# R-squared and F-distribution

I am trying to understand the relations between the coefficient of determination $$R^2$$ and the $$F$$-distributions.

Hereafter are the notations I use :

Assume a simple linear regression model $$y_i=ax_i+b + \varepsilon_i, 1\leq i\leq n$$, where the errors $$\varepsilon_i$$ follow a normal distribution with mean $$0$$ and constant variance $$\sigma^2$$. The regression line calculated by OLS is $$y=\hat{a}x+\hat{b}$$ and $$\hat{y}_i=\hat{a} x_i +\hat{b}$$ is the predicted value (at point $$x_i$$) and $$\hat{\varepsilon}_i=y_i-\hat{y}_i$$ is the residual.

Let $$SSR=\sum(\hat{y}_i - \bar{y})^2$$ be the sum of squares explained by the regression, $$SSE=\sum(y_i - \hat{y}_i)^2$$ the sum of squared errors and $$SST=\sum(y_i - \bar{y})^2$$ the total sum of squares.

Then $$R^2=\frac{SSR}{SST}=1-\frac{SSE}{SST}$$.

What I understood from documents about linear regression is that, under the null hypothesis that $$a=0$$ (the slope of the true regression line is zero), the two statistics $$\frac{SSR}{\sigma^2}$$ and $$\frac{SST}{\sigma^2}$$ follow (respectively) a chi-squared distribution with $$1$$ df and a chi-squared distribution with $$n-1$$ df, and that they are independent.

So, in my mind, the ratio $$\frac{\frac{SSR}{\sigma^2} / 1}{\frac{SST}{\sigma^2} / (n-1)}$$ should follow $$F(1, n-1)$$, a Fisher-Snedecor distribution with $$df$$ $$1$$ and $$n-1$$...

Instead of finding this result, what I see written everywhere is that $$(n-2) \frac{R^2}{1-R^2}$$ follows $$F(1, n-2)$$...

Why do people use this latter form instead of the former ? Are they equivalent ? Something escapes me…

In a model with a single explanatory variable and an intercept term you have regression degrees-of-freedom $$DF_R = 1$$ and residual degrees-of-freedom $$DF_E = n-2$$, so you get:
\begin{aligned} F \equiv \frac{MSR}{MSE} &= \frac{SSR / DF_R}{SSE / DF_E} \\[6pt] &= \frac{DF_E}{DF_R} \cdot \frac{SSR}{SSE} \\[6pt] &= \frac{DF_E}{DF_R} \cdot \frac{SSR}{SST-SSR} \\[6pt] &= (n-2) \cdot \frac{R^2}{1-R^2}. \\[6pt] \end{aligned}
In your own calculations, you appear to have used an incorrect value for the residual degrees-of-freedom ($$n-1$$ instead of $$n-2$$) so your result is not equivalent to the correct form.