Louis52 Junior 20180201 19:57 Posting: # 18328 Views: 1,270 

Hello, I'm new at this subject matter. I'm trying to learn more about BEQ studies, especially about how to get the CV(intra) when little or no (or not reliable) prior info available. I went over the "Biostatistics Sample Size Estimation for BE Studies" presentation (thank you) and at one point in one of the "Hints" slide it says that "All you need is the 90% geometric confidence interval and the sample size." My question is if I don't have no prior pilot study from which to get a CV intra, how would I use all this information? Am I understanding correctly that the CLlow and CLhigh would be the 90% confidence limits for a prior CV intra, or they refer to something else? I'm asking that because of the "...geometric confidence interval...". Why are we talking here about that? Thanks! 
Helmut Hero Vienna, Austria, 20180202 00:56 @ Louis52 Posting: # 18330 Views: 1,066 

Hi Louis, » […] how to get the CV(intra) when little or no (or not reliable) prior info available. I went over the "Biostatistics Sample Size Estimation for BE Studies" presentation […] and at one point in one of the "Hints" slide it says that "All you need is the 90% geometric confidence interval and the sample size." I guess you are referring to this presentation (slide 24), right? » Am I understanding correctly that the CLlow and CLhigh would be the 90% confidence limits for a prior CV intra… Essentially yes. <nitpick> The 100(1–2α) CI is calculated from
It’s not rocket science to perform the calculation “backwards” (get the unknown #4 from the known #1–#3 and the CI) and then CV_{intra}=√ℯ^{MSE}–1. If the GMR is not known, use √CL_{lower}×CL_{upper}. </nitpick> » I'm asking that because of the "...geometric confidence interval...". Why are we talking here about that? I wanted to make clear that we need for the calculations outlined in slides 25–28 the confidence interval derived from logtransformed data (which is required for most PK metrics like AUC, C_{max},…). Note that the calculation in the presentation is outlined for 2×2 crossover designs. It can be a little bit tricky to perform the calculations “manually” if you are unsure how the degrees of freedom in other designs are defined. I recommend the package PowerTOST for R (open source & free of costs). _{}Examples:
— Regards, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
Louis52 Junior 20180203 17:12 @ Helmut Posting: # 18337 Views: 906 

I guess I'm still a bit confused. Yes, it is about what is on the slide 25 'Hints' in the last bullet point: "All you need is the 90% geometric confidence interval and the sample size." If I am to look on slide 26 when the PE is derived it seems that the CLlo and CLhi are for the ratio for AUCs or Cmaxs as Test vs Ref from previous study. So maybe this is case when there is no other information about a previous study other than that. Because otherwise if we know the MSE for AUC or Cmax then it it easy to go get directly Cvintra. Am I correct on this? Or PE is something else? I'm asking because when delta(CL) is derived it seems that we take the logs. Not sure why... Thanks! 
Helmut Hero Vienna, Austria, 20180203 21:48 @ Louis52 Posting: # 18338 Views: 904 

Hi Louis, I don’t know which presentation you are referring to. Link? I’m referring to the one held in Moscow 2014 (slides 24–28). » If I am to look on slide 26 when the PE is derived it seems that the CLlo and CLhi are for the ratio for AUCs or Cmaxs as Test vs Ref from previous study. No. See slide 27: The confidence interval (CI) is given with 0.91–1.15. Hence, the lower confidence limit (CL_{lo}) was 0.91 and the upper one (CL_{hi}) 1.15. The point estimate (PE) aka geometric means ratio (GMR) was not given. It can be calculated by √CL_{lo}×CL_{hi}, i.e., √0.91×1.15=1.023. » So maybe this is case when there is no other information about a previous study other than that. Because otherwise if we know the MSE for AUC or Cmax then it it easy to go get directly Cvintra. Am I correct on this? Yes. This entire game is only necessary if the MSE is not given. » I'm asking because when delta(CL) is derived it seems that we take the logs. Not sure why... As I already wrote, all calculations are done in the logarithmic* domain. Did you ever wonder why the acceptance range of 80–125% is not symmetric around 100%? Most biologic variables follow a lognormal distribution. Makes sense because negative and even zero values are not possible (what’s a negative concentration?). It means also that logtransformed values follow a normal distribution which is one of the assumptions in the ANOVA. In BE we accept a deviation (Δ) of 20% as to be clinically not relevant. Consequently the lower limit of the acceptance range is 100(1–Δ)=80% and the upper one its reciprocal 100(1–Δ)^{–1}=125%. Transformed to logs we get –0.2231 and +0.2231 which are symmetric around zero. Bingo. We could apply the ANOVA. Remember some useful rules for working with logs:
CI=ℯ^{log(PE)∓t·SEΔ}. SE_{Δ} is calculated from the mean square error MSE and the total sample size (N) as √2·MSE∕N (slide 28). See slide 27 how the MSE is calculated from the difference between one of the CLs and PE, the subjects in both sequences (n_{1}, n_{2}), and the tvalue. Note that log(CL_{hi})–log(PE), log(PE)–log(CL_{lo}), or the logs of ratios give the same result: log(1.15)–log(1.023)=log(1.023)–log(0.91)=log(1.15∕1.023)=log(1.023∕0.91)=0.11702. Once we have the MSE, calculation of the CV is easy. If you do it by hand (I hope not), I suggest to calculate the CI afterwards. If you don’t get the original values, something went wrong.
— Regards, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
Louis52 Junior 20180204 14:38 @ Helmut Posting: # 18341 Views: 853 

Hi Helmut, Thank you for the explanation Probably, most of the times we would actually know the PE from a previous study together with the CI. That's why I would not see the need the recalculate the PE (as the GMR for either AUCs or Cmaxs). But I completely agree that if we get from a published article only the information about the GMR limits for AUC or Cmax, then this is a very clever way of deriving the CVintra. Couple of more questions:

Helmut Hero Vienna, Austria, 20180205 11:39 @ Louis52 Posting: # 18346 Views: 796 

Hi Louis, » 1. Presumably, n1 and n2 are form the previous study, right? As already explained these are the number of subjects / sequence (TR or RT). If known, fine. The calculated CV will be exact (to the numeric precision of the input). If unknown (rarely given in the literature) use the total number of subjects N and split it in such a way that the study is as balanced as possible. If the study was more unbalanced than assumed you will get a CV which is higher than the true one. Since this is conservative you will be one the safe side basing your sample size estimation on it (see there for an example). » 2. Where 0.91 and 1.15 are coming from? Did you select them just as an example or there is a particular reason? What do you think the word example in ■ Example: 90% CI [0.91 – 1.15], N 21 (n_{1} = 11, n_{2} = 10) in this presentation means?— Regards, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 