Suppose we have independent random samples of size n1= 604 and n2 = 490. The proportions of success in the two samples are p1= 0.43 and p2 = 0.51. Find the 95% confidence interval for the difference in the two population proportions.

[-0.1393, -0.0207]

Step-by-step explanation:

The formula for the confidence interval for the difference in the two population proportions =

(p1 - p2) ± z × √[p1 (1 - p1)/n1]+ [p2 (1 - p2)/n2]

From the question

p1= 0.43

p2 = 0.51

n1= 604

n2 = 490

z = z score for 95 % confidence interval = 1.96

Confidence Interval = (p1 - p2) ± z × √[p1 (1 - p1)/n1]+ [p2 (1 - p2)/n2]

= (0.43 - 0.51) ± 1.96 × √[0.43(1 - 0.43)/604]+ [0.51 (1 - 0.51)/490]

= -0.08 ± 1.96 × √[ 0.43 × 0.57/604] + [0.51 × 0.49/490]

= -0.08 ± 1.96 × √0.0004057947 + 0.00051

= -0.08 ± 1.96 × √0.0009157947

= - 0.08 ± 1.96 × 0.0302621001

= -0.08 ± 0.0593137161

Confidence Interval =

-0.08 - 0.00593137161 = - 0.1393137161

Approximately = -0.1393

-0.08 + 0.00593137161 = -0.0206862839

Approximately = -0.0207

The confidence Interval for the difference in the two population proportions is [-0.1393, -0.0207]

answer: 16

step-by-step explanation: first you would add 6+6 which equals 12 then add 2+2 which equals 4. then add 12 + 4 which is 16.

hope this ^_^