(This article was first published on

**Econometrics by Simulation**, and kindly contributed to R-bloggers)# Item equating is the practice of making the results from two

# different assessments equivalent. This can be done by either

# 1. having the same group take both assessments

# 2. having equivalent groups take the different assessments

# 3. having non-equivalent groups which use common items take the different

# assessments.

# In this post I will cover topic 1.

# For this code I will use the catR package to generate my assessments

# and responses.

library("catR")

# Let's attempt the first proceedure:

# First let's generate our item parameters for assessment 1.

# Let's create an item bank with 100 items for the assessments

nitems <- 100

bank1 <- createItemBank(model="2PL", items=nitems)$itemPar

bank2 <- createItemBank(model="2PL", items=nitems)$itemPar

# Now let's generate a 1000 person population sample to take our assessment

npeep <- 1000

theta <- rnorm(npeep)

# Calculate the score on both assessments

resp1 <- resp2 <- matrix(0, nrow=npeep, ncol=nitems)

for (i in 1:npeep) {

resp1[i,Pi(theta[i],bank1)$Pi(nitems)] <- 1 # Test 1

resp2[i,Pi(theta[i],bank2)$Pi(nitems)] <- 1 # Test 2

}

# To calculate total score on the tests we can sum the results of each row

score1 <- apply(resp1, 1, sum)

score2 <- apply(resp2, 1, sum)

# Since we know both forms of the test are parrellel we can check the correlation

# between scores on the different forms of the test as a measure of reliability.

cor(score1,score2)

# This gives me a correlation estimate of .95 which is very good.

# However, we are not interested in reliability right now. We would like to

# equate the two tests using the information thus far garnished.

# First let's estimate our parameters.

# I will use the ltm command in the package ltm for this.

library("ltm")

est2pl1 <- ltm(resp1~z1)

est2pl2 <- ltm(resp2~z1)

# Now we have two tests with the different items.

# We want to make sure the items are on the same scale

# so that it does not matter which test individuals take.

# Their expected score will be the same.

cbind(score1,score2)

# Because we simulated the generation of both tests which are very lengthy

# the tests are pretty much already equated by design which is frankly

# much easier to do by simulation and perhaps impossible to do with

# real assessments.

# Nevertheless, let's act as if our tests were not already parrellel and

# equate them.

# Let's first do some linear equating which is done by setting the standardized

# scores of the two exames equal to each other. (page 33)

# See http://ncme.org/linkservid/65DCF34D-1320-5CAE-6E896C93330B9C73/showMeta/

# X1 and X2 refers to total scores for each individual on exams 1 and 2

# (X1 - mean(X1))/Sd(X1) = (X2 - mean(X2))/Sd(X2)

# X1 = Sd(X1)/Sd(X2)*X2 + (mean(X1)-Sd(X1)/Sd(X2)*mean(X2)) = A*X2 + B

# Where A=Sd(x1)/Sd(x2) and B=mean(X1)-A*mean(X2)

(A <- sd(score1)/sd(score2))

# A being close to 1 indicates that the tests are scaled similary

(B <- mean(score1)-A*mean(score2))

# B being close to 0 indicates the tests are of similar difficulty

# Now let's calculate what score2 would be if scaled on assessment 1.

score2scaled <- A*score2 + B

summary(cbind(score1,score2,score2scaled))

# We can see that score2scaled looks slightly closer to score1.

# Let's try the same thing with slightly more interesting.

# Let's imagine that score1 is for ACT from 0 to 32 and score 2 is from SAT 200 to 800

# The standard deviation is 6 and average score around 20 for the ACT

ACT <- (score1-mean(score1))/sd(score1)*6 + 20

ACT[ACT>32] <- 32

ACT[ACT<4] <- 4

ACT <- round(ACT) # ACT rounds to whole numbers

# The standard deviation is 100 and average score around 500 for the ACT

SAT <- (score2-mean(score2))/sd(score2)*100 + 500

SAT[SAT>800] <- 800

SAT[SAT<200] <- 200

SAT <- round(SAT/10)*10 # SAT rounds to nearest 10

summary(cbind(SAT,ACT))

# Now let's see if we can transform our SAT scores to be on our ACT scale

(A <- sd(ACT)/sd(SAT))

(B <- mean(ACT)-A*mean(SAT))

SATscaled <- A*SAT + B

summary(cbind(SAT,ACT,SATscaled))

# The results from taking parrellel tests should fall on a linear form

plot(ACT,SATscaled, main="SAT results placed on ACT scale")

# This is the easiest method of equating two tests.

# However, it is not usually the most practical since it is costly to get the same

# group of individuals to take two different tests. In addition, there

# may be issues with fatigue which could be alleviated somewhat if for half of the

# group the first assessment was given first and for a different half

# the second assessment assessment was given first.

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