Chapter 12Choosing an Appropriate Statistical TestiStockph.docx

Chapter 12
Choosing an Appropriate Statistical Test
iStockphoto/ThinkstockLearning Objectives
After reading this chapter, you will be able to. . .
·
understand the importance of using the proper statistical analysi
s.
· identify the type of analysis based on four critical questions.
· use the decision tree to identify the correct statistical test.
Here we are in the final chapter that will pull all prior chapters t
ogether. Chapters 1 to 3 discussed descriptive statistics while th
e latterchapters, 4 to 11, discussed inferential statistics. Each of
the inferential chapters presented a statistical concept then con
ducted the appropriateanalysis to be able to test a hypothesis. T
he big question for students learning statistics is, "How do I kno
w if I'm using the correct statisticaltest?" For experienced statis
ticians this question is easy to answer as it is based on a few cri
teria. However, to a student just learning statisticsor to the novi
ce researcher, this question is a legitimate one. Many statistical
reference texts include a guide that asks specific questionsregar
ding the type of research question, design, number and scales of
measurement of variables, and statistical assumption of the dat
a thatallows you to use an elegant chart known as a decision tre
e. Based on the answers to these questions, the decision tree is u
sed to helpdetermine the type of analysis to be used for the rese
arch, thereby helping you answer this big question.
12.1 Considerations
To make the correct decisions based on the use of a decision tre
e, there are four specific questions that must be answered. Thes
e questions areas follows:
· What is your overarching research question?

·
How many independent, dependent, and covariate variables are
used in the study?
· What are the scales of measurement of each of your variables?
· Are there violations of statistical assumptions?
If you are able to answer these specific questions, then you will
be able to determine the proper analysis for your study. These q
uestions arecritically important, and if they cannot be answered,
then not enough thought has gone into the research. That said, l
et us discuss each ofthese questions so that they can be consider
ed and answered in the use of the decision tree.
What Is Your Overarching Research Question?Try It!
Derive your ownresearch question foryour Master's Thesisor Do
ctoralDissertation. Have a colleague orprofessor read it. What a
re theirthoughts or suggestions forimprovements?
Answering this question seems simple enough as all research ha
s an overarching research questionthat drives the study, especial
ly since this dictates the type of quantitative methodology. Ther
e arekey words in every research question that help determine th
e appropriate type of analysis. Forinstance, if the research quest
ion states, "What are the effects of job satisfaction on employee
productivity?" the keyword is "effects" as in the cause and effec
t of job satisfaction (theindependent variable) on productivity (t
he dependent variable). We know that to establish causeand effe
ct we would need to have at least two groups, the experimental
group and the controlgroup, and measure their levels of producti
vity at least two times based on an intervention ortraining progr
am that purports to promote increased job satisfaction. If there a
re significantdifferences in the two groups, before and after the
experiment, then a cause and effect has beensupported. As descr
ibed in Chapters 5 through 8, you know that differences betwee
n or withingroups involve either t-tests or ANOVAs.
Another research question may be, "What is the degree of associ
ation between levels of drugabuse and depression?" Here the op

erative word is "association" meaning a relationship orcorrelatio
n must be executed. Correlations are discussed at length in Chap
ter 9 and involve different types of relationships. Taking this re
searchquestion a step further, "Do higher depression scores pred
ict increased drug abuse?" Here the key word is "predict," whic
h is synonymous withregression. This statistical concept is disc
ussed in Chapter 10.
How Many Independent, Dependent, and Covariate Variables Ar
e Used?
This seems like an obvious question to answer in terms of the in
dependent and dependent variables, but it can get quite complex
whenadditional covariate variables are added. Covariates are ad
ditional variables that may have an influence on the independent
and dependentvariable relationship. These variables may be add
ed intentionally by the researcher or may be detected and contro
lled for as confoundingvariables (discussed in Chapter 8). The p
oint here is that if there are additional variables, then best resea
rch practices dictate that they must beadded to the analysis for i
mproved statistical conclusion validity of the research as in a re
gression or ANCOVA. Additionally, we know fromChapter 8 th
at if we have more than one dependent variable, then we cannot
run a regression or ANOVA, but instead MANOVAs andMANC
OVAs are appropriate. Last, when the independent variable is ca
tegorical (nominal or ordinal data) and the dependent variable is
continuous (interval or ratio), the number of groups in a betwee
n-group design or the treatment or times in a within-
group design willdetermine the methodology to be employed. T
herefore, the number of groups, times, or treatments will need t
o be known for the decisiontree.
What Are the Scales of Measurement of Each of Your Variables
?
Recall that the scales of measurement (nominal, ordinal, interva
l, and ratio) were discussed in Chapter 1. The relevance of discu
ssing thesedifferences in the type of data is that different analys

es are dictated by the type of data or scale of measurement. For
instance, we now knowthat if we are working with nominal data,
then a Pearson's correlation is not appropriate; instead, a Pears
on chi-
square test of independence isnecessary. Similarly, if we have c
ategorical data (nominal or ordinal data) for our independent va
riable with a continuous (interval or ratio)dependent variable, th
en a t-
test or factorial design is appropriate. If we know that both vari
ables are continuous, then a correlation orregression is appropri
ate. All of these statistical analyses were previously discussed a
nd should be familiar to you.
Are There Violations of Statistical Assumptions?Try It!
Identify the numberof variables in yourstudy, their scales ofmea
surement, andthe violations of assumptions ifthere are any. Can
you identifythe proper statistical analysis?(Hint: If you are havi
ng a difficulttime, use the decision treesprovided below.)
Throughout the previous chapters, we discussed both parametric
and nonparametric analysisbased on satisfying assumptions of t
esting. We reviewed the various tests of assumptions thatinclud
e normality, homogeneity of variance, sphericity, and linearity.
Normality is based on theGaussian distribution where significan
t deviation away from a normal distribution is a violation.This
may be attributed to small sample sizes, outliers, or both. Homo
geneity of variance violationsare significant differences in the v
ariances between groups. This can be problematic since theseflu
ctuations should be homogenous between groups. Sphericity is a
nother assumption similar tohomogeneity of variance, but this te
st of assumptions is within pairs of times/treatments. In otherwo
rds, significant differences in pairs of treatments should not occ
ur. Last, linearity must also betested as issues of significant cur
vilinearity can lead to erroneous statistical findings. In the end,
ifany or all of these are violated, then a nonparametric test is th
e proper option.

12.2 Decision Trees
Whether your research question focuses on the effects of the IV
on the DV (comparison groups), the relationships of the IV and
the DV(association between variables), or the prediction of the
DV based on the IV (prediction) you can determine the proper st
atistical analysis usingthe three statistical decision trees discuss
ed next. All inquiries start at the root of the tree, which is locat
ed at the top of the decision tree orflow chart, and then evaluate
the various branches depending upon what the research involve
s. The goal is to find the appropriate statisticaltest based on the
research question, the number of DVs and IVs, and the type of
DVs and IVs. Thus, to use each of the decision treessuccessfull
y, you must correctly identify the following items:
· The number and types of DVs and IVs.
· The number of groups or treatments included.
· Whether the data is normally distributed.
· If there are any covariates.
Recall from Chapter 1 that categorical variables have one or mo
re categories (such as ethnic groups), continuous variables allo
w for an possiblevalue and are not limited to whole numbers (ti
me to complete a problem), and dichotomous variables allow for
only two categories or levels(male or female). Figure 12.1 pres
ents a decision tree for research questions that involve comparin
g groups.
Figure 12.1: Statistical decision tree for the effects of IVs on D
Vs
Try It!
A A researchpsychologist wishes totest the effects of twobrands
ofantidepressant drugs bymeasuring the level ofdepression on t
wo test groupsand a control group using theBeck Depression Inv
entory.Determine the proper statisticalanalysis.
See answer here.
The following provides a basic example of how to use Figure 12
.1. For this example our goal is toexplore the differences betwee
n IQ test scores in males and females. Our IV is gender, and our

DVis IQ test scores. Based on this information, we know we on
ly have one IV (gender) and one DV(IQ test scores). For this ex
ample, gender is a categorical variable, so we evaluate the items
in thedecision tree and determine that we must identify the num
ber of IV groups. Since we have twogroups (male and female) a
nd no treatments, we would then need to consider whether oursa
mple means are non-
normally distributed (ND) or nonnormally distributed (NND). B
ecause IQscores generally have a mean of 100 and a standard de
viation of 15, IQ scores are considerednormally distributed. Sin
ce we did not include any covariates, the decision tree indicates
that anindependent t-
test is the appropriate statistical analysis. Now let's consider an
example usingFigure 12.2.
Figure 12.2: Statistical decision tree for the relationship among
IVsand DVs
Try It!
B A market researchanalyst wants toexamine consumerpurchasi
ng behaviors,based on gender differences, forthree different typ
es of mustard.Determine the proper statisticalanalysis.
See answer here.
Similar to our previous example, let's say we now want to invest
igate whether there is arelationship between IQ test scores and i
ndividuals' GPAs. In this example, IQ scores are the IVand GPA
s are the DV. Since we know GPA is a continuous variable and I
Q is a continuous variable,and that the sample means are normal
ly distributed, we would select the Pearson Correlation.Now let'
s consider an example using Figure 12.3.
Figure 12.3: Statistical decision tree forpredicting DVs based on
IVst!
C A forensicpsychologist wants toexplore the effect ofseveral pr
edictors(i.e., intelligence, levels ofdepression, race, and persona
litytype) on incidences of crimeacross state prisons. Determinet
he proper statistical analysis.

See answer here.
Say we want to investigate whether socioeconomic status (SES)
and IQ are predictive of GPAs. Inthis case, SES and IQ are the
predictors of independent variables and GPA is the dependentva
riable. As we know from the previous examples, GPA is a conti
nuous variable. Since we havetwo IVs, one that is continuous (I
Q) and one that is categorical (SES), we see that the mostapprop
riate statistical test is the Multiple Regression.
Summary
The chapter demonstrated the importance of performing the pro
per statistical analysis to ensure statistical conclusion validity (
Objective 1). Theproper analysis can be identified by answering
four key questions, (1) What is your overarching research quest
ion? (2) How many independent,dependent, and covariate variab
les are used in the study? (3) What are the scales of measureme
nt of your variables? and (4) Are thereviolations of statistical as
sumptions? (Objective 2) Based on these questions, the decision
tree can be used to identify the proper test, therebyavoiding crit
ical analysis mistakes for the student and novice researcher (Obj
ective 3).
In the end, the authors wish each student and statistical learner t
he very best. Happy analyzing!Key Terms
Click on the key term to see the definition.
decision treeChapter Exercises
Answers to Try It! Questions
The answers to all Try It! questions introduced in this chapter a
re provided below:
A. One-way ANOVA
B. Chi-square test of independence; Cramér's V
C. Multiple regression
Review Questions
The answers to the odd-
numbered items can be found in the answers appendix.
1.
Working backwards with the decision trees, a dependent/repeate
d t-test will involve how many IVs and DVs?

