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

General Structural EquationGeneral Structural Equation(LISREL) Models(LISREL) Models

Week 3 # 3Week 3 # 3MODELS FOR MEANS AND INTERCEPTSMODELS FOR MEANS AND INTERCEPTS

22

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

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 + E4L1

V1

E1

1

1

V2

E2

1

V3

E3

1

V4

E4

1

4


The covariance matrix upon which this model is based:

)3var()2,4cov()2,4cov()1,4cov(

)3var()2,3cov()1,3cov(

)2var()1,2cov(

)1var(

xxxxxxx

xxxxx

xxx

x

5


Simple replacements in this matrix:1. For any element, covariance replaced by

moment:

1/))((),(

1/))((),cov(

nyxYXmoment

nyyxxYX

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

Models for Means and Intercepts

Augmented moment matrix:

14321

)4()3)(4()2)(4()1)(4(

)3()2)(3()1)(3(

)2()1)(2(

)1(

2

2

2

2

xxxx

xxxxxxx

xxxxx

xxx

x

Each of the above divided by N-1

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


MEASUREMENT EQUATIONS NOW BECOME:

V1 = a1 + 1.0L1 + E1V2 = a2 + b1 L1 + E2V3 = a3 + b2 L1 + E3V4 = a4 + b3 L1 + E4

And there is a final equation for the mean of the latent variable:

L1 = a5

9


LV1X1 e11

X2 e2b2

X3 e3

b3Conventional Model:

X1 = 1.0 LV1 + e1

X2 = b2 LV1 + e2

X3 = b3 LV1 + e3

LV1X1 e11

X2 e2b2

X3 e3

b3

1a4 a1

a2 a3

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


LV1X1 e11

X2 e2b2

X3 e3

b3

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


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


LV1X1 e11

X2 e2b2

X3 e3

b3

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


LV1X1 e11

X2 e2b2

X3 e3

b3

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 + e1X2 = a2 + b2 L1 + e2X3 = a3 + b3 L1 + e3

14


L1

X1

e1

1

X2

e2

b1X3

e3

b2

L2

Y1

e4

Y2

e5

Y3

e6

1 b3 b4

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


L1

X1

e1

1

X2

e2

b1X3

e3

b2

L2

Y1

e4

Y2

e5

Y3

e6

1 b3 b4

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


L1

X1

e1

1

X2

e2

b1X3

e3

b2

L2

Y1

e4

Y2

e5

Y3

e6

1 b3 b4

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

L1

X1

e1

1

X2

e2

b1X3

e3

b2

L2

Y1

e4

Y2

e5

Y3

e6

1 b3 b4

b5

D2

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


L1

X1

e1

1

X2

e2

b1X3

e3

b2

L2

Y1

e4

Y2

e5

Y3

e6

1 b3 b4

b5

D2

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:

L1

X1

e1

1

X2

e2

b1X3

e3

b2

L2

Y1

e4

Y2

e5

Y3

e6

1 b3 b4

b5

D2

18


Same principle can be applied to multiple group models:

L1X1 e11

X2 e2b2

X3 e3

b3

L1X1 e11

X2 e2b2

X3 e3

b3

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) = a4We usually constrain measurement coefficients:

b2[1]=b2[2] & b3[1]=b3[2]

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Applications:

#1: A two-group model

L1

V1

E1

1

1

V2

E2

1

V3

E3

1

V4

E4

1

L1

V1

E1

1

1

V2

E2

1

V3

E3

1

V4

E4

1

Group 1 Group 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 Mean(L1) =a5

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Group 1 Group 2

L1

V1

E1

1

1

V2

E2

1

V3

E3

1

V4

E4

1

L1

V1

E1

1

1

V2

E2

1

V3

E3

1

V4

E4

1

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 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

A practical example:

Differences in religiosity, World Values Study 1990

In PRELIS, generate mean vectors as well as covariances

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

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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

LISREL TERMINOLOGY

Equations:X1 = τx1 + λ11ξ1 + δ1

X2 = τx2 + λ21ξ1 + δ2

X3 = τx3 + λ31ξ1 + δ3

X4 = τx4 + λ41ξ1 + δ4

New vector: Tau-X (TX)

