Bioequivalence Tests for Parallel Trial Designs with Log-Normal Data

In the following examples, we demonstrate the use of SimTOST for parallel trial designs with data assumed to follow a normal distribution on the log scale. We start by loading the package.

library(SimTOST)

Multiple Independent Co-Primary Endpoints

We here consider a bio-equivalence trial with 2 treatment arms and $m=5$ endpoints. The sample size is calculated to ensure that the test and reference products are equivalent with respect to all 5 endpoints. The true ratio between the test and reference products is assumed to be 1.05. It is assumed that the standard deviation of the log-transformed response variable is $\sigma = 0.3$ , and that all tests are independent ( $\rho = 0$ ). The equivalence limits are set at 0.80 and 1.25. The significance level is 0.05. The sample size is determined at a power of 0.8.

This example is adapted from Mielke et al. (2018), who employed a difference-of-means test on the log scale. The sample size calculation can be conducted using two approaches, both of which are illustrated below.

Approach 1: Using sampleSize_Mielke

In the first approach, we calculate the required sample size for 80% power using the sampleSize_Mielke() function. This method directly follows the approach described in Mielke et al. (2018), assuming a difference-of-means test on the log-transformed scale with specified parameters.

ssMielke <- sampleSize_Mielke(power = 0.8, Nmax = 1000, m = 5, k = 5, rho = 0, 
                              sigma = 0.3, true.diff = log(1.05), 
                              equi.tol = log(1.25), design = "parallel", 
                              alpha = 0.05, adjust = "no", seed = 1234, 
                              nsim = 10000)
ssMielke
#> power.a      SS 
#>  0.8196 68.0000

For 80% power, 68 subjects per sequence (136 in total) would have been required.

Approach 2: Using sampleSize

Alternatively, the sample size calculation can be performed using the sampleSize() function. This method assumes that effect sizes are normally distributed on the log scale and uses a difference-of-means test (ctype = "DOM") with user-specified values for mu_list and sigma_list. Unlike the first approach, this method allows for greater flexibility in specifying parameter distributions.

mu_r <- setNames(rep(log(1.00), 5), paste0("y", 1:5))
mu_t <- setNames(rep(log(1.05), 5), paste0("y", 1:5))
sigma <- setNames(rep(0.3, 5), paste0("y", 1:5))
lequi_lower <- setNames(rep(log(0.8), 5), paste0("y", 1:5))
lequi_upper <- setNames(rep(log(1.25), 5), paste0("y", 1:5))

ss <- sampleSize(power = 0.8, alpha = 0.05,
                 mu_list = list("R" = mu_r, "T" = mu_t),
                 sigma_list = list("R" = sigma, "T" = sigma),
                 list_comparator = list("T_vs_R" = c("R", "T")),
                 list_lequi.tol = list("T_vs_R" = lequi_lower),
                 list_uequi.tol = list("T_vs_R" = lequi_upper),
                 dtype = "parallel", ctype = "DOM", lognorm = FALSE, 
                 adjust = "no", ncores = 1, nsim = 10000, seed = 1234)
ss
#> Given a 80%  target power with 100(1-2*0.05)% confidence level.
#> 
#> The total required sample size to achieve 80.1% power is 136 sample units.
#> 
#>  n_drop   n_R   n_T n_total  power power_LCI power_UCI
#>   <num> <num> <num>   <num>  <num>     <num>     <num>
#>       0    68    68     136 0.8011 0.7931107 0.8088566

For 80% power, a total of 136 would be required.

Multiple Correlated Co-Primary Endpoints

In the second example, we have $k=m=5$ , $\sigma = 0.3$ and $\rho = 0.8$ . This example is also adapted from Mielke et al. (2018), who employed a difference-of-means test on the log scale. The sample size calculation can again be conducted using two approaches, both of which are illustrated below.

Approach 1: Using sampleSize_Mielke

ssMielke <- sampleSize_Mielke(power = 0.8, Nmax = 1000, m = 5, k = 5, rho = 0.8, 
                              sigma = 0.3, true.diff = log(1.05), 
                              equi.tol = log(1.25), design = "parallel", 
                              alpha = 0.05, adjust = "no", seed = 1234, 
                              nsim = 10000)
ssMielke
#> power.a      SS 
#>  0.8031 52.0000

For 80% power, 52 subjects per sequence (104 in total) would be been required.

Approach 2: Using sampleSize

mu_r <- setNames(rep(log(1.00), 5), paste0("y", 1:5))
mu_t <- setNames(rep(log(1.05), 5), paste0("y", 1:5))
sigma <- setNames(rep(0.3, 5), paste0("y", 1:5))
lequi_lower <- setNames(rep(log(0.8), 5), paste0("y", 1:5))
lequi_upper <- setNames(rep(log(1.25), 5), paste0("y", 1:5))

ss <- sampleSize(power = 0.8, alpha = 0.05,
                 mu_list = list("R" = mu_r, "T" = mu_t),
                 sigma_list = list("R" = sigma, "T" = sigma),
                 rho = 0.8, # high correlation between the endpoints
                 list_comparator = list("T_vs_R" = c("R", "T")),
                 list_lequi.tol = list("T_vs_R" = lequi_lower),
                 list_uequi.tol = list("T_vs_R" = lequi_upper),
                 dtype = "parallel", ctype = "DOM", lognorm = FALSE, 
                 adjust = "no", ncores = 1, k = 5, nsim = 10000, seed = 1234)
ss
#> Given a 80%  target power with 100(1-2*0.05)% confidence level.
#> 
#> The total required sample size to achieve 80.5% power is 108 sample units.
#> 
#>  n_drop   n_R   n_T n_total  power power_LCI power_UCI
#>   <num> <num> <num>   <num>  <num>     <num>     <num>
#>       0    54    54     108 0.8052 0.7972703 0.8128938

References

Mielke, Johanna, Byron Jones, Bernd Jilma, and Franz König. 2018. “Sample Size for Multiple Hypothesis Testing in Biosimilar Development.” Statistics in Biopharmaceutical Research 10 (1): 39–49. https://doi.org/10.1080/19466315.2017.1371071.

Thomas Debray

November 18, 2024

Multiple Independent Co-Primary Endpoints

Approach 1: Using sampleSize_Mielke

Approach 2: Using sampleSize

Multiple Correlated Co-Primary Endpoints

Approach 1: Using sampleSize_Mielke

Approach 2: Using sampleSize

References