Interconstruct Correlation Greater than 1

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Chris Castille

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Mar 28, 2017, 9:14:51 AM3/28/17

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Hi folks,

I’m testing a latent variable model that involves two latent personality factors (a ‘stability’ and a ‘plasticity’ factor) as underlying scores on a personality inventory that measures basic personality traits (the Big Five, or agreeableness, neuroticism, conscientiousness, openness, and extraversion). Basically, agreeableness, neuroticism, and conscientiousness are theorized as reflecting stability while extraversion and openness reflect plasticity. To ensure identification, I’ve constrained the factor loadings for extraversion and openness to be equal to one another (note: I’ve also tested free factor loadings and this doesn’t resolve my issue).

My issue is that the factor correlations (stability and plasticity) are greater than 1 when estimated freely. In other words, I have a non-positive definite matrix. I'm looking for advice for overcoming this issue.

To help diagnose this issue, I do have data on a measure of halo. Thinking that the two traits may be better represented as a general factor that could be interpreted as a halo factor, I correlated the two and, while this correlation is significant, it is too small to be a halo factor (r ~= .30).

Note that all indicators in the model are theta scores obtained from applying scoring algorithms to the scale scores as part of a separate aspect to this investigation.

My R script describing the model is below. Data are attached. Any help would be appreciated.

modl1 <- '

#Model meta-traits

Stability=~A+C+N

Plasticity=~v1*E+v1*O

#Model halo

A+C+N+E+O ~ bias

#Link Big Five traits to 10 Aspects

p+c ~ A

i+o ~ C

v+w ~ N

e+ass ~ E

int+a ~ O

‘

modl <- cfa(modl1, estimator = "MLM", data=HABM, std.lv = TRUE)

summary(modl, standardized = TRUE, fit.measures = TRUE)

The output is below:

lavaan WARNING: covariance matrix of latent variables is not positive definite; use inspect(fit,"cov.lv") to investigate.> summary(modl, standardized = TRUE, fit.measures = TRUE)

lavaan (0.5-20) converged normally after 151 iterations

Used Total

Number of observations 610 632

Estimator ML Robust

Minimum Function Test Statistic 615.868 407.954

Degrees of freedom 55 55

P-value (Chi-square) 0.000 0.000

Scaling correction factor 1.510

for the Satorra-Bentler correction

Model test baseline model:

Minimum Function Test Statistic 12198.485 8515.398

Degrees of freedom 120 120

P-value 0.000 0.000

User model versus baseline model:

Comparative Fit Index (CFI) 0.954 0.958

Tucker-Lewis Index (TLI) 0.899 0.908

Loglikelihood and Information Criteria:

Loglikelihood user model (H0) -7888.474 -7888.474

Loglikelihood unrestricted model (H1) -7580.540 -7580.540

Number of free parameters 95 95

Akaike (AIC) 15966.948 15966.948

Bayesian (BIC) 16386.227 16386.227

Sample-size adjusted Bayesian (BIC) 16084.623 16084.623

Root Mean Square Error of Approximation:

RMSEA 0.129 0.103

90 Percent Confidence Interval 0.120 0.139 0.095 0.110

P-value RMSEA <= 0.05 0.000 0.000

Standardized Root Mean Square Residual:

SRMR 0.077 0.077

Parameter Estimates:

Information Expected

Standard Errors Robust.sem

Latent Variables:

Estimate Std.Err Z-value P(>|z|) Std.lv Std.all

Stability =~

A 0.581 0.048 12.136 0.000 0.581 0.573

C 0.687 0.043 16.019 0.000 0.687 0.681

N -0.602 0.049 -12.388 0.000 -0.602 -0.558

Plasticity =~

E (v1) 0.458 0.044 10.402 0.000 0.458 0.439

O (v1) 0.458 0.044 10.402 0.000 0.458 0.490

Regressions:

Estimate Std.Err Z-value P(>|z|) Std.lv Std.all

A ~

bias 0.203 0.052 3.881 0.000 0.203 0.173

C ~

bias 0.432 0.043 10.105 0.000 0.432 0.372

N ~

bias -0.411 0.053 -7.824 0.000 -0.411 -0.331

E ~

bias 0.492 0.046 10.577 0.000 0.492 0.409

O ~

bias 0.392 0.045 8.653 0.000 0.392 0.365

p ~

A 0.711 0.025 28.878 0.000 0.711 0.757

c ~

A 0.708 0.053 13.427 0.000 0.708 0.770

i ~

C 0.847 0.021 40.336 0.000 0.847 0.927

o ~

C 0.814 0.020 41.592 0.000 0.814 0.838

v ~

N 0.884 0.021 41.347 0.000 0.884 0.904

w ~

N 0.880 0.017 53.161 0.000 0.880 0.914

e ~

E 0.902 0.016 57.865 0.000 0.902 0.909

ass ~

E 0.793 0.017 46.682 0.000 0.793 0.846

int ~

O 0.790 0.026 30.779 0.000 0.790 0.727

a ~

O 0.843 0.016 51.870 0.000 0.843 0.887

Covariances:

