It is true that PROC TTEST excludes the observation (subject 17) which has missing Cmax for one of the periods. Are you sure that you want a=0.1 as the significance level? Note that the confidence intervals involved in the TOST are (1- 2 a)*100% CIs, so these would be 80% CIs.Is this your focus and not the comparison of treatments? The first table ("result of Equivalence Analysis") shows the tests for the comparison of periods.The permutation of Cmax1 and Cmax2 in the VAR statement of PROC TTEST adds confusion and does not really compensate for the data issue because then the even-numbered subjects have their Cmax values incorrectly assigned.Most importantly, there are serious inconsistencies between your raw data files test-2.csv and unbalance.csv: For the odd-numbered subjects the Cmax values are assigned to different treatments (i.e., T and R have been interchanged).Įxample: Subject 1 had Cmax=23.7 for treatment T and 19.2 for treatment R according to test-2.csv, but (seemingly) vice versa according to unbalance.csv.Hello of all, I see several (potential) issues in your second post: Var Cmax1 Cmax2 / crossover=(drug1 drug2) Proc ttest data=trans dist=lognormal tost(0.8, 1.25) alpha=0.1 order=data
#Phoenix winnonlin training code
And your PROC TTEST code doesn't reflect the crossover design.įinally, the results from "Phoenix/Winnonlin" can be replicated (up to minor differences in the CI, also not converting ratios into percentages) with the code below: ods output statistics=stats1(where=(sequence=:'Both' & treatment=:'Ratio'))ĮquivTests=eqtst(where=(test ne: 'O' & treatment=:'R' & method=:'P')) However, I think this table is incorrect anyway because you specify the log-normal distribution for the logarithms of Cmax. The table "Equivalence Analysis" can be replicated with dataset TEMP1_1 as created above and your code, but, obviously, only after changing the TOST parameters to those in column "Null" of the output: (0.8, 1.25). Note that the "Difference Between Means" (named "Mean" in my output) was calculated for R − T (referring to lnCmax), whereas the ratio (in percent, for Cmax) was calculated in the opposite direction: T / R. Without the CROSSOVER option the confidence limits would be quite different. Var lowerclmean mean upperclmean point_esti lower upper ĬLMean Mean CLMean point_esti lower upper Var lnCmax1 lnCmax2 / crossover=(drug1 drug2) PROC PRINT output: Subject drug1 drug2 Cmax1 Cmax2 lnCmax1 lnCmax2Ħ R T 22.1 26.9 3.09558 3.29213 ods output statistics=stats0(where=(sequence=:'Both' & treatment=:'D' & method=:'P')) To replicate the results in your first screenshot I created a transposed dataset with a structure similar to Example 122.4 AB/BA Crossover Design in the PROC TTEST documentation (which I selected because obviously your data is from such a crossover trial): proc sort data=have out=tmp Input Subject Treatment $ Period Sequence $ Cmax Īpparently you created an additional variable lnCmax. Infile 'C:\Temp\test-2.csv' dsd firstobs=2 So, you started with a dataset created from test-2.csv: data have