# Arcsin CI for proportion

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.

• Show some data please. Commented Feb 9, 2015 at 16:40
• 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. Commented Feb 9, 2015 at 17:20
• You don't show us your exact calculations. Commented Feb 9, 2015 at 17:20
• 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! Commented Feb 9, 2015 at 17:23
• So are these durations? That isn't really a proportion, even if you are going to censor at 20 minutes. Commented Feb 9, 2015 at 21:36