1. General Structural Equation (LISREL) Models Week 3 # 3 MODELS FOR MEANS AND INTERCEPTS

2. 2 Refer to slides from previous class (Week 3 #2) if not covered in full on Tuesday. Refer to slides from previous class (Week 3 #2) if not covered in full on Tuesday.

3. 3 Models with Means and Intercepts Review of material from last class (detail of coverage to depend on progress from Tuesday’s class) Consider a measurement model: Equations: V1 = 1.0 L1 + E1 V2 = b1L1 + E2 V3 = b2L1 + E3 V4 = b3L1 + E4

4. 4 Models with Means and Intercepts The covariance matrix upon which this model is based:

5. 5 Models with Means and Intercepts Simple replacements in this matrix: 1. For any element, covariance replaced by moment: 2. And an “augmented moment matrix” is created by letting the first (or the last) element of the data matrix (the “X” in X’X) be a vector of 1’s

6. 6 Models for Means and Intercepts Augmented moment matrix: Each of the above divided by N-1

7. 7 Means and intercepts in SEM Models Working from this matrix instead of working from S, we can add intercepts back into equations (reproduce M instead of S).

8. 8 Models for Means and Intercepts MEASUREMENT EQUATIONS NOW BECOME: V1 = a1 + 1.0L1 + E1 V2 = a2 + b1 L1 + E2 V3 = a3 + b2 L1 + E3 V4 = a4 + b3 L1 + E4 And there is a final equation for the mean of the latent variable: L1 = a5

9. 9 Means and intercepts in SEM Models Conventional Model: X1 = 1.0 LV1 + e1 X2 = b2 LV1 + e2 X3 = b3 LV1 + e3 Extended to include intercepts: X1 = a1 + 1.0 LV1 + e1 X2 = a2 + b2 LV1 + e2 X3 = a3 + b3 LV1 + e3 [LV1 = a4] EQS calls this “V999”. Other programs do not explicitly model “1” as if it were a variable

10. 10 Means and intercepts in SEM Models Three new pieces of information: Means of X1, X2, X3 Equations: X1 = a1 + 1.0 L1 + e1 X2 = a2 + b2 L1 + e2 X3 = a3 + b3 L1 + e3 Other parameters:Var(e1) Var(e2) Var(e3) Var(L1) Mean(L1) One of the following parameters needs to be fixed: a1,a2,a3, mean(L1)

11. 11 Models for Means and Intercepts From the augmented moment matrix, 4 new pieces of information 5 new (possible) parameters: a1 through a5  cannot identify equation intercepts (under- identified)  but we can identify differences between intercepts.

12. 12 Means and intercepts in SEM Models Equations: X1 = a1 + 1.0 L1 + e1 X2 = a2 + b2 L1 + e2 X3 = a3 + b3 L1 + e3 Conventions: a1 = 0 Then Mean(L1) = Mean(X1) and a2 is difference between means X1,X2 (not usually of interest) a3 is difference between means X1, X3 (not usually of interest)

13. 13 Means and intercepts in SEM Models Conventions: Mean(L1) = 0 Then a1=mean of X1 a2 = mean of X2 a3 = mean of X3 Not particularly useful: means of LV’s by definition =0 Equations: X1 = a1 + 1.0 L1 + e1 X2 = a2 + b2 L1 + e2 X3 = a3 + b3 L1 + e3

14. 14 Means and intercepts in SEM Models In longitudinal case, more interesting possibilities: Constrain measurement models: b1=b3 b2=b4 Constrain intercepts: a1 = a4 a2 = a5 a3 = a6 Fix Mean(L1) to 0 Can now estimate parameter for Mean (L2) Equations: X1 = a1 + 1.0 L1 + e1 X2 = a2 + b1 L1 + e2 X3 = a3 + b2 L1 + e3 X4 = a4 + 1.0 L2 + e4 X5 = a5 + b3 L2 + e5 X6 = a6 + b4 L2 + e6

