Introduction to the Rasch and 2PL Model

The Rasch model, also known as the Rasch measurement model or the Rasch model for item response theory (IRT), is a widely used psychometric model for analyzing data in educational and psychological measurement. It provides a framework for assessing the relationship between respondents’ abilities and item difficulties.

The Rasch model assumes that \[Pr(U_{pi} = 1 | \theta_p, \beta_i)\], the probability of an individual \(p\) correctly solving a test item \(i\), is influenced by both the characteristics of the item (item parameters) and the person (person parameters). Specifically, the probability of a person with a math ability \(\theta_p\) answering an item correctly increases as their math ability becomes more pronounced. Conversely, as the difficulty of an item \(\beta_i\) increases, the probability of answering that item correctly decreases. This relationship can be represented by the following formula:

\[ Pr(U_{pi} = 1 | \theta_p, \beta_i) = \frac{e^{\theta_p - \beta_i}}{1 + e^{\theta_p - \beta_i}} \]

The Rasch model assumes a strict equality of item discrimination across all items. In general, the item discrimination parameter measures the extent to which an item is capable of distinguishing between individuals with different abilities. In analytical terms, the item discrimination parameter represents the slope of the item response function graph. A steeper slope indicates a stronger relationship between the ability (\(\theta\)) and a correct response, indicating how effectively the item discriminates among examinees along the ability scale continuum. A higher value of the item discrimination parameter suggests that the item is more effective in differentiating examinees. Practically, a higher discrimination parameter value means that the probability of a correct response increases more rapidly as the latent trait (\(\theta\)) increases. The Rasch model assumes this item discrimination to be equal among all items, which might not be the optimal fit for the data. Thus a second model, the 2PL model, was introduced.

The two-parameter logistic (2PL) model is a more flexible alternative and can be seen as an extension of the Rasch model. When the assumption of equal item discrimination is no good fit for the data, the 2PL model is preferred over the Rasch model. The 2PL model introduces a second item parameter, called the discrimination or slope parameter (denoted as \(\alpha\)), in addition to the difficulty parameter (denoted as \(\beta_i\)) for each item \(i\). Mathematically, the 2PL model can be described as follows:

\[ Pr(U_{pi} = 1 | \theta_p, \alpha_i, \beta_i) = \frac{e^{\alpha_i(\theta_p - \beta_i)}}{1 + e^{\alpha_i(\theta_p - \beta_i)}} \]

A high value of \(\alpha_i\) indicates that the item has a strong ability to distinguish among test takers. This means that the probability of giving a correct response increases more rapidly as the ability \(\theta\) increases.

When evaluating psychological and educational tests, researchers often face the question of which IRT model better describes the data. In this chapter, our specific hypothesis test aims to address whether the inclusion of the additional discrimination parameter in the 2PL model is essential for describing the data, or if the discrimination parameters are sufficiently similar for the Rasch model to adequately describe the data. Therefore, it becomes relevant to test the null hypothesis of equal slope parameters, as this may allow us to reject an incorrectly assumed Rasch model. Conducting a power analysis can help determine the minimum sample size needed to reject the simpler Rasch model. In the following sections, we will first perform an a priori power analysis and then a posteriori power analysis to test the Rasch model against the 2PL model.

Scenario

A school is interested in accurately assessing the math skills of their students to inform instructional strategies and curriculum development. They want to administer a math test to a group of students and collect their item response data. In the end the goal is to find an adequate model for the data to better understand how the math test assesses the math ability of students. We propose two modeling approaches: the Rasch model or the Two-Parameter Logistic (2PL) model.

The Rasch model assumes that an item only influences the response probability through item difficulty, disregarding variations in item discrimination. We are primarily interested in estimating the difficulty level of each math item to identify areas where students might struggle the most. The Rasch model can provide precise estimates of item difficulty, allowing the school to focus on improving those specific concepts or skills.

However, we also recognize that the 2PL model can offer additional insights by considering item discrimination. By incorporating discrimination parameters, the 2PL model captures the extent to which an item differentiates between students with high and low math abilities. This information can help identify items that are particularly effective at discriminating between students with varying skill levels.

To make an informed decision, we plan to compare the Rasch and 2PL by testing them against each other using a likelihood ratio test. Once we have data we could simply test if there is a statistically significant difference between the models. But this test is only reliable if there is enough data to begin with. To ensure that our planned test will be able to detect differences up to a margin of error, we decide to do an a-priori power analysis to determine the required sample size before we start collecting data. For this we use the mlpwr package.

