臨床試驗 Analysis of 2x2 Crossover Design 授課老師: 劉仁沛教授 國立台灣大學 與 國家衛生研究院 臨床試驗 Analysis of 2x2 Crossover Design 【本著作除另有註明外，採取創用CC「姓名標示－非商業性－相同方式分享」台灣3.0版授權釋出】

Inference for 2x2 Crossover Design References: Chow, SC, and Liu, JP (2014) Design and Analysis of Clinical Trials, 3rd Ed. Wiley Chow, SC and liu, JP (2008) Design and Analysis of Bioavailability and Bioequivalence, 3rd Ed., Chapman & Hall/ CRC, Chapter 3 Chow, SC and Liu, JP (1998) Design and Analysis of Animal Studies in Pharmaceutical Development, Marcel Dekker. PhRMA (2007) Drug Discovery and Development: understanding the R&D process, The Pharmaceutical Research and Manufactures of America

Inferences for the 2x2 Design 3.1 Introduction Data Structure Sequence Period I Period II 1 (RT) Reference Test Data: Yi11 Data:Yi21 2 (TR) Test Reference Data: Yi12 Data:Yi22

Inferences for the 2x2 Design The general model for the standard 2x2 crossover design Yijk =  + Sik + Pj + F(j,k) + C(j-1,k) + eijk where i(subject) = 1,…,nk, j(period) = 1,2, k(sequence) = 1,2.

Inferences for the 2x2 Design FR, if k=j F(j,k) = { k=1,2; j=1,2 FT, if kj CR, if k=1, j=2 C(j-1,k) = { CT, if k=2, j=2

Inferences for the 2x2 Design Fixed effects Sequence Period I Period II 1 (RT) 11=  +P1+FR 21 =  +P2+FT+CR 2 (TR) 12=  +P1+FT 22=  +P2+FR+CT where P1 + P2 = 0, FR + FT = 0, and CR + CT =0.

Inferences for the 2x2 Design Assumptions Sik iid ~ N(0, S2) eijk iid ~ N(0, e2) Sik and eijk are mutually independent

Inferences for the 2x2 Design 3.2 The Carryover Effects Subject totals: Uik = Yi1k + Yi2k 2 + CR, if sequence 1 E(Uik) = { 2 + CT, if sequence 2 V(Uik) = 2(e2 + 2s2) = u2 {U11…,Un11} and {U12…,Un22} are independent samples with the same variance u2

Inferences for the 2x2 Design Define C = CT - CR Ho: C = 0 vs. Ha: C  0  U.k = (1/nk)Uik, k=1,2. The MVUE OF C is given as       Ĉ =U.2 – U.1 = (Y.12+ Y.22) - (Y.11+ Y.21)

Inferences for the 2x2 Design V(Ĉ) = u2 [(1/n1) + (1/n2)]  V(Ĉ) = su2 [(1/n1) + (1/n2)], where  su2 =  (Uik - U.k)2 /(n1+n2 –2) Tc = c/v(c)

Inferences for the 2x2 Design Reject Ho if Tc > t(/2, n1+n2 –2) Confidence interval c  t(/2, n1+n2 –2)v(c) The carryover effect is confounded with the sequence effect

Inferences for the 2x2 Design 2.3 The Direct Drug Effect Period differences dik = (Yi2k – Yi1k)/2 [(P2 - P1) + (FT - FR) + CR]/2, if sequence = 1 E(dik) = { [(P2 - P1) + (FR - FT) + CT]/2, if sequence = 2 V (dik) = d2 = e2/2

Inferences for the 2x2 Design {d11…,dn11} and {d12…,dn22} are independent samples with the same variance d2 Sample means of period differences  d.k = (1/nk)dik, k=1,2 Define F = FT - FR   E(d.1 - d.2) = F – C/2.

