Honors Graduates last year had counts from public, private, home-schooled, and abroad. Is there evidence that the distribution of high school types among Honors Graduates differs from the distribution among the rest of the college students?

Honors Graduates last year had counts from public, private, home-schooled, and abroad. Is there evidence that the distribution of high school types among Honors Graduates differs from the distribution among the rest of the college students?

• A. Chi-square goodness of fit with 2 degrees of freedom
• B. Chi-square test of independence with 6 degrees of freedom
• C. ANOVA
• D. Z-test for proportions

Answer: (A)

The main idea is to test whether the distribution of a categorical variable—high school type (public, private, home-schooled, abroad)—is the same in Honors Graduates as it is in the rest of the college students. When you have counts in several categories and you want to see if one group follows a specified distribution (the distribution seen in the rest of the students), you use a chi-square goodness-of-fit test. This test compares the observed counts for Honors Graduates in each high school category to the expected counts you'd get if Honors Graduates followed the same distribution as the rest of the students. You compute those expected counts, calculate the chi-square statistic sum((observed − expected)² / expected), and use it to determine whether the observed pattern could reasonably arise if there were no real difference between the groups. If the statistic is large, it suggests the Honors group does not mirror the rest of the student body in high school type. Why not the other options? ANOVA is for comparing means of a continuous variable across groups, not counts across categories. A Z-test for proportions looks at a single proportion against a fixed value, not the entire distribution across multiple categories. A chi-square test of independence (the two-group, four-category contingency table) would also test compatibility of the two groups’ distributions, but the situation is typically framed as a goodness-of-fit check against the known distribution of the rest of the students, which is exactly what this test assesses.

The main idea is to test whether the distribution of a categorical variable—high school type (public, private, home-schooled, abroad)—is the same in Honors Graduates as it is in the rest of the college students. When you have counts in several categories and you want to see if one group follows a specified distribution (the distribution seen in the rest of the students), you use a chi-square goodness-of-fit test.

This test compares the observed counts for Honors Graduates in each high school category to the expected counts you'd get if Honors Graduates followed the same distribution as the rest of the students. You compute those expected counts, calculate the chi-square statistic sum((observed − expected)² / expected), and use it to determine whether the observed pattern could reasonably arise if there were no real difference between the groups. If the statistic is large, it suggests the Honors group does not mirror the rest of the student body in high school type.

Why not the other options? ANOVA is for comparing means of a continuous variable across groups, not counts across categories. A Z-test for proportions looks at a single proportion against a fixed value, not the entire distribution across multiple categories. A chi-square test of independence (the two-group, four-category contingency table) would also test compatibility of the two groups’ distributions, but the situation is typically framed as a goodness-of-fit check against the known distribution of the rest of the students, which is exactly what this test assesses.