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 version 4.1.3 (2022-03-10) and later.
All scripts were run on a Xeon E3-1245v3 @ 3.40GHz (1/4 cores) 16GB RAM with R 4.2.0 on Windows 7 build 7601, Service Pack 1, Universal C Runtime 10.0.10240.16390.

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 log-normal 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 \(\small{\alpha}\) (in BE commonly 0.05),
2. the BE-margins (commonly 0.80 – 1.25),
3. the desired (or target) power \(\small{\pi}\),
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.

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.⁵

Since generally the within-subject variability is lower than the between-subject variability, crossover studies are so popular. The efficiency of a crossover study compared to a parallel study is given by \(\small{\frac{\sigma_\textrm{intra}^2\;+\,\sigma_\textrm{inter}^2}{0.5\,\times\,\sigma_\textrm{intra}^2}}\). If, say, \(\small{\sigma_\textrm{intra}^2=0.5\times\sigma_\textrm{inter}^2}\) in a paralled study we need six times as many subjects than in a crossover to obtain the same power. On the other hand, in a crossover we have two measurements per subject, which makes the parallel study approximately three times more costly.

Of note, there is no relationship between \(\small{CV_\textrm{intra}}\) and \(\small{CV_\textrm{inter}}\).
An example are drugs which are subjected to polymorphic metabolism. For these drugs \(\small{CV_\textrm{intra}\ll CV_\textrm{inter}}\). On the other hand, some HVD(P)s show \(\small{CV_\textrm{intra}>CV_\textrm{inter}}\).

However, it is a prerequisite that no carryover from one period to the next exists. Only then the comparison of treatments will be unbiased. For details see another article.⁶

Subjects have to be in the same physiological state⁷ throughout the study – guaranteed by sufficiently long wash­out phases. Crossover studies can not only be performed in healthy volunteers but also in patients with a stable disease (e.g., asthma). Studies in patients with an instable disease (e.g., in oncology) must be performed in a pa­ral­lel design (covered in another article).
If crossovers are not feasible (e.g., for drugs with a very long half life), studies could be performed in a parallel design as well.

previous section ↩︎

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 four-period full replicate study with two 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")

# 
# +++++++++++ Equivalence test - TOST +++++++++++
#             Sample size estimation
# -----------------------------------------------
# Study design: 2x2x4 (4 period full replicate) 
# log-transformed data (multiplicative model)
# 
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25 
# True ratio = 0.95,  CV = 0.45
# 
# Sample size (total)
#  n     power
# 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 named x).

x <- 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 x:

names(x)
# [1] "Design"         "alpha"          "CV"            
# [4] "theta0"         "theta1"         "theta2"        
# [7] "Sample size"    "Achieved power" "Target power"

Now we can access the elements of x by their names. Note that double square brackets [[…]] have to be used.

x[["Sample size"]]
# [1] 42
x[["Achieved power"]]
# [1] 0.818228

Although you could access the elements by the number of the column(s), I don’t recommend that, since in various other functions of PowerTOST these numbers are different and hence, difficult to remember. If you insist in accessing elements by column-numbers, use single square brackets […].

x[7:8]
#   Sample size Achieved power
# 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)

# Unbalanced design. n(i)=21/20 assumed.

# [1] 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 ↩︎

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, believing¹⁴ 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 \(\small{\chi^2}\)-distribution of its error variance \(\small{\sigma^2}\) 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)

# lower CL upper CL 
#   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 \(\small{\chi^2}\)-distribution is skewed to the right, it is more likely to face a higher CV than a lower one in the pivotal study.

If we plan the study based on 0.45, 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)

#      CV      power
#  0.3223 0.57315300
#  0.4500 0.19171300
#  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)
x               <- data.frame(n = m, CV = 0.45,
                              l = NA, u = NA)
for (i in 1:nrow(df)) {
  x[i, 3:4] <- CVCL(CV = 0.45, df = m[i] - 2,
                    side = "2-sided",
                    alpha = 0.05)
}
x[, 3:4]      <- signif(x[, 3:4], 4)
names(x)[3:4] <- c("lower CL", "upper CL")
print(x, row.names = FALSE)

#   n   CV lower CL upper CL
#  12 0.45   0.3069   0.8744
#  18 0.45   0.3282   0.7300
#  24 0.45   0.3415   0.6685
#  NA   NA       NA       NA
#  NA   NA       NA       NA
#  NA   NA       NA       NA
#  NA   NA       NA       NA

