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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 and the online manual3 for details and a collection of other articles.

• The right-hand badges give the respective section’s ‘level’.

1. Basics about sample size methodology – requiring no or only limited statistical expertise.

1. These sections are the most important ones. They are – hopefully – easily comprehensible even for novices.

1. A somewhat higher knowledge of statistics and/or R is required. May be skipped or reserved for a later reading.

1. An advanced knowledge of statistics and/or R is required. Not recommended for beginners in particular.

1. If you are not a neRd or statistics afficionado, skipping is recommended. Suggested for experts but might be confusing for others.
• Click to show / hide R code.
Abbreviation Meaning
(A)BE (Average) Bioequivalence
ABEL Average Bioequivalence with Expanding Limits
CVinter Between-subject Coefficient of Variation
CVintra Within-subject Coefficient of Variation
CVwT, CVwR Within-subject Coefficient of Variation of the Test and Reference treatment
H0 Null hypothesis
H1 Alternative hypothesis (also Ha)
HVD(P) Highly Variable Drug (Product)
NTID Narrow Therapeutic Index Drug
RSABE Reference-scaled Average Bio­equivalence
TOST Two One-Sided Tests

# Introduction

What are the main statistical issues in planning a confirmatory experiment?

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.

top of section ↩︎

## Preliminaries

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 all sequences). Furthermore, the estimated sample size is the total number of subjects (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.

previous section ↩︎

## Terminology

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. The level of the test (in BE commonly 0.05),
2. the BE-margins (commonly 0.80 – 1.25),
3. the desired (or target) power,
4. the variance (commonly expressed as a coefficient of variation), and
5. the deviation of the test from the reference treatment.

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.

Realization: Observations (in a sample) of a random variable (of the population).

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 CVintra is lower than the between-subject variability CVinter, crossover studies are so popular. Of note, there is no relationship between CVintra and CVinter. An example are drugs which are subjected to polymorphic metabolism. For these drugs CVintraCVinter.

Carryover: A residual effect of a previous period.

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).

previous section ↩︎

# Power → Sample size

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, 2nd ed 2007.

Contrary to parallel and 2×2×2 crossover designs I will not give approximations for the sample size. Since there are numerous replicate designs possible (each of them with different degrees of freedom), the R code would be too long – even for my taste (Spaghetti Viennese).

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.

There is no need to re-invent the wheel. Let’s start 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 method10 for the first guess. You can unveil the course of iterations with details = TRUE.

sampleN.TOST(CV = 0.45, theta0 = 0.95,
targetpower = 0.80,
design = "2x2x4", details = TRUE)
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R>             Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2x4 (4 period full replicate)
R> Design characteristics:
R> df = 3*n-4, design const. = 1, 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 = 0.45
R>
R> Sample size search (ntotal)
R>  n     power
R> 40   0.799409
R> 42   0.818228
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.

$$\small{100(1-2\,\alpha)}$$ confidence interval of the point estimate within $$\small{\theta_1,\theta_2}$$.

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. 40    -2 -4.8% 0.803974
R>     central t-distribution 42    ±0 ±0.0% 0.817294
R>  noncentral t-distribution 42    ±0 ±0.0% 0.818228
R>    exact method (Owen’s Q) 42     –   –   0.818228

previous section ↩︎

# Examples

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.

## A Simple Case

We assume a CV of 0.45, a T/R-ratio of 0.95, a target a power of 0.80, and want to perform the study in a 4-period full replicate study with sequences.

Since 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 and design = "2x2x4".

sampleN.TOST(CV = 0.45, design = "2x2x4")
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R>             Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2x4 (4 period full replicate)
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 = 0.45
R>
R> Sample size (total)
R>  n     power
R> 42   0.818228

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).

df <- sampleN.TOST(CV = 0.45, design = "2x2x4",
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 df:

names(df)
R>  "Design"         "alpha"          "CV"
R>  "theta0"         "theta1"         "theta2"
R>  "Sample size"    "Achieved power" "Target power"

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).

df[["Sample size"]]
R>  42
df[["Achieved power"]]
R>  0.818228

If you insist in accessing elements by column-numbers, use single square brackets […].

df[7:8]
R>   Sample size Achieved power
R> 1          42       0.818228

With 42 subjects (21 per sequence) we achieve the power we desire.

