I've described a typical design for my experiments in this question. Well, 1-way RM ANOVA assumes a Gaussian distributed vector.
I try $y=\arcsin{\sqrt{x}}$. But for some data it works, for some doesn't... So the next step is to find the most powerful/general statistics. Freeman — Tukey's transformation with $y=\sqrt{x}+\sqrt{x+1}$ is the most powerful and seems to be appropriate in almost all my cases but again it doesn't give normal data sometimes.
What should I do this case? Box — Cox transformation?
Is it OK to transform data using different statistics within a paper?
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1$\begingroup$ Do you mean the arcsin(square root(x))? If not, that could be why it didn't work... $\endgroup$– Aaron - mostly inactiveCommented Jun 21, 2011 at 1:57
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1$\begingroup$ What do you mean by "But for some data it doesn't work"? In what sense did that transformation not work? Also, what do you mean by "the most powerful statistics"? Powerful in what sense? $\endgroup$– WolfgangCommented Oct 27, 2011 at 1:31
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$\begingroup$ @Wolfgang: thanks for your questions. Let say I have 2 cell types. For cell.type_1 the FT works i.e. it transforms to normality, for the other cell type it doesn't. I meant "powerful" as synonym for "powerful" (i.e. capable to transform "any" to normality) (Correct me please if I'm wrong?). Here I was recommended to use $\beta$ distribution but I can't understand what do I have to do in this case... :( $\endgroup$– abcCommented Oct 27, 2011 at 3:23
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$\begingroup$ Okay - that clarifies things a bit. But more questions: How are you determining whether a particular set of data are normally distributed? Also, since you mention the 1-way RM ANOVA: Are you looking at the data of each individual person or are you looking at the data for each separate condition/time-point? Finally, according to the other question you linked to, are you looking again at percentages? $\endgroup$– WolfgangCommented Oct 28, 2011 at 1:13
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2$\begingroup$ I see. What you really should be checking is the normality of the residuals after fitting the model, not the normality of the dependent variable itself. See, for example this question, which is quite related. $\endgroup$– WolfgangCommented Oct 31, 2011 at 6:33
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