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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
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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
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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
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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
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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
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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.
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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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Copyright 1995-2010 David W. A. Bourne (david@boomer.org)