A Bayesian method is implemented in PowerTOST, which takes the uncertainty of estimates into account. We can explore the uncertainty of the CV or the T/R-ratio and of both.

m      <- 16   # sample size of pilot
CV     <- 0.45
theta0 <- 0.95
x      <- expsampleN.TOST(CV = CV, theta0 = theta0,
                          targetpower = 0.80,
                          design = "2x2x4",
                          prior.parm = list(
                            m = m, design = "2x2x4"),
                          prior.type = "CV", # uncertain CV
                          details = FALSE)

# 
# ++++++++++++ Equivalence test - TOST ++++++++++++
#        Sample size est. with uncertain CV
# -------------------------------------------------
# Study design:  2x2x4 replicate crossover 
# log-transformed data (multiplicative model)
# 
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25 
# Ratio = 0.95
# CV = 0.45 with 44 df
# 
# Sample size (ntotal)
#  n   exp. power
# 42   0.800179

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

What about an uncertain T/R-ratio?

m      <- 16
CV     <- 0.45
theta0 <- 0.95
x      <- expsampleN.TOST(CV = CV, theta0 = theta0,
                          targetpower = 0.80,
                          design = "2x2x4",
                          prior.parm = list(
                            m = m, design = "2x2x4"),
                          prior.type = "theta0", # uncertain T/R-ratio
                          details = FALSE)

# 
# ++++++++++++ Equivalence test - TOST ++++++++++++
#      Sample size est. with uncertain theta0
# -------------------------------------------------
# Study design:  2x2x4 replicate crossover 
# log-transformed data (multiplicative model)
# 
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25 
# Ratio = 0.95
# CV = 0.45
# 
# Sample size (ntotal)
#  n   exp. power
# 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)

# [1] 0.7932

Explore 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 = FALSE)
print(comp2, row.names = FALSE)

#      CV    power
#  0.4500 0.818228
#  0.4725 0.783684
#  theta0    power
#  0.9500 0.818228
#  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.

• 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 = FALSE)
print(comp2, row.names = FALSE)

#    CV theta0  base  n  power
#  0.45 0.9500  TRUE 42 0.8182
#  0.45 0.9524 FALSE 42 0.8270
#  0.45 1.0500 FALSE 42 0.8270
#  0.45 1.0530 FALSE 42 0.8182
#    CV theta0  base  n  power
#  0.45 0.9500 FALSE 40 0.7994
#  0.45 0.9524 FALSE 40 0.8083
#  0.45 1.0500  TRUE 40 0.8083
#  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 power and sensitivity analyses, which are covered in other articles.
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)[2] <- "relative change"
print(pa, plotit = FALSE)
print(comp, row.names = FALSE)

# Sample size plan ABE
#  Design alpha   CV theta0 theta1 theta2 Sample size
#   2x2x4  0.05 0.45   0.95    0.8   1.25          42
#  Achieved power
#        0.818228
# 
# Power analysis
# CV, theta0 and number of subjects leading to min. acceptable power of ~0.7:
#  CV= 0.5247, theta0= 0.9248
#  n = 32 (power= 0.7005)
# 
#  parameter relative change
#         CV         +16.60%
#     theta0          -2.66%
#          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
x      <- expsampleN.TOST(CV = CV, theta0 = theta0,
                          targetpower = 0.80,
                          design = "2x2x4",
                          prior.parm = list(
                            m = m, design = "2x2x4"),
                          prior.type = "both", # uncertain CV and T/R-ratio
                          details = FALSE)

# 
# ++++++++++++ Equivalence test - TOST ++++++++++++
#   Sample size est. with uncertain CV and theta0
# -------------------------------------------------
# Study design:  2x2x4 replicate crossover 
# log-transformed data (multiplicative model)
# 
# alpha = 0.05, target power = 0.8
# BE margins = 0.8 ... 1.25 
# Ratio = 0.95 with 44 df
# CV = 0.45 with 44 df
# 
# Sample size (ntotal)
#  n   exp. power
# 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.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.

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)

#  theta0  n    power sigma.u  n assurance
#    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 C_max and AUC_0–t are mandatory (in some jurisdictions like the FDA additionally AUC_0–∞).

We don’t have to worry about multiplicity issues (inflated Type I Error) since if all tests must pass at level \(\small{\alpha}\), we are protected by the 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.

