Reading in the data

Here the data required for the analysis are entered. There are up to five objects to enter:
- a title for the analysis - the choice data for the 5000 examinees - the right answer key - a character string for each item or question to display in plots - a character string for each option choice within each item The two essential inputs are the choice data and the right answer key.

The choice data are an N by n matrix called U_5000 in the code. The rows of U_5000 correspond to examinees and the 80 columns to items or questions. Each number in U_5000 is the index of the option or answer that the examinee has chosen using the number 1 to M where M is the number of answers available for choice. We call this data structure index mode data, in contrast to score data where the number indicate pre-assigned scoring weights.

The choice indices may also be stored in a dataframe where all variables are of the factor type. We have avoided this level of sophistication since the only operations on U_5000 are summing and retrieving values. Given the potential size of the choice data, it can be worthwhile to place additional restrictions on the structure such being in integer mode.

The second data to read in are the indices of the right answers in the vector key. This object is used to assign scores of 0 to wrong answers and 1 to right answers.

The reader may find it helpful to consult man pages for the functions used using the command help("filename") to obtain more detail and see example code.

First we will set up a title for the data to be used in plots:

titlestr  <- "SweSAT-Q: 24 math analysis items and 1000 examinees"

Then we set up the path or connection to the location of the choice data.

The first five lines in the file “U_short_5000.txt” in TestGardener’s data folder are:

21323211244334312213413543333122422234448412231413431343313231533254433231121343 21223212144334242213413353433112123234441412221413411343113133513241441121231343 21423433234314141333433421444212314214443413231423432412313431332154123134242343 21344243214214312314223525223111142314324132124231431243313113534233123134444123 22143311341414332244143553222231121231431414343424432342313233555231143223121143

We use the basic `scan()’ command to read the 1000 lines in the file like this:

U <- scan("U_5000.txt","o")
N <- length(U) # Number of examinees

The command assumes that the working directory is the TestGardener folder, within which there is a “data” folder containing various data files. The second argument deals with the fact that there is no space between values, and the last argument declares each symbol to be a character value. The data are now 1000 strings, each 24 characters long, and the second line sets N, the number of examinees.

We will use the stringr package to break the each 80 character string into separate characters and then convert these characters to integers. The result is an integer vector of length 80 x 5000 = 40,000. The number of questions n is this length divided by the number of examinees. The third line reformats Uvector into an N by n matrix.

Uvector <- as.integer(unlist(stringr::str_split(U,"")))
n       <- length(Uvector)/N # Number of items
U       <- matrix(Uvector,N,n,byrow=TRUE)

However, we have also set up both U and key objects as .rda compressed files that are automatically loaded if library(TestGardener) has been entered.

There are many other ways in R to achieve the same result, of course, and this is unlikely to be viewed as the most elegant.

Inputting the right answer key object uses the same techniques, and we have in fact saved these also as a set of integer index values looking like the data themselves.

key   <- scan("key_5000.txt","o")
key <- as.integer(unlist(stringr::str_split(key,"")))

Selecting the 24 mathematical problem solving items and the subsample of 1000 examinees

itemindex <- c(1:12, 41:52)
examinees <- 4001:5000
U <- U[examinees,itemindex]
key <- key[itemindex]
N <- 1000
n <- 24

Setting up the analysis

Now we turn to computing objects using these data that are essential to the analysis of the data.

All questions for these 24 questions in the test have four answers, but we will add an additional category for answers that are missing or illegitimate.

noption <- rep(5,n)

The data have two integers for missing or illegitimate items: 8 and 9. We change these to the garbage category 5.

for (j in 1:N) {
  for (i in 1:n) {
    if (U[j,i] == 8 || U[j,i] == 9) U[j,i] = 5
  }
}

Now we turn to code that sets up objects that are essential to the analysis but that the analyst may want to set up using code rather than inputting. The first of these is a list vector of length n where each member contains a numeric vector of length M containing the designed scores assigned to each option. For multiple choice tests, that is easily done using the following code:

ScoreList <- list() # option scores
for (item in 1:n){
  scorei <- rep(0,noption[item])
  scorei[key[item]] <- 1
  ScoreList[[item]] <- scorei
}

There is also the possibility of inputting strings for each question and each option within each question. We won’t use this feature in this analysis.

