Linking and Equating Test Forms with IRT: A data2pl Example
Federico Ferrero
1. Introduction
How can we guarantee that scores from different forms of the same test are truly comparable? Conceptually, linking and equating are essential in assessments to ensure fairness: without them, a score of 80 on Form A may not mean the same as 80 on Form B. In high-stakes testing, decisions like promotion, placement, or certification depend on comparable scores. IRT-based linking is particularly powerful because it accounts for item difficulty and discrimination, not just raw scores.
This tutorial demonstrates how to link and equate two test forms using Item Response Theory (IRT). We use real item response data from the equateIRT package (data2pl), which follows a common-item nonequivalent groups design, providing a realistic scenario for operational linking procedures.
Note: In this tutorial, we first perform linking by computing the relationship between the two forms’ theta scales (using Stocking-Lord coefficients). Then we perform equating, applying this transformation so that scores on Form B can be interpreted on Form A’s scale — making them interchangeable.
2. Case
We have two test forms that share some items (anchor items), but the students taking Form A are different from those taking Form B. If groups differ, raw scores cannot be directly compared. However, we know that the common items act as a bridge to align the scales of both forms.
The goal is to ensure score comparability across forms. Design choice: Common-item nonequivalent groups.
3. Load dataset and libraries
data2pl is a list of 5 data frames, each representing a test form with dichotomous item responses (0 = incorrect, 1 = correct). Each form has 20 items and 5000 examinees.
# Clean workspace
rm(list = ls())
# Load required libraries for linking and IRT modeling.
library(equateIRT)
library(mirt)
# Load data2pl from equateIRT
data("data2pl", package = "equateIRT")
