2015 Duke-Industry Statistics Symposium. Sample Size Determination for a Three-arm Equivalence Trial of Poisson and Negative Binomial Data

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1 2015 Duke-Industry Statistics Symposium Sample Size Determination for a Three-arm Equivalence Trial of Poisson and Negative Binomial Data Victoria Chang Senior Statistician Biometrics and Data Management October 23 rd, 2015

2 Outline Equivalence Assessment: The Gold Standard Design Hypotheses based on Regression Model Numerical Results Summary

3 Equivalence Assessment: The Gold Standard Design In a three-arm clinical trial, an equivalence assessment may be carried out by demonstrating the following two superiority tests and one equivalence test Superiority Test H H 01 a1 (at α = 2.5%) : T 0 H P 02 : m R - m P 0 and : 0 H a2 : m R - m P > 0 T P Equivalence Test H H 03 a3 : : T * 1 R * 1 T R or * 2 (at α = 5%) T R * 2 or T m H03 : T 1 or ³ d 2 R m R : T H a R 3

4 Equivalence Test for Discrete Data Poisson and Negative Binomial Distributed Data Example: Frequency of adverse events, asthma, migraines Assume lower incidence rate is more favorable Poisson : Mean=Variance Negative Binomial: overdispersion parameter 4

5 Regression Model Let Y ij ~Pois μ j t ij and Z ij ~NB μ j t ij, k, where i = 1,2 n j, j = 0,1,2 Link Function: log(μ j t ij ) = log t ij + β 0 + β 1 x i1 + β 2 x i2 x i1 and x i2 are indicator functions Then μ 0 = e β 0 μ 1 = e β 0+β 1 μ 2 = e β 0+β 2 5

6 Hypotheses Superiority Test Equivalence Test H 01 : μ 2 1 H μ 02 : μ 2 δ 0 μ 1 or μ 2 δ 1 μ 2 1 α = 2.5% α = 5% H a1 : μ 2 μ 0 < 1 H a2 : δ 1 < μ 2 μ 1 < δ 2 H 01 : β 2 0 H 02 : β 2 β 1 δ 1 H 03 : β 2 β 1 δ 2 H a1 : β 2 < 0 H a2 : β 2 β 1 > δ 1 H a3 : β 2 β 1 < δ 2 where δ i = logδ i i = 1,2 Assume smaller incidence rate is more favorable 6

7 Test Statistics and Decision Rule Test Statistics: T 1 = β 2 V 01, T 2 = β 2 β 1 δ 1 V 02, T 3 = β 2 β 1 δ 2 V 03 where V 0i is the variance under H 0i, i = 1,2,3 Decision Rule: Claim equivalence of the test and the reference treatments by rejecting all null hypotheses if T 1 < Z 0.025, T 2 < Z 0.05 and T 3 > Z

8 Power Function Power (μ 0, μ 1, μ 2, δ 1, δ 2 ) P T 1 < Z 0.025, T 2 < Z 0.05, T 3 > Z 0.05 μ 0, μ 1, μ 2, δ 1, δ 2 P μ0,μ 1,μ 2,δ 1,δ 2 Z 1 < Z V 01 β 2 V a1, Z 0.05 V 02 β 2 β 1 + δ 1 V a2 < Z 2 < Z 0.05 V 03 β 2 β 1 + δ 2 V a2 Z 1 = β 2 β 2 V a1, Z 2 = β 2 β 1 (β 2 β 1 ) V a2 (Z 1, Z 2 )~N 0 0, 1 ρ with ρ = Var β 2 Cov(β 1,β 2 ) ρ 1 Var β 2 Var(β 2 β 1 ) V 01, V 02, V 03 are the variances of β 2 and β 2 β 1 under null V a1, V a2 are the variances of β 2 and β 2 β 1 under alternative space 8

