RE: ETABAR p-value
I think that only the mean of the mean posterior etabars, not the mean
of the post-hoc etabars are to be zero (at least it is the case when
using the MC-PEM algorithm). Within an MC-PEM framework, you get that
equality only when the individual posterior distributions have a mean
equal to the mode.
Serge Guzy
President, CEO, POP-PHARM,Inc.
Quoted reply history
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Mats Karlsson
Sent: Wednesday, February 21, 2007 12:32 AM
To: [EMAIL PROTECTED]
Cc: [email protected]
Subject: RE: [NMusers] ETABAR p-value
Dear Leonid,
In your case there is no reason to expect the mean of the etabar to be
zero, so a test of it does actually not make sense. You have pronounced
shrinkage in your posthoc etas and then you don't know what the
distribution should be (apart from expected median of zero). We have
noted mean posthoc eta significantly different from zero even when the
model is correct (see reference below for some more discussion on the
"uselessness" of posthoc etas).
http://www.aapspharmaceutica.com/search/abstract_view.asp?id=941&ct=06Ab
stra
cts
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 20:49
To: Mats Karlsson
Cc: [email protected]
Subject: Re: [NMusers] ETABAR p-value
Mats,
Thanks for your reply. I think I found where the problem was: to
estimate the probability of observed ETAbar, I used the distribution
with the variance estimated by the nonmem (0.97) while t-test uses
variance estimated from the data (0.11784 in this case). In this
particular case, variance estimated from the data is much lower than
OMEGA(3,3) because the POSTHOC ETA3 distribution is rather non-normal,
resulting in a small p-value.
Thanks
Leonid
Mats Karlsson wrote:
> 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
>
>
>
>
>
>
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