Statistics in bioequivalence Didier Concordet [email protected] NATIONAL VETERINARY S C H O O L T...

Statistics in bioequivalence

Didier [email protected]

NATIONALVETERINARYS C H O O L

T O U L O U S E

May 4-5 2004

2

Statistics in bioequivalencemay 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







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)

ExampleThe 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

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.

PracticallyThese 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

Parametric : normalitymay 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) distributionThe monotone transformation is the logarithmThe 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 dogsis a mixture of the N(3.5, 0.5²) and N(3.7, 0.5²) distributions

7

Parametric methodsmay 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

Non-parametric methodsmay 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







10







11

Transformations of parametersmay 4-5 2004

DataParametric

methodsAssumptions ?

yes

Transformation

no

yes

Non parametricmethods

no

12

Three fundamental assumptionsmay 4-5 2004

HomoscedasticityThe variance of the dependent variable is constant ; it does not vary with independent variables : formulation, animal, period.

IndependenceThe random variables implied in the analysis are independent.

NormalityThe random variables implied in the analysis are normally distributed

13

Fundamental assumptions : homoscedasticity

may 4-5 2004

HomoscedasticityThe 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 groupGroup 1 : Reference Group 2 : Test

AU

C

Ref Test

Ln

AU

C

Ref Test

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

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 checkHas to be assumed

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

Fundamental assumptions : Normality

may 4-5 2004

Normality • Not important in practice

when the sample size is large enough, the central limit theorem protects uswhen 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

In practice for bioequivalencemay 4-5 2004

Log transformationAUC : to stabilise the variance

to obtain a the symmetric distributionCMAX : to stabilise the variance

to obtain a the symmetric distributionTMAX (sometimes) : to obtain a the symmetric distribution

usually heteroscedasticity remainsWithout transformation

TMAX (sometimes)usually heteroscedasticity

19

The ln transformation : side effectmay 4-5 2004

If ),(~ln 2mNX

22

2

2

2

1)(

~)(

m

m

eeXVar

eXE

m is the pop. mean of lnX is the pop. median of X

1)(2

eXCV

After a logarithmic transformationbioequivalence methods compares the median (not the mean) of the parameters obtained with each formulation

20







21







22

Parallel and Cross-over designsmay 4-5 2004

parallel

Test

Ref. Seq

uenc

e 1

2

Period1 2

22 Cross-over

23

Parallel vs Cross-over designmay 4-5 2004

Advantages Drawbacks

Parallel

Cross-over

Easy to organiseEasy to analyseEasy to interpret

Comparison is carried-outbetween animals:not very powerful

Comparison is carried-outwithin animals:powerful

Difficult to organisePossible unequal carry-overDifficult to analyse

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 formulationthe variance of interest

between subjects for the parallel designwithin subject for the cross-over

Why ?

• To assess bioequivalence (student t-test or Fisher test)

NO

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

Analysis of parallel designsmay 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

0

50

100

150

200

250

AU

C

Examplemay 4-5 2004

AUC78.867.548.614.659.995.951.542.023.537.036.743.6198.570.433.127.150.126.938.3120.9

Test Ref

Tes

tR

ef

6.610ˆ 2 T

3.3008ˆ 2 R

Variances comparison : P = 0.026 Heteroscedasticity

28

1

10

100

1000

AU

C

Example on log transformed data may 4-5 2004

Test Ref

Tes

tR

ef

320.0ˆ 2 T

434.0ˆ 2 R

Variances comparison : P = 0.66 Homoscedasticity

ln AUC4.374.213.882.684.094.563.943.743.163.613.603.775.294.253.503.303.913.293.644.80

29

Examplemay 4-5 2004

Tes

tR

ef

320.0ˆ 2 T

434.0ˆ 2 R

Pooled variance

825.3TX

937.3RX

2

ˆ)1(ˆ)1(ˆ

222

RT

RRTT

nn

nn

377.021010

434.0)110(320.0)110(ˆ 2

ln AUC4.374.213.882.684.094.563.943.743.163.613.603.775.294.253.503.303.913.293.644.80

30

Another way to proceed : ANOVAmay 4-5 2004

Write an ANOVA model to analyse data useless here but useful to understand cross-overln AUC

4.374.213.882.684.094.563.943.743.163.613.603.775.294.253.503.303.913.293.644.80

Yij = ln AUC for the ith animal that received formulation i

Notations

formulation 1 = Test, formulation 2 = Ref

i = 1..2 ; j = 1..10Yij = µ + Fi + ij

y11=4.37µ = population meanFi = effect of the ith formulationij = indep random effects assumed to be drawn from N(0,²)

