

= number of the

plus signs of D j ,j

=

1,2,. .. ,n for HI:

) ji, +6, (3) ~ = r n i n ( ~ - , R +forH,: Sign Test Procedure for ji,

Step 1: State Ho and HI. Ho: Po = 6 , HI:

p, + 6 ,

for two-sided test ji, < 6, or ji, > 6, for one-sided test

p,

< 6,

15-6

Chapter 15

Sign Test Procedure for pD (cont.)

Nonparametric Statistics Step 2: Determine a test statistic and its value. R- = number of the minus signs of D j for upper-sided test Rf = number of the plus signs of D j


R = min(R-, R + )

for two-sided test

Step 3: Determine a critical value(s) for a. r,*,,,, for two-sided test; r , for one-sided test Step 4: Make a conclusion. Reject Ho if r 5 , for upper-sided test + * r I ran for lower-sided test for two-sided test

r5r

Example 15.2

II

pD ;Sign Test) Scores of 10 students for two exams are below. Test Ho: pD = 0 vs. H I : pD # 0 at a = 0.05. (Test on

No. Exam2(X2$

---em*

a

*ma

-am-

1

2

90

73*

-a m--sw---sw

3

4

5

--

98""-*- 80

**am"

6 92 *

7 72 *

ewma%m

8 78 -**

ex"v-

9

-85-

1

0

75 ---

92

*

-

< *a

w*

Step 1: State Ho and H I . Ho: p , = o HI:

I

I

p,

#O

Steo 2: Determine a test statistic and its value.

Sign m.*"we**-,".B,a

".,"m~w"ws,~"""w~.a,~"m.--#~*#~~""#

+

-

-.-*-

",a,~w.,"s."""*."ww~*".".".~-~~." s+-"*a"w

* Excluded since d j = 6,.

".>,"*w.B.w

..w.w

8..s",

".%"*.".""""*,"""-.a

.a.s

,-* ~ ~ . - - . ~ " ~ " ~ . ,",*"cw ~~~-~,*",

Step 3: Determine a critical value@) for a. r,*,,,, = r;,,,,, = 1 (Note) number of ties = I Step 4: Make a conclusion. Since r = 1 I r,*,,,, = 1 , reject Ho at a = 0.05.


The diameter of a ball bearing was measured by 12 inspectors. The measurements, corresponding differences, and their signs are presented in the table on next page. Test if there is a significant difference between the medians of the populations of measurements by using two different calipers. Use a = 0.05.

15-7

15-3 Wilcoxon Signed-Rank Test


15-3

Wilcoxon Signed-Rank Test

--

uswe -wm

a

*

a

*

*>w

" a -a

#

-

A #%*

a**--

a me

e *.d

*a

kw

mmVe-B

w

*

*e a*- mr-

rs

-

*".me

*

. --

Read the Wilcoxon signed-rank test table.

O Test a hypothesis on the mean ( p ) of a continuous distribution by the Wilcoxon signed-rank test. Apply the normal approximation to the signed-rank statistic when n is moderately large.

O Test a hypothesis on the mean of differences (b)for paired samples by the Wilcoxon signed-rank test.

Wilcoxon Signed-Rank Test

While the sign test uses only the plus and minus signs of the differences between the observations X, and the hypothesized median ii,,the Wilcoxon signed-rank test uses both the of the differences between the observations X, and the hypothesized mean po and the &of the absolute magnitudes of the differences. The signed-rank test assumes that the underlying distribution is svrnmetric and continuous, i.e., the mean equals the median. The Wilcoxon signed-rank statistics W- , W+ , and Ware determined by the following procedure: Step 1: Compute the differences X i- p,, i = 1,2,. .. , n.

