Biostatistics Case Studies
Peter D. Christenson
Biostatistician
http://gcrc.humc.edu/Biostat
Session 3: Missing Data in Longitudinal Studies
Case Study
Hall S et al: A comparative study of Carvedilol,
slow release Nifedipine, and Atenolol in the managementof essential hypertension.
J of Cardiovascular Pharmacology 1991;18(4)S35-38.
Case Study Outline
Subjects randomized to one of 3 drugs for controllinghypertension:
A: Carvedilol (new)B: Nifedipine (standard)C: Atenolol (standard)
Blood pressure and HR measured at baseline and 4 post-treatment periods.
Primary analysis is unclear, but changes over time in HRand bp are compared among the 3 groups.
Available Data: sitting dbp
Visit #
Week
Number of Subjects
Paper
Data
A
B
C
Screen
-1
Baseline
dbp1
0
100
93
95
Post 1
2
3
2
100
93
94
Post 2
3
4
4
94
91
94
Post 3
4
5
6
87
88
93
Post 4
5
6
8
83
84
91
Sitting dbp from Figure 2
Group A: Baseline and Final dbp
Week 0
Last Value:
Pre Week 8
Week 8
Final
Δ
Graph
N=100
103.04
± 0.52
N=83
90.43
± 0.96
N=83
90.43
± 0.96
12.61
± ?
Completers
N=83
102.99
± 0.53
N=83
90.43
± 0.96
N=83
90.43
± 0.96
12.55
± 1.10
LastObservation
Carried Forward
(LOCF)
N=100
103.04
± 0.52
N=17
97.47
± 3.47
N=83
90.43
± 0.96
N=100
91.63
± 1.02
11.41
± 1.11
Wanted: Use N=100 w/o LOCF
Combine:
Info on true 8 week change in 83 subjects.
Info on baseline only in 17 subjects.
Use week0-week8 correlation in 83 subjects.
More generally:
Suppose 9 subjects had only week 0 and 8 subjectshad only week 8.
Then, really 2 experiments, 1 paired (N=83) and 1unpaired (N1=9 and N2=8).
Combining involves weighting Δs from the 2experiments. Does not impute (substitute) values forthe 17 unknown values.
Generalize further to >2 time periods and >1 treatment, etc.
Mixed Models
Mixed models implement our need here.
“Mixed” means combination of fixed effects (e.g., drugs;want info on those particular drugs) and random effects(e.g., centers or patients; not interested in the particularones in the study).
AKA multilevel models, hierarchical models.
Very flexible, incorporate unequal patient variability,correlation, pairing, repeated values at multiple levels (e.g.,sitting and standing dbp in Fig 2, or if subjects wereclustered, say from the same family and genetics was anissue, etc), and data missing at random.
More assumptions required than typical analyses.
Data Structure for Software
Need:
patient week dbp
1 0 97
1 2 101
1 4 88
1 6 89
1 8 86
2 0 109
2 2 72
etc
Not:
patient wk0 wk2 wk4 wk6 wk8
1 97 101 88 89 86
2 109 72 . . .
Software
Need to use a mixed model module. Often, options areunclear. Use:
SPSS Analyze > Mixed
SAS proc mixed.
Repeated measures modules with options for randomfactors do not typically handle missing data, e.g.:
SPSS Analyze > GLM > Repeated > … Random
SAS proc glm; model …; random …;
are not in general OK, but will work with certain balancedpatterns of missing data.
Mixed Models in SPSS
Select Analyze > Mixed > Linear. First menu:
Mixed Models in SAS
Select Solutions > Analysis > Analyst >
Statistics > ANOVA > Mixed models
Alternatively, typical code is:
proc mixed;
class week patient;
model dbp=week/ddfm=satterthwaite;
lsmeans week/cl;
estimate 'Week Diff' week 1 -1;
repeated week/subject=patient type=un rcorr;
title 'Mixed Model N=100+83 Unstructured';
run;
Model 1 Results
Estimated Change:
Standard
Label Estimate Error DF t Value Pr > |t|
Week Diff 12.6058 1.0441 95.6 12.07 <.0001
So, Δ = 12.61±1.04 incorporates 100 + 83 observations.
Estimated Means:
Standard
Effect week Estimate Error
week 0 103.04 0.7059
week 8 90.43 0.7749
Group A: Baseline and Final dbp Update
Week 0
Last Value:
Pre Week 8
Week 8
Final
Δ
Graph
N=100
103.04
± 0.52
N=83
90.43
± 0.96
N=83
90.43
± 0.96
12.61
± 1.04
Completers
N=83
102.99
± 0.53
N=83
90.43
± 0.96
N=83
90.43
± 0.96
12.55
± 1.10
LastObservation
Carried Forward
(LOCF)
N=100
103.04
± 0.52
N=17
97.47
± 3.47
N=83
90.43
± 0.96
N=100
91.63
± 1.02
11.41
± 1.11
Is model appropriate? Depends on assumed covariance pattern.
