## Elementary Statistics (12th Edition)

$\mu_d$ is between 0.691 and 5.559.
The corresponding critical value using the table with df=8-1=7: $t_{\alpha/2}=t_{0.025}=2.365.$ The margin of error: $E=t_{\alpha/2}\frac{s_d}{\sqrt{n}}=2.365\frac{2.9114}{\sqrt{8}}=2.434.$ Hence the confidence interval $\mu_d$ is between $\overline{d}-E$=3.125-2.434=0.691 and $\overline{d}+E$=3.125+2.434=5.559. The interval doesn't contain 0, so it seems to be effective.