BEQool ☆ Slovenia, 20230524 23:05 (3 d 22:22 ago) Posting: # 23567 Views: 211 

Hello all! I have a question regarding sample size estimation for parallel designs with more than 2 treatment arms (with more than 2 groups). Package PowerTOST in R covers sample size estimation for parallel design for just 2 parallel groups ( known.designs() ) and not for more. Lets say we are conducting a (pilot) parallel study with 3 test formulations and 1 reference. How do we estimate sample size in this case? I would say that we just double the necessary sample size estimated (in R with sampleN.TOST) for 2 parallel groups (to get 4 independent groups for 3 test formulations and 1 reference). For example for CV=0.2 (parallel design and theta0=0.95) for 2 parallel groups we get sample size of 36 subjects (total sample size for 2 groups), so if we double it we get 72 subjects (for 4 groups?). Is this assumption and sample size estimation a correct one? On the other hand I have read this article regarding multiple treatments comparison in parallel design and I dont exactly understand why should we just control Type I Error and change alpha when estimating sample size with multiple treatments? So in our case (when using Bonferroni adjustment) we would get the total sample size (for 4 groups) of 50 subjects (parallel design, CV=0.2, theta0=0.95, alpha=0.0167)? Is this correct? What is more, 50 is not even dividable by 4. I am confused and I probably didnt get this explanation in the article right. Nevertheless I wouldnt mind this being the right way as the needed sample size is smaller Regarding alpha adjustment in the first case (for doubling sample size from 36 to 72) I would say that it is not necessary as it is a pilot study. On the other hand, if this was a pivotal study and we would want to show equivalence for just 1 test formulation (among 3) then we would have to adjust alpha (to 0.0167) and then I would say we would double the estimated sample size for 2 groups the same way as described above (so 50x2=100 subjects)? So which sample size estimation is a correct one? BEQool 
dshah ★★ India/United Kingdom, 20230525 11:42 (3 d 09:45 ago) @ BEQool Posting: # 23568 Views: 166 

Hello BEQool! When we refer to EMA BE guideline, the BE estimation shall be considered for the concerned treatment and the rest of the treatment shall be separated from calculation purpose. So if there are more than two treatments, why can't the sample size be considered separately and then take a overall decision? Consider below case. Case 1: Desired Power 80% r (note: nT*r=nR) 1 a 5% Total CV 40.00% Expected mT/mR Ratio 95.00% Lower Equivalence Limit 80.0% Upper Equivalence Limit 125.0% Required Sample Size 65 (nA)+65 (nB)=130 (Total) Case 2: Desired Power 80% r (note: nT*r=nR) 1 a 5% Total CV 40.00% Expected mT/mR Ratio 90.00% Lower Equivalence Limit 80.0% Upper Equivalence Limit 125.0% 1b 0.800076232 Required Sample Size 133 (nA)+133 (nB)=266 (Total) your case: Desired Power 80% r (note: nT*r=nR) 1 a 5% Total CV 20.00% Expected mT/mR Ratio 95.00% Lower Equivalence Limit 80.0% Upper Equivalence Limit 125.0% Required Sample Size 18 (nA)+18 (nB)=36 (Total) And for 4 arm> sample size is 72. But why the assumption are same? what is the purpose of additional arm, if the expected ratio is same? Regards, Divyen 
Helmut ★★★ Vienna, Austria, 20230527 10:26 (1 d 11:01 ago) @ BEQool Posting: # 23573 Views: 101 

Hi BEQool, ❝ I have a question regarding sample size estimation for parallel designs with more than 2 treatment arms (with more than 2 groups). Package PowerTOST in R covers sample size estimation for parallel design for just 2 parallel groups ( sampleN.TOST(..., design = "parallel") – like all functions of PowerTOST – gives the total sample size.❝ Lets say we are conducting a (pilot) parallel study with 3 test formulations and 1 reference. […] I would say that we just double the necessary sample size estimated (in R with sampleN.TOST) for 2 parallel groups (to get 4 independent groups for 3 test formulations and 1 reference). For example for CV=0.2 (parallel design and theta0=0.95) for 2 parallel groups we get sample size of 36 subjects (total sample size for 2 groups), so if we double it we get 72 subjects (for 4 groups?). Is this assumption and sample size estimation a correct one?
❝ […] I have read this article regarding multiple treatments comparison in parallel design and I dont exactly understand why should we just control Type I Error and change alpha when estimating sample size with multiple treatments? In a pilot study we don’t have to adjust \(\alpha\). Actually the CI is not relevant at all and therefore, we can use any (not too small) sample size. We are interested in the T/Rratios and CVs for selecting the ‘best’ candidate (see case 1 in this article about higherorder Xovers) and planning the pivotal study. ❝ So in our case (when using Bonferroni adjustment) we would get the total sample size (for 4 groups) of 50 subjects (parallel design, CV=0.2, theta0=0.95, alpha=0.0167)? Is this correct? adj < TRUE (potentially applicable in a pivotal study; see the explanations at the end):
❝ What is more, … ❝ … 50 is not even dividable by 4. ❝ I am confused and I probably didnt get this explanation in the article right. Nevertheless I wouldnt mind this being the right way as the needed sample size is smaller ❝ Regarding alpha adjustment in the first case (for doubling sample size from 36 to 72) I would say that it is not necessary as it is a pilot study. ❝ On the other hand, if this was a pivotal study and we would want to show equivalence for just 1 test formulation (among 3) then we would have to adjust alpha (to 0.0167) and then I would say we would double the estimated sample size for 2 groups the same way as described above (so 50x2=100 subjects)?
Furthermore, we don’t need an adjustment when comparing one test to comparators from two regions.
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