1

Step 2: Rank the absolute differences (xi - po in ascending order. For t k , use the average of the ranks that would be assigned if they differed. Step 3: Assign the signs of the differences to the corresponding ranks. Step 4: Calculate the sum of the signed ranks: W- = sum of the negative ranks W + =sum of the positive ranks

w = min(W-, W ' )

15-8

Chapter 15

Wilcoxon Signed-Rank Test Table

Nonparametric Statistics Like the sign test table, the Wilcoxon signed-rank test table (see Appendix Table VIII in MR) provides critical values ( wi,, ) of a signed-rank test statistic for various sample sizes (n) and levels of significance (a). (e.g.) Reading the signed-rank test table

Inference Context for p

(1)

w;.05,10 =

10

(2)

w;.025,20 = 52

(3)

w;.005.,5 =

15

Parameter of interest: p Test statistic of p: The appropriate signed-rank statistic is selected depending on the type of HI as follows: (I) W-=sumofthene.gativeranksof X i - p o , i = 1,2, ..., n forH1: p > p , (2) w + = sum of the positive ranks of

xi - po , i = 1,2,. .. , n

for HI: p < p0

(3) W=min(W-,W+) forHl: p # p 0

Wilcoxon Signed-Rank Test Procedure for p

Step 1: State HOand H I . Ho: C L = C L ~

HI: p # po for two-sided test p < po or p > p0 for one-sided test Step 2: Determine a test statistic and its value. W-= sum of the negative ranks of X , - po


W+= sum of the positive ranks of X i- po


W = min(W-, w + )

for two-sided test

Step 3: Determine a critical value(s) for a. w : , ~ , ~for two-sided test; w;,,

for one-sided test

Step 4: Make a conclusion. Reject Ho if for upper-sided test w- I w;,,

w+ Iw;,, w5w Approximation

,

,

for lower-sided test for two-sided test

If the sample size is moderately large (say, n > 20), the signed-rank statistic W + ( W- or)'+I has approximately a normal distribution with mean and variance n(n + 1)(2n + 1) - n(n + 1) Pw+ -- 4 and ow+= 24 Therefore, a hypothesis on the single mean by W+ can be tested by using the following z-statistic:

15-9


Normal Approximation (cont.)

r

Example 15.3

Test scores of 10 students in a statistics class are below. No. 1 2 3 4 5 6 7 8 9 Score(X$ 85 75 96 74 87 71 73 80 * *- * * -" * -* ww

ww-*

-*,-*w+

w-

a

#-w**

aww'

-4-*

weew+

w

#

1 75 -

->

*

0 90 -

-

1. (Test on p; Wilcoxon Signed-Rank Test) Test if the mean of the test scores is equal to 80 at a = 0.05. m Step 1: State Ho and H I . Ho: p = 80 H,: p + 8 0 Ster, 2: Determine a test statistic and its value.

xi-80

5

-5

16

-6

0

-5

10

(xi-801 Rank

5 2

5 2

16 9

* 6 7 9 7 4 5 . 5 7 5 . 5 *

5 2

10 8

*

-2

8

Signed 2 -2 9 rank * Excluded since xi = p, . w = min(w-, w')

-4

7

5.5

-9

-7

-7

-5.5

= min(20.5,24.5) = 20.5

Step 3: Determine a critical value@) for a. w : , ~ , ~=

~ i =, 5 ~(Note) ~ ~number , ~of ties = 1

Step 4: Make a conclusion. Since w = 20.5 $ w:,,,,

= 5 , fail to reject Ho at a = 0.05.

2. (Normal Approximation) Apply the normal approximation to the signed-rank statistic W and draw a conclusion at a = 0.05. k Step 1 : State Ho and H I . Ha: p = 80 H,: j,l # 80

je de

Step 2: Determine a test statistic and its value. 9x(9+1) n(n + 1) W-20.5 za = n(n+1)(2n+l) = 9 x ( 9 + 1 ) x ( 2 x 9 + 1 ) = -0.24

1sai pap!s-lamol.~oj

fa j o syum ah!l!sod aqljo urns = + A

lsal pap!s-laddn ~ o j 30 syum a~!@au ayljo urns = - A .anlaa s$! pua 3!p!$aqs qsaq s aupmalaa :Z days

