# PearsonsCorrelation.getCorrelationPValues() precision limited by machine epsilon

Rank to TopRank to BottomBulk Copy AttachmentsBulk Move AttachmentsVotersWatch issueWatchersConvert to sub-taskLinkCloneUpdate Comment AuthorReplace String in CommentUpdate Comment VisibilityDelete Comments
XMLWordPrintableJSON

#### Details

• Bug
• Status: Closed
• Major
• Resolution: Fixed
• 2.0
• None
• None

#### Description

Similar to the issue described in MATH-201, using PearsonsCorrelation.getCorrelationPValues() with many treatments results in p-values that are continuous down to 2.2e-16 but that drop to 0 after that.

In MATH-201, the problem was described as such:
> So in essence, the p-value returned by TTestImpl.tTest() is:
>
> 1.0 - (cumulativeProbability(t) - cumulativeProbabily(-t))
>
> For large-ish t-statistics, cumulativeProbabilty(-t) can get quite small, and cumulativeProbabilty(t) can get very close to 1.0. When
> cumulativeProbability(-t) is less than the machine epsilon, we get p-values equal to zero because:
>
> 1.0 - 1.0 + 0.0 = 0.0

The solution in MATH-201 was to modify the p-value calculation to this:
> p = 2.0 * cumulativeProbability(-t)

Here, the problem is similar. From PearsonsCorrelation.getCorrelationPValues():
p = 2 * (1 - tDistribution.cumulativeProbability(t));

Directly calculating the p-value using identical code as PearsonsCorrelation.getCorrelationPValues(), but with the following change seems to solve the problem:
p = 2 * (tDistribution.cumulativeProbability(-t));

#### People

Unassigned
Kevin Childs
0 Vote for this issue
Watchers:
0 Start watching this issue

#### Dates

Created:
Updated:
Resolved: