10.5 Estimating $2\times 2$ Contingency Table

To estimate the table for attribute $k$, we need to know $P({\alpha }_{ik}=a|y_i)$ and $I({\hat{\alpha }}_{ik}=b|y_i)$. 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 $${\hat{p}}_{00}=\frac{1}{n}\sum _{i=1}^{n}P({\alpha }_{ik}=0|y_i)I({\hat{\alpha }}_{ik}=0|y_i)$$ Verify that for attribute 1 (i.e., $k=1$) and student 8 (i.e., $i=8$), $I({\hat{\alpha }}_{ik}=0|y_i)=1$ and $P({\alpha }_{ik}=0|y_i)=$ 0.9959