9 Variance Derivation for Poisson Model 2 j=0 n j l θ; Y = μ j t ij + y ij log μ j t ij logy ij! i=1 Assume the exposure time of the subjects in each arm are the same and denoted as t 0, t 1 and t 2 I 1 = 1 p p+q 1 q 1 1 p+r Var β 1 = p 1 + q 1 = n 0 t 0 μ 0 n 1 t 1 μ 1 Var β 2 = p 1 + r 1 = n 0 t 0 μ 0 n 2 t 2 μ 2 Var β 2 β 1 = q 1 + r 1 = n 1 t 1 μ 1 n 2 t 2 μ 2 r where p = n 0 t 0 e β 0, q = n 1 t 1 e β 0+β 1 and r = n 2 t 2 e β 0+β 2 9

10 Variances under Null Derive V 01 under H 01 : β 2 = 0 μ 0 = μ 2 RMLE: μ 0 = μ 2 = μ 0n 0 t 0 +μ 2 n 2 t 2 n 0 t 0 +n 2 t 2, μ 1 = μ 1 V 01 = Var(β 2 ) H 01 = 1 n 0 t 0 μ n 2 t 2 μ 2 Derive V 02, V 03 under H 02, H 03 : β 2 β 1 = δ i μ 2 = e δ iμ 1 Apply Lagrange multiplier RMLE: μ 1 = μ 1n 1 t 1 +μ 2 n 2 t 2 e δ in 2 t 2 +n 1 t 1, μ 2 = e δ iμ 1 V 0i = Var(β 2 β 1 ) H 0i = 1 n 1 t 1 μ n 2 t 2 μ 2 where i = 1,2 10

11 Sample Size Example for Poisson Data Power = 80% δ 1, δ 2 = 0.8,1.25 n 1 = a n 0 and n 2 = b n 0 μ t μ 0 μ 1 μ 2 /μ 1 (a, b) n 0 Total (1,1) (1,2) (2,2) (1,1) (1,2) (2,2) /0.9 (1,1) (1,2) (2,2)

12 Sample Size Example for Poisson Data Power = 80% δ 1, δ 2 = 0.8,1.25 n 1 = a n 0 and n 2 = b n 0 μ t μ 0 μ 1 μ 2 /μ 1 (a, b) n 0 Total (1,1) (test better) (1,2) (2,2) (1,1) (1,2) (2,2) /0.9 (1,1) (reference better) (1,2) (2,2)

13 power power Power vs. sample size for the placebo arm of Poisson and negative binomial data for T = 3, μ 1 = 1. 8, μ 2 = 2, a = b = 2, and dispersion parameter k =1. Poisson mut=3_mu1=1.8_mu2=2_a=2_b=2 mut=3_mu1=1.8_mu2=2_a=2_b=2 Negative Binomial Mu0=3 μ 0 = 3 μmu0=2.2 0 = μmu0=2.1 0 = Mu0=3 μ 0 = 3 μmu0=2.2 0 = μmu0=2.1 0 = n0 n n0n 0 13

14 power power Power vs. placebo incidence rate of Poisson and negative binomial data for T = 1, μ 1 = μ 2 = 2, a = b = 2 and dispersion parameter k =1. Poisson Poisson_mut=1_Mu1=2_Mu2=2_a=2_b=2 Negative Binomial NB_Mut=1_Mu1=2_Mu2=2_a=2_b=2_k=1 nn.0=100 0 = nn.0=200 0 = nn.0=100 0 = nn.0=200 0 = mu0 μ μmu0 0 14

15 power power Power vs. incidence rate of test treatment of both Poisson and negative binomial data for T = 1, μ 0 = 3, μ 1 = 2, n 0 = 200 and k =1. Poisson Poisson_Mut=1,Mu0=3,Mu1=2,n0=200 Negative Binomial NB_mut=1_Mu0=3_Mu1=2_N0=200 (a,b)=(3,3) (a,b)=(1,5) (a,b)=(5,1) (a,b)=(3,3) (a,b)=(1,5) (a,b)=(5,1) μ μ μ 2 /μ 1 mu μ 2 /μ 1 Mu

