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9-8 CONTINGENCY TABLE TESTS 321

where u is the probability that a randomly selected element falls in row class i and v is the
i
j
probability that a randomly selected element falls in column class j. Now, assuming inde-
pendence, the estimators of u and v are
j
i
1 c
n a
u ˆ i O ij
j 1
1 r
n a
v ˆ O ij (9-40)
j
i 1
Therefore, the expected frequency of each cell is
1 c r
E nu ˆ v ˆ n a O ij a O ij (9-41)
i j
ij
j 1 i 1
Then, for large n, the statistic

r c 1O E 2 2
ij
ij
2
0 a a (9-42)
i 1 j 1 E ij
has an approximate chi-square distribution with (r 1)(c 1) degrees of freedom if the null
hypothesis is true. Therefore, we would reject the hypothesis of independence if the observed
2 2
value of the test statistic 0 exceeded ,(r 1)(c 1) .
EXAMPLE 9-14 A company has to choose among three pension plans. Management wishes to know whether
the preference for plans is independent of job classification and wants to use 0.05. The
opinions of a random sample of 500 employees are shown in Table 9-3.
To find the expected frequencies, we must first compute (340 500) 0.68, u ˆˆ u 1 2
(160 500) 0.32, v ˆ 1 (200 500) 0.40, v ˆ 2 (200 500) 0.40, and v ˆ 3 (100 500)
0.20. The expected frequencies may now be computed from Equation 9-41. For example, the
expected number of salaried workers favoring pension plan 1 is

nu ˆ v ˆ 50010.68210.402 136
E 11 1 1
The expected frequencies are shown in Table 9-4.
The eight-step hypothesis-testing procedure may now be applied to this problem.
1. The variable of interest is employee preference among pension plans.
2. H 0 : Preference is independent of salaried versus hourly job classification.

Table 9-3 Observed Data for Example 9-14 Table 9-4 Expected Frequencies for Example 9-14

Pension Plan Pension Plan
Job Classification 1 2 3 Totals Job Classification 1 2 3 Totals
Salaried workers 160 140 40 340 Salaried workers 136 136 68 340
Hourly workers 40 60 60 160 Hourly workers 64 64 32 160
Totals 200 200 100 500 Totals 200 200 100 500