# Han (2012) in the paper "An Efficiency Balanced Information Criterion # for Item Selection in Computerized Adaptive Testing" proposes a method # of evaluating potential items based on expected item potential information # as a function of maximum potential item information. # This method favors items which have lower a values to be initially # selected when there is greater uncertainty in the test but favors selection # of items with higher a parameters as the test progresses. # This small bit of code demonstrates how such a proceedure rescales # item information. # First we will define a few functions that we will use to construct our scale. # Birbaum approximates the theta which maximizes the information function at # a specific a, b, and c parameter level: tmax <- function(a,b,c,D=1.7) b+1/(D*a)+log((1+sqrt(1+8*c))/2) # For example: tmax(a=2,b=2,c=.2) # This is the item information function for a 3PL (3 parameter logistic) iinfo <- function(theta,a,b,c,D=1.7) ((D*a)^2*(1-c))/((c+exp(D*a*(theta-b)))* (1+exp(-D*a*(theta-b)))^2) iinfo(theta=0,a=1,b=0,c=.1) # Now we define a function which approximates the integration of function # "fun" from start to end. integ <- function(start,end, step, fun, ...) { x <- seq(start+step/2,end-step/2,step) sum(get(fun)(x, ...)*step) } # As step size goes to zero the integ function approaches true integration. # Of course that would mean infinite calculations which would be impossible # for any computer. Thus a larger step size is a worse approximation but # uses less machine time. # For example a <- function(x,y) x^y # Let's see integ(0,2,.00001, "a", y=0) integ(0,2,.00001, "a", y=1) # Looking good. # This is the big function that we are interested in: IE <- function(thetahat,SEE,a,b,c,D=1.7,step=.001) { # thetahat is the current estimate of ability # SSE is the current standard error of the estimate # step is the number of steps used to estimate the integral # We calculate the item information at the current thetahat ii <- iinfo(thetahat,a=a,b=b,c=c,D=D) # Now we calculate the "max" theta value for the item. thetamax <- tmax(a=a,b=b,c=c,D=D) # Now the max information for that item. maxI <- iinfo(thetamax,a=a,b=b,c=c,D=D) # The efficient information as defined by Han at the # current theta is: ie <- ii/maxI # einfo is the expected information for a particular # item integrated across the range thetahat-SEE to # thetahat+SEE. einfo <- integ(thetahat-SEE*2, thetahat+SEE*2, step=step, "iinfo", a=a,b=b,c=c,D=D) # Finally we can rescale the expected item information # by the maxI to find the expected item efficiency. eie <- einfo/maxI # This provides a list of returned values. list(eie=eie, ii=ii, ie=ie, maxI=maxI, thetamax=thetamax, einfo=einfo) } test <- IE(0,1,a=1,b=0,c=.1,step=.001) test # Let's see this criterion in action: theta <- seq(-3,3,.1) # Make a list of returns returns <- names(test) for(v in returns) assign(v,NULL) # Let's create one last function that returns a list of # mappings for each of the ability levels. mapping <- function(theta=seq(-3,3,.1), SEE=.5,a=1,b=0,c=.1,step=.001) { I1 <- list() for(i in 1:length(theta)) { res <- IE(theta=theta[i],SEE=SEE,a=a,b=b,c=c,step=step) for(v in returns) I1[[v]][i] <- res[[v]] } I1 } # Now let's imagine five different items I1 <- mapping(a=.5 , b=-1.5, c=.3, SEE=.5) I2 <- mapping(a=1 , b=-1 , c=.3, SEE=.5) I3 <- mapping(a=1.7, b=0 , c=.3, SEE=.5) I4 <- mapping(a=1 , b=1 , c=.3, SEE=.5) I5 <- mapping(a=1.5, b=1.5 , c=.3, SEE=.5) plot(theta , I3$ii, type="n", main="Item Information at ThetaHat SEE=.5", xlab="ThetaHat", ylab="Information") lines(theta, I1$ii, lwd=2, col="red") lines(theta, I2$ii, lwd=2, col="blue") lines(theta, I3$ii, lwd=2, col="green") lines(theta, I4$ii, lwd=2, col="purple") lines(theta, I5$ii, lwd=2, col="black") # We can see that some items have much more information # than other items such that they would almost never # be selected. Item 4 for instance is almost never expected # to yeild higher information. # If we are less sure of our theta estimate we may instead # calculate our expected information. plot(theta , I3$einfo, type="n", main="Expected Item Information at ThetaHat SEE=.5", xlab="ThetaHat", ylab="Information") lines(theta, I1$einfo, lwd=2, col="red") lines(theta, I2$einfo, lwd=2, col="blue") lines(theta, I3$einfo, lwd=2, col="green") lines(theta, I4$einfo, lwd=2, col="purple") lines(theta, I5$einfo, lwd=2, col="black") # In general this basically makes the peaks less extreme but # does not generally favor our items with lower a values.

