# $S_{xx}$ and $S_{xy}$ in linear regression

I am trying to prove that an equivalent way to perform the test for significance of regression in multiple linear regression is to base the test on $$R^2$$ as follows:

to test $$H_0:\beta_1 = \beta_2 = ... = \beta_k$$ versus $$H_1$$ : at least one $$\beta_j \neq 0$$, calculate $$F_0=\frac{R^2(n-p)}{k(1-R^2)}$$ and to reject $$H_0$$ if the computed value of $$F_0$$ exceeds $$F_{\alpha,k,n-p}$$, where $$p = k + 1$$.

Now, I am starting from the easier case of simple linear regression, and so I want to show that $$t^2=\frac{R^2(n-2)}{1-R^2}.$$ I did the following. $$t=\frac{\hat{\beta_1}}{se(\hat{\beta_1})}=\frac{S_{xy}}{S_{xx}}\sqrt{\frac{S_{xx}(n-2)}{SS_{res}}} \Rightarrow t^2= \frac{S^2_{xy}(n-2)}{S_{xx}SS_{res}}$$ and $$R^2=\frac{SS_{reg}}{SS_{tot}}=\frac{S^2_{xy}}{S_{xx}SS_{tot}} \Rightarrow 1-R^2=\frac{S_{xx}SS_{tot} - S^2_{xy}}{S_{xx}SS_{tot}} \Rightarrow \frac{R^2(n-2)}{1-R^2}=\frac{S^2_{xy}(n-2)}{S_{xx}SS_{tot} - S^2_{xy}}.$$

How do I conclude? Thanks!

• Welcome - would you maybe consider rethinking the title of your question? I feel it could better reflect your actual topic/question Mar 17 at 7:20

$$S_{xy}=\sum_{i=1}^{n}(x_i-\bar x)(y_i-\bar y)$$ and correspondingly for $$S_{xx}$$. $$SS_{tot}=\sum_{i=1}^{n}(y_{i}-\bar{y})^{2}$$, $$SS_{reg}=\sum_{i=1}^{n}(\widehat{y}_{i}-\bar{y})^{2}$$ and $$SS_{res}=\sum_{i=1}^{n}\hat{u}_{i}^{2}$$ stand for total (i.e., variation of the $$y_i$$), regression (i.e., variation of the fitted values) and residual sum of squares.
We have that $$R^2=\frac{S_{xy}^2}{S_{xx}SS_{tot}}$$ and, from $$SS_{tot}=SS_{reg}+SS_{res}$$, $$R^2=1-\frac{SS_{res}}{SS_{tot}}$$ so that $$SS_{tot}(1-R^2)=SS_{res}^2$$ and hence $$t^2= \frac{S^2_{xy}(n-2)}{S_{xx}SS_{tot}(1-R^2)}.$$ One could also have continued from your denominator $$S_{xx}SS_{tot} - S^2_{xy}$$ to get $$S_{xx}(SS_{reg}+SS_{res}) - S^2_{xy},$$ so that it would remain to show $$S_{xx}SS_{reg} - S^2_{xy}=0.$$ This follows from $$SS_{reg}=\frac{S^2_{xy}}{S_{xx}},$$ which, in turn, follows from $$\widehat{y}_{i}=a+bx_{i}$$ and $$\widehat{y}_{i}-\bar{y}=b(x_{i}-\bar{x})$$ so that $$SS_{reg}=b^{2}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}=b^{2}S_{xx}.$$ Plugging in $$b=S_{xy}/S_{xx}$$ gives $$SS_{reg}=\frac{S^2_{xy}}{S_{xx}}$$