- On 27 Sep 2005 at 12:32:46, "sulagna" (sdas.at.clinsearchlabs.com) sent the message

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Dear all

A Bioequivalence question ...................

Our guidance says that if Confidence interval is with in 80-125 %

then products are bioequivalent. Thus our conclusions are entirely

based on confidence interval.

But there can be cases where CI falls within prespecified limits

though ANOVA shows highly significant difference between formulations

( p = .00n). I have come across such cases many a

times ............... I

How would you conclude your results in such situations ? How

would you justify your decision based on CI ?

Regards

Sulagna

Associate Scientist ( Biostatistics) - On 27 Sep 2005 at 17:11:25, "Porzio, Stefano" (Stefano.Porzio.-a-.ZambonGroup.com) sent the message

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Dear Sulagna,

I'm extremely interested of a practical case where, as you say,

" ...CI falls within prespecified limits

though ANOVA shows highly significant difference between formulations

( p = .00n). I have come across such cases many a

times ............... "

Best regards

Stefano Porzio

Pharmacokinetic and Tox. Dept.

Inpharzam Ricerche SA - ZAMBON-GROUP

Taverne - Switzerland - On 27 Sep 2005 at 11:19:04, syngenor (syngenor.at.earthlink.net) sent the message

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The following message was posted to: PharmPK

I think you just answered your question. BE are based on CI limits

80-125% and not on the F test.

Regards,

Dominique Paccaly

Syngenor Pharma Consulting - On 27 Sep 2005 at 12:07:07, DDubins.-a-.allied-research.com sent the message

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Hello Sulagna, interesting question.

The bioequivalence limits were selected from a regulatory standpoint

rather than a statistical one. They are a criteria rather than a

statistical test. Thus two products need not demonstrate to be

similar in an ANOVA, only pass the bioequivalence limits. The

conclusions are based on the bioequivalence limits, not the p-values

of the ANOVA. Thus your conclusions would state that the 90%

confidence intervals of the parameters of interest fell within the

bioequivalence limits as defined by your regulatory body.

One corollary to what your saying is that you may have enough power

to determine a difference between a Cmax of 100 ug/mL (Test) and 102

ug/mL (Reference), but the question remains, is the difference

physiologically significant? Small differences between formulations

(p<0.05) may not be important. In addition, there may be considerable

variability between formulations precluding the ability to determine

a significant statistical difference, yet the ratios may well fall

outside the confidence limits. Large differences between formulations

(p>0.05) may be important.

If you have seen many studies then you know that there are

considerable intra- and inter-individual differences with the same

drug, and you have to draw lines somewhere. A problem with using the

confidence interval approach is that if you were a drug manufacturer

with infinite resources, as long as your Test/Reference ratios are

within 80-125, you can find enough subjects to get your drug to pass,

as the more subjects you have, the smaller the confidence intervals

will become. However, we reconcile this by the fact that it is not

terribly economically feasible to have football stadium sized Phase I

BE studies.

I found one article (Rani S and Pargal A., Indian J Pharmacol (2004);

36(4):209-216) which answers your question better than I could:

"In ANOVA, the ratio of the formulations' mean and sum of squares to

the error mean sum of squares gives an F-statistic to test the null

hypothesis Ho: uT=uR. This provides a test of whether the mean amount

of drug absorbed from the test formulation is identical to the mean

abount of drug absorbed from the reference. The test of this simple

null hypothesis of identity is of little interest in bioequivalence

studies, since the answer is always negative. This is because we

cannot expect the mean amounts of drug absorbed from two different

formulations or two different batches of the same formulation to be

identical. They may be very nearly equal, but not identical. Also, if

the trial is run under tightly controlled conditions (resulting in a

small error mean sum of squares in the analysis) and if the number of

subjects is large enough, no matter how small the difference between

the formulations, it will be detected as significant.

Thus the detection of the difference (which as indicated above,

will always exist) becomes simply a function of sample size, and

since the probable magnitude of the difference is the critical

factor, this gives rise to two anamalies:

1. A large difference between two formulations which is

nevertheless not statistically significant if error variability is

high and/or sample size not large enough.

