khaoula ★ Algeria, 20140606 14:04 Posting: # 13037 Views: 14,317 

Hi everybody , I have a question about impact of randomization in result of bioequivlence study: we have design of bioequivalence study cross over design with 16 subject (randomised), but at the firt period the 15 suject didn't come, after we collected the results of 15 subjects and we analyzed Kinetica without protocol for the management of dropouts we had abberants results: for Cmax CV = 0,03 % for hight variable drug, power: 50%, with subject effect and subject/sequence effect if does not take into account the exclusion of subject N 15 have impact in the result of the study? 
Helmut ★★★ Vienna, Austria, 20140606 14:47 @ khaoula Posting: # 13038 Views: 13,239 

Hi Khaoula, » […] we collected the results of 15 subjects and we analyzed Kinetica without protocol for the management of dropouts We recently have submitted a paper to the AAPS J (which currently is under review). It seems that Kinetica is not able to correctly deal with imbalanced studies. I strongly suggest to use another software (we got correct results in R, Phoenix/WinNonlin, EquivTest/PK, and SAS). » we had abberants results: for Cmax CV = 0,03 % for hight variable drug … I don’t understand what you mean here. The limit for HVDs/HVDPs is 30% CV of the reference obtained in a replicate design. CV_{intra} from a 2×2 serves only as a hint of a highly variable reference (since pooled from CV_{WR} and CV_{WT}). » power: 50% … A posteriori (aka posthoc) power is irrelevant in BE. Stop calculating it. Either the study passes, or not. » with subject effect and subject/sequence effect Subject effects are normal in a crossover. It tells you that subjects differ – well, they should… » […] the exclusion of subject N 15 have impact in the result of the study? If evaluated by Kinetica, yes. — Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
khaoula ★ Algeria, 20140606 16:03 @ Helmut Posting: # 13039 Views: 13,138 

Hi Helmut, » I don’t understand what you mean here. The limit for HVDs/HVDPs is 30% CV of the reference obtained in a replicate design. CV_{intra} from a 2×2 serves only as a hint of a highly variable reference (since pooled from CV_{WR} and CV_{WT}). this is a real study that was done in the institute were I do my post graduate disertation, and the drug is omeprazole, that is a HVDP (in bibliography I read a lot of bioequivalence stydy were CV > 30%) I know that the cross over isn't adapted for this, but CV C max of the study Was very low!!!!!!!! 
Helmut ★★★ Vienna, Austria, 20140607 13:52 @ khaoula Posting: # 13041 Views: 13,283 

Hi Khaoula, » […] the drug is omeprazole, that is a HVDP (in bibliography I read a lot of bioequivalence stydy were CV > 30%) Agree. » […] CV C max of the study Was very low!!!!!!!! The 0.03% you stated above are practically impossible. Even if you had only monozygotic multiples in your study (violating the “independence” of IID) there would still be variability of the bioanalytical method as a component of the residual variance. Exceptionally good methods still have a CV of ~2–3%. The lowest CV_{intra} of C_{max} I ever have seen was ~7% (more than two orders of magnitude higher than yours). Please check your calculations: \(CV\% = 100\sqrt{e^{MSE}1}\), where MSE is obtained from ANOVA/GLM of lntransformed data. Avoid Kinetica.— Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
jag009 ★★★ NJ, 20140609 15:36 @ khaoula Posting: # 13042 Views: 13,061 

Hi, I seriously doubt that you ended up with a 0.03% CV for Cmax, particularly for Omeprazole. Can you post a printout of the stats from Kinetica for Cmax? John 
khaoula ★ Algeria, 20140609 22:07 @ jag009 Posting: # 13044 Views: 13,036 

Patient Sequence CmaxT CmaxR LnCmaxT LnCmaxR I havnt randomisation of subject 15, and then they put the result in kinetica they excluded him, they havnt protocol for outlier etc thank you 
Helmut ★★★ Vienna, Austria, 20140610 00:37 @ khaoula Posting: # 13045 Views: 13,154 

