mahmoudteaima Regular Cairo, Egypt, 20160720 11:03 (edited by mahmoudteaima on 20160720 19:27) Posting: # 16496 Views: 3,705 

_{[/sub]Good morning all,
I want to know if bear 2.7.7 in R 3.3.1 running on macos 10.11.6 Elcaptin can perform ANOVA for lambda[sub]z}, t1/2 and tmax for test and reference products; also can it perform Kruskal Wallis analysis for tmax as these are non official requirements issued by the some members of the Bioequivalence committee at the central administration of pharmaceutical affairs (CAPA) at the ministry of health and population in egypt. If not, can these be considered in the coming updates of bear!!!!. Greetings for all. — Mahmoud Teaima, PhD. Assistant Professor at Department of Pharmaceutics and Industrial Pharmacy. Vicedirector at Center of Applied Research and Advanced Studies (CARAS). Faculty of Pharmacy, Cairo University, Cairo, Egypt. 
Helmut Hero Vienna, Austria, 20160720 12:25 @ mahmoudteaima Posting: # 16498 Views: 3,231 

Hi Mahmoud, we have to wait for Yungjin passing by when it comes to bear. I have some questions / remarks. » […] new requirements issued by the Bioequivalence committee at the central administration of pharmaceutical affairs (CAPA) at the ministry of health and population in egypt. Are these requirements available on the net? The latest version I know is of January 2010, which states: For t_{max} descriptive statistics should be given. If t_{max} is to be subjected to a statistical analysis this should be based on nonparametric methods and should be applied to untransformed data. » […] ANOVA for K (1st order elimination rate constant), t1/2 and tmax for test and reference products;
» […] also can it perform Kruskal Wallis analysis for tmax… Kruskal Wallis makes only sense for more than two treatments and will give only a global comparison. IMHO, not very helpful since posthoc tests for pairwise comparisons with a Bonferroniadjustment are needed. — Regards, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
mahmoudteaima Regular Cairo, Egypt, 20160720 14:12 @ Helmut Posting: # 16499 Views: 3,215 

Thanks Helmut for the prompt detailed fully scientific and statistically logic answers which i totally agree with you. Unfortunately things don't run fully according to book here. The official Egyptian guidelines to conduct BE studies totally agree with your words, but some members of the committee for evaluation of BE studies may insist on requests like the ones formentioned in my first post. that is why i see it will be helpful if bear can do such extra calculations. Herein below is an example of data submitted before: thalf I really don't know how these calculations were done, my guess it may be done by winnonlin!!!!!! Edit: Tabulators changed to spaces and BBcoded; see also this post #6. [Helmut] — Mahmoud Teaima, PhD. Assistant Professor at Department of Pharmaceutics and Industrial Pharmacy. Vicedirector at Center of Applied Research and Advanced Studies (CARAS). Faculty of Pharmacy, Cairo University, Cairo, Egypt. 
mittyri Senior Russia, 20160720 17:00 @ mahmoudteaima Posting: # 16503 Views: 3,156 

Hi Mahmoud, this doesn't look like Winnonlin/Phoenix output: 1. there is no Total row in Phoenix SS tables. Of course you can add it manually 2. As Helmut mentioned Kruskal Wallis test extends the Mann–Whitney U test when there are more than two groups. You won't find this test in Phoenix/Winnonlin, Mann–Whitney U test only By the way I don't see any difficulties to get these results in R using BEAR output as source data — Kind regards, Mittyri 
d_labes Hero Berlin, Germany, 20160720 16:05 @ Helmut Posting: # 16501 Views: 3,189 

Dear Helmut, » BTW, asking for a comparison of both k and t_{½} demonstrates a lack of understanding of statistics. They differ only by a factor of ln(2). Any comparison will give identical variance, CI, etc. I think here you err. t_{½} = ln(2)/k (some sort of 'inverse transformation').Lets consider for simplicity the GMR's (I consider the metric names are the geometric means here): t_{½T}/t_{½R} = (ln(2)/k_{T})/(ln(2)/k_{R}) = k_{R}/k_{T} That's not identical to the GMR for k IMHO . BTW: @Mahmoud. k or k_{e} will be better called lambda_{Z}  terminal rate constant. » » 1st order elimination rate constant is only correct within an onecompartment model. — Regards, Detlew 
Helmut Hero Vienna, Austria, 20160720 16:31 @ d_labes Posting: # 16502 Views: 3,161 

Dear Detlew, » » BTW, asking for a comparison of both k and t_{½} demonstrates a lack of understanding of statistics. They differ only by a factor of ln(2). Any comparison will give identical variance, CI, etc. » » I think here you err. I stand partly corrected. Just checked one of my studies (never compared this stuff before). λ_{z}: 99.66% (90% CI: 92.23% – 107.69%) t_{½}: 100.34% (90% CI: 92.86% – 108.42%) Within and betweensubject variances are identical. Hence, any conclusions (pass/fail) will be the same. Note the PEs close to 100% (“proving” the assumption?) and 1/1.0034 ≈ 0.9966 (and so are the CIs: 1/CL_{lower} of one metric = CL_{upper} of the other). Hence, testing both is nuts. — Regards, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
d_labes Hero Berlin, Germany, 20160721 14:42 @ Helmut Posting: # 16510 Views: 3,023 

