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

[Athanasius]

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?

[Anonymous Monk]

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