Modelos para datos longitudinales

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 6.374 7.319 7.796 10.386 /MEANS 18.034 25.819 35.255 46.593 /MATRIX 1 .809 1 .806 .850 1 .765 .831 .867 1 /LMtest /END

Single-factor model

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

Autoregressive model

Autoregressive Model * * * * V1 V2 * V4 V3 * *

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 6.374 7.319 7.796 10.386 /MEANS 18.034 25.819 35.255 46.593 /MATRIX 1 .809 1 .806 .850 1 .765 .831 .867 1 /LMtest /END

Estimates

Simplex model

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

EQS imput for simplex model

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

Restricting equality of effects across time

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

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)

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 .

college education n = 143. 1972. 1974. 1976. B. C. J. B. C. J. B. C college education n = 143 1972 1974 1976 B C J B C J B C J B 1 C .43 1 J .47 .29 1 B .79 .43 .48 1 C .39 .54 .38 .45 1 J .50 .28 .56 .56 .35 1 B .71 .37 .49 .78 .44 .59 1 C .27 .53 .18 .35 .60 .20 .34 1 J .47 .29 .49 .48 .32 .61 .53 .28 1 SD 2.03 1.84 1.67 1.76 1.68 1.48 1.74 1.83 1.54 B Busing C Criminals J Jobs

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

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*F1 + E2; V3 = 1*F1 + 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 2.03 1.84 1.67 1.76 1.68 1.48 1.74 1.83 1.54 /MATRIX 1 .43 1 .47 .29 1 .79 .43 .48 1 .39 .54 .38 .45 1 .50 .28 .56 .56 .35 1 .71 .37 .49 .78 .44 .59 1 .27 .53 .18 .35 .60 .20 .34 1 .47 .29 .49 .48 .32 .61 .53 .28 1 /END

Estimated Time effects F2 =F2 = .932*F1 + 1.000 D1 .102 9.106 F3 =F3 = 1.003*F2 + 1.000 D2 .085 11.800 CHI-SQUARE = 47.577 BASED ON 41 DEGREES OF FREEDOM PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS 0.22257

V1 =V1 = 1.114*F1 + 1.000 E1 .113 9.897 V2 =V2 = .839*F1 + 1.000 E2 .120 6.999 V3 =V3 = 1.005*F1 + 1.000 E3 .115 8.730 V4 =V4 = 1.000 F2 + 1.000 E4 V5 =V5 = .773*F2 + 1.000 E5 .131 5.922 V6 =V6 = .907*F2 + 1.000 E6 .147 6.174 V7 =V7 = 1.000 F3 + 1.000 E7 V8 =V8 = .552*F3 + 1.000 E8 .123 4.496 V9 =V9 = .836*F3 + 1.000 E9 .142 5.890

Random walk model

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

EQS code for random walk model

Estimates I F1 - F1 .806*I RESIDUAL COVARIANCE MATRIX (S-SIGMA) : ABIL6 ABIL7 ABIL9 ABIL11 V 1 V 2 V 3 V 4 ABIL6 V 1 0.060 ABIL7 V 2 0.003 -0.033 ABIL9 V 3 0.000 -0.049 -0.075 ABIL11 V 4 -0.041 -0.068 -0.074 -0.095 CHI-SQUARE = 6.867 BASED ON 5 DEGREES OF FREEDOM PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS 0.23072 I F1 - F1 .806*I I .087 I I 9.226 I I I I F2 - F2 .093*I I .030 I I 3.109 I I F3 - F3 .042*I I .022 I I 1.883 I I F4 - F4 .020*I I .028 I I .703 I

Estimates of meas. Error E1 -ABIL6 .134*I I .014 I I 9.376 I I I I E2 -ABIL7 .134*I I E3 -ABIL9 .134*I I E4 -ABIL11 .134*I I

Growth curve model

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

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 + F2 + E2; V3 = F1 + 3*F2 + 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 6.374 7.319 7.796 10.386 /MEANS 18.034 25.819 35.255 46.593 /MATRIX 1 .809 1 .806 .850 1 .765 .831 .867 1 /END

