# PharmPK Discussion - Tests of Normality

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• On 23 Aug 2003 at 20:50:33, Vlase Laurian (vlase.aaa.email.ro) sent the message
`The following message was posted to: PharmPKHi,Can anyone help me to interpret the results for these normality tests?I know, based on histograms that the distribution is not normal.Thank youVlase LaurianDept. Pharm. Technol. & BiopharmaceuticsFaculty of PharmacyCluj-NapocaRomaniaTests of Normality		Kolmogorov-Smirnov			Shapiro-Wilk			TRATAMEN   Statistic    df  Sig.  Statistic df  Sig.R_AUC   1.00    .383    18  .000    .601    18  .010         2.00    .316    18  .000    .650    18  .010R_CMAX  1.00    .231    18  .012    .775    18  .010         2.00    .344    18  .000    .695    18  .010**	This is an upper bound of the true significance.a	Lilliefors Significance Correction`
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• On 25 Aug 2003 at 23:26:45, =?ISO-8859-1?Q?"Helmut_Sch=FCtz"?= (helmut.schuetz.-at-.chello.at) sent the message
`The following message was posted to: PharmPKVlase Laurian wrote:> Hi,> Can anyone help me to interpret the results for these normality tests?> I know, based on histograms that the distribution is not normal.> Thank you> Vlase Laurian>[...]> Tests of Normality>            Kolmogorov-Smirnov       Shapiro-Wilk			> TRATAMEN   Statistic    df  Sig.  Statistic df  Sig.>> R_AUC   1.00    .383    18  .000    .601    18  .010>        2.00    .316    18  .000    .650    18  .010> R_CMAX  1.00    .231    18  .012    .775    18  .010>        2.00    .344    18  .000    .695    18  .010>> **	This is an upper bound of the true significance.> a	Lilliefors Significance Correction>Dear Vlase,the Kolmogorov-Smirnov test is poor in terms of statistical power withsample sizes generally applied in BE studies. The Shapiro-Wilk test isthe better choice. From the table you gave I guess that you applied thetests separatelly on test/reference treatments (1/2 ?).In my opinion this is not correct. One of the main assumptions of ANOVAis the normality (0,sigma) of errors, which are approximated in thesample by studentized residuals.Imagine the simple (and artificial) case of test and referencetreatments following both (the same) triangular distribution (whichclearly is not normal). If you take differences (or use the ratio...)you will end up with normally distributed residuals. And this is whatwe assumed in the model, so the ANOVA is valid.If you additionally test both treatments for normality this is fine,but what you have to test are the residuals...I would suggest the following procedure:o) lay down a statistical protocol describing what you plan to do (andadhere to that protocol!)o) run the ANOVA and calculate studentized intra- and (not soimportant) inter-subject residualso) test these residuals for normality (according to the protocol; onlyone test, otherwise you will run into problems with multiplicity andcontradictory results) and outliers (to be excluded only if you canjustify errors in the clinical and/or analytical conduct of the study)o) if non-normality of residuals is detected, use a nonparametricmethod (e.g., Wilcoxon-Mann-Whitney) instead of ANOVAGood luck!Helmut--Helmut Schütz          Biokinet GmbH / Dept BiostatisticsNeubaugasse 36/11      Nattergasse 4A-1070 Vienna/Austria  A-1170 Vienna/Austriatel/fax +43 1 9713935  tel +43 1 4856969 62no cell phone ;-)      fax +43 1 4856969 90http://www.goldmark.org/netrants/no-word/attach.html`
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• On 27 Aug 2003 at 08:50:47, kim.travis.at.syngenta.com sent the message
`The following message was posted to: PharmPKHelmut's advise is good.  A transformation of the data might be worthconsidering to help normalise the error, before resorting to anonparametricmethod,Kim`
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• On 27 Aug 2003 at 16:45:42, martin.schumacher.aaa.pharma.novartis.com sent the message
`Dear Vlase,In your recent mail you wrote:"Can anyone help me to interpret the results for these normality tests?I know, based on histograms that the distribution is not normal."Tests of Normality                Kolmogorov-Smirnov               Shapiro-WilkTRATAMEN  Statistic  df Sig. Statistic df Sig.R_AUC  1.00  .383  18 .000  .601  18 .010    2.00  .316  18 .000  .650  18 .010R_CMAX 1.00  .231  18 .012  .775  18 .010    2.00  .344  18 .000  .695  18 .010You applied 2 different statistical methods to test for normality. Theinterpretation of the results of the 2 methods is evident. Both methodsindicate with high confidence that the data of both treatments are notnormally distributed. This is what you already knew based on yourvisual inspection of the data.However, more important than applying a statistical test is a veryclear explanation of the data under consideration as well as precisestatement of the goal(s). It would be helpful to get this information.Additionally, the distribution of PK parameters like AUC and Cmax hasbeen addressed rather often in the literature and I feel that there's aconsensus that they are log-normally distributed. In this case youwould just take logarithms of the individual data before they enter anystatistical testing, without any tests regarding the distributionalform. Sequential statistical testing is a problematic task and shouldbe discouraged if possible.Best regards,MartinMartin M. Schumacher, Ph.D.Principal ScientistNovartis Pharma AGBioMarker Development (BMD)WKL-136.1.19CH-4002 BaselSwitzerlandTel. +41 - 61 696 1030 (direct)Fax  +41 - 61 696 6212Email martin.schumacher.aaa.pharma.novartis.com`
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• On 27 Aug 2003 at 22:26:02, =?ISO-8859-1?Q?"Helmut_Sch=FCtz"?= (helmut.schuetz.-at-.chello.at) sent the message
`The following message was posted to: PharmPKDear Kim, dear Martin,I think I have to be more precise. I agree that parameters like AUC andCmax are most likely log-normally distributed, so a transformationshould be the starting point.But I disagree with Martin's "... just take logarithms of theindividual data before they enter any statistical testing, without anytests regarding the distributional form." I know that this procedure(not testing for normality after transformation) is stated in therecent FDA guideline, but it's still very bad statistical practice. Wehave to test our assumptions (in ANOVA additivity of effects, normaldistribution of errors, independence of subjects, etc....) before weapply a statistical method, and if some of them are violated, we haveto take measures and not only close our eyes and continue...If we decide to go for a nonparametric method, this is not sequentialtesting, since we have not calculated the ANOVA confidence limits, andthe alpha-level is still intact.Best regards,HelmutHelmut Schütz          Biokinet GmbH / Dept BiostatisticsNeubaugasse 36/11      Nattergasse 4A-1070 Vienna/Austria  A-1170 Vienna/Austriatel/fax +43 1 9713935  tel +43 1 4856969 62no cell phone ;-)      fax +43 1 4856969 90http://www.goldmark.org/netrants/no-word/attach.html`
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• On 27 Aug 2003 at 16:57:03, "Daniel Sitar" (sitar.-at-.Ms.UManitoba.CA) sent the message
`	This suggestion is not logical, since the fundamentalassumptions of nonparametric statistics indicate that datadistribution is not part of the underlying principles of thetest(s).  Normalization of data sets only applies to the use ofparametric tests to try and achieve something resembling aGaussian distribution.   Bad data are not salvaged by statisticalcontortions.   If you keep transforming data by variousmathematical contortions, eventually you will find a way toget a statistically significant result, but the process is invalidbecause of the repeated analysis of the data withoutpartitioning the finding of a difference by chance.	The best suggestion is to redesign the experiment andto power it adequately to be able to detect a difference by astatistical approach that is prospective and scientificallyjustifiable.	Dan Sitar, University of ManitobaDaniel S. Sitar, PhDProfessor and HeadDepartment of Pharmacology and Therapeuticssitar.-a-.ms.UManitoba.caTel: 204-789-3532  FAX: 204-789-3932`
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• On 28 Aug 2003 at 09:04:29, kim.travis.aaa.syngenta.com sent the message
`The following message was posted to: PharmPKHaving suggested the possibility of a transformation I'd like tostronglyqualify it.  When you visualise the data (eg plot it) you should makesurethat you do it in units which makes most sense to you / most sensebiologically.  If you do not then your ability to judge quality of fitetcare severely compromised, and poor conclusions can result.  This oftenmeansthat you have to untransform your data in order to visualise it.People became accustomed to all kinds of transformations as a matter ofcourse in the old days, often so that a simple linear regression canyieldkey parameter estimates (pre computers).  Often this was doneirrespectiveof error distribution.  There is no excuse for this nowadays as theparameters can be fitted directly without transformation, but in manyapplications the habit dies hard.The good news is that ANOVA is remakably robust to most breaches of thekeyassumptions listed by Helmut.  But as Helmut indicates, historicalknowledgeof the likely distributional form of a variable is valuable.  In factit maybe at least as good a basis for deciding on a transformation as astatistical test of Normality, especially if sample numbers are low,Regards,Kim`
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