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Let's see if I understand this correctly. The normality assumption means that for each group I am testing the response within each group is normally distributed. So in order to check all the groups together would I calculate the residuals and then try to fit the combination of all residuals to a normal distribution?

Thanks

Edit: If normality in this case means tha: The responses in group A should be normally distributed with respect to group A. The responses in group B should be normally distributed with respect to group B. etc. Is the way of checking this to individually check each group? Or is it valid to calculate the residual of each group, for each group the mean value will be different. And then check whether the residuals are normally distributed in one go?

I'm concerned than doing it in one go could lead to the results showing normally distributed when within the group they're not normally distributed.

Thanks

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The answer to your question is discussed more completely at this Cross Validated thread.

But to answer your question more compactly, if you think about anova as being conducted in a general linear model framework, the model you are working with is something like $Y_{ij}=\mu_{j}+\sigma\epsilon_{ij}$. Here, an assumption to make the inferences valid is that the $\epsilon_{ij}$ piece, which is estimated by the residuals from the fitted model, has a normal distribution.

Practically speaking, the residuals from the model can be examined to see if they are reasonably normal, perhaps with a histogram or quantile-quantile plot.

There's no need to think about the residuals from the individual groups. For a one-way anova, you could calculate the difference between the mean and the values for each group, and then aggregate them. But that would be the same as the residuals of the linear model. And this method would get complicated for complicated designs.

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  • $\begingroup$ You could still look for residuals within groups to investigate the constancy of variance. $\endgroup$ – kjetil b halvorsen Nov 24 '19 at 17:34

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