Next, we set up a named list called optList that contains ScoreList and default values for the item and option strings:

optList <- list(itemLab=NULL, optLab=NULL, optScr=ScoreList)

We’re now finished with the data input and setup phase. From these objects we have a function make_dataList that completes the set up:

Math_dataList <- TestGardener::make.dataList(U, key, optList)
names(Math_dataList)
#  [1] "U"           "N"           "n"           "titlestr"    "optList"    
#  [6] "WfdList"     "key"         "grbg"        "WfdPar"      "noption"    
# [11] "nbin"        "scrrng"      "scrfine"     "scrvec"      "scrjit"     
# [16] "itmvec"      "percntrnk"   "thetaQnt"    "Wdim"        "PcntMarkers"
# [21] "titlestr"    "grbg"

The result is a named list with many members whose values may be required in the analysis phase. Consult the documentation using help("make.dataList.R").

One default option requires a bit of explanation. Function make.dataList() computes sum scores for each examinee, which in the multiple choice case are simply counts of the number of correct answer choices. The analysis phase does not use these sum scores directly. Instead, it uses the rank orders of the sum scores (ordered from lowest to highest) divided by N and multipled by 100, so that the scores now look like percentages, and we refer to them as “percent ranks.”

A problem is that many examinees will get the same rank value if we order these ranks directly. We can’t know whether there might be some source of bias in how these tied ranks are arranged. For example, suppose male examinees for each rank value are followed by female examinees with that rank value. We deal with this problem by default by adding a small random normally distributed number to each tied rank that is too small to upset the original rank order. This within-rank randomization is called “jittering” in the statistics community, and its tends to break up the influence of any source of bias. This is the default, but can be turned off if desired.

These jittered percent ranks are used to provide initial values for a cycled optimization procedure used by TestGardener.

Let’s make our first plot a display of the histogram and the probability density function for the sum scores.

hist(Math_dataList$scrvec, Math_dataList$scrrng[2], xlab="Sum Score",
     main=titlestr)

This histogram tells us a number of things. First, this is obviously a difficult test from the standpoint of the examinees at or below the mid-score of 40 questions correct. It is even difficult for the strongest examinees since no examinees obtained perfect scores. But the fact that no examinees had a sum score less tan 10 needs explaining, too. When completely lost, an examinee can always guess and choose one at random. For four-option question, this will strategy will produce a correct answer about 20 times and, for a test this long, is unlikely to produce a score less than 10. Guessing inflates sum scores over what we would desire, and our analyses will compensate for this by recognizing the signs of guessing and deflating the resulting score.

The next steps are optional. The analysis can proceed without them, but the analyst may want to look at preliminary results that are associated with the the objects in the set up phase.

First we compute item response or item characteristic curves for each response within each question. These curves are historically probability curves, but we much prefer to work with a transformation of probability, “surprisal = - log(probability),” for two reasons: (1) this greatly aids the speed and accuracy of computing the curves since they are positive unbounded values, and (2) surprisal is in fact a measure of information, and as such has the same properties as other scientific measures. It’s unit is the “bit” and what is especially important is that any fixed difference between two bit values means exactly the same thing everywhere on a surprisal curve.

The initial surprisal curves are achieved as follows using the important function Wbinsmth():

theta     <- Math_dataList$percntrnk
thetaQnt  <- Math_dataList$thetaQnt
WfdResult <- TestGardener::Wbinsmth(theta, Math_dataList)

One may well want to plot these initial surprisal curves in terms of both probability and surprisal, this is achieved in this way:

WfdList <- WfdResult$WfdList
binctr  <- WfdResult$aves
Qvec    <- c(5,25,50,75,95)
indfine <- seq(0,100,len=101)
TestGardener::ICC.plot(indfine, WfdList, Math_dataList, Qvec, binctr,  
                       Wrng=c(0,3), plotindex=1)

Here Ωand elsewhere we only plot the probability and surprisal curves for the first time in order to allow this vignette to be displayed compactly. There are comments on how to interpret these plots in the call to function ICC.plot() after the analysis step.

The bottom plot displays surprisal, the logarithm to the base M of the negative of probability. Surprisal is a valuable transform of probability. As the name suggests, when probability approach one, surprisal move close to zero, but probabilities near zero imply surprisals high surprisals. Surprisal is a measure if information having the “M-bit” as its unit. Most of us are familiar with the “2-bit”, associated with coin tosses. Surprisal is essentially the number of heads you would have to toss on the average to yield a particular probability. Probability 0.5 corresponds to 1 “2-bit”, and the famous probabilities 0.05 and 0.01 in statistics correspond to 4.3 and 6.1 “2-bits” respectively. These ideas generalize to “M-sided coins” such as the “6-bit” for a dice.

