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Examples were generated with
4.2.1
by the packages PowerTOST
^{1} and
randomizeBE
.^{2} More examples are given in the respective
vignette.^{3} See also the README on GitHub
for an overview and the Online manual^{4} for details.
Abbreviation  Meaning 

\(\small{\alpha}\)  Nominal level of the test, probability of Type I Error, patient’ risk 
\(\small{\beta}\)  Probability of Type II Error, producer’s risk (\(\small{1\pi}\)) 
BE  Bioequivalence 
BIBD  Balanced Incomplete Block Design 
\(\small{CV}\)  Coefficient of Variation 
\(\small{CV_\textrm{inter}}\)  Betweensubject \(\small{CV}\) 
\(\small{CV_\textrm{intra}}\)  Withinsubject \(\small{CV}\) (also ISCV) 
\(\small{H_0}\)  Null hypothesis 
\(\small{H_1}\)  Alternative hypothesis (also \(\small{H_\textrm{a}}\)) 
IBD  Incomplete Block Design 
NTID  Narrow Therapeutic Index Drug 
\(\small{\pi}\)  (Prospective) power 
TOST  Two OneSided Tests 
What are the main statistical issues in planning a confirmatory experiment?
For details about inferential statistics and hypotheses in equivalence see another article.
An ‘optimal’ study design is one which (taking all assumptions into account) has a reasonably high chance of demonstrating equivalence (power) whilst controlling the consumer risk.
A basic knowledge of R is
required. To run the scripts at least version 1.4.3 (20161101) of
PowerTOST
and at least version 0.3.3 (20170322) of
randomizeBE
is suggested. Any version of R would likely do. The current release of
PowerTOST
was only tested with R
4.1.3 (20220310) and later; the current release of
randomizeBE
was only tested with R 4.1.1 (20210810)) and later. Scripts were run on
a Xeon E31245v3 @ 3.40GHz (1/4 cores) 16GB RAM with R 4.2.1 on Windows 7 build 7601, Service Pack 1,
Universal C Runtime 10.0.10240.16390.
Note that in all functions of PowerTOST
– and scripts
using them – the arguments (say, the assumed T/Rratio
theta0
, the BElimits (theta1
,
theta2
), the assumed coefficient of variation
CV
, the desired (target) power targetpower
,
the anticipated dropoutrate do.rate
, etc.) have to be
given as ratios and not in percent.
sampleN.TOST()
gives balanced sequences (i.e.,
the same number of subjects is allocated to each of the sequences).
Furthermore, the estimated sample size is the total number of
subjects (not subjects per sequence – like in some other software
packages).
HigherOrder Designs with three and four treatments are supported.
design  internal name  df  common name 

"3x3"

3x3 crossover

2n4

3×3 Latin Square 
"3x6x3"

3x6x3 crossover

2n4

3×6×3 Williams’ design 
"4x4"

4x4 crossover

3n6

4×4 Latin Square or a Williams’ design 
"3x3"
denotes the Latin Square
(ABC
BCA
CAB
).
"3x6x3"
denotes the 6sequence Williams’ design
(ABC
ACB
BAC
BCA
CAB
CBA
).
"4x4"
denotes the Latin Square
(ABCD
BCDA
CDAB
DABC
)
or any of the six possible Williams’ designs with four
sequences:
library(randomizeBE)
< data.frame(option = NA_integer_,
williams4 seq.1 = NA_character_, seq.2 = NA_character_,
seq.3 = NA_character_, seq.4 = NA_character_)
for (i in 1:50) {
2:5] < williams(4)
williams4[i,
}< unique(williams4)
williams4 < 1:nrow(williams4)
options $option < 1:nrow(williams4)
williams4print(williams4, row.names = FALSE)
# option seq.1 seq.2 seq.3 seq.4
# 1 ABDC BCAD CDBA DACB
# 2 ADCB BCDA CABD DBAC
# 3 ACDB BDCA CBAD DABC
# 4 ABCD BDAC CADB DCBA
# 5 ACBD BADC CDAB DBCA
# 6 ADBC BACD CBDA DCAB
df
are the degrees of freedom, where n
is
the sample size.
Most examples deal with studies where the response variables likely
follow a lognormal
distribution, i.e., we assume a multiplicative model
(ratios instead of differences). We work with \(\small{\log_{e}}\)transformed data in
order to allow analysis by the ttest (requiring differences).
This is the default in most functions of PowerTOST
and
hence, the argument logscale = TRUE
does not need to be
specified.
HigherOrder Crossover Designs are particular because two approaches for the evaluation are possible. The sample size depends on the chosen approach.
“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.
Which approach you could / should use is discussed further down in the sections
It may sound picky but ‘sample size calculation’ (as used in most guidelines and alas, in some publications and textbooks) is sloppy terminology. In order to get prospective power (and hence, a sample size), we need five values:
1 – 2 are fixed by the agency,
3 is set by the sponsor
(commonly to 0.80 – 0.90), and
4 – 5 are just (uncertain!) assumptions.
In other words, obtaining a sample size is not an exact calculation like \(\small{2\times2=4}\) but always just an estimation.
“Power Calculation – A guess masquerading as mathematics.
As an aside, it is extremely unlikely that all assumptions will be exactly realized in a particular study. Hence, calculating retrospective (a.k.a. post hoc, a posteriori) power is not only futile but plain nonsense.^{7}
Since generally the withinsubject variability is lower than the betweensubject variability, crossover studies are so popular. The efficiency of a crossover study compared to a parallel study is given by \(\small{\frac{\sigma_\textrm{intra}^2\;+\,\sigma_\textrm{inter}^2}{0.5\,\times\,\sigma_\textrm{intra}^2}}\). If, say, \(\small{\sigma_\textrm{intra}^2=0.5\times\sigma_\textrm{inter}^2}\) in a paralled study we need six times as many subjects than in a crossover to obtain the same power. On the other hand, in a crossover we have two measurements per subject, which makes the parallel study approximately three times more costly.
Of note, there is no relationship between \(\small{CV_\textrm{intra}}\) and \(\small{CV_\textrm{inter}}\).
An example are drugs which are subjected to polymorphic metabolism. For
these drugs \(\small{CV_\textrm{intra}\ll
CV_\textrm{inter}}\). On the other hand, some
HVD(P)s show
\(\small{CV_\textrm{intra}>CV_\textrm{inter}}\).
However, it is a prerequisite that no carryover from one period to the next exists. Only then the comparison of treatments will be unbiased. For details see another article.^{8}
Subjects have to be in the same physiological state^{9} throughout the study –
guaranteed by sufficiently long washout phases. Crossover studies can
not only be performed in healthy volunteers but also in patients with a
stable disease (e.g., asthma). Studies in patients
with an instable disease (e.g., in oncology)
must be performed in a parallel design (covered in another
article).
If crossovers are not feasible (e.g., for drugs with a very
long half life), studies could be performed in a parallel design as
well.
The sample size cannot be
directly estimated,
only power calculated for an already given sample
size.
The power equations cannot be rearranged to solve for sample size.
“Power. That which statisticians are always calculating but never have.
If you are a geek out in the wild and have only your scientific calculator with you, use the large sample approximation (i.e., based on the normal distribution of \(\small{\log_{e}}\)transformed data) taking power and \(\small{\alpha}\) into account.^{11}
Convert the coefficient of variation to the variance \[\begin{equation}\tag{1} \sigma^2=\log_{e}(CV^2+1) \end{equation}\] Numbers needed in the following: \(\small{Z_{10.2}=0.8416212}\), \(\small{Z_{10.1}=1.281552}\), \(\small{Z_{10.05}=1.644854}\). \[\begin{equation}\tag{2a} \textrm{if}\;\theta_0<1:n\sim\frac{2\,\sigma^2(Z_{1\beta}Z_{1\alpha})^2}{\left(\log_{e}(\theta_0)\log_{e}(\theta_1)\right)^2} \end{equation}\] \[\begin{equation}\tag{2b} \textrm{if}\;\theta_0>1:n\sim\frac{2\,\sigma^2(Z_{1\beta}Z_{1\alpha})^2}{\left(\log_{e}(\theta_0)\log_{e}(\theta_2)\right)^2} \end{equation}\] \[\begin{equation}\tag{2c} \textrm{if}\;\theta_0=1:n\sim\frac{2\,\sigma^2(Z_{1\beta/2}Z_{1\alpha})^2}{\left(\log_{e}(\theta_2)\right)^2} \end{equation}\] Generally you should use \(\small{(2\textrm{a},\,}\)\(\small{2\textrm{b})}\). Only if you are courageous, use \(\small{(2\textrm{c})}\). However, I would not recommend that.^{12}
The sample size by \(\small{(2)}\) can be a real
number. We have to round it up to the next even
integer to obtain balanced sequences (equal number of subjects in
sequences TR
and RT
).
In \(\small{(1}\)\(\small{2)}\) we assume that the
population variance \(\small{\sigma^2}\) is known, which is never
the case. In practice is is uncertain because only its estimate (the
sample variance \(\small{s^2}\)) is known. Therefore, \(\small{(1}\)\(\small{2)}\) may
give a sample size which is too low:
For \(\small{CV=0.25}\), \(\small{\theta_0=0.95}\), \(\small{\alpha=0.05}\), \(\small{\beta=0.20,}\) and the conventional
limits we get 25.383 for the total sample size (rounded up to 27). As we
will see later, the correct sample size is also 27.
“Power: That which is wielded by the priesthood of clinical trials, the statisticians, and a stick which they use to beta their colleagues.
Since we evaluate studies based on the ttest, we have to
take the degrees of freedom (in the 3treatment designs \(\small{df=2\,n4}\) and in the 4treatment
designs \(\small{df=3\,n6}\)) into
account.
Power depends on the degrees of freedom, which themselves depend on the
sample size.
Hence, the power equations are used to estimate the sample size in an iterative way.
Example based on the noncentral tdistribution (warning: 114 lines of code).
