Bioequivalence and Bioavailability Forum

Hi BEQool,

❝ […]: why does the sample size estimation with R with package PowerTOST differ between sampleN.TOST and sampleN.scABEL when CV=30%?

Contrary to ABE, where power (and thus, the sample size) can be obtained by closed formulas, ABEL is a framework, which depends on the realized (observed) CV_wR, the upper constraint uc = 50%, and the point estimate.

Therefore, simulations are needed. At a true CV_wR = 30% we have – roughly* – a 50% chance that in the actual study CV_wR > 30%. Then we could expand the limits, gain power for a given sample size, or need less subjects for a certain power than we would need in ABE.

❝ So why do the sample size estimations differ? They have the same arguments (design="2x3x3", theta0=0.95 ...). CV is 30% so there should be no scaling (conventional BE limits, i.e., 80.00-125.00).

Nope. See above.

You could also remove all scaling conditions. Then you get the same sample size than with sampleN.TOST():
reg <- reg_const(regulator = "USER", r_const = 1, CVswitch = Inf, CVcap = Inf, pe_constr = FALSE)
sampleN.scABEL(CV = 0.3, theta0 = 0.95, design = "2x3x3", regulator = reg, details = FALSE)
+++++++++++ scaled (widened) ABEL +++++++++++
            Sample size estimation
   (simulation based on ANOVA evaluation)
---------------------------------------------
Study design: 2x3x3 (partial replicate)
log-transformed data (multiplicative model)
1e+05 studies for each step simulated.

alpha  = 0.05, target power = 0.8
CVw(T) = 0.3; CVw(R) = 0.3
True ratio = 0.95
ABE limits / PE constraint = 0.8 ... 1.25
USER defined regulatory settings
- CVswitch            = Inf
- no cap on scABEL
- regulatory constant = 1
- no pe constraint

Sample size
 n     power
30   0.8196

❝ Does it have to do anything with simulations? But even when I increase number of simulations, the differences aren't that big.

No again.

❝ Or if I reformulate the question, why dont the following powers match:

❝ a) power.TOST(CV=0.3, theta0=0.95,design="2x3x3", n=30)


❝ [1] 0.8204004

❝ b) power.scABEL(CV=0.3, theta0=0.95, design="2x3x3", n=30)


❝ [1] 0.85977

You asked the wrong question in b) because
power.scABEL(CV=0.3, theta0=0.95, design="2x3x3", n=27)
[1] 0.82566

Let’s consider an example where expanding the limits is less likely.

1. ABE (exact)
   sampleN.TOST(CV = 0.25, theta0 = 0.95, design = "2x3x3")
+++++++++++ Equivalence test - TOST +++++++++++
            Sample size estimation
-----------------------------------------------
Study design: 2x3x3 (partial replicate)
log-transformed data (multiplicative model)

alpha = 0.05, target power = 0.8
BE margins = 0.8 ... 1.25
True ratio = 0.95,  CV = 0.25

Sample size (total)
 n     power
21   0.814342
   
   
2. ABEL (simulations)
   Note that for the partial replicate design you should use subject simulations by the function sampleN.scABEL.sdsims() instead of simulating the associated statistics by the function sampleN.scABEL().
   sampleN.scABEL.sdsims(CV = 0.25, theta0 = 0.95, design = "2x3x3", details = FALSE)
+++++++++++ scaled (widened) ABEL +++++++++++
            Sample size estimation
   (simulation based on ANOVA evaluation)
---------------------------------------------
Study design: 2x3x3 (partial replicate)
log-transformed data (multiplicative model)
1e+05 studies for each step simulated.

alpha  = 0.05, target power = 0.8
CVw(T) = 0.25; CVw(R) = 0.25
True ratio = 0.95
ABE limits / PE constraint = 0.8 ... 1.25
Regulatory settings: EMA

Sample size
 n   power
21   0.8223

Now the sample sizes are identical. Power for ABEL is slightly larger than for ABE because there is a certain – while small – chance to expand the limits.

BTW, I would never ever assume theta = 0.95 for a HVD(P). That’s why in the reference-scaling functions of PowerTOST theta = 0.90 is the default.

---

• The simulated variance \(\small{s_\text{wR}^2}\) depends on its associated \(\small{\chi^2}\)-distribution with \(\small{n-2}\) degrees of freedom. The \(\small{\chi^2}\)-distribution is skewed to the right. Hence, when simulating any \(\small{CV_\text{wR}=x}\), you will obtain slightly more values \(\small{>x}\) than \(\small{<x}\).
  Try this to show the asymmetric distribution:
  n  <- 1e10
df <- 27 - 2
set.seed(1234567)
x  <- rchisq(n = 1e6, df = df)
y  <- hist(x, breaks = "FD", plot = FALSE)
plot(y, freq = FALSE, col = "lightblue", border = NA,
     xlim = c(0, max(y$mids)), cex.main = 1, las = 1,
     xlab = bquote("Number of random samples of "*chi^2 == .(n)),
     main = bquote(italic(df) == .(df)))
abline(v = y$mids[y$density == max(y$density)], col = "blue")
box()