Cycling through the estimation steps

Our approach to computing optimal estimates of surprisal curves and examinee percentile rank values involves alternating between:

1. estimating surprisal curves assuming the previously computed percentile ranks are known
2. estimating examinee percentile ranks assuming the surprisal curves are known.

This is a common strategy in statistics, and especially when results are not required to be completely optimal. We remind ourselves that nobody would need a test score that is accurate to many decimal places. One decimal place would surely do just fine from a user’s perspective.

We first choose a number of cycles that experience indicates is sufficient to achieve nearly optimal results, and then at the end of the cycles we display a measure of the total fit to the data for each cycle as a check that sufficient cycles have been carried. Finally, we choose a cycle that appears to be sufficiently effective. A list object is also defined that contains the various results at each cycle.

In this case the measure of total fit is the average of the fitting criterion across all examinees. The fitting criterion for each examinee is the negative of the log of the likelihood, or maximum likelihood estimation. Here we choose 10 cycles.

ncycle <- 10

Here is a brief description of the steps within each cycle

AnalyzeResult <- TestGardener::Analyze(theta, thetaQnt, Math_dataList, 
                                       ncycle, itdisp=FALSE, verbose=FALSE) 

parList      <- AnalyzeResult$parList
meanHsave    <- matrix(0,ncycle,1)
arclengthvec <- matrix(0,ncycle,1)
for (icycle in 1:ncycle) {
  meanHsave[icycle]    <- parList[[icycle]]$meanH
  arclengthvec[icycle] <- parList[[icycle]]$arclength
}

Displaying the results of the analysis cycles

cycleno <- 1:ncycle
plot(cycleno,    meanHsave[cycleno], type="b", lwd=2, xlab="Cycle Number", 
     ylab="Mean Fit")

cycleno <- 1:ncycle
plot(cycleno, arclengthvec[cycleno], type="b", lwd=2, xlab="Cycle Number",   
     ylab="Arclength")

Math_parListi  <- parList[[ncycle]]

WfdList    <- Math_parListi$WfdList
theta      <- Math_parListi$theta
Qvec       <- Math_parListi$Qvec
Hval       <- Math_parListi$Hval
DHval      <- Math_parListi$DHval
D2Hval     <- Math_parListi$D2Hval
binctr     <- Math_parListi$binctr
indfine    <- seq(0,100,len=101)

thetacheck  <- thetasearch(WfdList, U, theta, Hval, DHval, D2Hval)
theta       <- thetacheck$theta
Hval        <- thetacheck$Hval
DHval       <- thetacheck$DHval
D2Hval      <- thetacheck$D2Hval
changeindex <- thetacheck$changeindex
print(paste("The number of altered theta values is", length(changeindex)))
# [1] "The number of altered theta values is 183"

Math_infoList <- TestGardener::theta2arclen(theta, Qvec, WfdList, binctr)

arclength     <- Math_infoList$arclength
arclengthvec  <- Math_infoList$arclengthvec
arclengthfd   <- Math_infoList$arclengthfd
theta_al      <- Math_infoList$theta_al
thetafine_al  <- Math_infoList$thetafine.rng
Qvec_al       <- Math_infoList$Qvec_al
binctr_al     <- Math_infoList$binctr_al
Wfd_info      <- Math_infoList$Wfd_info
Wdim          <- Math_infoList$Wdim

indfine <- seq(0,100,len=101)
TestGardener::ICC.plot(indfine, WfdList, Math_dataList, Qvec, binctr,  
                       data_point=TRUE, plotType=c("P", "W"), Wrng=c(0,3), plotindex=1)

indfine <- seq(0,100,len=101)
TestGardener::ICC.plot(arclengthvec, WfdList, Math_dataList, Qvec_al, binctr_al,  
                       data_point=TRUE, plotType=c("P", "W"), Wrng=c(0,3), plotindex=1)

Plot the probability density of the score index and of arc length

The density of the percentile ranks that we initially used will no longer be a flat line.
This is because examinees tend to cluster at various score levels, no matter how the score is defined. Certain groups of items will tend to be correctly answered by the middle performance folks and other by only the top performers. Among the weakest examinees, there will still be a subset of questions that they can handle. We want to see these clusters emerge.

TestGardener::density_plot(theta, c(0,100), Qvec, xlabstr="Score index", 
                           titlestr="Theta Density", scrnbasis=15)

In fact, there are four distinct clusters of score index values below the 75% level. Within each of these clusters there are strong similarities in examinee’s choice patterns, whether right or wrong. We have only plotted score indices which are away from the two boundaries, because there are significant tendencies to have estimated score index values at 0 and 100.

