Answer
Fail to reject the null hypothesis.
Work Step by Step
$H_{0}:\mu_d=0,$ $H_{a}:\mu_d$ is more than 0. The test statistic is:$z=\frac{\overline{d}-\mu_d}{s_d/\sqrt{n}}=\frac{0.0.0588235-0}{9.604106/\sqrt{5}}=0.035713579.$ The P is the corresponding probability using the table, hence P=0.4858629. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.0.4858629 is more than $\alpha=0.05$, hence we fail to reject the null hypothesis.
