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 726 suspected criminals is drawn. Of these people, 232 were 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. previous answer if needed was .320

Answer 1

Solution :

Given that,

n = 726

x = 232

Point estimate = sample proportion = = x / n = 0.320

1 - = 0.680

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z/2 = Z0.01 = 2.326

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

= 2.326 * (((0.320 * 0.680) / 726)

= 0.040

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.320 - 0.040 < p < 0.320 + 0.040

0.280 < p < 0.360

( 0.280 , 0.360 )