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:$t=\frac{\overline{d}-\mu_d}{s_d/\sqrt{n}}=\frac{24.98-0}{109.2187/\sqrt{50}}=1.617.$ The P is the corresponding probability using the table with df=50-1=49, hence P is between 0.05 and 0.1. 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 more than$\alpha=0.05$, hence we fail to reject the null hypothesis. Hence the differences don't seem to be more than 0.