300 Chapter 8:Hypothesis Testing
≈ 1 − (.723)
≈.235
Therefore, if the message is sent 20 times, then there is a 76.5 percent chance that the
null hypothesisμ= 8 will be rejected when the true mean is 9.2. ■
ONE-SIDED TESTS
In testing the null hypothesis thatμ=μ 0 , we have chosen a test that calls for rejection
whenXis far fromμ 0. That is, a very small value ofXor a very large value appears to
make it unlikely thatμ(whichXis estimating) could equalμ 0. However, what happens
when the only alternative toμbeing equal toμ 0 is forμto be greater thanμ 0? That is,
what happens when the alternative hypothesis toH 0 :μ=μ 0 isH 1 :μ>μ 0? Clearly,
in this latter case we would not want to rejectH 0 whenXis small (since a smallXis more
likely whenH 0 is true than whenH 1 is true). Thus, in testing
H 0 :μ=μ 0 versus H 1 :μ>μ 0 (8.3.8)
we should rejectH 0 whenX, the point estimate ofμ 0 , is much greater thanμ 0. That is,
the critical region should be of the following form:
C={(X 1 ,...,Xn):X−μ 0 >c}
Since the probability of rejection should equalαwhenH 0 is true (that is, whenμ=μ 0 ),
we require thatcbe such that
Pμ 0 {X−μ 0 >c}=α (8.3.9)
But since
Z=
X−μ 0
σ/
√
n
has a standard normal distribution whenH 0 is true, Equation 8.3.9 is equivalent to
P
{
Z>
c
√
n
σ
}
=α
whenZis a standard normal. But since
P{Z>zα}=α
we see that
c=
zασ
√
n