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Hello,
I have a basic question about reporting an average
half life for animal data and giving appropriate
statistics.
For example: I can determine the first order
elimination rate constant (k) from the slope of plots
of ln(concentration) versus time. Then calculate the
half life, t1/2 = ln(2)/k. To get the following data:
rat1 k=0.22 t1/2 = 3.15
rat2 k=0.25 t1/2 = 2.77
rat3 k=0.32 t1/2 = 2.17
rat4 k=0.40 t1/2 = 1.73
k(average) = 0.2975
t1/2(average) = 2.46
Everything looks ok so far but if I calculate the t1/2
based on the k(average) I get a different value:
t1/2 = ln(2)/k(average) = 2.33
So what is right?
Also, what is the best way to give an indication of
variability (or confidence)in the "average" t1/2
value, whichever it may be?
Thanks,
Stephen Day
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Date: Fri, 26 Nov 1999 11:12:35 -0700
From: Stacey Tannenbaum
Organization: College of Pharmacy
To: PharmPK.-at-.boomer.org
Subject: Re: PharmPK Statistical treatment of PK data
Hi Stephen, you should probably be using harmonic mean to calculate the
average t1/2. You may want to read the following articles:
Estimation of Variance for Harmonic Mean Half-Lives, by Lam et al
J Pharm Sci, Vol 74 (2), p.229, Feb 1985
Comparison of Harmonic Mean versus Arithmetic Mean Clearance Values, by
Schaff et al
J Pharm Sci, Vol 75 (4), p.427, April 1986
Averaging Pharmacokinetic Parameter Estimates from Experimental Studies:
Statistical Theory and Application, by Roe et al
J Pharm Sci, Vol 86 (5), p.621, May 1997
Hope that these help!
Stacey Tannenbaum
University of Arizona College of Pharmacy
---
Date: Fri, 26 Nov 1999 10:13:34 -0800
From: "Dr. Michael Mayersohn"
Organization: College of Pharmacy
X-Sender: "Dr. Michael Mayersohn"
To: PharmPK.aaa.boomer.org
Subject: Re: PharmPK Statistical treatment of PK data
Stephen,
Use the harmonic mean and 'pseudo' standard deviation method outlined
by Lam et al., in order to average half-lives and express variation.
J. Pharmaceutical Sciences, 74, 229 (1985)
---
X-Sender: jelliffe.aaa.hsc.usc.edu
Date: Fri, 26 Nov 1999 10:54:56 -0800
To: PharmPK.aaa.boomer.org
From: Roger Jelliffe
Subject: Re: PharmPK Statistical treatment of PK data
Dear Dr. Day:
Why don't you make a population model of your 4 rats? Then
you will have a
real model, with both parameters, and you will be in a much better position
to say something about what you have.
Very best regards,
Roger Jelliffe
Roger W. Jelliffe, M.D. Professor of Medicine, USC
USC Laboratory of Applied Pharmacokinetics
2250 Alcazar St, Los Angeles CA 90033, USA
Phone (323)442-1300, fax (323)442-1302, email= jelliffe.-at-.hsc.usc.edu
Our web site= http://www.usc.edu/hsc/lab_apk
********************************************************************
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Hello Stephen,
Statistics is concerned with the analysis of observations, such
that the average or "mean" T1/2 would be the arithmetic average of the
individual observations:
Mean = Sum(i-n)Xi/(n)
Mean T1/2= (3.15+2.77+2.17+1.73)/4= 2.45
The T1/2 of .693/Ke(mean) would be a derived value, and would not be
the mean T1/2. This is since the T1/2 has a distribution of its own,
distinct from the Ke.
Measures of dispersion include variance and standard deviation.
Standard deviation is most commonly used, and the mean value is
frequently reported as Mean +/-SD.
Variance= S2= Sum(i-n)(Xi- Xmean)2/(n-1)
Standard deviation= S= square root of variance
S2= [(3.15-2.45)2 + (2.77-2.45)2 + (2.17-2.45)2 + (1.73-2.45)2/ (3)
= 0.3964000
S= 0.62960305
So, the mean+/-SD for your data is: 2.45 +/- 0.62960305
The 95% confidence interval is Mean +/- 2SD which is for your study:
2.45 +/- 1.25920610
That is, there is a 95% probability that a T1/2 measured in this
population will lie within the above confidence interval.
This is not to be confused with confidence interval for the Mean
T1/2 of your study, which involves the standard error of mean.
