Answer
Confidence interval: $0.829\lt p ̂\lt0.873$
We are 95% confident that the proportion of adults who always wear seat belts is between 0.829 and 0.873.
Work Step by Step
$p̂ =\frac{x}{n}=\frac{862}{1013}=0.851$
Required condition:
$1013\times0.851(1-0.851)=1013\times0.851(1-0.851)=128.4\gt10$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
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.
$Lower~bound=p ̂-z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.851-1.96\times\sqrt {\frac{0.851(1-0.851)}{1013}}=0.829$
$Upper~bound=p ̂+z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.851+1.96\times\sqrt {\frac{0.851(1-0.851)}{1013}}=0.873$
