# Adjusted R Squared formula - why is it that it's lower when it shold be higher given the calculation?

As I understand it, the formula for R2 is:

1 - (Var Residuals / Var Actuals)


The 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:

• Hi @SmallChess, I added a screenshot from the text book I was studying. Looking at the deduced formula on the right of the = sign, I understood the first component to be the R2 formula and then the right side to be the part I'm questioning. won't it always be greater than 1? Apr 5, 2018 at 2:34
• To me, the left part is not R2. Apr 5, 2018 at 2:39
• Check your order of operations. The RHS is not $R^2$ times a number greater than 1. Apr 5, 2018 at 3:09
• Maybe easier to memorize is the following formula for adjusted R-squared: 1 - (Var* Residuals / Var Actuals), where Var* uses n-number of parameters in the denominator and equals the square of the residual standard error. Apr 5, 2018 at 4:40

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 $R^2$.
with the inequality being strict when $k>0$ and $R^2<1$.