Parameter  Argument  Purpose  Default 

\(\small{\alpha}\)  alpha 
Nominal level of the test  0.05 
\(\small{\pi}\)  targetpower 
Minimum desired power  0.80 
logscale  logscale 
Analysis on logtransformed or original scale?  TRUE 
\(\small{\theta_0}\)  theta0 
‘True’ or assumed deviation of T from R  see below 
\(\small{\theta_1}\)  theta1 
Lower BE limit  see below 
\(\small{\theta_2}\)  theta2 
Upper BE limit  see below 
CV  CV 
CV  none 
design  design 
Planned design  "2x2" 
method  method 
Algorithm  "exact" 
robust  robust 
‘Robust’ evaluation (Senn’s basic estimator)  FALSE 
print 
Show information in the console?  TRUE 

details  details 
Show details of the sample size search?  FALSE 
imax  imax 
Maximum number of iterations  100 
Defaults depending on the argument logscale
:
Parameter  Argument  logscale=TRUE 
logscale=FALSE 

\(\small{\theta_0}\)  theta0 
0.95 
+0.05 
\(\small{\theta_1}\)  theta1 
0.80 
−0.20 
\(\small{\theta_2}\)  theta2 
1.25 
+0.20 
Arguments targetpower
, theta0
, theta1
, theta2
, and CV
have to be given as fractions, not in percent.
The CV is generally the within (intra) subject coefficient of variation. Only for design = "parallel"
it is the total (a.k.a. pooled) CV. ^{1}
The terminology of the argument design
follows this pattern: treatments x sequences x periods
. The conventional TRRT (a.k.a. ABBA) design can be abbreviated as "2x2"
. Some call the "parallel"
design a ‘onesequence’ design. The design "paired"
has two periods but no sequences and is the standard design for studying linear pharmacokinetics (where a single dose is followed by multiple doses). A profile in steady state (T) is compared to the one after the single dose (R). Note that the underlying model assumes no period effects.
Implemented exact algorithms are "exact"
/ "owenq"
(Owen’s Q function, default) ^{2} and "mvt"
(direct integration of the bivariate noncentral tdistribution). Approximations are "noncentral"
/ "nct"
(noncentral tdistribution) and "shifted"
/ "central"
(‘shifted’ central tdistribution).
"robust = TRUE"
forces the degrees of freedom to nseq
and is used only in higherorder crossover designs. It could be used if the evaluation was done with a mixedeffects model.
With sampleN.TOST(..., details = FALSE, print = FALSE)
results are provided as a data frame ^{3} with nine columns Design
, alpha
, CV
, theta0
, theta1
, theta2
, Sample size
, Achieved power
, and Target power
.
To access e.g., the sample size use either sampleN.TOST[1, 7]
or sampleN.TOST[["Sample size"]]
. We suggest to use the latter in scripts for clarity.
The estimated sample size gives always the total number of subjects (not subject/sequence in crossovers or subjects/group in a parallel design – like in some other software packages).
Designs with one (parallel) to four periods (replicates) are supported.
# design name df
# "parallel" 2 parallel groups n2
# "2x2" 2x2 crossover n2
# "2x2x2" 2x2x2 crossover n2
# "2x2x3" 2x2x3 replicate crossover 2n3
# "2x2x4" 2x2x4 replicate crossover 3n4
# "2x4x4" 2x4x4 replicate crossover 3n4
# "2x3x3" partial replicate (2x3x3) 2n3
# "2x4x2" Balaam’s (2x4x2) n2
# "2x2x2r" Liu’s 2x2x2 repeated xover 3n2
# "paired" paired means n1
Estimate the sample size for assumed intrasubject CV 0.30.
sampleN.TOST(CV = 0.30)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 2x2 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25
# True ratio = 0.95, CV = 0.3
#
# Sample size (total)
# n power
# 40 0.815845
To get only the sample size:
Note that the sample size is always rounded up to give balanced sequences (here a multiple of two). Since power is higher than our target, likely this was the case here. Let us check that.
Which power will we get with a sample size of 39?
Confirmed that with 39 subjects we will already reach the target power. That means also that one dropout will not compromise power. We could explore that further in a Power Analysis.
Maybe the clinical site has a limited capacity. Any study can also be performed in a replicate design and assessed for ABE. As a rule of thumb the total sample in a 3period replicate is ~¾ of the 2×2×2 design and the one of a 2sequence 4period replicate ~½ of the 2×2×2. The number of treatments and hence, of biosamples – which mainly drives the study’s cost – will be roughly the same.
designs < c("2x2x2", "2x2x3", "2x3x3", "2x2x4")
# data.frame of results
res < data.frame(design = designs, n = NA, power = NA, n.do = NA,
power.do = NA,
stringsAsFactors = FALSE) # this line for R <4.0.0
for (i in 1:4) {
# print = FALSE and details = FALSE suppress output to the console
# we are only interested in columns 78
# let's also calculate power for one dropout
res[i, 2:3] < sampleN.TOST(CV = 0.30, design = res$design[i],
print = FALSE, details = FALSE)[7:8]
res[i, 4] < res[i, 2]  1
res[i, 5] < suppressMessages(
power.TOST(CV = 0.30, design = res$design[i],
n = res[i, 4]))
}
print(res, row.names = FALSE)
# design n power n.do power.do
# 2x2x2 40 0.8158453 39 0.8056171
# 2x2x3 30 0.8204004 29 0.8068731
# 2x3x3 30 0.8204004 29 0.8063834
# 2x2x4 20 0.8202398 19 0.7991508
As expected – and as a bonus – we get a small gain in power, though in the 4period with one dropout power will be slightly compromised.
