Answer
$28.4\lt\mu\lt38.0$
Work Step by Step
Given $n=14, \bar X=33.2, s=8.3$, at a 95% confidence and $df=13$,
the critical t-value is $t_{\alpha/2}=2.16 $ (use table F)
The margin of error can be found as $E=2.16\times\frac{8.3}{\sqrt {14}}=4.8$
Thus, the interval of the true mean can be estimated as
$\bar X-E\lt\mu\lt\bar X+E$ which gives $28.4\lt\mu\lt38.0$
