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_bOct 21, 2013 at 1:06
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1$\begingroup$ 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. $\endgroup$– AureonOct 21, 2013 at 2:13
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1$\begingroup$ 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_bOct 21, 2013 at 2:42
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4$\begingroup$ 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, 2013 at 13:42
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$\begingroup$ 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, 2016 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.
answered Oct 20, 2013 at 23:38
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1$\begingroup$ Thank you, what I understood is that the criterion can be changed according to the data. $\endgroup$– AureonOct 21, 2013 at 2:15
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1$\begingroup$ I agree because statisticians may have just decided that 1.5 would be fit $\endgroup$– BobDec 13, 2016 at 3:40
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$\begingroup$ To be specific, 0.7% of the data. These commonly used limits make sure that '0.7%' of data is treated as an outlier; if there's any data point in that region. $\endgroup$ Nov 1, 2019 at 15:13