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I want to calculate unit changes (or percentile changes) from an intervention for a continuous outcome variable where I have the pre and post test means and standard deviations. From that I can calculate a Cohen's d, but I want to use the result for a CBA and therefore have to transform Cohen's d to unit changes. I am looking at symptoms for different mental disorders, and not specifically diagnosis/non-diagnosis.

An idea is to use a cut-off on the normal distribution, for instance 90th percentile, which then determines diagnosis/non-diagnosis. Is it a sound approach to (from a z-score table) start at the 90th percentile, see where I end up in the distribution after the given effect size (Cohen's d), and thereby find the percentile change from the intervention for the 90th percentile of the sample population?

If yes, can I then turn that percentile into percentages by dividing the percentile change by .5 and multiply it by 100?

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This seems to be similar to making three right turns instead of a left turn. Cohen's d gets in the way of understanding and assumes that the standard deviation is a relevant normalization factor and dispersion measure. SD is not appropriate for asymmetric distributions in my view.

I suggest you stick to actual data units for the entire analysis, also avoiding percentiling which can be manipulated by changes in your sampling scheme and which does not correspond to physical units either.

I do not know what CBA is.

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  • $\begingroup$ thanks @FrankHarrell. CBA = cost benefit analysis, so I need to be able to transform cohen's d into units in some way to be able to calculate the benefits from an intervention. I am measuring the variable on a scale, and only know the mean values that people have scored on that scale. And i thought that if i used some sort of cut-off, I can approximate how many people will be below/above that cut-off before and after a certain intervention. But I don't seem to find and sound way of transforming cohen's d into units and therefore don't know how to incorporate the answer to a CBA. $\endgroup$ – cam9876 Nov 23 '15 at 8:06
  • $\begingroup$ That seems to be assumption laden, and begs the question of why one would ever use Cohen's d in the first place. You need access to measurements on meaningful absolute scales. $\endgroup$ – Frank Harrell Nov 23 '15 at 13:54

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