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
x       <- data.frame(metric = metrics,
                      theta1 = theta1, theta2 = theta2,
                      CV = CV, theta0 = theta0, n = NA)
for (i in 1:nrow(x)) {
  x$n[i] <- sampleN.TOST(CV = CV[i],
                         theta0 = theta0[i],
                         theta1 = theta1[i],
                         theta2 = theta2[i],
                         targetpower = target,
                         design = "2x2x4",
                         print = FALSE)[["Sample size"]]
}
x$theta1 <- sprintf("%.4f", x$theta1)
x$theta2 <- sprintf("%.4f", x$theta2)
txt       <- paste0("Sample size based on ",
                    x$metric[x$n == max(x$n)], ".\n")
print(x, row.names = FALSE)
cat(txt)

#  metric theta1 theta2   CV theta0  n
#    Cmax 0.7500 1.3333 0.45   0.95 24
#    AUCt 0.8000 1.2500 0.35   1.04 24
#  AUCinf 0.8000 1.2500 0.37   1.06 32
# 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?

opt <- function(y) {
  power.TOST(theta0 = y, CV = x$CV[2],
             theta1 = theta1,
             theta2 = theta2,
             design = design,
             n = x$n[1]) - 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"
x       <- data.frame(metric = metrics, theta0 = theta0,
                      CV = CV, n = NA, power = NA)
for (i in 1:nrow(x)) {
  x[i, 4:5] <- sampleN.TOST(CV = CV[i],
                            theta0 = theta0,
                            theta1 = theta1,
                            theta2 = theta2,
                            targetpower = target,
                            design = design,
                            print = FALSE)[7:8]
}
x$power <- signif(x$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[2],
                 " and sample size ",
                 x$n[1], "\nachieved for theta0 ",
                 sprintf("%.4f", theta0s[1]), " or ",
                 sprintf("%.4f", theta0s[2]), ".\n")
print(x, row.names = FALSE)
cat(txt)

#  metric theta0   CV  n    power
#    Cmax   0.95 0.45 42 0.818228
#     AUC   0.95 0.30 20 0.820240
# Target power for AUC and sample size 42
# achieved for theta0 0.8959 or 1.1161.

That means, although we assumed for AUC the same T/R-ratio as for C_max – with the sample size of 42 required for C_max – for AUC it can be as low as ~0.90 or as high as ~1.12, 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 \(\small{\widehat{\lambda}_\textrm{z}}\) in some subjects and therefore, less data of AUC_0–∞ than of AUC_0–t). Again, it will not hurt because for the originally assumed T/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

metrics <- c("Cmax", "AUC")
CV      <- c(0.90, 0.30)
theta0  <- 0.95
target  <- 0.80
design  <- "2x2x4"
alpha   <- c(0.50, 0.05)
x       <- data.frame(metric = metrics, CV = CV, theta0 = theta0,
                      alpha = alpha, n = NA)
for (i in 1:nrow(x)) {
  x$n[i] <- sampleN.TOST(alpha = alpha[i],
                         CV = CV[i],
                         theta0 = theta0,
                         targetpower = target,
                         design = design,
                         print = FALSE)[["Sample size"]]
}
txt       <- paste0("Sample size based on ",
                     x$metric[x$n == max(x$n)], ".\n")
print(x, row.names = FALSE)
cat(txt)

#  metric  CV theta0 alpha  n
#    Cmax 0.9   0.95  0.50 22
#     AUC 0.3   0.95  0.05 20
# 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.

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 \[\begin{equation}\tag{7} \alpha_\textrm{adj}=\alpha\,/\,k \end{equation}\] where \(\small{k}\) is the number of simultaneous tests. The Type I Error \(\small{TIE}\) will always be controlled because \[\begin{equation}\tag{8} TIE=1-{(1-\alpha_\textrm{adj})}^k \end{equation}\] For \(\small{\alpha=0.05}\) and \(\small{k=2}\) we get \(\small{0.049375<\alpha}\).

