3 Power and sample size for ABE in the 2× 2 design Here we give formulae for the calculation of the power of the TOST procedure assuming that there are in total n subjects in the 2× 2 trial. Suppose that the null hypotheses given below are to be tested using a 100(1 − α)% two-sided confidence interval and a power of (1 − β) is required to reject these hypotheses when they are false. H :µ ≤− ln 1.25 H :µ ≥ ln 1.25. Let us define x to be a random variable that has a noncentral t-distribution with df degrees of freedom and noncentrality parameter nc, i.e., x ∼ t(df,nc). The cumulative distribution function of x is defined as CDF(t,df,nc) = Pr(x≤ t). Assume that the power is to be calculated using log(AUC). If σ is the common within-subject variance for T and R for log(AUC), and n/2 subjects are allocated to each of the sequences RT and TR, then 1−β = CDF(t , df,nc )−CDF(t , df,nc ) (7.1) where √ n(log(µ )− log(0.8)) nc = 2σ √ n(log(µ )− log(1.25)) nc = 2σ √ (n− 2)(nc −nc = ) df 2t and t is the 100(1 − α)% point of the central t-distribution on n− 2 degrees of freedom. Some SAS code to calculate the power for an ABE 2 × 2 trial is given below, where the required input variables are α, σ value of the ratio µ and n, the total number of subjects in the trial.