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First I'll start out with some background information. I conducted image analysis on 200 images to identify the area % of 7 different phases. I then averaged the results from each of for the 200 images to determine the average area % for each phase. For example phase 1 the average area is 20.1%.

Typically, I would like to report average results from a test with the mean +- 1.96 x SE, if it is normally distributed. Therefore, I conducted Anderson Darling tests on the area % data to determine if the data followed a normal, exponential, weibull, or gamma distribution, or could be transformed into one of those. However, all aforementioned distributions failed the Anderson Darling test with p-values that exceed 0.10.

Am I likely failing the distribution identification tests, because my average area % is a discrete variable and is not continuous (aka it only ranges from 0 to 100%)?

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  • $\begingroup$ So for each of the 200 images, you compute area % for 7 phases. And for each of the 7 phases, you compute an average over the 200 images? $\endgroup$ – DatamineR Jan 20 '16 at 18:29
  • $\begingroup$ That is correct @DatamineR. $\endgroup$ – Joshua Wright Jan 20 '16 at 21:55
  • $\begingroup$ 1. that a variable is on a bounded range doesn't make it discrete (it may be discrete but being on 0-100% isn't why -- continuous proportions are still bounded). 2. Likely the actual distribution won't be any common distribution, but I'd probably start with a beta model for one proportion (and Dirichlet for a larger set of proportions that add to one). 3. Failure to reject normality in a distributional test doesn't tell you that you have normality. 4. Why do you care whether the data were drawn from some particular distribution or some other distribution? $\endgroup$ – Glen_b Jan 20 '16 at 22:39
  • $\begingroup$ @Glen_b so that I can report a mean +- an error, aka for normal distribution mean +- 1.96 x SE (e.g. for Phase 1 20% +- 1%). I would like to give readers and idea of how accurate my image analysis technique is. $\endgroup$ – Joshua Wright Jan 21 '16 at 13:32
  • $\begingroup$ This aim should go in your question; it's important information that will (or rather, should) affect the answers you get. $\endgroup$ – Glen_b Jan 21 '16 at 16:41
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The answer depends on your data. If the standard deviation is very small compared to the distance from the mean to 0 and 100%, then you have a chance. Otherwise, it won't work.

With bounded data you might have a better luck with Beta or similar bounded distributions

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Normal distribution is probably not a good option in this case. The problem is not continuity. The % area cannot be lower than zero, and larger than 100%, while normally distributed variable is unbounded.

It is true that normal distribution may be reasonably used to model e.g. human height, but the standard deviation (few inches) is so large compared to the mean (few dozen inches) that the likelihood height, if it was really distributed normally, would be zero. The same may not apply to the % area.

Other issue is that most of the 7 distributions should be assymetric, with the mass concentrated closer to 0% than to 100%, but at the same time allowing for some images, the % area may be quite large.

I think that the best thing for you to do is to have a look at the histogram of the 7 areas. If you see there is a substantial number of images with areas close to 0% or 100%, but you otherwise see the middle parts of the "bell curve", you may try truncated normal. If the number of images with area close to 0 or 100% is very low and the distribution does not look symetric around its mean, you may look into log-normal distribution.

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