At one vehicle inspection station, 13 of 52 trucks and 11 of 88 cars failed the emissions test. Is this evidence of a difference in the percentages of all cars and trucks that are not in compliance with air quality regulations?

At one vehicle inspection station, 13 of 52 trucks and 11 of 88 cars failed the emissions test. Is this evidence of a difference in the percentages of all cars and trucks that are not in compliance with air quality regulations?

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

Answer: (B)

When you want to know if two independent groups have different failure rates, you compare their proportions with a two-proportion z-test. Here, trucks have 13 failures out of 52, and cars have 11 failures out of 88. The goal is to see if the true failure rates p1 and p2 differ. Compute the sample proportions: p1̂ = 13/52 = 0.25 and p2̂ = 11/88 = 0.125. Under the null hypothesis that the true proportions are equal, pool the successes and totals to get the pooled proportion p̂ = (13 + 11) / (52 + 88) ≈ 0.171. The standard error for the difference in proportions is sqrt[p̂(1 − p̂)(1/n1 + 1/n2)], which works out to about 0.066. The observed difference is 0.25 − 0.125 = 0.125, so the z-statistic is roughly 0.125 / 0.066 ≈ 1.9, corresponding to a two-tailed p-value around 0.06. This method is the best choice because it directly tests whether two independent proportions differ. A paired t-test wouldn’t apply since the groups aren’t matched. A one-proportion z-test would test a single group against a fixed value, not two groups. A chi-square test could be used on a 2×2 table, but the two-proportion z-test is the standard focused approach for comparing two independent proportions.

When you want to know if two independent groups have different failure rates, you compare their proportions with a two-proportion z-test. Here, trucks have 13 failures out of 52, and cars have 11 failures out of 88. The goal is to see if the true failure rates p1 and p2 differ.

Compute the sample proportions: p1̂ = 13/52 = 0.25 and p2̂ = 11/88 = 0.125. Under the null hypothesis that the true proportions are equal, pool the successes and totals to get the pooled proportion p̂ = (13 + 11) / (52 + 88) ≈ 0.171. The standard error for the difference in proportions is sqrt[p̂(1 − p̂)(1/n1 + 1/n2)], which works out to about 0.066. The observed difference is 0.25 − 0.125 = 0.125, so the z-statistic is roughly 0.125 / 0.066 ≈ 1.9, corresponding to a two-tailed p-value around 0.06.

This method is the best choice because it directly tests whether two independent proportions differ. A paired t-test wouldn’t apply since the groups aren’t matched. A one-proportion z-test would test a single group against a fixed value, not two groups. A chi-square test could be used on a 2×2 table, but the two-proportion z-test is the standard focused approach for comparing two independent proportions.