# One of the basic tasks in item response theory is estimating the ability of a test taker from the responses to a series of items (questions).

# Let's draw the same pool of items that we have used on several previous posts:

# First let's imagine we have a pool of 100 3PL items.

set.seed(101)

npool = 500

pool = as.data.frame(cbind(item=1:npool, a=abs(rnorm(npool)*.25+1.1), b=rnorm(npool), c=abs(rnorm(npool)/7.5+.1)))

summary(pool)

# Drawing on code from a previous post we can calculate several useful functions:

# (http://www.econometricsbysimulation.com/2012/11/matching-item-information.html)

# Each item has a item characteristic curve (ICC) of:

PL3 = function(theta,a, b, c) c+(1-c)*exp(a*(theta-b))/(1+exp(a*(theta-b)))

# Let's imagine that we have a single test taker and that test taker has taken the first 15 items from the pool.

items.count = 15

items.taken = pool[1:items.count,]

# And that the person has a latent theta ability of 1

theta = 1.3

# Let's calculate the cut points for each of the items.

items.cut = PL3(theta, items.taken$a, items.taken$b, items.taken$c)

# We can see how the cut point works by graphing

plot(0,0, type="n", xlim=c(-3,3),ylim=c(0,1), xlab=~theta,

ylab="Probability of Correct Response", yaxs = "i", xaxs = "i" , main="Item Characteristics Curves and Ability Level")

for(i in 1:items.count) {

lines(seq(-3,3,.1),PL3(seq(-3,3,.1), items.taken$a[i], items.taken$b[i], items.taken$c[i]), lwd=2)

abline(h=items.cut[i], col="blue")

}

abline(v=theta,col="red", lwd=3)

# Now let's draw a uniform draw that we will use to calculate whether each item as passed.

rdraw = runif(items.count)

# Finally, we will calculate item responses

item.responses = 0

item.responses = items.cut > rdraw

###############################################

# Done with Simulation - Time for Estimation

# We want to use the information we know about the items (the item parameters and the responses) in order to estimate a best guess at the true ability of the test taker.

# First we must check if the person got all of the items either correct or incorrect.

sum(item.responses)

# If this is either a 0 or a number equal to the number of items then we cannot esimate an interior maximum without additional assumptions.

# We will attempt to recover our theta value using the r command optim

# First we need to specify the function to optimize over.

MLE = function(theta) sum(log((item.responses==T)*PL3(theta, items.taken$a, items.taken$b, items.taken$c) +

(item.responses==F)*(1-PL3(theta, items.taken$a, items.taken$b, items.taken$c))))

# The optimization function takes as its argument the choice variables to be optimized (theta).

# The way the above optimization works is that you specify the probility of each response piecewise.

# If the response is correct, then you count the CDF of theta up to that point as contributing to the probability of observing a correct outcome. If the response is negative, then you count it as contributing the the probability of a incorrect outcome. You then choose the theta that produces the greatest total probability.

MLEval = 0

theta.range = seq(-3,3,.1)

for(i in 1:length(theta.range)) MLEval[i] =MLE(theta.range[i])

plot(theta.range, MLEval, type="l", main="Maximim Likelihood function", xlab= ~theta, ylab="Sum of Log Likelihood")

abline(v=theta, col="blue")

# We can visually see that the maximum of the slope will not be at the true value though it will be close.

optim(0,MLE, method="Brent", lower=-6, upper=6, control=list(fnscale = -1))

abline(v=optim(0,MLE, method="Brent", lower=-6, upper=6, control=list(fnscale = -1))$par, col="red")

# We can use R's optim function to find the ideal theta that would maximize the information from this test.

# Item information is:

PL3.info = function(theta, a, b, c) a^2 *(PL3(theta,a,b,c)-c)^2/(1-c)^2 * (1-PL3(theta,a,b,c))/PL3(theta,a,b,c)

# Notice, this is not the best way of defining the test information function since the items are not arguments.

test.info = function(theta) sum(PL3.info(theta, items.taken$a, items.taken$b, items.taken$c))

# Construct a vector to hold the test information

info = 0

for(i in 1:length(theta.range)) info[i]=test.info(theta.range[i])

plot(theta.range, info, type="l", main="Information Peaks Slightly Above 0", xlab= ~theta, ylab="Information")

abline(v=theta, col="blue")

# But we want to know about the test taker at theta

optim(0,test.info, method="Brent", lower=-6, upper=6, control=list(fnscale = -1))

# The person this test would be best suited to evaluate would have an ability rating of .19

abline(v=optim(0,test.info, method="Brent", lower=-6, upper=6, control=list(fnscale = -1))$par, col="red")

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