# PharmPK Discussion - Statistical treatment of PK data

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• On 26 Nov 1999 at 10:36:57, Stephen Day (shday.-at-.yahoo.com) sent the message
`Hello,I have a basic question about reporting an averagehalf life for animal data and giving appropriatestatistics.For example: I can determine the first orderelimination rate constant (k) from the slope of plotsof ln(concentration) versus time. Then calculate thehalf life, t1/2 = ln(2)/k. To get the following data:rat1  k=0.22  t1/2 = 3.15rat2  k=0.25  t1/2 = 2.77rat3  k=0.32  t1/2 = 2.17rat4  k=0.40  t1/2 = 1.73k(average) = 0.2975t1/2(average) = 2.46Everything looks ok so far but if I calculate the t1/2based on the k(average) I get a different value:t1/2 = ln(2)/k(average) = 2.33So what is right?Also, what is the best way to give an indication ofvariability (or confidence)in the "average" t1/2value, 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`
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• On 26 Nov 1999 at 14:45:20, David_Bourne (david.-a-.boomer.org) sent the message
`Date: Fri, 26 Nov 1999 11:12:35 -0700From: Stacey Tannenbaum Organization: College of PharmacyTo: PharmPK.-at-.boomer.orgSubject: Re: PharmPK Statistical treatment of PK dataHi Stephen, you should probably be using harmonic mean to calculate theaverage t1/2.  You may want to read the following articles:Estimation of Variance for Harmonic Mean Half-Lives, by Lam et alJ Pharm Sci, Vol 74 (2), p.229, Feb 1985Comparison of Harmonic Mean versus Arithmetic Mean Clearance Values, bySchaff et alJ Pharm Sci, Vol 75 (4), p.427, April 1986Averaging Pharmacokinetic Parameter Estimates from Experimental Studies:Statistical Theory and Application, by Roe et alJ Pharm Sci, Vol 86 (5), p.621, May 1997Hope that these help!Stacey TannenbaumUniversity of Arizona College of Pharmacy---Date: Fri, 26 Nov 1999 10:13:34 -0800From: "Dr. Michael Mayersohn" Organization: College of PharmacyX-Sender: "Dr. Michael Mayersohn" To: PharmPK.aaa.boomer.orgSubject: Re: PharmPK Statistical treatment of PK dataStephen,  Use the harmonic mean and 'pseudo' standard deviation method outlinedby Lam et al., in order to average half-lives and express variation.J. Pharmaceutical Sciences, 74, 229 (1985)---X-Sender: jelliffe.aaa.hsc.usc.eduDate: Fri, 26 Nov 1999 10:54:56 -0800To: PharmPK.aaa.boomer.orgFrom: Roger Jelliffe Subject: Re: PharmPK Statistical treatment of PK dataDear Dr. Day:	Why don't you make a population model of your 4 rats? Thenyou will have areal model, with both parameters, and you will be in a much better positionto say something about what you have.Very best regards,Roger JelliffeRoger W. Jelliffe, M.D. Professor of Medicine, USCUSC Laboratory of Applied Pharmacokinetics2250 Alcazar St, Los Angeles CA 90033, USAPhone (323)442-1300, fax (323)442-1302, email=  jelliffe.-at-.hsc.usc.eduOur web site=  http://www.usc.edu/hsc/lab_apk********************************************************************`
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• On 27 Nov 1999 at 15:30:21, ml11439.-a-.goodnet.com (Michael J. Leibold) sent the message
`Hello Stephen,     Statistics is concerned with the analysis of observations, suchthat the average or "mean" T1/2 would be the arithmetic average of theindividual 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 bethe 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 isfrequently 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 thispopulation will lie within the above confidence interval.      This is not to be confused with confidence interval for the MeanT1/2 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 standarderror of mean:     2.45 +/- (3.1825)(.314801520)= 2.45+/- 1.00185585     This means that there is a 95% probability that the populationmean lies within the above interval.     The coefficient of variation is useful for comparing the resultsobtained 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'sstudy relative to yours if the reported CV is >25.7%.     In comparing the variance in a measured variable such as T1/2 amongdifferent studies, it is important to distinguish between standard errorof mean, and standard devation. This is since the amount of variance indata can be underestimated if the standard error of mean is reported, ratherthan the standard deviation. As can be seen with the above example, thestandard error is smaller than the standard deviation.    Other caveats lie in the assumption of a normal distribution. Some kineticparameters are log normally distributed where the log of the variable isthe statistical variable being studied. Sometimes nonparametric statisticsare used, where the assumption of normal distribution is not necessary.     Good luck with your study!          Mike Leibold, PharmD, RPh          ML11439.-at-.goodnet.comReferences:1) Daniel, W.W., Biostatisitics, Wiley Interscience, New York 19792) Glantz, S.A., Primer of Biostatistics 4th edition, McGraw-Hill,     New York 19973) Martin, H.F. et al, Normal Values in Clinical Chemistry, Marcel Dekker,    New York 1975`