What happens if we have one dropout?

power.TOST(CV = 0.45, design = "2x2x4",
n = df[["Sample size"]] - 1)
R> Unbalanced design. n(i)=21/20 assumed.
R>  0.8088329

Still above the 0.80 we desire. However, with two dropouts (40 eligible subjects) we would slightly fall short (0.7994).

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.

previous section ↩︎

## Dropouts

We define two supportive functions:

1. Provide equally sized sequences, i.e., any total sample size n will be rounded up to achieve balance.
balance <- function(n, n.seq) {
return(as.integer(n.seq * (n %/% n.seq + as.logical(n %% n.seq))))
}
1. Provide the adjusted sample size based on the original sample size n and the anticipated droput-rate do.rate.
nadj <- function(n, do.rate, n.seq) {
return(as.integer(balance(n / (1 - do.rate), n.seq)))
}

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.

### Dropout-rate

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.11 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.

CV      <- 0.45 # within-subject CV
target  <- 0.80 # target (desired) power
theta0  <- 0.95 # assumed T/R-ratio
theta1  <- 0.80 # lower BE limit
theta2  <- 1.25 # upper BE limit
design  <- "2x2x4"
do.rate <- 0.10 # anticipated dropout-rate 10%
n.seq   <- as.integer(substr(design, 3, 3))
df      <- sampleN.TOST(CV = CV, theta0 = theta0,
theta1 = theta1,
theta2 = theta2,
targetpower = target,
design = design,
print = FALSE)
# calculate the adjusted sample size
# (decreasing) vector of eligible subjects
info    <- paste0("Assumed CV              : ",
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   : ",
df[["Sample size"]], " (",
df[["Sample size"]]/2, "/sequence)",
"\nAchieved power          : ",
signif(df[["Achieved power"]], 4),
"\n\n")
# explore the potential outcome for
# an increasing number of dropouts
eligible = n.elig,
do.act   = do.act,
power    = NA)
for (i in 1:nrow(df)) {
df$power[i] <- suppressMessages( power.TOST(CV = CV, theta0 = theta0, theta1 = theta1, theta2 = theta2, design = design, n = df$eligible[i]))
}
cat(info); print(round(df, 4), row.names = FALSE)
R> Assumed CV              : 0.45
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   : 42 (21/sequence)
R> Achieved power          : 0.8182
R> Adjusted sample size    : 48 (24/sequence)
R>  dosed eligible dropouts do.act  power
R>     48       48        0 0.0000 0.8645
R>     48       47        1 0.0208 0.8576
R>     48       46        2 0.0417 0.8506
R>     48       45        3 0.0625 0.8429
R>     48       44        4 0.0833 0.8352
R>     48       43        5 0.1042 0.8267
R>     48       42        6 0.1250 0.8182

In the worst case (6 dropouts) we end up with the originally estimated sample size of 42. 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 48 subjects like the 48 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 43 eligible subjects (power 0.8267). In this example we achieve still more than our target power but the loss might be relevant in other cases.

### Post hoc Power

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.

n.act <- 41
signif(power.TOST(CV = 0.45, n = n.act, design = "2x2x4"), 6)
R> Unbalanced design. n(i)=21/20 assumed.
R>  0.808833

Say, our study was more unbalanced. Let us assume that we dosed 48 subjects, the total number of subjects was also 41 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).

n.adj    <- 48
n.act    <- 41
n.s2     <- n.act - n.s1
post.hoc <- signif(power.TOST(CV  = 0.45,
n = c(n.s1, n.s2),
design = "2x2x4"), 6)
fmt      <- paste0("%", sig.dig, ".0f (%",  sig.dig, ".0f dropouts)")
"\nEligible      : ",
"\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: 48
R> Eligible      : 41 ( 7 dropouts)
R>   Sequence 1  : 24 ( 0 dropouts)
R>   Sequence 1  : 17 ( 7 dropouts)
R> Power         :  0.797608

Of course, in a particular study you will provide the numbers in the n vector directly.

