- On 26 Nov 1999 at 10:36:57, Stephen Day (shday.-at-.yahoo.com) sent the message

Back to the Top

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

_______________

Do You Yahoo!?

Thousands of Stores. Millions of Products. All in one place.

Yahoo! Shopping: http://shopping.yahoo.com - On 26 Nov 1999 at 14:45:20, David_Bourne (david.-a-.boomer.org) sent the message

Back to the Top

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

******************************************************************** - On 27 Nov 1999 at 15:30:21, ml11439.-a-.goodnet.com (Michael J. Leibold) sent the message

Back to the Top

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 - On 27 Nov 1999 at 22:08:50, Walter Wolf (wwolfw.-at-.hsc.usc.edu) sent the message

Back to the Top

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 |

====================================================================== - On 28 Nov 1999 at 23:17:07, "Edward F. O'Connor" (efoconnor.aaa.snet.net) sent the message

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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? - On 29 Nov 1999 at 19:30:56, David_Bourne (david.aaa.boomer.org) sent the message

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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 - On 30 Nov 1999 at 10:11:57, "Meyer Katzper 301-827-2514 FAX 301-827-2531" (KATZPER.aaa.cder.fda.gov) sent the message

Back to the Top

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 - On 30 Nov 1999 at 22:54:12, Roger Jelliffe (jelliffe.at.usc.edu) sent the message

Back to the Top

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

****************** - On 1 Dec 1999 at 21:57:43, "Joan Korth-Bradley" (KorthBJ.aaa.war.wyeth.com) sent the message

Back to the Top

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 - On 2 Dec 1999 at 21:42:07, David_Bourne (david.at.boomer.org) sent the message

Back to the Top

[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 - On 3 Dec 1999 at 21:05:57, Sander Vinks (A.A.T.M.M.Vinks.aaa.caiw.nl) sent the message

Back to the Top

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

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