Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

CHAPTER 10

Section 10-2

10-1 a) 1) The parameter of interest is the difference in means μ1 − μ 2 . Note that Δ0 = 0.

2) H0 : μ1 − μ 2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ 2 ≠ 0 or μ1 ≠ μ 2

4) The test statistic is

( x1 − x2 ) − Δ 0
z0 =
σ 12 σ 22
+
n1 n2

5) Reject H0 if z0 < −zα/2 = −1.96 or z0 > zα/2 = 1.96 for α = 0.05

6) x1 = 4.7 x2 = 7.8

σ1 = 10 σ2 = 5

n1 = 10 n2 = 15
(4.7 − 7.8)
z0 = = −0.9
(10) 2 (5) 2
+
10 15
7) Conclusion: Because –1.96 < –0.9 < 1.96, do not reject the null hypothesis. There is not sufficient
evidence to conclude that the two means differ at α = 0.05.

P-value = 2(1 − Φ(0.9)) = 2(1 − 0.815950) = 0.368

σ 12 σ 22 σ 12 σ 22
b) ( x1 − x2 ) − zα / 2 + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + zα / 2 +
n1 n2 n1 n2

(10)2 (5) 2 (10) 2 (5) 2

( 4.7 − 7.8) − 1.96 + ≤ μ1 − μ2 ≤ ( 4.7 − 7.8 ) + 1.96 +
10 15 10 15
−9.79 ≤ μ1 − μ 2 ≤ 3.59

With 95% confidence, the true difference in the means is between −9.79 and 3.59. Because zero is
contained in this interval, we conclude there is no significant difference between the means. We fail
to reject the null hypothesis.

10-1
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎜ Δ − Δ0 ⎟ ⎜ Δ − Δ0 ⎟
c) β = Φ ⎜ zα / 2 − ⎟ − Φ ⎜ − zα / 2 − ⎟
⎜ σ 12 σ 22 ⎟ ⎜ σ 12 σ 22 ⎟
⎜ + ⎟ ⎜ + ⎟
⎝ n1 n2 ⎠ ⎝ n1 n2 ⎠

⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
3 3
= Φ ⎜ 1.96 − ⎟ − Φ ⎜ −1.96 − ⎟
⎜ (10) (5) 2
2 ⎟ ⎜ (10) 2
(5) 2 ⎟
⎜⎜ + ⎟⎟ ⎜⎜ + ⎟⎟
⎝ 10 15 ⎠ ⎝ 10 15 ⎠
= Φ (1.08) − Φ ( −2.83) = 0.8599 − 0.0023 = 0.86

Power = 1 − 0.86 = 0.14

d) Assume the sample sizes are to be equal, use α = 0.05, β = 0.05, and δ = 3

(z + zβ ) (σ 12 + σ 22 ) (1.96 + 1.645) (10 + 52 )

2 2 2
α /2
n≅ = = 180.5
δ2 (3) 2

Use n1 = n2 = 181

10-2 a) 1) The parameter of interest is the difference in means μ1 − μ 2 . Note that Δ0 = 0.

2) H0 : μ1 − μ 2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ2 < 0 or μ1 < μ 2

4) The test statistic is

( x1 − x2 ) − Δ 0
z0 =
σ 12 σ 22
+
n1 n2

5) Reject H0 if z0 < −zα = −1.645 for α = 0.05

6) x1 = 14.2 x2 = 19.7

σ1 = 10 σ2 = 5

n1 = 10 n2 = 15
(14.2 − 19.7)
z0 = = −1.61
(10) 2 (5) 2
+
10 15
7) Conclusion: Because –1.61 > −1.645, do not reject the null hypothesis. There is not sufficient
evidence to conclude that the two means differ at α = 0.05.

P-value = Φ(−1.61) = 0.0537

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

σ 12 σ 22
b) μ1 − μ2 ≤ ( x1 − x2 ) + zα +
n1 n2

(10) 2 (5) 2
μ1 − μ2 ≤ (14.2 − 19.7 ) + 1.645 +
10 15
μ1 − μ 2 ≤ 0.12

With 95% confidence, the true difference in the means is less than 0.12. Because zero is contained in
this interval, we fail to reject the null hypothesis.

⎛ ⎞
⎜ ⎟
⎜ δ ⎟
c) β = 1 − Φ ⎜ − zα − ⎟
⎜ σ 1 σ 22
2
⎟
⎜ + ⎟
⎝ n1 n2 ⎠

⎛ ⎞
⎜ ⎟
−4
= 1 − Φ ⎜ −1.65 − ⎟
⎜ (10) 2 (5) 2 ⎟
⎜⎜ + ⎟⎟
⎝ 10 15 ⎠
= 1 − Φ ( −0.4789 ) = 0.316

Power = 1 − 0.316 = 0.684

d) Assume the sample sizes are to be equal, use α = 0.05, β = 0.05, and δ = Δ − Δ 0 = 4

(z + zβ ) (σ 12 + σ 22 ) (1.645 + 1.645) (10 + 52 )

2 2 2
α
n≅ = = 85
δ2 (4) 2

Use n1 = n2 = 85

10-3 a) 1) The parameter of interest is the difference in means μ1 − μ2 . Note that Δ0 = 0.

2) H0 : μ1 − μ 2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ 2 > 0 or μ1 > μ 2

4) The test statistic is

( x1 − x2 ) − Δ 0
z0 =
σ 12 σ 22
+
n1 n2

5) Reject H0 if z0 > zα = 2.325 for α = 0.01

6) x1 = 24.5 x2 = 21.3

σ1 = 10 σ2 = 5

n1 = 10 n2 = 15

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

(24.5 − 21.3)
z0 = = 0.937
(10) 2 (5) 2
+
10 15
7) Conclusion: Because 0.937 < 2.325, we fail to reject the null hypothesis. There is not sufficient
evidence to conclude that the two means differ at α = 0.01.

P-value = 1 − Φ(0.94) = 1 − 0.8264 = 0.1736

σ 12 σ 22
b) μ1 − μ2 ≥ ( x1 − x2 ) − zα +
n1 n2

(10) 2 (5) 2
μ1 − μ2 ≥ ( 24.5 − 21.3) − 2.325 +
10 15
μ1 − μ2 ≥ −4.74
With 99% confidence, the true difference in the means is greater than -4.74. Because 0 is contained
in this interval, we fail to reject the null hypothesis.

⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎜ δ ⎟ ⎜ 2.325 − 2 ⎟
c) β = Φ ⎜ zα − = Φ
2 ⎟ ⎜ 2 ⎟
⎜ σ 2
σ (10) 2
(5)
⎜
1
+ 2 ⎟ ⎜⎜ + ⎟⎟
⎝ n1 n2 ⎟⎠ ⎝ 10 15 ⎠

= Φ (1.74 ) = 0.959

Power = 1 − 0.96 = 0.04

d) Assume the sample sizes are to be equal, use α = 0.05, β = 0.05, and Δ = 3

(z ) (σ ) = (1.645 + 1.645) (10 ) = 339

2
+ zβ + σ 22 + 52
2 2 2
α 1
n≅
δ 2
(2) 2

Use n1 = n2 = 339

10-4 a) 1) The parameter of interest is the difference in fill volume μ1 − μ 2 . Note that Δ0 = 0.

2) H0 : μ1 − μ2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ 2 ≠ 0 or μ1 ≠ μ 2

4) The test statistic is

( x1 − x2 ) − Δ 0
z0 =
σ 12 σ 22
+
n1 n2

5) Reject H0 if z0 < −zα/2 = −1.96 or z0 > zα/2 = 1.96 for α = 0.05

6) x1 = 473.581 x2 = 473.324

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

σ 1 = 0.6 σ 2 = 0.75
n1 = 10 n2 = 10
(473.581 − 473.324)
z0 = = 0.85
(0.6) 2 (0.75) 2
+
10 10
7) Conclusion: Because –1.96 < 0.85 < 1.96, fail to reject the null hypothesis. There is not sufficient
evidence to conclude that the two machine fill volumes differ at α = 0.05.
P-value = 2(1 − Φ (0.85)) = 2(1 − 0.8023) = 0.395

σ 12 σ 22 σ 12 σ 22
b) ( x1 − x2 ) − zα / 2 + ≤ μ1 − μ 2 ≤ ( x1 − x2 ) + zα / 2 +
n1 n2 n1 n2

(0.6) 2 (0.75) 2 (0.6) 2 (0.75) 2

( 473.581 − 473.324 ) − 1.96 + ≤ μ1 − μ2 ≤ ( 473.581 − 473.324 ) + 1.96 +
10 10 10 10
−0.3383 ≤ μ1 − μ 2 ≤ 0.8523

With 95% confidence, we believe the true difference in the mean fill volumes is between −0.3383
and 0.8523. Because 0 is contained in this interval, we can conclude there is no significant
difference between the means.

⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎜ Δ − Δ0 ⎟ ⎜ Δ − Δ0 ⎟
c) β = Φ ⎜ zα / 2 − ⎟ − Φ ⎜ − zα / 2 − ⎟
⎜ σ 12 σ 22 ⎟ ⎜ σ 12 σ 22 ⎟
⎜ + ⎟ ⎜ + ⎟
⎝ n1 n2 ⎠ ⎝ n1 n2 ⎠

⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
1.2 1.2
= Φ ⎜1.96 − ⎟ − Φ ⎜ −1.96 − ⎟
⎜ (0.6) 2
(0.75) 2 ⎟ ⎜ (0.6) 2 (0.75) 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 − 0.9481 = 0.0519

d) Assume the sample sizes are to be equal, use α = 0.05, β = 0.05, and Δ = 0.04

(z + zβ ) (σ 12 + σ 22 ) (1.96 + 1.645 ) ( ( 0.60 ) + ( 0.75 ) ) = 8.326

2 2 2 2
α /2
n≅ =
δ 2
(1.2) 2

Use n1 = n2 = 9

10-5 a) 1) The parameter of interest is the difference in breaking strengths μ1 − μ 2 and Δ0 = 70

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

2) H0 : μ1 − μ 2 = 10

3) H1 : μ1 − μ 2 > 10

4) The test statistic is

( x1 − x2 ) − Δ 0
z0 =
σ 12 σ 22
+
n1 n2

5) Reject H0 if z0 > zα = 1.645 for α = 0.05

6) x1 = 1120 x2 = 1070 δ = 70

σ1 = 7 σ2 = 7

n1 = 10 n2 = 12
(1120 − 1070) − 70
z0 = = −6.67
(7) 2 (7) 2
+
10 12
7) Conclusion: Because –6.67 < 1.645 fail to reject the null hypothesis. There is insufficient evidence to
support the use of plastic 1 at α = 0.05.
P-value = 1 − Φ ( −6.67 ) = 1 − 0 = 1

σ 12 σ 22
b) μ1 − μ 2 ≥ ( x1 − x2 ) − zα +
n1 n2

(7) 2 (7) 2
μ1 − μ2 ≥ (1120 − 1070 ) − 1.645 +
10 12
μ1 − μ2 ≥ 45.07

⎛ ⎞
⎜ (84 − 70) ⎟
c) β = Φ ⎜ 1.645 − ⎟ = Φ ( −10.715 ) = 0
⎜ 7 7 ⎟
⎜ + ⎟
⎝ 10 12 ⎠
Power = 1 – 0 = 1

( zα 2 + z β ) 2 (σ 12 + σ 22 ) (1.645 + 1.645) 2 (1 + 1)
d) n = = = 5.41 ≅ 6
(Δ − Δ 0 ) 2
(12 − 10) 2

Yes, the sample size is adequate

10-6 a) 1) The parameter of interest is the difference in mean burning rate, μ1 − μ 2

2) H0 : μ1 − μ 2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ 2 ≠ 0 or μ1 ≠ μ 2

4) The test statistic is

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

( x1 − x2 ) − Δ 0
z0 =
σ 12 σ 22
+
n1 n2

5) Reject H0 if z0 < −zα/2 = −1.96 or z0 > zα/2 = 1.96 for α = 0.05

6) x1 = 18 x2 = 24

σ1 = 3 σ2 = 3
n1 = 20 n2 = 20
(18 − 24)
z0 = = −6.32
(3) 2 (3) 2
+
20 20
7) Conclusion: Because −6.32 < −1.96 reject the null hypothesis and conclude the mean burning rates
differ significantly at α = 0.05.
P-value = 2(1 − Φ (6.32)) = 2(1 − 1) = 0

σ 12 σ 22 σ 12 σ 22
b) ( x1 − x2 ) − zα / 2 + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + zα / 2 +
n1 n2 n1 n2

(3) 2 (3) 2 (3) 2 (3) 2

(18 − 24 ) − 1.96 + ≤ μ1 − μ2 ≤ (18 − 24 ) + 1.96 +
20 20 20 20
−7.86 ≤ μ1 − 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.

⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎜ Δ − Δ0 ⎟ ⎜ Δ − Δ0 ⎟
c) β = Φ ⎜ zα / 2 − ⎟ − Φ − z
⎜ α /2 − ⎟
⎜ σ 12 σ 22 ⎟ ⎜ σ 12 σ 22 ⎟
⎜ + +
⎝ n1 n2 ⎟⎠ ⎜
⎝ n1 n2 ⎟⎠

⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
2.5 2.5
= Φ ⎜ 1.96 − ⎟ − Φ ⎜ −1.96 − ⎟
⎜ (3) 2 (3) 2 ⎟ ⎜ (3) 2 (3) 2 ⎟
⎜⎜ + ⎟⎟ ⎜⎜ + ⎟⎟
⎝ 20 20 ⎠ ⎝ 20 20 ⎠

= Φ (1.96 − 2.64 ) − Φ ( −1.96 − 2.64 ) = Φ ( −0.68 ) − Φ ( −4.6 ) = 0.24825 − 0 = 0.24825

d) Assume the sample sizes are to be equal, use α = 0.05, β = 1-power = 0.1, and Δ = 4

( zα + zβ ) (σ 12 + σ 22 ) (1.96 + 1.28) (3 + 32 )
2 2 2
/2
n≅ = = 12
δ2 (4)2

Use n1 = n2 = 12
10-7 x1 = 89.6 x2 = 92.5

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

σ 12 = 1.5 σ 22 = 1.2
n1 = 15 n2 = 20

a) 1) The parameter of interest is the difference in mean road octane number μ1 − μ 2 and Δ0 = 0

2) H0 : μ1 − μ 2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ 2 < 0 or μ1 < μ 2

4) The test statistic is

( x1 − x2 ) − Δ 0
z0 =
σ 12 σ 22
+
n1 n2

5) Reject H0 if z0 < −zα = −1.645 for α = 0.05

6) x1 = 89.6 x2 = 92.5

σ 12 = 1.5 σ 22 = 1.2
n1 = 15 n2 = 20
(89.6 − 92.5)
z0 = = −7.25
1.5 1.2
+
15 20
7) Conclusion: Because −7.25 < −1.645 reject the null hypothesis and conclude the mean road octane
number for formulation 2 exceeds that of formulation 1 using α = 0.05.
P-value ≅ P( z ≤ −7.25) = 1 − P( z ≤ 7.25) = 1 − 1 ≅ 0

b) 95% confidence interval:

σ 12 σ 22 σ 12 σ 22
( x1 − x2 ) − zα / 2 + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + zα / 2 +
n1 n2 n1 n2

1.5 1.2 1.5 1.2

(89.6 − 92.5) − 1.96 + ≤ μ1 − μ2 ≤ ( 89.6 − 92.5 ) + 1.96 +
15 20 15 20
−3.684 ≤ μ1 − μ2 ≤ −2.116

With 95% confidence, the mean road octane number for formulation 2 exceeds that of formulation 1
by between 2.116 and 3.684.

c) 95% level of confidence, E = 1, and z0.025 = 1.96

2 2
⎛z ⎞
(
n ≅ ⎜ 0.025 ⎟ σ 12 + σ 22 = ⎜)
⎛ 1.96 ⎞
⎟ (1.5 + 1.2) = 10.37,
⎝ E ⎠ ⎝ 1 ⎠
Use n1 = n2 = 11

10-8 a) 1) The parameter of interest is the difference in mean batch viscosity before and after the process
change, μ1 − μ 2

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

2) H0 : μ1 − μ 2 = 10

3) H1 : μ1 − μ 2 < 10

4) The test statistic is

( x1 − x2 ) − Δ 0
z0 =
σ 12 σ 22
+
n1 n2

5) Reject H0 if z0 < −zα where z0.1 = −1.28 for α = 0.10

6) x1 = 750.2 x2 = 756.88 Δ0 = 10

σ 1 = 20 σ 2 = 20
n1 = 15 n2 = 8

(750.2 − 756.88) − 10
z0 = = −1.90
(20) 2 (20) 2
+
15 8
7) Conclusion: Because −1.90 < −1.28 reject the null hypothesis and conclude the process change has
increased the mean by less than 10.
P-value = P ( Z ≤ −1.90) = 1 − P( Z ≤ 1.90) = 1 − 0.97128 = 0.02872

b) Case 1: Before Process Change Case 2: After Process Change

μ1 = mean batch viscosity before change μ2 = mean batch viscosity after change
x1 = 750.2 x2 = 756.88

σ1 = 20 σ 2 = 20

n1 = 15 n2 = 8

90% confidence on μ1 − μ 2 , the difference in mean batch viscosity before and after process

change:

σ 12 σ 22 σ 12 σ 22
( x1 − x2 ) − zα / 2 + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + zα / 2 +
n1 n2 n1 n2

(20) 2 (20) 2 (20) 2 (20) 2

(750.2 − 756.88) − 1.645 + ≤ μ1 − μ 2 ≤ ( 750.2 − 756 / 88 ) + 1.645 +
15 8 15 8
−21.08 ≤ μ1 − μ2 ≤ 7.72

We are 90% confident that the difference in mean batch viscosity before and after the process
change lies within −21.08 and 7.72. Because 0 is contained in this interval we fail to detect a
difference in the mean batch viscosity from the process change.

c) Parts (a) and (b) conclude that the mean batch viscosity change is less than 10. This conclusion is
obtained from the confidence interval because the interval does not contain the value 10. The upper
endpoint of the confidence interval is only 7.72.

