Answer
Confidence interval: $0.903\lt p ̂\lt0.937$
Work Step by Step
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
$Lower~bound=p ̂-z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.920-1.96\sqrt {\frac{0.920(1-0.920)}{1001}}=0.903$
$Upper~bound=p ̂+z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.920+1.96\sqrt {\frac{0.920(1-0.920)}{1001}}=0.937$