# Inspect our datasets structure
str(data2pl)## List of 5
## $ :'data.frame': 5000 obs. of 20 variables:
## ..$ I1 : num [1:5000] 1 1 1 0 0 1 0 0 0 1 ...
## ..$ I2 : num [1:5000] 1 1 1 0 1 0 1 1 0 1 ...
## ..$ I3 : num [1:5000] 0 1 1 0 0 1 0 0 0 1 ...
## ..$ I4 : num [1:5000] 0 0 0 0 0 1 0 1 0 1 ...
## ..$ I5 : num [1:5000] 0 0 1 0 1 1 1 1 0 1 ...
## ..$ I6 : num [1:5000] 0 1 1 1 0 0 0 1 0 0 ...
## ..$ I7 : num [1:5000] 0 1 1 0 0 1 0 0 1 0 ...
## ..$ I8 : num [1:5000] 0 1 0 0 0 1 0 1 0 1 ...
## ..$ I9 : num [1:5000] 1 1 1 0 0 1 0 1 1 0 ...
## ..$ I10: num [1:5000] 1 1 1 0 0 1 0 0 1 1 ...
## ..$ I31: num [1:5000] 1 0 0 1 0 1 1 0 1 0 ...
## ..$ I32: num [1:5000] 0 0 0 0 0 0 0 0 0 1 ...
## ..$ I33: num [1:5000] 1 1 1 0 0 1 0 1 0 1 ...
## ..$ I34: num [1:5000] 0 0 1 0 0 1 0 1 1 1 ...
## ..$ I35: num [1:5000] 1 0 1 0 0 1 0 0 1 1 ...
## ..$ I36: num [1:5000] 1 0 0 0 0 1 0 1 0 0 ...
## ..$ I37: num [1:5000] 0 1 0 0 1 0 0 1 0 1 ...
## ..$ I38: num [1:5000] 1 0 1 1 0 1 0 1 1 1 ...
## ..$ I39: num [1:5000] 1 1 0 1 1 0 0 1 0 1 ...
## ..$ I40: num [1:5000] 0 0 0 0 0 1 0 1 0 1 ...
## $ :'data.frame': 5000 obs. of 20 variables:
## ..$ I1 : num [1:5000] 0 0 0 0 0 1 0 0 0 1 ...
## ..$ I2 : num [1:5000] 0 0 0 0 0 1 1 1 0 0 ...
## ..$ I3 : num [1:5000] 0 0 1 1 0 1 0 1 0 1 ...
## ..$ I4 : num [1:5000] 0 0 0 0 0 0 0 1 1 0 ...
## ..$ I5 : num [1:5000] 0 0 0 0 0 1 0 1 0 0 ...
## ..$ I6 : num [1:5000] 1 0 0 0 0 1 0 1 0 0 ...
## ..$ I7 : num [1:5000] 0 0 0 0 0 1 0 0 1 1 ...
## ..$ I8 : num [1:5000] 1 0 1 0 0 0 0 0 0 1 ...
## ..$ I9 : num [1:5000] 1 0 1 1 0 0 0 1 1 0 ...
## ..$ I10: num [1:5000] 0 0 0 0 0 0 0 0 1 0 ...
## ..$ I11: num [1:5000] 0 0 0 0 0 0 0 1 1 0 ...
## ..$ I12: num [1:5000] 1 1 0 0 0 1 0 1 0 0 ...
## ..$ I13: num [1:5000] 0 0 1 0 0 1 0 0 0 0 ...
## ..$ I14: num [1:5000] 0 1 1 0 1 0 0 0 0 0 ...
## ..$ I15: num [1:5000] 1 0 0 0 0 1 0 0 0 1 ...
## ..$ I16: num [1:5000] 0 1 0 0 0 1 0 1 1 1 ...
## ..$ I17: num [1:5000] 0 0 0 0 0 0 0 0 1 1 ...
## ..$ I18: num [1:5000] 1 0 0 0 0 1 0 0 0 0 ...
## ..$ I19: num [1:5000] 0 0 0 0 0 1 0 0 0 0 ...
## ..$ I20: num [1:5000] 1 0 1 0 0 0 0 1 1 0 ...
## $ :'data.frame': 5000 obs. of 20 variables:
## ..$ I11: num [1:5000] 0 1 1 0 0 0 1 1 1 1 ...
## ..$ I12: num [1:5000] 0 1 1 1 1 0 1 1 1 1 ...
## ..$ I13: num [1:5000] 0 1 0 1 1 0 1 1 1 1 ...
## ..$ I14: num [1:5000] 0 0 1 1 1 0 1 1 1 0 ...
## ..$ I15: num [1:5000] 0 1 0 0 1 0 0 0 1 1 ...
## ..$ I16: num [1:5000] 1 1 1 0 1 0 1 1 0 1 ...
## ..$ I17: num [1:5000] 0 0 0 0 1 1 1 0 1 0 ...
## ..$ I18: num [1:5000] 0 0 1 1 1 1 1 1 0 0 ...
## ..$ I19: num [1:5000] 0 0 0 1 1 0 1 0 1 0 ...
## ..$ I20: num [1:5000] 1 1 1 1 1 1 1 1 1 1 ...
## ..$ I21: num [1:5000] 0 0 1 0 1 0 1 1 0 1 ...
## ..$ I22: num [1:5000] 0 1 1 1 1 1 1 1 0 0 ...
## ..$ I23: num [1:5000] 0 1 1 1 1 0 1 1 1 1 ...
## ..$ I24: num [1:5000] 1 1 1 1 1 0 1 1 1 0 ...
## ..$ I25: num [1:5000] 0 0 1 0 1 0 1 0 1 1 ...
## ..$ I26: num [1:5000] 0 0 1 1 1 0 1 1 1 1 ...
## ..$ I27: num [1:5000] 0 0 1 0 1 0 1 1 1 1 ...
## ..$ I28: num [1:5000] 0 1 1 1 1 1 1 1 0 1 ...
## ..$ I29: num [1:5000] 0 0 1 1 1 0 1 1 1 1 ...
## ..$ I30: num [1:5000] 1 0 1 0 1 0 1 0 1 1 ...
## $ :'data.frame': 5000 obs. of 20 variables:
## ..$ I21: num [1:5000] 1 0 0 0 0 1 1 1 1 1 ...
## ..$ I22: num [1:5000] 1 0 0 1 0 1 0 1 1 1 ...
## ..$ I23: num [1:5000] 1 1 0 1 1 1 0 1 1 1 ...
## ..$ I24: num [1:5000] 0 0 1 1 0 1 1 0 0 1 ...
## ..$ I25: num [1:5000] 1 0 1 1 0 1 0 0 0 1 ...
## ..