But why is power in the replicate designs higher than in the 2×2×2? If residual variances are equal, the width of the confidence interval depends only on the tvalue and in particular on the degrees of freedom – which themselves depend on the design and the sample size.
# design name n formula df t.value
# 2x2x2 2x2x2 crossover 40 n2 38 1.686
# 2x2x3 2x2x3 replicate crossover 30 2*n3 57 1.672
# 2x3x3 partial replicate (2x3x3) 30 2*n3 57 1.672
# 2x2x4 2x2x4 replicate crossover 20 3*n4 56 1.673
If the capacity is 24 beds, we would opt for a 4period full replicate.
As another option (e.g., if the blood volume is limited and/or there are concerns about a higher dropoutrate in a multipleperiod study) we could stay with the 2×2×2 design but split the sample size into groups. In Europe (and for the FDA if certain conditions ^{4} are fulfilled), there are no problems pooling the data and use the conventional model.
sequence + subject(sequence) + period + treatment
However, some regulators prefer to incorporate groupterms in the model.
group + sequence + subject(group × sequence) +
period(group) + group × sequence + treatment
Since we have more terms in the model, we will loose some degrees of freedom. Let us explore in simulations how that would impact power. By default function power.TOST.sds()
performs 100,000 simulations.
grouping < function(capacity, n) {
# split sample size into >=2 groups based on capacity
if (n <= capacity) { # make equal groups
ngrp < rep(ceiling(n/2), 2)
} else { # at least one = capacity
ngrp < rep(0, ceiling(n / capacity))
grps < length(ngrp)
ngrp[1] < capacity
for (j in 2:grps) {
n.tot < sum(ngrp) # what we have so far
if (n.tot + capacity <= n) {
ngrp[j] < capacity
} else {
ngrp[j] < n  n.tot
}
}
}
return(ngrp = list(grps = length(ngrp), ngrp = ngrp))
}
CV < 0.30
capacity < 24 # clinical capacity
res < data.frame(n = NA, grps = NA, n.grp = NA, m.1 = NA, m.2 = NA)
x < sampleN.TOST(CV = CV, print = FALSE, details = FALSE)
res$n < x[["Sample size"]]
res$m.1 < x[["Achieved power"]]
x < grouping(capacity = capacity, n = res$n)
res$grps < x[["grps"]]
ngrp < x[["ngrp"]]
res$n.grp < paste(ngrp, collapse = "")
res$m.2 < power.TOST.sds(CV = CV, n = res$n, grps = res$grps,
ngrp = ngrp, gmodel = 2, progress = FALSE)
res$loss < 100*(res$m.2  res$m.1)/res$m.1
names(res)[2:6] < c("groups", "n/group", "pooled model",
"group model", "loss (%)")
res[1, 4:6] < sprintf("%6.4f", res[1, 4:6])
cat("Achieved power and relative loss\n"); print(res, row.names = FALSE)
# Achieved power and relative loss
# n groups n/group pooled model group model loss (%)
# 40 2 2416 0.8158 0.8120 0.4664
With ~0.5% the relative loss in power is practically negligible.
Estimate the sample size for equivalence of the ratio of two means with normality on original scale based on Fieller’s (‘fiducial’) confidence interval. ^{5} Crossover design, withinsubject CV_{w} 0.20, betweensubject CV_{b} 0.40.
sampleN.RatioF(CV = 0.20, CVb = 0.40)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# based on Fieller's confidence interval
# Sample size estimation
# 
# Study design: 2x2 crossover
# Ratio of means with normality on original scale
# alpha = 0.025, target power = 0.8
# BE margins = 0.8 ... 1.25
# True ratio = 0.95, CVw = 0.2, CVb = 0.4
#
# Sample size
# n power
# 28 0.807774
In this function the default \(\small{\alpha}\) is 0.025, since it is intended for studies with clinical endpoints, where the 95% confidence interval is usually used for equivalence testing. ^{6}
Estimate the sample size based on the results of a 2×2×2 pilot study in 16 subjects where we observed an intrasubject CV 0.20 and \(\small{\theta_0}\) 0.92.
If we believe [sic] that in the pivotal study both the \(\small{\theta_0}\) and CV will be exactly like in the pilot, this is a straightforward exercise. We simply provide the required arguments.
sampleN.TOST(CV = 0.20, theta0 = 0.92)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 2x2 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25
# True ratio = 0.92, CV = 0.2
#
# Sample size (total)
# n power
# 28 0.822742
This approach is called by some ‘carved in stone’ because it relies on – very strong – assumptions which likely are not justified. Although power curves are relatively flat close to unity (i.e., the impact on power is small when moving from say, \(\small{\theta_0}\) 1 to 0.95) but they are getting increasingly steep when moving away more from unity.
