Answer
Confidence interval: $82.58\lt x ̅\lt93.22$
We are 95% confident that adult Americans would like to live between 82.58 and 93.22 years.
Work Step by Step
$n=35$, so:
$d.f.=n-1=34$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.032$
(According to Table VI, for d.f. = 34 and area in right tail = 0.025)
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=87.9-2.032\times\frac{15.5}{\sqrt {35}}=82.58$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=87.9+2.032\times\frac{15.5}{\sqrt {35}}=93.22$
