8.3Tests Concerning the Mean of a Normal Population 297
will depend on the value ofμ, and so let us defineβ(μ)by
β(μ)=Pμ{acceptance ofH 0 }=Pμ{∣∣
∣∣
∣X−μ 0
σ/√
n∣
∣∣
∣
∣≤zα/2}=Pμ{
−zα/2≤X−μ 0
σ/√
n≤zα/2}The functionβ(μ) is called theoperating characteristic(or OC)curveand represents the
probability thatH 0 will be accepted when the true mean isμ.
To compute this probability, we use the fact thatXis normal with meanμand variance
σ^2 /nand so
Z≡X−μ
σ/√
n∼N(0, 1)Hence,
β(μ)=Pμ{
−zα/2≤X−μ 0
σ/√
n≤zα/2}=Pμ{
−zα/2−μ
σ/√
n≤X−μ 0 −μ
σ/√
n≤zα/2−μ
σ/√
n}=Pμ{
−zα/2−μ
σ/√
n≤Z−μ 0
σ/√
n≤zα/2−μ
σ/√
n}=P{
μ 0 −μ
σ/√
n−zα/2≤Z≤μ 0 −μ
σ/√
n+zα/2}=(
μ 0 −μ
σ/√
n+zα/2)
−(
μ 0 −μ
σ/√
n−zα/2)
(8.3.4)where is the standard normal distribution function.
For a fixed significance levelα, the OC curve given by Equation 8.3.4 is symmetric
aboutμ 0 and indeed will depend onμonly through (
√
n/σ)|μ−μ 0 |. This curve with
the abscissa changed fromμtod =(
√
n/σ)|μ−μ 0 |is presented in Figure 8.2 when
α=.05.
EXAMPLE 8.3c For the problem presented in Example 8.3a, let us determine the probability
of accepting the null hypothesis thatμ=8 when the actual value sent is 10. To do so,
we compute
√
n
σ
(μ 0 −μ)=−√
5
2× 2 =−√
5