2. Working backwards with the decision trees, a chi-
square test of
independence will involve what scales of measurement of thevar
iables?
3.
If a psychologist wanted to evaluate the effects of three therapy
sessions on her patients' level of depression, what analysis woul
d sherun?
4.
Using the decision trees, what is the main difference between pe
rforming a Cramér's V and executing a phi coefficient?
5.
If a researcher wishes to explore covariates and their influence
on a single IV and DV, what analysis will he perform?
6.
What is the main difference between performing an analysis of r
elationships and conducting an analysis of effects between twov
ariables?
7.
A student is working with two continuous variables that are not
normally distributed. What analysis should she perform?
Smart Lab Study
2
Smart Lab Study
Mary Garcia
RES 5400 Understanding, Interpreting & Applying Statistical
Concepts
Instructor: Kari Terzino
August 8, 2019

Population is the entire group of people, objects or events which
one wants to research. However, it not always feasible or
possible to study all member of a population. Therefore, it is
more effective to study a subset of the population. A sample is a
subset of a group of people, objects or events from a larger
population that one collects and study to make a conclusion. To
ensure that the population is adequately represented the sample
data must be random and must include a large amount of the
population being study. An example, one could count the
number of children with strawberry birthmarks in a random
sample then use a hypothesis test to estimate the percentage of
all children with strawberry birthmarks.
There are a number of different methods of random sampling.
These include simple, stratified, cluster, systemic, and multi-
stage sampling. The first type of random sampling is called
simple sampling Simple sampling is the most basic of the
random sampling. With simple sampling everyone has an equal
chance of being part of the sample. The 2nd method of random
sampling is stratified sampling. Stratified sampling is when the
population is divided into subgroups that do not overlapped.
Theses non-overlapping subgroup called strata. Simple random
samples are chosen from each stratum. Strata are group that are
similar due to some of the characteristic of the group members.

The next method of random sampling is cluster sampling. With
cluster sampling the populate is divided I into strata called
cluster. Clusters are randomly select and them all the
individuals in the cluster are contained in the sample. An
example of cluster sampling would be to select a sample of
student to answer a survey on the policy of late assignment. One
way to accomplish this would be to randomly select 4 classes
during the winter session. Surveying all the student in the
classes would be cluster sampling.
The next method of random sampling is systematic sample.
Systematic sample selects a random starting point from the
population. Then a sample is taken from a regular set interval of
the population. An example would be a local Head Start
program wants to form a systematic sample of 400 volunteers
from a population of 4000. By selecting of every 10th person in
the population would be a systematic sample. The final of
random sampling is Multi-stage sampling. Multi-stage sampling
can include of different stages of one kind of sampling. It can
however it can be a combination of sampling methods.
The non-random sample are convenience sampling and
volunteer sampling. Convenience sample is using research that
is already available. An example would be the researcher asking
people at the mall taking a poll. Volunteer sample is based on
individual’s decision to volunteer or not.
The next top this paper will explore is variables and
measurement. Variables is what you want to study. Anything
can be considered as a variable. A simple way to look at
variable is by their qualitative or quantitative properties.
Qualitative data is the conclusion of describing or classifying
qualities of a population that is neither measure nor counted.
Whereas, quantitative data is conclusion base on counting or
measuring qualities of a population.
The two main variable in a research is independent and
dependent. An independent variable is the variable that is
controlled or changes in research to test the properties of the

dependent variable. The dependent variable is the one being
tested a measure in the research. In other words, the dependent
relies on the independent variable.
The different kinds of variables are measure in different ways.
There are four scales of measurement. These four scales of
measurement are: nominal scales, ordinal scales, interval scales
and ratio scales. Nominal scales have no quantitative value and
are used for labeling variables. Ordinal scales the important
thing is the order of the values. However, the difference
between each value is not known. Interval scales are numeric
both the order and the exact different between the values are
known. Ratio scales tell the order, the exact value between the
elements and they have an absolute zero.
References:
Carruthers, M. W., Maggard, M. (2019). Smart Lab: A Statistics
Primer. San Diego, CA: Bridgepoint Education, Inc.

Research Question FOR WEEK ONE
Background
During this week you will brainstorm a list of research
questions you are interested in, which will help you work
towards your Week 1 Assignment. You are working towards
creating a list of at least 10 unique research questions that
encompass a variety of topics and types of variables. Think
about exploring relationships between variables, making
predictions for one variable using one or more other variables,
and determining differences between groups across one or two
variables. In future weeks, you will pull questions from this list
that might lend themselves to a particular statistical analysis,
thus saving valuable time in not needing to brainstorm research
ideas. During those weeks you will take the research question
and create a mini-research proposal that will help you consider
the application of a specific statistical analysis to that question.
Discussion Assignment Requirements
Initial Posting - To earn full participation points, include in
your initial posting at least 5 potential research questions by
Day 3. Have fun with these questions and choose topics you are
truly interested in, whether they are leadership, training, sports,
social media, politics, movies, or food. This will make the
research design process much more enjoyable. If you need help
coming up with ideas, ask your instructor for examples. Also,
feel free to post more than 5 research questions as it would be
useful to get feedback on as many questions as possible.
For each of the questions, provide the following:
· List the research question (be sure to phrase as a measurable
question)
· Identify the variables presented in the question
· Provide an operational definition for each variable

· Describe each variable’s scale of measurement (nominal,
ordinal, interval, or ratio) and characteristics (i.e., discrete vs.
continuous, numerical vs. categorical, etc.)
Replies - Though you may respond to your peers multiple times
during the week to provide support or feedback, students are
required to respond substantively to at least two of their
classmates’ postings by
ANSWER FOR DISCUSSION WEEK 1
Research discussion
Research questions one: How does leadership style affect
organizational performance?
In this research question, the independent variables are
leadership style, while the dependent variable is organizational
performance (Sukal, 2019). Leadership styles are techniques
used by organizations to run their activities to achieve their
objectives. Besides, organizational performance entails various
achievements of an entity that are accrued from its business
operations. An ordinal scale of measurement can be used in this
case.
Research questions two: Effects of technology on students'
performance?
In the case, technology is the independent variable while
students' performance is the dependent variable. Technology in
education is scientific knowledge used to improve the level of
education (Sukal, 2019). Student performance refers to how
students carry out their studies. An ordinal scale of
measurement is appropriate to measure how technology affects
students' performance.
Research questions Three: what are the effects of smoking on
human health?
Smoking is the independent variable, while human health is the
dependent variable. Smoking is the inhalation of tobacco
products, while human health is the well-being of the human
condition (Carruthers & Maggard, 2019). An ordinal scale of
measurement is used in this case.

Research questions four: Effects of training on employee
performance?
Training is the independent variable, while employee
performance is the dependent variable (Carruthers & Maggard,
2019). Training involves equipping employees with the
knowledge to perform their duties appropriately. Employee
performance is the output that is accrued from different
activities. An ordinal scale is used in this research question.
Research questions five: How does management styles affect
employee performance?
Management styles are the independent variable, while
employee performance is the dependent (Carruthers & Maggard,
2019). Management styles are techniques used by the
management to run business activities while employee
performance is output accrued from employees' actions. An
ordinal scale is used in this research question.
References
Carruthers, M. W., Maggard, M. (2019). Smart Lab: A Statistics
Primer. San Diego, CA: Bridge point Education, Inc.
Sukal, M. (2019). Research methods: Applying statistics in
research. San Diego, CA: Bridge point Education, Inc.
PROFFESSOR RESPOND:
Interesting questions!
Please be sure to include operational definitions of your DVs -
i.e. employee performance. How would you measure it? It
might be helpful to review the operational definition
announcement in the course. Remember, we need to include
enough detail about our methodology and variables so that
anyone could replicate our work.
Andreas Rentz/Getty Images

chapter 4
Applying z to Groups
Learning Objectives
After reading this chapter, you will be able to. . .
1. describe the distribution of sample means.
2. explain the central limit theorem.
3. analyze the relationship between sample size and confidence
in normality.
4. calculate and explain z-test results.
5. explain statistical significance.
6. calculate and explain confidence intervals.
7. explain how decision errors can affect statistical analysis.
8. calculate the z-test using Excel.
9. present results and draw conclusions based on z-tests.
10. interpret results of z-tests in APA format.
CN
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suk85842_04_c04.indd 103 10/23/13 1:16 PM
CHAPTER 4Section 4.1 The Distribution of Sample Means
Chapter 3 ended by noting that as interesting as it is to be able
to determine the percent-age of individuals below or above a
point or between two scores, we are more often
interested in groups than in individuals. A researcher is more
likely to investigate the
probability that a group of clients a psychologist has been
working with will score below
some point on a depression scale. In this chapter, what you have
learned in the first three
chapters will be applied to analyses of groups.
In Chapters 2 and 3, we discussed that many characteristics that
interest behavioral sci-
entists are normally distributed in a population. But, by
inference, that also means that
some characteristics are probably not normally distributed, and
in the moment it may not
be clear which are not. Relying on Table A in the Appendix to
reveal the proportions of the
entire population that fall in certain areas is appropriate only if
the data is normal; Table
A assumes a normal distribution.
4.1 The Distribution of Sample Means

So where do we go if we are suspicious about the normality of
the data? The answer is distribution of sample means, a
distribution made of the means of samples rather
than individual scores.
To this point, population has meant populations created by
sampling one subject at a time,
measuring each individual on some trait, and then plotting each
score in a frequency dis-
tribution. Consider an alternative. What if instead of selecting
each individual in a popu-
lation one-at-a-time, an analyst
1. selects a group with a specified size,
2. calculates the sample mean (M) for each group, and then
3. plots M (rather than the individual scores) in a frequency
distribution, and
4. continues doing this until the population is exhausted.
How would that affect the distribution? Would it still be a
population? The answer to the
second question is yes, it is still a population. Recall that, by
definition, a population is all
members of a defined group. Whether the members of the
population are measured indi-
vidually or as members of a group is incidental, as long as they
are all included.
If researchers want to know how dogmatic registered voters in
Brazos County, Texas,
are, they can measure each voter and then record the mean level
of dogmatism for each
group of 30. If the mean for groups of 30 are recorded until the
population is exhausted,
the result is still a population.

