1. Modelos para datos longitudinales

2. Data example
“We have four measures of hability, for a sample of 204 children, at ages 6,7,9 and 11 years (Osbourne and Suddick, 1972)”
(see Dunn et al. p. 100)
The Mean, Standard deviation, Correlation matrix:
/STANDARD DEVIATIONS
/MEANS
/MATRIX 1
.809 1
/LMtest
/END

3. Single-factor model

4. Single-factor Model * * * * V1 V2 V3 V4 * * * * F1

5. Autoregressive model

6. Autoregressive Model * * * * V1 V2 * V4 V3 * *

7. EQS code: /TITLE longitudinal model (Dunn et al. p. 108)
/SPECIFICATIONS
CAS=204 ; VAR=4; ME=ML;
ANAL = COV;
/LABELS
V1 = ABIL6; V2 = ABIL7; V3 = ABIL9; V4 = ABIL11;
/EQUATIONS
V2 = 1.0*V1+ E2;
V3 = 1.0*V2 + E3 ;
V4 = 1.0*V3 + E4;
/VARIANCES
V1 = 36*; E2 TO E4 = 1.0*;
/PRINT
EFFECT = YES;
/STANDARD DEVIATIONS
/MEANS
/MATRIX 1
.809 1
/LMtest
/END

8. Estimates

9. Simplex model

10. Simplex Model * * * * V4 V1 V2 V3 * * F4 F3 F1 * F2 * * * *

11. EQS imput for simplex model

12. Estimates, s.e., and chi2 g.o.f test

13. Restricting equality of effects across time

14. Exercise: using the cov. Matrix below, fit a simplex model

15. Dynamic factor model (Dunn et al. P. 140)
Judd and Milburn (1980) used a latent variable analysis to examine attitudes in a nation-wide sample of individuals who were surveyed on three occasions, in 1972, 1974 and 1976.
(Dunn et al. P. 140)

16. Part of the data involved recording attitudes on three topics: busing- a policy designed to achieve school integration; criminals - the protection for the legal rights of those accused of crimes; jobs- whether government should guarantee jobs and standard of living. The sample consisted of 143 individuals each with four years of college education, and 203 individuals who had no college education .

17. college education n = 143. 1972. 1974. 1976. B. C. J. B. C. J. B. C
college education n = B C J B C J B C J B 1 C J B C J B C J SD B Busing C Criminals J Jobs

18. Path diagram for effects across time* * * T1 * T3 T2 * V1 V2 V3 V4 V5 V6 V7 V8 V9

19. EQS code /TITLE liberalism-conservatism exmple
factor loadings and latent variable regression coefficients constrained
to be equal across groups
group 1 - four years of college education
/SPECIFICATIONS
GROUPS = 1; CAS=143; VAR=9; MATRIX = CORR; ANALYSIS = COV;
/EQUATIONS
V1 = 1*F1 + E1; ! F1 is liberalism in 1972
V2 = 1*F E2;
V3 = 1*F E3;
V4 = F2 + E4; !F2 is liberalism in 1974
!scale of F2 set to that of V4
!scale can not be set in /VARIANCE
! since F2 appears later as a depend. var
V5 = 1*F2 + E5;
V6 = 1*F2 + E6;
V7 = F3 + E7; !F3 is liberalism in 1976, again scale ..
V8 = 1*F3 + E8;
V9 = 1*F3 + E9;
F2 = 1*F1 + D1; !Regression of 1974 kuberakusn ib 1972
F3 = 1*F2 + D2; !Regression of 1976 liberalism on 1974
/VARIANCES
F1 = 1;
E1 TO E9 = 1*;
D1 TO D2 = .2*;
/COVARIANCES
E1,E4 = .5*;
E1,E7 = .5*;
E2,E5 = .5*;
E2,E8 = .5*;
E3,E6 = .5*;
E3,E9 = .5*;
E4,E7 = .5*;
E5,E8 =.5*;
E6,E9 = .5*;
/STANDARD DEVIATIONS
/MATRIX 1
.43 1
/END

20. Estimated Time effects
F2 =F2 = *F D1
.102
9.106
F3 =F3 = *F D2
.085
11.800
CHI-SQUARE = BASED ON DEGREES OF FREEDOM
PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS

21. V1 =V1 = *F E1
.113
9.897
V2 =V2 = *F E2
.120
6.999
V3 =V3 = *F E3
.115
8.730
V4 =V4 = F E4
V5 =V5 = *F E5
.131
5.922
V6 =V6 = *F E6
.147
6.174
V7 =V7 = F E7
V8 =V8 = *F E8
.123
4.496
V9 =V9 = *F E9
.142
5.890

22. Random walk model

23. Random Walk Model * * * * V4 V1 V2 V3 1 1 1 1 F4 F3 F1 F2 * * * *

24. EQS code for random walk model

25. Estimates I F1 - F1 .806*I RESIDUAL COVARIANCE MATRIX (S-SIGMA) :
ABIL ABIL ABIL ABIL11
V V V V 4
ABIL6 V
ABIL7 V
ABIL9 V
ABIL11 V
CHI-SQUARE = BASED ON DEGREES OF FREEDOM
PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS
I F F *I
I I
I I
I I
I F F *I
I I
I I
I F F *I
I I
I I
I F F *I
I I
I I

