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Statistical considerations Alfredo García – Arieta, PhD Training workshop: Training of BE assessors, Kiev, October 2009.

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Presentation on theme: "Statistical considerations Alfredo García – Arieta, PhD Training workshop: Training of BE assessors, Kiev, October 2009."— Presentation transcript:

1 Statistical considerations Alfredo García – Arieta, PhD Training workshop: Training of BE assessors, Kiev, October 2009

2 2 |2 | Outline Basic statistical concepts on equivalence 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

3 Training workshop: Training of BE assessors, Kiev, October 2009 3 |3 | Basic statistical concepts

4 Training workshop: Training of BE assessors, Kiev, October 2009 4 |4 | Type of studies Superiority studies –A is better than B (A = active and B = placebo or gold-standard) –Conventional one-sided hypothesis test Equivalence studies –A is more or less like B (A = active and B = standard) –Two-sided interval hypothesis Non-inferiority studies –A is not worse than B (A = active and B = standard with adverse effects) –One-sided interval hypothesis

5 Training workshop: Training of BE assessors, Kiev, October 2009 5 |5 | Hypothesis test Conventional hypothesis test H 0 :  =  1 H 1 :    1 (in this case it is two-sided) If P<0,05 we can conclude that statistical significant difference exists If P≥0,05 we cannot conclude –With the available potency we cannot detect a difference –But it does not mean that the difference does not exist –And it does not mean that they are equivalent or equal We only have certainty when we reject the null hypothesis –In superiority trials: H 1 is for existence of differences This conventional test is inadequate to conclude about “equalities” –In fact, it is impossible to conclude “equality”

6 Training workshop: Training of BE assessors, Kiev, October 2009 6 |6 | Null vs. Alternative hypothesis Fisher, R.A. The Design of Experiments, Oliver and Boyd, London, 1935 “The null hypothesis is never proved or established, but is possibly disproved in the course of experimentation. Every experiment may be said to exist only in order to give the facts a chance of disproving the null hypothesis” Frequent mistake: The absence of statistical significance has been interpreted incorrectly as absence of clinically relevant differences.

7 Training workshop: Training of BE assessors, Kiev, October 2009 7 |7 | Equivalence We are interested in verifying (instead of rejecting) the null hypothesis of a conventional hypothesis test We have to redefine the alternative hypothesis as a range of values with an equivalent effect The differences within this range are considered clinically irrelevant Problem: it is very difficult to define the maximum difference without clinical relevance for the Cmax and AUC of each drug Solution: 20% based on a survey among physicians

8 Training workshop: Training of BE assessors, Kiev, October 2009 8 |8 | Interval hypothesis or two one-sided tests Redefine the null hypothesis: How? Solution: It is like changing the null to the alternative hypothesis and vice versa. Alternative hypothesis test: Schuirmann, 1981 –H 01 :   1 H a1 :  1 <  –H 02 :   2 H a2 :  <  2. This is equivalent to: – H 0 :   1 or   2 H a :  1 <  <  2 It is called as an interval hypothesis because the equivalence hypothesis is in the alternative hypothesis and it is expressed as an interval

9 Training workshop: Training of BE assessors, Kiev, October 2009 9 |9 | Interval hypothesis or two one-sided tests The new alternative hypothesis is decided with a statistic that follows a distribution that can be approximated to a t- distribution To conclude bioequivalence a P value <0.05 has to be obtained in both one-sided tests The hypothesis tests do not give an idea of magnitude of equivalence (P<0001 vs. 90% CI: 0.95 – 1.05). That is why confidence intervals are preferred

10 Training workshop: Training of BE assessors, Kiev, October 2009 10 | Point estimate of the difference If T=R, d=T-R=0 If T>R, d=T-R>0 If T 0 Positive effect

11 Training workshop: Training of BE assessors, Kiev, October 2009 11 | Estimation with confidence intervals in a superiority trial d < 0 Negative effect d = 0 No difference d > 0 Positive effect Confidence interval 90% - 95% It is not statistically significant! Because the CI includes the d=0 value

12 Training workshop: Training of BE assessors, Kiev, October 2009 12 | Estimation with confidence intervals in a superiority trial d < 0 Negative effect d = 0 No difference d > 0 Positive effect Confidence interval 90% - 95% It is statistically significant! Because the CI does not includes the d=0 value

13 Training workshop: Training of BE assessors, Kiev, October 2009 13 | Estimation with confidence intervals in a superiority trial d < 0 Negative effect d = 0 No difference d > 0 Positive effect Confidence interval 90% - 95% It is statistically significant with P=0.05 Because the boundary of the CI touches the d=0 value

