Supplemental Examples For Illustrating Statistical Concepts Described in the VICH In Vivo 
Bioequivalence Draft Guidance GL52 
 
EXAMPLE: SAMPLE SIZE ESTIMATION: 
 
Scenario: The study will be conducted as a two-period, two-treatment, two-sequence crossover 
design where subjects receive a single dose of the test and reference products.  For the sake of 
this example, animals are sorted by identification  (ID) number and assigned to sequence 1 or 2 
completely at random.  The study design is as follows: 
 
Sequence 1: Period 1 = Test; Period 2 = Reference 
Sequence 2: Period 1 = Reference; Period 2 = Test 
 
To estimate the number of subjects needed in the study, a pilot crossover study was conducted.  
The expected ratio of the test and reference means = 1.05.  The anticipated within subject error 
= 15% Coefficient of variation (%CV).  The iterative equation used for estimating  the number 
of subjects is as follows: 
 
If = 1, then: n  >  [t(, 2n-2) + t(/2, 2n-2)]2 [CV/ ln1.25]2 
If 1< < 1.25, then: n  >  [t(, 2n-2) + t(, 2n-2)]2 [CV/ (ln 1.25  ln )]2 
If 0.8 < < 1, then: n  >  [t(, 2n-2) + t(, 2n-2)]2 [CV/ (ln 0.8  ln )]2 
 
Where 1- = the power of the study (80%);  = the Type 1 error for each side of the 90% 
confidence interval (= 0.05), n = the number of subjects per sequence (and the total number of 
subjects = N = 2n), and  = the anticipated ratio of the test/reference mean. 
 
Based upon this equation and the results of the pilot study, sample size estimation  procedure 
was as follows: 
 
If we begin our sample size estimation with n = 5 (N=10), the equation would be: 
 
5 > [1.860+0.889]2  * [0.15/(Ln 1.25- Ln 1.05)]2 = 5.59. 
 
Since this is not a true statement, we need to try the next highest value, n = 6 (N=12).  In this 
case, the calculation  is as follows: 
 
6 > [1.812+0.879]2  * [0.15/(Ln 1.25- Ln 1.05)]2 = 5.40. 
 
When we use n=6, the conditional  statement is now correct.  Therefore, our sample size 
estimate (the number of subjects per sequence) = 6 and the total number of subjects included in 
this study should be no less than 12. 
 
With N=12, the results of the simulated bioequivalence  trial are as follows:  EXAMPLE: BIOEQUIVALENCE DATA STATISTICAL ANALYSIS: 
 
Scenario: The study will be conducted as a two-period, two-treatment, two-sequence crossover 
design where subjects receive a single dose of the test and reference products.  For the sake of 
this example, subjects are sorted by identification  ID number and assigned to sequence 1 or 2 
completely at random.  The study design is as follows: 
 
Sequence 1: Period 1 = Test; Period 2 = Reference 
Sequence 2: Period 1 = Reference; Period 2 = Test 
 
With N=12, the results of the simulated trial were as follow (Table 1): 
 
Table 1: Dataset from a simulated bioequivalence  trial 
Animal  Sequence  Period  Treatment  Value 
1  1  2  Reference  86.76 
2  1  2  Reference  72.23 
3  1  2  Reference  102.10 
4  1  2  Reference  138.42 
5  1  2  Reference  120.67 
6  1  2  Reference  81.83 
7  2  1  Reference  84.91 
8  2  1  Reference  92.84 
9  2  1  Reference  114.42 
10  2  1  Reference  119.48 
11  2  1  Reference  95.32 
12  2  1  Reference  105.77 
1  1  1  Test  93.38 
2  1  1  Test  78.81 
3  1  1  Test  108.81 
4  1  1  Test  154.68 
5  1  1  Test  131.96 
6  1  1  Test  71.30 
7  2  2  Test  75.80 
8  2  2  Test  96.98 
9  2  2  Test  129.46 
10  2  2  Test  131.24 
11  2  2  Test  91.27 
12  2  2  Test  90.47 
 
Prior to analysis, all data were transformed to the natural logarithm.   The statistical model used 
in the analysis included sequence, period and treatment as fixed effects and animal-nested-
within-sequence as a random effect.  There are numerous statistical programs and program 
specifications that can be used.  All correctly programmed analyzed data should give the 
following  results (Tables 2 and 3). 
  Table 2: Test of fixed effects 
Effect  Numerator Degrees  Denominator  F-Value  Probability  of a 
of Freedom  Degrees of Freedom  Greater  F 
Sequence  1  10  <0.01  0.9527 
Period  1  10  0.89  0.3667 
Treatment  1  10  0.43  0.5274 
 