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 VarianceSource 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 : ANOVAmay 4-5 2004

TX

RX2̂

Does not give any information about bioeq

32

Analysis of cross-over designsmay 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

A model for the 22 crossover design

may 4-5 2004

Seq

uenc

e 1

Tes

t + R

efS

eque

nce

2R

ef +

Tes

t

PER 1 PER 278.8 125.367.5 94.648.6 66.314.6 26.959.9 82.695.9 91.851.5 72.442.0 74.323.5 41.737.0 58.236.7 24.943.6 42.5198.5 99.870.4 49.733.1 12.627.1 9.150.1 22.026.9 9.038.3 10.6120.9 67.3

AUClkjijljikjiljikji SANPSFAUC ,,,),(),,(,,

AUCij,k(i,j),l = AUC for the lth 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

A model for the 22 crossover design

may 4-5 2004

lkjijljikjiljikji SANPSFAUC ,,,),(),,(,,

Y1,1,1,1=78.8

µ = population meanFi = effect of the ith formulation

Sj = effect of the jth sequence

Pk(i,j) = effect of the kth period

Anl|Sj = random effect of the lth 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

Homoscedasticity ?may 4-5 2004

02040

6080

100120

140160

av

era

ge

AU

C/a

nim

al

Seq. 1 Seq. 2

Anl|Sj = assumed independent distrib according a N(0,²)In particular : Var(An|S1)=Var(An|S2)

Average AUC102.081.057.420.871.393.962.058.232.647.630.843.0149.260.022.918.136.018.024.494.1

Seq

uenc

e 1

Seq

uenc

e 2

Comparison of interindividual variances P = 0.038Usually this test is not powerful

36

Homoscedasticity ?may 4-5 2004

lkji ,,,̂

0 50 100 150 200ESTIMATE

-40

-30

-20

-10

0

10

20

30

40

RE

SID

UA

L

jljikji SNAPSF ˆˆˆˆˆ ),(

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

37

After a ln tranformation...may 4-5 2004

Seq

uenc

e 1

Tes

t + R

efS

eque

nce

2R

ef +

Tes

t

ln AUC PER 1 PER 24.37 4.834.21 4.553.88 4.192.68 3.294.09 4.414.56 4.523.94 4.283.74 4.313.16 3.733.61 4.063.60 3.213.77 3.755.29 4.604.25 3.913.50 2.543.30 2.203.91 3.093.29 2.203.64 2.364.80 4.21

lkjijljikjiljikji SANPSFAUC ,,,),(),,(,,ln

0

1

2

3

4

5

6

aver

age

ln A

UC

/ani

mal

Comparison of interindividual variances P = 0.137

Seq. 1 Seq. 22 3 4 5 6

ESTIMATE

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

RE

SID

UA

LHomoscedasticity seems reasonable

38

ANOVA tablemay 4-5 2004

Effects coding used for categorical variables in model.Categorical values encountered during processing are:FORMUL$ (2 levels) Ref, TestPERIOD (2 levels) 1, 2SEQUENCE (2 levels) 1, 2ANIMAL (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: 0.958440048Analysis of Variance

Source Sum-of-Squares df Mean-Square F-ratio P FORMUL$ 3.077972446 1 3.077972446 6.06297E+01 0.000000526PERIOD 0.293816162 1 0.293816162 5.787574765 0.027801823SEQUENCE 1.987295663 1 1.987295663 3.91456E+01 0.000008686ANIMAL(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


39

Period effectmay 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



7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20





41

Sequence effect : carryover effectmay 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

Testing the carryover effectmay 4-5 2004

Analysis of VarianceSource Sum-of-Squares df Mean-Square F-ratio P FORMUL$ 3.077972446 1 3.077972446 6.06297E+01 0.000000526PERIOD 0.293816162 1 0.293816162 5.787574765 0.027801823SEQUENCE 1.987295663 1 1.987295663 3.91456E+01 0.000008686ANIMAL(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: SEQUENCETest 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

Testing the carryover effectmay 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

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 22 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 animalsconsequently a poor randomisation can be wrongly interpreted as a carryover effect

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 22 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

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



7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20




P0.120875160


Inter animalsvariability

2̂

TX

RX

48

Balance sheetmay 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.

