Suggest you Google confidence interval on medians. Best wishes, David Booth

(1) With thousands of values you can nicely bootstrap the sampling distribution of any statistic (like a difference in medians, if you wish, but also a difference in means). From this distribution you can derive the confidence interval (see "bootstrapping" in google).

(2) If you really have thousands of values, you can rely on the central limit theore if you want to estimate mean differences.

(3) For these kinds of right-skewed distributions of seemingly strictly positive values, the difference or means or medians is often not a very useful measure. The ratio is often a much more sensible measure in such cases. As the log ratio is a difference of logs, you may simply get an estimate and a confidence interval using the log of your data. The distribution of the log-transformed data should be more symmetric, and additionally you can rely on the central limit theorem, even if the distribution of the transformed data is not perfectly symmetric.

I want to add a few things on top of what already has been said. You have a skewed distribution, but I do not think you have outliers (at least it is not obviously the case). My first-hand choice would be to transform the variable to attain approximate symmetry and compute the mean and use the conventional statistical machinery.

If you really insist on working in the original scale and use the median you can refer to:

Article Resistant outlier rules and the non-Gaussian case

and eq. (2.3) for the sampling distribution of the median statistics. But I really think that is overkill.

The msmedse function in WRS2 (Wair & Wilcox) in R estimates the standard error of the median (and has other functions, like mestse for the SE for M-estimators, and there are variations like bpmedse, which uses a different algo and is on https://dornsife.usc.edu/assets/sites/239/docs/Rallfun-v35.txt). For the CI, the functions msmedci on his web site calls msmedse or medcipb provides the bootstrap interval estimate. To activate these in R (and these should work in S+ too, it was his functions that got me using S+ in the 1990s) just:

source("https://dornsife.usc.edu/assets/sites/239/docs/Rallfun-v35.txt")

then

> msmedci(rchisq(100,3))

$test

[1] 13.37466

$ci.low

[1] 1.903511

$ci.hi

[1] 2.557197

$p.value

[1] 0

$median

[1] 2.230354

But, as Jochen Wilhelm says, given the sample size these methods should all yield similar results. If you are using the median for some of your statistics, Wilcox has several good papers on inference with medians and other robust measures. One question is why the median? My answer sometimes follows Galton's "One vote, one value" paper (https://www.nature.com/articles/075414a0). His argument was (say the measurement is people estimating some value) half the people will think the value is too low and half too high, but this doesn't speak to the statistical qualities of the estimator.

One further note of caution (making a kind of polyphony of caution at this point!) is that the difference between the medians of two distributions is not the same as the median difference between the observations, unlike the case for the mean.

If you have Stata, or even if you haven't, you should read Roger Newson's papers in the Stata journal. Start here: Article Parameters behind “Nonparametric” Statistics: Kendall's tau,...

First, I may be an outlier among those who follow this question, for many believe in Gaussian statistics, whereas I believe in Paretian statistics.

Second, I consider heavy skewed as a norm rather than something peculiar. In other words, heavy skewed is more normal than normal distribution; see my detailed arguments in this paper:

Article Geospatial Analysis Requires a Different Way of Thinking: Th...

The contrast between Gaussian statistics and Paretian statistics reflects that between Tobler's law (of more or less similar) and scaling law (of far more smalls than larges):

Article Editorial: Spatial Heterogeneity, Scale, Data Character, and...

Bin Jiang – I don't think that Gaussian statistics are a) a discipline or b) a belief system. Certainly I don't know any statisticians who believe in anything, though some Bayesians have told me that they believed in Bayesian statistics a bit more after having read the evidence…

But your input is most welcome here. I'm looking forward to reading your papers – many thanks!

Ronán Michael Conroy – Many thanks! My wording "believe in" may not be appropriately used here, but my point was that there are two camps: one dominated by Gaussian thinking, and the other dominated by Paretian thinking, or alternatively one on Tobler's law and the other on scaling law.

Hmmm, and I think the two camps are those who think about an appropriate model and those who don't. The latter is simply trained to use analyses assuming Gaussian distributions. Nothing else is taught (usually).

Jochen Wilhelm absolutely. I think that our first question should be a forensic one : what's the data generating function here? Data are like a crime scene. The analyst has to regard this motionless arrangement as the result of the forces that placed things where they are observed. And, like crime scenes, things are being moved by multiple forces. So how can we decompose those multiple forces into a family of data generating processes, rather like doing a deranged Fourier transform?

No need to quote Box in present company, is there?

It's the fun bit of the job, I find.

Janis Frisfelds You may use weighted mean instead of median since median does not consider all observations in a data set. Please check the following link. Thanks in advance. http://www.naturalspublishing.com/Article.asp?ArtcID=11225

Md. Moyazzem Hossain , please clarify what you mean by the "median does not consider all observations". Each point influences the location of the median (change the maximum to be the minimum, and the location of the median for continuous data changes). The median minimizes the sum of the least absolute values (as opposed to the least squares as the mean does). Consider at example:

x = 1,2,3,4

with a median of 2.5. Which observation can you remove and keep the median the same?

Recently I found on nice text on such topic, Equimodal Frequency Distributions, written by Edwin Mouzon Jr. (attached). Text is very old but useful. Basic concept is to build theoretical model by use of mode.

Daniel Wright Thanks. Actually I want to say that median is not affected by the outlier. For example, X=1,2,3,4,5 median=3; x=1,2,3,4,100 median=3,; x=1,2,3,2000,5000 median=3; x=1,1,3,10000,50000 median=3. So, in all cases median=3, however the scenario are different. Can I say median is a good measure?

Md. Moyazzem Hossain , yep, that makes sense. I wouldn't say any estimator is "good" in general, but certainly in some circumstances it is good. The textbook example is Tukey (1960) showing it can be better measure of the population mean than the population mean. Try:

vals

Md. Moyazzem Hossain It's not a question of whether the median is a 'good' measure or not. The median measures something, the mean measures something else. The mean is a function of the sum while the median is not.

If you are admitted to hospital and wonder how long you will be there, I could tell you that half of the patients are discharged at or before five days. And that only ten percent of patients stay longer than 21 days. Quantiles are flexible and useful distributional statistics.

But if you are the hospital manager and you are costing the service, then the mean occupancy is useful. If you know the cost of one bed-day, then you can estimate the cost by multiplying it by the mean length of stay times the number of patients. Although patients with long stays are rare, they contribute to the cost!

So the question is really what a descriptive statistic is good for, rather than whether it is good or not.

A very good point to consider, Ronán Michael Conroy . It is important to know the context when answering questions. The problem is that many times we focus on the question as it is, and it turns out that whoever asks the question knew only the mean and believes, many times wrongly, that the median is an option to measure the same. By the shape of the histogram, this is a situation where the mean and median measure and calculate completely different things. It is important to know if Janis is truly interested in the mean. (Suppose so.) On the other hand, let's don't forget that Janis wants to compare several groups with a control one, but shows only a histogram of all the data. Residual analysis is necessary because it shows different things. Among the possibilities is that a transformation of the data allows applying a Dunnett test, which would be appropriate to compare the groups with the control. I suggest starting with multiple boxplots which give initial ideas about what happens.

If x and y follow a non-symmetric distributions, sometimes (x-y) follows a symmetric distribution, so you can apply a Wilcoxon test.