1. Sample Size Calculations for the Rate of Changes in Repeated Measures Designs Chul Ahn, Ph.D. UT Southwestern Medical Center at Dallas (Joint work with Sinho Jung at Duke)

2. Normal outcomes 1. Univariate summary statistics Kirby et al. (1994) Overall and Doyle (1994) 2. Univariate split-plot ANOVA Bloch (1986) Lui and Cumberland (1992) 3. Hotelling’s T 2 Vonesch and Schork (1986) Rochon (1991)

3. 4. Multivariate ANOVA Muller and Barton (1989) Muller et al. (1992)

4. Binary Outcomes 1. Extension of univariate split-plot model Lui (1999) 2. Weighted least squares Rochon (1989) Lipsitz and Fitzmaurice (1994)

5. GEE Liu and Liang (1997): Score test, no closed form formula except for some special cases Rochon (1998), Wald test {Pan (2001), Z-test, a special case of Wald test, SAS and S-Plus use Wald test} Jung and Ahn (2003, 2004, 2005) Ahn and Jung (2003, 2005), Z-test Dahmen et al. (2004) Kim et al. (2005)

6. Other Approaches Hedeker et al. (1999) Yi and Panzarella (2002) Gastanaga et al. (2006) Tu et al. (2004, 2007)

7. Problem formulation Diggle et al. (2002) “Correlation between repeated observations affects the sample size estimates in a different way depending on the problem.”. Leon (2004) and Rochon (1998): As correlation increases, the required sample size increases when comparing group averages. Jung and Ahn (2003, 2005) and Ahn and Jung (2005) show that it may not be the case when comparing the rates of changes over time within subjects

8. GEE (Jung and Ahn, 2003, 2005) A closed form formula for sample size and power for comparing the rate of changes between two groups Sample size can be computed using a scientific calculator

9. GEE for continuous outcomes Let y ij be a continuous variable at measurement time t ij (j=1, …, K i ) for subject i. Let r i =0 for control group and r i =1 for experimental group.

10. We assume missing completely at random (MCAR) β 4 is the parameter of interest Let

11. S n (b)=0 where

12. is approximately normal with mean 0 and variance Σ n = A n -1 V n A n -1 and

13. Reject H 0 : β 4 = 0 if the absolute value of is larger than z 1-α/2 where is the (4,4)-component of Σ n

14. Sample size estimation Sample size estimate to detect H 1 : β 4 = β 40 with a two-sided α test and power 1-γ Assume that the visits are either made at scheduled times or missing, and the missing probability depends on measurement time only.

15. Let A and V denote the limits of A n and V n. Then, Σ n converges to Σ=A -1 V A -1 Let σ 4 2 denote the (4,4) component of Σ Then, the required sample size is

16. We need to derive the expression of A and V for σ 4 2 to calculate the sample size Let δ ij =0 for missing observation, and δ ij =1 otherwise Under MCAR ( δ i1, …, δ iK ) is independent of (y i1, …,y iK ) Let visit times be fixed (t 1, …,t K )

19. Let σ 2 = var (ε ij ), ρ jj’ = corr (ε ij, ε ij’ ), p j =E(δ ij )=p(observation at t j ) p jj’ =E(δ ij δ ij’ ) =p(observation at both t j and t j’ )

20. Where

22. Sample Size Formula σ 4 2 is the (4,4) component of Σ=A -1 V A -1 σ 4 2 =σ 2 s t 2 /(μ 0 2 σ r 2 σ t 4 ), where σ t 2 = μ 2 - μ 1 2

23. The required sample size is given by Note that we do not have to specify the true values for β 1, β 2, and β 3 in sample size calculation for testing β 4

24. Calculation of σ 4 2 requires projection of the missing probabilities and true correlation structure As a special case, we consider two missing patterns; independent missing (p jj’ = p j p j’ ) and monotone missing (p jj’ = p j’ for j<j’)

25. We can use any correlation structures. The commonly used correlation structures are AR(1) with ρ jj’ = ρ |j-j’|, and compound symmetry (exchangeable) with ρ jj’ = ρ for j≠j’ The sample size calculation can be done easily with a scientific calculator

26. Example Davis (1991, SIM) 83 women in labor were randomized to receive a pain medication (43 women) or placebo (40 women). The amount of pain was self-reported (0 = no pain, 100 = extreme pain) K=6, maximum number of measurements Monotone missing pattern

27. Sample size calculation From the data, we got σ 2 = 815.84 H 1 : β 40 =5.71 in a new study Assign equal number of subjects in each group: σ r 2 = 0.25 (=r(1-r)) Proportion of observed measurements (p 1, …, p 6 )=(1, 0.9, 0.78,, 0.67, 0.54, 0.41) From these, we get μ 0 =4.31, μ 1 =2.02, μ 2 =6.73, σ t 2 =2.65

28. Under CS, we get ρ=0.64 and s t 2 =8.30 from the data  We need n=67 to detect β 40 =5.71 with α=0.05 and 90% power Under AR(1), we get ρ=0.80 and s t 2 =13.73 from the data  We need n=111 to detect β 40 =5.71 with α=0.05 and 90% power

