I have measured Gaussian curvature data of 3D objects from two different groups, A and B. I would like to find out whether the objects differ in curvature.

The distribution of data values for each subject looks like an extreme value distribution:

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(This is expected since Gaussian curvature is computed as the product of the 2 principal curvature k1 and k2, which are the maximum and the minimum over all possible curvature lines.)

If the data were normally distributed, I would take the mean of the data for each subject and fit a linear model to the data, then look at the main effect of the group variable.

However, taking the mean of such a distribution does not seem to make much sense. What would be a suitable approach here?

Note that one could model positive and negative values separately, I guess. Would it be valid to apply a Box-Cox transformation (to the positive and negative parts separately) and then check for mean differences of the resulting distributions? (I somehow doubt it.) Or should I use a GAM?

  • $\begingroup$ You can always do Mann-whitney or kolmogorov-smirnov tests. But you say that the mean of the distribution does not make sense, is it because you know that the mean does not exist? And what would it help to treat positive and negative parts? $\endgroup$ – svendvn Apr 29 at 20:14
  • $\begingroup$ Could you pls explain looks like an extreme value distribution? $\endgroup$ – Yves Apr 30 at 6:28
  • $\begingroup$ @svendvn: I felt that looking at the mean does not make much sense because it is very close to zero, and that is the case because 1) the surface has lots of points with very low curvature, but also because 2) it has positive and negative curvature values that cancel each other out. The latter is also the reason why I thought it might be beneficial to look at positive and negative curvature separately. $\endgroup$ – John Silver May 27 at 13:38
  • $\begingroup$ @Yves: I took the positive part of the data and fitted various distributions to it in GNU R. The ones that matched best were extreme value distributions (I think Weibull had the best AIC). $\endgroup$ – John Silver May 27 at 13:48

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