- 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:currdataaldareading.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 4.084Residual Variances PIAT1 7.279 7.920 0.919 PIAT2 36.044 6.954 5.183 PIAT3 11.791 15.240 0.774