9-65

8) The P-value is greater than any acceptable significance level α. Therefore, we do not reject the null hypothesis. There is insufficient evidence to support the claim that the standard deviation is less than 20 microamps.

b) 7) n = 51, s = 20

χ02

250 157400

30 81= =( . ) .

P-value = ( )P χ2 3081< . ; 0 01 0 025. .< − <P value

8) The P-value is less than 0.05, therefore we reject the null hypothesis and conclude that the standard deviation is significantly less than 20 microamps. c) Increasing the sample size increases the test statistic χ0

2 and therefore decreases the P-value, providing more evidence against the null hypothesis. Mind Expanding Exercises 9-126 The parameter of interest is the true,μ. H0 : μ = μ0 H1 μ ≠ μ0

9-127 a) Reject H0 if z0 < -zα-ε or z0 > zε

)|//

()| //

( 000

000 μμ

σμ

σμμμ

σμ

σμ

=−

>−

+==−

−<−

nX

nXP

nX

nXP P

αεεαεεαεεα

=−−+−=Φ−+−Φ=>+−< −−

))1(1())(()(1)()()( 00 zzzzPzzP

b) β = P(zε ≤ X ≤ zε when d+= 01 μμ )

or β μ μ δα ε ε= − < < = +−P z Z z( | )0 1 0

β μ μ δα εμ

σε

α εδ

σε

δ

σ

εδ

σα ε

δ

σ

= − < < = +

= − − < < −

= − − − −

−−

−

−

P z z

P z Z z

z z

x

n

n n

n n

( | )

( )

( ) ( )

/

/ /

/ /

02

2 2

2 2

1 0

Φ Φ

9-128 1) The parameter of interest is the true mean number of open circuits, λ. 2) H0 : λ = 2 3) H1 : λ > 2 4) α = 0.05 5) Since n>30 we can use the normal distribution

z0 = n

X/λλ−

6) Reject H0 if z0 > zα where z0.05 =1.65 7) x= 1038/500=2.076 n = 500

9-66

z0 = 202.1500/2

2076.2=

−

8) Because 1.202 < 1.65, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true mean number of open circuits is greater than 2 at α = 0.01

9-129 a) 1) The parameter of interest is the true standard deviation of the golf ball distance σ. 2) H0: σ = 10 3) H1: σ < 10 4) α=0.05 5) Because n > 30 we can use the normal distribution

z0 = )2/(2

0

0

n

S

σ

σ−

6) Reject H0 if z0 < zα where z0.05 =-1.65 7) s= 13.41 n = 100

z0 = 82.4)200/(10

1041.132

=−

8) Since 4.82 > -1.65, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true standard deviation is less than 10 at α = 0.05

b) 95% percentile: σμθ 645.1+=

95% percentile estimator: SX 645.1ˆ +=θ From the independence

)2/(645.1/)ˆ( 222 nnSE σσθ +≅ The statistic S can be used as an estimator for σ in the standard error formula.

c) 1) The parameter of interest is the true 95th percentile of the golf ball distance θ. 2) H0: θ = 285 3) H1: θ < 285 4) α = 0.05 5) Since n > 30 we can use the normal distribution

z0 = )ˆ(ˆ

ˆ0

θθθ

ES−

6) Reject H0 if z0 < -1.65

7) θ̂ = 282.36 , s = 13.41, n = 100

z0 = 283.1200/41.13645.1100/41.13

28536.282222

−=+

−

8) Because -1.283 > -1.65, do not reject the null hypothesis. There is not sufficient evidence to indicate that the true θ is less than 285 at α = 0.05

9-130 1) The parameter of interest is the true mean number of open circuits, λ. 2) H0 : λ = λ0 3) H1 : λ ≠ λ0 4) α = 0.05 5) test statistic

9-67

∑

∑

=

=

−=

n

ii

n

ii

X

X

1

012

0

2

2

λ

λλχ

6) Reject H0 if 2

2,2/20 naχχ > or 2

2,2/120 na−< χχ

7) compute ∑=

n

iiX

12λ and plug into

∑

∑

=

=

−=

n

ii

n

ii

X

X

1

012

0

2

2

λ

λλχ

8) make conclusions alternative hypotheses

1) H0 : λ = λ0 H1 : λ > λ0

Reject H0 if 2

2,20 naχχ >

2) H0 : λ = λ0

H1 : λ < λ0

Reject H0 if 2

2,20 naχχ <