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