multcomp package
Thanks Evan and Gabriel for the reply. I think it might help me make the
question clearer if I show the data and the model here (I actually asked
questions related to this data before but I still need some help). The data
looks like the following:
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
I used the following random intercept and random slope model for this data.
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 common for all three methods;
T_k is the time corresponding to y_ijk;
e_ijk is the residual.
Here I used the following specification for the lme function
mod1 <- lme(fixed= reponse ~ method*time, random=~ 1 +time | individual,
data=one, weights= varIdent(form=~1|method),
control = lmeControl(opt = "optim"))
I did not add the method in random effects because here I assumed common
random slope for all three methods.
This model is used to initially check whether there fixed slopes are equal.
I wanted to further evaluate whether each fixed slope (beta_1, beta_2 and
beta 3) is significantly different from zero. I was hoping to evaluate this
based on the same model.
The output is as follows. Does the highlighted part below already gives the
result for testing beta_1=0; beta_2=0 and beta_3=0?
Thanks very much.
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
2016-06-06 11:21 GMT-04:00 Gabriel Baud-Bovy <baud-bovy.gabriel at hsr.it>:
On 06/06/2016 4:57 PM, li li wrote: Hi all, After fitting a random slope and random intercept model using lme function, I want to test whether each of the fixed slopes is equal to zero (The output of model is below). Can this be done (testing each individual slope) using multcomp package? I don't understand what you mean by testing individual slope ? For the fixed effects, you might test whether there is a method, time or interaction effect using one of the methods described below For the randome effects, according to your model specification, the time dependency might vary for each individual. the sd for the time (0.0001841179) is small. You might want to test whether to include randome slope by doing a LRT between a model with it and another model without. Whay not include method in the random effects ? Gabriel Thanks much for the help. Hanna To get p values: http://stats.stackexchange.com/questions/118416/getting-p-value-with-mixed-effect-with-lme4-package http://mindingthebrain.blogspot.it/2014/02/three-ways-to-get-parameter-specific-p.html Using lmerTest package https://cran.r-project.org/web/packages/lmerTest/lmerTest.pdf or using mixed in afex package http://rpackages.ianhowson.com/cran/afex/man/mixed.html both use pbkrtest packages https://cran.r-project.org/web/packages/pbkrtest/pbkrtest.pdf a faq http://glmm.wikidot.com/faq 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 > [[alternative HTML version deleted]]
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