To construct a 95% confidence interval for the population proportion of people who are captured after appearing on the 10 Most Wanted list, we can use the formula for a confidence interval for a population proportion:
\[ \hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
Where:
- \( \hat{p} = \frac{x}{n} \) is the sample proportion
- \( z \) is the z-score for a 95% confidence interval, which is 1.96
- \( n \) is the sample size, which is 531
- \( x \) is the number of captured individuals in the sample, which is 201
First, calculate the sample proportion:
\[ \hat{p} = \frac{201}{531} \approx 0.378 \]
Next, calculate the margin of error:
\[ ME = 1.96 \times \sqrt{\frac{0.378 \times 0.622}{531}} \approx 0.036 \]
Finally, construct the 95% confidence interval:
\[ 0.378 - 0.036 < p < 0.378 + 0.036 \]
\[ 0.342 < p < 0.414 \]
Therefore, the 95% confidence interval for the population proportion of people who are captured after appearing on the 10 Most Wanted list is approximately 0.342 to 0.414.
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 531
suspected criminals is drawn. Of these people, 201
were captured. Using the data, construct the 95%
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.
1 answer
1 answer