Statistics in bioequivalence

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Statistics in bioequivalence

• Didier Concordet
• d.concordet_at_envt.fr

May 4-5 2004
2
Statistics in bioequivalence
Parametric or non-parametric ?
Transformation of parameters
Experimental design parallel and crossover
Confidence intervals and bioequivalence
Sample size in bioequivalence trials
3
Statistics in bioequivalence
Parametric or non-parametric ?
Transformation of parameters
Experimental design parallel and crossover
Confidence intervals and bioequivalence
Sample size in bioequivalence trials
4
Parametric ?
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, s²)
The parameters m, s² defined the distribution of
AUC (its ln) that can be observed in this dog.
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Non parametric ?
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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Parametric normality
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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Parametric methods

• 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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Non-parametric methods

• 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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Statistics in bioequivalence
Parametric or non-parametric ?
Transformation of parameters
Experimental design parallel and crossover
Confidence intervals and bioequivalence
Sample size in bioequivalence trials
10
Statistics in bioequivalence
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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Transformations of parameters
Data
Parametric methods
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Three fundamental assumptions
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. Normalit
y The random variables implied in the analysis
are normally distributed
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Fundamental assumptions homoscedasticity
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
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Fundamental assumptions homoscedasticity

• 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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Fundamental assumptions Independence

• 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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Fundamental assumptions Normality

• 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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Fundamental assumptions Normality

• 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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In practice for bioequivalence
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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The ln transformation side effect
m is the pop. mean of lnX is the pop. median
of X
If
After a logarithmic transformation bioequivalence
methods compares the median (not the mean) of
the parameters obtained with each formulation
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Statistics in bioequivalence
Parametric or non-parametric ?
Transformation of parameters
Experimental design parallel and crossover
Confidence intervals and bioequivalence
Sample size in bioequivalence trials
21
Statistics in bioequivalence
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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Parallel and Cross-over designs
Period
1
2
Test
1
Sequence
2
Ref.
2?2 Cross-over
parallel
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Parallel vs Cross-over design
Drawbacks
Advantages
Easy to organise Easy to analyse Easy to interpret
Comparison is carried-out between animals not
very powerful
Parallel
Comparison is carried-out within animals powerful
Difficult to organise Possible unequal
carry-over Difficult to analyse
Cross-over
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Analysis of parallel and cross-over designs
Why ?

• 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

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

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Why ?
Classical hypotheses for student t-test and
Fisher test (ANOVA)
?T and ?R population mean for test and reference
formulation respectively
H 0 ?T ?R
H 1 ?T ? ?R
Hypotheses for the bioequivalence test
H 0 ?T - ?R gtD
bioinequivalence
H 1 ?T - ?R ? D
bioequivalence
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Analysis of parallel designs
Step 1 Check (at least graphically)
homoscedasticity
Transformation ?
Step 2 Estimate the mean for each formulation,
estimate the between subjects variance.
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Example
Test
Ref
Test
Ref
Heteroscedasticity
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Example on log transformed data
Test
Ref
Test
Ref
Homoscedasticity
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Example
Pooled variance
Test
Ref
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Another way to proceed ANOVA
Write an ANOVA model to analyse data useless
here but useful to understand cross-over
Notations
Yij ln AUC for the ith animal that received
formulation i
formulation 1 Test, formulation 2 Ref
i 1..2 j 1..10
Yij µ Fi eij
y114.37 µ population mean Fi effect of the
ith formulation eij indep random effects
assumed to be drawn from N(0,s²)
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Another way to proceed ANOVA
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
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Analysis of cross-over designs
Difficult to analyse by hand, especially when the
experimental design is unbalanced. Need of a
model to analyse data.

• Step 1 Write the model to analyse the
  cross-over

• Step 2 Check (at least graphically)
  homoscedasticity

Transformation ?

• Step 3 Check the absence of a carry-over effect

• Step 4 Estimate the mean for each formulation,
  
• estimate the within (intra) subjects variance.

