Answer
Confidence interval: $114.98\lt x ̅\lt126.02$
Work Step by Step
We want to estimate the mean using a sample whose size is greater than 30.
$n=40$, so:
$d.f.=n-1=39$
$level~of~confidence=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
$t_{\frac{α}{2}}=t_{0.005}=2.708$
(According to Table VI, for d.f. = 39 and area in right tail = 0.005)
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=120.5-2.708\times\frac{12.9}{\sqrt {40}}=114.98$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=120.5+2.708\times\frac{12.9}{\sqrt {40}}=126.02$
