Answer
$17.87\lt\mu\lt20.53$
this range is higher than the national average of $15.9$
Work Step by Step
Given $n=12, \bar X=19.2, s^2=4.41$, we have $s=2.1$ and
at a 95% confidence and $df=11$, the critical t-value is $t_{\alpha/2}=2.2 $ (use table F)
The margin of error can be found as $E=2.2\times\frac{2.1}{\sqrt {12}}=1.33$
Thus, the interval of the true mean can be estimated as
$\bar X-E\lt\mu\lt\bar X+E$ which gives $17.87\lt\mu\lt20.53$
It appears that this range is higher than the national average of $15.9$