mixed - Stata

mixed — Multilevel mixed-effects linear regression 33
The unstructured covariance matrix is the most general and contains many parameters. In this example,
we estimate a distinct residual variance for each day and a distinct covariance for each pair of days.
That there is positive covariance between all pairs of measurements is evident, but what is not as
evident is whether the covariances may be more parsimoniously represented. One option would be to
explore whether the correlation diminishes as the time gap between strength measurements increases
and whether it diminishes systematically. Given the irregularity of the time intervals, an exponential
structure would be more appropriate than, say, an AR or MA structure.
. estimates store unstructured
. mixed strength i.program##i.day || id:, noconstant
> residuals(exponential, t(day)) nolog nofetable
Mixed-effects ML regression Number of obs = 173
Group variable: id Number of groups = 37
Obs per group: min = 3
avg = 4.7
max = 5
Wald chi2(9) = 36.77
Log likelihood = -307.83324 Prob > chi2 = 0.0000
Random-effects Parameters Estimate Std. Err. [95% Conf. Interval]
id:
(empty)
Residual: Exponential
rho .9786462 .0051238 .9659207 .9866854
var(e) 11.22349 2.338371 7.460765 16.88389
LR test vs. linear regression: chi2(1) = 292.17 Prob > chi2 = 0.0000
Note: The reported degrees of freedom assumes the null hypothesis is not on
the boundary of the parameter space. If this is not true, then the
reported test is conservative.
In the above example, we suppressed displaying the main regression parameters because they
did not differ much from those of the previous model. While the unstructured model estimated 15
variance–covariance parameters, the exponential model claims to get the job done with just 2, a fact
that is not disputed by an LR test comparing the two nested models (at least not at the 0.01 level).
. lrtest unstructured .
Likelihood-ratio test LR chi2(13) = 22.50
(Assumption: . nested in unstructured) Prob > chi2 = 0.0481
Note: The reported degrees of freedom assumes the null hypothesis is not on
the boundary of the parameter space. If this is not true, then the
reported test is conservative.