< function(design) {
get.sequences < known.designs()[, c(2, 5)]
known < as.integer(known[known$design == design,
ns "steps"])
return(ns)
}< function(design, n) {
get.df < known.designs()[, 2:3]
known < known[known$design == design, "df"]
dfe < as.integer(eval(parse(text = dfe)))
df return(df)
}< function(n, ns) {# equally sized sequences
make.equal return(as.integer(ns * (n %/% ns + as.logical(n %% ns))))
}< 0.25 # withinsubject CV
CV < 0.95 # T/Rratio
theta0 < 0.80 # lower BElimit
theta1 < 1.25 # upper BElimit
theta2 < 0.80 # desired (target) power
target < 0.05 # level of the test
alpha < 1  target # producer's risk
beta < "3x3"
design < get.sequences(design)
ns < data.frame(method = character(),
df iteration = integer(),
n = integer(),
power = numeric())
< log(CV^2 + 1)
s2 < sqrt(s2)
s < qnorm(1  alpha)
Z.alpha < qnorm(1  beta)
Z.beta .2 < qnorm(1  beta / 2)
Z.beta# large sample approximation
if (theta0 == 1) {
< 2 * s2 * (Z.beta.2 + Z.alpha)^2
num < ceiling(num / log(theta2)^2)
n.seq else {
} < 2 * s2 * (Z.beta + Z.alpha)^2
num if (theta0 < 1) {
< (log(theta0)  log(theta1))^2
denom else {
}< (log(theta0)  log(theta2))^2
denom
}< ceiling(num / denom)
n.seq
}< make.equal(n.seq, ns)
n < pnorm(sqrt((log(theta0)  log(theta2))^2 * n /
power 2 * s2))  Z.alpha) +
(pnorm(sqrt((log(theta0)  log(theta1))^2 * n /
2 * s2))  Z.alpha)  1
(1, 1:4] < c("large sample", "\u2013",
df[sprintf("%.6f", power))
n, # check by central t approximation
# (since the variance is unknown)
< qt(1  alpha, get.df(design, n))
t.alpha if (theta0 == 1) {
< 2 * s2 * (Z.beta.2 + t.alpha)^2
num < ceiling(num / log(theta2)^2)
n.seq else {
} < 2 * s2 * (Z.beta + t.alpha)^2
num if (theta0 < 1) {
< (log(theta0)  log(theta1))^2
denom else {
}< (log(theta0)  log(theta2))^2
denom
}< ceiling(num / denom)
n.seq
}< make.equal(n.seq, ns)
n < qt(1  alpha, get.df(design, n))
t.alpha < pnorm(sqrt((log(theta0)  log(theta2))^2 * n /
power 2 * s2))  t.alpha) +
(pnorm(sqrt((log(theta0)  log(theta1))^2 * n /
2 * s2))  t.alpha)  1
(2, 1:4] < c("central t", "\u2013",
df[sprintf("%.6f", power))
n, < 1
i if (power < target) {# iterate upwards
repeat {
< sqrt(2 / n) * s
sem < c((log(theta0)  log(theta1)) / sem,
ncp log(theta0)  log(theta2)) / sem)
(< diff(pt(c(+1, 1) * qt(1  alpha,
power df = get.df(design, n)),
df = get.df(design, n), ncp = ncp))
+ 2, 1:4] < c("noncentral t", i,
df[i sprintf("%.6f", power))
n, if (power >= target) {
break
else {
}< i + 1
i < n + 2
n
}
}else { # iterate downwards
} repeat {
< sqrt(2 / n) * s
sem < c((log(theta0)  log(theta1)) / sem,
ncp log(theta0)  log(theta2)) / sem)
(< diff(pt(c(+1, 1) * qt(1  alpha,
power df = get.df(design, n)),
df = get.df(design, n), ncp = ncp))
+ 2, 1:4] < c("noncentral t", i,
df[i sprintf("%.6f", power))
n, if (power < target) {
< df[nrow(df), ]
df break
else {
}< i + 1
i < n  2
n
}
}
}print(df, row.names = FALSE)
# method iteration n power
# large sample – 27 0.813971
# central t – 27 0.805100
# noncentral t 1 27 0.803494
Now it is time to leave Base R behind and
continue with PowerTOST
. Makes our life much easier.
library(PowerTOST) # attach libraries
library(randomizeBE) # to run the examples
The sample size functions of PowerTOST
use a
modification of Zhang’s method^{14} based on the large sample approximation
as the starting value of the iterations. You can unveil the course of
iterations with details = TRUE
.
sampleN.TOST(CV = 0.25, theta0 = 0.95, design = "3x3",
targetpower = 0.80, details = TRUE)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 3x3 crossover
# Design characteristics:
# df = 2*n4, design const. = 2, step = 3
#
# logtransformed 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 search (ntotal)
# n power
# 24 0.753428
# 27 0.803494
#
# Exact power calculation with
# Owen's Q functions.
Two iterations. In many cases it hits the bull’s eye right away, i.e., already with the first guess.
Never (ever!) estimate the sample size based on the large
sample approximation \(\small{(2)}\).
We evaluate our studies based on the ttest, right?
The sample size based on the normal distribution may be too
small, thus compromising power.
# method n Delta RE power
# large sample approx. (2) 27 ±0 ±0.0% 0.803494
# noncentral tdistribution 27 ±0 ±0.0% 0.803494
# exact method (Owen’s Q) 27 – – 0.803494
In this example it works but it may fail in others.
top of section ↩︎ previous section ↩︎
Throughout the examples I’m referring to studies in a single center – not multiple groups within them or multicenter studies. That’s another pot of tea.
Most methods of PowerTOST
are based on pairwise
comparisons. It is up to you to adjust the level of the test
alpha
if you want to compare more (say, two test treatments
vs a reference or each of them against one of the others) in
order to avoid inflation of the familywise
error rate due to multiplicity. Details are given further down.
First of all we have have to consider the type of study, its purpose, and how we intend to evaluate it.
Let’s discuss the cases from above and whether we can evaluate the study by the common \(\small{\alpha=0.05}\) (i.e., with a \(\small{90\%}\) confidence interval) or have to adjust for multiplicity (\(\small{\alpha<0.05 \to\; >90\%}\) CI).
In this approach – which is preferred by some agencies, e.g., the EMA^{18} ^{19} ^{20} – all but two treatments are excluded. That means you obtain separate Incomplete Block Designs^{21} ^{22} (IBD), where the coding (treatments, sequences, periods) is kept and analyses of the respective pairs of interest are performed.
Williams’ designs with three and four treatments and the resulting IBDs (excluded treatments are denoted by \(\small{{\color{Red}\bullet}\,}\)): \[\small{\begin{array}{cccc} s/p & \text{I} & \text{II} & \text{III} \\\hline 1 & \text{A} & \text{B} & \text{C}\\ 2 & \text{A} & \text{C} & \text{B}\\ 3 & \text{B} & \text{A} & \text{C}\\ 4 & \text{B} & \text{C} & \text{A}\\ 5 & \text{C} & \text{A} & \text{B}\\ 6 & \text{C} & \text{B} & \text{A}\\ \end{array}{\color{Blue}\mapsto} \begin{array}{cccc} s/p & \text{I} & \text{II} & \text{III} \\\hline 1 & \text{A} & {\color{Red}\bullet} & \text{C}\\ 2 & \text{A} & \text{C} & {\color{Red}\bullet}\\ 3 & {\color{Red}\bullet} & \text{A} & \text{C}\\ 4 & {\color{Red}\bullet} & \text{C} & \text{A}\\ 5 & \text{C} & \text{A} & {\color{Red}\bullet}\\ 6 & \text{C} & {\color{Red}\bullet} & \text{A}\\ \end{array}{\color{Blue}\wedge} \begin{array}{cccc} s/p & \text{I} & \text{II} & \text{III} \\\hline 1 & {\color{Red}\bullet} & \text{B} & \text{C}\\ 2 & {\color{Red}\bullet} & \text{C} & \text{B}\\ 3 & \text{B} & {\color{Red}\bullet} & \text{C}\\ 4 & \text{B} & \text{C} & {\color{Red}\bullet}\\ 5 & \text{C} & {\color{Red}\bullet} & \text{B}\\ 6 & \text{C} & \text{B} & {\color{Red}\bullet}\\ \end{array}}\] \[\small{\begin{array}{ccccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & \text{A} & \text{C} & \text{B} & \text{D}\\ 2 & \text{B} & \text{A} & \text{D} & \text{C}\\ 3 & \text{C} & \text{D} & \text{A} & \text{B}\\ 4 & \text{D} & \text{B} & \text{C} & \text{A}\\ \end{array}{\color{Blue}\mapsto} \begin{array}{ccccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV} \\\hline 1 & \text{A} & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{D}\\ 2 & {\color{Red}\bullet} & \text{A} & \text{D} & {\color{Red}\bullet}\\ 3 & {\color{Red}\bullet} & \text{D} & \text{A} & {\color{Red}\bullet}\\ 4 & \text{D} & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{A}\\ \end{array}{\color{Blue}\wedge} \begin{array}{ccccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{B} & \text{D}\\ 2 & \text{B} & {\color{Red}\bullet} & \text{D} & {\color{Red}\bullet}\\ 3 & {\color{Red}\bullet} & \text{D} & {\color{Red}\bullet} & \text{B}\\ 4 & \text{D} & \text{B} & {\color{Red}\bullet} & {\color{Red}\bullet}\\ \end{array}{\color{Blue}\wedge} \begin{array}{ccccc} s/p & \text{I} & \text{II} & \text{III} & \text{IV}\\\hline 1 & {\color{Red}\bullet} & \text{C} & {\color{Red}\bullet} & \text{D}\\ 2 & {\color{Red}\bullet} & {\color{Red}\bullet} & \text{D} & \text{C}\\ 3 & \text{C} & \text{D} & {\color{Red}\bullet} & {\color{Red}\bullet}\\ 4 & \text{D} & {\color{Red}\bullet} & \text{C} & {\color{Red}\bullet}\\ \end{array}}\] With Latin Squares (\(\small{\text{ABC}\text{BCA}\text{CAB}}\), \(\small{\text{ABCD}\text{BCDA}\text{CDAB}\text{DABC}}\), etc.) balanced IBDs cannot be achieved.
