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 learningmplus posted on Saturday, October 22, 2011 - 3:44 pm
Hi Linda & Bengt!

I'm hoping you'll be able to help me. I am looking at the interaction between type of treatment (COND = family = 1; group = 0) on anxiety after treatment. In my model I have
Estimate P-Value

RCMASPPO ON
COND -0.444 0.471
DMARITAL 0.847 0.428
DMALE -0.516 0.460
D_ETHNIC -0.795 0.409
M_RCMAS 0.580 0.000
M_AGE -0.880 0.000
M_AGEXCOND 0.938 0.003


Intercepts
RCMASPPO 7.606 0.000


COND 1=family, 0=group
I'm thinking that the mean of change for group = 7.606; family = 7.606+(-.444) = 7.162.
As age goes up, change in anxiety goes down (i.e., anxiety changes less).
I am not sure how to interpret the product term. The reason I am asking this question here is because I have an additional question about missing data. I got this finding using FIML in MPlus. However, when I imputed the data in SPSS and used other tools to test the interaction I got a non-significant result. I am not sure if it is okay to use FIML when testing the above. Sorry if the questions sound basic, but whoever I asked in my program did not know and I need to figure it out for my dissertation. Thank you very much.
 Linda K. Muthen posted on Sunday, October 23, 2011 - 3:20 pm
FIML and imputation are asymptotically equivalent but may differ in practice with small samples. You may not have the same sample size in Mplus and SPSS. Mplus may delete observations with missing on x.

The interaction between age and cond is signficant in the prediction of anxiety. This means that treatment behaves differently with respect to age. For cond=1, the coefficient is -.880 + .938, a small positive value. For cond=0, it is -.880, a negative value.
 learningmplus posted on Sunday, October 23, 2011 - 6:31 pm
Thank you. I get the full sample in MPlus by bringing in the covariates x-y*; as advised here. I also get the full sample in SPSS when I impute the values, but I don't get the same results. Could clustering be the reason? MPlus accounts for clustering whereas SPSS doesn't. It would be interesting to know in what way, if any, clustering makes a difference in moderation analyses.

Based on the above -- would it be accurate if I say that for cond = 1 -- for every one unit that age goes up, anxiety goes up by .06. This is a small positive value.

for cond = 0 -- for every one unit that age goes up, anxiety goes down by -.880 units.

I'm not sure if the way I wrote this makes sense. I should say that I used

Anxiety Post on Anxiety Pre
Anxiety Post on M_Age
Anxiety Post ON M_AgeXCond as the syntax.

And one more thing -- I'm running some analyses with categorical moderators. How can I get the simple main effects and the means so I put them in a 2X2 table? It would make it easier for me to interpret the findings. I will then be able to report simple main effect contrasts and interaction contrasts.

Thanks very much! This helps me a lot toward my dissertation.
 Bengt O. Muthen posted on Tuesday, October 25, 2011 - 9:03 am
If you do Type = Complex in your Mplus run to deal with clustering, the parameter estimates are not affected, only their SEs. So that wouldn't be the source of the difference visavis SPSS.

Yes, your statements about how age affects anxiety are correct.

You should also have Cond on the right-hand side of your ON statements, which you did in your initial post.

You ask about means in a 2 x 2 table, so you must be asking about a binary moderator with your binary Cond variable. This is regression with two binary covariates. It is straightforward to get the 4 means by plugging in the 0's/1's for the two variables as in regular regression.
 Anke Schmitz posted on Wednesday, February 19, 2014 - 10:17 am
I have a signifikant moderation between two observed variables (one dichotomous and one continous). Where can I see how large the effect is for students with mean values, -1SD and +1 SD?
Regards Anke
 Bengt O. Muthen posted on Wednesday, February 19, 2014 - 11:34 am
See UG ex 3.18.
 Anke Schmitz posted on Thursday, February 20, 2014 - 7:07 am
Dear Bengt,
I have huge problems to apply ex 3.18 on my model.

