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General Structural Equations(LISREL)
Week 3 #4Mean Models Reviewed
Non-parallel slopesNon-normal data
2
Models for Means and Intercepts (continued)
Multiple Group Models:For “zero order” latent variable mean
differences: “free” individual measurement equation
intercepts but constrain them to equality across groups
Fix the latent variable means to 0 in group 1 Free the latent variable means in groups 2->k If the latent variables of interest are
endogenous and if there are exogenous latent variables in the model, constrain construct equation path coefficients to zero.
3
Models for Means and Intercepts (continued)
For “zero order” latent variable mean differences: “free” individual measurement equation intercepts but constrain them to
equality across groups Fix the latent variable means to 0 in group 1 Free the latent variable means in groups 2->k If the latent variables of interest are endogenous and if there are exogenous
latent variables in the model, constrain construct equation path coefficients to zero.
Individual LV mean parameters represent contrast with (differerence from) “reference group” (group with LV mean set to zero; LR tests requested for joint hypotheses (e.g, constrain means to zero in all groups vs. model with groups 2->k freed)
Check modification indices on measurement equation intercepts to verify “proportional indicator differences” assumption holds (or at least holds approximately)
4
AMOS Programming Check off “means and intercepts” Means and intercepts will now appear on
diagram. Where variances used to appear, there will now be two parameters (mean + variance); where the variable is dependent, one parameter (intercept) will appear.
Impose appropriate parameter constraints [insert brief demonstration here!]
5
Review yesterday’s slides from slide 52
Uses World Values Study 1990 data for an example
We’ll use an updated version (new data, some difference in countries) today
Refer to handout (slides not reproduced)
6
Means1a.LS8 - tau-x elements allowed to vary between
countries. Must fix kappa (mean of ksi’s) to 0 since otherwise
not identified. Chi-square=233.65 df=42
United States:
TAU-X
A006 F028 F066 F063 F118 F119
-------- -------- -------- -------- -------- --------
1.6191 3.6383 2.2287 8.5530 4.7504 2.9739
(0.0263) (0.0688) (0.0563) (0.0733) (0.0941) (0.0749)
61.4969 52.8937 39.5980 116.6334 50.4717 39.6838
TAU-X
F120 F121
-------- --------
4.3443 5.9000
(0.0883) (0.0757)
49.2263 77.9553
CANADA:
TAU-X
A006 F028 F066 F063 F118 F119
-------- -------- -------- -------- -------- --------
2.1202 4.7402 3.2042 7.4657 5.4974 3.3091
(0.0236) (0.0612) (0.0551) (0.0706) (0.0812) (0.0646)
89.9232 77.4453 58.1887 105.7780 67.7035 51.2445
TAU-X
F120 F121
-------- --------
4.4986 6.0079
(0.0713) (0.0636)
7
Means1b.ls8 Measurement model like means1a, but now we are expressing group 1 versus group 2 differences in means by 2 parameters (1 for each latent variable) as opposed to calculating them for each indicator using, e.g., TX 1 [1] – TX1 [2].
Chi-square=276.27 df=48KAPPA in Group 2 (Canada) [Kappa in Group 1 is fixed to zero] KSI 1 KSI 2 -------- -------- 1.0712 0.3236 (0.0731) (0.0948) 14.6538 3.4138Above provides significance tests for:Canada-U.S. differences in religiosity (z=14.6538, p<.001)Canada-U.S. differences in sex/morality attitudes (z=3.4138, p<.001)For a joint significance test to see if both the means for Religiosity and Sex/morality
are different (null hypothesis, differences both = 0), see program Means1c.ls8. Chi-square = 512.9661 df=50 for this model; subtract chi-squares (512-276) for test (df=2).
8
Diagnostics for this model: See Modification Indices for TX vectors:USA Modification Indices for TAU-X A006 F028 F066 F063 F118 F119 -------- -------- -------- -------- -------- -------- 0.6495 0.2995 2.8724 8.2808 27.0494 2.0749 Modification Indices for TAU-X F120 F121 -------- --------
12.1313 5.2727CANADAModification Indices for TAU-X A006 F028 F066 F063 F118 F119 -------- -------- -------- -------- -------- -------- 0.6495 0.2995 2.8725 8.2808 27.0495 2.0749 Modification Indices for TAU-X F120 F121 -------- -------- 12.1312 5.2728 Expected Change for TAU-X A006 F028 F066 F063 F118 F119 -------- -------- -------- -------- -------- -------- 0.0164 0.0238 0.0593 0.1261 0.3003 0.0637 Expected Change for TAU-X F120 F121 -------- -------- -0.1981 -0.1015
9
Means2a Model with exogenous single-indicator variables.