2̂TX RX

49







50







51

Additive bioequivalence test of hypotheses

may 4-5 2004

H 0 : T - R < or 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

Multiplicative bioequivalence test of hypotheses

21 or R

T

R

T

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])

21 R

T

53

Multiplicative bioequivalence test of hypotheses

211

210

:

or :

R

T

R

T

R

T

H

H

may 4-5 2004

Multiplicative hypotheses for the bioequivalence test

211

210

ln lnlnln:

lnlnlnor lnlnln:

RT

RTRT

H

H

bioequivalence

bioinequivalence

become additive after a ln transformation

54

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

Additive hypotheses for the bioequivalence test

bioequivalence

H 0 : T - R < or 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 H0

55

Decision rules for the two one-sided tests procedure

may 4-5 2004

1

2

1

11ˆ

df

RT

RT t

nnA

XX

First one-sided test

H 0 : T - R <

H 1 : T - R

Reject H0 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 µR and µT respectively

2̂TXRX

1dft 1

dft

A = 1 for parallel 0.5 for 22 crossover

56

Decision rules for the two one-sided tests procedure

may 4-5 2004

Reject H0 if

1

2

2

11ˆ

)(df

RT

RT t

nnA

XXH 0 : T - R >

H 1 : T - R

Second one-sided testWhere is the consumer risk (risk to wrongly conclude to bioequivalence)df is the degree of freedom of the variance and are estimates of µR and µT respectively

2̂TXRX

1dft 1

dft


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

RTdfRT

RTdfRT nn

AtXXnn

AtXX11

ˆ;11

ˆ 2121

Conclude to bioequivalence (with a risk ) if this interval is totally included in the equivalence interval [1 ; 2]


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

RTdfRT

RTdfRT nn

AtXXnn

AtXX11

ˆ;11

ˆ 22/122/1

Conclude to bioinequivalence (with a risk ) if this interval has no common point with the equivalence interval [1 ; 2]


59

Confidence intervals : summarymay 4-5 2004

1 2

Equivalence interval

90% CI Bioequivalence (=5%)

1- CI Bioinequivalence ()

1- CIBioinequivalence ()

No conclusion No conclusion

60

An example

.251ln lnln8.0ln:

25.1lnlnlnor 8.0lnlnln:

1

0

RT

RTRT

H

H

.251 8.0:

25.1or 8.0:

1

0

R

T

R

T

R

T

H

H

may 4-5 2004

Seq

uenc

e 1

Seq

uenc

e 2

ln AUC PER 1 PER 24.37 4.834.21 4.553.88 4.192.68 3.294.09 4.414.56 4.523.94 4.283.74 4.313.16 3.733.61 4.063.60 3.213.77 3.755.29 4.604.25 3.913.50 2.543.30 2.203.91 3.093.29 2.203.64 2.364.80 4.21


Homoscedasticity seems reasonableNo (differential) carryover effect

0.0508ˆ 2 3.51TX 08.4RX

nT=10 ; nR=10 ; df = nT+nR -2 = 18 734.195.018 t

61

An examplemay 4-5 2004

TX RXand respectively estimate ln µT and ln µR

90 % confidence interval for ln µT - ln µR

44.0;69.010

1

10

10508.05.0734.108.451.3;

10

1

10

10508.05.0734.108.451.3

11ˆ5.0;

11ˆ5.0 2121

RTdfRT

RTdfRT nn

tXXnn

tXX

90 % confidence interval for ln µT - ln µR = [-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

An examplemay 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

R

T

63

What is implicitly assumedmay 4-5 2004

The model can be written in an additive way via the ln transformation

211

210

:

or :

R

T

R

T

R

T

H

H

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

211

210

ln lnlnln:

lnlnlnor lnlnln:

RT

RTRT

H

H

This question translates in an additive way via the ln transformation

64

What is implicitly assumed

R

T

RT lnln

may 4-5 2004

What to do when the PK parameter (eg AUC) does meet the 3 fundamental assumptions without ln transformation ?

