- On 18 May 2006 at 11:32:45, =?ISO-8859-1?Q?Rog=E9rio_Antonio_de_Oliveira?= (rogerio.-a-.galenoresearch.com) sent the message

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The following message was posted to: PharmPK

Dear colleagues,

I would like to Know how WinNonlin calculates the power of ANOVA

which is

presented when we use the average bioequivalence procedures. For

instance,

we can get the results below.

Two One-Sided T-tests

Prob(< 80%)=0.0000 Prob(> 125%)=0.0589 Max=0.0589 Total=0.0589

Anderson-Hauck Procedure

A.H. p-value = 0.058886

Power of ANOVA for Confidence Level 90.00

Power at 20% = 0.998989

Which formula do we use to get this power of ANOVA? Do you can help me?

Thanks a lot.

--

Rogerio Antonio de Oliveira

Galeno Research Unit

www.galenoresearch.com - On 22 May 2006 at 16:03:20, "Support" (Support.-at-.Pharsight.com) sent the message

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The following message was posted to: PharmPK

Dear Rogerio,

In the WinNonlin BE Wizard, the following formulas are used to

calculate Power for Average Bioequivalence:

Definition: Power is the post-hoc probability of detecting a

difference greater than or equal to a specified percentage of the

reference least squares mean. In general,

Power = 1 - (probability of a type II error)

= probability of rejecting H0 when H1 is true.

In the bioequivalence calculations, the hypotheses being tested are:

H0: RefLSM = TestLSM

H1: RefLSM TestLSM

where: RefLSM = reference least squares mean (as computed by LinMix)

TestLSM = test least squares mean (as computed by LinMix)

For the no-transform case, the power is the probability of rejecting

H0 given:

|TestLSM - RefLSM | fractionToDetect

For ln-transform, and data already ln-transformed, this changes to:

|TestLSM - RefLSM | - ln (1 - fractionToDetect),

and similarly for log10-transform and data already log10-transformed.

For the sake of illustration, assume fractionToDetect = 0.2,

RefLSM>0, and no transform was done on the data. Also the maximum

probability of rejecting H0 occurs in the case of equality, |TestLSM

- RefLSM| = 0.2

Power = Pr (rejecting H0 at the alpha level given |Difference| = 0.2

= Pr (|Difference/DiffSE| > t (1-alpha/2), df given |Difference| = 0.2

= Pr (Difference/DiffSE > t (1-alpha/2),df given |Difference| = 0.2

+ Pr (Difference/DiffSE < -t (1-alpha/2),df given |Difference| = 0.2

Let:

t1 = t (1-alpha/2),df - 0.2

t2 = - t (1-alpha/2),df - 0.2

tstat = (Difference - 0.2

Then:

Power = Pr(tstat > t1 given |Difference| = 0.2

+ Pr(tstat < t2 given |Difference| = 0.2

Note that tstat is T-distributed. Let p1 be the p-value associated

with t1 (the area in the tail), and let p2 be the p-value associated

with t2 (the area in the tail; note that p2 may be negligible). Then:

Power = 1 - (p1 - p2)

Best Regards,

Pharsight Software Support

support.-at-.pharsight.com

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