Using the MPlus example data:
* Data for a two factor measurement model
infile x1-x6 using "Z:\PUBLIC_WEB\MPlus\Basics\CFA and MIMIC\ex5.1.dat"
// The congeneric model, where everything varies
sem (L1 -> x1-x3)
estimates store congeneric(500 observations read)
Endogenous variables
Measurement: x1 x2 x3
Exogenous variables
Latent: L1
Fitting target model:
Iteration 0: log likelihood = -2450.3813
Iteration 1: log likelihood = -2450.3813
Structural equation model Number of obs = 500
Estimation method = ml
Log likelihood = -2450.3813
( 1) [x1]L1 = 1
------------------------------------------------------------------------------
| OIM
| Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
Measurement |
x1 <- |
L1 | 1 (constrained)
_cons | -.0222186 .062788 -0.35 0.723 -.1452808 .1008436
-----------+----------------------------------------------------------------
x2 <- |
L1 | 1.125066 .0988703 11.38 0.000 .931284 1.318848
_cons | .0255831 .0624333 0.41 0.682 -.0967839 .1479501
-----------+----------------------------------------------------------------
x3 <- |
L1 | 1.019625 .0888617 11.47 0.000 .8454592 1.193791
_cons | .0346609 .0624935 0.55 0.579 -.0878241 .1571459
-------------+----------------------------------------------------------------
var(e.x1)| 1.063175 .0956754 .8912609 1.268249
var(e.x2)| .7996467 .0998985 .6259779 1.021498
var(e.x3)| 1.008739 .0952739 .8382702 1.213874
var(L1)| .9079904 .125112 .6930973 1.189511
------------------------------------------------------------------------------
LR test of model vs. saturated: chi2(0) = 0.00, Prob > chi2 = .And some example output:
// Tau-equivalent, all loadings equal
// a unit change on one measure is equal to that on another
sem (L1@a -> x1-x3)
// also written (L1 -> x1@a x2@a x3@a), or as three separate paths
estimates store tauequivalent
lrtest tauequivalent congenericEndogenous variables
Measurement: x1 x2 x3
Exogenous variables
Latent: L1
Fitting target model:
Iteration 0: log likelihood = -2452.1656
Iteration 1: log likelihood = -2451.4314
Iteration 2: log likelihood = -2451.4237
Iteration 3: log likelihood = -2451.4237
Structural equation model Number of obs = 500
Estimation method = ml
Log likelihood = -2451.4237
( 1) [x1]L1 - [x3]L1 = 0
( 2) [x2]L1 - [x3]L1 = 0
------------------------------------------------------------------------------
| OIM
| Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
Measurement |
x1 <- |
L1 | 1.049357 598.7102 0.00 0.999 -1172.401 1174.5
_cons | -.0222186 .063565 -0.35 0.727 -.1468037 .1023665
-----------+----------------------------------------------------------------
x2 <- |
L1 | 1.049357 598.7102 0.00 0.999 -1172.401 1174.5
_cons | .0255831 .0613638 0.42 0.677 -.0946877 .1458539
-----------+----------------------------------------------------------------
x3 <- |
L1 | 1.049357 598.7102 0.00 0.999 -1172.401 1174.5
_cons | .0346609 .0629133 0.55 0.582 -.0886469 .1579687
-------------+----------------------------------------------------------------
var(e.x1)| 1.019347 .0857791 .8643556 1.20213
var(e.x2)| .8818478 .0783523 .7409079 1.049598
var(e.x3)| .9781344 .0834668 .8274909 1.156202
var(L1)| .9089662 1037.221 . .
------------------------------------------------------------------------------
LR test of model vs. saturated: chi2(1) = 2.08, Prob > chi2 = 0.1488
Likelihood-ratio test LR chi2(1) = 2.08
(Assumption: tauequivalent nested in congeneric) Prob > chi2 = 0.1488