model specification using lme
Thanks Thierry for the reply. I think I now have a better understanding for the specification of the random effects when using lme function. Are my interpretations below correct? random=~ 1 | individual (same random intercept no random slope) random=~ 1 +method| individual (same random intercept and same random slope) random=~ 1 +method:time| individual (same random intercept and different random slope for different method) random=~ 1 +method + method:time| individual (different random intercept and different random slope for different method The summary results from the lme function shows whether the slopes for the three methods are equal (parallelism). I also wanted to test the hypotheses that each of the fixed slopes (corresponding to the three methods) equals 0, can I use multicomp package for that purpose? I am confused on how to make correct specifications in glht function to test these hypotheses. Hanna
summary(mod1)
Linear mixed-effects model fit by REML
Data: one
AIC BIC logLik
304.4703 330.1879 -140.2352
Random effects:
Formula: ~1 + time | individual
Structure: General positive-definite, Log-Cholesky parametrization
StdDev Corr
(Intercept) 0.2487869075 (Intr)
time 0.0001841179 -0.056
Residual 0.3718305953
Variance function:
Structure: Different standard deviations per stratum
Formula: ~1 | method
Parameter estimates:
3 1 2
1.00000 26.59750 24.74476
Fixed effects: reponse ~ method * time
Value Std.Error DF t-value p-value(Intercept)
96.65395 3.528586 57 27.391694 0.0000
method2 1.17851 4.856026 57 0.242689 0.8091
method3 5.87505 3.528617 57 1.664973 0.1014
*time 0.07010 0.250983 57 0.279301 0.7810 method2:time
-0.12616 0.360585 57 -0.349877 0.7277 method3:time -0.08010 0.251105 57
-0.318999 0.7509*
Correlation:
(Intr) methd2 methd3 time mthd2:
method2 -0.726
method3 -0.999 0.726
time -0.779 0.566 0.779
method2:time 0.542 -0.712 -0.542 -0.696
method3:time 0.778 -0.566 -0.779 -0.999 0.696
Standardized Within-Group Residuals:
Min Q1 Med Q3 Max
-2.67575293 -0.51633192 0.06742723 0.59706762 2.81061874
Number of Observations: 69
Number of Groups: 7 >
---------- Forwarded message ----------
From: Thierry Onkelinx <thierry.onkelinx at inbo.be>
Date: 2016-05-30 4:40 GMT-04:00
Subject: Re: [R] model specification using lme
To: li li <hannah.hlx at gmail.com>
Cc: r-help <r-help at r-project.org>
Dear Hanna,
None of the models are correct is you want the same random intercept for
the different methods but different random slope per method.
You can random = ~ 1 + time:method | individual
The easiest way to get alpha_0 and tau_i is to apply post-hoc contrasts.
That is fairly easy to do with the multcomp package.
alpha_0 = (m1 + m2 + m3) / 3
m1 = intercept
m2 = intercept + method2
m3 = intercept + method3
hence alpha_0 = intercept + method2/3 + method3/3
m1 = alpha_0 + tau_1
tau_1 = intercept - method2/3 - method3/3
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to say
what the experiment died of. ~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data. ~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of data.
~ John Tukey
2016-05-29 21:23 GMT+02:00 li li <hannah.hlx at gmail.com>:
Hi all,
For the following data, I consider the following random intercept and
random slope model. Denote as y_ijk the response value from *j*th
individual within *i*th method at time point *k*. Assume the following
model for y_ijk:
y_ijk= (alpha_0+ tau_i +a_j(i))+(beta_i+b_j(i)) T_k + e_ijk
Here alpha_0 is the grand mean;
tau_i is the fixed effect for ith method;
a_j(i) is random intercept corresponding to the *j*th individual
within *i*th method, assumed to be common for all three methods;
beta_i is the fixed slope corresponding to the ith method;
b_j(i) is the random slope corresponding to jth individual for
the ith method, assumed to be different for different methods;
T_k is the time corresponding to y_ijk;
e_ijk is the residual.
For this model, I consider the three specification using the lme function
as follows:
mod1 <- lme(fixed= reponse ~ method*time, random=~ 1 +time | individual,
data=one, weights= varIdent(form=~1|method),
control = lmeControl(opt = "optim"))
mod2 <- lme(fixed= reponse ~ method*time, random=~ 0 +time | individual,
data=one, weights= varIdent(form=~1|method),
control = lmeControl(opt = "optim"))
mod3 <- lme(fixed= reponse ~ method*time, random=~ method +time |
individual, data=one, weights= varIdent(form=~1|method),
control = lmeControl(opt = "optim"))
I think mod1 is the correct one. However, I am kind of confused with the
right usage of lme function. Can someone familiar with this give some help
here?
