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

Question

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. This means that the median in general is also unsuitable.

It may be assumed that each invidual measurement has normal distribution for the instrumental error of the measurement. Additional sources of error are impurities in methanol. The most common impurity is water which increases the refractive index, but authors usually dry methanol before measurement. Probably this is the reason why the density is higher for low values: other impurities such as some common organic solvents (benzene) and (possibly) diluted gases (methane, carbon oxide, carbon dioxide) lower the refractive index. From the other hand alcohols, aldehydes and ketones which may present in industrial methanol and some organic solvents (CCl4) increase the refractive index. So there at least two sources of error: one (instrumental) may be assumed normally distributed and even equal for the all measurements and the second (impurities) is probably asymmetrical (and may be even bimodal) and depends on how methanol was obtained and purified.

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.

Answer 1

What you're basically saying is that there are systematic errors in the measurements whose distributions you don't know. In that case, the answer to your question is that no amount of statistical analysis will allow you to estimate unknown systematic errors.

Answer 2

You appear to think that medians and means don't make sense unless you are certain that the underlying distribution is symmetric. That attitude would leave you without tools to describe most of the data that arise in science. Means and medians are what they are and are often cited for asymmetric distributions. (It can even be the case that mean and median are very close for such asymmetric distributions, as witness many Poisson distributions.)

We can say something about your example dataset.

There is one apparent outlier here at 1.335; your histogram omits it.

For 34 values some plot of the raw data is advisable as well as a histogram, especially as your bins are apparently subjectively chosen.

A very simple descriptive technique is trimmed means, i.e. we trim (or set aside) so many values in each tail before taking a mean. Once the outlier is trimmed, any summary between 1.3268 and 1.3266 is defensible. I conjecture that just about any fancier summary will give results around there. With a small dataset, it is easy to show all possible trimmed means. The table shows number trimmed in each tail, # used and the resulting trimmed mean; it thus spans the mean and median at either end.

  +----------------------------+
  | number    #   trimmed mean |
  |----------------------------|
  |      0   34         1.3270 |
  |      1   32         1.3268 |
  |      2   30         1.3268 |
  |      3   28         1.3268 |
  |      4   26         1.3267 |
  |      5   24         1.3267 |
  |      6   22         1.3267 |
  |      7   20         1.3267 |
  |      8   18         1.3266 |
  |      9   16         1.3266 |
  |     10   14         1.3266 |
  |     11   12         1.3266 |
  |     12   10         1.3266 |
  |     13    8         1.3266 |
  |     14    6         1.3266 |
  |     15    4         1.3266 |
  |     16    2         1.3266 |
  +----------------------------+

Nothing can be done about systematic errors without information on what they are.