A survey asks whether the population proportion favors the union more than 0.60. In a sample, 74 of 120 employees favor the union. Which inference method is appropriate to test whether the true proportion is greater than 0.60?

A survey asks whether the population proportion favors the union more than 0.60. In a sample, 74 of 120 employees favor the union. Which inference method is appropriate to test whether the true proportion is greater than 0.60?

• A. 2-proportion z-test
• B. 2-sample t-test
• C. Matched pairs t-test
• D. 1-proportion z-test

Answer: (D)

Testing a population proportion with a single sample against a specified value uses a one-proportion z-test. You have one sample of 120 employees and want to know if the true proportion who favor the union is greater than 0.60, so you compare the observed proportion to 0.60 using the normal approximation. Check the conditions: n p0 and n(1-p0) should be at least 5. Here p0 = 0.60 gives n p0 = 72 and n(1-p0) = 48, so the normal approximation is appropriate. Compute the sample proportion: p_hat = 74/120 ≈ 0.6167. The standard error under the null is sqrt(p0(1-p0)/n) = sqrt(0.6×0.4/120) ≈ 0.0447. The z statistic is (p_hat − p0)/SE ≈ (0.6167 − 0.60)/0.0447 ≈ 0.37. This yields a one-sided p-value of about 0.36, which is not significant at common alpha levels. The key point is that the correct method is a one-proportion z-test, not tests designed for two samples or for comparing means.

Testing a population proportion with a single sample against a specified value uses a one-proportion z-test. You have one sample of 120 employees and want to know if the true proportion who favor the union is greater than 0.60, so you compare the observed proportion to 0.60 using the normal approximation.

Check the conditions: n p0 and n(1-p0) should be at least 5. Here p0 = 0.60 gives n p0 = 72 and n(1-p0) = 48, so the normal approximation is appropriate.

Compute the sample proportion: p_hat = 74/120 ≈ 0.6167. The standard error under the null is sqrt(p0(1-p0)/n) = sqrt(0.6×0.4/120) ≈ 0.0447. The z statistic is (p_hat − p0)/SE ≈ (0.6167 − 0.60)/0.0447 ≈ 0.37.

This yields a one-sided p-value of about 0.36, which is not significant at common alpha levels. The key point is that the correct method is a one-proportion z-test, not tests designed for two samples or for comparing means.