Lab exercise 6

Please upload your .pdf file on Moodle by 23:59.

Suppose you want to create a function named my_obj that takes three inputs:

response data frame resp
a criterion score vector Y
weight \(w\), where the default weight is \(w = 0.5\).

and returns the weighted average of the reliability \(s_ir_{iX}\) and the item validity-to-reliability ratio \(\frac{s_ir_{iY}}{s_ir_{iX}}\), that is,

\[w\cdot s_ir_{iX}+(1-w)\frac{s_{i}r_{iY}}{s_ir_{iX}}\].

Complete the following code chunk:

my_obj <- function(resp, Y, w = 0.5){
  # item difficulty
  difficulty <- colMeans(resp)
  
  # item discrimination
  X <- rowSums(resp)
  discrimination <- numeric(ncol(resp))
  for (i in 1:ncol(resp)) {
    discrimination[i] <- cor(resp[, i], X)
  }

  # item SD
  item_sd <- 
  
  # item reliability
  item_rel <- 
  
  # item validity
  r_iy <- numeric(ncol(resp))
  for (i in 1:ncol(resp)) {
    r_iy[i] <- 
  }
  item_val <- 
  
  # weighted average
  obj <- w * item_rel + (1 - w) * (item_val / item_rel)
  
  return(obj)
}

order(my_obj(responses, Y), decreasing = TRUE)[1:10]

Try different values of w (0.0, 0.1, 0.2, …, 0.8, 0.9, 1.0). For each w values, select 10 items with the largest weighted average, \(w \cdot s_ir_{iX} + (1-w) \cdot \frac{s_{i} r_{iY}}{s_i r_{iX}}\).

hints:

ws <- seq(0, 1, 0.1) will create a vector from 0 to 1 with 0.1 increment.

ws <- seq(0, 1, 0.1)
ws

##  [1] 0.0 0.1 0.2 0.3 0.4 0.5
##  [7] 0.6 0.7 0.8 0.9 1.0

Create an output object which is a \(11 \times 10\) matrix.

output <- matrix(NA, nrow = length(ws), ncol = 10)

Write a for loop for (i in 1:length(ws)) {}
At each iteration in the for loop:
- set the new w as the \(i\)th element of ws
- obtain the weighted average using my_obj() function with the new weight w
- find the 10 item numbers that maximize obj
- save the 10 item numbers to the \(i\)th row of the output matrix