In a NEJM study of mild gestational diabetes, 958 women were randomized to usual prenatal care (473) or intensive treatment (485). Among controls, 68 babies were large for gestational age (LGA), and among the treatment group, 29 were LGA. Which test should be used to compare these two proportions?

In a NEJM study of mild gestational diabetes, 958 women were randomized to usual prenatal care (473) or intensive treatment (485). Among controls, 68 babies were large for gestational age (LGA), and among the treatment group, 29 were LGA. Which test should be used to compare these two proportions?

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

Answer: (A)

When you want to compare whether two independent groups have different proportions for a binary outcome, use a two-proportion test. In this study, being large for gestational age is a yes/no outcome, and the two groups (usual care vs intensive treatment) are independent. The question asks if the proportion with LGA differs between groups, so applying a two-proportion z-test is appropriate because it uses the normal approximation to compare the difference in proportions with large-sample counts (473 vs 485, with 68 vs 29 events). A chi-square test of homogeneity could also address this in a 2x2 table, and would give a similar result in large samples, but the two-proportion z-test is the most direct, specific method for this exact comparison. A paired t-test is for paired data with a continuous outcome, and ANOVA is for comparing means across groups with continuous outcomes, not binary ones. So the two-proportion z-test is the best choice.

When you want to compare whether two independent groups have different proportions for a binary outcome, use a two-proportion test. In this study, being large for gestational age is a yes/no outcome, and the two groups (usual care vs intensive treatment) are independent. The question asks if the proportion with LGA differs between groups, so applying a two-proportion z-test is appropriate because it uses the normal approximation to compare the difference in proportions with large-sample counts (473 vs 485, with 68 vs 29 events).

A chi-square test of homogeneity could also address this in a 2x2 table, and would give a similar result in large samples, but the two-proportion z-test is the most direct, specific method for this exact comparison. A paired t-test is for paired data with a continuous outcome, and ANOVA is for comparing means across groups with continuous outcomes, not binary ones.

So the two-proportion z-test is the best choice.