Introduction to Probability and Statistics for Engineers and Scientists

8.3Tests Concerning the Mean of a Normal Population 299

it follows, since is an increasing function, that
(
μ 0 −μ 1
σ/

√

n

−zα/2

)

≤ (−zα/2)=P{Z≤−zα/2}=P{Z≥zα/2}=α/2

Hence, we can take
(
μ 0 −μ 1
σ/

√

n

−zα/2

)

≈ 0

and so from Equation 8.3.5

β≈

(

μ 0 −μ 1

σ/

√

n

+zα/2

)

(8.3.6)

or, since

β=P{Z>zβ}=P{Z<−zβ}= (−zβ)

we obtain from Equation 8.3.6 that

−zβ≈(μ 0 −μ 1 )

√

n

σ

+zα/2

or

n≈

(zα/2+zβ)^2 σ^2

(μ 1 −μ 0 )^2

(8.3.7)

In fact, the same approximation would result whenμ 1 <μ 0 (the details are left as an
exercise) and so Equation 8.3.7 is in all cases a reasonable approximation to the sample
size necessary to ensure that the type II error at the valueμ=μ 1 is approximately equal
toβ.

EXAMPLE 8.3d For the problem of Example 8.3a, how many signals need be sent so that
the .05 level test ofH 0 :μ=8 has at least a 75 percent probability of rejection when
μ=9.2?

SOLUTION Sincez.025=1.96,z.25=.67, the approximation 8.3.7 yields

n≈

(1.96+.67)^2

(1.2)^2

4 =19.21

Hence a sample of size 20 is needed. From Equation 8.3.4, we see that withn= 20

β(9.2)=

(

−

1.2

√

20

2

+1.96

)

−

(

−

1.2

√

20

2

−1.96

)

= (−.723)− (−4.643)