# How to estimate true value and 95% bands when distribution is asymmetrical?

I have a set of results of independent measurements of some physical quantity. As an example I give here real expermental data on methanol refractive index at 25 degrees Celsius published in scientific literature from 1960 to 2011:

data={{1960, 1.32652}, {1961, 1.32662}, {1963, 1.32650}, {1963,
1.32750}, {1968, 1.32698}, {1968, 1.32890}, {1970, 1.32657}, {1970,
1.32660}, {1971, 1.3260}, {1971, 1.32610}, {1971, 1.32630}, {1971,
1.3350}, {1972, 1.32640}, {1972, 1.32661}, {1973, 1.32860}, {1975,
1.32515}, {1975, 1.32641}, {1976, 1.32663}, {1977, 1.32670}, {1980,
1.3250}, {1983, 1.32850}, {1985, 1.32653}, {1991, 1.32710}, {1995,
1.32621}, {1995, 1.32676}, {1996, 1.32601}, {1996, 1.32645}, {1996,
1.32715}, {1998, 1.32820}, {1999, 1.32730}, {1999, 1.32780}, {2001,
1.32634}, {2006, 1.32620}, {2011, 1.32667}};


The first number is the year of first publication, the second is the published value. The authors do not always provide their own estimate for error of the published value and even when they do this, the published estimate is often not accurate.

It is easy to see that the distribution is asymmetrical (other datasets I work with are even more asymmetrical). Here is a density histogram with bins selected by hands:

bins = {1.325, 1.3259, 1.3261, 1.3263, 1.3265, 1.3266, 1.3267, 1.3269,
1.3275, 1.328};
Histogram[dataFirst[[All, 2]], {bins}, "PDF"] Because of asymmetric feature of the distribution I cannot use the mean as an estimate for the true value of methanol refractive index.

The fact is that the probability to get smaller value than the true value in general is not equal to the probability to get higher value (the reasons for this are subtle and I will not explain them here). This means that the median is also unsuitable.

It may be assumed that each invidual measurement has normal distribution.

What is the best way to estimate the true value and its 95% confidence bands for the measured quantity in such cases?

P.S. References for relevant scientific papers will be appreciated.

P.S.2. The above code is for Mathematica system.

## 1 Answer

Sounds like a good situation for bootstrapping. You would still have to settle on a statistic, though, as I don't totally follow your logic behind ruling out both mean and median as measure of central tendency.

EDIT: See various answers to this question: Explaining to laypeople why bootstrapping works

• What do you mean by bootstrapping? Could you show an example? – Alexey Popkov May 7 '13 at 12:33
• Bootstrapping is a family of procedures that resample your data to estimate a parameter. For example, if you want a confidence interval for the median of your data (sample size n), you might take 5000 n-size samples (with replacement) from your data, calculate the median on each sample to create a distribution, and then calculate the 2.5 and 97.5 percentiles on that distribution to obtain the confidence interval for the median of your original sample. This, as I said in my answer, requires that you settle on a statistic. – Thomas May 7 '13 at 13:15
• Interesting. Could you provide a link where these procedures are explained in detail? – Alexey Popkov May 7 '13 at 14:06
• See link in edit. – Thomas May 7 '13 at 14:53
• Here I read: "Broadly speaking, the purpose of the bootstrap is to construct an approximate sampling distribution for the statistic of interest. It's not about actual estimation of the parameter." Is this what you mean when say that bootstrapping "requires that I settle on a statistic"? – Alexey Popkov May 7 '13 at 17:07