Introduction to Probability and Statistics for Engineers and Scientists

304 Chapter 8:Hypothesis Testing

and so thep-value is given by

p-value=P{Z<−.336}, Z∼N(0, 1)

=.368

Since this value is greater than .05, the foregoing data do not enable us to reject, at the .05
percent level of significance, the hypothesis that the mean nicotine content exceeds 1.6
mg. In other words, the evidence, although supporting the cigarette producer’s claim, is
not strong enough to prove that claim. ■

REMARKS

(a)There is a direct analogy between confidence interval estimation and hypothesis testing.
For instance, for a normal population having meanμand known varianceσ^2 , we have
shown in Section 7.3 that a 100(1−α) percent confidence interval forμis given by

μ∈

(

x−zα/2

σ

√

n

,x+zα/2

σ

√

n

)

wherexis the observed sample mean. More formally, the preceding confidence interval
statement is equivalent to

P

{

μ∈

(

X−zα/2

σ

√

n

,X+zα/2

σ

√

n

)}

= 1 −α

Hence, ifμ=μ 0 , then the probability thatμ 0 will fall in the interval

(

X−zα/2

σ

√

n

,X+zα/2

σ

√

n

)

is 1−α, implying that a significance levelαtest ofH 0 :μ=μ 0 versusH 1 :μ=μ 0 is
to rejectH 0 when

μ 0 ∈

(

X−zα/2

σ

√

n

,X+zα/2

σ

√

n

)

Similarly, since a 100(1−α) percent one-sided confidence interval forμis given by

μ∈

(

X−zα

σ

√

n

,∞

)

it follows that anα-level significance test ofH 0 :μ≤μ 0 versusH 1 :μ>μ 0 is to reject
H 0 whenμ 0 ∈(X−zασ/

√

n,∞) — that is, whenμ 0 <X−zασ/

√

n.