Suppose I am building a multiple regression model, perhaps with 5 explanatory variables. Suppose the residuals are not normally distributed (based on a Q-Q plot and a D'Agostino test, due to kurtosis). Suppose I have a sample size in the neighborhood of 100: small enough that I can trust the D'Agostino test to be telling me about a significant departure from normality, but large enough that I can lean on the Central Limit Theorem.

Can I trust the ANOVA F-statistic and the corresponding p-value given by R? In other words, is there a theoretical reason that, with a sufficiently large sample size (much bigger than 100 if you want), that the ANOVA F-statistic should be F-distributed even if the residuals are not normal?

This is claimed by Wooldridge on page 176 of his Introductory Econometrics book (4th edition). Wooldridge is usually fairly careful and gives theoretical justifications, but does not in this case. I am a mathematician, so I'd like to understand why the F-statistic should be F-distributed in this setting. I know that the t-statistics for the individual explanatory variables (assuming no multicollinearity) are approximately normally distributed by the Central Limit Theorem. I know that I can do a randomization based test and build an empirical distribution for the F-statistics, and it looks like an F-distribution. I just don't have a theoretical reason to believe it is one, and my students keep asking me this question. Thanks!

  • $\begingroup$ If all other assumptions are fulfilled, you are usually fine under non-normality and large $n$ as long as there are no high leverage points. But often under non-normality, we also have variance inequality which invalidates p-values directly. $\endgroup$
    – Michael M
    Commented Apr 14, 2016 at 11:36

1 Answer 1


I am not entirely sure this answers your question, but here we go. Asymptotically, $q\times F$, with $F$ an F-distributed r.v. with $q$ "numerator degrees of freedom corresponding to the number of restrictions tested in the F-test, converges in distribution to a $\chi^2_q$ r.v. as the denominator degrees of freedom (a function of $n$, the sample size) tend to infinity.

Rearranging this result gives that $F$ itself converges to a $\chi^2_q/q$-distributed r.v., or a $F_{q,\infty}$ r.v. This r.v. can be well approximated by a $F_{q,n}$ r.v. for $n$ "sufficiently large" (as with most asymptotic approximations, there is no general answer to what "sufficiently large" precisely is), just as a $t$-distributed r.v. can be well approximated by a standard normal when degrees of freedom of $t$ are sufficiently large.

The figure below shows that even for 25 denominator degrees of freedom, the approximation is already quite close, for $q=3$.

enter image description here


To (hopefully) address Glenb's valid comment: why would $F$ be approximately $\chi^2_q/q$? We may write $F$ as $$ F=\frac{(R'\hat{\beta}-r)'\left\{R'(X' X)^{-1}R\right\}^{-1}(R'\hat{\beta}-r)/q}{\hat{\sigma}^{2}}, $$ where $R$ and $r$ specify the null $H_0:R'\beta=r$ to be tested.

Let us take the leading case of $R=I$, so testing restrictions on the coefficients as opposed to, say, testing restrictions on the sum of coefficients. If the null is true, $r=\beta$, so that we may write $$ qF=\frac{(\hat{\beta}-\beta)'\left\{(X' X)^{-1}\right\}^{-1}(\hat{\beta}-\beta)}{\hat{\sigma}^{2}}=\sqrt{n}(\hat{\beta}-\beta)'\left\{\hat{\sigma}^{2}(X' X/n)^{-1}\right\}^{-1}\sqrt{n}(\hat{\beta}-\beta) $$ Now, we know that $\sqrt{n}(\hat{\beta}-\beta)$ is asymptotically normal with an asymptotic variance that may be estimated consistently by $\hat{\sigma}^{2}(X' X/n)^{-1}$. So, the middle term $\bigl\{\hat{\sigma}^{2}(X' X/n)^{-1}\bigr\}^{-1}$, loosely speaking, "standardizes" both asymptotic multivariate normal random vectors in the quadratic form by their "standard deviation", so that we (again asymptotically) obtain standard normal vectors.

In large but finite samples, this vector will only approximately be standard normal. Hence, the resulting sum of squares of the standard normal independent entries of the vector will also only approximately be $\chi^2_q$.

But for $n$ reasonably large, the approximation will be pretty good.

  • $\begingroup$ So now you just need the numerator to be a scaled chi-squared ... but why would that be the case? $\endgroup$
    – Glen_b
    Commented Apr 14, 2016 at 11:59
  • $\begingroup$ Was that where your question aimed at? $\endgroup$ Commented Apr 14, 2016 at 12:27
  • $\begingroup$ @ChristophHanck: Thanks for your answer! I'm a little bit confused. In the first paragraph you say $q \times F$ is approximately $\chi^2_q$-distributed, but in the edit you said $F/q$ is approximately $\chi^2_q$-distributed. Can you clarify which one is correct? Also, I am not 100% sure this answers my question. I have the ANOVA F test-statistic, but I don't know it's F-distributed, because I don't know the residuals are normal. Is there some reason to believe that the distribution of the F-statistics is close enough to the true F-distribution for this argument to work? Thanks! $\endgroup$ Commented Apr 14, 2016 at 15:26
  • $\begingroup$ Ups. Thanks for your careful reading - the first is correct, see my edits. As for whether the residuals are normal, you could test that. That the unobserved errors are normal (what you need for the F statistic to be F) would however need to be an assumption. My answer attempts to show that $F$ will be approximately $\chi^2/q$ for large $n$, which in turn is close to the $F$ distribution (see the picture). But there may indeed be better answers to come by other users! $\endgroup$ Commented Apr 14, 2016 at 16:21
  • 1
    $\begingroup$ I don't think you need to get this complicated. If there is only an intercept in the model your inference is for a mean. If the data come from a log normal distribution a sample size of $n=20,000$ can be far too small to trust asymptotic results, i.e., you can get horrible confidence interval coverage if you use the CLT. Also, the CLT assumes the standard deviation is a correct dispersion measure, which is often not the case. $\endgroup$ Commented Apr 14, 2016 at 16:31

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