10.5 Estimating \(2\times 2\) Contingency Table

To estimate the table for attribute \(k\), we need to know \(P(\alpha_{ik}=a|\mathbf{y}_i)\) and \(I(\hat{\alpha}_{ik}=b|\mathbf{y}_i)\). Let us take student 8 as an example. Run the following code:

Code

library(GDINA)
Y <- ecpe$dat
Q <- ecpe$Q
lcdm <- GDINA(Y, Q, model = "GDINA", verbose = 0)
mp <- personparm(lcdm, "mp")
mp.student8 <- mp[8, ]
mp.student8

##     A1     A2     A3 
## 0.0041 0.4801 0.9648

Recall that \[ \hat{p}_{00}=\frac{1}{n}\sum_{i=1}^nP(\alpha_{ik}=0|\mathbf{y}_i)I(\hat{\alpha}_{ik}=0|\mathbf{y}_i) \] Verify that for attribute 1 (i.e., \(k=1\)) and student 8 (i.e., \(i=8\)), \(I(\hat{\alpha}_{ik}=0|\mathbf{y}_i)=1\) and \(P(\alpha_{ik}=0|\mathbf{y}_i)=\) 0.9959