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

10-9 Catalyst 1 Catalyst 2

x1 = 65.22 x2 = 68.42

σ1 = 3 σ2 = 3
n1 = 10 n2 = 10

a) 95% confidence interval on μ1 − μ2 , the difference in mean active concentration

σ 12 σ 22 σ 12 σ 22
( x1 − x2 ) − zα / 2 + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + zα / 2 +
n1 n2 n1 n2

(3) 2 (3) 2 (3) 2 (3) 2

(65.22 − 68.42) − 1.96 + ≤ μ1 − μ2 ≤ ( 65.22 − 68.42 ) + 1.96 +
10 10 10 10
−5.83 ≤ μ1 − μ 2 ≤ −0.57

We are 95% confident that the mean active concentration of catalyst 2 exceeds that of catalyst
1 by between 0.57 and 5.83 g/l.

P-value:
( x1 − x2 ) − Δ 0 (65.22 − 68.42)
z0 = = = −2.38
σ 12 σ 22 32 32
+ +
n1 n2 10 10

Then P-value = 2(0.008656) = 0.0173

b) Yes, because the 95% confidence interval does not contain the value zero. We conclude that the
mean active concentration depends on the choice of catalyst.

⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
(5) (5)
c) β = Φ ⎜ 1.96 − ⎟ − Φ ⎜ −1.96 − ⎟
⎜ 3 32
2 ⎟ ⎜ 3 32
2 ⎟
⎜⎜ + ⎟⎟ ⎜⎜ + ⎟⎟
⎝ 10 10 ⎠ ⎝ 10 10 ⎠
= Φ ( −1.77 ) − Φ ( −5.69 ) = 0.038364 − 0
= 0.038364
Power = 1 − β = 1− 0.038364 = 0.9616.

d) Calculate the value of n using α and β.

( zα + z β ) (σ 12 + σ 22 ) (1.96 + 1.77 )
2
(9 + 9)
2
/2
n≅ = = 10.02,
( Δ − Δ0 )
2
(5) 2

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Therefore, 10 is only slightly too few samples. The sample sizes are adequate to detect the difference
of 5.
The data from the first sample n = 15 appear to be normally distributed.

99

95
90

80
70

Percent
60
50
40
30
20

10
5

700 750 800

The data from the second sample n = 8 appear to be normally distributed

99

95
90

80
70
Percent

60
50
40
30
20

10
5

700 750 800

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Plots for both samples are shown in the following figure.

99

95
90

80
70
Percent

60
50
40
30
20

10
5

55 65 75

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Section 10-2
10-10 a) x1 = 10.94 x2 = 12.15 s12 = 1.262 s22 = 1.992 n1 = 12 n2 = 16

(n1 − 1) s12 + (n2 − 1) s22 (12 − 1)1.262 + (16 − 1)1.992

sp = = = 1.7194
n1 + n2 − 2 12 + 16 − 2

Degree of freedom = n1 + n2 - 2 = 12 + 16 – 2 = 26.

( x1 − x2 ) − Δ 0 (−1.21)
t0 = = = −1.8428
1 1 1 1
sp + 1.7194 +
n1 n2 12 16

P-value = 2[P(t > 1.8428)] and 2(0.025) < P-value < 2(0.05) = 0.05 < P-value < 0.1
This is a two-sided test because the hypotheses are mu1 – mu2 = 0 versus not equal to 0.

b) Because 0.05 < P-value < 0.1 the P-value is greater than α = 0.05. Therefore, we fail to reject the
null hypothesis of mu1 – mu2 = 0 at the 0.05 and 0.01 levels of significance.

c) Yes, the sample standard deviations are somewhat different, but not excessively different.
Consequently, the assumption that the two population variances are equal is reasonable.

d) P-value = P (t < -1.8428) and 0.025 < P-value < 0.05

Because 0.025 < P-value < 0.05 the P-value is less than α = 0.05. Therefore, we reject the null
hypothesis of mu1 – mu2 = 0 at the 0.05 level of significance.

10-11 a) x1 = 54.73 x2 = 58.64 s12 = 2.132 s22 = 5.282 n1 = 15 n2 = 20

s12 s22 2 2.132 5.282 2

(+ ) ( + )
n1 n2 15 20
ν= 2 2
= 2 2
= 26.45 ≈ 26 (truncated)
( s1 ) 2
( s 2 ) 2 ( 2.13 ) 2
(5.28 ) 2

n1 n2 15 + 20
+ 15 − 1 20 − 1
n1 − 1 n2 − 1

The 95% upper one-sided confidence interval: t0.05,26 = 1.706

s12 s22
μ1 − μ2 ≤ ( x1 − x2 ) + tα ,ν +
n1 n2

( 2.13) ( 5.28)
2 2

μ1 − μ2 ≤ ( 54.74 − 58.64 ) + 1.706 +

15 20
μ1 − μ2 ≤ −1.6880
P-value = P(t < −3.00): 0.0025 < P-value < 0.005
This is one-sided test because the hypotheses are mu1 – mu2 = 0 versus less than 0.

b) Because 0.0025 < P-value < 0.005 the P-value < α = 0.05. Therefore, we reject the null hypothesis
of mu1 – mu2 = 0 at the 0.05 or the 0.01 level of significance.

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

c) Yes, the sample standard deviations are quite different. Consequently, one would not want to
assume that the population variances are equal.

d) If the alternative hypothesis were changed to mu1 – mu2 ≠ 0, then the P-value = 2 * P (t < −3.00)
and 0.005 < P-value < 0.01. Because the P-value < α = 0.05, we reject the null hypothesis of mu1 –
mu2 = 0 at the 0.05 level of significance.

10-12 a) 1) The parameter of interest is the difference in mean, μ1 − μ2

2) H0 : μ1 − μ2 = 0 or μ1 = μ2

3) H1 : μ1 − μ2 ≠ 0 or μ1 ≠ μ2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 < − tα / 2,n1+n2 −2 where −t0.025,28 = −2.048 or t0 > tα / 2, n + n − 2 where
1 2

t0.025,28 = 2.048 for α = 0.05

(n1 − 1) s12 + (n2 − 1) s22

6) x1 = 5.7 x2 = 8.3 sp =
n1 + n2 − 2

14(4) + 14(6.25)
s12 = 4 s22 = 6.25 = = 2.26
28
n1 = 15 n2 = 15
(5.7 − 8.3)
t0 = = −3.15
1 1
2.26 +
15 15

7) Conclusion: Because −3.15 < −2.048, reject the null hypothesis at α = 0.05.
P-value = P ( t > 3.15 ) < 2(0.0025), P-value < 0.005

b) 95% confidence interval: t0.025,28 = 2.048

1 1 1 1
( x1 − x2 ) − tα / 2, n + n − 2 ( s p )
1 2
+ ≤ μ1 − μ 2 ≤ ( x1 − x2 ) + tα / 2, n1 + n2 − 2 ( s p ) +
n1 n2 n1 n2

1 1 1 1
(5.7 − 8.3) − 2.048(2.26) + ≤ μ1 − μ2 ≤ (5.7 − 8.3) + 2.048(2.26) +
15 15 15 15
−4.29 ≤ μ1 − μ2 ≤ −0.91

Because zero is not contained in this interval, we are 95% confident that the means are different.

c) Δ = 3 Use sp as an estimate of σ:

10-14
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

μ2 − μ1 3
d= = = 0.66
2s p 2(2.26)

Using Chart VII (e) with d = 0.66 and n = n1 = n2 we obtain n* = 2n − 1 = 29 and α = 0.05. Therefore,
β = 0.1 and the power is 1 − β = 0.9

2 n ∗ +1
d) β = 0.05, d = = 0.44, therefore n* ≅ 75 then n = = 38, then n = n1 = n2 = 38
2(2.26) 2

10-13 a) 1) The parameter of interest is the difference in means, μ1 − μ2 , with Δ0 = 0

2) H0 : μ1 − μ2 = 0 or μ1 = μ2

3) H1 : μ1 − μ2 < 0 or μ1 < μ2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 < −tα , n1 + n2 − 2 where − t0.05,28 = −1.701 for α = 0.05

(n1 − 1) s12 + (n2 − 1) s22

6) x1 = 7.2 x2 = 7.9 sp =
n1 + n2 − 2

14(4) + 14(6.25)
s12 = 4 s22 = 6.25 = = 2.26
28
n1 = 15 n2 = 15
(7.2 − 7.9)
t0 = = −0.85
1 1
2.26 +
15 15
7) Conclusion: Because −0.85 > −1.701 we fail to reject the null hypothesis at the 0.05 level of
significance.
P-value = P ( t > 0.85) , 0.1 < P-value < 0.25

b) 95% confidence interval: t0.05,28 = 1.701

1 1
μ1 − μ2 ≤ ( x1 − x2 ) + tα , n + n − 2 ( s p )
1 2
+
n1 n2

1 1
μ1 − μ2 ≤ (7.2 − 7.9) + 1.701(2.26) +
15 15
μ1 − μ2 ≤ 0.704
Because zero is contained in this interval, we are 95% confident that μ1 > μ 2

c) Δ = 3 Use sp as an estimate of σ:

10-15
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

μ2 − μ1 3
d= = = 0.66
2s p 2(2.26)

Using Chart VII (g) with d = 0.66 and n = n1 = n2 we get n* = 2n − 1 = 29 and α = 0.05. Therefore, β

= 0.05 and the power is 1 − β = 0.95

2.5 n ∗ +1
d) β = 0.05, d = = 0.55. Therefore n* ≅ 40 and n = ≅ 21. Thus, n = n1 = n2 = 21
2(2.26) 2

10-14 a) 1) The parameter of interest is the difference in means, μ1 − μ2 , with Δ0 = 0

2) H0 : μ1 − μ2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ2 > 0 or μ1 > μ 2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 > tα , n1 + n2 − 2 where t 0.05,18 = 1.734 for α = 0.05

(n1 − 1) s12 + (n2 − 1) s22

6) x1 = 7.8 x2 = 5.6 sp =
n1 + n2 − 2

9(4) + 9(6.25)
s12 = 4 s22 = 6.25 = = 2.26
18
n1 = 10 n2 = 10
(7.8 − 5.6)
t0 = = 2.18
1 1
2.26 +
10 10
7) Conclusion: Because 2.18 > 1.714 reject the null hypothesis at the 0.05 level of significance.
P-value = P ( t > 2.18 ) and 0.025 < P-value < 0.05

b) 95% confidence interval:

1 1
μ1 − μ 2 ≥ ( x1 − x2 ) − tα , n + n − 2 ( s p ) +
1 2
n1 n2

1 1
μ1 − μ2 ≥ (7.8 − 5.6) − 1.734(2.26) +
10 10
μ1 − μ2 ≥ 0.45

Because zero is not contained in this interval, we are 95% confident that μ1 − μ 2 ≥ 0 .

c) Δ = 3 Use sp as an estimate of σ:

10-16
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

μ2 − μ1 3
d= = = 0.66
2s p 2(2.26)

Using Chart VII (g) with d = 0.66 and n = n1 = n2 = 10 we obtain n* = 2n – 1 = 19 and α = 0.05.

Therefore, β ≈ 0.12 and the power is 1 − β = 0.88

2.5 n ∗ +1
d) β = 0.05, d = = 0.55, therefore n* ≅ 40. Finally, n = ≅ 21, and n = n1 = n2 = 21
2(2.26) 2

10-15 a) 1) The parameter of interest is the difference in mean rod diameter, μ1 − μ 2

2) H0 : μ1 − μ 2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ 2 ≠ 0 or μ1 ≠ μ 2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 < −tα / 2, n1 + n2 − 2 where −t0.025,31 = −2.04 or t0 > tα / 2, n1 + n2 − 2 where t0.025,31 =

2.04 for α = 0.05

(n1 − 1) s12 + (n2 − 1) s22

6) x1 = 8.73 x2 = 8.68 sp =
n1 + n2 − 2

14(0.35) + 17(0.90)
s12 = 0.35 s22 = 0.90 = = 0.807
31
n1 = 15 n2 = 18
(8.73 − 8.68)
t0 = = 0.177
1 1
0.807 +
15 18

7) Conclusion: Because −2.04 < 0.177 < 2.04, we fail to reject the null hypothesis. There is insufficient
evidence to conclude that the two machines produce different mean diameters at α = 0.05.
P-value = 2P ( t > 0.177 ) > 2(0.40), P-value > 0.80

b) 95% confidence interval: t0.025,31 = 2.04

1 1 1 1
( x1 − x2 ) − tα / 2,n + n − 2 (s p ) + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + tα / 2, n1 + n2 − 2 ( s p ) +
1 2
n1 n2 n1 n2

1 1 1 1
(8.73 − 8.68) − 2.04(0.807) + ≤ μ1 − μ2 ≤ ( 8.73 − 8.68 ) + 2.04(0.807) +
15 18 15 18
−0.526 ≤ μ1 − μ 2 ≤ 0.626

Because zero is contained in this interval, there is insufficient evidence to conclude that the two
machines produce rods with different mean diameters.

10-17
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

10-16 a) Assume the populations follow normal distributions and σ 12 = σ 22 . The assumption of equal variances

may be relaxed in this case because it is known that the t-test and confidence intervals involving the t-
distribution are robust to this assumption of equal variances when sample sizes are equal.

Case 1: AFFF Case 2: ATC

μ1 = mean foam expansion for AFFF μ2 = mean foam expansion for ATC
x1 = 4.7 x 2 = 6.9

s1 = 0.6 s 2 = 0.8

n1 = 5 n2 = 5

4(0.602 ) + 4(0.802 )
95% confidence interval: t0.025,8 = 2.306 sp = = 0.7071
8

1 1 1 1
( x1 − x2 ) − tα / 2, n + n − 2 ( s p ) + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + tα / 2, n1 + n2 − 2 ( s p ) +
1 2
n1 n2 n1 n2

1 1 1 1
(4.7 − 6.9) − 2.306(0.7071) + ≤ μ1 − μ2 ≤ ( 4.7 − 6.9 ) + 2.306(0.7071) +
5 5 5 5
−3.23 ≤ μ1 − μ2 ≤ −1.17

b) Yes, with 95% confidence, the mean foam expansion for ATC exceeds that of AFFF by between
1.17 and 3.23 units.

10-17 a) 1) The parameter of interest is the difference in mean catalyst yield, μ1 − μ2 , with Δ0 = 0

2) H0 : μ1 − μ2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ2 < 0 or μ1 < μ2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 < where − t 0.01, 25 = −2.485 for α = 0.01

(n1 − 1) s12 + (n2 − 1) s22

6) x1 = 86 x2 = 89 sp =
n1 + n2 − 2

11(3) 2 + 14(2) 2
s1 = 3 s2 = 2 = = 2.4899
25
n1 = 12 n2 = 15
(86 − 89)
t0 = = −3.11
1 1
2.4899 +
12 15

10-18
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

7) Conclusion: Because −3.11 < −2.485, reject the null hypothesis and conclude that the mean yield of
catalyst 2 exceeds that of catalyst 1 at α = 0.01.

b) 99% upper confidence interval μ1 − μ 2 : t0.01,25 = 2.485

1 1
μ1 − μ2 ≤ ( x1 − x2 ) + tα / 2, n + n − 2 ( s p ) +
1 2
n1 n2

μ 1 − μ 2 ≤ (86 − 89 ) + 2 .485 ( 2 .4899 )

1 1
+
12 15
μ1 − μ 2 ≤ −0.603 or equivalently μ1 + 0.603 ≤ μ 2

We are 99% confident that the mean yield of catalyst 2 exceeds that of catalyst 1 by at least 0.603
units.

