## Elementary Statistics (12th Edition)

$\mu_d$ is between -7.3 and 6.3.
The corresponding critical value using the table with df=8-1=7: $t_{\alpha/2}=t_{0.005}=3.499.$ The margin of error: $E=t_{\alpha/2}\frac{s_d}{\sqrt{n}}=3.499\frac{5.4772}{\sqrt{8}}=6.8.$ Hence the confidence interval $\mu_d$ is between $\overline{d}-E$=-0.5-6.8=-7.3 and $\overline{d}+E$=-0.5+6.8=6.3. The interval contains 0, so it seems to be quite accurate.