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AP Statistics 3.11 Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Proportions- Exam Style Questions - MCQs - New Syllabus

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Question

Employees were randomly sampled from two large corporations. Out of 68 randomly selected employees from Corporation A, 58 were happy with their job. Out of 74 randomly selected employees from Corporation B, 52 were happy with their job. A 95 percent confidence interval was computed as follows:

\(\left(\frac{58}{68}-\frac{52}{74}\right)\pm1.96\sqrt{\frac{\frac{58}{68}\left(1-\frac{58}{68}\right)}{68}+\frac{\frac{52}{74}\left(1-\frac{52}{74}\right)}{74}}=[0.016,\;0.284]\)

Which of the following is an appropriate interpretation of the 95 percent confidence interval?

(A) We are 95% confident the proportion of employees happy with their job at both corporations is between 0.016 and 0.284.
(B) We are 95% confident the sample proportion of employees happy with their job at both corporations is between 0.016 and 0.284.
(C) We are 95% confident the difference in population proportions of employees happy with their job at Corporation A and Corporation B is between 0.016 and 0.284
(D) We are 95% confident the difference in the number of employees happy with their job at Corporation A and Corporation B is between 0.016 and 0.284.
(E) We are 95% confident the proportion of 68 employees happy with their job at Corporation A minus the proportion of 74 employees happy with their job at Corporation B is between 0.016 and 0.284.

▶️ Answer/Explanation

The confidence interval was constructed using:

\(\hat{p}_A-\hat{p}_B\)

which estimates the difference between the population proportions of employees who are happy with their jobs at the two corporations.

A confidence interval should always be interpreted in terms of the population parameter being estimated, not the sample statistics used to compute it.

Therefore, the correct interpretation is that we are 95% confident the true difference in population proportions of employees happy with their jobs at Corporation A and Corporation B lies between 0.016 and 0.284.

Because the entire interval is positive, it suggests that the proportion of happy employees is higher at Corporation A than at Corporation B.

✅ Answer: (C)

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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