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Topics in Clinical Trials (5 ) - 2012. J. Jack Lee, Ph.D. Department of Biostatistics University of Texas M. D. Anderson Cancer Center. Objectives of Phase II Trials. Initial assessment of drug’s efficacy (IIA) Refine drug’s toxicity profile
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Topics in Clinical Trials (5) - 2012 J. Jack Lee, Ph.D. Department of Biostatistics University of Texas M. D. Anderson Cancer Center
Objectives of Phase II Trials • Initial assessment of drug’s efficacy (IIA) • Refine drug’s toxicity profile • Compare efficacy among active agents and send the most promising one(s) to Phase III trials (IIB)
Key Elements of Phase II Trials • Patient Population • More homogeneous group with specific disease site, histology, and stage • Dose Level • At MTD or RPTD from Phase I trials • Dose adjustment may be needed • Endpoints • Short term efficacy endpoint • Response categories: CR, PR, NC, PD at, say, 4 wks • Response rate: proportion in CR + PR • Disease control rate: proportion in CR + PR + SD (for targeted agents) • Disease-free survival or progression-free survival
Available Phase II Designs • Phase IIA Trials • Gehan’s design (J. Chron. Dis. 1961) • Simon’s two-stage designs (Contr. Clin. Trials, 1989) • Other multi-stage designs • Predictive probability design • Phase IIB Trials • Simon et al.’s ranking and selection randomized phase II design (Cancer Treat. Rep. 1985) • Thall and Simon’s Bayesian phase IIB design (Biometrics, 1994)
Phase IIA Trials • Let the probability of response be p H0: p p0 H1: p p1 • p0 -- an uninteresting response rate • response rate from a standard treatment • p1 -- a desired response rate • target response rate • Control type I and II error rates • Type I error: false positive rate • Type II error: false negative rate
Hypothesis Testing Truth Ho H1 Ho H1 b Action a • Framework of hypothesis testing a: Type I error (level of significance) b: Type II error (1- b = Power) Sample Size Calculation: Find N s.t. to a and b are under control. Typically, compute N for a given a to yield (1-b)x100% power. For example, compute N for a = 0.05 to yield 80% power.
Power
Phase IIA Example • For the binary data (response or no response), the number of response follows a binomial distribution. • Under Ho: P=0.30, with N=30 observations
Under H0 : 14 responses out of 30 pts # of response 14 15 16 17 18 19 prob. 0.023 0.011 0.004 0.001 0 0 P value = 0.040, Two-sided 95% CI: (0.28, 0.66)
Prior Distribution: Prob(Resp.) ~ Beta(0.5, 0.5) Data: 14 out of 30 responded Posterior Distribution: Prob(Resp.) ~ Beta(14.5, 16.5) Likelihood(p0)=0.023, Likelihood(p1)=0.135, Bayes Factor=0.17 95% highest probability interval: (0.30, 0.63) Bayesian Analysis P(Resp. Rate > 0.3) = 0.974
Gehan’s Design • Assume p0 = 0 and p1 > 0 • A two-stage design • First stage: Enroll n1 patients, if no one responds to treatment, Reject H1, quit • Second stage: If there is at least one response, enroll n2 patients to refine the estimate of the response rate • What is the type I error rate? • How to choose type II error rate? • Small b (10% or less) is preferred • Don’t want to reject promising drugs
Simon’s Optimal 2-Stage Design • Assume p0 > 0 and p1 = p0 + d • Typically, d = 0.15 to 0.25 • A two-stage design • First stage: Enroll n1 patients, if r1 or less respond, Reject H1, quit • Second stage: If there is at least r1 + 1 responses, enroll n2 more patients • Final decision: If total number of responses is r or less, reject H1. Otherwise, reject H0 • Two types of design • Optimal: minimize the expected N under H0 • Minimax Design: minimize the maximum N
Simon’s Optimal 2-Stage Design n Rejection Region in # of Responses 21 0 ~ 2 50 3 ~ 7 H0: p p0 H1: p p1 • p0=0.10, p1=0.25, = = 0.10 E(N | p0) = 31.2 PET(p0) = 0.65
Comments on Simon’s Design • Suitable for p0 > 0 • Allow stopping early due to futility: Good • Does not allow stopping early due to initial efficacy: Good • Drawback: • No early stopping when a long string of failures are observed • Similar to all frequentist designs, when the conduct deviates from the design, all statistical properties no longer hold