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• On 27 Nov 1999 at 22:08:50, Walter Wolf (wwolfw.-at-.hsc.usc.edu) sent the message
`Stephen:I have read with interest some of the comments that have come back to yourquestion on the differences in the elimination rate constants you haveobserved in 4 different animals (rats? mice?).Let me suggest a different viewpoint.You have reported data on the elimination rate constant of a product (notspecified) in animals (again, species, age, sex not specified).You then asked how to treat those data statistically for that group.  What Isee are 4 individuals, and before I am willing to consider them a group, whatare their physiological characteristics? Are these animals all of the sameweight, age, sex; if elimination is renal, do they all have comparable renalfunction? If elimination is hepatic, have they all be dosed at a comparabletime in their feeding cycle? Hence, are they a true group, or 4 individualanimals sharing only some limited characteristics? In a word, does it makesense, from a pathophysiological viewpoint, to treat them as a group wherecomparisons make sense?In a study we published many years ago (Noninvasive Estimation of Bound andMobile 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 frommeasurements in a single individual, and the effect of physiologically basedprocesses on interanimal variability.Hence, my question to your data would be: is there any basis for thesesignificant differences in the rates of elimination measured, and isthere a pathophysiological reason for these interindividual differences,differences that could significantly affect drug pharmacokinetics andpharmacodynamics?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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• On 28 Nov 1999 at 23:17:07, "Edward F. O'Connor" (efoconnor.aaa.snet.net) sent the message
`The precision you list is greater for the t1/2 than for the k.   if yourun out the k to 3 sig figs, is the difference less?`
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• On 29 Nov 1999 at 19:30:56, David_Bourne (david.aaa.boomer.org) sent the message
`Date: Mon, 29 Nov 1999 01:32:53 -0700 (MST)X-Sender: ml11439.-at-.pop.goodnet.comTo: PharmPK.-at-.boomer.orgFrom: ml11439.-at-.goodnet.com (Michael J. Leibold)Subject: Re: PharmPK Re: Statistical treatment of PK dataHello 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 = 1.73T1/2 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 standarderror of mean:     .2975 +/- (3.1825)(.04007805)= .2975+/- .12754839      The coefficient of variation is useful for comparing the resultsobtained 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 standarderror 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 lessvariance in the data. This may really reflect computational aspectsof the much smaller numbers involved in the Ke calculations.          Mike Leibold, PharmD, RPh          ML11439.aaa.goodnet.comReferences:1) Daniel, W.W., Biostatisitics, Wiley Interscience, New York 19792) Glantz, S.A., Primer of Biostatistics 4th edition, McGraw-Hill,     New York 19973) 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 dataTo: PharmPK.at.boomer.org,   Multiple recipients of PharmPK - Sent by One value of t1/2 is the arithmetic mean, the other isthe harmonic mean (the one calculated from ke1). Thereis 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 dataTo: PharmPK.aaa.boomer.org,   Multiple recipients of PharmPK - Sent by Hello,Thank you for your comments.Actually, the values I used were fictional, just toillustrate a point. I am working in drug discoverywere relatively little (nothing) is known about theroute of elimination prior to dosing. Most of the timethe data is, thankfully, not as variable as theexample shown (hence the method of calculating t1/2doesn't matter much). Occasionally, however, there isconsiderable variation. As your comments suggest, thismay  be due to some physiological processes that isnot well controlled. I think that, even in thissituation, an attempt to find a t1/2 representative ofthe popuplation still makes sence (thus treating thetested animals as a group). I agree that, in the caseof variability caused by a pathophysiologicalcondition, finding reperesentative t1/2 is not veryuseful.Stephen DayMerck-Frosst Centre for Therapeutic ResearchKirkland, QC    CANADA`
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• 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