Here is my command:

DEFINE: int1 = t*g; !Interaction between continuous observed variable an binary observed categorical variable

model:

LV by lv4 lv7 lv9 lv10 lv11 lv13 lv14 lv17 lv21;

VW by vw2 vw3 vw5 vw6 vw7;

int2 | vw XWITH text;

LV on VW t g int1 int2;


There is a significant interaction of int1. I have to know how the effect is when the continuous varibale is -1SD, Mean and +1SD. I don't understand the command in ex. 3.18.
 Bengt O. Muthen posted on Friday, February 21, 2014 - 8:31 am
The LOOP plot is described in the UG on page 696. An application is shown on slides 21-24 in the V7Part1.pdf handout shown at

http://www.statmodel.com/v7workshops.shtml

See Part 1 under

Handouts for New Developments in Mplus Version 7
The handouts for the Mplus Version 7 workshops at Utrecht University on August 27-29, 2012 are posted here in 4-per-page format and in regular format:

Part 1: 4-per-page Regular
Part 2: 4-per-page Regular
Part 3: 4-per-page Regular

You also find a video of that there.
 Anke Schmitz posted on Saturday, February 22, 2014 - 6:35 am
Bengt, thank you very much! It worked quite well. I have one more question: How do I compare a model without interaction (just with main effects) with a model with interaction term?
There are no model criteria in the output. Just those information:
MODEL FIT INFORMATION

Number of Free Parameters 36

Loglikelihood

H0 Value -6874.044
H0 Scaling Correction Factor 1.0009
for MLR

Information Criteria

Akaike (AIC) 13820.088
Bayesian (BIC) 13987.030
Sample-Size Adjusted BIC 13872.714
(n* = (n + 2) / 24)


Can I do it with the formula on this website?

http://www.ats.ucla.edu/stat/mplus/faq/s_b_chi2.htm

Kind regards
Anke
 Bengt O. Muthen posted on Saturday, February 22, 2014 - 3:54 pm
Since it is only 1 parameter that differs between the 2 models you can simply use the z test that is given in the output. Using MLR, this takes the non-normality into account.
 Anke Schmitz posted on Thursday, March 06, 2014 - 7:39 am
Thanks for your replys on feb, 22. It worked quite good. Is there any chance to compare moderationsmodels between two observed groups (1 versus 2)? I just want to know if the interaction effects differ between two groups of students or if they are similar. I tried the "grouping is" command, but it did not work. Do I have to split the data file and generate two separate models?
Regards Anke
 Linda K. Muthen posted on Thursday, March 06, 2014 - 2:02 pm
Please send the output showing GROUPING not working and your license number to support@statmodel.com.
 Sara Namazi posted on Friday, June 01, 2018 - 11:38 am
Hi Dr. Muthen,

I have a question. I am doing moderation analysis with latent variables using Mplus. If my main predictor and outcome is not signification but my interaction term is, can I still interpret the significant interaction term?
 Bengt O. Muthen posted on Friday, June 01, 2018 - 5:40 pm
Yes, but better still is looking at the CI in the moderation plot in line with UG ex 3.18.
 Sara Namazi posted on Monday, June 04, 2018 - 1:29 pm
Thank you, Dr. Muthen. I have a quick question to ask:

I am reading an article by Maslowsky et al. In this article, it is indicated that when working with latent interaction terms, one must estimate a structural model in two steps: (1). A structural model without the latent interaction term (model 0) & (2). a structural model with latent interaction term (model 1).

Doing this in two steps allows for testing of model fit (Klein & Moosebrugger, 2002; Muthen, 2012).

My main question is that if I am looking at B1 as my moderator in the relationship between X and Y, do I include B1 as a predictor along with X in model 0 (that is the model that does not include the interaction term)?
 Bengt O. Muthen posted on Monday, June 04, 2018 - 2:01 pm
I would do that.

See also Section 1.2 of the FAQ on our website

http://www.statmodel.com/download/LVinteractions.pdf
 Sara Namazi posted on Tuesday, June 05, 2018 - 11:50 am
Thank you for your reply, Dr. Muthen.

I am trying to plot my latent interaction terms in excel. However, I am having a hard time trying to figure out where to find the intercept for my model.
 Bengt O. Muthen posted on Tuesday, June 05, 2018 - 5:04 pm
If not shown in the output, latent variables have zero intercepts.
 Sara Namazi posted on Tuesday, June 05, 2018 - 8:11 pm
Thank you, Dr. Muthen. Does the new version of Mplus produce standardized path coefficients when including an interaction term in a structural equation model with latent variables?
 Sara Namazi posted on Wednesday, June 06, 2018 - 11:06 am
Hi Dr. Muthen,

I wanted to follow-up with you. In my reading about latent interaction terms, I learned that Mplus does not produce standardized beta coefficient when including latent interaction terms in your model. However, I am able to get standardized beta coefficient using the following command in Mplus:

OUTPUT: STAND TECH1 tech4 sampstat PATTERNS RESIDUAL modindices ;

Not sure If I am missing something or whether my understanding is correct.

Thanks,
Sara
 Bengt O. Muthen posted on Wednesday, June 06, 2018 - 4:31 pm
You get standardized coefficients also with latent variable interactions in the current Mplus. This is described in the FAQ on our website:

Latent variable interactions
 alessandra monni posted on Monday, July 02, 2018 - 2:13 pm
Dear Dr Muthen,
I have a question regarding the interpretation of a moderation model output.
I have 2 indipendent latent factor: a, b and a dipendent measured variable c.
I ran a simple model:
c on a b;
and I had in the output a negative estimate for both:
c on a -0.300
c on b -0.100
I decided to run a moderation model because I assumed that variable a could interact with b and produce a stronger effect on c:
(high a and high b --> lesser c)
On the contrary I had in the output:
c on a -0.348
c on b -0.108
c on ab +0.250
In the interaction a and b the estimated relation with c is reversed. That does not make any sense in theory.
Am I interpreting the output in a wrong way?
 Bengt O. Muthen posted on Tuesday, July 03, 2018 - 2:45 pm
This is a good question for SEMNET.
 Sara Namazi posted on Tuesday, July 17, 2018 - 4:10 pm
Hi Dr, Muthen,

I have a question about latent moderated structural questions.
So to test interaction terms in Mplus, Maslowsky et al. 2015 article indicates that to estimate latent moderated structural equations, you should run a model without the interaction term (called Model 0) and then add a model with the interaction term (called Model 1) to compare model fit using the XWITH Command in Mplus.

So if I am looking at the relationship between a --> M --> b and a--b so a simple mediation, would that mean that my model 0 is:

M on a
b on M
b on a

and model 1 (v is my moderator pointing to the path going from a to M):

M on a
M on v
M on a*v
b on M
b on a

Thanks.
 alessandra monni posted on Wednesday, July 18, 2018 - 5:42 am
Dear Dr Muthen,
I'm running a moderation model and the moderation variable is a measured variable.
When I run the model it takes 10 minutes to complete.
When I used a moderation variable that is a latent variable in the same model it takes 3 minutes to complete.
Is that normal or am I doing in the wrong way?

I write you the model with the measured moderation variable below:

USEVARIABLES ARE

ZAAT2 ZAAT4 ZAAT5 ZAAT8 ZAAT10 ZAAT11
ZAAT1 ZAAT3 ZAAT6 ZAAT7 ZAAT9 ZAAT12

BSI20 BSI21 BSI22 BSI42

Zalfa1;

CATEGORICAL ARE

BSI20 BSI21 BSI22 BSI42;


ANALYSIS:
TYPE =RANDOM;
ALGORITHM=INTEGRATION;
INTEGRATION=MONTECARLO;

OUTPUT: standardized;


MODEL:

AATap by ZAAT2 ZAAT4 ZAAT5 ZAAT8 ZAAT10 ZAAT11;
AATav by ZAAT1 ZAAT3 ZAAT6 ZAAT7 ZAAT9 ZAAT12;

INT by BSI20 BSI21 BSI22 BSI42;


!!!---MODERATION FACTORS-----

AATapZalfa1 | AATap XWITH Zalfa1;
AATavZalfa1 | AATav XWITH Zalfa1;

!!!---MODERATION MODEL-----

INT on AATap Zalfa1 AATapZalfa1;
INT on AATav Zalfa1 AATavZalfa1;

Thank you in advance
 Bengt O. Muthen posted on Wednesday, July 18, 2018 - 7:02 am
Answer for Sarah:

We describe model fitting in section 1.2 of the FAQ on our website: Latent variable interactions.
 Bengt O. Muthen posted on Wednesday, July 18, 2018 - 8:15 am
Answer for Allessandra:

We need to see the 2 outputs - send to Support along with your license number.
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