##### Using the $C$% confidence interval for $\mu$ to perform right sided one sample $t$ test: full explanation

If $\mu_0$ is not in the $C$% confidence interval and the sample mean is larger than $\mu_0$ (as expected), the sample mean is significantly larger than $\mu_0$ at significance level $\alpha = \frac{1 \, - \, C/100}{2}$ If $\mu_0$ is not in the $C$% confidence interval and the sample mean is larger than $\mu_0$, the sample mean is significantly larger than $\mu_0$ at significance level $\alpha = \frac{1 \, - \, C/100}{2}$. In order to understand why, we have to be aware of the following three points, which are illustrated in the figure above: The lower bound of the $C$% confidence interval for $\mu$ is $\bar{y} - t^* \times SE$, the upper bound is $\bar{y} + t^* \times SE$. That is, we subtract $t^* \times SE$ from the sample mean, and we add $t^* \times SE$ to the sample mean. Here: $SE$ is the standard error of $\bar{y}$, which is $s / \sqrt{N}$ the critical value $t^*$ is the value under the $t$ distribution with area $C / 100$ between $-t^*$ and $t^*$, and therefore area $\frac{1\,-\,C/100}{2}$ to the right of $t^*$. In the figure above, the red dots represent sample means, the blue arrows represent the distance $t^* \times SE$. If we set the significance level $\alpha$ at $\frac{1 \, - \, C/100}{2}$, the critical value $t^*$ used for the significance test is the value under the $t$ distribution with area $\alpha = \frac{1\,-\,C/100}{2}$ to the right of $t^*$. This is the same critical value $t^*$ as the $t^*$ used for the $C$% confidence interval. We reject the null hypothesis at $\alpha = \frac{1\,-\,C/100}{2}$ if the $t$ value $t = \frac{\bar{y} - \mu_0}{SE}$ is at least as large as $t^*$. This means that we reject the null hypothesis at $\alpha = \frac{1\,-\,C/100}{2}$ if sample mean $\bar{y}$ is at least $t^* \times SE$ above $\mu_0$. In the figure above, the rejection region for $\bar{y}$ is represented by the green area. If $\mu_0$ is not in the $C$% confidence interval [$\bar{y} - t^* \times SE \,;\, \bar{y} + t^* \times SE$] and the sample mean $\bar{y}$ is larger than $\mu_0$, then $\bar{y}$ must be at least $t^* \times SE$ above $\mu_0$. In the figure above, the first confidence interval does not contain $\mu_0$ and the sample mean is larger than $\mu_0$, so the sample mean must be at least $t^* \times SE$ above $\mu_0$. The second confidence interval does contain $\mu_0$, so the sample mean must be less than $t^* \times SE$ above $\mu_0$. In sum, if $\mu_0$ is not in the $C$% confidence interval [$\bar{y} - t^* \times SE \,;\, \bar{y} + t^* \times SE$] and the the sample mean is larger than $\mu_0$, we know that the the sample mean is significantly larger than $\mu_0$ at significance level $\alpha = \frac{1 \, - \, C/100}{2}$.