Answer
$38.7\lt\mu\lt48.3$
The interval contains the national average of 44.7 cents.
Work Step by Step
1. Based on the data set we can find $\bar X=43.5, s=9.24 , n=12 $
2. At the 90% confidence and $df=11$, the critical t-value is $t_{\alpha/2}=1.8$ (use table F)
3. The margin of error can be found as $E=1.8\times\frac{9.24}{\sqrt {12}}=4.8$
4. Thus, the interval of the true mean can be estimated as $\bar X-E\lt\mu\lt\bar X+E$ which gives $38.7\lt\mu\lt48.3$
5. The interval contains the national average of 44.7 cents.
