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Examples in this article were generated with 4.0.5 by the package PowerTOST
.^{1}
More examples are given in the respective vignette.^{2} See also the README on GitHub for an overview, the online manual^{3} for details, and a collection of other articles.
Abbreviation | Meaning |
---|---|
BE | Bioequivalence |
CV | Coefficient of Variation |
CV_{inter} | Between-subject Coefficient of Variation |
CV_{intra} | Within-subject Coefficient of Variation |
H_{0} | Null hypothesis |
H_{1} | Alternative hypothesis (also H_{a}) |
NTID | Narrow Therapeutic Index Drug |
TOST | Two One-Sided Tests |
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 (2016-11-01) of PowerTOST
is suggested. Any version of R would likely do, though the current release of PowerTOST
was only tested with version 3.6.3 (2020-02-29) and later.
Note that in all functions of PowerTOST
the arguments (say, the assumed T/R-ratio theta0
, the BE-limits (theta1
, theta2
), the assumed coefficient of variation CV
, 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 the sequence TR
as to the sequence RT
). Furthermore, the estimated sample size is the total number of subjects, which is always an even number (not subjects per sequence – like in some other software packages).
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 t-test (requiring differences). This is the default in most functions of PowerTOST
and hence, the argument logscale = TRUE
does not need to be specified.
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.
— Stephen Senn, Guernsey McPearson’s Drug Development Dictionary.
Of note, 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.^{8}
Since generally the within-subject variability CV_{intra} is lower than the between-subject variability CV_{inter}, crossover studies are so popular. Of note, there is no relationship between CV_{intra} and CV_{inter}. An example are drugs which are subjected to polymorphic metabolism. For these drugs CV_{intra} ≪ CV_{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.^{9}
Crossover studies can be not only performed in healthy volunteers but also in patients with a stable disease (e.g., asthma).
If crossovers are not feasible (e.g., for drugs with a very long half-life or in patients with an instable disease), studies in a parallel design should be performed (covered in another article).
The sample size cannot be directly estimated,
only power calculated for an already given sample size.
The power equations cannot be re-arranged to solve for sample size.
“Power. That which statisticians are always calculating but never have.
— Stephen Senn, Statistical Issues in Drug Development. Wiley, 2^{nd} ed 2007.
If you are a nerd 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.^{10}
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: \(\small{Z_{1-0.2}=0.8416212}\), \(\small{Z_{1-0.1}=1.281552}\), \(\small{Z_{1-0.05}=1.644854}\). \[\begin{equation}\tag{2a} \textrm{if}\;\theta_0<1:n\sim\frac{2\,\sigma^2(Z_{1-\beta}-Z_{1-\alpha})^2}{\left(\log_{e}(\theta_0)-\log_{e}(\theta_1)\right)^2} \end{equation}\] \[\begin{equation}\tag{2b} \textrm{if}\;\theta_0>1:n\sim\frac{2\,\sigma^2(Z_{1-\beta}-Z_{1-\alpha})^2}{\left(\log_{e}(\theta_0)-\log_{e}(\theta_2)\right)^2} \end{equation}\] \[\begin{equation}\tag{2c} \textrm{if}\;\theta_0=1:n\sim\frac{2\,\sigma^2(Z_{1-\beta/2}-Z_{1-\alpha})^2}{\left(\log_{e}(\theta_2)\right)^2} \end{equation}\] Generally you should use \(\small{(2\textrm{a},2\textrm{b})}\). Only if you are courageous, use \(\small{(2\textrm{c})}\). However, I would not recommend that.^{11}
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 \(\small{(1-2)}\) we assume that the population variance \(\small{\sigma^2}\) is known, which is never the case. In practice is is uncertain because only its sample estimate \(\small{s^2}\) is known. Therefore, \(\small{(1-2)}\) will always give a sample size which is too low:
For \(\small{CV=0.20}\), \(\small{\theta_0=0.95}\), \(\small{\alpha=0.05}\), \(\small{\beta=0.20,}\) and the conventional limits we get 25.383 for the total sample size (rounded up to 26). As we will see later, the correct sample size is 28.
“Power: That which is wielded by the priesthood of clinical trials, the statisticians, and a stick which they use to beta their colleagues.
— Stephen Senn, Statistical Issues in Drug Development. Wiley, 2004.
Since we evaluate studies based on the t-test, we have to take the degrees of freedom (in 2×2×2 designs \(\small{df=n-2)}\) 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 t-distribution (warning: 100 lines of code).
function(n) { # equally sized sequences
make.even <-return(as.integer(2 * (n %/% 2 + as.logical(n %% 2))))
} 0.25 # within-subject CV
CV <- 0.95 # T/R-ratio
theta0 <- 0.80 # lower BE-limit
theta1 <- 1.25 # upper BE-limit
theta2 <- 0.80 # desired (target) power
target <- 0.05 # level of the test
alpha <- 1 - target # producer's risk
beta <- 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.even(n.seq)
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, n - 2)
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.even(n.seq)
n <- qt(1 - alpha, n - 2)
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 = n - 2),
df = n - 2, 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 = n - 2),
df = n - 2, 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)
R> method iteration n power
R> large sample – 26 0.799499
R> central t – 28 0.810646
R> noncentral t 1 28 0.807439
Now it’s high time to leave Base R behind and continue with PowerTOST
.
Makes our life much easier.
library(PowerTOST) # attach it to run the examples
Its sample size functions use a modification of Zhang’s method^{12} for the first guess. You can unveil the course of iterations with details = TRUE
.
sampleN.TOST(CV = 0.25, theta0 = 0.95,
targetpower = 0.80, details = TRUE)
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R> Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2 crossover
R> Design characteristics:
R> df = n-2, design const. = 2, step = 2
R>
R> log-transformed data (multiplicative model)
R>
R> alpha = 0.05, target power = 0.8
R> BE margins = 0.8 ... 1.25
R> True ratio = 0.95, CV = 0.25
R>
R> Sample size search (ntotal)
R> n power
R> 26 0.776055
R> 28 0.807439
R>
R> Exact power calculation with
R> 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 t-test, right?