Both \(\small{\theta_0}\) and CV (as every estimate) are uncertain to some extent, which depends on the degrees of freedom (sample size and design). Hence, it might not be good idea to perform very small pilot studies (e.g., in only six subjects). Although it might be possible that in the pivotal study the CV is indeed lower than the one we observed in the pilot, it would be even more risky than the ‘carved in stone’ approach to assume a lower one in planning the pivotal study.
With the function CVCL()
we can calculate confidence limits of the CV. It is advisable to use the upper confidence limit as a conservative approach. As a side effect – if the CV will be lower than assumed – we get a ‘safety margin’ for the T/R ratio.
df < 16  2 # degrees of freedom of the 2x2x2 crossover pilot
CVCL(CV = 0.20, df = df, side = "upper", alpha = 0.20)[["upper CL"]]
# [1] 0.2443631
I prefer \(\small{\alpha=0.20}\) in analogy to the producer’s risk \(\small{\beta=0.20}\) when planning for power \(\small{\pi=1\beta=0.80}\). Gould ^{7} proposed the more liberal \(\small{\alpha=0.25}\). Let us repeat the sample size estimation based on the upper CL of the CV.
CL.upper < CVCL(CV = 0.20, df = 16  2, side = "upper",
alpha = 0.20)[["upper CL"]]
res < sampleN.TOST(CV = CL.upper, theta0 = 0.92, print = FALSE)
print(res[7:8], row.names = FALSE)
# Sample size Achieved power
# 40 0.816919
Of course, this has a massive impact on the sample size, which increases from 28 to 40. It might be difficult to convince the management to invest ~40% more than with the ‘carved in stone’ approach.
However, we can also explore how power would be affected if our assumption is true and the study will nevertheless be performed with only 28 subjects.
CL.upper < CVCL(CV = 0.20, df = 16  2, side = "upper",
alpha = 0.20)[["upper CL"]]
power.TOST(CV = CL.upper, theta0 = 0.92, n = 28)
# [1] 0.679253
There will be a drop in power from the ~0.82 the management expects to only ~0.67. That’s just slightly higher than betting for two dozens in Roulette…
As mentioned above, if the CV turns out to be lower than assumed, we gain a ‘safety margin’ for the T/R ratio. Let us explore that. We perform the study with 40 subjects, the CV will be 0.22 (less than the ~0.24 we assumed), and the T/R with 0.90 will be worse than the 0.92 we assumed.
Below our target but still acceptable.
In the basic approach we concentrated mainly on the uncertainty of the CV. But this is not the end of the story. Clearly \(\small{\theta_0}\) is uncertain as well. With the function expsampleN.TOST()
we can dive deeper into this area. Let us start with the CV only.
expsampleN.TOST(CV = 0.20, theta0 = 0.92, prior.type = "CV",
prior.parm = list(m = 16, design = "2x2x2"))
#
# ++++++++++++ Equivalence test  TOST ++++++++++++
# Sample size est. with uncertain CV
# 
# Study design: 2x2 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25
# Ratio = 0.92
# CV = 0.2 with 14 df
#
# Sample size (ntotal)
# n exp. power
# 30 0.806069
Not that bad. The sample size increases fairly from the 28 of the ‘carved in stone’ approach to 30 but is substantially lower than the 40 we estimated based on the upper confidence limit of the CV.
Let us keep the CV ‘fixed’ and take only the uncertainty of \(\small{\theta_0}\) into account.
expsampleN.TOST(CV = 0.20, theta0 = 0.92, prior.type = "theta0",
prior.parm = list(m = 16, design = "2x2x2"))
#
# ++++++++++++ Equivalence test  TOST ++++++++++++
# Sample size est. with uncertain theta0
# 
# Study design: 2x2 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25
# Ratio = 0.92
# CV = 0.2
#
# Sample size (ntotal)
# n exp. power
# 46 0.805236
It starts to hurt. We saw already that power curves are getting steep if the T/R ratio is not close to unity. Our \(\small{\theta_0}\) 0.92 was not very nice but in the pivotal study it might be even lower as well – which has a larger impact on power than the CV.
Now for the ‘worst case scenario’, where we take both uncertainties into account.
expsampleN.TOST(CV = 0.20, theta0 = 0.92, prior.type = "both",
prior.parm = list(m = 16, design = "2x2x2"),
details = FALSE)
#
# ++++++++++++ Equivalence test  TOST ++++++++++++
# Sample size est. with uncertain CV and theta0
# 
# Study design: 2x2 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25
# Ratio = 0.92 with 14 df
# CV = 0.2 with 14 df
#
# Sample size (ntotal)
# n exp. power
# 54 0.802440
This sample size is almost twice the 28 your boss got from his/her ExcelSheet. ^{8} If you are not fired right away when suggesting such a study, take it as a warning what might happen.
