Calculating F-statistic, why is SS used instead of just $r^2$? The calculations and question are for a simple regression (one independent and one dependent variable).
SSres = SSy * (1 - $r^2$)
SSreg = SSy * $r^2$
F = (SSreg / DFreg)/(SSres / DFres)
So, I've figured out that the value of SSy does not matter for what F-value we get, and that's a good thing since otherwise the F-value is in part contingent upon the scale of our measure. So, in actuality, am I correct to assume that we could in fact simply replace SSres and SSreg with $r^2$ and $(1 - r^2)$ and ignore the in-between calculations of SSres and SSreg? Or is this due to that the F-statistic in this case comes from a simple regression, as opposed to multiple regression?
 A: If the degrees-of-freedom are taken as fixed, the F-statistic and the coefficient of determination are related as one-to-one functions:
$$F = \frac{MS_{Reg}}{MS_{Res}} = \frac{df_{Res}}{df_{Reg}} \cdot \frac{SS_{Reg}}{SS_{Res}} = \frac{df_{Res}}{df_{Reg}} \cdot \frac{R^2}{1-R^2}.$$
If you have already calculated the coefficient of determination then you can use this to obtain the F-statistic via the above formula.  From this equation we see that these two measures of goodness-of-fit are monotonicly related (a higher coefficient of determination goes with a higher F-statistic).
Incidentally, this means that a goodness-of-fit test could just as legitimately be conducted directly on the coefficient of determination instead of the F-statistic.  The latter is generally used becuase it has a simpler distribution, but they are both increasing monotonic functions of one another, so either statistic gives a proper ordinal measure of the evidence against the null hypothesis in a goodness-of-fit test.
A: We could replace SSres and SSreg with $r^2$ and $(1 - r^2)$. For multiple regression, you can replace SSres and SSreg with $R^2$ and $(1 - R^2)$, where $R^2$ is (unadjusted) R square.  
