Introduction to Probability and Statistics for Engineers and Scientists

494 Chapter 11:Goodness of Fit Tests and Categorical Data Analysis

compute the test statistic

T=

∑k

i= 1

(Xi−nˆpi)^2

npˆi

whereXiis, as before, the number ofYjthat fall in regioni,i =1,...,k, andpˆiis
the estimated probability of the event thatYjfalls in regioni, which is determined by

substitutingλˆforλin expression 11.3.1 forpi.
In general, this approach can be utilized whenever there are unspecified parameters in
the null hypothesis that are needed to compute the quantitiespi,i=1,...,k. Suppose
now that there aremsuch unspecified parameters and that they are to be estimated by the
method of maximum likelihood. It can then be proven that whennis large, the test statistic
Twill have, whenH 0 is true, approximately a chi-square distribution withk− 1 −m
degrees of freedom. (In other words, one degree of freedom is lost for each parameter that
needs to be estimated.) The test is, therefore, to

reject H 0 if T≥χα^2 ,k− 1 −m

accept H 0 otherwise

An equivalent way of performing the foregoing is to first determine the value of the test
statisticT, sayT=t, and then compute

p-value≈P{χk^2 − 1 −m≥t}

The hypothesis would be rejected ifα≥p-value.

EXAMPLE 11.3a Suppose the weekly number of accidents over a 30-week period is as
follows:

80013402125

18020193 45

33474012 12

Test the hypothesis that the number of accidents in a week has a Poisson distribution.

SOLUTION Since the total number of accidents in the 30 weeks is 95, the maximum
likelihood estimate of the mean of the Poisson distribution is

λˆ=^95

30

=3.16667

Since the estimate ofP{Y=i}is then

P{Y=i}=est

e−λˆλˆi

i!