If we want to test the hypothesis $H_0$: $\mu = 0$, where $H_a$: $\mu \neq 0$. Then the t statistic is $t=\bar{y}/(s/\sqrt{n})$ and 95% confidence interval is $\bar{y} \pm t_{.975}s/\sqrt{n}$ by t-test. If the sample mean $\bar{y} > 0$, why the rejection region is $t>t_{.975}$ instead of $t<-t_{.975}$?