Introduction to Probability and Statistics for Engineers and Scientists

326 Chapter 8:Hypothesis Testing

Suppose now that we want to test the null hypothesis thatpis equal to some specified
value; that is, we want to test

H 0 :p=p 0 versus H 1 :p=p 0

IfX, a binomial random variable with parametersnandp, is observed to equalx, then
a significance levelαtest would rejectH 0 if the valuexwas either significantly larger or
significantly smaller than what would be expected whenpis equal top 0. More precisely,
the test would rejectH 0 if either

P{Bin(n,p 0 )≥x}≤α/2 or P{Bin(n,p 0 )≤x}≤α/2

In other words, thep-value whenX=xis

p-value=2 min(P{Bin(n,p 0 )≥x},P{Bin(n,p 0 )≤x})

EXAMPLE 8.6c Historical data indicate that 4 percent of the components produced at
a certain manufacturing facility are defective. A particularly acrimonious labor dispute has
recently been concluded, and management is curious about whether it will result in any
change in this figure of 4 percent. If a random sample of 500 items indicated 16 defectives
(3.2 percent), is this significant evidence, at the 5 percent level of significance, to conclude
that a change has occurred?

SOLUTION To be able to conclude that a change has occurred, the data need to be strong
enough to reject the null hypothesis when we are testing

H 0 :p=.04 versus H 1 :p=.04

wherepis the probability that an item is defective. Thep-value of the observed data of 16
defectives in 500 items is

p-value=2 min{P{X≤ 16 },P{X≥ 16 }}

whereXis a binomial (500, .04) random variable. Since 500×.04=20, we see that

p-value= 2 P{X≤ 16 }

SinceXhas mean 20 and standard deviation

√
20(.96)=4.38, it is clear that twice the
probability thatXwill be less than or equal to 16 — a value less than one standard deviation
lower than the mean — is not going to be small enough to justify rejection. Indeed, it can
be shown that

p-value= 2 P{X≤ 16 }=.432

and so there is not sufficient evidence to reject the hypothesis that the probability of
a defective item has remained unchanged. ■