Suppose you compare the proportion of students showing school pride between two schools. Which test would you use to test for a difference in proportions between two independent groups?

Suppose you compare the proportion of students showing school pride between two schools. Which test would you use to test for a difference in proportions between two independent groups?

• A. 1-proportion z-test
• B. 2-sample t-test
• C. 2-proportion z-test
• D. Chi-square test

Answer: (C)

When you want to know if two independent groups have different proportions, you test whether the difference between their sample proportions is truly different in the population. The standard method is a two-proportion z-test. You compute the sample proportions from each group, p1 = x1/n1 and p2 = x2/n2, where x is the number showing school pride and n is the group size. Under the null hypothesis that the population proportions are the same (p1 = p2), you pool the data to get the combined proportion p_hat = (x1 + x2) / (n1 + n2). The standard error for the difference uses this pooled proportion: SE = sqrt[p_hat(1 - p_hat) * (1/n1 + 1/n2)]. The test statistic is z = (p1 - p2) / SE, which you compare to the standard normal distribution to decide if the observed difference is likely due to chance. This approach is specifically for comparing two independent proportions and relies on a large-sample approximation. Other tests fit different scenarios: a 1-proportion z-test compares a single proportion to a known value, a 2-sample t-test compares means rather than proportions, and a chi-square test can assess association in a contingency table (and, in large samples, aligns with the two-proportion test results).

When you want to know if two independent groups have different proportions, you test whether the difference between their sample proportions is truly different in the population. The standard method is a two-proportion z-test.

You compute the sample proportions from each group, p1 = x1/n1 and p2 = x2/n2, where x is the number showing school pride and n is the group size. Under the null hypothesis that the population proportions are the same (p1 = p2), you pool the data to get the combined proportion p_hat = (x1 + x2) / (n1 + n2). The standard error for the difference uses this pooled proportion: SE = sqrt[p_hat(1 - p_hat) * (1/n1 + 1/n2)]. The test statistic is z = (p1 - p2) / SE, which you compare to the standard normal distribution to decide if the observed difference is likely due to chance.

This approach is specifically for comparing two independent proportions and relies on a large-sample approximation. Other tests fit different scenarios: a 1-proportion z-test compares a single proportion to a known value, a 2-sample t-test compares means rather than proportions, and a chi-square test can assess association in a contingency table (and, in large samples, aligns with the two-proportion test results).