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 1313 suspected criminals is drawn. Of these people, 828 were not captured. Using the data, construct the 98% 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 = 1313

x = 828

Point estimate = sample proportion = = x / n = 828 / 1313 = 0.631

1 - = 1 - 0.631 = 0.369

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

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

= 2.326 * (((0.631 * 0.369) / 1313)

= 0.031

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.631 - 0.031 < p < 0.631 + 0.031

0.600 < p < 0.662

The 98% confidence interval for the population proportion p is : 0.600 , 0.662