Single indicator ksi-variables: gender, age, education.
Specification GA=IN in group 2 implies a parallel slopes model.
Thus, the AL parameters in group 2 can be interpreted as “group 1 vs. group 2
differences, controlling for differences in sex, education and age”.
TAU-X
GENDER AGE EDUC
-------- -------- --------
0.4217 42.3840 4.5365
(0.0146) (0.4750) (0.0413)
28.9360 89.2300 109.8409
ALPHA
ETA 1 ETA 2
-------- --------
1.2272 0.5898
(0.0714) (0.0954)
17.1899 6.1819
KAPPA
GENDER AGE EDUC
-------- -------- --------
-0.0196 3.3360 -0.4151
(0.0187) (0.6297) (0.0504)
-1.0482 5.2977 -8.2333
10
Diagnostics: Test of equal slopes (GA=IN) assumption:Modification Indices for GAMMA GENDER AGE EDUC -------- -------- -------- ETA 1 7.7083 6.9705 0.2122 ETA 2 3.1923 0.1765 9.3836
A global test will require the estimation of a separate model (Means2b) with GA=PS (parallel slopes assumption relaxed).
Chi-square dfCFI
Chi-square comparisons Means2a: 699.80790 .9635
Means2b: 669.59484 .9649
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Means2bALPHA CANADA (FIXED TO 0 IN US) ETA 1 ETA 2 -------- -------- 1.2545 0.6371 (0.0725) (0.0968) 17.3057 6.5809GAMMA - USA GENDER AGE EDUC -------- -------- -------- ETA 1 0.6845 -0.0170 0.0817 (0.1003) (0.0031) (0.0352) 6.8230 -5.5398 2.3209 ETA 2 0.0624 -0.0144 0.3074 (0.1462) (0.0045) (0.0520)GAMMA-Canada GENDER AGE EDUC -------- -------- -------- ETA 1 0.9597 -0.0308 0.1525 (0.0931) (0.0028) (0.0389) 10.3099 -11.1125 3.9173 ETA 2 -0.0936 -0.0246 0.5333 (0.1200) (0.0036) (0.0521)
12
Expressing effects when parallel slope assumption is relaxed:is pattern diverging, converging, crossover?
Equations:Eta1 = alpha1 + gamma1 Ksi 1 + gamma2 Ksi2 + gamma3 Ksi 3 + zeta1Hold constant at the 0 values of all Ksi variables except one. Not quite the overall
mean (Ksi=0 in group 1, but in group 2 it’s 0 + kappa), but close enough.In group 1, alpha1 = 0, equation is:Eta1 = gamma1 [1]Ksi1 [+alpha1=0 + gamma2 Ksi2=0 + gamma3 Ksi3=0 +
zeta1
where E(zeta1)=0
In group 2, alpha1 = alpha1[2]Eta1 = alpha1[2] + gamma1[2] Ksi1 [+ other terms =0]Now, the question is, at what values do we evaluate the equation?
1. Ksi1=0 This is the Ksi1 mean in group 1. (we could, alternativelyuse something like kappa1[2]/2, which is half way betweenthe group 1 and the group 2 mean of kappa1 … or even a weighted version)
2. Ksi1 = 0 + k standard deviations, where k can be any reasonable number1? 1.5? 2.0?
3. Ksi1 = 0 – k standard deviations.
13
How do we find the standard deviation of Ksi?Look at the PHI matrix to obtain variances, and take the
square root of these!PHI USA GENDER AGE EDUC -------- -------- -------- GENDER 0.2441 (0.0102) 23.9687 AGE -0.4381 259.2400 (0.2350) (10.8158) -1.8642 23.9687 EDUC 0.0251 1.7457 1.9599 (0.0204) (0.6670) (0.0818)
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For education, if we had a pooled estimate (Canada + US) we could use it, otherwise, we can be approximate 1.9599, 1.4733 ~ 1.72 sqrt(1.72) = 1.3. So we will want to evaluate at EDUC=0, EDUC=+1.3 (or perhaps +2.6?), EDUC=-1.3 (or perhaps -2.6?).
