Author: Brenda Gunderson, Ph.D., 2012
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Mind on Statistics
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Module 10: Chi-Square Tests
Objective: In this module, you will learn how to perform three Chi-square tests
(the test of goodness of fit, the test of independence, and the test of homogeneity)
that are used to analyze categorical responses.
Overview: There are three Chi-Square tests presented in this module: the tests of
goodness of fit, independence, and homogeneity. For all three tests, the data are
generally presented in the form of a contingency table (a rectangular array of
numbers in cells). All three tests are based on the Chi-Square statistic:
(O  Ei ) 2
, where Oi is the observed count and Ei is the expected count
2   i
Ei
under the corresponding null hypothesis.
Goodness of Fit Test: This test answers the question, “Do the data fit well
compared to a specified distribution?” It considers one categorical variable, and
assesses whether the proportion of sampled observations falling into each
category matches well enough to the null distribution for the given problem. The
null hypothesis for the goodness of fit test specifies this null distribution which
describes the population proportion of observations in each category.
Test of Homogeneity: This test answers the question, “Do two or more
populations have the same distribution for one categorical variable?” It considers
one categorical variable, and assesses whether this variable is distributed the same
in two (or more) different populations. The null hypothesis for the test of
homogeneity is that the distribution of the categorical variable is the same for the
two (or more) populations.
Test of Independence: This test answers the question, “Are two factors (or
variables) independent for a population under study?” It considers two categorical
variables, and assesses whether there is a relationship between these two
variables for a single population. The null hypothesis for the test of independence
is that the two categorical variables are independent (that is, they are not related)
for the population of interest.
There are a few properties of the Chi-square distribution that you might find
useful. The expected value of a Chi-square distribution is its degrees of freedom
(mean =   df ), and its variance is 2 times its degrees of freedom. Thus, its
standard deviation is the square root of 2 times the degrees of freedom (
 2  2 * df so   2 * df ). This frame of reference can help us assess if our
observed statistic is unusual under the null hypothesis or somewhat consistent
with the null hypothesis.
128
For details on how to enter cross-tabulated data and perform these tests in SPSS,
please refer to the How To Do It in SPSS section for Module 10 online in CTOOLS.
Output will be provided for all activities.
Formula Card:
Activity 1: Is There a Different Pattern in the Distribution
of Accidental Deaths in a Certain Region
Compared to the Pattern in the Entire United
States?
Background: According to the records of the National Safety Council, accidental
deaths in the United States during 2002 had the following distribution according to
the principal types of accidents.
Motor
Vehicle
45%
Falls
Drowning
Fire
Poison
Other
15%
4%
3%
16%
17%
Suppose that an accidental death data set from a particular geographical region
yielded the following frequency distribution for the principal types of accidents:
Motor
Falls
Drowning
Fire
Poison
Other
Vehicle
442
161
42
33
162
150
Do these data show a significantly different pattern in the distribution of accidental
deaths in the particular region compared to the pattern in the entire United
States? Use a 5% significance level.
(Source: National Safety Council Website, 2005)
129
Task: Perform a Chi-square goodness of fit test to assess whether the data fit well
with the model specified in the null hypothesis.
Recall: Write out the Five Steps for conducting a test of hypotheses
(reference page 53).
1.
2.
3.
4.
5.
Hypothesis Test: You can now implement the Five Steps for conducting a test of
hypotheses.
1. State the hypotheses:
a. Explain in one sentence why the test for this scenario is a goodness of fit
test.
b. State the null hypothesis H0: ________________________________ ,
where __________ represents:
Remember: Your hypotheses and parameter definition should always be a
statement about the population(s) under study.
2. Checking the Assumptions and Computing the Test Statistic
a. Find the expected counts for the different categories of accidental deaths,
and fill them in the table that has the null hypothesis proportions and the
observed counts.
130
Null %
Observed
Expected
Motor
Vehicle
45%
442
Falls
Drowning
Fire
Poison
Other
Total
15%
161
4%
42
3%
33
16%
162
17%
150
100%
990
b. Do all cells have expected counts greater than 5?
Yes
No
c. Complete the calculation of the test statistic based on your table above.
X2 
442  445.52  161  148.52  42  39.62  33  29.7 2  162  158.42  ..........................2
445.5
148.5
39.6
29.7
158.4
 3.66
3. Calculate the p-Value:
Based on the SPSS output, answer the following:
Test Statistics
Category
3.663
5
.599
Chi-Square
df
Asymp. Sig.
i.
The p-value is _____________ .
ii. The expected value of the test statistic assuming the null hypothesis is true
is ____________ .
iii. The large p-value is consistent with the fact that our observed test statistic
value is
greater than
less than
the expected test statistic value
(under the null hypothesis).
4. Decision:
What is your decision at a 5% significance level? Reject H0 Fail to Reject H0
Remember: Reject H0