## Lost in Assumptions

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, believing12 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 16 subjects and estimated the CV as 0.45.

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.

m  <- 16 # pilot study
ci <- CVCL(CV = 0.45, df = m - 2,
side = "2-sided", alpha = 0.05)
signif(ci, 4)
R> lower CL upper CL
R>   0.3223   0.7629

Surprised? Although 0.45 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 0.45, we would opt for 42 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.

m    <- 16
ci   <- CVCL(CV = 0.45, df = m - 2,
side = "2-sided", alpha = 0.05)
n    <- 28
comp <- data.frame(CV = c(ci[["lower CL"]], 0.45,
ci[["upper CL"]]),
power = NA)
for (i in 1:nrow(comp)) {
comp$power[i] <- power.TOST(CV = comp$CV[i],
n = n)
}
comp[, 1] <- signif(comp[, 1], 4)
comp[, 2] <- signif(comp[, 2], 6)
print(comp, row.names = FALSE)
R>      CV      power
R>  0.3223 0.57315300
R>  0.4500 0.19171300
R>  0.7629 0.00127429

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.

m               <- seq(12, 24, 6)
df              <- data.frame(n = m, CV = 0.45,
l = NA, u = NA)
for (i in 1:nrow(df)) {
df[i, 3:4] <- CVCL(CV = 0.45, df = m[i] - 2,
side = "2-sided",
alpha = 0.05)
}
df[, 3:4]      <- signif(df[, 3:4], 4)
names(df)[3:4] <- c("lower CL", "upper CL")
print(df, row.names = FALSE)
R>   n   CV lower CL upper CL
R>  12 0.45   0.3069   0.8744
R>  18 0.45   0.3282   0.7300
R>  24 0.45   0.3415   0.6685

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.

m      <- 16   # sample size of pilot
CV     <- 0.45
theta0 <- 0.95
df     <- expsampleN.TOST(CV = CV, theta0 = theta0,
targetpower = 0.80,
design = "2x2x4",
prior.parm = list(
m = m, design = "2x2x4"),
prior.type = "CV",
details = FALSE)
R>
R> ++++++++++++ Equivalence test - TOST ++++++++++++
R>        Sample size est. with uncertain CV
R> -------------------------------------------------
R> Study design:  2x2x4 replicate 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 = 0.45 with 44 df
R>
R> Sample size (ntotal)
R>  n   exp. power
R> 42   0.800179

Good that our pilot was in a fully replicated design (high degrees of freedom).

m      <- 16
CV     <- 0.45
theta0 <- 0.95
df     <- expsampleN.TOST(CV = CV, theta0 = theta0,
targetpower = 0.80,
design = "2x2x4",
prior.parm = list(
m = m, design = "2x2x4"),
prior.type = "theta0",
details = FALSE)
R>
R> ++++++++++++ Equivalence test - TOST ++++++++++++
R>      Sample size est. with uncertain theta0
R> -------------------------------------------------
R> Study design:  2x2x4 replicate 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 = 0.45
R>
R> Sample size (ntotal)
R>  n   exp. power
R> 128   0.800005

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.

m      <- 16
CV     <- 0.45
pe     <- 0.95
ci     <- round(CI.BE(CV = CV, pe = 0.95, n = m,
design = "2x2x4"), 4)
if (pe <= 1) {
cl <- ci[["lower"]]
} else {
cl <- ci[["upper"]]
}
print(cl)
R>  0.7932

Exlore the impact of a relatively 5% higher CV and a relatively 5% lower T/R-ratio on power for a given sample size.

