An exercise scientist claims that exercise is the closest thing we have to a miracle drug, especially when it comes to sleeping well. The scientist claims that 92% of adult runners sleep well at night. To investigate this claim, an independent study selects a random sample of 200 adult runners and finds that 95% of them sleep well at night. Do these data provide convincing evidence that the true proportion of all adult runners who sleep well at night is different than 92%? The z-table can be found here.What is the correct test statistic, P-value, and conclusion of this test?
Answer
A
z = –1.56, P-value = 0.118, reject H0. We have convincing evidence to conclude that the true proportion of all adult runners who sleep well at night is different than 92%.
B
z = 1.56, P-value = 0.059, reject H0. We have convincing evidence to conclude that the true proportion of all adult runners who sleep well at night is different than 92%.
C
z = 1.56, P-value = 0.118, fail to reject H0. We do not have convincing evidence to conclude that the true proportion of all adult runners who sleep well at night is different than 92%.
D
z = 0.95, P-value = 0.059, fail to reject H0. We do not have convincing evidence to conclude that the true proportion of all adult runners who sleep well at night is different than 92%.
Explanation
Test the null hypothesis that the true proportion is 0.92 against the two-sided alternative that it differs from 0.92. The standard error under the null is √(0.92 × 0.08 / 200) ≈ 0.0192, so z = (0.95 − 0.92) / 0.0192 ≈ 1.56. The two-sided P-value is about 2 × 0.059 = 0.118; this is not small enough to reject the null hypothesis. Thus, choice C is correct: there is not convincing evidence that the proportion differs from 92%.
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