# How to compute adjusted R-squared (in R)

See all solutions.

If we have fit a multivariate linear model, how can we compute the Adjusted $R^2$ for that model, to measure its goodness of fit?

## Solution

We assume you have already fit a multivariate linear model to the data, as in the code below. (If you’re unfamiliar with how to do so, see how to fit a multivariate linear model.) The data shown below is fake, and we assume you will replace it with your own real data if you use this code.

1
2
3
4
5
x1 <- c(2, 7, 4, 3, 11, 18, 6, 15, 9, 12)
x2 <- c(4, 6, 10, 1, 18, 11, 8, 20, 16, 13)
x3 <- c(11, 16, 20, 6, 14, 8, 5, 23, 13, 10)
y <- c(24, 60, 32, 29, 90, 45, 130, 76, 100, 120)
model <- lm(y ~ x1 + x2 + x3)


You can get a lot of information about your model from its summary.

1
summary(model)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Call:
lm(formula = y ~ x1 + x2 + x3)

Residuals:
Min      1Q  Median      3Q     Max
-25.031 -20.218  -8.373  22.937  35.640

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)   77.244     27.366   2.823   0.0302 *
x1            -2.701      2.855  -0.946   0.3806
x2             7.299      2.875   2.539   0.0441 *
x3            -4.861      2.187  -2.223   0.0679 .
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 30.13 on 6 degrees of freedom
Multiple R-squared:  0.5936,	Adjusted R-squared:  0.3904
F-statistic: 2.921 on 3 and 6 DF,  p-value: 0.1222


In particular, that printout contains the Adjusted $R^2$ value; it is the second value in the right-hand column, near the top.

You can also obtain it directly, as follows:

1
summary(model)$adj.r.squared  1  0.3903924  In this case, the Adjusted$R^2$is$0.3904\$.