in reply to How to best eliminate values in a list that are outliers

In D, .1 will be set as the "global min" but it is a mistake and 1500 should be set as the true global min.

To eleborate on ww’s response:

As LanX says, the standard test for outliers is Grubbs’s test. There is an online calculator for Grubbs’s test at http://www.graphpad.com/quickcalcs/grubbs2/, and entering your sample data for series “D” — with Alpha set to either 0.05 or 0.01 — produces the following result :

Row Value Z Significant Outlier?
1 0.1 1.471 Furthest from the rest, but not a significant outlier (P > 0.05).
2 1500.0 0.173
3 1700.0 0.000
4 2100.0 0.346
5 3200.0 1.298

So, why do you identify 0.1 as an outlier?

Athanasius <°(((>< contra mundum Iustus alius egestas vitae, eros Piratica,

  • Comment on Re: How to best eliminate values in a list that are outliers

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Re^2: How to best eliminate values in a list that are outliers
by Anonymous Monk on Nov 10, 2015 at 14:15 UTC

    Presumably, because his data does not represent some quantity to approximate normal distribution.

    Transforming the D with f = 1 / x yields:
    Row Value Z Significant Outlier?
    1 10.000000000000000000 1.78885438120805620000 Significant outlier. P < 0.05
    2 0.000666666666666667 0.44717876261411860000  
    3 0.000588235294117647 0.44719630129821775000  
    4 0.000476190476190476 0.44722135656121640000  
    5 0.000312500000000000 0.44725796073450363000  

    Transforming the D with f = log x gives:
    Row Value Z Significant Outlier?
    1 -2.30258509299405 1.7851028840345886 Significant outlier. P < 0.05
    2 7.31322038709030 0.3777123990128823  
    3 7.43838353004431 0.4058644616784443  
    4 7.64969262371151 0.4533927252416877  
    5 8.07090608878782 0.5481332981015742  

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