#### Answer

Reject the null hypothesis.

#### Work Step by Step

The differences: 22-44=-22, 37-41=-4, 28-62=-34, 63-52=11, 32-41=-9. $\overline{d}$ is the averages of the differences, hence: $\overline{d}=\frac{−22−4−34+11−9}{5}=-11.6.$
$s_d$ is the standard deviation of the differences, hence$s_d=\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(−22+11.6)^2+(−4+11.6)^2+(−34+11.6)^2+(11+11.6)^2+(−9+11.6)^2}{4}}=17.2134.$
The test statistic is:$t=\frac{\overline{d}-\mu_d}{s_d/\sqrt{n}}=\frac{-11.6-0}{17.2134/\sqrt{5}}=-1.507.$ t is the corresponding critical value using the table with df=5-1=4, hence $t_{\alpha/2}=t_{0.025}=\pm 2.776.$ If the value of the test statistic is in the rejection region, we reject the null hypothesis. -1.507 is between -2.776 and 2.776 hence we reject the null hypothesis.