Introduction to Probability and Statistics for Engineers and Scientists

330 Chapter 8:Hypothesis Testing

8.7 TESTS CONCERNING THE MEAN OF A

POISSON DISTRIBUTION

LetXdenote a Poisson random variable having meanλand consider a test of

H 0 :λ=λ 0 versus H 1 :λ=λ 0

If the observed value ofXisX=x, then a levelαtest would rejectH 0 if either

Pλ 0 {X≥x}≤α/2 or Pλ 0 {X≤x}≤α/2 (8.7.1)

wherePλ 0 means that the probability is computed under the assumption that the Poisson
mean isλ 0. It follows from Equation 8.7.1 that thep-value is given by

p-value=2 min(Pλ 0 {X≥x},Pλ 0 {X≤x})

The calculation of the preceding probabilities that a Poisson random variable with mean
λ 0 is greater (less) than or equal toxcan be obtained by using Program 5.2.

EXAMPLE 8.7a Management’s claim that the mean number of defective computer chips
produced daily is not greater than 25 is in dispute. Test this hypothesis, at the 5 percent
level of significance, if a sample of 5 days revealed 28, 34, 32, 38, and 22 defective chips.

SOLUTION Because each individual computer chip has a very small chance of being defec-
tive, it is probably reasonable to suppose that the daily number of defective chips is
approximately a Poisson random variable, with mean, say,λ. To see whether or not
the manufacturer’s claim is credible, we shall test the hypothesis

H 0 :λ≤25 versus H 1 :λ> 25

Now, underH 0 , the total number of defective chips produced over a 5-day period is
Poisson distributed (since the sum of independent Poisson random variables is Poisson)
with a mean no greater than 125. Since this number is equal to 154, it follows that the
p-value of the data is given by

p-value=P 125 {X≥ 154 }

= 1 −P 125 {X≤ 153 }

=.0066 from Program 5.2

Therefore, the manufacture’s claim is rejected at the 5 percent (as it would be even at the
1 percent) level of significance. ■