# Statistics in bioequivalence Didier Concordet NATIONAL VETERINARY S C H O O L T O U L O U S E May 4-5 2004.

## Presentation on theme: "Statistics in bioequivalence Didier Concordet NATIONAL VETERINARY S C H O O L T O U L O U S E May 4-5 2004."— Presentation transcript:

1 Statistics in bioequivalence Didier Concordet d.concordet@envt.fr NATIONAL VETERINARY S C H O O L T O U L O U S E May 4-5 2004

2 2 Statistics in bioequivalence may 4-5 2004 Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

3 3 Statistics in bioequivalence may 4-5 2004 Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

4 4 Parametric ? may 4-5 2004 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.

5 5 Non parametric ? may 4-5 2004 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.

6 6 Parametric : normality may 4-5 2004 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

7 7 Parametric methods may 4-5 2004 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

8 8 Non-parametric methods may 4-5 2004 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

9 9 Statistics in bioequivalence may 4-5 2004 Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

10 10 Statistics in bioequivalence may 4-5 2004 Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

11 11 Transformations of parameters may 4-5 2004 Data Parametric methods Assumptions ? yes Transformation no yes Non parametric methods no

12 12 Three fundamental assumptions may 4-5 2004 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

13 13 Fundamental assumptions : homoscedasticity may 4-5 2004 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

14 14 Fundamental assumptions : homoscedasticity may 4-5 2004 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.

15 15 Fundamental assumptions : Independence may 4-5 2004 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

16 16 Fundamental assumptions : Normality may 4-5 2004 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.

17 17 Fundamental assumptions : Normality may 4-5 2004 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…

18 18 In practice for bioequivalence may 4-5 2004 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

19 19 The ln transformation : side effect may 4-5 2004 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

20 20 Statistics in bioequivalence may 4-5 2004 Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

21 21 Statistics in bioequivalence may 4-5 2004 Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

22 22 Parallel and Cross-over designs may 4-5 2004 parallel Test Ref. Sequence 1 2 Period 12 2 2 Cross-over

23 23 Parallel vs Cross-over design may 4-5 2004 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

24 24 Analysis of parallel and cross- over designs may 4-5 2004 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

25 25 Why ? may 4-5 2004 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

26 26 Analysis of parallel designs may 4-5 2004 Step 1 : Check (at least graphically) homoscedasticity Step 2 : Estimate the mean for each formulation, estimate the between subjects variance. Transformation ?

27 27 Example may 4-5 2004 Test Ref Test Ref Variances comparison : P = 0.026 Heteroscedasticity

28 28 Example on log transformed data may 4-5 2004 Test Ref Test Ref Variances comparison : P = 0.66 Homoscedasticity

29 29 Example may 4-5 2004 Test Ref Pooled variance

30 30 Another way to proceed : ANOVA may 4-5 2004 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 = 1..10 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)

31 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: 0.095661810 Squared multiple R: 0.009151182 Analysis of Variance Source Sum-of-Squares df Mean-Square F-ratio P FORMUL\$ 0.062709687 1 0.062709687 0.166242589 0.688281535 Error 6.789922946 18 0.377217941 Least squares means LS Mean SE N FORMUL\$ =Ref 3.936724342 0.194220993 10 FORMUL\$ =Test 3.824733550 0.194220993 10 Another way to proceed : ANOVA may 4-5 2004 Does not give any information about bioeq

32 32 Analysis of cross-over designs may 4-5 2004 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.

33 33 A model for the 2 2 crossover design may 4-5 2004 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

34 34 A model for the 2 2 crossover design may 4-5 2004 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)

35 35 Homoscedasticity ? may 4-5 2004 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 = 0.038 Usually this test is not powerful

36 36 Homoscedasticity ? may 4-5 2004 i,j,k,l = indep random effects assumed to be drawn from N(0, ²)

37 37 After a ln tranformation... may 4-5 2004 Sequence 1 Test + Ref Sequence 2 Ref + Test ln AUC Comparison of interindividual variances P = 0.137 Seq. 1 Seq. 2 Homoscedasticity seems reasonable