n      <- 42
CV     <- 0.45
theta0 <- 0.95
comp1  <- data.frame(CV = c(CV, CV*1.05),
power = NA)
comp2  <- data.frame(theta0 = c(theta0, theta0*0.95),
power = NA)
for (i in 1:2) {
comp1$power[i] <- power.TOST(CV = comp1$CV[i],
theta0 = theta0,
design = "2x2x4",
n = n)
}
comp1$power <- signif(comp1$power, 6)
for (i in 1:2) {
comp2$power[i] <- power.TOST(CV = CV, theta0 = comp2$theta0[i],
design = "2x2x4",
n = n)
}
comp2$power <- signif(comp2$power, 6)
print(comp1, row.names = F); print(comp2, row.names = F)
R>      CV    power
R>  0.4500 0.818228
R>  0.4725 0.783684
R>  theta0    power
R>  0.9500 0.818228
R>  0.9025 0.564727

Interlude 1 Fig. 1 Power curves for n = 42, 2×2×4 design.

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>
• A common flaw in protocols is the phrase
»The sample size was calculated [sic] based on a T/R-ratio of 0.95 – 1.05…«
If you assume a deviation of 5% of the test from the reference and are not sure about its direction (lower or higer than 1), always use the lower T/R-ratio. If you would use the upper T/R-ratio, power would be only preserved down to 1/1.05 = 0.9524.
Given, sometimes you will need a higher sample size with the lower T/R-ratio. There’s no free lunch.
CV      <- 0.45
delta   <- 0.05 # direction unknown
design  <- "2x2x4"
theta0s <- c(1 - delta, 1 / (1 + delta),
1 + delta, 1 / (1 - delta))
n       <- sampleN.TOST(CV = CV, theta0 = 1 - delta,
design = design,
print = FALSE)[["Sample size"]]
comp1   <- data.frame(CV = CV, theta0 = theta0s,
base = c(TRUE, rep(FALSE, 3)),
n = n, power = NA)
for (i in 1:nrow(comp1)) {
comp1$power[i] <- power.TOST(CV = CV, theta0 = comp1$theta0[i],
design = design, n = n)
}
n       <- sampleN.TOST(CV = CV, theta0 = 1 + delta,
design = design,
print = FALSE)[["Sample size"]]
comp2   <- data.frame(CV = CV, theta0 = theta0s,
base = c(FALSE, FALSE, TRUE, FALSE),
n = n, power = NA)
for (i in 1:nrow(comp2)) {
comp2$power[i] <- power.TOST(CV = CV, theta0 = comp2$theta0[i],
design = design, n = n)
}
comp1[, c(2, 5)] <- signif(comp1[, c(2, 5)] , 4)
comp2[, c(2, 5)] <- signif(comp2[, c(2, 5)] , 4)
print(comp1, row.names = F); print(comp2, row.names = F)
R>    CV theta0  base  n  power
R>  0.45 0.9500  TRUE 42 0.8182
R>  0.45 0.9524 FALSE 42 0.8270
R>  0.45 1.0500 FALSE 42 0.8270
R>  0.45 1.0530 FALSE 42 0.8182
R>    CV theta0  base  n  power
R>  0.45 0.9500 FALSE 40 0.7994
R>  0.45 0.9524 FALSE 40 0.8083
R>  0.45 1.0500  TRUE 40 0.8083
R>  0.45 1.0530 FALSE 40 0.7994

</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:

