# PharmPK Discussion - Power and WinNonlin

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• On 18 May 2006 at 11:32:45, =?ISO-8859-1?Q?Rog=E9rio_Antonio_de_Oliveira?= (rogerio.-a-.galenoresearch.com) sent the message
`The following message was posted to: PharmPKDear colleagues,I would like to Know how WinNonlin calculates the power of ANOVAwhich ispresented when we use the average bioequivalence procedures. Forinstance,we can get the results below.                         Two One-Sided T-testsProb(< 80%)=0.0000 Prob(> 125%)=0.0589 Max=0.0589 Total=0.0589                        Anderson-Hauck Procedure                       A.H. p-value = 0.058886Power of ANOVA for Confidence Level 90.00Power at 20% = 0.998989Which 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`
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• On 22 May 2006 at 16:03:20, "Support" (Support.-at-.Pharsight.com) sent the message
`The following message was posted to: PharmPKDear Rogerio,In the WinNonlin BE Wizard, the following formulas are used tocalculate Power for Average Bioequivalence:Definition: Power is the post-hoc probability of detecting adifference greater than or equal to a specified percentage of thereference 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 = TestLSMH1: RefLSM TestLSMwhere: 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 rejectingH0 given:|TestLSM - RefLSM | fractionToDetectFor 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 maximumprobability of rejecting H0 occurs in the case of equality, |TestLSM- RefLSM| = 0.2Power = 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| = 0.2Let:t1 = t (1-alpha/2),df - 0.2t2 = - t (1-alpha/2),df - 0.2tstat = (Difference - 0.2Then:Power = Pr(tstat > t1 given |Difference| = 0.2+ Pr(tstat < t2 given |Difference| = 0.2Note that tstat is T-distributed. Let p1 be the p-value associatedwith t1 (the area in the tail), and let p2 be the p-value associatedwith t2 (the area in the tail; note that p2 may be negligible). Then:Power = 1 - (p1 - p2)Best Regards,Pharsight Software Supportsupport.-at-.pharsight.com`
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