Suppose that I have a randomly samplied group of people (-person_id-) about whom I have measured two difference characteristics (-measure- and -x-). I would like to compute the difference between the mean of two measures and model a heterogeneous variance for each separate type of measure. Clearly a t-test for independent groups is inappropriate.
To help, I've provided a small, reproducible dataset with models that I think do what I'm asking.
Code:
clear set seed 423 set obs 100 mat def M = (4, 7) mat def SD = (1.5, 0.75) mat def R = (1, 0.2 \ 0.2, 1) drawnorm x1 x2, mean(M) sd(SD) corr(R) gen int person_id = _n order person_id, first reshape long x , i(person_id) j(measure) compress
Code:
. mixed x ibn.measure, nocons || (person_id : , nocons), resid(ind, by(measure)) reml dfmethod(kroger) * output omitted Mixed-effects REML regression Number of obs = 200 Group variable: person_id Number of groups = 100 Obs per group: min = 2 avg = 2.0 max = 2 DF method: Kenward-Roger DF: min = 99.00 avg = 49.50 max = 99.00 F(2, 263.30) = 3910.89 Log restricted-likelihood = -297.96965 Prob > F = 0.0000 ------------------------------------------------------------------------------ x | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- measure | 1 | 3.912262 .1333588 29.34 0.000 3.647649 4.176875 2 | 7.102253 .0850039 83.55 0.000 6.935648 7.268857 ------------------------------------------------------------------------------ ------------------------------------------------------------------------------ Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval] -----------------------------+------------------------------------------------ person_id: (empty) | -----------------------------+------------------------------------------------ Residual: Independent, | by measure | 1: var(e) | 1.778458 .2527791 1.346043 2.349785 2: var(e) | .7225666 .1027011 .5468817 .9546901 ------------------------------------------------------------------------------ LR test vs. linear model: chi2(1) = 19.43 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.
Code:
. mixed x ibn.measure, nocons || (person_id : , nocons), resid(un, t(measure)) reml dfmethod(kroger) * output omitted Mixed-effects REML regression Number of obs = 200 Group variable: person_id Number of groups = 100 Obs per group: min = 2 avg = 2.0 max = 2 DF method: Kenward-Roger DF: min = 99.00 avg = 99.00 max = 99.00 F(2, 98.00) = 3463.37 Log restricted-likelihood = -293.14194 Prob > F = 0.0000 ------------------------------------------------------------------------------ x | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- measure | 1 | 3.912262 .1333588 29.34 0.000 3.647649 4.176875 2 | 7.102253 .0850039 83.55 0.000 6.933587 7.270919 ------------------------------------------------------------------------------ ------------------------------------------------------------------------------ Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval] -----------------------------+------------------------------------------------ person_id: (empty) | -----------------------------+------------------------------------------------ Residual: Unstructured | var(e1) | 1.778458 .2527791 1.346043 2.349785 var(e2) | .7225667 .1027011 .5468817 .9546901 cov(e1,e2) | .3455619 .1191073 .112116 .5790078 ------------------------------------------------------------------------------ LR test vs. linear model: chi2(2) = 29.09 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.
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