'SO'() = x, 1" Uo!Sn~3U03 a ~ s l pus p + A 3!ls!yels yual-pauZ!s aqlol uo!~sur!xo.~dds pm~.~ou aql klddv ' 2 .SO.O

= x, 1s wdd

s.2 usql35j s! ]anal Lpt~dur!aqljo usaw aqlj! IsaL '1

-

15-11


Example

(Test on p D ;Wilcoxon Signed-Rank Test) The exam scores in Exercise 14.2 is presented again below. Test Ho: p D = 0 VS.H I : p D # 0 at a = 0.05.

askiwjra 2 >~~&~~&~#@f&~j~#~~$~~~$~~j~;~~g&~$~$$~~#gjjj~~#$@&~~$ Examl(X1)

, ;%::."

'J

"-8

#-"

%~CW t

$i

&&ej8

85 90

h

'

n

75 73

*-navasw.

*

96 98

m*

74 80

& - # *

87 92

71 72

73 78

80 85

75 75

90 92-**

-

Step 1 : State Ho and H I . Ho: p D = O

H,: pD f O

I

I

I

6.5 3 3 Rank Signed -6.5 3 -3 rank ,~", * Excluded since d j = 6 , . ~

"s*sm.#*s"-

9

6.5

1

6.5

6.5

-9

-6.5

-1

-6.5

-6.5

* *

3 -3

----

%a%dw-,d,a,a,-a-.#-

.e-~~ " " " ~ , . " v " , " . ~ ' " . ~ ~ e ~ . ~ . ~ " ~ ~ . ~ w , ~ ~ ~ ~ ~ " ~ - ~,~c-ds . " ~a.aa""" ~ ' ~ m,,,m,*, , ~ ~ ,"""a,n"*~%v"*b " . e . " ~ ~ ~ ~ , ~ ~ s w ~ " . " ~ " e ~ . ~

*e#.~..--e~.*m-.",*a

Step 3: Determine a critical value(s) for a. w : ~ =~w, ;~ , ~=~5 ~ (Note) , ~ number of ties = 1

m

Step 4: Make a conclusion. Since w = 3 5 = 5, reject Ho at a = 0.05.

~


i

~

~

,

~

Reconsider the ball bearing diameter data in Exercise 15.2. The measurements, corresponding differences, and their signed ranks are presented below. Test Ho: p D = 0 vs. H1: p D # 0 at a = 0.05.

Measurement X2 XI

(XI-X2)

14

Rgok

Signed

15-12 Chapter 15

15-4


Wilcoxon Rank-Sum Test

0 Read the Wilcoxon rank-sum test table. Test a hypothesis on the difference in mean of two continuous populations (p1 - p2) by the Wilcoxon rank-sum test. 0 Apply the normal approximation to the rank-sum statistic when nl and n2 are moderately large. m . , ~ . ~ , " ~ ~ ~ B - - . m . ~ ~ ~ ~ ~ ~ ~ A ~ * * - . ~ e ' e w a ~ ~ * ~ : ~ - , * - - ~ ~ " ~ ~ ~ ~ ~ ~ ~ u s * ~ m - ~ ~ ~ . , # ~ ~ - ~ ~ . ~ * ~ ' , ~ B a w ~ - ~ . - ~ ~ ~ ~

Wilcoxon Rank-Sum Test

The Wilcoxon rank-sum test (often called Mann-Whitney test) uses the ranks of the observations from two independent continuous populations XI and X2 to test the difference in mean ( p -~p2). Let XI,, X12,...,X,,,, and X2, ,X2, ,...,XZn2denote two independent random samples of sizes nl and n2 (a1 5 n2) from the populations XI and X2. Then, the Wilcoxon rank-sum statistics Wl and W2 are determined by the following procedure: Step 1: Rank the n1 + n2 observations in ascending order. For t k , use the average of the ranks that would be assigned if they differed. Step 2: Calculate the sum of the ranks for each sample: W, = sum of the ranks in the smaller sample (size = nl) W2 = sum of the ranks in the larger sample (size = n2) Note that Wl and W2 will be nearly equal for both the samples after adjusting for the difference in sample size.

