Prove F test is equal to T test squared I need to show that F test is equal to T test squared, when the T test is for 2 independent groups and assuming variances are equal.
I know that $F=\frac{MSB}{MSW}=\frac{SSB/k-1}{SSW/N-K}$
and I know that $T=\frac{X-Y}{S_p \sqrt{\frac{1}{n}+\frac{1}{m}}}$, 
so $T^2=\frac{(X-Y)^2}{S_p^2 ({\frac{1}{n}+\frac{1}{m}})}$
I've seen this proof in Regression but here we're not using MSE and MSR, so i'm not sure how to connect between the two.
 A: Because one has $\boxed{T^2=F}$.
To show that, you have to check that (with $N=mn$):


*

*$SSW/(N-2)= S^2_p$ (the unbiaised estimate of $\sigma^2$)

*$SSB = {(\bar X- \bar Y)}^2/(\frac{1}{n}+\frac{1}{m})$
To show the second point you only have to use : 


*

*the elementary equality $SSB=m{(\bar x - \bar{x\cdot y})}^2+n{(\bar y - \bar{x\cdot y})}^2$

*the fact that the mean of the whole sample $x\cdot y=(x_1, \ldots, x_m, y_1, \ldots y_n)$ is the weighted mean $\frac{m \bar x + n \bar y}{m+n}$

*some elementary but a little tiedous calculations to conclude
Sorry for the strange notation $x\cdot y$ for the "whole sample", this was my first idea and I'm in a hurry now.
A: You can rewrite the equation as \begin{equation}\frac{SSB/\left(k-1\right)}{SSW/\left(N-k\right)}=\frac{SSB\left(k-1\right)/\sigma^{2}\left(k-1\right)^{2}}{SSW\left(N-k\right)/\sigma^{2}\left(N-k\right)^{2}} \end{equation}
For $k=2$ (two groups), \begin{equation}\frac{SSB\left(k-1\right)/\sigma^{2}\left(k-1\right)^{2}}{SSW\left(N-k\right)/\sigma^{2}\left(N-k\right)^{2}}=\frac{SSB/\sigma^{2}}{SSW\left(N-2\right)/\sigma^{2}\left(N-2\right)^{2}}. \end{equation} 
The numerator is a $\chi^{2}$  distribution with one degree of freedom. The denominator has the following distribution: \begin{equation} SSW\left(N-2\right)/\sigma^{2}\left(N-2\right)^{2}\sim\frac{\chi_{N-2}^{2}}{\left(N-2\right)^{2}}. \end{equation} Therefore, you have the ratio of two $\chi^{2}$  distributions. This ratio is equivalent to a $t$ distribution with $N-1$  degrees of freedom squared: \begin{equation}\frac{\chi_{1}^{2}}{\chi_{N-2}^{2}/\left(N-2\right)^{2}}\sim t_{N-2}^2. \end{equation}
