AMOS, like other SEM packages offers standardized and unstandardized estimates of the parameters. What is the difference? Are unstandardized estimates based on the covariance matrix and standarized on the correlation matrix? Or is it something else?


I don't think the currently accepted answer is correct. What it describes is identification, not standardization. The unstandardized coefficients is what comes directly out of the estimation procedure. The standardized coefficients recast regression coefficients and covariance in metrics of correlations, and unique variances, in terms of $R^2$. So if you have a confirmatory factor analysis model $$ y_j = \alpha_j + \lambda_j \xi + \delta_j $$ with ${\rm E}\delta_j=0$, ${\rm E}\delta_j^2 = \psi_j$, ${\rm E}\xi=0$, ${\rm E}\xi^2=\phi$ in the standard SEM notation, then we have ${\rm Var}[y_j] = \lambda_j^2 \phi + \psi_j$, and the standardized coefficients are: $\tilde\alpha_j-$irrelevant; $$ \tilde\lambda_j = {\rm Corr}(y_j,\xi) = \lambda_j {\rm Var}^{1/2}[\xi] {\rm Var}^{-1/2}[y_j]=\frac{\lambda_j \phi^{1/2}}{\sqrt{\lambda_j^2 \phi + \psi_j}} $$ $$ \tilde \psi_j = \frac{\psi_j}{\lambda_j^2 \phi + \psi_j} $$ Unlike the ``raw'' estimates, standardized estimates and standard errors do not depend on particular parameterization and the choice of the identifying scale parameter (i.e., whether the model is identified by setting $\phi=1$ or one of $\lambda_j=1$ -- and is then independent of the choice of the particular scaling variable, provided $\lambda_j\neq0$ in population).


Two weeks later, I see that no one has answered the question. However extensive Google searching revealed the following.

Assume multivariate data where the $p$ variables supposedly indicate the presence of one or more underlying factors. A simple measurement model involves relationships of the following type:

$$x_i=\lambda_i \xi + \epsilon_i$$

for common factor $\xi$ and uniqueness factor (i.e. noise) $\epsilon_i$. Several such relationships may exist, depending on the number of common factors. None of the "Greeks" are observable, but must be inferred from the data. The problem as stated does not have a unique solution. If correlation $\lambda_i$ and $\xi$ are both solutions, then so are the scaled quantities $a \lambda_i$ and $\xi / a$.

For identifiability, one of $\lambda_i$ and $\xi$ must be fixed -- typically to the value 1.

When the correlation $\lambda_i$ is set to 1, the solutions are said to be "unstandardized". When the common factor $\xi$ is set to one, the solutions are said to be "standardized."

Note that only one $\lambda_i$ need to standardized amongst the correlations associated with a given factor. So if in addition to the model above, I also have:

$$x_j=\lambda_j \xi + \epsilon_j$$

for another variable $x_j$, then only one of the lambda's need be set to 1 --- or equivalently, the common factor.

AMOS, by default, standardizes one of the correlation parameters, but one can request that the common factors be standardized instead.

  • $\begingroup$ +1 very nice. I had my own explanation for the terms, based on R packages, but you nailed it much more concretely. $\endgroup$ – Wayne Apr 1 '13 at 22:01

I think from my simple experience in Amos that unstandardized we request one of the parameters should be 1 to give as the result of factor loading to help determine which parameter needs to dropped first and to keep many parameter that need to be estimated. According to standardized: Here we got the real value of parameter.

  • $\begingroup$ Welcome to the site, @Najat. I do not understand your post. I tried to edit it for English. See if you can make your point clearer. $\endgroup$ – gung - Reinstate Monica Mar 8 '14 at 3:27
  • $\begingroup$ Najat, I think you've got unstandardized and standardized mixed up. Also, the difference shouldn't matter for model selection. I'll try to see if I can offer another answer here soon... $\endgroup$ – Nick Stauner Mar 8 '14 at 3:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.