Solve the problem.
-A study involves 634 randomly selected deaths, with 32 of them caused by accidents. Construct a 98% confidence interval for the true percentage of all deaths that are caused by accidents.
A)3.61%

Question

Solve the problem.
-A study involves 634 randomly selected deaths, with 32 of them caused by accidents. Construct a 98% confidence interval for the true percentage of all deaths that are caused by accidents.
A)3.61% < p < 6.48%
B)3.02% < p < 7.07%
C)2.80% < p < 7.29%
D)3.34% < p < 6.75%

Quizplus · Accepted Answer

To construct a 98% confidence interval for the proportion, we use the formula for the confidence interval of a proportion: $\hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $Z_{\alpha/2}$ is the Z-score corresponding to the confidence level, and $n$ is the sample size.Given:- $n = 634$- Number of successes (deaths by accidents) = 32- Therefore, $\hat{p} = \frac{32}{634}$For a 98% confidence interval, the Z-score ($Z_{\alpha/2}$) is approximately 2.33 (since the area in each tail is 1% for a 98% confidence level, and the corresponding Z-score for 0.01 in the tail is 2.33).First, calculate the sample proportion:$\hat{p} = \frac{32}{634} \approx 0.0505$Then, calculate the standard error:$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.0505(1-0.0505)}{634}} \approx 0.0087$Finally, calculate the margin of error:$ME = Z_{\alpha/2} \times SE = 2.33 \times 0.0087 \approx 0.0203$Thus, the confidence interval is:$\hat{p} \pm ME = 0.0505 \pm 0.0203$This gives us:Lower bound = $0.0505 - 0.0203 \approx 0.0302$ or 3.02%Upper bound = $0.0505 + 0.0203 \approx 0.0708$ or 7.08%Rounding slightly, the correct answer is 3.02% < p < 7.07%, which matches option B.

Solve the Problem. -A Study Involves 634 Randomly Selected Deaths, with 32 of with 32