jamovi

I would agree that, provided we can assume the variances to be the same, then the most appropriate approach is to check for normality in the residuals from all groups combined, after fitting the model (not in the original variable, where differences in group mean would make the distribution look les...

That's good to know - I'll try not to make any promises you can't keep.
Thanks for all of your hard work.

Hi jamovi developers I'm preparing a conference presentation to introduce members of my university to jamovi and I wondered if there are any major new features in the pipeline that I could mention. In particularly, any new data management features or anything that enhances the customisability of the...

Hi. jamovi's chi square contingency analyses aren't limited to 2 by 2. 3 by 4 works just fine. That was my first thought, but the original query was specifically about the "comparative measures" output (odds ratios, etc.) that are only available for 2x2 tables. I couldn't see any other wa...

You could try using Logistic regression, which will produce odds-ratios and other related results. For a 3x4, you would need to select N-outcomes (Multinomial), and put one variable (probably the 3-level) as the dependent and the other as a factor. Interpretation of the results will be tricky though...

Please could I request the addition of a mosaic plot (using mosaicplot(table(data)) or spineplot()) to the Frequencies>Contingency Tables analysis. I think it should be a fairly easy one to implement.

Thanks

Matt

Correction
The formulae for the two curves mentioned in my post should read:
y=a+c/(1+exp(-b*(x-m))) and
y=a+d*stim+c/(1+exp(-b*(x-m)))

The suggestions from @reason180 are both good. Using a two-way anova with interaction (response~stimulus+orientation+stimulus:orientation - assuming stimulus and orientation are both factors), the main effect for stimulus would provide a test for a difference in mean for the two level of stimulus, w...