A total of 150 households were asked if they had Internet access,of which 57 responded "Yes".Which of the following would represent a 98% confidence interval around this sample proportion?
A) $0.380 \mp 2.33$
B)$0.380 \mp 0.092$
C)$0.380 \mp 0.039$
D) $0.380 \mp 0.251$
E) $57 \pm 2.33$

Question

Quizplus · Accepted Answer

The correct confidence interval is calculated using the formula for a proportion's confidence interval: $\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $Z$ is the Z-score corresponding to the confidence level, and $n$ is the sample size. Here, $\hat{p} = \frac{57}{150} = 0.380$, $n = 150$, and for a 98% confidence level, $Z \approx 2.33$. Plugging these values into the formula gives a margin of error around 0.092, making option B correct.

A Total of 150 Households Were Asked If They Had 0.380∓2.330.380 \mp 2.330.380∓2.33

A Total of 150 Households Were Asked If They Had $0.380 \mp 2.33$