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
ETABAR p-value
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]
Quoted reply history
-----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
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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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=06Abstra
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]
Quoted reply history
-----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
>
>
>
>
>
>
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
>
>
>
>
>
>
--
The information contained in this email message may contain confidential or
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views of XOMA.
Dear NONMEM user,
I appreciate if someone provide me with suggestions regarding the problem
indicated below:
I have been trying to analyze PK data from a dosage form which releases it
content in proximal and distal part of GI tract. Before fitting the model
to my data, I tried to simulate the data. However, I am only getting a
single pulse of a drug with a lag time corresponding to the ALAG2 (lag
time for the second GI compartment). I am wondering if I have errors in
my codes or the data file. Any recommendations greatly appreciated. My
control stream and few lines from the input file are given below:
Control Stream:
;Model Desc: ONE Compartment Model - Double INPUT
;Project Name: XXXX
;Project ID: NO PROJECT DESCRIPTION
$PROB RUN# 1 YOUR TEXT
$INPUT C ID CMT DAT=DROP DATE TIME DV AMT MDV EVID SS II
$DATA POPPK_129V2.CSV IGNORE=C
$SUBROUTINES ADVAN6 TOL=3
$MODEL
COMP=(1) ;(DEPOT1, DEFDOSE)
COMP=(2) ;(DEPOT2)
COMP=(3) ;(CENTRAL,DEFOBS)
$PK
TVCL=THETA(1)
CL=TVCL*EXP(ETA(1))
TVV=THETA(2)
V=TVV*EXP(ETA(2))
TVKA1=THETA(3)
KA1=TVKA1*EXP(ETA(3))
TVKA2=THETA(4)
KA2=TVKA2*EXP(ETA(4))
TVF1=THETA(5)
F1=TVF1
TVLT1=THETA(6)
ALAG1=TVLT1*EXP(ETA(5))
TVLT2=THETA(7)
ALAG2=TVLT2*EXP(ETA(6))
K30=CL/V
IF (F1 .GT. 1) F1=0.999
IF (ALAG1 .LE.0) ALAG1=0
F2=1-F1
S3=V
$DES
RINA=KA1*A(1)
RINB=KA2*A(2)
C3=A(3)/V
DADT(1)=-RINA
DADT(2)=-RINB
DADT(3)=RINA+RINB-K30*A(3)
$ERROR
DEL=0
IF (F.EQ.0) DEL=1
IPRED=F
W=IPRED+DEL
IRES=DV-IPRED
IWRES=IRES/W
Y=F+W*ERR(1)+ERR(2)
$THETA
11000 ;[CL]
21000 ;[TVV]
1 ;[TVKA1]
0.5 ;[TVKA2]
0.25 FIXED ;[TVF1]
0.5 ;[TVLT1]
2.5 ;[TVLT2]
$OMEGA
0 FIXED ;[ETACL]
0 FIXED ;[ETAV]
0 FIXED ;[ETAKA1]
0 FIXED ;[ETAKA2]
0 FIXED ;[ETALT1]
0 FIXED ;[ETALT2]
$SIGMA
0 FIXED ;[ERR1]
0 FIXED ;[ERR2]
;$MSFI
;$EST MAXEVAL=9999 NSIG=3 MSF=1.msf PRINT=5 NOABORT METHOD=0 POSTHOC
;$COVARIANCE
$SIMULATION ONLYSIM (9215690)
$TABLE ID DATE DOSE TIME IPRED CL V KA1 KA2 F1
ALAG2 NOPRINT ONEHEADER FILE=1.TAB
;$TABLE ID CL VC Q VP KA IC50 NOPRINT ONEHEADER FILE=patab43
;$TABLE ID AGE CRCL WT ALB NOPRINT ONEHEADER FILE=cotab1
;$TABLE ID SEX NOPRINT ONEHEADER FILE=catab40
;$TABLE ID TIME IPRED NOPRINT ONEHEADER FILE=sdtab43
INPUT File:
Many Thank,
Majid Vakily, Ph.D.
Senior Research Investigator
Department of Drug Metabolism & Pharmacokinetics
Phone: (847) 582-2198
Fax; (847) 582-2388
The information contained in this communication is confidential and may
constitute non-public and/or "inside" information. It is intended only for the
use of the addressee and is the property of TAP Pharmaceutical Products Inc.
Unauthorized use, disclosure, or copying of this communication, or any part
thereof, is strictly prohibited and may be unlawful. If you received this
communication in error, please notify me immediately by return e-mail and
destroy this communication and all copies thereof, including all attachments.
gifqnhA9BUJWy.gif
Description:
GIF image
Dear Majid,
Since I recently made a similar mistake I can spot one reason for why you only
get a single pulse corresponding to ALAG2. In the INPUT file both does events
(CMT1 & CMT2) have EVID = 4. EVID = 4 leads to reset of the system and the
previous dose record is therefore not taken into account. Change to EVID = 1
and at least one problem is out of the way.
I don't know the nature of your formulation but I think that it in many cases
might also be possible to model this as one single dose and a chain of
absorption compartments. This might be a more physiological approach than to
have two fix doses into two separate absorption compartments.
