Degrees of freedom for missing data m...
Message/Author
 Anonymous posted on Tuesday, April 01, 2003 - 11:10 am
I have three annual assessments of two variables for husbands and wives from 71 couples: ratings of frequency of child problems (h8t-h10t for husbands and w8t-w10t for wives)and parenting satisfaction (h8psat-h10psat for husbands and w8psat-w10psat for wives). There are missing data. Because I know that there is no growth for either variable, I want to examine the intra- and cross-parent relation between averaged parenting satisfaction and averaged ratings of child problems for each parent over the 3 assessments.
I want errors for husbands' ratings of child problems and wives' ratings of child problems to correlate. Here is my MPLUS syntax:

USEVARIABLES ARE h8t-h10t h8psat-h10psat w8t-w10t w8psat-w10psat;
MISSING ARE ALL (-99);
ANALYSIS: TYPE = MISSING H1;
ITERATIONS=5000;
H1ITERATIONS=5000;
MODEL:
hit BY h8t-h10t@1;
hipsat BY h8psat-h10psat@1;
wit BY w8t-w10t@1;
wipsat BY w8psat-w10psat@1;
hit ON hipsat wipsat;
wit ON hipsat wipsat;

Here is selected output:
Chi-Square Test of Model Fit

Value 101.594
Degrees of Freedom 56
P-Value 0.0002

Chi-Square Test of Model Fit for the Baseline Model

Value 383.150
Degrees of Freedom 66
P-Value 0.0000

CFI/TLI

CFI 0.856
TLI 0.831

Loglikelihood

H0 Value -1968.303
H1 Value -1917.506

Information Criteria

Number of Free Parameters 34
Akaike (AIC) 4004.606
Bayesian (BIC) 4081.537
(n* = (n + 2) / 24)

RMSEA (Root Mean Square Error Of Approximation)

Estimate 0.107
90 Percent C.I. 0.073 0.140
Probability RMSEA <= .05 0.005

SRMR (Standardized Root Mean Square Residual)

Value 0.181

MODEL RESULTS

Estimates S.E. Est./S.E.

HIT BY
H8T 1.000 0.000 0.000
H9T 1.000 0.000 0.000
H10T 1.000 0.000 0.000

HIPSAT BY
H8PSAT 1.000 0.000 0.000
H9PSAT 1.000 0.000 0.000
H10PSAT 1.000 0.000 0.000

WIT BY
W8T 1.000 0.000 0.000
W9T 1.000 0.000 0.000
W10T 1.000 0.000 0.000

WIPSAT BY
W8PSAT 1.000 0.000 0.000
W9PSAT 1.000 0.000 0.000
W10PSAT 1.000 0.000 0.000

HIT ON
HIPSAT -1.442 0.497 -2.903
WIPSAT -0.007 0.682 -0.011

WIT ON
HIPSAT -0.343 0.390 -0.881
WIPSAT -0.856 0.563 -1.519

WIT WITH
HIT 13.997 5.449 2.569

WIPSAT WITH
HIPSAT 1.309 0.573 2.284

Intercepts
H8T 12.276 1.057 11.618
H9T 11.804 1.021 11.560
H10T 12.519 1.218 10.279
H8PSAT 17.535 0.331 52.950
H9PSAT 17.168 0.320 53.610
H10PSAT 17.272 0.307 56.250
W8T 11.343 0.773 14.665
W9T 12.915 0.958 13.483
W10T 12.392 1.076 11.520
W8PSAT 17.183 0.276 62.289
W9PSAT 17.148 0.267 64.280
W10PSAT 17.124 0.288 59.469

Variances
HIPSAT 4.840 1.015 4.769
WIPSAT 2.674 0.654 4.088

Residual Variances
H8T 22.196 6.302 3.522
H9T 15.799 5.485 2.881
H10T 37.567 8.939 4.202
H8PSAT 2.947 0.722 4.078
H9PSAT 1.969 0.557 3.534
H10PSAT 1.123 0.432 2.601
W8T 11.455 4.456 2.571
W9T 33.161 7.305 4.539
W10T 40.830 9.589 4.258
W8PSAT 2.729 0.638 4.279
W9PSAT 1.966 0.566 3.474
W10PSAT 2.239 0.615 3.640
HIT 41.801 9.210 4.538
WIT 23.935 5.776 4.144

