Displaying 9 results from an estimated 9 matches for "nnfi".
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anfi
2013 Apr 28
0
hierarchical confirmatory factor analysis with sem package
...1 and F2 are nested within F3.
Here is the code that I have, but it is giving me an error message "Warning
message:
In eval(expr, envir, enclos) : Negative parameter variances.
Model may be underidentified." and a further error "Error in
summary.objectiveML(cfa, fit.indices = c("NNFI", "CFI", "RMSEA")) :
coefficient covariances cannot be computed". I have run CFA before with no
issues. This is the first time I am running a nested model. Any help will be
greatly appreciated.
Regards,
Mat
cov.matrix<-cov(na.omit(df))
cfa.model<-specifyModel()
F...
2009 Mar 09
2
path analysis (misspecification?)
...esults;
Model Chisquare = 12.524 Df = 1 Pr(>Chisq) = 0.00040179
Chisquare (null model) = 812.69 Df = 3
Goodness-of-fit index = 0.98083
Adjusted goodness-of-fit index = 0.885
RMSEA index = 0.16545 90% CI: (0.09231, 0.25264)
Bentler-Bonnett NFI = 0.98459
Tucker-Lewis NNFI = 0.9573
Bentler CFI = 0.98577
SRMR = 0.027022
BIC = 6.4789
Parameter Estimates
Estimate Std Error z value Pr(>|z|)
x1-y1 -0.67833 0.033967 -19.970 0 y1 <--- x1
x2-x1 3.88384 0.293743 13.222 0 x1 <--> x2
x2-x2 12.05082 0.831569 14.492 0 x2...
2007 Mar 07
1
No fit statistics for some models using sem
...all the usual statistics:
Model Chisquare = 1303.7 Df = 1 Pr(>Chisq) = 0
Chisquare (null model) = 8526.8 Df = 6
Goodness-of-fit index = 0.95864
Adjusted goodness-of-fit index = 0.58639
RMSEA index = 0.30029 90% CI: (NA, NA)
Bentler-Bonnett NFI = 0.84711
Tucker-Lewis NNFI = 0.082726
Bentler CFI = 0.84712
BIC = 1294.1
My understanding is the you should always put in the correlation
between exogenous predictors, but when I do this I don't get fit
statistics. Can anyone help me understand what is happening here?
Thank you,
Ista
2007 Jun 27
1
SEM model fit
...rning messages with the output.
RESULTS:
Model Chisquare = 1374 Df = 185 Pr(>Chisq) = 0
Chisquare (null model) = 12284 Df = 210
Goodness-of-fit index = 0.903
Adjusted goodness-of-fit index = 0.88
RMSEA index = 0.0711 90% CI: (NA, NA)
Bentler-Bonnett NFI = 0.888
Tucker-Lewis NNFI = 0.888
Bentler CFI = 0.902
SRMR = 0.0682
BIC = 51.4
SYNTAX
rm(sem.enf.rq)
mdl.rq <- specify.model()
enf -> law2, NA, 1
enf -> law3, lam2, 1
enf -> law4, lam3, 1
enf <->...
2006 Aug 16
1
Specifying Path Model in SEM for CFA
...converges:
> summary(fit.sem)
Model Chisquare = 2147 Df = 10 Pr(>Chisq) = 0
Chisquare (null model) = 2934 Df = 15
Goodness-of-fit index = 0.4822
Adjusted goodness-of-fit index = -0.087387
RMSEA index = 0.66107 90 % CI: (NA, NA)
Bentler-Bonnett NFI = 0.26823
Tucker-Lewis NNFI = -0.098156
Bentler CFI = 0.26790
BIC = 2085.1
Normalized Residuals
Min. 1st Qu. Median Mean 3rd Qu. Max.
-5.990 -0.618 0.192 0.165 1.700 3.950
Parameter Estimates
Estimate Std Error z value Pr(>|z|)
l11 -0.245981 0.21863 -1.12510 0.26054748 X1 <--- F1
l21...
2007 Feb 20
0
Standardized residual variances in SEM
...t; summary(semModif.tmp, digits=2)
Model Chisquare = 601 Df = 229 Pr(>Chisq) = 0
Chisquare (null model) = 2936 Df = 253
Goodness-of-fit index = 0.81
Adjusted goodness-of-fit index = 0.78
RMSEA index = 0.08 90% CI: (0.072, 0.088)
Bentler-Bonnett NFI = 0.8
Tucker-Lewis NNFI = 0.85
Bentler CFI = 0.86
BIC = -667
Normalized Residuals
Min. 1st Qu. Median Mean 3rd Qu. Max.
-2.6300 -0.5640 -0.0728 -0.0067 0.5530 3.5500
Parameter Estimates
Estimate Std Error z value Pr(>|z|)
param1 0.78 0.073 10.7 0.0e+00 Q1 <--- G
param2...
2006 Aug 22
1
Total (un)standardized effects in SEM?
Hi there,
as a student sociology, I'm starting to learn about SEM. The course I
follow is based on LISREL, but I want to use the SEM-package on R
parallel to it.
Using LISREL, I found it to be very usable to be able to see the
total direct and total indirect effects (standardized and
unstandardized) in the output. Can I create these effects using R? I
know how to calculate them
2008 Dec 22
1
sem package fails when no of factors increase from 3 to 4
#### I checked through every 3 factor * 3 loading case.
#### While, 4 factor * 3 loading failed.
#### the data is 6 factor * 3 loading
require(sem);
cor18<-read.moments();
1
.68 1
.60 .58 1
.01 .10 .07 1
.12 .04 .06 .29 1
.06 .06 .01 .35 .24 1
.09 .13 .10 .05 .03 .07 1
.04 .08 .16 .10 .12 .06 .25 1
.06 .09 .02 .02 .09 .16 .29 .36 1
.23 .26 .19 .05 .04 .04 .08 .09 .09 1
.11 .13 .12 .03 .05 .03
2008 Dec 29
0
Serial Correlation Test for Short Time Series
...or18, 500))
Model Chisquare = 80.675 Df = 48 Pr(>Chisq) = 0.0021920
Chisquare (null model) = 1106.4 Df = 66
Goodness-of-fit index = 0.9747
Adjusted goodness-of-fit index = 0.95888
RMSEA index = 0.036935 90% CI: (0.022163, 0.050657)
Bentler-Bonnett NFI = 0.92708
Tucker-Lewis NNFI = 0.95682
Bentler CFI = 0.9686
SRMR = 0.032512
BIC = -217.63
Normalized Residuals
Min. 1st Qu. Median Mean 3rd Qu. Max.
-1.71000 -0.23300 -0.00337 0.08850 0.26700 2.13000
Parameter Estimates
Estimate Std Error z value Pr(>|z|)
TD11 0.30641 0.037053 8.2694 2....