10-18 a) According to the normal probability plots, the assumption of normality is reasonable because the
data fall approximately along straight lines. The equality of variances does not appear to be severely
violated either because the slopes are approximately the same for both samples.

10-19
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

b) 1) The parameter of interest is the difference in deflection temperature under load, μ1 − μ2 , with Δ0 = 0

2) H0 : μ1 − μ2 = 0 or μ1 = μ2

3) H1 : μ1 − μ2 > 0 or μ1 > μ2

4) The test statistic is

10-20
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 > tα ,n1 +n2 − 2 where t0.05, 28 = 1.701 for α = 0.05

6) Type 1 Type 2

(n1 − 1) s12 + (n2 − 1) s22

x1 = 91.47 x2 = 89.07 sp =
n1 + n2 − 2

14(5.93) 2 + 14(5.28) 2
s1 = 5.93 s2 = 5.28 sp = = 5.61
28
n1 = 15 n2 = 15
(91.47 − 89.07)
t0 = = 1.17
1 1
5.61 +
15 15
7) Conclusion: Because 1.17 < 1.701 we fail to reject the null hypothesis. There is insufficient evidence
to conclude that the mean deflection temperature under load for Type 1 exceeds the mean for Type 2
at the 0.05 level of significance.
P-value = P ( t > 1.17 ) , 0.1 < P-value < 0.25

c) Δ = 5 Use sp as an estimate of σ:
μ1 − μ2 5
d= = = 0.446
2s p 2(5.61)

Using Chart VII (g) with β = 0.10, d = 0.446 we get n* ≅ 40. Because n* = 2n − 1, n1 = n2 = 21.
Therefore, the sample sizes of 15 are not adequate to meet the given probability of detection.

10-19 a) According to the normal probability plots, the assumption of normality appears to be reasonable
because the data from both the samples fall approximately along a straight line. The equality of
variances does not appear to be severely violated either since the slopes are approximately the same for
both samples.

10-21
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

b) 1) The parameter of interest is the difference in mean etch rate, μ1 − μ 2 , with Δ0 = 0

2) H0 : μ1 − μ 2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ 2 ≠ 0 or μ1 ≠ μ 2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 < −tα / 2, n1 + n2 − 2 where − t 0.025,18 = −2.101 or t0 > t α/ 2,n1 + n 2 − 2 where

t 0.025,18 = 2.101 for α = 0.05

(n1 − 1) s12 + (n2 − 1) s22

6) x1 = 0.2533 x2 = 0.2642 sp =
n1 + n2 − 2

9(0.011) 2 + 9(0.006) 2
s1 = 0.011 s2 = 0.006 sp = = 0.0089
18
n1 = 10 n2 = 10

10-22
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

(0.2533 − 0.2642)
t0 = = −2.74
1 1
0.0089 +
10 10
7) Conclusion: Because −2.74 < −2.101 reject the null hypothesis and conclude the two machines mean
etch rates differ at α = 0.05.
P-value = 2P ( t < −2.74 ) 2(0.005) < P-value < 2(0.010) = 0.010 < P-value < 0.020

c) 95% confidence interval: t0.025,18 = 2.101

1 1 1 1
( x1 − x2 ) − tα / 2,n + n − 2 (s p ) + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + tα / 2, n1 + n2 − 2 ( s p ) +
1 2
n1 n2 n1 n2

1 1 1 1
(0.2533 − 0.2642) − 2.101(0.0089) + ≤ μ1 − μ2 ≤ (0.2533 − 0.2642) + 2.101(0.0089) +
10 10 10 10
−0.01926 ≤ μ1 − μ2 ≤ −0.00254

We are 95% confident that the mean etch rate for solution 2 exceeds the mean etch rate for solution
1 by between 0.00254 and 0.01926.

10-20 a) 1) The parameter of interest is the difference in mean impact strength, μ1 − μ2 , with Δ0 = 0

2) H0 : μ1 − μ2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ2 < 0 or μ1 < μ2

4) The test statistic is

( x1 − x 2 ) − Δ 0
t0 =
s12 s 22
+
n1 n 2

5) Reject the null hypothesis if t0 < −t α , ν where t 0.05, 23 = 1.714 for α = 0.05 since

2
⎛ s12 s22 ⎞
⎜ + ⎟
ν = ⎝ 1 2 2 ⎠ = 23.21
n n
⎛ s1 ⎞
2
⎛ s22 ⎞
⎜ ⎟ ⎜ ⎟
⎝ n1 ⎠ + ⎝ n2 ⎠
n1 − 1 n2 − 1
ν ≅ 23
(truncated)
6) x1 = 395 x2 = 435

s1 = 15 s2 = 30

n1 = 10 n2 = 16
(395 − 435)
t0 = = −4.51
152 302
+
10 16

10-23
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

7) Conclusion: Because −4.51 < −1.714 reject the null hypothesis and conclude that supplier 2 provides
gears with higher mean impact strength at the 0.05 level of significance.
P-value = P(t < −4.51): P-value < 0.0005

b) 1) The parameter of interest is the difference in mean impact strength, μ 2 − μ1

2) H0 : μ2 − μ1 = 25

3) H1 : μ2 − μ1 > 25 or μ2 > μ1 + 25

4) The test statistic is

( x2 − x1 ) − δ
t0 =
s12 s22
+
n1 n2

5) Reject the null hypothesis if t0 > t α , ν = 1.714 for α = 0.05 where

2
⎛ s12 s22 ⎞
⎜ + ⎟
ν = ⎝ 1 2 2 ⎠ = 23.21
n n
⎛ s1 ⎞
2
⎛ s22 ⎞
⎜ ⎟ ⎜ ⎟
⎝ n1 ⎠ + ⎝ n2 ⎠
n1 − 1 n2 − 1
ν ≅ 23
6) x1 = 395 x2 = 435 Δ 0 = 35 s1 = 15 s2 = 30 n1 = 10 n2 = 16

(435 − 395) − 35
t0 = = 0.563
152 302
+
10 16
7) Conclusion: Because 0.563 < 1.714, fail to reject the null hypothesis. There is insufficient evidence to
conclude that the mean impact strength from supplier 2 is at least 35 Nm higher than from supplier 1
using α = 0.05.

c) Using the information provided in part (a), and t0.025,25 = 2.069, a 95% confidence interval on the
difference μ 2 − μ1 is

s12 s22 s2 s2
( x2 − x1 ) − t0.025,25 + ≤ μ 2 − μ1 ≤ ( x2 − x1 ) + t0.025,25 1 + 2
n1 n2 n1 n2
40 − 2.069(8.874) ≤ μ 2 − μ1 ≤ 40 + 2.069(8.874)
21.64 ≤ μ 2 − μ1 ≤ 58.36

Because zero is not contained in the confidence interval, we conclude that supplier 2 provides gears
with a higher mean impact strength than supplier 1 with 95% confidence.

10-21 a) 1) The parameter of interest is the difference in mean melting point, μ1 − μ2 , with Δ0 = 0

2) H0 : μ1 − μ2 = 0 or μ1 = μ2

3) H1 : μ1 − μ2 ≠ 0 or μ1 ≠ μ2

10-24
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 < − t α/ 2,n1 + n 2 − 2 where − t 0.0025, 40 = −2.021 or t0 > t α/ 2,n1 + n 2 − 2 where

t 0.025, 40 = 2.021 for α = 0.05

(n1 − 1) s12 + (n2 − 1) s22

6) x1 = 215 x2 = 219 sp =
n1 + n2 − 2

20(2) 2 + 20(1.7)2
s1 = 2 s2 = 1.7 = = 1.856
40
n1 = 21 n2 = 21
(215 − 219)
t0 = = −6.984
1 1
1.856 +
21 21
7) Conclusion: Because −6.984 < −2.021 reject the null hypothesis. The alloys differ significantly in
mean melting point at α = 0.05.
P-value = 2P ( t < −6.984 ) P-value < 0.0010

| μ1 − μ2 | 1.7
b) d = = = 0.425
2σ 2(2)

Using the appropriate chart in the Appendix, with β = 0.10 and α = 0.05 we have n* = 75.

n* + 1
Therefore, n = = 38 , n1 = n2 = 38
2

10-22 a) 1) The parameter of interest is the difference in mean speed, μ1 − μ2 , Δ0 = 0

2) H0 : μ1 − μ2 = 0 or μ1 = μ2

3) H1 : μ1 − μ2 > 0 or μ1 > μ 2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 > t α, n1 + n 2 − 2 where t 0.10,14 = 1.345 for α = 0.10

6) Case 0: 65 mm Case 2: 0.5 mm

10-25
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

(n1 − 1) s12 + (n2 − 1) s22

x1 = 0.03 x2 = 0.027 sp =
n1 + n2 − 2

7(0.0028) 2 + 7(0.0023) 2
s1 = 0.0028 s2 = 0.0023 = = 2.56 × 10−3
14
n1 = 8 n2 = 8
(0.03 − 0.027)
t0 = = (0.03 − 0.027)2.34
1 1
2.56 × 10−3 +
8 8
7) Because 2.34 > 1.345 reject the null hypothesis and conclude that reducing the film thickness from
0.65 mm to 0.5 mm significantly increases the mean speed of the film at the 0.10 level of
significance (Note: an increase in film speed will result in lower values of observations).
P-value = P ( t > 2.34 ) 0.01 < P-value < 0.025

b) 95% confidence interval: t0.025,14 = 2.145

1 1 1 1
( x1 − x2 ) − tα / 2,n + n − 2 (s p ) + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + tα / 2, n1 + n2 − 2 ( s p ) +
1 2
n1 n2 n1 n2

1 1 1 1
(0.03 − 0.027) − 2.145(2.56 × 10−3 ) + ≤ μ1 − μ 2 ≤ ( 0.03 − 0.027 ) + 2.145(2.56 × 10−3 ) +
8 8 8 8
−0.00025 ≤ μ1 − μ2 ≤ 0.00575

We are 95% confident the difference in mean speed of the film is between 0.00025 and 0.00575
μJ/mm2.

10-23 a) 1) The parameter of interest is the difference in mean wear amount, μ1 − μ2 , with Δ0 = 0

2) H0 : μ1 − μ2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ2 ≠ 0 or μ1 ≠ μ2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
s12 s22
+
n1 n2

5) Reject the null hypothesis if t0 < − t 0.025, 26 or t0 > t 0.025, 26 where t 0.025, 26 = 2.056 for α = 0.05 because

2
⎛ s12 s22 ⎞
⎜ + ⎟
ν = ⎝ 1 2 2 ⎠ = 26.98
n n
⎛ s1 ⎞
2
⎛ s22 ⎞
⎜ ⎟ ⎜ ⎟
⎝ n1 ⎠ + ⎝ n2 ⎠
n1 − 1 n2 − 1
ν ≅ 26
(truncated)
6) x1 = 20 x2 = 15

10-26
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

s1 = 2 s2 = 8

n1 = 25 n2 = 25
(20 − 15)
t0 = = 3.03
(2) 2 (8) 2
+
25 25
7) Conclusion: Because 3.03 > 2.056 reject the null hypothesis. The data support the claim that the two
companies produce material with significantly different wear at the 0.05 level of significance.
P-value = 2P(t > 3.03), 2(0.0025) < P-value < 2(0.005), 0.005 < P-value < 0.010

b) 1) The parameter of interest is the difference in mean wear amount, μ1 − μ2

2) H0 : μ1 − μ2 = 0

3) H1 : μ1 − μ2 > 0

( x1 − x2 ) − Δ 0
4) The test statistic is t0 =
s12 s22
+
n1 n2

5) Reject the null hypothesis if t0 > t 0.05,27 where t 0.05, 26 = 1.706 for α = 0.05 since

6) x1 = 20 x2 = 15

s1 = 2 s2 = 8

n1 = 25 n2 = 25
(20 − 15)
t0 = = 3.03
(2) 2 (8) 2
+
25 25
7) Conclusion: Because 3.03 > 1.706 reject the null hypothesis. The data support the claim that the
material from company 1 has a higher mean wear than the material from company 2 at a 0.05 level
of significance.

c) For part (a) use a 95% two-sided confidence interval:

t0.025,26 = 2.056

s12 s22 s12 s22

( x1 − x2 ) − tα ,ν + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + tα ,ν +
n1 n2 n1 n2

(2) 2 (8) 2 (2) 2 (8) 2

(20 − 15) − 2.056 + ≤ μ1 − μ 2 ≤ (20 − 15) + 2.056 +
25 25 25 25
1.609 ≤ μ1 − μ 2 ≤ 8.391

For part (b) use a 95% lower one-sided confidence interval:

t 0.05, 26 = 1.706

s12 s22
( x1 − x2 ) − tα ,ν + ≤ μ1 − μ2
n1 n2

10-27
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

( 2) (8)
2 2

( 20 − 15) − 1.706 + ≤ μ1 − μ2
25 25
2.186 ≤ μ1 − μ 2
For part a) we are 95% confident the mean abrasive wear from company 1 exceeds the mean
abrasive wear from company 2 by between 1.609 and 8.391 mg/1000.

For part b) we are 95% confident the mean abrasive wear from company 1 exceeds the mean
abrasive wear from company 2 by at least 2.186 mg/1000.

10-24 a) 1) The parameter of interest is the difference in mean coating thickness, μ1 − μ2 , with Δ0 = 0.

2) H0 : μ1 − μ2 = 0

3) H1 : μ1 − μ2 > 0

4) The test statistic is

( x1 − x2 ) − δ
t0 =
s12 s22
+
n1 n2

5) Reject the null hypothesis if t0 > t0.01,18 where t0.01,18 = 2.552 for α = 0.01 since
2
⎛ s12 s22 ⎞
⎜ + ⎟
ν = ⎝ 1 2 2 ⎠ = 18.23
n n
⎛ s12 ⎞ ⎛ s22 ⎞
⎜ ⎟ ⎜ ⎟
⎝ n1 ⎠ + ⎝ n2 ⎠
n1 − 1 n2 − 1
ν ≅ 18
(truncated)
6) x1 = 2.65 x2 = 2.55

s1 = 0.25 s2 = 0.5 20.1

n1 = 11 n2 = 13
(2.65 − 2.55)
t0 = = 0.634
(0.25) 2 (0.5) 2
+
11 13
7) Conclusion: Because 0.634 < 2.552, fail to reject the null hypothesis. There is insufficient evidence
to conclude that increasing the temperature reduces the mean coating thickness at α = 0.01.
P-value = P(t > 0.634), 0.25 < P-value < 0.40

b) If α = 0.01, construct a 99% two-sided confidence interval on the difference in means. Here,
t0.005,19 = 2.878

10-28
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

s12 s22 s12 s22

( x1 − x2 ) − tα / 2,ν + ≤ μ1 − μ 2 ≤ ( x1 − x2 ) + tα / 2,ν +
n1 n2 n1 n2

(0.25) 2 (0.5) 2 (0.25) 2 (0.5) 2

(2.65 − 2.55) − 2.878 + ≤ μ1 − μ2 ≤ (2.65 − 2.55) − 2.878 +
11 13 11 13
−0.354 ≤ μ1 − μ2 ≤ 0.554

Because the interval contains zero, there is no significant difference in the mean coating thickness
between the two temperatures.

10-25 a) 1) The parameter of interest is the difference in mean width of the backside chip-outs for the single
spindle saw process versus the dual spindle saw process, μ1 − μ2

2) H0 : μ1 − μ2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ2 ≠ 0 or μ1 ≠ μ2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 < − tα / 2,n1 + n2 − 2 where − t 0.025, 28 = −2.048 or t0 > t α/ 2, n1 + n 2 − 2 where

t0.025, 28 = 2.048 for α = 0.05

(n1 − 1) s12 + (n2 − 1) s22

6) x1 = 66.385 x2 = 45.278 sp =
n1 + n2 − 2

14(7.895) 2 + 14(8.612) 2
s12 = 7.8952 s22 = 8.6122 = = 8.26
28
n1 = 15 n2 = 15
(66.385 − 45.278)
t0 = = 7.00
1 1
8.26 +
15 15
7) Conclusion: Because 7.00 > 2.048, we reject the null hypothesis at α = 0.05. P-value ≅ 0

b) 95% confidence interval: t0.025,28 = 2.048

1 1 1 1
( x1 − x2 ) − tα / 2,n + n − 2 (s p ) + ≤ μ1 − μ 2 ≤ ( x1 − x2 ) + tα / 2, n1 + n2 − 2 ( s p ) +
1 2
n1 n2 n1 n2

1 1 1 1
(66.385 − 45.278) − 2.048(8.26) + ≤ μ1 − μ2 ≤ (66.385 − 45.278) + 2.048(8.26) +
15 15 15 15
14.93 ≤ μ 1 − μ 2 ≤ 27.28
Because zero is not contained in this interval, we are 95% confident that the means are different.

