Introduction to Probability and Statistics for Engineers and Scientists

8.6Hypothesis Tests in Bernoulli Populations 325

should be rejected. Now

P.02{X≥ 10 }= 1 −P.02{X< 10 }

= 1 −

∑^9

i= 0

(

300

i

)

(.02)i(.98)^300 −i

=.0818 from Program 3.1

and so the manufacturer’s claim cannot be rejected at the 5 percent level of significance. ■

When the sample sizenis large, we can derive anapproximatesignificance levelαtest
ofH 0 :p≤p 0 versusH 1 :p>p 0 by using the normal approximation to the binomial. It
works as follows: Because whennis largeXwill have approximately a normal distribution
with mean and variance

E[X]=np, Var(X)=np(1−p)

it follows that

X−np

√

np(1−p)

will have approximately a standard normal distribution. Therefore, an approximate
significance levelαtest would be to rejectH 0 if

X−np 0

√

np 0 (1−p 0 )

≥zα

Equivalently, one can use the normal approximation to approximate thep-value.

EXAMPLE 8.6b In Example 8.6a,np 0 =300(.02)=6, and

√

np 0 (1−p 0 )=

√
5.88.
Consequently, thep-value that results from the dataX=10 is

p-value=P.02{X≥ 10 }

=P.02{X≥9.5}

=P.02

{

X− 6

√

5.88

≥

9.5− 6

√

5.88

}

≈P{Z≥1.443}

=.0745 ■