At Educ=0, Canada-US difference is 1.2545 (see alpha parameter, above) USA=0 Canada=1.2545
At Educ=-2.6, USA= 0 + (-2.6 * .0817) [usa gamma for educ = .0817]= -.2124Canada = 1.2545 + (-2.6 * .1525) [Canadian gamma for educ = .1525]
= 858At Educ = +2.6, USA = 0 + (2.6 * .0817) = .2124
Canada = 1.2545 + (2.6 * .1525) = 1.651
15
-0.5
0
0.5
1
1.5
2
-2.6 0 2.6
Education
USA
Canada
16
For age, approximate variance is sqrt (270) = 16.43. We could thus use 0 ± 16.43 or 0 ≠ 32.86 (or 0 ≠ (1.5 * 16.43) or if we knew that the mean was
approximately 42 (see tau-x parameter), we could simply do something like ± 20 years (more intuitive)
-0.5
0
0.5
1
1.5
2
-20 0 20
Age (0=42 years)
USA
Canada
17
Models for Four Groups• Canada• U.S.A.• Germany• U.K.
Means3a GA=PS Chi-square = 1892.25 df=180
Means3b GA=IN Chi-square = 1986.94 df=198
18
Value of USA Canada Educ USA Canada UK Germany
-2.6 -0.19188 0.70818 2.73732 1.56508 0 0 1.087 2.4339 1.8139
2.6 0.19188 1.46582 2.13048 2.06272
Formulas: USA: =0.0738*B8
Canada: =1.087+(B8*0.1457)
UK : =2.4339+(B8*-0.1167)
Germany: =1.8139+(B8*0.0957)
[B8 refers to the first education row. Formula becomes B9, B10
For rows below]
19
-0.5
0
0.5
1
1.5
2
2.5
3
-2.6 0 2.6
Education
USA
Canada
UK
Germany
Dealing with data that are not normally distributed within the traditional LISREL framework
Questions:-how bad is it if our data are not normally distributed?- what can we do about it?-are there easy “fixes”?
21
Non-Normal Data
How about just ignoring the problem?
Early 1980s: Robustness studies. Major findings:
In almost all cases, using LV models better than OLS even if data non-normal
(assumes multiple indicators available)
some discussion of conditions under which parameters might not be accurate (e.g., low measurement coefficient models)
22
Non-Normal Data
Early articles: A. Boomsa, On the Robustness of LISREL Johnson and Creech, American Sociological
Review, 48(3), 1983, 398-403 Henry, ASR, 47: 299-307 (related: Bollen and Barb, ASR, 46: 232-39)
See a good summary of early and later simulation studies: West, Finch and Curran in Hoyle.
23
Non-Normal Data See a good summary of early and later simulation studies:
West, Finch and Curran in Hoyle. Formal properties:
Consistent?
Asymp. Efficient?
Acov(θ)
X2
Multinormal (no kurtosis)
√ √ √ √
Elliptical √ √ X X
Arbitrary √ X X X
24
Non-Normal DataMany of the studies have involved CFA models
•E.g., Curran, West, Finch, Psych. Methods, 1(1), 1996.
• General findings (non-normal data):
• ML, GLS produce X2 values too high
•Overestimated by 50% in simulations
•GLS, ML produce X2 value slightly larger when sample sizes small, even when data are normally distributed
•Underestimation of NFI, TLI, CFI
•Also underestimated in small samples esp. NFI
•Moderate underestimation of std. errors (phi 25%, lambda 50%)
25
Non-Normality
Detection: ur = E(x – ur)r kurtosis 4th moment
Mean of 3 standardized: u4 / u22
Standardized 3rd moment u3/ (u2)3/2
Tests of statistical significance usually available (Bollen, p. 421) b1, b2 (skew,kurt)
N(0,1) test statistic for Kurtosis (H0: B2 – 3 = 0) Different tests (one approx. requires N>1000)
Joint test κ2 Approx. distr. as X2, df=2 Mardia’s multivariate test: skewedness, kurtosis, joint.
26
Non-NormalityNon-Normality
An alternative estimator:Fwls (also called Fagls):
[s – σ(θ)’ w-1 [s – σ(θ)]Browne, British Journal of Mathematical and Statistical Psychology, 41 (1988)
193ff.also 37 (1984), 62-83
Optimal weight matrix?asymptotic covariance matrix of sij
Acov(sij,sgh) = N-1 (σijgh - σij σgh)Sijgh = 1/N Σ (zi)(zj)(zg)(zh)
where zi is the mean-deviated value
If multinomial: σijgh = σij σgh + σjg σjh + σjh σjg (reduces to GLS)
W-1 is ½ * (k)(k+1) + ½ (k)(K+1)
27
Non-Normality
An alternative estimator:
Fwls (also called Fagls):[s – σ(θ)’ w-1 [s – σ(θ)]
W-1 is ½ * (k)(k+1) + ½ (k)(K+1)Computationally intense:
20 variables: 22,155 distinct elementsTo be non-singular,
N must be > p + ½ (p)(p+1)20 variables: minimum 23030 variables: minimum 495
Older versions of LISREL used to impose higher restrictions (refused to run until thresholds well above the minima shown above were reached)
28
Non-Normality
An alternative estimator:
Fwls (also called Fagls):[s – σ(θ)’ w-1 [s – σ(θ)]
W-1 is ½ * (k)(k+1) + ½ (k)(K+1)The AGLS estimator is commonly available in SEM
software LISREL 8 AMOS SAS-CALIS EQS
Be careful! Not really suitable for small N problems Good idea to have sample sizes in the thousands, not
hundreds.
