Interpretation of coefficients in gro... PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
 Jenny Oaklin posted on Monday, December 06, 2004 - 5:21 pm
What do the coefficients mean in this equation?

SALES = 2143.869 - 0.923(Winter) - 0.963(Spring/Summer) + 1.004(TREND)

Winter, Spring/summer are dummy variables coded 1 and 0 with the base season being fall and trend is the time period.

does SALES decrease by 0.923 in the winter and 0.963 in the spring/summer?

does the intercept 2143.869 have any significant meaning?

Thanks in advance
 bmuthen posted on Monday, December 06, 2004 - 5:34 pm
The intercept is the SALES value when all the predictors have value 0 (so for Fall with TREND=0). So, yes on your last question.

And yes on your first question, with some elaboration. E.g., -0.923 for Winter says that for any given TREND value, Winter has 0.923 less SALES value than Fall. Analogous for -0.963 for Spring/Summer.
 tonkin posted on Tuesday, April 19, 2005 - 5:03 pm
Three questions: First, I ran a two group growth model of continuous outcomes at four time points. I believe Dr. Muthen said I could do a t-test for the intercepts and slopes for the two groups to see if they differed signficantly. Is this so? If so, what is the formula? a Z-test for two samples? What do I put in the denominator?

Second, I have read several of your articles on gc and still do not really understand exactly what the latent intercept and slope are (technically). I know they are defined as the initial status (when timepoint is designated zero) and trend, but is it a factor score, or mean, or something else? When I look at the plots, they say "means" and are in the metric of my measures, but I think I need to know how the latent intercept and slope are calculated? Does this make sense?

Finally, Can I use censored variables in gc models, and particularly in three level models?

Thank You,
Peggy Tonkin
 BMuthen posted on Wednesday, April 20, 2005 - 9:18 am
Yes, you could do a t-test for two samples. Or, you could do two analyses where the growth factor means are held equal in one analysis and are not held equal in the other. The chi-square difference test can be used to determine whether the means are the same across groups.

The latent intercept and slope are latent variables, that is, scores on factors. We are interested in their means, variances, and covariances. The plots are for the model estimated means.

Yes, growth models can be estimated for censored outcomes.
 tonkin posted on Wednesday, April 20, 2005 - 2:10 pm
Thank you for your quick response you have been very helpful. One more question, can I treat my outcomes (continuous) as censored if I am doing a three level gc with covariates in the within model?
Thanks so much again,
Peggy Tonkin
 Linda K. Muthen posted on Wednesday, April 20, 2005 - 2:54 pm
A two-level growth model(three-level in HLM terms for example) cannot be estimated for a censored outcome. Only a one-level growth model (two-level in HLM terms) can be estimated for a censored outcome.
 Amy Tobler posted on Wednesday, September 15, 2010 - 11:29 am
We have a group randomized study and are trying to ascertain potential treatment effects on alcohol use trajectories.
iw sw | ALCYEAR2@0 ALCYEAR3@12 ALCYEAR4@24;
iw sw ON fam lunch;
ib sb | ALCYEAR2@0 ALCYEAR3@12 ALCYEAR4@24;
sb ON trt;

My question is: how do we interpret the beta of treatment regressed on sb? Is it the difference between the control group slope and the treatment group slope? If so how can we recover these slope from the output?
 Linda K. Muthen posted on Thursday, September 16, 2010 - 10:11 am
The regression coefficient in difference between the control and treatment group means of sb. The mean for the control group is the intercept. The mean for the treatment group is the intercept plus the slope.
 Amy Tobler posted on Thursday, September 16, 2010 - 11:05 am
Thanks for the response. One quick clarification. We get the following output:

TRT -0.037 0.022 -1.650 0.099
IB 0.000 0.000 999.000 999.000
SB 0.063 0.137 0.456 0.648

When you say the mean for the control group is the intercept, do you mean the mean of the intercept factor (in out output this would be 0)? or the intercept for the slope factor (in our output this would be 0.063)? Then the mean slope of the treatment group would be either the 0 or the 0.063 + (-0.037)? Thanks for your time.
 Linda K. Muthen posted on Thursday, September 16, 2010 - 1:38 pm
The mean for the control group is the intercept of sb.
 fred posted on Tuesday, October 25, 2016 - 1:22 am
I am running a LGC with one intercept and two slope of which the first one is negative (estimate= -12.477. SE=1.795). To achieve model fit this slope needs to be regressed on the intercept, which makes theoretical sense. The regression weight is 0.133, SE=0.021. I am somehow unsure if this can imply that the higher the intercept the "slower" or "faster" decrease as indicated by the slope since the regression weight is less than 1.
 Bengt O. Muthen posted on Tuesday, October 25, 2016 - 10:13 am
The higher that intercept is, the less negative is the slope. Try out different values of the intercept in that regresion and you will find this.
 Kelly Warmuth posted on Monday, April 22, 2019 - 8:08 am
Hi Drs. Muthén,

I created a growth model with 6 time points of a continuous dependent variable and a nominal time-invariant covariate with four categories. I created three dummy variables for the TIC (i s ON dum1 dum2 dum3) and am now trying to understand my output.

