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Statistics in

Criminal Justice

Fourth Edition

David Weisburd

Chester Britt

Statistics in Criminal Justice

Statistics in

Criminal Justice

Edition

David Weisburd

Hebrew University of Jerusalem, Jerusalem, Israel

Fourth

Chester Britt

Northeastern University, Boston, MA, USA

and

and George Mason University, Fairfax, VA, USA

David Weisburd Chester Britt

Institute of Criminology School of Criminology and Criminal Justice Faculty of Law Northeastern University

Hebrew University of Jerusalem Jerusalem, Israel

ISBN 978-1-4614-9169-9 ISBN 978-1-4614-9170-5 (eBook) DOI 10.1007/978-1-4614-9170-5 Springer New York Heidelberg Dordrecht London Library of Congress Control Number: © Springer Science+Business Media New York 2014 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. Exempted from this legal reservation are brief excerpts in connection with reviews or scholarly analysis or material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Duplication of this publication or parts thereof is permitted only under the provisions of the Copyright Law of the Publisher’s location, in its current version, and permission for use must always be obtained from Springer. Permissions for use may be obtained through RightsLink at the Copyright Clearance Center. Violations are liable to prosecution under the respective Copyright Law. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. While the advice and information in this book are believed to be true and accurate at the date of publication, neither the authors nor the editors nor the publisher can accept any legal responsibility for any errors or omissions that may be made. The publisher makes no warranty, express or implied, with respect to the material contained herein. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)

Boston, MA, USA

2013952914

and

George Mason University Fairfax, VA, USA

Department of Criminology, Law and Society

Additional material to this book can be downloaded from http://extras.springer.com

For Bryan, who made the desert bloom, used

sun to brighten the night, and brought such joy

to family and friends

D. W.

v

For my parents, Chester and Lila, who have been

a constant source of support

C. B.

iv C H A P T E R N U M B E R : C H A P T E R T I T L E

Contents

1

Statistics Are Used to Solve Problems 4

The Uses of Statistics 7

Measurement: The Basic Building Block of Research 13 Science and Measurement: Classification as a First Step in Research 14

Levels of Measurement 15

Relating Interval, Ordinal, and Nominal Scales: The Importance of Collecting Data

at the Highest Level Possible 22

What Is a Good Measure? 23

Representing and Displaying Data 36 What Are Frequency Distributions and Histograms? 37

Extending Histograms to Multiple Groups: Using Bar Charts 43

Using Bar Charts with Nominal or Ordinal Data 50

Pie Charts 51

Time Series Data 52

Describing the Typical Case: Measures of Central Tendency 65 The Mode: Central Tendency in Nominal Scales 66

The Median: Taking into Account Position 68

The Mean: Adding Value to Position 74

Statistics in Practice: Comparing the Median and the Mean 82

How Typical Is the Typical Case?: Measuring Dispersion 94

The Purpose of Statistics Is to Clarify 3

Measuring Dispersion in Interval Scales: The Range, Variance, and Standard Deviation 102

Introduction: Statistics as a Research Tool

P r e f a c e

C h a p t e r o n e

C h a p t e r t w o

C h a p t e r t h r e e

C h a p t e r f o u r

C h a p t e r f i v e

x i i i

vii

Basic Principles Apply Across Statistical Techniques 5

Measures of Dispersion for Nominal- and Ordinal-Level Data 95

The Logic of Statistical Inference: Making Statements About Populations from Sample Statistics 125

The Dilemma: Making Statements About Populations from Sample Statistics 126

The Research Hypothesis 129

The Null Hypothesis 131

Risks of Error in Hypothesis Testing 133

Risks of Error and Statistical Levels of Significance 135

Departing from Conventional Significance Criteria 1 7

Defining the Observed Significance Level of a Test: A Simple Example Using the Binomial Distribution 145

The Fair Coin Toss 147

Different Ways of Getting Similar Results 151

Solving More Complex Problems 154

The Binomial Distribution 155

Using the Binomial Distribution to Estimate the Observed Significance Level of a Test 159

Steps in a Statistical Test: Using the Binomial Distribution to Make Decisions About Hypotheses

Chi-Square: A Test Commonly Used for Nominal-Level Measures

Extending the Chi-Square Test to a Relationship Between Two Ordinal Variables: Identification with Fathers

The Normal Distribution and Its Application to Tests

171 The Problem: The Impact of Problem-Oriented Policing on Disorderly Activity at Violent-Crime Hot Spots 172

Assumptions: Laying the Foundations for Statistical Inference 174

Selecting a Sampling Distribution 180

Significance Level and Rejection Region 182

The Test Statistic 187

Making a Decision 187

197 Testing Hypotheses Concerning the Roll of a Die 198

Relating Two Nominal-Scale Measures in a Chi-Square Test 206

Extending the Chi-Square Test to Multicategory Variables: The Example of Cell Allocations in Prison 212

and Delinquent Acts 217

The Use of Chi-Square When Samples Are Small: A Final Note 222

of Statistical Significance 234 The Normal Frequency Distribution, or Normal Curve 235

Applying Normal Sampling Distributions to Nonnormal Populations 247

Comparing a Sample to an Unknown Population: The Single-Sample z-Test for Proportions 252

Comparing a Sample to an Unknown Population: The Single-Sample t-Test for Means 257

C h a p t e r s i x

C h a p t e r s e v e n

C h a p t e r e i g h t

C h a p t e r n i n e

C h a p t e r t e n

C O N T E N T Sviii

3

Distinguishing Statistical Significance and Strength of Relationship:

Measuring Association for Interval-Level Data: Pearson’s Correlation Coefficient

Testing the Statistical Significance of Pearson’s r

Testing the Statistical Significance of Spearman’s r

An Introduction to Bivariate Regression

Multivariate Regression

Comparing Means and Proportions in Two Samples 269 Comparing Sample Means 270

Comparing Sample Proportions: The Two-Sample t-Test for Differences of Proportions 282

The t-Test for Dependent Samples 288

A Note on Using the t-Test for Ordinal Scales 293

Comparing Means Among More Than Two Samples: Analysis of Variance Analysis of Variance 307

Defining the Strength of the Relationship Observed 328

Making Pairwise Comparisons Between the Groups Studied 331

A Nonparametric Alternative: The Kruskal-Wallis Test 334

Measures of Association for Nominal and Ordinal Variables 351

The Example of the Chi-Square Statistic 352

Measures of Association for Nominal Variables 355

Measures of Association for Ordinal-Level Variables 367

Choosing the Best Measure of Association for Nominal- and Ordinal-Level Variables 385

398 Measuring Association Between Two Interval-Level Variables 399

Pearson’s Correlation Coefficient 401

Spearman’s Correlation Coefficient 419

421

428

439 Estimating the Influence of One Variable on Another: The Regression Coefficient 440

Prediction in Regression: Building the Regression Line 445

Evaluating the Regression Model 453

The F-Test for the Overall Regression 467

481 The Importance of Correct Model Specifications 482

Correctly Specifying the Regression Model 494

C h a p t e r e l e v e n

C h a p t e r t w e l v e

C h a p t e r t h i r t e e n

C h a p t e r f o u r t e e n

C h a p t e r f i f t e e n

C h a p t e r s i x t e e n

C O N T E N T S ix

306

Logistic Regression

Multivariate Regression: Additional Topics Non-linear Relationships 516

Interaction Effects 522

An Example: Punishment Severity 533

An Example: Race and Punishment Severity 525

The Problem of Multicollinearity 534

548 Why Is It Inappropriate to Use OLS Regression for a Dichotomous Dependent Variable? 550

Logistic Regression 555

Interpreting Logistic Regression Coefficients 567

Comparing Logistic Regression Coefficients 577

Evaluating the Logistic Regression Model 583

Statistical Significance in Logistic Regression 587

C h a p t e r s e v e n t e e n

C h a p t e r e i g h t e e n

C O N T E N T Sx

514

C h a p t e r n i n e t e e n

Special Topics: Randomized Experiments 674

Sample Size, Equivalence, and Statistical Power 683

C h a p t e r t w e n t y o n e

Statistical Power

Examining Interaction Terms in Experimental Research 695

The Structure of a Randomized Experiment

The Main Advantage of Experiments: Isolating Causal Effects 677

Internal Validity 682

and Block Randomization 691

Using Covariates to Increase Statistical Power in Experimental Studies 693

C h a p t e r t w e n t y

Multilevel Regression Models 637 Variance Components Model 640

Random Intercept Model 646

Random Coefficient Model 655

Adding Cluster (Level 2) Characteristics 660

Special Topics: Confidence Intervals

Constructing Confidence Intervals

Confidence Intervals 704

C h a p t e r t w e n t y t w o

676

Multivariate Regression with Multiple Category Nominal or Ordinal Measures: Extending the Basic Logistic Regression Model 601

Multinomial Logistic Regression 603

Ordinal Logistic Regression 615

Substantive Example: Severity of Punishment Decisions 619

702

708

C O N T E N T S xi

C h a p t e r t w e n t y t h r e e

Critical Values of �2

Critical Value for P (Pcrit

Factorials 759

Distribution 760

Areas of the Standard Normal Distribution 761

Critical Values of Student’s t Distribution 762

Critical Values of the F-Statistic 763

), Tukey’s HSD Test 766

Critical Values for Spearman’s Rank-Order Correlation Coefficient 767

Fisher r-to-Z* Transformation 768

Glossary 770

Index 778

Appendix 1

Appendix 2

Appendix 3

Appendix 4

Appendix 5

Appendix 6

Appendix 7

Appendix 8

Special Topics: Statistical Power Statistical Power

Estimating Statistical Power and Sample Size for a Statistically Powerful Study 738

Summing Up: Avoiding Studies Designed for Failure 747

726 728

Components of Statistical Power 731

Preface

Oliver Wendell Holmes, the distinguished associate justice of the

Supreme Court, was noted for his forgetfulness. On a train leaving

Washington, D.C., he is said to have been approached by a

awkward moments, the conductor recognized the distinctive-

however, is said to have looked sternly at the conductor and

responded, “Young man, the problem is not where is my ticket;

the problem is where am I going.”

basic understanding of statistics in this field. In the first chapter, the main

themes of the text are outlined and discussed. This preface describes

how the text is organized.

The text takes a building-block approach. This means that each chap-

ter helps prepare you for the chapters that follow. It also means that the

level of sophistication of the text increases as the text progresses. Basic

concepts discussed in early chapters provide a foundation for the intro-

duction of more complex statistical issues later. One advantage to this

approach is that it is easy to see, as time goes on, how much you have

learned about statistics. Concepts that would have seemed impossible to

of the book now, you will see equations that are quite forbidding. How-

ever, when you come to these equations after covering the material in

earlier chapters, you will be surprised at how easy they are to under-

stand.

Throughout the text, there is an emphasis on comprehension and not

his case and his pockets, could not locate his pass. After a few

conductor who requested his ticket. Holmes, searching through

the rail company the ticket when he found it. Justice Holmes,

looking and well-known jurist and suggested that he just send

For the student of statistics, a textbook is like a train ticket. Not only does

it provide a pass the student can use for entering a new and useful area

of study; it also defines the route that will be taken and the goals that

are important to achieve. Different textbooks take different approaches

and emphasize different types of material. Statistics in Criminal Jus-

tice emphasizes the uses of statistics in research in crime and justice.

This text is meant for students and professionals who want to gain a

examine real-life criminal justice problems. In the opening chapters of the

sible but sophisticated understanding of statistics that can be used to

simple when you encounter them later on. If you turn to the final chapters

understand, had they been introduced at the outset, are surprisingly

xiii

computation. This approach is meant to provide readers with an acces-

book, basic themes and materials are presented. Chapter 1 provides an

introduction to how we use statistics in criminal justice and the problems

we face in applying statistics to real-life research problems. Chapters 2

through 5 introduce basic concepts of measurement and basic methods

build on the themes covered in these early chapters.

One of the fundamental problems researchers face is that they seek

to make statements about large populations (such as all U.S. citizens)

but are generally able to collect information or data on only a sample,

or smaller group, drawn from such populations. In Chapters 6 through

12, the focus is on how researchers use statistics to overcome this

special problems are encountered in criminal justice research, and how

should the researcher approach them? Some texts skip over the basics,

Having examined how we can make statements about populations

from information gained from samples, we turn to how we describe the

strength of association between variables. In the social sciences, it is

often essential not only to determine whether factors are related but also

to define the strength and character of those relationships. Accordingly,

in Chapters 13 and 14, we look at measures of association, and in Chap-

remember that the more advanced statistics presented in later chapters

Many of the statistics provided here will be familiar to you; however,

ing statements about populations based on samples? What are the

problem. What is the logic that underlies the statistics we use for mak-

for graphically representing data and using statistics to describe data.

different types of statistical procedures or tests that can be used? What

moving students from test to test before they understand the logic

behind the tests. The approach here is to focus in greater detail on

relatively simple statistical decisions before moving on to more com-

plex ones.

P R E F A C Exiv

In the concluding chapters, we look at three special topics. Chapter

estimates that you obtain from a sample. Because our emphasis is on

research in criminal justice, we conclude the text with a chapter that

examines methods for evaluating and improving the design of a research

project. The statistical concept that is central to Chapter 23—statistical

ters 15 through 20, we examine bivariate and different types of multi-

a method for assessing how much trust you can place in the specific

power—follows directly from the concepts developed in prior chapters.

it has become a central concern in criminal justice research and accord-

Statistical power is often ignored in introductory statistics texts. However,

While it is always difficult in statistics to decide where an introduc-

tory text should stop, with an understanding of these techniques you

variate regression. These are likely to be new topics for you, though they

are statistics commonly used in criminal justice.

criminal justice interventions. Chapter 22 describes confidence intervals,

21 focuses on the design of randomized experiments. Randomized experi-

ments allow criminal justice researchers to be confident about the causal

relationships between variables, and are often used in the evaluation of

in this text.ingly is given emphasis

point in learning statistics, but also to leave you with the confidence and

A working knowledge of computers is not required to understand

the statistical concepts or procedures presented in the text. However,

computers have become a very important part of research in statistics,

and thus we provide computer exercises for relevant chapters and a

tools to tackle more complex problems on your own. Each chapter starts

with a statement of the basic concepts and problems addressed and ends

with a full chapter summary. There is also a list of equations, when

relevant, at the end of the chapter. These materials should help you to

review what you have learned and to identify the basic knowledge you

need to move on to subsequent chapters.

All of the chapters contain a list of key terms with short definitions.

The key terms appear in boldface the first time they are mentioned in

the chapter. Sometimes a term may have been briefly explained in an

earlier chapter, but is designated as a key term in the chapter where

the concept is more central. A general glossary of key terms appears at

at the end. The questions are designed to make you think about the

subjects covered in the chapter. Sometimes they are straightforward,

following directly from the text. Sometimes they ask you to develop

ideas in slightly different ways than in the text. In constructing the ques-

tions, we sought to make working on statistical issues as much fun as

possible. In statistics, it is crucial to go over material more than once.

The questions are meant to reinforce the knowledge you have gained.

will have the basic tools to comprehend and conduct criminal justice

research. Of course, these tools constitute a building block for more

advanced methods. The goal of the text is not only to bring you to this

web site where you can access the data needed for those exercises

P R E F A C E xv

the end of the book. Chapters 2 through 23 each have a set of questions

Statistics in Criminal Justice will allow you to approach statistics in a

familiar context. It emphasizes the statistics and the problems that are

commonly encountered in criminal justice research. It focuses on under-

standing rather than computation. However, it takes a serious approach

to statistics, which is relevant to the real world of research in crime and

justice. The text is meant not only as an introduction for students but

also as a reference for researchers. The approach taken will help both

students who want an introduction to statistics and professionals who

seek a straightforward explanation for statistics that have become a

routine tool in contemporary criminal justice systems.

You are encouraged to use the web site. It will help you to see the

connection between the topics discussed in the chapters and statistical

computing.

(see the Computer Exercises at the end of Chapter 2 for details).

A c k n o w l e d g m e n t s

xvi

Professor Joseph Naus of the Department of Statistics of Rutgers Univer-

In the development of any academic enterprise, many students and

colleagues provide support and advice. We are particularly indebted to

of the new material for the third edition. A number of current and former

original edition of this book, and to Daniel Salem, a graduate of the Insti-

sity, who played a crucial advisory role in the preparation of the

Stephen Schnebly of the University of Illinios at Springfield, and Marc

work and provided comments over the previous editions, particularly the

tute of Criminology of the Hebrew University, who played a major role

of South Carolina was particularly helpful in helping us to refine some

in the production of the first edition. Robert Brame of the University

graduate students helped us in revising and editing the manuscript

across the four editions, including SueMing Yang and Kristen Miggans at

the University of Maryland, and Shomron Moyal, Rochelle Schnurr, Tal

Yonaton and Gali Weissman of the Hebrew University, and Arian Ferrer

We also want to thank the many scholars who have read and used our Emmanuelle Klossou, and Michael Rocque of Northeastern University.

comments and suggestions made by Todd Armstrong of Sam Houston

Swatt of Justice & Security Strategies, Inc. The final product reflects their

keen insights and thoughtful suggestions.

State University, Pamela Lattimore of the Research Triangle Institute,

xvii

A b o u t t h e A u t h o r s

David Weisburd is a Distinguished Professor of Criminology, Law and

Society at George Mason University (and Director of the Center for

Evidence Based Crime Policy) and Walter E. Meyer Professor of Law and

Criminal Justice at the Hebrew University. He also serves as a Senior

Fellow at the Police Foundation in Washington DC and is Chair of its

Research Advisory Committee. Professor Weisburd is an elected Fellow of

the American Society of Criminology and of the Academy of Experimental

Criminology. He is a member of the Science Advisory Board of the Office

of Justice Programs, the Steering Committee of the Campbell Crime and

Justice Group, the National Institute of Justice/Harvard Executive Session

in Policing, the Scientific Commission of the International Society of

Criminology, and the Committee on Law and Justice of the National

Research Council of the National Academy of Sciences (USA). Professor

Weisburd is one of the leading international researchers in crime and

justice. He is author or editor of more than twenty books and more than

100 scientific articles that cover a wide range of criminal justice research

topics, including crime at place, white collar crime, policing, illicit markets,

terrorism, criminal justice statistics and social deviance. Professor Weisburd

is the recipient of the 2010 Stockholm Prize in Criminology and the 2011

Klachky Family Prize for the Advancement of the Frontiers of Science.

Professor Weisburd is also the founding editor of the Journal of Experi-

mental Criminology.

Chester L. Britt is a researcher and scholar in criminology and criminal

from the University of Arizona, Professor Britt taught at the University

journals as Law and Society Review, Justice Quarterly, journal of Quanti-

and Journal of Research in Crime and Deliquency.

justice. He is Dean of the School of Criminology and Criminal Justice at

of Illinois, Penn State University, and Arizona State University. He served

Justice Quarterly from 2005 through 2007. He has also

co-edited a volume on criminological theory entitled Control Theories

of Crime and Delinquency. His research articles have appeared in such

Northeastern University in Boston. After receiving his Ph.D. in sociology

as the Editor of

tative Criminonology,

Introduction: Statistics as a Research Tool

Do Statisticians Have to Be Experts in Mathematics?

Are Computers Making Statisticians Redundant?

What Basic Principles Apply to Different Types of Statistics?

What Are the Different Uses of Statistics in Research?

C h a p t e r o n e

I n i t i a l h u r d l e s

K e y p r i n c i p l e s

What is Our Aim in Choosing a Statistic?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_1,

© Springer Science+Business Media New York 2014

tool for answering questions. It allows us to take large bodies of infor-

mation and summarize them with a few simple statements. It lets us

come to solid conclusions even when the realities of the research world

make it difficult to isolate the problems we seek to study. Without statis-

tics, conducting research about crime and justice would be virtually im-

possible. Yet, there is perhaps no other subject in their university studies

that criminal justice students find so difficult to approach.

A good part of the difficulty lies in the links students make between

statistics and math. A course in statistics is often thought to mean long

hours spent solving equations. In developing your understanding of sta-

tistics in criminal justice research, you will come to better understand the

formulas that underlie statistical methods, but the focus will be on con-

cepts and not on computations. There is just no way to develop a good

understanding of statistics without doing some work by hand. But in the

age of computers, the main purpose of doing computations is to gain a

deeper understanding of how statistics work.

Researchers no longer spend long hours calculating statistics. In the

1950s, social scientists would work for months developing results that

can now be generated on a computer in a few minutes. Today, you do

not need to be a whiz kid in math to carry out a complex statistical

analysis. Such analyses can be done with user-friendly computer pro-

grams. Why then do you need a course in statistics? Why not just leave it

to the computer to provide answers? Why do you still need to learn the

basics?

The computer is a powerful tool and has made statistics accessible to

a much larger group of criminal justice students and researchers. How-

ever, the best researchers still spend many hours on statistical analysis.

Now that the computer has freed us from long and tedious calculations,

what is left is the most challenging and important part of statistical analy-

sis: identifying the statistical tools that will best serve researchers in inter-

preting their research for others.

2

THE PURPOSE OF STATISTICAL ANALYSIS is to clarify and not confuse. It is a

T H E P U R P O S E O F S T A T I S T I C S I S T O 3

need to choose statistics for research and interpret them. It is meant for

students of criminology and criminal justice. As in other fields, there are

specific techniques that are commonly used and specific approaches that

have been developed over time by researchers who specialize in this

area of study. These are the focus of this text. Not only do we draw our

examples from crime and justice issues; we also pay particular attention

to the choices that criminal justice researchers make when approaching

statistical problems.

Before we begin our study of statistics in criminal justice, it is useful

to state some basic principles that underlie the approach taken in this

text. They revolve around four basic questions. First, what should our

purpose be in choosing a statistic? Second, why do we use statistics to

answer research questions? Third, what basic principles apply across

very different types of statistics? And finally, what are the different uses

for statistics in research?

guage. In this sense, statistics provide a way for the initiated to share

ideas and concepts without including the rest of us. Of course, it is nec- essary to use a common language to report research results. This is one

reason why it is important for you to take a course in statistics. But the

reason we use statistics is to make research results easier—not more dif-

would be just to tell us about your subjects. You could describe each

offender and his or her criminal history without creating any real

confusion. But what if you wanted to report on 20 offenders? It would

take quite a long time to tell us about each one in some detail, and it

is likely that we would have difficulty remembering who was who.

It would be even more difficult to describe 100 offenders. With

thousands of offenders, it would be just about impossible to take this

approach.

This is one example of how statistics can help to simplify and clarify

the research process. Statistics allow you to use a few summary state- ments to provide a comprehensive portrait of a large group of offenders.

For example, instead of providing the name of each offender and telling

us how many crimes he or she committed, you could present a single

ficult—to understand. For example, if you wanted to provide a description

of three offenders you had studied, you would not need to search for statis-

tics to summarize your results. The simplest way to describe your sample

The goal of this text is to provide you with the basic skills you will

It sometimes seems as if researchers use statistics as a kind of secret lan-

C L A R I F Y

T h e P u r p o s e o f S t a t i s t i c s I s t o C l a r i f y

4 C H A P T E R O N E : I N T R O D U C T I O N

statistic that described the average number of crimes committed by the

people you studied. You might say that, on average, the people you

studied committed two to three crimes in the last year. Thus, although it

might be impossible to describe each person you studied, you could, by

using a statistic, give your audience an overall picture. Statistics make it

possible to summarize information about a large number of subjects with

a few simple statements.

Given that statistics should simplify the description of research results,

it follows that the researcher should utilize the simplest statistics appro-

priate for answering the research questions that he or she raises.

Nonetheless, it sometimes seems as if researchers go out of their way to

identify statistics that few people recognize and even fewer understand.

This approach does not help the researcher or his or her audience.

There is no benefit in using statistics that are not understood by those in-

terested in your research findings. Using a more complex statistic when a

simpler one is appropriate serves no purpose beyond reducing the num-

ber of people who will be influenced by your work.

The best presentation of research findings will communicate results in

a clear and understandable way. When using complex statistics, the re-

searcher should present them in as straightforward a manner as possible.

The mark of good statisticians is not that they can mystify their audi-

ences, but rather that they can communicate even complex results in a

way that most people can understand.

S t a t i s t i c s A r e U s e d t o S o l v e P r o b l e m s

Statistics develop because of a need to deal with a specific type of ques-

tion or problem. In the example above, you were faced with the

dilemma that you could not describe each person in a very large study

without creating a good deal of confusion. We suggested that an average

might provide a way of using one simple statistic to summarize a charac-

teristic of all the people studied. The average is a statistical solution. It is

a tool for solving the problem of how to describe many subjects with a

short and simple statement.

As you will see in later chapters, statistics have been developed to

deal with many different types of problems that researchers face. Some

of these may seem difficult to understand at the outset, and indeed it is

natural to be put off by the complexities of some statistics. However, the

solutions that statisticians develop are usually based on simple common

sense. Contrary to what many people believe, statistics follow a logic

that you will find quite easy to follow. Once you learn to trust your com-

mon sense, learning statistics will turn out to be surprisingly simple. Indeed,

5

encounter in this text. Stating them at the outset will help you to see

how statistical procedures in later chapters are linked one to another. To

understand these principles, you do not need to develop any computa-

tions or formulas; rather, you need to think generally about what we are

trying to achieve when we develop statistics.

The first is simply that in developing statistics we seek to reduce the

level of error as much as possible. The purpose of research is to provide

answers to research questions. In developing these answers, we want to

be as accurate as we can. Clearly, we want to make as few mistakes as

possible. The best statistic is one that provides the most accurate state-

ment about your study. Accordingly, a major criterion in choosing which

statistic to use—or indeed in defining how a statistic is developed—is the

amount of error that a statistic incorporates. In statistics, we try to mini-

mize error whenever possible.

Unfortunately, it is virtually impossible to develop any description

without some degree of error. This fact is part of everyday reality. For

example, we do not expect that our watches will tell perfect time or that

our thermostats will be exactly correct. At the same time, we all know

that there are better watches and thermostats and that one of the factors

that leads us to define them as “better” is that they provide information

with less error. Similarly, although we do not expect our stockbroker to

possible. Fear of statistics is a greater barrier to learning than any of the com-

putations or formulas that we will use. It is difficult to learn anything when

you approach it with great foreboding. Statistics is a lot easier than you think.

The job of this text is to take you step by step through the principles and

ideas that underlie basic statistics for criminal justice researchers. At the

beginning, we will spend a good deal of time examining the logic behind

statistics and illustrating how and why statisticians choose a particular

that the solutions statisticians use make very good sense.

solution to a particular statistical problem. What you must do at the outset

is take a deep breath and give statistics a chance. Once you do, you will find

our experience is that students who have good common sense, even

if they have very little formal background in this area, tend to become

A few basic principles underlie much of the statistical reasoning you will

be correct all of the time, we are likely to choose the broker who we

believe will make the fewest mistakes.

the best criminal justice statisticians. But in order to be able to use com-

B A S I C P R I N C I P L E S A P P L Y A C R O S S S T A T I S T I C A L T E C H N I Q U E S

mon sense, it is important to approach statistics with as little fear as

B a s i c P r i n c i p l e s A p p l y A c r o s s S t a t i s t i c a l T e c h n i q u e s

6 C H A P T E R O N E : I N T R O D U C T I O N

In choosing a statistic, we also use a second principle to which we

will return again and again in this text: Statistics based on more informa-

tion are generally preferred over those based on less information. This

principle is common to all forms of intelligence gathering and not just

those that we use in research. Good decision making is based on infor-

mation. The more information available to the decision maker, the better

he or she can weigh the different options that are presented. The same

goes for statistics. A statistic that is based on more information, all else

being equal, will be preferred over one that utilizes less information.

There are exceptions to this rule, often resulting from the quality or form

of the information or data collected. We will discuss these in detail in the

text. But as a rule, the best statistic utilizes the maximum amount of

information.

Our third principle relates to a danger that confronts us in using statis-

tics as a tool for describing information. In many studies, there are cases

that are very different from all of the others. Indeed, they are so different

that they might be termed deviant cases or, as statisticians sometimes call

them, “outliers.” For example, in a study of criminal careers, there may

be one or two offenders who have committed thousands of crimes,

whereas the next most active criminal in the sample has committed only

choice of statistics and your presentation of results.

In almost every statistic we will study, outliers present a distinct and

troublesome problem. A deviant case can make it look as if your offend-

ers are younger or older than they really are—or less or more criminally

active than they really are. Importantly, deviant cases often have the

most dramatic effects on more complex statistical procedures. And it is

precisely here, where the researcher is often preoccupied with other sta-

tistical issues, that deviant cases go unnoticed. But whatever statistic is

used, the principle remains the same: Outliers present a significant prob-

lem in choosing and interpreting statistics.

The final principle is one that is often unstated in statistics, because it

is assumed at the outset: Whatever the method of research, the researcher

must strive to systematize the procedures used in data collection and

analysis. As Albert J. Reiss, Jr., a pioneer in criminal justice methodolo-

gies, has noted, “systematic” means in part “that observation and record-

ing are done according to explicit procedures which permit replication

and that rules are followed which permit the use of scientific inference.”1

1A. J. Reiss, Jr., “Systematic Social Observation of Social Phenomenon,” in Herbert

Costner (ed.), Sociological Methodology (San Francisco: Jossey Bass, 1971), pp. 3–33.

a few hundred crimes. Although such cases form a natural part of the

research process, they often have very significant implications for your

T H E U S E S O F S T A T I S T I C S 7

While Reiss’s comment will become clearer as statistical concepts are

defined in coming chapters, his point is simply that you must follow

clearly stated procedures and rules in developing and presenting statisti-

follow a consistent logic from start to finish. You should not jump from

statistic to statistic merely because the outcomes are favorable to the the-

sis you raise. In learning about statistics, it is also important to go step by

step—and to be well organized and prepared. You cannot learn statistics

by cramming in the last week of classes. The key to learning statistics is

to adopt a systematic process and follow it each week.

Statistical procedures are built on all of the research steps that pre-

cede them. If these steps are faulty, then the statistics themselves will be

faulty. In later chapters, we often talk about this process in terms of the

assumptions of the statistics that we use. We assume that all of the rules

of good research have been followed up to the point where we decide

on a statistic and calculate it. Statistics cannot be disentangled from the

larger research process that comes before them. The numbers that we

systematic approach is crucial not only to the statistical procedures that

you will learn about in this text but to the whole research process.

three ways in which statistics are used in criminal justice. The first is

called descriptive statistics, because it helps in the summary and de-

scription of research findings. The second, inferential or inductive sta-

tistics, allows us to make inferences or statements about large groups

from studies of smaller groups, or samples, drawn from them. Finally,

we introduce multivariate statistics toward the end of the text. Multi-

variate statistics, as the name implies, allow us to examine a series of

variables at one time.

Descriptive Statistics

We are all familiar in some way with descriptive statistics. We use them

often in our daily lives, and they appear routinely in newspapers and on

television. Indeed, we use them so often that we sometimes don’t think

cal findings. It is important to approach statistics in a systematic way. You

cannot be sloppy or haphazard, at least if the statistic is to provide a good

answer to the research question you raise. The choice of a statistic should

In the chapters that follow, we will examine three types of statistics or

employed. Very complex statistics cannot hide bad research methods. A

use are only as good as the data collection techniques that we have

T h e U s e s o f S t a t i s t i c s

8 C H A P T E R O N E : I N T R O D U C T I O N

of them as statistics at all. During an election year, everyone is con-

cerned about the percentage support that each candidate gains in the

primaries. Students at the beginning of the semester want to know what

proportion of their grades will be based on weekly exercises. In deciding

whether our salaries are fair, we want to know what the average salary is

for other people in similar positions. These are all descriptive statistics.

They summarize in one simple statement the characteristics of many

people. As discussed above in the example concerning criminal histories,

descriptive statistics make it possible for us to summarize or describe

large amounts of information.

In the chapters that follow, we will be concerned with two types of

descriptive statistics: measures of central tendency and measures of

dispersion. Measures of central tendency are measures of typicality.

They tell us in one statement what the average case is like. If we could

take only one person as the best example for all of the subjects we stud-

ied, who would it be? If we could choose only one level of criminal ac-

tivity to typify the frequency of offending of all subjects, what level

would provide the best snapshot? If we wanted to give our audience a

general sense of how much, on average, a group of offenders stole in a

year, what amount would provide the best portrait? Percentages, propor-

tions, and means are all examples of measures of central tendency that

we commonly use. In the coming chapters, you will learn more about

these statistics, as well as more complex measures with which you may

not be familiar, such as correlation and regression coefficients.

Having a statistic that describes the average case is very helpful in de-

scribing research results. However, we might also want to know how

typical this average case is of the subjects in our study. The answer to

this question is provided by measures of dispersion. They tell us to what

extent the other subjects we studied are similar to the case or statistic we

have chosen to represent them. Although we don’t commonly use mea-

sures of dispersion in our daily lives, we do often ask similar questions

without the use of such statistics.

For example, in deciding whether our income is fair, we might want

to know not only the average income of others in similar positions, but

also the range of incomes that such people have. If the range was very

small, we would probably decide that the average provides a fairly good

portrait of what we should be making. If the range was very large, we

might want to investigate more carefully why some people make so

much more or less than the average. The range is a measure of disper-

sion. It tells us about the spread of scores around our statistic. In the

chapters that follow, we will look at other measures of dispersion—for

example, the standard deviation and variance, which may be less famil-

iar to you. Without these measures, our presentation of research findings

T H E U S E S O F S T A T I S T I C S 9

would be incomplete. It is not enough simply to describe the typical

case; we must also describe to what degree other cases in our study are

different from or similar to it.

Inferential Statistics

Inferential statistics allow us to make statements about a population, or

the larger group of people we seek to study, on the basis of a sample

drawn from that population. Without this very important and powerful

tool, it would be very difficult to conduct research in criminal justice.

The reason is simple. When we conduct research, we do so to answer

questions about populations. But in reality we seldom are able to collect

information on the whole population, so we draw a sample from it. Sta-

tistical inference makes it possible for us to infer characteristics from that

sample to the population.

Why is it that we draw samples if we are really interested in making

statements about populations? In good part it is because gaining informa-

tion on most populations is impractical and/or too expensive. For exam-

ple, if we seek to examine the attitudes of U.S. citizens toward criminal

justice processing, we are interested in how all citizens feel. However,

studying all citizens would be a task of gigantic proportion and would

cost billions of dollars. Such surveys are done every few years and are

called censuses. The last census in the United States took many years to

prepare and implement and cost over $5 billion to complete. If every re-

search study of the American population demanded a census, then we

would likely cost millions of dollars to complete a simple study of their

attitudes. This is because the most inexpensive data collection can still

cost tens of dollars for each subject studied. When you consider that the

National Institute of Justice, the primary funder of criminal justice re-

search in the United States, provides a total of about $100 million a year

for all research, it is clear that criminal justice research cannot rely on

studies of whole populations.

It is easy to understand, then, why we want to draw a sample or sub-

set of the larger population to study, but it is not obvious why we should

believe that what we learn from that sample applies to the population

from which it is drawn. How do we know, for example, that the atti-

tudes toward criminal justice expressed by a sample of U.S. citizens are

similar to the attitudes of all citizens? The sample is a group of people

would have very few research projects indeed. Even when we are interested

in much smaller populations in the criminal justice system, examination of

the entire population is often beyond the resources of the criminal justice

researcher. For example, to study all U.S. prisoners, we would have to

study over 1 million people.

Even if we wanted to look at the 100,000 or so women prisoners, it

10 C H A P T E R O N E : I N T R O D U C T I O N

drawn from the population; it is not the population itself. How much

surveys that now form so much a part of public life or the studies that

you read about in your other college classes. When a news organization

sions about populations—whether of offenders, criminal justice agents,

crime-prone places, or criminal justice events—on samples. Statistical in-

ference provides a method for deciding to what extent you can have

faith in such results. It allows you to decide when the outcome observed

in a sample can be generalized to the population from which it was

drawn. Statistical inference is a very important part of statistics and one

we will spend a good deal of time discussing in this text.

Taking into Account Competing Explanations: Multivariate Statistics

Multivariate statistics allow us to solve a different type of problem in re-

search. It is often the case that the issue on which we want to focus is

confounded by other factors in our study. Multivariate statistics allow us

to isolate one factor while taking into account a host of others. For ex-

ample, a number of criminal justice studies examine the impact of im-

prisonment on the future criminal behavior of offenders. In general, they

compare offenders who are found guilty in court and sentenced to

once they are released into the community, is different from that of non-

prisoners. Researchers conducting these studies face a very difficult re-

search problem. Prisoners and nonprisoners are often very different

types of people, and some of these differences are likely to affect their

impact of imprisonment on future offending. If we discover that prison-

ers, once released into the community, are more likely than nonprison-

ers to commit a crime, how can we tell whether this was a result of the

can we rely on such estimates? And to what extent can we trust such statis-

tics? You have probably raised such issues already, in regard to either the

conducts a survey of 1,000 people to tell us how all voters will vote in the

next election, it is using a sample to make statements about a population.

The criminal justice studies you read about also base their conclu-

prison with those who are found guilty but do not receive a prison sanction.

Such studies focus on whether the criminal behavior of prisoners,

criminal behavior in the community. For example, prisoners are more likely

than nonprisoners to have been arrested before, since a prior arrest is often an important factor in the judge’s decision to incarcerate a convicted off-

ender in the first place. And we know from research about criminal careers

that people with a prior history of arrest are much more likely than people

without such a history to commit a crime in the future. Accordingly, prisoners

are more likely to commit a crime in the future, irrespective of the fact that

they have served a prison sentence. This makes it very difficult to assess the

C H A P T E R S U M M A R Y 11

experience of imprisonment? It might be due to the simple fact that pris-

oners are more likely than nonprisoners to commit crimes in the first

place. Their more serious arrest histories would predict this result.

The complex task facing the criminal justice researcher is to isolate

the specific impact of imprisonment itself from all of the other possible

explanations for differences in reoffending between prisoners and non-

prisoners. Multivariate analysis provides a statistical solution to this prob-

lem. It allows the criminal justice researcher to isolate the impact of one

factor—in this case, imprisonment—from those of other factors, such as

prior criminal history, that might confound the researcher’s conclusions.

C h a p t e r S u m m a r y

Statistics seem intimidating because they are associated with complex

mathematical formulas and computations. Although some knowledge of

math is required, an understanding of the concepts is much more im-

portant than an in-depth understanding of the computations. Today’s

computers, which can perform complex calculations in a matter of sec-

onds or fractions of seconds, have drastically cut the workload of the

researcher. They cannot, however, replace the key role a researcher

plays in choosing the most appropriate statistical tool for each research

problem.

The researcher’s aim in using statistics is to communicate findings in a

clear and simple form. As a result, the researcher should always choose

the simplest statistic appropriate for answering the research question.

ing principles apply to all types of statistics: (1) In developing statistics,

we seek to reduce the level of error as much as possible. (2) Statistics

based on more information are generally preferred over those based on

less information. (3) Outliers present a significant problem in choosing

and interpreting statistics. (4) The researcher must strive to systematize

the procedures used in data collection and analysis.

There are three principal uses of statistics discussed in this book. In

descriptive statistics, the researcher summarizes large amounts of in-

formation in an efficient manner. Two types of descriptive statistics that

go hand in hand are measures of central tendency, which describe

the characteristics of the average case, and measures of dispersion,

which tell us just how typical this average case is. We use inferential

statistics to make statements about a population on the basis of a sam-

ple drawn from that population. Finally, in multivariate statistics, we

isolate the impact of one factor from others that may distort our results.

Statistics offer commonsense solutions to research problems. The follow-

12 C H A P T E R O N E : I N T R O D U C T I O N

K e y T e r m s

descriptive statistics A broad area of sta-

tistics that is concerned with summarizing

large amounts of information in an efficient

manner. Descriptive statistics are used to

describe or represent in summary form the

characteristics of a sample or population.

inferential, or inductive, statistics A

broad area of statistics that provides the re-

searcher with tools for making statements

about populations on the basis of knowl-

edge about samples. Inferential statistics

allow the researcher to make inferences re-

garding populations from information

gained in samples.

measures of central tendency Descrip-

tive statistics that allow us to identify the

typical case in a sample or population.

Measures of central tendency are measures

of typicality.

measures of dispersion Descriptive sta-

tistics that tell us how tightly clustered or

dispersed the cases in a sample or popula-

tion are. They answer the question “How

typical is the typical case?”

multivariate statistics Statistics that

examine the relationships among vari-

ables while taking into account the possi-

variable from others that may distort his

or her results.

ble influences of other confounding factors.

Multivariate statistics allow the resear-

cher to isolate the impact of one

Measurement: The Basic Building Block

of Research

How Do Criminal Justice Researchers Develop Knowledge?

How Do the Different Scales Interconnect?

C h a p t e r t w o

C r i m i n a l j u s t i c e r e s e a r c h a s a s c i e n c e

T h e f o u r s c a l e s o f m e a s u r e m e n t

D e f i n i n g a g o o d m e a s u r e

What are They?

What are Their Characteristics?

Which is the Most Appropriate Scale of Measurement?

What is Meant by the “Validity” of a Measure?

What is Meant by the “Reliability” of a Measure?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_2,

© Springer Science+Business Media New York 2014

MEASUREMENT LIES AT THE HEART of statistics. Indeed, no statistic would

be possible without the concept of measurement. Measurement is also an

integral part of our everyday lives. We routinely classify and assign values

to people and objects without giving much thought to the processes that

underlie our decisions and evaluations. In statistics, such classification and

ordering of values must be done in a systematic way. There are clear rules

for developing different types of measures and defined criteria for decid-

ing which are most appropriate for answering a specific research question.

Although it is natural to focus on the end products of research, it is im-

portant for the researcher to remember that measurement forms the first

building block of every statistic. Even the most complex statistics, with

numbers that are defined to many decimal places, are only as accurate as

the measures upon which they are built. Accordingly, the relatively simple

rules we discuss in this chapter are crucial for developing solid research

findings. A researcher can build a very complex structure of analysis. But if

the measures that form the foundation of the research are not appropriate

for the analyses that are conducted, the findings cannot be relied upon.

We begin Chapter 2 by examining the basic idea of measurement in sci-

ence. We then turn to a description of the main types of measures in statis-

tics and the criteria used to distinguish among them. We are particularly

concerned with how statisticians rank measurement based on the amount

of information that a measure includes. This concept, known as levels of

measurement, is very important in choosing which statistical procedures are

appropriate in research. Finally, we discuss some basic criteria for defining

a good measure.

S c i e n c e a n d M e a s u r e m e n t : C l a s s i f i c a t i o n a s a F i r s t S t e p i n R e s e a r c h

Criminal justice research is a scientific enterprise that seeks to develop

knowledge about the nature of crimes, criminals, and the criminal justice

14

L E V E L S O F M E A S U R E M E N T 15

system. The development of knowledge can, of course, be carried out in

a number of different ways. Criminal justice researchers may, for exam-

ple, observe the actions of criminal justice agents or speak to offenders.

They may examine routine information collected by government or crim-

inal justice agencies or develop new information through analyses of the

content of records in the criminal justice system. Knowledge may be de-

veloped through historical review or even through examination of ar-

chaeological records of legal systems or sanctions of ancient civilizations.

The methods that criminal justice researchers use vary. What they

have in common is an underlying philosophy about how knowledge

may be gained and what scientific research can tell us. This philosophy,

which is predominant in scientific study in the modern world, is usually

called positivism.1 At its core is the idea that science is based on facts

and not values. Science cannot make decisions about the way the world

should be (although scientific observation may inform such decisions).

Rather, it allows us to examine and investigate the realities of the world

as we know it. The major tool for defining this reality in science is mea-

surement.

Measurement in science begins with the activity of distinguishing

groups or phenomena from one another. This process, which is gener-

ally termed classification, implies that we can place units of scientific

study—such as victims, offenders, crimes, or crime places—in clearly de-

fined categories. The classification process leads to the creation of vari-

ables. A variable is a trait, characteristic, or attribute that can be mea-

sured. What differentiates measurement in science from measurement in

our everyday lives is that there must be systematic criteria for determin-

ing both what each category of a variable represents and the boundaries

between categories. We now turn to a discussion of these criteria as they

relate to different levels of measurement.

L e v e l s o f M e a s u r e m e n t

Classification forms the first step in measurement. There are a number of

different ways we can classify the people, places, or phenomena we

wish to study. We may be content to simply distinguish one category

from another. But we may also be interested in how those categories re-

late to one another. Do some represent more serious crime or less seri-

ous crime? Can we rank how serious various crimes are in a clear and

1See D. Black, “The Boundaries of Legal Sociology,” in D. Black and M. Mileski (eds.),

The Social Organization of Law (New York: Seminar Press, 1973), pp. 41–47.

16 C H A P T E R T W O : M E A S U R E M E N T

defined order? Is it possible to define exactly how serious one crime is

relative to another?

As these types of questions suggest, measurement can be a lot more

complex than simply distinguishing one group from another. Recogniz-

ing this complexity, statisticians have defined four basic groups of mea-

sures, or scales of measurement, based on the amount of information

that each takes advantage of. The four are generally seen as occupying

different positions, or levels, on a ladder of measurement (see Figure

2.1). Following a principle stated in Chapter 1—that statistics based on

more information are generally preferred—measures that include more

information rank higher on the ladder of measurement.

Nominal Scales

At the bottom of the ladder of measurement are nominal scales. Nominal-

scale variables simply distinguish one phenomenon from another. Sup-

pose, for example, that you want to measure crime types. In your study,

you are most interested in distinguishing between violent crime and

other types of crime. To fulfill the requirements of a nominal scale, and

thus the minimum requirements of measurement, you need to be able to

take all of the crime events in your study and place them in one of two

categories: either violent crime or other crime. There can be no overlap.

In practice, you might come across many individual events that seem dif-

ficult to classify. For example, what would you do with a crime event in

which the offender first stole from his victim and then assaulted him?

This event includes elements of both violent and property crime. What

about the case where the offender did not assault the victim, but merely

Ratio

Interval

Ordinal

Nominal Categorization

Order + Categorization

True Zero + Set Intervals + Order + Categorization

Set Intervals + Order + Categorization

Ladder of MeasurementFigure 2.1

threatened her? Would you decide to include this in the category of

violent crime or other crime?

L E V E L S O F M E A S U R E M E N T 17

In criminology and criminal justice, we often make use of nominal-

scale variables. Many of these reflect simple dichotomies, like the distinc-

tion between violent and other crime. For example, criminologists often

seek to examine differences between men and women in their involve-

ment in criminality or treatment in the criminal justice system. It is com-

mon as well to distinguish between those who are sentenced to prison

and those who are not or those who commit more than one crime (“re-

cidivists”) and those who are only one-shot offenders.

It is often necessary to distinguish among multiple categories of a

nominal-level variable. For example, if you wanted to describe legal rep-

resentation in court cases, you would provide a very simplistic picture if

you simply distinguished between those who had some type of legal

representation and those who did not. Some of the offenders would be

likely to have private attorneys and others court-appointed legal repre-

sentation. Still others might gain help from a legal aid organization or a

public defender. In order to provide a full portrait of legal representa-

tion, you would likely want to create a nominal-scale variable with five

distinct categories: No attorney, Legal aid, Court appointed, Public de-

fender, and Private attorney. Table 2.1 presents a number of examples of

nominal-level scales commonly used in criminal justice.

Nominal-scale measures can include any number of different cate-

gories. The Uniform Crime Reporting system, which keeps track of ar-

rests in the United States, includes some 29 categories of crime. These

Nominal-Scale Variables Commonly Found in Criminal Justice Research

VARIABLE COMMON CATEGORIES

Gender Male, Female Race-Ethnicity Non-Hispanic Black, Non-Hispanic White, Hispanic (any race) Marital Status Single, Married, Separated, Divorced, Widowed Pretrial Release Status Detained, Released Type of Case Disposition Dismissed, Acquitted, Diverted, Convicted Method of Conviction Negotiated guilty plea, Nonnegotiated guilty plea, Bench trial, Jury trial Type of Punishment Incarceration, Nonincarceration

Table 2.1

In measurement, you must make systematic choices that can be applied

across events. You cannot decide one way for one event and another

way for another. In the situation described above, you might conclude

that the major issue in your study was the presence of violence. Thus,

all cases with any violent events would be placed in the violent cate-

gory. Similarly, you might conclude that violence had to include physical

victimization. Whatever your choice, to meet the requirements of mea-

surement you must define clearly where all events in your study are to

be placed.

18 C H A P T E R T W O : M E A S U R E M E N T

range from violent crimes, such as murder or robbery, to vagrancy and

vandalism. Although there is no statistical difficulty with defining many

categories, the more categories you include, the more confusing the de-

scription of the results is likely to be. If you are trying to provide a sense

of the distribution of crime in your study, it is very difficult to practically

describe 20 or 30 different crime categories. Keeping in mind that the

purpose of statistics is to clarify and simplify, you should try to use the

smallest number of categories that will accurately describe the research

problem you are examining.

At the same time, do not confuse collection of data with presentation

of your findings. You do not lose anything by collecting information in

the most detailed way that you can. If you collect information with a

large number of categories, you can always collapse a group of cate-

gories into one. For example, if you collect information on arrest events

utilizing the very detailed categories of the criminal law, you can always

combine them later into more general categories. But if you collect infor-

mation in more general categories (for example, just violent crime and

property crime), you cannot identify specific crimes such as robbery or

car theft without returning to the original source of your information.

Though nominal-scale variables are commonly used in criminology

and criminal justice, they provide us with very limited knowledge about

the phenomenon we are studying. As you will see in later chapters, they

also limit the types of statistical analyses that the researcher can employ.

In the hierarchy of measurement, nominal-scale variables form the low-

est step in the ladder. One step above are ordinal scales.

Ordinal Scales

What distinguishes an ordinal from a nominal scale is the fact that we as-

sign a clear order to the categories included. Now not only can we dis-

tinguish between one category and another; we also can place these

categories on a continuum. This is a very important new piece of infor-

mation; it allows us to rank events and not just categorize them. In the

case of crime, we might decide to rank in order of seriousness. In mea-

suring crime in this way, we would not only distinguish among cate-

gories, such as violent, property, and victimless crimes; we might also

argue that violent crimes are more serious than property crimes and that

victimless crimes are less serious than both violent and property crimes.

We need not make such decisions arbitrarily. We could rank crimes by

the amount of damage done or the ways in which the general popula-

tion rates or evaluates different types of crime.

Ordinal-scale variables are also commonly used in criminal justice

and criminology. Indeed, many important criminal justice concepts are

measured in this way. For example, in a well-known London survey of

L E V E L S O F M E A S U R E M E N T 19

2

Ranking crime by seriousness and measuring people’s fear of crime

tences, damage to victims, complexity of crime, or seriousness of prior

records of offenders, as illustrated in Table 2.2. What all of these vari-

ables have in common is that they classify events and order them along

a continuum. What is missing is a precise statement about how various

categories differ one from another.

Interval and Ratio Scales

Interval scales not only classify and order people or events; they also

define the exact differences between them. An interval scale requires

that the intervals measured be equal for all of the categories of the scale

examined. Thus, an interval-scale measure of prior record would not

simply rank prior record by seriousness; it would allow us to say how

much more serious one offender’s record was than another’s in a stan-

dard unit of measurement—for example, number of arrests, convictions,

or prison stays.

2See R. Sparks, H. Genn, and D. Dodd, Surveying Victims: A Study of the Measurement

of Criminal Victimization (New York: Wiley, 1977).

Ordinal Scale Variables Commonly Found in Criminal Justice Research

VARIABLE COMMON CATEGORIES

Level of Education Less than high school, Some high school, High school graduation, Some college or trade school, College graduate, Graduate/ professional school

Severity of Injury in an Assault None, Minor—no medical attention, Minor—medical attention required, Major—medical attention required with no hospitalization, Major—medical attention required with hospitalization

Attitude and Opinion Survey Strongly disagree, Disagree, No opinion, Agree, Strongly agree; Questions Very high, High, Moderate, Low, Very low

Bail-Release Decision Released on own recognizance, Released on bail, Detained— unable to post bail, Denied release

Type of Punishment Probation/community service, Jail incarceration, Prison incarceration, Death sentence

Table 2.2

victimization, fear of crime was measured using a simple four-level

ordinal scale. Researchers asked respondents: “Are you personally con-

cerned about crime in London as a whole? Would you say you are (1)

very concerned, (2) quite concerned, (3) a little concerned, or (4) not

concerned at all?”

are only two examples of the use of ordinal scales in criminal justice

research. We could also draw examples regarding severity of court sen-

20 C H A P T E R T W O : M E A S U R E M E N T

Most criminal justice variables that meet the criteria of an interval scale

also meet the criteria of a ratio scale. A ratio scale has all of the charac-

teristics of an interval scale but also requires that there be a non-arbitrary,

or true, zero value. This means simply that zero represents the absence

that the former has 15 more arrests than the latter. We have an important

piece of information that we would not have gained with an ordinal

scale. Now, not only can we say that the prior record of one offender is

more serious than that of another, but we can specify exactly how many

more arrests the offender has. This variable thus meets the requirements

of an interval scale. But it also meets the additional requirement of a

ratio scale that there be a true zero value, since we can state that some-

one with 20 arrests has 4 times as many arrests as someone with 5 ar-

rests. If the zero value were arbitrary, we could not make this statement.

This fact is best illustrated with an example. Suppose we alter our

measure of prior record to focus on the degree to which offenders ex-

ceed a specific threshold of prior offending. Let’s say that our threshold

is 4 prior arrests and we are interested only in offenders who have 4 or

more prior arrests. An offender with 5 arrests would gain a score of 1 on

this new measure, and an offender with 20 arrests would have a score of

16. An offender with 4 arrests would have a score of 0. This variable

meets the criteria of an interval scale because we can distinguish scores,

rank them, and define the exact difference between them. A score of 16

represents a more serious prior criminal record than a score of 1. In turn,

an offender with a score of 16 has 15 more arrests than an offender with

a score of 1. However, we cannot say that the offender with a score of

16 on this measure had 16 times as many prior arrests as the offender

with a score of 1. This is because the scale has an arbitrary zero point.

Zero represents not the absence of a prior record, but the fact that the

offender has 4 prior arrests. Thus, the scale is an interval scale but not a

ratio scale.

Nearly all the statistics that we use in criminal justice (and all those

that we describe in this text) are also appropriate for interval scales if

they are appropriate for ratio scales. For this reason, most statistics texts

do not differentiate between the scales in practice, even if they identify

how they differ in theory. We follow the same approach. For the rest of

the chapter and indeed the rest of this text, we will concentrate on the

differences among nominal, ordinal, and at least interval scales.

Criminal justice researchers use interval scales to present findings

about criminal justice agency resources, criminal sentences, and a whole

of the trait under study. To understand how interval scales differ from

ordinal scales and from ratio scales, it is useful to examine a concrete

example. We commonly measure prior offending in terms of the number

of arrests on an offender’s criminal history record. If we compare an

offender who has 20 arrests with one who has only 5 arrests, we know

L E V E L S O F M E A S U R E M E N T 21

host of other issues related to crimes and criminals. For example, we can

measure the amount spent by criminal justice agencies to pay the salaries

of police officers or to pay for the health care costs of prison inmates.

We can measure the financial costs of different types of crime by mea-

suring the amount stolen by offenders or the amount of time lost from

meet the requirements of at least an interval level of measurement.

Now that we have defined each step in the ladder of measurement,

we can summarize. As is illustrated in Table 2.4, as you move up the

ladder of measurement, the amount of information that is gained in-

creases. At the lowest level, you have only categorization. At the next

level, you add knowledge about the order of the categories included.

With interval scales, you not only classify and order your measure but

also define how much categories differ one from another. A ratio scale

requires all of these characteristics as well as a non-arbitrary, or true,

zero value.

Variables Commonly Found in Criminal Justice Research

That Are Measured on at Least Interval Scales

VARIABLE COMMON CATEGORIES

Age Years Education Years Income or Salary Dollars, etc. Number of Crimes in a

City/County State Nation Count Crime Rates for a

City/County/State/Nation Count of crimes, adjusted for the size of the population Self-Reported Delinquent Acts Count

Table 2.3

Summary of the Information Required for Each Level of Measurement

TRUE ZERO �

ORDER � SET INTERVALS � SET INTERVALS �

LEVEL OF CATEGOR- CATEGOR- ORDER � ORDER �

MEASUREMENT IZATION IZATION CATEGORIZATION CATEGORIZATION

Ratio X X X X

Interval X X X

Ordinal X X

Nominal X

Table 2.4

arrested. Table 2.3 provides examples of criminal justice variables that

of prison served or sentenced or the age at which offenders were first

work by violent crime victims. We can measure the number of years

22 C H A P T E R T W O : M E A S U R E M E N T

R e l a t i n g I n t e r v a l , O r d i n a l , a n d N o m i n a l S c a l e s : T h e I m p o r t a n c e o f C o l l e c t i n g D a t a a t t h e H i g h e s t L e v e l P o s s i b l e

Take, for example, the measurement of victimization. If you decided

to simply compare the types of victimization involved in a crime event,

you would measure victimization using a nominal scale. You might

choose the following categories: events involving loss of money or prop-

erty, events including physical harm, a combination of such events, and

all other events. But let us assume, for a moment, that at some time after

you collected your data, a colleague suggests that it is important to dis-

tinguish not only the type of event but also the seriousness of crimes

within each type. In this case, you would want to distinguish not only

whether a crime included monetary loss or violence but also the serious-

ness of each loss. However, because your variable is measured on a

nominal scale, it does not include information on the seriousness of loss.

Accordingly, from the information available to you, you cannot create an

ordinal-level measure of how much money was stolen or how serious

the physical harm was.

Similarly, if you had begun with information only on the order of

crime seriousness, you could not transform that variable into one that

defined the exact differences between categories you examined. Let’s

say, for example, that you received data from the police that ranked

($10,001 and above). If you decide that it is important to know not just

the general order of monetary harm but also the exact differences in

harm between crimes, these data are insufficient. Such information

would be available only if you had received data about harm at an in-

terval level of measurement. In this case, the police would provide in-

formation not on which of the four categories of harm a crime belonged

to, but rather on the exact amount of harm in dollars caused by each

crime.

is that you should measure variables in a study at the highest level of

One important lesson we can draw from the ladder of measurement

measurement your data allow. This is because each higher level of mea-

information at the outset, you may not be able to add it at the end of

surement cannot be transformed easily into measures higher on the

surement requires additional information. And if you fail to collect that

your study. In general, variables measured lower on the ladder of mea-

ladder. Conversely, variables measured higher on the ladder of mea-

surement can be transformed easily into measures lower on the ladder.

moderate monetary harm ($501–10,000), and serious monetary harm

monetary victimization for each crime into four ordinally scaled

categories: no monetary harm, minor monetary harm (up to $500),

W H A T I S A G O O D M E A S U R E ? 23

While you cannot move up the ladder of measurement, you can move

down it. Thus, for example, if you have information collected at an inter-

val level, you can easily transform that information into an ordinal-scale

measure. In the case of victimization, if you have information on the

exact amount of harm caused by a crime in dollars, you could at any

point decide to group crimes into levels of seriousness. You would sim-

ply define the levels and then place each crime in the appropriate level.

For example, if you defined crimes involving harm between $501 and

$10,000 as being of moderate victimization, you would take all of the

crimes that included this degree of victimization and redefine them as

falling in this moderate category. Similarly, you could transform this mea-

sure into a nominal scale just by distinguishing between those crimes

that included monetary harm and those that did not.

Beyond illustrating the connections among different levels of mea-

surement, our discussion here emphasizes a very important rule of

thumb for research. You should always collect information at the highest

level of measurement possible. You can always decide later to collapse

such measures into lower-level scales. However, if you begin by collect-

ing information lower on the ladder of measurement, you will not be

able to decide later to use scales at a higher level.

W h a t I s a G o o d M e a s u r e ?

In analysis and reporting of research results, measures that are of a

higher scale are usually preferred over measures that are of a lower

scale. Higher-level measures are considered better measures, based on

the principle that they take into account more information. Nonetheless,

this is not the only criterion we use in deciding what is a good variable

in research. The researcher must raise two additional concerns. First,

does the variable reflect the phenomenon to be described? Second, will

the variable yield results that can be trusted?

The first question involves what those who study research methods

call validity. Validity addresses the question of whether the variable

used actually reflects the concept or theory you seek to examine. Thus,

for example, collecting information on age in a sample is not a valid way

of measuring criminal history. Age, although related to criminal history,

is not a measure of criminal history. Similarly, work history may be re-

lated to criminality, but it does not make a valid measure of criminality.

But even if we restrict ourselves to variables that directly reflect criminal

history, there are often problems of validity to address.

Let’s say that you wanted to describe the number of crimes that of-

fenders committed over a one-year period. One option you might have

24 C H A P T E R T W O : M E A S U R E M E N T

The most valid measure of frequency of offending is the one that

most directly assesses how many crimes an individual has committed.

Associated with each of the three variables included on the rap sheet is

some degree of threat to validity. This means that each can be criticized

because it does not quite reflect the concept we wish to study. Incarcera-

tion, for example, is more a measure of seriousness of crime than fre-

quency of offending. This is because judges may impose a number of

different types of sanctions, and they are more likely to impose a prison

sentence for more serious crimes. Many crimes that result in a conviction

lead not to incarceration but rather to probation, fines, or community

service. Thus, if we use incarceration to measure frequency of offending,

we are likely to miss many crime events in an offender’s criminal record.

Accordingly, incarceration provides a biased picture of the number of of-

fenses committed by an offender. It is not a highly valid measure of this

concept.

Using this logic, criminologists have generally assumed that arrest is

the most valid measure of frequency of offending that can be gained

from official data sources, such as the FBI rap sheet. Arrests are much

closer in occurrence to the actual behavior we seek to study and are not

filtered by the negotiations found at later stages of the legal process.

While criminologists have assumed that arrests reflect criminal behavior

more accurately than convictions or incarceration, some legal scholars

contend that arrests are a less valid measure of criminality precisely be-

cause they come before the court reaches a conclusion regarding the in-

nocence or guilt of a defendant. They contend that someone has not

committed a crime until the legal system defines an act as such.

Self-report surveys are generally considered to provide the most valid

measure of frequency of offending. This is because an individual can be

asked directly how many crimes he or she has committed. But self-report

studies are often criticized in terms of another concern in measurement,

which is termed reliability.

Reliability addresses the question of whether a measure gains infor-

mation in a consistent manner. Will you get the same result if you repeat

measurement of the same case or person? If different people have similar

characteristics, will your measure reflect that similarity? Returning to the

above example of criminal history, we would ask not whether the mea-

sure reflects the concept of frequency of offending, but whether mea-

surement of the concept is reliable across different subjects.

is to examine their criminal history as it is recorded on the Federal

Bureau of Investigation’s (FBI) criminal history record, or rap sheet.

The rap sheet includes information on arrests, convictions, and incar-

cerations. Although each of these variables tells us something about a

person s criminal history, they are not all equally valid in terms of ans-

wering the research question we have proposed.

C H A P T E R S U M M A R Y 25

Self-reports, which allow us to ask valid questions about the number

of crimes that a person has committed, have been challenged on the basis

of their reliability. One problem is that people may lie about their crimi-

nal histories. Crime is a sensitive issue, and no matter what efforts the re-

searcher makes to assure subjects of confidentiality, people may be hesi-

tant to talk about crimes in their past. Accordingly, depending on the

degree of hesitancy of subjects, a researcher might gain different answers,

irrespective of a person’s actual criminal history. But even if a person is

willing to provide accurate responses to such questions, he or she may

not be able to. Some people have better memories than others, and the

reliability of this measure depends in part on a person’s ability to recall

events generally. Such issues of reliability have begun to be addressed di-

rectly by criminologists, who are trying to increase the reliability of self-

report methods by improving interview techniques and protocols.

Returning to the FBI rap sheets, we can also assess their reliability. In

general, not only is arrest assumed to be the most valid of official mea-

sures; it is also the measure most reliably recorded on the FBI rap sheets.

This is the case in good part because the rap sheets are built around fin-

gerprint records, which police agencies have come to routinely send to

the FBI. This helps the police agencies as well, because they often use

this information to check the identities of arrestees and to assess their

criminal histories. Other types of agencies are less consistent in their

transfer of information to the FBI, and as a result convictions and incar-

cerations are less reliably recorded.

The issues raised in connection with the validity and reliability of

criminal history information are good examples of the kinds of problems

you will encounter in assessing measures in criminal justice. You should

keep in mind that no variable is perfect. Some threat to validity is likely

velop or choose the best measure you can. The best measure is the one

that most closely reflects the concept you wish to study and assesses it in

a consistent and reliable way across subjects or events.

C h a p t e r S u m m a r y

In science, we use measurement to make accurate observations. All

measurement must begin with a classification process—a process

that in science is carried out according to systematic criteria. This

process implies that we can place units of scientific study in clearly

defined categories. The end result of classification is the development

of variables.

reliability is almost always present in measurement. Your task is to de-

to be encountered, no matter how careful you are. Some degree of un-

26 C H A P T E R T W O : M E A S U R E M E N T

There are four scales of measurement: nominal, ordinal, interval,

and ratio. With a nominal scale, information is organized by simple

classification. The aim is merely to distinguish between different phe-

nomena. There can be no overlap between categories nor can there be

cases that do not fit any one category. There is no theoretical limit to the

number of nominal categories possible. With an ordinal scale, not only

is information categorized, but these categories are then placed in order

of magnitude. An interval scale is one that, in addition to permitting the

processes of categorization and ordering, also defines the exact differ-

ence between objects, characteristics, or events. A ratio scale is an inter-

val scale for which a non-arbitrary, or true, zero value can be identified.

Data collected at a higher level of measurement may subsequently be

reduced to a lower level, but data collected at a lower level may not be

transformed to a higher one. For this reason, it is always advisable to

collect data at the highest level of measurement possible.

There are three separate factors that affect the quality of a measure.

The researcher should strive for a measure that has (1) a high scale of

measurement (one that uses the most information); (2) a high level of

validity (one that provides an accurate reflection of the concept being

studied); and (3) a high level of reliability (one that provides consistent

results across subjects or units of study).

K e y T e r m s

classification The process whereby data

are organized into categories or groups.

data Information used to answer a re-

search question.

interval scale A scale of measurement

that uses a common and standard unit and

enables the researcher to calculate exact

differences between scores, in addition to

categorizing and ordering data.

levels of measurement Types of mea-

surement that make use of progressively

larger amounts of information.

measurement The assignment of numeri-

cal values to objects, characteristics, or

events in a systematic manner.

nominal scale A scale of measurement

that assigns each piece of information to an

appropriate category without suggesting

any order for the categories created.

ordinal scale A scale of measurement that

categorizes information and assigns it an

order of magnitude without using a stan-

dard scale of equal intervals.

ratio scale A scale of measurement identi-

cal to an interval scale in every respect ex-

cept that, in addition, a value of zero on

the scale represents the absence of the

phenomenon.

reliability The extent to which a measure

provides consistent results across subjects

or units of study.

scale of measurement Type of catego-

rization used to arrange or assign values to

data.

E X E R C I S E S 27

validity The extent to which a variable

accurately reflects the concept being

measured.

variable A trait, characteristic, or attribute

of a person/object/event that can be mea-

sured at least at the nominal-scale level.

E x e r c i s e s

2.1 For each of the following examples of criminal justice studies, state whether the scale of measurement used is nominal, ordinal, or at least interval (i.e., interval or ratio). Explain your choice.

a. In a door-to-door survey, residents of a neighborhood are asked how many times over the past year they (or anyone in their house- hold) have been the victims of any type of crime.

b. Parole-board members rate inmate behavior on a scale with values ranging from 1 to 10; a score of 1 represents exemplary behavior.

c. One hundred college students are asked whether they have ever been arrested.

d. A researcher checks prison records to determine the racial back- ground of prisoners assigned to a particular cell block.

e. In a telephone survey, members of the public are asked which of the following phrases best matches how they feel about the perfor-

f. A criminologist measures the diameters (in centimeters) of the skulls of inmates who have died in prison, in an attempt to develop a biological theory of the causes of criminality.

g. Secretaries at a top legal firm are asked the following question: “Over the past year, have you been the victim of sexual harass-

2.2 You have been given access to a group of 12 jurors, with a mandate from your senior researcher to “go and find out about their prior jury experience.” Under each of the following three sets of restrictions, de- vise a question to ask the jurors about the number of experiences they have had with previous juries.

a. The information may be recorded only on a nominal scale of measurement.

indifferent, satisfied, or very satisfied. mance of their local police force: totally dissatisfied, dissatisfied,

follows: never, once, two or three times, more than three times, or ment—and if so, how many times?” Answers are categorized as

refused to answer.

28 C H A P T E R T W O : M E A S U R E M E N T

b. The information may be recorded on an ordinal scale but not on any higher scale of measurement.

c. The information may be recorded on an interval scale.

Your senior researcher subsequently informs you that she wishes to know the answers to the following five questions:

—How many of the jurors have served on a jury before?

—Who is the juror with the most prior experience?

—What is the sum total of previous jury experience?

—Is there anyone on the jury who has served more than three times?

—What is the average amount of prior jury experience for this group?

d. If you had collected data at the nominal level, which (if any) of the above questions would you be in a position to answer?

e. If you had collected data at the ordinal level, which (if any) of the above questions would you be in a position to answer?

f. If you had collected data at the interval level, which (if any) of the above questions would you be in a position to answer?

2.3 You have been asked to measure the public’s level of support for using the death penalty. Devise questions to gauge each of the following:

a. Overall support for using the death penalty.

b. Support for using the death penalty if there are other punishment options.

c. Support for using the death penalty if the chances of an innocent person being executed are

i. 1 in 1,000.

ii. 1 in 100.

iii. 1 in 10.

2.4 You are investigating the effects of a defendant’s prior record on vari- ous punishment decisions made by the court. One variable that you have access to in local court records is the total number of prior felony arrests for each defendant.

a. What kinds of questions would you be able to answer with prior record measured in this way?

b. Explain how you would recode this information on a nominal scale

c. Explain how you would recode this information on an ordinal scale

answer with prior record measured in this way?

answer with prior record measured in this way?

of measurement. What kinds of questions would you be able to

of measurement. What kinds of questions would you be able to

E X E R C I S E S 29

2.5 Because the Ministry of Transport (MOT) is concerned about the num- ber of road accidents caused by motorists driving too close together, it has, on an experimental 2-km stretch of road, painted “chevrons” (lane markings) every few meters in each lane. By the roadside it has erected a sign that reads: “KEEP YOUR DISTANCE: STAY AT LEAST 3 CHEVRONS FROM THE CAR IN FRONT!” The MOT has asked you to measure the extent to which this instruction is being followed. There are a number of possible measures at your disposal. Assess the relia- bility and validity of each approach suggested below. Which is the best measure?

a. Stand on a bridge over the experimental stretch of road and count how many of the cars passing below do not keep the required distance.

b. Compare police figures on how many accidents were recorded on that stretch of road over the periods before and after it was painted.

c. Study the film from a police camera situated 5 km farther down the same stretch of road (after the end of the experimental stretch) and count how many cars do not keep a safe distance.

2.6 The police are planning to introduce a pilot “community relations strategy” in a particular neighborhood and want you to evaluate whether it has an effect on the willingness of citizens to report crimes to the police. There are a number of possible measures at your dis- posal. Assess the reliability and validity of each approach suggested below. Which is the best measure?

a. Telephone every household and ask respondents to measure, on a scale of 1 to 10, how willing they are to report particular types of crime to the police. Repeat the experiment after the scheme has been in operation six months.

b. Compare a list of offenses reported by members of the neighbor- hood in the six months before introduction of the scheme with a similar list for the six months after introduction of the scheme. (It is standard procedure for the police to record the details of the com- plainant every time a crime is reported to them.)

2.7 You are comparing the psychological condition of three inmates serv- ing out long terms in different high-security prisons, and you are in- terested in the amount of contact each one has with the outside world. You wish to determine how many letters each one has sent over the past 12 months. No official records of this exist. There are a number of possible measures at your disposal. Assess the reliability and validity of each approach suggested below. Which is the best measure?

a. Ask each prisoner how many letters he or she sent over the past year.

30 C H A P T E R T W O : M E A S U R E M E N T

b. Check the rules in each of the prisons to see how many letters high security prisoners are allowed to send each year.

c. Check the records of the prison postal offices to see how many times each prisoner bought a stamp over the past year.

2.8 The government is interested in the link between employment and criminal behavior for persons released from prison. In a study de- signed to test for an effect of employment, a group of people released from prison are randomly assigned to a job training program, where they will receive counseling, training, and assistance with job place- ment. The other offenders released from prison will not receive any special assistance. There are a number of possible measures at your disposal. Assess the reliability and validity of each approach suggested below. Which is the best measure?

a. Eighteen months after their release from prison, interview all the of- fenders participating in the study and ask about their criminal activ- ity to determine how many have committed criminal acts.

2.9 In a recent issue of a criminology or criminal justice journal, locate a research article on a topic of interest to you. In this article, there should be a section that describes the data. A well-written article will describe how the variables were measured.

a. Make a list of the variables included in the article and how each was measured.

b. What is the level of measurement for each variable—nominal, ordi- nal, or at least interval? Explain why.

c. Consider the main variable of interest in the article. Assess its relia- bility and validity.

returned to prison within 18 months of release.

arrested for a new crime within 18 months of release.

b. Look at prison records to determine how many offenders were

c. Look at arrest records to determine how many offenders were

C o m p u t e r E x e r c i s e s

There are a number of statistical software packages available for data analysis.

Most spreadsheet programs will also perform the basic statistical analyses of the

kind described in this text through Chapter 17. The computer exercises included

in this text focus on the use of two different software programs: SPSS and Stata.

Most universities that we are aware of make at least one of these two statistical

programs available in student computer labs. There are also student versions of

each program that can be purchased separately, sometimes through a university

bookstore or other offices on campus—see each company’s Web site for details

(www.spss.com or www.stata.com). There are many excellent reference books

C O M P U T E R E X E R C I S E S 31

is not to repeat what is said in those books. Rather, our goal with the computer

exercises is to illustrate some of the power available to you in two widely used

packages. In real-world situations where you are perform some type of statistical

analysis, you rarely work through a problem by hand, especially if the number of

observations is large.

SPSS

To begin our exploration of SPSS, we will focus here on some of the data

management features available to users. After starting the SPSS program on

your computer, you will need to open the National Youth Survey data file from

the Web site (nys_1.sav). For those readers working with the Student Version

of SPSS, you are limited to data files with no more than 50 variables and 1,500

cases. We have also included a smaller version of the NYS data (nys_1_student.

sav) that contains a random sample of 1,000 cases (of the original 1,725).

After you start SPSS and open the data file, the raw data should appear in a

window that looks much like a spreadsheet. Each column represents a different

variable, while each row represents a different observation (individual, here). If

you scroll down to the end of the data file, you should see that there are 1,725

lines of data (or 1,000 if using the student version of the data file).

There are three direct ways to learn about the variables included in this data

file. First, notice the buttons in the lower center of the spreadsheet. One button

(which should appear as the darker shade of the two buttons) is labeled “Data

View,” and the other is labeled “Variable View.” The data view button presents

us with the spreadsheet of values for each observation and variable. If you click

on the button labeled “Variable View,” you should now see another spreadsheet,

in which variable names are listed in the first column and the other columns

contain additional information about each variable. For example, the first col-

umn provides the name of the variable, another column provides a label for the

Several SPSS and Stata files are available at the following Web address:

http://extras.springer.com. The data file we will use first represents a subset of

the data from the National Youth Survey, Wave 1. The sample of 1,725 youth

is representative of persons aged 11–17 years in the USA in 1976, when the first

wave of data was collected. While these data may seem old, researchers continue

to publish reports based on new findings and interpretations of these data. One

of the apparent strengths of this study was its design; the youth were inter-

viewed annually for 5 years from 1976 to 1980 and then were interviewed again

in 1983 and 1987. The data file on our Web site was constructed from the full

data source available at the Inter-University Consortium of Political and Social

Research, which is a national data archive. Data from studies funded by the

National Institute of Justice (NIJ) are freely available to anyone with an Internet

connection; go to http://www.icpsr.umich.edu/NACJD. All seven waves of data

from the National Youth Survey are available, for example.

on the use of either SPSS or Stata for statistical data analysis—our intent here

32 C H A P T E R T W O : M E A S U R E M E N T

variable (allowing us to add a more informative description of our variable), and

an additional column provides value labels. It is from this column that we will

be able to learn more about each variable. For example, click on the cell in this

column for the sex variable, and you should see a small gray box appear in the

cell. Now click on this small gray box and you will be presented with a new win-

dow that lists possible values for sex and the corresponding labels. Here, we see

that males have been coded as “1” and females as “2.” If you click on “OK” or

“Cancel,” the window disappears. You can then perform this same operation for

every other variable.

A second way of obtaining information about the variables in an SPSS data

file involves using the “Variables” command. To execute this command, click

on “Utilities” on the menu bar; then click on “Variables.” What you should see

is a list of variables on the left and another window on the right that presents

information about the highlighted variable. If you click on the sex variable, you

should see information on its coding and values in the window on the right. This

command is particularly useful if you are working with an SPSS data file and

simply need a reminder of how the variables are coded and what categories or

values are included. This feature is useful if you are working with a data set and

need to know what a particular variable refers to or how it is measured in order

to continue working.

A third way of obtaining information about the variables in an SPSS data file

involves the “Display Data File Information” command. To run this command,

click on “File” on the menu bar; then click on “Display Data File Information,”

and then select the option for “Working File” (the data you have already opened

up in SPSS). This command generates text for the output window in SPSS. This

output contains all the information SPSS has on every variable in a data file.

Executing this command is equivalent to executing the “Variables” command

for every variable in the data set and saving that information in another file. Be

aware that using this command on a data file with many variables will produce a

very large output file. This command is most useful when you are first working

with an SPSS data set that someone else has conveniently set up for you and you

need to verify the contents of the data set and the nature of the variables includ-

ed in the data set. Similar to what you are now doing with the NYS data file.

In subsequent chapters, the computer exercises will make use of syntax com-

mands and files. These are the command lines that each program uses to run a

particular procedure. To begin, open a syntax window in SPSS by clicking on

“File,” then “New” and “Syntax.” A blank window opens in which you may

enter command lines and save for future use. Alternatively, instead of open-

ing a blank syntax window, you could open one that had been saved, such as

Chapter_2.sps included on the text Web site.

Open the Chapter_2.sps syntax file. As a convention throughout the text, we

will use CAPS to denote required or really useful components to a command,

although you are free to use lowercase letters in any command that you run in

SPSS. The first line of this file begins with /*—this is a comment line. Any of

C O M P U T E R E X E R C I S E S 33

the SPSS syntax files that you open and begin with this line do not run com-

mands, but are included to help explain what it is you are reading in the file.

The first two command lines of Chapter_2.sps are:

The GET FILE command reads the NYS data file into SPSS. Note that you

will need to provide a directory name that contains your copy of the nys_1.sav

data file. The DATASET command gives you the option of naming the data

file something else internally, while you are working in SPSS. Here, the name

given is simply “nys.” The WINDOW = FRONT option is useful for those times

when you may have more than one data file opened in SPSS—this forces it to

keep this spreadsheet in the position where it is the first one you will see on your

computer screen.

It is important to note that each command line ends with a period—this

ensures that SPSS will know where the command ends and when to start execut-

ing the instruction.

To run these command lines—after you have edited the directory

information—highlight the two lines and do one of the following:

Click on the right-pointing triangle toward the top right of the syntax window.

Click on “Run” on the menu bar and then select the option for “Selection”—

The keystrokes for this same procedure will appear when you click on “Run.”

They are slightly different for Mac and Windows users, but may simplify your

To obtain the file information—the complete listing of variables and their char-

acteristics, run the third line of Chapter_2.sps:

DISPLAY DICTIONARY.

This will display all of the information we highlighted above in using the

point-and-click interactive method to display the file information.

Stata

Much like our discussion of SPSS, we begin our exploration of Stata by focusing

on some of the basic data management features available to users. After starting

the Stata program on your computer, you will see a window that has several dif-

ferent boxes. On the far left, the box labeled “Review” will contain a growing

list of commands that you run during a Stata session. The largest box (top and

center) is labeled “Results” and will present all of the output from various com-

mand that you have run as well as any error messages. Just below this box is the

“Command” box, where you will enter commands, if going one command at a

time. On the right side of the Stata window is a box labeled “Variables”—the

upper half will contain a complete list of all the variables in the data file, while

the lower half provides various details about the variable that is highlighted in the

upper half.

GET FILE = '[directory_name]nys_1.sav'.

DATASET NAME nys WINDOW = FRONT.

34 C H A P T E R T W O : M E A S U R E M E N T

Begin by opening the National Youth Survey data file obtained from the Web

site (nys_1.dta) using “File” from the menu bar and the “Open…” option listed.

Note that the “Review” box on the left now contains the Stata command that

was run to open the data file, the “Variables” box on the right contains the list-

ing of variables, and the “Results” window simply lists the Stata command that

was run.

If you would like to view the data in a spreadsheet format, click on the box at

the top center of the Stata window for “Data Browser” or “Data Editor”—the

“Browser” view will allow you to look at the data, while the “Editor” view will

allow you to make changes to the data. The window that opens will present each

variable as a separate column, each observation as a separate row, and the con-

tents of each cell will either be a numeric value or a value label that is associated

with a numeric value. If you click on a cell with a value label, the corresponding

numeric value appears in the white horizontal bar just above the spreadsheet.

There are several different ways of learning about the variables included in

this data file. As noted already, the “Variables” box on the right contains infor-

mation about each variable and is displayed as soon as the variable name is high-

lighted (i.e., clicked on) in the upper box.

A much more comprehensive method for obtaining information about the

variables in the data file is to use the “Codebook” command, which will list all

of the variables in the data file, their values, value labels, as well as other infor-

mation that will become useful in subsequent chapters of this text. To run the

“Codebook” command, click on “Data” on the menu bar, then the options for

“Describe data” and selecting the option for “Describe data contents (code-

book).” The “Results” window will generate a great deal of output that will list

each variable, the variable label, the possible (numeric) values when there are

fewer than 10 unique values, and any value labels. The other information pre-

sented will be discussed later.

Similar to the observation made in regard to the use of SPSS, in subsequent

chapters, the computer exercises will make use of syntax commands and files,

rather than the point-and-click options. Although Stata syntax is different from

SPSS syntax, how it works is very much the same: the command line(s) inform

Stata what procedure to run, whether it is opening a date file for use, or some

other statistical procedure. To open a new syntax window in Stata, where they

are called “do files,” start by clicking on “File,” then “New Do-file. A blank

window opens in which you may enter command lines and save for future use.

Alternatively, instead of opening a blank do file window, you could open one

that had been saved, such as Chapter_2.do included on the text Web site.

Stata commands must be in lowercase letters, unless otherwise noted in the

documentation for the command. As a convention throughout our text, we will

place Stata commands and useful or required options in boldface font.

Open the Chapter_2.do file. The first line of this file begins with /* and ends

with */—this is a comment line. In Stata, comments are bounded by these two

mirror sets of characters, so without the */, Stata would interpret everything

C O M P U T E R E X E R C I S E S 35

following the initial /* as a comment—so it is important to pay attention to your

use of these in any Stata do file.

The first command line of Chapter_2.do is:

The use command reads the NYS data file into Stata. Note that you will

need to provide a directory name that contains your copy of the nys_1.dta data

file. Also note that after the file name and location, there is a comma followed by

the clear option. Using this option tells Stata to clear out any existing data file

currently being used without saving any changes made to that data file.

To run this command line—after you have edited the directory information—

highlight the command line and do one of the following:

will run the command quietly, meaning there is no output in the “Results”

window, other than a line indicating a command was run.

will run the command and generate output in the “Results” window.

Keystroke options are available for both of these commands, to determine

what they are for your operating system (Mac or Windows), click on “View”

the keystroke shortcuts that can be used to run the command quietly or not.

To obtain the file information—the complete listing of variables and their char-

acteristics, run the second command line of Chapter_2.do:

codebook

This will display all of the information we highlighted above in using the point-

and-click interactive method to display the file information.

Problems

Using one of the ways described above, work through all the variables included

in the NYS data file:

1. Note the level of measurement for each variable and then briefly explain

why it is what it is. (You should not rely on the level of measurement

information given in the SPSS data file, especially if someone else has

constructed the SPSS data file for you.)

2. Describe the likely levels of reliability and validity for each variable and

explain why they are what they are.

use “[directory_name]nys_1.dta”, clear

Representing and Displaying Data

When Should a Pie Chart Be Used?

t i m e s e r i e s d a t a v i s u a l l y

How Can the Researcher Present Time Series Data Graphically?

C h a p t e r t h r e e

T h e f r e q u e n c y d i s t r i b u t i o n

U s i n g b a r c h a r t s t o r e p r e s e n t d a t a

U s i n g p i e c h a r t s t o r e p r e s e n t d a t a

R e p r e s e n t i n g h i s t o r i c a l o r

What is It?

When are Bar Charts Useful?

How are Bar Charts Created?

How are Pie Charts Created?

How are Data Represented Visually in a Histogram?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_3,

© Springer Science+Business Media New York 2014

THE GRAPHICAL DISPLAY OF DATA is an important tool for presenting sta-

tistical results in such a way that the key features or characteristics of an

analysis are highlighted. There are many different ways the same data

might be displayed. Indeed, many books have been written that focus

entirely on graphical presentation of data. In this chapter, we introduce

some common ways of representing data in graphical form, along with

suggestions for effectively presenting information in an accurate way.

We begin by discussing the most basic way of summarizing—and then

graphing—data: frequency distributions and histograms. Building on the

discussion of histograms, we move on to more general bar charts, noting

W h a t A r e F r e q u e n c y D i s t r i b u t i o n s a n d H i s t o g r a m s ?

When we array scores according to their value and frequency, we con-

struct what is called a frequency distribution. Let’s take the following

data on previous arrests of 100 known offenders as an example:

14 0 34 8 7 22 12 12 2 8 6 1 8 1 18 8 1 10 10 2

12 26 8 7 9 9 3 2 7 16 8 65 8 2 4 2 4 0 7 2 1 2 11 2 1 1 5 7 4 10

11 3 41 15 1 23 10 5 2 10 20 0 7 6 9 0 3 1 15 5 27 8 26 8 1 1 11 2 4 4 8 41 29 18 8 5 2 10 1 0 5 36 3 4 9 5 10 8 0 7

We first group all of the cases with the same value together. Accordingly,

we group together the cases with no prior arrests, one prior arrest, two

Finally, we examine how graphs can be used to represent a series of

the variety of information that can be presented in bar and pie charts.

observations over time.

37

prior arrests, and so forth, until we have covered all of the potential

scores in the distribution. Then we arrange these scores in order of mag-

nitude, as is done in Table 3.1. Looking at the data in this way allows us

to get a sense of the nature of the distribution of scores.

In practice, creating a frequency distribution is usually the first step a

researcher takes in analyzing the results of a study. Looking at the distri-

bution of scores not only provides a first glance at the results of a study;

it also allows the researcher to see whether there are scores that do not

make sense. For example, coding errors in the data set may have given

rise to impossible scores. In our example, a result of thousands of arrests

would be very unlikely and would thus lead the researcher to take an-

other look at that particular case.

Constructing frequency distributions by hand can be very time and

labor intensive. Researchers today seldom construct frequency distribu-

tions by hand. This task can be done simply and easily with packaged

statistical software, such as SPSS.

Frequency Distribution of Prior Arrests for 100 Known Offenders

VALUE FREQUENCY

0 6 1 11 2 11 3 4 4 6 5 6 6 2 7 7 8 12 9 4

10 7 11 3 12 3 14 1 15 2 16 1 18 2 20 1 22 1 23 1 26 2 27 1 29 1 34 1 36 1 41 2 65 1 Total 100

Table 3.1

38 C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A

If you decide to present a frequency distribution of the results of a

study, you must choose what particular format to use. For example, the

distribution of prior arrests could simply be presented as in Table 3.1. Or

the same information could be presented graphically in what is called a

histogram. To make a histogram, we take the scores and values from a

frequency distribution and represent them in pictorial form. In this case,

we use a bar to represent each value in the frequency distribution. The

x-axis (the horizontal axis) of the histogram represents the values of the

variable we are analyzing—here, the number of arrests. The y-axis (the

vertical axis) captures the height of the bars and indicates the number of

scores—the frequency—found in each category. A histogram of the data

on prior arrests is provided in Figure 3.1. The information presented in

the histogram is identical to the information presented in the frequency

distribution in Table 3.1, but the histogram conveys to the reader an im-

mediate sense of the range of values, the location of clusters of cases,

and the overall shape of the distribution—information that is not as eas-

ily obtainable from a frequency distribution.

It is important to note that the x-axis in Figure 3.1 correctly represents

the full range of values for the variable. In particular, note that there is a

large gap in the distribution from 41 arrests to 65 arrests. Depending on

F re

q u

en cy

14

12

10

8

6

4

2

0

Number of Arrests

0 5 10 15 20 25 30 35 40 45 50 55 60 65

Histogram of Frequency DistributionFigure 3.1

W H A T A R E F R E Q U E N C Y D I S T R I B U T I O N S A N D H I S T O G R A M S ? 39

the software used to generate a histogram, this information may or may

not be represented. Excluding this information and placing all the bars

adjacent to each other essentially means ignoring the level of measure-

ment of the variable and treating the variable as if it were nominal or or-

dinal. Why would you want to include information about those values

that have an observed frequency of zero? If you look at Figure 3.1, you

can see that there are relatively few observations for increasingly greater

numbers of arrests and that the distance between categories starts to in-

crease. If the bars were adjacent to each other and correctly labeled, it

would still be possible for the reader to discern the spread of observa-

tions. But when the values with an observed frequency of zero are por-

trayed as well, it is often easier to interpret the histogram.

Before we get into working with more complex graphs, it is worth

noting that frequency distributions of interval-level variables are often

dispersed across a large number of values. Thus, in presenting a fre-

people, we would likely not want to present the simple distribution of

income scores. If we did, we might end up with thousands of scores,

most of which would include only one or two cases. It would take pages

and pages to illustrate these data in the form of either a frequency distri-

bution or a histogram. The solution to this problem is to “group” data

together in larger categories—for example, by thousands or tens of thou-

sands of dollars in the case of incomes. Although there is no hard-and-

fast rule about how to create such larger groupings, it should be done in

a way that fairly represents the raw distribution of scores. Do not create

such a small group of categories that important variation in your data is

hidden.

A common source of confusion for students of statistics is the fact that

statisticians often represent distributions as “curves” rather than his-

tograms or frequency distributions. For example, Figure 3.2 uses a curve

to represent the 2001 SAT math scores of over 1.2 million college-bound

test takers.1 What is the relationship between a frequency distribution or

histogram and a distribution represented by a curve, such as the one in

Figure 3.2?

When we represent a distribution as a curve, it is usually a distribu-

tion with a very large number of cases, such as that of SAT math scores

of 2001 college-bound seniors. We can represent these distributions as

curves because, with a true interval-scale measure, as the number of

cases becomes very large, we can construct a histogram in such a way

1The College Board, 2001 College-Bound Seniors: A Profile of SAT Program Test Tak-

ers, accessed at http://www.collegeboard.com/sat/cbsenior/yr2001/pdf/NATL.pdf.

if we were looking at the incomes of a random sample of thousands of

together into categories that represent a range of values. For example,

quency distribution or histogram, it is often necessary to group scores

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A40

that it begins to approximate a curve. We do this by making the intervals

of scores smaller and smaller.

This process is illustrated in Figure 3.3. We begin with a histogram of

almost 10,000 cases, in which all of the scores are placed within 10

200 300 400 500 600 700 800

Smooth Curve Representation of a Histogram:

SAT Math Scores of 2001 College-Bound Seniors Figure 3.2

4,000

3,000

2,000

1,000

0

F re

q u

en cy

Std. dev. = 1.00 Mean = 0.00 N = 9,994.00

Values

Constructing a Distribution That Approximates a CurveFigure 3.3

(a) Distribution with 10 Intervals

W H A T A R E F R E Q U E N C Y D I S T R I B U T I O N S A N D H I S T O G R A M S ? 41

broad categories (see part a). Here each category is represented by one

large bar. When we increase the number of intervals, or categories, in

the histogram to 30 (part b), we can still see the individual bars but the

shape of the distribution is not as jagged. When the number of cate-

gories is increased to 650 (part c), the histogram looks more like a

smooth curve than a histogram, although if you look closely you will be

1,200

1,000

800

600

400

200

0

F re

q u

en cy

Std. dev. = 1.00 Mean = 0.00 N = 9,994.00

Values

(b) Distribution with 30 Intervals

50

40

30

20

10

0

F re

q u

en cy

Std. dev. = 1.00 Mean = 0.00 N = 9,994.00

Values

(c) Distribution with 650 Intervals

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A42

able to identify the bars that make up the curve. If our distribution had

included an even larger number of scores and categories, the curve

would have become even smoother.

E x t e n d i n g H i s t o g r a m s t o M u l t i p l e G r o u p s : U s i n g B a r C h a r t s

Although histograms provide a quick and easy way to present data from

a frequency distribution, they can be used to present the frequency dis-

tribution for only a single group. What happens when we have fre-

quency distributions for the same variable for more than one group?

There are many instances where an investigator will want to present re-

sults graphically for two or more groups, such as treatment and control

groups, males and females, or states. We could simply construct a his-

togram for each group we were interested in, but we would not be able

to make direct comparisons across the groups in the form and shape of

the distributions. A simple extension of the histogram to multiple groups

makes use of the bar chart. Bar charts allow us to present information

for multiple groups simultaneously. Bar charts are constructed in much

the same way as histograms. The x-axis generally represents the values

of the variable, and the y-axis the size of the bar.2 Most statistical soft-

ware and spreadsheet packages allow the user to construct bar charts to

depict patterns in the various groups. Two of the more common ap-

proaches involve placing the bars side by side and on top of each other.

Presenting data for each group in adjacent bars gives the reader a sense

of the distribution for each group and allows for immediate comparison

of distributions across groups.

Table 3.2 presents simulated frequency distributions for numbers of

prior convictions among 100 men and 100 women arrested for drug of-

fenses. The frequencies for male and female arrestees suggest that males

In addition to being able to incorporate data from more than one

group, a bar chart has other benefits. For example, bars may be pre-

2Most statistical software and spreadsheet packages allow for manipulation of many

characteristics of a bar chart, including color, shading, patterning, and dimensions

(two vs. three). While this allows for the construction of unique charts, the investiga-

tor should be wary of adding so much detail to a chart that the reader loses the point

the investigator is trying to make.

females. Figure 3.4 portrays the male and female frequency distributions

have had more prior contact with the criminal justice system than the

in a bar chart.

43E X T E N D I N G H I S T O G R A M S T O M U L T I P L E G R O U P S

sented either vertically, as in Figure 3.4, or horizontally, as in Figure 3.5.

The only difference between the bar charts in Figures 3.4 and 3.5 is that

in the latter the axes have been flipped: The y-axis now represents the

rules about which form of bar chart is better. Some people like vertical

bar charts because they can draw an imaginary horizontal line across the

graph to get a sense of which bars are larger and smaller. Alternatively,

other people prefer horizontal bars because looking at them mimics the

Frequency Distributions of Number of Prior Convictions

for Male and Female Drug Arrestees

FREQUENCY

Number of Prior Convictions Male Female

0 25 40 1 20 25 2 15 15 3 10 7 4 8 4 5 6 0 6 2 3 7 6 3 8 5 1 9 0 2

10 3 0 Total 100 100

Table 3.2

F re

q u

en cy

40

30

20

10

0

Number of Prior Convictions

0 1 2 3 4 5 6 7 8 9 10

Male

Female

Bar Chart for Male and Female Frequency DistributionsFigure 3.4

values of the variable (here, the number of prior convictions), and the

x-axis represents the size of the bar (the frequency). There are no specific

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A44

process of reading text from left to right. Since the research on visual

perception is mixed about whether vertical or horizontal bars are more

effective—there are benefits and drawbacks to both approaches—the

preferences of the investigator typically determine which approach is

used.3

A cautionary note is needed about the use of bar charts for comparing

multiple groups. In Figure 3.4, the number of individuals in each group

was equal, allowing us to make a direct comparison of each group’s fre-

quency distribution. An investigator would run into trouble, however, if

the two groups did not have the same number of cases. Say one group

had two or three times as many cases as a second group. If we simply

presented a bar chart of the observed frequencies, then we would be

limited to discussing the shape of each group’s distribution; we would

be unable to make valid comparisons across the two groups. For exam-

ple, in the frequency distributions of prior convictions in Table 3.2, sup-

Male

Female

10

9

8

7

6

5

4

3

2

1

0

0 10 20 30 40

Frequency

N u

m b er

o f

P ri

or C

on vi

ct io

n s

Horizontal Bar Chart for Male and Female Frequency DistributionsFigure 3.5

3For a good discussion of the benefits and drawbacks of various approaches, see Gary

T. Henry, Graphing Data: Techniques for Display and Analysis (Thousand Oaks, CA:

Sage, 1995).

4E X T E N D I N G H I S T O G R A M S T O M U L T I P L E G R O U P S 5

pose that we doubled the frequencies for the male arrestees, giving us a

total of 200 males in the sample. In Figure 3.6, the bars representing

male arrests are twice as large as they are in Figure 3.4, and one might

be tempted to say that more males have zero, one, or two arrests. Strictly

speaking, such an observation is correct—the frequency is larger for one

group than for the other group—but it misrepresents the relative sizes of

the two groups and the relative distribution of cases within each group,

which are unchanged by doubling the number of male arrests.

In research on crime and criminal justice, the groups we are inter-

ested in comparing rarely have an equal number of cases. For example,

Table 3.3 presents the frequency distributions of years of education for

capital offenders executed in the United States between 1977 and 1995,

distinguished by the recorded race of the offender. These data come

from a public data file archived by the Bureau of Justice Statistics.4 The

bar chart for these data is presented in Figure 3.7. It is important to note

F re

q u

en cy

50

40

30

20

10

0

Number of Prior Convictions

0 1 2 3 4 5 6 7 8 9 10

Male

Female

Bar Chart for 200 Male and 100 Female ArresteesFigure 3.6

4Capital Punishment in the United States, ICPSR Study #6956, available through the

National Archive of Criminal Justice Data (NACJD) at http://www.icpsr.umich.edu/

NACJD.

that whites outnumber African Americans at a rate of about 3 to 2. This tells

us that although we can evaluate the frequency distributions for whites and

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A46

Number of Years of Education Among Offenders Executed

in the United States, 1977 to 1995, by Race of Offender

RACE OF OFFENDER

Years of Education White African American

7 24 7 8 14 17 9 19 16

10 24 20 11 14 18 12 60 31 13 8 1 14 12 2 15 2 0 16 3 0 17 1 0 Total 181 112

Table 3.3

F re

q u

en cy

70

60

50

40

30

20

10

0

Years of Education

7 8 9 10 11 12 13 14 15 16 17

Whites

African Americans

Bar Chart for Number of Years of Education Among Offenders

Executed in the United States, 1977 to 1995, by Race of Offender Figure 3.7

4E X T E N D I N G H I S T O G R A M S T O M U L T I P L E G R O U P S 7

African Americans separately, we should avoid any direct comparisons

across the two groups.

How do we address the problem of unequal group sizes? The most

direct way is to convert the observed frequencies into proportions or

percentages. A proportion has a value between 0 and 1 and represents

the fraction of all observations that fall into a category. We calculate a

proportion as:

Equation 3.1

where Ncat refers to the number of observations in a given category and

Ntotal refers to the total number of observations. For example, to find the

proportion of whites who had 8 years of education, we would take the

number of white offenders who had 8 years of education (14) and divide

it by the total number of whites executed (181):

Proportion � Ncat

Ntotal

By convention, we generally round a proportion to the second decimal

place. (However, as we will discuss in more detail in the next chapter,

your decision as to how precisely to present a statistic should be based

on the specific research problem you are examining.) The proportion of

executed white offenders who had 8 years of education is thus about

0.08.

Sometimes researchers like to transform a proportion into a percent-

age, because percentages are more commonly used in our everyday

lives. We obtain the percentage simply by multiplying the proportion

by 100:

Equation 3.2Percentage � � Ncat

Ntotal � � 100

W orking It Out

� 0.077

� 14 181

Proportion � Ncat

Ntotal

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A48

For our example of executed white offenders with 8 years of education,

we multiply 14/181 by 100:

W orking It Out

� 7.7%

� � 14 181� � 100

Percentage � � Ncat

Ntotal � � 100

This tells us that about 8% of executed white offenders had 8 years of

education. Table 3.4 presents the observed frequencies and correspond-

ing percentages (calculated to the third decimal place) for every cell in

Table 3.3.

At this point, we can graph either the proportions or the percent-

ages in a bar chart. The selection of proportions or percentages does

not matter, as it will not have any bearing on the shape of the distribu-

tions. The bar chart in Figure 3.8 uses percentages to display years of

education among executed offenders in the United States, distinguished

by race.

Frequency Distributions and Percentages for Number of Years

of Education Among Executed Offenders, by Race

RACE OF OFFENDER

WHITE AFRICAN AMERICAN

Years of Education Freq. % Freq. %

7 24 13.260 7 6.250 8 14 7.735 17 15.179 9 19 10.497 16 14.286

10 24 13.260 20 17.857 11 14 7.735 18 16.071 12 60 33.149 31 27.679 13 8 4.420 1 0.893 14 12 6.630 2 1.786 15 2 1.105 0 0.000 16 3 1.657 0 0.000 17 1 0.552 0 0.000 Total 181 112

Table 3.4

4E X T E N D I N G H I S T O G R A M S T O M U L T I P L E G R O U P S 9

U s i n g B a r C h a r t s w i t h N o m i n a l o r O r d i n a l D a t a

The previous examples of bar charts and histograms have focused on

variables measured at the interval level. Bar charts are also quite useful

for visually representing nominally or ordinally measured variables. In

much of the research on public opinion about use of the death penalty in

the United States, there are sharp differences between the views of whites

and those of African Americans. Table 3.5 presents the level of support

for using the death penalty for first-degree murderers, distinguished by

P er

ce n

ta ge

35

30

25

20

15

10

5

0

Years of Education

7 8 9 10 11 12 13 14 15 16 17

Whites

African Americans

Percentages of Executed Offenders with Various Numbers of Years of Education, by RaceFigure 3.8

Level of Agreement with Use of the Death Penalty

for Convicted First-Degree Murderers

RACE OF RESPONDENT

WHITE AFRICAN AMERICAN

Level of Agreement Freq. % Freq. %

Strongly agree 567 51.266 57 37.748 Agree 290 26.221 32 21.192 Neutral/no opinion 76 6.872 24 15.894 Disagree 106 9.584 25 16.556 Strongly disagree 67 6.058 13 8.609 Total 1,106 151

Table 3.5

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A50

race of the respondent. The data come from the General Social Survey

(GSS), administered by the National Opinion Research Center at the Uni-

versity of Chicago. Respondents were asked: “How strongly do you agree

with the following statement: The death penalty should be used for per-

sons convicted of first-degree murder.” The responses were strongly

agree, agree, neutral/no opinion, disagree, and strongly disagree.5

To chart these data, we cannot graph the numbers of white and

African American respondents in each category and hope to make sense

of the patterns, since there are about seven times more white respon-

dents than African American respondents. Again, we construct a bar

chart using the proportion or percentage of respondents who fall into

each category. A bar chart presenting percentages of white and African

American respondents in each category appears in Figure 3.9.

P i e C h a r t s

Pie charts offer another way of displaying data graphically if the num-

ber of categories of a variable is relatively small. Each wedge in a pie

chart is a proportional representation of the number of cases in that cate-

gory. When we present data on the percentage or proportion of cases in

5The entire GSS database is publicly available and can be accessed at http://www.

icpsr.umich.edu/GSS. Data presented here are drawn from a 1991 study.

P er

ce n

ta ge

60

50

40

30

20

10

0

Use of Death Penalty

Whites

African Americans

Strongly Agree Agree Neutral Disagree Strongly Disagree

Bar Chart for Level of Agreement with Use of the Death Penalty

for Convicted First-Degree Murderers Figure 3.9

51P I E C H A R T S

a pie chart, the information contained in a pie chart is identical to that

We will use the death penalty opinion data in Table 3.5 to illustrate

The information presented in Figure 3.10 is identical to that in Figure

3.9. But the need for two separate pie charts to represent the data for the

two groups of respondents points out one of the limitations of using pie

charts for comparing multiple groups. In Figure 3.9, the bars for the two

groups of respondents are presented side by side. The reader is able to

quickly assess which group is larger, which group is smaller, and the mag-

nitude of the difference. With the two pie charts represented in parts a and

b of Figure 3.10, the reader has to go back and forth between the pies,

match up the category of interest, and then try to infer the magnitude of the

difference in size between the two wedges. Although pie charts are an easy

and effective way of representing the relative sizes of different categories of

a variable, we discourage the use of pie charts for any type of cross-group

comparison, to avoid the potential for confusing or misleading the reader.

T i m e S e r i e s D a t a

Study of many important issues in crime and criminal justice requires the

use of time series data. Time series data include measures on the same

variable for the same unit of analysis at more than one point in time. For

Strongly Disagree (6.1%)

Agree (26.2%)

Neutral (6.9%)

Disagree (9.6%) Strongly Agree (51.3%)

Strongly Disagree (8.6%)

Agree (21.2%)

Neutral (15.9%)

Disagree (16.6%)

Strongly Agree (37.7%)

Percentage of Agreement with Use of the Death Penalty for Convicted First-Degree MurderersFigure 3.10

(a) White Respondents (b) African American Respondents

aspects—size, shape, orientation, dimensions (two or three), position

of wedges, colors, shades, and so on.

and spreadsheet packages allow the researcher to manipulate various

presented in a bar chart. As with bar charts, most statistical software

responses for white and African American respondents, respectively.

the construction of pie charts. Parts a and b of Figure 3.10 present the

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A52

example, in a study of adolescents, the same individuals may be inter-

viewed two or more times. Or, in an examination of the effectiveness of

a treatment program, information on key characteristics of participants

may be collected before and after the treatment, to test for change.

An example of time series data more familiar to many students comes

from the Federal Bureau of Investigation’s annual Uniform Crime Re-

ports.6 Included in the reports are the “official” crime rates, which tell us

the total crime rate, the murder rate, the burglary rate, and so on. We

can easily locate a crime rate for the United States—as well as smaller

geopolitical units, such as states or counties—for some period of time.

Table 3.6 presents the total crime rate and murder rate for the United

States per 100,000 people for the years 1990 to 2000.

In a time series plot, data are graphed over time, with the measure

of time along the x-axis. For the data in Table 3.6, the measure of time is

years, but in other cases, we may have daily, weekly, monthly, or quar-

terly data. The y-axis then represents the value—here, the crime rate.

Figure 3.11 presents a plot of the total crime rate per 100,000 people in

the United States for the period 1990 to 2000. As you can see, the time

series plot provides a very powerful way to present data over time. The

“crime drop” in the United States during this period is clearly illustrated

in Figure 3.11.

It is also possible to plot more than one time series on the same

graph, but be careful about the values of the different series. If one se-

ries has values that are many times larger than those in the other series,

you run the risk of generating a plot in which the line of one group

looks misleadingly like a straight line. Figure 3.12, which shows total

Total Crime Rate and Murder Rate in the United States, 1990 to 2000

TOTAL CRIME RATE MURDER RATE

YEAR (PER 100,000) (PER 100,000)

1990 5,820.300 9.4000 1991 5,897.800 9.8000 1992 5,660.200 9.3000 1993 5,484.400 9.5000 1994 5,373.500 9.0000 1995 5,275.900 8.2000 1996 5,086.600 7.4000 1997 4,930.000 6.8000 1998 4,619.300 6.3000 1999 4,266.800 5.7000 2000 4,124.800 5.5000

Table 3.6

6Federal Bureau of Investigation, Crime in the United States, available at http://www.

fbi.gov/ucr/ucr.htm.

T I M E S E R I E S D A T A 53

crime rates and murder rates for the 1990 to 2000 period, illustrates this

problem. Since the total crime rate ranges between about 4,000 and

6,000 per 100,000 while the murder rate varies between about 5 and 10

per 100,000, the line representing the murder rate appears straight and

indeed can hardly be seen. In such a case, one solution is to construct a

R at

e

7,000

6,500

6,000

5,500

5,000

4,500

4,000

3,500

3,000

Year 1990

1991 1992

1993 1994

1995 1996

1997 1998

1999 2000

Total Crime Rate

Total Crime Rate (per 100,000 People) in the United States, 1990 to 2000Figure 3.11

R at

e

7,000

6,000

5,000

4,000

3,000

2,000

1,000

0

Year

1990 1991

1992 1993

1994 1995

1996 1997

1998 1999

2000

Total Crime Rate

Murder Rate

Total Crime Rate and Murder Rate (per 100,000 People)

in the United States, 1990 to 2000: Single Vertical Axis Figure 3.12

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A54

time series plot with a second y-axis on the right side of the chart. Figure

3.13 adds a second y-axis to account for the smaller values associated

with the murder rate, providing a better depiction of the annual variation

in both rates.

C h a p t e r S u m m a r y

A frequency distribution is an arrangement of data according to the

frequency with which each value occurs. The data may be represented

in a table or in graphic form in a histogram, which uses a bar to repre-

sent the frequency for each value.

Bar charts can be used to represent the frequencies, percentages,

or proportions of variables, regardless of whether they have been mea-

sured at the nominal, ordinal, or interval level. In a vertical bar chart, the

sizes of the bars are indicated along the y-axis and correspond to the fre-

quency, the percentage, or the proportion. The values, or categories, of

the variable are represented along the x-axis. When comparing two or

more groups in a bar chart, it is important to have the values represent

the percentage or proportion of cases, since the presentation of frequen-

cies could be misleading if the groups are of very different sizes.

Pie charts are another common way to represent variables with a

small number of categories or values. The size of each wedge in a pie

chart corresponds to the relative size of that category’s frequency count.

T ot

al C

ri m

e R

at e

M u

rd er

R at

e

7,000

6,000

5,000

4,000

3,000

2,000

1,000

0

Year

1990 1991

1992 1993

1994 1995

1996 1997

1998 1999

2000

Total Crime Rate

Murder Rate

12

10

8

6

4

2

0

Total Crime Rate and Murder Rate (per 100,000 People) in the United States,

1990 to 2000: Two Vertical Axes Figure 3.13

55C H A P T E R S U M M A R Y

Pie charts are better suited to describing data from a single sample or

group than data from multiple groups.

Time series data may be graphically displayed in a time series plot,

a line graph that portrays the values of a variable over some period of

time, such as days, weeks, months, or years. To allow for the compari-

son of multiple variables, additional lines can be easily incorporated into

the line graph. If the magnitudes of the variables differ by a large degree,

it is possible to add an additional y-axis so that the lines can be overlaid

and common or unique trends in the plotted variables discerned.

K e y T e r m s

bar chart A graph in which bars represent

frequencies, percentages, or proportions

for the categories or values of a variable.

frequency The number of times that a

score or value occurs.

frequency distribution An arrangement

of scores in order from the lowest to the

highest, accompanied by the number of

times each score occurs.

histogram A bar graph used to represent

a frequency distribution.

percentage A relation between two num-

bers in which the whole is accorded a

value of 100 and the other number is given

a numerical value corresponding to its

share of the whole.

pie chart A graph in which a circle (called

a pie) is cut into wedges to represent the

relative size of each category’s frequency

count.

proportion A relation between two num-

bers in which the whole is accorded a

value of 1 and the other number is given a

numerical value corresponding to its share

of the whole.

time series data Repeated measures of

the same variable over some regularly oc-

curring time period, such as days, months,

or years.

time series plot A line graph that con-

nects repeated measures of the same vari-

able over some regularly occurring time

period, such as days, months, or years.

S y m b o l s a n d F o r m u l a s

Ncat Number of cases in a category of a variable

Ntotal Total number of cases

To calculate the proportion of cases falling into a category:

Proportion � Ncat

Ntotal

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A56

To calculate the percentage of cases falling into a category:

E x e r c i s e s

3.1 A team of researchers observed the behavior of 20 children during breaktime on a school playground. The observers recorded how many times each child performed a “violent act”—be it a push, a kick, or a punch—against another child. The scores of the 20 children were as follows:

2 0 1 0 0 4 0 10 2 1 3 3 0 1 4 11 0 0 2 0

a. Construct a frequency distribution table for the above results.

b. Construct a histogram of the frequency distribution.

c. How might you interpret the results?

3.2 Workers at an inner-city rape crisis center asked all the callers on a given night for their ages. For a total of 50 callers, the results were as follows:

28 39 17 18 22 31 26 27 16 20

34 35 29 26 17 23 22 23 37 28

24 19 14 25 27 19 24 26 41 27

21 24 17 16 35 25 19 23 29 18

23 26 24 43 28 21 21 36 26 27

a. Construct a frequency distribution table for the above results.

b. Based on the frequency distribution, construct three different his- tograms, with the data grouped in 1-year intervals, 3-year intervals, and 10-year intervals.

3.3 A review of the records of 20 male and 20 female adjudicated delin- quents revealed the following numbers of prior arrests for violent crimes:

Male: 0 2 1 7 4 2 2 1 6 5 0 0 1 2 4 1 0 0 2 9

Female: 0 0 1 4 2 3 1 1 5 1 0 0 0 1 0 3 2 1 0 1

a. Construct three frequency distribution tables: one for males, one for females, and one for the male and female data combined.

b. Construct a histogram for each of the three frequency distributions.

c. How might you interpret the results?

Percentage � � Ncat

Ntotal � � 100

E X E R C I S E S 57

3.4 In a survey of 75 offenders convicted through guilty pleas, respon- dents were asked who had the greatest influence on their decision to plead guilty. Here are their responses: defense attorney (n � 35), spouse/partner (n � 25), prosecutor (n � 10), and judge (n � 5).

a. Graph these data on three bar charts using frequencies, propor- tions, and percentages.

b. Graph these data on a pie chart.

3.5 Suppose a researcher conducted a survey to assess the link between

researcher found that 150 thought the punishments were “too harsh,” 100 thought the punishments were “about right,” and 50 thought the punishments were “too lenient.”

b. Explain your choice of chart.

c. Describe the patterns that you find in this type of chart.

3.6 The burglary rates for two cities over a 10-year period are reported in the following table:

Year City A City B

1991 2,576 875

1992 2,675 966

1993 2,892 1,015

1994 3,014 1,325

1995 2,852 1,779

1996 2,651 1,954

1997 2,443 2,333

1998 2,519 2,121

1999 2,999 2,657

2000 2,840 2,005

a. Present this information in the way that you think is most informative.

b. Explain why you selected this type of chart.

c. Describe the patterns that you observe in this chart.

separately.

a. Present this information graphically in the way that you think is most informative—either for the full sample or for each gender

ments were “too harsh.” Among 300 female respondents, the

gender and attitudes about the criminal justice system. Among 850 male respondents, the researcher found that 500 thought the punishments given by the courts were “too lenient,” 200 thought the punish- ments were “about right,” and the remaining 150 thought the punish-

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A58

C o m p u t e r E x e r c i s e s

SPSS and Stata both contain a wide range of graphics tools that are comparable

to those found in many of the better spreadsheet packages. The following com-

puter exercises are intended to guide you through the basic steps in producing

graphs in the two different programs. In fact, the graphics commands we illus-

trate from SPSS are what are referred to as “legacy” commands in SPSS’s win-

dowed environment. The change in SPSS nomenclature is a consequence of the

graphics capabilities of the program having expanded greatly in recent years. The

graphics options in SPSS are now so complex that the best way to learn about

and work with the graphics commands is to do it interactively. The description

and discussion of each command is meant only as a brief overview of some of

the capabilities of either SPSS or Stata that will produce figures comparable to

those discussed in this chapter. We encourage you to experiment and to explore

the range of features available in the graphics commands. On the Web site for

the text, we have included syntax files that provide examples of each of the

graphics commands in either SPSS (Chapter_3.sps) or Stata (Chapter_3.do).

SPSS

Frequency Distributions and Histograms

Frequency distributions and histograms are produced quite easily in SPSS through

the use of the FREQUENCIES command (which can also be shortened as

FREQ):

Following VARIABLES = would be a list of the variables for which you wanted

SPSS to compute frequency distributions. The /HISTOGRAM option will gen-

erate a histogram for the corresponding variables included in the list. You should

note that the default in SPSS is to construct intervals for the values of the varia-

ble to be graphed. The size and the number of intervals can be modified by edit-

ing the chart in SPSS, which is accomplished by double-clicking on the chart in

the SPSS output window. This will open an additional window that allows you to

make numerous changes to your chart and customize its appearance. We encour-

age you to experiment with many of the options available to you.

The GRAPH command with the /HISTOGRAM option provides an

alternative way of obtaining just a histogram, without the corresponding frequen-

cy distribution. The simplest way of generating a histogram using SPSS syntax is

the following:

The histogram generated by this command will be identical to that produced by

the FREQUENCIES command.

C O M P U T E R E X E R C I S E S 59

FREQUENCIES VARIABLES = variable_names /HISTOGRAM.

GRAPH /HISTOGRAM = variable_name.

This will generate a bar chart where each bar reflects the number of cases with

that value, much like the histogram, but with one important difference. Bar

charts are not sensitive to the values of the variable being graphed—the values

are treated as ordinal and are all positioned along the x-axis at constant intervals.

Histograms make use of the level of measurement, so that a variable measured at

the interval level of measurement will be appropriately scaled on the x-axis.

To produce a bar chart that includes bars for multiple groups, there are sev-

eral different ways of doing this in SPSS. A simple bar chart for multiple groups

can be obtained by again executing the GRAPH command. To generate a mul-

tiple group bar chart that uses percentages as the point of comparison across

groups, use the following command:

To generate a multiple group bar chart that uses the number of cases in each

group, use the following command:

Note that the type of graph is now /BAR(GROUPED), the percentage of cases

by the PCT option, the number of cases by the COUNT option, the variable

we are interested in plotting is listed after the first BY and the grouping variable

after the second BY option.

For example, if we were interested in plotting a measure of delinquency by

sex of respondent in a self-report survey using percentages as the comparison,

we would use the following command:

The resulting graph would have two bars—one for each sex—at each level of the

delinquency measure.

Pie Charts

Pie charts are also very easy to produce in SPSS with the GRAPH command, but

changing the type of chart to /PIE=

Similar to the bar chart options, rather than the number of cases (here the

COUNT option), you can also ask for SPSS to use percentages (PCT option in

Bar Charts

To have SPSS produce bar charts, you will need to use the GRAPH command

again, but simply change the option to /BAR=

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A60

GRAPH /BAR = variable_name.

GRAPH /BAR(GROUPED) = PCT BY variable_name BY group_var_

name.

GRAPH /BAR(GROUPED) = COUNT BY variable_name BY

group_var_name.

GRAPH /BAR(GROUPED) = PCT BY delinquency BY sex.

GRAPH /PIE = COUNT BY variable_name.

the command line, replacing COUNT). As with every other chart in SPSS, you

can edit various dimensions of the chart after it has been created and presented

in the output window.

Line Graphs

The fourth type of chart we highlight here is the line graph. Again, the GRAPH

command is used and the type of chart is changed to /LINE

This command will generate a simple line graph that defaults to plotting the

number of cases at each value of the variable being plotted—equivalent to the

simple bar chart, but with a line connecting what would have been the top of

each bar.

To produce a time plot, such as that seen in Figure 3.11, we begin with the

same GRAPH command:

The VALUE option tells SPSS to plot the value of the variable included in the

parentheses. So, in Figure 3.11, we plotted a crime rate measure—we would

simply include the name of that variable here. After the BY option, we would

include the name of the time variable, such as month, quarter, year, etc.

Figure 3.12 includes two lines that represent two different crime rates. To

include additional lines in a time plot in SPSS, we would simply add the addi-

tional variable name to a list included in the parentheses following the VALUE

option:

NOTE: SPSS will not produce a line graph with two y-axes, such as that found

in Figure 3.13.

Stata

Frequency Distributions and Histograms

Frequency distributions are computed in Stata with the tab1 command and have

the following format:

If more than one variable name is listed, multiple frequency distributions will be

computed. Note that some of the other options for computing frequency dis-

tributions in Stata will produce more complicated tables should more than one

variable be listed (e.g., the tab command).

C O M P U T E R E X E R C I S E S 61

GRAPH /LINE(SIMPLE) = variable_name.

GRAPH /LINE(SIMPLE) = VALUE(variable_name)

BY time_variable.

GRAPH /LINE(SIMPLE) = VALUE(variable_name1

tab1 variable_name(s)

variable_name2) BY time_variable.

Histograms are also easy to produce in Stata with the histogram command:

Note that you are limited to one variable at a time in running the histogram

command.

Bar Charts

Although Stata has an impressive set of graphics commands, the ability to cre-

ate simple bar charts like those discussed in this chapter is limited. Rather than

rely on one of the built-in Stata commands, we will use a user-written command

called fbar that is available for installation on your local computer running Stata.

The command to type and run (one time only) is:

This command will install fbar from an archive maintained by Stata and will make it

available to you while you’re running Stata.1 Once the command has been installed,

the structure to the fbar command is:

Where everything appearing after the comma is an option. The default for fbar

is to produce a bar chart using frequencies. To have it compute percentages, use

the percent option (which can be abbreviated perc). If there is an interest in

comparing bar charts for different groups, then use the by option and include the

grouping variable’s name in the parentheses and use the option totalpc, which

will calculate percentages in the multiple group bar chart using all of the catego-

ries within each group as the reference, as we have in this chapter (see Figure 3.8).

Similar to our discussion of multiple bar charts in SPSS, if we were to

compare delinquency counts by sex, we would run the following command:

Pie Charts

Pie charts are very easy to produce in Stata using the graph command:

Where the over(…) option tells Stata how to classify the information into

categories.

1 Throughout the text, we will make use of user-written procedures for Stata. The process will

be very much the same every time we execute one of these commands to make a procedure

available within Stata. If you are working on a computer in a student computer lab setting, you

will need to install these kinds of procedures every time you move to a different machine and

run Stata.

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A62

histogram variable_name

ssc install fbar

fbar variable_name, by(group_variable_name)

percent totalpc

fbar delinquency, by(sex) totalpc

graph pie, over(variable_name)

Where the y_variable_name and x_variable_name refer to the placement of the

variable on the two axes of the plot.

To produce a time plot, such as that seen in Figure 3.11, we begin with the

same twoway command, and just make explicit reference to the x-axis being

the indicator of time:

When we want to add an additional line to the time series plot, we just include

another line graph and pair of variable names in parentheses:

Note that the time variable should be the same, if the comparison is between two

(or more) measures taken at the same points in time.

One of the options available in Stata, but not SPSS, is the ability to add a

second y-axis, which as we saw in Figure 3.13 can be useful for comparing trends

measured on different scales. The key is to include a reference to the axis within

each set of parentheses in the twoway command:

Where the axis(#) is inserted following a comma that follows the name of the

time variable. The Stata do file for Chapter 3 (Chapter_3.do) illustrates how to

use this command to reproduce Figures 3.11 through 3.13.

Problems

1. Enter the data from Exercise 3.1.

a. Produce a frequency distribution table and a histogram.

b. How does the histogram produced by the statistical software differ

from the one you constructed?

c. Experiment with the interval length to see if you can reproduce in the

statistical software the chart you constructed by hand.

2. Enter the data from Exercise 3.3.

a. Produce two frequency distribution tables and two histograms, one

each for males and for females.

Line Graphs

The final chart we highlight here is the line graph, many of which can be

produced with the twoway command in Stata, which for simple examples takes

on the following form:

C O M P U T E R E X E R C I S E S 63

twoway (line y_variable_name x_variable_name)

twoway (line y_variable_name time_variable_name)

twoway (line y_var_name1 time_variable_name)

(line y_var_name2 time_variable_name)

twoway (line y_var_name1 time_variable_name, axis(1)) (line

y_var_name2 time_variable_name, axis(2))

b. Produce a bar chart that presents the information for males and

females simultaneously, in terms of

i. Number of cases

ii. Percentage of cases.

3. Enter the data from Exercise 3.6. Produce three time plots:

a. One for City A

b. One for City B

c. One that contains lines representing both City A and City B.

4. Open the NYS data file (nys_1.sav, nys_1_student.sav, or nys_1.dta).

a. Produce a frequency distribution table and a histogram for each of the

delinquency variables.

b. Describe the shapes of the distributions produced in part a.

c. Choose one of the delinquency variables:

i. Produce a bar chart using percentages for each gender. Describe how

the patterns are similar and different across gender.

ii. Produce a bar chart using percentages for each race. Describe how

the patterns are similar and different across race.

C H A P T E R T H R E E : R E P R E S E N T I N G A N D D I S P L A Y I N G D A T A64

d. Select two of the nominal or ordinal variables (excluding gender and

race).

i. Produce pie charts for both variables.

ii. Produce pie charts for both variables, distinguished by gender.

iii. Produce pie charts for both variables, distinguished by race.

iv. Describe how the patterns are similar and different across gender and

race.

Describing the Typical Case:

Measures of Central Tendency

What Information Does It Use?

What Information Does It Use?

C h a p t e r f o u r

T h e m o d e

T h e m e d i a n

T h e m e a n

How is It Calculated?

What are Its Advantages and Disadvantages?

How is It Calculated?

What are Its Advantages and Disadvantages?

How is It Calculated?

What are Its Advantages and Disadvantages?

What are Its Other Unique Properties?

What Information Does It Use?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_4,

© Springer Science+Business Media New York 2014

THE NATURAL FIRST STEP in summarizing research is to provide a basic

portrait of the characteristics of a sample or population. What is the typ-

ical case? If the researcher could choose one case to represent all oth-

ers, which would it be? When a sample is very small, it is possible

merely to show the array of cases and let the reader decide. However,

as the number of cases grows, it becomes difficult to make a decision

about typicality from the distribution as a whole. This is the function of

measures of central tendency in statistics. They provide us with a simple

snapshot of our data that can be used to gain a picture of the average

case.

In this chapter, three commonly used measures of central tendency

are discussed and compared. The first, the mode, is used primarily with

nominal-level data. It is the simplest measure of central tendency, draw-

ing information only about the frequency of events in each category. The

second measure, the median, takes into account not only frequency but

also the order or ranking of study subjects. Finally, the mean adds the

additional factor of the exact scores associated with each subject studied.

As in the discussion of levels of measurement, we emphasize in this

chapter the benefits gained from statistics that use more information. But

we also illustrate the importance of looking carefully at the distribution

of cases in your study before deciding which measure of central ten-

dency is most appropriate.

T h e M o d e : C e n t r a l T e n d e n c y i n N o m i n a l S c a l e s

Faced with a nominal-scale measure, how would you define a typical

case? Take as an example Table 4.1. Here you have a nominal scale of

legal representation for a sample of offenders convicted of white-collar

crimes in U.S. federal courts. Offenders were placed into one of five cat-

egories, indicating the type of legal representation they had: no attorney

66

T H E M O D E : C E N T R A L T E N D E N C Y I N N O M I N A L S C A L E S 67

present, legal-aid attorney, court-appointed attorney, public defender,

and privately retained attorney. The number of individuals who fall in

each category—or, in statistical language, the N of cases—is reported.

Clearly, you have very limited information in this example on which

to base a choice about typicality. Here, as with other nominal-scale mea-

sures, you simply know how many cases fall into one category or an-

other. You would probably choose the category “private attorney” as

most representative of this sample, because it contains by far the most

cases (380). And indeed, this is precisely how statisticians define typical-

ity for nominal-level variables. We call the category with the largest N, or

number of cases, the mode. In this sample of white-collar offenders, the

modal category for type of representation is “private attorney.”

By defining one category as the modal category, we are able to pro-

vide a summary of the type of case that is typical of our sample or popu-

lation. Such statements are common in criminal justice research. We

often are interested in the racial category that appears most often in our

data or the type of offense that is most common. The modal category

can also provide a basis for making comparisons among samples. For

example, let’s say that a sample of offenders convicted of nonviolent

property crimes that would not ordinarily be defined as white collar was

compared to this larger sample of offenders convicted of white-collar

crimes. For the former group, as is apparent from Table 4.2, the modal

category is not “private attorney” but rather “court-appointed attorney.”

Although this comparison of the two samples is not a complex one, it

Legal Representation for White-Collar Crime

CATEGORY FREQUENCY (N)

No Attorney 20 Legal Aid 26 Court Appointed 92 Public Defender 153 Private Attorney 380 Total (�) 671

Table 4.1

Legal Representation for Common Crime

CATEGORY FREQUENCY (N)

No Attorney 40 Legal Aid 7 Court Appointed 91 Public Defender 22 Private Attorney 70 Total (�) 230

Table 4.2

68 C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

In general, we do not use the mode to describe central tendency with

ordinal or interval scales. The reason, in good part, is that the mode does

not take advantage of the additional information that such scales pro-

vide. The average case should not be chosen simply on the basis of the

frequency of events in a particular category, because higher-level scales

also provide information on the order or nature of the differences be-

tween categories.

Nonetheless, there are cases where researchers choose to use the

mode to describe ordinal- or interval-level measures. Generally this oc-

curs when there is a very large group of cases in one particular category.

Table 4.3, for example, provides an ordinal-level measure of the financial

harm caused by a sample of convicted offenders. Because almost two-

thirds of the individuals studied fall in the category “$101–$2,500,” you

might want to describe typicality in this case by saying that this category

is the modal category. Similarly, if you were examining prior arrests and

two-thirds of the offenders in your sample had no prior arrests, you

might want to report no arrests as the modal category. Even though this

measure is an interval measure, the mode provides a fairly good sum-

mary of the typical case in your sample.

T h e M e d i a n : T a k i n g i n t o A c c o u n t P o s i t i o n

In constructing the median, we utilize information not only on the num-

ber of cases found in a particular category, but also on the positions of

the categories. The median may be defined simply as the middle score in

a distribution. For ordinal scales, it is the category in which the middle

score lies. For interval scales, the median is the value that splits the dis-

tribution of scores in half.

There are two general steps in determining the median for a distribu-

tion of scores. First, the values need to be arranged from low to high

Financial Harm for a Sample of Convicted Offenders

CATEGORY FREQUENCY (N)

Less than $100 15 $101–$2,500 92 $2,501–$10,000 20 More than $10,000 19 Total (�) 146

Table 4.3

illustrates the different backgrounds of the two groups. White-collar

offenders are much more likely than common criminals to have the

resources to pay for private legal representation.

T H E M E D I A N : T A K I N G I N T O A C C O U N T P O S I T I O N 69

scores. As we saw in Chapter 3, a frequency distribution allows us to

represent our data in this way. Table 4.4 presents a frequency distribu-

tion of views of public drunkenness, drawn from a survey of students.

The students were presented with an ordinal-scale measure that allowed

them to rate the seriousness of a series of crimes. The ratings ranged

from “not serious at all” to “most serious.”

Second, we need to determine which observation splits the distribu-

tion. A simple formula, Equation 4.1, allows us to define which observa-

tion is the median when the number of observations in the distribution is

odd, as is the case with our example of views of public drunkenness.

Equation 4.1

In this case, we add 1 to the total number of observations in the sample

or population we are studying and then divide by 2. For the frequency

distribution in Table 4.4, the median observation is the 141st score:

Median observation � N � 1

2

Student Views on Public Drunkenness

CATEGORY FREQUENCY (N) CUMULATIVE N

Not serious at all 73 73 A bit serious 47 120 Somewhat serious 47 167 Serious 27 194 Very serious 26 220 Extremely serious 39 259 Most serious 22 281 Total (�) 281

Table 4.4

W orking It Out

� 141

� 281 � 1

2

Median observation � N � 1

2

However, because our variable, student views on drunkenness, is mea-

sured on an ordinal scale, it does not make sense to simply state that the

141st observation is the median score. To give a substantive meaning to

the median, it is important to define which category the median score

C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

falls in. The 141st observation in our distribution of ordered scores falls

in the category labeled “somewhat serious.”

The advantage of the median over the mode for describing ordinal

scales is well illustrated by our example of views of public drunkenness.

If we used the mode to describe typicality in student assessments of the

seriousness of public drunkenness, we would conclude that the typical

student did not see drunkenness as at all serious. But even though the

“not serious at all” category includes the largest number of cases, almost

three-quarters of the students rate this behavior more seriously. The me-

dian takes this fact into consideration by placing the typical case in the

middle of a distribution. It is concerned with not only the number of

cases in the categories, but also their position.

If the number of observations or cases in your distribution is even, then

you cannot identify a single observation as the median. While statisticians

recognize that the median is ambiguously defined in this case, by conven-

tion they continue to use Equation 4.1 to identify the median for an ordi-

nal-level distribution. In practice, this places the median score between two

observations. For example, consider the distribution of 146 scores in Table

4.3, representing financial harm in a sample of offenders. Here the number

1

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73rd observation: $101–$2,500

74th observation: $101–$2,500

� 73.5

� 146 � 1

2

Median observation � N � 1

2

1With this method, it is possible that the defined median value will fall between two

categories of an ordinally measured variable. In that case, you simply note that the

median falls between these two categories.

The median is sometimes used for defining typicality with interval

scales. For example, Table 4.5 presents the average number of minutes

of public disorder (per 70-minute period) observed in a sample of 31

of scores is even, and thus there is not a single observation that can be

defined as the median. Using Equation 4.1, we can see that the median is

defined as the halfway point between the 73rd and the 74th observation.

This means that the median falls in the category defined as $101 to $2,500.

70

T H E M E D I A N : T A K I N G I N T O A C C O U N T P O S I T I O N

“hot spots of crime,” or city blocks with high levels of crime. The hot

spots are arranged in ascending order on the basis of the number of

minutes of disorder observed. In this case, the distribution has an odd

number of observations, and thus the median is the score in the middle

of the distribution, or the 16th observation, which has a value of 2.12.

Hot Spots: Minutes of Public Disorder (A)

HOT SPOT

SCORE FREQUENCY (N) CUMULATIVE (N)

0.35 1 1 0.42 1 2 0.46 1 3 0.47 1 4 0.52 1 5 0.67 1 6 1.00 1 7 1.06 1 8 1.15 1 9 1.19 2 11 1.48 1 12 1.60 1 13 1.63 1 14 2.02 1 15 2.12 1 16 2.21 1 17 2.34 1 18 2.45 1 19 2.66 1 20 3.04 1 21 3.19 1 22 3.23 1 23 3.46 1 24 3.51 1 25 3.72 1 26 4.09 1 27 4.47 1 28 4.64 1 29 4.65 1 30 6.57 1 31 Total (�) 31 31

Table 4.5

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� 16

� 31 � 1

2

Median observation � N � 1

2

71

C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

Accordingly, using the median, we would describe the average hot spot as

having a little more than two minutes of disorder in each 70-minute period.

As noted above, when the number of observations in a distribution is

even, the median is ambiguously defined. Let’s, for example, delete the hot

spot with a score of 6.57 from Table 4.5. In this case, there is no single mid-

dle value for the array of cases in the table. If we use Equation 4.1 to define

the median observation, we get a value of 15.5.

2Sometimes the median for ordinal-level variables is also calculated using this method.

In such cases, the researcher should realize that he or she is treating the variable

under consideration as an interval-level measure. Only for an interval-level measure

can we assume that the units of measurement are constant across observations.

W orking It Out

� 15.5

� 30 � 1

2

Median observation � N � 1

2

But what is the value or score associated with an observation that lies

between two scores in an interval-level scale? If both the 15th and the

16th observation are in the same category, then the solution is easy. You

simply define the median as the score associated with both the 15th and

the 16th observation. However, it will sometimes be the case with an

interval-level variable that each of these observations will have a differ-

ent value on the scale, as we find here. There is no true median value

for this example. By convention, however, we define the median with

interval-level measures as the midpoint between the observation directly

below and the observation directly above the median observation. In our

example, this is the midpoint on our scale between the scores 2.02 and

2.12. The median in this case is defined as 2.07.2

W orking It Out

� 2.07

Median � 2.02 � 2.12

2

16th case � 2.12

15th case � 2.02

72

T H E M E D I A N : T A K I N G I N T O A C C O U N T P O S I T I O N

The median is generally more appropriate than the mode for assess-

ing central tendency for both ordinal- and interval-level measures. How-

ever, the median does not take advantage of all the information included

in interval-level scales. Although it recognizes the positions of the values

of a measure, it does not take into account the exact differences among

these values. In many cases, this can provide for a misleading estimate of

typicality for interval-level measures.

For example, let’s say that the distribution of disorder in hot spots is

that represented in Table 4.6. In this case, the median is 1.83. But is 1.83

a good estimate of central tendency for this distribution? The 17th score

is 3.34, which is not very close to 1.83 at all. The score of 1.83 is not an

the values of the 16th and 17th cases. This is because it looks only at

Hot Spots: Minutes of Public Disorder (B)

HOT SPOT

SCORE FREQUENCY (N) CUMULATIVE (N) CUMULATIVE %

0.35 1 1 3.2 0.42 1 2 6.5 0.46 1 3 9.7 0.47 1 4 12.9 0.52 1 5 16.1 0.67 1 6 19.4 1.00 1 7 22.6 1.06 1 8 25.8 1.15 1 9 29.0 1.19 2 11 35.5 1.48 1 12 38.7 1.60 1 13 41.9 1.63 1 14 45.2 1.73 1 15 48.4 1.83 1 16 51.6 3.34 1 17 54.9 3.44 1 18 58.1 3.45 1 19 61.3 3.66 1 20 64.5 4.04 1 21 67.7 4.19 1 22 71.0 4.23 1 23 74.2 4.46 1 24 77.4 4.51 1 25 80.6 4.72 1 26 83.9 5.09 1 27 87.1 5.47 1 28 90.3 5.64 1 29 93.5 5.65 1 30 96.8 5.57 1 31 100.0 Total (�) 31 31 100.0

Table 4.6

bution. The median is not sensitive to the gap in our measure between

ideal estimate of typicality, as it is far below half the scores in the distri-

73

C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

position and not at the size of the differences between cases. The me-

dian does not take advantage of all the information provided by interval-

level measures.

Another way to describe the median in interval-level measures is to

say that it is the 50th percentile score. A percentile score is the point or

score below which a specific proportion of the cases is found. The 50th

percentile score is the score below which 50% of the cases in a study lie.

For the data in Table 4.6, if we defined the median in this way, we again

choose 1.83 as the median minutes of disorder observed in the hot spots.

In this case, if we add the percentage of cases for all of the scores up

until the middle, or 16th, score, we come to a total (or cumulative per-

centage) of 51.6. At the 15th score, or 1.73, the cumulative percentage is

only 48.4, less than 50%.

T h e M e a n : A d d i n g V a l u e t o P o s i t i o n

The mean takes into account not only the frequency of cases in a cate-

gory and the positions of scores on a measure, but also the values of

these scores. To calculate the mean, we add up the scores for all of the

subjects in our study and then divide the total by the total number of

subjects. In mathematical language, the mean can be written as a short

equation:

Equation 4.2

Even though equations sometimes put students off, they are an im-

portant part of statistics. Indeed, equations are the language of statistics.

They show how a statistic is constructed and the method we use to cal-

culate it. Equations provide a short way of writing out what would often

take a number of sentences to describe in English. One of our tasks in

this text is to help you to translate such equations and to become more

comfortable with them.

In the case of the mean, we introduce what are for most students of

criminal justice some new symbols and concepts. First, to express the

mean, statisticians provide us with a shorthand symbol, —in English,

“X bar.” The equation also includes the summation symbol, �. Under the

symbol is i � 1, and above it is N. What this means is that you should

start summing your cases with the first subject in the sample and end

X

X � � N

i�1

Xi

N

74

T H E M E A N : A D D I N G V A L U E T O P O S I T I O N

with the last one (represented by N because, as we have already dis-

cussed, N is the number of cases in your sample). But what should you

sum? X represents the measure of interest—in the case of our example,

minutes of disorder. We use the subscript i to denote each of the obser-

vations of the variable X. If, for example, we wrote X3, we would be re-

ferring only to the 3rd observation of the variable. So Equation 4.2 says

that you should sum the scores for minutes of disorder from the first to

the last case in your study. Then you should divide this number by the

total number of cases.

Table 4.7 presents information about the total number of prior arrests

for a sample of 20 individuals arrested for felony offenses. To calculate

the mean, we first sum all of the scores, as shown in the numerator of

Equation 4.2:

Total Number of Prior Arrests

TOTAL NUMBER

OF ARRESTS FREQUENCY (N) CUMULATIVE (N)

0 4 4 1 1 5 2 2 7 4 3 10 5 3 13 7 4 17 8 2 19

10 1 20

Table 4.7

W orking It Out

� 86

� 5 � 5 � 5 � 7 � 7 � 7 � 7 � 8 � 8 � 10

� 0 � 0 � 0 � 0 � 1 � 2 � 2 � 4 � 4 � 4

� N

i�1

Xi � � 20

i�1

Xi

We then take the sum of the values, 86, and divide by the number of ob-

servations in the sample.

75

76 C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

The result, 4.3, tells us that in this sample the typical person arrested for

a felony has, on average, 4.3 prior arrests.

As another example, let’s take the data from Table 4.5 on minutes of

disorder in crime hot spots. According to Equation 4.2, the first step is to

sum all of the scores:

W orking It Out

� 4.3

� 86 20

X � � N

i�1

Xi

N

W orking It Out

� 71.56

� 3.51 � 3.72 � 4.09 � 4.47 � 4.64 � 4.65 � 6.57

� 2.21 � 2.34 � 2.45 � 2.66 � 3.04 � 3.19 � 3.23 � 3.46

� 1.15 � 1.19 � 1.19 � 1.48 � 1.60 � 1.63 � 2.02 � 2.12

� 0.35 � 0.42 � 0.46 � 0.47 � 0.52 � 0.67 � 1.00 � 1.06

� N

i�1

Xi � � 31

i�1

Xi

We then take this number, 71.56, and divide it by N, or 31, the number

of cases in our sample.

W orking It Out

� 2.308387097

� 71.56

31

X � � N

i�1

Xi

N

T H E M E A N : A D D I N G V A L U E T O P O S I T I O N 77

The result, 2.308387097 (rounded to the ninth decimal place), brings up

an issue that often arises in reporting statistics. Do you really need to

provide your audience with the level of precision that is given by your

statistic? In this case, for example, at what level of precision should min-

utes of disorder be presented?

A basic rule of thumb is to use your common sense in answering such

questions. Don’t provide statistics developed out to a large number of

decimal places just to impress others. In making this decision, you

should ask: What is the simplest presentation of my results that will pro-

vide the reader or listener with enough information to understand and

evaluate my work? Overall, criminal justice researchers seldom report the

mean to more than two decimal places. This is a good choice in our ex-

ample. Rounding to the second decimal place gives a mean of 2.31. Pro-

viding a more precise representation of the mean here would not add

important information for the reader.

In some cases, it is useful to develop estimates with much greater pre-

cision. In particular, if the values for the cases you are examining are

very small in the first place, you will want to present a more precise

mean. For example, Lawrence Sherman and his colleagues looked at the

mean daily rate of reported domestic violence in a study that compared

the impact of arrests versus warnings as a strategy for controlling spouse

abusers.3 Had they reported their findings only to the second decimal

place, as recommended above, they would have ended up with a mean

daily rate over the longest follow-up period (361–540 days) of 0.00 for

short arrest and 0.00 for warning. The difficulty here is that individuals

are unlikely to report cases of domestic violence on a very frequent

basis. Sherman et al. needed a much higher degree of precision to exam-

ine the differences between the two groups they studied. Accordingly,

they reported their results to the fourth decimal place. For arrests, the

rate was 0.0019, and for warnings it was 0.0009. These differences,

though small, were found to be meaningful in their research.

Comparing Results Gained Using the Mean and Median

Returning to the example from Table 4.5, we see that the mean for min-

utes of disorder, 2.31, is very similar to the median of 2.12 calculated

earlier. In this case, adding knowledge about value does not change our

portrait of the typical hot spot very much. However, we get a very differ-

ent sense of the average case if we use the data from Table 4.6. Here,

the median provided a less than satisfying representation of the average

3L. Sherman, J. D. Schmidt, D. Rogan, P. Gartin, E. G. Cohn, D. J. Collins, and A. R.

Bacich, “From Initial Deterrence to Long-Term Escalation: Short-Custody Arrest for

Poverty Ghetto Domestic Violence,” Criminology 29 (1991): 821–850.

78 C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

case. It was not sensitive to the fact that there was a large gap in the

scores between the 16th and 17th cases. Accordingly, the median, 1.83,

was very close in value to the first half of the cases in the sample, but

very far from those hot spots with higher values. The mean should pro-

vide a better estimate of typicality here, because it recognizes the actual

values of the categories and not just their positions. Let’s see what hap-

pens when we calculate the mean for Table 4.6.

Following our equation, we first sum the individual cases:

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� 84.41

� 4.51 � 4.72 � 5.09 � 5.47 � 5.64 � 5.65 � 5.77

� 3.34 � 3.44 � 3.45 � 3.66 � 4.04 � 4.19 � 4.23 � 4.46

� 1.15 � 1.19 � 1.19 � 1.48 � 1.60 � 1.63 � 1.73 � 1.83

� 0.35 � 0.42 � 0.46 � 0.47 � 0.52 � 0.67 � 1.00 � 1.06

� N

i�1

Xi � � 31

i�1

Xi

We then divide this number by the total number of cases:

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� 2.7229

� 84.41

31

X � � N

i�1

Xi

N

Here, we gain an estimate of typicality of 2.72 (rounding to the second

decimal place). As you can see, this score is much better centered in

our distribution than is the median. The reason is simple. The median

does not take into account the values of the categories. The mean does

take value into account and thus is able to adjust for the gap in the

distribution.

T H E M E A N : A D D I N G V A L U E T O P O S I T I O N 79

There are cases in which the sensitivity of the mean to the values of

the categories in a measure can give misleading results. For example,

let’s say that one case in your study is very different from the others. As

noted in Chapter 1, researchers call such a case an outlier, because it is

very much outside the range of the other cases you studied. Taking the

example of minutes of disorder from Table 4.5, let’s say that the last case

had 70 minutes of disorder (the maximum amount possible) rather than

6.57 minutes. When we calculate the mean now, the sum of the cases is

much larger than before:

W orking It Out

� 134.99

� 3.51 � 3.72 � 4.09 � 4.47 � 4.64 � 4.65 � 70.0

� 2.21 � 2.34 � 2.45 � 2.66 � 3.04 � 3.19 � 3.23 � 3.46

� 1.15 � 1.19 � 1.19 � 1.48 � 1.60 � 1.63 � 2.02 � 2.12

� 0.35 � 0.42 � 0.46 � 0.47 � 0.52 � 0.67 � 1.00 � 1.06

� N

i�1

Xi � � 31

i�1

Xi

Dividing this sum by the total number of cases provides us with a mean

of 4.35 (rounded to the second decimal place):

W orking It Out

� 4.3545

� 134.99

31

X � � N

i�1

Xi

N

The mean we calculated with the original score was 2.31 (see page 76).

Accordingly, merely by changing one score to an outlier, we have almost

doubled our estimate of typicality. In this case, the sensitivity of the

mean to an extreme value in the distribution led it to overestimate the

C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

average case. This illustrates the general principle that the mean is sensi-

tive to outliers. Because the mean is used to develop many other more

complex statistics, this principle is relevant not only to the mean itself

but also to a number of other important statistical techniques used by re-

searchers.

So what should you do if outliers lead to a misleading conclusion re-

garding typicality in your study? One solution is simply to exclude the

outliers from specific analyses and let your readers or audience know

that some cases have been excluded and why. If the number of extreme

cases is large enough, you may want to analyze these cases separately.

Other Characteristics of the Mean

Two other traits of the mean are important because they play a role in

how we develop other statistics. The first concerns what happens when

we look at deviations (or differences) from the mean. This will be-

come an issue in the next chapter, when we discuss measures of disper-

sion. The second, often termed the least squares property of the

mean, will become important to us in Chapter 15, when we discuss re-

gression.

If we take each score in a distribution, subtract the mean from it, and

sum these differences, we will always get a result of 0. In equation form,

this principle is represented as follows:

Equation 4.3

In English, this equation says that if we sum the deviations from the

mean, from the first to the last case, we will always get a result of 0. This

principle is illustrated in Table 4.8, using the data on minutes of public

disorder from Table 4.5. Here we have taken the 31 scores and sub-

tracted the mean from each one. We then added these differences. Be-

cause the positive scores balance out the negative ones, the result is 0.

This will always happen when we use the mean.

The second trait, the least squares property, is very important for un-

derstanding regression analysis (introduced in Chapter 15), a technique

commonly used for describing relationships among variables in criminal

justice. For the moment, it is enough to note this fact and that the issues

� N

i�1

(Xi � X ) � 0

Another solution is to transform the outliers. That is, you may want to

highest value that is not an outlier). In this way, you can include the

replace them with values closer to the rest of the distribution (e.g., the

cases, but minimize the extent to which they affect your estimate of

typicality. However, you should be cautious in developing such trans-

formations of your scores, keeping in mind that you are changing the

character of the distribution examined in your study.

80

T H E M E A N : A D D I N G V A L U E T O P O S I T I O N

we address early on in statistics are often the bases for much more com-

plex types of analysis. “Don’t forget the basics” is a good rule. Many mis-

takes that researchers make in developing more complex statistics come

from a failure to think about the basic issues raised in the first few chap-

ters of this text.

The least squares property is written in equation form as follows:

Equation 4.4

What this says in English is that if we sum the squared deviations from

the mean for all of our cases, we will get the minimum possible result.

That is, suppose we take each individual’s score on a measure, subtract

� N

i�1

(Xi � X )2 � minimum

Deviations from the Mean for Minutes of Public Disorder (A)

SCORE (X) DEVIATION FROM THE MEAN (Xi � )

0.35 0.35 � 2.31 � �1.96 0.42 0.42 � 2.31 � �1.89 0.46 0.46 � 2.31 � �1.85 0.47 0.47 � 2.31 � �1.84 0.52 0.52 � 2.31 � �1.79 0.67 0.67 � 2.31 � �1.64 1.00 1.00 � 2.31 � �1.31 1.06 1.06 � 2.31 � �1.25 1.15 1.15 � 2.31 � �1.16 1.19 1.19 � 2.31 � �1.12 1.19 1.19 � 2.31 � �1.12 1.48 1.48 � 2.31 � �0.83 1.60 1.60 � 2.31 � �0.71 1.63 1.63 � 2.31 � �0.68 2.02 2.02 � 2.31 � �0.29 2.12 2.12 � 2.31 � �0.19 2.21 2.21 � 2.31 � �0.10 2.34 2.34 � 2.31 � �0.03 2.45 2.45 � 2.31 � �0.14 2.66 2.66 � 2.31 � �0.35 3.04 3.04 � 2.31 � �0.73 3.19 3.19 � 2.31 � �0.88 3.23 3.23 � 2.31 � �0.92 3.46 3.46 � 2.31 � �1.15 3.51 3.51 � 2.31 � �1.20 3.72 3.72 � 2.31 � �1.41 4.09 4.09 � 2.31 � �1.78 4.47 4.47 � 2.31 � �2.16 4.64 4.64 � 2.31 � �2.33 4.65 4.65 � 2.31 � �2.34 6.57 6.57 � 2.31 � �4.26 Total (�) 0*

*Because of rounding error, the actual column total is slightly less than zero.

X

Table 4.8

81

C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

the mean from that score, and then square the difference. If we then

sum all of these values, the result we get will be smaller than the result

we would have gotten if we had subtracted any other number besides

the mean. You might try this by calculating the result for minutes of dis-

order using the mean. Then try other values and see if you can find

some other number of minutes that will give you a smaller result. The

least squares property says you won’t.

Using the Mean for Noninterval Scales

The mean is ordinarily used for measuring central tendency only with in-

terval scales. However, in practice, researchers sometimes use the mean

with ordinal scales as well. Is this wrong? In a pure statistical sense, it is.

However, some ordinal scales have a large number of categories and

thus begin to mimic some of the characteristics of interval-level mea-

sures.

This is particularly true in cases where the movements from one cate-

gory to another in an ordinal scale can be looked at as equivalent, no

matter which category you move from. Taking our example of student

attitudes toward public drunkenness in Table 4.4, a researcher might

argue that the difference between “somewhat serious” and “a bit serious”

is about equivalent to that between “very serious” and “extremely seri-

ous,” and so forth. Thus, the difference between these categories is not

just a difference of position; it is also a movement of equal units up the

scale. Taking this approach, we can say that this measure takes into ac-

count both position and value, although the values here are not as

straightforward as those gained from true interval scales such as number

of crimes or dollar amount stolen.

A researcher might argue that the mean is appropriate for presenting

findings on views of public drunkenness because this ordinal-scale mea-

sure of attitudes is like an interval-scale measure. Although it is easy to

see the logic behind this decision, it is important to note that such a de-

cision takes a good deal of justification. In general, you should be very

cautious about using the mean for ordinal-level scales, even when the

above criteria are met.

S t a t i s t i c s i n P r a c t i c e : C o m p a r i n g t h e M e d i a n a n d t h e M e a n

The general rule is that the mean provides the best measure of central

tendency for an interval scale. This follows a principle stated in Chap-

ter 1: In statistics, as in other decision-making areas, more information

is better than less information. When we use the mean, we take into

82

S T A T I S T I C S I N P R A C T I C E

tion, but also the values or scores of those categories. Because more in-

formation is used, the mean is less likely than other measures of central

tendency to be affected by changes in the nature of the sample that a re-

searcher examines. It is useful to note as well that the mean has some al-

gebraic characteristics that make it more easily used in developing other

types of statistics.

The mean is generally to be preferred, but when the distribution of a

variable is strongly skewed, the median provides a better estimate of

central tendency than the mean. “Skewed” means that the scores on the

cases. A distribution that has extreme values lower than the main cluster

of observations (i.e., there is a “tail” to the left in the distribution) is said

to be negatively skewed, while a distribution that has extreme values

greater than the main cluster of observations (i.e., there is a “tail” to the

right in the distribution) is said to be positively skewed.4

A good example of a skewed distribution in criminal justice is crimi-

nal history as measured by self-reports of prisoners. Horney and Mar-

shall, for example, reported results on the frequency of offending for a

sample of prisoners.5 As is apparent from Figure 4.1, most of the offend-

ers in their sample had a relatively low offending rate—between 1 and

20 offenses in the previous year. But a number of offenders had rates of

more than 100, and a fairly large group had more than 200. The mean

for this distribution is 175.

Clearly, 175 offenses provides a misleading view of typical rates of of-

fending for their sample. Because the mean is sensitive to value, it is in-

flated by the very high frequency scores of a relatively small proportion

of the sample. One solution suggested earlier to the problem of outliers

4A formal statistic for measuring the degree of skewness of a distribution is given by

the following equation:

In words, this equation tells us to take the deviation between a value and the mean

and cube it, then sum these values over all observations; the sum of the cubed devia-

tions is then divided by the sample size (N) multiplied by the standard deviation

cubed. The measure of skewness will have a value of 0 if the distribution is symmetri-

cal, a negative value if the distribution is negatively skewed, and a positive value if

the distribution is positively skewed. The greater the value of the measure, the greater

the degree of positive or negative skewness. 5J. Horney and I. H. Marshall, “An Experimental Comparison of Two Self-Report Meth-

ods for Measuring Lambda,” Journal of Research in Crime and Delinquency 29 (1992):

102–121.

skewness � (Xi � X )3

Ns3

account not only the frequency of events in each category and their posi-

variable are very much weighted to one side and that frequencies of

extreme values trail off in one direction away from the main cluster of

83

� N

i�1

C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

was to exclude such cases. But here, this would mean excluding almost

30% of the sample. Thus, these are not outliers in the traditional sense.

Another option mentioned earlier is to analyze the “outliers” separately.

But again, there is quite a spread of scores even if we look at those

above 50 or 100 separately, and the analysis of the outliers might in itself

provide a misleading view of central tendency. A common solution used

for describing this type of skewed interval-level distribution is to use the

median rather than the mean to describe central tendency. The median

for this distribution is 4, which is certainly more representative of the av-

erage case than is the mean. But even if you choose this solution, it is

very important to note to your audience that the distribution is skewed

and to tell them a bit about the nature of the distribution.

How should you decide when a distribution is so skewed that it is

preferable to use the median as opposed to the mean? You should begin

by comparing the mean and the median. When there is a very large dif-

ference between them, it may be the result of skewness. In such cases,

you should look at the distribution of the scores to see what is causing

the mean and median to differ widely. But there is no solid boundary

line to guide your choice.

cases where the mean and median provide relatively close estimates,

P er

ce n

t of

R es

p on

d en

ts

Frequency of Offending

1–20

0

20

40

60

80

21–40 41–60 61–80 81–100 101–120 121–140 141–160 161–180181–200 200+

Individual Frequency of Offending for a Sample of Offenders:

A Case Where the Mean Is a Misleading Measure of Central Tendency Figure 4.1

your choice will be clear. In the former case you would choose the

84

In extreme cases (such as that of criminal history in our example) or

C H A P T E R S U M M A R Y

where in between, you will have to use common sense and the experi-

ences of other researchers working with similar data as guidelines. What

seems to make sense? What have other researchers chosen to do? One

way of being fair to your audience is to provide results for both the

mean and the median, irrespective of which you choose as the best mea-

sure of typicality.

C h a p t e r S u m m a r y

The mode is calculated by identifying the category that contains the

greatest number of cases. It may be applied to any scale of measure-

ment. Because the mode uses very little information, it is rarely used

with scales of measurement higher than the nominal scale. It can occa-

sionally serve as a useful summary tool for higher-level scales, however,

when a large number of cases are concentrated in one particular cate-

gory.

The median is calculated by locating the middle score in a distribu-

tion and identifying in which category it falls. It is also known as the

50th percentile score, or the score below which 50% of the cases lie. The

information used includes both the number of cases in a particular cate-

gory and the positions of the categories. The median uses more informa-

tion than does the mode and requires a scale of measurement that is at

least ordinal in magnitude.

The mean is calculated by dividing the sum of the scores by the

number of cases. The information used includes not only the number of

cases in a category and the relative positions of the categories, but also

the actual value of each category. Such information normally requires at

least an interval scale of measurement. For this reason, the researcher

should be cautious about using the mean to describe an ordinal scale.

The mean uses more information than the mode and the median. It is,

however, sensitive to extreme cases—outliers. Faced with the distorting

effect of outliers, the researcher may choose to keep them, to transform

them to other values, or to delete them altogether. If a distribution of

scores is substantially skewed, then it may be more appropriate to use

the median than to use the mean.

The sum derived by adding each score’s deviation from the mean

will always be 0. If the deviation of each score from the mean is

squared, then the sum of these squares will be less than it would be if

any number other than the mean were used. This is called the least

squares property.

median, and in the latter the mean. However, when your results fall some-

85

86 C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

K e y T e r m s

deviation from the mean The extent to

which each individual score differs from

the mean of all the scores.

least squares property A characteristic

of the mean whereby the sum of all the

squared deviations from the mean is a

minimum—it is lower than the sum of the

squared deviations from any other fixed

point.

mean A measure of central tendency cal-

culated by dividing the sum of the scores

by the number of cases.

median A measure of central tendency cal-

culated by identifying the value or category

of the score that occupies the middle posi-

tion in the distribution of scores.

mode A measure of central tendency cal-

culated by identifying the score or category

that occurs most frequently.

outlier(s) A single or small number of ex-

ceptional cases that substantially deviate

from the general pattern of scores.

skewed Describing a spread of scores that

is clearly weighted to one side.

S y m b o l s a n d F o r m u l a s

X Individual score

Mean

N Number of cases

� Sum

To calculate the median observation:

To calculate the mean:

To show how the sum of the deviations from the mean equals 0:

To express the least squares property:

� N

i�1

(Xi � X)2 � minimum

� N

i�1

(Xi � X) � 0

X � � N

i�1

Xi

N

Median observation � N � 1

2

X

E X E R C I S E S 87

E x e r c i s e s

4.1 Drivers cited for moving violations are required by a state’s laws to take a driving safety course taught by the local police department. The sign-in sheet asks individuals to note why they received a ticket. The 14 participants at a recent class noted the following:

Speeding, Running a red light, Running a stop sign, Speeding, Speeding, Running a red light, Tailgating, Speeding, Running a red light, Recklessness, Speeding, Running a red light, Speeding, Running a stop sign

a. Categorize these data and calculate the mode.

b. Explain why the median would not be an appropriate measure of central tendency for these data.

4.2 Calculate the mode, median, and mean for the following data:

a. Number of previous employments held by 25 convicts:

3 3 1 1 0 1 0 2 1 0 8 4 3

1 2 1 9 0 1 7 0 7 2 0 1

b. Weeks of training undergone by 20 prison guards:

10 16 12 16 16 16 10 8 10 12

16 18 12 16 16 8 0 12 10 16

c. Height (in meters) of 30 convicts:

1.72 1.78 1.73 1.70 1.81 1.64 1.76 1.72 1.75 1.74

1.88 1.79 2.01 1.80 1.77 1.79 1.69 1.74 1.75 1.66

1.77 1.73 1.72 1.91 1.80 1.74 1.72 1.82 1.86 1.79

4.3 A researcher checked the response times of police to ten emergency telephone calls. The data below record the number of minutes that elapsed from when the telephone call ended to when the police arrived:

24 26 14 27 198 22 27 17 19 29

a. Calculate the mode, the median, and the mean.

b. Which of these measures is the most suitable for this particular case? Explain your choice.

4.4 Airport officials wished to check the alertness of their security officers over the two busiest weeks of the summer. During this period, they sent out 50 undercover staff carrying suspicious items of hand lug- gage. Five of them were stopped at the entrance to the airport. Six made it into the airport, but were stopped at check-in. Thirteen more got into the airport and through check-in, only to be stopped at the

88 C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

hand-luggage inspection point. Two passed the airport entrance, check-in, and hand-luggage inspection, but were stopped when presenting their boarding cards at the gate. Four people made it past every one of these stages, only to be stopped when boarding the plane. Twenty of the undercover staff were not detected at all.

a. Categorize the data and calculate the median category.

b. Is the median a good measure of central tendency in this case? Ex- plain your answer. If you think it is not, suggest an alternative and explain why.

4.5 On the first day of the term in a statistics course, the professor admin- istered a brief questionnaire to the students, asking how many statis- tics courses they had ever taken before the current term. Of the 33 students who answered the question, 17 said none, 9 said one, 3 said two, 2 said three, 1 said four, and 1 said five.

a. Calculate the mode, median, and mean for number of prior statis- tics classes.

b. Which one of these measures of central tendency best measures the typicality of these data?

4.6 As part of her undergraduate thesis, a criminal justice student asked ten other criminal justice majors to rate the fairness of the criminal jus- tice system. The students were asked to say whether they strongly agreed, agreed, were uncertain, disagreed, or strongly disagreed with the following statement: “The criminal justice system in our country treats all defendants fairly.” The ten responses were

Strongly agree, Strongly agree, Strongly disagree, Strongly disagree, Uncertain, Disagree, Disagree, Agree, Strongly disagree, Uncertain

a. Categorize these data and calculate an appropriate measure of cen- tral tendency.

b. Explain why this measure of central tendency best represents the typicality of these data.

4.7 There are five prisoners in the high-security wing in a prison—Albert, Harry, Charlie, Dave, and Eddie. Only Eddie’s biographical details have been lost. The information available is as follows:

Age Previous Convictions

Albert 23 1

Harry 28 4

Charlie 18 1

Dave 41 1

Eddie ? ?

E X E R C I S E S 89

a. Can we compute any of the following for previous convictions of the five prisoners: the mode, the median, or the mean? If any (or all) of these three measures may be calculated, what are their values?

b. If the mean number of previous convictions is 2.0, how many con- victions does Eddie have?

c. If we know that the median age for the five prisoners is 28, what does this tell us about Eddie’s age? Explain why.

d. If the mean age for the five prisoners is 28.2, how old is Eddie?

4.8 A researcher sat on a bench along the main shopping street of a city center on ten successive Saturdays from 11:00 A.M. to 2:00 P.M.—the three busiest shopping hours of the day—and recorded the number of times a police officer passed by. The results for the ten weeks are as follows:

Week No.: 1 2 3 4 5 6 7 8 9 10

No. of Police Officers Observed: 4 3 6 4 4 5 4 35 3 5

On week 8, local unionists held a demonstration in the city center, and the high number of observations for that week can be explained by the extra officers called in to police the rally.

b. Which measure of central tendency best represents typicality for these data? Discuss the issues involved in choosing the most appro- priate means of describing the data.

c. Imagine that the unionists had decided to hold a regular demon- stration in the city center on alternating weeks. The results recorded for the same study would be as follows:

Week No.: 1 2 3 4 5 6 7 8 9 10

No. of Police Officers Observed: 4 30 6 31 6 52 4 35 4 34

4.9 On a recent evening, a police crackdown on prostitution solicitation resulted in 19 arrests. The ages of the persons arrested were

17 18 24 37 32 49 61 20 21 21

25 24 24 26 30 33 35 22 19

a. Calculate an appropriate measure of central tendency.

b. Explain why this measure of central tendency best represents typi- cality for these data.

officers observed. a. Calculate the mode, median, and mean for the number of police

Would the measure of central tendency you recommended in part b still be the best measure of typicality? Explain why.

C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E

4.10 Using your answers from part a of Exercise 4.5, calculate

a. The sum of the deviations from the mean.

b. The sum of the squared deviations from the mean.

c. The sum of the squared deviations from the median.

d. The sum of the squared deviations from the mode.

e. Which of these sums of squared deviations has the smallest value?

90

C o m p u t e r E x e r c i s e s

Measures of central tendency are reported along with many other statistics in

many of the software programs you might encounter. The commands that we

describe below will be the same as those that we highlight at the end of

Chapter 5.

SPSS

There are two primary ways of obtaining measures of central tendency in SPSS.

The quickest way to obtain information on the mean for one or more variables

measured at the interval level of measurement is to use the DESCRIPTIVES

command

The output window will contain quite a bit of information for each variable

that you have named—much of it will make more sense after reading Chapter

5. It is important to note that in regard to measures of central tendency, the

DESCRIPTIVES command will report only the mean, not the mode or the

median. Since the only measure of central tendency this command will calculate

is the mean, this command is generally useful only for interval-level data. This

command is most useful when you are working with a data set that contains

almost exclusively interval-level variables.

An alternative to obtaining measures of central tendency is to again use the

FREQUENCIES command discussed at the end of Chapter 3. As you may recall

from the previous chapter’s computer exercises, this command will produce

frequency distributions for the variables whose names are included in the list of

variables. To obtain measures of central tendency for the variables of interest, we

simply need to add an option requesting these values:

DESCRIPTIVES VARIABLES = variable_names.

FREQUENCIES VARIABLES = variable_names

/FORMAT = NOTABLE

/STATISTICS = MEAN MEDIAN MODE.

C O M P U T E R E X E R C I S E S 91

Where the /STATISTICS = option lists the three measures of central tendency

we are interested in. The /FORMAT = NOTABLE option suppresses the print-

ing of the frequency distributions for all the variables included in the list. Should

you want the frequency distributions, just omit this line. Also, the separation of

the command across three lines is simply for ease of reading—all of this material

can be included on a single line in a syntax file. The formatting on this page of

the text would have made it difficult to read what was being done within SPSS.

After running this command, the output window will contain a box labeled

“Statistics.” Each column of this table refers to a separate variable. As you move

down the rows, you should see the reported values for the mode, the median,

and the mean for each variable.

Caution: The mode and the median are listed as numbers, even though the data

may be nominal or ordinal and you have entered value labels. To report correctly

the value of the mode or median, you need to report the category represented

by that number. For example, suppose you had analyzed the variable labeled

“gender,” where males were coded as 1 and females as 2, the mode would be

reported by SPSS as either 1 or 2, but it would be up to you to report correctly

whether the modal category was male or female–not a 1 or a 2.

Specific examples of the use of each of these commands are provided in the

accompanying SPSS syntax file for Chapter 4 (Chapter_4.sps).

Stata

In Stata, there are also two primary ways of obtaining median and mean—both

methods are fairly straightforward. The summarize command is

The basic output will include the mean and other measures on the variables

included in the list. To obtain the median, you will need to add the detail option:

You should note that the output will label the median as the 50th percentile.

Alternatively, if we want to avoid looking at a variety of other statistics that

may not be of interest, we can use the tabstat command and explicitly ask for

only the median and the mean:

The output will list the variables named across column and the median and the

mean will appear in separate rows of the table.

The mode is not reported in any of the standard Stata output, but is easily

determined by running the tab1 command to obtain a frequency distribution

that was described in the Computer Exercises at the end of Chapter 3.

Specific examples of the use of each of these commands are provided in the

accompanying Stata do file for Chapter 4 (Chapter_4.do).

summarize variable_names

summarize variable_names, detail

tabstat variable_names, statistics ( median mean)

C H A P T E R F O U R : D E S C R I B I N G T H E T Y P I C A L C A S E92

Recoding Variables

A common situation in statistical analysis is the need to recode the values for

some variable included in our data file. There are many reasons for why we may

need to recode, such as collapsing categories to simplify the categories of a vari-

able or defining some values as “missing” or inappropriate for our analysis.

The recode commands in SPSS and Stata are quite similar. In SPSS,

RECODE takes the following form:

The structure of RECODE will have a series of old and new values listed in

parentheses and can refer to specific values, ranges of values as well as missing

values. For the values that are not going to be changed, the (ELSE = COPY)

ensures that they are copied to the new variable. We would also like to empha-

size the importance of using new variable names to contain the recoded

values—it is good practice and helps to protect the data file that you are working

with. More than one researcher has made the mistake of sending the recodes

back into the original variable, realizing a mistake was made, and just damaged

the data file that was being used and the need to start over.

The EXECUTE command following the RECODE command is necessary to

force SPSS to perform the recodes now, rather than waiting for a call to another

procedure and performing the recodes at that time.

In Stata, the recode command takes the form

The recoding of old to new values occurs in however many parentheses are

required, just as in SPSS. There is no need to refer to the other values in the

original variable—they are automatically carried over to the new variable.

Of special use is the option of coding one or more values as missing. For

example, a person responding to a survey writes down an incorrect number that

is beyond the range of acceptable responses. In SPSS, system missing values are

noted by SYSMIS in the RECODE command. In Stata, a period (.) denotes a

missing value.

Problems

1. Open the NYS data file (nys_1.sav, nys_1_student.sav, or nys_1.dta).

Consider the level of measurement for each variable, and then compute

and report appropriate measures of central tendency for each variable.

RECODE variable_name (old = new)(old = new) (ELSE = Copy) INTO

new_variable_name.

EXECUTE.

recode variable_name (old = new), gen(new_variable_name)

Following all of the recodes in parentheses, the creation of a new variable is noted by

adding a comma to the command and then the gen(new_variable_name)

command to generate a new variable that contains the recoded values.

93

2. Do any of the variables included in the data file appear to have potential

outliers? (You may want to consult your histograms from the Chapter 3

computer exercises or create histograms now for the interval-level vari-

ables included in the data file.)

3. If you find one or more potential outliers, take the following steps to

investigate their effect on the measures of central tendency.

a. Use the recode command to create two new variables: one that recodes

the outliers as “System Missing” values and one that recodes the

outliers as the next smaller value.

Hints:

b. Using your variable that recodes the outliers as missing, report how

the values of the mean and the median change when potential outliers

are removed from the analysis.

c. Using your variable that recodes the outliers as the next smaller value,

report how the values of the mean and the median change when

potential outliers are made less extreme.

d. Which approach for handling potential outliers do you think is more

appropriate for analyzing these data? Explain why. Faced with potential

outliers, which measure of central tendency would you report?

C O M P U T E R E X E R C I S E S

� You will need to look at frequency distributions to determine the

maximum and next to maximum values.

� Keep in mind that SPSS will look for SYSMIS in the RECODE

command, while Stata will look for a period (.) in the recode command.

How Typical Is the Typical Case?:

Measuring Dispersion

What Do They Tell Us About Our Data?

a n d o r d i n a l s c a l e s : p r o p o r t i o n s ,

p e r c e n t a g e s , a n d t h e v a r i a t i o n r a t i o

r a n g e , v a r i a n c e , a n d s t a n d a r d d e v i a t i o n

C h a p t e r f i v e

M e a s u r e s o f d i s p e r s i o n

M e a s u r i n g d i s p e r s i o n i n n o m i n a l

M e a s u r i n g d i s p e r s i o n i n i n t e r v a l s c a l e s :

How are They Calculated?

What are Their Characteristics?

How are They Calculated?

What are Their Characteristics?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_5,

© Springer Science+Business Media New York 2014

MEASURES OF CENTRAL TENDENCY provide a snapshot of the typical

case; however, the same statistic may be obtained from samples or pop-

ulations that are in fact quite dissimilar. For example, a sample of police

recruits with a mean or median age of 23 is not likely to include people

younger than 18 or older than 30, because most police departments have

age requirements for incoming officers. A sample of offenders with a

mean or median age of 23, however, will include offenders younger than

18 and much older than 30. In both these samples, the average person

studied is 23 years old. But the sample of offenders will include more

younger and older people than the sample of police recruits. The ages of

the offenders are dispersed more widely around the average age.

Measures of dispersion allow us to fill a gap in our description of the

samples or populations we study. They ask the question: How typical is

the typical case? They tell us to what extent the subjects we studied are

similar to the case we have chosen to represent them. Are most cases

clustered closely around the average case? Or, as with the sample of of-

fenders above, is there a good deal of dispersion of cases both above

and below the average?

M e a s u r e s o f D i s p e r s i o n f o r N o m i n a l - a n d O r d i n a l - L e v e l D a t a

With nominal scales, we define the typical case as the category with

the largest number of subjects. Accordingly, in Chapter 4 we chose

“private attorney” as the modal category for legal representation for a

sample of white-collar offenders. But how would we describe to what

extent the use of a private attorney is typical of the sample as a whole?

Put another way, to what degree are the cases concentrated in the

modal category?

95

C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

The Proportion in the Modal Category

The most straightforward way to answer this question is to describe the

proportion of cases that fall in the modal category. Recall from Chapter 3

that a proportion is represented by the following equation:

Accordingly, we can represent the proportion of cases in the modal cate-

gory using Equation 5.1:

Equation 5.1

That is, we take the number of cases in the modal category and divide it

by the total number of cases in the sample.

Taking the example of legal representation, we divide the N of cases

in the modal category (private attorney) by the total N of cases in the

sample (see Table 5.1):

Proportion � Nmodal cat.

Ntotal

Proportion � Ncat

Ntotal

Legal Representation for White-Collar Crime

CATEGORY FREQUENCY (N )

No Attorney 20 Legal Aid 26 Court Appointed 92 Public Defender 153 Private Attorney 380 Total (�) 671

Table 5.1

W orking It Out

� 0.5663

� 380

671

Proportion � Nmodal cat.

Ntotal

Following our earlier suggestions regarding rounding to the second deci-

mal place, we say that the proportion of white-collar offenders in the

modal category was about 0.57.

96

M E A S U R E S O F D I S P E R S I O N F O R N O M I N A L A N D O R D I N A L D A T A

Table 5.2 presents information about the method of execution used

on the 683 persons executed in the United States from 1977 to 2000. The

modal category is lethal injection, so the proportion in the modal cate-

gory is found by dividing the N of cases in that category by the total N of

cases:

Method of Execution in the United States, 1977–2000

CATEGORY FREQUENCY (N )

Lethal Injection 518 Electrocution 149 Lethal Gas 11 Hanging 3 Firing Squad 2 Total (�) 683

Source: Tracy L. Snell, “Capital Punishment 2000,” Bureau of Justice Statistics Bulletin, 2001, p. 12.

Table 5.2

W orking It Out

� 0.7584

� 518

683

Proportion � Nmodal cat.

Ntotal

Of persons executed in the United States from 1977 to 2000, the propor-

tion killed through lethal injection was about 0.76.

The Percentage in the Modal Category

Alternatively, we may refer to the percentage in the modal category.

Most people find percentages easier to understand than proportions. Re-

call that a percentage is obtained by taking a proportion and multiplying

it by 100. Accordingly, we can take Equation 5.1 and multiply the result

by 100 to get the percentage of cases in the modal category.

Percentage � Nmodal cat.

Ntotal

� 100

97

C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

For our legal representation example,

W orking It Out

� 56.6319

� 380

671 � 100

Percentage � Nmodal cat.

Ntotal

� 100

That is, about 57% of the cases in the sample fall in the modal category.

Similarly, for the method of execution example, the percentage in the

modal category is

The Variation Ratio

Another way to describe the degree to which the modal category repre-

sents the cases in a sample is to use a statistic called the variation ratio

(VR). The variation ratio is based on the same logic as a proportion, but

it examines the extent to which the cases are spread outside the modal

category, rather than concentrated within it. The proportion of cases in

the modal category is subtracted from 1:

Equation 5.2

For the legal representation example, the variation ratio is

VR � 1 � �Nmodal cat.

Ntotal �

� 75.8419

Percentage � 518

683 � 100

W orking It Out

� 0.4337

� 1 � �380

671� VR � 1 � �Nmodal cat.

Ntotal �

involved the use of lethal injection.

About 76% of all executions in the United States from 1977 to 2000

98

M E A S U R E S O F D I S P E R S I O N F O R N O M I N A L A N D O R D I N A L D A T A 99

The variation ratio for legal representation in this sample of white-collar

offenders is about 0.43. But what does this say about the extent to

which cases in the sample are clustered around the typical case? Is a

variation ratio of 0.43 large or small? What rule can we use for deciding

more generally whether the distribution we are examining is strongly

clustered?

One approach is to define at the outset the upper and lower limits for

the variation ratio or proportion for a particular measure. Obviously, the

largest proportion, regardless of the study, is 1.0, which would mean that

all of the cases were in the modal category. Having all of the cases in

the modal category would lead to a variation ratio of 0, indicating no

dispersion.

The smallest proportion (or largest VR) depends, however, on the

number of categories in your measure. The mode is defined as the cate-

gory in your measure with the most cases, so it must have at least one

more case than any other category. If you have only two categories, then

the modal category must include one more than half of the cases in your

study. So, in the instance of two categories, the least possible concentra-

tion is just over 0.50 of the cases. The least possible dispersion, as mea-

sured by the variation ratio, would be 1 minus this proportion, or just

under 0.50. If you have four categories, the modal category must have

more than one-quarter of the cases. Accordingly, the smallest variation

ratio would be a bit smaller than 0.75.

What about our example of legal representation? We have five cate-

gories and 671 cases. The smallest number of cases the modal cate-

gory could have with these numbers is 135. In this instance, each of

the other four categories would have 134 cases. This is the maximum

amount of dispersion that could exist in this sample, and it amounts to

about 20.12% of the total number of cases in the sample, or a variation

ratio of 0.7988. As noted earlier, the greatest degree of concentration

in the modal category would yield a proportion of 1 and a variation

ratio of 0. The estimates we calculated for legal representation

(proportion � 0.57; VR � 0.43) lie somewhere between these two

extremes.

Is this dispersion large or small? As with many of the statistics we will

examine, the answer depends on the context in which you are working.

“Large” or “small” describes a value, not a statistical concept. Statistically,

you know that your estimate falls somewhere between the largest possi-

ble degree of concentration and the largest possible degree of disper-

sion. But whether this is important or meaningful depends on the prob-

lem you are examining and the results that others have obtained in prior

research.

For example, if, in a study of legal representation for white-collar

crime in England, it had been found that 90% of the cases were concen-

trated in the private attorney category, then we might conclude that our

C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

results reflected a relatively high degree of dispersion of legal represen-

tation in the United States. If, in England, only 25% of the cases had

been in the modal category, we might conclude that there was a rela-

tively low degree of dispersion of legal representation in the United

States.

The proportion and the variation ratio are useful primarily for describ-

ing dispersion with nominal-level measures. In some circumstances,

however, they can be useful for describing ordinal-level variables as

well. This is true primarily when there are just a few categories in a mea-

sure or when there is a very high degree of concentration of cases in

one category. The problem in using a simple proportion or variation

ratio for ordinal-level measures is that the mode, upon which these sta-

tistics are based, is often a misleading measure for ordinal scales. As dis-

cussed in Chapter 4, the mode does not take into account the positions

of scores in a measure, and thus it may provide a misleading view of the

average case.