10-29
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

15
c) For β < 0.01 and d = = 0.91, with α = 0.05 then using Chart VII (e) we find n* > 15. Then
2(8.26)

15 + 1
n> =8
2

10-26 a) 1) The parameter of interest is the difference in mean blood pressure between the test and control
groups, μ1 − μ2 , with Δ0 = 0

2) H0 : μ1 − μ2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ2 < 0 or μ1 < μ2

4) The test statistic is

( x1 − x 2 ) − Δ 0
t0 =
s12 s 22
+
n1 n 2

5) Reject the null hypothesis if t0 < − t α , ν where t0.05,12 = −1.782 for α = 0.05 since
2
⎛ s12 s22 ⎞
⎜⎜ + ⎟⎟
ν = ⎝ 1 2 2 ⎠ = 12
n n
⎛ s12 ⎞ ⎛ s22 ⎞
⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟
⎝ n1 ⎠ + ⎝ n2 ⎠
n1 − 1 n2 − 1
ν ≅ 12
(truncated)
6) x1 = 90 x2 = 115

s1 = 5 s2 = 10

n1 = 8 n2 = 9
(90 − 115)
t0 = = −6.63
(5) 2 (10) 2
+
8 9
7) Conclusion: Because −6.62 < −1.782 reject the null hypothesis and conclude that the test group has
higher mean arterial blood pressure than the control group at the 0.05 level of significance.
P-value = P(t < −6.62): P-value ≅ 0

b) 95% confidence interval: t0.05,12 = 1.782

s12 s22
μ1 − μ2 ≤ ( x1 − x2 ) + tα ,ν +
n1 n2

52 102
μ1 − μ2 ≤ (90 − 115) + 1.782 +
8 9
μ1 − μ2 ≤ −18.28
Because zero is not contained in this interval, we are 95% confident that mean 2 exceeds mean1.

10-30
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

c) 1) The parameter of interest is the difference in mean blood pressure between the test and control
groups, μ1 − μ2 , with Δ0 = −15

2) H0 : μ1 − μ 2 = −15 or μ1 = μ2

3) H1 : μ1 − μ 2 < −15 or μ1 < μ 2 − 15

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
s12 s22
+
n1 n2

5) Reject the null hypothesis if t0 < − t α , ν where t0.05,12 = −1.782 for α = 0.05 since
2
⎛ s12 s22 ⎞
⎜⎜ + ⎟⎟
ν = ⎝ 1 2 2 ⎠ = 12
n n
⎛ s1 ⎞
2
⎛ s22 ⎞
⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟
⎝ n1 ⎠ + ⎝ n2 ⎠
n1 − 1 n2 − 1
ν ≅ 12
(truncated)

6) x1 = 90 x2 = 115

s1 = 5 s2 = 10

n1 = 8 n2 = 9
(90 − 115) + 15
t0 = = −2.65
(5) 2 (10) 2
+
8 9
7) Conclusion: Because −2.65 < −1.782 reject the null hypothesis and conclude that the test group has
higher mean arterial blood pressure than the control group at the 0.05 level of significance.

d) 95% confidence interval: t0.05,12 = 1.782

s12 s22
μ1 − μ2 ≤ ( x1 − x2 ) + tα ,ν +
n1 n2

μ1 − μ2 ≤ −18.28

Because −15 is greater than the values in this interval, we are 95% confident that the mean for the
test group is at least 15 mmHg higher than the control group.

10-27 a) 1) The parameter of interest is the difference in mean number of periods in a sample of 200 trains for
two different levels of noise voltage, 100mv and 150mv μ1 − μ2 , with Δ0 = 0

2) H0 : μ1 − μ 2 = 0 or μ1 = μ 2

10-31
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

3) H1 : μ1 − μ2 > 0 or μ1 > μ 2
4) The test statistic is
( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 > tα , n1 + n2 − 2 where t0.01,198 = 2.326 for α = 0.01

(n1 − 1) s12 + (n2 − 1) s22

6) x1 = 7.9 x2 = 6.9 sp =
n1 + n2 − 2

99(2.6) 2 + 99(2.4) 2
s1 = 2.6 s2 = 2.4 = = 2.5
198
n1 = 100 n2 = 100
(7.9 − 6.9)
t0 = = 2.82
1 1
2.5 +
100 100
7) Conclusion: Because 2.82 > 2.326 reject the null hypothesis at the 0.01 level of significance.
P-value = P (t > 2.82) P-value ≅ 0.0025

b) 99% confidence interval: t0.01,198 = 2.326

1 1
μ1 − μ2 ≥ ( x1 − x2 ) − tα , n + n − 2 ( s p ) +
1 2
n1 n2

μ1 − μ2 ≥ 0.178
Because zero is not contained in this interval, we are 99% confident that mean 1 exceeds mean 2.

10-28 a) The probability plots below show that normality assumption is reasonable for both data sets.

10-32
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

b) 1) The parameter of interest is the difference in mean weight of two sheets of paper, μ1 − μ 2 . Assume

equal variances.
2) H0 : μ1 − μ 2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ 2 ≠ 0 or μ1 ≠ μ 2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 < − t α / 2, n1 + n2 − 2 where − t 0.025,28 = −2.048 or t0 > t α / 2, n1 + n2 − 2 where

t0.025,28 = 2.048 for α = 0.05

6) x1 = 3.472 x2 = 3.2494

s12 = 0.008312 s22 = .007142

n1 = 15 n2 = 15

(n1 − 1) s12 + (n2 − 1) s22 14(.00831) 2 + 14(0.00714) 2

sp = = = .00775
n1 + n2 − 2 28

t0 = 78.66

7) Conclusion: Because 78.66 >2.048, reject the null hypothesis at α = 0.05.

P-value ≅ 0

c) 1) The parameter of interest is the difference in mean weight of two sheets of paper, μ1 − μ2

2) H0 : μ1 − μ2 = 0 or μ1 = μ2

3) H1 : μ1 − μ2 ≠ 0 or μ1 ≠ μ 2

4) The test statistic is

10-33
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 < −tα / 2, n1 + n2 − 2 where −t0.05,28 = −1.701 or t0 > tα / 2, n1 + n2 − 2 where t0.05, 28

= 1.701 for α = 0.1

6) x1 = 3.472 x2 = 3.2494

s12 = 0.008312 s22 = .007142

n1 = 15 n2 = 15

(n1 − 1) s12 + (n2 − 1) s22 14(.00831) 2 + 14(0.00714) 2

sp = = = .00775
n1 + n2 − 2 28

t0 = 78.66

7) Conclusion: Because 78.66 > 1.701, reject the null hypothesis at α = 0.10. P-value ≅ 0

d) The answer is the same because the decision to reject the null hypothesis made in part (b) was at a
lower level of significance than the test in (c). Therefore, the decision would be the same for any value
of α larger than that used in part (b). Alternatively, the P-value from part (b) is essentially 0, meaning
that for any level of α greater than or equal to the P-value, the decision is to reject the null hypothesis.

e) 95% confidence interval for part (b): t0.025,28 = 2.048

1 1 1 1
( x1 − x2 ) − tα / 2,n + n − 2 (s p ) + ≤ μ1 − μ 2 ≤ ( x1 − x2 ) + tα / 2, n1 + n2 − 2 ( s p ) +
1 2
n1 n2 n1 n2

0.216 ≤ μ1 − μ2 ≤ 0.228

Because zero is not contained in this interval, we are 95% confident that the means are different.
90% confidence interval for part (c): t0.05,28 = 1.701

1 1 1 1
( x1 − x2 ) − tα / 2,n + n − 2 (s p ) + ≤ μ1 − μ 2 ≤ ( x1 − x2 ) + tα / 2, n1 + n2 − 2 ( s p ) +
1 2
n1 n2 n1 n2

0.217 ≤ μ1 − μ 2 ≤ 0.227
Because zero is not contained in this interval, we are 90% confident that the means are different.

10-29 a) The data appear to be normally distributed and the variances appear to be approximately equal. The
slopes of the lines on the normal probability plots are almost the same.

10-34
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

b) 1) The parameter of interest is the difference in mean overall distance, μ1 − μ2 , with Δ0 = 0

2) H0 : μ1 − μ2 = 0 or μ1 = μ2

3) H1 : μ1 − μ2 ≠ 0 or μ1 ≠ μ2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 < −tα / 2, n1 + n2 − 2 or t0 > tα / 2, n1 + n2 − 2 where t0.025,18 = 2.101 for α = 0.05

(n1 − 1) s12 + (n2 − 1) s22

6) x1 = 252.1 x2 = 242.6 sp =
n1 + n2 − 2

9(7.61) 2 + 9(9.26) 2
s1 = 7.61 s2 = 9.26 = = 8.04
20
n1 = 10 n2 = 10
(252.1 − 242.6)
t0 = = 2.642
1 1
8.04 +
10 10
7) Conclusion: Because 2.642 > 2.101 reject the null hypothesis. The data support the claim that the
means differ at α = 0.05.
P-value = 2P ( t > 2.642) P-value ≈ 2(0.01) = 0.02

1 1 1 1
c) ( x1 − x2 ) − tα ,ν s p + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + tα ,ν s p +
n1 n2 n1 n2

1 1 1 1
(252.1 − 242.6) − 2.101(8.04) + ≤ μ1 − μ2 ≤ (252.1 − 242.6) + 2.101(8.04) +
10 10 10 10
1.94 ≤ μ1 − μ2 ≤ 17.05

10-35
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

4.5
d) d = = 0.28 β = 0.95 Power = 1 − 0.95 = 0.05
2(8.04)

2.75
e) β = 0.25 d= = 0.171 n* = 100 Therefore, n = 51
2(8.04)

10-30 a) The data appear to be normally distributed and the variances appear to be approximately equal. The
slopes of the lines on the normal probability plots are almost the same.

Normal Probability Plot for Club1...Club2

ML Estimates - 95% CI

Club1
99
Club2

95
90

80
70
Percent

60
50
40
30
20

10
5

0.75 0.77 0.79 0.81 0.83 0.85 0.87 0.89

Data

b) 1) The parameter of interest is the difference in mean coefficient of restitution, μ1 − μ 2

2) H0 : μ1 − μ2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ2 ≠ 0 or μ1 ≠ μ 2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
1 1
sp +
n1 n2

5) Reject the null hypothesis if t0 < −tα / 2, n1 + n2 − 2 or t0 > tα / 2, n1 + n2 − 2 where t0.025,22 = 2.074 for α = 0.05

(n1 − 1) s12 + (n2 − 1) s22

6) x1 = 0.8161 x2 = 0.8271 sp =
n1 + n2 − 2

11(0.0217) 2 + 11(0.0175) 2
s1 = 0.0217 s2 = 0.0175 = = 0.01971
22
n1 = 12 n2 = 12
(0.8161 − 0.8271)
t0 = = −1.367
1 1
0.01971 +
12 12
7) Conclusion: Because –1.367 > –2.074 fail to reject the null hypothesis. The data do not support the
claim that there is a difference in the mean coefficients of restitution for club1 and club2 at α = 0.05
P-value = 2P ( t < −1.36) , P-value ≈ 2(0.1) = 0.2

c) 95% confidence interval

10-36
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

(x1 − x2 ) − tα ,ν s p 1
+
1
≤ μ1 − μ 2 ≤ ( x1 − x 2 ) + tα ,ν s p
1
+
1
n1 n2 n1 n2

1 1 1 1
(0.8161 − 0.8271) − 2.074(0.01971) + ≤ μ1 − μ2 ≤ (0.8161 − 0.8271) + 2.074(0.01971) +
12 12 12 12
−0.0277 ≤ μ1 − μ2 ≤ 0.0057

Because zero is included in the confidence interval there is not a significant difference in the mean
coefficients of restitution at α = 0.05.

0.2
d) d = = 5.07 β ≅ 0, Power ≅1
2(0.01971)

0.1 n * +1
e) 1 − β = 0.8 β = 0.2 d= = 2.53 n* = 4, n = = 2.5 n≅3
2(0.01971) 2

10-37
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Section 10-3

10-31 a) 1) The parameters of interest are the mean current (note: set circuit 1 equal to sample 2 so that Table X
can be used. Therefore μ1 = mean of circuit 2 and μ2 = mean of circuit 1)
2) H 0 : μ1 = μ2
3) H1 : μ1 > μ2

(n1 + n2 )(n1 + n2 + 1)
4) The test statistic is w2 = − w1
2
5) We reject H0 if w2 ≤ w0.025
*
= 51, because α = 0.025 and n1 = 8 and n2 = 9, Appendix A, Table X

gives the critical value.

6) w1 = 78and w2 = 75 and because 75 is less than 51 fail to reject H0
7) Conclusion, fail to reject H0. There is not enough evidence to conclude that the mean of circuit 1
exceeds the mean of circuit 2.

b) 1) The parameters of interest are the mean image brightness of the two tubes
2) H 0 : μ1 = μ2
3) H1 : μ1 > μ2

W1 − μ w1
4) The test statistic is z0 =
σw 1

5) We reject H0 if Z0 > Z0.025 = 1.96 for α = 0.025

6) w1 = 78, μ w1 = 72 and σ w21 = 108

78 − 72
z0 = = 0.58
10.39
Because Z0 < 1.96, fail to reject H0
7. Conclusion: fail to reject H0. There is not a significant difference in the heat gain for the heating
units at α = 0.05.
P-value = 2[1 − P(Z < 0.58 )] = 0.5619

10-32 1) The parameters of interest are the mean flight delays

2) H 0 : μ1 = μ2
3) H1 : μ1 ≠ μ 2

(n1 + n2 )(n1 + n2 + 1)
4) The test statistic is w2 = − w1
2
5) We reject H0 if w ≤ w0.01
*
= 23, because α = 0.01 and n1 = 6 and n2 = 6, Appendix A, Table X gives

the critical value.

6. w1 = 40 and w2 = 38 and because 40 and 38 are greater than 23, fail to reject H0
7. Conclusion: fail to reject H0. There is no significant difference in the flight delays at α = 0.01.

10-38
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

10-33 a) 1) The parameters of interest are the mean heat gains for heating units
2) H 0 : μ1 = μ2
3) H1 : μ1 ≠ μ 2

(n1 + n2 )(n1 + n2 + 1)
4) The test statistic is w2 = − w1
2
6) We reject H0 if w ≤ w0.01
*
= 78, because α = 0.01 and n1 = 10 and n2 = 10, Appendix A, Table X

gives the critical value.

7. w1 = 77 and w2 = 133 and because 77 is less than 78, we can reject H0
8. Conclusion: reject H0 and conclude that there is a significant difference in the heating units at α =
0.05.

b) 1) The parameters of interest are the mean heat gain for heating units
2) H 0 : μ1 = μ2
3) H1 : μ1 ≠ μ 2

W1 − μ w1
4) The test statistic is z0 =
σw 1

5) Reject H0 if |Z0| > Z0.025 = 1.96 for α = 0.05

6) w1 = 77, μw 1
= 105 and σ w21
= 175

77 − 105
z0 = = −2.12
13.23
Because |Z0 | > 1.96, reject H0
7. Conclusion: reject H0 and conclude that there is a difference in the heat gain for the heating units at
α = 0.05.
P-value = 2[1 − P(Z < 2.19 )] = 0.034

10-34 a) 1) The parameters of interest are the mean etch rates

2) H 0 : μ1 = μ2
3) H1 : μ1 ≠ μ 2

(n1 + n2 )(n1 + n2 + 1)
4) The test statistic is w2 = − w1
2
5) We reject H0 if w ≤ w0.05
*
= 78, because α = 0.05 and n1 = 10 and n2 = 10, Appendix A, Table X gives

the critical value.

