Solve the problem.
-Of 346 items tested, 12 are found to be defective. Construct the 98% confidence interval for the proportion of all such items that are defective.
A)0.0154

Question

Solve the problem.
-Of 346 items tested, 12 are found to be defective. Construct the 98% confidence interval for the proportion of all such items that are defective.
A)0.0154 < p < 0.0540
B)0.0118 < p < 0.0576
C)0.0345 < p < 0.0349
D)0.0110 < p < 0.0584

Quizplus · Accepted Answer

To construct a 98% confidence interval for the proportion of defective items, we first calculate the sample proportion ($p$) and then use the formula for the confidence interval for a proportion, which is $p \pm Z \sqrt{\frac{p(1-p)}{n}}$, where $Z$ is the Z-score corresponding to the confidence level (in this case, 98%), $p$ is the sample proportion, and $n$ is the sample size.Given:- Number of items tested ($n$) = 346- Number of defective items = 12Sample proportion ($p$) = Number of defective items / Number of items tested = 12 / 346 ≈ 0.034682For a 98% confidence level, the Z-score is approximately 2.33 (since the Z-score for 99% confidence level for a two-tailed test is about 2.576, and for 95% it's about 1.96, 98% would be in between, closer to 2.33).Plugging into the formula:$$0.034682 \pm 2.33 \sqrt{\frac{0.034682(1-0.034682)}{346}}$$Calculating the margin of error:$$2.33 \sqrt{\frac{0.034682(0.965318)}{346}} \approx 0.0229$$Thus, the confidence interval is approximately:$$0.034682 - 0.0229 < p < 0.034682 + 0.0229$$$$0.011782 < p < 0.057582$$Rounded to four decimal places, the confidence interval is:$$0.0118 < p < 0.0576$$Therefore, the correct answer is B) 0.0118 < p < 0.0576.

Solve the Problem. -Of 346 Items Tested, 12 Are Found to Be Defective