The sample size based on the normal distribution may be too small, thus compromising power.
R> method n Delta RE power
R> large sample approx. (2) 26 -2 -7.1% 0.776055
R> noncentral t-distribution 28 ±0 ±0.0% 0.807439
R> exact method (Owen’s Q) 28 – – 0.807439
Throughout the examples by I’m referring to studies in a single center – not multiple groups within them or multicenter studies. That’s another story.
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 v.s. a reference or each of them against one of the others) in order to avoid inflation of the family-wise error rate due to multiplicity. Examples are given further down.
We assume a CV of 0.25, a T/R-ratio of 0.95, and target a power of 0.80.
Since design = "2x2"
, theta0 = 0.95
, and targetpower = 0.80
are defaults of the function, we don’t have to give them explicitely. As usual in bioequivalence, alpha = 0.05
is employed (we will assess the study by a \(\small{100(1-2\,\alpha)=90\%}\) confidence interval) and the BE-limits 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, you need to specify only the CV
.
sampleN.TOST(CV = 0.25)
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R> Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2 crossover
R> log-transformed data (multiplicative model)
R>
R> alpha = 0.05, target power = 0.8
R> BE margins = 0.8 ... 1.25
R> True ratio = 0.95, CV = 0.25
R>
R> Sample size (total)
R> n power
R> 28 0.807439
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 df
).
sampleN.TOST(CV = 0.25, print = FALSE) df <-
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 df
:
names(df)
> [1] "Design" "alpha" "CV"
R> [4] "theta0" "theta1" "theta2"
R> [7] "Sample size" "Achieved power" "Target power" R
Now we can access the elements of df
by their names. Note that double square brackets [[…]]
have to be used (single ones are used to access elements by their numbers).
"Sample size"]]
df[[> [1] 28
R"Achieved power"]]
df[[> [1] 0.8074395 R
If you insist in accessing elements by column-numbers, use single square brackets […]
.
7:8]
df[> Sample size Achieved power
R> 1 28 0.8074395 R
With 28 subjects (14 per sequence) we achieve the power we desire.
What happens if we have one dropout?
power.TOST(CV = 0.25, n = df[["Sample size"]] - 1)
R> Unbalanced design. n(i)=14/13 assumed.
R> [1] 0.7918272
Still above the 0.80 we desire. However, with two dropouts (26 eligible subjects) we would slightly fall short (0.7761).
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.
We define two supportive functions:
n
will be rounded up to the next even number. function(n) {
up2even <-return(as.integer(2 * (n %/% 2 + as.logical(n %% 2))))
}
n
and the anticipated droput-rate do.rate
. function(n, do.rate) {
nadj <-return(as.integer(up2even(n / (1 - do.rate))))
}
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 theorise before one has data.
— Arthur Conan Doyle
The dropout-rate is calculated from the eligible and dosed subjects
or simply \[\begin{equation}\tag{3}
do.rate=1-n_\textrm{eligible}/n_\textrm{dosed}
\end{equation}\] 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 dropout-rate and rearrangement to find the adjusted number of dosed subjects \(n_\textrm{adj}\) we should use \[\begin{equation}\tag{4} n_\textrm{adj}=\;\upharpoonleft n\,/\,(1-do.rate) \end{equation}\] 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 dropout-rate according to \[\begin{equation}\tag{5} n_\textrm{adj}=\;\upharpoonleft n\times(1+do.rate) \end{equation}\]
If you used \((5)\) in the past – you are not alone. In a small survey a whooping 29% of respondents reported to use it.^{13} Consider changing your routine.
“There are no routine statistical questions,
only questionable statistical routines.
— David R. Cox
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 # within-subject CV
CV <- 0.80 # target (desired) power
target <- 0.95 # assumed T/R-ratio
theta0 <- 0.80 # lower BE limit
theta1 <- 1.25 # upper BE limit
theta2 <- 0.10 # anticipated dropout-rate 10%
do.rate <- sampleN.TOST(CV = CV, theta0 = theta0,
df <-theta1 = theta1,
theta2 = theta2,
targetpower = target,
print = FALSE)
# calculate the adjusted sample size
nadj(df[["Sample size"]], do.rate)
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 dropout-rate: ",
do.rate,"\nEstimated sample size : ",
"Sample size"]], " (",
df[["Sample size"]]/2, "/sequence)",
df[["\nAchieved power : ",
signif(df[["Achieved power"]], 4),
"\nAdjusted sample size : ",
" (", n.adj/2, "/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)
R> Assumed CV : 0.25
R> Assumed T/R ratio : 0.95
R> BE limits : 0.8…1.25
R> Target (desired) power : 0.8
R> Anticipated dropout-rate: 0.1
R> Estimated sample size : 28 (14/sequence)
R> Achieved power : 0.8074
R> Adjusted sample size : 32 (16/sequence)
R> dosed eligible dropouts do.act power
R> 32 32 0 0.0000 0.8573
R> 32 31 1 0.0312 0.8458
R> 32 30 2 0.0625 0.8343
R> 32 29 3 0.0938 0.8209
R> 32 28 4 0.1250 0.8074
In the worst case (4 dropouts) we end up with the originally estimated sample size of 28. Power preserved, mission accomplished. If we have less dropouts, splendid – we gain power.
If we would have adjusted the sample size acc. to (5) we would have dosed also 32 subjects like the 32 acc. to (4).
Cave: This might not always be the case… If the anticipated dropout rate of 10% is realized in the study, we would have also 28 eligible subjects (power 0.8074). 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.