 
Table 3: Difference and confidence interval 
Difference  Standard Error  Lower 90% Limit  Upper 90% Limit 
0.1958  0.02991  0.0346  0.0738 
 
Using the statistical output information,  we calculated the confidence bounds: 
 
Lower BE Bound = Exp(-0.0346) = 0.97 
Upper BE Bound = Exp(0.0738) = 1.08 
 
If both the lower and upper BE bounds are between 0.80 and 1.25, bioequivalence  is 
established for that value.  
  EXAMPLE:   SEQUENTIAL ANALYSIS: 
 
The use of a sequential analysis allows for an opportunity  to recalculate sample size based 
upon the observed study variance by altering  to accommodate an interim analysis of the 
dataset.  However, it is important to note that to avoid inflation  of the Type I error, the 
sequential analysis does not allow for sample size adjustments based upon incorrect 
assumptions of the ratio of treatment means. 
 
There are several types of sequential designs that can be used for bioequivalence  studies.  The 
following  is only one possible example of how this analysis may be executed.  The primary 
reference used in the writing of this example is Potvin et al., Pharm Stat, 7:245-262. 
 
Method B:  Test for bioequivalence  at the  = 0.0294 level, regardless of power.  At Stage 1, 
re-calculate necessary sample size for the entire study.  Confidence intervals can be estimated 
at both Stages 1 and 2.  A schematic diagram of the steps in the Method B version of the 
sequential analysis (based upon Potvin et al., 2008) is provided in Figure 1. 
 
 
 
Figure 1: Schematic diagram of the steps involved in Method B version of the sequential 
analysis. 
 
Evaluate BE at Stage 1 using an  level of 0.0294, regardless of the power achieved.  If the BE 
criteria are met or if the study power is equal to or greater than 80%, then no additional 
subjects should be tested.  Conversely, if the BE criteria are not met, the sample size necessary 
to achieve 80% power should be calculated based on the information  derived at Stage 1.  At 
Stage 2, the confidence intervals are re-calculated at a level of  = 0.0294 using data generated 
at Stages 1 and 2. 
 
Bioequivalence  evaluation does not extend beyond Stage 2, regardless of the outcome. 
 
  SCENARIO: For the sake of this example, we will use the following  Stage 1 assumptions: 
 
  We estimate a residual error of 15% CV and a ratio of the test/reference means of 0.90. 
  We will conduct Stage 1 with 20 subjects (10 per sequence). 
 
At Stage 1, we did not meet the BE criteria of 0.80 to 1.25.  By having pre-planned the use of a 
sequential analysis (using Method B), we can combine the data generated with the original  20 
subjects with that obtained from the additional  subjects at Stage 2.  So, the first question is how 
many total subjects will be needed to demonstrate product BE in this situation and how many 
additional  subjects will need to be included  in our trial at Stage 2? 
 
To answer that question, we need to plug these observed values into the equation for sample 
size: 
 
SAMPLE SIZE EQUATION:  
 
If 0.8 < < 1, then: n > [t(, 2n-2) + t(, 2n-2)]2 [CV/ (ln 0.8  ln )]2 
 
Our first estimate is that N = 40 (20 per sequence).  In that case, our calculation is as follows 
(keeping in mind that  is now set at 0.0294);  
 
[t(, 2n-2)  = 1.948, t(, 2n-2) = 0.851,  
[t(, 2n-2) + t(, 2n-2)]2 = 7.834 
[CV/ (ln 0.8  ln )]2 = [0.2/(ln 0.8  ln 0.90)]2 = 2.883 
n > [t(, 2n-2) + t(, 2n-2)]2 [CV/ (ln 0.8  ln )]2  = 22.587  
 
We repeat our sample size calculation,  this time using n = 23.  If we plug the corresponding t 
values into the equation, the results again indicate an n of at least 23 subjects per sequence (i.e., 
convergence is achieved).  Based upon this outcome, we conclude that the total number of 
study subjects (N) needed to meet the BE criteria with 20% CV and  = 0.90 (at  = 0.0294) 
will be 46.  Since we already have Stage 1 data generated with N=20 (i.e., 10 subjects per 
sequence), Stage 2 will require an additional  13 subjects per sequence (N=26) based upon our 
revised estimates. 
 
For the sake of comparison,  if we had instead opted to do a pilot study prior to executing the 
pivotal BE trial, the total number of subjects (N) that would have been needed (at  = 0.05 
rather than =0.0294) with the current estimates of  = 0.90 and CV = 0.20 would have been 
38 rather than 46.