29. Simulation study

31. With the same ρ value, sample size under AR(1) is larger than that under CS for testing the rates of changes between two groups  A conservative approach is to use AR(1) With the same ρ value, sample size under CS is larger than that under AR(1) when comparing marginal means between two groups  A conservative approach is to use CS (Rochon, 1998)

32. K group comparisons Jung and Ahn (2004) Two group comparisons can be extended to K (K≥3) group comparisons Use of non-central chi-square distribution

33. Increase n or m? Ahn and Jung (2004) Efficiency of the slope estimator in repeated measurements Relative benefit of adding subjects (n) versus adding measurements (m) on a specified fixed study period [0,T] n and m will affect the standard error of β 4 estimate

34. Given m, let g(m)=n 1/2 se(β) The effect of increase from m to (m+1) on se(β) is the same as that from n to n’, where n’ satisfies g(m+1)/ n 1/2 =g(m)/ (n’) 1/2 That is, n’=n{g(m)/g(m+1)} 2

35. True correlation, CS Under no missing, p j =p jj’ =1, σ m 2 = 12 σ 2 (1-ρ)m/{(m+1)(m+2)T 2 } σ m+1 /σ m does not depend on ρ in the complete data case, while it depends on ρ in the missing data case

36. Adding one more measurements in [0, T] is equivalent to adding n(m-1)/(m+1) 2 more subjects in the complete data case. That is, we can reduce n(m-1)/(m(m+3)) patients by adding one more assessments to achieve the same precision in the complete data case

37. Suppose that we increase the number of measurements from m to m+1, the relative reduction in standard error of slope is (se(β m )- se(β m+1 ))/se(β m )

39. Effect of dropout on sample size estimate Monotone missing Let N be the estimated total sample size under no missing data, and q be the proportion of dropout at the end of the study Can we estimate the sample size using N/(1-q)?

40. Dropout patterns

42. Binary Repeated Measurements Jung and Ahn (2005, SIM) g(p kij )= a k + b k t kij where g(p)=log{p/(1-p)} p kij (a k,b k )=g -1 (a k + b k t kij ) =exp(a k +b k t kij )/{1+ exp(a k +b k t kij )} Closed-form sample size formula can be derived in a similar way as we did for continuous outcomes

43. Sample size to test H 1 : |b 1 – b 2 |=d

45. Steps for sample size calculation 1. Choose type I error α and power 1-β 2. Schedule measurement times (t 1,…,t m ) 3. Choose allocation proportions r 1 and r 2 4. Given p k1 and p km, calculate (a k,b k ), and p kj Set d= b 2 - b 1 5. Specify non-missing proportions (δ 1,…,δ m ), and a missing pattern for δ jj’ 6. Specify the true correlation structure and the associated correlation parameter ρ 7. Calculate the variance v k and the sample size n

46. Example 75% of scleroderma patients do not have pulmonary fibrosis at baseline in the ongoing GENOSIS trial A new clinical trial will examine the effect of a new drug in preventing the occurrence of pulmonary fibrosis Presence or absence of pulmonary fibrosis will be assessed at baseline, and at months 6, 12, 18, 24 and 30.

47. Compare the occurrence of pulmonary fibrosis from baseline to 30 months for placebo versus a new drug Within-group correlation structure: AR(1) with ρ=0.8, ρ jj’ =0.8 |j-j’| Assign equal number of patients in each group, r 1 = r 2 =0.5 We project that proportion of subjects without pulmonary fibrosis is p 11 =0.75 at baseline, and p 16 =0.5 at 30 month in a placebo group

48. We assume that a new therapy will prevent further occurrence of pulmonary fibrosis That is, p 21 = p 26 =0.75 b 1 = {g(0.5)-g(0.75)}/(6-1)=-0.220 a 1 =g(0.75)=1.099 Similarly, we obtain (a 2,b 2 )=(1.099,0) So, d=0-(-0.220)=0.220

49. The probabilities of no pulmonary fibrosis can be estimated from the logistic regression equation (0.750, 0.707, 0.659, 0.608, 0.555, 0.500) for the placebo group (0.750, 0.750, 0.750, 0.750, 0.750, 0.750) for the treatment group The proportions of observed measurements are expected to be (δ 1,…,δ 6 )=(1.0, 0.95, 0.90, 0.85, 0.80, 0.75)

50. Suppose that we expect independent missing Now, we have all the parameters values to compute the sample size n From the parameters, we obtain v 1 =0.305 and v 2 =0.353 Finally, n=(1.96+0.84) 2 (0.305/0.5+0.353/0.5)/0.220 2 =214

51. Software for sample size estimate GEESIZE version 3.1 http://www.imbs.uni- luebeck.de/pub/Geesize/ “GEESIZE computes the minimum sample size in studies with correlated response data based on GEE. These correlated response data arise e.g. in repeated measurement designs, family studies or studies involving paired organs.”

52. RMASS2: Repeated measuers with attrition: sample size for 2 groups http://tigger.uic.edu/~hedeker/ml.html