Closed drveera closed 6 years ago
I guess its pnorm(abs(Z), lower.tail=FALSE)*2
.
Correct! In case of super large z stats (as often in genomics) you might want to use the log option too. Otherwise you end up with p=0.
Sure. thanks. Also regarding the lambda I didnt specify anything. Whats the default value the script uses? does it calculated from sample size from reference panel?
The default lambda (2/sqrt(n)) is fine. And yes, n is the number of individuals in the refpanel.
The newest version (just uploaded) has the P value column included.
awesome. thanks :)
Hi,
Could you tell how to calculate P values using R? For now I am doing like,
pnorm(abs(Z), lower.tail=FALSE)
Is this right? I couldnt understand the formula you described in the manual.Regards, Veera