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AP Statistics 3.9 Sampling Distributions for the Difference Between Sample Proportions- Exam Style Questions - MCQs - New Syllabus

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Question 

Suppose that 64% of students at Alpha High School and 70% of students at Beta High School favor Instagram as their preferred messaging app (both schools are large public schools). Independent random samples of 50 students from each school are selected and asked if they favor Instagram as their preferred messaging app. Let \( \hat{p}_A \) and \( \hat{p}_B \) represent the sample proportions of students at Alpha High School and Beta High School who favor Instagram, respectively.

Which of the following is the correct standard deviation of the sampling distribution of the difference in sample proportions, \( \hat{p}_A – \hat{p}_B \)?

(A) Standard Deviation:
\( \sqrt{\frac{0.64(0.36)}{50}+\frac{0.70(0.30)}{50}} \)

(B) Standard Deviation:
\( \sqrt{\frac{0.64(0.36)}{50}-\frac{0.70(0.30)}{50}} \)

(C) Standard Deviation:
\( \sqrt{\frac{0.64(0.36)}{50}}+\sqrt{\frac{0.70(0.30)}{50}} \)

(D) Standard Deviation:
\( \sqrt{\frac{0.64}{50^2}+\frac{0.70}{50^2}} \)

(E) Standard Deviation:
\( \sqrt{\frac{0.64}{50^2}}+\sqrt{\frac{0.70}{50^2}} \)

▶️ Answer/Explanation

For two independent sample proportions, the standard deviation of the sampling distribution of \( \hat{p}_A – \hat{p}_B \) is calculated using:

\( SD(\hat{p}_A-\hat{p}_B) \)
\( = \sqrt{\frac{p_A(1-p_A)}{n_A}+\frac{p_B(1-p_B)}{n_B}} \)

Substituting the given values:
\( = \sqrt{\frac{0.64(0.36)}{50}+\frac{0.70(0.30)}{50}} \)

This matches choice (A). The variances are added because the samples are independent, and the entire sum must remain under the square root sign.

✅ Answer: (A)

Question

A polling organization surveyed 2,002 randomly selected adults who are not scientists and 3,748 randomly selected adults who are scientists. Each adult was asked the question “Do you think that genetically modified foods are safe to eat?” Of those who are not scientists, 37 percent responded yes, and of those who are scientists, 88 percent responded yes. Which of the following is the standard error used to construct a confidence interval for the difference between the proportions of all adults who are not scientists and all adults who are scientists who would answer yes to the question?

(A) \(\sqrt{\frac{(0.37)(0.63)}{2,002}+\frac{(0.88)(0.12)}{3,748}}\)
(B) \(\sqrt{\frac{(0.37)(0.63)}{2,002}-\frac{(0.88)(0.12)}{3,748}}\)
(C) \(\sqrt{\frac{(0.37)(0.63)}{2,002}}+\sqrt{\frac{(0.88)(0.12)}{3,748}}\)
(D) \(\sqrt{\frac{(0.70)(0.30)}{2,002}}+\sqrt{\frac{(0.70)(0.30)}{3,748}}\)
(E) \(\frac{(0.37)(0.63)}{\sqrt{2,002}}+\frac{(0.88)(0.12)}{\sqrt{3,748}}\)

▶️ Answer/Explanation

For a confidence interval for the difference between two population proportions, the standard error is:

\(\text{SE}=\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\)

Substituting the sample proportions and sample sizes:

\(\text{SE}=\sqrt{\frac{(0.37)(0.63)}{2002}+\frac{(0.88)(0.12)}{3748}}\)

Notice that for a confidence interval we use the individual sample proportions, not a pooled proportion. The variances are added because the samples are independent.

Therefore, the correct standard error expression is given in choice (A).
✅ Answer: (A)

Question

In a random sample of 400 private high school students, 320 said they brought a smartphone to school that day, while in a random sample of 400 public high school students, 288 said they brought a smartphone to school that day. Which of the following represents a 98 percent confidence interval estimate for the difference (private high school minus public high school) between the proportions of all private high school and public high school students who bring a smartphone to school?

(A) \( (0.8-0.72)\pm1.96 \sqrt{ \frac{(0.8)(0.2)}{400} + \frac{(0.72)(0.28)}{400} } \)
(B) \( (0.8-0.72)\pm2.054 \sqrt{ \frac{(0.8)(0.2)}{400} + \frac{(0.72)(0.28)}{400} } \)
(C) \( (0.8-0.72)\pm2.326 \sqrt{ \frac{(0.8)(0.2)}{400} + \frac{(0.72)(0.28)}{400} } \)
(D) \( (0.8-0.72)\pm2.054 \sqrt{ (0.76)(0.24) \left( \frac{1}{400} + \frac{1}{400} \right) } \)
(E) \( (0.8-0.72)\pm2.326 \sqrt{ (0.76)(0.24) \left( \frac{1}{400} + \frac{1}{400} \right) } \)

▶️ Answer/Explanation
First calculate the sample proportions:
\(\hat{p}_1=\frac{320}{400}=0.80\)
\(\hat{p}_2=\frac{288}{400}=0.72\)
For a confidence interval for the difference of two population proportions, the formula is:
\((\hat{p}_1-\hat{p}_2) \pm z^* \sqrt{ \frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2} }\)
Since the confidence level is (98%), [ z^*=2.326 ] Substituting the sample values:
\((0.80-0.72) \pm 2.326 \sqrt{ \frac{(0.80)(0.20)}{400} + \frac{(0.72)(0.28)}{400} }\)
Notice that a pooled proportion is not used for confidence intervals; pooled proportions are used in hypothesis tests.
Therefore choices (D) and (E) are incorrect. The expression above matches choice (C).
✅ Answer: (C)
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