Index of Qualitative Variation

One measure of dispersion that is not based on the mode—and that

can be used for both nominal and ordinal scales—is the index of

qualitative variation (IQV). The IQV compares the amount of varia-

tion observed in a sample to the total amount of variation possible,

given the number of cases and categories in a study. It is a standard-

ized measure. This means that whatever the number of cases or cate-

gories, the IQV can vary only between 0 and 100. An IQV of 0 means

that there is no variation in the measure, or all of the cases lie in one

category. An IQV of 100 means that the cases are evenly dispersed

across the categories.

Equation 5.3

Equation 5.3 provides a guide for how to compute the IQV. You are

already familiar with the summation symbols within the parentheses.

Here we are summing not across cases, but across products of distinct

categories. Nobs represents the number of cases we observe within a cate-

gory in our study. Nexp represents the number of cases we would expect

in a category if the measure were distributed equally across the cate-

gories. That is, it is the N we would expect if there were the maximum

amount of dispersion of our cases. We use the subscripts i, j, and k as a

IQV � �� k�1

i�1

� k

j�i�1

Nobsi Nobsj

� k�1

i�1

� k

j�i�1 Nexpi

Nexpj � � 100

100

M E A S U R E S O F D I S P E R S I O N F O R N O M I N A L A N D O R D I N A L D A T A

shorthand way to say that we should multiply all of the potential pairs of

categories. Here’s how this works: k represents the total number of cate-

gories of a variable. In the legal representation example, k � 5. Sub-

scripts i and j index the categories of the variable. Use of the subscripts i

and j provides us with a way of keeping track and making sure that we

have multiplied all possible pairs of observed frequencies from each of

the categories.

For example, if a variable had three categories, then the numerator

(the measure of observed variation) would be equal to

If a variable had four categories, then the numerator would be equal to

A concrete example will make it much easier to develop this statistic

in practice. Let’s say that we wanted to describe dispersion of an ordinal-

scale measure of fear of crime in a college class of 20 students. The stu-

dents were asked whether they were personally concerned about crime

on campus. The potential responses were “very concerned,” “quite con-

cerned,” “a little concerned,” and “not concerned at all.” The responses

of the students are reported under the “N observed” column in Table 5.3.

As you can see, the cases are fairly spread out, although there are more

students in the “very concerned” and “quite concerned” categories than

in the “a little concerned” and “not concerned at all” categories. The ex-

pected number of cases in each category under the assumption of maxi-

mum dispersion is 5. That is, if the cases were equally spread across the

categories, we would expect the same number in each. Following Equa-

tion 5.3, we first multiply the number of cases observed in each category

by the number observed in every other category and then sum. We then

divide this total by the sum of the number expected in each category

Nobs1 Nobs2

� Nobs1 Nobs3

� Nobs1 Nobs4

� Nobs2 Nobs3

� Nobs2 Nobs4

� Nobs3 Nobs4

Nobs1 Nobs2

� Nobs1 Nobs3

� Nobs2 Nobs3

Fear of Crime Among Students

CATEGORY N OBSERVED N EXPECTED

Not Concerned at All 3 20/4 � 5 A Little Concerned 4 20/4 � 5 Quite Concerned 6 20/4 � 5 Very Concerned 7 20/4 � 5 Total (�) 20 20

Table 5.3

101

C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

multiplied by the number expected in every other category. This amount

is then multiplied by 100:

W orking It Out

� 96.6667

� �145 150� � 100

� �(3 � 4) � (3 � 6) � (3 � 7) � (4 � 6) � (4 � 7) � (6 � 7)

(5 � 5) � (5 � 5) � (5 � 5) � (5 � 5) � (5 � 5) � (5 � 5)� � 100

IQV � � k�1

i�1

� k

j�i�1

Nobsi Nobsj

� k�1

i�1

� k

j�i�1

Nexpi Nexpj

� 100

The observed variation is 145. The expected variation is 150, represent-

ing the maximum amount of dispersion possible for the measure. The

IQV for this measure is 96.67, meaning that the cases studied are very

dispersed among the categories of the measure.

M e a s u r i n g D i s p e r s i o n i n I n t e r v a l S c a l e s : T h e R a n g e , V a r i a n c e , a n d S t a n d a r d D e v i a t i o n

A common method of describing the spread of scores on interval or

higher scales is to examine the range between the highest and lowest

scores. Take, for example, the distribution of cases in Table 5.4. Let’s say

that this was a distribution of crime calls at hot spots over a one-year pe-

riod. In describing typicality in this distribution, we would report the

mean number of calls for the 12 places, which is 21.50. In describing

how dispersed the scores are, we would report that the scores range be-

tween 2 and 52, or that the range of scores is 50.

The range is very simple and easy to present. Its attraction lies pre-

cisely in the fact that everyone understands what a range represents.

However, the range is an unstable statistic because it uses very little of

the information available in interval-level scales. It bases its estimate of

dispersion on just two observations, the highest and lowest scores. This

means that a change in just one case in a distribution can completely

102

M E A S U R I N G D I S P E R S I O N I N I N T E R V A L S C A L E S

alter your description of dispersion. For example, if we changed the case

with the most calls in Table 5.4 from 52 to 502, the range would change

from 50 to 500.

One method for reducing the instability of the range is to examine

cases that are not at the extremes of your distribution. In this way, you

are likely to avoid the problem of having the range magnified by a few

very large or small numbers. For example, you might choose to look at

the range between the 5th and 95th percentile scores, rather than that

between the lowest and highest scores. It is also common to look at the

range between the 25th and 75th percentile scores or between the 20th

and 80th percentile scores. But however you change the points at which

the range is calculated, you still rely on just two scores in determining

the spread of cases in your distribution. The range provides no insight

into whether the scores below or above these cases are clustered to-

gether tightly or dispersed widely. Its portrait of dispersion for interval

scales is thus very limited.

How can we gain a fuller view of dispersion for interval scales? Re-

member that we became interested in the problem of dispersion because

we wanted to provide an estimate of how well the average case repre-

sented the distribution of cases as a whole. Are scores clustered tightly

around the average case or dispersed widely from it? Given that we have

already described the mean as the most appropriate measure of central

tendency for such scales, this is the natural place to begin our assess-

ment. Why not simply examine how much the average scores differ from

the mean?

In fact, this is the logic that statisticians have used to develop the

main measures of dispersion for interval scales. However, they are faced

with a basic problem in taking this approach. As we discussed in Chap-

ter 4, if we add up all of the deviations from the mean, we will always

Crime Calls at Hot Spots in a Year

HOT SPOT NUMBER NUMBER OF CALLS

1 2 2 9 3 11 4 13 5 20 6 20 7 20 8 24 9 27

10 29 11 31 12 52

Table 5.4

103

C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

come up with a value of 0. You can see this again by looking at the data

on crime calls at hot spots in Table 5.5. If we take the sum of the differ-

ences between each score and the mean, written in equation form as

the total, as expected, is 0.

As discussed in Chapter 4, when we add up the deviations above and

below the mean, the positive and negative scores cancel each other out.

In order to use deviations from the mean as a basis for a measure of dis-

persion, we must develop a method for taking the sign, or direction, out

of our statistic. One solution is to square each deviation from the mean.

Squaring will always yield a positive result because multiplying a posi-

tive number or a negative number by itself will result in a positive out-

come. This is the method that statisticians have used in developing the

measures of dispersion most commonly used for interval scales.

The Variance

When we take this approach, the variance (s 2) provides an estimate of

the dispersion around the mean. It is the sum of the squared deviations

from the mean divided by the number of cases. Written in equation

form, it is

Equation 5.4s 2 � � N

i�1

(Xi � X )2

N

� N

i�1

(Xi � X )

Deviations from the Mean for Crime Calls at Hot Spots in a Year

HOT SPOT NUMBER DEVIATIONS FROM

NUMBER OF CALLS THE MEAN (Xi � )

1 2 2 � 21.5 � �19.5 2 9 9 � 21.5 � �12.5 3 11 11 � 21.5 � �10.5 4 13 13 � 21.5 � �8.5 5 20 20 � 21.5 � �1.5 6 20 20 � 21.5 � �1.5 7 20 20 � 21.5 � �1.5 8 24 24 � 21.5 � 2.5 9 27 27 � 21.5 � 5.5

10 29 29 � 21.5 � 7.5 11 31 31 � 21.5 � 9.5 12 52 52 � 21.5 � 30.5

Total (�) � 0.0

X

Table 5.5

104

M E A S U R I N G D I S P E R S I O N I N I N T E R V A L S C A L E S

In practice, you must take the following steps to compute the variance

(as we do for our example in Table 5.6):

1. Take each case and subtract the mean from it, to get the deviation

from the mean. For our example of crime calls at hot spots, we first

take the case with 2 calls and subtract the mean of 21.5 from it, to

get a score of �19.5.

2. Square each of these scores. For the first case, our result is 380.25.

3. Sum the results obtained in step 2. For our example of hot spots of

crime, this yields a total of 1,839.

4. Finally, divide this result by the number of cases in the study. For our

12 cases, this leads to a variance of 153.25.1

Variance for Crime Calls at Hot Spots in a Year

HOT SPOT NUMBER

NUMBER OF CALLS (Xi � ) (Xi � )2

1 2 2 � 21.5 � �19.5 380.25 2 9 9 � 21.5 � �12.5 156.25 3 11 11 � 21.5 � �10.5 110.25 4 13 13 � 21.5 � �8.5 72.25 5 20 20 � 21.5 � �1.5 2.25 6 20 20 � 21.5 � �1.5 2.25 7 20 20 � 21.5 � �1.5 2.25 8 24 24 � 21.5 � 2.5 6.25 9 27 27 � 21.5 � 5.5 30.25

10 29 29 � 21.5 � 7.5 56.25 11 31 31 � 21.5 � 9.5 90.25 12 52 52 � 21.5 � 30.5 930.25

Total (�) � 0.0 Total (�) � 1,839.00

XX

Table 5.6

1If you are working with SPSS or another computer package, you will notice that the

result you get computing the variance by hand using this formula and the result pro-

vided by the computer package are slightly different. For example, SPSS computes a

variance of 167.18 for the distribution provided in Table 5.6. The difference develops

from the computer’s use of a correction for the bias of sample variances: 1 is sub-

tracted from the N in the denominator of Equation 5.4. The correction is used primar-

ily as a tool in inferential statistics and is discussed in Chapter 10. Though it is our

view that the uncorrected variance should be used in describing sample statistics,

many researchers report variances with the correction factor for sample estimates.

When samples are larger, the estimates obtained with and without the correction are

very similar, and thus it generally makes very little substantive difference which ap-

proach is used.

105

C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

amounts required for a group of 15 defendants. The mean bail amount is

$3,263.33. Following the same procedure as before, we subtract the

mean from each of the individual observations. These values are pre-

cases, we gain a variance of $6,984,155.56 for the dollar amount of bail.

W orking It Out

� 153.25

� 1,839

12

� � 12

i�1

(Xi � 21.5)2

12

s

2 � � N

i�1

(Xi � X )2

N

Variance for Bail Amounts for a Sample of Persons Arrested for Felonies

BAIL

DEFENDANT AMOUNT (Xi � ) (Xi � )2

1 500 �2,763.33 7,635,992.69 2 1,000 �2,263.33 5,122,662.69 3 1,000 �2,263.33 5,122,662.69 4 1,000 �2,263.33 5,122,662.69 5 1,200 �2,063.33 4,257,330.69 6 1,500 �1,763.33 3,109,332.69 7 2,500 �763.33 582,672.69 8 2,500 �763.33 582,672.69 9 2,500 �763.33 582,672.69

10 2,750 �513.33 263,507.69 11 5,000 1,736.67 3,016,022.69 12 5,000 1,736.67 3,016,022.69 13 5,000 1,736.67 3,016,022.69 14 7,500 4,236.67 17,949,372.69 15 10,000 6,736.67 45,382,722.69

Total (�) � 0.05 Total (�) � 104,762,333.33

XX

Table 5.7

sented in the third column. The squared deviations from the mean

appears at the bottom of the column. When we divide the total by the N of

appear in the fourth column, and the sum of the squared deviations

As another example, consider the data presented in Table 5.7 on bail

106

M E A S U R I N G D I S P E R S I O N I N I N T E R V A L S C A L E S

With the variance, we now have a statistic for computing dispersion

based on deviations from the mean. However, how can we interpret

whether the variance for a distribution is large or small? If you are having

trouble making sense of this from our two examples, you are not alone.

While squaring solves one problem (the fact that the raw deviations from

the mean sum to 0), it creates another. By squaring, we generally obtain

numbers that are much larger than the actual units in the distributions

we are examining.2

The Standard Deviation

Another measure of dispersion based on the variance provides a solution

to the problem of interpretation. This measure, the standard deviation,

is calculated by taking the square root of the variance. Accordingly, it re-

duces our estimate of dispersion, using a method similar to the one we

employed to solve the problem of positive and negative differences from

the mean adding to 0. The standard deviation (s) provides an estimate of

dispersion in units similar to those of our original scores. It is described

in equation form as

Equation 5.5s � �� N

i�1

(Xi � X )2

N

W orking It Out

� 6,984,155.56

� 104,762,333.33

15

� � 15

i�1

(Xi � 3,263.33)2

15

s

2 � � N

i�1

(Xi � X )2

N

2In the special case of a fraction, the result will be smaller numbers.

107

Although Equations 5.4 and 5.5 provide a useful way of conceptualizing and

measuring the variance and standard deviation, you can also use a com-

puting formula that has fewer steps and is less likely to result in computa-

tional error. In Table 5.6, we rounded the mean and then calculated squared

deviations based on values that were rounded at each step. In an attempt

to limit the amount of rounding, and consequently decrease the chances of

a mistake, an alternative equation that can be used for the variance is

And an alternative for the standard deviation is

Let’s reconsider the data in Table 5.6 on hot spots. The following table

illustrates the key calculations:

HOT SPOT NUMBER NUMBER OF CALLS (Xi)

1 2 4

2 9 81

3 11 121

4 13 169

5 20 400

6 20 400

7 20 400

8 24 576

9 27 729

10 29 841

11 31 961

12 52 2,704

Total (�) 258 7,386

X

2 i

s � �� N

i�1

X

2 i �

��N i�1

Xi�2

N

N

s

2 � � N

i�1

X

2 i �

�� N

i�1

Xi�2

N

N

Computational Equations for the Variance and Standard Deviation

The variance (s 2) is then calculated with the computational equation as

And the standard deviation is simply the square root of the variance:

Similarly, let’s revisit the bail data in Table 5.7 and compute the vari-

ance with the computational formula. The following table illustrates the

key calculations.

DEFENDANT BAIL AMOUNT (Xi )

1 500 250,000

2 1,000 1,000,000

3 1,000 1,000,000

4 1,000 1,000,000

5 1,200 1,440,000

6 1,500 2,250,000

7 2,500 6,250,000

8 2,500 6,250,000

9 2,500 6,250,000

10 2,750 7,562,500

11 5,000 25,000,000

12 5,000 25,000,000

13 5,000 25,000,000

14 7,500 56,250,000

15 10,000 100,000,000

Total (�) 48,950 264,502,500

The variance is

And the standard deviation is

s � �6,984,155.56 � 2,642.76

s

2 �

264,502,500 � (48,950)2

15 15

� 6,984,155.56

X

2 i

s � �153.25 � 12.38

s 2 �

7,386 � (258)2

12 12

� 153.25

110 C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

In calculating the standard deviation, we add one step to our calcula-

tion of variance: We take the square root of our result. For the example

of crime calls at 12 hot spots (where the variance equaled 153.25), we

obtain a standard deviation of .3 If you were to define,

on average, how much the scores differed from the mean just by looking

at these 12 cases, you would probably come to a conclusion close to that

provided by the standard deviation. Similarly, if we take the square root

of the variance for the bail example above, we come up with a figure

that makes much more intuitive sense than the variance. In this case, the

standard deviation is , or $2,642.76.

The standard deviation has some basic characteristics, which relate

generally to its use:

1. A standard deviation of 0 means that a measure has no variability.

For this to happen, all of the scores on a measure have to be the

same. For example, if you examine a group of first-time offenders,

there will be no variation in the number of offenses in their criminal

records. By definition, because they are all first-time offenders, the

standard deviation (and the variance) will be 0.

2. The size of the standard deviation (and the variance) is dependent

on both the amount of dispersion in the measure and the units of

analysis that are used. When cases are spread widely from the mean,

there is more dispersion and the standard deviation will be larger.

When cases are tightly clustered around the mean, the standard

deviation will be smaller.

Similarly, when the units of analysis in the measure are large, the

standard deviation will reflect the large units. For example, if you

report the standard deviation of police salaries in a particular city in

dollars, your standard deviation will be larger than if you reported

those salaries in units of thousands of dollars. If the standard

deviation is 3,350 in dollars, the standard deviation would be 3.35

using the unit of thousands of dollars.

3. Extreme deviations from the mean have the greatest weight in

constructing the standard deviation. What this means is that here, as

with the mean, you should be concerned with the problem of

outliers. In this case, the effect of outliers is compounded because

they affect not only the mean itself, which is used in computing the

standard deviation, but also the individual deviations that are

obtained by subtracting the mean from individual cases.

�6,984,155.56

�153.25 � 12.38

3As discussed in footnote 1, SPSS and many other computer packages would provide

a slightly different result, based on the use of a correction of �1 in the denominator.

M E A S U R I N G D I S P E R S I O N I N I N T E R V A L S C A L E S 111

The standard deviation is a useful statistic for comparing the extent to

which characteristics are clustered or dispersed around the mean in dif-

4

samples are very similar (61.05 for antitrust violators; 59.27 for bribery

offenders), but the standard deviation for those convicted of bribery is

about twice that of those convicted of antitrust violations.

Figure 5.1 illustrates why these two samples yield similar means but

very different standard deviations. The scores for most antitrust offend-

the distribution, including many more cases between 75 and 90 and

below 50. What this tells us is that the antitrust sample includes a fairly

homogeneous group of offenders, ranking on average relatively high

on the Duncan socioeconomic index. Bribery is a much more diverse

category. Although the means are similar, the bribery category includes

many more lower- and higher-status individuals than does the antitrust

category.

The Coefficient of Relative Variation

For the data on bribery and antitrust offenders in Table 5.8, in which the

means of the two groups are fairly similar, a direct comparison of stan-

dard deviations provides a good view of the differences in dispersion.

When the means of two groups are very different, however, this compar-

ison may not be a fair one. If the mean Duncan score for one group was

10 and for the other was 50, we might expect a larger standard deviation

in the latter group simply because the mean was larger and there was

Duncan SEI for Bribery and Antitrust Offenders

CATEGORY N s

Bribery 83 59.27 19.45 Antitrust 112 61.05 11.13 Total (�) 195

X

Table 5.8

4See Albert J. Reiss, Occupations and Social Status (New York: Free Press, 1961).

victed of antitrust violations is compared to a sample of offenders convicted

ferent samples. For example, in Table 5.8, a sample of offenders con-

of bribery. The characteristic examined is social status, as measured

by the interval-scale Duncan socioeconomic index (SEI). The index

is based on the average income, education, and prestige associated

with different occupations. The mean Duncan scores for these two

offenders, in contrast, the scores are much more widely spread across

ers are clustered closely within the range of 55 to 75. For bribery

112 C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

N u

m b er

o f

O ff

en d

er s

Std. dev. = 11.13 Mean = 61.05 N = 112.00

0 20 40 60 80

100

80

60

40

20

0

SEI

Std. dev. = 19.45 Mean = 59.27 N = 83.00

N u

m b er

o f

O ff

en d

er s

SEI

30

20

10

0 0 20 40 60 80

Socioeconomic IndicesFigure 5.1

(a) Antitrust Offenders

(b) Bribery Offenders

M E A S U R I N G D I S P E R S I O N I N I N T E R V A L S C A L E S 113

greater potential for dispersion. Similarly, if two measures use different

units of analysis—for example, dollars and number of offenses—a direct

comparison of standard deviations does not make sense.

One solution to this problem is to use the coefficient of relative

variation (CRV). The coefficient of relative variation looks at the size of

the standard deviation of a measure relative to the size of its mean:

Equation 5.6

In the example of the SEI for antitrust offenders, we divide the standard

deviation (11.13) by the mean (61.05) to obtain a CRV of 0.18, meaning

that the standard deviation is about one-fifth the size of the mean. Be-

cause the CRV expresses dispersion in a measure in a standardized form

relative to the mean, we can compare the CRV across measures that have

widely different means and standard deviations.

CRV � s

X

W orking It Out

� 0.1823

� 11.13

61.05

CRV � s

X

A Note on the Mean Deviation

The standard deviation allows us to measure dispersion in interval

scales, taking into account the deviation from the mean of each case in

our sample or population. But it is not the only measure that allows us

to do this. The mean deviation takes a similar approach, but relies on

absolute values, rather than squaring, to overcome the fact that the sum

of the deviations from the mean equals 0. When you take the absolute

value of a number, you ignore its sign. Accordingly, �8 and 8 both have

an absolute value of 8; in mathematical notation, � �8 � � � 8 � � 8.

The equation for the mean deviation is similar to that for the vari-

ance. The only difference is that we take the absolute value of the

antitrust offenders.

A measure that has a CRV of 1, for example, may be considered to

include much greater relative variation than is found in our sample of

114 C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

Mean Deviation for Crime Calls at Hot Spots in a Year

HOT SPOT NUMBER DEVIATIONS FROM

NUMBER OF CALLS THE MEAN �Xi � �

1 2 � 2 � 21.5 � � 19.5

2 9 � 9 � 21.5 � � 12.5

3 11 � 11 � 21.5 � � 10.5

4 13 � 13 � 21.5 � � 8.5

5 20 � 20 � 21.5 � � 1.5

6 20 � 20 � 21.5 � � 1.5

7 20 � 20 � 21.5 � � 1.5

8 24 � 24 � 21.5 � � 2.5

9 27 � 27 � 21.5 � � 5.5

10 29 � 29 � 21.5 � � 7.5

11 31 � 31 � 21.5 � � 9.5

12 52 � 52 � 21.5 � � 30.5

Total (�) � 111.0

X

Table 5.9

difference between each score and the mean, rather than the square of

the difference:

Using the data on crime calls in hot spots from Table 5.4, we take the

following steps to obtain the mean deviation. We first take the absolute

value of the difference between each score and the mean (see Table

5.9). We then sum up the 12 scores. Notice that we obtain a positive

number now (111), and not 0, because we are taking the absolute values

of the differences. Dividing this sum by the number of cases, N, we get a

mean deviation of 9.25.

Mean deviation � � N

i�1 �Xi � X �

N

W orking It Out

� 9.25

� 111 12

� � 12

i�1 �Xi � 21.5 �

N

Mean deviation � � N

i�1 �Xi � X �

N

Equation 5.7

C H A P T E R S U M M A R Y 115

The mean deviation and the standard deviation provide similar esti-

mates of dispersion, but the mean deviation here is a bit smaller than the

standard deviation of 12.38 that we calculated earlier. Which is the better

estimate of dispersion? In some sense, the mean deviation is more

straightforward. It simply looks at the average deviation from the mean.

In obtaining the standard deviation, we first must square the deviations;

then later, to return our result to units similar to those of the original dis-

tribution, we must take the square root of the variance.

Given our rule that we should use the least complex presentation

that is appropriate to answering our research question, you may won-

der why the standard deviation is almost always preferred over the

mean deviation in criminal justice research. As you will see in the next

few chapters, the answer is that the standard deviation is relevant to a

number of other statistics that we use in analyzing and describing

data.

C h a p t e r S u m m a r y

Measures of dispersion describe to what extent cases are distributed

around the measure of central tendency. They tell us just how typical the

typical case is.

There are several measures of dispersion for nominal and ordinal

scales. Proportions and percentages describe the extent to which cases

are concentrated in the modal category. The variation ratio (VR) de-

scribes the extent to which cases are spread outside the modal category.

A proportion of 1 (VR of 0) means that all the cases are in the modal cat-

egory. This represents the least possible amount of dispersion. The value

for the greatest possible dispersion can be determined by calculating the

minimum possible value of the modal category and then translating that

into a proportion or VR value. These measures can, in principle, be used

with ordinal-level data, but the results may be misleading, as they take

into account only the value of the mode. As an alternative, the index of

qualitative variation (IQV) is a standardized measure that takes into ac-

count variability across all the categories of a nominal- or ordinal-level

variable. An IQV of 0 means that there is no variation; an IQV of 100

means that there is maximum variation across the categories.

A different set of measures is used to measure dispersion for inter-

val and ratio scales. The range measures the difference between the

highest and lowest scores. It has the advantage of simplicity, but it

uses very little information (only two scores) and the scores used are

taken from the two extremes. It is also very sensitive to outliers. A

116 C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

95th and the 5th percentile. Such measures, however, are still based

on minimal information and thus are generally considered unstable

statistics. A more stable statistic for measuring dispersion in interval-

level scales is the variance. The variance is the sum of the squared

deviations of each score from the mean divided by the number of

cases. The standard deviation (s) is the square root of the variance.

The advantage of the standard deviation over the variance is that the

results are more easily interpreted. If all the scores in a sample are the

same, s will be 0. The more widely the scores are spread around the

mean, the greater will be the value of s. Outliers have a considerable

impact on the standard deviation.

Comparing the standard deviations of means is problematic when

the means are very different or when their units of measurement are

different. An alternative measure, the coefficient of relative varia-

tion (CRV), enables comparisons among samples with different

means. A less often used measure of dispersion for interval scales is

the mean deviation. The mean deviation is computed by taking the

sum of the absolute values of the deviations from the mean divided by

the number of cases.

K e y T e r m s

coefficient of relative variation A mea-

sure of dispersion calculated by dividing

the standard deviation by the mean.

index of qualitative variation A measure

of dispersion calculated by dividing the

sum of the possible pairs of observed

scores by the sum of the possible pairs of

expected scores (when cases are equally

distributed across categories).

mean deviation A measure of dispersion

calculated by adding the absolute deviation

of each score from the mean and then di-

viding the sum by the number of cases.

range A measure of dispersion calculated

by subtracting the smallest score from the

largest score. The range may also be calcu-

lated from specific points in a distribution,

such as the 5th and 95th percentile scores.

standard deviation A measure of disper-

sion calculated by taking the square root of

the variance.

variance (s2) A measure of dispersion cal-

culated by adding together the squared de-

viation of each score from the mean and

then dividing the sum by the number of

cases.

variation ratio A measure of dispersion

calculated by subtracting the proportion of

cases in the modal category from 1.

researcher may instead choose to measure the range between, say, the

S Y M B O L S A N D F O R M U L A S 117

S y m b o l s a n d F o r m u l a s

Nmodal cat. Number of cases in the modal category

Ntotal Total number of cases

Nobs Number of cases observed in each category

Nexp Number of cases expected in each category

s Standard deviation

s 2 Variance

To calculate the proportion of cases falling in the modal category:

To calculate the percentage of cases falling in the modal category:

To calculate the variation ratio:

To calculate the index of qualitative variation:

To calculate the variance:

To calculate the standard deviation:

s � �� N

i�1

(Xi � X )2

N

s

2 � � N

i�1

(Xi � X )2

N

IQV � �� k�1

i�1

� k

j�i�1

Nobsi Nobsj

� k�1

i�1

� k

j�i�1

Nexpi Nexpj

� � 100

VR � 1 � �Nmodal cat.

Ntotal �

Percentage � Nmodal cat.

Ntotal � 100

Proportion � Nmodal cat.

Ntotal

118 C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

To calculate the coefficient of relative variation:

To calculate the mean deviation:

E x e r c i s e s

5.1 Police records for 105 rape victims were analyzed to determine whether any prior relationship existed between the victim and the offender. The results were as follows:

Spouse 41

Family member other than spouse 14

Acquaintance 22

No prior relationship 28

a. Calculate the modal proportion and the variation ratio.

b. What are the minimum and maximum possible values for the varia- tion ratio?

c. Calculate the index of qualitative variation.

5.2 As part of a larger study on the influence of delinquent peers, a sam- ple of high school youth were asked how much they wanted to be like their best friend. The responses were coded as follows: in every way, 26; in most ways, 36; in some ways, 41; and not at all, 8.

a. Calculate the variation ratio for these data.

b. Calculate the index of qualitative variation for these data.

5.3 People convicted of minor traffic offenses who appeared in the magistrate’s court of a given locality on a given day were sentenced as follows: conditional discharge, 14; fine, 35; and license dis- qualification, 11.

a. Calculate the variation ratio.

b. Calculate the index of qualitative variation.

c. Why do these two results differ?

5.4 A sample of women was drawn from town A, and another sample was drawn from town B. All the women were asked how safe or unsafe

Mean deviation � � N

i�1 �Xi � X �

N

CRV � s

X

E X E R C I S E S 119

they felt walking alone at night in their neighborhoods. The results were recorded on a scale as follows: totally unsafe (town A: 40; town B: 25), quite unsafe (town A: 29; town B: 23), quite safe (town A: 10; town B: 15), and totally safe (town A: 21; town B: 17).

a. For each town, describe the typical case, using an appropriate mea- sure of central tendency. Explain why this is the best measure of central tendency for these data.

b. For each town, describe how typical the typical case is, using an appropriate measure of dispersion. Explain why this is the best measure of dispersion for these data.

c. In comparing the measures of central tendency and dispersion for the two towns, what conclusions may be drawn about the attitudes of the women?

5.5 For a sample of 12 offenders convicted of weapons violations, the length of prison sentence in months was recorded as:

6 6 2 12 36 48 60 24 24 20 18 15

a. Calculate the range for these data.

b. Calculate the mean and the variance for these data.

5.6 A group of 20 prisoners in a particular cell block were tested on their knowledge of the rules of the institution. The marks (out of a possible 70) were as follows:

31 28 27 19 18 18 41 0 30 27

27 36 41 64 27 39 20 28 35 30

a. Calculate the range.

b. Remove the largest and smallest scores. Calculate the range for the remaining cases.

c. How do you account for the difference between the values of the above two measures of dispersion?

5.7 Police crack a drug ring of 18 suppliers and discover that of the 18, only 4 have no previous convictions for drug- or theft-related offenses. Eight of those arrested have 1 previous conviction, and the others have 2, 3, 4, 5, 6, and 8, respectively.

a. Calculate the mean and the standard deviation of the 18 cases.

b. If each of the drug suppliers is convicted this time around, does the extra conviction on each of their criminal records affect the mean or the standard deviation in any way? Explain your answer.

120 C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

5.8 Use the data collected from tests of prisoners’ knowledge of institution rules in Exercise 5.6.

a. Calculate the mean and the standard deviation for these data.

b. If you remove the two most extreme scores, 0 and 64, what are the new mean and standard deviation?

c. How do you account for this effect?

5.9 When asked about how often in the last year they drank more than four beers in one evening, a sample of college students reported the following:

Number

of Times Frequency

0 187

1 213

2 162

3 94

4 71

5 55

6 39

7 12

8 9

9 5

10 13

a. Calculate an appropriate measure of dispersion for these data. Explain why this measure is most appropriate for these data.

b. Describe one way these data could be recoded to reduce the number of categories. Calculate an appropriate measure of dispersion for the recoded data and explain why this measure is most appropriate.

5.10 A researcher takes a sample of shop owners in Tranquiltown and a sample of shop owners in Violenceville and asks them to estimate the value of goods stolen from their shops in the past 12 months. The mean figure is $11.50 (s � $2.50) for Tranquiltown and $4,754.50 (s � $1,026.00) for Violenceville. When the study is published, the mayor of Violenceville protests, claiming that the mean sum for his town is a misleading figure. Because the standard deviation for Vio- lenceville is much bigger than that for Tranquiltown, he argues, it is clear that the mean from Violenceville is a much less typical descrip- tion of the sample than the mean from Tranquiltown.

a. What statistic might help the researcher to refute this criticism? Why?

C O M P U T E R E X E R C I S E S 121

b. Calculate this statistic for each town. What should the researcher conclude?

5.11 A researcher investigating differences in violence among preschool- age boys and girls found that the average number of violent acts per week was 7.6 (s � 4.8) for boys and 3.1 (s � 1.9) for girls.

a. Calculate the coefficient of relative variation for boys and for girls.

b. How can the coefficient of relative variation be used to compare these two groups? What does it tell you?

C o m p u t e r E x e r c i s e s

Similar to the measures of central tendency discussed in Chapter 4, there are

several ways to obtain measures of dispersion for interval-level variables in SPSS

and Stata (Neither program computes measures of dispersion, such as the index

of qualitative variation, for nominal and ordinal variables.).

SPSS

The same two commands that we used to obtain measures of central tendency—

DESCRIPTIVES and FREQUENCIES—will be used to obtain measures of

dispersion. Since you have already used these commands, our discussion here is

focused on the default measures of dispersion and additional options available

for measuring dispersion.

The DESCRIPTIVES command allows you to compute the standard devia-

tion and variance. The default is to compute the standard deviation, as well as

minimum and maximum values. To obtain the variance and/or range, add the

option /STATISTICS = to the command line:

The abbreviations should be clear, but STDDEV is the standard deviation, MIN

is the minimum value, and MAX is the maximum value.

In computing the variance, SPSS uses a correction for the bias of sample

measures of variance and dispersion: 1 is subtracted from the N in the denomi-

nator of Equations 5.4 and 5.5. The correction is used primarily as a tool in

inferential statistics and is discussed in greater detail in Chapter 10. Though it is

our view that the uncorrected variance and standard deviation should be used

in describing sample statistics, many researchers report these statistics with the

correction factor for sample estimates. When samples are larger, the estimates

obtained with and without the correction are very similar, and thus it generally

makes very little substantive difference which approach is used.

DESCRIPTIVES VARIABLES = variable_names

/STATISTICS = MEAN STDDEV VARIANCE RANGE MIN MAX.

122 C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

The FREQUENCIES command provides similar measures of disper-

sion through use of the /STATISTICS = option. You may request the stand-

ard deviation, variance, range, minimum, and maximum. In addition to these

measures of dispersion, the FREQUENCIES command has the option of com-

puting percentiles. Since calculation of percentiles by hand can be very difficult

(and error prone), this is a nice feature. There are three options for calculating

percentiles in SPSS: quartiles (25th, 50th, and 75th percentiles) or cut points for

some number of equally spaced groups (e.g., 10th, 20th, …, 90th percentiles)

by using the NTILES = option, as well as specific percentiles that you may be

interested in (e.g., 5th, 95th, 99th) by using the /PERCENTILES = option. The

general form, including all the various options, is:

If you are not interested in percentiles or equally spaced groups (e.g., quartiles,

deciles, etc.), then omit those lines from the command.

Specific examples of the use of each of these commands are provided in the

accompanying SPSS syntax file for Chapter 5 (Chapter_5.sps).

Stata

To obtain measures of dispersion in Stata, the same two commands used in

obtaining measures of central tendency will again be used. To obtain detailed

measures of dispersion, it is best to simply add the detail option to the

command line:

This will produce the variance, standard deviation, and numerous percentiles.

Stata makes the correction to the computation of the variance in the same way

that SPSS does by subtracting 1 from the total sample size.

The tabstat command is also useful in trimming out the unnecessary pieces

of information and printing only those items we want:

Where min is the minimum, max is the maximum, sd is the standard deviation,

var is the variance, cv is the coefficient of variation (not available in SPSS), and

FREQUENCIES VARIABLES = variable_names

/PERCENTILES = list_of_percentiles

/NTILES = number_of_equally_spaced_groups

/STATISTICS = STDDEV VARIANCE RANGE MINIMUM MAXIMUM

MEAN MEDIAN MODE.

summarize variable_names, detail

tabstat variable_names, statistics (median mean min max

range sd var cv p#)

C O M P U T E R E X E R C I S E S 123

in the 5th, 10th, 90th, and 95th percentiles, we would include p5 p10 p90 p95

on the command line (and within the parentheses).

Specific examples of the use of each of these commands are provided in the

accompanying Stata do file for Chapter 5 (Chapter_5.do).

Problems

1. Enter the data from Exercise 5.6 on the 20 prisoners’ test scores.

a. What is the range?

b. What are the 5th and 95th percentiles? What is the range between the

5th and 95th percentiles?

c. How does your answer to part b compare to your answer to part b in

Exercise 5.6?

2. Enter the data from Exercise 5.7. (Be sure that you have 18 lines of data,

since there are 18 observations listed in the question.)

a. What are the mean and the standard deviation? How does the standard

deviation differ from the value you calculated in Exercise 5.7?

b. To add 1 to each person’s number of prior convictions, you will need to

create a new variable.

In SPSS:

In Stata:

3. Open the NYS data file (nys_1.sav, nys_1_ student.sav, or nys_1.dta).

a. Choose five of the delinquency measures. What are the quartiles for

b. What is the range between the 25th and 75th percentiles for each

delinquency measure? (This difference is known as the “inter- quartile

range.”) What do the differences in inter-quartile ranges appear to

indicate about the dispersion of these different measures self-reported

delinquency?

p# refers to a specific percentile we may be interested in and included for as

many percentiles as we would like reported. For example, if we were interested

COMPUTE new_var_name = old_var_name + 1.

EXECUTE.

gen new_var_name = old_var_name +1

What are the mean and standard deviation for this new

variable? What has changed? What has remained the same?

124 C H A P T E R F I V E : H O W T Y P I C A L I S T H E T Y P I C A L C A S E ?

-

able (Remember, this measure is available in the tabstat command in

Stata, but will need to be calculated by hand if using SPSS.). What do

variable indicate about the relative dispersion of these variables?

c. What number of delinquent acts would mark the 15 % least delinquent

youth? The 20 % most delinquent youth?

The Logic of Statistical Inference:

Making Statements About Populations

from Sample Statistics

How Can a Sample Teach Us About a Population?

C h a p t e r s i x

S a m p l e d i s t r i b u t i o n s a n d p o p u l a t i o n d i s t r i b u t i o n s

A s k i n g t h e r e s e a r c h q u e s t i o n

A n s w e r i n g t h e r e s e a r c h q u e s t i o n

How are They Defined?

What Symbols are Used?

How are the Two Interrelated?

What Types of Error are Possible?

What is an “Acceptable” Risk of Error?

When Might It be Necessary to Accept a Different Level of Risk?

What are the Research and Null Hypotheses?

How are They Set Up?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_6,

© Springer Science+Business Media New York 2014

IN THIS CHAPTER, we look at an important dilemma that researchers face

in conducting criminal justice research. Although they seek to make

statements about populations, generally they collect data on samples

drawn from such populations. Statistical inference provides a solution to

about the characteristics of a population from data collected from a sam-

ple drawn from the population. We begin our discussion of statistical in-

ference by explaining the dilemma researchers face in making statements

about populations from samples. We then examine the logic of statistical

inference and the statistical risks associated with using this logic. You

will be introduced to how null and research hypotheses are set up, how

risks of error are assessed, and how levels of statistical significance are

used to limit this error.

T h e D i l e m m a : M a k i n g S t a t e m e n t s A b o u t P o p u l a t i o n s f r o m S a m p l e S t a t i s t i c s

In descriptive statistics, we are concerned with two basic types of distrib-

utions. One is the distribution of scores in the sample, or the sample

distribution. The second is the distribution of scores in the population

from which the sample is drawn. This is referred to as the population

distribution. One of the fundamental problems in research in criminal

justice, as in other fields, is that we want to make statements about the

characteristics of the population distribution, but we generally have in-

formation only about the distribution of sample scores. For example,

when we draw a sample of 2,000 voters in an election survey, we are

not interested per se in how those people will vote. Rather, we examine

their voting preference to learn something about how all people will

vote in the election. In statistical terms, we want to use information on

characteristics of the distribution of sample scores to make statements

about characteristics of the distribution of population scores.

126

this dilemma: it allows the researcher to make statements, or inferences,

T H E D I L E M M A 127

It is important to note at the outset that populations can be defined in

a number of different ways. There is, for example, the population of the

entire United States or the population of a particular state. There is the

population of all prisoners in the United States or the population of pris-

oners only in a specific state. Although the population of cases is fixed at

any particular time, actual populations are constantly changing across

time. For example, we can speak of the population of prisoners on any

particular day. But every day new people enter prisons and some prison-

ers are freed. Since the population of prisoners changes every day, the

population of prisoners at any one time is only a sample of the popula-

tion of prisoners across a longer period of time—for example, a year or

two.

Statisticians use different symbols to distinguish statistics on a popula-

tion from statistics on a sample (see Table 6.1). Population statistics, or

parameters, are defined using Greek letters. For example, the parame-

ter for the mean in a distribution of population scores is represented by

� and the standard deviation by �. Sample statistics are represented by

roman letters. We denote the mean in a distribution of sample scores as

and the standard deviation as s.

Why do we study sample statistics if we really want to say something

about population parameters? It certainly makes more sense to collect in-

formation on the population if that is what is of interest in the long run.

In practice, however, it is usually very difficult to gain information on the

universe, or total group of cases in the population. One reason is sim-

ply financial. As was pointed out in Chapter 1, to carry out just one sur-

vey of the U.S. population regarding their attitudes toward crime would

exhaust the budget of the National Institute of Justice (the major funder

of research in criminal justice in the United States) for many years. But

beyond the costs of such studies, there is the problem of their manage-

ment. A study of an entire population will often demand contact with

hundreds of thousands or even millions of people. Such an effort is

likely to be not just expensive, but difficult to manage and time consum-

ing to complete.

Because of the difficulty of gaining information on the characteristics

of an entire population, such parameters are generally unknown. How-

ever, when a parameter is available, there is no point in drawing statis-

tics from a sample. In recent years, advances in computer technology

X

Representing Population Parameters and Sample Statistics

MEAN VARIANCE STANDARD DEVIATION

Sample distribution s 2 s Population distribution � �2 �

X

Table 6.1

128 C H A P T E R S I X : T H E L O G I C O F S T A T I S T I C A L I N F E R E N C E

and recognition by public officials of the importance of data in making

policy decisions about the criminal justice system have led to the devel-

opment of a number of databases that include information on the popu-

lation of cases. For example, we now have population parameters on

characteristics of sentencing in the federal courts and basic demographic

characteristics of offenders held in jails and prisons. Information on the

population of arrests and emergency calls to the police is routinely col-

lected and computerized in most cities. One scholar suggests that this

trend means that criminologists in the future will have to pay less and

less attention to samples and the problems they create for researchers.1

However, whatever the future will bring, at present criminal justice re-

searchers must rely primarily on sample statistics in trying to say some-

thing about the characteristics of a population.

Given our reliance on sample statistics, it is important at the outset to

define how they might differ from population parameters. One obvious

difference is that sample statistics are generally known—put differently,

they can be defined by the researcher in the context of a research study.

In contrast, parameters are generally unknown, although, as we noted

above, there is a trend toward development of parameters about major

issues in criminal justice.

Even though parameters are often unknown, they are assumed to be

fixed. By that we mean that there is one true parameter for any measure.

For example, there is a true mean age at first arrest for the population of

all criminals in the United States at a specific time. In contrast, sample

statistics vary from sample to sample. For example, if you were to draw

10 samples from a population, using exactly the same method each time,

each sample would likely provide different sample statistics.

This is illustrated in Table 6.2. Ten random samples of 100 offenders

were drawn from a population of 1,940 offenders. Sample statistics are

presented for mean age and number of prior arrests. Although the sample

statistics obtained are generally similar to the population parameters, each

sample provides a somewhat different group of estimates, and in some

cases the differences are relatively large. In the case of sample 10, for ex-

ample, the average number of arrests for the sample is more than a third

lower than the population score. In sample 4, the average age is more

than two years older than the population parameter. This occurs despite

the fact that we drew each of the samples using the same technique and

from the same population of scores. You might want to try this yourself

by drawing a series of samples from your class or dormitory. Using the

same method, you will almost always obtain different sample statistics.

1M. Maltz, “Deviating from the Mean: The Declining Significance of Significance,”

Journal of Research in Crime and Delinquency 31 (1994): 434–463.

T H E R E S E A R C H H Y P O T H E S I S 129

This fact is one of the fundamental problems we face in statistics. We

want to make statements about populations, but we generally must rely

on sample statistics to do so. If sample statistics vary from sample to

sample, how can we use them to make reliable statements about the pa-

rameters associated with a population? Put differently, what is the use of

most studies in criminal justice, if they are based on samples rather than

populations? Fortunately, there is an area of statistics that provides us

with a systematic way to make decisions about population parameters

based on sample statistics. This area is called statistical inference, and

in the remaining sections of this chapter we focus on the logic that un-

derlies statistical inference.

T h e R e s e a r c h H y p o t h e s i s

Statistical inference begins with the definition of the questions that the

researcher seeks to answer in a research project. Sometimes research

questions in criminal justice are focused on specific agencies in the

criminal justice system. For example, we may want to learn more about

the police, the courts, or probation services. Other times research ques-

tions revolve around broad theoretical concerns that can be applied

across criminal justice agencies. We may, for example, seek to define

common features of criminal justice programs that lead to a reduction in

recidivism (reoffending). Sometimes our questions relate to offenders,

other times to victims of crime or criminal justice agents.

To answer a research question, we have to set up at least one and

sometimes several research hypotheses related to it. A hypothesis is a

proposed answer to our research question that we can then test in the

Ten Random Samples of 100 Offenders Drawn

from a Population of 1,940 Offenders

MEAN AGE MEAN ARRESTS

Population 39.7 2.72 Sample 1 41.4 2.55 Sample 2 41.2 2.19 Sample 3 38.8 2.09 Sample 4 42.1 3.45 Sample 5 37.9 2.58 Sample 6 41.1 2.62 Sample 7 39.2 2.79 Sample 8 39.2 2.48 Sample 9 37.8 2.55 Sample 10 37.7 1.72

Table 6.2

130 C H A P T E R S I X : T H E L O G I C O F S T A T I S T I C A L I N F E R E N C E

context of a study. Stating a research hypothesis does not mean that we

assume that the hypothesis is true. Rather, it focuses our research ques-

tion in such a way that it can be directly examined in the context of a

study. When the research hypothesis does not indicate a specific type of

outcome, stating only that there is a relationship or a difference, we say

that it is a nondirectional hypothesis. However, in those cases where

a researcher has a very clear idea of what to expect—based on prior re-

search evidence and/or theory—the research hypothesis may be more

Suppose we are interested in comparing the arrest records of drug-

involved offenders with those of offenders who do not use drugs. Our

research hypothesis might be simply that the arrest records of drug-

behavior among drug-involved offenders, we might want to state a direc-

tional hypothesis—that drug-involved offenders have more serious arrest

records than do non–drug-involved offenders. One problem with choos-

ing the latter option is that if we state our research hypothesis as a direc-

tional hypothesis, we are stating that we are not interested in outcomes

that fall in the opposite direction. In criminal justice research, we can

often be surprised by what we learn in a study. Accordingly, researchers

generally are cautious in defining a directional research hypothesis.

Having defined a research hypothesis, we want to examine whether it

is true for the population in which we are interested. For our example of

drug-involved offenders, if we could collect information about all offend-

ers, we could simply look at the parameters drawn from our study to see

whether they support the research hypothesis and, if so, to what degree.

In this case, we would not need to use the logic of statistical inference.

We would collect data directly on the population parameters. But ordi-

narily we cannot collect information on the population parameters and

must rely on the statistics drawn from a sample in making our decision.

Our problem is that we cannot come to an absolute conclusion regard-

ing the research hypothesis because we know that statistics vary from

sample to sample.

On the basis of a sample, we can never be sure of the true value of a

population parameter. Accordingly, we can never be absolutely certain

as to whether the research hypothesis is true. But does the fact that we

cannot be sure mean that we cannot come to a reasonable conclusion

regarding our hypotheses?

In fact, we often make decisions about hypotheses on the basis of

samples in our daily lives. For example, let’s say that you are deciding

involved offenders and offenders who do not use drugs are different

(a nondirectional hypothesis). But based on prior knowledge of criminal

precise. In this case, the researcher may specify the nature of the relation-

ship that is expected. Such a research hypothesis is called a directional

hypothesis. When a directional hypothesis is used, the researcher

states at the outset that he or she is interested in a specific type of

outcome—for example, that one group has more arrests than another.

T H E N U L L H Y P O T H E S I S 131

whether to sign up for a course taught by an instructor named Professor

Justice. One issue that you are particularly concerned about is the impact

the course will have on your grade point average. To make an informed

decision about the course, you might decide to ask friends of yours who

took the course last year how Professor Justice grades in comparison to

others at your college. Although you might not think of them quite in

this way, your friends represent your sample. In turn, the hypothesis that

the professor grades differently from other faculty members in your col-

lege is similar to a research hypothesis.

If your friends gave a mixed view or generally were unable to say

whether Professor Justice grades more harshly or more easily than other

professors, you would likely conclude that the course would not have

much impact on your grade point average. Put differently, you would

decide that the research hypothesis is probably false. If most of your

friends said that the professor is an easy grader or, conversely, that she

was a hard grader, you would take this as evidence that the research hy-

pothesis is most likely correct—that the professor grades differently and

that the course is likely to have an impact on your grade point average.

Once you have made the decision that Professor Justice is different

from others, you are likely to assess how she is different. If your friends

define the professor as a hard grader, you might decide to avoid the

course because you fear you would get a lower grade than is usual with

other professors. If they define the professor as an easy grader, you

might be encouraged to take the course, with the expectation that your

grade will be higher than usual.

In effect, you make a decision about the research hypothesis based

on information that you draw from your “sample” of friends. Your confi-

dence in making a decision will depend greatly on how reliable you be-

lieve your friends’ observations to be and to what degree they represent

other students in the class. This is very similar to the logic we use in

making statistical inferences from samples to populations. However, in

statistical inference, we test hypotheses not in reference to the research

hypothesis but in reference to a type of hypothesis that statisticians call

the null hypothesis.

T h e N u l l H y p o t h e s i s

The null hypothesis—or H0—gains its name from the fact that it usually

states that there is no relationship, or no difference. It is the flip side of

the research hypothesis (H1), which usually posits that there is a relation-

ship. In the example of the professor’s grading, the null hypothesis

would simply be that “there is no difference between the grading of Pro-

fessor Justice and that of others in the university.”

132 C H A P T E R S I X : T H E L O G I C O F S T A T I S T I C A L I N F E R E N C E

In practice, in statistics, we make decisions about hypotheses in rela-

tion to the null hypothesis rather than the research hypothesis. This is

because the null hypothesis states that the parameter in which we are in-

terested is a particular value. For example, returning to the comparison

of drug-involved and other offenders, your null hypothesis (H0) might be

that there is no difference between the two groups in the average num-

ber of crimes committed, or, put differently, that the difference is equal

to zero. In the case of Professor Justice’s grading, the null hypothesis

also states that there is no difference, or again that the difference be-

tween the average grade given by Professor Justice and the average

grade given by her colleagues is equal to zero.

In contrast, the research hypothesis is ordinarily not stated in exact

terms. A number of potential outcomes can satisfy the research hypothe-

sis. In the example of the professor’s grading, any average grade that is

different from that of other professors in the college is consistent with

the research hypothesis. But only one result, that the professor’s grading

is the same as that of others, is consistent with the null hypothesis. The

null hypothesis, accordingly, has the advantage of defining a specific

value for the population parameter.2

By stating the null and research hypotheses, we have taken a first

very important step in making statistical inferences. However, we still

have the problem of how to make decisions about these hypotheses on

the basis of sample statistics.

Whenever we rely on sample statistics to make statements about popu-

This means that when we test hypotheses in research, we generally

do not ask whether a hypothesis is true or false. To make such a state-

ment would require knowledge about the population parameters.

Rather, we ask whether we can make an inference, or draw a conclu-

sion, about our hypotheses based on what we know from a sample. In

statistical inference, we use sample statistics to infer to, or draw conclu-

sions about, population parameters.

2Some statisticians prefer to call the research hypothesis the “alternative” hypothesis,

because we can, in theory, choose any value as the null hypothesis, and not just the

value of zero or no difference. The alternative hypothesis, in this case, can be defined

as all other possible outcomes or values. For example, you could state in your null

hypothesis that the professor’s grades are, on average, five points higher than those of

other professors in the college. The alternative hypothesis would be that the profes-

sor’s grades are not, on average, five points higher than those of other professors.

lation parameters, we must always accept that our conclusions are

tentative. The only way to come to a definitive conclusion regarding the

population parameter is to actually examine the entire population. This

happens more and more with criminal justice data today. However, in most

research, we are still able to collect information only about sample statistics.

R I S K S O F E R R O R I N H Y P O T H E S I S T E S T I N G 133

In order to understand the logic of making inferences, it will help to

return to our example of drug-involved offenders. Let’s say we are inter-

ested in the number of crimes that offenders commit in a given year. We

decide to use arrests as our measure of criminal behavior. We might state

our null hypothesis as follows: “Drug-involved offenders and offenders

who do not use drugs have, on average, the same number of arrests in a

given year.” To test our hypothesis, we take a sample of drug-involved

As illustrated earlier in regard to Professor Justice’s grading, in every-

day life we make such decisions through a combination of intuition,

prior experience, and guesswork. In statistical inference, we take a sys-

tematic approach to this decision making, which begins with the recog-

nition that whatever decision we make has a risk of error.

R i s k s o f E r r o r i n H y p o t h e s i s T e s t i n g

What types of error do we risk when making a decision about a popula-

pothesis concerning arrests among drug-involved and non–drug-involved

offenders, there are only two possible scenarios for the population. In

the first case, the null hypothesis is true, meaning that there is no differ-

ence in average number of arrests between offenders in the population

who use drugs and those who do not. Alternatively, the null hypothesis

may be false, meaning that there is a difference in the population be-

Based on our sample statistic, we can, as well, come to only two pos-

sible conclusions regarding the null hypothesis. We can reject the null

hypothesis and infer that there is a difference in the average numbers of

arrests of drug-involved and other offenders in the population. Alterna-

tively, we can fail to reject the null hypothesis and state that our sample

does not provide sufficient evidence to conclude that there is a differ-

ence in the average numbers of arrests of drug-involved offenders and

offenders in the population who do not use drugs.

tion parameter from a sample? A simple way to examine this question

is to compare the potential decisions that can be made about a null

hypothesis with the value of the population parameter. For our null hy-

do offenders who do not use drugs.

offenders have, on average, either fewer or more arrests in a year than

tween drug-involved and other offenders. In this case, drug-involved

offenders and another sample of offenders who do not use drugs. We find

that drug-involved offenders in our sample have a mean of five arrests

per year, whereas offenders who do not use drugs have a mean of three

arrests per year. Should we reject the null hypothesis? Should we conclude

that there is a difference in the numbers of arrests in the population

based on results from our sample?

134 C H A P T E R S I X : T H E L O G I C O F S T A T I S T I C A L I N F E R E N C E

If we cross these two sets of possibilities, we define four possible situ-

ations, as represented in Figure 6.1. Two of these are desirable, because

they suggest that our decision about the null hypothesis is consistent

with the population parameter. In one case (box 1), we fail to reject the

null hypothesis, and it is in fact true. In the second (box 4), we reject the

null hypothesis on the basis of our sample results, and the null hypothe-

sis is false.

In the remaining situations, however, our decisions are not consistent

with the population parameter. In one (box 2), we fail to reject the null

hypothesis on the basis of our sample statistic, but it is in fact false. In

this case, we have made what statisticians call a Type II (or beta)

error. A Type II error would occur in our example of arrests among

offenders if we did not reject the null hypothesis on the basis of our

sample results when in fact the average numbers of arrests for drug-

involved and other offenders were different in the population to which

we want to infer. We make a Type I (or alpha) error (see box 3)

when we reject the null hypothesis on the basis of sample statistics but

H0 is true. In this case, we infer from our sample that drug offenders are

different from offenders who do not use drugs, when in fact they are

similar in the population.

Whenever we make a decision about a population parameter from a

sample statistic, we risk one of these two types of statistical error. If we

Box 1 ✔

Box 3 �

Type I error

Box 2 �

Type II error

Box 4 ✔

H0 = True H0 = False

Population

Reject H0

Decision

Fail to Reject H0

Types of Error in a Statistical TestFigure 6.1

S T A T I S T I C A L L E V E L S O F S I G N I F I C A N C E 135

fail to reject the null hypothesis, there is always the possibility that we

have failed to reject it when it was false (Type II error). If we reject the

null hypothesis, there is always the possibility that we have rejected it

when it was true (Type I error). Although we cannot avoid the possibil-

ity of error when we study samples, we can decide at the outset how

R i s k s o f E r r o r a n d S t a t i s t i c a l L e v e l s o f S i g n i f i c a n c e

In statistical inference, we assess the risk of making a wrong decision

about the population parameter in reference to Type I error. This is why

it is very important in statistical inference that we use the phrase “reject”

or “fail to reject” the null hypothesis, rather than the simpler statement

that we “reject” or “accept” the null hypothesis. Type I error is concerned

with rejection of the null hypothesis when it is true. It does not refer di-

rectly to the risk of accepting the null hypothesis when it is false (Type II

error). This latter problem will be the focus of a discussion in Chapter

make a decision either to “reject” or to “fail to reject” the null hypothesis

on the basis of the amount of Type I error we are willing to risk.

We define the amount of Type I error we are willing to risk as the

significance level of a test of statistical significance. In a test of statis-

tical significance, we make a decision to reject or to fail to reject the null

hypothesis on the basis of a sample statistic. The significance criterion,

or level, of a test of statistical significance is ordinarily represented by the

symbol �. The estimate of the risk of Type I error that is associated with

rejecting the null hypothesis in a test of statistical significance (based on

a sample statistic) is called the observed significance level and is ordi-

narily represented by the symbol p. In statistical inference, we first iden-

tify the amount of Type I error we are willing to risk, or the significance

level of a test. We then estimate the observed significance level from our

sample statistics. Finally, we compare the observed significance level

gained from our study with the criterion significance level we set at the

outset of our test of statistical significance. If the observed significance

level is less than the significance criterion, or level, that we set at the

outset of the test, we reject the null hypothesis. In the next chapter, we

examine how statisticians estimate the observed significance level of a

test. At this juncture, it is important to consider how we decide on the

amount of Type I error we are willing to risk. How do we choose the

significance level in a test of statistical significance?

hypotheses.

much risk or error we are willing to take in making decisions about

21. For now, it is important to remember that in statistical inference we

136 C H A P T E R S I X : T H E L O G I C O F S T A T I S T I C A L I N F E R E N C E

If we are willing to take a good deal of risk of Type I error, we set a

very lenient significance level. This means that we are willing to reject

the null hypothesis on the basis of our sample statistic, even if the risk of

a Type I error is fairly large. If we set a very strict significance level, this

means that we are unwilling to reject the null hypothesis unless we are

fairly certain that our decision is correct.

If we return to the example of Professor Justice’s grading, the impor-

tance of Type I error in making statistical inferences will become clearer.

Let’s say that every one of the friends you ask reports that the professor

is a much easier grader than other professors at your college. In this

case, you would probably conclude that the observed significance level,

or risk of a Type I error, in your study was very small. It is unlikely that

all of your friends would say that the professor was an easy grader if she

was in reality very similar to others in the college. Of course, your

friends could provide a mistaken view of the professor’s grading habits.

But you would probably assume that this is not very likely.

But what if your friends provided you with a mixed view of Professor

Justice’s grading? What if 40% of your friends say that Professor Justice

grades similarly to other professors and 60% say that she is easier? Would

you be so confident in rejecting the null hypothesis with these results?

Overall, the majority of your friends still say the professor is a relatively

easy grader. But in this case there is a substantial group reporting that

she grades much as other professors do. If you set a strict significance

level, you might not be willing to reject the null hypothesis based on

your observations (your observed significance level), and you might sim-

ply conclude that there is not enough evidence, on the basis of your

sample of friends, to say that Professor Justice grades differently.

It might seem at first that we would want to be fairly lenient in setting

our significance level. Often, in defining the research hypothesis, we are

expressing what we believe to be true. Why, then, would we want to

make it difficult to reject the null hypothesis? By rejecting the null hy-

pothesis that there is no difference, we are led to infer that the research

hypothesis is correct. This would seem in our best interest. In fact, by

convention we set a fairly strict significance level. In order to reject the

null hypothesis, we are expected to provide convincing evidence that

our conclusions reflect the true population parameters.

What is “convincing evidence”? How much risk of a Type I error

should we be willing to take in a test of statistical significance? In

criminal justice, and in most of the social sciences, a 5% level of statis-

tical significance is generally considered rigorous enough for tests of

hypotheses. This means that if the observed significance level of our

test is greater than 0.05, we will fail to reject the null hypothesis. If the

observed significance level is less than 0.05, we will reject the null

hypothesis.

D E P A R T I N G F R O M C O N V E N T I O N A L C R I T E R I A 137

D e p a r t i n g f r o m C o n v e n t i o n a l S i g n i f i c a n c e C r i t e r i a

In statistics, as in everyday life, it is simplest to follow the accepted con-

research may be better served by departing from established norms.

Sometimes this is the case because our criterion is not cautious enough

for the issue we are examining.

For example, a criminal justice agency making a decision about its

policies may want you to use a stricter standard than you would ordinar-

ily apply to your research. This may be the case if your conclusions

could lead to expensive or time-consuming changes in the structure or

activities of the agency. The agency will likely not want to make such

changes unless you are very confident that your conclusions are correct.

In this situation, you might decide to set your significance level at 0.01 or

0.001, meaning that you are willing to reject the null hypothesis only if

the observed significance level of a test is less than 1% or less than 0.1%.

Sometimes a researcher may decide to use a more lenient significance

level than 5%. This ordinarily occurs when the researcher is particularly

concerned with a Type II error rather than a Type I error. In the Min-

neapolis hot spots experiment, for example, which evaluated the impact

of police patrol on street blocks with high crime activity, the principal in-

vestigators discussed the dangers of a Type II error at the outset of their

study.3 They argued that conventional significance criteria might be too

strict for assessing the effectiveness of new police initiatives. Failure to

reject the null hypothesis of no program impact in this case, if it were

false, would lead the police not to pursue a potentially effective new

method of police patrol. The principal investigators, accordingly, de-

cided to use a 0.10, rather than a 0.05, significance level.

Why would a concern with a Type II error lead us to change the sig-

nificance level of our test of hypotheses? As we noted earlier, the signifi-

cance level is based on a Type I error, not a Type II error. However, the

two types of statistical error are related. When we increase the risk of a

Type I error, we reduce the risk of a Type II error. When we decrease

the risk of a Type I error, we increase the risk of a Type II error. This

3See Lawrence Sherman and David Weisburd, “General Deterrent Effects of Police

Patrol in Crime ‘Hot Spots’: A Randomized Study,” Justice Quarterly 12:4 (1995):

625–648.

ventions. Accordingly, most criminal justice researchers apply the 5%

significance level fairly automatically to the research questions they

consider. The problem with this approach is that the purposes of the

reject the null hypothesis in your study.

Whenever you use a significance level more stringent than 5%, it is

important to explain clearly why you have chosen to make it harder to

138 C H A P T E R S I X : T H E L O G I C O F S T A T I S T I C A L I N F E R E N C E

relationship is an obvious one, because setting a stricter significance

level (by decreasing the risk of a Type I error) naturally makes it less

likely that you will reject the null hypothesis. Similarly, when we make it

easier to reject the null hypothesis by increasing the risk of a Type I

error, it is that much harder in practice to fail to reject the null hypothe-

sis (and therefore to make a Type II error). The relationship between

Type I and Type II error is easy to understand, but it is not directly pro-

cept of “statistical power,” other factors in addition to the significance

level of a study affect the risk of a Type II error.

You should always consider carefully the implications of risks of error in

your research before setting the significance level of a test of statistical sig-

nificance. Even though you are likely in the end to rely on the conventional

norm of 5%, there are, as discussed above, some cases for which you might

want to consider stricter or more lenient levels of significance. If you do

choose a level other than 0.05, you must explain to your audience the fac-

tors that led you to depart from common practice in criminal justice.

The level of significance in a study should be defined at the outset,

and not after your results are in. If you wait until you have your sample

data, there will always be the temptation to adjust your significance level

to fit your sample statistics. This is particularly important if you decide to

use more stringent or more lenient criteria. If you use a more lenient cri-

terion, others might argue that you have made that decision in order to

allow rejection of the null hypothesis (and thus support for the research

hypothesis). If you use a stricter criterion, others might argue that you

are trying to avoid rejecting the null hypothesis. In many funded re-

search studies, researchers specify in their original proposals the signifi-

cance levels that they intend to apply to tests of hypotheses in order to

prevent such criticism later on.

In this chapter, we discussed the logic underlying statistical inference.

In the coming chapters, we will examine how statisticians define the risk

of Type I error associated with a specific outcome in a study, or the ob-

served significance level of a test. We also detail how such tests are ap-

plied to different types of statistics. You should not expect to have a full

understanding of statistical inference at this point. We have only begun

to develop these ideas and will return to them again and again.

C h a p t e r S u m m a r y

In descriptive statistics we are concerned with two types of distribu-

tions. The sample distribution is the distribution of scores in the sam-

ple. The population distribution is the distribution of scores in the

portional. As we will discuss in Chapter 21 when we examine the con-

C H A P T E R S U M M A R Y 139

population from which the sample is drawn. Population statistics are

known as parameters, and they have symbols different from those

which you have encountered so far. The mean in a population distribu-

tion is represented by � and the standard deviation by �. Parameters

are assumed to be fixed and are generally unknown. Sample statistics

are by definition known, but they vary from sample to sample. Statisti-

cians are faced with a fundamental dilemma in that they are usually in-

terested in making statements about population parameters, but they

generally study sample statistics. Statistical inference provides a solu-

tion to this dilemma.

Statistical inference begins with the definition of the research hy-

pothesis and the null hypothesis. These hypotheses are set up by the

researcher to answer the broader research question. The research hy-

pothesis is the proposed answer to a specific research question. When

the research hypothesis does not indicate a specific type of outcome,

stating only that there is a relationship or a difference, we say that it is a

nondirectional hypothesis. When the research hypothesis identifies

the nature of the relationship that is expected, it is called a directional

hypothesis. The null hypothesis generally posits that there is no such

relationship or no difference. The null hypothesis is stated in exact

terms. The research hypothesis encompasses a range of possible alterna-

Given the difficulties involved in collecting information on an entire

population, we are forced to work with samples. The tools of statistical

inference enable us to infer from a sample to a population by identifying

the risk of making a mistaken decision and determining the amount of

risk we are prepared to take. Two possible errors can be made when

making decisions about a population from a sample. A researcher who

rejects the null hypothesis when it is in fact true has made a Type I

error. A researcher who fails to reject the null hypothesis when it is in

fact false has made a Type II error.

In a test of statistical significance, we make a decision to reject or

to fail to reject the null hypothesis on the basis of a sample statistic. The

significance level defines the risk of Type I error that a researcher is

willing to take in a test of statistical significance. The risk of Type I error

associated with a specific sample statistic is the observed significance

level of a test. A commonly accepted standard significance level is 5%,

but researchers may choose to set a lower level if they want it to be

more difficult to reject the null hypothesis—or a higher level if they want

to make it easier. A researcher who wishes to depart from the accepted

standard should explain the reasons for such a decision at the outset.

Decreasing the risk of a Type I error increases the risk of a Type II error,

and vice versa.

hypothesis.

tives. It is thus easier to focus decisions on whether to reject the null

140 C H A P T E R S I X : T H E L O G I C O F S T A T I S T I C A L I N F E R E N C E

K e y T e r m s

directional hypothesis A research hy-

pothesis that indicates a specific type of

outcome by specifying the nature of the re-

lationship that is expected.

nondirectional hypothesis A research

hypothesis that does not indicate a specific

type of outcome, stating only that there is a

relationship or a difference.

null hypothesis A statement that reduces

the research question to a simple assertion

to be tested by the researcher. The null hy-

pothesis normally suggests that there is no

relationship or no difference.

observed significance level The risk of

Type I error associated with a specific sam-

ple statistic in a test. When the observed

significance level is less than the criterion

significance level in a test of statistical sig-

nificance, the researcher will reject the null

hypothesis.

parameter A characteristic of the pop-

ulation—for example, the mean number

of previous convictions for all U.S.

prisoners.

population The universe of cases that the

researcher seeks to study. The population

of cases is fixed at a particular time (e.g.,

the population of the United States). How-

ever, populations usually change across

time.

population distribution The frequency

distribution of a particular variable within a

population.

research hypothesis The antithesis of the

null hypothesis. The statement normally

answers the initial research question by

suggesting that there is a relationship or a

difference.

sample A set of actual observations or

cases drawn from a population.

sample distribution The frequency distri-

bution of a particular variable within a

sample drawn from a population.

sample statistic A characteristic of a sam-

ple—for example, the mean number of

previous convictions in a random sample

of 1,000 prisoners.

significance level The level of Type I

error a researcher is willing to risk in a sta-

tistical test of significance.

statistical inference The process of mak-

ing generalizations from sample statistics to

population parameters.

test of statistical significance A test in

which a researcher makes a decision to re-

ject or to fail to reject the null hypothesis

on the basis of a sample statistic.

Type I error Also known as alpha error.

The mistake made when a researcher re-

jects the null hypothesis on the basis of a

sample statistic (i.e., claiming that there is a

relationship) when in fact the null hypothe-

Type II error Also known as beta error.

The mistake made when a researcher fails

to reject the null hypothesis on the basis of

a sample statistic (i.e., failing to claim that

there is a relationship) when in fact the

null hypothesis is false (i.e., there actually

is a relationship).

universe The total population of cases.

by means of a study.

researcher hopes to be able to answer

research question The question the

relationship in the population).

sis is true (i.e., there is actually no such

E X E R C I S E S 141

S y m b o l s a n d F o r m u l a s

H0 Null hypothesis

H1 Research hypothesis

� Significance level of a test

� Population mean

p Observed significance level of a test

�2 Population variance

E x e r c i s e s

6.1 For each of the following random samples, describe the population to which the results could be generalized:

a. Subscribers to a magazine on hunting are asked about gun legislation.

b. Youth aged 15 to 19 years in the United States are asked about per- sonal drug use.

c. Registered voters in Los Angeles are asked whom they will vote for in the next election for mayor.

d. Visitors to a domestic violence shelter are assessed for psychologi- cal distress.

e. Members of the National Organization for Women (NOW) are asked about sexual harassment in the workplace.

f. Judges in New Jersey are asked about their sentencing philosophy.

6.2 A foundation sponsored a review of all studies carried out over the past 15 years into the link between smoking and juvenile delinquency. Eric, a criminologist commissioned by the foundation, unearthed five studies that sought to determine at what age delinquents who smoke began their habit. The five studies, conducted at the same time, use identical sampling techniques and sample sizes and draw their sam- ples from a fixed database of delinquents. The mean age, however, is different for each of the samples:

Study sample no. 1: mean age � 12.2

Study sample no. 2: mean age � 11.6

Study sample no. 3: mean age � 14.0

Study sample no. 4: mean age � 11.3

Study sample no. 5: mean age � 12.8

Overall computed mean � 12.38

142 C H A P T E R S I X : T H E L O G I C O F S T A T I S T I C A L I N F E R E N C E

a. During Eric’s presentation, one member of the foundation asks him to explain how it can be that each of the study samples produced a different result: Does this mean that there was something wrong with the sampling techniques used? How should Eric respond?

b. Another foundation member also seems confused. She asks if the overall computed mean is the mean age of the population. How should Eric respond?

6.3 A researcher collects data on the families of 20 delinquent and 20 nondelinquent children from the records of a school and checks how many children from each group come from broken homes.

a. State the null hypothesis.

b. State a nondirectional research hypothesis.

c. State a directional research hypothesis.

6.4 A researcher is interested in whether electronic monitoring of offend- ers on probation is an effective means of reducing crime. To test for an effect of electronic monitoring, probationers are randomly assigned either to a monitored group or to a control group that has no elec- tronic monitoring.

a. State the null hypothesis.

b. State a nondirectional research hypothesis.

c. State a directional research hypothesis.

6.5 A study published in a distinguished journal reported the results of a series of tests carried out on 50 convicted burglars. One of the claims of the investigators is that the average IQ of convicted burglars is 120.

a. From the following list of options, choose an appropriate null hy- pothesis and research hypothesis for testing this claim.

IQ � 120 IQ � 120 IQ � 120 IQ � 120

b. Explain your choice of null and research hypotheses.

6.6 A gang of criminals is planning to rob a supermarket. Eddy, the gang leader, reports that he “staked the store out” the day before and saw the store detective going for a 15-minute coffee break at 9.15 A.M. He sug- gests that this would be the best time to strike. Clive, a gang member, thinks that this plan is too risky—how do they know that the detective takes his break at the same time each day? Eddy, who is desperate for the money, thinks that the plan is safe enough and wants to carry out the robbery the next day. After an argument, they agree to compromise and watch the supermarket for three more days. On each of the three days, the store detective indeed takes his 15-minute break at 9:15 A.M. The gang decides to go ahead with the robbery on the fourth day.

The robbers can be seen as having set themselves a research ques- tion and having made a statistical decision based on a simple study.

E X E R C I S E S 143

a. How would you frame the robbers’ null hypothesis and research hypothesis?

b. Based on these hypotheses, what is their decision?

c. How would the robbers make a Type I error? How would the rob- bers make a Type II error? What type of statistical error ought the robbers to be most concerned with making? Explain why.

d. How does the argument between Eddy and Clive relate to the con- cept of statistical significance?

6.7 The government wishes to launch a pre-Christmas advertising cam- paign warning about the dangers of drunk driving. It suspects that dri- vers aged 18 to 21 are most likely to drive while under the influence of alcohol and is considering targeting the campaign specifically at this age group. A preliminary study gathers data on the ages of drunk dri- vers apprehended in a particular district over a six-month period.

b. How might the government make a Type I error?

c. How might the government make a Type II error?

d. The government accepts that targeting this specific age group in the advertising campaign will not cost any extra money. There is a feel- ing that the new campaign will be “worth a try,” even if the study doesn’t find enormous differences between the offending rate of 18- to 21-year-olds and that of other ages. How should these con- siderations affect the researchers’ decision on what level of signifi- cance to set?

6.8 The head of the police force in the city of Cheadle suspects that in- creasing the pay of his officers might increase their efficiency. A police researcher is assigned to test whether there is a difference between the crime-solving rates of a group of detectives who have been ran- domly awarded pay raises and a control group of detectives who have not been awarded pay raises. In writing up his report, the researcher concludes as follows:

The results show that the observed significance level is 0.14, mean- ing that rejecting the null hypothesis would run a 14% risk of a Type I error. Although a 5% significance level is considered stan- dard, in light of the potential benefits of salary increases for crime control rates, a higher 15% threshold is justified here, and the H0

may therefore be rejected.

a. What is the null hypothesis to which the researcher refers?

b. Explain why the researcher’s statistical reasoning is problematic.

research hypothesis is directional or nondirectional. a. What are the null and research hypotheses? Explain why the

144 C H A P T E R S I X : T H E L O G I C O F S T A T I S T I C A L I N F E R E N C E

6.9 A study explored whether there is a link between male aggression and climate. The researcher recorded her results as follows: 5% signifi- cance level set, H0 could not be rejected.

a. Explain these results in plain English.

b. Why is it important for the researcher to set the significance level at the beginning of the research and not at the end?

6.10 A private research foundation claims that increased regulation of handguns would reduce homicides in the United States. To examine this relationship, the foundation funds a study to assess the impact of handgun legislation on homicides in four states that recently passed laws restricting handgun ownership. The foundation expects that the reduced availability of handguns following the change in law will re- duce the number of opportunities for lethal violence. The researcher collects data on homicides in the four states for one year before and one year after the change in handgun laws.

b. Explain how the researcher could make a Type I error.

c. Explain how the researcher could make a Type II error.

dation wants the researcher to increase the significance level of the study to 10% and reject the null hypothesis. Should the researcher increase the significance level of the study? Why?

6.11 A group of researchers at a private think tank claims that the increased use of incarceration in the United States has not been harmful to the social fabric of the country. To support this claim, the researchers con- duct a study looking at rates of incarceration and rates of divorce in all fifty states for a 20-year period. The group of researchers tests for a relationship between rates of incarceration and rates of divorce, but does not expect to find a relationship between these two variables.

b. Explain how the research group could make a Type I error.

c. Explain how the research group could make a Type II error.

d. The observed significance level developed in their study is 0.03. The research group initially set a 5% risk of Type I error. One member of the research group suggests that they simply decrease the significance level of the study to 1% and fail to reject the null hypothesis. Should the research group decrease the significance level of the study? Why?

a. What are the null and research hypotheses? Explain why the research hypothesis is directional or nondirectional.

d. The results show an observed significance level of 0.06. The researcher wants to conclude that the null hypothesis cannot be rejected based on a 5% risk of Type I error. An official from the foun-

research hypothesis is directional or nondirectional. a. What are the null and research hypotheses? Explain why the

Defining the Observed Significance Level

of a Test: A Simple Example Using

the Binomial Distribution

What Role Do They Play in Inferential Statistics?

How Does One Calculate the Probability of a Given Outcome?

C h a p t e r s e v e n

S a m p l i n g d i s t r i b u t i o n s

P r o b a b i l i t i e s a n d p r o b a b i l i t y d i s t r i b u t i o n s

T h e b i n o m i a l d i s t r i b u t i o n

What are They?

What are Probability Distributions?

How are They Used?

What is It?

How is It Calculated?

What are Its Characteristics?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_7,

© Springer Science+Business Media New York 2014

WHEN WE MAKE INFERENCES to a population, we rely on a statistic in

our sample to make a decision about a population parameter. At the

heart of our decision is a concern with Type I error. Before we reject

our null hypothesis, we want to be fairly confident that it is in fact false

for the population we are studying. For this reason, we want the ob-

served risk of a Type I error in a test of statistical significance to be as

small as possible. But how do statisticians calculate that risk? How do

they define the observed significance level associated with the outcome

of a test?

The methods that statisticians use for calculating the observed sig-

nificance level of a test of statistical significance vary depending on

the statistics examined. Sometimes these methods are very complex.

But the overall logic that underlies these calculations is similar, irre-

spective of the statistic used. Thus, we can take a relatively simple

example and use it as a model for understanding how the observed

significance level of a test is defined more generally in statistics. This

is fortunate for us as researchers, because it means that we do not

have to spend all of our time developing complex calculations to de-

fine risks of error. Once we understand how risks of error are defined

for one problem, we can let statisticians calculate the risks for other

more complex problems. Our concern is not with the calculations

themselves, but with understanding the general logic that underlies

them.

We begin this chapter by discussing a very simple decision. When

should we begin to suspect that a coin used in a coin toss is unfair or bi-

ased? Ordinarily, we might come to a conclusion based on common

sense or intuition. In statistics, we take a more systematic approach, rely-

ing on the logic of hypothesis testing and a type of distribution called a

sampling distribution. Using this example of the coin toss and a sam-

pling distribution called the binomial distribution, we illustrate how sta-

tisticians use probability theory to define the observed significance level,

or risk of Type I error, for a test of statistical significance.

146

T H E F A I R C O I N T O S S 147

T h e F a i r C o i n T o s s

Imagine that you and your friends play a volleyball game each week

against a group of criminal justice students from another school. You al-

ways begin the game with a coin toss to decide who will serve the ball

first. Your opponents bring an old silver dollar, which you have agreed

to use for the toss. They choose heads and continue to choose heads

each time you play. At first, this does not seem like a problem. However,

each week you play, the coin comes up heads and they serve the ball.

Suppose that this happened for four straight weeks. Would you begin

to become suspicious? What if it went on for six weeks? How many

times in a row would they have to win the coin toss before you and

your team accused them of cheating? Would they have to win for ten or

twenty weeks? You might worry about accusing them too quickly, be-

cause you know that even if the coin is fair it sometimes happens that

someone is lucky and just keeps on winning. You would want to be

fairly certain that the coin was biased before concluding that something

was wrong and taking some action.

In everyday life, you are likely to make this decision based on intu-

ition or prior experience. If you ask your classmates, each one is likely

to come up with a slightly different number of coin tosses before he or

she would become suspicious. Some students may be willing to tolerate

only four or five heads in a row before concluding that they have

enough evidence to accuse their opponents of cheating. Others may be

unwilling to reach this conclusion even after ten or fifteen tosses that

come up heads. In part, the disagreement comes from personality differ-

ences. But more important, guesswork or common sense does not give

you a common yardstick for deciding how much risk you take in coming

to one conclusion or another.

Sampling Distributions and Probability Distributions

Statistical inference provides a more systematic method for making deci-

sions about risk. The coin toss can be thought of as a simple test of statisti-

cal significance. The research hypothesis is that the coin is biased in favor

of your opponents. The null hypothesis is that the coin is fair. Each toss of

the coin is an event that is part of a sample. If you toss the coin ten times,

you have a sample of ten tosses. Recall from Chapter 6 that Type I error

is the error of falsely rejecting the null hypothesis that the coin is fair. If

you follow the common norm in criminal justice, then you are willing to

reject the null hypothesis if the risk of a Type I error is less than 5%.

But how can we calculate the risk of a Type I error associated with a

specific outcome in a test of statistical significance, or what we generally

term the observed significance level of a test? One simple way to gain an

148 C H A P T E R S E V E N : D E F I N I N G T H E O B S E R V E D R I S K

estimate of the risk of unfairly accusing your friends is to check how

often a fair coin would give the same result as is observed in your series

of volleyball games. For example, let’s say that you played ten games in a

season and in all ten games the old silver dollar came up heads (meaning

that the opposing team won the toss). To check out how often this might

happen just by chance when the coin is in fact a fair one, you might go

to a laboratory with a fair coin and test this out in practice. One problem

you face is deciding how many samples or trials you should conduct. For

example, should you conduct one trial or sample by flipping your fair

coin just ten times and stopping? Or should you conduct multiple trials or

samples, each with ten tosses of the fair coin? Clearly, one trial or sample

of ten tosses will not tell you very much. Indeed, one of the reasons you

have gone to the laboratory is that you know it sometimes happens that a

fair coin will come out heads ten times in a row. What you want to know

is how rare an event this is. How often would you gain ten heads in a

row in a very large number of samples or trials of a fair coin?

The distribution that is gained from taking a very large number of

samples or trials is called a sampling distribution. In principle, one

could create a sampling distribution by drawing thousands and thou-

sands of samples from a population. For example, in the case of our

coin toss, we might conduct thousands of trials of ten flips of a fair coin.

If we recorded the outcome for each trial and placed our results in a fre-

quency distribution, we would have a sampling distribution for a sample

of ten tosses of a fair coin.

This sampling distribution would allow us to define the risk of a Type

I error we would face in rejecting the null hypothesis that the old silver

dollar is fair. For example, suppose that in the sampling distribution we

gained a result of ten heads in only 1 in 1,000 samples. If we reject the

null hypothesis in this case, our risk of making a Type I error, according

to the sampling distribution, is only 0.001. This is the observed signifi-

cance level of our test of statistical significance. In only 1 in 1,000 sam-

ples of ten tosses of a fair coin would we expect to gain a result of ten

heads. If the old silver dollar was indeed a fair coin, it would seem very

unlikely that on our one trial of ten tosses of the coin each toss would

come out heads. Of course, in making our decision we cannot be certain

that the silver dollar used in the volleyball toss is not a fair coin. While

the occurrence of ten heads in ten tosses of a fair coin is rare, it can hap-

pen about once in every 1,000 samples.

Building a sampling distribution provides a method for defining our

risk of a Type I error. However, it is very burdensome to create a sam-

pling distribution by hand or even in the laboratory. If you try out our

example of ten tosses of a fair coin, you will see that developing even

100 samples is not easy. If we had to actually construct a sampling distri-

bution every time we wanted to make a decision about a hypothesis, it

would be virtually impossible to make statistical inferences in practice.

T H E F A I R C O I N T O S S 149

Fortunately, there is another method we can use for creating sampling

distributions. This method relies on probability theory, rather than a bur-

densome effort to collect samples in the real world. Because we use

probabilities, the distributions that are created using this method are

called probability distributions. Importantly, though we rely on prob-

ability theory because it is very difficult to develop sampling distributions

in practice, we do not suffer for our approach. This is because probabil-

ity theory allows us to calculate the outcomes one would expect in a

perfect world. In the real world, we might flip the coin slightly differ-

The Multiplication Rule

In order to estimate the risk of a Type I error in the case of a series of

tosses of a fair coin, we can use the multiplication rule, a simple rule

about probabilities drawn from probability theory. The multiplication

rule tells you how likely we are to gain a series of events one after

another—in this case, a series of outcomes in a toss of a coin. It allows

us to estimate theoretically how often we would gain a specific series of

events if we drew an infinite number of samples. The multiplication rule

generally used to establish probabilities in statistics is based on the as-

sumption that each event in a sample is independent of every other

event. In the case of the coin toss, this means that the outcome of one

toss of a coin is unaffected by what happened on the prior tosses. Each

time you toss the coin, it is as if you started with a clean slate. That

would seem a fairly reasonable assumption for our problem. What wor-

ries us is that the coin is unfair overall, not that it is becoming less or

more unfair as time goes on.

An example of a series of events that are not independent is draws

from a deck of cards. Each time you draw a card, you reduce the num-

ber of cards left in the deck, thus changing the likelihood of drawing any

card in the future. For example, let’s say that on your first draw from a

deck of 52 cards you drew an ace of spades. On your second draw, you

cannot draw an ace of spades because you have already removed it from

the deck. The likelihood of drawing an ace of spades on the second

draw has thus gone from 1 in 52 to 0 in 51. You have also influenced the

likelihood of drawing any other card because there are now 51, not 52,

cards left in the deck. If you want a series of draws from a deck to be

independent of one another, you have to return each card to the deck

after you draw it. For example, if you returned the ace of spades to the

deck, the chance of choosing it (assuming the deck was mixed again)

would be the same as it was on the first draw. The chances of choosing

any other card would also be the same because you once again have all

52 cards from which to draw.

another, thus affecting the outcomes we gain. In probability theory, we

remove the imperfections of the real world from our estimates.

ently as we got tired or the coin might become worn on one side or

150 C H A P T E R S E V E N : D E F I N I N G T H E O B S E R V E D R I S K

The multiplication rule for four independent events is stated in Equa-

tion 7.1. It says that the likelihood of any series of events, represented as

A, B, C, and D, happening one after another is equal to the probability

of event A times the probability of event B times the probability of event

C times the probability of event D. The rule can be extended to as many

events as you like. We have chosen four here, because this was the

number of volleyball games we began with. If you wanted to extend the

rule, you would simply include the additional events on the left side of

the equation (for example, E and F ) and include the probability of each

on the right side of the equation [e.g., P (E ) • P (F )].

P (A & B & C & D) � P (A) • P (B) • P (C ) • P (D) Equation 7.1

Extending this logic to our example of the coin toss is straightforward.

The probability of A and B and C and D can represent the probability

that four tosses in a row come up heads. Our main problem is to estab-

lish what the probability is of a head coming up on any particular toss of

the coin. In this, we are helped by our null hypothesis, which states that

the coin is fair. If the coin is fair, then there should be an even chance of

a head or a tail coming up on any particular toss of the coin. Put differ-

ently, under the assumption of the null hypothesis that the coin is fair,

the likelihood of a head coming up is 0.50.

What, then, does the multiplication rule tell us about the chances of

getting four heads in a row on four tosses of a fair coin? In part a of

Table 7.1, we calculate that probability by multiplying 0.50 (the likeli-

hood of gaining a head on any toss of a fair coin) by itself four times, to

represent four tosses of an unbiased coin. The result is 0.0625. If you

had decided at the outset to make a decision about the null hypothesis—

that the coin is fair—after four tosses of the coin, then you have con-

ducted a type of test of statistical significance. For this test, the observed

significance level (or risk of a Type I error) of rejecting the null hypothe-

sis on the basis of four heads is 0.0625.

If you use the norms of criminal justice research, this is not enough,

however, for you to reject the null hypothesis that the coin is fair. Crimi-

nal justice researchers generally want the risk of falsely rejecting the null

hypothesis to be less than 5%. A bit over 6% is still more than the 5% sig-

nificance criterion that is used by convention. So if you had decided to

Probabilities Associated with Tosses of a Fair Coin

a. P (A & B & C & D) � P (A) • P (B) • P (C) • P (D) � (0.50)(0.50)(0.50)(0.50) � 0.0625

b. P (A & B & C & D & E) � P (A) • P (B) • P (C) • P (D) • P (E) � (0.50)(0.50)(0.50)(0.50)(0.50) � 0.0313

c. P (A & B & C & D & E & F & G & H & I & J) � P (A) • P (B) • P (C) • P (D) • P (E) • P (F ) • P (G) • P (H) • P (I ) • P (J) � (0.50)(0.50)(0.50)(0.50)(0.50)(0.50)(0.50)(0.50)(0.50)(0.50) � 0.0010

Table 7.1

D I F F E R E N T W A Y S O F G E T T I N G S I M I L A R R E S U L T S 151

make a decision about the fairness of the coin after four coin tosses, you

would probably not want to reject the null hypothesis that the coin is fair

and confront your opponents. Under this criterion, the likelihood of

falsely rejecting the null hypothesis, or the observed significance level of

your test, would have to be below 0.05.

What if you had decided at the outset to make a decision about the

null hypothesis after five tosses of a coin? Would a result of five heads in

a row lead you to reject the null hypothesis? As illustrated in part b of

Table 7.1, the multiplication rule tells you that the likelihood of getting

five heads in a row if the coin is fair is 0.0313. This is less than our

threshold of 0.05, and thus would lead you to reject the null hypothesis.

Is this consistent with your earlier commonsense conclusions? Students

are usually surprised at how quickly they reach the 0.05 significance

threshold in this example.

If you had decided at the outset that you would need ten or fifteen

heads in a row, you may want to reconsider, given what we have

learned from the multiplication rule. The likelihood of getting ten heads

in a row in ten tosses of a fair coin is only 1 in 1,000 (see part c of Table

7.1). The likelihood of getting fifteen heads in a row in fifteen tosses of a

fair coin is even lower: about 3 in 100,000. In both of these cases, you

would take a very small risk of a Type I error if you rejected the null hy-

pothesis. Nonetheless, the multiplication rule tells us that, even if the

coin is fair, it is possible to get ten or even fifteen heads in a row. It just

does not happen very often.

The multiplication rule allows us to estimate how often we would ex-

pect to get a series of specific outcomes in a very large number of trials

or samples, without actually going out and doing the hard work of con-

structing a sampling distribution in the real world. However, the problem

as examined so far assumes that the coin will come up heads every time.

What if the coin comes up heads generally, but not all the time? For ex-

ample, what if you play ten games and the coin comes up heads nine

times? The situation here is not as one-sided. Nonetheless, it still seems

unlikely that your opponents would win most of the time if the coin

were fair. The multiplication rule alone, however, does not allow us to

define how likely we are to get such a result.

D i f f e r e n t W a y s o f G e t t i n g S i m i l a r R e s u l t s

The multiplication rule allows us to calculate the probability of getting a

specific ordering of events. This is fine so far in our coin toss because in

each example we have chosen there is only one way to get our out-

come. For example, there is only one way to get five heads in five coin

152 C H A P T E R S E V E N : D E F I N I N G T H E O B S E R V E D R I S K

tosses or ten heads in ten coin tosses. In each case, your opponents

must toss a head before each game. This would be the situation as well

if your opponents tossed tails ten times in ten coin tosses. However, for

any outcome in between, there is going to be more than one potential

way to achieve the same result.

For example, if your opponents tossed nine heads in ten coin tosses,

they could win the coin toss nine times (with a head) and then lose the

toss in the tenth game (with a tail). Or they could lose the first toss (with

a tail) and then win the remaining nine. Similarly, they could lose the

second, third, fourth, fifth, sixth, seventh, eighth, or ninth coin toss and

win all the others. Each of these possible ordering of events is called an

arrangement. As is illustrated in Table 7.2, there are ten possible

arrangements, or different ways that you could get nine heads in ten

coin tosses. In the case of ten heads in ten coin tosses, there is only one

possible arrangement.

It is relatively simple to list all of the arrangements for our example of

nine heads in ten coin tosses, but listing becomes very cumbersome in

practice as the split of events becomes more even. For example, if we

were interested in how many ways there are of getting eight heads in ten

coin tosses, we would have to take into account a much larger number

of arrangements. As Table 7.3 illustrates, it takes a good deal of effort to

list every possible arrangement even for eight heads. In the case of a

more even split of events—for example, five heads in ten tosses—it be-

comes extremely cumbersome to list each arrangement one by one. Be-

cause of this, we generally use the formula in Equation 7.2 to define the

number of arrangements in any series of events.

Equation 7.2

On the left side of this equation we have N “choose” r, where N is the

number of events in the sample and r is the number of successes in the

�N

r� � N !

r !(N � r)!

Arrangements for Nine Successes in Ten Tosses of a Coin

Arrangement 1 � � � � � � � � � � Arrangement 2 � � � � � � � � � � Arrangement 3 � � � � � � � � � � Arrangement 4 � � � � � � � � � � Arrangement 5 � � � � � � � � � � Arrangement 6 � � � � � � � � � � Arrangement 7 � � � � � � � � � � Arrangement 8 � � � � � � � � � � Arrangement 9 � � � � � � � � � � Arrangement 10 � � � � � � � � � �

� � Head; � � Tail

Table 7.2

D I F F E R E N T W A Y S O F G E T T I N G S I M I L A R R E S U L T S 153

total number of events. In our case, N is the number of coin tosses and r

is the number of times that the coin comes up heads. Put together, this

statement establishes our question: How many ways are there of gaining

r heads in N tosses of a coin? To answer our question, we need to solve

the right side of the equation. Each of the terms in the equation is de-

fined as a factorial, indicated by the symbol !. When we take a factorial

of a number, we merely multiply it by all of the whole numbers smaller

than it. For example, 3! is equal to (3)(2)(1), or 6. Because factorials get

very large very quickly, a table of factorials is presented in Appendix 1.

Note that 0! � 1. Applied to our problem of nine heads in ten coin

tosses, Equation 7.2 is worked out below:

Arrangements for Eight Successes in Ten Tosses of a Coin

1: ���������� 16: ���������� 31: ���������� 2: ���������� 17: ���������� 32: ���������� 3: ���������� 18: ���������� 33: ���������� 4: ���������� 19: ���������� 34: ���������� 5: ���������� 20: ���������� 35: ���������� 6: ���������� 21: ���������� 36: ���������� 7: ���������� 22: ���������� 37: ���������� 8: ���������� 23: ���������� 38: ���������� 9: ���������� 24: ���������� 39: ����������

10: ���������� 25: ���������� 40: ���������� 11: ���������� 26: ���������� 41: ���������� 12: ���������� 27: ���������� 42: ���������� 13: ���������� 28: ���������� 43: ���������� 14: ���������� 29: ���������� 44: ���������� 15: ���������� 30: ���������� 45: ����������

� � Head; � � Tail

Table 7.3

W orking It Out

� 10

� 3,628,800

362,880(1)

� 10!

9! 1!

�10 9 � �

10! 9!(10 � 9)!

�N

r� � N !

r !(N � r)!

154 C H A P T E R S E V E N : D E F I N I N G T H E O B S E R V E D R I S K

Using this method, we get the same result as before. There are ten possi-

ble arrangements to get nine heads in ten tosses of a coin. When we

apply Equation 7.2 to the problem of five heads in ten coin tosses, its

usefulness becomes even more apparent. There are 252 different ways of

getting five heads in ten tosses. Listing each would have taken us consid-

erably longer than the calculation below.

W orking It Out

� 252

� 3,628,800

120(120)

� 10!

5! 5!

�10 5 � �

10! 5!(10 � 5)!

�N

r� � N !

r !(N � r)!

S o l v i n g M o r e C o m p l e x P r o b l e m s

Now that we have a method for calculating arrangements, we can return

to our original problem, which was to define the probability of your op-

ponents tossing the coin in ten games and getting heads nine times. Be-

cause there are ten different ways of getting nine heads in ten coin

tosses, you need to add up the probabilities associated with these ten se-

quences. This is what is done in Table 7.4. The multiplication rule is

used to calculate the probability for each sequence, or arrangement,

under the assumption of the null hypothesis that the coin is fair. Because

our null hypothesis states that the coin is fair, we can assume that the

chances of gaining a head and a tail are even. The probability of any

event, whether a head or a tail, is 0.50, and the probability of a sequence

of ten events is always the same. This makes our task easier. But it is im-

portant to note that if the null hypothesis specified an uneven split (for

example, 0.75 for a head and 0.25 for a tail), then each of the sequences

would have a different probability associated with it. In any case, the

likelihood of getting any one of these sequences is about 0.001, rounded

to the nearest thousandth. When we add together the ten sequences, we

get a probability of 0.010.

T H E B I N O M I A L D I S T R I B U T I O N 155

This means that we would expect to get nine heads in ten coin tosses

of a fair coin in only about 1 in 100 samples in a very large number of

trials of a fair coin. But is this the observed significance level of a test of

statistical significance in which we gain nine heads in ten tosses of a

coin? Or put in terms of Type I error, is this the total amount of risk we

face of falsely rejecting the null hypothesis when we gain nine heads?

The answer to this question is no, although it may be difficult at first to

understand why. If we are willing to reject the null hypothesis based on

an outcome of nine heads in ten trials, then we are, by implication, also

willing to reject the null hypothesis if our outcome is ten heads in ten tri-

als. In calculating our total risk of a Type I error, we must add together

the risk of all potential outcomes that would lead us to reject the null hy-

pothesis. This is why, when testing hypotheses, we generally do not

begin with an estimate of the specific probability associated with a single

outcome, but rather with the sampling distribution of probabilities of all

possible outcomes.

T h e B i n o m i a l D i s t r i b u t i o n

To construct a probability or sampling distribution for all of the possible

outcomes of ten coin tosses, we could continue to compute the number

of permutations and the likelihood of any particular arrangement. How-

ever, Equation 7.3 provides us with a more direct method for calculating

the probability associated with each of the potential outcomes in our

The Sum of Probabilities for All Arrangements of Nine Heads in Ten Tosses of a Fair Coin

PROBABILITY

Arrangement 1 � � � � � � � � � � 0.001 Arrangement 2 � � � � � � � � � � 0.001 Arrangement 3 � � � � � � � � � � 0.001 Arrangement 4 � � � � � � � � � � 0.001 Arrangement 5 � � � � � � � � � � 0.001 Arrangement 6 � � � � � � � � � � 0.001 Arrangement 7 � � � � � � � � � � 0.001 Arrangement 8 � � � � � � � � � � 0.001 Arrangement 9 � � � � � � � � � � 0.001 Arrangement 10 � � � � � � � � � � 0.001

Total Probability: 0.01

Probability of throwing each arrangement of 10 throws � P (A) • P (B) • P (C) • P (D) • P (E) • P (F ) • P (G) • P (H) • P (I ) • P (J ) � (0.50)(0.50)(0.50)(0.50)(0.50)(0.50)(0.50)(0.50)(0.50)(0.50) � 0.001

Table 7.4

156 C H A P T E R S E V E N : D E F I N I N G T H E O B S E R V E D R I S K

sample. Equation 7.3 is generally defined as the binomial formula, and

the distribution created from it is called the binomial distribution. As

the name suggests, the binomial distribution is concerned with events in

which there are only two possible outcomes—in our example, heads

and tails.

Equation 7.3

The binomial formula may look confusing, but most of it is familiar from

material already covered in this chapter. The left-hand side of the equa-

tion represents the quantity in which we are interested—the probability

of getting r successes (in our case, r heads) in a sample of N events (for

us, ten tosses of a coin). The first part of the equation provides us with

the number of arrangements for that number of heads. This quantity is

then multiplied by pr(1 � p)N�r, where p is the probability of a successful

outcome (a head) under the null hypothesis and r is the number of suc-

cesses. This formula gives us the probability associated with each

arrangement. Although this part of the equation looks somewhat differ-

ent from the multiplication rule we used earlier, it provides a shortcut for

getting the same result, as the example below illustrates.

We have already calculated the likelihood of getting nine or ten heads

in ten coin tosses if the coin is fair. To complete our sampling distribu-

tion, we need to compute probabilities associated with zero through

eight heads as well. Let’s begin with eight heads in ten coin tosses of an

unbiased coin:

Step 1: Calculating the number of arrangements

P �10 8 � �

10! 8!(10 � 8)!

(0.50)8(1 � 0.50)10�8

P �N

r� � N !

r !(N � r)! pr(1 � p)N�r

W orking It Out

� 45

� 3,628,800

40,320(2)

� 10!

8! 2!

�10 8 � �

10! 8!(10 � 8)!

T H E B I N O M I A L D I S T R I B U T I O N 157

In step 1 we simply follow the same method as we did earlier in estab-

lishing the number of ways of getting eight heads in ten tosses of a coin.

Our conclusion is that there are 45 different arrangements.

Step 2: Calculating the probability of any specific arrangement

W orking It Out

� 0.00098

� (0.50)10

� (0.50)8(0.50)2

pr(1 � p)N�r � (0.50)8(1 � 0.50)10�8

Step 2 provides us with the likelihood of getting any particular arrange-

ment under the assumption of the null hypothesis that the coin is fair. By

the null hypothesis, p is defined as 0.50, and r is the number of suc-

cesses (heads) in our example, or 8. So pr is (0.50)8, and (1 � p)N�r is

(1 � 0.50)10�8, or (0.50)2. The outcome of this part of the equation can

be simplified to (0.50)10. This in turn is the same outcome that we would

obtain using the multiplication rule, because the expression (0.50)10

means that we multiply the quantity 0.50 by itself 10 times. Using the

multiplication rule, we would have done just that.

Step 3: Combining the two outcomes

W orking It Out

� 0.0441

P �10 8 � � 45(0.00098)

P �N

r� � N !

r !(N � r)! pr(1 � p)N�r

of tossing eight heads in ten tosses of a fair coin is about 0.044. In

Combining the two parts of the equation, we find that the likelihood

Table 7.5, we calculate the probabilities associated with all the potential

158 C H A P T E R S E V E N : D E F I N I N G T H E O B S E R V E D R I S K

The probability or sampling distribution for ten tosses of a fair coin il-

lustrates how likely you are to get any particular outcome. All of the out-

comes together add up to a probability of 1.1 Put differently, there is a

100% chance that in ten tosses of a coin you will get one of these 11 po-

tential outcomes. This is obvious, but the sampling distribution allows

you to illustrate this fact. Following what our common sense tells us, it

also shows that an outcome somewhere in the middle of the distribution

Computation of Probability Distribution for Ten Tosses of a Fair Coin

0 heads

1 head

2 heads

3 heads

4 heads

5 heads

6 heads

7 heads

8 heads

9 heads

10 heads

� � 1.0*

*The total in the last column is in fact slightly greater than 100%. This is due to rounding the numbers to the nearest decimal place in order to make the calculation more manageable.

3,628,800

3,628,800(10 � 10)! �

3,628,800

3,628,800 � 1 1(0.00098) � 0.0010

3,628,800

362,880(10 � 9)! �

3,628,800

362,880 � 10 10(0.00098) � 0.0098

3,628,800

40,320(10 � 8)! �

3,628,800

80,640 � 45 45(0.00098) � 0.0441

3,628,800

5,040(10 � 7)! �

3,628,800

30,240 � 120 120(0.00098) � 0.1176

3,628,800

720(10 � 6)! �

3,628,800

17,280 � 210 210(0.00098) � 0.2058

3,628,800

120(10 � 5)! �

3,628,800

14,400 � 252 252(0.00098) � 0.2470

3,628,800

24(10 � 4)! �

3,628,800

17,280 � 210 210(0.00098) � 0.2058

3,628,800

6(10 � 3)! �

3,628,800

30,240 � 120 120(0.00098) � 0.1176

3,628,800

2(10 � 2)! �

3,628,800

80,640 � 45 45(0.00098) � 0.0441

3,628,800

1(10 � 1)! �

3,628,800

362,880 � 10 10(0.00098) � 0.0098

3,628,800

1(10 � 0)! �

3,628,800

3,628,800 � 1 1(0.00098) � 0.0010

�N r �pr (1 � p)N�r�N

r � � N !

r ! (N � r )!

Table 7.5

1Because of rounding error, the total for our example is actually slightly larger than 1

(see Table 7.5).

outcomes in this binomial distribution. The resulting sampling distribution

is displayed in Table 7.6.

U S I N G T H E B I N O M I A L D I S T R I B U T I O N 159

is most likely. If the coin is fair, then we should more often than not get

about an even split of heads and tails.

The largest proportion (0.247) in the sampling distribution is found at

five heads in ten tosses of a coin. As you move farther away from the

center of the distribution, the likelihood of any particular result declines.

The smallest probabilities are associated with gaining either all heads or

no heads. Like many of the distributions that we use in statistics, this dis-

tribution is symmetrical. This means that the same probabilities are asso-

ciated with outcomes on both sides.

In Chapter 6, we talked about the fact that samples vary from one an-

other. This is what makes it so difficult to make inferences from a sam-

ple to a population. Based on a sample statistic, we can never be sure

about the actual value of the population parameter. However, as illus-

trated in this sampling distribution, samples drawn from the same popu-

lation vary in a systematic way in the long run. It is very unlikely to draw

a sample with ten heads in ten tosses of a fair coin. On the other hand, it

is very likely to draw a sample with four, five, or six heads in ten tosses.

U s i n g t h e B i n o m i a l D i s t r i b u t i o n t o E s t i m a t e t h e O b s e r v e d S i g n i f i c a n c e L e v e l o f a T e s t

Using the sampling distribution, we can now return to the problem of

identifying the risks of error associated with rejecting the null hypothesis

that the coin brought by the other volleyball team is fair. Earlier we sug-

gested that you might want to use a 5% significance level for this test, in

part because it is the standard or conventional significance level used by

Probability or Sampling Distribution for Ten Tosses of a Fair Coin

0 heads 0.001 1 head 0.010 2 heads 0.044 3 heads 0.118 4 heads 0.206 5 heads 0.247 6 heads 0.206 7 heads 0.118 8 heads 0.044 9 heads 0.010 10 heads 0.001

Table 7.6

most criminal justice researchers. This means that in order to reject the null

hypothesis you would require that the observed significance level (p) of your

test (or the risk of making a Type1 error by incorrectly rejecting the null hypo-

to reject the null hypothesis that the coin is fair and confront your opponents?

thesis) be less than 5% (or p < .05). Using this level, when would you be willing

Applying the Binomial Distribution to Situations Where p � 0.5

The examples in the text focus on applying the binomial distribution to sit-

uations where the probability of a success is equal to 0.5. There are other

situations where we are interested in the probability of multiple suc-

cesses (or failures), but success and failure are not equally likely. For ex-

ample, many of the games of chance that a person might play at a casino

are constructed in such a way that winning and losing are not equally

likely—the chances of losing are greater than the chances of winning—

but use of the binomial distribution would allow for calculation of the

chances of winning over several plays of the game.

Consider the following more detailed example. Suppose that we have a

quiz with five questions and we are interested in the probability of a student

sponse on any single question is p � 1/2 � 0.5. We can then apply the bino-

mial in the same way as we have in the previous examples to determine the

probability of some number of correct answers. The following table pre-

sents the numbers of correct answers and the corresponding probabilities.

NUMBER OF CORRECT ANSWERS

0 correct

1 correct

2 correct

3 correct

4 correct

5 correct

Now suppose that the questions are worded as multiple-choice items

and the student has to choose one answer from four possibilities. For any

single question, the probability of guessing the correct answer is p � 1/4

� 0.25. Given that we have multiple questions, we can again calculate the

5!

5!(5 � 5)! 0.55 (1 � 0.5)5�5 � 0.03125

5!

4!(5 � 4)! 0.54 (1 � 0.5)5�4 � 0.15625

5!

3!(5 � 3)! 0.53 (1 � 0.5)5�3 � 0.3125

5!

2!(5 � 2)! 0.52 (1 � 0.5)5�2 � 0.3125

5!

1!(5 � 1)! 0.51 (1 � 0.5)5�1 � 0.15625

5!

0!(5 � 0)! 0.50 (1 � 0.5)5�0 � 0.03125

�N r �pr (1 � p)N�r

Computation

of Binomial

Probabilities for

Five True-False

Questions

answers are true or false, then the probability of guessing the correct re-

correctly guessing all of the answers on the quiz. If the only possible

probability for the number of correct responses using the binomial distri-

bution, but we need to replace p � 0.5 with p � 0.25 in the equations to

reflect the different probability of a correct answer. The following table

presents the numbers of correct responses and the corresponding prob-

abilities for the multiple-choice response set.

NUMBER OF CORRECT ANSWERS

0 correct

1 correct

2 correct

3 correct

4 correct

5 correct

It is important to note that the distribution presented in the second table

is no longer symmetrical, reflecting the fact that the probability of a cor-

rect response is no longer equal to the probability of an incorrect re-

sponse. For the true-false questions, where the probabilities of correct

and incorrect answers are the same, we see that the probabilities of zero

and five correct responses are equal, the probabilities of one and four cor-

rect responses are equal, and the probabilities of two and three correct

responses are equal. In contrast, when we look at the probabilities for

multiple-choice questions with four possible answers, there is no such

symmetry. The most likely outcome is one correct response, with a proba-

bility of 0.3955. The probability of guessing four or five correct multiple-

choice answers is much lower than the probability of guessing four or five

correct true-false answers. In general, the probabilities in the table show

that increasing the number of possible answers makes it much more diffi-

cult for the student to correctly guess all the answers and increases the

chances of getting no correct responses or only one correct response.

5!

5!(5 � 5)! 0.255 (1 � 0.25)5�5 � 0.0010

5!

4!(5 � 4)! 0.254 (1 � 0.25)5�4 � 0.0146

5!

3!(5 � 3)! 0.253 (1 � 0.25)5�3 � 0.0879

5!

2!(5 � 2)! 0.252 (1 � 0.25)5�2 � 0.2637

5!

1!(5 � 1)! 0.251 (1 � 0.25)5�1 � 0.3955

5!

0!(5 � 0)! 0.250 (1 � 0.25)5�0 � 0.2373

�N r �pr (1 � p)N�r

Computation

of Binomial

Probabilities

for Five

Multiple-

Choice

Questions

162 C H A P T E R S E V E N : D E F I N I N G T H E O B S E R V E D R I S K

At first glance, you might decide to reject the null hypothesis for out-

comes of zero, one, two, eight, nine, and ten heads. Each of these is

below the threshold of 0.05 that we have suggested. However, at the

outset we stated in our research hypothesis that we were concerned not

that the coin was biased per se, but that it was biased against your team.

This means that we set up our research hypotheses in such a way that

we would reject the null hypothesis only if the outcomes were mostly

heads. Although tossing zero, one, or two heads is just as unlikely as

tossing eight, nine, or ten heads, our research hypothesis states our in-

tention not to consider the former outcomes.

What about the risk of falsely rejecting the null hypothesis in the case

of eight, nine, or ten heads? As we noted earlier, in calculating the risk of

a Type I error, we must add up the probabilities associated with all the

outcomes for which we would reject the null hypothesis. So, for exam-

ple, if we want to know the risk of falsely rejecting the null hypothesis

on the basis of eight heads in ten coin tosses, we have to add together

the risks associated with eight, nine, and ten heads in ten tosses. The

question we ask is, What is the risk of falsely rejecting the null hypothe-

sis if we gain eight or more heads in ten coin tosses? The total risk, or

observed significance level, would be about 0.055 (that is, 0.044 � 0.010

� 0.001), which is greater than our threshold of 0.05 for rejecting the

null hypothesis. It is too large an outcome for you to confront your op-

ponents and accuse them of cheating.

In the case of nine heads, the outcome is well below the threshold of

a Type I error we have chosen. By adding together the probabilities as-

sociated with gaining nine or ten heads in ten coin tosses, we arrive at a

risk of 0.011 of falsely rejecting the null hypothesis. If we decided to re-

ject the null hypothesis that the coin is fair on the basis of an outcome of

nine heads, then the observed significance value for our test would be

0.011. For ten heads, as we noted earlier, the risk of a Type I error is

even lower (p � 0.001). Because there are no outcomes more extreme

than ten heads in our distribution, we do not have to add any probabili-

ties to it to arrive at an estimate of the risk of a Type I error.

You would take a very large risk of a Type I error if you decided in

advance to reject the null hypothesis that the coin is fair based on six

heads in ten tosses of a coin. Here, you would have to add the probabil-

ities associated with six (0.206), seven (0.118), eight (0.044), nine

(0.010), and ten heads (0.001).

As the coin toss example illustrates, sampling distributions play a very

important role in inferential statistics. They allow us to define the ob-

served significance level, or risk of a Type I error, we take in rejecting

the null hypothesis based on a specific outcome of a test of statistical

significance. Although most sampling distributions we use in statistics are

considerably more difficult to develop and involve much more complex

K E Y T E R M S 163

mathematical reasoning than the binomial distribution, they follow a

logic similar to what we have used here. For each distribution, statisti-

cians use probabilities to define the likelihood of gaining particular out-

comes. These sampling distributions provide us with a precise method

for defining risks of error in tests of statistical significance.

What you have learned here provides a basic understanding of how

sampling distributions are developed from probability theory. In later

chapters, we will rely on already calculated distributions. However, you

should keep in mind that steps similar to those we have taken here have

been used to construct these distributions.

C h a p t e r S u m m a r y

Whereas a sample distribution is the distribution of the results of one

sample, a sampling distribution is the distribution of outcomes of a

very large number of samples, each of the same size. A sampling distrib-

ution that is derived from the laws of probability (without the need to

take countless samples) may also be called a probability distribution.

A sampling distribution allows us to define the observed significance

level of a test of statistical significance, or the estimated risk of a Type I

error we take in rejecting the null hypothesis based on sample statistics.

To guide our decision as to whether to reject or fail to reject the null hy-

pothesis, we compare the observed significance level with the criterion

significance level set at the outset of the test of statistical significance.

By using the multiplication rule, we can calculate the probability of

obtaining a series of results in a specific order. The number of arrange-

ments is the number of different ways of obtaining the same result. The

total probability of obtaining any result is the individual probability mul-

tiplied by the number of different possible arrangements.

The binomial distribution is the sampling distribution for events

with only two possible outcomes—success or failure, heads or tails, etc.

It is calculated using the binomial formula. When deciding whether

the result achieved, or observed significance level, passes the desired

threshold for rejecting the null hypothesis, it is important to remember to

take a cumulative total of risk.

K e y T e r m s

arrangements The different ways events

can be ordered and yet result in a single

outcome. For example, there is only one

arrangement for gaining the outcome of ten

heads in ten tosses of a coin. There are,

however, ten different arrangements for

gaining the outcome of nine heads in ten

tosses of a coin.

164 C H A P T E R S E V E N : D E F I N I N G T H E O B S E R V E D R I S K

S y m b o l s a n d F o r m u l a s

! Factorial

r Number of successes

N Number of trials

p

To determine the probability of events A, B, C, and D occurring jointly

under the assumption of independence (the multiplication rule):

P (A & B & C & D) � P (A) • P (B) • P (C ) • P (D)

To determine the number of arrangements of any combination of

events:

To determine the probability of any binomial outcome occurring in all

its possible arrangements (the binomial formula):

P�N

r� � N !

r !(N � r)! pr(1 � p)N�r

�N

r� � N !

r !(N � r)!

binomial distribution The probability or

sampling distribution for an event that has

only two possible outcomes.

binomial formula The means of deter-

mining the probability that a given set of

binomial events will occur in all its possi-

ble arrangements.

factorial The product of a number and all

the positive whole numbers lower than it.

independent Describing two events when

the occurrence of one does not affect the

occurrence of the other.

multiplication rule The means for deter-

mining the probability that a series of

events will jointly occur.

probability distribution A theoretical

distribution consisting of the probabilities

expected in the long run for all possible

outcomes of an event.

sampling distribution A distribution

of all the results of a very large number

of samples, each one of the same size

and drawn from the same population

under the same conditions. Ordinarily,

sampling distributions are derived using

probability theory and are based on prob-

ability distributions.

The probability of a success in the binomial formula. It is also used as a

significance. symbol of the observed significance level of a test of statistical

E X E R C I S E S 165

E x e r c i s e s

7.1 Calculate the probability for each of the following:

a. Two tails in two tosses of a fair coin.

b. Three heads in three tosses of a fair coin.

c. Four heads in four tosses of an unfair coin where the probability of a head is 0.75.

d. Three sixes in three rolls of a fair die.

e. Five fours in five rolls of an unfair die where the probability of a four is 0.25.

7.2 All of Kate’s children are boys.

a. Intuitively, how many boys do you think Kate would have to have in succession before you would be willing to say with some cer- tainty that, for some biological reason, she is more likely to give birth to boys than girls?

b. Now calculate the number of successive births required before you could make such a decision statistically with a 5% risk of error.

c. How many successive boys would have to be born before you would be prepared to come to this conclusion with only a 1% risk of error?

7.3 The Federal Bureau of Investigation trains sniffer dogs to find explo- sive material. At the end of the training, Lucy, the FBI’s prize dog, is let loose in a field with four unmarked parcels, one of which contains Semtex explosives. The exercise is repeated three times, and on each occasion, Lucy successfully identifies the suspicious parcel.

a. What is the chance of an untrained dog performing such a feat? (Assume that the untrained dog would always approach one of the parcels at random.)

b. If there had been five parcels instead of four and the exercise had been carried out only twice instead of three times, would the chances of the untrained dog finding the single suspicious parcel have been greater or less?

7.4 Alex, an attorney, wishes to call eight witnesses to court for an impor- tant case. In his mind, he has categorized them into three “strong” wit- nesses and five “weaker” witnesses. He now wishes to make a tactical decision on the order in which to call the strong and the weaker witnesses.

166 C H A P T E R S E V E N : D E F I N I N G T H E O B S E R V E D R I S K

For example, one of his options is

Strong Weak Weak Strong Weak Weak Weak Strong

a. In how many different sequences can he call his strong and weaker witnesses?

b. If Alex decides that one of his three strong witnesses is in fact more suited to the weaker category, how many options does he now have?

7.5 In a soccer match held at a low-security prison, the inmates beat the guards 4 to 2.

a. How many different arrangements are there for the order in which the goals were scored?

b. What would your answer be if the final score were 5 to 1?

7.6 At the end of each year, the police force chooses its “Police Officer of the Year.” In spite of the fact that there are equal numbers of men and women on the force, in the last 15 years, 11 of the winners have been men and 4 have been women. Paul has been investigating whether women and men are treated differently in the police force.

a. Do these figures provide Paul with a reasonable basis to suspect that the sex of the officer is an active factor? Explain your answer.

b. Looking back further into the records, Paul discovers that for the three years before the 15-year span initially examined, a woman was chosen each time. Does this affect his conclusion? Explain your answer.

7.7 Use the binomial distribution to calculate each of the following probabilities:

a. Three heads in eight tosses of a fair coin.

b. Six tails in thirteen tosses of a fair coin.

c. Four fives in five rolls of a fair die.

d. Two ones in nine rolls of a fair die.

e. Five sixes in seven rolls of a fair die.

7.8 Tracy, a teacher, gives her class a ten-question test based on the homework she assigned the night before. She strongly suspects that Mandy, a lazy student, did not do the homework. Tracy is surprised to see that of the ten questions, Mandy answers seven correctly. What is the probability that Mandy successfully guessed seven of the ten an- swers to the questions if

a. The questions all required an answer of true or false?

b. The questions were all in the multiple-choice format, with students having to circle one correct answer from a list of five choices?

E X E R C I S E S 167

7.9 After a supermarket robbery, four eyewitnesses each report seeing a man with glasses fleeing from the scene. The police suspect Eddy and make up an identity parade of five men with glasses. Eddy takes his place in the parade alongside four randomly chosen stooges. Of the four eyewitnesses who are brought in, three identify Eddy and the fourth points to one of the stooges. The detective in charge decides that there is enough evidence to bring Eddy to trial.

a. The detective’s superior wishes to know the probability that Eddy would have been chosen by three out of the four eyewitnesses if each witness had chosen a member of the identity parade entirely at random. What is the probability?

b. What is the probability of Eddy being chosen at random by only two of the four witnesses?

7.10 A gang of five child thieves draws straws each time before they go shoplifting. Whoever draws the short straw is the one who does the stealing. By tradition, Anton, the leader, always draws first. On the four occasions that the gang has performed this ritual, Anton has drawn the short straw three times.

a. Construct a table to illustrate the binomial distribution of Anton’s possible successes and failures for each of the four draws.

b. Should he accuse his fellow gang members of rigging the draw if

i. He is willing to take a 5% risk of falsely accusing his friends?

ii. He is willing to take only a 1% risk of falsely accusing his friends?

7.11 Baron, a gambler, plays 11 rounds at a casino roulette wheel, each time placing a $100 note on either black or red.

a. Construct a table to illustrate the binomial distribution of Baron’s possible successes and failures for each of the 11 rounds.

b. The casino croupiers have been told to inform the management if a client’s winning streak arouses suspicion that he might be cheating. The threshold of suspicion is set at 0.005. How many successes does Baron need on 11 trials to arouse the management’s suspicion?

7.12 Nicola is playing roulette on an adjacent table. On 12 successive spins of the wheel, she places a $100 note on either the first third (numbers 1–12), the second third (numbers 13–24), or the final third (numbers 25–36).

a. Construct a table to illustrate the binomial distribution of Nicola’s possible successes and failures for each of the 12 spins.

b. How many times out of the 12 would Nicola need to win in order to arouse the suspicion of the casino manager that she was cheat- ing, if the management policy is to limit the risk of falsely accusing a customer to 0.001?

7.13 A security consultant hired by store management thinks that the prob- ability of store security detecting an incident of shoplifting is 0.1. Sup - pose the consultant decides to test the effectiveness of security by try - ing to steal an item ten different times.

a. Construct a table to illustrate the binomial distribution of possible detections for each of the ten attempted thefts.

b. Store management claims that the chances of detection are greater than 0.1. If the consultant set the threshold for detection at 0.05, how many times would she have to be detected to increase the probability of detection?

7.14 In a crime spree, Joe commits six robberies.

a. If the probability of arrest for a single robbery is 0.7, what is the probability that Joe will be arrested for three of the robberies?

b. If the probability of detection for a single robbery is 0.4, what is the probability that Joe will not be arrested for any of his crimes?

7.15 The arrest histories for a sample of convicted felons revealed that, with ten previous arrests, the probability of a drug arrest was 0.25. If an offender has been arrested ten times, what is the probability of

a. two drug arrests?

b. five drug arrests?

c. seven drug arrests?

C H A P T E R S E V E N : D E F I N I N G T H E O B S E R V E D R I S K168

C o m p u t e r E x e r c i s e s

The computation of binomial probabilities by hand can be quite tedious and time

consuming. Spreadsheet packages typically include a function that can be inserted

into a cell that will allow for the computation of a binomial probability. To com-

pute the probability correctly, you will need to enter three pieces of information:

the number of “successes” (i.e., events of interest), the total number of trials,

and the probability of success on a single trial. Another item that you will need

to pay attention to is whether the binomial function you are using computes a

cumulative probability—the default in many spreadsheet packages. Throughout

this chapter, we have not computed cumulative probabilities, but rather, what are

labeled, “probability densities” in many spreadsheet packages.

Although not quite as flexible as a spreadsheet package, the computation of

binomial probabilities in both SPSS and Stata is not complicated. We illustrate

how the various commands work in each program.

The computation of binomial probabilities in SPSS requires the use of the

COMPUTE command that was noted in the Chapter 5 Computer Exercises. As

a means of illustrating the computation of binomial probabilities in SPSS, we will

SPSS

169C O M P U T E R E X E R C I S E S

a new data set in SPSS that contains one variable: the number of success. This

new variable will have values ranging from 0 to 10—enter them in order for ease

of interpretation later. (After you enter these data, you should have 11 values for

your new variable. For ease of illustration below, rename this variable to “suc-

cesses”—without the quotation marks.)

To compute the binomial probabilities, the general form of the COMPUTE

command will be:

The PDF.BINOM function will compute binomial probabilities for any given

combination of successes (q), number of trials (n), and probability of success on

a single trial (p). Prior to executing this command, we need to insert values for

each item in the PDF.BINOM function.

of successes—this is the variable that you created with values ranging from 0

to 10. Enter the variable name here.

The value for “n” (the second value referenced in the parentheses) is the total

number of trials. For our example, enter the number 10.

The value for “p” (the third value referenced in the parentheses) is the

probability of success for a single trial. For our example, enter the value 0.5.

Assuming that you named the new variable successes, the COMPUTE command

would look like:

We have named the new variable “binom_prob” and once this command has

been run, you should see a new variable in the second column of the SPSS data

window that contains binomial probabilities. With the exception of rounding

differences, these values are identical to those presented in Table 7.5.

Stata

The computation of binomial probabilities in Stata is nearly identical to that in

SPSS and requires the use of the gen command that was noted in the Chapter

5 Computer Exercises. We walk through the same illustration of the computa-

tion of binomial probabilities as we did in SPSS, and will reproduce the binomial

probabilities listed in Table 7.5 with the commands in Stata.

To create a new data set in Stata, click on the “Data Editor” button at the

top center of the Stata window. You should see a spreadsheet layout, much like

in SPSS. Again, we will begin by creating a new variable representing the number

of success. Enter values for this new variable that range from 0 to 10. (After you

reproduce the binomial probabilities listed in Table 7.5. To begin, we will create

COMPUTE new_var_name = PDF.BINOM(q,n,p).

EXECUTE.

COMPUTE binom_prob = PDF.BINOM(successes,10,0.5).

EXECUTE.

C H A P T E R S E V E N : D E F I N I N G T H E O B S E R V E D R I S K170

enter these data, you should have 11 values for your new variable. For ease of

illustration below, rename this variable to “successes”—without the quotation

marks.)

To compute the binomial probabilities, the general form of the gen com-

mand will be:

Where the binomialp function will compute binomial probabilities for any

given combination of number of trials (n), of number of successes (k), and

probability of success on a single trial (p). Prior to executing this command, we

need to insert values for each item in the binomialp function.

The value for n -

ber of trials. For our example, enter the number 10.

The value for k (the second value referenced in the parentheses) is the number

of successes—this is the variable that you created with values ranging from 0

to 10. Enter the variable name here.

The value for p (the third value referenced in the parentheses) is the

probability of success for a single trial. For our example, enter the value 0.5.

Assuming that you named the new variable successes, the gen command would

look like:

The new variable is named “binom_prob” and once this command has been run,

you should see a new variable in the second column of the data window that

contains binomial probabilities. Similar to the analysis with SPSS, these values

are identical to those presented in Table 7.5, with the exception of rounding

differences.

Problems

2. Verify the probabilities you calculated for any of the Exercises you worked

through at the end of Chapter 7.

3. Construct a table of binomial probabilities for each of the following

combinations:

a. Number of trials = 10, probability of success = 0.2.

b. Number of trials = 10, probability of success = 0.7.

c. Number of trials = 15, probability of success = 0.3.

d. Number of trials = 15, probability of success = 0.5.

e. Number of trials = 15, probability of success = 0.8.

gen new_var_name = binomialp(n,k,p)

gen binom_prob = binomialp(10,successes,0.5)

1. Reproduce the tables of binomial probabilities on pages 160 and 161 in

the box applying binomial probabilities when p ≠ 0.5.

Steps in a Statistical Test:

Using the Binomial Distribution

to Make Decisions About Hypotheses

Are Assumptions Made About the Population Distribution?

C h a p t e r e i g h t

S t a t i s t i c a l a s s u m p t i o n s

S a m p l i n g d i s t r i b u t i o n

S i g n i f i c a n c e l e v e l

T e s t s t a t i s t i c a n d d e c i s i o n

What Type of Measurement is Being Used?

What Sampling Method is Being Used?

What are the Hypotheses?

Which Sampling Distribution is Appropriate?

What is the Rejection Region?

Should a One-Tailed or a Two-Tailed Test be Used?

What is the Test Statistic?

How is a Final Decision Made?

Where is It Placed?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_8,

© Springer Science+Business Media New York 2014

IN THE PREVIOUS CHAPTER, you saw how probability theory is used to

identify the observed significance level in a test of statistical significance.

But you cannot simply rely on mathematical calculations to determine

whether to reject the null hypothesis. You must make sure at the outset

that the methods used are appropriate to the problem examined. You

must clearly state the assumptions made. You must define the specific

hypotheses to be tested and the specific significance criteria to be used.

It is best to take a careful step-by-step approach to tests of statistical sig-

nificance. Using this approach, you will be much less likely to make seri-

ous mistakes in developing such tests.

In this chapter, we introduce the basic elements of this step-by-step

approach. To place this approach in context, we illustrate each step with

a specific research problem that can be addressed using the binomial

distribution. Although we use the binomial distribution as an example,

you should not lose sight of the fact that our purpose here is to establish

a general model for presenting tests of statistical significance, which can

be used whichever sampling distribution is chosen.

T h e P r o b l e m : T h e I m p a c t o f P r o b l e m - O r i e n t e d P o l i c i n g o n D i s o r d e r l y A c t i v i t y a t V i o l e n t - C r i m e H o t S p o t s

In Jersey City, New Jersey, researchers developed a problem-oriented

policing program directed at violent-crime hot spots.1 Computer map-

ping techniques were used to identify places in the city with a very high

level of violent-crime arrests or emergency calls to the police. Jersey City

police officers, in cooperation with staff of the Rutgers University Center

for Crime Prevention Studies, developed strategies to solve violent-crime

1See Anthony Braga, “Solving Violent Crime Problems: An Evaluation of the Jersey

City Police Department’s Pilot Program to Control Violent Crime Places,” unpublished

dissertation, Rutgers University, Newark, NJ, 1996.

172

T H E P R O B L E M

problems at a sample of 11 places. The strategies followed a problem-

oriented policing (POP) approach, in which police collect a wide variety

of information about each hot spot, analyze that information to identify

the source of the problem, develop tailor-made responses to do some-

thing about the problem, and finally assess whether their approach actu-

ally had an impact.2

The evaluation involved a number of different components. One part

of the research sought to identify whether “disorderly” activity at the hot

spots had declined during the period of the study. For example, the re-

searchers wanted to see whether the number of loiterers or homeless

people had been reduced as a result of the efforts of the police. The

treatment areas were compared to a matched group, or control group, of

similar but untreated violent-crime places. Table 8.1 presents the overall

2Problem-oriented policing is an important new approach to police work formulated

by Herman Goldstein of the University of Wisconsin Law School. See H. Goldstein,

Problem-Oriented Policing (New York: McGraw-Hill, 1990).

Results at Treatment and Control Locations Derived from Observations

of Disorderly Behavior Before and After Intervention

TRIAL PLACE OUTCOME

1 Journal Square East �

Newport Mall

2 Stegman & Ocean �

Clerk & Carteret

3 Glenwood & JFK �

Journal Square West

4 Bergen & Academy �

Westside & Duncan

5 Westside & Clendenny �

Franklin & Palisade

6 Belmont & Monticello �

MLK & Wade

7 MLK & Atlantic �

Neptune & Ocean

8 MLK & Armstrong �

Ocean & Eastern

9 Westside & Virginia �

JFK & Communipaw

10 Park & Prescott �

Dwight & Bergen

11 Old Bergen & Danforth �

Bramhall & Arlington

Note: Experimental or treatment hot spots are listed in boldface type. � � Relative improvement in experimental locations � � Relative improvement in control locations

Table 8 .1

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C H A P T E R E I G H T : S T E P S I N A S T A T I S T I C A L T E S T

results of pre- and posttest comparisons of outcomes for the 11 matched

pairs of locations. In 10 of the 11 pairs, the experimental hot spots (those

receiving POP intervention) improved relative to the control locations.

The research question asked by the evaluator was whether the POP

approach has an impact on disorderly activity at violent-crime hot spots.

The statistical problem faced is that the 11 comparisons are only a sample

of such comparisons. What conclusions can the researcher make regard-

A s s u m p t i o n s : L a y i n g t h e F o u n d a t i o n s f o r S t a t i s t i c a l I n f e r e n c e

The first step in a test of statistical significance is to establish the as-

sumptions on which the test is based. These assumptions form the

foundation of a test. No matter how elegant the statistics used and the

approach taken, if the assumptions on which they are built are not solid,

then the whole structure of the test is brought into question.

Level of Measurement

Our first assumption is related to the type of measurement used. Differ-

ent types of tests of statistical significance demand different levels of

measurement.

Accordingly, it is important to state at the outset the type of measure-

ment required by a test. For the binomial test, which is based on the bi-

nomial distribution, a nominal-level binary measure is required. A binary

measure has only two possible outcomes, as was the case with the coin

toss example in Chapter 7. The type of outcome measure used to evalu-

ate the impact of problem-oriented policing on disorderly activity—

whether the treatment hot spot improved (or got worse) relative to the

control location—fits this assumption. In stating our assumptions (as is

done at the end of this section), we include a specific definition of the

level of measurement required:

Level of Measurement: Nominal binary scale.

Shape of the Population Distribution

The second assumption refers to the shape of the population distribu-

tion. In statistical inference, we are generally concerned with two types

of tests. In the first type—termed parametric tests—we make an as-

sumption about the shape of the population distribution. For example, in

a number of tests we will examine in later chapters, there is a require-

appropriate for our problem is based on the binomial sampling distribution.

question, we use a test of statistical significance. The specific test that is

ing the larger population of violent-crime hot spots? To answer this

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A S S U M P T I O N S : L A Y I N G T H E F O U N D A T I O N S

ment that for the population to which you infer, the scores on the vari-

able be normally distributed.

The second type of test of statistical significance does not make a spe-

cific assumption regarding the population distribution. These tests are

called nonparametric tests or distribution-free tests. The advantage

of nonparametric tests is that we make fewer assumptions. The disad-

vantage is that nonparametric tests do not allow us to analyze data at

higher levels of measurement. They are generally appropriate only for

nominal and ordinal scales. The binomial test is a nonparametric test. Ac-

Population Distribution: No assumption made.

Sampling Method

The third assumption concerns the sampling method used. When we

conduct a test of statistical significance, we want our sample to be a

good representation of the population from which it is drawn. Put in sta-

tistical terms, we want our study to have high external validity.

Let’s suppose you are interested in attitudes toward the death penalty.

Would a sample of your friends provide an externally valid sample of all

Americans? Clearly not, because a sample of only your friends is not

likely to include age or ethnic or class differences that typify the U.S.

population. Even if we used your friends as a sample of U.S. college stu-

dents, we could still identify threats to the external validity of the study.

Colleges have differing criteria for admission, so it is not likely that one

college will be representative of all colleges. Even as a sample of stu-

dents at your college, your friends may not provide a valid sample. They

may be drawn primarily from a specific year of college or have other

characteristics that make them attractive as friends but also mean that

they are a poor representation of others in the college.

How can we draw a representative sample? The most straightfor-

ward approach is to choose cases at random from the population. This

type of sampling is called random sampling. Random samples are as-

sumed to have high external validity compared with what may be termed

convenience samples. A convenience sample consists of whatever sub-

jects are readily available to the researcher. Your friends form a conve-

nience sample of students at your college or of all college students.

It is important to note that convenience samples are not always bad

samples. For example, if you choose to examine prisoners in one prison on

the assumption that prisoners there provide a cross section of the different

types of prisoners in the United States, you might argue that it is a repre-

sentative sample. However, if you use a convenience sample, such as pris-

oners drawn from a single prison, you must always be wary of potential

threats to external validity. Convenience samples are prone to systematic

biases precisely because they are convenient. The characteristics that make

175

cordingly, in stating our assumptions we write:

C H A P T E R E I G H T : S T E P S I N A S T A T I S T I C A L T E S T

them easy for the researcher to define are likely as well to differentiate

them in one way or another from the population the researcher seeks to

study.

Statistical tests of significance generally assume that the researcher has

used a type of random sampling called independent random sampling.

Independent random sampling requires not only that cases be identified at

random, but also that the selection of cases be independent. As discussed

in the previous chapter, two events are statistically independent when the

occurrence of one does not affect the occurrence of the other. In sam-

pling, this means that the choice of one case or group of cases will not

have any impact on the choice of another case or group of cases. This is a

useful assumption in assuring the external validity of a study because it

prevents biases that might be brought into the process of sampling.

For example, suppose you want to select 1,000 prisoners from the

population of all prisoners in the United States. Each time you select a

prisoner for your sample, you use a random method of selection. How-

ever, prison officials have told you that if you select one prisoner from a

cell then you cannot select any other prisoner from that cell. Accord-

In order to ensure independent random sampling, the same popula-

the deck. If we didn’t replace the card, we would influence the likeli-

hood of a specific card being chosen on the next draw from the deck.

For example, if we started with a full deck of 52 cards, the likelihood of

getting the queen of spades would be 1 in 52. However, if we drew, say,

a jack of hearts and didn’t return it to the deck, what would be the likeli-

hood of getting a queen of spades on our next draw? This time we

would have only 51 cards to draw from, so our likelihood would change

to 1 in 51. In order to gain a fully independent random sample, we must

use a method of sampling called sampling with replacement. This

means that we must use the same population each time we select a case.

For every selection, the sampling frame must remain exactly the same. In

this way, we can ensure that the choice of one case cannot have any im-

pact on the choice of another.

Though this method ensures independence, it also means that a par-

ticular case may be selected more than once. For example, suppose you

choose a particular prisoner as case number five in your sample. Be-

cause you must use the same sampling frame each time you select a

case, that prisoner is returned to the sampling frame after selection. Later

ingly, after each selection of a prisoner, you must remove all of his

cellmates from your sampling frame, or universe of eligible cases. The

result is that there are now systematic reasons why you might suspect

that your sample is not representative of the population.

tion of cases must be used in drawing each case for a sample. As we

discussed in Chapter 7, if we want each draw from a deck of cards to be

independent, we have to return the card chosen on any specific draw to

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A S S U M P T I O N S : L A Y I N G T H E F O U N D A T I O N S

in your study, you might choose that prisoner again. Accordingly, while

sampling with replacement, or returning sampled cases to the sampling

frame after each selection, makes statistical sense, it often does not make

practical sense when you are carrying out research in the real world. If

you are conducting an interview study, for example, independent ran-

dom sampling would allow individuals to be interviewed more than

once. It is likely that subjects would find it strange to be reinterviewed

using the same interview schedule. Moreover, their responses would

likely be influenced by their knowledge of the survey. Similarly, if a sub-

ject or place is chosen twice in a study that involves a specific treatment

or intervention, then that subject or place should be given the treatment

after each selection. Here there is the difficulty that it may be harmful to

provide the treatment more than once.

Even when there are no specific practical barriers to sampling with re-

placement, it is difficult to explain to practitioners or even many re-

searchers why an individual may appear twice in the same sample. As a

result, many, if not most, criminal justice studies do not replace individu-

als in the sampling frame once they have been selected. Although this

represents a formal violation of the assumptions of your test, in most

cases its impact on your test result is negligible. This is because samples

are generally very small relative to populations, and thus in practice

there is little chance of selecting a case more than once even when sam-

pling with replacement. If, however, your sample reaches one-fifth or

more of the size of your population, you may want to include a correc-

tion factor in your test.3

For this test of statistical significance, we assume that researchers in

the Jersey City POP study sampled cases randomly from a large popula-

tion of hot spots during the sample selection month. Because it would

not have been practical to implement treatments more than once at any

site, the researchers did not sample with replacement.

3The correction factor adjusts your test to account for the fact that you have not al-

lowed individuals to be selected from the population more than once. Not including a

correction factor makes it more difficult to reject the null hypothesis. That is, the in-

clusion of a correction factor will make it easier for you to reject the null hypothesis.

One problem criminal justice scholars face in using a correction factor is that they

often want to infer to populations that are beyond their sampling frame. For example,

a study of police patrol at hot spots in a particular city may sample 50 of 200 hot

spots in the city during a certain month. However, researchers may be interested in

making inferences to hot spots generally in the city (not just those that exist in a par-

ticular month) or even to hot spots in other places. For those inferences, it would be

misleading to adjust the test statistic based on the small size of the sampling frame.

For a discussion of how to correct for sampling without replacement, see Paul S. Levy

and Stanley Lemeshow, Sampling of Populations: Methods and Applications (New

York: Wiley, 1991).

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C H A P T E R E I G H T : S T E P S I N A S T A T I S T I C A L T E S T

The binomial test, however, like most tests of statistical significance

examined in this book, assumes independent random sampling. Accord-

ingly, in stating our assumptions, it is important to note both the require-

ment for this test and our failure to meet that requirement. Therefore we

state our assumption:

Sampling Method: Independent random sampling (no replacement; sam-

ple is small relative to population).

Throughout this text, we state the assumptions of a test and then place

any violations of assumptions in parentheses. This is good practice, as it

will alert you to the fact that in many studies there are violations of one

type or another of assumptions. Some of these violations are not impor-

tant. For example, not sampling with replacement in this study does not

affect the test outcome because the population of hot spots is assumed

to be very large relative to the sample. However, you will sometimes

find more serious violations of assumptions. In those cases, you will

have to take a more critical view of the results of the test. It is good practice to define not only the sampling method used but

also the sampling frame of your study. In our example, we can make in-

ferences based on our random sample to the population of hot spots in

Jersey City during the month of sample selection. Accordingly, we state

in our assumptions:

Sampling Frame: Hot spots of violent crime in one month in Jersey

City.

Our sampling frame reminds us of the specific population to which our

sample infers. However, researchers usually want to infer beyond the

specific population identified by their sampling frame. For example, the

population of interest for the POP study is likely to be hot spots through-

out the year, not just those in a specific month. Researchers may even

want to infer to violent-crime hot spots generally, not just those in Jersey

City.

We cannot assume that our sample is a representative sample for

these inferences based on our sampling method, since these populations

did not constitute our sampling frame. However, we can ask whether

our sample is likely to provide valid inferences to those populations. In

the case of hot spots in Jersey City, we would need to question whether

there is any reason to suspect that hot spots chosen in the month of

study were different from those that would be found in other months of

the year. For inferences to the population of hot spots in other locations,

we would have to assume that Jersey City hot spots are similar to those

in other places and would respond similarly to POP interventions. In

making any inference beyond your sampling frame, you must try to

identify all possible threats to external validity.

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The Hypotheses

The final assumptions we make in a test of statistical inference refer to

the hypotheses of our study. As discussed in Chapter 6, hypotheses are

developed from the research questions raised in a project. Hypotheses

must be stated before the researcher collects outcome data for a study. If

hypotheses are stated only after data have been collected and analyzed,

the researcher might be tempted to make changes in the hypotheses that

unfairly affect the tests of statistical significance that are conducted.

As discussed in Chapter 6, the researcher ordinarily begins by defin-

ing the research hypothesis. In the problem-oriented policing study, we

might state our research hypothesis in three different ways:

Hypothesis 1. Incivilities in treatment hot spots decline relative to incivil-

ities in control hot spots after POP intervention.

Hypothesis 2. Incivilities in treatment hot spots increase relative to inci-

vilities in control hot spots after POP intervention.

Hypothesis 3. The level of incivilities in treatment hot spots relative to

incivilities in control hot spots changes after POP intervention.

hypotheses. The first two research hypotheses are directional hypotheses

because they specify the direction, or type of relationship, that is ex-

pected. For example, hypothesis 1 is concerned only with whether the

POP program is successful in reducing incivilities. If the researcher

adopts this hypothesis, then he or she is stating that the statistical test

employed will not be concerned with the second hypothesis—that the

intervention makes matters worse and increases incivilities. The third hy-

pothesis is a nondirectional hypothesis. In this case, the researcher is in-

terested in testing the possibility that the intervention improves hot spots

or makes them worse.

In the POP study, researchers wanted to assess both positive and neg-

ative outcomes. Although they believed that problem-oriented policing

should reduce incivilities at violent-crime hot spots, they did not want to

preclude at the outset a finding that the program actually made matters

worse. Accordingly, they used a nondirectional research hypothesis:

“The level of incivilities in treatment hot spots relative to incivilities in

control hot spots changes after POP intervention.” The null hypothesis is

“The level of incivilities in treatment hot spots does not change relative

to incivilities in control hot spots after POP intervention.”

In practice, the null hypothesis may be stated in terms of probabili-

ties, just as we could state the coin toss hypothesis in the last chapter

in terms of probabilities. In this study, the researchers examined (for

each matched pair of hot spots) whether the hot spot that received the

problem-oriented policing intervention improved or worsened relative to

Recall from Chapter 6 that we distinguish directional from nondirectional

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the control location. The null hypothesis suggests that the treatment and

control hot spots are equally likely to improve. Put in terms of probabili-

ties, there is a 0.50 chance of success (P � 0.50) for the intervention

under the null hypothesis. The research hypothesis represents all other

possible outcomes (P � 0.50). Remember that our hypotheses are state-

ments about the populations examined. Accordingly, in stating the hy-

potheses, we use symbols appropriate for population parameters—in

this case P rather than p. Stating our assumptions, we write

Hypotheses:

H0: The level of incivilities in treatment hot spots does not change rela-

tive to incivilities in control hot spots after POP intervention, P � 0.50.

H1: The level of incivilities in treatment hot spots relative to incivilities in

control hot spots changes after POP intervention, P � 0.50.

Stating All of the Assumptions

Our assumptions may be stated as follows:

Assumptions:

Level of Measurement: Nominal binary scale.

Population Distribution: No assumption made.

Sampling Method: Independent random sampling (no replacement; sam-

ple is small relative to population).

Sampling Frame: Hot spots of violent crime in one month in Jersey City.

Hypotheses:

H0: The level of incivilities in treatment hot spots does not change rela-

tive to incivilities in control hot spots after POP intervention, P � 0.50.

H1: The level of incivilities in treatment hot spots relative to incivilities in

control hot spots changes after POP intervention, P � 0.50.

S e l e c t i n g a S a m p l i n g D i s t r i b u t i o n

In stating our hypotheses, we already noted the specific requirements of

the binomial sampling distribution. Now we must state why we have

chosen the binomial distribution and identify the specific characteristics

of the sampling distribution that will be used to assess the risk of falsely

rejecting the null hypothesis in our problem-oriented policing example.

Choosing a sampling distribution is one of the most important decisions

that researchers make in statistical inference. As we will show in later

chapters, there are a number of different types of sampling distributions.

Moreover, as with the binomial distribution, a single type of sampling

distribution may have different forms depending on the problem

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S E L E C T I N G A S A M P L I N G D I S T R I B U T I O N

Because our measure is nominal and binary (see assumptions), we se-

lected the binomial distribution for our test. The specific distribution that we

use is based on our null hypothesis and the size of our sample. As illus-

trated in Chapter 7, the binomial distribution provides the likelihood of gain-

ing a particular number of successes (heads in the example of the coin toss)

in a fixed number of trials. In order to assess that likelihood, we also need

to know what the probability of a success or failure is on any particular trial.

In our example, there are 11 trials, or 11 matched comparisons. Our

null hypothesis states that the likelihood of a success for any comparison

is 0.50. To build our sampling distribution, we apply the binomial for-

mula to each of the 12 possible outcomes that could be gained in our

study, under the assumption that P � 0.50. This is done in Table 8.2.

The resulting distribution is presented in Table 8.3.

Computation of Sampling Distribution of Success or Failure in 11 Trials

0 successes 1(0.00049) � 0.00049

1 success 11(0.00049) � 0.00537

2 successes 55(0.00049) � 0.02686

3 successes 165(0.00049) � 0.08057

4 successes 330(0.00049) � 0.16113

5 successes 432(0.00049) � 0.22638*

6 successes 432(0.00049) � 0.22638*

7 successes 330(0.00049) � 0.16113

8 successes 165(0.00049) � 0.08057

9 successes 55(0.00049) � 0.02686

10 successes 11(0.00049) � 0.00537

11 successes 1(0.00049) � 0.00049

*Probabilities contain rounding error.

39,916,800

39,916,800(11 � 11)! �

39,916,800

39,916,800 � 1

39,916,800

3,628,800(11 � 10)! �

39,916,800

3,628,800 � 11

39,916,800

362,880(11 � 9)! �

39,916,800

725,760 � 55

39,916,800

40,320(11 � 8)! �

39,916,800

241,920 � 165

39,916,800

5,040(11 � 7)! �

39,916,800

120,960 � 330

39,916,800

720(11 � 6)! �

39,916,800

86,400 � 462

39,916,800

120(11 � 5)! �

39,916,800

86,400 � 462

39,916,800

24(11 � 4)! �

39,916,800

120,960 � 330

39,916,800

6(11 � 3)! �

39,916,800

241,920 � 165

39,916,800

2(11 � 2)! �

39,916,800

725,760 � 55

39,916,800

1(11 � 1)! �

39,916,800

3,628,800 � 11

39,916,800

1(11 � 0)! �

39,916,800

39,916,800 � 1

�N r �pr (1 � p)N�r�N

r � � N !

r !(N � r )!

Table 8.2

research problem examined, then the conclusion reached will be suspect.

examined. If the sampling distribution used is inappropriate for the

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S i g n i f i c a n c e L e v e l a n d R e j e c t i o n R e g i o n

Having selected the distribution that will be used to assess Type I error,

we are ready to define the outcomes that will lead us to reject the null

hypothesis. Our first step is to choose the significance level of our test.

As described in Chapter 6, the significance level of a test is the amount

of Type I error we are willing to risk in rejecting the null hypothesis. By

convention, criminal justice researchers use a 5% significance threshold.

But, as discussed in Chapter 6, we should consider at the outset whether

a more lenient or more stringent significance level is appropriate for our

study.

As researchers in the problem-oriented policing study do not present

any special reason for altering conventionally accepted levels of signifi-

cance, we will set a 5% significance threshold for our test of statistical

significance. As noted in Chapter 6, in articles and books the significance

level is often expressed by the Greek letter �. For our test, � � 0.05.

The significance level defines the Type I error we are willing to risk in

our test. But it does not tell us directly what outcomes in our sample

would lead us to reject the null hypothesis. For this, we need to turn to

our sampling distribution and define an area within it called a rejection

region. The rejection region of a test is the area in the sampling distribu-

tion that includes those outcomes that would lead to rejection of the null

hypothesis. If the observed significance level of a test, or the p value of

the test, falls within the rejection region, then the researcher rejects the

null hypothesis and concludes that the outcome is statistically significant.

The area covered by the rejection region is equivalent to the significance

level of a test. The point at which the rejection region begins is called

the critical value because it is the point at which the test becomes criti-

cal and leads the researcher to reject the null hypothesis.

Sampling Distribution of Success or Failure in 11 Trials

OUTCOME OF TRIALS OVERALL PROBABILITY

0 successes 0.00049 1 success 0.00537 2 successes 0.02686 3 successes 0.08057 4 successes 0.16113 5 successes 0.22559 6 successes 0.22559 7 successes 0.16113 8 successes 0.08057 9 successes 0.02686

10 successes 0.00537 11 successes 0.00049

Table 8.3

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S I G N I F I C A N C E L E V E L A N D R E J E C T I O N R E G I O N

In the problem-oriented policing example, the rejection region in-

cludes 5% of the sampling distribution. Our initial problem is to define

which 5%. Should we define the rejection region to be in the middle of

the distribution represented in Table 8.3—for example, at 5 or 6 suc-

cesses in 11 comparisons? Or should we look only at the extreme values

on the positive side of the distribution, where there are mostly successes?

Or should we include the area on the negative side of the distribution,

where there are no successes?

Choosing a One-Tailed or a Two-Tailed Rejection Region

The answer to our questions comes in part from common sense and in

part from our assumptions. It just would not make sense to place the re-

jection region in the middle of the sampling distribution. We are trying

to decide whether the outcomes observed in our sample are very differ-

ent from the outcomes that would be expected if problem-oriented

policing had no impact. Putting the rejection region in the middle of the

distribution would place it among those outcomes that are most likely

under the null hypothesis. Clearly, we want the rejection region to be on

the edges of the distribution, or in what statisticians call the tails of the

distribution. These are the unlikely events—those that we would not

expect if the null hypothesis were true. As indicated in our sampling dis-

tribution in Table 8.3, we would expect to get 11 successes in a row in

about 5 of 10,000 samples if the program had no impact on the popula-

tion. This is a very unlikely event and one that would lead us to reject

the null hypothesis.

But zero successes is also an unlikely event, with the same probability

of occurrence as 11 successes. Should we include only one tail of the dis-

tribution in our rejection region—the tail that assesses whether the pro-

gram was a success? Or should we also include the opposite side of the

distribution, which suggests that the program led to more disorder? Our

answer is drawn from the research hypothesis that we stated in our as-

sumptions. We chose a nondirectional research hypothesis, meaning that

we are interested in evaluating both the possibility that the experimental

sites improved relative to the control hot spots and the potential outcome

that they got worse relative to the control hot spots. In terms of the sam-

pling distribution, our research hypothesis suggests that the rejection re-

gion for our test should be split between both tails of the distribution.

This type of test is called a two-tailed test of significance. If we had

stated a directional research hypothesis, we would be concerned with out-

comes on only one side of the sampling distribution. Such a test is called a

one-tailed test of significance. For example, if our research hypothesis

were that incivilities in treatment hot spots decrease relative to incivilities

in control hot spots after POP intervention, we would be concerned only

with outcomes on the side of the distribution that shows program success.

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The choice of a one-tailed or two-tailed test of statistical significance

has important implications for the types of study outcomes that will lead

to rejection of the null hypothesis. Because our test is a two-tailed test,

the rejection region must be divided between both sides of the sampling

distribution. This means in practice that the total significance level of

0.05 must be divided in half. Half of the rejection region, or 0.025, is

found in the tail associated with success of the program, and half, or

0.025, in the tail associated with failure.

What outcomes would lead to rejection of the null hypothesis in our

example? When we add 0 and 1 successes or 10 and 11 successes, we

gain a probability value of 0.00586 (in each tail of the distribution,

0.00049 � 0.00537). This is less than the 0.025 value that we have de-

fined as the rejection region for each tail of our test. Accordingly, an out-

come of 0, 1, 10 or 11 would lead to an observed significance level less

than the significance level of 0.05 that we have set, and thus we would

reject the null hypothesis (p � 0.05). However, including 2 or 9 suc-

cesses, each of which has a probability value of 0.027, increases the area

of the distribution to 0.066. This area is larger than our rejection region.

An outcome of 9 or 2 would result in an observed significance level

greater than 0.05, and thus we would fail to reject the null hypothesis.

Figure 8.1 presents the binomial probabilities for our example and high-

lights the two tails of the distribution that are used to test our nondirec-

tional hypothesis.

P ro

b ab

il it

y

0.25

0.2

0.15

0.1

0.05

0

Number of Successes

0 1 2 3 4 5 6 7 8 9 10 11

0.0005 0.005

0.03 0.03

0.08 0.08

0.16 0.16

0.23 0.23

0.005 0.0005

Outcomes in the rejection region

Outcomes in the rejection region

Outcomes That Would Lead to Rejecting the Null Hypothesis

for a Two-Tailed Test of Significance (� � 0.05) Figure 8.1

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S I G N I F I C A N C E L E V E L A N D R E J E C T I O N R E G I O N

But what if we state a directional research hypothesis? How does

this affect our rejection region? In this case, we calculate the area of the

rejection region on only one side of the sampling distribution. Parts a

and b of Figure 8.2 present the binomial probabilities for our two dis-

tinct directional hypotheses and highlight the tail of the distribution

that is potentially of interest. For example, if our research hypothesis is

that incivilities in treatment hot spots decline relative to incivilities in

control hot spots after POP intervention, we look at outcomes only on

the tail of the distribution that shows program success (Figure 8.2b).

Because we are concerned only about these outcomes, all 5% of the re-

jection region is placed in this one tail of the distribution. We do not

have to split the area of the rejection region. In this example, outcomes

of 9, 10, and 11 successes are all within the rejection region, because

adding their probabilities results in a value of 0.033 (0.00049 � 0.00537

� 0.02686). An outcome of 9, 10, or 11 results in an observed signifi-

cance level that is less than the 5% significance threshold of our test

(see Figure 8.2b). Adding the probability of 8 successes (or 0.08057)

puts us above that threshold. If our research hypothesis is that incivili-

ties increase in treatment hot spots relative to control hot spots, then

we look at outcomes only on the opposite tail of the distribution (Fig-

ure 8.2a). In this case, outcomes of 0, 1, and 2 successes lead us to re-

ject the null hypothesis.

This example reinforces a rule described earlier: It is important to

state the research hypothesis before you gain study outcomes. What if

the problem-oriented policing hot spots improved relative to control

locations in nine comparisons? With a one-tailed test, the result would

fall within our rejection region and lead to rejection of the null hypoth-

esis. With a two-tailed test, the result would be outside our rejection re-

gion. The choice of a directional or nondirectional research hypothesis

can have an important impact on our conclusions. Merely by stating

the research hypothesis a bit differently, we can change the outcome of

the test.

A one-tailed test makes it easier to reject the null hypothesis based on

outcomes on one side of a sampling distribution because it precludes re-

jection of the null hypothesis based on outcomes on the opposite side.

The price of a larger rejection region in one-tail of the sampling distribu-

tion is no rejection region in the other tail. Similarly, the price of being

able to examine outcomes on both sides of the distribution, as is the

case with a two-tailed test, is that the rejection region will be smaller on

each side. The benefit is that you can assess results in both directions. If

you already know the outcomes of a test, you might be tempted to ad-

just the direction of the test according to the observed outcomes of a

study. Taking such an approach unfairly adjusts the rejection region to

your advantage.

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P ro

b ab

il it

y

0.25

0.2

0.15

0.1

0.05

0

Number of Successes

0 1 2 3 4 5 6 7 8 9 10 11

0.0005 0.005

0.03 0.03

0.08 0.08

0.16 0.16

0.23 0.23

0.005 0.0005

Outcomes in the rejection region

P ro

b ab

il it

y

0.25

0.2

0.15

0.1

0.05

0

Number of Successes

0 1 2 3 4 5 6 7 8 9 10 11

0.0005 0.005

0.03 0.03

0.08 0.08

0.16 0.16

0.23 0.23

0.005 0.0005

Outcomes in the rejection region

Outcomes That Would Lead to Rejecting the Null Hypothesis

for a One-Tailed Test of Significance (� � 0.05) Figure 8.2

(b) Focus on Program Successes

(a) Focus on Program Failures

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M A K I N G A D E C I S I O N

T h e T e s t S t a t i s t i c

In most tests of statistical significance, it is necessary to convert the spe-

cific outcome of a study to a test statistic. A test statistic expresses the

value of your outcome in units of the sampling distribution employed in

your test. For the binomial distribution, the units are simply the number

of successes in the total number of trials. The test statistic for our POP

intervention example is 10.

M a k i n g a D e c i s i o n

The final step in a test of statistical significance is making a decision. If

you have laid out all of the steps discussed above, then your choice

should be easy. If your test statistic falls within the rejection region, then

you reject the null hypothesis. This means in practice that the observed

significance level of your test is less than the criterion significance level

that you set when you defined the significance level and rejection region

for your test. If the test statistic does not fall in the rejection region, you

cannot reject the null hypothesis. In our example, the test statistic (10)

does fall in the rejection region, which includes 0, 1, 10, and 11 suc-

cesses. In this case, our observed significance level is less than the 0.05

threshold we set earlier. Our decision, then, is to reject the null hypothe-

sis that incivilities in treatment hot spots do not change relative to incivil-

ities in control hot spots after POP intervention. We conclude that the

differences observed are statistically significant.

But what does this mean? When we say that a result is statistically

significant, we are not claiming that it is substantively important. The

importance of a result depends on such issues as whether the research

affects real-life criminal justice decision making or whether it con-

tributes new knowledge to a specific area of criminology or criminal

justice. We also are not stating that we are certain that the null hypothe-

sis is untrue for the population. Without knowledge of the population

parameter, we cannot answer this question with certainty. Statistical sig-

nificance has a very specific interpretation. The fact that an outcome is

statistically significant means that it falls within the rejection region of

your test. This happens when the observed significance level for a test

is smaller than the significance criterion, or significance level, set at the

outset of the test. A statistically significant result is one that is unlikely if

the null hypothesis is true for the population. Whenever we make a

statement that a result is statistically significant, we do it with the recog-

nition that we are risking a certain level of Type I error. In this test, as

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C H A P T E R E I G H T : S T E P S I N A S T A T I S T I C A L T E S T

in most tests of statistical significance in criminal justice, we were will-

ing to take a 5% risk of falsely rejecting the null hypothesis.

C h a p t e r S u m m a r y

The first stage in a test of statistical significance is to state one’s as-

sumptions. The first assumption is about the type of measurement

used. The second assumption concerns the shape of the population

distribution. A parametric test is one that makes assumptions about

the shape of the population distribution. A nonparametric test

makes no such assumptions. Although nonparametric tests have the

advantage of making fewer assumptions, they are generally used only

for nominal and ordinal scales. The third assumption relates to the

sampling method. A random sample is generally considered to be

pling, it is in theory necessary to return the subject to the sampling

frame after selection. Sampling with replacement creates practical

problems, however, and is generally not required if the sample is

small relative to the population. The fourth assumption states the null

and research hypotheses. Care should be taken in framing them and in

deciding whether the research hypothesis should be directional.

The second stage is to select an appropriate sampling distribution.

The third stage is to select a significance level. The significance level

determines the size of the rejection region and the location of the

critical values of the test. If a test result falls within the rejection re-

gion, the researcher is prepared to reject the null hypothesis. This

means that the observed significance level of the test is less than the

significance level the researcher set at the outset of the test. If the hy-

potheses are directional, then the researcher will be concerned only

with one tail of the distribution, and the entire rejection region will

be placed on one side of the distribution (a one-tailed test of signifi-

cance). If the hypotheses are nondirectional, then the researcher is

concerned with results in both tails, and the rejection region will be

divided equally between both sides of the distribution (a two-tailed

test of significance).

The fourth stage involves calculating a test statistic. The study re-

sult is now converted into the units of the sampling distribution. Fi-

nally, a decision is made: The null hypothesis will be rejected if the

test statistic falls within the rejection region. When such a decision can

be made, the results are said to be statistically significant.

accepted form of sampling. To ensure the independence of the sam-

more representative, or to have greater external validity, than a

convenience sample. Independent random sampling is the most

188

K E Y T E R M S

assumptions Statements that identify the

requirements and characteristics of a test of

statistical significance. These are the foun-

dations on which the rest of the test is built.

convenience sample A sample chosen not

at random, but according to criteria of expe-

dience or accessibility to the researcher.

critical value The point at which the re-

jection region begins.

distribution-free tests Another name for

nonparametric tests.

external validity The extent to which a

study sample is reflective of the population

from which it is drawn. A study is said to

have high external validity when the sam-

ple used is representative of the population

to which inferences are made.

independent random sampling A form

of random sampling in which the fact that

one subject is drawn from a population in

no way affects the probability of drawing

any other subject from that population.

nonparametric tests Tests of statistical

significance that make no assumptions as

to the shape of the population distribution.

one-tailed test of significance A test of

statistical significance in which the region

for rejecting the null hypothesis falls on

only one side of the sampling distribution.

One-tailed tests are based on directional re-

search hypotheses.

parametric tests Tests of statistical signifi-

cance that make assumptions as to the

shape of the population distribution.

random sampling Drawing samples from

the population in a manner that ensures

every individual in that population an

equal chance of being selected.

rejection region The area of a sampling

distribution containing the test statistic val-

ues that will cause the researcher to reject

the null hypothesis.

representative sample A sample that

reflects the population from which it is

drawn.

sampling frame The universe of eligible

cases from which a sample is drawn.

sampling with replacement A sampling

method in which individuals in a sample

are returned to the sampling frame after

they have been selected. This raises the

possibility that certain individuals in a pop-

ulation may appear in a sample more than

once.

statistically significant Describing a test

statistic that falls within the rejection region

defined by the researcher. When this oc-

curs, the researcher is prepared to reject the

null hypothesis and state that the outcome

or relationship is statistically significant.

tails of the distribution The extremes on

the sides of a sampling distribution. The

events represented by the tails of a sam-

pling distribution are those deemed least

likely to occur if the null hypothesis is true

for the population.

test statistic The outcome of the study,

expressed in units of the sampling distribu-

tion. A test statistic that falls within the re-

jection region will lead the researcher to re-

ject the null hypothesis.

two-tailed test of significance A test of

statistical significance in which the region

for rejecting the null hypothesis falls on

both sides of the sampling distribution.

Two-tailed tests are based on nondirec-

tional research hypotheses.

K e y T e r m s

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E x e r c i s e s

8.1 Answer the following conceptual questions:

a. Is it better to have more or fewer assumptions at the beginning of a test of statistical significance? Explain your answer.

b. Why is it important to state all of the assumptions at the outset of the test?

c. In what sense can stating the null and research hypotheses be seen as making assumptions?

8.2 Gatley University is an elite university of 1,000 students. Nadia, a stu- dent studying Chinese at the university, wishes to determine the aver- age IQ of students at Gatley. She has decided that her sample size will be 50, and she is considering several different sampling methods. For each method, state the sampling frame and discuss whether the sam- pling method is random and whether it is independent.

a. Nadia chooses 50 names at random from the list of language stu- dents at the university.

b. Nadia asks 50 of her acquaintances at the university if they would mind taking an IQ test.

c. Nadia chooses the first two students from the alphabetical list of each of the 25 university departments.

d. Nadia takes all 1,000 names and puts them into a hat. She draws out a name, writes it down, and then puts it back in the hat and draws again. This procedure is repeated 50 times.

8.3 Hale Prison is renowned for its poor internal discipline. The new prison governor wants to tackle this problem and decides to investi- gate whether removing prisoners’ visiting privileges will act as a deter- rent against future misbehaving. From 100 prisoners who recently took part in a violent prison riot, he selects the 25 inmates with the worst disciplinary records, removes their visiting privileges, and begins to monitor their progress relative to the others.

a. Does this method meet the criteria of independent random sampling?

b. Is independent sampling possible in this case?

c. Describe a more appropriate sampling method.

8.4 For each of the following hypotheses, state whether a one-tailed or a two-tailed test of statistical significance would be appropriate. In each case, explain your choice.

a. H1: Citizens over the age of 50 are more likely to be the victims of assault than citizens under the age of 50.

190

E X E R C I S E S

b. H1: Children raised by adopted parents have rates of delinquency different from those of children raised by their biological parents.

c. H1: The experience of imprisonment has an impact on the chances of an ex-convict reoffending.

d. H1: Women are more likely than men to support increased sentences for rapists.

e. H1: Persons who are not victims of assault have lower levels of anger than persons who have been victims of assault.

f. H1: White offenders are less likely to be sentenced to prison than Hispanic offenders.

g. H1: Teenagers have rates of crime that are different from adult rates of crime.

h. H1: Defendants charged with property crimes have different rates of pretrial misconduct than defendants charged with violent crimes.

i. H1: Male defendants are more likely to be held on bail than female defendants.

j. H1: Women are more supportive of capital punishment than men.

k. H1: States with higher unemployment rates have higher rates of property crime.

l. H1: The level of poverty in a neighborhood affects the neighbor- hood’s crime rate.

m. H1: Democrats are less supportive of cutting taxes than Republicans.

n. H1: Graduates from private law schools are more likely to become federal judges than graduates from state law schools.

8.5 In Chapter 7, we constructed a binomial distribution showing the chances of success and failure for ten tosses of a fair coin. The distri- bution was as follows:

0 heads 0.001

1 head 0.010

2 heads 0.044

3 heads 0.118

4 heads 0.206

5 heads 0.247

6 heads 0.206

7 heads 0.118

8 heads 0.044

9 heads 0.010

10 heads 0.001

191

C H A P T E R E I G H T : S T E P S I N A S T A T I S T I C A L T E S T

Consider the following alternative hypotheses:

Alternative 1: H0: The coin is fair. H1: The coin is biased.

Alternative 2: H0: The coin is fair. H1: The coin is biased in favor of heads.

a. Would a one-tailed or a two-tailed test be more appropriate for a researcher who chose alternative 1? Explain why.

b. For a sequence of ten throws, what results would cause a re- searcher operating under the hypotheses listed under alternative 1 to reject the null hypothesis at a significance level of 5%?

c. Would a one-tailed or a two-tailed test be more appropriate for a researcher who chose alternative 2? Explain why.

d. For a sequence of ten throws, what results would cause a re- searcher operating under the hypotheses listed under alternative 2 to reject the null hypothesis at a significance level of 5%?

8.6 Use the following binomial distribution showing the chances of suc- cess and failure for 12 trials.

Number of Successes Probability

0 successes 0.00118

1 success 0.01065

2 successes 0.04418

3 successes 0.11110

4 successes 0.18857

5 successes 0.22761

6 successes 0.20032

7 successes 0.12953

8 successes 0.06107

9 successes 0.02048

10 successes 0.00463

11 successes 0.00064

12 successes 0.00004

Using a significance level of 0.05, what outcomes would lead you to reject the null hypothesis for each of the following pairs of hypotheses?

a. H0: P � 0.50 H1: P � 0.50

b. H0: P � 0.50 H1: P � 0.50

192

E X E R C I S E S

c. H0: P � 0.50 H1: P � 0.50

d. If you changed the significance level to 0.01, how would your an- swers to parts a, b, and c change?

8.7 Use the following binomial distribution showing the chances of suc- cess and failure for 15 trials.

Number of Successes Probability

0 successes 0.00000

1 success 0.00000

2 successes 0.00001

3 successes 0.00006

4 successes 0.00042

5 successes 0.00228

6 successes 0.00930

7 successes 0.02928

8 successes 0.07168

9 successes 0.13650

10 successes 0.20051

11 successes 0.22313

12 successes 0.18210

13 successes 0.10288

14 successes 0.03598

15 successes 0.00587

Using a significance level of 0.05, what outcomes would lead you to reject the null hypothesis for each of the following pairs of hypotheses?

a. H0: P � 0.50 H1: P � 0.50

b. H0: P � 0.50 H1: P � 0.50

c. H0: P � 0.50 H1: P � 0.50

d. If you changed the significance level to 0.01, how would your an- swers to parts a, b, and c change?

8.8 Locate a research article in a recent issue of a criminology or criminal justice journal.

a. State the research hypotheses tested by the researcher(s).

b. Describe the sampling method, the sample, and the sampling frame used by the researcher(s).

193

C H A P T E R E I G H T : S T E P S I N A S T A T I S T I C A L T E S T194

C o m p u t e r E x e r c i s e s

SPSS and Stata both have the capability to test hypotheses using the binomial

distribution. As discussed in each subsection below, there are sample syntax files

in both SPSS (Chapter_8.sps) and Stata (Chapter_8.do) that illustrate the com-

mands for testing against the binomial distribution.

SPSS

The NPTESTS command will use a two-tailed test to calculate an observed

significance level for a binary variable (e.g., success vs. failure) and compare the

probabilities observed to a binomial distribution (among many other options

in the NPTESTS command). The default probability of a single success in the

BINOMIAL command is p = 0.50, meaning that this command tests the follow-

ing hypotheses:

After executing this command, the output window presents a table of results

that indicates the null hypothesis being tested and the observed significance level

(labeled “Exact Significance” in the table). You will see from this output window

that the observed significance level is 0.012, which is identical to the value calcu-

lated on p. 184 in the text.

As we illustrate below, the probability can be changed easily in the NPTESTS

command line.

To try out the NPTESTS command and apply it to a binomial distribution,

open the SPSS syntax file Chapter_8.sps. A small data file (ex_8_1.sav) will be

read into SPSS when you execute the first two lines of command syntax. This

small data file contains the data from Table 8.1 in the text. Relative decreases in

post-intervention crime are indicated by a value of 1, and relative increases in

post-intervention crime are indicated by a value of 0.

The structure to the NPTESTS command for a comparison to a binomial

distribution is:

Where /ONESAMPLE indicates that we have data from only one sample that

we are going to compare with a hypothesized population. The TEST(crime)

statement indicates that the variable “crime” (the only one in the data file) is

to be the focus of the test. Within the BINOMIAL option—which tells SPSS

to compare “crime” against a binomial distribution—the TESTVALUE is the

hypothesized value of the probability of a success (i.e., P in the hypotheses

above) and SUCCESSCATEGORICAL = LIST(1) tells SPSS that we have coded

our measure so that a value of 1 is a success. All other values for a variable

would be interpreted as failures.

H 0 :P=0.50

H 1 :P≠0.50

NPTESTS /ONESAMPLE TEST (crime) BINOMIAL(TESTVALUE = 0.5

SUCCESSCATEGORICAL = LIST(1)).

C O M P U T E R E X E R C I S E S 195

Stata

The binomial test in Stata is remarkably simple in form:

The output from running the bitest command will be a table indicating the

observed number of success and then both one-tail and the two-tail tests of the

hypothesized probability. Should you be interested in testing a different

hypothesized probability, just alter the value from a presumed 0.5 to the value

you want to test.

Open the file with the Stata do file Chapter_8.do to reproduce the results

from Table 8.1 in the text. The first command line opens the data file, which is

identical to that referred to in the discussion of SPSS above.

The binomial test of the crime variable is then simply:

The results for the two-tail test show a significance level of 0.012 (rounded),

exactly the same as reported above.

Problems

1. The director of a special drug treatment program claims to have found

a cure to drug addiction. As supporting evidence, the director produces

information on a random sample of 13 former clients who were followed

for 12 months after completing the program. Here is how the director

classified each former client:

Success, Failure, Success, Success, Success, Success, Success, Failure,

Success, Success, Failure, Success, Success

Enter these data into SPSS.

a. State all the assumptions of the hypothesis test.

b. What is the test statistic?

c. What decision can be made about the null hypothesis? (Assume that the

d. Can the director conclude that the program is effective? Explain why.

2. A group of researchers wanted to replicate previous research on hot

spot interventions in another city, using a sample of 25 hot spots. When

comparing post-intervention crime levels, they classified the 25 locations

as follows:

Decrease, Decrease, Increase, Decrease, Decrease, Increase, Increase,

Decrease, Decrease, Decrease, Decrease, Decrease, Decrease, Decrease, Decrease, Increase, Increase, Decrease, Decrease, Decrease,

Decrease, Decrease, Increase, Decrease, Decrease

bitest variable_name == hypothesized_probability

bitest crime == 0.5

C H A P T E R E I G H T : S T E P S I N A S T A T I S T I C A L T E S T196

Enter these data into SPSS.

a. State all the assumptions of the hypothesis test.

b. What is the test statistic?

c. What decision can be made about the null hypothesis? (Assume that

d. Did this study show a post-intervention change in crime?

come to the same conclusion? Explain why.

Chi-Square: A Test Commonly Used

for Nominal-Level Measures

How Do Degrees of Freedom Affect the Distribution?

How Does One Interpret a Chi-Square Statistic?

Two Nominal-Scale Variables?

Can the Chi-Square Test Be Used to Examine the Relationship Between

Ordinal-Level Variables?

Nominal-Scale Measure?

C h o o s i n g t h e c h i - s q u a r e d i s t r i b u t i o n

C a l c u l a t i n g t h e c h i - s q u a r e s t a t i s t i c

S u b s t a n t i v e e x a m p l e s u s i n g t h e c h i - s q u a r e t e s t

C h a p t e r n i n e

When is the Chi-Square Distribution Appropriate?

What are Degrees of Freedom?

How is the Chi-Square Statistic Calculated?

How is the Chi-Square Test Carried Out When There Is Only One

How is the Chi-Square Test Applied in the Case of the Relationship Between

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_9,

© Springer Science+Business Media New York 2014

THE BINOMIAL TEST provides a good introduction to the problem of sta-

tistical inference because it examines relatively simple statistical deci-

sions. Using the binomial test, we illustrated how statisticians build a

sampling distribution from probabilities. But the binomial distribution

can be applied only to a single binary variable. In this chapter, we look

at a more commonly used nonparametric test of statistical significance

for nominal-level measures: chi-square. The chi-square test allows the re-

searcher to examine multicategory nominal-level variables as well as the

relationship between nominal-level measures.

We begin our discussion of chi-square with an example similar to the

one used to introduce the binomial distribution in Chapter 7. In this

case, we examine the problem of a fair roll of a die. We then turn to ap-

plications of the chi-square test in criminal justice.

T e s t i n g H y p o t h e s e s C o n c e r n i n g t h e R o l l o f a D i e

In Chapter 7, we examined how you might make a decision about

whether to challenge the fairness of a coin used to decide who would

serve first in a weekly volleyball match. But what if you had the same

question regarding a die used in a friendly game of chance at a local

club? Each week, you and a few friends go down to the club and play a

game of chance that involves the toss of a die. Let’s say that the house

(the club) wins whenever you roll a two or a six. You win whenever

you roll a three or a four, and no one wins when you roll a one or a

five. Over the month, you have played the game 60 times. Of the 60 rolls

of the die, you have lost 24, rolling six 20 times and rolling two 4 times

(see Table 9.1). You have won 10 times in total, rolling three 6 times and

rolling four 4 times. The remaining 26 rolls of the die were split, with 16

ones and 10 fives.

198

As in the case of the coin toss, you and your friends have begun to be

suspicious. Does it make sense that there should be such an uneven split

in the outcomes of the game if the die is fair? Should you raise this issue

with the club and suggest that they change their die? You don’t want to

appear to be a sore sport. Nonetheless, if the distribution of rolls of the

die that you observed is very unlikely given a fair die, you would be

willing to make a protest.

The Chi-Square Distribution

You cannot use the binomial distribution to assess the fairness of the die

because the binomial distribution assumes that there are only two poten-

tial outcomes for each event—for example, a head or tail on each toss of

a coin. For the die, there are six potential outcomes: a roll of one, two,

three, four, five, or six. In such cases, you can make use of another sam- 2

ducted, the chi-square distribution varies from problem to problem.

However, the chi-square distribution varies not according to the number

of trials that are conducted but according to the number of degrees of

freedom (df) associated with a test. The number of degrees of freedom

refers to how much a mathematical operation is free to vary, or take on

any value, after an agreed-upon set of limitations has been imposed.

In the chi-square distribution, these limitations are associated with

the number of categories, or potential outcomes, examined. To define

the degrees of freedom of a chi-square test, we ask how many cate-

gories would have to be known for us to predict the remaining cate-

gories with certainty. For example, if we know that there are 60 rolls of

the die and we also know the precise number of events that fall in five

of the six categories, we will be able to predict the sixth category sim-

ply by subtracting from the total number of events (60) the number in

the five known categories (see Table 9.2). If two categories are blank,

we can predict the total of both, but not the exact split between them.

Accordingly, the number of degrees of freedom for this example is 5.

Once we know the number of events or observations in five categories,

Frequency Distribution for 60 Rolls of a Die

1 No winner 16 2 You lose 4 3 You win 6 4 You win 4 5 No winner 10 6 You lose 20

60

Table 9.1

pling distribution, called the chi-square (� ) distribution. Like the

binomial distribution, which varies depending on the number of trials con-

�Total ( )

T E S T I N G H Y P O T H E S E S C O N C E R N I N G T H E R O L L O F A D I E 199

we can predict the number in the sixth with certainty. More generally,

you can identify the degrees of freedom for a one-variable chi-square

distribution using the equation df � k � 1, where k equals the number

of categories in your measure (for our example, 6 � 1 � 5).

Figure 9.1 shows how chi-square distributions vary according to the

number of degrees of freedom. The height of the distribution represents

the proportion of cases found at any specific value of the chi-square

statistic. As the number of degrees of freedom grows, the height of the

chi-square distribution decreases, with a longer and longer tail to the

right. This means that the proportion of cases found above higher values

of the chi-square statistic grows as the number of degrees of freedom in-

creases. To understand what this means substantively, as well as how the

chi-square distribution is used in making decisions about hypotheses, it

is important to see how the chi-square statistic is calculated.

Frequency Distribution for 60 Rolls of a Die with Information Missing

1 No winner 16 2 You lose 4 3 You win 6 4 You win 4 5 No winner 10 6 You lose ?

60

Frequency of category 6 � (total frequency) � (sum of categories 1 to 5) 20 � 60 � 40

Table 9.2

F re

q u

en cy

Values of χ2

df = 1

df = 2 df = 6

df = 15

Chi-Square Distributions for Various Degrees of FreedomFigure 9.1

�Total ( )

C H A P T E R N I N E : C H I - S Q U A R E200

Calculating the Chi-Square Statistic

The formula for the chi-square statistic is presented in Equation 9.1.

