I have four weights. I want to test if there is an outlier within those four weights.

To me, the number seems quite small to detect outliers.

I thought that the Median Absolute Deviation (MAD) may be a way to detect outliers.

I used it in R with

D[which((abs(D-median(D))/mad(D)) >2)]

where D is those numbers:

[1,] 1.2492
[2,] 1.2885
[3,] 1.1707
[4,] 1.3207

The result was numeric (0). This has to do with the >2 term (which is supposed to mean 2 MAD -SDs away from the middle mad or so). Does that mean there is no outlier within this group?

Is this actually a sound way to check for an outlier in this specific case?



Yes, that would mean there is no outlier, using that definition of outlier.

Whether that is a good way to detect outliers is not clear. I am leery of any automated method or formula. They may be necessary when you have a lot of variables and too little time, but I'd say they are dangerous.

Why are you trying to detect outliers? What will you do if you find them? What are the data?

In general, I'd say the way to detect outliers (if you need to do so at all) is using substantive knowledge.

  • 1
    $\begingroup$ Thank you! I do this actually in my Masters' course, sso it's also a thing of practice. In this special case, outliers would be those weights that are unlikely to have had occured by what I research, i.e. weight loss due to decomposition; where for instance weight was lost because the samples (sachets buried in the earth) were ripped and lost their contents. But maybe yes, I'll just throw them out when they're overly deviating. Thank you for the answer! $\endgroup$ – Vera Marya Jun 11 '17 at 13:57
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    $\begingroup$ @VeraMarya: If you found this answer helpful, then please consider upvoting and/or accepting it. $\endgroup$ – Stephan Kolassa Apr 25 '19 at 15:07

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