Question

Consider two independent random samples with the following results: n = 532 n2 = 730 X = 390 X2 = 139 Use this data to find the 90 % confidence interval for the true difference between the population proportions. Show supporting work for each ste that follows Step 2 of 3: Find the margin of error. Round your answer to six decimal places. Answer 5 Points Tables Keypad

Answer 1

Pooled proportion $\hat{p}$ = ( x1 + x2) / ( n1 + n2 ) = ( 390 + 139 ) / ( 532 + 730 ) = 0.4192

$\hat{p}$1 = 390 / 532 = 0.7331 , $\hat{p}$ = 139 / 730 = 0.1904

Margin of error = Z$\alpha$/2 * sqrt [ $\hat{p}$ ( 1 - $\hat{p}$ ) * ( 1 / n1 + 1 / n2 ) ]

= 1.6449 * sqrt [ 0.4192 * ( 1 - 0.4192) * ( 1 / 532 + 1 / 730 ) ]

= 0.046267