Likan Zhan

Why squared-t-statistic and F-statistic are identical

Likan · 2017-05-23

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,

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

2. The logic behind the prove

\[ F = \frac{MSR}{MSE} \]

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,

\[ MSE = \frac{SS_1 + SS_2}{df_1 + df_2} \]

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

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.