# MAD equivalent for standard error

As far as I know, one can calculate the relative standard error from the standard deviation of a data sample. I am looking for the Median Absolute Deviation equivalent for standard error.

Does one exist? Also which methods exist to calculate the relative standard error, that do not require the standard deviation?

• What properties do you want of a "standard error" (of the mean, presumably) that is a "MAD equivalent"? – whuber Jun 7 '12 at 16:53
• What do you mean by equivalent? Are you looking for a formula that applies generally for distributions that relates MAD to se? – Michael Chernick Jun 7 '12 at 16:54
• The goal is to find the statistical significance of a certain metric. Lots of data is involved, so best to come up with an efficient method. I started looking into MAD, because of possible outliers due to data problems ... – Navi Jun 7 '12 at 20:25

When you say standard error, you should be talking about the standard error of something, such as the standard error of the sample mean.

Similarly you could for example talk about the median absolute deviation of the sample median. It is possible to calculate this, at least as an approximation for large samples.

It is well known that for a continuous random variable with population median $m$, continuous probability density of the median $f(m)$ and a large odd sample size $n$, the sample median is approximately normally distributed with median $m$ and variance $\frac{1}{4 n f(m)^2}$, i.e. with median absolute deviation approximately $\dfrac{\Phi^{-1}\left(\frac34 \right)}{2 \sqrt{n} f(m)}$ where $\frac{\Phi^{-1}\left(\frac34 \right)}{2} \approx 0.337$.

If you want to have this as a relative median absolute deviation of the sample median, then presumably you divide by $m$.

• Did you notice that "MAD" is a median absolute deviation? It is used as a robust statistic. Its utility in the role of a putative SE is something to be doubted because it does not enjoy the asymptotic properties of a standard error, nor does it have a comparable interpretation. – whuber Jun 7 '12 at 20:04
• @whuber: I had typed mean absolute deviation when I meant median absolute deviation. The median absolute deviation of the sample median has the asymptotic property of being roughly proportional to $\frac{1}{\sqrt{n}}$, similar to the standard error of the sample mean. – Henry Jun 8 '12 at 6:22
• @whuber: For large samples the absolute difference between the population median and the sample median has a roughly 95% chance of being less than 2.906 times the expression I gave, similar to the 1.96 property of the standard error. – Henry Jun 8 '12 at 6:30
• @whuber: though I accept the point that the median absolute deviation of the sample median depends on the underlying population distribution near the median, and this may not be so immediately easy to estimate from the sample. – Henry Jun 8 '12 at 8:17
• Thanks for the extended analysis--clear and to the point. (+1) – whuber Jun 8 '12 at 12:29

An alternative approach is to look at the quantiles of the the sample: order the observations so $x_1 \le x_2 \le x_3 \le \cdots \le x_n$ and (for reasonably large samples from a continuous distribution) there is a 95% probability that the interval $$\left[x_{\frac{n}{2} - 0.98 \sqrt{n}}, x_{\frac{n}{2} + 0.98 \sqrt{n}}\right]$$ contains the population median.

Clearly ${\frac{n}{2} \pm 0.98 \sqrt{n}}$ may not be an integer: you can round outwards to be conservative or use one of the many possibilities for interpolating quantiles: you are now looking for $Q_p$ where $p = {\frac{1}{2} \pm 0.98 /\sqrt{n}}$.

Again for relative, you can divide by the median.