A college alumni fund compares two contact methods (phone versus email) for a random sample of alumni. In the phone group 40% donated and in the email group 30% donated, with 150 alumni in each group. The fund wishes to estimate the difference in donation proportions between the two contact methods. Which inference method is appropriate?

A college alumni fund compares two contact methods (phone versus email) for a random sample of alumni. In the phone group 40% donated and in the email group 30% donated, with 150 alumni in each group. The fund wishes to estimate the difference in donation proportions between the two contact methods. Which inference method is appropriate?

• A. 2-Proportion Z-Interval
• B. 1-Proportion Z-Interval
• C. Paired T-Test
• D. Chi-Square Test

Answer: (A)

Estimating the difference in proportions from two independent groups requires a two-proportion z-interval. Here you have two independent samples: alumni reached by phone and alumni reached by email. The interest is in p1 − p2, where p1 is the true donation rate for the phone method and p2 for the email method. With the data, p1̂ = 0.40, p2̂ = 0.30, and n1 = n2 = 150. A two-proportion z-interval uses the standard error based on the individual sample proportions: SE = sqrt[p1̂(1 − p1̂)/n1 + p2̂(1 − p2̂)/n2] Plugging in the numbers: SE ≈ sqrt[(0.40)(0.60)/150 + (0.30)(0.70)/150] ≈ sqrt[(0.24 + 0.21)/150] ≈ sqrt[0.003] ≈ 0.0548. The difference in sample proportions is 0.40 − 0.30 = 0.10. A (roughly) 95% confidence interval would be: 0.10 ± 1.96 × 0.0548 ≈ 0.10 ± 0.107, which gives about (−0.008, 0.207). This interval reflects the range of plausible values for the true difference in donation proportions between the two contact methods, based on the data. Why not the other options? A single-proportion interval is for one group, not two. A paired t-test is for matched or paired data and uses differences within pairs, not independent samples with a proportion outcome. A chi-square test is for testing associations in a contingency table or goodness-of-fit; it’s a test, not an interval estimate of the difference in proportions. The two-proportion z-interval is the appropriate method for estimating p1 − p2 with independent samples.

Estimating the difference in proportions from two independent groups requires a two-proportion z-interval. Here you have two independent samples: alumni reached by phone and alumni reached by email. The interest is in p1 − p2, where p1 is the true donation rate for the phone method and p2 for the email method.

With the data, p1̂ = 0.40, p2̂ = 0.30, and n1 = n2 = 150. A two-proportion z-interval uses the standard error based on the individual sample proportions:

SE = sqrt[p1̂(1 − p1̂)/n1 + p2̂(1 − p2̂)/n2]

Plugging in the numbers:

SE ≈ sqrt[(0.40)(0.60)/150 + (0.30)(0.70)/150] ≈ sqrt[(0.24 + 0.21)/150] ≈ sqrt[0.003] ≈ 0.0548.

The difference in sample proportions is 0.40 − 0.30 = 0.10. A (roughly) 95% confidence interval would be:

0.10 ± 1.96 × 0.0548 ≈ 0.10 ± 0.107, which gives about (−0.008, 0.207).

This interval reflects the range of plausible values for the true difference in donation proportions between the two contact methods, based on the data.

Why not the other options? A single-proportion interval is for one group, not two. A paired t-test is for matched or paired data and uses differences within pairs, not independent samples with a proportion outcome. A chi-square test is for testing associations in a contingency table or goodness-of-fit; it’s a test, not an interval estimate of the difference in proportions. The two-proportion z-interval is the appropriate method for estimating p1 − p2 with independent samples.