Equation 9.1

The summation symbol in the body of the equation has i � 1 below it

and k above it. This means that we sum the quantity that follows the

summation symbol for each category from the first to the k th, or last,

category. Since there are six categories in our example, we will have to

carry out the same calculation six times, once for each of the six poten-

tial outcomes of the roll of a die.

The quantity that follows the summation symbol includes two sym-

bols, fo and fe. The symbol fo represents the frequency of the events ob-

served in a category, or the observed frequencies. For example, in 20

of the 60 trials, a six was rolled (see Table 9.1). The observed frequency

for a roll of six is 20. The symbol fe represents the expected frequency

of a category. The expected frequencies are ordinarily defined by the

null hypothesis. In our example, they represent the number of events

that would be expected in each category in the long run if the die were

fair. Because a fair die is one for which there is an equal chance of ob-

taining any of the six potential outcomes, we divide the 60 observations

evenly across the six categories. This leads to an expected frequency of

10 for each potential outcome. Table 9.3 shows the expected and ob-

served frequencies for our example.

To calculate the chi-square statistic, Equation 9.1 tells us first to sub-

tract the expected frequency from the observed frequency in each cate-

gory. We then square the result and divide that quantity by the expected

frequency of the category. For example, for a roll of six, we subtract 10

(the expected frequency) from 20 (the observed frequency). We then

square that quantity (to get 100) and divide the result by 10. This gives

us 10 for a roll of six. After carrying out this computation for each cate-

gory, as is done in Table 9.4, we add up the results for all six categories

�2 � � k

i�1

( fo � fe)

2

fe

Expected and Observed Frequencies for 60 Rolls of a Fair Die

fe fo

1 10 16 2 10 4 3 10 6 4 10 4 5 10 10 6 10 20

60 60

Table 9.3

�Total ( )

T E S T I N G H Y P O T H E S E S C O N C E R N I N G T H E R O L L O F A D I E 201

to obtain the total chi-square statistic. The chi-square statistic for this ex-

ample is 22.4.

The chi-square statistic measures how much the observed distribution

differs from that expected under the null hypothesis. If the observed fre-

quencies are similar to the expected frequencies, the chi-square statistic

is small. If the observed frequencies are the same as the expected fre-

quencies, the chi-square statistic equals 0. The more the observed fre-

quencies differ from the expected frequencies, the larger the chi-square

statistic will be. What does this mean in terms of making a decision

about the fairness of the die? To find out, we have to turn to a table of

probabilities associated with the chi-square distribution.

Linking the Chi-Square Statistic to Probabilities: The Chi-Square Table

In Chapters 7 and 8, we used the binomial formula to calculate the prob-

ability associated with each of the possible outcomes in our sample. For

other tests of statistical significance, including chi-square, we can take

advantage of already calculated probability distributions. Appendix 2

presents a table of probabilities associated with chi-square distributions

with degrees of freedom from 1 to 30. The chi-square table does not give

us the probability associated with every possible outcome, but rather

provides probabilities and then lists the chi-square statistics associated

with them.

As illustrated in the chi-square table in Appendix 2, a larger chi-

square statistic is associated with a smaller significance level, or � value.

For example, under one degree of freedom, a statistic of 2.706 is associ-

ated with a significance level of 0.10, a statistic of 3.841 with an � value

of 0.05, and a statistic of 10.827 with an � value of 0.001. This also

means that the larger the chi-square statistic obtained in a test, the less

likely it is that the observed distribution is drawn from the expected dis-

tribution. This logic makes good common sense. For our example of the

roll of a die, it is reasonable to become more suspicious about the fair-

ness of the die as the number of events in the different categories

Computation of Chi-Square for 60 Rolls of a Die

OUTCOME A fo fe (fo � fe) (fo � fe)2

1 16 10 6 36 3.6 2 4 10 �6 36 3.6 3 6 10 �4 16 1.6 4 4 10 �6 36 3.6 5 10 10 0 0 0 6 20 10 10 100 10.0

� � 22.4

(fo � fe)2

fe

Table 9.4

C H A P T E R N I N E : C H I - S Q U A R E202

becomes more uneven. If we expect 10 events in each category and ac-

tually get one with 20, one with 16, and two others with only 4, this

should begin to make us suspicious. If one or two categories have 25

cases and two or three have none, it seems even more likely that the die

is not a fair one. But if each category has about 10 cases, which is to be

expected in the long run with 60 rolls of a fair die, both common sense

and chi-square give us little reason to suspect a biased die.

Notice as well in Appendix 2 that as the number of degrees of free-

dom gets larger, a larger chi-square statistic is needed to arrive at the

same probability value. For example, with one degree of freedom, a chi-

square statistic of 3.841 is associated with an � value of 0.05. With 30 de-

grees of freedom, a statistic of 43.773 is needed to achieve the same

threshold. This reflects the difference in the shape of chi-square distribu-

tions with different degrees of freedom and makes good sense if you

consider how the chi-square statistic is calculated. A separate addition is

made to the chi-square statistic for each possible category. Accordingly,

it makes sense to demand a larger statistic as the number of categories in

the test increases.

What about our decision regarding the roll of the die? Looking at Ap-

pendix 2, we can see that with five degrees of freedom a chi-square sta-

tistic of 11.070 is associated with a significance level of 0.05. This means

that in the long run we would expect to obtain a chi-square statistic of

11.070 in only 5 in 100 samples if the die is fair. In fact, we obtained a

chi-square statistic of 22.4. This number is even larger than that needed

for a significance level of 0.001. Accordingly, the observed significance

level for this test is less than 0.001 ( p � 0.001). If the die were fair, the

probability of getting a distribution like the one observed in our 60 rolls

of a die would be less than 1 in 1,000. Given this result, we would likely

come to the conclusion that the die was not a fair one and call for the

club to use a new one.

A Substantive Example: The Relationship Between Assault

Victims and Offenders

We can illustrate the chi-square test for a single variable by considering

the responses from a random sample survey of Illinois residents.1 One of

the primary purposes of the survey was to examine the effect of victim-

ization on the physical and mental health of adults. Each respondent was

asked about a variety of possible victimization experiences. When the

person claimed to have experienced a crime, a series of follow-up ques-

tions were asked about the circumstances of the event. Table 9.5 pre-

sents the frequency distribution of the responses to a question about the

1See Chester L. Britt, “Health Consequences of Criminal Victimization,” International

Review of Victimology 8 (2001): 63–73 for a description of the study.

T E S T I N G H Y P O T H E S E S C O N C E R N I N G T H E R O L L O F A D I E 203

relationship between the victim and the offender for those persons who

claimed to have been assaulted.

A simple research question using these data might focus on whether

the victim-offender relationship was unevenly distributed among the

population of assault victims. To answer our research question, we fol-

low the form of a statistical test introduced in Chapter 8.

We begin by stating the assumptions of our chi-square test. The level of

measurement required for chi-square is nominal. We make no specific as-

sumptions regarding the shape of the population distribution, as the chi-

square test is a nonparametric test of statistical significance. Although the

chi-square test ordinarily requires a fully independent random sample, this

sample was selected without replacement.2 This is not a serious violation

of our assumptions because the sample is very small relative to the popu-

lation of interest. Note that our null hypothesis is that the victim-offender

relationship in the population is evenly or randomly distributed across the

categories examined. The research hypothesis is that the victim-offender

relationship is not randomly or evenly distributed in the population.

Assumptions:

Level of Measurement: Nominal scale.

Population Distribution: No assumption made.

Sampling Method: Independent random sampling (no replacement; sam-

ple is small relative to population).

Sampling Frame: Persons aged 18 and over in the state of Illinois.

Hypotheses:

H0: The type of victim-offender relationship for assault victims is ran-

domly distributed.

H1: The type of victim-offender relationship for assault victims is not ran-

domly distributed.

Relationship Between Assault Victim and Offender

CATEGORY FREQUENCY (N )

Stranger 166 Acquaintance 61 Friend 35 Boyfriend/girlfriend 38 Spouse 66 Other relative 44 Total (�) 410

Table 9.5

2There are certain specific situations in which the chi-square test does not require

sampling with replacement; see B. S. Everitt, The Analysis of Contingency Tables (Lon-

don: Chapman and Hall, 1997).

C H A P T E R N I N E : C H I - S Q U A R E204

The Sampling Distribution Since we are analyzing the distribution of

cases for a nominal variable, the chi-square distribution provides an ap-

propriate means of assessing whether the observations are randomly dis-

tributed across the six categories of victim-offender relationships. For a

single nominal variable, the number of degrees of freedom for the chi-

square test is df � k � 1 � 6 � 1 � 5.

Significance Level and Rejection Region Since we have no reason to im-

pose a stricter or more lenient level of statistical significance on our analy-

sis, we will use a significance level (�) of 0.05. Given that the number of

degrees of freedom associated with this chi-square test is 5 and the signifi-

cance level is 0.05, we see from Appendix 2 that the corresponding critical

value of the chi-square distribution is 11.070. Accordingly, if the calculated

value of the chi-square statistic in our example is greater than 11.070, we

will reject the null hypothesis and conclude that type of victim-offender re-

lationship among assault victims is not randomly distributed.

The Test Statistic Equation 9.1 provides the formula for calculating the chi-

square statistic to test for random assignment of cases to each category or

value. We begin by calculating the expected frequency (fe) for each cell in

the table. Again, as in the example of the die, we would expect under the

null hypothesis that there would be an equal number of cases in each of

the categories examined. To calculate the expected frequency mathemati-

cally, we divide the total N of cases by the number of categories. This is

done below, where we get an expected value for each category of 68.333:

After calculating the expected frequency, we can proceed to calculate the

chi-square statistic. Table 9.6 presents the observed and expected frequen-

cies for each cell and the appropriate calculations for determining the value

of the chi-square statistic. We find the value of the test statistic to be 178.85.

fe � N k

� 410 6

� 68.333

Computation of Chi-Square for Type of Victim-Offender Relationship

CATEGORY fo fe (fo � fe) (fo � fe)2

Stranger 166 68.333 97.667 9,538.843 139.593 Acquaintance 61 68.333 �7.333 53.773 0.787 Friend 35 68.333 �33.333 1,111.089 16.260 Boyfriend/girlfriend 38 68.333 �30.333 920.091 13.465 Spouse 66 68.333 �2.333 5.443 0.080 Other relative 44 68.333 �24.333 592.095 8.665

� � 178.849

(fo � fe)2

fe

Table 9.6

T E S T I N G H Y P O T H E S E S C O N C E R N I N G T H E R O L L O F A D I E 205

The Decision The critical value for our test of statistical significance was

11.070, meaning that a calculated chi-square statistic greater than this

critical value would lead to rejection of the null hypothesis. The value of

our test statistic is 178.85, which is much larger than our critical chi-

square value. Accordingly, the observed significance level of our test is

less than the significance criterion we set at the outset (p � 0.05). On

the basis of this outcome, we reject the null hypothesis and conclude

that type of victim-offender relationship among assault victims is not ran-

domly distributed. Of course, we cannot be certain that the null hypothe-

sis is false for the population we are examining. We make our decision

with a set risk of a Type I error defined at the outset of our test.

R e l a t i n g T w o N o m i n a l - S c a l e M e a s u r e s i n a C h i - S q u a r e T e s t

In criminal justice and criminology, we seldom examine research issues like

the fairness of a die or the randomness of type of victim-offender relation-

ship, which are concerned with outcomes on only one measure. More often,

we are interested in describing the relationships among two or more vari-

ables. For example, we may want to assess whether men and women are

likely to be placed in different types of treatment facilities or whether differ-

ent ethnic groups receive different types of sanctions. For each of these ex-

amples, two measures must be assessed at the same time. In the former, we

examine both gender and type of treatment facility. In the latter, we exam-

ine type of sentence and ethnicity. Below, we use the example of a study of

white-collar criminals to illustrate the use of chi-square in making inferences

about the relationship between two variables: recidivism and sanction type.

A Substantive Example: Type of Sanction and Recidivism

Among Convicted White-Collar Criminals

In a study of white-collar offenders, data on reoffending from FBI records

over a ten-year period were examined. The sample included offenders

from seven U.S. district courts, convicted of eight different white-collar

crimes (antitrust violations, securities fraud, mail and wire fraud, false

claims and statements, credit and lending institution fraud, bank embez-

zlement, income tax fraud, and bribery). The sample was chosen ran-

domly without replacement.3 The research question concerned whether

imprisonment of white-collar offenders impacted upon reoffending.

3In this case, a stratified random sample was selected in order to ensure a broad sampling

of white-collar offenders. For our example here, we treat the sample as a simple random

sample. See David Weisburd, Elin Waring, and Ellen Chayet, “Specific Deterrence in a

Sample of Offenders Convicted of White Collar Crimes,” Criminology 33 (1995): 587–607.

C H A P T E R N I N E : C H I - S Q U A R E206

The likelihood of rearrest for a group of offenders who received a

prison sanction was compared with that of a matched group who did not

receive a prison sanction. The researchers found that 33.0% of the prison

group (N � 100) was rearrested during the follow-up period, in contrast

to 28.4% of the no-prison group (N � 67). What conclusions can we

come to concerning white-collar criminals generally?

To answer our research question, we follow the standard format of a

test of statistical significance. We begin by stating our assumptions. Re-

member that to state the assumptions you must choose the type of test

you will use. In this case, we have chosen a chi-square test for relating

two nominal-level measures.

Assumptions:

Level of Measurement: Nominal scales.

Population Distribution: No assumption made.

Sampling Method: Independent random sampling (no replacement; sam-

ple is small relative to population).

Sampling Frame: Offenders from seven federal judicial districts convicted

of eight different white-collar crimes.

Hypotheses:

H0: There is no difference in the likelihood of rearrest among similar

white-collar offenders sentenced or not sentenced to prison. (Or, likeli-

hood of rearrest and imprisonment are independent.)

H1: There is a difference in the likelihood of rearrest among similar

white-collar offenders sentenced or not sentenced to prison. (Or, likeli-

hood of rearrest and imprisonment are not independent.)

The level of measurement required for a chi-square test is nominal.

Our example includes two variables: rearrest and type of sanction. Each

is measured as a binary nominal variable. For rearrest, we examine those

rearrested versus those not rearrested in the follow-up period. For type

of sanction, we differentiate between those who were sentenced to

prison and those who did not receive a prison sanction. In regard to the

population distribution, chi-square is a nonparametric test and therefore

requires no specific assumption.

The sample was selected randomly, but as is the case with most crim-

inal justice studies, the investigators did not sample with replacement. At

the same time, the population from which the sample was drawn is very

large relative to the sample examined, and thus we have no reason to

suspect that this violation of the assumptions will affect our test result.

The sampling frame includes offenders from seven federal judicial dis-

tricts. As discussed in Chapter 8, it is necessary to explain why your

tricts convicted of eight different white-collar crimes. Accordingly, our

inferences relate directly to the population of those offenses and those

R E L A T I N G T W O N O M I N A L - S C A L E M E A S U R E S 207

sample is representative of a broader population of cases if you want to

make inferences beyond your sampling frame. In this study, the seven dis-

tricts examined were seen as providing a sample with geographic spread

throughout the United States, and the selected white-collar offenses were

defined as offering a “broad sampling of white-collar offenders.”

In most applications of the chi-square test, one cannot assign a direc-

tional research hypothesis. This is because chi-square requires a nominal

ables, as examined here, the researcher can choose between a direc-

tional and a nondirectional research hypothesis. In our example, a

directional hypothesis would be that the prison group is more likely than

the no-prison group to be rearrested or that the no-prison group is more

likely than the prison group to be rearrested. The research hypothesis

stated by the investigators in this study was nondirectional. It stated sim-

ply that the two groups (prison and no-prison) differ in terms of likeli-

hood of rearrest during the follow-up period. The null hypothesis was

that there is no difference between the prison and no-prison groups.

Researchers often state the hypotheses of a chi-square test in terms of

the independence of the variables that are examined. Stated this way, the

null hypothesis would be that prison group (prison or no-prison) is inde-

pendent, or unrelated to rearrest, in the follow-up period. The research

hypothesis is that prison group is not independent. While this method of

stating the hypotheses for your test sounds different, it leads to the same

conclusions. If the two groups differ in terms of likelihood of rearrest in

the follow-up period, then prison group and recidivism are related and

thus not independent. If there is no difference, then prison group is un-

related to, or independent of, recidivism in the follow-up period.

The Sampling Distribution Because we are examining the relationship

between two nominal-scale variables, the chi-square distribution pro-

vides an appropriate sampling distribution for our test. However, our de-

cision about degrees of freedom is not as straightforward as that in the

example of a roll of a die. In this case, we must take into account the

Recidivism Among 167 White-Collar Criminals According to Whether They Did or Did Not Receive Prison Sentences

Imprisoned Not imprisoned

Subsequently arrested Cell A Cell B 52 33 19

Not subsequently arrested Cell C Cell D 115 67 48

100 67 167

Table 9.7

catagories examined. Nonetheless, in the special case of two binary vari-

level of measurement, which does not assign order or value to the

Row total

Column total

C H A P T E R N I N E : C H I - S Q U A R E208

joint distribution of our measures. This is illustrated in Table 9.7, which

shows two potential outcomes for the prison variable and two potential

outcomes for the arrest variable. We have four cells, or four possible

combined outcomes. Cell A is for offenders who received a prison sanc-

tion and were arrested in the follow-up period. Cell B is for offenders

who did not receive a prison sanction and were arrested in the follow-up

period. Cell C is for offenders who received a prison sanction and were

not arrested in the follow-up period. Cell D is for offenders who did not

receive a prison sanction and were not arrested in the follow-up period.

If we sum across and down the cells, we gain two row marginals and

two column marginals. The row marginals represent the totals for the rows:

52 for those arrested and 115 for those not arrested. The column marginals

represent the totals for the columns: N � 100 for the prison group, and N �

67 for the no-prison group. If you know the row and column marginals, as

is assumed in computing the degrees of freedom for chi-square, you can

predict with certainty the remaining cells, once the value of any one cell is

known (see Table 9.8). Degrees of freedom for a two-variable chi-square

can be gained more simply through the formula df � (r � 1)(c � 1), where

r represents the number of rows and c the number of columns. For our ex-

ample, there are two rows and two columns, so df � (2 � 1)(2 � 1) � 1.

Significance Level and Rejection Region We stated no reason at the outset

for choosing for our example a stricter or more lenient significance thresh-

old than is used by convention. Accordingly, we use a significance level of

0.05 for our test. Our rejection region is defined by the chi-square table

(see Appendix 2). Importantly, the chi-square distribution is not concerned

with the direction of outcomes in a test. It tells us to what extent the ob-

served frequencies in our example differ from those that would be ex-

pected under the null hypothesis of no difference. Whether they differ in

one direction or another, the chi-square statistic will always be positive.

Predicting the Missing Cells in a Two-Variable Chi-Square Test

Row

Cell B 52 33 ?

Cell C Cell D 115 ? ?

67 167

Given that cell A � 33: Cell B � (52 � 33) � 19 Cell C � (100 � 33) � 67 Cell D � (115 � 67) � 48

Table 9.8

Imprisoned Not imprisoned total

Column total 100

Subsequently arrested

Not subsequently arrested

Cell A

R E L A T I N G T W O N O M I N A L - S C A L E M E A S U R E S 209

The terms “directional” and “nondirectional” are very tenuous ones in

a chi-square test. Chi-square assumes nominal-scale variables, which by

definition do not provide information about the order of values in a

measure. If we cannot specify the order of two measures, we cannot

speak of the direction of their relationship. As noted earlier, in most situ-

ations a directional hypothesis is not appropriate for a chi-square test. In

the special case of two binary variables, however, researchers do some-

times use chi-square to examine directional research hypotheses. For ex-

ample, we might have stated our research hypothesis as “The likelihood

of arrest in the follow-up period for white-collar offenders sentenced to

prison is lower than that of similar white-collar offenders not sentenced

to prison.”

However, our research hypothesis is nondirectional, as is the table of

chi-square values. To define our rejection region, we turn to the row of

the table associated with one degree of freedom. Under a significance

level (�) of 0.05, we see a score of 3.841. For us to reject the null hy-

pothesis, our test statistic will have to be greater than this value.4

The Test Statistic To apply chi-square to the two-variable case, we need

to adapt our original equation. The formula for the chi-square statistic for

relating two measures is presented in Equation 9.2.5

Equation 9.2

The only difference between Equation 9.2 and Equation 9.1 is that we

have an additional summation symbol. In this case, we do not sum

simply across the categories of one measure; rather, we sum across

each row (r) and column (c) of the joint distribution of two measures.

Accordingly, Equation 9.2 reminds us that we must examine the ex-

pected and observed frequencies for every potential outcome we can

observe—or, in terms of the chi-square equation, for every cell in our

table.

�2 � � r

i�1

� c

j�1

( fo � fe)

2

fe

4What if we had defined a directional research hypothesis? In this case, we look to the

column of the table for twice the value of the desired significance level, since we now

have placed all risk of falsely rejecting the null hypothesis in only one direction. For

example, for a 0.05 significance level, we turn to the test statistic for a 0.10 level. 5When a chi-square test has only one degree of freedom, it is recommended that a

correction factor be added if the expected frequencies of any cell fall below 20. The

correction provides a more conservative, or smaller, chi-square statistic:

�2 � � r

i�1

� c

j�1

(� fo � fe � � 0.5)2

fe

C H A P T E R N I N E : C H I - S Q U A R E210

For our example, this means we must sum across cells A, B, C, and D.

As before, we want to compare the observed frequency to the expected

frequency in each cell. The observed frequencies are those gained in our

research. The expected frequencies are defined through the null hypoth-

esis. The null hypothesis states that there is no difference in arrest rates

between the prison and no-prison groups. If this is true, then we should

expect the same proportion of arrests in both groups. To calculate the

expected frequencies, accordingly, we first need to define the overall

proportion of offenders arrested in the follow-up period.

The proportion of offenders arrested overall in the sample in the follow-

up period is obtained by dividing the total number of offenders arrested

(Ncat � 52) by the total number of offenders in the sample (Ntotal � 167):

To get the expected frequency for cell A, we multiply this proportion by

the marginal total of 100 ( fe � 31.14). For the no-prison group, we have

a total of 67 offenders. Applying the proportion of the total sample to

this group, we multiply 0.3114 by 67 and get an expected frequency of

20.86 for cell B. In practice, we do not need to compute the expected

frequencies for the remaining two cells, C and D. Indeed, we could have

assigned all of the cells expected frequencies based on knowledge of

only one cell. This is what the number of degrees of freedom for this ex-

ample tells us. If you know the number of cases in one cell, you can

predict with certainty the rest. The expected and observed frequencies

for our example are shown in Table 9.9.

Now that we have calculated the observed and expected frequencies

for each potential outcome, or cell, we can calculate the chi-square sta-

Proportion � Ncat

N total �

52

167 � 0.3114

Expected and Observed Frequencies of Recidivism and Nonrecidivism for White-Collar Offenders According to Whether They Received Prison Sentences

Imprisoned Not imprisoned

Subsequently arrested Cell A Cell B 52 fo � 33 fo � 19

fe � 31.14 fe � 20.86

Not subsequently arrested Cell C Cell D 115 fo � 67 fo � 48

fe � 68.86 fe � 46.14

100 67 167

Table 9.9

tistic. To do this, we first square the difference of the observed and

Row total

Column total

R E L A T I N G T W O N O M I N A L - S C A L E M E A S U R E S 211

expected frequency of the cell:

This is done in Table 9.10 for each of the four cells in our problem.

Using cell A as an example, we first subtract the expected frequency of

31.14 from the observed frequency of 33. We then square this quantity

(1.86), obtaining a result of 3.4596. Dividing this result by the expected

frequency in the cell (31.14) gives us 0.1111. The sum of all four cells,

0.4021, is our test statistic.

The Decision Our rejection region was defined as including any chi-

square statistic greater than 3.841. The test statistic for our example is

only 0.402. Accordingly, we choose not to reject the null hypothesis. The

observed significance level for our test is greater than the significance

level, or threshold, we set at the outset (p � 0.05). We conclude that

there is no significant difference in the likelihood of recidivism between

white-collar offenders who have and have not been sentenced to prison.

Our inferences are made directly to the specific offenses and judicial dis-

tricts defined in the sampling frame.

E x t e n d i n g t h e C h i - S q u a r e T e s t t o M u l t i c a t e g o r y V a r i a b l e s : T h e E x a m p l e o f C e l l A l l o c a t i o n s i n P r i s o n

The previous example illustrates the use of chi-square in the case of two

binary variables. We now turn to an extension of the chi-square test to

an example including a multicategory nominal-level variable. Our exam-

ple is drawn from a study of the relationship between prisoners’ race

and their cell assignments in a large state prison in the northeastern

( fo � fe) 2

fe

Computation of Chi-Square for 167 White-Collar Criminals

CELL fo fe (fo � fe) (fo � fe)2

A 33 31.14 1.86 3.4596 0.1111 B 19 20.86 �1.86 3.4596 0.1658 C 67 68.86 �1.86 3.4596 0.0502 D 48 46.14 1.86 3.4596 0.0750

� � 0.4021

(fo � fe)2

fe

Table 9.10

expected frequencies for each cell, and then we divide this quantity by the

C H A P T E R N I N E : C H I - S Q U A R E212

United States.6 We examine the placement of non-Hispanic white and

“nonwhite” inmates (including Hispanics) into seven cell blocks. The

sample includes all prisoners in the general prison population for a sin-

gle day. The distribution of cases is presented in Table 9.11.

If cell assignments were made on considerations unrelated to race, we

would expect to find the proportion of non-Hispanic whites in each cell

block roughly equivalent to the proportion of non-Hispanic whites in the

general prison population (19.9%; see the marginal for non-Hispanic

whites in Table 9.11). Such equivalence is not evident. In block G, for ex-

ample, non-Hispanic whites constituted 12.2% of the inmates. In block H,

they comprised 60.7%. Do results for this sample allow us to conclude that

Assumptions:

Level of Measurement: Nominal scales.

Population Distribution: No assumption made.

Sampling Method: Independent random sampling (the entire sampling

frame is examined).

Sampling Frame: All prisoners in the general prison population on a

specific day.

6See Douglas McDonald and David Weisburd, “Segregation and Hidden Discrimina-

tion in Prisons: Reflections on a Small Study of Cell Assignments,” in C. Hartchen

(ed.), Correctional Theory and Practice (Chicago: Nelson Hall, 1991).

Proportions of Non-Hispanic White Prisoners in Seven Cell Blocks

Non-Hispanic

whites Nonwhites

Cell block C 48 208 256 18.7% 81.3% 100%

Cell block D 17 37 54 31.5% 68.5% 100%

Cell block E 28 84 112 25.0% 75.0% 100%

Cell block F 32 79 111 28.8% 71.2% 100%

Cell block G 37 266 303 12.2% 87.8% 100%

Cell block H 34 22 56 60.7% 39.3% 100%

Cell block I 44 268 312 14.1% 85.9% 100%

240 964 1,204 19.9% 80.1% 100%

Table 9.11

Row

total

total

Column

there is disparity in cell-block assignments throughout the year in the prison.

E X T E N D I N G T H E T E S T T O M U L T I C A T E G O R Y V A R I A B L E S 213

Hypotheses:

H0: Cell-block assignment and race are independent.

H1: Cell-block assignment and race are not independent.

As in our previous example, we assume a nominal level of measure-

ment for our test and do not make assumptions regarding the form of

the population distribution. Prisoner race is measured at the binary nom-

inal level, and cell block is a multicategory nominal scale.

The sample includes all cases in the sampling frame. Accordingly, we

do not need to use statistical inference to make statements about that

population. However, the study was designed not only to describe

prison-cell allocations on that day, but also to make more general state-

ments about cell allocations in the prison studied throughout the year.

This is not an uncommon scenario in criminal justice research, in good

part because the realities of the criminal justice system often preclude

sampling beyond specific institutions or outside of specific time frames.

This means, however, that the researchers seek to make inferences be-

yond their sampling frame.

If cell allocations on the day examined in this study are representative

of cell allocations more generally throughout the year, then the infer-

ences made on the basis of the test will be reliable. If not, then the test

will not provide for valid inferences. In our example, the investigators

argue:

There was no reason to suspect that the cell assignments of prisoners

on that day differed substantially from assignments on other days.

Moreover, these cell assignments represented the results of decisions

made over the course of months and perhaps years prior to the date

of drawing the sample. There was every reason to believe, conse-

quently, that cell assignments on that date constituted a valid repre-

sentation of cell assignment decisions made during the several months

prior to that day.

Our research question asks whether we would be likely to obtain the

distribution we observe in our sample if assignment to cell blocks were

colorblind in the population. Stated in the form of hypotheses, we ask

whether race and cell-block assignment are independent. If they are in-

dependent, as proposed in our null hypothesis, then we would expect

about the same proportion of nonwhite and non-Hispanic white prison-

ers in each cell block. Our research hypothesis is nondirectional. It states

that race and cell-block assignment are not independent. In this exam-

ple, as in most chi-square tests, use of nominal-scale measures, which do

not assign order or value to categories, means that one cannot define a

directional research hypothesis.

C H A P T E R N I N E : C H I - S Q U A R E214

The Sampling Distribution Because we are examining the relationship

between two nominal variables, one binary and one multicategory, we

use the chi-square sampling distribution. The number of degrees of free-

dom for our problem is defined as in the previous example:

W orking It Out

� 6

� (7 � 1)(2 � 1)

df � (r � 1)(c � 1)

In this case, we have seven categories for our row variable (cell block)

and two categories for our column variable (ethnicity). The number of

degrees of freedom for our sampling distribution is six.

Significance Level and Rejection Region As we have no reason to pro-

pose more lenient or stricter significance criteria than are used by con-

vention, we will set a 0.05 significance level. To define our rejection re-

gion, we turn to the row of the chi-square table associated with six

degrees of freedom. Under the 0.05 column, a chi-square statistic of

12.592 is listed. If the test statistic is greater than this critical value, then it

falls within the rejection region of the test.

The Test Statistic To calculate the test statistic in this multicategory ex-

ample, we follow the same procedure used for the two-by-two table in

the previous section. Our first task is to define the expected frequency

for each cell of the table. We do this, as before, by dividing a marginal of

the table by the total proportion of cases. Taking the overall number of

non-Hispanic whites in the sample, we obtain a proportion of 0.1993:

W orking It Out

� 0.199335

� 240

1,204

Proportion � Ncat

N total

E X T E N D I N G T H E T E S T T O M U L T I C A T E G O R Y V A R I A B L E S 215

To calculate the expected frequency in each cell in the non-Hispanic

whites column, we multiply this proportion by the marginal total for

each row. So, for example, for cell block C, we multiply 256 by

0.199335, leading to an expected frequency for non-Hispanic whites of

51.030. We then replicate this procedure for each of the six other cells in

the non-Hispanic whites column. To calculate the expected frequencies

for the nonwhites column, we simply subtract the expected frequency

for the non-Hispanic whites column from the row marginal. So, for ex-

ample, for nonwhites in cell block C, the expected frequency is 256 (the

marginal total) minus 51.030 (the expected frequency for non-Hispanic

whites for that cell block), or 204.970. Table 9.12 includes the expected

and observed frequencies for the 14 cells in our example.

To obtain the test statistic, we use Equation 9.2, which may be ap-

plied to any two-variable chi-square problem:

Again we begin by subtracting the expected frequency from the observed

frequency in each cell and squaring the result. This quantity is then di-

vided by the expected frequency of the cell. The chi-square statistic is

found by summing the result across all 14 cells. The full set of calculations

for the test statistic is presented in Table 9.13. The chi-square score for our

example is 88.3610.

�2 � � r

i�1

� c

j�1

( fo � f e)

2

fe

Observed Frequencies and Expected Frequencies for Non-Hispanic White and Nonwhite Prisoners in Seven Cell Blocks

Non-Hispanic

whites Nonwhites

Cell block C f � 48 fo � 208 256 fe � 51.030 fe � 204.970

Cell block D fo � 17 fo � 37 54 fe � 10.764 fe � 43.236

Cell block E fo � 28 fo � 84 112 fe � 22.326 fe � 89.674

Cell block F fo � 32 fo � 79 111 fe � 22.126 fe � 88.874

Cell block G fo � 37 fo � 266 303 fe � 60.399 fe � 242.601

Cell block H fo � 34 fo � 22 56 fe � 11.163 fe � 44.837

Cell block I fo � 44 fo � 268 312 fe � 62.193 fe � 249.807

240 964 1,204

Table 9.12

Row

total

Colum total

o

C H A P T E R N I N E : C H I - S Q U A R E216

The Decision The outcome of 88.3610 is much greater than the critical

value for our test of 12.592. Accordingly, we reject the null hypothesis

that race and cell-block allocation are independent (using a 5% signifi-

cance level). We conclude that there is a statistically significant relation-

ship between the distribution of prisoners across cell blocks and their

race.

E x t e n d i n g t h e C h i - S q u a r e T e s t t o a R e l a t i o n s h i p B e t w e e n T w o O r d i n a l V a r i a b l e s : I d e n t i f i c a t i o n w i t h F a t h e r s a n d D e l i n q u e n t A c t s

The examples of the application of the chi-square test presented so far

have used only nominal-scale variables. This is consistent with the as-

sumptions of the chi-square test. But in practice researchers sometimes

use chi-square to test for independence when one or both of the vari-

ables have been measured at the ordinal level of measurement. This test

for independence can provide important information to the researcher.

However, because the chi-square test assumes a nominal scale of mea-

surement, it does not pay attention to the order of the categories in an

ordinal scale. This means that a statistically significant finding can tell the

researcher only that the distribution of scores observed is different from

that expected had there been no relationship. It cannot test for whether

the values of one variable increase as the values of the other increase or,

conversely, whether the scores on one measure increase as those on the

Computation of Chi-Square for Non-Hispanic White (W) and Nonwhite (NW) Prisoners in Seven Cell Blocks

CELL BLOCK RACE fo fe (fo � fe) (fo � fe) 2

C W 48 51.030 �3.030 9.1809 0.1799 C NW 208 204.970 3.030 9.1809 0.0448 D W 17 10.764 6.236 38.8877 3.6128 D NW 37 43.236 �6.236 38.8877 0.8994 E W 28 22.326 5.674 32.1943 1.4420 E NW 84 89.674 �5.674 32.1943 0.3590 F W 32 22.126 9.874 97.4959 4.4064 F NW 79 88.874 �9.874 97.4959 1.0970 G W 37 60.399 �23.399 547.5132 9.0649 G NW 266 242.601 23.399 547.5132 2.2568 H W 34 11.163 22.837 521.5286 46.7194 H NW 22 44.837 �22.837 521.5286 11.6317 I W 44 62.193 �18.193 330.9852 5.3219 I NW 268 249.807 18.193 330.9852 1.3250

� � 88.3610

(fo � fe) 2

fe

Table 9.13

E X T E N D I N G T H E T E S T T O O R D I N A L V A R I A B L E S 217

other decrease. When you use the chi-square test for ordinal-scale vari-

ables, the test itself treats the variables as if they were simply composed

of a group of nominal categories.

Table 9.14 presents data from the Richmond Youth Survey report on

the relationship between number of delinquent acts and affectional iden-

tification with one’s father. The distribution of cases presented refers

only to the white males who responded to the survey.7 The sample was

a random sample (drawn without replacement) for all high school–age

white males in Richmond, California, in 1965. The size of the sample is

small relative to the sampling frame.

If delinquency were unrelated to attachment to one’s family—here in-

dicated by the level of affectional identification with one’s father—we

would expect to find the distribution of cases for each level of delin-

quency to be roughly equal across levels of identification. The distribu-

tion of cases provides some indication that these variables are not, in

fact, independent. For example, among the youths who wanted to be

like their father in every way, 63% reported that they had not committed

a delinquent act. This was true for only 41% of those who did not want

to be at all like their fathers. Our question is whether the differences we

observe in our sample are large enough for us to conclude that identifi-

cation with one’s father and delinquency are related in the population

from which our sample has been drawn.

7David F. Greenberg, “The Weak Strength of Social Control Theory,” Crime and Delin-

quency 45:1 (1999): 66–81.

Affectional Identification with Father by Number of Delinquent Acts

AFFECTIONAL D E L I N Q U E N T A C T S IDENTIFICATION WITH FATHER None One Two or More

In every way 77 25 19 121 63.636% 20.661% 15.702% 100%

In most ways 263 97 44 404 65.099% 24.010% 10.891% 100%

In some ways 224 97 66 387 57.881% 25.065% 17.054% 100%

In just a few ways 82 52 38 172 47.674% 30.233% 22.093% 100%

Not at all 56 30 52 138 40.580% 21.739% 37.681% 100%

702 301 219 1,222 57.447% 24.632% 17.921% 100%

Table 9.14

Row total

Column total

C H A P T E R N I N E : C H I - S Q U A R E218

Assumptions:

Level of Measurement: Nominal scales (our study examines two ordinal-

scale measures).

Population Distribution: No assumption made.

Sampling Method: Independent random sampling (no replacement;

sample is small relative to population).

Sampling Frame: High school–age white males in Richmond, California,

in 1965.

Hypotheses:

H0: Affectional identification with father and delinquency are independent.

H1: Affectional identification with father and delinquency are not

independent.

The Sampling Distribution Although we are using two ordinal-scale mea-

sures rather than two nominal-scale measures, we have chosen to use

the chi-square sampling distribution to test for a relationship. This viola-

tion of the nominal-scale assumption for the chi-square test is accept-

able. However, by placing the violation of the assumption in parentheses

next to the test requirement of a nominal level of measurement, we re-

mind ourselves that chi-square is not concerned with the order of the

categories in the measures examined. It treats the two ordinal-scale mea-

sures as if they were nominal-scale measures and simply tests for

whether the distributions among the categories depart from what we

would expect under an assumption of independence. As we did in the

two previous examples, we calculate degrees of freedom as follows:

W orking It Out

� 8

� (5 � 1)(3 � 1)

df � (r � 1)(c � 1)

Significance Level and Rejection Region There is no reason to propose a

more lenient or stricter significance level for this analysis, so we will

stick with a 0.05 significance level. Given that we have eight degrees of

freedom and a significance level of 0.05, we can consult the chi-square

table and determine that our critical value of the chi-square statistic is

15.507. If the test statistic is greater than this value, it falls in the rejection

E X T E N D I N G T H E T E S T T O O R D I N A L V A R I A B L E S 219

region of the test, and we can conclude that delinquency is significantly

related to affectional identification.

The Test Statistic To determine the expected frequency for each cell in

the table, we follow the same format we have used in the previous two

examples. As before, we start with the calculation of the marginal for no

delinquent acts and divide by the total number of cases, which gives us

a value of 0.574468:

W orking It Out

� 0.574468

� 702

1,222

Proportion � Ncat

N total

To calculate the expected frequency for each cell in the no delinquent

acts column, we take this proportion and multiply it by the marginal total

for the row. For the first row, we multiply 0.574468 by 121, which gives

us an expected frequency of 69.511. Similarly, for the second row, we

multiply 0.574468 by 404, giving us an expected frequency of 232.085.

We continue this procedure for the remaining three rows in the no delin-

quent acts column.

For the second column, we need to determine the marginal proportion

for those cases with one delinquent act. Since there are 301 cases in the

marginal for one delinquent act, the corresponding proportion is

(301/1,222) � 0.246318. To obtain the expected frequencies for this sec-

ond column, we multiply 0.246318 by the corresponding row marginal.

So, for the first row of the second column, the expected frequency is ob-

tained by multiplying 0.246318 by 121, which gives us 29.804. This proce-

dure is repeated to complete the remaining cells in the second column.

Finally, to determine the expected frequencies for the cells in the

third column, we simply add the expected frequencies for the first two

columns and subtract that sum from the row marginal. For example, in

the first row, the two expected frequencies obtained thus far are 69.511

and 29.804. If we add these two values (69.511 � 29.804 � 99.315) and

subtract this sum from the row marginal (121), we find that the ex-

pected frequency for the cell in the third column of the first row is equal to

(121 � 99.315) � 21.685. To complete the table of expected frequencies,

C H A P T E R N I N E : C H I - S Q U A R E220

we repeat this operation for the remaining cells in the third column.

Table 9.15 contains all the observed and expected frequencies.

To obtain the test statistic, we again use Equation 9.2, which may be

applied to any two-variable chi-square problem:

Again we begin by subtracting the expected from the observed fre-

quency in each cell and squaring the result. This quantity is then divided

by the expected frequency of the cell. The chi-square statistic is found

by summing the result across all cells. The full set of calculations neces-

sary for obtaining the value of the chi-square test statistic appears in

Table 9.16. The chi-square statistic for our test has a value of 61.532.

The Decision

The calculated chi-square statistic of 61.532 is much larger than the criti-

cal value of 15.507 for the chi-square distribution with eight degrees of

freedom. This means that the observed significance level for our test is

less than the criterion significance level we set at the outset (p � 0.05).

Thus, we reject the null hypothesis that affectional identification with fa-

ther is not related to number of delinquent acts (at a 5% significance

level). In turn, we conclude that for adolescent males there is a statisti-

cally significant relationship between delinquency and affectional identi-

fication with father. Importantly, this statistical inference refers directly to

our sampling frame: high school–age white males in Richmond, Califor-

nia, in 1965.

�2 � � r

i�1

� c

j�1

( fo � fe)

2

fe

Observed and Expected Frequencies for Affectional Identification with Father and Number of Delinquent Acts

AFFECTIONAL D E L I N Q U E N T A C T S IDENTIFICATION WITH FATHER None One Two or More

In every way fo � 77 fo � 25 fo � 19 121 fe � 69.511 fe � 29.804 fe � 21.685

In most ways fo � 263 fo � 97 fo � 44 404 fe � 232.085 fe � 99.512 fe � 72.403

In some ways fo � 224 fo � 97 fo � 66 387 fe � 222.319 fe � 95.325 fe � 69.355

In just a few ways fo � 82 fo � 52 fo � 38 172 fe � 98.809 fe � 42.367 fe � 30.824

Not at all fo � 56 fo � 30 fo � 52 138 fe � 79.277 fe � 33.992 fe � 24.731

702 301 219 1,222

Table 9.15

Row total

Column total

E X T E N D I N G T H E T E S T T O O R D I N A L V A R I A B L E S 221

T h e U s e o f C h i - S q u a r e W h e n S a m p l e s A r e S m a l l : A F i n a l N o t e

The chi-square test is often used by criminal justice researchers. How-

ever, it has a very important limitation in its application to studies with

small or highly skewed samples. When more than one in five (20%) of

the cells in your table has an expected frequency of five or less, it is gen-

erally considered inappropriate to use a chi-square test. In such situa-

tions, it is recommended that you combine categories of your variables

until you meet the minimum expected-frequencies requirement.

C h a p t e r S u m m a r y

Whereas the binomial distribution is relevant only for binary variables,

the chi-square distribution can be used to examine a variable with

more than two categories.

The shape of the chi-square distribution chosen depends on the

degrees of freedom associated with the test. The formula for degrees of

freedom defines how many categories would have to be known for us to

the number of degrees of freedom, the flatter the distribution. In practi-

cal terms, as the number of degrees of freedom increases, a larger chi-

square statistic is required to reject the null hypothesis.

Computation of Chi-Square for Affectional Identification with Father and Delinquency

IDENTIFICATION DELINQUENCY fo fe (fo � fe) (fo � fe)2

Every way None 77 69.511 7.489 56.085 0.807 Every way One 25 29.804 �4.804 23.078 0.774 Every way Two or more 19 21.685 �2.685 7.209 0.332 Most ways None 263 232.085 30.915 955.737 4.118 Most ways One 97 99.512 �2.512 6.310 0.063 Most ways Two or more 44 72.403 �28.403 806.730 11.142 Some ways None 224 222.319 1.681 2.826 0.013 Some ways One 97 95.325 1.675 2.806 0.029 Some ways Two or more 66 69.355 �3.355 11.256 0.162 Few ways None 82 98.809 �16.809 282.543 2.859 Few ways One 52 42.367 9.633 92.795 2.190 Few ways Two or more 38 30.824 7.176 51.495 1.671 Not at all None 56 79.277 �23.277 541.819 6.835 Not at all One 30 33.992 �3.992 15.936 0.469 Not at all Two or more 52 24.731 27.269 743.598 30.067

� � 61.532

(fo � fe )2

fe

Table 9.16

8

8Another alternative solution is to use another group of non-parametric tests, defined as

be able to predict the remaining categories with certainty. The greater

‘e xact ’ tests, to estimate the observed significance level (see Alan Agresti, Categorical Data Analysis, New York, John Wiley, 1990). Such tests (e.g., Fisher ’s Exact Test) which develop

C H A P T E R N I N E : C H I - S Q U A R E222

The chi-square test of statistical significance is a nonparametric test. To

calculate the test statistic, the researcher must first identify the observed

frequency and the expected frequency of each category. The expected

frequencies are those one would expect under the assumption of the null

hypothesis. They are distributed in the same proportions as the marginal

frequencies. The chi-square formula is then applied to each category, or

cell, in the table. If the observed frequencies differ substantially from the

square statistic will be small. If the two frequencies are the same, the statis-

tic will be 0. The larger the statistic (and the smaller the number of degrees

of freedom), the easier it will be to reject the null hypothesis. The chi-

square statistic is always positive. Because the chi-square test relies on nom-

inal nonordered data, it is not concerned with the direction of outcomes.

K e y T e r m s

cells The various entries in a table, each of

which is identified by a particular row and

column. When we use a table to compare

two variables, it is convenient to refer to

each combination of categories as a cell.

chi-square distribution A sampling distri-

bution that is used to conduct tests of sta-

tistical significance with binary or multicat-

egory nominal variables. The distribution is

nonsymmetrical and varies according to de-

grees of freedom. All the values in the dis-

tribution are positive.

chi-square statistic The test statistic re-

sulting from applying the chi-square for-

mula to the observed and expected fre-

quencies for each cell. This statistic tells

us how much the observed distribution

differs from that expected under the null

hypothesis.

degrees of freedom A mathematical

index that places a value on the extent to

which a particular operation is free to vary

after certain limitations have been imposed.

Calculating the degrees of freedom for a

chi-square test determines which chi-

square probability distribution we use.

expected frequency The number of ob-

servations one would predict for a cell if

the null hypothesis were true.

marginal The value in the margin of a

table that totals the scores in the appropri-

ate column or row.

observed frequency The observed result

of the study, recorded in a cell.

S y m b o l s a n d F o r m u l a s

�2 Chi-square

df Degrees of freedom

observed frequencies are similar to the expected frequencies, then the chi-

expected frequencies, then the chi-square statistic will be large. If the

begun to provide exact test options for larger tables.

a sampling distribution for each problem examined, have been made more practical with

Computational options for cross tabulations. A few statistical programs (e.g., SAS) have the advent of powerful Computers. SPSS provides exact tests for two by two tables as

S Y M B O L S A N D F O R M U L A S 223

fo Observed frequency

fe Expected frequency

c Number of columns

r Number of rows

k Number of categories

To determine the degrees of freedom for a chi-square test including only

one variable:

df � k � 1

To determine the degrees of freedom for a chi-square test including two

variables:

df � (r � 1)(c � 1)

To determine the chi-square statistic for one variable:

To determine the chi-square statistic for two variables:

E x e r c i s e s

9.1 Local community leaders are concerned about the distribution of homicides in their small town. The local police department broke the city into six recognizable neighborhoods of the same size and discov- ered the following distribution of homicides:

Neighborhood Number of Homicides

A 14

B 9

C 17

D 3

E 7

F 10

�2 � � r

i�1

� c

j�1

( fo � fe)

2

fe

�2 � � k

i�1

( fo � fe)

2

fe

C H A P T E R N I N E : C H I - S Q U A R E224

Community leaders would like to know whether the homicides are randomly distributed across these six neighborhoods.

a. Use a 5% level of significance and outline each of the steps re- quired in a test of statistical significance.

b. What can you conclude about the distribution of homicides across these six neighborhoods?

9.2 Sergeant Bob is in charge of the duty roster at Gatley police station. Every week, it is his responsibility to randomly assign the five beat of- ficers, including his son Bob Jr., to patrol in each of the five zones that make up the city of Gatley. Zones A and D are favored by all the officers because they are usually quiet. Of the others, Zone C is noto- riously dangerous. The officers have recently begun to suspect Sergeant Bob of favoritism toward his son. In the last 30 weeks, Bob Jr. has been assigned to Zone A 12 times, Zone B and Zone C 2 times each, Zone D 9 times, and Zone E 5 times.

a. Do the other officers have reason to believe that Sergeant Bob is not assigning zones in a random manner? Use a 5% level of signifi- cance and outline each of the steps required in a test of statistical significance.

b. Would your answer be any different if a 1% level of significance were used?

9.3 In the past 100 years, there have been more than 250 successful breakouts from Didsbury Prison. Mike is a researcher who has been hired by the prison governor to investigate the phenomenon. Details are available only for those breakouts that took place in the past ten years—a total of 30. Using the records of these 30 breakouts as a sam- ple, Mike decides to break the figures down to see whether breakouts were more common in certain wings of the prison than in others. It transpires that of the 30 breakouts, 4 have been from A-Wing, 8 from B-Wing, 15 from C-Wing, and 3 from D-Wing.

a. Does Mike have enough evidence to conclude that, over the 100- year period, breakouts were more (or less) likely to occur from cer- tain wings than from others? Use a 5% level of significance and out- line each of the steps required in a test of statistical significance.

b. Would your answer be any different if a 1% level of significance were used?

c. Are there any problems with Mike’s choice of a sample? Explain your answer.

9.4 A study of death penalty cases (all first-degree murder charges with aggravating circumstances) revealed the following relationship be- tween the victim’s race and the chances the offender was sentenced to death: In 100 cases involving white victims, 20 offenders were sen-

E X E R C I S E S 225

tenced to death. In 100 cases involving black victims, 10 offenders were sentenced to death.

a. Is there a relationship between the race of the victim and the likeli- hood an offender was sentenced to death? Use a 5% level of signifi- cance and outline each of the steps required in a test of statistical significance.

b. Would your answer be different if a 1% level of significance were used?

9.5 At a local school, 46 children were accused of cheating on exams over the course of a semester. In an innovation, the principal decided that every second child accused of cheating would be brought before a “peer jury” to decide guilt or innocence. In all other cases, the deci- sion would be made by the examiners as usual. Of the 30 children who were adjudged guilty over the course of the semester, 18 were convicted by the peer jury, and the rest were convicted by the exam- iners. Of the children who were adjudged not guilty, 5 were acquitted by their peers.

a. The principal is mainly interested in the educational value of the experiment, but he will discontinue it if it becomes clear that the peer jury and the examiners make different decisions to a statisti- cally significant degree. He is willing to take a 5% risk of error. Should the scheme be continued? Outline each of the steps of a test of statistical significance.

b. Could the principal base the test on a directional hypothesis? If so, what would that hypothesis be, and would it make a difference in his final decision?

9.6 In the course of a year, Jeremy, a law student, observed a total of 55 cases in which an accused male pleaded guilty to a serious traffic of- fense. He observed that of the 15 who were sentenced to prison, 6 wore a shirt and tie in court. Of the 40 who were not sentenced to prison, 8 wore a shirt and tie in court.

a. Can Jeremy conclude that there is a link between the physical ap- pearance of the accused and whether he is imprisoned? Use a 5% level of significance and outline each of the steps required in a test of statistical significance.

b. What level of significance would be required for his decision to be reversed?

9.7 Sasha was interested in the extent to which people are prepared to in- tervene to help a stranger and whether the race of the stranger is rele- vant to the likelihood of intervention. She hired four male actors: one of African ancestry, one of Asian ancestry, one of European ancestry, and one of Indian ancestry. The actors were each told to fake a fall in a busy shopping street and pretend to be in some pain. Sasha

C H A P T E R N I N E : C H I - S Q U A R E226

observed from nearby and recorded whether, within five minutes of the actor’s fall, anyone had stopped to see if he was okay. Each actor repeated the experiment 40 times.

The results were as follows:

Ancestry � �

African 4 36

Asian 0 40

European 20 20

Indian 8 32

(� � Intervention within 5 mins; � � no intervention)

a. Can Sasha conclude that there is a link between race of victim and readiness to intervene? Use a 5% level of significance and outline each of the steps required in a test of statistical significance.

b. Would your answer be any different if a 1% level of significance were used?

9.8 Dave takes a random sample of the speeches, interviews, and official statements given by the prime minister and the interior minister of a particular country over the course of a year in which reference is made to “prison policy.” He analyzes the content of the statements in his sample and discovers five different types of justification for the government’s prison policy. Dave then records each time the prime minister or interior minister refers to any of the five justification types. The results are as follows:

Justification Type Prime Minister Interior Minister

Incapacitation or protecting society 6 16

Specific deterrence 2 14

General deterrence 4 20

Rehabilitation 0 15

Retribution 13 10

a. Is there a statistically significant difference between the policy state- ments of the prime minister and those of the interior minister? Use a 5% level of significance and outline each of the steps required in a test of statistical significance.

b. Would your answer be any different if a 1% level of significance were used?

9.9 The Television Complaints Board monitors the standards of morality for a nation’s TV channels. It has recently set up a telephone hotline for viewers who wish to complain about sex, violence, or foul lan- guage on any of the nation’s three TV channels. In its first month of operation, the board received the following complaints:

E X E R C I S E S 227

Channel 1 Channel 2 Channel 3

Sex 2 8 10

Violence 10 12 10

Foul language 3 10 15

a. Which of the following questions would a chi-square test of these results seek to answer?

i. Is there a statistically significant difference between the number of complaints made against each channel?

ii. Is there a statistically significant difference between the number of each type of complaint made?

iii. Is there a statistically significant difference between the types of different complaints received about the three different stations?

b. Answer the question you chose in part a by running a chi-square test at a 5% level of significance. Should the null hypothesis be overturned?

9.10 A survey of public opinion about the criminal justice system asked re- spondents to complete the following statement: “The criminal justice system treats offenders. . . .” The researchers found the following dis- tribution of responses by gender of the respondent:

Gender of Respondent Too Lenient About Right Too Harsh

Female 15 50 35

Male 40 35 25

a. Is there a relationship between the gender of the respondent and per- ceptions of punishment severity? Use a 5% level of significance and outline each of the steps required in a test of statistical significance.

b. Would your answer be different if the significance level were 0.01?

c. What can you conclude about the relationship between gender and perceptions of punishment severity?

9.11 A researcher is interested in the link between the type of offense a de- fendant is charged with and the manner in which a conviction is ob- tained. An examination of court records of a random sample of con- victed offenders reveals the following distribution of cases:

How Conviction Was Obtained

Type of Charge Offense Jury Trial Bench Trial Guilty Plea

Violent 19 13 67

Property 5 8 92

Drug 8 11 83

Other 10 6 74

C H A P T E R N I N E : C H I - S Q U A R E228

a. Is there a relationship between type of charge offense and method of conviction? Use a 5% level of significance and outline each of the steps required in a test of statistical significance.

b. Would your answer be any different if a 1% level of significance were used?

c. What can you conclude about the relationship between type of charge offense and method of conviction?

C o m p u t e r E x e r c i s e s

Up to this point, all of our computer examples have relied on individual-level

data, where each line of data in the spreadsheet view, whether SPSS or Stata,

referred to one case. However, there are many other times where we have been

given data in tabular form, say from a government publication, and we wish to

perform some analysis on that data. In these instances, we may have data on

several hundred, or even thousand, cases, but the data appear in aggregated form.

All of the examples in this chapter take that form, too.

Prior to discussing how to obtain chi-square values to test for a relationship

between two categorical variables, we illustrate how to enter tabular data into the

software program.

SPSS

Entering Tabular Data

The most direct way of entering tabular data is to think of the rows and columns

in a table as two separate variables and the number of cases in each cell of the

table as a third variable (representing a count of the number of cases with the

combination of the values for the two variables). For example, consider the data

presented in Table 9.7:

Imprisoned Not Imprisoned Row Margin

Subsequently Arrested 33 19 52

Not Subsequently Arrested 67 48 115

Column Margin 100 67 167

We can enter the information from this table as three variables: the row,

the column, and the count. Note that as you enter data into SPSS, the default

variable names will be VAR00001, VAR00002, and VAR00003. There are two

equivalent ways of renaming these variables in SPSS. We will rename the row

variable from VAR00001 to arrested (to represent whether or not the person was

subsequently arrested), the column variable from VAR00002 to prison (to repre-

sent whether or not the person had been imprisoned), and the number of cases

in each cell from VAR00003 to count. The equivalent syntax for renaming these

variables is:

C O M P U T E R E X E R C I S E S 229

After entering these data into SPSS and changing the variable names, you would

have three variables and four lines of data like the following (with the variable

names appearing at the top of each column of data):

arrest prison count

1.0000 1.0000 33.0000

1.0000 2.0000 19.0000

2.0000 1.0000 67.0000

2.0000 2.0000 48.0000

If you were to begin working with SPSS at this point, you would not

obtain the correct results, since SPSS will treat these data as representing

only four observations (You may want to confirm this for yourself—run the

FREQUENCIES command on either arrest or prison and see how many obser-

vations SPSS thinks it is working with.).

To have SPSS recognize the 167 observations represented by these four lines,

you need to tell SPSS to weight the cases by the “count” variable. This is done

with the WEIGHT command:

For our example, this would be:

The WEIGHT command tells SPSS that you have entered data in tabular form.

Obtain the Chi-Square Statistic

To continue our example using the data from Table 9.7, we would run the fol-

lowing command:

C H A P T E R N I N E : C H I - S Q U A R E230

The chi-square statistic is obtained by using the CROSSTABS command. The

basic format of the CROSSTABS command is

RENAME VARIABLES VAR00001 = arrest VAR00002 = prison

VAR00003 = count.

Or,

RENAME VARIABLES (VAR00001 VAR00002 VAR00003 = arrest

prison count).

WEIGHT BY variable_name.

WEIGHT BY count.

CROSSTABS

/TABLES = row_variable BY column_variable

/STATISTICS = CHISQ.

CROSSTABS

/TABLES = arrest BY prison

/STATISTICS = CHISQ.

The output produced by executing this command will contain a cross-tabulation

of the data that should be identical to Table 9.7. Immediately below this table

will be another table labeled “Chi-square tests.” The “Pearson Chi-Square” is the

name of the chi-square statistic that you have learned to calculate in this chap-

ter. The value reported by SPSS is 0.403, which differs from the value reported

above (0.402) by 0.001, which can be attributed to rounding error above.

All of the commands discussed above are illustrated in the SPSS syntax file

Chapter_9.sps.

Stata

A similar process is used in Stata to enter the data from Table 9.7. Recall that to

enter data, you will need to click on the “Data Editor” button at the top center

of the Stata window. Note that as you enter the data, the default variable

names are var1, var2, and var3. To change these to arrest, prison, and count,

respectively, enter the following commands:

Alternatively, you could edit the variable name in the far right box in the Stata

window (be sure to click on the lock symbol to “unlock” the variable properties

and allow you to make changes to the variable name).

In contrast to SPSS, there is no direct weighting of the data directly. Rather,

Stata includes an option in nearly all of its statistical procedures that will allow

the user to include a variable that represents frequency counts, as we have in our

example.

To obtain the chi-square statistic and a two-way cross-tabulation, we enter the

following command:

Note that we have assumed we are not working with tabular data in this example,

but individual-level data as we have in all previous chapters. The addition of the

chi2 option after the comma requests the chi-square statistic.

For our example, we would enter the following command:

We have added the fweight = option [in required brackets] to indicate that we

have a frequency weight variable named count that indicates the number of cases

with each characteristic of the two variables.

The output from running this command includes a cross-tabulation that

should be identical to Table 9.7. Just below the table is the chi-square statistic,

C O M P U T E R E X E R C I S E S 231

rename var1 arrest

rename var2 prison

rename var3 count

tabulate row_variable column_variable, chi2

tabulate arrest prison [fweight = count], chi2

reported as “Pearson chi2(1) = 0.4031.” The value “1” in the parentheses refers

to the number of degrees of freedom for the chi-square statistic. This is a com-

mon reporting format in Stata output for a chi-square statistics—the degrees of

freedom associated with the test will be listed in the adjacent parentheses.

All of these commands are illustrated in the Stata do file Chapter_9.do.

Problems

1. Input the data on race and cell-block assignment from Table 9.11.

Compute the value of the chi-square statistic for these data. How does it

compare to the value reported in the text?

2. Input the data on affectional identification with father and delinquency

from Table 9.14. Compute the value of the chi-square statistic for these

data. How does it compare to the value reported in the text?

3. Enter the data from Exercise 9.7. Compute the value of the chi-square

statistic for these data. How does it compare with the value that you have

calculated for this exercise?

4. Enter the data from Exercise 9.11. Compute the value of the chi-square

statistic for these data. How does it compare with the value that you have

calculated for this exercise?

5. Open the NYS data file (nys_1.sav, nys_1_ student.sav, or nys_1.dta). Use

a 5% level of significance and outline each of the steps required in a test

of statistical significance for each of the following possible relationships:

a. Is ethnicity related to grade point average?

b. Is marijuana use among friends related to the youth’s attitudes about

marijuana use?

c. Is the importance of going to college related to the importance of

having a job?

d. Is grade point average related to the importance of having a job?

e. Is the sex of the youth related to the importance of having friends?

f. Is the importance of having a job related to the youth’s attitudes about

marijuana use?

g. SPSS notes at the bottom of each cross-tabulation the number and

percentage of all cells that had expected frequencies less than 5. For

parts a through f, are there any cross-tabulations that produce expected

frequencies of less than 5 for 20 % or more of all cells in the table? If

so, what are the consequences for interpreting the chi-square statistic?

Explain how the categories of one or more variables could be com-

bined to produce a table that has fewer cells with expected frequencies

of less than 5.

C H A P T E R N I N E : C H I - S Q U A R E232

6. Open the data file pcs_98.sav. These data represent a random sample

of 1,300 offenders convicted of drug, property, or violent offenses and

sentenced in Pennsylvania in 1998. The full data file contains information

on all offenders (more than 40,000) who were convicted of misdemea-

nor and felony offenses and sentenced to some form of punishment in

Pennsylvania in 1998. The full data file is available through the National

Archive of Criminal Justice Data at http://www.icpsr.umich.edu/NACJD

(Prior to answering the following questions, you may find it helpful to

review the list of variables and how each is coded.). Use a 5% level of

significance and outline each of the steps required in a test of statistical

significance for each of the following relationships:

a. Is the sex of the offender related to the method of conviction (i.e.,

plea, bench trial, or jury trial)?

b. Is the race-ethnicity of the offender related to whether the offender was

incarcerated or not?

c. Is the method of conviction related to the type of punishment received

(i.e., probation, jail, or prison)?

d. Is the type of conviction offense (i.e., drug, property, or violent) related

to the method of conviction?

C O M P U T E R E X E R C I S E S 233

The Normal Distribution

and Its Application

to Tests of Statistical Significance

w h e n p o p u l a t i o n p a r a m e t e r s a r e u n k n o w n

How Can We Make Assumptions About an Unknown Population?

When Can It Be Used?

How Can We Define a Sampling Distribution

When the Parameters Are Unknown?

p o p u l a t i o n d i s t r i b u t i o n

P a r a m e t r i c t e s t s f o r a n o r m a l

U s i n g t h e n o r m a l s a m p l i n g d i s t r i b u t i o n

T w o e x a m p l e s

C h a p t e r t e n

What are the Characteristics of the Normal Frequency Distribution?

What is the z-Score?

When can We Use the Normal Sampling Distribution?

What are the Assumptions of the One-Sample z-Test for Means?

What is the Central Limit Theorem?

What is the z-Test for Proportions?

What is the t-Test for Means?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_10,

© Springer Science+Business Media New York 2014

IN CHAPTERS 8 AND 9, tests of statistical significance were presented that

did not make assumptions about the population distribution of the char-

acteristics studied. We now turn to a different type of test of statistical

significance in which the researcher must make certain assumptions

about the population distribution. These tests, called parametric tests, are

widely used in criminal justice and criminology because they allow the

researcher to test hypotheses in reference to interval-level scales.

inference is to make statements about populations from what is known

about samples. However, parametric tests require that we make assump-

tions about the population at the outset. If population parameters are

generally unknown, how can we make assumptions about them? In this

chapter, we examine this dilemma in the context of two types of para-

metric tests that are based on the normal distribution.

T h e N o r m a l F r e q u e n c y D i s t r i b u t i o n , o r N o r m a l C u r v e

In Chapter 3, we noted that frequency distributions may take many dif-

ferent forms. Sometimes there is no pattern to a distribution of scores.

This is the case for the example in Figure 10.1, in which the frequency of

scores goes up and down without consistency. But often a distribution

begins to take a specific shape. For example, Floyd Allport suggested

more than half a century ago that the distribution of deviant behavior is

shaped like a J.1 His J curve, represented in Figure 10.2, fits many types

1F. H. Allport, “The J-Curve Hypothesis of Conforming Behavior,” Journal of Social

Psychology 5 (1934): 141–183.

We begin by introducing the normal sampling distribution and its

application to tests of significance for measures that are normally dis-

researchers in the use of parametric tests. The purpose of statistical

tributed in the population. We then turn to a basic dilemma faced by

235

of rule-breaking behavior and suggests a theory of deviance in which so-

cial control leads most people to conform more or less to societal rules.

Allport fit a J curve to behaviors as diverse as parking violations, confor-

mity to religious rituals in church, and stopping at a stop sign.

The most widely utilized distributional form in statistics is what is de-

fined as the normal frequency distribution or normal curve. The

normal distribution is the basis for a number of parametric statistical

Random Frequency DistributionFigure 10.1

The J CurveFigure 10.2

Conformity Deviance

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N236

tests. This is the case in good part because of a set of special characteris-

tics associated with the normal curve.

Characteristics of the Normal Frequency Distribution

A normal distribution is always symmetrical and bell shaped. By that we

mean that it is shaped exactly the same on both sides of its mean. If you

represent a normal distribution as a curve, you can fold it over at its

mean and gain two half-curves that are exactly alike. Of course, there

are many different potential bell-shaped curves that are symmetrical, as

illustrated in Figure 10.3. The curve in part a of Figure 10.3, for example,

is fairly flat. What this means is that the scores are fairly widely spread

around the mean. The curve in part b, in contrast, is very peaked. Here,

the scores are tightly clustered around the mean. In the statistical lan-

guage developed in earlier chapters, we can say that the standard devia-

tion of the first distribution is much larger than that of the second.

In a true normal distribution, the mean, mode, and median are always

the same. This can be seen in the normal curves in Figure 10.3. If the

distribution is completely symmetrical, then the 50th percentile score, or

the median, must be right in the middle of the distribution. In turn, since

Two Examples of Normal CurvesFigure 10.3

(a) Normal Curve with a Large Standard

Deviation

(b) Normal Curve with a Small Standard

Deviation

T H E N O R M A L F R E Q U E N C Y D I S T R I B U T I O N , O R N O R M A L C U R V E 237

the middle of the distribution represents its highest peak, and thus the

largest frequency of scores, it is also the location of the mode for the

normal distribution. Finally, given that there is an exactly equal distribu-

tion of scores below and above that peak, the same value must also be

the mean for a normal distribution.

All of these traits help to define a normal distribution. However, the

most useful characteristic of a normal distribution develops from the fact

that the percentage of cases between its mean and points at a measured

distance from the mean is always fixed. The measure in this case is the

standard deviation unit. A standard deviation unit is simply the stan-

dard deviation for the particular distribution being examined. For exam-

ple, let’s say that you were examining the results of a standardized test

for assessing adjustment of prisoners and that the distribution obtained

was a normal distribution. You obtained a mean score of 90 and a stan-

dard deviation of 10 for your sample. The standard deviation unit of this

distribution would be 10. That is, if you measured one standard devia-

tion unit from the mean in either direction, you would move 10 points

from the mean, to 100 and 80. If you measured two standard deviation

units from the mean, you would move 20 points, to 110 and 70.

In a normal distribution, 68.26% of the cases in the distribution are

found within one standard deviation unit above and below the mean

(see Figure 10.4). Because the normal distribution is symmetrical, this

means that 34.13% of the cases lie within one standard deviation unit to

either the right (positive side) or the left (negative side) of the mean.

Percentage of Cases Under Portions of the Normal CurveFigure 10.4

Standard Deviations

–4 –3 –2 –1 0 +1 +2 +3 +4

0.13% 2.14% 13.59% 34.13% 34.13% 13.59% 2.14% 0.13%

68.26%

95.46%

σσσσ σ σ σ σ

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N238

Fully 95.46% of the cases are found within two standard deviation units

above and below the mean. Virtually all of the cases in a distribution

with a normal form are within three standard deviation units of the

mean, although in theory the tails (or extremes) of this distribution go on

forever. For the sample of inmates discussed above, we thus know that

slightly more than two-thirds have adjustment scores of between 80 and

100 (one standard deviation unit above and below the mean). Very few

members of the sample have adjustment scores above 120, which repre-

sents a score that is three standard deviation units from the mean.

z-Scores

scores in our sample or population to z-scores, which represent stan-

dard deviation units for the standard normal distribution. This distribu-

tion has a mean of 0 and a standard deviation unit of 1. The formula for

converting a specific score to a z-score is represented by Equation 10.1.

Equation 10.1

For this equation, we take the score of interest and subtract from it the

mean score for the population distribution (represented by �). We then

divide that result by the standard deviation of the population distribution

we are examining (represented by �). In practice, what this formula does

is allow us to convert any specific score in any normal distribution to a

z-score in a standard normal distribution. We can then use a standard-

ized table to identify the location of that score. A concrete example will

make this conversion easier to understand.

Intelligence quotient (IQ) scores are normally distributed in the U.S.

population, with a mean of 100 and a standard deviation of about 15.

Suppose a probation officer is writing a report on a young offender. She

finds that the young man has an IQ of 124. She wants to give the sen-

tencing judge a good sense of what this means in terms of how this

young man compares to others. She can do this by transforming the

mean IQ of the offender (124) to a z-score and then identifying where

this z-score fits in the standard normal distribution. We use Equation

10.1 for this purpose.

As shown in the numerator of the equation, we subtract the popula-

tion mean (�) of IQ scores, which we already noted was 100, from the

score of 124. By doing this we shift the position of our score. We now

have its location if the mean of our distribution were 0—the mean of a

standard normal distribution.

z � Xi � �

Using a simple equation, we can convert all normal distributions, irrespec-

normal distribution. This distribution can then be used to identify

the exact location of any score. We do this by converting the actual

tive of their particular mean or standard deviation, to a single standard

T H E N O R M A L F R E Q U E N C Y D I S T R I B U T I O N , O R N O R M A L C U R V E 239

If the mean were 0, then the score for this offender would be 24 (and

not 124). As a second step, we divide this result by 15, the standard devi-

ation (�) of IQ scores in the U.S. population. This is equivalent to con-

verting our sample standard deviation unit to 1, the standard deviation of

the standard normal distribution, since each score of 15 is equivalent to

one z standard deviation unit. The result is 1.60.

Our final step is to compare this z-score to an already prepared table

of the standard normal distribution, provided in Appendix 3. You will

notice that the z table goes up to only 0.50. This is because it provides

us with only half of the normal curve. On this half of the normal curve,

W orking It Out

� 1.60

� 124 � 100

15

z � Xi � �

IQ Score of Young Prisoner Compared to Average IQ Score of the General PopulationFigure 10.5

z = 1.60 score for prisoners

94.52% 5.48% 44.52%

µ

our z-score of 1.60 is equivalent to 0.4452, meaning that 44.52% of

the scores lie between 0 and �1.60 standard deviations from 0. In

Figure 10.5, our result is illustrated in the context of the normal curve.

Because our result is a positive score, we place the value on the right-

hand side of the normal distribution.

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N240

To identify the percentage of people in the general population with

IQ scores higher than that of the young offender, we subtract our result

of 0.4452 from 0.50 (the proportion of cases in this half of the curve).

Our result of 0.0548 means that only a bit more than 5% of the general

population has higher IQ scores than this offender. Conversely, almost

95% of the population has lower IQ scores than this offender. By con-

verting the offender’s score to a score on the standard normal distribu-

tion, we are able to place his intelligence in context. From our finding,

we can see that he is indeed a highly intelligent young man, based on

his IQ score.

Developing Tests of Statistical Significance Based on the Standard Normal

Distribution: The Single-Sample z-Test for Known Populations

The normal distribution can also be used as a sampling distribution.

population characteristics for all Americans are, as discussed above,

known. The mean score for the population is 100, and the standard devi-

ation of the population mean is 15. You conduct a study of 125 prisoners

selected through an independent random sampling procedure from the

population of American prisoners. You find that the mean IQ in your

sample is 90.2 This mean is different from the mean of the American

population. But we know that samples vary, and thus you might get a

mean of 90 even if the mean for American prisoners were the same as

that for the general population. What we want to know is how likely we

are to get such an outcome in our sample if the distribution of American

prisoners is the same as that of the general American population.3 Be-

2Our hypothesized results mirror those found in prior studies; see R. J. Hernstein,

“Some Criminogenic Traits of Offenders,” in J. Q. Wilson (ed.), Crime and Public Pol-

icy (San Francisco: Institute for Contemporary Studies, 1983). Whether these differ-

ences mean that offenders are, on average, less intelligent than nonoffenders is an

issue of some controversy in criminology, in part because of the relationship of IQ to

other factors, such as education and social status. 3By implication, we are asking whether it is reasonable to believe that our sample of

prisoners was drawn from the general population. For this reason, the z-test can also

be used to test for random sampling. If you have reason to doubt the sampling meth-

ods of a study, you can conduct this test, comparing the observed characteristics of

your sample with the known parameters of the population from which your sample

was drawn.

distribution of sample means will also be distributed normally. This means

When a population of scores is distributed normally, the sampling

in practice that we can use the normal distribution as our sampling

This is the case for the IQ test, so we will continue to use it as an example.

Let’s say that you were interested in whether American prisoners

differ from Americans generally in terms of average IQ scores. The

that the variable of interest is normally distributed in the population.

distribution for a test of statistical significance if we know at the outset

T H E N O R M A L F R E Q U E N C Y D I S T R I B U T I O N , O R N O R M A L C U R V E 241

cause the population parameters of the American population are known,

a single-sample z-test for known populations is appropriate.

We set up our test of statistical significance the same way we did

other tests in previous chapters.

Assumptions:

Level of Measurement: Interval scale.

Population Distribution: Normal distribution.

Sampling Method: Independent random sampling.

Sampling Frame: The American prison population.

Hypotheses:

H0: The mean IQ of the population from which our sample of prisoners

was drawn is the same as the mean IQ of the general population of

Americans (� � 100).

H1: The mean IQ of the population from which our sample of prisoners

was drawn is not the same as the mean IQ of the general population of

Americans (� � 100).

As required by the single-sample z-test for known populations, IQ

scores are measured at an interval level. As already noted, IQ is also nor-

mally distributed in the general population, meaning that it is appropri-

ate to use a normal sampling distribution to conduct our test of statistical

significance. Our sample, as required by our test, is drawn randomly

pothesis is that the mean IQ of American prisoners is the same as the

mean IQ of the general American population (� � 100). Our research

hypothesis is that the mean IQ of prisoners is different from that of the

The Sampling Distribution

population distribution and apply it to the sampling distribution for our

test, because the standard deviation of the sampling distribution is influ-

enced by the number of observations in a sample. This is illustrated in

Figure 10.6, which presents three different sampling distributions for the

same population distribution of scores. In the first, there are only 10

cases in the samples from which the sampling distribution is developed.

In the second, there are 25 cases in each sample. Finally, in the third

that the mean of the sampling distribution is the same as the mean of the

However, we cannot simply take the standard deviation of scores for the

ledge of the standard deviation of the population distribution of scores.

The mean of the sampling distribution we use

average American (� � 100).

with replacement from the American prison population. Our null hy-

for our test of statistical inference is defined, as in other tests, by our null

hypothesis. In statistical tests using the normal distribution we can assume

population distribution. In this case, it is 100, or the mean IQ for the American population. The standard deviation is drawn from our know-

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N242

that the spread of scores is reduced as the size of samples in the distrib-

ution increases. This fact is a very important one in statistics and follows

samples increases. Put in lay terms, larger samples are more trustworthy

or more likely to reflect the true population score, all else being equal,

than are smaller samples.

In order to differentiate between the standard deviation of a popula-

tion distribution of scores and that of a sampling distribution, statisticians

call the standard deviation of a sampling distribution the standard

error. Using Equation 10.2, we adjust our standard error for the fact that

the dispersion of sample means decreases as sample size increases. In

order to distinguish the standard deviation (�) from the standard error in

this text, we will use the subscripts sd (for sampling distribution) when-

ever we refer to the standard error of a sampling distribution. Accord-

ingly, the standard error of a sampling distribution is represented as �sd

in Equation 10.2.

Equation 10.2

For our example, we find the standard error of the sampling distribu-

tion by dividing the population standard deviation of IQ, 15, by the

square root of our sample N. The result is 1.342.

Standard error � �sd � �

�N

Normal Distribution of Scores from Samples of Varying Sizes: N � 10, N � 25, and N � 100Figure 10.6

distribution, there are 100 cases in each sample. What is clear here is

what our common sense tells us: Our sampling distribution becomes

more tightly clustered around the mean as N increases. This implies,

in practice, that we are less likely to draw deviant samples (those far

from the mean of the sampling distribution) as the N of cases in our

T H E N O R M A L F R E Q U E N C Y D I S T R I B U T I O N , O R N O R M A L C U R V E 243

Significance Level and Rejection Region Given that no special concerns

have been stated in regard to the risk of either a Type I or a Type II

error, we use a conventional 0.05 significance threshold. As our research

hypothesis is nondirectional, we use a two-tailed test. What this means

for our rejection region is illustrated in Figure 10.7. On the right-hand

side of the distribution are outcomes greater than the average American

IQ of 100. On the left-hand side of the distribution are outcomes less

than the average. Because our research hypothesis is not directional, we

split our total rejection region of 5% between both tails of the distribu-

tion. This is represented by the shaded area. Each shaded area repre-

sents half the total rejection region, or 0.025.

W orking It Out

� 1.342

� 15

�125

Standard error � �

�N

Rejection Region on a Normal Frequency Distribution for a 0.05 Two-Tailed

Significance Test Figure 10.7

z = –1.96

Total Rejection Region, = 0.025 + 0.025 = 0.05

Rejection Region = 0.025

Rejection Region = 0.025

z = +1.96

Outcomes Supporting Lower IQs for Prisoners

Outcomes Supporting Higher IQs for Prisoners

µ

α

α

α

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N244

To define the z-score that corresponds with our rejection region,

we must turn to the table of probability values associated with the z

distribution in Appendix 3. As discussed earlier in the chapter, the z

table represents only half of the normal curve. We look at the value

associated with 0.4750 (0.5000 � 0.0250) in the table, which is 1.96. If

we observe a test statistic either greater than 1.96 or less than �1.96,

we will reject the null hypothesis of our test (see Figure 10.7). In this

case, our observed significance level would be less than the 0.05 crite-

rion for our test.

If we had stated a directional research hypothesis, we would place

the entire rejection region (� � 0.05) in one of the two tails of the nor-

mal distribution. In this case, we would conduct a one-tailed statistical

test. Parts a and b of Figure 10.8 represent the rejection regions for two

different one-tailed tests of statistical significance. If our research hypoth-

esis stated that average IQs for prisoners were less than those for the

U.S. population, we would place the entire rejection region of 0.0500 in

the left tail of the distribution (see Figure 10.8a). We again consult the z

table in Appendix 3 to identify the z-score associated with a value of

0.4500 (0.5000 � 0.0500). We observe that 0.4500 falls exactly halfway

between two values in the table —0.4495 and 0.4505—corresponding to

z-scores of �1.64 and �1.65, respectively. How do we determine the

value of z in such a case? The most accurate value for z would be found

by interpolating between �1.64 and �1.65, which would give �1.645,

The Test Statistic To calculate our test statistic, we can use the same for-

mula we did in examining the relative position of a score in the standard

normal distribution, with two important differences. In this case, we

have to take into account the fact that sampling distributions become

more tightly spread around their mean as the N of sample cases becomes

larger. As discussed in defining the sampling distribution above, we need

to adjust the standard deviation of the population distribution by divid-

ing it by the square root of the N of our sample. This provides us with

the standard error (�sd) for our distribution. We also need to subtract the

average IQ for prisoners is less than the U.S. average. If our research

than �1.645, then we reject our null hypothesis and conclude that the

portions reported in the z table. In this case, if our test statistic is less

since the value we are looking for is halfway between the two pro-

hypothesis stated that the average IQ of prisoners was greater than the

U.S. average, we would place the rejection region on the right side

of the distribution (see Figure 10.8b). In such a case, our critical value

would be �1.645, meaning that if our test statistic was greater than

1.645, we would reject the null hypothesis and conclude that the ave-

rage IQ for prisoners was greater than the U.S. average.

T H E N O R M A L F R E Q U E N C Y D I S T R I B U T I O N , O R N O R M A L C U R V E 245

Rejection Region on a Normal Frequency Distribution for a 0.05 One-Tailed Significance TestFigure 10.8

z = –1.645 µ

Rejection Region = 0.05

Outcomes Supporting Lower IQs for Prisoners

Outcomes Supporting Higher IQs for Prisoners

α

Rejection Region = 0.05

z = +1.645

Outcomes Supporting Lower IQs for Prisoners

Outcomes Supporting Higher IQs for Prisoners

µ

α

(a) Lower IQs for Prisoners

(b) Higher IQs for Prisoners

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N2 64

mean (�) of the population score from , rather than Xi. These adjust-

ments are made in Equation 10.3.

Equation 10.3

Inserting into our equation the mean value of our sample and its N of

cases and the mean and standard deviation for the population of scores,

we obtain a z-test statistic of �7.453.

z � X � �

�sd �

X � �

�/�N

X

W orking It Out

� �7.453

� 90 � 100

15/�125

z � X � �

�/�N

The Decision Because our test statistic is less than our negative critical

value (�7.453 � �1.96) and falls in the rejection region, we reject the

null hypothesis. We conclude on the basis of our study that the mean IQ

of the population from which our sample was drawn is different from

that of the general American population.

A p p l y i n g N o r m a l S a m p l i n g D i s t r i b u t i o n s t o N o n n o r m a l P o p u l a t i o n s

The example of IQ presents a case where the single-sample z-test can be

used to test hypotheses involving interval-scale measures. However, it re-

quires that the population distribution for the measure be normal. In

some fields in the social sciences, measures are constructed in such a way

that they are normally distributed in practice.4 But in criminology, there

4In principle, any distribution may be arranged in such a way that it conforms to a

normal shape. This can be done simply by ranking scores and then placing the appro-

priate number within standard deviation units appropriate for constructing a standard

normal distribution.

A P P L Y I N G N O R M A L S A M P L I N G D I S T R I B U T I O N S 247

has been much less use of distributions that are standardized in normal

form, in part because the distributions of the behaviors and populations

that we confront do not often conform to the shape of a normal curve.

Even measures that do begin to approximate the shape of the normal dis-

tribution seldom meet all the requirements of a true normal distribution.

How, then, can parametric tests based on the normal distribution be

widely used to make statistical inferences? Not only do they demand that

we make an assumption about a population we usually know little

about, but the assumption we are being asked to make does not make

very much sense for criminal justice measures. The answer may be

found in an important distinction between population distributions on

the one hand and sampling distributions on the other. While we have

every reason to be hesitant in assuming that the population distribution

of scores is normal for criminal justice measures, we can assume with a

good deal of confidence that the sampling distributions for such mea-

sures are approximately normal. Using the toss of a fair coin as an exam-

ple, we can provide a simple illustration of this fact.

In Figure 10.9, we overlay the distribution of scores for a population

of 1,000 tosses of a fair coin over the normal distribution. As is apparent,

outcomes in a coin toss are not distributed normally. This makes good

sense, since there are only two possible scores for the coin toss: heads

and tails. No matter what the outcome, it is impossible for a coin toss to

approximate the form of the normal distribution.

But let’s now turn to a sampling distribution for the coin toss. In

this case, we want to know the likelihood of gaining a specific num-

ber of heads in a set number of coin tosses. This is the logic we used

Distribution of 1,000 Tosses of a Fair Coin Contrasted to the Normal DistributionFigure 10.9

Heads Tails

F re

q u

en cy

Population Distribution of 1,000 Tosses of a Fair Coin

Normal Distribution

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N248

in developing the binomial probability distribution in Chapter 7. Fig-

ure 10.10 presents the binomial distribution for different-size samples

of the coin toss under the null hypothesis that the coin is fair.

For a sample size of 1 (Figure 10.10a), the shape of the sampling dis-

tribution is the same as the shape of the population distribution of

scores. However, notice what happens as the size of the samples used to

construct the sampling distributions grows. For a sample of 10 (Figure

10.10b), the histogram for the distribution of scores is still jagged, but it

has begun to take a shape similar to the normal distribution. Importantly,

for a sample of 10, we do not have two potential outcomes, which

Sampling Distribution of Coin TossesFigure 10.10

Number of Heads

F re

q u

en cy

Normal Curve

Number of Heads

F re

q u

en cy

Normal Curve

(a) 1 Toss of a Fair Coin

(b) 10 Tosses of a Fair Coin

A P P L Y I N G N O R M A L S A M P L I N G D I S T R I B U T I O N S 249

would make a normal shape impossible, but 11 potential outcomes (no

heads, one head, two heads, three heads, four heads, . . . to ten heads).

This is the case because we are flipping the coin ten times for each sam-

ple. The sampling distribution is telling us the number of times we

would expect to gain a specific number of heads in ten tosses of a fair

coin in a very large number of trials. For a sample of 100 flips of a fair

coin (Figure 10.10c), the sampling distribution even more closely ap-

proximates the normal curve. By the time we get to a sample of 400 flips

of the coin (Figure 10.10d), the sampling distribution of a fair coin is al-

most indistinguishable from the normal curve.

Number of Heads

F re

q u

en cy

Normal Curve

Number of Heads

F re

q u

en cy

Normal Curve

(c) 100 Tosses of a Fair Coin

(d) 400 Tosses of a Fair Coin

Sampling Distribution of Coin Tosses (cont.)Figure 10.10

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N250

The population distribution of scores for a fair coin is very far from a

normal form. Yet sampling distributions for the same population begin

to approximate the normal distribution as the size of the sample of coin

tosses grows. This remarkable fact is summarized in a very important

theorem, or statement, about sampling distributions called the central

limit theorem. The central limit theorem allows us to overcome our ini-

tial dilemma because it says that under many circumstances we can use

a normal sampling distribution for making inferences about a population

that is not normal in shape.

Central Limit Theorem

If repeated independent random samples of size N are drawn from a

population, then as N grows large, the sampling distribution of sample

means will be approximately normal.

The central limit theorem tells us that when the number of cases in a

sample is large, we can assume that the sampling distribution of sample

means is approximately normal even if the population distribution itself

is not normal. This is what is meant by the statement “then as N grows

large, the sampling distribution of sample means will be approximately

normal.” However, the theorem does not provide us with a clear state-

ment about how large the number of cases in a sample must be before

we can make this assumption.

One reason for this ambiguity is that the number of cases needed be-

fore the sampling distribution begins to approximate normality depends

in part on the actual distribution of the measure examined in the popula-

tion. As can be seen from the example of the coin toss, even when the

population distribution departs markedly from the normal distribution,

the sampling distribution fits fairly closely to the normal curve with a

sample size of 100. For this reason, you will find wide agreement that a

normal sampling distribution can be assumed for samples of 100 or

more, irrespective of the distribution of scores in a population.

There is much less agreement about what to do when a sample is

smaller than 100 cases. Some statisticians argue that with 50 cases you

can be fairly confident that the central limit theorem applies in most cir-

cumstances. Others apply this yardstick to 25 or 30 cases, and still others

argue that under certain circumstances—for example, when prior studies

suggest a population distribution fairly close to normality—only 15 cases

is enough. In conducting research in criminal justice, you should recog-

nize that there is no hard and fast rule regarding sample size and the

central limit theorem. In practice, in criminal justice, researchers gener-

ally assume that 30 cases is enough for applying the central limit theo-

rem. However, when a distribution strongly departs from normality, as is

the case with a proportion, it is safer to require more than 100 cases.

A P P L Y I N G N O R M A L S A M P L I N G D I S T R I B U T I O N S 251

While the central limit theorem solves a major problem in applying

normal distribution tests to criminological questions, we are still faced

with a barrier in actually carrying out such tests. As we saw earlier (see

Equation 10.3), the standard error of the z sampling distribution is gained

from knowledge about the standard deviation of the population distribu-

tion. How can we identify the standard error of a sampling distribution if

we do not know the standard deviation of the population distribution? In

the following sections, we illustrate two methods for defining � for an

unknown population. In the first, we take advantage of a special rela-

tionship between the mean and the standard deviation of a proportion.

In the second, we estimate the unknown parameter based on informa-

tion gained in our sample.

C o m p a r i n g a S a m p l e t o a n U n k n o w n P o p u l a t i o n : T h e S i n g l e - S a m p l e z - T e s t f o r P r o p o r t i o n s

One implication of the central limit theorem is that we can use a normal

begins to approximate a normal distribution when the number of cases

for the sample becomes large. The central tendency of this distribution

and its dispersion are measured by the mean and standard error, just as in

distributions that develop from interval-level data. Accordingly, although

it would be inappropriate to use the mean and standard deviation to de-

scribe a sample or population distribution of a proportion, the mean and

standard error are appropriate statistics for describing the normal sam-

pling distribution that is associated with the same proportion.

Computing the Mean and Standard Deviation

for the Sampling Distribution of a Proportion

How do we compute the mean and standard deviation of a proportion?

One way to do this would be to simply apply the formula for the mean

and the standard deviation to the scores associated with a proportion.

However, there is a simpler way to arrive at the same result. It turns out

that the mean of a proportion is equal to the proportion itself. This is il-

lustrated in Table 10.1, which shows an example in which the mean and

proportion are calculated for five heads in ten tosses of a coin.

sampling distribution to test hypotheses involving proportions. This might

seem strange at first, since we estimate the shape of a normal distribu-

tion through knowledge of its mean and standard deviation. As discus-

sed in Chapter 4, the mean and standard deviation are not appropriate

statistics to use with a nominal-level measure such as a proportion.

bution of a proportion—in our example, the coin toss (see Figure 10.10)—

Nonetheless, as illustrated in the previous section, the sampling distri-

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N252

For the numerator of the mean, we sum the scores on the ten trials

(five ones and five zeros) and get 5. The numerator of a proportion is

the N of cases in the category of interest. If the category is heads, then

we also get a result of 5. The denominators for both equations are the

same (10), and thus the outcomes are also the same. As a general rule,

we state that for a proportion � � P.

What about the standard deviation of a proportion? It turns out that

we can calculate the standard deviation with knowledge of only the pro-

portion itself. This is illustrated in Table 10.2.

Taking the sum of the squared deviations from the mean and divid-

ing it by N, we get a result of 0.25. But we can get this same result by

multiplying the proportion of heads (P) by the proportion of tails

(Q)—in our case, multiplying 0.5 by 0.5. Accordingly, we can substitute

P • Q for

� N

i�1

(Xi � X)2

N

Calculating the Mean and Proportion of 5 Heads in 10 Tosses of a Coin

CALCULATING THE PROPORTION

CALCULATING THE MEAN FOR FIVE HEADS FOR FIVE HEADS

Proportion � Nsuccesses

Ntotal

� 5

10 � 0.5X �

� N

i�1

Xi

N �

1 � 1 � 1 � 1 � 1 � 0 � 0 � 0 � 0 � 0

10 � 0.5

Table 10.1

Calculating the Standard Deviation of 5 Heads in 10 Tosses of a Coin

CALCULATING THE STANDARD DEVIATION CALCULATING THE STANDARD

FROM THE RAW SCORES DEVIATION FROM P AND Q

� �0.25 � 0.5

� �0.25 � 0.25 � 0.25 � 0.25 �0.25 � 0.25 �0.25 � 0.25 �0.25 � 0.25

10

� � �PQ � �(0.5)(0.5) � �0.25 � 0.5 � � �� N

i�1

(Xi � X)2

N

Table 10.2

Note: head � 1� tail � 0

T H E S I N G L E - S A M P L E z - T E S T F O R P R O P O R T I O N S 253

in the equation for the standard deviation for the mean:

Equation 10.4

Because of this relationship between the mean and the standard devi-

ation of a proportion, when we state the proportion of successes ex-

pected under the null hypothesis, we also state by implication the mean

and the standard deviation for the population distribution of scores. So if

we state in the null hypothesis that the proportion of successes in the

population is 0.50, we know that the mean of the population distribution

of scores for our test of the null hypothesis is 0.50 and its standard devia-

tion is .

What this means in practice is that we need not have any a priori

knowledge of the shape of the population distribution to construct a

sampling distribution for our test of proportions. With a large N, we can

assume a normal sampling distribution, irrespective of the actual form of

the population distribution. Through our null hypothesis, we can define

both the mean and the standard deviation of the population distribution

for our test. We are now ready to use the normal distribution to test hy-

potheses about unknown population parameters.

Testing Hypotheses with the Normal Distribution:

The Case of a New Prison Program

Suppose that you were asked to evaluate a new prison education pro-

ported by the foundation. Managers of the program claim that the suc-

cess rate is actually much greater than the criteria set by the foundation.

However, a recent newspaper exposé claims that the success rate of the

program is actually much below 75%. You are able to collect information

on 150 prisoners, selected using independent random sampling. You

find that 85% of your sample successfully completed the course. What

conclusions can you make, based on your sample results, about the

claims of managers and the newspaper exposé?

Assumptions:

Level of Measurement: Interval scale (program success is measured as a

proportion).

Population Distribution: Normal distribution (relaxed because N is

large).

0.25 (� � �PQ � �(0.50)(0.50) � �0.25 � 0.50)

� � �� N

i�1

(Xi � X)2

N � �PQ

gram. The foundation sponsoring the effort sought to achieve a pro-

gram success rate of 75% among the 100,000 prisoners enrolled in the

program. Success was defined as completion of a six-month course sup-

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N254

Sampling Method: Independent random sampling.

Sampling Frame: 100,000 prisoners in the program.

Hypotheses:

H0: The success rate of the program is 0.75 (P � 0.75).

H1: The success rate of the program is not 0.75 (P � 0.75).

Because the number of cases in our sample is greater than the thresh-

old of 100 suggested for invoking the central limit theorem in the case of

a proportion, we can ignore—or, in statistical terms, relax—assump-

tions regarding the shape of the population distribution. In the special

case of a proportion, we can also relax the assumption of an interval

scale of measurement.5 Our sample, as assumed by our test, is drawn

randomly with replacement from the sampling frame of 100,000 prison-

ers in the program.

Our research hypothesis is nondirectional. Managers of the program

claim that the program has a success rate of greater than 0.75 (P � 0.75).

The newspaper exposé claims that the success rate is much below 75%

(P � 0.75). Accordingly, we want to be able to examine both of these

potential outcomes in our test. The null hypothesis is that the rate of suc-

cess for the program is 0.75 (P � 0.75).

The Sampling Distribution In calculating the mean and standard devia-

tion or standard error for our sampling distribution, we rely on our null

hypothesis. Our null hypothesis states that the proportion of successes in

the population is 75%. This means that the mean of the sampling distrib-

ution is also 0.75. We can calculate the standard error of the sampling

5It would not make sense, however, to use a normal distribution test for nominal-scale

measures with more than two categories. The normal distribution assumes scores

above and below a mean. The sampling distribution of a proportion follows this

pattern because it includes only two potential outcomes, which then are associated

with each tail of the distribution. In a multicategory nominal-scale measure, we have

more than two outcomes and thus cannot fit each outcome to a tail of the normal

curve. Because the order of these outcomes is not defined, we also cannot place them

on a continuum within the normal distribution. This latter possibility would suggest

that the normal distribution could be applied to ordinal-level measures. However, be-

cause we do not assume a constant unit of measurement between ordinal categories,

the normal distribution is often considered inappropriate for hypothesis testing with

ordinal scales. In the case of a proportion, there is a constant unit of measurement

between scores simply because there are only two possible outcomes (e.g., success

and failure).

T H E S I N G L E - S A M P L E z - T E S T F O R P R O P O R T I O N S 255

distribution by adjusting Equation 10.2 to the case of a proportion, as il-

lustrated in Equation 10.5:

Equation 10.5

Applying this equation to our problem, we obtain a standard error of

0.035 for the normal sampling distribution associated with our null

hypothesis:

�sd � �

�N �

�PQ

�N � �PQ

N

W orking It Out

� 0.0353

� 0.433 12.25

� �0.1875

�150

� �(0.75)(0.25) 150

�sd � �PQ

N

In order to test our hypothesis, we will convert this sampling distribu-

tion, with mean 0.75 and standard error 0.035, to the standard normal

distribution (or z), which has a mean of 0 and a standard deviation or

standard error of 1. This calculation is done when we calculate the test

statistic below.

Significance Level and Rejection Region Given that no special concerns

have been stated in regard to the risk of either a Type I or a Type II

error, we use a conventional 0.05 significance threshold. As our research

hypothesis is nondirectional, we use a two-tailed test. As our level of sig-

nificance is the same as in our previous problem, we follow the same

procedure and arrive at a critical value of 1.96. If we observe a test statis-

tic either greater than 1.96 or less than �1.96, we will reject the null hy-

pothesis of our test.

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N256

The Test Statistic We can rely on the same formula used in the single-

sample z-test for known populations, presented earlier in Equation 10.3.

However, in Equation 10.6, we express the formula with proportions

rather than means:

Equation 10.6

The mean of the sample (p) is 0.85, since this is the outcome of the

study. The mean of the sampling distribution P (0.75) is taken from our

null hypothesis. The standard error of the sampling distribution (�sd) was

calculated earlier based on our null hypothesis that the proportion of

successes was 0.75. Our result is a z-score of 2.833.

z � X � �

�sd �

X � �

�/�N �

p � P

�PQ/N

W orking It Out

� 2.8329

� 0.10

0.0353

� 0.85 � 0.75

�(0.75)(0.25)/150

z � p � P

The Decision Our test statistic is well within the rejection region of our

test (which includes scores greater than 1.96 or less than �1.96), meaning

that our observed significance level is less than the significance level we

set for our test at the outset (p � 0.05). We therefore reject the null hy-

pothesis at a 0.05 significance level. We come out on the side of the man-

agers of the program. Our sample results support their position that the

overall program has exceeded the criterion for success of the foundation.

C o m p a r i n g a S a m p l e t o a n U n k n o w n P o p u l a t i o n : T h e S i n g l e - S a m p l e t - T e s t f o r M e a n s

The proportion provides us with a special case in which we can calcu-

late the standard error of our sampling distribution based on our null hy-

pothesis. But this is not possible when our null hypothesis relates to a

mean of an interval-level measure. In this case, there is not one specific

variance or standard deviation associated with a mean but an infinite

�PQ/N

T H E S I N G L E - S A M P L E t - T E S T F O R M E A N S 257

number of potential variances or standard deviations. How, then, can we

test hypotheses about unknown parameters in the case of the mean?

One obvious method is to simply use the variance of our sample as a

“guesstimate” of the variance of the population distribution. The problem

with this solution is that the variance of a sample is a somewhat biased

estimate of the variance of the population. By this we mean that the av-

erage of repeated observations of the variance (s2) tends in the long run

not to be equivalent to the value of �2. We can transform s2 to a better

estimate of �2 through a very small correction to the equation for the

variance. This new statistic (expressed as since it is an estimate of �2)

is represented in Equation 10.7.6 An estimate of the standard deviation

can be gained by taking the square root of this value.

Equation 10.7

In order to use this new statistic to test hypotheses, we must also use

a slightly different sampling distribution, called the t distribution. It is

sometimes called Student’s t because its inventor, W. S. Gossett

(1876–1936), first described the distribution under the pen name Student

in 1908.

The t distribution (Appendix 4) is very similar to the z distribution

(Appendix 3). However, as with the chi-square test, the shape of the t

distribution is dependent on the number of degrees of freedom. The

number of degrees of freedom for a single-sample t-test is defined as

N � 1. When the number of cases in a sample is greater than 500, then

the t and z distributions are virtually identical. However, as the number

of cases in a sample gets smaller and smaller, and so accordingly does

the number of degrees of freedom, the t distribution becomes flatter and

a larger and larger test statistic is needed to reject the null hypothesis.

This fact can be illustrated by looking at the t table in Appendix 4. As

you can see, the t table lists the critical values associated with six signifi-

cance thresholds for both one- and two-tailed tests. Let’s focus on the

fourth column, which is the critical value associated with a two-tailed,

5% significance level. When the number of degrees of freedom is 500,

the critical value for the t-statistic is the same as for the z distribution:

1.960. At 120, the t value needed to reject the null hypothesis is still al-

most the same: 1.980. At 100, the value is 1.982: at 50, it is 2.008; and at

25, it is 2.060. The largest differences come for even smaller degrees of

freedom.

�̂2 � � N

i�1

(Xi � X)2

N � 1

(�̂)

�̂2

6As noted on page 105 (footnote 1), computerized statistical analysis packages, such as

SPSS, use this corrected estimate in calculating the variance and standard deviation for

sample estimates.

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N258

The t distribution presents a new problem as well in making infer-

ences to unknown populations. Relaxing the assumption of normality is

generally considered more risky in a t-test than in a z-test. This makes

good sense because we are now using an estimate of � rather than the

actual population parameter. As the number of cases increases, our con-

fidence in this estimate grows.7 How large should N be before you are

willing to use a t-test? With samples of more than 30 cases, your statisti-

cal conclusions are not likely to be challenged. However, the t distribu-

tion is particularly sensitive to outliers. Conclusions based on smaller

samples should be checked carefully to make sure that one or two ob-

servations are not the cause of a very large statistical outcome.

Testing Hypotheses with the t Distribution

We are now ready to turn to a practical example. Suppose that the study

described earlier also examined the average test scores for those prison-

ers who had completed the program. The foundation set a standard of

success of 65 on the test. Program managers say that prisoners who have

completed the program achieve average scores much higher than this.

The newspaper exposé again claims that the average scores are consid-

erably lower than those expected by the foundation. In this case, you are

able to take an independent random sample of 51 prisoners who have

completed the test. You find that the test mean for the sample is 60, and

the standard deviation is 15. What conclusions about the larger popula-

tion of prisoners can you come to based on your sample results?

Assumptions:

Level of Measurement: Interval scale.

Population Distribution: Normal distribution (relaxed because

N is large).

Sampling Method: Independent random sampling.

Sampling Frame: Prisoners who have completed the program.

Hypotheses:

H0: The mean test score for prisoners who have completed the program

is 65 (� � 65).

H1: The mean test score for prisoners who have completed the program

is not 65 (� � 65).

7Our statistical problem is that we assume that � and � are independent in developing

the t distribution. When a distribution is normal, this is indeed the case. However, for

other types of distributions, we cannot make this assumption, and when N is small, a

violation of this assumption is likely to lead to misleading approximations of the ob-

served significance level of a test.

T H E S I N G L E - S A M P L E t - T E S T F O R M E A N S 259

Following the assumptions of our test, we use an interval scale (the

mean of test scores) and an independent random sampling method. We

relax the assumption of normality because N is larger than the minimum

threshold of 30 recommended for interval-level measures. Our research hy-

pothesis is once again nondirectional so that we can examine the positions

of both the managers of the program and the newspaper exposé. The null

hypothesis is that the mean test score for the population of prisoners com-

pleting the program is 65 (the foundation standard), or that � � 65.

The Sampling Distribution Because � is unknown and cannot be deduced

from our null hypothesis, we will use the t distribution. The number of

degrees of freedom for our example is defined as N � 1, or 51 � 1 � 50.

Significance Level and Rejection Region Again, we have no reason in this

example to depart from the 0.05 significance threshold. Because our re-

search hypothesis is not directional, we use a two-tailed test. Turning to

the t table, we find that a t-score of 2.008 is associated with a two-tailed,

5% significance threshold (at 50 degrees of freedom). This means that we

will reject our null hypothesis if we obtain a test statistic greater than

2.008 or less than �2.008. For these observed values of our test statistic,

the observed significance level of our test is less than the criterion of

0.05 that we have selected.

The Test Statistic The test statistic for the t distribution is similar to that

for the z distribution. The only difference is that we now use an estimate

of the standard deviation rather than � itself.

Equation 10.8

Although we can get an estimate of � by adjusting the calculation for

s, the formula for t may also be written in a way that allows us to calcu-

late t from the unadjusted sample standard deviation.

� X � �

�� N

i�1

(Xi � X )2

N �N � 1

t � X � �

�̂/�N �

X � �

�� N

i�1

(Xi � X )2

N � 1 �N

t � X � �

�sd �

X � �

�̂/�N

(�̂)

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N260

This means that we can simplify the equation for the t-test as follows:

Equation 10.9

Applying the t formula to our example, we use the mean of the sample,

60, as ; � is defined by the null hypothesis as 65; s is our sample stan-

dard deviation of 15; and N is the number of cases for our sample (51).

X

t � X � �

s/�N � 1

W orking It Out

� �2.3570

� �5

2.1213

� �5

15/�50

� 60 � 65

15/�51 � 1

t � X � �

s/�N � 1

The Decision Because the test statistic of �2.3570 is less than �2.008,

we reject the null hypothesis and conclude that the result is significantly

different from the goal set by the foundation. In this case, our decision is

on the side of the newspaper exposé. We can conclude from our sample

(with a 5% level of risk of falsely rejecting the null hypothesis) that the

test scores in the population of prisoners who have completed the pro-

gram are below the foundation goal of 65.

C h a p t e r S u m m a r y

Parametric tests of statistical significance allow us to make inferences

about a population from samples using interval-level data. In a paramet-

ric test, we make certain assumptions about the shape of the population

distribution at the outset.

The normal distribution, or normal curve, is widely used in statis-

tics. It is symmetrical and bell shaped. Its mean, mode, and median are

always the same. There will always be a set number of cases between

C H A P T E R S U M M A R Y 261

the mean and points a measured distance from the mean. The measure

of this distance is the standard deviation unit. All normal distributions,

irrespective of their mean or standard deviations, can be converted to a

single standard normal distribution by converting the actual scores in the

sample or population to z-scores. To use a normal sampling distribution

for a test of statistical significance, we must assume that the characteristic

studied is normally distributed in the population.

An important dilemma in statistical inference is created by this as-

sumption. How can we make assumptions about the population distribu-

tion when its characteristics are generally unknown? The central limit

theorem describes an important fact that allows us to solve this prob-

lem. As stated in the theorem, when the number of cases in a sample is

large, the sampling distribution will be approximately normal in shape,

even if the population distribution itself is not. In the field of criminal

justice, it is generally assumed that the central limit theorem can be ap-

plied where the sample size is 30 or greater. When dealing with propor-

tions, though, it is safer to require a sample size of at least 100. In such

circumstances, we may relax the assumption of normality. We can

now make inferences using a normal sampling distribution, even though

the shape of the population distribution is unknown.

In order to define the sampling distribution, we need information

about the population parameters—information that is not usually avail-

able. In the case of a test involving proportions, the null hypothesis can

be used to define both the mean and the standard error of the popula-

tion distribution. Once the population parameters have been defined by

the null hypothesis, we can apply the formula for the z-test of statistical

significance. In the case of a test of means, the null hypothesis cannot be

used directly to define the standard error. We may, however, use the t

K e y T e r m s

central limit theorem A theorem that

states: “If repeated independent random

samples of size N are drawn from a

population, as N grows large, the

sampling distribution of sample means

will be approximately normal.” The

central limit theorem enables the researcher

to make inferences about an unknown

population using a normal sampling

distribution.

normal curve A normal frequency distrib-

ution represented on a graph by a continu-

ous line.

normal frequency distribution A bell-

shaped frequency distribution, symmetrical

in form. Its mean, mode, and median are

always the same. The percentage of cases

between the mean and points at a mea-

sured distance from the mean is fixed.

sampling distribution, which relies on an estimate of the standard error.

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N262

S y m b o l s a n d F o r m u l a s

p Proportion of successes (sample)

P Proportion of successes (population)

Q Proportion of failures (population)

�sd

t t-score

Estimate of �

To determine the z-score for a single observation:

To determine the standard error of a sampling distribution:

�sd � �

�N

z � Xi � �

�̂

relaxing an assumption Deciding that

we need not be concerned with that as-

sumption. For example, the assumption

that a population is normal may be relaxed

if the sample size is sufficiently large to in-

voke the central limit theorem.

single-sample t-test A test of statistical

significance that is used to examine

whether a sample is drawn from a specific

population with a known or hypothesized

mean. In a t-test, the standard deviation of

the population to which the sample is

being compared is unknown.

single-sample z-test A test of statistical sig-

nificance that is used to examine whether a

sample is drawn from a specific population

with a known or hypothesized mean. In a

z-test, the standard deviation of the popula-

tion to which the sample is being compared

either is known or—as in the case of a pro-

portion— is defined by the null hypothesis.

standard deviation unit A unit of mea-

surement used to describe the deviation of

a specific score or value from the mean in

a z distribution.

standard error The standard deviation of

a sampling distribution.

standard normal distribution A normal

frequency distribution with a mean of 0

and a standard deviation of 1. Any normal

frequency distribution can be transformed

into the standard normal distribution by

using the z formula.

z-score Score that represents standard

deviation units for a standard normal

distribution.

Standard error of a normal distribution

S Y M B O L S A N D F O R M U L A S 263

To determine the z-score for a sample mean:

To determine the standard deviation of a proportion:

To determine the z-score for a sample proportion:

To estimate the value of � from data in a sample:

To determine the value of t:

E x e r c i s e s

10.1 In which of the following circumstances would a researcher be justi- fied in using a normal sampling distribution? Explain how or why for each case.

a. A sample of 10 subjects is drawn to study a variable known to be normally distributed in the population.

b. A sample of 50 subjects is drawn to study a variable known to be normally distributed in the population.

c. A sample of 10 subjects is drawn to study a variable. The shape of the distribution of this variable in the population is unknown.

d. A sample of 50 subjects is drawn to study a variable. The shape of the distribution of this variable in the population is unknown.

e. A sample of 50 subjects is drawn to study a proportion. The shape of the distribution of this proportion in the population is un- known.

t � X � �

s/�N � 1

�̂ � �� N

i�1

(Xi � X)2

N � 1

z � X � �

�sd �

X � �

�/�N �

p � P

�PQ /N

� � �PQ

z � X � �

�sd �

X � �

�/�N

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N264

10.2 A team of psychologists has created an index they claim measures an individual’s “ability to control anger.” The index is calculated from the answers to a detailed questionnaire and is normally distributed among U.S. adult males, with a mean of 100 and a standard deviation of 30. Researchers assess a group of ten prisoners, all of whom have been convicted for violent rapes. They discover that the mean score for the group is 50.8.

a. What percentage of U.S. adult males would be expected to obtain a score equal to or less than that of the rapists?

b. The psychologists who constructed the index consider the bottom 10% of U.S. adult males on their distribution to be “strongly in- clined to use violence to solve social problems.” Albert is a re- spectable businessman who scores 60.6 on the scale. Is Albert in- cluded in this category? Explain why.

c. What percentage of U.S. adult males would be expected to score between 110 and 120 on the “anger index”?

10.3 A teacher gives the following assignment to 200 students: Check the local newspaper every morning for a week and count how many times the word “gun” is mentioned on the “local news” pages. At the end of the week, the students report their totals. The mean result is 85, with a standard deviation of 8. The distribution of scores is normal.

a. How many students would be expected to count fewer than 70 cases?

b. How many students would be expected to count between 80 and 90 cases?

c. Karen is a notoriously lazy student. She reports a total of 110 cases at the end of the week. The professor tells her that he is convinced she has not done the assignment, but has simply made up the number. Are his suspicions justified?

10.4 The professors who teach the Introduction to Psychology course at State University pride themselves on the normal distributions of exam scores. After the first exam, the current professor reports to the class that the mean for the exam was 73, with a standard deviation of 7.

a. What proportion of student would be expected to score above 80?

b What proportion of students would be expected to score between 55 and 75?

c. What proportion of students would be expected to score less than 65?

d. If the top 10% of the class receive an A for the exam, what score would be required for a student to receive an A?

e. If the bottom 10% of the class fail the exam, what score would earn a student a failing grade?

E X E R C I S E S 265

10.5 A noted criminologist, Leslie Wilkins, has suggested that the distribu- tion of deviance in the population follows a normal bell-shaped curve, with “sinners” at one extreme, “saints” at the other, and most of us falling somewhere in between the two. Working on the basis of this theory, a researcher constructs a detailed self-report survey whereby individuals are given a score based on the offenses they have commit- ted in the past year, with the score weighted according to the relative triviality or seriousness of each offense. The lower the score, the nearer the individual approximates “sinner” status, and the higher the score, the closer he or she is to being a “saint.” From his initial sample of 100 adults in a specific state, the researcher computes a mean score of 30, with a standard deviation of 5.

a. If the researcher’s model is correct, below which score should he expect to find the 5% of U.S. society with the greatest propensity to deviance?

b. In his sample of 100, the researcher is surprised to discover that 50 subjects score greater than 35 on the deviance test. How many cases would be expected under the assumption of a normal distrib- ution of saints and sinners? What does this suggest about the origi- nal theory?

10.6 An established test measuring “respect for authority” has a mean among U.S. adults of 73 and a standard error of 13.8. Brenda gives the test to 36 prison inmates and finds the mean score to be 69.

a. Is this enough evidence to suggest that the prisoners belong to a population that has significantly less respect for authority than the general U.S. adult population?

b. Assuming there is enough information, test whether this sample dif- fers significantly from the population. Use a significance level of 5% and outline each of the stages of a test of statistical significance.

10.7 The governor of Stretford Prison has a biographical record of all the inmates. The mean age of all the inmates is 22, with a standard devia- tion of 7.5. A recent survey by a hostile researcher makes damaging criticisms of the educational standards in the prison. The prison gover- nor suspects that the 50 prisoners interviewed for the study were not chosen at random. The mean age of the prisoners chosen is 20. Show how a test for statistical significance can be used by the governor to cast doubt on the sampling method of the survey. Use a significance level of 5% and outline each of the stages of a test of statistical significance.

10.8 A hundred years ago, an anonymous scientist wrote a famous indict- ment of a notoriously cruel prison somewhere in the United States. Without ever referring to the prison by name, the scientist checked the records of all those who were imprisoned over its 50-year history and found that 15% of those who entered died within. Henry, a historian,

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N266

is intrigued by the old report and publishes an article in a historical journal in which he states his conviction that the report was referring to Grimsville Prison, which existed about that time. In a subsequent issue of the journal, a rival historian claims that Henry has shown no evidence to support his theory.

Henry finds the records from Grimsville, and from a sample of 80 prisoner records he discovers that 11% of the prisoners died inside. Can he use this information to substantiate his claim that the object of the report is indeed Grimsville? Use a significance level of 5% and out- line each of the stages of a test of statistical significance.

10.9 Every pupil at Foggy Lane College was asked a series of questions, which led to an overall score grading “satisfaction” with the college’s discipline procedures. The overall mean score was 65. Roger suspects that the black students at the college feel differently. He takes a ran- dom sample of 25 black students from the college and finds that their mean satisfaction score is 61, with a standard deviation of 8.

Are the black students’ views on discipline significantly different from those of the general student population? Use a significance level of 1% and outline each of the stages of a test of statistical significance.

10.10 A special police unit has spent several years tracking all the members of a large child-abuse ring. In an interview with a daily newspaper, a junior detective on the unit claims that the ringleaders have been tracked down and will shortly be arrested. In response to questions from the interviewer about the makeup of the child-abuse ring, the detective replies, “We have gathered details on every last member of this criminal group—they come from very varied backgrounds and their average age is 36.”

X is the chairperson of a charitable club, which is in fact a front for a substantial child-abuse circle. He reads the newspaper article and fears that it might refer to him and his group. He looks through the club’s membership files and draws a sample of 50 members, finding an average age of 40 with a standard deviation of 9.

Can X be confident that the detective interviewed in the newspaper was not referring to his criminal group?

10.11 A civil rights group is concerned that Hispanic drug offenders are being treated more severely than all drug offenders in Border State. A state government web site reports that all drug offenders were sen- tenced to an average of 67 months in prison. The group conducts a small study by taking a random sample of public court records. For the 13 Hispanic drug offenders in the sample, the average sentence was 72 months (s = 8.4). Use a 5% significance level and test whether Hispanic drug offenders in Border State are sentenced more severely. Be sure to outline the steps in a test of statistical significance.

10.12 A researcher believes that offenders who are arrested for committing homicides in her city are younger than the national average. A review

E X E R C I S E S 267

of FBI arrest statistics for recent years indicates that the mean age of homicide offenders is 18.7. The researcher collects information on a random sample of 25 persons arrested for homicide in her city and finds the mean age to be 16.8, with a standard deviation of 4.1. Can the researcher conclude that homicide offenders in her city are younger than the national average? Use a significance level of 0.05. Be sure to outline the steps in a test of statistical significance.

10.13 Following a revolution, the new leadership of the nation of Kippax decides to hold a national referendum on whether the practice of cap- ital punishment should be introduced. In the buildup to the referen- dum, a leading army general wishes to gauge how the people are likely to vote so that he can make a public statement in line with pop- ular feeling on the issue. He commissions Greg, a statistician, to carry out a secret poll of how people expect to vote. The results of Greg’s poll are as follows: The sample proportion in favor of introducing capital punishment is 52%.

Do the results indicate that the majority of the population favors in- troducing capital punishment? Use a significance level of 5% and out- line each of the stages of a test of statistical significance.

10.14 The Silver Star Treatment Center claims to be effective at reducing drug addiction among the persons who go through its treatment regi- men. As evidence of the effectiveness of the Silver Star treatment, the director claims that 63% of all drug users nationally have a relapse within 12 months of treatment, but in a random sample of 91 cases treated by Silver Star, only 52% had a relapse within 12 months of completing the treatment. Use a 1% level of significance to test whether Silver Star’s treatment is effective at reducing drug use. Be sure to outline the steps in a test of statistical significance.

10.15 A federal judge issues an opinion claiming that nonviolent drug of- fenders should make up no more than 20% of the local jail popula- tion. If a jail is found to have more than 20% nonviolent drug offend- ers, the jail will fall under court order and be required to release inmates until the composition of the jail population conforms to the judge’s standard. The local sheriff draws a random sample of 33 in- mates and finds that 23% have been convicted of nonviolent drug of- fenses. Should the sheriff be concerned about the jail coming under court supervision? Use a significance level of 0.05. Be sure to outline the steps in a test of statistical significance.

C H A P T E R T E N : T H E N O R M A L D I S T R I B U T I O N268

Comparing Means and Proportions

in Two Samples

t h a t a r e n o t i n d e p e n d e n t

C h a p t e r e l e v e n

C o m p a r i n g s a m p l e m e a n s

C o m p a r i n g s a m p l e p r o p o r t i o n s

C o m p a r i n g m e a n s i n s a m p l e s

What is the Two-Sample t-Test for Means?

What are the Assumptions of the Test?

How is the Test Carried Out?

What is the Two-Sample t-Test for Proportions?

What are the Assumptions of the Test?

How is the Test Carried Out?

What is the t-Test for Dependent Samples?

What are the Assumptions of the Test?

How is the Test Carried Out?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_11,

© Springer Science+Business Media New York 2014

IN CHAPTER 10, we used parametric significance tests to compare the

mean or proportion of a single sample with a population goal or para-

meter. In this chapter, we turn to a more commonly used application of

parametric tests of statistical significance: comparisons between samples.

Let’s say, for example, that you are interested in whether there is a differ-

ence in the mean salaries of male and female police officers or in the

proportions of African Americans and others arrested last year. Your

question in either of these cases is not whether the population parame-

ters have particular values, but whether the parameters for the groups

examined in each case are different. This involves comparing means and

proportions for two populations. If you take samples from these popula-

tions, you can make inferences regarding the differences between them

by building on the normal distribution tests covered in Chapter 10.

C o m p a r i n g S a m p l e M e a n s

The Case of Anxiety Among Police Officers and Firefighters

In a study conducted by University of Washington researchers, police of-

ficers were compared to firefighters in terms of the amount of stress and

anxiety they experienced on the job.1 One measure the researchers used

was derived by creating an interval-scale index from questions about the

occurrence on the job of symptoms of anxiety, such as sweating and “the

jitters.” The researchers drew a sample of police officers by going to po-

lice stations and asking officers to be paid participants in their study. For

firefighters, the researchers randomly selected subjects. The final sample,

1Michael Pendleton, Ezra Stotland, Philip Spiers, and Edward Kirsch, “Stress and Strain

among Police, Firefighters, and Government Workers: A Comparative Analysis,” Crim-

inal Justice and Behavior 16 (1989): 196–210.

270

all drawn from one city, included 127 firefighters and 197 police officers.

For this sample, the researchers found that the mean anxiety-on-the-job

score for police officers was 12.8 (s1 � 2.76), whereas that for firefighters

was 8.8 (s2 � 2.85). What conclusions regarding the larger populations

of firefighters and police officers can the researchers draw from these

sample statistics?

As in other problems involving comparisons between the means of

two groups, we are not able to define the standard deviations of the

population distributions for our test. Indeed, we conduct a test of statisti-

cal significance for the differences between the two samples precisely

because we do not have information on the population parameters. Ac-

cordingly, we turn again to the t-test introduced in Chapter 10. In this

case, we use a two-sample t-test for means.

Assumptions:

Level of Measurement: Interval scale.

Population Distribution: Normal distribution in both populations (re-

laxed because N is large).

Sampling Method: Independent random sampling (a nonrandom sam-

pling technique was used for police officers; random sampling without

replacement was used for firefighters).

Sampling Frame: All police officers and firefighters in one city.

Hypotheses:

H0: The mean anxiety-on-the-job score for the population of police offi-

cers is the same as that for the population of firefighters (�1 � �2).

H1: The mean anxiety-on-the-job score for the population of police offi-

cers is different from that for the population of firefighters (�1 � �2).

The assumptions for the two-sample t-test are similar to those for the

one-sample t-test. An interval level of measurement is assumed, and in-

deed the characteristic being examined, anxiety on the job, is measured

at the interval-scale level. The two-sample t-test also requires that both

population distributions be normal in form. When this is the case, the

sampling distribution of the difference between means—the focus of our

test—is also normally distributed. Even when the populations examined

are not normally distributed, the sampling distribution of the difference

between the sample means will be normally distributed if the N of cases

for both samples is large.

The definition of how large samples must be to invoke the central

limit theorem is again a matter of debate. In Chapter 10, we noted that a

sample size of 30 or more was generally large enough to apply the cen-

tral limit theorem in a single-sample test for means. For a two-sample

test, we need a minimum of 30 cases in each sample. In our example,

C O M P A R I N G S A M P L E M E A N S 271

both samples include a much larger number of subjects, and thus we can

relax the assumption of normality.

As with other tests we have examined, here we are required to use an

independent random sampling method. For a two-sample t-test, we must

assume that both samples are independent random samples. In practice,

researchers do not ordinarily use separate sampling procedures to iden-

tify the samples representing each population of interest. Rather, they

draw a random sample from all members of a population and then as-

sume that specific samples within the larger sample are also independent

and random. For example, researchers interested in attitudes toward

crime in the United States generally draw an independent random sam-

ple of all U.S. residents. They may, however, also have an interest in

comparing attitudes of men and women or of college graduates and

non–college graduates. If the larger sample has been drawn as an inde-

pendent random sample, the subsamples are also independent random

samples.2

The one practical difficulty with this assumption arises when the num-

ber of subjects in a particular subpopulation is small. For example, in a

survey of U.S. residents, a very small group of Jews or Muslims is likely

to be sampled when researchers draw a simple independent random

sample. Thus, even though such a subsample will still be independent

and random (if the larger sample is independent and random), re-

searchers may not end up with many cases because such a group repre-

sents a small proportion of the U.S. population. When there is interest in

a subpopulation that is small, researchers often identify such groups for

special attention and attempt to draw larger samples from them.

For the firefighters in our example, the researchers used a random

sampling method, but they did not sample with replacement. This viola-

tion of assumptions is not serious because the sample of firefighters

drawn was small relative to the number of subjects in the sampling

frame. The method of sampling for police officers represents a more seri-

ous violation of the assumptions of the two-sample t-test. The re-

searchers did not draw a random sample. Nonetheless, they argued that

2The logic here follows simple common sense. If you select each case independently

and randomly from a population, on each selection you have an equal probability of

choosing any individual, whether male or female, college-educated or not, and so on.

From the perspective of a particular group—for example, males—each time you

choose a man, the method can be seen as independent and random. That is, the like-

lihood of choosing any male from the sample is the same each time you draw a case.

Of course, sometimes you will draw a female. However, within the population of

males, each male has an equal chance of selection on each draw. And if the sampling

method is independent, then each male has an equal chance of being selected every

time a case is selected.

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S272

their sample was still representative of the population of police officers

in the city:

Participant officers were compared with nonparticipant officers on avail-

able data (which were acquired by the police department independently

of the study). These data included entrance psychological tests, current

departmental physical fitness tests, age, sex, and so on. . . . The partici-

pant and nonparticipant groups did not differ significantly on 25 com-

parison variables.

generally. For this inference to be justified, the researchers would have

to explain why firefighters and police officers in this city are representa-

tive of firefighters and police officers in other cities.

1

(�2) is the same (�1 � �2). The research hypothesis was that there is a

difference (�1 � �2). The researchers did not define the direction of this

difference. Their research hypothesis allows the possibility that police of-

ficers experience more anxiety at work than firefighters as well as the op-

tion that firefighters experience more anxiety at work than police officers.

The Sampling Distribution For a difference of means test, we use the t

sampling distribution. The number of degrees of freedom for the distri-

bution is obtained by adding the numbers of cases in the two samples

and subtracting 2: df � N1 � N2 � 2. For our example, the number of

degrees of freedom is 322.

The mean of the sampling distribution is defined, as in the case of a

difference of proportions test, by the null hypothesis. It is represented by

�1 � �2, or the hypothesized difference between the means of the two

populations studied. Since the null hypothesis states that �1 � �2, the

mean of the sampling distribution is 0.

� 322

� 197 � 127 � 2

df � N1 � N2 � 2

The validity of our inferences to the larger population of police officers

in the city depends on how persuasive we find the researchers’ claims

that their sample was representative. But irrespective of the generali-

zability of these samples to the population of police officers and fire-

fighters in the city, the researchers also want to infer their findings

beyond their sampling frame to police officers and firefighters more

The null hypothesis for a difference of means test is generally that there

is no difference, and this was the case in the University of Washington

research. The null hypothesis stated simply that the mean anxiety-on-the-

) and firefightersjob score for the populations of police officers (�

C O M P A R I N G S A M P L E M E A N S 273

In defining the standard error of the sampling distribution for compar-

ing two samples, we take into account the variances of the two popula-

tions. Accordingly, the standard error of a sampling distribution of the

difference of sample means is the square root of the sum of

the two sample variances, each divided by its sample N:

Equation 11.1

In calculating this standard error, we can use either of two ap-

proaches. The first assumes that the two population distributions not

only have equal means but also have equal variances. In this case, we

are assuming that the two population distributions are the same. This is

often called the pooled variance method. The assumption we make in

this approach, a common one in statistical tests, is often referred to as

homoscedasticity (from the Greek for “same scatter [or spread]”). It can

be written in mathematical form as follows:

A second approach, called the separate variance method, does not

make a specific assumption that the variances of the two populations are

equal. You should note, however, that when samples are very small or

one sample is much larger than the other, the simple estimate of degrees

of freedom noted above must be corrected if the separate variance

method is used. The correction commonly employed involves a good

deal of computation.3 For our problem, which involves large samples of

relatively similar size, it is unnecessary to take this approach.

Given that the pooled variance method requires an additional assump-

tion, that of homoscedasticity, you might question why researchers would

choose this approach to analyze the statistical significance of their study

results. One advantage of the pooled variance method is that you will

generally get a more efficient estimate of the standard error of your sam-

pling distribution. This means that the pooled variance method often

leads to a larger t-statistic (though this is not always the case, as illustrated

later in the chapter). But should you take advantage of this method if it

means that you add the risk of violating an additional assumption?

The separate variance method should be used in most circumstances.

As a general rule, it is better to make fewer assumptions, because this cre-

ates less potential for violating them and coming to a mistaken conclusion.

Nonetheless, sometimes your sample results or prior research suggests

�2 1 � �2

2 � �2 or �1 � �2 � �

�sd(X1 � X2) � ��2

1

N1 �

�2 2

N2

(�sd(X1 � X2) )

3See H. M. Blalock, Social Statistics (New York: McGraw-Hill, 1979), p. 231.

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S274

strongly that an assumption of equal variances can be made. For example,

if there is little difference in the standard deviations you find in your sam-

ples, you may be able to conclude with confidence that the population

standard deviations do not differ.4 If, in turn, prior studies show that the

standard deviations between the groups studied are very similar, this

might also lead you to apply this assumption in your test. Most statistical

How are the two methods different in practice? Let’s start with the

pooled variance approach.

If we assume that the two populations of interest have equal vari-

ances, we can simplify Equation 11.1, which defines the standard error

of the sampling distribution for a difference of means test.

plified formula is given in Equation 11.2.

Equation 11.2

Because we do not know the actual value of �, we rewrite the equation,

substituting an estimate of �, or , as shown in Equation 11.3.

Equation 11.3

This, of course, creates another problem for us. How do we calculate ?

We now have two estimates of the sample variance, one from each sample.

And we also need to take into account the bias associated with using s 2 to

estimate , as discussed in Chapter 10. Our solution to the former problem

is to weight the two sample variances by the N of cases in each sample.

This is only fair, because the larger sample is likely to provide a better

estimate of the joint standard deviation than the smaller sample. We in-

clude a correction for bias of the sample variances directly in our esti-

mate of � by subtracting 2 (1 for each sample) in the denominator of the

equation, as shown in Equation 11.4.

Equation 11.4�̂ � � N1s 2 1 � N2s

2 2

N1 � N2 � 2

�̂

�̂

�̂sd(X1 � X2) � �̂�N1 � N2

N1N2

�̂

�sd(X1 � X2) � ��N1 � N2

N1N2

(�sd(X1 � X2) )

4A test of statistical significance may be performed to assess differences in variances. It

is based on the F distribution, which is discussed in detail in Chapter 12. The test

takes a ratio of the two variances being examined:

F � �̂2

larger variance

�̂2 smaller variance

analysis computer programs provide test outcomes for both of these

options with the correct degrees of freedom applied.

This simplification process is outlined in the box on page 264. The sim-

C O M P A R I N G S A M P L E M E A N S 275

To work out the pooled variance method for our example, we first es-

timate the pooled standard deviation for the two populations, which pro-

vides a result of 2.8043. We then calculate the standard error (

for our sampling distribution. Our result is 0.3191.

�̂sd(X1 � X2) )

�sd(X1 � X2) � ��2

1

N1

� �2

2

N2

� �� 1 N1

� 1

N2

� ��N1 � N2

N1N2

Simplification of the Equation for the Standard Error of the Sampling Distribution for Two Samples (�1 � �2)

W orking It Out

� 2.8043

� �7.86405

� �2,532.22

322

� �(197)(2.76)2 � (127)(2.85)2

197 � 127 � 2

�̂ � � N1s 2 1 � N2s

2 2

N1 � N2 � 2

W orking It Out

� 0.3191

� 2.804�0.01295

� 2.804�197 � 127

(197)(127)

�̂sd(X1 � X2) � �̂�N1 � N2

N1N2

How does the pooled variance method differ from the separate vari-

ance method? We once again begin with Equation 11.1. Because and

are unknown, we use and to gain an estimate of . In

turn, as before, the variances of our samples are not considered unbi-

ased estimates of the variances of the population distributions. Accord-

ingly, in order to obtain an unbiased estimate of the standard error using

this method, we need once more to adjust the equations—in this case,

by subtracting 1 from the denominator of each variance estimate, as

shown in Equation 11.5.

Equation 11.5

Based on the sample variances of police officers and firefighters in our

example, we get 0.321 as an estimate of the standard error of the two-

sample t-text using the separate variance method.

�̂sd(X1 � X2) � � s

2 1

N1 � 1 �

s

2 2

N2 � 1

sd(X1 � X2) s2

2s2 1�2

2

�2 1

As you can see, the result found using the pooled variance method

(0.319) is very similar to that found using the separate variance method

(0.321). This will often be the case, especially when samples are rela-

tively large or evenly divided between the two groups. Nonetheless,

even small differences can sometimes affect the conclusions you reach in

a two-sample t-test.

Significance Level and Rejection Region The University of Washington

researchers used a 0.05 significance level and a two-tailed significance

test. The two-tailed test was based on their nondirectional research hy-

job scores between firefighters and police officers.

W orking It Out

� 0.3214

� � (2.76)2

197 � 1 �

(2.85)2

127 � 1

�̂sd(X1 � X2) � � s

2 1

N1 � 1 �

s

2 2

N2 � 1

pothesis, which stated simply that there is a difference in anxiety-on-the-

C O M P A R I N G S A M P L E M E A N S 277

Interpolating from the t table (see Appendix 4), we find that a

t-value of about 1.97 is associated with a two-tailed 5% significance

threshold (with 322 degrees of freedom). This means that a test

statistic greater than 1.97 or less than �1.97 is needed to reject the

null hypothesis.

The Test Statistic To define the t-score appropriate for our test, we must

alter the single-sample t-test equation to account for the fact that we are

comparing two samples. First, we must adjust the numerator to reflect

our comparisons of the differences in the means observed in our study

with those defined in the null hypothesis. Second, we must adjust the

denominator to reflect the standard error of the difference between

means. Because we now have two methods for defining the standard

error of the sampling distribution, we have two separate equations. The

first reflects the difference of means test using a separate variance esti-

mate (11.6a), and the second the difference of means test using a pooled

variance estimate (11.6b).

Both Equation 11.6a and Equation 11.6b have two quantities in the

numerator. The first is the difference between the two sample means

(represented by ). The second is the difference between the two

population means (�1 � �2) as defined by the null hypothesis. Because

the null hypothesis is that the two populations are equal, this quantity is

equal to 0.

In the denominator in each equation, the standard error used for a

sampling distribution when comparing a sample mean to a population

mean has been replaced with the standard error used when comparing

sample means drawn from two populations. This quantity was defined in

our discussion of the sampling distribution.

The t-score for this problem is 12.461 using the separate variance esti-

mate and 12.539 using the pooled variance estimate. As recommended

(although the differences are small in this case), we use the separate

variance method in making our decision.

X1 � X2

Equation 11.6b

Pooled Variance Methodt � (X1 � X2) � (�1 � �2)

�̂�N1 � N2

N1N2

Equation 11.6a

Separate Variance Methodt � (X1 � X2) � (�1 � �2)

� s

2 1

N1 � 1 �

s

2 2

N2 � 1

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S278

The Decision Because our test statistic of 12.4456 is larger than the criti-

cal value of our rejection region of 1.97, we reject the null hypothesis

that there is no difference in anxiety-on-the-job scores between the pop-

ulations of police officers and firefighters to which our test infers. For

our test, the observed significance level is less than the significance

threshold we set at the outset (p � 0.05).

W orking It Out Separate Variance

� 12.4456

� 4

0.3214

� (12.8 � 8.8) � 0

� (2.76)2

197 � 1 �

(2.85)2

127 � 1

t � (X1 � X2) � (�1 � �2)

� s 2 1

N1 � 1 �

s 2 2

N2 � 1

W orking It Out Pooled Variance

� 12.5353

� 4

0.3191

� (12.8 � 8.8) � 0

�(197)(2.76)2 � (127)(2.85)2

197 � 127 � 2 �197 � 127

(197)(127)

t � (X1 � X2) � (�1 � �2)

�N1s 2 1 � N2s

2 2

N1 � N2 � 2 �N1 � N2

N1N2

t � (X1 � X2) � (�1 � �2)

�̂�N1 � N2

N1N2

, where �̂ � �N1s 2 1 � N2s

2 2

N1 � N2 � 2

C O M P A R I N G S A M P L E M E A N S 279

Bail in Los Angeles County: Another Example

of the Two-Sample t-Test for Means

As a second example of the two-sample t-test for means, we will

examine a study of bail setting in Los Angeles County in the 1990s.

The State Court Processing Statistics database represents a random

sample of felony defendants from more than 50 urban court districts

in the United States. Since Los Angeles County participated in the

study throughout the 1990s, data are available for 1990, 1992, 1994,

and 1996.5

An important issue in criminal justice decision making has been the

impact of the defendant’s race or ethnicity on the type of decision

made. We can focus on the amount of bail set as one way to begin to

test for racial or ethnic differences in criminal case processing. In Los

Angeles County in the 1990s, a sample of 1,121 African Americans were

required to post a mean bail amount of $50,841 (s � 115,565). A sample

of 1,798 Hispanics of any race were required to post a mean bail

amount of $66,552 (s � 190,801). Although the difference in mean bail

amounts for these two samples of defendants appears to be large (ap-

proximately $16,000), can we conclude that this difference is statistically

significant?

Assumptions

Level of Measurement: Interval scale.

Population Distribution: Normal distribution in both populations (re-

laxed because N is large).

Sampling Method: Independent random sampling.

Sampling Frame: All felony arrestees in Los Angeles County in

the 1990s.

Hypotheses:

H0: The mean bail amount set for the population of African American

felony defendants is the same as the mean bail amount set for the popu-

lation of Hispanic felony defendants of any race (�1 � �2).

H1: The mean bail amount set for the population of African American

felony defendants is different from the mean bail amount set for the

population of Hispanic felony defendants of any race (�1 � �2).

5These data are available through the National Archive of Criminal Justice Data and

can be accessed at http://www.icpsr.umich.edu/NACJD.

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S280

The Sampling Distribution Because we are interested in comparing

means and � for the population distributions is unknown, we use a t-test

for means. Since the number of cases in each sample is large, we can

relax the normality assumption for this test. The number of degrees of

freedom for the test is df � N1 � N2 � 2 � 1,121 � 1,798 � 2 � 2,917.

Significance Level and Rejection Region Let’s assume that we want to set

a fairly strict level of statistical significance for this test—say, 0.01. We

might argue that we are particularly concerned with a Type I error in this

example, since concluding that there are racial differences may have

very important implications for the criminal justice system. At the same

time, there is no stated reason for expecting one group to have higher

bail amounts than the other group, so we use a two-tailed test. Given

that we have 2,917 degrees of freedom, a significance level of 0.01, and

a two-tailed test, we can consult the t distribution table and determine

that our critical values for this analysis are �2.576. If the test statistic is

greater than �2.576 or less than �2.576, then it falls into the rejection re-

gion for the test, and we will conclude that bail amounts set are not

equal across the two felony groups of defendants.

The Test Statistic We calculate the test statistic using both the separate

variance and the pooled variance methods. As we demonstrate below,

the t-score is �2.7694 using the separate variance method and �2.4863

using the pooled variance method. Following the earlier recommenda-

tion, we will use the separate variance method in making our decision

about the null hypothesis.

W orking It Out Separate Variance Method

� �2.7694

� �15,711

5,673.02

� (50,841 � 66,552) � 0

� 115,5652

1,121 � 1 �

190,8012

1,798 � 1

t � (X1 � X2) � (�1 � �2)

� s2 1

N1 � 1 �

s2 2

N2 � 1

C O M P A R I N G S A M P L E M E A N S 281

The Decision Because our test statistic of �2.7694 is less than the criti-

cal value of �2.576, we reject the null hypothesis that there is no differ-

ence in bail amounts set for African Americans and Hispanics of any race

in Los Angeles County. In this case, it is interesting to note that if we had

used the pooled variance method, we would have failed to reject the

null hypothesis. This points to the importance of making your assump-

tions before you see the study results.

C o m p a r i n g S a m p l e P r o p o r t i o n s : T h e T w o - S a m p l e t - T e s t f o r D i f f e r e n c e s o f P r o p o r t i o n s

As we noted in Chapter 10, one implication of the central limit theorem

portion, the sampling distribution of a proportion begins to approximate

comes large. The central tendency of this distribution and its dispersion

are measured by the mean and standard error, just as for distributions

that develop from interval-level data. In a difference of proportions test,

our interest is in the difference between the populations studied. This

W orking It Out Pooled Variance Method

� �2.4863

� �15,711

6,319.07

� (50,841 � 66,552) � 0

�(1,121)(115,565)2 � (1,798)(190,801)2

1,121 � 1,798 � 2 �1,121 � 1,798

(1,121)(1,798)

� (X1 � X2) � (�1 � �2)

� N1s 2 1 � N2s

2 2

N1 � N2 � 2 �N1 � N2

N1N2

t � (X1 � X2) � (�1 � �2)

�̂�N1 � N2

N1N2

, � where �̂ � � N1s 2 1 � N2s

2 2

N1 � N2 � 2

a normal distribution when the number of cases for the sample be-

is that we can use a normal sampling distribution to test hypotheses

involving proportions. While the mean and standard deviation are not

appropriate statistics to use with a nominal-level measure such as a pro-

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S2 28

difference is also a proportion. Though it would be inappropriate to use

the mean and standard deviation to describe the sample or population

distribution of this proportion, the mean and standard error are appro-

At the same time, we generally cannot use the z-test for conducting a

difference of proportions test. Rather, as in the previous examples, we

rely on the t distribution to test our null hypothesis. You may wonder

why we use the t-test rather than the z-test for making statistical infer-

ences in the case of a difference of proportions. A t-test is used when the

standard deviation of the population distribution is unknown and must

be estimated. In Chapter 10, we noted that when we stated the propor-

tion of successes expected under the null hypothesis, we also stated by

implication the mean and the standard deviation of the population distri-

bution of scores and thus the mean and the standard error of the sam-

In fact, we again define the mean of the sampling distribution for

such a test through the null hypothesis. For a difference of proportions

test, the null hypothesis is ordinarily that there is no difference between

the proportions of the two populations to which we seek to infer. This

null hypothesis defines the mean of our sampling distribution: no differ-

ence, or zero.

As noted earlier in the chapter, in defining the standard error of our

sampling distribution, we take into account the variances of the two

populations. Our problem is that defining the standard error requires

knowing the values of P and Q for each of the two populations we are

interested in. (This is the case because we obtain the variance of a pro-

the null hypothesis states only that the proportions of the two popula-

tions are equal; it does not tell us the value of those proportions. Be-

cause of this, when testing for differences of proportions, the researcher 6

6In practice, many statistics texts use the z-test for examples involving proportions.

Generally this is done because a difference of proportions test is appropriate only for

larger samples, and with larger samples, there is substantively little difference between

the outcomes of these two normal distribution tests. We illustrate a difference of pro-

portions problem using a t-test because it follows the logic outlined in Chapter 10.

That is, in the case where � is unknown, a t-test should be used. Moreover, most

packaged statistical programs provide outcomes only in terms of t-tests.

associated with this proportion.

priate statistics for describing the normal sampling distribution that is

pling distribution for our test. Why can’t we just rely on the same logic

to produce the mean and the standard error for a test comparing two

proportions?

must apply a t rather than a z distribution to his or her test. If the

standard deviations for each distribution were known, it would not be

portion by taking the product of P and Q; see Chapter 10, page 253.) But

C O M P A R I N G S A M P L E P R O P O R T I O N S 283

The Case of Drug Testing and Pretrial Misconduct

In a study conducted in Maricopa County, Arizona, criminal justice

researchers examined whether drug testing of defendants released

before trial had an impact on pretrial misconduct.7 They compared

two groups of defendants who were released before trial. The first

group was monitored through drug testing twice a week. The second

group was released without subsequent drug testing. The sample

was chosen over a six-month period. The researchers identified sub-

jects for the study through identification numbers kept in a computer-

ized case-management system. Defendants with odd identification

numbers were placed in the drug-testing group. Defendants with even

identification numbers were placed in the control, or no-drug-testing,

group. The drug-testing group had 118 subjects. The control group

had 116 subjects.

The researchers followed up on both of these groups of defendants

for 90 days. One measure of pretrial misconduct was failure to appear at

a hearing during the follow-up period. A total of 38% of the control

group and 30% of the drug-testing group failed to appear at a hearing

during this period. The question the researchers wanted to answer was

whether they could infer from the difference between these two samples

that there was in fact a difference in pretrial misconduct between the

populations these samples represent. A two-sample t-test is an appropri-

ate statistical test with which to answer this question.

Assumptions:

Level of Measurement: Interval scale (failure to appear is measured as a

proportion).

Population Distribution: Normal distribution in both populations (re-

laxed because N is large).

Sampling Method: Independent random sampling (all cases for six

months are selected).

Sampling Frame: Defendants released before trial for a six-month period

in Maricopa County, Arizona.

7See Chester Britt III, Michael Gottfredson, and John S. Goldkamp, “Drug Testing and

Pretrial Misconduct: An Experiment on the Specific Deterrent Effects of Drug Monitor-

ing Defendants on Pretrial Release,” Journal of Research in Crime and Delinquency 29

(1992): 62–78.

necessary for the researcher to conduct a statistical test of significance

at all. In this case, the value of the proportion for each of the two popu-

lations would be known by implication.

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S284

Hypotheses:

H0: The two populations do not differ in terms of the proportion who

fail to appear for a pretrial hearing (P1 � P2).

H1: Defendants subject to drug testing will be more likely to appear for

a pretrial hearing (P1 � P2).

The two-sample t-test requires an interval level of measurement, as

well as a normal population distribution for each of the two samples ex-

amined. The actual level of measurement for our example (as stated in

parentheses in our assumptions) is nominal—we compare two propor-

tions. As with a single-sample test of proportions, when the sample sizes

are large, we can relax assumptions regarding the level of measurement

used and the shape of the populations examined. Because we have two

samples and not just one, the central limit theorem applies only if both

samples are large. The definition of how large samples must be to in-

voke the central limit theorem in the case of a difference of proportions

test is a matter of debate. However, when each sample includes more

than 100 cases, as is true for the Arizona study, there will be little argu-

ment regarding the use of this parametric test for proportions.

We must again assume that both samples are independent random

samples. In our example, the researchers did not draw an independent

random sample for either the drug-testing group or the control group.

Rather, as was the case with the cell-allocation study examined in Chap-

ter 9, they sampled all defendants released before trial in Maricopa

County for a specific period of time—in this case, six months. In order to

create the two samples, the researchers assigned the defendants to the

groups according to their identification numbers: Even-numbered sub-

jects were assigned to the control group, and odd-numbered subjects to

the drug-testing group.

In making statistical inferences, the researchers were clearly interested

in inferring beyond their sampling frame (defendants released before

8

8In fact, although we do not examine their findings here, Britt and colleagues con-

ducted their study in two Arizona counties.

Maricopa County but also to other jurisdictions and other programs

trial during the six-month period), not only to other time periods in

likely to apply to other “sophisticated and experienced pretrial services

agencies. They also noted that it “is reasonable to assume that the

programs that were implemented are comparable to the programs that

are likely to be implemented in similar agencies.” When drawing con-

clusions from this research, we would have to consider whether the sample

used can in fact be seen as representative of these larger populations.

similar to the one they studied. They argued that their findings were

C O M P A R I N G S A M P L E P R O P O R T I O N S 285

Our final assumptions relate to the hypotheses. The null hypothesis,

as for earlier tests, is that there is no difference. It assumes that those

monitored through drug testing and those not so monitored (the control

group) will have the same proportion of defendants who fail to appear.

Another way of expressing this is to say that the proportion of failures to

appear in the drug-tested population (P1) of released defendants is the

same as that in the population that is not drug-tested (P2), or that P1 �

P2. The researchers chose a directional research hypothesis because they

were concerned only with the possibility that the program decreased the

likelihood that offenders would fail to appear for a pretrial hearing.

Accordingly, the research hypothesis was stated as P1 � P2. The re-

searchers were interested in testing whether drug testing would increase

compliance.

The Sampling Distribution Because N is large for both samples, we can

use a t distribution as the sampling distribution for testing the difference

between proportions. The number of degrees of freedom for the distri-

bution is obtained by adding the numbers of cases in the two samples

N1 � N2

As we noted earlier, the null hypothesis defines the mean of the sam-

pling distribution we will use for our test. The mean of the sampling dis-

tribution is P1 � P2, or simply 0, because the null hypothesis states that

the two population proportions are the same.

In defining the standard error of the sampling distribution, we can

rely on the pooled variance approach. This is always the case when we

are examining differences between proportions. When we assume in the

null hypothesis that the two population proportions are the same, then

by implication we also assume that the two standard deviations for these

population distributions are also equal. This fact derives from the

method by which the population variances are calculated. As noted in

Chapter 10, the variance of a proportion is

�2 � PQ

and the standard deviation of a proportion is

Accordingly, if P is the same for two populations, then we can also as-

sume that the variances of those populations are equal. In statistical

� � �PQ

� 2. In our example, the number of and subtracting 2: df �

degrees of freedom equals 118 � 116 � 2, or 232.

terms, as we noted earlier, this is defined as the assumption of homos-

cedasticity.

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S286

Significance Level and Rejection Region The researchers in the Mari-

copa County study decided to use “conventional levels” of statistical

significance—that is, a rejection region of � � 0.05. Following their re-

search hypothesis, they also used a one-tailed test of statistical signifi-

cance. Given that we have 232 degrees of freedom, a significance level

of 0.05, and a one-tailed test, we can consult the t distribution table

and determine that the critical value for this analysis is �1.654. If the

this test.

The Test Statistic To define the t-score appropriate for our test, we

must alter the formula for a t-test of means to take into account the

fact that we are examining sample proportions. Accordingly, we re-

place the difference between the sample means with the dif-

ference between the sample proportions (p1 � p2). We also replace

the assumed differences between the population means with (P1 �

P2), as stated by the null hypothesis. Because the null hypothesis

states that the two populations are equal, this quantity is equal to 0.

Equation 11.7 presents a modified formula for calculating the t-statistic

for proportions.

Equation 11.7

Below we calculate the t-statistic for our test. Note that we must

first calculate the variance for each of the sample proportions. To cal-

culate the variance of a sample proportion, we use the formula

s2 � pq. This formula is identical to that presented above for the

variance of a population proportion. The only difference here is the

use of the symbols p, q, and s to represent sample rather than popula-

tion estimates. For our example, � (0.3)(0.7) � 0.21 and �

(0.38)(0.62) � 0.2356.

After inserting the values for the two sample variances and into

Equation 11.7, we obtain a t-statistic equal to �1.29.

s

2 2)(s

2 1

s

2 2s

2 1

� (p1 � p2) � (P1 � P2)

� N1s

2 1 � N2s

2 2

N1 � N2 � 2 �N1 � N2

N1N2

t � (p1 � p2) � (P1 � P2)

�̂�N1 � N2

N1N2

, where �̂ � � N1s

2 1 � N2s

2 2

N1 � N2 � 2

(X1 � X2)

test statistic is less than �1.654, then it falls in the rejection region for

C O M P A R I N G S A M P L E P R O P O R T I O N S 287

The Decision Because our test statistic of �1.29 is greater than the criti-

cal value of �1.654, we fail to reject the null hypothesis that there is no

difference in failure to appear at hearings for the drug testing and control

populations. Based on these and other similar results, the researchers in

this study concluded that “systematic drug testing and monitoring in the

pretrial setting, in programs such as those described above [i.e., exam-

ined in this research], is not likely to achieve significant” change in pre-

trial misconduct.

T h e t - T e s t f o r D e p e n d e n t S a m p l e s

One of the requirements of the two-sample t-test is that the samples ex-

W orking It Out Pooled Variance Method

� �1.29

� �0.08 0.062

� (0.30 � 0.38) � 0

�(118)(0.21) � (116)(0.2356)

118 � 116 � 2 �118 � 116

(118)(116)

� (p1 � p2) � (P1 � P2)

� N1s 2 1 � N2s

2 2

N1 � N2 � 2 �N1 � N2

N1N2

t � (p1 � p2) � (P1 � P2)

�̂�N1 � N2

N1N2

, � where �̂ � � N1s 2 1 � N2s

2 2

N1 � N2 � 2

amined be independent. However, sometimes criminal justice resear-

chers examine samples that are not independent. For example, subjects

may be matched and placed in like pairs based on such characteristics

as social status, education, gender, age, and IQ. Dependent samples

will also result when a researcher takes measurements on the same

subject or unit of analysis over time. For example, a researcher may

examine the attitudes of juvenile delinquents before and after partici-

pation in a specific program or may study changes at specific crime

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S288

hot spots before and after some type of police intervention. Sometimes

the same individuals are compared at different ages or stages in their de-

velopment. Even though in such instances the researcher has two sam-

ples of observations—for example, before and after the program—the

samples are not independent.

The t-test for dependent samples is commonly used in such situa-

tions.9 It focuses on the differences between the pairs in developing the

sampling distribution of the test statistic. Each pair in a t-test for depen-

dent samples is considered a single observation.

The Effect of Police Presence Near High-Crime Addresses

Let’s suppose that a police department took an independent random

sample of 35 high-crime addresses from all high-crime addresses in a

city. The department then assigned a police officer to walk the beat on

each block where one of the addresses was located for a full month. As-

sume we are asked to assess whether the strategy was effective in reduc-

ing calls for police service. We have data on the number of emergency

calls for police service for the month before the officer walked the beat

and for the month during which the officer walked the beat. These data

are given in Table 11.1. The mean number of calls for service the month

before was 30, and the mean for the month when the officer was present

was 20. Can we conclude from this that the program would be effective

in reducing calls for service if applied generally to high-crime addresses?

Assumptions:

Level of Measurement: Interval scale.

Population Distribution: Normal distribution (relaxed because N is large).

Sampling Method: Independent random sampling.

Sampling Frame: All high-crime addresses in the city.

Hypotheses:

H0: There is no difference in the number of calls for service at high-

crime addresses whether an officer walks the beat or not (�1 � �2).

1

1 � �2).

9Here we examine the t-test for dependent samples only in reference to mean differ-

ences for interval-level data. However, this test may also be used for dichotomous

nominal-level data. Suppose you were assessing the absence or presence of some char-

acteristic or behavior at two points in time. If each observation were coded as 0 or 1,

then you would calculate the mean difference and the standard deviation of the

difference (sd) using the same equations as in this section. The only difference from the

example discussed in the text is that you would work only with zeros and ones.

(Xd)

H : There are fewer calls for service at high-crime addresses when an

officer walks the beat (�

T H E t - T E S T F O R D E P E N D E N T S A M P L E S 289

Number of calls for service is an interval-scale measure, as required

by the t-test. The test also requires that the population of differences be-

tween pairs be normally distributed. Because our sample is large (greater

than 30), we are able to relax this assumption for our test.

Our null hypothesis states that there is no difference in the number of

calls for police service at high-crime addresses whether a police officer

walks the beat or not. Because the police department is concerned only

with whether the presence of a police officer walking the beat is effec-

tive in reducing emergency calls for service, we use a directional re-

search hypothesis. It states that the number of calls for police service will

be lower when a police officer walks the beat.

Emergency Calls to Police for the Month Before and the Month During Which an Officer Walked the Beat

CALLS CALLS CALLS CALLS

LOCATION BEFORE DURING LOCATION BEFORE DURING

1 29 14 19 18 22 2 50 28 20 27 24 3 14 8 21 42 16 4 16 6 22 31 14 5 11 20 23 51 30 6 31 17 24 28 8 7 33 4 25 26 11 8 37 22 26 14 19 9 21 20 27 29 21

10 40 27 28 39 26 11 30 29 29 40 20 12 22 30 30 30 20 13 30 18 31 26 11 14 36 20 32 30 28 15 30 22 33 27 13 16 29 26 34 33 20 17 24 19 35 35 34 18 41 33 � 1,050 700

30 20 s 9.21 7.52 X

Table 11.1

A t-test for dependent samples requires that the pairs examined be

selected randomly and independently from the target population of

dependent (i.e., they are related to one another), the pairs themselves

are independent from one another. Because we began with an inde-

pendent random sample of high-crime addresses, we can assume that

the paired observations taken before and during the police interven-

tion are random and independent.

pairs. Accordingly, although the scores for the subjects in the pairs are

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S290

The Sampling Distribution The number of degrees of freedom for a t-test

for dependent samples is obtained by taking the number of pairs studied

and subtracting 1: df � N � 1. In our example, which involves paired

observations for the same subjects over two time periods, the number of

degrees of freedom is 35 � 1, or 34. If we had examined subjects

The mean of the sampling distribution is defined by the null hypothe-

sis. It is represented by �d, or the mean of the population of differences

between crime calls when a police officer is and is not walking the beat.

Because the null hypothesis states that there is no difference in crime

calls during these time periods, �d for the sampling distribution is 0. The

estimated standard error of the sampling distribution is found using

Equation 11.8.

Equation 11.8

Using the estimates of variance calculated in Table 11.2, we find that the

estimated standard error for the sampling distribution is 1.559.

�̂sd � � s2 d

N � 1

(�̂sd)

Significance Level and Rejection Region Following conventional norms,

we use a 0.05 level of statistical significance. However, our research hy-

pothesis suggests a directional test, which means that we place the entire

rejection region on one side of the t distribution. Because the research

hypothesis states that the period during which a police officer is walking

the beat will have a lower number of crime calls, we are interested in the

negative area of the t distribution below the mean. Looking at the t table

(Appendix 4) under a one-tailed significance test with 34 degrees of free-

dom, we see that a critical value of t less than �1.691 is needed to place

the score in the rejection region.

W orking It Out

� 1.5595

� �82.6857 35 � 1

�̂sd �� s2 d

N � 1

matched by common traits, the degrees of freedom would also be

obtained by taking the number of pairs of subjects and subtracting 1.

T H E t - T E S T F O R D E P E N D E N T S A M P L E S 291

The Test Statistic To define the t-score appropriate for our test, we alter

Equation 10.9 (used for the single-sample test of means in Chapter 10) to

take into account the fact that we are now concerned with the mean differ-

ence between pairs of observations. Accordingly, instead of comparing the

mean of one sample to the hypothesized population parameter ,

we now compare the mean of the observed differences between the pairs

with the hypothesized difference between the pairs based on the null hy-

pothesis . As discussed in the section on the sampling distribu-

tion, the hypothesized difference is 0. We also adjust the denominator of

the equation to reflect the standard error of the differences between the

pairs of observations. The revised formula is presented in Equation 11.9.

Equation 11.9t � Xd � �d

� s2 d

N � 1

(Xd � �d)

(X � �)

Calculation of the Standard Deviation for the Differences Between the Two Sets of Dependent Observations

DIFFER- DIFFER-

LOCATION DURING BEFORE ENCE Xi LOCATION DURING BEFORE ENCE Xi

1 14 29 �15 25 19 22 18 �4 196 2 28 50 �22 144 20 24 27 �3 49 3 8 14 �6 16 21 16 42 �26 256 4 6 16 �10 0 22 14 31 �17 49 5 20 11 �9 361 23 30 51 �21 121 6 17 31 �14 16 24 8 28 �20 100 7 4 33 �29 361 25 11 26 �15 25 8 22 37 �15 25 26 19 14 �5 225 9 20 21 �1 81 27 21 29 �8 4

10 27 40 �13 9 28 26 39 �13 9 11 29 30 �1 81 29 20 40 �20 100 12 30 22 �8 324 30 20 30 �10 0 13 18 30 �12 4 31 11 26 �15 25 14 20 36 �16 36 32 28 30 �2 64 15 22 30 �8 4 33 13 27 �14 16 16 26 29 �3 49 34 20 33 �13 9 17 19 24 �5 25 35 34 35 �1 81 18 33 41 �8 4 Totals (�) �350 2,894

� 82.6857� �10

� 2,894

35 �

�350

35

s2 d �

� N

i�1

(Xi � Xd)2

N Xd �

� N

i�1

Xi

N

(Xi � Xd)2(Xi � Xd)2

Table 11.2

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S2 29

Substituting the values from our example, we obtain a t-score of �6.412.

10See Chapter 12 for an example of a rank-order test (the Kruskal-Wallis one-way

analysis of variance).

W orking It Out

� �6.4123

� �10 � 0

�82.6857 35 � 1

t � Xd � �d

� s2 d

N � 1

The Decision Because our test statistic of �6.4123 is less than the crit-

ical value of our rejection region, �1.691, we reject the null hypothe-

sis. The observed significance level of our test is less than the criterion

significance level we set when defining the significance level and

rejection region for our test (p � 0.05). We can conclude that there

is a statistically significant decrease in the number of calls for police

services at high-crime addresses when a police officer is walking

the beat.

A N o t e o n U s i n g t h e t - T e s t f o r O r d i n a l S c a l e s

Ordinal scales create a special problem in conducting tests of statistical

significance. Most tests we have examined so far assume either a nomi-

nal or an interval level of measurement. There are nonparametric tests

for ordinal-level measures; however, these generally assume that the re-

searcher can rank order all scores in a distribution.10 Typically, with ordi-

nal measures, there are a limited number of ordinal categories and many

observations, so such tests are not appropriate.

There is no simple guideline for deciding which test to use for ordinal-

scale variables. In practice, when there are a number of categories in an

T H E t - T E S T E 293 O N U S I N G F O R O R D I N A L S C A L E S A N O T

ordinal scale, researchers use the t-test for means to calculate statistical

significance. When N is large and the number of categories is more than

five, this approach is generally accepted. However, you should keep

in mind when you use this approach that the t-test assumes not only

that the categories in the scale are ordered but also that the intervals

represented by the categories are equal for all of the categories. To

the extent that this can be assumed, you will be on more solid ground

using the t-test for ordinal scales. When the number of categories is less

than five, it may be better to use the chi-square statistic, discussed in

Chapter 9. In Chapter 13, we will discuss other descriptive statistics and

associated significance tests that are often appropriate for ordinal-level

scales.

C h a p t e r S u m m a r y

The two-sample t-test is a parametric test of statistical significance that

may be used to test for equality of two population means or propor-

tions. Although the test requires an interval level of measurement and

normal population distributions, it is nonetheless appropriate for un-

Like other tests of statistical significance examined in previous chap-

ters, the two-sample t-test requires independent random sampling. The

null hypothesis states that the population means or proportions for the

two samples studied are the same. A critical value for the test is identi-

fied on the t sampling distribution, after first determining the degrees

of freedom. The mean of the sampling distribution is again defined

with reference to the null hypothesis. There are two options for calcu-

lating the standard error of the sampling distribution for a difference of

means test. The first is termed the pooled variance method; it gener-

ally provides a more efficient statistical estimate but requires the addi-

tional assumption of homoscedasticity—that the standard deviations

variance method because it does not make an assumption about the

equality of variances between the two population distributions. For a

difference of proportions test, the pooled variance method is always

used.

When the two samples examined are not independent, the t-test for

dependent samples should be used. The calculation of this statistic is

based on the mean difference between pairs of samples and the standard

deviation of the differences between the pairs.

of the two populations are the same. The second is called the separate

both samples.

known populations and binary variables when N is sufficiently large for

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S294

K e y T e r m s

homoscedasticity A statement that the

variances and standard deviations of two or

more populations are the same.

pooled variance A method of obtaining

the standard error of the sampling distribu-

tion for a difference of means test. The

pooled variance method requires an as-

sumption of homoscedasticity.

separate variance A method of obtaining

the standard error of the sampling distribu-

tion for a difference of means test. The

separate variance method does not require

an assumption of homoscedasticity.

t-test for dependent samples A test of

statistical significance that is used when

two samples are not independent.

two-sample t-test A test of statistical sig-

nificance that examines the difference ob-

served between the means or proportions

of two samples.

S y m b o l s a n d F o r m u l a s

Estimate of the standard deviation of a population

Estimate of the standard error of a sampling distribution

To calculate degrees of freedom for the two-sample t-test:

df � N1 � N2 � 2

To calculate an unbiased estimate of the standard error for the sampling

distribution in a two-sample t-test (separate variance method):

To estimate a pooled joint standard deviation of two populations for the

pooled variance method:

To estimate the standard error for the sampling distribution in a two-

sample t-test (pooled variance method):

�̂sd(X1 � X2) � �̂�N1 � N2

N1N2

�̂ � � N1s 2 1 � N2s

2 2

N1 � N2 � 2

�̂sd(X1 � X2) � � s2

1

N1 � 1 �

s2 2

N2 � 1

�̂sd

�̂

S Y M B O L S A N D F O R M U L A S 295

To calculate the two-sample t-test statistic for means (separate variance

method):

To calculate the two-sample t-test statistic for means (pooled variance

method):

To calculate the variance of a sample proportion:

s2 � pq

To calculate the two-sample t-test statistic for proportions:

To calculate the two-sample t-test statistic for means of dependent

samples:

E x e r c i s e s

11.1 Test the following pairs of hypotheses using the information given. Assume that the variable has been measured at the interval level and the cases have been selected at random. For each test, answer the fol- lowing questions:

— Does the test require a one-tailed or a two-tailed test of statistical significance?

— What is (are) the critical value(s) for the stated level of significance?

— What is the value of the test statistic?

— What is the decision regarding the null hypothesis?

t � Xd � �d

� s2 d

N � 1

t � (p1 � p2) � (P1 � P2)

� N1s 2 1 � N2s

2 2

N1 � N2 � 2 �N1 � N2

N1N2

t � (X1 � X2) � (�1 � �2)

�̂�N1 � N2

N1N2

t � (X1 � X2) � (�1 � �2)

� s2 1

N1 � 1 �

s2 2

N2 � 1

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S296

— Does the sample size suggest the need for caution in drawing conclusions?

a. H0: �1 � �2

H1: �1 � �2

� � 0.05

s1 � 4 s2 � 6 N1 � 14 N2 � 18

b. H0: �1 � �2

H1: �1 � �2

� � 0.01

s1 � 8 s2 � 10 N1 � 41 N2 � 41

c. H0: �1 � �2

H1: �1 � �2

� � 0.05

s1 � 6 s2 � 6 N1 � 122 N2 � 215

d. H0: �1 � �2

H1: �1 � �2

� � 0.05

s1 � 2 s2 � 3 N1 � 29 N2 � 33

e. H0: �1 � �2

H1: �1 � �2

� � 0.01

s1 � 35 s2 � 25 N1 � 513 N2 � 476

f. H0: �1 � �2

H1: �1 � �2

� � 0.05

s1 � 1 s2 � 2 N1 � 85 N2 � 93

11.2 Greg wishes to investigate whether there is any difference in the amount of violent crowd behavior that supporters of two soccer teams report having seen in one season. He distributes questionnaires at ran- dom to season-ticket holders at United and at City. The mean number

X2 � 4X1 � 2

X2 � 32X1 � 45

X2 � 6X1 � 15

X2 � 28X1 � 33

X2 � 20X1 � 10

X2 � 30X1 � 24

E X E R C I S E S 297

of matches at which the sample of 110 United fans remember seeing violent incidents is 15 (s � 4.7). For the sample of 130 City fans, the mean number of such matches is 8 (s � 4.2).

a. Can Greg conclude that there are differences in the amount of violent crowd behavior observed between the two populations of season-ticket holders? Outline all the steps required in the test of sta- tistical significance. Choose an appropriate level of significance and calculate the t-test statistic according to the separate variance method.

b. Would Greg’s conclusion be any different if he were to use the pooled variance method?

c. Which of the two methods is preferred in this case?

11.3 To see if there is truth in the claim by a prominent graduate of the police academy that white officers are awarded more promotions than African American officers, an independent random sample is drawn from the 1,000 police officers in Bluesville who graduated from the academy ten years earlier. For the 42 white officers sam- pled, the mean number of promotions received in the ten years since graduation was 3.2 (s � 0.8). For the 20 African American officers sampled, the mean number of promotions received was 2.8 (s � 0.65).

a. From these data, can you conclude that white officers who gradu- ated ten years ago have been awarded more promotions than their African American counterparts? Use the separate variance method and set a 5% significance level.

b. Would your answer be any different if you used the pooled vari- ance method?

c. If the level of significance had been set at 1%, would there be any difference in the decisions you would make based on the separate variance and pooled variance methods?

d. Does the sample size have any relevance to the extent to which you can rely on the results?

11.4 By surveying a random sample of 100 students from Partytime High School and 100 students from Funtime High School, a researcher learns that those from Partytime High School have smoked marijuana an average of 9.8 times (s � 4.2) in the last six months, while those from Funtime High School have smoked marijuana an average of 4.6 times (s � 3.6) in the last six months. Can the researcher conclude that use of marijuana differs between Partytime and Funtime high schools? Use the separate variance method and set a significance level of 0.01. Be sure to state the assumptions of the statistical test.

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S298

11.5 Test the following pairs of hypotheses using the information given. Assume that the variable has been measured at the nominal level, the value reported is the proportion, and the cases have been selected at random. For each test, answer the following questions:

— Does the test require a one-tailed or a two-tailed test of statistical significance?

— What is (are) the critical value(s) for the stated level of significance?

— What is the value of the test statistic?

— What is the decision regarding the null hypothesis?

— Does the sample size suggest the need for caution in drawing conclusions?

a. H0: P1 � P2

H1: P1 � P2

� � 0.05 p1 � 0.80 p2 � 0.60 N1 � 6 N2 � 8

b. H0: P1 � P2

H1: P1 � P2

� � 0.01 p1 � 0.73 p2 � 0.75 N1 � 211 N2 � 376

c. H0: P1 � P2

H1: P1 � P2

� � 0.05 p1 � 0.46 p2 � 0.54 N1 � 86 N2 � 76

d. H0: P1 � P2

H1: P1 � P2

� � 0.01 p1 � 0.28 p2 � 0.23 N1 � 192 N2 � 161

e. H0: P1 � P2

H1: P1 � P2

� � 0.01 p1 � 0.12 p2 � 0.10 N1 � 57 N2 � 45

f. H0: P1 � P2

H1: P1 � P2

� � 0.05 p1 � 0.88 p2 � 0.94 N1 � 689 N2 � 943

E X E R C I S E S 299

11.6 After a long political battle, certain categories of prisoners in Rainy State have been given the right to vote in upcoming local elections. Carolyn wishes to know whether there is any difference between the proportion of eligible prisoners and the proportion of eligible nonpris- oners in Rainy State who will take advantage of their right to vote. She draws two random independent samples—one of 125 prisoners, and the other of 130 nonprisoners. The samples are drawn from the entire eligible prisoner and nonprisoner populations of Rainy State. She finds that 60% of her prisoner sample and 44% of her nonprisoner sample intend to vote.

a. Why is a statistical test of significance necessary here?

b. Carry out a test of statistical significance, remembering to outline each step carefully. Can Carolyn conclude that the two populations are different in terms of their respective members’ intentions to vote?

11.7 Eric takes a random sample of 200 offenders convicted of bribery and a random sample of 200 offenders convicted of robbery over the past five years in Sunny State. By checking court records, he finds that 9% of the bribery offenders and 1% of the robbery offenders in his sam- ples have university educations.

a. By using a two-tailed test with a significance level of 0.01, can Eric conclude that the differences he observes are statistically significant?

b. What steps would you recommend that Eric take if he wishes to ex- tend his conclusions to the prisoner population of neighboring Rainy State?

11.8 Three hundred prisoners, all convicted of violent crimes against persons, have enrolled in a six-month course in anger control. A random sample of 41 of the prisoners are chosen to complete the same questionnaire on two separate occasions—once during the first lesson and once during the last lesson. The questionnaire measures how likely respondents are to resort to violence to solve problems. The results are translated into an index from 0 to 10, with higher scores indicating a higher tendency to seek nonviolent solutions to problems. The 41 prisoners’ scores are shown in the

a. What is the mean change in scores?

b. What is the standard deviation for the differences between scores?

c. Carry out a test of statistical significance, remembering to outline all of the steps required by the test. Can you reject the null hypothesis for the test on the basis of the differences observed?

table on page 301.

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S300

First Last First Last

Subject Lesson Lesson Subject Lesson Lesson

1 1 2 22 6 6

2 2 4 23 4 4

3 1 6 24 9 6

4 3 2 25 9 7

5 4 5 26 0 1

6 7 9 27 1 4

7 6 6 28 1 4

8 4 3 29 3 3

9 4 7 30 2 2

10 1 1 31 2 4

11 2 1 32 1 1

12 3 4 33 0 3

13 4 9 34 4 5

14 6 7 35 4 5

15 7 8 36 6 7

16 2 2 37 6 6

17 2 7 38 7 7

18 3 3 39 3 6

19 1 4 40 1 1

20 6 4 41 1 2

21 2 4

11.9 A random sample of adults in a midwestern state were interviewed twice over a period of two years. Each time, as part of the survey, they were asked how many times their home had been burglarized in the previous 12 months. The numbers of burglaries reported by 27 of the respondents in each interview are shown in the table on

a. What is the mean change in scores?

b. What is the standard deviation for the differences between scores?

c. Carry out a test of statistical significance. Use a significance level of 5% and outline all of the steps required by the test. Can you reject the null hypothesis for the test on the basis of the differences observed?

d. Would your answer have been any different if you had used a sig- nificance level of 1%? Explain why.

page 302.

E X E R C I S E S 301

Respondent First Interview Second Interview

1 0 2

2 1 1

3 2 0

4 1 0

5 0 1

6 0 0

7 3 0

8 0 0

9 0 1

10 5 1

11 2 2

12 2 2

13 1 0

14 1 0

15 0 0

16 0 2

17 0 1

18 0 1

19 1 3

20 0 1

21 1 0

22 2 0

23 0 0

24 0 0

25 2 1

26 0 2

27 1 0

C o m p u t e r E x e r c i s e s

The commands for computing independent (two-sample) and dependent samples

t-tests are generally straightforward in both SPSS and Stata. As in previous

chapters, we have included syntax files for both SPSS (Chapter_11.sps) and Stata

(Chapter_11.do) that contain examples of the commands we illustrate below.

SPSS

Independent Samples t-Test

The two-sample t-test for differences in group means is performed by using the

T-TEST command:

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S302

T-TEST GROUPS = grouping_variable(#1 #2)

/VARIABLES = variable_name(s).

where the GROUPS= indicates the variable containing information on the

groups being tested, which are then listed within the parentheses. A nice fea-

ture of this option is presented when a categorical variable has three or more

categories, say a measure of race–ethnicity, and the interest is in comparing any

two of the groups, such as the 3rd and the 5th groups. We would then insert

the values “3” and “5” into the parentheses to designate these as the groups. All

other cases would be dropped from that particular run of the T-TEST command.

The /VARIABLES= option is where you would list all of the interval-level

variables that you are interested in testing for differences across the two groups.

The output generated by the T-TEST command will contain two lines in the

main results table. The first line will present the t-test results using the pooled

variance method, while the second line will present the t-test results using the

separate variance method.

Dependent Samples t-Test

A dependent samples t-test is obtained using the T-TEST command but using

the PAIRS option to specify the type of t-test to be performed:

The output generated by executing a dependent samples t-test will show the

the standard deviation of the difference, and the corresponding t-statistic.

Stata

Independent Samples t-Test

This command assumes that the variances are equal. To run the ttest command

assuming unequal variance, add the option unequal to the command line:

In the output generated by this command, Stata will show the results for both

one-tail as well as two-tail t-test for differences in the group means.

Dependent Samples t-Test

The syntax for a dependent samples t-test in Stata is remarkably simple:

C O M P U T E R E X E R C I S E S 303

The order the variables are listed in the PAIRS= option is important and should

match exactly the order of the variables as they appear in your null and research

hypotheses.

The two-sample t-test for differences in group means is performed by using the

ttest command:

T-TEST PAIRS = var_1_name WITH var_2_name (PAIRED).

ttest variable_name, by(grouping_variable)

ttest variable_name, by(grouping_variable) unequal

ttest var_1_name == var_2_name

The key to the command syntax is the double equal sign (==) between the

two variable names. Similar to the SPSS command, the order of the variables

is important, and they should be listed here in exactly the same order as they

appear in the null and research hypotheses.

Problems

1. Open the LA bail data (labail.sav or labail.dta). This data file

contains the values used for the bail example in the text. There are only

two variables included in this data set: bail amount and race (coded as

t-test command to

t-test command to test the hypotheses listed below. For each hypothesis test,

use a significance level of 5 %, state each of the assumptions, and explain

whether there is a statistically significant difference between the two groups.

a. The number of times a youth has stolen something valued at $5 to $50

is different for whites and African Americans.

b. The number of times a youth has smoked marijuana is greater for males

than for females.

c. The number of times a youth has physically attacked another student is

different for males and females.

d. The number of times a youth has hit his or her parents is greater for

e. The number of times a youth has cheated on schoolwork is greater for

students earning mostly C’s than for students earning mostly A’s.

dta). Use a t-test command to test the hypotheses listed below. For each

hypothesis test, use a significance level of 5%, state each of the

assumptions, and explain whether there is a statistically significant

difference between the two groups.

a. The length of incarceration sentence is shorter for female than for male

offenders.

offenders.

c. The length of incarceration sentence is longer for offenders convicted

in a jury trial than for offenders who plead guilty.

d. The length of incarceration sentence is different for offenders

convicted of drug and property offenses.

C H A P T E R E L E V E N : C O M P A R I N G M E A N S A N D P R O P O R T I O N S304

reproduce the results on pages 281 and 282 in the text.

paired samples t

paired samples t

C O M P U T E R E X E R C I S E S 305

Comparing Means Among More

Than Two Samples: Analysis of Variance

How Can the Strength of the Relationship Be Defined?

How Does One Make Comparisons Between the Groups?

C h a p t e r t w e l v e

A n a l y s i s o f v a r i a n c e ( A N O V A )

T h e K r u s k a l - W a l l i s t e s t

What is the Logic Underlying ANOVA?

What are the Assumptions of the Test?

How is the Test Carried Out?

When is the Test Used?

How is the Test Carried Out?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_12,

© Springer Science+Business Media New York 2014

IN CHAPTER 11, we used the t distribution to test hypotheses about

means from two independent samples. But what if we are interested in

looking at more than two samples at a time? This is a common problem

in criminology and criminal justice, where many important questions

can be raised across a number of different samples. For example, race is

a central concern in criminal justice and criminology, and often it does

not make sense to restrict comparisons involving race to just two

groups. Similarly, in many criminal justice studies, a number of interven-

tions are compared simultaneously. In such studies, researchers want to

compare not just two but three or more means in the context of one

statistical test.

Analysis of variance (ANOVA) is a commonly used parametric test of

statistical significance that allows the researcher to compare multiple

groups on specific characteristics. ANOVA also provides an opportunity

to introduce several important statistical concepts used in more complex

types of analysis. In this chapter, we examine in this context the con-

cepts of explained and unexplained variation and consider how they re-

late to the total variation found in a specific measure. This chapter also

introduces a nonparametric test, the Kruskal-Wallis test, which can be

used for comparisons across multiple groups when the assumptions un-

derlying ANOVA are difficult to meet.

A n a l y s i s o f V a r i a n c e

Analysis of variance is based on a simple premise. As the differences

between the means of samples become larger relative to the variability

of scores within each sample, our confidence in making inferences

grows. Why is this the case? Certainly it makes sense that the more the

mean differs from one sample to another, the stronger the evidence

307

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

supporting the position that there are differences between the popula-

tion means. All else being equal, the larger the differences between the

samples, the more confident we are likely to be in rejecting the position

that the population means are equal.

But we are faced with a problem in making such an inference. How

much confidence can we place in the observed means of our samples?

As we have stated many times in this book, sample estimates vary from

sample to sample. If the sample means are likely to vary considerably,

then we want to be cautious in drawing strong conclusions from our

study. If the sample means are not likely to vary greatly, we can have

more confidence in conclusions drawn from them. Analysis of variance

uses the variability within the observed samples to come to conclusions

about this variability.

Suppose, for example, that you are examining two separate studies,

each including three samples. In the first study, the scores are widely

dispersed around the mean for each group. In contrast, in the second

study, the scores are tightly clustered around the group means. If you

take the variability you observe in these samples as an indication of the

variability in the populations from which they were drawn, you are

likely to have more confidence in estimates gained from the second

study than from the first. Those estimates appear to be more stable, evi-

dencing much less variability.

