1 3 Statistical considerations

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					   Statistical considerations

                                 Alfredo García – Arieta, PhD




Training workshop: Training of BE assessors, Kiev, October 2009
                                                    Outline

             Basic statistical concepts on equivalence

             How to perform the statistical analysis of a
              2x2 cross-over bioequivalence study

             How to calculate the sample size of a 2x2
              cross-over bioequivalence study

             How to calculate the CV based on the 90% CI
              of a BE study



2|   Training workshop: Training of BE assessors, Kiev, October 2009
                     Basic statistical concepts




3|   Training workshop: Training of BE assessors, Kiev, October 2009
                                      Type of studies

 Superiority studies
     – A is better than B (A = active and B = placebo or gold-standard)
     – Conventional one-sided hypothesis test

 Equivalence studies
     – A is more or less like B (A = active and B = standard)
     – Two-sided interval hypothesis

 Non-inferiority studies
     – A is not worse than B (A = active and B = standard with
       adverse effects)
     – One-sided interval hypothesis


4|   Training workshop: Training of BE assessors, Kiev, October 2009
                                      Hypothesis test
 Conventional hypothesis test
 H0:  = 1              H1:   1                (in this case it is two-sided)
 If P<0,05 we can conclude that statistical significant difference exists
 If P≥0,05 we cannot conclude
     – With the available potency we cannot detect a difference
     – But it does not mean that the difference does not exist
     – And it does not mean that they are equivalent or equal

 We only have certainty when we reject the null hypothesis
     – In superiority trials: H1 is for existence of differences

 This conventional test is inadequate to conclude about “equalities”
     – In fact, it is impossible to conclude “equality”


5|   Training workshop: Training of BE assessors, Kiev, October 2009
            Null vs. Alternative hypothesis

 Fisher, R.A. The Design of Experiments, Oliver and Boyd,
  London, 1935

 “The null hypothesis is never proved or established, but is
  possibly disproved in the course of experimentation.
  Every experiment may be said to exist only in order to
  give the facts a chance of disproving the null hypothesis”

 Frequent mistake: The absence of statistical significance
  has been interpreted incorrectly as absence of clinically
  relevant differences.


6|   Training workshop: Training of BE assessors, Kiev, October 2009
                                           Equivalence

 We are interested in verifying (instead of rejecting) the null
  hypothesis of a conventional hypothesis test

 We have to redefine the alternative hypothesis as a range of values
  with an equivalent effect

 The differences within this range are considered clinically irrelevant

 Problem: it is very difficult to define the maximum difference without
  clinical relevance for the Cmax and AUC of each drug

 Solution: 20% based on a survey among physicians




7|   Training workshop: Training of BE assessors, Kiev, October 2009
Interval hypothesis or two one-sided tests
  Redefine the null hypothesis: How?
  Solution: It is like changing the null to the alternative hypothesis
   and vice versa.
  Alternative hypothesis test: Schuirmann, 1981
      – H01:  1             Ha1: 1<
      – H02:  2             Ha2: < 2.

  This is equivalent to:
      – H0:  1 or  2                    Ha: 1<<2

  It is called as an interval hypothesis because the equivalence
   hypothesis is in the alternative hypothesis and it is expressed as an
   interval


 8|   Training workshop: Training of BE assessors, Kiev, October 2009
Interval hypothesis or two one-sided tests

  The new alternative hypothesis is decided with a statistic
   that follows a distribution that can be approximated to a t-
   distribution

  To conclude bioequivalence a P value <0.05 has to be
   obtained in both one-sided tests

  The hypothesis tests do not give an idea of magnitude of
   equivalence (P<0001 vs. 90% CI: 0.95 – 1.05).