Inferences for the 2x2 Design Unless C = 0, no unbiased estimator for F based on the data from both period exists       f = d.1 – d.2 = (Y.21 - Y.11) - (Y.22 - Y.12)     = (Y.21+ Y.12) - (Y.11+ Y.12)   = Y.T - Y.R f is a linear combination of the sequence-by-period means and the least squares estimator of F

Inferences for the 2x2 Design V(f) = d2 [(1/n1) + (1/n2)] v(f) = sd2 [(1/n1) + (1/n2)], where  sd2 =  (dik - d.k)2 /(n1+n2 –2) Ho: F = 0 vs. Ha: F  0 Test Statistic Td = f/v(f)

Inferences for the 2x2 Design Under the assumption of C=0 Reject Ho if TF > t(/2, n1+n2 –2) Confidence interval f  t(/2, n1+n2 –2)v(f)

Inferences for the 2x2 Design When C  0, f is not unbiased for F. However, unbiased estimator can be obtained as the difference of sample means of the first period between the two sequences:   f C = Y.11 - Y.12 V(f C) = (e2 + s2) [(1/n1) + (1/n2)] V(f C) – V(f) = (e2/2 + s2) [(1/n1) + (1/n2)]

Inferences for the 2x2 Design 3.4 The Period Effect Crossover difference dik , if sequence = 1 Oik = { -dik , if sequence = 2

Inferences for the 2x2 Design [(P2 - P1) + (FT - FR) + CR]/2, if sequence = 1 E(Oik) = { [(P1 – P2) + (FT - FR) - CT]/2, if sequence = 2 V (Oik) = d2 = e2/2 {O11…,On11} and {O12…,On22} are independent samples with the same variance d2. The inference for the period effect can be performed as that for the carryover effects and direct effect. (homework)

Inferences for the 2x2 Design 3.5 The Analysis of Variance  SStotal =  (Yijk - Y…)2    =  (Yijk - Yi.k + Yi.k - Y... )2    =  (Yijk - Yi.k)2 + (Yi.k - Y... )2 = SSwithin + SSbetween

Inferences for the 2x2 Design SSbetween = SScarry + SSinter     SScarry = [2n1n2/(n1+n2)][(Y.12+ Y.22) - (Y.11+ Y.21)]2 df = 1 SSinter = Y2i.k/2 - Y2..k/2nk, df = n1+n2-2 E(MScarry) = [2n1n2/(n1+n2)](CT - CR)2 + e2 + 2s2 E(MSinter) = e2 + 2s2

Inferences for the 2x2 Design SSwithin = SSdrug + SSperiod + SSintra     SSdrug = [2n1n2/(n1+n2)]{1/2[(Y.21 - Y.11) - (Y.22+ Y.12)]}2 df =1 SSperiod = [2n1n2/(n1+n2)]{1/2[(Y.21 - Y.11) - (Y.12+ Y.22)]}2 df = 1 SSintra = Y2ijk - Y2i.k/2 - Y2.jk /nk - Y2..k/2nk, df = n1+n2-2

Inferences for the 2x2 Design E(MSdrug)=[2n1n2/(n1+n2)][(FT - FR)+(CT - CR)/2]2+e2 E(MSperiod) = [2n1n2/(n1+n2)](P2 – P1)2 + e2 E(MSintra) = e2

Inferences for the 2x2 Design Carryover effect: Fc = MScarry/MSinter Direct Drug Effect: Fd = MSdrug/MSintra Period Effect : FP = MSperiod/MSintra Intersubject variability Fv = MSinter/MSintra

Inferences for the 2x2 Design 3.6 Example (Chow and Liu, 2008) Sequence Sequence Period I Period II Mean 1 (RT) Y.11= 85.82 Y.21= 81.80 83.81 2 (TR) Y.12= 78.74 Y.22= 79.30 79.02 Period mean 82.28 80.55 81.42

Inferences for the 2x2 Design Summary of Inference for the Fixed Effects Estimated Effect MVUE Variance 95% C.I. T p-value Carryover -9.59 245.63 (-42.10, 22.91) -0.612 0.5468 Direct Drug -2.29 13.97 (-10.03, 5.46) -0.613 0.5463 Period -1.73 13.97 ( -9.47, 6.01) -0.464 0.6474

