One sided confidence interval for the mean

I want to test the following hypothesis: $$H_0: \mu \geq \mu_0 \\ H_a: \mu < \mu_0.$$

I use one sample t-test: $$t_0 = \frac{\bar{X} - \mu_0}{S/\sqrt{n}}.$$ Since under the alternative we expect that $\bar{X}$ is small, we should reject for small values of $t_0$. Thus, given that the size of the test is equal to $\alpha$, we reject if $$t_0 \leq -t_{1-\alpha}(n-1).$$ My aim is to construct a confidence interval for $\mu$. Given my results, can I construct a confidence interval for $\mu$ based on the fact that we don't reject if $$t_0 > -t_{1-\alpha}(n-1),$$ thus the confidence interval should be constructed by solving the following inequality $$\frac{\bar{X} - \mu}{S/\sqrt{n}} > -t_{1-\alpha}(n-1)?$$ I am not sure that my way of reasoning is consistent since I use t-test (with $\mu_0$) in order to construct the rejection region and I use $\frac{\bar{X} - \mu}{S/\sqrt{n}}$ (with $\mu$ instead of $\mu_0$) in order to determine the confidence interval.

• You may find this answer of some use. – Glen_b Nov 29 '13 at 21:36