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 S.Arunachalam posted on Monday, August 26, 2013 - 3:55 pm
Respected Prof. Muthen,

I am trying to get significance for each value of x (moderator) in my moderated-mediation model using

MODEL CONSTRAINT:
new(ind1-ind21);
loop (x,-1,1,.1);
do (1,21) ind# = b1 * (a1 + a2 * x);

However the results report the same value for all the 21 indirect effect as
IND1 0.047 0.024 1.997 0.046
IND2 0.047 0.024 1.997 0.046
IND3 0.047 0.024 1.997 0.046
.... ........
IND21 0.047 0.024 1.997 0.046

While I can use PLOT(ind) to get the plot, I need the significance of the indirect effect for every legitimate value of X ( similar to Johnson-Neuman technique) which I am not getting using the PLOT(ind). My apologies if I am missing something very obvious, please guide.
 Bengt O. Muthen posted on Monday, August 26, 2013 - 8:31 pm
The DO function is not connected to LOOP. Instead follow page 16 of the May 2013 UConn Workshop handout on our website which shows how LOOP is connected with PLOT to do what you want:

MODEL CONSTRAINT:
PLOT(indirect direct);
! let moderate represent the range of the agg1 moderator
LOOP(moderate, -2, 2, 0.1);
indirect = beta1*(gamma1+gamma3*moderate);
direct = beta2+beta4*moderate;
 S.Arunachalam posted on Monday, August 26, 2013 - 10:28 pm
Dear Prof. Muthen, Yes I did use the PLOT to get the graph. However the PLOT doesn't print the values of the confidence limits (in the .out file) for the different values of the moderator. I need those to be reported in my article to show the points where the moderated-mediation is significant at 95% level. Is there a way to get those confidence limits which PLOT uses to be printed in the .out file? Thank you.
 Thuy Nguyen posted on Monday, August 26, 2013 - 10:56 pm
You can use the “Save plot data” option under the Plot menu to save the values plotted in the loop plot. The file will contain 4 columns of data. The first column has the x-values; the second column has the estimates; the third column has the lower confidence limit; and the fourth column has the upper confidence limit.
 S.Arunachalam posted on Monday, August 26, 2013 - 11:01 pm
Thank you so much, Thuy! I truly appreciate it.
Respectfully, Arun
 Joe posted on Thursday, September 25, 2014 - 2:59 pm
I have a question about the LOOP Plot for a growth model. (Note I am using BAYES estimator here.) I used the LOOP command to get the 95% credible interval for the quadratic growth function. I also ran my model t times, so that each timepoint t served as the intercept.

Using the LOOP function, the CIs get larger as t gets larger. For the multiple models, each centered at each timepoint, the CIs remain small.

Can you shed some light on the difference here? My primary purpose is to find the time point at which the CIs DO NOT overlap. I get different results for my different methods.

Thank you.
 Bengt O. Muthen posted on Thursday, September 25, 2014 - 9:51 pm
I don't think I understand - it doesn't sounds like you are looking at CI for estimated means of outcomes, but for something else - what is it?
 Joe posted on Thursday, September 25, 2014 - 10:16 pm
I apologize for being unclear. I am running a growth model with 18 occasions, and would like CI for estimates of each measurement occasion.

Method 1 - I center the growth model at time 1. I then create a NEW parameter for times 2 through 17. For each parameter I get a CI, and I notice the the CI increases as t increases.

est lower_2.5ci upper_2.5ci
85.29 83.82 86.40
97.68 96.78 98.47
109.11 108.52 109.78
...
167.46 154.11 171.68
164.98 149.09 170.15
161.55 142.86 167.67

Method 2 - I run 18 models (one for each measurement occasion), in which each occasion serves as the center (intercept). The CI for occasion 18 here is not equal to the CI of the estimated parameter for occasion 18 in Method 1.

est lower_2.5ci upper_2.5ci
85.20 83.74 86.37
97.63 96.75 98.43
109.09 108.47 109.76
...
171.95 171.18 172.63
170.53 169.73 171.26
168.24 167.39 168.99

From what I can tell, this difference is because in Method 1, the Posterior S.D. increases as t increases, but this is not the case in Method 2. Can you explain why?
 Bengt O. Muthen posted on Thursday, September 25, 2014 - 11:14 pm
For what is the CI? For the estimated mean of the outcome?
 Joe posted on Friday, September 26, 2014 - 1:23 am
Yes. Each of the repeated outcomes.
 Joe posted on Friday, September 26, 2014 - 3:46 am
I should be more accurate.

The CI for Method 1 is for the intercept and then each estimated parameter (i+time*l+(time**2*q).

The CI for Method 2 is for the intercept of each of the 18 models.

I apologize for the double post.
 Bengt O. Muthen posted on Friday, September 26, 2014 - 6:13 pm
First check that Method 1 and Method 2 give the same number of parameters and the same loglikelihood.

CIs widening over time would happen if the estimated variance of the outcomes increases over time - you find that in the RESIDUAL output.
 Joe posted on Friday, September 26, 2014 - 7:17 pm
Although the estimated variance of the outcomes generally increase over time, the increase is not monotonic, but the widening of the CIs is monotonic.

The Number of Free parameters is the same, but the Estimated Number of Parameters (pD) is different. In fact, pD is different across all 18 models for Method 2, where the only change is the center of the intercept. Might this be related to the collinearity of the linear and quadratic factor loadings?

Finally, which method would you prefer to find the point at which credible intervals (CI) overlap: (1) using the LOOP plot function; or (2) Multiple growth models so that each measurement occasion (T = 18) is allowed to serve as the intercept, and comparing CIs of 18 intercepts?

Thank you very much for your time and help, they are greatly appreciated.
 Bengt O. Muthen posted on Friday, September 26, 2014 - 7:33 pm
The Bayesian pD isn't behaved like regular parameter counting, and in fact I would suggest that you study this first using ML to explore in a more familiar context. With ML you can check that you get the same logL which you can't with Bayes. And then see if you get the same CI behavior with ML as you saw with Bayes.

I would work with LOOP.
 Joe posted on Friday, September 26, 2014 - 8:25 pm
Thank you very, very much.

Best,
Joe
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