Inferences for the 2x2 Design The Analysis of Variance Table SOV df SS MS F p-value Intersubject Carryover 1 276.00 276.00 0.37 0.5468 Residual 22 16211.49 736.89 4.41 0.0005 Intrasubject Direct Drug 1 62.79 62.79 0.38 0.5463 Period 1 35.97 35.97 0.22 0.6474 Residual 22 3679.43 167.25 Total 47 20265.68

Inferences for the 2x2 Design

Inferences for the 2x2 Design

臨床試驗 Evaluation of Equivalence and Non-inferiority 授課老師: 劉仁沛教授 國立台灣大學 與 國家衛生研究院 【本著作除另有註明外，採取創用CC「姓名標示－非商業性－相同方式分享」台灣3.0版授權釋出】

Design and Analysis of Bioequivalence Studies References: Chow, S.C. and Liu, J.P. (2008)Design of Bioavailability and Bioequivalence Studies, 3rd Ed., CRC/Chapman & Hall, Chapter 4 and 5 FDA (2003) Guidance on Bioavailability and Bioequivalence Studies for Orally Administered Drug Products – General Considerations FDA(2001) Guidance on Statistical Approaches to Establishing Bioequivalence

Methods for Average Bioavailability 4.1 Introduction 4.2 Interval Hypothesis Testing 4.3 The Confidence Interval Approach 4.4 Log-transformation 4.4 Discussion

Methods for Average Bioavailability 4.1 Introduction The general model for the 2x2 crossover design without carryover effect: Yijk =  + Sik + Pj + F(j,k) + eijk where i(subject) = 1,…,nk, j(period), k(sequence) = 1,2. Define T =  + FT and R =  + FR       Y.R = (Y.11+ Y.22) and Y.T = (Y.21+ Y.12)   E(Y.T) = T and E(Y.R) = R

Methods for Average Bioavailability Issues on Assessment of ABE Selection of ABE measures difference T - R vs ratio T/R Raw Data vs Log-transformed Data Determination of ABE limits 20/20 in difference or 80/125 in ratio 4. Hypotheses Differences vs equivalence

Methods for Average Bioavailability Raw Data: L < T - R < U 20/20 rule: L= -0.2R and U = 0.2R, R is unknown L < T/R < U  (L – 1)R < T - R < (U – 1)R, R is unknown. Log-transformed data: L < T/R < U  ln(L)< ln(T) – ln(R) < log(U) Limits are not functions of unknown R

Methods for Average Bioavailability 4.2 Interval Hypotheses Testing Difference: Ho: T - R  L or T - R  U vs. Ha: L < T - R < U Ratio: Ho: T/R  L or T/R  U vs. Ha: L < T/R < U

Methods for Average Bioavailability Two one-sided hypotheses HoL: T - R  L vs. HaL: T - R > L and HoU: T - R  U vs. HaU:T - R < U The parameter space of Ho is the union of the parameter spaces of HoLand HoU. The parameter space of Ha is the intersection of the parameter spaces of HaLand HaU.

Methods for Average Bioavailability 4.2.2 Schuirmann’s Two One-sided Tests (TOST) Procedure Conclude ABE if TL = (f - L)/v(f) > t(, n1+n2 –2) and TU = (f - U)/v(f) < -t(, n1+n2 –2). TOST procedure control the type I error rate at its nominal  level. Nonparametric TOST is also available.

Methods for Average Bioavailability Example 4.2.1:   Y.R = 82.559, and Y.T = 80.272 t(0.05, 22) = 1.717 and sd = 9.145 U = -L = 0.2(82.559) = 16.51 TL = [(80.272-82.559) +16.51]/9.145sqrt(1/6) = 3.810 TU = [(80.272-82.559) -16.51]/9.145sqrt(1/6) = -5.036 Issue: U = -L is actually an estimate.

Methods for Average Bioavailability 4.3 Confidence Interval Approach Data: Raw and untransformed data If a (1-2)100% confidence interval for the difference T - R or the ratio T/R is within the acceptance limits as recommended by the regulatory agency, then accept the test formulation; otherwise reject it. The confidence interval approach is operationally equivalent to the TOST procedure based on the difference T - R  = 5%  90% C.I. T - R: 20% T/R: (80%, 120%).