The standard error of mean= S/square root(n)= 0.62960305/(2)=
0.31480152
The 95% confidence interval for the population mean is:
Mean +/- t(.975)SEm
That is, the sample mean +/- the t value for 95% times the standard
error of mean:
2.45 +/- (3.1825)(.314801520)= 2.45+/- 1.00185585
This means that there is a 95% probability that the population
mean lies within the above interval.
The coefficient of variation is useful for comparing the results
obtained by different investigators involving the same variable:
C.V.= [S/(Xmean)](100)
C.V.= [0.62960305/(2.45)](100)= 25.698082%
Such that the variance may be greater in another investigator's
study relative to yours if the reported CV is >25.7%.
In comparing the variance in a measured variable such as T1/2 among
different studies, it is important to distinguish between standard error
of mean, and standard devation. This is since the amount of variance in
data can be underestimated if the standard error of mean is reported, rather
than the standard deviation. As can be seen with the above example, the
standard error is smaller than the standard deviation.
Other caveats lie in the assumption of a normal distribution. Some kinetic
parameters are log normally distributed where the log of the variable is
the statistical variable being studied. Sometimes nonparametric statistics
are used, where the assumption of normal distribution is not necessary.
Good luck with your study!
Mike Leibold, PharmD, RPh
ML11439.-at-.goodnet.com
References:
1) Daniel, W.W., Biostatisitics, Wiley Interscience, New York 1979
2) Glantz, S.A., Primer of Biostatistics 4th edition, McGraw-Hill,
New York 1997
3) Martin, H.F. et al, Normal Values in Clinical Chemistry, Marcel Dekker,
New York 1975
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Stephen:
I have read with interest some of the comments that have come back to your
question on the differences in the elimination rate constants you have
observed in 4 different animals (rats? mice?).
Let me suggest a different viewpoint.
You have reported data on the elimination rate constant of a product (not
specified) in animals (again, species, age, sex not specified).
You then asked how to treat those data statistically for that group. What I
see are 4 individuals, and before I am willing to consider them a group, what
are their physiological characteristics? Are these animals all of the same
weight, age, sex; if elimination is renal, do they all have comparable renal
function? If elimination is hepatic, have they all be dosed at a comparable
time in their feeding cycle? Hence, are they a true group, or 4 individual
animals sharing only some limited characteristics? In a word, does it make
sense, from a pathophysiological viewpoint, to treat them as a group where
comparisons make sense?
In a study we published many years ago (Noninvasive Estimation of Bound and
Mobile Platinum Compounds in the Kidney Using a Radiopharmacokinetic Model.
R. R. Brechner, D. Z. D'Argenio, R. Dahalan and W. Wolf, J. Pharm. Sci., 53,
873-877, 1986) we document the importance of estimating parameters from
measurements in a single individual, and the effect of physiologically based
processes on interanimal variability.
Hence, my question to your data would be: is there any basis for these
significant differences in the rates of elimination measured, and is
there a pathophysiological reason for these interindividual differences,
differences that could significantly affect drug pharmacokinetics and
pharmacodynamics?
PLEASE NOTE NEW ZIP-CODE AT USC
======================================================================
| Professor Walter Wolf, Ph.D. E-Mail: wwolfw.at.hsc.usc.edu |
| Distinguished Professor of Pharmaceutical Sciences |
| Director, Pharmacokinetic Imaging Program |
| Department of Pharmaceutical Sciences, School of Pharmacy |
| University of Southern California Telephone:323-442-1405|
| 1985 Zonal Ave., Los Angeles, CA 90089-9121 Fax: 323-442-9804|
| |
|Center for Noninvasive Pharmacology, Los Angeles Oncologic Institute|
| MRI at St. Vincent Medical Center Telephone: 213-484-7235 |
| 2131 Third St., Los Angeles, CA 90057 Fax: 213-484-7447 |
======================================================================
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The precision you list is greater for the t1/2 than for the k. if you
run out the k to 3 sig figs, is the difference less?