Dear Helmut, » ... Hence, testing both is nuts. Totally correct IMHO. That'also my opinion. My concern was [nitpicking] "Any comparison will give identical variance, CI, etc." [/nitpicking]. But I must confess that I couldn't convince all sponsors I worked with of this opinion. Especially one from UpperBavaria was resistant. — Regards, Detlew 
Helmut Hero Vienna, Austria, 20160721 14:54 @ d_labes Posting: # 16511 Views: 3,025 

Dear Detlew, » But I must confess that I couldn't convince all sponsors I worked with of this opinion. Especially one from UpperBavaria was resistant. Haha! They once forced me to test AUC_{0–τ} and C_{average}… „Doppelt genäht hält besser” — Regards, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
yjlee168 Senior Kaohsiung, Taiwan, 20160720 21:10 @ mahmoudteaima Posting: # 16505 Views: 3,152 

Dear Mahmoud, No Kruskal Wallis analysis for tmax with bear at this moment. Sorry about that. Probably you can do it manually or using other statistical package for now. » ...these are non official requirements issued by the some members of the Bioequivalence committee at the central administration of pharmaceutical affairs (CAPA) at the ministry of health and population in egypt. I see. sound interesting. » If not, can these be considered in the coming updates of bear!!!!. I will consider your suggestion. BTW, thanks for letting me know that bear works well on the new MacOS. — All the best, Yungjin Lee bear v2.8.3: created by Hsinya Lee & Yungjin Lee Kaohsiung, Taiwan http://pkpd.kmu.edu.tw/bear Download link (updated) > here 
mahmoudteaima Regular Cairo, Egypt, 20160720 22:40 @ yjlee168 Posting: # 16506 Views: 3,116 

thanks Yungjin Lee for your reply, here is the results of kruskal wallis test of tmax data performed in R 3.3.1 on macos 10.11.6 Elcaptin. KruskalWallis rank sum test of tmax ref = c(1, 0.5, 2, 0.75, 1, 1.5, 2, 2, 1, 1.5, 0.75, 1.5, 0.75, 0.75, 1, 1, 0.75, 1.5, 1, 3, 0.75, 0.75, 1, 0.75) so no significant difference in tmax values between reference and test products. thanks for your advise in that. waiting your advise on how to make the remaining statistical analysis of lambdaz and t1/2?!!!!!. Greetings. — Mahmoud Teaima, PhD. Assistant Professor at Department of Pharmaceutics and Industrial Pharmacy. Vicedirector at Center of Applied Research and Advanced Studies (CARAS). Faculty of Pharmacy, Cairo University, Cairo, Egypt. 
yjlee168 Senior Kaohsiung, Taiwan, 20160721 11:11 (edited by yjlee168 on 20160721 11:44) @ mahmoudteaima Posting: # 16509 Views: 3,091 

Dear Mahmoud, » ... » KruskalWallis rank sum test of tmax ... » KruskalWallis chisquared = 0.0054846, df = 1, » pvalue = 0.941 Great. much better than SPSS, right? » waiting your advise on how to make the remaining statistical analysis of lambdaz and t1/2? I guess the descriptive statistics for λz and t_{1/2} should be good enough based on Egypt's BE GL, as mentioned previously by Helmut. I am not sure if the challenge was due to that you inappropriately submitted ANOVA for λz and t_{1/2} to the regulatory before. If so, you may consider to remove ANOVA stuffs and just present descriptive statistics for λz and t_{1/2}. — All the best, Yungjin Lee bear v2.8.3: created by Hsinya Lee & Yungjin Lee Kaohsiung, Taiwan http://pkpd.kmu.edu.tw/bear Download link (updated) > here 
mahmoudteaima Regular Cairo, Egypt, 20160727 12:13 @ yjlee168 Posting: # 16520 Views: 2,868 

This is the results of tmax for my BE project from both R3.3.3 and spss20 on macos 10.11.6 from spss: NPAR TESTS /KW=Tmax BY formulation(1 2) /STATISTICS DESCRIPTIVES /MISSING ANALYSIS. NPar Tests Syntax NPAR TESTS /KW=Tmax BY formulation(1 2) /STATISTICS DESCRIPTIVES /MISSING ANALYSIS. Descriptive Statistics N Mean Std. Deviation Minimum Maximum Tmax 48 1.1271 .50010 .50 3.00 treatment 48 1.50 .505 1 2 KruskalWallis Test Ranks treatment N Mean Rank Tmax ref 24 24.35 test 24 24.65 Total 48 Test Statisticsa,b Tmax ChiSquare .005 df 1 Asymp. Sig. .941 a Kruskal Wallis Test b Grouping Variable: treatment From R: KruskalWallis rank sum test of tmax > ref = c(1, 0.5, 2, 0.75, 1, 1.5, 2, 2, 1, 1.5, 0.75, 1.5, 0.75, 0.75, 1, 1, 0.75, 1.5, 1, 3, 0.75, 0.75, 1, 0.75) > test = c(1, 0.5, 2, 0.8, 0.8, 1.5, 1, 1.5, 1, 2, 0.5, 1.5, 0.8, 1, 1, 0.8, 1, 1, 1, 1.5, 0.8, 0.8, 1, 0.8) > dati = list(g1=ref, g2=test) > kruskal.test(dati) KruskalWallis rank sum test data: dati KruskalWallis chisquared = 0.0054846, df = 1, pvalue = 0.941 Can anyone see a difference? — Mahmoud Teaima, PhD. Assistant Professor at Department of Pharmaceutics and Industrial Pharmacy. Vicedirector at Center of Applied Research and Advanced Studies (CARAS). Faculty of Pharmacy, Cairo University, Cairo, Egypt. 