6. w1 = 74 and w2 = 136 and because 74 is less than 78, we reject H0
7. Conclusion: reject H0 and conclude that there is a significant difference in the mean etch rate at α =
0.05.

b) 1) The parameters of interest are the mean temperatures

10-39
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

2) H 0 : μ1 = μ2
3) H1 : μ1 ≠ μ 2

W1 − μ w1
4) The test statistic is z0 =
σw 1

5) We reject H0 if |Z0| > Z0.025 = 1.96 for α = 0.05

6) w1 = 74, μ w1 = 105 and σ w21 = 175

74 − 105
z0 = = −2.343
13.229
Because |Z0| > 1.96, it rejects H0
7) Conclusion: It rejects H0. There is a difference in the pipe deflection temperatures at α = 0.05.
P-value = 2[P(Z < −2.343)] = 0.019

10-35 a) 1) The parameters of interest are the mean temperatures

2) H 0 : μ1 = μ 2

3) H1 : μ1 ≠ μ 2

(n1 + n2 )(n1 + n2 + 1)
4) The test statistic is w2 = − w1
2
5) We reject H0 if w ≤ w0.05
*
= 185, because α = 0.05 and n1 = 15 and n2 = 15, Appendix A, Table X

gives the critical value.

6) w1 = 259 and w2 = 206 and because both 259 and 206 are greater than 185, we fail to reject H0
7) Conclusion: fail to reject H0. There is not a significant difference in the mean pipe deflection
temperature at α = 0.05.

b) 1) The parameters of interest are the mean etch rates

2) H 0 : μ1 = μ2
3) H1 : μ1 ≠ μ 2

W1 − μ w1
4) The test statistic is z0 =
σw 1

5) We reject H0 if |Z0| > Z0.025 = 1.96 for α = 0.05

6) w1 = 259, μ w1 = 232.5 and σ w21 = 581.25

259 − 232.5
z0 = = 1.1
24.11
Because |Z0| < 1.96, do not reject H0
7) Conclusion: Fail to reject H0. There is not a significant difference between the mean etch rates.
P-value = 0.2713

10-36 a) The data are analyzed in ascending order and ranked as follows:

10-40
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Group Distance Rank

2 223 1.0
2 236 2.0
2 238 3.0
1 240 4.5
2 240 4.5
2 242 6.0
1 244 7.0
2 245 8.0
2 247 9.0
1 248 11.0
1 248 11.0
2 248 11.0
2 250 13.0
1 251 14.5
1 251 14.5
1 255 16.0
2 257 17.0
1 259 18.0
1 262 19.0
1 263 20.0

The sum of the ranks for group 1 is w1 = 135.5 and for group 2, w2 = 74.5. Because w2 is less than

w0.05 = 78 , we reject the null hypothesis that both groups have the same mean.

b) When the sample sizes are equal it does not matter which group we select for w1
10(10 + 10 + 1)
μW1 = = 105
2
10*10(10 + 10 + 1)
σ W2 1 = = 175
12
135.5 − 105
Z0 = = 2.31
175
Because z0 > z0.025 = 1.96, reject H0 and conclude that the sample means for the two groups are
different.
When z0 = 2.31, P-value = 0.021

10-41
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Section 10-4

10-37 a) d = 0.2738 sd = 0.1351, n = 9

95% confidence interval:
⎛s ⎞ ⎛s ⎞
d − tα / 2, n −1 ⎜ d ⎟ ≤ μd ≤ d + tα / 2, n −1 ⎜ d ⎟
⎝ n⎠ ⎝ n⎠

⎛ 0.1351 ⎞ ⎛ 0.1351 ⎞
0.2738 − 2.306 ⎜ ⎟ ≤ μd ≤ 0.2738 + 2.306 ⎜ ⎟
⎝ 9 ⎠ ⎝ 9 ⎠

0.1699 ≤ μd ≤ 0.3776
With 95% confidence, the mean shear strength of Karlsruhe method exceeds the mean shear strength of
the Lehigh method by between 0.1699 and 0.3776. Because zero is not included in this interval, the
interval is consistent with rejecting the null hypothesis that the means are equal.

The 95% confidence interval is directly related to a test of hypothesis with 0.05 level of significance and
the conclusions reached are identical.

b) It is only necessary for the differences to be normally distributed for the paired t-test to be
appropriate and reliable. Therefore, the t-test is appropriate.
Normal Probability Plot

.999
.99
.95
Probability

.80
.50
.20
.05
.01
.001

0.12 0.22 0.32 0.42 0.52

diff
Average: 0.273889 Anderson-Darling Normality Test
StDev: 0.135099 A-Squared: 0.318
N: 9 P-Value: 0.464

10-38 a) 1) The parameter of interest is the difference between the mean parking times, μd.
2) H0 : μ d = 0

3) H1 : μ d ≠ 0

4) The test statistic is

d
t0 =
sd / n

5) Reject the null hypothesis if t0 < −t 0.05,13 where −t 0.05,13 = −1.771 or t0 > t 0.05,13 where t 0.05,13 = 1.771

for α = 0.10

10-42
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

6) d = 1.21
sd = 12.68

n = 14
1.21
t0 = = 0.357
12.68 / 14
7) Conclusion: Because −1.771 < 0.357 < 1.771 fail to reject the null. The data fail to support the claim
that the two cars have different mean parking times at the 0.10 level of significance.

b) The result is consistent with the confidence interval constructed because zero is included in the 90%
confidence interval.
c) According to the normal probability plots, the assumption of normality does not appear to be violated
because the data fall approximately along a straight line.

Normal Probability Plot

.999
.99
Pr .95
ob .80
abi .50
lity .20
.05
.01
.001

-20 0 20
diff
Average: 1.21429 Anderson-Darling Normality Test
StDev: 12.6849 A-Squared: 0.439
N: 14 P-Value: 0.250

10-39 d = 868.375 sd = 1290, n = 8 where di = brand 1 − brand 2

95% confidence interval:
⎛s ⎞ ⎛s ⎞
d − tα / 2, n −1 ⎜ d ⎟ ≤ μd ≤ d + tα / 2, n −1 ⎜ d ⎟
⎝ n⎠ ⎝ n⎠

⎛ 1290 ⎞ ⎛ 1290 ⎞
868.375 − 2.365 ⎜ ⎟ ≤ μd ≤ 868.375 + 2.365 ⎜ ⎟
⎝ 8 ⎠ ⎝ 8 ⎠
−210.26 ≤ μd ≤ 1947.01
Because this confidence interval contains zero, there is no significant difference between the two brands
of tire at a 5% significance level.

10-40 a) According to the normal probability plots, the assumption of normality does not appear to be violated
because the data fall approximately along a straight line.

10-43
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Normal Probability Plot

.999
.99
.95

Probability
.80
.50
.20
.05
.01
.001

-5 0 5
diff
Average: 0.666667 Anderson-Darling Normality Test
StDev: 2.96444 A-Squared: 0.315
N: 12 P-Value: 0.502

b) d = 0.667 sd = 2.964, n = 12
95% confidence interval:

⎛s ⎞ ⎛s ⎞
d − t α / 2, n −1 ⎜⎜ d ⎟⎟ ≤ μ d ≤ d + t α / 2, n −1 ⎜⎜ d ⎟⎟
⎝ n ⎠ ⎝ n⎠
⎛ 2.964 ⎞ ⎛ 2.964 ⎞
0.667 − 2.201⎜ ⎟ ≤ μ d ≤ 0.667 + 2.201⎜ ⎟
⎝ 12 ⎠ ⎝ 12 ⎠
−1.216 ≤ μd ≤ 2.55
Because zero is contained within this interval, there is no significant indication that one design
language is preferable at a 5% significance level

10-41 a) 1) The parameter of interest is the difference in blood cholesterol level, μd

where di = Before − After.
2) H0 : μd = 0

3) H1 : μd > 0

4) The test statistic is

d
t0 =
sd / n

5) Reject the null hypothesis if t0 > t 0.05,14 where t 0.05,14 = 1.761 for α = 0.05

6) d = 26.867
sd = 19.04

n = 15
26.867
t0 = = 5.465
19.04 / 15
7) Conclusion: Because 5.465 > 1.761 reject the null hypothesis. The data support the claim that the
mean difference in cholesterol levels is significantly less after diet and an aerobic exercise program
at the 0.05 level of significance.

10-44
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

P-value = P(t > 5.565) ≅ 0

b) 95% confidence interval:

⎛s ⎞
d − tα , n −1 ⎜ d ⎟ ≤ μd
⎝ n⎠

⎛ 19.04 ⎞
26.867 − 1.761⎜ ⎟ ≤ μd
⎝ 15 ⎠
18.20 ≤ μd
Because the lower bound is positive, with 95% confidence the mean difference in blood cholesterol level
is significantly less after the diet and aerobic exercise program.

10-42 a) 1) The parameter of interest is the mean difference in natural vibration frequencies, μd where di = finite
element − Equivalent Plate.
2) H0 : μd = 0

3) H1 : μd ≠ 0

4) The test statistic is

d
t0 =
sd / n

5) Reject the null hypothesis if t0 < −t0.025,6 or t0 > t0.025, 6 where t 0.005,6 = 2.447 for α = 0.05

6) d = −5.49
sd = 5.924

n=7
−5.49
t0 = = −2.45
5.924 / 7
7) Conclusion: Because −2.45 < −2.447, reject the null hypothesis. The two methods have different mean
values for natural vibration frequency at the 0.05 level of significance.

b) 95% confidence interval:

⎛s ⎞ ⎛s ⎞
d − tα / 2, n −1 ⎜ d ⎟ ≤ μd ≤ d + tα / 2, n −1 ⎜ d ⎟
⎝ n⎠ ⎝ n⎠

⎛ 5.924 ⎞ ⎛ 5.924 ⎞
−5.49 − 2.447 ⎜ ⎟ ≤ μd ≤ −5.49 + 2.447 ⎜ ⎟
⎝ 7 ⎠ ⎝ 7 ⎠
−10.969 ≤ μd ≤ −0.011
With 95% confidence the mean difference between the natural vibration frequency from the equivalent
plate method and the finite element method is between −10.969 and −0.011 cycles.

10-43 a) 1) The parameter of interest is the difference in mean weight, μd where di = Weight Before − Weight
After.

10-45
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

2) H0 : μd = 0

3) H1 : μd > 0

4) The test statistic is

d
t0 =
sd / n

5) Reject the null hypothesis if t0 > t 0.05,9 where t 0.05,9 = 1.833 for α = 0.05

6) d = 8
sd = 3.2

n = 10
8
t0 = = 7.906
3.2 / 10
7) Conclusion: Because 7.906 > 1.833 reject the null hypothesis and conclude that the mean weight
loss is significantly greater than zero. That is, the data support the claim that this particular diet
modification program is effective in reducing weight at the 0.05 level of significance.

b) 1) The parameter of interest is the difference in mean weight loss, μd where di = Before − After.
2) H0 : μd = 10

3) H1 : μd > 10

4) The test statistic is

d − Δ0
t0 =
sd / n

5) Reject the null hypothesis if t0 > t 0.05,9 where t 0.05,9 = 1.833 for α = 0.05

6) d = 8
sd = 3.2

n = 10
8 − 4.5
t0 = = 3.46
3.2 / 10
7) Conclusion: Because 3.46 > 1.833 reject the null hypothesis. There is evidence to support the claim
that this particular diet modification program is effective in producing a mean weight loss of at least
4.5 kg at the 0.05 level of significance.

c) Use sd as an estimate for σ:

⎛ ( zα + zβ ) σ d
2
⎞ (1.645 + 1.29)3.2 ⎞
2
n=⎜ ⎟ = ⎛⎜ ⎟ = 0.88 , n = 1
⎜ 10 ⎟ ⎝ 10 ⎠
⎝ ⎠
Yes, the sample size of 10 is adequate for this test.

10-46
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

10-44 a) 1) The parameter of interest is the mean difference in impurity level, μd where di = Test 1 − Test 2.
2) H0 : μ d = 0

3) H1 : μ d ≠ 0

4) The test statistic is

d
t0 =
sd / n

5) Reject the null hypothesis if t0 < − t 0.005, 7 or t0 > t 0.005, 7 where t 0.005, 7 = 3.499 for α = 0.01

6) d = −0.2125
sd = 0.1727

− 0.2125
n=8 t0 = = −3.48
0.1727 / 8
7) Conclusion: Because −3.499 < −3.48 < 3.499 fail to reject the null hypothesis. There is not sufficient
evidence to conclude that the tests generate different mean impurity levels at α = 0.01.

b) 1) The parameter of interest is the mean difference in impurity level, μd where di = Test 1 − Test 2.
2) H0 : μd + 0.1 = 0

3) H1 : μ d + 0.10 < 0

4) The test statistic is

d + 0.1
t0 =
sd / n

5) Reject the null hypothesis if t0 < − t0.05, 7 where t0.05, 7 = 1.895 for α = 0.05

6) d = −0.2125
sd = 0.1727

n=8
− 0.2125 + 0.1
t0 = = −1.8424
0.1727 / 8
7) Conclusion: Because −1.895 < −1.8424 fail to reject the null hypothesis at the 0.05 level of
significance.

c) β = 1 − 0.9 = 0.1
0.1
d= = 0.579
0.1727
n = 8 is not an adequate sample size. From the chart VII (g), n ≈ 30

10-45 a) The probability plot below show that normality assumption is reasonable.

10-47
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

b) d = −0.015
sd = 0.5093

n = 10

t0.025,9 = 2.262
95% confidence interval:
⎛s ⎞ ⎛s ⎞
d − tα / 2, n −1 ⎜ d ⎟ ≤ μd ≤ d + tα / 2, n −1 ⎜ d ⎟
⎝ n⎠ ⎝ n⎠
−0.379 ≤ μ d ≤ 0.3493

Because zero is contained in the confidence interval there is not sufficient evidence that the mean IQ
depends on birth order.

c) β = 1 − 0.9 = 0.1
δ 1
d= = = 1.96
σ sd
Thus 6 ≤ n would be enough.

10-46 a) Let x12 = x 2 − x1 and x 23 = x3 − x 2 and x d = x23 − x12

1) The parameter of interest is the mean difference in circumference μd where x d = x23 − x12

2) H0 : μd = 0
3) H1 : μd ≠ 0
4) The test statistic is

d
t0 =
sd / n

5) Reject the null hypothesis if t0 < − t0.05, 4 or t0 > t0.05, 4 where t0.05, 4 = 2.132 for α = 0.10

10-48
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

6) d = 8.6
sd = 7.829

n=5 8.6
t0 = = 2.456
7.829 / 5
7) Conclusion: Because 2.132 < 2.456 reject the null hypothesis. The means are significantly different at
α = 0.1.

b) Let x67 = x7 − x6

Let x d = x12 − x67

1) The parameter of interest is the mean difference in circumference μd where xd = x12 − x67

2) H0 : μ d = 0

3) H1 : μ d > 0

4) The test statistic is

d
t0 =
sd / n

5) Reject the null hypothesis if t0 > t0.05, 4 where t0.05, 4 = 2.132 for α = 0.05

6) d = −24.4
sd = 7.5

n=5 − 24.4
t0 = = 7.27
7.5 / 5
7) Conclusion: Because 7.27 > 2.132 reject the null hypothesis. The means are significantly different at α
= 0.1.
P-value = P(t > 7.27) ≈ 0

c) No, the paired t test uses the differences to conduct the inference.

10-47 1) Parameters of interest are the median cholesterol levels for two activities.
2) H 0 : μ D = 0 2) H 0 : μ1 − μ 2 = 0
or
3) H1 : μ D > 0 3) H1 : μ1 − μ 2 > 0

4) r−

5) Because α = 0.05 and n = 15, Appendix A, Table VIII gives the critical value of r0.05
*
= 3 . We

reject H0 in favor of H1 if r− ≤ 3.
6) The test statistic is r− = 2.

Observation Before After Difference Sign

1 265 229 36 +

10-49
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

2 240 231 9 +
3 258 227 31 +
4 295 240 55 +
5 251 238 13 +
6 245 241 4 +
7 287 234 53 +
8 314 256 58 +
9 260 247 13 +
10 279 239 40 +
11 283 246 37 +
12 240 218 22 +
13 238 219 19 +
14 225 226 −1 -
15 247 233 14 +

15
⎛15 ⎞
P-value = P(R+ ≥ r+ = 14 | p = 0.5) = ∑ ⎜⎜ r ⎟⎟(0.5) r
(0.5) 20−r = 0.00049
r =13 ⎝ ⎠
7) Conclusion: Because the P-value = 0.00049 is less than α = 0.05, reject the null hypothesis. There
is a significant difference in the median cholesterol levels after diet and exercise at α = 0.05.

10-48 1) The parameters of interest are the median cholesterol levels for two activities.
H0 : μD = 0 H 0 : μ1 − μ2 = 0
2) and 3) or
H1 : μ D > 0 H1 : μ1 − μ2 > 0

4) w−
5) Reject H0 if w − ≤ w0*.05,n =15 = 30 for α = 0.05

6) The sum of the negative ranks is w− = (1 + 15) = 16.