15. 15 Means and intercepts in SEM Models Constrain measurement models: b1=b3 b2=b4 Constrain intercepts: a1 = a4 a2 = a5 a3 = a6 Fix Mean(L1) to 0 Can now estimate parameter for Mean (L2) Equations: X1 = a1 + 1.0 L1 + e1 X2 = a2 + b1 L1 + e2 X3 = a3 + b2 L1 + e3 Y1 = a4 + 1.0 L2 + e4 Y2 = a5 + b3 L2 + e5 Y3 = a6 + b4 L2 + e6 Example: X1 X2 X3 X4 X5 X6 Means: 2 3 2.5 3 4 3.5 Y1 = a4 + 1.0 L2 + e4 (E(L2)=a7 Estimate: a7=1.0 Y1 = 2 + 1.0*1 + 0 (expected value of L2=1.0) Y2 = 3 + b3*1 + 0 (expected value of L2 = 1.0) New parameter:a7

16. 16 Means and intercepts in SEM Models Equations: X1 = a1 + 1.0 L1 + e1 X2 = a2 + b1 L1 + e2 X3 = a3 + b2 L1 + e3 Y1 = a4 + 1.0 L2 + e4 Y2 = a5 + b3 L2 + e5 Y3 = a6 + b4 L2 + e6 There can be a construct equation intercept parameter in causal models L2 = a7 + b1 L1 + D2 If mean(L1) fixed to 0 E(L2) = a7 + b1*0 = a7 As before, a7 represents the expected difference between the mean of L1 and the mean of L2

17. 17 Means and intercepts in SEM Models L2 = a7 + b1 L1 + D2 If mean(L1) fixed to 0 E(L2) = a7 + b1*0 = a7 In practice, if L1 and L2 represent time 1 and time 2 measures of the same thing, we would expect correlated errors:

18. 18 Means and intercepts in SEM Models Same principle can be applied to multiple group models: Group 1 Group 2 X1 = a1 + 1.0 L1 + e1 X2 = a2 + b2 L1 + e2 X3 = a3 + b3 L1 + e3 X1 = a1 + 1.0 L1 + e1 X2 = a2 + b2 L1 + e2 X3 = a3 + b3 L1 + e3 a1[1] = a1[2] a2[1]=a2[2] a3[1]=a3[2] Mean(L1)=0 Mean(L1) = a4 We usually constrain measurement coefficients: b2[1]=b2[2] & b3[1]=b3[2]

19. 19 Models for Means and Intercepts Applications: #1: A two-group model Group 1Group 2 Group 1 V1 = a1 + 1.0L1 + E1 V2 = a2 + b1 L1 + E2 V3 = a3 + b2 L1 + E3 V4 = a4 + b3 L1 + E4 Group 2 V1 = a1 + 1.0 L1 + E1 V2 = a2 + b1 L1 + E2 V3 = a3 + b2 L1 + E3 V4 = a4 + b3 L1 + E4 Mean(L1) =a5

20. 20 Models for Means and Intercepts Group 1Group 2 Group 1 V1 = a1 + 1.0L1 + E1 V2 = a2 + b1 L1 + E2 V3 = a3 + b2 L1 + E3 V4 = a4 + b3 L1 + E4 Group 2 V1 = a1 + 1.0 L1 + E1 V2 = a2 + b1 L1 + E2 V3 = a3 + b2 L1 + E3 V4 = a4 + b3 L1 + E4 Mean(L1) =a5 Constraints: 1. Measurement model 2. intercepts: a1[1] = a1[2] ; a2[1] = a2[2] etc. 3. a5[1] = 0 THIS MEANS THAT a5[2] represents between- group mean differences.

21. 21 A practical example: Differences in religiosity, World Values Study 1990 In PRELIS, generate mean vectors as well as covariances

22. 22 Looking at item means Means: U.S.: Means v9 v147 v175 v176 -------- -------- -------- -------- 1.700 3.854 1.401 8.126 Canada: Means high = less religious except for V176 v9 v147 v175 v176 -------- -------- -------- -------- 2.193 4.811 1.750 7.005

23. 23 Factor Means: We cannot establish a factor mean for each group, but we CAN get a coefficient representing the difference between the factor means (factor mean in each group can be established trivially as equal to the mean of one of the indicators – not particularly helpful though).

