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Statistics in bioequivalence Didier Concordet NATIONAL VETERINARY S C H O O L T O U L O U S E May

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2 Statistics in bioequivalence may Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

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3 Statistics in bioequivalence may Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

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4 Parametric ? may A statistical property of the distribution of data All data are drawn from distribution that can be completely described by a finite number of parameters (refer to sufficiency) Example The ln AUC obtained in a dog for a formulation is a figure drawn from a N(m, ²) The parameters m, ² defined the distribution of AUC (its ln) that can be observed in this dog.

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5 Non parametric ? may The distribution of data is not defined by a finite number of parameters. It is defined by its shape, number of modes, regularity….. The number of parameters used to estimate the distribution with n data increases with n. Practically These distributions have no specific name. The goal of a statistical study is often to show that some distributions are/(are not) different. It suffice to show that a parameter that participate to the distribution description (eg the median) is not the same for the compared distributions.

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6 Parametric : normality may Usually, the data are assumed to be drawn from a (mixture) of gaussian distribution(s) up to a monotone transformation Example : The ln AUC obtained in a dog for a formulation is drawn from a N(3.5, 0.5²) distribution The monotone transformation is the logarithm The ln AUC obtained in another dog for the same formulation is drawn from a N(3.7, 0.5²) distribution The distribution of the data that are observable on these 2 dogs is a mixture of the N(3.5, 0.5²) and N(3.7, 0.5²) distributions

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7 Parametric methods may Methods designed to analyze data from parametric distributions Standard methods work with 3 assumptions (detailed after) homoscedasticity independence normality Practically for bioequivalence studies AUC and CMAX : parametric methods

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8 Non-parametric methods may Used when parametric methods cannot be used (e.g. heteroscedasticity) Usually less powerful than their parametric counterparts (it is more difficult to show bioeq. when it holds) Lie on assumptions on the shape, number of modes, regularity….. Practically for bioequivalence studies The distribution of (ln) TMAX is assumed to be symmetrical

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9 Statistics in bioequivalence may Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

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10 Statistics in bioequivalence may Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

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11 Transformations of parameters may Data Parametric methods Assumptions ? yes Transformation no yes Non parametric methods no

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12 Three fundamental assumptions may Homoscedasticity The variance of the dependent variable is constant ; it does not vary with independent variables : formulation, animal, period. Independence The random variables implied in the analysis are independent. Normality The random variables implied in the analysis are normally distributed

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13 Fundamental assumptions : homoscedasticity may Homoscedasticity The variance of the dependent variable is constant, does not vary with independent variables : formulation, animal, period. Example : Parallel group design, 2 groups, 10 dogs by group Group 1 : Reference Group 2 : Test AUC Ref Test Ln AUC Ref Test

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14 Fundamental assumptions : homoscedasticity may Homoscedasticity Maybe the most important assumption Analysis of variance is not robust to heteroscedasticity More or less easy to check in practice : - graphical inspection of data (residuals) - multiple comparisons of variance (Cochran, Bartlett, Hartley…). These tests are not very powerful Crucial for the bioequivalence problem : the width of the confidence interval mainly depends on the quality of estimation of the variance.

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15 Fundamental assumptions : Independence may Independence (important) The random variables implied in the analysis are independent. In a parallel group : the (observations obtained on) animals are independent. In a cross-over : the animals are independent. the difference of observations obtained in each animals with the different formulations are independent. In practice : Difficult to check Has to be assumed

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16 Fundamental assumptions : Normality may Normality The random variables implied in the analysis are normally distributed. In a parallel group : the observations of each formulation come from a gaussian distribution. In a cross-over : - the "animals" effect is assumed to be gaussian (we are working on a sample of animals) - the observations obtained in each animal for each formulation are assumed to be normally distributed.

