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AP Statistics 5.5 Sampling Distributions for Sample Proportions - MCQs - Exam Style Questions

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Question

In a certain county, 84 percent of high school graduates obtained a driver’s license before graduation. Which of the following samples is more likely to have more than 90 percent of high school graduates who obtained a driver’s license before graduation?
• Sample A: A random sample of 200 high school graduates from the county
• Sample B: A random sample of 400 high school graduates from the county
(A) Sample A, because there is less sampling variability in the sampling distribution for samples of size 200 than samples of size 400
(B) Sample A, because there is more sampling variability in the sampling distribution for samples of size 200 than samples of size 400
(C) Sample B, because there is more sampling variability in the sampling distribution for samples of size 400 than samples of size 200
(D) Sample B, because there is less sampling variability in the sampling distribution for samples of size 400 than samples of size 200
(E) Neither, because the mean of the sampling distribution is the same for both samples
▶️ Answer/Explanation
Detailed solution
For a sample proportion, the standard deviation is \(\sqrt{\dfrac{p(1-p)}{n}}\).
A smaller \(n\) → larger spread of the sampling distribution → more chance to see values far from \(p=0.84\) (e.g., \(>0.90\)).
Therefore the smaller sample (n=200) is more likely to exceed \(0.90\).
✅ Answer: (B) Sample A, because there is more sampling variability for n = 200 than for n = 400.

Question

A large movie theater franchise rewards Movie Club members who attend a show during the month of their birthday with a discount on items purchased from the theater’s snack bar. Assume that \(52.5\%\) of all Movie Club members will use the birthday reward. Suppose a manager at the franchise will survey a random sample of \(40\) Movie Club members who are celebrating a birthday next month. The manager will ask each member whether they will take advantage of the birthday reward. Let \(\hat p\) be the proportion of members in the sample who say they will take advantage of the birthday reward during the month of their birthday. The franchise wants to verify the normality of the sampling distribution of \(\hat p\).
Which of the following statements provides the most appropriate justification for the approximate normality of the sampling distribution of \(\hat p\)?
(A) The sample size of \(40\) Movie Club members is less than \(\dfrac{1}{10}\) of the assumed minimum population size of more than \(400\) Movie Club members.
(B) The sample size of Movie Club members is greater than \(30\), so the central limit theorem applies.
(C) The Movie Club members who participated in the survey were chosen from a random sample.
(D) In the random sample of \(40\) Movie Club members, \(21\) are expected to use the birthday reward and \(19\) are expected not to use the birthday reward. Both expected counts are greater than \(10\).
(E) The parameter \(52.5\%\) is close to \(50\%\), resulting in a near-symmetric sampling distribution.
▶️ Answer/Explanation
Step 1: Identify the correct normality condition for proportions

For a sampling distribution of a proportion \(\hat p\) to be approximately normal, we need the success–failure condition (a.k.a. large counts):
\(np \ge 10\) and \(n(1-p) \ge 10\), where \(n\) is the sample size and \(p\) is the population proportion.

Step 2: Compute the expected counts

Here, \(n=40\) and \(p=0.525\). Then:
\(np = 40(0.525) = 21 \ge 10\)
\(n(1-p) = 40(0.475) = 19 \ge 10\).
Both conditions are satisfied, so the sampling distribution of \(\hat p\) is approximately normal.

Step 3: Evaluate the choices

(A) Addresses the independence condition (10% condition), not normality. ❌
(B) \(n>30\) CLT is for means, not proportions. ❌
(C) Random sampling is important, but by itself doesn’t ensure normality. ❌
(D) Uses the success–failure counts \(np=21\) and \(n(1-p)=19\), both \(\ge 10\). ✔️
(E) “Close to \(50\%\)” is not a formal criterion. ❌

✅ Answer: (D)

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