Regression and $R^2$ I have a question: Regression results are reported from a data set satisfying the classical regression model(with homoskedastic errors) and $N=506$ observations. The authors report that the null hypothesis of "all five slope coefficients equal zero" is not rejected at an approximate 5% level. Assuming the authors used the "homoskedasticity-only" version of the relevant test statistic, is $R^2=0.04$ consistent with the authors' conclusion? 
 A: Yes, that makes perfect sense.  If your regression has an R Square of essentially zero it means your regression really does not estimate the actual data well.  Given that it makes good sense that none of your regression coefficients are statistically significant at the 0.05 level.  I suspect several are not even statistically significant at the 0.10 level or even higher. 
A: I don't get that. Putting in $n=506$ and $r=0.2$ in the Vassastats calculator, I get $t=4.583$ or $p<0.0001$. Now, if I goof up and put in $r=0.04$ then I get not significant. I think the authors made a boo-boo, namely that $(\pm 0.2)*(\pm 0.2)=0.04$ and the calculator takes $r$ not $r^2$.
Why? The symmetric $\pm$t-statistic is from $t=r \sqrt{\frac{n-2}{1-r^2}}$, and is applied to the single and doubled right tail CDF Student's-t distribution with $n-2$ degrees of freedom. Note that as $n$ increases the magnitude of the t-statistic increases such that in the limit, a vanishingly small magnitude $r$-value can be made significant for an $n$ sufficiently large. Now, with respect to the current question, an $n=506$ is larger than for most patient series, and as such, a rather smaller magnitude $r$-value becomes significant than we are "accustomed" to while reading about smaller patient series. I hypothesize the authors of the unnamed paper may have fallen victim to their expectations of a significance-value as opposed to a proper calculation of same.
