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Question 2 of 10, Step 2 of 2
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The FBI wants to determine the effectiveness of their 10 Most Wanted list. To do so, they need to find out the fraction of people who appear on the list that are actually caught.
Step 2 of 2 : Suppose a sample of 334 suspected criminals is drawn. Of these people, 86 were captured. Using the data, construct the $90\mathrm{\%}$ confidence interval for the population proportion of people who are captured after appearing on the 10 Most Wanted list. Round your answers to three decimal places.
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Step 1 ：The question is asking for a 90% confidence interval for the population proportion of people who are captured after appearing on the 10 Most Wanted list. The sample size is 334 and the number of successes (people captured) is 86.

Step 2 ：The formula for a confidence interval for a population proportion is given by: $\hat{p}\pm Z\sqrt{(\hat{p}(1-\hat{p}))/n}$ where $\hat{p}$ is the sample proportion, Z is the Z-score corresponding to the desired level of confidence, and n is the sample size.

Step 3 ：First, we need to calculate the sample proportion ( $\hat{p}$ ), which is the number of successes divided by the sample size.

Step 4 ：Then, we need to find the Z-score corresponding to a 90% confidence level. The Z-score for a 90% confidence level is 1.645.

Step 5 ：Finally, we can substitute these values into the formula to find the confidence interval.

Step 6 ：Let's calculate this: n = 334, x = 86, $\hat{p}=0.257$, Z = 1.645, SE = 0.024, CI_lower = 0.218, CI_upper = 0.297.

Step 7 ：The 90% confidence interval for the population proportion of people who are captured after appearing on the 10 Most Wanted list is approximately $\boxed{{\displaystyle [0.218,0.297]}}$.