— Russell V. Lenth, Two Sample-Size Practices that I Don’t Recommend.
However, sometimes we are interested in it for planning the next study.
If you give and odd total sample size n
, power.TOST()
will try to keep sequences as balanced as possible and show in a message how that was done.
27
n.act <-signif(power.TOST(CV = 0.25, n = n.act), 6)
R> Unbalanced design. n(i)=14/13 assumed.
R> [1] 0.791827
Say, our study was more unbalanced. Let us assume that we dosed 32 subjects, the total number of subjects was also 27 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 the TR
or RT
sequence).
32
n.adj <- 27
n.act <- n.adj / 2
n.s1 <- n.act - n.s1
n.s2 <- signif(power.TOST(CV = 0.25,
post.hoc <-n = c(n.s1, n.s2)), 6)
nchar(as.character(n.adj))
sig.dig <- paste0("%", sig.dig, ".0f (%", sig.dig, ".0f dropouts)")
fmt <-cat(paste0("Dosed subjects: ", sprintf("%2.0f", n.adj),
"\nEligible : ",
sprintf(fmt, n.act, n.adj - n.act),
"\n Sequence 1 : ",
sprintf(fmt, n.s1, n.adj / 2 - n.s1),
"\n Sequence 1 : ",
sprintf(fmt, n.s2, n.adj / 2 - n.s2),
"\nPower : ", post.hoc, "\n"))
R> Dosed subjects: 32
R> Eligible : 27 ( 5 dropouts)
R> Sequence 1 : 16 ( 0 dropouts)
R> Sequence 1 : 11 ( 5 dropouts)
R> Power : 0.778224
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/R-ratio are only assumptions. Whatever their origin might be (literature, previous studies) they carry some degree of uncertainty. Hence, believing^{14} that they are the true ones may be risky.
Some statisticians call that the ‘Carved-in-Stone’ 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 \(\chi^2\)-distribution of its error variance \(\sigma^2\) with \(\small{n-2}\) degrees of freedom. \[\begin{matrix}\tag{6} s^2=\log_{e}(CV^2+1)\\ L=\frac{(n-1)\,s^2}{\chi_{\alpha/2,\,n-2}^{2}}\leq\sigma^2\leq\frac{(n-1)\,s^2}{\chi_{1-\alpha/2,\,n-2}^{2}}=U\\ \left\{lower\;CL,\;upper\;CL\right\}=\left\{\sqrt{\exp(L)-1},\sqrt{\exp(U)-1}\right\} \end{matrix}\]Let’s calculate the 95% confidence interval of the CV to get an idea.
12 # pilot study
m <- CVCL(CV = 0.25, df = m - 2,
ci <-side = "2-sided", alpha = 0.05)
signif(ci, 4)
R> lower CL upper CL
R> 0.1733 0.4531
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 \(\chi^2\)-distribution is skewed to the right, it is more likely to get a higher CV than a lower one in the planned study.
If we plan the study based on 0.25, we would opt for 28 subjects like in the examples before (not adjusted for the dropout-rate).
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, df = m - 2,
ci <-side = "2-sided", alpha = 0.05)
28
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],
compn = n)
}1] <- signif(comp[, 1], 4)
comp[, 2] <- signif(comp[, 2], 6)
comp[, print(comp, row.names = FALSE)
R> CV power
R> 0.1733 0.976850
R> 0.2500 0.807439
R> 0.4531 0.184866
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 = m[i] - 2,
df[i, side = "2-sided",
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)
R> n CV lower CL upper CL
R> 6 0.25 0.1483 0.8060
R> 8 0.25 0.1597 0.5846
R> 10 0.25 0.1675 0.4992
R> 12 0.25 0.1733 0.4531
R> 14 0.25 0.1779 0.4238
R> 16 0.25 0.1817 0.4034
A Bayesian method is implemented in PowerTOST
, which takes the uncertainty of estimates into account. We can explore the uncertainty of the CV, the T/R-ratio, and both.
12 # sample size of pilot
m <- 0.25
CV <- 0.95
theta0 <- expsampleN.TOST(CV = CV, theta0 = theta0,
df <-targetpower = 0.80,
design = "2x2",
prior.parm = list(
m = m, design = "2x2"),
prior.type = "CV",
details = FALSE)
R>
R> ++++++++++++ Equivalence test - TOST ++++++++++++
R> Sample size est. with uncertain CV
R> -------------------------------------------------
R> Study design: 2x2 crossover
R> log-transformed data (multiplicative model)
R>
R> alpha = 0.05, target power = 0.8
R> BE margins = 0.8 ... 1.25
R> Ratio = 0.95
R> CV = 0.25 with 10 df
R>
R> Sample size (ntotal)
R> n exp. power
R> 34 0.802586
Nasty. 21% more subjects required.
What about an uncertain T/R-ratio?
12
m <- 0.25
CV <- 0.95
theta0 <- expsampleN.TOST(CV = CV, theta0 = theta0,
df <-targetpower = 0.80,
design = "2x2",
prior.parm = list(
m = m, design = "2x2"),
prior.type = "theta0",
details = FALSE)
R>
R> ++++++++++++ Equivalence test - TOST ++++++++++++
R> Sample size est. with uncertain theta0
R> -------------------------------------------------
R> Study design: 2x2 crossover
R> log-transformed data (multiplicative model)
R>
R> alpha = 0.05, target power = 0.8
R> BE margins = 0.8 ... 1.25
R> Ratio = 0.95
R> CV = 0.25
R>
R> Sample size (ntotal)
R> n exp. power
R> 72 0.800974
Terrible! The sample size more than doubles.
