Answer
Confidence interval: $99.39\lt µ\lt110.21$
We are 99% confident that the population mean is between 99.39 and 110.21
Work Step by Step
$n=23$, so:
$d.f.=n-1=22$
$level~of~confidence=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
$t_{\frac{α}{2}}=t_{0.005}=2.819$
(According to Table VI, for d.f. = 22 and area in right tail = 0.005)
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=104.8-2.819\times\frac{9.2}{\sqrt {23}}=99.39$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=104.8+2.819\times\frac{9.2}{\sqrt {23}}=110.21$
