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| Questions of second order model |
 
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| Daisy Chang posted on Tuesday, November 08, 2005 - 10:21 am
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Hi,
I am trying to do a second order factor model, but I constantly receiving an error message saying that THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL.
PROBLEM INVOLVING PARAMETER 10.
I have tried different ways including changing the starting value and fixing the loading of bpress to be 1, but none of it worked. I would really appreicate your input. Thanks. Here is my program.
title: project
data: file is u:\bio.dat;
variable: names are HDL_CHO HB_A1C DHEA_S CORTIS1 NOREPI1 EPINEP1
BP_S BP_D WAISTHIP TOTHDL age control edu;
usevariables are lncor lnnor lntot sqep sbp_s sbp_d lnhb sqdh sqep ;
missing are all (9999);
define: lncor= log(cortis1) if cortis1 > 0 then lncor = log(cortis1);
lntot= log(tothdl) if tothdl > 0 then lntot= log(tothdl);
lnnor= log(norepi1) if norepi1 > 0 then lnnor = log(norepi1);
lnhb = log(hb_a1c) if hb_a1c > 0 then lnhb = log(hb_a1c);
sqep = sqrt(epinep1) if epinep1 >=0 then sqep = sqrt(epinep1);
sqdh = sqrt(dhea_s) if dhea_s >= 0 then sqdh = sqrt(dhea_s);
sbp_s = (bp_s)/50 if bp_s > 0 then sbp_s = (bp_s)/50;
sbp_d = (bp_d)/50 if bp_d > 0 then sbp_d = (bp_d)/50;
analysis: type = missing h1;
convergence=1000;
model:bpress by sbp_s sbp_d;
hpa by sqdh lncor;
metabo by lnhb lntot waisthip;
nerv by sqep lnnor;
age by hpa* bpress metabo nerv;
age@1; |
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| You would need to send your input, data, output, and license number to support@statmodel.com to get an answer to this question. |
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Hi,
we are establishing a 2nd order model. We've done CFA for seven tests separately and now try to model them in one SEM.
The structure is as follows:
f1 BY x1 x2 x3;
f2 BY y1 y2 y3;
...
f7 BY z1 z2 z3;
g BY f1 f2 ...f7;
The model fits our data well. However, loadings of all items from the last test on its latent factor are negative (all z are negative). Hence, f7 loads negatively on the second order factor g when all other loadings (f1 to f6)are positive.
What could be the reason for this result? Does it have implications for further analyses?
Thank you very much for your help. |
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| If all are negative, they can all be changed to positive. I would try positive starting values. |
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