ratnakar1811 ★ India, 2012-03-19 13:49 (4412 d 07:36 ago) Posting: # 8295 Views: 7,227 |
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Dear All, This is to confirm what all primary and secondary PK parameters and criteria for BE should be considered for steady state study for a modified released tablet for EU submission? As per current EU guideline it only mentions about AUC0-tau to be considered as the additional parameter for bioequivalence. Whereas for Steady State study for immediate release formulations it mentions parameters like AUC0-tau, Cmax,ss and Tmax,ss also. But as per FDA we do consider Cmin, Cav, degree of fluctuation [(Cmax-Cmin)/Cav], and swing [(Cmax-Cmin)/Cmin]. Thanking you for your guidance in advance. Ratnakar |
Helmut ★★★ Vienna, Austria, 2012-03-19 15:18 (4412 d 06:08 ago) @ ratnakar1811 Posting: # 8296 Views: 6,521 |
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Dear Ratnakar! ❝ […] what all primary and secondary PK parameters and criteria for BE should be considered for steady state study for a modified released tablet for EU submission? As per current EU guideline it only mentions about AUC0-tau to be considered as the additional parameter for bioequivalence. Unfortunately the MR NfG is quite old (1999) and somewhat ambiguous: 4.1 Bioavailability studies
4.1.1. Rate and extent of absorption, fluctuation Interesting points:
❝ Whereas for Steady State study for immediate release formulations it mentions parameters like AUC0-tau, Cmax,ss and Tmax,ss also. Tmax – where? ❝ But as per FDA we do consider Cmin, Cav, degree of fluctuation [(Cmax-Cmin)/Cav], and swing [(Cmax-Cmin)/Cmin]. Not relevant here. BTW, I can’t imagine that FDA rejects a study based on failing Swing if all other metrics pass. Swing is a lousy metric, since you divide by the one concentration with the largest variability of the entire profile. — Dif-tor heh smusma 🖖🏼 Довге життя Україна! Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
Jaime_R ★★ Barcelona, 2012-03-19 17:26 (4412 d 03:59 ago) @ Helmut Posting: # 8298 Views: 6,257 |
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Dear Helmut! ❝ ● Fluctuation: No equivalence to IR, but non-superiority! Oh, I missed that! Being less than a statistical amateur, does that require the upper one-sided 95% CL included in 125% – or, if the statistical software (e.g. WinNonlin) does not give this result directly – assessing the upper limit of the 90% CI only? How can we calculate the sample size? — Regards, Jaime |
Helmut ★★★ Vienna, Austria, 2012-03-19 17:59 (4412 d 03:26 ago) @ Jaime_R Posting: # 8299 Views: 6,259 |
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Dear Jaime! ❝ […] does that require the upper one-sided 95% CL included in 125% – or […] – assessing the upper limit of the 90% CI only? Both will work. ❝ How can we calculate the sample size? Ouch; good question! Steven Julious (Sample Sizes for Clinical Trials, 2010) gives a method for cross-over studies (actually the other way ’round: non-inferiority) but only for normal distributed data. Have to sleep over it. — Dif-tor heh smusma 🖖🏼 Довге життя Україна! Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
d_labes ★★★ Berlin, Germany, 2012-03-20 16:48 (4411 d 04:38 ago) (edited by d_labes on 2012-03-21 11:35) @ Helmut Posting: # 8305 Views: 6,855 |
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Dear Helmut, dear Jaime! ❝ ❝ […] does that require the upper one-sided 95% CL included in 125% – or […] – assessing the upper limit of the 90% CI only? ❝ ❝ Both will work. With the exception that the ICH E9 guidance suggests for one sided CIs an alpha=0.025 I think. Or am I wrong . That would require 95% CI's and assessing their upper limit only. ❝ ❝ How can we calculate the sample size? ❝ ❝ Ouch; good question! Steven Julious (Sample Sizes for Clinical Trials, 2010) gives a method for cross-over studies (actually the other way ’round: non-inferiority) but only for normal distributed data. Have to sleep over it. Non-superiority is a rather unusual term . The other way round - non-inferiority - would mean that we formulate our problem as non-inferiority. It's easy. Instead of the hypotheses H0: µT/µR > 1.25 ('superiority') we formulate H0: µR/µT < 0.8 ('inferiority') In case of log-normal distributed data the hypotheses change as usual to hypotheses of differences H0: log(µR) - log(µT) < -0.2231436 ('inferiority') The corresponding test is a one-sided t-test which has power for a 2x2 crossover (Julious "Sample Sizes for Clinical Trials", 2010, equation 6.22) 1-beta = pt(t1-alpha,n-2,df=n-2,tau) Edit: I can only reproduce PASS values for power and sample size if I use power=1-pt(...) although Julious termed (1-beta) power . where pt() is the cumulative distribution function of the non-central t-distribution with non-centrality parameter tau= abs((log(µT)-log(µR) - d)*sqrt(n)/sqrt(2*MSE)) and d=log(R0) with R0 the null or 'true' ratio. BTW: PASS 2008 has a module "Noninferiority & superiority -> Two means in a 2x2 crossover -> specify using ratios" that will do the calculations for you. It contains also an option in that module which allows the calculations for the 'non-superiority' case named "Higher is bad". If you don't own PASS here a quick shot in R: (design is in the moment only a place holder, nothing other then 2x2 crossover is implemented) # power function For a targetpower=0.8, CV=0.3, margin=0.8 the last function will give the sample sizes alpha= The whole discussion up to here depends on the assumption of log-normality. If this is a reasonable assumption for measures of fluctuation like PTF or swing is left to you. — Regards, Detlew |
d_labes ★★★ Berlin, Germany, 2012-03-22 11:20 (4409 d 10:06 ago) @ Helmut Posting: # 8312 Views: 6,172 |
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Dear All! In my previous post I have noticed that I could reproduce the power and sample size for non-inferiority trials only if I used 1-beta = 1 - pt(t1-alpha,n-2,df=n-2,tau) Seems even the pope can err . The formulas given in the book are in error here, not only for the non-inferiority case (chapter 6) but also for the superiority tests (chapter 3/4). Compare them to those in the paper
where the formulas are yet given with power=1-probt(...). Compare further the results of the normal approximation formulas (f.i. formula 6.21 1-beta = pnorm(tau-t1-alpha,n-2 )to see that 1-probt(...) is correct. With tau=2.5, df=7, alpha=0.05 (in R syntax):
(6.21) pnorm(2.5 - qt(1-0.05, df=7)) =0.72755 BTW: Would it desirable to have non-inferiority power / sample size in PowerTOST (although it is not based on TOST but on OOST - one one-sided t-test)? I could ask the author if he had some spare time . — Regards, Detlew |
Helmut ★★★ Vienna, Austria, 2012-03-30 16:13 (4401 d 06:12 ago) @ d_labes Posting: # 8353 Views: 6,037 |
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Dear Detlew! ❝ I could ask the author if he had some spare time. Obviously he had. Released 2012-03-26:
power.noninf() , sampleN.noninf() .THX for PowerOOST !— Dif-tor heh smusma 🖖🏼 Довге життя Україна! Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |