Should the formula for test statistic have an absolute value? I have been given a formula for calculating a t test statistic as 
$$t_{n-1} = \frac{ \bar{x}- \mu_0}{\frac{S}{\sqrt{n}}} $$
Where $\bar{x}$ is the sample mean, $\mu_0$ is the hypothetical true mean, $S$ is the sample standard deviation and $n$ is the sample count.
But when I plug my values in, the outcome is negative.
I don't see how to look up a negative value on the T-table.
Should I be taking the absolute value?
 A: The support of $\mathrm t$ distribution is $\mathbb R. $ So, it is still a legit case if $t_{n-1}<0.$ And don't forget the probability curve is symmetric about $0.$
A: As User1865345 states, the t statistic can take negative or positive values.
Your reference to the table depends on your hypotheses. The common null hypothesis is $H_0: \mu = \mu_0$. The alternative hypotheses can be one sided ($H_1: \mu \le \mu_0$ or $H_1: \mu \ge \mu_0$) or two sided ($H_1: \mu \neq \mu_0$).
For a one-sided alternative $H_1: \mu \ge \mu_0$, at a level of significance $\alpha$, you reject $H_0$ if the calculated value, $t_{calc}$(without taking absolute value) is greater than $t_{n-1, \alpha}$ from the table.
For a one-sided alternative $H_1: \mu \le \mu_0$, at a level of significance $\alpha$, you reject $H_0$ if the calculated value, $t_{calc}$(without taking absolute value) is lesser than $-t_{n-1, \alpha}$ from the table. (Again, as pointed out, the t distribution is symmetric about the t = 0 and the left tail probability at level of significance $\alpha$ is $-t_{n-1, \alpha}$)
For a two-sided alternative $H_1: \mu \neq \mu_0$, the tail probability or the rejection region is split into two halves, and you reject $H_0$ if the calculated value, $|t_{calc}|$ is greater than $t_{n-1, \alpha/2}$ from the table. (Note the absolute value and $\alpha/2$).
It's best to remember your alternatives, visualize the graph of the distribution for critical region, and glance at the table before concluding the test.
A: In a previous question you were looking at 
and wanted the column labelled cum.prob $t_{.975}$ or one-tail $0.025$ or two-tails $0.05$
If the table had been extended to the left, then by the symmetry of the $t$ distribution,

*

*there would have been a column labelled cum.prob $t_{.025}$ or one-tail $0.025$ or two-tails $0.05$

*and it would have had the values $-12.71, -4.303,-3.182,\ldots$
and similarly with the other columns.
Doing this would double the size of the table without adding useful information beyond the change of signs.
