I have recently come across this post on Median Absolute Deviation (MAD). The Wikipedia article here, by the article as an estimator Standard deviation of the distribution is 'k' times MAD, where the value of 'k' is dependent on the distribution. For a normal distribution the value is approximately '1.4826' for k. Now, my question is if MAD is calculated for a data set and then standard deviation is calculated and the standard deviation is divided by MAD and it does not come out to be approximately 1.4826 e.g. if it comes out as 6.0, can it be stated that the data distribution is not likely to be normally distributed.

  • $\begingroup$ You mean minimum absolute deviation which is a measure of spread just like the standard deviation. However it does not provide enough information to distinguish the normal distribution from other unimodal symmetric distributions. $\endgroup$ Feb 13, 2017 at 14:53
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    $\begingroup$ From what I understand, MAD is calculated by first calculating the median of the data set and then absolute difference between the median and each of the data point is calculated then the median of the differences is taken to be the MAD $\endgroup$
    – Ironluca
    Feb 13, 2017 at 16:56
  • $\begingroup$ @MichaelChernick I've never met minimum absolute deviation: on the face of it it might be highly uninformative about spread and could be (near) zero, regardless of where it is measured from (median, mean, something else). All flavours of (mean, median) absolute deviation from the (mean, median) seem in use somewhere. $\endgroup$
    – Nick Cox
    Feb 14, 2017 at 10:21
  • $\begingroup$ if the distribution is not normal how to use median absolute deviation build in function in R .what should we give the value of constant in case of normal distribution mad(x, center = median(x), constant = 1.4826, na.rm = FALSE, low = FALSE, high = FALSE) $\endgroup$
    – user366766
    Sep 1, 2022 at 11:36

1 Answer 1

  1. Considered as a formal test of normality: If $M$ = (sample) median absolute deviation from the median and $s$ = standard deviation, then you could indeed use a measure like $R = M/s$ (or its reciprocal) as a test statistic for a test of normality.

    Note however, that such tests cannot tell you something is normal, only - sometimes - that it isn't.

    To make it a test, all you'd need is the distribution of/a table of percentiles of the distribution of the ratio under the null (i.e. at normality) for various sample sizes. This can be obtained by simulation, for example -- though it might also be possible to obtain it analytically.

    It's actually a close kin to an old test statistic proposed by Geary[1], which was the ratio of mean deviation to standard deviation, sometimes referred to as Geary's $a$ test (because he proposed a number of test statistics it's necessary to distinguish them, and he used $a$ - and later, $a_1$, to denote this ratio of mean deviation to standard deviation).

    Geary's $a$ test has quite good power compared to the Shapiro-Wilk test in small to moderate sized samples for a wide range of symmetric alternatives, beating it in a number of situations. To my recollection is has quite good power against heavier tailed cases like the logistic and Laplace. Your proposal should have somewhat similar properties.

    Indeed I think that the likelihood ratio test for normality against a Laplace alternative would correspond to looking at the ratio of mean deviation from the median to standard deviation (which would be a third statistic a bit more like Geary's than yours).

    [My guess is that Geary's $a$ test statistic would have better power against something like a logistic alternative than yours, but yours might be more competitive with even peakier-and-heavier-tailed alternatives than the Laplace -- an example of an alternative that I'd expect it to do especially well against would be the location-scale family based off the distribution of the product of two independent standard normals. It might also do fairly well against something like a t-distribution with low d.f. It would be interesting to see if such guesses hold up, and whether it does well in other situations.]

    Against general alternatives, the power may sometimes be poor, however - for example we should anticipate relatively low power against lightish-tailed, skew alternatives (at least ones that have similar population ratios of median absolute deviation to standard deviation), compared to widely used omnibus tests. However, many skew alternatives of interest are also heavy-tailed, so it may still do fairly well against some of those.

    It wouldn't be suitable in every situation but might work very well if you anticipate the kind of alternatives against which it should have reasonably good power.

    There are a number of papers that have investigated Geary's test but off the top of my head I don't recall any for your proposed statistic. I'd bet that it has been looked at but I didn't find any papers on it with a quick search.

    The closest I came was Gel et al [2] which discusses a test based on the ratio of standard deviation to mean deviation from the median (which they call MAAD), which would be a version of the test I suggested for a Laplace alternative above. They say that compared to the test based on the MAAD, the MAD has lower power against heavy tailed alternatives (which they say is due to the higher variance of MAD at the normal) but they don't give further details (however, they do say that MAD is better for diagnostic displays, which relates to my point 2. below). Aside from that brief passing mention I haven't found anything else on power comparisons.

    One big advantage of these kinds of tests is their simplicity; they don't require specialized routines to compute the statistic and are amenable to hand computation in small samples, even for beginning students. In the case of Geary's test there's a normal approximation (D'Agostino, 1970 [3]) for $n>40$; there's likely to be a suitable normal approximation in medium-to-large samples here as well. That they can also have good power in situations we might actually care to identify may make them worth considering -- certainly it could be worth a bit of time investigating the power properties more closely and some investigation to find any previous investigations of the test.

  2. As a diagnostic tool. Rather than a formal test (which may answer a question we already know the answer to instead of one we'd be better to answer), we could use the ratio as a diagnostic -- a measure of how far from normality we might be (in effect as a kind of raw "effect size" of a particular kind of non-normality).

    For example, if we're particularly concerned about how heavy-tailed our distribution might be this sort of ratio might be worth considering as a diagnostic measure for that situation, rather than computing something like kurtosis, say.

* (i.e. has relatively high power in that situation)

[1] Geary, R. C. 1935. "The ratio of mean deviation to the standard deviation as a test of normality." Biometrika 27: 310-332

[2] Gel, Y. R., Miao, W., and Gastwirth, J. L. (2007) Robust Directed Tests of Normality Against Heavy Tailed Alternatives. Computational Statistics and Data Analysis 51, 2734-2746.

[3] D'Agostino, Ralph B. (1970),
"Simple compact portable test of normality: Geary's test revisited"
Psychological Bulletin, Vol 74(2), Aug, 138-140


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