Answer
$\mu_d$ is between -66.8 and 49.8
Work Step by Step
The corresponding critical value using the table with df=10-1=9: $t_{\alpha/2}=t_{0.005}=3.25.$ The margin of error: $E=t_{\alpha/2}\frac{s_d}{\sqrt{n}}=3.25\frac{56.6789}{\sqrt{10}}=58.3.$ Hence the confidence interval $\mu_d$ is between $\overline{d}-E$=-8.5-58.3=-66.8 and $\overline{d}+E$=-8.5+58.3=49.8.