An example to get a randomization list and extract the IBDs. We assume a CV of 0.25, a T/Rratio of 0.95, target a power of 0.80, and plan the evaluation for the ‘Two at a Time’ approach.
< function(n, ns) {# equally sized sequences
make.equal return(as.integer(ns * (n %/% ns + as.logical(n %% ns))))
}< function(n, do.rate, ns) {
nadj return(as.integer(make.equal(n / (1  do.rate), ns)))
}< 4 # number of treatments (at least 3)
ntmt < 0.25 # assumed CV
CV < 0.95 # assumed T/Rratio
theta0 < 0.80 # lower and
theta1 < 1.25 # upper BE limit
theta2 < 0.80 # target (desired) power
target < FALSE # TRUE if the study should have balanced sequences
bal < 0 # anticipated dropoutrate
do.rate # sample size for evaluation by ‘Two at a Time’
# note that the design is specified as "2x2"
< sampleN.TOST(CV = CV, theta0 = theta0, targetpower = target,
n theta1 = theta1, theta2 = theta2,
design = "2x2", print = FALSE)[["Sample size"]]
# sequences of a Williams’ design
< williams(ntmt)
seqs < 2 # because 2x2!
ns # adjust for dropoutrate
< nadj(n, do.rate, ns)
n # eventual uprounding if balance is requested
if (bal) n < make.equal(n, length(seqs))
< paste("Sequences:", paste(seqs, collapse = ", "), "\n")
txt1 < paste0(seqs, collapse = "")
s # the last treatment in lexical order will act as the reference
< paste(sort(unlist(strsplit(s, ""))), collapse = "")
s < substr(s, nchar(s), nchar(s))
ref # extract treatment codes
< unique(c(strsplit(split = "", seqs), recursive = TRUE))
trts # extract tests in increasing order
< sort(trts[!trts == ref])
tests < 0
calls repeat {# call randomizations until all IBDs are balanced
# get a randomization list
< calls + 1
calls < suppressWarnings( # not relevant for TaT
tmp RL4(nsubj = n, seqs = seqs, randctrl = FALSE))
< tmp$rl
rand # generate columns for the IBDs
for (j in seq_along(tests)) {
paste0("IBD.", j)]] < NA
rand[[
}# build the IBDs
for (j in 1:nrow(rand)) {
for (k in seq_along(tests)) {
< tests[!tests == tests[k]]
excl < paste0("[", paste(excl, collapse = ", "), "]")
excl 3 + k] < gsub(excl, "•", rand$sequence[j])
rand[j,
}
}# check IBDs for balance
< rep(NA, length(tests))
checks for (j in seq_along(tests)) {
< gsub("[^[:alnum:], ]", "", rand[3 + j])
ibd < gsub("c", "", ibd)
ibd < gsub(" ", "", unlist(strsplit(ibd, ",")))
ibd < length(ibd[ibd == unique(ibd)[1]]) ==
checks[j] length(ibd[ibd == unique(ibd)[2]])
}if (sum(checks) == length(tests)) break
}< names(rand)[4:ncol(rand)]
ibd names(rand)[4:ncol(rand)] < gsub("\\.", " ", ibd)
< paste0("All extracted IBDs are balanced.",
txt2 "\nRandomized : ",
format(tmp$date, usetz = TRUE),
"\nSeeed : ", tmp$seed,
"\nCalls until balance was achieved: ", calls, "\n")
cat(txt1)
print(rand, row.names = FALSE)
cat(txt2)
# Sequences: ACBD, BADC, CDAB, DBCA
# subject seqno sequence IBD 1 IBD 2 IBD 3
# 1 1 ACBD A••D ••BD •C•D
# 2 4 DBCA D••A DB•• D•C•
# 3 4 DBCA D••A DB•• D•C•
# 4 2 BADC •AD• B•D• ••DC
# 5 2 BADC •AD• B•D• ••DC
# 6 3 CDAB •DA• •D•B CD••
# 7 1 ACBD A••D ••BD •C•D
# 8 3 CDAB •DA• •D•B CD••
# 9 3 CDAB •DA• •D•B CD••
# 10 2 BADC •AD• B•D• ••DC
# 11 4 DBCA D••A DB•• D•C•
# 12 3 CDAB •DA• •D•B CD••
# 13 2 BADC •AD• B•D• ••DC
# 14 1 ACBD A••D ••BD •C•D
# 15 4 DBCA D••A DB•• D•C•
# 16 1 ACBD A••D ••BD •C•D
# 17 4 DBCA D••A DB•• D•C•
# 18 4 DBCA D••A DB•• D•C•
# 19 2 BADC •AD• B•D• ••DC
# 20 3 CDAB •DA• •D•B CD••
# 21 3 CDAB •DA• •D•B CD••
# 22 1 ACBD A••D ••BD •C•D
# 23 2 BADC •AD• B•D• ••DC
# 24 1 ACBD A••D ••BD •C•D
# 25 3 CDAB •DA• •D•B CD••
# 26 2 BADC •AD• B•D• ••DC
# 27 1 ACBD A••D ••BD •C•D
# 28 4 DBCA D••A DB•• D•C•
# All extracted IBDs are balanced.
# Randomized : 20220711 12:42:25 CEST
# Seeed : 3965897
# Calls until balance was achieved: 1
In this example we have seven subjects in all sequences,
i.e., the design is balanced. This will not be so if the sample
size is not a multiple of the number of sequences. However, that’s not
relevant because we aim at balance of the
IBDs. If you desire
balance for any reason, specify bal < TRUE
, which will
require more subjects in such a case.
Since we are working in a regulated environment, perhaps you will be asked by an inspector how you performed the randomization. You need to provide the sample size, all sequences and the seed to reproduce it.
library(randomizeBE)
RL4(n = 28, seqs = c("ACBD", "BADC", "CDAB", "DBCA"), seed = 3965897)
Although there is just one Williams’ design for three treatments,
there are six for four treatments, twelve for five, 120 for six, and 360
for seven. The function williams()
of
randomizeBE
will arbitrarily select one of them. Hence, if
you run the script, the chance to get the same sequences like in the
example is ^{1}/_{120}.
Apart from the regulatory recommendation I suggest this approach, since the ‘All at Once’ approach may lead to biased estimates and an inflated Type I Error.^{5} ^{26}
In this approach we assume homogenicity in the ANOVA of pooled data (equal variances in the pairwise comparisons) and obtain one residual variance.
To plan for this approach specify one of the design arguments given
above, i.e., "3x3"
,
"3x6x3"
, or "4x4"
.
Whilst popular (and hence, used in most of the examples), I don’t
recommend it. Note that more than four treatments are not implemented in
sampleN.TOST()
. Consider the ‘Two
at a Time’ approach instead (see the examples below).
top of section ↩︎ previous section ↩︎
We assume a CV of 0.25, a T/Rratio of 0.95, and target a power of 0.80.
Since theta0 = 0.95
and targetpower = 0.80
are defaults of the function, we don’t have to give them explicitly. As
usual in BE,
alpha = 0.05
is employed (we will assess the study by a
\(\small{100\,(12\,\alpha)=90\%}\)
confidence interval). The BElimits
are theta1 = 0.80
and theta2 = 1.25
. Since
they are also defaults of the function, we don’t have to give them as
well. Hence, we need to specify only the CV
and the
design
.
sampleN.TOST(CV = 0.25, design = "3x3") # All at Once
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 3x3 crossover
# logtransformed 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
# 27 0.803494
sampleN.TOST(CV = 0.25, design = "2x2") # Two at a Time
#
# +++++++++++ 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 = 0.25
#
# Sample size (total)
# n power
# 28 0.807439
Don’t get confused by Study design: 2x2 crossover
in the
output. It only means that we will evaluate the study by the
‘Two at a Time’ approach as
IBDs. In the ‘All at Once’
approach we have more degrees of freedom (\(\small{2\times274=50}\)) than in the ‘Two
at a Time’ approach (\(\small{282=26}\)). An example to get the
correct sample size is given in the Interlude
further down.
Sometimes we are not interested in the entire output and want to use
only a part of the results in subsequent calculations. We can suppress
the output by the argument print = FALSE
and assign the
result to a data frame (here named x
).
< sampleN.TOST(CV = 0.25, design = "3x3",
x print = FALSE)
Although you could access the elements by the number of the column(s), I don’t recommend that, since in various functions these numbers are different and hence, difficult to remember.
Let’s retrieve the column names of x
:
names(x)
# [1] "Design" "alpha" "CV"
# [4] "theta0" "theta1" "theta2"
# [7] "Sample size" "Achieved power" "Target power"
Now we can access the elements of x
by their names. Note
that double square brackets [[…]]
have to be used.
"Sample size"]]
x[[# [1] 27
"Achieved power"]]
x[[# [1] 0.8034938
Although you could access the elements by the number of the
column(s), I don’t recommend that, since in various other functions of
PowerTOST
these numbers are different and hence, difficult
to remember. If you insist in accessing elements by columnnumbers, use
single square brackets […]
.
7:8]
x[# Sample size Achieved power
# 1 27 0.8034938
With 27 subjects (nine per sequence) we achieve the power we desire.
What happens if we have one dropout?
power.TOST(CV = 0.25, design = "3x3",
n = x[["Sample size"]]  1)
# Unbalanced design. n(i)=9/9/8 assumed.
# [1] 0.7868869
Slightly below the 0.80 we desire.
Interlude 1
As we have seen above, when we plan to
evaluate the study by the ‘Two at a Time’ approach, we will get
an even number of subjects from
sampleN.TOST(design = "2x2")
. If we want balanced sequences
in our design (which will be a multiple of three in the
"3x3"
design, a multiple of six in the "3x6x3"
design, and a multiple of four in the "4x4"
design),
sometimes we have to round the sample size up.