Model 1 Covariance Pattern: Compound Symmetry
Software Output
Estimated R Correlation
Matrix for patient 4
Row Col1 Col2
1 1.0000 0.008760
2 0.008760 1.0000
Covariance ParameterEstimates
Cov Parm Subject Estimate
CS patient 0.4366
Residual 49.3989
Output Interpretation
Estimated Covariance Pattern:
Week 0 8
0 (7.06)2 0.44
8 0.44 (7.06)2
(7.06)2 = 49.3989 + 0.4366
Note that this model assumes thatvariability among subjects is thesame at each week, and thatthere is a correlation between theweeks (estimated at 0.00876).
But: Week 0 SD = 5.2
Week 8 SD = 8.8
Model 2 Covariance Pattern: Unstructured
Software Output
Estimated R Correlation
Matrix for patient 4
Row Col1 Col2
1 1.0000 0.01129
2 0.01129 1.0000
Covariance ParameterEstimates
Cov Parm Subject Estimate
UN(1,1) patient 27.1700
UN(2,1) patient 0.5169
UN(2,2) patient 77.2008
Output Interpretation
Estimated Covariance Pattern:
Week 0 8
0 (5.21)2 0.44
8 0.44 (8.79)2
(5.21)2 = 27.17
This model allows differentvariability among subjects at eachweek, and a correlation betweenthe weeks (estimated at 0.011).
This better models the SDs:
Week 0 SD = 5.2
Week 8 SD = 8.8
Model 3 Covariance: Heterogeneous Uncorrelated
Software Output
Estimated R Correlation
Matrix for patient 4
Row Col1 Col2
1 1.0000
2 1.0000
Covariance ParameterEstimates
Cov Parm Subject Estimate
UN(1,1) patient 27.1701
UN(2,1) patient 0
UN(2,2) patient 77.1998
Output Interpretation
Estimated Covariance Pattern:
Week 0 8
0 (5.21)2 0
8 0 (8.79)2
(5.21)2 = 27.17
This model allows differentvariability among subjects at eachweek, but no correlation betweenthe two weeks.
Matches: Week 0 SD = 5.2
Week 8 SD = 8.8
Choice of Covariance Pattern
Model
Covariance Pattern
-2 Log Likelihood
1: Comp Sym
1: Corr & = SDs
1230.2
2: Unstructured
2: Corr & ≠ SDs
1206.0
3: Heterog Uncorr
3: 0 Corr & ≠ SDs
1206.0
Use likelihood ratio test to test whether a more complex modelsignificantly improves fit of the data. Models must be “nested”.
Is model 2 significantly better than model 1?
Χ2 = 1230.2-1206.0 = 24.2 has Χ2 distribution with d.f.=difference in # of estimated parameters (here 3-2) if model 2 isnot an improvement. P-value=Prob(Χ2 >24.2) <0.0001, somodel 2 is needed. Final choice: model 3.
Model 3 Results
Estimated Change:
Standard
Label Estimate Error DF t Value Pr > |t|
Week Diff 12.6063 1.0963 128 11.50 <.0001
Thus, use Δ = 12.61±1.10 from 100 + 83 observations.
Estimated Means:
Standard
Effect week Estimate Error DF
week 0 103.04 0.5212 99
week 8 90.43 0.9644 82
Conclusions for Group A Week 0 to Week 8 dbp Δ
Last observation carried forward overestimates dbp at week 8.
Essentially 0 correlation between residual week 0 and week 8dbp.
Use mixed model with heterogeneous uncorrelated covariancepattern.
This mixed model is equivalent to a 2-sample t-test withunequal variance using Satterthwaite’s weighting. This wouldnot happen if either (1) some subjects only had dbp at week 8,or (2) correlation was stronger between weeks 0 and 8, whichusually happens.
Generalize: Group A with all 5 Time Periods
Covariance Pattern
Parameters
-2 Log Likelihood
Compound Symmetry
2
3193.7
Heterogeneous Uncorrelated
5
3245.4
Toeplitz
5
3172.0
Heterogeneous Toeplitz
9
3141.4
Unstructured
15
3111.7
Since LR = 3141.4 - 3111.7 = 30.7 is large for a Χ26 , thereis substantial unstructured correlation over weeks.
Conclusions: Repeated Measures with Mixed Models
Very useful for missing data.
Requires more than usual assumptions.
Mild deviations from assumed covariance pattern do not have alarge influence.
Software can be intimidating due to specifying many modelassumptions, since the method is so general and flexible.
May be difficult to apply unbiasedly in clinical trials where theprimary analysis needs to be specifically detailed.