Wilcoxon Rank-Sum Test Table

The Wilcoxon rank-sum test table (see Appendix Table IX in MR) provides critical values ( w;,,,, ) of a rank-sum statistic for various sample sires (nl and n2 where nl I n2) and levels of significance (a = 0.05 and 0.01) for a two-sided test. (e.g.) Reading the rank-sum test table for a two-sided test (1) w;.05i2,10,15 = w;;.025,10,15= 94

(2) w;.01i2,5,,0 = w~.005,5,10= 19 Inference Context for PI - CLZ

Wilcoxon Rank-Sum Test Procedure for PI - Pz

Parameter of interest: p1 Test statistic of p1- p2: The appropriate rank-sum statistic is selected depending on the type of HI as follows: (1) W,=sumoftheranksof X , j , j = 1 , 2 ,..., nl forH1: p , - p 2 60 or p1 - p2 < 8o for one-sided test

15-13

15-4 Wilcoxon Rank-Sum Test

Wilcoxon Rank-Sum Test Procedure (cont.)

Step 2: Determine a test statistic and its value. W, = sum of the ranks of X , ; for lower-sided test 'J

W2 = sum of the ranks of XZj for upper-sided test Wl or W2

for two-sided test

Step 3: Determine a critical value(s) for a. *

for two-sided test;

W a I 2,n, ,n2

w

2

for one-sided test

Step 4: Make a conclusion. Reject Hoif 1

Wa,nl ,n2


w2 5 w:,nI,n2


W1

*

W1

Normal Approximation

5 W a I 2,n, ,n2 Or

W2

* 5 Wa/2,n,,n2

for two-sided test

When nl and n2 are moderately large (say, nl and n2 > 8), the rank-sum statistic Wl (or W2 ) has approximately a normal distribution with mean and variance n, (n1 + n2 + 1) 2 n1nz(n1+n2+1) and owl= 2 12 Therefore, a hypothesis on the difference in mean by Wl can be tested by using the z-statistic

Pw, =

Y

Example 15.5

The life lengths (unit: hour) of INFINITY(XI) and FOREVER(X2) light bulbs are under study. Suppose that no information is available about the underlying distributions of XI and X2. A random sample selected from each brand results in the following:

F O R E V E R ( X ~795 L

-am.m

a.m--

wwaMMM

w-

780

r*a-w--m-

820

770

810 765

ma wawam-m-*-

wM--s-s

800 790

- 750

---Maww#%w-m-

730

*#"B#a*m wwa-s.8 'r

1. (Test on p1- p2; Wilcoxon Rank-Sum Test) Test if the mean life lengths of the two brands are the same at a = 0.05. o- Step 1: State Ho and H I . Ho: p1- p 2 = 0 H1: P1 - p 2 2 0 a test statistic and its value.

%9pfpB" ##@dk46 146

XI Rank X2 Rank

"mu

760 810 775 680 9 17.5 12 1 795 780 820 770 15 13 19 11 w l = 6 1 and w2= 129 ~a*-e"--ams.#m*---

730 5.5 810 17.5

725 4 765 10

740 7 800 16

700 2 790 14

$ (8~d -

& $ g81 C ~ j 4 s g g .

720 3 750 8 -.#*-

730 5.5-

* w - ~ # - ~ ~ " ~ a " ~ - ~ ~ - ~ * * ~ w w w - w m M " ~ - w ~ ~ # ~ " w ~ " ~ - ~*aw'"**" -

15-14 Chapter 15


Nonparametric Statistics Step 3: Determine a critical value(s) for a. ~;/2,n,,n,

- ~ ~ . 0 5 1 2 , 9 , 1=0 65 -

Step 4: Make a conclusion. Since w, = 61 I

~ i / =~65,,reject ~ Ho , at~a ~= 0.05.

