jamovi

Is the problem simply that, for at least one pair of the y columns, the difference scores are all the same, so the variance is zero, and ANOVAs don't work when zero becomes a denominator of an equation? See below.

This repeated-measures ANOVA (2 by 3 by 6, with eight participants) does work, and includes sphericity tests (jamovi 2.5.6 for Windows).

Are there just a few <min and >max values? If so, then what you're seeing as a problem (with this pseudo-Winsorization I've proposed) may be a benefit in that it prevents outliers from distorting your results. Alternatively, if there are lots of <min and >max values, then you really do have a proble...

To me, a straightforward and convincing approach would be to convert each instance of "max" to equal the value of the largest non-max value in your data set. Likewise, convert each "min" the the value of the smallest non-min value you measured. (This approach is very similar [tho...

The edit tab is for formatting any textual annotations you might add to the analysis output. It's not for formatting the spreadsheet. You can't format the styling of the spreadsheet cells.

If you're willing to constrain the statistical question a little: Instead of asking whether "the rank ordering between the two groups is significantly different," you could ask the following four questions. Is there an effect of group on the ranked value of Animal 1? Is there an effect of ...

Thanks! (Also, I meant to write: "why wouldN'T the denominator for Cohen's d suffer from the same complication/ambiguity")

I wonder, why would the denominator for Cohen's d suffer from the same complication/ambiguity that other LMM effect-size measures suffer from? Why would the denominator for Cohen's d need to somehow account for variances explained by random coefficients?

How about this approach? https://cran.r-project.org/web/packages ... tics.html

I pretty sure you can't do that in jamovi (unless you want to use the Rj interface to R and do the permutation test in R).