Bioequivalence and Bioavailability Forum

Dear Alice!

❝ […] I'm sorry if my english is not really good

No problem - we have just a few native speakers of English here.

❝ I've got some difficulties to understand tost power.

Power=1-beta, where in the framework of BE beta is the producer's risk to fail in demonstrating bioequivalence with a true bioequivalent formulation. Most companies try to plan for a sample size of ~90% (optimistic case: all assumptions on the CV, T/R-ratio hold, no drop-outs) in order to get ~80% if assumptions are violated (higher CV, T/R deviating more from unity, drop-outs).
You cannot calculate sample size directly, but only power based on fixed values: CV, T/R-ratio, alpha (generally 0.05), acceptance range (generally 0.80-1.25), sample size. For any combination of these values you get a power value. Now you increase the sample size until the calculated power is > the target power.
Example: alpha 0.05, beta 20% (target power: 1-beta=80%), T/R 0.95, CV_intra 20%. You start the iterative search with a sample size of 16 subjects and obtain:
  n    power
 16   73.54%
 17   76.51%
 18   79.12%
 19   81.43%
 20   83.47%
With 19 subjects you already exceed the target power of 80%. In a TR/RT 2×2×2 cross-over you will start with a balanced design (equal number of subjects in each sequence) - therefore you round up to the next even number 20 [N = n_TR (10) + n_RT (10); power 83.47%].

❝ Maybe have you some reading for help me?

Maybe you find one of my presentations useful. References are given at the end.

❝ I've looked at D.Labes R packages, but I don't understand how this function run and so I can't writte it in my report.

Have you tried help(PowerTOST) after loading the package?
The example above would be coded by means of sampleN.TOST()

sampleN.TOST(alpha = 0.05, targetpower = 0.8, logscale = TRUE,
theta1 = 0.8, theta2 = 1.25, diff = 0.95, CV = 0.2, design = "2x2",
exact = TRUE, print = TRUE, details = TRUE)

resulting in

+++++++++ Equivalence test - TOST +++++++++
          Sample size estimation
-------------------------------------------
Study design:  2x2 crossover
Design characteristics:
df = n-2, design const. = 2, step = 2

log-transformed data (multiplicative model)

alpha = 0.05, target power = 0.8
BE margins        = 0.8 ... 1.25
Null (true) ratio = 0.95,  CV = 0.2

Sample size search
 n     power
16   0.735413
18   0.791240
20   0.834680

Exact power calculation with
Owen's Q functions.

Package sampleN.TOST() gives samples for balanced designs only (therefore no values for 17 and 19).


At CV 20% we plan the study with 20 subjects (red diamonds). Power (green lines) is 83.47%. If CV increases, we still can go with 20 subjects (although power decreases), until we reach CV 20.98%. Power would be 79.99% and we have to increase the sample size in order to stay >80%.
Another interesting point: The minimum sample size in most regulations is 12. This translates to a CV of 15.63% (power 80.02%). If we keep the sample size at 12 and the CV is even lower, it becomes more and more likely that we get a significant treatment effect (confidence interval does not include 100%). Might be problematic in Denmark. For formulations with very low variability (yes, I've seen a CV of 6%), 4 subjects would be enough (power 80.52%). If we run the study in 12, power will be 99.99993%.

According to ICH-E9 you should perform a sensitivity analysis in study planning. In that case power.TOST() helps. Let's assume that you planned the study with 20 subjects and want to know the power if T/R is 0.90 instead of 0.95...
power.TOST(alpha = 0.05, logscale = TRUE,
theta1 = 0.8, theta2 = 1.25, diff = 0.90,
CV = 0.2, n = 20, design = "2x2", exact = TRUE)

We get
[1] 0.5649986

Oh, that's bad.
Let's keep the T/R-ratio at 0.95 and increase the CV to 0.25 instead - we get
[1] 0.6430574

Generally power functions are quite flat on the top (~±5% from 100%, example plot), but drop off quite fast if we move away from 100%. The impact of CV is not so important. Drop outs have the least impact (we have seen already above that with 16 subjects power will still be 73.54%).