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AP Statistics 3.2 Sampling Distributions for Sample Proportions- Exam Style Questions - MCQs - New Syllabus

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Question 

The director of a marketing department wants to estimate the proportion of people who purchase a certain product online. The director originally planned to obtain a random sample of 2,500 people who purchased the product. However, because of budget concerns, the sample size will be reduced to 1,500 people. Which of the following describes the effect of reducing the number of people in the sample?

(A) The variance of the sample will increase.
(B) The variance of the population will decrease.
(C) The variance of the sampling distribution of the estimator will increase.
(D) The variance of the sampling distribution of the estimator will decrease.
(E) The variance of the sampling distribution of the estimator will remain the same.

▶️ Answer/Explanation

The estimator for a population proportion is the sample proportion \(\hat{p}\). The variance of its sampling distribution is:

\(\mathrm{Var}(\hat{p})=\frac{p(1-p)}{n}\)

When the sample size \(n\) decreases from 2,500 to 1,500, the denominator becomes smaller. As a result, the variance of the sampling distribution becomes larger. The population variance does not change simply because a smaller sample is taken.

Therefore, reducing the sample size increases the variance of the sampling distribution of the estimator.
✅ Answer: (C)

Question 

According to data from the United States Elections Project, only 36 percent of eligible voters voted in the 2014 elections. For random samples of size 40, which of the following best describes the sampling distribution of \(\hat{p}\), the sample proportion of people who voted in the 2014 elections?

(A) The sampling distribution is skewed to the left, with mean 0.36 and standard deviation 0.076.
(B) The sampling distribution is skewed to the right, with mean 0.64 and standard deviation 0.006.
(C) The sampling distribution is approximately normal, with mean 0.36 and standard deviation 0.076.
(D) The sampling distribution is approximately normal, with mean 0.36 and standard deviation 0.006.
(E) The sampling distribution is approximately normal, with mean 0.64 and standard deviation 0.076.

▶️ Answer/Explanation

For a sampling distribution of a sample proportion:

\(\mu_{\hat{p}}=p=0.36\)

Check the Large Counts Condition:

\(np=(40)(0.36)=14.4\)
\(n(1-p)=(40)(0.64)=25.6\)

Both values are at least 10, so the sampling distribution is approximately normal.

The standard deviation is:

\(\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}\)
\(\sigma_{\hat{p}}=\sqrt{\frac{(0.36)(0.64)}{40}}\approx0.0759\approx0.076\)

Therefore, the sampling distribution is approximately normal with mean 0.36 and standard deviation 0.076.
✅ Answer: (C)

Question 

A 2017 survey of 1077 registered voters found that 83 percent of the voters surveyed oppose the government’s plan to repeal its net neutrality rules for internet providers. The poll’s margin of error was ±3 percentage points at a 99% confidence level. Which of the following best describes what is meant by the poll having a margin of error of ±3 percentage points?

(A) Three percent of those surveyed were unavailable to participate in the poll.
(B) It would not be unexpected for 3% of the population to readily agree to the net neutrality rules.
(C) Between 862 and 926 of the 1077 voters surveyed responded that they would oppose the government’s plan to repeal its net neutrality rules for internet providers.
(D) The poll used a method that gets an answer within 3% of the truth about the population approximately 99% of the time.
(E) Between 80% and 86% of all registered voters oppose the government’s plan to repeal its net neutrality rules for internet providers.

▶️ Answer/Explanation
A margin of error of \(\pm 3%\) at a \(99%\) confidence level means that the sampling method is expected to produce an estimate within \(3\) percentage points of the true population proportion about \(99%\) of the time.
The confidence interval for this survey would be:
\(83% \pm 3%\)
\((80%, 86%)\)
However, the confidence level refers to the reliability of the method, not to a guarantee that the true population value is definitely between \(80%\) and \(86%\).
The best interpretation is that the procedure used will capture the true population proportion within the margin of error approximately \(99%\) of the time when repeated many times.
Choice (E) incorrectly interprets the confidence interval as a certainty rather than a confidence statement about the method.
✅ Answer: (D)
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