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## STATISTICS 542 Introduction to Clinical Trials SAMPLE SIZE ISSUES

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**STATISTICS 542Introduction to Clinical TrialsSAMPLE SIZE**ISSUES Ref: Lachin, Controlled Clinical Trials 2:93-113, 1981.**Sample Size Issues**• Fundamental Point Trial must have sufficient statistical power to detect differences of clinical interest • High proportion of published negative trials do not have adequate power Freiman et al, NEJM (1978) 50/71 could miss a 50% benefit**Example: How many subjects?**• Compare new treatment (T) with a control (C) • Previous data suggests Control Failure Rate (Pc) ~ 40% • Investigator believes treatment can reduce Pc by 25% i.e. PT = .30, PC = .40 • N = number of subjects/group?**Estimates only approximate**• Uncertain assumptions • Over optimism about treatment • Healthy screening effect • Need series of estimates • Try various assumptions • Must pick most reasonable • Be conservative yet be reasonable**Statistical Considerations**Null Hypothesis (H0): No difference in the response exists between treatment and control groups Alternative Hypothesis (Ha): A difference of a specified amount () exists between treatment and control Significance Level (): Type I Error The probability of rejecting H0 given that H0 is true Power = (1 - ): ( = Type II Error) The probability of rejecting H0 given that H0 is not true**Standard Normal Distribution**Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.**Standard Normal Table**Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.**Distribution of Sample Means (1)**Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.**Distribution of Sample Means (2)**Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.**Distribution of Sample Means (3)**Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.**Distribution of Sample Means (4)**Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.**Distribution of Test Statistics**• Many have this common form • Testing a population parameter (eg difference in means) • Θ = sample estimate of a population parameter • Then • Z = [Θ – E(Θ)]/√V(Θ) • And then Z has a Normal (0,1) distribution for large sample size**If statistic z is large enough (e.g. falls into red area of**scale), we believe this result is too large • to have come from a distribution with mean O (i.e. Pc - Pt = 0) • Thus we reject H0: Pc - Pt = 0, claiming that their exists 5% chance this result could have come from distribution with no difference**Normal Distribution**Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.**Two Groups**or OR**Test of Hypothesis**• Two sided vs. One sided e.g. H0: PT = PC H0: PT< PC • Classic test za = critical value If |z| > z If z > z Reject H0 Reject H0 = .05 , z = 1.96 = .05, z = 1.645 where z = test statistic • Recommend zbe same value both cases (e.g. 1.96) two-sided one-sided = .05 or = .025 z = 1.96 1.96**Typical Design Assumptions (1)**1. = .05, .025, .01 2. Power = .80, .90 Should be at least .80 for design 3. = smallest difference hope to detect e.g. = PC - PT = .40 - .30 = .10 25% reduction!**Typical Design Assumptions (2)**Two Sided Significance Level Power**Sample Size Exercise**• How many do I need? • Next question, what’s the question? • Reason is that sample size depends on the outcome being measured, and the method of analysis to be used**Simple Case - Binomial**1. H0: PC = PT 2. Test Statistic (Normal Approx.) 3. Sample Size Assume • NT = NC = N • HA:= PC - PT**Sample Size Formula (1)Two Proportions**Simple Case • Za = constant associated with a P {|Z|> Za } = a two sided! (e.g. a = .05, Za =1.96) • Zb = constant associated with 1 - b P {Z< Zb} = 1- b (e.g. 1- b = .90, Zb =1.282) • Solve for Zb(1- b) or D**Sample Size Formula (2)Two Proportions**• Za = constant associated with a P {|Z|> Za } = a two sided! (e.g. a = .05, Za =1.96) • Zb = constant associated with 1 - b P {Z< Zb} = 1- b (e.g. 1- b = .90, Zb =1.282)**Sample Size Formula**Power • Solve for Zb1- b Difference Detected • Solve for D**Simple Example (1)**• H0: PC = PT • HA: PC = .40, PT = .30 = .40 - .30 = .10 • Assume a = .05 Za = 1.96 (Two sided) 1 - b = .90 Zb = 1.282 • p = (.40 + .30 )/2 = .35**Simple Example (2)**Thus a. N = 476 2N = 952 b. 2N = 956 N = 478**Approximate* Total Sample Size for Comparing Various**Proportions in Two Groups with Significance Level (a) of 0.05 and Power (1-b) of 0.80 and 0.90**Comparison of Means**• Some outcome variables are continuous • Blood Pressure • Serum Chemistry • Pulmonary Function • Hypothesis tested by comparison of mean