Hi Khaoula, I’m too lazy to check with my old installation of Kinetica. But there are some points to note: » SOURCE D.F SS MS F p » Period 1 0.0264787 0.0264787 0.56186 0.4669 NS » Subject(Seq) 13 4.8517 0.373208 7.91934 0.0003379 *** » Formulation 1 0.211625 0.211625 4.4906 0.05391 NS » Sequence 1 0.686301 0.686301 14.5631 0.002141 *** » Error 13 0.61264 0.0471261 » Total 29 6.38875 Old story. Kinetica calculates the sequence effect wrong – since ages. See this thread. Instead of F=0.686301/0.0471261=14.5631 (which is highly significant with p=0.002141) the correct test is against Subject(Seq) : F=0.686301/0.373208=1.8389 (which is not significant with p=0.1982).» Root Mean Square Error = 0.217086 The RMSE is a crude estimate of the CV and should be avoided. » ; CV = 0.0347883 That’s nonsense! \(CV\% = 100\sqrt{e^{0.0471261}1} = 21.97\%\) » Power of the test = 0.500777 » 1  ( Power of the test ) = 0.499223 Wrong. If you insist in post hoc power: 0.09885. With a T/Rratio close to the upper limit of the acceptance range would you really expect a ~50% chance to pass BE in 15 subjects? » around the ratio:[test form]/[ref form])=[1.028, 1.3612] Yes, but around which ratio? I would guess: \(\sqrt{1.028 \times 1.3612} = {\color{Red} {1.829\ldots}}\) The geometric mean of Test is 559.01 and the one of Reference 461.21. As a first guess of the T/Rratio we get 461.21/559.01=1.2121. Since this is an imbalanced dataset, we have to use the least squares (or adjusted) means instead, which are 552.58 and 456.58. Therefore, T/R=1.2103. What does Kinetica “calculate” here? Due to the bug in Kinetica this is likely wrong. In Phoenix/WinNonlin I got: Hypothesis DF SS MS F_stat P_value BTW: Parameter Estimate Check: CV% = 100√(ℯ^{0.053184} – 1) = 23.37%. Low for omeprazole, but possible. If you want an independent evaluation by noncommercial software: In R (you find the script in my lectures) I got: 121.02% [104.22–140.53%] CV 23.37% » I havnt randomisation of subject 15, and then they put the result in kinetica they excluded him, The randomization of an excluded subject is not required anyhow. — Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
ElMaestro ★★★ Denmark, 20140610 01:00 @ Helmut Posting: # 13046 Views: 12,962 

Hi Khaoula and Helmut, » » I havnt randomisation of subject 15, and then they put the result in kinetica they excluded him, » » The randomization of an excluded subject is not required anyhow. I think what was meant was that the allocated sequence for subject 15 is unknown/missing/evaporated/gone/lost? If aiming for balance which is standard you may be able to guess it pretty well provided the sequences of the other 8+7 subjects are correct Anyways, guessing isn't science, and I just wonder how such an event could happen  could you comment, Khaoula? How could info on randomisation for a single subject be missing? — I could be wrong, but... Best regards, ElMaestro 
Helmut ★★★ Vienna, Austria, 20140610 14:53 @ ElMaestro Posting: # 13050 Views: 12,887 

Hi ElMaestro & Khaoula, » I think what was meant was that the allocated sequence for subject 15 is unknown/missing/evaporated/gone/lost? If aiming for balance which is standard you may be able to guess it pretty well provided the sequences of the other 8+7 subjects are correct Since we have seven subjects in sequence RT and eight in sequence TR, we might guess that #15 was planned for sequence RT. » Anyways, guessing isn't science, Yep. » and I just wonder how such an event could happen […] Khaoula wrote in her OP: » » we have design of bioequivalence study cross over design with 16 subject (randomised), but at the firt period the 15 suject didn't come […] » How could info on randomisation for a single subject be missing? Good question. » » without protocol for the management of dropouts I wouldn’t call #15 a “dropout” but rather a “noshow”. — Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
khaoula ★ Algeria, 20140611 00:17 @ Helmut Posting: # 13056 Views: 12,883 

Hi ElMaestro & Helmut I think that it's important to have randomisation of subject 15 (I dont now the basis of a software (kinetica) and how tu use it), I havnt randomisation because they dont give me it, they asked me to try to understand where is the error, so when I search I found that oeprazole is a HVDP and I try to understand what FDA and EMA do, I concluded that our design is't appropriate (must have replicate design), but with Phoenix/WinNonlin CV=23.37%, so the problem of our study is number of subject, with hightest number we can conclude to biequivalence (the test drug is really bioequivalent of réference, this study was done to learn how to do biequivalennce study) I calculated CV with error = 0.0471261 and I had CV% = 100√(ℯ^{0.0471261} – 1) = 21.97% but I thought that I was wrong 
Helmut ★★★ Vienna, Austria, 20140611 02:01 @ khaoula Posting: # 13057 Views: 13,177 

Hi Khaoula! » […] I havnt randomisation because they dont give me it, they asked me to try to understand where is the error, […] May I ask: Who are “they”? » […] our design is't appropriate (must have replicate design), […] Not necessarily. Only if you want to widen the acceptance range (EMA) or go with referencescaling (FDA). But if CV_{WR} <30% scaling is not allowed. » […] the problem of our study is number of subject, […] Mainly. If we assume the T/Rratio and CV “as carved in stone” – which I would not recommend anyway – you would have needed the following sample sizes in order to demonstrate BE with 80% power: design method sample size Note that referencescaling would not help with such a low CV. You have a lot of problems arising from the study:
» the test drug is really bioequivalent of réference, […] Why do you think so? You would need hundreds of subjects to squeeze the CI within the acceptance range. » this study was done to learn how to do biequivalennce study Can you briefly summarize what you have learned? A drug is that substance which, when injected into a rat, will produce a scientific report. Anonymous Pharmacokinetics: one of the magic arts of divination whereby needles are stuck into dummies in an attempt to predict profits. Stephen Senn » I calculated CV with error = 0.0471261 and I had CV% = 100√(ℯ^{0.0471261} – 1) = 21.97% but I thought that I was wrong So instead of following a formula given in numerous papers and textbooks you rather trusted in the output of a piece of software? — Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
khaoula ★ Algeria, 20140724 13:36 @ Helmut Posting: # 13306 Views: 12,485 

