Chi-Square Analysis Test of Homogeneity

Sometimes we compare samples of different populations for certain characteristics. This data is often presented in a two way table. The test we use is a chi-square test of homogeneity. This is one of two similar chi-square tests, the other being a test for independence.

Relapse No Relapse Yes Total Desipramine Lithium61824 Placebo42024 Total The data represent random samples of cocaine addicts treated with different drugs to reduce the likelyhood of a relapse. Are the proportion of addicts who relapse the same for each category?

Step 1: H a : The proportion of cocaine addicts who relapse is not the same for all treatments. H 0 : The proportion of cocaine addicts who relapse is the same for the different treatments. Step 2: Assumptions: Our data are counts. We have random samples of the populations, as stated.

To find expected counts we enter the data in [A], run the test, and check [B]. All expected counts are more than 5.

Step 3: Degrees of freedom are the number of rows minus 1 times the number of columns minus 1.

Step 4: This graph does not really show that a tiny tail region should be shaded. Step 5:

Step 6: Reject H 0, a test statistic this large will occur by chance alone less than 1% of the time. Step 7: We have strong evidence that the proportion of cocaine addicts who relapse is not the same for every treatment. Further if we examine the actual contributions from each cell, we may be able to see the reason for our positive results.

In this example, we see that addicts treated with desipramine contributed the most to the test statistic. We can carry out a specific comparison of desipramine and placebo using a 2- proportion Z test, if we want to make an exact comparison.

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