Another question is regarding the fixed effect tau_1, tau_2 and tau_3
(corresponding to the three methods). One main question I am interested in
is whether each of them are statistically different from zero. In the
summary results below (shaded part), it looks that the result for method 2
and 3 are given with reference to method 1). Is there a way to obtain
specific result separately for alpha_0 (the overall mean) and also tau_1,
tau_2 and tau3?
Thanks very much for the help!
Hanna
summary(mod1)Linear mixed-effects model fit by REML
Data: one
AIC BIC logLik
304.4703 330.1879 -140.2352
Random effects:
Formula: ~1 + time | individual
Structure: General positive-definite, Log-Cholesky parametrization
StdDev Corr
(Intercept) 0.2487869075 (Intr)
time 0.0001841179 -0.056
Residual 0.3718305953
Variance function:
Structure: Different standard deviations per stratum
Formula: ~1 | method
Parameter estimates:
3 1 2
1.00000 26.59750 24.74476
Fixed effects: reponse ~ method * time
Value Std.Error DF t-value p-value(Intercept)
96.65395 3.528586 57 27.391694 0.0000
method2 1.17851 4.856026 57 0.242689 0.8091
method3 5.87505 3.528617 57 1.664973 0.1014time
0.07010 0.250983 57 0.279301 0.7810
method2:time -0.12616 0.360585 57 -0.349877 0.7277
method3:time -0.08010 0.251105 57 -0.318999 0.7509
Correlation:
(Intr) methd2 methd3 time mthd2:
method2 -0.726
method3 -0.999 0.726
time -0.779 0.566 0.779
method2:time 0.542 -0.712 -0.542 -0.696
method3:time 0.778 -0.566 -0.779 -0.999 0.696
Standardized Within-Group Residuals:
Min Q1 Med Q3 Max
-2.67575293 -0.51633192 0.06742723 0.59706762 2.81061874
Number of Observations: 69
Number of Groups: 7 >
one response individual time method
1 102.9 3 0 3
2 103.0 3 3 3
3 103.0 3 6 3
4 102.8 3 9 3
5 102.2 3 12 3
6 102.5 3 15 3
7 103.0 3 18 3
8 102.0 3 24 3
9 102.8 1 0 3
10 102.7 1 3 3
11 103.0 1 6 3
12 102.2 1 9 3
13 103.0 1 12 3
14 102.8 1 15 3
15 102.8 1 18 3
16 102.9 1 24 3
17 102.2 2 0 3
18 102.6 2 3 3
19 103.4 2 6 3
20 102.3 2 9 3
21 101.3 2 12 3
22 102.1 2 15 3
23 102.1 2 18 3
24 102.2 2 24 3
25 102.7 4 0 3
26 102.3 4 3 3
27 102.6 4 6 3
28 102.7 4 9 3
29 102.8 4 12 3
30 102.5 5 0 3
31 102.4 5 3 3
32 102.1 5 6 3
33 102.3 6 0 3
34 102.3 6 3 3
35 101.9 7 0 3
36 102.0 7 3 3
37 107.4 3 0 1
38 101.3 3 12 1
39 92.8 3 15 1
40 73.7 3 18 1
41 104.7 3 24 1
42 92.6 1 0 1
43 101.9 1 12 1
44 106.3 1 15 1
45 104.1 1 18 1
46 95.6 1 24 1
47 79.8 2 0 1
48 89.7 2 12 1
49 97.0 2 15 1
50 108.4 2 18 1
51 103.5 2 24 1
52 96.4 4 0 1
53 89.3 4 12 1
54 112.6 5 0 1
55 93.3 6 0 1
56 99.6 7 0 1
57 109.5 3 0 2
58 98.5 3 12 2
59 103.5 3 24 2
60 113.5 1 0 2
61 94.5 1 12 2
62 88.5 1 24 2
63 99.5 2 0 2
64 97.5 2 12 2
65 98.5 2 24 2
66 103.5 4 0 2
67 89.5 5 0 2
68 87.5 6 0 2
69 82.5 7 0 2
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