Local depedence for polytomous data

261 views
Skip to first unread message

Nikolaos Tsigilis

unread,
May 27, 2020, 3:14:26 AM5/27/20
to mirt-package
Dear Phil,

I am fitting a unidimentional questionnaire applying the graded response model (itemtype="graded").  I am checking the local depedency assumption using the residuals(mydata, type = "LD", table=TRUE).  However, I cannot understand the output.  Moreover, in Paek & Cole textbook "Using R for  Item Response Theory Model Applications" it is suggested that we can standardize the residuals using (abs(res)-(k-1)*(m-1))/sqrt(2*(k-1)*(m-1)) formula. Do you suggest doing so? 

Thanks 
Nikos

Phil Chalmers

unread,
May 27, 2020, 10:18:37 AM5/27/20
to Nikolaos Tsigilis, mirt-package
That's one version of a standardization, but I prefer using the signed Cramer's V standardization instead since it doesn't require making asymptotic normal transformation approximations for chi-squared distributions. That's what mirt's residual() function prints in the upper diagonal portion. Also, table = TRUE is not a supported argument to itemfit(), so I'm not sure what you're seeing.

Phil


--
You received this message because you are subscribed to the Google Groups "mirt-package" group.
To unsubscribe from this group and stop receiving emails from it, send an email to mirt-package...@googlegroups.com.
To view this discussion on the web visit https://groups.google.com/d/msgid/mirt-package/77c15d12-53c4-4354-bcd3-a0919151dad3%40googlegroups.com.

Phil Chalmers

unread,
May 27, 2020, 10:24:56 AM5/27/20
to Nikolaos Tsigilis, mirt-package
Sorry, I had the wrong documentation open (multiple window problem). In residuals(), tables = TRUE gives the observed and expected counts for the tables, as well as the standardized residuals based on the usual chi-squared residual form (O - E) / √E

Phil

Nikolaos Tsigilis

unread,
May 28, 2020, 10:11:22 AM5/28/20
to mirt-package
Thank you very mauch Phil for your response.  It is very much appreciated.  From your exprerience or readings is there a particular value over which Cramer's V indicates a local depedency problem, or you interpret Cramer's V values relatively?

Nikos 


Τη Τετάρτη, 27 Μαΐου 2020 - 5:24:56 μ.μ. UTC+3, ο χρήστης Phil Chalmers έγραψε:
Sorry, I had the wrong documentation open (multiple window problem). In residuals(), tables = TRUE gives the observed and expected counts for the tables, as well as the standardized residuals based on the usual chi-squared residual form (O - E) / √E

Phil


On Wed, May 27, 2020 at 10:17 AM Phil Chalmers <rphilip...@gmail.com> wrote:
That's one version of a standardization, but I prefer using the signed Cramer's V standardization instead since it doesn't require making asymptotic normal transformation approximations for chi-squared distributions. That's what mirt's residual() function prints in the upper diagonal portion. Also, table = TRUE is not a supported argument to itemfit(), so I'm not sure what you're seeing.

Phil


On Wed, May 27, 2020 at 3:14 AM Nikolaos Tsigilis <ntsi...@gmail.com> wrote:
Dear Phil,

I am fitting a unidimentional questionnaire applying the graded response model (itemtype="graded").  I am checking the local depedency assumption using the residuals(mydata, type = "LD", table=TRUE).  However, I cannot understand the output.  Moreover, in Paek & Cole textbook "Using R for  Item Response Theory Model Applications" it is suggested that we can standardize the residuals using (abs(res)-(k-1)*(m-1))/sqrt(2*(k-1)*(m-1)) formula. Do you suggest doing so? 

Thanks 
Nikos

--
You received this message because you are subscribed to the Google Groups "mirt-package" group.
To unsubscribe from this group and stop receiving emails from it, send an email to mirt-p...@googlegroups.com.

Phil Chalmers

unread,
Jun 14, 2020, 9:48:43 PM6/14/20
to Nikolaos Tsigilis, mirt-package
I'm hesitant to report hard cut-offs as there are many things that come into play here (degree of multidimensional misspecification, directionality of the violation, number of 'packets' showing residual covariation, etc). Interpreting the values relatively is fine, but the values themselves do follow a continuum of severity, so in that sense anything different from values close to 0 could be problematic depending on the application.

Phil


To unsubscribe from this group and stop receiving emails from it, send an email to mirt-package...@googlegroups.com.
To view this discussion on the web visit https://groups.google.com/d/msgid/mirt-package/f4ece047-d953-49cc-9659-f19e219cf978%40googlegroups.com.

lee...@umich.edu

unread,
Nov 25, 2020, 12:22:54 AM11/25/20
to mirt-package
Hello Phil, 

I have been using Cramer's V standardization to evaluate local independence for a stress scale (all items intended to capture stress). I have, however, come across an interesting issue in that many of the Cramer's V coefficients range from .10-.20 (negative or positive) -- which some folks have said were too high, requiring a multidimensional model. What's strange is when I do other tests (parallel analysis, CFA, bifactor EFA), the results seem to suggest that there is a dominating single factor with some, but a marginal amount of multidimensionality (e.g., ECV in the bifactor CFA was near .90). I am not sure how to proceed and would deeply appreciate your insights. Thank you for developing MIRT and for your wisdom in this field.

Dan

Phil Chalmers

unread,
Dec 1, 2020, 2:58:50 PM12/1/20
to lee...@umich.edu, mirt-package
Hi Dan,

In the limiting case, all tests are multidimensional in that each item really represents its own unique factor, though this is useless for understanding the empirical world. On the other hand, a one factor model effectively tries as hard as possible to force all items to be independent after removing one common axis of variation, which could fail if the items have more in common than just a single dominating factor. The local dependence measures like Cramer's V really are just present to detect whether there are any bivariate residual relationships present between items that could be a problem, and might be worth modeling. There's no guarantee that there is more 'in common' over and above bivariate comparisons, but if a common structure can be identified empirically and rationally then a multidimensional model should be considered more plausible. Note that lack of detecting more than 1 factor in the data does suggest that there probably are no other dominant traits that should be modeled (largely a function of statistical/measurement power), but it doesn't patch the problem of residual dependencies between items. 

In your case, I might consider fitting a bifactor model with specific bivariate testlet effects (with slopes constrained to be equal to each other for identification) to account for the covariation between residuals, and thereby avoid the bias that could be invoked by these bivariate dependencies in the general trait of interest. At the very least this would address the residual covariation in an attempt to unbias the quasi-unidimensional trait, though of course doing so is not always free (costs degrees of freedom, and suggests a specific, modelable relationship between residuals). HTH. 

Phil


Reply all
Reply to author
Forward
0 new messages