- On 23 Aug 2003 at 20:50:33, Vlase Laurian (vlase.aaa.email.ro) sent the message

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The following message was posted to: PharmPK

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

Dept. Pharm. Technol. & Biopharmaceutics

Faculty of Pharmacy

Cluj-Napoca

Romania

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 - On 25 Aug 2003 at 23:26:45, =?ISO-8859-1?Q?"Helmut_Sch=FCtz"?= (helmut.schuetz.-at-.chello.at) sent the message

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The following message was posted to: PharmPK

Vlase 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 with

sample sizes generally applied in BE studies. The Shapiro-Wilk test is

the better choice. From the table you gave I guess that you applied the

tests separatelly on test/reference treatments (1/2 ?).

In my opinion this is not correct. One of the main assumptions of ANOVA

is the normality (0,sigma) of errors, which are approximated in the

sample by studentized residuals.

Imagine the simple (and artificial) case of test and reference

treatments following both (the same) triangular distribution (which

clearly is not normal). If you take differences (or use the ratio...)

you will end up with normally distributed residuals. And this is what

we 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 (and

adhere to that protocol!)

o) run the ANOVA and calculate studentized intra- and (not so

important) inter-subject residuals

o) test these residuals for normality (according to the protocol; only

one test, otherwise you will run into problems with multiplicity and

contradictory results) and outliers (to be excluded only if you can

justify errors in the clinical and/or analytical conduct of the study)

o) if non-normality of residuals is detected, use a nonparametric

method (e.g., Wilcoxon-Mann-Whitney) instead of ANOVA

Good luck!

Helmut

--

Helmut Schütz Biokinet GmbH / Dept Biostatistics

Neubaugasse 36/11 Nattergasse 4

A-1070 Vienna/Austria A-1170 Vienna/Austria

tel/fax +43 1 9713935 tel +43 1 4856969 62

no cell phone ;-) fax +43 1 4856969 90

http://www.goldmark.org/netrants/no-word/attach.html - On 27 Aug 2003 at 08:50:47, kim.travis.at.syngenta.com sent the message

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The following message was posted to: PharmPK

Helmut's advise is good. A transformation of the data might be worth

considering to help normalise the error, before resorting to a

nonparametric

method,

Kim - On 27 Aug 2003 at 16:45:42, martin.schumacher.aaa.pharma.novartis.com sent the message

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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-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

You applied 2 different statistical methods to test for normality. The

interpretation of the results of the 2 methods is evident. Both methods

indicate with high confidence that the data of both treatments are not

normally distributed. This is what you already knew based on your

visual inspection of the data.

However, more important than applying a statistical test is a very

clear explanation of the data under consideration as well as precise

statement of the goal(s). It would be helpful to get this information.

Additionally, the distribution of PK parameters like AUC and Cmax has

been addressed rather often in the literature and I feel that there's a

consensus that they are log-normally distributed. In this case you

would just take logarithms of the individual data before they enter any

statistical testing, without any tests regarding the distributional

form. Sequential statistical testing is a problematic task and should

be discouraged if possible.

Best regards,

Martin

Martin M. Schumacher, Ph.D.

Principal Scientist

Novartis Pharma AG

BioMarker Development (BMD)

WKL-136.1.19

CH-4002 Basel

Switzerland

Tel. +41 - 61 696 1030 (direct)

Fax +41 - 61 696 6212

Email martin.schumacher.aaa.pharma.novartis.com - On 27 Aug 2003 at 22:26:02, =?ISO-8859-1?Q?"Helmut_Sch=FCtz"?= (helmut.schuetz.-at-.chello.at) sent the message

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The following message was posted to: PharmPK

Dear Kim, dear Martin,

I think I have to be more precise. I agree that parameters like AUC and

Cmax are most likely log-normally distributed, so a transformation

should be the starting point.

But I disagree with Martin's "... just take logarithms of the

individual data before they enter any statistical testing, without any

tests regarding the distributional form." I know that this procedure

(not testing for normality after transformation) is stated in the

recent FDA guideline, but it's still very bad statistical practice. We

have to test our assumptions (in ANOVA additivity of effects, normal

distribution of errors, independence of subjects, etc....) before we

apply a statistical method, and if some of them are violated, we have

to take measures and not only close our eyes and continue...

If we decide to go for a nonparametric method, this is not sequential

testing, since we have not calculated the ANOVA confidence limits, and

the alpha-level is still intact.

Best regards,

Helmut

Helmut Schütz Biokinet GmbH / Dept Biostatistics

Neubaugasse 36/11 Nattergasse 4

A-1070 Vienna/Austria A-1170 Vienna/Austria

tel/fax +43 1 9713935 tel +43 1 4856969 62

no cell phone ;-) fax +43 1 4856969 90

http://www.goldmark.org/netrants/no-word/attach.html - On 27 Aug 2003 at 16:57:03, "Daniel Sitar" (sitar.-at-.Ms.UManitoba.CA) sent the message

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This suggestion is not logical, since the fundamental

assumptions of nonparametric statistics indicate that data

distribution is not part of the underlying principles of the

test(s). Normalization of data sets only applies to the use of

parametric tests to try and achieve something resembling a

Gaussian distribution. Bad data are not salvaged by statistical

contortions. If you keep transforming data by various

mathematical contortions, eventually you will find a way to

get a statistically significant result, but the process is invalid

because of the repeated analysis of the data without

partitioning the finding of a difference by chance.

The best suggestion is to redesign the experiment and

to power it adequately to be able to detect a difference by a

statistical approach that is prospective and scientifically

justifiable.

Dan Sitar, University of Manitoba

Daniel S. Sitar, PhD

Professor and Head

Department of Pharmacology and Therapeutics

sitar.-a-.ms.UManitoba.ca

Tel: 204-789-3532 FAX: 204-789-3932 - On 28 Aug 2003 at 09:04:29, kim.travis.aaa.syngenta.com sent the message

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The following message was posted to: PharmPK

Having suggested the possibility of a transformation I'd like to

strongly

qualify it. When you visualise the data (eg plot it) you should make

sure

that you do it in units which makes most sense to you / most sense

biologically. If you do not then your ability to judge quality of fit

etc

are severely compromised, and poor conclusions can result. This often

means

that you have to untransform your data in order to visualise it.

People became accustomed to all kinds of transformations as a matter of

course in the old days, often so that a simple linear regression can

yield

key parameter estimates (pre computers). Often this was done

irrespective

of error distribution. There is no excuse for this nowadays as the

parameters can be fitted directly without transformation, but in many

applications the habit dies hard.

The good news is that ANOVA is remakably robust to most breaches of the

key

assumptions listed by Helmut. But as Helmut indicates, historical

knowledge

of the likely distributional form of a variable is valuable. In fact

it may

be at least as good a basis for deciding on a transformation as a

statistical test of Normality, especially if sample numbers are low,

Regards,

Kim

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