At least, if the pivotal study is performed in a lower sample size and fails, you know why.
If you are adventurous consider an Adaptive TwoStage Sequential Design with sample size reestimation. Various methods are supported in the package Power2Stage. ^{9}
An alternative to working with an assumed \(\small{\theta_0}\) is using statistical assurance as proposed by Ring et al. ^{10} This concept uses the distribution of T/Rratios and assumes an uncertainty parameter \(\small{\sigma_\textrm{u}}\). A natural assumption is \(\small{\sigma_\textrm{u}=1\theta_0}\), i.e., for the commonly applied \(\small{\theta_0=0.95}\) one can use the argument sem = 0.05
of the function expsampleN.TOST()
where the argument theta0
has to be kept at 1. The following example reproduces Table 1 of the paper.
CV < 0.214
res < data.frame(target = c(rep(0.8, 5), rep(0.9, 5)),
theta0 = rep(c(1, seq(0.95, 0.92, 0.01)), 2),
n.1 = NA, power = NA,
sigma.u = rep(c(0.0005, seq(0.05, 0.08, 0.01)), 2),
n.2 = NA, assurance = NA)
for (i in 1:nrow(res)) {
res[i, 3:4] < sampleN.TOST(CV = CV, targetpower = res$target[i],
theta0 = res$theta0[i], details = FALSE,
print = FALSE)[7:8]
res[i, 6:7] < expsampleN.TOST(CV = CV, targetpower = res$target[i],
theta0 = 1, prior.type = "theta0",
prior.parm = list(sem = res$sigma.u[i]),
details = FALSE, print = FALSE)[9:10]
}
res < signif(res, 3)
res[, 5] < sprintf("%.2f", res[, 5])
names(res)[c(3, 6)] < "n"
print(res, row.names = FALSE)
# target theta0 n power sigma.u n assurance
# 0.8 1.00 18 0.833 0.00 18 0.833
# 0.8 0.95 22 0.824 0.05 22 0.833
# 0.8 0.94 24 0.817 0.06 22 0.800
# 0.8 0.93 26 0.801 0.07 26 0.819
# 0.8 0.92 30 0.802 0.08 28 0.803
# 0.9 1.00 22 0.916 0.00 22 0.916
# 0.9 0.95 28 0.904 0.05 28 0.904
# 0.9 0.94 32 0.909 0.06 32 0.903
# 0.9 0.93 36 0.905 0.07 38 0.902
# 0.9 0.92 42 0.908 0.08 48 0.902
One caveat: Assuming no uncertainty (\(\small{\sigma_\textrm{u}=0}\)) would fail because the level of technical success is zero. Here a small value of 0.0005 was used instead.
Estimate the sample size for a study of two blood pressure lowering drugs in a 2×2 design assessing the difference in means of untransformed data (raw, linear scale; i.e., specifying logscale = FALSE
). In this setup everything has to be given with the same units (i.e., here \(\small{\theta_0}\) –5 mm Hg, \(\small{\theta_1}\) –15 mm Hg, \(\small{\theta_2}\) +15 mm Hg systolic BP).
Assuming a residual standard deviation 20 mm Hg.
planned < "2x2x2"
logscale < FALSE
theta0 < 5
theta1 < 15
theta2 < +15 # if not given, theta1 is used
SD.resid < 20 # residual standard deviation
sampleN.TOST(CV = SD.resid, theta0 = theta0, theta1 = theta1,
theta2 = theta2, logscale = logscale, design = planned)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 2x2 crossover
# untransformed data (additive model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 15 ... 15
# True diff. = 5, CV = 20
#
# Sample size (total)
# n power
# 52 0.807468
Assuming a standard deviation of the difference T – R 28 mm Hg.
known < known.designs()[, c(2, 6)] # extract relevant information
bk < known[known$design == planned, "bk"] # retrieve design constant
cat(paste0("The design constant for design \"", planned, "\" is ", bk, "\n"))
# The design constant for design "2x2x2" is 2
SD.delta < 28 # standard deviation of the difference
SD.resid < SD.delta / sqrt(bk) # convert to residual SD
sampleN.TOST(CV = SD.resid, theta0 = theta0, theta1 = theta1,
theta2 = theta2, logscale = logscale, design = planned)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 2x2 crossover
# untransformed data (additive model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 15 ... 15
# True diff. = 5, CV = 19.79899
#
# Sample size (total)
# n power
# 50 0.800423
Note that other software packages (e.g., PASS, nQuery, StudySize,…) require the standard deviation of the difference as input.
Designs with three and four treatments/periods are supported.
# design name df
# "3x3" 3x3 crossover 2n4
# "3x6x3" 3x6x3 crossover 2n4
# "4x4" 4x4 crossover 3n6
"3x3"
denotes the Latin Square (ABCBCACAB
), "3x6x3"
the 6sequence Williams’ design (ABCACBBACBCACABCBA
), "4x4"
the Latin Square (ABCDBCDACDABDABC
) or any of the possible Williams’ designs with four periods (ADBCBACDCBDADCAB
, ADCBBCDACABDDBAC
, ACDBBDCACBADDABC
, ACBDBADCCDABDBCA
, ABDCBCADCDBADACB
, ABCDBDACCADBDCBA
).
