# Adjusted $R^2$ calculations

I'm struggling to figure out how these adjusted $$R^2$$ values for linear regression were calculated with $$n=8$$ observations:

Footnote 124 says that for a model with just an intercept, $$RSS$$ (residual sum of squares) equals $$TSS$$ (total sum of squares). So using $$R^2=1-\frac{RSS}{TSS}$$, we get $$R^2=0$$ for the model with just an intercept. Then I use the formula $$R^2_{adj} = 1-\left((1-R^2)\frac{n-1}{n-k-1}\right)$$ where $$n$$ is the number of observations (here $$n=8$$), and $$k$$ is the number of slopes (not including the intercept). So for the model with just the intercept, I get $$R^2_{adj} = 1-\left((1-0)\frac{8-1}{8-1}\right) = 0$$ whereas the book has $$0.4077$$.

I get a different answer for the other models as well. For instance, for the model only using $$X_2$$ I get $$R^2_{adj} = 1-\left(\frac{6981.58}{10693.5}\cdot \frac{8-1}{8-2}\right)=0.2383.$$

For the model with $$X_1$$ and $$X_2$$: $$R^2_{adj} = 1-\left(\frac{915.375}{10693.5}\cdot\frac{8-1}{8-3}\right) = 0.8802$$

For the model using all three predictors: $$R^2_{adj} = 1-\left(\frac{908.166}{10693.5}\cdot\frac{8-1}{8-4}\right) = 0.8514.$$

What am I missing or doing wrong?

• Are you sure that $r^2$ is the residual sos divided by the total? Mar 6, 2021 at 16:47
• Could you please edit the question to include a citation of and link to the book you are quoting from?
– EdM
Mar 6, 2021 at 17:02
• @EdM The book is not freely available online. It is Howard Mahler's Guide to Statistical Learning: howardmahler.com/Teaching/MAS-1.html
– kccu
Mar 6, 2021 at 17:57
• @mdewey No, $R^2$ is the model sum of squares divided by the total sum of squares. Since $TSS=MSS+RSS$, $R^2 = \frac{MSS}{TSS}=1-\frac{RSS}{TSS}$. The formula for $R^2_{adj}$ uses $1-R^2$, which is equal to $\frac{RSS}{TSS}$. All these formulas are on the Wikipedia page for coefficient of determination, and in the textbook as well: en.wikipedia.org/wiki/Coefficient_of_determination
– kccu
Mar 6, 2021 at 18:00
• There seem to be two different models with X1 and X3 as predictors, with different coefficients. Seems a typo; the third one likely involved X2 and X3. Mar 7, 2021 at 2:18

Not a definitive answer but from what I gathered, there are different formulas for calculating the adjusted R-squared. The adjusted R-squared tries to express the proportion of variance explained by a model on a population level. Since this is not an easy thing to estimate, there have been different proposals for calculating the adjusted R-squared. Some of the different versions include:

• Wherry’s formula: $$1-(1-R^2)\frac{(n-1)}{(n-v)}$$
• McNemar’s formula: $$1-(1-R^2)\frac{(n-1)}{(n-v-1)}$$
• Lord’s formula: $$1-(1-R^2)\frac{(n+v-1)}{(n-v-1)}$$
• Stein's formula: $$1-\big[\frac{(n-1)}{(n-k-1)}\frac{(n-2)}{(n-k-2)}\frac{(n+1)}{n}\big](1-R^2)$$

An often cited study in this context is Yin and Fan (2001), which is a comparison study of different R-squared versions based on simulated data. See also these three questions about this issue:

What is the adjusted R-squared formula in lm in R and how should it be interpreted?

What is an unbiased estimate of population R-square?

For your specific example I did not get the shown solutions with any of the above listed formulas, but I guess it is possible that the author used yet another formula? Perhaps footnot/reference 125 in your passage gives some indication of what was used?

Reference:

• Yin, P., & Fan, X. (2001). Estimating $$R^2$$ shrinkage in multiple regression: A comparison of different analytical methods. The Journal of Experimental Education, 69(2), 203-224.
• All of these formulas should still yield an adjusted $R^2$ of $0$ for the model with just an intercept though, correct? I don't understand how the text has a nonzero adjusted $R^2$ for that model.
– kccu
Mar 7, 2021 at 15:39
• Also the formula I used is the only one presented in the text the screenshot is from.
– kccu
Mar 7, 2021 at 15:40
• Wherry's formula above does not yield 0 for the model with just an intercept...But it doesn't result in the value of your passage. Perhaps its easiest to write the author and ask directly what software/formula he used for his calculations. Mar 8, 2021 at 20:40
• Ah, you are correct about Wherry's formula (and Stein's as well?). But the adjusted $R^2$ of $0.4077$ still seems entirely too large for a model with just an intercept.
– kccu
Mar 16, 2021 at 18:06

The text is almost certainly in error. This is a danger with self-published works, as this appears to be. Without peer review and editors, mistakes easily creep in.

The adjusted $$R^2$$ can be negative, and its value will always be less than or equal to that of $$R^2$$.
There is no way that an intercept-only model can have an $$R^2$$ other than 0, so for the text to claim a substantial positive adjusted $$R^2$$ for that model (0.4077) must be an error.
We can also examine the implications of the claim that the 3-predictor model has an adjusted $$R^2$$ of 0.912. That is lower than the unadjusted $$R^2$$ of 0.915 so it can't be completely ruled out. But what would that mean for the relationship between $$n$$ and $$p$$? With those values and the adjusted $$R^2$$ formula, I get:
$$\frac{n-1}{n-p-1}= 1.035$$
or $$n \approx 1 +30p$$. That's not compatible with the assumed $$n=8$$, particularly not if $$p=3$$.