Two dog parks in a city are studied. Dogs are categorized by size: toy, small, medium, large, and giant. Is the breed size distribution different between the two parks?

Two dog parks in a city are studied. Dogs are categorized by size: toy, small, medium, large, and giant. Is the breed size distribution different between the two parks?

• A. ANOVA
• B. Paired t-test
• C. 2-proportion z-test
• D. Chi-square test for homogeneity

Answer: (D)

When you want to know if the distribution of a categorical variable is the same across two independent groups, you use a chi-square test for homogeneity. Here, dogs are categorized by size into five groups, and you’re comparing how those size categories are distributed between the two parks. You’d put the counts in a 2-by-5 contingency table (parks by size categories), test the null that the distributions are identical across parks, and look at the chi-square statistic with degrees of freedom (2−1)×(5−1) = 4. A significant result means the size distribution differs between parks; a non-significant result suggests the distributions are similar. This approach is appropriate because it analyzes multiple categorical outcomes simultaneously across two groups. ANOVA is for continuous outcomes, so it wouldn’t fit here. A paired t-test requires related or paired data, which isn’t the case since the parks are independent. A 2-proportion z-test compares only a single proportion between two groups and doesn’t assess the full distribution across all five size categories, whereas the chi-square test for homogeneity evaluates the entire distribution at once.

When you want to know if the distribution of a categorical variable is the same across two independent groups, you use a chi-square test for homogeneity. Here, dogs are categorized by size into five groups, and you’re comparing how those size categories are distributed between the two parks. You’d put the counts in a 2-by-5 contingency table (parks by size categories), test the null that the distributions are identical across parks, and look at the chi-square statistic with degrees of freedom (2−1)×(5−1) = 4. A significant result means the size distribution differs between parks; a non-significant result suggests the distributions are similar.

This approach is appropriate because it analyzes multiple categorical outcomes simultaneously across two groups. ANOVA is for continuous outcomes, so it wouldn’t fit here. A paired t-test requires related or paired data, which isn’t the case since the parks are independent. A 2-proportion z-test compares only a single proportion between two groups and doesn’t assess the full distribution across all five size categories, whereas the chi-square test for homogeneity evaluates the entire distribution at once.