2. A small difference, probably of no therapeutic importance

whatsoever, that is shown to be statistically significant if error

variability is minimal and/or sample size adequately large.

The first case suggests a lack of sensitivity in the analysis,

and the second an excess of it. Consequently, any practice that

increases the variability of the study (sloppy designs, assay

variability and within formulation variability) would reduce the

chances of finding a significant difference and hence improve the

cances of concluding bioequivalence."

David Dubins, B.A.Sc., Ph.D.

Pharmacokinetic Scientist - On 27 Sep 2005 at 14:01:24, Angusmdmclean.aaa.aol.com sent the message

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Sulagna: we are asked to use Ln transformed data: it is my

understanding that the confidence intervals approach is an effort to

quantitative the magnitude of the difference between the LS means

ratio (test vs reference) of the exposure parameter of interest.

Therefore this is the criterion fro BE with the ranges quoted

(0.8-1.25, after back transformation) that is to be followed not the

p value of the F test from the ANOVA.

Hope above helps,

Angus McLean, Ph.D.

8125 Langport Terrace,

Suite 100,

Gaithersburg,

MD 20877

Tel 301-869-1009

fax 301-869-5737

BioPharm Global Inc. - On 29 Sep 2005 at 10:26:14, "sulagna" (sdas.at.clinsearchlabs.com) sent the message

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Dear Stefano

I might have overstated by saying that " ANOVA shows highly significant

difference between formulations " . Actually p values for significance

were slightly less than p=.05 at times even though CI was well within

0.80-1.25 .

I think, if p value (for formulations only) is highly

significant

then CI is bound to fall outside limits. This is because we take the

ratio

of formulations during calculation of CI. However this will not be

true for

other effects like period and sequence. Inspite of being highly

significant CI can lie within 80-125%.

Period and Sequence effects only add to the total residual

variability and give us an explaination as to what contributed to the

total

error.

Regards

Sulagna

Associate Scientist ( Biostatistics) - On 29 Sep 2005 at 17:36:49, "Hans Proost" (j.h.proost.at.rug.nl) sent the message

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Dear Sulagna,

You wrote:

> I think, if p value (for formulations only) is highly significant

> then CI is bound to fall outside limits. This is because we take

> the ratio of formulations during calculation of CI.

This is not correct. The p-value and CI have a completely different

background and a completely different meaning.

The p-value refers to the test of the null hypothesis that both

formulations

are identical, i.e that their ratio is 1. If p<0.05, this null

hypothesis is

rejected, and we conclude that both formulations are significantly

different. If the null hypothesis cannot be rejected, one cannot make a

conclusion. Note that one should not conclude that the formulations are

bioequivalent: this would be a pertinently wrong conclusion.

The confidence interval is used as a statistic to test the null

hypothesis

that the formulations are bioinequivalent, i.e. that their true ratio is

outside the interval 0.8 to 1.25 (either below 0.8, or above 1.25;

hence the

'two one-sided tests', hence the 90% confidence interval). If the entire

confidence interval is within the interval 0.8 to 1.25, the null

hypothesis

is rejected, and the formulations are considered bioequivalent. If

the null

hypothesis cannot be rejected, one cannot make a conclusion. Actually

one

should not conclude that the formulations are bioequivalent (e.g. a

larger

samples might have resulted in a confidence interval within the

interval 0.8

to 1.25). It is better to say that the formulation failed the test on

bioequivalence.

Please note that the first test is not relevant in bioequivalence

testing.

It is really not important whether or not two formulations are

significantly

different or not. The second test gives what we want to know: that a

product

passing the test will have a bioavailability within 0.8 and 1.25 of

that of

the reference product.

Period and sequence effects does not necessarily play a role here.