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17 Fundamental assumptions : Normality may Normality Not important in practice when the sample size is large enough, the central limit theorem protects us when the sample size is small, the tests use to detect non normality are not powerful (they do not detect non normality) The analysis of variance is robust to non normality Difficult to check : - graphical inspection of the residuals : Pplot (probability plot) - Kolmogorov-Smirnov, Chi-Square test…

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18 In practice for bioequivalence may Log transformation AUC : to stabilise the variance to obtain a the symmetric distribution CMAX : to stabilise the variance to obtain a the symmetric distribution TMAX (sometimes) : to obtain a the symmetric distribution usually heteroscedasticity remains Without transformation TMAX (sometimes) usually heteroscedasticity

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19 The ln transformation : side effect may If m is the pop. mean of lnX is the pop. median of X After a logarithmic transformation bioequivalence methods compares the median (not the mean) of the parameters obtained with each formulation

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20 Statistics in bioequivalence may Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

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21 Statistics in bioequivalence may Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

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22 Parallel and Cross-over designs may parallel Test Ref. Sequence 1 2 Period Cross-over

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23 Parallel vs Cross-over design may Advantages Drawbacks Parallel Cross-over Easy to organise Easy to analyse Easy to interpret Comparison is carried-out between animals: not very powerful Comparison is carried-out within animals: powerful Difficult to organise Possible unequal carry-over Difficult to analyse

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24 Analysis of parallel and cross- over designs may To check whether or not the assumptions (especially homoscedasticity) hold To check there is no carry-over (cross-over design) To obtain a good estimate for the mean of each formulation the variance of interest between subjects for the parallel design within subject for the cross-over Why ? To assess bioequivalence (student t-test or Fisher test) NO

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25 Why ? may H 0 : | T - R | > H 1 : | T - R | Classical hypotheses for student t-test and Fisher test (ANOVA) T and R population mean for test and reference formulation respectively Hypotheses for the bioequivalence test H 0 : T = R H 1 : T R bioequivalence bioinequivalence

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26 Analysis of parallel designs may Step 1 : Check (at least graphically) homoscedasticity Step 2 : Estimate the mean for each formulation, estimate the between subjects variance. Transformation ?

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27 Example may Test Ref Test Ref Variances comparison : P = Heteroscedasticity

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28 Example on log transformed data may Test Ref Test Ref Variances comparison : P = 0.66 Homoscedasticity

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29 Example may Test Ref Pooled variance

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30 Another way to proceed : ANOVA may Write an ANOVA model to analyse data useless here but useful to understand cross-over Y ij = ln AUC for the i th animal that received formulation i Notations formulation 1 = Test, formulation 2 = Ref i = 1..2 ; j = Y ij = µ + F i + ij y 11 =4.37 µ = population mean F i = effect of the i th formulation ij = indep random effects assumed to be drawn from N(0, ²)

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31 Effects coding used for categorical variables in model. Categorical values encountered during processing are: FORMUL$ (2 levels) Ref, Test Dep Var: LN_AUC N: 20 Multiple R: Squared multiple R: Analysis of Variance Source Sum-of-Squares df Mean-Square F-ratio P FORMUL$ Error Least squares means LS Mean SE N FORMUL$ =Ref FORMUL$ =Test Another way to proceed : ANOVA may Does not give any information about bioeq

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32 Analysis of cross-over designs may Step 1 : Write the model to analyse the cross-over Step 3 : Check the absence of a carry-over effect Transformation ? Difficult to analyse by hand, especially when the experimental design is unbalanced. Need of a model to analyse data. Step 2 : Check (at least graphically) homoscedasticity Step 4 : Estimate the mean for each formulation, estimate the within (intra) subjects variance.