The Central Limit Theorem
The first question—how would the distribution be affected?—is
a little more involved, but
it is very important to nearly everything we do in statistical
analysis. The answer requires
introducing the central limit theorem, which holds that
• if a population is sampled an infinite number of times using
sample size n and
• the mean (M) of each sample is determined,
H1
TX_DC
BLF
TX
BL
BLL
suk85842_04_c04.indd 104 10/23/13 1:16 PM
• the multiple Ms will take on the characteristics of a normal
distribution whether or
not the original population of individuals is normal.
Take a minute to absorb this. A population of an infinite

number of sample means drawn
from one population will reflect a normal distribution whatever
the nature of the original
distribution. A healthy skepticism prompts at least two
questions:
1. How would we know whether this is true given that an
infinite number of
samples is out of everyone’s reach?
2. How can sampling in groups rather than as individuals affect
normality?
Although prove is too strong a word, we can at least provide
evidence for the effect of the
central limit theorem with an example. Perhaps a psychologist
is working with 10 people
who are very resistant to change; they are highly dogmatic.
Technically, because 10 is the
number in the entire group, the population is N 5 10 (the
uppercase N signifies the popu-
lation). A small population does not change the fact that there
still cannot be an infinite
number of samples, of course, but for the sake of the illustration
let us assume that
• dogmatism scores are available for each of the 10 people;
• the data is interval scale;
• the scores range from 1 to 10; and
• each person receives a different score.
So with N 5 10, the scores are
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Figure 4.1 is a frequency distribution of those 10 scores.

Figure 4.1: A frequency distribution for the scores 1–10, with
each score
occurring once
S
c
o
re
F
re
q
u
e
n
c
y
Score Values
10
9
8
7
6
5

4
3
2
1
10987654321
suk85842_04_c04.indd 105 10/23/13 1:17 PM
The distribution in Figure 4.1 is many things, perhaps, but it is
not normal. With
range 5 10 2 1 5 9 and s 5 3.028 (a calculation worth checking),
it is extremely platykur-
tic, with no apparent mode (or 10 modes). We can illustrate the
workings of the central
limit theorem with the following procedure:
a. We will use samples of just n 5 2.
b. Rather than an infinite number of samples, we will make the
example
manageable by using one sample for each possible combination
of scores in
samples of n 5 2 from the population.
All the possible combinations of two scores from values 1–10
are listed in Table 4.1. There
are 90 possible combinations of the 10 dogmatism scores.

Table 4.1: All possible combinations of the integers 1–10
1, 2 2, 1 3, 1 4, 1 5, 1 6, 1 7, 1 8, 1 9, 1 10, 1
1, 3 2, 3 3, 2 4, 2 5, 2 6, 2 7, 2 8, 2 9, 2 10, 2
1, 4 2, 4 3, 4 4, 3 5, 3 6, 3 7, 3 8, 3 9, 3 10, 3
1, 5 2, 5 3, 5 4, 5 5, 4 6, 4 7, 4 8, 4 9, 4 10, 4
1, 6 2, 6 3, 6 4, 6 5, 6 6, 5 7, 5 8, 5 9, 5 10, 5
1, 7 2, 7 3, 7 4, 7 5, 7 6, 7 7, 6 8, 6 9, 6 10, 6
1, 8 2, 8 3, 8 4, 8 5, 8 6, 8 7, 8 8, 7 9, 7 10, 7
1, 9 2, 9 3, 9 4, 9 5, 9 6, 9 7, 9 8, 9 9, 8 10, 8
1, 10 2, 10 3, 10 4, 10 5, 10 6, 10 7, 10 8, 10 9, 10 10, 9
If we calculate a mean and plot the value in a frequency
distribution as a test of the central
limit theorem for each possible pair of scores, the result is
Figure 4.2. Because the entire
distribution is based on sample means, Figure 4.2 is a
distribution of sample means.
suk85842_04_c04.indd 106 10/23/13 1:17 PM
Figure 4.2: A frequency distribution of the means of all possible
pairs of
scores 1–10

The Mean of the Distribution of Sample Means
The symbol used for a population mean to this point, m, is
actually the symbol for a
population mean formed from one score at a time. To
distinguish between the mean of
the population of individual scores and the mean of the
population of sample means,
we will subscript m with an M: mM. This symbol indicates a
population mean based on
sample means.
With a distribution of just 90 sample means, this is nothing like
an infinite number, of
course, but the resulting figure is instructive nevertheless.
• The mean of the scores 1–10 is 5.5: m 5 5.5.
Study Figure 4.2 for a moment. What is the mean of that
distribution?
• The mean of that distribution of sample means is also 5.5: mM
5 5.5.
The point is this: When the same data is used to create two
distributions, one a population
based on individual scores and the other a distribution of
sample means, the two popula-
tion means will have the same value,
• m 5 mM.
Describing the distribution as “normal” is a stretch, but Figure
4.2 is certainly much more
like a normal distribution than Figure 4.1 is. For one thing,

rather than the perfectly flat
distribution that occurs when all the scores have the same
frequency, mean scores near
S
c
o
re
F
re
q
u
e
n
c
y
Sample Means
10
9
8
7
6
5
4

3
2
1
6.05.55.04.54.03.53.02.5 9.08.58.07.57.0 9.56.52.01.5
suk85842_04_c04.indd 107 10/23/13 1:17 PM
A Why is there
less variability
in the distribution of
sample means than in a
distribution of individual
scores?
Try It!
the middle of the distribution in Figure 4.2 occur more
frequently than
means at the extreme right or left. Why are extreme scores less
likely
than scores near the middle of the distribution? It is because
many
combinations of scores can produce the mean values in the
middle of
the distribution, but comparatively few combinations can
produce the
values in the tails. With repetitive sampling, the mean scores
that can
be produced by multiple combinations increase in frequency and

the
more extreme scores occur only occasionally; this tendency is
illus-
trated in the next section.
Variability in the Distribution of Sample Means
In the original distribution of 10 scores (Figure 4.1), what is the
probability that someone
could randomly select one score (x) that happens to have a
value of 1? Because there are 10
scores, and just one score of 1, the probability is p 5 1/10 5 .1,
right? For the same reason,
what is the probability of selecting x 5 10? It is the same, p 5
.1.
Now, moving to the distribution based on the 90 scores, what is
the probability of select-
ing a sample of n 5 2 that will have M 5 1.0? Is there any
probability of selecting two scores
out of the 10 that will have M 5 1.0? Because there is only a
single 1, the answer is no. As
soon as a score of 1 is averaged with any other score in the
group, M . 1 because all other
scores are greater than 1. That’s why the lowest possible mean
score in Figure 4.2 is 1.5,
which can occur only when 1 and 2 are in the same sample.
The same thing occurs in the upper end of the distribution. The
probability of selecting a
group of n 5 2 with M 5 10 is also zero (p 5 0) because all other
scores are lower than 10.
In the 90 possible combinations, the highest possible mean
score is 9.5, which can occur
only when the 10 and the 9 happen to be in the same sample.
There will always be less

variability in a distribution of sample means than in a
distribution of scores sampled one
at a time. As the size of the sample increases, the impact of the
most extreme scores like-
wise diminishes as they are included in samples with less
extreme scores.
The Standard Error of the Mean
The sigma, s, which indicates a population standard deviation,
is specific to a population
based on individual scores. The symbol for the standard
deviation of the sample means is
sM. The formal name for this value is standard error of the
mean.
Note the difference between the language of statistics and
everyday language. The error
part in “standard error of the mean” has nothing to do with
making a mistake. In statistics,
there are actually several different kinds of “standard errors,”
and they all have one thing
in common: They are all measures of data variability. The size
of the statistic indicates the
amount of variability in whatever the particular standard error is
gauging.
Earlier in this chapter, we noted that whether it is the
distribution of individual scores or
the distribution of sample means, the means of the two
distributions will always be equal,
m 5 mM. Does the same hold true for the measures of
variability? Will s 5 sM? Actually,
we answered that question when noting that there is always
more variability in the distri-
bution of individual scores than in the distribution of sample

means. Put more succinctly,
s . sM. We can check this conclusion with our data.
suk85842_04_c04.indd 108 10/23/13 1:17 PM
The standard deviation of the 10 original scores (1, 2, 3, 4, 5, 6,
7, 8, 9, 10) is
s 5 2.872
It is a lot of data to enter, but it is a good idea to check this. For
this calculation and for the
calculation of the standard error of the mean that follows, use
the formula for the popula-
tion rather than the sample standard deviation (N rather than n 2
1 in the denominator).
Elsewhere in this presentation, it will always be n 2 1. Here we
get into the degrees of
freedom (df ) as discussed in Chapter 1, which is a theoretical
and mathematical adjust-
ment for the use of samples. Keep in mind that if we want to
look at a population param-
eter, we would use N or the population size, but because we are
using samples, the adjust-
ment would be the sample size minus 1, or (n 2 1).
The standard error of the mean can be calculated by taking the
standard deviation of the
mean scores of each of those 90 samples from which the
distribution of sample means was
constituted. It is a little laborious, and happily this is not a
pattern that must be followed

later, but the value is
sM 5 1.915
This too is a parameter, so it involves the N rather than n 2 1 in
the denominator. As
predicted, s has a larger value than sM. It reflects the
moderating influence that the less
extreme scores have on the more extreme scores when they
occur in the same sample.
Sampling Error
Although the standard error of the mean does not refer to a
mistake per se, another kind
of error, the sampling error, does refer to a mistake in sampling
that causes an error. In
inferential statistics, samples are important for what they reveal
about populations. This is
effective only when the sample accurately represents the
population. The degree to which
the sample does not represent the population is the degree of
sampling error.
Samples tend to accurately reflect the population when two
important prerequisites are
satisfied:
• the sample must be relatively large, and
• the sample must be based on random selection.
The safety of large samples is explained by the law of large
numbers. According to this
mathematical principle, errors diminish as a proportion of the
whole as the number of
data points increases. The potential for serious sampling error

