analysis of a cross over design

Alfredo García – Arieta, PhD

Analysis of a cross-over design

Assessment of Bioequivalence Data

23rd June 2015

Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training

Outline

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

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

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


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


Statistical Analysis of BE studies

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

In the past, it was possible to find statistical analyses performed with incorrect software.– Calculations based on arithmetic means, instead of

Least Square Means, give biased results in unbalanced studies

• Unbalance: different number of subjects in each sequence– Calculations for replicate designs are more

complex and prone to mistakes


The statistical analysis is not so complex

2x2 BE trial

N=12

Period 1Period 2

Sequence 1 (BA)

BA is RT

Y11Y12

Sequence 2 (AB)

AB is TR

Y21Y22


We don’t need to calculate an ANOVA table

Sources of variation d. f. SS MS F P Inter-subject 23 16487,49 716,85 4,286 Carry-over 1 276,00 276,00 0,375 0,5468 Residual / subjects 22 16211,49 736,89 4,406 0,0005

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

Total 47 20265,68


With complex formulae

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And really complex formulae

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Given the following data, it is simple

2x2 BE trial

N=12

Period 1Period 2

Sequence 1 (BA)Y11

75, 95, 90, 80, 70, 85

Y12

70, 90, 95, 70, 60, 70

Sequence 2 (AB)Y21

75, 85, 80, 90, 50, 65

Y22

40, 50, 70, 80, 70, 95


First, log-transform the data

2x2 BE trial

N=12

Period 1Period 2

Sequence 1 (BA)Y11

4.3175, 4.5539, 4.4998, 4.3820, 4.2485, 4.4427

Y12

4.2485, 4.4998, 4.5539, 4.2485, 4.0943, 4.2485

Sequence 2 (AB)Y21

4.3175, 4.4427, 4.3820, 4,4998, 3,9120, 4.1744

Y22

3.6889, 3,9120, 4,2485, 4.3820, 4.2485, 4.5539


Second, calculate the arithmetic mean of each period and sequence

2x2 BE trial

N=12

Period 1Period 2

Sequence 1 (BA)Y11 = 4.407Y12 = 4.316

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


Note the difference between Arithmetic Mean and Least Square Mean

The arithmetic mean (AM) of T (or R) is the mean of all observations with T (or R) irrespective of its group or sequence

– All observations have the same weight

The LSM of T (or R) is the mean of the two sequence by period means

– In case of balanced studies AM = LSM – In case of unbalanced studies observations in sequences with

less subjects have more weight– In case of a large unbalance between sequences due to drop-

outs or withdrawals the bias of the AM is notable


Third, calculate the LSM of T and R

2x2 BE trial

N=12

Period 1Period 2

Sequence 1 (BA)Y11 = 4.407Y12 = 4.316

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

B = 4.2898 A = 4.3018


Fourth, calculate the point estimate

F = LSM Test (A) – LSM Reference (B)

F = 4.30183 – 4.28985 = 0.01198

Fifth step! Back-transform to the original scale

Point estimate = eF = e0.01198 = 1.01205

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


Now we need to calculate the variability!

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

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

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

Step 4: Sum these squared differences

Step 5: Divide it by (n1+n2-2), where n1 and n2 is the number of subjects in each sequence. In this example 6+6-2 = 10

– This value multiplied by 2 is the MSE– CV (%) = 100 x √eMSE-1


This can be done easily in a spreadsheet!

I II Step 1 Step 1 Step 3 Step 3 Step 4R T P2-P1 (P2-P1)/2 d - mean d squared Sum = 0,23114064

4,31748811 4,24849524 -0,06899287 -0,03449644 0,01140614 0,00013014,55387689 4,49980967 -0,05406722 -0,02703361 0,01886897 0,00035604 Step 54,49980967 4,55387689 0,05406722 0,02703361 0,07293619 0,00531969 Sigma2(d) = 0,023114064,38202663 4,24849524 -0,13353139 -0,0667657 -0,02086312 0,00043527 MSE= 0,046228134,24849524 4,09434456 -0,15415068 -0,07707534 -0,03117276 0,00097174 CV = 21,75162184,44265126 4,24849524 -0,19415601 -0,09707801 -0,05117543 0,00261892