24. 24 LISREL TERMINOLOGY Equations: X 1 = τx 1 + λ 11 ξ 1 + δ 1 X 2 = τx 2 + λ 21 ξ 1 + δ 2 X 3 = τx 3 + λ 31 ξ 1 + δ 3 X 4 = τx 4 + λ 41 ξ 1 + δ 4 New vector: Tau-X (TX) Normally, λ 11 = 1.0 (reference indicator) Variances, covariances, means: VAR(δ 1 ), VAR(δ 2 ), VAR(δ 3 ), VAR(δ 4 ), MEAN(ξ 1 ) New vector: Kappa (vector of means of ξ’s)

25. 25 LISREL TERMINOLOGY Constraints: Group 1Group 2 TX(1)= TX(1) TX(2)=TX(2) TX(3)=TX(3) TX(4)=TX(4) Kappa1 = 0Kappa1 = free*

26. 26 LISREL TERMINOLOGY Constraints: Group 1Group 2 TX(1)= TX(1) TX(2)=TX(2) TX(3)=TX(3) TX(4)=TX(4) Kappa1 = 0Kappa1 = free* Tau-X : vector of manifest variable intercepts Kappa: vector of latent (exogenous) variable means PROGRAMMING: Group 1: TX=FR KA=FI Group 2: TX=IN KA=FR

27. 27 LISREL TERMINOLOGY Equivalent for Y-variables: Tau-Y: intercepts for manifest variable eq’s Alpha: intercepts for construct equations Eta1 = alpha1 + gamma ksi + zeta Important Note: When gammas are constrained to equality across groups, alphas represent a between-group differences in means controlling for differences in Ksi.

28. 28 Factor Mean differences Variances: PHI USA Canada USA KSI 1 KSI 1 -------- 2.751 3.268 (0.187) (0.162) TAU-X TAU-X is constrained to equality (both groups) v9 v147 v175 v176 -------- -------- -------- -------- 1.715 3.828 1.428 8.197 (0.023) (0.058) (0.016) (0.065) 74.484 65.924 86.556 127.025 KAPPA Kappa is zero in group 1 KSI 1 Lambda-X V9.458 -------- V147 1.00 1.005 V175.276 (0.072) V176 -1.289 13.927

29. 29 Models for Means and Intercepts Testing assumptions we have assumed that the pattern of differences between corresponding measurement equation intercepts can be expressed by a single coefficient V1 = a1 + 1.0 L1 + e1 V2 = a2 + b2 L1 + e2 V3 = a3 + b3 L1 + e3 V4 = a4 + b4 L1 + e4 L1=a5 We constrain a1,a2,a3,a4 to equality across groups and estimate a5 to represent between-group differences

30. 30 Models for Means and Intercepts What if the pattern is: Group 1Group 2 v13.24.2 v22.23.2 v31.92.8 v42.01.5 a5 will be positive, but the fact that the group1-group2 difference on V4 is not consistent will lead to poorer fit Could estimate model with a1[1]=a1[2], a2[1]=a2[2], a3[1]=a3[2] BUT a4[1]  a4[2]

31. 31 LISREL TERMINOLOGY LISREL: Equations: X 1 = τx 1 + λ 11 ξ 1 + δ 1 X 2 = τx 2 + λ 21 ξ 1 + δ 2 X 3 = τx 3 + λ 31 ξ 1 + δ 3 X 4 = τx 4 + λ 41 ξ 1 + δ 4 Normally, τx 1 = 1.0 (reference indicator) Variances, covariances, means: VAR(δ 1 ), VAR(δ 2 ), VAR(δ 3 ), VAR(δ 4 ), MEAN(ξ 1 )

32. 32 Models for Means and Intercepts Modification Indices for TAU-X v9 v147 v175 v176 -------- -------- -------- -------- 4.941 0.613 13.006 16.218 We could estimate a model with TX 4 free (not essential; would be more important if chi-square really large) Expected Change for TAU-X v9 v147 v175 v176 -------- -------- -------- -------- 0.045 -0.036 0.063 0.236 TAU-X (repeated from previous slide): v9 v147 v175 v176 -------- -------- -------- -------- 1.715 3.828 1.428 8.197 (0.023) (0.058) (0.016) (0.065) 74.484 65.924 86.556 127.025