< function(design) {
get.sequences < known.designs()[, c(2, 5)]
known < as.integer(known[known$design == design,
ns "steps"])
return(ns)
}< function(n, ns) {
make.equal return(as.integer(ns * (n %/% ns + as.logical(n %% ns))))
}< "3x3"
design < 0.25
CV < get.sequences(design)
ns < sampleN.TOST(CV = CV, design = "2x2",
n.TaT print = FALSE)[["Sample size"]]
< sampleN.TOST(CV = CV, design = design,
n.AaO print = FALSE)[["Sample size"]]
< make.equal(n = n.TaT, ns = ns)
n < data.frame(design = design,
df approach = c("\u2018All at Once\u2019",
"\u2018Two at a Time\u2019"),
n = c(n.AaO, n), power = NA)
1, 4] < power.TOST(CV = CV, design = design,
df[n = n.AaO)
2, 4] < power.TOST(CV = CV, design = "2x2",
df[n = n)
print(df, row.names = FALSE)
# design approach n power
# 3x3 ‘All at Once’ 27 0.8034938
# 3x3 ‘Two at a Time’ 30 0.8342518
This uprounding from the sample size of the ‘All at Once’ approach protects us also against loss of power due to dropouts.
However, when planning for the ‘Two at a Time’ approach, it is not necessary to have balanced sequences – only the resulting IBDs have to be. Hence, we could plan it as a conventional 2×2 crossover.
< "3x3"
design < 0.25
CV < data.frame(design = design,
df approach = "\u2018Two at a Time\u2019",
n = NA, power = NA)
1, 3:4] < sampleN.TOST(CV = CV, design = "2x2", # for TaT use "2x2"
df[print = FALSE)[7:8]
print(df, row.names = FALSE)
# design approach n power
# 3x3 ‘Two at a Time’ 28 0.8074395
Since dropouts are common, it makes sense to include / dose more subjects in order to end up with a number of eligible subjects which is not lower than our initial estimate.
Let us explore that in the next section.
top of section ↩︎ previous section ↩︎
We define two supportive functions:
n
will be rounded up to the next multiple of the
number of sequences ns
.< function(n, ns) {
make.equal return(as.integer(ns * (n %/% ns + as.logical(n %% ns))))
}
n
, the anticipated droputrate do.rate
, and
the number of sequences ns
.< function(n, do.rate, ns) {
nadj return(as.integer(make.equal(n / (1  do.rate), ns)))
}
In order to come up with a suggestion we have to anticipate a (realistic!) dropout rate. Note that this not the job of the statistician; ask the Principal Investigator.
“It is a capital mistake to theorize before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts.
The dropoutrate is calculated from the eligible and
dosed subjects
or simply \[\begin{equation}\tag{3}
do.rate=1n_\textrm{eligible}/n_\textrm{dosed}
\end{equation}\] Of course, we know it only after the
study was performed.
By substituting \(n_\textrm{eligible}\) with the estimated sample size \(n\), providing an anticipated dropoutrate and rearrangement to find the adjusted number of dosed subjects \(n_\textrm{adj}\) we should use \[\begin{equation}\tag{4} n_\textrm{adj}=\;\upharpoonleft n\,/\,(1do.rate) \end{equation}\] where \(\upharpoonleft\) denotes rounding up to the next even number as implemented in the functions above.
An all too common mistake is to increase the estimated sample size \(n\) by the drop outrate according to \[\begin{equation}\tag{5} n_\textrm{adj}=\;\upharpoonleft n\times(1+do.rate) \end{equation}\]
If you used \(\small{(5)}\) in the past – you are not alone. In a small survey a whopping 29% of respondents reported to use it.^{28} Consider changing your routine.
“There are no routine statistical questions, only questionable statistical routines.
In the following I specified more arguments to make the function more
flexible.
Note that I wrapped the function power.TOST()
in
suppressMessages()
. Otherwise, the function will throw for
any odd sample size a message telling us that the design is
unbalanced. Well, we know that.
< 0.25 # withinsubject CV
CV < 0.80 # target (desired) power
target < 0.95 # assumed T/Rratio
theta0 < 0.80 # lower BE limit
theta1 < 1.25 # upper BE limit
theta2 < "3x3"
design < 3
ns < 0.15 # anticipated dropoutrate 10%
do.rate < sampleN.TOST(CV = CV, theta0 = theta0,
df theta1 = theta1,
theta2 = theta2,
targetpower = target,
design = design,
print = FALSE)
# calculate the adjusted sample size
< nadj(df[["Sample size"]], do.rate, ns)
n.adj # (decreasing) vector of eligible subjects
< n.adj:df[["Sample size"]]
n.elig < paste0("Assumed CV : ",
info
CV,"\nAssumed T/R ratio : ",
theta0,"\nBE limits : ",
paste(theta1, theta2, sep = "\u2026"),
"\nTarget (desired) power : ",
target,"\nAnticipated dropoutrate: ",
do.rate,"\nEstimated sample size : ",
"Sample size"]], " (",
df[["Sample size"]]/ns, "/sequence)",
df[["\nAchieved power : ",
signif(df[["Achieved power"]], 4),
"\nAdjusted sample size : ",
" (", n.adj/ns, "/sequence)",
n.adj, "\n\n")
# explore the potential outcome for
# an increasing number of dropouts
< signif((n.adj  n.elig) / n.adj, 4)
do.act < data.frame(dosed = n.adj,
df eligible = n.elig,
dropouts = n.adj  n.elig,
do.act = do.act,
power = NA)
for (i in 1:nrow(df)) {
$power[i] < suppressMessages(
dfpower.TOST(CV = CV,
theta0 = theta0,
theta1 = theta1,
theta2 = theta2,
design = design,
n = df$eligible[i]))
}cat(info)
print(round(df, 4), row.names = FALSE)
# Assumed CV : 0.25
# Assumed T/R ratio : 0.95
# BE limits : 0.8…1.25
# Target (desired) power : 0.8
# Anticipated dropoutrate: 0.15
# Estimated sample size : 27 (9/sequence)
# Achieved power : 0.8035
# Adjusted sample size : 33 (11/sequence)
#
# dosed eligible dropouts do.act power
# 33 33 0 0.0000 0.8745
# 33 32 1 0.0303 0.8641
# 33 31 2 0.0606 0.8536
# 33 30 3 0.0909 0.8430
# 33 29 4 0.1212 0.8300
# 33 28 5 0.1515 0.8168
# 33 27 6 0.1818 0.8035
In the worst case (six dropouts) we end up with the originally estimated sample size of 27. Power preserved, mission accomplished. If we have less dropouts, splendid – we gain power.
If we would have adjusted the sample size acc. to \(\small{(5)}\) we
would have dosed also 33 subjects like the 33 acc. to \(\small{(4)}\).
Cave: This might not always be the case… If the anticipated dropout rate
of 15% is realized in the study, we would have also 28 eligible subjects
(power 0.8168). In this example we achieve still more than our target
power but the loss might be relevant in other cases.
As said in the preliminaries, calculating post hoc power is futile.
“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.
However, sometimes we are interested in it for planning the next study.
If you give a total sample size n
with unbalanced
sequences, power.TOST()
will try to keep sequences as
balanced as possible and show in a message how that was done.
< 26
n.act signif(power.TOST(CV = 0.25, design = "3x3",
n = n.act), 6)
# Unbalanced design. n(i)=9/9/8 assumed.
# [1] 0.786887
Say, our study was more unbalanced. Let us assume that we dosed 33
subjects, the total number of subjects was also 26 but all dropouts
occured in one sequence (unlikely but possible).
Instead of the total sample size n
we can give the number
of subjects of each sequence as a vector (the order is not relevant,
i.e., it does not matter which element refers to which
sequence).
< head(df$dosed, 1)
n.adj < tail(df$eligible, 1)
n.act < tail(df$dropouts, 1)
do < rep(n.adj / ns, ns)
n.eli 1] < n.adj / ns  do
n.eli[< signif(power.TOST(CV = 0.25,
post.hoc n = n.eli,
design = design), 6)
< nchar(as.character(n.adj))
sig.dig < paste0("%", sig.dig, ".0f (%",
fmt ".0f dropouts)")
sig.dig, < paste0("Dosed subjects: ",
info sprintf("%2.0f", n.adj),
"\nEligible : ",
sprintf(fmt, n.act,
 n.act))
n.adj for (i in seq_along(n.eli)) {
< paste(info, "\n Sequence ", i, ":",
info sprintf(fmt, n.eli[i],
/ ns  n.eli[i]))
n.adj
}< paste(info, "\nPower : ",
info "\n")
post.hoc, cat(info)
# Dosed subjects: 33
# Eligible : 27 ( 6 dropouts)
# Sequence 1 : 5 ( 6 dropouts)
# Sequence 2 : 11 ( 0 dropouts)
# Sequence 3 : 11 ( 0 dropouts)
# Power : 0.747186
Of course, in a particular study you will provide the numbers in the
n
vector directly.
As stated already in the Introduction, the
CV and the T/Rratio are only assumptions. Whatever
their origin might be (literature, previous studies) they carry some
degree of uncertainty. Hence, believing^{30} that they are the
true ones may be risky.
Some statisticians call that the ‘CarvedinStone’ approach.
Say, we performed a pilot study in 12 subjects and estimated the CV as 0.25.
The \(\alpha\) confidence interval of the CV is obtained via the \(\small{\chi^2}\)distribution of its error variance \(\small{\sigma^2}\) with \(\small{n2}\) degrees of freedom. \[\eqalign{\tag{6} s_\text{w}^2&=\log_{e}(CV_\text{w}^2+1)\\ L=\frac{(n1)\,s_\text{w}^2}{\chi_{\alpha/2,\,n2}^{2}}&\leq\sigma_\text{w}^2\leq\frac{(n1)\,s_\text{w}^2}{\chi_{1\alpha/2,\,n2}^{2}}=U\\ \left\{lower\;CL,\;upper\;CL\right\}&=\left\{\sqrt{\exp(L)1},\sqrt{\exp(U)1}\right\} }\] Let’s calculate the 95% confidence interval of the CV to get an idea.
< 12 # pilot study
m < CVCL(CV = 0.25, df = get.df(design, m),
ci side = "2sided", alpha = 0.05)
signif(ci, 4)
# lower CL upper CL
# 0.1901 0.3671
Surprised? Although 0.25 is the best estimate for planning the next study, there is no guarantee that we will get exactly the same outcome. Since the \(\small{\chi^2}\)distribution is skewed to the right, it is more likely to face a higher CV than a lower one in the pivotal study.
