1. 1 Chapter 6 SAMPLE SIZE ISSUES Ref: Lachin, Controlled Clinical Trials 2:93-113, 1981.

2. 2 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

3. 3 Example: How many subjects? Compare new treatment (T) with a control (C) Previous data suggests Control Failure Rate (P c ) ~ 40% Investigator believes treatment can reduce P c by 25% i.e. P T =.30, P C =.40 N = number of subjects/group?

4. 4 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

5. 5 Statistical Considerations Null Hypothesis (H 0 ): No difference in the response exists between treatment and control groups Alternative Hypothesis (H a ): A difference of a specified amount (  ) exists between treatment and control Significance Level (  ): Type I Error The probability of rejecting H 0 given that H 0 is true Power = (1 -  ): (  = Type II Error) The probability of rejecting H 0 given that H 0 is not true

6. 6 Standard Normal Distribution Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.

7. 7 Standard Normal Table Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.

8. 8 Distribution of Sample Means (1) Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.

9. 9 Distribution of Sample Means (2) Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.

10. 10 Distribution of Sample Means (3) Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.

11. 11 Distribution of Sample Means (4) Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.

12. 12 Distribution of Test Statistics Many have a common form = population parameter (eg difference in means) = sample estimate Then –Z =[ – E( )]/SE( ) And then Z has a Normal (0,1) distribution

13. 13 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. P c - P t = 0) Thus we reject H 0 : P c - P t = 0, claiming that there exists 5% chance this result could have come from distribution with no difference

14. 14 Normal Distribution Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.

15. 15 Two Groups ORor

16. 16 Test Statistics

17. 17 Test of Hypothesis Two sidedvs.One sided e.g. H 0 : P T = P C H 0 : P T < P C Classic testz  = critical value If |z| > z  If z > z  Reject H 0  =.05, z  = 1.96  =.05, z  = 1.645 where z = test statistic Recommend z  be same value both cases (e.g. 1.96) two-sided one-sided   =.05 or =.025 z  = 1.96 1.96

18. 18 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.  = P C - P T =.40 -.30 =.1025% reduction!

19. 19 Typical Design Assumptions (2) Two Sided PowerSignificance Level

20. 20 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

21. 21 One Sample Test for Mean H 0 :  =  0 vs. H A :  =  0 +  z α = constant associated with a P {|Z|> z α } = α two sided

22. 22 One Sample Test for Mean Under the alternative hypothesis that  =  0 + , the test statistic follows a standard normal variable.

23. 23 One Sample Test for Mean

24. 24 Simple Case - Binomial 1.H 0 :P C = P T 2.Test Statistic (Normal Approx.) 3.Sample Size Assume N T = N C = N H A :  = P C - P T

25. 25 Sample Size Formula (1) Two Proportions Simpler Case Z  = constant associated with  P {|Z|> Z  } =  two sided! (e.g.  =.05, Z  =1.96) Z  = constant associated with 1 -  P {Z< Z  } = 1-  (e.g. 1-  =.90, Z  =1.282) Solve for Z  (  1-  ) or 

26. 26 Sample Size Formula (2) Two Proportions Z  = constant associated with  P {|Z|> Z  } =  two sided! (e.g.  =.05, Z  =1.96) Z  = constant associated with 1 -  P {Z< Z  } = 1-  (e.g. 1-  =.90, Z  =1.282)

27. 27 Sample Size Formula Power Solve for Z    1-  Difference Detected Solve for 

28. 28 Simple Example (1) H 0 : P C = P T H A : P C =.40, P T =.30  =.40 -.30 =.10 Assume  =.05Z  = 1.96 (Two sided) 1 -  =.90Z  = 1.282 p = (.40 +.30 )/2 =.35

29. 29 Simple Example (2) Thus a. N = 476 2N = 952 b. N = 478 2N = 956

30. 30 Approximate* Total Sample Size for Comparing Various Proportions in Two Groups with Significance Level (  ) of 0.05 and Power (1-  ) of 0.80 and 0.90

31. 31

32. 32 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

33. 33 Comparison of Two Means H 0 :  C =  T  C -  T = 0 H A :  C -  T =  Test statistic for sample means ~ N (  ) Let N = N C = N T for design Power ~N(0,1) for H 0

34. 34 Example e.g. IQ  = 15  = 0.3x15 = 4.5 Set 2  =.05  = 0.10 1 -  = 0.90 H A :  = 0.3    /  = 0.3 Sample Size N = 234  2N = 468

35. 35

36. 36 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

37. 37 Assuming Exponential Survival Distributions Then define the effect size by Standard difference

38. 38 Time to Failure (1) Use a parametric model for sample size Common model - exponential –S(t) = e - t = hazard rate –H 0 : I = C –Estimate N George & Desu (1974) Assumes all patients followed to an event (no censoring) Assumes all patients immediately entered

39. 39 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.

40. 40 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) =

41. 41 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

42. 42 1.Instant Recruitment Study Censored At Time T 2.Continuous Recruiting (O,T) & Censored at T 3.Recruitment (O, T 0 ) & Study Censored at T (T > T 0 )

43. 43 Example Assume  =.05 (2-sided) & 1 -  =.90 C =.3 and I =.2 T = 5 years follow-up T 0 = 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 T 0 N = 233

44. 44 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

45. 45 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

46. 46 Comparing Two Proportions –Assumes event rates will be altered by non ‑ compliance –Define P T * = adjusted treatment group rate P C * = adjusted control group rate If P T < P C, 0 PTPT PCPC P T * P C * 1.0

47. 47 Simple Model - Compute unadjusted N –Assume no dropins –Assume dropout proportion R –ThusP C * = P C P T * = (1-R) P T + R P C –Then adjust N –Example R1/(1-R) 2 % Increase.1 1.23 23%.25 1.78 78% Adjusted Sample Size

48. 48 Sample Size Adjustment for Non-Compliance Dropouts & dropins (R 0, R I ) –Example R 0 R 1 1/(1- R 0 - R 1 ) 2 % Increase.1.1 1.5656%.25.25 4.04 times%

49. 49 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

50. 50 Sample Size Summary Ethically, the size of the study must be large enough to achieve the stated goals with reasonable probability (power) Sample size estimates are only approximate due to uncertainty in assumptions Need to be conservative but realistic

51. 51 Demo of Sample Size Program www.biostat.wisc.edu/ Program covers comparison of proportions, means, & time to failure Can vary control group rates or responses, alpha & power, hypothesized differences Program develops sample size table and a power curve for a particular sample size