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Date: Mon, 29 Nov 1999 01:32:53 -0700 (MST)
X-Sender: ml11439.-at-.pop.goodnet.com
To: PharmPK.-at-.boomer.org
From: ml11439.-at-.goodnet.com (Michael J. Leibold)
Subject: Re: PharmPK Re: Statistical treatment of PK data
Hello Edward,
Data:
rat1 k=0.22 t1/2 = 3.15
rat2 k=0.25 t1/2 = 2.77
rat3 k=0.32 t1/2 = 2.17
rat4 k=0.40 t1/2 = 1.73
T1/2 Data:
Mean = Sum(i-n)Xi/(n)
Mean T1/2= (3.15+2.77+2.17+1.73)/4= 2.455
Mean Ke= (.22+.25+.32+.40)/4= .2975
Variance= S2= Sum(i-n)(Xi- Xmean)2/(n-1)
Standard deviation= S= square root of variance
S2= [(.22-.2975)2 + (.25-.2975)2 + (.32-.2975)2 + (.4-.2975)2/ (3)
= .006425
S= 0.08015610
So, the mean+/-SD for your data is: .2975 +/-.08015610
The 95% confidence interval is Mean +/- 2SD which is for your study:
.2975 +/- .16031220
The standard error of mean= S/square root(n)= 0.0801561/(2)=
0.04007805
The 95% confidence interval for the population mean is:
Mean +/- t(.975)SEm
That is, the sample mean +/- the t value for 95% times the standard
error of mean:
.2975 +/- (3.1825)(.04007805)= .2975+/- .12754839
The coefficient of variation is useful for comparing the results
obtained by different investigators involving the same variable:
C.V.= [S/(Xmean)](100)
C.V.= [0.0801561/(.2975)](100)= 26.943227%
Ke Data:
Mean = Sum(i-n)Xi/(n)
Mean T1/2= (3.15+2.77+2.17+1.73)/4= 2.455
Variance= S2= Sum(i-n)(Xi- Xmean)2/(n-1)
Standard deviation= S= square root of variance
S2= [(3.15-2.455)2 + (2.77-2.455)2 + (2.17-2.455)2 + (1.73-2.455)2/ (3)
= 0.39636667
S= 0.62957658
So, the mean+/-SD for your data is: 2.455 +/- 0.62957658
The 95% confidence interval is Mean +/- 2SD which is for your study:
2.455 +/- 1.25915316
The standard error of mean= S/square root(n)= 0.62957658(2)=
0.31478829
The 95% confidence interval for the population mean is:
Mean +/- t(.975)SEm
That is, the sample mean +/- the t value for 95% times the standard
error of mean:
2.455 +/- (3.1825)(.31478829)= 2.455+/- 1.00181373
C.V.= [S/(Xmean)](100)
C.V.= [0.62957658/(2.455)](100)= 25.644667%
Summary:
The CV is slightly less for the T1/2 data, suggesting less
variance in the data. This may really reflect computational aspects
of the much smaller numbers involved in the Ke calculations.
Mike Leibold, PharmD, RPh
ML11439.aaa.goodnet.com
References:
1) Daniel, W.W., Biostatisitics, Wiley Interscience, New York 1979
2) Glantz, S.A., Primer of Biostatistics 4th edition, McGraw-Hill,
New York 1997
3) Martin, H.F. et al, Normal Values in Clinical Chemistry, Marcel Dekker,
New York 1975
---
Date: Mon, 29 Nov 1999 05:51:52 -0800 (PST)
From: Stephen Day
Subject: Re: PharmPK Re: Statistical treatment of PK data
To: PharmPK.at.boomer.org,
Multiple recipients of PharmPK - Sent by
One value of t1/2 is the arithmetic mean, the other is
the harmonic mean (the one calculated from ke1). There
is a difference, regardless of sig figs.
Stephen Day
---
Date: Mon, 29 Nov 1999 07:31:42 -0800 (PST)
From: Stephen Day
Subject: Re: PharmPK Re: Statistical treatment of PK data
To: PharmPK.aaa.boomer.org,
Multiple recipients of PharmPK - Sent by
Hello,
Thank you for your comments.
Actually, the values I used were fictional, just to
illustrate a point. I am working in drug discovery
were relatively little (nothing) is known about the
route of elimination prior to dosing. Most of the time
the data is, thankfully, not as variable as the
example shown (hence the method of calculating t1/2
doesn't matter much). Occasionally, however, there is
considerable variation. As your comments suggest, this
may be due to some physiological processes that is
not well controlled. I think that, even in this
situation, an attempt to find a t1/2 representative of
the popuplation still makes sence (thus treating the
tested animals as a group). I agree that, in the case
of variability caused by a pathophysiological
condition, finding reperesentative t1/2 is not very
useful.
Stephen Day
Merck-Frosst Centre for Therapeutic Research
Kirkland, QC CANADA
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Stephen Day's basic question is best addressed by looking at the the underlying
pharmacokinetic derivation of the halflife relationship. For a one compartment
model concentration = (Initial concentration)*exp(-k*t).