26. Estimates of meas. Error
E1 -ABIL *I I
.014 I I
9.376 I I
I I
E2 -ABIL *I I
E3 -ABIL *I I
E4 -ABIL *I I

27. Growth curve model

28. Growth curve model * * * * V1 V2 V3 V4 * 1 * 1 * D1 1 * D2 1 Slope F2
Intercept F1 * 1 *

29. growth curve model /TITLE
Dependence between ability scores at 6,7,9 and 11
Growth curve model of latent ability
/SPECIFICATIONS
CAS=204 ; VAR=4; ANALYSIS=MOMENT; MATRIX = CORR;
ANAL = COV;
/LABELS
V1 = ABIL6; V2 = ABIL7; V3 = ABIL9; V4 = ABIL11;
F1 = intercept; F2 = slope;
/EQUATIONS
V1 = F1 + E1;
V2 = F1 + F E2;
V3 = F1 + 3*F E3;
V4 = F1 + 5*F2 + E4;
F1 = 18*V999 + D1;
F2 = 7*V999 + D2;
/VARIANCES
D1 = 25*; D2 = 1*; E1 TO E4 = 1.0*;
/CONSTRAINTS
(E1,E1 ) = (E2,E2) = (E3,E3) = (E4,E4);
!/PRINT
!EFFECT = YES;
/STANDARD DEVIATIONS
/MEANS
/MATRIX 1
.809 1
/END

30. Moment Matrix RESIDUAL COVARIANCE/MEAN MATRIX (S-SIGMA) :
4 VARIABLES (SELECTED FROM 4 VARIABLES), BASED ON CASES.
ABIL ABIL ABIL ABIL V999
V V V V V999
ABIL6 V
ABIL7 V
ABIL9 V
ABIL11 V
V999 V
RESIDUAL COVARIANCE/MEAN MATRIX (S-SIGMA) :
ABIL6 V
ABIL7 V
ABIL9 V
ABIL11 V
V999 V
CHI-SQUARE = BASED ON DEGREES OF FREEDOM
PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS LESS THAN 0.001

31. Estimates
ABIL6 =V1 = F E1
ABIL7 =V2 = F F E2
ABIL9 =V3 = F *F E3
.069
31.739
ABIL11 =V4 = F *F E4
.119
30.755
INTERCEP=F1 = *V D1
.470
38.417
SLOPE =F2 = *V D2
.305
25.612

32. Estimates of variances
E D
E1 -ABIL *I D1 -INTERCEP *I
.625 I I
I I
I I
E2 -ABIL *I D2 -SLOPE *I
.625 I I
I I
E3 -ABIL *I I
.625 I I
I I
E4 -ABIL *I I

33. Model with cov. Betw. F1 and F2
CHI-SQUARE = BASED ON DEGREES OF FREEDOM
PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS
COVARIANCES AMONG INDEPENDENT VARIABLES
E D
I D2 -SLOPE *I
I D1 -INTERCEP I
I I
CORRELATIONS AMONG INDEPENDENT VARIABLES
I D2 -SLOPE *I
I D1 -INTERCEP I

34. /TITLE
Dependence between ability scores at 6,7,9 and 11
Growth curve model of latent ability
/SPECIFICATIONS
CAS=204 ; VAR=4; ANALYSIS=MOMENT; MATRIX = CORR;
ANAL = COV;
/LABELS
V1 = ABIL6; V2 = ABIL7; V3 = ABIL9; V4 = ABIL11;
F1 = intercept; F2 = slope;
/EQUATIONS
V1 = F1 + E1;
V2 = F1 + F E2;
V3 = F1 + 3*F E3;
V4 = F1 + 5*F2 + E4;
F1 = 18*V999 + D1;
F2 = 7*V999 + D2;
/VARIANCES
D1 = 25*; D2 = 1*; E1 TO E4 = 1.0*;
/COVARIANCES
D1,D2 = .4*;
/CONSTRAINTS
(E1,E1 ) = (E2,E2) = (E3,E3) = (E4,E4);
!/PRINT
!EFFECT = YES;
/STANDARD DEVIATIONS
/MEANS
/MATRIX 1
.809 1
/END

36. Exercice: One reason why the ability data might not show uniform linear growth is that there are practice effects in the testing; in particular, the initial test might give relatively low scores in comparison to the later tests because it was new to the children. The addition of a direct link from V999 to V1 would allow for such an effect. Does the linear growth model fit better when allowing for such a practice effect ?.

37. CHI-SQUARE = 14.266 BASED ON 5 DEGREES OF FREEDOM
/EQUATIONS
V1 = *V999 + F1 + E1;
V2 = F1 + F E2;
V3 = F1 + 3*F E3;
V4 = F1 + 5*F2 + E4;
F1 = 18*V999 + D1;
F2 = 7*V999 + D2;
MEASUREMENT EQUATIONS WITH STANDARD ERRORS AND TEST STATISTICS
ABIL6 =V1 = F *V E1
1.852
-1.626
ABIL7 =V2 = F F E2
ABIL9 =V3 = F *F E3
.730
4.000
ABIL11 =V4 = F *F E4
1.624
3.262
CHI-SQUARE = BASED ON DEGREES OF FREEDOM
PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS