[R-sig-ME] Why am I getting a Variance of 0 for my random effect
Kevin E. Thorpe
kevin.thorpe at utoronto.ca
Wed Aug 11 21:18:31 CEST 2010
On 08/11/2010 02:25 PM, Douglas Bates wrote:
> On Wed, Aug 11, 2010 at 1:13 PM, Kevin E. Thorpe
> <kevin.thorpe at utoronto.ca> wrote:
>> Hello.
>
>> I'm getting a variance of 0 on a random effect and I don't know why.
>
> It's not a bug - it's a feature. ML estimates or REML estimates of
> variance components can be zero. This simply indicates that the
> variability in the response associated with the factor, RN in your
> case, is not sufficient to warrant the additional complexity in the
> model.
>
>> I suspect I've not set the model up correctly. My transcript is below
>> with my own comments sprinkled in for time to time.
>>
>> A little bit about the data (which I will provide off-list if requested).
>> We have nurses managing an aspect of patient care
>> according to different algorithms. Interest focuses on of the
>> algorithms result in different outcomes. I have restricted this
>> to only nurses who did each algorithm twice (in case my problem
>> was being caused by some nurses doing only one algorithm, possibly
>> only one time).
>>
>> I figured that since I have multiple observations per nurse, I
>> should treat nurse as a random effect, but maybe I confused myself
>> again.
>
> You are quite correct that it is appropriate to allow for the
> possibility of RN having an effect on the response and that it should
> be incorporated as a random effect if it were in the model but the
> results indicate that RN does not have sufficient effect on the
> response.
Thanks Doug. Your response is helpful, as always. If RN does not
contribute a random effect, would it be appropriate to revert to a
standard regression model, or is it best to leave the unimportant
random effect? In the case of ordinary regression, dropping
variables based on their p-values compromises inference. Does the
same apply with dropping a random effect with no variance?
Kevin
>
>>
>> R version 2.11.1 Patched (2010-07-21 r52598)
>> Copyright (C) 2010 The R Foundation for Statistical Computing
>> ISBN 3-900051-07-0
>>
>> R is free software and comes with ABSOLUTELY NO WARRANTY.
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>>
>> Natural language support but running in an English locale
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>> Type 'contributors()' for more information and
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>> Type 'demo()' for some demos, 'help()' for on-line help, or
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>> Type 'q()' to quit R.
>>
>>> library(lattice)
>>> library(lme4)
>>
>>> str(data1)
>> 'data.frame': 72 obs. of 3 variables:
>> $ RN : int 1 1 2 3 7 7 9 9 15 15 ...
>> $ Assignment: Factor w/ 2 levels "E","N": 1 1 1 1 1 1 1 1 1 1 ...
>> $ AUChr : num 12.26 7.23 9.26 4.04 10.31 ...
>>> tmp1<- with(data1,aggregate(AUChr,list(RN=RN,Assigment=Assignment),mean))
>>> names(tmp1)[3]<- "Mean"
>>>
>>> tmp2<- with(data1,aggregate(AUChr,list(RN=RN,Assignment=Assignment),var))
>>> names(tmp2)[3]<- "Variance"
>>>
>>> meanvar<- merge(tmp1,tmp2)
>>
>> The point of this is to show that the means are not all the same,
>> nor are the variances.
