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CHAPTER 10
Section 10-2
10-1. a)
1) The parameter of interest is the difference in fill volume,
µ
1
− µ
2
( note that
∆
0
=0)
2) H
0
:
3) H
1
:
µ
1
−
µ
2
=
0
µ
1
−
µ
2
≠
0
or
or
µ
1
=
µ
2
µ
1
≠
µ
2
z
0
=
( x
1
−
x
2
)
− ∆
0
2
σ
1
σ
2
+
2
n
1
n
2
4)
α
= 0.05
5) The test statistic is
6) Reject H
0
if z
0
<
−z
α/2
=
−1.96
or z
0
> z
α/2
= 1.96
7)
x
1
=
16.015 x
2
=
16.005
σ
1
=
0.02
n
1
= 10
σ
2
=
0.025
n
2
= 10
z
0
=
(16.015
−
16.005)
(0.02)
(0.025)
+
10
10
2
2
=
0.99
8) since -1.96 < 0.99 < 1.96, do not reject the null hypothesis and conclude there is no evidence that the
two machine fill volumes differ at
α
= 0.05.
b)
P-value
=
2(1
− Φ
(0.99))
=
2(1
−
0.8389)
=
0.3222
c) Power = 1
− β
, where
β = Φ
z
α
/ 2
−
∆ − ∆
0
2
σ
1
σ
2
+
2
n
1
n
2
− Φ −
z
α
/ 2
−
∆ − ∆
0
2
σ
1
σ
2
+
2
n
1
n
2
(0.02)
2
(0.025)
2
(0.02)
2
(0.025)
2
+
+
10
10
10
10
=
Φ
(
1.96
−
3.95
)
− Φ
(
−
1.96
−
3.95
)
= Φ
(
−
1.99
)
− Φ
(
−
5.91
)
= 0.0233
−
0
= 0.0233
Power = 1
−0.0233
= 0.9967
d)
=
Φ
1.96
−
0.04
− Φ −
1.96
−
0.04
(
x
1
−
x
2
)
−
z
α
/ 2
2
2
σ
1
σ
2
σ
1
σ
2
2
+
≤ µ
1
− µ
2
≤
(
x
1
−
x
2
)
+
z
α
/ 2
+
2
n
1
n
2
n
1
n
2
(
16.015
−
16.005
)
−
1.96
(0.02)
2
(0.025)
2
(0.02)
2
(0.025)
2
+
≤ µ
1
− µ
2
≤
(
16.015
−
16.005
)
+
1.96
+
10
10
10
10
−
0.0098
≤ µ
1
− µ
2
≤
0.0298
With 95% confidence, we believe the true difference in the mean fill volumes is between
−0.0098
and
0.0298. Since 0 is contained in this interval, we can conclude there is no significant difference between
the means.
e) Assume the sample sizes are to be equal,
use
α
= 0.05,
β
= 0.05, and
∆
= 0.04
10-1
(
z
n
≅
α
/2
2
+
z
β
)
σ
12
+
σ
2
2
(
δ
2
)
=
(
1.96
+
1.645
)
(
(0.02)
2
2
+
(0.025)
2
(0.04)
2
)
=
8.35,
n = 9,
use n
1
= n
2
= 9
10-2.
1) The parameter of interest is the difference in breaking strengths,
µ
1
− µ
2
and
∆
0
= 10
2) H
0
:
µ
1
− µ
2
=
10
3) H
1
:
µ
1
− µ
2
>
10
4)
α
= 0.05
5) The test statistic is
z
0
=
( x
1
−
x
2
)
− ∆
0
2
σ
1
σ
2
+
2
n
1
n
2
6) Reject H
0
if z
0
> z
α
= 1.645
7) x
1
=
162.5 x
2
=
155.0
δ
= 10
σ
1
=
1.0
n
1
= 10
σ
2
=
1.0
n
2
= 12
z
0
=
(162.5
−
155.0)
−
10
2
2
= −
5.84
(1.0)
(10)
.
