Thank you SteveDenham. All the best, Skillside ... View more

Dear Steve, thank you for your quick response, your questions are crucial. However, I have a feeling that in this case we too much focus on a dependent variable, while the predictor can change over time as well. The predictor in this case is ordinal expressed on Lickert's scale, Is there any statistical test that focuses on the variability of the predictor as well? I hope you understood my idea. ... View more

Thank you for your all answers FreelanceReinhard, Of course if the H 0 : p 1 =p 2 is a true assumption we are probably won't ever know where is the truth before the study. So in whole disccusion i don't see the point of calculating samplesize for noninferiority/superiority study, isn't it better to calculate samplesize for test which gonna be used at the end? In that case Chi square? May I ask you, where do you work? Best, ... View more

Great to hear that, i regret that i am not on your level. I was pretty sure that I have some skills in sas, but after a discussion with you I feel like you brought me down on earth. Thanks. So let's keep the optimistic variant, and let me keep the sample size for noninferiorityty. So at the end of the study what kind of test you would recommend using for testing the hypothesis that there is a difference between frequencies in groups? Chi-square assumes no difference, what is the alternative for that? and isn't it give a higher probability of committing a Beta error in that case? ... View more

And thank you for providing the link to documentation, but unfortunately, i don't understand most of the equations, i am not mathematician ... View more

Thank You for your answer, and i am sorry for delay in resposne. Tough day, today 😉 So if i understood it correctly the following proc will be correct for noninferiority sample size calculation, so i cannot set nullpdiff below 2% proc power; twosamplefreq groupps=(0.97 0.95) alpha=0.025 nullpdiff=0.02 power=0.9 test=fm sides=1 npergroup=.; run; That gives 535 and for nullpdiff is more optimistic 356. But then when i calcualte power for Chi-square test it doesn't give the sufficient power.   I wrote: the drugs do not differ between each other, i either believe that this one with 95% success rate(SR) is superior to drug with 97% SR Well, they differ between endpoints but i assume that in real life scenario and on the bigger sample it will be the same as first drug or even better. I meant that i assume the margin which will show equality or even superiority. Could you provide some simulations, please?   ... View more

Thank you for your support. This discussion is exciting for me. But hold on..... seems like i misunderstood something. Once again, let's use a two scenarios (S1 and S2). S1: I would like to conduct samplesize estimation in a noninferiority study. The drug 2 is not worse than drug 1, where the success rate of drug 1 is 97% and the success rate for drug 2 is 95%. So difference in the endpoints is 2%, i suggested margin (probably nullpdiff) 1% or 3% or 4%. I was sure that in this case the assumption for margin = 1 is that drug 2 will have success rate between 94 and 96%. S2: let's assume that drug 2 to will be superior to drug 1. Still the succes rates are the same for both drugs, but then i have to expect margin at least 3? Other questions: if i have set the nullpdiff option as positive value then swaping the order in the groupps option changes the calculation from superior to noninferiority samplesize calculation. And if i won't swap order in groupps option and i will change the value of nullpdiff for negative then i will as well change from superiority to noninferiority study? What should i do to decrease the samplesize? the drugs do not differ between each other, i either believe that this one with 95% success rate(SR) is superior to drug with 97% SR. Please, suggest a solution for that case. The study will take too much time if i will have 1500 participants per arm. In my time zone it after midnight, 00:22. Have a nice night! Thanks again, and i hope to talk to you tomorrow. ... View more

Thank you for your answer. So let me show you what i've done. Superiority for drug 1 vs drug 2 proc power; twosamplefreq groupps=(0.97 0.95) alpha=0.025 nullpdiff=0.01 power=0.8 test=fm sides=1 npergroup=.; run; The result is 689. Then the outcome - endpoints (binary variables) will be tested via Chi-square test. So i checked the power of the Chi-square. The hypothesis in Chi-square test assumes no difference: proc power; twosamplefreq test=pchi groupproportions = (.97 .95) nullproportiondiff = 0.00 power = . npergroup = 698; run; The result show power at level 0.4... Thank you for your answer. Could you please explain me what nullproportiondiff is for in chi-square while the it assumes no differences? ... View more

I have an addtional question. So i calculated the required sample size and i am going to conduct a study. Probably, at the end of the study, i will asses the differences (of outcomes) with Chi square test and when i implement the results from sample size estimation (it was set on 0.9 power) the power of the Chi-square test is around 0.6. I was sure that after the calculation of the required sample size the tests will be more powerful, could you please provide me an explanation of this situation? ... View more

Thank you for your explanation. ... View more

Thank you for your replay. Is it mean that 0.90 in your code is the ref group? and experimental group is 0.96 so you calculate the superiority? and if you swap the order then you will have non-inferiroirty sample size calculation? thank you ... View more

Thank you, it seems that it works! Hope that journal editor won't request different format. Regards, Skillside ... View more