Dear Leonid,
The etabar test is a t-test of the mean of the posthoc etas. I would not
discard a model just because of this not being the case as there may be
other reasons than misspecification for a no-zero mean of posthoc etas. Only
when data are very rich and there is no shrinkage (or when eta shrinkage is
identically large for both positive and negative etas) would we expect the
mean of posthoc etas to be zero.
Best regards,
Mats
Mats Karlsson, PhD
Professor of Pharmacometrics
Div. of Pharmacokinetics and Drug Therapy
Dept. of Pharmaceutical Biosciences
Faculty of Pharmacy
Uppsala University
Box 591
SE-751 24 Uppsala
Sweden
phone +46 18 471 4105
fax +46 18 471 4003
[EMAIL PROTECTED]
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Leonid Gibiansky
Sent: Tuesday, February 20, 2007 18:25
To: [email protected]
Subject: [NMusers] ETABAR p-value
Dear All,
Could anyone help me to interpret ETAbar p value? I have:
TOT. NO. OF OBS RECS: 1486
TOT. NO. OF INDIVIDUALS: 213
ETABAR: -0.52E-02 -0.27E-01 -0.93E-01
P VAL.: 0.90E+00 0.24E+00 0.81E-04
ETA1 ETA2 ETA3
ETA1 4.08E-01
ETA2 2.26E-01 2.31E-01
ETA3 0.00E+00 0.00E+00 9.70E-01
which looks too low for me for eta3. I checked that p of abs(mean(eta)) >
0.093 is about 0.17 for
normally distributed variable with SD=sqrt(0.97) and about 200 patients.
> sum1 <- 0
> for(i in 1:1000000) if(abs(mean(rnorm(213,0,sqrt(0.97))))> 0.093) sum1 <-
sum1+1
> print(sum1/1000000)
[1] 0.168624
How exactly this p-value is computed (NONMEM V) ?
Thanks
Leonid