That should not take us by surprise. We don’t know where the true T/R-ratio 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, n = m,
ci <-design = "2x2"), 4)
if (pe <= 1) {
ci[["lower"]]
cl <-else {
} ci[["upper"]]
cl <-
}print(cl)
R> [1] 0.7918
Exlore the impact of a relatively 5% higher CV and a relatively 5% lower T/R-ratio on power for a given sample size.
28
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,
n = n)
}$power <- signif(comp1$power, 6)
comp1for (i in 1:2) {
$power[i] <- power.TOST(CV = CV,
comp2theta0 = comp2$theta0[i],
n = n)
}$power <- signif(comp2$power, 6)
comp2print(comp1, row.names = F); print(comp2, row.names = F)
R> CV power
R> 0.2500 0.807439
R> 0.2625 0.769438
R> theta0 power
R> 0.9500 0.807439
R> 0.9025 0.554599
Interlude 1
Note the log-scale of the x-axis. 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> 0.30
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 <-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)
} sampleN.TOST(CV = CV, theta0 = 1 + delta,
n <-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)) {
$power[i] <- power.TOST(CV = CV,
comp2theta0 = comp2$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 = F); print(comp2, row.names = F)
R> CV theta0 base n power
R> 0.3 0.9500 TRUE 40 0.8158
R> 0.3 0.9524 FALSE 40 0.8246
R> 0.3 1.0500 FALSE 40 0.8246
R> 0.3 1.0530 FALSE 40 0.8158
R> CV theta0 base n power
R> 0.3 0.9500 FALSE 38 0.7953
R> 0.3 0.9524 FALSE 38 0.8043
R> 0.3 1.0500 TRUE 38 0.8043
R> 0.3 1.0530 FALSE 38 0.7953
</nitpick>
Since power is much more sensitive to the T/R-ratio than to the CV, the results obtained with the Bayesian method should be clear now.
Essentially this leads to the murky waters of prospective sensitivity analyses, which will be covered in another article.
An appetizer to show the maximum deviations (CV, T/R-ratio 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)
100*(tail(pa$paCV[["CV"]], 1) -
change.CV <- pa$plan$CV) / pa$plan$CV
100*(head(pa$paGMR$theta0, 1) -
change.theta0 <- pa$plan$theta0) /
pa$plan$theta0
100*(tail(pa$paN[["N"]], 1) -
change.n <- pa$plan[["Sample size"]]) /
pa$plan[["Sample size"]]
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)
R> Sample size plan ABE
R> Design alpha CV theta0 theta1 theta2 Sample size
R> 2x2 0.05 0.25 0.95 0.8 1.25 28
R> Achieved power
R> 0.8074395
R>
R> Power analysis
R> CV, theta0 and number of subjects leading to min. acceptable power of =0.7:
R> CV= 0.2843, theta0= 0.9268
R> n = 23 (power= 0.7173)
R> parameter relative change
R> CV +13.70%
R> theta0 -2.44%
R> n -17.86%
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/R-ratio:
12
m <- 0.25
CV <- 0.95
theta0 <- expsampleN.TOST(CV = CV, theta0 = theta0,
df <-targetpower = 0.80,
design = "2x2",
prior.parm = list(
m = m, design = "2x2"),
prior.type = "both",
details = FALSE)
R>
R> ++++++++++++ Equivalence test - TOST ++++++++++++
R> Sample size est. with uncertain CV and theta0
R> -------------------------------------------------
R> Study design: 2x2 crossover
R> log-transformed data (multiplicative model)
R>
R> alpha = 0.05, target power = 0.8
R> BE margins = 0.8 ... 1.25
R> Ratio = 0.95 with 10 df
R> CV = 0.25 with 10 df
R>
R> Sample size (ntotal)
R> n exp. power
R> 108 0.800126
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/R-ratio is based on statistical assurance.^{15} This concept uses the distribution of T/R-ratios 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/R-ratio 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,
details = FALSE,
print = FALSE)[7:8]
5:6] <- expsampleN.TOST(CV = CV,
comp[theta0 = 1, # fixed!
targetpower = target,
design = "2x2",
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)
R> theta0 n power sigma.u n assurance
R> 0.95 28 0.807439 0.05 28 0.817016
It is not unusal 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 \(\alpha\), we are protected by the intersection-union principle.^{16} ^{17}
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/R-ratios.
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)) {
$n[i] <- sampleN.TOST(CV = CV[i],
dftheta0 = theta0[i],
theta1 = theta1[i],
theta2 = theta2[i],
targetpower = target,
print = FALSE)[["Sample size"]]
}$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)
R> metric theta1 theta2 CV theta0 n
R> Cmax 0.7500 1.3333 0.25 0.95 16
R> AUCt 0.8000 1.2500 0.19 1.04 16
R> AUCinf 0.8000 1.2500 0.20 1.06 20
R> Sample size based on AUCinf.
Even if we assume the same T/R-ratio 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/R-ratio 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/R-ratio of AUC.
Which are the extreme T/R-ratios (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]) - 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,
print = FALSE)[7:8]
}$power <- signif(df$power, 6)
dfif (theta0 < 1) {
uniroot(opt, tol = 1e-8,
res <-interval = c(theta1 + 1e-4, theta0))
else {
} uniroot(opt, tol = 1e-8,
res <-interval = c(theta0, theta2 - 1e-4))
} 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)
R> metric theta0 CV n power
R> Cmax 0.95 0.25 28 0.807439
R> AUC 0.95 0.20 20 0.834680
R> Target power for AUC and sample size 28
R> achieved for theta0 0.9158 or 1.0920.
That means, although we assumed for AUC the same T/R-ratio as for C_{max} – with the sample size of 28 required for C_{max} – for AUC it can be as low as ~0.92 or as high as ~1.09, which is a soothing side-effect.