Setting concentration=(Initial concentration)/2 and solving for the time yields
t1/2=ln(2)/k .
If it is meaningful to take an average of k over a number of individual people
or animals then the average half live must be taken as
t1/2(average=ln(2)/k(average)
otherwise averaging over the individual t1/2 values is averaging over a
nonlinear transformation of k which (as he discovered for himself) will not
equal t1/2(average).
If this pseudo-average is plugged back into the original equation for
concentration it will not yield (Initial concentration)/2. Generously we might
call this a non-consistent estimator. The stronger term is false.
Unfortunately it is very easy and natural to carry out a naiive
average in cases
where such an operation is not permissible. Similar problems are
found hidden in
complex calcultions (an issue I plan to address at the forthcoming
Simulation in
the Health Sciences conference.)
Meyer Katzper
katzper.-at-.cder.fda.gov
Addendum to Stephen Day's averaging question.
The harmonic mean provides the correct answer as it essentially derives the
value of k(average) and so is equivalent to ln(2)/k(average).
katzper.-at-.cder.fda.gov
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About the issue of the effective half time and the business of getting the
average or whatever:
I have seen a lot of ideas expressed about this now. My
question is - what
will you DO with the information? Why do we want to find a single number
that best describes the distribution anyway? I think that the reason we
make models is to act most intelligently on the results found, and to me
that means developint the best dosage regimen, and that means the regimen
that will achieve a desired target goal with the greatest precision, or one
which maximizes the probability of the concentrations being within a
desired window, for example. The separation principle [1] states that
whenever one seeks to control a system and separates that process into the
2 phases, of:
1. Getting single point estimates for the parameters in the
model, and then,
2. Using these estimates to control the system,
that the job is usually done supoptimally. It is simply assumed that the
target goal of the dosage regimen will be achieved exactly. This is
apparently what makes the control suboptimal. There is no way to evaluate
the precision with which the target goal is predicted to be achieved.
However, this is the approach generally taken by the PK community, as
witness the current discussion.
However, Multiple Model dosage design [2] gets around the
problems posed
by the separation principle, and specifically permits prediction of the
precision with which a given regimen will achieve a desired target goal.
Instead of using a single value for each PK parameter, it uses many values.
For example, it could, if one wished, use the entire collection of the MAP
Bayesian posterion parameter values of the subjects in a population.
However, the most likely set of parameter values, given the raw population
data, is the nonparametric joint parameter density as given with the NPML
method of Mallet [3] or the NPEM method of Schumitzky [4]. This gives
essentially one set of parameter values for each subject in the population,
and an estimate of the probability of each parameter set. Thus there are
multiple parameter sets in the population model, essentially one for each
subject stidied in the past. This constitutes the most likely solution to
the population analysis problem. It provides the richest, and most likely,
Bayesian prior to use for the design of the initial dosage regimen for the
next patient.
Instead of having only one set of parameter values, there are now many
with the NP models. Instead of finding the regimen to achieve the desired
target goal exactly, as one must do when there is only 1 model (impossible
when there is any diversity at all in the population), a candidate regimen
can be given to each of the models in the nonparametric joint density.
Multiple predictions can be made of the concentrations calculated to be
present at a desired time, and their distance from the desired garget goal
at that time can be found. One thus can compute the weighted squared error
of the failure to achieve the desired goal. This can then be worked on
further, and the regimen can be found which specifically minimizes that
error. In this way, we can now design dosage regimens that spacifically
achieve a desired goal with the greatest possible precision [2]. Similar
approaches have been used to maximize the probability of being within a
desired window [5].
Why don't we use parametric population modeling methods to
take advantage
of the real strengths of each? Let's use parametric methods to
separate inter from intra individual variability, (and from that of the
assay error itself), and then put that information into nonparametric
methods to get the final entire discrete joint parameter density? Then we
can develop dosage regimens that are maximally precise. This, I think, is
what is really important, not discussing what kind of a mean is the best
estimator of a density that is usually not normal, not lognormal, but
somethine else anyway, and which is not infrequently multimodal, so that we
then can DO something maximally useful with the population raw data.
1. Bertzekas D: Dynamic Programming: Deterministic and Stochastic Models
Englewood Cliffs NJ: Prentice-Hall, 1987, pp 144-146.
2. Jelliffe R, Schumitzky A, Bayard D, Milman M, Van Guilder M, Wang X,
Jiang F, Barbaut X, and Maire P: Model-Based, Goal-Oriented, Individualised
Drug Therapy - Linkage of Population Modelling, New Multiple Model Dosage
Design, Bayesian Feeback, and Individualized Target Goals.
