Question

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 1753 suspected criminals is drawn. Of these people, 701 were captured. Using the data, construct the 99% 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.

Answer 1

Solution :

Given that,

n = 1753

x = 701

Point estimate = sample proportion = = x / n = 701 / 1753 = 0.400

1 - = 0.600

Z_/2 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.400 * 0.600) / 1753)

= 0.030

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.400 - 0.030 < p < 0.400 + 0.030

0.370 < p < 0.430

(0.370 , 0.430)