Best regards,

Hans Proost

Johannes H. Proost

Dept. of Pharmacokinetics and Drug Delivery

University Centre for Pharmacy

Antonius Deusinglaan 1

9713 AV Groningen, The Netherlands

tel. 31-50 363 3292

fax 31-50 363 3247

Email: j.h.proost.at.rug.nl - On 29 Sep 2005 at 14:08:18, "Mike Davenport" (Mike.Davenport.-at-.richmond.ppdi.com) sent the message

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The following message was posted to: PharmPK

The usual test of the null hypothesis is Ho: B=0, a p-value less than

0.05 would imply that you have rejected the null hypothesis at the

alpha-level of significance of 0.05.

The corresponding 95% CI would not include the value of zero in the

interval.

So, if you have a UMP test then you can construct a corresponding UMA

CI.

Therefore, its quite possible to have a significant treatment difference

p < 0.05 and still claim bioequivalence because the confidence interval

could be within .80 to 1.25, but not include one.

Note that, the two-one sided t-test for BE is computationally equivalent

to a 90% confidence interval.

mike - On 29 Sep 2005 at 14:50:52, Angusmdmclean.-at-.aol.com sent the message

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Johannes: I agree with your previous message, but I would to add one

detail to a sentence in your message: please see addition in

parenthesis below.

The confidence interval is used as a statistic to test the null

hypothesis (p value set to 0.05) that the formulations are

bioinequivalent, i.e., that their true ratio is outside the interval

0.8 to 1.25 (either below 0.8, or above 1.25; hence the 'two one-

sided tests', hence the 90% confidence interval). If the entire

confidence interval is within the interval 0.8 to 1.25, the null

hypothesis.

In other words for the second null hypothesis the p value is set to

0.05.

Do you agree with this addition?

Angus McLean, Ph.D.

8125 Langport Terrace,

Suite 100,

Gaithersburg,

MD 20877

Tel 301-869-1009

fax 301-869-5737

BioPharm Global Inc. - On 30 Sep 2005 at 08:02:24, dengcq.-a-.netscape.net sent the message

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It is possible to obtain a 90% confidence interval (for the ratio of

geometric means) within the ranges like 85-95% or 105-120%. Under

these situations, we can claim the bioequivalence (since the 90% CI

falls entirely within the range of 80-125%), but the p-value will be

<0.05 (significant, ratio does not include 1).

This is why we need to rely on the confidence interval (not p-value)

to make the judgment in BE study.

Thanks,

CQ Deng, PhD

http://www.talecris.com - On 30 Sep 2005 at 14:37:44, Rajkumar Radhakrishnan (cryier.-at-.gmail.com) sent the message

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What if your 90% CI ranges from for eg. say 75% - 130%? It includes

80%-125% and also includes mean ratio 1 - On 2 Oct 2005 at 20:28:37, "A.J. Rossini" (blindglobe.-a-.gmail.com) sent the message

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On 9/29/05, Hans Proostwrote:

> Dear Sulagna,

> You wrote:

> > I think, if p value (for formulations only) is highly significant

> > then CI is bound to fall outside limits. This is because we take

> > the ratio of formulations during calculation of CI.

> This is not correct. The p-value and CI have a completely different

> background and a completely different meaning.

Be careful with this one -- p-values and CI's are mirror images of

each other, for appropriately matched computations. Modulo regularity

conditions on the hypothesis test (which some common ones fail), any

hypothesis test (through its p-values) can be used to generate

confidence intervals, and any confidence interval can be used to

create a hypothesis test and corresponding p-value.

It does get tricky, though -- for instance, take the wilcoxon rank sum

(2-sample) and signed-rank tests. They don't qualify, because they

don't satisfy a transitivity property. It is possible to find real

data sets such that A > B (hypothesis of one being larger is

evidenced) and B > C, and C > A.

For tests like the t-test, it's easy to show that you can match

confidence intervals with p-values. When they don't match, then you

generally are using a better formula for one than the other (i.e. CI

for difference of 2 means under the assumption of non-equal variances,

vs. a T-test assuming equal variances, or similar mis-matches).

