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Chi-Square Analysis Test of Homogeneity. Sometimes we compare samples of different populations for certain characteristics. When we do so we are performing.

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Presentation on theme: "Chi-Square Analysis Test of Homogeneity. Sometimes we compare samples of different populations for certain characteristics. When we do so we are performing."— Presentation transcript:

1 Chi-Square Analysis Test of Homogeneity

2 Sometimes we compare samples of different populations for certain characteristics. When we do so we are performing a test of homogeneity. The following data come from an article “Heavy Drinking and Problems Among Wine Drinkers” (Journal of Studies on Alcohol (1999): 467-471) analyzed drinking problems among Canadians. The article gave the percentage of drinkers who never drank to intoxication for seven different types of drinkers. We have the following data:

3 Never Intoxicated Sometimes Intoxicated Beer Only515741 Wine Only875232 Spirits Only387372 Beer and Wine774560 Beer and Spirits312727 Wine and Spirits708349 Beer, Wine, and Spirits10321119 The data represent seven random samples of drinkers of the different categories. Are the proportion of drinkers who drink (sometimes) to intoxication the same for each category?

4 Step 1: H a : The proportion of drinkers who sometimes drink to intoxication is not the same for the different groups. H 0 : The proportion of drinkers who sometimes drink to intoxication is the same for the different groups. Step 2: Assumptions: Our data are counts. We have random samples of the seven populations, as stated.

5 To find expected counts we enter the data in [A], run the  2 test, and check [B]. All expected counts are more than 5.

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

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

8 Step 6: Reject H 0, a value this extreme will rarely occur by chance alone. Step 7: We have strong evidence that the proportion who drink (sometimes) to intoxication varies with the group. Further if we examine the actual  2 contributions from each cell, we may be able to see the reason for our positive results.

9 As we examine the observed and expected counts, it is evident that several groups have counts different from expected. It should be noted that the sample sizes are large, and with  2 analysis the  2 statistic gets larger and larger as the sample size increases, if the the null hypothesis is false. This means that the exact  2 statistic is very dependent upon the sample size and you should not conclude so much about the size of the p-value, if it the test is significant. As sample sizes increase it becomes less and less likely that we will see no deviation from an expected pattern. This is a weakness in  2 analysis. If you have a significant test, you may gain important information by examining the  2 contributions of each cell. This type of analysis is most often done by computer.

10 We will stop the analysis here, as we do not have the component chi-square contributions for each cell, and the calculation is fairly tedious.

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