# [Fwd: Re: [R] degrees of freedom for t-tests in lme]

Han-Lin Lai Han-Lin.Lai at noaa.gov
Fri Apr 26 18:18:26 CEST 2002

Sorry, by mistake I sent this to Professor Bates instead of r-help.
Han

-------- Original Message --------
Subject: Re: [R] degrees of freedom for t-tests in lme
Date: Thu, 25 Apr 2002 09:16:16 -0700
From: Han-Lin Lai <Han-Lin.Lai at noaa.gov>
To: Douglas Bates <bates at stat.wisc.edu>
References: <3CC6E87F.5400277D at noaa.gov>
<6rg01lottu.fsf at franz.stat.wisc.edu>

Thank you very much for the answer.  I find that there is couple typo in
my
data.  Now I get the correct results from fitting the model:
lme(y~x+log(den)+sex+dep,data=lwd,random= list(group=~x))

numDF denDF      F-value         p-value
(Intercept)             1  3209     17660.85      <.0001
x                 1  3209       6411.56      <.0001
sex                1  3209          12.07      0.0005
dep               3    22            25.10      <.0001

Fixed effects: y ~ x + sex + dep
Value        Std.Error        DF   t-value
p-value
(Intercept)     -11.24380     0.1725474     3209 -65.16353  <.0001
x        3.05784      0.0382305     3209  79.98440  <.0001
sex       0.01958     0.0057823      3209   3.38548  0.0007
depD27       0.01382     0.0593448          22   0.23293  0.8180
depD35      -0.06537     0.0550606         22  -1.18723  0.2478
depD50      -0.28581     0.0557979         22  -5.12220  <.0001

However, my real questions is:  How DF = 3209 and 22 are calculated?  I
follow
the book by Pinheiro and Bates (2000, p.91),
Q = 1
mo=1 because I have intercept in model
m1=26 groups
m2=3237 observations
p0=1
p1=3, (I have 4 levels of dep's, therefore, it is 4-1. Isn't it?)
p2=??

Then, DF for dep is 26-(1+3)=22.  Following this way, p2 = 2 so the DF
for
intercept, x, and sex is 3237-(26+2)=3209.  How is p2 determined?

Thanks for the help.
Han

Douglas Bates wrote:

> Han-Lin Lai <Han-Lin.Lai at noaa.gov> writes:
>
> > I have trouble to figure out how the df is derived in LME.  Here is my
> > model,
> >
> >    lme(y~x+log(den)+sex+dep,data=lwd,random= list(group=~x))
> >
> > Number of total samples (N) is 3237
> > number of groups (J) is 26
> > number of level-1 variables (Q1) is 3, i.e., x, log(den) and sex
> > number of level-2 variables (Q2) is 1, i.e., dep
> > x and den are continuous variable
> > sex is associated with individual samples and has 2 levels
> > dep is associated with group has 4 levels: depD15, depD27, and depD35,
> > and depD35.
> >
> > I got the results:
> >
> > Fixed effects: y ~ x + log(den) + sex + dep
> >                          Value       Std.Error   DF   t-value p-value
> > (Intercept) -11.29271     0.1681915 3206 -67.14200  <.0001
> >               x     3.05937     0.0367970 3206  83.14182  <.0001
> >    log(den)     0.00898     0.0022357 3206   4.01732  0.0001
> >            sex     0.01980     0.0057675 3206   3.43216  0.0006
> >     depD27   -0.01505     0.0560142 3206  -0.26872  0.7882
> >     depD35   -0.06102     0.0548647 3206  -1.11227  0.2661
> >     depD50   -0.29123     0.0567132     24  -5.13511  <.0001
> >
> > Are these coefficients are the level-2, and thus, DF for testing dep's
> > should be (J-Q2_1=26-1-1=24).  Why I get the number 3206, especially for
> > depD27 depD35 and depD50.
> >
> > Thanks
> > Han
> > Han-Lin.Lai at noaa.gov
>
> By saying that dep is a level-2 variable do you mean that it does not
> change within groups?  Have you checked that, say by
>
> gsummary(mydata, form = ~ group, invariants = TRUE)
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