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33 A model for the 2 2 crossover design may Sequence 1 Test + Ref Sequence 2 Ref + Test AUC AUC ij,k(i,j),l = AUC for the l th animal of the seq. j when it received formulation i at period k(i,j) Notations formulation 1 = Test, formulation 2 = Ref i = 1..2 ; j = 1..,2 ; k(1,1) = 1 ; k(1,2) = 2 ; k(2,1) = 2 ; k(2,2) = 1 ; l=1..10

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34 A model for the 2 2 crossover design may Y 1,1,1,1 =78.8 µ = population mean F i = effect of the i th formulation S j = effect of the j th sequence P k(i,j) = effect of the k th period An l |S j = random effect of the l th animal of sequence j, they are assumed independent distrib according a N(0, ²) i,j,k,l = indep random effects assumed to be drawn from N(0, ²)

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35 Homoscedasticity ? may Seq. 1 Seq. 2 An l |S j = assumed independent distrib according a N(0, ²) In particular : Var(An|S 1 )=Var(An|S 2 ) Average AUC Sequence 1 Sequence 2 Comparison of interindividual variances P = Usually this test is not powerful

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36 Homoscedasticity ? may i,j,k,l = indep random effects assumed to be drawn from N(0, ²)

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37 After a ln tranformation... may Sequence 1 Test + Ref Sequence 2 Ref + Test ln AUC Comparison of interindividual variances P = Seq. 1 Seq. 2 Homoscedasticity seems reasonable

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38 ANOVA table may Effects coding used for categorical variables in model. Categorical values encountered during processing are: FORMUL$ (2 levels) Ref, Test PERIOD (2 levels) 1, 2 SEQUENCE (2 levels) 1, 2 ANIMAL (20 levels) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 Dep Var: LN_AUC N: 40 Multiple R: Squared multiple R: Analysis of Variance Source Sum-of-Squares df Mean-Square F-ratio P FORMUL$ E PERIOD SEQUENCE E ANIMAL(SEQUENCE) E E Error Least squares means LS Mean SE N FORMUL$ =Ref FORMUL$ =Test Does not give any information about bioeq

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39 Period effect may Does not invalidate a crossover design Does affect in the same way the 2 formulations Origin : environment, equal carry-over Period effect significant

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40 ANOVA table may Effects coding used for categorical variables in model. Categorical values encountered during processing are: FORMUL$ (2 levels) Ref, Test PERIOD (2 levels) 1, 2 SEQUENCE (2 levels) 1, 2 ANIMAL (20 levels) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 Dep Var: LN_AUC N: 40 Multiple R: Squared multiple R: Analysis of Variance Source Sum-of-Squares df Mean-Square F-ratio P FORMUL$ E PERIOD SEQUENCE E ANIMAL(SEQUENCE) E E Error Least squares means LS Mean SE N FORMUL$ =Ref FORMUL$ =Test Does not give any information about bioeq

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41 Sequence effect : carryover effect may Differential carryover effect significant ? For all statistical softwares, the only random variables of a model are the residuals The ANOVA table is built assuming that all other effects are fixed However We are working on a sample of animals Independent random variables

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42 Testing the carryover effect may Analysis of Variance Source Sum-of-Squares df Mean-Square F-ratio P FORMUL$ E PERIOD SEQUENCE E ANIMAL(SEQUENCE) E E Error The test for the carryover (sequence) effect has to be corrected Test for effect called: SEQUENCE Test of Hypothesis Source SS df MS F P Hypothesis Error E The good P value

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43 Testing the carryover effect may The test for a carryover effect should be declared significant when P<0.1 In the previous example P=0.12 : the carryover effect is not significant

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44 How to interpret the (differential) carryover effect ? may A carryover effect is the effect of the drug administrated at a previous period (pollution). In a 2 2 crossover, it is differential when it is not the same for the sequence TR and RT. A non differential carryover effect translates into a period effect It is confounded with the groups of animals consequently a poor randomisation can be wrongly interpreted as a carryover effect

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45 What to do if the carryover effect is significant ? may The kinetic parameters obtained in period 2 are unequally polluted by the treatment administrated at period 1. In a 2 2 crossover, it is not possible to estimate the pollution When the carryover effect is significant the data of period 2 should be discarded. In such a case, the design becomes a parallel group design.

