testing for significant difference in item difficulty

[Balazs Klein]

How could I test if the difficulty of two items (in a 2 parameter model) is significantly different?

The situation: I made some changes in some items and trialled it with a number of test takers. The item difficulty changed. How can I test if this change was significant?

Thanks for the help.

Balázs

[Phil Chalmers]

Use anova(mod1, mod2), where one model is the nested version of the other (i.e., more constrained).

Phil

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[Balazs Klein]

Sorry, I don't think I understood your answer

Could you give an example?

library(mirt)

data <- expand.table(LSAT7)

mod1 <- mirt(data, 1)

coef(mod1)$Item.1

coef(mod1)$Item.3

Item1 -seems- to be more difficult than Item3 (d1=1.856, d3=1.804).

How do I know if this difference was significant?

Thanks for the help.

B.

On Wednesday, August 17, 2016 at 1:19:11 PM UTC+2, Phil Chalmers wrote:

Use anova(mod1, mod2), where one model is the nested version of the other (i.e., more constrained).

Phil

On Wed, Aug 17, 2016 at 6:39 AM, Balazs Klein <balazs...@gmail.com> wrote:

How could I test if the difficulty of two items (in a 2 parameter model) is significantly different?

The situation: I made some changes in some items and trialled it with a number of test takers. The item difficulty changed. How can I test if this change was significant?

Thanks for the help.

Balázs

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[Seongho Bae]

Estimating SEs may help to know differences among difficulty.

mod1 <- mirt(data, 1, SE = T)

coef(mod1)$Item.1

coef(mod1)$Item.3

Seongho

2016년 8월 17일 수요일 오후 10시 29분 23초 UTC+9, Balazs Klein 님의 말:

[Phil Chalmers]

Seongho has the right idea here, though if you want a formal test with the ACOV matrix you can us wald(). Otherwise, you can apply equality constraints to do a likelihood-ratio test by fitting another model and comparing with anova(). See below.

#-------------

# Wald test

dat <- expand.table(LSAT7)

mod1 <- mirt(dat, 1, SE = T)

wald(mod1) #see parameters and location

# setup test matrix (test whether 2 and 6 are equal)

L <- matrix(0, 1, 10)

L[1,2] <- 1

L[1,6] <- -1

wald(mod1, L)

# ------------

# LR test 

# Constrain intercepts 1 and 3 to be equal, and test whether this makes the model fit worse

mod2 <- mirt(dat, 'F = 1-5

                   CONSTRAIN = (1, 3, d)')

anova(mod1, mod2)

Phil

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[Balazs Klein]

Many thanks for this.
B.