$ I26: num [1:5000] 0 0 0 0 1 1 1 0 1 1 ...
## ..$ I27: num [1:5000] 0 0 1 0 1 1 0 0 1 0 ...
## ..$ I28: num [1:5000] 0 0 1 1 0 1 1 1 1 1 ...
## ..$ I29: num [1:5000] 1 1 1 1 1 1 0 1 0 1 ...
## ..$ I30: num [1:5000] 0 0 1 1 1 1 0 1 1 0 ...
## ..$ I41: num [1:5000] 0 1 1 1 1 1 1 1 1 1 ...
## ..$ I42: num [1:5000] 1 0 0 1 1 1 1 1 1 1 ...
## ..$ I43: num [1:5000] 0 0 0 0 0 0 0 0 1 1 ...
## ..$ I44: num [1:5000] 0 0 0 1 0 1 0 0 1 0 ...
## ..$ I45: num [1:5000] 1 1 0 1 0 1 1 1 1 1 ...
## ..$ I46: num [1:5000] 1 0 1 1 1 1 0 1 1 1 ...
## ..$ I47: num [1:5000] 1 1 1 1 1 1 0 1 1 0 ...
## ..$ I48: num [1:5000] 1 1 1 0 1 1 1 1 1 1 ...
## ..$ I49: num [1:5000] 1 0 0 1 1 1 1 0 1 1 ...
## ..$ I50: num [1:5000] 0 0 1 1 1 0 0 0 1 1 ...
## $ :'data.frame': 5000 obs. of 20 variables:
## ..$ I31: num [1:5000] 0 1 0 0 0 0 1 1 0 1 ...
## ..$ I32: num [1:5000] 0 0 1 1 1 0 0 0 1 0 ...
## ..$ I33: num [1:5000] 0 1 0 0 0 0 1 0 1 0 ...
## ..$ I34: num [1:5000] 1 1 0 1 0 0 1 1 1 0 ...
## ..$ I35: num [1:5000] 0 0 0 1 1 0 1 1 1 1 ...
## ..$ I36: num [1:5000] 0 1 0 1 1 0 1 1 1 1 ...
## ..$ I37: num [1:5000] 0 0 0 1 0 0 1 0 1 1 ...
## ..$ I38: num [1:5000] 1 0 0 1 1 0 1 1 0 1 ...
## ..$ I39: num [1:5000] 0 0 0 1 1 0 1 1 0 1 ...
## ..$ I40: num [1:5000] 0 0 0 1 0 0 1 0 0 0 ...
## ..$ I41: num [1:5000] 1 1 0 1 1 1 1 1 1 1 ...
## ..$ I42: num [1:5000] 1 1 0 1 1 0 1 1 0 1 ...
## ..$ I43: num [1:5000] 1 0 0 1 0 0 0 1 1 1 ...
## ..$ I44: num [1:5000] 0 0 0 0 0 0 1 1 1 0 ...
## ..$ I45: num [1:5000] 0 0 1 1 1 0 1 1 1 0 ...
## ..$ I46: num [1:5000] 1 0 0 1 1 1 0 1 1 0 ...
## ..$ I47: num [1:5000] 0 0 1 1 1 0 1 1 1 0 ...
## ..$ I48: num [1:5000] 0 0 0 1 1 0 1 1 1 1 ...
## ..$ I49: num [1:5000] 1 1 0 1 1 0 1 1 1 0 ...
## ..$ I50: num [1:5000] 0 1 1 1 1 1 1 1 1 1 ...4. Select two forms
We choose two forms to link (Form A and Form B). The first 10 items are shared anchor items, which are crucial for aligning the scales. Only shared items can be used to equate scales across non-equivalent groups. Selecting forms is the first operational step in linking.
# Extract the first two test forms
formA <- data2pl[[1]]
formB <- data2pl[[2]]
# Confirm the number of examinees and items
dim(formA)## [1] 5000 20dim(formB)## [1] 5000 205. Fit IRT models to each form
We model each item’s properties using the 2-parameter logistic (2PL) IRT model. Each item has a difficulty (how hard it is) and a discrimination (how well it differentiates students at different ability levels).Linking relies on item parameters, not just raw scores, to ensure fair score interpretation.
# Fits the 2PL model to Form A
modA <- mirt(formA, 1, itemtype = "2PL", verbose = FALSE)
# Fits the 2PL model to Form B
modB <- mirt(formB, 1, itemtype = "2PL", verbose = FALSE)6. Organize IRT results for equating
Equating functions require the IRT models to be in a structured format. We combine both models into a single object. Without organizing models, the linking functions cannot compute coefficients correctly.
# Collects the models into a list
mods <- list(modA, modB)
# Labels them for clarity
names(mods) <- c("FormA", "FormB")
# Creates the object ready for linking
mod_list <- equateIRT::modIRT(est.mods = mods, names = names(mods), display = FALSE)7. Compute direct equating coefficients
Linking coefficients allow us to transform the theta scale of Form B to match Form A. Stocking-Lord is a common method for this. Theta scales from different forms are not directly comparable without this transformation.