Set Up

To simulate data for an IRT task like the one described above, we will use the mirt package. For the subsequent power simulation we will use the mlpwr package. If it is your first time using them, you have to install them like so:

install.packages("mlpwr")
install.packages("mirt")

Now the packages are permanently installed on your computer. To use them in R you need to (re-)load them every time you start a session.

library(mlpwr)
library(mirt)

Data Preparation

To simulate the data for our scenario, we need the following specifications:

1. In discussion with the schools experts we define the discriminability of the items a, and the item difficulty d. This corresponds to slopes and intercepts in the 2PL model.
2. For illustration purposes we simulate N = 100 observations.
3. The data items should be simulated from a 2PL model with binary response format (0 = incorrectly solved, 1 = correctly solved)

# Defining intercepts and slopes
a <- c(1.04, 1.2, 1.19, 0.61, 1.31, 0.83, 1.46, 1.27, 0.51, 0.81)
d <- c(0.06, -1.79, -1.15, 0.88, -0.2, -1.87, 1.23, -0.08, -0.71, 0.6)

# Setting number of observations 
N <- 100

# Itemtype 
itemtype <- "2PL"

Afterwards we can simulate a dataset using the function simdata in the package mirt.

# Simulate Data
sim_data <- simdata(a = a, d = d, N = N, itemtype = itemtype)

# First 5 rows if simulated data
sim_data[1:5,]

##      Item_1 Item_2 Item_3 Item_4 Item_5 Item_6 Item_7 Item_8 Item_9 Item_10
## [1,]      1      0      0      0      0      0      1      0      1       1
## [2,]      1      0      1      0      0      0      0      0      0       1
## [3,]      1      0      0      1      1      0      0      1      0       1
## [4,]      0      0      0      0      0      0      1      0      0       0
## [5,]      0      0      1      1      0      0      0      0      1       1

Model fit

After discussions with the schools experts we deem items 1-4 the most important for their test. Thus we want to ensure a good model fit for these items. Both a Rasch model or 2PL model are plausible for these items, thus we fit them both then administrate a likelihood ratio test to check which model fits better. The rest of the items are fit with a 2PL model.

# Fit 2PL model
mod <- mirt(sim_data)

# Rasch contsraint for items 1-4
constrained <- "F = 1-4
          CONSTRAIN = (1-4, a1)"
# Fit constrained model
mod_constrained <- mirt(sim_data, constrained)  # Fit 2PL with equal item discrimination

# Compare model fit
res <- anova(mod_constrained, mod)  

We rely on the p-value of the likelihood ratio test to compare the models. We deem the 2PL model significantly better if the p-value is below 0.01 and extract this in the form of a logical TRUE/FALSE response.

res$p[2] < 0.01  # extract significance

## [1] TRUE

Power Analysis

The previous section showed how we can perform a data simulation test with a subsequent hypothesis test. Now we know the test’s items approximately follow the specifications above and we want to make sure that the school recruits enough participants to take their test in order to later select the correct model. Thus we perform a power simulation with mlpwr.

A power simulation can be done using the find.designfunction. For our purpose we submit 4 parameters to it:

1. simfun: a simulation function, that generates the data and does the hypothesis test, outputting a logical value. We use the function described before.
2. boundaries: the boundaries of the design space that are searched. This should be set to a large enough value in order to explore the design space sufficiently.
3. power: the desired power level. We set 0.95 as the desired power level.
4. evaluations (optional): this optional parameter makes the computation faster by limiting the number of reevaluations. But this also makes the computation less stable and precise.

Note: additional specifications are possible (see documentation) but not necessary. For most parameters like the choice of surrogate function the default values should already be good choices. As we are working with only 1 design parameter here (N = sample size) we don’t need to submit a cost function that weighs the design parameters.

The find.design function needs to reiterate a data simulation multiple times. For this purpose it expects a data generating function (DGF) as its main input. The DGF takes an observation number N as input and must return a logical value of whether the hypothesis test was TRUE or FALSE. We can formulate the above simulation process in a function like so:

simfun_irt1 <- function(N) {

    # generate data
    dat <- simdata(a = c(1.04, 1.2, 1.19, 0.61, 1.31,
        0.83, 1.46, 1.27, 0.51, 0.81), d = c(0.06,
        -1.79, -1.15, 0.88, -0.2, -1.87, 1.23, -0.08,
        -0.71, 0.6), N = N, itemtype = "2PL")

    # test hypothesis
    mod <- mirt(dat)  # Fit 2PL Model
    constrained <- "F = 1-4
          CONSTRAIN = (1-4, a1)"
    mod_constrained <- mirt(dat, constrained)  # Fit 2PL with equal slopes

    res <- anova(mod_constrained, mod)  # perform model comparison
    res$p[2] < 0.01  # extract significance
}