Simon’s Minimax/Optimal 2-Stage Design Stage 1: Enroll n1 patients, If r1responses, stop trial, reject H1; Otherwise continue to Stage 2; Stage 2: Enroll Nmax - n1patients, If r responses, reject H1 PET(p0) = Prob(Early Termination | p0) = pbinom(r1, n1 ,p0) E(N |p0) = n1 + [1 - PET(p0)] (Nmax- n1) Minimax Design: Smallest Nmax Optimal Design: Smallest E(N|p0)
Example n n Rejection Region in # of Responses Rejection Region in # of Responses 19 17 0 ~ 3 0 ~ 3 36 37 4 ~ 10 4 ~ 10 • Simon’s Optimal 2-Stage Design = 0.095 = 0.097 E(N | p0) = 26.02 PET(p0) = 0.55 • p0=0.20, p1=0.40, = = 0.10 • Simon’s MiniMax 2-Stage Design = 0.086 = 0.098 E(N | p0) = 28.26 PET(p0) = 0.46
Predictive Probability Design Setting: p0=0.20, p1=0.40 Prior for p = Beta(0.2, 0.8) N1=10, Nmax =36 (search 27 ~ 42) Cohort size=1 (continuously monitoring) Constraints: a = 0.1, b = 0.1 Goal: Find L , T , and Nmax to satisfy the constraints Lee JJ and Liu DD. A predictive probability design for phase II cancer clinical trials. Clinical Trials 5:93-106, 2008
Bayesian Approach Prior distribution of response rate: p ~ beta(a0, b0) f0(p)=(a0 + b0)/[(a0)(b0)] pa0-1 (1-p) b0-1 Enroll n patients (maximum accrual=Nmax) Observed responses X ~ bin(n, p) Likelihood function: Lx(p) px (1-p)n-x Posterior distribution: beta(a0 + x,b0+ n x) f(p|x) = f0(p) Lx(p) p a0+ x – 1 (1-p) b0+ n – x 1
Predictive Probability (PP) Design • PP is the probability of rejecting H0 at the end of study should the current trend continue, • i.e., given the current data, the chance of declaring a “positive” result at the end of study. • If PP is very large or very small, essentially we know the answer can stop the trial now and draw a conclusion. • # of future patients: m=Nmax n • # of future responses: Y ~ beta-binomial(m, a0 + x, b0 + n x) • For each Y=i: f(p|x, Y=i)= beta(a0 + x + i,b0 + Nmax x i)
Reject H0 Y=i Prob(Y=i | x) Bi=P[p>p0|p ~ f(p|x,Y=i), x, Y=i] Ii(Bi > T) 0 Prob(Y=0 | x) B0= B0(Y=0) 0 1 Prob(Y=1 | x) B1= B1(Y=1) 0 … … … … m-1 Prob(Y=m-1| x) Bm-1= Bm-1(Y=m-1) 1 m Prob(Y=m | x) Bm= Bm(Y=m) 1 Predictive Probability (PP) = {Prob(Y=i | x) [Prob(p > p0 | p ~ f(p|x,Y=i) , x, Y=i) >T ]} = [Prob(Y=i | x) Ii(Bi >T)]
PP Decision Rule At Each Stage, If PP = [Prob(Y=i | x) Ii(Bi > T)] <L , Stop Trial, accept H0 Otherwise, Continue to the Next Stage until Nmax Goal: Find L , T , and Nmaxto satisfying the constraint of type I and type II error rates (NOTE: In a phase IIA trial, we typically don’t want to stop early due to efficacy; can treat more patients and learn more about the treatment’s efficacy and toxicity.)
Predictive Probability Searching Process Nmax =36 L= 0.001 T= 0.852 ~ 0.922 L T
Power Nmax =36 L= 0.001 T= 0.852 ~ 0.922 L T
Operating characteristics of designs with type I and type II error rates 0.10, prior for p = beta(0.2,0.8), p0=0.2, p1=0.4
Stopping Boundaries for p0=0.20, p1=0.40, = b= 0.10 prior for p = beta(0.2,0.8) • = 0.088 • = 0.094 E(N | p0) = 27.67 PET(p0) = 0.86 Simon’s MiniMax: = 0.086 = 0.098 E(N | p0) = 28.26 PET(p0) = 0.46 Simon’s Optimal: = 0.095 = 0.097 E(N | p0) = 26.02 PET(p0) = 0.55
PP Boundaries by varying the prior for p = 0.088, = 0.094 E(N|p0)=27.7, PET(p0) = 0.86 = 0.089, = 0.091 E(N|p0)=28.9, PET(p0) = 0.86 = 0.089, = 0.091 E(N|p0)=30.6, PET(p0) = 0.85
Case 1: 1st stage: 0 of 10 responses, Nmax=36, prior of p=beta(0.2, 0.8), L=0.001, T =0.900 Will you stop the trial? PP=0.0008 < L , Stop the trial
0/10 responses Case 1 PP=0.0008 < L , Stop the trial
0/10 responses STOP Declare H0 Case 1 Prob(p > 0.2) = 0.0086
Case 2: 1st stage: 2 of 10 responses, Nmax=36, prior of p=beta(0.2, 0.8), L=0.001, T =0.900 Will you stop the trial? PP=0.1766 > L , Continue the trial
Case 2: 1st stage: 2 of 10 responses, followed by 0 out of 12 patients responsesNmax=36, prior of p=beta(0.2, 0.8), L=0.001, T =0.900 Will you stop the trial? PP= 0.0002 < L , Stop the trial
2/10 responses GO Case 2 PP=0.1766 > L
0/12 responses Case 2 PP=0.0002 < L, Stop the trial
STOP Declare H0 Case 2 Prior Dist=Beta(2.2, 8.8) Posterior Dist=Beta(2.2, 20.8) 8 8 Prob(p > 0.2) = 0.063 6 6 Density 4 Density 4 Prob(p>0.2)=0.063 2 2 0 0 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Prob of Response Prob of Response
2/10 responses GO Case 3