33. 33 LISREL PROGRAMMING CODE FOR PREVIOUS EXAMPLE: 2 group model for relig 1:USA DA NG=2 NI=23 NO=1456 CM FI=g:\Means&Intercepts\usa.cov ME FI=g:\Means&Intercepts\usa.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 / MO NX=4 NK=1 LX=FU,FI PH=SY,FR TD=SY C TX=FR KA=FI VA 1.0 LX 2 1 FR LX 1 1 LX 3 1 LX 4 1 OU ME=ML SE TV MI SC ND=3 Group 2: Canada DA NI=23 NO=1474 CM FI=g:\Means&Intercepts\cdn.cov ME FI=g:\Means&Intercepts\cdn.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 / MO LX=IN PH=PS TD=PS KA=FR TX=IN OU ME=ML SE TV MI SC ND=3 New Do not include MA=CM

34. 34 Doing it in AMOS: Add this

35. 35 AMOS: For each exogenous variable, the Object Properties box will now have Mean and Variance

36. 36 AMOS: For each endogenous variable, the Object Properties box will now have an Intercept For all indicators, type in a parameter name here. For all indicators, click “all groups” to impose equality constraint.

37. 37 AMOS Constraints: Group 1Group 2 b1 = b1 b2 =b2 b3 = b3 a1 =a1 a2 = a2 a3 =a3 a4 =a4 a5=0 a5 free (parameter for mean differences)

38. 38 AMOS Group: Canada Means EstimateS.E.C.R.PLabel RELIG1.0050.07213.9310.000a5 Group: United States Intercepts EstimateS.E.C.R.PLabel V91.7150.02374.5120.000a1 V1473.8280.05865.9470.000a2 V1751.4280.01686.5850.000a3 V1768.1970.065127.0680.000a4 REFER TO: Model2.amw for more extended example

39. 39 Moving to Y, Eta and adding a 3 rd country 2 group model for relig 1:USA DA NG=3 NI=23 NO=1456 CM FI=H:\Means&Intercepts\usa.cov ME FI=H:\Means&Intercepts\usa.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 / MO NY=4 NE=1 LY=FU,FI PS=SY,FR TE=SY C TY=FR AL=FI VA 1.0 LY 2 1 FR LY 1 1 LY 3 1 LY 4 1 OU ME=ML SE TV MI SC ND=3 Group 2: Canada DA NI=23 NO=1474 CM FI=H:\Means&Intercepts\cdn.cov ME FI=H:\Means&Intercepts\cdn.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 / MO LY=IN PS=PS TE=PS AL=FR TY=IN OU ME=ML SE TV MI SC ND=3 Group 3: Netherlands DA NI=23 NO=909 CM FI=H:\Means&Intercepts\neth.cov ME FI=H:\Means&Intercepts\neth.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 / MO LY=IN PS=PS TE=PS AL=FR TY=IN OU ME=ML SE TV MI SC ND=3

40. 40 Mean comparisons USA=0 ALPHA Canada ETA 1 -------- 0.896 (0.064) 14.077 ALPHA Netherlands ETA 1 -------- 2.069 (0.087) 23.889 Chi-square = 280.733, df=18 With AL(1)=AL(1)=AL(1) 3 groups (i.e., AL=0 in all three groups) Chi-square = 888. 794 df=20

41. 41 Mean comparisons USA=0 In USA Modification Indices for TAU-Y v9 v147 v175 v176 -------- -------- -------- -------- 0.855 6.130 13.121 2.882 In Canada: Modification Indices for TAU-Y v9 v147 v175 v176 -------- -------- -------- -------- 20.003 18.756 3.873 69.008 In the Netherlands: Modification Indices for TAU-Y v9 v147 v175 v176 -------- -------- -------- -------- 19.044 60.570 4.629 62.667

42. 42 Models for Means and Intercepts: Interpreting Mean differences with exogenous variables GROUP 1 GROUP 2 Equations: L3 = a1 + b1 L1 + b2 L 2 + D3 In group 1, we will hold a1 fixed to 0. In group 2, a1 will be free. IF b1 group 1 = b1 group 2 AND b2 group 1 = b2 group 2 THEN a1 is the between-group difference in L3, controlling for the effects of L1 and L2

