Interpretation and significance of gr...
Message/Author
 Marja Aartsen, Vrije Universiteit Amsterdam, The Netherlands posted on Thursday, June 20, 2002 - 3:21 am
Dear Linda and Bengt Muthen,

I am trying to understand the output of a latent growth model, estimated with Mplus. I therefore estimated the same model with LISREL 8.12 and find almost the same estimates for the parameters. However, there are two things that I don't understand:

1: What is the meaning of the estimated Intercepts in the first section of the Model results? The growth factor 'Intercept' is about two times higher than what I expect on the basis of the raw data. In the Tech4 output the estimated Intercepts are in line with what I expected, but no significance (s.e, z-value or variances) are printed there. 2. What can I do see whether these parameters are significant?

Both LISREL and Mplus inputfiles are provided below.
Thank you very much in advance. Marja Aartsen

title: linear and quadratic growth in delayed recall, three timepoints total time span 6 years,
plus the effect of two explanatory variables age and sex.
residuals are homoscedastic and not correlated over time.
latent growth factors are not correlated
data: file is c:\data\bcddr1.dat;
format is 6f6.2;
variable: names are respnr bmwtdr cmwtdr dmwtdr age sex;
usevariables bmwtdr cmwtdr dmwtdr age sex;
analysis: type=meanstructure;
model: i by bmwtdr-dmwtdr@1;
s by bmwtdr@0 cmwtdr@3 dmwtdr@6;
q by bmwtdr@0 cmwtdr@3 dmwtdr@9;
[bmwtdr-dmwtdr@0 i s q];
i s q on age;
i s q on sex;
bmwtdr cmwtdr dmwtdr (2);
i with s@0;
i with q@0;
s with q@0;
output: tech4 tech3;

(Lisrel input)
DA NG=1 NI=6 NO=999
RA FI=C:\DATA\BCDDR1.DAT FO
(6F6.2)
LA
RESPNR BMWTDR CMWTDR DMWTDR AGE SEX
SE
2 3 4 5 6/
MO NY=3 TY=ZE NE=3 TE=SY,FI AL=FR BE=ZE PS=SY,FI NX=2 NK=2 TX=FR TD=ZE c
GA=FU,FI PH=SY,FR
LE
INTERCEPT SLOPE SLOPE2
MA LY
1 0 0
1 3 3
1 6 9
FR TE(1,1) TE(2,2) TE(3,3)
!FR TE(1,2) TE(2,3)
EQ TE(1,1) TE(2,2) TE(3,3)
FR PS(1,1) PS(2,2) PS(3,3)
LK
AGE SEX
MA LX
1 0
0 1
FR GA(1,1) GA(2,1) GA(3,1)
FR GA(1,2) GA(2,2) GA(3,2)
PATH DIAGRAM
OU ML SE TV RS ND=6 IT=1000.
 Linda K. Muthen posted on Thursday, June 20, 2002 - 8:29 am
If you send your Mplus output and data if possible to support@statmodel.com, I will try to answer your questions.
 Anonymous posted on Wednesday, June 16, 2004 - 7:39 pm
I'm using v2.14 to estimate a random effects model. I have measures of student performances at five periods. What I would like to do is fix the intercept at some point just after the second period and run the model. Saving the intercept coefficient, I would like to rerun the model fixing the intercept at some point prior to the third period. Then I would like to run the model a third time replacing the intercept with the value I saved from the first run and perform a chi-square test on the difference. My question is, is the Loglikelihood value and number of free parameters actual (or pseudo) values that would permit me to calculate a difference test?

