# Testing the difference of two means with unknown variances

Let $X\sim N(\mu_1,\sigma_1^2)$ and $Y\sim N(\mu_2,\sigma_2^2)$ be two populations with unknown means and variances.

Suppose $X_1,...,X_n$ and $Y_1,...,Y_m$ are iid samples from $X$ and $Y$ respectively.

How can I test the hypothesis that $\mu_1=\mu_2$?

Can I use the t-test with the static

$$t=\frac{\bar X-\bar Y}{([S_1^2/(n-1)]+[S_2^2/(m-1)])^{1/2} }$$ to use t-test?

If so, which number should I use for the degree of freedom. What would be the mathematical justification?

If not, how may I test the hypothesis?

I am stuck with this for hours. Any hint would be a great help. Thanks!

Since both $X$ and $Y$ are normal distributed, they are as well $i.i.d.$ and you can use the t-test to test the difference in mean. If they would be dependent on each other, you could not use the formula provided because the standard error $SE(\bar{X}-\bar{Y})$ would depend on the covariance of $\bar{Y}$ and $\bar{X}$.
For $X$ and $Y$ you are using data to estimate the random variables $\bar{X}$ and $\bar{Y}$. This means you are losing one degree of freedom as stated in your formula. In some textbooks, only the large sample formula is provided without adjusting for $dof$. The small and large sample formulas should come close to one another when $n,m$ are each larger than $90$. Then, the Central limit theorem should ensure that your sample averages follow a normal distribution.
As for the critical value from the t-distribution, you should use $t_{n+m-2,1-\alpha/2}$ for two-sided tests and $t_{n+m-2,1-\alpha}$ for one-sided tests, where $\alpha$ is your significance level.