Furthermore, sometimes we have less data of AUC than of C_{max} (samples at the end of the profile missing or unreliable estimation of \(\small{\lambda_\textrm{z}}\) in some subjects and therefore, less data of AUC_{0–∞} than of AUC_{o–t}). Again, it will not hurt because for the originally assumed T/R-ratio we need only 20 subjects.
Since – as a one-point 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.^{18}
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)) {
$n[i] <- sampleN.TOST(alpha = alpha[i],
dfCV = CV[i],
theta0 = theta0,
targetpower = target,
print = FALSE)[["Sample size"]]
} paste0("Sample size based on ",
txt <-$metric[df$n == max(df$n)], ".\n")
dfprint(df, row.names = FALSE); cat(txt)
R> metric CV theta0 alpha n
R> Cmax 0.6 0.95 0.50 24
R> AUC 0.2 0.95 0.05 20
R> 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}.
In the most simple case one may compare two test treatments to one reference and want to demonstrate equivalence of both. The well-known Bonferroni-adjustment can be used \[\begin{equation}\tag{6} \alpha_\textrm{adj}=\alpha\,/\,k \end{equation}\] where \(\small{k}\) is the number of simultaneous tests. The type I error \(\small{TIE}\) will always be controlled because \[\begin{equation}\tag{7} TIE=1-{(1-\alpha_\textrm{adj})}^k \end{equation}\] 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 \(\small{100(1-2\,\alpha_\textrm{adj})}\) 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)
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R> Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2 crossover
R> log-transformed data (multiplicative model)
R>
R> alpha = 0.025, target power = 0.8
R> BE margins = 0.8 ... 1.25
R> True ratio = 0.95, CV = 0.25
R>
R> Sample size (total)
R> n power
R> 36 0.816081
The sample size increased by ~29% from the 28 subjects required in a single comparison.
If all treatments should be tested for equivalence (say, T_{1} v.s. R, T_{2} v.s. R, and T_{2} v.s. T_{1}), naturally the sample size increases further.
3 # comparisons
k <- 0.05 / k # Bonferroni
alpha.adj <-sampleN.TOST(alpha = alpha.adj, CV = 0.25,
theta0 = 0.95, targetpower = 0.80)
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R> Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2 crossover
R> log-transformed data (multiplicative model)
R>
R> alpha = 0.01666667, target power = 0.8
R> BE margins = 0.8 ... 1.25
R> True ratio = 0.95, CV = 0.25
R>
R> Sample size (total)
R> n power
R> 40 0.812356
With an increasing number of comparisons, the Bonferroni-adjustment 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.
So far we employed the common (and hence, default) BE-limits theta1 = 0.80
and theta2 = 1.25
. In some jurisdictions tighter limits of 90.00 – 111.11% have to be used.^{19}
Generally NTIDs show a low within-subject variability (though the between-subject CV can be much higher – these drugs require quite often dose-titration).
You have to provide only the lower BE-limit 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)
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R> Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2 crossover
R> log-transformed data (multiplicative model)
R>
R> alpha = 0.05, target power = 0.8
R> BE margins = 0.9 ... 1.111111
R> True ratio = 0.95, CV = 0.125
R>
R> Sample size (total)
R> n power
R> 68 0.805372
Not so nice. If you would exhaust the capacity of the clinical site, you may consider a replicate design.
Interlude 2
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></nitpick>
Hence, for Health Canada you have to set both limits.
sampleN.TOST(CV = 0.125, theta1 = 0.90, theta2 = 1.12)
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R> Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2 crossover
R> log-transformed data (multiplicative model)
R>
R> alpha = 0.05, target power = 0.8
R> BE margins = 0.9 ... 1.12
R> True ratio = 0.95, CV = 0.125
R>
R> Sample size (total)
R> n power
R> 68 0.805372
However, only rarely a bit lower than with the common limits for NTIDs (here we get the same sample size).
The FDA requires for NTIDs tighter batch-release specifications (±5% instead of ±10%). Let’s hope that your product complies and the T/R-ratio will be closer to 1:
sampleN.TOST(CV = 0.125,
theta0 = 0.975, # ‘better’ T/R-ratio
theta1 = 0.90)
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R> Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2 crossover
R> log-transformed data (multiplicative model)
R>
R> alpha = 0.05, target power = 0.8
R> BE margins = 0.9 ... 1.111111
R> True ratio = 0.975, CV = 0.125
R>
R> Sample size (total)
R> n power
R> 32 0.800218
Substantially lower sample and doable.
Sometimes we are interested in assessing differences of responses and not 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,
logscale = FALSE)
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R> Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2 crossover
R> untransformed data (additive model)
R>
R> alpha = 0.05, target power = 0.8
R> BE margins = -15 ... 15
R> True diff. = -5, CV = 20
R>
R> Sample size (total)
R> n power
R> 52 0.807468
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 a 2×2×2 design it’s \(\small{SD/\sqrt{2}}\).
36
SD.delta <-sampleN.TOST(CV = SD.delta / sqrt(2), theta0 = -5,
theta1 = -15, theta2 = +15,
logscale = FALSE)
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R> Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2 crossover
R> untransformed data (additive model)
R>
R> alpha = 0.05, target power = 0.8
R> BE margins = -15 ... 15
R> True diff. = -5, CV = 25.45584
R>
R> Sample size (total)
R> n power
R> 82 0.801691
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 reference-means.
In such a case Fieller’s (‘fiducial’) confidence interval^{20} 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.^{21} Note that we need additionally the between-subjct CV (here arbitrarily twice the CV).
0.25
CV <- 2 * CV
CVb <-sampleN.RatioF(CV = CV, CVb = CVb)
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R> based on Fieller's confidence interval
R> Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2 crossover
R> Ratio of means with normality on original scale
R> alpha = 0.025, target power = 0.8
R> BE margins = 0.8 ... 1.25
R> True ratio = 0.95, CVw = 0.25, CVb = 0.5
R>
R> Sample size
R> n power
R> 42 0.802153
“He who seeks for methods without having a definite problem in mind seeks in the most part in vain.