38 38 ANOVA table may 4-5 2004 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: 0.978999514 Squared multiple R: 0.958440048 Analysis of Variance Source Sum-of-Squares df Mean-Square F-ratio P FORMUL\$ 3.077972446 1 3.077972446 6.06297E+01 0.000000526 PERIOD 0.293816162 1 0.293816162 5.787574765 0.027801823 SEQUENCE 1.987295663 1 1.987295663 3.91456E+01 0.000008686 ANIMAL(SEQUENCE) 1.34946E+01 18 0.749700010 1.47676E+01 0.000000479 Error 0.863034164 18 0.050766716 Least squares means LS Mean SE N FORMUL\$ =Ref 4.077673811 0.050381899 20 FORMUL\$ =Test 3.507676363 0.053107185 20 Does not give any information about bioeq

39 39 Period effect may 4-5 2004 Does not invalidate a crossover design Does affect in the same way the 2 formulations Origin : environment, equal carry-over Period effect significant

40 40 ANOVA table may 4-5 2004 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: 0.978999514 Squared multiple R: 0.958440048 Analysis of Variance Source Sum-of-Squares df Mean-Square F-ratio P FORMUL\$ 3.077972446 1 3.077972446 6.06297E+01 0.000000526 PERIOD 0.293816162 1 0.293816162 5.787574765 0.027801823 SEQUENCE 1.987295663 1 1.987295663 3.91456E+01 0.000008686 ANIMAL(SEQUENCE) 1.34946E+01 18 0.749700010 1.47676E+01 0.000000479 Error 0.863034164 18 0.050766716 Least squares means LS Mean SE N FORMUL\$ =Ref 4.077673811 0.050381899 20 FORMUL\$ =Test 3.507676363 0.053107185 20 Does not give any information about bioeq

41 41 Sequence effect : carryover effect may 4-5 2004 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

42 42 Testing the carryover effect may 4-5 2004 Analysis of Variance Source Sum-of-Squares df Mean-Square F-ratio P FORMUL\$ 3.077972446 1 3.077972446 6.06297E+01 0.000000526 PERIOD 0.293816162 1 0.293816162 5.787574765 0.027801823 SEQUENCE 1.987295663 1 1.987295663 3.91456E+01 0.000008686 ANIMAL(SEQUENCE) 1.34946E+01 18 0.749700010 1.47676E+01 0.000000479 Error 0.863034164 18 0.050766716 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 1.987295663 1 1.987295663 2.650787831 0.120875160 Error 1.34946E+01 18 0.749700010 The good P value

43 43 Testing the carryover effect may 4-5 2004 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

44 44 How to interpret the (differential) carryover effect ? may 4-5 2004 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

45 45 What to do if the carryover effect is significant ? may 4-5 2004 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.

46 46 How to avoid a carryover effect ? may 4-5 2004 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

47 47 ANOVA table may 4-5 2004 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: 0.978999514 Squared multiple R: 0.958440048 Analysis of Variance Source Sum-of-Squares df Mean-Square F-ratio P FORMUL\$ 3.077972446 1 3.077972446 6.06297E+01 0.000000526 PERIOD 0.293816162 1 0.293816162 5.787574765 0.027801823 SEQUENCE 1.987295663 1 1.987295663 3.91456E+01 0.000008686 ANIMAL(SEQUENCE) 1.34946E+01 18 0.749700010 1.47676E+01 0.000000479 Error 0.863034164 18 0.050766716 Least squares means LS Mean SE N FORMUL\$ =Ref 4.077673811 0.050381899 20 FORMUL\$ =Test 3.507676363 0.053107185 20 P 0.120875160 Does not give any information about bioeq Inter animals variability

48 48 Balance sheet may 4-5 2004 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.

49 49 Statistics in bioequivalence may 4-5 2004 Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

50 50 Statistics in bioequivalence may 4-5 2004 Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

51 51 Additive bioequivalence test of hypotheses may 4-5 2004 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

52 52 Multiplicative bioequivalence test of hypotheses may 4-5 2004 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])

53 53 Multiplicative bioequivalence test of hypotheses may 4-5 2004 Multiplicative hypotheses for the bioequivalence test bioequivalence bioinequivalence become additive after a ln transformation

54 54 The two one-sided tests (Schuirman) may 4-5 2004 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

55 55 Decision rules for the two one-sided tests procedure may 4-5 2004 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

56 56 Decision rules for the two one-sided tests procedure may 4-5 2004 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

57 57 Same procedure with confidence intervals may 4-5 2004 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

58 58 Same procedure with confidence intervals may 4-5 2004 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

