Question

Consider two independent random samples with the following results:n 1 =165 x 1 =104 n 2 =83 x 2 =9 Use this data to find the 98% confidence interval for the true difference between the population proportions.

Answer 1

Solution :- Given That :- X1 =104 Favourable case: Sample Size: X2 = 9 n2 = 83 n1 = 165 Sample proportion 1: @1 = x1 / n1 = 104 / 165 = 0.6303 Sample proportion 2: f2 = x2 / n2 = 9/83 = 0.1084 At CI= 98% Confidence Level'Z' is, By using standard normal'z'table a = 1 - 98% a = 1 - 0.98 a = 0.02 a/2 = 0.01 za/2 = 2.326 At 98% Confidence interval for difference between two Population proportion is, CÔ 1 - ô2) + margin of error P: (1 – Ô1) 02 (1 - 02) (6 -P2){za/2 - + M N2 (0.6303 - 0.1084) + (2.326)* sqrt(0.6303 * ( 1-0.6303)/165 + 0.1084 *(1-0.1084)/8 0.5219 + (0.1182) (0.4037, 0.6401 ) At, 98% Confidence interval for difference between two Population proportion is, (0.4037, 0.6401 )