Question

A researcher wants to estimate the difference in population proportions for two populations using a 98% confidence interval. In a sample of 900 persons from the first population, 288 possessed the characteristic of interest. In a sample of 800 from the second population, 240 possessed the characteristic of interest. A 98% confidence interval for the difference in the population proportions is:

{0.0323, 0.0723}

{-0.0323, 0.0723}

{0.0323, -0.0723}

{0.0223, 0.0923}

{0.0423, 0.0723}

Answer 1

1 = 288 / 900 = 0.32

2 = 240 / 800 = 0.30

Pooled proportion = (x1+ x2) / (n1 + n2)

= (288 + 240 ) / ( 900 + 800)

= 0.3106

98% confidence interval for p1 - p2 is

(1 - 2) - Z * sqrt( ( 1 - ) ( 1 / n1 + 1 / n2) ) < p1 - p2 < (1 - 2) + Z * sqrt( ( 1 - ) ( 1 / n1 + 1 / n2) )

(0.32-0.3) - 2.326 * sqrt( 0.3106 * 0.6894 ( 1 / 900 + 1 / 800) ) < p1 - p2 ) <

(0.32-0.3) + 2.326 * sqrt( 0.3106 * 0.6894 ( 1 / 900 + 1 / 800) )

-0.0323 < p1 - p2 < 0.0723

98% confidence interval for p1 - p2 is ( -0.0323 , 0.0723)