Normally, λ11 = 1.0 (reference indicator)

Variances, covariances, means:

VAR(δ1), VAR(δ2), VAR(δ3), VAR(δ4), MEAN(ξ1)

New vector: Kappa (vector of means of ξ’s)

25

LISREL TERMINOLOGY

Constraints:Group 1 Group 2

TX(1) = TX(1)

TX(2) = TX(2)

TX(3) = TX(3)

TX(4) = TX(4)

Kappa1 = 0 Kappa1 = free*

26

LISREL TERMINOLOGY

Constraints:Group 1 Group 2

TX(1) = TX(1)TX(2) = TX(2)TX(3) = TX(3)TX(4) = TX(4)

Kappa1 = 0 Kappa1 = free*

Tau-X : vector of manifest variable interceptsKappa: vector of latent (exogenous) variable meansPROGRAMMING:Group 1: TX=FR KA=FIGroup 2: TX=IN KA=FR

27

LISREL TERMINOLOGY

Equivalent for Y-variables:

Tau-Y: intercepts for manifest variable eq’sAlpha: 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

Factor Mean differences

Variances: PHI USA CanadaUSA 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

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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

Models for Means and InterceptsWhat if the pattern is:

Group 1 Group 2v1 3.2 4.2v2 2.2 3.2v3 1.9 2.8v4 2.0 1.5

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

LISREL TERMINOLOGY

LISREL:Equations:

X1 = τx1 + λ11ξ1 + δ1

X2 = τx2 + λ21ξ1 + δ2

X3 = τx3 + λ31ξ1 + δ3

X4 = τx4 + λ41ξ1 + δ4

Normally, τx1 = 1.0 (reference indicator)

Variances, covariances, means:

VAR(δ1), VAR(δ2), VAR(δ3), VAR(δ4), MEAN(ξ1)

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Models for Means and InterceptsModification 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

LISREL PROGRAMMING CODE FOR PREVIOUS EXAMPLE:

2 group model for relig 1:USADA NG=2 NI=23 NO=1456 CM FI=g:\Means&Intercepts\usa.covME FI=g:\Means&Intercepts\usa.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 2 4 5 /MO NX=4 NK=1 LX=FU,FI PH=SY,FR TD=SY CTX=FR KA=FIVA 1.0 LX 2 1 FR LX 1 1 LX 3 1 LX 4 1 OU ME=ML SE TV MI SC ND=3Group 2: CanadaDA NI=23 NO=1474CM FI=g:\Means&Intercepts\cdn.covME FI=g:\Means&Intercepts\cdn.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 2 4 5 /MO LX=IN PH=PS TD=PS KA=FR TX=INOU ME=ML SE TV MI SC ND=3

New

Do not include MA=CM

34

Doing it in AMOS:

Add this

35

AMOS:For each exogenous variable, the Object Properties box will now have

Mean and

Variance

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

AMOS

Constraints:

Group 1 Group 2

b1 = b1

b2 = b2

b3 = b3

a1 = a1

a2 = a2

a3 = a3

a4 = a4

a5=0 a5 free (parameter for mean

differences)

