I was wondering if there is a relationship between $R^2$ and a F-Test.

Usually $$R^2=\frac {\sum (\hat Y_t - \bar Y)^2 / T-1} {\sum( Y_t - \bar Y)^2 / T-1}$$ and it measures the strength of the linear relationship in the regression.

An F-Test just proves a hypothesis.

Is there a relationship between $R^2$ and a F-Test?

  • 2
    $\begingroup$ The formula for $R^2$ looks incorrect, not just because it's missing some characters in the denominator: those "$-1$" terms don't belong. The correct formula looks much more like an $F$ statistic :-). $\endgroup$
    – whuber
    Apr 22, 2013 at 17:28
  • $\begingroup$ See stats.stackexchange.com/questions/58107/… $\endgroup$ May 21, 2013 at 9:31

4 Answers 4


If all the assumptions hold and you have the correct form for $R^2$ then the usual F statistic can be computed as $F = \frac{ R^2 }{ 1- R^2} \times \frac{ \text{df}_2 }{ \text{df}_1 }$. This value can then be compared to the appropriate F distribution to do an F test. This can be derived/confirmed with basic algebra.

  • 3
    $\begingroup$ could you please define df1 and df2? $\endgroup$
    – bonobo
    Jan 20, 2017 at 14:49
  • 1
    $\begingroup$ @bonobo, df1 is the numerator degrees of freedom (based on the number of predictors) and df2 is the denominator degrees of freedom. $\endgroup$
    – Greg Snow
    Jan 20, 2017 at 17:23
  • 4
    $\begingroup$ To clarify further about the degrees of freedom: df1=k, where k is number of predictors. df1 is called the "numerator degrees of freedom," even though it's in the denominator in this formula. df2=n−(k+1), where n is the number of observations and k is the number of predictors. df2 is called the "denominator degrees of freedom," even though it's in the numerator in this formula. $\endgroup$
    – Tim Swast
    Feb 11, 2018 at 17:38
  • 8
    $\begingroup$ @GregSnow could you consider adding the definitions for the degrees of freedom to the answer? I suggested such a change at stats.stackexchange.com/review/suggested-edits/175306 but it was rejected. $\endgroup$
    – Tim Swast
    Feb 11, 2018 at 17:45

Recall that in a regression setting, the F statistic is expressed in the following way.

$$ F = \frac{(TSS - RSS)/(p-1)}{RSS/(n-p)} $$

where TSS = total sum of squares and RSS = residual sum of squares, $p$ is the number of predictors (including the constant) and $n$ is the number of observations. This statistic has an $F$ distribution with degrees of freedom $p-1$ and $n-p$.

Also recall that $$ R^2 = 1 - \frac{RSS}{TSS} = \frac{TSS - RSS}{TSS} $$

simple algebra will tell you that $$ R^2 = 1 - (1 + F \cdot \frac{p-1}{n-p})^{-1} $$

where F is the F statistic from above.

This is the theoretical relationship between the F statistic (or the F test) and $R^2$.

The practical interpretation is that a bigger $R^2$ lead to high values of F, so if $R^2$ is big (which means that a linear model fits the data well), then the corresponding F statistic should be large, which means that that there should be strong evidence that at least some of the coefficients are non-zero.


Intuitively, I like to think that the result of the F-ratio first gives a yes-no response to the the question, 'can I reject $H_0$?' (this is determined if the ratio is much larger than 1, or the p-value < $\alpha$).

Then if I determine I can reject $H_0$, $R^2$ then indicates the strength of the relationship between.

In other words, a large F-ratio indicates that there is a relationship. High $R^2$ then indicates how strong that relationship is.


Also, quickly:

R2 = F / (F + n-p/p-1)

Eg, The R2 of a 1df F test = 2.53 with sample size 21, would be:

R2 = 2.53 / (2.53+19) R2 = .1175


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.