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)
Dec 23, 2015
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 statisticsKirby et al. (1994) Overall and Doyle (1994)
2. Univariate split-plot ANOVABloch (1986)Lui and Cumberland (1992)
3. Hotelling’s T2
Vonesch and Schork (1986)Rochon (1991)
4. Multivariate ANOVA
Muller and Barton (1989)
Muller et al. (1992)
Binary Outcomes
1. Extension of univariate split-plot model
Lui (1999)
2. Weighted least squares
Rochon (1989)
Lipsitz and Fitzmaurice (1994)
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)
Other Approaches
Hedeker et al. (1999)Yi and Panzarella (2002)Gastanaga et al. (2006)Tu et al. (2004, 2007)
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
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
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.
We assume missing completely at random (MCAR)
β4 is the parameter of interest
Let
Let
Sn(b)=0
where
is approximately normal with mean 0 and variance Σn = An
-1 Vn An-1
and
Reject H0 : β4 = 0
if the absolute value of is larger than z1-α/2
where is the (4,4)-component of Σn
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.
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
We need to derive the expression of A and V for σ4
2 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)
Let σ2 = var (εij), ρjj’ = corr (εij, εij’),
pj=E(δij)=p(observation at tj)
pjj’=E(δij δij’)
=p(observation at both tj and tj’)
Where
Sample Size Formula
σ42 is the (4,4) component of Σ=A-1 V A-1
σ42 =σ2st
2 /(μ02 σr
2 σt4),
where σt2 = μ2 - μ1
2
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
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’)
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
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 measurementsMonotone missing pattern
Sample size calculation
From the data, we got σ2 = 815.84H1 : β40 =5.71 in a new study
Assign equal number of subjects in each group: σr
2 = 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, σt
2=2.65
Under CS, we get ρ=0.64 and st2=8.30
from the dataWe need n=67 to detect β40 =5.71 with
α=0.05 and 90% powerUnder 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
Simulation study
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)
K group comparisons
Jung and Ahn (2004)Two group comparisons can be
extended to K (K≥3) group comparisonsUse of non-central chi-square
distribution
Increase n or m?
Ahn and Jung (2004)Efficiency of the slope estimator in
repeated measurementsRelative 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
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
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
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
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)
Effect of dropout on sample size estimate
Monotone missingLet 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)?
Dropout patterns
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
Sample size to test H1 : |b1 – b2 |=d
Steps for sample size calculation
1. Choose type I error α and power 1-β
2. Schedule measurement times (t1,…,tm)
3. Choose allocation proportions r1 and r2
4. Given pk1 and pkm, calculate (ak,bk), and pkj Set d= b2 - b1
5. Specify non-missing proportions (δ1,…,δm), and a missing pattern for δjj’
6. Specify the true correlation structure and the associated correlation parameter ρ
7. Calculate the variance vk and the sample size n
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.
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
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
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)
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
Software for sample size estimate
GEESIZE version 3.1http://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.”
RMASS2: Repeated measuers with attrition: sample size for 2 groups
http://tigger.uic.edu/~hedeker/ml.html