>>
>>> meanvar
>> RN Assignment Mean Variance
>> 1 1 E 9.745 12.65045
>> 2 1 N 7.185 1.36125
>> 3 15 E 10.605 15.07005
>> 4 15 N 10.385 4.41045
>> 5 16 E 8.175 0.00845
>> 6 16 N 8.420 1.03680
>> 7 2 E 7.300 7.68320
>> 8 2 N 6.950 1.00820
>> 9 21 E 9.670 9.41780
>> 10 21 N 10.535 2.44205
>> 11 22 E 7.720 2.04020
>> 12 22 N 7.930 1.21680
>> 13 24 E 9.555 10.35125
>> 14 24 N 9.330 0.38720
>> 15 25 E 8.240 0.92480
>> 16 25 N 9.485 0.00125
>> 17 27 E 8.635 0.08405
>> 18 27 N 7.745 3.72645
>> 19 28 E 9.635 8.61125
>> 20 28 N 8.315 10.35125
>> 21 3 E 6.005 7.72245
>> 22 3 N 11.435 55.44045
>> 23 31 E 9.590 9.94580
>> 24 31 N 10.570 16.70420
>> 25 35 E 9.055 0.32805
>> 26 35 N 9.925 14.41845
>> 27 36 E 9.040 2.08080
>> 28 36 N 7.395 1.14005
>> 29 5 E 8.430 3.38000
>> 30 5 N 17.385 139.94645
>> 31 6 E 6.930 0.24500
>> 32 6 N 8.330 1.72980
>> 33 7 E 10.650 0.23120
>> 34 7 N 7.375 0.09245
>> 35 9 E 8.885 7.56605
>> 36 9 N 8.405 0.73205
>>
>> Model with "Assignment" (algorithm).
>>
>>> lmer(AUChr~Assignment+(1|RN),data=data1,REML=FALSE)
>> Linear mixed model fit by maximum likelihood
>> Formula: AUChr ~ Assignment + (1 | RN)
>> Data: data1
>> AIC BIC logLik deviance REMLdev
>> 365.7 374.8 -178.8 357.7 356.9
>> Random effects:
>> Groups Name Variance Std.Dev.
>> RN (Intercept) 0.0000 0.0000
>> Residual 8.4152 2.9009
>> Number of obs: 72, groups: RN, 18
>>
>> Fixed effects:
>> Estimate Std. Error t value
>> (Intercept) 8.7703 0.4835 18.14
>> AssignmentN 0.5131 0.6837 0.75
>>
>> Correlation of Fixed Effects:
>> (Intr)
>> AssignmentN -0.707
>>
>>
>> Model without the algorithm variable.
>>
>>> lmer(AUChr~(1|RN),data=data1,REML=FALSE)
>> Linear mixed model fit by maximum likelihood
>> Formula: AUChr ~ (1 | RN)
>> Data: data1
>> AIC BIC logLik deviance REMLdev
>> 364.3 371.1 -179.1 358.3 358.5
>> Random effects:
>> Groups Name Variance Std.Dev.
>> RN (Intercept) 0.000 0.0000
>> Residual 8.481 2.9122
>> Number of obs: 72, groups: RN, 18
>>
>> Fixed effects:
>> Estimate Std. Error t value
>> (Intercept) 9.0268 0.3432 26.3
>>>
>>> sessionInfo()
>> R version 2.11.1 Patched (2010-07-21 r52598)
>> Platform: i686-pc-linux-gnu (32-bit)
>>
>> locale:
>> [1] LC_CTYPE=en_US LC_NUMERIC=C LC_TIME=en_US
>> [4] LC_COLLATE=C LC_MONETARY=C LC_MESSAGES=en_US
>> [7] LC_PAPER=en_US LC_NAME=C LC_ADDRESS=C
>> [10] LC_TELEPHONE=C LC_MEASUREMENT=en_US LC_IDENTIFICATION=C
>>
>> attached base packages:
>> [1] stats graphics grDevices utils datasets methods base
>>
>> other attached packages:
>> [1] lme4_0.999375-34 Matrix_0.999375-42 lattice_0.18-8
>>
>> loaded via a namespace (and not attached):
>> [1] grid_2.11.1 nlme_3.1-96 stats4_2.11.1
>>>
>>> proc.time()
>> user system elapsed
>> 3.488 0.056 3.536
--
Kevin E. Thorpe
Biostatistician/Trialist, Knowledge Translation Program
Assistant Professor, Dalla Lana School of Public Health
University of Toronto
email: kevin.thorpe at utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016
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