Estimate Std.Err Z-value P(>|z|) Std.lv Std.all

Stability ~~

Plasticity 1.158 0.087 13.295 0.000 1.158 1.158

p ~~

c -0.051 0.020 -2.496 0.013 -0.051 -0.137

i 0.001 0.009 0.147 0.883 0.001 0.006

o 0.007 0.014 0.462 0.644 0.007 0.020

v -0.001 0.014 -0.062 0.950 -0.001 -0.003

w -0.023 0.018 -1.247 0.212 -0.023 -0.087

e 0.020 0.012 1.674 0.094 0.020 0.073

ass -0.031 0.017 -1.821 0.069 -0.031 -0.094

int 0.028 0.018 1.514 0.130 0.028 0.064

a -0.018 0.011 -1.635 0.102 -0.018 -0.069

c ~~

i 0.009 0.011 0.863 0.388 0.009 0.045

o -0.011 0.016 -0.683 0.494 -0.011 -0.034

v -0.042 0.013 -3.150 0.002 -0.042 -0.154

w 0.027 0.011 2.441 0.015 0.027 0.108

e -0.014 0.010 -1.369 0.171 -0.014 -0.054

ass -0.011 0.014 -0.779 0.436 -0.011 -0.034

int -0.054 0.020 -2.691 0.007 -0.054 -0.131

a 0.029 0.011 2.732 0.006 0.029 0.119

i ~~

o -0.146 0.011 -13.363 0.000 -0.146 -0.791

v 0.012 0.007 1.674 0.094 0.012 0.079

w -0.008 0.008 -1.111 0.266 -0.008 -0.057

e -0.034 0.008 -4.115 0.000 -0.034 -0.231

ass 0.039 0.009 4.607 0.000 0.039 0.219

int 0.068 0.012 5.821 0.000 0.068 0.285

a -0.023 0.007 -3.060 0.002 -0.023 -0.161

o ~~

v -0.023 0.014 -1.638 0.101 -0.023 -0.095

w 0.016 0.013 1.210 0.226 0.016 0.071

e 0.048 0.010 4.716 0.000 0.048 0.207

ass -0.055 0.014 -4.044 0.000 -0.055 -0.196

int -0.107 0.020 -5.277 0.000 -0.107 -0.288

a 0.045 0.012 3.730 0.000 0.045 0.206

v ~~

w -0.125 0.014 -8.631 0.000 -0.125 -0.656

e -0.007 0.012 -0.626 0.531 -0.007 -0.037

ass 0.035 0.013 2.605 0.009 0.035 0.147

int 0.047 0.016 3.044 0.002 0.047 0.151

a -0.013 0.008 -1.523 0.128 -0.013 -0.070

w ~~

e 0.005 0.010 0.512 0.608 0.005 0.028

ass -0.020 0.015 -1.333 0.183 -0.020 -0.092

int -0.026 0.015 -1.674 0.094 -0.026 -0.087

a 0.009 0.008 1.135 0.256 0.009 0.050

e ~~

ass -0.152 0.017 -8.970 0.000 -0.152 -0.676

int -0.116 0.014 -8.201 0.000 -0.116 -0.387

a 0.057 0.008 6.696 0.000 0.057 0.321

ass ~~

int 0.137 0.019 7.144 0.000 0.137 0.377

a -0.068 0.010 -6.473 0.000 -0.068 -0.317

int ~~

a -0.225 0.019 -12.138 0.000 -0.225 -0.789

Intercepts:

Estimate Std.Err Z-value P(>|z|) Std.lv Std.all

A 0.097 0.041 2.400 0.016 0.097 0.096

C 0.203 0.038 5.334 0.000 0.203 0.202

N -0.151 0.041 -3.647 0.000 -0.151 -0.140

E 0.062 0.038 1.621 0.105 0.062 0.059

O 0.060 0.036 1.680 0.093 0.060 0.064

p -0.013 0.025 -0.504 0.614 -0.013 -0.013

c -0.010 0.021 -0.453 0.651 -0.010 -0.010

i 0.012 0.013 0.973 0.330 0.012 0.014

o 0.028 0.021 1.310 0.190 0.028 0.029

v -0.054 0.018 -2.997 0.003 -0.054 -0.051

w 0.026 0.017 1.492 0.136 0.026 0.025

e 0.044 0.017 2.545 0.011 0.044 0.043

ass -0.059 0.021 -2.766 0.006 -0.059 -0.060

int -0.071 0.028 -2.509 0.012 -0.071 -0.070

a 0.051 0.016 3.117 0.002 0.051 0.058

Stability 0.000 0.000 0.000

Plasticity 0.000 0.000 0.000

Variances:

