Answer
Confidence interval: $49.52\lt x ̅\lt60.08$
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
$n=30$, so:
$d.f.=n-1=29$
$level~of~confidence=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
$t_{\frac{α}{2}}=t_{0.005}=2.756$
(According to Table VI, for d.f. = 29 and area in right tail = 0.005)
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=54.8-2.756\times\frac{10.5}{\sqrt {30}}=49.52$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=54.8+2.756\times\frac{10.5}{\sqrt {30}}=60.08$