TX RXand respectively estimate µT and µR

butR

T

X

Xdoes not estimate

RT XX lnln does not estimate

Another method is needed to build the 90 % confidence interval of

R

T

65

Confidence interval for µT/µR

2̂

95.0dft

may 4-5 2004

TX RXand respectively estimate µT and µR

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

0ˆ

2ˆ 2

95.022

95.022

TdfTTR

RdfR n

AtXXXx

n

AtXx


66

Second degree equation

RdfR

TdfT

RdfRTRTR

RdfR

TdfT

RdfRTRTR

TdfTTR

RdfR

nA

tX

nA

tXn

AtXXXXX

x

nA

tX

nA

tXn

AtXXXXX

x

n

AtXXXx

n

AtXx

295.02

295.02

295.0222

2

295.02

295.02

295.0222

1

295.02

295.022

ˆ

ˆˆ

ˆ

ˆˆ

0ˆ

2ˆ

may 4-5 2004

The two solutions x1, x2 give the 90% confidence interval [x1 ; x2 ]

67

Example : parallel groups design

78.867.548.614.659.995.951.542.023.537.036.743.6

114.470.433.127.150.158.585.495.4

may 4-5 2004AUC

Tes

tR

ef

60.610ˆ 2 T

55.851ˆ 2 R

Variances comparison : P = 0.62

0

20

40

60

80

100

120

140

AU

C

Test Ref

06.73118

55.851960.6109ˆ 2

51.93TX 46.61RX

68

Example

734.195.018 t

may 4-5 2004

090.256961.638309.3651

0ˆ

2ˆ

2

295.02

295.022

xx

ntXXXx

ntXx

TdfTTR

RdfR

nR =10 ; nT = 10

x1 =0.63 ; x2 = 1.12

The 90% confidence interval of µT/µR 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







70







71


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

• 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 µT/µR (or µT-µR)

What one need to know to determine the sample size ?

21 RT

21 R

T

may 4-5 2004

additive

multiplicative

2

73


• The equivalence interval : [, 1.25]

• The experimental design : crossover (22) 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 µT/µR

The most common situation

25.18.0 R

T

may 4-5 2004

multiplicative

2

74

The most common situationmay 4-5 2004

• An estimate of : intra for the log transformed data

• An idea about the true value of µT/µR

It remains to know2

1)(2

eXCV

We have already seen that if ),(~ln 2mNX

then

Different scenarios for CV and µT/µR can be simulated

75

Sample sizemay 4-5 2004

µT/µR

CV % 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.205.0% 12 6 4 4 4 6 8 227.5% 22 8 6 6 6 8 12 4410.0% 36 12 8 6 8 10 20 7612.5% 54 16 10 8 10 14 30 11815.0% 78 22 12 10 12 20 42 168

Number of animal per sequence for a 22 crossover, log transformation, equivalence interval : [0.8, 1.25], =5%, = 20%

76

More generallymay 4-5 2004


• 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 µT/µR

21 R

T

77

More generallymay 4-5 2004

Iterative procedure; N = number of animals per sequence

• if µT/µR=1 2

21

2122

122 ln;lnmax

CV

ttN NN

• if 1<µT/µR<2

2

2

2122

122 /lnln

RTNN

CVttN

• if 1<µT/µR<

2

1

2122

122 /lnln

RTNN

CVttN

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






Sample size in bioequivalence trialsSynthesis e

xercise

79

Exercisemay 4-5 2004

You have to design a bioequivalence trial for a generic of a reference formulation.This trial should allow to check if 25.1 8.0

R

G

where µG and µR are the median for the generic and the referenceformulation respectively.From the Freedom of Information, one knows that the intra individualCV 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

Exercisemay 4-5 2004

The consumer risk is set to 5%. You chose a power 1-=80%.

You have planned a 22 crossover design with a washout period of

about 48 hours.

You expect the ratio µG/µR 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=nR=nG= 12 animals have been allocated randomly within the two

sequences.

81

Sample sizemay 4-5 2004

µT/µR

CV % 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.205.0% 12 6 4 4 4 6 8 227.5% 22 8 6 6 6 8 12 4410.0% 36 12 8 6 8 10 20 7612.5% 54 16 10 8 10 14 30 11815.0% 78 22 12 10 12 20 42 168

Number of animal per sequence for a 22 crossover, log transformation, equivalence interval : [0.8, 1.25], =5%, = 20%

82

Resultsmay 4-5 2004

Seq

uenc

e 1

Ref

.+ G

en.

Seq

uenc

e 2

Gen

. + R

ef.

What to do next ?107.1 134.6144.5 189.0113.8 186.487.2 130.2

155.1 213.4115.2 153.0188.5 266.5127.3 163.8132.4 173.9216.2 302.2108.3 147.3155.2 219.3107.8 191.194.1 170.2

108.7 170.8137.0 221.3120.7 203.6102.3 168.289.8 175.292.8 150.2

103.2 159.8119.6 201.5138.1 215.9135.5 251.9

AUC

Homoscedasticity : inter individuals ?

