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Mplus Discussion > Growth Modeling of Longitudinal Data >
 Larry Kurdek posted on Saturday, February 08, 2003 - 7:21 am
Using married couples as the unit of analysis, I am estimating growth curves (2 parameters, intercepts and slopes) for 3 variables (self-perception, spouse-perception, and marital distress) for each spouse, for a total of 12 latent variables. I have 4 annual assessments for each variable for a total of 24 observed variables.

I am especially interested in links among the latent variables. I can estimate the overall model in Mplus very easily with good fit, but am wondering how I can assess some specific contrasts in the overall model.

For example, looking at relations among the slopes, the link between change in self perception and change in distress is .02 for husbands and .03 for wives. The link between change in spouse perception and change in distress is .08 for husbands and .07 for wives. All 4 links are significant. (The self and spouse assessments are on the same scale, so relations involving them can be compared directly.)

I want to test whether the difference between the spouse-distress and the self-distress links differs between husbands and wives. That is, is the difference between the 2 links for husbands (.08 versus .02 = .06) different from the parallel 2 links for wives (.07 versus .03 = .04), or is .06-.04 = .02 reliably different from 0? In ANOVA terms, this is equivalent to asking whether the interaction between two within-subjects effects is significant (type of perception, self vs. other x spouse, husband vs. wife). Can I perform this test in Mplus?
 Linda K. Muthen posted on Saturday, February 08, 2003 - 7:42 am
You can use chi-square difference testing to accomplish this. Run a model with the husband and wife coefficients held equal and another model with them not held equal.
 Larry Kurdek posted on Saturday, February 08, 2003 - 12:35 pm
Thanks for the quick reply. I have run chi-square tests, but I do not think they get at the 'double difference' question I have. That is, is the difference between two coefficients different for husbands and wives?

Here is my Mplus model statement where h=husband, w=wife, i = intercept, l=linear change, ns=negative self, np=negative partner, and ds=distress:

MODEL: hins BY h1ns-h4ns@1;
hlns BY h1ns@0 h2ns@1 h3ns@2 h4ns@3;
hinp BY h1np-h4np@1;
hlnp BY h1np@0 h2np@1 h3np@2 h4np@3;
hids BY h1dis-h4dis@1;
hlds BY h1dis@0 h2dis@1 h3dis@2 h4dis@3;
wins BY w1ns-w4ns@1;
wlns BY w1ns@0 w2ns@1 w3ns@2 w4ns@3;
winp BY w1np-w4np@1;
wlnp BY w1np@0 w2np@1 w3np@2 w4np@3;
wids BY w1dis-w4dis@1;
wlds BY w1dis@0 w2dis@1 w3dis@2 w4dis@3;

[h1ns-h4ns@0 h1np-h4np@0 h1dis-h4dis@0
w1ns-w4ns@0 w1np-w4np@0 w1dis-w4dis@0
Adding the following constraint statements
allows me to test whether the set of coefficients involving change differs between spouses (by comparing the difference in chi-square values for this model and the original model):

hlds WITH hlns (1);
hlds WITH hlnp (1);
wlds WITH wlns (1);
wlds WITH wlnp (1);

However, what I am interested in is whether the difference between the coefficients for the first two statements (ds with ns versus ds with np for husbands) is the same as the difference between the coefficients for the last two statements (ds with ns versus ds with np for wives).

Is there a set of equality constraints that will give me this double-difference test?

Thank you so much for your assistance.
 bmuthen posted on Sunday, February 09, 2003 - 9:19 am
A general approach to this is to draw on the estimated covariance matrix of the parameter estimates obtained from Tech3, which can be saved using SAVEDATA and FILE(TECH3). This gives the standard deviation in the denominator of a z test such as:

(a - b)/sd(a - b)

where sd(a-b) involves the variances and covariance of a and b found in Tech3.

To save the computational work of the above approach, there are probably several ways of parameterizing the model so that the double-difference test can be done through chi-square difference testing. One that comes to mind is to add a second-order factor that influences your slopes with unit coefficients, so that the factor variance captures the presumably positive covariance (the link you talk about). For each gender you are interested in the difference in a covariance difference - so for one gender you would have:

hf by hlds@1 hlns@1 hlnp@1;
! this gives the covariance between each of the
! 3 pairs as the variance of hf.
hlds with hlnp;
! this residual covariance gives the extra
! covariance - the difference

Then you test equality of the "hlds with hlnp" residual covariance and the "wlds with hlnp" residual covariance. - Note that I have not tested this thinking, so there may be a glitch, but this gives the general idea.
 Larry Kurdek posted on Sunday, February 09, 2003 - 3:34 pm
Thanks for the very helpful reply. I will check it out. (By the way, the support you provide for your program is simply incredible!).
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