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AP Statistics 6.5 Interpreting p-Values- MCQs - Exam Style Questions

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Question

After reading a report that indicates that 67 percent of people in a country use a certain social media platform, David decides to investigate whether the percent is greater for people in the city where he lives. From a random sample of 500 people from the city, 72 percent indicated they use the social media platform. David then tests the following hypotheses for the proportion \(p\) of all people in the city who use the social media platform: \(H_0:p=0.67\) and \(H_a:p>0.67\).
David’s one-sample z test for a population proportion produced a test statistic of 2.38 and a p-value of 0.009. Which of the following statements is true?
(A) The probability of obtaining a test statistic equal to 2.38 from a random sample of 500 people in the city is 0.009, if the actual percent is 67%.
(B) The probability of obtaining a test statistic of 2.38 or greater from a random sample of 500 people in the city is 0.009, if the actual percent is 67%.
(C) The probability is 0.009 that the percent of people in the city who use the social media platform is greater than 67%.
(D) The probability is 0.009 that the percent of people in the city who use the social media platform is equal to 67%.
(E) The probability is 0.009 that the percent of people in the city who use the social media platform is at most 67%.
▶️ Answer/Explanation
Detailed solution
The p-value is the probability, assuming \(H_0\) is true (here \(p=0.67\)), of getting a test statistic as large as or larger than the observed value.
So it is \(P(Z\ge2.38\mid p=0.67, n=500)\approx 0.009\).
That matches option (B). Options (C)–(E) misinterpret the p-value as a probability statement about the parameter.
✅ Answer: (B)

Question

A researcher conducted a \(t\)-test of the hypotheses \(H_{0}:\,\mu=38\) versus \(H_{a}:\,\mu\ne 38\). The sample mean was \(35\) and the \(p\)-value for the test was \(0.0627\). What would the \(p\)-value have been if the researcher had used \(H_{a}:\,\mu<38\) as the alternative hypothesis?
(A) \(1-0.0627\)
(B) \(1-2(0.0627)\)
(C) \(1-\left(\dfrac{1}{2}\right)(0.0627)\)
(D) \(2(0.0627)\)
(E) \(\dfrac{1}{2}(0.0627)\)
▶️ Answer/Explanation
Detailed solution

The original test was two-sided: \(H_a:\,\mu\ne 38\). Its \(p\)-value \(0.0627\) equals the combined area in both tails beyond \(\pm t_{\text{obs}}\).
Changing to the left-tailed alternative \(H_a:\,\mu<38\) uses only the left tail (same observed \(t\) since \(\bar x=35<38\)).
Therefore the one-sided \(p\)-value is half of the two-sided value: \(\dfrac{1}{2}\times 0.0627=0.03135\).
✅ Answer: (E)

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