If we plan the study based on 0.25, we would opt for 27 subjects like
in the examples before (not adjusted for the dropoutrate).
If the CV will be lower, we gain power. Fine.
But what if it will be higher? Of course, we will loose power. But how
much?
Let’s explore what might happen at the confidence limits of the CV.
< 12
m < CVCL(CV = 0.25, get.df(design, m),
ci side = "2sided", alpha = 0.05)
< 27
n < data.frame(CV = c(ci[["lower CL"]], 0.25,
comp "upper CL"]]),
ci[[power = NA)
for (i in 1:nrow(comp)) {
$power[i] < power.TOST(CV = comp$CV[i],
compdesign = design,
n = n)
}1] < signif(comp[, 1], 4)
comp[, 2] < signif(comp[, 2], 6)
comp[, print(comp, row.names = FALSE)
# CV power
# 0.1901 0.951526
# 0.2500 0.803494
# 0.3671 0.417982
Ouch!
What can we do? The larger the previous study was, the larger the degrees of freedom and hence, the narrower the confidence interval of the CV. In simple terms: The estimate is more certain. On the other hand, it also means that very small pilot studies are practically useless.
< seq(6, 16, 2)
m < data.frame(n = m, CV = 0.25,
df l = NA, u = NA)
for (i in 1:nrow(df)) {
3:4] < CVCL(CV = 0.25,
df[i, df = get.df(design, m[i]),
side = "2sided",
alpha = 0.05)
}3:4] < signif(df[, 3:4], 4)
df[, names(df)[3:4] < c("lower CL", "upper CL")
print(df, row.names = FALSE)
# n CV lower CL upper CL
# 6 0.25 0.1675 0.4992
# 8 0.25 0.1779 0.4238
# 10 0.25 0.1849 0.3883
# 12 0.25 0.1901 0.3671
# 14 0.25 0.1940 0.3528
# 16 0.25 0.1973 0.3425
A Bayesian
method is implemented in PowerTOST
, which takes the
uncertainty of estimates into account. We can explore the uncertainty of
the CV or the T/Rratio and of both.
< 12 # sample size of pilot
m < 0.25
CV < 0.95
theta0 < expsampleN.TOST(CV = CV, theta0 = theta0,
df targetpower = 0.80,
design = design,
prior.parm = list(
m = m, design = design),
prior.type = "CV",
details = FALSE)
#
# ++++++++++++ Equivalence test  TOST ++++++++++++
# Sample size est. with uncertain CV
# 
# Study design: 3x3 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25
# Ratio = 0.95
# CV = 0.25 with 20 df
#
# Sample size (ntotal)
# n exp. power
# 30 0.804060
Not that bad. 11% more subjects required.
What about an uncertain T/Rratio?
< 12
m < 0.25
CV < 0.95
theta0 < expsampleN.TOST(CV = CV, theta0 = theta0,
df targetpower = 0.80,
design = design,
prior.parm = list(
m = m, design = design),
prior.type = "theta0",
details = FALSE)
#
# ++++++++++++ Equivalence test  TOST ++++++++++++
# Sample size est. with uncertain theta0
# 
# Study design: 3x3 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25
# Ratio = 0.95
# CV = 0.25
#
# Sample size (ntotal)
# n exp. power
# 183 0.881163
Terrible! What a nightmare.
That should not take us by surprise. We don’t know where the true T/Rratio lies but we can calculate the lower 95% confidence limit of the pilot study’s point estimate to get an idea about a worst case.
< 12
m < 0.25
CV < 0.95
pe < round(CI.BE(CV = CV, pe = 0.95,
ci n = m, design = design), 4)
if (pe <= 1) {
< ci[["lower"]]
cl else {
} < ci[["upper"]]
cl
}print(cl)
# [1] 0.7988
Exlore the impact of a relatively 5% higher CV and a relatively 5% lower T/Rratio on power for a given sample size.
< 27
n < 0.25
CV < 0.95
theta0 < data.frame(CV = c(CV, CV*1.05),
comp1 power = NA)
< data.frame(theta0 = c(theta0, theta0*0.95),
comp2 power = NA)
for (i in 1:2) {
$power[i] < power.TOST(CV = comp1$CV[i],
comp1theta0 = theta0,
design = design,
n = n)
}$power < signif(comp1$power, 6)
comp1for (i in 1:2) {
$power[i] < power.TOST(CV = CV,
comp2theta0 = comp2$theta0[i],
design = design,
n = n)
}$power < signif(comp2$power, 6)
comp2print(comp1, row.names = FALSE)
print(comp2, row.names = FALSE)
# CV power
# 0.2500 0.803494
# 0.2625 0.765085
# theta0 power
# 0.9500 0.803494
# 0.9025 0.550795
Interlude 2
Note the logscale of the xaxis. It demonstrates that power curves are symmetrical around 1 (\(\small{\log_{e}(1)=0}\), where \(\small{\log_{e}(\theta_2)=\left\log_{e}(\theta_1)\right}\)) and we will achieve the same power for \(\small{\theta_0}\) and \(\small{1/\theta_0}\) (e.g., for 0.95 and 1.0526). Furthermore, all curves intersect 0.05 at \(\small{\theta_1}\) and \(\small{\theta_2}\), which means that for a true \(\small{\theta_0}\) at one of the limits \(\small{\alpha}\) is strictly controlled.
<nitpick>< 0.25
CV < 0.05 # direction unknown
delta < c(1  delta, 1 / (1 + delta),
theta0s 1 + delta, 1 / (1  delta))
< sampleN.TOST(CV = CV, theta0 = 1  delta,
n design = design,
print = FALSE)[["Sample size"]]
< data.frame(CV = CV, theta0 = theta0s,
comp1 base = c(TRUE, rep(FALSE, 3)),
n = n, power = NA)
for (i in 1:nrow(comp1)) {
$power[i] < power.TOST(CV = CV,
comp1theta0 = comp1$theta0[i],
n = n, design = design)
}< sampleN.TOST(CV = CV, theta0 = 1 + delta,
n design = design,
print = FALSE)[["Sample size"]]
< data.frame(CV = CV, theta0 = theta0s,
comp2 base = c(FALSE, FALSE, TRUE, FALSE),
n = n, power = NA)
for (i in 1:nrow(comp2)) {
$power[i] < power.TOST(CV = CV,
comp2design = design,
theta0 = comp2$theta0[i],
n = n)
}c(2, 5)] < signif(comp1[, c(2, 5)] , 4)
comp1[, c(2, 5)] < signif(comp2[, c(2, 5)] , 4)
comp2[, print(comp1, row.names = FALSE)
print(comp2, row.names = FALSE)
# CV theta0 base n power
# 0.25 0.9500 TRUE 27 0.8035
# 0.25 0.9524 FALSE 27 0.8124
# 0.25 1.0500 FALSE 27 0.8124
# 0.25 1.0530 FALSE 27 0.8035
# CV theta0 base n power
# 0.25 0.9500 FALSE 27 0.8035
# 0.25 0.9524 FALSE 27 0.8124
# 0.25 1.0500 TRUE 27 0.8124
# 0.25 1.0530 FALSE 27 0.8035
</nitpick>
Since power is much more sensitive to the T/Rratio than to the CV, the results obtained with the Bayesian method should be clear now.
Essentially this leads to the murky waters of prospective power
analysis, which is covered in another
article.
An appetizer to show the maximum deviations (CV, T/Rratio and
decreased sample size due to dropouts) which give still a minimum
acceptable power of ≥ 0.70:
< 0.25
CV < 0.95
theta0 < 0.80
target < 0.70
minpower < pa.ABE(CV = CV, theta0 = theta0,
pa targetpower = target,
minpower = minpower,
design = design)
< 100*(tail(pa$paCV[["CV"]], 1) 
change.CV $plan$CV) / pa$plan$CV
pa< 100*(head(pa$paGMR$theta0, 1) 
change.theta0 $plan$theta0) /
pa$plan$theta0
pa< 100*(tail(pa$paN[["N"]], 1) 
change.n $plan[["Sample size"]]) /
pa$plan[["Sample size"]]
pa< data.frame(parameter = c("CV", "theta0", "n"),
comp change = c(change.CV,
change.theta0,
change.n))$change < sprintf("%+.2f%%", comp$change)
compnames(comp)[2] < "relative change"
print(pa, plotit = FALSE)
print(comp, row.names = FALSE)
# Sample size plan ABE
# Design alpha CV theta0 theta1 theta2 Sample size
# 3x3 0.05 0.25 0.95 0.8 1.25 27
# Achieved power
# 0.8034938
#
# Power analysis
# CV, theta0 and number of subjects leading to min. acceptable power of ~0.7:
# CV= 0.2827, theta0= 0.9276
# n = 22 (power= 0.7106)
#
# parameter relative change
# CV +13.09%
# theta0 2.36%
# n 18.52%
Confirms what we have seen above. Interesting that the sample size is the least sensitive one. Many overrate the impact of dropouts on power.
If you didn’t stop reading in desperation (understandable), explore both uncertain CV and T/Rratio:
< 12
m < 0.25
CV < 0.95
theta0 < expsampleN.TOST(CV = CV, theta0 = theta0,
df targetpower = 0.80,
design = design,
prior.parm = list(
m = m, design = design),
prior.type = "both",
details = FALSE)
#
# ++++++++++++ Equivalence test  TOST ++++++++++++
# Sample size est. with uncertain CV and theta0
# 
# Study design: 3x3 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25
# Ratio = 0.95 with 20 df
# CV = 0.25 with 20 df
#
# Sample size (ntotal)
# n exp. power
# 198 0.866206
I don’t suggest that you should use it in practice. AFAIK, not even the main author of this function (Benjamin Lang) does. However, it serves educational purposes to show that it is not that easy and why even properly planned studies might fail.
An alternative to assuming a T/Rratio is based on ‘Statistical
Assurance’.^{31} 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 T/Rratio of 0.95 one can use
the argument sem = 0.05
in the function
expsampleN.TOST()
, where the argument theta0
has to be fixed at 1.