This is precisely the approach taken in analysis of variance. The vari-

ability between the groups studied is contrasted with the variability

within these groups to produce a ratio:

The larger this ratio—the larger the differences between the groups

relative to the variability within them—the more confidence we can

have in a conclusion that the population means are not equal. When

the ratio is smaller, meaning that the differences between the groups

are small relative to the variability within them, we have less reason to

conclude that differences exist in the populations to which we want to

infer.

Developing Estimates of Variance Between and Within Groups

The first step in analysis of variance is to define the variability be-

tween and within the groups studied. To make this process more con-

Table 12.1).

Variability between groups

Variability within groups

crete, let’s use a hypothetical study of depression among 12 prison

inmates drawn from high-, moderate-, and low-security prisons (see

308

A N A L Y S I S O F V A R I A N C E

Between-group variability is measured by first subtracting the grand

mean, or overall mean, of the three samples from the mean of each

sample. This difference must then be adjusted to take into account the

number of scores or observations in each sample. Equation 12.1 repre-

sents this process in mathematical language. The sample (or category)

means are represented by , the overall mean (or grand mean) is repre-

sented by , Nc represents the number of scores or observations in the

sample (or category), and tells us to sum the results from the first

sample (or category) mean (c � 1) to the last sample (or category) mean

(c � k).

Equation 12.1

As illustrated in Table 12.1, the overall, or grand, mean is found by

adding up all the scores in the three samples and dividing by the total

number of scores (N � 12). To calculate the amount of between-group

variability for our example, we take just three quantities—the mean de-

pression score of high-security inmates minus the overall mean,

the mean depression score of moderate-security inmates minus

the overall mean, and the mean depression score of low-security inmates

minus the overall mean—and multiply each by its sample size.

Within-group variability is identified by summing the difference be-

tween each subject’s score and the mean for the sample in which the

subject is found. In Equation 12.2, Xi represents a score from one of the

(X � 4)

(X � 8)

(X � 9)

� k

c�1

[Nc(Xc � Xg)]

� k

c�1

Xg

Xc

Depression Scores for 12 Prison Inmates Drawn from High-, Moderate-, and Low-Security Prisons

LOW SECURITY MODERATE SECURITY HIGH SECURITY

(GROUP 1) (GROUP 2) (GROUP 3)

3 9 9 5 9 10 4 8 7 4 6 10

To calculate the grand mean: Xg � � N

i�1

Xi

N �

84

12 � 7

X � 9X � 8X � 4

� � 36� � 32� � 16

Table 12.1

309

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

three samples and, as before, represents the mean for that sample.

Here we sum from i � 1 to N, or from the first to the last score in the

overall study.

Equation 12.2

For within-group variability, we have four calculations to carry out for

each sample (group). For the first subject in the low-security prison sam-

ple, for example, we subtract from the subject’s score of 3 the mean

score of low-security inmates in the study (4). The same is done for each

of the other three members of this sample. For the moderate-security

sample, we repeat the process, starting with the first subject, with a score

of 9, and using the mean of 8 for the group as a whole. The same is

done for the high-security sample.

When we add up the deviations between the groups and those within

them, as is done in Tables 12.2 and 12.3, we find that both are 0. This

does not mean that there is an absence of variability either within or be-

tween the samples we are examining. Rather, this outcome reflects a rule

stated in Chapter 4: The sum of the deviations from a mean equals 0.

Clearly, we cannot use the sum of the deviations from the mean as an in-

dicator of variation. As in other similar problems, it makes sense to

square the deviations from the mean. The squares of all the deviations

from the mean will be positive or 0.

The result when we square these quantities and then add them is

commonly referred to as a sum of squares. The variability between

groups measured in this way is called the between sum of squares

(BSS). It is calculated by taking the sum of the squared deviation of each

sample mean from the overall mean multiplied by the number

of cases (Nc) in that sample.

Equation 12.3BSS � � k

c�1

[Nc(Xc � Xg) 2]

(Xg)(Xc )

� N

i�1

(Xi � Xc )

Xc

Summing the Deviations of the Group Means from the Grand Mean for the Three Groups in the Inmate Depression Study

N c

4 7 �3 �12 8 7 1 4 9 7 2 8

� � 0� � 0

(Xc � Xg)(Xc � Xg)XgXc

Table 12.2

310

A N A L Y S I S O F V A R I A N C E

To calculate the between sum of squares for our hypothetical study,

we take the same approach as is shown in Table 12.2. The one differ-

ence is that after we subtract the overall mean from a category mean, we

square the result. Our final result is 56.

Summing the Deviations of the Scores from the Group Means Within the Three Groups in the Inmate Depression Study

X i

3 4 �1 5 4 1 4 4 0 4 4 0 9 8 1 9 8 1 8 8 0 6 8 �2 9 9 0

10 9 1 7 9 �2

10 9 1

� � 0

(Xi � Xc)Xc

Table 12.3

W orking It Out

� 56

� 4(4 � 7)2 � 4(8 � 7)2 � 4(9 � 7)2

BSS � � k

c�1

[Nc(Xc � Xg) 2]

is defined as the within sum of squares (WSS). The within sum of

squares is obtained by taking the sum of the squared deviation of each

score from its category mean, as represented in Equation 12.4. As before,

we first take the score for each subject and subtract from it the sample

mean. However, before adding these deviations together, we square

each one. The within sum of squares for our hypothetical example is

equal to 14.

Equation 12.4WSS � � N

i�1

(Xi � Xc )2

When we measure variability within groups using this method, the result

311

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

Partitioning Sums of Squares

We can also calculate a third type of variability for our inmate depres-

sion study: total sum of squares (TSS). The total sum of squares

takes into account all of the variability in our three samples. It is

calculated by taking the sum of the squared deviation of each score

from the overall mean of scores for the three groups, as shown in

Equation 12.5.

Equation 12.5

In practice, we first take the deviation of a score from the overall

mean and then square it. For example, the first subject in the low-secu-

rity prison sample has a score of 3. We subtract from this score the over-

all mean of 7 and then square the result (�4), to obtain a value of 16. To

arrive at the total sum of squares, we do this for each of the 12 scores in

the study and then sum the results.

TSS � � N

i�1

(Xi � Xg) 2

W orking It Out

� 14

�� (7 � 9)2 � (10 � 9)2

�� (9 � 8)2 � (8 � 8)2 � (6 � 8)2 � (9 � 9)2 � (10 � 9)2

� (3 � 4)2 � (5 � 4)2 � (4 � 4)2 � (4 � 4)2 � (9 � 8)2

WSS � � N

i�1

(Xi � Xc )2

W orking It Out

� 70

� (10 � 7)2 � (7 � 7)2 � (10 � 7)2

� (9 � 7)2 � (9 � 7)2 � (8 � 7)2 � (6 � 7)2 � (9 � 7)2

� (3 � 7)2 � (5 � 7)2 � (4 � 7)2 � (4 � 7)2

TSS � � N

i�1

(Xi � Xg) 2

312

A N A L Y S I S O F V A R I A N C E

squares and the within sum of squares. That is, the total variability across

all of the scores is made up of the variability between the samples and

the variability within the samples. More generally, the three types of vari-

ability discussed so far may be expressed in terms of a simple formula

that partitions the total sum of squares into its two component parts: the

between sum of squares and the within sum of squares.

Total sum of squares � between sum of squares

� within sum of squares Equation 12.6

(For our example, 70 � 56 � 14.)

Another way to express the relationship among the three types of

sums of squares is to partition the total sum of squares into explained

and unexplained components:

Total sum of squares � explained sum of squares

� unexplained sum of squares Equation 12.7

In this equation, the between sum of squares is represented by the ex-

plained sum of squares (ESS) because the between sum of squares

represents the part of the total variability that is accounted for by the dif-

ferences between the groups. For our hypothetical example, this is the

proportion of the total variability in depression that is “explained” by the

type of prison in which the subject is incarcerated.

The within sum of squares is represented in Equation 12.7 by the un-

explained variability, or the unexplained sum of squares (USS). This

is the part of the total variability that differences between the groups do

not explain. We usually do not know the cause of this variability.

Developing Estimates of Population Variances

So far we have defined the sums of squares associated with between-

group and within-group variability. But analysis of variance, as its name

implies, is concerned with variance, not just variability. Accordingly, we

have to adjust our sums by taking into account the appropriate number

of degrees of freedom. In Chapter 5, when we developed estimates of

variance, we divided the squared deviations from the mean by the num-

ber of cases in the sample or population. For analysis of variance, we di-

vide the between and within sums of squares by the appropriate degrees

of freedom.

For the between-group estimate of variance, we define the number of

degrees of freedom as k � 1, where k is the number of samples or cate-

gories examined. If we are comparing three sample means, the number

of degrees of freedom associated with the between-group estimate

of variance is thus 2. As illustrated by Equation 12.8, an estimate of the

The quantity obtained is equivalent to the sum of the between sum of

313

The ANOVA model is often presented using a different set of statistical

notation than that used in this text. In this book, we define the total sum of

squares as equal to the sum of the within sum of squares and the be-

tween sum of squares:

In many other statistics texts, the following equation is used for the de-

composition of the total sum of squares:

The double summation symbols indicate that we need to first sum over

each group—denoted by the subscript j—and then sum over each indi-

vidual observation—denoted by the subscript i. In terms of this book’s no-

tation, .

Although we have reduced the number of summation signs to one for

each sum of squares, the calculations with the two equations are identi-

cal for the total sum of squares and the within sum of squares. The one

difference between the two equations lies in the calculation of the be-

tween sum of squares. The notation we offer simplifies this calculation

by taking into account the fact that all of the individual scores in a single

group (Nc) will have the same value. Rather than repeat the same calcu-

lation for all the individuals in the same group, we produce the identical

answer by multiplying the squared difference of the sample mean and

the overall mean by the number of observations in the corresponding

sample.

X • • � Xg and X •j � Xc

� N

i�1 �

k

j�1

(Xij � X••)2 � � N

i�1 �

k

j�1

(Xij � X•j) 2 � �

N

i�1 �

k

j�1

(X•j � X••)2

� N

i�1

(Xi � Xg)2 � � N

i�1

(Xi � Xc)2 � � k

c�1

[Nc(Xc � Xg)2]

in Different Formats Representing Sums of Squares: ANOVA Notation

A N A L Y S I S O F V A R I A N C E

between-group variance is obtained by dividing the between sum

of squares by k � 1.

Equation 12.8

The number of degrees of freedom for the within-group estimate of

variance is N � k, or the number of cases in the sample minus the num-

ber of samples or categories examined. The within-group variance esti-

mate is calculated by dividing the within sum of squares by N � k

(see Equation 12.9).

Equation 12.9

A Substantive Example: Age and White-Collar Crimes

Now that we have defined the two types of variance that are compared

in analysis of variance, let’s look at a substantive problem. Table 12.4

presents data on age for three samples of offenders convicted of white-

collar crimes in seven federal district courts over a three-year period.1

�̂2 wg �

� N

i�1

(Xi � Xc) 2

N � k

(�̂2 wg)

�̂2 bg �

� k

c�1

[Nc(Xc � Xg) 2]

k � 1

(�̂2 bg)

Ages of 30 White-Collar Criminals Convicted of Three Different Offenses

OFFENSE 1 OFFENSE 2 OFFENSE 3

BANK EMBEZZLEMENT BRIBERY ANTITRUST VIOLATION

19 28 35 21 29 46 23 32 48 25 40 53 29 42 58 30 48 61 31 58 62 35 58 62 42 64 62 49 68 75

s � 8.98 s � 13.99 s � 10.54

Grand mean � 1333/30 � 44.43(Xg)

X � 56.2X � 46.7X � 30.4

Table 12.4

1The data are drawn from S. Wheeler, D. Weisburd, and N. Bode, Sanctioning of

White Collar Crime, 1976–1978: Federal District Courts (Ann Arbor, MI: Inter-

University Consortium for Political and Social Research, 1988).

315

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

The first sample was drawn from offenders convicted of bank embezzle-

ment, the second from offenders convicted of bribery, and the third from

offenders convicted under criminal antitrust statutes. Ten subjects were

drawn randomly from each of these populations.

The values listed in Table 12.4 represent the ages of the sampled of-

fenders. The mean age of the bank embezzlers is 30.4 years; of the

bribery offenders, 46.7 years; and of the antitrust offenders, 56.2 years.

Can we conclude from the differences found among these samples that

there are differences among the means of the populations from which

these samples were drawn?

Assumptions:

Level of Measurement: Interval scale.

Population Distribution: Normal distribution for each population (the

shape of the population distributions is unknown, and the sizes of the

samples examined are small).

Sampling Method: Independent random sampling (no replacement;

sample is small relative to population).

Sampling Frame: All white-collar offenders convicted of the

crimes examined in seven federal judicial districts over a three-year

period.

Population variances are equal .

Hypotheses:

H0: Population means of age for bank embezzlers, bribery offenders,

and antitrust offenders are equal (�1 � �2 � �3).

H1: Population means of age for bank embezzlers, bribery offenders,

and antitrust offenders are not equal (�1 � �2 � �3).

ance requires that an interval level of measurement be used for the

scores to be examined. Our example meets this assumption because

age is an interval measure. Some statisticians add an assumption of

nominal measurement because a comparison of means across samples

requires that we define categories (or samples) for comparison. For

example, in the case of age and white-collar crime, the three samples

represent three categories of offenses. Our interest in this case is in

the relationship between age (an interval-scale variable) and category

of crime (a nominal-scale variable). In the hypothetical study dis-

cussed earlier in this chapter, we were interested in the relationship

between depression (measured at an interval level) and type of prison

(a nominal-scale variable).

(�2 1 � �2

2 � �2 3)

Like other parametric tests of statistical significance, analysis of vari-

316

A N A L Y S I S O F V A R I A N C E

Analysis of variance also requires that the populations underlying the

samples examined be normally distributed. For our example, this is the

most troubling assumption. We do not have evidence from prior studies

that age is distributed normally within categories of crime. Nor are our

samples large enough to allow us to invoke the central limit theorem.

For ANOVA, as for the two-sample t-test, we want to have at least 30

cases per sample in order to relax the normality assumption. Because the

computations for analysis of variance are complex, having only ten cases

in each sample makes it easier to learn about ANOVA. However, our test

will provide valid results only if the population distributions we infer to

are in fact normally distributed.

We must also assume that the samples being compared were drawn

randomly and independently. In practice, as we discussed in Chapter

11, researchers often make comparisons between groups within a sin-

gle larger sample. For example, we might draw an independent ran-

dom sample of white-collar offenders and then compare the means

found in this larger sample for bank embezzlement, bribery, and an-

titrust offenders. As explained in Chapter 11, if the larger sample was

drawn as an independent random sample, then we may assume that

subsamples consisting of offenders convicted of different types of

crimes are also independent random samples. In this study, random

samples were drawn independently from each category of crime. Al-

districts studied.

For analysis of variance, we must also assume that the population

variances of the three groups are equal. This assumption of homoscedas-

ticity is similar to that introduced for the two-sample t-test (using the

pooled-variance method) in Chapter 11. However, in contrast to the

2

One reason researchers, as opposed to statisticians, are not very

concerned about the assumption of homoscedasticity is that even large

2For example, see G. Hornsnell, “The Effect of Unequal Group Variances on the

F Test for Homogeneity of Group Means,” Biometrika 40 (1954): 128–136; G. E. P.

Box, “Some Theorems on Quadratic Forms Applied in the Study of Analysis of

Variance Problems. I. Effect of Inequality of Variance in the One Way Classification,”

Annals of Mathematical Statistics 25 (1954): 290–302.

though the researchers did not sample with replacement, we can assume

that this violation of assumptions is not serious because the sample

drawn is very small relative to the population of offenders in the

t-test, ANOVA has no alternative test if we cannot assume equal variances

between the groups. Although this seems at first to be an important

barrier to using analysis of variance, in practice it is generally accepted

that violations of this assumption must be very large before the results

of a test come into question.

317

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

violations of this assumption affect the estimates of statistical signifi-

cance to only a small degree. Sometimes it is suggested that you simply

define a more conservative level of statistical significance when you

are concerned with a serious violation of the homoscedasticity assump-

tion.3 Accordingly, you might select a 1% significance threshold as

opposed to the more conventional 5% threshold. In general, large devi-

ations in variance are not likely to occur across all of the groups stud-

ied. If one group is very different from the others, you might choose to

conduct your test two ways—both including the group that is very dif-

ferent from the others and excluding it—and compare your results. In

our example, the variances do not differ widely one from another (see

Table 12.4).

Our final assumptions for this analysis relate to the null and research

hypotheses. For analysis of variance, the null hypothesis is that the

means of the groups are the same. In our example, the null hypothesis

is that the mean ages of offenders in the populations of the three crime

categories are equal. Our research hypothesis is that the means are not

equal. As is true for ANOVA more generally, our research hypothesis is

nondirectional. If we are making inferences to three or more popula-

tions, it is not possible to define the direction of the differences among

them.4

The Sampling Distribution The sampling distribution used for making de-

cisions about hypotheses in analysis of variance is called the F distribu-

tion, after R. A. Fisher, the statistician who first described it. The shape of

the F distribution varies, depending on the number of degrees of free-

dom of the variance estimates being compared. The number of degrees

3Most packaged statistical programs provide a test for equivalence of variances as an

option with their ANOVA program. However, be careful not to automatically reject

use of analysis of variance on the basis of a statistically significant result. In smaller

studies, with samples of less than 50 per group, a finding of a statistically significant

difference should make you cautious about using analysis of variance. In such a

study, you may want to adjust the significance level, as suggested here, or consider al-

ternative nonparametric tests (discussed later in the chapter). With larger samples,

a statistically significant result at conventional significance levels should not necessar-

ily lead to any adjustments in your test. For such adjustments to be made, the differ-

ence should be highly significant and reflect large actual differences among variance

estimates. 4However, in the special case of analysis of variance with only two samples, the re-

searcher can use a directional research hypothesis. This will sometimes be done in ex-

perimental studies when the researcher seeks to examine differences across experi-

mental and control groups, taking into account additional factors [e.g., see L. W.

Sherman and D. Weisburd, “General Deterrent Effects of Police Patrol in Crime ‘Hot

Spots.’ A Randomized Study,” Justice Quarterly 12 (1995): 625–648.]

318

A N A L Y S I S O F V A R I A N C E

of freedom is represented by k � 1 for the between-group variance and

N � k for the within-group variance:

W orking It Out

df for between-group variance � k � 1 � 3 � 1 � 2

df for within-group variance � N � k � 30 � 3 � 27

Because we need to take into account two separate degrees of freedom

at the same time, a different table of probability estimates is given for

each significance threshold. Accordingly, Appendix 5 provides F tables

for 0.05, 0.01, and 0.001 significance levels.

Each table provides the F-scores, adjusted for degrees of freedom,

that correspond to the particular significance threshold. Thus, for

example, in the table for � � 0.05, the values given are the critical

values for the test. For all tests using the F distribution, we need to ob-

tain an F-score greater than this critical value to reject the null hypothesis

of equal means. Looking at the table for � � 0.05, we can identify two

interesting characteristics of the F distribution.

First, the F distribution is unidirectional, consisting only of positive

values. Consistent with the fact that the research hypothesis in analysis

of variance with three or more population means states simply that the

means are not equal, the F distribution is concerned only with the ab-

solute size of the statistic obtained.

Second, as the number of degrees of freedom associated with the

within-group variance grows, the F-value needed to reject the null hy-

pothesis gets smaller. Remember that the number of degrees of freedom

for the within-group variance is equal to N � k. Accordingly, as the

number of cases in the sample gets larger, the number of degrees of

freedom also gets larger. Why should the F-value needed to reject the

null hypothesis be related to the size of the sample? As with a t-test, as

the number of cases increases, so too does our confidence in the esti-

mate we obtain from an F-test.5

Significance Level and Rejection Region Given that no special concerns

have been stated in regard to the risk of either a Type I or a Type II

error, we use a conventional 0.05 significance level. Looking at the F

table for � � 0.05 (Appendix 5), with 2 and 27 degrees of freedom,

5Indeed, note that the values of F with 1 degree of freedom for the between sum of

squares are simply the values of t squared.

319

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

respectively, we find a critical value of 3.35. This tells us that we need an

F-score greater than 3.35 to reject our null hypothesis of no difference

between the population means. An observed F-score greater than 3.35

means that the observed significance level for our test is less than the

0.05 criterion level we have set.

The Test Statistic To calculate the F-ratio, we must compute estimates of

the between-group and within-group population variances based on our

three samples. Computing the between-group variance is relatively easy.

As noted earlier, the formula for between-group variance is

Applying this formula to our example, we first take the mean for each

group, subtract from it the overall mean of the three groups (44.43),

square the result, and multiply by 10—the number of observations in

each group. After this process has been carried out for each of the three

groups, the totals are then added together and divided by the degrees of

freedom for the between-group variance (2). These calculations are illus-

trated below. The between sum of squares for our example is 3,405.267.

Dividing it by the number of degrees of freedom (2) results in a

between-group variance estimate of 1,702.634.

�̂2 bg �

� k

c�1

[Nc(Xc � Xg) 2]

k � 1

(�̂2 bg)

W orking It Out

� 1,702.6335

� 3,405.267

2

� 10(30.4 � 44.43)2 � 10(46.7 � 44.43)2 �10(56.2 � 44.43)2

3 � 1

�̂2 bg �

� k

c�1

[Nc(Xc � Xg) 2]

k � 1

Applying the formula for within-group variance is more difficult in large

part because the calculation of a within-group sum of squares demands a

good deal of computation even for small samples. For that reason, some

texts provide an alternative estimating technique for the within-group sum

assume that you will turn to statistical computing packages when conduct-

ing research in the future and the purpose here is to gain a better under-

320

of squares (see the box on p. 322). However, because it is probably safe to

A N A L Y S I S O F V A R I A N C E

standing of analysis of variance, we will focus on the raw computation. Al-

though cumbersome, it illustrates more directly the logic behind ANOVA.

As discussed earlier, the formula for within-group variance is

For our example, we first take each individual score as illustrated in Table

12.5, and subtract from it the mean for its group: . We then

square this quantity: . This is done for all 30 individual scores,

and the results are then summed. The within-group sum of squares is

3,874.1. When we divide this quantity by the correct degrees of freedom

(N � k, or 27), we obtain a within-group variance estimate of 143.485.

To obtain the F-statistic for our example, we simply calculate the ratio

of the between- and within-group variances (see Equation 12.10), obtain-

ing 11.866.

Equation 12.10F � between-group variance

within-group variance

(X � Xc )2

(X � Xc )

�̂2 wg �

� N

i�1

(Xi � Xc )2

N � k

(�̂2 wg)

Calculating the Within-Group Sum of Squares

OFFENSE 1 OFFENSE 2 OFFENSE 3

BANK EMBEZZLEMENT BRIBERY ANTITRUST VIOLATION

� 30.4 � 46.7 � 56.2

X X X

19 �11.4 129.96 28 �18.7 349.69 35 �21.2 449.44 21 �9.4 88.36 29 �17.7 313.29 46 �10.2 104.04 23 �7.4 54.76 32 �14.7 216.09 48 �8.2 67.24 25 �5.4 29.16 40 �6.7 44.89 53 �3.2 10.24 29 �1.4 1.96 42 �4.7 22.09 58 1.8 3.24 30 �0.4 0.16 48 1.3 1.69 61 4.8 23.04 31 0.6 0.36 58 11.3 127.69 62 5.8 33.64 35 4.6 21.16 58 11.3 127.69 62 5.8 33.64 42 11.6 134.56 64 17.3 299.29 62 5.8 33.64 49 18.6 345.96 68 21.3 453.69 75 18.8 353.44

� � 3,874.1

(Xi � Xc ) 2(Xi � Xc )(Xi � Xc ) 2(Xi � Xc )(Xi � Xc ) 2(Xi � Xc )

XXX

Table 12.5

W orking It Out

� 11.8663

F � 1,702.6335

143.4852

F � between-group variance

within-group variance

321

i

i

While it is important for you to understand the concepts underlying the

equations used in the computations for ANOVA, the actual calculation of

the within-group sum of squares can be quite tedious. Since it is often

easier to calculate the total sum of squares and the between-group sum

of squares, the simplest way of obtaining the within-group sum of squares

is to rely on the relationship stated in Equation 12.6:

Total sum of squares � between-group sum of squares

� within-group sum of squares

This equation can be rearranged and solved for the within-group sum of

squares:

Within-group sum of squares � total sum of squares

� between-group sum of squares

A formula for computing the total sum of squares is

This equation tells us to square the value of each observation and add the

resulting squared values together . From this quantity, we then

subtract the square of the sum of all the values divided by the total num-

ber of observations .

For an illustration of the use of this formula, we can turn to the data on

the ages of white-collar criminals in Table 12.4. In the following table, we

take each offender’s age (X ) and square it (X 2). We then sum each column.

To calculate the total sum of squares, we just enter the two sums into

the computational formula:

� 7,279.367

� 66,509 � (1,333)2

30

TSS � � N

i �1

X

2 i �

(� N

i�1

X i) 2

N

�(� N

i�1

X i) 2/N�

��N i�1

X

2 i�

TSS � � N

i �1

X

2 i �

(� N

i�1

X i) 2

N

Computational Formulas for the Within-Group Sum of Squares

Age (X ) Age Squared (X 2)

19 361

21 441

23 529

25 625

29 841

30 900

31 961

35 1,225

42 1,764

49 2,401

28 784

29 841

32 1,024

40 1,600

42 1,764

48 2,304

58 3,364

58 3,364

64 4,096

68 4,624

35 1,225

46 2,116

48 2,304

53 2,809

58 3,364

61 3,721

62 3,844

62 3,844

62 3,844

75 5,625

At this point, since we know the total sum of squares (7,279.367) and have

already calculated the between-group sum of squares (3,405.267), we can

easily see that the within-group sum of squares is 3,874.1:

Within-group sum of squares � total sum of squares

� between-group sum of squares

� 7,279.367 � 3,405.267

� 3,874.1

� � 66,509� � 1,333

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

The Decision Because the test statistic for our example (11.866) is larger

than 3.35, the critical value of the rejection region, our result is statisti-

cally significant at the 0.05 level. Accordingly, we reject the null hypothe-

sis of no difference between the population means and conclude (with a

conventional level of risk of falsely rejecting the null hypothesis) that the

average age of offenders differs across the three types of crime exam-

ined. However, given our concern about violating the assumption of

normality, our conclusion will be valid only if age is indeed normally

distributed in the three populations studied.

Another ANOVA Example: Race and Bail Amounts

Among Felony Drug Defendants

Table 12.6 presents data on bail amounts set for three samples of

felony drug defendants in large urban court districts in the United

States in the 1990s.6 The first sample is taken from non-Hispanic

whites, the second sample from non-Hispanic African Americans, and

the third sample from Hispanics of any race. Twelve defendants were

drawn at random from the population of each group. The mean bail

amount is $4,833.33 for non-Hispanic whites, $8,833.33 for non-

Hispanic African Americans, and $30,375.00 for Hispanics of any race. Do

Bail Amounts (in Dollars) for 36 Felony Drug Defendants

NON-HISPANIC NON-HISPANIC HISPANICS

WHITES BLACKS OF ANY RACE

1,000 1,000 1,000 1,000 1,000 2,000 1,500 2,000 4,000 2,000 2,500 5,000 2,500 3,000 10,000 3,000 4,000 12,500 5,000 5,000 25,000 7,000 10,000 25,000 7,500 12,500 25,000 7,500 20,000 40,000

10,000 20,000 65,000 10,000 25,000 150,000

s � 3,287.18 s � 8,201.46 s � 42,028.74

g

X � 30,375.00X � 8,833.33X � 4,833.33

Table 12.6

6The data are taken from State Court Processing Statistics: 1990, 1992, 1994, 1996 and

can be accessed through the National Archive of Criminal Justice Data web site at

http://www.icpsr.umich.edu/NACJD.

324

Grand mean (X ) � 528500/36 = 14,680.56

Most statistical analysis software presents the results of an analysis of

variance in the form of an ANOVA table. An ANOVA table provides a com-

pact and convenient way to present the key elements in an analysis of

variance. In addition to ensuring that the researcher has all the neces-

sary information, it also allows the researcher to reproduce the estimates

of the variance and the F-statistic. The general form of an ANOVA table is

as follows:

Source df Sum of Squares Mean Square F

Between k � 1

Within N � k

Total N � 1

Each row gives the pieces of information needed and the formulas for the

calculations. For example, the “Between” row gives the corresponding

degrees of freedom and the formulas for calculating between-group vari-

ability, between-group variance (in the “Mean Square” column), and the

F-statistic. The “Within” row gives the corresponding degrees of freedom

and the formulas for calculating within-group variability and within-group

variance.

Following is the ANOVA table for the results of our calculations using

the data on the ages of white-collar criminals:

Source df Sum of Squares Mean Square F

Between 2 3,405.267 1,702.6335 11.8663

Within 27 3,874.100 143.4852

Total 29 7,279.367

� N

i�1

(Xi � Xg)2

WSS

N � k� N

i�1

(Xi � Xc)2

�̂2 bg

�̂2 wg

BSS

k � 1� k

c�1

[Nc(Xc � Xg)2]

The ANOVA Table

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

the differences in sample means indicate that there are differences in

the population means?

Assumptions:

Level of Measurement: Interval scale.

Population Distribution: Normal distribution for each population (the

shape of the population distributions is unknown, and the sizes of the

samples examined are small).

Sampling Method: Independent random sampling (no replacement; sam-

ples are small relative to populations).

Sampling Frame: Felony drug defendants in large urban court districts in

the United States in the 1990s.

Population variances are equal .

Hypotheses:

H0: Population means of bail amounts set for felony drug defendants

who are non-Hispanic whites, non-Hispanic African Americans, and His-

panics of any race are equal (�1 � �2 � �3).

H1: Population means of bail amounts set for felony drug defendants

who are non-Hispanic whites, non-Hispanic African Americans, and His-

panics of any race are not equal (�1 � �2 � �3).

The Sampling Distribution We again use the F distribution to test for dif-

ferences among our three sample means. Recall that we need two indi-

cators of degrees of freedom: one for the between-group variance and

one for the within-group variance.

df for between-group variance � k � 1 � 3 � 1 � 2

df for within-group variance � N � k � 36 � 3 � 33

Significance Level and Rejection Region As in the preceding example, we

do not have any particular concerns about the risk of a Type I or Type II

error, so we can use the conventional 0.05 significance level. Given that

we have degrees of freedom equal to 2 and 33 with a 0.05 significance

level, the critical value of the F-statistic is about 3.29. If our calculated

F-statistic is greater than 3.29, then we will reject our null hypothesis of

equal means.

The Test Statistic We begin our calculation of the F-statistic by comput-

ing estimates of the between-group variance and the within-group vari-

to be 2,264,840,905.56.�̂2 bg

(�2 1 � �2

2 � �2 3)

ance. Applying the formula for between-group variance, we find the

estimate of

326

A N A L Y S I S O F V A R I A N C E

W orking It Out

� 2,264,840,905.56

� 4,529,681,811.11

2

� � 12(4,833.33 � 14,680.56)2

� 12(8,833.33 � 14,680.56)2 � 12(30,375.00 � 14,680.56)2� 3 � 1

�̂2 bg �

� k

c�1

[Nc(Xc � Xg) 2]

k � 1

The value of the within-group variance is 617,193,813.13. Table

12.7 presents the calculation of the within-group sum of squares, which

turns out to be equal to 20,367,395,833.33. We then divide the value of

the within-group sum of squares by the corresponding degrees of free-

dom (N � k � 36 � 3 � 33), which gives us an estimate for the within-

group variance of 617,193,813.13.

(�̂2 wg)

W orking It Out

� 617,193,813.13

� 20,367,395,833.33

36 � 3

�̂2 wg �

� N

i�1

(Xi � Xc) 2

N � k

The value of the F-statistic is obtained by dividing the estimate of

between-group variance by the estimate of within-group variance. For

our example, F is found to be 3.67.

W orking It Out

� 3.67

� 2,264,840,905.56

617,193,813.13

F � between-group variance

within-group variance

327

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

The Decision Since our test statistic of 3.67 is greater than the critical

value of 3.29 for the F distribution, we reject the null hypothesis of equal

population means; we conclude that there is a statistically significant re-

lationship between bail amount and race of defendant. However, as in

our example concerning age and white-collar crime, we began our test

with strong doubts about whether we could meet a core assumption of

analysis of variance. The samples examined are not large, and thus we

cannot relax the normality assumption for our test. In turn, we do not

have strong reason to believe that the populations to which we want to

infer actually meet the criteria of this assumption. As has been noted

throughout the text, statistical conclusions are only as solid as the as-

sumptions the researchers make. In this case, our statistical conclusions

clearly do not stand on solid ground.

D e f i n i n g t h e S t r e n g t h o f t h e R e l a t i o n s h i p O b s e r v e d

Even though analysis of variance is concerned with comparing means

from independent samples, in practice the samples are usually defined

as representing a multicategory nominal-level variable. For example, as

noted earlier, in comparing three samples of white-collar criminals, we

could define each as one category in a nominal-scale measure of type of

white-collar crime. Similarly, in our example concerning the relationship

between bail amount and race, we spoke about differences among three

samples of offenders: non-Hispanic whites, non-Hispanic blacks, and

Calculating the Within-Group Sum of Squares

NON-HISPANIC NON-HISPANIC HISPANICS

WHITES BLACKS OF ANY RACE

X X X

1,000 14,694,418.89 1,000 61,361,058.89 1,000 862,890,625.00 1,000 14,694,418.89 1,000 61,361,058.89 2,000 805,140,625.00 1,500 11,111,088.89 2,000 46,694,398.89 4,000 695,640,625.00 2,000 8,027,758.89 2,500 40,111,068.89 5,000 643,890,625.00 2,500 5,444,428.89 3,000 34,027,738.89 10,000 415,140,625.00 3,000 3,361,098.89 4,000 23,361,078.89 12,500 319,515,625.00 5,000 27,778.89 5,000 14,694,418.89 25,000 28,890,625.00 7,000 4,694,458.89 10,000 1,361,118.89 25,000 28,890,625.00 7,500 7,111,128.89 12,500 13,444,468.89 25,000 28,890,625.00 7,500 7,111,128.89 20,000 124,694,518.89 40,000 92,640,625.00

10,000 26,694,478.89 20,000 124,694,518.89 65,000 1,198,890,625.00 10,000 26,694,478.89 25,000 261,361,218.89 150,000 14,310,140,625.00

� � 20,367,395,833.33

(Xi � Xc ) 2(Xi � Xc )

2(Xi � Xc ) 2

Table 12.7

328

D E F I N I N G T H E S T R E N G T H O F T H E R E L A T I O N S H I P O B S E R V E D

Hispanics of any race. Nonetheless, these three samples can be seen as

three groups in a nominal-level measure of race of defendant.

ences between the means of the samples. With just three samples, we can

get a pretty good sense of the strength of a relationship using this method.

But even with three samples, it is difficult to summarize the extent of the

relationship observed because we must look at three separate compar-

isons (that between group 1 and group 2, that between group 2 and group

3, and that between group 3 and group 1). With four samples, the number

of comparisons is six; for seven samples, there are 21 comparisons.

Clearly, it is useful, especially as the number of samples grows, to have a

single statistic for establishing the strength of the observed relationship.

A commonly used measure of association for ANOVA is a statistic

called eta (�). Eta relies on the partialing of sums of squares to establish

the relationship, or correlation, between the interval-level variable in

ANOVA and the nominal-level variable. To calculate eta, we simply take

the square root of the ratio of the between sum of squares to the total

sum of squares (see Equation 12.11).

Equation 12.11

Although it might not seem so at first glance, this measure makes good

sense. Understanding eta, however, will be easier if we start with an-

other statistic, eta squared (�2), which is sometimes referred to as the

percent of variance explained (see Equation 12.12).

Equation 12.12

Eta squared is the proportion of the total sum of squares that is accounted

for by the between sum of squares. As previously noted, the between

sum of squares is also defined as the explained sum of squares because it

represents the part of the total variation that is accounted for by the dif-

ferences between the samples. Eta squared thus identifies the proportion

of the total sum of squares that is accounted for by the explained sum of

squares—hence its identification as the percent of variance explained.

The larger the proportion of total variance that is accounted for by the

between sum of squares, the stronger the relationship between the nom-

inal- and interval-level variables being examined. When the means of the

samples are the same, eta squared will be 0. This means that there is no

relationship between the nominal- and interval-level measures the study

is examining. The largest value of eta squared is 1, meaning that all of

�2 � BSS TSS

� � �BSS TSS

identified?” The simplest way to answer this question is to look at the differ-

significant result is “How strong is the overall relationship we have

Accordingly, one question we might ask after finding a statistically

329

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

the variability in the samples is accounted for by the between sum of

squares. In practice, as eta squared increases in value between 0 and 1,

the relationship being examined gets stronger.

The square root of eta squared is a measure more sensitive to small

relationships. For example, a value for eta squared of 0.04 is equal to a

value for eta of 0.20, and a value for eta squared of 0.1 is equivalent to a

value for eta of 0.32, as shown in Table 12.8. In criminal justice, where

the relationships examined are often not very large, measures such as

this one, which allow us to distinguish relatively small values more

clearly, are particularly useful.

Turning to our example concerning age and white-collar crime, we

can see that the differences between the groups account for a good deal

of the variability in the total sum of squares. Taking the between sum of

squares for that example and dividing it by the total sum of squares

gives a value for eta squared of 0.468.

Comparing Eta Squared with Eta

� 2 �

0.00 0.00 0.01 0.10 0.02 0.14 0.03 0.17 0.04 0.20 0.05 0.22 0.10 0.32 0.25 0.50 0.50 0.71 0.75 0.87 1.00 1.00

Table 12.8

W orking It Out

� 0.4678

� 3,405.267

7,279.367

�2 � BSS TSS

� 7,279.367

� 3,405.267 � 3,874.100

TSS � BSS � WSS

BSS � 3,405.267

330

M A K I N G P A I R W I S E C O M P A R I S O N S B E T W E E N T H E G R O U P S S T U D I E D

By taking the square root of this value, we obtain a correlation coeffi-

cient, or eta, of 0.684.

W orking It Out

� 0.6840

� �0.4678

� � �BSS TSS

� �3,405.267

7,279.367

Does an eta of 0.684 signify a large or small relationship? To some ex-

tent, differentiating between “large” and “small” in this context is a value

judgment rather than a statistical decision. We might decide whether a

particular value of eta is large or small based on results from other stud-

ies in other areas of criminal justice or perhaps similar studies that drew

different samples. There is no clear yardstick for making this decision.

One psychologist suggests that any value for eta greater than 0.371 rep-

resents a large effect.7 A moderate-size effect is indicated by a value of

0.243. Using this criterion, we would define the relationship between age

and type of white-collar crime as very strong. However, in this example,

we should be cautious about relying on the results obtained. With small

samples, the values of eta are not considered very reliable.8

M a k i n g P a i r w i s e C o m p a r i s o n s B e t w e e n t h e G r o u p s S t u d i e d

Once you have established through an analysis of variance that there is a

statistically significant difference across the samples studied, you may

want to look at differences between specific pairs of the samples. To do

this, you make comparisons between two sample means at a time. Such

comparisons within an analysis of variance are often called pairwise

comparisons.

7See Jacob Cohen, Statistical Power Analysis for the Behavorial Sciences (Hillsdale, NJ:

Lawrence Erlbaum, 1988), pp. 285–287. 8Once again, there is no universally accepted definition of what is “small.” There will

be little question regarding the validity of your estimate of eta if your samples meet

the 30 cases minimum defined for invoking the central limit theorem. Some statisti-

cians suggest that you will gain relatively reliable estimates even for samples as small

as 10 cases.

331

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

It would seem, at first glance, that you could simply apply the two-

sample t-test discussed in Chapter 11 to test hypotheses related to such

comparisons. However, you are faced with a very important statistical

problem. If you run a number of t-tests at the same time, you are un-

fairly increasing your odds of obtaining a statistically significant finding

along the way. For example, let’s say that you have conducted an analy-

sis of variance comparing seven samples and obtained a statistically sig-

nificant result. You now want to look at the pairwise comparisons to see

which of the specific comparisons are significantly different from one an-

other. There are a total of 21 separate comparisons to make (see Table

12.9). For each test, you set a significance level of 0.05, which means

that you are willing to take a 1 in 20 chance of falsely rejecting the null

hypothesis. Thus, if you run 20 tests, you might expect to get at least one

statistically significant result just by chance.

Here a finding of a significant result could simply be attributed to the

fact that you have run a large number of tests. Accordingly, to be fair, you

should adjust your tests to take into account the change in the probabili-

ties that results from looking at a series of pairwise comparisons. A num-

ber of different tests allow you to do this; many are provided in standard

statistical packages.9 One commonly used test is the honestly significant

difference (HSD) test developed by John Tukey (see Equation 12.13).

Equation 12.13

HSD defines the value of the difference between the pairwise compar-

isons that is required to reject the null hypothesis at a given level of sta-

tistical significance.

HSD � Pcrit � �̂ 2 wg

Nc

The 21 Separate Pairwise Comparisons To Be Made for an Analysis of Variance with Seven Samples (Categories)

SAMPLE 1 2 3 4 5 6 7

1 2 � 3 � � 4 � � � 5 � � � � 6 � � � � � 7 � � � � � �

Table 12.9

9For a discussion of pairwise comparison tests, see A. J. Klockars and G. Sax, Multiple

Comparisons (Quantitative Applications in the Social Science, Vol. 61) (London: Sage,

1986).

332

M A K I N G P A I R W I S E C O M P A R I S O N S B E T W E E N T H E G R O U P S S T U D I E D

For our white-collar crime example, with a conventional 5% signifi-

cance threshold, we first identify the critical value (Pcrit) associated with

that significance threshold by looking at Appendix 6. With three samples

and 27 degrees of freedom in the within sum of squares estimate, the

critical value is about 3.51. We then multiply this value by the square

root of the within-group variance estimate ( ) divided by the number

of cases in each sample (Nc)—in our example, 10.10 Our result is 13.296,

meaning that the absolute value of the difference in mean age between

the pairwise comparisons must be greater than 13.296 to reject the null

hypothesis of no difference (using a 5% significance threshold).

�̂ 2 wg

10Most pairwise comparison tests, including Tukey’s HSD test, require that the sample

sizes of the groups examined be equal. While most statistical software packages pro-

vide adjustments of these tests to account for unequal sample sizes, there is still de-

bate over whether the estimates gained can be relied upon [e.g., see Robert R. J. Sokal

and F. J. Rohlf, Biometry: The Principles and Practice of Statistics in Biological Re-

search, 3rd ed. (New York: W. H. Freeman, 1995), Chap. 9]. Irrespective of this de-

bate, when unequal sample sizes are examined, the adjusted estimates are to be pre-

ferred over the unadjusted estimates.

W orking It Out

� 13.296

� 3.51 �143.49 10

HSD � Pcrit � �̂ 2 wg

Nc

Table 12.10 shows the absolute differences found for the three com-

parisons between means. Two of the three comparisons are statistically

significant at the 5% level—the absolute differences are greater than

13.296 for the difference between bank embezzlers and bribery offend-

ers and for that between bank embezzlers and antitrust offenders. How-

have to conclude that our sample results do not provide persuasive evi-

dence for stating that the mean ages of bribery and antitrust offenders

are different in the larger populations from which these two samples

were drawn.

result just misses the value needed to reject the null hypothesis. We would

ever, for the difference between bribery and antitrust offenders, our

333

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

The comparisons we have made so far have been based on a statis-

tically significant overall result for ANOVA. Should such comparisons

be made if the overall differences across the means are not statistically

significant? In general, it is not a good idea to look for pairwise com-

parisons if the overall analysis of variance is not statistically signifi-

cant. This is a bit like going fishing for a statistically significant result.

However, sometimes one or another of the pairwise comparisons is of

pothesis for a pairwise comparison, it is acceptable to examine it, irre-

spective of the outcomes of the larger test. In such circumstances, it is

also acceptable to use a simple two-sample t-test to examine group

differences.

A N o n p a r a m e t r i c A l t e r n a t i v e : T h e K r u s k a l - W a l l i s T e s t

For studies where you cannot meet the parametric assumptions of the

analysis of variance test, you may want to consider a nonparametric

rank-order test. In performing a rank-order test, you lose some

crucial information because you focus only on the order of scores and

not on the differences in values between them. However, such tests

have the advantage of not requiring assumptions about the population

distribution.

One rank-order test is the Kruskal-Wallis test. As a nonparametric

test of statistical significance, it requires neither a normal distribution

nor equal variances between the groups studied. The test asks simply

whether the distribution of ranked scores in the three groups is what

would be expected under a null hypothesis of no difference. When the

Results of the Pairwise Comparison Tests

OFFENSE 1 OFFENSE 3

BANK OFFENSE 2 ANTITRUST

EMBEZZLEMENT BRIBERY VIOLATION

Offense 1 Bank Embezzlement

Offense 2 16.300* Bribery

Offense 3 25.800* 9.500 Antitrust Violation

*p � 0.5

Table 12.10

develop your analyses. However, if you do start off with a strong hy-

particular interest. Such interest should be determined before you

334

A N O N P A R A M E T R I C A L T E R N A T I V E

number of cases in each group is greater than 5, the sampling distribu-

tion of the Kruskal-Wallis test score, denoted H, is approximately chi-

square.

As an illustration, let’s examine whether this nonparametric test sug-

gests significant differences in terms of age across the white-collar crime

categories of our earlier example.

Assumptions:

Level of Measurement: Ordinal scale.

Population Distribution: No assumption made.

Sampling Method: Independent random sampling (no replacement; sam-

ple is small relative to population).

Sampling Frame: All white-collar offenders convicted of the crimes ex-

amined in seven federal judicial districts over a three-year period.

Hypotheses:

H0: The distribution of ranked scores is identical in the three populations.

H1: The distribution of ranked scores differs across the three populations.

the sample. To obtain this measure, we simply rank the 30 subjects stud-

ied according to age, with the youngest offender given a rank of 1 and

the oldest a rank of 30 (see Table 12.11). In the case of ties, subjects

share a rank. For example, the two subjects aged 29 share the rank of

6.5 (the average of ranks 6 and 7), and the three subjects aged 62 share

the rank of 26 (the average of ranks 25, 26, and 27).

White-Collar Offenders Ranked According to Age

OFFENSE 1 OFFENSE 3

BANK OFFENSE 2 ANTITRUST

EMBEZZLEMENT BRIBERY VIOLATION

Age Rank Age Rank Age Rank

19 1 28 5 35 11.5 21 2 29 6.5 46 16 23 3 32 10 48 17.5 25 4 40 13 53 20 29 6.5 42 14.5 58 22 30 8 48 17.5 61 24 31 9 58 22 62 26 35 11.5 58 22 62 26 42 14.5 64 28 62 26 49 19 68 29 75 30

� � 219� � 167.5� � 78.5

Table 12.11

In this test, we use an ordinal-level measure: the rank order of ages in

335

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

Our null hypothesis is that the distribution of ranked scores in the

three populations is identical. Our research hypothesis is that it is not

identical across the three populations.

Sampling Distribution The sampling distribution H is distributed approx-

imately according to chi-square because the number of cases in each

group is greater than 5. The number of degrees of freedom for the distri-

bution is defined as k � 1, where k refers to the number of samples (or

categories). Because our example involves three samples, the number of

degrees of freedom for the chi-square distribution is 3 � 1, or 2.

Significance Level and Rejection Region Consistent with our earlier

choice of a 0.05 significance threshold, we turn to the 0.05 value with 2

degrees of freedom in the chi-square table (see Appendix 2). The critical

value identified is 5.991.

The Test Statistic The formula for H given in Equation 12.14 looks

complex. However, it is relatively simple to compute if broken into

pieces.

Equation 12.14

There is only one complex term in the equation. It is

This term tells us to take the sum of the ranks in each sample, square it,

and divide it by the number of cases in the sample; then we sum these

values for all the samples.11 The H-score obtained for our problem is

13.038.

� k

c�1

��

Nc

i�1

Ri�2

Nc

H � � 12 N(N � 1)� ��

k

c�1

��

Nc

i�1

Ri�2

Nc � � 3(N � 1)

11Most statistical computing packages provide an alternative calculation that adjusts for

ties. In practice, the differences between using this correction procedure and perform-

ing the unadjusted test are generally small. For our example, where there are a large

number of ties relative to the sample size (14/30), the difference in the observed sig-

nificance level is only 0.0001.

336

C H A P T E R S U M M A R Y

The Decision As with the F-test, our score exceeds the critical value

needed to reject the null hypothesis of no difference. The observed sig-

nificance level of our test is thus less than the criterion significance level

we set at the outset (p � 0.05). From the Kruskal-Wallis test, we can

again conclude that there is a statistically significant relationship between

type of white-collar crime and age of offender. This time, however, we

can have more confidence in our conclusion because the assumptions of

the test are met more strictly.

C h a p t e r S u m m a r y

ANOVA is a parametric test of statistical significance that allows a researcher

to compare means across more than two groups. It takes into account not

only variability between groups but also variability within groups. The larger

the differences between the groups relative to the variability within them,

the more confidence the researcher can have in a conclusion that differ-

ences exist among the population means. Between-group variability is mea-

sured by the between sum of squares (or explained sum of squares).

Within-group variability is measured by the within sum of squares (or un-

explained sum of squares). The total sum of squares is equal to the

sum of the between and within sums of squares. To develop estimates of

population variances, the sums of squares are divided by the appropriate

degrees of freedom. ANOVA requires the following assumptions: interval

scales, normal population distributions, independent random sampling, and

homoscedasticity. The sampling distribution for ANOVA is denoted as F.

W orking It Out

� 13.038

� � 12 930�8,217.95 � 3(31)

� � 12 30(31)��(78.5)2

10 �

(167.5)2

10 �

(219)2

10 � � 3(31)

H � � 12 N(N � 1)���

k

c�1

��

Nc

i�1

Ri�2

Nc � � 3(N � 1)

337

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

The F-value needed to reject the null hypothesis gets smaller as the

within-group degrees of freedom grows. The F-statistic is calculated by

dividing the between-group variance by the within-group variance.

The strength of the relationship observed is measured by the statistic

eta squared, or percent of variance explained. Eta squared is the

ratio of the between sum of squares to the total sum of squares. An eta

squared value of 0 indicates that there is no relationship between the

nominal and interval variables (i.e., the means are the same). An eta

squared value of 1 represents a perfect relationship between the nominal

A researcher who wishes to compare means of pairs of specific sam-

ples within a larger test makes a pairwise comparison. Running a se-

ries of two-sample t-tests, however, unfairly increases the odds of getting

a statistically significant result. The honestly significant difference

(HSD) test is a pairwise comparison test that corrects for this bias.

When the assumptions underlying ANOVA are difficult to meet, the

researcher may choose a nonparametric alternative—the Kruskal-Wallis

test. This test does not require an assumption of normal population dis-

tributions or homoscedasticity. As a rank-order test, however, it does

not use all of the information available from interval-level data.

K e y T e r m s

analysis of variance (ANOVA) A para-

metric test of statistical significance that

assesses whether differences in the means

of several samples (groups) can lead the

researcher to reject the null hypothesis

that the means of the populations from

which the samples are drawn are the

same.

between sum of squares (BSS) A mea-

sure of the variability between samples

(groups). The between sum of squares is

calculated by taking the sum of the

squared deviation of each sample mean

from the grand mean multiplied by the

number of cases in that sample.

correlation A measure of the strength of a

relationship between two variables.

eta A measure of the degree of correlation

between an interval-level and a nominal-

level variable.

eta squared The proportion of the total

sum of squares that is accounted for by the

between sum of squares. Eta squared is

sometimes referred to as the percent of

variance explained.

explained sum of squares (ESS) Another

name for the between sum of squares. The

explained sum of squares is the part of the

total variability that can be explained by

visible differences between the groups.

grand mean The overall mean of every

single case across all of the samples.

honestly significant difference (HSD)

test A parametric test of statistical signifi-

338

and interval variables. The correlation coefficient, or eta, is obtained

by taking the square root of eta squared.

S Y M B O L S A N D F O R M U L A S

S y m b o l s a n d F o r m u l a s

Xi Individual subject or score

Sample or category mean

Grand or overall mean

Nc Number of cases in each sample

k Number of categories or samples

� Correlation coefficient eta

�2 Percent of variance explained; eta squared

Pcrit Critical value for HSD test

Ri Individual rank of score

To calculate the between sum of squares:

BSS � � k

c�1

[Nc(Xc � Xg) 2]

Xg

Xc

cance, adjusted for making pairwise compar-

isons. The HSD test defines the difference

between the pairwise comparisons required

to reject the null hypothesis.

Kruskal-Wallis test A nonparametric test

of statistical significance for multiple

groups, requiring at least an ordinal scale

of measurement.

overall mean See grand mean.

pairwise comparisons Comparisons

made between two sample means ex-

tracted from a larger statistical analysis.

percent of variance explained The pro-

portion of the total sum of squares that is

accounted for by the explained sum of

squares; eta squared.

rank-order test A test of statistical

significance that uses information relating to

the relative order, or rank, of variable scores.

sum of squares The sum of squared devia-

tions of scores from a mean or set of means.

total sum of squares (TSS) A measure of

the total amount of variability across all of

the groups examined. The total sum of

squares is calculated by summing the

squared deviation of each score from the

grand mean.

unexplained sum of squares (USS) An-

other name for the within sum of squares.

The unexplained sum of squares is the part

of the total variability that cannot be ex-

plained by visible differences between the

groups.

within sum of squares (WSS) A measure

of the variability within samples (groups).

The within sum of squares is calculated by

summing the squared deviation of each

score from its sample mean.

339

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

To calculate the within sum of squares:

To calculate the total sum of squares:

To partition the total sum of squares:

Total sum of squares � between sum of squares

� within sum of squares

TSS � BSS � WSS

To estimate between-group variance:

To estimate within-group variance:

To calculate F:

To calculate eta:

To calculate eta squared:

To perform the HSD test:

To perform the Kruskal-Wallis test:

H � � 12 N(N � 1)���

k

c�1

��

Nc

i�1

Ri�2

Nc � � 3(N � 1)

HSD � Pcrit � �̂2 wg

Nc

�2 � BSS TSS

� � �BSS TSS

F � between-group variance

within-group variance

�̂2 wg �

� N

i�1

(Xi � Xc) 2

N � k

�̂2 bg �

� k

c�1

[Nc(Xc � Xg) 2

k � 1

TSS � � N

i�1

(Xi � Xg) 2

WSS � � N

i�1

(Xi � Xc) 2

340

E X E R C I S E S

E x e r c i s e s

12.1 Dawn, a criminal justice researcher, gives 125 pretrial defendants scores based on a questionnaire that assesses their ability to under- stand the court process. The defendants were selected from five sepa- rate counties. Dawn took an independent random sample of 25 defen- dants from each county. The scores for the five populations are normally distributed. Dawn runs an ANOVA test on her results, which produces a test statistic of 3.35.

a. Would Dawn be able to reject her null hypothesis that there is no difference between the populations in their ability to comprehend the court process if she were to set a 5% significance level?

b. Would she be able to reject the null hypothesis using a 1% signifi- cance level?

c. Would your answer to either part a or part b be different if Dawn’s sample had consisted of five equally sized groups of 200 subjects each?

12.2 Random samples of individuals were drawn from three neighborhoods by a policing research foundation to study the level of public support for the local police department. The research foundation constructed a complicated interval-level measure of police support, in which higher values indicated more support. The researchers found the following pattern across the three neighborhoods: The mean level of support in neighborhood A was 3.1 (N � 15); in neighborhood B, it was 5.6 (N � 17); and in neighborhood C, 4.2 (N � 11). The measure of be- tween-group variance was 4.7, and the measure of within-group vari- ance was 1.1.

a. If the significance level is 0.05, can the research foundation con- clude that there are different levels of support for the police depart- ment across neighborhoods? Write out all of the steps of a test of statistical significance, including any violations of assumptions.

b. What if the significance level is 0.01?

12.3 Random sampling of individuals with four different majors at a univer- sity found the following grade point averages (GPAs) for the four groups:

Major A: GPA � 3.23 (N � 178)

Major B: GPA � 2.76 (N � 64)

Major C: GPA � 2.18 (N � 99)

Major D: GPA � 3.54 (N � 121)

If the between-group variance is 5.7 and the within-group variance is 1.5, are the GPAs different for the four majors? Use a significance level of 0.01, and write out all of the steps of a test of statistical significance, including any violations of assumptions.

341

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

12.4 Random sampling offenders convicted of minor drug possession in Border County found the average jail sentence for white offenders to be 86 days (N � 15), for African American offenders to be 99 days (N � 10), and for Hispanic offenders to be 72 days (N � 7). Further analysis of jail sentence lengths by race found the between sum of squares to be 250 and the within sum of squares to be 1,300. Are the jail sentence lengths significantly different across race categories?

a. Use a significance level of 0.05. Write out all of the steps of a test of statistical significance, including any violations of assumptions.

b. Would the conclusion be any different if the significance level had been set at 0.01?

12.5 Listed below is a set of data identifying previous convictions for any

Robbery Rape Murder Drug Dealing

1 1 0 5

0 1 0 3

2 1 0 7

6 0 6 4

4 0 2 8

5 2 7 0

3 2 1 6

1 1 4 2

5 0 2 1

3 2 3 4

Calculate the following values:

a.

b. df for between-group variance

c. df for within-group variance

d. the four values of

e. the total sum of squares

f. the between sum of squares

g. the within sum of squares

12.6 Convicted drug dealers held in Grimsville Prison are placed in cell block A, B, or C according to their city of origin. Danny (who has little knowledge of statistics) was once an inmate in the prison. Now re-

Xc

Xg

342

murder, and drug dealing. offense of 40 inmates serving prison sentences for robbery, rape,

E X E R C I S E S

leased, he still bears a grudge against the prison authorities. Danny wishes to make up a series of statistics to show that the convicts in the various blocks are treated differently. According to his fictitious sam- ple, the mean number of hours of exercise per week given to the in- mates is 10 hours for block A offenders, 20 hours for block B offend- ers, and 30 hours for block C offenders. Shown below are two fictitious sets of results.

Fictitious study 1:

Block A Block B Block C

9 21 30

10 19 29

9 20 31

11 19 29

11 21 31

Fictitious study 2:

Block A Block B Block C

18 16 37

16 18 36

10 2 7

2 31 41

4 33 29

a. From simply looking at the numbers, without running any statistical tests, which of the two fictitious studies would you expect to provide stronger backing for Danny’s claim? Explain your answer.

b. Calculate the between sum of squares and the within sum of squares for study 1.

c. Calculate the between sum of squares and the within sum of squares for study 2.

d. Calculate the value of eta for each study. How do you account for the difference?

12.7 A researcher takes three independent random samples of young pick- pockets and asks them how old they were when they first committed the offense. The researcher wishes to determine whether there are any differences among the three populations from which the samples were drawn—those who have no siblings, those who have one or two sib- lings, and those with three or more siblings.

X � 30X � 20X � 10

X � 30X � 20X � 10

343

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S

Age at first theft:

0 Siblings 1 or 2 Siblings 3� Siblings

10 14 15

8 15 15

16 15 10

14 13 13

7 12 16

8 9 15

a. Show that the total sum of squares is equal to the between sum of squares plus the within sum of squares.

b. What is the value of eta?

c. Can the researcher reject the null hypothesis on the basis of the dif- ferences observed? Run an F-test using a 5% significance level. Re- member to outline all of the steps of a test of statistical significance, including any violations of assumptions.

12.8 Using independent random sampling, Sophie draws samples from three different populations: psychologists, police officers, and factory workers. She gives each subject a hypothetical case study of a drug dealer who has been found guilty and awaits sentencing. The subjects are then asked to suggest how many years the drug dealer should serve in prison. The results are presented below:

Psychologists Police Factory Workers

2 3 5

1 2 6

0 3 4

0 3 8

1 4 7

2.5 1 7

2 1.5 6

1.5 0 2

4 0.5 3

1 7 2

a. Can Sophie conclude that the three populations are different in terms of their attitudes toward punishing convicted drug dealers? Run an F-test using a 5% significance level. Remember to outline all of the steps of a test of statistical significance, including any viola- tions of assumptions.

b. Would Sophie’s decision be any different if she chose a 1% or a 0.1% level of significance?

344

C O M P U T E R E X E R C I S E S

c. Calculate the value of eta for the results above. Is the relationship a strong one?

12.9 For the data in Exercise 12.4, run a Kruskal-Wallis test using a 5% level of statistical significance. Remember to outline all of the steps of a test of statistical significance, including any violations of assump- tions. Are you able to reject the null hypothesis?

345

C o m p u t e r E x e r c i s e s

In the following discussion for estimating one-way analysis of variance models

in SPSS and Stata, you may find it useful to open the data file presented in Table

12.4 in the text (ex12_1.sav or ex12_1.dta) or the corresponding syntax file

(Chapter_12.sps or Chapter_12.do).

SPSS

ANOVA

To compute a one-way ANOVA model in SPSS, you will use the ONEWAY

command:

The output will present the ANOVA table (discussed in the box on p. 325). If

you run this command using the data from Table 12.4 (i.e., ex12_1.sav), it would

be

Tukey’s HSD

The ONEWAY command will perform a wide range of additional calculations

on a data file, including Tukey’s HSD statistic. To obtain Tukey’s HSD statistic,

execute the ONEWAY command with the /POSTHOC = TUKEY option:

The value of the F-test reported by SPSS matches that reported on page 321.

ANOVA results can also be obtained with the MEANS command:

where the /CELLS option will generate a table of results that lists the mean,

number of cases, and standard deviation for the variable(s) of interest by group

category. The /STATISTICS option with ANOVA will generate the ANOVA

table produced by ONEWAY (and discussed on p. 325). The ANOVA option

will also compute the value of Eta and Eta-squared, which are not available in

the ONEWAY command.

ONEWAY variable_name(s) BY grouping_variable.

ONEWAY age BY crime.

MEANS TABLES = variable_name(s) BY grouping_variable

/CELLS MEAN COUNT STDDEV

/STATISTICS ANOVA.

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S346

The output presented will contain the ANOVA table that you have already seen

and an additional table that lists all possible comparisons of group means.

If you run this command using the data from Table 12.4, the command

would be

Executing this command will reproduce the results from Table 12.10. The three

major rows in this table represent the three samples of offenders. Within each

major row are two smaller rows that represent contrasts between the groups. So,

for example, in the first major row (the embezzlement sample), there are calcula-

tions for the mean of this group minus the mean of the second group (the bribery

sample) in the first line, followed by calculations for the mean of the first group

minus the mean of the third group (the antitrust sample) in the second line. The

values for Tukey’s HSD reported in the first major row match those reported

in Table 12.10. In the second major row (the bribery sample), the second line

represents the difference between this group’s mean and the mean for the third

group (the antitrust sample), and the value for Tukey’s HSD again matches that

reported in Table 12.10.

Sometimes the labels in the table of results for Tukey’s HSD can be

confusing, so you will need to pay attention to the lines you are working with.

Keep in mind that the variable listed in the first column of each major row has

the mean for every other group (listed in the second column) subtracted

from its mean.

Kruskal–Wallis Test

The Kruskal–Wallis test is available in SPSS through the use of the NPAR

command with the /K-W option:

ONEWAY variable_name BY grouping_variable

/POSTHOC = TUKEY.

ONEWAY age BY crime

/POSTHOC = TUKEY.

Note the parentheses following the name of the grouping variable. The values to

be included here are the values representing the minimum (minrange) and maxi-

mum (maxrange) for the grouping variable. Returning to the use of the data in

Table 12.4, we would enter the following command:

where 1 and 3 represent the bounds of the grouping variable.

NPAR TESTS

/K-W = variable_name BY grouping_variable (minrange, maxrange).

NPAR TESTS

/K-W = age BY crime (1,3).

C O M P U T E R E X E R C I S E S 347

The output generated by the NPAR TESTS command will contain two small

tables. The first table lists each group or category and its average rank. The sec-

ond table presents the results for the Kruskal–Wallis test. Note that the value of

the test statistic reported by SPSS differs slightly from that reported in the text

(SPSS: 13.073; text: 13.038). The reason for this difference was noted in footnote

11: SPSS corrects the calculation of the test statistic by adjusting for ties in rank,

and the formula in the text does not make such a correction.

Stata

ANOVA

Similar to many other statistical packages, there are multiple ways of obtaining

one-way ANOVA results in Stata. The two most direct commands are oneway

and anova. The oneway command provides sufficient information for most

purposes. If you are interested in obtaining Tukey’s HSD, however, you will need

to use the anova command. To compute a one-way ANOVA model in Stata,

you will use the oneway command:

The output will present the ANOVA table. If the tabulate option is included

on the command line, Stata will generate group means, standard deviations,

and counts for the number of cases on the variable of interest. If you omit the

tabulate option, oneway will simply generate an ANOVA table.

To run this command using the data from Table 12.4 (i.e., ex12_1.dta) enter

the following:

The value of the F-test reported by SPSS matches that reported on page 321.

The format for the anova command is identical:

Not surprisingly, the output from this command will be the ANOVA table.

Tukey’s HSD

Tukey’s HSD statistic is not an option in any of the Stata ANOVA commands.

To obtain Tukey’s HSD statistic, we must first install a pair of user-written pro-

cedures that will use the results from an anova command and then compute

Tukey’s HSD.

To install these user-written procedures, enter the following two commands

(one time only for each one):

oneway variable_name grouping_variable, tabulate

oneway age crime, tabulate

anova variable_name grouping_variable

net install tukeyhsd, from (http://www.ats.ucla.edu/stat/

stata/ado/analysis)

net install sg101, from (http://www.stata.com/stb/stb47)

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S348

The tukeyhsd command will be the procedure we use, but its calculations are

based on the other procedure installed (sg101).

To obtain Tukey’s HSD, we first run a one-way ANOVA using the anova

command, followed by the tukeyhsd command:

For the data in Table 12.4 that we have been working with, the two commands

would be

The output from the running of the tukeyhsd command will show only three

comparisons, 1 v. 2, 1 v. 3, and 2 v. 3, making the interpretation of the output

somewhat simpler than in SPSS, where all possible comparisons are presented.

The results are identical to the results from Table 12.10.

Kruskal–Wallis Test

The Kruskal–Wallis test is available in Stata with the kwallis command:

Returning to the use of the data in Table 12.4, we would run the following

command:

The output generated by the kwallis command will contain one small table

listing the group, number of cases in that group, and rank sum. Below the table,

there are chi-square test statistics for both of the methods we have noted: with

and without a correction for ties. Consequently, Stata reproduces the value in the

text (13.038) that does not correct for ties and the value also estimated by SPSS

(13.073) that does correct for ties.

Problems

1. Input the data from Table 12.6 as two variables: bail amount and race

(use 1 = non-Hispanic white, 2 = non-Hispanic African American, and

3 = Hispanic of any race).

a. Reproduce the ANOVA results in the text.

b. Compute the HSD for these data. What can you conclude about the

pairwise comparisons across race categories?

c. Perform the Kruskal–Wallis test. How do the results from the Kruskal–

Wallis test compare to the ANOVA results in part a? Do the results

from the Kruskal–Wallis test alter the conclusions obtained through the

use of ANOVA?

anova variable_name grouping_variable

tukeyhsd grouping_variable

anova age crime

tukeyhsd crime

kwallis variable_name, by(grouping_variable)

kwallis age, by(crime)

C O M P U T E R E X E R C I S E S 349

2. Enter the data from Exercise 12.5. Use one of the ANOVA commands to

test for differences in group means.

a. Write out the assumptions of the test, critical value of the test statistic,

value of the computed test statistic, and decision regarding the null

hypothesis.

b. Compute the HSD for each of the group comparisons. What can you

conclude about pairwise comparisons for each group?

c. Use the Kruskal–Wallis test to test for differences in rank order across

groups. Write out the assumptions of the test, critical value of the test

statistic, value of the computed test statistic, and decision regarding the

null hypothesis.

3. Enter the data from Exercise 12.6. Use one of the ANOVA commands to

test for differences in group means.

a. Write out the assumptions of the test, critical value of the test statistic,

value of the computed test statistic, and decision regarding the null

hypothesis.

b. Compute the HSD for each of the group comparisons. What can you

conclude about pairwise comparisons for each group?

c. Use the Kruskal–Wallis test to test for differences in rank order across

groups. Write out the assumptions of the test, critical value of the test

statistic, value of the computed test statistic, and decision regarding the

null hypothesis.

4. Open the NYS data file (nys_1.sav, nys_1_student.sav, or nys_1.dta). Carry

out the following statistical analyses for each of the research questions in

parts a through e:

a. Does the mean number of thefts valued at $5–$50 vary across academic

ability?

b. Does the mean number of times drunk vary across race?

c. Does the level of marijuana use vary across amount of contact with

delinquent peers?

d. Does the mean number of attacks on other students vary across victim-

ization experience?

e. Does the mean number of times cheating on schoolwork vary across

grade point average?

Use ANOVA to test for differences in group means. For each

hypothesis test, write out the assumptions of the test, critical value

of the test statistic, value of the computed test statistic, and decision

regarding the null hypothesis.

C H A P T E R T W E L V E : M O R E T H A N T W O S A M P L E S350

Compute the HSD for each of the pairwise comparisons. What can

you conclude about pairwise comparisons for each research question?

Use the Kruskal–Wallis test to test for differences in rank order

across groups. For each hypothesis test, write out the assumptions

of the test, critical value of the test statistic, value of the computed

test statistic, and decision regarding the null hypothesis.

5. Open the Pennsylvania Sentencing data file (pcs_ 98.sav or pcs_98.dta).

Carry out the following statistical analyses for each of the research

questions in parts a through c:

a. Does the length of incarceration sentence vary across race?

b. Does the length of incarceration sentence vary across method of conviction?

c. Does the length of incarceration sentence vary by type of conviction offense?

� Use ANOVA to test for differences in group means. For each

hypothesis test, write out the assumptions of the test, critical value

of the test statistic, value of the computed test statistic, and decision

regarding the null hypothesis.

� Compute the HSD for each of the pairwise comparisons. What can

you conclude about pairwise comparisons for each research question?

� Use the Kruskal–Wallis test to test for differences in rank order

across groups. For each hypothesis test, write out the assumptions

of the test, critical value of the test statistic, value of the computed

test statistic, and decision regarding the null hypothesis.

Measures of Association

for Nominal and Ordinal Variables

What Do Nominal Measures of Association Describe?

What Do Ordinal Measures of Association Describe?

C h a p t e r t h i r t e e n

M e a s u r i n g t h e s t r e n g t h o f a r e l a t i o n s h i p

M e a s u r e s o f a s s o c i a t i o n f o r n o m i n a l v a r i a b l e s

M e a s u r e s o f a s s o c i a t i o n f o r o r d i n a l v a r i a b l e s

When are These Measures Used?

What is the Test of Statistical Significance for These Measures?

When are These Measures Used?

What is the Test of Statistical Significance for These Measures?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_13,

© Springer Science+Business Media New York 2014

CHAPTER 12 INTRODUCED eta (�) and the more general concept of mea-

sures of association. Eta is a descriptive statistic that allows us to define

how strongly the categorical variable or sample in an analysis of variance

is related to the interval-level variable or trait we examined across the

samples. But there are many other useful measures of association that

allow us to define relationships among variables. Over the next few

chapters, we will focus on some of these that are particularly useful in

studying criminal justice. We will still be concerned with statistical signif-

icance in these chapters, but we will examine not only whether a mea-

sure is statistically significant but also how strong the relationship is.

In this chapter, our focus is on nominal- and ordinal-level measures of

association. We begin with a discussion of why it is important to distin-

guish between statistical significance and strength of association. While

statistical significance can tell us whether we can make reliable state-

ments about differences in a population from observations made from

samples, it does not define the size of the relationship observed. It is im-

portant to define the strength of the relationship between variables being

D i s t i n g u i s h i n g S t a t i s t i c a l S i g n i f i c a n c e a n d S t r e n g t h o f R e l a t i o n s h i p : T h e E x a m p l e o f t h e C h i - S q u a r e S t a t i s t i c

In Chapter 9, we explored the chi-square statistic as a way to determine

whether there was a statistically significant relationship between two

nominal-level variables. The chi-square statistic is useful as a way of test-

ing for such a relationship, but it is not meant to provide a measure of the

strength of the relationship between the variables. It is tempting to look at

the value of the chi-square statistic and the observed significance level as-

sociated with a particular chi-square value and infer from these statistics

the strength of the relationship between the two variables. If we follow

such an approach, however, we run the risk of an interpretive error.

results that are statistically significant are also substantively important.

examined because that puts us in a better position to decide whether

352

S T A T I S T I C A L S I G N I F I C A N C E A N D S T R E N G T H

The problem with using the chi-square statistic—or outcomes of other

tests of statistical significance—in this way is that the size of the test sta-

tistic is influenced not only by the nature of the relationship observed

but also by the number of cases in the samples examined. As we have

noted a number of times in the text, this makes good sense. Larger sam-

ples, all else being equal, are likely to be more trustworthy. Just as we

feel more confident in drawing inferences from a sample of 10 or 20

coin tosses than from a sample of 2 or 3 tosses, our confidence in mak-

ing a decision about the null hypothesis grows as the sizes of the sam-

ples examined using a chi-square statistic increase.

The following example will help to illustrate this problem. Suppose we

have a sample of 200 cases that cross-tabulate experimental condition with

an outcome measure, as shown in Table 13.1. We see that 60% of those in

the treatment group have an outcome classified as a success, while only

40% of those in the control group have an outcome classified as a success.

Our calculated value of chi-square for these data is 8.00 with df � 1, which

has an observed significance level less than 0.01 (see Appendix 2). See

Table 13.2 for detailed calculations for obtaining the chi-square statistic.

Observed Frequencies ( fo) and Expected Frequencies ( fe) for Two Outcomes of an Experimental Condition with 200 Cases

OUTCOME

EXPERIMENTAL

CONDITION Failure Success Total

Treatment f � 40 fo � 60 100 fe � 50 fe � 50

Control fo � 60 fo � 40 100 fe � 50 fe � 50

Total 100 100 200

Table 13.1

Calculations for Obtaining Chi-Square Statistic for the Example in Table 13.1

EXPERIMENTAL

CONDITION OUTCOME f fe fo � fe (fo � fe)2

Treatment Failure 40 50 �10 100 2 Treatment Success 60 50 10 100 2 Control Failure 60 50 10 100 2 Control Success 40 50 �10 100 2

� � 8.0

(fo � fe)2

fe

Table 13.2

353

o

o

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

Without changing the proportional distribution of cases for this exam-

ple—keeping success at 60% for the treatment group and 40% for the

control group—suppose we multiply the number of cases by 10. We

now have 2,000 total observations, as shown in Table 13.3, but the rela-

tionship between experimental condition and outcome is the same. Our

calculated chi-square statistic, however, now has a value of 80.00 (see

Table 13.4) with df � 1, and the observed significance level is less than

0.0001. So, simply by increasing the size of the sample, we increase the

value of chi-square and decrease the corresponding observed signifi-

cance level.

This feature of the chi-square statistic applies to all tests of statistical

significance. Irrespective of the observed relationship between measures,

as the sample size increases, the observed significance level associated

The rule does not raise any new questions regarding the meaning of sta-

tistical significance. It simply reminds us that, all else being equal, we can

be more confident in making statistical inferences from larger samples. It

also emphasizes the importance of distinguishing between statistical sig-

nificance and the size or strength of a relationship between variables.

To allow researchers to define the strength of a relationship among

nominal-level or ordinal-level variables, statisticians have developed a vari-

ety of measures of association. Some of these measures are based on the

value of the chi-square statistic; others are based on unique transforma-

tions of the counts or distributions of cases within a table. All the measures

of association that we discuss share a standardized scale: A value of 0 is

interpreted as no relationship, and a value of 1.0 (or, in the case of ordinal

scales, �1 or �1) is interpreted as a perfect relationship between the two

variables. The discussion that follows describes some of the more fre-

quently used measures of association for nominal and ordinal variables.

Observed Frequencies ( fo) and Expected Frequencies ( fe) for Two Outcomes of an Experimental Condition with 2,000 Cases

OUTCOME

EXPERIMENTAL

CONDITION Failure Success Total

Treatment fo � 400 fo � 600 1,000 fe � 500 fe � 500

Control fo � 600 fo � 400 1,000 fe � 500 fe � 500

Total 1,000 1,000 2,000

Table 13.3

relationship between statistical significance and sample size will be exam-

with that relationship will also increase. This simple rule regarding the

354

ined in more detail in the discussion of statistical power in Chapter 21.

M E A S U R E S O F A S S O C I A T I O N F O R N O M I N A L V A R I A B L E S

M e a s u r e s o f A s s o c i a t i o n f o r N o m i n a l V a r i a b l e s

Measures of Association Based on the Chi-Square Statistic

The preceding example illustrated how the chi-square statistic is affected

by sample size. With a 2 � 2 table (i.e., two rows and two columns),

one straightforward way of measuring the strength of a relationship be-

tween two variables that adjusts for the influence of sample size is to

transform the value of the chi-square statistic by adjusting for the total

number of observations. One measure of association that does this is

phi (�). Phi is obtained simply by dividing the chi-square statistic by the

total number of observations (N) and taking the square root of this value

(see Equation 13.1).

Equation 13.1

Phi will have a value of 0 if the value of the chi-square statistic is 0 and

there is no relationship between the two variables. Phi will have a value

of 1 if the chi-square statistic takes on a value equal to the sample size,

which can occur only when there is a perfect relationship between two

categorical variables. It is important to note that phi is appropriate only

eliminating the possibility of any kind of meaningful interpretation.

Consider the two chi-square statistics that we calculated above for the

data in Tables 13.1 and 13.3: 8.00 and 80.00, respectively. If we insert

these values for chi-square and the sample size, we find that the value of

phi for both tables is 0.20.

� � ��2

N

Calculations for Obtaining Chi-Square Statistic for the Example in Table 13.3

EXPERIMENTAL

CONDITION OUTCOME fo fe fo � fe (fo � fe)2

Treatment Failure 400 500 �100 10,000 20 Treatment Success 600 500 100 10,000 20 Control Failure 600 500 100 10,000 20 Control Success 400 500 �100 10,000 20

� � 80.0

(fo � fe)2

fe

Table 13.4

355

exceeds two, then it is possible for phi to take on values greater than 1.0,

for analyses that use a 2 � 2 table. If the number of rows or columns

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

We now have a measure of association that is not influenced by sample

size. For both of our examples, in which the proportion of cases in each

group was similar, we have the same phi statistic. However, is the relation-

ship large or small? As noted in Chapter 12, defining “large” and “small” is

a matter of judgment and not statistics. In judging the importance of a re-

sult, researchers can compare it with other findings from prior studies. Or

they may examine the importance of the policy implications that could be

drawn from the result. For example, a very small change in rates of heart

attacks in the population could save many lives, and thus a small relation-

ship may still be important. According to a standard measure of effect size

suggested by Jacob Cohen, a phi of 0.10 is considered to indicate a small

relationship, one of 0.30 a medium relationship, and one of 0.50 a large

relationship.1

Our examples suggest why we might be misled if we used the chi-

square statistic and its corresponding significance level as an indicator of

the strength of the relationship between two variables. If we had tried to

infer the strength of the relationship between experimental condition

and outcome from the value of the chi-square statistic, we would have

been tempted to conclude that Table 13.3 showed a stronger relationship

than Table 13.1. However, once we take into account the size of the

sample, we see that the two tables reflect the same relationship between

the two variables. The data in Table 13.3 lead to a higher observed sig-

nificance level because the samples examined are larger. However, the

strength of the relationship observed in the two tables is the same.

For tables with more than two rows or two columns, we cannot use

phi. Instead, we use a measure of association known as Cramer’s V,

which is also based on the value of the chi-square statistic but makes an

adjustment for the number of categories in each variable. Equation 13.2

presents the formula for calculating Cramer’s V.

Equation 13.2

In Equation 13.2, the chi-square statistic (�2) is divided by the product of

the total number of observations (N ), and the smaller of two numbers,

V � � �2

N � min(r � 1, c � 1)

1See Jacob Cohen, Statistical Power Analysis for the Behavioral Sciences (Hillsdale, NJ:

Lawrence Erlbaum, 1988), pp. 215–271.

W orking It Out

and � � �80.00 2,000

� 0.20� � �8.00 200

� 0.20

356

M E A S U R E S O F A S S O C I A T I O N F O R N O M I N A L V A R I A B L E S

r � 1 or c � 1 (i.e., the minimum of these two values), where r is the

number of rows in the table and c is the number of columns. For exam-

ple, if we had a table with two rows and three columns, we would have

r � 1 � 2 � 1 � 1 and c � 1 � 3 � 1 � 2. The value for r � 1 is the

smaller of these two numbers, so we would use that value (1) for

min(r � 1, c � 1) in the denominator of the formula. If we were work-

ing with a larger table with, say, five rows and four columns, we would

have r � 1 � 5 � 1 � 4 and c � 1 � 4 � 1 � 3. Since 3 is less than 4,

we would use the value 3 for min(r � 1, c � 1) in the denominator.

Let’s consider an example. Table 13.5 reproduces the data from Table

9.9 on cell-block assignment and race of prisoner. Recall from Chapter 9

that the chi-square statistic for this cross-tabulation was 88.2895, and

with df � 6, the observed significance level was less than 0.001. We can

use the data in this table to illustrate the calculation of V. The table has

seven rows (r � 7) and two columns (c � 2), meaning that r � 1 � 7 �

1 � 6 and c � 1 � 2 � 1 � 1. The smaller of these two values is 1,

which we substitute for min(r � 1, c � 1) in the denominator of the for-

mula for V. After inserting the other values into Equation 13.2, we find

that V � 0.2708.

Assignment of Non-Hispanic White and Nonwhite Prisoners in Seven Prison Cell Blocks

NON-HISPANIC

CELL BLOCK WHITES NONWHITES

C 48 208 1,256 D 17 37 1, 54 E 28 84 1,112 F 32 79 1,111 G 37 266 1,303 H 34 22 1, 56 I 44 268 1,312

240 964 1,204

Table 13.5

W orking It Out

� 0.2708

� � 88.2895

(1,204)(1)

V � � �2

N � min(r � 1, c � 1)

357

Column total

ROW TOTAL

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

Proportional Reduction in Error Measures: Tau and Lambda

Some measures of association that are appropriate for nominal-level vari-

ables are based on the idea of proportional reduction in error, or

PRE. Such measures indicate how much knowledge of one variable

helps to reduce the error we make in defining the values of a second

variable. If we make about the same number of errors when we know

the value of the first variable as when we don’t, then we can conclude

that the PRE is low and the variables are not strongly related. However,

if knowledge of one variable helps us to develop much better predic-

tions of the second variable, then we have a high PRE and the variables

may be assumed to be strongly related.

Two of the more common measures of association between nominal

variables, Goodman and Kruskal’s tau (�) and lambda (�) are both

PRE measures. Both of these measures require that we identify at the

outset which variable is the dependent variable and which variable is

the independent variable. A dependent variable is an outcome vari-

able—it represents the phenomenon that we are interested in explaining.

It is “dependent” on other variables, meaning that it is influenced—or we

expect it to be influenced—by other variables. Any variable that affects,

or influences, the dependent variable is referred to as an independent

variable. The values of Goodman and Kruskal’s tau (�) and lambda ( )

For most research projects, a body of prior research and/or theory

will indicate which variables are dependent and which are independent.

For example, for the study in Table 13.1, the independent variable is the

experimental condition: the treatment or the control group. Whether the

person participated in the treatment or the control group is generally

theorized to influence outcome success or failure, which is the depen-

dent variable. In other words, the experiment tests whether success or

failure is due, at least in part, to participation in some kind of treatment.

PRE measures of association, such as tau and lambda, require the use

of two decision rules. The first decision rule—the naive decision rule—

involves making guesses about the value of the dependent variable with-

out using any information about the independent variable. The second

decision rule—the informed decision rule—involves using information

about how the cases are distributed within levels or categories of the in-

dependent variable. The question becomes “Can we make better predic-

tions about the value of the dependent variable by using information

about the independent variable?” Will the informed decision rule provide

358

relationship between cell-block assignment and race of prisoner.

The value of Cramer’s V may be interpreted in the same way as that

of phi. Accordingly, a value for V of 0.2708 is suggestive of a moderate

dependent variable and which as the independent variable.

will generally differ depending on which variable is identified as the

M E A S U R E S O F A S S O C I A T I O N F O R N O M I N A L V A R I A B L E S

better predictions than the naive decision rule? PRE measures of associa-

tion have a value of 0 when there is no relationship between the two

variables and a value of 1 when there is a perfect relationship between

the two variables. Table 13.6 presents two hypothetical distributions il-

lustrating PRE measures showing no relationship (part a) and a perfect

relationship (part b). In part a, we see that knowledge of one variable

does not help us make predictions about the second variable, since the

A key advantage to PRE measures of association is the interpretation of

values between 0 and 1. Any value greater than 0 may be interpreted as a

proportionate reduction in error achieved by using information on the in-

dependent variable. Alternatively, we can multiply the PRE measure by

100 and interpret the value as the percent reduction in errors. For exam-

For an illustration of the calculation of tau and lambda, consider the

data presented in Table 13.7. These data come from responses to a sur-

Hypothetical Distribution of 200 Cases for Two Nominal Variables

(a) PRE Measure of Association � 0.0

VARIABLE 2

VARIABLE 1 Category 1 Category 2

Category 1 50 50 100 Category 2 50 50 100

100 100 200

(b) PRE Measure of Association � 1.0

VARIABLE 2

VARIABLE 1 Category 1 Category 2

Category 1 0 100 100 Category 2 100 0 100

100 100 200

Table 13.6

2For a description of the study, see Chester L. Britt, “Health Consequences of Criminal

Victimization,” International Review of Victimology, 8 (2001): 63–73.

359

errors of 50% when information about the independent variable is used.

ple, a PRE measure of 0.50 indicates a percent reduction in prediction

Row total

Column total

Row total

Column total

knowledge of one variable determines, without error, the value of the

cases are evenly distributed across all possible cells of the table (e.g., of

the 100 cases in Category 1 of Variable 1, exactly 50 cases each fall into

2 of Variable 2).

Category 1 and 2 of Variable 2). In the perfect relationship shown in part b,

second variable (e.g., all cases in Category 1 of Variable 1 fall into Category

vey by adult residents of the state of Illinois.2 Respondents who reported

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

questions about the most recent event. Two of these questions addressed

the relationship between the victim and the offender and the location of

the assault. Here we have classified the victim-offender relationship into

four categories: stranger, acquaintance/friend, partner (includes spouse

and boyfriend or girlfriend), and relative. Location of the assault is also

classified into four categories: home, neighborhood, work, and someplace

else. For this analysis, we assume that the victim-offender relationship is

the independent variable and the location of the assault is the dependent

variable. Our research question is “What is the strength of the relationship

between victim-offender relationship and location of an assault?”

Goodman and Kruskal’s tau uses information about the marginal dis-

tributions of the two variables to test whether knowledge of the inde-

pendent variable reduces prediction errors for the dependent variable.

The first step in computing this statistic is to ask how many errors we

would expect to make, on average, if we did not have knowledge about

the victim-offender relationship. This is our naive decision rule, where

we are effectively trying to guess what category of the dependent vari-

able an observation might belong to, without using any information

about the independent variable. For our example, we begin by looking

at the column totals in Table 13.7, which reflect the categories of the de-

pendent variable. Of the 410 assaults, we see that 139 occurred in the

home, 77 in the neighborhood, 30 at work, and 164 someplace else. We

use these column totals to help us determine the average number of er-

rors we would expect to make if we assigned cases without any informa-

tion about the victim-offender relationship.

Let’s begin with assaults in the home. Of the 410 total assaults, 139

belong in the assaulted-in-the-home category, while 271 do not belong

in this category (i.e., the assault occurred elsewhere). Proportionally,

0.6610 (271 of 410) of the cases do not belong in the assaulted-in-the-

Data on Victim-Offender Relationship and Location of Assault

VICTIM- LOCATION OF ASSAULT

OFFENDER

RELATIONSHIP Home Neighborhood Work Someplace Else

Stranger 10 49 18 89 166 Acquaintance/friend 21 22 7 46 96 Partner 77 5 3 19 104 Relative 31 1 2 10 44

139 77 30 164 410

Table 13.7

360

assaulted-in-the-home category, we would expect 0.6610 of these 139

home category. If we randomly assigned 139 of the 410 cases to the

that they had experienced an assault were asked a series of follow-up

Row total

Column total

M E A S U R E S O F A S S O C I A T I O N F O R N O M I N A L V A R I A B L E S

incorrectly—the number of prediction errors—we multiply the propor-

tion of cases not in the category by the number of cases assigned to that

category. For assaulted in the home, this is (0.6610 � 139) � 92. The

value 92 represents the number of prediction errors we would expect to

make, on average, in assigning cases to the assaulted-in-the-home cate-

gory without any knowledge of the victim-offender relationship.3

Turning to assaults in the neighborhood, we see that 77 cases belong in

this category, and the remaining 333 do not belong in this category. As a

proportion, 0.8122 of the cases (333 of 410) do not belong in the as-

saulted-in-the-neighborhood category. This means that we would expect

to make 0.8122 � 77 � 63 prediction errors, on average, in assigning

cases to this category without any knowledge of the victim-offender rela-

tionship. For assaults at work, 30 cases belong in this category and 380 do

not, meaning that we would expect to make (380/410) � 30 � 28 predic-

tion errors, on average, in assigning cases to the assaults-at-work category

without any information about the victim-offender relationship. There are

164 cases that belong to the assaulted-someplace-else category, meaning

that 246 cases do not belong in this category. We would expect to make

(246/410) � 164 � 98 prediction errors, on average, in assigning cases to

this category without any information about the victim-offender relation-

ship. To determine the total number of prediction errors we would make

without any knowledge of the victim-offender relationship, we add these

four values together: 92 � 63 � 28 � 98 � 281 total prediction errors.

If we then use information about the victim-offender relationship—

whether the victim and offender were strangers, acquaintances/friends,

partners, or relatives—we can test whether this information improves our

ability to predict the location of the assault. This reflects the use of our

informed decision rule: Does our assignment of cases to categories of

the dependent variable improve when we use information about the cat-

egory of the independent variable? In other words, does knowing the

category of the independent variable (victim-offender relationship) re-

duce the number of prediction errors we make about the category of the

dependent variable (location of assault)? To the extent that the indepen-

dent variable has a relationship with the dependent variable, the number

of prediction errors should decrease.

The logic behind calculating the prediction errors is the same as before,

except that we focus on the row totals in the table, rather than the total

number of cases in each category of the dependent variable. We start with

the first category of the independent variable (i.e., the first row of Table

13.7) and note that 166 cases involved offenders who were strangers to

3For all calculations of prediction errors, we have rounded the result to the nearest

integer.

361

cases to be assigned incorrectly. To obtain the number of cases assigned

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

the victim. In a process similar to our earlier analysis, we begin by noting

the placement of cases within this row: 10 assaults occurred at home, 49 in

the neighborhood, 18 at work, and 89 someplace else. Starting with the as-

signment of cases to assaulted-in-the-home, we note that 10 cases belong

in this category and 156 do not belong in this category. As a proportion,

0.9398 of the cases (156 of 166) do not belong in the assaulted-in-the-

gory, we would expect to make 0.9398 � 10 � 9 prediction errors, on

average. Turning to the assaulted-in-the-neighborhood category, we note

that 49 cases belong in this category and 117 do not belong in this cate-

gory, which means that we would expect to make (117/166) � 49 � 35

prediction errors. For the assaulted-at-work category, we would expect to

make (148/166) � 18 � 16 prediction errors, and for the assaulted-

someplace-else category, we would expect to make (77/166) � 89 � 41

prediction errors. The total number of prediction errors in assigning cases

involving offenders who were strangers is 101 (that is, 9 � 35 � 16 � 41).

To determine the prediction errors for each of the remaining cate-

gories of the independent variable (assaults involving offenders who

were acquaintances/friends, partners, or relatives), we use the same ap-

proach with the three remaining rows of Table 13.7. Table 13.8 presents

all the calculations of prediction errors necessary for obtaining tau.

We obtain the total number of prediction errors made using informa-

tion about the victim-offender relationship by summing the errors across

each category of relationship. For cases involving an offender who was a

tion errors; for cases involving partners, 44 prediction errors; and for

cases involving a relative, 20 prediction errors (see the bottom row of

Table 13.8). Altogether, we would expect to make 228 (that is, 101 � 63 �

Calculations of Prediction Errors for Obtaining Tau for a Relationship Between Victim-Offender Relationship and Location of Assault

PREDICTION PREDICTION PREDICTION PREDICTION PREDICTION

ERRORS: ERRORS: ERRORS: ERRORS: ERRORS:

No Knowledge of Offender Offender Was an Offender Offender

LOCATION OF Victim-Offender Was a Acquaintance Was a Was a

ASSAULT Relationship Stranger or a Friend Partner Relative

Home 139(271/410) � 92 10(156/166) � 9 21(75/96) � 16 77(27/104) � 20 31(13/44) � 9 Neighborhood 77(333/410) � 63 49(117/166) � 35 22(74/96) � 17 5(99/104) � 5 1(43/44) � 1 Work 30(380/410) � 28 18(148/166) � 16 7(89/96) � 6 3(101/104) � 3 2(42/44) � 2 Someplace else 164(246/410) � 98 89(77/166) � 41 46(50/96) � 24 19(85/104) � 16 10(34/44) � 8 Total � � 281 � � 101 � � 63 � � 44 � � 20

Table 13.8

362

assigned 10 of the 166 cases in this row to the assaulted-in-the-home cate-

home category when the offender is a stranger. Thus, if we randomly

involving an acquaintance or friend, we would expect to make 63 predic-

stranger, we would expect to make 101 prediction errors; for cases

M E A S U R E S O F A S S O C I A T I O N F O R N O M I N A L V A R I A B L E S

Goodman and Kruskal’s tau is a measure of the reduction in predic-

tion errors achieved by using knowledge of the independent variable—

which, again, in our example is the victim-offender relationship. Equa-

tion 13.3 presents the general formula for calculating tau.

Equation 13.3

For our example, tau is equal to 0.1886. If we multiply this proportion by

100%, we can discern that knowledge of the victim-offender relationship

reduced our prediction errors by 18.86%, which implies a weak to mod-

erate relationship between victim-offender relationship and location of

assault.

� �

�number of errors without knowledge of independent variable � � �number of errors

with knowledge of independent variable�

number of errors without knowledge of independent variable

W orking It Out

� � 281 � 228

281 � 0.1886

Lambda ( ) is a measure of association that is conceptually very simi-

lar to Goodman and Kruskal’s tau in that it is a PRE measure. However,

rather than using the proportional distribution of cases to determine pre-

diction errors, lambda uses the mode of the dependent variable. We

begin with the naive decision rule, placing all possible observations in

the modal category of the dependent variable and counting as errors the

number of cases that do not belong in that modal category. We then use

information about the value of the independent variable (the informed

decision rule), making assignments of cases based on the mode of the

dependent variable within each category of the independent variable.

Equation 13.4 shows that lambda is calculated in a manner similar to

that used to calculate tau.

�number of errors

dependent variable� � � number of errors using mode of dependent variable by level of independent variable

363

relationship to predict location of assault.

44 � 20) prediction errors using information about the victim-offender

Equation 13.4

using mode of

number of errors using mode of dependent variable

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

The calculation of lambda is less tedious, since we use only informa-

tion on the modal category overall and then within each level of the

independent variable. Without knowledge of the victim-offender rela-

tionship, we would assign all 410 cases to the assaulted-someplace-

else category, resulting in 410 � 164 � 246 classification errors.

What about the number of classification errors when we use knowl-

someplace-else category, resulting in 166 � 89 � 77 classification

errors. For assaults where the offender was an acquaintance or friend,

we would assign all 96 cases to the assaulted-someplace-else category,

resulting in 96 � 46 � 50 classification errors. All 104 partner offend-

ers and 44 relative offenders would both be assigned to the home cat-

use knowledge of the victim-offender relationship, compared to 246

prediction errors made without any knowledge of the victim-offender

relationship. The value of lambda is 0.3211, meaning that knowledge

of the modal location of assault for each type of victim-offender rela-

tionship reduces our errors in predicting location of assault by 32.11%.

W orking It Out

� 246 � 167

246 � 0.3211

As can be seen from our example, different measures of association

may lead to somewhat different interpretations of the relationship be-

tween two variables. This occurs because different measures use differ-

ent strategies in coming to a conclusion about that relationship. Which

is the best measure of association for assessing the strength of the rela-

tionship between two nominal-level variables? Researchers often prefer

the two PRE measures—tau and lambda—over phi and V, since PRE

measures have direct interpretations of values that fall between 0 and

1. However, to use PRE measures, a researcher must assume that one

measure (the independent variable) affects a second (the dependent

variable). Of tau and lambda, tau is often defined as the better measure

able, lambda will have a value of 0, implying that there is no relation-

ship between the two variables. Since tau relies on the marginal

distributions of observations both overall and within each category of

364

offender was a stranger, we would assign all 166 cases to the assaulted-

edge of the victim-offender relationship? For assaults where the

errors, respectively. We have a sum of 167 prediction errors when we

egory, resulting in 104 � 77 � 27 and 44 � 31 � 13 classification

dependent variable is the same for all categories of the independent vari-

of association for two reasons. First, if the modal category of the

M E A S U R E S O F A S S O C I A T I O N F O R N O M I N A L V A R I A B L E S

the independent variable, tau can still detect a relationship between the

marginal totals (i.e., row or column totals). When row or column totals

are not approximately equal, the value of lambda may be artificially

high or low. The reliance on marginal distributions in the calculation of

tau allows that measure of association to account for the size of the

marginal totals directly and causes it not to be as sensitive to differ-

ences in marginal totals.

Statistical Significance of Measures of Association for Nominal Variables

The statistical significance of each of the nominal measures of associa-

tion just discussed can be assessed with the results of a chi-square test.

When the chi-square statistic has a value of 0, each of the four coeffi-

cients will also have a value of 0. The null hypothesis for each of the

four coefficients is simply that the coefficient is equal to 0. The research

hypothesis is simply that the coefficient is not equal to 0.

We illustrate the steps of a hypothesis test for tau and lambda, using

the data on victim-offender relationship and location of assault.

Assumptions:

Level of Measurement: Nominal scale.

Population Distribution: No assumption made.

Sampling Method: Independent random sampling.

Sampling Frame: Adults aged 18 years and older in the state of

Illinois.

Hypotheses:

H0: There is no association between victim-offender relationship and

location of assault (�p � 0).

H1: There is an association between victim-offender relationship and

location of assault (�p � 0).

or

H0: There is no association between victim-offender relationship and

location of assault ( p � 0).

H1: There is an association between victim-offender relationship and

location of assault ( p � 0).

The Sampling Distribution Since we are testing for a relationship be-

tween two nominal-level variables, we use the chi-square distribution,

where degrees of freedom � (r � 1)(c � 1) � (4 � 1)(4 � 1) � 9.

365

related to the marginal distributions, the value of lambda is sensitive to

independent and the dependent variables. Second, and this is again

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

Observed Frequencies and Expected Frequencies for Victim-Offender Relationship and Location of Assault

LOCATION OF ASSAULT

VICTIM-

OFFENDER Someplace

RELATIONSHIP Home Neighborhood Work Else

Stranger fo � 10 fo � 49 fo � 18 fo � 89 166 fe � 56.2780 fe � 31.1756 fe � 12.1463 fe � 66.4000

Acquaintance/ fo � 21 fo � 22 fo � 7 fo � 46 96 friend fe � 32.5463 fe � 18.0293 fe � 7.0244 fe � 38.4000

Partner fo � 77 fo � 5 fo � 3 fo � 19 104 fe � 35.2585 fe � 19.5317 fe � 7.6098 fe � 41.6000

Relative fo � 31 fo � 1 fo � 2 fo � 10 44 fe � 14.9171 fe � 8.2634 fe � 3.2195 fe � 17.6000

139 77 30 164 410

Table 13.9

Calculations of Chi-Square for Victim-Offender Relationship and Location of Assault

VICTIM-

OFFENDER LOCATION

RELATIONSHIP OF ASSAULT fo fe fo � fe (fo � fe)2

Stranger Home 10 56.2780 �46.2780 2141.6578 38.0549 Stranger Neighborhood 49 31.1756 17.8244 317.7089 10.1909 Stranger Work 18 12.1463 5.8537 34.2653 2.8210 Stranger Someplace else 89 66.4000 22.6000 510.7600 7.6922 Friend Home 21 32.5463 �11.5463 133.3180 4.0963 Friend Neighborhood 22 18.0293 3.9707 15.7667 0.8745 Friend Work 7 7.0244 �0.0244 0.0006 0.0001 Friend Someplace else 46 38.4000 7.6000 57.7600 1.5042 Partner Home 77 35.2585 41.7415 1742.3498 49.4164 Partner Neighborhood 5 19.5317 �14.5317 211.1705 10.8117 Partner Work 3 7.6098 �4.6098 21.2499 2.7924 Partner Someplace else 19 41.6000 �22.6000 510.7600 12.2779 Other relative Home 31 14.9171 16.0829 258.6605 17.3399 Other relative Neighborhood 1 8.2634 �7.2634 52.7572 6.3844 Other relative Work 2 3.2195 �1.2195 1.4872 0.4619 Other relative Someplace else 10 17.6000 �7.6000 57.7600 3.2818

� � 168.0005

(fo � fe)2

fe

Table 13.10

366

Row total

Column total

M E A S U R E S O F A S S O C I A T I O N F O R O R D I N A L - L E V E L V A R I A B L E S

Significance Level and Rejection Region We use the conventional 5% sig-

nificance level for this example. From Appendix 2, we see that the criti-

cal value of chi-square associated with a significance level of 5% and

df � 9 is 16.919. If the calculated chi-square statistic is greater than

16.919, we will reject the null hypotheses and conclude that the associa-

tion between victim-offender relationship and location of assault is statis-

tically significant.

The Test Statistic The chi-square statistic for the data in Table 13.7 is

168.001. See Table 13.9 for the expected and observed frequencies and

Table 13.10 for the detailed calculations.

The Decision Since our calculated chi-square statistic of 168.001 is much

M e a s u r e s o f A s s o c i a t i o n f o r O r d i n a l - L e v e l V a r i a b l e s

The preceding discussion described several measures of association for

nominal variables, where there is no rank ordering of the categories of

each variable. With ordinal-level variables, we can use the ordering of the

categories to measure whether there is a positive or a negative relation-

ship between two variables. A positive relationship would be indicated

by higher ranks on one variable corresponding to higher ranks on a sec-

ond variable. A negative relationship would be indicated by higher ranks

on one variable corresponding to lower ranks on a second variable. The

measures of association for ordinal-level variables all have values that

value of 0.0 indicates no relationship between the two variables.

Table 13.11 illustrates these variations in the strength of the relation-

ship between two ordinal variables with a hypothetical distribution of

450 cases. Part a presents a pattern of no association between the two

variables. Since the cases are evenly distributed across all the cells of the

table, knowledge of the level of one ordinal variable does not provide

any information about the level of the second ordinal variable. Parts b

and c show perfect negative and positive relationships, respectively,

where knowledge of the level of one ordinal variable determines, with-

out error, the level of the second ordinal variable.

The calculation of ordinal measures of association is tedious to perform

by hand. When doing data analysis, you would likely rely on a statistical

367

larger than our critical chi-square of 16.919, we reject the null hypotheses

and conclude that there is a statistically significant relationship between

victim-offender relationship and location of assault.

relationship, a value of �1.0 indicates a perfect positive relationship, and a

range from �1.0 to �1.0. A value of �1.0 indicates a perfect negative

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

software package to perform the calculations for you. Most common statis-

tical software packages will compute the measures of association for ordi-

nal variables described here (see, for example, the computer exercises at

the end of this chapter). The following discussion is intended to help you

understand how these various measures are calculated.

There are four common measures of association for ordinal variables:

gamma (�), Kendall’s b, Kendall’s c, and Somers’ d. Common to all

four is the use of concordant pairs and discordant pairs of observa-

tions. The logic behind using concordant and discordant pairs of obser-

vations is that we take each possible pair of observations in a data set

and compare the relative ranks of the two observations on the two vari-

ables examined. Concordant pairs are those pairs of observations for

which the rankings are consistent: One observation is ranked high on

both variables, while the other observation is ranked low on both vari-

Hypothetical Distribution of 450 Cases for Two Ordinal Variables

(a) Measure of Association � 0.0

VARIABLE 2

VARIABLE 1 Low Medium High

Low 50 50 50 150 Medium 50 50 50 150 High 50 50 50 150

150 150 150 450

(b) Measure of Association � �1.0

VARIABLE 2

VARIABLE 1 Low Medium High

Low 0 0 150 150 Medium 0 150 0 150 High 150 0 0 150

150 150 150 450

(c) Measure of Association � �1.0

VARIABLE 2

VARIABLE 1 Low Medium High

Low 150 0 0 150 Medium 0 150 0 150 High 0 0 150 150

150 150 150 450

Table 13.11

368

ables. For example, one observation is ranked 1 (of five ranked categories)

Row total

Column total

Column total

Row total

Row total

Column total

� �

M E A S U R E S O F A S S O C I A T I O N F O R O R D I N A L - L E V E L V A R I A B L E S

ond variable, while the other observation is ranked 4 on the first variable

and 3 on the second variable. Discordant pairs refer to those pairs of ob-

servations for which the rankings are inconsistent: One observation is

ranked high on the first variable and low on the second variable, while

the other observation is ranked low on the first variable and high on the

second variable. For example, one observation is ranked 1 on the first

variable and 5 on the second variable, while the other observation is

called a tied pair of observations (tie).4

How do we decide whether a pair of observations is a concordant

pair, a discordant pair, or a tied pair? Let’s look at the determination of

concordant, discordant, and tied pairs for the data presented in Table

13.12, which represents a cross-tabulation of two ordinal variables, each

with three categories: low, medium, and high.

4All the measures of association for ordinal variables that we discuss here are for

grouped data that can be represented in the form of a table. In Chapter 14, we discuss

another measure of association for ordinal variables—Spearman’s r (rs)—that is most

useful in working with ungrouped data, such as information on individuals. The diffi-

culty we confront when using Spearman’s r on grouped data is that the large number

of tied pairs of observations complicates the calculation of this measure of association.

Spearman’s r is a more appropriate measure of association when we have ordinal

variables with a large number of ranked categories for individual cases or when we

take an interval-level variable and rank order the observations (see Chapter 14).

Cross-Tabulation of Two Ordinal Variables

DEPENDENT VARIABLE

INDEPENDENT

VARIABLE Low Medium High

Low Cell A Cell B Cell C 12 4 3

Medium Cell D Cell E Cell F 5 10 6

High Cell G Cell H Cell I 3 5 14

Table 13.12

on the first variable and 2 (of five ranked categories) on the sec-

369

ranked 4 on the first variable and 2 on the second variable. A pair of

observations that has the same rank on one or both of the variables is

the dependent and the independent variables is required. The value of d

four measures of association for ordinal variables for which specification of

Somers’ d is the only one of the

depending on which variable is specified as the dependent variable. To simp-

lify the following discussion, the examples we present in the next section de-

fine one variable as the dependent and the other as the independent variable.

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

We begin by determining the concordant pairs—those pairs of observa-

since the cases in the same row or column will have the same ranking on

the independent and dependent variables, respectively, and thus represent

ties. We then look for cases located below and to the right of the cell of in-

terest. For Cell A, the cells we will use to determine concordant pairs are

of observations that are concordant for observations in Cell A, we begin

by summing the number of observations in Cells E, F, G, and I: 10 � 6 �

5 � 14 � 35. This tells us that for a single observation in Cell A, there are

35 concordant pairs of observations. Since there are 12 observations in

Cell A, we multiply the number of cases in Cell A (12) by the sum of the

cases in Cells E, F, H, and I. For Cell A, there are 420 concordant pairs.

W orking It Out

12(10 � 6 � 5 � 14) � 420

Continuing to work across the first row of Table 13.12, we move to Cell

B. The cells located below and to the right of Cell B are Cells F and I, so

the number of concordant pairs is 4(6 � 14) � 80. When we move to Cell

C, we see there are no cells below and to the right, so we drop down to

the next row and start with Cell D. The cells located below and to the right

of Cell D are Cells H and I, so the number of concordant pairs is 5(5 � 14)

� 95. Moving to Cell E, we see that only Cell I is below and to the right, so

the number of concordant pairs is 10(14) � 140. The remaining cells in the

table—F, G, H, and I—have no other cells located below and to the right,

so they are not used in the calculation of concordant pairs. After calculating

concordant pairs for all cells in the table, we sum these values to get the

number of concordant pairs for the table. For Table 13.12, the total number

of concordant pairs is 735 (that is, 420 � 80 � 95 � 140).

W orking It Out

Sum � 420 � 80 � 95 � 140 � 735

Cell A: Cell B: Cell D: Cell E:

12(10 � 6 � 5 � 14) � 420 4(6 � 14) � 80 5(5 � 14) � 95

10(14) � 140

370

tions that have consistent relative rankings. Let’s start with Cell A. We

Cells E, F, H, and I, since the ranks are consistently lower on both the

independent and the dependent variables. To determine the number of pairs

remove from consideration the row and column that Cell A is located in,

M E A S U R E S O F A S S O C I A T I O N F O R O R D I N A L - L E V E L V A R I A B L E S

To calculate discordant cells, we begin in the upper right corner of

Table 13.12 (Cell C), locate cells that are positioned below and to the

left of the cell of interest, and perform calculations similar to those for

concordant pairs. Beginning with Cell C, we multiply the number of

cases in Cell C by the sum of cases in Cells D, E, G, and H, which are

located below and to the left of Cell C. The number of discordant pairs

for Cell C is 69.

W orking It Out

3(5 � 10 � 3 � 5) � 69

Moving from right to left in the top row of Table 13.12, we shift our

attention to Cell B. The discordant pairs for Cell B are calculated by mul-

tiplying the number of cases in Cell B by the sum of cases in Cells D and

G. We find the number of discordant pairs for Cell B to be 4(5 � 3) �

32. Since there are no cells located below and to the left of Cell A, it is

not used to calculate discordant pairs, and we move on to Cell F. The

cells located below and to the left of Cell F are Cells G and H, so the

is 10(3) � 30. There are no cells located below and to the left of Cells D,

G, H, and I, so no further calculations are performed. As with the con-

cordant pairs, we sum our discordant pairs for Table 13.12 and find the

sum to be 179 (that is, 69 � 32 � 48 � 30).

W orking It Out

Sum � 69 � 32 � 48 � 30 � 179

Cell C: Cell B: Cell F: Cell E:

3(5 � 10 � 3 � 5) � 69 4(5 � 3) � 32 6(3 � 5) � 48

10(3) � 30

To calculate ties in rank for pairs of observations, we have to con-

sider the independent and dependent variables separately. We denote

ties on the independent variable as TX and ties on the dependent vari-

able as TY. Since the independent variable is represented by the rows

371

number of discordant pairs is 6(3 � 5) � 48. For Cell E, the only cell

located below and to the left is Cell G, so the number of discordant pairs

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

in Table 13.12, the pairs of observations that will be defined as ties on

the independent variable will be those cases located in the same row

of Table 13.12. To calculate the number of ties in each row, we use

Equation 13.5.

Equation 13.5

where TX is the number of ties on the independent variable and Nrow is

the row total. Equation 13.5 tells us to calculate the product of the num-

ber of observations in a row and the number of observations in a row

minus 1 for all rows. We then sum the products calculated for each row

and multiply the sum by .

For Table 13.12, the three row totals are 19 (row 1), 21 (row 2), and

22 (row 3). When we insert these values into Equation 13.5, we find the

number of ties on the independent variable to be 612.

1 \2

TX � 1 2

� Nrow(Nrow � 1)

W orking It Out

� 612

� 1 2

(342 � 420 � 462) � 1 2

(1,224)

� 1 2

[(19)(18) � (21)(20) � (22)(21)]

� 1 2

[(19)(19 � 1) � (21)(21 � 1) � (22)(22 � 1)]

TX � 1 2

� Nrow(Nrow � 1)

The ties on the dependent variable are found in a similar manner. Since

the dependent variable is represented in the columns, we perform the

same type of calculation, but using column totals rather than row totals.

Equation 13.6 presents the formula for calculating ties on the dependent

variable.

Equation 13.6

In Equation 13.6, TY is the number of ties on the dependent variable and

Ncol is the total number of observations in the column.

TY � 1 2

� Ncol(Ncol � 1)

372

M E A S U R E S O F A S S O C I A T I O N F O R O R D I N A L - L E V E L V A R I A B L E S

Gamma

Once we have calculated the numbers of concordant pairs and discor-

dant pairs, gamma ( ) is the simplest of the ordinal measures of associa-

tion to calculate, since it does not use information about ties in rank.

Gamma has possible values that range from �1.0 to �1.0. Gamma may

also be interpreted as a PRE measure: We can interpret the value of

gamma as indicating the proportional reduction in errors in predicting

the dependent variable, based on information about the independent

variable.

Equation 13.7 presents the formula for calculating gamma. Gamma is

the difference between the number of concordant (C ) and discordant

(D) pairs, (C � D), divided by the sum of the concordant and discordant

pairs, (C � D).

Equation 13.7

For the data in Table 13.12, gamma is equal to 0.6083. The positive value

of gamma tells us that as we move from lower ranked to higher ranked

categories on the independent variable, the category of the dependent

variable also tends to increase. In regard to the relative strength of the

relationship, a value of 0.6083 suggests a strong relationship between the

independent and dependent variables, since knowledge of the indepen-

dent variable reduces our errors in predicting the dependent variable by

60.83%.

� C � D C � D

W orking It Out

� 614

� 1 2

(380 � 342 � 506) � 1 2

(1,228)

� 1 2

[(20)(19) � (19)(18) � (23)(22)]

� 1 2

[(20)(20 � 1) � (19)(19 � 1) � (23)(23 � 1)]

TY � 1 2

� Ncol(Ncol � 1)

In Table 13.12, the column totals are 20 (column 1), 19 (column 2),

and 23 (column 3). After inserting these values into Equation 13.6, we

find the number of ties on the dependent variable to be 614.

373

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

Kendall’s �b and �c

Kendall’s tau measures—�b and �c—also assess the strength of associa-

tion between two ordinal variables.5 The two measures are conceptually

very similar in that they use information about concordant and discor-

dant pairs of observations. But they also utilize information about tied

pairs on both the independent and the dependent variables. Both tau

measures have possible values ranging from �1.0 to �1.0. There are two

important differences between �b and �c: First, �b should be applied only

to a table where the number of rows is equal to the number of columns;

�c should be applied to a table where the number of rows is not equal to

the number of columns. When the number of rows is equal to the num-

ber of columns, �c will have a value close to that of �b. Second, �b may

be interpreted as a PRE measure, but �c may not. The differences in the

application and interpretation of each measure suggest that knowing the

dimensions of the table is important in deciding which measure is most

appropriate.

Equations 13.8 and 13.9 present the formulas for calculating �b and �c,

respectively.

Equation 13.8

In Equation 13.8, C and D represent the concordant and the discordant

pairs, respectively; N represents the total number of cases; TX represents

the number of ties on the independent variable; and TY represents the

number of ties on the dependent variable.

�b � C � D

�[N(N � 1)/2 � TX][N (N � 1)/2 � TY]

W orking It Out

� 0.6083

� 556 914

� 735 � 179 735 � 179

� C � D C � D

5These two tau measures are different from Goodman and Kruskal’s tau, which mea-

sures the strength of association between two nominal variables.

374

M E A S U R E S O F A S S O C I A T I O N F O R O R D I N A L - L E V E L V A R I A B L E S

Let’s return to the data presented in Table 13.12. We have already

calculated the number of concordant pairs to be 735, the number of

discordant pairs to be 179, the total number of cases to be 62, the

number of ties on the independent variable to be 612, and the number

of ties on the dependent variable to be 614. After inserting these val-

ues into Equation 13.6, we find �b to be 0.4351. This indicates that

knowledge of the independent variable reduces our prediction errors

by 43.51%.

W orking It Out

� 0.4351

� 556

�(1,279)(1,277)

� 556

�[1,891 � 612][1,891 � 614]

� 735 � 179

�[62(62 � 1)/2 � 612][62(62 � 1)/2 � 614]

�b � C � D

�[N (N � 1)/2 � TX ][N (N � 1)/2 � TY ]

Equation 13.9 presents the formula for calculating �c. We do not calcu-

late �c for Table 13.12, since the number of rows is equal to the number

of columns. We do, however, illustrate its calculation below with another

example.

Equation 13.9

In Equation 13.9, C and D represent the concordant and the discordant

pairs, respectively; N represents the total number of cases; and m is the

smaller of the number of rows (r) and the number of columns (c). Sup-

pose, for example, that we had a table with five rows (r � 5) and four

columns (c � 4). The number of columns is smaller than the number of

rows, so m would be 4.

where m � min(r, c)

�c � C � D

1 2

N

2[(m � 1)/m]

375

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

Somers’ d

The fourth measure of association for ordinal variables that we present

ables, Somers’ d uses information on ties on only the independent vari-

able. It is important to remember that the statistic you get for Somers’ d

may vary, depending on which variable is defined as the dependent vari-

able. The formula for calculating Somers’ d is given in Equation 13.10.

Equation 13.10

In Equation 13.10, where C, D, N, and TX represent the concordant pairs,

the discordant pairs, the total number of cases, and the number of ties

For Table 13.12, we have already calculated values for C, D, N, and

TX. After inserting these values into Equation 13.10, we find Somers’ d to

be 0.4347.

dYX � C � D

N (N � 1)/2 � TX

W orking It Out

� 0.4347

� 556

1,279

� 556

1,891 � 612

� 735 � 179

62(62 � 1)/2 � 612

dYX � C � D

N (N � 1)/2 � TX

A Substantive Example: Affectional Identification

with Father and Level of Delinquency

Table 9.14 presented a cross-tabulation of two ordinal variables: affec-

much they wanted to grow up and be like their fathers. The responses

were classified into five ordered categories: in every way, in most

376

information about ties on both the independent and the dependent vari-

here—Somers’ d—is similar to the tau measures, but instead of using

on the independent variable, respectively. The subscript YX on d

denotes the dependent and the independent variables, in order.

father was determined by the youth’s responses to a question about how

tional identification with father and delinquency. Identification with

M E A S U R E S O F A S S O C I A T I O N F O R O R D I N A L - L E V E L V A R I A B L E S

ways, in some ways, in just a few ways, and not at all. Delinquent acts

were classified into three ordered categories: none, one, and two or

more. The data came from the Richmond Youth Survey report, and the

distribution of cases presented refers only to the white males who re-

sponded to the survey.6 We reproduce this cross-tabulation in Table

13.13.

In our earlier analysis of the data in this table (see Chapter 9), we

found a statistically significant relationship between affectional identifica-

tion with father and delinquency. However, the chi-square statistic told

us nothing about the direction of the effect or the strength of the rela-

tionship between these two variables. We can use the measures of asso-

We begin by calculating the numbers of concordant pairs, discordant

pairs, and tied pairs of observations. The number of concordant pairs of

observations is 201,575; the number of discordant pairs is 125,748; the

number of pairs tied on the independent variable is 187,516; and the

number of pairs tied on the dependent variable is 315,072.

Affectional Identification with Father by Number of Delinquent Acts

AFFECTIONAL

IDENTIFICATION NUMBER OF DELINQUENT ACTS

WITH FATHER None One Two or more

In every way Cell A Cell B Cell C 121 77 25 19

In most ways Cell D Cell E Cell F 404 263 97 44

In some ways Cell G Cell H Cell I 387 224 97 66

In just a few ways Cell J Cell K Cell L 172 82 52 38

Not at all Cell M Cell N Cell O 138 56 30 52

702 301 219 1,222

Table 13.13

6David F. Greenberg, “The Weak Strength of Social Control Theory,” Crime and Delin-

quency 45:1 (1999): 66–81.

377

between identification with father and level of delinquency.

ciation for ordinal variables to test the strength of the relationship

Row total

Column total

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

W orking It Out

Concordant Pairs:

Cell A: 77(97 � 44 � 97 � 66 � 52 � 38 � 30 � 52) � 36,652

Cell B: 25(44 � 66 � 38 � 52) � 5,000

Cell D: 263(97 � 66 � 52 � 38 � 30 � 52) � 88,105

Cell E: 97(66 � 38 + 52) � 15,132

Cell G: 224(52 � 38 � 30 � 52) � 38,528

Cell H: 97(38 � 52) � 8,730

Cell J: 82(30 � 52) � 6,724

Cell K: 52(52) � 2,704

Sum � 36,652 � 5,000 � 88,105 � 15,132 � 38,528 � 8,730

� 6,724 � 2,704

� 201,575

Discordant Pairs:

Cell C: 19(263 � 97 � 224 � 97 � 82 � 52 � 56 � 30) � 17,119

Cell B: 25(263 � 224 � 82 � 56) � 15,625

Cell F: 44(224 � 97 � 82 � 52 � 56 � 30) � 23,804

Cell E: 97(224 � 82 � 56) � 35,114

Cell I: 66(82 � 52 � 56 � 30) � 14,520

Cell H: 97(82 � 56) � 13,386

Cell L: 38(56 � 30) � 3,268

Cell K: 52(56) � 2,912

Sum � 17,119 � 15,625 � 23,804 � 35,114 � 14,520 � 13,386

� 3,268 + 2,912

� 125,748

378

M E A S U R E S O F A S S O C I A T I O N F O R O R D I N A L - L E V E L V A R I A B L E S

After calculating the concordant pairs, discordant pairs, and pairs

tied on the independent and dependent variables, we can calculate

the measures of association for ordinal variables. We find the value of

gamma to be 0.2317. Don’t be confused by the fact that for affectional

identification movement from lower to higher ordered categories rep-

resents movement from more to less identification with the father.

Substantively, what this value of gamma tells us is that as the level of

affectional identification with father decreases (i.e., as we move down

the rows of the table), the youth are likely to report higher levels of

delinquency. The value of gamma also indicates that we reduce our

prediction errors about level of delinquency by 23.17% when we use

information about the level of affectional identification with father. If

affectional identification in this example had been measured from less

to more identification with father (rather than more to less identifica-

tion), gamma would have been negative. As a general rule, it is impor-

tant to look carefully at the ordering of the categories of your measure

in order to make a substantive interpretation of your result.

Pairs Tied on the Independent Variable:

Pairs Tied on the Dependent Variable:

� 315,072

� (1 \2)(630,144)

� (1 \2)(492,102 � 90,300 � 47,742)

TY � (1 \2)[(702)(701) � (301)(300) � (219)(218)]

� 187,516

� (1 \2)(375,032)

� (1 \2)(14,520 � 162,812 � 149,382 � 29,412 � 18,906)

�� (172)(171) � (138)(137)]

TX � (1 \2)[(121)(120) � (404)(403) � (387)(386)

379

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

Recall that there are two tau measures: �b and �c. If the number of

rows were equal to the number of columns, then we would use �b. Since

the number of rows is different from the number of columns in Table

13.13, we use �c. For the data presented in Table 13.13, �c has a value of

0.1523, meaning that as the level of affectional identification with father

decreases, the level of delinquency increases. However, since �c

W orking It Out

� 0.2317

� 75,827

327,323

� 201,575 � 125,748

201,575 � 125,748

� C � D C � D

W orking It Out

� 0.1523

� 75,827

497,761.3333

� 201,575 � 125,748

1 2

(1,222)2 �3 � 1 3 �

�c � C � D

1 2

N

2�m � 1 m �

, � where m � min(r, c) � min(5, 3) � 3

Our third measure of association for ordinal variables, Somers’ d, has

a value of 0.1358. The interpretation is the same as that for gamma and

�c: Lower levels of affectional identification with father are associated

level of delinquency by 13.58%.

380

is not

reduction in error.

a PRE measure, we cannot interpret this result in terms of proportional

affectional identification with father reduces our prediction errors about

with higher levels of delinquency. In this case, knowledge of level of

M E A S U R E S O F A S S O C I A T I O N F O R O R D I N A L - L E V E L V A R I A B L E S

Note on the Use of Measures of Association for Ordinal Variables

As illustrated in our example, the values for gamma, Kendall’s tau mea-

sures, and Somers’ d will generally not be the same. The difference in val-

ues can be attributed primarily to whether the measure accounts for tied

pairs of observations. Gamma does not account for tied pairs of observa-

pairs of observations tied on the independent variable, while Kendall’s tau

measures account for tied pairs of observations on both variables.

Which of these measures is best to use in which situations? As in

our discussion of measures of association for nominal variables, to

begin to address this question, we need to consider the dimensions of

the table and our desire for a PRE measure. If the number of rows is

equal to the number of columns, then �b is likely the best overall mea-

sure of association for two reasons: First, it has a PRE interpretation,

meaning that values falling between 0 and 1 have direct interpreta-

tions in terms of reduction of error. Second, since �b accounts for pairs

of observations tied on both the independent and the dependent vari-

ables, it will provide a more conservative estimate than gamma. If the

number of rows is not equal to the number of columns, Somers’ d is

sometimes considered a better measure of association than �c, since it

has a PRE interpretation and �c does not. Somers’ d offers the addi-

tional advantage of being an appropriate measure of association for

Statistical Significance of Measures of Association for Ordinal Variables

Each of the four measures of association for ordinal variables can

be tested for statistical significance with a z-test. The general formula

for calculating the z-score is given in Equation 13.11, where we divide

W orking It Out

� 0.1358

� 75,827

558,515

� 201,575 � 125,748

[(1,222)(1,222 � 1)/2] � 187,516

dYX � C � D

N (N � 1)/2 � TX

381

association between two ordinal variables. Somers’ d accounts for only the

tions and thus is sometimes criticized for overestimating the strength of

dependent variables.

those situations where we have clearly defined independent and

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

the measure of association by the standard error of the measure of

association.

Equation 13.11

What will differ for each of the measures of association for ordinal vari-

ables is the calculation of the standard error. Equations 13.12, 13.13, and

13.14 present approximate standard errors for gamma, Kendall’s tau

measures, and Somers’ d, respectively.7

In all three equations, N is the total number of observations, r is the

number of rows, and c is the number of columns in the table.

Assumptions:

Level of Measurement: Ordinal scale.

Population Distribution: Normal distribution for the relationship exam-

ined (relaxed because N is large).

Sampling Method: Independent random sampling.

Sampling Frame: High school–age white males in Richmond, California,

in 1965.

Hypotheses:

H0: There is no association between affectional identification with father

and delinquency ( p � 0).

H1: There is an association between affectional identification with father

and delinquency ( p � 0).

Equation 13.14

Approximate Standard Error for

Somers’ d

�̂d � �4(r 2 � 1)(c � 1)

9Nr

2(c � 1)

Equation 13.13

Approximate Standard Error for

Kendall’s Tau Measures

�̂� � �4(r � 1)(c � 1) 9Nrc

Equation 13.12

Approximate Standard Error for

Gamma

�̂ � � 4(r � 1)(c � 1)

9N (r � 1)(c � 1)

z � measure of association

standard error of measure of association

7For a more detailed discussion of these issues, see Jean Dickson Gibbons, Nonpara-

metric Measures of Association (Newbury Park, CA: Sage, 1993).

382

M E A S U R E S O F A S S O C I A T I O N F O R O R D I N A L - L E V E L V A R I A B L E S

or

H0: There is no association between affectional identification with father

and delinquency (�c (p) � 0).

H1: There is an association between affectional identification with father

and delinquency (�c (p) � 0).

or

H0: There is no association between affectional identification with father

and delinquency (dp � 0).

H1: There is an association between affectional identification with father

and delinquency (dp � 0).

The Sampling Distribution We use the normal distribution to test

whether the measures of ordinal association differ significantly from 0.

As with our earlier examples using a normal sampling distribution, the N

of cases must be large in order for us to relax the normality assumption.

When examining the relationship between two ordinal-level variables,

we recommend a sample of at least 60 cases.

Significance Level and Rejection Region We use the conventional 5% sig-

nificance level for our example. From Appendix 3, we can determine

that the critical values for z are �1.96. If the calculated z-score is greater

than 1.96 or less than �1.96, we will reject the null hypotheses and con-

clude that the measure of association between affectional identification

with father and delinquency is significantly different from 0.

The Test Statistic Since we have three different measures of associa-

tion— , �c, and d—we need to calculate three separate test statistics. We

first need to calculate the approximate standard error for gamma, using

Equation 13.12. We find the standard error for gamma to be 0.0330.

W orking It Out

� 0.0330

� �0.00109

� � 96

87,984

� � 4(5 � 1)(3 � 1)

(9)(1,222)(5 � 1)(3 � 1)

�̂ � � 4(r � 1)(c � 1)

9N (r � 1)(c � 1)

383

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

Using the standard error for gamma, we then calculate the z-score

using Equation 13.11. In our example, we find the z-score for gamma to

be 7.0212.

W orking It Out

� 7.0212

� 0.2317 0.0330

z �

�̂

Turning to �c, we calculate the standard error using Equation 13.13.

For our example, the standard error for �c is 0.0241.

W orking It Out

� 0.0241

� �0.00058

� � 96

164,970

� �4(5 � 1)(3 � 1)

(9)(1,222)(5)(3)

�̂� � �4(r � 1)(c � 1) 9Nrc

Using the standard error for �c

W orking It Out

� 6.3195

� 0.1523

0.0241

z � �c

�̂�

384

to be 6.3195.

and Equation 13.11, we find the z-score

C H O O S I N G T H E B E S T M E A S U R E O F A S S O C I A T I O N

For Somers’ d, we follow the same process, calculating the standard

error for d and then using the standard error to calculate the z-score for

d. For our example, the standard error for d is 0.0264 and the corre-

sponding z-score is 5.1439.

W orking It Out

� 0.0264

� �0.00070

� � 384

549,900

� � 4(52 � 1)(3 � 1)

(9)(1,222)(52)(3 � 1)

�̂d � �4(r

2 � 1)(c � 1)

9Nr

2(c � 1)

W orking It Out

� 5.1439

� 0.1358

0.0264

z � d

�̂d

The Decision All three of the calculated z-scores are greater than 1.96,

meaning that we reject the null hypotheses and conclude in the case of

each test that there is a statistically significant relationship between affec-

tional identification with father and delinquency.

C h o o s i n g t h e B e s t M e a s u r e o f A s s o c i a t i o n f o r N o m i n a l - a n d O r d i n a l - L e v e l V a r i a b l e s

Because we have covered so many different measures in this chapter,

we thought it would be useful to recap them in a simple table that can

be used in deciding which measure of association is appropriate

385

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

for which specific research problem. Table 13.14 presents summary

information on the measures of association for nominal and ordinal

variables discussed in this chapter. The first column of Table 13.14

gives the measure of association, the second column notes the appro-

priate level of measurement for the two variables, the third column

tells whether the measure of association is also a PRE measure, and the

fourth column lists any restrictions on the size of the table used in the

analysis. Thus, for any given pair of nominal or ordinal variables, you

should be able to determine which measure of association best suits

your needs.

C h a p t e r S u m m a r y

Measures of association for nominal and ordinal variables allow

researchers to go beyond a simple chi-square test for independence

between two variables and assess the strength of the relationship.

The measures of association discussed in this chapter are the most

commonly used measures of association for nominal and ordinal

variables.

Two of the measures of association for nominal variables are based

on the value of the chi-square statistic. Phi (�) adjusts the value of chi-

square by taking into account the size of the sample, but is useful only

for 2 � 2 tables. Cramer’s V is also based on the value of the chi-square

statistic, but makes an additional adjustment for the numbers of rows

and columns in the table. One of the difficulties with the interpretation

Summary of Measures of Association for Nominal and Ordinal Variables

MEASURE OF LEVEL OF DIMENSIONS OF TABLE

ASSOCIATION MEASUREMENT PRE MEASURE? (ROWS BY COLUMNS)

� Nominal No 2 � 2 V Nominal No Any size Nominal Yes Any size � Nominal Yes Any size Ordinal Yes Any size �b Ordinal Yes Number of rows �

Number of columns �c Ordinal No Number of rows �

Number of columns d Ordinal Yes Any size

Table 13.14

386

C H A P T E R S U M M A R Y

of phi and V is that a value that falls between 0 and 1 does not have a

precise interpretation. We can infer that as values approach 0, there is a

weak relationship between the two variables. Similarly, as values ap-

proach 1, there is a strong (or near perfect) relationship between the two

variables.

Goodman and Kruskal’s tau and lambda are measures of associa-

tion that are not based on the value of the chi-square statistic and instead

use different decision rules for classifying cases. Tau relies on the pro-

portional distribution of cases in a table, while lambda relies on the

modal values of the dependent variable overall and within each level or

category of the independent variable. Tau and lambda offer an improve-

ment over phi and V in that a value between 0 and 1 can be interpreted

directly as the proportional reduction in errors made by using informa-

tion about the independent variable. More generally, this characteristic is

called proportional reduction in error, or PRE. PRE measures tell us

how much knowledge of one measure helps to reduce the errors we

make in defining the values of a second measure. Both measures require

that we define at the outset which variable is the dependent variable

and which variable is the independent variable. The dependent vari-

able is the outcome variable—the phenomenon that we are interested in

explaining. As it is dependent on other variables, it is influenced—or we

expect it to be influenced—by other variables. The variables that affect,

or influence, the dependent variable are referred to as the independent

variables.

There are four common measures of association for ordinal vari-

ables: gamma (�), Kendall’s �b and �c, and Somers’ d. Measures of

association for ordinal variables are all based on concordant pairs

and discordant pairs of observations. Concordant pairs are pairs of

observations that have consistent rankings on the two variables (e.g.,

high on both variables or low on both variables), while discordant

pairs are those pairs of observations that have inconsistent rankings on

the two variables (e.g., high on one variable and low on the other vari-

able). Gamma uses information only on the concordant and discordant

pairs of observations. The remaining measures of association—

Kendall’s tau measures and Somers’ d—use information about pairs of

observations that have tied rankings. All four of the measures of associ-

ation for ordinal variables discussed in this chapter have values ranging

from �1.0 to 1.0, where a value of �1.0 indicates a perfect negative re-

lationship (i.e., as we increase the value of one variable, the other vari-

able decreases), a value of 1.0 indicates a perfect positive relationship

(i.e., as we increase the value of one variable, the other variable also

increases), and a value of 0.0 indicates no relationship between the

two variables. Gamma ( ), Kendall’s �b, and Somers’ d all have PRE

interpretations.

387

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

K e y T e r m s

concordant pairs of observations Pairs

of observations that have consistent rank-

ings on two ordinal variables.

Cramer’s V A measure of association for

two nominal variables that adjusts the chi-

square statistic by the sample size. V is ap-

propriate when at least one of the nominal

variables has more than two categories.

dependent variable The outcome vari-

able; the phenomenon that we are inter-

ested in explaining. It is dependent on

other variables in the sense that it is influ-

enced—or we expect it to be influenced—

by other variables.

discordant pairs of observations Pairs

of observations that have inconsistent rank-

ings on two ordinal variables.

gamma (�) PRE measure of association

for two ordinal variables that uses informa-

tion about concordant and discordant

pairs of observations within a table.

Gamma has a standardized scale ranging

from �1.0 to 1.0.

Goodman and Kruskal’s tau (�) PRE

measure of association for two nominal

variables that uses information about the

proportional distribution of cases within a

table. Tau has a standardized scale ranging

from 0 to 1.0. For this measure, the re-

searcher must define the independent and

dependent variables.

independent variable A variable as-

sumed by the researcher to affect or influ-

ence the dependent variable.

Kendall’s �b PRE measure of association

for two ordinal variables that uses informa-

tion about concordant pairs, discordant

pairs, and pairs of observations tied on

both variables examined. �b has a standard-

ized scale ranging from �1.0 to 1.0 and is

appropriate only when the number of rows

equals the number of columns in a table.

Kendall’s �c A measure of association for

two ordinal variables that uses information

about concordant pairs, discordant pairs,

and pairs of observations tied on both vari-

ables examined. �c has a standardized scale

ranging from �1.0 to 1.0 and is appropriate

when the number of rows is not equal to

the number of columns in a table.

lambda (�) PRE measure of association

for two nominal variables that uses infor-

independent variable. Lambda has a stan-

dardized scale ranging from 0 to 1.0.

phi (�) A measure of association for two

nominal variables that adjusts the chi-

square statistic by the sample size. Phi is

appropriate only for nominal variables that

each have two categories.

proportional reduction in error (PRE)

The proportional reduction in errors made

when the value of one measure is pre-

dicted using information about the second

measure.

Somers’ d PRE measure of association for

two ordinal variables that uses information

about concordant pairs, discordant pairs,

and pairs of observations tied on the inde-

pendent variable. Somers’ d has a standard-

ized scale ranging from �1.0 to 1.0.

tied pairs of observations (ties) Pairs of

observation that have the same ranking on

two ordinal variables.

388

dependent variable for each category of the

mation about the modal category of the

S Y M B O L S A N D F O R M U L A S

S y m b o l s a n d F o r m u l a s

C Number of concordant pairs of observations

D Number of discordant pairs of observations

Nrow Total number of observations for each row

Ncol Total number of observations for each column

TX Number of pairs of observations tied on the independent

variable

TY Number of pairs of observations tied on the dependent

variable

� Phi; measure of association for nominal variables

V Cramer’s V; measure of association for nominal variables

Lambda; measure of association for nominal variables

� Goodman and Kruskal’s tau; measure of association for nominal

variables

gamma; measure of association for ordinal variables

�b Kendall’s �b; measure of association for ordinal variables

�c Kendall’s �c; measure of association for ordinal variables

d Somers’ d; measure of association for ordinal variables

To calculate phi (�):

To calculate Cramer’s V:

V � � � 2

N � min(r � 1, c � 1)

� � �� 2

N

389

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

To calculate Goodman and Kruskal’s tau:

To calculate lambda:

To calculate the number of tied pairs of observations on the

independent variable:

To calculate the number of tied pairs of observations on the dependent

variable:

To calculate gamma:

To calculate �b:

To calculate �c:

To calculate Somers’ d:

To calculate the z-score:

z � measure of association

standard error of measure of association

dYX � C � D

N (N � 1)/2 � TX

�c � C � D

1 2

N

2[(m � 1)/m] , where m � min(r, c)

�b � C � D

�[N (N � 1)/2 � TX][N (N � 1)/2 � TY ]

� C � D C � D

TY � 1 2 � Ncol(Ncol � 1)

TX � 1 2 � Nrow(Nrow � 1)

�number of errors

dependent variable� � � number of errors using mode of dependent variable by level of independent variable

� �

�number of errors without knowledge of independent variable � � �number of errors

with knowledge of independent variable�

number of errors without knowledge of independent variable

390

using mode of

number of errors using mode of dependent variable

E X E R C I S E S

To calculate the standard error for gamma:

To calculate the standard error for Kendall’s tau measures:

To calculate the standard error for Somers’ d:

E x e r c i s e s

13.1 A researcher studies the link between race of offender and death sen- tence decision in a state by selecting a random sample of death penalty cases over a 20-year period. The researcher finds the follow- ing distribution of death sentence decisions by race:

Race Sentenced to Death Not Sentenced to Death

White 8 73

African American 16 52

a. Calculate phi for these data.

b. Calculate Goodman and Kruskal’s tau for these data.

c. Using the values that you calculated for phi and tau, how strongly related are the race of the offender and receiving a death sentence?

13.2 Silver Bullet Treatment Services claims to have an effective system for treating criminal offenders. As evidence for the effectiveness of its pro- gram, a spokesperson from the organization presents information on rearrest within one year for 100 individuals randomly assigned to the treatment program and for 100 individuals randomly assigned to a control group. The distribution of cases follows:

Experimental Condition Not Rearrested Rearrested

Treatment group 75 25

Control group 40 60

a. Calculate phi for these data.

b. Calculate Goodman and Kruskal’s tau for these data.

�̂d � �4(r

2 � 1)(c � 1)

9Nr

2(c � 1)

�̂� � �4(r � 1)(c � 1) 9Nrc

�̂ � � 4(r � 1)(c � 1)

9N (r � 1)(c � 1)

391

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

c. Calculate lambda for these data.

d. Based on these three measures of association, what can you con- clude about the strength of the relationship between the treatment and rearrest?

13.3 A graduate student is interested in the relationship between the gen- der of a violent crime victim and the victim’s relationship to the of- fender. To study this relationship, the student analyzes survey data collected on a random sample of adults. Among those persons who had been victims of violent crimes, the student finds the following dis- tribution of cases by gender:

Relationship of Offender to Victim

Gender Stranger Friend Partner

Male 96 84 21

Female 55 61 103

a. Calculate V for these data.

b. Calculate Goodman and Kruskal’s tau for these data.

c. Calculate lambda for these data.

d. Based on these three measures of association, what can you con- clude about the strength of the relationship between gender and the victim’s relationship to a violent offender?

13.4 In an attempt to explore the relationship between type of legal repre- sentation and method of case disposition, a student working on a re- search project randomly selects a small sample of cases from the local court. The student finds the following distribution of cases:

Method of Case Disposition

Type of Legal Convicted by Convicted by Guilty

Representation Trial Plea Acquitted

Privately retained 10 6 4

Public defender 3 17 2

Legal aid 3 13 1

a. Calculate V for these data.

b. Calculate Goodman and Kruskal’s tau for these data.

c. Calculate lambda for these data.

d. Based on these three measures of association, what should the stu- dent conclude about the relationship between type of legal repre- sentation and method of case disposition?

392

E X E R C I S E S

13.5 A researcher interested in the link between attacking other students and being bullied by other students at school used data from a self- report survey administered to a random sample of teenagers. The dis- tribution of responses was as follows:

Attacked Another Student

Bullied Never Once Two or More Times

Never 59 22 19

Once 31 44 52

Two or more times 25 29 61

a. Calculate gamma for these data.

b. Calculate �b for these data.

c. Calculate Somers’ d for these data.

d. Interpret each of the three measures of association. What can you conclude about the relationship between being bullied and attack- ing other students?

13.6 In response to an increasing reluctance of individuals to serve on ju- ries, a study is commissioned to investigate what might account for the public’s change of heart. Wondering whether prior jury experience has any effect on how favorably the jury system is viewed, a re- searcher constructs the following table:

“How would you rate the current jury system?”

Very Very

Served on a jury Unfavorable Unfavorable Favorable Favorable

Never 22 20 21 26

Once 11 19 12 13

Two or three times 18 23 9 6

Four or more times 21 15 7 4

a. Calculate gamma for these data.

b. Calculate �b for these data.

c. Calculate Somers’ d for these data.

d. Interpret each of the three measures of association. What can you conclude about the relationship between serving on a jury and atti- tudes about the jury system?

13.7 A researcher interested in the relationship between attitudes about school and drug use analyzed data from a delinquency survey admin- istered to a random sample of high school youth. The researcher was

393

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S

particularly interested in how well the youth liked school and their use of marijuana. A cross-tabulation of responses revealed the follow- ing distribution of cases:

Smoked Marijuana in the Last Year

Once or Three or

I Like School Never Twice More Times

Strongly agree 52 20 12

Agree 48 26 20

Disagree 31 32 33

Strongly disagree 35 45 50

a. Calculate gamma for these data.

b. Calculate �c for these data. Explain why �b is not appropriate for these data.

c. Calculate Somers’ d for these data.

d. Interpret each of the three measures of association. What can you conclude about the relationship between liking school and smoking marijuana?

13.8 A public opinion poll asked respondents whether punishments for

whether their political views were liberal, moderate, or conservative. A cross-tabulation of the responses to these two questions shows the fol- lowing distribution of cases:

Criminal punishments should be . . .

Political

Views More Severe About the Same Less Severe

Liberal 8 54 79

Moderate 35 41 37

Conservative 66 38 12

a. Calculate gamma for these data.

b. Calculate �c for these data. Explain why �b is not appropriate for these data.

c. Calculate Somers’ d for these data.

d. Interpret each of the three measures of association. What can you conclude about the relationship between views about politics and attitudes about criminal punishments?

394

convicted criminals should be made more severe, made less severe, or kept about the same. The respondents were also asked to state

C O M P U T E R E X E R C I S E S 395

C o m p u t e r E x e r c i s e s

Many of the measures of association discussed in this chapter are available in

common statistical packages. There are variations in coverage, however, as we

note below. There are also sample files containing examples of syntax for both

SPSS (Chapter_13.sps) and Stata (Chapter_13.do).

SPSS

Each measure of association discussed in this chapter is available in SPSS

with the CROSSTABS command discussed in Chapter 9. Recall from that

discussion that the computation of the chi-square statistic was obtained

through the /STATISTICS= option. To obtain all the nominal and ordinal

measures of association we have discussed in this chapter, we would simply

add to the list of association measures:

where CHISQ is the chi-square, D is Somers’ d, and BTAU and CTAU are

Kendall’s Tau-b and Tau-c, respectively. PHI, LAMBDA, and GAMMA are

self-explanatory. Although it is not listed in the command line, Goodman and

Kruskal’s tau is obtained with the LAMBDA option. Note that you will not need

all of these measures for every comparison, and you should pay some attention

to the level of measurement and select only those measures that make sense

for your data.

In the output generated by this command, you will be presented with the

cross-tabulation of the two variables, followed by additional tables that give

the various measures of association. Depending on which measures you

have requested, you may have three measures of lambda and two measures of

Goodman and Kruskal’s tau reported. The key to reading the correct values for

lambda and tau is to know which variable is the dependent (or outcome) variable.

To illustrate this process, enter the data from Table 13.5 on race and cell

block (follow the same process used in the exercises in Chapter 9). Recall in

the discussion of the cell block assignment data that we treated cell block as

the dependent variable. The value reported for lambda in the line for cell block

as the dependent variable will match the value reported in the text. The value

reported for Goodman and Kruskal’s tau in the line for cell block as the

dependent variable will differ slightly from that reported in the text because

SPSS does not round the prediction errors to the nearest integer; instead, it

records prediction errors with digits after the decimal.

CROSSTABS

/TABLES = row_variable BY column_variable

/STATISTICS = CHISQ PHI LAMBDA GAMMA D BTAU CTAU.

C H A P T E R T H I R T E E N : N O M I N A L A N D O R D I N A L V A R I A B L E S396

Stata

Measures of association are available through the use of the tabulate

command by adding either a list of association measures desired or using all.

Unfortunately, Stata does not compute a wide range of association measures

for nominal and ordinal measures.

The command for computing all of the possible (in Stata) measures of

association is

The output will contain chi-square, Cramer’s V, gamma, and Kendall’s Tau-b.

Cramer’s V will apply to nominal data, while gamma and Kendall’s Tau-b will

apply to ordinal data.

Problems

The first four problems are likely done more effectively in SPSS, since Stata has

limited abilities to compute these measures of association.

2. Enter the data from Table 13.13 into SPSS. Compute the values of

gamma, τ c , and Somers’ d for these data. How do the values of these

measures of association compare to those reported in the text? Test the

statistical significance of each of the measures of association.

3. Enter the data from Exercise 13.2 into SPSS. Compute the values of phi,

tau, and lambda. How do these measures of association compare to the

values that you calculated for this exercise? Test the statistical significance

of each of the measures of association.

4. Enter the data from Exercise 13.6 into SPSS. Compute the values of

gamma, τ c , and Somers’ d for these data. How do these measures of asso-

ciation compare to the values that you calculated for this exercise?

5. Open the NYS data file (nys_1.sav, nys_1_student.sav, or nys_1.dta). Each

pair of variables listed below was tested for a relationship using the

chi-square test in the computer exercises at the end of Chapter 9. For each

pair of variables, determine the level of measurement (nominal or ordinal)

and the dependent and the independent variables; then compute appropri-

ate measures of association to the extent that you are able. Interpret each

of the measures of association that you have computed. Test the statistical

significance of each of the measures of association. What can you con-

clude about the relationship between each pair of variables?

a. What is the relationship between ethnicity and grade point average?

b. What is the relationship between marijuana use among friends and the

youth’s attitudes about marijuana use?

tabulate row_variable column_variable, all

1. Enter the data from Table 13.7 into SPSS. Compute the values of

Cramer’s V, tau, and lambda for these data. How do the values of these

measures of association compare to those reported in the text?

C O M P U T E R E X E R C I S E S 397

c. What is the relationship between the importance of going to college

and the importance of having a job?

d. What is the relationship between grade point average and the impor-

tance of having a job?

e. What is the relationship between the youth’s sex and the importance of

having friends?

f. What is the relationship between the importance of having a job and

the youth’s attitudes about having a job?

6. Open the Pennsylvania Sentencing data file (pcs_98.sav or pcs_98.dta).

Each pair of variables listed below was tested for a relationship using the

chi-square test in the computer exercises at the end of Chapter 9. For each

pair of variables, determine the level of measurement (nominal or ordinal)

and the dependent and the independent variables; then compute appropri-

ate measures of association. Interpret each of the measures of association

that you have computed. Test the statistical significance of each of the

measures of association. What can you conclude about the relationship

between each pair of variables?

a. Is the sex of the offender related to the method of conviction?

b. Is the race–ethnicity of the offender related to whether the offender

was incarcerated or not?

c. Is the method of conviction related to the type of punishment

received?

d. Is the type of conviction offense related to the method of conviction?

Measuring Association for Interval-Level

Data: Pearson’s Correlation Coefficient

What Does a Correlation Coefficient Describe?

When Might Pearson’s r Provide Misleading Results?

C h a p t e r f o u r t e e n

T h e l i n e a r c o r r e l a t i o n c o e f f i c i e n t

T e s t i n g f o r s t a t i s t i c a l s i g n i f i c a n c e

What are the Characteristics of Pearson’s r?

What are the Characteristics of Spearman’s r?

What is the Test of Statistical Significance for Pearson’s r?

What is the Test of Statistical Significance for Spearman’s r?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_14,

© Springer Science+Business Media New York 2014

THIS CHAPTER INTRODUCES the linear correlation coefficient, a widely

used descriptive statistic that enables the researcher to describe the rela-

tionship between two interval-level measures. This situation is encoun-

tered often in criminal justice research. For example, researchers may

want to establish whether number of prior arrests is related to age, edu-

cation, or monthly income. Similarly, it is common in criminal justice re-

search to ask whether the severity of a sanction measured on an interval

scale (e.g., number of years sentenced to imprisonment or amount of a

fine) is related to such variables as the amount stolen in an offense or

the number of prior arrests or convictions of a defendant. We also exam-

ine an alternative rank-order measure of association that may be used

when the linear correlation coefficient will lead to misleading results.

M e a s u r i n g A s s o c i a t i o n B e t w e e n T w o I n t e r v a l - L e v e l V a r i a b l e s

example, that we are presented with the data in Table 14.1. Can we find a

simple way of expressing the relationship between these two variables?

For each of the 15 young offenders in our sample, we have informa-

tion regarding age and number of arrests over the last year. The mean

age of the sample overall is 17.1 years. The mean number of arrests is

4.9. These statistics describe the characteristics of our sample overall,

One way to understand this relationship is to change one of these

measures into a categorical variable. For example, we might divide the

399

of examining a new statistic to describe the relationship between two

It may not be intuitively obvious why we need to go to the trouble

interval-level measures. Why can’t we just use the means, as we did when

we examined interval-level measures in Chapters 11 and 12? Suppose, for

but, importantly, they do not help us to understand the relationship

between age and arrests in the study.

offenders into two groups—one consisting of offenders under age 18

and the other of offenders 18 and older. Then we could use the same

approach taken in earlier chapters and simply compare the means for

the younger and older groups, as shown in Table 14.2. On average, the

older offenders appear to have more arrests than the younger offenders

( versus ).

Similarly, we could divide arrests into categories and compare the

mean age of offenders in each category. For example, Table 14.3 divides

arrests into three categories: low number of arrests (less than 3), moder-

ate number of arrests (3–8), and high number of arrests (9 and above).

This table again shows that, on average, older offenders have more ar-

rests than younger ones. In this case, the mean age for the high-arrest

X � 2.500X � 7.571

Age and Number of Arrests over the Last Year for 15 Young Offenders

SUBJECT NUMBER OF ARRESTS AGE

1 0 14 2 1 13 3 1 15 4 2 13 5 2 14 6 3 14 7 3 17 8 4 19 9 4 21

10 6 19 11 8 16 12 9 18 13 9 20 14 10 21 15 11 22

X � 17.0667X � 4.8667

Table 14.1

Mean Numbers of Arrests for Offenders Under Age 18 versus Those Age 18 and Older

NUMBER OF ARRESTS NUMBER OF ARRESTS

(UNDER AGE 18) (AGE 18 AND OLDER)

0 4 1 4 1 6 2 9 2 9 3 10 3 11 8

X � 7.5714X � 2.5000

Table 14.2

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A400

group was 20.3 and those for the moderate- and low-arrest groups were

17.7 and 13.8, respectively.

Although this approach allowed us to come to a general conclusion

regarding the relationship between age and arrests in our sample, it

forced us to convert one measure from an interval- to a nominal-level

variable. In each example, we had to take a step down the ladder of

measurement, which means that we did not use all of the information

provided by our data. This, of course, violates one of the general princi-

ples stated earlier in the text: Statistics based on more information are

generally preferred over those based on less information.

But how can we describe the relationship between two interval-level

variables without converting one to a nominal scale? A logical solution to

this dilemma is provided by a coefficient named after Karl Pearson, a

noted British statistician who died in 1936. Pearson’s r estimates the

correlation, or relationship, between two measures by comparing how

specific individuals stand relative to the mean of each measure. Pear-

son’s correlation coefficient (r) has become one of the most widely

used measures of association in the social sciences.

P e a r s o n ’ s C o r r e l a t i o n C o e f f i c i e n t

Pearson’s r is based on a very simple idea. If we use the mean of each

distribution as a starting point, we can then see how specific individuals

in the sample stand on each measure relative to its mean. If, in general,

people who are above average on one trait are also above average on

another, we can say that there is a generally positive relationship be-

tween the two traits. That is, being high, on average, on one trait is re-

lated to being high, on average, on the other. If, in contrast, people who

are higher, on average, on one trait tend to be low, on average, on the

Mean Ages for Offenders with Low, Moderate, and High Numbers of Arrests

LOW NUMBER OF ARRESTS MODERATE NUMBER OF ARRESTS HIGH NUMBER OF ARRESTS

(0–2) (3–8) (9�)

14 14 18 13 17 20 15 19 21 13 21 22 14 19

16 X � 20.2500X � 17.6667X � 13.8000

Table 14.3

P E A R S O N ’ S C O R R E L A T I O N C O E F F I C I E N T 401

other, then we conclude that there is a negative relationship between

those traits.

To illustrate these relationships, let’s use the data presented in Table

14.1. If we put a plus next to each subject whose average age or number

of arrests is above the mean for the sample overall and a minus next to

those whose average is below the mean, a pattern begins to emerge (see

Table 14.4). When a subject is above average in number of arrests, the

subject is also generally above average in age. This is true for five of the

six subjects above average in number of arrests (subjects 10, 12, 13, 14,

and 15). Conversely, when a subject is below average in number of ar-

rests, the subject is generally below the mean age for the sample. This is

true for seven of the nine subjects below average in number of arrests

(subjects 1 through 7).

Accordingly, for this sample, subjects generally tend to stand in the

same relative position to the mean for both age and arrests. When indi-

viduals in the sample have a relatively high number of arrests, they also

tend to be relatively older. When they have fewer arrests, they tend to

be younger than average for the sample. A simple mathematical way to

express this relationship is to take the product of the signs. By doing

this, we find that for 12 of the 15 subjects, the result is a positive value

(see Table 14.4). Put simply, 12 of the cases move in the same direction

relative to the mean. The relationship observed in this case is generally

positive.

A Positive Relationship Between Age and Number of Arrests for 15 Young Offenders Relative to the Means

ABOVE OR ABOVE OR

NUMBER OF BELOW THE BELOW THE PRODUCT OF

SUBJECT ARRESTS MEAN? AGE MEAN? THE SIGNS

1 0 � 14 � �

2 1 � 13 � �

3 1 � 15 � �

4 2 � 13 � �

5 2 � 14 � �

6 3 � 14 � �

7 3 � 17 � �

8 4 � 19 � �

9 4 � 21 � �

10 6 � 19 � �

11 8 � 16 � �

12 9 � 18 � �

13 9 � 20 � �

14 10 � 21 � �

15 11 � 22 � �

X � 17.0667X � 4.8667

Table 14.4

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A402

A generally negative relationship can be illustrated by reversing the

scores for arrests in Table 14.4. That is, the first subject does not have 0

arrests, but 11; the second does not have 1 arrest, but 10; and so forth. If

we now indicate each subject’s placement relative to the mean, we ob-

tain the set of relationships listed in Table 14.5. In this table, subjects

who are above average in number of arrests are generally below average

in age, and subjects who are below average in number of arrests are

generally above average in age. The products of these signs are mostly

negative. Put differently, the scores generally move in opposite direc-

tions relative to the mean. There is still a relationship between age and

number of arrests, but in this case the relationship is negative.

This is the basic logic that underlies Pearson’s r. However, we need to

take into account two additional pieces of information to develop this

correlation coefficient. The first is the values of scores. Using plus (�)

and minus (�) divides the scores into categories and thus does not take

full advantage of the information provided by interval-level measures.

Accordingly, instead of taking the product of the signs, we take the

product of the difference between the actual scores and the sample

means. This measure is termed the covariation of scores and is ex-

pressed mathematically in Equation 14.1.

Equation 14.1Covariation of scores � � N

i�1

(X1i � X1)(X2i � X2)

A Negative Relationship Between Age and Number of Arrests for 15 Young Offenders Relative to the Means

ABOVE OR ABOVE OR

NUMBER OF BELOW THE BELOW THE PRODUCT OF

SUBJECT ARRESTS MEAN? AGE MEAN? THE SIGNS

1 11 � 14 � �

2 10 � 13 � �

3 9 � 15 � �

4 9 � 13 � �

5 8 � 14 � �

6 6 � 14 � �

7 4 � 17 � �

8 4 � 19 � �

9 3 � 21 � �

10 3 � 19 � �

11 2 � 16 � �

12 2 � 18 � �

13 1 � 20 � �

14 1 � 21 � �

15 0 � 22 � �

X � 17.0667X � 4.8667

Table 14.5

P E A R S O N ’ S C O R R E L A T I O N C O E F F I C I E N T 403

Table 14.6 illustrates what we gain by including the values of the

scores. We now have not only a measure of the subjects’ placement on

both variables relative to the mean—the sign of the relationship— but

also an estimate of how strongly the scores vary from the mean. In

general, for this distribution, the stronger the deviation from the mean

on one variable, the stronger the deviation on the second variable. For

example, if we look at the scores most distant in value from the mean

in terms of number of arrests over the last year, we also find the scores

most distant in terms of age. Those subjects with either zero or one ar-

rest are not just younger, on average, than other subjects; they are

among the youngest offenders overall in the sample. Similarly, those

with the most arrests (10 or 11) are also the oldest members of the

sample (ages 21 and 22).

The covariation of scores provides an important piece of information

for defining Pearson’s r. However, the size of the covariation between

two measures depends on the units of measurement used. To permit

comparison of covariation across variables with different units of

measurement, we must standardize the covariation between the two

Covariation of Number of Arrests (X1) and Age (X2) for 15 Young Offenders

NUMBER

OF ARRESTS AGE

SUBJECT X1 X2

1 0 �4.8667 14 �3.0667 14.9247

2 1 �3,8667 13 �4.0667 15.7247

3 1 �3.8667 15 �2.0667 7.9913

4 2 �2.8667 13 �4.0667 11.6580

5 2 �2.8667 14 �3.0667 8.7913

6 3 �1.8667 14 �3.0667 5.7246

7 3 �1.8667 17 �0.0667 0.1245

8 4 �0.8667 19 1.9333 �1.6756

9 4 �0.8667 21 3.9333 �3.4090

10 6 1.1333 19 1.9333 2.1910

11 8 3.1333 16 �1.0667 �3.3423

12 9 4.1333 18 0.9333 3.8576

13 9 4.1333 20 2.9333 12.1242

14 10 5.1333 21 3.9333 20.1908

15 11 6.1333 22 4.9333 30.2574

� N

i�1

(X1i � X1)(X2i � X2) � 125.1332X2 � 17.0667X1 � 4.8667

(X1i � X1)(X 2i � X 2)X 2i � X 2X1i � X1

Table 14.6

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A404

variables according to the variability within each. This is done by taking

the square root of the product of the sums of the squared deviations

from the mean for the two variables. Pearson’s r is then the ratio be-

tween the covariation of scores and this value (see Equation 14.2). The

numerator of the equation is the covariation of the two variables. The

Equation 14.2

ables is negative (i.e., when subjects’ scores vary in opposite directions

relative to the mean). The ratio will be largest when there is a good deal

of covariation of the variables and when the variability of scores around

each mean is small. The ratio will be smallest when there is little covaria-

tion and a good deal of variability in the measures. The range of possible

values of r is between �1 and �1.

The Calculation

Calculating Pearson’s r by hand takes a good deal of work. For that rea-

son, in the future you will probably enter the data into a computer and

then use a packaged statistical program to calculate correlation coeffi-

cients. But it will help you to understand r better if we take the time to

calculate an actual example. We will use the data on number of arrests

and age presented in Table 14.1. The calculations needed for Pearson’s r

are shown in Table 14.7.

To calculate the numerator of Equation 14.2, we must first take the

simple deviation of each subject’s score from the mean number of arrests

(Table 14.7, column 3) and multiply it by the deviation of the subject’s

age from the mean age of the sample (column 6). The result, the covari-

ation between the measures, is presented in column 8. So, for the first

subject, the product of the deviations from the means is 14.9247; for the

second, it is 15.7247; and so on. The covariation for our problem,

125.1332, is gained by summing these 15 products.

To obtain the denominator of the equation, we again begin with the

deviations of subjects’ scores from the mean. However, in this case we

Pearson’s r � � N

i�1

(X1i � X1)(X2i � X2)

��� N

i�1

(X1i � X1) 2���

N

i�1

(X2i � X2) 2�

denominator of the equation standardizes this outcome according to

the square root of the product of the variability found in each of the

two distributions, again summed across all subjects.

This ratio will be positive when the covariation between the variables

is positive (i.e., when subjects’ scores vary in the same direction relative

to the mean). It will be negative when the covariation between the vari-

P E A R S O N ’ S C O R R E L A T I O N C O E F F I C I E N T 405

do not multiply the two scores for each subject. Rather, we first square

the deviations from each mean (columns 4 and 7) and then sum the

squared deviations for each variable. The sum of the squared deviations

of each score from the mean number of arrests is equal to 187.7333; the

sum of the squared deviations of each score from the mean age is equal

to 138.9326. Next we take the product of those deviations, and finally we

take the square root of that product.

W orking It Out

� 161.5001

� �26,082.2755

��� N

i�1

(X1i � X1) 2���

N

i�1

(X2i � X2) 2� � �(187.7333)(138.9326)

Calculations for the Correlation of Number of Arrests (X1) and Age (X2) for 15 Young Offenders

NUMBER OF ARRESTS AGE

SUBJECT X1 X2

(1) (2) (3) (4) (5) (6) (7) (8)

1 0 �4.8667 23.6848 14 �3.0667 9.4046 14.9247 2 1 �3.8667 14.9514 13 �4.0667 16.5380 15.7247 3 1 �3.8667 14.9514 15 �2.0667 4.2712 7.9913 4 2 �2.8667 8.2180 13 �4.0667 16.5380 11.6580 5 2 �2.8667 8.2180 14 �3.0667 9.4046 8.7913 6 3 �1.8667 3.4846 14 �3.0667 9.4046 5.7246 7 3 �1.8667 3.4846 17 �0.0667 0.0044 0.1245 8 4 �0.8667 0.7512 19 1.9333 3.7376 �1.6756 9 4 �0.8667 0.7512 21 3.9333 15.4708 �3.4090

10 6 1.1333 1.2844 19 1.9333 3.7376 2.1910 11 8 3.1333 9.8176 16 �1.0667 1.1378 �3.3423 12 9 4.1333 17.0842 18 0.9333 0.8710 3.8576 13 9 4.1333 17.0842 20 2.9333 8.6042 12.1242 14 10 5.1333 26.3508 21 3.9333 15.4708 20.1908 15 11 6.1333 37.6174 22 4.9333 24.3374 30.2574

X2 � 17.0667X1 � 4.8667

� N

i�1

(X1i � X1)(X2i � X2) � 125.1332� N

i�1

(X2i � X 2) 2 � 138.9326�

N

i�1

(X1i � X1) 2 � 187.7333

(X1i � X1)(X 2i � X 2)(X2i � X2) 2X 2i � X 2(X1i � X1)

2X1i � X1

Table 14.7

This leaves us with a value of 161.5001 for the denominator of our

equation.

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A406

Our correlation is about 0.77, meaning that the correlation between age

and number of arrests is a positive one. As number of arrests increases,

so does the average age of the offenders in our sample. But what is the

strength of this relationship? Is it large or small? As discussed in Chapter

12 when we examined the correlation coefficient eta, whether something

is large or small is in good measure a value judgment. The answer de-

pends in part on how the result compares to other research in the same

area of criminal justice. For example, if other studies produced correla-

tions that were generally much smaller, we might conclude that the rela-

tionship in our sample was a very strong one. Jacob Cohen suggests that

a correlation of 0.10 may be defined as a small relationship; a correlation

of 0.30, a moderate relationship; and a correlation of 0.50, a large rela-

tionship.1 On this yardstick, the relationship observed in our example is

a very strong one.

A Substantive Example: Crime and Unemployment in California

W orking It Out

� 0.7748

� 125.1332

161.5001

� 125.1332

�(187.7333)(138.9326)

Pearson’s r � � N

i�1

(X1i � X1)(X2i � X2)

��� N

i�1

(X1i � X1) 2���

N

i�1

(X2i � X2) 2�

We are now ready to calculate Pearson’s r for our example. We sim-

ply take the covariation of 125.1332 and divide it by 161.5001, to get

0.7748:

1See Jacob Cohen, Statistical Power Analysis for the Behavioral Sciences (Hillsdale, NJ:

statisticians develop standardized estimates of “effect size.”

An area of study that has received extensive attention from criminologists

is the relationship between crime rates and other social or economic

indicators, such as unemployment. An example of such data is provided

in Table 14.8, which presents the burglary rate and the unemployment

Lawrence Erlbaum, 1988), pp. 79–80. In Chapter 21, we discuss in greater detail how

P E A R S O N ’ S C O R R E L A T I O N C O E F F I C I E N T 407

Unemployment Rate and Burglary Rate for 58 California Counties in 1999

UNEMPLOYMENT RATE BURGLARY RATE (PER 100,000)

COUNTY (X1) (X2)

Alameda 3.5 837.89 Alpine 9.1 2,037.49 Amador 4.6 818.55 Butte 6.8 865.04 Calaveras 6.9 989.76 Colusa 15.9 520.06 Contra Costa 3.0 664.73 Del Norte 8.0 1,200.91 El Dorado 3.9 509.87 Fresno 13.4 924.10 Glenn 11.2 845.29 Humboldt 6.4 1,027.79 Imperial 23.4 1,526.40 Inyo 5.7 511.12 Kern 11.4 960.18 Kings 13.1 649.22 Lake 7.7 1,333.21 Lassen 7.0 361.24 Los Angeles 5.9 610.28 Madera 11.5 929.32 Marin 1.9 526.98 Mariposa 7.4 775.92 Mendocino 6.7 843.92 Merced 13.3 1,214.69 Modoc 8.5 325.08 Mono 6.7 957.95 Monterey 9.6 570.14 Napa 3.3 477.54 Nevada 4.1 455.37 Orange 2.6 464.52 Placer 3.2 646.12 Plumas 9.0 1,030.58 Riverside 5.4 1,049.18 Sacramento 4.2 925.61 San Benito 8.0 845.75 San Bernadino 4.8 883.02 San Diego 3.1 539.82 San Francisco 3.0 744.81 San Joaquin 8.8 896.85 San Luis Obispo 3.2 540.79 San Mateo 2.0 355.82 Santa Barbara 3.9 444.07 Santa Clara 3.0 347.57 Santa Cruz 6.3 647.73 Shasta 7.0 823.95 Sierra 9.2 699.71 Siskiyou 10.3 575.09 Solano 4.6 769.30 Sonoma 2.7 555.44 Stanislaus 10.5 1,057.99 Sutter 13.0 859.11 Tehama 6.7 816.55 Trinity 11.5 676.23 Tulare 16.5 1,047.32 Tuolumne 6.5 908.79 Ventura 4.8 491.86 Yolo 4.3 591.28 Yuba 11.6 1,366.76

Table 14.8

Calculations for the Correlation of Unemployment Rate (X1) and Burglary Rate (X2) for 58 California Counties

UNEMPLOYMENT RATE BURGLARY RATE

X1i X2i

3.5 15.2639 837.89 2,208.9342 �183.6216 9.1 2.8666 2,037.49 1,554,009.8148 2,110.6173 4.6 7.8787 818.55 765.0369 �77.6369 6.8 0.3683 865.04 5,498.1187 �45.0012 6.9 0.2569 989.76 39,548.9985 �100.8068

15.9 72.1327 520.06 73,349.2681 �2,300.1922 3.0 19.4208 664.73 15,916.5222 555.9776 8.0 0.3518 1,200.91 168,115.8264 243.1824 3.9 12.2983 509.87 78,972.6338 985.5115

13.4 35.9172 924.10 17,744.7176 798.3367 11.2 14.3876 845.29 2,959.2838 206.3420

6.4 1.0138 1,027.79 56,121.2783 �238.5339 23.4 255.7792 1,526.40 540,973.9304 11,763.0738

5.7 2.9135 511.12 78,271.6446 477.5406 11.4 15.9448 960.18 28,658.8671 675.9891 13.1 32.4114 649.22 20,070.5872 �806.5455

7.7 0.0859 1,333.21 294,110.2232 158.9538 7.0 0.1656 361.24 184,599.7240 174.8249 5.9 2.2707 610.28 32,620.2250 272.1623

11.5 16.7535 929.32 19,162.6711 566.6050 1.9 30.3259 526.98 69,648.8576 1,453.3298 7.4 0.0000 775.92 224.1219 0.1033 6.7 0.4997 843.92 2,812.1067 �37.4864

13.3 34.7286 1,214.69 179,605.8467 2,497.4917 8.5 1.1949 325.08 216,979.6082 �509.1777 6.7 0.4997 957.95 27,908.8097 �118.0942 9.6 4.8097 570.14 48,730.8716 �484.1284 3.3 16.8666 477.54 98,188.6612 1,286.9000 4.1 10.9356 455.37 112,574.1401 1,109.5334 2.6 23.1063 464.52 106,517.8338 1,568.8313 3.2 17.6980 646.12 20,958.5556 609.0359 9.0 2.5380 1,030.58 57,450.9605 381.8490 5.4 4.0276 1,049.18 66,713.3625 �518.3608 4.2 10.2842 925.61 18,149.2898 �432.0313 8.0 0.3518 845.75 3,009.5428 32.5371 4.8 6.7959 883.02 8,487.8079 �240.1719 3.1 18.5494 539.82 63,036.4964 1,081.3364 3.0 19.4208 744.81 2,123.4309 203.0730 8.8 1.9407 896.85 11,227.3733 147.6119 3.2 17.6980 540.79 62,550.3601 1,052.1486 2.0 29.2346 355.82 189,286.5140 2,352.3838 3.9 12.2983 444.07 120,284.5979 1,216.2655 3.0 19.4208 347.57 196,533.2430 1,953.6700 6.3 1.2252 647.73 20,494.9860 158.4646 7.0 0.1656 823.95 1,092.9173 �13.4518 9.2 3.2152 699.71 8,313.9201 �163.4961

10.3 8.3700 575.09 46,569.9421 �624.3330 4.6 7.8787 769.30 466.1583 60.6029 2.7 22.1549 555.44 55,437.0321 1,108.2429

10.5 9.5673 1,057.99 71,342.0361 826.1648 13.0 31.2828 859.11 4,653.8729 381.5574

(X1i � X1)(X 2i � X 2)(X2i � X2) 2(X1i � X1)

2

Table 14.9

(continued on next page)

P E A R S O N ’ S C O R R E L A T I O N C O E F F I C I E N T 409

Table 14.9 presents the detailed calculations for the unemployment

and burglary data from California. The sum of the covariations is

37,128.9297, the sum of the squared deviations for the unemployment

rate is 1,010.3570, and the sum of the squared deviations for the bur-

glary rate is 5,659,402.5114. After inserting these values into Equation

14.2, we find that r � 0.4910. The positive correlation between the un-

employment rate and the burglary rate means that counties with higher

rates of unemployment also tended to have higher rates of burglary,

while counties with lower rates of unemployment tended to have lower

rates of burglary.

W orking It Out

� 0.4910

� 37,128,9297

�(1,010.3570)(5,659,402.5114)

r � � N

i�1

(X1i � X1)(X2i � X2)

�� � N

i�1

(X1i � X1) 2�� �

N

i�1

(X2i � X2) 2�

Calculations for the Correlation of Unemployment Rate (X1) and Burglary Rate (X2) for 58 California Counties (Continued)

UNEMPLOYMENT RATE BURGLARY RATE

X1i X2i

6.7 0.4997 816.55 658.3997 �18.1386 11.5 16.7535 676.23 13,147.0761 �469.3177 16.5 82.6845 1,047.32 65,755.9859 2,331.7373

6.5 0.8225 908.79 13,900.2449 �106.9229 4.8 6.7959 491.86 89,419.3595 779.5431 4.3 9.6528 591.28 39,844.4316 620.1705

11.6 17.5821 1,366.76 331,625.4507 2,414.6776

X2 � 790.8907X1 � 7.4069

� N

i�1

� 37,128.9297� N

i�1

� 5,659,402.5114� N

i�1

� 1,010.3570

(X1i � X1)(X 2i � X 2)(X2i � X2) 2(X1i � X1)

2

Table 14.9

rate for all 58 counties in California in 1999. The burglary rate represents

the number of burglaries per 100,000 population, and the unemployment

rate represents the percentage of persons actively looking for work who

have not been able to find employment.

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A410

Nonlinear Relationships and Pearson’s r

Pearson’s r allows us to assess the correlation between two interval-level

measures, taking into account the full amount of information that these

measures provide. However, it assesses the strength of only a linear re-

lationship. If the correlation between two variables is not linear, then

Pearson’s r will give very misleading results.

A simple way to see this is to look at scatterplots, or scatter dia-

grams, representing different types of relationships. A scatterplot posi-

tions subjects according to their scores on both variables being examined.

Figure 14.1 represents the subjects in our example concerning age and

number of arrests. The first case (age � 14, arrests � 0) is represented

by the dot with a 1 next to it. The overall relationship in this example is

basically linear and positive. That is, the dots move together in a positive

direction (as age increases, so too do arrests). A scatterplot of the data in

Table 14.5, where the highest number of arrests is found among younger

rather than older subjects, is presented in Figure 14.2. In this case, the

scatterplot shows a negative relationship (as age increases, number of ar-

rests decreases).

But what would happen if there were a curvilinear relationship be-

tween age and number of arrests? That is, what if the number of arrests

for both younger and older subjects was high, and the number for those

of average age was low? This relationship is illustrated in Figure 14.3. In

Scatterplot Showing a Positive Relationship Between Age

and Number of Arrests for 15 Subjects Figure 14.1

0

12

10

8

6

4

2

12 14 16 18 20 22 24

Age

N u

m b

er o

f A

rr es

ts

1

P E A R S O N ’ S C O R R E L A T I O N C O E F F I C I E N T 411

Scatterplot Showing a Negative Relationship

Between Age and Number of Arrests for 15 Subjects Figure 14.2

0

12

10

8

6

4

2

12 14 16 18 20 22 24

Age

N u

m b er

o f

A rr

es ts

Scatterplot Showing a Curvilinear Relationship

Between Age and Number of Arrests for 15 Subjects Figure 14.3

0

12

10

8

6

4

2

12 14 16 18 20 22 24

Age

N u

m b er

o f

A rr

es ts

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A412

this scatterplot, the relationship is clear: The number of arrests declines

until age 17 and then increases. However, Pearson’s r for these data is

close to 0. If there is a relationship, why does this happen? Table 14.10

shows why. Subjects who are either much above or much below the

mean for age have large numbers of arrests. The covariance for these

subjects is accordingly very high. However, for those below the mean in

age, the covariance is negative, and for those above the mean, the co-

variance is positive. If we add these scores together, they cancel each

other out. As a result, Pearson’s r for this example is close to 0.

W orking It Out

� 0.0066

� 0.8668

�(123.7338)(138.9326)

Pearson’s r � � N

i�1

(X1i � X1)(X2i � X2)

��� N

i�1

(X1i � X1) 2���

N

i�1

(X2i � X2) 2�

Curvilinear Relatonship: Calculations for the Correlation of Number of Arrests (X1) and Age (X2) for 15 Young Offenders

NUMBER OF ARRESTS AGE

SUBJECT X1 X2

1 9 1.8667 3.4846 14 �3.0667 9.4046 �5.7246 2 11 3.8667 14.9514 13 �4.0667 16.5380 �15.7247 3 6 �1.1333 1.2844 15 �2.0667 4.2712 2.3422 4 10 2.8667 8.2180 13 �4.0667 16.5380 �11.6580 5 8 0.8667 0.7512 14 �3.0667 9.4046 �2.6579 6 7 �0.1333 0.0178 14 �3.0667 9.4046 0.4088 7 2 �5.1333 26.3508 17 �0.0667 0.0044 0.3424 8 5 �2.1333 4.5510 19 1.9333 3.7376 �4.1243 9 9 1.8667 3.4846 21 3.9333 15.4708 7.3423

10 6 �1.1333 1.2844 19 1.9333 3.7376 �2.1910 11 2 �5.1333 26.3508 16 �1.0667 1.1378 5.4757 12 4 �3.1333 9.8176 18 0.9333 0.8710 �2.9243 13 7 �0.1333 0.0178 20 2.9333 8.6042 �0.3910 14 10 2.8667 8.2180 21 3.9333 15.4708 11.2756 15 11 3.8667 14.9514 22 4.9333 24.3374 19.0756

X2 � 17.0667X1 � 7.1333

� N

i�1

(X1i � X1)(X2i � X2)� 0.8668� N

i�1

(X2i � X2) 2 � 138.9326�

N

i�1

(X1i � X1) 2 � 123.7338

(X1i � X1)(X 2i � X 2)(X2i � X2) 2X 2i � X 2(X1i � X1)

2X1i � X1

Table 14.10

P E A R S O N ’ S C O R R E L A T I O N C O E F F I C I E N T 413

Pearson’s r will provide a good estimate of correlation when the

relationship between two variables is approximately linear. However,

a strong nonlinear relationship will lead to a misleading correlation

coefficient. Figure 14.4 provides examples of a number of nonlinear

relationships. These examples illustrate why it is important to look at

the scatterplot of the relationship between two interval-level measures

to establish that it is linear before calculating Pearson’s correlation

coefficient. Linear relationships are much more common in criminal

justice than nonlinear ones. But you would not want to conclude,

based on r, that there was a very small relationship between two vari-

ables when in fact there was a very strong nonlinear correlation be-

tween them.

What can you do if the relationship is nonlinear? Sometimes the so-

lution is simply to break up the distribution of scores. For example,

Figure 14.3 shows a nonlinear relationship that results in an r of 0.007.

If we break this distribution at the point where it changes direction,

we can calculate two separate Pearson’s correlation coefficients,

each for a linear relationship. The first would provide an estimate

of the relationship for younger offenders (which is negative), and the

second an estimate of the relationship for older offenders (which is

positive).