diminishes as the size of
the sample grows.
Random selection refers to a situation where every member of
the population has an
equal probability of being selected. A random sample of n 5 5
could be created from the
10 people being treated for dogmatic behavior by
• assigning each person a number,
• placing the 10 numbers into a paper bag,
suk85842_04_c04.indd 109 10/23/13 1:17 PM
CHAPTER 4Section 4.2 The z-Test
• shaking the bag well, and
• without looking, drawing out five numbers.
The result would be a randomly selected sample. When they are
randomly selected,
samples differ from populations only by chance. Such
randomization of subjects can
be achieved through a random number generator (e.g., in SPSS)
used for experimental
purposes.
If the sample should fail to capture some important
characteristic of the population other
than its size, there is a sampling-error problem. The important
characteristic might be the
mean, for example, and if M ? m, this indicates a sampling
error. Actually, there is always
some sampling error because a sample can never exactly

duplicate all the descriptive
characteristics of the population, but the sampling error will
usually be minor if samples
are relatively large and randomly selected. Statistical analysis
procedures tolerate minor,
random sampling error, but systematic sampling error is another
matter. Systematic sam-
pling error occurs when the same mistake is made time after
time.
In 1936, the publishers of The Literary Digest, a prominent
publication of the time, decided
to predict the outcome of that year’s presidential election in the
United States. To ensure
that the sample size would not be a problem, they sent out
millions of postcards to reg-
istered voters. It would seem that they at least met the
requirement for a relatively large
sample, because the Harris and Gallup organizations typically
get very accurate results
with a few thousand, and sometimes just a few hundred,
responses. Fatefully, they
decided to use telephone books and automobile registrations to
locate those who would
be polled. Consider the historical setting. At the height of the
Great Depression, voters
were identified by two indicators of relative prosperity: a
telephone in the home and a
currently registered car. The study was disastrous for the
magazine’s reputation. This
misprediction was directly challenged by George Gallup
(founder of the American Institute
of Public Opinion), who in fact predicted that FDR would win
(using quota sampling) and
that The Literary Digest poll was false. The results indicated
that Alf Landon would win,

but of course Franklin D. Roosevelt was elected in a landslide
to a second term, carrying
every state in the union except Maine and Vermont. Since
Gallup was correct, the Gallup
poll gained credibility and went on to become one of the most
recognized and used polling
systems in public opinion polling.
The problem encountered by The Literary Digest poll was
systematic sampling error. The
voters were consistently and nonrandomly selected from groups
not representative of the
entire population. If they had been randomly selected, chances
are that with the large
sample size, the study would have predicted the election results
very accurately, but the
sample size alone was not enough to salvage the effort.
4.2 The z-Test
To summarize, the distribution of sample means is a distribution
based not on indi-vidual scores but on the means of samples of
the same size repeatedly drawn from a
population. The central limit theorem indicates that when a
population is based on sam-
ples rather than individual scores, the resulting population will
be normal regardless of
how the population of individual scores was distributed.
suk85842_04_c04.indd 110 10/23/13 1:17 PM
Because the central limit theorem provides assurance of a

normal distribution, if the z
score formula from Chapter 3 is adjusted to accommodate
groups rather than individual
scores, Table A will answer all the same questions about groups
that were initially asked
about individuals in Chapter 3. Recall that the z score formula
(3.1) had the following
form:
z 5
x 2 M
s
If the following substitutions are made:
M for x so that the focus is on the mean of the group rather than
the individual,
mM for M to shift from the sample mean to the mean of the
distribution of sample
means, and
sM for s so that the measure of variability is for the distribution
rather than the
sample, the result is the z-test:
z 5
M 2 mM
sM
Formula 4.1
The z-test produces a z value for groups rather than individual
scores which indicates
how distant a particular sample mean is from the mean of the

distribution of sample
means. The procedure is the same as it was for individual
scores: Calculate a value of z
and then use the table to interpret that value.
Note the similarities between Formula 3.1 and Formula 4.1:
• Both formulas produce values of z.
• Both numerators call for subtractions that result in difference
scores.
• Both denominators measure data variability.
Calculating the z-Test
When calculating z scores, as you saw in Chapter 3, everything
that is needed (x, M, and
s) can be determined from the sample:
z 5
x 2 M
s
What is needed for the z-test, however, is often not as easy to
determine. Because mM 5 m,
one of those two parameters must be provided. The standard
error of the mean (sM) can
also be a problem. No one wishing to complete a z-test is going
to have the mean scores
for that infinite number of samples that make up the distribution
of sample means. So
calculating the standard deviation of those means, which is what
the standard error of
the mean represents, is not an option. Nevertheless, there is a
way to determine this value
that lets us skip the tedium of calculating a standard deviation

for who-knows-how-many
scores. If sM is not given (“And the standard error of the mean
is . . .”), but the population
standard deviation (s) is provided, sM can be found as follows:
sM 5
s
"N
Formula 4.2
suk85842_04_c04.indd 111 10/23/13 1:17 PM
Where
sM 5 the standard error of the mean
s 5 the population standard deviation
N 5 the number in the group
So for a group of 100 with the value of s as 15, then sM is
sM 5
s
"N
sM 5
15
"100

5
15
10
5 1.5
This is only a partial solution, however, because it still requires
at least s .
You will learn a way around the problem of mM in Chapter 5,
but in the meantime, let us
take the following example. A marriage and family counselor
has access to some national
data on the frequency of negative verbal comments exchanged
between divorcing cou-
ples. The counselor finds that
• couples in troubled marriages tend to have 11 negative
exchanges per week, with
a standard deviation of 4.755, and
• a study of 45 couples who have filed for divorce in the
counselor’s county reveals
that the mean number of negative comments per week is 12.865.
• Given the national data, the counselor wants to know the
probability that a ran-
domly selected group of couples from that population will have
as many nega-
tive exchanges as the counselors’ clients, or more.
Although the question is about groups rather than individuals,
the problem is much like
a z score problem. Here is the information that is available:

m and therefore mM 5 11.0
s 5 4.755
N 5 45
M 5 12.865
1. The standard error of the mean is
sM 5
s
"N
5
4.755
"45
5 0.709
2. And z is
z 5
M 2 mM
sM
5
12.865 2 11.0
0.798
5 2.630
suk85842_04_c04.indd 112 10/23/13 1:17 PM

Comparing M to mM indicates that the counselor’s group has a
higher number of negative
verbal exchanges per week than the number nationally among
couples with troubled mar-
riages: 12.865 is a higher value than 11.0. What else can be
determined from the analysis?
Interpreting the Value from the z-Test
This is a value of z just like those that were calculated in
Chapter 3, except that the value
indicates how much a sample mean (M) differs from the mean of
a population of samples
(mM) rather than how an individual (x) differs from either a
sample mean (M) or a popula-
tion mean (m). The Table A value indicates that 0.4957 out of
0.5 occurs between a value
for z 5 2.63 and the mean of the distribution.
• So among the population of couples with troubled marriages,
49.57% will have
negative verbal exchanges somewhere between the level of this
group (12.865 per
week) and the mean of the population (11.0) per week.
• But the question is the probability that a group of clients
selected at random
would have 12.865 negative comments per week, or more.
• Because 49.57% will have 12.865 or fewer negative exchanges
per week, just
0.43% (50% 2 49.57%) will have 12.865 negative comments per

week or more.
Stated as a probability, p 5 .0043, a group of individuals in
troubled marriages
will have 12.865 negative exchanges per week or more.
This result is depicted in Figure 4.3.
Figure 4.3: The probability of selecting a sample with M 5
12.865 or
higher from a population with MM 5 11.0
The probability of selecting a sample with M 5 12.865 or higher
is indicated by determin-
ing the z equivalent of a sample with M 5 12.865 and then
determining the proportion of
the distribution at that point and higher in the population.
There are some important differences between this z and those
calculated in Chapter 3.
p = 0.5 p = 0.4957 p = 0.0043
0 z = 2.63
z-value
suk85842_04_c04.indd 113 10/23/13 1:17 PM
• Note that the difference between the mean of the population
(mM 5 11.0) and the
sample mean (M 5 12.865) is really quite modest, but the z
value (z 5 2.630) is

comparatively extreme. Recall that 62z includes 95% of the
distribution, and at
z 5 2.630 we are substantially beyond that.
• The reason for the rather large value of z is the quite small
standard error of the
mean, 0.709.
• Because variability between group means tends to be small
relative to the vari-
ability between individuals, it does not take much of a
difference between the
sample mean (M) and the mean of the distribution of sample
means (mM) to pro-
duce an extreme value of z.
Apply It!
Confidence in the Claim
A parent is looking at private high schools for his child. A
particular high school
claims that last year their students performed above average in
math and ver-
bal SAT scores. The parent, who knows statistics, decides to
test this claim. The parent finds
the nationwide results for last year’s SAT scores. The mean
math SAT score was 520, with a
standard deviation of 110. The mean verbal score was 508, with
a standard deviation of 98.
The parent asks to see the high school’s study.
The high school looked at SAT scores from a random sample of
40 students for that same
year. The mean math score was 535 and the mean verbal score

was 540.The parent would
like to test if the high school scores come from a different
population than the national
scores. The z-test will give him a way to determine this. If the
value of z could occur by
chance with a probability p 5 .05 or less, the parent will view
this as a nonrandom occur-
rence. First, he looked at the math scores.
Math Scores
m, and therefore mM, 5 520
s 5 110
N 5 40
M 5 535
Calculate the standard error of the mean:
sM 5
s
"N
5
110
"40
5 17.39
Then determine z:
z 5
M 2 mM
sM

