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Sorry, my previous question was not very clear I deleted it and am trying again.

Question: What options do I have for two-factor non-parametric version of ANOVA with repetition? What is the follow-up?

Relevant Details: Friedman's Test doesn't handle repetiton from my understanding, and Agricolae package in R has a method of handling this with a modified Friedman's but is not well suited to my data. I cannot find anything else to utilize against this.

Non-parametric is required due to normality assumption failure caused by small sample size. Perhaps there is a way to correct this? Transformations have failed to correct the normality.

Faugaun
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    In comments on your question that you deleted (your choice, naturally) I made the point that your data are binomial and so suggest a generalized linear model. Tests of normality are thus irrelevant and sample size has nothing to do with the issue. Naturally, again, nothing stops you asking a different question. – Nick Cox Oct 21 '14 at 15:36
  • So something like this will allow me to ignore normality violation? http://stats.stackexchange.com/questions/60352/comparing-levels-of-factors-after-a-glm-in-r – Faugaun Oct 21 '14 at 15:45
  • What I tell you three times is true. If you use models relevant for binomial responses, you are not assuming normality of your data in any sense. In fact you are making different assumptions. (In fact normality is the least important assumption in linear models; I am not to blame for a large literature implying otherwise.) – Nick Cox Oct 21 '14 at 15:50
  • http://en.wikipedia.org/wiki/Generalized_linear_model Reading this it looks like exactly what I need to use. Now to learn binomial glm ... Thanks a Bunch Nick Cox :) and sorry if I'm not very clear :( – Faugaun Oct 21 '14 at 16:02
  • Nick could I more simply just build the CI for each treatment (12 cells) For instance 15 of 20 = 0.75 +- X vs the other groups, CI's that don't overlap are different, maybe use multcomp package tukeyHSD boxplots to create significance groups from the raw data. Then do it for each factor cell also, since 2x6 these would be x of 120 vs y of 120 or x of 40, y of 40 ... – Faugaun Oct 21 '14 at 16:17
  • never mind you can ignore me I am thinking out loud, thanks a million mate you got me thinking about this the correct way now :) – Faugaun Oct 21 '14 at 16:20
  • Sorry, but this is a new question that requires **much more explanation**. You won't get serious, detailed help via comments. A different problem is that you're assuming that I use R and can advise, but I only very occasionally use R, and certainly don't advise on code. Box plots don't work well with discrete data; I made that point on your previous question. – Nick Cox Oct 21 '14 at 16:21
  • disregard mate it's not an r question and I got it figured out :) you are the man! I just was thinking incorrectly on how to approach the data and need to use binomial confidence interval comparisons, simple :) Should have this complete in half an hour now. – Faugaun Oct 21 '14 at 16:28

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