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A model for the 2?2 crossover design
AUC
Notations
Sequence 1 Test Ref
formulation 1 Test, formulation 2 Ref
AUCij,k(i,j),l AUC for the lth animal of the
seq. j when it received formulation i at period
k(i,j)
Sequence 2 Ref Test
i 1..2 j 1..,2 k(1,1) 1 k(1,2) 2
k(2,1) 2 k(2,2) 1 l1..10
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A model for the 2?2 crossover design
Y1,1,1,178.8 µ population mean Fi effect of
the ith formulation Sj effect of the jth
sequence Pk(i,j) effect of the kth
period AnlSj random effect of the lth animal
of sequence j, they are assumed independent
distrib according a N(0,?²) ei,j,k,l indep
random effects assumed to be drawn from N(0,s²)
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Homoscedasticity ?
AnlSj assumed independent distrib according a
N(0,?²)
In particular Var(AnS1)Var(AnS2)
Average AUC
Sequence 1
Sequence 2
Comparison of interindividual variances P
0.038 Usually this test is not powerful
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Homoscedasticity ?
ei,j,k,l indep random effects assumed to be
drawn from N(0,s²)
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After a ln tranformation...
ln AUC
Sequence 1 Test Ref
Sequence 2 Ref Test
Seq. 1
Seq. 2
Comparison of interindividual variances P 0.137
Homoscedasticity seems reasonable
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ANOVA table
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 6.06297E01
0.000000526 PERIOD 0.293816162
1 0.293816162 5.787574765 0.027801823 SEQUENCE
1.987295663 1 1.987295663
3.91456E01 0.000008686 ANIMAL(SEQUENCE)
1.34946E01 18 0.749700010 1.47676E01
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
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Period effect
Period effect significant

• Does not invalidate a crossover design
• Does affect in the same way the 2 formulations
• Origin environment, equal carry-over

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ANOVA table
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 6.06297E01
0.000000526 PERIOD 0.293816162
1 0.293816162 5.787574765 0.027801823 SEQUENCE
1.987295663 1 1.987295663
3.91456E01 0.000008686 ANIMAL(SEQUENCE)
1.34946E01 18 0.749700010 1.47676E01
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
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Sequence effect carryover effect
Differential carryover effect significant ?

• For all statistical softwares, the only random
  variables of a model are the residuals e
• The ANOVA table is built assuming that all other
  effects are fixed

However
We are working on a sample of animals
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Testing the carryover effect
The test for the carryover (sequence) effect has
to be corrected
Analysis of Variance Source
Sum-of-Squares df Mean-Square F-ratio
P FORMUL 3.077972446 1
3.077972446 6.06297E01 0.000000526 PERIOD
0.293816162 1 0.293816162 5.787574765
0.027801823 SEQUENCE 1.987295663
1 1.987295663 3.91456E01 0.000008686 ANIMAL(SEQU
ENCE) 1.34946E01 18 0.749700010
1.47676E01 0.000000479 Error
0.863034164 18 0.050766716
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 1.34946E01 18
0.749700010
The good P value
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Testing the carryover effect

• The test for a carryover effect should be
  declared significant when Plt0.1
• In the previous example P0.12 the carryover
  effect is not significant

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How to interpret the (differential) carryover
effect ?

• 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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What to do if the carryover effect is significant
?

• 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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How to avoid a carryover effect ?

• 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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ANOVA table
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 6.06297E01
0.000000526 PERIOD 0.293816162
1 0.293816162 5.787574765 0.027801823 SEQUENCE
1.987295663 1 1.987295663
3.91456E01 0.000008686 ANIMAL(SEQUENCE)
1.34946E01 18 0.749700010 1.47676E01
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
Inter animals variability
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Balance sheet

