testlet.marginalized {sirt} | R Documentation |
Marginal Item Parameters from a Testlet (Bifactor) Model
Description
This function computes marginal item parameters of a general factor if item parameters from a testlet (bifactor) model are provided as an input (see Details).
Usage
testlet.marginalized(tam.fa.obj=NULL,a1=NULL, d1=NULL, testlet=NULL,
a.testlet=NULL, var.testlet=NULL)
Arguments
tam.fa.obj |
Optional object of class |
a1 |
Vector of item discriminations of general factor |
d1 |
Vector of item intercepts of general factor |
testlet |
Integer vector of testlet (bifactor) identifiers (must be integers
between 1 to |
a.testlet |
Vector of testlet (bifactor) item discriminations |
var.testlet |
Vector of testlet (bifactor) variances |
Details
A testlet (bifactor) model is assumed to be estimated:
P(X_{pit}=1 | \theta_{p}, u_{pt} )=
invlogit( a_{i1} \theta_p + a_t u_{pt} - d_{i} )
with Var( u_{pt} )=\sigma_t^2
. This multidimensional
item response model with locally independent items is
equivalent to a unidimensional IRT model with locally
dependent items (Ip, 2010). Marginal item parameters a_i^\ast
and d_i^\ast
are obtained according to the response
equation
P(X_{pit}=1 | \theta_{p}^\ast )=
invlogit( a_{i}^\ast \theta_p^\ast - d_{i}^\ast )
Calculation details can be found in Ip (2010).
Value
A data frame containing all input item parameters and
marginal item intercept d_i^\ast
(d1_marg
) and
marginal item slope a_i^\ast
(a1_marg
).
References
Ip, E. H. (2010). Empirically indistinguishable multidimensional IRT and locally dependent unidimensional item response models. British Journal of Mathematical and Statistical Psychology, 63, 395-416.
See Also
For estimating a testlet (bifactor) model see
TAM::tam.fa
.
Examples
#############################################################################
# EXAMPLE 1: Small numeric example for Rasch testlet model
#############################################################################
# Rasch testlet model with 9 items contained into 3 testlets
# the third testlet has essentially no dependence and therefore
# no testlet variance
testlet <- rep( 1:3, each=3 )
a1 <- rep(1, 9 ) # item slopes first dimension
d1 <- rep( c(-1.25,0,1.5), 3 ) # item intercepts
a.testlet <- rep( 1, 9 ) # item slopes testlets
var.testlet <- c( .8, .2, 0 ) # testlet variances
# apply function
res <- sirt::testlet.marginalized( a1=a1, d1=d1, testlet=testlet,
a.testlet=a.testlet, var.testlet=var.testlet )
round( res, 2 )
## item testlet a1 d1 a.testlet var.testlet a1_marg d1_marg
## 1 1 1 1 -1.25 1 0.8 0.89 -1.11
## 2 2 1 1 0.00 1 0.8 0.89 0.00
## 3 3 1 1 1.50 1 0.8 0.89 1.33
## 4 4 2 1 -1.25 1 0.2 0.97 -1.21
## 5 5 2 1 0.00 1 0.2 0.97 0.00
## 6 6 2 1 1.50 1 0.2 0.97 1.45
## 7 7 3 1 -1.25 1 0.0 1.00 -1.25
## 8 8 3 1 0.00 1 0.0 1.00 0.00
## 9 9 3 1 1.50 1 0.0 1.00 1.50
## Not run:
#############################################################################
# EXAMPLE 2: Dataset reading
#############################################################################
library(TAM)
data(data.read)
resp <- data.read
maxiter <- 100
# Model 1: Rasch testlet model with 3 testlets
dims <- substring( colnames(resp),1,1 ) # define dimensions
mod1 <- TAM::tam.fa( resp=resp, irtmodel="bifactor1", dims=dims,
control=list(maxiter=maxiter) )
# marginal item parameters
res1 <- sirt::testlet.marginalized( mod1 )
#***
# Model 2: estimate bifactor model but assume that items 3 and 5 do not load on
# specific factors
dims1 <- dims
dims1[c(3,5)] <- NA
mod2 <- TAM::tam.fa( resp=resp, irtmodel="bifactor2", dims=dims1,
control=list(maxiter=maxiter) )
res2 <- sirt::testlet.marginalized( mod2 )
res2
## End(Not run)