# 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?

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.

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 $$0.$$

• Thank you, but then how do I look up the value using ttable.org ? Commented Feb 4, 2023 at 9:12
• As Mewbacca comprehensively wrote, it depends on what you are testing. As for how to use a table, hint symmetry. Commented Feb 4, 2023 at 9:29
• Thank you, but the formula I have been given does not ask what I am testing. I can just plug in the null-hypothetical mean . So is the formula missing something? Commented Feb 4, 2023 at 9:35
• @Kirsten, check Henry's answer which I hope would clear your impending confusion. This is again symmetry at work. Commented Feb 4, 2023 at 18:18

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.