1) Under “Intercepts”, how should the significance of the estimates for I and S be interpreted? I understand that this is the baseline group and the estimates are the intercept and slope for that group, but I don’t know what to make of the two-tailed p-value. Does it indicate that the intercept and slope of the baseline group differ significantly from zero or that they differ from the other three groups?

2) For the results regarding the dummy coded TIC, if the slope for “s ON dum1” is significant, does that mean that group’s slope differs significantly from just the baseline group or all other groups?

Thank you for your help and for this wonderful resource!
 Bengt O. Muthen posted on Monday, April 22, 2019 - 5:07 pm
1) intercept and slope of the baseline group differ significantly from zero (that's the only case where all dummies are zero).

2) all other groups (when dum1=1, the other dummies are all zero).

This is discussed in regression books which also discuss other coding schemes answering other questions.
 Tor Neilands posted on Tuesday, March 10, 2020 - 2:35 pm

Apologies in advance for what is probably a very basic question or FAQ, but in years of fitting SEMs I have yet to encounter this particular issue. I have helped an investigator at our institution fit an LGM in Mplus. The LGM has a random intercept and a random slope, with the latter being defined by the loading coefficients -3, -2, -1, and 0 so that the intercept reflects the last time point.

The random intercepts and slopes are explained by multiple covariates (uncentered). One continuous covariate has a significant positive coefficient .06 on the random slope term. The investigators desire to interpret this coefficient substantively. From reviewing a nice video by Patrick Curran ( near 17:50 in the video), it seems that one recommended way to do that would be to pick points of the continuous covariate - say, first quartile, median, and third quartile - and generate trajectory plots plus significance tests of the resulting simple slopes.

Would you agree this is a good way to go for interpreting the meaning of a significant coefficient of the regression of a latent slope onto a continuous covariate or is there another approach you would recommend instead? Assuming this is the approach you would recommend, can you point me to an example of how to generate the simple slopes and plot in Mplus? Thanks much, Tor Neilands
 Bengt O. Muthen posted on Tuesday, March 10, 2020 - 3:02 pm
I think it is a good idea to plot estimated lines for the outcome as a function of time where the lines correspond to different covariate values. You can do that with the Model Constraint command options LOOP and PLOT. 3 lines corresponding to the covariate values can be shown in the same plot with CIs. You express each estimated line like we show in our Short Course Topic 3 slides 97-98.
 Tor Neilands posted on Sunday, March 15, 2020 - 12:47 pm
Thanks, Bengt, this is very helpful. I set up MODEL CONSTRAINT code to emulate slide 97 of Short Course Topic 3. Alpha and beta are intercepts of the random intercepts and slopes, respectively. i1-i13 and s1-s13 are the slopes for the regressions of the random intercept and slope terms onto the covariates, respectively. m1-m13 are the means for the covariates, estimated in the model to address covariate missing values.

I am getting quite different values for the model constraint intercept and slope terms (52.969 and -6.332) whereas Tech4 shows they are 12.994 and -.443, respectively. The Tech 4 output is more consistent with the sample means for the four time points, which are 14.243, 14.007, 13.433, and 12.997 (LGM loadings are coded -3, -2, -1, 0 so the intercept reflects the last time point). What am I doing incorrectly?

Model Constraint:
New int_mean slp_mean T1_mean T2_mean T3_mean T4_mean ;
int_mean = alpha + i1*m1 + i2*m2 + i3*m3 + i4*m4 + i5*m5 + i6*m6 + i7*m7 +
i8*m8 + i9*m9 + i10*m10 + i11*m11 + i12*m12 + i13*m13 ;
slp_mean = beta + s1*m1 + s2*m2 + s3*m3 + s4*m4 + s5*m5 + s6*m6 + s7*m7 +
s8*m8 + s9*m9 + s10*m10 + s11*m11 + s12*m12 + s13*m13 ;
 Bengt O. Muthen posted on Sunday, March 15, 2020 - 4:51 pm
That's odd. Why don't you send your output to Support so we can look at the whole picture.
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