- On 23 Jan 2005 at 06:44:38, "Someswara Rao" (korlas.-a-.rediffmail.com) sent the message

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Dear all

What is the difference between Conventional 90% confidence intervals

and Westlake's 90% confidence intervals? If you provide formula for the

Westlake's 90% CI that would be greatful.

Someswara Rao.K - On 23 Jan 2005 at 19:15:42, dengcq.aaa.netscape.net sent the message

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The following message was posted to: PharmPK

Hi, Someswara,

It is better to check the book:

Design and Analysis of Bioavailability and Bioequivalence Studies

Author: Shein-Chung Chow Jen-Pei Liu

ISBN: 0824775724

It has all information you need. Other than the conventional and

westlake's CI, other approaches for obtaining the CI are also discussed

in this book in detail.

Thanks,

CQ Deng - On 24 Jan 2005 at 13:23:05, =?ISO-8859-1?Q?Helmut_Sch=FCtz?= (helmut.schuetz.at.bebac.at) sent the message

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The following message was posted to: PharmPK

Dear Someswara,

Westlake's symetrical confidence limits are obsolete and

should not be used any more.

The conventional confidence limit is symetrical about the

point estimate, whereas Westlake's is symetrical about unity.

In order to construct the conventional 90% confidence interval,

we use twice the 95% coverage of the t-distribution (subtracting

from / adding to the point estimate).

Westlake assumed that 'clinicians' have difficulties

understanding something like '86% - 102% of the reference' and

prefer a statement like '100% +/- 10%'.

In a mathematical sense it's possible to construct a 90%

confidence interval which is symetrical about unity by using

*asymetrical* lower and upper coverages of the t-distribution.

E.g., subtracting t99% and adding t91% we get a 90% confidence

interval which is narrower *below* unity, and wider *above*

unity:

0.8-------0.9-------1.0-------1.1-------1.2

| | | | |

conventional: |-------|-------|

Westlake: |-------------------|

There is no explicit formula, you have to apply an

iterative process (calculating the interval, looking whether

it's symetrical, adjusting the lower and upper percentages

of the t-distribution,.... until you're satisfied).

Westlake's method should be avoided, because it obscures

information.

Imagine:

Study 1: 86% - 102%, point estimate 94%

Study 2: 90% - 110%, point estimate 100%

give the *same* Westlake confidence limits (obscuring two

facts: point estimate and variability is lower in study 1

than in study 2)

Anyhow, if you are interested in some background

information, please have a look at the original reference

W.J. WESTLAKE;

Symetrical Confidence Intervals for Bioequivalence Trials

Biometrics 32, 741-744 (1976)

Symetrical confidence limits are computed by modern

statistical packages, but if you want to get some 'feeling'

about your data and Wetlake's method, you can do the

calculation 'manually' using:

SPRIET, A. and D. BEILER;

Table to Facilitate Determination of Symetrical Confidence Intervals in

Bioavailability Trials with Westlake's Method

Europ. J. Drug Metabol. Pharmacokin. 2, 129-132 (1978)

Both references are given for linear models [0.8-1.2]

and 95% confidence intervals.

Best regards,

Helmut

--

Helmut Schutz

BEBAC

Consultancy Services for Bioequivalence and Bioavailability Studies

Neubaugasse 36/11

A-1070 Vienna/Austria

tel/fax +43 1 2311746

http://BEBAC.at Bioequivalence/Bioavailability Forum at

http://forum.bebac.at

http://www.goldmark.org/netrants/no-word/attach.html - On 25 Jan 2005 at 19:36:55, =?ISO-8859-1?Q?Helmut_Sch=FCtz?= (helmut.schuetz.at.bebac.at) sent the message

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The following message was posted to: PharmPK

<>Since yesterday's mail might have been a little confusing

(blanks were just 'swallowed' somewhere within the net),

I will give it a second try. Today I will use dots instead

of blanks as 'spacers':

......

In a mathematical sense it's possible to construct a 90%

confidence interval which is symetrical about unity by using

*asymetrical* lower and upper coverages of the t-distribution.

E.g., subtracting t99% and adding t91% we get a 90% confidence

interval which is narrower *below* unity, and wider *above*

unity:

<><>.............0.8-------0.9-------1.0-------1.1-------1.2

...............|........ |........ |........ |....... |

<>conventional:........ |=======x=======|...............

