A CDC report found arthritis among adults: 51 of 100 men and 80 of 782 women. To estimate the difference in proportions with 95% confidence, which method would you use?

A CDC report found arthritis among adults: 51 of 100 men and 80 of 782 women. To estimate the difference in proportions with 95% confidence, which method would you use?

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

Answer: (A)

Estimating the difference between two independent proportions with a confidence interval. When you want a 95% confidence interval for p1 − p2 based on two independent samples, you use a two-proportion z-interval. Take the sample proportions and compute the standard error as SE = sqrt[p1̂(1−p1̂)/n1 + p2̂(1−p2̂)/n2]. Then use the 1.96 multiplier (for 95% confidence) to form the interval: (p1̂ − p2̂) ± 1.96 × SE. Here, p1̂ = 51/100 = 0.51, p2̂ = 80/782 ≈ 0.1023, and the difference is about 0.4077. The standard error is SE ≈ sqrt(0.51×0.49/100 + 0.1023×0.8977/782) ≈ sqrt(0.002499 + 0.000118) ≈ 0.0512. So the 95% confidence interval for the difference is approximately 0.4077 ± 0.1004, i.e., (0.307, 0.508). This means we’re estimating that arthritis is higher among men by about 30.7 to 50.8 percentage points in this sample. The 2-proportion z-interval is the standard method for this kind of estimate; the 2-proportion z-test would be used for testing whether the two proportions are equal, not for a confidence interval, and the other options are for different goals.

Estimating the difference between two independent proportions with a confidence interval. When you want a 95% confidence interval for p1 − p2 based on two independent samples, you use a two-proportion z-interval. Take the sample proportions and compute the standard error as SE = sqrt[p1̂(1−p1̂)/n1 + p2̂(1−p2̂)/n2]. Then use the 1.96 multiplier (for 95% confidence) to form the interval: (p1̂ − p2̂) ± 1.96 × SE.

Here, p1̂ = 51/100 = 0.51, p2̂ = 80/782 ≈ 0.1023, and the difference is about 0.4077. The standard error is SE ≈ sqrt(0.51×0.49/100 + 0.1023×0.8977/782) ≈ sqrt(0.002499 + 0.000118) ≈ 0.0512. So the 95% confidence interval for the difference is approximately 0.4077 ± 0.1004, i.e., (0.307, 0.508).

This means we’re estimating that arthritis is higher among men by about 30.7 to 50.8 percentage points in this sample. The 2-proportion z-interval is the standard method for this kind of estimate; the 2-proportion z-test would be used for testing whether the two proportions are equal, not for a confidence interval, and the other options are for different goals.