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The analysis of variance usually requires the assumption of normality. However I always read different approaches on how to test this assumption (between different books but also here) especially in the case of repeated measures anova. Mostly regarding what to check for normality (e.g. dependent variable, residuals) and on what level (eg. per cell, between cells or per sample). Normality on the cell level would imply residual normality on the sample level, but not the other way around. But is it necessary? Residual normality on the sample level is much easier to check (i.e. faster) than anything on the cell level.

My question is twofold. First, what would be the exact normality assumption we have to check for each of the following designs? Second, what would be the a) most rigorous method (i.e. in a perfect world) b) the standard method (but maybe not perfect due to effort vs time constraints) method to check the specific assumption. Please assume that each independent variable has 3 levels.

The designs are the following:

A) One-way anova

B) One-way repeated measures anova

C) two-way anova

D) two-way repeated-measures anova

E) two-way mixed anova

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There are pluses and minuses to residual vs cell-wise checks for normality. On the plus side for residuals, there is more data, so the character of the error distribution is estimated better. On the minus side, the test is confounded with heteroscedasticity, so even when the cell-wise distributions are precisely normal, the residuals will indicate non-normality.

On the plus side for cell-wise checking, the assumption actually refers to normality within cells, so it is a direct assessment, and will be excellent with large within-cell sample sizes. On the minus side, small within-cell sample sizes render such assessments nearly useless, and there is also the practical problem of having to perform and collate results of many assessments (one for each cell).

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  • $\begingroup$ That's a nice and straightforward answer. Thank you, if nobody comes up with a more elegant/elaborate answer, I will accept this answer. $\endgroup$
    – Mr Pi
    Oct 3, 2023 at 13:19

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