My question involves the stated 56 degrees of freedom for the chi-square test of model fit.
There are 12 observed scores, giving (12x13)/2 = 78 parameters. The output says I have 34 free parameters, and the tech1 output confirms this:

PARAMETER SPECIFICATION

NU
H8T H9T H10T H8PSAT H9PSAT
________ ________ ________ ________ ________
1 1 2 3 4 5

NU
H10PSAT W8T W9T W10T W8PSAT
________ ________ ________ ________ ________
1 6 7 8 9 10

NU
W9PSAT W10PSAT
________ ________
1 11 12

LAMBDA
HIT HIPSAT WIT WIPSAT
________ ________ ________ ________
H8T 0 0 0 0
H9T 0 0 0 0
H10T 0 0 0 0
H8PSAT 0 0 0 0
H9PSAT 0 0 0 0
H10PSAT 0 0 0 0
W8T 0 0 0 0
W9T 0 0 0 0
W10T 0 0 0 0
W8PSAT 0 0 0 0
W9PSAT 0 0 0 0
W10PSAT 0 0 0 0

THETA
H8T H9T H10T H8PSAT H9PSAT
________ ________ ________ ________ ________
H8T 13
H9T 0 14
H10T 0 0 15
H8PSAT 0 0 0 16
H9PSAT 0 0 0 0 17
H10PSAT 0 0 0 0 0
W8T 0 0 0 0 0
W9T 0 0 0 0 0
W10T 0 0 0 0 0
W8PSAT 0 0 0 0 0
W9PSAT 0 0 0 0 0
W10PSAT 0 0 0 0 0

THETA
H10PSAT W8T W9T W10T W8PSAT
________ ________ ________ ________ ________
H10PSAT 18
W8T 0 19
W9T 0 0 20
W10T 0 0 0 21
W8PSAT 0 0 0 0 22
W9PSAT 0 0 0 0 0
W10PSAT 0 0 0 0 0

THETA
W9PSAT W10PSAT
________ ________
W9PSAT 23
W10PSAT 0 24

ALPHA
HIT HIPSAT WIT WIPSAT
________ ________ ________ ________
1 0 0 0 0

BETA
HIT HIPSAT WIT WIPSAT
________ ________ ________ ________
HIT 0 25 0 26
HIPSAT 0 0 0 0
WIT 0 27 0 28
WIPSAT 0 0 0 0

PSI
HIT HIPSAT WIT WIPSAT
________ ________ ________ ________
HIT 29
HIPSAT 0 30
WIT 31 0 32
WIPSAT 0 33 0 34

I would think this model would give me 78-34=44 degrees of freedom, but I get 56 df. P. 36 of the manual says that MEANSTRUCTURE is included by default in MISSING analyses. Am I correct in assuming that the discrepancy between the reported 56 df and the expected 44 df is that the intercept parameters (NU 1-12) are not really included as parameters?

I ran the problem in LISREL 8.53 and get nearly identical results with 56 df and no intercept parameters.
 Linda K. Muthen posted on Tuesday, April 01, 2003 - 2:36 pm
The means are both part of your sample statistics and also part of your estimates. Note that there are twelve free parameters in the nu matrix. So your sample statistis are 78 + 12 = 90 - 34 = 56. If you ran the problem without means, you would have 78 - 22 = 56. Please let me know if this does not answer your question.
 Anonymous posted on Wednesday, April 02, 2003 - 10:32 am
I can easily get the analysis I want (without means) by dropping the TYPE=MISSING entry in the ANALYSIS section. How can I use the MISSING option and run the problem without means? I have tried setting the intercepts and means of the latent variables to 0, get the identification of the estimated parameters I want, but the estimates themselves are off. Is this a scaling issue?
 Linda K. Muthen posted on Wednesday, April 02, 2003 - 10:41 am
You cannot run the MISSING analysis without means. Means are required for this estimation. However, having unstructured means as part of the model does not affect the results. These means are estimated as the sample values. You can try this out without missing to show yourself. Fixing the means to zero is not correct.
 Anonymous posted on Wednesday, April 02, 2003 - 2:27 pm
Thank you. I now see that the structure of the output from Mplus and LISREL differs because Mplus requires the means for estimation. Parameter estimates themselves are nearly identical in the two programs.