Yura Junior Belarus, 20170117 09:30 Posting: # 16963 Views: 1,730 

Dear all For calculation of a confidence interval of a difference of test and reference pharmaceuticals for two sequences (RTTR) the formula is used: What formula is used for calculation of a confidence interval of a difference of test and reference pharmaceuticals for three sequences (TRRRTRRRT)? Kindest regards 
BEproff Senior Russia, 20170117 12:33 @ Yura Posting: # 16965 Views: 1,577 

Hi Yura, Certainly I am not an expert in math but it seems that your formula uses arithmetic means while BE operates with geometric means... 
Yura Junior Belarus, 20170117 14:13 @ BEproff Posting: # 16966 Views: 1,555 

It is clear that log transformation becomes. I'm interested in the boundaries of the interval. 
d_labes Hero Berlin, Germany, 20170118 14:36 @ BEproff Posting: # 16974 Views: 1,432 

Dear BEproff! » Certainly I am not an expert in math ... Whereof one cannot speak, thereof one must be silent. Ludwig Wittgenstein SCRN. — Regards, Detlew 
d_labes Hero Berlin, Germany, 20170118 14:30 @ Yura Posting: # 16973 Views: 1,443 

Dear Yura! » For calculation of a confidence interval of a difference of test and reference pharmaceuticals for two sequences (RTTR) the formula is used: » what you have delivered here is not the formula for a 2x2 design , but the formula for a parallel group design with 2 groups. Replace sigma^{2} with 0.5*sigma^{2} and you get the correct formula. » What formula is used for calculation of a confidence interval of a difference of test and reference pharmaceuticals for three sequences (TRRRTRRRT)? The formula for the partial replicate is very similar to the above formula: Replace sigma^{2} with (1/6)*sigma^{2} and use (1/n_{1}+1/n_{2}+1/n_{3}) under the squareroot and you got it . The n_{i} are the number of subjects in the 3 sequence groups. Check out the function CI.BE() in package PowerTOST which can do the calculations for you. Given you have the n_{i}, the point estimate in the original scale (i.e. sloppy spoken the GMR) and the residual CV. The formula for (TRRRTRRRT) clearly depends on the statistical model. I gave you the one which is obtained if the same model for replicate studies is used as in case of a 2x2 crossover: all effects fixed, data without missings. That way is what the EMA wants us to go. A mixed model analysis is a different cattle of fish . Moreover I strongly recommend not to use an explicit formula for the CI but let the software do that job. Most if not all statistical software has tools for that. For instance estimate statement in SAS Proc GLM or confint() function in R just to name two of them. Doing so circumvents all the bells and whistles with missing data, unbalancedness and so on.Hope this helps. — Regards, Detlew 
Yura Junior Belarus, 20170119 05:43 @ d_labes Posting: # 16976 Views: 1,340 

Dear d_labes thank you very much opened their eyes Regards 
Mahmoud Junior Jordan, 20170119 07:01 @ Yura Posting: # 16977 Views: 1,328 

» What formula is used for calculation of a confidence interval of a difference of test and reference pharmaceuticals for three sequences (TRRRTRRRT)? Dear Sir ===== Lower= EXP(Mean(T)Mean(R)SQRT(var(intra)*1/6(1/n1+1/n2+1/n3))) Upper= EXP(Mean(T)Mean(R)+ SQRT(var(intra)*1/6(1/n1+1/n2+1/n3))) Mahmoud AbdelMohsen Ph.D Edit: Full quote removed. Please delete everything from the text of the original poster which is not necessary in understanding your answer; see also this post #5! [Helmut] 
Yura Junior Belarus, 20170119 08:16 @ Mahmoud Posting: # 16978 Views: 1,312 

Dear Mahmoud Where tst? 
Mahmoud Junior Jordan, 20170119 08:55 @ Yura Posting: # 16979 Views: 1,293 

» Where tst? Dear Sir ===== This is the exact formula for find the CIs for this design 
Yura Junior Belarus, 20170120 06:28 @ Mahmoud Posting: # 16984 Views: 1,192 

Dear Mahmoud Using a formula in an example of the working group of EMA, results don't turn out. Regards 
Mahmoud Junior Jordan, 20170120 09:11 @ Yura Posting: # 16986 Views: 1,178 

Dear Sir ===== Lower= EXP(Mean(T)Mean(R) t(.05,2(n1+n2+n3)3)*SQRT(var(intra)*1/6(1/n1+1/n2+1/n3))) Upper= EXP(Mean(T)Mean(R)+ t(.05,2(n1+n2+n3)3)*SQRT(var(intra)*1/6(1/n1+1/n2+1/n3))) Edit: Full quote removed. Please delete everything from the text of the original poster which is not necessary in understanding your answer; see also this post #5! Yesterday I even deleted a post of yours which contained only a lengthy quote of yourself (i.e., without any additional text). First warning posted on the 16^{th}. Since you repeatedly did not follow the Forum’s Policy, I blocked your account for three weeks. [Helmut] 
Yura Junior Belarus, 20170120 11:33 @ Mahmoud Posting: # 16988 Views: 1,154 

Dear Mahmoud now yes 