Using the $C$% confidence interval for $\mu_1 - \mu_2$ to perform left sided two sample $t$ test (equal variances not assumed): full explanation

If 0 is not in the $C$% confidence interval and the difference between the two sample means is smaller than 0 (as expected), the difference between the two sample means is significantly smaller than 0 at significance level $\alpha = \frac{1 \, - \, C/100}{2}$


confidence interval as significance test

If 0 is not in the $C$% confidence interval and the difference between the two sample means is smaller than 0, the difference between the two sample means is significantly smaller than 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:
  1. The lower bound of the $C$% confidence interval for $\mu_1 - \mu_2$ is $\bar{y} - t^* \times SE$, the upper bound is $\bar{y} + t^* \times SE$. That is, we subtract $ t^* \times SE$ from the difference between the two sample means, and we add $ t^* \times SE$ to the difference between the two sample means. Here:
    • $ SE$ is the standard error of $\bar{y}_1 - \bar{y}_2$, which is $\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}$
    • 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^*$ and to the left of $ -t^*$.
    In the figure above, the red dots represent differences between two sample means, the blue arrows represent the distance $ t^* \times SE$.
  2. 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 left of $ t^*$. This is the same critical value $ t^*$ as the (negative) $ 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}_1 - \bar{y}_2) - 0}{SE}$ is at least as small as the negative $ t^*$. This means that we reject the null hypothesis at $\alpha = \frac{1\,-\,C/100}{2}$ if the difference between the two sample means $\bar{y}_1 - \bar{y}_2$ is at least $ t^* \times SE$ below 0. In the figure above, the rejection region for $\bar{y}_1 - \bar{y}_2$ is represented by the green area.
  3. If 0 is not in the $C$% confidence interval [$\bar{y} - t^* \times SE \,;\, \bar{y} + t^* \times SE$] and the difference between the two sample means $\bar{y}_1 - \bar{y}_2$ is smaller than 0, then $\bar{y}_1 - \bar{y}_2$ must be at least $ t^* \times SE$ below 0. In the figure above, the first confidence interval does not contain 0 and the difference between the two sample means is smaller than 0, so the difference between the two sample means must be at least $ t^* \times SE$ below 0. The second confidence interval does contain 0, so the difference between the two sample means must be less than $ t^* \times SE$ below 0.
In sum, if 0 is not in the $C$% confidence interval [$\bar{y} - t^* \times SE \,;\, \bar{y} + t^* \times SE$] and the difference between the two sample means is smaller than 0, we know that the difference between the two sample means is significantly smaller than 0 at significance level $\alpha = \frac{1 \, - \, C/100}{2}$.