Question

A random sample found that forty-two percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 92% confidence interval for the true proportion of Americans who were satisifed with the gun control laws in 2017. Fill in the blanks appropriately. A 92% for the true proportion of Americans who were satisfied with the gun control laws in 2017 is(____)(_____)(Keep 3 decimal places)

Answer 1

$\alpha$ = 1 - 0.92 = 0.08

From Z table,

Z$\alpha$/2 = 1.7507 (For two tailed confidence interval)

92% confidence interval for p is

[ $\hat{p}$ - Z$\alpha$/2 * sqrt ( $\hat{p}$ ( 1 - $\hat{p}$ ) / n ] < p < [ $\hat{p}$ + Z$\alpha$/2 * sqrt ( $\hat{p}$ ( 1 - $\hat{p}$ ) / n ]

0.42 - 1.7507 * sqrt [ 0.42 * ( 1 - 0.42) / 100] < p < 0.42 + 1.7507 * sqrt [ 0.42 * ( 1 - 0.42) / 100]

0.334 < p < 0.506

92% CI is ( 0.334 , 0.506 )