Bioequivalence and Bioavailability Forum

Hi BE-proff,

❝ library(PowerTOST)

❝ sampleN.scABEL.ad(CV=0.398, theta0=1.15, targetpower=0.80, design="2x2x4")

THX for being concerned about the consumer risk! With a CV_wR of 39.8% we can expect a slight inflation of the Type I Error (TIE).

❝ Is it correct that:

❝ - at least 48 subjects to be randomized

Yes.

❝ - CI for Cmax is 0.74 - 1.34

No. For C_max the expanded limits are 0.7472...1.3383. The confidence interval has to lie entirely within these limits and additionally the PE must lie within 0.8–1.25.
Note that you have to use an adjusted α. Only if the CV_wR in the study would be exactly as assumed and there will be no dropouts, you could used the adjusted α from the sample size estimation above and calculate the 100(1–2×0.04444)=91.112% CI.

If you observe a different CV_wR in the study (likely) and there were dropouts you have to check whether the original iteratively adjusted α still suits and – if not – get a new one based on the study’s data. Examples for one dropout in the first sequence and two in the other:

• CV_wR 38% (lower than expected):
  
  library(PowerTOST)
scABEL.ad(CV=0.38, theta0=1.15, design="2x2x4", n=c(23, 22), alpha.pre=0.0444)

+++++++++++ scaled (widened) ABEL +++++++++++
         iteratively adjusted alpha
---------------------------------------------
Study design: 2x2x4 (RTRT|TRTR)
log-transformed data (multiplicative model)
1,000,000 studies in each iteration simulated.

CVwR 0.38, n(i) 23|22 (N 45)
Nominal alpha                 : 0.05, pre-specified alpha 0.0444
True ratio                    : 1.1500
Regulatory settings           : EMA (ABEL)
Empiric TIE for alpha 0.0444  : 0.05485
Power for theta0 1.1500       : 0.761
Iteratively adjusted alpha    : 0.03910
Empiric TIE for adjusted alpha: 0.05000
Power for theta0 1.1500       : 0.745
  
  Since the CV_wR is closer to the critical range (maximum TIE at CV_wR 30%) you have to use the adjusted α 0.03910 (which is lower than the planned one). Hence you loose some power. Assess ABEL based on the 100(1–2×0.03910)=92.18% CI.
  
  
• CV_wR 42% (higher than expected):
  
  scABEL.ad(CV=0.42, theta0=1.15, design="2x2x4", n=c(23, 22), alpha.pre=0.0444)

+++++++++++ scaled (widened) ABEL +++++++++++
         iteratively adjusted alpha
---------------------------------------------
Study design: 2x2x4 (RTRT|TRTR)
log-transformed data (multiplicative model)
1,000,000 studies in each iteration simulated.

CVwR 0.42, n(i) 23|22 (N 45)
Nominal alpha                 : 0.05, pre-specified alpha 0.0444
True ratio                    : 1.1500
Regulatory settings           : EMA (ABEL)
Empiric TIE for alpha 0.0444  : 0.04491
Power for theta0 1.1500       : 0.811
TIE not > nominal alpha; the chosen pre-specified alpha is justified.
  
  Since no inflation of the TIE is expected with α 0.0444, drop it from the arguments (i.e., try with the nominal α 0.05 only).
  
  scABEL.ad(CV=0.42, theta0=1.15, design="2x2x4", n=c(23, 22))

+++++++++++ scaled (widened) ABEL +++++++++++
         iteratively adjusted alpha
---------------------------------------------
Study design: 2x2x4 (RTRT|TRTR)
log-transformed data (multiplicative model)
1,000,000 studies in each iteration simulated.

CVwR 0.42, n(i) 23|22 (N 45)
Nominal alpha                 : 0.05
True ratio                    : 1.1500
Regulatory settings           : EMA (ABEL)
Empiric TIE for alpha 0.0500  : 0.04803
Power for theta0 1.1500       : 0.823
TIE not > nominal alpha; no adjustment of alpha is required.
  
  Perfect. Use the nominal α 0.05 and assess ABEL based on the 100(1–2×0.05)=90% CI. You gain power.

❝ - CIs for AUCs are 0.80-1.25

The acceptance limits.