CV       <- 0.45
theta0   <- 0.95
target   <- 0.80
minpower <- 0.70
pa       <- pa.ABE(CV = CV, theta0 = theta0,
targetpower = target,
minpower = minpower,
design = "2x2x4")
change.CV     <- 100*(tail(pa$paCV[["CV"]], 1) - pa$plan$CV) / pa$plan$CV change.theta0 <- 100*(head(pa$paGMR$theta0, 1) - pa$plan$theta0) / pa$plan$theta0 change.n <- 100*(tail(pa$paN[["N"]], 1) -
pa$plan[["Sample size"]]) / pa$plan[["Sample size"]]
comp     <- data.frame(parameter = c("CV", "theta0", "n"),
change = c(change.CV,
change.theta0,
change.n))
comp$change <- sprintf("%+.2f%%", comp$change)
names(comp) <- "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>   2x2x4  0.05 0.45   0.95    0.8   1.25          42
R>  Achieved power
R>        0.818228
R>
R> Power analysis
R> CV, theta0 and number of subjects leading to min. acceptable power of =0.7:
R>  CV= 0.5247, theta0= 0.9248
R>  n = 32 (power= 0.7005)
R>  parameter relative change
R>         CV         +16.60%
R>     theta0          -2.66%
R>          n         -23.81%

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:

m      <- 16
CV     <- 0.45
theta0 <- 0.95
df     <- expsampleN.TOST(CV = CV, theta0 = theta0,
targetpower = 0.80,
design = "2x2x4",
prior.parm = list(
m = m, design = "2x2x4"),
prior.type = "both",
details = FALSE)
R>
R> ++++++++++++ Equivalence test - TOST ++++++++++++
R>   Sample size est. with uncertain CV and theta0
R> -------------------------------------------------
R> Study design:  2x2x4 replicate 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 44 df
R> CV = 0.45 with 44 df
R>
R> Sample size (ntotal)
R>  n   exp. power
R> 142   0.800556

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.13 This concept uses the distribution of T/R-ratios and assumes an uncertainty parameter $$\sigma_\textrm{u}$$. A natural assumption is $$\sigma_\textrm{u}=1-\theta_0$$, 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.

CV        <- 0.45
theta0    <- 0.95
target    <- 0.80
sigma.u   <- 1 - theta0
comp      <- data.frame(theta0 = theta0,
n.1 = NA, power = NA,
sigma.u = sigma.u,
n.2 = NA, assurance = NA)
comp[2:3] <- sampleN.TOST(CV = CV,
targetpower = target,
theta0 = theta0,
design = "2x2x4",
details = FALSE,
print = FALSE)[7:8]
comp[5:6] <- expsampleN.TOST(CV = CV,
theta0 = 1, # fixed!
targetpower = target,
design = "2x2x4",
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 42 0.818228    0.05 40   0.80949

previous section ↩︎

## Multiple Endpoints

It is not unusal that equivalence of more than one endpoint has to be demonstrated. In bioequivalence the pharmacokinetic metrics Cmax and AUC0–t are mandatory (in some jurisdictions like the FDA additionally AUC0–∞).

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.14 15

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 Cmax are acceptable. Let’s explore that with different CVs and T/R-ratios.

metrics <- c("Cmax", "AUCt", "AUCinf")
CV      <- c(0.45, 0.35, 0.37)
theta0  <- c(0.95, 1.04, 1.06)
theta1  <- c(0.75, 0.80, 0.80)
theta2  <- 1 / theta1
target  <- 0.80
df      <- data.frame(metric = metrics,
theta1 = theta1, theta2 = theta2,
CV = CV, theta0 = theta0, n = NA)
for (i in 1:nrow(df)) {
df$n[i] <- sampleN.TOST(CV = CV[i], theta0 = theta0[i], theta1 = theta1[i], theta2 = theta2[i], targetpower = target, design = "2x2x4", print = FALSE)[["Sample size"]] } df$theta1 <- sprintf("%.4f", df$theta1) df$theta2 <- sprintf("%.4f", df$theta2) txt <- paste0("Sample size based on ", df$metric[df$n == max(df$n)], ".\n")
print(df, row.names = FALSE); cat(txt)
R>  metric theta1 theta2   CV theta0  n
R>    Cmax 0.7500 1.3333 0.45   0.95 24
R>    AUCt 0.8000 1.2500 0.35   1.04 24
R>  AUCinf 0.8000 1.2500 0.37   1.06 32
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 Cmax. 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?