values between groups, or comparison of mean changes**Comparison of Two Means**• H0: C = TC - T = 0 • HA: C - T = • Test statistic for sample means ~ N (, ) • Let N = NC = NT for design ~N(0,1) for H0**Comparison of Means**• Power Calculation**Example**e.g. IQ = 15 = 0.3x15 = 4.5 • Set 2 = .05 = 0.10 1 - = 0.90 • HA: = 0.3 / = 0.3 • Sample Size • N = 234 2N = 468**Comparing Time to Event Distributions**• Primary efficacy endpoint is the time to an event • Compare the survival distributions for the two groups • Measure of treatment effect is the ratio of the hazard rates in the two groups = ratio of the medians • Must also consider the length of follow-up**Assuming Exponential Survival Distributions**• Then define the effect size by • Standard difference**Time to Failure (1)**• Use a parametric model for sample size • Common model - exponential • S(t) = e-t = hazard rate • H0: I = C • Estimate N George & Desu (1974) • Assumes all patients followed to an event (no censoring) • Assumes all patients immediately entered**Assuming Exponential Survival Distributions**• Simple case • The statistical test is powered by the total number of events observed at the time of the analysis, d.**Converting Number of Events (D) to Required Sample Size (2N)**• d = 2N x P(event) 2N = d/P(event) • P(event) is a function of the length of total follow-up at time of analysis and the average hazard rate • Let AR = accrual rate (patients per year) A = period of uniform accrual (2N = AR x A) F = period of follow-up after accrual complete A/2 + F = average total follow-up at planned analysis = average hazard rate • Then P(event) = 1 – P(no event) =**Time to Failure (2)**• In many clinical trials 1. Not all patients are followed to an event (i.e. censoring) 2. Patients are recruited over some period of time (i.e. staggered entry) • More General Model (Lachin, 1981) where g() is defined as follows**1. Instant Recruitment Study Censored At Time T**2. Continuous Recruiting (O,T) & Censored at T 3. Recruitment (O, T0) & Study Censored at T (T > T0)**Example**Assume = .05 (2-sided) & 1 - = .90 C = .3 and I = .2 T = 5 years follow-up T0 = 3 0. No Censoring, Instant Recruiting N = 128 1. Censoring at T, Instant Recruiting N = 188 2. Censoring at T, Continual Recruitment N = 310 3. Censoring at T, Recruitment to T0 N = 233**Sample Size Adjustment for Non-Compliance (1)**• References: 1. Shork & Remington (1967) Journal of Chronic Disease 2. Halperin et al (1968) Journal of Chronic Disease 3. Wu, Fisher & DeMets (1988) Controlled Clinical Trials • Problem Some patients may not adhere to treatment protocol • Impact Dilute whatever true treatment effect exists**Sample Size Adjustment for Non-Compliance (2)**• Fundamental Principle Analyze All Subjects Randomized • Called Intent-to-Treat (ITT) Principle • Noncompliance will dilute treatment effect • A Solution Adjust sample size to compensate for dilution effect (reduced power) • Definitions of Noncompliance • Dropout: Patient in treatment group stops taking therapy • Dropin: Patient in control group starts taking experimental therapy**Comparing Two Proportions**• Assumes event rates will be altered by non‑compliance • Define PT* = adjusted treatment group rate PC* = adjusted control group rate If PT < PC, 1.0 0 PC PT PC * PT ***Adjusted Sample Size**Simple Model - Compute unadjusted N • Assume no dropins • Assume dropout proportion R • Thus PC* = PC PT* = (1-R) PT + R PC • Then adjust N • Example R1/(1-R)2% Increase .1 1.23 23% .25 1.78 78%**Sample Size Adjustment for Non-Compliance**Dropouts & dropins (R0, RI) • Example R0 R1 1/(1- R0- R1)2% Increase .1 .1 1.56 56% .25 .25 4.0 4 times%**Sample Size Adjustments**• More Complex Model Ref: Wu, Fisher, DeMets (1980) • Further Assumptions • Length of follow-up divided into intervals • Hazard rate may vary • Dropout rate may vary • Dropin rate may vary • Lag in time for treatment to be fully effective**Example: Beta-Blocker**Heart Attack Trial (BHAT) (1) • Used complex model • Assumptions 1. = .05 (Two sided) 1 - = .90 2. 3 year follow-up 3. PC = .18 (Control Rate) 4. PT = .13 Treatment assumed 28% reduction 5. Dropout 26% (12%, 8%, 6%) 6. Dropin 21% (7%, 7%, 7%)**Example: Beta-Blocker**Heart Attack Trial (BHAT) (2) UnadjustedAdjusted PC = .18 PC* = .175 PT = .13 PT* = .14 28% reduction 20% reduction N = 1100 N* = 2000 2N = 2200 2N* = 4000**Multiple Response Variables**• Many trials measure several outcomes (e.g. MILIS, NOTT) • Must force investigator to rank them for importance • Do sample size on a few outcomes (2-3) • If estimates agree, OK If not, must seek compromise