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The Borenstein (2008) textbook I'm using doesn't make any references to estimating a population median. Searching the internet I find lots of references to using the median and range to estimate the population mean, but what if I want to estimate the population median instead?

The only suggestion I've found that addresses this question is in this comment here, where the commenter mentions never having seen a method that allows the estimation of the population median.

You can assume that I could have the Mean, SD, interquartile range, and range, among other statistics from the original studies.

Borenstein, M. (2008). Introduction to metaanalysis.

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    $\begingroup$ Can you assume that the raw data within the original studies have a (roughly) symmetric distribution? Or even a normal distribution? $\endgroup$ – Wolfgang Jul 13 '16 at 8:05
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    $\begingroup$ Yes, I can assume it has a roughly symmetric distribution. In some cases I will be able to test this by looking at the raw data. $\endgroup$ – user1205901 - Reinstate Monica Jul 13 '16 at 8:10
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    $\begingroup$ If the distribution is symmetric, then the mean is equal to the median. So why not meta-analyze the means (which is more efficient), since that will then also give you an estimate of the median? $\endgroup$ – Wolfgang Jul 13 '16 at 9:52
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    $\begingroup$ That is a good idea. I will leave the question as is, though, because it is inevitable that in future I will encounter cases in which I can't assume my datasets are symmetrically distributed. I am still interested to know what to do in those cases. $\endgroup$ – user1205901 - Reinstate Monica Jul 13 '16 at 11:00
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    $\begingroup$ You might want to look at my recent answer to this question stats.stackexchange.com/questions/88813/… which has some recent references and a link to an R package. $\endgroup$ – mdewey Oct 17 '18 at 14:59
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If the distribution is symmetric, then the mean is equal to the median. So then you can meta-analyze the means, since that also gives you an estimate of the median. In addition, it is more efficient.

It is also possible to meta-analyze medians directly. The large-sample variance of a sample median ($m$) from a normal distribution is $$\mbox{Var}[m] = \frac{\pi \sigma^2}{2n}.$$ An estimate of the sampling variance can be obtained by replacing $\sigma^2$ with the observed sample variance. So, given multiple medians and corresponding variances thereof, one can easily proceed with a meta-analysis of these values. But again, it would be more efficient to meta-analyze the means then.

If the raw data did not come from a normal distribution (and especially if the distribution is not symmetric), then things are different. Then meta-analyzing means and medians are really different things. However, if the data are not normally distributed, but follow some other distribution with density function $f(x)$, then the variance equation above is not correct. The more general equation for the large-sample variance is $$\mbox{Var}[m] = \frac{1}{4nf(M)^2},$$ so $f(M)$ is the density at the true median (see, for example, https://en.wikipedia.org/wiki/Median#Sampling_distribution or https://stats.stackexchange.com/a/45143/1934). Note that the density of the normal distribution at the median (= mean) is $1/\sqrt{2 \sigma^2 \pi}$, so just plug that into the equation and you get the variance equation given earlier.

So if you want to meta-analyze medians, you first need to have some kind of idea what density function would be applicable for a given dataset. Then you can compute (or rather: estimate) the sampling variance and then again proceed with a standard meta-analysis.

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In systematic reviews and meta-analysis, standard routines summarize results in terms of sample mean and standard deviation. Several trials report the median with minimum/maximum or the inter quartile range. A practical and sound approach is to estimate the sample mean and standard deviation for such trials. In this paper of Wang and colleagues are reported formulas to implement such approach. The paper has been cited about 500 times according to Google Scholar and I have applied such approach. Furthermore it would be interesting to see the cost benefit ratio of having more imprecisioni with more power.

The median is very similar to the mean when the distribution of the data is symmetrical, and so occasionally can be used directly in meta-analyses.  However, means and medians can be very different from each other if the data are skewed, and medians are often reported because the data are skewed (see Chapter 9, Section 9.4.5.3).

The Cochrane handbook does not encourage the meta_analysis of medians and interquartile ranges, except when sample sizes are large and the distribution of the outcome is similar to the normal distribution, which can be roughly tested by checking if the width of the interquartile range is approximately 1.35 standard deviations. The Cochrane handbook pointed out that the "use of interquartile ranges rather than standard deviations can often be taken as an indicator that the outcomes distribution is skewed".

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  • $\begingroup$ You might have an unfinished sentence ("Furthermore it would be interesting...") $\endgroup$ – Candamir Nov 6 '18 at 22:46
  • $\begingroup$ Yes, I have completed the sentence $\endgroup$ – paoloeusebi Nov 6 '18 at 22:49

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