• 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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Statistics in bioequivalence
Parametric or non-parametric ?
Transformation of parameters
Experimental design parallel and crossover
Confidence intervals and bioequivalence
Sample size in bioequivalence trials
50
Statistics in bioequivalence
Parametric or non-parametric ?
Transformation of parameters
Experimental design parallel and crossover
Confidence intervals and bioequivalence
Sample size in bioequivalence trials
51
Additive bioequivalence test of hypotheses
?T and ?R population mean for test and reference
formulation respectively
D1 D2 Absolute equivalence interval
Additive hypotheses for the bioequivalence test
H 0 ?T - ?R lt D1 or ?T - ?R gt D2
bioinequivalence
H 1 D1 ? ?T - ?R ? D2
bioequivalence
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Multiplicative bioequivalence test of hypotheses
?T and ?R population median for test and
reference formulation respectively
D1 D2 Relative equivalence interval where
0lt D1 lt1lt D2 (eg 0.8 1.25)
Multiplicative hypotheses for the bioequivalence
test
H 0
bioinequivalence
H 1
bioequivalence
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Multiplicative bioequivalence test of hypotheses
Multiplicative hypotheses for the bioequivalence
test
bioinequivalence
bioequivalence
become additive after a ln transformation
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The two one-sided tests (Schuirman)
Additive hypotheses for the bioequivalence test
bioequivalence
H 0 ?T - ?R gt D2
H 0 ?T - ?R lt D1
H 1 ?T - ?R ? D2
H 1 D1 ? ?T - ?R
second one-sided test
First one-sided test
Bioequivalence when the 2 tests reject H0
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Decision rules for the two one-sided tests
procedure
Reject H0 if
A 1 for parallel 0.5 for 2?2 crossover
First one-sided test
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Decision rules for the two one-sided tests
procedure
H 0 ?T - ?R gt D2
Reject H0 if
H 1 ?T - ?R ? D2
A 1 for parallel 0.5 for 2?2 crossover
Second one-sided test
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Same procedure with confidence intervals
Build a 1-2a (90 for a consumer risk a 5)
confidence interval for ?T - ?R
A 1 for parallel 0.5 for 2?2 crossover
Conclude to bioequivalence (with a risk a) if
this interval is totally included in the
equivalence interval D1 D2
58
Same procedure with confidence intervals
Build a 1-b (95 for the drug company risk b
5) confidence interval for ?T - ?R
A 1 for parallel 0.5 for 2?2 crossover
Conclude to bioinequivalence (with a risk b) if
this interval has no common point with the
equivalence interval D1 D2
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Confidence intervals summary
No conclusion
No conclusion
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An example
ln AUC
Homoscedasticity seems reasonable No
(differential) carryover effect
Sequence 1
nT10 nR10 df nTnR -2 18
Sequence 2
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An example
and
respectively estimate ln µT and ln µR
90 confidence interval for ln µT - ln µR
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
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An example
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
Actually, the 90 confidence interval has no
common point with the equivalence interval
Conclude to bioinequivalence (risklt10)
63
What is implicitly assumed
The model can be written in an additive way via
the ln transformation
Assume that the question is formulated in a
multiplicative way
bioinequivalence
bioequivalence
This question translates in an additive way via
the ln transformation
It is implicitly assumed that the PK parameters
(eg AUC) has to be ln transformed to meet the 3
fundamental assumptions
64
What is implicitly assumed
What to do when the PK parameter (eg AUC) does
meet the 3 fundamental assumptions without ln
transformation ?
does not estimate
but
does not estimate
Another method is needed to build the 90
confidence interval of
65
Confidence interval for µT/µR
estimate the between (parallel) or within
(crossover) subjects variance
Critical value of a student distribution with df
degrees of freedom
degree of freedom for
df
Solve the second degree equation
A 1 for parallel 0.5 for 2?2 crossover
66
Second degree equation
The two solutions x1, x2 give the 90 confidence
interval x1 x2
67
Example parallel groups design
AUC
Test
Ref
Variances comparison P 0.62
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Example
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
Statistics in bioequivalence
Parametric or non-parametric ?
Transformation of parameters
Experimental design parallel and crossover
Confidence intervals and bioequivalence
Sample size in bioequivalence trials
70
Statistics in bioequivalence
Parametric or non-parametric ?
Transformation of parameters
Experimental design parallel and crossover
Confidence intervals and bioequivalence
Sample size in bioequivalence trials
71
Sample size in bioequivalence trials
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
What one need to know to determine the sample
size ?