— David Hilbert
With a few exceptions (i.e., simulation-based methods), in PowerTOST
the default method = "exact"
implements Owen’s Q function^{22} which is also used in SAS’ Proc Power
.
Other implemented methods are "mvt"
(based on the bivariate non-central t-distribution), "noncentral"
/ "nct"
(noncentral t-distribution), and "shifted"
/ "central"
(shifted central t-distribution). Although "mvt"
is also exact, it may have a somewhat lower precision compared to Owen’s Q and has a much longer run-time.
Let’s compare them.
c("exact", "mvt", "noncentral", "central")
methods <- data.frame(method = methods,
df <-power = rep(NA, 4))
for (i in 1:nrow(df)) {
$power[i] <- power.TOST(CV = 0.25,
dfn = 28,
method = df$method[i])
}$power <- round(df$power, 5)
dfprint(df, row.names = FALSE)
R> method power
R> exact 0.80744
R> mvt 0.80744
R> noncentral 0.80744
R> central 0.80303
Power approximated by the shifted central t-distribution is generally slightly lower compared to the others. Hence, if used in sample size estimations, occasionally two more subjects are ‘required’ – which is not correct as demonstrated by the exact method.
Here an example for CV 0.28.
c("exact", "mvt", "noncentral", "central")
methods <- data.frame(method = methods)
df <-for (i in 1:nrow(df)) {
sampleN.TOST(CV = 0.28, theta0 = 0.95,
tmp <-method = df$method[i],
targetpower = 0.80, print = FALSE)
$n[i] <- tmp[["Sample size"]]
df$power[i] <- tmp[["Achieved power"]]
df$power.exact[i] <- power.TOST(CV = 0.28,
dftheta0 = 0.95,
n = df$n[i],
method = "exact")
}print(df, row.names = FALSE)
R> method n power power.exact
R> exact 34 0.8017690 0.8017690
R> mvt 34 0.8017720 0.8017690
R> noncentral 34 0.8017690 0.8017690
R> central 36 0.8210282 0.8242676
Particularly for ‘nice’ combinations of the CV and T/R-ratio the relative increase in sample sizes (and hence, study costs) might be relevant.
set.seed(123456)
1e4
nsims <- signif(runif(nsims, min = 0.13, max = 0.40), 2)
CV <- signif(runif(nsims, min = 0.85, max = 0.95), 2)
theta0 <- data.frame(CV = CV, theta0 = theta0,
df <-exact = NA, shifted = NA,
diff = NA, delta.n = FALSE)
unique(df)
df <-for (i in 1:nrow(df)) {
$exact[i] <- sampleN.TOST(CV = df$CV[i],
dftheta0 = df$theta0[i],
design = "2x2",
method = "exact",
print = FALSE)[["Sample size"]]
$shifted[i] <- sampleN.TOST(CV = df$CV[i],
dftheta0 = df$theta0[i],
design = "2x2",
method = "shifted",
print = FALSE)[["Sample size"]]
if (df$shifted[i] != df$exact[i]) df$delta.n[i] <- TRUE
}$diff <- 100 * (df$shifted - df$exact) / df$exact
df df[with(df, order(-diff, theta0, CV)), ]
df <-$diff <- sprintf("%+.3f%%", df$diff)
dfprint(df[which(df$delta.n == TRUE), 1:5], row.names = FALSE)
R> CV theta0 exact shifted diff
R> 0.14 0.92 14 16 +14.286%
R> 0.17 0.95 14 16 +14.286%
R> 0.22 0.95 22 24 +9.091%
R> 0.19 0.92 24 26 +8.333%
R> 0.14 0.88 28 30 +7.143%
R> 0.19 0.91 28 30 +7.143%
R> 0.18 0.90 30 32 +6.667%
R> 0.28 0.95 34 36 +5.882%
R> 0.21 0.90 40 42 +5.000%
R> 0.27 0.93 40 42 +5.000%
R> 0.27 0.92 46 48 +4.348%
R> 0.33 0.95 46 48 +4.348%
R> 0.34 0.94 54 56 +3.704%
R> 0.36 0.95 54 56 +3.704%
R> 0.30 0.92 56 58 +3.571%
R> 0.33 0.93 58 60 +3.448%
R> 0.36 0.94 60 62 +3.333%
R> 0.36 0.93 68 70 +2.941%
R> 0.27 0.89 78 80 +2.564%
R> 0.25 0.88 84 86 +2.381%
R> 0.23 0.87 92 94 +2.174%
R> 0.38 0.91 102 104 +1.961%
R> 0.25 0.87 108 110 +1.852%
R> 0.36 0.90 110 112 +1.818%
R> 0.40 0.91 112 114 +1.786%
R> 0.36 0.89 134 136 +1.493%
R> 0.32 0.86 232 234 +0.862%
R> 0.34 0.86 260 262 +0.769%
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: About the acceptability of Base R see ‘A Guidance Document for the Use of R in Regulated Clinical Trial Environments’.
The authors of PowerTOST
tried to do their best to provide reliable and valid results. The ‘NEWS’-file on CRAN documents the development of the package, bug-fixes, and introduction of new methods.
Validation of any software (yes, of SAS as well…) lies in the hands of the user. The package contains data.frames of sample size from the literature^{23} ^{24} (show them by ct5.1
, ct5.2
, ct5.3
, and ct5.4.1
). Execute the script test_2x2.R
which can be found in the /tests
sub-directory of the package to reproduce them. If results don’t agree, something went wrong. Do you use a version <0.9.3 of 2012-02-13? If yes, update to the current one.
