Why is the statistic $t=r \sqrt{\frac{n-2}{1-r ^2}} \approx t (n-2)$? Let $r$ be the sample correlation for two random variables $X,Y$ based on a random sample $(X_1, Y_1), (X_2, Y_2), \dots (X_n, Y_n)$. According to Wikipedia, under the null hypothesis of zero correlation, the test statistic $t=r \sqrt{\frac{n-2}{1-r ^2}}$ approximately follows a t-distribution with $n-2$ degrees of freedom when the number of observations $n$ is large enough. Is there an easy way to prove this? So far, I tried to rewrite the formula for $r$ in such a way that I can apply the Central Limit Theorem, but I was unable to make something out of it.

In Wackerly et al. this is actually given as a problem 11.55:
Testing the null hypothesis $H_0:\beta_1 = 0$, the statistic
$$T = \frac{\hat \beta_1 - 0}{\frac{S}{\sqrt{S_{xx}}}}$$ possesses a $t$ distribution with $n-2$ degrees of freedom if the null hypothesis is true. Show that the equation for T can also be written as $$T = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$$
So you want to start from the first equation and convert it to the second.