43. 43 Lisrel model for mean comparisons with controls 2 group model for relig 1:USA DA NG=3 NI=23 NO=1456 CM FI=H:\Means&Intercepts\usa.cov ME FI=H:\Means&Intercepts\usa.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 12 13 14 / MO NY=4 NX=3 NK=3 FIXEDX NE=1 LY=FU,FI PS=SY,FR TE=SY C TY=FR AL=FI GA=FU,FR KA=FI TX=FR VA 1.0 LY 2 1 FR LY 1 1 LY 3 1 LY 4 1 OU ME=ML SE TV MI SC ND=3 Group 2: Canada DA NI=23 NO=1474 CM FI=H:\Means&Intercepts\cdn.cov ME FI=H:\Means&Intercepts\cdn.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 12 13 14 / MO LY=IN PS=PS TE=PS AL=FR TY=IN FIXEDX GA=IN KA=FR TX=IN OU ME=ML SE TV MI SC ND=3 Group 3: Netherlands DA NI=23 NO=909 CM FI=H:\Means&Intercepts\neth.cov ME FI=H:\Means&Intercepts\neth.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 12 13 14 / MO LY=IN PS=PS TE=PS AL=FR TY=IN FIXEDX GA=IN KA=FR TX=IN OU ME=ML SE TV MI SC ND=3

44. 44 Lisrel model for mean comparisons with controls Group 3: Netherlands DA NI=23 NO=909 CM FI=H:\Means&Intercepts\neth.cov ME FI=H:\Means&Intercepts\neth.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 12 13 14 / MO LY=IN PS=PS TE=PS AL=FR TY=IN FIXEDX GA=IN KA=FR TX=IN OU ME=ML SE TV MI SC ND=3 GROUP #1 SPECIFICATION: MO NY=4 NX=3 NK=3 FIXEDX NE=1 LY=FU,FI PS=SY,FR TE=SY C TY=FR AL=FI GA=FU,FR KA=FI TX=FR Exogenous variable mean =0 in group 1 Exogenous variable mean reflects difference from group 1 TX parameters constrained to =

45. 45 Lisrel model for mean comparisons with controls Group 3: Netherlands DA NI=23 NO=909 CM FI=H:\Means&Intercepts\neth.cov ME FI=H:\Means&Intercepts\neth.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 12 13 14 / MO LY=IN PS=PS TE=PS AL=FR TY=IN FIXEDX GA=IN KA=FR TX=IN OU ME=ML SE TV MI SC ND=3 GROUP #1 SPECIFICATION: MO NY=4 NX=3 NK=3 FIXEDX NE=1 LY=FU,FI PS=SY,FR TE=SY C TY=FR AL=FI GA=FU,FR KA=FI TX=FR GA matrix fixed to invariance (ksi- variables have same effect in each group) Alpha zero in group 1 Group m coefficient represents differences from group 1

46. 46 Mean differences, with controls ALPHA CANADA ETA 1 -------- 0.854 (0.062) 13.822 KAPPA v355 v356 sex -3.465 -0.391 0.008 (0.621) (0.085) (0.018) -5.579 -4.592 0.410 ALPHA ETA 1 -------- 2.080 (0.086) 24.324 KAPPA v355 v356 sex -------- -------- -------- -4.017 -0.908 -0.059 (0.703) (0.111) (0.021) -5.711 -8.179 -2.812

47. 47 Models for Means and Intercepts GROUP 1 GROUP 2 L3 = a1[1] + b1[1]L1 + b2[1]L2 + D3 L3 = a1[2] + b1[2]L1 + b2[2]L2 + D3 Models/constraints: {1} a1[1]=0 (always) {2} b1[1] = b1[2] and b2[1]=b2[2] (normally; parallel slopes) a1[2]=0 vs. a1[2]  0 under {2}: mean diff’s controlling for L1,L2 a1[2]=0 vs. a1[2]  0 under b1=b2=0: mean diff’s without controls

48. 48 Models for Means and Intercepts If slopes of all exogenous variables (L1 and L2 in this example) are parallel, a1 is the mean difference controlling for exog. var’s a1 b1

49. 49 Models for Means and Intercepts What if slopes are not parallel? L1 L3 A1 only represents between-group difference when L1=0 Between-group difference contingent upon value of L1

50. 50 Models for Means and Intercepts #2 A longitudinal model Fix measurement model intercepts to equality Equations: LVTime2 = a6 + b5*LVTime1 + D2 LVTime1 = a5 b5 Fix a5=0; a6 represents change in level over time (We would also normally fix measurement model b coefficients to equality)