• The hypotheses to be tested
• The equivalence interval D1, D2
• The experimental design parallel or crossover
• The consumer risk (a 5) risk to wrongly
  conclude to bioeq
• The drug company risk (b 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)

additive
multiplicative
73
The most common situation

• The hypotheses to be tested
• The equivalence interval 0.8, 1.25
• The experimental design crossover (2?2) with
  the same number of
• animals per sequence N
• The consumer risk (a 5)
• The drug company risk (b 20)
• A ln transformation is required
• An estimate of (intra for the log
  transformed data)
• An idea about the true value of µT/µR

multiplicative
74
The most common situation
It remains to know

• An estimate of intra for the log
  transformed data
• An idea about the true value of µT/µR

We have already seen that if
then
Different scenarios for CV and µT/µR can be
simulated
75
Sample size
Number of animal per sequence for a 2?2
crossover, log transformation, equivalence
interval 0.8, 1.25, a5, b 20
76
More generally

• The hypotheses to be tested
• The equivalence interval D1, D2
• The experimental design crossover
• The consumer risk (a) risk to wrongly conclude
  to bioeq
• The drug company risk (b) 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

77
More generally
Iterative procedure N number of animals per
sequence

• if µT/µR1

• if 1ltµT/µRltD2

• if D1ltµT/µRlt1

D. Hauschke coll. Sample size determination for
bioequivalence assessment using a multiplicative
model. J. Pharmacokin. Biopharm. 20557-561 (1992)
For additive hypotheses
K.F. Phillips. Power of the two one-sided tests
procedure in bioequivalence. J. Pharmacokin.
Biopharm. 18137-144 (1990)
78
Statistics in bioequivalence
Parametric or non-parametric ?
Transformation of parameters
Synthesis exercise
Experimental design parallel and crossover
Confidence intervals and bioequivalence
Sample size in bioequivalence trials
79
Exercise
You have to design a bioequivalence trial for a
generic of a reference formulation. This trial
should allow to check if
where µG and µR 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
Exercise
The consumer risk a is set to 5. You chose a
power 1-b80. You have planned a 2?2 crossover
design with a washout period of about 48
hours. You expect the ratio µG/µR to be within
the range 0.91.15. If the difference in the
population is larger, the two formulations will
not be declared bioequivalent. NnRnG 12
animals have been allocated randomly within the
two sequences.
81
Sample size
Number of animal per sequence for a 2?2
crossover, log transformation, equivalence
interval 0.8, 1.25, a5, b 20
82
Results
AUC
What to do next ?
Sequence 1 Ref . Gen.
Sequence 2 Gen. Ref.
83
After a ln transformation
ln AUC
Mean/animal
Homoscedasticity inter individuals ?
Sequence 1 Ref . Gen.
Seq. 2
Seq. 1
Sequence 2 Gen. Ref.
P (Fisher)0.083
Homoscedasticity inter individuals
84
ANOVA
Homoscedasticity intra individuals ?
What to do now ?
Homoscedasticity
85
ANOVA Table
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 Source Sum-of-Sq
uares df Mean-Square F-ratio
P PERIOD 2.1441351 1 2.144135 800.613948
0.000000 SEQUENCE 0.076541 1 0.076541
28.580350 0.000023 FORMUL 0.124074
1 0.124074 46.328848 0.000001 ANIMAL(SEQUEN
CE) 1.938929 22 0.088133 32.908674
0.000000 Error 0.058918 22 0.002678 Least
squares means LS Mean SE N FORMUL Gene 4.96294
3 0.010564 24 FORMUL Ref 5.064627 0.010564 24
86
Need to correct the test for the carryover effect
Test for effect called SEQUENCE Test of
Hypothesis Source SS df MS F P Hypothesis 0.076
541 1 0.076541 0.868475 0.361493 Error 1.938929
22 0.088133
No significant (differential) effect carryover
87
Confidence interval
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 Source Sum-of-Sq
uares df Mean-Square F-ratio
P PERIOD 2.1441351 1 2.144135 800.613948
0.000000 SEQUENCE 0.076541 1 0.076541
28.580350 0.000023 FORMUL 0.124074
1 0.124074 46.328848 0.000001 ANIMAL(SEQUEN
CE) 1.938929 22 0.088133 32.908674
0.000000 Error 0.058918 22 0.002678 Least
squares means LS Mean SE N FORMUL Gene 4.96294
3 0.010564 24 FORMUL Ref 5.064627 0.010564 24
88
Confidence interval
Build a 90 (for a consumer risk a 5)
confidence interval for ln ?G ln ?R
nG12 nR10 df nTnR -2 22
90 confidence interval for ln µG ln µR -
0.12565 - 0.07435
89
Conclusion
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