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I was looking at the outlier detection formula which uses the IQR and I wonder why it should be multiplied by 1.5? Can the constant be increased i.e 3 or 6 to be more "acid" if so under what criteria?

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  • $\begingroup$ The criteria are yours to choose. Why would you declare any observation to be an outlier? $\endgroup$ – Glen_b Oct 21 '13 at 1:06
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    $\begingroup$ Ok thanks. I just looked at some text books and found that the IQR is one of the myriad of methods to spot an outlier. $\endgroup$ – Aureon Oct 21 '13 at 2:13
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    $\begingroup$ It's really your model (at least in the loose sense) that makes an outlier an outlier (because what is an outlier but something that 'doesn't fit'?) $\endgroup$ – Glen_b Oct 21 '13 at 2:42
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    $\begingroup$ John Tukey, who invented the approach from which this method appears derived, used two multipliers: He set one "fence" at 1.5 times (an analog of) the IQR away from each quartile and another fence at 3 times the IQR from each quartile. Values beyond the first fence were "out" and values beyond the second were considered "far out" (those who remember the '60s will understand this terminology). If you think you need more extreme fences, then most likely you should consider re-expressing your data rather than changing the fences. $\endgroup$ – whuber Oct 21 '13 at 13:42
  • $\begingroup$ Question was also asked here: mathoverflow.net/questions/26434/why-1-5iqr-whiskers-in-boxplot and one of the answers is that such rule in case of normal distribution marks cases that appear with less then probability smaller then 1%. $\endgroup$ – Tim Oct 5 '16 at 11:32
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Certainly you can change the criterion.

The 1.5 multiplier is so that a certain proportion of the sample in a normal population will be outside it. But there is nothing sacred about it.

However, I would caution against any automatic method of selecting outliers.

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    $\begingroup$ Thank you, what I understood is that the criterion can be changed according to the data. $\endgroup$ – Aureon Oct 21 '13 at 2:15
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    $\begingroup$ I agree because statisticians may have just decided that 1.5 would be fit $\endgroup$ – Bob Dec 13 '16 at 3:40

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