2. (Normal Approximation) Apply the normal approximation to the rank-sum statistic Wl and draw a conclusion at a = 0.05. Step 1: State HOand H I . Ho: p1 -p2 = o H1: p1 - p 2 + 0

(


Step 3: Determine a critical value(s) for a. Z,/2 = Z0,025 = 1.96

I

The manufacturer of a hot tub is interested in testing two different types of heating elements. The heating element that produces a larger heat gain (unit: OF) after 15 minutes will be preferable. Ten observations of the heat gain for each heating element are obtained as follows:


(

I 15-5

Step 4: Make a conclusion. Since lzol = 2.37 > Z0,025 = 1.96 ,reject HO at a = 0.05.

1. Test if one heating element is superior to the other at a = 0.05. 2. Apply the normal approximation to the rank-sum statistic Wl and draw a conclusion at a = 0.05.

Nonparametric Methods in the Analysis of Variance

O Perform the Kruskal-Wallis test in the single-factor analysis of variance (ANOVA). Explain use of ranks in ANOVA.

KruskalWallis Test

The Kruskal-Wallis test is a nonparametric alternative to the parametric F-test of the single-factor analysis of variance (ANOVA). While the parametric ANOVA model, Yi,. = p + Z~ + E~ ,i = 1,2,. .. ,a and j = 1,2,... ,n i , assumes the error

15-5 Nonparametric Methods in the Analysis of Variance

KruskalWallis Test (cont.)

15-15

terms ei, are normally distributed with mean zero and variance 02, the nonparametric ANOVA method assumes that the eYhave the same continuous distribution for all the factor levels i = 1,2, ... ,a. The Kruskal-Wallis statistic H i s determined by the following procedure: Step 1: Rank all N (=

za r=l

ni ) observations in ascending order. For t k , use

the average of the ranks that would be assigned if they differed. Step 2: Calculate the sum ( R,,) of the ranks RYin the ithtreatment, i = I, 2, ... , a. Step 3: Calculate the Kruskal-Wallis statistic l2 - 3(N + 1) for observations without ties H= N(N + 1) i=l ni

gc

4 *i

where S 2 =-!-[T~R~ N - 1 i=l j=1

x2

Approximation Inference Context

1

for observations with ties

I

N ( N + 112 -

4

The Kruskal-Wallis static approximates to a X2-distributionwith a - 1 degrees of freedom when the sample sizes ni are moderately large (ni 2 6 for a = 3; ni 2 5 for a > 3).

Parameters of interest: 2, ,2,, ...,z, Test statistic of T,,2,, ...,2, (1) Case 1: Observations with ties

(2) Case 2: Observations without ties

KruskalWallis Test Procedure

Step 1: State HOand H I . Ho: 7, = z 2 =...=z,

=o

(pl =p2 =.-.=Po)

HI: zi + 0 foratleastone i,i=1 , 2,..., a Step 2: Determine a test statistic and its value. l2 - 33(N+ 1) for observations without ties H= N(N + 1) ni

gg

4

for observations with ties

4

I

15-16 Chapter 15

KruskalWallis Test Procedure (cont.)



Step 4: Make a conclusion. Reject Ho if 2

h 2 %a,,-,

If the ordinary F-test were to be applied to the ranks instead of the original data in the single-factor ANOVA, the F-test statistic would be H /(a - 1) F, = (N-1-H)l(N-a) Note that Fois closely related to the Kruskal-Wallis statistic H, which indicates that the Kruskal-Wallis test is approximately equivalent to the parametric ANOVA to the ranks. Accordingly, the F-test on the ranks can be applied when a nonparametric alternative to the analysis of variance is unavailable.

