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Sample Size Calculations for the Rate of Changes in Repeated Measures Designs

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). Normal outcomes. 1. Univariate summary statistics Kirby et al. (1994) Overall and Doyle (1994)

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Sample Size Calculations for the Rate of Changes in Repeated Measures Designs

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  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 T2 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 yij be a continuous variable at measurement time tij (j=1, …, Ki) for subject i. Let ri =0 for control group and ri =1 for experimental group.

  10. We assume missing completely at random (MCAR) • β4 is the parameter of interest • Let

  11. Let • Sn(b)=0 • where

  12. is approximately normal with mean 0 and variance Σn = An-1 Vn An-1 • and

  13. Reject H0 : β4= 0 if the absolute value of is larger than z1-α/2 where is the (4,4)-component of Σn

  14. Sample size estimation • Sample size estimate to detect H1 : β4 = β40 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 An and Vn. Then, Σn converges to Σ=A-1 V A-1 • Let σ42 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 σ42 to calculate the sample size • Let δij =0 for missing observation, and δij =1 otherwise • Under MCAR (δi1, …,δiK) is independent of (yi1, …,yiK) • Let visit times be fixed (t1, …,tK)

  17. Let σ2 = var (εij), ρjj’ = corr (εij, εij’), • pj=E(δij)=p(observation at tj) • pjj’=E(δijδij’) =p(observation at both tj and tj’)

  18. Where

  19. Sample Size Formula σ42 is the (4,4) component of Σ=A-1 V A-1 σ42 =σ2st2 /(μ02σr2σt4), where σt2 = μ2 - μ12

  20. The required sample size is given by • Note that we do not have to specify the true values for β1, β2, and β3 in sample size calculation for testing β4

  21. Calculation of σ42 requires projection of the missing probabilities and true correlation structure • As a special case, we consider two missing patterns; independent missing (pjj’ = pj pj’) and monotone missing (pjj’ = pj’ for j<j’)

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

  23. 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

  24. Sample size calculation • From the data, we got σ2 = 815.84 • H1 : β40 =5.71 in a new study • Assign equal number of subjects in each group: σr2 = 0.25 (=r(1-r)) • Proportion of observed measurements (p1 , …, p6 )=(1, 0.9, 0.78,, 0.67, 0.54, 0.41) • From these, we get μ0=4.31, μ1=2.02, μ2=6.73, σt2=2.65

  25. Under CS, we get ρ=0.64 and st2=8.30 from the data • We need n=67 to detect β40 =5.71 with α=0.05 and 90% power • Under AR(1), we get ρ=0.80 and st2=13.73 from the data • We need n=111 to detect β40 =5.71 with α=0.05 and 90% power

  26. Simulation study

  27. 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)

  28. 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

  29. 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 β4 estimate

  30. Given m, let g(m)=n1/2 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)/ n1/2 =g(m)/ (n’)1/2 That is, n’=n{g(m)/g(m+1)}2

  31. True correlation, CS • Under no missing, pj=pjj’=1, σm2 = 12 σ2(1-ρ)m/{(m+1)(m+2)T2 } σm+1/σm does not depend on ρ in the complete data case, while it depends on ρ in the missing data case

  32. 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

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

  34. 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)?

  35. Dropout patterns

  36. Binary Repeated Measurements • Jung and Ahn (2005, SIM) • g(pkij )= ak + bk tkij where g(p)=log{p/(1-p)} • pkij (ak,bk)=g-1(ak + bk tkij) =exp(ak+bk tkij)/{1+ exp(ak+bk tkij)} • Closed-form sample size formula can be derived in a similar way as we did for continuous outcomes

  37. Sample size to test H1 : |b1 – b2 |=d

  38. Steps for sample size calculation • Choose type I error α and power 1-β • Schedule measurement times (t1,…,tm) • Choose allocation proportions r1 and r2 • Given pk1 and pkm, calculate (ak,bk), and pkj Set d= b2 - b1 • Specify non-missing proportions (δ1,…,δm), and a missing pattern for δjj’ • Specify the true correlation structure and the associated correlation parameter ρ • Calculate the variance vk and the sample size n

  39. 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.

  40. 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, ρjj’=0.8|j-j’| • Assign equal number of patients in each group, r1= r2=0.5 • We project that proportion of subjects without pulmonary fibrosis is p11 =0.75 at baseline, and p16 =0.5 at 30 month in a placebo group

  41. We assume that a new therapy will prevent further occurrence of pulmonary fibrosis • That is, p21 = p26 =0.75 • b1 = {g(0.5)-g(0.75)}/(6-1)=-0.220 a1 =g(0.75)=1.099 • Similarly, we obtain (a2,b2)=(1.099,0) • So, d=0-(-0.220)=0.220

  42. 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 (δ1,…,δ6)=(1.0, 0.95, 0.90, 0.85, 0.80, 0.75)

  43. Suppose that we expect independent missing • Now, we have all the parameters values to compute the sample size n • From the parameters, we obtain v1 =0.305 and v2 =0.353 • Finally, n=(1.96+0.84)2 (0.305/0.5+0.353/0.5)/0.2202=214

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