Two tail t-critical value for margin of error for one tail t-test I have done a one-tailed one-sample t-test with the following setup:
$H_0:\overline{x} = \mu = \$151$
$H_A: \overline{x} < \mu $
$n = 24$
For $\alpha = 0.05$, the t-critical value is $-1.711$ (from the t-table, using degrees of freedom $= 24$), and I calculate $\overline{x} = \$126$ (skipping the details of the calculation). I get t-statistic $=-2.5$ so I reject the $H_0$. So far, so good. 
Now, the problem asks to calculate the confidence interval using $\alpha = 0.05$. 
I know that the margin of error $\overline{x} = \pm t{\text -}critical \times standard \ error$. 
The answer to the problem uses the two-tailed t-critical value, which is $2.064$, for $95\%$ confidence interval.
I am confused about why they use the two-tailed t-critical value rather than the one-tailed one? 
Is it because calculating the confidence interval is a separate task, and has nothing to do with the original hypothesis, and they are interested to see the deviation from the mean in both directions? 
 A: Your use of notation is strange.  The hypothesis should be written $$H_0 : \mu = 151 \quad \text{vs.} \quad H_a : \mu < 151.$$  The hypothesis is a statement about the value of the unknown parameter, in this case the mean $\mu$.  We make an inference about its value based on the data, from which we calculate a test statistic, which under the null hypothesis is $$T \mid H_0 = \frac{\bar x - 151}{s/\sqrt{n}},$$ where $\bar x$ is the sample mean and $s$ the sample standard deviation.  A statistic is never included in the hypothesis because it is something that is calculated from the data; there is no uncertainty about the value we get for it.
Given that $\bar x = 126$ and the value of your test statistic is $T = -2.5$, I get $s \approx 48.9898$ (the exact value of which may differ due to the limited precision you reported for the test statistic).
As for the calculation of a confidence interval, this need not be one-sided even if the hypothesis is one-sided.  The two concepts are related, but a confidence interval is nothing more than an interval estimate.  So, rather than using the sample mean as a simple point estimate for the true mean, we can incorporate the variability observed in the data to give a more sophisticated estimate of the true mean.  You can calculate a two-sided interval or a one-sided interval.  The question as stated is not specific about which one is intended.
A two-sided $95\%$ interval is computed as $$\bar x \pm t_{n-1,\alpha/2}^* \frac{s}{\sqrt{n}},$$ where $s/\sqrt{n}$ is the standard error of the mean, and $t_{n-1,\alpha/2}^* $ is the upper $\alpha/2$ quantile of the student $t$ distribution.  In your case, it is $$t_{23,0.025}^* = 2.06866.$$  Thus the two-sided interval is $$[105.31, 146.69].$$  The one-sided interval that is in the same direction as the hypothesis test is $$\left(-\infty, \bar x + t_{n-1,\alpha}^* \frac{s}{\sqrt{n}}\right].$$  Here, $t_{23,0.95}^* = 1.71387$ so the upper confidence limit is $143.14$, which is smaller than the upper confidence limit of the two-sided interval.  This makes sense because we selected the one-sided upper confidence limit in such a way that the probability the interval does not contain the true mean is $\alpha$; i.e., the upper tail probability of the student $t$ distribution is the full $0.05$, rather than the equal-tailed, two-sided interval.