  That is why confidence intervals are preferred


 9|   Training workshop: Training of BE assessors, Kiev, October 2009
             Point estimate of the difference


                                     If T=R, d=T-R=0
                                     If T>R, d=T-R>0
                                     If T<R, d=T-R<0




        d<0                                     d=0                      d>0
        Negative effect                     No difference                Positive effect


10 |   Training workshop: Training of BE assessors, Kiev, October 2009
       Estimation with confidence intervals
               in a superiority trial

                                          It is not statistically significant!
                                       Because the CI includes the d=0 value


                                   Confidence interval 90% - 95%




        d<0                                     d=0                      d>0
        Negative effect                     No difference                Positive effect


11 |   Training workshop: Training of BE assessors, Kiev, October 2009
       Estimation with confidence intervals
               in a superiority trial

                                              It is statistically significant!
                                   Because the CI does not includes the d=0 value


                                   Confidence interval 90% - 95%



        d<0                                     d=0                      d>0
        Negative effect                     No difference                Positive effect


12 |   Training workshop: Training of BE assessors, Kiev, October 2009
       Estimation with confidence intervals
               in a superiority trial

                          It is statistically significant with P=0.05
                     Because the boundary of the CI touches the d=0 value


                                   Confidence interval 90% - 95%



        d<0                                     d=0                      d>0
        Negative effect                     No difference                Positive effect


13 |   Training workshop: Training of BE assessors, Kiev, October 2009
                                   Equivalence study



                                            Region of
                                             clinical
                                           equivalence


                                 -d                                      +d
        d<0                                     d=0                           d>0
        Negative effect                     No difference                     Positive effect


14 |   Training workshop: Training of BE assessors, Kiev, October 2009
                        Equivalence vs. difference

                              Region of clinical equivalence
Equivalent?                                                                     Different?
                    ?                                                          ?
          No                                                                      Yes
                    ?                                                          Yes
         Yes                                                                      Yes
                  Yes                                                          ?
         Yes                                                                      Yes
                    ?                                                          Yes
         No                                                                       Yes
                                  -d                                      +d
         d<0                                     d=0                            d>0
         Negative effect                     No difference                      Positive effect


 15 |   Training workshop: Training of BE assessors, Kiev, October 2009
                               Non-inferiority study

                      Inferiority limit
Inferior?
                                                                                   ?
                                   Yes
                                           ?
                                               No
                                                            No
                                                                         No
                                                                              No
                                                                                   No
                                 -d
        d<0                                     d=0                                    d>0
        Negative effect                     No difference                              Positive effect


16 |   Training workshop: Training of BE assessors, Kiev, October 2009
                               Superiority study (?)

                                                             Superiority limit
                                                                             Superior?
                                                                         ?
                                 No
                                         No
                                               No
                                                    No, not clinically and ? statistically
                                                           No, not clinically, but yes statistically
                                                                ?, but yes statistically
                                                                      Yes, statistical & clinically
                                                                                   Yes, but only the
                                                            +d
                                                                                    point estimate
        d<0                                     d=0                     d>0
        Negative effect                     No difference               Positive effect


17 |   Training workshop: Training of BE assessors, Kiev, October 2009
How to perform the statistical analysis of a
  2x2 cross-over bioequivalence study




 18 |   Training workshop: Training of BE assessors, Kiev, October 2009
                     Statistical Analysis of BE studies


               Sponsors have to use validated software
                     – E.g. SAS, SPSS, Winnonlin, etc.

               In the past, it was possible to find statistical
                analyses performed with incorrect software.
                     – Calculations based on arithmetic means, instead of
                       Least Square Means, give biased results in
                       unbalanced studies
                        • Unbalance: different number of subjects in each sequence
                     – Calculations for replicate designs are more
                       complex and prone to mistakes


19 |   Training workshop: Training of BE assessors, Kiev, October 2009
The statistical analysis is not so complex

          2x2 BE trial                                    Period 1       Period 2

                N=12


       Sequence 1 (BA)                                         Y11         Y12

             BA is RT


       Sequence 2 (AB)                                        Y21          Y22

             AB is TR



20 |   Training workshop: Training of BE assessors, Kiev, October 2009
We don’t need to calculate an ANOVA table


 Sources of variation                   d. f.                  SS          MS       F           P

 Inter-subject                                   23            16487,49    716,85       4,286
   Carry-over                                     1              276,00    276,00       0,375   0,5468
   Residual / subjects                           22            16211,49    736,89       4,406   0,0005

 Intra-subject                                                  3778,19
   Formulation                                    1               62,79     62,79       0,375   0,5463
   Period                                         1               35,97     35,97       0,215   0,6474
   Residual                                      22             3679,43    167,25