Clin.Pharmacokinet. 1998, 34: 57-77, 1998.
3. Mallet A: A Maximum Likelihood Estimation Method for Random Coefficiant
Regression Models. Biometrika 73: 645-656, 1986.
4. Schumitzky A: Nonparametric EM Algorithms for Estimating Prior
Distributions. App. Math. Comput. 45: 143-157, 1991.
5. Taright N, Mentre F, Mallet A, and Jouvent R: Nonparametric Estimation
of Population characteristics of the Kinetics of Lithium from
Observational and Experimental Data: Individualization of Chromic Dosing
Regimen using a new Bayesian approach. Ther. Drug Monit. 16: 258-269, 1994.
Very warmly to all,
Roger Jelliffe
Roger W. Jelliffe, M.D. Professor of Medicine, USC
USC Laboratory of Applied Pharmacokinetics
2250 Alcazar St, Los Angeles CA 90033, USA
Phone (323)442-1300, fax (323)442-1302, email= jelliffe.aaa.hsc.usc.edu
Our web site= http://www.usc.edu/hsc/lab_apk
******************
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Roger as always provides considerable food
for thought.
How can a pkist in an industrial setting best
communicate his results so that the end user
has the information they need to optimize
drug therapy?
What would the clinical pharmacology section
of a product monograph look like?
Kind regards,
Joan K-B
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[Two replies - db]
From: "Jones, Mike"
To: "'PharmPK.at.boomer.org'"
Subject: RE: PharmPK Re: Statistical treatment of PK data
Date: Thu, 2 Dec 1999 09:20:11 -0700
This is a very good question.
I for one would like to see pharmacokinetic parameters in product monographs
linked to physiologic parameters.
For example it would be nice to see something like:
Total Drug Clearance = (Renal Drug Clearance * Creatinine Clearance) +
(Hepatic Clearance * Body Weight) + ... + Constant
Volume of distribution should have a similar relationship.
For clinical ease of use these parameters should be based one the one
compartment model.
Multi-compartment model parameters would also be useful and I would like to
see those as well, but the minimum set for clinical use should be the one
compartment model parameters.
Thanks,
Mike Jones, PharmD
Micromedex, Inc.
6200 S. Syracuse Way, Suite 300
Englewood, CO 80111-4740
(303) 486-6723
(800) 525-9083 ext. 6723
mike.jones.aaa.mdx.com
---
Date: Thu, 02 Dec 1999 11:59:01 -0500
From: "Bryan Facca"
To:
Subject: PharmPK Re: Statistical treatment of PK data
Joan,
my suggestion would be to work with the drug info people at your
particular setting. If the PK info is with a product, you also might
try to include it with the dosing guidelines for the particular
product. If its general info on P-K, set your target for the
audience; ie: pharmacist, physicians and publish in respective
journals for those groups.
Bryan Facca RPh PharmD
Metropolitan Hospital Grand Rapids, MI
faccabf.-a-.metrogr.org
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A very good question! For those working in TDM initial parameter estimates
of different patient groups can be extremely useful. At present these
values are very difficult to find in the literature. What we need is
information on the PK constants defining the model: i.e. for a 2
compartment model estimates of Vc, K12, K21 ans K10 [+ dispersion
factors]. Preferably, relationships with relevant patient parameters need
to be reported as well: e.g. Vc as a function of body weight, K10 as a
function of creatinine clearance (or clearance parameters during
intermittent and continuous hemodialysis), cardiac output etc. With these
initial estimates candidate dosage regimes can be simulated. Subsequently,
the model can be individualized based on feed-back such as measured serum
concentrations. This is the way we currently perform TDM for a small number
of drugs intelligently. We need to expand this type of clinically very
useful dosage optimization for many more drugs in patient populations with
changing PK (and PD). For instance in the setting of the ICU, oncology and
HIV treatment.
best regards,
Sander Vinks
Alexander A.T.M.M. Vinks, Pharm.D., Ph.D.
Director, TDM & Clinical Toxixology Laboratory
The Hague Central Hospital Pharmacy
P.O. Box 43100
NL-2504 AC The Hague, The Netherlands
Tel: +31-70-3217-217
Fax: +31-70-3217-156
email: vinks.aaa.caiw.nl
PharmPK Discussion List Archive Index page
Copyright 1995-2010 David W. A. Bourne (david@boomer.org)