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Modelos para datos longitudinales

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

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Modelos para datos longitudinales

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  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 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

  3. Single-factor model

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

  5. Autoregressive model

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

  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 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 JB 1C .43 1J .47 .29 1B .79 .43 .48 1C .39 .54 .38 .45 1J .50 .28 .56 .56 .35 1B .71 .37 .49 .78 .44 .59 1C .27 .53 .18 .35 .60 .20 .34 1J .47 .29 .49 .48 .32 .61 .53 .28 1SD 2.03 1.84 1.67 1.76 1.68 1.48 1.74 1.83 1.54 B BusingC CriminalsJ 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*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

  20. 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

  21. 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

  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 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 I I F3 - F3 .042*I I .022 I I 1.883 I I I I F4 - F4 .020*I I .028 I I .703 I

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

  27. Growth curve model

  28. Growth curve model * * * * V1 V2 V3 V4 * 1 * * 1 1 D1 * 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 + 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

  30. Moment Matrix 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 ABIL7 ABIL9 ABIL11 V999 V 1 V 2 V 3 V 4 V999 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

  31. 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

  32. 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 I I E3 -ABIL9 8.907*I I .625 I I 14.248 I I I I E4 -ABIL11 8.907*I I .625 I I 14.248 I I

  33. 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 --------------------------------------- E D --- --- I D2 -SLOPE .561*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 + 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

  35. 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 ?.

  36. /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

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