Introduction to Probability and Statistics for Engineers and Scientists

324 Chapter 8:Hypothesis Testing

If we letXdenote the number of defects in the sample of sizen, then it is clear that
we wish to rejectH 0 whenXis large. To see how large it need be to justify rejection at the
αlevel of significance, note that

P{X≥k}=

∑n

i=k

P{X=i}=

∑n

i=k

(

n

i

)

pi(1−p)n−i

Now it is certainly intuitive (and can be proven) thatP{X≥k}is an increasing function
ofp— that is, the probability that the sample will contain at leastkerrors increases in the
defect probabilityp. Using this, we see that whenH 0 is true (and sop≤p 0 ),

P{X≥k}≤

∑n

i=k

(

n

i

)

p 0 i(1−p 0 )n−i

Hence, a significance levelαtest ofH 0 :p≤p 0 versusH 1 :p>p 0 is to rejectH 0 when

X≥k∗

wherek∗is the smallest value ofkfor which

∑n

i=k

(n

i

)

p 0 i(1−p 0 )n−i≤α. That is,

k∗=min

{

k:

∑n

i=k

(

n

i

)

pi 0 (1−p 0 )n−i≤α

}

This test can best be performed by first determining the value of the test statistic —
say,X=x— and then computing thep-value given by

p-value=P{B(n,p 0 )≥x}

=

∑n

i=x

(

n

i

)

p 0 i(1−p 0 )n−i

EXAMPLE 8.6a A computer chip manufacturer claims that no more than 2 percent of the
chips it sends out are defective. An electronics company, impressed with this claim, has
purchased a large quantity of such chips. To determine if the manufacturer’s claim can be
taken literally, the company has decided to test a sample of 300 of these chips. If 10 of
these 300 chips are found to be defective, should the manufacturer’s claim be rejected?

SOLUTION Let us test the claim at the 5 percent level of significance. To see if rejection
is called for, we need to compute the probability that the sample of size 300 would
have resulted in 10 or more defectives whenpis equal to .02. (That is, we compute the
p-value.) If this probability is less than or equal to .05, then the manufacturer’s claim