14 Training workshop: Training of BE assessors, Kiev, October 2009 14 | Equivalence study d < 0 Negative effect d = 0 No difference d > 0 Positive effect -- ++ Region of clinical equivalence

15 Training workshop: Training of BE assessors, Kiev, October 2009 15 | Equivalence vs. difference d < 0 Negative effect d = 0 No difference d > 0 Positive effect -- ++ Region of clinical equivalence Equivalent?Different? No ? Yes ? ? ? ? No

16 Training workshop: Training of BE assessors, Kiev, October 2009 16 | Non-inferiority study d < 0 Negative effect d = 0 No difference d > 0 Positive effect -- Inferiority limit Inferior? Yes ? No ?

17 Training workshop: Training of BE assessors, Kiev, October 2009 17 | Superiority study (?) d < 0 Negative effect d = 0 No difference d > 0 Positive effect ++ Superiority limit Superior? No, not clinically, but yes statistically ?, but yes statistically Yes, statistical & clinically No No, not clinically and ? statistically ? Yes, but only the point estimate

18 Training workshop: Training of BE assessors, Kiev, October 2009 18 | How to perform the statistical analysis of a 2x2 cross-over bioequivalence study

19 Training workshop: Training of BE assessors, Kiev, October 2009 19 | 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

20 Training workshop: Training of BE assessors, Kiev, October 2009 20 | The statistical analysis is not so complex Period 2Period 12x2 BE trial N=12 Y 12 Y 11 Sequence 1 (BA) BA is RT Y 22 Y 21 Sequence 2 (AB) AB is TR

21 Training workshop: Training of BE assessors, Kiev, October 2009 21 | We don’t need to calculate an ANOVA table

22 Training workshop: Training of BE assessors, Kiev, October 2009 22 | With complex formulae

23 Training workshop: Training of BE assessors, Kiev, October 2009 23 | More complex formulae

24 Training workshop: Training of BE assessors, Kiev, October 2009 24 | And really complex formulae

25 Training workshop: Training of BE assessors, Kiev, October 2009 25 | Given the following data, it is simple Period 2Period 12x2 BE trial N=12 Y 12 70, 90, 95, 70, 60, 70 Y 11 75, 95, 90, 80, 70, 85 Sequence 1 (BA) Y 22 40, 50, 70, 80, 70, 95 Y 21 75, 85, 80, 90, 50, 65 Sequence 2 (AB)

26 Training workshop: Training of BE assessors, Kiev, October 2009 26 | First, log-transform the data Period 2Period 12x2 BE trial N=12 Y 12 4.2485, 4.4998, 4.5539, 4.2485, 4.0943, 4.2485 Y 11 4.3175, 4.5539, 4.4998, 4.3820, 4.2485, 4.4427 Sequence 1 (BA) Y 22 3.6889, 3,9120, 4,2485, 4.3820, 4.2485, 4.5539 Y 21 4.3175, 4.4427, 4.3820, 4,4998, 3,9120, 4.1744 Sequence 2 (AB)

27 Training workshop: Training of BE assessors, Kiev, October 2009 27 | Second, calculate the arithmetic mean of each period and sequence Period 2Period 12x2 BE trial N=12 Y 12 = 4.316Y 11 = 4.407Sequence 1 (BA) Y 22 = 4,172Y 21 = 4.288Sequence 2 (AB)

28 Training workshop: Training of BE assessors, Kiev, October 2009 28 | 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

29 Training workshop: Training of BE assessors, Kiev, October 2009 29 | Third, calculate the LSM of T and R Period 2Period 12x2 BE trial N=12 Y 12 = 4.316Y 11 = 4.407Sequence 1 (BA) Y 22 = 4,172Y 21 = 4.288Sequence 2 (AB) B = 4.2898 A = 4.3018

30 Training workshop: Training of BE assessors, Kiev, October 2009 30 | 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 = e F = e 0.01198 = 1.01205 Five very simple steps to calculate the point estimate!!!