This function is also implemented the same way in the mlpwr package and can be accessed using:

example.simfun("irt1")

With the specified parameters we can perform the power analysis. The boundaries were set based on an educated guess from experience, but if the analysis fails one can always readjust them and do the analysis again. We set a random seed for reproducibility reasons.

set.seed(123)
res <- find.design(simfun = simfun_irt1, boundaries = c(40,
    100), power = .95, evaluations = 2000)

Now we can summarize the results using the summary command.

summary(res)

## 
## Call:
## find.design(simfun = simfun_irt1, boundaries = c(40, 100), power = 0.95, 
##     evaluations = 2000)
## 
## Design: N = 56
## 
## Power: 0.95271,  SE: 0.00601
## Evaluations: 2000,  Time: 655.23,  Updates: 16
## Surrogate: Logistic regression

As we can see the calculated sample size for the desired power of 0.95 is 56. The estimated power for this sample size is 0.95271 with a standard error of 0.00601.The summary additionally reports the number of simulation function evaluations, the time until termination in seconds, and the number of surrogate model updates. See Zimmer & Debelak (2023) for more details. We can also plot our power simulation and look at the calculated function using the plot function.

plot(res)

Confidence Intervals (gray) are printed in addition to the estimated power curve (black), so we can get a feel for how the design parameter (here sample size N) influences the power level and also where the prediction is more or less uncertain. The black dots show us the simulated data.

We finished our power analysis and report back to the school that a sample size of 56 should be sufficient to determine an appropriate model for their IRT problem.

Scenario

A school is developing a math test. They want to ensure that the test items are fair and do not favor or disadvantage any specific group of students based on their first language. The school’s experts suspect that especially the first test item could be problematic. Thus, they plan to investigate the presence of DIF in the first item of the test in a study. The school will need to recruit native English speakers and non-native English speakers for this study, but they don’t know how many. Additionally their time resources for recruitment are limited. To determine an appropriate sample size under the given cost constraint (limited time restraints) we use the mlpwr package for the power analysis.

Set Up

To simulate data for an IRT task like the one described above, we will use the mirt package. For the subsequent power simulation we will use the mlpwr package. If you did not run the first part of the script, the mlpwr and mirt packages have to be installed and loaded.

library(mlpwr)
library(mirt)

Data Preparation

The data simulation follows a similar approach to that in the Rasch vs. 2PL section. Thus we will directly set up the simulation function.

simfun_irt2 <- function(N1, N2) {
  # generate data
  a1 <- a2 <- c(1.04, 1.2, 1.19, 0.61, 1.31, 0.83,
                1.46, 1.27, 0.51, 0.81)
  d1 <- d2 <- c(0.06, -1.79, -1.15, 0.88, -0.2, -1.87,
                1.23, -0.08, -0.71, 0.6)
  a2[1] <- a2[1] + 0.3
  d2[1] <- d2[1] + 0.5
  dat1 <- simdata(a = a1, d = d1, N = N1, itemtype = "2PL")
  dat2 <- simdata(a = a2, d = d2, N = N2, itemtype = "2PL")
  dat <- as.data.frame(rbind(dat1, dat2))
  group <- c(rep("1", N1), rep("2", N2))
  
  # Fit model
  model <- multipleGroup(dat, 1, group = group)
  # Perform DIF for item 1
  dif <- DIF(model, which.par = c("a1", "d"), items2test = c(1))
  #Extract significance
  dif$p[1] < 0.05
}

The simulation consists of the following steps:

1. Data Generation: The function generates simulated data using the simdata function. It defines two sets of item parameters, a1 and d1, for the first group, and a2 and d2 for the second group. These parameters represent slope and intercept (equates to item discrimination and item difficulty), respectively. The N1 and N2 parameters are the design parameters and specify the sample sizes for the first and second groups, respectively. We assume DIF for item 1, meaning native English speakers (group == 1) have an easier time solving this item than non-native English speakers (group == 2) Thus, the parameters for the first item differ between the groups (higher difficulty and discrimination for non-native English speakers), while all the others are the same.

2. Data Preparation: The generated data for both groups are combined into a single data frame dat, and a vector group is created to indicate the group membership (native vs. non-native) of each observation.

3. Fit model: The multipleGroup function is used to fit a multiple-group IRT model to the combined data (dat) with one latent trait dimension. The group information is provided using the group argument.

