## Elementary Statistics (12th Edition)

$H_{0}:\mu_d=0,$ $H_{a}:\mu_d$ is more than 0. The test statistic is:$t=\frac{\overline{d}-\mu_d}{s_d/\sqrt{n}}=\frac{18.6667-0}{10.0935/\sqrt{12}}=6.406.$ The P is the corresponding probability using the table with df=12-1=11, hence P is less than 0.005. 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 less than $\alpha=0.01$, hence we reject the null hypothesis. Hence we can say that there is evidence to support that Captopril is effective.