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I'm evaluating two different techniques of identifying the representation of the motor network within the brain. Both techniques identify the center of gravity of the motor network in MRI coordinates based on physiological output.

So what I have is 8 datasets of coordinates for both techniques. The 2 techniques clearly identify different centers of gravity and I would like to quantify this. However, it is hard to determine whether the coordinates/distances are normally distributed due to the limited sample size.

I am currently running a t-test on the x coordinate (which is most relevant) and a one-sided t test on the distance (which I'm not sure is accurate). Any advise would be very helpful.

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  • $\begingroup$ Why would you expect that the coordinates and distances would be normally distributed? Why would you care? Why look at one coordinate at a time when you have spatial data? You need to use a method that uses all three coordinates at once. Distance might do that, but then you would be testing whether the distance is far from 0 and you could use a nonparametric test. But maybe some sort of kernel smoother might be good. Analysis of MRI data is a whole subspecialty. I used to know a bit about it, but it's been 10+ years since I did this. You might want to consult a specialist. $\endgroup$ – Peter Flom Feb 23 '18 at 13:18
  • $\begingroup$ I'm experienced with analysis of fMRI data. That is not the issue. So I figured I'd do statistics on the distance like you suggested, but what nonparametric test would you suggest? $\endgroup$ – Jord Feb 23 '18 at 13:50
  • $\begingroup$ Normality of the data is largely immaterial, as @Peter indicates. However, what does matter is normality of the sampling distribution of the statistics, such as the center of gravity. That very well could be close enough to Normal to make a (multivariate) T test applicable. Testing the distances is a little delicate because you are in an ANOVA kind of situation with 8 datasets to compare (for 28 total comparisons), but the distances will have chi-squared distributions rather than Normal distributions. This is the main issue. (+1) $\endgroup$ – whuber Feb 23 '18 at 15:26

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