For some nonlinear relationships, you may want to consider using al-

ternative statistics. For example, it may be worthwhile to break up your

Examples of Nonlinear RelationshipsFigure 14.4

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A414

sample into a number of groups, or categories, and then look at the

means for each. In some cases, it may be possible to change the form of

the variables and, in doing so, increase the linearity of the relationship

being examined. Although such transformations are beyond the scope of

this text, you should be aware that they provide one solution to prob-

lems of nonlinearity.2

Beware of Outliers

For Pearson’s r, as for other statistics based on deviations from the mean,

outliers can have a strong impact on results. For example, suppose we

add to our study of age and number of arrests (from Table 14.1) one

subject who was very young (12) but nonetheless had an extremely large

number of arrests over the last year (25), as shown in the scatterplot in

Figure 14.5. If we take the covariation for this one relationship (see sub-

2For a discussion of this issue, see. J. Fox, Linear Statistical Models and Related Meth-

ods (New York: Wiley, 1994).

Scatterplot Showing the Relationship Between Age and Number of Arrests for 16 Subjects,

Including an Outlier Figure 14.5

0

30

20

10

10 12 14 16 18 20 22 24

Age

N u

m b er

o f

A rr

es ts

P E A R S O N ’ S C O R R E L A T I O N C O E F F I C I E N T 415

ject 16 in Table 14.11), we see that it is very large relative to that of other

subjects in our analysis. Because it is negative, it cancels out the positive

covariation produced by the other subjects in the sample. Indeed, with

this subject included, the correlation decreases from 0.77 to 0.10.

Calculations for the Correlation of Number of Arrests (X1) and Age (X2) for 16 Young Offenders

NUMBER OF ARRESTS AGE

SUBJECT X1 X2

1 0 �6.1250 37.5156 14 �2.7500 7.5625 16.84375

2 1 �5.1250 26.2656 13 �3.7500 14.0625 19.21875

3 1 �5.1250 26.2656 15 �1.7500 3.0625 8.96875

4 2 �4.1250 17.0156 13 �3.7500 14.0625 15.46875

5 2 �4.1250 17.0156 14 �2.7500 7.5625 11.34375

6 3 �3.1250 9.7656 14 �2.7500 7.5625 8.59375

7 3 �3.1250 9.7656 17 0.2500 0.0625 �0.78125

8 4 �2.1250 4.5156 19 2.2500 5.0625 �4.78125

9 4 �2.1250 4.5156 21 4.2500 18.0625 �9.03125

10 6 �0.1250 0.0156 19 2.2500 5.0625 �0.28125

11 8 1.8750 3.5156 16 �0.7500 0.5625 �1.40625

12 9 2.8750 8.2656 18 1.2500 1.5625 3.59375

13 9 2.8750 8.2656 20 3.2500 10.5625 9.34375

14 10 3.8750 15.0156 21 4.2500 18.0625 16.46875

15 11 4.8750 23.7656 22 5.2500 27.5625 25.59375

16 25 18.8750 356.2656 12 �4.7500 22.5625 �89.65625

� 567.7496 � 163.000 � 29.5000

� N

i�1

(X1i � X1)(X2i � X2)� N

i�1

(X2i � X2) 2X2 � 16.7500�

N

i�1

(X1i � X1) 2X1 � 6.1250

(X1i � X1)(X 2i � X 2)(X2i � X2) 2X 2i � X 2(X1i � X1)

2X1i � X1

Table 14.11

W orking It Out

� 0.0970

� 29.5000

�(567.7496)(163)

Pearson’s r � � N

i�1

(X1i � X1)(X2i � X2)

��� N

i�1

(X1i � X1) 2���

N

i�1

(X2i � X2) 2�

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A416

What should you do when faced with outliers? When you have just a

few deviant cases in your sample, the best decision may be to exclude

them from your analysis. If you take this approach, it is important to

clearly state that certain cases have been excluded and to explain why.

Before excluding outliers, however, you should compare the correlations

with and without them. When samples are large, deviant cases may have

a relatively small impact; thus, including them may not lead to mislead-

ing results.

When there are a relatively large number of outliers that follow the

general pattern of relationships in your data, it may be better to choose

an alternative correlation coefficient rather than exclude such cases. For

example, suppose we add to our study of age and arrests three different

subjects for whom the relationships are similar to those noted previously,

but the number of arrests and the average age are much higher (see sub-

jects 16, 17, and 18 in Table 14.12). These data are depicted in the scat-

Calculations for the Correlation of Number of Arrests (X1) and Age (X2) for 18 Young Offenders

NUMBER OF ARRESTS AGE

SUBJECT X1 X2

1 0 �11.2778 127.1888 14 �8.5556 73.1983 96.4883

2 1 �10.2778 105.6332 13 �9.5556 91.3095 98.2105

3 1 �10.2778 105.6332 15 �7.5556 57.0871 77.6549

4 2 �9.2778 86.0776 13 �9.5556 91.3095 88.6549

5 2 �9.2778 86.0776 14 �8.5556 73.1983 79.3771

6 3 �8.2778 68.5220 14 �8.5556 73.1983 70.8215

7 3 �8.2778 68.5220 17 �5.5556 30,8647 45.9881

8 4 �7.2778 52.9664 19 �3.5556 12.6423 25.8769

9 4 �7.2778 52.9664 21 �1.5556 2.4199 11.3213

10 6 �5.2778 27.8552 19 �3.5556 12.6423 18.7675

11 8 �3.2778 10.7440 16 �6.556 42.9759 21.4879

12 9 �2.2778 5.1884 18 �4.5556 20.7535 10.3767

13 9 �2.2778 5.1884 20 �2.5556 6.5311 5.8211

14 10 �1.2778 1.6328 21 �1.5556 2.4199 1.9877

15 11 �0.2778 0.0772 22 �0.5556 0.3087 0.1543

16 36 24.7222 611.1872 40 17.4444 304.3071 431.2639

17 40 28.7222 824.9648 50 27.4444 753.1951 788.2635

18 54 42.7222 1,825.1864 60 37.4444 1,402.0831 1,5999.7071

� 4,065.6116 � 3,050.4446 � 3,472.2214

� N

i�1

(X1i � X1)(X2i � X2)� N

i�1

(X2i � X2) 2X2 � 22.5556�

N

i�1

(X1i � X1) 2X1 � 11.2778

(X1i � X1)(X 2i � X 2)(X2i � X2) 2X 2i � X 2(X1i � X1)

2X1i � X1

Table 14.12

P E A R S O N ’ S C O R R E L A T I O N C O E F F I C I E N T 417

terplot in Figure 14.6. In such situations, Pearson’s r is likely to give a

misleading view of the relationship between the two variables. For our

example, the correlation changes from 0.77 to 0.99.

Scatterplot Showing the Relationship Between Age and Number of Arrests

for 18 Subjects, Including Three Outliers Who Follow the General Pattern

Figure 14.6

0

60

40

20

50

30

10

10 20 30 40 50 60 70

Age

N u

m b er

o f

A rr

es ts

W orking It Out

� 0.9859

� 3,472.2214

�(4,065.6116)(3,050.4446)

Pearson’s r � � N

i�1

(X1i � X1)(X2i � X2)

��� N

i�1

(X1i � X1) 2���

N

i�1

(X2i � X2) 2�

In such situations, you may want to use a rank-order correlation coef-

ficient called Spearman’s r. Pearson’s r is generally more appropriate

for interval-level data. However, where a number of outliers are found in

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A418

the distribution, Spearman’s correlation coefficient can provide a use-

ful alternative.

S p e a r m a n ’ s C o r r e l a t i o n C o e f f i c i e n t

Spearman’s r compares the rank order of subjects on each measure

rather than the relative position of each subject to the mean. Like Pear-

son’s r, its range of possible values is between �1 and �1. It is calcu-

lated using Equation 14.3.

Equation 14.3

Let’s calculate rs for our original example with 15 cases (from Table

14.1), and for the example with the additional three outliers (from Table

14.12). To carry out the calculation, we must first rank order the cases, as

rs � 1 �

6� N

i�1

D

2 i

N (N

2 � 1)

Calculation of Difference in Rank (D) for Spearman’s r for 15 Young Offenders

NUMBER OF RANK ARRESTS RANK AGE D

SUBJECT ARRESTS Rk1 AGE Rk2 (Rk1 � Rk2) D 2

1 0 1 14 4 �3 9

2 1 2.5 13 1.5 1 1

3 1 2.5 15 6 �3.5 12.25

4 2 4.5 13 1.5 3 9

5 2 4.5 14 4 0.5 0.25

6 3 6.5 14 4 2.5 6.25

7 3 6.5 17 8 �1.5 2.25

8 4 8.5 19 10.5 �2 4

9 4 8.5 21 13.5 �5 25

10 6 10 19 10.5 �0.5 0.25

11 8 11 16 7 4 16

12 9 12.5 18 9 3.5 12.25

13 9 12.5 20 12 0.5 0.25

14 10 14 21 13.5 0.5 0.25

15 11 15 22 15 0 0

� N

i�1

D

2 i � 98X2 � 17.0667X1 � 4.8667

Table 14.13

S P E A R M A N ’ S C O R R E L A T I O N C O E F F I C I E N T 419

shown in Tables 14.13 and 14.14. We then take the squared difference in

ranks for each subject on the two measures and sum it across all the

cases in our example. This value is multiplied by 6, and then divided by

N(N 2 � 1). The final figure is then subtracted from 1.

Calculation of Difference in Rank (D) for Spearman’s r for 18 Young Offenders

NUMBER OF RANK ARRESTS RANK AGE D

SUBJECT ARRESTS Rk1 AGE Rk2 (Rk1 � Rk2) D 2

1 0 1 14 4 �3 9

2 1 2.5 13 1.5 1 1

3 1 2.5 15 6 �3.5 12.25

4 2 4.5 13 1.5 3 9

5 2 4.5 14 4 0.5 0.25

6 3 6.5 14 4 2.5 6.25

7 3 6.5 17 8 �1.5 2.25

8 4 8.5 19 10.5 �2 4

9 4 8.5 21 13.5 �5 25

10 6 10 19 10.5 �0.5 0.25

11 8 11 16 7 4 16

12 9 12.5 18 9 3.5 12.25

13 9 12.5 20 12 0.5 0.25

14 10 14 21 13.5 0.5 0.25

15 11 15 22 15 0 0

16 36 16 40 16 0 0

17 40 17 50 17 0 0

18 54 18 60 18 0 0

� N

i�1

D

2 i � 98X2 � 22.5556X1 � 11.2778

Table 14.14

W orking It Out No Outliers

� 0.8250

� 1 � 0.1750

� 1 � 588

3360

� 1 � (6)(98)

(15)(224)

rs � 1 �

6� N

i�1

D

2 i

N(N

2 � 1)

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A420

The correlation coefficients for our two distributions (without outliers

and with outliers) are similar. In the case without the outliers, rs � 0.83;

in the case with them, rs � 0.90. The outliers here do not have as much

of an impact because Spearman’s correlation coefficient does not take

into account the actual values of the scores, but only their ranks in the

distribution. Note that the correlation coefficient obtained here for the 15

cases, rs � 0.83, is a bit larger than, although similar to, r � 0.77. Which

is the better estimate of the correlation between these two variables? In

the case without the outliers, Pearson’s r would be preferred because it

takes into account more information (order as well as value). In the case

with the outliers, however, Spearman’s r would be preferred because it

is not affected by the extreme values of the three outliers, but only by

their relative positions in the distributions.

T e s t i n g t h e S t a t i s t i c a l S i g n i f i c a n c e o f P e a r s o n ’ s r

As in Chapter 13, our emphasis in this chapter has been not on statistical

inference but rather on statistical description. Our concern has been to

describe the strength or nature of the relationship between two interval-

level variables. Nonetheless, it is important here, as before, to define

whether the differences observed in our samples can be inferred to the

populations from which they were drawn.

Statistical Significance of r : The Case of Age and Number of Arrests

We can use the t-distribution introduced in Chapter 10 to test for the sig-

nificance of Pearson’s r. We begin by conducting a test of statistical sig-

nificance for our example of the correlation between age and number of

arrests.

W orking It Out With Outliers

� 0.8989

� 1 � 0.1011

� 1 � 588 5814

� 1 � (6)(98)

(18)(323)

rs � 1 �

6� N

i�1

D

2 i

N (N

2 � 1)

S T A T I S T I C A L S I G N I F I C A N C E O F P E A R S O N ’ S r 421

Assumptions:

Level of Measurement: Interval scale.

Population Distribution: Normal distribution of Y around each value of

X (must be assumed because N is not large).

Homoscedasticity.

Linearity.

Sampling Method: Independent random sampling.

Sampling Frame: Youth in one U.S. city.

Hypotheses:

H0: There is no linear relationship between age and number of arrests in

the population of young offenders (rp � 0).

H1: There is a linear relationship between age and number of arrests in

the population of young offenders (rp � 0).

The t-test for Pearson’s r assumes that the variables examined are

measured on an interval scale. In practice, researchers sometimes use

ordinal-scale measures for calculating these coefficients, particularly

when an interval-level measure is related to an ordinal-scale variable.

There is no simple answer to the question of which statistic is most ap-

propriate in such cases, and Pearson’s r is often considered a good solu-

tion. Nonetheless, you should keep in mind that Pearson’s r, like other

statistics that require an interval level of measurement, assumes that the

categories are not only ranked but also equal to one another. When

there are clear differences between the categories, the meaning of r be-

comes ambiguous. For example, suppose you were interested in the re-

lationship between amount stolen in robberies and age, where amount

stolen in robberies was measured on an ordinal scale, with the first cat-

egory as $1–50, the second as $51–200, and subsequent intervals also of

unequal size. If the real relationship between amount stolen and age

was truly linear, with every year of age related to a measured increase

in amount stolen, you would likely get a misleading correlation coeffi-

cient. In this case, an interval-scale measurement would allow you to

represent the linear relationship between amount stolen and age. The

ordinal scale we have described might mask or misrepresent that rela-

tionship. In practice, you should also be wary of using Pearson’s corre-

lation coefficient when the number of categories is small (for example,

less than 5). While r is sometimes used when the researcher wants to

represent the relationship between an ordinal-scale measure and an in-

terval-level variable, it should not be used to define the relationship be-

tween two ordinal-level measures or when nominal-scale measurement

is involved. As noted in earlier chapters, other statistics are more appro-

priate for measuring such relationships.

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A422

Because our test of the statistical significance of r is a parametric test

of significance, we must also make assumptions regarding the popula-

tion distribution. For tests of statistical significance with r, we must

assume a normal distribution. However, in this case, it is useful to think

of this distribution in terms of the joint distribution of scores between

X1 and X2. Assume that the scatterplot in Figure 14.7 represents the

relationship between age and number of arrests for ages 16–19 for the

population of scores. The relationship, as in our sample, is linear. Notice

how the points in the scatterplot are distributed. Suppose we put an

imaginary line through the scatter of points (shown as a real line in Fig-

ure 14.7). There is a clustering of points close to the line, in the middle

Scatterplot Showing Normal Distribution and HomoscedasticityFigure 14.7

16 17 18 19

Age

A

Cross Sections:

Arrests

Age = 16 A

N u

m b er

o f

A rr

es ts

F re

q u

en cy

1 case 5 cases 10 cases 25 cases 50 cases 100 cases

B

C

D

Arrests

Age = 17 B

F re

q u

en cy

Arrests

Age = 18 C

F re

q u

en cy

Arrests

Age = 19 D

F re

q u

en cy

S T A T I S T I C A L S I G N I F I C A N C E O F P E A R S O N ’ S r 423

of the distribution. As we move away from the center of the distribution

of number of arrests for each age (represented by the imaginary line),

there are fewer points. The distribution of number of arrests for each

value of age is basically normal in form, as illustrated in the cross section

for each of the four ages examined. This imaginary population distribu-

tion meets the normality assumption of our test.

One problem in drawing conclusions about our assumptions is that

they relate to the population and not to the sample. Because the popula-

tion distribution is usually unknown, we generally cannot come to solid

conclusions regarding our assumptions about the population. In the case

of an assumption of normality, the researcher is most often aided by the

central limit theorem. When the number of cases in a sample is greater

than 30, the central limit theorem can be safely invoked. For our exam-

ple, we cannot invoke the central limit theorem. Accordingly, our test re-

sults cannot be relied on unless the assumption of a normal distribution

is true for the population to which we infer.

For our t-test, we must also assume that the variances of the joint dis-

tribution of scores are equal. In our example, this means that the spread

of number of arrests around each value of age should be about the

same. This is the assumption of homoscedasticity. To visualize this as-

sumption, it is useful to look again at the scatterplot in Figure 14.7.

We can see that for each age examined, the variance in the distribu-

tion of scores for number of arrests is about equal. That is, the spread of

the scores around our imaginary line for each value of age in this popu-

lation distribution is about equal, whether we look at the cases associ-

ated with the youngest subjects (on the left side of the scatterplot), those

associated with average-age subjects (in the middle of the scatterplot), or

those associated with the oldest offenders (on the right side of the scat-

terplot). With regard to the assumption of homoscedasticity, researchers

generally use the scatterplot of sample cases as an indication of the form

of the population distribution. As with analysis of variance, we are gen-

erally concerned with only marked violations of the homoscedasticity as-

sumption. Given the small number of cases in our sample distribution of

scores, it is very difficult to examine the assumption of homoscedasticity.

Nonetheless, if you look back at Figure 14.1, it seems reasonable to con-

clude that there are no major violations of this assumption.

What would a joint distribution of the relationship between arrests

and age look like if both the normality and the homoscedasticity as-

the points for each age category are not clustered in the center of the

distribution of scores for number of arrests. Indeed, they form a type of

bimodal distribution, with peaks above and below our imaginary line

(see the cross section for each age group). Heteroscedasticity (or un-

equal variances), rather than homoscedasticity, is also represented in

the scatterplot in Figure 14.8. For subjects aged 17 and 19, the scores

sumption were violated? In the scatterplot in Figure 14.8 (page 425),

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A424

are scattered widely. For subjects aged 16 and 18, however, the scores

are tightly clustered around the imaginary line. If this were the popula-

tion distribution of the variables under study, you would want to be very

cautious in applying the t-test to your correlation coefficient.

As with other tests of statistical significance, we assume independent

random sampling. Pearson’s correlation coefficient adds one new as-

sumption to our test—that of linearity. Our null hypothesis is simply that

there is no linear relationship between age and number of arrests, or that

the population correlation coefficient (rp) is equal to 0. The research hy-

pothesis is nondirectional; that is, we simply test for a linear relationship

Scatterplot Showing Nonnormal Distribution and HeteroscedasticityFigure 14.8

16 17 18 19

Age

A

Cross Sections:

Arrests

Age = 16 A

N u

m b er

o f

A rr

es ts

F re

q u

en cy

100 cases 50 cases 25 cases 10 cases 5 cases 1 case B

C

D

Arrests

Age = 17 B

F re

q u

en cy

Arrests

Age = 18 C

F re

q u

en cy

Arrests

Age = 19 D

F re

q u

en cy

S T A T I S T I C A L S I G N I F I C A N C E O F P E A R S O N ’ S r 425

between age and number of arrests (i.e., rp � 0). However, we might

have proposed a directional research hypothesis—for example, that

there is a positive relationship between age and number of arrests in the

population, or that rp � 0.

The Sampling Distribution The sampling distribution is t, with N � 2 de-

grees of freedom. For our example, df � 15 � 2 � 13.

Significance Level and Rejection Region With a two-tailed 0.05 signifi-

cance threshold, the critical value for the t-test (with 13 degrees of free-

dom) is 2.160 (see Appendix 4). We will reject the null hypothesis if the

t-score is greater than 2.160 or less than �2.160. In these cases, the ob-

served significance level of the test will be less than the criterion signifi-

cance level we have chosen.

The Test Statistic It is important to note that there is more than one way

to test statistical significance for r. Equation 14.4 provides a straightfor-

ward estimate of t, based on our calculation of r.

Equation 14.4

Inserting our sample estimates, we calculate that the t-statistic for r is

4.4188:

t � r �N � 2

1 � r

2

W orking It Out

� 4.4188

� 0.7748�32.5256

� 0.7748� 15 � 2

1 � (0.7748)2

t � r �N � 2

1 � r

2

The Decision Since 4.419 is greater than our critical value of t (2.160),

we reject the null hypothesis and conclude that there is a statistically sig-

nificant relationship between age and number of arrests. However, be-

cause we cannot strongly support the assumption of normality in this test

or relax that assumption because N is large, we cannot place strong re-

liance on our test result.

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A426

Statistical Significance of r :

Unemployment and Crime in California

In our analysis of unemployment and burglary rates in California coun-

ties, we found r � 0.4910. We can test the statistical significance of this

result by following the same approach we used in the previous example.

We start by outlining our assumptions and hypotheses.

Assumptions:

Level of Measurement: Interval scale.

Population Distribution: Normal distribution of Y (i.e., burglary rate)

around each value of X (i.e., unemployment rate) (relaxed because N is

large).

Homoscedasticity.

Sampling Method: Independent random sampling (the 58 counties repre-

sent all counties in California in 1999).

Sampling Frame: California counties.

Linearity.3

Hypotheses:

H0: There is no linear relationship between unemployment rate and bur-

glary rate (rp � 0).

H1: There is a linear relationship between unemployment rate and bur-

glary rate (rp � 0).

might question why we would choose to conduct a statistical test of sig-

nificance. Why do we need to make inferences? We already have the

population of scores. One reason might be that we want to look at data

for the year observed as a sample of the relationships that occur over a

number of years. Similarly, we might want to use the data in California

to represent other states. For either of these inferences, we would need

to explain why this sample was representative of the population. An-

other reason we might choose to conduct a statistical test of significance

is to see whether the correlation observed is likely to be a chance occur-

rence. We would expect differences across the counties simply as a

product of the natural fluctuations that occur in statistics. A significance

test in this case can tell us whether the relationship observed is likely to

be the result of such chance fluctuations or whether it is likely to repre-

sent a real relationship between the measures examined.

3It is good practice to examine the sample scatterplot of scores to assess whether this

assumption is likely to be violated. We find no reason to suspect a violation of the as-

sumption when we examine this scatterplot (see Chapter 15, Figure 15.2).

Since we have data for all counties in California for a given year, you

S T A T I S T I C A L S I G N I F I C A N C E O F P E A R S O N ’ S r 427

The Sampling Distribution The sampling distribution is t, with N � 2 de-

grees of freedom. For this example, df � 58 � 2 � 56.

Significance Level and Rejection Region With a two-tailed 0.05 signifi-

cance level and 56 degrees of freedom, interpolation yields estimates of

�2.003 and �2.003 for the critical values of the t-test.4

The Test Statistic We again use Equation 14.4 to calculate a t-value to

test the significance of r. Inserting the values for our data, we find the

value of the t-statistic to be 4.2177:

The Decision Since 4.218 is greater than our critical value of 2.003, we

reject the null hypothesis and conclude that there is a statistically sig-

nificant relationship between the unemployment rate and the burglary

rate.

T e s t i n g t h e S t a t i s t i c a l S i g n i f i c a n c e o f S p e a r m a n ’ s r

For Spearman’s r, we use a nonparametric statistical test. With N 30,

we use an exact probability distribution constructed for the distribution

of differences between ranked pairs (see Appendix 7). For larger sam-

ples, a normal approximation of this test is appropriate. It is con-

structed by taking the difference between the observed value of rs and

the parameter value under the null hypothesis (rs(p)). This value is then

divided by 1 divided by the square root of N � 1, as shown in Equa-

tion 14.5.

Equation 14.5z � rs � rs (p)

1

�N � 1

� 4.2177

� 0.491�73.789165

� 0.491� 58 � 2

1 � (0.491)2

t � r �N � 2

1 � r

2

4The table does not list a t-value for df � 56. We therefore interpolate from the values

of df � 55 (2.004) and df � 60 (2.000).

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A428

Because we are examining less than 15 cases, we will use the exact

probability table presented in Appendix 7.

Assumptions:

Level of Measurement: Ordinal scale.

Population Distribution: No assumption made.

Sampling Method: Independent random sampling.

Sampling Frame: Youth in one U.S. city.

Hypotheses:

H0: There is no linear relationship between the rank order of scores in

the population (rs(p) � 0).

H1: There is a linear relationship between the rank order of scores in the

population (rs(p) � 0).

The Sampling Distribution Because N is small, we use the exact probabil-

ity distribution constructed for Spearman’s r in Appendix 7.

Significance Level and Critical Region As earlier, we use the conventional

0.05 significance threshold. Since our research hypothesis is not direc-

tional, we use a two-tailed rejection region. From Appendix 7, under a

two-tailed 0.05 probability value and an N of 15, we find that an rs

greater than or equal to 0.525 or less than or equal to �0.525 is needed

to reject the null hypothesis.

The Test Statistic In the case of the exact probability distribution, the

test statistic is simply the value of rs. As calculated earlier in this chapter

s equals 0.825.

The Decision Because the observed rs is larger than 0.525, we reject the

null hypothesis and conclude that there is a statistically significant linear

relationship between ranks of age and number of arrests in the popula-

tion. The observed significance level of our test is less than the criterion

significance level we set at the outset (p � 0.05).

Because we use a nonparametric test, we do not need to make

assumptions regarding the population distribution. The null hypothesis

is the same as for the correlation coefficient r ; however, it is concerned

with ranks rather than raw scores.

(see page 420), r

S T A T I S T I C A L S I G N I F I C A N C E O F S P E A R M A N ’ S r 429

C h a p t e r S u m m a r y

Linear correlation coefficients describe the relationship between two

interval-level measures, telling us how strongly the two are associated.

Pearson’s r is a widely used linear correlation coefficient. It examines

the placement of subjects on both variables relative to the mean and

estimates how strongly the scores move together or in opposite direc-

tions relative to the mean. The covariation, which is the numerator of

the Pearson’s r equation, is positive when both scores vary in the

same direction relative to the mean and negative when they vary in

opposite directions. Dividing the covariation by the denominator of

the Pearson’s r equation serves to standardize the coefficient so that it

varies between �1 and �1. Pearson’s r will produce a misleading cor-

relation coefficient if there is a nonlinear relationship between the

variables.

Outliers have a strong impact on Pearson’s r. If there are several out-

liers that follow the general pattern of relationships in the data, Spear-

man’s r may provide less misleading results. Spearman’s r also varies

between �1 and �1. It compares the rank order of subjects on each

measure.

The t distribution may be used to test significance for Pearson’s

correlation coefficient. It is assumed that the variables examined

are measured on an interval scale. There is also an assumption of

normality and a requirement of homoscedasticity. These assumptions re-

late to the joint distribution of X1 and X2. The researcher must also as-

sume linearity. For Spearman’s r, a nonparametric test of statistical signif-

icance is used.

K e y T e r m s

covariation A measure of the extent to

which two variables vary together relative

to their respective means. The covariation

between the two variables serves as the

numerator for the equation to calculate

Pearson’s r.

curvilinear relationship An association

between two variables whose values may

be represented as a curved line when plot-

ted on a scatter diagram.

heteroscedasticity A situation in which

the variances of scores on two or more

variables are not equal. Heteroscedasticity

violates one of the assumptions of the

parametric test of statistical significance for

the correlation coefficient.

linear relationship An association be-

tween two variables whose joint distribu-

tion may be represented in linear form

when plotted on a scatter diagram.

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A430

S y m b o l s a n d F o r m u l a s

r Pearson’s correlation coefficient

rs Spearman’s correlation coefficient

D Difference in rank of a subject on two variables

To calculate the covariation of scores for two variables:

To calculate Pearson’s correlation coefficient:

To calculate Spearman’s correlation coefficient:

To test statistical significance for Pearson’s r :

To test statistical significance for Spearman’s r where N is large:

z � rs � rs (p)

1

�N � 1

t � r �N � 2

1 � r

2

rs � 1 �

6� N

i�1

D

2 i

N (N

2 � 1)

Pearson’s r � � N

i�1

(X1i � X1)(X2i � X2)

��� N

i�1

(X1i � X1) 2���

N

i�1

(X2i � X2) 2�

Covariation of scores � � N

i�1

(X1i � X1)(X2i � X2)

Pearson’s correlation coefficient See

Pearson’s r.

Pearson’s r A commonly used measure of

association between two variables. Pear-

son’s r measures the strength and direction

of linear relationships on a standardized

scale from �1.0 to 1.0.

scatter diagram See scatterplot.

scatterplot A graph whose two axes are

defined by two variables and upon which a

point is plotted for each subject in a sam-

ple according to its score on the two vari-

ables.

Spearman’s correlation coefficient See

Spearman’s r.

Spearman’s r (rs) A measure of associa-

tion between two rank-ordered variables.

Spearman’s r measures the strength and di-

rection of linear relationships on a stan-

dardized scale between �1.0 and 1.0.

S Y M B O L S A N D F O R M U L A S 431

E x e r c i s e s

14.1 A researcher draws four random samples of ten offenders, aged be- tween 30 and 35 years, all of whom are currently serving out a term of imprisonment and all of whom have been in prison before. For each sample, she compares the subjects on the following pairs of variables:

Sample 1 1 2 3 4 5 6 7 8 9 10

X1: Number of convictions 3 5 1 7 6 2 4 9 10 8

X2: Average sentence 2 2.5 0.5 3 3 1 2 4.5 5 3.5

Sample 2 1 2 3 4 5 6 7 8 9 10

X1: Years of education 9 12 17 16 9 14 10 17 17 9

X2: Age at first offense 14 17 14 16 10 17 16 10 12 12

Sample 3 1 2 3 4 5 6 7 8 9 10

X1: Age at first offense 13 17 10 16 14 11 18 19 15 12

X2: Number of convictions 7 3 10 4 6 9 1 1 6 8

Sample 4 1 2 3 4 5 6 7 8 9 10

X1: Age at first offense 11 16 18 12 15 17 13 20 20 13

X2: Average sentence 3 5 1.5 1 1 4 4.5 5 3 2.5

a. Calculate the mean scores of both variables for Samples 1, 2, 3, and 4.

b. Display the data for the four samples in four frequency distribution tables. For each score, add a positive or negative sign to indicate the direction in which the score differs from the mean (as done in Tables 14.4 and 14.5). Add an extra column in which you record a plus or a minus for the product of the two signs.

c. Draw four scatterplots, one for each sample distribution. State whether each scatterplot shows a positive relationship, a negative relationship, a curvilinear relationship, or no relationship between the two variables.

d. Would you advise against using Pearson’s correlation coefficient as a measure of association for any of the four samples? Explain your answer.

14.2 Jeremy, a police researcher, is concerned that police officers may not be assigned to areas where they are needed. He wishes to find out whether there is a connection between the number of police officers assigned to a particular block and the number of violent incidents reported on that block during the preceding week. For ten different blocks (designated A

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A432

through J), the number of patrolling officers assigned and the number of prior violent incidents reported are as follows:

A B C D E F G H I J

X1: Violent incidents 7 10 3 9 8 0 4 4 2 8

X2: Officers assigned 6 9 3 10 8 1 4 5 2 7

a. Calculate the covariance for the data recorded above.

b. Calculate the value of Pearson’s r for the data recorded above.

c. On an 11th block—block K—there are no police officers patrolling, yet in the previous week 11 violent incidents were reported there. What effect would it have on Pearson’s r if Jeremy included block K in his calculations?

d. How do you explain this difference?

14.3 Seven subjects of different ages are asked to complete a questionnaire measuring attitudes about criminal behavior. Their answers are coded into an index, with scores ranging from 1 to 15. The subjects’ scores are as follows:

X1: Age 12 22 10 14 18 20 16

X2: Score 6 3 3 9 9 6 13

a. Calculate Pearson’s correlation coefficient for the two variables listed above.

b. Illustrate the sample distribution on a scatterplot.

c. Divide the scatterplot into two sections, and calculate the value of Pearson’s r for each section.

d. Explain the difference between the r values you obtained in parts a and c.

14.4 Eight homeowners in the inner-city neighborhood of Moss Tide are asked how long they have been living in the neighborhood and how many times during that period their house has been burglarized. The results for the eight subjects are listed below:

X1: Years in neighborhood 2 1.5 3.5 28 1 5 20 3

X2: Number of burglaries 2 1 5 55 0 4 10 3

a. Calculate Pearson’s r for the two variables recorded above.

b. Calculate Spearman’s r for the same data.

c. Illustrate the sample distribution on a scatterplot.

d. Which of the two correlation coefficients is more appropriate, in your opinion, for this case? Refer to the scatterplot in explaining your answer.

E X E R C I S E S 433

14.5 Eleven defendants arrested for violent offenses were all required to post bail. The judge said that the amount of bail assigned was related to the total number of prior arrests. The results for the 11 defendants are as follows:

X1: Number of

prior arrests 0 3 9 13 2 7 1 4 7 20 5

X2: Amount of

bail assigned 100 500 2,500 10,000 1,000 10,000 100 7,500 5,000 100,000 4,000

a. Calculate Pearson’s r for the two variables recorded above.

b. Calculate Spearman’s r for the same data.

c. Illustrate the sample distribution on a scatterplot.

d. Which of the two correlation coefficients is more appropriate, in your opinion, for this case? Refer to the scatterplot in explaining your answer.

14.6 Researchers looking at age and lifetime assault victimization inter- viewed nine adults and found the following values:

X1: Age 18 20 19 25 44 23 67 51 33

X2: Number of times

assaulted in lifetime 1 4 8 0 6 2 9 3 10

a. Calculate Pearson’s r for the two variables recorded above.

b. Calculate Spearman’s r for the same data.

c. Illustrate the sample distribution on a scatterplot.

d. Which of the two correlation coefficients is more appropriate, in your opinion, for this case? Refer to the scatterplot in explaining your answer.

14.7 In a study looking at the relationship between truancy and theft, a sample of ten youth were asked how many times in the last year they had skipped school and how many times they had stolen something worth $20 or less. Their responses were

X1: Number of times

skipped school 9 2 4 0 0 10 6 5 3 1

X2: Number of thefts

valued at $20 or less 25 10 13 0 2 24 31 20 1 7

a. Calculate Pearson’s r for the two variables recorded above.

b. Use a 5% level of significance and outline each of the steps re- quired in a test of statistical significance of r.

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A434

14.8 A study investigating the link between child poverty and property crime rates gathered information on a random sample of 13 counties. The values for the percentage of children under 18 living in poverty and property crime rates (given per 100,000) are

a. Calculate Pearson’s r for the two variables recorded above.

b. Use a 5% level of significance and outline each of the steps re- quired in a test of statistical significance of r.

X1: Percentage of

children living

in poverty 10 8 43 11 27 18 15 22 17 17 20 25 35

X2: Property crime

rate 1,000 2,000 7,000 4,000 3,000 4,500 2,100 1,600 2,700 1,400 3,200 4,800 6,300

C o m p u t e r E x e r c i s e s

The process for obtaining correlation coefficients and scatterplots is similar in

SPSS and Stata. Similar to other chapters, we have made available example

syntax files that illustrate the commands below for both SPSS (Chapter_14.sps)

and Stata (Chapter_14.do).

SPSS

Correlation Coefficients

Both correlation coefficients discussed in this chapter—Pearson’s r and

Spearman’s r—can be obtained with similar commands. To obtain Pearson’s r,

use the command CORRELATIONS:

The /STATISTICS DESCRIPTIVES line is optional and would simply

generate a table of descriptive statistics, much like that produced with the

DESCRIPTIVES command in Chapter 4. Although it is much more efficient

to have SPSS compute all the correlations simultaneously for however many

variables you have included in the command, rather than selectively picking

out specific pairs that you might be most interested in, bear in mind that

the output may be rather unwieldy if too many variables are listed in the

The output from the CORRELATIONS command will consist of a

matrix (grid) of correlations for all the variables whose names appear in the

C O M P U T E R E X E R C I S E S 435

/VARIABLES= line.

/VARIABLES= line. You should also note that this matrix of correlations is sym-

CORRELATIONS

/VARIABLES = list_of_variable_names

/STATISTICS DESCRIPTIVES.

metric; running from the upper left to the lower right corner of the matrix is a

diagonal that is made up of 1s (the correlation of the variable with itself). The cor-

relations that appear above the diagonal will be a mirror image of the correlations

that appear below the diagonal. Thus, to locate the value of the correlation coeffi-

cient you are most interested in, you simply find the row that corresponds to one

of the variables and the column that corresponds to the other variable. It does not

matter which variable you select for the row or for the column; the correlation

coefficient reported in the matrix will be the same.

The command for obtaining Spearman’s r is NONPAR CORR

(nonparametric correlation in SPSS):

The output will again be a matrix of correlations for the variables listed

on the /VARIABLES= line of the command. The /PRINT = SPEARMAN

option ensures that SPSS will compute the correct correlation.

Scatterplots

The Computer Exercises section of Chapter 3 illustrated a wide range of graphics

commands available in SPSS. One of the simple, but powerful, graphs that SPSS

can produce is a scatterplot, which is obtained with the GRAPH command

and /SCATTERPLOT(BIVAR) option:

The graph produced by this command will be a simple scatterplot that can be

edited to suit your needs. As we noted in Chapter 3, the interactive graphical

options in SPSS are extensive, and we encourage you to explore some of the

additional possibilities for creating novel and informative scatterplots.

Stata

Correlation Coefficients

Obtaining Pearson’s r in Stata is also simple using the pwcorr command

(pairwise correlation in Stata):

where the options obs and sig will print in the correlation matrix the number

of observations used in the calculation of each correlation and the observed

significance level, making the output comparable to that in SPSS. The pwcorr

command does not require these options; they are included here simply to

provide more comprehensive and informative output.

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A436

(nonparametric correlation in SPSS):

NONPAR CORR

/VARIABLES = list_of_variable_names

/PRINT = SPEARMAN.

GRAPH

/SCATTERPLOT(BIVAR) = x_axis_var WITH y_axis_var.

pwcorr list_of_variable_names, obs sig

There is one notable difference in the Stata output compared to that in

SPSS. Rather than printing a full symmetric matrix of correlations (see previous

discussion of SPSS output), Stata prints only the diagonal (where r = 1.0 and the

variable is correlated with itself) and the correlations appearing below the

diagonal (i.e., lower left half of the correlation matrix). This makes the output

less overwhelming and easier to sort through a large number of correlations.

Spearman’s r is obtained with the spearman command:

where the stats(rho obs p) option will print the correlation (rho), the

number of observations used in calculating the correlation (obs), and the

observed significance level of the correlation (p). The default is to simply print

the value of Spearman’s r without information about the number of observations

or observed significance level. The pw option ensures that Stata uses all avail-

able cases in computing the correlation rather than deleting an observation if it is

missing information on any one of the variables included in the list of variables.

The output will be a matrix of Spearman’s r correlations.

Scatterplots

In Chapter 3’s Computer Exercises section we illustrated the use of the twoway

command for the creation of other graphs. It can also be used for the creation of

simple scatterplots by changing the graph-type option to scatter:

As with all other graphs in Stata, this scatterplot may be edited and customized

to suit your purpose.

Problems

1. Open the California UCR data file (caucr_ 99.sav or caucr_99.dta). These

are the data presented in Table 14.8. Compute Pearson’s r for these data

and note that it matches the value reported in the text.

a. Compute Spearman’s r for these data.

b. How does the value of Pearson’s r compare to that for Spearman’s r?

What might account for this?

c. Generate a scatterplot for these data.

2. Enter the data from Exercise 14.2.

a. Compute both Pearson’s r and Spearman’s r for these data.

b. Add the extra case presented in part c of Exercise 14.2. Recompute

Pearson’s r and Spearman’s r for these data.

c. Compare your answers from parts a and b. How do you explain this

pattern?

d. Generate a scatterplot for the 11 cases.

C O M P U T E R E X E R C I S E S 437

spearman list_of_variable_names, stats(rho obs p) pw

twoway (scatter y_var_name x_var_name)

3. Enter the data from Exercise 14.8.

a. Compute Spearman’s r for these data.

b. How does the value of Pearson’s r compare to that for Spearman’s r ?

What might account for this?

c. Generate a scatterplot for these data.

4. Open the NYS data file (nys_1.sav, nys_1_student.sav, or nys_1.dta). Use a

5% level of significance, and outline each of the steps required in a test of

statistical significance for each of the following relationships:

a. Age and number of thefts valued at less than $5 in the last year.

b. Number of times drunk and number of thefts valued at $5 to $50 in

the last year.

c. Frequency of marijuana use and number of times the youth has hit

other students in the last year.

d. Number of times the youth has hit a parent and number of thefts val-

ued at more than $50 in the last year.

e. Number of times the youth has been beaten up by parent and number

of times the youth has hit a teacher in the last year.

5. Generate a scatterplot for each pair of variables listed in Computer

Exercise 4.

6. Open the Pennsylvania Sentencing data file (pcs_ 98.sav or pcs_98.dta).

Use a 5% level of significance, and outline each of the steps required in a

test of statistical significance for each of the following relationships:

a. Age of offender and length of incarceration sentence.

b. Severity of conviction offense and length of incarceration sentence.

c. Prior criminal history score and age.

d. Length of incarceration sentence and prior criminal history score.

7. Generate a scatterplot for each pair of variables listed in Computer

Exercise 6.

C H A P T E R F O U R T E E N : I N T E R V A L - L E V E L D A T A438

An Introduction to Bivariate Regression

What Does a Regression Coefficient Describe?

How Can We Predict the Value of Y ?

2 Statistic?

Coefficient b?

C h a p t e r f i f t e e n

T h e r e g r e s s i o n c o e f f i c i e n t

P r e d i c t i o n : b u i l d i n g t h e r e g r e s s i o n l i n e

E v a l u a t i n g t h e r e g r e s s i o n l i n e

S t a t i s t i c a l i n f e r e n c e i n r e g r e s s i o n

How is the Regression Coefficient b Expressed?

How is the Regression Line Built?

What are the Limits of Prediction?

What is the R

What is the Test of Statistical Significance for the Regression

What is the F-Test for the Overall Regression?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_15,

© Springer Science+Business Media New York 2014

IN THE PREVIOUS CHAPTER, we asked how strongly two interval-level vari-

ables were correlated. In this chapter, we ask a related question that is

central to much statistical analysis in criminology and criminal justice: Can

we predict the level of one interval-level variable from the value of an-

other? As in the description of specific measures of association for nominal

and ordinal variables in Chapter 13, in regression analysis we must define

which variable is the independent variable, or predictor variable, and

which variable is the dependent variable, or the variable being predicted.

We also introduce the idea of regression modeling in this chapter. Our dis-

cussion focuses on how regression analysis is used to create a prediction

model and the statistics that researchers use to evaluate such models.

E s t i m a t i n g t h e I n f l u e n c e o f O n e V a r i a b l e o n A n o t h e r : T h e R e g r e s s i o n C o e f f i c i e n t

By making only a slight change to the equation for Pearson’s correlation

coefficient, we can construct the regression coefficient b—a statistic that

estimates how much one variable influences another. As with Goodman

and Kruskal’s tau, lambda, and Somers’ d (see Chapter 13), in developing

this measure we again make a very important assumption about the rela-

tionship between the variables examined. We assume that the indepen-

dent variable (X) influences or predicts the dependent variable (Y).

The regression coefficient b asks how much impact one variable (the

independent variable) has on another (the dependent variable). It an-

swers this question not in standardized units, but in the specific units of

the variables examined. The specific interpretation of a regression coeffi-

cient will depend on the units of measurement used. Nonetheless, b has

a general interpretation in terms of X, the symbol for the independent

variable, and Y, the symbol for the dependent variable:

A change of one unit in X produces a change of b units in the estimated

value of Y.

440

E S T I M A T I N G T H E I N F L U E N C E O F O N E V A R I A B L E O N A N O T H E R

Let’s take a concrete example. Suppose that you are studying the rela-

tionship between education and reoffending. You assume that education

has an impact on reoffending and thus define years of education as the

independent variable, X, and number of arrests as the dependent vari-

able, Y. You calculate the regression coefficient b and find that it has a

value of �2. You can interpret this coefficient as meaning that a one-

year change (or increase) in education produces, on average, a two-unit

change—in this case, reduction—in number of rearrests. If b had been

positive, we would have concluded that a one-year increase in education

produces, on average, an increase of two arrests.

This interpretation of the regression coefficient reminds us that we are

once again concerned with linear relationships. If we say that a unit

change in X results in b units of change in Y, we are also saying that

You can see from this example why regression analysis is such a

widely used tool in criminal justice and criminology. Contrary to what

students often fear, regression coefficients are very easy to understand

and make good intuitive sense. The regression coefficient in this case

Hypothetical Regression Line of the Relationship

Between Number of Arrests and Years of Education Figure 15.1

N u

m b er

o f

A rr

es ts

20

15

10

5

0

Years of Education

8 9 10 11 12 13 14 15 16

441

In our example, the regression coefficient predicts that a one-year increase

in education produces a reduction of two arrests, irrespective of how

education an offender already has. If the change is constant, then the

result is a linear relationship. This is illustrated in Figure 15.1. Importantly,

linearity is a central assumption of regression analysis. As with linear

correlation coefficients, you should always examine the scatterplot of

the relationship between the variables to be sure that the relationship

you are estimating is a linear one.

the change produced by X is constant. By “constant” we mean that the

change produced in Y is always the same for each unit change in X.

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

tells us that education reduces reoffending. It also tells us by how much.

For each year of education, there is a reduction of about two arrests.

Often, criminologists use regression coefficients to estimate how much

1

Calculating the Regression Coefficient

The calculation for the regression coefficient b (Equation 15.1) is very

similar to that for the correlation coefficient r (Equation 14.2). Once

again, the covariation between the two variables examined is in the nu-

merator, although the two variables are expressed as X and Y rather than

X1 and X2. The difference is in the denominator of the equation. Instead

of standardizing this value according to the variability found in both

measures, we now contrast the covariation of the relationship of X and Y

with the variability found only in X.

Equation 15.1

By taking the example of age and number of arrests over the last year

from Table 14.1, we can see how a regression coefficient is calculated in

practice (see Table 15.1). Our first step, as noted above, is to examine

the scatterplot of points to make sure that the relationship we are esti-

mating is linear. This was done in Chapter 14, so we will not repeat that

(Y ). Our assumption here is that age influences the number of arrests on

an offender’s record over the previous year. As with the correlation coef-

ficient, we first calculate the covariation between age (X ) and number of

arrests (Y ), as shown in Equation 15.2.

Covariation of X and Y � Equation 15.2� N

i�1

(Xi � X)(Yi � Y)

b � � N

i�1

(Xi � X)(Yi � Y)

� N

i�1

(Xi � X)2

1Jihong Zhao and Quint Thurman, “A National Evaluation of the Effect of COPS Grants

on Crime from 1994–1999,” unpublished manuscript, University of Nebraska, Decem-

ber, 2001.

442

crime-control benefit is gained from a particular strategy or from the

step here. The next step is to define X and Y. In this case, age is the inde-

pendent variable (X ), and number of arrests is the dependent variable

addition of criminal justice resources. In a report to the federal govern-

ment, for example, a group of researchers from the University of Nebraska

used regression coefficients to estimate how much each dollar spent on

additional police in the United States during the 1990s influenced crime

rates in American cities. Clearly, a statistic that can tell us how much

change in one variable influences another can be very useful in deciding

on criminal justice strategies or policies.

E S T I M A T I N G T H E I N F L U E N C E O F O N E V A R I A B L E O N A N O T H E R

Our result (calculated in column 6 in Table 15.1) is the same as the

simply take the sum of the squared deviations of each subject’s age from

the mean for age. The result, calculated in column 3 of Table 15.1, is

138.9326. Our regression coefficient is obtained by dividing these two

sums.

Calculations for the Regression Coefficient of Age and Number of Arrests for 15 Young Offenders

AGE ARRESTS

X Xi � (Xi � )2 Y Yi � (Xi � )(Yi � ) (1) (2) (3) (4) (5) (6)

14 �3.0667 9.4046 0 �4.8667 14.9247 13 �4.0667 16.5380 1 �3.8667 15.7247 15 �2.0667 4.2712 1 �3.8667 7.9913 13 �4.0667 16.5380 2 �2.8667 11.6580 14 �3.0667 9.4046 2 �2.8667 8.7913 14 �3.0667 9.4046 3 �1.8667 5.7246 17 �0.0667 0.0044 3 �1.8667 0.1245 19 1.9333 3.7376 4 �0.8667 �1.6756 21 3.9333 15.4708 4 �0.8667 �3.4090 19 1.9333 3.7376 6 1.1333 2.1910 16 �1.0667 1.1378 8 3.1333 �3.3423 18 0.9333 0.8710 9 4.1333 3.8576 20 2.9333 8.6042 9 4.1333 12.1242 21 3.9333 15.4708 10 5.1333 20.1908 22 4.9333 24.3374 11 6.1333 30.2574

� 125.1332� N

i�1 (Xi � X)(Yi � Y)�

N

i�1

(Xi � X)2 � 138.9326

Y � 4.8667X � 17.0667

YXYXX

Table 15.1

W orking It Out

� 0.9007

� 125.1332

138.9326

b � � N

i�1

(Xi � X)(Yi � Y)

� N

i�1

(Xi � X)2

This result of 0.9007 can be interpreted as meaning that a one-year in-

crease in age produces, on average, a 0.90 increase in number of arrests.

443

denominator, however, involves less work than before. In this case, we

result for the correlation coefficient: 125.1332. The calculation of the

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

A Substantive Example: Unemployment and Burglary in California

In Chapter 14, in an analysis of unemployment and burglary rates for

counties in the state of California in 1999, we reported that the correla-

tion between unemployment rates and burglary rates was r � 0.4910.

We can continue to work with these data to provide another illustration

practice in regression to start by examining a scatterplot (see Figure

15.2). Our interest is in confirming that the relationship between unem-

ployment rates and burglary rates in California counties is linear. The

scatterplot in Figure 15.2 provides no strong evidence to suggest that the

relationship is not linear.

In Table 14.9, we reported the two results needed to calculate the re-

gression coefficient b. The covariation between the unemployment rate

and the burglary rate is 37,128.9297, and the variability of the unemploy-

ment rate is 1,010.3570 (see the bottom of Table 14.9). Inserting these

values into Equation 15.1, we find that b � 36.7483.

� 36.7483

� 37128.9297 1010.3570

b � � N

i�1

(Xi � X)(Yi � Y)

� N

i�1

(Xi � X)2

Scatterplot of Unemployment Rates and Burglary Rates in California, 1999Figure 15.2

B u

rg la

ry R

at e

3,000

2,000

1,000

0

Unemployment Rate

0 10 20 30

444

relationship between unemployment rates and burglary rates, it is good

of regression analysis. While the correlation of 0.49 suggests a linear

P R E D I C T I O N I N R E G R E S S I O N

This coefficient tells us that, on average, for every one-unit increase in

the unemployment rate, the burglary rate increases by about 36.75

crimes (per 100,000).

P r e d i c t i o n i n R e g r e s s i o n : B u i l d i n g t h e R e g r e s s i o n L i n e

The regression coefficient provides a method for estimating how

change in an independent variable influences change in a dependent

variable. However, by itself, it does not allow the researcher to predict

the actual values of the dependent variable. For example, we found in

our examination of age and number of arrests that a one-year increase

in age in our sample of young offenders was associated with a 0.9007

increase in number of arrests over the last year. Accordingly, based on

our analysis, we would predict that a 14-year-old would have about

0.9 more arrests than a 13-year-old. Someone 15 years old would be

expected to have about 1.8 more arrests over the last year than some-

one 13 years old. This is because our regression coefficient suggests

expect a county with an unemployment rate of 5.0 to have a burglary

rate that was about 36.75 (per 100,000) higher than that of a county

with an employment rate of 4.0. A county with an unemployment rate

of 6.0 would be expected to have a burglary rate that was about 36.75

(per 100,000) higher than that of the county with an unemployment

rate of 5.0.

But this still does not tell us how many arrests overall a person 13, 14,

or 15 years old would be expected to have over the last year. Nor can

we define the rate of burglary in a county with an unemployment rate of

4.0, 5.0, or 6.0. To answer these questions, we need another piece of in-

formation: a starting point from which to calculate change. That starting

point is provided by a statistic called the Y-intercept.

The Y-Intercept

The Y-intercept, or b0, is the expected value of Y when X equals 0.2 It is

calculated by taking the product of b and the mean of X, and subtracting

it from the mean of Y (Equation 15.3).

Equation 15.3b0 � Y � bX

2Note that there is no single accepted convention for representing the Y-intercept.

Some researchers use the symbol � (alpha), while others prefer to use a.

445

that for each year increase in age, we can expect about 0.9 more arrests.

Similarly, for the unemployment rate and burglary rate analysis, we would

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

Let’s begin with our example of age and number of arrests. We get

the value of b0 by first taking the product of b (0.9007) and the mean

for age (17.0667; see Table 15.1) and then subtracting that value

(15.3720) from 4.8667, the mean for arrests in the sample. The result is

�10.5053.

W orking It Out

� �10.5053

� 4.8667 � 15.3720

� 4.8667 � (0.9007)(17.0667)

b0 � Y � bX

The Regression Line

By looking at a scatterplot (see Figure 15.3), we can see how the Y-

intercept helps in developing predictions of number of arrests (Y ) from

age (X). If we put the value �10.5053 on the line where the value of age

is 0, we can then use the regression coefficient b to draw a line of pre-

Scatterplot and Regression Line Showing the Relationship

Between Age and Number of Arrests for 15 Subjects Figure 15.3

–10

20

0,0

10

10 20 30

Age

N u

m b

er o

f A

rr es

ts

Y-intercept

446

P R E D I C T I O N I N R E G R E S S I O N

0 � 0.9007, or �10.5053 � 0.9007 � �9.6046. For

0 � 0.9007 � 0.9007, and

so forth. By plotting these values, we can draw the regression line for

our example. This is done in Figure 15.3, which also includes the scatter-

plot of the 15 sample scores (from Table 15.1).

The predicted values of Y, designated , can also be found through a

simple equation. In this case, is equivalent to the Y-intercept plus the

regression coefficient times the value of X (Equation 15.4).

Equation 15.4

For our example, this equation may be represented as shown in Equa-

tion 15.5.

Predicted number of arrests � �10.5053 � (0.9007)(age) Equation 15.5

We now have a method for predicting the number of arrests based on

the age of subjects in our sample. For example, looking at our regression

line, we would predict that someone 13 years old would have had about

one arrest over the last year. To obtain the exact prediction, we simply

insert age 13 in Equation 15.5. The result is 1.2038:

Ŷ � b0 � bX

W orking It Out

� 1.2038

� �10.5053 � (0.9007)(13)

Predicted number of arrests � �10.5053 � (0.9007)(age)

W orking It Out

� 3.9059

� �10.5053 � (0.9007)(16)

Predicted number of arrests � �10.5053 � (0.9007)(age)

For someone 16 years old, we would predict about four arrests.

447

us that for each increase of one year in age, there is a corresponding

diction, called the regression line. The regression coefficient tells

bof arrests should be

2 years of age, the number of arrests should be b

increase of 0.9007 arrest. This means that when age is about 1, number

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

Predictions Beyond the Distribution Observed in a Sample

Are our predictions good ones? We will turn a bit later in the chapter to

methods of evaluating how well the regression line predicts or fits the

data in our sample. Nonetheless, at this point it is useful to look at Fig-

ure 15.3 and evaluate on your own how well we have done. I think you

will agree that we have done a fairly good job of drawing a line through

the data points in our sample. For the most part, the points are close to

the line, meaning that our prediction of number of arrests based on age

of the offender is fairly close to the actual scores in our sample. But what

about predicting beyond our sample scores? Can we use our regression

line to predict number of arrests for those 30 or 40 years old? What about

for offenders younger than 13?

This approach should be used with caution. In order to predict be-

yond the data points in your sample, you must assume that the relation-

ships are similar for offenders your study has not examined. As is always

the case when you try to make inferences beyond your sampling frame,

example, we would have to explain why we think that the relationship

3

W orking It Out

� 7.5087

� �10.5053 � (0.9007)(20)

Predicted number of arrests � �10.5053 � (0.9007)(age)

For someone 20 years old, we would predict between seven and eight

arrests.

3For a discussion of the relationship between age and crime, see David F. Farrington,

“Age and Crime,” in Crime and Justice: An Annual Review of Research, Vol. 7

(Chicago: University of Chicago Press, 1986), pp. 189–250.

448

your conclusions cannot be justified directly from the statistics you develop.

You must explain why you expect the relationships observed to con-

tinue for populations that have not been represented. In our case, for

between age and number of arrests continues for both older and youn-

ger offenders. What we know about offending in criminology suggests

that this approach would not be a good one to use for older offenders.

A number of studies have shown that offending rates often decrease

as offenders get older. Certainly our predictions of number of arrests

for those younger than age 12 are to be viewed with caution. Our

P R E D I C T I O N I N R E G R E S S I O N

regression line predicts negative values of arrests for these young offend-

ers. This example illustrates a very important limitation of regression

analysis. You should be very cautious about predicting beyond the distri-

bution observed in your sample. As a general rule, regression can pro-

vide good estimates of predictions for the range of offenders represented

by the data; however, it often does not provide a solid basis for predic-

tions beyond that range.

Predicting Burglary Rates from Unemployment Rates in California

What would our prediction model look like for the case of the unemploy-

ment rate and burglary rate data from counties in California? First, we

need to define the Y-intercept by using Equation 15.3. The value of b is

36.7483, the mean unemployment rate is 7.4069, and the mean burglary

rate is 790.8907 (see Table 14.9). The Y-intercept has a value of 518.6997.

W orking It Out

� 518.6997

� 790.8907 � (36.7483)(7.4069)

b0 � Y � bX

Once we know the Y-intercept, we can develop the regression line. For

our unemployment and burglary example, we write the regression equa-

tion as shown in Equation 15.6.

Burglary rate � 518.6997 � 36.7483(unemployment rate) Equation 15.6

Equation 15.6 can be used to calculate the predicted value of the bur-

glary rate, given a value for the unemployment rate. For example, if a

county had an unemployment rate of 2.5, we would predict from our

model that the burglary rate would be about 610.57.

W orking It Out

� 610.5705

� 518.6997 � (36.7483)(2.5)

Burglary rate � 518.6997 � 36.7483(unemployment rate)

449

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

For a county with an unemployment rate of 4.2, we would predict a bur-

glary rate of about 673.04.

W orking It Out

� 673.0426

� 518.6997 � (36.7483)(4.2)

Burglary rate � 518.6997 � 36.7483(unemployment rate)

And for a county that had an unemployment rate of 7.5, we would pre-

dict a burglary rate of about 794.31.

W orking It Out

� 794.3120

� 518.6997 � (36.7483)(7.5)

Burglary rate � 518.6997 � 36.7483(unemployment rate)

How well our regression line fits the data is illustrated in Figure 15.4,

which presents the regression line in the scatterplot of the data. Overall,

it looks as though the regression line captures the relationship between

unemployment rates and burglary rates reasonably well—the values for

burglary rate tend to cluster near the regression line.

Scatterplot and Regression Line Showing the Relationship Between

Unemployment Rate and Burglary Rate for California Counties, 1999 Figure 15.4

B u

rg la

ry R

at e

3,000

2,000

1,000

0

Unemployment Rate

0 10 20 30

450

P R E D I C T I O N I N R E G R E S S I O N

Choosing the Best Line of Prediction Based on Regression Error

One question we might ask is whether the regression line we identify is

the best one that could be drawn, given the data available to us. In order

to answer this question, we must first decide on the criteria that we will

use for defining the best line. In regression, as in many of the statistical

techniques we have examined, we use the criterion of minimizing error.

Regression error (e) is defined as the difference between the actual

values of Y and the predicted values of Y, or , as shown in Equation 15.7.

Equation 15.7

In Table 15.2, the actual values of Y and the predicted values ( ) are

predicted value for arrests based on our regression equation is 6.6080,

so the error in this case is �2.6080. In other words, the actual value of

Y is made up of both our prediction and some amount of error. The

equation form of this relationship gives the basic regression model for

our example. This model may be expressed either in terms of a theoret-

ical model describing the relationships in the population or in terms of

the observed relationships found in our sample. For a population

Y � �0 � �(age) � �

e � Y � Ŷ

Contrast of the Predicted Values for Y, or , and the Actual Values for Y

NUMBER OF

AGE ARRESTS

SUBJECT X Y Yi �

1 14 0 2.1046 �2.1046 2 13 1 1.2039 �0.2039 3 15 1 3.0053 �2.0053 4 13 2 1.2039 0.7961 5 14 2 2.1046 �0.1046 6 14 3 2.1046 0.8954 7 17 3 4.8066 �1.8066 8 19 4 6.6080 �2.6080 9 21 4 8.4093 �4.4093

10 19 6 6.6080 �0.6080 11 16 8 3.9060 4.0940 12 18 9 5.7073 3.2927 13 20 9 7.5086 1.4914 14 21 10 8.4093 1.5907 15 22 11 9.3100 1.6900

� N

i�1

(Yi �Ŷi ) � 0.0000

ŶiŶ

Ŷ Table 15.2

451

arrests. Subject 8, for example, had four arrests over the last year. The

contrasted for the 15 subjects in our example of age and number of

model, we use Greek letters to represent the parameters. So for our

example, the population model would be

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

As we see from Table 15.2, this subject’s actual value for Y is also 1 ar-

rest over the last year.

In using error as a criterion for choosing the best line, we are forced

to base our decision not on the sum of errors in our equation, but on the

sum of the squared errors. This is because the deviations above and

below the regression line cancel each other out (see Table 15.2). In re-

gression, as in deviations from the mean, the sum of the deviations of

from Y are always equal to 0.

Squaring the deviations of from Y provides estimates with only pos-

itive signs and allows us to assess the amount of error found. The regres-

sion line we have constructed is the best line in terms of the criteria of

squared deviations of from Y. In mathematical language, the regression

line is the line for which the sum of the squared errors is a minimum

(Equation 15.8).

Equation 15.8

For this reason, we call this approach ordinary least squares regres-

sion analysis, or OLS regression. We hope you remember that we first

introduced the concept of the least squares property when we discussed

the mean in Chapter 4. As is often the case in statistics, ideas learned early

on continue to be important in understanding more complex statistics.

� N

i�1

(Yi � Ŷ )2 � minimum

W orking It Out

� 1

� �10.5053 � 13.5105 � (�2.0053)

� �10.5053 � 0.9007(15) � (�2.0053)

Y � b0 � b(age) � e

The model for the example would be expressed as

Y � b0 � b(age) � e

By looking at one of our 15 subjects, we can see this relationship in

practice. Subject 3’s age is 15. The difference between the predicted

value for arrests and the actual number of arrests, or the error (e), for

Subject 3 is �2.0053. If we add the Y-intercept, the subject’s age times

the coefficient b, and the error, we obtain a value of 1.

452

E V A L U A T I N G T H E R E G R E S S I O N M O D E L

E v a l u a t i n g t h e R e g r e s s i o n M o d e l

Having noted that OLS regression provides the best line in terms of the

least squares criteria for error, we may still ask how well this line pre-

dicts the dependent variable. Does the regression model add to our abil-

ity to predict number of arrests in our sample? Researchers commonly

use a measure called the percent of variance explained to answer this

question.

Percent of Variance Explained

The percent of variance explained, or R2, in regression is analogous to

eta squared in analysis of variance. With eta squared, we examine the

proportion of the total sum of squares accounted for by the between (or

explained) sum of squares. In the case of regression, the explained sum

of squares (ESS) is calculated from the difference between the predicted

value of Y, or , and the mean of Y, or : ( i � )2. The total sum of

squares (TSS) is represented by the difference between Y and the mean of

Y: (Yi � )2. R 2 for regression, like eta squared for analysis of variance, is

the ratio of the explained to the total sum of squares (Equation 15.9).

Equation 15.9

Why do we define the explained and total sums of squares in terms

of the mean? If we did not have our regression model, but had only the

raw data in our sample, our best single prediction of Y would be the

mean of Y. The question answered by R 2 is “How much additional

)

).

The explained deviation thus represents the improvement in predicting

Y that the regression line provides over the mean.

To calculate the explained sum of squares for our example, we sim-

ply take the difference between the predicted score for arrests and the

mean for arrests, square that value, and sum the outcomes across the 15

subjects. This is done in column 5 of Table 15.3, where our final result is

112.7041. For the total sum of squares, we subtract the mean of arrests

for the sample from each subject’s actual number of arrests. This value is

Y

ˆ

R

2 � ESS TSS

� � N

i�1

(Ŷi � Y )2

� N

i�1

(Yi � Y )2

Y

YŶYŶ

453

knowledge have we gained by developing the regression line?” This is

illustrated in Figure 15.5, where we take one subject from our analysis

of the effect of age on number of arrests, Subject 13, and plot that subject’s

. The distance from the mean of Y to the explained deviation, (Y � Y

actual score for the subject is the total deviation from the mean, (Y �

between the predicted value of Y and the mean of Y represents the

score relative to the regression line and the mean of Y. The distance

i

i

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

squared, and the 15 outcomes are added together (see Table 15.3, col-

umn 7). The total sum of squares for our example is 187.7333.

To obtain the percent of variance explained, or R 2, we take the ratio

of these two values. The percent of variance explained beyond the mean

in our regression model is 0.60.4

The Explained, Unexplained, and Total Deviations

from the Mean for Subject 13 (Age � 20; Arrests � 9) Figure 15.5

12

10

8

6

4

2

0

12 14 16 18 20 22 24

Unexplained Variation

Explained Variation Total Variation

Age

N u

m b er

o f

A rr

es ts

X13

Y13

Y – Y

Y – Y

Y – Y

Y

Y

4In the case of multivariate regression analysis (described in detail in Chapter 16),

where there are multiple independent variables, some statisticians advise using an

adjusted measure of R2. Commonly referred to as adjusted R2, it is routinely provided

by most statistical software programs. Adjusted R2 is calculated using the following

formula:

Adjusted R2 � 1 � (1 � R2)

where k equals the number of independent variables and N is the number of observa-

tions in the sample.

Adjused R2 can prevent a misleading interpretation of the strength of prediction of a

model, because it offsets the artificial inflation in the statistic that is created with every

additional variable added to a regression model. Nonetheless, the adjustment repre-

sents a transformation of the R2 value and thus alters the simple meaning of the statis-

tic. We do not advise using the adjusted R2 value unless it differs substantially from

the simple R2 value.

� N � 1 N � k � 1�

454

i

i

i

i

E V A L U A T I N G T H E R E G R E S S I O N M O D E L

Is an R 2 of 0.6003 large or small? As noted in earlier chapters, deter-

mining whether a relationship is large or small inevitably involves a

value judgment. In deciding on the strength of your prediction, you

would likely compare your results to those of other research on the

same or related topics. As a general rule in criminal justice, regression

models seldom result in R 2 values greater than 0.40. If your R 2 is larger

than 0.40, you can usually assume that your prediction model is a pow-

erful one. The percent of variance explained in our model accordingly

suggests a very high level of prediction. Conversely, when the percent of

variance explained is less than 0.15 or 0.20, the model is likely to be

viewed as relatively weak in terms of prediction.

W orking It Out

� 0.6003

� 112.7041 187.7333

R

2 � ESS TSS

� � N

i�1

(Ŷi � Y )2

� N

i�1

(Yi � Y )2

Calculations for R2 Values for 15 Young Offenders

X Y i � ( i � )2 Yi � (Yi � )2

(1) (2) (3) (4) (5) (6) (7)

14 0 2.1046 �2.7621 7.6292 �4.8667 23.6848 13 1 1.2039 �3.6628 13.4158 �3.8667 14.9514 15 1 3.0053 �1.8614 3.4649 �3.8667 14.9514 13 2 1.2039 �3.6628 13.4158 �2.8667 8.2180 14 2 2.1046 �2.7621 7.6292 �2.8667 8.2180 14 3 2.1046 �2.7621 7.6292 �1.8667 3.4846 17 3 4.8066 �0.0601 0.0036 �1.8667 3.4846 19 4 6.6080 1.7413 3.0320 �0.8667 0.7512 21 4 8.4093 3.5426 12.5501 �0.8667 0.7512 19 6 6.6080 1.7413 3.0320 1.1333 1.2844 16 8 3.9060 �0.9607 0.9230 3.1333 9.8176 18 9 5.7073 0.8406 0.7066 4.1333 17.0842 20 9 7.5086 2.6419 6.9798 4.1333 17.0842 21 10 8.4093 3.5426 12.5501 5.1333 26.3508 22 11 9.3100 4.4433 19.7427 6.1333 37.6174

� 17.0667 � 112.7041 � 187.7333� N

i�1

(Yi � Y )2� N

i�1

(Ŷi � Y )2X

YYYŶYŶŶ

Table 15.3

455

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

Percent of Variance Explained:

Unemployment Rates and Burglary Rates in California

Returning to our example of unemployment rates and burglary rates, we

can similarly examine how well unemployment rates help to explain

burglary rates in California counties. The two key pieces of information

we need are the explained sum of squares and the total sum of squares.

The first column of Table 15.4 presents the burglary rates, and the sec-

ond column presents the predicted value of the burglary rate for each

county. The third and fourth columns present the calculations to obtain

the explained sum of squares. Column 3 displays the difference be-

tween the predicted value of the burglary rate ( ) and the mean of the

burglary rate ( ), while column 4 presents the squared difference be-

tween the predicted value and the mean. The explained sum of squares,

which has a value of 1,364,425.7358 (see the bottom of column 4), is

the sum of the squared differences between the predicted value of the

burglary rate and the mean of the burglary rate. The calculations for the

total sum of squares are presented in columns 5 and 6. The fifth column

represents the difference between the observed burglary rate (Yi ) and

the mean of the burglary rate ( ), while the sixth column contains the

squared differences between the observed burglary rate and the mean

of the burglary rate. The total sum of squares, which has a value of

5,659,404.5114 (located at the bottom of column 6), is the sum of the

squared differences between the observed burglary rate and the mean

of the burglary rate.

To obtain the value of R 2, we again use Equation 15.9, inserting our

values for the explained and total sums of squares.

Y

Y

Ŷi

W orking It Out

� 0.2411

� 1,364,425.7358

5,659,404.5114

R

2 � ESS TSS

� � N

i�1

(Ŷi � Y )2

� N

i�1

(Yi � Y )2

Our R 2 of 0.2411 indicates that about 24% of the variance in bur-

glary rates in California counties is explained by the unemployment

rate.

456

Calculations for R2 Values for Unemployment Rates and Burglary Rates in California Counties

Y i � ( i � )2 Yi � (Yi � )2

(1) (2) (3) (4) (5) (6)

837.89 647.3188 �143.5720 20,612.9048 46.9993 2,208.9342 2,037.49 853.1092 62.2185 3,871.1455 1,246.5993 1,554,009.8148

818.55 687.7419 �103.1488 10,639.6791 27.6593 765.0369 865.04 768.5881 �22.3026 497.4042 74.1493 5,498.1187 989.76 772.2630 �18.6277 346.9923 198.8693 39,548.9985 520.06 1,102.9977 312.1070 97,410.7607 �270.8307 73,349.2681 664.73 628.9446 �161.9461 26,226.5393 �126.1607 15,916.5222

1,200.91 812.6861 21.7954 475.0395 410.0193 168,115.8264 509.87 662.0181 �128.8726 16,608.1548 �281.0207 78,972.6338 924.10 1,011.1269 220.2362 48,503.9926 133.2093 17,744.7176 845.29 930.2807 139.3900 19,429.5609 54.3993 2,959.2838

1,027.79 753.8888 �37.0019 1,369.1391 236.8993 56,121.2783 1,526.40 1,378.6099 587.7192 345,413.8816 735.5093 540,973.9304

511.12 728.1650 �62.7257 3,934.5122 �279.7707 78,271.6446 960.18 937.6303 146.7396 21,532.5161 169.2893 28,658.8671 649.22 1,000.1024 209.2117 43,769.5480 �141.6707 20,070.5872

1,333.21 801.6616 10.7709 116.0125 542.3193 294,110.2232 361.24 775.9378 �14.9529 223.5892 �429.6507 184,599.7240 610.28 735.5147 �55.3760 3,066.5047 �180.6107 32,620.2250 929.32 941.3052 150.4145 22,624.5068 138.4293 19,162.6711 526.98 588.5215 �202.3692 40,953.3053 �263.9107 69,648.8576 775.92 790.6371 �0.2536 0.0643 �14.9707 224.1219 843.92 764.9133 �25.9774 674.8248 53.0293 2,812.1067

1,214.69 1,007.4521 216.5614 46,898.8356 423.7993 179,605.8467 325.08 831.0603 40.1695 1,613.5927 �465.8107 216,979.6082 957.95 764.9133 �25.9774 674.8248 167.0593 27,908.8097 570.14 871.4834 80.5927 6,495.1801 �220.7507 48,730.8716 477.54 639.9691 �150.9216 22,777.3324 �313.3507 98,188.6612 455.37 669.3677 �121.5230 14,767.8322 �335.5207 112,574.1401 464.52 614.2453 �176.6454 31,203.6044 �326.3707 106,517.8338 646.12 636.2943 �154.5964 23,900.0593 �144.7707 20,958.5556

1,030.58 849.4344 58.5437 3,427.3648 239.6893 57,450.9605 1,049.18 717.1405 �73.7502 5,439.0891 258.2893 66,713.3625

925.61 673.0426 �117.8481 13,888.1841 134.7193 18,149.2898 845.75 812.6861 21.7954 475.0395 54.8593 3,009.5428 883.02 695.0915 �95.7992 9,177.4791 92.1293 8,487.8079 539.82 632.6194 �158.2713 25,049.7949 �251.0707 63,036.4964 744.81 628.9446 �161.9461 26,226.5393 �46.0807 2,123.4309 896.85 842.0847 51.1940 2,620.8297 105.9593 11,227.3733 540.79 636.2943 �154.5964 23,900.0593 �250.1007 62,550.3601 355.82 592.1963 �198.6944 39,479.4646 �435.0707 189,286.5140 444.07 662.0181 �128.8726 16,608.1548 �346.8207 120,284.5979 347.57 628.9446 �161.9461 26,226.5393 �443.3207 196,533.2430 647.73 750.2140 �40.6767 1,654.5947 �143.1607 20,494.9860 823.95 775.9378 �14.9529 223.5892 33.0593 1,092.9173 699.71 856.7841 65.8934 4,341.9349 �91.1807 8,313.9201 575.09 897.2072 106.3165 11,303.1960 �215.8007 46,569.9421 769.30 687.7419 �103.1488 10,639.6791 �21.5907 466.1583 555.44 617.9201 �172.9706 29,918.8250 �235.4507 55,437.0321

1,057.99 904.5569 113.6661 12,919.9937 267.0993 71,342.0361 859.11 996.4276 205.5369 42,245.4173 68.2193 4,653.8729 816.55 764.9133 �25.9774 674.8248 25.6593 658.3997 676.23 941.3052 150.4145 22,624.5068 �114.6607 13,147.0761

1,047.32 1,125.0467 334.1560 111,660.1989 256.4293 65,755.9859 908.79 757.5637 �33.3270 1,110.6923 117.8993 13,900.2449 491.86 695.0915 �95.7992 9,177.4791 �299.0307 89,419.3595 591.28 676.7174 �114.1733 13,035.5447 �199.6107 39,844.4316

1,366.76 944.9800 154.0893 23,743.5062 575.8693 331,625.4507

� 1,364,425.7358 � 5,659,404.5114� N

i�1

(Yi � Y )2� N

i�1

(Ŷi � Y )2

YYYŶYŶŶ

Table 15.4

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

Statistical Significance of the Regression Coefficient:

The Case of Age and Number of Arrests

In assessing the statistical significance of the regression coefficient, we

once again use the t distribution, introduced in Chapter 10.

Assumptions:

Level of Measurement: Interval scale.

Population Distribution: Normal distribution of Y around each value of

X (must be assumed because N is not large).

Homoscedasticity.

Linearity.

Sampling Method: Independent random sampling.

Sampling Frame: Youth in one U.S. city.

Hypotheses:

H0: Age does not predict the number of arrests over the last year in the

population of young offenders ( � � 0).

H1: Age does predict the number of arrests over the last year in the pop-

ulation of young offenders ( � � 0).

To test the significance of a regression coefficient, we assume that

both variables are measured at the interval level. Because this is a para-

metric test of significance, we must also make assumptions regarding

the population distribution. For tests of statistical significance with b,

we must assume that for each value of X the scores of Y are normally

distributed around the regression line. We must also assume that the

variances of the distribution of Y scores around each value of X are

equal. This is the assumption of homoscedasticity, described in Chapter

14. Researchers generally rely on the central limit theorem to relax as-

sumptions of normality. As with analysis of variance and correlation,

To visualize these assumptions in the case of regression, it is once

again useful to look at a scatterplot. Suppose Figure 15.6 represents the

scatterplot of the population of scores for age and number of arrests in

joint distribution between X and Y that both is normal in form and meets

the assumption of homoscedasticity. But here, our imaginary line is actu-

ally the regression line. When we examine points above and below the

regression line, we see that there is a clustering of points close to the

458

we are generally concerned only with marked violations of the homos-

cedasticity assumption.

assumptions in terms of the correlation coefficient, this figure illustrates a

the city examined. Like Figure 14.7, which we used to examine these

line. Farther from the line, there are fewer points. This distribution is

basically normal in that the scores of Y for every value of X form a bell

E V A L U A T I N G T H E R E G R E S S I O N M O D E L

shape that is highest on the regression line and then slopes down in nor-

mal form away from the line. This is illustrated in the cross section for

each age group.

Also, the points around the regression line have about equal variances

(homoscedasticity). That is, the spread of the Y scores around each X in

this distribution is about equal, whether we look at the cases associated

with the youngest subjects (on the left side of the scatterplot), those asso-

ciated with average-age subjects (in the middle of the scatterplot), or those

associated with the oldest offenders (on the right side of the scatterplot).

Scatterplot Showing Normal Distribution and HomoscedasticityFigure 15.6

16 17 18 19

Age

A

Cross Sections:

Arrests

Age = 16 A

N u

m b

er o

f A

rr es

ts

F re

q u

en cy

1 case 5 cases 10 cases 25 cases 50 cases 100 cases

B

C

D

Arrests

Age = 17 B

F re

q u

en cy

Arrests

Age = 18 C

F re

q u

en cy

Arrests

Age = 19 D

F re

q u

en cy

Regression Line

459

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

In contrast, the scatterplot in Figure 15.7 shows a case that violates—

rather than meets—the assumptions for our t-test. In this scatterplot, the

points are not normally distributed around the regression line. Indeed,

they form a type of bimodal distribution with peaks above and below

the regression line (see the cross section for each age group). As

discussed in Chapter 14, heteroscedasticity refers to unequal variances

around the regression line. The scatterplot in Figure 15.7 also shows a

distribution with unequal variances. For subjects aged 17 and 19, the

scores of Y are scattered widely around the regression line. For subjects

Scatterplot Showing Nonnormal Distribution and HeteroscedasticityFigure 15.7

16 17 18 19

Age

A

Cross Sections:

Arrests

Age = 16 A

N u

m b

er o

f A

rr es

ts

F re

q u

en cy

100 cases 50 cases 25 cases 10 cases 5 cases 1 case B

C

D

Arrests

Age = 17 B

F re

q u

en cy

Arrests

Age = 18 C

F re

q u

en cy

Arrests

Age = 19 D

F re

q u

en cy

Regression Line

460

E V A L U A T I N G T H E R E G R E S S I O N M O D E L

aged 16 and 18, however, the scores of Y are tightly clustered around

the regression line.

In our example, we have only 15 cases, and thus we cannot invoke

the central limit theorem. As in the case of Pearson’s correlation coeffi-

cient, we recommend that the N of cases be at least 30 before this is

We also must assume linearity, which we examined in our scatterplot

of the relationship between age and number of arrests . Finally, our null

hypothesis is that age does not influence number of arrests over the last

year for the population of young offenders from which our sample was

drawn ( � � 0). Our research hypothesis is that age does influence num-

ber of arrests in that population ( � � 0).

The Sampling Distribution The sampling distribution is t, with N � 2 de-

grees of freedom. For our example, df � 15 � 2 � 13.

Significance Level and Rejection Region With a two-tailed 0.05 signifi-

cance threshold, the critical value for the t-test (with 13 degrees of free-

dom) is 2.160 (see Appendix 4). We will reject the null hypothesis if the

t-score is greater than 2.160 or less than �2.160.

The Test Statistic The t-test for the significance of the regression coeffi-

cient is performed by taking the difference between the observed value

of b and the hypothesized value � and then dividing that result by the

standard error of b, or . The formula for the t-statistic is shown in

Equation 15.10.

Equation 15.10

where b is the estimated regression coefficient, � is the hypothesized

population value, and is the standard error of b. In practice, this for-

mula can simply be written as b divided by the standard error of b, since

the null hypothesis for � ordinarily is that it is equal to 0.

�̂b

t � b � �

�̂b

�̂b

461

done. Accordingly, our test results cannot be relied on unless the assump-

tion of a normal distribution of Y around each value of X is true for the

population to which we infer. Of course, we cannot make this assump-

indication of the form of the population distribution. Based on the scat-

terplot shown in Figure 14.1, we concluded earlier that there was not

evidence of serious violations of this assumption.

tion without some prior knowledge of the population distribution of

scores. With regard to the assumption of homoscedasticity, as we noted in

Chapter 14, researchers generally use the scatterplot of sample cases as an

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

To determine the standard error of b in a bivariate regression

model (one with one dependent and one independent variable), we use

Equation 15.11.5

Equation 15.11

i

Table 15.5 presents the values for age and number of arrests for 15 of-

fenders and illustrates the calculations necessary for obtaining the standard

error of b. In the fourth column, we have calculated the difference between

the observed value for number of arrests (Yi ) and the predicted value ( ).Ŷi

ˆ i

�̂b � �� N

i�1

(Yi � Ŷi ) 2/(N � 2)

� N

i�1

(Xi � X )2

Calculations for the Standard Error of b for Age and Number of Arrests for 15 Young Offenders

X Y Yi � i (Yi � i) 2 Xi � (Xi � )2

(1) (2) (3) (4) (5) (6) (7)

14 0 2.1046 �2.1046 4.4294 �3.0667 9.4046 13 1 1.2039 �0.2039 0.0416 �4.0667 16.5380 15 1 3.0053 �2.0053 4.0211 �2.0667 4.2712 13 2 1.2039 0.7961 0.6337 �4.0667 16.5380 14 2 2.1046 �0.1046 0.0109 �3.0667 9.4046 14 3 2.1046 0.8954 0.8017 �3.0667 9.4046 17 3 4.8066 �1.8066 3.2639 �0.0667 0.0044 19 4 6.6080 �2.6080 6.8015 1.9333 3.7376 21 4 8.4093 �4.4093 19.4420 3.9333 15.4708 19 6 6.6080 �0.6080 0.3696 1.9333 3.7376 16 8 3.9060 4.0940 16.7612 �1.0667 1.1378 18 9 5.7073 3.2927 10.8419 0.9333 0.8710 20 9 7.5086 1.4914 2.2242 2.9333 8.6042 21 10 8.4093 1.5907 2.5303 3.9333 15.4708 22 11 9.3100 1.6900 2.8562 4.9333 24.3374

� 17.0667 � 75.0293 � 138.9326� N

i�1

(Xi � X )2� N

i�1

(Yi � Ŷi) 2X

XXŶŶŶ

Table 15.5

5It is important to note that this equation for the standard error of b is appropriate

only if we have a bivariate regression model. If we have two or more independent

variables, then a modified equation is necessary to calculate the standard error of the

regression coefficients.

462

the observed value of the dependent variable (Y ) and the predicted

value of the dependent variable (Y ) divided by the number of observa-

tions (N) minus 2. The denominator is the measure of variability for

the independent variable, X.

The numerator represents the sum of the squared differences between

E V A L U A T I N G T H E R E G R E S S I O N M O D E L

Once we have calculated the standard error of b, we can return to Equa-

tion 15.10 and calculate a t-score for b to obtain our test statistic. The test

statistic is 4.4195 for our example.6

W orking It Out

� 0.2038

� �75.0293/(15 � 2)

138.9326

�̂b � �� N

i �1

(Yi � Yi ˆ )2/(N � 2)

� N

i �1

(Xi � X )2

Column 5 shows the squares of the differences, which are then summed at

i

value (

appear in column 7. The sum of the squared deviations between age and

mean age is 138.9326 (see the bottom of column 7). We insert these values

into Equation 15.11 to calculate the standard error for b, which has a value

of 0.2038.

Ŷi

W orking It Out

� 4.4195

� 0.9007 0.2038

t � b � �

�̂b

6Except for rounding error, this result is the same as the one we obtained in testing

the significance of the correlation coefficient for this relationship (4.4195 vs. 4.4188).

In practice, you could use the correlation coefficient significance test result for defin-

ing the statistical significance of the regression coefficient. Indeed, in many texts only

one formula is provided for both coefficients.

463

the bottom of the column. We find the sum of the squared deviations bet-

ween the observed value for number of arrests (Y ) and the predicted

appear in column 6, while the squared deviations between age and mean age

) to be 75.0293. The deviations between age and mean age

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

The Decision As 4.4195 is greater than our critical value of t (2.160), we

reject the null hypothesis that age does not predict number of arrests.

We conclude that there is a statistically significant relationship between

age and number of arrests in the population of young offenders. How-

ever, because we cannot strongly support the assumption of normality in

this test or relax that assumption because N is large, we cannot place

strong reliance on our test result.

Testing the Statistical Significance of the Regression Coefficient for

Unemployment Rates and Burglary Rates in California

Let’s assess the statistical significance of the regression coefficient in our

example concerning unemployment rates and burglary rates from Cali-

fornia counties. We again begin by stating the assumptions and hypothe-

ses of our test.

Assumptions:

Level of Measurement: Interval scale.

Population Distribution: Normal distribution of Y around each value of

X (relaxed because N is large).

Homoscedasticity.

Linearity.

Sampling Method: Independent random sampling (all counties are in-

cluded in one year).

Sampling Frame: Counties in California.

Hypotheses:

H0: Unemployment rates do not influence burglary rates in California

counties ( � � 0).

H1: Unemployment rates do influence burglary rates in California coun-

ties ( � � 0).

The Sampling Distribution The sampling distribution is the t distribution,

with df � 58 � 2 � 56.

Significance Level and Rejection Region Since the research hypothesis is

nondirectional, we use a two-tailed test of statistical significance and set the

significance level at 0.05. The critical values for the t-test with 56 degrees of

freedom are about �2.003 and 2.003, meaning that we will reject the null

hypothesis if the test statistic is less than �2.003 or greater than 2.003.

The Test Statistic To test the statistical significance of our regression

coefficient b, we again use Equation 15.11 to determine the standard

error of b and Equation 15.10 to calculate the t-score. Table 15.6

464

Calculations for the Standard Error of b for Unemployment Rates and Burglary Rates for 58 California Counties

X Y Yi � i (Yi � i) 2 Xi � (Xi � )2

(1) (2) (3) (4) (5) (6) (7)

3.5 837.89 647.3188 190.5712 36,317.4013 �3.9069 15.2639 9.1 2,037.49 853.1092 1,184.3808 1,402,757.8083 1.6931 2.8666 4.6 818.55 687.7419 130.8081 17,110.7643 �2.8069 7.8787 6.8 865.04 768.5881 96.4519 9,302.9613 �0.6069 0.3683 6.9 989.76 772.2630 217.4970 47,304.9581 �0.5069 0.2569

15.9 520.06 1,102.9977 �582.9377 339,816.3271 8.4931 72.1327 3.0 664.73 628.9446 35.7854 1,280.5949 �4.4069 19.4208 8.0 1,200.91 812.6861 388.2239 150,717.7965 0.5931 0.3518 3.9 509.87 662.0181 �152.1481 23,149.0352 �3.5069 12.2983

13.4 924.10 1,011.1269 �87.0269 7,573.6848 5.9931 35.9172 11.2 845.29 930.2807 �84.9907 7,223.4123 3.7931 14.3876

6.4 1,027.79 753.8888 273.9012 75,021.8564 �1.0069 1.0138 23.4 1,526.40 1,378.6099 147.7901 21,841.9077 15.9931 255.7792

5.7 511.12 728.1650 �217.0450 47,108.5364 �1.7069 2.9135 11.4 960.18 937.6303 22.5497 508.4881 3.9931 15.9448 13.1 649.22 1,000.1024 �350.8824 123,118.4797 5.6931 32.4114

7.7 1,333.21 801.6616 531.5484 282,543.6909 0.2931 0.0859 7.0 361.24 775.9378 �414.6978 171,974.2653 �0.4069 0.1656 5.9 610.28 735.5147 �125.2347 15,683.7226 �1.5069 2.2707

11.5 929.32 941.3052 �11.9851 143.6438 4.0931 16.7535 1.9 526.98 588.5215 �61.5415 3,787.3525 �5.5069 30.3259 7.4 775.92 790.6371 �14.7171 216.5936 �0.0069 0.0000 6.7 843.92 764.9133 79.0067 6,242.0571 �0.7069 0.4997

13.3 1,214.69 1,007.4521 207.2379 42,947.5513 5.8931 34.7286 8.5 325.08 831.0603 �505.9803 256,016.0134 1.0931 1.1949 6.7 957.95 764.9133 193.0367 37,263.1637 �0.7069 0.4997 9.6 570.14 871.4834 �301.3434 90,807.8327 2.1931 4.8097 3.3 477.54 639.9691 �162.4291 26,383.2093 �4.1069 16.8666 4.1 455.37 669.3677 �213.9977 45,795.0284 �3.3069 10.9356 2.6 464.52 614.2453 �149.7253 22,417.6595 �4.8069 23.1063 3.2 646.12 636.2943 9.8257 96.5452 �4.2069 17.6980 9.0 1,030.58 849.4344 181.1456 32,813.7284 1.5931 2.5380 5.4 1,049.18 717.1405 332.0395 110,250.2163 �2.0069 4.0276 4.2 925.61 673.0426 252.5674 63,790.3117 �3.2069 10.2842 8.0 845.75 812.6861 33.0639 1,093.2215 0.5931 0.3518 4.8 883.02 695.0915 187.9285 35,317.1061 �2.6069 6.7959 3.1 539.82 632.6194 �92.7994 8,611.7342 �4.3069 18.5494 3.0 744.81 628.9446 115.8654 13,424.7909 �4.4069 19.4208 8.8 896.85 842.0847 54.7653 2,999.2337 1.3931 1.9407 3.2 540.79 636.2943 �95.5043 9,121.0637 �4.2069 17.6980 2.0 355.82 592.1963 �236.3763 55,873.7552 �5.4069 29.2346 3.9 444.07 662.0181 �217.9481 47,501.3612 �3.5069 12.2983 3.0 347.57 628.9446 �281.3746 79,171.6655 �4.4069 19.4208 6.3 647.73 750.2140 �102.4840 10,502.9682 �1.1069 1.2252 7.0 823.95 775.9378 48.0122 2,305.1713 �0.4069 0.1656 9.2 699.71 856.7841 �157.0741 24,672.2603 1.7931 3.2152

10.3 575.09 897.2072 �322.1172 103,759.4841 2.8931 8.3700 4.6 769.30 687.7419 81.5581 6,651.7269 �2.8069 7.8787 2.7 555.44 617.9201 �62.4801 3,903.7641 �4.7069 22.1549

10.5 1,057.99 904.5569 153.4332 23,541.7315 3.0931 9.5673 13.0 859.11 996.4276 �137.3176 18,856.1233 5.5931 31.2828

6.7 816.55 764.9133 51.6367 2,666.3478 �0.7069 0.4997 11.5 676.23 941.3052 �265.0752 70,264.8351 4.0931 16.7535 16.5 1,047.32 1,125.0467 �77.7267 6,041.4321 9.0931 82.6845

6.5 908.79 757.5637 151.2263 22,869.4089 �0.9069 0.8225 4.8 491.86 695.0915 �203.2315 41,303.0589 �2.6069 6.7959 4.3 591.28 676.7174 �85.4374 7,299.5476 �3.1069 9.6528

11.6 1,366.76 944.9800 421.7800 177,898.3853 4.1931 17.5821

� 7.4069 � 4,294,978.7756 � 1,010.3570� N

i�1

(Xi � X )2� N

i�1

(Yi � Ŷi) 2X

XXŶŶŶ

Table 15.6

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

presents the calculations necessary for calculating the standard error

of b. Column 4 of Table 15.6 provides the difference between the ob-

served burglary rate (Yi) and the predicted burglary rate ( i). Column

5 shows the square of each of the differences; the squares are then

summed at the bottom of the column. We find the sum of the squared

deviations between the observed burglary rate and the predicted bur-

glary rate to be 4,294,978.7756. The deviations between observed un-

employment (Xi ) and mean unemployment appear in Column 6, while

the squared deviations between unemployment and mean unemploy-

ment appear in Column 7. The sum of the squared deviations between

age and mean age is 1,010.3570 (see the bottom of Column 7). After

inserting these values into Equation 15.11, we calculate a value of

8.7126 for the standard error for b.

W orking It Out

� 8.7126

� �4,294,978.7756/(58 � 2)

1,010.3570

�̂b � �� N

i�1

(Yi � Ŷi ) 2/(N � 2)

� N

i�1

(Xi � X )2

The t-statistic is then calculated by inserting our values for b and the

standard error of b into Equation 15.10. For our example, the test statistic

is 4.2178.

W orking It Out

� 4.2178

� 36.7483 � 0

8.7126

t � b � �

�̂b

466

T H E F - T E S T F O R T H E O V E R A L L R E G R E S S I O N

The Decision Our test statistic of 4.2178 is greater than our critical

t-value of 2.003, leading us to reject the null hypothesis that unemploy-

T h e F - T e s t f o r t h e O v e r a l l R e g r e s s i o n

In regression analysis, we can carry out a second type of test of statistical

significance to evaluate whether the overall regression model contributes

particularly useful when we have more than one independent variable in

our model, a situation we will examine in Chapter 16. For this second

test of statistical significance, we draw on the logic we used in develop-

ing the measure of percent of variance explained, or R 2.

Percent of variance explained tells how much our model improves

our predictions beyond what can be learned from the mean. But another

question is whether we can conclude from our sample R 2 that R 2 is in

fact different from 0 in the population. This is the test of statistical signif-

icance for the regression model overall. The assumptions for this test are

the same as those described in the previous section.

To test this hypothesis, we use analysis of variance, which was intro-

Equation 15.12

The explained sum of squares was discussed above. The unexplained

sum of squares is simply the sum of the squared errors of the regression:

Age and Number of Arrests

The total sum of squares can be partitioned into its explained and unex-

plained components. The total sum of squares for our example of age

USS � � N

i �1

(Yi � Y i ˆ )2

F � ESS/df USS/df

467

conclude that there is a statistically significant relationship between

unemployment rates and burglary rates.

ment rates do not influence burglary rates in California counties and

duced in Chapter 12. Again, the F-test is based on a ratio of the explained

variance to the unexplained variance. The explained and unexplained

variance estimates are obtained by dividing the explained sum of squares

(ESS) and unexplained sum of squares (USS) by their appropriate degrees

of freedom, as shown in Equation 15.12.

7 This test issignificantly to our understanding of the dependent variable .

7

discussed in the prior section. We assume an interval scale, a normal distribution (relaxed

While we do not state the assumption of the tests below formally, they follow those

when N is large), homoscedasticity, linearity, and independent random sampling.

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

sum of squares (112.7041) plus the unexplained sum of squares, or error

sum of squares of the regression (75.0292).

The number of degrees of freedom for the ESS (df1) is k, or the num-

ber of variables in the regression. In our example, the regression in-

cludes only one independent variable—age—and thus the number of

degrees of freedom for the ESS is 1. For the USS, the number of degrees

of freedom (df2) is equal to N � k � 1. In the case of our example of

number of arrests and age, it is equal to 15 � 1 � 1, or 13. The F-statistic

for our regression is thus calculated by dividing the ratio of the ex-

plained variance (112.7041/1) by that of the unexplained variance

(75.0292/13), obtaining F � 19.53.

W orking It Out

� 19.5278

� 112.7041 5.7715

� 112.7041/1 75.0292/13

F � ESS/df USS/df

Setting a 5% significance threshold, we can see in the F table in Ap-

pendix 5 that the critical value associated with 1 and 13 degrees of free-

dom is 4.67. If our F-statistic is larger than this value, then our observed

significance level is less than the criterion significance level we set for

our test. Our test statistic of 19.53 is much larger than this value, and

W orking It Out

187.7333 � 112.7041 � 75.0292

� N

i�1

(Yi � Y )2 � � N

i�1

(Yi ˆ � Y )2 � �

N

i�1

(Yi � Ŷ

i) 2

TSS � ESS � USS

468

and number of arrests is 187.7333, which is equivalent to the explained

T H E F - T E S T F O R T H E O V E R A L L R E G R E S S I O N

plained by the regression line in the population to which we infer is 0.

Unemployment Rates and Burglary Rates in California

The F-test may also be used to assess the overall regression for the rela-

tionship between unemployment rates and burglary rates in California.

We reported in our calculations of R 2 that the value for the total sum of

squares is 5,659,404.5114 and the value for the explained sum of squares

is 1,364,425.7358, which means that the value for the unexplained sum

of squares is 4,294,978.7756.

We have likely violated the normality assumption of our test because we do not

have knowledge about the shape of the joint distribution of age and number of arrests

in the population and N � 15 cases is not enough to safely invoke the central limit

theorem.

W orking It Out

5,659,404.5114 � 1,364,425.7358 � 4,294,978.7756

TSS � ESS � USS

The number of degrees of freedom for the ESS (df1) is k � 1, and the

number of degrees of freedom for the USS (df2) is N � k � 1, which is

58 � 1 � 1 � 56. The F-statistic for our regression of burglary rates on

unemployment rates is calculated by dividing the ratio of the explained

variance (1,364,425.7358/1) by that of the unexplained variance

(4,294,978.7756/56), which gives us F � 17.7900.

W orking It Out

� 17.7900

� 1,364,425.7358

76,696.0496

� 1,364,425.7358/1

4,294,978.7756/56

F � ESS/df USS/df

469

thus we would reject the null hypothesis that the percent of variance ex- 8

8

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

If we set a 5% significance threshold, we see in the F table (Appendix 5)

that the critical value associated with 1 and 56 degrees of freedom is not

given in the table. Interpolating from the values for 40 degrees of free-

dom (4.008) and 60 degrees of freedom (4.000) for df2, we estimate a

critical value of 4.002. Our test statistic of 17.79 is much larger than the

critical value, so we reject the null hypothesis that the percent of vari-

ance explained by the regression line in the population to which we

infer is 0.

C h a p t e r S u m m a r y

The regression coefficient b tells us how much one variable (the in-

dependent variable, X) influences another variable (the dependent

variable, Y ). The regression coefficient is expressed in specific units of

the dependent variable and is interpreted as follows: A change of one

unit in X produces a change of b units in the estimated value of Y.

A researcher cannot predict values of Y using the regression coeffi-

cient alone. The additional piece of information required is the Y-

intercept (b0). The b0 coefficient may be interpreted as the expected

value of Y when X � 0. The predicted value of Y for other values of

X can be calculated by adding b0 to the product of the regression

coefficient and X. Regression error is the difference between the

predicted and actual values of Y. The regression line is the line

for which the sum of the squared errors is at a minimum—hence

the name ordinary least squares regression (OLS). OLS regression

is a solid basis for prediction within, but not beyond, the sample

range.

The R2 statistic is the proportion of the total sum of squares (Y � )2

accounted for by the explained sum of squares . This propor-

tion represents the improvement in predicting Y that the regression line

provides over the mean of Y.

sured on an interval scale. There is also an assumption of normality and

a requirement of homoscedasticity. These assumptions relate to the dis-

tribution of Y around each value of X. The researcher must also assume

2 that R 2

(Ŷ � Y )2

Y

470

regression coefficient b. It is assumed that the variables examined are mea-

linearity. The F-test for the overall regression determines whether the

The t distribution may be used to test statistical significance for the

in the population.

researcher can conclude from the sample R is different from 0

i

i

S Y M B O L S A N D F O R M U L A S

K e y T e r m s

bivariate regression A technique for pre-

dicting change in a dependent variable

using one independent variable.

dependent variable (Y) The variable as-

sumed by the researcher to be influenced

by one or more independent variables.

independent variable (X) A variable

assumed by the researcher to have an impact

on the value of the dependent variable, Y.

OLS regression See ordinary least squares

regression analysis.

ordinary least squares regression

analysis A type of regression analysis in

which the sum of squared errors from the

regression line is minimized.

percent of variance explained (R2) A

measure for evaluating how well the re-

gression model predicts values of Y. It rep-

resents the improvement in predicting Y

that the regression line provides over the

mean of Y.

regression coefficient b A statistic used

to assess the influence of an independent

variable, X, on a dependent variable, Y.

The regression coefficient b is interpreted

as the estimated change in Y that is associ-

ated with a one-unit change in X.

regression error (e) The difference be-

tween the predicted value of Y and the

regression line The line predicting values

of Y. The line is plotted from knowledge of

the Y-intercept and the regression coeffi-

cient.

regression model The hypothesized

statement by the researcher of the factor or

factors that define the value of the depen-

dent variable, Y. The model is normally

Y-intercept (b0) The expected value of Y

when X � 0. The Y-intercept is used in

predicting values of Y.

S y m b o l s a n d F o r m u l a s

b0 Y-intercept

�0 Y-intercept for population model

b Regression coefficient

� Regression coefficient for the population

X Independent variable

Y Dependent variable

e Error

� Error for population model

ESS Explained sum of squares

471

actual value of Y.

expressed in equation form.

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

USS Unexplained sum of squares

TSS Total sum of squares

k Number of variables in the overall regression model

Standard error of the regression coefficient

To determine the value of the Y-intercept:

b0 �

To predict values of the dependent variable, Y:

To identify the regression error:

e � Y �

To show that the sum of squared error in an OLS regression line is a

minimum:

� minimum

A bivariate regression model for a sample:

Y � b0 � bX � e

A bivariate regression model for a population:

Y � �0 � �X � �

To calculate the percent of explained variance:

R 2 �

To calculate the value of t for the regression coefficient b:

t �

To calculate the standard error of the regression coefficient:

�̂b � �� N

i�1

(Yi � Ŷi ) 2/(N � 2)

� N

i�1

(Xi � X )2

b � �

�̂b

ESS TSS

� � N

i�1

(Ŷi � Y )2

� N

i�1

(Yi � Y )2

� N

i�1

(Yi � Ŷi) 2

Ŷ � b0 � bX

Y � bX

�̂b

472

E X E R C I S E S

To calculate the value of F for the overall regression:

F �

E x e r c i s e s

15.1 A researcher carries out a series of regression analyses for different studies. The results of three of the studies are given below. In each case, explain what the results mean in plain English.

a. X � number of prior driving offenses; Y � fine in dollars imposed by magistrate; b � 72.

b. X � number of times a household has been broken into prior to purchase of first burglar alarm; Y � amount of money in dollars spent by homeowner on first burglar alarm; b � 226.

c. X � number of times subject has been involved in a car accident; Y � estimated average speed of subject when driving on a freeway in miles per hour; b � �8.5.

15.2 Nine adolescents are interviewed about the number of hours per week they work and the number of times they have smoked marijuana in the last year. The results are recorded as follows:

X: Number of hours worked per week 0 10 10 15 5 30 20 40 15

Y: Number of times smoked marijuana in the last year 1 3 2 5 0 13 10 20 25

a. Calculate the regression coefficient b, and explain what it means in plain English.

b. Calculate the value of the Y-intercept, b0.

c. Calculate a test of statistical significance for the regression coeffi- cient b.

15.3 A study of sentencing decisions hypothesized that judges would become increasingly lenient with drug offenders as they accumulated more years of experience. To test this hypothesis, researchers gathered data on a sample of 12 drug offenders. The data included information on number of years on the bench for the judge and

ESS/df USS/df

� � N

i�1

(Ŷi � Y )2/k

� N

i�1

(Yi � Ŷi )2/(N � k � 1)

473

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

length of the sentence (in months). The results were recorded as follows:

X: Number of years as a judge 3 1 0 1 2 5 9 13 17 0 6 2

Y: Length of sentence (months) 14 22 24 20 15 12 3 6 18 18 10 18

a. Calculate the regression coefficient b, and explain what it means in plain English.

b. Calculate the value of the Y-intercept, b0.

c. Calculate a test of statistical significance for the regression coeffi- cient b.

15.4 Ten police officers are asked how many promotions they have re- ceived and how many years they have served on the force. The results are recorded below:

X: Years on the force 7 1 5 3 12 2 4 1 9 6

Y: Number of promotions 5 1 3 1 8 1 2 0 7 2

a. Calculate the regression coefficient b.

b. Calculate the value of the Y-intercept, b0.

c. How many promotions would you predict for an officer who had served 10 years on the force?

d. What is the regression error in predicting the number of promotions for an officer who has served 12 years on the force?

15.5 Ten prosecutors were asked what percentage of their cases ended in guilty pleas and how many years of experience each had as a prose- cutor. The results were recorded as

X: Number of years of experience as a prosecutor 10 12 8 1 0 2 7 20 5

Y: Percentage of cases resulting in a guilty plea 93 90 87 72 70 70 82 97 94

a. Calculate the regression coefficient b, and explain what it means in plain English.

b. Calculate the value of the Y-intercept, b0.

c. Calculate a test of statistical significance for the regression coeffi- cient b.

d. If a prosecutor had six years of experience, what would be the pre- dicted percentage of guilty pleas?

15.6 A study exploring the link between aggression and crime reported ag- gression scores and number of arrests for six individuals who partici-

474

E X E R C I S E S

pated in the study. The values for aggression and number of arrests are as follows:

X: Aggression score 92 63 77 29 51 10

Y: Number of arrests 6 2 3 1 2 0

a. Calculate the regression coefficient b, and explain what it means in plain English.

b. Calculate the value of the Y-intercept, b0.

c. Calculate a test of statistical significance for the regression coefficient b.

d. What would be the predicted number of arrests for a person with an aggression score of 75?

e. What would be the predicted number of arrests for a person with an aggression score of 25?

15.7 For the last ten convicts released from Wilmslow Prison, Joan recorded the percentage of their initial sentence from which they were excused. She also recorded the number of times each convict was called before a disciplinary committee over the course of his sentence. The scores of each subject on these two variables are listed below:

X: Number of disciplinary hearings 0 5 2 1 6 4 4 0 5 3

Y: Percentage of sentence not served 33 5 18 32 0 10 5 30 0 17

a. Calculate the regression coefficient b.

b. Calculate the value of the Y-intercept, b0.

c. Using the data provided, show that the sum of the error on either side of the regression line equals 0:

d. Using the data provided, show that

is less than

e. Explain in plain English the meaning of what you showed in part d for the regression model.

15.8 In running a small pilot study for a large-scale research project, George gathers data on the average number of homicides monthly for

� N

i�1

(Yi � Y )2

� N

i�1

(Yi � Ŷ )2

� N

i�1

(Yi � Y ̂) � 0

475

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N

five U.S. cities. While he is looking for a good predictor of the differ- ent homicide rates, he stumbles across the following set of data on the number of theaters in each of the cities:

X: Number of theaters 1 3 6 7 8

Y: Homicides monthly 10 14 23 26 32

a. Calculate the regression coefficient b.

b. Calculate the value of the Y-intercept, b0.

c. According to this regression model, how many homicides would a city with ten theaters expect per month?

d. Why is this model misleading?

15.9 Lee is investigating six recent cases of vandalism in the local shopping mall. She compares the amount of damage done in each case with the number of vandals involved in each incident. Her findings are as follows:

X: Number of vandals 3 2 6 4 1 2

Y: Damage done ($) 1,100 1,850 3,800 3,200 250 1,200

a. Calculate the regression coefficient b.

b. Calculate the value of the Y-intercept, b0.

c. Plot the scores on a scatterplot, and draw the regression line where you think it should go.

d. Calculate the value of R 2. What does this tell you about the model?

15.10 Refer to the data from Exercise 15.2.

a. Calculate the value of R 2. What does it tell you about the model?

b. Run an F-test for the overall regression. Remember to outline all of the steps required in a test of statistical significance, including any violations of your assumptions. Can you conclude that the percent of explained variance (R 2) is different from 0 for the population?

15.11 Refer to the data from Exercise 15.5.

a. Calculate the value of R 2. What does it tell you about the model?

b. Run an F-test for the overall regression. Remember to outline all of the steps required in a test of statistical significance, including any violations of your assumptions. Can you conclude that the percent of explained variance (R 2) is different from 0 for the population?

15.12 Refer to the data from Exercise 15.9.

a. Calculate the value of R 2. What does it tell you about the model?

b. Run an F-test for the overall regression. Remember to outline all of the steps required in a test of statistical significance, including any violations of your assumptions. Can you conclude that the percent of explained variance (R 2) is different from 0 for the population?

476

C O M P U T E R E X E R C I S E S 477

C o m p u t e r E x e r c i s e s

Basic OLS regression analysis is the core of many statistical packages, and SPSS

and Stata have features that make obtaining basic results straightforward but have

also added a number of options and features for obtaining much more complex

results. We discuss some of these features and options in subsequent chapters.

As before, there are example syntax files that illustrate the commands below for

both SPSS (Chapter_15.sps) and Stata (Chapter_15.do).

SPSS

Ordinary least squares regression analyses are performed with the

REGRESSION command and is generally self-explanatory:

The output from executing this command will present three tables of results that

contain the statistics discussed in this chapter. The first table of results, labeled

“Model Summary,” presents the value for R2. The second table of results, labeled

“ANOVA,” contains the ANOVA table that was originally discussed in Chapter

12. In this table, you will find the values for the explained, unexplained, and

total sums of squares; the F-statistic; and the observed significance level of F.

Please note, however, that the labels for the sums of squares are different. The

explained sum of squares is labeled “Regression,” and the unexplained sum of

squares is labeled “Residual.” The third table of results presents the regression

coefficients, standard errors of the coefficients, t-statistic for each coefficient,

and observed significance level for each regression coefficient. Note, too, that

there is a distinction between unstandardized and standardized regression coef-

ficients. The unstandardized coefficients are the regression coefficients presented

in this chapter. We discuss standardized regression coefficients in Chapter 16.

It is also possible to have SPSS compute predicted values and residuals for

each observation in the data file. To obtain one or both of these values, you will

need to add a /SAVE line to the command line:

where /SAVE PRED RESID will save the predicted values (PRED) and the

residuals (RESID) to the data file, where they will appear as new variables

(i.e., additional columns of data on the far right of the spreadsheet). The default

REGRESSION

/DEPENDENT dep_var_name

/METHOD = ENTER indep_var_name.

REGRESSION

/DEPENDENT dep_var_name

/METHOD = ENTER indep_var_name

/SAVE PRED RESID.

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N478

variable name for the predicted values is PRE_1, and default variable name

for the residuals is RES_1. We will return to this command in the computer

exercises for Chapter 16.

Stata

The estimation of OLS regression models uses the regress command:

The output generated by the regress command is compact but includes an

ANOVA table, a listing of model summary statistics (e.g., R2), and a table of

results presenting the coefficients, standard errors, t-statistics for the coeffi-

cients, an observed significance level, and something labeled a “95 % Confidence

Interval.” We discuss confidence intervals in Chapter 20.

To compute predicted values and residuals in Stata, you will use the predict

command immediately following a regress command. The default output for pre-

dict is to compute the predicted value; the residuals are designated with an option.

To request predicted values:

In contrast to SPSS, which will automatically generate a new variable name to

contain the predicted values or residuals, Stata requires that the user enter a new

variable name. If you wanted to use the same convention as SPSS, this command

would appear as

To request residuals, run another predict command immediately following the

previous line for predicted values:

The option r that appears after the comma indicates to Stata to compute the

residuals rather than some other quantity. Again, if you wanted to use the same

convention as in SPSS for naming these variables, this command would look like

It is important to bear in mind that if you run several different models, you will

need to continually modify the variable names for the predicted values and

residuals. Stata will not allow you to overwrite existing variables.

Problems

1. Open the California UCR data file (caucr_99.sav or caucr_99.dta),

which contains the data presented in Table 14.8. Run the linear regres-

sion command in SPSS, using burglary rate as the dependent variable

and unemployment rate as the independent variable. Note that the values

reported in the three tables of results match those reported in the text.

regress dep_var_name indep_var_name

predict new_variable_name

predict PRE_1

predict new_variable_name, r

predict RES_1, r

C O M P U T E R E X E R C I S E S 479

2. Enter the data from Exercise 15.2.

a. Run the linear regression command, using number of times smoked

marijuana as the dependent variable and number of hours worked per

week as the independent variable.

b. Compare your answers to Exercises 15.2 and 15.10 with the results

produced by SPSS.

3. Enter the data from Exercise 15.3.

a. Run the linear regression command, using sentence length as the

dependent variable and number of years as a judge as the independent

variable.

b. What is the value of R2?

c. Perform an F-test for the overall regression. Outline all of the steps

your assumptions. Can you conclude that the percent of explained

variance (R2) is different from 0 for the population?

4. Enter the data from Exercise 15.7.

a. Run the linear regression command, using percentage of sentence not

served as the dependent variable and number of disciplinary hearings as

the independent variable.

b. What is the value of R2?

c. Perform an F-test for the overall regression. Outline all of the steps

your assumptions. Can you conclude that the percent of explained

variance (R2) is different from 0 for the population?

5. Open the NYS data file (nys_1.sav, nys_1_student.sav, or nys_1.dta). Use

the linear regression command, and run regression analyses for the pairs

of variables listed below. Do the following for each pair of variables:

Perform a t

Report the value of R2.

Perform an F-test for the overall regression.

a. Age (X) and number of thefts valued at less than $5 in the last year (Y).

b. Number of times drunk (X) and number of thefts valued at $5–$50

in the last year (Y).

c. Frequency of marijuana use (X) and number of times the youth has

hit other students in the last year (Y).

C H A P T E R F I F T E E N : B I V A R I A T E R E G R E S S I O N480

d. Number of times the youth has hit a parent (X) and number of

thefts valued at more than $50 in the last year (Y).

e. Number of times the youth has been beaten up by a parent (X) and

number of times the youth has hit a teacher in the last year (Y).

6. Open the Pennsylvania Sentencing data file (pcs_98.sav or pcs_98.dta).

Use the linear regression command, and run regression analyses for the

pairs of variables listed below. Do the following for each pair of variables:

Perform a t

Report the value of R2.

Perform an F-test for the overall regression.

a. Age (X) and length of incarceration sentence (Y).

b. Prior criminal history score (X) and length of incarceration

sentence (Y).

c. Offense severity score (X) and length of incarceration sentence (Y).

Multivariate Regression

How Does the Researcher Try to Correctly Specify the Model?

How Can You Include Nominal and Ordinal Variables in a Regression Model?

How Do You Compare Regression Coefficients?

C h a p t e r s i x t e e n

S p e c i f i c a t i o n : b u i l d i n g a m u l t i v a r i a t e m o d e l

M o d e l b u i l d i n g

What is a Correctly Specified Model?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_16,

© Springer Science+Business Media New York 2014

ONE OF THE STATISTICAL TOOLS most commonly used in criminal justice

and criminology is regression modeling. A regression model allows the

researcher to take a broad approach to criminological research problems.

It is based not simply on understanding the relationships among vari-

ables, but on specifying why changes occur and what factors are directly

responsible for these changes. In a regression model, the researcher tries

to disentangle the various potential factors that have an impact on the

dependent variable, in order to provide an accurate picture of which

variables are in fact most important in causing change.

In this chapter, we discuss why it is generally necessary to take into

account more than just one independent variable in building a regres-

sion model. Previous chapters have focused on bivariate statistical analy-

sis, in which we relate two variables—nominal, ordinal, or interval—to

T h e I m p o r t a n c e o f C o r r e c t M o d e l S p e c i f i c a t i o n s

The most important assumption we make in regression modeling is that the

model we have estimated is specified correctly. A correctly specified re-

gression model is one in which the researcher has taken into account all

of the relevant predictors of the dependent variable and has measured

them accurately. This requirement of regression modeling is the most diffi-

cult one that researchers face. Its importance is linked both to prediction of

the dependent variable and to correct estimation of regression coefficients.

Errors in Prediction

Predictions of Y in regression are based on the factors that are included

in a regression model. So far, we have examined bivariate regression

models, in which one independent variable is used to predict values of

Y. But in the real world it is unlikely that only one variable will influence

the dependent measure you are examining. Most often, it will be neces-

sary to take into account a number of independent variables. Regression

statistical model.

researcher takes into account a series of independent variables within one

each other. This chapter introduces multivariate analysis, in which the

482

analysis that takes into account more than one independent variable is

called multivariate regression analysis. The regression model we have

discussed so far can be extended to the multivariate case simply by

adding a term for each new variable. For example, to include years of

education in the model predicting number of arrests presented earlier,

we would express our regression equation as follows:

Yarrests � b0 � b1(age) � b2(education) � e

The population model for this equation would be written as

Yarrests � �0 � �1(age) � �2(education) � �

Sometimes, when examining multiple independent variables, re-

searchers find it tedious to include the names of the variables in sub-

scripts. Accordingly, they will often use a general form of the regression

equation and then define each variable in a table or in their description

of results. For example, the above equation could be expressed in terms

of the population parameters as

Model 1: Y � �0 � �1X1 � �2X2 � �

where Y � arrests

X1 � age

X2 � years of education

In theory, you could define all relevant predictors of Y and include them

all in your regression model. This correctly specified model would also

provide the most accurate predictions of Y. Conversely, a misspecified

model, or one that does not include all relevant predictors, will provide

biased predictions of Y.

Let’s say, for example, that family median income is also an important

predictor of arrests. In this case, the corrected population regression

equation would be written as follows:

Model 2: Y � �0 � �1X1 � �2X2 � �3X3 � �

where Y � arrests

X1 � age

X2 � years of education

X3 � family median income

By adding this additional variable, we improve our predictions of Y over

those provided by Model 1. Because we have taken into account the in-

fluence of family income on arrests, we have added to our ability to cor-

rectly predict the dependent variable. By implication, our predictions of

Y will be less trustworthy when we do not include a factor that influ-

ences the dependent variable.

T H E I M P O R T A N C E O F C O R R E C T M O D E L S P E C I F I C A T I O N S 483

Sometimes, statisticians express this fact in terms of an assumption

about the error term in the population regression model. The error term,

�, should represent only random fluctuations that are related to the out-

residuals, since they are in theory what is left over once you have taken

into account all systematic causes of Y. However, if you fail to include an

important predictor of Y as an independent variable, then by implication

it moves to your error term. The error term now is not made up only of

random—or what statisticians sometimes call stochastic—variation in Y,

but rather includes the systematic variation that can be attributed to the

excluded variable. For example, in Model 1, the effect of median family

income is not taken into account, and thus the systematic relationship

between median family income and number of arrests is found in the

error term for that regression equation. When a model is not correctly

specified, the error term will not represent only random or stochastic

variation, as is required by the assumptions of regression analysis; it will

be systematically related to the dependent variable.

Correctly Estimating the Effect of b

dent variables. Suppose, for example, that a bivariate regression is de-

fined in which number of years in prison is identified as influencing

number of arrests after prison:

Yrearrests � b0 � b1(years in prison) � e

In estimating this relationship from the data presented in Table 16.1, we

find that the regression coefficient based on this model is 1.709. That is,

every additional year of imprisonment produces about a 1.709 increase

in our prediction of number of subsequent arrests.

W orking It Out

� 1.7089

� 31.7 18.55

b � � N

i�1

(Xi � X )(Yi � Y )

� N

i�1

(Xi � X )2

Our model for subsequent arrests states that the only causal factor in-

fluencing arrests is years of imprisonment. This, of course, is a question-

able statement, because common sense tells us that this model is not

researcher to present biased estimates of the effects of specific indepen-

Failure to correctly specify a regression model may also lead the

comes (Y ) that you are examining. For this reason, we also call the errors

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N484

correctly specified. There are certainly other factors that influence arrests.

Some of those factors, in turn, may also be related to the number of

years that an offender serves in prison. If this is true—that relevant fac-

tors related to years of imprisonment have been omitted from the

model—then the regression coefficient may provide a very misleading

estimate of the effect of imprisonment on arrests.

Judges, for example, are likely to impose longer prison sentences on

offenders with more serious prior records. Using the sample data in

Table 16.2, we can look at the correlations among these three variables

(see Table 16.3). The number of prior arrests is strongly related (r �

0.63) to the length of prison term served. Prior arrests are even more

strongly related to subsequent arrests (r � 0.76). This suggests, first of

all, that prior record is a relevant factor that should be included if our

model is to be correctly specified. But it also raises a very important con-

cern: How do we know that our finding that “years in prison” increases

reoffending is not simply a result of the fact that those who serve longer

prison terms generally have more serious prior records of offending?

Number of Rearrests (Y ) and Years Spent in Prison (X ) for 20 Former Inmates

YEARS SPENT

REARRESTS IN PRISON

SUBJECT Y Yi � X Xi � (Xi � )2 (Xi � )(Yi � )

1 0 �3.1 2 �1.15 1.3225 3.565 2 0 �3.1 3 �0.15 0.0225 0.465 3 1 �2.1 1 �2.15 4.6225 4.515 4 1 �2.1 2 �1.15 1.3225 2.415 5 1 �2.1 3 �0.15 0.0225 0.315 6 1 �2.1 3 �0.15 0.0225 0.315 7 2 �1.1 4 0.85 0.7225 �0.935 8 2 �1.1 2 �1.15 1.3225 1.265 9 2 �1.1 2 �1.15 1.3225 1.265

10 3 �0.1 3 �0.15 0.0225 0.015 11 3 �0.1 3 �0.15 0.0225 0.015 12 3 �0.1 3 �0.15 0.0225 0.015 13 4 0.9 3 �0.15 0.0225 �0.135 14 4 0.9 4 0.85 0.7225 0.765 15 4 0.9 4 0.85 0.7225 0.765 16 4 0.9 4 0.85 0.7225 0.765 17 5 1.9 4 0.85 0.7225 1.615 18 6 2.9 4 0.85 0.7225 2.465 19 7 3.9 5 1.85 3.4225 7.215 20 9 5.9 4 0.85 0.7225 5.015

� 18.55

Bivariate Regression Model: Dependent Variable: Subsequent Rearrests Independent Variable: Years in Prison Regression Coefficient: b (years in prison) � 31.7/18.55 � 1.7089

� N

i�1

(Xi � X)(Yi � Y )� N

i�1

(Xi � X )2X � 3.15Y � 3.1

YXXXY

Table 16.1

� 31.7

485T H E I M P O R T A N C E O F C O R R E C T M O D E L S P E C I F I C A T I O N S

would be made with subjects who were otherwise similar. That is, we

would want to be sure that the offenders with longer and shorter prison

sentences were comparable on other characteristics, such as the serious-

ness of prior records. In this case, there would be no relationship be-

tween prior arrests and length of imprisonment, and thus we would not

have to be concerned with the possibility that the effect of length of im-

In criminal justice, this approach is taken in the development of

randomized experiments.1 A randomized study of the impact of

Number of Rearrests, Years Spent in Prison, and Number of Prior Arrests for 20 Former Inmates

SUBJECT REARRESTS YEARS IN PRISON PRIOR ARRESTS

1 0 2 4 2 0 3 2 3 1 1 2 4 1 2 3 5 1 3 3 6 1 3 2 7 2 4 3 8 2 2 3 9 2 2 1

10 3 3 2 11 3 3 3 12 3 3 3 13 4 3 4 14 4 4 3 15 4 4 4 16 4 4 5 17 5 4 4 18 6 4 5 19 7 5 5 20 9 4 6

= 3.10 = 3.15 = 3.35 s � 2.300 s � 0.9631 s � 1.2360

XXY

Table 16.2

Correlation Coefficients for the Variables Years in Prison, Prior Arrests, and Subsequent Rearrests Based on Data from 20 Former Inmates

YEARS IN PRISON PRIOR ARRESTS

Prior Arrests r � 0.6280 Subsequent Rearrests r � 0.7156 r � 0.7616

Table 16.3

1

Wadsworth, 1995). For a comparison of experimental and nonexperimental methods,

see D. Weisburd, C. Lum, and A. Petrosino, “Does Research Design Affect Study Out-

comes in Criminal Justice?” The Annals 578 (2001): 50–70.

In an ideal world, our comparisons of the impact of imprisonment

M. Maxfield, The Practice of Social Research in Criminal Justice (Belmont, CA:

prisonment actually reflects the influence of prior arrests on reoffending.

For a discussion of experimental methods in criminal justice, see E. Babbie and

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N486

487T H E I M P O R T A N C E O F C O R R E C T M O D E L S P E C I F I C A T I O N S

length of imprisonment on reoffending would be one in which the re-

searcher took a sample of offenders and assigned them to treatment

and control conditions at random. For example, the researcher might

define a sentence of 6 months as a control condition and a sentence of

1 year as an experimental condition. In this case, the researcher could

of subjects to treatment and control conditions allows the researcher

to assume that other traits, such as prior record, are randomly scattered

across the treatment and control conditions. Our problem in criminal

justice is that it is often impractical to develop experimental research

designs. For example, it is highly unlikely that judges would allow a

researcher to randomly allocate prison sanctions. The same is true for

many other research problems relating to crime and justice.

Fortunately for criminal justice researchers, a correctly specified re-

gression model will take into account and control for relationships that

exist among the independent variables included in the model. So, for

example, the inclusion of both length of imprisonment and prior ar-

rests in one regression model will provide regression coefficients that

reflect the specific impact of each variable, once the impact of the

other has been taken into account. This is illustrated in Equation 16.1,

which describes the calculation of a multivariate regression coefficient

in the case of two independent variables (X1 and X2). Equation 16.2

applies Equation 16.1 to the specific regression model including both

length of imprisonment and prior arrests. The model can be described

as follows:

Y � b0 � b1X1 � b2X2 � e

where Y � subsequent rearrests

X1 � years in prison

X2 � prior arrests

Here we calculate the multivariate regression coefficient b1 for length of

imprisonment.

Equation 16.1

Equation 16.2

In Equations 16.1 and 16.2, the bivariate correlations among the

three measures examined, as well as the standard deviations of years in

bX1 � �rY,YP � (rY,PArYP,PA)

1 � r 2 YP,PA

�� sY

s �

bX1 � �rY,X1

� (rY,X2 rX1,X2

)

1 � r 2 X1,X2

�� sY

s �

rearrests without concern about the confounding influences of other vari-

examine the effects of a longer versus a shorter prison sentence on

YP

X1

ables. In Chapter 21, we focus more directly on the analysis of experi-

mental data. But it is important to note here that random allocation

prison and rearrests, are used to calculate the multivariate regression

coefficients. The three correlations for our specific example are (1) rY,YP,

or the correlation between subsequent rearrests and years in prison;

(2) rY,PA, or the correlation between subsequent rearrests and prior ar-

rests; and (3) rYP,PA, or the correlation between years in prison and

prior arrests.

What is most important to note in Equation 16.2 is that the numer-

ator (in the first part) takes into account the product of the relation-

ship between prior arrests and subsequent rearrests and that of prior

arrests and years in prison. This relationship is subtracted from the

simple correlation between years in prison and subsequent arrests. In

this way, multivariate regression provides an estimate of b that takes

into account that some of the impact of years in prison may be due to

the fact that longer prison terms are associated with more serious

prior records. This estimate is now purged of the bias that was intro-

duced when prior record was not included in the regression model.

The multivariate regression coefficient for years in prison when prior

record is included in the regression model (0.936) is considerably

smaller than the estimate calculated earlier in the bivariate regression

(1.709).

W orking It Out

� 0.9358

� �0.2373152

0.605616 �(2.388122)

� �0.7156 � (0.7616)(0.6280)

1 � (0.6280)2 �� 2.300

0.9631�

bX1 � �rY,YP � (rY,PArYP,PA)

1 � r 2 YP,PA

�� sY

sYP�

With the same information, we can calculate the multivariate regres-

sion coefficient for prior arrests. We find the value for b2 to be 0.9593 (see

sults. As you can see, the value of b when we take into account years in

prison (0.96) is much smaller than that in the bivariate case (1.4).

working it out, page 490). The bivariate regression coefficient for prior ar-

rests is calculated in the box on page 489 so that you can compare the re-

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N488

PRIOR

REARRESTS ARRESTS

SUBJECT Y Yi � X Xi � (Xi � )2 (Xi � )(Yi � )

1 0 �3.1 4 0.65 0.4225 �2.015

2 0 �3.1 2 �1.35 1.8225 4.185

3 1 �2.1 2 �1.35 1.8225 2.835

4 1 �2.1 3 �0.35 0.1225 0.735

5 1 �2.1 3 �0.35 0.1225 0.735

6 1 �2.1 2 �1.35 1.8225 2.835

7 2 �1.1 3 �0.35 0.1225 0.385

8 2 �1.1 3 �0.35 0.1225 0.385

9 2 �1.1 1 �2.35 5.5225 2.585

10 3 �0.1 2 �1.35 1.8225 0.135

11 3 �0.1 3 �0.35 0.1225 0.035

12 3 �0.1 3 �0.35 0.1225 0.035

13 4 0.9 4 0.65 0.4225 0.585

14 4 0.9 3 �0.35 0.1225 �0.315

15 4 0.9 4 0.65 0.4225 0.585

16 4 0.9 5 1.65 2.7225 1.485

17 5 1.9 4 0.65 0.4225 1.235

18 6 2.9 5 1.65 2.7225 4.785

19 7 3.9 5 1.65 2.7225 6.435

20 9 5.9 6 2.65 7.0225 15.635

= 3.10 = 3.35

� 30.55 � 43.30

Bivariate Regression Model: Dependent Variable: Subsequent Rearrests Independent Variable: Prior Arrests Regression Coefficient: b (Prior Arrests) � 43.30/30.55 � 1.4173

� N

i�1

(Xi � X )(Yi � Y )� N

i�1

(Xi � X )2XY

YXXXY

Calculation of Bivariate Regression Coefficient for Number of Rearrests (Y) and Number of Prior Arrests (X) for 20 Former Inmates

The fact that the results are different when we examine the effects of

years in prison and prior arrests in the multivariate regression model

shows that the bivariate regression coefficients were indeed biased. In

both cases, the estimate of the effect of b provided by the bivariate re-

gression coefficient was much too high. These differences also reflect a

difference in interpretation between the multivariate regression coeffi-

duced by a one-unit change in X. In the multivariate case, b represents

the estimated change in Y associated with a one-unit change in X when

all other independent variables in the model are held constant. Holding

prior arrests constant leads to a reduction in the impact of years in

prison. Holding years in prison constant leads to a reduction in the esti-

mate of the effect of prior arrests. These differences may be seen as the

bias introduced by misspecifying the regression model through the ex-

clusion of prior arrests.

We can also identify this bias in terms of assumptions related to the

error term in regression. It is assumed not only that the errors in the re-

gression are stochastic, but also that there is no specific systematic rela-

tionship between the error term and the independent variables included

in the regression. If there is such a relationship, the regression coefficient

will be biased. While this may seem like a new concept, it is really a re-

statement of what you learned above.

Let’s use our model predicting rearrest as a substantive example. We

saw that if we estimated the regression coefficient for years in prison

without taking into account prior arrests, the regression coefficient

would be biased—in this case, overestimated. What happens in theory to

the error term in this case? As we discussed earlier in the chapter, when

we exclude an independent variable, the effect of that variable moves to

W orking It Out

� 0.9593

� �0.3122032

0.605616 �(1.8608)

� �0.7616 � (0.7156)(0.6280)

1 � (0.6280)2 �� 2.300

1.2360�

bX2 � �rY,PA � (rY,YPrYP,PA)

1 � r 2 YP,PA

�� sY

sPA�

regression coefficient represents the estimated change in Y that is pro-

cient and the bivariate regression coefficient. In the bivariate case, the

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N490

the error term. In our case, the population model including both inde-

pendent variables may be stated as follows:

Y � �0 � �1X1 � �2X2 � �

where Y � subsequent rearrests

X1 � years in prison

X2 � prior arrests

When we take into account only one independent variable, the model

includes only the term X1:

Y � �0 � �1X1 � �

where Y � subsequent rearrests

X1 � years in prison

In the latter model, number of prior arrests is included by implication in

the error term. But what does this mean regarding the relationship in this

model between the error term and years in prison? Our sample data sug-

gest that the number of prior arrests is related to years in prison (as was

By looking at bias in terms of the error term, we can also specify when

excluding an independent variable will not lead to bias in our estimates

of the regression coefficients of other variables. If the excluded variable is

unrelated to other variables included in the regression, it will not cause

bias in estimates of b for those specific variables. This is the case because

when there is no systematic relationship between the excluded variable

and the included variable of interest, its exclusion does not lead to a sys-

tematic relationship between the error term and the variable of interest.

For example, if years in prison and prior arrests were not systemati-

cally related (e.g., the correlation between the variables was 0), it would

not matter whether we took into account prior arrests in estimating the re-

gression coefficient for years in prison.2 In this case, the exclusion of prior

arrests would not lead to a systematic relationship between the error term

2It is important to note that bias can be caused by a nonlinear relationship between

the excluded and the included variable. The assumption is that there is no systematic

relationship of any form.

shown in Table 16.3). By implication, since number of prior arrests is

related to years in prison as well. Accordingly, if we leave prior arrests

out of our equation, then we violate the assumption that there is no

varibles in the equation.

now found in the error term, the error term can now be assumed to be

systematic relationship between the error term and the independent

491T H E I M P O R T A N C E O F C O R R E C T M O D E L S P E C I F I C A T I O N S

and years in prison, because there is no systematic relationship between

prior arrests and years in prison. However, it is important to remember

that the exclusion of prior arrests will still cause bias in our estimate of Y.

In this situation, we continue to violate the assumption that the error term

is stochastic. It now includes a systematic predictor of Y, prior arrests.

Comparing Regression Coefficients Within a Single

Model: The Standardized Regression Coefficient

A multivariate regression model allows us to specify the impact of a spe-

cific independent variable while holding constant the impact of other in-

dependent variables. This is a very important advantage of multivariate

regression analysis over bivariate regression analysis. However, when we

include multiple variables in the same model, it is natural to want to

compare the impact of the different variables examined. For example, in

our case, does years in prison have a stronger effect on subsequent rear-

rests than number of prior arrests does? Or does number of prior arrests

have a stronger effect than years in prison? The ordinary regression coef-

ficient b does not allow us to answer this question, since it reports the

effect of a variable in its original units of measurement. Accordingly, the

regression coefficient for years in prison reports the predicted change in

subsequent rearrests for each year change in years in prison. The regres-

sion coefficient for number of prior arrests reports the predicted change

in subsequent rearrests for each change in number of prior arrests.

Though the interpretation of the regression coefficients in these cases is

straightforward, we cannot directly compare them.

Another statistic, called the standardized regression coefficient or

Beta, allows us to make direct comparisons. Beta weights take the re-

gression coefficients in an equation and standardize them according to

the ratio of the standard deviation of the variable examined to the stan-

dard deviation of the dependent variable. Beta is expressed mathemati-

cally in Equation 16.3:

Equation 16.3

The interpretation of the standardized coefficient is similar to that of b

(the unstandardized coefficient), except that we change the units. We in-

terpret Beta as the expected amount of change in the standard deviation

of the dependent variable, given a one-unit change in the standard devi-

ation of the independent variable.

For years in prison in our example, we take the regression coefficient

of 0.9358 and multiply it by the ratio of the standard deviation of years in

prison (0.9631) and subsequent rearrests (2.3000). The result is 0.3919,

Beta � b�sX

sY�

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N492

which tells us that an increase of one standard deviation in years in

prison is expected to result in an increase of 0.392 standard deviation in

rearrests.

W orking It Out

� 0.3919

� 0.9358�0.9631 2.3000�

Beta � b�sX

sY�

For prior arrests, we begin with our regression coefficient of 0.9593.

Again, we standardize our estimate by taking the ratio of the standard

deviation of prior arrests (1.2360) and subsequent rearrests (2.3000). Our

estimate of Beta here is 0.5155, which indicates that an increase of one

standard deviation in prior arrests is expected to result in an increase of

0.516 standard deviation in rearrests.

W orking It Out

� 0.5155

� 0.9593�1.2360 2.3000�

Beta � b�sX

sY�

In our example, the Beta weight for prior arrests is larger than that for

years in prison. According to this estimate, the number of prior arrests

has a greater impact on subsequent rearrests than the number of years in

prison does. The standardized regression coefficient thus provides us

with an answer to our original question regarding which of the indepen-

dent variables examined has the most influence on the dependent vari-

able. As you can see, the standardized regression coefficient is a useful

tool for comparing the effects of variables measured differently within a

493T H E I M P O R T A N C E O F C O R R E C T M O D E L S P E C I F I C A T I O N S

single regression model. However, because standardized regression coef-

ficients are based on the standard deviations of observed samples, they

are generally considered inappropriate for making comparisons across

samples.

C o r r e c t l y S p e c i f y i n g t h e R e g r e s s i o n M o d e l

The previous section illustrated the importance of a correctly specified

regression model. If a regression model is not correctly specified, then

the predictions that are made and the coefficients that are estimated may

provide misleading results. This raises important theoretical as well as

practical questions for criminal justice research.

In criminal justice research, we can seldom say with assurance that

the models we develop include all relevant predictors of the dependent

variables examined. The problem is often that our theories are not pow-

erful enough to clearly define the factors that influence criminal justice

questions. Criminal justice is still a young science, and our theories for

explaining crime and justice issues often are not well specified. This fact

has important implications for the use of criminal justice research in de-

veloping public policy. When our predictions are weak, they do not

form a solid basis on which to inform criminal justice policies.3

One implication of our failure to develop strongly predictive models

in criminal justice is that our estimates of variable effects likely include

some degree of bias. We have stressed in this chapter the importance of

controlling for relevant predictors in regression modeling. The cost of

leaving out important causes is not just weaker prediction but also esti-

mates of variable effects that include potentially spurious components.

This fact should make you cautious in reporting regression analyses and

critical in evaluating the research of others. Just because regression coef-

ficients are reported to the fifth decimal place on a computer printout

does not mean that the estimates so obtained are solid ones.

The fact that regression models often include some degree of mis-

specification, however, should not lead you to conclude that the regres-

3Mark Moore of Harvard University has argued, for example, that legal and ethical

dilemmas make it difficult to base criminal justice policies about crime control on

models that still include a substantial degree of statistical error. See M. Moore, “Pur-

blind Justice: Normative Issues in the Use of Prediction in the Criminal Justice Sys-

tem,” in A. Blumstein, J. Cohen, A. Roth, and C. A. Visher (eds.), Criminal Careers

and “Career Criminals,” Vol. 2 (Washington, DC: National Academy Press, 1986).

sion approach is not useful for criminal justice researchers. As in any

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N494

available. The researcher’s task in developing regression models is to im-

prove on models that were developed before. With each improvement,

the results we gain provide a more solid basis for making decisions

about criminal justice theory and policy. This, of course, makes the prac-

tical task of defining the correct model for the problem you are examin-

ing extremely important. How then should you begin?

Defining Relevant Independent Variables

Importantly, model specification does not begin with your data. Rather,

it starts with theory and a visit to the library or other information sys-

tems. To build a regression model, you should first identify what is al-

ready known about the dependent variable you have chosen to study. If

your interest, for example, is in the factors that influence involvement in

criminality, you will need to carefully research what others have said and

found regarding the causes of criminality. Your regression model should

take into account the main theories and perspectives that have been

raised by others.

If you do not take prior research and theory into account, then those

reviewing your work will argue that your predictions and your estimates

of variable effects are biased in one way or another. Just as the exclusion

of prior record from our example led to a misleading estimate of its im-

pact on length of imprisonment, so too the exclusion of relevant causal

factors in other models may lead to bias. The only way to refute this po-

tential criticism is to include such variables in your regression model.

Taking into account the theories and perspectives of others is the first

step in building a correctly specified regression model. However, in most

research we seek to add something new to existing knowledge. In regres-

sion modeling, this usually involves the addition of new variables. Some-

times, such new variables are drawn from an innovative change in theory.

Other times, they involve improvements in measurement. Often, the find-

ing that these new or transformed variables have an independent impact

above and beyond those of variables traditionally examined by researchers

leads to important advances in criminal justice theory and policy.

Taking into Account Ordinal- and Nominal-Scale

Measures in a Multivariate Regression

Until now, we have assumed that ordinary least squares regression

analysis requires an interval level of measurement, both for the depen-

dent and for the independent variables. However, criminal justice re-

searchers will sometimes use this approach with ordinal-level dependent

variables when there are a number of categories and there is good rea-

son to assume that the intervals for the categories are generally similar.

In practice, you should be cautious in using OLS regression when your

science, the task is to continue to build on the knowledge that is presently

C O R R E C T L Y S P E C I F Y I N G T H E R E G R E S S I O N M O D E L 495

What about the inclusion of non–interval-level independent variables?

Such variables often are important in explaining criminal justice out-

comes. If we are required to include all relevant causes of Y in order to

correctly specify our model, how can we exclude ordinal- and nominal-

level measures? Fortunately, we do not have to. In multivariate regres-

sion, it is acceptable to include ordinal- and nominal-level independent

variables as long as there is at least one interval-level independent vari-

able also included in the analysis.

But even though you can include ordinal- and nominal-level variables,

you need to take into account the specific interpretation used by regres-

sion analysis for interpreting the effects of one variable on another. In-

cluding an ordinal-level measure in a multivariate regression is relatively

straightforward. This is done in Table 16.4, which presents a standard

SPSS printout for a regression analysis. The data used are drawn from a

national sample of police officers developed by the Police Foundation.4

The dependent variable in this analysis is hours worked per week. There

are two independent variables. One, years with the department, is mea-

sured at the interval level. The second, level of education, is on an ordi-

SPSS Printout for Regression Analysis of the Police Officer Example

Coefficients

Unstandardized Standardized Coefficients Coefficients t Sig.

Model B Std. Error Beta

1 (Constant) 44.968 .749 60.031 .000

YEARS WITH �7.354E-02 .026 �.092 �2.816 .005

DEPARTMENT

LEVEL OF .456 .173 .086 2.636 .009 EDUCATION

a Dependent Variable: HOURS PER WEEK WORKED

Table 16.4

4David Weisburd et al., The Abuse of Authority: A National Study of Police Officers’ At-

titudes (Washington, DC: The Police Foundation, 2001).

nal scale with eight levels, ranging from some high school to doctoral

regression cannot be met in the case of ordinal dependent variables, you

dependent variable is not interval level. When the assumptions of OLS

should use ordinal regression (see Chapter 19). As will be explained in Chapter 18, the use of OLS regression in the case of a binary dependent

variable is inappropriate.

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N496

SPSS table) are less than the conventionally applied significance level of

0.05 we would likely use in this case. This result, as in most statistical

packages, is calculated for a two-tailed test of statistical significance (gen-

erally the default option). For years with the department, we can see that

the impact is negative. When we control for the impact of level of educa-

tion, each year with the department is associated with an average de-

But what is the meaning of the effect of level of education? Here, what

we have is not an interval scale but a group of ordered categories. For the

regression, this ordinal-level scale is treated simply as an interval-level

scale. It is assumed that the categories must be roughly similar in value, or

that each level increase in that scale is related in a linear manner to the de-

pendent variable. Thus, our interpretation of this regression coefficient is

that for every one-level increase in education level, there is, on average, a

0.456 increase in the number of hours worked (once we have taken into

account years with the department). In this case, the standardized regres-

sion coefficient is very useful. It appears from the size of the coefficients

This example illustrates how we can include an ordinal-level variable

in a multivariate regression. The inclusion of an ordinal variable is

straightforward, and its interpretation follows that of an interval-level in-

dependent variable. But when we include a nominal-level variable, we

have to adjust our interpretation of the regression coefficient.

SPSS Printout for Regression Analysis with an Interval-Level and Nominal-Level Variable

Coefficients

Unstandardized Standardized Coefficients Coefficients t Sig.

Model B Std. Error Beta

1 (Constant) 48.550 .977 49.672 .000

YEARS WITH �7.737E-02 .026 �.097 �2.943 .003

DEPARTMENT

RESPONDENT �1.669 .803 �.068 �2.077 .038 GENDER

a Dependent Variable: HOURS PER WEEK WORKED

Table 16.5

hours worked per week—the observed significance levels (“Sig.” in the

degree. We can see that both of these variables have a significant impact on

crease of about 0.074 hours in number of hours worked each week.

that the overall effect of years with the department is much less than that

of level of education. However, the standardized regression coefficients

(represented by Beta) show that the difference between the two variables

is not large.

C O R R E C T L Y S P E C I F Y I N G T H E R E G R E S S I O N M O D E L 497

dependent variable and data examined in our prior example. To make

our example easier, we include only two measures in predicting number

of hours worked. Again, we have an interval-level measure, years with

the department. A binary independent variable, gender, is also included.

In regression analysis, a binary nominal-level independent variable is

generally called a dummy variable. Our first problem is to give num-

bers to this dummy variable. Multivariate regression analysis does not

recognize qualitative categories. By convention, we give one category a

value of 0 and the other a value of 1. It is generally good practice to give

the category with the largest number of cases a value of 0 because, as

we will illustrate in a moment, that category becomes the reference cate-

gory. Since this sample included many more men than women, we as-

signed men the value 0 and women the value 1.

Again, we can see that both variables have a statistically significant im-

pact on hours worked per week. The observed significance level for years

with the department is 0.003, and that for gender is 0.038. But how can

we interpret the dummy variable regression coefficient of �1.669? One

way to gain a better understanding of the interpretation of dummy vari-

able regression coefficients is to see how they affect our regression equa-

tion. Let’s begin by writing out the regression equation for our example:

Y � b0 � b1X1 � b2X2

where Y � hours worked per week

X1 � years with the department

X2 � gender of officer

As a second step, let’s insert the coefficients gained in our regression

analysis:

Y � 48.550 � (�0.077)X1 � (�1.669)X2

or

Y � 48.550 � 0.077X1 � 1.669X2

What happens if we try to write out the regression equations for men

and women separately? For men, the regression equation is

Y � 48.550 � 0.077X1 � 1.669(0)

Table 16.5 reports the results of a regression with a single interval-

level variable and a binary nominal-level variable. We use the same

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N498

Because men are coded as 0, the second term of the equation falls out.

But what about for women? The second term in the equation is a con-

stant because all of the women have a value of 1. If we write it out, we

have the following result:

Y � 48.550 � 0.077X1 � 1.669(1) or

Y � 48.550 � 0.077X1 � 1.669

We can simplify this formula even more, because the two constants at

the beginning and the end of the equation can be added together:

Y � 46.881 � 0.077X1

What then is the difference between the regression equations for men

and women? In both cases, the slope of the regression line is given by

the term �0.077X1. The difference between the two equations lies in the

Y-intercept, as illustrated in Figure 16.1. As you can see, men and

women have parallel regression lines. However, the women’s line inter-

sects the Y-axis about 1.7 hours lower than the men’s line. This provides

us with the interpretation of our coefficient. Women police officers, on

Regression Lines for Men and Women Police OfficersFigure 16.1

H ou

rs W

or ke

d p

er W

ee k

50

49.5

49

48.5

48

47.5

47

46.5

46

Years with Department (X1)

0 2 4 6 8 10 12 14 16 18 20

Males

Females

Y � 48.550 � 0.077X1

or

C O R R E C T L Y S P E C I F Y I N G T H E R E G R E S S I O N M O D E L 499

average, work about 1.669 hours a week less than men police officers,

taking into account years with the department.

This example also suggests why it is generally recommended that you

place the category with the largest number of cases as the 0 category of

a binary dummy variable. The category men, in this case, is the reference

category, meaning that the coefficient for gender gives us the estimate of

the female category in reference to the male category. We want our ref-

erence category to be as stable as possible, and a large number of cases

makes this category more stable.

But how can we assess the impact of a nominal variable that has

multiple categories? In fact, multiple-category nominal variables create

a good deal more complexity for the researcher than do ordinal or

binary nominal variables. In this case, you must create a separate

variable for each category in your analysis. For example, the Police

Foundation study divided the United States into four regions: North

Central, Northeast, South, and West. In practice, you would need to

create a separate variable for each of these regions. In other words,

you would define a variable North Central, which you would code 1

for all those officers in the North Central region and 0 for all other of-

ficers. You would repeat this process for each of the other regional

categories.

As with the binary independent variable, you must choose one of

the categories to be a reference category. In this case, however, the

reference category is excluded from the regression. Again, it is gener-

ally recommended that you choose as the reference category the cate-

gory with the largest number of cases.5 In our example, the largest

number of officers is drawn from the South. Suppose that we include

only one interval-level variable in our equation: years with the depart-

ment. Table 16.6 presents the results from an analysis in which years

with the department and region are used to predict number of hours

worked.

In this example, we included in the regression a single interval-level

variable and three region measures: North Central, Northeast, and West.

5There may be times when you want to choose a category that does not include the

largest number of cases as the reference. For example, if you wanted to compare a se-

ries of treatments to a no-treatment, or control, condition, it would make sense to

have the control condition as the excluded category, even if it did not include the

largest N. However, if the excluded category has a small number of cases, it may lead

to instability in the regression estimates.

While South is not included as a variable, it is in fact the reference

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N500

Y � b0 � b1X1 � b2X2 � b3X3 � b4X4

where Y � hours worked per week

X1 � years with department

X2 � North Central

X3 � Northeast

X4 � West

Using the results for this model presented in Table 16.6, we can write

the results in equation form as follows:

Y � 47.654 � 0.061X1 � 2.335X2 � 1.758X3 � 0.846X4

In this case, we can also write out a separate regression equation for

each of the four regions. Since the South is our reference category,

those from the South are coded 0 on the three included variables.

Thus, our equation is simply the Y-intercept and the variable years

with the department. For officers from the North Central region,

the equation includes the Y-intercept, b1X1, and b2X2. The other para-

meters are set to 0, since those in the North Central region have 0 val-

ues on X3 and X4. Similarly, for both the Northeast and the West, only

SPSS Printout for Regression Analysis with Multiple-Category Nominal Variable

Coefficients

Unstandardized Standardized Coefficients Coefficients t Sig.

Model B Std. Error Beta

1 (Constant) 47.654 .470 101.435 .000

YEARS WITH �6.138E-02 .026 �.077 �2.352 .019

DEPARTMENT

NORTH CENTRAL �2.335 .610 �.141 �3.825 .000

NORTHEAST �1.758 .573 �.114 �3.067 .002

WEST �.846 .616 �.050 �1.372 .170

a Dependent Variable: HOURS PER WEEK WORKED

Table 16.6

category. If we again write out our regression equation, we can see why

this is the case:

C O R R E C T L Y S P E C I F Y I N G T H E R E G R E S S I O N M O D E L 501

one of the three dummy variables is included. For each equation,

we can once again add the constant for the dummy variable to the

Y-intercept:

Officers from the South:

Y � 47.654 � 0.061X1

Officers from the North Central:

Y � 47.654 � 0.061X1 � 2.335X2 or Y � 45.319 � 0.061X1

Officers from the Northeast:

Y � 47.654 � 0.061X1 � 1.758X3 or Y � 45.896 � 0.061X1

Officers from the West:

Y � 47.654 � 0.061X1 � 0.846X4 or Y � 46.808 � 0.061X1

Once again, we can gain a better conceptual understanding of our re-

sults if we plot them, as in Figure 16.2. In this case, each of the included

categories is found to have a Y-intercept lower than that of the excluded

Plot of Hours Worked and Years with Department, by RegionFigure 16.2

H ou

rs W

or ke

d p

er W

ee k

48

47.5

47

46.5

46

45.5

45

Years with Department (X1)

0 2 4 6 8 10 12 14 16 18 20

South

West

Northeast

North Central

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N502

category, the South. This means that, on average, officers work fewer

hours in all of the other regions. The least number of hours worked per

week is found in the North Central region. Here, officers work, on aver-

age, about 2.335 hours less than they do in the South, once we have

taken into account years with the department. In the Northeast, officers

work about 1.758 hours less and in the West about 0.846 hour less a

week.

Are these differences statistically significant? It is important to note

that the significance statistic reported for each coefficient tells us only

whether the category is significantly different from the reference cate-

gory. This is one reason it is so important to be clear about the definition

of the reference category. In our example, the North Central and North-

east regions are significantly different from the South, using a 5% signifi-

cance threshold and a two-tailed significance test (the default option in

SPSS). The West, however, is not significantly different from the South,

using this threshold.

If you wanted to determine whether region overall as a variable

had a statistically significant impact on hours worked per week, you

would have to run an additional significance test based on the F-test

for the regression model, introduced in Chapter 15. The F-test for mul-

tiple-category dummy variables in regression compares the R 2 statistic

gained with the dummy variables included in the regression to the R 2

statistic gained without those variables. In practice, you must run two

separate regressions, although most computer programs now provide

this statistic directly. First, you calculate the regression without the

new dummy variable categories (referred to as the reduced model)

and identify its R 2. In our case, the regression without the dummy

variables produces an R 2 of only 0.008. You then compute the regres-

sion with the dummy variables, as we did earlier (referred to as the

full model). In this case, R 2 is 0.023. The F-test formula is presented in

Equation 16.4.

Equation 16.4

To apply Equation 16.4 to our example, we first subtract the R 2 of the

reduced model from the R 2 of the full model and then divide

this quantity by the number of variables in the full model (k fm) minus the

number of variables in the reduced model (k rm), which is 3. The denom-

inator is found by subtracting the R 2 of the full model from 1, and then

dividing this quantity by N � k fm � 1. For this sample, N is 923 and k fm

is 4. Our final result is F � 4.55. Looking at the F-distribution (see

(R 2 fm)(R 2

rm)

F � (R 2

f m � R 2 rm)/(k fm � k rm)

(1 � R 2 fm)/(N � k fm � 1)

C O R R E C T L Y S P E C I F Y I N G T H E R E G R E S S I O N M O D E L 503

Appendix 5) with 3 and 918 degrees of freedom, we can see that our re-

sult is statistically significant at the 0.05 level.

W orking It Out

� 4.5455

� (0.015/3)

(0.977/918) �

0.005 0.0011

� (0.023 � 0.008)/(4 � 1)

(1 � 0.023)/(923 � 4 � 1)

F � (R 2

fm � R 2 rm)/(k fm � k rm)

(1 � R 2 fm)/(N � k fm � 1)

One final question we might ask is whether we can use the standard-

ized regression coefficient to compare dummy variables to ordinal- and

interval-level measures. In general, statisticians discourage such use of

standardized regression coefficients, since they are based on standard de-

viations and the standard deviation is not an appropriate statistic for a

nominal-level variable. Additionally, for a nominal-level variable, the

standardized regression coefficient refers only to the difference between

the reference category and the dummy variable category examined. This

may sometimes make sense in the case of a binary dummy variable, since

we can say that one category is Beta standard deviations higher or lower

on the dependent variable. But it can be extremely misleading in the case

The size of the standardized regression coefficient, like the size of the co-

efficient itself, will depend on which category is excluded. In general,

you should exercise caution when interpreting standardized regression

coefficients for dummy variables in a multivariate regression analysis.

of multi-category nominal-level variables, such as region in our example.

C h a p t e r S u m m a r y

In a bivariate regression model, there is only one independent vari-

able, and it must be an interval-level measure. Importantly, the re-

searcher can rarely be sure that the change observed in the dependent

variable is due to one independent variable alone. And if variables that

have an impact on the dependent measure are excluded, the predictions

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N504

related to the included factor, then the estimate of b for the included fac-

tor will also be biased. Randomized experiments, which scatter differ-

ent traits at random, offer a solution to the latter problem, but they are

often impractical in criminal justice research. A statistical solution that

enables us to correct for both types of bias is to create a multivariate

regression model.

In a multivariate regression model, there may be several independent

variables, only one of which needs to be interval level. Such a model

considers the effect of each independent variable, while holding all the

other variables constant. A binary nominal-level variable included in a

regression model is called a dummy variable. Regression coefficients

measured using different scales may be compared with a standardized

regression coefficient (Beta). A regression model is correctly speci-

fied if the researcher has taken into account and correctly measured all

of the relevant predictors of the dependent variable. Existing literature

and prior research are suitable places to start.

K e y T e r m s

biased Describing a statistic when its esti-

mate of a population parameter does not

center on the true value. In regression

analysis, the omission of relevant indepen-

dent variables will lead to bias in the esti-

mate of Y. When relevant independent vari-

ables are omitted and those measures are

related to an independent variable included

in regression analysis, then the estimate of

the effect of that variable will also be biased.

correctly specified regression model A

regression model in which the researcher

has taken into account all of the relevant

predictors of the dependent variable and

has measured them correctly.

dummy variable A binary nominal-level

variable that is included in a multivariate

regression model.

multivariate regression A technique for

predicting change in a dependent variable,

using more than one independent variable.

randomized experiment A type of study

in which the effect of one variable can be

examined in isolation through random allo-

cation of subjects to treatment and control,

or comparison, groups.

standardized regression coefficient

(Beta) Weighted or standardized estimate

of b that takes into account the standard

deviation of the independent and the de-

pendent variables. The standardized regres-

sion coefficient is used to compare the ef-

fects of independent variables measured on

different scales in a multivariate regression

analysis.

of Y gained in a regression will be biased. If the excluded variables are

K E Y T E R M S 505

S y m b o l s a n d F o r m u l a s

k Number of independent variables in the overall regression model

Correlation coefficient for Y and X1

Correlation coefficient for Y and X2

Correlation coefficient for X1 and X2

sY Standard deviation for Y

Standard deviation for X1

R 2 obtained for the full regression model

R 2 obtained for the reduced regression model

k fm Number of independent variables in the full regression model

k rm Number of independent variables in the reduced regression model

To calculate a multivariate regression coefficient for two independent

variables:

and

A sample multivariate regression model with three independent variables:

0 � b1X1 � b2X2 � b3X3 � e

A population multivariate regression model with three independent

variables:

Y � �0 � �1X1 � �2X2 � �3X3 � �

To calculate the standardized coefficient (Beta):

Beta � b�sX

sY�

b � �r )

1 � �� sY

s �

b 1 � �rY,X1

� (rY,X2 rX1,X2

)

1 � r 2 X1,X2

�� sY �

R

2 rm

R

2 fm

sX1

rX1,X2

rY,X2

rY,X1

Y,

Y � b

Y,

x

x 2

� (r r, X1,X2

r 2 X1,X2

s

X1

2X

X1

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N506

To calculate an F-test on a subset of variables in a regression model:

E x e r c i s e s

16.1 Consider the following regression model, which purports to predict the length of sentence given to convicted thieves:

Y � b0 � bX � e

where Y � length of sentence

X � number of prior sentences

a. List the variables you might wish to include in a more comprehen- sive model. Include a brief statement about why each additional variable should be included.

b. Present your model in equation form.

16.2 In an article in the newspaper, a researcher claims that low self- esteem is the cause of crime. Upon closer inspection of the results in the paper, you learn that the researcher has computed a bivariate model using self-reported theft as the dependent variable (Y ) and self- esteem as the one independent variable (X ).

a. List the variables you might wish to include in a more comprehen- sive model. Include a brief statement about why each additional variable should be included.

b. Present your model in equation form.

16.3 A researcher has built a multivariate regression model to predict the effect of prior offenses and years of education on the length of sen- tence received by 100 convicted burglars. He feeds the data into a computer package and obtains the following printout:

Dependent Variable (Y): Length of Sentence (months)

Independent Variable (X1 ): Number of Prior Offenses

Independent Variable (X2 ): Years of Education

F sig � 0.018

R Square � 0.16

X1 : b � �0.4 Sig t � 0.023

X2 : b � �0.3 Sig t � 0.310

Evaluate the results, taking care to explain the meaning of each of the statistics produced by the computer.

F � (R 2

fm � R 2 rm)/(k fm � k rm)

(1 � R 2 fm)/(N � k fm � 1)

E X E R C I S E S 507

16.4 An analysis of the predictors of physical violence at school produced the following results:

Independent Variable b Beta

Age (Years) 0.21 0.05

Sex (Female � 1, Male � 0) �3.78 0.07

�1.34 0.06

1.96 0.33

3.19 0.24

2.05 0.27

Explain what each regression coefficient (b) and standardized regres- sion coefficient (Beta) means in plain English.

16.5

explain variations in the amount of drugs seized per month and runs a regression analysis to check the effect of his independent variable—the total number of customs officers on duty for each

coefficient is �4.02. Danny is worried, however, that his bivariate model might not be correctly specified, and he decides to add another variable—the number of ships that arrive at the port each month. He calculates the correlations between the three pairs of variables, and the results are as follows:

Y (drugs seized), X1 (customs officers): �0.55

Y (drugs seized), X2 (ships arriving): �0.60

X1 (customs officers), X2 (ships arriving): �0.80

The standard deviations for the three variables are 20 kg (quantity of drugs seized per month), 1.6 (number of customs officers on duty), and 22.5 (number of ships arriving).

a. Calculate the regression coefficient for customs officers.

b. Calculate the regression coefficient for ships arriving.

c. How do you account for the difference between your answer to part a and the regression coefficient of �4.02 that Danny obtained earlier?

16.6 A study of prison violence examined the effects of two independent variables—percent of inmates sentenced for a violent crime (X1) and average amount of space per inmate (X2)—on the average number of violent acts per day (Y ). All variables were measured for a random

Race (White � 1, Non-white � 0)

month—on the quantity of drugs seized. The resulting regression

Number of Friends Arrested

Danny has obtained figures on the amount of drugs seized per

Number of Times Attacked by Others

month at a seaport over the course of two years. He wishes to

Number of Times Hit by Parents

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N508

selection of cell blocks in three prisons. The researcher reported the following results:

a. Calculate the regression coefficients for the effects of X1 and X2 on Y. Explain what these coefficients mean in plain English.

b. Calculate the standardized regression coefficients for the effects of X1 and X2 on Y. Explain what these coefficients mean in plain English.

c. Which one of the variables has the largest effect on prison vio-

16.7 A study of recidivism classified offenders by type of punishment re- ceived: prison, jail, probation, fine, or community service. A re- searcher interested in the effects of these different punishments ana- lyzes data on a sample of 967 offenders. She computes two regression models. In the first, she includes variables for age, sex, race, number of prior arrests, severity of the last conviction offense, and length of punishment. The R 2 for this model is 0.27. In the second model, she adds four dummy variables for jail, probation, fine, and community service, using prison as the reference category. The R 2 for this model is 0.35. Explain whether the type of punishment had an effect on re- cidivism (assume a 5% significance level).

16.8 A public opinion poll of 471 randomly selected adult respondents asked about their views on the treatment of offenders by the courts. Expecting race/ethnicity to be related to views about the courts, a re- searcher classifies respondents as African American, Hispanic, and white. To test for the effect of race/ethnicity, he computes one regres- sion using information about the age, sex, income, and education of the respondents and finds the R 2 for this model to be 0.11. In a sec- ond regression, he adds two dummy variables for African American and Hispanic, using white as the reference category. The R 2 for this second model is 0.16. Explain whether the race/ethnicity of the re- spondent had a statistically significant effect on views about the courts (assume a 5% significance level).

sX2 � 2.64

sX1 � 10.52

sY � 0.35

rX1,X2

rY,X2

rY,X1 � 0.20

� 0.20

lence? Explain why.

� 0.20

E X E R C I S E S 509

C o m p u t e r E x e r c i s e s

In Chapter 15, we explored the basic features of the regression commands in

SPSS and Stata in the computation of a bivariate regression model. To compute

a multivariate regression model, we simply add additional independent variable

names to the list of independent variables on the command line. The following

exercises illustrate some of the additional features of the regression command.

Please see the appropriate SPSS (Chapter_16.sps) or Stata (Chapter_16.do)

syntax file for specific examples.

SPSS

Standardized Regression Coefficients (Betas)

The standardized regression coefficients (Betas) are part of the standard

output for SPSS’s linear regression command. In the table of results presenting

the coefficients, the standardized coefficients are located in the column

following those presenting the values for the regression coefficients (b) and the

standard errors of b. Nothing else is required to obtain the standardized

coefficients.

F-Test for a Subset of Variables

The computation of an F-test for a subset of variables requires a little

planning in setting up a multivariate linear regression model. When think-

ing about your regression model and a test of one or more subsets of

variables, you will need to enter these independent variables on separate

interested in, we would use the following syntax:

The trick here is to keep track of all the independent variables in your regression

model and determine whether they belong to the first or the second group—the

second group would be the subset of interest. For example, suppose we had an

interest in looking at whether demographic characteristics of offenders affected

punishment severity. In this case, we would then list the demographic character-

istics (however measured) in the second block (i.e., /METHOD = ENTER line).

We have also added the /STATISTICS option line to the REGRESSION

command. The reason for this is to force SPSS to compute the F-test on the

subset of variables and to simultaneously report all of the other results in a

linear regression analysis that it usually reports. Specifically, the items on the

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N510

/METHOD = ENTER lines. In general, if we have one subset that we are

/STATISTICS line request the coefficient table (COEFF), model summary (R),

REGRESSION

/STATISTICS COEFF R ANOVA CHANGE

/DEPENDENT dep_var_name

/METHOD = ENTER list_of_variables_NOT_in_subset

/METHOD = ENTER list_of_variables_in_subset.

ANOVA table (ANOVA), and change in R2 when the second block of variables

is added to the regression model (CHANGE). The F-test on the subset of

variables is produced with the CHANGE option.

The output from running this command is nearly identical to what you have

viewed previously. The major difference is that there will be two major rows of

results for all of the tables viewed in the output before—one row will be labeled

Model 1 and the other row Model 2. In other words, there will be a row for the

“reduced” model (Model 1 in SPSS) that contains only those variables included

in the first block of variables and a second row for the “full” model that includes

all variables (Model 2 in SPSS).

The F-test for the subset of variables can be found in the “Model Summary”

table of results under the columns labeled “Change Statistics.” For Model 2, the

F-statistic for the subset of variables appears in the column labeled “F Change.”

the next column, providing you with all the information you need to test whether

the subset of variables makes a statistically significant contribution to the overall

regression model.

C O M P U T E R E X E R C I S E S 511

Since the description of the various pieces may be confusing, we encourage

you to open and run the accompanying SPSS syntax file for this chapter

(Chapter_16.sps).

Residual Plot

It is also possible with the regression command to analyze residuals in ways ranging

from simple to complex. Perhaps the most straightforward way of analyzing residu-

als is graphically, through the use of a residual plot that SPSS can produce. There

are many different kinds of residual plots that SPSS could create—we highlight

only one simple example here. A histogram of the residuals from a regression analy-

sis with a normal curve overlaid on the histogram is obtained as follows:

where the /RESIDUALS line will request a plot of residuals—the

HISTOGRAM(ZRESID) option specifies a histogram of what are known as “stand-

ardized residuals.” A histogram of the residuals with the overlaid normal curve will

give you some idea of how closely the residuals approximate a normal distribution

(which is what is to be expected). If the residuals do not resemble a normal distribu-

tion, this is often an indication of a problem with the regression model, such as one

or more relevant independent variables having been omitted from the analysis.

REGRESSION

/DEPENDENT dep_var_name

/METHOD = ENTER list_of_indep_vars

/RESIDUALS HISTOGRAM(ZRESID).

The value for the numerator degrees of freedom (df 1 ) will appear in the next

column to the right and will equal the number of independent variables included

in the subset. The value for the denominator degrees of freedom (df 2 ) appears in

Stata

Standardized Regression Coefficients (Betas)

To request the standardized regression coefficients (Betas) in a multivariate

linear regression model in Stata, you will need to add the option b to a regress

command:

The last column of output in the coefficient table will then report the

standardized coefficients.

F-Test for a Subset of Variables

In contrast to the cumbersome syntax in SPSS for testing a subset of

variables, the syntax required in Stata involves two steps: (1) estimate the full

regression model with the regress command and (2) test the subset of variables

using the testparm command. The form of the testparm command is simply

C H A P T E R S I X T E E N : M U L T I V A R I A T E R E G R E S S I O N512

The output from running this command is an F-test on the set of variables listed

on the testparm command line.

Residual Plot

We illustrated at the end of Chapter 15 the process for computing residuals from

a linear regression analysis. To create a histogram of the residuals, we would use

the histogram command (discussed at the end of Chapter 3), and if we wanted

to overlay a normal curve, we use the normal option:

Problems

1. Enter the data from Table 16.2. Run the regression command to

reproduce the unstandardized and standardized regression coefficients

presented in this chapter.

a. Compute two bivariate regression models, using years in prison as the

independent variable in one regression and prior arrests as the inde-

pendent variable in the second regression. Generate a histogram of the

residuals for each regression model. What does the pattern of results in

this plot suggest to you about the distribution of error terms?

b. Compute the multivariate model, and generate a histogram of the

residuals for this regression model. How has the pattern of error terms

changed relative to the two histograms produced in part a?

to do Exercises 2 through 5.

regress dep_var_name indep_var_names, b

testparm list_of_subset_variables

regress dep_var_name indep_var_names

predict RES_1, r

histogram RES_1, normal

2. Compute a multivariate regression model, using number of times the

student hit other students as the dependent variable. From the variables

included in the data file, select at least five independent variables that you

think have some relationship to hitting other students.

-

b. Generate a histogram of the residuals for this regression model. What

does the pattern of results in this plot suggest to you about the distri-

bution of error terms?

3. Compute a multivariate regression model, using number of times

something worth $5 or less has been stolen as the dependent variable.

From the variables included in the data file, select at least five independ-

ent variables that you think have some relationship to stealing something

worth $5 or less.

b. Generate a histogram of the residuals for this regression model.

What does the pattern of results in this plot suggest to you about the

distribution of error terms?

4. Compute a multivariate regression model, using number of times the

student cheated on exams as the dependent variable. From the variables

included in the data file, select at least five independent variables that you

think have some relationship to cheating on exams.

b. Generate a histogram of the residuals for this regression model. What

does the pattern of results in this plot suggest to you about the

distribution of error terms?

5. Compute a multivariate regression model, using number of times drunk as

the dependent variable. Use age, sex, race, employment status, hours spent

studying per week, grade point average, and number of friends who use

alcohol as the independent variables.

a. Use an F-test to test whether demographic characteristics—age, sex,

and race—affect drinking behavior.

b. Use an F-test to test whether academic characteristics—hours spent

studying per week and grade point average—affect drinking behavior.

C O M P U T E R E X E R C I S E S 513

494 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

C h a p t e r s e v e n t e e n

Multivariate Regression:

Additional Topics

N o n - l i n e a r r e l a t i o n s h i p s

What are They?

How are They Included in Regression Models?

How are They Interpreted?

I n t e r a c t i o n e f f e c t s

What are They?

How are They Included in Regression Models?

How are They Interpreted?

M u l t i c o l l i n e a r i t y

When Does It Arise?

How is It Diagnosed?

How is It Treated?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_17,

© Springer Science+Business Media New York 2014

IN THE PREVIOUS CHAPTER we extended the bivariate regression approach

by showing how we could include multiple independent variables simulta-

neously in a single model. We illustrated how we incorporate variables

measured not only at the interval level of measurement, but also nominal

and ordinal independent variables, into a regression model. While the mod-

els we have examined so far allow us to approximate linear relationships

between various independent variables and the dependent variable, in the

real world we are sometimes confronted with more complex research ques-

tions that require us to make additional modifications to our model.

For example, in the OLS regression model, our interpretation of the coef-

ficients is based on the notion that there is a linear relationship between the

independent and the dependent variable. But what if we find evidence of

a curvilinear relationship? Or, theory suggests that there may be a non-linear

relationship between two variables? Although the OLS regression model is

based on the assumption of a linear relationship between the dependent

and each of the independent variables, non-linear relationships can be

incorporated into an OLS regression model in a straightforward manner.

Another issue in the application of OLS regression is that the interpre-

tation of the coefficients is based on the idea that each independent vari-

able has a constant effect irrespective of the levels of other independent

variables. For example, if we include a dummy variable we assume that

the effect of every other independent variable is the same for men and

women. But what if there was a good theoretical or policy reason to sus-

pect that the effect of some variable was different for men and women?

How would we incorporate that into our model? In the statistical litera-

ture, these are known as interaction effects, and they allow us to test

whether the effects of specific independent variables in a regression model

vary by the level of other independent variables.

In this chapter we also introduce an important problem that

researchers sometimes face when estimating multivariate regression mod-

els. We have emphasized so far that researchers must include all relevant

independent variables in a model if it is to be correctly specified. Correct

515

model specification, in turn, is necessary to avoid bias in regression models.

But sometimes the inclusion of multiple independent variables can lead

to a problem we term multicollinearity which is likely to lead to esti-

mation of unstable regression coefficients. Multicollinearity refers to the

situation where independent variables are very highly correlated with

each other, which then makes it very difficult for OLS regression to deter-

mine the unique effects of each independent variable.

N o n - l i n e a r R e l a t i o n s h i p s

Policy-oriented research focused on the severity of punishment for convict-

ed offenders illustrates that as the severity of an offender’s prior record and

the severity of the conviction offense increase, the severity of the punish-

ment tends to increase.1 In the interpretation of OLS regression results, we

would say something, for example, about how each one unit increase in

the severity of an offender’s prior record results in the length of a sentence

increasing by some fixed time period (e.g., 8 months). Key to the interpre-

tation of OLS regression coefficients is the idea that the level of the inde-

pendent variable does not matter – each unit change is expected to result

in the same change in the dependent variable regardless of whether we are

looking at small or large values on the independent variable. What if this is

not always the case?

For example, an assumption often made in research on sentencing out-

comes, but rarely examined, is the idea that first-time offenders (i.e., those

with no prior record) or those offenders convicted of relatively minor

forms of crime will be punished much more leniently than other offenders.

Then, as the severity of prior record or of conviction offense increase,

there is an expectation of an increasingly punitive response by the crimi-

nal justice system. Put another way, there is an expectation of a non-linear

relationship between the severity of the conviction offense or the offender’s

prior record and the severity of punishment – changes in the level of the

dependent variable may vary by the level of the independent variable.

Figure 17.1 presents a hypothetical plot for punishment severity and

prior criminal history that reflects increasingly harsher punishments for

offenders with more extensive criminal records.

As can be seen in the figure, there is a gradual increase in the severi-

ty of the punishment as the severity of the prior record increases. Then,

the increases in the severity of the punishment become larger for the same

unit increase in prior criminal history.

The range of potential non-linear relationships is limitless and is

bounded only by the imagination and creativity of the researcher and the

theoretical basis for conducting the research. Yet, while there may be a

wide range of possible non-linear relationships, most researchers will con-

fine their analyses to a relatively limited group of non-linear possibilities,

C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

1 Michael H. Tonry, Sentencing Matters (New York, Oxford University Press, 1996).

516

N O N - L I N E A R R E L A T I O N S H I P S

some of which are displayed in Figure 17.2. Panel (a) presents what is

referred to as a quadratic equation. All that this means is that a squared

term has been added to the equation to give it a form such as Y = X + X2.

The quadratic equation is one of the more commonly used transforma-

tions in criminology and criminal justice, and has had frequent applica-

tion in the study of age-related behavior.

Panel (b) presents an inverse function of the form Y = 1/X. This kind

of transformation helps to capture relationships where there is a decreasing

negative effect of the independent variable on the dependent variable.

Prior Criminal History

P u

n is

h m

e n

t S

e v e ri

ty

Figure 17.1 Hypothetical Non-linear Relationship Between Punishment Severity and Prior Criminal

History

X

Y

Figure 17.2 Common Non-linear Relationships Used in Criminal Justice Research

(a) Quadratic

517

C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

Panel (c) presents a square root transformation of the form Y = X. This

kind of transformation is useful when there is a diminishing positive

impact of the independent variable on the dependent variable.

Finding a Non-linear Relationship: Graphical Assessment

Perhaps the most straightforward way of exploring data for a non-linear

relationship is to use a line graph (see Chapter 3). A simple scatterplot

(discussed in Chapter 14) will often contain so many data points that it is

difficult, if not impossible, to discern any pattern in the data. A line graph

that plots the mean of the dependent variable against the value of the

X

Y

(b) Inverse

Figure 17.2 Continued

X

Y

(c) Square Root

518

0

20

40

60

80

100

120

140

160

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Offense Severity

L e

n g

th o

f S

e n

te n

c e

Mean Length of Sentence OLS Regression Line

Figure 17.3 Plot for Mean Length of Sentence by Offense Severity for Offenders in Pennsylvania

independent variable will likely provide a rough indication of the nature of

the bivariate relationship between the two variables. For example, Figure 17.3

presents the mean for length of sentence against the severity of the convic-

tion offense for over 20,000 offenders sentenced in Pennsylvania in 1998.2

As you look at Figure 17.3, you can see that there is a gradual, linear

increase in length of sentence as offense severity increases to about level

6 to 7. At that point, the increases in sentence length become larger for each

additional increase in offense severity. To highlight the curvilinear nature of

the relationship between offense severity and length of punishment, the

OLS regression line for these data is overlayed in Figure 17.3, indicating that

the straight-line relationship does not capture the relationship between

length of sentence and severity of offense particularly well.

Incorporating Non-linear Relationships into an OLS Model

Assuming that we have good reason for assuming that a non-linear

relationship exists between the dependent variable and one or more of

the independent variables, how do we incorporate this information into

2 These data are available through the National Archive of Criminal Justice Data and can be accessed at http://www.icpsr.umich.edu/NACJD

519N O N - L I N E A R R E L A T I O N S H I P S

the OLS regression model? The first step, as noted above, is to try and

gain a sense of the relationship graphically. In most circumstances, if the-

ory suggests or if we find evidence of a curvilinear relationship, the most

straightforward approach is to add a quadratic term – the squared value of

the independent variable – such as that in Panel (a) of Figure 17.2. More

formally, a quadratic regression equation would have the following form:

Y = b 0

+ b 1 X

1 + b

2 X

1 2

where Y represents the dependent variable

X 1

represents the independent variable

In our example presented in Figure 17.3, we have evidence of a curvi-

linear relationship that might be accounted for by adding a squared term

for offense severity to a regression equation. We begin by noting that the

OLS regression line portrayed in Figure 17.3 is

Y = −9.85 + 4.62 X 1

where Y represents length of sentence (in months)

X 1

represents offense severity

To incorporate a non-linear relationship, we begin by transforming the

variable – in this case offense severity – and then add this transformed

variable to the regression equation. In most statistical software packages

this would simply involve the creation of a new variable which represents

the original variable squared. When we square offense severity and add

it to the regression equation, we obtain the following equation:

Y = b 0

+ b 1 X

1 + b

2 (X

1 * X

1 ) = b

0 + b

1 X

1 + b

2 X

1 2.

If we then estimate this new regression equation that includes both the

original measure of offense severity and the squared value of offense

severity, we obtain the following results:

Y = 7.07 − 3.99 X 1

+ 0.77 X 1 2

Substantively, this regression equation captures the curvilinear rela-

tionship between offense severity and sentence length much better than

a straight-line relationship, since there are increasingly larger increases in

sentence length for each unit change in offense severity. Figure 17.4

presents the mean sentence length by offense severity (similar to that in

Figure 17.3) along with the new regression line based on including the

quadratic term.

Interpreting Non-linear Coefficients

In many practical applications of adding non-linear terms to OLS regres-

sion models, there is often less emphasis on the interpretation of the indi-

vidual coefficients that represent transformed variables. The reason for

520 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

this is the difficulty in making sense of individual coefficients. For example,

in the example using the data on offenders sentenced in Pennsylvania, the

coefficient for the squared value of offense severity is given as 0.77. Like

any other OLS coefficient, this coefficient can be interpreted in the con-

text of one unit changes of the transformed variable:

For each one unit increase in the squared value of offense severity, the length

of sentence is expected to increase by 0.77 months.

Unfortunately, this kind of interpretation tends not to have much intu-

itive meaning for most people – researchers and others. Consequently, the

description of results using non-linear transformations in OLS regression

models will often focus on the general pattern of results, rather than on

the specific coefficients. This is not entirely satisfactory, however, because

it still leaves the reader wondering whether the transformation added

much to the analysis. Our suggestion is to use graphs, such as that pre-

sented in Figure 17.4, which do provide an effective way to convey evi-

dence of a non-linear relationship between the dependent and inde-

pendent variables. What makes this kind of plot particularly useful is that

it conveys both the pattern in the observed data and the predicted values

based on the estimated regression model.