Which design
argument in studies with more than two periods you should use depends on the planned evaluation.
Suppose we have a bioequivalence study with three treatments – A, B, and C – and the objective of the study is to make pairwise comparisons among the treatments. Suppose further that treatment C is different in kind from A and B, so that the assumption of homogeneous variance among the three treatments is questionable. One way to do the analyses, under normality assumptions, is Two at a Time – e.g., to test hypotheses about A and B, use only the data from A and B. Another way is All at Once – include the data from all three treatments in a single analysis, making pairwise comparisons within this analysis. If the assumption of homogeneous variance is correct, the All at Once approach will provide more d.f. for estimating the common variance, resulting in increased power. If the variance of C differs from that of A and B, the All at Once approach may have reduced power or an inflated type I error rate, depending on the direction of the difference in variances.
All at Once
In this approach you assume homogenicity in the ANOVA of pooled data and get one residual variance.
To plan for this approach specify one of the design
arguments given above.
Two at a Time
In this approach – which is preferred by some agencies, e.g., the EMA ^{11} – you exclude one treatment and perform the analysis on the remaining two. That means you obtain two separate Incomplete Block Designs (IBD).
One example is comparing a test (T
) and to references from two different regions (R1
, R2
). You get not only two point estimates (like in the ANOVA) but also two withinsubject variances (in the comparisons T
vs R1
and T
vs R2
).
Another example is a pilot study with two candidate treatments (C1
, C2
) and one reference (R
). You would select the candidate with \(\small{\min \left \{ \left  \log_{e}\theta_\textrm{C1}\right ,\left  \log_{e}\theta_\textrm{C2}\right  \right \}}\) for the pivotal study. If \(\small{ \left  \log_{e}\theta_\textrm{C1}\right \sim \left  \log_{e}\theta_\textrm{C2}\right  }\), select the candidate with lower withinsubject variance.
To plan for this approach specify design = "2x2x2"
.
I suggest to opt for the latter since the former may lead to biased estimates and an inflated type I error. ^{12}
Three treatments intended for evaluation ‘All at Once’ or ‘Two at a Time’.
CV < 0.20
res < data.frame(design = c("3x6x3", "2x2x2"), n = NA, power = NA,
stringsAsFactors = FALSE)
for (i in 1:2) {
res[i, 2:3] < sampleN.TOST(CV = CV, design = res$design[i],
details = FALSE, print = FALSE)[7:8]
}
print(res, row.names = FALSE)
# design n power
# 3x6x3 18 0.8089486
# 2x2x2 20 0.8346802
Four treatments (e.g., Test fasting and fed, Reference fasting and fed).
CV < 0.20
res < data.frame(design = c("4x4", "2x2x2"), n = NA, power = NA,
stringsAsFactors = FALSE)
for (i in 1:2) {
res[i, 2:3] < sampleN.TOST(CV = CV, design = res$design[i],
details = FALSE, print = FALSE)[7:8]
}
print(res, row.names = FALSE)
# design n power
# 4x4 20 0.8527970
# 2x2x2 20 0.8346802
Let us first recap the hypotheses in bioequivalence.
The ‘Two OneSided Tests Procedure’ (TOST) ^{13} \[\small{H_\textrm{0L}:\frac{\mu_\textrm{T}}{\mu_\textrm{R}} \leq \theta_1\:vs\:H_\textrm{1L}:\frac{\mu_\textrm{T}}{\mu_\textrm{R}}>\theta_1}\] \[\small{H_\textrm{0U}:\frac{\mu_\textrm{T}}{\mu_\textrm{R}} \geq \theta_2\:vs\:H_\textrm{1U}:\frac{\mu_\textrm{T}}{\mu_\textrm{R}}<\theta_2}\]
The confidence interval inclusion approach \[\small{H_0:\frac{\mu_\textrm{T}}{\mu_\textrm{R}}\notin \left [ \theta_1, \theta_2 \right]\:vs\:H_1:\theta_1<\frac{\mu_\textrm{T}}{\mu_\textrm{R}}<\theta_2}\]
Note that the null hypotheses imply bioinequivalence where \(\small{[\theta_1,\theta_2]}\) are the lower and upper limits of the bioequivalence range.
TOST provides two \(\small{p}\) values (where \(\small{H_0}\) is not rejected if \(\small{\max}\,[p_\textrm{L},p_\textrm{U}]>\alpha\)) and is of historical interest only because the CI inclusion approach is preferred in regulatory guidelines.
From a regulatory perspective the outcome of a comparative BA study is dichotomous. Either the study demonstrated bioequivalence (confidence interval entirely within \(\small{[\theta_1,\theta_2]}\)) or not. ^{14} Only if the CI lies entirely outside \(\small{[\theta_1,\theta_2]}\), the null hypothesis is not rejected and further studies not warranted.
