Re: [R] p values for a GEE model

2006-04-11 Thread Prof Brian Ripley
On Mon, 10 Apr 2006, Tarca, Adi wrote:

 Hi all,

 I have a dataset in which the output Y is observed on two groups of
 patients (treatment factor T with 2 levels).

 Every subject in each group is observed three times (not time points but
 just technical replication).

 I am interested in estimating the treatment effect and take into account
 the fact that I have repeated measurements for every subject.

 If I do this with repeated measures ANOVA (in which the patient is
 considered a random effect) I got the following results:

  library(nlme)

 data-read.table(http://146.9.88.18/uploads/dataGEE.txt,header=TRUE)

   res-lme(Y~T,random=~1|P,data=data)
   summary(res)

 So the p-value for significance of the treatment effect is 0.069.

 I would like to use also as a variant analysis a Generalized Estimation
 Equation Model, like

library(gee)
summary(gee(Y~T,id=P,data=data))

 Questions:

 A) Is the gee approach suitable in this case with the model formulae I
 use?

Yes, but your choice of working correlation probably is not.  Your results 
are basically least-squares here.  I think you want corstr=exchangeable 
or perhaps some form of AR if the repeated observations have a clear time 
base.

 B) Can I obtain a p-value for the fixed effect T ?

Well, the output is

Coefficients:
  Estimate Naive S.E.   Naive z Robust S.E.  Robust z
(Intercept) 14.781968  0.3686055 40.102410   0.3608597 40.963207
TPTLD1.857166  0.5259206  3.531266   0.8570370  2.166961

and if you read up on the background you will know that the 'z' scores are 
approximately normally distributed.

There are detailed comparisons of examples of the lme and gee approaches 
in MASS chapter 10.

-- 
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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Re: [R] p values for a GEE model

2006-04-11 Thread Renaud Lancelot
2006/4/10, Tarca, Adi [EMAIL PROTECTED]:
 Hi all,

 I have a dataset in which the output Y is observed on two groups of
 patients (treatment factor T with 2 levels).

 Every subject in each group is observed three times (not time points but
 just technical replication).

 I am interested in estimating the treatment effect and take into account
 the fact that I have repeated measurements for every subject.

 If I do this with repeated measures ANOVA (in which the patient is
 considered a random effect) I got the following results:



   library(nlme)

 data-read.table(http://146.9.88.18/uploads/dataGEE.txt,header=TRUE)

res-lme(Y~T,random=~1|P,data=data)

summary(res)



 So the p-value for significance of the treatment effect is 0.069.

 I would like to use also as a variant analysis a Generalized Estimation
 Equation Model, like

 library(gee)

 summary(gee(Y~T,id=P,data=data))

Beware: the default within-group correlation structure is independence
in the gee function (see argument corstr). I think you want an
exchangeable correlation structure, i.e. the same within-group
correlation for all the measurements:

 Data - read.table(http://146.9.88.18/uploads/dataGEE.txt;, header = TRUE)

 library(gee)
 fm1 - gee(Y ~ T, id = P, data = Data, corstr = exchangeable)
[1] Beginning Cgee S-function, @(#) geeformula.q 4.13 98/01/27
[1] running glm to get initial regression estimate
[1] 14.781968  1.857166
 summary(fm1)

 GEE:  GENERALIZED LINEAR MODELS FOR DEPENDENT DATA
 gee S-function, version 4.13 modified 98/01/27 (1998)

Model:
 Link:  Identity
 Variance to Mean Relation: Gaussian
 Correlation Structure: Exchangeable

Call:
gee(formula = Y ~ T, id = P, data = Data, corstr = exchangeable)

Summary of Residuals:
Min  1Q  Median  3Q Max
-3.42512726 -0.99762726 -0.04887726  0.48282733  7.73737274


Coefficients:
 Estimate Naive S.E.   Naive z Robust S.E.  Robust z
(Intercept) 14.853839  0.6069154 24.474317   0.3855683 38.524536
TPTLD1.713788  0.8586943  1.995807   0.8423924  2.034429

Estimated Scale Parameter:  3.945557
Number of Iterations:  2

Working Correlation
 [,1] [,2] [,3]
[1,] 1.00 0.897846 0.897846
[2,] 0.897846 1.00 0.897846
[3,] 0.897846 0.897846 1.00

 Questions:

 A) Is the gee approach suitable in this case with the model formulae I
 use?