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

T R4,3175 3,6889 -0,62860866 -0,31430433 -0,25642187 0,065752184,4427 3,9120 -0,53062825 -0,26531413 -0,20743167 0,04302794,3820 4,2485 -0,13353139 -0,0667657 -0,00888324 7,8912E-054,4998 4,3820 -0,11778304 -0,05889152 -0,00100906 1,0182E-063,9120 4,2485 0,33647224 0,16823612 0,22611858 0,051129614,1744 4,5539 0,37948962 0,18974481 0,24762727 0,06131926

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

PERIOD


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

I II Step 1 Step 1R T P2-P1 (P2-P1)/2

4,31748811 4,24849524 -0,06899287 -0,034496444,55387689 4,49980967 -0,05406722 -0,027033614,49980967 4,55387689 0,05406722 0,027033614,38202663 4,24849524 -0,13353139 -0,06676574,24849524 4,09434456 -0,15415068 -0,077075344,44265126 4,24849524 -0,19415601 -0,09707801

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

T R4,3175 3,6889 -0,62860866 -0,314304334,4427 3,9120 -0,53062825 -0,265314134,3820 4,2485 -0,13353139 -0,06676574,4998 4,3820 -0,11778304 -0,058891523,9120 4,2485 0,33647224 0,168236124,1744 4,5539 0,37948962 0,18974481

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

PERIOD


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

I II Step 1 Step 1R T P2-P1 (P2-P1)/2

4,31748811 4,24849524 -0,06899287 -0,034496444,55387689 4,49980967 -0,05406722 -0,027033614,49980967 4,55387689 0,05406722 0,027033614,38202663 4,24849524 -0,13353139 -0,06676574,24849524 4,09434456 -0,15415068 -0,077075344,44265126 4,24849524 -0,19415601 -0,09707801

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

T R4,3175 3,6889 -0,62860866 -0,314304334,4427 3,9120 -0,53062825 -0,265314134,3820 4,2485 -0,13353139 -0,06676574,4998 4,3820 -0,11778304 -0,058891523,9120 4,2485 0,33647224 0,168236124,1744 4,5539 0,37948962 0,18974481

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

PERIOD


I II Step 1 Step 1 Step 3 Step 3R T P2-P1 (P2-P1)/2 d - mean d squared

4,31748811 4,24849524 -0,06899287 -0,03449644 0,01140614 0,00013014,55387689 4,49980967 -0,05406722 -0,02703361 0,01886897 0,000356044,49980967 4,55387689 0,05406722 0,02703361 0,07293619 0,005319694,38202663 4,24849524 -0,13353139 -0,0667657 -0,02086312 0,000435274,24849524 4,09434456 -0,15415068 -0,07707534 -0,03117276 0,000971744,44265126 4,24849524 -0,19415601 -0,09707801 -0,05117543 0,00261892

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

T R4,3175 3,6889 -0,62860866 -0,31430433 -0,25642187 0,065752184,4427 3,9120 -0,53062825 -0,26531413 -0,20743167 0,04302794,3820 4,2485 -0,13353139 -0,0667657 -0,00888324 7,8912E-054,4998 4,3820 -0,11778304 -0,05889152 -0,00100906 1,0182E-063,9120 4,2485 0,33647224 0,16823612 0,22611858 0,051129614,1744 4,5539 0,37948962 0,18974481 0,24762727 0,06131926

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

PERIOD

Step 3: Squared differences


Step 3 Step 4squared Sum = 0,23114064

0,00013010,00035604 Step 50,00531969 Sigma2(d) = 0,023114060,00043527 MSE= 0,046228130,00097174 CV = 21,75162180,00261892

0,065752180,04302797,8912E-051,0182E-060,051129610,06131926

Step 4: Sum these squared differences


Step 3 Step 4squared Sum = 0,23114064

0,00013010,00035604 Step 50,00531969 Sigma2(d) = 0,023114060,00043527 MSE= 0,046228130,00097174 CV = 21,75162180,00261892

0,065752180,04302797,8912E-051,0182E-060,051129610,06131926

Step 5: Divide the sum by n1+n2-2


Calculate the confidence interval withpoint estimate and variability

Step 11: In log-scale

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

F has been calculated before

The t value is obtained in t-Student tables with 0.1 alpha and n1+n2-2 degrees of freedom

– Or in MS Excel with the formula =DISTR.T.INV(0.1; n1+n2-2)

Sigma2(d) has been calculated before.


