5.3 Likelihood of an individual response vector (Cont’d)

Suppose these items are DINA-type items and assume that guessing and slip parameters are .1 and .2 respectively for all of them.

Recall that for the DINA model, the conditional success probability of student $i$ to item $j$ is $$\begin{equation} P(Y_{ij}=1|\mathbf{\alpha}_{c})= \begin{cases} g_j & \quad \text{if } \mathbf{\alpha}_{c}^T\mathbf{q}_j< \mathbf{q}_{j}^T\mathbf{q}_j\\ 1-s_j & \quad \text{otherwise} \end{cases} \end{equation}$$

Calculate the conditional success probability $P(Y_{j}=1|\mathbf{\alpha}_c)$, which can be saved in a matrix with each item (in rows) and each attribute profile (i.e., 00,10,01 and 11 in columns):

Code

g <- 0.1
s <- 0.2
J <- 5
K <- 2
N <- 10
p.y.given.alpha <- matrix(NA, nrow = J, ncol = 2^K)
attributeprofiles <- GDINA::attributepattern(2)
for (j in 1:J) {
    for (a in 1:2^K) {
        if (sum(attributeprofiles[a, ] * Q[j, ]) < sum(Q[j, ] * Q[j, ])) {
            p.y.given.alpha[j, a] <- g
        } else {
            p.y.given.alpha[j, a] <- 1 - s
        }
    }
}

	00	10	01	11
item 1	0.1	0.8	0.1	0.8
item 2	0.1	0.8	0.1	0.8
item 3	0.1	0.1	0.8	0.8
item 4	0.1	0.1	0.8	0.8
item 5	0.1	0.1	0.1	0.8

Let $\alpha_i$ be the attribute profile of student $i$. Under the assumption of local independence, the likelihood of observing a response vector $\mathbf{Y}_i$ for student $i$ is

$$\begin{equation} L( {\mathbf{Y}_i}|{\alpha_i})=\prod_{j = 1}^J P ({Y_{ij}}=1|\mathbf{\alpha}_i)^{Y_{ij}}[1 - P(Y_{ij}=1|\mathbf{\alpha}_i)]^{1 - {Y_{ij}}} \end{equation}$$

Exercise 5.1 Verify that for the first student, $L( {\mathbf{Y}_1}|{\mathbf{\alpha} _i=10})=.00144$.

What is $L( {\mathbf{Y}_3}|{\mathbf{\alpha} _i=01})$?

Click for Answer

$L( {\mathbf{Y}_i}|{\mathbf{\alpha} _c})$ can be presented in a matrix of $N\times 2^K$:

Code

Li.given.alpha.c <- matrix(NA, nrow = 10, ncol = 4)
for (i in 1:10) {
    # for each student for each latent class
    for (cc in 1:4) {
        Li.given.alpha.c[i, cc] <- prod((p.y.given.alpha[, cc]^y[i, ]) *
            (1 - p.y.given.alpha[, cc])^(1 - y[i, ]))
    }
}
colnames(Li.given.alpha.c) <- c("00", "10", "01", "11")
rownames(Li.given.alpha.c) <- paste("student", 1:10)
Li.given.alpha.c

##                 00      10      01     11
## student 1  0.00081 0.00144 0.00144 0.0205
## student 2  0.00729 0.01296 0.01296 0.0051
## student 3  0.00081 0.00004 0.05184 0.0205
## student 4  0.00081 0.05184 0.00144 0.0205
## student 5  0.00729 0.01296 0.00036 0.0051
## student 6  0.00729 0.00036 0.01296 0.0051
## student 7  0.00729 0.01296 0.01296 0.0051
## student 8  0.00009 0.00016 0.00576 0.0819
## student 9  0.00081 0.05184 0.00144 0.0205
## student 10 0.00081 0.00144 0.00144 0.0205