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I'm trying to replicate a study. I have 165 observations of a proportion variable (the distribution is skewed to the right as many proportions are low). I need to report the CI. I've applied arcsin(sqrt), computed the mean and SE of the transformed proportions, got the CI, and transformed back to the original scale.

Now, the thing is, that the total proportion of the sample does not fall into that interval. (I feel it has to do with the lower bound of the binomial, that's why the study is stretching those values by arcsin(sqr)).

But how should I properly state this result?

The following article comments on this issue:

Comment: The logarithm is the only non-linear transformation that produces results that can be cleanly expressed in terms of the original data. Other transformations, such as the square root, are sometimes used, but it is difficult to restate their results in terms of the original data.

the raw data for one of the variables:

Eser 0 0 0 0 2 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 10 0 1 0 0 1 0 0 0 0 0 0 0 1 1 0 6 0 0 0 0 0 0 10 0 1 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 9 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 9 0 0 0 1 0 1 0 0 0 6 0 2 0 0 1 0 4 0 1 2 0 0 1 0 0 10 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 6 1 0 0 0 10 10 3 1 2 1 0 0 0 0 0 0 0 0 0 2 0 0 0 0 3 0

It's 165 proportions (between 0 and 1) computed as those counts out of n=20 for each case.

I want to compute CI for that proportion for whole sample.

I tried comparing sample variance, skewness and kurtosis to that of the theoretical binomial with the same p. My data seem to be somewhat underdispersed and the transformation did not normalize it anyway. So what would you suggest? I need to get some kind of confidence interval for the mean proportion.

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  • $\begingroup$ Show some data please. $\endgroup$ – Nick Cox Feb 9 '15 at 16:40
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    $\begingroup$ Here 127/165 points are exact zeros. No transformation can make such a sample non-spikey. asin(sqrt()) doesn't make it much better behaved. I would want to know about the data generation process. $\endgroup$ – Nick Cox Feb 9 '15 at 17:20
  • $\begingroup$ You don't show us your exact calculations. $\endgroup$ – Nick Cox Feb 9 '15 at 17:20
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    $\begingroup$ I don't understand Jerry Dallal's comment on the site you cite. So long as a transformation is invertible, you can always show results on the original scale. There are many special features of logarithms, but they are not the only invertible transformation; most of the others are! $\endgroup$ – Nick Cox Feb 9 '15 at 17:23
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    $\begingroup$ So are these durations? That isn't really a proportion, even if you are going to censor at 20 minutes. $\endgroup$ – gung - Reinstate Monica Feb 9 '15 at 21:36
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Without seeing all your calculations, it is hard to be sure, but...

I think you are (implicitly anyway) assuming that at each minute interval an event either happens or it doesn't happen, and that the probability of it happening in any interval is constant. If that were true, X/20 would be a proportion that follows a binomial distribution, so your method of computing a confidence interval would make sense.

Assuming you did the calculations correctly, the fact that the confidence interval is obviously wrong suggests to me that your assumption is wrong. Perhaps, the probability of an event happening in any minute interval is not always the same. If the binomial assumption of consistent probability of the event occurring is wrong, the variable is not binomial, so computing a confidence interval based on the binomial distribution wouldn't be helpful.

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  • $\begingroup$ I see the interval does not contain the sample mean because the angle transform stretches the lower values. In the raw data the skew pulls the mean up. Now, what's the point of using transformations in such situations, anyway? The study I'm replicating uses the same transform, but it's not that much skewed as my data. $\endgroup$ – Germaniawerks Feb 9 '15 at 19:35
  • $\begingroup$ @Germaniawerks, I don't quite follow your comment. But I think you need to first make it clear whether the binomial assumption applies to your data. If not, then methods based on it are sure to give misleading or inconsistent results. $\endgroup$ – Harvey Motulsky Feb 9 '15 at 20:51

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