MAD (Median Absolute Deviation) is:

$\text{MAD} = M_i(|x_i-M_j(x_j)|)$

where $M()$ is the median operator ($M_i(x_i) = \text{median}(x_1,...,x_n)$).

I'd like to scale the MAD in such a way as to include (say) 95% of a distribution around the median, the way that that 95% of a normal distribution is within $1.96\sigma$ of the mean.

That is, if $m = M_i(x_i)$ and $d = \text{MAD}_i(x_i)$, make an interval like $m \pm b\cdot d$ (where $b$ depends on the distribution you are dealing with) that includes 95% of the distribution.

Can this be done?

  • $\begingroup$ There is no universal relationship (although there are some bounds): that's why people don't use MAD to compute confidence intervals! Incidentally, "sigma" usually refers to a sample standard deviation (and so does your tag) but you probably intended to refer to a standard error because that's what the width of a confidence interval depends on in the Normal distribution situation. BTW, what are "Mi", "Mj", "xi", and "xj"? Please consider using $\TeX$ markup (enclose it between dollar signs $\$$). $\endgroup$
    – whuber
    Commented Aug 13, 2013 at 21:26
  • $\begingroup$ GK89 - I'll start you off with adding markup, but you should definitely define all your terms. You seek a confidence limit -- but for which quantity? Are you asking for an interval under an assumption of normality, or were you hoping for the existence of some distribution-free limit? $\endgroup$
    – Glen_b
    Commented Aug 13, 2013 at 23:03
  • $\begingroup$ If you check the Wikipedia article, MAD has no "$b$" at all. You should start with what you're trying to achieve, and put $b$ where it goes (as a scaling factor in your interval) rather in your definition of MAD (where it doesn't belong). Your question doesn't explain enough about what you're trying to do. $\endgroup$
    – Glen_b
    Commented Aug 13, 2013 at 23:11
  • $\begingroup$ I believe what you're trying to do is use the MAD (without a $b$) to estimate the standard deviation, or a multiple of it, like $1.96\sigma$ or $3.84\sigma$ (by multiplying MAD by $b$). If your data normally distributed, I believe the asymptotic scaling constant is about 1.48 -- that is, in sufficiently large samples (possibly very large $n$ may be needed for everything to work well) you would estimate $\sigma$ by $1.48 \text{MAD}$-- but this number is different for different distributions; and at the same time the 1.96 or 3.84 also changes from distribution to distribution. $\endgroup$
    – Glen_b
    Commented Aug 14, 2013 at 23:44
  • $\begingroup$ GK89 I have attempted to clarify your question. If you disagree with the edits, revert or edit to make it say what you want. $\endgroup$
    – Glen_b
    Commented Aug 15, 2013 at 0:06

2 Answers 2


I know the original post is over a year old, but I would like some more information on this topic. I currently run a proficiency program for manure testing and soil testing laboratories. A colleague, who knows much more about statistics than I do, suggested the following to get a 95% confidence interval using the MAD and median.

  1. Calculate the median and MAD values.
  2. Remove results exceeding plus or minus 4.0 MAD units from the median as outliers.
  3. Recalculate the median and MAD values on the reduced data set.
  4. Results exceeding plus or minus 2.9 MAD units from the second median are outside the 95% confidence interval.

There is one other kicker. I use the statistical program R. When calculating MAD I use the following:

mad(x, constant = 1)

The default in R is: constant = 1.4826.

Typically, we have from 140 to 200 datapoints for each analysis. Often, the results are right skewed, occasionally left skewed, and rarely normally distributed. After removing the 4.0 MAD outliers, we have a much more normally distributed histogram. I suspect at that point we might be able to use mean and SD to calculate the confidence interval.

For a number of years, we just ran the data one time. Labs were flagged for accuracy if their results deviated by more than 2.5 MAD units from the median. I have compared both methods, and usually 2.5 MAD units from the median (just one calculation) is quite close to the two-step method using 2.9 MAD units from the median after removing the 4.0 outliers.

I hope this method gives us a 95% confidence interval. But, if anyone has a better suggestion, I'd like to hear it.

  • 2
    $\begingroup$ Welcome to the site, @JerryFloren. This is an answer to the OP's question, but you also seem to have some (related) questions of your own. Be aware that CV is not a discussion forum--it is a pure Q&A site. If you also want to ask follow-up questions, please do so by clicking the gray ASK QUESTION at the top & ask there, not here. Since you're new here, you may want to take our tour, which has information for new users. $\endgroup$ Commented Oct 14, 2014 at 20:28
  • 2
    $\begingroup$ Please allow me to extend @gung's welcome to you as well, Jerry. I apologize for the deceiving title: neither the original question nor, as far as I can tell, your post are about confidence: they seem to be about something called a tolerance interval, which is a way of estimating (say) the middle 95% of a population. I am hoping you will post your question in a separate thread so it can be properly addressed, but when you do so you might want to contemplate what kind of interval you really want and perhaps write a few words about what it means to you: what will you use it for? $\endgroup$
    – whuber
    Commented Oct 14, 2014 at 20:46
  • $\begingroup$ @JerryFloren, I'm curious for the reasoning in choosing 4 for the removal of outliers and 2.9 for being outside 95% $\endgroup$
    – GK89
    Commented Oct 15, 2014 at 13:06

Yes it can! In this rather recent preprint authors defined asymptotic CI of Median absolute deviation (MAD) as

$$ \left[L,U\right]_{MAD} = \left[ \widehat{MAD}_X ± \textit{z}_{1 - \alpha / 2} \frac{ \widehat{ASD}(\mathcal{MAD}, F_n)) }{ \sqrt{n} } \right] $$

where ASD is a squared root of asymptotic variation, the $z_{1 − \alpha/2}$ is the $(1 − \alpha/2)×100$ percentile of the standard normal distribution. But then getting into how to calculate asymptotic variation< I got a bit lost, but all the math is in the preprint.

Futhermore, this CI was also implemented in R in DescTools package.


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