< 0.25
CV < 0.95
theta0 < 0.80
target < 1  theta0
sigma.u < data.frame(theta0 = theta0,
comp n.1 = NA, power = NA,
sigma.u = sigma.u,
n.2 = NA, assurance = NA)
2:3] < sampleN.TOST(CV = CV,
comp[targetpower = target,
theta0 = theta0,
design = design,
details = FALSE,
print = FALSE)[7:8]
5:6] < expsampleN.TOST(CV = CV,
comp[theta0 = 1, # fixed!
targetpower = target,
design = design,
prior.type = "theta0",
prior.parm =
list(sem = sigma.u),
details = FALSE,
print = FALSE)[9:10]
names(comp)[c(2, 5)] < rep("n", 2)
print(signif(comp, 6), row.names = FALSE)
# theta0 n power sigma.u n assurance
# 0.95 27 0.803494 0.05 27 0.813331
top of section ↩︎ previous section ↩︎
It is not unusual that equivalence of more than one endpoint has to be demonstrated. In bioequivalence the pharmacokinetic metrics C_{max} and AUC_{0–t} are mandatory (in some jurisdictions like the FDA additionally AUC_{0–∞}).
We don’t have to worry about multiplicity issues (inflated Type I Error) since if all tests must pass at level \(\small{\alpha}\), we are protected by the intersectionunion principle.^{32} ^{33}
We design the study always for the worst case combination, i.e., based on the PK metric requiring the largest sample size. In some jurisdictions wider BE limits for C_{max} are acceptable. Let’s explore that with different CVs and T/Rratios.
< c("Cmax", "AUCt", "AUCinf")
metrics < c(0.25, 0.19, 0.20)
CV < c(0.95, 1.04, 1.06)
theta0 < c(0.75, 0.80, 0.80)
theta1 < 1 / theta1
theta2 < 0.80
target < data.frame(metric = metrics,
df theta1 = theta1, theta2 = theta2,
CV = CV, theta0 = theta0, n = NA)
for (i in 1:nrow(df)) {
$n[i] < sampleN.TOST(CV = CV[i],
dftheta0 = theta0[i],
theta1 = theta1[i],
theta2 = theta2[i],
targetpower = target,
design = design,
print = FALSE)[["Sample size"]]
}$theta1 < sprintf("%.4f", df$theta1)
df$theta2 < sprintf("%.4f", df$theta2)
df< paste0("Sample size based on ",
txt $metric[df$n == max(df$n)], ".\n")
dfprint(df, row.names = FALSE)
cat(txt)
# metric theta1 theta2 CV theta0 n
# Cmax 0.7500 1.3333 0.25 0.95 18
# AUCt 0.8000 1.2500 0.19 1.04 15
# AUCinf 0.8000 1.2500 0.20 1.06 21
# Sample size based on AUCinf.
Even if we assume the same T/Rratio for two PK metrics, we will get a wider margin for the one with lower variability.
Let’s continue with the conditions of our previous examples, this time assuming that the CV and T/Rratio were applicable for C_{max}. As common in PK, the CV of AUC is lower, say, only 0.20. That means, the study is ‘overpowered’ for the assumed T/Rratio of AUC.
Which are the extreme T/Rratios (largest deviations of T from R) giving still the target power?
< function(x) {
opt power.TOST(theta0 = x, CV = df$CV[2],
theta1 = theta1, theta2 = theta2,
n = df$n[1], design = design)  target
}< c("Cmax", "AUC")
metrics < c(0.25, 0.20) # Cmax, AUC
CV < 0.95 # both metrics
theta0 < 0.80
theta1 < 1.25
theta2 < 0.80
target < data.frame(metric = metrics, theta0 = theta0,
df CV = CV, n = NA, power = NA)
for (i in 1:nrow(df)) {
4:5] < sampleN.TOST(CV = CV[i],
df[i, theta0 = theta0,
theta1 = theta1, theta2 = theta2,
targetpower = target,
design = design, print = FALSE)[7:8]
}$power < signif(df$power, 6)
dfif (theta0 < 1) {
< uniroot(opt, tol = 1e8,
res interval = c(theta1 + 1e4, theta0))
else {
} < uniroot(opt, tol = 1e8,
res interval = c(theta0, theta2  1e4))
}< unlist(res)
res < c(res[["root"]], 1/res[["root"]])
theta0s < paste0("Target power for ", metrics[2],
txt " and sample size ",
$n[1], "\nachieved for theta0 ",
dfsprintf("%.4f", theta0s[1]), " or ",
sprintf("%.4f", theta0s[2]), ".\n")
print(df, row.names = FALSE)
cat(txt)
# metric theta0 CV n power
# Cmax 0.95 0.25 27 0.803494
# AUC 0.95 0.20 18 0.808949
# Target power for AUC and sample size 27
# achieved for theta0 0.9164 or 1.0912.
That means, although we assumed for AUC the same T/Rratio as for C_{max} – with the sample size of 27 required for C_{max} – for AUC it can be as low as ~0.92 or as high as ~1.09, which is a soothing sideeffect.
Furthermore, sometimes we have less data of AUC than of C_{max} (samples at the end of the profile missing or unreliable \(\small{\widehat{\lambda}_\textrm{z}}\) in some subjects and therefore, less data of AUC_{0–∞} than of AUC_{0–t}). Again, it will not hurt because for the originally assumed T/Rratio we need only 18 subjects.
Since – as a onepoint metric – C_{max} is
inherently more variable than AUC, Health Canada does not
require assessment of its confidence interval – only the point estimate
has to lie within 80.0 – 125.0%. We can explore that by setting
alpha = 0.5
for it.^{34}
< c("Cmax", "AUC")
metrics < c(0.60, 0.20)
CV < 0.95
theta0 < 0.80
target < c(0.50, 0.05)
alpha < data.frame(metric = metrics, CV = CV, theta0 = theta0,
df alpha = alpha, n = NA)
for (i in 1:nrow(df)) {
$n[i] < sampleN.TOST(alpha = alpha[i],
dfCV = CV[i],
theta0 = theta0,
targetpower = target,
design = design,
print = FALSE)[["Sample size"]]
}< paste0("Sample size based on ",
txt $metric[df$n == max(df$n)], ".\n")
dfprint(df, row.names = FALSE)
cat(txt)
# metric CV theta0 alpha n
# Cmax 0.6 0.95 0.50 24
# AUC 0.2 0.95 0.05 18
# Sample size based on Cmax.
Only with a relatively high CV (compared to the one of AUC) you may face a situation where you have to base the sample size on C_{max}.
top of section ↩︎ previous section ↩︎
First of all we have have to consider the purpose of the study and how we will evaluate it (see above).
The wellknown Bonferroniadjustment can be used \[\begin{equation}\tag{7} \alpha_\textrm{adj}=\alpha\,/\,k \end{equation}\] where \(\small{k}\) is the number of simultaneous tests. The Type I Error \(\small{TIE}\) will always be controlled because \[\begin{equation}\tag{8} TIE=1{(1\alpha_\textrm{adj})}^k \end{equation}\] For \(\small{\alpha=0.05}\) and \(\small{k=2}\) we get \(\small{0.049375<\alpha}\).
Of course, all comparisons in the study should be done by \(\small{100\,(12\,\alpha_\textrm{adj})}\) confidence intervals.
Extending the basic example and using the
argument alpha
:
< 2 # comparisons
k < 0.05 / k # Bonferroni
alpha.adj sampleN.TOST(alpha = alpha.adj, CV = 0.25,
theta0 = 0.95, targetpower = 0.80,
design = design)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 3x3 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.025, target power = 0.8
# BE margins = 0.8 ... 1.25
# True ratio = 0.95, CV = 0.25
#
# Sample size (total)
# n power
# 36 0.827798
The sample size increased by ~33% from the 27 subjects required in a single comparison.
With an increasing number of comparisons, the Bonferroniadjustment quickly becomes overly conservative. Another issue is that the tests are not strictly independent (at least when we compare different tests to the same reference treatment some – unknown – correlation exists). More powerful methods (Holm, Hochberg, …) are out of scope of this article.
Three treatments intended for evaluation ‘All at Once’ or ‘Two at a Time’. Equivalence is seeked for both (say, a tablet and a capsule vs a reference).
< 0.25
CV < data.frame(design = c("3x3", "3x6x3",
df "2x2x2"),
n = NA, power = NA)
for (i in 1:nrow(df)) {
2:3] < sampleN.TOST(alpha = 0.05 / 2,
df[i, CV = CV,
design = df$design[i],
print = FALSE)[7:8]
}$power < signif(df$power, 4)
dfprint(df, row.names = FALSE)
# design n power
# 3x3 36 0.8278
# 3x6x3 36 0.8278
# 2x2x2 36 0.8161
Four treatments (e.g., Test fasting and fed state, Reference fasting and fed state). Two comparisons are confirmatory (Test vs Reference fasting, Test vs Reference fed) and two are exploratory (Test fed vs fasting, Reference fed vs fasting). We have to adjust only for the former since the latter are only ‘nice to know’.
< 0.25
CV < data.frame(design = c("4x4", "2x2x2"),
df n = NA, power = NA)
for (i in 1:nrow(df)) {
2:3] < sampleN.TOST(alpha = 0.05 / 2,
df[i, CV = CV,
design = df$design[i],
print = FALSE)[7:8]
}$power < signif(df$power, 4)
dfprint(df, row.names = FALSE)
# design n power
# 4x4 36 0.8315
# 2x2x2 36 0.8161
top of section ↩︎ previous section ↩︎
So far we employed the common (and hence, default)
BElimits theta1 = 0.80
and theta2 = 1.25
. In some jurisdictions for Narrow
Therapeutic Index Drugs tighter limits of 90.00 – 111.11% have to be
used.^{35}
Generally NTIDs show a low withinsubject variability (though the betweensubject CV can be much higher – these drugs require quite often dosetitration).
You have to provide only the lower
BElimit theta1
(the
upper one will be automatically calculated). In the examples I used a
lower CV of 0.125.
sampleN.TOST(CV = 0.125, theta1 = 0.90,
design = design)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 3x3 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 0.9 ... 1.111111
# True ratio = 0.95, CV = 0.125
#
# Sample size (total)
# n power
# 69 0.813991
Not so nice. If you would exhaust the capacity of the clinical site, you may consider a replicate design.