Estimate Std.Err Z-value P(>|z|) Std.lv Std.all

A 0.661 0.072 9.167 0.000 0.661 0.642

C 0.406 0.043 9.447 0.000 0.406 0.398

N 0.673 0.054 12.527 0.000 0.673 0.579

E 0.696 0.068 10.164 0.000 0.696 0.640

O 0.547 0.043 12.761 0.000 0.547 0.627

p 0.388 0.038 10.168 0.000 0.388 0.427

c 0.355 0.051 6.911 0.000 0.355 0.407

i 0.119 0.014 8.532 0.000 0.119 0.140

o 0.286 0.020 14.086 0.000 0.286 0.298

v 0.204 0.021 9.753 0.000 0.204 0.183

w 0.178 0.032 5.591 0.000 0.178 0.165

e 0.186 0.025 7.575 0.000 0.186 0.174

ass 0.272 0.039 6.953 0.000 0.272 0.284

int 0.485 0.030 16.031 0.000 0.485 0.471

a 0.168 0.017 10.102 0.000 0.168 0.213

Stability 1.000 1.000 1.000

Plasticity 1.000 1.000 1.000

HABM.csv

Terrence Jorgensen

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Mar 29, 2017, 9:12:57 AM3/29/17

to lavaan

My issue is that the factor correlations (stability and plasticity) are greater than 1 when estimated freely. In other words, I have a non-positive definite matrix. I'm looking for advice for overcoming this issue.

Have you considered the possibility that your hypothesis is wrong? The fact that the estimated correlation exceeds 1 indicates that at least one pair of first-order factors (indicators from different second-order factors) are more correlated than the simple second-order structure allows for. You could look at the latent covariance matrix from a model with only first-order factors for clues.

Model misfit is only one possible cause of a Heywood case. Another is sampling error. At smaller sample sizes, you can expect Heywood cases if the true parameter is near a boundary. You have a pretty big sample size, though. But you can check whether the null hypothesis is plausible that the true parameter is an acceptable value. Check your CI, and you will notice it includes (barely) includes some plausible correlations below 1, indicating the data are also consistent with a true correlation of, e.g., 0.995. So although you cannot disregard sampling error as the cause of the Heywood case (nor model misfit as the cause), the CI still indicates that if your model is correct, your data are only consistent with a true correlation very near the boundary, in which case you could try a 1-factor solution as well. But I would recommend inspecting the first-order factor correlations for clues as to how your hypothesized 2-factor structure might not be appropriate.

Terrence D. Jorgensen

Postdoctoral Researcher, Methods and Statistics

Research Institute for Child Development and Education, the University of Amsterdam

UvA web page: http://www.uva.nl/profile/t.d.jorgensen

Chris Castille

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Mar 29, 2017, 11:40:07 AM3/29/17

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Hi Terrence,

Thanks for taking this on. Your feedback was helpful.

I have considered this possibility (that the hypothesis is incorrect) and have run a one-factor model. The fit of a one factor model is largely identical, granted there is no Heywood case issue. Huzzah.

One more point, I’m not sure what you mean by 'checking out the first-order factor correlations.’ My indicators are not correlated so strongly as to suggest redundancy and maybe I was not so clear in my first explanation. The indicators in my data are trait scores - there are no item-level indicators in the model. At the lowest level in the hierarchy are trait scores corresponding to 10 Aspects of the Big Five, the latter of which are at the next level in the hierarchy. The intercorrelations appear fine at their respective levels. High intercorrelation across levels is expected because the item responses informing those trait scores are shared.

Still, I’m skeptical, and I think more context would explain why. Research from more rigorous investigations using multi-method/measurement approaches have largely discredited the single-factor as an explanation for common Big Five variance. Through these investigations, it was revealed that the two higher-order factors are positively correlated when ratings are provided by single informants. When ratings are provided by multiple informants, they are typically relatively uncorrelated. When that single factor emerges, multifaceted method variance seems to be the explanation. Now for my data...My data are single-source and self-reported. While I have controlled for halo effects, I think that other sources of method variance that were not captured by my design (e.g., response style) are resulting in the emergence of a single factor. Bummer.

Chris

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Terrence Jorgensen

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Mar 30, 2017, 5:46:00 AM3/30/17

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One more point, I’m not sure what you mean by 'checking out the first-order factor correlations.’ My indicators are not correlated so strongly as to suggest redundancy and maybe I was not so clear in my first explanation.

OK, so it is only a first-order model. In that case correlation residual might point out which relationships your 2-factor solution does not adequately account for:

resid(fit, type = "cor")

Regarding context and lingering skepticism, you might find some personality researchers on SEMNET who could offer useful discussionL