# 1. Notations

Let’s suppose two independent samples,

Sample 1 Sample 2
$$X_{11}$$ $$X_{12}$$
$$X_{21}$$ $$X_{22}$$
$$X_{31}$$ $$X_{32}$$
$$\vdots$$ $$\vdots$$
$$X_{n1}$$ $$X_{n2}$$

where,

• $$n_1$$ is the size of sample one;
• $$n_2$$ is the size of sample two;
• $$M_1$$ is the mean of sample one;
• $$M_2$$ is the mean of sample two;
• $$M_g$$ is the weighted mean of the two samples, defined as follows

$M_g = \frac{n_1 \cdot M_1 + n_2 \cdot M_2}{n_1 + n_2}$

# 2. The logic behind the prove

• Given the two samples, the $$F$$-statistic is,

$F = \frac{MSR}{MSE}$

There are only two conditions, so $$k = 2$$ and the sum of squares ($$SSR$$) and mean sum of squares ($$MSR$$) are the same,

$MSR = SSR = n_1 \cdot (M_1 - M_g)^2 + n_2 \cdot (M_2 - M_g)^2$

And the $$MSE$$ in the denominator of the $$F$$-ratio is,

$MSE = \frac{SS_1 + SS_2}{df_1 + df_2}$

• Given the same two samples, the squared $$t$$-statistic is

$\begin{equation} \begin{split} t^2 & = \Bigg(\frac{M_1 - M_2}{\sqrt{\frac{sp^2}{n_1} +\frac{sp^2}{n_2}}}\Bigg)^2\\ & = \frac{(M_1 - M_2) ^2}{\frac{sp^2}{n_1} + \frac{sp^2}{n_2}}\\ & = \frac{\frac{(M_1 - M_2) ^2}{\frac{n_1 + n_2}{n_1 \cdot n_2}}}{sp^2} \end{split} \end{equation}$

As we can see, the $$sp^2$$ in the $$t$$-statistic and $$MSE$$ in the $$F$$-statistic are the same,

$SSE = sp^2 = \frac{SS_1 + SS_2}{df_1 + df_2}$

So to prove $$t^2 = F$$, we need to prove the numerator in the squared-t-statistic and $$MSR$$ are the same, i.e.,

$n_1 \cdot (M_1 - M_g)^2 + n_2 \cdot (M_2 - M_g)^2 = \frac{(M_1 - M_2) ^2}{\frac{n_1 + n_2}{n_1 \cdot n_2}}$

# 3. The process of the prove

To the previous equation, we first need to replace the $$M_g$$ in $$MSR$$ with its definition, i.e., the weigheted mean of $$M_1$$ and $$M_2$$,

$\begin{equation} \begin{split} MSR & = n_1 \cdot (M_1 - M_g)^2 + n_2 \cdot (M_2 - M_g)^2\\ & = n_1 \cdot \Bigg(M_1 - \frac{n_1 \cdot M_1 + n_2 \cdot M_2}{n_1 + n_2}\Bigg)^2 + n_2 \cdot \Bigg(M_2 - \frac{n_1 \cdot M_1 + n_2 \cdot M_2}{n_1 + n_2}\Bigg)^2\\ & = n_1 \cdot \Bigg(\frac{n_1 \cdot M_1 + n_2 \cdot M_1 - n_1 \cdot M_1 - n_2 \cdot M_2}{n_1 + n_2}\Bigg)^2 + n_2 \cdot \Bigg(\frac{n_1 \cdot M_2 + n_2 \cdot M_2 - n_1 \cdot M_1 - n_2 \cdot M_2}{n_1 + n_2}\Bigg)^2\\ & = n_1 \cdot \Bigg(\frac{n_2 \cdot M_1 - n_2 \cdot M_2}{n_1 + n_2}\Bigg)^2 + n_2 \cdot \Bigg(\frac{n_1 \cdot M_2 - n_1 \cdot M_1 }{n_1 + n_2}\Bigg)^2\\ & = \Bigg( \frac{M_1 - M_2}{n_1 + n_2} \Bigg)^2 \cdot (n_1 \cdot n_2^2 + n_2 \cdot n_1^2)\\ & = \frac{(M_1 - M_2)^2}{(n_1 + n_2)^2} \cdot (n_1 \cdot n_2 ) \cdot (n_1 + n_2)\\ & = \frac{(M_1 - M_2)^2}{n_1 + n_2} \cdot (n_1 \cdot n_2 ) \\ & = \frac{(M_1 - M_2) ^2}{\frac{n_1 + n_2}{n_1 \cdot n_2}}\\ \end{split} \end{equation}$

# 4. Reference

O’Brien, & Francis J., Jr. (1982). A Proof That t2 and F are Identical: The General Case.