Methods for Average Bioavailability 4.3.1 The Classical Confidence Interval:   L1= (Y.T - Y.R) - t(, n1+n2 –2)sd2 [(1/n1) + (1/n2)],   U1= (Y.T - Y.R) + t(, n1+n2 –2)sd2 [(1/n1) + (1/n2)]. and   L2= (L1/Y.R + 1)x100%, and U2= (U1/Y.R + 1)x100% Conclude ABE if (1) (L1, U1)  (L, U), L,= -0.2R and U = 0.2R or (2) (L2, U2)  (80%, 120%)

Methods for Average Bioavailability Example 4.3.1   Y.R = 82.559, and Y.T = 80.272 t(0.05, 22) = 1.717 and sd = 9.145 U= -L = 0.2(82.559) = 16.51 (L1, U1) = (-8.698, 4.123) (-16.51, 16.51) (L1, U1) = (89.46%, 104.99%) (80%, 120%) Issue: U = -L is actually an estimate.

Methods for Average Bioavailability 4.3.2 CI Based on Fieller’s Theorem 1. For ratio T/R based on raw data 2. Take variability of estimator of R into account  = T/R = ( + FT)/( + FR) Define (Yi21 - Yi11) , if sequence 1 U*ik = { (Yi12 - Yi22) , if sequence 2

Methods for Average Bioavailability {U*11,…, U*n11} are iid normal with mean (P2 - P1) and variance 2, {U*12,…, U*n22} are iid normal with mean -(P2 - P1) and variance 2, where 2 = (T2+S2) - 2S2+ 2(R2+S2).   T = (U*.1 + U*.2)/sqrt(wS2U)  t(n1+n2 –2).     U*.1 + U*.2 = Y.T - Y.R, w = [(1/n1) + (1/n2)]/4 S2U = S2TT - 2S2TR + 2S2RR.

Methods for Average Bioavailability   S2RR = [1/(n1+n2-2)][(Yi11- Y.11)2 + (Yi22- Y.22)2]   S2TT = [1/(n1+n2-2)][(Yi21- Y.21)2 + (Yi12- Y.12)2]   SSTR = [1/(n1+n2-2)][(Yi11- Y.11)(Yi21 - Y.21) +   (Yi12- Y.12)(Yi22 - Y.22)]

Methods for Average Bioavailability   T = (Y.T - Y.R.)/sqrt[w(S2TT - 2STR + 2S2RR)] A (1-2)100% CI for  is the set that {| T2 < t2(, n1+n2 –2)}. To solve the following quadratic equation (Y.T - Y.R.)2 - t2(,n1+n2 –2)[w(S2TT - 2STR + 2S2RR)] Conditions for the limits are positive real numbers:  Y.R/sqrt(wS2RR) > t(,n1+n2 –2) and Y. T/sqrt(wS2TT) > t(,n1+n2 –2).

Methods for Average Bioavailability Example 4.3.2:  Y.R/sqrt(wS2RR) = 19.27 > t(0.05, 22) = 1.717 Y. T/sqrt(wS2TT) = 18.73 > t(0.05, 22) = 1.717 90% CI for  is given as (89.78%, 105.19%)

4.4 Log-transformation The distributions of PK measures such as AUC and Cmax are skewed. FDA 2003 guidance suggest that statistical evaluation for BE should be performed based on the log-transformation of AUC and Cmax.

Log-transformation The multiplicative (log-transformed) model Xijk = ’S’ikP’jF’(j,k)C’(j-1,k)e’ijk or Yijk = ln(Xijk) = +Sik+Pj+F(j,k)+C(j-1,k)+eijk Xijk = exp( + Sik + Pj + F(j,k) + C(j-1,k) + eijk)

Log-transformation (Xi1k, Xi2k) follows a bivariate lognormal distn. Sequence-by-period means and medians in log-transformed model Seq Period I Period II 1 Mean exp[(+P1+FR)+(R2+S2)/2] exp[(+P2+FT)+(T2+S2)/2] Median exp(+P1+FR) exp(+P2+FT) 2 Mean exp[(+P1+FT)+(T2+S2)/2] exp[(+P2+FR)+(R2+S2)/2] Median exp(+P1+FT) exp(+P2+FR)

Log-transformation BE measures 1 Ratio of Medians  = ’T/’R = (’ + F’T)/(’ + F’R) = exp[( + FT)/( + FR)] = exp[(FT - FR)] = exp(F) 2. Ratio of Means M = exp[F +(T2-R2)/2].

