How to calculate R-square from adjusted r-square, n, and p?

Question

Let $\bar{R}^2$ denote the adjusted coefficient of determination.

I have $\bar{R}^2 = 0.9199$ with 15 cases. Now I am trying to find $R^2$ given the results below.

I found the formula for $R^2$ but did not understand it. How do you calculate $R^2$ from $\bar{R}^2$?

$\bar{R}^2 = 1-\dfrac{(n-1)(1- R^2)}{n-p-1}$

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0.9199=1-(15-i)(1-RSQ)/(15-p), like this? But p unknown and i unknown?! — hhh, May 24 '13 at 09:48
The wikipedia page says that the $i$ should be a $1$ and that $p$ is the number of regressors not counting the constant term. This would be $2$ in your case. Now, you should be able to isolate $R^2$. — Stefan Hansen, May 27 '13 at 05:43

score 2 · Accepted Answer · answered May 27 '13 at 05:43

Given the equation for $\bar{R}^2$ you have:

$\bar{R}^2 = 1-\dfrac{(n-1)(1- R^2)}{n-p-1}$

You have $3$ regressors and a sample of $15$, thus substituting these and $\bar{R}2$ into the equation yields:

$0.9199 = 1 - \dfrac{(15-1)(1-R^2)}{15-3-1}$

Rearranging this expression and solving for $R^2$ gives:

$R^2 = 0.9371$

score 0 · Answer 2 · edited Oct 12 '19 at 18:26

As was stated above, the answer by GovEcon is wrong.

Wiki defines p in the above formula as: "where p is the total number of explanatory variables in the model (not including the constant term), and n is the sample size."

The parameters three parameters. Excluding the intercept (constant/beta0) p = 2.

That being said it would be easier to calculate $R^2$ as follows.

The formula for $R^2$ adjusted can be given as:

$R^2_{adj} = 1 - (n-1){MSE \over SST}, MSE = {SSE \over (n-p-1)}$

$= 1 - (n-1)[({SSE \over n-p-1})/SST] = 1 - [{(n-1)\over(n-p-1)}]*({SSE \over SST})$

If you recall from the definition that $R^2$ adjusted controls for increase in $R^2$ due to increase in parameters then it makes sense that removing [{(n-1) \over (n-p-1)}$] should give you $R^2$

$R^2 = 1 - {SSE \over SST}$

Check by plugging in:

${SSE \over SST} = 1 - R^2$

Original formula:

$R^2_{adj} = 1 - [(n-1)/(n-p-1)]*(1-R^2)$ = original formula

Then, $R^2_{adj} = 1 - (n-1)*(MSE/SST) = 1- (15-1)(8224/1436706) = ~.9198 $

$R^2 = 1- SSE/SST = 1- 98690/1436706 = ~.931$ NOT $.9371$

Note: the anova table is already rounded MSE = SSE/ (n-p-1) = 98690/(15-2-1) = 8224.16666 = ~8224. So discrepancies with table arise from here.

Aside: I don't know how to format equations and don't have time to do it now, but I do not want the current answer to mislead more people.

score 0 · Answer 3 · edited Nov 03 '15 at 10:20

0

You must also take into consideration what does constant mean. If it is Intercept, then you have two independent variables, which means $k=2$, not $3$.

edited Nov 03 '15 at 10:20

Harish Chandra Rajpoot

37,450

answered Nov 03 '15 at 10:02

Giorgi Shubladze

1

score 0 · Answer 4 · edited Apr 26 '14 at 20:18

Well, I think there is a little mistake in the equation. $p$ stands for the number of parameters, which means the number of predictors plus one for the constant.

$k$ stands for the number of predictors. Thus, \begin{align} p=k+1\tag{1}. \end{align}

The equation using $p$ is as follows: \begin{align} 1-\frac{(n-1)(1-R^2)}{n-p} \tag{2}. \end{align}

If you substitute equation $(1)$ in $(2)$ you got $$ 1-\frac{(n-1)(1-R^2)}{n-(k+1)}=1-\frac{(n-1)(1-R^2)}{n-k-1}. $$ The equation using $k$ is as follows: $$ 1-\frac{(n-1)(1-R^2)}{n-k-1}. $$ I know it is very common mistake. The degree of freedom for the SSR is $p-1$. For simple regression, the $\mathrm{df}$ of SSR is either $k=p-1=1$.

How to calculate R-square from adjusted r-square, n, and p?

4 Answers4