opt <- function(x) {
power.TOST(theta0 = x, CV = df$CV, theta1 = theta1, theta2 = theta2, design = design, n = df$n) - target
}
metrics <- c("Cmax", "AUC")
CV      <- c(0.45, 0.30) # Cmax, AUC
theta0  <- 0.95          # both metrics
theta1  <- 0.80
theta2  <- 1.25
target  <- 0.80
design  <- "2x2x4"
df      <- data.frame(metric = metrics, theta0 = theta0,
CV = CV, n = NA, power = NA)
for (i in 1:nrow(df)) {
df[i, 4:5] <- sampleN.TOST(CV = CV[i],
theta0 = theta0,
theta1 = theta1,
theta2 = theta2,
targetpower = target,
design = design,
print = FALSE)[7:8]
}
df$power <- signif(df$power, 6)
if (theta0 < 1) {
res <- uniroot(opt, tol = 1e-8,
interval = c(theta1 + 1e-4, theta0))
} else {
res <- uniroot(opt, tol = 1e-8,
interval = c(theta0, theta2 - 1e-4))
}
res     <- unlist(res)
theta0s <- c(res[["root"]], 1/res[["root"]])
txt     <- paste0("Target power for ", metrics,
" and sample size ",
df$n, "\nachieved for theta0 ", sprintf("%.4f", theta0s), " or ", sprintf("%.4f", theta0s), ".\n") print(df, row.names = FALSE); cat(txt) R> metric theta0 CV n power R> Cmax 0.95 0.45 42 0.818228 R> AUC 0.95 0.30 20 0.820240 R> Target power for AUC and sample size 42 R> achieved for theta0 0.8959 or 1.1161. That means, although we assumed for AUC the same T/R-ratio as for Cmax – with the sample size of 42 required for Cmax – for AUC it can be as low as ~0.90 or as high as ~1.12, which is a soothing side-effect. $$\small{\lambda_\textrm{z}}$$: Apparent terminal elimination rate – con­stant. Furthermore, sometimes we have less data of AUC than of Cmax (samples at the end of the profile missing or unreliable estimation of $$\lambda_\textrm{z}$$ in some subjects and therefore, less data of AUC0–∞ than of AUCo–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 – Cmax 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.16 metrics <- c("Cmax", "AUC") CV <- c(0.90, 0.30) theta0 <- 0.95 target <- 0.80 design <- "2x2x4" alpha <- c(0.50, 0.05) df <- data.frame(metric = metrics, CV = CV, theta0 = theta0, alpha = alpha, n = NA) for (i in 1:nrow(df)) { df$n[i] <- sampleN.TOST(alpha = alpha[i],
CV = CV[i],
theta0 = theta0,
targetpower = target,
design = design,
print = FALSE)[["Sample size"]]
}
txt       <- paste0("Sample size based on ",
df$metric[df$n == max(df$n)], ".\n") print(df, row.names = FALSE); cat(txt) R> metric CV theta0 alpha n R> Cmax 0.9 0.95 0.50 22 R> AUC 0.3 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 Cmax. previous section ↩︎ ## Multiple Treatments 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: k <- 2 # comparisons alpha.adj <- 0.05 / k # Bonferroni sampleN.TOST(alpha = alpha.adj, CV = 0.45, theta0 = 0.95, targetpower = 0.80, design = "2x2x4") R> R> +++++++++++ Equivalence test - TOST +++++++++++ R> Sample size estimation R> ----------------------------------------------- R> Study design: 2x2x4 (4 period full replicate) 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 = 0.45 R> R> Sample size (total) R> n power R> 52 0.813541 The sample size increased by ~10% from the 42 subjects required in a single comparison. If all treatments should be tested for equivalence (say, T1 v.s. R, T2 v.s. R, and T2 v.s. T1), naturally the sample size increases further. k <- 3 # comparisons alpha.adj <- 0.05 / k # Bonferroni sampleN.TOST(alpha = alpha.adj, CV = 0.45, theta0 = 0.95, targetpower = 0.80, design = "2x2x4") R> R> +++++++++++ Equivalence test - TOST +++++++++++ R> Sample size estimation R> ----------------------------------------------- R> Study design: 2x2x4 (4 period full replicate) 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 = 0.45 R> R> Sample size (total) R> n power R> 58 0.812372 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. previous section ↩︎ ## NTIDs 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.17 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, design = "2x2x4") R> R> +++++++++++ Equivalence test - TOST +++++++++++ R> Sample size estimation R> ----------------------------------------------- R> Study design: 2x2x4 (4 period full replicate) 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> 34 0.807718 Doable. Interlude 2 Health Canada requires for NTIDs (termed by HC ‘critical dose drugs’) that the confidence interval of Cmax lies within 80.0–125.0% and the one of AUC within 90.0–112.0%. <nitpick> • 100(1/0.9) ≠ 112.0, right? I always thought that it’s 111.11… On the average generic products will be approved with 100√0.9×1.12 ≈ 100.4% When asked, the reply was: »These numbers are more easy to remember.« </nitpick> Hence, for Health Canada you have to set both limits. sampleN.TOST(CV = 0.125, theta1 = 0.90, theta2 = 1.12, design = "2x2x4") R> R> +++++++++++ Equivalence test - TOST +++++++++++ R> Sample size estimation R> ----------------------------------------------- R> Study design: 2x2x4 (4 period full replicate) 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> 34 0.807718 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, design = "2x2x4") R> R> +++++++++++ Equivalence test - TOST +++++++++++ R> Sample size estimation R> ----------------------------------------------- R> Study design: 2x2x4 (4 period full replicate) 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> 16 0.805921 Substantially lower sample size. That’s nice. previous section ↩︎ ## Difference of Means 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: • As a difference of means relative to the same (underlying) reference mean. • In units of the difference of means. 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, design = "2x2x4") R> R> +++++++++++ Equivalence test - TOST +++++++++++ R> Sample size estimation R> ----------------------------------------------- R> Study design: 2x2x4 (4 period full replicate) 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> 26 0.810576 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. SD.delta <- 36 design <- "2x2x4" # extract relevant information and retrieve design constant known <- known.designs()[7:11, c(2, 6, 9)] bk <- known[known$design == design, "bk"] #
sampleN.TOST(CV = SD.delta / sqrt(bk), theta0 = -5,
theta1 = -15, theta2 = +15,
logscale = FALSE,
design = "2x2x4")
R>
R> +++++++++++ Equivalence test - TOST +++++++++++
R>             Sample size estimation
R> -----------------------------------------------
R> Study design: 2x2x4 (4 period full replicate)
R>
R> alpha = 0.05, target power = 0.8
R> BE margins = -15 ... 15
R> True diff. = -5,  CV = 36
R>
R> Sample size (total)
R>  n     power
R> 82   0.805692