 B) Can I obtain a p-value for the fixed effect T ?

You can extract the fixed-effect coefficients with coef:

 coef(summary(fm1))
 Estimate Naive S.E.   Naive z Robust S.E.  Robust z
(Intercept) 14.853839  0.6069154 24.474317   0.3855683 38.524536
TPTLD1.713788  0.8586943  1.995807   0.8423924  2.034429

Then, get the P values using a normal approximation for the distribution of z:

 2 * pnorm(abs(coef(summary(fm1))[,5]), lower.tail = FALSE)
(Intercept)   TPTLD
 0.  0.04190831

Best,

Renaud




 Thanks,



 Laurentiu Tarca






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Re: [R] p values for a GEE model

2006-04-11 Thread Thomas Lumley
On Tue, 11 Apr 2006, Renaud Lancelot wrote:

 2006/4/10, Tarca, Adi [EMAIL PROTECTED]:
 Hi all,

 I have a dataset in which the output Y is observed on two groups of
 patients (treatment factor T with 2 levels).

 Every subject in each group is observed three times (not time points but
 just technical replication).

 I am interested in estimating the treatment effect and take into account
 the fact that I have repeated measurements for every subject.

Snip
 I would like to use also as a variant analysis a Generalized Estimation
 Equation Model, like

 library(gee)

 summary(gee(Y~T,id=P,data=data))

 Beware: the default within-group correlation structure is independence
 in the gee function (see argument corstr). I think you want an
 exchangeable correlation structure, i.e. the same within-group
 correlation for all the measurements:


He has a linear model with the same number of observations for each person 
and no covariates that vary within a person. The independence and 
exchangeable working correlations will give identical answers.


-thomas

Thomas Lumley   Assoc. Professor, Biostatistics
[EMAIL PROTECTED]   University of Washington, Seattle

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Re: [R] p values for a GEE model

2006-04-11 Thread Prof Brian Ripley
On Tue, 11 Apr 2006, Thomas Lumley wrote:

 On Tue, 11 Apr 2006, Renaud Lancelot wrote:

 2006/4/10, Tarca, Adi [EMAIL PROTECTED]:
 Hi all,

 I have a dataset in which the output Y is observed on two groups of
 patients (treatment factor T with 2 levels).

 Every subject in each group is observed three times (not time points but
 just technical replication).

 I am interested in estimating the treatment effect and take into account
 the fact that I have repeated measurements for every subject.

 Snip
 I would like to use also as a variant analysis a Generalized Estimation
 Equation Model, like

 library(gee)

 summary(gee(Y~T,id=P,data=data))

 Beware: the default within-group correlation structure is independence
 in the gee function (see argument corstr). I think you want an
 exchangeable correlation structure, i.e. the same within-group
 correlation for all the measurements:


 He has a linear model with the same number of observations for each person

Not so: some have 3 and some have 2, and the two levels of T are not quite 
balanced (29/28).

 and no covariates that vary within a person. The independence and
 exchangeable working correlations will give identical answers.

They do not in this case (nor should they, I believe).

-- 
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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Re: [R] p values for a GEE model

2006-04-11 Thread Thomas Lumley
On Tue, 11 Apr 2006, Prof Brian Ripley wrote:

 On Tue, 11 Apr 2006, Thomas Lumley wrote:
 
 He has a linear model with the same number of observations for each person

 Not so: some have 3 and some have 2, and the two levels of T are not quite 
 balanced (29/28).

 and no covariates that vary within a person. The independence and
 exchangeable working correlations will give identical answers.

 They do not in this case (nor should they, I believe).


You are right, of course.  I believed the message rather than looking at 
the data.

-thomas

Thomas Lumley   Assoc. Professor, Biostatistics
[EMAIL PROTECTED]   University of Washington, Seattle

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