Elena777 Junior Belarus, 20180120 11:13 (edited by Elena777 on 20180121 08:46) Posting: # 18229 Views: 1,156 

Dear participants of forum, I am relatively new to R and RStudio. And I really need your advise to clear some questions up which are important for my selfeducation and professional skills. 1. A BE study of Sotalol HCl 160 mg tablets (Mylan) and Betapace 160 mg tablets (Berlex Laboratories) with standard 2x2 crossover design was described in one of the applications on FDA (application number 75725). They planned to enroll 24 subjects. But in fact that study was conducted with 23 subjects. I would like to calculate CV from CI using data of that study. What data should I write for arguments "n" and "design"? If I input such data: CVfromCI(pe = 0.99, lower = 0.87, upper = 1.13, n = 23, design = "2x2", alpha = 0.05, robust=FALSE) RStudio gives me a result but with the following note: CVfromCI(pe = 0.99, lower = 0.87, upper = 1.13, n = 23, design = "2x2", alpha = 0.05, robust=FALSE) Can I use this result for futher calculations of CVpooled by means of using PowerTOST or should I correct my data? 2. While calculating CV from CI with data from another study I`ve got the following result: CVfromCI(pe = 0.96, lower = 0.9075, upper = 1.0089, n = 24, design = "2x2", alpha = 0.05, robust=FALSE) My input is correct because numbers are taken from that application not from my head. Should I pay attention to this warning message and can I use this result for futher calculations of CVpooled by PowerTOST? 3. While calculating CV from CI is it prefered to use CI for logtransformed parameters (for example LCmax) or for nontransformed parameters(Cmax)? Big thanks in advance. 
Helmut Hero Vienna, Austria, 20180121 17:56 @ Elena777 Posting: # 18241 Views: 877 

Dear Elena, » 1. A BE study […] planned to enroll 24 subjects. But in fact that study was conducted with 23 subjects. […] If I input such data: » » CVfromCI(pe=0.99, lower=0.87, upper=1.13, n=23, design="2x2", alpha=0.05, robust=FALSE) » » RStudio gives me a result but with the following note: » » CVfromCI(pe=0.99, lower=0.87, upper=1.13, n=23, design="2x2", alpha=0.05, robust=FALSE) » Unbalanced 2x2 design. n(i)= 12/11 assumed. » » Can I use this result for futher calculations of CVpooled by means of using PowerTOST or should I correct my data? The function does not “know” how many subjects in each of the sequences were dosed. The function tries to keep a 2×2 study with an odd number of subjects as balanced as possible (here 12 subjects in one sequence and 11 in the other) and throws this message. CVfromCI(pe=0.99, lower=0.87, upper=1.13, n=23, design="2x2") If you know the subjects per sequence and specify them in the argument n , you get the same result but without a message.CVfromCI(pe=0.99, lower=0.87, upper=1.13, n=c(12, 11), design="2x2") But the study might have been even more unbalanced. Let’s try 14 subjects in one sequence and 9 in the other: CVfromCI(pe=0.99, lower=0.87, upper=1.13, n=c(14, 9), design="2x2") Given that if you don’t know the subjects / sequence the code’s attempt to keep the sequences as balanced as possible gives you the highest (i.e., most conservative) estimate. For the background see this presentation (slides 25–29). » 2. While calculating CV from CI with data from another study I`ve got the following result: » » CVfromCI(pe=0.96, lower=0.9075, upper=1.0089, n=24, design="2x2", alpha=0.05, robust=FALSE) » [1] 0.1071475 » » My input is correct because numbers are taken from that application not from my head. Please check it again. The PE is given by √lower × upper. In your case that’s sqrt(0.9075*1.0089) or 0.9568577. The function checks the input for plausibility. Hence, the message is correct since your 0.96 is different to 0.9568577. » Should I pay attention to this warning message Yes. Check the PE. » and can I use this result for futher calculations of CVpooled by PowerTOST? I would (after checking the data for correctness) suggest the ones with the highest numeric precision. » 3. While calculating CV from CI is it prefered to use CI for logtransformed parameters (for example LCmax) or for nontransformed parameters(Cmax)? The former. According to all guidelines the analysis for C_{max} (and AUC as well) is done on logtransformed data. — Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
Elena777 Junior Belarus, 20180127 10:02 @ Helmut Posting: # 18295 Views: 747 

Dear Helmut, thanks, I really appreciate it. 