Confirmatory Factor Analysis of Longitudinal Data

David A. Kenny

December 23. 2013

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TaskSame set of measures that form a latent variables are measured at two or more times on the same sample.

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Example DataDumenci, L., & Windle, M.  (1996).

Multivariate Behavioral Research, 31, 313-330. Depression with four indicators (CESD)

              PA: Positive Affect (lack thereof)               DA: Depressive Affect

    SO: Somatic Symptoms               IN: Interpersonal Issues Four times separated by 6 months 433 adolescent females Age 16.2 at wave 1  

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Equal Loadings Over Time• Want to test that the factor loadings

are the same at all times.• If the loadings are the same, then it

becomes more plausible to argue that one has the same construct at each time.

• Many longitudinal models requires temporally invariant loadings.

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Correlated Measurement Error

• Almost always with longitudinal data, the errors of measurement of an indicator (technically called uniquenesses) should be correlated.

• To be safely identified, at least, three indicators are needed.

• Identified with just two indicators, but must assume the loadings are equal (i.e., both set to one).

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Equal Error Variances• Another possibility is that error

variance of the same measure at different times are equal.

• As in the example with four indicators at four times, each indicator would have one error variance for each of the four times, a total of 12 constraints.

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Equal Loadings

Correlated Errors

Equal Error Variances

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Latent Variable Measurement Models

Model ² df RMSEA ² diffdf

diff pComparison

Model

I No Correlated Errors 856.729 98 0.135       

II Correlated Errors (CE) 107.718 74 0.032 749.010 24 >.001 I

III CE and Equal Loadings (EL) 123.657 83 0.034 15.938  9  .068 II

IVCE, EL, and Equal Error Variances 143.645 95 0.034 19.998 12  .067 III

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Conclusion• Definitely need correlated errors in the

model (something that will almost always be the case).

• Forcing equal loadings, while worsening the fit some, seems reasonable in this case.

• Equal error variances is also reasonable.

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Means of a Latent Variable

• Fix the intercept of the marker variable at each time to zero.

• Free the other intercepts but set them equal over time; a total of (k – 1)(T – 1) constraints.– 9 constraints for the example dataset

• Free factor means and see if the model fits.

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Equality of the Means of a Latent Variable

• Assuming good fit of a model with latent means, fix the factor means (m1 = m2 = m3 = m4) to be equal to test the equality of factor means; T – 1 df.

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Equal Intercepts

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Example Means: Latent Variable, Base Model

• Model with No Constraints on the Means– 2(83) = 123.66, p = .003– RMSEA = 0.034; TLI = .985

• Model with Latent Means and Constraints on the Intercepts– 2(92) = 157.49, p = .003– RMSEA = 0.041; TLI = .979

• Fit is worse with the constraints, but the model fit (RMSEA and TLI) are acceptable.

• Can test if means differ.

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Example Means: Latent Variable

• The four means: 25.34, 25.82, 21.72, 20.09 • Base Model

– 2(92) = 157.49, p = .003– RMSEA = 0.041; TLI = .979

• Equal Latent Means– 2(95) = 182.94, p < .001– RMSEA = 0.046; TLI = .973

• Test of the null hypothesis of equal variance:2(3) = 25.44, p < .001

• Conclusion: Means differ.

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Equal Variance: Latent Variable

• Fix the T factor variances to be equal (s1 = s2 = s3 = s4).

• Compare this model to a model in which factor variances are free to vary with T – 1 df.

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Example: Latent Variable• The four latent variances: 25.34, 25.82, 21.72, 20.09 • Base Model

– 2(83) = 123.66, p = .003– RMSEA = 0.034; TLI = .985

• Equal Variances– 2(86) = 133.43, p = .001– RMSEA = 0.036; TLI = .984

• Test of the null hypothesis of equal variance:– 2(3) = 9.76, p = .021

• Variances significantly different, but model fit is not all that different from the base model.