Introduction to Probability and Statistics for Engineers and Scientists

11.4Tests of Independence in Contingency Tables 495

we obtain, after some computation, that with the five regions as given in the beginning of
this section,

ˆp 1 =.04214

ˆp 2 =.13346

ˆp 3 =.43434

ˆp 4 =.28841

ˆp 5 =.10164

Using the data valuesX 1 = 6,X 2 = 5,X 3 = 8,X 4 = 6,X 5 = 5, an additional
computation yields the test statistic value

T=

∑^5

i= 1

(Xi− 30 ˆpi)^2

30 ˆpi

=21.99156

To determine thep-value, we run Program 5.8.1a. This yields

p-value≈P{χ 32 >21.99}

= 1 −.999936

=.000064

and so the hypothesis of an underlying Poisson distribution is rejected. (Clearly,
there were too many weeks having 0 accidents for the hypothesis that the underlying
distribution is Poisson with mean 3.167 to be tenable.) ■

11.4 Tests of Independence in Contingency Tables

In this section, we consider problems in which each member of a population can
be classified according to two distinct characteristics — which we shall denote as the
X-characteristic and theY-characteristic. We suppose that there arerpossible values for
theX-characteristic andsfor theY-characteristic, and let

Pij=P{X=i,Y=j}

fori=1,...,r,j =1,...,s. That is,Pijrepresents the probability that a randomly
chosen member of the population will haveX-characteristiciandY-characteristicj.