#### Answer

Reject the null hypothesis.

#### Work Step by Step

The differences: 98-98=0, 97-97.6=-0.6, 98.6-98.8=-0.2, 97.4-98=-0.6. $\overline{d}$ is the averages of the differences, hence: $\overline{d}=\frac{0-0.6-0.2-0.6}{4}=-0.35.$
$s_d$ is the standard deviation of the differences, hence$s_d=\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(0+0.35)^2+(-0.6+0.35)^2+(-0.2+0.35)^2+(-0.6+0.35)^2}{3}}=0.3.$
The test statistic is:$t=\frac{\overline{d}-\mu_d}{s_d/\sqrt{n}}=\frac{-0.35-0}{0.3/\sqrt{4}}=-2.333.$
The corresponding critical value using the table with df=4-1=3: $\pm t_{\alpha/2}=\pm t_{0.025}=\pm 3.182.$
If the value of the test statistic is in the rejection region, we reject the null hypothesis. -2.333 is between -3.182 and 3.182 hence we reject the null hypothesis.