51. 51 Models for Means and Intercepts An example: MODEL 3A a1 to a11 = between groups Measurement (b1 to b9) = between groups Latent var. intercepts: 0 in group 1; free in group 2 Coeff’s b13-b17 = between groups

52. 52 LISREL EXAMPLES Group 1 specification: DA NG=2 NI=23 NO=1456 CM FI=h:\icpsr2003\Week3Examples\usacov1.cov ME FI=h:\icpsr2003\Week3Examples\usa.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 6 7 8 9 10 11 / MO NY=10 NE=2 LY=FU,FI PS=SY,FR C TE=SY TY=FR AL=FI FR LY 2 1 LY 3 1 LY 4 1 VA 1.0 LY 1 1 LY 5 2 FR LY 6 2 LY 7 2 LY 8 2 LY 9 2 LY 10 2 OU ME=ML SE TV MI SC ND=3

53. 53 LISREL EXAMPLES Group 2 specification: Group 2: Canada DA NI=23 NO=1474 CM FI=h:\icpsr2003\Week3Examples\cdncov1.cov ME FI=H:\ICPSR2003\Week3Examples\Cdn.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 6 7 8 9 10 11 / MO NY=10 LY=IN PS=PS TE=PS TY=IN AL=FR OU ME=ML SE TV MI SC ND=3

54. 54 LISREL EXAMPLES Summary: DA NG=2 NI=23 NO=1456 CM FI=h:\icpsr2003\Week3Examples\usacov1.cov ME FI=h:\icpsr2003\Week3Examples\usa.mn … MO NY=10 NE=2 LY=FU,FI PS=SY,FR C TE=SY TY=FR AL=FI … OU…. Group 2: Canada DA NI=23 NO=1474 CM FI=h:\icpsr2003\Week3Examples\cdncov1.cov ME FI=H:\ICPSR2003\Week3Examples\Cdn.mn … MO NY=10 LY=IN PS=PS TE=PS TY=IN AL=FR

55. 55 Group 2 printout TAU-Y v9 v147 v175 v176 v304 v305 -------- -------- -------- -------- -------- -------- 1.715 3.827 1.429 8.195 1.908 2.246 (0.023) (0.058) (0.016) (0.064) (0.040) (0.045) 74.370 65.859 86.728 127.177 47.806 49.412 TAU-Y v307 v308 v309 v310 -------- -------- -------- -------- 3.034 2.431 3.921 4.792 (0.065) (0.053) (0.065) (0.058) 46.933 45.617 60.368 82.751 ALPHA ETA 1 ETA 2 -------- -------- 0.460 0.555 (0.032) (0.045) 14.279 12.277 PSI ETA 1 ETA 2 -------- -------- ETA 1 0.688 (0.034) 20.416 ETA 2 0.525 1.153 (0.036) (0.088) 14.790 13.134

56. 56 Group 2 printout ALPHA ETA 1 ETA 2 -------- -------- 0.460 0.555 (0.032) (0.045) 14.279 12.277 PSI ETA 1 ETA 2 -------- -------- ETA 1 0.688 (0.034) 20.416 ETA 2 0.525 1.153 (0.036) (0.088) 14.790 13.134 Modification Indices for TAU-Y v9 v147 v175 v176 v304 v305 -------- -------- -------- -------- -------- -------- 4.900 0.666 13.544 16.089 11.448 16.359 Modification Indices for TAU-Y v307 v308 v309 v310 -------- -------- -------- -------- 0.006 6.567 11.094 18.489 Expected Change for TAU-Y v9 v147 v175 v176 v304 v305 -------- -------- -------- -------- -------- -------- 0.044 -0.037 0.064 0.235 0.161 0.234 Expected Change for TAU-Y v307 v308 v309 v310 -------- -------- -------- -------- -0.005 0.141 -0.194 -0.217 V176 lambda is -ve

57. 57 We can perform block tests (both latent variables at a time: MODEL 1: Group 1: TY=FR AL=FI Group 2: TY=IN AL=FR MODEL 2: Group 1: TY=FR AL=FI Group 2: TY=IN AL=FI

58. 58 With exogenous variables: 3 ksi variables (single indicator)  2 eta variables KSI variables: There is insufficient information to separately estimate latent variable means differences using Tau equality constraints as was done previously. We CAN fix the mean of Ksi’s to the mean of the manifest (single-indicator) variable, as follows: TX=FI(i.e., fixed to TX=0, both groups) KA=FR (Group 1) (Will register as mean of corresponding X-variable) KA=FR (Group 2) Important note: When “controlling” for the effects of the X-variables, we certainly want to allow between-group differences in X (Ksi) variables. Hence we usually impose no equality constraint.