For the 2×2×4 design the CV equals the SD because the design constant is 1. However, that’s not true for all replicate design. Show them:

R>  design  bk                      name
R>   2x2x3 1.5 2x2x3 replicate crossover
R>   2x2x4 1.0 2x2x4 replicate crossover
R>   2x4x4 1.0 2x4x4 replicate crossover
R>   2x3x3 1.5 partial replicate (2x3x3)
R>   2x4x2 8.0          Balaam's (2x4x2)

Note that other software packages (e.g., PASS, nQuery, StudySize, …) require the standard deviation of the difference as input.

previous section ↩︎

# Methods

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., simulation-based methods), in PowerTOST the default method = "exact" implements Owen’s Q function18 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.

methods <- c("exact", "mvt", "noncentral", "central")
df      <- data.frame(method = methods,
power = rep(NA, 4))
for (i in 1:nrow(df)) {
df$power[i] <- power.TOST(CV = 0.45, n = 42, design = "2x2x4", method = df$method[i])
}
df$power <- round(df$power, 5)
print(df, row.names = FALSE)
R>      method   power
R>       exact 0.81823
R>         mvt 0.81823
R>  noncentral 0.81823
R>     central 0.81729

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’.
Therefore, I recommend to use "method = shifted" / "method = central" only for comparing with old results (literature, own studies).

previous section ↩︎

# Q & A

• 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.

