The important thing for a 3way ANOVA to be valid is residuals being normal and homoscedastic, not the data. If your residuals satisfy these assumptions you do not need to worry.

It is sad that you did not explain at all what your response variable is, so it is almost impossible to give you any specific/gelpful advise. It is much much more important to know what kind of variable the response is, and it is almost irrelevant to know the information you provided in so great detail (the three experimental factors and their levels).

Just to give some possible directions:

One remark at the start: not the response values need to be normal distributed but the conditional distribution of the response must be (approximately!) normal. "Conditional" means conditional on the respective value of the linear predictor function. Since here -in the normal probability model- the mean and the variance are independent, this conditional distribution can be expressed as the distribution of the residuals (difference between the measured and the predicted value), which then should approximate N(0,s²). 

If your response is some kind of ...

...count value, a normal probability model is inappropriate and should be substituted by a Poisson or a negative binomial model.

... concentration, a log-normal or a gamma model might be appropriate.

... proportions based on fixed and known numbers of observations, a binomial ("logistic") model would be in place.

... proportions in (0,1) a beta model could be suited.

And if you have a rather large amount of data you can always bootstrap the sampling distribution of whatever statistic you think is relevant.

Another possible "soulution" (I would not favour, though) is to find some transformation of the response that will give you approximately normal distributed residuals. This could be a log- sqrt- inverse- logit- power- or some other transformation. The Box-Cox transformation will find a transformation from the log/power family that will result in a residual distribution with the best possible approximation to the normal distribution (that is possible within this family of transformations).

Thanks for all the useful suggestions.

My dependent variable is continous and ranges from 0 to virtually infinite.

I have about 1000 counts per condition (approx 8000 counts overall)

Thanks again

Stefano

Hello Stephano,

How did you see that your dependent variable seems to be not normally distributed ?

In addition of previous answers concerning assumptions for ANOVA, I let you consult the link below which will give you (I hope) interesting information.

Moreover, concerning your initial question, you can try to perform a data transformation (log, square root etc.) and see if your data "look more normal" (as suggested by Jochen).

You can also maybe try the ANOVA with the value rank... (looking at the number of repetition you have). But you should check ANOVA assumption first. You will loose power (for sure), but it may give interesting results...

code : anova(lm(rank(response)~factor1*factor2*factor3,data=your_table)) with R

Regards

http://www.statmethods.net/stats/anovaAssumptions.html

Hi Cyril,

I know that my dep var is not normally distributed because normality test failed in all my groups.

Moreover, it was expected as the variable I measured reflects a particular response (many cells present low value response, whereas few cells exhibit high value response).

Therefore, I just have to find the correct way to analyze my datasets and try to reveal main effects and interactions.

Thanks again

Stefano

Hi Stephano,

When you said "normality test failed in all my groups" there is two things I want to underline :

"(...) in all my groups"

First of all, be sure that you check normality of the residual of your model. This is the assumption requested concerning normality for linear model (and therefore for ANOVA).

"normality test failed (...)"

Then, let me recommend you two topics very close to yours :

• https://www.researchgate.net/post/Why_the_assumption_of_normality_of_residuals_ANOVA_is_still_violated_after_the_log_transformation?_tpcectx=qa_overview_following&_trid=39sElIn0D1ExawsCck9QJkaT_
• https://www.researchgate.net/post/Homogeneity_of_variance_is_one_of_assumption_for_Repeated_Measures_ANOVA_How_to_test_it_Is_the_Mauchlys_is_enough_or_should_I_use_Boxs_test_too

It is well explained that, sometimes (well... most of the time !), it is better to have a look at data (by doing as many graphs as you can) than testing them... Specially with the number of count you have.

Finally, you said "Therefore, I just have to find the correct way to analyze my datasets and try to reveal main effects and interactions."

Maybe not, I highly recommend you to follow the instruction I gave you in the previous link (even if you are not working with R, I think you can find easily the same commands in your own stat software) and have look at your data, residual etc. Maybe, you won't need any transformation otherwise you'll see better what data transformation you'll need.

Regards

Ok, Cyril. I'll try to do what you all suggested. Maybe I am complicating what is not so complex.

Thanks you all, again

Stefano

Stefano I guess so.

Dear Stefano,

I think with your sample size you do not have to worry about violation of the assumption of normal distiribution because the central limit theorem applies here.

regards

Thanks a lot, Claas. 

Stefano

I disagree. What Claas says is irrelevant.

As I understood  the relevance of the central limit theorem the greater the sample size, the greater the approximation of the distribution of the respective sample statistic to a normal distribution. So I certainly did not want to say that a test on normal distribution of the residuals would be wrong but most probably unnecessary when having 1000 cases per condition. A parametric ANOVA should lead to robust and correct inferences in any case with samples of this magnitude.

https://en.wikipedia.org/wiki/Central_limit_theorem

- I do not see why this does not apply here.

regards

Claas

Hello Claas,

At the question : "When checking normality of residual prior linear model is unnecessary ?" The answer is never !

Please find good literature on that point here :

https://www.researchgate.net/post/When_can_we_consider_ANOVA_robust_enough_in_front_of_normality_and_homoscedasticity_rejection

Regards.

@ Claas: somehow you are right and somehow not. For example Rand Wilcox shows very illustrative in his books, that the central limit theorem does not always hold, especially if the distribution in the population is very heavy tailed. So what is a reasonable sample size to rely on the central limit theorem, 20, 30 or 100 samples? He showed that even with larger sample sizes it does not always hold.

Dear Rainer,

thanks for the literature recommendation (and for the assurance that I have not been totally on the wrong track here). This discussion is a good reminder (at least is has been for me) to not forget to always be skeptical about recommendations of basic statistical textbooks.

In the last days I had fruitful conversations with my collegues about possibilities to plan a final data analysis after (and not before) a look at the data (to graphically check for normality) without letting in too much "researcher degrees of freedom". 

Regards.

Dear Stefano,

I've just been reading a paper about the robustness of 2-way ANOVA to violations of its assumptions. Apparently, non-normality is not as much of a problem as unequal variances (even with equal group sizes). The paper is:

Harwell, M.R., Rubinstein, E.N., Hayes, W.S. and Olds, C.C., 1992. Summarizing Monte Carlo results in methodological research: The one-and two-factor fixed effects ANOVA cases. Journal of Educational and Behavioral Statistics, 17(4), pp. 315-339.

I'm not sure whether this translates to 3-way ANOVA.

Thanks again to all those who contributed to this thread.

I am collecting lots of information to have elements to take a decision.

Best regards

Stefano

Notice that the DV runs from 0 to N. Normal distributions run from -inf to +inf. Thus the DV and hence the residuals can't be approx. normal. This leads us back to Prof Wilhelm's answer. Best, DB