Log-transformation Ho: ’T/’R  L or ’T/’R  U vs. Ha: L < ’T/’R < U Ho: ln(’T) – ln(’R) ln(L) or ln(’T)-ln(’R) ln(U) vs. Ha: ln(L) < ln(’T) - ln(’R) < ln(U) Ho: T - R  L or T - R  L vs. Ha: L < T - R < U L and U are known nonrandom constants, i.e., ±0.2231 for 80/125.

Log-transformation The Confidence interval approach and Schuirmann’s TOST can be directly applied to the log-transformed data for assessment of ABE. The period difference on the log-scale is the period ratio on the original scale dik = (Yi2k – Yi1k)/2 = ln (Xi2k/Xi1k)/2 = ln(rik)/2.

Log-transformation         f = d.1 – d.2 = Y.T - Y.R     f = d.1 – d.2 = Y.T - Y.R The MLE under the normality assumption for the log-transformed data is given as     f’ = exp(f) = exp(d.1 – d.2) = exp(Y.T - Y.R)

Log-transformation Let Rk be the geometric mean of period ratios obtained from sequence k, k =1,2. f’ = (R1/R2)1/2 f’ follows an univariate log-normal distribution with mean exp(md2/2). f’is biased for  The (1-2)100% CI for  is given as [exp(L1), exp(U1)]

Log-transformation Under 80/125 rule, ABE is concluded if exp(L1) > 0.8 and exp(U1) < 1.25, or L1 > -0.2231 and U1 < 0.2231 Reasons of selection of 80% and 125% as ABE limits 1. Symmetric about 0 on log-scale. 2. Maximal power occurs when ratio is 1. 3. Maximal power for 80/120 limit occurs at 0.98.

Log-transformation The MVUE for  is give as f’MVUE = f’(-mSSD), where SSD is the pooled SS of period differences on the log-transformed data,  is df, (-mSSD)={(/2)/[(/2+j)j!}[(-m/4)SSD]j and (.) is gamma function. E[(cSSD)] = exp[(c/2)d2] E{[(cSSD)]2} = exp[cd2](c d4)

Log-transformation Theorem Bias(f’) = [exp(md2/2) –1]. MSE(f’) = 2[exp(md2/2) –1]2 + 2exp(md2/2)[exp(md2/2) –1] V(f’MVUE) = 2{exp(md2/2)[(md2)2] – 1} v(f’MVUE) = exp(2f){[(-mSSD)]2 - (-4mSSD)}

Log-transformation Example Data: p181 of Chow and Liu (2008) ANOVA Table based on log-transformed data SS df SS MS F p-value Intersubject Carryover 1 1.3053 1.3053 2.50 0.1332 Residuals 16 8.3434 0.5215 1.69 Intrasubject Formulation 1 1.0470 1.0470 3.39 0.0843 Period 1 0.l959 0.1959 0.63 0.4376 Residual 16 4.9440 0.3090

Log-transformation   Y.T = 1.1798, Y.R= 1.5209, f = -0.3411 SSD = 2.472, sd2 = 0.1545 (L1, U1) = (-0.6646, -0.0176) f’ = exp(-0.3411) = 0.7111 [exp(-0.6646), exp(-0.0176)] = (0.5145, 0.9826) Bias (f’) = 0.0123, MSE(f’) = 0.0183 + 0.01232 =0.0184 f’MVUE = (0.7110)(0.9831) = 0.6990 v(f’) = 0.0184 v(f’MVUE) = 0.71102[(0.98312 – 0.9354] = 0.0157 Eff(f’MVUE, f’) = 0.0184/0.0157 = 117.39%

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