You can also reproduce Table 7.1 of Julious (2010, p 110–11).^{25}
paste("True Mean Ratios 0.80\u20131.20",
txt <-"\n90% power by the noncentral t-distribution",
"\n Levels of Bioequivalence",
"\n 10% -> 90.00\u2013111.11%",
"\n 15% -> 85.00\u2013117.65%",
"\n 20% -> 80.00\u2013125.00%",
"\n 25% -> 75.00\u2013133.33%",
"\n 30% -> 70.00\u2013142.86%",
"\nSample sizes (per arm)\n")
function(expr) {
tryCatch.W.E <-##' Catch *and* save both errors and warnings,
##' and in the case of a warning, also keep the
##' computed result.
##'
##' @title tryCatch both warnings (with value) and
##' errors
##' @param expr an \R expression to evaluate
##' @return a list with 'value' and 'warning', where
##' 'value' may be an error caught.
##' @author Martin Maechler;
##' Copyright (C) 2010-2012 The R Core Team
NULL
W <- function(w) { # warning handler
w.handler <- w
W <<-invokeRestart("muffleWarning")
}list(value = withCallingHandlers(
tryCatch(expr, error = function(e) e),
warning = w.handler),
warning = W)
} function(CV, theta0, theta1) {
fun <-return(
sampleN.TOST(CV = CV, theta0 = theta0,
theta1 = theta1,
targetpower = 0.90,
method = "noncentral",
print = FALSE)[["Sample size"]])
} seq(0.10, 0.45, 0.05)
CV <- seq(0.80, 1.20, 0.05)
ratio <- seq(0.90, 0.70, -0.05)
theta1 <- data.frame(CV = rep(CV, each = length(ratio)),
df <-Ratio = ratio, n10 = NA, n15 = NA,
n20 = NA, n25 = NA, n30 = NA)
for (i in 1:nrow(df)) {
for (j in 3:ncol(df)) {
# needed to continue if ratio at or outside limits
# (otherwise, stops with error)
tryCatch.W.E(fun(CV = df$CV[i],
x <-theta0 = df$Ratio[i],
theta1 = theta1[j-2]))
if (is.numeric(x$value)) {
x$value
df[i, j] <-# check whether we get the target with one
# subject less
suppressMessages(
power <-power.TOST(CV = df$CV[i],
theta0 = df$Ratio[i],
theta1 = theta1[j-2],
n = df[i, j] - 1,
method = "noncentral"))
if (power >= 0.90) df[i, j] <- df[i, j] - 1
# remove high ones like in the table
if (j == 3 & df[i, j] > 1082) df[i, j] <- NA
if (j == 4 & df[i, j] > 969) df[i, j] <- NA
if (j == 5 & df[i, j] > 1897) df[i, j] <- NA
if (j == 6 & df[i, j] > 612) df[i, j] <- NA
if (j == 7 & df[i, j] > 179) df[i, j] <- NA
}
}
}for (j in 3:ncol(df)) {
formatC(df[, j], format = "f",
df[, j] <-digits = 0, big.mark=",")
}== "NA")] <- ""
df[(df $CV <- df$CV*100
dfnames(df)[c(1, 3:7)] <- c("CV (%)",
paste0(seq(10, 30, 5), "%"))
cat(txt); print(df, row.names = FALSE)
R> True Mean Ratios 0.80–1.20
R> 90% power by the noncentral t-distribution
R> Levels of Bioequivalence
R> 10% -> 90.00–111.11%
R> 15% -> 85.00–117.65%
R> 20% -> 80.00–125.00%
R> 25% -> 75.00–133.33%
R> 30% -> 70.00–142.86%
R> Sample sizes (per arm)
R> CV (%) Ratio 10% 15% 20% 25% 30%
R> 10 0.80 43 12
R> 10 0.85 48 13 7
R> 10 0.90 54 14 8 6
R> 10 0.95 60 16 8 6 5
R> 10 1.00 21 10 7 5 5
R> 10 1.05 55 15 8 6 5
R> 10 1.10 40 13 7 5
R> 10 1.15 331 26 10 6
R> 10 1.20 104 17 8
R> 15 0.80 93 23
R> 15 0.85 106 26 12
R> 15 0.90 119 29 14 8
R> 15 0.95 132 33 15 9 7
R> 15 1.00 45 20 12 8 6
R> 15 1.05 121 31 15 9 7
R> 15 1.10 86 25 12 8
R> 15 1.15 738 57 19 10
R> 15 1.20 231 36 15
R> 20 0.80 163 40
R> 20 0.85 185 45 20
R> 20 0.90 207 50 22 13
R> 20 0.95 232 56 25 14 10
R> 20 1.00 78 34 19 12 9
R> 20 1.05 212 54 24 14 10
R> 20 1.10 151 43 20 12
R> 20 1.15 99 33 16
R> 20 1.20 405 62 24
R> 25 0.80 251 60
R> 25 0.85 284 68 30
R> 25 0.90 320 77 33 18
R> 25 0.95 357 86 37 21 14
R> 25 1.00 120 52 28 18 12
R> 25 1.05 326 82 36 21 14
R> 25 1.10 232 65 30 17
R> 25 1.15 151 49 24
R> 25 1.20 625 95 36
R> 30 0.80 356 85
R> 30 0.85 403 96 41
R> 30 0.90 454 108 46 25
R> 30 0.95 507 121 52 29 18
R> 30 1.00 170 73 39 25 17
R> 30 1.05 463 116 51 28 18
R> 30 1.10 329 92 42 24
R> 30 1.15 214 69 33
R> 30 1.20 888 135 50
R> 35 0.80 477 113
R> 35 0.85 540 128 54
R> 35 0.90 608 145 61 33
R> 35 0.95 679 162 69 38 24
R> 35 1.00 227 97 52 32 22
R> 35 1.05 620 155 67 37 24
R> 35 1.10 440 123 55 31
R> 35 1.15 287 92 44
R> 35 1.20 1,190 180 67
R> 40 0.80 612 144
R> 40 0.85 694 164 69
R> 40 0.90 780 185 78 42
R> 40 0.95 871 207 88 48 30
R> 40 1.00 291 124 66 41 27
R> 40 1.05 796 198 86 48 30
R> 40 1.10 565 157 71 39
R> 40 1.15 368 118 56
R> 40 1.20 1,527 231 86
R> 45 0.80 179
R> 45 0.85 861 203 86
R> 45 0.90 969 230 97 52
R> 45 0.95 1,082 257 109 60 37
R> 45 1.00 361 153 82 50 33
R> 45 1.05 989 246 106 59 37
R> 45 1.10 701 195 87 48
R> 45 1.15 456 146 69
R> 45 1.20 1,897 286 106
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 28 eligible subjects as desired. Why is the dropout-rate ~12% and not the anticipated 10%?