Note on Statistical Significance

Estimating statistical significance for a non-linear term does not present any

new problem to our understanding of multivariate regression. The statisti-

cal significance of both the individual coefficients and the overall model

in an OLS regression model incorporating non-linear terms is determined

521

0

20

40

60

80

100

120

140

160

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Offense Severity

L e

n g

th o

f S

e n

te n

c e

Mean Length of Sentence Regression Line (Quadratic)

Figure 17.4 Plot for Mean Length of Sentence by Offense Severity for Offenders in Pennsylvania with

Quadratic Regression Line

N O N - L I N E A R R E L A T I O N S H I P S

522 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

in the same way as for any other OLS regression model. For individual

coefficients, we use the t-test and for the overall model, we use the F-test.

Summary

How does one know whether to include a non-linear term in a regression

model? In light of the many different non-linear relationships that are pos-

sible – we could transform any number of our independent variables in

an OLS regression model – how do we settle on an appropriate model?

The single best guide for the researcher is prior theory and research. If a

theoretical perspective claims a non-linear relationship or prior research

has established a non-linear relationship between two variables, then the

researcher may want to examine a non-linear relationship in the regres-

sion analysis. Without the guidance of theory and prior research, the

researcher is better off using an OLS model without any non-linear

relationships included. If subsequent analyses, such as a residual analysis

(discussed in Chapter 16) indicate a non-linear relationship, then some

kind of transformation of an independent variable may be in order.

I n t e r a c t i o n E f f e c t s

A number of different theories of crime and delinquency make statements

about how the effect of one variable will vary by the level of some other

variable. A perspective known as general strain theory hypothesizes that

the effects of psychological strain (e.g., having one’s parents file for

divorce) on delinquency will vary by the ability of a youth to adapt to

strain.3 For example, if an individual characteristic, such as self-esteem,

helps individuals to adapt to various forms of strain, then the effect of that

strain may vary by the level of self-esteem: as the level of self-esteem

increases, the effect of strain on the chances of delinquency may become

smaller. Alternatively, research on criminal justice decision-making has

suggested that the effects of offender characteristics, such as the offend-

er’s age, may differentially affect the severity of punishment across differ-

ent racial or ethnic categories.4

Assuming that we have a rationale for including an interactions effect,

how do we incorporate it into our regression model? Let us begin with a

simple regression model that has two independent variables X 1

and X 2 .

Y = b 0

+ b 1 X

1 + b

2 X

2 + e

3 R. Agnew, 1992, Foundation for a general strain theory of crime and delinquency, Criminology, 30, 47-87. 4 D. Steffensmeier, J. Kramer, and J. Ulmer, 1995, Age differences in sentencing, Justice Quarterly, 12, 583-602.

To add an interaction effect to a regression model, all that we need to

do is to compute the product of the two variables: X 1 *X

2 = X

3 . We then

add this term to the regression equation:

Y = b 0

+ b 1 X

1 + b

2 X

2 + b

3 X

3 + e

Where X 3

= X 1 *X

2

Note that we now have an additional regression coefficient (b 3 ) in the

model that represents the interaction of the variables X 1 and X

2 , which we

will need to interpret. The interpretation of interaction effects can be quite

complicated, with the degree of complexity based on the level of meas-

urement of the two variables.

Interaction of a Dummy Variable and Interval-Level Variable

To illustrate the process of interpreting interaction effects, it is useful to

begin with a relatively simple case: the interaction of a dummy variable with

a variable measured at the interval level of measurement. In the regression

model above, let us assume that X 2

is a dummy variable, where the

two categories are coded as either 0 or 1.

It is now possible to work through a series of regression equations,

much like we did in the previous chapter in our discussion of dummy

variables, by inserting different values for X 2 . Note that the key difference

is we now have more than one place where we need to insert values for

the dummy variable.

If we set the value for X 2 = 1, we have the following regression equation:

Y = b 0

+ b 1 X

1 + b

2 X

2 + b

3 (X

1 *X

2 ) = b

0 + b

1 X

1 + b

2 (1) + b

3 (X

1 *1)

Which reduces to:

Y = b 0

+ b 1 X

1 + b

2 + b

3 X

1

By rearranging our terms, we can rewrite the regression equation as:

Y = (b 0

+ b 2 ) + b

1 X

1 + b

3 X

1 = (b

0 + b

2 ) + (b

1 + b

3 )X

1

As in the previous chapter, we see that when we focus our attention

on the category with the value 1, the model intercept changes by the

value of the coefficient for that variable (i.e., b 2 ). What is different in the

above equation is that the effect of variable X 1 is now the sum of two dif-

ferent regression coefficients: the original coefficient for X 1

(i.e., b 1 ) and

the coefficient for the interaction term (i.e., b 3 ).

How do we now interpret the effect of X 1 ? After summing the two

regression coefficients b 1

and b 3 , we would say that for cases that had a

value of 1 on X 2

(i.e., the cases were in Group 1), for each one unit

increase in X 1 , Y is expected to change by b

1 + b

3 units.

I N T E R A C T I O N E F F E C T S 523

When we set the value for X 2 = 0 (the reference category for our dummy

variable – Group 0), we now have the following regression equation:

Y = b 0

+ b 1 X

1 + b

2 (0) + b

3 (X

1 *0)

Which reduces to:

Y = b 0

+ b 1 X

1

This indicates that the model intercept (b 0 ) and coefficient for X

1 (b

1 )

represent the intercept for the reference category on X 2

and the effect of

X 1

for cases in the reference category, respectively.

To make the example more concrete, suppose that after estimating this

regression equation, we find the following results:

b 0

= 2.5

b 1

= 3.2

b 2

= 1.9

b 3

= 1.3

By inserting the values for the regression coefficients into the regres-

sion equation, we have the following:

Y = 2.5 + 3.2 X 1

+ 1.9 X 2

+ 1.3 (X 1 *X

2 )

For X 2

= 1, we have the following:

Y = 2.5 + 3.2 X 1

+ 1.9 (1) + 1.3 (X 1 *1)

= (2.5 + 1.9) + (3.2 + 1.3) X 1

= 4.4 + 4.5 X 1

And for X 2

= 0, we have:

Y = 2.5 + 3.2 X 1

+ 1.9 (0) + 1.3 (X 1 *0) = 2.5 + 3.2 X

1

The interpretation of the effect of X 1 is straightforward, but we need to

make sure that we are clear about the group for which we are interpret-

ing the effect of X 1 . Thus, for X

2 = 1, for each one unit increase in X

1 , we

expect Y to increase by 4.5 units. When X 2 = 0, for each one unit increase

in X 1 , Y is expected to increase by 3.2 units. Substantively, this type of

result would allow a researcher to say that the effect of X 1

varied across

the groups measured in X 2 . As a visual aid to understanding these results,

X 2

= 0 is represented by the dashed line and group X 2

= 1 is represented

by the solid line.

Up to this point, we have assumed that all of the coefficients are positive.

Figure 17.6 presents several additional possibilities for various combinations

524 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

we have presented the two regression lines in Figure 17.5, where group

A N E X A M P L E : R A C E A N D P U N I S H M E N T S E V E R I T Y 525

X1

Y

X2 = 0

X2 = 1

Figure 17.5 Regression Lines for the Interaction of X 1

and X 2

of positive and negative values for b 1

and b 3

(we assumed that b 0

and b 2

were positive in each plot). (Keep in mind that b 1 represents the effect of X

1

for the reference category of the dummy variable and b 3 represents the value

of the interaction effect.) Panel (a) is comparable to the preceding example,

where b 1 and b

3 are both positive. Panel (b) illustrates a hypothetical exam-

ple when b 1

is positive and b 3

is negative. Panels (c) and (d) illustrate pos-

sible patterns when b 1

is negative and b 3

is positive (panel (c)) or negative

(panel (d)). Clearly, there are many other possibilities, but we wanted to pro-

vide a few illustrations for different patterns that researchers have had to

address in their analyses.

A n E x a m p l e : R a c e a n d P u n i s h m e n t S e v e r i t y

Suppose that we are interested in testing whether the severity of a crimi-

nal offense differentially affects the length of time offenders are sentenced

to prison by race. Put another way, does the severity of the conviction

offense affect the severity of punishment in the same way for offenders

of different races? We again use data on the sentences of over 20,000

offenders sentenced to prison in Pennsylvania in 1998 to illustrate the

test for an interaction effect between severity of offense and race of

offender. To simplify our model here, we measure race as a dummy vari-

able (0 = white, 1 = African American). Offense severity is scored by the

Pennsylvania Sentencing Commission and has values ranging from 1 to 14

526 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

X1

Y

X2=0

X2=1

Figure 17.6 Hypothetical Interaction Effects for Different Combinations of Positive and Negative

Interaction Effects

(a) b 1

and b 3

are both positive

(b) b 1

is positive and b 3

is negative

X1

Y

X2=0

X2=1

527

Figure 17.6 Continued

1

Y

X2=0

X2=1

X1

Y

X2=0

X2=1

(c) b 1

is negative and b 3

is positive

(d) b 1

and b 3

are both negative

X

A N E X A M P L E : R A C E A N D P U N I S H M E N T S E V E R I T Y

and sentence length is measured in months sentenced to prison. The

regression model we set out to test can be written as:

Y = b 0

+ b 1

X 1

+ b 2

X 2

+ b 3

X 1 *X

2

where Y represents length of sentence (in months)

X 1

represents offense severity

X 2

represents race

When we estimate this regression model, we produce the following set

of results:

Y = −8.14 + 4.12 X 1

− 5.41 X 2

+ 1.31 X 1 *X

2

Using the same approach as above, we begin by focusing on African

Americans (X 2

= 1):

Y = −8.14 + 4.12 X 1

− 5.41 (1) + 1.31 (X 1 *1)

= (−8.14 − 5.41) + (4.12 + 1.31) X 1

= −13.55 + 5.43 X 1

For whites, the equation is:

Y = −8.14 + 4.12 X 1

− 5.41 (0) + 1.31 (X 1 *0)

= −8.14 + 4.12 X 1

Substantively, we can now directly interpret the effect of offense sever-

ity for white and African American offenders separately. Among white

offenders, each one unit increase in offense severity, is expected to

increase sentence length by 4.12 months, while for African American

offenders, each one unit increase in offense severity is expected to

increase sentence length by 5.43 months. More succinctly, these results

suggest that the effect of offense severity on punishment severity is greater

for African American offenders than for white offenders. These results are

presented graphically in Figure 17.7. The dashed line reflects the slope for

offense severity on sentence length for white offenders, while the solid line

reflects the effect for African American offenders. As one moves further out

the x-axis to greater values for offense severity, there is an increasingly

greater effect for African American offenders compared to white offenders.

Interaction Effects between Two Interval-level Variables

Up to this point, our attention has been focused on interaction effects

involving one variable measured at the interval level of measurement and

one dummy variable measured at the nominal level of measurement.

The inclusion of an interaction effect between two interval-level variables

in a regression model is done in exactly the same way – we compute a

product of the two variables and add the product to the regression equa-

tion. The interpretation of the interaction effect is much more complex,

528 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

however, since we are no longer able to simplify the regression equation

to represent the effect of one variable for two different groups.

In some cases we may not be concerned with a specific interpretation

of the interaction term. For example, we may want to simply identify

whether the interaction between two measures is statistically significant.

A statistically significant interaction term between two measures would

suggest that their effect cannot be measured only by the additive effects

of each measure in the model, but rather there is an additional effect that

is measured by the interaction term.

It may help to conceptualize this issue if we turn to a substantive exam-

ple. Many sociologists have suggested that extra-legal variables such as

income and social status impact upon sentencing outcomes.5 In a simple

additive model each of these factors would have some defined independ-

ent effect on the severity of a sentence measured in months of imprison-

ment. This model is illustrated in equation form below.

Y = b 0

+ b 1

X 1

+ b 2

X 2

where Y represents length of sentence (in months)

X 1

represents income

X 2

represents social status

529

−20

−10

0

10

20

30

40

50

60

70

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Offense Severity

L e n

g th

o f

S e n

te n

c e (

M o

n th

s )

White African American

Figure 17.7 Regression Lines for the Effect of Offense Severity on Sentence Length by Race of

Offender

5 Donald J Black, The Behavior of Law (New York; Academic Press, 1976).

A N E X A M P L E : R A C E A N D P U N I S H M E N T S E V E R I T Y

530 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

But what if the researcher believed that the effect of income and social sta-

tus was not simply additive but also multiplicative, meaning that there was

an added effect that was due to the interaction between the two. This the-

ory is represented in the equation below:

Y = b 0

+ b 1

X 1

+ b 2

X 2

+ b 3

X 1 * X

2

where Y represents length of sentence (in months)

X 1

represents income

X 2

represents social status

In this case the researcher is hypothesizing that there is not only the inde-

pendent effect of income and of social class, but that there is an additional

interaction effect that is measured by multiplying income by social class.

What if this effect is statistically significant? What interpretation can the

researcher draw? To illustrate this we take a hypothetical example of regres-

sion results as reported below:

Y = 7.2 − 2.4 X 1

− 1.6 X 2

− 1.1 X 1 * X

2

The additional interaction term in this case suggests that there is an addi-

tional benefit beyond that of the additive independent effects of income

and social status that must be taken into account. In a very simple inter-

pretation, we can say that not only does a high income high status indi-

vidual receive a benefit from their income and status, but when the

individual is high on both measures simultaneously they gain an added

benefit above and beyond that provided by each characteristic on its own.

While models specified in this way can help us to identify additional

impacts that come from the interactions between interval level variables,

it is very hard to develop interpretations beyond what we have noted

above. But the researcher can adjust such models to develop a more easily

interpretable understanding of interaction terms.

Conceptually, when we have an interaction between two interval-

level variables, we are testing the idea that the effect of one interval-level

variable varies by the level of the second interval-level variable. For

example, in the example noted above from general strain theory, the

hypothesis is that the effect of strain varies by the level of self-esteem. In

practice, the difficulty we often have in the interpretation of interaction

effects between two interval-level variables is in choosing values for one

variable to represent the effect of the other variable. In effect, we have

already done this in our example of an interaction between the dummy

variable and the interval-level variable. Recall that when we include an

interaction effect between a dummy variable and an interval-level variable,

we set the value of the dummy variable to either 0 or 1 and then inter-

pret the effect of the interval-level variable for each group represented in

the dummy variable.

In trying to determine how to interpret the interaction between two

interval-level variables, we would encourage you to first consider which

variable is of key importance for a study. The second variable would then

be set at a limited number of values, which allows the researcher to see

how the effect of the key variable changes across levels of the second

variable. For example, if we again refer to the interaction between strain

and self-esteem, the key theoretical variable is strain. Following these

guidelines, we would then want to interpret the effect of strain for several

designated values of self-esteem. Clearly, we could interpret the interaction

effect the other way: the effect of self-esteem for specified levels of strain,

but this is not a key piece of the theory.

What values do we use for the second interval-level variable? For any

given interval-level variable, there may be hundreds or thousands of real-

istic possible values that we could use. We think that a useful place to start

is to use the mean, one standard deviation above and below the mean,

and two standard deviations above and below the mean. This will cover

a wide range of possible values of the variable we are using and should

be ample for understanding how our key variable changes across values

of the second variable. In other cases, where there may be meaningful

values on the second independent variable that have more intuitive mean-

ing to the reader, these values should be used. For example, if we were

to fix years of education, we might use 8, 12, and 16 to reflect the com-

pletion of junior high school, high school, and undergraduate collegiate

education, respectively.

For example, suppose that we have estimated a regression model with

two interval-level variables X 1

and X 2

and the interaction of X 1

and X 2 :

Y = 2.3 + 1.7 X 1

+ 2.0 X 2

+ 0.5 (X 1 * X

2 )

For the purpose of this example, we will consider X 1

the key variable.

We find the mean and standard deviation of X 2

to be 3.2 and 1.2,

respectively.

The values that are one or two standard deviations above and below

the mean of X 2

are:

531

Two standard deviations above: 3.2 + 2 * 1.2 = 3.2 + 2.4 = 5.6

One standard deviation above: 3.2 + 1.2 = 4.4

One standard deviation below: 3.2 − 1.2 = 2.0

Two standard deviations below: 3.2 − 2 * 1.2 = 3.2 − 2.4 = 0.8

We can now input these values for X 2

to determine the effect of X 1

on Y:

Effect of X 1

at the mean of X 2 :

Y = 2.3 + 1.7 X 1

+ 2.0 * (3.2) + 0.5 (X 1 * 3.2)

= 2.3 + 1.7 X 1

+ 6.4 + 1.6 X 1

= (2.3 + 6.2) + (1.7 + 1.6) X 1

= 8.5 + 3.3 X 1

A N E X A M P L E : R A C E A N D P U N I S H M E N T S E V E R I T Y

If we wanted to interpret the effect of X 1

directly, then we would state

that at the mean for X 2 , each one-unit increase in X

1 is expected to

increase Y by 3.3 units.

Effect of X 1

at one standard deviation above the mean of X 2 :

Y = 2.3 + 1.7 X 1

+ 2.0 * (4.4) + 0.5 (X 1 * 4.4)

= 2.3 + 1.7 X 1

+ 8.8 + 2.2 X 1

= (2.3 + 8.8) + (1.7 + 2.2) X 1

= 11.1 + 3.9 X 1

If we wanted to interpret the effect of X 1

directly, then we would state

that at one standard deviation above the mean for X 2 , each one-unit

increase in X 1

is expected to increase Y by 3.9 units.

Effect of X 1

at one standard deviation below the mean of X 2 :

Y = 2.3 + 1.7 X 1

+ 2.0 * (2.0) + 0.5 (X 1 * 2.0)

= 2.3 + 1.7 X 1

+ 4.0 + 1.0 X 1

= (2.3 + 4.0) + (1.7 + 1.0) X 1

= 6.3 + 2.7 X 1

If we wanted to interpret the effect of X 1

directly, then we would state

that at one standard deviation below the mean for X 2 , each one-unit

increase in X 1

is expected to increase Y by 2.7 units.

Effect of X 1

at two standard deviations above the mean of X 2 :

Y = 2.3 + 1.7 X 1

+ 2.0 * (5.6) + 0.5 (X 1 * 5.6)

= 2.3 + 1.7 X 1

+ 11.2 + 2.8 X 1

= (2.3 + 11.2) + (1.7 + 2.8) X 1

= 13.5 + 4.5 X 1

If we wanted to interpret the effect of X 1

directly, then we would state

that at two standard deviations above the mean for X 2 , each one-unit

increase in X 1

is expected to increase Y by 4.5 units.

Effect of X 1

at two standard deviations below the mean of X 2 :

Y = 2.3 + 1.7 X 1

+ 2.0 * (0.8) + 0.5 (X 1 * 0.8)

= 2.3 + 1.7 X 1

+ 1.6 + 0.4 X 1

= (2.3 + 1.6) + (1.7 + 0.4) X 1

= 3.9 + 2.1 X 1

If we wanted to interpret the effect of X 1

directly, then we would state

that at two standard deviations below the mean for X 2 , each one-unit

increase in X 1

is expected to increase Y by 2.1 units.

Aside from making direct interpretations of the effect of X 1 at these five

values of X 2 , what we see is that as the value of X

2 increases, the effect

of X 1

on Y increases.

532 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

A n E x a m p l e : P u n i s h m e n t S e v e r i t y

We again use the data on the sentencing of offenders in Pennsylvania in

1998 and modify our regression model slightly. We continue to use length

of sentence as the dependent variable and severity of the offense as an

independent variable. Our second independent variable is a prior crimi-

nal history score that is computed by the Pennsylvania Sentencing

Commission and can take on values ranging from 0 to 8; larger values for

prior criminal history reflect both a greater number of prior offenses as

well as more serious prior offenses. For the purposes of this example, we

have added an interaction effect between severity of the offense and prior

criminal history and are interested in how the effect of offense severity

varies across levels of prior criminal history.

After estimating this model, we obtain the following regression equation:

Y = −7.83 + 3.58 X 1

− 1.21 X 2

+ 0.62 X 1 * X

2

Where Y represents length of prison sentence (in months)

X 1

represents offense severity

X 2

represents prior criminal history

Since we are primarily interested in the effect of offense severity across

levels of criminal history, we have computed the mean and standard devi-

ation of the criminal history variable to be 1.51 and 1.97, respectively.

Following the same procedure as in our hypothetical example, we calcu-

late the effect of offense severity at the mean of prior criminal history.

Effect of offense severity at the mean of prior criminal history:

Y = −7.83 + 3.58 X 1

– 1.21 (1.51) + 0.62 (X 1 * 1.51)

= −7.83 + 3.58 X 1

– 1.83 + 0.94 X 1

= (−7.83 – 1.83) + (3.58 + 0.94) X 1

= −9.66 + 4.52 X 1

If we wanted to interpret the effect of offense severity directly, then we

would state that at the mean for prior criminal history, each one-unit

increase in offense severity is expected to increase sentence length by

4.52 months.

Since the standard deviation of prior criminal history is greater than the

mean, we cannot sensibly compute values at 1 or two standard deviations

below the mean, since the values would be negative, and prior record

score is constrained to have a value ranging from 0 to 8. In such a case,

we can choose other values for the independent variable that carry sub-

stantive meaning. For example, a prior criminal history score of 0 implies

little or no prior criminal activity, while a prior criminal history score of 8

implies an extensive history of serious and likely violent offending. By

using the minimum and maximum values for prior criminal history, along

A N E X A M P L E : P U N I S H M E N T S E V E R I T Y 533

with the mean, we can gain a good sense of how the effect of offense

severity varies by level of prior criminal history.

Effect of offense severity at a prior criminal history score of 0:

Y = –7.83 + 3.58 X 1

– 1.21 (0) + 0.62 (X 1 * 0)

= –7.83 + 3.58 X 1

A direct interpretation of the effect of offense severity at a prior criminal

history score of 0 indicates that each one-unit increase in offense severity

is expected to increase sentence length by 3.58 months.

Similarly, if we take the maximum value for prior criminal history score

of 8:

Y = –7.83 + 3.58 X 1

– 1.21 (8) + 0.62 (X 1 * 8)

= –7.83 + 3.58 X 1

– 9.68 + 4.96 X 1

= (–7.83 – 9.68) + (3.58 + 4.96) X 1

= –17.51 + 8.54 X 1

Thus, among the historically most serious offenders, these results suggest

that each one-unit increase in offense severity is expected to increase sen-

tence length by 8.54 months.

Much like the hypothetical example earlier, we see that as the value

for prior criminal history increases, the effect of offense severity increas-

es. To appreciate the change in the effect of offense severity, we have

plotted these three regression equations in Figure 17.8. We see that at a

prior criminal history score of 0, the regression line is positive, but shows

modest increases over the range of offense severity. As we move to the

mean of prior criminal history and then to the maximum value for prior

criminal history, we see the slope of the regression line become steeper,

reflecting increasingly greater effects on sentence length for any given

increase in prior criminal history.

T h e P r o b l e m o f M u l t i c o l l i n e a r i t y

The use of interaction terms is very likely to create a problem in regres-

sion analyses that can lead to difficulty in estimation of the regression

equation. This is because we are likely to be interested not only in the

interaction between two variables, but also the simple additive effects of

each independently. When we include the original variables as well as the

term for their interaction, the three terms are likely to be very highly cor-

related. Even though multivariate regression was developed in part to take

into account the interrelationships among variables that predict Y, when

independent variables in a regression model are too strongly related to

one another, regression estimates become unstable. This problem is called

Multicollinearity.

534 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

In criminal justice, the independent variables examined are generally

multicollinear, or correlated with one another. Indeed, this correlation is

one reason it is so important to use multivariate techniques in criminal jus-

tice research. When variables are intercorrelated, as in the case of our

example of years in prison and prior arrests discussed in Chapter 16, it is

important to control for the potential confounding influences of one vari-

able on the other. Failure to do so is likely to lead to bias in our estimates

of the effects of specific regression coefficients. However, the irony of mul-

ticollinearity is that when variables become too correlated, or highly mul-

ticollinear, the regression estimates become unreliable. This may happen

when models do not include interaction terms, but is a particularly serious

concern when interactions between variables are specified in a model.

Multicollinearity can be identified in one of two ways. A common

method is to look at the intercorrelations among the independent variables

included in your model. Very high correlations between independent

variables are likely to lead to multicollinearity problems. What is considered

a very high correlation? As with many other definitions in statistics, there

is no absolute number at which multicollinearity is considered serious. As

a general rule, a correlation between two independent variables of greater

than 0.80 should be seen as a warning that serious multicollinearity may

be evident in your model.

Multicollinearity between two variables occurs less often than multi-

collinearity across a series of variables. To diagnose this type of multicolli-

nearity, we use a statistic that is usually defined as tolerance. Tolerance

T H E P R O B L E M O F M U L T I C O L L I N E A R I T Y 535

−20

0

20

40

60

80

100

120

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Offense Severity

L e n

g th

o f

S e n

te n

c e (

M o

n th

s )

Prior Criminal History = 0 Mean of Prior Criminal History

Prior Criminal History = 8

Figure 17.8 Regression Lines for the Effect of Offense Severity on Sentence Length by Level of Prior

Criminal History Score

measures the extent of the intercorrelations of each independent variable

with all other independent variables. It is defined as 1 minus the percent

of variance in X explained by the other independent variable examined

(Equation 17.1).

Tolerance = 1 − R X

2

Calculation of tolerance is generally provided as an option in standard sta-

tistical computing packages, but it also can be calculated by taking each

independent variable as the dependent variable in a regression that

includes all other independent variables. This value is then subtracted

from 1. For example, let’s say we defined a model for explaining sentence

length among offenders in Pennsylvania that included three independent

variables:

Y length

= b 0

+ b 1

(age) + b 2

(offense severity)

+ b 3

(prior criminal history) + e

The R2 X

for age would be estimated by calculating a regression in which

age was the dependent variable and offense severity and prior criminal

history were the independent variables. You would then take this R2 and

subtract it from 1. Similarly, to get R2 X

for offense severity, you would

regress age and prior criminal history on offense severity and then sub-

tract the resulting R2 from 1. Table 17.1 presents the Tolerance statistics

for this regression model.

How do we know if multicollinearity is negatively effecting our model

estimates based on the tolerance statistics? A very small tolerance statistic

suggests that the model is likely to include a high level of multicollinear-

ity. Again, there is no clear yardstick for defining a level of tolerance that

is likely to lead to estimation problems. In general, however, a tolerance

level of less than 0.20 should be taken as a warning that serious multi-

collinearity may exist in your model. We see from the results in Table 17.1

that the smallest tolerance statistic has a value of 0.94, which does not

indicate any serious multicollinearity in this regression model.

Beyond these diagnostic procedures for multicollinearity, there are

warning signs that can be observed in the regressions that are estimated.

Sometimes when multicollinearity is present, the percent of explained

536 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

Table 17.1 Tolerance Statistics for Regression of Sentence Length for Pennsylvania

Offenders

INDEPENDENT VARIABLE TOLERANCE

Age 0.940 Offense Severity 0.931 Prior Criminal History 0.975

Equation 17.1

variance in a model is high, but the regression coefficients overall fail to

reach conventional thresholds of statistical significance. Sometimes multi-

collinearity inflates coefficients to unrealistic sizes or produces coefficients

in a direction contrary to conventional wisdom. One problem in diagnos-

ing multicollinearity is that it may have such varied effects in your model

that you may have difficulty distinguishing a misleading result that is due

to multicollinearity from one that represents a new and interesting finding.

When there are indications of serious multicollinearity, you can take a

number of alternative corrective measures. The simplest is to exclude the

variable or variables that are contributing most to multicollinearity. In the

case of interaction terms this might require that one of the terms included

in the interaction be dropped from the model. In the case where a small

group of measures are highly collinear, you might choose to exclude the

one variable that appears to present the most serious problem (i.e., that has

the lowest tolerance value). The drawback of this approach is that the

exclusion of such measures is likely to lead to model misspecification and

may result in biased estimates of other regression coefficients that remain

in the model. This approach makes sense only when other variables that

remain in the model measure the same concept or theory. An approach that

achieves a similar result, without excluding specific measures, is to create

new indices from clusters of variables that are multicollinear. For example,

if a group of measures all relating to social status are multicollinear, you

may decide to create a new composite measure defined as social status and

use it as an independent variable in subsequent regressions.

C h a p t e r S u m m a r y

Non-linear relationships refer to the effect of the independent variable

on the dependent variable not being a straight-line (linear) relationship. A

linear relationship implies that each one unit increase in the independent

variable will result in the dependent variable increasing or decreasing by

some fixed amount, regardless of the level of the independent variable.

A non-linear relationship implies that each one unit increase in the inde-

pendent variable does not result in the same amount of change in the

dependent variable – it may be larger or smaller and will vary by the level

of the independent variable. A non-linear relationship can be incorporated

into an OLS regression equation by transforming the independent variable.

Interaction effects reflect the varying effect of one independent vari-

able on the dependent variable across the levels or values of a second

independent variable. When we have an interaction effect between a

dummy variable and an interval-level variable, we can directly interpret the

effect of the interval-level variable on the dependent variable for each

group measured by the dummy variable. Interpretation of an interaction of

two interval level independent variables is much more difficult. One way

C H A P T E R S U M M A R Y 537

538 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

of simplifying interpretation is to designate values for one variable, such as

the mean, one standard deviation above/below the mean, two standard

deviations above/below the mean, and so on, as fixed points to compute

the effect of the other interval-level variable on the dependent variable.

Multicollinearity occurs when independent variables in a regression

model are too strongly related. It leads to unstable results. The problem

may be diagnosed by checking the bivariate correlations between the

variables and by measuring tolerance. Multicollinearity may be dealt with

either by excluding specific variables altogether or by merging several

similar variables into one composite index.

K e y T e r m s

interaction effect An interaction effect is present when the effect of one independent variable on the dependent variable is conditional on the level of a second independent variable.

multicollinearity Condition in a multi- variate regression model in which independent variables examined are very strongly intercorrelated. Multicollinearity leads to unstable regression coefficients.

non-linear relationship Relationship between the dependent and the independent variable that is not captured by a straight line (linear) relationship.

tolerance A measure of the extent of the intercorrelations of each independent vari- able with all other independent variables. Tolerance may be used to test for multicollinearity in a multivariate regression model.

S y m b o l s a n d F o r m u l a s

R 2 X

R2 obtained when an independent variable is treated as a dependent variable in a test for tolerance

To calculate tolerance:

Tolerance = 1 – R 2 X

E x e r c i s e s

17.1 An analysis of shoplifting frequency among youth and young adults included a quadratic term for age of the individual and produced the following results:

INDEPENDENT VARIABLE B

Age (in years) 0.35 Age2 (in years2) −0.01

Interpret the effect of age on the frequency of shoplifting.

17.2 An analysis linking level of informal social control to frequency of delinquency produced the following results:

INDEPENDENT VARIABLE B

Age (in years) −0.12 Sex (1=Female, 0=Male) −1.50 Race (1=White, 0=Non-white) 0.27 Informal Social Control (1=Low, 10=High) −0.83

After plotting the mean level of delinquency by level of informal social con-

trol, the researcher observed what appeared to be an inverse relationship

(1/X) between delinquency and informal social control. After transforming

the measure of informal social control, the researcher estimated a new

regression and produced the following results:

INDEPENDENT VARIABLE B

Age (in years) −0.11 Sex (1=Female, 0=Male) −1.61 Race (1=White, 0=Non-white) 0.32 Inverse of Informal Social Control (1=Low, 10=High) 2.45

a. Interpret the effect of the inverse of informal social control.

b. Sketch the relationship between delinquency and informal social control using the coefficient for the inverse of informal social control.

17.3 A researcher wanted to test the hypothesis that adolescent females were more affected by parental supervision than adolescent males. In a regression analysis incorporating an interaction effect between sex and supervision, the researcher produced the following set of results:

INDEPENDENT VARIABLE B

Sex (1=Female, 0=Male) −2.7 Supervision (1=Low, 10=High) −1.3 Sex * Supervision −0.5

Interpret the effect of supervision for adolescent females and males.

17.4 A study of attitudes about punishment used a scale of punitiveness ranging in value from 1 (Low) to 100 (High). The researcher was particularly interested in whether there was an interaction effect between age and political conservatism. A regression analysis pro- duced the following results:

INDEPENDENT VARIABLE B MEAN

Age (years) 1.67 44.95 Political Conservatism (1=Low, 10=High) 0.92 6.5 Age * Political Conservatism 0.56

a. What is the effect of political conservatism at the mean age of the sample? Interpret this effect.

b. What is the effect of age at the mean level of political conser- vatism for the sample? Interpret this effect.

E X E R C I S E S 539

540 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

c. What is the effect of political conservatism at each of the following ages?

– 20 – 30 – 50 – 60

Describe how the effect of political conservatism changes as age

increases.

d. What is the effect of age at each of the following values of political conservatism?

– 0 – 2 – 5 – 8 – 10

Describe how the effect of age changes as the level of political con-

servatism increases.

17.5 A study of violence in prison cell blocks was concerned about the amount of space available to each inmate and the proportion of inmates identified as gang members who had been identified as gang members. The researcher tested the hypothesis of an interac- tion effect between space available and the proportion of inmates identified as gang members. A regression analysis produced the following results:

INDEPENDENT VARIABLE B MEAN

Space available (square feet per inmate) −0.25 10.0 Proportion gang members 0.77 0.77 Space available * Proportion gang members −0.05

a. What is the effect of space available at the mean proportion of gang members for the sample of cell blocks? Interpret this effect.

b. What is the effect of proportion of gang members at the mean level of space available for the sample of cell blocks? Interpret this effect.

c. What is the effect of space available at each of the following proportions of gang membership?

– 0.2 – 0.4 – 0.6 – 0.8

Describe how the effect of space available changes as proportion of

gang membership increases.

d. What is the effect of proportion of gang membership at each of the following values of space available?

– 3 – 6 – 12 – 15

Describe how the effect of proportion of gang membership changes as

the level of space available increases.

17.6 Rachel collects police data on a series of burglaries and wishes to determine the factors that influence the amount of property stolen in each case. She creates a multivariate regression model and runs a test of tolerance for each of the independent variables. Her results are as follows, where Y = Amount of property stolen ($):

INDEPENDENT VARIABLE SCALE TOLERANCE

X 1 : Time of robbery (AM or PM) Nominal 0.98

X 2 : Accessibility of property Ordinal 0.94

X 3 : Number of rooms in house Interval 0.12

X 4 : Size of house Interval 0.12

X 5 : Joint income of family Interval 0.46

Would you advise Rachel to make any changes to her model? Explain

your answer.

17.7 A researcher examining neighborhood crime rates computes a regression model using the following variables:

Y = crime rate (per 100,000)

X 1

= percent living in poverty

X 2

= percent unemployed

X 3

= median income

X 4

= percent of homes being rented

The researcher finds the F-statistic for the overall model to be statistically

significant (with α = 0.05), but the results for each variable are as follows:

INDEPENDENT VARIABLE B SIG. TOLERANCE

X 1 : Percent living in poverty 52.13 0.17 0.15

X 2 : Percent unemployed 39.95 0.23 0.07

X 3 : Median income 22.64 0.12 0.19

X 4 : Percent of homes being rented 27.89 0.33 0.05

a. Explain why the researcher found a statistically significant regression model, but no significant regression coefficients

b. What would you recommend the researcher do in this case?

C O M P U T E R E X E R C I S E S 541

C o m p u t e r E x e r c i s e s

In Chapter 16, we illustrated the use of the regression command in SPSS and

Stata to estimate multivariate regression models. The analyses described in this

chapter—nonlinear terms, interaction effects, and a test for multicollinearity—are

accomplished with the same regression command in each program. The

542 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

following exercises should help to illustrate how to perform these analyses, as

will the sample syntax files for SPSS (Chapter_17.sps) and Stata (Chapter_17.do).

SPSS

Computing Nonlinear and Interaction Terms

To include a nonlinear or an interaction term in a multivariate regression model,

it is necessary to first compute the nonlinear or the interaction term. This com-

putation is done with the COMPUTE command that we briefly discussed in the

Computer Exercises in Chapter 4. The general format for the COMPUTE com-

mand is

In SPSS, the double asterisk (**) indicates that we want to take a variable to a

power. In this example, we want to square AGE, so we add the value 2. If, for

some reason, we had wanted to cube AGE, then we would have typed AGE**3.

Also, recall that the addition of the EXECUTE command forces SPSS to per-

form this calculation immediately.

An alternative that accomplishes the same thing is

where we simply multiply the variable AGE by itself.

The calculation can be a function of one or more variables, which we illustrate

below.

Nonlinear Terms

Suppose we wanted to compute a squared term for a variable AGE. We might

name this new variable AGE_SQ. The COMPUTE command would look like

In either case, once the COMPUTE command has been executed, the new

variable will appear in the data file.

Interaction Terms

The computation of an interaction term is as direct as the equations given in this

chapter. We again use the COMPUTE command, but our calculation involves

multiplying the two variables of interest. We have found that it is often helpful to

make the name of the new variable representing an interaction term a combination

of fragments from both of the original variables being used in the calculation.

For example, suppose that we want to create an interaction term for two

variables that we have named EDUCATION and INCOME. We might call the

interaction variable EDUC_INC:

COMPUTE new_var_name = calculation

COMPUTE AGE_SQ = AGE**2.

EXECUTE.

COMPUTE AGE_SQ = AGE * AGE.

EXECUTE.

COMPUTE EDUC_INC = EDUCATION * INCOME.

EXECUTE.

543

Estimating the Regression Model

model—identical to how we presented these terms in the discussion in this

chapter. For situations where we are using nonlinear terms, we may need to drop

the original variable. Prior research and theory indicating that a nonlinear term

was appropriate will often be the best guide on what the regression equation

should look like. In the case of an interaction term, keep in mind that we must

include both of the original variables and the interaction term; otherwise, it will be

nearly impossible to interpret the coefficients that we do obtain from a regression

C O M P U T E R E X E R C I S E S

After computing the nonlinear or the interaction term, we then simply treat the

created variable as an additional variable added to our multivariate regression

analysis.

Collinearity Diagnostics

SPSS’s regression command will produce a wide assortment of collinearity

statistics, including the tolerance statistic discussed above. To obtain the

collinearity diagnostics, we include the /STATISTICS option in our

REGRESSION command:

where TOL requests the tolerance statistics for each independent variable and

COLLIN requests all other collinearity measures. The tolerance statistics are

presented in the coefficients table, where you will also find the regression

coefficients. Recall from the discussion in the chapter that a tolerance of less

than about 0.20 is indicative of collinearity problems in a regression model.

Stata

Computing Nonlinear and Interaction Terms

To include a nonlinear or an interaction term in a multivariate regression model,

it is necessary to first compute the nonlinear or the interaction term. This com-

putation is done with the gen command (short for generate) that we briefly

discussed in the Computer Exercises in Chapter 4. The general format for the

gen command is

The calculation can be a function of one or more variables, which we illustrate

below. (Note that the structure of the discussion is nearly identical to that in the

SPSS section, with slight changes for the Stata syntax.)

Nonlinear Terms

Suppose we wanted to compute a squared term for a variable AGE. We might

name this new variable AGE_SQ. The gen command would look like

REGRESSION

/STATISTICS COEFF R ANOVA COLLIN TOL

/DEPENDENT dep_var_name

/METHOD = ENTER list_of_indep_vars.

gen new_var_name = calculation

gen AGE_SQ = AGE^2

544 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

In Stata, the upward pointing arrow (^) indicates that we want to take a vari-

able to a power. In this example, we want to square AGE, so we add the value

2. If, for some reason, we had wanted to cube AGE, then we would have typed

AGE^3. An alternative that accomplishes the same thing is

where we simply multiply the variable AGE by itself.

In either case, once the gen command has been executed, the new variable

will appear in the data file.

Interaction Terms

The computation of an interaction term is as direct as the equations given in this

chapter. We again use the gen command, but our calculation involves multiply-

ing the two variables of interest. We have found that it is often helpful to make

the name of the new variable representing an interaction term a combination of

fragments from both of the original variables being used in the calculation.

For example, suppose that we want to create an interaction term for two

variables that we have named EDUCATION and INCOME. We might call the

interaction variable EDUC_INC:

Estimating the Regression Model

After computing the nonlinear or the interaction term, we then simply treat the

created variable as an additional variable added to our multivariate regression

model—identical to how we presented these terms in the discussion in this

chapter. For situations where we are using nonlinear terms, we may need to drop

the original variable. Prior research and theory indicating that a nonlinear term

was appropriate will often be the best guide on what the regression equation

should look like. In the case of an interaction term, keep in mind that we must

include both of the original variables and the interaction term; otherwise,

it will be nearly impossible to interpret the coefficients that we do obtain

from a regression analysis.

Collinearity Diagnostics

Within Stata, only the VIF is directly available, but there is a user-written pack-

age—collin—that will compute the full set of collinearity statistics, including the

tolerance. To install collin on your system, run the following command (one time

only—see Chapter_17.do):

This will make the collin command available for use.

In order to use the collin command, you must first run the regress

command. Immediately following the regress command, run the collin using

the same set of independent variables:

gen AGE_SQ = AGE * AGE

gen EDUC_INC = EDUCATION * INCOME

net install collin, from(http://www.ats.ucla.edu/stat/stata/

ado/analysis)

545

The output will be similar to that in SPSS and will produce a wide assortment

of collinearity statistics, including the tolerance statistic discussed above. Each

measure of collinearity is presented in a table that lists each independent variable

(by row) and collinearity statistic across columns.

Problems

Open the NYS data file (nys_1.sav, nys_1_student.sav, or nys_1.dta) to complete

Exercises 1 through 4.

C O M P U T E R E X E R C I S E S

regress dep_var list_of_independent_variables

collin list_of_independent_variables

1. Using one of the line graph commands (described in Chapter 3), generate

a plot for the mean of a delinquency measure by age.

In SPSS, this should look something like

2. Compute a squared term for the age variable as described above. Estimate

a multivariate regression model using one of the measures of delinquency

as the dependent variable. Use age and age squared as the independent

variables. As noted earlier in the chapter, this will provide a quadratic

equation that will test for a nonlinear relationship between age and the

measure of delinquency that you have selected.

Report the values of the regression coefficients for age and age

squared and whether or not the coefficients are statistically significant

(with α = 0.05). Interpret the effect of age on this measure of delinquency.

GRAPH /LINE(SIMPLE) = MEAN(delinquency_variable)

BY age .

In Stata, the corresponding syntax is a bit more complicated, but use the

following lines to accomplish a similar graph as in SPSS:

egen mean_delinq1 = mean(delinquency_variable), by(age)

sort age

twoway (line mn_delinq1 age, connect(ascending))

The resulting line graphs from the syntax given above will plot the mean

level of delinquency—for the variable you picked—for each age recorded

in the NYS sample. Try this command with other measures of delinquency

and see if you notice any variations in the shapes of the lines. (NOTE: If

you are using Stata, you will need to change the output variable name in the

egen command line, perhaps most simply by changing the number at the

end.)

Do they imply a linear relationship between age and delinquency? A

nonlinear relationship? If nonlinear, what kind of nonlinear relationship?

(You may want to refer back to Figure 17.2 for some approximations of

different curves.)

546 C H A P T E R S E V E N T E E N : M U L T I V A R I A T E R E G R E S S I O N

3. Compute a multivariate regression model using number of times drunk as

the dependent variable. Include the measures for age, sex, race (recoded

as a dummy variable), and at least two other variables that you think are

related to the number of times drunk. This will be the “baseline model” in

the following questions.

a. Compute the tolerance statistic for each of the independent variables in

the baseline model. Does it appear that there is a problem with

collinearity in the regression model? Explain why.

b. Compute an interaction term for sex with age. Add this term to the

baseline model, and rerun the regression command. Is the effect of age

effect of age on the number of times drunk for males and females.

i. Compute the tolerance statistic for each of the independent variables

in this model. Does it appear that there is a problem with collinearity

in this model? Explain why.

c. Compute an interaction term for race (which should be coded as a

dummy variable) and age. Add this term to the baseline model, and

rerun the regression command. (The interaction effect from part

(a) should no longer be included in the analysis.) Is the effect of age

groups? If so, interpret the effect of age on the number of times drunk

for each race group.

i. Compute the tolerance statistic for each of the independent variables

in this model. Does it appear that there is a problem with collinearity

in this model? Explain why.

d. If one of the additional variables that you have added to your

regression model is measured at the interval level of measurement,

compute an interaction term between this variable and either the sex

or the race variable. Add this term to the baseline model (there should

be no other interaction terms included in this analysis), and rerun the

regression command. Is the effect of this variable on number of times

effect of this variable for each group.

i. Compute the tolerance statistic for each of the independent variables

in this model. Does it appear that there is a problem with collinearity

in this model? Explain why.

4. Compute a multivariate regression model using number of times cheated

at school as the dependent variable. Include the measures for age, sex,

race (recoded as a dummy variable), grade point average, and amount of

time spent studying as the independent variables. This will be the baseline

model in the following questions.

547

a. Compute an interaction term for grade point average and time spent

studying and add this term to the regression model. Prior to rerunning

the regression command, check the item for descriptive statistics

available through the regression command window. Report the

b. Compute the tolerance statistic for each of the independent variables

in this model. Does it appear that there is a problem with collinearity in

this model? Explain why.

c. What is the effect of grade point average on number of times cheated

at the mean level of time spent studying? (You will need to use the

mean reported in the results for part (a).) Interpret this effect.

i. How does the effect of grade point average change as the value for

time spent studying increases or decreases?

d. What is the effect of time spent studying on number of times cheated

at the mean grade point average? Interpret this effect.

i. How does the effect of time spent studying change as the value for

grade point average increases or decreases?

C O M P U T E R E X E R C I S E S

Logistic Regression

f o r e x a m i n i n g a d i c h o t o m o u s d e p e n d e n t v a r i a b l e

Dependent Variable?

c o e f f i c i e n t s w i t h i n a s i n g l e m o d e l

Regression Coefficients?

Interpreted?

2

Model?

C h a p t e r e i g h t e e n

L o g i s t i c r e g r e s s i o n a s a t o o l

I n t e r p r e t i n g l o g i s t i c r e g r e s s i o n c o e f f i c i e n t s

C o m p a r i n g l o g i s t i c r e g r e s s i o n

E v a l u a t i n g t h e l o g i s t i c r e g r e s s i o n m o d e l

T e s t i n g f o r s t a t i s t i c a l s i g n i f i c a n c e

Why is It Inappropriate to Use OLS Regression for a Dichotomous

How is the Outcome Altered in a Logistic Regression Model?

Why is It Difficult to Interpret the Logistic Regression Coefficient?

How Can Probability Estimates be Used to Compare the Strength of Logistic

What is the Standardized Logistic Regression Coefficient and How is It

How is the Percent of Correct Predictions Interpreted?

What is Pseudo R and How is It Interpreted?

What is the Test of Statistical Significance for the Overall Logistic Regression

What is the Test of Statistical Significance for the Logistic Regression Coefficient?

What is the Test of Statistical Significance for a Multicategory Nominal Variable?

What is an Odds Ratio and How is It Interpreted?

What Shape Does the Logistic Model Curve Take?

What is the Derivative at Mean and How is It Interpreted?

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_18,

© Springer Science+Business Media New York 2014

ORDINARY LEAST SQUARES REGRESSION is a very useful tool for identi-

fying how one or a series of independent variables affects an interval-

level dependent variable. As noted in Chapter 16, this method may

also be used—though with caution—to explain dependent variables

that are measured at an ordinal level. But what should the researcher

do when faced with a binary or dichotomous dependent variable?

Such situations are common in criminology and criminal justice. For

example, in examining sentencing practices, the researcher may want

to explain why certain defendants get a prison sentence while others

do not. In assessing the success of a drug treatment program, the re-

searcher may be interested in whether offenders failed a drug test or

whether they returned to prison within a fixed follow-up period. In

each of these examples, the variable that the researcher seeks to ex-

plain is a simple binary outcome. It is not appropriate to examine bi-

nary dependent variables using the regression methods that we have

reviewed thus far.

This chapter introduces a type of regression analysis that allows us to

examine a dichotomous dependent variable. Called logistic regression

analysis, it has become one of the analysis tools most frequently used in

crime and justice research. We begin the chapter by explaining why the

OLS regression approach described in Chapters 15 and 16 is not appro-

trate the interpretation of logistic regression statistics in the context of a

substantive criminal justice research example. In this chapter, as in

Chapter 16, our focus will be more on explaining how logistic regres-

sion can be used in research than on describing the mathematical prop-

erties that underlie the computations used to develop logistic regression

statistics.

logistic regression approach and the logic underlying it. Finally, we illus-

priate when the dependent variable is binary. We then describe the

549

f o r a D i c h o t o m o u s D e p e n d e n t V a r i a b l e ?

In Chapter 16, you saw that we could use not only interval-level vari-

ables, but also ordinal- and nominal-level variables, as independent

measures in a multivariate ordinary least squares regression. While

we emphasized that the assumptions regarding measurement of the

dependent variable are much more stringent in OLS regression, even

in the case of the dependent variable the researcher may sometimes

decide to use ordinal- as well as interval-level variables. But applying

the OLS regression approach is inappropriate when the dependent

variable is nominal, as is the case with a binary or dichotomous depen-

dent variable.

Why do we state this rule so unequivocally? One reason is that the

logic underlying our explanation of a dichotomous dependent variable is

at odds with the models that we build using the OLS regression ap-

proach. In order to expand on this idea, we need to return to how pre-

dictions are developed using OLS regression. In the simple linear

model—the OLS model—we predict the value of Y based on an equation

that takes into account the values of a Y-intercept (b0) and one or a se-

ries of independent variables (e.g., b1X1). This model is represented

below for a bivariate regression example in which we seek to explain

the yearly budget of police departments based on the number of officers

employed.

where

This is an additive model, in which we predict the value of Y—in this

case, the yearly police department budget in dollars—by adding the

value of the Y-intercept to the value of the regression coefficient times

the value of X1 (the number of sworn officers in a department).

Let’s say that a representative sample of police agencies were sur-

veyed and analysis of the responses yielded the following regression

equation:

Y � 100,000 � 100,000X1

This equation suggests that for each additional officer employed, the de-

partment budget is expected to increase by $100,000. For a police

agency with 100 officers, we would expect a budget of about

$10,100,000:

X1 � number of sworn officers

Y � yearly police department budget in dollars

Y � b0 � b1X1

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N550

W h y i s I t I n a p p r o p r i a t e t o U s e O L S R e g r e s s i o n

For a police agency with 1,000 officers, we would expect a budget of

about $100,100,000:

W orking It Out

� 10,100,000

� 100,000 � 10,000,000

� 100,000 � 100,000(100)

Y � 100,000 � 100,000X1

W orking It Out

� 100,100,000

� 100,000 � 100,000,000

� 100,000 � 100,000(1,000)

Y � 100,000 � 100,000X1

This model, like other OLS regression models, assumes that there is

no real limit to the value that the dependent variable can attain. With

each additional officer comes an expected increase in the departmental

budget. Our model suggests that the increase is about $100,000 for each

additional officer. While this logic makes very good sense when we are

speaking about interval-scale measures, such as the budget of a police

agency, does it make sense when we are dealing with a dichotomous

dependent variable, such as whether a parolee has failed a drug test?

Let’s say that we surveyed 30 parolees who had been tested for drug

use. Our independent variable is prior drug arrests. The data in Table

ber of prior drug arrests for each parolee. We have coded a failed drug

gression results for our example. The OLS regression suggests a very

strong relationship between prior drug arrests and failing the drug test.

But if we look more closely at the regression model, we can see that this

approach may lead to outcomes that are not consistent with the

processes we seek to understand.

18.1 report whether a parolee failed the drug test and also give the num-

test as 1 and a successful drug test as 0. Table 18.2 provides the OLS re-

Figure 18.1 shows the data points for our example in a scatterplot, as

well as the regression line drawn from the outcomes in Table 18.2. It is

I S I T I N A P P R O P R I A T E T O U S E O L S R E G R E S S I O N ?W H Y 551

Drug Testing Results and Prior Drug Arrests for 30 Parolees

DRUG TEST RESULT DRUG TEST SCORE NUMBER OF DRUG ARRESTS

Pass 0 0 Pass 0 0 Pass 0 0 Pass 0 0 Pass 0 0 Fail 1 0 Pass 0 1 Pass 0 1 Pass 0 1 Pass 0 1 Pass 0 1 Pass 0 1 Fail 1 2 Fail 1 2 Fail 1 2 Fail 1 2 Fail 1 2 Fail 1 3 Fail 1 3 Pass 0 3 Fail 1 4 Fail 1 4 Fail 1 4 Fail 1 5 Fail 1 5 Fail 1 5 Fail 1 6 Fail 1 6 Fail 1 7 Fail 1 8

OLS Regression Results for Example of Drug Testing and Prior Drug Arrests

Unstandardized Standardized Coefficients Coefficients t Sig.

Model B Std. Error Beta

1 (Constant) .211 .104 2.025 .052

Drug Arrests 0.148 .030 .681 4.921 .000

a Dependent Variable: FAILURE ON DRUG TEST

Table 18.1

Table 18.2

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N552

clear that the OLS regression approach leads to predicted outcomes that

are not possible, given our dependent variable. For example, for a parolee

with six drug arrests, our model predicts that Y will have a value of 1.099:

Scatterplot of Example of Drug Testing with the Predicted Regression Line

D ru

g T

es t

Fa il

u re

s

2.0

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0.0

–0.2

Number of Drug Arrests

–2 0 2 4 6 8 10

W orking It Out

� 1.099

� 0.211 � 0.888

� 0.211 � 0.148(6)

Y � 0.211 � 0.148X

For a parolee with eight drug arrests, our model predicts that Y will have

a value of 1.395:

W orking It Out

� 1.395

� 0.211 � 1.184

� 0.211 � 0.148(8)

Y � 0.211 � 0.148X

Note: In this scatterplot (produced using SPSS), the points sometimes represent more than one observation.

Figure 18.1

553I S I T I N A P P R O P R I A T E T O U S E O L S R E G R E S S I O N ?W H Y

But in our example, the predicted value of Y should logically be no

greater than 1 or no less than 0. A value of 1 means that the parolee

failed the drug test, and a value of 0 means that the parolee passed the

drug test. Predicting values greater than 1 or less than 0 just does not

make sense given the possible outcomes of a binary dependent variable.

This example, then, illustrates a logical problem in using OLS meth-

ods to gain estimates for cases where the dependent variable is dichoto-

mous. The OLS approach assumes that there is no limit to the predicted

value of the dependent variable. But in the case of a dichotomous de-

pendent variable, there are limits—the values 0 and 1. While this as-

sumption of predictions within the limits of the possible outcomes of the

dependent variable is also violated when OLS regression is used for an

ordinal-level dependent variable and sometimes when it is applied to

specific interval-level measures, the violation is most extreme in the case

of a binary dependent variable, such as drug testing failures. It does not

make sense to analyze such situations with a model that allows the value

of Y to increase at a constant rate for each change in the value of X. For

our analysis to be consistent with the problem we are examining, it must

provide predictions that are constrained to values between 0 and 1.

method in a case where the dependent variable is dichotomous. In our

discussion of excluded variables in Chapter 16, we noted that a central

assumption of the regression approach is that there is no systematic rela-

tionship between the error term and the independent variables included

in the regression. When a systematic relationship exists, estimates of the

regression coefficients are likely to be biased. But if you look at parolees

for whom the value of prior drug arrests is greater than 5 (see Figure

gression error and the independent variable. Because the actual value of

Y cannot be greater than 1, and the predicted values continue to increase

in a linear fashion (as evidenced by the regression line), the regression

error increases in the negative direction as the number of prior drug ar-

rests increases. This means that as the number of prior drug arrests gets

larger and larger, we will make larger and larger negative errors in pre-

diction. When OLS regression is used to examine a binary dependent

variable, we are very likely to have a systematic relationship between the

independent variable and the errors we make in predicting Y, because

Y-values are constrained to 0 and 1 and predicted values of Y have no

limit.

pendent variable, we violate assumptions important to making statistical

inferences with OLS regression. We noted in Chapter 15 that two para-

metric assumptions of our tests of statistical significance in regression are

that the values of X are normally distributed around the regression line

Figure 18.1 illustrates additional problems that arise in using the OLS

18.1), you can see that there is a consistent relationship between the re-

Figure 18.1 also illustrates why, when we examine a dichotomous de-

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N554

and that they meet an assumption of homoscedasticity (equal variances

around the regression line). The normality assumption is clearly violated

the shape of the distribution of X around Y will be bimodal because the

observed values of our dependent variable are constrained in practice to

0 and 1. However, if our sample is large enough, we generally allow vio-

lations of this assumption. Our problem in regard to homoscedasticity is

more serious. We noted in Chapter 15 that violations of the assumption

of homoscedasticity must be large before they become a concern. In the

case of a binary dependent variable, heteroscedasticity (violation of the

homoscedasticity assumption) is likely to be large. As shown in Figure

scores of X around the regression line is likely to vary widely in form,

depending on the scores of the independent variable.

L o g i s t i c R e g r e s s i o n

While the application of OLS regression methods to a dichotomous de-

pendent variable raises a number of substantive and statistical concerns,

there are many advantages to the basic form of the regression approach

introduced in previous chapters. For example, the effect of each b was

constant. That is, we could define a single constant effect for each vari-

able in the regression model. That effect took into account the other

variables in the model. And we could add all of these effects and the Y-

intercept to get a predicted value for Y. Because of the utility of the re-

gression approach, statisticians have developed alternative methods for

conducting regression analysis with dichotomous dependent variables

that do not violate basic assumptions but allow us to continue to use the

overall regression approach. Perhaps the most widely used of these

methods is logistic regression analysis.1 Logistic regression analysis is

1A method called generalized least squares might also be used to deal with violations

of our assumptions, though logistic regression analysis is generally the preferred

method. See E. A. Hanushek and J. E. Jackson, Statistical Methods for Social Scientists

(New York: Academic Press, 1977) for a comparison of these approaches. See also

David W. Hosmer and Stanley Lemeshow, Applied Regression Analysis, 2nd ed. (New

York: Wiley, 2000). Another method, probit regression analysis, is very similar to that

presented here, though it is based on the standard normal distribution rather than the

logistic model curve. The estimates gained from probit regression are likely to be very

similar to those gained from logistic regression. Because logistic regression analysis

has become much more widely used and is available in most statistical software pack-

ages, we focus on logistic regression in this chapter.

when we have a dichotomous dependent variable. As Figure 18.1 shows,

18.1 for our distribution of drug testing failures, the distribution of the

L O G I S T I C R E G R E S S I O N 555

based on a transformation of the regression model that allows the out-

comes of the regression equation to vary without limit, but constrains the

predictions of the dependent variable to values between 0 and 1. At the

same time, the inferential statistics used in logistic regression do not rely

on assumptions regarding the population distribution of scores.

In fitting the data that are analyzed, logistic regression analysis uses

the logistic model curve for the probability that Y � 1. While the logis-

tic regression curve follows the linear model in the middle of its distribu-

tion, it does not allow values below 0 or above 1. Indeed, as the logistic

curve approaches 0 or 1, it begins to flatten, so it keeps coming closer

to—but never actually reaches—either of these two values. The logistic

curve thus satisfies our primary objection to the linear OLS regression

method. That is, it does not allow predictions greater than or less than the

actual values of the distribution of scores that we are trying to predict.

The logistic model curve provides a solution to the problem of pre-

dictions beyond the observed distribution. However, in order to gain the

desired property of outcomes between 0 and 1, we have to alter the

form of our regression equation. Using OLS regression, we represent our

equation for the prediction of Y with one independent variable as

follows:

Y � b0 � b1X1

As noted earlier, this approach may yield values that are greater than 1

or less than 0, as was the case in our drug testing example.

The Logistic Model Curve

P ro

b ab

il it

y (Y

= 1

)

1.0

0.8

0.6

0.4

0.2

0

Xb

–5 –4 –3 –2 –1 0 1 2 3 4 5

the logic of a curve rather than that of a straight line. Figure 18.2 shows

Figure 18.2

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N556

In logistic regression, we alter the form of what we are trying to pre-

dict. Rather than predicting Y, as in OLS regression, we now predict the

natural logarithm (ln) of the odds of getting a 1 on the dependent vari-

able. Although this sounds very imposing, it is simply a transformation of

equation for a bivariate logistic regression:

There is no change on the right side of this equation. We have the

constant b0 and the regression coefficient b1 that reflects a constant effect

for the independent variable examined. Moreover, the outcome of this

formula has no limit. But on the left side of the equation, we now have

the natural logarithm of the odds of Y, or what statisticians call the

logit of Y. A logarithm, or log, is the exponent of the power to which

a fixed number (called a base) must be raised to produce another num-

ber. So, for example, if the base is 10, the logarithm of 100 is 2. That is, if

we take 10 to the 2nd power (102), we get 100. In logistic regression, we

do not use a base of 10, which is associated with what is called the com-

mon logarithm; rather, we use a base of 2.71828, which is associated

with what is called the natural logarithm and is represented in symbol

form as ln. The base of the natural logarithm, 2.71828, is also known as

Euler’s constant and is denoted by the symbol e. What this means is that

ln(x) is the power to which e must be raised to get x.

This represents the odds of getting an outcome of 1, rather than 0, on

the dependent variable. The odds are determined by dividing the proba-

bility of getting a 1 [P(Y � 1)] by the probability of not getting a 1 [1 �

P(Y � 1)]. In our drug testing example, this would be the odds of failing

a drug test divided by those of not failing the test. If an individual had an

80% predicted likelihood of failing the drug test, then the odds would be

0.80/0.20, or 4 to 1.

ln � P(Y � 1)

1 � P(Y � 1)� � ln �P(Y � 1)

P(Y � 0)� � b0 � b1X1

W orking It Out

� 4.0

� 0.80 0.20

Odds � P(Y � 1)

1 � P(Y � 1)

the equation presented above. Equation 18.1 represents the prediction

Equation 18.1

What about the notation P(Y � 1)/[1�P(Y � 1)] in Equation 18.1?

557L O G I S T I C R E G R E S S I O N

If we transform this equation further, we see that it gives us the prop-

erty we are looking for. That is, the predicted values of Y produced by

our regression equation will vary between 0 and 1, despite the fact that

the outcomes in our regression equation can reach any value between

plus and minus infinity. In the box above, we show how to transform

the equation so that the outcome is the probability that Y will be 1. The

We begin with the specification of the logistic regression model:

To simplify, we let

and

Using these simplifications, we can rewrite the logistic regression equa-

tion as

If we exponentiate both sides of the equation (i.e., take the value of e to

the power of both sides of the equation), we obtain

Then, ln is the power to which we must raise e to get ; that is,

e ln[P/(1�P)] � P

1 �P

P 1 � P� P

1 � P� e ln[P/(1�P)] � e Xb

ln� P

1 � P� � Xb

P � P (Y � 1) ⇒ 1 � P (Y � 1) � 1 � P

Xb � b0 � b1X1

ln� P (Y � 1)

1 � P (Y � 1)� � b0 � b1X1

Derivation of P (Y � 1) from the Cumulative Logistic Probability Function

end result is a simple equation that can be calculated on a hand calcula-

tor with a natural log function. This equation is often called the cumula-

tive logistic probability function.

P(Y � 1) � 1

1 � e�(b0�b1X1)

This leads to rewriting the logistic regression equation as

We multiply both sides of the equation by (1 � P):

Then we add PeXb to both sides of the equation:

Next we rewrite the equation to pull out the common factor, P :

Now we divide both sides of the equation by (1 + e Xb) to solve for P :

Since, as noted above, P � P (Y � 1),

P (Y � 1) � 1

1 � e�Xb

� 1

1 � e�Xb

P � e Xb

1 � e Xb �

1

�1 � e Xb

e Xb � �

1

� 1

e Xb � 1�

P (1 � e Xb) � e Xb

P � Pe Xb � e Xb

P � e Xb(1 � P) � e Xb � Pe Xb

P

1 � P � e Xb

Equation 18.2

By using the term Xb to represent the right side of the regression equa-

any number of independent variables:

What this equation does is divide 1 by the sum of 1 and e (the value

2.71828) taken to the �Xb power. The process of taking a number to

some power is referred to as exponentiation. Here we exponentiate e to

the power �Xb. Exponentiation may also be familiar to you as the an-

tilog or inverse log.2

P(Y � 1) � 1

1 � e�Xb

Illustration of the Fact That P(Y � 1) Will Not Exceed 1 or Be Less Than 0

Xb P (Y � 1)

�25 0.000000000014 �20 0.000000002061 �15 0.000000305902 �10 0.000045397669 �5 0.006692850924 �4 0.017986209962 �3 0.047425873178 �2 0.119202922022 �1 0.268941421370

0 0.500000000000 1 0.731058578630 2 0.880797077978 3 0.952574126822 4 0.982013790038 5 0.993307149076

10 0.999954602131 15 0.999999694098 20 0.999999997939 25 0.999999999986

2Your calculator likely has a button labeled “e x,” which performs this operation. If

there is no ex button, then you should be able to locate a button labeled “INV” and

another for the natural logarithm, ln. By pushing “INV” and then “ln” (the inverse or

antilog), you will be able to perform this operation.

Equation 18.3

tion, we may write Equation 18.2 more generally to take into account

Table 18.3

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N560

Importantly, whatever the value associated with Xb, the value of

P(Y � 1) will always be between 0 and 1. The value of P(Y � 1) can get

closer and closer to 1 or to 0, but it will never exceed that number. This

tive values of Xb. As the values get very large, the gain for each increase

in Xb becomes smaller and smaller. Logistic regression, then, allows us

to use the traditional regression format in which the outcome, Xb, can

achieve any size without limit. However, since we have converted what

we are predicting to the logit of Y, our predictions of Y are bounded by

0 and 1.

comes of Y are constrained between 0 and 1. But what does a prediction

between the values 0 and 1 mean? As we have already noted, the ob-

served outcomes for a dichotomous dependent variable have a score of

either 0 or 1. For example, in our drug testing example, either parolees

failed the drug test (coded as 1) or they passed the test (coded as 0).

When we examined the regression approach in previous chapters, the

pendent variable, our interpretation must be different. The predicted

value of Y in this case is the predicted probability of getting an outcome

of 1. So, for example, a value of 0.50 in our drug testing example would

mean that, according to our model, an individual was predicted to have

about an equal chance of failing and not failing drug testing. A value of

0.90 would suggest that an individual was highly likely to have a drug

testing failure.

To estimate the coefficients of a logistic regression, we use a much

more complex mathematical process than was used in OLS regression.

It is based on maximum likelihood estimation (MLE) techniques.

Using these techniques, we try to maximize the probability that our re-

gression estimates will produce a distribution similar to that of the ob-

served data. With this approach, we do not simply derive a single

mathematical solution for obtaining the regression estimates.3 Rather,

we begin by identifying a tentative solution, which we then try to im-

prove upon. Our criterion for improvement is termed a likelihood func-

tion. A likelihood function measures the probability of observing the

3It should be noted, however, that maximum likelihood techniques do not always re-

quire an iterative process.

is illustrated in Table 18.3, where we take very large negative and posi-

allowed us to develop a regression model in which the predicted out-

Use of the natural logarithm of the odds of Y, or the logit of Y, has

bution of scores on our interval-level measure. With a dichotomous de-

predicted value of Y was simply one of the possible values in the distri-

561L O G I S T I C R E G R E S S I O N

results in the sample, given the coefficient estimates in our model. By

convention in logistic regression, we use �2 times the natural loga-

rithm of the likelihood function (or �2LL), which is defined as the log

likelihood function. We repeat this process again and again, until the

change in the likelihood function is considered negligible. Each time

we repeat the process and reestimate our coefficients is called an itera-

tion. Logistic regression is said to be an iterative procedure, because it

tries a number of solutions before arriving at a final result—or, in statis-

tical terms, converging.

Most packaged statistical programs set a default limit on the num-

ber of iterations that can be tried. In SPSS, that limit is 20 iterations.

Lack of convergence in a standard number of iterations may indi-

cate some type of problem in the regression model. Often, it occurs

when the number of variables examined is large relative to the

number of cases in the study. John Tukey, a noted statistician who

taught at Princeton University, has suggested a rule for logistic regres-

sion: that there be at least five cases and preferably at least ten in

the smaller category of the dependent variable for each independent

variable examined.4 Whatever the cause, if you receive a message

from a packaged statistical analysis program that your regression has

failed to converge, you should look carefully at your model and your

measures.

We have now looked at the basic logic of the logistic regression

model. While the logistic regression model differs from the OLS re-

gression model in the outcome predicted, the basic form of the addi-

tive linear model has been maintained. The right side of the equation

remains an additive function of the Y-intercept and the independent

variables (multiplied by their associated regression coefficients). The

effect of each independent variable remains its independent effect,

with the other variables in the model controlled. We also continue

to be constrained by the same regression assumptions regarding cor-

rect model specification. Excluding variables from the regression will

lead to bias, either in our prediction of Y or in our estimates of

specific regression coefficients. The models are also sensitive to prob-

lems of multicollinearity. These concepts were reviewed in Chapter

4See John Tukey, Report to the Special Master, p. 5; Report to the New Jersey Supreme

Court 27 (1997).

logistic regression.

16, but you should remember that they apply to our discussion of

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N562

A Substantive Example: Adoption of Compstat in U.S. Police Agencies

Application of logistic regression to a substantive problem will help you

to understand the use of logistic regression, as well as the different coef-

ficients associated with the technique. The example we use is drawn

from a Police Foundation survey of U.S. police agencies, begun in 1999 5

stat, a management system first developed in New York City in order to

reduce crime and improve quality of life, had been widely adopted in

some form by other U.S. police agencies. It was theorized that Compstat

would be much more likely to be adopted by larger police agencies.

Using logistic regression analysis, we will examine whether this hypothe-

sis is supported by the Police Foundation data.

The dependent variable in our analysis is dichotomous, measuring

whether the department claimed to have adopted a “Compstat-like pro-

gram.” The main independent variable is the number of sworn officers

serving in the department at the time of survey.6 We also include, as a

second independent variable, region, which divides the country into four

regions: South, West, North Central, and Northeast. For this multicate-

gory nominal variable, we use three dummy variables to represent re-

gion and define the North Central region as the reference, or excluded,

X4 � West

X3 � South

X2 � Northeast

where X1 � number of sworn officers

Xb � b0 � b1X1 � b2X2 � b3X3 � b4X4

5For a description of this study, see David Weisburd, Stephen Mastrofski, Ann Marie

McNally, and Rosann Greenspan, Compstat and Organizational Change (Washington,

DC: The Police Foundation, 2001). 6Departments with 1,300 or more officers were coded in our example as 1,300 officers.

This transformation was used in order to take into account the fact that only 5% of the

departments surveyed had more than this number of officers and their totals varied very

widely relative to the overall distribution. Another solution that could be used to address

the problem of outliers is to define the measure as the logarithm of the number of sworn

officers, rather than the raw scores. We relied on the former solution for our example be-

cause interpretation of the coefficients is more straightforward. In an analysis of this

problem, a researcher would ordinarily want to compare different transformations of the

dependent variable in order to define the one that best fit the data being examined.

police agencies with more than 100 sworn officers (N � 515) and got a

The Police Foundation surveyed all and completed in the year 2000.

response rate of 86%. A main concern of the survey was whether Comp-

category. Our regression model (Xb) is represented in Equation 18.4:

Equation 18.4

563L O G I S T I C R E G R E S S I O N

Summary of the Logistic Regression Coefficients Using SPSS’s Logistic Regression Program

Variables in the Equation

B S.E. Wald df Sig. Exp(B)

Step 1 NORTHEAST .359 .372 .931 1 .335 1.432

SOUTH .805 .332 5.883 1 .015 2.237

WEST .428 .367 1.360 1 .244 1.534

#SWORN .002 .000 24.842 1 .000 1.002

Constant �1.795 .311 33.378 1 .000 .166

a Variable(s) entered on step 1: NORTHEAST, SOUTH, WEST, #SWORN.

As you can see, it took only three iterations to achieve convergence. The

convergence criterion used in this SPSS run was that the log likelihood

function declined by less than 0.010%. As noted earlier, we use �2 times

the natural logarithm of the likelihood function (�2LL) to define the log

likelihood function. The final coefficients listed in this table are the same

as the regression coefficients (B) reported in the summary of the regres-

Iteration History Using SPSS’s Logistic Regression Program

Iteration History

�2 Log Coefficients likelihood

Iteration Constant NOREAST SOUTH WEST NMSWORN

Step 1 1 493.418 �1.555 .258 .629 .308 .001

2 492.515 �1.783 .351 .795 .419 .002

3 492.513 �1.795 .359 .805 .428 .002

a Method: Enter b Constant is included in the model. c Initial �2 Log Likelihood: 528.171 d Estimation terminated at iteration number 3 because log likelihood decreased by less than .010 percent.

Table 18.4

Table 18.4 shows the iteration history for estimating this regression.

Table 18.5

sion results provided in Table 18.5.

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N564

We can now express our regression equation in terms of the outcomes

Inserting the values from our regression analysis, we can express the

equation as follows (note that the constant in the SPSS printout is the

Y-intercept, b0):

Xb � �1.75 � 0.002X1 � 0.359X2 � 0.805X 3 � 0.428X 4

We can also develop predictions of Y from this model, as in the case

of the OLS model. However, as explained above, our predictions of Y

are not the direct outcome of our additive regression model. Rather, the

lative logistic probability function:

For example, let’s say that we want to predict the probability of

Because North Central is the reference category, the equation contains

only the Y-intercept and the effect of the number of sworn officers:

P(Y � 1) � 1

1 � e�Xb

X 4 � West

X 3 � South

X2 � Northeast

where X1 � number of sworn officers

Xb � b0 � b1X1 � b2X2 � b3X 3 � b4X 4

W orking It Out

X 4 � West

X 3 � South

X2 � Northeast

where X1 � number of sworn officers

� 0.205

� �1.795 � 2

� �1.795 � 0.002(1000)

� �1.795 � 0.002(1000) � 0.359(0) � 0.805(0) � 0.428(0)

Xb � b0 � b1X1 � b2X2 � b3X 3 � b4X4

of our analysis. Above, we defined our regression model using the term Xb :

probability of Y occurring was expressed in Equation 18.3 for the cumu-

a Compstat-like program in a department with 1,000 officers in the

North Central region. Our first task is to define the value of Xb. We do

that by applying coefficients gained in our logistic regression analysis.

565L O G I S T I C R E G R E S S I O N

For a police department in the South with 1,000 officers, the predicted

probability of having a Compstat-like program is fully 73%:

W orking It Out

And,

� 0.7330

� 1

1 � 0.3642

� 1

1 � e�1.01

P(Y � 1) � 1

1 � e�Xb

� 1.01

� �1.795 � 2 � 0.805

� �1.795 � 0.002(1000) � 0.805(1)

� �1.795 � 0.002(1000) � 0.359(0) � 0.805(1) � 0.428(0)

Xb � b0 � b1X1 � b2X2 � b3X 3 � b4X 4

Applying this result to our equation, we see that, according to our re-

gression model, the probability of having a Compstat-like program in

such a department is about 55%.

W orking It Out

� 0.55

� 1

1 � 0.8146

� 1

1 � e�0.205

P(Y � 1) � 1

1 � e�Xb

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N566

I n t e r p r e t i n g L o g i s t i c R e g r e s s i o n C o e f f i c i e n t s

in our model. In this case, the logistic regression coefficients are listed in

the column labeled B. As expected, the coefficient for the number of

sworn officers (#SWORN) is positive; that is, as the number of sworn offi-

cers increases, the likelihood of having a Compstat-like program also in-

category. This means that in the Police Foundation sample, police depart-

ments in the Northeast, West, and South regions were more likely to claim

to have a Compstat-like program than those in the North Central region.

What about the exact interpretation of the logistic regression coefficient?

Here, we can see the price we pay for developing a regression model in

W orking It Out

And,

� 0.3122

� 1

1 � 2.2034

� 1

1 � e�(�0.79)

P(Y � 1) � 1

1 � e�Xb

� �0.79

� �1.795 � 0.2 � 0.805

� �1.795 � 0.002(100) � 0.805(1)

� �1.795 � 0.002(100) � 0.359(0) � 0.805(1) � 0.428(0)

Xb � b0 � b1X1 � b2X2 � b3X 3 � b4X 4

If we apply our prediction model to smaller departments, we see that

they are less likely, according to our estimates, to have a Compstat-like

program. For example, our model suggests that a police agency with

only 100 officers from the South would have a probability of only 31% of

having a Compstat-like program:

increases. The three dummy variables included for region also have a

positive impact relative to the North Central region, which is the excluded

which the predictions of the probability of Y are constrained between

Using Table 18.5, we can also define the specific effects of the variables

I N T E R P R E T I N G L O G I S T I C R E G R E S S I O N C O E F F I C I E N T S 567

increase in Y (all other included independent variables held constant).

The interpretation of the logistic regression coefficient is not as

straightforward. Our regression equation is predicting not Y, but the loga-

rithm of the odds of getting a 1—or, in our example, the log of the odds of

having a Compstat-like program. In a multivariate logistic regression, b rep-

resents the estimated change in the log of the odds of Y occurring when all

other independent variables are held constant. The coefficient for number

of sworn officers is 0.002, meaning that each additional officer increases by

0.002 the log of the odds of having a Compstat-like program. While some

researchers may have an intuitive understanding of the change in the log of

the odds, the transformation of the outcome measure in the logistic regres-

sion model has made the regression coefficients very difficult to explain or

interpret in a way that nonstatisticians will understand.

The Odds Ratio

To make results easier to understand, statisticians have developed other

methods of interpreting logistic regression coefficients. An approach

commonly used is to report the regression coefficient in terms of its odds

ratio. The odds ratio, sometimes called the exponent of B, is reported

the impact of a one-unit change in X on the ratio of the probability of an

event occurring to the probability of the event not occurring. Equation

events separated by a change of one unit in X:

An odds ratio greater than 1 indicates that the odds of getting a 1 on

the dependent variable increase when the independent variable increases.

An odds ratio less than 1 indicates that the odds of getting a 1 on the de-

pendent variable decreases when the independent variable increases. For

our example, an odds ratio greater than 1 indicates that as the indepen-

dent variable increases, the odds of having a Compstat-like program also

increase. If the odds ratio were 3, for example, then a one-unit change in

where P(Y � 1) � 1

1 � e�Xb

Odds ratio � P(Y � 1)

1 � P(Y � 1) X

P(Y � 1)

1 � P(Y � 1) X � 1

0 and 1. In the OLS regression case, the interpretation of b is in reference

to units of measurement of Y. In the multivariate case, b represents the

estimated change in Y associated with a unit change in X, when all other

independent variables in the model are held constant. So a b of 2 in an OLS

regression suggests that a unit increase in X is associated with a two-unit

as Exp(B) in the SPSS printout in Table 18.5. The odds ratio represents

18.5 defines the odds ratio in terms of the calculation of the odds for two

Equation 18.5

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N568

X would make the event Y about three times as likely to occur. An odds

ratio less than 1 would suggest that the likelihood of having a Compstat-

like program decreased as the independent variable increased.

To calculate the odds ratio, we need to define the probability of get-

ting a 1 [i.e., P(Y � 1)] on our dependent variable at two values, X and

X � 1. We can choose any two consecutive values of the independent

variable; our odds ratio will be the same, no matter what consecutive val-

ues we choose. Let’s start with number of sworn officers. For simplicity,

we will take 101 and 100 as X and X � 1 and we will calculate the proba-

bilities when a department is in the North Central region. We first need to

work out the odds of getting a Compstat-like program from our model for

the case of 101 sworn officers. As shown below, the result is 0.2034.

W orking It Out Number of Sworn Officers � 101

Step 1: Defining the probability of Y � 1.

Step 2: Defining P(Y � 1).

Step 3: Defining the odds.

� 0.2034

P(Y � 1)

1 � P(Y � 1) X

� 0.1690 0.8310

� 0.1690

� 1

1 � 4.9185

� 1

1 � e�(�1.593)

P(Y � 1) � 1

1 � e�Xb

X 4 � West

X 3 � South

X2 � Northeast

where X1 � number of sworn officers

� �1.593

� �1.795 � 0.202

� �1.795 � 0.002(101)

� �1.795 � 0.002(101) � 0.359(0) � 0.805(0) � 0.428(0)

Xb � b0 � b1X1 � b2X2 � b3X 3 � b4X 4

569I N T E R P R E T I N G L O G I S T I C R E G R E S S I O N C O E F F I C I E N T S

We then need to follow the same procedure for 100 sworn officers. As

shown below, the odds in the case of 100 sworn officers in the North

Central region are 0.2029.

W orking It Out Number of Sworn Officers � 100

Step 1: Defining the probability of Y � 1.

Step 2: Defining P (Y � 1).

Step 3: Defining the odds.

� 0.2029

P(Y � 1)

1 � P(Y � 1) X�1

� 0.1687 0.8313

� 0.1687

� 1

1 � 4.9283

� 1

1 � e�(�1.595)

P(Y � 1) � 1

1 � e�Xb

X 4 � West

X 3 � South

X2 � Northeast

where X1 � number of sworn officers

� �1.595

� �1.795 � 0.2

� �1.795 � 0.002(100)

� �1.795 � 0.002(100) � 0.359(0) � 0.805(0) � 0.428(0)

Xb � b0 � b1X1 � b2X2 � b3X 3 � b4X 4

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N570

Finally, using these two odds, we can estimate our odds ratio, which is

simply the ratio of these two numbers, or 1.002:

W orking It Out

� 1.002

� 0.2034 0.2029

Odds ratio � P(Y � 1)

1 � P(Y � 1) X

P(Y � 1)

1 � P(Y � 1) X�1

Of course, it would be a lot easier to just look at your computer print-

out, which provides the same outcome. You will probably not calculate

odds ratios by hand outside your statistics class. But working out the

odds ratio gives you a better understanding of what it is and where it

comes from.

ber of sworn officers. Instead of working through the three steps

above, we can move directly from the logistic regression coefficient to

the odds ratio by exponentiating the value of the coefficient b. As

noted earlier in the chapter, when we exponentiate the value of the co-

efficient b, we take e—the value 2.71828—to the power of the coeffi-

cient b. For number of sworn officers, it is e (0.002) � 1.002. What this

means is that for any logistic regression analysis, all we need to do is

exponentiate the logistic regression coefficient to calculate the odds

ratio. SPSS, like most other statistical software, will automatically report

the odds ratios for each of the independent variables included in the

analysis.

It is important to keep in mind that the odds ratio provides an esti-

mate for only a single one-unit increase in the independent variable. The

odds ratio is not a linear function of the coefficients; thus, we cannot say

that for each one-unit increase in the independent variable, the odds in-

crease by some amount. If we are interested in a change of more than

one unit in our independent variable—say 2, 5, 10, or 100 units— we

multiply that number by our coefficient b and then exponentiate that

value. For example, returning to the number of sworn officers, suppose

we are interested in the odds of adopting a Compstat-like program for a

In Table 18.5, we see that b � 0.002 and Exp(B) � 1.002 for num-

571I N T E R P R E T I N G L O G I S T I C R E G R E S S I O N C O E F F I C I E N T S

department that added 100 officers. We multiply our coefficient of 0.002

by 100, getting a value of 0.2, and then take e to the power of 0.2, which

gives us a value of 1.2214.

W orking It Out

Odds ratio � e (0.002)(100) � e (0.2) � 1.2214

This odds ratio tells us that the odds of adopting a Compstat-like

program increase by a factor of 1.22 for a department with 100 addi-

tional officers. As an exercise, take the odds of adopting a Compstat-

like program for a department with 100 officers in the North Central

with 200 officers. Then take the ratio of these two odds—it will equal

1.2214.

Our focus on the number of sworn officers illustrates another fea-

ture of logistic regression coefficients that is easily overlooked. There

are times—usually for an interval-level independent variable— when

the logistic regression coefficient will appear to have a small value.

Yet, when we begin to account for the range of the independent vari-

able and start to look at increases of 10, 100, or even 1,000 in the inde-

pendent variable, we may find that the odds increase by a substantial

amount.

For our regional dummy variables, it should be remembered that the

three measures are compared to the reference category, the North Cen-

tral region. Because working out the odds ratio is tedious, we will carry

out the calculations only for the South. According to the results pre-

meaning that being in the South region of the country, as opposed to the

North Central region, more than doubles the odds of having a Compstat-

like program. As with our number of sworn officers coefficient, we get a

value of 2.2367 by taking e to the power of 0.805, which is the logistic

regression coefficient for the South region.

W orking It Out

Odds ratio � e (0.805) � 2.2367

sented in Table 18.5, the South has an associated odds ratio of 2.237,

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N

region (0.2029; see page 570) and calculate the odds for a department

572

Alternatively, we can work through the calculation of the odds ratio

to arrive at the same conclusion. Setting the number of sworn officers at

100, we will calculate the odds ratio of a Compstat-like program for the

case where a department is in the South versus the case where it is in

the North Central region.

W orking It Out Departments in the South

Step 1: Defining the probability of Y � 1.

Step 2: Defining P(Y � 1).

Step 3: Defining the odds.

� 0.4539

P(Y � 1)

1 � P(Y � 1) X

� 0.3122

0.6878

� 0.3122

� 1

1 � 2.2033

� 1

1 � e�(�0.790)

P(Y � 1) � 1

1 � e�Xb

X 4 � West

X 3 � South

X2 � Northeast

where X1 � number of sworn officers

� �0.790

� �1.795 � 0.200 � 0.805

� �1.795 � 0.002(100) � 0.805(1)

� �1.795 � 0.002(100) � 0.359(0) � 0.805(1) � 0.428(0)

Xb � b0 � b1X1 � b2X2 � b3X 3 � b4X 4

573I N T E R P R E T I N G L O G I S T I C R E G R E S S I O N C O E F F I C I E N T S

W orking It Out Departments in the North Central Region

Step 1: Defining the probability of Y � 1.

Step 2: Defining P (Y � 1).

Step 3: Defining the odds.

� 0.2029

P(Y � 1)

1 � P(Y � 1) X�1

� 0.1687 0.8313

� 0.1687

� 1

1 � 4.9283

� 1

1 � e�(�1.595)

P(Y � 1) � 1

1 � e�Xb

X 4 � West

X 3 � South

X2 � Northeast

where X1 � number of sworn officers

� �1.595

� �1.795 � 0.2

� �1.795 � 0.002(100)

� �1.795 � 0.002(100) � 0.359(0) � 0.805(0) � 0.428(0)

Xb � b0 � b1X1 � b2X2 � b3X 3 � b4X 4

W orking It Out

� 2.2367

� 0.4539 0.2029

Odds ratio � P(Y �1)

1 �P (Y �1) X

P (Y � 1)

1 � P(Y � 1) X�1

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N574

Turning to the odds comparing the West and Northeast regions with

the North Central region, we can see that the differences are smaller (see

meaning that departments in these regions have a higher likelihood of

reporting a Compstat-like program. The odds ratios for both regions are

about 1.5. A police department in these regions is about 1.5 times as

likely to have a Compstat-like program as a department in the North

Central region.

The Derivative at Mean

tic regression coefficient into a simple linear regression coefficient.

Accordingly, it has the advantage of having the same interpretation as

the result would have had if OLS regression had been appropriate to the

problem. The disadvantage of the derivative at mean is that it calculates

the regression coefficient as if it had a constant effect over the entire dis-

tribution of predicted values of Y, based on the change observed when

the predicted value of Y is at its mean. In fact, the logistic curve in Figure

derivative at mean will be largest when the mean of the dependent vari-

able is close to the middle of the logistic curve. As the mean of the distri-

The interpretation of the derivative at mean is similar to that of the

OLS regression coefficient. The derivative at mean may be defined as the

change in Y associated with a unit change in X at the mean value of

the dependent variable. The derivative at mean is defined mathemati-

where is the mean of the dependent variable (i.e., the proportion of

cases having a value of 1 for the dependent variable).

in our regression model. Since about 33% of the sample claimed to have

implemented a Compstat-like program, the derivative at mean is calcu-

lated for a mean of Y of 0.33. If we look at the derivative at mean for

the dummy variables associated with region, we can see the advantage

of this approach. Taking the South region, for which the difference

Y

DM � Y (1 � Y )bi

regression coefficient (but that is not reported in SPSS) is the derivative

at mean (DM). The derivative at mean converts the nonlinear logis-

18.2 shows that the impact of the parameters will change in absolute

cally in Equation 18.6:

Table 18.6 provides the derivative at mean for each of the coefficients

Equation 18.6

terms, depending on where in the distribution they are calculated. The

bution moves closer to the tails of the logistic curve, the derivative will

be smaller.

Table 18.5). Like the South region statistic, these coefficients are positive,

Another measure that sometimes makes it easier to understand the logistic

575I N T E R P R E T I N G L O G I S T I C R E G R E S S I O N C O E F F I C I E N T S

from the excluded category is largest, we calculate the derivative at

mean below:

W orking It Out

� 0.1778

� (0.33)(0.67)(0.805)

� (0.33)(1 � 0.33)(0.805)

DM � Y (1 � Y )bi

We can interpret this coefficient much as we interpreted the dummy

variable regression coefficients in Chapter 16. If a police department is

located in the South as opposed to the North Central region, its outcome

on the dependent variable is about 0.1778 unit higher. Since the depen-

dent variable has values ranging between 0 and 1, we can interpret this

coefficient in terms of percentages. Departments in the South have, on

average, about an 18 percentage-point higher chance of claiming to have

a Compstat-like program when Y is at its mean.

The derivative at mean for number of sworn officers is about 0.0004.

This suggests that for each additional officer, there is a 0.0004 increase in

the value of Y. According to the derivative at mean, an increase in 100

officers would lead to a 4 percentage-point increase in the likelihood of

having a Compstat-like program. An increase of 1,000 officers would

lead to a 40 percentage-point increase.

W orking It Out

� 0.0004

� (0.33)(0.67)(0.002)

� (0.33)(1 � 0.33)(0.002)

DM � Y (1 � Y )bi

Derivative at Mean for Each of the Regression Coefficients in the Compstat Example

VARIABLE b

Northeast 0.359 (0.33)(0.67)(0.359) � 0.0794 South 0.805 (0.33)(0.67)(0.805) � 0.1778 West 0.428 (0.33)(0.67)(0.428) � 0.0946 Number of Sworn Officers 0.002 (0.33)(0.67)(0.002) � 0.0004

DM � Y (1 � Y )bi

Table 18.6

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N576

C o m p a r i n g L o g i s t i c R e g r e s s i o n C o e f f i c i e n t s

In Chapter 16, you saw how standardized regression coefficients could

be used to compare the magnitude of regression coefficients measured

on different scales. There is no widely accepted method for comparing

the magnitude of the coefficients in logistic regression. When variables

are measured on the same scale, we can rely on comparisons of the sta-

tistics we have reviewed so far. For example, if our model includes two

binary dummy variables, we can easily gain a sense of the impact of

each variable by comparing the size of each odds ratio.

Let’s say that we are interested in predicting the likelihood of getting a

prison sentence for a sample of convicted burglars. We include two bi-

nary dummy variables in our analysis. The odds ratio for the first variable,

gender (0 � female; 1 � male), is 1.5. The odds ratio for the second,

whether a gun was used in the burglary (0 � no; 1 � yes), is 2.0. In this

case, we could say that use of a weapon has a larger effect on the likeli-

hood of getting a prison sentence than does gender. In the case of gen-

der, being a male as opposed to a female increases the odds of getting a

prison sentence by about 50%. However, according to these estimates,

using a gun in the burglary doubles the odds of getting a prison sentence.

Using Probability Estimates to Compare Coefficients

If variables are measured on very different scales, comparing the magni-

tude of effects from one variable to another is often difficult. One easily

understood and transparent method for doing this is to rely on the pre-

dicted probabilities of Y. In a study using logistic regression, Wheeler,

Weisburd, and Bode were confronted with a large number of statistically

significant independent variables measured on very different scales.7

They decided to calculate probability estimates for measures at selected

intervals when the scores of all other predictors were held at their mean.

They also calculated a range of predictions computed from the 5th to

95th percentile scores for the measure of interest. The table they devel-

The study examined factors that explained whether or not white-

collar offenders convicted in federal courts were sentenced to prison.

The table gives the reader a sense of how changes in the independent

variable affect changes in the dependent variable, as well as a general

idea (using the range) of the overall influence of the measure examined.

For example, the amount of “dollar victimization” in an offense (variable

2) and “role in offense” (variable 13) are both ordinal-level variables but

are measured with a different number of categories. Looking at the table,

7See Stanton Wheeler, David Weisburd, and Nancy Bode, “Sentencing the White Col-

lar Offender: Rhetoric and Reality,” American Sociological Review 47 (1982): 641–659.

oped is reproduced in Table 18.7.

C O M P A R I N G L O G I S T I C R E G R E S S I O N C O E F F I C I E N T S 577

Selected Probability Estimates and Calculated Range for Significant Variables in Wheeler, Weisburd, and Bode’s Study of White-Collar-Crime Sentencing

PROBABILITY PROBABILTY

INDEPENDENT VARIABLES ESTIMATESa RANGEb INDEPENDENT VARIABLES ESTIMATESa RANGEb

I. Act-Related Variables 1) Maximum Exposure to

Prison 44 1 day–1 year 32 1 year & 1 day–2 years 35 4 years & 1 day–5 years 45 14 years & 1 day–15 years 76

2) Dollar Victimization 41 $101–$500 27 $2,501–$5,000 38 $10,001–$25,000 47 $25,001–$100,000 51 over $2,500,000 68

3) Complexity/Sophistication 27 4 32 6 38 8 45

10 52 12 59

4) Spread of Illegality 21 Individual 40 Local 47 Regional 54 National/International 61

II. Actor-Related Variables 7) Social Background:

Duncan S.E.I. 29 15.1 28 49.4 41 62.0 47 66.1 49 84.0 57

8) Social Background: Impeccability 17

7 54 11 49 14 45 17 42 21 37

9) Criminal Background: Number of Arrests 22

0 37 1 43 2 45 5 51 9 59

10) Criminal Background: Most Serious Prior Conviction 20

None 37 Minor Offense 46 Low Felony 52 Moderate Felony 57

13) Role in Offense 24 Minor 26 Missing 33 Single/Primary 50

III. Legal Process Variables 16) Statutory Offense Category 39

Antitrust Violations 28 Bribery 30 Bank Embezzlement 36 False Claims 36 Postal Fraud 38 Lending/Credit Fraud 45 SEC Violations 65 Tax Violations 69

IV. Other Variables 30 17) Sex

Male 50 Female 20

18) Age ——c

22 42 30 48 39 50 48 46 61 32

21) District 28 Northern Georgia 34 Southern New York 34 Central California 43 Western Washington 43 Maryland 50 Northern Illinois 53 Northern Texas 62

aEstimated likelihood of imprisonment when scores on all other variables are held at their mean. bRange computed from 5th to 95th percentile score. cBecause of the curvilinear effect measured here, the range is not relevant.

Table 18.7

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N578

we can see that a person playing a minor role in an offense had a pre-

dicted probability of imprisonment of about 26%, while someone playing

a primary role had a 50% likelihood of imprisonment, according to the

model estimated (and holding all other independent variables constant at

their mean). A crime involving less than $500 in victimization led to an

estimated likelihood of imprisonment of 27%. A crime netting over

$2,500,000 led to an estimated likelihood of imprisonment of 68%. If we

compare the range of predicted values between the 5th and 95th per-

centile scores for each variable, our calculation suggests that dollar vic-

timization (with a range of 41%) has a much larger impact than role in

an offense (with a range of 24%). Of course, the choice of the 5th and

95th percentiles is arbitrary. And this method also arbitrarily holds every

other independent variable to its mean. Nonetheless, the advantage of

this approach is that it provides a method of comparison that is straight-

forward and easy for the nonstatistician to understand.

To apply this method to our data, we need information on the mean

for each independent variable. For our data, the means are

Northeast: 0.225

South: 0.373

West: 0.229

Number of sworn officers: 334.784

scribes the results. Using this table, we can see that there are very large

differences in the predicted probabilities of a Compstat-like program for

departments of varying size. This illustrates a point made earlier, when

we noted that the odds ratio for each change in number of sworn officers

was small. Though the change per unit change in X is small in this case

(because departments differ widely in size), the predicted change can be

very large. Under this approach, the range variable suggests a larger im-

pact for number of sworn officers than for region of country.

Table of Selected Probability Estimates and Range for the Compstat Model

VARIABLE PROBABILITY ESTIMATE RANGE

Number of sworn officers: 100 (5th percentile) 0.2468 0.53 500 0.4217 1,300 (95th percentile) 0.7831

Northeast 0.4090 ——

South 0.4647 ——

West 0.4216 ——

Table 18.8

579C O M P A R I N G L O G I S T I C R E G R E S S I O N C O E F F I C I E N T S

to the method employed by Wheeler, Weisburd, and Bode. Table 18.8 de-

In the box on pages 580 581, the calculations are carried out according and

“Standardized” Logistic Regression Coefficients

Some statistical software programs list the standardized logistic regres-

sion coefficient Beta, which is analogous to the standardized regression

coefficient. Like the standardized regression coefficient, the standardized

logistic regression coefficient can be interpreted relative to changes

For all of the following calculations,

P (Y � 1) � 1

1 � e�[�1.3159 � (0.002)(1300)] �

1

1 � e�(1.2841) � 0.7831

P (Y � 1) � 1

1 � e�[�1.3159 � (0.002)(500)] �

1

1 � e�(�0.3159) � 0.4217

P (Y � 1) � 1

1 � e�[�1.3159 � (0.002)(100)] �

1

1 � e�(�1.1159) � 0.2468

P (Y � 1) � 1

1 � e�(�1.3159�0.002X1)

� �1.3159 � 0.002X1

� �1.795 � 0.002X1 � 0.0808 � 0.3003 � 0.0980

� �1.795 � 0.002X1 � 0.359(0.225) � 0.805(0.373) � 0.428(0.229)

Xb � b0 � b1 X1 � b2 X2 � b3 X3 � b4 X4

X 4 � West

X 3 � South

X2 � Northeast

X1 � number of sworn officers

Calculating Selected Probability Estimates and Range for the Compstat Model

Probability estimate for number of sworn officers:

P (Y � 1) for 100 officers:

P (Y � 1) for number of sworn officers:

P (Y � 1) for 500 officers:

P (Y � 1) for 1,300 officers:

(measured in standard deviation units) in the independent variable. The

magnitude of the standardized logistic regression coefficient allows us to

compare the relative influence of the independent variables, since a

larger value for the standardized coefficient means that a greater change

in the log of the odds is expected. In contrast to the standardized regres-

Probability estimate for Northeast:

Probability estimate for South:

Probability estimate for West:

P (Y � 1) � 1

1 � e�(�0.7443 � 0.428X4) �

1

1 � e 0.3163 � 0.4216

P (Y � 1) for West:

� �0.7443 � 0.428X4

� �1.795 � 0.6696 � 0.0808 � 0.3003 � 0.428X4

� �1.795 � 0.002(334.784) � 0.359(0.225) � 0.805(0.373) � 0.428X4

Xb � b0 � b1X1 � b2 X2 � b3 X 3 � b4 X 4

P (Y � 1) � 1

1 � e�(�0.9466 � 0.805X3) �

1

1 � e 0.1416 � 0.4647

P (Y � 1) for South:

� �0.9466 � 0.805X 3

� �1.795 � 0.6696 � 0.0808 � 0.805X3 � 0.0980

� �1.795 � 0.002(334.784) � 0.359(0.225) � 0.805X3 � 0.428(0.229)

Xb � b0 � b1X1 � b2 X2 � b3 X 3 � b4 X 4

P (Y � 1) � 1

1 � e�(�0.7271 � 0.359X2) �

1

1 �e 0.3681 � 0.4090

P (Y � 1) for Northeast:

� �0.7271 � 0.359X2

� �1.795 � 0.6696 � 0.359X2 � 0.3003 � 0.0980

� �1.795 � 0.002(334.784) � 0.359X2 � 0.805(0.373) � 0.428(0.229)

Xb � b0 � b1X1 � b2 X2 � b3 X3 � b4X 4

gistic regression models does not fall between 0 and 1, but can take on

any value.8 Some statisticians warn that such coefficients should be inter-

preted with caution.9 Nonetheless, they can provide a method for gain-

ing a general sense of the strength of coefficients in logistic regression.

The standardized logistic regression coefficient is calculated using

where bi is the unstandardized coefficient for variable i from the origi-

nal logistic regression model and si is the standard deviation for vari-

able i. We interpret Beta as the change in the log of the odds of

P (Y � 1) relative to changes (measured in standard deviation units)

in the independent variable. For example, a Beta of 0.4 implies that

for a one-standard-deviation change in the independent variable,

the log of the odds is expected to increase by 0.4. Alternatively, if

Beta � �0.9, a one-standard-deviation change in the independent

variable is expected to result in a decrease of 0.9 in the log of the

odds that P (Y � 1).

Returning to our example using the Compstat data, we find the stan-

dardized coefficient for number of sworn officers to be 0.6616.

Betai � bisi

8Some researchers have proposed alternative ways of calculating standardized logistic re-

gression coefficients that allow for interpretations related to changes in probabilities. See,

for example, Robert L. Kaufman, “Comparing Effects in Dichotomous Logistic Regression:

A Variety of Standardized Coefficients,” Social Science Quarterly 77 (1996): 90–109. 9For example, see Andy Field, Discovering Statistics Using SPSS for Windows (London:

Sage Publications, 2000).

W orking It Out

� 0.6616

� (0.002)(330.797)

Beta � bisi

model and present the accompanying unstandardized logistic regression

coefficients. Though the unstandardized logistic regression coefficient for

the South (0.805) seems very large relative to that for number of sworn

Equation 18.7:

Equation 18.7

In Table 18.9, we calculate Beta for all four of the coefficients in the

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N

sion coefficients for linear regression models, the Beta calculated for log-

582

officers (0.002), the reverse relationship is found when we look at the

standardized coefficients. The standardized coefficient for number of

sworn officers is 0.662, while that for the South is 0.390. This is consis-

tent with our analysis of the probability estimates. Both of these methods

take into account the fact that the scales of measurement for these mea-

sures differ widely. While you should use caution in relying on standard-

coefficients within regression models.10

E v a l u a t i n g t h e L o g i s t i c R e g r e s s i o n M o d e l

In OLS regression, to assess how well our model explains the data, we

use a straightforward measure of the percent of variance explained be- 2

Nonetheless, a number of measures have been proposed for assessing

how well a model predicts the data.

Percent of Correct Predictions

One widely accepted method for assessing how well a logistic regression

model predicts the dependent variable is to compare the values of Y pre-

dicted by the model to those that would be obtained simply by taking

the observed distribution of the dependent variable. This statistic is com-

Beta and Associated Logistic Regression Coefficients for the Compstat Model

STANDARD

VARIABLE DEVIATION b BETA

Number of sworn officers 330.797 0.002 (330.797)(0.002) � 0.6616

Northeast 0.416 0.359 (0.416)(0.359) � 0.1493

South 0.484 0.805 (0.484)(0.805) � 0.3896

West 0.421 0.428 (0.421)(0.428) � 0.1802

10As with standardized regression coefficients in OLS regression, you should not com-

pare standardized logistic regression coefficients across models. Moreover, while we

report standardized regression coefficients for the dummy variables included in the

model, you should use caution in interpreting standardized coefficients for dummy

Table 18.9

provide a general yardstick for comparing the relative strength of

ized logistic regression coefficients, here, as in other cases, they can

). There is no equivalent statistic in logistic regression.yond the mean (R

monly described as the percent of correct predictions. Table 18.10

E V A L U A T I N G T H E L O G I S T I C R E G R E S S I O N M O D E L

variables. See Chapter 16, pages 493–494, for a discussion of this problem.

583

shows the percent of correct predictions for our Compstat example. The

Percent of correct predictions �

The observed predictions in our example represent the observed pro-

portion of departments that report having a Compstat-like program. As

we noted before, this number is about 0.33 (or 33%). We add the 106

vide this number by the total number of cases in the analysis (N � 419).

order to compare these predicted values with the observed values, we

must assign each case a 0 or a 1. The decision as to whether to define

the predicted value as a 1 or a 0 is based on a set cut-off point. Herein

lies the main drawback of this approach: The point at which you deter-

mine that the prediction is a 1 is arbitrary. In SPSS, as in other standard

software packages, 0.50 is used as a natural cut-off point. That is, if we

get a predicted probability of 0.50 or greater for a case in our study, it

will be counted as a prediction of 1. Remember that a 1 in our case

means that the department has a Compstat-like program. In this analysis,

if the prediction is 0.495, the case is given a 0. Clearly, by using a single

and arbitrary cut-off point, we are losing a good deal of information

about how well the model fits the data.

The proportion of correct predictions is worked out below, using

for which the actual and predicted values are the same. In 264 cases, the

�Ncorrrect predictions

Ntotal � � 100

Percent of Correct Predictions for the Logistic Regression of the Compstat Data

Classification Table

Predicted

COMPSTAT Percentage Correct

Observed .00 1.00

Step 1 COMPSTAT .00 264 19 93.3

1.00 106 30 22.1

Overall 70.2

a The cut value is .500.

Table 18.10

formula for percent of correct predictions is presented in Equation 18.8.

Equation 18.8

and 30 cases in Table 18.7 where the observed value is 1 and then di-

The predicted values are drawn from Equation 18.3. But, importantly, in

Equation 18.8. The N of correct predictions is found by taking each case

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N584

actual and predicted values are 0. In only 30 cases are the actual and

predicted values equal to 1. These are the correct predictions in our

analysis, so the total number of correct predictions is 264 � 30 � 294.

The percent of correct predictions is 70.2. This seems like a very high

level of prediction. However, to interpret this statistic, we must compare

it with the level we would have reached if we had not used our regres-

sion model. In that case, we would have had information only on the

split in the dependent variable. As noted earlier, 33% of the departments

claim to have implemented a Compstat-like program. Knowing only this,

our best bet would have been to predict that every department did not

have a Compstat-like program. If we did this, we would be correct about

67% of the time. Thus, our model did not improve our prediction very

much over what we would have predicted with knowledge of only the

outcomes of the dependent variable.

W orking It Out

� 70.17

� (0.7017) � 100

� �294

419� � 100

Percent of correct predictions � �Ncorrect predictions

Ntotal � � 100

Pseudo R 2

While there is no direct R2 measure for logistic regression, a number of

what may be termed pseudo R2 measures have been proposed. Like stan-

dardized logistic regression coefficients, these measures are not well ac-

cepted and must be used with caution. Nonetheless, by providing a general

sense of the prediction level of a model, they can add information to other

statistics, such as the percent of correct predictions. A commonly used

pseudo R2 measure is Cox and Snell’s R2.11 As with other pseudo R2 sta-

tistics, a main component of this measure is the log likelihood function

(�2LL). It makes good sense to rely on the log likelihood function, since it

measures the degree to which a proposed model predicts the data exam-

ined. In this case, we compare the difference between the �2LL estimate

11D. R. Cox and E. J. Snell, The Analysis of Binary Data, 2nd ed. (London: Chapman

and Hall, 1989).

585E V A L U A T I N G T H E L O G I S T I C R E G R E S S I O N M O D E L

obtained when no independent variables are included (the null model) and

the �2LL estimate obtained when all the independent variables are in-

cluded (the full model). The �2LL value for the null model (528.171) is

vides the method of calculation for Cox and Snell’s R2.

While this equation looks intimidating, it can be solved in two easy

steps. First, we calculate the number that appears above e, or the expo-

nent of the natural log:

R

2 � 1 � e�[(�2LLnull model) � (�2LLfull model)]/N

Pseudo R2 Statistics as Reported in SPSS

Model Summary

Step �2 Log Cox & Snell R Nagelkerke R likelihood Square Square

1 492.513 .082 .114

W orking It Out Step 1

� �0.085

� �35.658/419

� �[(528.171) � (492.513)]/419

�[(�2LLnull model) � (�2LLfull model)]/N

We then take e to the power of �0.085, which, as we noted earlier, can

be done simply on a basic scientific calculator. We next subtract this

number from 1:

W orking It Out Step 2

� 0.0816

� 1 � 0.9185

Cox and Snell’s R 2 � 1 � e�0.085

Table 18.11

given in Table 18.4. The �2LL value for the full model (492.513) is given in

the model summary statistics provided in Table 18.11. Equation 18.9 pro-

Equation 18.9

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N586

Rounding 0.0816 to three decimal places gives a result of 0.082, which is

identical to that produced in the SPSS printout.

Like the percent of correct predictions, Cox and Snells R 2 suggests

that our model does not provide for a very strong level of prediction.

SPSS produces another R 2 statistic: the Nagelkerke R 2. This statistic cor-

rects for the fact that Cox and Snell’s estimate, as well as many other

pseudo R 2 statistics, often have a maximum value of less than 1 (which

2 2

12 Other pseudo

R 2

these values should be seen as an exact representation of the percent of

variance explained in your model. But they can give you a rough sense

of how well your model predicts the outcome measure.

S t a t i s t i c a l S i g n i f i c a n c e i n L o g i s t i c R e g r e s s i o n

Statistical significance for a logistic regression can be interpreted in much

the same way as it was for the regression models discussed in Chapters

15 and 16. However, a chi-square distribution is used, and thus we do

not have to be concerned with assumptions regarding the population

distribution in our tests. For the overall model, there is a general test,

based on the difference between the �2LL statistics for the full and null

models. The chi-square formula for the overall model in logistic regres-

Model chi-square � (�2LLnull model) � (�2LLfull model)

next page). The number of degrees of freedom is determined by the

number of independent variables included in the model estimated. In

our case, there are three regression coefficients for the variable region

and the measure number of sworn officers. The number of degrees of

freedom thus equals 4. Looking at Appendix 2, we can see that a chi-

square statistic of greater than 18.465 is needed for a statistically signifi-

cant result at the 0.001 level. Because our chi-square statistic is much

larger than this, our observed significance level is less than 0.001. Using

12See N. J. D. Nagelkerke, “A Note on a General Definition of the Coefficient of Deter-

mination, Biometrika 78 (1991): 691–692.

erke’s R is thus generally larger than Cox and Snell’s R , which—

especially with large values—will be too conservative.

explained by the independent variables included in the model). Nagelk-

would indicate that all of the variance in the dependent variable was

statistics will give estimates similar to those produced here. None of

sion is represented in Equation 18.10.

Equation 18.10

S T A T I S T I C A L S I G N I F I C A N C E I N L O G I S T I C R E G R E S S I O N 587

For our example, the model chi-square is 35.658 (see working it out,

conventional significance criteria, we would reject the null hypothesis

and conclude that the model estimated provides significant improvement

over that without any independent variables.

W orking It Out

� 35.658

� 528.171 � 492.513

Model chi-square � (�2LLnull model) � (�2LLfull model)

In testing the statistical significance of individual parameters, statistical

This statistic

also has a chi-square distribution, and so the statistical significance of a

result may be checked in a chi-square table. The Wald statistic takes the

ratio of the logistic regression coefficient to its standard error (see Equa-

the South and North Central regions (the latter being the excluded cate-

gory), we take the logistic regression coefficient of 0.805 and divide it by

the reported standard error of 0.332. To get the Wald statistic, we square

this number. The result is 5.879.

W

2 � � b SEb

�2

13

W orking It Out South Region Measure

� 5.879

� �0.805 0.332�

2

W

2 � � b SEb

�2

tion 18.11). The standard error of the logistic regression coefficient is

provided in the SPSS printout (see Table 18.12). For the comparison of

Equation 18.11

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N

13software packages ordinarily provide the Wald statistic.

14

error.

We discuss the Wald statistic in detail here, because it is the most common test of

14

chapter and in more detail in Chapter 19 offers an alternative test for statistical significance

for Categorical and Limited Dependent Variables (Thousand Oaks, CA, Sage, 1997). that is appropriate to both small and large samples (see J. Scott Long, Regression Models

The difference between our result and that shown in Table 18.12 is due to rounding

statistical significance reported in many statistical software applications. it should be noted,

however, that some researchers have noted that the Wald statistic is sensitive to small sample sizes (e.g., less than 100 cases). The likelihood-ratio test discussed later in this

588

To determine whether this coefficient is statistically significant, we can

refer to the chi-square table for 1 degree of freedom. The number of de-

grees of freedom for an individual variable in a logistic regression will al-

ways be 1. Looking at Appendix 2, we see that a chi-square of 10.827 is

required for a result to be statistically significant at the 0.001 level. A chi-

square of 6.635 is required at the 0.01 level, and a chi-square of 5.412 at

the 0.02 level. Our observed significance level can therefore be defined

as falling between 0.01 and 0.02. The SPSS printout gives the exact ob-

served significance level as 0.015. Using conventional levels of statistical

significance, we would conclude that we can reject the null hypothesis

that there is no difference in the reported implementation of Compstat-

like programs in the South versus the North Central region.

that the number of sworn officers is also statistically significant—in this

case, at greater than the 0.001 level. It is important to note that the statistics

reported in this table, as well as in most statistical software, are for two-

tailed significance tests. We mentioned at the outset that there was a strong

hypothesis that larger departments are more likely to report a Compstat-like

program. If we wanted to use a directional test of statistical significance, we

would simply divide the observed significance level in our test by 2.

Looking at the other region dummy variables, we can see that there is

not a statistically significant difference between the Northeast and North

Central regions or between the West and North Central regions. But, as

noted in Chapter 16, it is important to ask whether the variable region

overall contributes significantly to the regression. To test this hypothesis,

we can conduct a likelihood ratio chi-square test, which compares the

SPSS Printout with B, SE of B, and Wald Statistics

Variables in the equation

B S.E. Wald df Sig. Exp(B)

Step 1 NORTHEAST .359 .372 .931 1 .335 1.432

SOUTH .805 .332 5.883 1 .015 2.237

WEST .428 .367 1.360 1 .244 1.534

#SWORN .002 .000 24.842 1 .000 1.002

Constant �1.795 .311 33.378 1 .000 .166

a Variable(s) entered on step 1: NORTHEAST, SOUTH, WEST, #SWORN.

Looking at the significance statistics column in Table 18.12, we can see

This printout is identical to that in Table 18.5. It is reproduced here for easy reference

as you work through the computations presented in this section.

Table 18.12

589

15

15

S T A T I S T I C A L S I G N I F I C A N C E I N L O G I S T I C R E G R E S S I O N

log likelihood function of the model with the multicategory nominal vari-

able (the full model) with the log likelihood function of the model without

details the likelihood ratio chi-square test. The number of degrees of free-

dom is defined as the number of dummy variables added by the multicate-

gory nominal variable. In our case, it is 3 for the three included regions.

Likelihood ratio chi-square test � (�2LLreduced model) � (�2LLfull model)

We can get the statistics for the test by running two separate regres-

sions. The reduced model regression excludes the dummy variables as-

sociated with region. The �2LL for this model is shown in the model

the full model—the model we have been using throughout the chapter,

with the region dummy variables included. The model statistics were re-

Below, we work out the likelihood ratio chi-square using these two

estimates. The likelihood ratio chi-square for the region variable is 6.934,

Compstat-like program at the 0.05 significance threshold. Because our

chi-square statistic is smaller than this number, we cannot conclude that

there is overall a statistically significant relationship between region and

claimed development of a Compstat-like program.

W orking It Out

� 6.934

� 499.447 � 492.513

Likelihood ratio chi-square test � (�2LLreduced model) � (�2LLfull model)

Model Summary for the Reduced Model

Model Summary

Step �2 Log Cox & Snell R Nagelkerke R likelihood Square Square

1 499.447 .066 .092

Table 18.13

the multicategory nominal variable (the reduced model). Equation 18.12

Equation 18.12

summary from an SPSS printout in Table 18.13. The second regression is

ported in Table 18.11.

with 3 degrees of freedom (the number of dummy variable measures

included in the model). Looking at Appendix 2, we can see that with 3

degrees of freedom, a chi-square of 7.815 would be needed to reject

the null hypothesis of no relationship between region and a reported

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N590

C h a p t e r S u m m a r y

Ordinary least squares regression is not an appropriate tool for analyzing

a problem in which the dependent variable is dichotomous. In such

cases, OLS regression is likely to predict values that are greater than 1

and less than 0 and thus outside the observed distribution of Y. Using

the OLS approach in this case will also lead to violations of parametric

assumptions required for associated statistical tests. Logistic regression

analysis uses a logistic model curve, rather than a straight line, to pre-

dict outcomes for Y in the case of a dichotomous dependent variable.

This constrains predictions to between 0 and 1.

While the logistic model curve provides a solution to predictions be-

yond the observed distribution, the outcome variable is transformed into

the natural logarithm of the odds of Y, or the logit of Y. Through use

of the cumulative logistic probability function, the logistic regression

equation may be used to predict the likelihood of Y occurring. Maxi-

mum likelihood techniques are used to estimate the coefficients in a

logistic regression analysis. In this approach, we begin by identifying a

tentative solution, which we then try to improve upon. Our criterion for

improvement is termed the log likelihood function (�2LL). We repeat

this process again and again until the change in the likelihood function

is considered negligible. Each time we repeat the process and reestimate

our coefficients is called an iteration. Lack of convergence in a stan-

dard number of iterations indicates some type of problem in the regres-

sion model that is being estimated.

The multivariate logistic regression coefficient, b, may be inter-

preted as the increase in the log of the odds of Y associated with a one-

unit increase in X (with all other independent variables in the model

held constant). The odds ratio, or Exp(B), and the derivative at

mean, DM, provide more easily interpreted representations of the logis-

change in Y associated with a unit change in X. The DM will change de-

pending on the mean value of Y in the problem examined.

There is no widely accepted method for comparing logistic regression

coefficients measured on different scales. One method is to compare

probability estimates at selected intervals. Standardized regression coeffi-

cients have been suggested for logistic regression, though they should be

interpreted with caution. There is no single widely accepted statistic for

assessing how well the logistic regression model predicts the observed

data. An approach commonly used is to calculate the percent of cor-

rect predictions. This method establishes an arbitrary decision point

regression coefficient, the derivative at mean may be interpreted as the

tic regression coefficient. The odds ratio represents the impact of a one-

unit change in X on the ratio of the probability of Y. Like an ordinary

C H A P T E R S U M M A R Y 591

(usually 0.50) for deciding when a predicted value should be set at 1.

These predictions are then compared to the observed data. Pseudo R2

statistics have also been developed, though they remain a subject of

debate.

Statistical significance for the overall logistic regression model is as-

sessed through computation of the model chi-square. Statistical signifi-

cance for individual regression coefficients is evaluated with the Wald

statistic. A likelihood ratio chi-square test can be used to calculate

the statistical significance of a multicategory nominal variable.

K e y T e r m s

Cox and Snell’s R2 A commonly used

pseudo R 2 measure whose main compo-

nent, as in other pseudo R2 statistics, is the

log likelihood function (�2LL).

cumulative logistic probability func-

tion A transformation of the logistic

probability function that allows

computation of the probability that

Y will occur, given a certain combination

of characteristics of the independent

variables.

derivative at mean (DM) A measure that

converts the nonlinear logistic regression

coefficient to a simple linear regression co-

efficient, which may be interpreted as the

change in Y associated with a unit change

in X.

iteration Each time we identify another

tentative solution and reestimate our logis-

tic regression coefficients.

lack of convergence Failure of a logistic

regression analysis to reach a result that

meets the criterion of reduction in the log

likelihood function.

likelihood ratio chi-square test A test

for statistical significance that allows the re-

searcher to examine whether a subset of

independent variables in a logistic regres-

sion is statistically significant. It compares

�2LL for a full model to �2LL for a re-

duced model.

log likelihood function A measure of the

probability of observing the results in the

sample, given the coefficient estimates in

the model. In logistic regression, the log

likelihood function (�2LL) is defined as �2

times the natural logarithm of the likeli-

hood function.

logarithm The power to which a fixed

number (the base) must be raised to pro-

duce another number.

logistic model curve The form of the pre-

dicted outcomes of a logistic regression

analysis. Shaped like an S, the logistic

curve begins to flatten as it approaches

never actually reaches—either of these two

values.

logistic regression analysis A type of re-

gression analysis that allows the researcher

to make predictions about dichotomous de-

pendent variables in terms of the log of the

odds of Y.

logistic regression coefficient The coef-

ficient b produced in a logistic regression

0 or 1, so it keeps coming closer to—but

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N

analysis. It may be interpreted as the

592

S y m b o l s a n d F o r m u l a s

e Base of the natural logarithm

ln Natural logarithm

�2LL �2 times the log likelihood function

The natural logarithm of the odds of P(Y � 1) to P(Y � 0):

ln � P(Y � 1)

1 � P(Y � 1)� � ln �P(Y � 1)

P(Y � 0)� � b0 � b1X1

ated with a one-unit increase in X.

maximum likelihood estimation A tech-

nique for estimating the parameters or co-

efficients of a model that maximizes the

probability that the estimates obtained will

produce a distribution similar to that of the

observed data.

model chi-square The statistical test used

to assess the statistical significance of the

overall logistic regression model. It com-

pares the �2LL for the full model with the

�2LL calculated without any independent

variables included.

Nagelkerke R2 A pseudo R 2 statistic that

corrects for the fact that Cox and Snell’s es-

timates, as well as many other pseudo R 2

statistics, often have a maximum value of

less than 1.

natural logarithm of the odds of Y (logit

of Y) The outcome predicted in a logistic

regression analysis.

odds ratio [Exp(B)] A statistic used to in-

terpret the logistic regression coefficient. It

represents the impact of a one-unit change

in X on the ratio of the probability of Y.

percent of correct predictions A

statistic used to assess how well a logistic

regression model explains the observed

data. An arbitrary decision point (usually

0.50) is established for deciding when a

predicted value should be set at 1, and

then the predictions are compared to the

observed data.

pseudo R2 The term generally used for a

group of measures used in logistic regres-

sion to create an approximation of the OLS

regression R 2. They are generally based on

comparisons of �2LL for a full model and

a null model (without any independent

variables).

standardized logistic regression

coefficient A statistic used to compare

logistic regression coefficients that use

different scales of measurement. It is

meant to approximate Beta, the

standardized regression coefficient in

OLS regression.

Wald statistic A statistic used to assess the

statistical significance of coefficients in a

logistic regression model.

S Y M B O L S A N D F O R M U L A S 593

change in the log of the odds of Y associ-

To calculate the probability that Y � 1:

To calculate the odds ratio for P(Y � 1), given a one-unit change in the

independent variable X:

� Exp(B)

To calculate the derivative at mean:

To calculate the standardized logistic regression coefficient:

Betai = bisi

To calculate the percent of correct predictions:

Percent correct �

To calculate Cox and Snell’s R 2:

To calculate the model chi-square:

Model chi-square � (�2LLnull model) � (�2LLfull model)

To calculate the Wald statistic:

To calculate the likelihood ratio chi-square statistic for a subset of

independent variables:

Likelihood ratio chi-square test � (�2LLreduced model) � (�2LLfull model)

W 2 � � b SEb

�2

R 2 � 1 � e�[(�2LLnull model) � (�2LLfull model)]/N

�Ncorrrect predictions

Ntotal � � 100

DM � Y (1 � Y )bi

Odds ratio � P(Y � 1)

1 � P(Y � 1) X

P(Y � 1)

1 � P(Y � 1) X�1

P(Y � 1) � 1

1 � e�Xb

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N594

E x e r c i s e s

As part of a research project for a class, a student analyzed data on a sample of adults who had been asked about their decision to report being assaulted to the police. Their decision was coded as 1 � assault

age (in years), sex (0 � male, 1 � female), and race (0 � white, 1 � nonwhite). The student reported the regression results as

Variable b

Age 0.01

Sex 0.5

Race �0.2

Constant �0.1

a. Calculate the predicted values for each of the following persons:

— A 65-year-old white female

— A 25-year-old nonwhite male

— A 40-year-old white male

— A 30-year-old nonwhite female

b. Should any of the student’s predicted values lead the student to question the use of ordinary least squares regression? Explain why.

A research institute concerned with raising public attention about the use of force by school children calculates the following effects on the likeli- hood of hitting another child at school, using logistic regression analysis:

Variable b

Sex (0 � girl, 1 � boy) 0.7

Grade in school �0.1

Constant �0.4

Hitting another child was coded as 1; no hitting was coded as 0.

a. Interpret the effects of sex and grade in school on the log of the odds that P(Y � 1).

b. Calculate and interpret the odds ratios for the effects of sex and grade in school on use of force.

Supervision of defendants on pretrial release is thought to reduce the chance that defendants will flee the community. A government agency funds a small study to examine whether supervision affects pretrial

E X E R C I S E S

student used ordinary least squares regression to estimate the effects of reported to the police, 0 � assault not reported to the police. The

595

18.1

18.2

18.3

flight (flight � 1, no flight � 0) and reports the following logistic regression results:

Variable b Standard Error of b

Age (years) �0.01 0.02

Sex ( 1 � male, 0.67 0.25

0 � female)

Severity of offense 0.21 0.03

scale (0 to 10)

Number of prior 0.35 0.09

felony convictions

Number of contacts with �0.13 0.03

supervision caseworker

Constant �0.52

a. Calculate and interpret the odds ratio for each of the independent variables.

b. Can the government agency conclude that supervision in the form of contact with a caseworker has a statistically significant effect on pretrial flight? Explain why.

c. If the agency reports that the �2LLnull model is 653.2 and the �2LLfull model is 597.6, can it conclude that the model is statistically significant? Explain why.

A survey of adolescents indicated that 17% had used marijuana in the last year. In addition to standard demographic predictors of drug use, a researcher expects that school performance also affects the likeli- hood of marijuana use. The researcher’s table of results follows.

Standard Standard

Variable Mean Deviation b Error of b

Age (years) 14.6 3.1 0.07 0.03

Sex (1 � male, 0 � female) 0.55 0.50 0.36 0.15

Race (1 � white, 0 � nonwhite) 0.75 0.43 �0.42 0.30

Grade point average 2.76 1.98 �0.89 0.24

Think of self as a good 0.59 0.49 �0.65 0.33

student (1 � yes, 0 � no)

Constant �0.87

a. Calculate the predicted probability of marijuana use for each of the following persons:

— A 14-year-old white male who does not think of himself as a good student and has a GPA of 3.07.

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N

18.4

596

— A 17-year-old nonwhite female who thinks of herself as a good student and has a GPA of 3.22.

— A 15-year-old white female who thinks of herself as a good stu- dent and has a GPA of 2.53.

b. Calculate the standardized coefficient for each of the independent variables in the model. Which variable appears to have the largest effect on marijuana use?

c. Calculate the derivative at mean for each of the independent vari-

d. Compare your answers for parts b and c. How do you explain this pattern?

After losing a court battle over a requirement that it reduce its jail population, a county conducted an analysis to predict which offenders would pose the greatest threat of committing a violent offense if re- leased early. A random sample of 500 inmates released from the jail in the last three years was analyzed to see what factors predicted arrest for a violent crime in the 12 months after release. For the final model, which included five predictors of violent arrest, the county reported the following statistics:

Predicted Predicted

No Violent Arrest Violent Arrest

Observed No Violent Arrest 439 19

Observed Violent Arrest 27 15

a. Calculate the percent correctly predicted for this model. What does this statistic indicate about the county’s prediction model?

b. Calculate the model chi-square for this model. Interpret this statistic.

c. Calculate Cox and Snell’s R 2 for this model. Interpret this statistic.

d. How do you explain the difference in the results for parts a through c?

Hopeful that media attention to wrongful convictions has increased public opinion in favor of abolishing the death penalty, an abolitionist organization conducts a study to assess public support for abolishing the death penalty. Overall, the organization finds that 35% would sup- port abolishing the death penalty if offenders could be sentenced to life without the option of parole (coded as 1 � abolish the death penalty, 0 � do not abolish the death penalty). In a logistic regression

�2LLfull model � 861.3

�2LLnull model � 876.5

effect on marijuana use? ables in the model. Which variable appears to have the largest

597

18.5

18.6

E X E R C I S E S

model examining the effects of respondent characteristics on support, the organization finds the following:

Standard Standard

Variable Mean Deviation b Error of b

Age (years) 41.2 15.4 �0.01 0.01

Sex (1 � male, 0 � female) 0.44 0.50 �0.42 0.19

Race (1 � white, 0 � nonwhite) 0.76 0.43 �0.24 0.09

Political conservative 0.33 0.47 �1.12 0.22

(1 � yes, 0 � no)

Region of Country:

South 0.23 0.42 �0.19 0.11

West 0.31 0.46 �0.09 0.04

North 0.27 0.44 0.27 0.12

(omitted � Central)

Constant 0.11

a. Which variable has a relatively greater impact on support for abol- ishing the death penalty? Explain why.

b. If �2LLreduced model � 376.19 and �2LLfull model � 364.72 when region variables are omitted from the analysis, do the region variables have a statistically significant effect on support for abolishing the death penalty?

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N

C o m p u t e r E x e r c i s e s

In both SPSS and Stata, the general format for many of the multivariate statistical

models is much the same. So, while the models may become increasingly

complex, the syntax necessary to run these models is oftentimes very

similar. We have included examples in both the SPSS (Chapter_18.sps) and Stata

(Chapter_18.do) syntax files.

SPSS

Logistic regression analyses are performed in SPSS with the LOGISTIC

REGRESSION command:

The structure to this command is identical to the REGRESSION command

discussed in previous chapters. Much of the output from running this command

has been discussed in this chapter. The difference between the linear regression

command and the logistic regression command is that the output from a logistic

regression presents information for the model that includes only the intercept

and is labeled “Block 0” in the SPSS output. The next section of output is

598

LOGISTIC REGRESSION VARIABLES dep_var_name

/METHOD = ENTER list_of_indep_vars.

C O M P U T E R E X E R C I S E S 599

labeled “Block 1,” and it contains the results discussed above: Omnibus Tests of

Model Coefficients, Model Summary, Classification Table, and Variables in the

Equation.

It is possible to have SPSS calculate the predicted probability and residual for

each observation in the data file. To obtain one or both of these values, insert

the /SAVE PRED RESID line, just as in the linear regression command—the

naming convention in the LOGISTIC REGRESSION is the same as that used

in the REGRESSION command.

Stata

Logistic regression analyses are performed in Stata with the logit command:

The output from running the logit command parallels that in Stata’s regress

command. The model summary results appear at the top of the output, followed

by a table that presents the coefficients, their standard errors, z-scores, and

confidence intervals (we discuss these in Chapter 20). Note that rather than the

Wald statistic for each variable, Stata computes the z-score. To obtain the Wald

statistic, simply square the z-score:

which will result in the same substantive conclusion (i.e., a significant Wald statis-

tic will also be a significant z-score).

To obtain the odds ratios, we need to add the option or to the logit

command:

Note that the table of coefficients will report the odds ratios rather than the

coefficients interpretable as the log of the odds.

In the model summary statistics, Stata reports the test for the overall model

as “LR chi2(#)” and what is labeled as “Pseudo R2.” Recall that in a chi-square

test in Stata, the degrees of freedom associated with a test are included within the

parentheses following “chi2.” The “Pseudo R2” value is the Cox and Snell R2.

Saving predicted values and residuals following the logit command is

identical to saving these same values after running the regress command:

which will then add two additional variables to the working data file.

Problems

1. Open the Compstat data file (compstat.sav or compstat.dta). These are

the data analyzed in this chapter. Use one of the binary logistic regres-

sion commands with Compstat as the dependent variable and number of

logit dev_var list_of_indep_vars

logit dev_var list_of_indep_vars, or

predict PRE_1

predict RES_1, r

Wald = z2

C H A P T E R E I G H T E E N : L O G I S T I C R E G R E S S I O N

sworn officers, Northeast, South, and West as the independent variables.

Note that the values reported in the software output match those reported

in the text in Tables 18.4, 18.5, 18.10, and 18.11.

2. Open the Pennsylvania Sentencing data file (pcs_98.sav or pcs_98.dta).

Use one of the binary logistic regression commands with incarceration as

the dependent variable and age, race, sex, offense severity score, and prior

criminal history score as the independent variables.

b. Explain the odds ratios for each independent variable in plain English.

c. Interpret the results of the Wald statistic for each of the logistic regres-

d. Interpret the value of Cox and Snell’s R2.

e. Perform a chi-square test for the overall regression model.

3. Open the NYS data file (nys_1.sav, nys_1_student.sav, or nys_1.dta). Using

one of the binary logistic regression commands, run an analysis for each

of the measures of delinquency below. As in the Computer Exercises in

Chapters 16 and 17, you will need to select a set of independent variables

that you think are related to the dependent variable. Note that each of the

delinquency items will need to be recoded as 0 or 1 to represent whether

or not the act was committed (see Chapter_18.sps and Chapter_18.do for

examples). Do the following for each analysis:

Explain the odds ratios in plain English.

Interpret the results of the Wald statistic for each of the logistic

Interpret the value of Cox and Snell’s R2.

Perform a chi-square test for the overall regression model.

a. Number of thefts valued at less than $5 in the last year; convert to

any thefts in the last year.

b. Number of times drunk in the last year; convert to any times drunk

in the last year.

c. Number of times the youth has hit other students in the last year;

convert to any times the youth has hit other students in the last year.

d. Number of times the youth has hit a parent in the last year; convert

to any times the youth has hit a parent in the last year.

600

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_19,

C h a p t e r n i n e t e e n

Multivariate Regression with Multiple Category Nominal or Ordinal Measures: Extending the Basic Logistic Regression Model

© Springer Science+Business Media New York 2014

H o w d o w e a n a l y z e a d e p e n d e n t v a r i a b l e w i t h

m o r e t h a n t w o c a t e g o r i e s ?

M u l t i n o m i a l l o g i s t i c r e g r e s s i o n

How Do We Interpret the Multinomial Logistic Regression Model?

Does the Reference Category Make a Difference?

What is the Test of Statistical Significance for Single Coefficients?

What is the Test of Statistical Significance for Multiple Coefficients?

What are the Practical Limits of the Multinomial Logistic Regression

Model?

O r d i n a l l o g i s t i c r e g r e s s i o n

How Do We Interpret the Ordinal Logistic Regression Model?

How Do We Interpret Cumulative Probabilities?

How are Cumulative Probabilities Related to Odds Ratios?

How Do We Interpret Ordinal Logistic Coefficients?

What is the Test of Statistical Significance for Coefficients?

What are the Parallel Slopes Tests?

How is the Score Test Computed?

How is the Brant Test Computed?

What is the Partial Proportional Odds Model?

How Do We Interpret the Results from the Partial Proportional Odds Model?

IN THE PREVIOUS CHAPTER, we examined how to analyze data in a binary

logistic regression model that included a dependent variable with two cat-

egories. This allowed us to overcome problems associated with using

Ordinary Least Squares Regression in cases where the variable that is

being explained is measured as a simple dichotomy. Accordingly, we

have now described tools that allow the researcher to develop explana-

tory models with either an interval dependent variable or a dichotomous

dependent variable.

But there are many situations in which researchers are faced with ana-

lyzing dependent variables that include more than two categories, or that

are measured on an ordinal scale. For example, we may want to identify

the factors that lead to a dismissal, a guilty plea conviction, or a trial con-

viction in court. The methods we have covered so far do not allow us to

examine this problem in a single multivariate statistical model. We have

also not discussed how a researcher should deal with dependent vari-

ables such as fear of crime, which are measured on an ordinal scale. As

we noted in Chapter 16, the assumptions of Ordinary Least Squares

variables.

regression are not likely to be met when using ordinal scale dependent

602

Fortunately, we can extend our discussion of the logistic regression

model to consider such dependent variables. However, such logistic

regression models need to be modified to take into account these new sit-

uations. In this chapter we provide an introduction to multinomial and

ordinal logistic regression. Though these topics are generally not included

in an introductory statistics text, we think the problems that these

approaches address are becoming very common in criminal justice

research and even for a basic understanding of statistical methods in

criminal justice it is important to be able to understand and apply them.

M U L T I N O M I A L L O G I S T I C R E G R E S S I O N

M u l t i n o m i a l L o g i s t i c R e g r e s s i o n

Multinomial logistic regression is used to examine problems where

there are more than two nominal categories in the dependent variable. We

have already mentioned the case where a researcher wants to explain

why convicted offenders are sentenced to prison, probation or fines, but

there are many situations in criminal justice in which dependent variables

include multiple nominal categories. For example, a researcher may want

to explain why certain offenders tend to specialize in either violent crime,

property crime, or white collar crime. Multinomial regression is particu-

larly useful when researchers create categorizations for groups of offenders

and then want to explain why certain people fall into those groups. This

is common, for example, in recent studies in developmental criminology

where offenders are placed into a small number of groups that evidence

different crime trajectories.1 It is then natural to ask why offenders fall

into those groups. Multinomial regression provides a very useful tool for

conducting multivariate analyses in such situations.

Multinomial regression is conceptually a straightforward extension of

the binary logistic regression model that we discussed in the previous

chapter. Recall that in the binary logistic regression model, we designated

one of the two outcome categories as the presence of a given trait and

the second as the absence of that trait. For example, we compared police

departments that had adopted Compstat (Y=1) versus those that did not

adopt the Compstat program (Y=0). In logistic regression the left side of

the regression equation is the natural logarithm (ln) of the odds of hav-

ing a 1 on the dependent variable (Y=1) as opposed to having a 0 (Y=0).

This transformation, illustrated in Equation 19.1, allowed us to develop a

prediction model in which the predictions of the regression equation are

constrained to fall between 0 and 1. We called this transformation the

logit of Y:

( ) ( )

ln P Y

P Y b b X

0 1

0 1 1 =

= = +e o Equation 19.1

603

What happens when the outcome variable has more than two cate-

Y in Equation 19.1 requires that

there be only the absence (Y=0) or the presence (Y=1) of a trait. This sit-

uation is not appropriate when you have the possibility of the presence of

more than one positive outcome (e.g. dismissal, a guilty plea conviction,

gories? The problem here is that we do not have a simple change in the odds

for one outcome, as we did with the example of Compstat in the prior

chapter. Here we have to take into account changes in the odds in relation

to more than two categories. The logit of

1See, for example, D. Nagin, Group-based Modeling of Development (Cambridge, MA: Harvard University Press, 2005).

C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

or a trial conviction) and you want to distinguish among them. Of course,

you could simply argue that you are only interested for example, in

whether individuals received a dismissal. However, you would not have

the possibility in the simple logistic model to predict why they received

alternatively a guilty plea conviction, or a trial conviction. This is the prob-

lem that multinomial regression seeks to solve.

Suppose that our outcome variable has three categories (C1, C2, and

C3) with the number of observations in each being represented by N C1

,

N C2

, and N C3

. We could begin by estimating three binary logistic regression

models that would allow for all possible comparisons of the outcome cate-

gories – the logits for C1 and C2, C2 and C3, and C1 and C3. The logit of

Y for each regression could be written simply as:

( ) ( )

, ( ) ( )

, ( ) ( )

ln ln ln P Y C

P Y C

P Y C

P Y C

P Y C

P Y C and

2 1

3 2

3 1

=

=

=

=

=

=e e eo o o for each comparison,

respectively.

Interestingly, these three logits can be linked in what can be defined as

an identity equation that illustrates how knowledge of any two logits can

produce the values of the third. The identity equation2 can be stated as:

( ) ( )

( ) ( )

( ) ( )

ln ln ln P Y C

P Y C

P Y C

P Y C

P Y C

P Y C

2 1

3 2

3 1

=

= +

=

= =

=

=e e eo o o

If we were to estimate these three separate logits, the coefficients would

be interpreted in the same way as we described in Chapter 18. While this

approach would allow us to make comparisons of the likelihood of sub-

jects falling in each of the three categories examined as compared to each

2You can verify this identity by using the fact that the logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator: ln(x/y) = ln(x) − ln(y). Specifically, for this equality, we note that

( )

( ) ( ( ) ) ( ( ) )ln ln ln

P Y C

P Y C P Y C P Y C

2

1 1 2

=

= = = - =e o

and

( )

( ) ( ( ) ) ( ( ) )ln ln ln

P Y C

P Y C P Y C P Y C

3

2 2 3

=

= = = - =e o .

When we put these two pieces together in a single equation, we have

[ ( ( ) ) ( ( ) )] [ ( ( ) ) ( ( ) )]

( ( ) ) ( ( ) ) ( ( ) ) ( ( ) )

( ( ) ) ( ( ) )

( ( ) )

( ( ) )

ln ln ln ln

ln ln ln ln

ln ln

ln ln

ln

P Y C P Y C P Y C P Y C

P Y C P Y C P Y C P Y C

P Y C P Y C

P Y C

P Y C

1 2 2 3

1 2 2 3

1 3

3

1

= - = + = - =

= = - = + = - =

= = - =

= =

=e o Which establishes the equality. We explain the practical implication of this equality below in our discussion of the interpretation of the coefficients from a multinomial logistic regression model.

604

other, it would require us to run three separate regressions. Moreover, and

more importantly from a statistical point of view, we would likely be

working with three completely different samples in each of the three

analyses: N C1

+ N C2

, N C2

+ N C3

, and N C1

+ N C3

. This is because the cases on

the dependent variable are unlikely to be distributed evenly. For exam-

ple, we would not expect sentences for 300 offenders to be distributed

with exactly one hundred in each group (e.g. dismissal, a guilty plea con-

viction, or a trial conviction). Given this, each of our comparisons would

be based on different samples. In comparing N C 1

to N C2

we would have

only defendants who had outcomes C1 and C2. Defendants that had out-

come C3 would not be included in that comparison. But what we are

really interested in is the choice among the three outcomes and how this

choice is distributed in our entire sample. The statistical problem here is

that the varying sample sizes would then result in incorrect standard

errors for the coefficients, leading to inaccurate tests of statistical signifi-

cance. The multinomial logistic regression model simultaneously accounts

for these different sample sizes, ensuring a more valid estimate of signif-

icance levels. It also has the benefit of allowing us to conduct our analysis

using only one regression model.

A Substantive Example: Case Dispositions in California

The State Court Processing Statistics database includes information on ran-

dom samples of individuals arrested for felony offenses in the largest court

districts in the United States. To illustrate the application of the multino-

mial logistic regression model, we focus on a random sample of 10,230

felony arrestees in California in the 1990s. A question of both policy and

theoretical relevance is the study of the factors that affect the type of case

disposition (outcome) – whether a dismissal, a guilty plea conviction, or a

trial conviction.3

A first step in a multinomial regression is to define a “reference cate-

gory.” This is necessary because we need to decide which category we

want to use as a baseline. It is an arbitrary decision about which category

is designated the reference category, but to the extent that we can make a

choice that has some theoretical relevance or makes the interpretation of

the results simpler, that would be the preferred choice. For case disposi-

tion, suppose that we choose dismissal as the reference category, which

then allows us to make two comparisons between a type of conviction –

guilty plea or trial – and dismissal. More directly, our multinomial logistic

3While it may appear odd at first glance that we have not included those individuals who were acquitted at a trial, there were very few individuals who fell into this cate- gory. Like most jurisdictions, courts in California acquit relatively few individuals through a trial – it was about 1% of all cases in the 1990s. What this means is that once the prosecutor has filed charges against a defendant, rather than dismiss the case, it will likely result in the conviction of the defendant through either a guilty plea or a trial conviction. This also implies that a dismissal of the case functions much like an acquittal, but one made by the prosecuting attorney rather than a judge or jury.

605M U L T I N O M I A L L O G I S T I C R E G R E S S I O N

C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

regression results will indicate (1) the relative likelihood of a guilty plea

conviction compared to a dismissal and (2) the relative likelihood of a trial

conviction compared to a dismissal. The one comparison we did not men-

tion was the relative likelihood of a guilty plea conviction compared to a

trial conviction. In the multinomial logistic regression model, this compar-

ison is not directly estimated, but as we illustrate shortly, the results can be

obtained very simply from the results for the comparison of each convic-

tion type to a dismissal.

In equation form, the multinomial model can be written as either a

probability model or an odds ratio model. In Equation 19.2 we provide

an example of the probability equation.

Probability Equation

)j

( ) (

( )

exp

exp P Y m

Xb

Xb

j

J

m

1

= =

= /

In this equation, m refers to the outcome category of interest and has val-

ues ranging from 1 to J (the last category). The numerator to the equa-

tion tells us to exponentiate the value of Xb for category m. The

denominator, in turn, tells us that we need to exponentiate the value of

Xb for all categories and then sum these values together. Since there is a

redundancy built into the values of the coefficients in a multinomial

logistic model, the values for one set of coefficients are set at 0 (e.g.,

b 1 = 0). This is the reference category and leads to the estimation of

coefficients for the total number of categories minus 1 ( i.e., J − 1).

For our three-category case disposition variable, m = 1, 2, or 3. Writing

out the probability equations for each outcome leads to the following for-

mulations of the probability of each of the three outcomes in our example.

For m = 1,

( ) ( ) ( ) ( )

( )

( ) ( )exp exp exp

exp

exp exp P Y

X Xb Xb

X

Xb Xb 1

0

0

1 1

2 3 2 3

= = + +

= + +

For m = 2 and m = 3, we have

( ) ( ) ( )

( )

( ) ( ) ( )

( )

exp exp

exp

exp exp

exp

P Y Xb Xb

Xb

P Y Xb Xb

Xb

2 1

3 1

2 3

2

2 3

3

= = + +

= = + +

Odds Ratio Equation

)

)

)

( ) ( )

(

( )

(

(

( )

( )

exp

exp

exp

exp

exp

exp OR

P Y n

P Y m

Xb

Xb

Xb

Xb

Xb

Xb m n

jj

J

n

jj

J

m

n

m

1

1 =

=

= = =e

=

=

/

/

Equation 19.2

Equation 19.3

606

= 0 b 1

We provide the odds ratio equation in Equation 19.3. Although this equa-

tion may look much more complicated, it uses the information in Equation

19.2 for the probabilities of each category. The equation reduces to some-

thing less complex, because the denominators in the fraction in the mid-

dle of the equation cancel each other out. The only notable difference here

is the notation of a second category by the subscript n. Thus, for any odds

ratio that we may be interested in between categories m and n, Equation

19.3 illustrates that it can be obtained from the respective probabilities.

If we are interested in computing the odds ratio for a comparison

between any category and the reference category (m = 1), where b 1

= 0,

we obtain

( ) ( )

( )

( ) ( )

exp

exp expOR

P Y

P Y m

X1 0/m n m

m= =

= = =

Xb Xb

This last result confirms how we are then to interpret the coefficients from

the multinomial logistic regression model. Since the coefficients for the ref-

erence category have been fixed at 0, the coefficients for each of the

remaining outcome categories will compare the relative likelihood of that

category compared to the reference category.4

In practice, what these equations tell us is that we will have J − 1 sets

of coefficients from a multinomial logistic regression model that can be

interpreted in the same way as binary logistic coefficients, where we com-

pare each outcome (m) to the reference category (m = 1) for the outcome

variable. In our example for case disposition, where we have designated

dismissal as the reference category, one set of coefficients will give us the

log of the odds or the odds ratios comparing the likelihood of a guilty

plea conviction relative to a dismissal, while the second set of coefficients

will give us the log of the odds or the odds ratios comparing the likelihood

of a trial conviction relative to a dismissal.

Table 19.1 presents the results from our application of the multinomial

logistic regression model. We have kept the multivariate model simple and

used age, sex (males = 1, females = 0), race (non-white = 1, white = 0),

Table 19.1 Multinomial Logistic Regression Coefficients

INDEPENDENT TRIAL CONVICTION V. GUILTY PLEA CONVICTION V. VARIABLES DISMISSAL DISMISSAL

Age .016 .004 Male 1.123 −.013 Non-white .043 −.266 Violent Offense .657 −.525 Number of Charges .325 .192 Intercept −4.767 1.318

4It is worth pointing out that the binary logistic regression model presented in Chapter 18 is a special case of the multinomial logistic regression model, where m = 2. If you work through both Equations 19.2 and 19.3 above assuming that m = 2, you will be able to replicate the equations in the previous chapter.

607M U L T I N O M I A L L O G I S T I C R E G R E S S I O N

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type of crime (violent =1, non-violent=0), and total number of charges as

predictors of the type of case disposition for the sample of 10,230 arrestees

in California in the 1990s. The first column lists all the independent vari-

ables, while the second and third columns present the coefficients for each

of the two comparisons: Column 2 presents the comparison of trial con-

viction to dismissal, while Column 3 presents the comparison of guilty plea

conviction to dismissal.

The results in Column 2 show that as age and the number of charges

increase, the likelihood of a trial conviction increases relative to a dis-

missal. Similarly, defendants who are male, non-white, and charged with

a violent offense are also more likely to be convicted at trial than to have

their case dismissed. As in the previous chapter, we can also interpret

each of these coefficients more directly as odds ratios. (Recall from the

previous chapter that the exponentiation of the coefficient provides us

with the odds ratio given a one-unit change in the independent variable.)

● If age is increased by one year, the odds of a trial conviction versus a

dismissal increase by a factor of exp(.016) = 1.016, controlling for all

other variables in the model. ● The odds of a trial conviction versus a dismissal are exp(1.123) = 3.074

times higher for male than for female defendants, controlling for all

other variables in the model. ● The odds of a trial conviction versus a dismissal are exp(.043) = 1.044

times higher for non-white than white defendants, controlling for all

other variables in the model. ● The odds of a trial conviction versus a dismissal are exp(.657) = 1.929

times higher for defendants charged with a violent rather than a non-

violent offense, controlling for all other variables in the model. ● If the number of charges is increased by one, the odds of a trial con-

viction versus a dismissal increase by a factor of exp(.325) = 1.384, con-

trolling for all other variables in the model.

We can similarly interpret the results in Column 3, which show that as

age and number of charges increase, the likelihood of a guilty plea con-

viction relative to a dismissal increases. We also see from these results that

defendants who are male, non-white and charged with a violent offense

will be less likely to be convicted with a guilty plea than to have their

cases dismissed. Again, the direct interpretations of the coefficients would

be the following:

● If age is increased by one year, the odds of a guilty plea conviction ver-

sus a dismissal increase by a factor of exp(.004) = 1.004, controlling for

all other variables in the model. ● The odds of a guilty plea conviction versus a dismissal are exp(−.013)

= .987 times smaller for male than for female defendants, controlling for

all other variables in the model.

608

● The odds of a guilty plea conviction versus a dismissal are exp(−.266)

= .766 times smaller for non-white than white defendants, controlling

for all other variables in the model. ● The odds of a guilty plea conviction versus a dismissal are exp(−.525)

= .592 times smaller for defendants charged with a violent rather than a

non-violent offense, controlling for all other variables in the model. ● If the number of charges is increased by one, the odds of a guilty plea

conviction versus a dismissal increase by a factor of exp(.192) = 1.212,

controlling for all other variables in the model.

The Missing Set of Coefficients

As we noted earlier, when we estimate a multinomial logistic regression

model, we obtain coefficients for all contrasts but one. In the example of

case disposition, we are missing the contrast between guilty plea convic-

tion and trial conviction. Based on the identity relationship of multiple

logits that we described earlier in the chapter (see, also, footnote 2), for

all possible comparisons of the outcome categories, the most direct way

of obtaining the missing coefficients is to simply subtract one set of coef-

ficients from another set of coefficients. In Table 19.1, the results in

Column 2 represent the logit for Trial Conviction and Dismissal, while

those in Column 3 represent the logit for Guilty Plea Conviction and

Dismissal.

Since the logarithm of a fraction can be rewritten as the subtraction of

the logarithm of the denominator from the logarithm of the numerator, the

logits can be rewritten as

( ) ( )

( ( ))

( ( ))

ln ln

ln

P Y Dismissal

P Y Trial Conviction P Y Trial Conviction

P Y Dismissal

=

= = =

- =

e o

and

( )

( ) ( ( ))

( ( ))

ln ln

ln

P Y Dismissal

P Y Guilty Plea Conviction P Y Guilty Plea Conviction

P Y Dismissal

=

= = =

- =

e o

By performing simple subtractions of the logits, we can generate addi-

tional contrasts between the outcome categories. To obtain the missing

coefficients for the comparison of Guilty Plea Conviction to Trial

Conviction, we subtract the logit for Trial Conviction and Dismissal from the

logit for Guilty Plea Conviction and Dismissal:

( )

( )

( ) ( )

( ( )) ( ( ))

( ( )) ( ( ))

ln ln

ln ln

ln ln

P Y Dismissal

P Y Guilty Plea Conviction

P Y Dismissal

P Y Trial Conviction

P Y Guilty Plea Conviction P Y Dismissal

P Y Trial Conviction P Y Dismissal

=

= -

=

=

= = - =

- = - =

e eo o 7 7

A A

( ( )) ( ( ))ln lnP Y Guilty Plea Conviction P Y Trial Conviction= = - =

609M U L T I N O M I A L L O G I S T I C R E G R E S S I O N

C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

( )

( ) ln

P Y Trial Conviction

P Y Guilty Plea Conviction =

=

=e o In other words, what this algebraic manipulation of logits shows us is

that we can obtain the coefficients for the omitted contrast simply by sub-

tracting one set of coefficients from another set of coefficients.

When applied to our case disposition coefficients, we obtain the results

presented in Table 19.2. Here, we see that as age and the number of

charges increase, the likelihood of a guilty plea conviction relative to a trial

conviction decrease. Similarly, defendants who are male, non-white and

charged with a violent offense will be less likely to be convicted through a

guilty plea than through a trial.

In regard to the direct interpretation of the coefficients, we have the

following:

● If age is increased by one year, the odds of a guilty plea conviction ver-

sus a trial conviction decrease by a factor of exp(−.012) = .988, con-

trolling for all other variables in the model. ● The odds of a guilty plea conviction versus a trial conviction are

exp(−1.136) = .321 times smaller for male than for female defendants, con-

trolling for all other variables in the model. ● The odds of a guilty plea conviction versus a trial conviction are

exp(−.309) = .734 times smaller for non-white than white defendants,

controlling for all other variables in the model. ● The odds of a guilty plea conviction versus a trial conviction are

exp(−1.182) = .307 times smaller for defendants charged with a violent

rather than a non-violent offense, controlling for all other variables in

the model. ● If the number of charges is increased by one, the odds of a guilty plea

conviction versus a trial conviction increase by a factor of exp(−.133) =

875, controlling for all other variables in the model.

A second way to obtain the coefficients for the comparison of guilty

plea conviction to trial conviction would be to simply redefine our statis-

tical model so that trial conviction was chosen as the reference category

and re-estimate our multinomial model. Upon rerunning the multinomial

logistic regression model, we obtain the results presented in Table 19.3.

Table 19.2 Coefficients for the Omitted Contrast of Guilty Plea Conviction v.Trial

Conviction through Subtraction of Coefficients

INDEPENDENT VARIABLES GUILTY PLEA CONVICTION V. TRIAL CONVICTION

Age .004 –.016 = −.012 Male −.013 – 1.123 = −1.136 Non-white −.266 –.043 = −.309 Violent Offense −.525 –.657 = −1.182 Number of Charges .192 –.325 = −.133 Intercept 1.318 – (−4.767) = 6.085

610

Note that Column 2 contains the coefficients for the contrast between dis-

missal v. trial conviction, which substantively gets at the same compari-

son that appears in Table 19.1, Column 2, except for the order of the

comparison (trial conviction v. dismissal). The only difference between

the results presented in Column 2 of both Tables 19.1 and 19.3 are the

signs of the coefficients, reflecting the order of comparison of the out-

come categories. Again, what this indicates to us is that the selection of

reference categories is arbitrary and that we will obtain the same sub-

stantive results, regardless of which category is selected. At the same time,

we need to be aware of the selection of categories so that we correctly

interpret our results.

Column 3 presents the contrast for guilty plea conviction relative to trial

conviction. The results in this column are identical to those appearing in

column 2 of Table 19.2, which were based on simply subtracting one set of

coefficients from another.

Had we been interested in the contrast of Trial Conviction relative to Guilty

Plea Conviction, we would have reversed the original order of subtraction

(i.e., the coefficients in Column 3 would have been subtracted from the

coefficients in Column 2). Then, the only difference that we would have seen

in Table 19.2 is that the signs of the coefficients would have been reversed.

Otherwise, the substantive interpretation of the results would be identical.

Statistical Inference

Single Coefficients

The results from a multinomial logistic regression analysis complicate

slightly tests of statistical significance. Since we now have multiple coef-

ficients for each independent variable, there are questions about how to

discern whether an independent variable has an effect on the dependent

variable. Specifically, there are two issues of statistical inference that are

important for interpreting the results from a multinomial logistic regres-

sion analysis. For each coefficient we can estimate the statistical signifi-

cance of each category compared to the reference category. But we also

can estimate the overall significance of the independent variable in

predicting the multi-category dependent variable.

Table 19.3 Multinomial Logistic Regression Coefficients Using Trial Conviction as the

Reference Category from Re-estimated Model

GUILTY PLEA CONVICTION INDEPENDENT VARIABLES DISMISSAL V. TRIAL CONVICTION V. TRIAL CONVICTION

Age −.016 −.012 Male −1.123 −1.136 Non-white −.043 −.309 Violent Offense −.657 −1.182 Number of Charges −.325 −.133 Intercept 4.767 6.085

611M U L T I N O M I A L L O G I S T I C R E G R E S S I O N

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To test the effect of each individual coefficient in comparison to the

reference category, we would again use the Wald statistic described in

Chapter 18, with degrees of freedom equal to 1. As noted in Chapter 18,

the Wald statistic has a chi-square distribution, and so the statistical sig-

nificance of a result may be checked in the chi–square table. Table 19.4

presents the Multinomial Logistic Coefficients from the original model

along with the standard errors (se) of the coefficients and value of the

Wald statistic (W).

If we set the significance level at 5%, the critical value of the Wald sta-

tistic with df = 1, is 3.841 (see Appendix 2). We can see that the violent

offense charge and number of charges have statistically significant effects

for both pairs of outcomes. The demographic characteristics have varying

effects, where age and sex have statistically significant effects on the like-

lihood of a trial conviction compared to a dismissal, but race has a statis-

tically significant effect on the likelihood of a guilty plea conviction

compared to a dismissal.

Multiple Coefficients

Note in Table 19.4 that there are two coefficients for each independent

variable. As we noted above, the number of coefficients from a multino-

mial logistic regression model for each independent variable will be one less

than the number of categories on the dependent variable (i.e., J – 1). How

do we assess the overall effect of each independent variable on the depend-

ent variable? There are two key ways of doing this – one is a likelihood

ratio test similar to the test we discussed in the previous chapter on binary

logistic regression. The other test is an extension of the Wald test we have

also already used. Regardless of the statistical software package one uses

to estimate a multinomial logistic regression model, one of these two meth-

ods will be reported to test the overall effect of each independent variable.

The likelihood ratio test involves estimating the full multinomial logistic

regression equation with all variables and then estimating reduced models

that eliminate one independent variable from each analysis. The difference

in the −2 log-likelihood function for each equation will then allow for the

test of each independent variable.

Table 19.4 Multinomial Logistic Regression Coefficients, Standard Errors, and Wald Test

Results

GUILTY PLEA CONVICTION V. TRIAL CONVICTION V. DISMISSAL DISMISSAL

INDEPENDENT VARIABLES B SE W B SE W

Age .016 .008 4.181 .004 .003 2.110 Male 1.123 .373 11.350 −.013 .071 .032 Non-white .043 .158 .072 −.266 .053 24.827 Violent Offense .657 .160 16.834 −.525 .060 76.287 Number of Charges .325 .043 58.524 .192 .021 82.642 Intercept −4.767 .438 118.726 1.318 .124 112.754

612

For example, in the case dismissal analysis, the value of the −2 log-like-

lihood for the full model is 3625.670. When we estimate the same model,

tions is 3653.501 – 3625.670 = 27.831. By eliminating the variable for

non-white, we have removed two coefficients from the analysis. If you

refer again to Table 19.4, you will see that each independent variable

appears twice to indicate its overall effect on case disposition – once to

represent the effect on trial conviction versus dismissal and once more

to represent the effect on guilty plea conviction versus dismissal. The degrees

of freedom for the test will be df = 2 to reflect the removal of the two coef-

ficients. At a significance level of 5%, the critical value of the chi-square is

5.991. This means that we would conclude that the race of the defendant

has a statistically significant effect on the type of case disposition. Table

19.5 presents the vales of the −2 log-likelihood function for each of

the reduced models and the value of the likelihood ratio test for each inde-

pendent variable. Based on the critical value of the chi-square of 5.991, we

see that race and sex of defendant, violent offense charge, and number of

charges all have statistically significant effects on type of case disposition,

while age of defendant does not have a statistically significant effect.

An alternative test of each independent variable is to use the Wald sta-

tistic. Up to this point, we have used the Wald statistic to test the statistical

significance of a single coefficient, but it can also be used to test the group

of coefficients representing the effect of any given independent variable.

Recall that the Wald test statistic for a single coefficient is computed by

dividing the coefficient by its standard error and then squaring this value.

The Wald statistic for a group of coefficients involves an analogous calcu-

lation, but requires the use of matrix algebra – a topic beyond the scope

of our text. That said, many statistical software packages will generate the

results for the Wald test as part of the standard output, and our attention

here is focused more on illustrating the interpretation of the results, rather

than the actual calculation. In most applications, the value of the Wald

statistic will be very similar to the value of the LR test.5

Table 19.5 Likelihood Ratio and Wald Statistic Results for the Overall Effect of Each

Independent Variable

−2 LOG-LIKELIHOOD FOR INDEPENDENT VARIABLE DF THE REDUCED MODEL LR TEST STATISTIC WALD

Age 2 3630.528 4.859 4.94 Male 2 3642.342 16.672 11.98 Non-white 2 3653.501 27.831 27.77 Violent Offense 2 3747.168 121.498 125.95 Number of Charges 2 3737.877 112.207 99.08

5Recall from footnote # 13 in Chapter 18 that the Wald statistic is sensitive to small samples (e.g., less than 100), while the LR test is not.

613M U L T I N O M I A L L O G I S T I C R E G R E S S I O N

but eliminate the variable non-white from the analysis, the value of the −2

log-likelihood is 3653.501. The difference of the two log-likelihood func-

To test the overall effect of an independent variable with the Wald sta-

tistic, we continue to use a chi-square distribution with degrees of free-

dom equal to the number of coefficients being tested – the number of

outcome categories minus 1 (i.e., df = J – 1).

In our case disposition example, we have three outcome categories

( J = 3), so the degrees of freedom will be 3 – 1 = 2, which again corre-

sponds to the number of sets of coefficients that have been estimated. The

values of the Wald test for each of the independent variables included in

our analysis are presented in Table 19.5.

Using a significance level of 5%, we see from Appendix 2 that the crit-

ical chi-square statistic has a value of 5.991. Based on this value, we see

that all the independent variables, except for age of defendant, have sta-

tistically significant effects on type of case disposition. The substance of

these results is identical to that using the LR test.

How should we address mixed results? For example, it is not uncom-

mon for a researcher to find that the overall effect of an independent

variable is not statistically significant, but one of the individual coeffi-

cients does have a significant effect on a comparison of two outcome

categories. Alternatively, the Wald test for the overall effect of an inde-

pendent variable may show it to have a statistically significant effect, but

there may be individual coefficients representing the effect of that inde-

pendent variable on a specific comparison that are not statistically

significant.

This kind of difficulty is illustrated in the results presented in Tables

19.4 and 19.5. Age of defendant does not have a statistically significant

effect on case disposition overall, yet age does have a statistically signifi-

cant effect on the comparison of trial conviction to dismissal. In such a

case, the researcher should carefully note the pattern of results, explain-

ing that overall age does not affect case disposition but that there appears

to be a statistically significant impact of age on gaining a trial conviction

as compared to a dismissal.

Alternatively, we also see that the overall effect of race is statistically

significant, but the individual coefficient for race on the comparison

between trial conviction and dismissal is not statistically significant. The

safest approach for the researcher in this type of situation is to note the

significance of the overall effect of the independent variable, but again

to clearly explain the pattern of results for the individual coefficients. In

this case, our model suggests that race has an overall impact on case

disposition but our data do not allow us to conclude, despite this, that

race has a significant effect on gaining a trial conviction as opposed to a

dismissal.

Our suggestion is to use caution in interpreting the results and to be

as clear as possible in explaining the nature and type of effect that is sta-

tistically significant. In multinomial regression, a number of statistical out-

comes are included and the researcher should be careful not to draw

selectively from the results gained.

C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N614

O R D I N A L L O G I S T I C R E G R E S S I O N

Overall Model

In addition to testing the statistical significance of the individual coeffi-

cients, we are also often interested in assessing the statistical significance

of the overall model. To assess the statistical significance of the full multi-

nomial regression model, we compute a model chi–square statistic that is

identical in form to that used for the binary logistic regression model dis-

cussed in Chapter 18. Recall that the model chi–square is computed as:

Model chi–square = (−2LL null model) – (−2LL

full model)

For our case disposition analysis, the −2LL null model

= 3923.540 and the

−2LL full model

= 3625.670, resulting in a model chi-square of 3923.540 –

3625.670 = 297.870. We have already noted that a total of 10 coefficients

have been estimated (2 for each of the 5 independent variables), which

gives us a degrees of freedom value for this test equal to 10. Looking at

Appendix 2, we see that at a significance level of 5%, we see that a chi–

square statistic greater than 18.307 is needed to reject the null hypothesis

that the model has no statistically significant effect on case disposition.

Since our model chi-square is larger than the critical value of the chi–

square, we conclude that the overall model has a statistically significant

effect on case disposition.

A Concluding Observation about Multinomial Logistic Regression Models

In our substantive example, we selected a dependent variable with only

three categories. Realistic applications of multinomial logistic regression

models with more than three categories can quickly become unwieldy in

regard to the number of contrasts that are being analyzed. For example, if

we had a dependent variable with four categories, we would have three

sets of coefficients to represent a total of six different contrasts (C1 and C2,

C1 and C3, C1 and C4, C2 and C3, C2 and C4, and C3 and C4). If we

increased the number of outcome categories to five, we would have four

sets of coefficients to represent a total of ten different contrasts, at which

point the results from a multinomial logistic regression analysis likely

become too difficult for most researchers to summarize in a coherent and

concise way.

O r d i n a l L o g i s t i c R e g r e s s i o n

Multinomial regression provides a solution to the important problem of

predicting multiple nominal category dependent variables. But in our dis-

cussions so far we have not examined how to analyze ordinal level

dependent variables. For many years researchers simply used Ordinary

Least Squares Regression to deal with this type of analysis. There are times

when this approach makes sense, and is unlikely to lead to estimation

615

C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

problems that are important. For example, if a researcher is examining an

ordinal level variable that is measured as ten ascending categories, and

can easily assume that the interval each category represents is the same

as the prior interval, OLS estimates are likely to be satisfactory from a

statistical perspective. In this case, the ordinal level variable of interest can

be assumed to have characteristics close to that of an interval level

dependent variable.

However, because until recently other estimation approaches for ordi-

nal dependent variables were not easily accessible, researchers often

made assumptions regarding an ordinal dependent variable that was

examined that were clearly not appropriate. For example, when we exam-

ine fear of crime measured as a series of categories from “very fearful” to

“not fearful at all” it is very hard to assume that there are equal intervals

between these qualitative responses. Today, with the availability of statis-

tical software packages that will estimate ordinal regression models,

researchers should be cautious in applying OLS regression to ordinal level

measures. In this section, we present the ordinal logistic regression

model that explicitly takes into account an ordered categorical dependent

variable.

In order to set up the application and interpretation of the ordinal

logistic model, we need to reconsider what a variable measured at the

ordinal level tells us. Recall from Chapter 2 that an ordinal variable has

ranked categories that are assumed to represent an underlying continuum.

For example, when respondents to a survey are presented with a state-

ment that has as response choices Strongly Agree, Agree, Disagree, and

Strongly Disagree, the variable is assumed to represent an underlying con-

tinuum of agreement-disagreement with some issue. Yet, we know that

however an individual responds to the question, any two individuals

falling in the same category may not mean exactly the same thing. For

example, if we randomly selected two individuals who had responded

Strongly Disagree with a policy statement and we were able to ask more

in-depth follow-up questions, we would likely discover that there were

degrees of how strongly each disagreed.

If we assume that an ordinal variable’s categories represent an under-

lying continuum, we can think of thresholds as those points where an

individual may move from one ordinal category to another (adjacent) cat-

egory. In the example above, we could make note of the thresholds

between Strongly Agree and Agree, Agree and Disagree, and Disagree

and Strongly Disagree. Figure 19.1 illustrates the link between the

underlying continuum and the variable measured at the ordinal level. In

Figure 19.1, each dot represents the “true value” for an individual’s attitudes

about a given issue – but this true value cannot be measured directly, and

we are left with the four response choices indicating degree of agreement

or disagreement. Each of the vertical lines marks the point between one

of the possible response choices and indicates the threshold for each

response category.

616

The ordinal logistic regression model represents something of a hybrid

of the binary logistic and multinomial logistic regression models. Similar

to the multinomial logistic regression model’s estimation of multiple

model intercepts, the ordinal logistic model estimates multiple intercepts

that represent the values of the thresholds. Comparable to the binary

logistic model, the ordinal logistic model estimates one coefficient for the

effect of each independent variable on the dependent variable. In part,

this is due to the added information contained in an ordinal variable,

rather than a multi-category nominal variable. The interpretation of the

results from the ordinal logistic model is also potentially much simpler

than the results from the multinomial logistic model.

One of the key differences between the ordinal logistic model and

other logistic models is that rather than estimating the probability of a sin-

gle category as in the binary and multinomial logistic models, the ordinal

logistic model estimates a cumulative probability – the probability that

the outcome is equal to or less than the category of interest. In equation

format:

Figure 19.1 Hypothetical Ordinal Variable and Underlying Continuum

617O R D I N A L L O G I S T I C R E G R E S S I O N

C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

Cumulative Probability

( ) ( )P Y m P Y j j

m

1

# = =

=

/

In Equation 19.4, m is the category of interest and can take on values rang-

ing from 1 to J − 1, while j denotes each individual category. The summa-

tion sign tells us that we are to add the probabilities for each individual

outcome from the first category (i.e., Y = 1) to the last category of interest

(i.e., Y = m). For example, using the four response categories above would

mean that J = 4, and we could compute a total of J − 1 = 4 − 1 = 3 cumu-

lative probabilities. If we define a Strongly Agree response as 1 and

Strongly Disagree response as 4, we could then compute probabilities for

P(Y ≤ 1), P(Y ≤ 2), and P(Y ≤ 3), representing P(Y ≤ Strongly Agree),

P(Y ≤ Agree), and P(Y ≤ Disagree), respectively. We would not include the

final category (i.e., P(Y ≤ 4) or P(Y ≤ Strongly Disagree)), since it would

have to be equal to 1 (or 100%) – all possible values have to fall in one of

the four response categories.

Using the cumulative probabilities, we can then compute odds ratios to

represent the effects of the independent variables on the dependent vari-

able. Again, there is a slight variation to the construction and interpretation

of the odds ratios, since we are using cumulative probabilities:

Odds Ratio Using Cumulative Probabilities

( ) ( )

( > ) ( )

OR P Y m

P Y m

P Y m

P Y m

1m #

# # =

- =

Substantively, the odds ratio presented in Equation 19.5 indicates the odds

of an outcome less than or equal to category m versus the odds of a cat-

egory greater than m. In the context of our four response choices, the three

odds ratios that we could make reference to would be the following:

● Odds of a Strongly Agree response versus the combined outcomes of

Agree, Disagree, and Strongly Disagree. ● Odds of the combined outcomes of Strongly Agree and Agree response

versus the combined outcomes of Disagree and Strongly Disagree. ● Odds of the combined outcomes of Strongly Agree, Agree, and Disagree

versus Strongly Disagree.

We can take Equation 19.5 for the odds ratio and rewrite it as a linear

model, similar to that for the binary and multinomial logistic models.

( > ) ( )

( )expOR P Y m

P Y m Xbm m

# = = -x .

The general form for this equation is very similar to that for either the binary

or multinomial logistic model, except that we have introduced a new term

(τ m ) and now have a negative sign to the left of Xb. The (τ

m ) represent the

Equation 19.4

Equation 19.5

618

threshold parameters, which function as intercepts in the model and will

take on values for j = 1 to J − 1.

By taking the natural logarithm of the odds ratio equation, we produce

the logit Equation 19.6:

Ordinal Logit Equation

( > ) ( )

[ ( )]ln ln exp P Y m

P Y m Xb Xbm m

# = - = -x xe o .

Equation 19.6 forms the basis for estimating ordinal logistic regression

models.

Interpretation of Ordinal Logistic Regression Coefficients

In our discussion of the binary logistic regression model, we illustrated

how a one-unit increase in the independent variable would modify the

odds of the different outcomes by a factor of exp(b). Since the form of

the ordinal logistic equation is slightly different, we cannot simply expo-

nentiate b to obtain the effect of the independent variable.

Thus, to interpret the effect of a one-unit change in the independent vari-

able in an ordinal logistic regression model, we will need to exponentiate

the negative value of the estimated coefficient. We can then interpret the

coefficient as indicating the odds of an outcome less than or equal to cate-

gory m versus the odds of a category greater than m.

S u b s t a n t i v e E x a m p l e : S e v e r i t y o f P u n i s h m e n t D e c i s i o n s

Using the State Court Processing data for California in the 1990s resulted

in a sample of 8,197 individuals being convicted for some type of crime.

The primary punishment outcomes – community-based, jail, and prison –

represent a continuum of punishment severity with prison the most severe

sentence. Again, if we keep our set of independent variables limited by

using the same variables as in our case disposition example above, we

have measures of age, sex (males = 1, females = 0), race (non-white = 1,

white = 0), type of crime (violent =1, non-violent=0), and total number of

charges as predictors of severity of punishment. Table 19.6 presents the

results of our ordinal logistic regression model. You will notice that ordi-

nal regression, like multinomial regression uses the Wald statistic to assess

the statistical significance of individual parameters.

S U B S T A N T I V E E X A M P L E : S E V E R I T Y O F P U N I S H M E N T D E C I S I O N S

Equation 19.6

619

To illustrate the modification, suppose we have two values of X: X and

+ 1. The odds ratio for X and X + 1 would be

OR X+1

OR X = exp X+1 b

exp Xb =exp X X+1 b =m

m

m

m

( ) ( )

( )( ) ( )

( )éë ùû( ) t

t eexp b .( )

-

- - -

X

Interpreting the Coefficients

While ordinal regression accounts for the fact that the categories in the

dependent variable are ranked, for example in our case from less to more

severe sanctions, the interpretation of the coefficients is similar to that

used in multinomial regression. In this case, we can compare lower cate-

gories to the categories ranked above them. For example, in the case of

sentences, we can compare either community-based punishment to jail

and prison sentences, or community-based punishment and jail sentences

to prison sentences. In both these cases, the exponent of the negative of

the coefficient provides the odds ratio for change. Since age, for example,

is measured at the interval level of measurement, we would note that

for a one year increase in age, the odds ratio changes by a factor of

exp(−.004) = .996, controlling for the other variables in the model. We can

write out the interpretations as follows:

● The odds of receiving a community-based punishment versus a jail and

prison punishment decrease by a factor of .996 for a one year increase

in age, controlling for all other variables in the model. ● The odds of receiving a community-based and a jail punishment versus

a prison punishment decrease by a factor of .996 for a one year increase

in age, controlling for all other variables in the model.

We see that the coefficient for male is .821. By exponentiating the negative

of .821 (exp(−.821) = .440), we see that males are likely to receive more

severe punishments than females, controlling for the other independent

variables in the model. More concretely, we can state the following about

the punishment of male offenders:

● The odds of receiving a community-based punishment versus a jail and

prison punishment are .440 times smaller for males than for females,

controlling for all other variables in the model. ● The odds of receiving a community-based and jail punishment versus a

prison punishment are .440 times smaller for males than for females,

controlling for all other variables in the model.

The effect of race on punishment severity is exp(−.166) = .847. Writing out

direct interpretations of this coefficient leads to the following statements:

C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

Table 19.6 Ordinal Logistic Regression Results for Severity of Punishment

INDEPENDENT VARIABLE B SE WALD

Age .004 .002 4.000 Male .821 .058 202.991 Non-white .166 .043 14.862 Violent Offense .328 .053 38.312 Number of Charges .014 .014 1.000 τ

1 −.881 .099 79.175

τ 2

1.720 .101 291.980

620

S U B S T A N T I V E E X A M P L E : S E V E R I T Y O F P U N I S H M E N T D E C I S I O N S

● The odds of receiving a community-based punishment versus a jail and

prison punishment are .847 times smaller for non-whites than for

whites, controlling for all other variables in the model. ● The odds of receiving a community-based and jail punishment versus a

prison punishment are .847 times smaller for non-whites than for

whites, controlling for all other variables in the model.

The effect of a violent offense charge is exp(−.328) = .720, indicating that

a violent offense is likely to result in more severe forms of punishment

(as we would expect):

● The odds of receiving a community-based punishment versus a jail and

prison punishment are .720 times smaller for individuals charged with a

violent offense rather than a non-violent offense, controlling for all

other variables in the model. ● The odds of receiving a community-based and jail punishment versus a

prison punishment are .720 times smaller for individuals charged with a

violent offense rather than a non-violent offense, controlling for all

other variables in the model.

Finally, the effect of a one unit increase in the number of charges is exp

(−.014) = .986. In practice, we would not spend much time interpreting this

coefficient, since the Wald statistic indicates it is not significantly different

from 0.6 However, as another illustration for how to interpret coefficients

from an ordinal logistic regression model, it is useful to write out the inter-

pretations of this coefficient:

● The odds of receiving a community-based punishment versus a jail and

prison punishment decrease by a factor of .986 for a one unit increase

in the number of charges, controlling for all other variables in the model. ● The odds of receiving a community-based and a jail punishment versus

a prison punishment decrease by a factor of .986 for a one unit increase

in the number of charges, controlling for all other variables in the model.

Note, too, that there are two threshold parameters representing the thresh-

old points between each of the ordered categories (i.e., community-based

and jail punishments and then between jail and prison punishments).

Statistical Significance

As we noted above, the test of statistical significance for each individual

coefficient is a Wald statistic that is computed and is interpreted in exactly

the same way as the Wald statistic for binary logistic and multinomial logistic

regression models. Table 19.6 reports the values of the Wald statistic for

6You should verify the statistical significance of each coefficient presented in Table 19.6 using a Wald test statistic with df = 1.

621

622 C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

each independent variable. The statistical significance of the overall model

is based on a model chi-square statistic that is also computed and inter-

preted in exactly the same way as for the binary logistic and multinomial

logistic regression models. In our punishment example, the −2LL null model

=

5883.113 and the −2LL full model

= 5601.386, resulting in a model chi–square

of 5883.113 – 5601.386 = 281.727. Since a total of 5 coefficients have been

estimated (one for each independent variable), the degrees of freedom

value for this test is equal to 5. Looking at Appendix 2, we see that at a

significance level of 5%, a chi–square statistic greater than 11.070 is needed

to reject the null hypothesis that the model had no effect on punishment

severity. Since our model chi–square is larger than the critical value of

the chi–square, we conclude that the overall model had a statistically

significant effect on punishment severity.

Parallel Slopes Tests

As we noted earlier, the proportional odds model assumes that the effects of the

independent variables are constant across all categories of the dependent variable,

which is analogous to our interpretation of coefficients in a multivariate linear

regression model. Regardless of the level (or category) of the dependent variable,

we expect the independent variable to exert a constant (i.e., proportional) effect

on the dependent variable. The constant effect of each independent variable should

have also been clear in the direct interpretations of the coefficients noted in the pre-

vious section. This is known more generally as the parallel slopes assumption.

Most statistical packages include a score test of this assumption that informs the

user of the appropriateness of the ordinal logistic model. Somewhat less common

is the Brant test, which tests for parallel slopes in the overall model and in each

independent variable.

Score Test

Conceptually, the parallel slopes score test is based on the idea that we could esti-

mate a series of J – 1 binary logistic regression models (i.e., one model less than the

number of ordered categories in the dependent variable) of the form P(Y ≤ m) that

allowed the effects for all K independent variables to vary by outcome category on

the dependent variable. The test would then focus on whether a single coefficient

or multiple coefficients best represented the effect of the independent variables

on the dependent variable. Technically, the score test uses information about the

log-likelihood for the ordinal logistic regression model and assesses how much it

would change by allowing the coefficients for all the independent variables to vary

by the outcome category on the dependent variable. The degree of change in the

likelihood function then indicates whether the parallel slopes assumption is met.

