Read this chapter, which introduces you to the three major uses of the chi-squared distribution: the goodness-of-fit test, the test of independence, and the test of a single variance. Attempt the practice problems and homework at the end of the chapter.
Test for Homogeneity
The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence.
Note
The expected value inside each cell needs to be at least five in order for you to use this test.
Hypotheses
H0: The distributions of the two populations are the same.
Ha: The distributions of the two populations are not the same.
Test Statistic
Use a χ2 test statistic. It is computed in the same way as the test for independence.
Degrees of Freedom (df)
df = number of columns - 1
Requirements
All values in the table must be greater than or equal to five.
Common Uses
Comparing two populations. For example: men vs. women, before vs. after, east vs. west. The variable is categorical with more than two possible response values.
Example 11.11
Problem
Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are shown in Table 11.18. Do male and female college students have the same distribution of living arrangements?
Dormitory | Apartment | With Parents | Other | |
Males | 72 | 84 | 49 | 45 |
Females | 91 | 86 | 88 | 35 |
H0: The distribution of living arrangements for male college students is the same as the distribution of living arrangements for female college students.
Ha: The distribution of living arrangements for male college students is not the same as the distribution of living arrangements for female college students.
Degrees of Freedom (df):
df = number of columns – 1 = 4 – 1 = 3
Distribution for the test:
Calculate the test statistic:

Make a decision: Because the calculated test statistic is in the tail we cannot accept H0. This means that the distributions are not the same.
Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the distributions of living arrangements for male and female college students are not the same.
Try It 11.11
Sport | Sedan | Hatchback | Truck | Van/SUV | |
---|---|---|---|---|---|
Family | 5 | 15 | 35 | 17 | 28 |
Single | 45 | 65 | 37 | 46 | 7 |
Try It 11.11
Application type accepted | Brown | Columbia | Cornell | Dartmouth | Penn | Yale |
---|---|---|---|---|---|---|
Regular | 2,115 | 1,792 | 5,306 | 1,734 | 2,685 | 1,245 |
Early decision | 577 | 627 | 1,228 | 444 | 1,195 | 761 |
Table 11.20
We want to know if the number of regular applications accepted follows the same distribution as the number of early applications accepted. State the null and alternative hypotheses, the degrees of freedom and the test statistic, sketch the graph of the χ2 distribution and show the critical value and the calculated value of the test statistic, and draw a conclusion about the test of homogeneity.