RE: ETABAR p-value
Leonid,
As I understand OMEGA(3,3) is the variance of the individual eta3's
whereas the t-test uses the variance of the average eta3, i.e. something
closer to OMEGA(3,3)/n where n is the number of subjects.
Br
Lars
Quoted reply history
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Leonid Gibiansky
Sent: 20 February 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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