It can now be proved that the sum of the relative difference between number of
observed weeks (O) and estimated weeks (E) is chi-square distributed:
2
calc
¼
X
N
O EðÞ
2
E
The distribution applies for positive values and has the parameter , which denotes the
number of degrees of freedom and is given by:
¼ðN 1Þ1
where N is the number of observations and the second 1 is the consequence of
introducing an estimate for in the computations.
If our hypothesis that the number of weeks is given by the Poisson distribution
is true (H
0
), the calculated value of the chi-square criterion should be less than
the critical value. We are then able to test the assumption of a Poisson distribution as
follows:
from Table 5.3:
2
calc
¼ 3:2328
degrees of freedom: ¼(4 1) – 1 ¼2
assuming significance level: ¼0.95
tabulated value (from handbook):
2
2,0:95
¼ 5:99
It can be concluded that the Poisson distribution is valid as the calculated value (3.23) is
less than the tabulated value.
5. 5 CHOOSING AMONG ALTERNATIVE TR AINING PROGR AMS
The chi-square test can also be useful for testing other models. Let us take the following
case described by ReVelle and Stephenson (1995). A company has tried out training
programmes of different duration: 1, 3, 5 and 10 days. The attending crew member s were
subject to a rating by their supervisors 6 months after the training session. The supervisor
used the follo wing ranking: excellent, good or poor.
The result of the assessment is shown in the upper part of Table 5.4. Observation of
the data may support the suspicion that there is no clear relationship between course
duration and rating. It is interesting to note the low number of ‘excellent’ ratings for the
participants in the 10-day program.
Against this background, it may be interesting to test the following null-hypothesis:
H
0
¼ No correlation between Duration and Rating
5.5 CHOOSING AMONG ALTERNATIVE TRAINING PR OG RAMS 127