I have several sets of data, unfortunately the data comes to me in a "summary" form. My job is to consolidate the several data sources into one general summary. I'm currently using the median to summarise the data, but I don't know if this is statistically sound. Here's a description of my problem:

There are $N_P$ samples, each with varying sample sizes, but all from a single population. Neither the sample size or the standard variation are known. Each sample can be divided into $N_Q$ disjoint groups (or qualities). From each sample, the only data that is known is what percent of the sample falls within a group (or category). For example, population $A$ contains, $x\%$ of $a$, $y\%$ of $b$ and $z\%$ of $c$.

The different samples are not disjoint, so a single item might be in several of the samples; but I don't know how much overlapping there is. There are 5-8 different samples with 5-7 categories. An example (smaller) table is the following.

            cat. a    cat. b    cat. c    
sample A    47.34%    30.05%    11.92%
sample B    41.60%    29.90%    11.90%
sample c    47.74%    29.67%    12.69%
--------    ------    ------    ------
median      47.34%    29.90%    11.92%

Now is it statistically sound to create this "median" summary, which takes each group from the different samples and finds the median? Maybe I should be using the mean? The problem I'm seeing is the "median sample" usually sums to less than 100%, even though the percentages from each sample sum to 100%. Should this matter?

Sample sizes: 100k - 100m
Population size: ~1 billion
  • $\begingroup$ do you know the sample sizes... the range of them? Small sample sizes wreak havoc on the mean... moreso than the median. $\endgroup$
    – John
    Dec 6 '10 at 0:40
  • $\begingroup$ the sample sizes range from 1-100 million and the population size is upwards to 1 billion. But I have no knowledge of how disjoint the samples are from the total population. $\endgroup$
    – Justin
    Dec 6 '10 at 3:33

What you are doing does not makes sense if your goal is to categorize what proportion of the entire population (sample A + sample B + sample C) is in category a, b, and c. Consider the following contingency table:

   a  b  c             a    b    c
A  8; 1; 1         A  .8;  .1;  .1
B  7; 2; 1         B  .7;  .2;  .1
C  1; 13; 16       C  .03; .43; .53

Then, for example, the median of the category a probabilities is 0.7 and the mean is 0.51, but only 16/50 = 0.32 of the all the observations are in column a. Likewise, the median of the category c probabilities would be 0.1, but only 0.36 of the observations are in column c. Does the "median summary" you propose tell you anything meaningful in a situation such as this one? Unless you have the marginal counts of either the samples or the categories, or you are willing to make some assumptions about them, I don't think there is a whole lot you can do in this case.

Do you have any specific goals in mind? Also, how many categories and samples do you have?

Edit: Your sample/population phrasing is slightly confusing. It's better to say you "have 3 samples, each which be sub-divided into 3 categories a,b, and c." The phrase "sample population" is troublesome, as is your reference to two different "populations."

  • $\begingroup$ ok, what I'm really interested is if it's statistically sound to include the median row in the summary table at en.wikipedia.org/wiki/Usage_share_of_web_browsers I can go into more details if you want. $\endgroup$
    – Justin
    Dec 6 '10 at 4:52
  • $\begingroup$ In that case you should be okay. You can consider each of the samples A,B,C to be experimental units, and each of the categories a,b,c to be some measurement taken on each sample (with the restriction the measurements sum to 1.) This could be helpful when you have a lot of samples, not so much when you have 3 or 5. $\endgroup$
    – HairyBeast
    Dec 6 '10 at 5:40
  • $\begingroup$ It's amusing that that site calculates the median (and goes to the trouble of labeling it as such and suggesting that they know the difference between the mean and the median) on just two points of data which results, actually, in the mean of the two values! $\endgroup$
    – Brett
    Dec 6 '10 at 23:33
  • $\begingroup$ The table with 2 data sources, is one of the things I'm trying to correct. Some of the editors of the page are handling the data in a way that's statistically unsound, or at least I'm trying to prove that. $\endgroup$
    – Justin
    Dec 8 '10 at 3:10

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