+
10
12
8) Since -5.84 < 1.645 do not reject the null hypothesis and conclude there is insufficient evidence to support
the use of plastic 1 at
α
= 0.05.
10-3
β
=
Φ
1.645
−
(12
−
10)
= Φ
(
−
3.03
)
=
0.0012
, Power = 1 – 0.0012 = 0.9988
1 1
+
10 12
(1.645
+
1.645)
2
(1
+
1)
=
=
5.42
≅
6
2
(12
−
10)
1
n
=
2
(
z
α
2
+
z
β
)
2
(
σ
12
+
σ
2
)
(
∆ − ∆
0
)
2
Yes, the sample size is adequate
10-4.
a) 1) The parameter of interest is the difference in mean burning rate,
µ
1
− µ
2
2) H
0
:
µ
1
− µ
2
=
0 or
µ
1
= µ
2
10-2
3) H
1
:
µ
1
− µ
2
≠
0 or
µ
1
≠ µ
2
4)
α
= 0.05
5) The test statistic is
z
0
=
( x
1
−
x
2
)
− ∆
0
2
σ
1
σ
2
+
2
n
1
n
2
6) Reject H
0
if z
0
<
−z
α/2
=
−1.96
or z
0
> z
α/2
= 1.96
7) x
1
=
18 x
2
=
24
σ
1
=
3
n
1
= 20
σ
2
=
3
n
2
= 20
z
0
=
(18
−
24)
(3)
(3)
+
20
20
2
2
= −
6.32
8) Since
−6.32
<
−1.96
reject the null hypothesis and conclude the mean burning rates differ
significantly at
α
= 0.05.
b) P-value =
2(1
− Φ
(6.32))
=
2(1
−
1)
=
0
∆ − ∆
0
2
σ
1
σ
2
+
2
n
1
n
2
c)
β = Φ
z
α
/ 2
−
∆ − ∆
0
2
σ
1
σ
2
+
2
n
1
n
2
− Φ −
z
α
/ 2
−
=
Φ
1.96
−
2.5
(3)
2
(3)
2
+
20
20
− Φ −
1.96
−
2.5
(3)
2
(3)
2
+
20
20
=
Φ
(
1.96
−
2.64
)
− Φ
(
−
1.96
−
2.64
)
= Φ
(
−
0.68
)
− Φ
(
−
4.6
)
= 0.24825
−
0
= 0.24825
d)
(
x
1
−
x
2
)
−
z
α
/ 2
(
18
−
24
)
−
1.96
2
2
σ
1
σ
2
σ
1
σ
2
2
+
≤ µ
1
− µ
2
≤
(
x
1
−
x
2
)
+
z
α
/ 2
+
2
n
1
n
2
n
1
n
2
(3)
2
(3)
2
(3)
2
(3)
2
+
≤ µ
1
− µ
2
≤
(
18
−
24
)
+
1.96
+
20
20
20
20
−
7.86
≤ µ
1
− µ
2
≤ −
4.14
We are 95% confident that the mean burning rate for solid fuel propellant 2 exceeds that of propellant 1 by
between 4.14 and 7.86 cm/s.
10-5.
x
1
=
30.87
x
2
=
30.68
10-3
σ
1
=
0.10
n
1
= 12
σ
2
=
0.15
n
2
= 10
a) 90% two-sided confidence interval:
(
x
1
−
x
2
)
−
z
α
/ 2
σ
12
n
1
+
2
σ
2
n
2
≤
µ
1
−
µ
2
≤
(
x
1
−
x
2
)
+
z
α
/ 2
σ
12
n
1
+
2
σ
2
n
2
(0.10)
2
(0.15)
2
(0.10)
2
(0.15)
2
+
≤
µ
1
−
µ
2
≤
(
30.87
−
30.68
)
+
1.645
+
(30.87
−
30.68)
−
1.645
12
10
12
10
0.0987
≤
µ
1
−
µ
2
≤
0.2813
We are 90% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between
0.0987 and 0.2813 fl. oz.