Westlake:...............|===================|.........

Best regards,

Helmut

--

Helmut Schutz

BEBAC

Consultancy Services for Bioequivalence and Bioavailability Studies

Neubaugasse 36/11

A-1070 Vienna/Austria

tel/fax +43 1 2311746

http://BEBAC.at Bioequivalence/Bioavailability Forum at

http://forum.bebac.at

http://www.goldmark.org/netrants/no-word/attach.html - On 4 Feb 2005 at 22:04:04, Wojciech Jawien (mfjawien.-at-.cyf-kr.edu.pl) sent the message

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Dear Someswara Rao,

Westlake's symmetrical CI can be calculating as

[mean-SD*tinv(alpha-phi) , mean+SD*tinv(phi)]

tinv is an inverse t-distribution function and phi has to satisfy

tinv(phi)-tinv(alpha-phi)=2*mean/SD

The 1-alpha is the nominal confidence level, in fact this interval

always has coverage greater than this value.

The above statements do not give an open-form formula for CI, since

such doesn't exist. The solution can be obtained, eg. using Excel with

its Solver.

I have included an appropriate worksheet [Not included on the listsev -

db]. For large absolute values of mean/SD there may be numerical

problems in finding solution, however.

I think you may also benefit from original Wstlake's paper (Biometrics

32,741-744 (1976)) as well as from some critics on it: (Biometrics

33,759-760; Biometrics 37,589-591) and Westlake's responses in the same

Biometrics issues. Also take into account the nearly 3 decades passed

after Westlake's paper. His ideas may be seen as an intuitive approach

to now well founded procedures of judging bioequivalence, like

Schuirmann's two one-sided tests. In other words, you should answer

yourself the question: Do I need Westlake's symmetrical CI?

Regards

Wojciech Jawien

Jagiellonian University

Krakow

Poland - On 9 Feb 2005 at 10:59:39, "Someswara Rao" (korlas.at.rediffmail.com) sent the message

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Dear Wojciech Jawien

I want to know the alpha(level of of significance) value in the

calculation of Westlake's CI, like a fixed value (0.05)in the

conventional CI.

If alpha=0.025 that will give more closure width of CI.If this is the

value in calculation of Westlake's CI, then we will get more closure

width of CI.

My idea is that, for any product weastlake's CI passes then it will

automatically passess the 90% conventional CI.

Thanks for your valuable information. - On 9 Feb 2005 at 22:39:21, Wojciech Jawien (mfjawien.-at-.cyf-kr.edu.pl) sent the message

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The following message was posted to: PharmPK

Dear Someswara,

First of all, your email causeed me to reveal bad mistake in my previous

answer.

Namely, the condition for phi should read:

tinv(phi)-tinv(alpha-phi)= - 2*mean/SD (the minus before 2 was

omitted). Apologies.

Now ad rem.

You wrote:

> I want to know the alpha(level of of significance) value in the

> calculation of Westlake's CI, like a fixed value (0.05)in the

> conventional CI.

If you choose nominal significance level alpha=0.05, i.e. nominal

confidence level 1- alpha=0.95 (as Westlake himself did) you will get CI

containing true value with probability greater than 0.95.

The actual coverage depends on true value (of mu_T-mu_R). If it is

exactly 0, it is always contained in Westlake's CI , thus conf.level is

exactly 1=100%. The nominal value is approached as abs(mu_T-mu_R) tends

to infinity.

This is not the case for conventional CI, which conf.level is always

exactly 1-alpha.

> If alpha=0.025 that will give more closure width of CI.If this is the

> value in calculation of Westlake's CI, then we will get more closure

> width of CI.

> My idea is that, for any product weastlake's CI passes then it will

> automatically passess the 90% conventional CI.

If mean>0 then tinv(phi)alpha/2 and right bound

of Westlake's CI is lower tahn that of conventional CI. And analogously

for mean<0 and left bound. Thus conventional CI is not fully contained

in Westlake's CI (and your idea is not necessarily correct).

In a book Chow and Liu, CQ Deng recommended, 90% CI's are used, because

of relation (at least in the case of the conventional CI) with the

Schuirmann two one-sided tests procedure.

It changes nothing but alpha value in above conclusions. Also, there is

no clear reason to use Westlake's CI instead of conventional in this

context.

Regards

Wojciech Jawien

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