a5,

RELIG

a1

V9

0,

E1

1

1

a2

V147

0,

E2

b1

1

a3

V175

0,

E3

b2

1

a4

V176

0,

E4

b3

1

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AMOS

Group: Canada

Means

Estimate S.E. C.R. P Label

RELIG 1.005 0.072 13.931 0.000 a5

Group: United States

Intercepts

Estimate S.E. C.R. P Label

V9 1.715 0.023 74.512 0.000 a1

V147 3.828 0.058 65.947 0.000 a2

V175 1.428 0.016 86.585 0.000 a3

V176 8.197 0.065 127.068 0.000 a4

REFER TO: Model2.amw for more extended example

39

Moving to Y, Eta and adding a 3rd

country

2 group model for relig 1:USADA NG=3 NI=23 NO=1456 CM FI=H:\Means&Intercepts\usa.covME FI=H:\Means&Intercepts\usa.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 2 4 5 /MO NY=4 NE=1 LY=FU,FI PS=SY,FR TE=SY CTY=FR AL=FIVA 1.0 LY 2 1 FR LY 1 1 LY 3 1 LY 4 1 OU ME=ML SE TV MI SC ND=3Group 2: CanadaDA NI=23 NO=1474CM FI=H:\Means&Intercepts\cdn.covME FI=H:\Means&Intercepts\cdn.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 2 4 5 /MO LY=IN PS=PS TE=PS AL=FR TY=INOU ME=ML SE TV MI SC ND=3Group 3: NetherlandsDA NI=23 NO=909CM FI=H:\Means&Intercepts\neth.covME FI=H:\Means&Intercepts\neth.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 2 4 5 /MO LY=IN PS=PS TE=PS AL=FR TY=INOU ME=ML SE TV MI SC ND=3

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

Mean comparisons USA=0

In USAModification 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

Models for Means and Intercepts: Interpreting Mean differences with exogenous variables

L3

1

1 1 1

L1

1

1 1 1

L21

111

b1

b2

L3

1

1 1 1

L1

1

1 1 1

L21

111

b1

b2

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

Lisrel model for mean comparisons with controls

2 group model for relig 1:USADA NG=3 NI=23 NO=1456 CM FI=H:\Means&Intercepts\usa.covME FI=H:\Means&Intercepts\usa.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 2 4 5 12 13 14 /MO NY=4 NX=3 NK=3 FIXEDX NE=1 LY=FU,FI PS=SY,FR TE=SY CTY=FR AL=FI GA=FU,FR KA=FI TX=FRVA 1.0 LY 2 1 FR LY 1 1 LY 3 1 LY 4 1 OU ME=ML SE TV MI SC ND=3Group 2: CanadaDA NI=23 NO=1474CM FI=H:\Means&Intercepts\cdn.covME FI=H:\Means&Intercepts\cdn.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 2 4 5 12 13 14 /MO LY=IN PS=PS TE=PS AL=FR TY=IN FIXEDX GA=IN KA=FR TX=INOU ME=ML SE TV MI SC ND=3Group 3: NetherlandsDA NI=23 NO=909CM FI=H:\Means&Intercepts\neth.covME FI=H:\Means&Intercepts\neth.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 2 4 5 12 13 14 /MO LY=IN PS=PS TE=PS AL=FR TY=IN FIXEDX GA=IN KA=FR TX=INOU ME=ML SE TV MI SC ND=3

44


Group 3: NetherlandsDA NI=23 NO=909CM FI=H:\Means&Intercepts\neth.covME FI=H:\Means&Intercepts\neth.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 2 4 5 12 13 14 /MO LY=IN PS=PS TE=PS AL=FR TY=IN FIXEDX GA=IN KA=FR TX=INOU 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 CTY=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


Group 3: NetherlandsDA NI=23 NO=909CM FI=H:\Means&Intercepts\neth.covME FI=H:\Means&Intercepts\neth.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 2 4 5 12 13 14 /MO LY=IN PS=PS TE=PS AL=FR TY=IN FIXEDX GA=IN KA=FR TX=INOU 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 CTY=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

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


L3

1

1 1 1

L1

1

1 1 1

L21

111

b1

b2L3

1

1 1 1

L1

1

1 1 1

L21

111

b1

b2

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


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

b1

49


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


#2 A longitudinal model

Fix measurement model intercepts to equality

LVTime1

1

1 1 1

LVTime2

1

1 1 1

D2

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


An example:

relig

0

v9

a1

e1

0,

1

1

v147

a2

e2

0,

b1

1

v151

a3

e3

0,

b2

1

v175

a4

e4

0,

b3

1

v176

a5

e5

0,

b4

1

sexmrl

0

v310

a6

e11

0,

1

1

v309

a7

e10

0,

b5

1

v308

a8

e9

0,

b6

1

v307

a9

e8

0,

b7

1

v305

a10

e7

0,

b8

1

v304

a11

e6

0,

b9

1

d1

0,

d2

0,

1

1

sex

v355

v356

b13

b14

b15

b16

b17

b18

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

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LISREL EXAMPLES

Group 1 specification:DA NG=2 NI=23 NO=1456 CM FI=h:\icpsr2003\Week3Examples\usacov1.covME FI=h:\icpsr2003\Week3Examples\usa.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355

v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8

occup9SE1 2 4 5 6 7 8 9 10 11 /MO NY=10 NE=2 LY=FU,FI PS=SY,FR CTE=SY TY=FR AL=FIFR LY 2 1 LY 3 1 LY 4 1 VA 1.0 LY 1 1 LY 5 2FR LY 6 2 LY 7 2 LY 8 2 LY 9 2 LY 10 2 OU ME=ML SE TV MI SC ND=3

