Stat Computing > MPlus > Textbook Examples > Applied Longitudinal Data Analysis

Mplus Textbook Examples
Applied Longitudinal Data Analysis: Modeling Change and Event Occurrence
by Judith D. Singer and John B. Willett
Chapter 5:  Treating time more flexibly

• Supplement 5.1 - Relaxing assumption of equality of variances across time.

Supplement 5.1 - Relaxing assumption of equality of variances across time.

This example uses the data file reading.dat.  AGEGRPi-6.5 is used as a temporal predictor, called cagegrpi (i.e., cagegrp1, cagegrp2 and cagegrp3).  These were created before making the data file.

• On page 145, in Table 5.2 Model A, the assumption is made that the residual variances are the same across all 3 time points.  This example shows how you can relax that assumption, allowing them to vary freely.
• Using a likelihood ratio test, we compared this model to the more restricted model (shown in Chapter 5) by taking 2 times the differences in the LogLikelihoods, yielding  2*(909.978 - 906.176) = 7.604.  When compared to a chi-square table with 2 degrees of freedom, the p value was .022.

Title: 
  Table 5.2, Model A.
Data:  
  File is G:\currdata\alda\reading.dat ;
Variable: 
  Names are 
     id agegrp1 agegrp2 agegrp3 age1 age2 age3 piat1 piat2 piat3 cage1
     cage2 cage3 cagegrp1 cagegrp2 cagegrp3;
  Missing are all (-999999999) ; 
  Usevariables are
     piat1 piat2 piat3 cagegrp1 cagegrp2 cagegrp3;
  Tscores cagegrp1-cagegrp3 ;
Analysis: 
  Type = random ;
  estimator = ml;
Model:
  i s | piat1-piat3 at cagegrp1-cagegrp3 ;
  i with s;
  ! piat1-piat3 (1) ; ! By commenting this line, the variances are not constrained to be
                      ! the same across the 3 time points.  

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TESTS OF MODEL FIT

Loglikelihood
          H0 Value                        -906.176
Information Criteria
          Number of Free Parameters              8
          Akaike (AIC)                    1828.353
          Bayesian (BIC)                  1848.262
          Sample-Size Adjusted BIC        1823.015
            (n* = (n + 2) / 24)
MODEL RESULTS
                   Estimates     S.E.  Est./S.E.
 I        WITH
    S                 -3.156    2.644     -1.193
 Means
    I                 20.772    0.610     34.077
    S                  5.053    0.305     16.551 Intercepts
    PIAT1              0.000    0.000      0.000
    PIAT2              0.000    0.000      0.000
    PIAT3              0.000    0.000      0.000
 Variances
    I                 24.024    8.484      2.832
    S                  6.587    1.613      4.084
 Residual Variances
    PIAT1              7.279    7.920      0.919
    PIAT2             36.044    6.954      5.183
    PIAT3             11.791   15.240      0.774

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