Observation Before After Difference Signed Rank

14 225 226 −1 −1
2 240 231 9 3
5 251 238 13 4.5
9 260 247 13 4.5
15 247 233 14 6

10-50
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

13 238 219 19 7
12 240 218 22 8
3 258 227 31 9
1 265 229 36 10
11 283 246 37 11
10 279 239 40 12
7 287 234 53 13
4 295 240 55 14
8 314 256 58 15
6 245 241 4 2

7) Conclusion: Because w− = 1 is less than the critical value w0*.05,n =15 = 30 , reject the null hypothesis.

There is a significant difference in the mean cholesterol levels after diet and exercise at α = 0.05.

Exercise10-47 tests the difference in the median cholesterol levels after diet and exercise while
Exercise 10-48 tests the difference in the mean cholesterol levels after diet and exercise.

10-51
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Section 10-5

1 1
10-49 a) f0.25,5,10 = 1.59 d) f0.75,5,10 = = = 0.529
f0.25,10,5 189
.

1 1
b) f0.10,24,9 = 2.28 e) f0.90,24,9 = = = 0 . 525
f 0 . 10 , 9 , 24 1 . 91

1 1
c) f0.05,8,15 = 2.64 f) f0.95,8,15 = = = 0.311
f0.05,15,8 3.22

10-50 a) f0.25,7,15 = 1.47 d) f0.75,7,15 = 1 1

= = 0 . 596
f 0 .25 ,15 , 7 1 . 68

b) f0.10,10,12 = 2.19 e) f0.90,10,12 = 1 1

= = 0 . 438
f 0 . 10 ,12 ,10 2 . 28

1 1
c) f0.01,20,10 = 4.41 f) f0.99,20,10 = = = 0.297
f0.01,10,20 3.37

10-51 1) The parameters of interest are the standard deviations σ 1 , σ 2

2) H0 : σ 12 = σ 22
3) H1 : σ 12 < σ 22
4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f .0.95, 4,9 = 1 / f .0.05,9, 4 = 1/6 = 0.1666 for α = 0.05

6) n1 = 5 n2 = 10 s12 = 23.2 s22 = 28.8

(23.2)
f0 = = 0.806
(28.8)

7) Conclusion: Because 0.1666 < 0.806 do not reject the null hypothesis.
95% confidence interval:

σ 12 ⎛ s12 ⎞
≤ ⎜ ⎟ f1−α ,n −1,n −1
σ 22 ⎜⎝ s 22 ⎟⎠ 2 1

σ12 ⎛ 23.2 ⎞ σ 12 σ
≤⎜
⎝ ⎟⎠ * f 0.05,9,4 where f.05,9,4 = 6 ≤ 4.83 or 1 ≤ 2.20
σ22
28.8 σ 22 σ2
Because the value one is contained within this interval, there is no significant difference in the variances.

10-52 1) The parameters of interest are the standard deviations, σ1 , σ 2

2) H0 : σ12 = σ 22

10-52
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

3) H1 : σ 12 > σ 22

4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 > f.0.01,19,7 ≅ 6.16 for α = 0.01

6) n1 = 20 n2 = 8 s12 = 4.5 s22 = 2.3

4.5
f0 = = 1.956
2.3
7) Conclusion: Because 6.16 > 1.956 fail to reject the null hypothesis.
95% confidence interval:

⎛ s12 ⎞ σ2
⎜⎜ 2 ⎟⎟ f 0.99,n2 −1,n1 −1 ≤ 12
⎝ s2 ⎠ σ2
σ12 σ 12
1.956 *1/ 6.16 ≤ 0.318 ≤
σ 22 σ 22
Because the value one is contained within this interval, there is no significant difference in the
variances.

10-53 a) 1) The parameters of interest are the standard deviations, σ1 , σ 2

2) H0 : σ 12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f.0.975,14,14 = 0.33 or f0 > f 0.025,14,14 = 3 for α = 0.05

6) n1 = 15 n2 = 15 s12 = 2.5 s22 = 2.2

2.5
f0 = = 1.14
2.2
7) Conclusion: Because 0.333 < 1.14 < 3, we fail to reject the null hypothesis. There is no sufficient
evidence that there is a difference in the standard deviation.
95% confidence interval:

⎛ s12 ⎞ σ12 ⎛ s12 ⎞

⎜⎜ 2 ⎟⎟ f1−α / 2, n2 −1,n1 −1 ≤ 2 ≤ ⎜⎜ 2 ⎟⎟ fα / 2, n2 −1,n1 −1
⎝ s2 ⎠ σ 2 ⎝ s2 ⎠

σ 12 σ12
(1.14)0.333 ≤ ≤ (1.14)3 0.38 ≤ ≤ 3.42
σ 22 σ 22
Because the value one is contained within this interval, there is no significant difference in the variances.

10-53
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

σ1
b) λ = =2
σ2
n1 = n2 = 5

α = 0.05
Chart VII (o) we find β = 0.35 then the power 1 − β = 0.65

σ1
c) β = 0.05 and σ 2 = σ 1 / 2 so that = 2 and n ≈ 31
σ2

10-54 1) The parameters of interest are the variances of concentration, σ12 , σ 22

2) H0 : σ12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f 0.975,9,15 where f 0.975,9,15 = 0.265 or f0 > f0.025,9 ,15 where

f 0.025,9,15 = 3.12 for α = 0.05

6) n1 = 10 n2 = 16

s1 = 4.7 s2 = 5.8

(4.7) 2
f0 = = 0.657
(5.8) 2

7) Conclusion: Because 0.265 < 0.657 < 3.12 fail to reject the null hypothesis. There is insufficient
evidence to conclude that the two population variances differ at the 0.05 level of significance.

10-55 a) 1) The parameters of interest are the time to assemble standard deviations, σ1 , σ 2

2) H0 : σ 12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f 1−α / 2, n1 −1, n2 −1 = 0.365 or f0 > f α / 2, n1 −1, n2 −1 = 2.86 for α = 0.02

6) n1 = 25 n2 = 21 s1 = 0.98 s2 = 1.02

(0.98) 2
f0 = = 0.923
(1.02) 2

10-54
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

7) Conclusion: Because 0.365 < 0.923 < 2.86 fail to reject the null hypothesis. There is not sufficient
evidence to support the claim that men and women differ in repeatability for this assembly task at the
0.02 level of significance.

ASSUMPTIONS: Assume that the two populations are independent and normally distributed.

b) 98% confidence interval:

⎛ s12 ⎞ σ 2 ⎛ s2 ⎞
⎜ 2 ⎟ f 1−α / 2, n1 −1, n2 −1 ≤ 12 ≤ ⎜ 12 ⎟ f α / 2, n1 −1, n2 −1
⎜s ⎟ σ 2 ⎜⎝ s 2 ⎟⎠
⎝ 2 ⎠
σ 12
(0.923)0.365 ≤ ≤ (0.923)2.86
σ 22

σ 12
0.3369 ≤ ≤ 2.640
σ 22
Because the value one is contained within this interval, there is no significant difference between the
variance of the repeatability of men and women for the assembly task at a 2% significance level.

10-56 a) 90% confidence interval for the ratio of variances:

⎛ s12 ⎞ σ 2 ⎛ s2 ⎞
⎜ 2 ⎟ f 1−α / 2, n1 −1, n2 −1 ≤ 12 ≤ ⎜ 12 ⎟ f α / 2, n1 −1, n2 −1
⎜s ⎟ σ 2 ⎜⎝ s 2 ⎟⎠
⎝ 2 ⎠

⎛ 0 .6 2 ⎞ σ 2 ⎛ 0 .6 2 ⎞ σ 12
⎜⎜ 2 ⎟⎟0.156 ≤ 12 ≤ ⎜⎜ 2 ⎟⎟6.39 0.08775 ≤ ≤ 3.594
⎝ 0 .8 ⎠ σ 2 ⎝ 0 .8 ⎠ σ 22

b) 95% confidence interval:

⎛ s12 ⎞ σ 2 ⎛ s2 ⎞
≤ 1 ≤ 1 f
⎜⎜ s 2 ⎟⎟ 1−α / 2,n1 −1, n2 −1 σ 2 ⎜⎜ s 2 ⎟⎟ α / 2,n1 −1, n2 −1
f
⎝ 2⎠ 2 ⎝ 2⎠

⎛ (0.6) 2 ⎞ σ 12 ⎛ (0.6) 2 ⎞ σ 12
⎜⎜ ⎟
2 ⎟
0.104 ≤ ≤⎜ ⎟ 9.60 0.0585 ≤ ≤ 5.4
⎝ (0.8) ⎠ σ 22 ⎜⎝ (0.8) 2 ⎟⎠ σ 22
The 95% confidence interval is wider than the 90% confidence interval.

c) 90% lower-sided confidence interval:

⎛ s12 ⎞ σ12
⎜⎜ 2 ⎟⎟ f1−α , n1 −1,n2 −1 ≤ 2
⎝ s2 ⎠ σ2

⎛ (0.6) 2 ⎞ σ 12 σ 12
⎜⎜ ⎟⎟ 0.243 ≤ 2 0.137 ≤
⎝ (0.8)
2
⎠ σ2 σ 22
σ
A 90% lower confidence bound on σ1 is given by 0.370 ≤ 1
σ2 σ2

10-55
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

10-57 a) 90% confidence interval for the ratio of variances:

⎛ s12 ⎞ σ 12 ⎛ s12 ⎞
⎜ 2 ⎟ f1−α / 2, n1 −1, n2 −1 ≤ 2 ≤ ⎜ 2 ⎟ fα / 2, n1 −1, n2 −1
⎝ s2 ⎠ σ 2 ⎝ s2 ⎠

⎛ (0.35) ⎞ σ 12 ⎛ (0.35) ⎞ σ 12 σ
⎜ ⎟ 0.412 ≤ ≤⎜ ⎟ 2.33 0.1602 ≤ ≤ 0.9061 0.4002 ≤ 1 ≤ 0.9519
⎝ (0.90) ⎠ σ 22 ⎝ (0.90) ⎠ σ 22 σ2

b) 95% confidence interval:

⎛ s12 ⎞ σ 12 ⎛ s12 ⎞
⎜ 2 ⎟ f1−α / 2, n1 −1, n2 −1 ≤ 2 ≤ ⎜ 2 ⎟ fα / 2, n1 −1, n2 −1
⎝ s2 ⎠ σ 2 ⎝ s2 ⎠

⎛ (0.35) ⎞ σ 12 ⎛ (0.35) ⎞ σ 12 σ1
⎜ ⎟ 0.342 ≤ 2 ≤ ⎜ ⎟ 2.82 0.133 ≤ ≤ 1.097 0.3647 ≤ ≤ 1.047
⎝ (0.90) ⎠ σ 2 ⎝ (0.90) ⎠ σ 22 σ2
The 95% confidence interval is wider than the 90% confidence interval.

c) 90% lower-sided confidence interval:

⎛ s12 ⎞ σ 12
⎜ 2 ⎟ f1−α , n1 −1, n2 −1 ≤ 2
⎝ s2 ⎠ σ2

⎛ (0.35) ⎞ σ 12 σ 12 σ1
⎜ ⎟ 0.500 ≤ 2 0.194 ≤ 0.441 ≤
⎝ (0.90) ⎠ σ2 σ 22 σ2

10-58 1) The parameters of interest are the strength variances, σ 12 , σ 22

2) H0 : σ 12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f0.975,9,15 where f0.975,9,15 = 0.265 or f0 > f0.025,9,15 where f0.025,9,15 =

3.12 for α = 0.05

6) n1 = 10 n2 = 16

s1 = 15 s2 = 30

(15) 2
f0 = = 0.25
(30)2

7) Conclusion: Because 0.25 < 0.265 reject the null hypothesis. The population variances differ at the
0.05 level of significance for the two suppliers.

10-59 1) The parameters of interest are the melting variances, σ 12 , σ 22

10-56
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

2) H0 : σ 12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f0.975,20,20 where f0.975,20,20 = 0.4058 or f0 > f0.025,20,20 where

f0.025,20,20 = 2.46 for α = 0.05

6) n1 = 21 n2 = 21

s1 = 2 s2 = 1.7

(2) 2
f0 = = 1.384
(1.7) 2

7) Conclusion: Because 0.4058 < 1.384 < 2.46 fail to reject the null hypothesis. The population variances
do not differ at the 0.05 level of significance.

10-60 1) The parameters of interest are the thickness variances, σ12 , σ 22

2) H0 : σ 12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f 0.995,10,12 where f 0.995,10,12 = 0.1766 or f0 > f 0.005,10,12 where

f 0.005,10,12 = 5.0855 for α = 0.01

6) n1 = 11 n2 = 13

s1 = 0.25 s2 = 0.5

(0.25) 2
f0 = = 0.25
(0.5) 2

7) Conclusion: Because 0.1766 < 0.25 < 5.0855 fail to reject the null hypothesis. The thickness variances
are not significantly different at the 0.01 level of significance.

10-61 1) The parameters of interest are the overall distance standard deviations, σ 1 , σ 2

2) H0 : σ 12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

10-57
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f .0.975,9,9 = 0.248 or f0 > f 0.025,9,9 = 4.03 for α = 0.05

6) n1 = 10 n2 = 10 s1 = 7.61 s2 = 9.26

(7.61) 2
f0 = = 0.6754
(9.26) 2

7) Conclusion: Because 0.248 < 0.6754 < 4.04 fail to reject the null hypothesis. There is not sufficient
evidence that the standard deviations of the overall distances of the two brands differ at the 0.05 level
of significance.
95% confidence interval:

⎛ s12 ⎞ σ 12 ⎛ s12 ⎞
⎜ 2 ⎟ f1−α / 2, n1 −1, n2 −1 ≤ 2 ≤ ⎜ 2 ⎟ fα / 2, n1 −1, n2 −1
⎝ s2 ⎠ σ 2 ⎝ s2 ⎠

σ 12 σ 12
(0.6754)0.248 ≤ ≤ (0.6754)4.03 0.168 ≤ ≤ 2.723
σ 22 σ 22

σ1
A 95% lower confidence bound on the ratio of standard deviations is given by 0.41 ≤ ≤ 1.65
σ2
Because the value one is contained within this interval, there is no significant difference in the
variances of the distances at a 5% significance level.

10-62 1) The parameters of interest are the time to assemble standard deviations, σ 1 , σ 2

2) H0 : σ 12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f.0.975,11,11 = 0.288 or f0 > f 0.025,11,11 = 3.474 for α = 0.05

6) n1 = 12 n2 = 12 s1 = 0.0217 s2 = 0.0175

(0.0217) 2
f0 = = 1.538
(0.0175) 2

7) Conclusion: Because 0.288 < 1.538 < 3.474 fail to reject the null hypothesis. There is not sufficient
evidence that there is a difference in the standard deviations of the coefficients of restitution between the
two clubs at the 0.05 level of significance.
95% confidence interval:

⎛ s12 ⎞ σ 12 ⎛ s12 ⎞
⎜ 2 ⎟ f1−α / 2, n1 −1, n2 −1 ≤ 2 ≤ ⎜ 2 ⎟ fα / 2, n1 −1, n2 −1
⎝ s2 ⎠ σ 2 ⎝ s2 ⎠

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

σ 12 σ 12
(1.538)0.288 ≤ ≤ (1.538)3.474 0.443 ≤ ≤ 5.343
σ 22 σ 22

A 95% lower confidence bound the ratio of standard deviations is given by 0.666 ≤ σ 1 ≤ 2.311
σ2
Because the value one is contained within this interval, there is no significant difference in the variances
of the coefficient of restitution at a 5% significance level.

10-63 1) The parameters of interest are the variances of the weight measurements between the two sheets of

paper, σ12 , σ 22

2) H0 : σ 12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f.0.975,14,14 = 0.33 or f0 > f 0.025,14,14 = 3 for α = 0.05

6) n1 = 15 n2 = 15 s12 = 0.008312 s22 = 0.007142

f 0 = 1.35

7) Conclusion: Because 0.333 < 1.35 < 3 fail to reject the null hypothesis. There is not sufficient evidence
that there is a difference in the variances of the weight measurements between the two sheets of paper.
95% confidence interval:

⎛ s12 ⎞ σ 12 ⎛ s12 ⎞
⎜ 2 ⎟ 1−α / 2, n2 −1, n1 −1
f ≤ ≤ ⎜ ⎟ fα / 2, n2 −1, n1 −1
⎝ s2 ⎠ σ 22 ⎝ s22 ⎠

σ 12 σ 12
(1.35)0.333 ≤ ≤ (1.35)3 0.45 ≤ ≤ 4.05
σ 22 σ 22

Because the value one is contained within this interval, there is no significant difference in the variances.