4. Perform DIF for item 1: The DIF function is applied to the fitted model (model) to perform DIF analysis. The which.par argument specifies the parameters to test for DIF, including discrimination (a) and difficulty (d). The items2test argument specifies the specific item(s) to test for DIF, with item 1 selected in this case.

5. Extract significance: The significance of the DIF analysis for item 1 is extracted by comparing the p-value (dif$p[1]) with a significance threshold of 0.05. If the p-value is less than 0.05, the function returns TRUE, indicating the presence of DIF for item 1. Otherwise, it returns FALSE, suggesting no significant DIF for item 1. In other words if the function returns true, there is evidence that the difficulty of item 1 varies for native vs. non-native speakers of the same ability.

Following is one example of a dataframe simulated by the function. The group argument corresponds to native vs. non-native speaker, the rows to individual students, the columns to the 10 items in the math test.

##    group Item_1 Item_2 Item_3 Item_4 Item_5 Item_6 Item_7 Item_8 Item_9 Item_10
## 1      1      1      1      0      1      0      0      1      1      1       1
## 2      1      0      1      1      0      1      1      1      1      0       1
## 3      1      1      0      0      1      0      0      1      0      0       0
## 4      1      0      1      0      0      1      0      0      0      0       0
## 5      1      1      0      0      0      1      1      0      0      0       1
## 6      2      0      0      0      1      1      0      1      1      0       1
## 7      2      1      0      1      1      1      0      0      1      1       0
## 8      2      1      1      1      1      1      0      0      1      1       1
## 9      2      0      0      1      1      0      0      1      0      0       0
## 10     2      1      0      1      1      0      0      1      1      1       1

Cost function

Now that the data generation and hypothesis test is done, we also need to specify a cost function because we have more than one design parameter. For this artificial scenario we assume that recruiting non-native speakers is harder than recruiting native English speakers, because the school has to spend more time recruiting. Put differently the cost for recruiting non-native participants is higher. We include this into the power analysis by defining a cost function like so:

costfun_irt2 <- function(N1, N2) 5 * N1 + 7 * N2

This assumes that the school spends on average 5 minutes recruiting a native speaker vs. 7 minutes for non-native speakers. Contrary to the previous example, the school now has a set budget of time to recruit participants. They want to know how they should allocate their resources in order to gain maximum power.

Power Analysis

Now that all components are defined, a power analysis using find.design from mlpwr is possible. We need the following inputs,

1. simfun: a simulation function, that generates the data and does the hypothesis test, outputting a logical value. We use the function described before.
2. boundaries: the boundaries of the design space that are searched. This should be put to large intervals in order to ensure a sufficient portion of the design space is explored by the algorithm. We have to submit this as a list of named vectors, so the function knows which boundaries correspond to what design parameter.
3. cost: the maximum cost (= time budget of the school) that the school has. The algorithm will maximize for power given the budget constraint.
4. evaluations (optional): this optional parameter makes the computation faster by limiting the number of reevaluations. But this also makes the computation less stable and precise.

Note 1: additional specifications are possible (see documentation) but not necessary. For most parameters like the choice of surrogate function the default values should already be good choices.

With the specified parameters we can perform the power analysis. We set a random seed for reproducibility reasons.

set.seed(111)
res <- find.design(simfun = simfun_irt2, boundaries = list(N1 = c(100,
    700), N2 = c(100, 700)), costfun = costfun_irt2,
    cost = 4000, evaluations = 1000)

Now we can summarize the results using the summary command.

summary(res)

## 
## Call:
## find.design(simfun = simfun_irt2, boundaries = list(N1 = c(100, 
##     700), N2 = c(100, 700)), evaluations = 1000, costfun = costfun_irt2, 
##     cost = 4000)
## 
## Design: N1 = 363, N2 = 312
## 
## Power: 0.63819,  SE: 0.01743,  Cost: 3999
## Evaluations: 1000,  Time: 232.59,  Updates: 32
## Surrogate: Gaussian process regression

As we can see the calculated sample sizes for a cost of max. 4000 are 363 for native speakers and 312 for the non-native speakers. The estimated power for the sample sizes is 0.63819 with a standard error of 0.01743. The summary additionally reports the number of simulation function evaluations, the time until termination in seconds, and the number of surrogate model updates. The details of the surrogate modeling algorithm are described in a paper.

We can also plot our power simulation and look at the calculated function using the plot function.

plot(res)

The black dots show us the simulated data. The red line corresponds to the cost constraint. The violet cross corresponds to the optimal design. The power level is indicated by the blue color.

We conclude our power analysis with the before stated results: 363 for native speakers and 312 for the non-native speakers and let the school take over the recruitment for their study.