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