Answer
$\mu_d$ is between -4.16 and 2.16.
Work Step by Step
The corresponding critical value using the table with df=12-1=11: $t_{\alpha/2}=t_{0.005}=3.106.$ The margin of error: $E=t_{\alpha/2}\frac{s_d}{\sqrt{n}}=3.106\frac{3.5196}{\sqrt{11}}=3.16.$ Hence the confidence interval $\mu_d$ is between $\overline{d}-E$=-1-3.16=-4.16 and $\overline{d}+E$=-1+3.16=2.16. The interval contains 0, hence there is no big difference.