16 power power Misuse of a Poisson Model Power vs. k for different assigned parameter values with required sample size calculated based on the Poisson Model. Power vs. k for different assigned parameter values. Each power curve is generated under the condition of total sample size = 500. k 0 Poisson T = mut=3_mu1=2.4 3, μ 1 = 2.4 T = 3, μ 1 = mut=3_mu1=2.4_n0= , n 0 = 100, a, b = 2,2 μmu1/mu2=1, 2 /μ 1 = 1, (a,b)=(2,2) (a,b)=(2,2) μmu1/mu2=1, 2 /μ 1 = 1, (a,b)=(1,2) (a,b)=(1,2) μmu1/mu2=0.9, 2 /μ 1 = (a,b)=(2,2) (a,b)=(2,2) μmu1/mu2=0.9, 2 /μ 1 = (a,b)=(1,2) (a,b)=(1,2) μmu1/mu2=1/0.9, 2 /μ 1 = 1/0.9, (a,b)=(2,2) (a,b)=(2,2) μmu1/mu2=1/0.9, 2 /μ 1 = 1/0.9, (a,b)=(1,2) μ 2 Mu1/Mu2=1 /μ 1 = 1 μ 2 Mu1/Mu2=0.9 /μ 1 = μ 2 Mu1/Mu2=1/0.9 /μ 1 = 1/ k k 16

17 Summary Poisson and Negative Binomial Data Power is affected by μ 0 when it is very close to μ 2, otherwise the power is dominated by the equivalence test. When sample size for the test arm and the reference arm are equal, it gives the highest power. Assign more subjects to the better treatment gives a higher power for Poisson Model. A wrong model may cause a very low power and incorrect interpretation. 17

18 Reference Pigeot, I., Schafer, J., Rohmel, J., & Hauschke, D. (2003). Assessing non-inferiority of a new treatment in a three-arm clinical trial including a placebo. Stat Med, 22(6), Rohmel, J., & Pigeot, I. (2010). A comparison of multiple testing procedures for the gold standard non-inferiority trial. J Biopharm Stat, 20(5), Esinhart, J. D., & Chinchilli, V. M. (1994). Extension to the use of tolerance intervals for the assessment of individual bioequivalence. J Biopharm Stat, 4(1), EMEA (2001). Note for Guidance on the Investigation of Bioavailability and Bioequivalence, The European Medicines Agency Evaluation of Medicines for Human Use. EMEA/EWP/QWP/1401/98, London, United Kingdom. FDA (2001). Guidance on Statistical Approaches to Establishing Bioequivalence, Center for Drug Evaluation and Research, the US Food and Drug Administration, Rockville, Maryland, USA. FDA (2003a). Guidance on Bioavailability and Bioequivalence Studies for Orally Administrated Drug Products General Considerations, Center for Drug Evaluation and Research, the US Food and Drug Administration, Rockville, Maryland, USA. FDA (2003b). Guidance on Bioavailability and Bioequivalence Studies for Nasal Aerosols and Nasal Sprays for Local Action, Center for Drug Evaluation and Research, the US Food and Drug Administration, Rockville, Maryland, USA. Friede, T., Mitchell, C., & Müller Velten, G. (2007). Blinded Sample Size Reestimation in Non Inferiority Trials with Binary Endpoints. Biometrical Journal, 49(6), Generic drug (2013). from Hsieh, T-C., Chow, S.-C., Liu, J-P., Hsiao, C-F., Chi, E. (2010). Statistical test for evaluation of biosimilarity in variability of follow-on biologics. J. Biopharm. Stat. 20(1): Hauschke, D., Kieser, M., Diletti, E., and Burke, M. (1999). Sample Size Determination for Proving Equivalence Based on the Ratio of Two Means for Normally Distributed Data. Stat Med,18: Hilton, J. F. (2006). Designs of superiority and noninferiority trials for binary responses are noninterchangeable. Biometrical journal, 48(6),

19 Thank you! Comments and Questions