 Interpreting standardized thresholds    Message/Author  Jim Prisciandaro posted on Sunday, May 09, 2010 - 11:18 pm
Hello,

I was wondering how one interprets the values of standardized thresholds (i.e., makes a descriptive statement about a given threshold). I would specifically like to be able to interpret them in terms of probability and/or in relation to item difficulty parameters in IRT.

For this example, I'm estimating a bifactor model with MLR. The thresholds I'm looking at are in the stdyx section.

Thanks,
Jim  Linda K. Muthen posted on Monday, May 10, 2010 - 8:02 am
The raw thresholds are used to compute probabilities and for translation to IRT parameters. See the IRT technical information on the website. Computing probabilities is described in Chapter 14.  Jim Prisciandaro posted on Monday, May 10, 2010 - 10:27 am
Hi Linda,

I'm afraid I'm going to need a little more handholding here.

I looked at the resources you suggested. In terms of the IRT technical information, all I saw was how thresholds are parameterized in IRT when there is a single factor. Does this generalize to the case where there is 1 general factor and 3 specific factors (i.e., bifactor model). If so, where in my output would I find the parameters mentioned in the formula (formula 19, page 5, MPlusIRT1.pdf); I'm afraid I don't know what parameters the symbols are referring to in general?

In terms of chapter 14, is the information I am looking for in "calculating probabilities from logistic regression coefficients" (p. 441)? I also noticed the section on pp. 412-413 ("latent class indicators") might be relevant; but I am looking for probabilities corresponding to threshold values, not descriptive explanations for the thresholds (I'm guessing the descriptives are referring to unstandardized thresholds, but I'm not sure).

A general question: can I get the information I am looking for using standardized thresholds or can I only use unstandardized thresholds? Do the standardized thresholds have any inherent meaning?

A worked through example would be tremendously helpful, but any guidance is appreciated.

Thanks,
Jim  Linda K. Muthen posted on Tuesday, May 11, 2010 - 9:52 am
The formulas can be generalized. Lambda are factor loadings. Tau are thresholds. Psi are the variances and covariances of the factors. Alpha are the factor means.

Yes, you should look at calculating probabilities from logistic regression coefficients. It shows how to compute probabilities using information from the output.

You should use the unstandardized thresholds. The standardized thresholds are not useful. Note that thresholds and intercepts differ only by their sign.  Jim Prisciandaro posted on Friday, May 14, 2010 - 2:56 pm
To help understand your response I searched through the forum some more and found Bengt's comment from 10/09:

"when you run Mplus with the factor standardized to zero mean and unit variance, a comparison of (1) and (2) gives
(3) a_j = lambda_j/D,
(4) b_jk = tau_jk/lambda_j"

and Tom's comment from 10/2008:

"The MPLUS loadings under MLR were the same as those estimated by multi-log. To get the thresholds, the transformation is MPLUS Threshold/factor loading=Multi-log threshold."

I am working with a bifactor model using MLR, does the standardized output set the factor mean to zero and the variance to 1 as per Bengt's comment?

It appears from these above comments that it is unnecessary to incorporate information regarding factor means, variances and covariances (tech4) when translating MPlus factor loadings and thresholds into discrimination and difficulty parameters; that is that one can translate these parameters just using the loadings and thresholds in the standard output. Am I mistaken?  Jim Prisciandaro posted on Friday, May 14, 2010 - 2:57 pm
(continued from above)....

1) Is the loading for a given item in Mplus the same as the IRT discrimination parameter for that item as Tom suggests? If so, is it the unstandardized or the standardized loading that is the same as the discrimination parameter in IRT?

2) Bengt suggested we have to divide by 1.7 to obtain the discrimination parameter whereas Tom suggested we do not. Should we divide the loading by 1.7 or not to obtain the discrimination parameter?

Thanks,
Jim  Linda K. Muthen posted on Saturday, May 15, 2010 - 8:10 am
Slides 93 and 94 of the Topic 2 course handout gives the formulas and shows how they relate to Mplus. I think these should answer your questions.  Jim Prisciandaro posted on Saturday, May 15, 2010 - 9:31 am
Thanks Linda,

1) My understanding now is that if I explicitly set the factor means to 0 and the factor variances to 1 in my input, I can use the simplified formulas Bengt provided above using unstandardized output.

2) If instead I freely estimate the factor means and variances, I have to apply the formulas that include Psi and Alpha on the unstandardized output (and obtain that info via Tech4).

Is my understanding correct on points 1 and 2 above?

3) If I'm understanding these points correctly, my final point of confusion is that you noted that Psi represents both factor variances and covariances. How then do I obtain a single number for Psi (variances and covariances) to plug into the formula to obtain the difficulty parameter for a given item?

Jim  Linda K. Muthen posted on Sunday, May 16, 2010 - 10:08 am
1. Yes.

2. Yes.

3. Ask for TECH4 in the output command. The variances are on the diagonal and the covariances are on the off-diagonal.  Jim Prisciandaro posted on Sunday, May 16, 2010 - 1:30 pm
Thanks Linda,

Almost there...

1) Regarding point 3, my question was: the formula for the difficulty parameter calls for calculation of the square root of psi, but psi isn't a single number, right?

2) Are we supposed to calculate the square root of the psi matrix (the matrix of variances and covariances) for that formula?

3) If yes, is it possible to calculate the square root of psi in mplus?

Thanks,
Jim  Linda K. Muthen posted on Monday, May 17, 2010 - 9:58 am
With one factor, psi is a single number, the variance of the factor. I think you are best off freeing all factor loadings and fixing the factor variance to one. Then you don't have to deal with psi. This parametrization is the one used most often in IRT.  Jim Prisciandaro posted on Monday, May 17, 2010 - 11:13 am
This makes sense. As a final checking in:

1) If I fixed the variances of the factors to 1, my formulas would be:

a_j = lambda_j/D
b_jk = tau_jk-(lambda_j*alpha)/lambda_j

2) If I fixed the variances of the factors to 1 AND the means of the factors to 0, my formulas would be:

a_j = lambda_j/D,
b_jk = tau_jk/lambda_j

3) And I would use the unstandardized output for all of these transformations, right?

Another very important question to me:

4) If one compares a model with a) factor variances freely estimated and the loading of the first item per factor constrained to "1" and b) the same model with factor variances set to "1" and all loadings freely estimated, would the fit of these 2 models (and AIC/BIC) be identical to one another or could they be different?

Thanks again,
Jim  Jim Prisciandaro posted on Monday, May 17, 2010 - 11:20 am
Sorry to put this on a separate post, but I was just doing further reading of the forum and:

5) Does Mplus automatically fix the factor means to "0"?

6) If so, then if I manually fix the factor variances to 1, the formulas I use should be:

a_j = lambda_j/D,
b_jk = tau_jk/lambda_j

because the factor means are "0" by default, right?  Linda K. Muthen posted on Tuesday, May 18, 2010 - 9:02 am
Yes. This all sounds correct.    Topics | Tree View | Search | Help/Instructions | Program Credits Administration