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AP Statistics 6.8 Confidence Intervals for the Difference of Two Proportions- MCQs - Exam Style Questions

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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, \(37\%\) responded yes, and of those who are scientists, \(88\%\) 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) \(\displaystyle \sqrt{\frac{(0.37)(0.63)}{2{,}002}+\frac{(0.88)(0.12)}{3{,}748}}\)
(B) \(\displaystyle \sqrt{\frac{(0.37)(0.63)}{2{,}002}-\frac{(0.88)(0.12)}{3{,}748}}\)
(C) \(\displaystyle \sqrt{\frac{(0.37)(0.63)}{2{,}002}}+\sqrt{\frac{(0.88)(0.12)}{3{,}748}}\)
(D) \(\displaystyle \sqrt{\frac{(0.70)(0.30)}{2{,}002}}+\sqrt{\frac{(0.70)(0.30)}{3{,}748}}\)
(E) \(\displaystyle \frac{(0.37)(0.63)}{\sqrt{2{,}002}}+\frac{(0.88)(0.12)}{\sqrt{3{,}748}}\)
▶️ Answer/Explanation
Detailed solution

For a confidence interval for a difference of proportions \((p_1-p_2)\) using two independent samples, the standard error is
\[ \mathrm{SE}=\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}. \]
Here, group \(1\): not scientists \(\Rightarrow \hat p_1=0.37,\; n_1=2{,}002\).
Group \(2\): scientists \(\Rightarrow \hat p_2=0.88,\; n_2=3{,}748\).
Therefore
\[ \mathrm{SE}=\sqrt{\frac{(0.37)(0.63)}{2{,}002}+\frac{(0.88)(0.12)}{3{,}748}} \approx \sqrt{0.0001164+0.0000282} \approx \sqrt{0.0001446} \approx 0.012. \]
This matches option (A).
✅ Answer: (A)

Question

In a large school district, 16 of 85 randomly selected high school seniors play a varsity sport. In the same district, 19 of 67 randomly selected high school juniors play a varsity sport. A 95 percent confidence interval for the difference between the proportion of high school seniors who play a varsity sport in the school district and high school juniors who play a varsity sport in the school district is to be calculated. What is the standard error of the difference?
(A) 0.0347
(B) 0.0695
(C) 0.1362
(D) 0.9800
(E) 1.6900
▶️ Answer/Explanation
Detailed solution

1. Calculate Sample Proportions:
\(\hat{p}_1 = \frac{16}{85} \approx 0.1882\) (seniors)
\(\hat{p}_2 = \frac{19}{67} \approx 0.2836\) (juniors)

2. Standard Error Formula:
\(SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\)

3. Calculate:
\(SE = \sqrt{\frac{0.1882(0.8118)}{85} + \frac{0.2836(0.7164)}{67}}\)
\(= \sqrt{\frac{0.1527}{85} + \frac{0.2032}{67}}\)
\(= \sqrt{0.001796 + 0.003033}\)
\(= \sqrt{0.004829} \approx 0.0695\)

✅ Answer: (B)

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