Discrete Time Survival Analysis plots PreviousNext
Mplus Discussion > Categorical Data Modeling >
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 Amy Hartl posted on Saturday, September 22, 2012 - 3:22 pm
Greetings,

I'm running a DTSA and would like to view the survival and hazard plots. I specify Plot2 in the output command, but it says the only plots available are:

Histograms (sample values, estimated factor scores)
Scatterplots (sample values, estimated factor scores)
Item characteristic curves
Information curves

Are survival plots only available for continuous time SA? If not, how can I obtain these?

I am following example 6.19 from the manual and am running version 6.11.

Thank you!

Amy
 Linda K. Muthen posted on Saturday, September 22, 2012 - 3:40 pm
You need to use the DSURVIVAL option of the VARIABLE command to specify which variables are the discrete-time survival variables. Then you will get the plots.
 Amy Hartl posted on Sunday, September 23, 2012 - 9:13 am
Great, thank you! This yielded the estimated baseline survival curves. Is there a way I can get the estimated baseline hazard curves? It won't let me select it when I'm the plot menu.

Below is the syntax in case that's useful.

Thank you!


TITLE: DTSA time and fship freq only
DATA: FILE IS rec7to12.dat;

VARIABLE: NAMES ARE
dyadn
f1
f2
freqF1
freqF2
u7
u8
u9
u10
u11
u12
;

USEVARIABLES=
freqF1 freqF2 u8-12
;
CATEGORICAL = u8-u12;
dsurvival= u8-u12;
MISSING ARE ALL (9999.00);

ANALYSIS:
ESTIMATOR = MLR;

MODEL:
f BY u8-u12@1;
f ON freqF1 freqF2;
f@0;

OUTPUT: sampstat; stdyx;
PLOT: TYPE IS Plot3;
 Linda K. Muthen posted on Sunday, September 23, 2012 - 6:09 pm
That plot is for continuous-time survival.
 Amy Hartl posted on Wednesday, September 26, 2012 - 9:09 am
I see. Okay, thank you!
 Amy Hartl posted on Wednesday, September 26, 2012 - 11:37 am
I see that loading all of the indicators @1 enforces the proportional odds assumption. How can I test the constant hazard rate assumption, i.e., how can I constrain the hazard rate to be equal across time?

Can I do this without using type=mixture and a latent class design?

Thank you for your help!
 Bengt O. Muthen posted on Wednesday, September 26, 2012 - 5:39 pm
You can run the model saying e.g.

u1-u5 on x;

versus

u1-u5 on x (1);

The latter approach is the same as saying

f by u1-u5@1;
f on x;

Twice the loglikelihood difference for the 2 models gives a chi-square test.
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