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 403 suspected criminals is drawn. Of these people, 149 were captured. Using the data, construct the 80% 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.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

For this case the estimated proportion of interest would be [tex]\hat p =\frac{149}{403}= 0.370[/tex]

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by [tex]\alpha=1-0.80=0.2[/tex] and [tex]\alpha/2 =0.1[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.28, t_{1-\alpha/2}=1.28[/tex]

The confidence interval for the mean is given by the following formula:

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.37 - 1.28\sqrt{\frac{0.37(1-0.37)}{403}}=0.339[/tex]

[tex]0.37 + 1.28\sqrt{\frac{0.37(1-0.37)}{403}}=0.401[/tex]

The 80% confidence interval would be given by (0.339;0.401)