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Please consider these data points as an example of many more, they represent single studies with multiple samples which have been done during the last decades.

7.7 +/-3.3) mg/L, n=10
3.6 +/-0.2 mg/L, n=22

The publication states that 7.7 resp. 3.6 represents mean OR median, without specifying which has been used where, and there is no way to find out.

Is there a way to get something like an average out of those data points, or is too much crucial information missing?

EDIT: This question should be deleted. I realized I misread the data in question: values with a standard deviation are means, values without are median. Thank you for your help, and sorry for bothering you.

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  • $\begingroup$ It is probably a rather sloppy way to indicate that mean and median coincide up to the stated precision. $\endgroup$
    – Michael M
    Commented Jul 27, 2015 at 7:45
  • $\begingroup$ I'm not sure that it is necessarily true. They are both measures of central tendency, but mean is moved by skew. $\endgroup$
    – EngrStudent
    Commented Jul 27, 2015 at 9:57
  • $\begingroup$ @engrstudent: Not in general of course (sorry for not being precise), but in this particular data example. $\endgroup$
    – Michael M
    Commented Jul 27, 2015 at 11:43
  • $\begingroup$ MichaelM, I think it's sloppy, too. The entire publication refers to "mean/median" without any further explanation. Perhaps I should discard the data for being too unclear. $\endgroup$
    – poshtad
    Commented Jul 27, 2015 at 12:22
  • $\begingroup$ I apologize for being a sloppy reader, see my edit. $\endgroup$
    – poshtad
    Commented Jul 27, 2015 at 14:28

1 Answer 1

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The answer depends on nature of your data. If you can assume that your data comes from an unimodal and symmetric distribution, then mean should be close to median, otherwise they could differ (see here for reading more on relations between mean and median). Below you can find a quick example:

x <- rexp(1000) # Exponential distribution (skewed)
mean(x)
## [1] 0.9910864
median(x)
## [1] 0.6553773

y <- rnorm(1000) # Normal distribution (symmetric)
mean(y)
## [1] -0.03853657
median(y)
## [1] -0.05458903

enter image description here

enter image description here

As I wrote in answer to your previous question, making some assumptions about distribution of your data would make it possible to make some educated guesses about your data (e.g. about mean given only a range, but it would be also possible to infer about mean given only a median using similar approach).

Also notice that if your sample sizes are small (n=10, n=22 as in your example), then you can expect the differences to be even bigger since a single outlier can possibly influence the mean.

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  • $\begingroup$ bootstrap resampling might provide a way to look at "what if" without knowing. $\endgroup$
    – EngrStudent
    Commented Jul 27, 2015 at 9:58
  • $\begingroup$ @EngrStudent but with single data point bootstrap won't help much ;) $\endgroup$
    – Tim
    Commented Jul 27, 2015 at 10:05
  • $\begingroup$ I was thinking in terms of Pearsons mode-median-mean expression. Maybe we can find a version that is a decent way to combine indicators of central tendency. We don't know the nature of the distribution or the label applied to the statistic. There is a tolerance and a sample size. Maybe we could find cases of GMM that result in mean and median or mean&median given. $\endgroup$
    – EngrStudent
    Commented Jul 27, 2015 at 11:19
  • $\begingroup$ Dear Tim, I finally commented your answer to my previous question - thank you very much for your work. I am uncertain how symmetric the data is, it's plasmatic blood levels of chemical compounds under certain medical conditions. From a pure biochemical point of view, I would say that distribution should be along a gaussian curve, but there are so many other, unknown factors, such as age or medical preconditions. The goal of my question is to actually answer the question "How high is too high?". $\endgroup$
    – poshtad
    Commented Jul 27, 2015 at 12:18
  • $\begingroup$ Continued: That number/average doesn't have to be precise, just an attempt at a best-effort approach with the poor data available. The link to Karl Pearsons approximation looks promising, do you think I could use this? $\endgroup$
    – poshtad
    Commented Jul 27, 2015 at 12:19

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