Interlude 3
Health Canada requires for NTIDs (termed by HC ‘critical dose drugs’) that the confidence interval of C_{max} lies within 80.0 – 125.0% and the one of AUC within 90.0 – 112.0%.
<nitpick>
\(\small{100\frac{1}{0.900}\,{\color{Red}\neq}\,112.0\%}\),
right? I always thought that it’s \(\small{111.1\dot{1}\%}\)… Consequently, on
the average generic products will be approved with \(\small{\sqrt{90.0\times112.0}\approx100.4\%}\).
When asked, the reply was:
»These numbers are more easy to remember.«
</nitpick>
Hence, for Health Canada you have to specify both limits.
sampleN.TOST(CV = 0.125, theta1 = 0.90, theta2 = 1.12,
design = design)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 3x3 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 0.9 ... 1.12
# True ratio = 0.95, CV = 0.125
#
# Sample size (total)
# n power
# 69 0.813991
Here we get the same sample size than with the common limits for NTIDs.
Sometimes – by the size of one sequence – smaller:
< function(design) {
get.sequences < known.designs()[, c(2, 5)]
known < as.integer(known[known$design == design,
ns "steps"])
return(ns)
}< function(n, ns) {
make.equal return(as.integer(ns * (n %/% ns + as.logical(n %% ns))))
}< "3x3"
design < 0.80
target < 0.975
theta0 < 0.075
CV.min < 0.20
CV.max < 1000
CV.step < seq(CV.min, CV.max, length.out = CV.step)
CV < get.sequences(design)
ns < data.frame(CV = CV, n.EMA = NA_integer_,
res n.HC = NA_integer_, less = FALSE)
for (i in 1:nrow(res)) {
< sampleN.TOST(CV = CV[i], theta0 = theta0,
n.TaT targetpower = target, design = "2x2",
theta1 = 0.90, theta2 = 1/0.90,
print = FALSE)[["Sample size"]]
$n.EMA[i] < make.equal(n = n.TaT, ns = ns)
res< sampleN.TOST(CV = CV[i], theta0 = theta0,
n.TaT targetpower = target, design = "2x2",
theta1 = 0.90, theta2 = 1.12,
print = FALSE)[["Sample size"]]
$n.HC[i] < make.equal(n = n.TaT, ns = ns)
resif (res$n.HC[i] < res$n.EMA[i]) res$less[i] = TRUE
}cat("design :", design,
"(intended for \u2018Two at a Time\u2019 approach)",
"\ntarget power :", target,
"\ntheta0 :", theta0,
"\nCV :", CV.min, "\u2013", CV.max,
"\nSample size for HC < common :",
sprintf("%.2f%%", 100 * sum(res$less) / CV.step),
"of cases.\n")
# design : 3x3 (intended for ‘Two at a Time’ approach)
# target power : 0.8
# theta0 : 0.975
# CV : 0.075 – 0.2
# Sample size for HC < common : 12.90% of cases.
The FDA requires for NTIDs tighter batchrelease specifications (±5% instead of ±10%). Let’s hope that your product complies and the T/Rratio will be closer to 1:
sampleN.TOST(CV = 0.125,
theta0 = 0.975, # ‘better’ T/Rratio
theta1 = 0.90,
design = design)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 3x3 crossover
# logtransformed data (multiplicative model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 0.9 ... 1.111111
# True ratio = 0.975, CV = 0.125
#
# Sample size (total)
# n power
# 33 0.820725
Substantially lower sample and doable.
top of section ↩︎ previous section ↩︎
Sometimes we are interested in assessing differences of responses
instead of their ratios. In such a case we have to set
logscale = FALSE
. The limits theta1
and
theta2
can be expressed in the following ways:
Note that in the former case the units of CV
, and
theta0
need also to be given relative to the reference mean
(specified as ratio).
Let’s estimate the sample size for an equivalence trial of two blood pressure lowering drugs assessing the difference in means of untransformed data (raw, linear scale). In this setup everything has to be given with the same units (i.e., here the assumed difference –5 mm Hg and the lower / upper limits –15 mm Hg / +15 mm Hg systolic blood pressure). Furthermore, we assume a CV of 25 mm Hg.
sampleN.TOST(CV = 20, theta0 = 5,
theta1 = 15, theta2 = +15,
design = design,
logscale = FALSE)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 3x3 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
# 51 0.805426
Sometimes in the literature we find not the CV but the standard deviation of the difference. Say, it is given with 36 mm Hg. We have to convert it to a CV. In all implemented HigherOrder Crossover designs: \(\small{CV=SD/\sqrt{2}}\).
< 36
SD.delta sampleN.TOST(CV = SD.delta / sqrt(2), theta0 = 5,
theta1 = 15, theta2 = +15,
design = design,
logscale = FALSE)
#
# +++++++++++ Equivalence test  TOST +++++++++++
# Sample size estimation
# 
# Study design: 3x3 crossover
# untransformed data (additive model)
#
# alpha = 0.05, target power = 0.8
# BE margins = 15 ... 15
# True diff. = 5, CV = 25.45584
#
# Sample size (total)
# n power
# 81 0.800354
Note that other software packages (e.g., PASS, nQuery, StudySize, …) require the standard deviation of the difference as input.
It would be a fundamental flaw if responses are assumed to a follow a normal distribution and not – like above – assess their difference and at the end instead giving an estimate of the difference, calculate the ratio of the test and referencemeans.
In such a case Fieller’s (‘fiducial’) confidence interval^{36} has to
be employed.
Note that in the functions sampleN.RatioF()
and
power.RatioF()
alpha = 0.025
is the default,
since it is intended for studies with clinical endpoints, where the 95%
confidence interval is usually used for equivalence testing.^{37} Only
the ‘Two at a Time’ approach is implemented, i.e., you have to
specify design = "2x2"
. Sometimes we have to round the
sample size up (see the interlude). Note that
we need additionally the betweensubject CV (here arbitrarily
twice the CV).
< "3x3"
planned < 0.25
CV < 2 * CV
CVb < function(design) {
get.sequences < known.designs()[, c(2, 5)]
known < as.integer(known[known$design == design,
ns "steps"])
return(ns)
}< function(n, ns) {
make.equal return(as.integer(ns * (n %/% ns + as.logical(n %% ns))))
}< get.sequences(planned)
ns < sampleN.RatioF(CV = CV, CVb = CVb, design = "2x2",
n print = FALSE)[["Sample size"]]
< make.equal(n = n, ns = ns)
n < data.frame(Design = planned, Eval = "2x2",
df alpha = 0.025, CV = CV, CVb = CVb, n = n)
$power < suppressMessages(
dfpower.RatioF(CV = CV, CVb = CVb,
design = "2x2", n = n))
names(df)[6:7] < c("Sample size", "Achieved power")
print(df, row.names = FALSE)
# Design Eval alpha CV CVb Sample size Achieved power
# 3x3 2x2 0.025 0.25 0.5 42 0.8021525
“He who seeks for methods without having a definite problem in mind seeks in the most part in vain.
With a few exceptions (i.e., simulationbased methods), in
PowerTOST
the default method = "exact"
implements Owen’s Q function^{38} which is also used in SAS’
Proc Power
.
Other implemented methods are "mvt"
(based on the
bivariate noncentral tdistribution),
"noncentral"
/ "nct"
(noncentral
tdistribution), and "shifted"
/
"central"
(shifted central tdistribution).
Although "mvt"
is also exact, it may have a somewhat lower
precision compared to Owen’s Q and has a much longer runtime.
Let’s compare them.
< 0.25
CV < c("exact", "mvt", "noncentral", "shifted")
methods < c("‘Two at a Time’", "‘All at Once’")
appr < data.frame(approach = rep(appr, each = 4), method = methods,
df n = NA_integer_, power = NA_real_)
for (i in 1:nrow(df)) {
if (df$approach[i] == "‘Two at a Time’") {
3:4] < sampleN.TOST(CV = CV, design = "2x2",
df[i, method = df$method[i], print = FALSE)[7:8]
else {
}3:4] < sampleN.TOST(CV = CV, design = "3x3",
df[i, method = df$method[i], print = FALSE)[7:8]
}
}$power < round(df$power, 5)
dfprint(df, row.names = FALSE, right = FALSE)
# approach method n power
# ‘Two at a Time’ exact 28 0.80744
# ‘Two at a Time’ mvt 28 0.80744
# ‘Two at a Time’ noncentral 28 0.80744
# ‘Two at a Time’ shifted 28 0.80303
# ‘All at Once’ exact 27 0.80349
# ‘All at Once’ mvt 27 0.80349
# ‘All at Once’ noncentral 27 0.80349
# ‘All at Once’ shifted 27 0.80113
Power approximated by the shifted central tdistribution is
generally slightly lower compared to the others. Hence, if used in
sample size estimations, occasionally one more complete
sequence is ‘required’.
Therefore, I recommend to use "method = shifted"
/
"method = central"
only for comparing with old results
(literature, own studies).
Q: Can we use R in a
regulated environment and is PowerTOST
validated?
A: See this document^{39} about the
acceptability of Base R
.
R
is updated every couple of
months with documented changes^{40} and maintaining a bugtracking system.^{41} I
recommed to use always the latest release.
The authors of PowerTOST
tried to do their best to provide
reliable and valid results. The package’s NEWS
file
documents the development of the package, bugfixes, and introduction of
new methods. The ultimate responsibility of validating any
software (yes, of SAS as well…) lies in the hands of the user.^{42} ^{43}
Q: Shall we throw away our sample size
tables?
A: Not at all. File them in your archives to collect
dust. Maybe in the future you will be asked by an agency how you arrived
at a sample size. But: Don’t use them any more. What you should not do
(and hopefully haven’t done before): Interpolate. Power and therefore,
the sample size depends in a highly nonlinear fashion on the five
conditions listed above, which makes
interpolation of values given in table a nontrivial job.
Q: Which of the methods should we use in our
daily practice?
