Lagrange (i.e., score) test to test whether parameters should be freed from a more constrained baseline model.
lagrange(mod, parnum, SE.type = "Oakes", type = "Richardson", ...)an estimated model
a vector, or list of vectors, containing one or more parameter
locations/sets of locations to be tested.
See objects returned from mod2values for the locations
type of information matrix estimator to use. See mirt for
further details
type of numerical algorithm passed to numerical_deriv to
obtain the gradient terms
additional arguments to pass to mirt
Chalmers, R., P. (2012). mirt: A Multidimensional Item Response Theory Package for the R Environment. Journal of Statistical Software, 48(6), 1-29. doi:10.18637/jss.v048.i06
# \donttest{
dat <- expand.table(LSAT7)
mod <- mirt(dat, 1, 'Rasch')
(values <- mod2values(mod))
#> group item class name parnum value lbound ubound est
#> 1 all Item.1 dich a1 1 1.0000000 -Inf Inf FALSE
#> 2 all Item.1 dich d 2 1.8680718 -Inf Inf TRUE
#> 3 all Item.1 dich g 3 0.0000000 0e+00 1 FALSE
#> 4 all Item.1 dich u 4 1.0000000 0e+00 1 FALSE
#> 5 all Item.2 dich a1 5 1.0000000 -Inf Inf FALSE
#> 6 all Item.2 dich d 6 0.7909134 -Inf Inf TRUE
#> 7 all Item.2 dich g 7 0.0000000 0e+00 1 FALSE
#> 8 all Item.2 dich u 8 1.0000000 0e+00 1 FALSE
#> 9 all Item.3 dich a1 9 1.0000000 -Inf Inf FALSE
#> 10 all Item.3 dich d 10 1.4608233 -Inf Inf TRUE
#> 11 all Item.3 dich g 11 0.0000000 0e+00 1 FALSE
#> 12 all Item.3 dich u 12 1.0000000 0e+00 1 FALSE
#> 13 all Item.4 dich a1 13 1.0000000 -Inf Inf FALSE
#> 14 all Item.4 dich d 14 0.5214399 -Inf Inf TRUE
#> 15 all Item.4 dich g 15 0.0000000 0e+00 1 FALSE
#> 16 all Item.4 dich u 16 1.0000000 0e+00 1 FALSE
#> 17 all Item.5 dich a1 17 1.0000000 -Inf Inf FALSE
#> 18 all Item.5 dich d 18 1.9927710 -Inf Inf TRUE
#> 19 all Item.5 dich g 19 0.0000000 0e+00 1 FALSE
#> 20 all Item.5 dich u 20 1.0000000 0e+00 1 FALSE
#> 21 all GROUP GroupPars MEAN_1 21 0.0000000 -Inf Inf FALSE
#> 22 all GROUP GroupPars COV_11 22 1.0219437 1e-04 Inf TRUE
#> prior.type prior_1 prior_2
#> 1 none NaN NaN
#> 2 none NaN NaN
#> 3 none NaN NaN
#> 4 none NaN NaN
#> 5 none NaN NaN
#> 6 none NaN NaN
#> 7 none NaN NaN
#> 8 none NaN NaN
#> 9 none NaN NaN
#> 10 none NaN NaN
#> 11 none NaN NaN
#> 12 none NaN NaN
#> 13 none NaN NaN
#> 14 none NaN NaN
#> 15 none NaN NaN
#> 16 none NaN NaN
#> 17 none NaN NaN
#> 18 none NaN NaN
#> 19 none NaN NaN
#> 20 none NaN NaN
#> 21 none NaN NaN
#> 22 none NaN NaN
# test all fixed slopes individually
parnum <- values$parnum[values$name == 'a1']
lagrange(mod, parnum)
#> X2 df p
#> 1 0.36714266 1 0.54456590
#> 5 0.04789243 1 0.82677204
#> 9 4.69717572 1 0.03021223
#> 13 1.44960175 1 0.22859187
#> 17 2.55408012 1 0.11000983
# compare to LR test for first two slopes
mod2 <- mirt(dat, 'F = 1-5
FREE = (1, a1)', 'Rasch')
coef(mod2, simplify=TRUE)$items
#> a1 d g u
#> Item.1 1.158017 1.9393429 0 1
#> Item.2 1.000000 0.7850362 0 1
#> Item.3 1.000000 1.4501782 0 1
#> Item.4 1.000000 0.5175801 0 1
#> Item.5 1.000000 1.9787221 0 1
anova(mod, mod2)
#> AIC SABIC HQ BIC logLik X2 df p
#> mod 5341.802 5352.192 5352.994 5371.248 -2664.901
#> mod2 5343.264 5355.386 5356.321 5377.618 -2664.632 0.538 1 0.463
mod2 <- mirt(dat, 'F = 1-5
FREE = (2, a1)', 'Rasch')
coef(mod2, simplify=TRUE)$items
#> a1 d g u
#> Item.1 1.0000000 1.8746626 0 1
#> Item.2 0.9464702 0.7810629 0 1
#> Item.3 1.0000000 1.4661198 0 1
#> Item.4 1.0000000 0.5234068 0 1
#> Item.5 1.0000000 1.9997347 0 1
anova(mod, mod2)
#> AIC SABIC HQ BIC logLik X2 df p
#> mod 5341.802 5352.192 5352.994 5371.248 -2664.901
#> mod2 5343.720 5355.842 5356.777 5378.075 -2664.860 0.081 1 0.775
mod2 <- mirt(dat, 'F = 1-5
FREE = (3, a1)', 'Rasch')
coef(mod2, simplify=TRUE)$items
#> a1 d g u
#> Item.1 1.000000 1.8101786 0 1
#> Item.2 1.000000 0.7649578 0 1
#> Item.3 1.848191 1.7774734 0 1
#> Item.4 1.000000 0.5042544 0 1
#> Item.5 1.000000 1.9316302 0 1
anova(mod, mod2)
#> AIC SABIC HQ BIC logLik X2 df p
#> mod 5341.802 5352.192 5352.994 5371.248 -2664.901
#> mod2 5335.130 5347.252 5348.187 5369.485 -2660.565 8.671 1 0.003
# test slopes first two slopes and last three slopes jointly
lagrange(mod, list(parnum[1:2], parnum[3:5]))
#> X2 df p
#> 1.5 0.4591775 2 0.79486042
#> 9.13.17 9.1527190 3 0.02732783
# test all 5 slopes and first + last jointly
lagrange(mod, list(parnum[1:5], parnum[c(1, 5)]))
#> X2 df p
#> 1.5.9.13.17 9.861713 5 0.07924974
#> 1.17 2.898233 2 0.23477762
# }