In any case, calculation of post hoc (a.k.a. a posteriori, retrospective) power is futile. ^{15}
There is simple intuition behind results like these: If my car made it to the top of the hill, then it is powerful enough to climb that hill; if it didn’t, then it obviously isn’t powerful enough. Retrospective power is an obvious answer to a rather uninteresting question. A more meaningful question is to ask whether the car is powerful enough to climb a particular hill never climbed before; or whether a different car can climb that new hill. Such questions are prospective, not retrospective.
If a study passes – despite lower than desired power – there is no reason to reject the study. It only means that assumptions (‼) in sample size estimation were not realized. The CV might have been higher, and/or the T/Rratio worse, and/or the dropoutrate higher than anticipated. On the other hand, if post hoc power is higher than desired, it does not further support a study which already demonstrated BE.
Nevertheless, exploring power is useful when trying to understand why a study failed and to plan another study. Let us continue with the example from above. Ignoring our concerns, the management decided to perform the pivotal study with 28 subjects. The T/Rratio was slightly worse (0.90), the CV higher (0.25), and we had one dropout in the first sequence and two in the second. The function CI.BE()
comes handy.
n < c(14  1, 14  2) # 14 dosed in each sequence
round(100*CI.BE(pe = 0.90, CV = 0.25, n = n), 2)
# lower upper
# 79.87 101.42
The study failed although by a small margin. One might be tempted to repeat the study with an – only slightly – higher sample size. But what was the post hoc power of the failed study?
Actually the chance of passing was worse than flipping a coin.
NB, in calculating post hoc power the observed \(\small{\theta_0}\) has to be used. In some statistical reports high ‘power’ is given even for a failed study, which is not even wrong. Unfortunately \(\small{\theta_0=1}\) is still the default in some software packages.
power.TOST(CV = 0.20, theta0 = 0.92, n = 28) # assumed in planning
# [1] 0.822742
power.TOST(CV = 0.25, theta0 = 1, n = c(13, 12)) # observed but wrong T/Rratio
# [1] 0.8558252
Since all estimates were worse than assumed, how could one get a ‘power’ even higher than desired, despite the fact that the study failed to demonstrate bioequivalence? That’s nonsense, of course. \(\small{\theta_0=1}\) gives the ‘power to detect a significant difference of 20%’ – a flawed concept which was abandoned after Schuirmann’s paper of 1987.
When planning the next study one can use the entire arsenal from above. Since we have more accurate estimates (from 25 subjects instead of the 16 of the pilot) the situation is more clear now.
As a further step we can take the information of both studies into account with the function CVpooled()
.
CVs < (" CV  n  design  study
0.20  16  2x2x2  pilot
0.25  25  2x2x2  pivotal")
txtcon < textConnection(CVs)
data < read.table(txtcon, header = TRUE, sep = "",
strip.white = TRUE, as.is = TRUE)
close(txtcon)
print(CVpooled(data, alpha = 0.20), digits = 4, verbose = TRUE)
# Pooled CV = 0.2322 with 37 degrees of freedom
# Upper 80% confidence limit of CV = 0.2603
Before pooling, variances are weighted by the degrees of freedom. Hence, the new estimate is with ~0.23 closer to the 0.25 of the larger study. Note also that the upper confidence limit is with ~0.26 higher than the one of the pilot study with ~0.24.
Don’t pool data blindly. In the ideal situation you know the entire background of all studies (clinical performance, bioanalytics). Even if all studies were performed at the same CRO, more things are important. One purpose of a pilot study is to find a suitable sampling schedule. If the sampling schedule of the pilot was not ideal (e.g., C_{max} was not sufficiently enough described), pooling is not a good idea. It might well be that in the pivotal study – with a ‘better’ sampling schedule – its CV is more reliable. On the other hand, AUC is less sensitive to different sampling schedules.
Pooling data from the literature should be done with great caution (if at all). Possibly critical information is missing. Consider using a CV from the upper end of values instead. Common sense helps.
An example where pooling could be misleading: C_{max} data of pilot and pivotal studies in five different designs with 11 to 39 subjects, fasting/fed, three different bioanalytical methods (GC/ECD, LCMS/MS, GC/MS), chiral and achiral (which is not relevant for this drug since the active enantiomer is ~95% of the total drug and there is no in vivo interconversion). Note that most pivotal studies were ‘overpowered’.
This is an applesandoranges comparison. Red squares show CVs which were above the upper CL of the pooled CV. Given, only in two studies (#1, #6) their lower CL did not overlap the upper one of the pooled CV.
Which side of the great divide are you on? Do you believe that meta is better or do you hold instead that pooling is fooling? Well, to nail my colours to the mast, I belong to the former school. It seems to me that there is no other topic in medical statistics, with the possible exceptions of crossover trials, bioequivalence and nof1 studies, which has the same capacity as this one to rot the brains.
