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P8157 Homework 3 solved

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Question 1:

In this question you are going to analyze the ‘wtloss’ data, available on the course website.
Briefly, the data come from a weight loss trial in which K=120 patients were randomized to three
treatment arms: dietary counseling at baseline (diet=0), dietary counseling at all sudy visits
(diet=1), and dietary counseling at all visits plus free access to an exercise facility (diet=2).
Each patient visited the study clinic monthly, for up to 12 months; at each visit their weight was
measured.

(a) Use lme() to fit two linear mixed effects models, both including a main effect for diet, a main
effect for time, and a diet by time interaction:
(i) random intercepts only
(ii) random intercepts and random slopes

Report the results in a table that would be suitable for a clinical journal, and provide precise
interpretations of the fixed effects and variance components from model (ii)
(b) Consider conducting a test for whether the random intercepts/slopes model provides a significantly better fit to the data than the random intercept model. Write down the null and
alternative hypotheses. In class we learned that this test is non-standard testing scenario,
and the likelihood ratio test statistic under the null is a mixture of χ
2
1
and χ
2

2 distributions. In
the lme help file look up simulate.lme. Use this function to simulate the null distribution,
setting n.sim=1000 and seed=1504. Plot this distribution, along with the χ
2
1 distribution
and the χ
2
2 distribution, highlighting the 95th percentiles. What do you conclude about the
adequacy of the random intercept model as compared to random intercept/slopes model.

(c) Conduct a residual analysis of model (i) from part (a). Report your results in a concise
manner and briefly summarize what you conclude, including whether the results from this
analysis are consistent with what you concluded from part (b).
(d) Use geeglm() to fit the same mean model from part (a) using GEE 1.5, based on: (i) working
independence (GEE-I), (ii) working exchangeable (GEE-E), and (iii) working AR-1 (GEEAR1). Report the results in a table that would be suitable for a clinical journal, and provide
precise interpretations of the regression parameter estimates from GEE-E.
1

(e) State the null and alternative hypothesis for the test of whether the rate of weight loss differs
for the treatment groups. Conduct the test for the GEE-E estimator and describe the results
using language that would be suitable to a non-biostatistician collaborator
(f) Assuming the mean model is correctly specified, comment on the consistency of the point
estimates reported in parts (a) and (d), as well as on the validity of the standard error
estimates.
2
where the weight matrix Wk is equal to V
−1
k

. For simplicity, we’ll consider the special case where
the response is continuous and the variance is constant (i.e. homogeneous). In that case, the GEE
estimating equation is given by:
X
K
k=1
DT
k V

−1
k
(Yk − µk
) = X
K
k=1
XT
k Wk(Yk − µk
).

Define the total weight given to cluster k as
Wtot,k =
Xnk
i=1
Xnk
j=1
wk,ij ,
where wk,ij is the (i, j)
th element of Wk.

(a) Assuming an exchangeable correlation structure with correlation parameter ρ, calculate Wtot,k
as a function of nk and ρ using the identity:
(aIm + b1m)
−1 =
1
a

Im −
b
a + mb1m

.

where Im denotes the m × m identity matrix and 1m denotes the m × m matrix of 1’s.
(b) From this, derive the form of the relative weight of a person with nk=10 to one where
nk=5.Calculate this value for ρ=0.9, ρ=0.5, and ρ=0.1 and comment on the trend that you
observe.

(c) What do the results in part (b) say about the weight for each subject when working independence is used?
(d) The per-observation weight (per single observation within a person) can be thought of as
Wtot,k/nk. Using the results from part (b), comment on the trend in the per observation
weight received.
3

Question 2 (Optional):

In this question we are going to try to understand how much each cluster (or subject) and
each observation per cluster (or subject) is weighted by GEE. That is, even though Vk is called a
’working covariance matrix,’ it might be more natural to think of it as a working weighting matrix,