Thank you.
 bmuthen posted on Wednesday, June 16, 2004 - 9:04 pm
It sounds like you want to consider the intercept factor mean at different time points by centering at these different time points. The intercept factor at the time point of centering is the (systematic part of) the growth function at that time point. (Different centering points give the same number of parameters and same chi-square test of fit value.) It sounds like you want to test for differences in the growth function at different time points. If I am understanding you correctly so far, it seems like you can conduct such a test by testing the slope mean. With a linear model with time scores say 0, 1, 2, ... the mean difference at different time points is the slope mean. Let me know if I am not getting the point of your question.
 Anonymous posted on Wednesday, June 16, 2004 - 9:28 pm
The wrinkle is that I'm including a time^2 slope as well because many of the individual plots show that a quadrilateral function works best.
 bmuthen posted on Thursday, June 17, 2004 - 8:14 am
With a quadratic growth function your mean difference would be due to the means of both the linear and quadratic slopes. A test of this difference would involve the variance of a sum involving those means which implies that you need the SE of each and also their covariance from Tech3, using the Delta method. Off hand, I can't think of a trick to get this directly by modeling, but it is quite likely it is possible.
 Anonymous posted on Thursday, June 17, 2004 - 8:32 am
Thank you for your response. If I change the intercepts for each of the two different runs, wouldn't the means of the linear and quad. slopes be 0 for each run and the only mean calculated is that of the intercept?
 bmuthen posted on Thursday, June 17, 2004 - 9:20 am
Yes, you are right of course - the time score multiplying the slope means is zero; hence no contribution from the slope means (these are not zero, however).
 Anonymous posted on Monday, June 21, 2004 - 7:22 am
If I may clarify one last point. If only the intercept is changing and the time scores * the two slopes is zero, can I fix the intercept in a second run and use the difference in the log likelihood as a test of significance?
 bmuthen posted on Monday, June 21, 2004 - 4:58 pm
I don't think that will do it. I think you should do the test in a single run. In that run you should consider the growth curve mean at two time points and that will then actually require the considerations I mentioned above:

With a quadratic growth function your mean difference would be due to the means of both the linear and quadratic slopes. A test of this difference would involve the variance of a sum involving those means which implies that you need the SE of each and also their covariance from Tech3, using the Delta method. Off hand, I can't think of a trick to get this directly by modeling, but it is quite likely it is possible.
 Carolin posted on Wednesday, September 07, 2011 - 1:17 am
Hi,

I'm analyzing a GMM with 4 timepoints and I have a question concerning the interpretation of the slope factor.
In my model, in one class the slope factor is not significant, but has a mean of 0.123. Does this mean, that there is no significant change although the slope indicates increase? Can I interprete that trajectory as a stable one (I think not).
In another class the slope is signifiacnt, but the value is small (0.045). Does this mean here I have a significant linear change, although the increase is that small and seems rather stable?

Thank a lot!
 Linda K. Muthen posted on Wednesday, September 07, 2011 - 7:56 am
I am not clear on your question. The slope growth factor has a mean and variance. If the mean is significant, it means that the trajectory is not flat. If the variance is significant, it means there is variability around the mean.
 Carolin posted on Thursday, September 08, 2011 - 12:10 am
I wonder if the mean of the slope is significant, but very small (0.045), so that it looks like a flat trajectory you nevertheless say there is significant increase??
And on the other side, if the mean isn't significant, but has a mean of e.g. 0.123 and it really looks like an increase, than you say the trajectory is flat??
Hope I am now clearer?!? Thanks a lot!!
 Linda K. Muthen posted on Thursday, September 08, 2011 - 6:01 am
The significance test is whether the mean of the slope growth factor is significantly different from zero. If the value is deemed to be not significantly different from zero, then the slope is flat. Whether the value is significantly different from zero depends not only on the parameter estimate but also the standard error of the parameter estimate.
 Carolin posted on Friday, September 09, 2011 - 6:13 am
Thanks a lot! Just for my understanding: I could say that the slope factor mean is significantly different from zero, but is interpreted as flat (because of small value) and in the other case I could say that the trajectory shows an increasing course but the mean of slope does not significantly differ from zero. Am I right?
 May Yang posted on Thursday, April 18, 2013 - 3:25 pm
I am attempting to re-create a growth model that was unfortunately conducted in AMOS, the reviewers are requesting the same statistics outputted from that program which is the issue since I am using Mplus.

In a growth model with covariates, I understand that intercepts are estimated instead of means and residual variances are estimated instead of variances for the latent growth measures (intercept and slope). I am also aware that we can request estimated means via TECH4.

My questions are:
1) can I obtain significance testing for the estimated means from tech4?
2) is there a way to obtain estimated variances which are not outputted in tech4?

Thank you.
 Bengt O. Muthen posted on Thursday, April 18, 2013 - 3:54 pm
1) This is available in the forthcoming Mplus Version 7.1.

2) You can use MODEL CONSTRAINT to compute estimates and SEs for any quantity that you express in terms of the model estimates.
 May Yang posted on Sunday, April 21, 2013 - 7:39 pm
Thank you Bengt,
In response to my 4/18 post, I have attempted to calculate the estimated means and variances of the slope and intercept latent variables but can not seem to figure out translating the formula (from short course slides) into the model constraint statements.