12.1984ˆ 21

46.629ˆ 22 Seq. 1 Seq. 2

Mean/animal120.9166.7150.1108.7184.2134.1227.5145.6153.2259.2127.8187.3149.4132.1139.8179.1162.1135.3132.5121.5131.5160.6177.0193.7

0

50

100

150

200

250

300

AU

C

P (Fisher)=0.038

Heteroscedasticity : inter individuals

83

After a ln transformationmay 4-5 2004

Seq

uenc

e 1

Ref

.+ G

en.

Seq

uenc

e 2

Gen

. + R

ef.

ln AUC

Homoscedasticity : inter individuals ?

0660.0ˆ 21

0221.0ˆ 22 Seq. 1 Seq. 2

Mean/animal

P (Fisher)=0.083

Homoscedasticity : inter individuals

P1 P24.67 4.904.97 5.244.73 5.234.47 4.875.04 5.364.75 5.035.24 5.594.85 5.104.89 5.165.38 5.714.69 4.995.04 5.394.68 5.254.54 5.144.69 5.144.92 5.404.79 5.324.63 5.134.50 5.174.53 5.014.64 5.074.78 5.314.93 5.374.91 5.53

4.795.114.984.675.204.895.414.975.025.544.845.224.974.844.915.165.054.884.834.774.865.055.155.22

33.5

44.5

55.5

66.5

7

me

an

ln A

UC

/an

ima

l

84

ANOVAmay 4-5 2004


What to do now ? Homoscedasticity : intra individuals ?

4.5 5.0 5.5 6.0ESTIMATE

-0.10

-0.05

0.00

0.05

0.10

RE

SID

UA

L Homoscedasticity

85

ANOVA Tablemay 4-5 2004

Effects coding used for categorical variables in model. Categorical values encountered during processing are:PERIOD (2 levels) 1, 2SEQUENCE (2 levels) 1, 2FORMUL$ (2 levels) Gene, RefANIMAL (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

Source Sum-of-Squares df Mean-Square F-ratio PPERIOD 2.1441351 1 2.144135 800.613948 0.000000SEQUENCE 0.076541 1 0.076541 28.580350 0.000023FORMUL$ 0.124074 1 0.124074 46.328848 0.000001ANIMAL(SEQUENCE) 1.938929 22 0.088133 32.908674 0.000000Error 0.058918 22 0.002678 Least squares means

LS Mean SE NFORMUL$ Gene 4.962943 0.010564 24FORMUL$ Ref 5.064627 0.010564 24

86

Need to correct the test for the carryover effect

may 4-5 2004

Test for effect called: SEQUENCE Test of Hypothesis

Source SS df MS F PHypothesis 0.076541 1 0.076541 0.868475 0.361493Error 1.938929 22 0.088133

No significant (differential) effect carryover

87

Effects coding used for categorical variables in model. Categorical values encountered during processing are:PERIOD (2 levels) 1, 2SEQUENCE (2 levels) 1, 2FORMUL$ (2 levels) Gene, RefANIMAL (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

Source Sum-of-Squares df Mean-Square F-ratio PPERIOD 2.1441351 1 2.144135 800.613948 0.000000SEQUENCE 0.076541 1 0.076541 28.580350 0.000023FORMUL$ 0.124074 1 0.124074 46.328848 0.000001ANIMAL(SEQUENCE) 1.938929 22 0.088133 32.908674 0.000000Error 0.058918 22 0.002678 Least squares means

LS Mean SE NFORMUL$ Gene 4.962943 0.010564 24FORMUL$ Ref 5.064627 0.010564 24

Confidence intervalmay 4-5 2004

2̂

RX

GX

88

Confidence intervalmay 4-5 2004

002678.0ˆ 2 064627.5RX962943.4GX

Build a 90% (for a consumer risk = 5%)

confidence interval for ln G – ln R

RGdfRG

RGdfRG nn

tXXnn

tXX11

ˆ5.0;11

ˆ5.0 2121

nG=12 ; nR=10 ; df = nT+nR -2 = 22 717.195.022 t

90% confidence interval for ln µG – ln µR = [ - 0.12565 ; - 0.07435]

89

Conclusionmay 4-5 2004

The 90% confidence interval for µG/µR = [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

Statistics in bioequivalence Didier Concordet [email protected] NATIONAL VETERINARY S C H O O L T...

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Transcript of Statistics in bioequivalence Didier Concordet [email protected] NATIONAL VETERINARY S C H O O L T...