5
535 2 520
17.39
5 0.8625
The table value for z 5 0.8625 is 0.3051.
(continued)
suk85842_04_c04.indd 114 10/23/13 1:17 PM
Apply It! (continued)
The percentage of the population of sampling means scoring
535 or more can be determined
by 0.50 2 0.3051 5 0.1949. About 19.5% of the samples of
student scores selected at ran-
dom will have a mean score higher than 535. That is almost a 1
in 5 chance. This result is not
statistically significant at the p 5 .05 level, so the hypothesis
that these students are better
is not supported by the results. In other words, a sample of
student scores with M 5 535
might well have been drawn from a population with mean scores
of mM 5 520.
The parent then looked at verbal scores.
Verbal Scores

m, and therefore mM, 5 508
s 5 98
N 5 40
M 5 540
Calculate the standard error of the mean:
sM 5
s
"N
5
98
"40
5 15.5
Then determine z:
z 2
M 2 mM
sM
5
540 2 508
15.5
5 2.06
The percentage of this group scoring 540 or more can be
determined by
0.50 2 0.4803 5 0.0197. About 2% of the samples of student

scores selected at random
will have a mean score higher than 540. This result is therefore
statistically significant at the
p 5 .05 level, so the null hypothesis can be rejected, and the
alternative (research) hypoth-
esis that these students have better verbal scores is supported.
At less than p 5 .05, the
outcome is 95% unlikely to have occurred by chance.
Using his knowledge of statistics, the parent was able to test the
high school’s claim of bet-
ter SAT scores. The parent rejects the claim of better math
scores and accepts the claim of
better verbal scores.
Apply It! boxes written by Shawn Murphy
suk85842_04_c04.indd 115 10/23/13 1:17 PM
CHAPTER 4Section 4.3 The Concept of Statistical Significance
Another z-Test
Interested in a possible connection between explicit
reinforcement and performance in the
workplace, researchers gather sales data for a group of 30 sales
associates whose manag-
ers provide daily verbal reinforcement. The mean level of sales
for this group in a particu-
lar month is $23,300. For people nationally in this type of retail
sales, the mean is $22,538
with a standard deviation of $5,822. What percentage of all
randomly selected groups will
have mean sales of $23,300 or higher?

m, and therefore mM , 5 22,538
s 5 5,822
N 5 30
M 5 23,300
First, calculate the standard error of the mean:
sM 5
s
"N
5
5,822
"30
5 1,063
Then determine z:
z 5
M 2 mM
sM
5
23,300 2 22,538
1,063
5 0.72

The question is, what percentage of the distribution of sample
means will have mean sales
as high as this group’s sales or higher? The proportion at
$22,538 or lower would be 0.50
(for the lower half of the distribution) plus the 0.2642 of the
distribution between the mean
and the z value for 23,300.
0.50 1 0.2642 5 0.7642
The percentage above M 5 23,300 5 1 2 0.7642 5 0.2358 3 100
(to convert the proportion
to a percentage) 5 23.58%. About 24% of the samples of sales
associates selected at ran-
dom will have mean sales of $23,300 or higher.
4.3 The Concept of Statistical Significance
Like the z score problems in Chapter 3, the z-test is a ratio of
the difference (the numera-tor) compared to data variability (the
denominator). When the ratio is large, it indi-
cates that the score (in the z score problem) or the sample mean
(in the case of the z-test) is
quite distant from the means to which it is compared.
Is there a point at which the sample mean (M) becomes so
different from the mean of
the distribution of sample means (reflected in a large value of z)
that it is more likely to
be characteristic of some other distribution? In the first z-test
problem, we proceeded as
suk85842_04_c04.indd 116 10/23/13 1:17 PM

though the sample of those who had filed for divorce was a
subgroup of all couples with
troubled marriages. What if the sample is actually more
characteristic of some other dis-
tribution, say, a population of couples for whom divorce is
imminent? Can large values of
z reflect the fact that the sample actually represents a
population different from the one to
which it was compared?
Consider another example before we answer these questions.
Those in a college honors
program are probably adults. If researchers are interested in
studying intelligence, would
it be reasonable to expect that the members of this group
represent what is characteristic
of all adults? From the standpoint of age (and the absence of
child prodigies), those honors
students are probably all adults, but in terms of intelligence,
they probably are not typical.
Perhaps they are more representative of the population of
intellectually gifted adults than
of adults in general.
The individuals in every sample belong to many different
populations. The couples on the
verge of divorce belong to
• the population of married people,
• the population of adults,
• the population of adults in the particular state,
• the population of adults in the particular county,
• the population of couples with troubled marriages, and so on.

One of the questions the z-test helps answer is whether a
particular sample is most charac-
teristic of the population to which it is compared, or whether
the sample is more like some
other population. The magnitude of the z value is the key to the
answer.
Statistical Significance and Probability
In the case of the z-test, an outcome is statistically significant
when
• it is so unlike the population to which it is compared that it
can be presumed to
reflect some other population;
• said another way, an outcome is statistically significant when
the value of M is
distant enough from mM that it probably was not randomly
selected from that
particular distribution of sample means.
So, at what point is an outcome nonrandom? Ronald A. Fisher
(1932), who coined the term
statistically significant, made the answer a matter of
probability. If the probability that an
outcome (in our case, the value of z) occurred by chance is p 5
.05 or less, the outcome is
probably not random; it is a statistically significant occurrence.
Although p 5 .05 is probably the most common, other
probability levels have also been
used to indicate statistical significance. Reviewing journal
articles will indicate statistical
testing done at p 5 .01, p 5 .001, and occasionally, even p 5 .1.

It is up to the person doing
the analysis to state the level chosen to indicate statistical
significance (before conducting
the test, by the way). That probability value is also called the
alpha (a) level for reasons
we will get to later.
suk85842_04_c04.indd 117 10/23/13 1:17 PM
Because we can use the z-test and the z score table to calculate
the probability of an occur-
rence (in addition to the other things we can do to determine the
percentage of the popu-
lation above a point, below a point, and between points), we can
determine statistical
significance. In the first z-test we completed,
• we compared the mean number of negative verbal exchanges
in a sample of
couples on the verge of divorce to the mean level of negative
exchanges among
those identified as the population of couples with “troubled”
marriages and
found that z 5 2.630.
• The table value indicates that the probability of randomly
selecting a sample
of couples that would have M 5 12.865 or more negative verbal
exchanges
per week was p 5 .0043. At less than p 5 .05, that outcome is
unlikely to have
occurred by chance. It is statistically significant.

In the second z-test,
• the issue was whether explicit reinforcement affects
sales performance.
• For that problem, z 5 0.717, and the table value for
z 5 0.72 is 0.2642.
• That means the probability of earning $23,300 or
more can be determined by taking the upper half of
the distribution, which is 0.50 2 0.2642. The differ-
ence is 0.2358 (Figure 4.4).
• The probability that a group of sales associates selected at
random would have
mean sales of $23,300 or higher is p 5 .2358. That is a
probability of occurrence of
almost 1 chance in 4. It is too likely to have occurred by chance
to be statistically
significant.
• A sample of sales associates with M 5 $23,300 sales for the
month might well
have been drawn from a population with mean monthly sales of
mM 5 $22,538.
Figure 4.4: The probability of selecting a sample with sales of
M 5 $23,300
or higher from a population with mean sales of MM 5 $22,538
B What does the
term statistically
significant mean?
Try It!

p = 0.5 p = 0.2358
p =
0.2642
0
z-value
suk85842_04_c04.indd 118 10/23/13 1:17 PM
Determining Significance Without the Table
Remember that 6z 5 1.0 includes about 68% of the z
distribution, so the probability of
randomly selecting an outcome that occurs in the 6z 5 1.0 area
is p 5 .68. Nothing in that
region is going to be statistically significant because those z
values indicate results that are
very characteristic of the distribution as a whole. It is the
uncharacteristic events that are
significant, and Fisher’s standard of p 5 .05 indicates that the
key is a z value that occurs
at a point where only the most extreme 5% of the distribution is
excluded.
Recall that normal distributions are symmetrical. That 5%
exclusion means that the most
extreme 2.5% of outcomes in the lower tail and the most
extreme 2.5% of outcomes in
the upper tail are statistically significant. Because Table A
provides proportions for only

the upper half of the distribution, the z value, which includes all
but the extreme 2.5% of
outcomes, will be the point at which results become statistically
significant. If 2.5% needs
to be the percentage excluded, 47.5% is the percentage
included. As a proportion, 47.5%
is 0.475.
• From Table A, what z value includes 0.475 of the distribution
back to the mean of
the distribution?
• Because z 5 1.96 includes 0.475 of the distribution, 6 that
value will include 0.95
of the distribution (2 3 0.475 5 0.95).
• Anytime a z-test produces a z 5 1.96 or greater, the result is
statistically signifi-
cant at p 5 .05.
Another View of Significance
Whether p 5 .05, .01, .001, or some other amount, the particular
standard for statistical
significance is somewhat arbitrary. Fisher picked a point and
said essentially, “Anything
beyond this level of probability is not likely to have occurred
by chance.” Yet another
debatable issue in statistics is that not everyone agrees that
there has to be such a stan-
dard. One approach was to calculate the probability that an
event could occur by chance,
and then let consumers make their own decision about whether
it is significant. Another
approach is to accompany the significance level with the effect
size or the magnitude of

the relationship or effect. This is an additional reporting value
without solely relying on
the significance values. Effect sizes are discussed using
Cohen’s (1988, pp. 145–153) effect
size values starting in Chapter 5.
Another traditional approach to hypothesis testing is calculating
statistical values (e.g., z
values) and then comparing them to the appropriate critical
value found in their respec-
tive tables (usually found in appendices of statistical
references). Seldom do researchers
deal with critical values of a test statistic; with modern
computing power, it is easy to get
the actual probability value for the test statistic from the data
and then to compare this
probability to the desired critical alpha level (e.g., a 5 .05). The
latter approach is most
suitably used in ongoing analyses for the remainder of this text.
suk85842_04_c04.indd 119 10/23/13 1:17 PM
Sampling Error as an Explanation of Difference
In virtually every z-test, there will be some difference between
M and mM, which means
that z will have some value other than 0. When the differences
fall short of statistical
significance (z , 1.96), how are they explained? The answer is
sampling error. Because
no sample can exactly emulate the population, most samples in
the distribution of sam-

ple means will have a sample mean different from the
population mean. In the second
example dealing with the explicit reinforcement of sales
associates, those who received
explicit reinforcement actually did better than the population of
all sales associates, but at
z 5 0.717 the difference is not large enough to be statistically
significant. Such a difference
might reflect the fact that those selected for the sample group
just happened to be gener-
ally above the mean of the distribution. In the first example on
the number of negative
exchanges, some of the difference reflects sampling error, but
that factor alone is not an
adequate explanation of the difference between M and mM.
More Confidence in the Sample
The foregoing underscores the importance of having confidence
in the sample to begin
with. Even though samples can never mirror populations
exactly, we noted that large,
randomly selected samples minimize sampling error. It can be
difficult to define “large,”
however. One approach to determining the optimal sample size
is based on the answers
to two questions:
1. How much certainty must there be that the sample is like the
population?
2. How much error can be tolerated?
The formula is as follows:
n 5 a
1z2 1s2

variation from s
b
2
Formula 4.3
Where
n 5 the required sample size
z 5 the value of z that corresponds to how certain we wish to be
of the result.
Because 6z 5 1.96 includes the middle 95% of the distribution,
using that
value in the formula provides p 5 .95 that the sample emulates
the popula-
tion. If .99 certainty is required, z 5 2.58.
s 5 the standard deviation of the population. If the population
standard devia-
tion is not available, a sample standard deviation (s) can be
substituted,
although the estimate will lose some precision.
variation from s 5 the amount we are willing to allow s to vary
from s
An instructor wants to gauge the impact that a service learning
course has on students’
attitudes toward community service. The university research
office has surveyed stu-
dents’ interest in service learning and from the scores on the
instrument has determined

suk85842_04_c04.indd 120 10/23/13 1:17 PM
a standard deviation of 8.294. The instructor is willing for the
sample
data to digress from university-wide data by 2 points and wishes
to be
.95 confident of the result.
n 5 a
1z2 1s2
variation from s
b
2
With
s 5 8.294
z 5 1.96
n 5 a
11.962 18.2942
2.0
b
2
5 approximately 66

With p 5 .95, a random sample of about 66 people will provide a
sample within 2 points
of the population standard deviation.
Changing the conditions can dramatically affect the required
sample size. If the instructor
needs to be within 1 point of the population standard deviation
and wishes for p 5 .99,
note the impact on the result:
n 5 a
12.582 18.2942
1.0
b
2
5 approximately 458 people
There is constant tension between how certain and how precise
we need to be on the
one hand and the size of the needed sample on the other. The
results illustrate that both
increasing the level of certainty or requiring less error
necessitates larger sample sizes,
but at least Formula 4.3 can help us strike a balance. Samples
that are very large can be
time-consuming and expensive to work with. Samples that are
very small may not reflect
the essential characteristics of the population, making
generalizing the results a problem.
Decision Errors
Statistical significance is based on the probability that an event

could occur by chance, and
interpreting outcomes based on probabilities carries a risk.
• Is it not possible, however unlikely, that a researcher could
accidentally sample
the couples in the distribution who have the most negative
exchanges? Maybe
they do not belong to a distinct population at all. Maybe they
are just from the
most extreme portion of the population of all married couples.
• On the other hand, is it not also possible that those sales
associates who were
explicitly reinforced actually did belong to a population of
higher-performing
salespeople, but because they were accidentally sampled from
the lowest region
in the population of all sales associates, their differences
appeared to be not
significant?
Because statistical decisions are based on probabilities rather
than certainties, any sta-
tistical decision can result in two decisions: a correct decision
and a decision error. The
two examples above represent the two types of decision errors,
and they are mutually
C If a result is
not statistically
significant, how is the
difference between M
and mM explained?
Try It!

suk85842_04_c04.indd 121 10/23/13 1:17 PM
exclusive. Any statistical decision involves the risk of one or
the other, but never both in
the same analysis. As these decisions are discussed in the next
section, refer to Figure 4.5
to aid in your understanding of correct decisions and type I and
type II errors.
Type I Errors
Type I errors in statistical testing occur when an outcome is
determined to be statistically
significant, but further research and testing would indicate that
it is not. In other words,
the first, errant conclusion is an anomaly that fails to hold up
under further scrutiny. The
probability of this error is defined by the level at which the
testing occurs. If the criterion
for statistical significance is p 5 .05 or a 5 .05, and the result is
deemed statistically sig-
nificant, the probability of a type I error is .05. Because type I
error is also called alpha (a)
error, the significance level of a test is sometimes noted in
terms of the risk of alpha error,
a 5 .05, rather than p 5 .05. It means the same thing, except that
the author has chosen
to indicate the probability of type I error rather than referring
directly to the criterion for
statistical significance. At a 5 .05, for every 100 times someone
concludes that a result is
statistically significant; there will be a type I error an average

of five times.
Note that what Fisher did was arbitrarily exclude the most
extreme 5% of the distribu-
tion as atypical of the most likely outcomes. Although we agree
that those most extreme
outcomes are the least likely to occur, that most extreme 5% of
the distribution is still part
of the distribution in question. Outcomes in that area of the
distribution hold the greatest
potential for a type I error.
• The only time a type I error is possible is when a result is
deemed statistically
significant. If there is no statistically significant outcome, there
is no potential for
a type I (a) error.
• In a particular significant finding, there is not any way to
know whether a type I
error has occurred. Gathering new data and repeating the
analysis is the only
way to check, which is why replication studies are so important.
Type II Errors
In a z-test, type II errors occur when the sample is actually
characteristic of some popula-
tion other than the distribution of sample means to which it was
compared, but the statisti-
cal testing (z , 1.96) suggests no significant difference. This
type of decision error is also
called a beta (b) error.
There would be little problem with type II errors if the
populations

involved were completely separate, but often there are
important sim-
ilarities. The population of all sales associates probably bears a
num-
ber of similarities to the population of sales associates who
receive
explicit reinforcement. The more the populations involving
sales asso-
ciates overlap, the more likely decision errors become.
Although the level at which the statistical test is conducted
(often
or a 5 .05) defines the likelihood of a type I error, the
probability of
a type II error is more elusive, and in fact we never know the
exact
probability of committing this error, although some statistical
tests are more prone to it
than others.
D An analysis
results in a finding
that is statistically
significant at p 5 .05.
What is the probability
of a type II error?
Try It!
suk85842_04_c04.indd 122 10/23/13 1:17 PM

• The only time a type II error can occur is when a result is
determined not to be
statistically significant.
• In a particular analysis where the result appears to be not
significant, there is not
any way to know whether a decision has resulted in a type II
error.
• See Tables 4.2 and 4.3 for a look as to how these are
interrelated.
Table 4.2: Correct decisions, type I, and type II errors
Reality
R
es
ea
rc
h
The null hypothesis
(Ho) is True
The alternative hypothesis
(Ha) is True
The null hypothesis (Ho) is
True
Accurate
p 5 1 2 a

Type II error
p 5 b
The alternative hypothesis
(Ha) is True
Type I error
p 5 a
Accurate
p 5 1 2 b
Table 4.3: Pregnancy test results
Result When Ho (No Baby) Is True When Ha (Baby) Is True
Not Pregnant Correct pregnancy test
“Whew! Parental planning effective!”
Type II error
False Negative pregnancy test
“Oops! Baby on the way!”
Pregnant Type I error
False Positive pregnancy test
“Where’s the baby?”
Correct pregnancy test
“As planned, baby on the way!”
Decision Errors in Summary
Is one error more damaging than the other? Do analysts have a
preference for one type of
error? The answer, of course, depends upon circumstances and
especially on the impact

that a decision error has on the people involved.
Perhaps a committee is evaluating certification programs for
mental health professionals,
and it deems the program at University A to be significantly
better than the competing
programs. If the result is that, the graduates from University A
receive preferential hiring
but the difference among programs really is not statistically
significant after all, then there
has been a type I error.
On the other hand, perhaps a client has a serious illness and
comes to a health profes-
sional for a diagnosis. If the health professional fails to
recognize that the client is not
suk85842_04_c04.indd 123 10/23/13 1:17 PM
G* Power 3 is a
free, online power
analysis tool available
via the Institute for
Experimental Psychology
at the Heinrich Heine
University Düsseldorf.
Follow the link below
to learn more about
the package, with
accompanying manuals
and articles, by developers
Buchner, Erdfelder, and

Faul (1996).
http://www.psycho.uni
-duesseldorf.de/abtei
lungen/aap/gpower3
Try It!
healthy and misses the condition that is affecting the client’s
well-
being, there has been a type II error.
Which type of error is the more serious depends upon
circumstances,
but statisticians may have their own bias. Power in statistical
testing is
described in terms of the likelihood of a type II error. The most
power-
ful tests are associated with the fewest beta errors. The power
of a sta-
tistical test is symbolically indicated this way: 1 2 b (refer to
the lower
right quadrant “correct decision” in Figure 4.5). Power is
indicated as
a probability of rejecting the null hypothesis and therefore the
higher
the power (1 2 b) the greater the likelihood of supporting the
alter-
native hypothesis. For instance, a power of 1 2 0.2(b) 5 0.8
indicates
an 80% chance of rejecting the null hypothesis or simply stated,
an
80% power. As a researcher, increasing this likelihood or
probability
is imperative to finding a statistically significant difference or
rela-
tionship in hypothesis testing, and this is most commonly

affected by
sample size. As a result, researchers will conduct a Power
analysis to
calculate a minimum sample size based on effect sizes and
statistical
significance criteria using appropriate software such as
G*Power 3 or
SPSS Sample Power 3.
On “Picking Your Poison”
Although type I and type II errors cannot both occur in the same
analysis, the probability
of one affects the likelihood of the other. Mental health
professionals ordinarily must pass
some sort of licensing requirement, perhaps in the form of a
test. Like most professional
licensing tests, it probably is a reasonably good, but certainly
not perfect, indicator of who
is competent.
• If test results indicate that an individual is competent, but the
individual actually
lacks the skills and knowledge required, there has been a type I
error.
• If test results indicate that an individual is not sufficiently
competent to be
licensed, but the person actually is, there has been a type II
error.
If a type I error is thought to be the greater problem, the
licensing body might simply raise
the required test score. This would probably reduce the number
of incompetent people
who are licensed, but the companion problem is that it would

also exclude more of those
who actually are competent but, because they do not test well,
fail to demonstrate their
competence on the required measure. This inherent connection
between the two kinds
of decision errors is why someone has to decide which is the
more damaging of the two.
With airline pilots and surgeons, it is straightforward. Usually
the decision is in favor of
excluding some who are competent (and therefore committing a
type II error) rather than
risk licensing some who are not competent (committing a type I
error). The potential cost
to the well-being of others is too great to do otherwise. In other
circumstances, the greatest
good is less clear.
suk85842_04_c04.indd 124 10/23/13 1:17 PM
http://www.psycho.uni-duesseldorf.de/abteilungen/aap/gpower3
CHAPTER 4Section 4.4 A Confidence Interval for the Mean of
the Population
4.4 A Confidence Interval for the Mean of the Population
When the results from a z-test are statistically significant, the
sample best represents a population other than the one to which
it was compared. In these instances, the
mean of the sample (M) is in fact an estimate of the value of
that other population mean.
Because it is a discrete value, M is called a “point estimate” of

the population mean mM.
For instance, a 95% confidence interval would provide a range
of values within which this
population mean mM would fall. As depicted in Figure 4.5, the
confidence interval pro-
vides a way to determine how precisely M estimates or predicts
the population mean, mM.
Figure 4.5: The confidence interval based on a normal
distribution
With regard to hypothesis testing, when a z value is significant,
the confidence interval
for the mean can produce a range of values within which mM is
likely to occur. If the value
produced is not within this confidence interval range, then this
is not a probable estimate
of mM. Conversely, if the z-test is not significant, there is no
need for the confidence interval
because our analysis indicates that the sample M belongs to the
population described, and
there is no need to estimate the value of mM. Calculating the
confidence interval involves
values from the z-test. The formula is
CI 5 6z(sM) 1 M Formula 4.4
Where
CI 5 the interval within which the population mean is expected
to occur
z 5 the table value that reflects the level at which the z-testing
was conducted
(for p 5 .05, z 5 1.96)

sM 5 the value of the standard error of the mean from the z-test
M 5 the value of the sample mean
�3�4 �2 �1 0 1 2 3 4
Prediction of the population mean
z-value
suk85842_04_c04.indd 125 10/23/13 1:17 PM
the Population
For the negative-verbal-exchanges problem, the result was
significant, indicating that the
sample probably belongs to some distribution of sample means
other than the one to
which it was compared. The confidence interval will establish a
range of values within
which the mean for that other population probably occurs.
CI 5 6z(sM) 1 M
CI 5 61.96(0.709) 1 12.865
CI 5 61.390 1 12.865 5 11.475, 14.255
With .95 probability, the population which the sample
represents has a mean (mM) value
somewhere between 11.475 and 14.255.
Note that the level of probability is one of the factors affecting

the size of the confidence
interval. If we wish to be more certain of capturing the
population mean, a .99 confidence
interval can be used instead of .95, and z 5 2.58 substituted for
z 5 1.96 in the formula.
Recalculating the confidence interval for p 5 .99,
CI 5 6z(sM) 1 M
CI 5 62.58(0.709) 1 12.865
CI 5 61.829 1 12.865 5 11.036, 14.694
A greater level of certainty of capturing the true mean of the
distribution represented by
the sample mean requires a wider confidence interval.
The other factor that affects the width of the confidence interval
is
the standard error of the mean, sM, which measures the amount
of
variability in the distribution of sample means. More data
variability
translates into a larger standard error of the mean, which makes
the
confidence interval larger.
There is no need for a confidence interval unless the z-test
results are statistically signifi-
cant. The reason can be illustrated by completing a confidence
interval for the nonsignifi-
cant salesperson problem. Recall that sM 5 1,063 and M 5
23,300 for that example.
CI 5 6z(sM) 1 M

CI 5 61.96(1,063) 1 23,300
CI 5 62,083 1 23,300 5 21,217, 25,383
Note that this confidence interval includes within its range the
value of the original popu-
lation, 22,538. That is because with a nonsignificant z value,
the conclusion is that the
population that the sample represented is likely the same
population to which it was
compared.
A nonsignificant z-test value will always produce a confidence
interval that includes
the original population mean. Note that neither the .95 nor the
.99 confidence intervals
included the original population mean from the negative-verbal-
exchanges problem.
E What does the
confidence interval
for z determine?
Try It!
suk85842_04_c04.indd 126 10/23/13 1:17 PM
the Population
Apply It!
Quality Control Revisited

Let us return to the example of the bottling company that uses
an automated
machine to fill 4-liter plastic containers with orange juice. The
recalibrated
machine fills the containers to a mean of 4.05 liters, with a
standard deviation of 0.09 liters.
The equipment engineer now wants to use the same machine to
fill the 4-liter containers
with apple juice. He would like to know if changing to apple
juice will affect the machine’s
performance. Will the mean fill amount still be 4.05 liters with
a standard deviation of
0.09 liters?
To find an answer, the engineer measures 20 of the apple juice
containers. The sample mean
fill is 3.99 liters. What is the probability that a randomly
selected group of 20 apple juice
containers would have a mean fill of 3.99 liters or less? Has
changing from orange juice to
apple juice affected the machine? The engineer decides to use a
value of p 5 .01 to indicate
statistical significance.
m, and therefore mM, 5 4.05 liters
s 5 0.09 liters
N 5 20
M 5 3.99 liters
First, calculate the standard error of the mean:
sM 5
s

"N
5
0.09
"20
5 0.02 liters
Then determine z:
z 5
M 2 mM
sM
5
3.99 2 4.05
0.02
5 23.0
Note that even though the difference between the mean of the
population (mM 5 4.05
liters) and the sample mean (M 5 3.99 liters) is small, the z
value is very large because of
the small standard error of the mean (sM 5 0.02 liters).
The table value for z 5 –3.0 is 0.4987.
The percentage of containers filled to 3.99 liters or less can be
determined by
0.50 2 0.4987 5 .0013. About .13% of the samples selected at
random will have a mean
fill level lower than 3.99 liters. This result is therefore
statistically significant at the p 5 .01
level. By switching from orange juice to apple juice, the mean

fill level has changed. The
machine controls will have to be adjusted to account for these
differences if a mean fill
amount of 4.05 liters is to be achieved when filling apple juice
containers.
(continued)
suk85842_04_c04.indd 127 10/23/13 1:17 PM
CHAPTER 4Section 4.5 The z-Test Using Excel
4.5 The z-Test Using Excel
A social worker’s caseload includes 8 people with annual
incomes as follows (in thou-sands of dollars):
13.5, 18, 22.375, 25.240, 26, 29.331, 30, 30
Is the mean income of the social worker’s clients significantly
different from the mean
income of all social workers’ clients for whom the average
annual income is 19.500 thou-
sand dollars with a standard deviation of 4.525 thousand
dollars? We will proceed in
Excel as follows:
1. Enter the income data into a spreadsheet in cells A1–A8.
2. Have Excel calculate the mean by entering the formula 5
average(A1:A8) in cell
A9.
3. In cell A11, determine the standard error of the mean by
dividing the population

standard deviation by the square root of the number. The
command in Excel is
54.525/sqrt(8).
4. Determine the z value in cell A13 by entering the command
5(A9 2 19.5)/A11.
The part in parentheses is the numerator in the z ratio: M (cell
A9) 2 mM. Figure 4.6
is a screenshot of what your display will look like just before
you press Enter.
The result is z 5 3.004. Testing at p 5 .05 (for which z 5 1.96),
these eight people have sig-
nificantly different incomes than the population of all social
workers’ clients.
Apply It! (continued)
Because the results from the z-test are statistically significant,
the sample mean of 3.99 best
represents the value of the population mean. The engineer next
computes the confidence
interval for p 5 .01 to determine the range of values within
which mM is likely to occur.
CI 5 6z(sM) 1 M
Where
z 5 the table value that reflects the level at which the z-testing
was conducted
For p 5 .01, z 5 2.58
CI 5 62.58(0.02) 1 3.99
CI 5 6.05 1 3.99 5 3.94, 4.04

Therefore, there is a 99% probability that the mean fill value is
between 3.94 and 4.04
liters when using the machine with apple juice.
Apply It! boxes written by Shawn Murphy
suk85842_04_c04.indd 128 10/23/13 1:17 PM
CHAPTER 4Section 4.6 Presenting Results
Figure 4.6: Calculating a z-test in Excel
4.6 Presenting Results
Using the data from Figure 4.6, the mean income of the social
worker’s caseload is 24.31 (in thousands). The population
average is 19.50 (SD 5 4.53). During the z-test,
the population average is standardized at 0, and the sample
mean is calculated as a z score
to determine its difference or distance from the population. In
this case, the sample mean
results in a z score of 3.00. The sample mean is 3 standard
deviations above the population
mean. We only need the z score to be as high as 1.96 in order
for the difference to be sta-
tistically significant at p 5 .05. In this case, we met this
criterion and can conclude that the
social worker’s caseload has a significantly higher income than
the population average.
suk85842_04_c04.indd 129 10/23/13 1:17 PM

CHAPTER 4Section 4.7 Interpreting Results
It is important to note in your interpretations the population
mean, sample mean, z score,
and significance level. Be sure to discuss whether the difference
is statistically significant
or not and whether or not the difference means the sample mean
is higher or lower than
the population mean.
4.7 Interpreting Results
Though you should refer to the most recent edition of the APA
manual for specific detail on formatting statistics, Table 4.4
may be used may be used as a quick guide in pre-
senting the statistics covered in this chapter.
Table 4.4: Guide to APA formatting of z test scores
Abbreviation or Term Description
CI Confidence interval; presented as CI [lowest, highest]
p Probability or significance level
If statistically significant, report p , .05 or .01
If not statistically significant, report p 5 “calculated p level”
SEM Standard error of the mean; standard error of measurement
z z-test statistic or score
Source: Publication Manual of the American Psychological
Association, 6th edition. ©2009 American Psychological
Association, pp. 119–122.