The null hypothesis of the score test is parallel (equal) slopes. The research hypoth-

esis is that the slopes are not parallel (equal). The value of the score test (reported

by most statistical software) is distributed as a chi-square with K( J – 2) degrees of

freedom.

S U B S T A N T I V E E X A M P L E : S E V E R I T Y O F P U N I S H M E N T D E C I S I O N S 623

For our severity of punishment example, we have K = 5 (i.e., five independent

variables) and J = 3 (i.e., three outcome categories on the dependent variable). The cor-

critical values for the chi-square reported in Appendix 2, the critical chi-square for a

significance level of 5% is 11.070. The value of the score test for our model is 57.890,

which indicates that we should reject our null hypothesis of parallel slopes and con-

clude that our model does not meet the parallel slopes assumption.

Brant Test

Similar to the score test, the Brant test is a Wald test that assesses whether all the

coefficients in a proportional odds model satisfy the parallel slopes assumption.

The computation of the Brant test is based on the values of the coefficients and

their respective variances. In addition to providing an overall test for the parallel

slopes assumption, the Brant test can be decomposed into values for each of the

independent variables in the ordinal logistic regression model to test whether each

independent variable meets the parallel slopes assumption.

The Brant test for the overall model is distributed as a chi-square with K( J-2)

degrees of freedom (same as in the score test). Each independent variable’s test

statistic is distributed as a chi-square with J – 2 degrees of freedom.

The results of the Brant test for our severity of punishment example appear

in Table 19.7. As expected, the overall test again indicates that the parallel slopes

assumption is violated for our model. The chi-square is computed as 65.34, and

with df = 5 and a critical chi-square of 11.070, we reject the null hypothesis of

parallel slopes for the full model. For each independent variable, the critical chi-

3.841. We see that age (chi-square = 40.25), male (chi-square = 4.72), and violent

offense charge (chi-square = 15.82) would lead us to reject the null hypothesis of

parallel slopes for each of these variables, since all have chi-square values greater

than 3.841. This means that the effects of these three independent variables are

not proportional (constant) across the levels of severity of punishment. In contrast,

the effects of non-white (chi-square = 2.28) and total number of charges (chi-

square = 2.14) have chi-square values less than 3.841, leading us to fail to reject the

parallel slopes assumption and conclude that the effects of these two independent

variables are proportional across the levels of severity of punishment.

Brant, Rollin. 1990. “Assessing proportionality in the proportional odds model for ordinal

logistic regression.” Biometrics 46: 1171–1178.

Brant Test Results for Severity of Punishment

VARIABLE χ2 df p

Overall 65.34 5 0.000 Age 40.25 1 0.000 Male 4.72 1 0.030 Non-white 2.28 1 0.131 Violent 15.82 1 0.000 Total number of charges 2.14 1 0.143

Table 19.7

7

7

responding degrees of freedom for our score test is equal to 5(3 – 2) = 5. Based on the

624 C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

Partial Proportional Odds

In much of the research in criminology and criminal justice, it is quite common for

the parallel slopes assumption not to be met in practice. Historically, when researchers

have been confronted with results from the score test indicating that the model failed

to satisfy the parallel slopes assumption, they were left with a choice of fitting the

proportional odds model and violating a key assumption of the model or of fitting a

multinomial logistic regression model and ignoring the ordinal nature of the dependent

variable, complicating the interpretation of the results through the increased number

of coefficients. Recently, a class of models referred to as partial proportional odds or

generalized ordinal logistic regression models has received increasing attention. The

logic to the partial proportional odds model is to allow some or all of the coefficients

of the independent variables to vary by the level of the dependent variable, much like

we see in the application of multinomial logistic regression, but to constrain other coef-

ficients to have a single value, as in the proportional odds model.

We obtain the partial proportional odds model by generalizing the proportional

odds equation from Section 4.1 to allow the coefficients (the b m

) to vary by level

of the dependent variable (m):

Without any further constraints on the coefficients, the total number of coef-

ficients estimated will be identical to that obtained from a multinomial logistic

regression analysis. It is important to note, however, that the coefficients do not

mean the same thing. Recall from our discussion above that multinomial logistic

regression coefficients refer to comparisons between a given category and the

reference category. As noted in the equation above, the logit in the partial propor-

tional odds model is identical to that in the proportional odds model and refers to

the odds of a category less than or equal to m versus a category greater than m.

Due to the potentially large number of coefficients in a fully generalized ordi-

nal logit model, most researchers will want to limit the number of variables with

nonconstant effects. The results from the Brant test are useful for determining

which independent variables, if any, appear to have varying effects on the differ-

ent categories of the dependent variable (i.e., the slopes are not parallel). If the

overall Brant test result is not statistically significant, it implies that the parallel

slopes assumption is met for the full model. In this case, there is likely little to be

gained by relaxing the parallel slopes assumption for a single variable—the results

become unnecessarily complicated and will not add much statistically to the model.

There are a number of sources interested readers can consult, although most of these are much more technical than the

material presented in this text. See, for example, Fu, Vincent. 1998. “Estimating generalized ordered logit models.” Stata

Technical Bulletin 8:160–164. Lall, R., Walters, S.J., Morgan, K., and MRC CFAS Co-operative Institute of Public Health.

2002. “A review of ordinal regression models applied on health-related quality of life assessments.” Statistical Methods

in Medical Research 11:49–67. O’Connell, Ann A. 2006. Logistic Regression Models for Ordinal Response Variables.

Thousand Oaks, CA: Sage. Peterson, Bercedis and Harrell, Jr, Frank E. 1990. “Partial proportional odds models for ordi-

nal response variables.” Applied Statistics 39: 205–217. Williams, Richard. 2006. “Generalized Ordered Logit/ Partial

Proportional Odds Models for Ordinal Dependent Variables.” The Stata Journal 6(1):58–82.

8

8

ln ln exp . P Y

P Y Xb Xb

£( ) >( )

æ

è çç

ö

ø ÷÷ = ( )éë ùû =

m

m m m m mt t- -

S U B S T A N T I V E E X A M P L E : S E V E R I T Y O F P U N I S H M E N T D E C I S I O N S 625

In those cases where the overall Brant test result is statistically significant, then

the Brant test results for individual variables will point to those variables with the

greatest divergence from the parallel slopes assumption and the best candidates for

allowing the effects to vary across the different ordinal logits.

All other features of the partial proportional odds model—tests for statistical

significance, interpretation of the coefficients, and the like—are the same as found

in the proportional odds model.

Severity of Punishment Example

In our application of the proportional odds model to the severity of punishment

data from California, we also noted that the parallel slopes assumption was not

satisfied for the overall model. In particular, the effects of age, male, and violent

offense charge violated the parallel slopes assumption, while those of non-white

and total number of charges did not (see Table 19.9).

To illustrate the application and interpretation of the partial proportional odds

model, we begin by allowing all five independent variables to have different effects

on the two ordinal logits:

ln 1

> 1 = ln

= Probation

= Jail or Prison

P Y

P Y

P Y

P Y

£( ) ( )

æ

è çç

ö

ø ÷÷

( ) ( )

æ

èè ç ç

ö

ø ÷ ÷

= 1 1t X b

and

ln ln Pr

Pr

P Y

P Y

P Y

P Y

£( ) >( )

æ

è çç

ö

ø ÷÷ =

=( ) =( )

æ

è

2

2

obation or Jail

ison çç ç

ö

ø ÷ ÷ = -t 2 2Xb .

Since there are two different ordinal logits being estimated, there are two full

sets of unique coefficients to interpret that illustrate the different effects the inde-

pendent variables have on the two different ordered logits. These coefficients are

presented in Table 19.8.

Some highlights found in the results are presented in Table 19.8:

Age:

Table 19.8 Partial Proportional Odds Model for Severity of Punishment—All Coefficients

Allowed to Vary

VARIABLE

PROBATION VS. JAIL AND/OR PRISON

PROBATION AND/ OR JAIL VS. PRISON

COEFFICIENT (SE) COEFFICIENT (SE)

Age −0.010(0.003) 0.011(0.003) Male 0.728(0.073) 0.916(0.074) Non-white 0.103(0.064) 0.190(0.048) Violent offense 0.101(0.081) 0.410(0.057) Total number of charges −0.006(0.020) 0.021(0.015) Constant (τ) −1.496(0.133) 2.073(0.116)

-

– The odds of probation versus jail or prison increase by a factor of exp

other variables in the model. For a ten-year increase in age, the odds of

626 C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

Steffensmeier, D., Ulmer, J. and Kramer, J. (1998), The Interaction of Race, Gender, and Age in

Criminal Sentencing: The Punishment Cost of Being Young, Black, and Male. Criminology, 36:

763–798.

Partial Proportional Odds Model for Severity of Punishment— Selected

Coefficients Allowed to Vary

Table 19.9

9

Substantively, these results indicate that offenders charged with a violent offense

are less likely to be treated leniently in the form of receiving either a probation or

a jail sentence and are more likely to receive a prison sentence. The results for age

suggest that while older offenders are more likely to receive a probation sentence

rather than a jail or a prison sentence, they are less likely to receive a probation or

a jail sentence rather than a prison sentence. These results may seem contradictory,

but one way of interpreting the pattern is that older offenders are more likely to

be sentenced to probation or prison, depending on crime, criminal history, and so

on, but less likely to receive jail sentences. This kind of variability in the effect of

age on sentencing has also been found in prior research.

Since the results of the Brant test indicated that there were only three of the

five independent variables that did not satisfy the parallel slopes assumption, we

have rerun the partial proportional odds model allowing only the effects for age,

male, and violent offense charge to vary. Table 19.9 presents these results. Since

the results presented here are nearly identical to those presented in Table 19.8 for

the coefficients allowed to vary, we limit our discussion here to the effects of male

(variable) and non-white (constrained):

9

0.010) = 1.105, controlling for all other variables in the model.

(0.011)) = 0.989 for a one-unit increase in age, controlling for all other vari-

ables in the model. For a ten-year increase in age, the odds of probation or

controlling for all other variables in the model.

� Violent offense charge:

smaller for offenders charged with a violent offense than for offenders charged

with a nonviolent offense, controlling for all other variables in the model.

smaller for offenders charged with a violent offense than for offenders charged

with a miscellaneous offense, controlling for all other variables in the model.

VARIABLE

CONSTRAINED PROBATION VS. JAIL AND/OR PRISON

PROBATION AND/ OR JAIL VS. PRISON

COEFFICIENT (SE) COEFFICIENT (SE) COEFFICIENT (SE)

Age 0.011 (0.003) Male 0.916 (0.074) Non-white 0.164 (0.043) Violent offense 0.415 (0.057) Total number of charges 0.013 (0.014) Constant (τ) 2.043 (0.115)

−0.010 (0.003) 0.731 (0.073)

0.096 (0.081)

−1.430 (0.128)

627

C h a p t e r S u m m a r y

In this chapter, we have examined two different multivariate statistical

models to be used when we are confronted with a categorical dependent

variable that has more than two categories. When the dependent variable

has three or more categories, we can use the multinomial logistic

regression model. The multinomial logistic regression model allows for

the computation of probabilities and odds ratios that indicate the effects

of the independent variables on the relative likelihood of the different

outcome categories.

Since the multinomial logistic regression model estimates a set of coef-

ficients for each independent variable, we have two issues of statistical

significance to assess: the individual coefficients and the overall effect of

the independent variable on the dependent variable. For the individual

coefficients, we continue to use the Wald statistic, which is distributed as

a chi-square statistic with one degree of freedom. For the overall effect of

the independent variable on the dependent variable, where we are testing

multiple coefficients, we can use the likelihood ratio (LR) test or the Wald

statistic. Both test statistics are distributed as a chi-square with J

of freedom.

When the dependent variable is measured at the ordinal level of meas-

urement, we can use the ordinal logistic regression model (or

proportional odds model). The ordinal logistic regression model also

C H A P T E R S U M M A R Y

� Male:

smaller for male offenders than for female offenders, controlling for all oth-

er variables in the model.

smaller for male offenders than for female offenders, controlling for all oth-

er variables in the model.

� Non-white:

smaller for non-white offenders than for white offenders, controlling for all

other variables in the model.

smaller for non-white offenders than for white offenders, controlling for all

other variables in the model.

Substantively, these results indicate that male and non-white offenders are less

likely to receive more lenient punishments (either probation or jail) and more likely

to receive a prison sentence. As noted above, this kind of pattern is consistent with

much of the prior research on punishment severity.

628 C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

allows for the computation of probabilities and odds ratios, but the focus

is on the likelihood of increasing or decreasing categories on the ordered

dependent variable. The ordinal logistic model assumes that the effects of

the independent variables are constant across the categories of the depend-

ent variable, which can be tested with the parallel slopes test that is

commonly reported in the output of most statistical programs. The parallel

slopes test statistic is distributed as a chi-square with K( J

freedom. The null hypothesis in such a test is that the slopes are parallel,

while the research hypothesis is that the slopes are not parallel.

When there is evidence that the parallel slopes assumption is not satis-

fied, we can use the partial proportional odds model that allows one

or more of the effects of the independent variable to vary across the levels

of the ordinal dependent variable. The interpretation of the results and the

tests for statistical significance work the same way in the partial proportional

odds model as they do in the ordinal logistic regression model.

K e y T e r m s

multinomial logistic regression A statistical

technique to predict the value of a dependent vari-

able with three or more categories measured at the

nominal level of measurement.

ordinal logistic regression (proportional

odds model) A statistical technique to predict the

value of a dependent variable with three or more cat-

egories measured at the ordinal level of measurement.

parallel slopes assumption In an ordinal

logistic regression model, the effect of each inde-

pendent variable is assumed to be constant across

all categories of the dependent variable.

partial proportional odds model An ordi-

nal logistic regression model that allows the effects

of one or more of the independent variables to

vary across the levels of the ordinal dependent

variable. Useful when the parallel slopes assump-

tion is violated.

thresholds Points that mark the limits of the under-

lying continuum measured by an ordinal variable.

S y m b o l s a n d F o r m u l a s

To calculate the probability that Y =m:

exp

exp P Y m

Xb

Xb

jj

J

m

1

= =

= /

_ ` _i j

i

To calculate the odds ratio in a multinomial logistic regression model for

P (Y =m) relative to P (Y =n), given a one-unit change in an independent

variable:

exp

exp

exp

exp

exp

exp OR

P Y n

P Y m

Xb

Xb

Xb

Xb

Xb

Xb |m n

jj

J

n

jj

J

m

n

m

1

1 =

=

= = =

=

=

/

/ _ _

` _

` _

_ _

i i

j i

j i

i i

629

To calculate the cumulative probability of P (Y ≤ m):

P Y m P Y j j

m

1

# = =

=

/_ _i i To calculate the odds ratio in an ordinal logistic regression model using

cumulative probabilities:

> OR

P Y m

P Y m

P Y m

P Y m

1m #

# # =

- =_

_ _ _

i i

i i

Ordinal logit equation:

> ln ln exp

P Y m

P Y m Xb Xbm m

# = - = -x x

J

L

K K _

_ _ N

P

O Oi

i i8 B

E x e r c i s e s

19.1 A large survey of adults asked about violent victimization experi-

ences. A question of particular interest to one researcher was the

location of the victimization event – home, work, or elsewhere.

She computed a multinomial logistic regression model that pro-

duced the following results:

INDEPENDENT VARIABLE HOME V. WORK ELSEWHERE V. WORK

Age (years) 0.01 0.05 Sex (1=Male, 0=Female) −0.19 0.22 Married (1=Yes, 0=No) 0.37 −0.13 Number of nights out per week for leisure 0.07 0.16

a. Calculate the odds ratio for each coefficient and explain what

each odds ratio means.

b. Calculate the coefficients and the odds ratios for the omitted

comparison and explain what each odds ratio means.

19.2 In an attempt to better understand how non-incarcerative punish-

ments were being used by judges, Blue State funded an evaluation

study of misdemeanor punishment decisions. The evaluators classi-

fied non-incarcerative sentences in the following four categories:

fine, restitution, community service, and electronic monitoring. The

researchers’ final analysis produced the following results:

COMMUNITY FINE V. RESTITUTION V. SERVICE V. ELECTRONIC ELECTRONIC ELECTRONIC

INDEPENDENT VARIABLE MONITORING MONITORING MONITORING

Any prior criminal record (1=Yes, 0=No) −0.06 −0.07 −0.10 Severity of offense −0.10 −0.12 −0.14 Employed (1=Yes, 0=No) 0.25 0.23 0.36

E X E R C I S E S

630 C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

a. Calculate the odds ratios for the effect of any prior record and

explain what each odds ratio means.

b. Calculate the odds ratios for the effect of severity of offense and

explain what each odds ratio means.

c. Calculate the odds ratios for the effect of employed and explain

what each odds ratio means.

d. Calculate the coefficients and the odds ratios for the comparison

between Fine and Community Service. Explain what each odds

ratio means.

19.3 Criminological theory has attempted to explain both the fre-

quency of delinquency as well as the type of delinquency an

individual is likely to commit. A longitudinal study of adolescents

tested for the effects of several background characteristics on the

likelihood an individual would commit a drug, property, violent,

or public order offense. The researchers used a multinomial

logistic regression model and found the value of the −2 log-

likelihood for the full model to be 5263.1. The values for the −2

log-likelihood for each of the independent variables was

reported as:

INDEPENDENT VARIABLE −2 LOG-LIKELIHOOD

Age 5264.7 Sex 5322.5 Race 5271.1 Grade point average 5267.9 Employment status 5414.6 Parental supervision 5272.3 Number of friends who had been arrested 5459.4

Calculate the LR Test statistic for each independent variable and

state whether this variable has a statistically significant effect on

type of crime (assume α = 0.05).

19.4 In response to public perceptions that the police in Riverside City

were too prone to use physical force on suspects, a study was

commissioned to examine the factors related to when police did

use physical force. After a year of data collection, the researchers

classified police use of force into the following three categories:

regression model of only demographic characteristics produced the

following results:

INDEPENDENT VARIABLE B

Age of officer (years) −0.02 Sex of officer (1 = Male, 0 = Female) 0.18 Race of officer (1 = White, 0 = Non-white) 0.13 Age of suspect (years) −0.03 Sex of suspect (1 = Male, 0 = Female) 0.33 Race of suspect (1 = White, 0 = Non-white) −0.11

Calculate the odds ratio for each coefficient and explain what each

odds ratio means.

None, Mild restraint, and Complete restraint. The ordinal logistic

631

19.5 A survey of adults in the US asked a series of questions about sup-

port for various policies related to the treatment of criminal

offenders. One question focused on the level of support for the

use of the death penalty – whether the respondent was opposed

to its use, neutral, or favored its use. An ordinal logistic regression

model that included age (years), sex (1 = male, 0 = female), race

(1 = African American, 0 = white), education (number of years

completed), and degree of political liberalism (1 = low, 10 = High)

produced the following results:

19.6 In a study of community perceptions of the local police department, indi-

viduals were asked a series of questions about their perceptions of police

behavior when interacting with local residents. Of particular interest to

the researchers was a question about trust that residents had in the police:

“Would you say that your level of trust in the police is …” The responses

were limited to Very Low, Low, Moderate, High, and Very High. The

researchers estimated an ordinal logistic regression model and Brant test

and found the following:

Independent Variable B Brant Test

Age (in years) 0.02 4.372

Sex (1 = male, 0 = female) −0.38 8.914

Race (1 = non-white, 0 = white) −0.42 12.695

Ever arrested? (1 = yes, 0 = no) −0.67 2.720

Ever reported a crime to the police? (1 = yes, 0 = no) −0.26 5.661

Total 34.362

b. Test the parallel slopes assumption for the full model and each coef-

recommendation be to the researchers about the use of the ordinal

19.7 A longitudinal study of delinquent and criminal behavior classified a

cohort of males (all the same race–ethnicity) into one of the three cat-

egories based on patterns of illegal behavior throughout adolescence:

Nondelinquent, Low-rate delinquent, and High-rate delinquent. On the

basis of Brant test results, the researchers estimated a partial proportional

odds model using a small subset of background characteristics to predict

delinquent group:

Independent variable

Nondelinquent vs. low and high rate

Nondelinquent and/or low rate vs. high rate

Coefficient Coefficient

Academic performance (1 = low to 10 = high) −0.12 −0.08

Risk scale (1 = low to 20 = high) 0.23 0.33

Parent arrested? (1 = yes, 0 = no) 0.47 0.85

Number of friends arrested 0.17 0.29

Level of parental supervision (1 = low, 10 = high) −0.09 −0.11

Interpret and explain these results.

E X E R C I S E S

632 C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

C o m p u t e r E x e r c i s e s

We will rely on two new commands to estimate multinomial logistic regression

estimate OLS regression models and binary logistic regression models. The data

file used to illustrate the application of the multinomial and ordinal models in

this chapter can be found in either SPSS (ca_scps9098.sav) or Stata (ca_scps9098.

dta) format. The illustration of the commands below assumes that you have

opened one of these files into SPSS or Stata and can also be found in the sample

syntax files in both SPSS (Chapter_19.sps) and Stata (Chapter_19.do) format.

SPSS

Multinomial Logistic Regression

To estimate a multinomial logistic regression model, you will need to use the

NOMREG command:

As in previous illustrations of SPSS commands, everything can be issued in upper

or lowercase, but we have used uppercase lettering to highlight the key components

of the command. The /PRINT= option forces SPSS to print all of the model and

individual coefficient results. Also take note that the default reference category in

SPSS is the category with the highest number (i.e., category value). To force a par-

ticular category as the reference, use the “base” option in parentheses.

To reproduce our results in Table 19.1, enter the following command:

Similarly, to reproduce the results in Table 19.3, where “Trial” was used as the

reference category, use the following command:

Note that the only difference between these two commands is changing the base

from 1 to 3.

Ordinal Logistic Regression

To estimate an ordinal logistic regression model in SPSS, use the PLUM

command:

and ordinal logistic regression models in both SPSS and Stata, but the basic format

of each of the new commands is nearly identical to previous commands used to

NOMREG depvar (BASE = #) WITH indepvars

/PRINT = PARAMETER SUMMARY LRT CPS MFI.

NOMREG casedisp (BASE = 1) WITH age male nonwhite violent

total_charges

/PRINT = PARAMETER SUMMARY LRT CPS MFI.

NOMREG casedisp (BASE = 3) WITH age male nonwhite violent

total_charges

/PRINT = PARAMETER SUMMARY LRT CPS MFI.

PLUM depvar WITH indepvars

/LINK = LOGIT

/PRINT = FIT PARAMETER SUMMARY TPARALLEL.

633

Since there are other types of models for ordinal regression, the /LINK= option

forces SPSS to estimate an ordinal logistic regression model. The /PRINT=

option forces SPSS to generate a full set of output that is consistent with the

items we have discussed in this chapter.

To reproduce the results in Table 19.6, enter the following command:

At the time of this writing, SPSS does not have the option of computing the

Brant test or estimating partial proportional odds models.

Stata

Multinomial Logistic Regression

To estimate a multinomial logistic regression model in Stata, we use the mlogit

command:

where baseoutcome refers to the category that should be used as the reference

category. The default in Stata is to use the category with the greatest number of

cases.

To reproduce our results in Table 19.1, enter the following command:

Note that the baseoutcome(#) option was used to force Stata into using “dis-

missal” as the reference category. If this option had been omitted, “plea” would

have been used as the reference category.

Similarly, to reproduce the results in Table 19.3, the command would be

Note, too, that the output provided in Stata in regard to statistical significance

is a z-score rather than a Wald statistic. This is not problematic, since there is a

simple relationship between the Wald and z-score:

Consequently, if you square the values of the z-scores in the Stata output, it

will reproduce the Wald statistics reported in the text (with some rounding error).

Ordinal Logistic Regression

To estimate an ordinal logistic regression model in Stata, use the ologit command:

To reproduce the results in Table 19.6, enter the following command:

C O M P U T E R E X E R C I S E S

PLUM ord_punishmentWITH age total_charges nonwhite male

violent

/LINK = LOGIT

/PRINT = FIT PARAMETER SUMMARY TPARALLEL.

mlogit depvar indepvars, baseoutcome(#)

mlogit casedisp age male nonwhite violent total_charges,

baseoutcome(1)

mlogit casedisp age male nonwhite violent total_charges,

baseoutcome(3)

ologit depvar indepvars

ologit ord_punishment age male nonwhite violent total_charges

Wald=z2.

634 C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

The Brant test requires downloading and installing a set of commands written by

Long and Freese. To install this package of user-written commands, enter the

following command (one time, only):

After this command has been run, the Brant test will be available and only

requires entering the command brant in the line after running the ologit

command. The following commands will reproduce the results for Tables 19.6

and then 19.7:

Partial Proportional Odds

Before running a partial proportional odds model, we again need to download

and install a user-written procedure called gologit2. The process here is the

same as before. Enter the following command (one time only):

This command will install gologit2 from an archive of procedures housed and

maintained by Stata. Once this command has been run, the basic structure of the

gologit2 command is

Long, J.S. and J. Freese, 2006, Regression Models for Categorical Dependent Variables

Using Stata, 2 ed., College Station, TX: Stata Press.

This will estimate a fully unconstrained model, where all of the independ-

ent variables are allowed to have variable effects across the levels of the ordinal

dependent variable.

To estimate a partial proportional odds model that constrains some independ-

ent variables to have the same effect and allows others to have variable effects,

use the pl(constrained_indepvars) option:

The variable names listed inside the parentheses with the pl option will be

constrained. Any other independent variable listed in the gologit2 command

line will be allowed to have variable effects.

To reproduce the results in Table 19.8, enter the following command:

To reproduce the results in Table 19.9, which constrains the effects of non-white

and number of charges, enter the following command:

10

10

net install spost9_ado,

from(http://www.indiana.edu/~jslsoc/stata)

ologit ord_punishment age male nonwhite violent total_charges

brant

ssc install gologit2

gologit2 depvar indepvars

gologit2 depvar indepvars, pl(constrained_indepvars)

gologit2 ord_punishment age male nonwhite violent

total_charges

gologit2 ord_punishment age male nonwhite violent total_charges,

pl(nonwhite total_charges)

635

Problems

C O M P U T E R E X E R C I S E S

1. As a first step in working with these two commands, open either the SPSS

(Chapter_19.sps) or the Stata (Chapter_19.do) files that will run all of

the commands described above. If you prefer to work directly with the

California data file, then open either the SPSS (ca_scps9098.sav) or the

Stata version (ca_scps9098.dta). The syntax and data files contain the fel-

ony arrest cases used in the examples in this chapter and will allow you to

reproduce the results in this chapter. Follow the commands as described

above or run the appropriate syntax file.

2. Compute a multinomial logistic regression model using employment status

(full-time, part-time, and not employed) as the dependent variable. From

the remaining variables included in the NYS data file, select at least five

variables that you think might have some relationship to an adolescent’s

employment status. Calculate the odds ratio for each coefficient, and

explain what each odds ratio means in plain English.

3. Compute an ordinal logistic regression model using the same set of

dependent and independent variables that you used in Question 2.

odds ratio means in plain English.

b. Test the parallel slopes assumption. If you have access to Stata, use the

c. If you estimate a different model, report the results, and explain what

4. Select one of the measures of delinquency and recode it into three cat-

egories representing no delinquency (a score of 0), one delinquent act, and

two or more delinquent acts. Compute an ordinal logistic regression model

using this recoded measure of delinquency as the dependent variable.

From the remaining variables in the data file, select at least five variables

that you think might have some relationship with this measure of

delinquency.

ratio means in plain English.

(Use the Brant test, if available.)

c. If you have access to Stata, estimate a partial proportional odds model

using the same dependent and independent variables that takes into

account your results from part (b). If you only have access to SPSS,

estimate a multinomial logistic regression model using the same

dependent and independent variables. Calculate the odds ratio for each

Questions 2 through 4 use the NYS data (nys_1.sav or nys_1_student.sav for SPSS and nys_1.dta for Stata).

636 C H A P T E R N I N E T E E N : M U L T I V A R I A T E R E G R E S S I O N

d. How does the substance of the results from the initial ordinal

logistic regression model compare to the substance of the results from

either the partial proportional odds model or the multinomial logistic

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_20,

© Springer Science+Business Media New York 2014

C h a p t e r t w e n t y

Multilevel Regression Models

H o w d o w e a n a l y z e c l u s t e r e d o r m u l t i l e v e l d a t a ?

V a r i a n c e c o m p o n e n t s m o d e l

What are Fixed and Random Effects?

How Do We Interpret the Results from a Variance Components Model?

What is the Intraclass Correlation and How is it Interpreted?

R a n d o m i n t e r c e p t m o d e l

How is the Random Intercept Model Different from the Variance Components

Model?

How is Explained Variance Computed and Interpreted?

What are Between and Within Effects?

How Do We Test for Differences in Between and Within Effects?

R a n d o m c o e f f i c i e n t m o d e l

How is the Random Coefficient Model Different from the Random Intercept

Model?

How Do We Test for Differences Between the Random Coefficient Model and

the Random Intercept Model?

A d d i n g c l u s t e r ( l e v e l - 2 ) c h a r a c t e r i s t i c s

How are Cluster (Level-2) Characteristics Added to the Random Coefficient

Model?

How Do We Interpret the Effects of Cluster (Level-2) Characteristics?

happens if there are clusters of cases in our data? For example, it is common in

community-based surveys to first select a set of neighborhoods from a larger

population of neighborhoods. Then, within these neighborhoods, a sample of

individuals is selected to respond to the survey questions. A second example may

involve researchers collecting data from a survey of youth by first sampling schools

and then administering the survey to students within the selected schools, which

may be further complicated by selecting a limited number of classrooms within

each school. A third example might involve an experiment with multiple treat-

ment and control sites, but the statistical analysis only distinguishes treatment from

control group, ignoring information about site. This kind of design, where there

are different levels of sampling, has the potential to allow the researcher to look at

the effects of the larger sampling unit (e.g., classroom, neighborhood, school, or

research site) on individual responses or outcomes.

Clustered data is also conceptualized as multilevel data (and hence, the

name of the statistical models we discuss in this chapter). In thinking about clus-

tered data as multilevel data, we would define the cluster—neighborhood, school,

treatment site above—as the level 2 data. The unique observations—typically,

individual cases—would be defined as the level 1 data. We are also not limited to

thinking of our data as having only two levels and could conceivably work with

data that have three or more levels. An example of three-level data might involve

a study that begins with a sample of schools (level 3), followed by a sample of

classrooms within each of those schools (level 2), and then the individual students

within each of the selected classrooms (level 1). Put in terms of clustered data, we

have students clustered within classrooms that are clustered within schools. The

more general point to describing clustered data as multilevel data is that the lowest

level of data will represent the total number of observations in our data—whatever

these observations happen to represent. Each additional level of clustering then

reflects a higher level of data.

Why does the clustering of data matter? There are both statistical and theoretical

reasons for why we may want to pay attention to clustered data. Statistically, obser-

vations within a cluster will tend to be more similar to each other than to observa-

638

hroughout the discussion of regression models—both OLS and logistic—T we have assumed a single sample of cases that represents a single sample frame. What

tions from different clusters. For example, survey respondents within a neighbor-

hood will tend to be more alike on key individual characteristics when compared to

survey respondents from another neighborhood, regardless of whether that other

neighborhood is across town or across the nation. The increased similarity of cases

within a cluster has consequences for our statistical tests, making it more likely that

we will find statistically significant results, since cases within a cluster will tend to

exhibit a similar pattern of association and consequently smaller standard errors.1

Theoretically, we may also have an interest in the multilevel structure of the

data that points to important effects of the cluster on relationships observed at

the individual level. For example, how might characteristics of a neighborhood—

such as poverty rate or unemployment rate—affect the relationship we might

observe between a respondent’s gender and fear of crime? If we find that female

respondents express higher levels of fear of crime, then we could ask the ques-

tion about whether this statistical relationship is the same across neighborhoods.

Does the effect of gender on fear of crime change across neighborhood? If the

effect is essentially the same across neighborhood, it tells us that neighborhood

may be unimportant for understanding fear of crime. In contrast, if we find the

effect of gender does vary across neighborhood, we may then want to investigate

why the effect varies. Is it due to other characteristics of the neighborhood, such

as poverty, unemployment, vacant houses, and the like? Multilevel data are struc-

tured in such a way that the clustering of cases presents both a challenge and an

opportunity to test for the effects of different independent variables measured for

different units of analysis.

In this chapter, we provide a brief introduction to what are known as multilevel

models2 that account for the clustering of cases—the multilevel structure of the

data—and can tell us interesting things about the nature of the statistical relation-

ships we are studying. We take as a given that there is something informative

or interesting about the multilevel structure of the data—that the clustering of

observations is not simply a statistical nuisance to be corrected. In the discussion

that follows, we restrict our attention to the analysis of dependent variables meas-

ured at the interval or ratio level of measurement. We also limit our discussion to

two-level models: we have individual-level data (level 1) nested within one set of

clusters (level 2). There is an extensive and growing literature on increasingly more

sophisticated multilevel models that account for dependent variables measured

at the nominal and the ordinal levels of measurement as well as multilevel data

with three or more levels.3 These models are too complex to examine in this brief

introductory treatment of multilevel models.

1If our concern is primarily in statistically accounting for clustered observations, we can use

what are referred to as robust standard errors and is available as an option in most statistical

packages. We do not discuss these standard errors in this chapter, but encourage curious read-

ers to consult Angrist, J.D., & Pischke, J. (2009). Mostly harmless econometrics: An empiri-

cist’s companion. Princeton, NJ: Princeton University Press. 2 These models are also known as mixed models, random effects models, and hierarchical linear

models. Since these phrases all take on different meanings across the social and behavioral sci-

ences, we use multilevel models, since that phrase seems to hold less potential for confusion

across disciplinary boundaries. 3 See, for example, Raudenbush, S., & Bryk, A. (2002). Hierarchical linear models:

Applications and data analysis methods, 2nd edn. Thousand Oaks, CA: Sage.

639M U L T I L E V E L R E G R E S S I O N M O D E L S

V a r i a n c e C o m p o n e n t s M o d e l

We begin our discussion of multilevel models by starting with the simplest case,

that of assessing how much cluster (i.e., group) means vary from each other. We

find that the most straightforward building block for accomplishing this is the sim-

ple one-way analysis of variance model. Recall from our discussion of analysis of

variance in Chap. 12 that our presentation emphasized how the decomposition of

the dependent variable’s total variance into two parts—between-group and within-

group—could be used to test whether group had any ability to “explain” (statisti-

cally) the total variation in the dependent variable. Fundamentally, we were trying

to assess whether the group means were significantly different from each other.

As a linear statistical model similar to OLS regression, we note that a one-way

analysis of variance can be written in equation form analogous to an OLS regres-

sion equation:

Y ij = b

0 + b

j X

ij + ϵ

ij ,

where Y ij is the value of the dependent variable for individual i in group j, X

ij rep-

resents the group (j) that an individual (i) belongs to, b 0 is a mean of the dependent

variable, b j a measure of the distance between each group and b

0 ij is the indi-

vidual residual. The specific meaning of b 0 and each b

j depends on how the X

ij have

been coded, which we explain below. As in the OLS regression model, the error term

( ij ) is assumed to have a normal distribution with a mean of 0 and variance of

e 2.

There are two primary ways that the X ij are typically coded to estimate this

model—regression coding and contrast coding. Regression coding refers to creat-

ing a series of dummy variables coded 0–1 for all groups except one, which is used

as the reference category. This procedure is identical to that discussed in Chap. 16

on the use of multiple dummy variables to represent a multi-category nominal

independent variable in OLS regression analysis. If regression coding is used, we

could rewrite the above equation as

Y ij = b

0 + b

1 D

1i + b

2 D

2i + . . . + b

(j−1) D

(j−1)i + ϵ

ij ,

where each D j represents a dummy variable indicator for up to j j groups

(since one group does not have an indicator variable and functions as the refer-

ence category). In this model, b 0 represents the mean for the omitted group and

each b j measures the difference between the mean for group j and the mean for

the omitted group.

Contrast coding works in a similar way to dummy variable coding, with the

difference being the reference category in the regression coding scheme (i.e., the

Contrast coding ensures that the sum of all the estimated effects (i.e., the b j ) is 0,

meaning that we can always determine the value for the reference category. We

could rewrite the one-way ANOVA equation using contrast coding as

Y ij = b

0 + b

1 C

1i + b

2 C

2i + . . . + b

(j−1) C

(j−1)i + ϵ

ij ,

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S640

ϵ

ϵ

A Regression Approach to Analysis of Variance

C 1 + C

2 ). Suppose that we esti-

mated this model and found the following:

b 0 = 2. 3, C

1 = 0. 3, C

2 = 0. 6:

Mean for Treatment 1: 2. 3 + 0. 3 = 2. 6

Mean for Treatment 2: 2. 3 + 0. 6 = 2. 9

Several decades of research have shown that defendants who have been released

during the pretrial period—the time between arrest and disposition (conclusion)

of a case—will tend to receive more lenient punishments if convicted. Those

defendants who remain in jail during the pretrial period, due to the judge deny-

ing release altogether or to the judge requiring a bail amount the defendant could

not pay, will typically be more likely to go to prison and to receive slightly longer

sentences if sentenced to prison. Importantly, these effects hold, even after tak-

ing into account other characteristics of the defendant, such as prior record and

severity of the offense.

As part of a larger project exploring judicial decision-making in the bail and

release decision, John Goldkamp and Michael Gottfredson conducted two stud-

ies in Philadelphia—the first a pilot study to examine the factors that influenced

the level of bail judges required and the second a test of whether the use of what

were called bail guidelines made the decision-making more consistent and equita-

ble across defendants. In this chapter, we focus our attention on data from the

first study. Goldkamp and Gottfredson selected a random sample of 20 judges to

where each C j represents the contrast coded indicator for up to j j

groups. In this model, b 0 represents the overall sample mean and the b

j measure of

the distance between each group mean and the overall sample mean. The distance

negative of the sum of the other effects.

For example, suppose we have an experiment with three conditions: Treatment

1, Treatment 2, and Control Group. If we designate the Control Group as the ref-

erence category, the contrast coding scheme would look like the following:

C 1

C 2

Treatment 1 Treatment 2 Control Group

1 0

−1

0 1

−1

Policy guidelines for bail: An experiment in

court reform. Phildelphia, PA: Temple University Press.

641

A Substantive Example: Bail Decision-Making Study

One of the most important decisions in the criminal process is the bail and

release decision made by the judge shortly after the arrest of most individuals.

V A R I A N C E C O M P O N E N T S M O D E L

C j is simply the

cases clustered evenly across the 20 judges. Put in the terminology of levels of

level 2 data.

We can consider each judge as a separate experimental condition—cases were

randomly assigned to each judge, ensuring broad similarity of the cases and thereby

creating the opportunity to assess how similarly or differently judges would process

these cases. Our attention in the example that follows is the bail decision for each

case that was indicated by the dollar amount the judge set for the person’s release.

Our dependent variable is the common logarithm of bail amount. Of the original

sample for the following analyses.

Table 20.1 presents the means and standard deviations for bail amount (in dol-

lars) and logged bail amount for each of the 20 judges. The average bail amount

required by each judge varies considerably. For example, the average bail amount

values for logged bail are much smaller and have a more limited range. This is due

to the fact that the logarithm used here—base 10—reflects the exponent for the

number of times 10 would be multiplied by itself to reproduce the bail amount

(e.g., log(100) = 2, log(1000) = 3, and so on). Consequently, a difference of 1.0 on

the logarithmic scale used here is equivalent to a ten-fold increase in bail amount.

To help establish a baseline for the multilevel models discussed in the remain-

der of this chapter, it will be useful to present the results from a one-way ANOVA,

where we treat each judge as a type of experimental condition and test for differ-

ences in mean bail amounts. Table 20.2 presents the results for logged bail amount

(since this will be the outcome measure we rely on in subsequent analyses in this

chapter). Note that for the coefficients reported in Table 20.2, we have used

dummy variable coding, using Judge 1 as the reference category. Consequently, the

model intercept represents the mean for Judge 1 and each of the reported coef-

ficients represents the difference between that judge’s mean and that for Judge 1.

6 The test of the null hypothesis of equality of means

across judge gives us an F-test value of F df 1 = 19, df

2

p < 0. 001. We then conclude that the mean logged bail amount across this sample

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S642

of 20 judges is significantly different.

Fixed and Random Effects

In the analysis of variance model as we have presented it above, the b j are referred

to as fixed effects, meaning that they represent a constant effect of the group

for all of the cases within that group. In experimental research, this implies that the

treatment received by each individual assigned to a particular condition is assumed

There was a 10 % rule in effect in Philadelphia at the time of the study, meaning that defend-

ants would only need to post 10 % of the dollar amount requested by the judge in order to

ensure their freedom during the pretrial period. 6 Due to rounding, some of the judge means estimated with the coefficients in Table 20.2 will

differ at the second decimal when compared to the means reported in Table 20.1.

to be the same. However, this assumption ignores the fact that there are often

differences in the kind of treatment each case within a particular condition may

receive—known as treatment effect heterogeneity. Although the lead researcher

may have designed a protocol that minimizes variations in the treatments received

by participants, the requirement for many treatments to be administered by anoth-

er human being introduces the possibility of differences in multiple administrations

of the treatment. For example, an individual police officer may think that he/she is

meeting the expectations of the researcher, but events and circumstances unique to

that officer, that day, that site, may result in slight differences in how a treatment

is administered.

If we assume that there are no systematic differences in how an individual

administers a treatment within a group, then the analysis of variance model can

be modified to incorporate random effects. These random effects allow for

Table 20.1 Means and Standard Deviations of Bail Amounts by Judge in Philadelphia

BAIL(DOLLARS) BAIL(LOGGED) BAIL(DOLLARS) BAIL(LOGGED)

JUDGE MEAN SD MEAN SD JUDGE MEAN SD MEAN SD

1 2 3 4 5 6 7 8 9 10

2076.30 4784.88 1901.68 1830.43 2204.58

17486.78 1842.76 2117.76 1652.86 3627.04

4513.85 8522.53 3414.13 2808.85 3890.20

60861.05 3320.16 3583.45 2099.00

10960.82

3.04 3.28 3.05 2.97 2.98 3.51 3.00 3.01 2.95 2.96

0.45 0.56 0.41 0.46 0.50 0.73 0.42 0.47 0.46 0.57

11 12 13 14 15 16 17 18 19 20

4576.19 7299.55 3385.96 7945.31 4944.00 8747.40 2184.62 4158.82 1956.30 6246.90

7170.58 16270.59 6310.05

24293.52 11726.94 19082.44 3182.93 7765.01 3053.54

23285.99

3.38 3.34 3.21 3.32 3.28 3.57 3.11 3.24 3.07 3.23

0.47 0.63 0.50 0.65 0.51 0.50 0.41 0.51 0.38 0.58

JUDGE COEFFICIENT (b J ) STD. ERROR

2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 Intercept

0.24 0.01

−0.08 −0.06

0.47 −0.05 −0.04 −0.09 −0.08

0.33 0.30 0.16 0.28 0.24 0.53 0.07 0.19 0.03 0.18 3.04

0.07 0.06 0.07 0.06 0.06 0.06 0.07 0.06 0.07 0.07 0.07 0.07 0.06 0.06 0.07 0.07 0.07 0.06 0.06 0.04

Table 20.2 Analysis of Variance of Logged Bail Amount

643V A R I A N C E C O M P O N E N T S M O D E L

variation within a group or condition, which acknowledges that there will be

differences in the treatments individuals in each group or condition receive. A par-

whatever the group represents, as a sample of all possible conditions within the

group.

The random effects model can be written as:

Y bij j ij= + +0 z ,

where Y ij , b

0 , and

ij j (Greek letter zeta) are the

random effects and represent the difference in mean for group j (as sampled) and

the overall sample mean b 0 j

are assumed to have a normal distribution with

a mean of 0 and variance of z

2.

Var( )Y = +s z e 2 2s

These are what are known as the variance components that can be used to

assess whether there is variation in the dependent variable across the group means.

How do we know when to choose a fixed or random effects model? Of pri-

mary consideration is whether the effect of the group is viewed as being consistent

across all cases within the group or whether the effect of the group represents a

sampling of all possible effects of the group. To the extent the effect of the group

is viewed as consistent across cases, then a fixed effects model is the optimal

choice and we would estimate a standard analysis of variance model. Alternatively,

if the effect of the group is expected to vary among cases within that group, then

a random effects model is the more appropriate choice.

From a practical standpoint, there are no firm rules about the sample sizes

needed to estimate models with fixed and random effects. The total sample size (N)

is used to estimate the fixed effects and much like estimating any linear regression

model, relatively modest sample sizes (100–200 cases) are often adequate. That

same guideline holds for multilevel models. Since the random effects are estimated

at the level of the cluster, it is unclear just how many clusters are necessary to esti-

mate a multilevel model, although 10–20 clusters provide a lower bound .

Intraclass Correlation and Explained Variance

Given the two measures of variance— z

2 and e 2—we can compute a measure of

explained variance (ρ):

r s

s s =

+ z

z e

2

2 2( ) ,

Rabe-Hesketh, S., & Skrondal, A. (2012). Multilevel and longitudinal modeling using

stata, volume I: Continuous responses, 3rd edn. College Station, TXL: Stata Press.

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S644

Equation 20.1

Equation 20.2

Equation 20.3

Note that we now have two measures of variance— z

2 and e 2—that reflect

variation between groups ( z 2 and within groups (

e 2) and combined represents the

total variation in the dependent variable:

)

allel way of considering random effects is to think of the conditions of the group,

ϵ

ϵ

where ρ has values ranging from 0 to 1 and measures the proportion of total

variation in the dependent variable that is due to the group. At ρ = 0, the group

explains none of the variation in the dependent variable, while at ρ = 1, the group

explains all of the variation in the dependent variable.

An alternative interpretation of ρ is as the intraclass correlation, which

indicates the level of absolute agreement of values within each group. By absolute

agreement, we’re trying to assess the extent to which the values within a group

correlation coefficients, we were assessing relative agreement in cases. For the

Pearson correlation, it was the relative agreement in values of two variables, while

for the Spearman correlation, it was the relative agreement in ranks of values of

two variables.

The intraclass correlation provides a measure that can be viewed in two differ-

ent ways. In part, it provides a measure of intergroup heterogeneity by measuring

how much of the total variation is due to the group. At the same time, it provides

a measure of within group homogeneity by measuring how similar the values are

within each group.

Statistical Significance

A natural question to arise in the application of random effects models is whether

the random effect—the estimate of variance z

2—is statistically significant.

Substantively, this is a question about whether allowing for random variation

around the overall mean adds anything statistically to the model over and above a

fixed effects model.

To test the statistical significance of z

2, we rely on a likelihood-ratio test,

similar to that used in previous chapters. To compute the LR test for the vari-

ance component, we need two values of the log-likelihood: (1) log-likelihood for

the ANOVA model and (2) log-likelihood for the random effects model (REM).

The LR test is computed as

c 2 2= - -( ( ) ( )).LL LLANOVA REM

The likelihood-ratio test statistic has a 2 sampling distribution with 1 degree of

freedom. We then divide the observed level of statistical significance for the com-

puted 2 by 2, since it is a test of variances, which can only take on positive values

and effectively truncates the sampling distribution to positive values.

Bail Decision-Making Study

We return to our example from the Bail Decision-Making Study and present the

results for a variance components model in Table 20.3. The model intercept is

variance of the groups ( z

2) is estimated to be 0.031, indicating the degree to which

the group (i.e., judge) means vary around the full sample mean. The unexplained

error variance ( e 2

To what extent does the judge making the decision about bail affect the

required amount? The intraclass correlation provides an indicator of the influence

645V A R I A N C E C O M P O N E N T S M O D E L

Equation 20.4

of the judge and is estimated to be 0.10 for logged bail. The intraclass correlation

can also be obtained from the two variance components estimates:

r s

s s =

+ =

+ =z

z e

2

2 2

0 03

0 03 0 27 0 10

( )

.

. . . .

The value of the intraclass correlation suggests that the decision-making judge only

accounts for about 10 % of the variation in the logged bail amounts.

In regard to statistical significance, we find that the log-likelihood for the

c 2 2 1871 73 1777 35 188 76= - - - =(( . ) ( . )) .

Based on 1 degree of freedom, we find the critical 2, assuming a p 2 p

before dividing it by 2, meaning the variance components model represents a sig-

nificant improvement over the standard one-way ANOVA model. Substantively,

these results indicate that the decision-making judge is important to understanding

bail amount requested.

R a n d o m I n t e r c e p t M o d e l

We can extend the basic variance components model to include independent

variables to estimate what is known as a random intercept model (RIM).

Alternatively, we could start with an OLS regression model and allow the intercept to

vary randomly across cluster. Either way, we estimate a model that takes on the form

Y b b Xij ij j= + + +0 1 1 z

where Y and X 1 represent the dependent and independent variables, respectively,

b 0 and b

1 represent the model intercept and the effect of X

1 j represents the

ij represents the random

error term for each individual observation.

In the random intercept model, the regression coefficients are interpreted in

the same way as discussed previously—a unit change in the independent variable

VARIABLE COEFFICIENT se z-SCORE

Fixed Effect: Intercept 3.17 0.04 77.73

Random Effects:

Intercept ( z

2) Error (

e 2)

ρ

0.031 0.270 0.104

Table 20.3 Variance Components Results for Logged Bail Amount

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S646

Equation 20.5

ϵ

ijϵ

is expected to result in a change in the dependent variable equal to b 1 and the

intercept is the expected value of the dependent variable if X 1 has a value of 0.

Explained Variance

With a random intercept model, we have three variations on explained variance

that help us to understand the patterns of association in our multilevel data. Recall

that the total variance in the dependent variable is Var(Y ) = z

2 + e 2. Following the

estimation of a random intercept model, we can compute the explained variance

(R2) of the dependent variable with the following equation:

R z e z e

z e

2 0 2

0 2

1 2

1 2

0 2

0 2

= + - +

+

( ) ( ) ,

s s s s s s

where z0

2 and z1

2 represent the variance of the random effects for the intercept

in the variance components model (subscripted with a 0) and the random intercept

model (subscripted with a 1), respectively. The error variance (unexplained vari-

ance) in the variance components model and the random intercept are indicated by

e0 2 and

e1 2, respectively. Consistent with previous interpretations of R2, a value of

0 indicates none of the variance in the dependent variable was explained, while a

value of 1 would indicate that all of the variance was explained.

We can further decompose the total explained variance into each of the two

levels of data: (a) explained variance at the level of the cluster (level 2) and (b)

unexplained variance at the level of the individual observations (level 1). The

explained variance at level 2 is:

Rz z z

z

2 0 2

1 2

0 2

= -s s s

.

The explained variance at level 1 is:

Re e e

e

2 0 2

1 2

0 2

= -s s s

.

What do these level-specific measures of explained variance tell us? The level 2

explained variance (R z

2) informs us how much of the random variation in cluster

means found in the variance components model is due to the individual level char-

acteristics of the observations (i.e., the set of independent variables we included in

the random intercept model). A value of 0 indicates that none of the cluster-level

variation was explained by the characteristics of the individuals included in the

data. Conversely, a value of 1 indicates that all of the cluster-level variation was due

to the characteristics of the individuals included in the data. Generally, the level 2

explained variance will tell us how much of the observed cluster-level variance is

due to the composition of the clusters.

The level 1 explained variance (R e 2) tells us how much of the error (residual)

variance—the “unexplained” variance—was reduced by adding in a set of inde-

pendent variables thought to be related to the dependent variable. A value of

0 indicates that none of the error variance was explained by the inclusion of

the independent variables, while a value of 1 would indicate that all of the error

R A N D O M I N T E R C E P T M O D E L 647

Equation 20.6

Equation 20.7

Equation 20.8

variance was explained by the independent variables. The explained variance at

level 1 is directly analogous to our discussion of R2 in linear regression models in

statistically explains the values on the dependent variable.

Statistical Significance

The results from a random intercept model will lead to testing the statistical signifi-

cance of both the effects of the independent variables included in the model and

the use of the random intercept model over an OLS linear regression model. In

regard to testing for the statistical significance of the effects of the independent var-

iables (e.g., b 1 in the equation above), we would use the following familiar equation:

z b

se b =

( ) ,

where se(b) is the standard error of the coefficient and z the test statistic assumed

to have a normal distribution. Similar to our discussion of testing the statistical

and 19, the coefficients in a random intercept model are also expected to be nor-

mally distributed. The maximum likelihood estimation procedure for the random

intercept model estimates standard errors that are adjusted for the clustered nature

of the data. Depending on the particular data we are working with, the standard

errors for the coefficients in a random intercept model will always be at least as

large, but more likely larger, than those estimated in an OLS linear regression

model with the same variables, since the clustered nature of the data has been

taken into account during the estimation process.

The test of the random intercept model against the OLS regression model

involves the use of the same type of likelihood-ratio test that we used in the test

of the variance components model against a one-way ANOVA model. The goal of

this test is to assess whether allowing the model intercept to vary randomly across

clusters improves the fit of the statistical model. The test is:

c 2 2= - -( ( ) ( )).LL LLOLS RIM

As before, the likelihood-ratio test statistic has a 2 sampling distribution with 1

degree of freedom for the one variance estimate of the random effects. We then

divide the observed level of statistical significance by 2, since it is a test of vari-

ances, which can only take on positive values and effectively truncates the sampling

distribution to positive values. Substantively, this test will indicate whether the

addition of a random intercept to a linear regression model makes a statistically

significant contribution.

Centering Independent Variables

There are many instances where the interpretation of the model intercept is

important to understand the implications of the estimated results. When we are

confronted with an independent variable that has no meaningful zero-point, the

meaning of the intercept is difficult to explain. One of the straightforward ways of

dealing with this issue is to center the independent variable. What do we mean by

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S648

Equation 20.9

Equation 20.10

center a variable? In general, centering refers to subtracting a measure of central

tendency (e.g., a mean) from each raw score.

The two types of centering of independent variable that are often most useful

for estimating multilevel models are (1) grand or overall mean centering and (2)

cluster or group-based centering. In grand-mean centering, we subtract the overall

sample mean of the independent variable from each observation

X Xij - ..

Note that the two periods in the subscript of X indicate that the mean is the same

for each individual observation i and group j.

In cluster-based centering, we subtract the relevant cluster (group) mean from

each observation

X Xij j- .

Note that there is a period in place of the i subscript to indicate the value of X j. is

the same for each observation i in a given group j.

How does the inclusion of a centered variable, instead of the original

independent variable, affect the interpretation of the results? The regression coef-

ficient for the centered value of the independent variable is interpreted in exactly

the same way: a unit change in X is expected to result in a change of b units in the

dependent variable. Keep in mind that all we have done by centering a variable is

shifted its location; its scale is unchanged and so the slope is unchanged (this was

The difference in interpretation that comes from using centered variables is in

the model’s intercept. In the grand-mean centering case, the model intercept now

represents the expected mean for a case that has a value equal to the overall mean

in the sample for X (which is equal to a value of zero on a grand-mean centered

variable). For the cluster-based centering, the model intercept now represents a

weighted average of the cluster means.

The other implication for interpreting the results is focused on the variance

component z

2. When grand-mean centering has been used, z

2 represents varia-

tion in the group means around the overall mean, identical to the case where no

centering has been used. With cluster-based centering, z

2 is interpreted as the vari-

ation of group means around the weighted average of cluster means.

When should centering be used? In general, centering an independent variable

measured at the interval or ratio level of measurement aids in the estimation of

multilevel models, particularly some of the more complex models that involve esti-

mating interaction effects across levels of data (which we discuss below). Centering

should not change the substance of the statistical results, since all that centering

accomplishes is a shifting of the independent variable so that a value of 0 repre-

sents either the overall mean for the sample or the cluster mean for each group.

Dummy variables may also be centered. In this case, the overall or group means

simply represent the proportion of cases in the full sample or the cluster that has

the characteristic measured by the dummy variable.

R A N D O M I N T E R C E P T M O D E L 649

Bail Decision-Making Study

To illustrate the application of random intercept models and the use of center-

ing, we return to the Bail Decision-Making Study. In the interest of keeping the

statistical model simple, we include only one independent variable predicting bail

amount. Clearly, there are numerous characteristics of defendants and their cases

that affect the bail decision. Our goal here is to illustrate how one would go about

estimating and interpreting a random intercept model with and without a centered

independent variable. We will develop a more complex model in a later section.

In making assessments about bail, the judge is expected to consider both the

chances of the defendant fleeing the community and the potential threat to public

safety. One indicator of a defendant’s overall risk is the number of prior drug

offenses in the person’s criminal history record. In general, the greater the evidence

of prior drug offending, the higher the perceived risk of some kind of pretrial mis-

conduct and consequently higher bail amounts being requested from defendants.

model (RIM) with no centering, RIM with grand mean centering, and RIM with

cluster mean centering. The OLS model ignores the clustering of cases by judge

and simply reports the effect of number of prior drug offenses on the amount of

bail requested. We see from the results in Column 1 that a one-unit increase in the

number of prior drug offenses increases logged bail by 0.12 units. Since talking

about changes in logged units likely makes little intuitive sense, a more meaningful

interpretation of this coefficient is in terms of percentage change in the logged out-

come variable. Our observed coefficient of 0.12 for number of prior drug offenses

can alternatively be interpreted as expecting bail amount to increase by 12 % for

each additional prior drug offense.

respectively) as does the effect of number of prior drug offenses (0.12 v. 0.11,

respectively). For the two RIMs with centering, the estimates for the intercept and

the effect of number of prior drug offenses are the same through two significant

digits (but not beyond). Note that the estimate of the intercept is the overall mean

Table 20.4

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RIM RIM

VARIABLE OLS RIM GRAND MEAN CENTERING CLUSTER CENTERING

Fixed Effects:

Intercept (se) (t or z-score)

2.69 (0.019)

(141.22)

2.71 (0.040)

(67.11)

3.17 (0.037)

(85.18)

3.17 (0.041)

(77.62)

Number of Prior

Drug Offenses (se) (t or z-score)

0.12 (0.004)

(29.87)

0.11 (0.004)

(29.87)

0.11 (0.004)

(29.87)

0.11 (0.004)

(29.80)

Random Effects:

Intercept ( z

2) Intercept (

e 2)

0.026 0.192

0.026 0.192

0.031 0.192

Regression Results for Logged Bail Amount on Number of Prior Drug Offenses

for logged bail amount reported earlier, which is expected under the use of grand

mean centering. The use of cluster mean centering estimates a model intercept that

is the weighted average of the group means. The reason the estimates for the inter-

cept are so similar in the two models using different types of centering is an artifact

of the study design that used a balanced approach to select the same number of

cases for each judge. The small variation in the number of cases per judge used in

these analyses accounts for the minor differences that do appear in the intercept

and the effect of number of prior drug offenses.

In regard to the effect of number of prior drug offenses, please note that the

effect is the same for all three RIMs, regardless of whether no centering (Column

used. This is to be expected, as centering an independent variable only shifts the

distribution of cases and does not alter anything else about the values, meaning

that the coefficient representing the effect of the independent variable should stay

the same.

As we did with the variance components model, we can test whether the RIM

offers a statistically significant improvement over the OLS regression model. This

again requires a chi-square test relying on the difference in the log-likelihood

values for the OLS model and the RIM. The log-likelihood for the RIM model

c 2 2 1513 37 1400 06 226 62= - - - - =(( . ) ( . )) .

Based on 1 degree of freedom, we find the critical 2

2 of 226.62 has a p

regardless of whether it is divided by 2. This result indicates that the RIM offers a

substantial improvement in the statistical model over the traditional linear regres-

sion model.

The explained variance for the RIM is based on variance estimates presented in

R2 to be:

R 2 0 031 0 270 0 026 0 192

0 031 0 270 0 276=

+ - + +

= ( . . ) ( . . )

. . . .

This shows that the inclusion of a single independent variable—number of prior

bail.

When we decompose R2 by level of data, we find the explained variance at level

2 (the judge) to be:

Rz 2 0 031 0 026

0 031 0 161=

− =

. .

. . .

Meaning that 16.1 % of the variation across judges is due to the composition of

the cases in their courtroom. Put another way, about 16 % of the variation across

judges is due to the number of drug offenses that defendants in their courtroom

possess prior to the current arrest.

R A N D O M I N T E R C E P T M O D E L 651

Although we have only added a single independent variable, we find the

explained variance at level 1 (error variance) to be:

Re 2 0 270 0 192

0 270 0 289=

− =

. .

. . .

What this means is a single independent variable—number of prior drug

offenses—explains nearly 29 % of the variation in the error variance.

What we have not yet addressed is how to determine which type of centering

to use, or whether to use any centering at all. We turn our attention to answering

this question in the next section.

Between and Within Effects

In our discussion of centering, we noted that centering variables can assist in

estimating multilevel models. What this means is that the algorithms used by vari-

ous statistical packages to estimate multilevel models perform better when using

centered independent variables. Although the explanation for how these algo-

rithms work goes beyond the focus of our text, we note that statistical packages

that estimate multilevel models often require multiple iterations to come to a

solution—the estimates of the intercept and the other coefficients.

We are still left with the question, then, of which method of centering to use.

How do we make this determination? One of the issues that naturally arises in the

study of clustered or multilevel data is whether the effects of the independent vari-

ables are the same across group as they are within group. For example, in the analy-

sis of judicial bail decision-making, we might wonder whether the effect of number

of prior drug offenses across judge—what we will call a between effect—is the

same as the effect of number of drug offenses processed by each judge—what

we will call a within effect. Conceptually, what we are attempting to get at is

whether a regression model for each judge (the within regression for each judge) is

parallel to a single regression line based on the means for each judge (the between

regression for all judges). To the extent the slopes (coefficients) are parallel, there

are similar between and within effects, meaning that each judge uses information

on number of prior drug offenses in approximately the same way. To the extent

the slopes differ, there are different between and within effects, indicating that

judges weight information about number of prior drug offenses differently.

Figures 20.1 and 20.2 present a way of thinking hypothetically about similar

and different effects. In each figure, the solid line represents the overall regression

slope for the effect of X 1 on Y . The dashed lines represent the regression lines

within each of the five clusters plotted. In Fig. 20.1, the between and within effects

are parallel to each other. The different placement of the dashed lines represents

j ), where two clusters have positive random

effects (and appear above the solid line), and three clusters have negative random

effects (and appear below the solid line). In Fig. 20.2, the between and within

effects are different—one slope is positive, one slope is essentially flat, while the

remaining three slopes are negative. Although these figures are informative in

highlighting similarities and differences in the between and within effects, it is

impractical to plot out regression lines for many groups, and will instead rely on a

statistical test for differences in the between and within effects.

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S652

Testing for Between and Within Effects

The most direct way of testing for a difference in the between and the within

effects is with the addition of the cluster mean for each independent variable

already included in the random intercept model. There are two equivalent ways

of testing for differences in the between and the within effects—both include an

estimate for the cluster mean, but differ in whether the original raw score or the

centered variable is included in the model. The following discussion illustrates the

differences and equivalences between the two approaches.

First, we start with a simple random intercept model that has only a single

independent variable:

Y b b Xij ij j= + + +0 1 1 z

We then add the cluster mean for X 1 and estimate the following equation:

Y b b X b Xij a ij b j j= + + + +0 1 1 1 1. z

Fig. 20.1 Parallel Between and Within Effects

Fig. 20.2 Different Between and Within Effects

R A N D O M I N T E R C E P T M O D E L 653

Y

X

Y

X

Equation 20.11

ij .ϵ

ij .ϵ

The coefficient b 1b

estimates the magnitude of difference in the between and the

within effects and captures any possible divergence in slope, such as that portrayed

in Fig. 20.2. If b 1b

is not significantly different from zero, then the between and

the within effects are the same, and the slopes for each cluster parallel those of

the overall effect. If b 1b

is significantly different from zero, then the between and

within effects diverge and the slopes are not parallel.

In the alternative, but fully equivalent, approach, we estimate a model that

includes the cluster mean centered value for X 1 and the cluster mean for X

1 as two

separate independent variables:

Y b b X X b Xij a ij j b j j= + - + + +0 1 1 1 1 1( ). . z

log( ) .BAIL b b DRUGOFF b DRUGOFFij a ij b j j= + + + +0 1 1 1 1 z

number of drug offenses, which represents the difference of the within and the

between effects. The within effects estimate has a value of b = 0. 11. The estimate

of the difference in the two effects has a value of b

effect is greater than the within effect, which confirms that the between and within

effects of number of drug offenses are significantly different from each other.

More importantly, in the context of understanding judicial decision-making, these

results imply that the 20 judges in this study differentially weight information about

number of prior drug offenses.

If we were interested in estimating both the within and between effects directly, we

could use the second approach described above and estimate the following equation:

log( ) ( ). .BAIL b b DRUGOFF DRUGOFF b DRUGOFFij a ij j b j j= + - + +0 1 1 1 1 1 z ++

VARIABLE ESTIMATE se z-SCORE

Intercept Number of Drug Offenses Number of Drug Offenses (Cluster means)

1.29 0.11 0.35

0.523 0.004 0.128

2.47 29.80 2.72

Table 20.5 Test of Between and Within Effects Similarity

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In this model, the coefficient b 1a

directly estimates the within effect and the coef-

ficient b 1b

directly estimates the between effect of X 1 . To obtain the difference in

between and within effects, we would simply subtract b 1a

from b 1b

.

Bail Decision-Making Study

In the Bail Decision-Making Study, one of the key areas of attention was a ques-

tion about whether judges weighted information about defendants in similar or

different ways. A direct test of this is provided by a test for similarity of between

and within effects. If we continue the example started previously using number of

drug offenses (DRUGOFF ) as the independent variable, we estimate the following

model:

Equation 20.12

ij .ϵ

.ijϵ

ij .ϵ

Table 20.6 presents the results from this analysis.

As expected, given the previous set of results, the between effect of number

of drug offenses is greater than the within effect of number of drug offenses,

again confirming that these 20 judges differentially used information about

drug offending when making bail decisions. There are two additional findings in

Table 20.6 worth noting. First, the effect of cluster deviations is the same as the

since this is just the effect for the cluster-mean centered number of drug offenses.

which is the estimate for the difference obtained directly using the first method

R a n d o m C o e f f i c i e n t M o d e l

A straightforward extension of the random intercept model involves thinking

about the effects of one or more of the independent variables in a multilevel model

also varying across cluster. Put another way, we may have justification, based on

prior research and theory, to expect the slope coefficients for a key variable to vary

across cluster. For example, in a study of fear of crime across neighborhoods, we

might expect the effect of gender to vary by neighborhood. Similarly, in our study

of judicial decision-making, we might expect judges to weight information about

cases and defendants differently, suggesting that we will find different slopes for

key predictors of the outcome variable.

The development of the random coefficient model (RCM) begins with

the random intercept model (here we have included only a single independent

variable X 1 ):

Y b b Xij ij j= + + +0 1 1 0z

where all terms are defined as above, except that there is now a 0 included in the

b 0 ).

For a random coefficient model, we add a random effect for the slope coefficient

in question. In our example, to estimate a model in which b 1 is allowed to vary

1j :

Y b b X Xij ij j j ij= + + + +0 1 1 0 1 1z z

VARIABLE ESTIMATE se z-SCORE

Intercept Number of Drug Offenses (Cluster deviations) Number of Drug Offenses (Cluster means)

1.29 0.11

0.46

0.523 0.004

0.128

2.47 29.80

3.61

Table 20.6 Test of Between and Within Effects Similarity

R A N D O M C O E F F I C I E N T M O D E L 655

Equation 20.13ij ,ϵ

ij ,ϵ

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S656

We can rewrite this equation to more directly link the random effects with the

proper slope coefficient:

Y b b X X

b b X

ij j ij j ij

j j ij

= + + + +

= + + + +

( ) ( )

( ) ( )

0 0 1 1 1 1

0 0 1 1 1

z z

z z

In this model, we estimate fixed effects for the model intercept (b 0 ) and the slope

coefficient for X 1 (b

1 ) and simultaneously estimate random effects for both the

0j 1j ).

Conceptually, what the random coefficient model does is analogous to estimat-

ing a regression model for each cluster and then examining whether the intercepts

and slope coefficients vary in any meaningful way across the clusters. We now turn

to a more formal examination of variance estimates from the random coefficient

model.

Variance Estimates

Similar to the variance components and random intercept models, the level 1 error

variance continues to be represented by e 2. The variance of the random effects

Variance of the intercept across cluster: z00 2

Variance of the slope coefficient across cluster: z11 2

We are now confronted with another choice in regard to the estimation of the

estimate the covariance of the random effects for the intercept and the slope,

which we can label either z01 2 or

z10 2 . What does this estimate of the covari-

0 1 assess? In general, it will indicate whether the magnitude of

the random effect for the intercept covaries with the magnitude of the random

effect for the slope coefficient. More directly, a positive value of 2 z01

indicates that

clusters with larger intercepts will tend to have larger values for the slope coeffi-

cient. Conversely, a negative covariance would suggest that smaller values of the

intercept are associated with larger values of the slope coefficient, and vice versa.

It is important to note that we cannot make direct comparisons of the variance

components 2

z00 , 2

z11 , and 2

z01 . Since the variance estimates reflect the different

metrics of the variables being analyzed, simply by changing the scale of one or

another variable, we could drastically alter the variance or covariance estimate. For

example, by expanding the scale of a variable, say from (0,1) to (0,10), it would

inflate the values of each variance estimate without changing the substantive inter-

pretation of the results.

Note on Explained Variance

In contrast to the variance components model and the random intercept model,

we are no longer able to compute estimates of explained variance in the random

coefficient model. The reason for this is the distribution of the residuals is heter-

Specifically, the total residual variance is now proportional to the value of the

independent variable with the random effect. For example, if we use the equation

ijϵ

ijj .ϵ

0j ),

ij ), and the product of X

1ij and the random effect

for X 1 1j

):

z z0 1 1j j ijX+ +

It is the product of X 1ij 1j

that creates the heteroskedastic error variance, since

the total residual will depend on the magnitude of X 1ij

. In short, as X 1ij

increases or

decreases, the magnitude of the residual will change.

Bail Decision-Making Study

To gain an appreciation of the random coefficient model, we begin our analysis of

the bail decision-making data by estimating regression equations for each of the

20 judges included in the sample. We continue to rely on the same simple model

of logged bail as the dependent variable and number of prior drug offenses as the

only independent variable in our model:

log( ) ( )BAIL b b DRUGOFFij j j ij= + +0 1 1

Note from the subscripts to the intercept (b 0 ) and the slope coefficient (b

1 ) that

there will be unique estimates for each value across the judges/clusters (j). Rather

than present the results from 20 regression analyses in a table, the results are pre-

sented graphically in Fig. 20.3, where each line represents the regression line for a

single judge. Clearly, the results show variation in the intercept across judge—note

the vertical placement of each regression line that reflects larger or smaller values

of the judge-specific intercept. We also note that there is variation in the regres-

sion slopes across the 20 judges—some of the slopes are steeper, some are flatter.

One way of starting to assess how similar or different the intercepts and

slope coefficients are for each judge can be viewed in a scatterplot of the inter-

graph represents one of the 20 judges included in the analysis. The spread of cases

Fig. 20.3 Judge-specific Regression Lines

R A N D O M C O E F F I C I E N T M O D E L 657

2 .5

3 3 .5

4 4 .5

5

L o

g (B

a il)

0 5 10 15 20

Number of Prior Drug Offenses

ij .ϵ

.ijϵ

Fig. 20.4 Scatterplot of Judge-specific Intercepts and Slope Coefficients

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S658

.0 5

.1 .1

5 .2

S lo

p e

2.4 2.6 2.8 3

Intercept

across both axes confirms what we viewed in Fig. 20.3—there is variation in the

a negative association: larger values of the intercept tend to have smaller slope

coefficients, while smaller values of the intercept tend to have larger slope coef-

-

tern suggests that judges with large intercepts—their cases receive relatively larger

bail amounts on average—tend to place less weight on the defendant’s number

of prior drug offenses. In contrast, for judges with smaller intercepts—their cases

tend to receive relatively lower bail amounts on average—place greater weight on

the defendant’s number of prior drug offenses.

The use of the random coefficient model offers a more efficient way of assess-

ing the similarities and differences across the judges, but most importantly, will

allow us to determine whether the variations in the traditional regression model

are statistically meaningful or reflect random variation in the values for each judge.

log( ) ( + )+( + X )+

( + )+(

0 0 1 1 1 1

0 0

Bail b b DRUGOFF

b

ij j ij j ij

j

=

=

z z

z bb DRUGOFFj ij1 1 1+ ) +z

to provide a ready point of comparison. Column 2 displays the results for the ran-

results using grand mean centering and cluster mean centering, respectively. The

estimates of the intercept and the slope coefficient in each of the columns are iden-

( 2 z11

) and the covariance of the intercept and slope random effects ( 2 z01

). As

before, we can test whether the addition of these random effects represents an

improvement in the statistical model over the previous model.

Recall from above that the test of statistical significance for the addition of a

random effect is a chi-square test that compares the log-likelihood values from two

ij

ij .

ϵ

ϵ

different models. In the present case, we can make two comparisons: (1) RIM v.

RCM with random effect for slope coefficient and (2) RCM with random effects

for the intercept and the slope v. RCM with the additional covariance of the ran-

dom effects. The first comparison assesses whether the basic RCM that adds a

random effect for number of drug offenses is an improvement over the RIM. The

second comparison tests whether the addition of the covariance of the random

effects adds to the model’s improvement over and above the basic RCM.

c 2 2 1391 74 1391 67 0 14= - - - =(( . ) ( . )) .

Based on 1 degree of freedom, we find the critical 2, assuming a p 2 p

of whether it is divided by 2. This result indicates that the RCM with the covari-

ance of the random effects does not improve the statistical model and could be

dropped from the analysis.

What do the results of these two comparisons mean? By finding the RCM

makes a significant improvement in the statistical model, we know that the

Table 20.7

R A N D O M C O E F F I C I E N T M O D E L 659

Regression Results for Logged Bail Amount on Number of Prior Drug Offenses

RIM RIM

VARIABLE OLS RIM GRAND MEAN CENTERING CLUSTER CENTERING

Fixed Effects:

Intercept (se) (t or z-score)

2.69 (0.019)

(141.22)

2.71 (0.037)

(72.95)

3.17 (0.037)

(85.54)

3.17 (0.041)

(77.72)

Number of Prior

Drug Offenses (se) (t or z-score)

0.12 (0.004)

(29.87)

0.11 (0.006)

(18.37)

0.11 (0.006)

(18.37)

0.11 (0.006)

(18.27)

Random Effects:

Intercept ( 2 z00

) Drug Offenses ( 2

z11 )

Covariance ( 2 z01

)

0.0209 0.0005

−0.0004

0.0258 0.0005 0.0016

0.0317 0.0005 0.0015

Comparison 1:

c 2 2 1400 06 1391 74 16 64= - - - =(( . ) ( . )) .

Based on 1 degree of freedom, we find the critical 2, assuming a p 2 p

before dividing it by 2. This result indicates that the RCM with a random effect

for the slope coefficient (number of prior drug offenses) represents a significant

improvement in the statistical model.

Comparison 2:

intercepts and the effects of number of prior drug offenses vary across the 20

judges in making bail decisions. The finding that including the covariances of the

random effects does not improve the model means the judge-specific intercepts

and coefficients for number of prior drug offenses are not correlated with each

between judge-specific intercepts and coefficients, there was no statistical evidence

of such a relationship once we more formally tested the model.

A d d i n g C l u s t e r ( L e v e l 2 ) C h a r a c t e r i s t i c s

Thus far in our discussion of multilevel models, we have focused strictly on char-

acteristics of the individual observations in the data—the level 1 characteristics.

One of the great strengths of multilevel models is the ability to include cluster-level

characteristics that will indicate how the effects of the independent variables may

vary across levels of a cluster characteristic. In the example we have used thus far

regarding judges and bail decision-making, we might hypothesize that characteris-

tics of judges would affect how each would weigh information about defendants in

making bail decisions. For example, gender of judge may alter the relationship that

we have observed between number of prior drug offenses and bail amount. Or,

years of service as a judge may affect the observed relationship between number

of prior drug offenses and bail amount. These are the kinds of questions to which

we now turn.

When considering adding cluster-level characteristics to a multilevel analy-

sis, the researcher is confronted with two important questions about a cluster

characteristic:

1. Is there an expectation that the cluster characteristic will directly affect the

dependent variable?

2. Is there an expectation that the effect of an independent variable will vary by

the level of the cluster characteristic?

Both of these questions force us to consider prior theory and research in thought-

fully developing our multilevel model. The first question is the more straightfor-

ward of the two questions and will often be a reflection of prior research showing

that the cluster characteristic is likely important to the dependent variable being

analyzed. For example, there is research indicating that gender of judge affects

how criminal defendants and offenders are treated. We would have justification for

hypothesizing that gender of judge would affect bail amount. Similarly, if we were

studying fear of crime across a large sample of different neighborhoods, we would

have justification for hypothesizing that official crime rates in the neighborhoods

may have a direct affect on an individual’s fear of crime.

The second question requires considerable care in developing the model,

especially since there is likely to be less evidence and/or theory on which to base

a hypothesis of an independent variable’s (level 1) effect varying by the level of

the cluster characteristic. In the literature on multilevel models, this kind of rela-

tionship is often referred to as a cross-level interaction, since they imply the

effect of one variable (the level 1 independent variable) changes across the levels of

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S660

another variable (the level 2 cluster characteristic). For example, in considering fear

of crime, we may hypothesize that official crime rates may interact with the effect

of age of a resident on fear of crime. Older individuals are more fearful of being

crime victims in general, and we could hypothesize that as the neighborhood crime

rate increased, there was a multiplier effect on the fear of crime among elderly

residents. At the same time, neighborhood crime rates may not affect the level of

fear of younger individuals.

Although the interpretation of cluster-level characteristics in a multilevel model

may become complicated, their inclusion in the statistical model is not compli-

cated. For the situation where we expect the cluster characteristic to have a direct

effect on the dependent variable, we simply include it as an additional independent

variable (denoted with a W in the following discussion) in our random intercept or

random coefficient model. In the form of a simple random coefficient model, we

would estimate the following model:

Y b b X bW Xij ij j j j ij= + + + + +0 1 1 3 1 0 1 1z z

Note that our cluster characteristic W 1j has only a j subscript, indicating that the

values of W 1 vary by cluster j but will be the same for all cases within that cluster.

Like any other variable we might include in a linear regression model, the cluster

characteristic can be a dummy variable or an interval level variable. The interpre-

tation of the cluster characteristic’s effect (b 3 ) is no different than that for other

independent variables included in the model: a unit change in W 1 is expected to

change the value of the dependent variable by b 3 .

If we expect the effect of one of our level 1 independent variables to vary by

level of a cluster characteristic, we include an interaction term between the two

variables—our cross-level interaction—and add it to the model. Continuing the

same random coefficient model from above, we would include an interaction

between X 1 and W

1 :

Y b b X bW b X W Xij ij j ij j j j ij= + + + + + +0 1 1 3 1 4 1 1 0 1 1z z

How do you interpret these results and make sense of the interaction effect? In

-

ferent ways: we fix the value of one variable and assess the effect of the second

variable. In a multilevel model, we will always fix the level (value) of the cluster

characteristic first and then interpret the effect of the level 1 independent variable.

In the equation above, the effect of X 1 can be written as

b X b X Wij ij j1 1 4 1 1+ .

If W 1 is a dummy variable, then for W

1 = 0, we have

b X b X b Xij ij ij1 1 4 1 1 10+ =( ) ,

meaning the effect of X 1 at W

1 = 0 is simply b

1 . In contrast, if W

1 = 1, we have

b X b X b X b X b b Xij ij ij ij ij1 1 4 1 1 1 4 1 1 4 11+ = + = +( ) ( ) ,

A D D I N G C L U S T E R ( L E V E L 2 ) C H A R A C T E R I S T I C S 661

Equation 20.14

Equation 20.15

ij .ϵ

ij .ϵ

meaning the effect of X 1 at W

1 = 1 is b

1 + b . For cluster-level characteristics

measured on an interval scale of measurement, we would typically pick out some

meaningful values and highlight the effect of X 1 at those values.

A Substantive Example: Race and Sentencing Across Pennsylvania Counties

In an analysis of sentencing decisions in Pennsylvania in the 1990s, Britt used

a multilevel model to assess the effects of various social, economic, and crime

measures on punishment severity decisions for offenders . Of particular interest

in Britt’s analysis was the effect of these kinds of community characteristics on

the effect of offender’s race on punishment severity. For example, were black

offenders punished more severely in those counties with higher rates of crime?

Alternatively, were black offenders punished less severely in those counties with

proportionally larger black populations? The theoretical rationale for these differ-

ent hypotheses is presented in the original paper.

In what follows, we highlight a few of his key findings as examples of the power

-

In the context of a multilevel model, the sentence length decisions represent the

individual-level data (i.e., level 1) and the counties in which the sentences were

given represent the cluster-level data (level 2). We focus our discussion on the

following variables:

Since our intention here is to illustrate the use and interpretation of multilevel

showing only those elements focused on the effect of county characteristics on

overall sentence length and county characteristics interacting with race of the

offender. Omitted from the table are numerous case and offender characteristics

relevant to predicting punishment severity, such as offense severity, criminal his-

tory, plea bargaining, and the like.

To begin the interpretation of the results, note that percentage of the popula-

tion classified as black, difference in white and black per capita income, percent-

age living in urban areas, and trend in unemployment all have direct effects on

Britt, C.L. (2000). Social context and racial disparities in punishment decisions. Justice

Quarterly

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S662

Individual-level (Level 1)

Sentence length: Months sentenced to jail or prison.

Black: Coded as 1 if offender was black, 0 if offender was white (all other

cases were excluded from analysis).

County-level (Level 2)

Percentage of population living in an urban area.

Average violent crime rate.

Table 20.8

A D D I N G C L U S T E R ( L E V E L 2 ) C H A R A C T E R I S T I C S 663

black offender and the interaction terms for black offender with percentage of the

population classified as black and the average violent crime rate. The interpretation

of the results at this point is no different than the interpretations we made in the

linear regression model. For the direct effects on sentence length:

As the percentage of blacks in a county increases, the average sentence length

decreases.

As the percentage of a county’s population living in an urban area increases,

the average sentence length increases.

As the county-level unemployment rate increased over time, the average

sentence length increases.

For the cross-level interaction effects of offender race with percentage of black

residents in a county and average crime rate, the interpretations follow the logic

to any other interaction effect. In general, what we find is that as the percentage

of a county’s population classified as black increases, the effect of being black

decreases, meaning that black offenders received significantly shorter sentences

than white offenders overall, but the magnitude of this difference increases as the

percentage of blacks in a county increases. Conversely, in counties where the aver-

age violent crime rate was higher, the effect of being black increases, meaning

that the punishments received by black offenders were more severe than those for

white offenders in counties with higher violent crime rates.

The following hypotheticals will help to illustrate these patterns. Suppose that

we have four different counties with the following characteristics:

County A: Percentage Black = 10, Violent Crime Rate = 100

County B: Percentage Black = 20, Violent Crime Rate = 100

County C: Percentage Black = 10, Violent Crime Rate = 200

County D: Percentage Black = 20, Violent Crime Rate = 200

The equation for the effect of being a black offender on sentence length is:

- - ´ + ´2 277 0 315 0 009. . .Black Black PercentBlack Black ViolentCrimeeRate.

The effect of being a black offender in County A:

− − × × + × × = −2 277 0 315 10 0 009 100 4 527. . . . .Black Black Black Black

VARIABLE ESTIMATE se z-SCORE

Intercept Percentage Black Percentage Urban Trend in Unemployment

Black Black × Percentage Black Black × Violent Crime Rate

13.968 −0.161

0.026 0.820

−2.277 −0.315

0.009

0.574 0.021 0.007 0.228

0.618 0.092 0.003

24.334 −7.750 3.910 3.596

−3.684 −3.292 3.000

Multilevel Regression Results for Sentence Length

The effect of being a black offender in County B:

− − × × + × × = −2 277 0 315 20 0 009 100 7 677. . . . .Black Black Black Black

The effect of being a black offender in County C:

− − × × + × × = −2 277 0 315 10 0 009 200 3 627. . . . .Black Black Black Black

The effect of being a black offender in County D:

− − × × + × × = −2 277 0 315 20 0 009 200 6 777. . . . .Black Black Black Black

In all cases, the effect of being a black offender resulted in a shorter sentence, 9 For the two pairs coun-

ties with matching percentage black populations, the increase in the violent crime

rate resulted in a shrinking of the effect of being black and moving the coefficient

closer to 0, where there is no difference between black and white offenders. For

the two pairs of counties with matching violent crime rates, as the percentage of

the black population increased, the sentence disparity increased further, with black

offenders receiving even more lenient sentences.

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S664

C h a p t e r S u m m a r y

9 While this finding often strikes many criminal justice students as counterintuitive, it is

consistent with much of the research done on the race effects on sentencing outcomes. What

is not shown here, but is included in Britt’s article, is the effect of race on the likelihood of

being incarcerated, where black offenders were much more likely than white offenders to be

sentenced to prison. The findings here just highlight that once sentenced to incarceration, the

length of time is shorter for black offenders compared to white offenders. In Pennsylvania, this

has typically taken the form of more black offenders being sentenced to local jails for relatively

short periods of time, while white offenders who have received incarceration sentences will be

sent to state prisons for relatively longer stays.

Multilevel models offer an important extension to traditional linear regres-

sion models by statistically accounting for possible clustering in a sample of data.

Observations that come from the same cluster (e.g., multiple survey respondents

within the same neighborhood, multiple cases processed by a judge or prosector,

and so on) will tend to be more similar to each other than to observations from

different clusters. This results in an increased likelihood of finding statistically

significant effects, since many of the cases within a cluster will exhibit a similar

pattern of association. Multilevel models account for clustering by allowing for

random variation in the intercepts and possibly the coefficients of the independ-

ent variables. Models that allow for variation in the model intercept are referred to

as random intercept models (RIMs), while models that contain a random

intercept and at least one random slope coefficient are referred to as random

coefficient models (RCMs).

Variation in both the intercept and the coefficient for an independent vari-

able are measured with what are called variance components—measures of

how much the intercept and slope may vary across cluster. These are also called

665K E Y T E R M S

the random effects. We test the significance of the variance estimates with a

chi-square test that compares the model with the random effects against a linear

regression model without any random effects.

In estimating multilevel models, we may also center the values of the independ-

ent variables. Centering can take on two forms: grand-mean or cluster-mean center-

ing. Centering has no effect on the interpretation of the slope coefficients, but will

alter the substantive meaning of the model intercept. In grand-mean centering,

the model intercept represents the overall sample mean for the dependent variable.

In cluster-mean centering, the model intercept represents a weighted sample

mean for the dependent variable that is conditioned on the number of cases per

cluster. The more balanced the size of the clusters, the more similar the two esti-

mates of the model intercept will be. In general, centering the values of the inde-

pendent variables will tend to simplify the estimation of the overall multilevel model.

Centering also allows for the testing of different between and within

group effects of the independent variables. Much of the research in criminology

and criminal justice assumes the between and within effects are the same without

ever testing for similarity. By estimating models that include the cluster means as

independent variables, it is possible to assess directly how or whether the between

cluster effects are the same as the within cluster effects. If the results of these

tests indicate the effects are the same, then grand-mean centering is appropriate.

Alternatively, if the between and within effects are different, then cluster-mean

centering is a more appropriate technique.

K e y T e r m s

between effect Effect of an independent

variable on the dependent variable using the

cluster as the unit of analysis—a regression of

cluster-level averages across all the clusters

included in the analysis.

cluster-mean centering Computed difference

between the observed raw score on some variable

for each observation in the sample and the cluster

mean for that variable.

contrast coding A method for recoding a

multi-category nominal variable into multiple

indicator variables (one less than the total number

of categories), where the indicator category is

and all other categories are coded as 0. Contrast

coding ensures that the sum of all the estimated

effects for the indicator variable is equal to 0.

cross-level interaction An interaction effect

included in a multilevel model between a level 1

independent variable and a level 2 cluster

characteristic.

fixed effects A descriptive label for the

regression coefficients (b k ) estimated in a model

with random effects. Fixed effects represent the

average effects of the independent variables on

the dependent variable across all individuals and

clusters in a multilevel model.

grand-mean centering Computed difference

between the observed raw score on some variable

for each observation in the sample and the overall

sample mean for that variable.

intraclass correlation A measure of

association that measures the level of absolute

agreement of values within each cluster.

multilevel data Sample data where individual

observations (level 1 data) are clustered within a

higher-level sampling unit (level 2 data).

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S666

random coefficient model A linear regres-

sion model that allows the intercept and the effect

of at least one independent variable to vary

randomly across cluster—random effects are

included for the model intercept and at least one

independent variable.

random effects A descriptive label for the

random error terms included in a multilevel

model that allow for variation across cluster

from the sample average estimated in the

fixed effects. Random effects are assumed

to be normally distributed in most multilevel

models.

random intercept model A linear regression

model that allows the intercept to vary randomly

across cluster—random effects are included for

the model intercept.

regression coding A method for recoding a

multi-category nominal variable into multiple

indicator dummy variables (one less than the

total number of categories), where the indicator

category is coded as 1 and all other categories

are coded as 0. The reference category does not

have an indicator variable and is coded as a 0 on

all the indicator dummy variables.

variance components model A one-way anal-

ysis of variance model that includes random effects

for each cluster that assesses whether there is

random variation in the mean of the dependent

variable across the clusters included in the analysis.

within effect Effect of an independent

variable on the dependent variable within each

cluster and then averaged across all clusters or

groups included in the analysis.

S y m b o l s a n d F o r m u l a s

0j Random effect for the model intercept b

0

kj Random effect for the regression coefficient for the independent variable k

(bk)

2 z0

variance of the random effect for the model intercept in a variance compo-

nents model

2 z1

variance of the random effect for the model intercept in a random intercept

model

2 e0

error variance (unexplained variance) in a variance components model

2 z1

error variance (unexplained variance) in a random intercept model

2 z00

Variance of the random effect for the model intercept in a random coef-

ficient model

2 zkk

Variance of the random effect for the independent variable k in a random

coefficient model

2 z0k

Covariance of the random effects for the model intercept and for the inde-

pendent variable k in a random coefficient model

General equation for the variance components model:

Y bij j= + +0 z

General equation for the random intercept model with one independent variable:

Y b b Xij ij j= + + +0 1 1 0z

ijϵ

ijϵ

General equation for the random coefficient model with one independent variable:

Y b b Xij j j ij= + + + +( ) ( )0 0 1 1 1z z

Likelihood-ratio test for variance components:

c 2 2 1 2= - -( ( ) ( ))LL LLModel Model

Equation for computing the intraclass correlation (ρ):

r s

s s =

+ z

z e

2

2 2( ) ,

Equation for grand-mean centering:

X Xij - ..

Equation for cluster-mean centering:

X Xij j− .

Equation for explained variance in a random intercept model:

R z e z e

z e

2 0 2

0 2

1 2

1 2

0 2

0 2

= + - +

+

( ) ( )s s s s s s

Equation for the explained variance at level 2 of a random intercept model:

Rz z z

z

2 0 2

1 2

0 2

= -s s s

Equation for the explained variance at level 1 of a random intercept model:

Re e e

e

2 0 2

1 2

0 2

= -s s s

E x e r c i s e s

1. Researchers interested in the possible effects of neighborhood poverty on

patterns of intimate partner violence (IPV) gathered interview data from

-

lishing neighborhood variability in IPV, the researchers estimated a vari-

ance components model and obtained the following results:

- = -2 2000LL( )ANOVA

- = -2 1900LL( )REM

E X E R C I S E S 667

ijϵ

s z 2 0 10= .

s e 2 0 40= .

b. To what extent does neighborhood affect the prevalence of IPV? Cal-

meaning.

self-reported delinquent behavior, which was measured with a scale that

researchers were particularly interested in the effects of academic perfor-

mance on delinquent behavior and estimated a random intercept model,

obtaining the following results:

VARIABLE COEFFICIENT

Intercept GPA Educational Aspirations Father’s Education Mother’s Education

0.50 − 0. 30 − 0. 05 − 0. 20 − 0. 35

2 z 2 e

0.15 0.95

components model and explain its substantive meaning.

model and explain its substantive meaning.

between the variance components and random intercept models.

covariates were added to the model? Describe how the statistical

3. A study of anti-social behavior among children collected information on

study. Based on prior research, the investigators expected the

within-family and between-family effects of parental attachment to be

different. The investigators found the following effects:

b Attachment

se

b ClusterMeanofAttachment

se = 0. 03

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S668

a. Explain whether the investigators found evidence of different within-

family and between-family effects of parental attachment.

b. Which type of centering would be most appropriate for these data if

the investigators simply want to estimate a single effect for parental at-

tachment that ignores the between and the within effects?

-

istics on individuals’ perceptions of fear of crime victimization selected a

within each neighborhood. Using a fear of crime scale as the dependent

variable and demographic characteristics as covariates, the researchers

estimated a series of regression models to test for random effects across

neighborhood. The results appear in the following table (assume all fixed

effects are statistically significant with p

result.

a random sample of 200 cases among those involving sentences to jail or

prison. Sentence length was measured as the number of months sentenced

to incarceration. To assess the impact of legal characteristics on sentence

length decisions, the researchers developed measures of severity of the

conviction crime and of criminal history. After establishing that the inter-

cept varied across judge, the researchers investigated a series of random

coefficient models that examined whether there were correlations of the

random effects for the intercept and the two covariates. The following

table presents their results:

VARIABLE OLS RANDOM INTERCEPT MODEL

RANDOM COEFFICIENT MODEL

Fixed Effects:

Intercept Age Female Black

4.50 0.07 0.78 0.64

4.40 0.06 0.80 0.52

4.50 0.05 0.90 0.48

Random Effects:

Intercept ( 00 2 )

Age ( 11 2 )

Female ( 22 2 )

Black ( 33 2 )

0.03 0.03 0.01 0.12 0.05

Model Information:

− 2L L −3176 −2273 −2095

E X E R C I S E S 669

your result.

b. Test whether the addition of the random effect covariances is statisti-

C o m p u t e r E x e r c i s e s

We have noted throughout this chapter that multilevel models can become com-

plex very quickly and so we have tried to keep our focus on the basic elements

of multilevel models. The essential syntax required to estimate these models is

generally straightforward and does not become complicated until we start cus-

tomizing the statistical model. The examples below and in the accompanying

syntax files for SPSS (Chapter_20.sps) and Stata (Chapter_20.do) illustrate key

components to the multilevel commands without getting bogged down in too

many of the options and details.

SPSS

In SPSS, random intercept models and random coefficient models are both esti-

mated with the MIXED command:

where the first line has the same structure as many of the other commands in SPSS.

The /PRINT=SOLUTION TESTCOV option requests SPSS to print out the

coefficient table (SOLUTION) and to test the random effect variances and covari-

ances for statistical significance (TESTCOV). The /METHOD=ML option forces

SPSS to estimate the models with maximum likelihood, while the FIXED = option

VARIABLE RIM RCM 1 RCM 2

Fixed Effects:

Intercept Severity of Offense Criminal History

5.19 2.72 5.31

5.23 2.65 5.62

7.98 1.95 4.97

Random Effects:

Intercept ( 00 2 )

Severity of Offense ( 11 2 )

Criminal History ( 22 2 )

Intercept-Severity ( 01 2 )

Intercept-History ( 02 2 )

Severity-History ( 12 2 )

0.29 0.26 0.16 0.21

0.21 0.12 0.18 0.06 0.04 0.11

Model Information:

− 2LL −2984 −2781 −2779

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S670

MIXED dep_var WITH list_of_indep_vars

/PRINT = SOLUTION TESTCOV

/FIXED = INTERCEPT list_of_indep_vars

/RANDOM = INTERCEPT random_indep_var(s) |

SUBJECT(cluster_variable) COVTYPE(UN) .

line should include INTERCEPT and all of the independent variables included in

the model. The /RANDOM = option will then determine whether a RIM or RCM

will be estimated. If a RIM is to be estimated, the /RANDOM line simplifies to:

For an RCM that estimates random effects covariances and variances you will

need to add the COVTYPE(UN) option to the /RANDOM line, since the

default in SPSS is to estimate an RCM without random effect covariances:

The output from running any of these commands is the same and includes

tables of coefficients (the fixed effects), of covariances and variances for the ran-

dom effects, and of model summary statistics.

Stata

Random Intercept Models

To estimate random coefficient models in Stata, you will have access to two dif-

ferent commands: xtreg and xtmixed. The structure to the xtreg command is:

The basic format is similar to the regress command—the difference follows the

comma, where the i() option indicates which variable provides information on

the cluster. In this chapter, the cluster was the judge identifier. The var option

is included to force Stata to estimate the variance of the random intercept—the

default output in Stata is to report the square root of the variance (i.e., the standard

deviation of the random effect). Finally, the mle option forces Stata to compute

the maximum likelihood estimates for the RIM.

The use of the xtmixed command is similar:

where instead of a comma immediately following the list of independent variables,

we have two vertical bars (||) that are followed by the cluster variable with a colon

(:) appended, and then the comma and request for maximum likelihood estimates.

As we explain in the next section regarding the estimation of random coefficient

models, the vertical bars will be useful for designating random coefficients.

Random Coefficient Models

The xtreg command cannot be used for random coefficient models, while

xtmixed can be used. To use xtmixed for an RCM, we simply add the name of

the independent variable(s) after the : that we want to estimate random effects for:

C O M P U T E R E X E R C I S E S 671

/RANDOM = INTERCEPT | SUBJECT(cluster_variable) .

For an RCM that only estimates random effects variances:

/RANDOM = INTERCEPT random_indep_var(s) |

SUBJECT(cluster_variable) .

/RANDOM = INTERCEPT random_indep_var(s) |

SUBJECT(cluster_variable) COVTYPE(UN) .

xtreg depvar list_of_indep_vars, i(cluster_variable) mle var

xtmixed depvar list_of_indep_vars || cluster_variable:,

mle var

The structure to the rest of the command is the same—all we do is include one or

more independent variable names after the : and before the comma. Note that the

default RCM in Stata is to estimate a model with no covariances of the random

effects. If we want to estimate the covariances of the random effects, we add the

option cov(unstructured) to the command line:

The output will then contain the variances for the intercept, any independent vari-

ables with effects allowed to vary across cluster, and all possible covariances of the

random effects.

Problems

1. Open the Bail Decision-Making data file (bail-data-example.sav or bail-

data-example.dta). The sample syntax files for this chapter include the

syntax required to reproduce most of the tables in this chapter. Work your

way through one of the syntax files and make sure you understand how it

works.

2. Using the Bail Decision-Making data file, use logbail as the dependent

-

ing models:

a. Random intercept model:

estimate the covariances of the random effects.

i. Did anything change in regard to the effects of these independent

-

mate the random effect covariances.

i. Did anything change in regard to the effects of these independent

C H A P T E R T W E N T Y : M U L T I L E V E L R E G R E S S I O N M O D E L S672

xtmixed depvar list_of_indep_vars || cluster_variable:

random_indep_var(s), mle var

xtmixed depvar list_of_indep_vars || cluster_variable:

random_indep_var(s), mle var cov(unstructured)

3. Continue to use the Bail Decision-Making data file, change the dependent

variable to the bail amount requested and estimate the same set of models

with the same independent variables as in Question 2.

a. Explain how the results are similar. Different. Focus on the values

effects.

b. What might account for these differences? (Hint: You may want to

generate histograms for both dependent variables as a starting point.)

C O M P U T E R E X E R C I S E S 673

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_21,

© Springer Science+Business Media New York 2014

C h a p t e r t w e n t y o n e

Special Topics: Randomized Experiments

What does a randomized experiment look like?

What are the advantages of randomized experiments?

How do randomized experiments maximize internal validity?

What is a block randomized trial?

How does block randomization help increase statistical power?

How can covariates be used to help increase statistical power?

Many of the descriptive and inferential statistical approaches we have

examined in this text are appropriate for analyzing randomized experiments.

For example, in Chapter 11 we used the two-sample test of proportions to ana-

lyze data from the Maricopa County Drug Testing Experiment. In this chapter

we want to focus more specifically on randomized studies. We begin the chapter

by showing why randomized experiments provide a very strong ability to make

causal inferences without concern for confounding. Indeed, many scholars have

taken the position that only randomized experiments can provide valid conclu-

sions regarding the impacts of treatments and programs.1 Joan McCord argued,

for example, that crime and justice evaluations should employ random assign-

ment “whenever possible.” 2

Once we have established why randomized experiments provide distinct advan-

tages, we will focus on some specific approaches for strengthening analysis of

1 See Robert Boruch, Brooke Snyder, and Dorothy DeMoya, “The Importance of Randomized

Field Trials,” Crime & Delinquency 46 (2000):156–180; Donald Campbell and Robert

F. Boruch, “Making the Case for Randomized Assignment to Treatments by Considering

the Alternatives: Six Ways in Which Quasi-Experimental Evaluations in Compensatory

Education Tend to Underestimate Effects,” In Carl A. Bennett and Arthur A. Lumsdaine

(eds.), Evaluation and Experiment: Some Critical Issues in Assessing Social Programs,

(New York: Academic Press, 1975: 195–296); Thomas Cook and Donald Campbell, Quasi-

Experimentation: Design and Analysis Issues for Field Settings, (Boston, MA: Houghton

Mifflin Harcourt, 1979); Lynette Feder, Annette Jolin, and William Feyerherm, “Lessons

from Two Randomized Experiments in Criminal Justice Settings,” Crime & Delinquency 46

(2000): 380–400; Brian R. Flay and J. Allen Best, “Overcoming Design Problems in Evaluating

Health Behavior Programs,” Evaluation and the Health Professions, 5 (1982): 43–69; David

Weisburd, “Randomized Experiments in Criminal Justice Policy: Prospects and Problems,”

Crime & Delinquency, 46 (2000): 181–193; David Weisburd, Cynthia Lum, and Anthony

Petrosino, “Does Study Design Affect Research Outcomes in Criminal Justice?” The Annals

of the American Academy of Political and Social Sciences 578 (2001): 50–70; Leland

Wilkinson and Task Force on Statistical Inference, APA Board of Scientific Affairs, “Statistical

Methods in Psychology Journals: Guidelines and Explanations,” American Psychologist 54

(1999): 594–604. 2 See Joan McCord, “Cures that Harm: Unanticipated Outcomes of Crime Prevention

Programs,” Annals of the American Academy of Political and Social Science 587 (2003):

16–30, p. 29.

675

randomized studies. While the basic tools we have already described for analyzing

data are also appropriate for experimental data, there are specific problems that

researchers might encounter in experimental research. One of these derives from

the fact that experiments are “carried out” by the researcher and practitioner in

the field. Unlike “observational” research studies that observe programs and draw

data from them, in a randomized experiment the researcher makes the scene by

randomly allocating subjects or places to treatment and control conditions. This

means that it is often difficult to gain enough cases for a statistically powerful

research design (see Chapter 23). In this chapter we examine two methods for

increasing the statistical power of experimental research programs. We also dis-

cuss a related problem that develops from the focus of experimental research on

specific research problems. Some scholars have criticized randomized experiments

T h e S t r u c t u r e o f a R a n d o m i z e d E x p e r i m e n t 4

The general structure of experiments in criminology is usually similar in design

regardless of the area or the question of interest. Generally, experiments in crimi-

nology start with an eligibility pool, randomization, group allocation, and posttest

measures relevant to the dependent variable of interest (Figure 21.1).

The eligibility pool is made up of those participants or units that are eligible for the

experiment. Units of analysis can be individuals or aggregated groups or other enti-

ties that often are found in clusters. For example, in an experiment that evaluates the

impact of increased foot patrol on crime rates, the eligibility pool may be individual

years of experience. Or the unit of analysis could be the geographic area or the “beat”

that will be assigned to different conditions. The eligibility pool thus comprises those

patrol officers (or patrol “beats”) that meet the criteria for inclusion in the study.

Next, researchers randomly assign members from this pool of eligible partici-

pants or units to the study conditions—often a treatment group and a control or

a comparison group. Historically, randomization was carried out through a simple

coin flip. There are now many ways to randomize subjects, but most often research-

ers rely on computerized statistical software to carry out randomization. Some

researchers may simply use the rule of odds and evens—that is, assigning every

other case to one particular group. This is often referred to as alternation and is

3 James J. Heckman and Jeffrey A. Smith, “Assessing the Case for Social Experimentation,”

Journal of Economic Perspectives, 9 (1995): 85–110. 4 For this section and the section on internal validity we draw heavily from David Weisburd,

Anthony Petrosino, Trevor Fronius (2013). Randomized Experiments in Criminology

and Criminal Justice. In David Weisburd and Gerben Bruinsma (Eds.). Encyclopedia of

Criminology and Criminal Justice. New York: Springer Verlaag.

C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S676

because they generally do not take into account “interactions” between the variable

or the treatment of interest and other relevant variables.3 For example, in assess-

ing a rehabilitation program in prison we may be interested not only in the general

impacts of the program but also the differential impacts across men and women,

or older or younger offenders. We will discuss how a researcher can do this in the

context of an experimental research model.

officers who “walk the beat” in a specific area, and who have a specific number of

considered quasi-random assignment as the assignment of the numbers used is not

actually random. The most critical factor in randomization, however, is that each

case has the same probability or likelihood of being selected for the control group

as the experimental group and that the assignment is based purely on chance.

In the usual criminological experiment, eligible cases are randomly assigned to

one of the two groups—treatment or control. Experiments in criminology may

have more than two groups. But typically, an experiment comprises a group that

receives the treatment or the intervention and a control or a comparison group that

does not. It is also quite common in a criminological experiment for the control

group to actually receive something rather than nothing. For example, in a foot

patrol experiment, the control group may receive treatment as usual or the same

number of foot patrol officers as typically employed.

Experimental designs can include any number of outcomes. If the randomiza-

tion was implemented with fidelity, it should produce two equivalent groups on

the pretest or baseline measures related to the outcome of interest. The researcher

then conducts analyses to determine if the intervention had any impact on the

posttest or follow-up measures of the outcomes of interest.

T h e M a i n A d v a n t a g e o f E x p e r i m e n t s : I s o l a t i n g C a u s a l E f f e c t s

In identifying whether a variable has a causal influence, or evaluating the out-

comes of treatments or programs, the key issue for researchers is getting an

unbiased estimate of the treatment or the intervention effect. Without that any

other considerations, such as the ability to generalize results, are superfluous. For

example, suppose an evaluator was asked to assess whether an intervention for

drug involved offenders provided an effective deterrent to future offending. In the

study employed, the treatment group was found to be half as likely to recidivate as

the control condition not receiving the intervention. This would ordinarily lead the

evaluator to report that the intervention was a success. But what if it was difficult

to “believe” the result that was gained in the study? Suppose that the design of the

study did not allow the evaluator to assume that the observed effect was actually

the result of the intervention. In this scenario, the evaluator could not be sure that

Diagram of the Typical Criminological ExperimentFigure 21.1

T H E M A I N A D V A N T A G E O F E X P E R I M E N T S 677

Cases Then

Followed to

Determine Impact

Experimental Outcome

Measures Eligibility Pool

Created

Eligible Cases Randomized Control

it was the intervention that caused the change or something else that was common

to the treatment group but not to the control group. In such a situation, it does not

do much good to argue about whether the results can be generalized to a specific

population of interest. The results themselves are not believable.

The main problem researchers face in producing believable results is that treat-

ments are often confounded with other factors. For example, suppose that the

reason for the outcome observed above was that the evaluator had not taken into

account the fact that the treated drug offenders were volunteers. Volunteers in turn

are more likely to be highly motivated to succeed in such programs than individu-

als who are not volunteers.5 This is often termed “creaming” in the identification

of the subjects in the treatment condition. The reason why the intervention group

had lower recidivism rates in this case could easily be explained by the fact that

they were on average more motivated to be rehabilitated than drug offenders in

the control condition.

All research studies that seek to establish causation between a specific variable

or treatment and an outcome must deal with this problem of confounding, and

it stands as the major barrier to drawing believable conclusions in criminological

studies. Non-experimental methods, such as regression techniques using obser-

vational data that we reviewed in Chapters 15–20, and quasi-experiments using

approaches such as matching of subjects rely on a similar logic to solve the prob-

lem of confounding. The logic is easily stated: if we know what the factors are that

confound treatment (or the variable of interest) we can take them into account.

In other words, non-experimental methods, as we noted in Chapter 16, rely on a

“knowledge solution” to the problem of confounding.

But how does knowledge solve the problem of confounding? Let us take the

example of regression analyses as described in Chapter 16 using observational data

examining the question of the effect of a drug intervention program on recidivism.

Figure 21.2a shows the effect of the intervention on recidivism using a standard-

ized regression coefficient approach (see pages 492–493). Here we have the simple

result suggesting that the intervention decreases recidivism. However, when we

include in our analysis the “confounding” factor—level of motivation—the rela-

tionship between treatment and recidivism changes (see Figure 21.2b). Taking into

account the effect of level of motivation, the effect of the intervention becomes 0

in this illustration. The observed effect was not due to the intervention but rather

due to the confounding of the intervention with motivation of offenders.

Notice that two extra coefficients are included in Figure 21.2b. The first repre-

sents the relationship between treatment (the variable of interest) and motivation

(the confounding variable). This standardized coefficient is .50 and represents the

extent to which treatment and motivation are related or confounded. The second

5 See George De Leon, Gerald Melnick, George Thomas David Kressel, and Harry K. Wexler,

“Motivation for Treatment in a Prison-Based Therapeutic Community,” American Journal

of Drug and Alcohol Abuse, 26 (2000): 33–46.; Faye Taxman, Reducing Recidivism through

a Seamless System of Care: Components of Effective Treatment, Supervision, and

Transition Services in the Community, (Washington, DC: Office of National Drug

Control Policy, 1998).; Robert Rosenthal, “The Volunteer Subject,” Human Relations

18 (1965): 389–406.

C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S678

and recidivism. Together these two relationships detail the extent of confound-

ing that is clouding our ability to estimate the treatment or the program effect.

Confounding in this context takes into account the extent to which the confound-

ing factor is related to the outcome of interest, and the degree to which it is related

to or confounded with the treatment variable. And indeed, we can estimate the

value of the simple relationship observed in Figure 21.2a. This can be defined as

the degree of bias. In this case, the degree of bias is equal to the observed effect.

Had we not taken motivation into account we would have erroneously concluded

that the treatment leads to lower rates of recidivism, when in fact it is the motiva-

The way in which multivariate regression approaches allow us to control for

confounding is illustrated in Equation 21.1 and described in Chapter 16. Here we

show the computation of the regression coefficient b for a treatment variable (Tr)

controlling for a confounding factor (CC, in this case motivation):

Equation 21.1

Example of the Bias in the Estimate of a Treatment Effect Caused by the Exclusion of an

Unknown or an Unmeasured Factor (X j ). (a) Estimate of B

1 in the Case Where the Factor

(X j ) Is Unmeasured and Excluded from the Model. Estimate of B

1 is −.25. (b) Estimate of

B 1 in the Case Where the Factor (X

j ) Is Included in the Model. Estimate of B

1 is .00.

Figure 21.2

T H E M A I N A D V A N T A G E O F E X P E R I M E N T S 679

b = r r r

r

S

S Tr

Y Tr Y CC TrCC

TrCC

Y

Tr

1 1

1 2

− ( ) −

  

  

 



a

b

Treatment

Outcome

Xj

B = .50

B1 = .00

B2 = -.50

B1 = -.25

Treatment

tion of offenders (represented by ) which is responsible for this result.Xj

The key part of the equation for our interest is the numerator in the first part of the

equation: r Y1Tr

r Y1CC

r TrCC

). Note that it includes the simple correlation between

the treatment variable and the outcome measure (r Y1Tr

). Subtracted from that is

the product of the correlation between the outcome measure and the confound-

ing variable (r Y1CC

) and the treatment and confounding variable (r TrCC

)—the two

components of confounding we have just described. In this context we can statis-

tically “un-confound” our estimate of the treatment if we have knowledge of the

This solution is also key to the myriad of approaches that have been developed

for other types of non-experimental approaches. All of them rely on knowledge

about confounding. For example, matching of subjects on known characteristics,

or the more sophisticated propensity score matching approach,6 begins with the

basic assumption that we have enough knowledge to create equivalence of units in

the treatment and control conditions. Because the subjects in the matched groups

are assumed to be alike, we make an assumption that confounders are not influ-

encing our observations of a treatment effect. Note that in this case we are trying

to statistically control for such confounding factors, by making the treatment and

control groups alike on these factors. In such a case, the correlation between treat-

ment and confounding variables is assumed to be 0. When it is, as illustrated in

Equation 21.2 for the bivariate regression coefficient (see Chapter 15), the effect

of the treatment breaks down to the simple correlation between treatment and

outcome:

b = r S

S Tr Y Tr

Y

Tr 1

( )

 

 Equation 21.2

The problem with these methods is that if we want to get an unbiased estimate of

treatment we would in theory have to identify all “confounding causes.” Using the

regression approach, which in some sense provides the most transparent form of non-

experimental methods, we would need to identify all confounding factors that also

have meaningful impacts on the outcome measure and include them in the regres-

sion. This would mean both that we would have to have knowledge about all such

confounding factors and that we would be able to measure them in a research study.

Randomized experiments start with a different logic. If we cannot control out

for confounding, we can make it irrelevant for the problem at hand. This is done

through the process of randomizing treatment. If treatment is randomized then

there is no reason to suspect systematic biases. This can be illustrated by return-

ing to the simple path diagrams we used earlier. In Figure 21.3a we show the

simple relationship between a treatment and outcome. In Figure 21.3b we include

a potential confounding variable. Note that the confounding factor has a strong

6 See Paul R. Rosenbaum and Dennis R. Rubin, “The Central Role of the Propensity Score in

Observational Studies for Causal Effects,” Biometrika 70 (1983): 41–55.; Paul R. Rosenbaum

and Dennis R. Rubin, “Reducing Bias in Observational Studies Using Subclassification on the

Propensity Score,” Journal of the American Statistical Association 79 (1984): 516–524.

C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S680

Figure 21.3 Example of the Lack of Confounding in the Treatment Effect When the Treatment (V

1 )

and Potential Confounder (V 2 ) Have No Relationship. (a) The Model Excluding a Potential

Confounder, V 2 . (b) The Model Including a Potential Confounder (V

2 ) but No Relationship

Between the Treatment (V 1 ) and the Confounder Because of Randomization. If V

2 is

Excluded, the Bias = B * B 2 = .00 * .50 = .00

T H E M A I N A D V A N T A G E O F E X P E R I M E N T S 681

standardized relationship with the outcome variable (B = .50). However, using the

theory of randomization we can assume that there is no systematic relationship

between the treatment and the confounder. This is the case because treatment has

been randomly allocated. In theory it is not going to be related systematically to

other factors such as gender, race, age, and attitudes.

What this means is that the relationship between any potential confounder

and the treatment can be assumed to be 0. By chance, fluctuations will occur,

and there will be systematic relationships observed, but these can in this case be

assumed to balance out in the long run. Or at least there is no reason to assume

that they will not. And if the relationship between the confounder and the treat-

ment is 0, then when we multiply this by the large relationship between the

confounder and the outcome we will also gain 0. The effect of treatment is not

confounded. This is also illustrated in Equation 21.2 presented earlier, though in

this case the assumption that treatment and the confounder have a 0 correlation

is more believable.

a

b

Y

V1

V2

B = .00

B1 = .50

B2 = .50

V1

Y

B1 = .50

I n t e r n a l V a l i d i t y

Our discussion so far is often subsumed under the heading of “internal validity”

in methodological texts in criminology. A research design in which the impact of

the intervention can be clearly distinguished from other observed factors is known

as having high internal validity. If there are confounding factors involved in the

impact of the intervention, then the evaluation design is considered to have low

internal validity. Shadish, Cook, and Campbell, among others, have identified the

most common threats to internal validity7:

1. Selection: The preexisting differences between treatment and control subject

or units.

2. History: An external event occurring at the same time of the study that may

influence impact.

3. Maturation: Changes in subjects or units between measurements of the

dependent variable. These changes may be of natural evolution (e.g., aging) or

due to time-specific incidences (e.g., fatigue, illness).

4. Testing: Measurement at pretest impacts measurement at posttest.

5. Instrumentation: Changes to the instrument or the method of measure-

ment in posttest measures.

6. Regression to the mean: Natural trends may cause extreme subjects or units

who score extremely high or low during the pretest to score closer to the

mean at posttest.

7. Differential attrition: The differential loss of subjects or units from the

treatment group compared to the control group.

8. Causal order: The certainty that the intervention did in fact precede the out-

come of interest.8

To further illustrate the importance of internal validity, let us suppose a researcher

internal validity is considered high if, at the end of the evaluation, the researcher

can show that the change in juvenile recidivism among the intervention group is

due only to the intervention (i.e., youth court) and no other confounding factors

were at play. The researcher must show through either research design or analytical

procedures that all confounding factors are accounted for in the measurement of

outcomes. If the researcher is unable to account for other factors such as serious-

ness of first offense or the maturation of the study population, he or she must

note that the observed effects may be due to other factors. If threats to validity (or

potential confounding factors) are not accounted for, the internal validity of the

study would be considered low.

7 See William R. Shadish, Thomas D. Cook, and Donald T. Campbell, Experimental and

Quasi-Experimental Designs for Generalized Causal Inference, (Boston: Houghton-

Mifflin, 2002) 8David P. Farrington and Brandon C. Welsh, “Randomized Experiments in Criminal Justice:

What Have we Learned in the Past Two Decades?” Journal of Experimental Criminology

1 (2005): 9–38.

C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S682

is interested in evaluating the impact of a youth court on juvenile recidivism. The

Generally speaking, a randomized experiment has the highest possible internal

validity, because as we illustrated above, this approach allows the researcher to assume

that other confounding causes of the outcome of interest, known and unknown, are

not systematically influencing the study results. High internal validity in randomized

experiments is gained through the process of randomly allocating the treatment or

the intervention to the experimental and control or comparison groups. Through

random assignment, the researcher is not just randomizing the treatment. He or she

is randomizing all other factors that may influence the outcome of the treatment.

Thus, there is no systematic bias that increases the odds of one unit’s assignment to

the treatment group and another unit’s assignment to the control or the comparison

group. This is not to imply that the groups are the same on every characteristic—it is

very possible that differences may occur; however, these differences can be assumed

to be randomly distributed and are accounted for in the probability distributions that

underlie statistical tests of significance. Regardless, neither the treatment group nor

the control group should have an advantage over the other on the basis of known

or unknown variables. Thus, randomized experiments are the only design that allows

the researcher to assume statistically unbiased effects.9

S A M P L E S I Z E , E Q U I V A L E N C E , A N D S T A T I S T I C A L P O W E R 683

The goal of most randomized experiments in criminology and criminal justice,

as in other social science fields, is to disentangle the impact of the treatment or

the intervention from the impact of other factors on the outcomes that are to be

tested. A randomized experiment allows the researcher to attribute differences

between the groups from pretest to posttest to the treatments or the interventions

that are applied. At the conclusion of the study the researcher is able to assert, with

confidence, that the differences are likely a result of the treatment and not due to

other confounding factors. It is more difficult for non-randomized studies, even

a high-quality quasi-experimental design, to make this assertion. This advantage is

underscored by Farrington10:

The unique advantage of randomized experiments over other methods is high internal validity. There are many threats to internal validity that are eliminated in randomized experiments but are serious in non-experimental research. In particular, selection effects, owing to differences between the kinds of persons in one condition and those in another, are eliminated.

S a m p l e S i z e , E q u i v a l e n c e , a n d S t a t i s t i c a l P o w e r 1 1

Despite the distinct advantages of randomized studies, it is often difficult to gain a

large number of cases in a randomized experiment. Sometimes this is true because

it is difficult to identify a large number of subjects who can be made eligible for

9 See Robert F. Boruch, Randomized Experiments for Planning and Evaluation:

A Practical Guide, (Thousand Oaks, CA: Sage, 1997). 10 See David P. Farrington “Randomized Experiments on Crime and Justice,” In Michael Tonry

(ed.), Crime and Justice: A Review of Research, 4 (1983): 257–308, p. 260. 11 Our discussion in this section relies heavily on David Weisburd and Charlotte Gill, “Block

Randomized Trials at Places: Rethinking the Limitations of Small N Experiments,” Journal of

Quantitative Criminology (2013).

randomization into treatment and control conditions. Sometimes this is the case

because treatment conditions or data collection are expensive, and each new case

will increase the cost of the study. These problems are particularly acute in place-

based randomized trials since the number of places with a specific crime problem

is generally limited.12 Moreover, place-based trials ordinarily demand significant

treatment resources per site, and accordingly it is expensive for agencies to “treat”

a large number of sites at one time.13

Farrington14 argues that small N studies are not likely to achieve realistic pretest

balance across measured and unmeasured covariates.15 This of course undermines

the main advantage of experimental studies—that the control and treatment

groups can be assumed to be equivalent and differ only in the receipt or the non-

receipt of treatment.

While Farrington and colleagues have focused primarily on the problem of

equivalence, a related criticism of the small N sizes of many randomized experi-

ments can also be raised. If the sample sizes used for such studies are small,

then their statistical power under traditional assumptions is also likely to be low.

Statistical power (see Chapter 23) is a particularly important component of evalu-

ation studies, because it assesses whether the study will provide a “fair test” of

the interventions examined. Sample size is one of the key elements of statistical

power, and experiments with very small samples are also likely to have low statisti-

cal power.

A design approach called “block randomization” provides a potential solution

to the risk of unbalanced samples in small N studies as well as a valid method for

increasing the statistical power of such studies. Block randomized experiments take

12 See Anthony A. Braga, David L. Weisburd, Elin J. Waring, Lorraine Green Mazerolle,

William Spelman, and Francis Gajewski, Problem-Oriented Policing in Violent Crime Places:

A Randomized Controlled Experiment,” Criminology 37 (1999): 541–580; Lawrence W.

Sherman, Patrick R. Gartin, and Michael E. Buerger, “Hot Spots of Predatory Crime: Routine

Activities and the Criminology of Place,” Criminology 27 (1989): 27–56. 13 See Robert Boruch, Henry May, Herbert Turner, Julia Lavenberg, Anthony Petrosino,

Dorothy De Moya, Jeremy Grimshaw, and Ellen Foley, “Estimating the Effects of

Interventions that are Deployed in Many Places: Place-Randomized Trials,” American

Behavioral Scientist 47 (2004): 608–633; David Weisburd, “Hot Spots Policing Experiments

and Criminal Justice Research: Lessons from the Field,” Annals of the American Academy

of Political and Social Science 599 (2005): 220–245. 14See note 8; See also David P. Farrington, Lloyd E. Ohlin, and James Q. Wilson,

Understanding and Controlling Crime: Toward a New Research Strategy, (New York:

Springer-Verlag, 1986); David P. Farrington and Maria M. Ttofi, “School-Based Programs

to Reduce Bullying and Victimization,” Campbell Systematic Reviews, 6(6), 2009; Darrick

Jolliffe and David P. Farrington, A Rapid Evidence Assessment of the Impact of Mentoring

on Reoffending, (London: Home Office Online Report, 2007). 15 Farrington notes in this regard, “(t)o understand why randomization ensures closer equiva-

lence with larger samples, imagine drawing samples of 10, 100, or 1,000 unbiased coins. With

10 coins, just over 10 percent of the samples would include 2 or less, or 8 or more, heads.

With 100 coins, just over 10 percent of the samples would include 41 or less, or 59 or more,

heads. With 1,000 coins, just over 10 percent of the samples would include 474 or less,

or 526 or more, heads. It can be seen that, as the sample size increases, the proportion of heads

in it fluctuates in a narrower and narrower band around the mean figure of 50 percent”

(see note 8 p. 263n).

C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S684

advantage of prior knowledge about the distribution of units in an experimental

study to maximize equivalence of treatment and control conditions.

We begin our discussion below by focusing on statistical theory, detailing how

block randomized trials maximize equivalence of experimental studies and pro-

vide valid methods for increasing statistical power in small sample experiments.

We then turn to an empirical illustration of these arguments, drawing from data

used in the Jersey City Drug Market Analysis Experiment (Jersey City Experiment

(JCE)).16 Using data from the JCE and simulation methods, we illustrate the advan-

tages of block randomization approaches over simple or “naïve” randomization in

developing equivalent groups. We then illustrate the overall increase in statistical

power provided by the block randomized statistical modeling approach.

Statistical Foundations for Block Randomization

While criminological researchers are often unfamiliar with block randomized

studies, the first randomized experiment in crime and justice, the Cambridge-

Somerville Youth Study, used a complete or a fully blocked randomized design.17

In that study, problem youths were paired on age, social background, biological

somatotype, and temperament. The researchers used this paired or fully blocked

design because the experimental treatment was lengthy and complex, so they

sought to maximize the equivalence of the comparisons they could make. Their

design is illustrated in Figure 21.4. In practice, the researchers matched the youths

into pairs on these characteristics and then randomly allocated them within the

pairs into treatment and control conditions. In the fully blocked randomized

design, each subject in a pair has an equal probability of being assigned to treat-

ment or control conditions, but randomization is restricted in that one subject

from each pair must be assigned to treatment and one subject to control.

16 See David Weisburd and Lorraine Green, “Policing Drug Hot Spots: The Jersey City Drug

Market Analysis Experiment,” Justice Quarterly 12 (1995): 711–735. 17 See Edwin Powers and Helen Witmer, An Experiment in the Prevention of Delinquency,

(New York: Columbia University Press, 1951).

Figure 21.4 Fully Blocked (Matched Pairs) Random Assignment

S A M P L E S I Z E , E Q U I V A L E N C E , A N D S T A T I S T I C A L P O W E R 685

Sample of units (100)

Units paired by similarity on relevant

characteristics

Pair 1

Treatment (1) Control (1)

Pair 2

Treatment (1) Control (1)

. . . Pair 50

Treatment (1) Control (1)

What advantage does this approach give over naïve or simple randomization?

Naïve randomization (illustrated in Figure 21.5), which is the most common

approach in crime and justice experiments, assigns the total sample under study to

treatment or control conditions without restrictions. Every subject in this case has

an equal probability of being assigned to either the treatment or the control condi-

tion. Naïve randomization relies on the assumption that there are no systematic

reasons for the treatment and control subjects to differ (since every subject had an

mental studies in the first place. But it does not guarantee equivalence, simply that

there is no reason for non-equivalence. When samples are large, this assumption

is reasonable because large differences between the groups are unlikely by chance.

subjects or the units in an experiment that can help us create equivalence on fac-

tors that are related to the outcomes observed. Age and social background were

considered key predictors of delinquency by the Cambridge-Somerville research-

ers, and their introduction as factors to match the youths in the study was seen as

a direct way of making sure that the treatment and control conditions were similar

on important influences of treatment success.

However, the benefit of equivalence gained through fully blocked randomiza-

tion comes at a statistical price. For each limitation on randomization the study

must “pay a fine” in terms of degrees of freedom. For example, in the Cambridge-

Somerville Youth Study 650 boys were matched into pairs. In a naïve randomiza-

tion design with 325 cases per group, the study would have had 648 degrees of

freedom for statistical tests of significance (N 1

N 2

design, the degrees of freedom of the tests declined to 324 (N pairs

The loss of degrees of freedom is meaningful because it changes the distribu-

tion of the test statistic. For example, as illustrated by Equation 21.3 (dependent

samples) and 21.4 (independent samples) below, in a t-test the estimated standard

Figure 21.5 Naïve (Balanced) Random Assignment

C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S686

Sample of units (100)

Treatment (50) Control (50)

equal probability of assignment to each condition), a key raison d’etre for experi-

But why shouldn’t we increase equivalence if we can, especially in small studies

where chance differences between control and treatment groups might be large in

the case of naïve randomization? Fully blocked randomized designs like the

Cambridge-Somerville Youth Study assume that we have knowledge about the

deviations are divided by the degrees of freedom. This means that as the degrees

of freedom of a test get smaller the t-value observed also gets smaller:

t = X -

s

df

d d

d

2

m Equation 21.3

t = X - X - -

N s + N s

df

N + N

N N

1 2 1 2

2

2 2

2

1 2

1 2

1 1

( ) ( )m m Equation 21.4

In turn, the value of the statistic needed to achieve statistical significance will be

larger as the degrees of freedom for a test get smaller (see the t distribution exam-

ple in Table 21.1 below). This difference is not meaningful in the case of relatively

large sample studies. For example, in the Cambridge-Somerville experiment, with

324 degrees of freedom, the critical value of the t-test (with standard criteria of

p < .05 and a nondirectional test) is about 1.967, almost the same as the 1.960 in

the z normal distribution without adjustment. But when the degrees of freedom

are reduced to 100, the critical value for the t-test becomes 1.984 and at 50 degrees

of freedom, 2.009.

The balance between loss of degrees of freedom and greater equivalence is

weighted toward the goal of equivalence in disciplines where the causal processes

underlying the impacts of treatment are well understood. This is the case because

the benefits of the complete or the fully blocked randomized design are greatest

when each loss of degrees of freedom is accompanied by a gain in the equivalence

of the treatment and control conditions on factors that are related to treatment

outcomes. If treatment outcomes are conditioned by such factors, then blocking

will decrease the heterogeneity of outcomes in the study. Looking at Equation 21.3

and Equation 21.4 above, this would mean that the numerators of the standard

errors are made smaller and accordingly the t-values observed are larger. This

makes intuitive sense because if the groups are more similar in terms of what

would have been expected absent treatment, then it should be easier to identify a

treatment outcome. In statistical terms, there is likely to be less noise in identifying

that outcome. In the case of a fully blocked design in which treatment outcomes

DEGREES OF FREEDOM CRITICAL VALUE

10 20 50 100 200 324 500 648

2.228 2.086 2.009 1.984 1.972 1.967 1.965 1.964

Table 21.1 Critical Values for the t Distribution (Two-Tailed, α = .05)

S A M P L E S I Z E , E Q U I V A L E N C E , A N D S T A T I S T I C A L P O W E R 687

were not related to the blocking factors, the standard deviations would remain the

same as in a naïve design, while there would be a substantial loss of degrees of

freedom. This would mean that a large price was paid for the fully blocked design

without a corresponding benefit.

And here lies the primary argument against the use of fully blocked randomized

designs in crime and justice. The level of knowledge of the causal processes under-

lying crime and justice research simply does not allow us to parse randomization

with sufficient distinctions to allow us to gain a benefit from a fully blocked rand-

omized design. This is one of the main reasons that matched pair designs are not

common more generally in criminology, though we suspect that criminologists are

often uninformed about the benefits of fully blocked randomized designs.

To say that criminologists do not have a full understanding of causal processes

in place-based studies does not mean that they do not have sufficient knowledge to

improve experimental designs using block randomization methods. A compromise

approach between the fully blocked and naïve randomization approach is a partially

blocked or termed here simply, block randomized design. A block randomized

design makes no assumptions regarding the number of blocks or groups identi-

fied at the outset. Rather, the number of blocks is determined by the researcher’s

assessment of the ability of known data to group units in ways that maximize

their similarities on key causal variables. Cases are placed within the specified ran-

domization blocks and then randomized within those blocks (see Figure 21.6). The

blocks do not have to be of equal size, but the number of cases in each block must

be even to allow for equal randomization and balance within blocks.

An Example: Jersey City Experiment

At this juncture it is worthwhile to introduce the substantive example we will

draw from for illustrating the benefits of a block randomized design for place-

based studies, the Jersey City Drug Market Analysis Experiment. The JCE evalu-

ated an innovative drug enforcement strategy involving police crackdowns along

with citizen and local business engagement in controlling crime at drug markets.

A total of 56 high drug activity hot spots were randomly assigned in equal num-

bers to receive either the experimental program or regular, unsystematic enforce-

ment on an ad hoc basis. Most of the drug market hot spots included fewer than

four street segments and intersections, though two places included more than

ten street segments. Police emergency calls for service for a variety of crime and

Figure 21.6 Partially Blocked Random Assignment (Jersey City Drug Market Analysis Experiment)

C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S688

Hot spots (56)

Very High

(5 sets of matched pairs)

T (5) C (5)

High (8)

T (4) C (4)

Moderate (26)

T (13) C (13)

Low (12)

T (6) C (6)

disorder-related issues were measured for 7-month pre- and post-intervention

periods. We focus below on three main outcome measures for disorder measured

in the study: suspicious persons, public morals, and police assistance.

Knowing that there was considerable variation in criminal activity even across

the sample of hot spots, the study authors were concerned at the outset that the

prior level of crime would influence the effect of treatment. Given the small sam-

ple of drug hot spots that could be identified in Jersey City, the authors were also

concerned that naïve randomization might lead to non-equivalent groups. At the

same time, there was concern that each loss of degrees of freedom in the experi-

ment would substantially impact the results, since the total N of available cases

was only 56. The solution in the JCE was to examine the distribution of both

emergency calls for service and arrests and then to identify natural cutting points.

In this way the researchers believed that they could gain greater equivalence

between the groups without a large loss of degrees of freedom (28) that would

have ensued if the fully blocked randomized design was adopted. The assumption

here was that prior crime and disorder would have a general impact on the effects

of treatment but would not be specific enough to distinguish sites in a way that

would justify a complete randomized block design. The researchers identified eight

statistical blocks for randomization. The ten highest activity hot spots were rand-

omized in pairs because of large gaps between them; these five pairs represented

the five “very-high-activity” statistical blocks. Of the rest of the sample of hot

spots, 8 were grouped into a “high-activity” block, 26 hot spots were classified as

a medium-activity block, and 12 were classified as a low-activity block.

The Benefits of Block Randomized Trials

One approach to examining the contribution of statistical blocking to equivalence

in the Jersey City Drug Market Analysis Experiment is to compare the equivalence

gained between the treatment and control conditions on key baseline (pretest)

measures. However, the Jersey City study is only one specific draw of randomi-

zation. By definition any specific draw of a sample is going to be different from

another draw. The statistical concern is whether on average, a draw using the

block randomization procedure is likely to produce a more equivalent outcome

than a draw using a simple randomization procedure. To examine this question we

develop 10,000 simulations of both naïve randomization and block randomization

using the Jersey City data.18 We focus on baseline calls for service for the three key

disorder outcomes in the study (suspicious persons, public morals, and assistance).

Table 21.2 reports the baseline information from the original study, the simula-

tion results for the blocking approach, and the simulation outcomes of a naïve

18 Stata programs were developed to run a randomization sequence (blocked or naïve) on the

JCE dataset and then run a t-test comparing the treatment and control group means at baseline

on the three outcomes of interest. Stata’s simulation function was then used to run each

program 10,000 times and create a dataset containing the group means, t-values, p-values, an

indicator showing whether or not the two groups were significantly different at baseline for

each iteration, and the absolute average mean group difference across all iterations. We are

grateful to David B. Wilson for developing the programs and simulation syntax.

S A M P L E S I Z E , E Q U I V A L E N C E , A N D S T A T I S T I C A L P O W E R 689

Table 21.2

19 Of course, this is about what we would have expected given a .10 significance threshold

and a fair randomization procedure. But the important point is that the block randomization

approach allows us to do better.

C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S690

Calls for Service at Baseline in Block and Naïve Randomizations of JCE Data

SUSPICIOUS PERSONS PUBLIC MORALS ASSISTANCE

Original JCE block randomized data

Treatment mean (SD) Control mean (SD) Absolute mean difference (SE)

17.00 (16.15) 17.93 (21.16)

.93 (5.03)

9.32 (10.58) 9.11 (13.36)

.21 (3.22)

43.86 (43.40) 42.11 (43.05) 1.75 (11.55)

10,000 simulations Block randomized data (N = 56)

Average absolute mean difference across all samples (SD) Number of samples with significant difference

at baseline (p ≤ 0.10)

2.67 (1.95) 93

1.29 (0.94) 4

7.11 (5.05) 218

10,000 simulations Naïve randomized data (N = 56)

Average absolute mean difference across all samples (SD) Number of samples with significant difference

at baseline (p ≤ 0.10)

3.98 (2.89) 955

2.56 (1.86) 989

9.13 (6.71) 966

randomization approach. In the case of the simulations, we report the number of

samples that have significant differences at baseline and the overall absolute mean

difference found in the 10,000 simulation samples.

Table 21.2 suggests the importance of the simulation approach. The specific

draw in the Jersey City study produced unusually equivalent groups on the three

baseline measures examined. The absolute mean difference for all three measures

is substantially lower than the average absolute mean difference produced in our

simulations. This does not mean that the Jersey City randomization was flawed but

rather that the investigators by chance gained one of the more equivalent randomi-

zations from the sampling distribution of all possible randomizations.

But despite the fact that the Jersey City randomization was a relatively “lucky

draw,” it is clear from Table 21.2 that the procedure used was likely to produce

much more equivalent groups than a simple randomization procedure. In the

10,000 simulations of the JCE block randomization procedure only 93 simulations

produced significantly different outcomes (p < .10) for treatment and control con-

ditions at baseline for suspicious persons calls, 4 for public moral calls, and 218

for assistance calls. In contrast, using the simple randomization approach on the

same 56 cases, 955 samples produced significant differences for suspicious person

calls, 989 for public moral calls, and 966 for assistance calls.19 These differences

are of substantial magnitude and are also reflected in the average absolute mean

difference across all of the simulation samples. For suspicious persons the mean

differences were almost 50 % larger in the naïve randomization sample, for public

morals about twice as large, and for assistance almost a third larger.

S t a t i s t i c a l P o w e r a n d B l o c k R a n d o m i z a t i o n

As we noted earlier, the benefits of a block randomized design are dependent

on the assumption that the blocking factors are related to the study outcomes.

This cannot be assessed ordinarily because knowledge about treatment outcomes

is unknown until the experiment is complete. However, we are able to exam-

ine this assumption using post-experiment data from Jersey City. For all three

outcomes of interest discussed here, the correlations between pre- and posttest

outcomes were significant and had greater than a moderate size coefficient.20

For suspicious persons the correlation was .44 (p < .10),21 for public morals .52

(p < .01), and for assistance .63 (p < .001). These results suggest that Weisburd and

Green’s assumption that there would be a strong relationship between the blocking

factors and the final study outcomes was correct.

In turn, the statistical model benefits of the identification of block variability

can be observed directly in these data. A simple or a naïve randomized experi-

ment presents a model for understanding outcomes where systematic variation is

determined only by treatment. Accordingly, the model can be expressed in terms

of sums of squares (SS; see Chapter 12) by Equation 21.5:

SS SS + SS + SS total intercept group error

= Equation 21.5

The total variability of the study in this case is broken down to the influence of the

treatment (SS group

) and the overall variability in the data (SS error

), with the intercept

(SS intercept

) completing the linear model. Remember that the statistical denominator

for the statistical significance test is going to be the error term, meaning that as the

error term gets smaller the significance of the test will get larger.

With the introduction of a blocking factor, an additional source of variability is

taken into account in the model—SS block

—as illustrated in Equation 21.6:

SS = SS + SS + SS + SS total intercept group block error Equation 21.6

would expect the overall size of SS error

to decline. This is the case because the block

randomized model limits any relationship between SS group

(i.e., the treatment com-

ponent of the model) and SS block

, meaning that the two components of variability

are constrained to be independent (because they are independent, the inclusion of

the blocking factor in the model will not impact the size of the treatment effect).

Accordingly, any SS block

effect will be drawn out of the error term for the model

(SS total

is a fixed quantity irrespective of the model defined). Since SS error

is a key

21 We calculated the correlation between the blocking factor and the three disorder outcome

measures by running a GLM with only the blocking factor included. The correlation is based

on taking the square root of the overall R2 of the model. We use a one-tailed test of

significance following the assumption that the correlation between the blocking factor

and the outcome is positive.

20 See Jacob Cohen, Statistical Power Analysis for the Behavioral Sciences, 2nd ed.,

(Hillsdale, NJ: Lawrence Erlbaum, 1988).

S T A T I S T I C A L P O W E R A N D B L O C K R A N D O M I Z A T I O N 691

When the blocking factor (SS block

) is strongly related to the outcome (SS total

) we

Table 21.3

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element of the denominator of the test statistic (the degrees of freedom being a

second key element), its reduction without a proportionate reduction in SS group

(and a large decrease in degrees of freedom) will lead to a more significant outcome

(i.e., a more powerful statistical outcome) than a naïve model.

As an illustration of these assumptions, we estimated univariate ANOVA

models using just the treatment factor, and separately with the treatment and

blocking factors as fixed effects (Table 21.3).22 In order to simplify our example,

we do not estimate the full model which could include a treatment by block inter-

action (see later).23 Following our assumptions, SS total

and SS group

are the same in

both models. The total variability in the model is constant irrespective of model

specification, and the effect of treatment is not influenced because of the balanced

randomization of cases within blocks. However, SS error

declines in the analyses that

include blocking as a factor. For suspicious persons the decline is from 6,373.643

to 5,021.566, for public morals from 2,884.107 to 2,080.761, and for assistance

from 14,875.821 to 8,788.761. Note as well that there is a corresponding decrease

in the degrees of freedom of SS error

in the block randomized design (from 54 to

47 in all the models), reflecting the “price” of this approach. But, following our

examination of the correlation between the blocking factor and the outcomes, the

loss of statistical power generated by the reduction of degrees of freedom of the

22 In this case the effects are fixed because we are assuming analysis of specific categories and

do not assume that those categories are representative of the population of cases. For example,

the analysis looks at the specific blocks of hot spots in the experiment; it does not assume that

we have a representative sample of all possible “blocks” of hot spots. If the effects were ran-

dom, the blocks observed would be seen as a representative sample of blocks of hot spots. 23 Where the interaction between treatment and block is significant, Fleiss recommends including

an interaction term in the model. When the blocking factor represents a substantively important

variable, the introduction of a block by treatment interaction can also add knowledge about

the differential effects of treatment across values of the blocking variable. See Joseph L. Fleiss,

The Design and Analysis of Clinical Experiments, (New York: John Wiley and Sons, 1986);

David Weisburd and Faye Taxman, “Developing a Multicenter Randomized Trial in Criminology:

The Case of HIDTA,” Journal of Quantitative Criminology 16 (200): 315–340.

UNIVARIATE ANALYSIS OF VARIANCE MODELS SUSPICIOUS PERSONS PUBLIC MORALS ASSISTANCE

Treatment-only model

SS intercept

(df) SS

group (df)

SS error

(df) SS

TOTAL (df)

F (p) for group effect

480.286 (1) 516.071 (1)**

6,373.643 (54) 7,370.000 (56)

4.372 (.041**)

21.875 (1) 129.018 (1)

2,884.107 (54) 3,035.000 (56)

2.416 (.126)

1,512.161 (1) 355.018 (1)

14,875.821 (54) 16,743.000 (56)

1.289 (.261)

Treatment and block model

SS intercept

(df) SS

group (df)

SS block

(df) SS

error (df)

SS TOTAL

(df) F (p) for group effect

510.998 (1) 516.071 (1)**

1,352.076 (7) 5,021.566 (47) 7,370.000 (56)

4.830 (.033**)

145.484 (1) 129.018 (1)* 803.346 (7)

2,080.761 (47) 3,035.000 (56)

2.914 (.094*)

3,250.491 (1) 355.018 (1)

6,087.060 (7) 8,788.761 (47)

16,743.000 (56) 1.899 (.175)

Notes: *p < .10 **p < .05.

Univariate Analysis of Variance for Treatment and Treatment–Block Effects (JCE)

U S I N G C O V A R I A T E S T O I N C R E A S E S T A T I S T I C A L P O W E R 693

error term is less than the gain from the inclusion of the blocking term. When we

combine treatment and block effects in the model, all three comparisons show

larger F-statistics. The observed p-value for suspicious persons declines from .041

to .033, for public morals from .126 to .094, and for assistance from .261 to .175.

U s i n g C o v a r i a t e s t o I n c r e a s e S t a t i s t i c a l P o w e r i n E x p e r i m e n t a l S t u d i e s

Another technique for increasing statistical power in experimental studies follows

the statistical logic of block randomization but does not balance the blocking

characteristics at the outset. It relies heavily on the logic of randomization that we

have already described. As noted earlier, if the cases are randomized to treatment

and control conditions then we can assume that there is no correlation between

treatment and possible confounding factors (see Figure 21.2). That means that

the inclusion of additional covariates in an analysis will not, in theory, affect the

estimate of the treatment effect. Since that is the case, we should be able to include

covariates without creating any bias in our assessment of the influence of the

experimental variable.

However, we do gain a direct benefit in calculating the statistical significance

of the test. Let us use again the approach of examining the sums of squares of our

equation. Suppose we convert the JCE to a simple naïve randomization sequence.

In this case our model includes treatment and error as the only variables (see

Equation 21.5).

If we add covariates the error term for the model should decline, because as we

noted earlier we have no reason to expect that the effect of treatment (i.e., group)

will change (see Equation 21.7). As an example, let us add as covariates variables

that should be related to the dependent variables: the pre-experiment calls for ser-

vice for robbery and aggravated assault (collectively the baseline violent crime calls

for service) and the baseline calls for service for each respective outcome measure.

Thus, for each outcome, we include three covariates: robbery at baseline, aggra-

vated assault at baseline, and the outcome at baseline (e.g., suspicious person calls

at baseline for the suspicious person outcome):

SS total

= SS intercept

+ SS group

+ SS covariates

+ SS error

Equation 21.7

In Table 21.4 we show the results using the simple naïve design as well as

the results we would gain for taking into account the three covariates for each

outcome. As can be seen from the table, the statistical significance of the results

including the covariates is considerably lower than when no covariates are includ-

ed. For public morals, for example, the p-value for the group effect has dropped

from a non-statistically significant .126 in the naïve example to a significant .041

when including the covariates. For all three outcomes, we have substantially low-

ered the SS error

by adding the covariates. Even though we paid a price in degrees of

freedom for using three covariates, the benefit in terms of reducing the error and

increasing the significance of our group findings outweighs the cost. We should

Table 21.4

C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S694

be cautious in including these covariates, however. Note that we would expect the

effect of treatment to remain similar between the simple model and the model

with covariates in terms of the sums of squares explained. This is largely true for

suspicious persons and public morals where the SS group

remains fairly similar in

both sections of Table 21.4. For assistance, however, there is a large increase in

the SS group

potentially suggesting that we may have introduced some level of bias

into the model with our choice of covariates.

Importantly, if we had the data available, we could add many different covari-

ates to the models. Those covariates could be related to the characteristics of the

units of randomization, for example the social characteristics of the hot spots.

In a study of individuals we might include gender or age or race to the analysis.

In some sense, the larger the group of covariates that are related to the outcome

measure that are included, the greater the benefit. This is because each additional

variable included that is relevant to the prediction of the outcome will decrease the

error term for the significance test. The rule does not apply if there is no relation-

ship between the covariate and the outcome. In that case, the researcher will lose

a degree of freedom (for every additional variable or parameter) in the analysis

without an additional benefit in reduction of the error variance.

As is apparent, there is much to be gained by including covariates in an experi-

mental analysis. However, as in other statistical procedures, covariates can be

manipulated in ways that affect the validity of your results. Randomization allows

us to assume that there is no relationship between the covariate and the treatment

or the variable interest. But this does not mean that there is not in the sample

of interest a spurious relationship that is observed. In any randomization there

are likely to be some measures that by chance are related to the treatment. What

randomization guarantees is that such bias will be random, and it is likely in the

long run that whenever there is a spurious correlation it is likely to be balanced

off with another correlation in the opposite direction. But what if the researcher

UNIVARIATE ANALYSIS OF VARIANCE MODELS SUSPICIOUS PERSONS PUBLIC MORALS ASSISTANCE

Treatment-only model

SS intercept

(df)

SS group

(df)

SS error

(df)

SS TOTAL

(df)

F (p) for group effect

480.286 (1)

516.071 (1)**

6,373.643 (54)

7,370.000 (56)

4.372 (.041**)

Treatment and covariate model

SS intercept

(df)

SS group

(df)

SS covariate (pre - outcome)

(df)

SS covariate (pre - robbery)

(df)

SS covariate (pre - assault)

(df)

SS error

(df)

SS TOTAL

(df)

F (p) for group effect

188.025 (1)

559.666 (1)**

678.443 (1)**

752.979 (1)**

55.281 (1)

5,334.804 (51)

7,370.000 (56)

5.350 (.025**)

Notes: *p < .10 **p < .05.

Univariate Analysis of Variance for Treatment and Treatment–Covariate Effects (JCE)

21.875 (1)

129.018 (1)

2,884.107 (54)

3,035.000 (56)

2.416 (.126)

1,512.161 (1)

355.018 (1)

14,875.821 (54)

16,743.000 (56)

1.289 (.261)

34.122 (1)

162.019 (1)**

751.986 (1)**

160.373 (1)**

15.301 (1)

1,875.149 (51)

3,035.000 (56)

4.407 (.041**)

84.815 (1)

638.146 (1)*

305.731 (1)

2403.850 (1)**

405.114 (1)

9596.472 (51)

16,743.000 (56)

3.391 (.071*)

chooses specific variables that have relationships in the dataset with the treatment

or the variable of interest? In this case, the error term will be reduced if these

measures are correlated with the outcome, but so will the estimate of the treatment

effect. Again, here as in regression analyses more generally, the researcher can “go

fishing” until the result they are looking for is gained.

The dangers of influencing the validity of the treatment effect should lead to

caution in using covariates to reduce error variance in the analysis of experimental

studies. A general rule that will protect you from the danger of manipulation of

results is for the researcher to define at the outset which covariates will be used

in analyzing the outcomes. In this way, the researcher cannot manage results post

facto on the basis of knowledge of sample characteristics. Clearly, one should not

run a large number of regressions with different covariates included until a “good”

result is gained. The process of selecting variables before the results of an experi-

ment are known is in our view a good rule to follow. But more generally, if an

experiment has sufficient statistical power, the researcher should use the simple

analysis approach, in which covariates are not included. This is the only way to

guarantee that the results are not being manipulated in a way that might lead to

spurious findings.

E x a m i n i n g I n t e r a c t i o n T e r m s i n E x p e r i m e n t a l R e s e a r c h

The best way to examine interactions in an experimental study is to define the

contextual factor at the outset. For example, in the JCE, the block randomiza-

tion procedure was based on the level of crime observed in the baseline year. For

illustration purposes we use four blocks in our analysis: very high, high, moderate,

and low. One might ask whether the level of crime was related to the effective-

ness of treatment. Importantly, because of the use of block randomization the

cases are equally divided between the blocks, meaning that the design of the study

ensures that there is no relationship between block and treatment. There are an

equal number of very high, high, moderate, and low baseline crime hot spots in

the experimental and control groups. We add to our equation an additional term

beyond treatment or group and block—that of the interaction between treatment

and block as illustrated in Equation 21.8:

SS total

=SS intercept

+SS group

+SS block

+SS group×block

+SS error

Equation 21.8

E X A M I N I N G I N T E R A C T I O N T E R M S I N E X P E R I M E N T A L R E S E A R C H 695

Some scholars have criticized experimental studies because in their simplest form

they do not allow us to examine contextual factors that might influence treatments

or outcomes. For example, the average treatment effect observed in an experiment

will tell you whether the treatment is effective or not for the entire sample. But

what if we are interested in understanding whether the treatment effect differs for

men and women or younger versus older subjects. Most experimental studies in

criminology have not looked at these “interaction effects” between treatment and

other factors. But this does not mean that interactions cannot be observed in a

valid way in experimental studies.

Table 21.5

Table 21.6

C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S696

In Table 21.5 we report the results using this approach. The addition of

another term has again reduced the error variance. But more importantly, given

our discussion, we now have an answer to the question of whether the treatment

differs in effect across the groups. The significance statistic for the interaction

between group and block is significant only for suspicious persons. Accordingly,

we now have an experimental result that tells us that the effect of treatment on

suspicious person calls for service is conditioned by the level of crime in the

baseline year (i.e., the blocking factor). Table 21.6 shows these results producing

the effect of treatment for each of the four blocks for suspicious persons. The

table suggests that the effect is consistent in three of the four blocks represent-

ing higher crime rates and that the overall effect of the treatment is much larger

in the highest rate blocks. However, in the low rate blocks, which represented

much less serious crime areas, the observed treatment effect is in the opposite

direction. These results would likely lead the researchers to recommend the hot

spot policing strategy tested in the Jersey City study for higher rate drug market

hot spots, at least in regard to influencing suspicious persons. It might also raise

questions regarding why the treatment did not have the desired effect in the lower

crime areas.

BLOCK TREATMENT MEAN (SD) CONTROL MEAN (SD)

Very high (n = 5) High (n = 4) Moderate (n = 13) Low (n = 6) Total (n = 28)

−4.20 (6.979) −12.00 (8.246)

3.15 (9.353) 4.17 (6.145) −.11 (9.689)

18.40 (19.995) 8.25 (5.188) 5.69 (5.991)

−5.33 (5.785) 5.96 (11.924)

Mean Change in Suspicious Person Calls for Service by Block and Group (JCE)

UNIVARIATE ANALYSIS OF VARIANCE MODEL SUSPICIOUS PERSONS PUBLIC MORALS ASSISTANCE

Treatment, block, and group × block model

SS intercept

(df)

SS group

(df)

SS block

(df)

SS group × block

(df)

SS error

(df)

SS TOTAL

(df)

F (p) for group x block effect

236.940 (1)

928.490 (1)**

564.676 (3)*

1,893.588 (3)**

3,915.378 (48)

7,370.000 (56)

7.738 (<.001**)

55.582 (1)

150.470 (1)*

581.946 (3)**

88.137 (3)

2,214.024 (48)

3,035.000 (56)

0.637 (.595)

1,729.492 (1)

693.174 (1)

1,865.060 (3)*

559.706 (3)

12,451.055 (48)

16,743.000 (56)

0.719 (.545)

Notes: * p < .10 ** p < .05.

Univariate Analysis of Variance for Treatment, Block, and Group by Block

Effects (JCE)

Some scholars suggest that interaction terms can be added without a block

randomized design.24 This approach is based on the assumption we noted earlier

that randomization is likely to lead to balance between the groups on characteristics

that might be measured. The problem of course is, as we noted earlier, that there

may be chance relationships between variables in your observed data and this

might lead to spurious results. It seems reasonable to us that the inclusion of

interaction terms without block randomization requires very large samples. Only

in such cases can you rely on randomization providing equivalence across a large

number of measures. We recommend samples larger than 300 per group before

examining interactions in the data without block randomization. Moreover, the

researchers should examine whether the data are balanced (and thus there is no

relationship between treatment and covariate) in each specific case. With smaller

samples, this approach should be carried out with caution, and the interactions

observed should be identified at the outset. In any event, only a discrete number

of interactions should be observed.

C h a p t e r S u m m a r y

Randomized experiments provide higher levels of internal validity than observa-

tional studies in terms of determining the impacts of a treatment or an interven-

tion. Through the process of determining an eligibility pool, randomizing the

eligible participants or units, and assigning them to treatment or control groups,

researchers can better deal with the problem of confounding in posttest measures

relevant to the dependent variable of interest. Randomized experiments have the

highest possible internal validity as they allow us to assume that confounding

causes of the dependent variable are not a concern—this since treatment has been

allocated randomly and we can assume that possible confounding factors are not

systematically related to treatment.

For concerns of statistical power related to smaller sample experiments,

blocked randomized trials can maximize equivalence of experimental studies.

Another way of increasing the statistical power in experimental studies is with the

inclusion of covariates; however, covariates should be used cautiously as they can

allow the researcher to manipulate the results of the study.

One criticism of experimental research is that in its simplest form it is unable to

examine contextual factors that may influence treatments or outcomes. Although

many experimental studies in criminology have not looked at interaction effects

between treatments and other factors that should not be taken to mean that inter-

actions cannot be observed with experimental designs. The best way to examine

such interactions is to define them at the outset and use block randomization

techniques.

24 See Barak Ariel and David Farrington, “Block Randomized Trials,” In Alex R. Piquero and

David Weisburd (eds.), Handbook of Quantitative Criminology, (New York: Springer, 2010:

437–454).

C H A P T E R S U M M A R Y 697

S y m b o l s a n d F o r m u l a s

To compute the regression coefficient b for a treatment variable (Tr) controlling

for a confounding factor (CC):

b = r - r r

- r

S

S Tr

Y Tr Y CC TrCC

2

TrCC

Y

Tr

1 1

1

( )

  

  

 



With the basic assumption that we have enough knowledge to create equivalence

of units in the treatment and control conditions:

b = r S

S Tr Y Tr

Y

Tr 1

( )

 



C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S698

K e y T e r m s

block randomization A type of

randomization whereby cases are first sorted

into like groups and then afterwards randomly

allocated into treatment and control

conditions.

group allocation In criminological experiments,

eligible cases are randomly assigned to two or

more groups—typically treatment or control.

treatment group One group that eligible cases

are randomly assigned to which receives the treat-

ment or the intervention being evaluated.

control group The group that eligible cases

are randomly assigned to which does not receive

the treatment or the intervention being evaluated.

In many criminological experiments the control

group may receive existing interventions in

contrast to the innovative treatment.

posttest measure Analyses conducted by the

researcher to determine if the intervention had any

impact on the outcome measures of interest.

internal validity Whether the research design

has allowed for the impact of the intervention

or the treatment to be clearly distinguished from

other factors.

alternation A type of quasi-random assignment

in which researchers assign every other case to

one particular group.

confounding factors Variables associated

with treatments and/or outcomes that can

bias overall results if not controlled for

statistically.

eligibility pool Participants or units that are

eligible for an experiment.

randomization The process of randomly

assigning members from the pool of eligible

participants or units to the study conditions—

often a treatment group and a control or a

comparison group.

T-value for dependent samples:

t = X

s

df

d d

d

- m 2

T-value for independent samples:

t = X - X - -

N s + N s

df

N + N

N N

1 1

1 1

1

2 2

1

2

2 2

2

2

2

( ) ( )m m

Sum of squares for a simple or a naïve randomized experiment:

SS total

= SS intercept

+ SS group

+ SS error

With the introduction of a blocking factor:

SS total

=SS intercept

+SS group

+SS block

+SS error

With the introduction of covariates:

SS total

= SS intercept

+ SS group

+ SS covariates

+ SS error

Accounting for interaction between the treatment and block:

SS total

= SS intercept

+ SS group

+ SS block

+ SS group×block

+ SS error

E x e r c i s e s

21.1 Danny randomly allocates 30 students to either a treatment group that

receives new instructional program or a control group that receives

standard instruction. After randomization, he compares the characteristics

of the treatment and control groups and finds that the treatment group

is significantly different from the control group in two characteristics

(age, reading level).

group indicate that randomization failed?

21.2 Mark finds in a bivariate regression analysis that a drug treatment program

has a significant impact on reducing the likelihood that patients will

E X E R C I S E S 699

the drug treatment program is effective. Brent, however, argues that

Mark’s results are confounded because he did not account for patients’

level of motivation. Brent notes that motivation and likelihood of relapse

are highly related (r = 0.50). He reruns the regression results controlling

for the level of motivation and finds that the impact of treatment has

a. Diagram the impact of treatment on likelihood of relapse based on

Mark’s initial result.

b. Diagram the impact of treatment on likelihood of relapse using Brent’s

analyses.

c. What is the level of bias Mark has introduced by not including this

confounder? What is the estimated r between the level of motivation

and treatment?

21.3 Darcy wants to test the effectiveness of a new police training program on

domestic violence. She identifies the officers with the least knowledge of

domestic violence and administers the training to this group because she

believes that it will be most worthwhile since they have the most to learn.

She tests this group on domestic violence knowledge before and after the

training. She also tests a comparison group of officers who did not receive

the training. She finds a major jump in knowledge in the trained officers

compared to the non-trained officers and concludes that her training

program was effective.

a. Are Darcy’s conclusions warranted? Are there any threats to internal

validity in her research design?

b. Design an alternative study to test the effectiveness of the training

program that has a higher level of internal validity than Darcy’s study.

21.4 Adrian is designing a randomized trial to examine the effectiveness of a

program designed to reduce recidivism in offenders. He has a sample of

200 prisoners that will all be released from prison on the same day and

can be randomly allocated to a treatment group receiving the program or a

control group that does not receive the program.

a. If Adrian uses a naïve randomization procedure, how many prisoners

will be in each group? What will be the total degrees of freedom for the

research design?

b. If Adrian uses a fully blocked randomization procedure, how many

pairs of prisoners will be randomized? What will be the total degrees of

freedom for the research design?

c. If Adrian wants to use a partially blocked randomization procedure,

what might be one prisoner characteristic he uses to create statistical

blocks? What are the statistical consequences if this prisoner character-

istic does not end up being related to the effectiveness of the program?

21.5 Logan is reexamining data from a policing experiment to assess whether

using blocking provided a statistical benefit. Results from a “treatment-

only” model that did not include the blocking factor and a “treatment and

block model” that did include the blocking factor are provided below.

C H A P T E R T W E N T Y O N E : S P E C I A L T O P I C S700

a. What is the total sum of squares in each model? Are these values the

same? Why or why not?

b. Why is the SS group

the same in both models?

c. How many total blocks were used in the treatment and block model?

21.6 Sharon is analyzing data from a large randomized trial of the impact

of after-school programs on juvenile delinquency. After completing the

experiment she has been considering adding a number of different

covariates to her overall analysis to minimize the error and improve her

ability to identify a treatment effect.

a. Do you have any concerns about the approach Sharon is taking to

analyzing the experimental data? If so, what would be a better

approach?

b. If Sharon has chosen good covariates, what should happen to the

SS total

in the model? What should happen to the SS error

? What should

happen to the SS group

?

c. With the large sample size, the statistical power in Sharon’s experiment

is estimated to be about 0.9. Does this affect whether she should

consider using covariates?

21.7 Refer to Table 21.6 that provides the mean change in suspicious person

calls for service by block and group.

a. This chapter described what the differences by block suggested for the

effectiveness of the treatment across the statistical blocks. What do the

results for the control group show?

b. What does the total change versus the change in each block suggest

about the importance of block by group interaction effects?

E X E R C I S E S 701

UNIVARIATE ANALYSIS OF VARIANCE MODELS CALLS FOR SERVICE

TREATMENT-ONLY MODEL

SS intercept

(df)

SS group

(df)

SS error

(df)

p-value for group effect

260.50 (1)

300.65 (1) *

10,875.75 (80)

.038*

Treatment and block model

SS intercept

(df)

SS group

(df)

SS block

(df)

SS error

(df)

p-value for group effect

500.25 (1)

300.65 (1) *

3,614.44 (5)*

7,021.56 (75)

.022*

*p < .05

Special Topics: Confidence Intervals

f o r d i f f e r e n t s t a t i s t i c s

Coefficient Calculated?

C h a p t e r t w e n t y t w o

D e s c r i b i n g a c o n f i d e n c e i n t e r v a l

D e f i n i n g c o n f i d e n c e i n t e r v a l s

What is a Confidence Interval?

How is It Constructed?

How is the Confidence Interval for Sample Means Calculated?

How is the Confidence Interval for Sample Proportions Calculated?

How is the Confidence Interval for a Difference of Means Calculated?

How is the Confidence Interval for a Correlation Coefficient Calculated?

How is the Confidence Interval for a Regression Coefficient Calculated?

How is the Confidence Interval for a Logistic Regression

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_22,

© Springer Science+Business Media New York 2014

ONE OF THE MAIN CONCERNS of this text has been to define how we

make inferences from samples to populations. This is also one of the

main concerns of researchers, since in most cases they must make deci-

sions about population parameters on the basis of sample statistics. Our

approach has been to use the logic of statistical inference, which begins

with the creation of a null hypothesis. Importantly, the logic of statistical

inference we have reviewed so far is concerned primarily with the deci-

sion as to whether to reject or fail to reject the null hypothesis. This

means in practice that we have relied on a logic that allows us to make a

statement about where the population parameter is not.

This approach has enabled us to come to concrete decisions about

population parameters on the basis of sample statistics. When we reject

the null hypothesis on the basis of a statistical test, we conclude that the

relationship we have examined is statistically significant. For example,

when we reject the null hypothesis on the basis of our sample statistics

in a statistical test of the difference of means, we have confidence that

there is a difference between the means of the two populations. When

we reject the null hypothesis that there is not a linear correlation be-

tween two variables, we have confidence that there is in fact a linear

correlation between these two variables in the population. But the logic

we have used so far does not allow us to zero in on the value of the

population parameter. When we find that the relationship between two

variables in a sample is statistically significant, we conclude that there is

likely to be a relationship in the population from which the sample was

drawn. But this decision does not provide us with an estimate of the size

of that relationship in the population.

In this chapter, we turn to an approach to statistical inference that

leads us to make specific statements about population parameters from

sample statistics. The logic used in this approach is very similar to that

described in earlier chapters. However, we do not make a single deci-

sion about the null hypothesis. Rather, we create an interval of values

within which we can be fairly confident that the true parameter lies. Of

703

course, without data on the population itself, we can never be certain of

the value of the population parameter. This interval is generally called a

confidence interval. In this chapter, we begin by explaining the logic be-

hind confidence intervals and how they are used. We then illustrate how

confidence intervals are constructed for the main statistics reviewed in

this text.

C o n f i d e n c e I n t e r v a l s

In the statistical tests presented in earlier chapters, we began by setting a

null hypothesis. Our null hypothesis made a statement about the value

of the population parameter. In practice, the null hypothesis generally

stated that a statistic had a value of 0. For example, for the difference of

means test, the null hypothesis generally stated that the difference be-

tween two population means was 0; for the correlation coefficient, that

the population correlation had a value of 0; or for the regression coeffi-

cient, that the population regression coefficient had a value of 0. When

the results of our statistical test indicated that we should reject the null

hypothesis, we concluded that it was unlikely that the population para-

meter had the value stated in the null hypothesis. Since the null hypothe-

sis was generally 0 or no difference, we rejected the hypothesis that the

population parameter had this null value.

We can use similar logic to make a very different statement about

population parameters. In this case, we ask where the population para-

meters are likely to be found. In statistics, the interval of values around

the sample statistic within which we can be fairly confident that the

true parameter lies is called a confidence interval. A confidence in-

terval makes it possible for us to state where we think the population

parameter is likely to be—that is, the range of values within which we

feel statistically confident that the true population parameter is likely to

be found. Importantly, the fact that we are confident does not mean

that the population parameter actually lies in that range of values. As

in tests of statistical significance, we rely on probabilities in making

our decisions.

One common illustration of confidence intervals comes from newspa-

results typically make reference to a range of values. For example, a poll

might indicate that 60% of adults in the United States favor using the

death penalty for convicted murderers, �4% (plus or minus 4 percent).

per articles and television news programs reporting the results from public

opinion polls. In addition to stating that some percentage of the population

supports a particular political candidate in an upcoming election or a

particular policy, more thorough accounts of these kinds of survey

C H A P T E R T W E N T Y T W O : S P E C I A L T O P I C S704

The value of 60% is often described in statistics as a point estimate. Ab-

sent knowledge of the population parameter, the statistic we obtain for

our sample is generally used as an estimate—in statistical terms, a point

estimate—of the population parameter. The range of values represented

by �4% is sometimes described as the margin of error of a poll. In sta-

tistics, we prefer to call this margin of error a confidence interval. Based

on a very specific set of statistical assumptions, it is the interval within

which the true population value is expected to fall.

Confidence intervals are based on the same statistical logic as tests of

statistical significance. It will be easier to understand the relationship be-

tween tests of statistical significance and the construction of confidence

intervals if we start with an example that—although it is very unusual—

allows us to make a straightforward link between these two concepts.

Let’s say that we gather data on attitudes toward the death penalty using

an interval-scale measure that has both positive values, indicating sup-

port for the death penalty, and negative values, indicating opposition to

the death penalty. We use an independent random sampling method to

draw our sample from the population of all adult Americans. After com-

pleting our study, we find that the mean score for attitudes toward the

death penalty is 0.

In determining a confidence interval, we rely on the same basic as-

sumptions that we use for tests of statistical significance. If we were

going to compare the mean in our sample to some hypothesized popula-

tion mean, we would use a t-test as our test of statistical significance.

This means that the t-test also underlies our confidence interval. Accord-

ingly, we have to assume an interval level of measurement and make

parametric assumptions regarding the population distribution. Let’s as-

sume that our sample is very large, so we can relax the assumption of a

normal population distribution. We have already noted that the sampling

method meets the requirements of a t-test.

If we intended to conduct a test of statistical significance for this ex-

ample, we would have stated a null hypothesis and an alternative hy-

pothesis and set a level of statistical significance. Let’s say that the null

hypothesis is that Americans are neutral regarding the death penalty.

This means that H0 for our example will be 0.0, as the scale is divided

for our sample that is greater than 1.96 or less than �1.96. Since our ob-

served value of the measure is 0, the value of t will also be 0. Clearly, we

would not reject the null hypothesis in this case.

between positive attitudes greater than 0 and negative attitudes less than 0.

There is no reason to posit a directional research hypothesis, so our test

will be two-tailed. We will use a standard 0.05 significance level.

Figure 22.1 illustrates the t-test for this example. The rejection region

begins at a t- value of 1.96 either above or below the null hypothesis of 0.

In order to reject the null hypothesis, we need an observed value of t

C O N F I D E N C E I N T E R V A L S 705

But what would a confidence interval for this example look like?

With a confidence interval, we are not concerned about whether the

population parameter is not at a specific value (for example, the null

hypothesis); rather, we are concerned about specifying a range of val-

ues within which we can be fairly confident (though not certain) that

the population parameter lies. How do we choose this range of values?

Clearly, we want to make the interval large enough that, given the ob-

served statistic in our sample, the population parameter is unlikely to

lie outside it. As in tests of statistical significance, our choice is based

on convention. With a test of statistical significance, it is common to set

a threshold of 5% for the risk we are willing to take of falsely rejecting

the null hypothesis. With confidence intervals, we define the width of

the interval so that we can be very confident that the true population

value lies within it. The confidence interval most commonly used is a

terval for our example. As you can see, the confidence interval extends

until the rejection region begins. It is, in this sense, the flip side of the

rejection region.

The 5% Rejection Region and 95% Confidence Interval

on a Normal Frequency Distribution (where and H0 � 0)X

95% Confidence Interval

Rejection Region 0.025

Rejection Region 0.025

Rejection Region for Two-Tailed Test = 0.05

t = –1.96 t = +1.960

α

Figure 22.1

95% confidence interval. Figure 22.1 illustrates the 95% confidence in-

Thus, a 95% confidence interval and a 5% significance level are directly

related. In our example, the 5% significance rejection region represents

values far enough away from the null hypothesis that we are confident

in rejecting it. The 95% confidence interval represents values close enough

C H A P T E R T W E N T Y T W O : S P E C I A L T O P I C S706

Of course, in practical examples it is very unlikely that our observed

sample statistic will be the same as the population parameter hypothe-

sized by the null hypothesis. A more common situation is that of the

opinion poll described earlier. What would the confidence interval look

like for our opinion poll? We have all the information we need to illus-

trate that example, except that the level of confidence of the interval was

not specified. Let’s assume that a 95% confidence interval was used in ar-

riving at the margin of error. The observed statistic, or point estimate, of

60% will be the mean of the distribution. We would use a z-test rather

than a t-test because we are concerned with only a single proportion.

Let’s assume that the other assumptions of the test were met. The margin

shows the confidence interval relative to the z distribution. As you can

see, the interval ranges between 56% and 64%.

But how does this confidence interval relate to a test of statistical sig-

nificance? First, we need to identify a null hypothesis. Suppose we make

the null hypothesis for our test that the population is evenly divided in

their attitudes toward the death penalty. In this case, the H0 takes on a

value of 0.50, meaning that about half of the population to which the

sample infers are for and half against the use of the death penalty for

95% Confidence Interval for the Public Opinion Poll Example

Test of Statistical Significance for Opinion Poll

95% Confidence for

Opinion Poll

95% Confidence

Interval

Rejection Region 0.025

Rejection Region 0.025

P = 0.50 H0

p = 0.600.56 0.64 –1.96 1.96

to our observed statistic, or point estimate, that we are confident that the

population parameter lies within that interval.

of error of the test, or size of the confidence interval, is 4%. Figure 22.2

Figure 22.2

C O N F I D E N C E I N T E R V A L S 707

convicted murderers. Note that this value is very far outside the confi-

dence interval that we have defined for our example.

can see, our point estimate of 0.60 falls much to the right of the critical

value (t � �1.96) of our test of statistical significance. As a general rule,

if the null hypothesis for a test of statistical significance lies outside the

confidence interval for the statistic (and the confidence interval and the

significance level represent opposite parts of the same criterion—for ex-

ample, 0.95 and 0.05; 0.99 and 0.01), then you may assume that the re-

sult is statistically significant. This again points to the close relationship

between tests of statistical significance and confidence intervals.

While we use the logic of confidence intervals to define where a pop-

ulation parameter is likely to be found, the confidence interval has a

very specific statistical interpretation. Were we to draw repeated samples

of a specific sample size from the same population, using a 95% confi-

dence interval, we would expect that in 95% of these samples the confi-

dence interval would include the population parameter. That is, we

C o n s t r u c t i n g C o n f i d e n c e I n t e r v a l s

Confidence intervals for many different sample statistics can be constructed

using the same basic equation. To illustrate how we construct a confidence

interval, we use the example of a t-statistic. The most general equation for

calculating a t-statistic for a sample statistic is written as follows:

To construct a confidence interval, we adjust this equation so that we

can solve it for the population parameter. We can do this through simple

algebra. Solving for the population parameter produces the following

equation:

Population parameter � �sample statistic� � t � standard error of

sampling distribution�

t � sample statistic � population parameter

standard error of sampling distribution

Figure 22.2 shows the sampling distribution for our example. As you

would expect that in only 5 out of every 100 samples would the parameter

lie outside the confidence interval. As in tests of statistical significance,

we must be aware at the outset that we are only making an informed

decision about the value of the population parameter. Using a 95%

confidence interval, we will make the wrong decision about 5 in a 100

times.

C H A P T E R T W E N T Y T W O : S P E C I A L T O P I C S708

In setting the boundaries for the confidence interval, we will use the

positive and negative values associated with a two-tailed t-test to provide

account for the positive and negative t-values, our confidence interval is

The t-value in the equation coincides with the level of confidence we re-

quire (i.e., the critical t-value). Following our earlier logic, this t-value is

the flip side of the significance threshold. For a 95% confidence interval,

we use a t-value associated with a two-tailed 0.05 significance level. For

a 99% confidence interval, we use a t-value associated with a two-tailed

0.01 significance level. In general, if � is our significance level for a

two-tailed test, then we can construct a confidence interval for 100 �

(1 � �) using the same critical t-values.

Confidence Intervals for Sample Means

Let’s start by constructing a confidence interval for a sample mean ( ). If

where is the sample mean, s is the sample standard deviation, N is the

sample size, and t is the critical t-value associated with a given signifi-

cance level. To determine our critical t, we use df � N � 1, as in the

single-sample t-test (see Chapter 10).

For an illustration of the calculation of a confidence interval for a

sample mean, consider a recent study of fear of crime among Korean

Americans living in the Chicago area.1 The investigators constructed a

fear of crime instrument that was measured on an interval scale and

ranged in value from 11.00 to 110.00. The mean fear of crime score for

the 721 respondents was 81.05, with a standard deviation of 23.41.

To calculate a 99% confidence interval for the fear of crime instrument,

we use the t-value associated with a 0.01 significance level and 720 de-

grees of freedom (df � 721 � 1). Using the last line of the t distribution

table in Appendix 4, we find the corresponding critical t-value to be 2.576.

X

Confidence limit � X � t � s

�N � 1 �

X

Confidence limit � �sample statistic� � t � standard error of

sampling distribution�

1Min Sik Lee and Jeffery T. Ulmer, “Fear of Crime Among Korean Americans in

Chicago Communities,” Criminology 38:4 (2000): 1173–1206.

the upper and lower boundaries, respectively (see Figure 22.1). After we

given by Equation 22.1.

Equation 22.1

we rewrite Equation 22.1 to replace the general terms with the mean and

the standard error, we have Equation 22.2.

Equation 22.2

C O N S T R U C T I N G C O N F I D E N C E I N T E R V A L S 709

W orking It Out

� 81.05 � 2.25

� 81.05 � 2.576� 23.41

�721 � 1 �

Confidence limit � X � t � s

�N � 1 �

The result of �2.25 indicates that the 99% confidence interval includes

values ranging from a low of 78.80 (81.05 � 2.25) to a high of 83.30

(81.05 � 2.25). By using a 99% confidence interval, we can be very con-

fident that the population mean lies somewhere between 78.80 and

83.30. In statistical terms, if we were to observe repeated samples of this

size drawn from this population and calculate a confidence interval for

each of them, only about 1 in 100 would fail to include the true popula-

tion parameter.

Confidence Intervals for Sample Proportions We can apply the same type of

logic to calculating a confidence interval for a sample proportion, modi-

value. To calculate a confidence interval for a sample proportion, we use

where p is the sample proportion, q is 1 � p, N is the sample size, and

z is the critical z-value associated with a given significance level.

In their study of fear of crime among Korean Americans in the

Chicago area, the investigators also included a question about the re-

spondent’s victimization experiences. Specifically, respondents were

asked whether they had experienced any kind of victimization in the

past three years. Included in this global indicator of victimization were

violent as well as property crime victimizations. The investigators re-

ported that 27% of the 721 respondents had experienced some form of

victimization during this time period.

Confidence limit � p � z��pq

N �

A 99% confidence interval for the fear of crime instrument has the

following values:

fying Equation 22.1 by replacing the critical t-value with the critical z-

Equation 22.3:

Equation 22.3

C H A P T E R T W E N T Y T W O : S P E C I A L T O P I C S710

Knowing that the sample proportion is 0.27 and the z-score is 1.96,

we can calculate a 95% confidence interval for this proportion. We insert

W orking It Out

� 0.27 � 0.03

� 0.27 � 1.960(0.0165)

� 0.27 � 1.960�(0.27)(1 � 0.27) 721

Confidence limit � p � z �pq

N

The 95% confidence interval is �3% around the sample mean of 27%. It

suggests that we can be confident that the percentage of Korean Ameri-

cans living in and around Chicago who experienced some form of crimi-

nal victimization within the three-year period lies between 24% and 30%.

Confidence Intervals for a Difference of Sample Means In Chapter 11, we

discussed calculating t-statistics to test for significant differences between

two sample means. Another way of calculating a confidence interval for

place the sample mean with the difference of sample means and insert

the appropriate standard error for the difference of two sample means.

Recall from Chapter 11, however, that there are two methods for calcu-

lating the standard error of the sampling distribution: the separate vari-

interval for a difference of two sample means, using either the separate

variance method or the pooled variance method.

Confidence limit � (X1 � X2) � t �� N1s 2 1 � N2s

2 2

N1 � N2 � 2 �N1 � N2

N1N2 �

Confidence limit � (X1 � X2) � t � s 2 1

N1 � 1 �

s 2 2

N2 � 1

our values for p, q, and N into Equation 22.3.

Equations 22.4a and 22.4b present formulas for calculating a confidence

Equation 22.4a Separate Variance Method

Equation 22.4b Pooled Variance Method

the difference of two sample means is by modifying Equation 22.1 to re-

ance method and the pooled variance method (see pages 274–279).

C O N S T R U C T I N G C O N F I D E N C E I N T E R V A L S 711

In both equations, represent the two sample means,

represent the two sample variances, N1 and N2 represent the two sample

sizes, and t is the critical t-value associated with a given significance level.

As with the two-sample t-test (see Chapter 11), the number of degrees of

freedom for determining the critical t-value will be df � N1 � N2 � 2.