`Stephen Day's basic question is best addressed by looking at the the underlyingpharmacokinetic derivation of the halflife relationship. For a one compartmentmodel concentration = (Initial concentration)*exp(-k*t).Setting concentration=(Initial concentration)/2 and solving for the time yieldst1/2=ln(2)/k .If it is meaningful to take an average of k over a number of individual peopleor animals then the average half live must be taken ast1/2(average=ln(2)/k(average)otherwise averaging over the individual t1/2 values is averaging over anonlinear transformation of k  which (as he discovered for himself) will notequal t1/2(average).If this pseudo-average is plugged back into the original equation forconcentration it will not yield (Initial concentration)/2. Generously we mightcall this a non-consistent estimator. The stronger term is false.Unfortunately it is very easy and natural to carry out a naiiveaverage in caseswhere such an operation is not permissible. Similar problems arefound hidden incomplex calcultions (an issue I plan to address at the forthcomingSimulation inthe Health Sciences conference.)Meyer Katzperkatzper.-at-.cder.fda.govAddendum to Stephen Day's averaging question.The harmonic mean provides the correct answer as it essentially derives thevalue of k(average) and so is equivalent to ln(2)/k(average).katzper.-at-.cder.fda.gov`
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• On 30 Nov 1999 at 22:54:12, Roger Jelliffe (jelliffe.at.usc.edu) sent the message
`About the issue of the effective half time and the business of getting theaverage or whatever:	I have seen a lot of ideas expressed about this now. Myquestion is - whatwill you DO with the information? Why do we want to find a single numberthat best describes the distribution anyway? I think that the reason wemake models is to act most intelligently on the results found, and to methat means developint the best dosage regimen, and that means the regimenthat will achieve a desired target goal with the greatest precision, or onewhich maximizes the probability of the concentrations being within adesired window, for example. The separation principle [1] states thatwhenever one seeks to control a system and separates that process into the2 phases, of:1.	Getting single point estimates for the parameters in themodel, and then,2.	Using these estimates to control the system,that the job is usually done supoptimally. It is simply assumed that thetarget goal of the dosage regimen will be achieved exactly. This isapparently what makes the control suboptimal. There is no way to evaluatethe precision with which the target goal is predicted to be achieved.However, this is the approach generally taken by the PK community, aswitness the current discussion.	However, Multiple Model dosage design [2] gets around theproblems posedby the separation principle, and specifically permits prediction of theprecision 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 MAPBayesian posterion parameter values of the subjects in a population.However, the most likely set of parameter values, given the raw populationdata, is the nonparametric joint parameter density as given with the NPMLmethod of Mallet [3] or the NPEM method of Schumitzky [4]. This givesessentially one set of parameter values for each subject in the population,and an estimate of the probability of each parameter set. Thus there aremultiple parameter sets in the population model, essentially one for eachsubject stidied in the past. This constitutes the most likely solution tothe population analysis problem. It provides the richest, and most likely,Bayesian prior to use for the design of the initial dosage regimen for thenext patient.	Instead of having only one set of parameter values, there are now manywith the NP models. Instead of finding the regimen to achieve the desiredtarget goal exactly, as one must do when there is only 1 model (impossiblewhen there is any diversity at all in the population), a candidate regimencan be given to each of the models in the nonparametric joint density.Multiple predictions can be made of the concentrations calculated to bepresent at a desired time, and their distance from the desired garget goalat that time can be found. One thus can compute the weighted squared errorof the failure to achieve the desired goal. This can then be worked onfurther, and the regimen can be found which specifically minimizes thaterror. In this way, we can now design dosage regimens that spacificallyachieve a desired goal with the greatest possible precision [2]. Similarapproaches have been used to maximize the probability of being within adesired window [5].	Why don't we use parametric population modeling methods totake advantageof the real strengths of each?  Let's use parametric methods             toseparate inter from intra individual variability, (and from that of theassay error itself), and then put that information into  nonparametricmethods to get the final entire discrete joint parameter density? Then wecan develop dosage regimens that are maximally precise. This, I think, iswhat is really important, not discussing what kind of a mean is the bestestimator of a density that is usually not normal, not lognormal, butsomethine else anyway, and which is not infrequently multimodal, so that wethen can DO something maximally useful with the population raw data.1.	