analysis of a cross over design
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Transcript of 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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
How to perform the statistical analysis of a 2x2 cross-over bioequivalence study
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
With complex formulae
22
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Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
More complex formulae
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Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
And really complex formulae
2
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Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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!!!
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
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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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
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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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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.
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
How to calculate the sample size of a 2x2 cross-over bioequivalence study
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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%)
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
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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)
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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%
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
2121
2
25.12
LnZZs
N w 22 1 CVLnsw
CV expressed as 0.3 for 30%
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
Example of calculation assuming thattreatments are equal: Effect of power
If we desire a 90% power, Z(1-(/2)) = -1.644853627
Consumer risk always 5%, Z(1-) = -1.644853627
The equation becomes: N = 434.686167 x S2
Given a CV of 30%, S2 = 0.086177696
Then N = 37.46
We have to round up to the next pair number: 38
Plus e.g. 4 extra subject in case of drop-outs
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
Example of calculation assuming thattreatments are equal: Effect of CV
If we desire a 90% power, Z(1-(/2)) = -1.644853627
Consumer risk always 5%, Z(1-) = -1.644853627
The equation becomes: N = 434.686167 x S2
Given a CV of 25%, S2 = 0.06062462
Then N = 26.35
We have to round up to the next pair number: 28
Plus e.g. 4 extra subject in case of drop-outs
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
Calculation assuming thattreatments are not equal
2
211
2
25.12
LnLnZZs
NRT
w
1RT
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
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
Example of calculation assuming thattreatments are 5% different
If we desire a 90% power, Z(1-) = -1.28155157
Consumer risk always 5%, Z(1-) = -1.644853627
If we assume thatT/R=1.05
The equation becomes: N = 563.427623 x S2
Given a CV of 40 %, S2 = 0.14842001
Then N = 83.62
We have to round up to the next pair number: 84
Plus e.g. 8 extra subject in case of drop-outs
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
Example of calculation assuming thattreatments are 5% different: Effect of power
If we desire a 80% power, Z(1-) = -0.84162123
Consumer risk always 5%, Z(1-) = -1.644853627
If we assume thatT/R=1.05
The equation becomes: N = 406.75918 x S2
Given a CV of 40 %, S2 = 0.14842001
Then N = 60.37
We have to round up to the next pair number: 62
Plus e.g. 6 extra subject in case of drop-outs
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
Example of calculation assuming thattreatments are 5% different: Effect of CV If we desire a 80% power, Z(1-) = -0.84162123
Consumer risk always 5%, Z(1-) = -1.644853627
If we assume thatT/R=1.05
The equation becomes: N = 406.75918 x S2
Given a CV of 20 %, S2 = 0.03922071
Then N = 15.95
We have to round up to the next pair number: 16
Plus e.g. 2 extra subject in case of drop-outs
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
Example of calculation assuming thattreatments are 10% different
If we desire a 80% power, Z(1-) = -0.84162123
Consumer risk always 5%, Z(1-) = -1.644853627
If we assume thatT/R=1.11
The equation becomes: N = 876.366247 x S2
Given a CV of 20 %, S2 = 0.03922071
Then N = 34.37
We have to round up to the next pair number: 36
Plus e.g. 4 extra subject in case of drop-outs
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
How to calculate the CVbased on the 90% CI of a BE study
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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)
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
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 %
Assessment of Bioequivalence Data22nd June 2015, Addis Ababa. EFMHACA Training
Thank you very much for your attention!