Question

Question 8 of 10 (2 points) Attempt 1 of 3 10.2 Section Exercise 11 Construct the 99.5% confidence interval for the difference Pi-p2 when x,-58, n,-83, x,-72, answer to at least three decimal places. n2-159. Round the 囤 A 99.5% confidence interval for the difference between the two proportions is P1-P2 alo

Answer 1

SOLUTION

given : .T1 = 58. n! = 83. T2= 72. n-= 159

1 58 n1 83 sample proportion (p) = 0.6988

sample proportion ()-2-150 T2 72 n2 159 -= 0.4528

for 99.5% confidence level. 2 za-2.807

$\\\mathbf{formula:}\; Confidence \;interval \;for\; the \;difference \;in\; proportion,\;p_1-p_2$

p (-pp(1 -p2 01) n1 7l

$=\left ( 0.6988-0.4528 \right )\pm2.807*\left \{ \sqrt{\frac{0.6988*(1-0.6988)}{83}+\frac{0.4528*(1-0.4528)}{159}} \right \}$

0.6988 0.30120.4528 * 0.5472 0.2460 ± 2.807 * 159 83

0.246 0.180

Lower Limit-0.246-0.180 0.066

Upper Limit 0.246 0.180 0.426

0.066< p1 -P2 <0.426