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46 How to avoid a carryover effect ? may Its origin is a too short washout period The washout period should be taken long enough to ensure that no drug is present at the next period of the experiment

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47 ANOVA table may Effects coding used for categorical variables in model. Categorical values encountered during processing are: FORMUL$ (2 levels) Ref, Test PERIOD (2 levels) 1, 2 SEQUENCE (2 levels) 1, 2 ANIMAL (20 levels) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 Dep Var: LN_AUC N: 40 Multiple R: Squared multiple R: Analysis of Variance Source Sum-of-Squares df Mean-Square F-ratio P FORMUL$ E PERIOD SEQUENCE E ANIMAL(SEQUENCE) E E Error Least squares means LS Mean SE N FORMUL$ =Ref FORMUL$ =Test P Does not give any information about bioeq Inter animals variability

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48 Balance sheet may The fundamental assumptions hold There is no carryover (crossover design) Estimate the mean for each formulation, estimate the between (parallel) or within (crossover) subjects variance.

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49 Statistics in bioequivalence may Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

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50 Statistics in bioequivalence may Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

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51 Additive bioequivalence test of hypotheses may H 0 : T - R H 1 : T - R T and R population mean for test and reference formulation respectively Additive hypotheses for the bioequivalence test bioequivalence bioinequivalence [ 1 ; 2 ] Absolute equivalence interval

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52 Multiplicative bioequivalence test of hypotheses may H 0 : H 1 : T and R population median for test and reference formulation respectively Multiplicative hypotheses for the bioequivalence test bioequivalence bioinequivalence [ 1 ; 2 ] Relative equivalence interval where 0< 1 <1< 2 (eg [0.8 ; 1.25])

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53 Multiplicative bioequivalence test of hypotheses may Multiplicative hypotheses for the bioequivalence test bioequivalence bioinequivalence become additive after a ln transformation

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54 The two one-sided tests (Schuirman) may Additive hypotheses for the bioequivalence test bioequivalence H 0 : T - R H 1 : T - R H 0 : T - R < H 1 : T - R H 0 : T - R > H 1 : T - R First one-sided test second one-sided test Bioequivalence when the 2 tests reject H 0

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55 Decision rules for the two one-sided tests procedure may First one-sided test H 0 : T - R < H 1 : T - R Reject H 0 if Where is the consumer risk (risk to wrongly conclude to bioequivalence) df is the degree of freedom of the variance and are estimates of µ R and µ T respectively A = 1 for parallel 0.5 for 2 2 crossover

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56 Decision rules for the two one-sided tests procedure may Reject H 0 if H 0 : T - R > H 1 : T - R Second one-sided test Where is the consumer risk (risk to wrongly conclude to bioequivalence) df is the degree of freedom of the variance and are estimates of µ R and µ T respectively A = 1 for parallel 0.5 for 2 2 crossover

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57 Same procedure with confidence intervals may Build a 1-2 (90% for a consumer risk = 5%) confidence interval for T - R Conclude to bioequivalence (with a risk ) if this interval is totally included in the equivalence interval [ 1 ; 2 ] A = 1 for parallel 0.5 for 2 2 crossover

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58 Same procedure with confidence intervals may Build a 1- (95% for the drug company risk = 5%) confidence interval for T - R Conclude to bioinequivalence (with a risk ) if this interval has no common point with the equivalence interval [ 1 ; 2 ] A = 1 for parallel 0.5 for 2 2 crossover

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59 Confidence intervals : summary may Equivalence interval 90% CI Bioequivalence ( =5%) 1- CI Bioinequivalence ( ) 1- CI Bioinequivalence ( ) No conclusion

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60 An example may Sequence 1 Sequence 2 ln AUC Homoscedasticity seems reasonable No (differential) carryover effect n T =10 ; n R =10 ; df = n T + n R -2 = 18