Although we suggest to explore the various options shown in Example 3, it is worthwhile to have a first look with the function pa.ABE()
.
pa.ABE(CV = 0.20, theta0 = 0.92)
# Sample size plan ABE
# Design alpha CV theta0 theta1 theta2 Sample size Achieved power
# 2x2 0.05 0.2 0.92 0.8 1.25 28 0.822742
#
# Power analysis
# CV, theta0 and number of subjects which lead to min. acceptable power of at least 0.7:
# CV= 0.2377, theta0= 0.9001
# n = 21 (power= 0.7104)
This exercise confirms what we already know. The most critical parameter is \(\small{\theta_0}\), whereas dropouts are the least important.
More details are given in the vignette Power Analysis.
As we have seen, the impact of dropouts on power is rather limited. Regularly CROs suggest additional subjects to ‘compensate for the potential loss in power’. IMHO, milking sponsors to make wealthy CROs richer. Note that the dropoutrate is based on dosed subjects. Hence, the correct formula for the adjusted sample size \(\small{n{}'}\) based on the estimated one \(\small{n}\) is \(\small{n{}'=n / (1{dropout\:rate})}\), and not \(\small{n{}'=n \times (1+{dropout\:rate})}\). An example for studies with three periods:
balance < function(x, y) {
return(y * (x %/% y + as.logical(x %% y)))
}
do < 0.15 # anticipated dropoutrate 15%
seqs < 3
n < seq(12, 96, 12)
res < data.frame(n = n,
adj1 = balance(n / (1  do), seqs), # correct
elig1 = NA, diff1 = NA,
adj2 = balance(n * (1 + do), seqs), # wrong
elig2 = NA, diff2 = NA)
res$elig1 < floor(res$adj1 * (1  do))
res$diff1 < sprintf("%+i", res$elig1  n)
res$elig2 < floor(res$adj2 * (1  do))
res$diff2 < sprintf("%+i", res$elig2  n)
invisible(
ifelse(res$elig1  n >=0, res$optim < res$elig1, res$optim < res$elig2))
res$diff < sprintf("%+i", res$optim  n)
names(res)[c(2, 5)] < c("n'1", "n'2")
res$diff1[which(res$diff1 == "+0")] < "\u00B10"
res$diff2[which(res$diff2 == "+0")] < "\u00B10"
res$diff[which(res$diff == "+0")] < "\u00B10"
print(res, row.names = FALSE)
# n n'1 elig1 diff1 n'2 elig2 diff2 optim diff
# 12 15 12 ±0 15 12 ±0 12 ±0
# 24 30 25 +1 30 25 +1 25 +1
# 36 45 38 +2 42 35 1 38 +2
# 48 57 48 ±0 57 48 ±0 48 ±0
# 60 72 61 +1 69 58 2 61 +1
# 72 87 73 +1 84 71 1 73 +1
# 84 99 84 ±0 99 84 ±0 84 ±0
# 96 114 96 ±0 111 94 2 96 ±0
With the wrong formula – especially for high dropout rates – you might end up with less eligible subjects (elig2
) than planned thus compromising power. On the other hand, with the correct one (due to rounding up to get balanced sequences) you might end up with slightly too many (elig1
). Of course, if you want to be one the safe side, you can select the ‘best’ (column optim
).
Sometimes the CV is not given in the literature. By means of the function CVfromCI()
we can calculate it from the confidence interval, the design, and the sample size. ^{16}
The method is exact if the subjects/sequence are known. In the literature quite often only the total sample size is given and the function tries to keep sequences as balanced as possible. What if the study was imbalanced?
A total sample size of 26 was reported. The study was either balanced or imbalanced to an unknown degree:
n < 26
CV.est < CVfromCI(lower = 0.8323, upper = 1.0392, design = "2x2x4", n = 26)
n.est < sampleN.TOST(CV = CV.est, design = "2x2x4", print = FALSE,
details = FALSE)[["Sample size"]]
n1 < balance(seq(n, 18, 1), 2) / 2
n2 < n  n1
nseqs < unique(data.frame(n1 = n1, n2 = n2, n = n))
res < data.frame(n1 = nseqs$n1, n2 = nseqs$n2, CV.true = NA,
CV.est = CV.est, n.true = NA, n.est = n.est)
for (i in 1:nrow(res)) {
res$CV.true[i] < CVfromCI(lower = 0.8323, upper = 1.0392, design = "2x2x4",
n = c(res$n1[i], res$n2[i]))
res$n.true[i] < sampleN.TOST(CV = res$CV.true[i], design = "2x2x4",
print = FALSE, details = FALSE)[["Sample size"]]
res$n.est[i] < sampleN.TOST(CV = CV.est, design = "2x2x4",
print = FALSE, details = FALSE)[["Sample size"]]
}
print(signif(res, 5), row.names = FALSE)
# n1 n2 CV.true CV.est n.true n.est
# 13 13 0.34986 0.34986 26 26
# 12 14 0.34876 0.34986 26 26
# 11 15 0.34546 0.34986 26 26
# 10 16 0.33988 0.34986 24 26
# 9 17 0.33196 0.34986 24 26
The true CV of any imbalanced study might have been lower than what we assumed. That means, if we use the estimated CV – falsely assuming balanced sequences – our sample size estimation will always be conservative.
