Introduction to Probability and Statistics for Engineers and Scientists

302 Chapter 8:Hypothesis Testing

can be obtained as follows:

β(μ)=Pμ

{

X≤μ 0 +zα

σ

√

n

}

=P

{

X−μ

σ/

√

n

≤

μ 0 −μ

σ/

√

n

+zα

}

=P

{

Z≤

μ 0 −μ

σ/

√

n

+zα

}

, Z∼N(0, 1)

where the last equation follows since

√
n(X−μ)/σhas a standard normal distribution.
Hence we can write

β(μ)=

(

μ 0 −μ

σ/

√

n

+zα

)

Since , being a distribution function, is increasing in its argument, it follows thatβ(μ)
decreases inμ; which is intuitively pleasing since it certainly seems reasonable that the
larger the true meanμ, the less likely it should be to conclude thatμ≤μ 0. Also since
(zα)= 1 −α, it follows that

β(μ 0 )= 1 −α
The test given by Equation 8.3.10, which was designed to testH 0 :μ=μ 0 versus
H 1 :μ>μ 0 can also be used to test, at level of significanceα, the one-sided hypothesis

H 0 :μ≤μ 0

versus

H 1 :μ>μ 0

To verify that it remains a levelαtest, we need show that the probability of rejection is
never greater thanαwhenH 0 is true. That is, we must verify that

1 −β(μ)≤α for allμ≤μ 0

or

β(μ)≥ 1 −α for allμ≤μ 0

But it has previously been shown that for the test given by Equation 8.3.10,β(μ) decreases
inμandβ(μ 0 )= 1 −α. This gives that

β(μ)≥β(μ 0 )= 1 −α for allμ≤μ 0

which shows that the test given by Equation 8.3.10 remains a levelαtest forH 0 :μ≤μ 0
against the alternative hypothesisH 1 :μ≤μ 0.