31 Training workshop: Training of BE assessors, Kiev, October 2009 31 | 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: d 1 and d 2 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 (n 1 +n 2 -2), where n 1 and n 2 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 √e MSE -1

32 Training workshop: Training of BE assessors, Kiev, October 2009 32 | This can be done easily in a spreadsheet!

33 Training workshop: Training of BE assessors, Kiev, October 2009 33 | Step 1: Calculate the difference between periods for each subject and divide it by 2: (P2-P1)/2

34 Training workshop: Training of BE assessors, Kiev, October 2009 34 | Step 2: Calculate the mean of these differences within each sequence to obtain 2 means: d1 & d2

35 Training workshop: Training of BE assessors, Kiev, October 2009 35 | Step 3: Squared differences

36 Training workshop: Training of BE assessors, Kiev, October 2009 36 | Step 4: Sum these squared differences

37 Training workshop: Training of BE assessors, Kiev, October 2009 37 | Step 5: Divide the sum by n 1 +n 2 -2

38 Training workshop: Training of BE assessors, Kiev, October 2009 38 | Calculate the confidence interval with point estimate and variability Step 11: In log-scale 90% CI: F ± t (0.1, n1+n2-2) -√((Sigma 2 (d) x (1/n 1 +1/n 2 )) F has been calculated before The t value is obtained in t-Studient tables with 0,1 alpha and n 1 +n 2 -2 degrees of freedom –Or in MS Excel with the formula =DISTR.T.INV(0.1; n 1 +n 2 -2) Sigma 2 (d) has been calculated before.

39 Training workshop: Training of BE assessors, Kiev, October 2009 39 | Final calculation: the 90% CI Log-scale 90% CI: F±t (0.1, n1+n2-2) -√((Sigma 2 (d)·(1/n 1 +1/n 2 )) F = 0.01198 t (0.1, n1+n2-2) = 1.8124611 Sigma 2 (d) = 0.02311406 90% CI: LL = -0.14711 to UL= 0,17107 Step 12: Back transform the limits with e LL and e UL e LL = e -0.14711 = 0.8632 and e UL = e 0.17107 = 1.1866

40 Training workshop: Training of BE assessors, Kiev, October 2009 40 | How to calculate the sample size of a 2x2 cross-over bioequivalence study

41 Training workshop: Training of BE assessors, Kiev, October 2009 41 | 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

42 Training workshop: Training of BE assessors, Kiev, October 2009 42 | 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 (1 st edition) and 2000 (2 nd 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)

43 Training workshop: Training of BE assessors, Kiev, October 2009 43 | 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%)

44 Training workshop: Training of BE assessors, Kiev, October 2009 44 | Calculation assuming that treatments 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%  CV expressed as 0.3 for 30%

45 Training workshop: Training of BE assessors, Kiev, October 2009 45 | Example of calculation assuming that treatments 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 S 2 Given a CV of 30%, S 2 = 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

46 Training workshop: Training of BE assessors, Kiev, October 2009 46 | Example of calculation assuming that treatments are equal 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 S 2 Given a CV of 25%, S 2 = 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

47 Training workshop: Training of BE assessors, Kiev, October 2009 47 | Calculation assuming that treatments are not equal 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

48 Training workshop: Training of BE assessors, Kiev, October 2009 48 | Example of calculation assuming that treatments are 5% different If we desire a 90% power, Z(1-  ) = -1.28155157 Consumer risk always 5%, Z(1-  ) = -1.644853627 If we assume that  T /  R =1.05 The equation becomes: N = 563.427623 x S 2 Given a CV of 40 %, S 2 = 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

49 Training workshop: Training of BE assessors, Kiev, October 2009 49 | Example of calculation assuming that treatments are 5% different If we desire a 80% power, Z(1-  ) = -0.84162123 Consumer risk always 5%, Z(1-  ) = -1.644853627 If we assume that  T /  R =1.05 The equation becomes: N = 406.75918 x S 2 Given a CV of 20 %, S 2 = 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

50 Training workshop: Training of BE assessors, Kiev, October 2009 50 | Example of calculation assuming that treatments are 10% different If we desire a 80% power, Z(1-  ) = -0.84162123 Consumer risk always 5%, Z(1-  ) = -1.644853627 If we assume that  T /  R =1.11 The equation becomes: N = 876.366247 x S 2 Given a CV of 20 %, S 2 = 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

51 Training workshop: Training of BE assessors, Kiev, October 2009 51 | How to calculate the CV based on the 90% CI of a BE study

52 Training workshop: Training of BE assessors, Kiev, October 2009 52 | 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 e 4.5654 = 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)

53 Training workshop: Training of BE assessors, Kiev, October 2009 53 | 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: 0.0891 2 = 0,0079 and divide it by 2 = 0,0040 Multiply it by the sample size: 0.0040x24 = 0,0953 = MSE CV (%) = 100 x √(e MSE -1) = 100 x √(e 0.0953 -1) = 31,63 %


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