Statistical Methods for Psychology

You can think of very much the way you thought of the treatment effect ( ) in the
analysis of variance. It is the contribution of. But for the analysis of variance,
was the amount that was addedto the grand mean to obtain the column mean. Here, on
the other hand, is the amount by which we multiply to obtain the column’s expected
frequency. 5 1.6062 says that the column one expected frequency is 1.6062 times
larger than the overall mean—or 160.62% of it. For the Not Guilty column,

Then we can show that for this model

For cell 11, we would have 80.3119 3 1.6062 5 129, which has reproduced the expected
frequency that we used in Table 17.2. The other expected cell frequencies follow because
rows and columns must sum to row and column totals; we have 1 df.
We have a similar model when we consider just the Fault variable instead of just the
Verdict variable. Here we have

To go one step further, we can consider the independence model (Table 17.1), which
contained both Fault and Verdict effects but not their interaction. Here we will need
both and to account for both Verdict and Fault. Working with the expected frequen-
cies from the independence model we have:

Then, for example,

which agrees, within rounding error, with the actual expected value for the independence
model. You should verify for yourself that in the general case, for the independence model,
the expected frequency for cellijis

I have led you through the last few paragraphs to make a simple but very important point.
In the analysis of variance we wrote an additivelinear model for observations in each cell as

With log-linear models of categorical data, we have seen that we can write the multi-
plicativeindependence model for expected cell frequenciesas

Fij=NttNiVtNjF

Xijk=m1ai1bj1abij 1 eijk

Fij=NttNiVtNjF

F 11 =tNNt 1 VtNF 1 =80.3069 3 1.6062 3 0.9889=127.557

Nt 2 F=

1 (130.441)(50.559)

tN

=

81.2094

80.3069

=1.0112

Nt 1 F=

(^1) (127.559)(49.441)
tN

=

79.4144

80.3069

=0.9889

Nt 2 V=

(^1) (49.441)(50.559)
tN

=

49.9969

80.3069

=0.6226

Nt 1 V=

(^1) (127.559)(130.441)
tN

=

128.9920

80.3069

=1.6062

tN= 14 (127.559)(49.441)(130.441)(50.559)=80.3069

tVj tFi

Fij=NttNiF

Fij=NttNjV

Nt 2 V=

1 (50)(50)

tN

=

50

80.3119

=0.6226

tNjV

NtjV tN

columnj bj

NtjV bj

Section 17.2 Model Specification 637