Answer
a) -11.6
b)17.2134
c)-1.507
d)$\pm 2.776$
Work Step by Step
a) 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: $\overline{d}=\frac{-22-4-34+11-9}{5}=-11.6.$
b)$s_d$ is the standard deviation of the differences, hence$s_d=\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(-22+11.6)^2+(-4+11.6)^2+(-34+11.6)^2+(11+11.6)^2+(-9+11.6)^2}{4}}=17.2134.$
c) The test statistic is:$t=\frac{\overline{d}-\mu_d}{s_d/\sqrt{n}}=\frac{-11.6-0}{17.2134/\sqrt{5}}=-1.507.$
d) The corresponding critical value using the table with df=5-1=4: $\pm t_{\alpha/2}=\pm t_{0.025}=\pm 2.776.$