Chi-Square Test of Homogeneity

This lesson explains how to conduct a chi-square 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 chi-square 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 Chi-Square Test for Homogeneity

The test procedure described in this lesson is appropriate when the following conditions are met:

For each population, the sampling method is simple random sampling.
The variable under study is categorical.
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 chi-square test for homogeneity to determine whether observed sample frequencies differ significantly from expected frequencies specified in the null hypothesis. The chi-square 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 P-value 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 chi-square random variable (Χ²) defined by the following equation.
Χ² = 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.
P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a chi-square, use the Chi-Square 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 P-value to the significance level, and rejecting the null hypothesis when the P-value 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	Total
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 chi-square test for homogeneity.
Analyze sample data. Applying the chi-square test for homogeneity to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
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

Χ² = Σ[ (O_r,c - E_r,c)² / E_r,c ]
Χ² = (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
Χ² = (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
Χ² = 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 P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 19.39.
We use the Chi-Square Distribution Calculator to find P(Χ² > 19.39) = 0.0000. (The actual P-value, of course, is not exactly zero. If the Chi-Square Distribution Calculator reported more than four decimal places, we would find that the actual P-value is a very small number that is less than 0.00005 and greater than zero.)
Interpret results. Since the P-value (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.