If I check for the assumptions of a repeated ANOVA (one IV with 3 measurements) I need to check for the normality of differences.

From t-test for paired data I know, I calculate the difference and run and hist(), qqplot() or shapiro.test() over the differences. Now, in ANOVA it is the same assumption. However, everybody seems to check the normality of residuals instead. As I asked in class I got the answer: it is faster. My question outside the speed-argument is, why is it the same?

Normally I would calculate 3 differences for 3 groups and then check the normality of each difference. Residuals are the deviation of the value to the mean(s). How is this the same?

Thanks for your time. I hope for helpful comments.


1 Answer 1


An ANOVA is a general linear model of the form (this is just one way of writing the model called the cell-means model, and I do so without loss of generality):


where $\mu_i$ are parameters (treatment means), $Y_{ij}$ is the value of the response variable for the $j$th trial for the $i$th treatment, and $i=1,...,r$ and $j=1..., n_i$, and $\epsilon_{ij}$ are independent $N(0, \sigma^2)$ as specified by the Gauss-Markov theorem.

Note that the error term is assumed to be Normal with mean 0 and variance $\sigma^2$ and that $\mu_i$ is considered a constant. One can validate the "normality assumption" by two methods:

  1. One can directly examine the errors $\epsilon_{ij}$ (residulas in the fitted model) which represent the error terms in the above model to verify that they are in fact distributed as $N(0, \sigma^2)$; or
  2. One can look at the $Y_{ij}$ terms directly since $Y_{ij}-\mu_i=\epsilon_{ij}.$ Because $\mu_{ij}$ is a constant:

\begin{eqnarray*} E(Y_{ij}) & = & E(\mu_{ij}-\epsilon_{ij})\\ & = & E(\mu_{ij})-E(\epsilon_{ij})\\ & = & \mu_{ij}-0\\ & & \mu_{ij} \end{eqnarray*}


\begin{eqnarray*} V(Y_{ij}) & = & V(\mu_{ij}-\epsilon_{ij})\\ & = & V(\mu_{ij})+V(\epsilon_{ij})\\ & = & 0+V(\epsilon_{ij})\\ & & \sigma^{2} \end{eqnarray*}

And because $Y_{ij}$ is a linear function of a normally distributed random variable, $\epsilon_{ij}$, $Y_{ij}$ itself is a normally distributed random variable (Recall that if $X$ is a random variable, then $Z=a+cX$ is normally distributed with mean $a+cE(X)$ and variance $c^2Var(X)$, given that $a$ and $c$ are constants.).

So, this implies $Y_{ij}$ are independent $N(\mu_{ij}, \sigma^2)$ in the cell-means ANOVA model, and, as a result, implies that one can verify normality be examining either (1) the residuals or by (2) examining the $Y_{ij}$ since they too are assumed normal.

So to summarize:

Because the $\epsilon_{ij}$ term is Normal and any linear combination of $\epsilon_{ij}$ is Normal, then the residuals must also be normal as well as the $Y_{ij}$ themselves since they are a linear combination of the $\epsilon_{ij}$.

  • $\begingroup$ Thanks. May I summarize in easiear words what I understood from your explanation: Yij−μi = ϵij Residuals have to be independent from each other and normal distributed. This I learned before. So the other side from the equation must be this as well. And this other side are my dv in different groups. Therefore I check normality in this groups. $\endgroup$
    – Daniela
    Jan 6, 2019 at 22:35
  • $\begingroup$ I did not get E, Z, a, c, V(prob Variance), X (prob the IV) $\endgroup$
    – Daniela
    Jan 6, 2019 at 22:41
  • $\begingroup$ That's was a general theorem from elementary probability. It simply states that any linear combination of a Normal variable is also Normal. $\endgroup$ Jan 6, 2019 at 22:43
  • $\begingroup$ Thanks for your fast answers. :) Now I'm good for tomorrow. $\endgroup$
    – Daniela
    Jan 6, 2019 at 22:53
  • 1
    $\begingroup$ I accepted. However, I'm not allowed yet to visibly vote. $\endgroup$
    – Daniela
    Jan 6, 2019 at 23:05

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