Answer
Confidence interval: $0.496\lt p ̂\lt0.548$
We are 90% confident that the proportion of adults Americans who are worried about having enough money to live comfortably in retirement is between 0.496 and 0.548.
Work Step by Step
$p̂ =\frac{x}{n}=\frac{526}{1008}=0.522$
Required condition:
$1008\times0.522(1-0.522)=1008\times0.522(1-0.522)=251.5\gt10$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.9=1-α$
$α=0.1$
$z_{\frac{α}{2}}=z_{0.05}$
If the area of the standard normal curve to the right of $z_{0.05}$ is 0.05, then the area of the standard normal curve to the left of $z_{0.05}$ is $1−0.05=0.95$
According to Table V, there are 2 z-scores which give the closest value to 0.95: 1.64 and 1.65. So, let's find the mean of these z-scores: $\frac{1.64+1.65}{2}=1.645$
$Lower~bound=p ̂-z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.522-1.645\times\sqrt {\frac{0.522(1-0.522)}{1008}}=0.496$
$Upper~bound=p ̂+z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.522+1.645\times\sqrt {\frac{0.522(1-0.522)}{1008}}=0.548$