Hi Helmut I made a big mistake when I copied the results of Cmax of our study and publish them here, I'm really sorry and confused, can you use the right results and recalculate it with your software kinetica v5 and Phoenix/WinNonlin? I don't want to waste your time and I am so grateful for your help, thank you 
Helmut ★★★ Vienna, Austria, 20140727 01:37 @ khaoula Posting: # 13317 Views: 12,697 

Hi Khaoula! » […] can you use the right results and recalculate it with your software kinetica v5 and Phoenix/WinNonlin? Meanwhile I retired the machine where Kinetica is installed on. Results from Phoenix/WinNonlin6.3: Hypothesis DF SS MS F_stat P_value Checks: T/R = 100(551.40/467.64) = 100ℯ^{0.1647} = 117.91% 90% CI = 100ℯ^{0.1647±1.7709√0.047816(1/(2×8)+1/(2×7))} = 102.33–135.86% CV% = 100√ℯ^{0.047816} – 1 = 21.13% — Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
Helmut ★★★ Vienna, Austria, 20140611 19:51 @ khaoula Posting: # 13059 Views: 12,957 

Dear Khaoula, I recalculated your C_{max}data in Kinetica v5.0.10. I didn’t import the “all missing” row of subject #15 and asked for logtransformed analysis. Now I’m even more confused. Below my results:
 Still wrong in many places, but closer to Phoenix/WinNonlin and R. PE 90% CI MSE CV — Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
khaoula ★ Algeria, 20140612 00:41 @ Helmut Posting: # 13061 Views: 12,791 

Dear Helmut, we have (Kinetica Version 4.4.1) I sent you a private mail, can you read it? thank you Edit: Look at your inbox. I have a 4.x version of Kinetica installed on one of my old machines. Maybe I will check – can’t promise. [Helmut] 
khaoula ★ Algeria, 20140725 00:49 @ Helmut Posting: # 13310 Views: 12,506 

thank you verry much Helmut !!!! so, what I saw: the sequence effect with kinetica V5 is wrong (like V4): seq effect = MS seq/ MS error(within), but the right mathematc formula is seq effect MS seq / MS subj(seq)(between) (I understand that thanks to you) calculation of CV: the software dont take this rule: but this one: (it's false) but for GMR, I'm really confused, It's an unbalenced cross over so you told me to use the least squares (or adjusted) means instead I read this post and use this formula: I have'nt found the same result (1,21), but 1,18 like kinetica v4... where is the problem? 
Helmut ★★★ Vienna, Austria, 20140727 02:23 @ khaoula Posting: # 13318 Views: 12,547 

HI Khaoula, » the software […] take this rule: (it's false) Yes, it’s wrong. Where did you find this formula? You would get a negative value for the root. Maybe Kinetica uses (also wrongly) CV% = 100√ℯ^{MSE²} – 1 which would give 4.72% – at least closer to your original 3.47% – no idea. It’s not my job to unveil all potential bugs in a software I neither seriously used nor validated myself… » but for GMR, I'm really confused, It's an unbalenced cross over so you told me to use the least squares (or adjusted) means instead I read this post » » I have'nt found the same result (1,21), but 1,18 like kinetica v4... where is the problem? Again, no idea. How did you calculate a GMR of 1.21? Your corrected logtransformed data gives a LSM for the reference of 6.1477 and for the test 6.3125. These values are derived exactly as I stated in my old post. subj seq form per logCmax subj seq form per logCmax subj seq form per logCmax subj seq form per logCmax Mean of R in sequence RT is 6.0203 and in sequence TR is 6.2751. LSM = (x_{RT}+x_{TR})/2 = 6.1477. ✔ Similarily we get the LSM of the test 6.3125. ✔ Backtransformed LSMs are ℯ^{6.1477} = 467.64 (R) and ℯ^{6.3125} = 551.40 (T). Therefore, the GMR is 551.40/467.64 = 1.1791, agreeing with what PHX/WNL reports. PS: Sometimes in study reports the geometric means of treatments and results of the ANOVA are given in the same table (without the LSMs). Then – if the study was not balanced – people are confused, since the ratio of geometric means does not match the point estimate (calculated from the LSMs). Here the backtransformed geometric means are 471.64 (R) and 557.90 (T). Wrongly calculating the ratio 557.90/471.64 = 1.1829 ≠ 1.1791… — Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
khaoula ★ Algeria, 20140825 21:41 @ Helmut Posting: # 13430 Views: 11,990 

Hi Helmut, thank you very much for your answer, I'm very sorry to reply so late, at first I did not have internet, then I fell very ill. so I was wrong, I thought that after I sent the new results you answered me and GMR = 1.21 again !!! so now it's OK there is no problem between kinetica and unbalanced cross over !!! two problems only:
thank you very much I am very happy to have met you, you help me a lot !!!!! 