F-test on Ranks

r

It is recommended that the F-test be performed on both the original data and the ranks when the normality assumption andlor the effect of an outlier(s) are concerned. If both the test results are similar, use of the original data is satisfactory; otherwise, use of the ranks is preferred. Example 15.6

1 (Nonpararnetric ANOVA; Kruskal-Wallis Test) Maximum power grip forces (unit: lb.) of a 45-year-old male at four elbow angles (0°, 45", 90°, and 135") are I obtained below. Test if elbow angle has a significant effect on maximum power 1

grip strength at a = 0.05.

0" (YI) 45" (y2) 90" (y3)

., r3XLri)

mmw-B,

h

* ,-,

65 70 67 58 -*-Bww--B

68 72 63 60 *-

- -

ms--wmB

62 69 68 59 , -**

-

m

=

66 75 64 63 ,. *--

-

ev

*

69 68 67 60 .-

mw mMw

Step 1: State HOand H I .


Hbanv Wzk

Ri.

135" bj) Rank (r4J

-**-"Mwwm

a#

58 1

60 3.5

" * ~ - ~ 6 w * - . m ~ " - ~ r a B ~ ~ " ~ " ~ " ~ w , w ~ ~ a ~ a v - - -

59 2

-w~w,-e"6a--*"M,vw--*~-,%"v

63 6.5

60 3.5

6---*---m"eMm*-m--w---

16.5

--

15-5 Nonparametric Methods in the Analysis of Variance

15-17


Step 3: Determine a critical value(s) for a. 2 - 2 2 Xa,a-1 - X O . O ~ , ~= - 1X 0 . 0 5 ~= 7.81

Step 4: Make a conclusion. Since h = 14.52 >

xi.,-,


= 7.81, reject Ho at a = 0.05.

The tensile strengths (unit: 1b.1in.~)of cements prepared by four different mixing techniques (A, B, C, and D) are measured below. Test if the difference in mixing technique affects the tensile strength at a = 0.05.

15-18 Chapter 15


MINITABApplications Example 15.1

on ; Sign Test) I (Inference (1) Choose File * New, click Minitab Project, and click OK.

I

Score

75-

96

*

Choose Stat Nonparametrics > 1-Sample Sign. In Variables select Score. Check Test median, enter '80' (hypothetical median), and in

I

Sign Test for Medim Slgn test of median = 80.00 versus

not =

80.00

............................................................................................... N 12

Below r- = 5

Equal 1

Above

P

v+ = 6

1.0000

j

................................. ........................................................: rr..

Median 82.501

3: "

I

Example 15.2

(Inference on

E D ;Sign Test)

(1) Choose File


(2) Enter the score data of the two exams on the worksheet.

(3) Choose Calc > Calculator. In Store result in variable, enter C3 as a column to store calculation results. In Expression, click Exam 1, minus

.I

1

(~xam 1 ( X I ) Exam 2 (x2) 85

- .

XI - x2

I

-4

-5 ..

15-20 Chapter 15



I

(5) Choose Stat > Nonparametrics > 1-Sample Sign. In Variables select x l - x2. Check Test median, enter '0' (hv~otheticalmedian of the score

I

Sign Test for Median

............................................................................................... Sign t e s t of median = 0.00000 v e r s u s

1x1

-

x2

N

Below

10

r- = 8

Equal 1

not =

Ahove r+=l

0.00000

P 0.0391

.............................................................................................

.i

Example 15.3

i

Hedian

1-3.500 :

.".. -

(Inference on p.;Wilcoxon Signed-Rank Test) (1) Choose File > New, click Minitab Project, and click OK.

