8.14:
Test for Homogeneity
Test for homogeneity is another chi-square-based test determining if two or more populations have similar distributions of a categorical variable.
In contrast, the other two chi-square tests, the goodness-of-fit and the test for independence, deal with the data from a single population.
Suppose researchers wish to study the susceptibility of carriers of sickle cell anemia to malaria infection and compare it with people with normal RBCs.
The null hypothesis states that people with normal RBCs and carriers of sickle cell anemia are equally susceptible to malaria infection, while the alternate hypothesis states the contrary.
Since the infection rates of two independent populations are compared, a test for homogeneity is required instead of a test for independence.
This test uses calculations similar to other chi-square-based methods to determine the chi-square value and P-value.
The null hypothesis is rejected as the chi-square value exceeds the critical value.
This means that there is sufficient evidence to conclude that people with normal RBCs and carriers of the sickle cell anemia trait are not equally susceptible to malaria.
8.14:
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 conclude 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. The hypotheses for the test for homogeneity can be stated as follows:
H0: The distributions of the two populations are the same.
H1: The distributions of the two populations are not the same.
This test uses a chi-square test Statistic and is computed in the same way as the test for independence. The degrees of freedom for this test are given as df = number of columns – 1
The most common use for this test is comparing two populations. For example, men vs. women, before vs. after, and east vs. west. The variable is categorical, with more than two possible response values.
The expected frequency values for this test are required to be at least 5, similar to the chi-square-based tests. However, if any value is below five, one can use a Fischer Exact Test. It is helpful for all chi-square-based tests and provides an exact P-value. However, since the calculations involved in this test are complex, computer software such as Minitab and STATDISK are used.
This text is adapted from Openstax, Introductory Statistics, Section 11.4 Test for Homogeneity.