Moment Matrix RESIDUAL COVARIANCE/MEAN MATRIX (S-SIGMA) : 4 VARIABLES (SELECTED FROM 4 VARIABLES), BASED ON 204 CASES. ABIL6 ABIL7 ABIL9 ABIL11 V999 V 1 V 2 V 3 V 4 V999 ABIL6 V 1 40.628 ABIL7 V 2 37.741 53.568 ABIL9 V 3 40.052 48.500 60.778 ABIL11 V 4 50.643 63.169 70.200 107.869 V999 V999 18.034 25.819 35.255 46.593 1.000 RESIDUAL COVARIANCE/MEAN MATRIX (S-SIGMA) : ABIL6 V 1 -4.237 ABIL7 V 2 1.322 6.190 ABIL9 V 3 3.633 6.587 3.895 ABIL11 V 4 14.225 17.565 13.692 29.420 V999 V999 -0.008 -0.046 0.104 -0.050 0.000 CHI-SQUARE = 47.294 BASED ON 7 DEGREES OF FREEDOM PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS LESS THAN 0.001

Estimates ABIL6 =V1 = 1.000 F1 + 1.000 E1 ABIL7 =V2 = 1.000 F1 + 1.000 F2 + 1.000 E2 ABIL9 =V3 = 1.000 F1 + 2.187*F2 + 1.000 E3 .069 31.739 ABIL11 =V4 = 1.000 F1 + 3.656*F2 + 1.000 E4 .119 30.755 INTERCEP=F1 = 18.042*V999 + 1.000 D1 .470 38.417 SLOPE =F2 = 7.822*V999 + 1.000 D2 .305 25.612

Estimates of variances E D --- --- E1 -ABIL6 8.907*I D1 -INTERCEP 31.568*I .625 I 3.725 I 14.248 I 8.474 I I I E2 -ABIL7 8.907*I D2 -SLOPE 2.028*I .625 I .361 I 14.248 I 5.621 I E3 -ABIL9 8.907*I I .625 I I 14.248 I I E4 -ABIL11 8.907*I I

Model with cov. Betw. F1 and F2 CHI-SQUARE = 16.468 BASED ON 6 DEGREES OF FREEDOM PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS 0.01145 COVARIANCES AMONG INDEPENDENT VARIABLES --------------------------------------- E D --- --- I D2 -SLOPE 4.490*I I D1 -INTERCEP .811 I I 5.538 I CORRELATIONS AMONG INDEPENDENT VARIABLES I D2 -SLOPE .561*I I D1 -INTERCEP I

/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 + F2 + E2; V3 = F1 + 3*F2 + 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 6.374 7.319 7.796 10.386 /MEANS 18.034 25.819 35.255 46.593 /MATRIX 1 .809 1 .806 .850 1 .765 .831 .867 1 /END

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 ?.

CHI-SQUARE = 14.266 BASED ON 5 DEGREES OF FREEDOM /EQUATIONS V1 = *V999 + F1 + E1; V2 = F1 + F2 + E2; V3 = F1 + 3*F2 + E3; V4 = F1 + 5*F2 + E4; F1 = 18*V999 + D1; F2 = 7*V999 + D2; MEASUREMENT EQUATIONS WITH STANDARD ERRORS AND TEST STATISTICS ABIL6 =V1 = 1.000 F1 +-3.011*V999 + 1.000 E1 1.852 -1.626 ABIL7 =V2 = 1.000 F1 + 1.000 F2 + 1.000 E2 ABIL9 =V3 = 1.000 F1 + 2.918*F2 + 1.000 E3 .730 4.000 ABIL11 =V4 = 1.000 F1 + 5.296*F2 + 1.000 E4 1.624 3.262 CHI-SQUARE = 14.266 BASED ON 5 DEGREES OF FREEDOM PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS 0.01400