Note that p, SEM, and z are italicized, whereas CI is not. The
following are some examples
of how to present results using these abbreviations, though you
may use different
combinations of results. These examples utilize the data
presented in Section 4.5.
• The average annual income for the social worker’s caseload
was significantly
higher (M 5 24.31) than the population average income (M 5
19.5; z 5 3.00,
p , .05).
• The annual income for the social worker’s caseload is
statistically different from
the population average income, z 5 3.00, SEM 5 1.60, p , .05.
Using the data from Apply It! Quality Control Revisited, we
could present the results in
the following way:
• The probability of an apple juice container being filled with
3.99 liters or less is
statistically significant at 0.0013, z 5 –3.00, SEM 5 0.02, p ,
.01, 99% CI [3.94,
4.04].
• The difference between the population mean (m 5 4.05) and
sample mean
(M 5 3.99) is statistically significant at p , .01, 99% CI [3.94,
4.04] (z 5 –3.00,
SEM 5 0.02).
suk85842_04_c04.indd 130 10/23/13 1:17 PM

CHAPTER 4Summary
Summary
The z-test provides a good introduction to formal statistical
testing. It is an uncomplicated
test that involves many of the same issues that come up in the
more advanced tests. In
general, behavioral researchers are much more interested in
analyzing the performance of
groups than of single individuals. We have many reasons to
wonder whether this or that
group truly represents the population to which they are
compared. The z-test provides a
mechanism for comparing one group for whom we have data to
an identified population.
The z-test is based on the distribution of sample means
(Objective 1), a population of the
means of samples rather than of individuals’ scores. The central
limit theorem indicates
that such a distribution will be normal even if the distribution
of individual scores is not
(Objective 2). The normality allows the use of the z table to
analyze how groups compare
to populations (Objectives 4 and 6). Because the sample data we
analyze sometimes does
not fit well with the population presumed to be the source, the
z-test provides a way to
determine whether the sample belongs to some other population,
an outcome related to
the concept of statistical significance (Objective 5).
When the sample is determined to represent some other
population, the sample mean is
a point estimate of the value of that other mM, but it is only an

estimate. The confidence
interval provides a range of values within which the mean of
that other population will
occur with a specified probability (Objective 7). In doing so,
the confidence interval gives
an indication of the precision with which M estimates mM.
Inferential statistical analysis involves the risk of making an
incorrect decision. Occasion-
ally, results that appear significant in one test will not hold up
when the study is repeated
with new data. On the other hand, sometimes a nonsignificant
finding will be overturned on
further analysis. These type I and type II errors, respectively,
are a reminder that statistical
decisions are based on probabilities rather than certainties
(Objective 8). In addition, how to
present results (Objective 9) and interpret them in APA format
(Objective 10) as they relate
to describing z-test results are important pieces of utilizing and
writing about statistical data.
Small samples, no matter how carefully selected, cannot mirror
all the relevant character-
istics of complex populations, and populations involving people
are invariably complex.
For this reason, a procedure for determining the size of the
sample needed to emulate the
important characteristics of the population has some utility
(Objective 3). Formula 4.3
meets that need. Statistical significance is a very important
concept in educational analy-
sis. When new programs or strategies are instituted, we often
look for ways to determine
whether the program makes a difference. The z-test helps
answer some of these questions.

As important as the z-test is as an introduction, it has
limitations in that it requires access
to both mM and sM. Although population means can usually be
figured out, the standard
error of the mean sometimes just is not accessible. The t-tests in
Chapter 5 will provide a
way around this difficulty.
This summary is probably a good barometer of your grasp of
Chapters 1–4. Although
some of the material has probably been familiar, many of the
ideas are likely new. If the
review here makes sense, that is excellent. If there are some
holes, it is a good idea to take
some time to go back to the relevant sections and review.
Statistical analysis is incremen-
tal, as we have stressed before, so it is important to understand
what has been presented
before continuing. Working the examples below, sometimes
repeatedly, will help.
suk85842_04_c04.indd 131 10/23/13 1:17 PM
CHAPTER 4Key Terms
Key Terms
central limit theorem Proposition that
holds that if a population is sampled an
infinite number of times using sample size
n and the mean of each sample is deter-
mined, the multiple means (Ms) will take
on the characteristics of a normal distribu-
tion whether or not the original population

of individuals was normal.
confidence interval (CI 5 6z(sM) 1 M)
Provides a way to determine how precisely
M estimates mM.
decision errors The two types of decision
errors are type I errors and type II errors.
distribution of sample means A distribu-
tion made of the means of samples rather
than individual scores.
law of large numbers The mathematical
principle that errors diminish as the num-
ber of data points increases.
power In statistical testing, the likelihood
of a type II error. The power of a statistical
test is indicated as 1 2 b.
random selection The selection of a sam-
ple from a population where every member
has an equal probability of being selected.
sampling error Error that is reflected in
the degree to which the characteristics of
the sample, such as the mean and standard
deviation, vary from those populations.
standard error of the mean The standard
deviation of the sample means (sM).
statistically significant An outcome so
unlike the population to which it is com-
pared that it can be presumed to reflect

some other population. Said another way,
the value of M is distant enough from mM
that it probably was not randomly selected
from the distribution of sample means. If
the probability that an outcome occurred
by chance is p 5 .05 or less, the outcome is
statistically significant.
systematic sampling error Sampling error
that occurs because the same mistake in
selecting a sample of a population is made
repeatedly.
type I errors Also alpha (a) errors; type
of decision errors made when a result is
judged to be statistically significant, but
further research and testing would show
that it is not.
type II errors Also beta (b) errors; type of
decision errors that occur when the sample
is characteristic of some population other
than the distribution of samples means to
which it was compared but the statistical
testing suggests no significant difference.
z-test Test that indicates how distant a
sample mean is from the mean of the dis-
tribution of sample means, in units of the
standard error of the mean. When the value
of z is 1.96 or greater, there is a probability
of p 5 .05 or less that the sample belongs to
the population.
suk85842_04_c04.indd 132 10/23/13 1:17 PM

CHAPTER 4Chapter Exercises
Chapter Exercises
Answers to Try It! Questions
The answers to all Try It! questions introduced in this chapter
are provided below.
A. There is less variability in the distribution of sample means
than in a distribu-
tion of individual scores because sample means moderate the
effect of extreme
scores. The larger the sample, the more extreme scores are
minimized as factors
in data variability.
B. “Statistically significant” means that the calculated value, z
in this case, is large
enough that it is not likely to have occurred by chance; it is
probably not a ran-
dom outcome.
C. When the difference between M and mM in a z-test is not
significant, the differ-
ence is attributed to random variability; the value of M is one of
the possible
values of samples drawn at random from the distribution of
sample means.
D. A type II, or beta, error can occur only when a result is
determined not statisti-
cally significant. When the result is significant, the probability
of b 5 0.

E. Calculated only for a statistically significant result, a
confidence interval for the
value of z indicates a range of scores within which the
population mean that the
sample does probably represent occurs.
Review Questions
The answers to the odd-numbered items can be found in the
answers appendix.
1. If all the psychologists working at a state mental hospital
have an average age of
47.5 years, what will be the value of mM if it were created from
such a population?
2. The standard deviation of the psychologists’ ages in Exercise
1 is calculated. If the
standard error of the mean for the distribution of sample means
is also calculated
for the same data, which will have the greater value? Why is
there a difference?
3. The assistant vice president for personnel at a college has
job-performance scores
for all clerical staff, with a mean value of 32.956 and a standard
error of the mean
of 5.924. What is the probability of randomly selecting a sample
with a job satisfac-
tion mean of 35.0 or higher?
4. If a group with M 5 35.0 is selected, are they significantly
different from the
population?
5. The clerical staff in a large law office has the following job
performance scores:

25, 37, 38, 43, 44, 48, 51
If the mean level of performance for all clerical staff is
33.255 with a standard error
of the mean of 3.248, are those in the law office characteristic
of that population? Test
at p 5 .05.
suk85842_04_c04.indd 133 10/23/13 1:17 PM
6. The standard deviation for a major intelligence test is s 5
15.0. If in a given year
the test is administered to 347 people, what is the value of the
standard error of the
mean?
7. An exclusive graduate program requires GRE Quantitative
scores of 500 or bet-
ter. This year’s entering class has n 5 16 and M 5 625. Are they
characteristic of a
national population of graduate students for whom m 5 500 with
s 5 100? What
is the probability that a group of 16 applicants selected at
random would have
s 5 100 and M 5 525 or better?
8. If a researcher wishes to gather a sample of people who have
intelligence scores
that differ from the national standard deviation of 15 by no
more than 3 points,
with .95 confidence, how large must the sample be? How large

must the sample be
if it is to vary from the national standard deviation by no more
than 2 points?
9. A group of social workers takes a measure of optimism and
scores as follows:
11, 14, 14, 16, 19, 20, 22, 23, 27, 30
If the population standard deviation is 4.554,
a. What is the value of the standard error of the mean?
b. What is the z value for a z-test with this group if mM 5
26.0?
c. If 26.0 is the mean for all employed adults, is this group of
social workers sig-
nificantly different?
d. Complete this problem on Excel. Refer to Figure 4.6 for
help.
10. If a z-test result is not significant, why will a confidence
interval for the popula-
tion mean contain the value of the population mean to which the
sample was
compared?
11. What factors will reduce the size of a confidence interval?
12. If someone is testing at p 5 .01 and the result is statistically
significant, what is the
probability of a type I error? What is the probability of a type II
error?
Analyzing the Research
Review the article abstract provided below. You can then access

the full article via your
university’s online library portal to answer the critical thinking
questions. Answers can be
found in the answers appendix.
Using Normative Data for a Neuropsychology Study
Crawford, J. R., & Garthwaite, P. H. (2008). On the ‘optimal’
size for normative samples
in neuropsychology: Capturing the uncertainty when normative
data are used to
quantify the standing of a neuropsychological test score. Child
Neuropsychology,
14(2), 99–117. doi:10.1080/09297040801894709
suk85842_04_c04.indd 134 10/23/13 1:17 PM
Article Abstract
Bridges and Holler (2007) have provided a useful reminder that
normative data are fal-
lible. Unfortunately, however, their paper misleads
neuropsychologists as to the nature
and extent of the problem. We show that the uncertainty
attached to the estimated z score
and percentile rank of a given raw score is much larger than
they report and that it varies
as a function of the extremity of the raw score. Methods for
quantifying the uncertainty
associated with normative data are described and used to
illustrate the issues involved.
A computer program is provided that, on entry of a normative

sample mean, standard
deviation, and sample size, provides point and interval
estimates of percentiles and
z scores for raw scores referred to these normative data. The
methods and program pro-
vide neuropsychologists with a means of evaluating the
adequacy of existing norms and
will be useful for those planning normative studies.
Critical Thinking Questions
1. The article states that Johnny has a z score of 21.66,
assuming p , .05. Is there a
significant difference from the mean of the distribution?
2. If the neuropsychological test has a m 5 50 and s 5 10, what
is the z score of someone
who received a 45 on the test?
3. Why would the psychologist want to convert their scores
from a neuropsychological
test to a z score?
suk85842_04_c04.indd 135 10/23/13 1:17 PM
suk85842_04_c04.indd 136 10/23/13 1:17 PM
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