Final calculation: the 90% CI

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

F = 0.01198

t(0.1, n1+n2-2) = 1.8124611

Sigma2(d) = 0.02311406

90% CI: LL = -0.14711 to UL= 0.17107

Step 12: Back transform the limits with eLL and eUL

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


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


Factors affecting the sample size

The error variance (CV%) of the primary PK parameters– Published data– Pilor study

The significance level derired (5%): consumer’s risk

The statistical power desired (>80%): producer’s risk

The mean deviation from comparator compativle with BE

The acceptance criteria: (usually 80-125% or ±20%)


Reasons for a correct calculation of the sample size

Too many subjects– It is unethical to disturb more subjects than necessary– Some subjects at risk and they are not necessary– It is an unnecessary waste of some resources ($)

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

Minimum number of subjects: 12


Frequent mistakes

To calculate the sample size required to detect a 20% difference assuming that treatments are e.g. equal

– Pocock, Clinical Trials, 1983

To use calculation based on data without log-transformation

– Design and Analysis of Bioavailability and Bioequivalence Studies, Chow & Liu, 1992 (1st edition) and 2000 (2nd edition)

Too many extra subjects. Usually no need of more than 10%. Depends on tolerability

– 10% proposed by Patterson et al, Eur J Clin Pharmacol 57: 663-670 (2001)


Methods to calculate the sample size

Exact value has to be obtained with power curves

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

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

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

CV=15%

CV=30%


Calculation assuming thattreatments are equal

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

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

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

2

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CV expressed as 0.3 for 30%


Example of calculation assuming thattreatments are equal

If we desire a 80% power, Z(1-(/2)) = -1.281551566

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

The equation becomes: N = 343.977655 x S2

Given a CV of 30%, S2 = 0.086177696

Then N = 29.64

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

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


Example of calculation assuming thattreatments are equal: Effect of power




Given a CV of 30%, S2 = 0.086177696

Then N = 37.46




Example of calculation assuming thattreatments are equal: Effect of CV




Given a CV of 25%, S2 = 0.06062462

Then N = 26.35




Calculation assuming thattreatments are not equal

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

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

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


Example of calculation assuming thattreatments are 5% different

If we desire a 90% power, Z(1-) = -1.28155157


If we assume thatT/R=1.05


Given a CV of 40 %, S2 = 0.14842001

Then N = 83.62




Example of calculation assuming thattreatments are 5% different: Effect of power





Given a CV of 40 %, S2 = 0.14842001

Then N = 60.37




Example of calculation assuming thattreatments are 5% different: Effect of CV If we desire a 80% power, Z(1-) = -0.84162123




Given a CV of 20 %, S2 = 0.03922071

Then N = 15.95




Example of calculation assuming thattreatments are 10% different





Given a CV of 20 %, S2 = 0.03922071

Then N = 34.37




How to calculate the CVbased on the 90% CI of a BE study


Example of calculation of the CV based on the 90% CI

Given a 90% CI: 82.46 to 111.99 in BE study with N=24

Log-transform the 90% CI: 4.4123 to 4.7184

The mean of these extremes is the point estimate: 4.5654

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

The width in log-scale is 4.7184 – 4.5654 = 0.1530

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


Example of calculation of the CV based on the 90% CI

Given a N = 24, the degrees of freedom are 22

t = DISTR.T.INV(0.1;n-2) = 1.7171

Standard error of the difference (SE(d)) = Width / t-value = 0.1530 / 1.7171 = 0.0891

Square it: 0.08912 = 0,0079 and divide it by 2 = 0.0040

Multiply it by the sample size: 0.0040x24 = 0.0953 = MSE

CV (%) = 100 x √(eMSE-1) = 100 x √(e0.0953-1) = 31.63 %


Thank you very much for your attention!

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analysis of a cross over design

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