• 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 42 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 (46.67…) to the next even number (48).
If you manage it to dose fractional subjects (I can’t) your dropout rate would indeed equal the anticipated one: 100(1 – 42/46.67…) = 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(x, y), where x and y are the number of subjects per sequence.

• Q: In other articles you mention a method for the ‘Ratio of Means’. I miss it here.
A: Only the 2×2×2 Crossover Design is implemented in sampleN.RatioF(). For a crude [sic] estimate half the obtained sample size for 4-period full replicate designs. However, I definitely don’t recommend that.

• 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).19

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:

CV    <- 0.45
des   <- "2x2x4"
n     <- sampleN.TOST(CV = CV, design = des,
print = FALSE)[["Sample size"]]
nsims <- c(1e5, 5e5, 1e6, 5e6, 1e7, 5e7, 1e8)
exact <- power.TOST(CV =CV, n = n, design = des)
df    <- data.frame(simulations = nsims,
exact = rep(exact, length(nsims)),
simulated = NA, SE = NA, RE = NA)
for (i in 1:nrow(df)) {
start.time      <- proc.time()[]
df$simulated[i] <- power.TOST.sim(CV = CV, n = n, design = des, nsims = nsims[i]) df$secs[i]      <- proc.time()[] - 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",
digits = 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 0.81823 0.81711 -0.1366000 -0.1366%   0.09
R>      500,000 0.81823 0.81728 -0.1164000 -0.1164%   0.52
R>    1,000,000 0.81823 0.81822 -0.0004899 -0.0005%   0.99
R>    5,000,000 0.81823 0.81842  0.0231700 +0.0232%   5.03
R>   10,000,000 0.81823 0.81816 -0.0077130 -0.0077%  10.03
R>   50,000,000 0.81823 0.81824  0.0014850 +0.0015%  50.84
R>  100,000,000 0.81823 0.81824  0.0019680 +0.0020% 100.81 Fig. 2 Empiric power for n = 28.

• Q: I still have questions. How to proceed?
A: The preferred method is to register at the BEBA Forum and post your question there (please read its Policy first).
You can contact me at [email protected]. Be warned – I will charge you for anything beyond most basic questions.

previous section ↩︎

Footnotes and References

1. Labes D, Schütz H, Lang B. PowerTOST: Power and Sample Size for (Bio)Equivalence Studies. 2021-01-18. CRAN.↩︎

2. Schütz H. Average Bioequivalence. 2021-01-18. CRAN.↩︎

3. Labes D, Schütz H, Lang B. Package ‘PowerTOST’. January 18, 2021. CRAN.↩︎

4. 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.↩︎

5. U.S. FDA, CDER. Guidance for Industry. Statistical Approaches Establishing Bioequivalence. January 2001. APPENDIX C.↩︎

6. 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.«↩︎

7. 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.↩︎

8. 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.↩︎

9. 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 Cmax can by excluded from the comparison if stated in the protocol. For details see another article.↩︎

10. 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.↩︎

11. Schütz H. Sample Size Estimation in Bioequivalence. Evaluation. 2020-10-23. ↩︎

12. Quoting my late father: »If you believe, go to church.«↩︎

13. 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↩︎

14. Berger RL, Hsu JC. Bioequivalence Trials, Intersection-Union Tests and Equivalence Confidence Sets. Stat Sci. 1996; 11(4): 283–302. JSTOR:2246021.↩︎

15. With alpha = 0.5 the test effectively reduces to a pass/fail assessment.↩︎

16. 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.↩︎

17. Owen DB. A special case of a bivariate non-central t-distribution. Biometrika. 1965; 52(3/4): 437–46. doi:10.2307/2333696.↩︎

18. 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.↩︎