TestGardener::density_plot(theta_al, c(0,arclength), Qvec_al, xlabstr="Arclength", 
                           titlestr="Arc length Density",  scrnbasis=15)

Compute expected test scores and plot their density function for all examinees

The expected score for an examinee are determined by his value theta for the question and by the score values assigned by the test designer to each option. We use a plotting interval defined by the range of the sum scores, but as a rule the density of expected scores is positive only over a shorter interval.

indfine <- seq(0,100,len=101)
mufine <- TestGardener::testscore(indfine, WfdList, optList)
TestGardener::mu.plot(mufine, Math_dataList$scrrng, titlestr)

Note that the bottom 5% examinees get on the average sum scores of about 5, which is 20% of the largest possible score of 24. This is because, even though they know nearly nothing about math, they can still score right often by just guessing or randomly choosing.

But at the top end, the best examinees lose heavily in average score terms. There are no average scores above 21, even though there are plenty of examinees with score indices of nearly or exactly 100. This effect is due to the fact that a few questions are faulty in the sense of not getting close to probability one or surprisal 0 at score index 100. We found, for example, that one question did not have any answer that could be judged correct by serious mathematical standards. And another question seemed to suffer from some teachers having overlooked covering the topic of “percent increase.”

It is typical that lower expected test scores are above the diagonal line and higher ones below. At the lower or left end, expected test scores are too high because of the possibility of getting question right by guessing. At the upper or right end the scores are below the line because of questions that fail to reach probability one for the right answer for even the strongest examinees.

The process of computing an expected score compresses the range of scores relative to that of the observed test scores.

Compute the arc length of the test effort curve and plot the curve

We have 412 curves simultaneously changing location simultaneously in both probability and surprisal continua as the score index moves from 0 to 100. We can’t visual such a thing, but we are right to think of this as a point moving along a curve in these two high dimensional spaces. In principle this curve has twists and turns in it, but we show below that they are not nearly as complex as one might imagine.

What we can study, and use in many ways, is the distance within the curve from its beginning to either any fixed point along it, or to its end. Thße curve is made especially useful because it can be shown that any smooth warping of the score index continuum will have no impact on the shape of this curve. Distance along the curve can be measured in bits, and a fixed change in bits has the same meaning at every point in the curve. The bit is a measure information, and we call this curve the test information curve.

The next analysis and plot displays the length of the curve and how it changes relative to the score index theta associated with any point on the curve.

print(paste("Arc length =", round(arclength,2)))
# [1] "Arc length = 39.18"
TestGardener::ArcLength.plot(arclength, arclengthvec, titlestr)

The relationship is quite linear, except for some curvature near 0 and 100. We can say of the top examinees that they acquire nearly 90 2-bits of information represented in the test. That is, the probability of aceing this test by chance alone is that of tossing 90 heads in a row. (We convert 5-bits into 2-bits by multiplying 39 by 2.32])

Display test information curve projected into its first two principal components

The test information curve is in principle an object of dimension Wdim, but in most cases almost all of its shape can be seen in either two dimensions or three. Here we display it in two dimensions as defined by a functional principal component analysis.

Result <- TestGardener::Wpca.plot(arclength, WfdList, Math_dataList$Wdim, titlestr=titlestr)

The change in direction of this curve between the 25% and 50% marker points indicates that the information acquired initially is qualitatively different than that mastered at the upper end of the curve. This seems obvious, since math moves from the concrete to the abstract over a secondary school program.

Investigate the status of score index estimate by plotting H(theta) and its derivatives

An optimal fitting criterion for modelling test data should have these features: (1) at the optimum value of theta fit values at neighbouring values should be larger than the optimum; (2) the first derivative of the fitting criterion should be zero or very nearly so, and (3) the second derivative should have a positive value.

But one should not automatically assume that there is a single unique best score index value that exists for an examinee or a ratings scale respondent. It’s not at all rare that some sets of data display more than one minimum. After all, a person can know some portion of the information at an expert level but be terribly weak for other types of information. By tradition we don’t like to tell people that there are two or more right answers to the question, “How much do you know?” But the reality is otherwise, and especially when the amount of data available is modest.

If an estimated score index seems inconsistent with the sum score value or is otherwise suspicious, the function Hfuns.plot() allows us to explore the shape of the fitting function H(theta) as well as that of its second derivative.

Here we produce these plots for the first examinee.

TestGardener::Hfuns.plot(theta, WfdList, U, plotindex=1)

Use simulated data to estimate the accuracy of test scores