10.8 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

EXERCISES

• Write your code to estimate $p_{00}$ and $p_{11}$ for attribute 1.
• Compare the classification accuracy of attribute 1 from the analytic solution with the one from Monte Carlo approach.