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Chi-Square Test for Homogeneity of Proportions (or Homogeneity of Populations) for AP Statistics

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By — McGraw-Hill Professional
Updated on Feb 4, 2011

Practice problems for these concepts can be found at:

In this lesson we use the chi-square statistic but will investigate whether or not the values of a single categorical variable are proportional among two or more populations. In the previous section, we considered a situation in which a sample of 36 students was selected and were then categorized according to gender and political party preference. We then asked if gender and party preference are independent in the population. Now suppose instead that we had selected a random sample of 20 males from the population of males in the school and another, independent, random sample of 16 females from the population of females in the school. Within each sample we classify the students as Democrat, Republican, or Independent. The results are presented in the following table, which you should notice is exactly the same table we presented earlier when gender was a category.

Because "Male" and "Female" are now considered separate populations, we do not ask if gender and political party preference are independent in the population of students. We ask instead if the proportions of Democrats, Republicans, and Independents are the same within the populations of Males and Females. This is the test for homogeneity of proportions (or homogeneity of populations). Let the proportion of Male Democrats be p1; the proportion of Female Democrats be p2; the proportion of Male Republicans be p3; the proportion of Female Republicans be p4; the proportion of Independent Males be p5; and the proportion of Independent Females be p6. Our null and alternative hypotheses are then

H0: p1 = p2, p3 = p4, p5 = p6.

HA: Not all of the proportions stated in the null hypothesis are true.

It works just as well, and might be a bit easier, to state the hypotheses as follows.

H0: The proportions of Democrats, Republicans, and Independents are the same among Male and Female students.

HA: Not all of the proportions stated in the null hypothesis are equal.

For a given two-way table the expected values are the same under a hypothesis of homogeneity or independence.

example: A university dean suspects that there is a difference between how tenured and nontenured professors view a proposed salary increase. She randomly selects 20 nontenured instructors and 25 tenured staff to see if there is a difference. She gets the following results.

Do these data provide good statistical evidence that tenured and nontenured faculty differ in their attitudes toward the proposed salary increase?

solution:

X2 = z2

The example just completed could have been done as a two-proportion z-test where p1 and p2 are defined the same way as in the example (that is, the proportions of tenured and nontenured staff that favor the new plan). Then

Computation of the z-test statistics for the two-proportion z-test yields z = 1.333. Now, z2 = 1.78. Because the chi-square test and the two-proportion z-test are testing the same thing, it should come as little surprise that z2 equals the obtained value of X2 in the example. For a 2 × 2 table, the X2 test statistic and the value of z2 obtained for the same data in a two-proportion z-test are the same. Note that the z-test is somewhat more flexible in this case in that it allows for the possibility of a one-sided test, whereas the chi-square test is strictly two sided (that is, it is simply a rejection of H0). However, this advantage only holds for a 2 × 2 table since there is no way to use a simple z-test for situations with more than two rows and two columns.

Practice problems for these concepts can be found at:

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