Sixty senior account executives are classified into three groups, A, B, and C. Testing whether the population proportions in these groups are equal uses which test?

Sixty senior account executives are classified into three groups, A, B, and C. Testing whether the population proportions in these groups are equal uses which test?

• A. Chi-square goodness-of-fit with 2 df
• B. Chi-square test of independence
• C. 1-proportion z-test
• D. ANOVA

Answer: (A)

You’re testing how a single categorical variable with three categories is distributed across those categories. Under the null, each category would have the same proportion, 1/3, of the total. A chi-square goodness-of-fit test compares the observed counts to these expected counts and uses the formula sum((O−E)²/E) across the categories. For three categories, the degrees of freedom are 3−1 = 2, so the test statistic follows a chi-square distribution with 2 degrees of freedom. This approach is appropriate here because you’re assessing whether the population proportions in the three groups are equal, not whether two variables are related (which would call for a chi-square test of independence), not a single proportion (which would use a 1-proportion z-test), and not comparing means (which would use ANOVA). For 60 executives, the expected counts would be 20 in each group, which satisfies the rule of thumb for the chi-square test.

You’re testing how a single categorical variable with three categories is distributed across those categories. Under the null, each category would have the same proportion, 1/3, of the total. A chi-square goodness-of-fit test compares the observed counts to these expected counts and uses the formula sum((O−E)²/E) across the categories. For three categories, the degrees of freedom are 3−1 = 2, so the test statistic follows a chi-square distribution with 2 degrees of freedom. This approach is appropriate here because you’re assessing whether the population proportions in the three groups are equal, not whether two variables are related (which would call for a chi-square test of independence), not a single proportion (which would use a 1-proportion z-test), and not comparing means (which would use ANOVA). For 60 executives, the expected counts would be 20 in each group, which satisfies the rule of thumb for the chi-square test.