ChiSquare Test of Homogeneity
This lesson explains how to conduct a
chisquare test of homogeneity.
The test is applied to a single
categorical variable from two or more different populations.
It is used to determine whether frequency counts are distributed
identically across different populations.
For example, in a survey of TV viewing preferences, we might ask
respondents to identify their favorite program. We might ask the
same question of two different populations, such as males and
females. We could use a chisquare test for
homogeneity to determine whether male viewing preferences
differed significantly from female viewing preferences. The
sample problem at the end of the lesson
considers this example.
When to Use ChiSquare Test for Homogeneity
The test procedure described in this lesson is appropriate when the
following conditions are met:
 If sample data are displayed in a
contingency table
(Populations x Category levels),
the expected frequency count for each cell of the table is
at least 5.
This approach consists of four steps: (1) state the hypotheses,
(2) formulate an analysis plan, (3) analyze sample data, and
(4) interpret results.
State the Hypotheses
Every hypothesis test requires the analyst
to state a
null hypothesis
and an
alternative hypothesis. The hypotheses are stated in such
a way that they are mutually exclusive. That is, if one is
true, the other must be false; and vice versa.
Suppose that data were sampled from r populations, and
assume that the categorical variable had c levels.
At any specified level of the categorical variable,
the null hypothesis states that each population has the same
proportion of observations. Thus,
H_{o}: P_{level 1 of pop 1}
= P_{level 1 of pop 2} = . . .
= P_{level 1 of pop r}
H_{o}: P_{level 2 of pop 1}
= P_{level 2 of pop 2} = . . .
= P_{level 2 of pop r}
. . .
H_{o}: P_{level c of pop 1}
= P_{level c of pop 2} = . . .
= P_{level c of pop r}

The alternative hypothesis (H_{a}) is that at least
one of the null hypothesis statements is false.
Formulate an Analysis Plan
The analysis plan describes
how to use sample data to accept or reject the null
hypothesis. The plan should specify the following elements.
 Significance level. Often, researchers choose
significance levels
equal to
0.01, 0.05, or 0.10; but any value between 0 and
1 can be used.
 Test method. Use the
chisquare test for homogeneity
to determine whether observed sample frequencies differ
significantly from expected frequencies specified in the
null hypothesis. The chisquare test for homogeneity is
described in the next section.
Analyze Sample Data
Using sample data from the contingency tables, find the
degrees of freedom, expected frequency counts,
test statistic, and the Pvalue associated with the test statistic.
The analysis described in this section is illustrated in the
sample problem at the end of this lesson.
 Degrees of freedom. The
degrees of freedom (DF) is equal to:
DF = (r  1) * (c  1)
where r is the number of populations, and
c is the number of levels for the categorical
variable.
 Expected frequency counts. The expected frequency counts
are computed separately for each population
at each level of the categorical variable, according to the
following formula.
E_{r,c} = (n_{r} * n_{c}) / n
where
E_{r,c} is the expected frequency count for
population r
at level c of the categorical variable,
n_{r} is the total number of observations from
population r,
n_{c} is the total number of observations at
treatment level c, and
n is the total sample size.
 Test statistic. The test statistic is a chisquare random variable
(Χ^{2}) defined by
the following equation.
Χ^{2} =
Σ [ (O_{r,c}  E_{r,c})^{2} / E_{r,c} ]
where
O_{r,c} is the observed frequency count in population r
for level c of the categorical variable, and
E_{r,c} is the expected frequency count in
population r for level c of the categorical
variable.
 Pvalue. The Pvalue is the probability of observing a
sample statistic as extreme as the test statistic. Since the
test statistic is a chisquare, use the
ChiSquare Distribution Calculator
to assess the probability associated with the test statistic. Use
the degrees of freedom computed above.
Interpret Results
If the sample findings are unlikely, given
the null hypothesis, the researcher rejects the null hypothesis.
Typically, this involves comparing the Pvalue to the
significance level,
and rejecting the null hypothesis when the Pvalue is less than
the significance level.
Test Your Understanding
Problem
In a study of the television viewing habits of children, a
developmental psychologist selects a random sample of
300 first graders  100 boys and 200 girls. Each child is asked
which of the following TV programs they like best: The Lone Ranger,
Sesame Street, or The Simpsons. Results are shown in the
contingency table below.