10-64 a) 1) The parameters of interest are the thickness variances, σ 12 , σ 22

2) H0 : σ 12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f0.99,7,7 where f0.99,7,7 = 0.143 or f0 > f0.01,7,7 where f0.01,7,7 = 6.99 for

α = 0.02
6) n1 = 8 n2 = 8

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

s1 = 0.0028 s2 = 0.0023

(0.0028) 2
f0 = = 1.48
(0.0023) 2

7) Conclusion: Because 0.143 < 1.48 < 6.99 fail to reject the null hypothesis. The thickness variances
do not significantly differ at the 0.02 level of significance.

b) If one population standard deviation is to be 50% larger than the other, then λ = 2. Using n = 8, α = 0.01
and Chart VII (p), we obtain β ≅ 0.85. Therefore, n = n1 = n2 = 8 is not adequate to detect this difference
with high probability.

10-65 a) 1) The parameters of interest are the etch-rate variances, σ 12 , σ 22 .

2) H0 : σ 12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22

5) Reject the null hypothesis if f0 < f0.975,9 ,9 = 0.248 or f0 > f0.025,9 ,9 = 4.03 for α = 0.05

6) n1 = 10 n2 = 10

s1 = 0.011 s2 = 0.006

(0.011) 2
f0 = = 3.361
(0.006) 2
7) Conclusion: Because 0.248 < 3.361 < 4.03 fail to reject the null hypothesis. There is not sufficient
evidence that the etch rate variances differ at the 0.05 level of significance.

b) With λ = 2 = 1.4 β = 0.10 and α = 0.05, we find from Chart VII (o) that n1* = n2* = 100.

Therefore, the samples of size 10 would not be adequate.

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Section 10-6

10-66 a) This is a two-sided test because the hypotheses are p1 – p2 = 0 versus not equal to 0.

b) pˆ = 54 = 0.216 60 54 + 60
pˆ 2 = = 0.207 pˆ = = 0.2111
1
250 290 250 + 290

Test statistic is pˆ1 − pˆ 2 where pˆ = x1 + x 2

z0 = n1 + n 2
⎛1 1⎞
pˆ (1 − pˆ )⎜⎜ + ⎟⎟
⎝ n1 n2 ⎠
0.0091
z0 = = 0.2584
⎛ 1 1 ⎞
(0.2111)(1 − 0.2111)⎜ + ⎟
⎝ 250 290 ⎠
P-value = 2[1 − P(Z < 0.2584)] = 2[1 − 0.6020] = 0.796

c) The P-value is greater than α = 0.05, so we fail to reject the null hypothesis of the difference of p1 – p2
= 0 at the 0.05 level of significance.
d) 90% two-sided confidence interval on the difference:

pˆ1 (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 ) pˆ1 (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 )

( pˆ1 − pˆ 2 ) − zα / 2 + ≤ p1 − p2 ≤ ( pˆ1 − pˆ 2 ) + zα / 2 +
n1 n2 n1 n2

0.216(1 − 0.216) 0.207(1 − 0.207) 0.216(1 − 0.216) 0.207(1 − 0.207)

(0.0091) − 1.65 + ≤ p1 − p2 ≤ (0.0091) + 1.65 +
250 290 250 290
− 0.0491 ≤ p1 − p2 ≤ 0.0673

10-67 a) This is one-sided test because the hypotheses are p1 – p2 = 0 versus greater than 0.

b) pˆ = 188 = 0.752 245 188 + 245

pˆ 2 = = 0 .7 pˆ = = 0.7217
1
250 350 250 + 350

Test statistic is pˆ1 − pˆ 2 where pˆ = x1 + x 2

z0 = n1 + n 2
⎛1 1⎞
pˆ (1 − pˆ )⎜⎜ + ⎟⎟
⎝ n1 n2 ⎠
0.052
z0 = = 1.4012
⎛ 1 1 ⎞
(0.7217)(1 − 0.7217)⎜ + ⎟
⎝ 250 350 ⎠
P-value = [1 − P(Z < 1.4012)] = 1 − 0.9194 = 0.0806
95% lower confidence interval on the difference:

pˆ1 (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 )
( pˆ1 − pˆ 2 ) − zα + ≤ p1 − p2
n1 n2

0.752(1 − 0.752) 0.7(1 − 0.7)

(0.052) − 1.65 + ≤ p1 − p2
250 350
− 0.0085 ≤ p1 − p2

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

c) The P-value = 0.0806 is less than α = 0.10. Therefore, we reject the null hypothesis that p1 – p2 = 0 at
the 0.1 level of significance. If α = 0.05, the P-value = 0.0806 is greater than α = 0.05 and we fail to
reject the null hypothesis.

10-68 a) 1) The parameters of interest are the proportion of successes of surgical repairs for different tears, p1
and p2
2) H0 : p1 = p2

3) H1 : p1 > p2

4) Test statistic is pˆ1 − pˆ 2 where pˆ = x1 + x 2

z0 = n1 + n 2
⎛1 1⎞
pˆ (1 − pˆ )⎜⎜ + ⎟⎟
⎝ n1 n2 ⎠

5) Reject the null hypothesis if z0 > z0.05 where z0.05 = 1.65 for α = 0.05
6) n1 = 18 n2 = 30

x1 = 14 x2 = 22

p̂1 = 0.78 p̂2 = 0.73 14 + 22

pˆ = = 0.75
18 + 30
0.78 − 0.73
z0 = = 0.387
⎛1 1 ⎞
0.75(1 − 0.75)⎜ + ⎟
⎝ 18 30 ⎠
7) Conclusion: Because 0.387 < 1.65 we fail to reject the null hypothesis at the 0.05 level of
significance.
P-value = [1 − P(Z < 0.387)] = 1 − 0.6517 ≈ 0.35

b) 95% confidence interval on the difference:

pˆ 1 (1 − pˆ 1 ) pˆ 2 (1 − pˆ 2 )
( pˆ 1 − pˆ 2 ) − zα + ≤ p1 − p2
n1 n2

0.78(1 − 0.78) 0.73(1 − 0.73)

(0.78 − 0.73) − 1.65 + ≤ p1 − p2
18 30
− 0.159 ≤ p1 − p2
Because this interval contains the value zero, there is not enough evidence to conclude that the in the
success rate p1 exceeds p2.

10-69 a) 1) The parameters of interest are the proportion of voters in favor of Bush vs those in favor of
Kerry, p1 and p2
2) H0 : p1 = p2

3) H1 : p1 ≠ p2

10-62
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

4) Test statistic is pˆ1 − pˆ 2 where pˆ = x1 + x 2

z0 = n1 + n 2
⎛1 1⎞
pˆ (1 − pˆ )⎜⎜ + ⎟⎟
⎝ n1 n2 ⎠
5) Reject the null hypothesis if z0 < −z0.005 or z0 > z0.005 where z0.005 = 2.58 for α = 0.01
6) n1 = 2020 n2 = 2020

x1 = 1071 x2 = 930

p̂1 = 0.53 p̂2 = 0.46 1071 + 930

pˆ = = 0.495
2020 + 2020
0.53 − 0.46
z0 = = 4.45
⎛ 1 1 ⎞
0.495(1 − 0.495)⎜ + ⎟
⎝ 2020 2020 ⎠
7) Conclusion: Because 4.45 > 2.58 reject the null hypothesis and conclude yes there is a significant
difference in the proportions at the 0.05 level of significance.
P-value = 2[1 − P(Z < 4.45)] ≈ 0

b) 99% confidence interval on the difference:

pˆ1 (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 ) pˆ1 (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 )

( pˆ1 − pˆ 2 ) − zα / 2 + ≤ p1 − p2 ≤ ( pˆ1 − pˆ 2 ) + zα / 2 +
n1 n2 n1 n2

0.029 ≤ p1 − p2 ≤ 0.11

Because this interval does not contain the value zero, we are 99% confident there is a difference in
the proportions.

10-70 a) 1) The parameters of interest are the proportion of defective parts, p1 and p2
2) H0 : p1 = p2

3) H1 : p1 ≠ p2

x1 + x2
4) Test statistic is z = pˆ1 − pˆ 2 where pˆ =
0 n1 + n2
⎛1 1⎞
pˆ (1 − pˆ ) ⎜ + ⎟
⎝ n1 n2 ⎠

5) Reject the null hypothesis if z0 < −z0.025 or z0 > z0.025 where z0.025 = 1.96 for α = 0.05

6) n1 = 300 n2 = 300

x1 = 20 x2 = 10

20 + 10
p̂1 = 0.067 p̂2 = 0.033 pˆ = = 0.05
300 + 300
0.067 − 0.033
z0 = = 1.91
⎛ 1 1 ⎞
0.05(1 − 0.05) ⎜ + ⎟
⎝ 300 300 ⎠

10-63
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

7) Conclusion: Because −1.96 < 1.91 < 1.96, we fail to reject the null hypothesis. There is no
significant difference in the fraction of defective parts produced by the two machines at the 0.05
level of significance.
P-value = 2[1 − P(Z < 1.91)] = 0.05613

b) 95% confidence interval on the difference:

pˆ1 (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 ) pˆ1 (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 )

( pˆ1 − pˆ 2 ) − zα / 2 + ≤ p1 − p2 ≤ ( pˆ1 − pˆ 2 ) + zα / 2 +
n1 n2 n1 n2

0.067(1 − 0.067) 0.033(1 − 0.033) 0.067(1 − 0.067) 0.033(1 − 0.033)

(0.067 − 0.033) − 1.96 + ≤ p1 − p2 ≤ (0.067 − 0.033) + 1.96 +
300 300 300 300
−0.00077 ≤ p1 − p2 ≤ 0.06877

Because this interval contains the value zero, there is no significant difference in the fraction of
defective parts produced by the two machines. We have 95% confidence that the difference in
proportions is between −0.00077 and 0.06877.

c) Power = 1 − β
⎛ ⎛ 1 1 ⎞ ⎞ ⎛ ⎛ 1 1 ⎞ ⎞
⎜z p q ⎜⎜ + ⎟⎟ − ( p1 − p 2 ) ⎟ ⎜− z p q ⎜⎜ + ⎟⎟ − ( p1 − p 2 ) ⎟
β= ⎜ ⎟ ⎜ ⎟
α /2 α /2
⎝ n1 n 2 ⎠ ⎝ n1 n 2 ⎠
Φ⎜ ⎟ − Φ⎜ ⎟
⎜ σˆ pˆ 1 − pˆ 2 ⎟ ⎜ σˆ pˆ 1 − pˆ 2 ⎟
⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟
⎝ ⎠ ⎝ ⎠
300(0.05) + 300(0.01)
p= = 0.03 q = 0.97
300 + 300

0.05(1 − 0.05) 0.01(1 − 0.01)

σˆ pˆ − pˆ = + = 0.014
1 2
300 300

⎛ ⎛ 1 1 ⎞ ⎞ ⎛ ⎛ 1 1 ⎞ ⎞
⎜ 1.96 0.03(0.97) ⎜ + ⎟ − ( 0.05 − 0.01) ⎟ ⎜ −1.96 0.03(0.97) ⎜ + ⎟ − ( 0.05 − 0.01) ⎟
⎝ 300 300 ⎠ ⎝ 300 300 ⎠
β = Φ⎜ ⎟ −Φ⎜ ⎟
⎜ 0.014 ⎟ ⎜ 0.014 ⎟
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= Φ ( −0.91) − Φ ( −4.81) = 0.18141 − 0 = 0.18141

Power = 1 − 0.18141 = 0.81859

2
⎛
⎜ zα / 2
( p1 + p2 )( q1 + q2 ) + z ⎞
p1q1 + p2 q2 ⎟
β
⎜ 2 ⎟
d) n = ⎝ ⎠
( p1 − p2 )
2

2
⎛
⎜1.96
( 0.05 + 0.01)( 0.95 + 0.99 ) + 1.29 ⎞
0.05(0.95) + 0.01(0.99) ⎟
⎜ 2 ⎟
=⎝ ⎠ = 382.11
( 0.05 − 0.01)
2

n = 383

10-64
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

⎛ ⎛1 1⎞ ⎞ ⎛ ⎛1 1⎞ ⎞
⎜ zα / 2 pq ⎜ + ⎟ − ( p1 − p2 ) ⎟ ⎜ − zα / 2 pq ⎜ + ⎟ − ( p1 − p2 ) ⎟
⎜ ⎝ n1 n2 ⎠ ⎟ ⎜ ⎝ n1 n2 ⎠ ⎟
e) β = Φ ⎜ ⎟ −Φ⎜ ⎟
⎜ σˆ pˆ1 − pˆ 2 ⎟ ⎜ σˆ pˆ1 − pˆ 2 ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠

p = 300(0.05) + 300(0.02) = 0.035 q = 0.965

300 + 300

0.05(1 − 0.05) 0.02(1 − 0.02)

σˆ pˆ − pˆ = + = 0.015
1 2
300 300

⎛ ⎛ 1 1 ⎞ ⎞ ⎛ ⎛ 1 1 ⎞ ⎞
⎜ 1.96 0.035(0.965) ⎜ + ⎟ − ( 0.05 − 0.02 ) ⎟ ⎜ −1.96 0.035(0.965) ⎜ + ⎟ − ( 0.05 − 0.02 ) ⎟
⎝ 300 300 ⎠ ⎝ 300 300 ⎠
β = Φ ⎜⎜ ⎟ − Φ⎜
⎟ ⎜
⎟
⎟
0.015 0.015
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= Φ ( −0.04 ) − Φ ( −3.96 ) = 0.48405 − 0.00004 = 0.48401

Power = 1 − 0.48401 = 0.51599

2
⎛
⎜ zα / 2
( p1 + p2 )( q1 + q2 ) + z ⎞
p1q1 + p2 q2 ⎟
β
⎜ 2 ⎟
f) n=⎝ ⎠
( p1 − p2 )
2

2
⎛
⎜1.96
( 0.05 + 0.02 )( 0.95 + 0.98 ) + 1.29 ⎞
0.05(0.95) + 0.02(0.98) ⎟
⎜ 2 ⎟
=⎝ ⎠ = 790.67
( 0.05 − 0.02 )
2

n = 791

10-71 a) 1) The parameters of interest are the proportion of satisfactory lenses, p1 and p2

2) H0 : p1 = p2

3) H1 : p1 ≠ p2

4) Test statistic is
x1 + x2
pˆ1 − pˆ 2 where pˆ =
z0 = n1 + n2
⎛1 1⎞
pˆ (1 − pˆ )⎜⎜ + ⎟⎟
⎝ n1 n2 ⎠

5) Reject the null hypothesis if z0 < − z 0.005 or z0 > z 0.005 where z 0.005 = 2.58 for α = 0.01
6) n1 = 300 n2 = 300

x1 = 253 x2 = 196

253 + 196
p̂1 = 0.843 p̂2 = 0.653 pˆ = = 0.748
300 + 300

10-65
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

0.843 − 0.653
z0 = = 5.36
⎛ 1 1 ⎞
0.748(1 − 0.748)⎜ + ⎟
⎝ 300 300 ⎠
7) Conclusion: Because 5.36 > 2.58 reject the null hypothesis and conclude yes there is a significant
difference in the fraction of polishing-induced defects produced by the two polishing solutions at the
0.01 level of significance.
P-value = 2[1 − P(Z < 5.36)] ≈ 0

b) By constructing a 99% confidence interval on the difference in proportions, the same question can
be answered by whether or not zero is contained in the interval.

10-72 a) 1) The parameters of interest are the proportion of residents in favor of an increase, p1 and p2
2) H0 : p1 = p2

3) H1 : p1 ≠ p2

4) Test statistic is pˆ1 − pˆ 2 where pˆ = x1 + x 2

z0 = n1 + n 2
⎛1 1⎞
pˆ (1 − pˆ )⎜⎜ + ⎟⎟
⎝ n1 n2 ⎠
5) Reject the null hypothesis if z0 < −z0.025 or z0 > z0.025 where z0.025 = 1.96 for α = 0.05

6) n1 = 500 n2 = 400

x1 = 385 x2 = 267

p̂1 = 0.77 p̂2 = 0.6675 385 + 267

pˆ = = 0.724
500 + 400
0.77 − 0.6675
z0 = = 3.42
⎛ 1 1 ⎞
0.724(1 − 0.724)⎜ + ⎟
⎝ 500 400 ⎠
7) Conclusion: Because 3.42 > 1.96 reject the null hypothesis and conclude yes there is a significant
difference in the proportions of support for increasing the speed limit between residents of the two
counties at the 0.05 level of significance.
P-value = 2[1 − P(Z < 3.42)] = 0.00062

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

b) 95% confidence interval on the difference:

pˆ1 (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 ) pˆ1 (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 )

( pˆ1 − pˆ 2 ) − zα / 2 + ≤ p1 − p2 ≤ ( pˆ1 − pˆ 2 ) + zα / 2 +
n1 n2 n1 n2

0.77(1 − 0.77) 0.6675(1 − 0.6675) 0.77(1 − 0.77) 0.6675(1 − 0.6675)

(0.77 − 0.6675) − 1.96 + ≤ p1 − p2 ≤ (0.77 − 0.6675) + 1.96 +
500 400 500 400
0.0434 ≤ p1 − p2 ≤ 0.1616

We are 95% confident that the difference in proportions is between 0.0434 and 0.1616. Because the
interval does not contain zero there is evidence that the counties differ in support of the change.