Now the interpretations are different (though related) and the

backgrounds/history are different, but if you look at them from a big

enough picture (and compute enough of them, in a sense -- i.e. for

different level sets, or different hypotheses "close" to the original,

you get the same results).

best,

-tony

blindglobe.-at-.gmail.com

Muttenz, Switzerland. - On 3 Oct 2005 at 09:02:29, "Hans Proost" (j.h.proost.at.rug.nl) sent the message

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The following message was posted to: PharmPK

Dear Tony,

On my comment to Sulagna:

> > This is not correct. The p-value and CI have a completely

different

> > background and a completely different meaning.

You wrote:

> Be careful with this one -- p-values and CI's are mirror images of

> each other, for appropriately matched computations. Modulo

regularity

> conditions on the hypothesis test (which some common ones fail), any

> hypothesis test (through its p-values) can be used to generate

> confidence intervals, and any confidence interval can be used to

> create a hypothesis test and corresponding p-value.

Thank you for pointing to my badly phrased statement. I fully agree with

your comments.

What I meant was: 'The statistical inference from a p-value obtained

from

ANOVA (i.e. aiming at the difference between the two formulations) is

completely different from the statistical inference from the 90%

confidence

interval in bioequivalence testing'. This difference is due to the

different

aim of the test, resulting in a different null hypothesis, as

explained in

my earlier message.

Two additional notes:

1) The 90% confidence interval is not appropriate to test the null

hypothesis that both formulations are equal with a p-value of 0.05;

to this

purpose a 95% confidence interval should be used.

2) In bioequivalence testing (i.e. null hypothesis: the formulations are

bioinequivalent, i.e. their true ratio is

outside the interval 0.8 to 1.25) the 90% confidence interval does not

restrict the type I error (the risk of falsely rejecting the null

hypothsis,

i.e. concluding bioequivalence whereas the true ratio is outside the

interval 0.8 and 1.25) to 0.05 in all cases; in some cases the true

value is

slightly higher. As far as I remember the 90% confidence interval was

adopted by FDA for practical reasons as a simple alternative to 'exact'

procedures.

Best regards,

Hans Proost

Johannes H. Proost

Dept. of Pharmacokinetics and Drug Delivery

University Centre for Pharmacy

Antonius Deusinglaan 1

9713 AV Groningen, The Netherlands

tel. 31-50 363 3292

fax 31-50 363 3247

Email: j.h.proost.-at-.rug.nl - On 3 Oct 2005 at 09:20:19, "Hans Proost" (j.h.proost.at.rug.nl) sent the message

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The following message was posted to: PharmPK

Dear Rajkumar,

You wrote:

> What if your 90% CI ranges from for eg. say 75% - 130%? It includes

> 80%-125% and also includes mean ratio 1

Null hypothesis of bioinequivalence cannot be rejected. The test

formulation

failed the test on bioequivalence.

Best regards,

Hans Proost

Johannes H. Proost

Dept. of Pharmacokinetics and Drug Delivery

University Centre for Pharmacy

Antonius Deusinglaan 1

9713 AV Groningen, The Netherlands

tel. 31-50 363 3292

fax 31-50 363 3247

Email: j.h.proost.aaa.rug.nl - On 3 Oct 2005 at 09:28:42, "Hans Proost" (j.h.proost.-at-.rug.nl) sent the message

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The following message was posted to: PharmPK

Dear Angus,

You added to my comment:

> (p value set to 0.05)

> In other words for the second null hypothesis the p value is set to

> 0.05.

> Do you agree with this addition?

You are right. I tacitly assumed a p-value of 0.05. To be even more

correctly, it is better to say either 'p < 0.05' or 'alpha = 0.05', to

discriminate between the cut-off value for the type I error (alpha =

0.05)

and the actual value (p-value, p < 0.05) of the probability of falsely

rejecting the null hypothesis.

Best regards,

Hans Proost

Johannes H. Proost

Dept. of Pharmacokinetics and Drug Delivery

University Centre for Pharmacy

Antonius Deusinglaan 1

9713 AV Groningen, The Netherlands

tel. 31-50 363 3292

fax 31-50 363 3247

Email: j.h.proost.at.rug.nl

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