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.gabr...@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]] > > _______________________________________________r-sig-mixed-mod...@r-project.org > mailing listhttps://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models > . > > > > > -- > --------------------------------------------------------------------- > Gabriel Baud-Bovy tel.: (+39) 02 2643 4839 (office) > UHSR University (+39) 02 2643 3429 (laboratory) > via Olgettina, 58 (+39) 02 2643 4891 (secretary) > 20132 Milan, Italy fax: (+39) 02 2643 4892 > --------------------------------------------------------------------- > > [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.