#### Answer

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:$t=\frac{\overline{d}-\mu_d}{s_d/\sqrt{n}}=\frac{7278.5-0}{6913.2593/\sqrt{6}}=2.579.$ The P is the corresponding probability using the table with df=6-1=5, hence P is between 0.01 and 0.025. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P is lessthan $\alpha=0.05$, hence we reject the null hypothesis.