 Total                                           47            20265,68




  21 |   Training workshop: Training of BE assessors, Kiev, October 2009
                           With complex formulae


                                                                                 
                                                      2       2      nk           2

                            SSTotal   Yijk  Y···
                                                    k 1 j 1 i 1



                                                                                     
                                                        2       2        nk               2

                            SSWithin   Yijk  Yi ·k
                                                      k 1 j 1 i 1



                                                                                 
                                                               2         nk           2

                            SS Between  2 Yi ·k  Y···
                                                             k 1 i 1


22 |   Training workshop: Training of BE assessors, Kiev, October 2009
                          More complex formulae

SS Between  SSCarry  SSint er

SSCarry                
                          2n1n2
                                 Y·12  Y·22   Y·11  Y·21 2

                         n1  n2
                               2        nk            2                  2     2
                                Y        Y
SS Inter                                       i ·k                     ·· k

                       k 1 i 1 2  k 1 2nk


23 |   Training workshop: Training of BE assessors, Kiev, October 2009
                  And really complex formulae

       SSWithin  SS Drug  SS Period  SS Intra
                                                                                                2
                                           1                                 
                                            Y·21  Y·11   Y·22  Y·12 
                          2n1n2
       SS Drug         
                         n1  n2           2                                 
                                                                                                    2
                                              1                                 
                                               Y·21  Y·11   Y·12  Y·22 
                            2n1n2
       SS Period         
                           n1  n2            2                                 
                             2     2     nk
                                                  Y          2     nk2     2
                                                                         Y··2   2   2   Y· 2
                        Yijk                              k
                                       2                                 i ·k              jk
       SS Intra
                        k 1 j 1 i 1   k 1 i 1 2  k 1 j 1 nk  k 1 2nk



24 |   Training workshop: Training of BE assessors, Kiev, October 2009
       Given the following data, it is simple

          2x2 BE trial                                    Period 1         Period 2

                N=12


       Sequence 1 (BA)                                         Y11           Y12

                                             75, 95, 90, 80, 70, 85 70, 90, 95, 70, 60, 70


       Sequence 2 (AB)                                        Y21            Y22

                                             75, 85, 80, 90, 50, 65 40, 50, 70, 80, 70, 95



25 |   Training workshop: Training of BE assessors, Kiev, October 2009
                   First, log-transform the data

          2x2 BE trial                                    Period 1          Period 2

                N=12


       Sequence 1 (BA)                                   Y11                  Y12
                                                   4.3175, 4.5539,       4.2485, 4.4998,
                                                   4.4998, 4.3820,       4.5539, 4.2485,
                                                   4.2485, 4.4427        4.0943, 4.2485
       Sequence 2 (AB)                                  Y21                   Y22
                                                   4.3175, 4.4427,       3.6889, 3,9120,
                                                   4.3820, 4,4998,       4,2485, 4.3820,
                                                   3,9120, 4.1744        4.2485, 4.5539


26 |   Training workshop: Training of BE assessors, Kiev, October 2009
Second, calculate the arithmetic mean of
      each period and sequence
          2x2 BE trial                                    Period 1        Period 2

                N=12


       Sequence 1 (BA)                                 Y11 = 4.407       Y12 = 4.316




       Sequence 2 (AB)                                 Y21 = 4.288       Y22 = 4,172




27 |   Training workshop: Training of BE assessors, Kiev, October 2009
 Note the difference between Arithmetic
     Mean and Least Square Mean
 The arithmetic mean (AM) of T (or R) is the mean of all
  observations with T (or R) irrespective of its group or
  sequence
       – All observations have the same weight

 The LSM of T (or R) is the mean of the two sequence by
  period means
       – In case of balanced studies AM = LSM
       – In case of unbalanced studies observations in sequences with
         less subjects have more weight
       – In case of a large unbalance between sequences due to drop-
         outs or withdrawals the bias of the AM is notable


28 |   Training workshop: Training of BE assessors, Kiev, October 2009
       Third, calculate the LSM of T and R

          2x2 BE trial                                    Period 1        Period 2

                N=12


       Sequence 1 (BA)                                 Y11 = 4.407       Y12 = 4.316

                                               B = 4.2898                    A = 4.3018


       Sequence 2 (AB)                                 Y21 = 4.288       Y22 = 4,172




29 |   Training workshop: Training of BE assessors, Kiev, October 2009
       Fourth, calculate the point estimate

 F = LSM Test (A) – LSM Reference (B)

 F = 4.30183 – 4.28985 = 0.01198

 Fifth step! Back-transform to the original scale

 Point estimate = eF = e0.01198 = 1.01205

 Five very simple steps to calculate the point estimate!!!