59 59 Confidence intervals : summary may 4-5 2004 1 2 Equivalence interval 90% CI Bioequivalence ( =5%) 1- CI Bioinequivalence ( ) 1- CI Bioinequivalence ( ) No conclusion

60 60 An example may 4-5 2004 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

61 61 An example may 4-5 2004 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] = [-0.223 ; +0.223], Cannot conclude to bioequivalence

62 62 An example may 4-5 2004 90 % 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

63 63 What is implicitly assumed may 4-5 2004 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

64 64 What is implicitly assumed may 4-5 2004 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

65 65 Confidence interval for µ T /µ R may 4-5 2004 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

66 66 Second degree equation may 4-5 2004 The two solutions x 1, x 2 give the 90% confidence interval [x 1 ; x 2 ]

67 67 Example : parallel groups design may 4-5 2004 AUC Test Ref Variances comparison : P = 0.62 Test Ref

68 68 Example may 4-5 2004 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 ?)

69 69 Statistics in bioequivalence may 4-5 2004 Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

70 70 Statistics in bioequivalence may 4-5 2004 Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials

71 71 Sample size in bioequivalence trials may 4-5 2004 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

72 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 4-5 2004 additive multiplicative

73 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 4-5 2004 multiplicative

74 74 The most common situation may 4-5 2004 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

75 75 Sample size may 4-5 2004 Number of animal per sequence for a 2 2 crossover, log transformation, equivalence interval : [0.8, 1.25], =5%, = 20%

76 76 More generally may 4-5 2004 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

77 77 More generally may 4-5 2004 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. 20:557-561 (1992) K.F. Phillips. Power of the two one-sided tests procedure in bioequivalence. J. Pharmacokin. Biopharm. 18:137-144 (1990) For additive hypotheses

78 78 Statistics in bioequivalence may 4-5 2004 Parametric or non-parametric ? Transformation of parameters Experimental design : parallel and crossover Confidence intervals and bioequivalence Sample size in bioequivalence trials Synthesis exercise

79 79 Exercise may 4-5 2004 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 ?

80 80 Exercise may 4-5 2004 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.

81 81 Sample size may 4-5 2004 Number of animal per sequence for a 2 2 crossover, log transformation, equivalence interval : [0.8, 1.25], =5%, = 20%

82 82 Results may 4-5 2004 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

83 83 After a ln transformation may 4-5 2004 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

84 84 ANOVA may 4-5 2004 What to do now ? Homoscedasticity : intra individuals ? Homoscedasticity

85 85 ANOVA Table may 4-5 2004 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: 0.993193 Squared multiple R: 0.986432 Analysis of Variance SourceSum-of-Squares dfMean-Square F-ratio P PERIOD2.1441351 12.144135 800.613948 0.000000 SEQUENCE0.076541 10.076541 28.580350 0.000023 FORMUL\$0.124074 10.124074 46.328848 0.000001 ANIMAL(SEQUENCE)1.938929 220.088133 32.908674 0.000000 Error0.058918 220.002678 Least squares means LS MeanSEN FORMUL\$Gene4.9629430.01056424 FORMUL\$Ref5.0646270.01056424

86 86 Need to correct the test for the carryover effect may 4-5 2004 Test for effect called: SEQUENCE Test of Hypothesis SourceSSdfMSFP Hypothesis0.07654110.0765410.8684750.361493 Error1.938929220.088133 No significant (differential) effect carryover

87 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: 0.993193 Squared multiple R: 0.986432 Analysis of Variance SourceSum-of-Squares dfMean-Square F-ratio P PERIOD2.1441351 12.144135 800.613948 0.000000 SEQUENCE0.076541 10.076541 28.580350 0.000023 FORMUL\$0.124074 10.124074 46.328848 0.000001 ANIMAL(SEQUENCE)1.938929 220.088133 32.908674 0.000000 Error0.058918 220.002678 Least squares means LS MeanSEN FORMUL\$Gene4.9629430.01056424 FORMUL\$Ref5.0646270.01056424 Confidence interval may 4-5 2004

88 88 Confidence interval may 4-5 2004 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 = [ - 0.12565 ; - 0.07435]

89 89 Conclusion may 4-5 2004 The 90% confidence interval for µ G /µ R = [exp( - 0.12565) ;exp(- 0.07435)] = [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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