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, |
In reply to Re: How to best eliminate values in a list that are outliers
by Athanasius
in thread How to best eliminate values in a list that are outliers
by Anonymous Monk
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