59. 59 With exogenous variables: TX=FI(i.e., fixed to TX=0) KA=FR (Group 1) (Will register as mean of corresponding X-variable) KA=FR (Group 2) Important note: When “controlling” for the effects of the X-variables, we certainly want to allow between-group differences in X (Ksi) variables. Hence we usually impose no equality constraint. To test for significance of differences of individual X/Ksi variables in a 2- group model, we can run another model that sets KA in group 2 = to KA in group 1 (Group 2: KA=IN). We would not keep this equality constraint in place when testing alpha parameters for equivalence.

60. 60 With exogenous variables: TX=FI(i.e., fixed to TX=0) KA=FR (Group 1) (Will register as mean of corresponding X-variable) KA=FR (Group 2) An alternative specification: TX=FR KA=FI (set to zero in group 1) Group 2TX=IN KA=FR(represents between-group differences in the exogenous single-indicator variables)

61. 61 With exogenous variables: 2 group mean model for relig & sexual moral group 1:USA ! adding exogenous variables DA NG=2 NI=23 NO=1456 CM FI=h:\icpsr2003\Week3Examples\usacov1.cov ME FI=h:\icpsr2003\Week3Examples\usa.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 6 7 8 9 10 11 12 13 14 / MO NY=10 NX=3 NE=2 NK=3 LX=ID TD=ZE PH=SY,FR LY=FU,FI PS=SY,FR C GA=FU,FR TY=FR AL=FI KA=FR TX=FI TE=SY GA=FU,FR FR LY 2 1 LY 3 1 LY 4 1 VA 1.0 LY 1 1 LY 10 2 FR LY 7 2 LY 8 2 LY 9 2 LY 6 2 ly 5 2 FR TE 2 1 TE 10 9 TE 6 5 OU ME=ML SE TV MI SC ND=3 Group 2: Canada DA NI=23 NO=1474 CM FI=h:\icpsr2003\Week3Examples\cdncov1.cov ME FI=H:\ICPSR2003\Week3Examples\Cdn.mn LABELS v9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9 SE 1 2 4 5 6 7 8 9 10 11 12 13 14 / MO NY=10 NX=3 LY=IN LX=IN PS=PS PH=PS TD=IN TE=PS C GA=IN LX=ID TD=ZE PH=SY,FR TX=FI KA=FR TY=IN AL=FR OU ME=ML SE TV MI SC ND=3

62. 62 With exogenous variables MO NY=10 NX=3 NE=2 NK=3 LX=ID TD=ZE PH=SY,FR LY=FU,FI PS=SY,FR C GA=FU,FR TY=FR AL=FI KA=FR TX=FI TE=SY GA=FU,FR Group 2: MO NY=10 NX=3 LY=IN LX=IN PS=PS PH=PS TD=IN TE=PS C GA=IN LX=ID TD=ZE PH=SY,FR TX=FI KA=FR TY=IN AL=FR Alternative specification in program MMODEL3.ls8 yields same estimates for alpha (but different estimates for kappa) Group 1: KA=FI TX=FR Group 2: KA=FR TX=IN

63. 63 With exogenous variables: Group 2 (Canada) results: ALPHA ETA 1 ETA 2 -------- -------- 0.426 0.839 (0.031) (0.064) 13.857 13.082 KAPPA v355 v356 sex -------- -------- -------- 43.035 7.383 0.498 (0.421) (0.063) (0.013) 102.251 117.894 38.210 Covariance Matrix of ETA and KSI ETA 1 ETA 2 v355 v356 sex -------- -------- -------- -------- -------- ETA 1 0.615 ETA 2 0.701 2.411 v355 -2.678 -5.809 260.920 v356 0.340 1.028 -12.609 5.776 sex 0.062 0.009 0.014 0.012 0.250