Kind regards,
Martin Bergstrand, MSc, PhD student
-----------------------------------------------
Division of Pharmacokinetics and Drug Therapy
Department of Pharmaceutical Biosciences
Uppsala University
-----------------------------------------------
P.O. Box 591
SE-751 24 Uppsala
Sweden
-----------------------------------------------
[EMAIL PROTECTED]
-----------------------------------------------
Work: +46 18 471 4639
Mobile: +46 709 994 396
Fax: +46 18 471 4003
Quoted reply history
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED]
Sent: 22 februari 2007 23:08
To: [email protected]
Subject: RE: [NMusers] ETABAR p-value
Dear NONMEM user,
I appreciate if someone provide me with suggestions regarding the problem
indicated below:
I have been trying to analyze PK data from a dosage form which releases it
content in proximal and distal part of GI tract. Before fitting the model to
my data, I tried to simulate the data. However, I am only getting a single
pulse of a drug with a lag time corresponding to the ALAG2 (lag time for the
second GI compartment). I am wondering if I have errors in my codes or the
data file. Any recommendations greatly appreciated. My control stream and few
lines from the input file are given below:
Control Stream:
;Model Desc: ONE Compartment Model - Double INPUT
;Project Name: XXXX
;Project ID: NO PROJECT DESCRIPTION
$PROB RUN# 1 YOUR TEXT
$INPUT C ID CMT DAT=DROP DATE TIME DV AMT MDV EVID SS II
$DATA POPPK_129V2.CSV IGNORE=C
$SUBROUTINES ADVAN6 TOL=3
$MODEL
COMP=(1) ;(DEPOT1, DEFDOSE)
COMP=(2) ;(DEPOT2)
COMP=(3) ;(CENTRAL,DEFOBS)
$PK
TVCL=THETA(1)
CL=TVCL*EXP(ETA(1))
TVV=THETA(2)
V=TVV*EXP(ETA(2))
TVKA1=THETA(3)
KA1=TVKA1*EXP(ETA(3))
TVKA2=THETA(4)
KA2=TVKA2*EXP(ETA(4))
TVF1=THETA(5)
F1=TVF1
TVLT1=THETA(6)
ALAG1=TVLT1*EXP(ETA(5))
TVLT2=THETA(7)
ALAG2=TVLT2*EXP(ETA(6))
K30=CL/V
IF (F1 .GT. 1) F1=0.999
IF (ALAG1 .LE.0) ALAG1=0
F2=1-F1
S3=V
$DES
RINA=KA1*A(1)
RINB=KA2*A(2)
C3=A(3)/V
DADT(1)=-RINA
DADT(2)=-RINB
DADT(3)=RINA+RINB-K30*A(3)
$ERROR
DEL=0
IF (F.EQ.0) DEL=1
IPRED=F
W=IPRED+DEL
IRES=DV-IPRED
IWRES=IRES/W
Y=F+W*ERR(1)+ERR(2)
$THETA
11000 ;[CL]
21000 ;[TVV]
1 ;[TVKA1]
0.5 ;[TVKA2]
0.25 FIXED ;[TVF1]
0.5 ;[TVLT1]
2.5 ;[TVLT2]
$OMEGA
0 FIXED ;[ETACL]
0 FIXED ;[ETAV]
0 FIXED ;[ETAKA1]
0 FIXED ;[ETAKA2]
0 FIXED ;[ETALT1]
0 FIXED ;[ETALT2]
$SIGMA
0 FIXED ;[ERR1]
0 FIXED ;[ERR2]
;$MSFI
;$EST MAXEVAL=9999 NSIG=3 MSF=1.msf PRINT=5 NOABORT METHOD=0 POSTHOC
;$COVARIANCE
$SIMULATION ONLYSIM (9215690)
$TABLE ID DATE DOSE TIME IPRED CL V KA1 KA2 F1
ALAG2 NOPRINT ONEHEADER FILE=1.TAB
;$TABLE ID CL VC Q VP KA IC50 NOPRINT ONEHEADER FILE=patab43
;$TABLE ID AGE CRCL WT ALB NOPRINT ONEHEADER FILE=cotab1
;$TABLE ID SEX NOPRINT ONEHEADER FILE=catab40
;$TABLE ID TIME IPRED NOPRINT ONEHEADER FILE=sdtab43
INPUT File:
Many Thank,
Majid Vakily, Ph.D.
Senior Research Investigator
Department of Drug Metabolism & Pharmacokinetics
Phone: (847) 582-2198
Fax; (847) 582-2388
The information contained in this communication is confidential and may
constitute non-public and/or "inside" information. It is intended only for the
use of the addressee and is the property of TAP Pharmaceutical Products Inc.
Unauthorized use, disclosure, or copying of this communication, or any part
thereof, is strictly prohibited and may be unlawful. If you received this
communication in error, please notify me immediately by return e-mail and
destroy this communication and all copies thereof, including all attachments.
ATT3923554.gif
Description:
ATT3923554.gif