# If we want to see how our expected efficiency item # information value will do we can see that as well. # However, before we do that imagine first each of these # information functions divided by it's peak value. plot(c(0,theta) , c(0,I1$eie), type="n", main="Expected Efficiency Item Information at ThetaHat SEE=.5", xlab="ThetaHat", ylab="Information") lines(theta, I1$eie, lwd=2, col="red") lines(theta, I2$eie, lwd=2, col="blue") lines(theta, I3$eie, lwd=2, col="green") lines(theta, I4$eie, lwd=2, col="purple") lines(theta, I5$eie, lwd=2, col="black") # Now we can see that item 1 (red) and 4 (purple) are favored by # this algorithm, though by standard item maximization or by # expected item maximization they would almost never have been # chosen.

# The authors suggest a summing or the Efficiency Information # and that of expected information might yeild a good solution. plot(c(0,theta) , c(0,I3$eie+I3$einfo), type="n", main="Expected Efficiency Item Information at ThetaHat SEE=.5", xlab="ThetaHat", ylab="Information") lines(theta, I1$eie+I1$einfo, lwd=2, col="red") lines(theta, I2$eie+I2$einfo, lwd=2, col="blue") lines(theta, I3$eie+I3$einfo, lwd=2, col="green") lines(theta, I4$eie+I4$einfo, lwd=2, col="purple") lines(theta, I5$eie+I5$einfo, lwd=2, col="black") # The argument is that as SEE gets small the information begins # to look much more like that of Item Information which is # appropropriate for later in the test. I1 <- mapping(a=.5 , b=-1.5, c=.3, SEE=.15) I2 <- mapping(a=1 , b=-1 , c=.3, SEE=.15) I3 <- mapping(a=1.7, b=0 , c=.3, SEE=.15) I4 <- mapping(a=1 , b=1 , c=.3, SEE=.15) I5 <- mapping(a=1.5, b=1.5 , c=.3, SEE=.15) plot(c(0,theta) , c(0,I3$eie), type="n", main="Expected Efficiency Item Information at ThetaHat SEE=.15", xlab="ThetaHat", ylab="Information") lines(theta, I1$eie, lwd=2, col="red") lines(theta, I2$eie, lwd=2, col="blue") lines(theta, I3$eie, lwd=2, col="green") lines(theta, I4$eie, lwd=2, col="purple") lines(theta, I5$eie, lwd=2, col="black") # Now we can see that item 1 (red) and 4 (purple) are favored by # this algorithm, though by standard item maximization or by # expected item maximization they would almost never have been # chosen. # The authors suggest a summing or the Efficiency Information # and that of expected information might yeild a good solution. plot(c(0,theta) , c(0,I3$eie+I3$einfo), type="n", main="Expected Efficiency Item Information at ThetaHat SEE=.15", xlab="ThetaHat", ylab="Information") lines(theta, I1$eie+I1$einfo, lwd=2, col="red") lines(theta, I2$eie+I2$einfo, lwd=2, col="blue") lines(theta, I3$eie+I3$einfo, lwd=2, col="green") lines(theta, I4$eie+I4$einfo, lwd=2, col="purple") lines(theta, I5$eie+I5$einfo, lwd=2, col="black") # We can see that item 1 is still favored though we expected # it to give us very little information. Overall, the # method seems interesting but not yet ideal.

## No comments :

## Post a Comment