Bertzekas D: Dynamic Programming: Deterministic and Stochastic ModelsEnglewood 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, IndividualisedDrug Therapy - Linkage of Population Modelling, New Multiple Model DosageDesign, Bayesian Feeback, and Individualized Target Goals.Clin.Pharmacokinet. 1998, 34: 57-77, 1998.3.	Mallet A: A Maximum Likelihood Estimation Method for Random CoefficiantRegression Models. Biometrika 73: 645-656, 1986.4.	Schumitzky A: Nonparametric EM Algorithms for Estimating PriorDistributions. App. Math. Comput. 45: 143-157, 1991.5.	Taright N, Mentre F, Mallet A, and Jouvent R: Nonparametric Estimationof Population characteristics of the Kinetics of Lithium fromObservational and Experimental Data: Individualization of Chromic DosingRegimen using a new Bayesian approach. Ther. Drug Monit. 16: 258-269, 1994.Very warmly to all,Roger JelliffeRoger W. Jelliffe, M.D. Professor of Medicine, USCUSC Laboratory of Applied Pharmacokinetics2250 Alcazar St, Los Angeles CA 90033, USAPhone (323)442-1300, fax (323)442-1302, email=  jelliffe.aaa.hsc.usc.eduOur web site=  http://www.usc.edu/hsc/lab_apk******************`
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• On 1 Dec 1999 at 21:57:43, "Joan Korth-Bradley" (KorthBJ.aaa.war.wyeth.com) sent the message
`Roger as always provides considerable foodfor thought.How can a pkist in an industrial setting bestcommunicate his results so that the end userhas the information they need to optimizedrug therapy?What would the clinical pharmacology sectionof a product monograph look like?Kind regards,Joan K-B`
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• On 2 Dec 1999 at 21:42:07, David_Bourne (david.at.boomer.org) sent the message
`[Two replies - db]From: "Jones, Mike" To: "'PharmPK.at.boomer.org'" Subject: RE: PharmPK Re: Statistical treatment of PK dataDate: Thu, 2 Dec 1999 09:20:11 -0700This is a very good question.I for one would like to see pharmacokinetic parameters in product monographslinked 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) + ... + ConstantVolume of distribution should have a similar relationship.For clinical ease of use these parameters should be based one the onecompartment model.Multi-compartment model parameters would also be useful and I would like tosee those as well, but the minimum set for clinical use should be the onecompartment model parameters.Thanks,Mike Jones, PharmDMicromedex, Inc.6200 S. Syracuse Way, Suite 300Englewood, CO 80111-4740(303) 486-6723(800) 525-9083 ext. 6723mike.jones.aaa.mdx.com---Date: Thu, 02 Dec 1999 11:59:01 -0500From: "Bryan Facca" To: Subject: PharmPK Re: Statistical treatment of PK dataJoan,my suggestion would be to work with the drug info people at yourparticular setting.  If the PK info is with a product, you also mighttry to include it with the dosing guidelines for the particularproduct.  If its general info on P-K, set your target for theaudience; ie: pharmacist, physicians and publish in respectivejournals for those groups.Bryan Facca RPh PharmDMetropolitan Hospital Grand Rapids, MIfaccabf.-a-.metrogr.org`
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• On 3 Dec 1999 at 21:05:57, Sander Vinks (A.A.T.M.M.Vinks.aaa.caiw.nl) sent the message
`A very good question! For those working in TDM initial parameter estimatesof different patient groups can be extremely useful. At present thesevalues are very difficult to find in the literature. What we need isinformation on the PK constants defining the model: i.e. for a 2compartment model estimates of Vc, K12, K21 ans K10 [+ dispersionfactors]. Preferably, relationships with relevant patient parameters needto be reported as well: e.g. Vc as a function of body weight, K10 as afunction of creatinine clearance (or clearance parameters duringintermittent and continuous hemodialysis), cardiac output etc. With theseinitial estimates candidate dosage regimes can be simulated. Subsequently,the model  can be individualized based on feed-back such as measured serumconcentrations. This is the way we currently perform TDM for a small numberof drugs intelligently. We need to expand this type of clinically veryuseful dosage optimization for many more drugs in patient populations withchanging PK (and PD). For instance in the setting of the ICU, oncology andHIV treatment.best regards,Sander VinksAlexander A.T.M.M. Vinks, Pharm.D., Ph.D.Director, TDM & Clinical Toxixology LaboratoryThe Hague Central Hospital PharmacyP.O. Box 43100NL-2504 AC The Hague, The NetherlandsTel:   +31-70-3217-217Fax:   +31-70-3217-156email: vinks.aaa.caiw.nl`
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