If you are unsure about the direction of the deviation of T from R (lower or higher) always assume \(\small{\theta_0<1}\).
CV < 0.21
d < 0.05 # delta 5%, direction unknown
n < sampleN.TOST(CV = CV, theta0 = 1  d, print = FALSE,
details = FALSE)[["Sample size"]]
res1 < data.frame(CV = CV, theta0 = c(1  d, 1 / (1  d)),
n = n, power = NA)
for (i in 1:nrow(res1)) {
res1$power[i] < power.TOST(CV = CV, theta0 = res1$theta0[i], n = n)
}
n < sampleN.TOST(CV = CV, theta0 = 1 + d, print = FALSE,
details = FALSE)[["Sample size"]]
res2 < data.frame(CV = CV, theta0 = c(1 + d, 1 / (1 + d)),
n = n, power = NA)
for (i in 1:nrow(res1)) {
res2$power[i] < power.TOST(CV = CV, theta0 = res2$theta0[i], n = n)
}
res < rbind(res1, res2)
print(signif(res[order(res$n, res$theta0), ], 4), row.names = FALSE)
# CV theta0 n power
# 0.21 0.9524 20 0.8081
# 0.21 1.0500 20 0.8081
# 0.21 0.9500 22 0.8374
# 0.21 1.0530 22 0.8374
If you use 1.05 (sample size 20) power will be maintained all the way down to its reciprocal (0.9524) but not to 0.95 (where you would already need a sample size of 22). On the other hand, 0.95 preserves power up to 1.053. The statement ‘sample size based on a deviation of ±5%’ seen in many protocol does not make sense.
It is common a misconception that in a parallel design one obtains the between (inter) subject variability. The fact that products are administered only once does not mean that the withinsubject variability does not exist (we observed only one occasion). Hence, the variability consists of both between and withincomponents. We can estimate the betweensubject variability only in a crossover design. Referring to betweensubject variability in a parallel design is sloppy terminology at least.↩︎
Owen DB. A special case of a bivariate noncentral tdistribution. Biometrika. 1965: 52(3/4); 437–46. doi:10.2307/2333696.↩︎
Schütz H. MultiGroup Studies in Bioequivalence. To pool or not to pool? BioBridges. Prague, 26–27 September, 2018. Slide 7: Review of Guidelines.↩︎
Fieller EC. Some Problems In Interval Estimation. J Royal Stat Soc B. 1954; 16(2): 175–85. JSTOR:2984043.↩︎
EMEA, CPMP. Points to Consider on Switching between Superiority and NonInferiority. London, 27 July 2000. CPMP/EWP/482/99.↩︎
Gould AL. Group Sequential Extension of a Standard Bioequivalence Testing Procedure. J Pharmacokin Biopharm. 1995: 23(1); 57–86. doi:10.1007/BF02353786.↩︎
Dubins D. My Tiny Contribution to Clinical Research: FARTSSIE. 20201013. FARTSSIE – Free Analysis Research Tool for Sample Size Iterative Estimation.↩︎
Labes D, Lang B, Schütz H. Power2Stage: Power and SampleSize Distribution of 2Stage Bioequivalence Studies. 20190420. Rpackage Version 0.5.2.↩︎
Ring A, Lang B, Kazaroho C, Labes D, Schall R, Schütz H. Sample size determination in bioequivalence studies using statistical assurance. Br J Clin Pharmacol. 2019; 85(10): 2369–77. doi:10.1111/bcp.14055.↩︎
EMA, CHMP. Guideline on the Investigation of Bioequivalence. London, 20 January 2010. CPMP/EWP/QWP/1401/98 Rev. 1/ Corr **.↩︎
D’Angelo P. Testing for Bioequivalence in Higher‐Order Crossover Designs: Two‐at‐a‐Time Principle Versus Pooled ANOVA. 2^{nd} Conference of the Global Bioequivalence Harmonisation Initiative. Rockville, 15–16 September, 2016.↩︎
Schuirmann DJ. A Comparison of the Two OneSided Tests Procedure and the Power Approach for Assessing the Equivalence of Average Bioavailability. J Pharmacokin Biopharm. 1987; 15(6): 657–80. doi:10.1007/BF01068419.↩︎
If the CI overlaps \(\small{[\theta_1,\theta_2]}\), the outcome is indecisive (called ‘the gray zone’ by some). As long as \(\small{\theta_0}\) lies within \(\small{[\theta_1,\theta_2]}\) you can hope to show BE in a larger sample size. However, once \(\small{\theta_0}\) approaches one of the limits, if might be futile – even with a very small CV. Try
sampleN.TOST(CV = 0.075, theta0 = 0.81)
and then
sampleN.TOST(CV = 0.075, theta0 = 0.80)
to get an idea.↩︎
Hoenig JM, Heisey DM. The Abuse of Power: The Pervasive Fallacy of Power Calculations for Data Analysis. Am Stat. 2001; 55(1): 19–24↩︎
Schütz H. Sample Size Estimation for BE Studies. Workshop. Bucarest, 19 March 2013. Slide 26: Algebra…↩︎