29
Non-Normality
An alternative estimator:
Fwls (also called Fagls):
[s – σ(θ)’ w-1 [s – σ(θ)]
W-1 is ½ * (k)(k+1) + ½ (k)(K+1)
The AGLS estimator is commonly available in SEM software LISREL 8: ME=WL in OU statement; must also provide
asymptotic covariance matrix generated by PRELIS AC FI= statement follows CM FI= statement
AMOS: check box on analysis options
Again, the problem is that this estimator can be unstable given the size of the matrix (acov) that needs to be inverted (especially in moderate sample sizes)
30
Non-Normality
Sample program in LISREL with adf estimator:
LISREL model for religiosity and moral conservatism
Part 2: ADF estimation
DA NI=14 NO=1456
CM FI=h:\icpsr2003\Week4Examples\nonnormaldata\relmor1.cov
ACC FI=h:\icpsr2003\Week4Examples\nonnormaldata\relmor1.acc
SE
1 2 3 4 5 6 7 8 9 10 11 12 13 14/
MO NY=11 NX=3 NE=2 Nk=3 fixedx ly=fu,fi ga=fu,fr c
ps=sy,fr te=sy
va 1.0 ly 1 1 ly 8 2
fr ly 2 1 ly 3 1 ly 4 1 ly 5 1
fr ly 6 2 ly 7 2 ly 9 2 ly 10 2 ly 11 2
fr te 2 1 te 11 10 te 7 6
ou me=ml se tv sc nd=3 mi
31
Non-Normality
Generating asymptotic covariance matrix in PRELIS
32
Non-Normality
Generating asymptotic covariance matrix in PRELIS
Resultant matrix will be much larger than covariance matrix
33
Non-Normality ADF estimation
LISREL model for religiosity and moral conservatism
Part 2: ADF estimation
DA NI=14 NO=1456
CM FI=h:\icpsr99\nonnorm\relmor1.cov
ACC FI=h:\icpsr99\nonnorm\relmor1.acc
SE
1 2 3 4 5 6 7 8 9 10 11 12 13 14/
MO NY=11 NX=3 NE=2 Nk=3 fixedx ly=fu,fi ga=fu,fr c
ps=sy,fr te=sy
va 1.0 ly 1 1 ly 8 2
fr ly 2 1 ly 3 1 ly 4 1 ly 5 1
fr ly 6 2 ly 7 2 ly 9 2 ly 10 2 ly 11 2
fr te 2 1 te 11 10 te 7 6
ou me=wl se tv sc nd=3 mi
34
Non-Normality ML, scaled statistics
LISREL model for religiosity and moral conservatism
Part 2: ADF estimation
DA NI=14 NO=1456
CM FI=h:\icpsr2003\Week4Examples\nonnormaldata\relmor1.cov
ACC FI=h:\icpsr2003\Week4Examples\nonnormaldata\relmor1.acc
SE
1 2 3 4 5 6 7 8 9 10 11 12 13 14/
MO NY=11 NX=3 NE=2 Nk=3 fixedx ly=fu,fi ga=fu,fr c
ps=sy,fr te=sy
va 1.0 ly 1 1 ly 8 2
fr ly 2 1 ly 3 1 ly 4 1 ly 5 1
fr ly 6 2 ly 7 2 ly 9 2 ly 10 2 ly 11 2
fr te 2 1 te 11 10 te 7 6
ou me=ml se tv sc nd=3 mi
35
Non-Normality
Low tech solutions:For variables that are continuous,
TRANSFORMATION See classic regression texts such as Fox
Common transformations: X log(X) (usually natural log) X sqrt (X) X X2
X 1/ X (even harder to interpret since this will result in sign reversal)
Transforming to remove skewedness often/usually removes kurtosis, but this is not guaranteed
“Normalization” as an extreme option (e.g., map rank-ordered data onto N(0,1) distribution).
