Answer
Confidence interval: $95.49\lt x ̅\lt113.11$
Compared to the results obtained in part (a), the increase in the level of confidence caused the confidence interval to become wider.
Work Step by Step
$n=15$, so:
$d.f.=n-1=14$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.145$
(According to Table VI, for d.f. = 14 and area in right tail = 0.025)
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=104.3-2.145\times\frac{15.9}{\sqrt {15}}=95.49$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=104.3+2.145\times\frac{15.9}{\sqrt {15}}=113.11$
