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Solutions to the exercises of
Chapter 10
• Exercise 10.1
The challenge in this exercise is that you must see that a Rasch
model has to be fit to the replicated data sets too in order to be able
to compute the Q3 index.
• Exercise 10.2
Call δ2 the pairwise interaction parameter and δ3 the three-way interaction parameter. Then,

0
yp1
yp1 + yp2
yp1 + yp2 + yp3




Wp = 



 yp1 + yp2 + yp3 + yp4
Define t1 =
i−1
j=1
ypj and t2 =
0
0
yp1 yp2
yp1 yp2 + yp1 yp3 + yp2 yp3
yp1 yp2 + yp1 yp3 + yp1 yp4
+yp2 yp3 + yp2 yp3 +
yp2 yp4 + yp3 yp4
i−2 i−1
j=1
j<k
k=1





.




ypj ypk . Interpretation:
Holding everything else constant, the odds of giving a 1-response
∗
change with factor eδ2 (t1 −t1 ) if t1 instead of t∗1 previous successes
are achieved. For δ3 a similar interpretation holds, but then for an
increase in previous number of pairs of successes. Note however that
the “holding everything constant” requirement is somewhat artificial
here because if t1 changes, t2 is likely to change as well (but not
automatically).
• Exercise 10.3
For two items:
Pr(Y1 = 1) =
1
Pr(Y1 = 1, Y2 = y)
y=0
=
• Exercise 10.4
exp (θp − β1 ) + exp (2θp − β1 − β2 + δ)
1 + exp (θp − β1 ) + exp (θp − β2 ) + exp (2θp − β1 − β2 + δ)
2
1. Solutions to the exercises of Chapter 10
Pr(Y1 = 1, Y2 = y) =
Pr(Y1 = 1, Y2 = y|θp )dF (θp )
=
Pr(Y1 = 1|θp )Pr(Y2 = y|θp )dF (θp )
1
Pr(Y1 = 1) =
Pr(Y1 = 1, Y2 = y)
y=0
=
Pr(Y1 = 1|θp )dF (θp )
• Exercise 10.5

with ρ1 =
1
 ..
 .

 ρ1
R=
 ρ2

 .
 ..
···
..
.
···
···
..
.
ρ1
..
.
ρ2
..
.
1
ρ2
..
.
ρ2
1
..
.
···
..
.
···
···
..
.
ρ2
···
ρ2
ρ3
···
σ2 +τ 2
σ2 +τ 2 +1 ,
ρ2 =
σ2 √
√
,
σ2 +τ 2 +1 σ2 +1

ρ2
.. 
. 

ρ2 
,
ρ3 

.. 
. 
1
and ρ3 =
σ2
σ2 +1
• Exercise 10.6
Pr(Y1 = y1 ) =
Pr(Y2 = y2 |y1 ) =
exp(y1 (θ − β1 ))
1 + exp(θ − β1 )
exp(y2 (θ − β2 + y1 δ1 ))
1 + exp(θ − β2 + y1 δ1 )
and
Pr(Y3 = y3 , Y4 = y4 |y1 , y2 ) =
exp(y3 (θ − β3 + sδ1 ) + y4 (θ − β4 + sδ1 ) + y3 y4 δ2 )
1 + exp(θ − β3 + sδ1 ) + exp(θ − β4 + sδ1 ) + exp(2θ − β3 − β4 + 2sδ1 + δ2 )
where s = y1 + y2 .
• Exercise 10.7
Design matrix for

1
 0

X=
 0
 0
0
the normal child:
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
y1
0
y1 + y2
0 y1 + y + 2 + y3
1 y1 + y2 + y3 + y4
0
0
0
0
0






1. Solutions to the exercises of Chapter 10
Design matrix for

1
 0

X =
 0
 0
0
3
the disabled child:
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
y1
y1 + y2
y1 + y2 + y3
y1 + y2 + y3 + y4






• Exercise 10.8
The model can be constructed by simply adding θp1 to the exponents in all cumulative probability equations pertaining to the items
involved in the testlet.