b) 95% two-sided confidence interval:
(
x
1
−
x
2
)
−
z
α
/ 2
2
σ
1
σ
2
σ
2
σ
2
+
2
≤ µ
1
− µ
2
≤
(
x
1
−
x
2
)
+
z
α
/ 2 1
+
2
n
1
n
2
n
1
n
2
(30.87
−
30.68)
−
1.96
(0.10)
2
(0.15)
2
(0.10)
2
(0.15)
2
+
≤ µ
1
− µ
2
≤
(
30.87
−
30.68
)
+
1.96
+
12
10
12
10
0.0812
≤
µ
1
−
µ
2
≤
0.299
We are 95% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between
0.0812 and 0.299 fl. oz.
Comparison of parts a and b:
As the level of confidence increases, the interval width also increases (with all other values held constant).
c) 95% upper-sided confidence interval:
µ
1
− µ
2
≤
(
x
1
−
x
2
)
+
z
α
2
σ
1
σ
2
+
2
n
1
n
2
µ
1
− µ
2
≤
(
30.87
−
30.68
)
+
1.645
µ
1
− µ
2
≤
0.2813
(0.10)
2
(0.15)
2
+
12
10
With 95% confidence, we believe the fill volume for machine 1 exceeds the fill volume of machine 2 by
no more than 0.2813 fl. oz.
10-4
10-6.
a) 1) The parameter of interest is the difference in mean fill volume,
µ
1
− µ
2
2) H
0
:
µ
1
− µ
2
=
0 or
µ
1
= µ
2
3) H
1
:
µ
1
− µ
2
≠
0 or
µ
1
≠ µ
2
4)
α
= 0.05
5) The test statistic is
z
0
=
( x
1
−
x
2
)
− ∆
0
2
σ
1
σ
2
+
2
n
1
n
2
6) Reject H
0
if z
0
<
−z
α/2
=
−1.96
or z
0
> z
α/2
= 1.96
7) x
1
=
30.87 x
2
=
30.68
σ
1
=
0.10
n
1
= 12
σ
2
=
0.15
n
2
= 10
z
0
=
(30.87
−
30.68)
(0.10)
(0.15)
+
12
10
2
2
=
3.42
8) Since 3.42 > 1.96 reject the null hypothesis and conclude the mean fill volumes of machine 1 and
machine 2 differ significantly at
α
= 0.05.
b) P-value = 2(1
− Φ
(3.42))
=
2(1
−
0.99969)
=
0.00062
c) Assume the sample sizes are to be equal,
2
use
α
= 0.05,
β
= 0.10, and
∆
= 0.20
n = 9, use n
1
= n
2
= 9
(
z
α
/ 2
+
z
β
)
(
σ
12
+ σ
2
)
=
(
1.96
+
1.28
)
2
(
(0.10)
2
+
(0.15)
2
)
=
8.53,
2
n
≅
(
∆ − ∆
0
)
2
(
−
0.20)
2
10-7.
x
1
=
89.6
2
σ
1
=
1.5
n
1
= 15
x
2
=
92.5
σ
2
=
1.2
2
n
2
= 20
a) 95% confidence interval:
(
x
1
−
x
2
)
−
z
α
/ 2
2
σ
1
σ
2
σ
2
σ
2
+
2
≤ µ
1
− µ
2
≤
(
x
1
−
x
2
)
+
z
α
/ 2 1
+
2
n
1
n
2
n
1
n
2
(89.6
−
92.5)
−
1.96
15 1.2
.
15 1.2
.
+
≤ µ
1
− µ
2
≤
(
89.6
−
92.5
)
+
1.96
+
15 20
15 20
−
3.684
≤ µ
1
− µ
2
≤ −
2.116
With 95% confidence, we believe the mean road octane number for formulation 2 exceeds that of
formulation 1 by between 2.116 and 3.684.
b)
1) The parameter of interest is the difference in mean road octane number,
µ
1
− µ
2
and
∆
0
= 0
10-5

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