Wald statistics for mirt models
Compute a Wald test given an L vector or matrix of numeric contrasts. Requires that the
model information matrix be computed (by passing SE = TRUE when estimating the model). Use
wald(model) to observe how the information matrix columns are named, especially if
the estimated model contains constrained parameters (e.g., 1PL).
wald(object, L, C = NULL)
object |
estimated object from |
L |
a coefficient matrix with dimensions |
C |
a constant vector of population parameters to be compared along side L, where
The following description is borrowed from Alternatively, the hypothesis can be specified symbolically as a character vector with one or more elements, each of which gives either a linear combination of coefficients, or a linear equation in the coefficients (i.e., with both a left and right side separated by an equals sign). Components of a linear expression or linear equation can consist of numeric constants, or numeric constants multiplying coefficient names (in which case the number precedes the coefficient, and may be separated from it by spaces or an asterisk); constants of 1 or -1 may be omitted. Spaces are always optional. Components are separated by plus or minus signs. Newlines or tabs in hypotheses will be treated as spaces. See the examples below." |
Phil Chalmers rphilip.chalmers@gmail.com
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
## Not run:
#View parnumber index
data(LSAT7)
data <- expand.table(LSAT7)
mod <- mirt(data, 1, SE = TRUE)
coef(mod)
# see how the information matrix relates to estimated parameters, and how it lines up
# with the parameter index
(infonames <- wald(mod))
index <- mod2values(mod)
index[index$est, ]
#second item slope equal to 0?
L <- matrix(0, 1, 10)
L[1,3] <- 1
wald(mod, L)
# same as above using character syntax input
infonames
wald(mod, "a1.5 = 0")
#simultaneously test equal factor slopes for item 1 and 2, and 4 and 5
L <- matrix(0, 2, 10)
L[1,1] <- L[2, 7] <- 1
L[1,3] <- L[2, 9] <- -1
L
wald(mod, L)
# Again, using more efficient syntax
infonames
wald(mod, c("a1.1 = a1.5", "a1.13 = a1.17"))
#log-Liklihood tests (requires estimating a new model)
cmodel <- 'theta = 1-5
CONSTRAIN = (1,2, a1), (4,5, a1)'
mod2 <- mirt(data, cmodel)
#or, equivalently
#mod2 <- mirt(data, 1, constrain = list(c(1,5), c(13,17)))
anova(mod2, mod)
#####
# test equality of means in multi-group model:
# H0: (mu1 - mu2) = (mu3 - mu4)
set.seed(12345)
a <- matrix(abs(rnorm(15,1,.3)), ncol=1)
d <- matrix(rnorm(15,0,.7),ncol=1)
itemtype <- rep('2PL', nrow(a))
N <- 500
dataset1 <- simdata(a, d, N, itemtype)
dataset2 <- simdata(a, d, N, itemtype, mu = .5)
dataset3 <- simdata(a, d, N, itemtype, mu = -1)
dataset4 <- simdata(a, d, N, itemtype, mu = -.5)
dat <- rbind(dataset1, dataset2, dataset3, dataset4)
group <- factor(rep(paste0('D', 1:4), each=N))
levels(group)
models <- 'F1 = 1-15'
# 3 means estimated
mod_free <- multipleGroup(dat, models, group = group, SE=TRUE,
invariance=c('slopes', 'intercepts', 'free_var','free_means'))
wald(mod_free) # obtain parameter names
# View(mod2values(mod_free))
# reference group mean = 0 by default
wald(mod_free, c("0 - MEAN_1.123 = MEAN_1.185 - MEAN_1.247"))
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