Ben,
I doubt the very small difference in log likelihood gives much, if any
information about which model is a better fit. Even if we overlook the
limited precision of the estimate of the REML criterion, the difference
is so small as to me of minimal importance.
John
John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and
Geriatric Medicine
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
On May 26, 2015, at 8:03 PM, Ben Bolker bbol...@gmail.com wrote:
These actually aren't terribly different from each other. I suspect
that lmer is slightly closer to the correct answer, because lme
reports a log-likelihood (really -1/2 times the REML criterion) of
49.30021, while lmer reports a REML criterion of -98.8 - slightly
better fit at -R/2 = 49.4. The residual sds are 0.0447 (lme) vs.
0.0442 (lmer); the intercept sd estimate is 0.016 vs 0.0089,
admittedly a bit low, and both month sds are very small. lmer
indicates a singular fit (correlation of -1).If you look at the
confidence intervals on these estimates (confint(fitted_model) in
lme4; intervals(fitted_model) in lme) I think you'll find that the
confidence intervals are much wider than these differences (you may
even find that lme reports that it can't give you the intervals
because the Hessian [curvature] matrix is not positive definite).
Both should be comparable to SAS PROC MIXED results, I think, if
you get the syntax right ...
On Tue, May 26, 2015 at 7:09 PM, li li hannah@gmail.com wrote:
Hi all,
I am fitting a random slope and random intercept model using R. I
used both lme and lmer funciton for the same model. However I got
different results as shown below (different variance component
estimates and so on). I think that is really confusing. They should
produce close results. Anyone has any thoughts or suggestions. Also,
which one should be comparable to sas results?
Thanks!
Hanna
## using lme function
mod_lme - lme(ti ~ type*months, random=~ 1+months|lot,
na.action=na.omit,
+ data=one, control = lmeControl(opt = optim))
summary(mod_lme)
Linear mixed-effects model fit by REML
Data: one
AIC BIC logLik
-82.60042 -70.15763 49.30021
Random effects:
Formula: ~1 + months | lot
Structure: General positive-definite, Log-Cholesky parametrization
StdDev Corr
(Intercept) 8.907584e-03 (Intr)
months 6.039781e-05 -0.096
Residual4.471243e-02
Fixed effects: ti ~ type * months
Value Std.Error DF t-value p-value
(Intercept) 0.25831245 0.016891587 31 15.292373 0.
type0.13502089 0.026676101 4 5.061493 0.0072
months 0.00804790 0.001218941 31 6.602368 0.
type:months -0.00693679 0.002981859 31 -2.326329 0.0267
Correlation:
(Intr) typPPQ months
type -0.633
months -0.785 0.497
type:months 0.321 -0.762 -0.409
Standardized Within-Group Residuals:
MinQ1 MedQ3 Max
-2.162856e+00 -1.962972e-01 -2.771184e-05 3.749035e-01 2.088392e+00
Number of Observations: 39
Number of Groups: 6
###Using lmer function
mod_lmer -lmer(ti ~ type*months+(1+months|lot), na.action=na.omit,
data=one)
summary(mod_lmer)
Linear mixed model fit by REML t-tests use Satterthwaite
approximations to
degrees of freedom [merModLmerTest]
Formula: ti ~ type * months + (1 + months | lot)
Data: one
REML criterion at convergence: -98.8
Scaled residuals:
Min 1Q Median 3Q Max
-2.1347 -0.2156 -0.0067 0. lot (Intercept) 2.870e-04 0.0169424
months 4.135e-07 0.0006431 -1.00
Residual 1.950e-03 0.0441644
Number of obs: 39, groups: lot, 6
Fixed effects:
Estimate Std. Errordf t value Pr(|t|)
(Intercept) 0.258312 0.018661 4.82 13.842 4.59e-05 ***
type 0.135021 0.028880 6.802000 4.675 0.00245 **
months 0.008048 0.001259 11.943000 6.390 3.53e-05 ***
type:months -0.006937 0.002991 28.91 -2.319 0.02767 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Correlation of Fixed Effects:
(Intr) typPPQ months
type -0.646
months -0.825 0.533
type:month 0.347 -0.768 -0.421
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