This exercise illustrates the joint effect of the parameter values under the threeparameter model.

(a) Set the values of the item parameters to a D 1:0, b D 2:0, and c D 0:10.

Type in the following command line:

> icccal(b=-2.0, a=1.0, c=.10)

(b) The computer will display the table of computations. Study the table for a few minutes to see the relation between the probability of correct response and the ability scores.

(c) The item characteristic curve will be displayed on the screen by typing in the following command line:

> icc(b=-2.0, a=1.0, c=.10)

(d) The item characteristic curve is located in the low-ability end of the scale and it is moderately steep.

(e) Next, we want to put another item characteristic curve on the same graph.

Type in the following command line:

> par(new=T)

(f) Now set the values of the item parameters to b D 0:0, a D 1:5, and c D 0:20.

Then repeat steps a through c.

(g) This will place a second curve on the graph.

(h) Now repeat steps e through f using the values of b D 2:0, a D 0:5, and c D 0:30.

(i) At this point you should have three item characteristic curves displayed on the graph. Again the values of b locate the items along the ability scale. But the ability level at which P. / D 0:5 does not correspond to the value of b but is slightly lower. Recall that under the three-parameter model, b is the point on the ability scale where the probability of correct response is .1 C c/=2 rather than 0:5. The slopes of the curves at b reflect the values of

a. The lower tails of the three curves approach their values of c at the lowest levels of ability. However, this is not apparent for the curve with b D 2:O as the values of P. / are still rather large at D 3:0.

(j) At this juncture Exercise 3 is completed.