Two treatments with five possible responses; test whether the distributions across treatments are the same. Which test should be used and what are the degrees of freedom?

Two treatments with five possible responses; test whether the distributions across treatments are the same. Which test should be used and what are the degrees of freedom?

• A. Chi-square test of independence with 4 degrees of freedom
• B. Chi-square test of homogeneity with 4 degrees of freedom
• C. ANOVA
• D. Kruskal-Wallis test

Answer: (B)

When you want to compare how a categorical outcome is distributed across different groups, you use a chi-square test of homogeneity. Here you have two treatments (two groups) and five possible response categories, so you’d lay out a 2-by-5 contingency table showing how many observations in each treatment fall into each category. The null hypothesis is that the distribution of responses is the same for both treatments. The chi-square statistic checks whether the observed counts differ from what we'd expect if the distributions were identical across treatments. The degrees of freedom come from (rows minus one) times (columns minus one): (2−1) × (5−1) = 4. So this is a chi-square test of homogeneity with four degrees of freedom. The other tests aren’t appropriate here: ANOVA and Kruskal-Wallis compare means or central tendencies of a numeric outcome, not distributions of a categorical one; the chi-square test of independence would test association between two categorical variables in a single population, which is a related idea but framed differently from asking whether distributions are the same across independent groups.

When you want to compare how a categorical outcome is distributed across different groups, you use a chi-square test of homogeneity. Here you have two treatments (two groups) and five possible response categories, so you’d lay out a 2-by-5 contingency table showing how many observations in each treatment fall into each category. The null hypothesis is that the distribution of responses is the same for both treatments. The chi-square statistic checks whether the observed counts differ from what we'd expect if the distributions were identical across treatments. The degrees of freedom come from (rows minus one) times (columns minus one): (2−1) × (5−1) = 4. So this is a chi-square test of homogeneity with four degrees of freedom. The other tests aren’t appropriate here: ANOVA and Kruskal-Wallis compare means or central tendencies of a numeric outcome, not distributions of a categorical one; the chi-square test of independence would test association between two categorical variables in a single population, which is a related idea but framed differently from asking whether distributions are the same across independent groups.