Monday, August 5, 2013

Item Equating with same Group - SAT, ACT example

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