Why can we use Student t distribution for testing small populations with same variance? Student's t distribution is defined as: $\frac{\bar{X} - \mu }{ S/\sqrt{n}}$.
When we are testing normal populations with equal variances, to calculate the critical region we can:


*

*for two random samples of sizes $n$ and $m$ from  $X\sim N(\mu_1,\sigma^2)$ and $X\sim N(\mu_1,\sigma^2)$.

*and hypothesis $H_0: \mu_1 - \mu_2 = 0, H_1: \mu_1 - \mu_2 > 0$
$\alpha = P(\text{Reject $H_0$} |\text{$H_0$ is true})$
$\alpha = P(\bar{X} - \bar{Y} < a | \mu_1 - \mu_2 = 0)$
$\alpha = P(\frac{\bar{X} - \bar{Y}}{S_p \sqrt{1/n + 1/m}} < \frac{a}{S_p \sqrt{1/n + 1/m}} | \mu_1 - \mu_2 = 0)$
My question is, given that the difference of two t Student random variables is not t-distributed, why does the test approximates $\frac{\bar{X} - \bar{Y}}{S_p \sqrt{1/n + 1/m}} < \frac{a}{S_p \sqrt{1/n + 1/m}}$ as t-distributed with $n+m -2$ degrees of freedom?
 A: [Your opening sentence doesn't define a t-distribution, it defines a t-statistic -- which random variable does have a t-distribution under the right conditions. This is something that can be shown, rather than being a matter of definition.]
It's true that the difference of two independent t-distributed random variates is not itself t-distributed.
It's also true that you could write the (usual equal-variance) two sample t-statistic as the difference of two terms that are t-distributed (albeit with scale $\neq 1$).
However, we don't have independence here; there's a very particular kind of dependence because of the shared denominator - exactly of the form needed to make the original statistic t-distributed.
If you put the two terms back together again, the numerator is a difference of two sample means whose individual observations are each mutually independent (both within and across those groups). That difference of sample means will be normally distributed.
The denominator is the standard error of the difference in means. With samples from a normal distribution, sample means and variances are independent, and because of cross-sample independence the numerator and denominator of the two-sample t-statistic are in turn independent. 
Note further that we can rewrite the statistic as 
$$\frac{\bar{X}-\bar{Y}}{\sqrt{\sigma^2(\frac{1}{n_x}+\frac{1}{n_y})}}\cdot\frac{1}{\sqrt{\frac{S_p^2}{\sigma^2}}}$$
Under the usual assumptions for the equal-variance two sample t-test, the first term is distributed as a standard normal. The second term is independent of the first and its denominator is the square root of (a chi-squared random variate divided by its degrees of freedom).
A standard normal divided by the square root of (a chi-square divided by its d.f.), where those two terms are independent of each other has a t-distribution.
This makes the whole statistic t-distributed (with d.f. equal to the d.f. of the chi-square in the second term of the expression I wrote above).
So (given the usual assumptions), the two-sample t-statistic actually does have a t-distribution.