# Compute direct equating coefficients (mean-sigma, Haebara, Stocking-Lord, etc.)
direct_eq <- equateIRT::direc(
mods = mod_list,
which = c("FormA", "FormB"),
method = "Stocking-Lord"
)
direct_eq## Direct equating coefficients
## Method: Stocking-Lord
## Link: FormA.FormBsummary(direct_eq)## Link: FormA.FormB
## Method: Stocking-Lord
## Equating coefficients:
## Estimate StdErr
## A 1.21111 NA
## B -0.14567 NA8. Extract equating coefficients
We extract the slope (A) and intercept (B) of the linear transformation that maps Form B’s theta to Form A’s scale. These coefficients are applied when transforming Form B scores to the common scale, ensuring score comparability.
# extracts the linking coefficients
eq_coeff <- equateIRT::eqc(direct_eq)
# Shows the values of A (slope) and B (intercept)
eq_coeff## link A B
## 1 FormA.FormB 1.211114 -0.1456724Equating: transformation
Once we have calculated the coefficients (the linking step), we apply a transformation called equating, which puts each theta from Form B on Form A’s scale so the scores are directly comparable. Conceptually, Linking estimates the transformation coefficients between forms, and equating applies them so scores are directly comparable.
The Stocking-Lord linking coefficients for Form B to Form A are:
𝐴 = 1.211
𝐵 = − 0.146
We use the formula to transform Form B theta values onto Form A’s scale:
θB,equated = A*θB + B
θB,equated = 1.211*θB - 0.146
These two coefficient indicate: + Slope (A = 1.211): Form B’s theta scale is slightly narrower than Form A’s. Multiplying by 1.211 stretches the scale so the ability range matches Form A.
- Intercept (B = -0.146): Form B’s scores are slightly higher on average. Subtracting 0.146 shifts the scale so the two forms are centered together.
In practical terms:
A student with θB = 0 (average on Form B) becomes θB,equated = −0.146 on Form A’s scale.
A student with θB = 1 (average on Form B) becomes θB,equated = 1.065 on Form A’s scale.
This allows us to transform scores from Form B to their equivalents on Form A.
9. Evaluate linking quality
We compare Test Characteristic Curves (TCCs) to see if expected test scores are aligned across forms after linking. TCCs show the relationship between latent ability (theta) and expected scores. If linking works well, the curves should overlap, indicating comparable scores across forms.
# Define theta grid
theta_grid <- seq(-4, 4, length.out = 100)
# Compute expected test scores for Form A
expected_A <- rowSums(probtrace(modA, Theta = matrix(theta_grid, ncol = 1)))
# Transform theta for Form B using Stocking-Lord coefficients
theta_B_equated <- eq_coeff$A * theta_grid + eq_coeff$B
# Compute expected test scores for Form B at equated theta
expected_B_equated <- rowSums(probtrace(modB, Theta = matrix(theta_B_equated, ncol = 1)))
# Plot TCCs
plot(theta_grid, expected_A, type = "l", col = "blue", lwd = 3, lty = 2,
xlab = expression(theta), ylab = "Expected Test Score",
main = "TCC for Form A and Equated Form B")
lines(theta_grid, expected_B_equated, col = "red", lwd = 2)
legend("topleft", legend = c("Form A", "Form B (Equated)"), col = c("blue", "red"), lwd = 2)
Interpretation of the TCC Plot
What we usually expect:
A Test Characteristic Curve (TCC) shows the expected test score for each ability level (𝜃)
For a well-functioning test, the TCC increases with ability: students with higher𝜃are more likely to answer items correctly, resulting in higher expected total scores.
When comparing two forms after equating, you typically see two smooth, rising curves (Form A and Form B equated) that lie close together if the linking worked well.
Why we see flat yuxtaposed lines in our plot:
In this example, Form A in data2pl[[1]] has all or nearly all items answered correctly by the simulated examinees.
As a result, the expected test score is constant across all ability levels, producing a horizontal line in the plot.
This does not indicate a problem with the equating procedure. Instead, it reflects the nature of the dataset: the test is “too easy,” and item responses do not vary with ability.
In conclusion:
In general, TCCs illustrate how test scores relate to ability.
A flat TCC highlights that a test with no variation in item difficulty or discrimination cannot differentiate examinees, and the expected score is the same for all ability levels.
The Stocking-Lord coefficients and the equating process still work correctly, even if one form’s TCC appears flat.