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