30 |   Training workshop: Training of BE assessors, Kiev, October 2009
Now we need to calculate the variability!

 Step 1: Calculate the difference between periods for each subject
  and divide it by 2: (P2-P1)/2

 Step 2: Calculate the mean of these differences within each
  sequence to obtain 2 means: d1 and d2

 Step 3:Calculate the difference between “the difference in each
  subject” and “its corresponding sequence mean”. And square it.

 Step 4: Sum these squared differences

 Step 5: Divide it by (n1+n2-2), where n1 and n2 is the number of
  subjects in each sequence. In this example 6+6-2 = 10
       – This value multiplied by 2 is the MSE
       – CV (%) = 100 x √eMSE-1


31 |   Training workshop: Training of BE assessors, Kiev, October 2009
This can be done easily in a spreadsheet!
         PERIOD
      I             II          Step 1          Step 1        Step 3        Step 3                   Step 4
      R             T           P2-P1         (P2-P1)/2     d - mean d     squared      Sum =     0,23114064
 4,31748811    4,24849524    -0,06899287    -0,03449644    0,01140614    0,0001301
 4,55387689    4,49980967    -0,05406722    -0,02703361    0,01886897    0,00035604                  Step 5
 4,49980967    4,55387689    0,05406722      0,02703361    0,07293619    0,00531969   Sigma2(d) = 0,02311406
 4,38202663    4,24849524    -0,13353139     -0,0667657    -0,02086312   0,00043527      MSE=     0,04622813
 4,24849524    4,09434456    -0,15415068    -0,07707534    -0,03117276   0,00097174      CV =     21,7516218
 4,44265126    4,24849524    -0,19415601    -0,09707801    -0,05117543   0,00261892

   Step 2       Mean d1 =    -0,09180516 -0,04590258
                  n1 =            6

      T              R
   4,3175         3,6889     -0,62860866    -0,31430433    -0,25642187   0,06575218
   4,4427         3,9120     -0,53062825    -0,26531413    -0,20743167   0,0430279
   4,3820         4,2485     -0,13353139     -0,0667657    -0,00888324   7,8912E-05
   4,4998         4,3820     -0,11778304    -0,05889152    -0,00100906   1,0182E-06
   3,9120         4,2485     0,33647224      0,16823612    0,22611858    0,05112961
   4,1744         4,5539     0,37948962      0,18974481    0,24762727    0,06131926

   Step 2       Mean d2 =    -0,11576491 -0,05788246
                  n2 =            6




32 |   Training workshop: Training of BE assessors, Kiev, October 2009
Step 1: Calculate the difference between periods
  for each subject and divide it by 2: (P2-P1)/2
                                                 PERIOD
                                            I               II            Step 1        Step 1
                                            R               T             P2-P1       (P2-P1)/2
                                       4,31748811      4,24849524     -0,06899287   -0,03449644
                                       4,55387689      4,49980967     -0,05406722   -0,02703361
                                       4,49980967      4,55387689      0,05406722    0,02703361
                                       4,38202663      4,24849524     -0,13353139    -0,0667657
                                       4,24849524      4,09434456     -0,15415068   -0,07707534
                                       4,44265126      4,24849524     -0,19415601   -0,09707801

                                          Step 2        Mean d1 =     -0,09180516 -0,04590258
                                                          n1 =             6

                                             T               R
                                          4,3175          3,6889      -0,62860866   -0,31430433
                                          4,4427          3,9120      -0,53062825   -0,26531413
                                          4,3820          4,2485      -0,13353139    -0,0667657
                                          4,4998          4,3820      -0,11778304   -0,05889152
                                          3,9120          4,2485       0,33647224    0,16823612
                                          4,1744          4,5539       0,37948962    0,18974481

                                          Step 2        Mean d2 =     -0,11576491 -0,05788246
                                                          n2 =             6



  33 |   Training workshop: Training of BE assessors, Kiev, October 2009
 Step 2: Calculate the mean of these differences
within each sequence to obtain 2 means: d1 & d2
                                                 PERIOD
                                            I               II            Step 1        Step 1
                                            R               T             P2-P1       (P2-P1)/2
                                       4,31748811      4,24849524     -0,06899287   -0,03449644
                                       4,55387689      4,49980967     -0,05406722   -0,02703361
                                       4,49980967      4,55387689      0,05406722    0,02703361
                                       4,38202663      4,24849524     -0,13353139    -0,0667657
                                       4,24849524      4,09434456     -0,15415068   -0,07707534
                                       4,44265126      4,24849524     -0,19415601   -0,09707801

                                          Step 2        Mean d1 =     -0,09180516 -0,04590258
                                                          n1 =             6

                                             T               R
                                          4,3175          3,6889      -0,62860866   -0,31430433
                                          4,4427          3,9120      -0,53062825   -0,26531413
                                          4,3820          4,2485      -0,13353139    -0,0667657
                                          4,4998          4,3820      -0,11778304   -0,05889152
                                          3,9120          4,2485       0,33647224    0,16823612
                                          4,1744          4,5539       0,37948962    0,18974481

                                          Step 2        Mean d2 =     -0,11576491 -0,05788246
                                                          n2 =             6



  34 |   Training workshop: Training of BE assessors, Kiev, October 2009
                   Step 3: Squared differences
                           PERIOD
                       I              II             Step 1        Step 1        Step 3        Step 3
                       R              T              P2-P1        (P2-P1)/2    d - mean d     squared
                  4,31748811     4,24849524      -0,06899287    -0,03449644   0,01140614    0,0001301
                  4,55387689     4,49980967      -0,05406722    -0,02703361   0,01886897    0,00035604
                  4,49980967     4,55387689       0,05406722    0,02703361    0,07293619    0,00531969
                  4,38202663     4,24849524      -0,13353139     -0,0667657   -0,02086312   0,00043527
                  4,24849524     4,09434456      -0,15415068    -0,07707534   -0,03117276   0,00097174
                  4,44265126     4,24849524      -0,19415601    -0,09707801   -0,05117543   0,00261892

                    Step 2        Mean d1 =      -0,09180516 -0,04590258
                                    n1 =              6

                       T               R
                    4,3175          3,6889       -0,62860866    -0,31430433   -0,25642187   0,06575218
                    4,4427          3,9120       -0,53062825    -0,26531413   -0,20743167   0,0430279
                    4,3820          4,2485       -0,13353139     -0,0667657   -0,00888324   7,8912E-05
                    4,4998          4,3820       -0,11778304    -0,05889152   -0,00100906   1,0182E-06
                    3,9120          4,2485        0,33647224    0,16823612    0,22611858    0,05112961
                    4,1744          4,5539        0,37948962    0,18974481    0,24762727    0,06131926

                    Step 2        Mean d2 =      -0,11576491 -0,05788246
                                    n2 =              6



35 |   Training workshop: Training of BE assessors, Kiev, October 2009
 Step 4: Sum these squared differences
                             Step 3                                            Step 4
                            squared                                 Sum =   0,23114064
                          0,0001301
                          0,00035604                                           Step 5
                          0,00531969                            Sigma2(d) = 0,02311406
                          0,00043527                               MSE=     0,04622813
                          0,00097174                               CV =     21,7516218
                          0,00261892




                          0,06575218
                          0,0430279
                          7,8912E-05
                          1,0182E-06
                          0,05112961
                          0,06131926


36 |   Training workshop: Training of BE assessors, Kiev, October 2009
          Step 5: Divide the sum by n1+n2-2
                         Step 3                                             Step 4
                        squared                                Sum =     0,23114064
                      0,0001301
                      0,00035604                                          Step 5
                      0,00531969                           Sigma2(d) = 0,02311406
                      0,00043527                              MSE=     0,04622813
                      0,00097174                              CV =     21,7516218
                      0,00261892




                      0,06575218
                      0,0430279
                      7,8912E-05
                      1,0182E-06
                      0,05112961
                      0,06131926


37 |   Training workshop: Training of BE assessors, Kiev, October 2009
   Calculate the confidence interval with
       point estimate and variability
 Step 11: In log-scale

 90% CI: F ± t(0.1, n1+n2-2)-√((Sigma2(d) x (1/n1+1/n2))

 F has been calculated before

 The t value is obtained in t-Studient tables with 0,1 alpha
  and n1+n2-2 degrees of freedom
       – Or in MS Excel with the formula =DISTR.T.INV(0.1; n1+n2-2)

 Sigma2(d) has been calculated before.


38 |   Training workshop: Training of BE assessors, Kiev, October 2009
                  Final calculation: the 90% CI

 Log-scale 90% CI: F±t(0.1, n1+n2-2)-√((Sigma2(d)·(1/n1+1/n2))

 F = 0.01198

 t(0.1, n1+n2-2) = 1.8124611

 Sigma2(d) = 0.02311406

 90% CI: LL = -0.14711 to UL= 0,17107

 Step 12: Back transform the limits with eLL and eUL

 eLL = e-0.14711 = 0.8632 and eUL = e0.17107 = 1.1866

39 |   Training workshop: Training of BE assessors, Kiev, October 2009
How to calculate the sample size of a 2x2
   cross-over bioequivalence study




40 |   Training workshop: Training of BE assessors, Kiev, October 2009
Reasons for a correct calculation of the
             sample size
 Too many subjects
       – It is unethical to disturb more subjects than necessary
       – Some subjects at risk and they are not necessary
       – It is an unnecessary waste of some resources ($)

 Too few subjects
       – A study unable to reach its objective is unethical
       – All subjects at risk for nothing
       – All resources ($) is wasted when the study is inconclusive




41 |   Training workshop: Training of BE assessors, Kiev, October 2009
                                   Frequent mistakes

 To calculate the sample size required to detect a 20%
  difference assuming that treatments are e.g. equal
       – Pocock, Clinical Trials, 1983

 To use calculation based on data without log-
  transformation
       – Design and Analysis of Bioavailability and Bioequivalence
         Studies, Chow & Liu, 1992 (1st edition) and 2000 (2nd edition)

 Too many extra subjects. Usually no need of more than
  10%. Depends on tolerability
       – 10% proposed by Patterson et al, Eur J Clin Pharmacol 57:
         663-670 (2001)

42 |   Training workshop: Training of BE assessors, Kiev, October 2009
       Methods to calculate the sample size

 Exact value has to be obtained with power curves

 Approximate values are obtained based on formulae
        – Best approximation: iterative process (t-test)
        – Acceptable approximation: based on Normal distribution

 Calculations are different when we assume products are
  really equal and when we assume products are slightly
  different

 Any minor deviation is masked by extra subjects to be
  included to compensate drop-outs and withdrawals (10%)


43 |   Training workshop: Training of BE assessors, Kiev, October 2009
                      Calculation assuming that
                        treatments are equal

            2  s  Z1b                         Z1a 
                                                                                                  
                      2                                              2
                                             2
N
                      w
                                                                           s  Ln 1  CV
                                                                             2                 2

                             Ln1.25               2                        w
                                                                         CV expressed as 0.3 for 30%



 Z(1-(b/2)) = DISTR.NORM.ESTAND.INV(0.05) for 90% 1-b

 Z(1-(b/2)) = DISTR.NORM.ESTAND.INV(0.1) for 80% 1-b

 Z(1-a) = DISTR.NORM.ESTAND.INV(0.05) for 5% a

44 |   Training workshop: Training of BE assessors, Kiev, October 2009
   Example of calculation assuming that
          treatments are equal
 If we desire a 80% power, Z(1-(b/2)) = -1.281551566