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61 An example may and respectively estimate ln µ T and ln µ R 90 % confidence interval for ln µ T - ln µ R 90 % confidence interval for ln µ T - ln µ R = [-0.69 ; -0.44] is not totally included within the (ln transformed) equivalence interval = [ln 0.8 ; ln 1.25] = [ ; ], Cannot conclude to bioequivalence

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62 An example may % confidence interval for = [exp(-0.69) ;exp(-0.44)] =[0.50 ; 0.64] is not totally included within the equivalence interval = [0.8 ; 1.25] Cannot conclude to bioequivalence Conclude to bioinequivalence (risk<10%) Actually, the 90 % confidence interval has no common point with the equivalence interval

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63 What is implicitly assumed may The model can be written in an additive way via the ln transformation Assume that the question is formulated in a multiplicative way bioequivalence bioinequivalence It is implicitly assumed that the PK parameters (eg AUC) has to be ln transformed to meet the 3 fundamental assumptions This question translates in an additive way via the ln transformation

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64 What is implicitly assumed may What to do when the PK parameter (eg AUC) does meet the 3 fundamental assumptions without ln transformation ? and respectively estimate µ T and µ R but does not estimate Another method is needed to build the 90 % confidence interval of

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65 Confidence interval for µ T /µ R may and respectively estimate µ T and µ R estimate the between (parallel) or within (crossover) subjects variance Critical value of a student distribution with df degrees of freedom df degree of freedom for Solve the second degree equation A = 1 for parallel 0.5 for 2 2 crossover

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66 Second degree equation may The two solutions x 1, x 2 give the 90% confidence interval [x 1 ; x 2 ]

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67 Example : parallel groups design may AUC Test Ref Variances comparison : P = 0.62 Test Ref

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68 Example may n R =10 ; n T = 10 x1 =0.63 ; x 2 = 1.12 The 90% confidence interval of µ T /µ R is [0.63 ; 1.12] This interval is not totally included in the equivalence interval [0.8 ; 1.25] Cannot conclude to bioequivalence (lack of power ?)

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69 Statistics in bioequivalence may Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

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70 Statistics in bioequivalence may Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

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71 Sample size in bioequivalence trials may The sample size is only an issue for the drug company Small sample size unable to prove bioequivalence Sample size calculation useful to design the experiment

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72 The hypotheses to be tested The equivalence interval : [ 1, 2 ] The experimental design : parallel or crossover The consumer risk ( = 5%) risk to wrongly conclude to bioeq The drug company risk ( = 20% ?) risk to wrongly conclude to bioineq or of no conclusion A ln transformation will be required ? An estimate of (inter individual for parallel, intra for crossover) An idea about the true value of µ T /µ R (or µ T -µ R ) What one need to know to determine the sample size ? may additive multiplicative

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73 The hypotheses to be tested The equivalence interval : [, 1.25] The experimental design : crossover (2 2) with the same number of animals per sequence N The consumer risk ( = 5%) The drug company risk ( = 20%) A ln transformation is required An estimate of (intra for the log transformed data) An idea about the true value of µ T /µ R The most common situation may multiplicative

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74 The most common situation may An estimate of : intra for the log transformed data An idea about the true value of µ T /µ R It remains to know We have already seen that if then Different scenarios for CV and µ T /µ R can be simulated

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75 Sample size may Number of animal per sequence for a 2 2 crossover, log transformation, equivalence interval : [0.8, 1.25], =5%, = 20%

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76 More generally may The hypotheses to be tested The equivalence interval : [ 1, 2 ] The experimental design : crossover The consumer risk ( ) risk to wrongly conclude to bioeq The drug company risk ( ) risk to wrongly conclude to bioineq or of no conclusion A ln transformation is required An estimate of CV % An idea about the true value of µ T /µ R