(2) Enter the test score data on the worksheet. (3) Choose Stat > Nonparametrics > 1-Sample Wilcoxon. In Variables select Score. Check Test median, enter '80' (hypothetical mean), and in

15-21


Example 15.3 (cont.) k -Wilcoxon Signed Rank Test Test of medzan = 80.00 versus medlan not = 80.00

....................................................................................... N for N

Test 9

Wilcoxon Statlstlc w + =24.5

i Estimated P

i

i ....................................................................................... iscore

Example 15.4

10

0.859

Nedian 80.001

71 %I

(Inference on p, ;Wilcoxon Signed-Rank Test) (1) Choose File

, ,

New, click Minitab Project, and click OK.

(2) Enter the score data of the two exams on the worksheet. (3) Choose Calc Calculator. In Store result in variable, enter C3 as a column to store calculation results. In Expression, click Exam 1, minus sign, and Exam 2 in sequence. Then click OK.

(4) Name the C3 column x l - x2.

,

,

(5) Choose Stat Nonparametrics 1-Sample Wilcoxon. In Variables select X I - x2. Check Test median, enter '0' (hypothetical mean of the

Wilcoxon Signed Rank Test

I Test of medlan

= 0.000000 versus median not = 0.000000

N for

Wilcoxon Statistic wf = 3.0

i~stimated Median -3.000 0.024 i

; :..................................................................................... . jxl - x2

N

Test

10

9

P

15-22 Chapter 15

Example 15.5

Nonpararnetric Statistics

(Inference on yl- CL;?;Wilcoxon Rank-Sum Test) (1) Choose File New, click Minitab Project, and click OK.

*

bulb brands on the worksheet.

J.

1 3

lnfin~ty / Forever I 760 p1 795 8101 7801

I

'

*

*

(3) Choose Stat Nonparametrics Mann-Whitney. In First Sample select Infinity, in Second Sample select Forever, and in Alternative

Mann-Whitney Confidence Interval and Test .......................................................................................

-

i

Infinity N = 9 Bedian 730.00 Forever N = 10 Median = 785.00 i Point estimate for ETA1-ETA2 is -47.50 I 95.5 Percent C I for ETA1-ETA2 is (-79.99,-10.01)i

............................................................... i

Test of ETA1 = ETA2 v s ETA1 not = ETA2 is significant at 0.0200 The test 1s significant at 0.0199 (ad]usted for ties11

$5

43

MINITABApplications

Example 15.6

I

15-23

(Nonparametric A N 0 VA; Kruskal-Wallis Test) (1) Choose File


(2)

ksheet.

I--, Force

. I

I.; ;I r

1

Angle

-

*

01 +--- -65 -

6ar -----. . 0,- ............. 62

...

..........

:hoose Stat > Nonparametrics ) Kruskal-Wallis. In R.esponse select

I

I

Kruskal-Wallis Test Kruskal-Wallis Test on Force Angle

N

Median

Ave Rank

Z

0

5

45

5

90

5 5 20

66.00 70.00 67.00 60.00

10.9 17.5 10.3 3.3 10.5

0.17 3.06 -0.09 -3.14

135

Overall

15-24 Chapter 15



1. (Test on

I

p ;Sign Test)

Step 1: State Ho and HI. Ho: ji =2.5 H I : p < 2.5 Step 2: Determine a test statistic and its value.

11

2 " ' - . 3 'B,'?:

'.-:..?;'.

.$

-

* Excluded since xi = P o . Step 3: Determine a critical value@) for a.

ri,, = r,',,,,,

I

I I

= 5 (Note) number of ties = 2

Step 4: Make a conclusion. Since r+ = 2 I r& = 5 ,reject Ho at a = 0.05. 2. (P-value Calculation)

Since P = 0.0002 < a = 0.05, reject HOat a = 0.05. 3. (Normal Approximation) Step 1: State Ho and HI. Ho: ji =2.5 H I : ji < 2.5

I


Step 3: Determine a critical value@) for a. Z a = Zo,05 = z0,05= 1.65

I

Step 4: Make a conclusion. Since z , = -3.58 < -z0,,, = -1.65, reject Ho at a = 0.05.

