When you do x1g$fdr, you adjusted for a total of 15568 tests, so FDR is high
for those first four entries. When you do p.adjust(pval,method="BH"), you
assumed there were only a total of 4 multiple tests in your experiment, so FDR
is low.
Ding
-Original Message-
From: R-help [mailto:r
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 1.000 1.000
If all of your t-tests are related, say by testing the location values
for each row against a fi
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 0.478278
Dear Sebastian,
I was able to replicate the R^2 value of 0.993019 using the function provided
as the second answer (by Julien Massardier) to the following question -
https://stats.stackexchange.com/questions/83826/is-a-weighted-r2-in-robust-linear-model-meaningful-for-goodness-of-fit-analys
Tha
Dear David
I have updated R version (3.6.2) as suggested by John and now Rcmdr
works :-)
Thanks again for your time and great help
Best regards
Toufik
On 06.01.2020 17:31, David Winsemius wrote:
On 1/6/20 5:16 AM, tzahaf wrote:
Dear
I have a problem when trying to use Rcmdr. This is the
Dear John, Jeff
I have updated R version (3.6.2) as suggested and now Rcmdr works :-)
Thanks again for your time et great help
Best regards
Toufik
On 06.01.2020 23:59, Fox, John wrote:
Dear Toufik,
-Original Message-
From: Toufik Zahaf
Sent: Monday, January 6, 2020 4:07 PM
To: Je
Hi,
recently I posted a question about how R computes the value for the
"r.squared" statistic internally for weighted least squares linear
models on crossvalidated.com (see:
https://stats.stackexchange.com/questions/439590/how-does-r-compute-r-squared-for-weighted-least-squares)
including a m
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