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77 More generally may Iterative procedure; N = number of animals per sequence if µ T /µ R =1 if 1<µ T /µ R < 2 if 1 <µ T /µ R < D. Hauschke & coll. Sample size determination for bioequivalence assessment using a multiplicative model. J. Pharmacokin. Biopharm. 20: (1992) K.F. Phillips. Power of the two one-sided tests procedure in bioequivalence. J. Pharmacokin. Biopharm. 18: (1990) For additive hypotheses

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78 Statistics in bioequivalence may Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials Synthesis exercise

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79 Exercise may You have to design a bioequivalence trial for a generic of a reference formulation. This trial should allow to check if where µ G and µ R are the median for the generic and the reference formulation respectively. From the Freedom of Information, one knows that the intra individual CV of AUC for the reference formulation is about 7%. The half life of the reference formulation is about 6 hours. What kind of experimental design do you choose ? How many animals do you include in the trial ?

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80 Exercise may The consumer risk is set to 5%. You chose a power 1- =80%. You have planned a 2 2 crossover design with a washout period of about 48 hours. You expect the ratio µ G /µ R to be within the range [0.9;1.15]. If the difference in the population is larger, the two formulations will not be declared bioequivalent. N=n R =n G = 12 animals have been allocated randomly within the two sequences.

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81 Sample size may Number of animal per sequence for a 2 2 crossover, log transformation, equivalence interval : [0.8, 1.25], =5%, = 20%

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82 Results may Sequence 1 Ref.+ Gen. Sequence 2 Gen. + Ref. What to do next ? AUC Homoscedasticity : inter individuals ? Seq. 1Seq. 2 Mean/animal P (Fisher)=0.038 Heteroscedasticity : inter individuals

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83 After a ln transformation may Sequence 1 Ref.+ Gen. Sequence 2 Gen. + Ref. ln AUC Homoscedasticity : inter individuals ? Seq. 1 Seq. 2 Mean/animal P (Fisher)=0.083 Homoscedasticity : inter individuals

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84 ANOVA may What to do now ? Homoscedasticity : intra individuals ? Homoscedasticity

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85 ANOVA Table may Effects coding used for categorical variables in model. Categorical values encountered during processing are: PERIOD (2 levels) 1, 2 SEQUENCE (2 levels) 1, 2 FORMUL$ (2 levels) Gene, Ref ANIMAL (24 levels) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24 Dep Var: LN_AUC N: 48 Multiple R: Squared multiple R: Analysis of Variance SourceSum-of-Squares dfMean-Square F-ratio P PERIOD SEQUENCE FORMUL$ ANIMAL(SEQUENCE) Error Least squares means LS MeanSEN FORMUL$Gene FORMUL$Ref

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86 Need to correct the test for the carryover effect may Test for effect called: SEQUENCE Test of Hypothesis SourceSSdfMSFP Hypothesis Error No significant (differential) effect carryover

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87 Effects coding used for categorical variables in model. Categorical values encountered during processing are: PERIOD (2 levels) 1, 2 SEQUENCE (2 levels) 1, 2 FORMUL$ (2 levels) Gene, Ref ANIMAL (24 levels) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24 Dep Var: LN_AUC N: 48 Multiple R: Squared multiple R: Analysis of Variance SourceSum-of-Squares dfMean-Square F-ratio P PERIOD SEQUENCE FORMUL$ ANIMAL(SEQUENCE) Error Least squares means LS MeanSEN FORMUL$Gene FORMUL$Ref Confidence interval may

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88 Confidence interval may Build a 90% (for a consumer risk = 5%) confidence interval for ln G – ln R n G =12 ; n R =10 ; df = n T + n R -2 = 22 90% confidence interval for ln µ G – ln µ R = [ ; ]

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89 Conclusion may The 90% confidence interval for µ G /µ R = [exp( ) ;exp( )] = [0.88 ; 0.93] [0.8 ; 1.25] is totally included within the equivalence interval. The generic and reference formulations are bioequivalent

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