RESEARCH METHOD COHEN ok

582 MULTIDIMENSIONAL MEASUREMENT
by first determining the degrees of freedom for the
marginals.
Each of the variables in our example (sex,
voting preference and social class) contains two
categories. It follows therefore that we have
(2 − 1) degrees of freedom for each of them, given
that the marginal for each variable is fixed. Since
the grand total of all the marginals (i.e. the sample
size) is also fixed, it follows that one more degree
of freedom is also lost. We subtract these fixed
numbers from the total number of cells in our
contingency table. In general therefore:
degrees of freedom (df) = the number of cells in the
table −1 (forN)− the number of cells fixed by the
hypothesis being tested.
Thus, where r = rows, c = columns and l = layers:
df = rcl(r − 1) − (c − 1) − (1 − 1) − 1
= rcl − r − c − l + 2
that is to say df = rcl − r − c − l + 2whenweare
testing the hypothesis of the mutual independence
of the three variables.
In our example:
df = (2)(2)(2) − 2 − 2 − 2 + 2 = 4
From chi-square tables we see that the critical
value of χ 2 with four degrees of freedom is 9.49
at p = 0.05. Our obtained value greatly exceeds
that number. We reject the null hypothesis and
conclude that sex, voting preference and social
class are significantly interrelated.
Having rejected the null hypothesis with respect
to the mutual independence of the three variables,
the researcher’s task now is to identify which
variables cause the null hypothesis to be rejected.
We cannot simply assume that because our
chi-square test has given a significant result,
it therefore follows that there are significant
associations between all three variables. It may
be the case, for example, that an association exists
between two of the variables while the third is
completely independent. What we need now is
atestof‘partialindependence’.Whiteley(1983)
shows the following three such possible tests in
respect of the data in Box 25.17. First, that sex is
independent of social class and voting preference:
(1) p ijk = (p i )(p jk )
Second, that voting preference is independent of
sex and social class
(2) p ijk = (p j )(p ik )
Third, that social class is independent of sex and
voting preference
(3) p ijk = (p k )(p ij )
The following example shows how to construct
the expected frequencies for the first hypothesis.
We can determine the probability of an individual
being, say, woman, Labour, and working-class,
assuming hypothesis (1), as follows:
p 222 = (p 2++ )(p +22 ) = (n 2++) (n +22 )
(N) (N)
p 222 = (270) (240)
(550) (550) = 0.214
E 222 = N(p 2++ )(p +22 ) = 550 (270) (240)
(550) (550) = 117.8
That is to say, assuming that sex is independent
of social class and voting preference, the expected
number of female, working-class Labour supporters
is 117.8.
When we calculate the expected frequencies
for each of the cells in our contingency table in
respect of our first hypothesis (p ijk ) = (p i )(p jk ), we
obtain the results shown in Box 25.20.
χ 2 = ∑ (O − E) 2
= 5.71
E
Box 25.20
Expected frequencies assuming that sex is
independent of social class and voting preference
Middle class
Working class
Conservative Labour Conservative Labour
Men 91.6 25.5 40.7 122.2
Women 88.4 24.5 39.3 117.8
Source:adaptedfromWhiteley1983