Lecture 46 Section 14.5 Wed, Apr 13, 2005 Test of Independence Lecture 46 Section 14.5 Wed, Apr 13, 2005

Independence Only one sample is taken. For each subject in the sample, two observations are made (i.e., two variables are measured). We wish to determine whether there is a relationship between the two variables. The two variables are independent if there is no relationship between them.

Example Suppose a university researcher suspects that a student’s SAT-M score is related to his performance in Statistics. At the end of the semester, he compares each student’s grade to his SAT-M score for all Statistics classes at that university. He wants to know whether the student’s with the higher SAT-M scores got the higher grades.

Example Does there appear to be a difference between the rows? Or are the rows independent? Grade A B C D F 400 - 500 2 5 13 20 500 – 600 8 47 18 22 600 – 700 10 32 6 700 - 800 15 1 SAT-M

The Test of Homogeneity The null hypothesis is that the variables are independent. The alternative hypothesis is that the variables are not independent. H0: The variables are independent. H1: The variables are not independent.

The Test Statistic The test statistic is the chi-square statistic, computed as The question now is, how do we compute the expected counts?

Expected Counts Under the assumption of independence (H0), the rows should exhibit the same proportions. This is the same as when testing for homogeneity. Therefore, we may calculate the expected counts in the same way.

Expected Counts A B C D F 400 - 500 2 (6) 5 (14.4) 13 (18) 20 (10.8) 500 – 600 (10) 8 (24) 47 (30) 18 22 600 – 700 10 32 6 700 - 800 (3) 15 (7.2) (9) 1 (5.4)

Validity of the Test This test assumes that proportions are normally distributed, as we discussed in Chapter 9. That will be the case provided np is at least 5 for each cell. np represented the expected count. Therefore, for the test to be valid, it ought to be the case that every expected count be at least 5. In this example, one cell has expected count 3.

The Test Statistic The value of 2 is 112.323.

df = (no. of rows – 1)  (no. of cols – 1). Degrees of Freedom The degrees of freedom are the same as before df = (no. of rows – 1)  (no. of cols – 1). In our example, df = (4 – 1)  (5 – 1) = 12.

The p-value To find the p-value, calculate 2cdf(112.323, E99, 12) = 2.075  10-18. The results are significant.

Let’s Do It! Let’s Do It! 14.7, p. 884 – Hiring Based on Race? Simpson’s paradox.

TI-83 – Test of Independence The test for independence on the TI-83 is identical to the test for homogeneity.

Practice The following matrix shows height-weight data for this class. Enter the matrix into your TI-83 and perform the test. 145 – 160 165 – 180 185 – 280 65 – 69 5 4 1 70 – 71 2 3 72 – 77

Practice Suppose that we had similar results using 10 times as many students. Enter the matrix into your TI-83 and perform the test. 145 – 160 165 – 180 185 – 280 65 – 69 50 40 10 70 – 71 20 30 72 – 77