Answer
$\mu_d$ is between 1 and 12
Work Step by Step
The corresponding critical value using the table with df=12-1=11: $t_{\alpha/2}=t_{0.05}=1.796.$ The margin of error: $E=t_{\alpha/2}\frac{s_d}{\sqrt{n}}=1.796\frac{10.6301}{\sqrt{12}}=5.5.$ Hence the confidence interval $\mu_d$ is between $\overline{d}-E$=6.5-5.5=1 and $\overline{d}+E$=6.5+5.5=12.