64. 64 With exogenous variables Modification Indices for TAU-Y v9 v147 v175 v176 v304 v305 -------- -------- -------- -------- -------- -------- 9.272 0.057 13.759 29.114 6.016 9.526 Modification Indices for TAU-Y v307 v308 v309 v310 -------- -------- -------- -------- 2.967 2.309 1.943 5.638

65. 65 With exogenous variables So far, we have assumed GA=IN What if this assumption is unreasonable? Modification index for Gamma: Modification Indices for GAMMA v355 v356 sex -------- -------- -------- ETA 1 11.943 1.530 4.826 ETA 2 0.097 0.036 0.950 Rerun model with FR GA 1 1 in group 2 BUT: alpha will no longer represent the between-group difference in eta1, eta2, controlling for age, educ, sex. … alpha 1 will be the INTERCEPT (when V355=0)

66. 66 With exogenous variables Rerun model with FR GA 1 1 in group 2 BUT: alpha will no longer represent the between-group difference in eta1, eta2, controlling for age, educ, sex. … alpha 1 will be the INTERCEPT (when V355=0) Interpretation will depend on coding of KA and TX If we specified TX=FI and KA=FR in groups 1 & 2, then V355 measured in YEARS so we would work out the equation at Ksi=20 Ksi=40 Ksi=40 If we specified TX=FR and KA=FI, we have effectively mean centred in group 1 and have centred the data at the value of the between-group difference in group 2. Would work out equation at Ksi=-20 Ksi=0 Ksi=+20 (still using same age metric)

67. 67 With exogenous variables With GA(1,1) free: ALPHA ETA 1 ETA 2 -------- -------- 0.657 0.840 (0.077) (0.064) 8.501 13.091 GAMMA group 1 v355 v356 sex -------- -------- -------- ETA 1 -0.006 0.039 0.246 (0.001) (0.007) (0.030) -5.401 5.746 8.205 GAMMA group 2 GAMMA v355 v356 sex -------- -------- -------- ETA 1 -0.011 0.039 0.246 (0.001) (0.007) (0.030) -8.666 5.746 8.205 US EQUATION: Eta1 = 0 -.006*Age CDN EQUATION Eta1 =.657 -.001*Age

68. 68 With exogenous variables US EQUATION: Eta1 = 0 -.006*Age CDN EQUATION Eta1 =.657 -.001*Age In this model, AGE is expressed in the same metric as the manifest variable (years).KA=FR TX=FI (Different model, age would be mean deviated [ group1] and centred on kappa, mean difference from group 1[in gr. 2] ) KA=FI TX=FR Group 1 KA=FR TX=IN Group 2

69. 69 With exogenous variables US EQUATION: Eta1 = 0 -.006*Age CDN EQUATION Eta1 =.657 -.011*Age In this model, AGE is expressed in the same metric as the manifest variable (years). KA=FR TX=FI (Different model, age would be mean deviated [ group1] and centred on kappa, mean difference from group 1[in gr. 2] ) KA=FI TX=FR Group 1 KA=FR TX=IN Group 2 At Ksi1= 20 (age= average age + 20): US: 0 -.006*20 = -.120 Cdn:.657 -.011*20 =.657 -.220 = +.437 (difference of.557 from US) At Ksi1 = 0 (age = average age: US: 0 Cdn:.657 (.657 difference from US) At Ksi1= -20 (average age – 20): US: 0 -.006*-20 = +.120 Cdn:.657 -.011*-20 =.657 +.220 =.877 (difference of.757 from US)

70. 70 With exogenous variables: MMODEL4.lS8 Less relig. Cdn. USAge At Ksi1= 20 (age= average age + 20): US: 0 -.006*20 = -.120 Cdn:.657 -.011*20 =.657 -.220 = +.437 (difference of.557 from US) At Ksi1 = 0 (age = average age: US: 0 Cdn:.657 (.657 difference from US) At Ksi1= -20 (average age – 20): US: 0 -.006*-20 = +.120 Cdn:.657 -.011*-20 =.657 +.220 =.877 (difference of.757 from US)

71. 71 LAST SLIDE