Confidence interval for sum of coefficients

Is there an issue with using the variance sum law and the var-covar matrix  to sum two parameters and estimate the variance of the sum? i.e., add their variances and covariances as expressed in the variance covariance matrix of the parameter estimates, probably obtained with vcov(modelObj).

Or is this too simplistic for a mixed model?

-Nate

--
Nathan J. Doogan, Ph.D.  | College of Public Health
Post-Doctoral Researcher | The Ohio State University



-----Original Message-----
From: r-sig-mixed-models-bounces at r-project.org [mailto:r-sig-mixed-models-bounces at r-project.org] On Behalf Of Michael Cone
Sent: Thursday, September 25, 2014 9:26 AM
To: r-sig-mixed-models at r-project.org
Subject: Re: [R-sig-ME] Confidence interval for sum of coefficients

Dear list,

Lorenz has pointed out to me Ben's suggestion to bootstrap the sums (or any linear combiantion) of coefficients I'm interested in. This may be the general approach, but I struggle to see why it would be illegitimate to simply change the reference level for the treatment contrast coding, fit the model again and run confint() a second time (and do so again for
MachineC):

Machines$Machine <- relevel(Machines$Machine, 'B')
fm2 <- lmer(score ~ Machine + (Machine | Worker), data = Machines)
summary(fm2)

Fixed effects:
             Estimate Std. Error t value
(Intercept)   60.322      3.529  17.096
MachineA      -7.967      2.421  -3.291
MachineC       5.950      2.446   2.432

confint(fm2)

(Intercept)  52.8500103 67.7944456
MachineA    -13.0931710 -2.8401544
MachineC      0.7692323 11.1307757

Now the CI of the intercept is the confidence interval for the overall score of MachineB. Adding lower and upper bounds from fm1 would have given somewhat similar, but somewhat wider intervals.
(I probably have a lack of understanding as to how CIs can be calculated with. Is there an inuitive explanation for why the bounds don't add?)

Machines$Machine <- relevel(Machines$Machine, 'C')
fm3 <- lmer(score ~ Machine + (Machine | Worker), data = Machines)
summary(fm3)

Fixed effects:
             Estimate Std. Error t value
(Intercept)   66.272      1.806   36.69
MachineB      -5.950      2.446   -2.43
MachineA     -13.917      1.540   -9.04

confint(fm3)

(Intercept)  62.4471752  70.0972752
MachineB    -11.1307677  -0.7692243
MachineA    -17.1780524 -10.6552759

Thanks, and best wishes
Michael

Am 25.09.2014 14:11 schrieb Michael Cone:

Hello,

I suspect this to be simple, but I can't figure it out.

library(lme4)
data(Machines)
fm1 <- lmer(score ~ Machine + (Machine | Worker), data = Machines)
summary(fm1)

Fixed effects:
            Estimate Std. Error t value
(Intercept)   52.356      1.681  31.151
MachineB       7.967      2.421   3.291
MachineC      13.917      1.540   9.036

confint(fm1)

                 2.5 %     97.5 %
[...]
(Intercept) 48.7964047 55.9147119
MachineB     2.8401623 13.0931789
MachineC    10.6552809 17.1780575

[and 14 warnings, but it's just an example:
In optwrap(optimizer, par = start, fn = function(x) dd(mkpar(npar1, 
... :
convergence code 1 from bobyqa: bobyqa -- maximum number of function 
evaluations exceeded ...
In profile.merMod(object, signames = oldNames, ...) : non-monotonic 
profile]

I'd like to have confidence intervals for the overall score of 
MachineA, MachineB, and MachineB. MachineA is easy (CI of the 
intercept), but how do I combine the CI of the intercept with the CI 
of the MachineB parameter, and likewise the CI of the intercept with 
the parameter of MachineC? Can I simply add the lower and upper bounds 
of the two intervals or is this naive?

Thank you for your time,

Michael

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