Of course, all comparisons in the study should be done by \(\small{100\,(1-2\,\alpha_\textrm{adj})}\) 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")

# 
# +++++++++++ Equivalence test - TOST +++++++++++
#             Sample size estimation
# -----------------------------------------------
# Study design: 2x2x4 (4 period full replicate) 
# log-transformed data (multiplicative model)
# 
# alpha = 0.025, target power = 0.8
# BE margins = 0.8 ... 1.25 
# True ratio = 0.95,  CV = 0.45
# 
# Sample size (total)
#  n     power
# 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")

# 
# +++++++++++ Equivalence test - TOST +++++++++++
#             Sample size estimation
# -----------------------------------------------
# Study design: 2x2x4 (4 period full replicate) 
# log-transformed data (multiplicative model)
# 
# alpha = 0.01666667, target power = 0.8
# BE margins = 0.8 ... 1.25 
# True ratio = 0.95,  CV = 0.45
# 
# Sample size (total)
#  n     power
# 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.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, design = "2x2x4")

# 
# +++++++++++ Equivalence test - TOST +++++++++++
#             Sample size estimation
# -----------------------------------------------
# Study design: 2x2x4 (4 period full replicate) 
# log-transformed data (multiplicative model)
# 
# alpha = 0.05, target power = 0.8
# BE margins = 0.9 ... 1.111111 
# True ratio = 0.95,  CV = 0.125
# 
# Sample size (total)
#  n     power
# 34   0.807718

Doable.

sampleN.TOST(CV = 0.125, theta1 = 0.90, theta2 = 1.12,
             design = "2x2x4")

# 
# +++++++++++ Equivalence test - TOST +++++++++++
#             Sample size estimation
# -----------------------------------------------
# Study design: 2x2x4 (4 period full replicate) 
# log-transformed data (multiplicative model)
# 
# alpha = 0.05, target power = 0.8
# BE margins = 0.9 ... 1.12 
# True ratio = 0.95,  CV = 0.125
# 
# Sample size (total)
#  n     power
# 34   0.807718

target  <- 0.80
theta0  <- 0.975
CV.min  <- 0.075
CV.max  <- 0.20
CV.step <- 1000
CV      <- seq(CV.min, CV.max, length.out = CV.step)
res     <- data.frame(CV = CV, n.EMA = NA_integer_,
                      n.HC = NA_integer_, less = FALSE)
for (i in 1:nrow(res)) {
  res$n.EMA[i] <- sampleN.TOST(CV = CV[i], theta0 = theta0,
                        targetpower = target, design = "2x2x4",
                        theta1 = 0.90, theta2 = 1/0.90,
                        print = FALSE)[["Sample size"]]
  res$n.HC[i]  <- sampleN.TOST(CV = CV[i], theta0 = theta0,
                        targetpower = target, design = "2x2x4",
                        theta1 = 0.90, theta2 = 1.12,
                        print = FALSE)[["Sample size"]]
  if (res$n.HC[i] < res$n.EMA[i]) res$less[i] = TRUE
}
cat("target power                :", target,
    "\ntheta0                      :", theta0,
    "\nCV                          :", CV.min, "\u2013", CV.max,
    "\nSample size for HC < common :",
    sprintf("%.2f%%", 100 * sum(res$less) / CV.step),
    "of cases.\n")

# target power                : 0.8 
# theta0                      : 0.975 
# CV                          : 0.075 – 0.2 
# Sample size for HC < common : 9.40% of cases.

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")

# 
# +++++++++++ Equivalence test - TOST +++++++++++
#             Sample size estimation
# -----------------------------------------------
# Study design: 2x2x4 (4 period full replicate) 
# log-transformed data (multiplicative model)
# 
# alpha = 0.05, target power = 0.8
# BE margins = 0.9 ... 1.111111 
# True ratio = 0.975,  CV = 0.125
# 
# Sample size (total)
#  n     power
# 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 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")

# 
# +++++++++++ Equivalence test - TOST +++++++++++
#             Sample size estimation
# -----------------------------------------------
# Study design: 2x2x4 (4 period full replicate) 
# untransformed data (additive model)
# 
# alpha = 0.05, target power = 0.8
# BE margins = -15 ... 15 
# True diff. = -5,  CV = 20
# 
# Sample size (total)
#  n     power
# 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")

# 
# +++++++++++ Equivalence test - TOST +++++++++++
#             Sample size estimation
# -----------------------------------------------
# Study design: 2x2x4 (4 period full replicate) 
# untransformed data (additive model)
# 
# alpha = 0.05, target power = 0.8
# BE margins = -15 ... 15 
# True diff. = -5,  CV = 36
# 
# Sample size (total)
#  n     power
# 82   0.805692

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

#  design  bk                      name
#   2x2x3 1.5 2x2x3 replicate crossover
#   2x2x4 1.0 2x2x4 replicate crossover
#   2x4x4 1.0 2x4x4 replicate crossover
#   2x3x3 1.5 partial replicate (2x3x3)
#   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 ↩︎