A: That’s due to rounding up the calculated adjusted sample size (31.11…) to the next even number (32).
If you manage it to dose fractional subjects (I can’t) your dropout rate would indeed equal the anticipated one: 100(1 – 28/31.11…) = 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
and bar
are the number of subjects per sequence (the order is not important).
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 \(\small{n-2}\) degrees of freedom and \(\small{\log_{e}(\theta_0)}\) follows a normal distribution.^{26}
Convergence takes a while. Empiric power, its standard error, its relative error compared to the exact power, and execution times on a Xeon E3-1245v3 3.4 GHz, 8 MB cache, 16 GB RAM, 64 bit R 4.0.5 on Windows 7:
0.25
CV <- sampleN.TOST(CV = CV,
n <-print = FALSE)[["Sample size"]]
c(1e5, 5e5, 1e6, 5e6, 1e7, 5e7, 1e8)
nsims <- power.TOST(CV = CV, n = n)
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,
dfnsims = nsims[i])
$secs[i] <- proc.time()[[3]] - start.time
df$SE[i] <- sqrt(0.025 / i)
df$RE <- 100 * (df$simulated - exact) / exact
df
}$exact <- signif(df$exact, 5)
df$SE <- signif(df$RE, 4)
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)
R> sim’s exact sim’d SE RE secs
R> 100,000 0.80744 0.80814 0.0867600 +0.0868% 0.17
R> 500,000 0.80744 0.80699 -0.0551700 -0.0552% 0.79
R> 1,000,000 0.80744 0.80766 0.0279300 +0.0279% 1.63
R> 5,000,000 0.80744 0.80751 0.0090080 +0.0090% 8.05
R> 10,000,000 0.80744 0.80733 -0.0136200 -0.0136% 16.17
R> 50,000,000 0.80744 0.80743 -0.0009096 -0.0009% 81.81
R> 100,000,000 0.80744 0.80744 0.0001147 +0.0001% 162.40
License
Helmut Schütz 2021
1^{st} version March 10, 2021.
Rendered 2021-04-16 16:10:22 CEST by rmarkdown in 2.37 seconds.
Footnotes and References
Labes D, Schütz H, Lang B. PowerTOST: Power and Sample Size for (Bio)Equivalence Studies. 2021-01-18. CRAN.↩︎
Labes D, Schütz H, Lang B. Package ‘PowerTOST’. January 18, 2021. CRAN.↩︎
If we plan all studies for 80% power, treatments are equivalent, and all assumptions are exactly realized, one out of five studies will fail by pure chance.
Science is a cruel mistress.↩︎
U.S. FDA, CDER. Guidance for Industry. Statistical Approaches Establishing Bioequivalence. January 2001. APPENDIX C.↩︎
According to the WHO:
»The a posteriori power of the study does not need to be calculated. The power of interest is that calculated before the study is conducted to ensure that the adequate sample size has been selected. […] The relevant power is the power to show equivalence within the pre-defined acceptance range.«↩︎
Fuglsang A. Pilot and Repeat Trials as Development Tools Associated with Demonstration of Bioequivalence. AAPS J. 2015; 17(3): 678–83. doi:10.1208/s12248-015-9744-6.↩︎
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.↩︎
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 pre-dose concentrations > 5% of their C_{max} can by excluded from the comparison if stated in the protocol. For details see another article.↩︎
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 allows a dose-adjustment in the evaluation of the study.↩︎
Zhang P. A Simple Formula for Sample Size Calculation in Equivalence Studies. J Biopharm Stat. 2003; 13(3): 529–538. doi:10.1081/BIP-120022772.↩︎
Schütz H. Sample Size Estimation in Bioequivalence. Evaluation. 2020-10-23. BEBA Forum.↩︎
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↩︎
Berger RL, Hsu JC. Bioequivalence Trials, Intersection-Union 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.↩︎
With alpha = 0.5
the test effectively reduces to a pass/fail assessment.↩︎
The FDA recommends Reference-Scaled Average Bioequivalence (RSABE) and a comparison of the within-subject variances of treatments, which requires a fully replicated design. It will be 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 Non-Inferiority. London, 27 July 2000. CPMP/EWP/482/99.↩︎
Owen DB. A special case of a bivariate non-central t-distribution. Biometrika. 1965; 52(3/4): 437–46. doi:10.2307/2333696.↩︎
Hauschke D, Steinijans VW, Pigeot I. Bioequivalence Studies in Drug Development. Chichester: John Wiley; 2007. Tables 5.1–5.3 (p 113–118).↩︎
Chow SC, Liu J-p. Design and Analysis of Bioavailability and Bioequivalence Studies. Boca Raton: CRC Press; 3^{rd} edition 2009. Table 5.4.1 (p 158).↩︎
Note that Table 7.1 gives sample sizes even if they are unbalanced. Julious did not report extreme sample sizes and therefore, I dropped them from the output as well.↩︎
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.↩︎