AP Statistics 3.3 Constructing a Confidence Interval for a Population Proportion- Exam Style Questions - MCQs - New Syllabus
▶️ Answer/Explanation
The 95% confidence interval for the 2022 proportion of single-person households is:
\((0.2813,\;0.3387)\)
The 2021 Census value was:
\(p=0.29\)
Notice that:
\(0.2813 < 0.29 < 0.3387\)
Since the 2021 value of 0.29 lies within the 95% confidence interval, it is a plausible value for the 2022 population proportion. Therefore, the data do not provide sufficient evidence that the proportion of single-person households changed between 2021 and 2022.
A confidence interval that contains the hypothesized value corresponds to failing to reject the null hypothesis in a two-sided significance test at the matching significance level.
Therefore, there is not sufficient evidence at the 0.05 significance level to conclude that a significant difference exists.
✅ Answer: (C)
▶️ Answer/Explanation
First combine the data from both days:
Total successes:
\( 93 + 26 = 119 \)
Total sample size:
\( 151 + 36 = 187 \)
Sample proportion:
\( \hat{p} = \frac{119}{187} \approx 0.636 \)
Standard error:
\( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \)
\( = \sqrt{\frac{(0.636)(0.364)}{187}} \)
\( \approx 0.0352 \)
Margin of error for a 95% confidence interval:
\( ME = 1.96(SE) \)
\( = 1.96(0.0352) \)
\( \approx 0.069 \)
Converting to a percentage:
\( 0.069 = 6.9\% \)
Therefore, the margin of error is approximately 6.9%.
✅ Answer: (C)
▶️ Answer/Explanation
For a large-sample confidence interval for a population proportion, the success-failure condition must be satisfied:
\( n\hat{p} \ge 10 \)
and
\( n(1-\hat{p}) \ge 10 \)
Here, only 4 of the 100 residents answered “yes,” so:
\( \hat{p}=\frac{4}{100}=0.04 \)
Therefore,
\( n\hat{p}=100(0.04)=4 \)
Since \(4 < 10\), the success-failure condition is not met. As a result, the large-sample z-confidence interval is not appropriate.
✅ Answer: (D)