Answer
Confidence interval: $104.578\lt x ̅\lt111.422$
Compared to the results obtained in part (a), the margin of error has decreased.
Work Step by Step
$n=25$, so:
$d.f.=n-1=24$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.9=1-α$
$α=0.1$
$t_{\frac{α}{2}}=t_{0.05}=1.711$
(According to Table VI, for d.f. = 24 and area in right tail = 0.05)
$Lower~bound=x ̅ -t_{\frac{α}{2}}.\frac{s}{\sqrt n}=108-1.711\times\frac{10}{\sqrt {25}}=104.578$
$Upper~bound=x ̅ +t_{\frac{α}{2}}.\frac{s}{\sqrt n}=108+1.711\times\frac{10}{\sqrt {25}}=111.422$
