adjusted p value, fdr
Hi Anamarija, In the first calculation you are implicity using n=15568 as the default as the number of p values tested.
p.adjust(pval,method="BH",n=15568)
[1] 0.4782784 0.5852553 1.0000000 1.0000000 If all of your t-tests are related, say by testing the location values for each row against a fixed value like zero, then this is an appropriate way to discover whether any differ significantly from your reference value. In general if you only want to test a subset of the rows rather than all, you should have a good reason for doing so. Jim
On Wed, Jan 8, 2020 at 9:15 AM Ana Marija <sokovic.anamarija at gmail.com> wrote:
Hello,
I have a data set which has 15568 entries
I am trying to calculate adjusted p values using p value via:
head(x1g)
t P.Value
adj.P.Val B fdr
3TXADqtSkhV1IXRHlg 4.468671 3.072189e-05 0.4782784 1.5253151 0.4782784
34lG83aZ6.WLFnge6s 4.217037 7.518696e-05 0.5852553 0.8660534 0.5852553
oS67lguv5HpU4i6Pvg -3.939182 1.959413e-04 0.9994637 0.1600655 0.9994637
Qpc2sX7gp.W5E2rRS4 3.732914 3.900822e-04 0.9994637 -0.3471331 0.9994637
NqsVOy_yos30cOQRSE -3.673551 4.737810e-04 0.9994637 -0.4902001 0.9994637
B3SDpegGjpdxnvU0S0 -3.665765 4.859528e-04 0.9994637 -0.5088638 0.9994637
x1g$fdr=p.adjust(x1g$P.Value,method="BH")
the fdr values seem unusually high
if I just do it for the first few p values from my x1g data frame:
pval=c(3.072189e-05,7.518696e-05,1.959413e-04,3.900822e-04) p.adjust(pval,method="BH")
[1] 0.0001228876 0.0001503739 0.0002612551 0.0003900822 Can someone please explain what might be the issue. Thanks Ana
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