Viewing Preferences 
Total 
Lone RangerLone Ranger 
Sesame StreetSesame Street 
The SimpsonsThe Simpsons 
Boys 
50 
30 
20 
100 
Girls 
50 
80 
70 
200 
Total 
100 
110 
90 
300 
Do the boys' preferences for these TV programs differ
significantly from the girls' preferences? Use a 0.05
level of significance.
Solution
The solution to this problem takes four steps:
(1) state the hypotheses, (2) formulate an analysis plan,
(3) analyze sample data, and (4) interpret results.
We work through those steps below:
State the hypotheses. The first step is to
state the null hypothesis and an alternative hypothesis.
 Null hypothesis: The null hypothesis states that the proportion
of boys who prefer the Lone Ranger is identical to the
proportion of girls. Similarly, for the other programs. Thus,
H_{o}: P_{boys like Lone Ranger}
= P_{girls like Lone Ranger}
H_{o}: P_{boys like Sesame Street}
= P_{girls like Sesame Street}
H_{o}: P_{boys like Simpsons}
= P_{girls like Simpsons}

 Alternative hypothesis: At least one of the null
hypothesis statements is false.
 Formulate an analysis plan. For this analysis,
the significance level is 0.05. Using sample data, we will
conduct a
chisquare test for homogeneity.
 Analyze sample data. Applying the chisquare
test for homogeneity to sample data, we compute
the degrees of freedom,
the expected frequency counts, and
the chisquare test statistic. Based on the
chisquare statistic and the
degrees of freedom, we determine the
Pvalue.
DF = (r  1) * (c  1)
DF = (r  1) * (c  1) = (2  1) * (3  1) = 2
E_{r,c} = (n_{r} * n_{c}) / n
E_{1,1} = (100 * 100) / 300 = 10000/300 = 33.3
E_{1,2} = (100 * 110) / 300 = 11000/300 = 36.7
E_{1,3} = (100 * 90) / 300 = 9000/300 = 30.0
E_{2,1} = (200 * 100) / 300 = 20000/300 = 66.7
E_{2,2} = (200 * 110) / 300 = 22000/300 = 73.3
E_{2,3} = (200 * 90) / 300 = 18000/300 = 60.0
Χ^{2} = Σ[ (O_{r,c}  E_{r,c})^{2} / E_{r,c} ]
Χ^{2} =
(50  33.3)^{2}/33.3 + (30  36.7)^{2}/36.7 +
(20  30)^{2}/30 + (50  66.7)^{2}/66.7 +
(80  73.3)^{2}/73.3 + (70  60)^{2}/60
Χ^{2} =
(16.7)^{2}/33.3 + (6.7)^{2}/36.7 +
(10.0)^{2}/30 + (16.7)^{2}/66.7 +
(3.3)^{2}/73.3 + (10)^{2}/60
Χ^{2} = 8.38 + 1.22 + 3.33 + 4.18 + 0.61 + 1.67 = 19.39
where
DF is the degrees of freedom,
r is the number of populations,
c is the number of levels of the categorical variable,
n_{r} is the number of observations from population r,
n_{c} is the number of observations from level c
of the categorical variable,
n is the number of observations in the sample,
E_{r,c} is the expected frequency count in population
r for level c, and
O_{r,c} is the observed frequency count in population
r for level c.
The Pvalue is the probability that a chisquare statistic
having 2 degrees of freedom is more extreme than 19.39.
We use the
ChiSquare Distribution Calculator
to find P(Χ^{2} > 19.39) = 0.0000. (The actual Pvalue,
of course, is not exactly zero. If the ChiSquare Distribution
Calculator
reported more than four decimal places, we would find that the
actual Pvalue is a very small number that is less than 0.00005 and
greater than zero.)
 Interpret results. Since the Pvalue (0.0000) is
less than the significance level (0.05), we reject the
null hypothesis.
Note: If you use this approach on an exam, you may also want to mention
why this approach is appropriate. Specifically, the approach is
appropriate because the sampling method was simple random sampling, the
variable under study was categorical, and the expected frequency count
was at least 5 in each population at each level of the categorical
variable.