A: method = exact
. Full stop. Why rely on
approximations? Since it is the default in sampleN.TOST()
and power.TOST()
, you don’t have to give this argument
(saves keystrokes).
Q: I fail to understand your example about
dropouts. We finish the study with 27 eligible subjects as desired. Why
is the dropoutrate ~18% and not the anticipated 10%?
A: That’s due to rounding up the calculated adjusted
sample size (31.76…) to the next complete sequence (33). If you
manage it to dose fractional subjects (I can’t) your dropout rate would
indeed equal the anticipated one: 100 × (1 – 27/31.76…) = 10%. ⬜
Q: Do we have to worry about unbalanced
sequences?
A: sampleN.TOST()
will always give the
total number of subjects for balanced sequences.
If you are interested in post hoc power, give the sample size
as a vector, i.e.,
power.TOST(..., n = c(foo, bar, ...)
, where
foo
, bar
, ...
are the number of
subjects in each sequence.
Q: Is it possible to simulate power of
studies?
A: That’s not necessary, since the available methods
provide analytical solutions. However, if you don’t trust them,
simulations are possible with the function
power.TOST.sim()
, which employs the distributional
properties:
\(\small{\sigma^2}\) follows a \(\small{\chi^2}\)distribution with \(\small{n2}\) degrees of freedom and \(\small{\log_{e}(\theta_0)}\) follows a normal distribution).^{44}
Convergence takes a while. Empiric power, its standard error, its relative error compared to the exact power, and execution times on a Xeon E31245v3 @ 3.40GHz (1/4 cores) 16GB RAM with R 4.2.1 on Windows 7 build 7601, Service Pack 1, Universal C Runtime 10.0.10240.16390:
< 0.25
CV < "3x3"
des < sampleN.TOST(CV = CV, design = des,
n print = FALSE)[["Sample size"]]
< c(1e5, 5e5, 1e6, 5e6, 1e7, 5e7, 1e8)
nsims < power.TOST(CV =CV, n = n, design = des)
exact < data.frame(simulations = nsims,
df exact = rep(exact, length(nsims)),
simulated = NA, SE = NA, RE = NA)
for (i in 1:nrow(df)) {
< proc.time()[[3]]
start.time $simulated[i] < power.TOST.sim(CV = CV, n = n,
dfdesign = des,
nsims = nsims[i])
$secs[i] < proc.time()[[3]]  start.time
df$SE[i] < sqrt(0.5 * df$simulated[i] / nsims[i])
df$RE < 100 * (df$simulated  exact) / exact
df
}$exact < signif(df$exact, 5)
df$SE < formatC(df$SE, format = "f", digits = 5)
df$RE < sprintf("%+.4f%%", df$RE)
df$simulated < signif(df$simulated, 5)
df$simulations < formatC(df$simulations, format = "f",
dfdigits = 0, big.mark=",")
names(df)[c(1, 3)] < c("sim\u2019s", "sim\u2019d")
print(df, row.names = FALSE)
# sim’s exact sim’d SE RE secs
# 100,000 0.80349 0.80222 0.00200 0.1585% 0.12
# 500,000 0.80349 0.80333 0.00090 0.0209% 0.58
# 1,000,000 0.80349 0.80338 0.00063 0.0145% 1.19
# 5,000,000 0.80349 0.80353 0.00028 +0.0040% 5.92
# 10,000,000 0.80349 0.80342 0.00020 0.0094% 11.86
# 50,000,000 0.80349 0.80353 0.00009 +0.0046% 59.28
# 100,000,000 0.80349 0.80350 0.00006 +0.0007% 119.53
With an increasing number of simulations values become more precise (the standard error decreases) as well as more accurate (the relative error decreases).
In the Monte Carlo method values approach the exact one asymptotically.
Acknowledgment
Member Shuanghe of the BEBAForum for discovering a bug in my script for extracting IBDs (they were not necessarily balanced).
Licenses
Helmut Schütz 2022
R
and packages GPL 3.0,
pandoc
GPL 2.0.
1^{st} version March 17, 2021. Rendered September 08, 2022 15:32
CEST by rmarkdown via pandoc in 2.08 seconds.
Footnotes and References
Labes D, Schütz H, Lang B. PowerTOST: Power and Sample Size for (Bio)Equivalence Studies. Package version 1.5.4. 20220221. CRAN.↩︎
Labes D. randomizeBE: Create a Random List for Crossover Studies. Package version 0.3.5. 20190824. CRAN.↩︎
Labes D, Schütz H, Lang B. Package ‘PowerTOST’. February 21, 2022. CRAN.↩︎
Schuirmann DJ. Two at a Time? Or All at Once? International Biometric Society, Eastern North American Region, Spring Meeting. Pittsburgh, PA. March 28 – 31, 2004. Abstract.↩︎
Senn S. Guernsey McPearson’s Drug Development Dictionary. 21 April 2020. Online.↩︎
Hoenig JM, Heisey DM. The Abuse of Power: The Pervasive Fallacy of Power Calculations for Data Analysis. Am Stat. 2001; 55(1): 19–24. doi:10.1198/000313001300339897. Open Access.↩︎
In short: There is no statistical method to ‘correct’ for unequal carryover. It can only be avoided by design, i.e., a sufficiently long washout between periods. According to the guidelines subjects with predose concentrations > 5% of their C_{max} can by excluded from the comparison if stated in the protocol.↩︎
Especially important for drugs which are autoinducers or inhibitors and biologics.↩︎
Senn S. Statistical Issues in Drug Development. Wiley, 2^{nd} ed 2007.↩︎
Julious SA. Sample Sizes for Clinical Trials. Chapter 7.2. Boca Raton: CRC Press; 2010.↩︎
Common batch release specifications are ±10% of the declared content. Of course, you will try to find a batch of the reference product which is as close as possible to the test (the EMA recommends ≤ 5%). However, keep analytical (in)accuracy and (im)precision in mind. Furthermore, the analytical method was only validated for the test. Try to get a CoA from the originator… Good luck. You never can be sure. Only Health Canada may (!) accept a doseadjustment in the evaluation of the study.↩︎
Senn S. Statistical Issues in Drug Development. Chichester: John Wiley; 2004.↩︎
Zhang P. A Simple Formula for Sample Size Calculation in Equivalence Studies. J Biopharm Stat. 2003; 13(3): 529–38. doi:10.1081/BIP120022772.↩︎
Let us assume that the – whilst unknown – withinsubject \(\small{CV\textrm{s}}\) of the two reference treatments are different \((\small{CV_{\textrm{w}_{\textrm{R}1}}\neq CV_{\textrm{w}_{\textrm{R}2}}\,}\)). Since in the ‘All at Once’ approach the pooled variance is used, the confidence interval of the treatment comparisons may be wider than in the ‘Two at a Time’ approach.↩︎
Since e.g., in the comparison of products from different regions the populations differ and the risk for both is still \(\small{\leq \alpha}\) – apart tourists buying the product in another region…↩︎
Whilst I used the former argument for many years, nowadays I’m strongly leaning towards the latter.↩︎
EMA, CHMP. Guideline on the Investigation of Bioequivalence. CPMP/EWP/QWP/1401/98 Rev. 1/Corr ** London. 20 January 2010. Online.↩︎
David Brown. Presentation at the 3^{rd} EGA Symposium on Bioequivalence. London. June 2010. Slides.↩︎
EGA. Revised EMA Bioequivalence Guideline. Questions & Answers. Online.↩︎
Cochran WG, Cox GM. Experimental Designs. New York: Wiley; 2^{nd} ed. 1957. Chapter 9. p. 376–95.↩︎
Dean A, Voss D, Dragulić D. Design and Analysis of Experiments. New York: Springer; 2^{nd} ed. 2017. Chapter 11. p. 349–397.↩︎
FDA, CDER. Guidance for Industry. Statistical Approaches to Establishing Bioequivalence. Rockville. January 2001. Download.↩︎
Council of the EEC. On the approval of the rules for conducting research of bioequivalence of drugs within the framework of the EEU. November 3, 2016; amended September 4, 2020.↩︎
Mandal BN. ibd: Incomplete Block Designs. Package version 1.5. 20190424. CRAN.↩︎
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, MD. 15–16 September, 2016.↩︎
Doyle AC. The Adventures of Sherlock Holmes. A Scandal in Bohemia. 1892. p. 3.↩︎
Schütz H. Sample Size Estimation in Bioequivalence. Evaluation. BEBA Forum. 20201023.↩︎
Lenth RV. Two SampleSize Practices that I Don’t Recommend. 24 October 2000. Online.↩︎
Quoting my late father: »If you believe, go to church.«↩︎
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. Open Access.↩︎
Berger RL, Hsu JC. Bioequivalence Trials, IntersectionUnion Tests and Equivalence Confidence Sets. Stat Sci. 1996; 11(4): 283–302. JSTOR:2246021.↩︎
Zeng A. The TOST confidence intervals and the coverage probabilities with R simulation. March 14, 2014. Online.↩︎
With \(\small{\alpha=0.5}\) the width of the CI is zero and the test effectively reduces to a passfail assessment.↩︎
The U.S. FDA and China’s CDE recommend ReferenceScaled Average Bioequivalence (RSABE) and a comparison of the withinsubject variances of treatments, which requires a fully replicated design. It is covered in another article.↩︎
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. CPMP/EWP/482/99. London. 27 July 2000. Online.↩︎
Owen DB. A special case of a bivariate noncentral tdistribution. Biometrika. 1965; 52(3/4): 437–46. doi:10.2307/2333696.↩︎
The R Foundation for Statistical Computing. A Guidance Document for the Use of R in Regulated Clinical Trial Environments. Vienna. October 18, 2021. Online.↩︎
FDA. Statistical Software Clarifying Statement. May 6, 2015. Internet Archive.↩︎
WHO. Guidance for organizations performing in vivo bioequivalence studies. Geneva. May 2016. Technical Report Series No. 996, Annex 9. Section 4. Online.↩︎
Zheng C, Wang J, Zhao L. Testing bioequivalence for multiple formulations with power and sample size calculations. Pharm Stat. 2012; 11(4): 334–41. doi:10.1002/pst.1522.↩︎