Introduction to Structural Equation Modeling with Latent Variables

The path diagram for the one-factor model with parallel tests is shown in Figure 29.

Figure 29: H1: One-Factor Model with Parallel Tests for Lord Data

The hypothesis differs from in that F1 and F2 have a perfect correlation in . This is indicated by the fixed value 1.0 for the double-headed path that connects F1 and F2 in Figure 29. Again, you need only minimal modification of the preceding specification for to specify the path diagram in Figure 29, as shown in the following statements:

proc calis data=lord;
   path
      W <=== F1   = beta1,
      X <=== F1   = beta1,
      Y <=== F2   = beta2,
      Z <=== F2   = beta2;
   pvar
      F1  = 1.0,
      F2  = 1.0,
      W X = 2 * theta1,
      Y Z = 2 * theta2;
   pcov
      F1 F2 = 1.0;
run;

The only modification of the preceding specification is in the PCOV statement, where you put a constant 1 for the covariance between F1 and F2. An annotated fit summary is displayed in Figure 30.

The chi-square value is 37.3337 (df=6, p<0.0001). This indicates that you can reject the hypothesized model H1 at the 0.01 -level. The standardized root mean square error (SRMSR) is 0.0286, the adjusted GFI (AGFI) is 0.9509, and Bentler’s comparative fit index is 0.9785. All these indicate good model fit. However, the RMSEA is 0.0898, which does not support an acceptable model for the data.

The estimation results are displayed in Figure 31.

The goodness-of-fit tests for the four hypotheses are summarized in the following table.

	Number of		Degrees of
Hypothesis	Parameters		Freedom	p-value
	4	37.33	6	< .0001	1.0
	5	1.93	5	0.8583	0.8986
	8	36.21	2	< .0001	1.0
	9	0.70	1	0.4018	0.8986

Recall that the estimates of for and are almost identical, about 0.90, indicating that the speeded and unspeeded tests are measuring almost the same latent variable. However, when was set to 1 in and (both one-factor models), both hypotheses were rejected. Hypotheses and (both two-factor models) seem to be consistent with the data. Since is obtained by adding four constraints (for the requirement of parallel tests) to (the full model), you can test versus by computing the differences of the chi-square statistics and their degrees of freedom, yielding a chi-square of 1.23 with four degrees of freedom, which is obviously not significant. In a sense, the chi-square difference test means that representing the data by would not be significantly worse than representing the data by . In addition, because offers a more precise description of the data (with the assumption of parallel tests) than , it should be chosen because of its simplicity. In conclusion, the two-factor model with parallel tests provides the best explanation of the data.