Chapter 11 presented a test for differences in bail amounts required of

African American and Hispanic defendants in Los Angeles County. A

sample of 1,121 African Americans were required to post a mean bail

amount of $50,841 (s � 115,565), while a sample of 1,798 Hispanics

were required to post a mean bail amount of $66,552 (s � 190,801). The

difference in the two sample means is $15,711, where Hispanics were re-

quired to post higher bail amounts, on average.

interval for this difference of sample means. For both equations, we use

the t-value associated with a 0.05 significance level and 2,917 degrees of

freedom. From the last line of the t distribution table in Appendix 4, we

find that critical t � 1.960.

s 2 1 and s 2

2X1 and X2

W orking It Out Separate Variance Method

� �15,711 � 11,119.12

� �15,711 � 1.960(5,673.02)

� (50,841 � 66,552) � 1.960 � 115,5652

1,121 � 1 �

190,8012

1,798 � 1

Confidence limit � (X1 � X2) � t � s 2 1

N1 � 1 �

s 2 2

N2 � 1

W orking It Out Pooled Variance Method

� �15,711 � 12,385.38

� �15,711 � 1.960(6,319.07)

� 1.960 ��(1,121)(115,5652) � (1,798)(190,8012)

1,121 � 1,798 � 2 �1,121 � 1,798

(1,121)(1,798)� � (50,841 � 66,552)

Confidence limit � (X1 � X2) � t�� N1s 2 1

� N2s 2 2

N1 � N2 � 2 �N1 � N2

N1N2 �

Using Equations 22.4a and 22.4b, we can calculate a 95% confidence

C H A P T E R T W E N T Y T W O : S P E C I A L T O P I C S712

Using the separate variance method, we find that the confidence in-

terval is �11,119.12 around the difference of sample means of 15,711.

This interval suggests that we can be confident, based on our sample

findings, that the average bail amounts posted in Los Angeles by African

Americans were from $4,591.88 to $26,830.12 less than the average bail

amounts required of Hispanics. The pooled variance method provides

very similar results, indicating that the confidence interval is �12,385.38

around the difference of sample means. Again, this interval suggests that

we can be fairly confident that African Americans were required to post

average bail amounts from $3,325.62 to $28,096.38 less than those re-

quired of Hispanics.

Confidence Intervals for Pearson’s Correlation Coefficient, r

The calculation of confidence intervals for Pearson’s correlation coeffi-

cient, r, relies on a similar logic, but requires an additional step. In

contrast to that for sample means, sample proportions, or differences

of means, the sampling distribution for Pearson’s r is not normal or

even approximately normal.2 Consequently, we need to convert r into

3

equation, we take the natural logarithm of 1 plus r divided by 1 minus r

and multiply this value by .

Values for Z* for correlation coefficients ranging in value from 0.000 to

1.000 are given in Appendix 8. Note that the correlations given in the ap-

pendix are all positive. If r is negative, then Z* will also be negative.

In Chapter 14, we reported that the correlation between unemploy-

ment rates and burglary rates for 58 counties in California was 0.491. If

Z * � 1 2

� ln �1 � r 1 � r�

1 \2

2The sampling distribution for r will generally be normal and symmetric only for the

case where r � 0, which is what allowed us to use the t distribution to test whether

rp � 0 (i.e., the null hypothesis) in Chapter 14. When r � 0, the sampling distribution

is not symmetric around r, so we cannot calculate a confidence interval for r in the

same way we did for sample means or the difference of sample means. 3Ronald A. Fisher, Statistical Methods for Research Workers, 14th ed. (New York:

Hafner, 1970).

another statistic, Z *, that does have a normal distribution. The conversion

After calculating of r is known as the Fisher r-to-Z * transformation.

a confidence interval for Z *. Since the values for Z * are not directly

interpretable, we will then convert the confidence limits back into

values of r.

The Fisher r-to-Z* transformation is given in Equation 22.5. In this

the standard error for Z *, we can then modify Equation 22.1 to calculate

Equation 22.5

C O N S T R U C T I N G C O N F I D E N C E I N T E R V A L S 713

we locate r � 0.491 in Appendix 8, we find Z* to be 0.5374. We obtain

W orking It Out

� 0.5374

� 1 2

� ln �1 � 0.491

1 � 0.491�

Z * � 1 2

� ln �1 � r 1 � r�

The standard error of Z*, which is based on the size of the sample

In our example concerning unemployment rates and burglary rates for

California counties, we have 58 observations, so the standard error of Z*

is 0.1348.

�sd(Z*) � 1

�N � 3

W orking It Out

� 0.1348

� 1

�58 � 3

�sd(Z*) � 1

�N � 3

tistic, a critical z-value (since Z* is approximately normally distributed),

and the equation for the standard error of Z*. The formula for the confi-

Confidence limit � Z * � z � 1

�N � 3 �

the same value for Z* if we use Equation 22.5.

Equation 22.6

(N), is presented in Equation 22.6.

dence interval for Z* is given in Equation 22.7.

Equation 22.7

We can now modify Equation 22.1 by inserting Z* as the sample sta-

C H A P T E R T W E N T Y T W O : S P E C I A L T O P I C S714

where Z* is based on the Fisher r -to-Z* transformation, N is the sample

size, and z is the critical z-value associated with a given significance

level.

Continuing our example for Z* � 0.5374 and N � 58, we calculate a

95% confidence interval for Z* by using critical z � 1.960 and inserting

W orking It Out

� 0.5374 � 0.2642

� 0.5374 � 1.960 � 1

�58 �3 �

Confidence limit � Z * � z � 1

�N � 3 �

The confidence interval is �0.2642 around Z* � 0.5374, indicating

that the range for Z* is 0.2732 to 0.8016. Since we are unable to directly

interpret values of Z*, we should convert the values of Z* back to values

of r, using Appendix 8. The conversion of Z* back to r will provide us

with the confidence interval for r. For Z* � 0.2732, we find that r �

0.267. For Z* � 0.8016, we find that r � 0.665. For both values of Z*, we

used the closest Z*-value reported in Appendix 8 to determine the val-

ues for r, since an exact match could not be found. These results suggest

that we can be confident that the population value for the correlation co-

efficient falls between 0.267 and 0.665. Note that the upper and lower

confidence limits are not symmetric around r —the lower limit is farther

away from r � 0.491 than is the upper limit.

Confidence Intervals for Regression Coefficients

Confidence intervals for regression coefficients are nearly identical in form

to confidence intervals for sample means. The formula for calculating con-

where b is the regression coefficient, is the standard error of b, and t

is the critical t -value associated with a given level of significance. The

number of degrees of freedom for the critical t will be equal to N � k �

1, where N is the sample size and k is the number of independent vari-

ables in the regression model.

�̂b

ˆ b

the values into Equation 22.7.

fidence intervals for regression coefficients is given in Equation 22.8.

Equation 22.8Confidence limit � b � � �t �

C O N S T R U C T I N G C O N F I D E N C E I N T E R V A L S 715

In Chapter 15, we reported that the regression coefficient representing

the effect of unemployment rates on burglary rates in California was

calculate a 99% confidence interval, the number of degrees of freedom

will be 56 (df � 58 � 1 � 1 � 56), so the critical t we will use is 2.669

(see Appendix 4).

W orking It Out

� 36.7483 � 23.2539

� 36.7483 � 2.669(8.7126)

Confidence limit � b � t �̂b

The result of �23.2539 indicates that the 99% confidence interval in-

cludes values ranging from a low of 13.4944 to a high of 60.0022. The

99% confidence interval suggests that we can be very confident that the

population value of the regression coefficient representing the effect of

unemployment rates on burglary rates lies somewhere between 13.4944

and 60.0022.

Confidence Intervals for Logistic Regression Coefficients and Odds Ratios

Confidence intervals for logistic regression coefficients are calculated in

The number of degrees of freedom for determining the critical t-value

also equals N � k � 1.

In addition to being able to calculate confidence intervals for the orig-

formed coefficients, which are difficult to interpret.

An illustration of the use of confidence intervals for logistic regression

coefficients is provided by a recent study examining the link between

mental disorders and violent victimization for a sample of 747 adults.4

The dependent variable measured whether the individual had reported a

4 Eric Silver, “Mental Disorder and Violent Victimization: The Mediating Role of In-

volvement in Conflicted Social Relationships,” Criminology 40 (2002): 191–212.

exactly the same way as logistic regression coefficients (Equation 22.8).

inal logistic regression coefficients, we can also refer to confidence inter-

regression coefficients into odds ratios by exponentiating the coefficient b.

This means that we can take the lower and upper limits of our confi-

dence interval for b and convert them to odds ratios. We can then

discuss the confidence interval relative to the odds, rather than the untrans-

36.7483 and the standard error for b was 8.7126 (see page 466). If we

vals for odds ratios. As noted in Chapter 18, we can convert our logistic

� �

C H A P T E R T W E N T Y T W O : S P E C I A L T O P I C S716

violent victimization in the preceding ten weeks. One of the nine inde-

pendent variables used by the researcher was the level of neighborhood

disadvantage, which was an interval-level instrument that combined eco-

nomic indicators, such as poverty rate, unemployment rate, and income.

The effect of neighborhood disadvantage was positive (b � 0.33), mean-

ing the greater the level of neighborhood disadvantage, the more likely

the individual was to have experienced a violent victimization. The stan-

dard error for b was reported to be 0.09.

To calculate a 99% confidence interval for b, we use critical t � 2.576,

W orking It Out

� 0.33 � 0.23

� 0.33 � 2.576(0.09)

Confidence limit � b � t �̂b

The result of �0.23 tells us that the 99% confidence interval includes

values ranging from a low of 0.10 to a high of 0.56. If we exponentiate

The lower limit of the confidence interval for the odds ratio is 1.105

[Exp(0.10)], and the upper limit of the confidence interval for the odds

ratio is 1.751 [Exp(0.56)]. These results suggest that we can be very confi-

dent that the population value of the odds ratio lies somewhere between

1.105 and 1.751. If we took repeated random samples of the size exam-

ined here and calculated a confidence interval for each, then in only

about 1 in 100 cases would that interval fail to include the true odds

ratio.

C h a p t e r S u m m a r y

In tests of statistical significance, we make a statement about where the

population parameter is not. In this chapter, we turned to an approach

to statistical inference that leads us to make a very different type of state-

ment about population parameters. The logic used in this approach is

similar to that described in earlier chapters. However, we do not make a

single decision about the null hypothesis. Rather, we create an interval

since df � 747 � 9 � 1 � 737, and insert the values into Equation 22.8.

the lower and upper limits of the confidence interval for b, we will have the

lower and upper limits of the confidence interval for the odds ratio.

� �

C H A P T E R S U M M A R Y 717

of values within which we can be fairly confident that the true parameter

lies—although, without data on the population itself, we can never be

certain of the value of the population parameter. This interval is gener-

ally called a confidence interval.

A confidence interval makes it possible for us to say where we

think the population parameter is likely to be—that is, the range of

values within which we feel statistically confident that the true popula-

tion parameter is likely to be found. A confidence interval is generally

constructed around the observed statistic of interest, commonly called

a point estimate. Absent knowledge of the population parameter, the

statistic we obtain for our sample is generally used as an estimate—in

statistical terms, a point estimate—of the population parameter. The

size of the confidence interval is often referred to as the margin of

error.

Confidence intervals may be constructed at any level of confidence.

By convention, we use 95% and 99% confidence levels, which are

based on 5% and 1% significance thresholds. While it is commonly

said, when using a confidence interval, that the researcher is confident

that the true parameter lies in the interval defined, confidence inter-

vals have a specific statistical interpretation. Suppose we find, using a

95% or 99% criterion, that a confidence interval is of a certain size. If

we were to draw repeated samples of the same size, using the same

methods, and calculate a confidence interval for each sample, then in

only 5 in 100 (for a 95% interval) or 1 in 100 (for a 99% interval) of

these samples would the interval fail to include the true population

parameter.

K e y T e r m s

confidence interval An interval of values

around a statistic (usually a point estimate).

If we were to draw repeated samples and

calculate a 95% confidence interval for each,

then in only 5 in 100 of these samples would

the interval fail to include the true popula-

tion parameter. In the case of a 99% confi-

dence interval, only 1 in 100 samples would

fail to include the true population parameter.

margin of error The size of the confi-

dence interval for a test. A margin of error

of �3% in an opinion poll means that the

confidence interval ranged between 3%

above and 3% below the point estimate or

observed statistic.

point estimate An estimate of the

population parameter. Absent knowledge

of the population parameter, the statistic

we obtain for a sample is generally used

as an estimate—or, in statistical terms,

a point estimate—of the population

parameter.

718 C H A P T E R T W E N T Y T W O : S P E C I A L T O P I C S

S y m b o l s a n d F o r m u l a s

To calculate the confidence interval for a sample mean:

To calculate the confidence interval for a sample proportion:

To calculate the confidence interval for a difference of sample means,

using the separate variance method:

To calculate the confidence interval for a difference of sample means,

using the pooled variance method:

To convert r to Z* (Fisher r -to-Z* transformation):

To calculate the confidence interval for Z*:

To calculate the confidence interval for a regression or logistic

regression coefficient:

E x e r c i s e s

In a study of self-reported marijuana use, a sample of high school stu- dents were asked how many times they had smoked marijuana in the last month. Researchers reported that the average for the sample was

Confidence limit � b � t �̂b

Confidence limit � Z * � z � 1

�N � 3 �

Z * � 1 2

� ln �1 � r 1 � r�

Confidence limit � (X1 � X2) � t �� N1s 2 1 � N2s

2 2

N1 � N2 � 2�N1 � N2

N1N2 �

Confidence limit � (X1 � X2) � t� s 2 1

N1 � 1 �

s 2 2

N2 � 1

Confidence limit � p � z ��pq

N �

Confidence limit � X � t � s

�N � 1 �

22.1

� �

E X E R C I S E S 719

2.4 times, with a 95% confidence interval of �1.3. Explain what this result means in plain English.

Following a revolution, the new leadership of the nation of Kippax intends to hold a national referendum on whether the practice of capital punishment should be introduced. In the buildup to the refer- endum, a leading army general wishes to gauge how the people are likely to vote so that he can make a public statement in line with popular feeling on the issue. He commissions Greg, a statistician, to carry out a secret poll of how people plan to vote. The results of Greg’s poll are as follows: The sample proportion in favor of intro- ducing capital punishment is 52%; the sample has a 95% confidence interval of �10%. How should Greg explain these results to the army general?

Concerned that taxpayers were not reporting incomes honestly, a state department of revenue commissioned an independent study to esti- mate the number of times people had cheated on their tax returns in the last five years. The researchers interviewed a random sample of 121 adults and found that the mean number of times they had cheated on their income taxes in the last five years was 2.7, with a standard deviation of 1.1.

a. Calculate a 95% confidence interval for this sample mean.

b. Explain what this result means.

The country of Mifflin is preparing for an upcoming presidential elec- tion. A random sample of 200 likely voters in Mifflin indicates that 57% are going to vote for the Hawk Party candidate, while the remain- ing 43% are planning on voting for the Gopher Party candidate.

a. Calculate a 95% confidence interval for the proportion voting for the Hawk Party candidate.

b. Calculate a 99% confidence interval for the proportion voting for the Hawk Party candidate.

c. Which of the two confidence intervals provides a better indicator

election?

A long-running disagreement between science and humanities profes- sors at Big Time University focuses on which department has the

grade point average for a random sample of 322 recent science gradu- ates was 3.51 (s � 1.2). Asserting that there is no meaningful differ- ence, a history professor shows that the mean grade point average for a sample of 485 recent humanities graduates was 3.36 (s � 1.6). Con-

22.2

22.3

22.4

22.5

of who will win the election? Who do you predict will win the

smarter students. As evidence supportive of the contention that

explain which professor appears to be more correct.

science students are smarter, a physics professor shows that the mean

struct a 99% confidence interval for this difference of means, and

C H A P T E R T W E N T Y T W O : S P E C I A L T O P I C S720

Interested in the effects of income and poverty on robbery rates, a stu- dent selected a random sample of 125 cities and correlated average in- come and percentage of persons living in poverty with the robbery rate. She reported the following correlations:

Income and robbery: r � �0.215

Poverty and robbery: r � 0.478

a. Calculate a 95% confidence interval for each correlation.

b. Explain what these results mean.

adolescents about their behavior. The researchers estimated a regres-

Variable b Standard Error

Intercept �0.21 0.15

Age �0.02 0.01

Number of friends arrested 2.56 0.73

Number of hours per week studying �0.17 0.08

Number of hours per week working 0.09 0.03

Self-esteem �1.05 0.51

a. Calculate a 95% confidence interval for each of the independent variable regression coefficients.

b. Explain what these results mean.

searchers recoded delinquency as 0 � no delinquency and 1 � one or more delinquent acts. They estimated a logistic regression model and found the following:

Variable b Standard Error

Intercept 0.05 0.04

Age �0.12 0.05

Number of friends arrested 1.86 0.57

Number of hours per week studying �0.23 0.09

Number of hours per week working 0.44 0.17

Self-esteem �0.79 0.38

a. Calculate a 95% confidence interval for each of the independent variable regression coefficients.

b. Explain what these results mean.

22.6

22.7

22.8 In a follow-up to the analysis reported in Exercise 22.7, the re-

dependent variable. The table of results follows:

Delinquency researchers at DP Institute interviewed a sample of 96

sion model, using number of delinquent acts in the last year as the

E X E R C I S E S 721

C H A P T E R T W E N T Y T W O : S P E C I A L T O P I C S722

C o m p u t e r E x e r c i s e s

All of the statistical packages that we are familiar with allow for the

straightforward computation of confidence intervals. SPSS and Stata both

allow for easy reporting of confidence intervals, as we illustrate below. Four

of the confidence intervals discussed in this chapter—sample mean, difference

of means, regression, and logistic regression—are the focus of our discussion.

There are no options in either program for computing confidence intervals for

Pearson’s r. Sample syntax in both SPSS (Chapter_22.sps) and Stata

(Chapter_22.do) illustrate each of the following commands.

SPSS

To obtain the confidence interval for a sample mean, use the T-TEST

command, but use the /TESTVAL option to test the value specified in the

null hypothesis (e.g., 0):

T-TEST

/TESTVAL = 0

/VARIABLES = variable_name

/CRITERIA = CI(.95).

where the /TESTVAL = 0 implements the null hypothesis (H 0

T-TEST command will test whether the sample mean is different from 0.

The /CRITERIA = CI(.95) requests a 95 % confidence interval. If the

/CRITERIA line is omitted from the command, the default output will still

contain the 95 % confidence interval. Where you may want to include the

/CRITERIA line is in the situation where you are interested in a different

confidence interval. For example, a 90 % confidence interval would be requested

with /CRITERIA = CI(.90).

In Chapter 11, we discussed how to compute an independent samples t-test

in SPSS with the T-TEST command. Recall the basic format for the command is

T-TEST GROUPS = grouping_variable(category_1 category_2)

/VARIABLES = variable_name.

Similar to the one-sample t-test, the default output from executing the T-TEST

for independent samples will include the 95% confidence interval. Should

you be interested in a confidence interval of a different size, insert the

/CRITERIA = CI(.##) option as explained above.

We discussed various features and option of the linear regression command

in SPSS (REGRESSION) in Chapters 15 through 17. Confidence intervals

are obtained by adding the option CI(##) to the /STATISTICS line in the

REGRESSION command:

C O M P U T E R E X E R C I S E S 723

REGRESSION

/STATISTICS COEFF CI(95) R ANOVA

/DEPENDENT dep_var

/METHOD = ENTER list_of_indep_vars.

where we have inserted a request for a 95 % confidence interval by adding

CI(95) to the /STATISTICS line. Should you want a different confidence inter-

val, simply change the values inside the parentheses. As we noted in previous

discussions of the REGRESSION command, when you desire some additional

output, it is necessary to also request all of the standard output from the com-

mand (i.e., COEFF R ANOVA). Note that the specification of the confidence

interval in the REGRESSION command does not require a decimal point in the

CI(##) option.

Chapter 18’s Computer Exercises focused on the use of the LOGISTIC

REGRESSION command. To obtain the confidence intervals for the estimated

coefficients, we need to add the /PRINT = CI(##) option line to the command:

LOGISTIC REGRESSION VARIABLES dep_var

/METHOD = ENTER list_of_indep_vars

/PRINT = CI(95).

The confidence intervals will appear as the far right columns in the table of

coefficients. Should you want a different confidence interval, all you need to

do is change the 95 in /PRINT = CI(95) to the value of interest. NOTE: The

confidence intervals computed by SPSS in the LOGISTIC REGRESSION com-

mand are for the odds ratios [Exp(B)], not the original coefficients (B). Similar

to requesting confidence intervals in REGRESSION, no decimal point is used in

the CI(##) option in the LOGISTIC REGRESSION command.

Stata

The ttest command is used to compute a one-sample ttest, where we compare a

sample mean to a hypothesized value:

ttest variable_name == hypothesized_value

In most cases, the hypothesized value will be 0. By default, the ttest command

will compute 95% confidence intervals. If you are interested in a different value

for the confidence interval, then add the level(##) option to the command line:

ttest variable_name == hypothesized_value, level(##)

For example, if we were interested in testing the hypothesis that GPA in the

NYS data was equal to 0 and use 90% confidence intervals, we would enter the

following command:

C H A P T E R T W E N T Y T W O : S P E C I A L T O P I C S724

ttest gpa == 0, level(90)

We discussed the independent samples t-test in Chapter 11’s Computer Exercises

section using the ttest command. The default output from the ttest command

is a 95% confidence interval. Should we be interested in a different range, we

would add the level(##) option:

ttest variable_name, by(grouping_variable) level(##)

In previous chapters, we have discussed the use of both the regress and logit

commands to estimate linear regression and binary logistic regression models,

respectively. Both of those commands report 95% confidence intervals in the

table of coefficients output by default. To request a different confidence interval,

add the level(##) option to the end of the command line (following a comma):

regress dep_var list_of_indep_vars, level(##)

and

logit dep_var list_of_indep_vars, level(##)

Consistent with all other output, the confidence intervals appear in the far right

column of the coefficients table.

Problems

Open the NY data file (nys_1.sav, nys _1_student.sav, or nys_1.dta) to answer

questions 1 through 4.

1. For each of the following measures of delinquency, compute a 95%

confidence interval and explain what it means.

a. Number of times the youth has stolen something valued at less than $5.

b. Number of times the youth has cheated on exams at school.

c. Number of times the youth has been drunk.

2. For each of the following difference of means tests, compute a 95% con-

fidence interval and explain what it means.

a. Does the number of times the youth has taken something valued at less

b. Does the number of times the youth has hit his or her parents differ for

c. Does the number of times the youth has cheated on exams differ for

C O M P U T E R E X E R C I S E S 725

3. Rerun two of the regression models you estimated in the Computer

Exercises in Chapter 16. For each model, compute 95% confidence inter-

vals for the regression coefficients and explain what each result means.

4. Rerun two of the logistic regression models you estimated in the

Computer Exercises in Chapter 18. For each model, compute 95% con-

fidence intervals for the odds ratios and explain what each result means.

Open the Pennsylvania Sentencing data file (pcs_98.sav or pcs_98.dta)

to answer questions 5 and 6.

5. Run a regression model using length of incarceration sentence as the

dependent variable and age, race, sex, offense severity score, and prior

criminal history score as the independent variables. Compute 99% confi-

dence intervals for the regression coefficients and explain what each result

means.

Run a binary logistic regression model using incarceration as the

dependent variable and age, race, sex, offense severity score, and prior

criminal history score as the independent variables (this is the model you

estimated in the Computer Exercises in Chapter 18). Compute 99%

confidence intervals for the odds ratios and explain what each result

means.

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5_23,

© Springer Science+Business Media New York 2014

Special Topics: Statistical Power

C h a p t e r t w e n t y t h r e e

A s s e s s i n g t h e r i s k o f T y p e I I e r r o r

How is Statistical Power Defined?

How Do Significance Criteria Influence Statistical Power?

How Does Effect Size Influence Statistical Power?

How Does Sample Size Influence Statistical Power?

E s t i m a t i n g s t a t i s t i c a l p o w e r

How Do We Define the Significance Criteria and Effect Size in a

Statistical Power Analysis?

How Do We Determine the Sample Size Needed to Ensure a Statistically

Powerful Study?

As we have seen in earlier chapters, criminal justice researchers place a

premium on statistical inference and its use in making decisions about population

parameters from sample statistics. In assessing statistical significance, the focus

hypothesis. Paying attention to the statistical significance of a finding should keep

researchers honest, because it provides a systematic approach for deciding when

the observed statistics are convincing enough for the researcher to state that they

reflect broader processes or relationships in the general population from which the

sample was drawn. If the threshold of statistical significance is not met, then the

researcher cannot reject the null hypothesis and cannot conclude that a relation-

ship exists.

Another type of error that most criminal justice researchers are aware of, but

pay relatively little attention to, is Type II, or beta (

failing to reject the null hypothesis that we originally introduced in Chapter 6.

A study that has a high risk of Type II error is likely to mistakenly conclude that

treatments are not worthwhile or that a relationship does not exist when in fact it

does. Understanding the risk of a Type II error is crucial to the development of a

research design that will give the researcher a good chance of finding a treatment

effect or a statistical relationship, if those effects and relationships exist in the

population. This is fundamentally what we mean by statistical power—given the

-

cally significant effects and relationships?

Although researchers in criminal justice have placed much more emphasis

on the statistical significance than on the statistical power of a study, research in

fields such as medicine and psychology routinely reports estimates of statistical

power.1

require research proposals to estimate how powerful the proposed research design

will be. The purpose of this chapter is to present an introductory discussion of

the key components in an assessment of the statistical power of a research design

727

1 See, for example, S. E. Maxwell, K. Kelley, and J. R. Rausch “Sample Size Planning for

Accuracy in Parameter Estimation,” Annual Review of Psychology

728 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

and to explain why it is important for criminal justice researchers to have a basic

understanding of the importance of statistical power in designing and evaluating

criminal justice research.

S t a t i s t i c a l P o w e r

Statistical power measures the probability of rejecting the null hypothesis when it

is false, but it cannot be measured directly. Rather, statistical power is calculated

by subtracting the probability of a Type II error—the probability of falsely failing

.

For many sample statistics, the Type II error can be estimated directly from the

sampling distributions commonly assumed for most test statistics. In contrast to a

traditional test of statistical significance, which identifies for the researcher the risk

-

cal power measures how often one would fail to identify a relationship that in fact

does exist in the population. For example, a study with a statistical power level

of 0.90 has only a 10% probability of falsely failing to reject the null hypothesis.

Alternatively, a study with a statistical power estimate of 0.40 has a 60% probability

of falsely failing to reject the null hypothesis. Generally as the statistical power of a

proposed study increases, the risk of making a Type II error decreases.

graphically. Suppose that we are interested in a difference in group means, say

between a control and treatment group in a criminal justice experiment, and

based on prior research and theory, we expect to find a positive difference in the

outcome measure. We would test for a difference in the group means by using

a one-tailed t-test. If we have 100 cases in each group, then the critical t-value

by the solid line represents the t-distribution—the sampling distribution—under

t-value.

Figure 23.1 Graphical Representation of Type I and Type II Errors in a Difference of Means Test

(100 Cases Per Sample)

−4 −2 0 2 4 6

0 .0

0 .1

0 .2

0 .3

0 .4

αβ

Critical t=1.653

S T A T I S T I C A L P O W E R 729

represents the hypothesized sampling distribution based on prior research and

theory and our expectations for the expected differences in the two group means.

The hypothesized sampling distribution is also a t-distribution, but it is known as

a non-central t-distribution—we illustrate below how this distribution is used to

compute statistical power. The probability of making a Type II error (

in the figure and is the cumulative probability in the distribution on the right up to

the critical t-value (i.e., t

test is represented in the figure by the area under the dashed line that falls to the

right of the critical value—the difference between 1 and —and represents t-values

the null hypothesis.

It is important to note that our estimate of is fully dependent on our estimate

illustrates the differences for two alternative effect sizes while assuming that the

sample sizes remain fixed at 100 cases per group. For example, if we expect the dif-

ference of means to be smaller, we would shift the hypothesized sampling distribu-

expect a larger difference, we would shift the hypothesized sampling distribution

to the right, reducing the estimate of

If the statistical power of a research design is high and the null hypothesis is false

for the population under study, then it is very likely that the researcher will reject

the null hypothesis and conclude that there is a statistically significant finding. If the

statistical power of a research design is low, it is unlikely to yield a statistically sig-

nificant finding, even if the research hypothesis is in fact true. Studies with very low

statistical power are sometimes described as being “designed for failure,” because a

study that is underpowered is unlikely to yield a statistically significant result, even

when the outcomes observed are consistent with the research hypothesis. 2

Figure 23.2 Graphical Representation of Type I and Type II Errors in a Difference of Means

Test—Changing the Difference in Mean Values. (a) Smaller Difference in Means—Fixed

Sample Size. (b) Larger Difference in Means—Fixed Sample Size

2 For an extended discussion of this, see D. Weisburd “Design Sensitivity in Criminal Justice

Experiments,” Crime and Justice

−4 −2 0 2 4

0 .0

0 .1

0 .2

0 .3

0 .4

α β

Critical t=1.653 a b

−4 −2 0 2 4 6 8

0 .0

0 .1

0 .2

0 .3

0 .4

αβ

Critical t=1.653

730 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

Consider the implications for theory and practice in criminal justice of a study

that has low statistical power. Suppose that a promising new program has been

developed for dealing with spouse assault. If that program is evaluated with a study

that has low statistical power, then the research team will likely fail to reject the null

hypothesis based on the sample statistics, even if the program does indeed have the

potential for affecting spouse assault. Although the research team is likely to say

that the program does not have a statistically significant impact on spouse assault,

this is not because the program is not an effective one, but because the research

team designed the study in such a way that it was unlikely to be able to identify

program success. Conceptually, this same problem occurs in the analysis of other

types of data when trying to establish whether a relationship exists between two

theoretically important variables. The relationship may exist in the population of

interest, but a study with low statistical power will be unlikely to conclude that the

relationship is statistically significant.

One might assume that researchers in criminal justice would work hard to

develop statistically powerful studies, because such studies are more likely to sup-

port the research hypothesis proposed by the investigators. Unfortunately, statisti-

cal power is often ignored altogether by criminal justice researchers, which results

in many criminal justice studies having a low level of statistical power.

Setting the Level of Statistical Power

What is a desirable level of statistical power? There is no single correct answer to

this question, since it depends on the relative importance of Type I and Type II

errors for the researcher. That said, one of the more common suggestions in the

statistical power literature has been that studies should attempt to achieve a power

level of 0.80, meaning that the chances of a Type II error are = 0.20. There are

many ways in which this is an arbitrary threshold. At the same time, it implies a

straightforward gauge for the relative importance of both types of error. If we use

If the target level of statistical power is 0.90, then = 0.10, and the ratio of prob-

abilities decreases to 0.10/0.05 = 2.0. What this means is that for a fixed level

chances of a Type II error (

to 1.0, where the chances of both types of error are viewed as equally important.

What happens if we reduce the desired level of statistical significance? For

example, suppose we were particularly concerned about our chances of mak-

See S. E. Brown “Statistical Power and Criminal Justice Research,” Journal of Criminal

Justice

Experiments,” Crime and Justice

of 0.80, this would imply that we are willing to accept a probability of making a

Type II error that is 20 times greater than the probability of a Type I error. If

we simultaneously increase the level of statistical power to 0.90 at the same time

we reduce the significance level, the

a much greater likelihood of a Type II error. If we wanted to keep the ratio

of error probabilities at 4.0, we would need a study with a power level of 0.96

-

neously increase our chances of failing to reject a false null hypothesis unless we

have a more powerful study.

C o m p o n e n t s o f S t a t i s t i c a l P o w e r

The level of statistical power associated with any given test of a sample statistic is

� Sample size

� Effect size

The level of statistical significance and sample size are assumed to be within the

control of the researcher, while the estimated effect size is not. The following

discussion briefly describes the links between each element and the statistical

power of any given test.

Statistical Significance and Statistical Power

The most straightforward way to increase the statistical power of a test is to change

the significance level used. As we reduce the chances of making a Type I error by

reducing the level of statistical significance from 0.10 to 0.05 to 0.01, it becomes

increasingly difficult to reject the null hypothesis. Simultaneously, the power of the

test is reduced. A significance level of 0.05 results in a more powerful test than a

significance level of 0.01, because it is easier to reject the null hypothesis using the

more lenient significance criteria. Conversely, a 0.10 level of significance would

make it even easier to reject the null hypothesis.

z-scores required to reject the

null hypothesis for several levels of statistical significance using a two-tailed test.

It would take a z

hypothesis with p = 0.10, a z

p = 0.05, and a z p = 0.01. Clearly,

α 0.20 0.10 0.05 0.01 0.001

z-score ±1.282 ±1.645 ±1.960 2.576 3.291

Table 23.1 z-Scores Needed to Reject the Null Hypothesis in a Two-Tailed Test of Statistical

Significance by Level of α

C O M P O N E N T S O F S T A T I S T I C A L P O W E R 731

732 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

it is much easier to reject the null hypothesis with a 0.10 significance threshold than

with a 0.01 significance threshold.

This method for increasing statistical power is direct, but it means that any

benefit we gain in reducing the risk of a Type II error is offset by an increase in

the risk of a Type I error. By setting a more lenient significance threshold, we

do indeed gain a more statistically powerful research study. However, the level

of statistical significance of our test also declines. Since a 0.05 significance level

has become the convention in much of the research in criminology and criminal

statistical significance is used.

Directional Hypotheses

A related method for increasing the statistical power of a study is to limit the

direction of the research hypothesis to either a positive or a negative outcome,

which implies the use of a one-tailed statistical test. A one-tailed test will provide

greater statistical power than a two-tailed test for the same reason that a less strin-

gent level of statistical significance provides more power than a more stringent

one. By choosing a one-tailed test, the researcher reduces the absolute value of the

test statistic needed to reject the null hypothesis by placing all of the probability of

making a Type I error in a single tail of the distribution.

z-scores

needed to reject the null hypothesis in one- and two-tailed tests for five different

levels of statistical significance. (For the sake of simplicity, we assume in the one-

tests, the test statistic required to reject the null hypothesis is smaller in the case

of a one-tailed test. For example, at p = 0.05, a z-score greater than or equal to

two-tailed test. In the one-tailed test, the z-score needs only to be greater than or

equal to 1.645. When we reduce the significance level to p = 0.01, a z-score greater

hypothesis in the two-tailed test, but in the one-tailed test, the z-score needs only

Although the researcher can increase the statistical power of a study by using

a directional, as opposed to a nondirectional, research hypothesis, there is a price

for shifting the rejection region to one side of the sampling distribution. Once a

one-directional test is defined, a finding in the direction opposite to that originally

predicted cannot be recognized. To do otherwise would bring into question the

integrity of the assumptions of the statistical test used in the analysis.

0.20 0.10 0.05 0.01 0.001

z-score (one-tail test) z-score (two-tail test)

−0.842 or 0.842 ±1.282

−1.282 or 1.282 ±1.645

−1.645 or 1.645 ±1.960

−2.326 or 2.326 2.576

−3.090 or 3.090 3.291

Table 23.2 z-Scores Needed to Reject the Null Hypothesis in One-Tailed and Two-Tailed

Tests of Statistical Significance

Sample Size and Statistical Power

The method used most often to change the level of statistical power in social

science research is to vary the size of the sample. Similar to specifying the level of

statistical significance, sample size can be controlled by the researcher. Modifying

the size of the sample is typically a more attractive option for increasing statistical

power than modifying the level of statistical significance, since the risk of a Type I

error remains fixed—presumably at the conventional p = 0.05.

The relationship between statistical power and sample size is straightforward.

All else being equal, larger samples provide more stable estimates of the popula-

tion parameters than do smaller samples. Assuming that we are analyzing data

from random samples of a population, the larger sample will have smaller standard

errors of the coefficients than will the smaller sample. As the number of cases in

a sample increases, the standard error of the sampling distribution (for any given

-

ard error for a single-sample t-test is computed as

s s

se N

= -1

.

As N

se

decreases, the likelihood of achieving statistical significance grows, because the

test statistic for a test of statistical significance is calculated by taking the ratio of

the difference between the observed statistic and the value proposed in the null

is held constant, then as the sample size increases, the standard error decreases,

and a larger test statistic is computed, making it easier to reject the null hypothesis.

The effect of sample size on statistical power for a t-test of the difference of

100 two-sample t-tests in which a mean difference of two arrests between groups

hypothesis changes substantially with each increase in sample size, even though all

SCENARIO SAMPLE SIZE(PER GROUP) μ1 − μ2 σ EXPECTED SIGNIFICANT OUTCOMES

1 2 3 4

35 100 200

1,000

0.2 0.2 0.2 0.2

1 1 1 1

13 29 51 99

Table 23.3 Number of Statistically Significant Outcomes Expected in 100 Two-Sample

t-Tests for Four Scenarios

733C O M P O N E N T S O F S T A T I S T I C A L P O W E R

734 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

other characteristics are held constant across the four scenarios. Under the first

tests. In the second scenario, 29 significant outcomes would be expected and in

the third, 51. In the final scenario of samples of 1,000, nearly every test (99 out of

statistical significance of a study.

In most cases, researchers maximize the statistical power of a study by increas-

ing sample size. The concern with sample size is also reflected in the number of

publications focused on advising researchers in all behavioral and social science

fields on how to determine the appropriate sample size for a proposed research

study.4

Although sample size should be under the control of the researcher, it is

important to be aware of the unanticipated consequences of simply increasing

sample size may have on other factors that influence statistical power, particularly

in evaluation research.5 For example, suppose a researcher has developed a com-

plex and intensive method for intervening with high-risk youth. The impact of the

treatment is dependent on each subject receiving the “full dosage” of the treatment

for a six-month period. If the researcher were to increase the sample size of this

study, it might become more difficult to deliver the treatments in the way that was

originally intended by the researcher. More generally, increasing the sample size

of a study can decrease the integrity or the dosage of the interventions that are

applied and result in the study showing no effect of the treatment. Increasing the

size of a sample may also affect the variability of study estimates in other ways. For

example, it may become more difficult to monitor implementation of treatments as

a study grows. It is one thing to make sure that 100 subjects receive a certain inter-

vention but quite another to ensure consistency of interventions across hundreds

or thousands of subjects. Also, studies are likely to include more heterogeneous

groups of subjects as sample size increases. For example, in a study of intensive

probation, eligibility requirements were continually relaxed in order to meet project

goals regarding the number of participants.6 As noted earlier, as the heterogeneity

of treatments or subjects in a study grows, it is likely that the standard deviations

of the outcomes examined will also get larger. This, in turn, leads to a smaller effect

size for the study and thus a lower level of statistical power.

4 For a range of examples, see P. Dattalo Determining Sample Size

How Many Subjects: Statistical

Power Analysis in Research

Myors Statistical Power Analysis 5 D. Weisburd “Design Sensitivity in Criminal Justice Experiments,” Crime and Justice

6

Evaluation Review

Effect Size and Statistical Power

for statistical significance used in a test. Effect size measures the difference

between the actual parameters in the population and those hypothesized in the null

hypothesis. In computing effect size, it is important to take into account both the

raw differences between scores and the degree of variability found in the measures

examined. Taking into account variability in effect size is a method of standardiza-

tion that allows for the comparison of effects across studies that may have used

different scales or slightly different types of measures. It has also allowed for the

standardization of estimates of statistical power across a wide range of studies and

types of analyses.

Generally, effect size is defined as

ES Parameter H

= -

.0

s Equation 23.1

The relationship between effect size and statistical power should be clear. When

the standardized population parameters differ substantially from those proposed

in the null hypothesis, the researcher should be more likely to observe a significant

examined. Effect size will increase when the difference between the population

parameter and the hypothesized parameter increases and the standard error is

held constant or when the difference is held constant and the standard error is

decreased, perhaps through the use of a larger sample of cases.

A difference of means test for two independent samples provides a simple

illustration for these relationships. In the difference of means test, effect size

would be calculated by first subtracting the population difference as stated in the

null hypothesis (H 0 μ

1 – H

0 μ

2

population (μ 1 – μ

2

as the pooled or the common standard deviation of the outcome measures in the

ES = -( )- -( )m m m m

s 1 2 0 1 0 2H H

. Equation 23.2

Effect size can also be calculated for observed differences in a study. This is a common

approach in meta-analysis, where a large group of studies are summarized in a single analysis.

For example, in calculating effect size for a randomized experiment with one treatment and

one control group, the researcher would substitute the outcome scores for both groups in the

numerator of the ES equation and the pooled standard deviation for the two outcome

measures in the denominator. For a more detailed discussion of effect size and its use generally

for comparing effects across different studies, see M. Lipsey and D. Wilson Practical

Meta-Analysis Meta-Analytic

Procedures for Social Research

735C O M P O N E N T S O F S T A T I S T I C A L P O W E R

736 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

Since the null hypothesis for a difference of means test is ordinarily that the two

population means are equal (i.e., H 0 μ

1 H

0 μ

2

ES = -( )m m

s 1 2

. Equation 23.3

Thus, the ES for a difference of means test may be defined simply as the raw

difference between the two population parameters, divided by their common

standard deviation. To reiterate an earlier comment, when the difference between

the population means is greater, the ES for the difference of means will be larger.

Also, as the variability of the scores of the parameters grows, as represented by the

standard deviation of the estimates, the ES will get smaller.

number of statistically significant outcomes expected in 100 t-tests (using a 0.05

significance threshold and a nondirectional research hypothesis, resulting in a two-

In the first three scenarios, the mean differences between the two populations

are varied and the standard deviations for the populations are held constant. In

the last three scenarios, the mean differences are held constant and the standard

deviations differ.

is expected in either the comparisons with the largest differences between mean

scores or the comparisons with the smallest standard deviations. As the differ-

likelihood of obtaining a statistically significant result. Conversely, as the popula-

expected number of significant outcomes decreases.

As this exercise illustrates, there is a direct relationship between the two com-

ponents of effect size and statistical power. Studies that examine populations in

which there is a larger effect size will, all else being equal, have a higher level of

statistical power. Importantly, the relationship between effect size and statistical

power is unrelated to the significance criteria we use in a test. In this sense, effect

SCENARIO μ1 μ2 σ EXPECTED SIGNIFICANT OUTCOMES

(a) Means differ; standard deviations constant

1 2 3

0.3 0.3 0.3

0.5 0.9 1.3

2 2 2

10 56 94

(b) Means constant; standard deviations differ

4 5 6

0.3 0.3 0.3

0.5 0.5 0.5

0.5 1 2

80 29 10

Table 23.4 Number of Statistically Significant Outcomes Expected in 100 Two-Sample

t-Tests for Six Different Scenarios (100 Cases in Each Sample)

size allows for increasing the statistical power of a study (and thus reducing the risk

-

Although effect size is often considered the most important component of

statistical power, it is generally very difficult for the researcher to manipulate in

a specific study.8 Ordinarily, a study is initiated in order to determine the type

and magnitude of a relationship that exists in a population. In many cases, the

researcher has no influence at all over the raw differences or the variability of the

scores on the measures examined. For example, a researcher who is interested in

identifying whether male and female police officers have different attitudes toward

corruption may have no idea prior to the execution of a study the nature of these

attitudes or their variability. It is then not possible for the researcher to estimate

the nature of the effect size prior to collecting and analyzing data—the effect size

may be large or small, but it is not a factor that the researcher is able to influence.

In contrast, evaluation research—in which a study attempts to assess a specific

program or intervention—the researcher may have the ability to influence the

effect size of a study and thus minimize the risk of making a Type II error. There is

growing recognition, for example, of the importance of ensuring the strength and

integrity of criminal justice interventions.9 Moreover, many criminal justice evalu-

ations fail to show a statistically significant result simply because the interventions

are too weak to have the desired impact or the outcomes are too variable to allow

a statistically significant finding.10

Statistical power suggests that researchers should be concerned with the effect

size of their evaluation studies if they want to develop a fair test of the research

hypothesis. First, the interventions should be strong enough to lead to the expect-

ed differences in the populations under study. Of course, the larger the differences

expected, the greater the statistical power of an investigation. Second, interven-

tions should be administered in ways that maximize the homogeneity of outcomes.

For example, interventions applied differently to each subject will likely increase

the variability of outcomes and thus the standard deviation of those scores. Finally,

researchers should recognize that the heterogeneity of the subjects studied (and

the statistical power of their tests. Different types of people are likely to respond

in different ways to treatment or interventions. If they do respond differently, the

variability of outcomes will be larger, and thus the likelihood of making a Type II

error will increase.

As a caution, we note that a wide range of research in criminology and crimi-

nal justice has increasingly made use of archival data sets that result in research-

ers analyzing populations rather than samples. Examples of this would include

studies that rely on archival data on all punishment decisions made in the US

8 M. Lipsey Design Sensitivity: Statistical Power for Experimental Research,

9

Evaluation Review 10 D. Weisburd “Design Sensitivity in Criminal Justice Experiments,” Crime and Justice

737C O M P O N E N T S O F S T A T I S T I C A L P O W E R

738 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

Federal District Courts or census data on all prisoners in a state on a specific date.

which populations are analyzed calls into question many of the assumptions about

performing tests for statistical significance.11 Put simply, the analysis of popula-

tion data implies no need for statistical significance testing, since the researcher is

not trying to generalize from a sample to a population. Clearly, issues of statistical

makes little sense, the number of cases in the data set is as large as it possibly can

E s t i m a t i n g S t a t i s t i c a l P o w e r a n d S a m p l e S i z e f o r a S t a t i s t i c a l l y P o w e r f u l S t u d y

A number of texts have been written that provide detailed tables for defining the

statistical power of a study.12 All of these texts also provide a means for computing

the size of the sample needed to achieve a given level of statistical power. In both

cases—the estimation of statistical power or the estimation of necessary sample

size—assumptions will need to be made about effect size and level of statistical

significance desired. The following discussion provides a basic illustration for

how to compute estimates of statistical power. (The computations reported in the

following discussion have been performed with a variety of statistical software

tools, several of which are freely available. More detail on several easily accessible

resources to compute power estimates is provided in the computer problems sec-

The most common application of statistical power analysis in criminology and

criminal justice research has been to compute the sample size needed to achieve a

to be cautious about simply increasing the size of the sample, since a larger sample

can affect other important features of statistical power. Thus, in using increased

sample size to minimize Type II error, we must consider the potential conse-

quences that larger samples might have on the nature of interventions or subjects

studied, particularly in evaluation research. Nonetheless, sample size remains the

tool most frequently used for adjusting the power of studies, because it can be

manipulated by the researcher and does not require changes in the significance

criteria of a test.

To define how many cases should be included in a study, we must conduct

power analyses before the study is begun, generally referred to as prospective or

11

Journal of Research in Crime and Delinquency 12 Among some of the more widely used examples are J. Cohen Statistical Power Analysis

for the Behavioral Sciences

and S. Thiemann How Many Subjects: Statistical Power Analysis in Research, (Newbury

Design Sensitivity: Statistical Power for Experimental

Research Statistical

Power Analysis

a priori power analysis, and where our attention has been focused thus far in this

chapter. Some authors have advocated the use of power analysis to evaluate wheth-

er studies already conducted have acceptable levels of statistical power, based

on the sample statistics, referred to as retrospective or post hoc power analy-

sis. Although there is much agreement about the utility of prospective power

analysis, there is little consensus about the appropriateness of retrospective power

analysis. The widespread use of secondary data sources in the study of crime and

criminal justice further complicates the interpretation of results from a statistical

power analysis. Since it is not possible for researchers to augment the original

that the results will indicate to the researchers using these data sources what the

archived data set can and cannot tell them about the statistical relationships they

may be most interested in.

To define the sample size needed for a powerful study, we must first clearly

define each of the components of statistical power other than sample size. These

1. The statistical test

2. The significance level

4. The effect size

The first three of these elements should be familiar, since they are based on com-

mon assumptions made in developing any statistical test. The statistical test is cho-

sen based on the type of measurement and the extent to which the study can meet

certain assumptions. For example, if we want to compare three sample means, we

will likely use analysis of variance as our test. If we are comparing means from two

samples, we will likely use a two-sample t-test. If we are interested in the unique

effects of a number of independent variables on a single interval-level dependent

variable, we will likely use OLS regression and rely on t-tests for the individual

coefficients and F-tests for either the full regression model or a subset of variables

from the full model.

To calculate statistical power, we must also define the significance level of a

test and its research hypothesis. By convention, we generally use a 0.05 significance

threshold, and thus we are likely to compute statistical power estimates based

on this criterion. The research hypothesis defines whether a test is directional or

nondirectional. When the statistical test allows for it, we will typically choose a

nondirectional test to take into account the different types of outcomes that can

be found in a study.14 If we were evaluating an existing study, we would use the

For an example, see the exchange between J. P. Hayes and R. J. Steidl “Statistical Power

Analysis and Amphibian Population Trends,” Conservation Biology

and L. Thomas “Retrospective Power Analysis,” Conservation Biology 14 J. Cohen Statistical Power Analysis for the Behavioral Sciences

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740 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

The fourth element, defining effect size, is perhaps the most difficult

component. If we are trying to estimate the magnitude of a relationship in the

population that has not been well examined in the past, how can we estimate the

effect size in the population? It may be useful to reframe this criterion. The pur-

pose of a power analysis is to see whether our study is likely to detect an effect

of a certain size. Usually, we define that effect in terms of what is a meaningful

outcome in a study. A power analysis, then, tells us whether our study is designed

in a way that is likely to detect that outcome (i.e., reject the null hypothesis on the

sometimes defined as design sensitivity.15 It assesses whether our study is designed

with enough sensitivity to be likely to reject the null hypothesis if an effect of a

certain size exists in the population under study.

The task of defining effect size has been made easier by identifying broad

have been the most widely adopted by other researchers and simply refer to classi-

fying effect sizes as small, medium, and large.16 The numeric value associated with

an effect size classified as small, medium, or large is contingent on the specific sta-

tistical test being considered. For example, if our focus is on a difference of means

test for two independent samples, the standardized effect size estimate is known

as d

effect if it is 0.5, and a large effect if it is 0.8. In contrast, if we are considering the

statistical power of an OLS regression model, the standardized effect size estimate

is known as f2 and is considered to be a small effect if it is 0.02, a medium effect

attempted to define similar types of standardized effects for more complex statisti-

The following illustration turns to a discussion of the computation of statistical

difference of means test, ANOVA, correlation, and OLS regression—all of which

have been the focus of previous chapters.

The computation of statistical power estimates requires the comparison of a

sampling distribution under the null hypothesis with a sampling distribution under

sampling distribution under the research hypothesis is referred to as a non-central

-

tion under the null hypothesis is the t-distribution, while the sampling distribution

under the research hypothesis is the non-central t-distribution.

The non-central sampling distribution is computed based on a “non-centrality”

parameter, which in all cases is a function of the standardized effect for the statisti-

cal test under consideration. For each of the statistical tests discussed below, we

describe both the standardized effect and the non-centrality parameter and explain

15 M. Lipsey Design Sensitivity: Statistical Power for Experimental Research, (Newbury

16 J. Cohen Statistical Power Analysis for the Behavioral Sciences

how to use these values to estimate the statistical power of a sample as well as the

size of sample needed to meet a target level of statistical power.

Difference of Means Test

Throughout this chapter, we have pointed to the difference of means test as an

example for many of the points we wanted to make about statistical power. More

directly, the standardized effect size d is

d = -m m s

1 2 ,

which is identical to the equation noted earlier for computing a standardized dif-

ference of means for two independent samples. Recall that represents the pooled,

or common, standard deviation for the difference of means.

The non-centrality parameter for the t-distribution is

d = d N

4 , Equation 23.4

where N = n 1 + n

2 when there are equal numbers of cases in each group (i.e.,

n 1 = n

2 n

1 n

2 , the non-centrality parameter is

d = = +

d N

where N n n

n n

H

H 2

2 1 2

1 2

, . Equation 23.5

To illustrate the computation of a statistical power estimate, suppose that we want

to assess the effectiveness of a treatment program for drug offenders. Our design

calls for random assignment of 100 cases to each group. We expect the program

to be effective at reducing recidivism in the treatment group and so can assume a

one-tailed t-test with a significance level of 5%. What is the statistical power of our

design for detecting standardized effects at the small (d d

and large (d

For all three scenarios, the critical t

test with a significance level of 0.05 and df = N – 2 = 198. For a small effect,

an estimate for risk of making a Type II error of

hypothesis when it is false. The corresponding estimate of statistical power is

in each group, our probability of rejecting the null hypothesis when it is false is

< 0.0001, and power > 0.9999.

It is not possible to include copies of non-central t-distribution tables in the same way that

t-distribution in Appendix 4. We will illustrate in the Computer

Exercises at the end of this chapter how to work with the non-centrality parameter to obtain

estimates of from various statistical packages.

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Putting these results together indicates that our design with 100 cases assigned to

each group provides a high level of statistical power for detecting medium effects

and larger but an inadequate level of power for detecting small effects.

Alternatively, we may be interested in determining the sample size needed to

provide us with a statistical power estimate of 80% for each of the three effect

to reject the null hypothesis when it is false about 80% of the time. To achieve a

power estimate of 80% for a medium effect, we only need 102 cases (51 in each

ANOVA

For a simple ANOVA, where we are looking only at fixed effects and assume equal

sample sizes across groups, the standardized effect size f is defined as

f m= s s

, Equation 23.6

where sm

i

k im m

k =

-( ) = å

1

2

, k is the number of groups, m is the grand mean, and

m i represents each of the group means with n

1 = n

2 = … = n

k .

The non-centrality parameter for the F-distribution is

=f 2N, Equation 23.7

where f2 refers to the square of the standardized effect size (f N refers to the

total sample size.

As an illustration of the calculation of statistical power estimates for a fixed-

effects ANOVA model, assume that we have three groups, each with 100 cases

participating in an experiment aimed at reducing recidivism among violent offend-

significance level has been set at 5%. What is the level of statistical power of our

design for detecting standardized effects at the small (f f

and large (f

For each of the three scenarios, the critical value of the F

(df 1 = 2, df

2

of = 0.681, suggesting that we have a probability of 68.1% of making a Type II

error and fail to reject the null hypothesis when it is false. The corresponding esti-

chance of rejecting the null hypothesis when it is false. This result is presented

F-distribution, and the dashed

line the non-central F-distribution. Below the two curves, represented by two dif-

ferent shades of grey, alpha is indicated by the darker shading in the right tail of

the F-distribution beyond the critical value, and beta is represented by the lighter

shaded area to the left of the critical value and under the non-central F-distribution.

For the medium and large effect size analyses, the F-distribution remains the

same, but the non-central F-distribution is shifted further to the right. For the

medium effect size,

has = 48, < 0.0001, and power > 0.9999. Similar to the previous analysis compar-

ing the means for only two groups, our research design with 100 cases assigned

to each of the three groups provides a high level of statistical power for detecting

medium and large effects but an inadequate level of power for detecting small

effects.

If our concern is focused on the size of the sample needed for a power level of

80% for each of the three effect sizes—small, medium, and large—then we would

again proceed in the same way as in the two-sample t-test. To have an 80% chance

of detecting a small effect (f

Correlation

To test the statistical power of a correlation coefficient, we can use either the

correlation coefficient (r r-to-Z transformation of the correlation

coefficient (r Z

power will not be identical, they will tend to be very close, typically differing only

at the second or the third decimal.

d = -

´ r

r N

2

21 , Equation 23.8

where r is either the sample correlation coefficient (r

(r Z

N is the sample size.

We can again illustrate the calculation of statistical power for correlations by

assuming that we have 100 observations that would allow us to compute a correla-

tion between two variables. For example, suppose we interview a random sample

of police officers and are interested in the correlation between the number of

Figure 23.3 Graphical Representation for Power Analysis in a One-Way ANOVA

E S T I M A T I N G S T A T I S T I C A L P O W E R A N D S A M P L E S I Z E 743

0 1 2 3 4 5 6 7 0 .0

0 .2

0 .4

0 .6

0 .8

1 .0

α β

Critical F=3.026

744 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

years on the police force and a scale that measured hostility toward judges. We

might expect that more years on the police force will have a positive correlation

with hostility toward judges, implying that we can conduct a one-tailed t-test of

statistical significance. As with the preceding examples, assume that the level of

statistical significance is 5%. What is the level of statistical power of our design

for detecting standardized effects at the small (r r

(r

The critical t-value for all three scenarios is 1.661, based on df = N

a small effect size (r

us with an estimate for risk of making a Type II error of

the null hypothesis when it is false. The corresponding estimate of statistical power

is 0.259, indicating that we would only reject the null hypothesis when it was false

-

cal power analysis of the medium effect indicates that

where < 0.0001, and power > 0.9999.

The sample size required to detect each of the three effect sizes—small,

medium, and large—with a statistical power of 80% again requires the use of the

t-distribution. To achieve a power level of 80% for a small effect (r -

ple of 614 cases would be needed. For the medium effect (r

number of cases drops to 64, while for the large effect (r

required to have an 80% chance of rejecting the null hypothesis when it is false.

Least-Squares Regression

The statistical power analysis of least-squares regression can take two different, but

related, forms. One question asks about the ability to detect whether a regression

model—a single dependent variable and two or more independent variables—has

a statistically significant effect on the dependent variable. This means that the null

hypothesis is focused on whether the regression model in its entirety has an effect

on the dependent variable. A second question asks about the ability to detect the

effect of a single variable or a subset of variables added to a regression model.

This addresses the more common substantive question in much of the published

Figure 23.4 Graphical Representation for Power Analysis of a Correlation

−4 −2 0 2 4 0 .0

0 .1

0 .2

0 .3

0 .4

αβ

Critical t=1.661

taken into account statistically, does variable X add anything to the overall model?

Whether we are analyzing the full model or a subset of the full model, the stand-

ardized effect size (denoted as f 2 R2 for the full model or

the partial R2 for the subset of variables we are interested in analyzing. Specifically,

R f

f

2 2

21 =

+ . Equation 23.9

To provide some context to these values, an f 2 value of 0.02 corresponds to an R2

of 0.02, while f 2 = 0.15 implies that R2 f 2 R2 = 0.26.

Statistical power analysis for least-squares regression uses the F-distribution.

As noted in the discussion of statistical power analysis for ANOVA models, the

non-centrality parameter for the F-distribution is

= f 2N.

To assess the statistical power for the full regression model consider the follow-

ing simple example. Suppose that we are interested in the effects of various case

and defendant characteristics on the amount of bail required by a court. Typical

analyses of bail decisions would consider some of the following characteristics (as

-

whether the defendant was under criminal justice supervision at the time of the

of the defendant. This provides us with a regression model with eight independent

variables.

As a point of illustration, we may want to estimate the statistical power of the

regression model assuming that we have a sample of only 100 cases and have set

a significance level of 5%, giving us a crucial F-value of 2.024. For the small effect

size ( f 2

find

a Type II error of just under 88%. Alternatively, the estimate of statistical power

is 0.124, meaning that we have a probability of only 12.4% of rejecting the null

hypothesis when it is false. The results for the medium effect ( f 2

= 15.0,

an inadequate level of power but is much closer to the target of 80%. For the large

effect (f 2

desired level of 80%.

For a regression model with eight independent variables, what sample size is

required to achieve a statistical power level of 80% for detecting effects at the small

(f2 f2 f2

-

um and large effects, we would require samples of 109 and 52 cases, respectively.

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The number of cases required to detect a statistically significant effect at either the

medium or the large effect level may strike many readers as small. It is important to

keep in mind that we have only been assessing the full model—the number of cases

required for detecting individual effects will tend to be different than the number of

cases required for detecting whether the full model is significant.

The assessment of statistical power for a single independent variable or a small

subset of independent variables proceeds in much the same way as the analysis for

the full model. The key difference is in the degrees of freedom required for the

F-distribution. In the case of a single independent variable, the numerator df = 1,

while the denominator df remains the same as in the full model. For a subset of

independent variables, the numerator df = the number of variables in the subset

(the denominator df

If we return to the bail example above, the analysis of statistical power for

any one of the independent variables will be identical. We continue to keep the

sample size at 100 cases, the level of statistical significance at 5%, and the defini-

tion of small, medium, and large effects the same as before. For the small effect

( f 2 = 2.0,

able to reject the null hypothesis of no relationship between the independent and

dependent variables about 28.8% of the time. For the medium effect ( f 2

= 15.0, f 2

< 0.0001, and power > 0.9999.

Similarly, we may be interested in assessing the statistical power of a subset

of variables. For example, in the bail example, the subset of demographic char-

theory predicting differential treatment of defendants within the courts. We find

a similar pattern to the results. For the small effect ( f 2 = 2.0, = 0.814,

and power = 0.186, again indicating a low level of statistical power for detecting a

statistically significant relationship between demographic characteristics and bail

amount. For the medium effect ( f 2 = 15.0, = 0.095, and power = 0.905,

while for the large effect ( f 2 = 0.001, and power = 0.999.

Sample size calculations work in the same way as for the full model. If we

hope to achieve a power level of 80%, what size sample is necessary to detect

Figure 23.5 Graphical Representation for Power Analysis of a Regression Model (with Eight Independent

Variables)

0 1 2 3 4 5 0 .0

0 .2

0 .4

0 .6

0 .8

1 .0

αβ

Critical F=2.042

small, medium, and large effects for either single variables or subsets of variables?

Continuing the bail example, we assume that there are eight independent variables.

For the single variable, the number of cases required to detect a small effect with

effect requires only 26 cases. It is worth noting that sample size calculations for

single variable effects are not affected by the number of variables included in the

full regression model.

In practice, many of the individual effects that researchers are trying to assess in

their multivariate models will tend toward the small effect size. For example, much

survey research aimed at trying to explain attitudes toward a particular topic will

often incorporate 10–20 independent variables and have a full model R2 typically

between 0.15 and 0.20. This implies that many of the effects of individual variables

will tend to be quite small in magnitude. In order for an analysis to detect a statisti-

cally significant relationship, a much large sample becomes necessary.

S u m m i n g U p : A v o i d i n g S t u d i e s D e s i g n e d f o r F a i l u r e

The statistical power of a test can be compared to the sensitivity of a radiation

meter. A very sensitive meter will be able to identify even the smallest deposits of

radioactivity. A meter that is not very sensitive will often miss such small deposits,

although it likely will detect very large radiation signals from areas rich in radio-

activity. Similarly, a statistically sensitive study will be able to identify even small

effects. This is usually because the researcher has increased the sample size of

the study to make it more statistically powerful. Conversely, a study that has little

sensitivity is unlikely to yield a statistically significant result even when relatively

large differences or program impacts are observed. Such studies may be seen as

“designed for failure,” not because of inadequacies in the theories or the programs

evaluated, but because the investigator failed to consider statistical power at the

outset of the study.

You might question why we would even bother to define the size of the sam-

ple needed for statistically powerful studies. Why not just collect 1,000 or more

cases in every study and be almost assured of a statistically powerful result? The

simple answer is that although you should try to sample as many cases as you can

in a study, there are generally constraints in developing samples. These constraints

may be monetary, related to time, or associated with access to subjects. It is often

important to know the minimum number of cases needed to achieve a certain

threshold of statistical power so that you can try, within the constraints of the

research setting, to reach an adequate level of statistical power in your study. It is

also important to be able to assess whether studies that you read or evaluate were

designed in such a way that they are reasonable tests of the hypotheses presented.

If such studies are strongly underpowered, then you should have much less confi-

dence in findings that do not support the research hypothesis.

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C h a p t e r S u m m a r y

A statistically powerful test is one for which there is a low risk of making a Type II

error. Statistical power can be defined as 1 minus the probability of falsely accept-

ing the null hypothesis. A test with a statistical power of 0.90 is one for which

there is only a 10% probability of making a Type II error. If the power of a test is

0.10, the probability of a Type II error is 90%. A minimum statistical power level

of at least 0.50 is recommended. However, it is generally accepted that in better

studies, the level of statistical power will be at least 0.80. A study with a low level

of statistical power can be described as “designed for failure,” as it is unlikely to

produce a statistically significant result even if the expected effect exists in the

population under study.

There are several ways in which statistical power can be maximized. First, we

may raise the significance threshold. Doing so, however, also increases the risk of

a Type I error. Second, we may limit the direction of the research hypothesis and

conduct a one-tailed test. Doing so, though, will necessarily ignore outcomes in the

opposite direction. Third, we may try to maximize the effect size. The greater the

differences betwen the populations and the smaller the variability of those differ-

ences, the larger the population effect size will be. Effect size, however, is usually

beyond the control of the researcher. Fourth, we may increase the sample size. A

larger sample produces a smaller standard error for the sampling distribution and

a larger test statistic. The larger the sample, all else being equal, the greater the

chance of rejecting the null hypothesis.

Sample size is generally the most useful tool for maximizing statistical power.

A power analysis before a study is begun will define the number of cases needed

to identify a particular size effect—small, medium, or large. A power analysis of an

existing study will help to identify whether it was well designed to assess the ques-

tions that were examined. To identify a small effect size, the overall sample must

be very large. For a large effect size, a much smaller sample will suffice.

K e y T e r m s

design sensitivity The statistical power

of a research study. In a sensitive study design,

statistical power will be maximized, and the

statistical test employed will be more capable

of identifying an effect.

effect size (ES) A standardized measure

derived by taking the effect size (e.g., the

measured in the raw units of the outcome

measure examined, and dividing it by the

pooled or common standard deviation of the

outcome measure.

statistical power One minus the probability of

a Type II error. The greater the statistical power

of a test, the less chance there is that a researcher

will mistakenly fail to reject the null hypothesis.

S y m b o l s a n d F o r m u l a s

D

F

N

n i

i

t-distribution

F-distribution (used for ANOVA and OLS

ES Parameter H

= - 0

s .

ES H H

= -( )- -( )m m m m

s 1 2 0 1 0 2

, which simplifies to ES = -( )m m

s 1 2 .

To calculate the non-centrality parameter for the t-distribution for difference

d = d N

4 .

d = d NH

2 , where N

n n

n n H =

+

2 1 2

1 2

f m= s s

, where sm i

k im m

k =

-( ) = å

1

2

, k is the number of groups, m is the

grand mean, and m i represents each of the group means with n

1 = n

2 = ⋯ = n

k .

To calculate the non-centrality parameter for the F

l = f N2 .

To calculate the non-centrality parameter for the t-distribution for correlation

d = -

´ r

r N

2

21 .

S Y M B O L S A N D F O R M U L A S 749

750 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

To calculate R2 for OLS regression using the standardized effect size f as defined

R f f2 2 21= +( ).

C o m p u t e r E x e r c i s e s

In contrast to many of the other computer exercises in this text, the computation

of statistical power estimates is not easily performed in any of the large stand-alone

statistical packages. There are a variety of software packages available for comput-

ing statistical power as well as a number of websites that host power calculators for

a wide range of statistical tests. All of the analyses presented in this chapter were

-

load from the Institut fur Experimentelle Psychologie at Universitat Dusseldorf

is a specialized package devoted to statistical power estimation and offers a wide

creation of powerful graphs that will plot power estimates across a range of sample

sizes, effect sizes, and statistical significance levels. The figures presented in this

18.

Power and Precision v. 2.0 is a commercially available software package

designed to compute power estimates for a wide range of statistical models in a

user-friendly environment.19 As a commercial software package, its range of capa-

all of the output—text and graphs—can be easily exported to other programs.

In the case that one simply wants to compute a small number of power esti-

mates without bothering to learn a new software package, a reasonably compre-

org/#Power. The list of websites hosting power calculators is categorized by the

type of statistical test that the user is searching for—one-sample t-test, two-sample

t-test, correlation, regression, and so on.

On a technical note, it is worth highlighting that there will be slight differences

across statistical software packages and power calculators in the estimated sample

sizes needed to achieve a given level of statistical power. The primary reason for

this appears to be focused on rounding the estimated sample size to an integer,

since we cannot sample a fraction of a case in any research study. Some packages

round up so that the estimated statistical power as always at least as great as the

target entered into the computation. Other packages and calculators will round

estimate of statistical power may be slightly less than the initial target.

18

Analysis Program for the Social, Behavioral, and Biomedical Sciences,” Behavior Research

Methods 19 M. Borenstein, H. Rothstein, and J. Cohen Power and Precision

Stata

Two-Sample Difference of Means Test

In Stata, one- and two-sample difference of means tests are performed with the

sampsi

sampsi Mean1 Mean2 , sd1(#) sd2(#) n1(#) n2(#) power(#)

onesided

where Mean1 and Mean2 refer to the expected population means for the two

samples being compared, sd1(#) and sd2(#) refer to the expected standard

n1(#) and

n2(#)

in each sample, power(#) is a designated level of power for sample size

onesided indicates that a

we are trying to estimate power and assume constant standard deviations and

sample sizes across the two samples, this can be simplified to

sampsi Mean1 Mean2, sd(#) n(#)

Upon entering the command, the output will list all of the assumptions (alpha

cases each, a difference of population means of 0.2, a common standard devia-

sampsi command, we would

sampsi 0 0.2 , sd(1) n(35)

The use of 0 and 0.2 for the two sample means is a convenient way to represent

the difference. It would make no difference what two numbers we inserted here

C O M P U T E R E X E R C I S E S 751

752 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

respectively.

If our interest is in estimating the sample size required to achieve a given

level of statistical power, we would alter the sampsi command by omitting the

sample size values (n(#)

difference of means power analysis, we assumed a small standardized effect (i.e.,

d

sampsi 0 .2 , sd(1) power(.8) onesided

sample size estimates for the medium and large effects, simply increase the value

of the second mean from 0.2 to 0.5 and 0.8.

ANOVA

Unfortunately, there is no built-in command in Stata to compute power in a

simple one-way ANOVA. Although there are several user-written commands

that can be installed and used, it is often difficult to obtain the kind of infor-

mation we may be most interested in gleaning from a power analysis. Based on

existing procedures in Stata, as well as other statistical packages, we have written

anova_pwr, which is a modest Stata procedure that you can use to replicate

the results in this chapter and perform other simple one-way ANOVA estimates

of power and sample size. To install this command on your copy of Stata, type

net install anova_pwr,

from(http://myfiles.neu.edu/c.britt/stata/ado/power)

The basic components to the anova_pwr

anova_pwr, ngp(#) f(#) min(#) max(#)

where ngp(#) represents the number of groups to be compared, f (#)

represents the standardized effect size (the default is f min(#) is the

max(#) is the

byvalue(#)

represents a way to control how much output is generated by the command

For our example above, we computed the power of a one-way ANOVA

design with three groups (k

effect sizes f

The anova_pwr command to compute the power estimate for the small

anova_pwr, ngp(3) f(.1) min(100) max(100)

Since we specified the minimum and maximum group sizes to be the same (100

If you rerun this command, but change the value for f to reflect the

medium and strong effect sizes, the power estimates reported above will also

be reproduced.

It is not possible to estimate directly the sample size required for a designated

level of statistical power in an ANOVA using Stata. The anova_pwr command

can be used to represent a range of group sizes through the use of the min(#)

and max(#) options that will estimate the power associated with a given effect

size. If our goal is to achieve a power of 0.80, then we might start by estimating

the power for a wide range of sample sizes and then narrowing down the range

on a second run. For example, if we are interested in determining the sample size

required to detect a small effect (f

enter the following command using 100 cases as the minimum, since we already

anova_pwr, ngp(3) f(.1) min(100) max(500) byvalue(10)

753C O M P U T E R E X E R C I S E S

754 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

The output from this run is

As we move through the values in the output, we see that a power level of 0.8

would be required to achieve a minimum power of 0.80. The table of results also

illustrates how sample size estimates may vary across programs to compute sta-

0.80, which some programs would round to 0.80. At the same time, it is

that threshold.

There is one user-written procedure that we are aware of for computing power

estimates of correlation coefficients in Stata. The command is sampsi_rho,

which bases power calculations on converting the correlation coefficient with

the Fisher z formula and then using the normal distribution (instead of a

t

ssc install sampsi_rho

The basic structure of the sampsi_rho command is

sampsi_rho , null(#) alt(#) n(#) power(#) solve() alpha(#)

onesided

755C O M P U T E R E X E R C I S E S

Correlation

756 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

where null(#) specifies the value of the correlation for the null hypothesis

alt(#) specifies the alternative hypothesis value of the correlation

n(#) power(#) indi-

solve() notes whether to solve

alpha(#) specifies the alpha

To replicate the values above in our analysis of power estimates for correla-

tion coefficients for a sample size of 100, we would enter the following com-

sampsi_rho, solve(power) n(100) alt(0.1) onesided

The estimated power is 0.260, very nearly the same as the estimate produced

using the correlation coefficient and the t-distribution. If you were interested in

reproducing the power estimates for the medium and strong effects, you would

just need to change the value of alt(#) to alt(0.3) and alt(0.5), respectively.

In a similar way, we can estimate the sample size needed to achieve a desig-

nated level of statistical power for a hypothesized effect size by making just a

few changes to the sampsi_rho command. For example, if we wanted to esti-

mate the sample size needed to detect a medium correlation (i.e., r

power level of 0.80, we would omit the sample size and solve options but insert

power(0.8)

sampsi_rho, alt(0.3) power(0.8) onesided

-

tion of a case, we would typically round up to 68 in this case. The rationale, as

we noted above, in rounding up is to ensure that a power level of no less than

entirely due to the use of the Fisher-transformed value of the correlation and use

of the normal distribution and is to be expected.

OLS Regression

Similar to computing power with ANOVA in Stata, it is necessary to rely on the

user-written command powerreg

net install powerreg, from(http://www.ats.ucla.edu/stat/

stata/ado/analysis)

The basic structure to the powerreg command is

powerreg, r2f(value) r2r(value) nvar(#) ntest(#) alpha(value)

where r2f(value) is the hypothesized value of R2 expected, r2r(value) is the

R2 nvar(#) refers to the total number of inde-

pendent variables included in the regression model, and ntest(#) refers to the

number of independent variables being tested. Alpha is assumed to be 0.05, and

nvar and ntest are both set at a default of 1.

To reproduce the results reported above for power in OLS regression for a

weak effect (i.e., R2

powerreg, r2f(.02) r2r(0) n(100) nvar(8) ntest(8)

Note that the value for r2r is entered as 0—this is the expected value of R2

without any of the independent variables included in the analysis. The power

above. Results for the moderate (R2 R2

are obtained by simply altering the value of r2f in the powerreg command.

Note that the power estimates reported by Stata vary slightly from those reported

To compute the sample size needed to achieve a designated level of statistical

power, we would omit the n(#) option but insert an option for power(#)

powerreg, r2f(.02) r2r(0) power(0.8) nvar(8) ntest(8)

We find that the estimated sample size needed to detect a weak effect (R2

the calculation of the standardized effect (f2

effect size20—the values for the medium and large effect sizes are nearly identical

and differ by only 1 case.21

Problems

1. Compute the estimates of statistical power for each of the four

scenarios in Exercise 21.1. Which scenario has the highest level of

statistical power? Explain why.

2. Compute the estimates of statistical power for each of the four scenarios

Explain why.

in Exercise 21.6. Was this study designed to have a high level of statistical

power to identify small and medium effects? Explain why.

4. Compute the estimates of statistical power for each of the following

one-way ANOVA studies. (For all scenarios, assume that the researcher is

Which scenario would have the highest level of statistical power?

Explain why.

20 If the value of r2r(#) in the command is changed to r2r(0.0196), the resulting estimate

21 The reason for this difference is that the powerreg command computes sample size esti-

sample size that ensures that the power level is at least 0.80 and so the estimates reported above

are greater by 1.

757C O M P U T E R E X E R C I S E S

758 C H A P T E R T W E N T Y T H R E E : S P E C I A L T O P I C S

5. In attempting to design a correlation study looking at academic

performance and delinquency, a researcher expects a small-to-moderate

correlation among a population of adolescents he or she will sample from.

a. If he or she computes estimates of statistical power assuming a

two-tail test, what size sample would he or she need to detect a small

correlation? Medium correlation?

b. Do you think he or she could justify a one-tail test of the correlation?

If a one-tail test was used, how does the estimated sample size change

for both the small and medium correlations?

6. A research team is preparing to launch a statewide survey to gauge public

sentiment about the incarceration of juvenile offenders, focusing primarily

on support for more lenient punishments. Consistent with much public

opinion research, expectations are that a combination of ten independent

variables is likely to explain about 15% of the variation in views about

juvenile punishment.

a. What size sample would the researchers need to have to achieve a

power of 0.80? 0.90?

b. Of particular interest to the researchers is the effect of three different

measures of experience with the justice system, but their expectation is

that the overall effect of these three measures will be small. What size

sample would the researchers need to achieve a power of 0.80? 0.90?

c. What size sample should the researchers try to obtain? Explain why.

Appendix 1 Factorials

0! � 1

1! � 1

2! � 2

3! � 6

4! � 24

5! � 120

6! � 720

7! � 5,040

8! � 40,320

9! � 362,880

10! � 3,628,800

11! � 39,916,800

12! � 479,001,600

13! � 6,227,020,800

14! � 87,178,291,200

15! � 1,307,674,368,000

16! � 20,922,789,888,000

17! � 355,687,428,096,000

18! � 6,402,373,705,728,000

19! � 121,645,100,408,832,000

20! � 2,432,902,008,176,640,000

21! � 51,090,942,171,709,440,000

22! � 1,124,000,727,777,607,680,000

23! � 25,852,016,738,884,976,640,000

24! � 620,448,401,733,239,439,360,000

25! � 15,511,210,043,330,985,984,000,000

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5,

© Springer Science+Business Media New York 2014

759

Appendix 2 Critical Values of �2 Distribution

df 0.20 0.10 0.05 0.02 0.01 0.001

1 1.642 2.706 3.841 5.412 6.635 10.827 2 3.219 4.605 5.991 7.824 9.210 13.815 3 4.642 6.251 7.815 9.837 11.341 16.268 4 5.989 7.779 9.488 11.668 13.277 18.465 5 7.289 9.236 11.070 13.388 15.086 20.517 6 8.558 10.645 12.592 15.033 16.812 22.457 7 9.803 12.017 14.067 16.622 18.475 24.322 8 11.030 13.362 15.507 18.168 20.090 26.125 9 12.242 14.684 16.919 19.679 21.666 27.877

10 13.442 15.987 18.307 21.161 23.209 29.588 11 14.631 17.275 19.675 22.618 24.725 31.264 12 15.812 18.549 21.026 24.054 26.217 32.909 13 16.985 19.812 22.362 25.472 27.688 34.528 14 18.151 21.064 23.685 26.873 29.141 36.123 15 19.311 22.307 24.996 28.259 30.578 37.697 16 20.465 23.542 26.296 29.633 32.000 39.252 17 21.615 24.769 27.587 30.995 33.409 40.790 18 22.760 25.989 28.869 32.346 34.805 42.312 19 23.900 27.204 30.144 33.687 36.191 43.820 20 25.038 28.412 31.410 35.020 37.566 45.315 21 26.171 29.615 32.671 36.343 38.932 46.797 22 27.301 30.813 33.924 37.659 40.289 48.268 23 28.429 32.007 35.172 38.968 41.638 49.728 24 29.553 33.196 36.415 40.270 42.980 51.179 25 30.675 34.382 37.652 41.566 44.314 52.620 26 31.795 35.563 38.885 42.856 45.642 54.052 27 32.912 36.741 40.113 44.140 46.963 55.476 28 34.027 37.916 41.337 45.419 48.278 56.893 29 35.139 39.087 42.557 46.693 49.588 58.302 30 36.250 40.256 43.773 47.962 50.892 59.703

Source: From Table IV of R. A. Fisher and F. Yates, Statistical Tables for Biological, Agricultural and Medical Research (London: Longman Group Ltd., 1974). (Previously published by Oliver & Boyd, Edinburgh.) Reprinted by permission of Pearson Education Ltd.

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5,

© Springer Science+Business Media New York 2014

760

Appendix 3 The entries in this table are the proportion of the cases in a standard

normal distribution that lie between 0 and z.

SECOND DECIMAL PLACE IN z

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879

0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389

1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319

1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767

2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936

2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986

3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990 3.1 0.4990 0.4991 0.4991 0.4991 0.4992 0.4992 0.4992 0.4992 0.4993 0.4993 3.2 0.4993 0.4993 0.4994 0.4994 0.4994 0.4994 0.4994 0.4995 0.4995 0.4995 3.3 0.4995 0.4995 0.4995 0.4996 0.4996 0.4996 0.4996 0.4996 0.4996 0.4997 3.4 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4998

3.5 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 3.6 0.4998 0.4998 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 3.7 0.4999 4.0 0.49997 4.5 0.499997 5.0 0.4999997

Source: R. Johnson, Elementary Statistics (Belmont, CA: Duxbury Press, 1996).

Areas of the Standard Normal Distribution

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5,

© Springer Science+Business Media New York 2014

761

Appendix 4 ONE-TAILED VALUE

Degrees 0.25 0.10 0.05 0.025 0.01 0.005

of TWO-TAILED VALUE

Freedom 0.50 0.20 0.10 0.05 0.02 0.01

1 1.000 3.078 6.314 12.706 31.821 63.657 2 0.816 1.886 2.920 4.303 6.965 9.925 3 0.765 1.638 2.353 3.182 4.541 5.841 4 0.741 1.533 2.132 2.776 3.747 4.604 5 0.727 1.476 2.015 2.571 3.365 4.032 6 0.718 1.440 1.943 2.447 3.143 3.707 7 0.711 1.415 1.895 2.365 2.998 3.499 8 0.706 1.397 1.860 2.306 2.896 3.355 9 0.703 1.383 1.833 2.262 2.821 3.250

10 0.700 1.372 1.812 2.228 2.764 3.169 11 0.697 1.363 1.796 2.201 2.718 3.106 12 0.695 1.356 1.782 2.179 2.681 3.055 13 0.694 1.350 1.771 2.160 2.650 3.012 14 0.692 1.345 1.761 2.145 2.626 2.977 15 0.691 1.341 1.753 2.131 2.602 2.947 16 0.690 1.337 1.746 2.120 2.583 2.921 17 0.689 1.333 1.740 2.110 2.567 2.898 18 0.688 1.330 1.734 2.101 2.552 2.878 19 0.688 1.328 1.729 2.093 2.539 2.861 20 0.687 1.325 1.725 2.086 2.528 2.845 21 0.686 1.323 1.721 2.080 2.518 2.831 22 0.686 1.321 1.717 2.074 2.508 2.819 23 0.685 1.319 1.714 2.069 2.500 2.807 24 0.685 1.318 1.711 2.064 2.492 2.797 25 0.684 1.316 1.708 2.060 2.485 2.787 26 0.684 1.315 1.706 2.056 2.479 2.779 27 0.684 1.314 1.703 2.052 2.473 2.771 28 0.683 1.313 1.701 2.048 2.467 2.763 29 0.683 1.311 1.699 2.045 2.462 2.756 30 0.683 1.310 1.697 2.042 2.457 2.750 31 0.682 1.309 1.696 2.040 2.453 2.744 32 0.682 1.309 1.694 2.037 2.449 2.739 33 0.682 1.308 1.692 2.035 2.445 2.733 34 0.682 1.307 1.691 2.032 2.441 2.728 35 0.682 1.306 1.690 2.030 2.438 2.724 40 0.681 1.303 1.684 2.021 2.423 2.704 45 0.680 1.301 1.680 2.014 2.412 2.690 50 0.680 1.299 1.676 2.008 2.403 2.678 55 0.679 1.297 1.673 2.004 2.396 2.669 60 0.679 1.296 1.671 2.000 2.390 2.660 70 0.678 1.294 1.667 1.994 2.381 2.648 80 0.678 1.293 1.665 1.989 2.374 2.638 90 0.678 1.291 1.662 1.986 2.368 2.631

100 0.677 1.290 1.661 1.982 2.364 2.625 120 0.677 1.289 1.658 1.980 2.358 2.617

�500 0.674 1.282 1.645 1.960 2.326 2.576

Source: “Table D, The t Table” adapted from SCIENTIFIC TABLES, published by Ciba-Geigy, in WAYS AND MEANS OF STATISTICS by Leonard Tashman and Kathleen Lamborn, Copyright � 1979 by Harcourt Brace & Company, reprinted by permission of Harcourt Brace & Company.

Critical Values of Student’s t Distribution

One-tailed value

Two-tailed value

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5,

© Springer Science+Business Media New York 2014

762

Appendix 5 Critical Values of the F-Statistic

(� � 0.05)

NUMERATOR DEGREES OF FREEDOM

df1

df2 1 2 3 4 5 6 8 12 24 �

1 161.4 199.5 215.7 224.6 230.2 234.0 238.9 243.9 249.0 254.3

2 18.51 19.00 19.16 19.25 19.30 19.33 19.37 19.41 19.45 19.50

3 10.13 9.55 9.28 9.12 9.01 8.94 8.84 8.74 8.64 8.53

4 7.71 6.94 6.59 6.39 6.26 6.16 6.04 5.91 5.77 5.63

5 6.61 5.79 5.41 5.19 5.05 4.95 4.82 4.68 4.53 4.36

6 5.99 5.14 4.76 4.53 4.39 4.28 4.15 4.00 3.84 3.67

7 5.59 4.74 4.35 4.12 3.97 3.87 3.73 3.57 3.41 3.23

8 5.32 4.46 4.07 3.84 3.69 3.58 3.44 3.28 3.12 2.93

9 5.12 4.26 3.86 3.63 3.48 3.37 3.23 3.07 2.90 2.71

10 4.96 4.10 3.71 3.48 3.33 3.22 3.07 2.91 2.74 2.54

11 4.84 3.98 3.59 3.36 3.20 3.09 2.95 2.79 2.61 2.40

12 4.75 3.88 3.49 3.26 3.11 3.00 2.85 2.69 2.50 2.30

13 4.67 3.80 3.41 3.18 3.02 2.92 2.77 2.60 2.42 2.21

14 4.60 3.74 3.34 3.11 2.96 2.85 2.70 2.53 2.35 2.13

15 4.54 3.68 3.29 3.06 2.90 2.79 2.64 2.48 2.29 2.07

16 4.49 3.63 3.24 3.01 2.85 2.74 2.59 2.42 2.24 2.01

17 4.45 3.59 3.20 2.96 2.81 2.70 2.55 2.38 2.19 1.96

18 4.41 3.55 3.16 2.93 2.77 2.66 2.51 2.34 2.15 1.92

19 4.38 3.52 3.13 2.90 2.74 2.63 2.48 2.31 2.11 1.88

20 4.35 3.49 3.10 2.87 2.71 2.60 2.45 2.28 2.08 1.84

21 4.32 3.47 3.07 2.84 2.68 2.57 2.42 2.25 2.05 1.81

22 4.30 3.44 3.05 2.82 2.66 2.55 2.40 2.23 2.03 1.78

23 4.28 3.42 3.03 2.80 2.64 2.53 2.38 2.20 2.00 1.76

24 4.26 3.40 3.01 2.78 2.62 2.51 2.36 2.18 1.98 1.73

25 4.24 3.38 2.99 2.76 2.60 2.49 2.34 2.16 1.96 1.71

26 4.22 3.37 2.98 2.74 2.59 2.47 2.32 2.15 1.95 1.69

27 4.21 3.35 2.96 2.73 2.57 2.46 2.30 2.13 1.93 1.67

28 4.20 3.34 2.95 2.71 2.56 2.44 2.29 2.12 1.91 1.65

29 4.18 3.33 2.93 2.70 2.54 2.43 2.28 2.10 1.90 1.64

30 4.17 3.32 2.92 2.69 2.53 2.42 2.27 2.09 1.89 1.62

40 4.08 3.23 2.84 2.61 2.45 2.34 2.18 2.00 1.79 1.51

60 4.00 3.15 2.76 2.52 2.37 2.25 2.10 1.92 1.70 1.39

120 3.92 3.07 2.68 2.45 2.29 2.17 2.02 1.83 1.61 1.25

�500 3.84 2.99 2.60 2.37 2.21 2.09 1.94 1.75 1.52 1.00

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(� � 0.01)

NUMERATOR DEGREES OF FREEDOM

df1

df2 1 2 3 4 5 6 8 12 24 �

1 4052 4999 5403 5625 5764 5859 5981 6106 6234 6366

2 98.49 99.01 99.17 99.25 99.30 99.33 99.36 99.42 99.46 99.50

3 34.12 30.81 29.46 28.71 28.24 27.91 27.49 27.05 26.60 26.12

4 21.20 18.00 16.69 15.98 15.52 15.21 14.80 14.37 13.93 13.46

5 16.26 13.27 12.06 11.39 10.97 10.67 10.27 9.89 9.47 9.02

6 13.74 10.92 9.78 9.15 8.75 8.47 8.10 7.72 7.31 6.88

7 12.25 9.55 8.45 7.85 7.46 7.19 6.84 6.47 6.07 5.65

8 11.26 8.65 7.59 7.01 6.63 6.37 6.03 5.67 5.28 4.86

9 10.56 8.02 6.99 6.42 6.06 5.80 5.47 5.11 4.73 4.31

10 10.04 7.56 6.55 5.99 5.64 5.39 5.06 4.71 4.33 3.91

11 9.65 7.20 6.22 5.67 5.32 5.07 4.74 4.40 4.02 3.60

12 9.33 6.93 5.95 5.41 5.06 4.82 4.50 4.16 3.78 3.36

13 9.07 6.70 5.74 5.20 4.86 4.62 4.30 3.96 3.59 3.16

14 8.86 6.51 5.56 5.03 4.69 4.46 4.14 3.80 3.43 3.00

15 8.68 6.36 5.42 4.89 4.56 4.32 4.00 3.67 3.29 2.87

16 8.53 6.23 5.29 4.77 4.44 4.20 3.89 3.55 3.18 2.75

17 8.40 6.11 5.18 4.67 4.34 4.10 3.79 3.45 3.08 2.65

18 8.28 6.01 5.09 4.58 4.25 4.01 3.71 3.37 3.00 2.57

19 8.18 5.93 5.01 4.50 4.17 3.94 3.63 3.30 2.92 2.49

20 8.10 5.85 4.94 4.43 4.10 3.87 3.56 3.23 2.86 2.42

21 8.02 5.78 4.87 4.37 4.04 3.81 3.51 3.17 2.80 2.36

22 7.94 5.72 4.82 4.31 3.99 3.76 3.45 3.12 2.75 2.31

23 7.88 5.66 4.76 4.26 3.94 3.71 3.41 3.07 2.70 2.26

24 7.82 5.61 4.72 4.22 3.90 3.67 3.36 3.03 2.66 2.21

25 7.77 5.57 4.68 4.18 3.86 3.63 3.32 2.99 2.62 2.17

26 7.72 5.53 4.64 4.14 3.82 3.59 3.29 2.96 2.58 2.13

27 7.68 5.49 4.60 4.11 3.78 3.56 3.26 2.93 2.55 2.10

28 7.64 5.45 4.57 4.07 3.75 3.53 3.23 2.90 2.52 2.06

29 7.60 5.42 4.54 4.04 3.73 3.50 3.20 2.87 2.49 2.03

30 7.56 5.39 4.51 4.02 3.70 3.47 3.17 2.84 2.47 2.01

40 7.31 5.18 4.31 3.83 3.51 3.29 2.99 2.66 2.29 1.80

60 7.08 4.98 4.13 3.65 3.34 3.12 2.82 2.50 2.12 1.60

120 6.85 4.79 3.95 3.48 3.17 2.96 2.66 2.34 1.95 1.38

�500 6.64 4.60 3.78 3.32 3.02 2.80 2.51 2.18 1.79 1.00

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764 A P P E N D I C E S

(� � 0.001)

NUMERATOR DEGREES OF FREEDOM

df1

df2 1 2 3 4 5 6 8 12 24 �

1 405284 500000 540379 562500 576405 585937 598144 610667 623497 636619

2 998.5 999.0 999.2 999.2 999.3 999.3 999.4 999.4 999.5 999.5

3 167.5 148.5 141.1 137.1 134.6 132.8 130.6 128.3 125.9 123.5

4 74.14 61.25 56.18 53.44 51.71 50.53 49.00 47.41 45.77 44.05

5 47.04 36.61 33.20 31.09 29.75 28.84 27.64 26.42 25.14 23.78

6 35.51 27.00 23.70 21.90 20.81 20.03 19.03 17.99 16.89 15.75

7 29.22 21.69 18.77 17.19 16.21 15.52 14.63 13.71 12.73 11.69

8 25.42 18.49 15.83 14.39 13.49 12.86 12.04 11.19 10.30 9.34

9 22.86 16.39 13.90 12.56 11.71 11.13 10.37 9.57 8.72 7.81

10 21.04 14.91 12.55 11.28 10.48 9.92 9.20 8.45 7.64 6.76

11 19.69 13.81 11.56 10.35 9.58 9.05 8.35 7.63 6.85 6.00

12 18.64 12.97 10.80 9.63 8.89 8.38 7.71 7.00 6.25 5.42

13 17.81 12.31 10.21 9.07 8.35 7.86 7.21 6.52 5.78 4.97

14 17.14 11.78 9.73 8.62 7.92 7.43 6.80 6.13 5.41 4.60

15 16.59 11.34 9.34 8.25 7.57 7.09 6.47 5.81 5.10 4.31

16 16.12 10.97 9.00 7.94 7.27 6.81 6.19 5.55 4.85 4.06

17 15.72 10.66 8.73 7.68 7.02 6.56 5.96 5.32 4.63 3.85

18 15.38 10.39 8.49 7.46 6.81 6.35 5.76 5.13 4.45 3.67

19 15.08 10.16 8.28 7.26 6.61 6.18 5.59 4.97 4.29 3.52

20 14.82 9.95 8.10 7.10 6.46 6.02 5.44 4.82 4.15 3.38

21 14.59 9.77 7.94 6.95 6.32 5.88 5.31 4.70 4.03 3.26

22 14.38 9.61 7.80 6.81 6.19 5.76 5.19 4.58 3.92 3.15

23 14.19 9.47 7.67 6.69 6.08 5.65 5.09 4.48 3.82 3.05

24 14.03 9.34 7.55 6.59 5.98 5.55 4.99 4.39 3.74 2.97

25 13.88 9.22 7.45 6.49 5.88 5.46 4.91 4.31 3.66 2.89

26 13.74 9.12 7.36 6.41 5.80 5.38 4.83 4.24 3.59 2.82

27 13.61 9.02 7.27 6.33 5.73 5.31 4.76 4.17 3.52 2.75

28 13.50 8.93 7.19 6.25 5.66 5.24 4.69 4.11 3.46 2.70

29 13.39 8.85 7.12 6.19 5.59 5.18 4.64 4.05 3.41 2.64

30 13.29 8.77 7.05 6.12 5.53 5.12 4.58 4.00 3.36 2.59

40 12.61 8.25 6.60 5.70 5.13 4.73 4.21 3.64 3.01 2.23

60 11.97 7.76 6.17 5.31 4.76 4.37 3.87 3.31 2.69 1.90

120 11.38 7.31 5.79 4.95 4.42 4.04 3.55 3.02 2.40 1.56

�500 10.83 6.91 5.42 4.62 4.10 3.74 3.27 2.74 2.13 1.00

Source: From Table IV of R. A. Fisher and F. Yates, Statistical Tables for Biological, Agricultural and Medical Research (London: Longman Group Ltd., 1974). (Previously published by Oliver & Boyd, Edinburgh.) Reprinted by permission of Pearson Education Ltd.

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A P P E N D I C E S 765

Appendix 6 Critical Value for P (Pcrit), Tukey’s HSD Test

LEVEL OF SIGNIFICANCE (� � 0.05)

k � THE NUMBER OF MEANS OR

NUMBER OF STEPS BETWEEN ORDERED MEANS

dfw 2 3 4 5 6 7 8 9 10 12 15 20

1 17.97 26.98 32.82 37.08 40.41 43.12 45.40 47.36 49.07 51.96 55.36 59.56

2 6.08 8.33 9.80 10.88 11.74 12.44 13.03 13.54 13.99 14.75 15.65 16.77

3 4.50 5.91 6.82 7.50 8.04 8.48 8.85 9.18 9.46 9.95 10.52 11.24

4 3.93 5.04 5.76 6.29 6.71 7.05 7.35 7.60 7.83 8.21 8.66 9.23

5 3.64 4.60 5.22 5.67 6.03 6.33 6.58 6.80 6.99 7.32 7.72 8.21

6 3.46 4.34 4.90 5.30 5.63 5.90 6.12 6.32 6.49 6.79 7.14 7.59

7 3.34 4.16 4.68 5.06 5.36 5.61 5.82 6.00 6.16 6.43 6.76 7.17

8 3.26 4.04 4.53 4.89 5.17 5.40 5.60 5.77 5.92 6.18 6.48 6.87

9 3.20 3.95 4.41 4.76 5.02 5.24 5.43 5.59 5.74 5.98 6.28 6.64

10 3.15 3.88 4.33 4.65 4.91 5.12 5.30 5.46 5.60 5.83 6.11 6.47

11 3.11 3.82 4.26 4.57 4.82 5.03 5.20 5.35 5.49 5.71 5.98 6.33

12 3.08 3.77 4.20 4.51 4.75 4.95 5.12 5.27 5.39 5.61 5.88 6.21

13 3.06 3.73 4.15 4.45 4.69 4.88 5.05 5.19 5.32 5.53 5.79 6.11

14 3.03 3.70 4.11 4.41 4.64 4.83 4.99 5.13 5.25 5.46 5.71 6.03

15 3.01 3.67 4.08 4.37 4.59 4.78 4.94 5.08 5.20 5.40 5.65 5.96

16 3.00 3.65 4.05 4.33 4.56 4.74 4.90 5.03 5.15 5.35 5.59 5.90

17 2.98 3.63 4.02 4.30 4.52 4.70 4.86 4.99 5.11 5.31 5.54 5.84

18 2.97 3.61 4.00 4.28 4.49 4.67 4.82 4.96 5.07 5.27 5.50 5.79

19 2.96 3.59 3.98 4.25 4.47 4.65 4.79 4.92 5.04 5.23 5.46 5.75

20 2.95 3.58 3.96 4.23 4.45 4.62 4.77 4.90 5.01 5.20 5.43 5.71

24 2.92 3.53 3.90 4.17 4.37 4.54 4.68 4.81 4.92 5.10 5.32 5.59

30 2.89 3.49 3.85 4.10 4.30 4.46 4.60 4.72 4.82 5.00 5.21 5.47

40 2.86 3.44 3.79 4.04 4.23 4.39 4.52 4.63 4.73 4.90 5.11 5.36

60 2.83 3.40 3.74 3.98 4.16 4.31 4.44 4.55 4.65 4.81 5.00 5.24

120 2.80 3.36 3.68 3.92 4.10 4.24 4.36 4.47 4.56 4.71 4.90 5.13

� 2.77 3.31 3.63 3.86 4.03 4.17 4.29 4.39 4.47 4.62 4.80 5.01

Source: From Comprehending Behavioral Statistics, by R. T. Hurlburt, Copyright � 1994, Brooks/Cole Publishing Company, Pacific Grove, CA 93950, a division of International Thomson Publishing Inc. By permission of the publisher. Adapted from Biometrika Tables for Statisticians, vol. 1, 3rd ed., E. S. Pearson and H. O. Hartley (eds.). Copyright � 1966, Cambridge University Press for Biometrika Trust. By permission of the Biometrika Trust.

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5,

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Appendix 7 Critical Values for Spearman’s Rank-Order Correlation Coefficient

LEVEL OF SIGNIFICANCE (�) FOR ONE-TAILED TEST

0.05 0.025 0.01 0.005

LEVEL OF SIGNIFICANCE (�) FOR TWO-TAILED TEST

n 0.10 0.05 0.02 0.01

5 0.900 — — — 6 0.829 0.886 0.943 — 7 0.714 0.786 0.893 0.929 8 0.643 0.738 0.833 0.881 9 0.600 0.700 0.783 0.833

10 0.564 0.648 0.745 0.794 11 0.536 0.618 0.709 0.818 12 0.497 0.591 0.703 0.780 13 0.475 0.566 0.673 0.745 14 0.457 0.545 0.646 0.716 15 0.441 0.525 0.623 0.689 16 0.425 0.507 0.601 0.666 17 0.412 0.490 0.582 0.645 18 0.399 0.476 0.564 0.625 19 0.388 0.462 0.549 0.608 20 0.377 0.450 0.534 0.591 21 0.368 0.438 0.521 0.576 22 0.359 0.428 0.508 0.562 23 0.351 0.418 0.496 0.549 24 0.343 0.409 0.485 0.537 25 0.336 0.400 0.475 0.526 26 0.329 0.392 0.465 0.515 27 0.323 0.385 0.456 0.505 28 0.317 0.377 0.448 0.496 29 0.311 0.370 0.440 0.487 30 0.305 0.364 0.432 0.478

Source: From Comprehending Behavioral Statistics, by R. T. Hurlburt, Copyright � 1994, Brooks/Cole Publishing Company, Pacific Grove, CA 93950, a division of International Thomson Publishing Inc. By permission of the publisher.

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5,

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767

Appendix 8 Fisher r-to-Z* Transformation

The entries in this table are the Z-values for correlations ranging from 0.000 to 0.999.

THIRD DECIMAL PLACE IN r

r 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009

0.00 0.0000 0.0010 0.0020 0.0030 0.0040 0.0050 0.0060 0.0070 0.0080 0.0090 0.01 0.0100 0.0110 0.0120 0.0130 0.0140 0.0150 0.0160 0.0170 0.0180 0.0190 0.02 0.0200 0.0210 0.0220 0.0230 0.0240 0.0250 0.0260 0.0270 0.0280 0.0290 0.03 0.0300 0.0310 0.0320 0.0330 0.0340 0.0350 0.0360 0.0370 0.0380 0.0390 0.04 0.0400 0.0410 0.0420 0.0430 0.0440 0.0450 0.0460 0.0470 0.0480 0.0490 0.05 0.0500 0.0510 0.0520 0.0530 0.0541 0.0551 0.0561 0.0571 0.0581 0.0591 0.06 0.0601 0.0611 0.0621 0.0631 0.0641 0.0651 0.0661 0.0671 0.0681 0.0691 0.07 0.0701 0.0711 0.0721 0.0731 0.0741 0.0751 0.0761 0.0772 0.0782 0.0792 0.08 0.0802 0.0812 0.0822 0.0832 0.0842 0.0852 0.0862 0.0872 0.0882 0.0892 0.09 0.0902 0.0913 0.0923 0.0933 0.0943 0.0953 0.0963 0.0973 0.0983 0.0993 0.10 0.1003 0.1013 0.1024 0.1034 0.1044 0.1054 0.1064 0.1074 0.1084 0.1094 0.11 0.1104 0.1115 0.1125 0.1135 0.1145 0.1155 0.1165 0.1175 0.1186 0.1196 0.12 0.1206 0.1216 0.1226 0.1236 0.1246 0.1257 0.1267 0.1277 0.1287 0.1297 0.13 0.1307 0.1318 0.1328 0.1338 0.1348 0.1358 0.1368 0.1379 0.1389 0.1399 0.14 0.1409 0.1419 0.1430 0.1440 0.1450 0.1460 0.1471 0.1481 0.1491 0.1501 0.15 0.1511 0.1522 0.1532 0.1542 0.1552 0.1563 0.1573 0.1583 0.1593 0.1604 0.16 0.1614 0.1624 0.1634 0.1645 0.1655 0.1665 0.1676 0.1686 0.1696 0.1706 0.17 0.1717 0.1727 0.1737 0.1748 0.1758 0.1768 0.1779 0.1789 0.1799 0.1809 0.18 0.1820 0.1830 0.1841 0.1851 0.1861 0.1872 0.1882 0.1892 0.1903 0.1913 0.19 0.1923 0.1934 0.1944 0.1955 0.1965 0.1975 0.1986 0.1996 0.2007 0.2017 0.20 0.2027 0.2038 0.2048 0.2059 0.2069 0.2079 0.2090 0.2100 0.2111 0.2121 0.21 0.2132 0.2142 0.2153 0.2163 0.2174 0.2184 0.2195 0.2205 0.2216 0.2226 0.22 0.2237 0.2247 0.2258 0.2268 0.2279 0.2289 0.2300 0.2310 0.2321 0.2331 0.23 0.2342 0.2352 0.2363 0.2374 0.2384 0.2395 0.2405 0.2416 0.2427 0.2437 0.24 0.2448 0.2458 0.2469 0.2480 0.2490 0.2501 0.2512 0.2522 0.2533 0.2543 0.25 0.2554 0.2565 0.2575 0.2586 0.2597 0.2608 0.2618 0.2629 0.2640 0.2650 0.26 0.2661 0.2672 0.2683 0.2693 0.2704 0.2715 0.2726 0.2736 0.2747 0.2758 0.27 0.2769 0.2779 0.2790 0.2801 0.2812 0.2823 0.2833 0.2844 0.2855 0.2866 0.28 0.2877 0.2888 0.2899 0.2909 0.2920 0.2931 0.2942 0.2953 0.2964 0.2975 0.29 0.2986 0.2997 0.3008 0.3018 0.3029 0.3040 0.3051 0.3062 0.3073 0.3084 0.30 0.3095 0.3106 0.3117 0.3128 0.3139 0.3150 0.3161 0.3172 0.3183 0.3194 0.31 0.3205 0.3217 0.3228 0.3239 0.3250 0.3261 0.3272 0.3283 0.3294 0.3305 0.32 0.3316 0.3328 0.3339 0.3350 0.3361 0.3372 0.3383 0.3395 0.3406 0.3417 0.33 0.3428 0.3440 0.3451 0.3462 0.3473 0.3484 0.3496 0.3507 0.3518 0.3530 0.34 0.3541 0.3552 0.3564 0.3575 0.3586 0.3598 0.3609 0.3620 0.3632 0.3643 0.35 0.3654 0.3666 0.3677 0.3689 0.3700 0.3712 0.3723 0.3734 0.3746 0.3757 0.36 0.3769 0.3780 0.3792 0.3803 0.3815 0.3826 0.3838 0.3850 0.3861 0.3873 0.37 0.3884 0.3896 0.3907 0.3919 0.3931 0.3942 0.3954 0.3966 0.3977 0.3989 0.38 0.4001 0.4012 0.4024 0.4036 0.4047 0.4059 0.4071 0.4083 0.4094 0.4106 0.39 0.4118 0.4130 0.4142 0.4153 0.4165 0.4177 0.4189 0.4201 0.4213 0.4225 0.40 0.4236 0.4248 0.4260 0.4272 0.4284 0.4296 0.4308 0.4320 0.4332 0.4344 0.41 0.4356 0.4368 0.4380 0.4392 0.4404 0.4416 0.4428 0.4441 0.4453 0.4465 0.42 0.4477 0.4489 0.4501 0.4513 0.4526 0.4538 0.4550 0.4562 0.4574 0.4587 0.43 0.4599 0.4611 0.4624 0.4636 0.4648 0.4660 0.4673 0.4685 0.4698 0.4710 0.44 0.4722 0.4735 0.4747 0.4760 0.4772 0.4784 0.4797 0.4809 0.4822 0.4834 0.45 0.4847 0.4860 0.4872 0.4885 0.4897 0.4910 0.4922 0.4935 0.4948 0.4960 0.46 0.4973 0.4986 0.4999 0.5011 0.5024 0.5037 0.5049 0.5062 0.5075 0.5088 0.47 0.5101 0.5114 0.5126 0.5139 0.5152 0.5165 0.5178 0.5191 0.5204 0.5217 0.48 0.5230 0.5243 0.5256 0.5269 0.5282 0.5295 0.5308 0.5321 0.5334 0.5347

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THIRD DECIMAL PLACE IN r

r 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009

0.49 0.5361 0.5374 0.5387 0.5400 0.5413 0.5427 0.5440 0.5453 0.5466 0.5480 0.50 0.5493 0.5506 0.5520 0.5533 0.5547 0.5560 0.5573 0.5587 0.5600 0.5614 0.51 0.5627 0.5641 0.5654 0.5668 0.5682 0.5695 0.5709 0.5722 0.5736 0.5750 0.52 0.5763 0.5777 0.5791 0.5805 0.5818 0.5832 0.5846 0.5860 0.5874 0.5888 0.53 0.5901 0.5915 0.5929 0.5943 0.5957 0.5971 0.5985 0.5999 0.6013 0.6027 0.54 0.6042 0.6056 0.6070 0.6084 0.6098 0.6112 0.6127 0.6141 0.6155 0.6169 0.55 0.6184 0.6198 0.6213 0.6227 0.6241 0.6256 0.6270 0.6285 0.6299 0.6314 0.56 0.6328 0.6343 0.6358 0.6372 0.6387 0.6401 0.6416 0.6431 0.6446 0.6460 0.57 0.6475 0.6490 0.6505 0.6520 0.6535 0.6550 0.6565 0.6580 0.6595 0.6610 0.58 0.6625 0.6640 0.6655 0.6670 0.6685 0.6700 0.6716 0.6731 0.6746 0.6761 0.59 0.6777 0.6792 0.6807 0.6823 0.6838 0.6854 0.6869 0.6885 0.6900 0.6916 0.60 0.6931 0.6947 0.6963 0.6978 0.6994 0.7010 0.7026 0.7042 0.7057 0.7073 0.61 0.7089 0.7105 0.7121 0.7137 0.7153 0.7169 0.7185 0.7201 0.7218 0.7234 0.62 0.7250 0.7266 0.7283 0.7299 0.7315 0.7332 0.7348 0.7365 0.7381 0.7398 0.63 0.7414 0.7431 0.7447 0.7464 0.7481 0.7498 0.7514 0.7531 0.7548 0.7565 0.64 0.7582 0.7599 0.7616 0.7633 0.7650 0.7667 0.7684 0.7701 0.7718 0.7736 0.65 0.7753 0.7770 0.7788 0.7805 0.7823 0.7840 0.7858 0.7875 0.7893 0.7910 0.66 0.7928 0.7946 0.7964 0.7981 0.7999 0.8017 0.8035 0.8053 0.8071 0.8089 0.67 0.8107 0.8126 0.8144 0.8162 0.8180 0.8199 0.8217 0.8236 0.8254 0.8273 0.68 0.8291 0.8310 0.8328 0.8347 0.8366 0.8385 0.8404 0.8423 0.8441 0.8460 0.69 0.8480 0.8499 0.8518 0.8537 0.8556 0.8576 0.8595 0.8614 0.8634 0.8653 0.70 0.8673 0.8693 0.8712 0.8732 0.8752 0.8772 0.8792 0.8812 0.8832 0.8852 0.71 0.8872 0.8892 0.8912 0.8933 0.8953 0.8973 0.8994 0.9014 0.9035 0.9056 0.72 0.9076 0.9097 0.9118 0.9139 0.9160 0.9181 0.9202 0.9223 0.9245 0.9266 0.73 0.9287 0.9309 0.9330 0.9352 0.9373 0.9395 0.9417 0.9439 0.9461 0.9483 0.74 0.9505 0.9527 0.9549 0.9571 0.9594 0.9616 0.9639 0.9661 0.9684 0.9707 0.75 0.9730 0.9752 0.9775 0.9798 0.9822 0.9845 0.9868 0.9892 0.9915 0.9939 0.76 0.9962 0.9986 1.0010 1.0034 1.0058 1.0082 1.0106 1.0130 1.0154 1.0179 0.77 1.0203 1.0228 1.0253 1.0277 1.0302 1.0327 1.0352 1.0378 1.0403 1.0428 0.78 1.0454 1.0479 1.0505 1.0531 1.0557 1.0583 1.0609 1.0635 1.0661 1.0688 0.79 1.0714 1.0741 1.0768 1.0795 1.0822 1.0849 1.0876 1.0903 1.0931 1.0958 0.80 1.0986 1.1014 1.1042 1.1070 1.1098 1.1127 1.1155 1.1184 1.1212 1.1241 0.81 1.1270 1.1299 1.1329 1.1358 1.1388 1.1417 1.1447 1.1477 1.1507 1.1538 0.82 1.1568 1.1599 1.1630 1.1660 1.1692 1.1723 1.1754 1.1786 1.1817 1.1849 0.83 1.1881 1.1914 1.1946 1.1979 1.2011 1.2044 1.2077 1.2111 1.2144 1.2178 0.84 1.2212 1.2246 1.2280 1.2315 1.2349 1.2384 1.2419 1.2454 1.2490 1.2526 0.85 1.2562 1.2598 1.2634 1.2671 1.2707 1.2745 1.2782 1.2819 1.2857 1.2895 0.86 1.2933 1.2972 1.3011 1.3050 1.3089 1.3129 1.3169 1.3209 1.3249 1.3290 0.87 1.3331 1.3372 1.3414 1.3456 1.3498 1.3540 1.3583 1.3626 1.3670 1.3714 0.88 1.3758 1.3802 1.3847 1.3892 1.3938 1.3984 1.4030 1.4077 1.4124 1.4171 0.89 1.4219 1.4268 1.4316 1.4365 1.4415 1.4465 1.4516 1.4566 1.4618 1.4670 0.90 1.4722 1.4775 1.4828 1.4882 1.4937 1.4992 1.5047 1.5103 1.5160 1.5217 0.91 1.5275 1.5334 1.5393 1.5453 1.5513 1.5574 1.5636 1.5698 1.5762 1.5826 0.92 1.5890 1.5956 1.6022 1.6089 1.6157 1.6226 1.6296 1.6366 1.6438 1.6510 0.93 1.6584 1.6658 1.6734 1.6811 1.6888 1.6967 1.7047 1.7129 1.7211 1.7295 0.94 1.7380 1.7467 1.7555 1.7645 1.7736 1.7828 1.7923 1.8019 1.8117 1.8216 0.95 1.8318 1.8421 1.8527 1.8635 1.8745 1.8857 1.8972 1.9090 1.9210 1.9333 0.96 1.9459 1.9588 1.9721 1.9857 1.9996 2.0139 2.0287 2.0439 2.0595 2.0756 0.97 2.0923 2.1095 2.1273 2.1457 2.1649 2.1847 2.2054 2.2269 2.2494 2.2729 0.98 2.2976 2.3235 2.3507 2.3796 2.4101 2.4427 2.4774 2.5147 2.5550 2.5987

0.99 2.6467 2.6996 2.7587 2.8257 2.9031 2.9945 3.1063 3.2504 3.4534 3.8002

Note: Values were computed using the equation for the Fisher r-to-Z* transformation.

A P P E N D I C E S 769

Glossary

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5,

© Springer Science+Business Media New York 2014

770

analysis of variance (ANOVA) A parametric test of statistical significance that assesses whether differences in the means of several samples (groups) can lead the researcher to reject the null hypothesis that the means of the popula- tions from which the samples are drawn are the same.

arrangements The different ways events can be ordered and yet result in a single outcome. For example, there is only one arrangement for gaining the outcome of ten heads in ten tosses of a coin. There are, however, ten different arrangements for gaining the outcome of nine heads in ten tosses of a coin.

assumptions Statements that identify the require- ments and characteristics of a test of statistical significance. These are the foundations on which the rest of the test is built.

bar chart A graph in which bars represent frequen- cies, percentages, or proportions for the cate- gories or values of a variable.

between sum of squares (BSS) A measure of the variability between samples (groups). The be- tween sum of squares is calculated by taking the sum of the squared deviation of each sample mean from the grand mean multiplied by the number of cases in that sample.

biased Describing a statistic when its estimate of a population parameter does not center on the true value. In regression analysis, the omission of relevant independent variables will lead to bias in the estimate of Y. When relevant inde- pendent variables are omitted and those mea- sures are related to an independent variable included in regression analysis, then the esti- mate of the effect of that variable will also be biased.

binomial distribution The probability or sampling distribution for an event that has only two possi- ble outcomes.

binomial formula The means of determining the probability that a given set of binomial events will occur in all its possible arrangements.

bivariate regression A technique for predicting change in a dependent variable using one inde- pendent variable.

cells The various entries in a table, each of which is identified by a particular row and column. When we use a table to compare two variables, it is convenient to refer to each combination of categories as a cell.

central limit theorem A theorem that states: “If re- peated independent random samples of size N are drawn from a population, as N grows large, the sampling distribution of sample means will be approximately normal.” The central limit the- orem enables the researcher to make inferences about an unknown population using a normal sampling distribution.

chi-square distribution A sampling distribution that is used to conduct tests of statistical signifi- cance with binary or multicategory nominal vari- ables. The distribution is nonsymmetrical and varies according to degrees of freedom. All the values in the distribution are positive.

chi-square statistic The test statistic resulting from applying the chi-square formula to the observed and expected frequencies for each cell. This statis- tic tells us how much the observed distribution dif- fers from that expected under the null hypothesis.

classification The process whereby data are orga- nized into categories or groups.

coefficient of relative variation A measure of dis- persion calculated by dividing the standard devi- ation by the mean.

concordant pairs of observations Pairs of obser- vations that have consistent rankings on two or- dinal variables.

confidence interval An interval of values around a statistic (usually a point estimate). If we were to draw repeated samples and calculate a 95% con- fidence interval for each, then in only 5 in 100 of these samples would the interval fail to in- clude the true population parameter. In the case of a 99% confidence interval, only 1 in 100 sam- ples would fail to include the true population parameter.

alternation A type of quasi-random assignment in which researchers assign every other case to one particular group.

between effect Effect of an independent variable on the dependent variable using the cluster as the unit of analysis—a regression of cluster-level averages across all the clusters included in the analysis.

cluster-mean centering Computed difference between the observed raw score on some vari- able for each observation in the sample and the cluster mean for that variable.

block randomization A type of randomization whereby cases are first sorted into like groups and then afterwards randomly allocated into treatment and control conditions.

confounding factors Variables associated with treatments and/or outcomes that can bias overall results if not controlled for statistically.

contrast coding A method for recoding a multi- category nominal variable into multiple indicator

G L O S S A R Y 771

Cramer’s V A measure of association for two nomi- nal variables that adjusts the chi-square statistic by the sample size. V is appropriate when at least one of the nominal variables has more than two categories.

critical value The point at which the rejection re- gion begins.

cumulative logistic probability function A trans- formation of the logistic probability function that allows computation of the probability that Y will occur, given a certain combination of character- istics of the independent variables.

curvilinear relationship An association between two variables whose values may be represented as a curved line when plotted on a scatter dia- gram.

data Information used to answer a research ques- tion.

degrees of freedom A mathematical index that places a value on the extent to which a particu- lar operation is free to vary after certain limita-

determines which chi-square probability distrib- ution we use.

distribution-free tests Another name for nonpara- metric tests.

dummy variable A binary nominal-level variable that is included in a multivariate regression model.

effect size (ES) A standardized measure derived by taking the effect size (e.g., the difference be- tween two populations), measured in the raw units of the outcome measure examined, and di- viding it by the pooled or common standard de- viation of the outcome measure.

eta A measure of the degree of correlation between an interval-level and a nominal-level variable.

eta squared The proportion of the total sum of squares that is accounted for by the between sum of squares. Eta squared is sometimes re- ferred to as the percent of variance explained.

expected frequency The number of observations one would predict for a cell if the null hypothe- sis were true.

explained sum of squares (ESS) Another name for the between sum of squares. The explained sum of squares is the part of the total variability that can be explained by visible differences be- tween the groups.

tions have been imposed. Calculating the degrees of freedom for a chi-square test

cross-level interaction An interaction effect included in a multilevel model between a level 1 independent variable and a level 2 cluster characteristic.

convenience sample A sample chosen not at ran- dom, but according to criteria of expedience or accessibility to the researcher.

correctly specified regression model A regres- sion model in which the researcher has taken into account all of the relevant predictors of the dependent variable and has measured them cor- rectly.

correlation A measure of the strength of a relation- ship between two variables.

covariation A measure of the extent to which two variables vary together relative to their respec- tive means. The covariation between the two variables serves as the numerator for the equa- tion to calculate Pearson’s r.

Cox and Snell’s R2 A commonly used pseudo R 2

measure whose main component, as in other pseudo R 2 statistics, is the log likelihood func- tion (�2LL).

variables (one less than the total number of cate- gories), where the indicator category is coded as 1, the reference category is coded as − 1, and all other categories are coded as 0. Contrast coding ensures that the sum of all the estimated effects for the indicator variable is equal to 0.

control group The group that eligible cases are randomly assigned to which does not receive the treatment or the intervention being evalu- ated. In many criminological experiments the control group may receive existing interventions in contrast to the innovative treatment.

eligibility pool Participants or units that are eligible for an experiment.

dependent variable (Y) The variable assumed by the researcher to be influenced by one or more independent variables; the outcome variable; the phenomenon that we are interested in ex- plaining. It is dependent on other variables in the sense that it is influenced—or we expect it to be influenced—by other variables.

derivative at mean (DM) A measure that converts the nonlinear logistic regression coefficient to a simple linear regression coefficient, which may be interpreted as the change in Y associated with a unit change in X.

descriptive statistics A broad area of statistics that is concerned with summarizing large amounts of information in an efficient manner. Descriptive statistics are used to describe or represent in summary form the characteristics of a sample or population.

design sensitivity The statistical power of a re- search study. In a sensitive study design, statisti- cal power will be maximized, and the statistical test employed will be more capable of identify- ing an effect.

deviation from the mean The extent to which each individual score differs from the mean of all the scores.

directional hypothesis A research hypothesis that indicates a specific type of outcome by specify- ing the nature of the relationship that is ex- pected.

discordant pairs of observations Pairs of obser- vations that have inconsistent rankings on two ordinal variables.

G L O S S A R Y772

external validity The extent to which a study sam- ple is reflective of the population from which it is drawn. A study is said to have high external validity when the sample used is representative of the population to which inferences are made.

factorial The product of a number and all the posi- tive whole numbers lower than it.

frequency The number of times that a score or value occurs.

frequency distribution An arrangement of scores in order from the lowest to the highest, accompa- nied by the number of times each score occurs.

gamma (�) PRE measure of association for two or- dinal variables that uses information about con- cordant and discordant pairs of observations within a table. Gamma has a standardized scale ranging from �1.0 to 1.0.

Goodman and Kruskal’s tau (�) PRE measure of association for two nominal variables that uses information about the proportional distribution of cases within a table. Tau has a standardized scale ranging from 0 to 1.0. For this measure, the researcher must define the independent and dependent variables.

grand mean The overall mean of every single case across all of the samples.

heteroscedasticity A situation in which the vari- ances of scores on two or more variables are not equal. Heteroscedasticity violates one of the assumptions of the parametric test of statistical significance for the correlation coefficient.

histogram A bar graph used to represent a fre- quency distribution.

fixed effects A descriptive label for the regres- sion coefficients (b

k ) estimated in a model with

random effects. Fixed effects represent the aver- age effects of the independent variables on the dependent variable across all individuals and clusters in a multilevel model.

grand-mean centering Computed difference between the observed raw score on some vari- able for each observation in the sample and the overall sample mean for that variable.

group allocation In criminological experiments, eligible cases are randomly assigned to two or more groups—typically treatment or control.

homoscedasticity A statement that the variances and standard deviations of two or more popula- tions are the same.

honestly significant difference (HSD) test A parametric test of statistical significance, ad- justed for making pairwise comparisons. The HSD test defines the difference between the pairwise comparisons required to reject the null hypothesis.

independent Describing two events when the oc- currence of one does not affect the occurrence of the other.

independent random sampling A form of ran- dom sampling in which the fact that one subject is drawn from a population in no way affects the probability of drawing any other subject from that population.

independent variable (X) A variable assumed by the researcher to have an impact on or influence the value of the dependent variable, Y.

index of qualitative variation A measure of dis- persion calculated by dividing the sum of the possible pairs of observed scores by the sum of the possible pairs of expected scores (when

inferential, or inductive, statistics A broad area of statistics that provides the researcher with tools for making statements about populations on the basis of knowledge about samples. Infer- ential statistics allow the researcher to make in- ferences regarding populations from information gained in samples.

interval scale A scale of measurement that uses a common and standard unit and enables the re- searcher to calculate exact differences between scores, in addition to categorizing and ordering data.

iteration Each time we identify another tentative solution and reestimate our logistic regression coefficients.

Kendall’s �b PRE measure of association for two or- dinal variables that uses information about con-

b

has a standardized scale ranging from �1.0 to 1.0 and is appropriate only when the number of rows equals the number of columns in a table.

Kendall’s �c A measure of association for two ordi- nal variables that uses information about con-

c

has a standardized scale ranging from �1.0 to 1.0 and is appropriate when the number of rows is not equal to the number of columns in a table.

cordant pairs, discordant pairs, and pairs of observations tied on both variables examined. �

cordant pairs, discordant pairs, and pairs of observations tied on both variables examined. �

variable on the dependent variable is conditional on the level of a second independent variable.

cases are equally distributed across categories).

sent when the effect of one independent interaction effect An interaction effect is pre-

internal validity Whether the research design has allowed for the impact of the intervention or the treatment to be clearly distinguished from other factors.

intraclass correlation A measure of association that measures the level of absolute agreement of values within each cluster.

Kruskal-Wallis test A nonparametric test of statisti- cal significance for multiple groups, requiring at least an ordinal scale of measurement.

G L O S S A R Y 773

lack of convergence Failure of a logistic regression analysis to reach a result that meets the criterion of reduction in the log likelihood function.

lambda (�) PRE measure of association for two nominal variables that uses information about

least squares property A characteristic of the mean whereby the sum of all the squared devia- tions from the mean is a minimum—it is lower than the sum of the squared deviations from any other fixed point.

levels of measurement Types of measurement that make use of progressively larger amounts of information.

likelihood ratio chi-square test A test for statisti-

ables in a logistic regression is statistically significant. It compares �2LL for a full model to �2LL for a reduced model.

linear relationship An association between two variables whose joint distribution may be repre- sented in linear form when plotted on a scatter diagram.

log likelihood function A measure of the proba- bility of observing the results in the sample, given the coefficient estimates in the model. In logistic regression, the log likelihood function (�2LL) is defined as �2 times the natural loga- rithm of the likelihood function.

logarithm The power to which a fixed number (the base) must be raised to produce another number.

logistic model curve The form of the predicted outcomes of a logistic regression analysis. Shaped like an S, the logistic curve begins to flatten as it approaches 0 or 1, so it keeps com- ing closer to—but never actually reaches—either of these two values.

logistic regression analysis A type of regression analysis that allows the researcher to make pre- dictions about dichotomous dependent variables in terms of the log of the odds of Y.

logistic regression coefficient The coefficient b produced in a logistic regression analysis. It may be interpreted as the increase in the log of the odds of Y associated with a one-unit in-

margin of error The size of the confidence inter- val for a test. A margin of error of �3% in an opinion poll means that the confidence interval ranged between 3% above and 3% below the point estimate or observed statistic.

cal significance that allows the researcher to examine whether a subset of independent vari-

crease in X.

the modal category of the dependent variable for each category of the independent variable. Lambda has a standardized scale ranging from 0 to 1.0.

maximum likelihood estimation A technique for estimating the parameters or coefficients of a model that maximizes the probability that the

estimates obtained will produce a distribution similar to that of the observed data.

mean A measure of central tendency calculated by dividing the sum of the scores by the number of cases.

mean deviation A measure of dispersion calcu- lated by adding the absolute deviation of each score from the mean and then dividing the sum by the number of cases.

measurement The assignment of numerical values to objects, characteristics, or events in a system- atic manner.

measures of central tendency Descriptive statis- tics that allow us to identify the typical case in a sample or population. Measures of central ten- dency are measures of typicality.

measures of dispersion Descriptive statistics that tell us how tightly clustered or dispersed the cases in a sample or population are. They answer the question “How typical is the typical case?”

median A measure of central tendency calculated by identifying the value or category of the score that occupies the middle position in the distribu- tion of scores.

mode A measure of central tendency calculated by identifying the score or category that occurs most frequently.

model chi-square The statistical test used to assess the statistical significance of the overall logistic regression model. It compares the �2LL for the full model with the �2LL calculated without any independent variables included.

multicollinearity Condition in a multivariate re- gression model in which independent variables examined are very strongly intercorrelated. Mul- ticollinearity leads to unstable regression coeffi- cients.

marginal The value in the margin of a table that to- tals the scores in the appropriate column or row.

multilevel data Sample data where individual observations (level 1 data) are clustered within a higher-level sampling unit (level 2 data).

probability that a series of events will jointly occur.

multivariate regression A technique for predict- ing change in a dependent variable, using more than one independent variable.

multivariate statistics Statistics that examine the relationships among variables while taking into account the possible influences of other con- founding factors. Multivariate statistics allow the researcher to isolate the impact of one variable from others that may distort his or her results.

multiplication rule The means for determining the

multinomial logistic regression A statistical te-

variable with three or more categories measured chnique to predict the value of a dependent

at the nominal level of measurement.

Nagelkerke R2 A pseudo R 2 statistic that corrects for the fact that Cox and Snell’s estimates, as well as many other pseudo R 2 statistics, often have a maximum value of less than 1.

G L O S S A R Y774

natural logarithm of the odds of Y (logit of Y) The outcome predicted in a logistic regression analysis.

nominal scale A scale of measurement that assigns each piece of information to an appropriate cat- egory without suggesting any order for the cate- gories created.

nondirectional hypothesis A research hypothesis that does not indicate a specific type of out- come, stating only that there is a relationship or a difference.

nonparametric tests Tests of statistical signifi- cance that make no assumptions as to the shape of the population distribution.

normal curve A normal frequency distribution rep- resented on a graph by a continuous line.

normal frequency distribution A bell-shaped fre- quency distribution, symmetrical in form. Its mean, mode, and median are always the same. The percentage of cases between the mean and points at a measured distance from the mean is fixed.

null hypothesis A statement that reduces the re- search question to a simple assertion to be tested by the researcher. The null hypothesis normally suggests that there is no relationship or no difference.

observed frequency The observed result of the study, recorded in a cell.

observed significance level The risk of Type I error associated with a specific sample statistic in a test. When the observed significance level is less than the criterion significance level in a test of statistical significance, the researcher will re- ject the null hypothesis.

odds ratio [Exp(B)] A statistic used to interpret the logistic regression coefficient. It represents the impact of a one-unit change in X on the ratio of the probability of Y.

OLS regression See ordinary least squares regres- sion analysis.

one-tailed test of significance A test of statistical significance in which the region for rejecting the null hypothesis falls on only one side of the sampling distribution. One-tailed tests are based on directional research hypotheses.

dependent and the independent variable that is

ordinal logistic regression A statistical technique

level of measurement.

non-linear relationship Relationship between the

not captured by a straight line (linear) relationship.

three or more categories measured at the ordinal to predict the value of a dependent variable with

ordinal scale A scale of measurement that catego- rizes information and assigns it an order of mag- nitude without using a standard scale of equal intervals.

parameter A characteristic of the population—for example, the mean number of previous convic- tions for all U.S. prisoners.

parametric tests Tests of statistical significance that make assumptions as to the shape of the population distribution.

Pearson’s correlation coefficient See Pearson’s r.

tionships on a standardized scale from �1.0 to 1.0.

percent of correct predictions A statistic used to assess how well a logistic regression model ex- plains the observed data. An arbitrary decision point (usually 0.50) is set for deciding when a predicted value should be set at 1, and then the predictions are compared to the observed data.

percent of variance explained (1) R 2, a measure for evaluating how well the regression model predicts values of Y; it represents the improve- ment in predicting Y that the regression line provides over the mean of Y. (2) �2, the propor- tion of the total sum of squares that is ac- counted for by the explained sum of squares.

percentage A relation between two numbers in which the whole is accorded a value of 100 and the other number is given a numerical value corresponding to its share of the whole.

phi (�) A measure of association for two nominal variables that adjusts the chi-square statistic by the sample size. Phi is appropriate only for nominal variables that each have two categories.

pie chart A graph in which a circle (called a pie) is cut into wedges to represent the relative size of each category’s frequency count.

point estimate An estimate of the population para- meter. Absent knowledge of the population pa- rameter, the statistic we obtain for a sample is

r Pearson’s r A commonly used measure of asso-

ciation between two variables. Pearson’s measures the strength and direction of linear rela-

parallel slopes assumption In an ordinal logistic

ordinary least squares regression analysis A type of regression analysis in which the sum of squared errors from the regression line is mini- mized.

outlier(s) A single or small number of exceptional cases that substantially deviate from the general pattern of scores.

overall mean See grand mean. pairwise comparisons Comparisons made be-

tween two sample means extracted from a larger statistical analysis.

categories of the dependent variable.

regression model, be effect of each independent variable is assumed to be constant across all

generally used as an estimate—or, in statistical terms, a point estimate—of the population para- meter.

G L O S S A R Y 775

the other number is given a numerical value corresponding to its share of the whole.

proportional reduction in error (PRE) The pro- portional reduction in errors made when the value of one measure is predicted using infor- mation about the second measure.

pseudo R2 The term generally used for a group of measures used in logistic regression to create an approximation of the OLS regression R 2. They are generally based on comparisons of �2LL for a full model and a null model (without any in- dependent variables).

random sampling Drawing samples from the pop- ulation in a manner that ensures every individ- ual in that population an equal chance of being selected.

randomized experiment A type of study in which the effect of one variable can be examined in isolation through random allocation of subjects to treatment and control, or comparison, groups.

pooled variance A method of obtaining the standard error of the sampling distribution for a difference of means test. The pooled variance method re- quires an assumption of homoscedasticity.

population The universe of cases that the re- searcher seeks to study. The population of cases is fixed at a particular time (e.g., the population of the United States). However, populations usually change across time.

population distribution The frequency distribu- tion of a particular variable within a population.

probability distribution A theoretical distribution consisting of the probabilities expected in the long run for all possible outcomes of an event.

proportion A relation between two numbers in which the whole is accorded a value of 1 and

posttest measure Analyses conducted by the researcher to determine if the intervention had any impact on the outcome measures of interest.

random coefficient model A linear regression model that allows the intercept and the effect of at least one independent variable to vary randomly across cluster—random effects are included for the model intercept and at least one independent variable.

random effects A descriptive label for the random error terms included in a multilevel model that allow for variation across cluster from the sample average estimated in the fixed effects. Random effects are assumed to be normally distributed in most multilevel models.

random intercept model A linear regression model that allows the intercept to vary randomly across cluster—random effects are included for the model intercept.

randomization The process of randomly assigning members from the pool of eligible participants or units to the study conditions—often a treatment group and a control or a comparison group.

range A measure of dispersion calculated by sub- tracting the smallest score from the largest score. The range may also be calculated from specific points in a distribution, such as the 5th and 95th percentile scores.

rank-order test A test of statistical significance that uses information relating to the relative order, or rank, of variable scores.

ratio scale A scale of measurement identical to an interval scale in every respect except that, in ad- dition, a value of zero on the scale represents the absence of the phenomenon.

regression coefficient b A statistic used to assess the influence of an independent variable, X, on a dependent variable, Y. The regression coefficient b is interpreted as the estimated change in Y that is associated with a one-unit change in X.

regression error (e) The difference between the predicted value of Y and the actual value of Y.

regression coding A method for recoding a multi- category nominal variable into multiple indicator dummy variables (one less than the total number of categories), where the indicator category is coded as 1 and all other categories are coded as 0. The reference category does not have an indicator variable and is coded as a 0 on all the indicator dummy variables.

representative sample A sample that reflects the population from which it is drawn.

research hypothesis The antithesis of the null hy- pothesis. The statement normally answers the initial research question by suggesting that there is a relationship or a difference.

research question The question the researcher hopes to be able to answer by means of a study.

sample A set of actual observations or cases drawn from a population.

sample distribution The frequency distribution of a particular variable within a sample drawn from a population.

sample statistic A characteristic of a sample—for example, the mean number of previous convic- tions in a random sample of 1,000 prisoners.

regression line The line predicting values of Y. The line is plotted from knowledge of the Y-intercept and the regression coefficient.

regression model The hypothesized statement by the researcher of the factor or factors that define the value of the dependent variable, Y. The model is normally expressed in equation form.

rejection region The area of a sampling distribu- tion containing the test statistic values that will

relaxing an assumption Deciding that we need not be concerned with that assumption. For ex- ample, the assumption that a population is nor- mal may be relaxed if the sample size is suffi- ciently large to invoke the central limit theorem.

reliability The extent to which a measure provides consistent results across subjects or units of study.

cause the researcher to reject the null hypothesis.

G L O S S A R Y776

sampling distribution A distribution of all the re- sults of a very large number of samples, each one of the same size and drawn from the same population under the same conditions. Ordinar- ily, sampling distributions are derived using probability theory and are based on probability distributions.

sampling frame The universe of eligible cases from which a sample is drawn.

sampling with replacement A sampling method in which individuals in a sample are returned to the sampling frame after they have been se- lected. This raises the possibility that certain in-

scale of measurement Type of categorization used to arrange or assign values to data.

scatter diagram See scatterplot. scatterplot A graph whose two axes are defined by

two variables and upon which a point is plotted for each subject in a sample according to its score on the two variables.

separate variance A method of obtaining the stan- dard error of the sampling distribution for a dif-

or—as in the case of a proportion—is defined by the null hypothesis.

skewed Describing a spread of scores that is clearly weighted to one side.

Somers’ d PRE measure of association for two ordi- nal variables that uses information about con-

Somers’ d has a standardized scale ranging from �1.0 to 1.0.

Spearman’s correlation coefficient See Spear- man’s r.

Spearman’s r (rs) A measure of association be- tween two rank-ordered variables. Spearman’s r measures the strength and direction of linear re- lationships on a standardized scale between �1.0 and 1.0.

standard deviation A measure of dispersion calcu- lated by taking the square root of the variance.

standard deviation unit A unit of measurement used to describe the deviation of a specific score or value from the mean in a z distribution.

standard error The standard deviation of a sam- pling distribution.

standard normal distribution A normal fre- quency distribution with a mean of 0 and a stan- dard deviation of 1. Any normal frequency dis- tribution can be transformed into the standard normal distribution by using the z formula.

standardized logistic regression coefficient A statistic used to compare logistic regression coef- ficients that use different scales of measurement. It is meant to approximate Beta, the standardized regression coefficient in OLS regression.

standardized regression coefficient (Beta) Weighted or standardized estimate of b that takes into account the standard deviation of the independent and the dependent variables. The standardized regression coefficient is used to compare the effects of independent variables

dividuals in a population may appear in a sample more than once.

observations tied on the independent variable. cordant pairs, discordant pairs, and pairs of

ference of means test. The separate variance method does not require an assumption of ho- moscedasticity.

significance level The level of Type I error a re- searcher is willing to risk in a test of statistical significance.

single-sample t-test A test of statistical significance that is used to examine whether a sample is drawn from a specific population with a known or hypothesized mean. In a t-test, the standard deviation of the population to which the sample is being compared is unknown.

single-sample z-test A test of statistical signifi- cance that is used to examine whether a sample is drawn from a specific population with a known or hypothesized mean. In a z-test, the standard deviation of the population to which the sample is being compared either is known tails of the distribution The extremes on the sides

of a sampling distribution. The events repre- sented by the tails of a sampling distribution are those deemed least likely to occur if the null hy- pothesis is true for the population.

test of statistical significance A test in which a researcher makes a decision to reject or to fail to reject the null hypothesis on the basis of a sam- ple statistic.

test statistic The outcome of the study, expressed in units of the sampling distribution. A test sta- tistic that falls within the rejection region will lead the researcher to reject the null hypothesis.

tied pairs of observations (ties) Pairs of observa- tions that have the same ranking on two ordinal variables.

tresholds Points that mark the limits of the under

measured on different scales in a multivariate re- gression analysis.

statistical inference The process of making gener- alizations from sample statistics to population parameters.

statistical power One minus the probability of a Type II error. The greater the statistical power of a test, the less chance there is that a re- searcher will mistakenly fail to reject the null hypothesis.

statistically significant Describing a test statistic that falls within the rejection region defined by the researcher. When this occurs, the researcher is prepared to reject the null hypothesis and state that the outcome or relationship is statisti- cally significant.

sum of squares The sum of squared deviations of scores from a mean or set of means.

lying continuum measured by an ordinal variable.

G L O S S A R Y 777

time series data Repeated measures of the same

time series plot A line graph that connects re- peated measures of the same variable over some regularly occurring time period, such as days, months, or years.

tolerance A measure of the extent of the intercor- relations of each independent variable with all other independent variables. Tolerance may be used to test for multicollinearity in a multivariate regression model.

total sum of squares (TSS) A measure of the total amount of variability across all of the groups ex- amined. The total sum of squares is calculated by summing the squared deviation of each score from the grand mean.

t-test for dependent samples A test of statistical significance that is used when two samples are not independent.

two-sample t-test A test of statistical significance that examines the difference observed between the means or proportions of two samples.

two-tailed test of significance A test of statistical significance in which the region for rejecting the null hypothesis falls on both sides of the sam- pling distribution. Two-tailed tests are based on nondirectional research hypotheses.

Type I error Also known as alpha error. The mis- take made when a researcher rejects the null hy- pothesis on the basis of a sample statistic (i.e., claiming that there is a relationship) when in fact the null hypothesis is true (i.e., there is ac- tually no such relationship in the population).

Type II error Also known as beta error. The mis- take made when a researcher fails to reject the null hypothesis on the basis of a sample statistic (i.e., failing to claim that there is a relationship) when in fact the null hypothesis is false (i.e., there actually is a relationship).

unexplained sum of squares (USS) Another name for the within sum of squares. The unex- plained sum of squares is the part of the total variability that cannot be explained by visible differences between the groups.

universe The total population of cases. validity The extent to which a variable accurately

reflects the concept being measured. variable A trait, characteristic, or attribute of a per-

son/object/event that can be measured at least at the nominal-scale level.

variance (s2) A measure of dispersion calculated by adding together the squared deviation of each score from the mean and then dividing the sum by the number of cases.

variation ratio A measure of dispersion calculated by subtracting the proportion of cases in the modal category from 1.

Wald statistic A statistic used to assess the statisti- cal significance of coefficients in a logistic re- gression model.

within sum of squares (WSS) A measure of the vari- ability within samples (groups). The within sum of squares is calculated by summing the squared de- viation of each score from its sample mean.

Y-intercept (b0) The expected value of Y when X � 0. The Y-intercept is used in predicting values of Y.

z-score Score that represents standard deviation units for a standard normal distribution.

period, such as days, months, or years. variable over some regularly occurring time

treatment group One group that eligible cases are randomly assigned to which receives the treat- ment or the intervention being evaluated.

variance components model A one-way analysis of variance model that includes random effects for each cluster that assesses whether there is random variation in the mean of the depend- ent variable across the clusters included in the analysis.

within effect Effect of an independent variable on the dependent variable within each cluster and then averaged across all clusters or groups included in the analysis.

I n d e x

D. Weisburd and C. Britt, Statistics in Criminal Justice, DOI 10.1007/978-1-4614-9170-5,

© Springer Science+Business Media New York 2014

778

A

alpha error. See Type I error

analysis of variance (ANOVA), 307–337

arrangements, 152–147, 163

equation for, 152

association, measures of, 354–355, 359, 363–364,

367, 369, 382–383, 386–388

Cramer’s V, 356, 358, 386, 388–389

387–388

b c ), 368, 374, 381,

387–388, 390–391

lambda ( ), 358, 387–388, 390

nominal measures of statistical significance, 365

ordinal measures of statistical significance,

367–368

phi ( ), 364, 388–389

Somers’ d, 365, 368–369, 376, 380, 387–388

assumptions, for statistical inference, 174–180, 214

B

b (regression coefficient), 440–447, 458, 461,

464, 624

calculation of, 442

statistical significance of, 458–467

b 0 (Y-intercept), 445–447, 449, 470–472

equation for, 445, 447

bar charts, 43–44, 55–56

horizontal, 43–45

for nominal and ordinal data, 50

vertical, 43–44

beta error. See Type II error

between effect, 652

testing of, 654

between sum of squares (BSS), 310–311, 313–315,

320, 329–330, 337–338

equation for, 310, 339

binomial distribution, 155–164

equation for, 156, 164

bivariate regression, 439–473

block randomization, 685–688

benefits of, 689–690

statistical power, 691–693

Brant test, 623

BSS. See between sum of squares

C

cells, 209, 223

centering, 648–649

central limit theorem, 251–252, 262

central tendency, 8, 66, 73

chi-square ( 2) distribution, 199–202, 223, 612, 614,

622, 623, 628

chi-square ( 2 ) statistic, 201–203, 223, 614, 627

for relating two measures, 210

chi-square ( 2) table, 202–203, 612

chi-square ( 2 ) test, 203–205

with multicategory variables, 212–217

with ordinal variables, 217–222

with small samples, 222

classification, 14–15, 26

clustered data, 638–639

coefficient of relative variation (CRV), 111–112

equation for, 113, 118

coin toss, fair, 147–151

column marginals, 209

confidence intervals, 702–718

compared to rejection region, 706–707

for logistic regression coefficient, 716, 719

for Pearson’s r, 713

for regression coefficient, 716, 719

for sample mean, 711–713

for sample proportion, 710–713

confidence limit equations, 709–711, 714–715, 719

convenience samples, 175, 189

correctly specified model, 483–484

correlation, 361 (see also eta ( ), Pearson’s r)

equation for, 403, 431

Spearman’s r covariation, 403–404

correct model specification, 482–484

Cramer’s V, 354, 386

statistical significance of, 363–364

critical region. See rejection region

critical value, 182, 189, 612, 613, 615, 622, 623

cross-level interaction, 660–662

CRV. See coefficient of relative variation

curvilinear relationship, 43, 411–412, 430

I N D E X 779

D

data, 22–23, 26, 602, 614, 619, 625. See also decisions

about data

by comparing sample means, 270–279

clustered, 638–639

with chi-square test, 199–202

collection of, 18

multilevel, 638–639

with chi-square test, 201–203

by comparing sample means, 270–275, 278

decisions about data, making, 180

making analysis of, 5–11

with Kruskal-Wallis test, 334–335

with logistic regression coefficients, 567–576

making analysis of, 6–11, 367–368

with Pearson’s r, 401, 403–405, 407

with regression coefficients, 456–462

with single-sample t test, 257–261

with Spearman’s r, 420–421

with t-test for dependent samples, 289–291,

294–295

with z-test for proportions, 252–257

degrees of freedom (df), 199–200, 202–203, 205, 209,

321, 365, 426, 428, 461, 612–615, 622, 623

dependent variable (Y), 440, 462, 603, 605, 611, 612,

615–618, 622–624, 627, 628

derivative at mean, 575–576, 591, 594

descriptive statistics, 7–8, 11

design sensitivity, 740, 748

deviation from the mean, 81, 85–86, 104–105,

114–115, 310, 395, 452

equation for, 80–82, 254

differenes from the mean. See deviation

from the mean

directional hypotheses, 179, 185

dispersion measures, 8

distribution

binomial, 156–158, 192–193, 199

normal, 235–241

normal frequency, 236–237, 244, 246, 262–263

predictions beyond observed, 448–449, 551

probability, 147–149, 158, 163–164

rejection region of, 182–183

sampling, 146–148, 164, 174, 180–183

skewed, 83

standard normal, 239–241, 245, 256, 262–263

tails of, 183–184, 189, 244–245

types of, 126–127, 138

distribution-free tests. See nonparametric tests

E

e (regression error), 451–453, 470

effect size (ES), 735–745

error

in prediction, 440

reduction of, 5, 381

risk of in hypothesis testing, 133–135

standard, 243–245, 605, 612, 613

Type I, 134–135, 139–140

Type II, 134–135, 137–140

ESS, 313, 338

equation for, 329–330, 338 2), 329–330, 338. See also percent

of variance explained

equation for, 329–330, 338

expected frequency, 201, 205, 211, 212, 216

experiments

advantages of, 677–681

interactions and, 695–697

internal validity of, 682–683

explained sum of squares (ESS), 313, 329, 337–338

explained variance, 313–314, 453–455, 647–648

external validity, 175, 178, 188

F

failure, studies designed for, 747

F equation, 319, 337

factorials (!), 153

frequency distribution, 37–38, 56

frequency, expected, 201, 205, 222, 223

studies designed for, 747–748

-test for overall regression, 467–470

fixed effects, 641–644

G

statistical significance of, 381

387–388

grand mean, 309–310, 333

statistical significance of, 365

group variability, 309–310, 313, 337

H

H 0 (null hypothesis), 131–133

H 1 (research hypothesis), 129–131

heteroscedasticity, 426–427, 430

histogram, 39, 43, 56

I N D E X780

homoscedasticity, 274, 295

honestly significant difference (HSD), 332–334, 338

equation for, 332, 340

hypotheses, 129–133, 622, 628

and risk of error, 133–136

I

independence, defined, 164

independent random sampling, 176–178, 189

independent variable (X ), 358–365, 388, 390, 608,

611–615, 617–620, 622–628

index of qualitative variation (IQV), 100–102, 116

equation for, 100, 117

inferential statistics, 9–10

interaction effects, 522–525, 528–532, 537–538

cross-level, 660–662

interpretation of dummy variable

and interval-level variable, 498–499

interpretation of two interval-level variables, 530–532

interval scales, 19–21, 26

measuring dispersion in, 102–111

intraclass correlation, 644–645

IQV. See index of qualitative variation

K

b c ), 368, 374, 381,

386–387, 390–391

statistical significance of, 381–383

Kruskal-Wallis test, 334–337

equation for, 336, 339–340

L

statistical significance of, 363

level-2 characteristics, 660–662

levels of measurement, 15–21, 26

likelihood ratio chi-square test, 589–590, 592, 594, 613

linear relationship, 411, 430

logistic model curve, 556, 592

logistic regression, 549–594, 602–607, 609–612, 615,

617, 619, 621–624, 627–629

compared to OLS regression, 550–555

estimating coefficients for, 550, 554

statistical significance for, 555–558

logistic regression coefficients, 567–568, 572,

580–583, 592, 594, 624

comparing, 577–583

confidence interval for, 582

statistical significance of, 587–590, 613

M

marginals, 209, 223

maximum likelihood estimation (MLE), 561, 593

mean, 74–81, 85–86

comparing to median, 77–80

computing, 90–91

confidence interval for, 711

equation for, 74, 87

grand, 307

for noninterval scales, 82

overall, 309

mean deviation, 113–115

measurement

concept of, 14–15, 26

ladder of, 16

levels of, 15–16, 26

scales of, 16

measures

of association, 354–385

of central tendency, 8, 66–77

of dispersion, 9, 95–111

median, 66–74, 86

comparing to mean, 77–80

mode, 66–68, 86

multicollinearity, 516, 534–538

multilevel data, 638–639

multiplication rule, 149–151, 164

multivariate regression, 481–507

multivariate statistics, 7, 10–12

N

National Archive of Criminal Justice Data

(NACJD), 31

nominal-level data, 16–18

nominal-level variable, 17

nominal scales

bar charts with, 50

central tendency in, 66–68

chi-square test with, 198–202

measures of association for, 355–365

nondirectional hypotheses, 179

relating to other scales, 22–23

nondirectional hypotheses, 179

non-linear relationship, 516–522

coefficients, 519–521

detecting with line graphs, 517–519

statistical significance, 522

nonparametric tests, 175, 179. See also

Kruskal-Wallis test

I N D E X 781

normal curve, 235–247, 262

normal frequency distribution, 235–247, 262

null hypothesis (H 0 ), 131–133

O

observed frequencies, 201–202, 216

odds ratio, 568–569, 571–572, 607, 608, 618, 619,

620, 627–629, 716–717

OLS regression, 452, 470

one-tailed rejection region, 183

one-tailed test of significance, 183, 186, 188, 287

ordinal-level data, 18–19

ordinal logistic regression, 602, 616, 617, 619, 620,

622–624, 627, 628

parallel slopes assumption, 622–627

partial proportional odds, 624–625

ordinal scales, 18–20, 26, 602

bar charts with, 50

measures of association for, 367–369, 381

ordinary least squares regression analysis, 452, 495–496

outlier, 6, 79–80, 83–86

overall mean, 309, 338

P

pairwise comparisons, 331–334, 339

between groups studied, 331–334

parallel slopes test, 622–623

Brant test, 623

score test, 622–623

parameters, 127, 139–140, 619, 621

parametric tests, 174, 235

partial proportional odds, 624–625

Pearson, Karl, 401

Pearson’s correlation coefficient. See Pearson’s r

Pearson’s r, 401–405, 430–431

calculation of, 405–407

confidence interval for, 713–715

equation for, 440–442

for nonlinear relationships, 411–415

statistical significance of, 421–426

percentage, 48–49, 56

equation for, 48–49, 57

percent of variance explained (R2), 453–455, 467,

470. See also eta squared ( 2 ) equation for

phi ( ), 353, 386

pie charts, 51–52, 55–56

pooled variance method, 274, 276–277, 282, 288,

294–295

equation for, 282

population, 140

population distribution, 126, 140

positivism, 15

probability distributions, 147–149, 164

problem-oriented policing (POP) approach, 173

proportion, 48, 56

confidence interval for, 710–711

equation for, 48, 56

pseudo R2, 585–587, 592

R

r. See Pearson’s r

r s . See Spearman’s r

R2. See percent of variance explained

random coefficient model, 655–660

level-2 characteristics, 660–662

statistical significance, 636–637

random effects, 642–644

model, 642–644

randomized experiments, 486, 505

random intercept model, 646–653

statistical significance, 648–649

random sampling, 175, 189

range, 102–103, 116

rank order test, 334, 338, 339. See also

nonparametric tests

ratio scales, 19–21, 26

regression

approach to ANOVA, 640–641

bivariate, 439, 462, 471–472, 482, 484, 485, 488

F-test, 467–470

regression coefficient (b), 440–444, 464, 470, 471, 624

calculation of, 440–443

confidence interval for, 715

correctly estimating, 482–492

statistical significance of, 458–461

regression error (e), 451–452, 471

regression line, 446–450

regression model, 449–450, 469

multivariate, 482, 490, 492, 494, 505, 605, 622

predictions in, 445–449, 603

Reiss, Albert J., Jr., 6

rejection region, 182–186, 188

in ANOVA, 319–321

with chi-square test, 202–203

compared to confidence interval, 706–707

with Kruskal-Wallis test, 334

for Pearson’s r, 421–426

for regression coefficients, 464

I N D E X782

for Spearman’s r, 428

with t-test for dependent samples, 289–291

with two-sample t-test, 277–278

with z-test for proportions, 252–257

relationship

defining strength of, 328–331

linear, 411, 414

relaxing assumptions, 259, 263

relevant independent variables, defining, 495

reliability, 24–25

representative sample, 175, 189

research

measurement in, 14–15

purpose of, 15

research hypothesis (H 1 ), 129–131, 140,

622, 628

research questions, 129, 139–140

equation for, 449, 470

row marginals, 209

S

s. See standard deviation

sample distribution, 126, 140

sample statistics, 127–129, 140

sampling, 175–178

random, 175–176

with replacement, 176, 189

sampling distribution, 147–149, 164

for chi-square test, 208–209

for comparing sample means, 273–277

for Kruskal-Wallis test, 336

for nonnormal populations, 247–252

for proportion, 252–254

selecting, 180–181

for single-sample t-test, 258–260

for Spearman’s r, 429

for t-test for dependent samples, 291

sampling frame, 176, 189

scales, measurement, 16, 26

interval, 19–21, 26

mean and, 82

nominal, 16–18, 26

ordinal, 18–19, 26, 602

ratio, 19–21, 26

scatter diagrams, 411–415, 431

scatterplots, 411–415, 431

science and measurement, 14–15

score test, 622–623

separate variance method, 67–278, 295

significance criteria, 137–138

significance level

in ANOVA, 319–320

with chi-square test, 202–203, 205

with Kruskal-Wallis test, 336

for logistic regression coefficients, 557–558

for nominal measures of association, 365–367

for ordinal measures of association, 381–383

for Pearson’s r, 426

for regression coefficients, 462

and rejection region, 182–183

for Spearman’s r, 429

with t-test for dependent samples, 291

with two-sample t-test, 277–278

with z-test for proportions, 244

skewed, 83–86, 222

Somers’ d, 368–369, 376, 380, 385, 387–391

statistical significance of, 381–382

Spearman’s correlation coefficient. See Spearman’s r

Spearman’s r, 369, 418–421, 430–431

equation for, 419, 431

statistical significance of, 428–429

SPSS, 31–33

standard deviation (s), 107–111, 116

equation for, 107, 117

estimate of, 258, 264

of population distribution, 239–243

of sampling distribution of a proportion, 252–254

standard deviation sample, 140

standard deviation unit, 238–240, 263

standard error, 243, 252, 263, 605, 612, 613

standard normal distribution, 239, 263

Stata, 33–35

statistical error, 134–138

statistical inference, 129–138, 611–615

statistical power, 728–730, 748

components of, 726

use of covariates, 693–695

statistical significance, 135, 140, 605, 611–613, 615,

619, 621–622, 625, 627

test of, 135, 140

statistics

descriptive, 7–9

faulty, 7

fear of, 5

inductive, 7

inferential, 9–10

multivariate, 10–12

problem solving with, 4–5

purpose of, 3–4

techniques of, 5–6

I N D E X 783

Student’s. See t distribution

studies designed for failure, 729, 747

sum of deviations from the mean, equation for, 80, 86

sums of squares, 312–313, 337

partitioning of, 312–313

T

t distribution, 258–260, 281, 283, 286–287, 291, 307,

421, 423, 430, 458, 464, 470, 709

test statistic, 187–189, 205, 210, 215–216, 220, 245,

257, 260, 281, 627, 628

in ANOVA, 319, 324

for chi-square test, 201

for comparing sample means, 278, 287

for Kruskal-Wallis test, 335, 337–340, 345–346

for logistic regression coefficients, 564, 567–568,

572, 580, 582–583, 585, 591, 716

for Pearson’s r, 405, 431

for regression coefficients, 477

for sample proportions, 287

for Spearman’s r, 428

for t-test for dependent samples, 288–290,

294–295

for z-test for proportions, 252

time series data, 52–53, 56

time series plot, 53–56

tolerance

equation for, 536, 538

total sum of squares ( TSS), 312, 339, 453

equation for, 312–314, 322, 330, 340, 468

t-test

for dependent samples, 288–291, 294–296

for means, 271, 280–281, 294, 296

for ordinal scales, 294

tails of the distribution, 183–184, 189, 244

two-sample t-test, 271, 277, 280, 284, 285, 294, 295

equations for, 277, 295, 296

two-tailed rejection region, 181

two-tailed test of significance, 183–185, 429

Type I error, 134–140

Type II error 1, 134–140

U

unbiased estimate of standard error equation, 259,

277, 295

unexplained sum of squares (USS), 313, 337, 339,

467–469, 472

universe, 127, 140, 176, 189

unknown population, 252, 254, 259, 262

V

Validity, 23–27, 32, 175–176, 178, 188–189,

273, 331

Variable, 16–17, 19–22, 31–32, 613, 631

Variance, 104–107, 116

equation for, 104, 117

between and within groups, 313

pooled, 272, 274–275, 278, 294

separate, 274, 295

variation ratio, 98–100, 115–117

variance components, 640–646

statistical significance of, 645

W

Wald statistic, 588, 589, 612–614, 619,

621, 627

within effect, 652

testing of, 653–654

within group variance equation, 315, 340

equation for, 315

within sum of squares (WSS), 311, 337, 339

equation for, 311–313, 340

equation for, 313

X

X (independent variable), 440–441, 471

Y

Y (dependent variable), 440–441, 471

Z

z-score, 239–241

z-test for proportions, 252–257

  • Contents
  • Preface
  • A c k n o w l e d g m e n t s
  • A b o u t t h e A u t h o r s
  • C h a p t e r o n e
    • Introduction: Statistics as a Research Tool
      • T h e P u r p o s e o f S t a t i s t i c s I s t o C l a r i f y
      • S t a t i s t i c s A r e U s e d t o S o l v e P r o b l e m s
      • B a s i c P r i n c i p l e s A p p l y A c r o s s S t a t i s t i c a l T e c h n i q u e s
        • Descriptive Statistics
      • T h e U s e s o f S t a t i s t i c s
        • Inferential Statistics
        • Taking into Account Competing Explanations: Multivariate Statistics
      • C h a p t e r S u m m a r y
      • K e y T e r m s
  • C h a p t e r t w o
    • Measurement: The Basic Building Block of Research
    • S c i e n c e a n d M e a s u r e m e n t : C l a s s i f i c a t i o n a s a F i r s t S t e p i n R e s e a r c h
    • L e v e l s o f M e a s u r e m e n t
      • Nominal Scales
      • Ordinal Scales
      • Interval and Ratio Scales
    • R e l a t i n g I n t e r v a l , O r d i n a l , a n d N o m i n a l S c a l e s : T h e I m p o r t a n c e o f C o l l e c t
    • W h a t I s a G o o d M e a s u r e ?
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
  • C h a p t e r t h r e e
    • Representing and Displaying Data
      • W h a t A r e F r e q u e n c y D i s t r i b u t i o n s a n d H i s t o g r a m s ?
      • E x t e n d i n g H i s t o g r a m s t o M u l t i p l e G r o u p s : U s i n g B a r C h a r t s
      • U s i n g B a r C h a r t s w i t h N o m i n a l o r O r d i n a l D a t a
      • P i e C h a r t s
      • T i m e S e r i e s D a t a
      • C h a p t e r S u m m a r y
      • K e y T e r m s
      • S y m b o l s a n d F o r m u l a s
      • E x e r c i s e s
      • Computer Exercises
        • SPSS
        • Stata
  • C h a p t e r f o u r
    • Describing the Typical Case: Measures of Central Tendency
    • T h e M o d e : C e n t r a l T e n d e n c y i n N o m i n a l S c a l e s
    • T h e M e d i a n : T a k i n g i n t o A c c o u n t P o s i t i o n
    • T h e M e a n : A d d i n g V a l u e t o P o s i t i o n
      • Comparing Results Gained Using the Mean and Median
      • Other Characteristics of the Mean
      • Using the Mean for Noninterval Scales
    • S t a t i s t i c s i n P r a c t i c e : C o m p a r i n g t h e M e d i a n a n d t h e M e a n
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
      • Recoding Variables
  • C h a p t e r f i v e
    • How Typical Is the Typical Case?: Measuring Dispersion
    • M e a s u r e s o f D i s p e r s i o n f o r N o m i n a l a n d O r d i n a l L e v e l D a t a
      • The Proportion in the Modal Category
      • The Percentage in the Modal Category
      • The Variation Ratio
      • Index of Qualitative Variation
    • M e a s u r i n g D i s p e r s i o n i n I n t e r v a l S c a l e s : T h e R a n g e , V a r i a n c e , a n d S t a n d a r
      • The Variance
      • The Standard Deviation
      • The Coefficient of Relative Variation
      • A Note on the Mean Deviation
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
  • C h a p t e r s i x
    • The Logic of Statistical Inference: Making Statements About Populations from Sample Statistics
    • T h e D i l e m m a : M a k i n g S t a t e m e n t s A b o u t P o p u l a t i o n s f r o m S a m p l e S t a t i s t i c s
    • T h e R e s e a r c h H y p o t h e s i s
    • T h e N u l l H y p o t h e s i s
    • R i s k s o f E r r o r i n H y p o t h e s i s T e s t i n g
    • R i s k s o f E r r o r a n d S t a t i s t i c a l L e v e l s o f S i g n i f i c a n c e
    • D e p a r t i n g f r o m C o n v e n t i o n a l S i g n i f i c a n c e C r i t e r i a
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
  • C h a p t e r s e v e n
    • Defining the Observed Significance Level of a Test: A Simple Example Using the Binomial Distribution
    • T h e F a i r C o i n T o s s
      • Sampling Distributions and Probability Distributions
      • The Multiplication Rule
    • D i f f e r e n t W a y s o f G e t t i n g S i m i l a r R e s u l t s
    • S o l v i n g M o r e C o m p l e x P r o b l e m s
    • T h e B i n o m i a l D i s t r i b u t i o n
    • U s i n g t h e B i n o m i a l D i s t r i b u t i o n t o E s t i m a t e t h e O b s e r v e d S i g n i f i c a n c e L e v
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
  • C h a p t e r e i g h t
    • Steps in a Statistical Test: Using the Binomial Distribution to Make Decisions About Hypotheses
    • T h e P r o b l e m : T h e I m p a c t o f P r o b l e m O r i e n t e d P o l i c i n g o n D i s o r d e r l y A c t i v i t
    • A s s u m p t i o n s : L a y i n g t h e F o u n d a t i o n s f o r S t a t i s t i c a l I n f e r e n c e
      • Level of Measurement
      • Shape of the Population Distribution
      • Sampling Method
      • The Hypotheses
      • Stating All of the Assumptions
    • S e l e c t i n g a S a m p l i n g D i s t r i b u t i o n
    • S i g n i f i c a n c e L e v e l a n d R e j e c t i o n R e g i o n
      • Choosing a One-Tailed or a Two-Tailed Rejection Region
    • T h e T e s t S t a t i s t i c
    • M a k i n g a D e c i s i o n
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
  • C h a p t e r n i n e
    • Chi-Square: A Test Commonly Used for Nominal-Level Measures
    • T e s t i n g H y p o t h e s e s C o n c e r n i n g t h e R o l l o f a D i e
      • The Chi-Square Distribution
      • Calculating the Chi-Square Statistic
      • Linking the Chi-Square Statistic to Probabilities: The Chi-Square Table
      • A Substantive Example: The Relationship Between Assault Victims and Offenders
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • R e l a t i n g T w o N o m i n a l S c a l e M e a s u r e s i n a C h i S q u a r e T e s t
      • A Substantive Example: Type of Sanction and Recidivism Among Convicted White-Collar Criminals
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • E x t e n d i n g t h e C h i S q u a r e T e s t t o M u l t i c a t e g o r y V a r i a b l e s : T h e E x a m p l e o f C e
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • E x t e n d i n g t h e C h i S q u a r e T e s t t o a R e l a t i o n s h i p B e t w e e n T w o O r d i n a l V a r i a b l
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • T h e U s e o f C h i S q u a r e W h e n S a m p l e s A r e S m a l l : A F i n a l N o t e
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
  • C h a p t e r t e n
    • The Normal Distribution and Its Application to Tests of Statistical Significance
    • T h e N o r m a l F r e q u e n c y D i s t r i b u t i o n , o r N o r m a l C u r v e
      • Characteristics of the Normal Frequency Distribution
      • Scores
      • Developing Tests of Statistical Significance Based on the Standard Normal Distribution: The Single-Sample
      • Test for Known Populations
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • A p p l y i n g N o r m a l S a m p l i n g D i s t r i b u t i o n s t o N o n n o r m a l P o p u l a t i o n s
    • C o m p a r i n g a S a m p l e t o a n U n k n o w n P o p u l a t i o n : T h e S i n g l e S a m p l e
    • T e s t f o r P r o p o r t i o n s
      • Computing the Mean and Standard Deviation for the Sampling Distribution of a Proportion
      • Testing Hypotheses with the Normal Distribution: The Case of a New Prison Program
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • C o m p a r i n g a S a m p l e t o a n U n k n o w n P o p u l a t i o n : T h e S i n g l e S a m p l e
    • T e s t f o r M e a n s
      • Testing Hypotheses with the
      • Distribution
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
  • C h a p t e r e l e v e n
    • Comparing Means and Proportions in Two Samples
    • C o m p a r i n g S a m p l e M e a n s
      • The Case of Anxiety Among Police Officers and Firefighters
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
      • Bail in Los Angeles County: Another Example of the Two-Sample
      • Test for Means
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • C o m p a r i n g S a m p l e P r o p o r t i o n s : T h e T w o S a m p l e
    • T e s t f o r D i f f e r e n c e s o f P r o p o r t i o n s
      • The Case of Drug Testing and Pretrial Misconduct
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • T h e
    • T e s t f o r D e p e n d e n t S a m p l e s
      • The Effect of Police Presence Near High-Crime Addresses
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • A N o t e o n U s i n g t h e
    • T e s t f o r O r d i n a l S c a l e s
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
  • C h a p t e r t w e l v e
    • Comparing Means Among More Than Two Samples: Analysis of Variance
    • A n a l y s i s o f V a r i a n c e
      • Developing Estimates of Variance Between and Within Groups
      • Partitioning Sums of Squares
      • Developing Estimates of Population Variances
      • A Substantive Example: Age and White-Collar Crimes
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
      • Another ANOVA Example: Race and Bail Amounts Among Felony Drug Defendants
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • D e f i n i n g t h e S t r e n g t h o f t h e R e l a t i o n s h i p O b s e r v e d
    • M a k i n g P a i r w i s e C o m p a r i s o n s B e t w e e n t h e G r o u p s S t u d i e d
    • A N o n p a r a m e t r i c A l t e r n a t i v e : T h e K r u s k a l W a l l i s T e s t
      • Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
  • C h a p t e r t h i r t e e n
    • Measures of Association for Nominal and Ordinal Variables
    • D i s t i n g u i s h i n g S t a t i s t i c a l S i g n i f i c a n c e a n d S t r e n g t h o f R e l a t i o n s h i p : T
    • M e a s u r e s o f A s s o c i a t i o n f o r N o m i n a l V a r i a b l e s
      • Measures of Association Based on the Chi-Square Statistic
      • Proportional Reduction in Error Measures: Tau and Lambda
      • Statistical Significance of Measures of Association for Nominal Variables
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • M e a s u r e s o f A s s o c i a t i o n f o r O r d i n a l L e v e l V a r i a b l e s
      • Gamma
      • Kendall’s
      • and
      • Somers’
      • A Substantive Example: Affectional Identification with Father and Level of Delinquency
      • Note on the Use of Measures of Association for Ordinal Variables
      • Statistical Significance of Measures of Association for Ordinal Variables
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • C h o o s i n g t h e B e s t M e a s u r e o f A s s o c i a t i o n f o r N o m i n a l a n d O r d i n a l L e v e l V a r i
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
  • C h a p t e r f o u r t e e n
    • Measuring Association for Interval-Level Data: Pearson’s Correlation Coefficient
    • M e a s u r i n g A s s o c i a t i o n B e t w e e n T w o I n t e r v a l L e v e l V a r i a b l e s
    • P e a r s o n ’ s C o r r e l a t i o n C o e f f i c i e n t
      • The Calculation
      • A Substantive Example: Crime and Unemployment in California
      • Nonlinear Relationships and Pearson’s
      • Beware of Outliers
    • S p e a r m a n ’ s C o r r e l a t i o n C o e f f i c i e n t
    • T e s t i n g t h e S t a t i s t i c a l S i g n i f i c a n c e o f P e a r s o n ’ s
      • Statistical Significance of
      • : The Case of Age and Number of Arrests
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
      • Statistical Significance of
      • Unemployment and Crime in California
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • T e s t i n g t h e S t a t i s t i c a l S i g n i f i c a n c e o f S p e a r m a n ’ s
      • The Sampling Distribution
      • Significance Level and Critical Region
      • The Test Statistic
      • The Decision
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
  • C h a p t e r f i f t e e n
    • An Introduction to Bivariate Regression
    • E s t i m a t i n g t h e I n f l u e n c e o f O n e V a r i a b l e o n A n o t h e r : T h e R e g r e s s i o n C o e f f i
      • Calculating the Regression Coefficient
      • A Substantive Example: Unemployment and Burglary in California
    • P r e d i c t i o n i n R e g r e s s i o n : B u i l d i n g t h e R e g r e s s i o n L i n e
      • The
      • Intercept
      • The Regression Line
      • Predictions Beyond the Distribution Observed in a Sample
      • Predicting Burglary Rates from Unemployment Rates in California
      • Choosing the Best Line of Prediction Based on Regression Error
    • E v a l u a t i n g t h e R e g r e s s i o n M o d e l
      • Percent of Variance Explained
      • Percent of Variance Explained: Unemployment Rates and Burglary Rates in California
      • Statistical Significance of the Regression Coefficient: The Case of Age and Number of Arrests
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
      • Testing the Statistical Significance of the Regression Coefficient for Unemployment Rates and Burglary Rates in California
      • The Sampling Distribution
      • Significance Level and Rejection Region
      • The Test Statistic
      • The Decision
    • T h e
    • T e s t f o r t h e O v e r a l l R e g r e s s i o n
      • Age and Number of Arrests
      • Unemployment Rates and Burglary Rates in California
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
  • C h a p t e r s i x t e e n
    • Multivariate Regression
    • T h e I m p o r t a n c e o f C o r r e c t M o d e l S p e c i f i c a t i o n s
      • Errors in Prediction
      • Correctly Estimating the Effect of
      • Comparing Regression Coefficients Within a Single Model: The Standardized Regression Coefficient
    • C o r r e c t l y S p e c i f y i n g t h e R e g r e s s i o n M o d e l
      • Defining Relevant Independent Variables
      • Taking into Account Ordinaland Nominal-Scale Measures in a Multivariate Regression
    • C h a p t e r S u m m a r y
    • K e y T e r m s
    • S y m b o l s a n d F o r m u l a s
    • E x e r c i s e s
    • Computer Exercises
      • SPSS
      • Stata
  • C h a p t e r s e v e n t e e n
    • Multivariate Regression: Additional Topics
      • N o n l i n e a r R e l a t i o n s h i p s
        • Finding a Non-linear Relationship: Graphical Assessment
        • Incorporating Non-linear Relationships into an OLS Model
          • Interpreting Non-linear Coefficients
          • Note on Statistical Significance
        • Summary
      • I n t e r a c t i o n E f f e c t s
        • Interaction of a Dummy Variable and Interval-Level Variable
      • A n E x a m p l e : R a c e a n d P u n i s h m e n t S e v e r i t y
        • Interaction Effects between Two Interval-level Variables
      • A n E x a m p l e : P u n i s h m e n t S e v e r i t y
      • T h e P r o b l e m o f M u l t i c o l l i n e a r i t y
      • C h a p t e r S u m m a r y
      • K e y Te r m s
      • S y m b o l s a n d F o r m u l a s
      • E x e r c i s e s
      • Computer Exercises
        • SPSS
        • Stata
  • C h a p t e r e i g h t e e n
    • Logistic Regression
      • W h y i s I t I n a p p r o p r i a t e t o U s e O L S R e g r e s s i o n f o r a D i c h o t o m o u s D e p e n d e n t V a r i a b l e ?
      • L o g i s t i c R e g r e s s i o n
        • A Substantive Example: Adoption of Compstat in U.S. Police Agencies
      • I n t e r p r e t i n g L o g i s t i c R e g r e s s i o n C o e f f i c i e n t s
        • The Odds Ratio
        • The Derivative at Mean
      • C o m p a r i n g L o g i s t i c R e g r e s s i o n C o e f f i c i e n t s
        • Using Probability Estimates to Compare Coefficients
        • “Standardized” Logistic Regression Coefficients
      • E v a l u a t i n g t h e L o g i s t i c R e g r e s s i o n M o d e l
        • Percent of Correct Predictions
        • Pseudo R2
      • S t a t i s t i c a l S i g n i f i c a n c e i n L o g i s t i c R e g r e s s i o n
      • C h a p t e r S u m m a r y
      • K e y T e r m s
      • S y m b o l s a n d F o r m u l a s
      • E x e r c i s e s
      • Computer Exercises
        • SPSS
        • Stata
  • Ch a p t e r n i n e t e e n
    • Multivariate Regression with Multiple Category Nominal or Ordinal Measures: Extending the Basic Logistic Regression Model
    • M u l t i n o m i a l L o g i s t i c R e g r e s s i o n
      • A Substantive Example: Case Dispositions in California
      • Probability Equation
      • Odds Ratio Equation
      • The Missing Set of Coefficients
      • Statistical Inference
        • Single Coefficients
        • Multiple Coefficients
        • Overall Model
      • O r d i n a l L o g i s t i c R e g r e s s i o n
        • Cumulative Probability
        • Odds Ratio Using Cumulative Probabilities
        • Ordinal Logit Equation
        • Interpretation of Ordinal Logistic Regression Coefficients
      • S u b s t a n t i v e E x a m p l e : S e v e r i t y o f P u n i s h m e n t D e c i s i o n s
        • Interpreting the Coefficients
        • Statistical Significance
        • Parallel Slopes Tests
        • Score Test
        • Brant Test
        • Partial Proportional Odds
        • Severity of Punishment Example
      • C h a p t e r S u m m a r y
      • K e y Te r m s
      • S y m b o l s a n d F o r m u l a s
      • E x e r c i s e s
      • Computer Exercises
        • SPSS
        • Stata
          • Problems
  • C h a p t e r t w e n t y
    • Multilevel Regression Models
      • V a r i a n c e C o m p o n e n t s M o d e l
        • A Substantive Example: Bail Decision-Making Study
        • Fixed and Random Effects
        • Intraclass Correlation and Explained Variance
        • Statistical Significance
        • Bail Decision-Making Study
      • R a n d o m I n t e r c e p t M o d e l
        • Explained Variance
        • Statistical Significance
        • Centering Independent Variables
        • Bail Decision-Making Study
        • Between and Within Effects
        • Testing for Between and Within Effects
      • R a n d o m C o e f f i c i e n t M o d e l
        • Variance Estimates
        • Note on Explained Variance
        • Bail Decision-Making Study
      • A d d i n g C l u s t e r ( L e v e l 2 ) C h a r a c t e r i s t i c s
        • A Substantive Example: Race and Sentencing Across Pennsylvania Counties
      • C h a p t e r S u m m a r y
      • K e y Te r m s
      • S y m b o l s a n d F o r m u l a s
      • E x e r c i s e s
      • Computer Exercises
        • SPSS
        • Stata
          • Random Intercept Models
          • Random Coefficient Models
          • Problems
  • C h a p t e r t w e n t y o n e
    • Special Topics: Randomized Experiments
      • T h e S t r u c t u r e o f a R a n d o m i z e d E x p e r i m e n t
      • T h e M a i n A d v a n t a g e o f E x p e r i m e n t s : I s o l a t i n g C a u s a l E f f e c t s
      • n t e r n a l V a l i d i t y
      • S a m p l e S i z e , E q u i v a l e n c e , a n d S t a t i s t i c a l P o w e r
        • Statistical Foundations for Block Randomization
        • An Example: Jersey City Experiment
        • The Benefits of Block Randomized Trials
      • S t a t i s t i c a l P o w e r a n d B l o c k R a n d o m i z a t i o n
      • U s i n g C o v a r i a t e s t o I n c r e a s e S t a t i s t i c a l P o w e r i n E x p e r i m e n t a l S t u d i e s
      • E x a m i n i n g I n t e r a c t i o n T e r m s i n E x p e r i m e n t a l R e s e a r c h
      • C h a p t e r S u m m a r y
      • K e y Te r m s
      • S y m b o l s a n d F o r m u l a s
      • E x e r c i s e s
  • C h a p t e r t w e n t y t w o
    • Special Topics: Confidence Intervals
      • C o n f i d e n c e I n t e r v a l s
      • C o n s t r u c t i n g C o n f i d e n c e I n t e r v a l s
        • Confidence Intervals for Sample Means
        • Confidence Intervals for Sample Proportions
        • Confidence Intervals for a Difference of Sample Means
        • Confidence Intervals for Pearson’s Correlation Coefficient, r
        • Confidence Intervals for Regression Coefficients
        • Confidence Intervals for Logistic Regression Coefficients and Odds Ratios
      • C h a p t e r S u m m a r y
      • K e y T e r m s
      • S y m b o l s a n d F o r m u l a s
      • E x e r c i s e s
      • Computer Exercises
        • SPSS
        • Stata
          • Problems
  • C h a p t e r t w e n t y t h r e e
    • Special Topics: Statistical Power
      • S t a t i s t i c a l P o w e r
        • Setting the Level of Statistical Power
      • C o m p o n e n t s o f S t a t i s t i c a l P o w e r
        • Statistical Significance and Statistical Power
          • Directional Hypotheses
        • Sample Size and Statistical Power
        • Effect Size and Statistical Power
        • E s t i m a t i n g S t a t i s t i c a l P o w e r a n d S a m p l e S i z e f o r a S t a t i s t i c a l l y P o w e r f u l S t u d y
          • Difference of Means Test
          • ANOVA
          • Correlation
          • Least-Squares Regression
        • S u m m i n g U p : A v o i d i n g S t u d i e s D e s i g n e d f o r F a i l u r e
        • C h a p t e r S u m m a r y
        • K e y Te r m s
        • S y m b o l s a n d F o r m u l a s
        • Computer Exercises
          • Stata
            • Two-Sample Difference of Means Test
            • ANOVA
          • Correlation
            • OLS Regression
            • Problems
  • Appendix 1
    • Factorials
  • Appendix 2
    • Critical Values of X2 Distribution
  • Appendix 3
    • Areas of the Standard Normal Distribution
  • Appendix 4
    • Critical Values of Student’s t Distribution
  • Appendix 5
    • Critical Values of the F-Statistic
  • Appendix 6
    • Critical Value for P (Pcrit), Tukey’s HSD Test
  • Appendix 7
    • Critical Values for Spearman’s Rank-Order Correlation Coefficient
  • Appendix 8
    • Fisher r-to-Z* Transformation
  • Glossary
  • I n d e x