Test for Homogeneity

10.13 Test for Homogeneity

When we tested for independence in Section 10.12, a random sample of 1000 vot- ers was selected and the row and column totals for our contingency table were determined by chance. Another type of problem for which the method of Section

10.12 applies is one in which either the row or column totals are predetermined. Suppose, for example, that we decide in advance to select 200 Democrats, 150 Republicans, and 150 Independents from the voters of the state of North Carolina and record whether they are for a proposed abortion law, against it, or undecided. The observed responses are given in Table 10.8.

Table 10.8: Observed Frequencies Political Affiliation

Abortion Law Democrat Republican Independent Total

150 500 Now, rather than test for independence, we test the hypothesis that the popu- lation proportions within each row are the same. That is, we test the hypothesis

that the proportions of Democrats, Republicans, and Independents favoring the abortion law are the same; the proportions of each political affiliation against the law are the same; and the proportions of each political affiliation that are unde- cided are the same. We are basically interested in determining whether the three categories of voters are homogeneous with respect to their opinions concerning the proposed abortion law. Such a test is called a test for homogeneity.

Assuming homogeneity, we again find the expected cell frequencies by multi- plying the corresponding row and column totals and then dividing by the grand

10.13 Test for Homogeneity 377 total. The analysis then proceeds using the same chi-squared statistic as before.

We illustrate this process for the data of Table 10.8 in the following example. Example 10.14: Referring to the data of Table 10.8, test the hypothesis that opinions concerning

the proposed abortion law are the same within each political affiliation. Use a 0.05 level of significance.

Solution :

1. H 0 : For each opinion, the proportions of Democrats, Republicans, and Inde- pendents are the same.

2. H 1 : For at least one opinion, the proportions of Democrats, Republicans, and Independents are not the same.

4. Critical region: χ 2 > 9.488 with v = 4 degrees of freedom.

5. Computations: Using the expected cell frequency formula on page 375, we need to compute 4 cell frequencies. All other frequencies are found by sub- traction. The observed and expected cell frequencies are displayed in Table

Table 10.9: Observed and Expected Frequencies

Political Affiliation

Abortion Law Democrat Republican Independent Total

6. Decision: Do not reject H 0 . There is insufficient evidence to conclude that the proportions of Democrats, Republicans, and Independents differ for each stated opinion.

Testing for Several Proportions

The chi-squared statistic for testing for homogeneity is also applicable when testing the hypothesis that k binomial parameters have the same value. This is, therefore, an extension of the test presented in Section 10.9 for determining differences be- tween two proportions to a test for determining differences among k proportions. Hence, we are interested in testing the null hypothesis

H 0 :p 1 =p 2 =···=p k

378 Chapter 10 One- and Two-Sample Tests of Hypotheses against the alternative hypothesis, H 1 , that the population proportions are not all

equal. To perform this test, we first observe independent random samples of size n 1 ,n 2 ,...,n k from the k populations and arrange the data in a 2 × k contingency table, Table 10.10.

Table 10.10: k Independent Binomial Samples Sample:

Depending on whether the sizes of the random samples were predetermined or occurred at random, the test procedure is identical to the test for homogeneity or the test for independence. Therefore, the expected cell frequencies are calculated as before and substituted, together with the observed frequencies, into the chi-squared statistic

χ 2 (o i −e i ) = 2 ,

with

v = (2 − 1)(k − 1) = k − 1

degrees of freedom. By selecting the appropriate upper-tail critical region of the form χ 2 >χ 2 α , we

can now reach a decision concerning H 0 .

Example 10.15: In a shop study, a set of data was collected to determine whether or not the proportion of defectives produced was the same for workers on the day, evening, and night shifts. The data collected are shown in Table 10.11.

Table 10.11: Data for Example 10.15 Shift:

Day Evening Night

Defectives

870 Use a 0.025 level of significance to determine if the proportion of defectives is the

same for all three shifts. Solution : Let p 1 ,p 2 , and p 3 represent the true proportions of defectives for the day, evening, and night shifts, respectively.

1. H 0 :p 1 =p 2 =p 3 .

2. H 1 :p 1 ,p 2 , and p 3 are not all equal.

4. Critical region: χ 2 > 7.378 for v = 2 degrees of freedom.

10.14 Two-Sample Case Study 379

5. Computations: Corresponding to the observed frequencies o 1 = 45 and o 2 =

= 57.0 and e 2 =

All other expected frequencies are found by subtraction and are displayed in Table 10.12.

Table 10.12: Observed and Expected Frequencies Shift:

Night Total Defectives

6. Decision: We do not reject H 0 at α = 0.025. Nevertheless, with the above P -value computed, it would certainly be dangerous to conclude that the pro- portion of defectives produced is the same for all shifts.

Often a complete study involving the use of statistical methods in hypothesis testing can be illustrated for the scientist or engineer using both test statistics, complete with P -values and statistical graphics. The graphics supplement the numerical diagnostics with pictures that show intuitively why the P -values appear as they do, as well as how reasonable (or not) the operative assumptions are.