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LISREL EXAMPLES

Group 2 specification:

Group 2: CanadaDA NI=23 NO=1474CM FI=h:\icpsr2003\Week3Examples\cdncov1.covME FI=H:\ICPSR2003\Week3Examples\Cdn.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 2 4 5 6 7 8 9 10 11 /MO NY=10 LY=IN PS=PS TE=PS TY=IN AL=FROU ME=ML SE TV MI SC ND=3

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LISREL EXAMPLES

Summary:

DA NG=2 NI=23 NO=1456 CM FI=h:\icpsr2003\Week3Examples\usacov1.covME FI=h:\icpsr2003\Week3Examples\usa.mn…

MO NY=10 NE=2 LY=FU,FI PS=SY,FR CTE=SY TY=FR AL=FI…OU….Group 2: CanadaDA NI=23 NO=1474CM FI=h:\icpsr2003\Week3Examples\cdncov1.covME FI=H:\ICPSR2003\Week3Examples\Cdn.mn…

MO NY=10 LY=IN PS=PS TE=PS TY=IN AL=FR

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Group 2 printoutTAU-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

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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


v307 v308 v309 v310 -------- -------- -------- -------- 0.006 6.567 11.094 18.489Expected 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

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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

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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.

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TX=FI (i.e., fixed to TX=0)



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.

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TX=FI (i.e., fixed to TX=0)



KA=FR (Group 2)

An alternative specification:

TX=FR

KA=FI (set to zero in group 1)

Group 2 TX=IN

KA=FR (represents between-group differences in the

exogenous single-indicator variables)

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With exogenous variables:2 group mean model for relig & sexual moral group 1:USA! adding exogenous variablesDA NG=2 NI=23 NO=1456 CM FI=h:\icpsr2003\Week3Examples\usacov1.covME FI=h:\icpsr2003\Week3Examples\usa.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 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 CGA=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 2FR 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=3Group 2: CanadaDA NI=23 NO=1474CM FI=h:\icpsr2003\Week3Examples\cdncov1.covME FI=H:\ICPSR2003\Week3Examples\Cdn.mnLABELSv9 v147 v151 v175 v176 v304 v305 v307 v308 v309 v310 v355 v356 sex occup1 occup2 occup3 occup4 occup5 occup6 occup7 occup8 occup9SE1 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 CGA=IN LX=ID TD=ZE PH=SY,FR TX=FI KA=FR TY=IN AL=FROU ME=ML SE TV MI SC ND=3

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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 CGA=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 CGA=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

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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

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v9 v147 v175 v176 v304 v305 -------- -------- -------- -------- -------- -------- 9.272 0.057 13.759 29.114 6.016 9.526


v307 v308 v309 v310 -------- -------- -------- -------- 2.967 2.309 1.943 5.638

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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)

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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)

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With GA(1,1) free:

ALPHA

ETA 1 ETA 2 -------- -------- 0.657 0.840 (0.077) (0.064) 8.501 13.091GAMMA 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 2GAMMA

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

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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

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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)

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With exogenous variables: MMODEL4.lS8

Less relig.

Cdn.

US Age

At Ksi1= 20 (age= average age + 20):US: 0 - .006*20 = -.120Cdn: .657 - .011*20 = .657 - .220 = + .437 (difference of .557 from US)

At Ksi1 = 0 (age = average age:US: 0Cdn: .657 (.657 difference from US)

At Ksi1= -20 (average age – 20):US: 0 - .006*-20 = +.120Cdn: .657 - .011*-20 = .657 + .220 = .877 (difference of .757 from US)

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

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