# Confidence interval for the median

I have a set of values ${x_i}, i=1, \dots ,N$ of which I calculate the median M. I was wondering how I could calculate the error on this estimation.

On the net I found that it can be calculated as $1.2533\frac{\sigma}{\sqrt{N}}$ where $\sigma$ is the standard deviation. But I did not find references about it. So I do not understand why.. Could someone explain it to me?

I was thinking that I could use bootstrap to have an estimate of the error but I would like to avoid it because it would slow down a lot my analysis.

Also I was thinking to calculate the error on the median in this way $$\delta M = \sqrt{ \frac{\sum_i(x_i - M)^2}{N-1} }$$

Does it make sense?

• Do you know with absolute certainty that the data are normally distributed? Feb 15, 2016 at 12:48
• they are lognormal Feb 15, 2016 at 12:52
• Bootstrap should work and it could not take a long time. Either you have a complete enough dataset and no need to do a bootstrap, just take the median of your variable as a good estimation of the real median. Or you have a rather small dataset and you could use bootstrap to estimate a median with your margin error in no excessive time.
– YCR
Feb 15, 2016 at 12:52
– Tim
Feb 15, 2016 at 12:56
• Extensive information about the distribution of the median appears in my post at stats.stackexchange.com/a/86804/919. It develops the theory needed for both nonparametric and normal-approximation confidence intervals.
– whuber
Feb 15, 2016 at 19:53

To directly deal with error on the median you can use the exact nonparametric confidence interval for the median, which uses order statistics. If you want something different, i.e., a measure of dispersion, consider Gini's mean difference. Code is here for the median's confidence interval.

• I was actually considering to use an analog of the Gini coefficient: $S_n=c * med_j (med_j |x_i-x_j|)$ as defined by Rousseeuw and Croux (web.ipac.caltech.edu/staff/fmasci/home/astro_refs/…). Feb 15, 2016 at 13:48
• The median must have an asymmetric error if the data distribution is asymmetric. Feb 15, 2016 at 14:19

As pointed out in the other answer, there is a non-parametric CI for the median using the order statistics. That CI is better in many aspects than what you found on the net.

Now, if you must know where the $1.2533\frac{\sigma}{\sqrt{N}}$ factor comes from, the answer is from the asymptotic distribution of the median. If we denote the sample median by $\tilde{\theta}$ and the population median by $\theta$ then it can be shown that

$$\sqrt{n} \left( \tilde{\theta} - \theta \right) \xrightarrow{L} \mathcal{N} \left(0, \frac{1}{4 \left[f \left( \theta \right) \right]^2} \right)$$

where $f$ is the distribution of your sample. The result is not as universal as the CLT because the asymptptic distribution still depends on the underlying distribution of your sample (through the term $\left[f \left( \theta \right) \right]^2$). You can, however, make the drastic simplication that your sample comes from a normal distribution with mean -and median- $\theta$ and variance $\sigma^2$. Evaluating $f$ at its point of symmetry then yields

$$\left[f \left( \theta \right) \right]^2 = \frac{1}{2\pi \sigma^2}$$

and so the asymptotic variance becomes

$$\frac{2\pi}{4} \sigma^2$$.

Divide by $N$ and take the square root of that to arrive at your standard error $1.2533\frac{\sigma}{\sqrt{N}}$.