If the estimated intercept mean = est intercept + estimated slope (X1) * sample mean (X1);

Here is a sample of my program:

MODEL:
i s | ls1@0 ls2@1 ls3@2 ls4@3;
i on age_c (b);
i(a);

MODEL CONSTRAINT:
new (mean_i);
mean_i= a + b*age_c;

This gives an 'unknown parameter' error for age_c. I am having trouble figuring out how the "sample mean X1" part of the equation should be programmed. Can you help? I need to report the significant of the means so cannot simply report the mean from TECH4.
 Linda K. Muthen posted on Monday, April 22, 2013 - 9:47 am
MODEL:
i s | ls1@0 ls2@1 ls3@2 ls4@3;
i on age_c (b);
i(a);
[age_c] (c);

MODEL CONSTRAINT:
new (mean_i);
mean_i= a + b*c;
 May Yang posted on Monday, April 22, 2013 - 4:51 pm
Thanks Linda. I am getting model
non-covergence issues after implementing [age_c] when the model converged before. And the mean estimate for the intercept (mean_i) does not equal to what is outputted under TECH4 for intercept mean. Any pointers?
 Linda K. Muthen posted on Monday, April 22, 2013 - 5:28 pm
 Michael T Weaver posted on Thursday, July 17, 2014 - 11:26 am
I estimated a parallel process latent growth model of the following form:
Model:
i1 s1 | se0@0 se3@3 se6@6 se9@9 se12@12 se15@15 se18@18;
i2 s2 | dmgmt0@0 dmgmt3@3 dmgmt6@6 dmgmt9@9 dmgmt12@12 dmgmt15@15 dmgmt18@18;
i1 on hgba1c0;
i2 ON i1;
s2 ON i1 s1;
hgba1c0 on i2;

I used bootstrapped estimates, and would like to get either an inferential test or confidence intervals for the slope parameters.

I can get inferential test out of TECH4 using MLR estimate, but did not see corresponding test or CI when using BOOTSTRAPPED estimates.

Michael
 Linda K. Muthen posted on Friday, July 18, 2014 - 1:17 pm
We don't give those with the BOOTSTRAP option.
 Michael T Weaver posted on Friday, July 18, 2014 - 2:38 pm
Thanks.

Would it be legitimate to use the TECHNICAL 4 output estimates for the variance in the latent variable covariance matrix to construct a confidence interval?

Michael
 Bengt O. Muthen posted on Friday, July 18, 2014 - 4:14 pm
Do you mean using TECH4 SEs?
 Michael T Weaver posted on Saturday, July 19, 2014 - 7:42 pm
Bengt:

Since I used BOOTSTRAP (RESIDUAL), there are no SEs in the TECH4 output.

I was wondering if the variances provided in TECH4 output "ESTIMATED COVARIANCE MATRIX FOR THE LATENT VARIABLES" could be used to produce the SEs required to calculate a confidence interval.

I saw SEs and tests of null that parameter=0 provided in TECH 4 using MLR estimate, but wanted to use bootstrapped estimates.

Thanks...

Michael
 Bengt O. Muthen posted on Sunday, July 20, 2014 - 6:27 am
The TECH4 results cannot be used for SEs.
 Michael T Weaver posted on Sunday, July 20, 2014 - 8:31 am
Thanks - I was afraid of that.

If there is a way to estimate those SEs or confidence intervals, I'd welcome any suggestions/pointing to references.

I appreciate the time you and Linda spent answering my questions.

Michael
 Chris Cambron posted on Tuesday, April 26, 2016 - 4:24 pm
Hi,
I am estimating an unconditional latent growth curve with count data and am getting a negative intercept. This occurs with any count specification I use (zero-inflated negative binomial is the most appropriate). I've rechecked the data numerous times and there are no errors that I can find. Any thoughts of what else I can check would be much appreciated.

MODEL:
i s | ALC1@0 ALC2@3 ALC3@6 ALC4@9 ALC5@12 ALC6@18;
 Linda K. Muthen posted on Tuesday, April 26, 2016 - 5:29 pm
The growth model for a count variable is on the log rate scale. See Slide 139 of the Topic 6 course handout.