As I understand it, the formula for R2 is:
1 - (Var Residuals / Var Actuals)
The free book Open Intro Statistics has this example:
Guided Practice 8.10 The variance of the residuals for the model
given in Guided Practice 8.7 is 23.34, and the variance of the total
price in all the auctions is 83.06. Calculate R2 for this model.
1 - (VarResiduals / VarActuals) =
1 - (23.34 / 83.06) = 0.72
So far so good.
However, if the adjusted R2 is R2 * (n - 1 / n - k - 1) where n is sample size and k is number of predictors, then the denominator in the brackets will always be lower than the numerator (n - 1 > n - k -1). Therefore, the result will be a number greater than 1 and therefore shouldn't adjusted R2 always be greater than R2?
Adjusted R2 = R2 * [a number greater than 1]
My understanding was that adjusted R2 is intended to be a tougher evaluator of model fit. But if I'm reading the formula correct, shouldn't it always be a value greater than R2 since the right hand side of the multiplication sign in R2 * (n - 1 / n - k - 1) will always be greater than 1, because the numerator will always be greater than the denominator.
I don't follow this calculation, any guidance appreciated.
Here's the formula from page 377 of Open Intro Statistics:
Best Answer
There's no mistake, it's quite clear that for $k>0$ and $R^2<1$ that $R^2_\text{adj}$ is smaller than $R^2$.
Look at the formula closely. You multiply $1-R^2$ by a number greater than $1$ and then subtract that larger quantity from $1$, resulting in a smaller quantity than $R^2$.
Or working through it algebraically:
\begin{eqnarray} R^2_\text{adj} &=& 1-(1-R^2) \times \frac{n-1}{n-k-1}\\ &=& R^2 \times \frac{n-1}{n-k-1}-\frac{k}{n-k-1}\\ &=& R^2-\frac{k}{n-k-1}(1-R^2)\\ &\leq& R^2 \end{eqnarray}
with the inequality being strict when $k>0$ and $R^2<1$.