Fail to Reject H0 
Results statistically significant
Results not statistically significant
5. Conclusion:
What is your conclusion in the context of the problem?
131
Activity 2: Are Angry People More Likely to Have Heart
Disease?
Background: “People who get angry easily tend to be more likely to have heart
disease.” That is the conclusion of a study that followed a random sample of
12,986 people from three locations over about four years. All subjects were free
of heart disease at the beginning of the study. The subjects took the Spielberger
Trait Anger Scale, which measures how prone a person is to sudden anger. The
8,474 people in the sample who had normal blood pressure were classified
according to whether they had coronary heart disease (CHD) or not and whether
they had low anger, moderate anger, or high anger according to the Anger Scale.
The classification summary is given.
Low Anger
Moderate Anger
CHD
53
110
No CHD
3057
4621
(Source: Moore, 2001, page 476)
High Anger
27
606
Task: Perform a Chi-square test of independence at the 10% level to assess if
Anger classification and Heart Disease Status are related.
Hypothesis Test: You can now implement the Five Steps for conducting a test of
hypotheses.
1. State the hypotheses:
a. Explain in one sentence why the test for this scenario is a test of
independence.
b. State the null hypothesis H0:
______________________________________________________
c. Intuition about what the conclusion may be can be derived from looking at
some basic probabilities.
 What proportion of sampled subjects had CHD?

What proportion of High anger subjects had CHD?

What proportion of Moderate anger subjects had CHD?

What proportion of Low anger subjects had CHD?
If the null hypothesis were true, we would expect the above four numbers
to be _________.
132
2. Checking the Assumptions and Computing the Test Statistic
CHD * TEMPER Crosstabulation
CHD
CHD
No CHD
Total
Count
Expected Count
Count
Expected Count
Count
Expected Count
Low anger
53
69.7
3057
3040.3
3110
3110.0
TEMPER
Moderate
anger
110
106.1
4621
4624.9
4731
4731.0
High anger
27
14.2
606
618.8
633
633.0
Total
190
190.0
8284
8284.0
8474
8474.0
Chi-Square Te sts
Value
16.077 a
13.999
Pearson Chi-Square
Lik elihood Ratio
Linear-by-Linear
As soc iation
N of Valid Cases
2
2
As ymp. Sig.
(2-sided)
.000
.001
1
.000
df
13.184
8474
a. 0 c ells (.0% ) have expected count less than 5. The
minimum expected count is 14. 19.
a. Briefly, explain what the “expected counts” are.
b. Based on the table, do the assumptions appear to be met to perform the
test? (Are all expected counts greater than 5?) Yes
No
c. The expected value of the test statistic assuming the null hypothesis is true
is _________.
d. What is the value of the test statistic? ____ = ____________
3. Calculate the p-Value:
Based on the SPSS output, answer the following:
i. The p-value is _____________ .
ii. The small p-value is consistent with the fact that our observed test statistic
value is
greater than
less than
the expected test statistic value
(under the null hypothesis).
4. Decision:
What is your decision at a 5% significance level? Reject H0 Fail to Reject H0
Remember: Reject H0

Fail to Reject H0 
Results statistically significant
Results not statistically significant
133
5. Conclusion:
What is your conclusion in the context of the problem?
Activity 3: Comparison of the Distribution of Academic
Degrees: Males Versus Females
Background: How do women and men compare in the pursuit of academic
degrees? The table below present counts (in thousands) from the Statistical
Abstract of degrees earned in 1996 categorized by the level of the degree and the
sex of the recipient.
Female
Male
Bachelor
642
522
Master
227
179
Professional
32
45
Doctorate
18
27
Task: Perform a Chi-square test of homogeneity. Use a 1% significance level.
Hypothesis Test: You can now implement the Five Steps for conducting a test of
hypotheses.
1. State the hypotheses: State the null hypothesis
H0: __________________________________________________________
__________________________________________________________
2. Checking the Assumptions and Computing the Test Statistic
134
a. Show how the expected count 531.8 (first cell for males) was computed.
b. Based on the table, do the assumptions appear to be met to perform the
test? (Are all expected counts greater than 5?) Yes
No
c. The expected value of the test statistic assuming the null hypothesis is true
is _________.
d. What is the value of the test statistic? ____ = ____________
Chi-Square Te sts
Pearson Chi-Square
Lik elihood Ratio
Linear-by-Linear
As soc iation
N of Valid Cases
Value
9.514a
9.485
5.099
3
3
As ymp. Sig.
(2-sided)
.023
.023
1
.024
df
1692
a. 0 c ells (.0% ) have expected count less than 5. The
minimum expected count is 20. 56.
3. Calculate the p-Value:
Based on the SPSS output, the p-value is _____________ .
4. Decision:
What is your decision at a 5% significance level? Reject H0
Remember: Reject H0

Fail to Reject H0 
Fail to Reject H0
Results statistically significant
Results not statistically significant
135
5. Conclusion:
What is your conclusion at a 1% significance level
in the context of the problem?
Would your decision and conclusion change if the significance level was:
 5% instead of 1%?
 3% instead of 1%?
 2.3% instead of 1%?
 2% instead of 1%?
Based on your answers above, you can see that the p-value represents the
_________________ significance level at which the results would be
statistically significant.
Check Your Understanding:
Fill in the blank with the most appropriate Chi-square test to address the research
question.
A researcher wants to determine if scoring high or low on an artistic ability test
depends on being right or left-handed.
___________________________________________
A national organization wants to compare the distribution of level of highest
education completed (high school, college, masters, doctoral) for Republicans
versus Democrats.
___________________________________________
A preservation society has the percentages of five main types of fish in the river
from 10 years ago. After noticing an imbalance recently, they add some fish
from hatcheries to the river. How can they determine if they restored the
ecosystem from a new sample of fish?
___________________________________________
136
Example Exam Question on Chi-Square Tests
A study was performed to examine the attendance pattern and exam performance
of students in an intro statistics course. A sample of 96 students enrolled in an
intro stat course was selected (these can be considered a random sample of such
students). These 96 students were classified by their attendance status (regularly
attend lecture or not) and their performance on the midterm exam, classified as
low (below 50%), middle (50% to 80%), and high (above 80%). The data and SPSS
output are provided.
Regularly Attend? * Exam Performance Crosstabulation
Count
Low
Regularly
Attend?
Yes
No
Total
Exam Performance
Middle
High
12
20
28
16
12
8
28
32
36
Total
60
36
96
Chi-Square Tests
Pears on Chi-Square
Likelihood Ratio
Linear-by-Linear As sociation
N of Valid Cases
Value
8.195 a
8.298
8.067
96
df
2
2
1
Asymp. Sig.
.017
.016
.005
a. 0 cells (.0%) have expected count les s than 5. The minimum
expected count is 10.50.
Give the name of the chi-square test for assessing if there is a relationship
between attendance status and exam performance.
_______________________________________________________________
b. Based on the above data, what proportion of regular attendees performed
high on the exam?
Final answer: ________________________
c. Assuming there is no relationship between attendance status and exam
performance, how many regular attendees would you expect to perform high
on the exam? Show all work.
Final answer: ________________________
137
d. Assuming there is no relationship between attendance status and exam
performance, what is the distribution of the test statistic? (Include all relevant
information.)
Final answer: ___________________________________
e. Use a level of 0.05 to assess if there is a significant relationship between
attendance status and exam performance.
Test Statistic Value: ________________
Thus, there (circle your answer):
does
p-value: ______________________
does not
appear to be
an association between attendance status and exam performance.
138
139