13.2 The Chi Square Test for Homogeneity of Populations The setting: Used to compare distribution of proportions in two or more populations. Data is organized in a two way table Explanatory variable (Treatments) are columns. Response variable levels are rows. Must have separate and independent random samples from each population. Each individual is classified based on one categorical variable. Each column (Treatment) can be considered a separate sample. H 0 : The distribution of the response variable is the same in all populations. H a : The distribution of the response variable is not the same in all populations.
72x100 divide by 300 = 24 expected count for cell one Step 1: I am interested in finding if the three cough suppressants act differently in suppressing coughs Step 2: I will perform a Chi Square Test of Homogeneity. The 300 subjects must be randomly allocated into the three treatment groups. The expected counts are greater than 5. Each subject is independent.
Chi square Test of Independence The chi square test of independence is probably the most frequently used hypothesis test in the social sciences. Two variable are associated or independent of one another. E.g. association between smoking and lung cancer. SRS or random allocation for experiments
I am interested in finding if political party is independent of opinion on a tax proposal. OR H 0 : There is no association between property tax reform and political party. H a : There is an association between property tax reform and political party.
or Since the p value is small (0.005 < p<0.01), there is evidence to suggest that opinion on tax reform and political party are associated (there is a relationship between opinion on tax reform and political party). ( ) ( ) 332.49 1447.51 1780 ( 418.22) 1820.78 ( ) 2239 ( 253.29) ( 1102.71) 1356 1004 4371 5375 or or There is no association between parent's smoking habits and child's smoking habits. There is an association between parent's smoking habits and child's smoking habits. H o : Parent's smoking habit is independent of child's smoking habit. H a : Parent's smoking habit is dependent on child's smoking habit. H o : There is not a relationship between a parent's smoking habit and a student's smoking habit. H a : There is a relationship between a parent's smoking habit and a student's smoking habit. I want to find if there is a relationship between a parent's smoking habit and a student's smoking habit.
Step 2: The conditions I will perform a chi-square test of independence. Both the parents and students are independent samples. The two samples need to be considered as an SRS. All expected counts are greater than one and at least 80% are more than 5. (20% under 5). Step 3: Calculations (400-332.49) 2 (1168-1102.71) 2 +... + 332.49 1102.71 = 13.7 + 3.15 + 0.01 + 0.003 + 16.83+ 3.87 = 37.563 df: ( 2-1) (2-1) Step 4 : Conclusion Since the p-value is less than any significance level, say alpha at 0.05, then we have significant statistical evidence to reject the null hypothesis. Therefore, we can conclude that parent's smoking habits and student's smoking habits are related.
Try:
Step 1: H o : There is no association between headache status and social status. H a : There is an association between headache status and social status. Step 2: To use the chi-square test of association/independence, we must check all expected cell counts are at least 1, and that no more than 20% are less than 5. Step 3: Calculations degrees of freedom - Step 4: Interpret your results in context.