36
Non-Normality
Generally, if kurtosis between +1 and -1, not considered too problematic
(See Bollen, 1989)From this…….
37
Transformations
AMOS: Transformations must be performed on SPSS dataset. Save new dataset, and work from
this. (e.g, COMPUTE X1 = LOG(X1).)
LISREL: Transformations can be performed
in PRELIS.
PRELIS
already provides distribution
information on variables as a
“check”PRELIS “compute” dialogue box under transformations
Remember to SAVE the Prelis dataset after each
transformation. Use of stat package (SPSS, Stata,
SAS) may be preferable
38
Transformations
All the usual caveats apply:All the usual caveats apply:1. If a variable only has 4-5 values, transformation
will not normalize a variable (at the very least, will still have tucked-in tails) – though it could help bring it closer to within the +1 -1 range (Kurtosis)
2. If a categorized variable has one value with a majority of cases, then no transformation will work
3. If the variable has negative values, make sure to add a constant (“offset”) before logging
39
Other solutions:
1. Robust test statistics (Bentler) Implementation: EQS, LISREL
2. Muthen has recently developed a WLSM (mean-adjusted) and WLSMV (mean and variance adjusted) estimator
Implementation: MPLUS only3. Bootstrapping
Implementation: AMOS (easy to use) LISREL (awkward)
4. CATEGORICAL VARIABLE MODELS (CVM).
40
Bootstrapping
Computationally intensive Sampling with replacement; from
resampling space R draw bootstrap sample S*
n,j where j=# of samples, n=bootstrap n
Typically, bootstrap N = sample N Repeat resampling B times, get set of
values Issue: what if, across 200 resamples, 2 of them
have ill-defined matrices? Usually, these are discarded
41
Bootstrapping
Computationally intensive Sampling with replacement; from resampling
space R draw bootstrap sample S*n,j where j=# of samples,
n=bootstrap n
Typically, bootstrap N = sample N Repeat resampling B times, get set of values
Issue: what if, across 200 resamples, 2 of them have ill-defined matrices?
Usually, these are discarded Tests: 5% confidence intervals (want large # of
samples… confidence intervals do not need to be symmetric (can look to value at 95th percentile and at 5th among bootstrapped samples).
More common to compute standard errors
42
Bootstrapping
Overall model X2 correction (available in AMOS).. Bollen and Stine.
Yang and Bentler (chapter in Marcoulides & Schumacker): “faith” in bootstrap based on its
appropriateness in other app’s Simulation study, 1995, if explor. factor
analysis … rotated solutions close, but not so with unrotated solutions
“It seems that in the present stage of development, the use of the bootstrap estimator in covariance structure analysis is still limited. It is not clear whether one can trust the bias estimates.”
43
Bootstrapping
Ichikawa and Konishi, 1995 When data multinormal, bootstrap se’s not as
good as ML Bootstrap doesn’t seem to work when N<150
consistent overestimation (at N=300, not a problem though).
44
The Categorical Variable Model
Conceptual background:
We observe y interested in latent y*with C discrete valuesYi = Ci – 1 if vi,ci-1 < yi* where v is a threshhold
Yi = Ci – 2 if vi,ci-2 < yi* ≤ vi,ci-1
Yi = Ci – 3 if vi,ci-3 < yi* ≤ ≤ vi,ci-2
…..1 If v1,1 if vi,1 < yi
* ≤ vi,2
0 if yi* ≤ vi,1 v’s are threshhold parameters
to be estimated.
45
The Categorical Variable Model
Observed and Latent CorrelationsX-variable scale y-variable scale Observed correl. Latent corr.
Continuous continuous pearson pearson
Contiuous categorical pearson polyserial
Continuous dichtoomous point-biserial biserial
Categorical categorical pearson polychoric
Dichotomous dichotomous phi tetrachoric
If it is reasonable to assume that continuous and normally distributed y* variables underlie the categorical y variables… a variety of latetn correlations can be specified.
46
The Categorical Variable Model
If it is reasonable to assume that continuous and normally distributed y* variables underlie the categorical y variables… a variety of latetn correlations can be specified.
First step: estimate thresholds using MLSecond step: latent correlations estimatedThird step: obtain a consistent estimator of the asymptotic
covariance matrix of the latent correlations (for use in a weighted least squares estimator in the SEM model).
Extreme case: ability to recover y* model when variables split into 25%/75% dichotomies: promising (though X2 underestimated)