Answer
Confidence interval: $0.932\lt p ̂\lt0.964$
We are 99% confident that the proportion of Americans who live in neighborhoods with acceptable levels of carbon monoxide is between 0.932 and 0.964.
Work Step by Step
$level~of~confidence=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
$z_{\frac{α}{2}}=z_{0.005}$
If the area of the standard normal curve to the right of $z_{0.005}$ is 0.005, then the area of the standard normal curve to the left of $z_{0.005}$ is $1−0.005=0.995$
According to Table V, there are 2 z-scores which give the closest value to 0.995: 2.57 and 2.58. So, let's find the mean of these z-scores: $\frac{2.57+2.58}{2}=2.575$
$Lower~bound=p ̂-z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.948-2.575\times\sqrt {\frac{0.948(1-0.948)}{1201}}=0.932$
$Upper~bound=p ̂+z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.948+2.575\times\sqrt {\frac{0.948(1-0.948)}{1201}}=0.964$
