A policeman believes that more than 40% of older drivers speed on highways, but a confidential survey found that 49 of 88 randomly selected older drivers admitted speeding. Is this strong evidence that the policeman was wrong?

A policeman believes that more than 40% of older drivers speed on highways, but a confidential survey found that 49 of 88 randomly selected older drivers admitted speeding. Is this strong evidence that the policeman was wrong?

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

Answer: (D)

The idea is to test a single population proportion against a specific value. Here you’re checking whether the true proportion of older drivers who speed is greater than 0.40, so you set up a one-proportion z-test with p0 = 0.40. Compute the sample proportion: p-hat = 49/88 ≈ 0.557. Under the null, the standard error is sqrt(p0(1−p0)/n) = sqrt(0.4×0.6/88) ≈ 0.0522. The z-statistic is (p-hat − p0)/SE ≈ (0.557 − 0.40)/0.0522 ≈ 3.0. A one-sided p-value for z ≈ 3.0 is about 0.0013, which is far below common significance levels (like 0.05). So you reject the null in favor of p > 0.40, meaning there is strong evidence that more than 40% of older drivers speed. This supports the policeman’s belief rather than showing he was wrong. The other tests aren’t appropriate here: a two-proportion z-test compares two independent proportions, a chi-square test for proportions is an alternative for categorical data but not needed when testing a single proportion, and a paired t-test is for comparing means of paired measurements.

The idea is to test a single population proportion against a specific value. Here you’re checking whether the true proportion of older drivers who speed is greater than 0.40, so you set up a one-proportion z-test with p0 = 0.40.

Compute the sample proportion: p-hat = 49/88 ≈ 0.557. Under the null, the standard error is sqrt(p0(1−p0)/n) = sqrt(0.4×0.6/88) ≈ 0.0522. The z-statistic is (p-hat − p0)/SE ≈ (0.557 − 0.40)/0.0522 ≈ 3.0. A one-sided p-value for z ≈ 3.0 is about 0.0013, which is far below common significance levels (like 0.05).

So you reject the null in favor of p > 0.40, meaning there is strong evidence that more than 40% of older drivers speed. This supports the policeman’s belief rather than showing he was wrong.

The other tests aren’t appropriate here: a two-proportion z-test compares two independent proportions, a chi-square test for proportions is an alternative for categorical data but not needed when testing a single proportion, and a paired t-test is for comparing means of paired measurements.