#gS

fd


Exercise 15.2

(Test on

15-25

PD ;Sign Test)

Step 1: State Ho and H I . Ho: PD = 0 H,: jiD# O Step 2: Determine a test statistic and its value. r=min(r-,r') =min(2,6)=2 Step 3: Determine a critical value(s) for a. ri12,n= r;025,8 = 0 (Note) number of ties = 4 Step 4: Make a conclusion. Since r = 2 g = 0 , fail to reject Ho at a = 0.05.

riI2,,

Exercise 15.3

1. (Test on p; Wilcoxon Signed-Rank Test) Step 1: State Ho and H I . Ho: p = 2.5 HI: p < 2.5


#xl-ls) .

No,

3 -q-2.3

12 13 14 15

2.3 2.0 1.8 1.3

-0.2 -0.5 -0.7 -1.2

0.2 0.5 0.7 1.2

5.5 8.0 14.0 19.5

-5.5 -8.0 -14.0 -19.5

17 18 19 20

2.0 1.9 2.3 1.9

-0.5 -0.6 -0.2 -0.6

0.5 0.6 0.2 0.6

8.0 11.5 5.5 11.5

-8.0 -1 1.5 -5.5 -11.5

17.5

- -17.5 -

22 1.6 -0.9 * Excluded since x, = p, .

A m - - s - - - " # - - d . v - - P

*'.

? " .*

0.9

Mm -m

&&

"+a%%* * e--*--

.

9i@J@d fa&

15-26 Chapter 15



Step 3: Determine a critical value(s) for a. w:,~ = u ' ( ; , ~=~60 , ~ ~(Note) number of ties = 2

Step 4: Make a conclusion. Since w* = 5 I w:.,

= 60, reject Hoat

a = 0.05.

2. (Normal Approximation) Step 1: State Ho and H I . Ho: p =2.5 HI: p < 2.5 Step 2: Determine a test statistic and its value.

1

Step 3: Determine a critical value(s) for a. Z, = z0,05= 1.65 Step 4: Make a conclusion. Since zo = -3.73 < -z,,,,

Exercise 15.4

= -1.65

,reject Hoat a = 0.05.

(Test on y, ;Wilcoxon Signed-Rank Test) Step 1: State Ho and H I . Ho: y, = o H I : y, # O Step 2: Determine a test statistic and its value. w = min(w-, w') = min(14.5,21.5) = 14.5 Step 3: Determine a critical value(s) for a. w:/~,, = ~

: , ~ = ~ 3 ~ , (Note) 8

Step 4: Make a conclusion. Since w = 14.5 $ w:,,,, Exercise 15.5

I

number of ties = 4

= 3 , fail to reject Hoat

1. (Test on pl - p2; Wilcoxon Rank-Sum Test) Step 1 : State Ho and H I . Ho: -~ 2 = 0 H I : p1 -y2.;tO

a = 0.05.

15-27


Exercise 15.5 (cont.) Rank X2

2 31

4 33

5.5 32

9.5 35

7.5 34

3 29

1 38

11.5 35

Step 3: Determine a critical value(s) for a. wi12,n,,n2 -w ~ , ~ = 78~ for ~ two-sided ~ , ~ test ~ , Step 4: Make a conclusion. Since w, = 77 I w;,,

,,,,,,= 78 ,reject Ho at a

~ ~

= 0.05.

2. (Normal Approximation) Step 1 : State Ho and H I . H,: p, - p2 = o H I : p1 - p 2 + 0 Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. z a I 2= z ~= 1.96 , ~ ~ ~ Step 4: Make a conclusion. Since lz, = 2.12 > z,,,,,

1

Exercise 15.6

= 1.96, reject Ho at

(Nonparametric ANOVA; Kruskal-Wallis Test) Step 1 : State Ho and H I . H,: 7 , = 7 , = ... = 7 , = 0 H I : zi # 0 for at least one i, i = 1, 2, 3, 4

13.5 37

a = 0.05.

19.5 30

ZP'E I =

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