 Consumer risk always 5%, Z(1-a) = -1.644853627

 The equation becomes: N = 343.977655 x S2

 Given a CV of 30%, S2 = 0,086177696

 Then N = 29,64

 We have to round up to the next pair number: 30

 Plus e.g. 4 extra subject in case of drop-outs

45 |   Training workshop: Training of BE assessors, Kiev, October 2009
   Example of calculation assuming that
          treatments are equal
 If we desire a 90% power, Z(1-(b/2)) = -1.644853627

 Consumer risk always 5%, Z(1-a) = -1.644853627

 The equation becomes: N = 434.686167 x S2

 Given a CV of 25%, S2 = 0,06062462

 Then N = 26,35

 We have to round up to the next pair number: 28

 Plus e.g. 4 extra subject in case of drop-outs

46 |   Training workshop: Training of BE assessors, Kiev, October 2009
                      Calculation assuming that
                       treatments are not equal

                   2  s  Z1 b  Z1a 
                           2                                2

   N                      w
                                                                         T  R  1
             LnT              R   Ln1.25                 2




 Z(1-b) = DISTR.NORM.ESTAND.INV(0.1) for 90% 1-b

 Z(1-b) = DISTR.NORM.ESTAND.INV(0.2) for 80% 1-b

 Z(1-a) = DISTR.NORM.ESTAND.INV(0.05) for 5% a

47 |   Training workshop: Training of BE assessors, Kiev, October 2009
   Example of calculation assuming that
       treatments are 5% different
 If we desire a 90% power, Z(1-b) = -1.28155157

 Consumer risk always 5%, Z(1-a) = -1.644853627

 If we assume that T/R=1.05

 The equation becomes: N = 563.427623 x S2

 Given a CV of 40 %, S2 = 0,14842001

 Then N = 83.62

 We have to round up to the next pair number: 84

 Plus e.g. 8 extra subject in case of drop-outs


48 |   Training workshop: Training of BE assessors, Kiev, October 2009
   Example of calculation assuming that
       treatments are 5% different
 If we desire a 80% power, Z(1-b) = -0.84162123

 Consumer risk always 5%, Z(1-a) = -1.644853627

 If we assume that T/R=1.05

 The equation becomes: N = 406.75918 x S2

 Given a CV of 20 %, S2 = 0,03922071

 Then N = 15.95

 We have to round up to the next pair number: 16

 Plus e.g. 2 extra subject in case of drop-outs


49 |   Training workshop: Training of BE assessors, Kiev, October 2009
   Example of calculation assuming that
      treatments are 10% different
 If we desire a 80% power, Z(1-b) = -0.84162123

 Consumer risk always 5%, Z(1-a) = -1.644853627

 If we assume that T/R=1.11

 The equation becomes: N = 876.366247 x S2

 Given a CV of 20 %, S2 = 0,03922071

 Then N = 34.37

 We have to round up to the next pair number: 36

 Plus e.g. 4 extra subject in case of drop-outs


50 |   Training workshop: Training of BE assessors, Kiev, October 2009
            How to calculate the CV
        based on the 90% CI of a BE study




51 |   Training workshop: Training of BE assessors, Kiev, October 2009
                    Example of calculation of
                   the CV based on the 90% CI
 Given a 90% CI: 82.46 to 111.99 in BE study with N=24

 Log-transform the 90% CI: 4.4123 to 4.7184

 The mean of these extremes is the point estimate: 4.5654

 Back-transform to the original scale e4.5654 = 96.08

 The width in log-scale is 4.7184 – 4.5654 = 0,1530

 With the sample size calculate the t-value. How?
       – Based on the Student-t test tables or a computer (MS Excel)


52 |   Training workshop: Training of BE assessors, Kiev, October 2009
                    Example of calculation of
                   the CV based on the 90% CI
 Given a N = 24, the degrees of freedom are 22

 t = DISTR.T.INV(0.1;n-2) = 1.7171

 Standard error of the difference (SE(d)) = Width / t-value =
  0.1530 / 1.7171 = 0,0891

 Square it: 0.08912 = 0,0079 and divide it by 2 = 0,0040

 Multiply it by the sample size: 0.0040x24 = 0,0953 = MSE

 CV (%) = 100 x √(eMSE-1) = 100 x √(e0.0953-1) = 31,63 %

53 |   Training workshop: Training of BE assessors, Kiev, October 2009

				
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