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Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Supplemental Exercises

10-73 a) SE Mean1 = s1 = 2.23 = 0.50

n1 20

x1 = 11.87 x2 = 12.73 s12 = 2.232 s22 = 3.19

2
n1 = 20 n2 = 20

Degrees of freedom = n1 + n2 − 2 = 20 – 20 − 2 = 38.

(n1 − 1) s12 + (n2 − 1) s22 (20 − 1)2.232 + (20 − 1)3.192

sp = = = 2.7522
n1 + n2 − 2 20 + 20 − 2

( x1 − x2 ) − Δ 0 (−0.86)
t0 = = = −0.9881
1 1 1 1
sp + 2.7522 +
n1 n2 20 20

P-value = 2 [P(t < −0.9881)] and 2(0.10) <P-value < 2(0.25) = 0.20 <P-value < 0.5

The 95% two-sided confidence interval: tα / 2,n + n −2 = t 0.025,38 = 2.024

1 2

(x1 − x2 ) − tα / 2,n + n −2 s p 1 1
+ ≤ μ1 − μ 2 ≤ (x1 − x2 ) + t α / 2 , n1 + n2 − 2 s p
1 1
+
1 2
n1 n2 n1 n2

1 1 1 1
( −0.86 ) − (2.024)(2.7522) + ≤ μ1 − μ2 ≤ ( −0.86 ) + (2.024)(2.7522) +
20 20 20 20
− 2.622 ≤ μ1 − μ 2 ≤ 0.902

b) This is two-sided test because the alternative hypothesis is μ1 – μ2 not = 0.

c) Because the 0.20 < P-value < 0.5 and the P-value > α = 0.05 we fail to reject the null hypothesis at
the 0.05 level of significance. If α = 0.01, we also fail to reject the null hypothesis.

10-74 a) This is one-sided test because the alternative hypothesis is μ1 – μ2 < 0.

b)
2 2
⎛ s12 s22 ⎞ ⎛ 2.982 5.362 ⎞
⎜ + ⎟ ⎜ + ⎟
ν= ⎝ n1 n2 ⎠ = ⎝ 16 25 ⎠
= 38.44 ≈ 38 (truncated)
( ) ( )
2 2 2 2
⎛ s12 ⎞ ⎛ s22 ⎞ 2.982 5.362
⎜ n⎟ ⎜ n ⎟ 16 25
⎝ 1⎠
+⎝
2⎠ +
n1 − 1 n2 − 1 16 − 1 25 − 1

Degree of freedom = 38.

P-value = P(t < −1.65) and 0.05 < P-value < 0.1

10-68
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

c) Because 0.05 < P-value < 0.1 and the P-value > α = 0.05 we fail to reject the null hypothesis of μ1 –
μ2 = 0 at the 0.05 level of significance. If α = 0.1, we reject the null hypothesis because the P-value
< 0.1.

d) The 95% upper one-sided confidence interval: t0.05,38 = 1.686

s12 s22
μ1 − μ2 ≤ ( x1 − x2 ) + tα ,ν +
n1 n2

μ1 − μ2 ≤ (− 2.16) + 1.686
(2.98)2 + (5.36)2
16 25
μ1 − μ2 ≤ 0.0410

10-75 a) Assumptions that must be met are normality, equality of variance, and independence of the
observations. Normality and equality of variances appear to be reasonable from the normal probability
plots. The data appear to fall along straight lines and the slopes appear to be the same. Independence of
the observations for each sample is obtained if random samples are selected.
.
Normal Probability Plot Normal Probability Plot

.999 .999
.99 .99
.95 .95
Probability

Probability

.80 .80
.50 .50
.20 .20
.05 .05
.01 .01
.001 .001

14 15 16 17 18 19 20 8 9 10 11 12 13 14 15
9-hour 1-hour
Average: 16.3556 Anderson-Darling Normality Test Average: 11.4833 Anderson-Darling Normality Test
StDev: 2.06949 A-Squared: 0.171 StDev: 2.37016 A-Squared: 0.158
N: 9 P-Value: 0.899 N: 6 P-Value: 0.903

b) x1 = 16.36 x2 = 11.483

s1 = 2.07 s2 = 2.37

n1 = 9 n2 = 6

99% confidence interval: t α / 2, n1 + n 2 − 2 = t 0.005,13 where t 0.005,13 = 3.012

8(2.07) 2 + 5(2.37) 2
sp = = 2.19
13

( x1 − x2 ) − tα / 2,n + n − 2 ( s p ) ≤ μ1 − μ 2 ≤ ( x1 − x2 ) + tα / 2, n1 + n2 − 2 ( s p )
1 1 1 1
1 2
+ +
n1 n2 n1 n2

1 1 1 1
(16.36 − 11.483) − 3.012 ( 2.19 ) + ≤ μ1 − μ2 ≤ (16.36 − 11.483) + 3.012 ( 2.19 ) +
9 6 9 6
1.40 ≤ μ1 − μ 2 ≤ 8.36

c) Yes, we are 99% confident the results from the first test condition exceed the results of the second
test condition because the confidence interval contains only positive values.

10-69
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

d) 95% confidence interval for σ 12 / σ 22

σ 12
95% confidence interval on :
σ 22
1 1
f 0.975,8,5 = = = 0.2075 , f 0.025,8,5 = 6.76
f 0.025,5,8 4.82

s12 σ 12 s12
f ≤ ≤ f 0.025,8,5
σ 22 s22
0.975,8,5
s22
⎛ 4.283 ⎞ σ 12 ⎛ 4.283 ⎞
⎜ ⎟ ≤ ≤
σ 22 ⎜⎝ 5.617 ⎟⎠
(0.2075) (6.76)
⎝ 5.617 ⎠
σ 12
0.1582 ≤ ≤ 5.157
σ 22

e) Because the value one is contained within this interval, the population variances do not differ at a
5% significance level.

10-76 a) Assumptions that must be met are normality and independence of the observations. Normality appears
to be reasonable from the normal probability plots. The data appear to fall along straight lines. Because
the slopes appear to be the same, it appears the population standard deviations are similar (so that we
expect that we will fail to reject H0). Independence of the observations for each sample is obtained if
random samples are selected.

Normal Probability Plot Normal Probability Plot

.999 .999
.99 .99
.95 .95
Probability

Probability

.80 .80
.50 .50
.20 .20
.05 .05
.01 .01
.001 .001

97 98 99 100 101 102 103 104 100 101 102 103 104 105 106 107 108 109
vendor 1 vendor 2
Average: 99.576 Anderson-Darling Normality Test Average: 105.069 Anderson-Darling Normality Test
StDev: 1.52896 A-Squared: 0.315 StDev: 1.96256 A-Squared: 0.376
N: 25 P-Value: 0.522 N: 35 P-Value: 0.394

b) 1) The parameters of interest are the variances of resistance of products, σ 12 , σ 22

2) H0 : σ 12 = σ 22

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22

10-70
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

1 1
5) Reject H0 if f0 < f0.975,24 ,34 where f0.975,24,34 = = = 0.459 for α = 0.05
f 0.025,34, 24 2.18

or f0 > f0.025,24,34 where f0.025,24,34 = 2.07 for α = 0.05

6) s1 = 1.53 s2 = 1.96
n1 = 25 n2 = 35
(1.53) 2
f0 = = 0.609
(1.96) 2

7) Conclusion: Because 0.459 < 0.609 < 2.07, fail to reject H0. There is not sufficient evidence to
conclude that the variances are different at α = 0.05.

10-77 a) 1) The parameter of interest is the mean weight loss, μd where di = Initial Weight − Final Weight.
2) H0 : μ d = 1.5

3) H1 : μ d > 1.5

4) The test statistic is

d − Δ0
t0 =
sd / n

5) Reject H0 if t0 > tα,n-1 where t0.05,7 = 1.895 for α = 0.05.

6) d = 1.875
sd = 0.641
n=8
1.875 − 1.5
t0 = = 1.655
0.641/ 8
7) Conclusion: Because 1.655 > 1.895, fail to reject the null hypothesis and conclude the average
weight loss is significantly less than 1.5 at α = 0.05.

b) 2) H0 : μ d = 1.5

3) H1 : μ d > 1.5

4) The test statistic is

d − Δ0
t0 =
sd / n

5) Reject H0 if t0 > tα,n-1 where t0.01,7 = 2.998 for α = 0.01.

6) d = 1.875
sd = 0.641
n=8

10-71
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

1.875 − 1.5
t0 = = 1.655
0.641/ 8
7) Conclusion: Because 1.655 < 2.998, fail to reject the null hypothesis. The average weight loss is
not significantly greater than 1.5 at α = 0.01.

c) 2) H0 : μd = 2.2

3) H1 : μd > 2.2

4) The test statistic is

d − Δ0
t0 =
sd / n

5) Reject H0 if t0 > tα,n-1 where t0.05,7 = 1.895 for α = 0.05

6) d = 1.875
sd = 0.641
n=8
1.875 − 2.2
t0 = = −1.434
0.641/ 8
7) Conclusion: Because −1.434 < 1.895, fail to reject the null hypothesis and conclude that the average
weight loss is not significantly greater than 2.2 at α = 0.05.

d) 2) H0 : μd = 2.2

3) H1 : μd > 2.2

4) The test statistic is

d − Δ0
t0 =
sd / n

5) Reject H0 if t0 > tα,n-1 where t0.01,7 = 2.998 for α = 0.01.

6) d = 1.875
sd = 0.641
n=8
1.875 − 2.2
t0 = = −1.434
0.641/ 8
7) Conclusion: Because −1.434 < 2.998, fail to reject the null hypothesis and conclude the average
weight loss is not significantly greater than 2.2 at α = 0.01.

(x1 − x2 ) − zα / 2 σ 1 + σ 2 σ 12 σ 22
2 2
10-78 ≤ μ1 − μ 2 ≤ (x1 − x2 ) + zα / 2 +
n1 n2 n1 n2

a) 90% confidence interval: zα / 2 = 1.65

10-72
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

352 302 352 302

( 600 − 625) − 1.65 + ≤ μ1 − μ2 ≤ ( 600 − 625 ) + 1.65 +
20 20 20 20
−42.01 ≤ μ1 − μ 2 ≤ −7.99

Yes, the data indicate that the mean breaking strength of the yarn of manufacturer 2 exceeds that of
manufacturer 1 by between 42.01 and 7.99 with 90% confidence.

b) 98% confidence interval: zα / 2 = 2.33

352 302 352 302

( 600 − 625) − 2.33 + ≤ μ1 − μ 2 ≤ ( 600 − 625 ) + 2.33 +
20 20 20 20
−49.02 ≤ μ1 − μ2 ≤ 0.98

Yes, we can again conclude that yarn of manufacturer 2 has greater mean breaking strength than that of
manufacturer 1 by between 49.02 and 0.98 with 98% confidence.

c) The results of parts (a) and (b) are same although the confidence level used is different. The
appropriate interval depends upon the level of confidence considered acceptable.

10-79 a) 1) The parameters of interest are the proportions of children who contract polio, p1, p2
2) H0 : p1 = p2
3) H1 : p1 ≠ p2
4) The test statistic is
pˆ1 − pˆ 2
z0 =
⎛1 1⎞
pˆ (1 − pˆ )⎜⎜ + ⎟⎟
⎝ n1 n2 ⎠
5) Reject H0 if z0 < −zα /2 or z0 > zα /2 where zα /2 = 1.96 for α = 0.05

x1 110 x1 + x2
6) pˆ1 = = = 0.00055 (Placebo) pˆ = = 0.000356
n1 201299 n1 + n2

x2 33
pˆ 2 = = = 0.00016 (Vaccine)
n2 200745

0.00055 − 0.00016
z0 = = 6.55
⎛ 1 1 ⎞
0.000356(1 − 0.000356) ⎜ + ⎟
⎝ 201299 200745 ⎠
7) Because 6.55 > 1.96 reject H0 and conclude the proportion of children who contracted polio is
significantly different at α = 0.05.

b) α = 0.01
Reject H0 if z0 < −zα /2 or z0 > zα /2 where zα /2 = 2.58. Still z0 = 6.55.

Because 6.55 > 2.58, reject H0 and conclude the proportion of children who contracted polio is
significantly different at α = 0.01.

10-73
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

c) The conclusions are the same because z0 is so large it exceeds zα/2 in both cases.

10-80 a) α = 0.10 zα / 2 = 1.65

( zα / 2 ) 2 (σ12 + σ 22 ) (1.65) 2 (1225 + 900)

n≅ ≅ = 57.85, n = 58
( E )2 (10) 2

b) α = 0.02 zα / 2 = 2.33

( zα / 2 ) 2 (σ12 + σ 22 ) ( 2.33) 2 (1225 + 900)

n≅ ≅ = 115.36, n = 116
( E )2 (10)2
c) As the confidence level increases, sample size also increases.
d) α = 0.10 zα / 2 = 1.65

( zα / 2 ) 2 (σ12 + σ 22 ) (1.65) 2 (1225 + 900)

n≅ ≅ = 231.41, n = 232
( E )2 (5) 2

α = 0.02 zα / 2 = 2.33

( zα / 2 ) 2 (σ12 + σ 22 ) ( 2.33) 2 (1225 + 900)

n≅ ≅ = 461.5, n = 462
( E )2 (5) 2
e) As the error decreases, the required sample size increases.

x1 387 x2 310
10-81 pˆ1 = = = 0.258 pˆ 2 = = = 0.2583
n1 1500 n2 1200

pˆ1 (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 )
( pˆ1 − pˆ 2 ) ± zα / 2 +
n1 n2

a) zα / 2 = z0.025 = 1.96

0.258(0.742) 0.2583(0.7417)
( 0.258 − 0.2583) ± 1.96 +
1500 1200
− 0.0335 ≤ p1 − p2 ≤ 0.0329
Because zero is contained in this interval, there is no significant difference between the proportions of
unlisted numbers in the two cities at a 5% significance level.

b) zα / 2 = z0.05 = 1.65

0.258(0.742) 0.2583(0.7417)
( 0.258 − 0.2583) ± 1.65 +
1500 1200
−0.0282 ≤ p1 − p2 ≤ 0.0276

The proportions of unlisted numbers in the two cities do not significantly differ at a 5% significance
level.

x1 774
c) pˆ1 = = = 0.258
n1 3000

10-74
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

x2 620
p 2 = = = 0.2583
n2 2400

95% confidence interval:

0.258(0.742) 0.2583(0.7417)
( 0.258 − 0.2583) ± 1.96 + −0.0238 ≤ p1 − p2 ≤ 0.0232
3000 2400
90% confidence interval:

0.258(0.742) 0.2583(0.7417)
( 0.258 − 0.2583) ± 1.65 + −0.0201 ≤ p1 − p2 ≤ 0.0195
3000 2400
Increasing the sample size decreased the width of the confidence intervals, but did not change the
conclusions drawn. The conclusion remains that there is no significant difference.

10-82 a) 1) The parameters of interest are the proportions of those residents who wear a seat belt regularly, p1,
p2
2) H0 : p1 = p2
3) H1 : p1 ≠ p2
4) The test statistic is
pˆ1 − pˆ 2
z0 =
⎛1 1⎞
pˆ (1 − pˆ ) ⎜ + ⎟
⎝ n1 n2 ⎠
5) Reject H0 if z0 < −zα /2 or z0 > zα / 2 where z0.025 = 1.96 for α = 0.05

x1 165 x +x
6) pˆ1 = = = 0.825 pˆ = 1 2 = 0.807
n1 200 n1 + n2

x2 198
pˆ 2 = = = 0.792
n2 250

0.825 − 0.792
z0 = = 0.8814
⎛ 1 1 ⎞
0.807(1 − 0.807) ⎜ + ⎟
⎝ 200 250 ⎠

7) Conclusion: Because −1.96 < 0.8814 < 1.96 fail to reject H0. There is not sufficient evidence that
there is a difference in seat belt usage at α = 0.05.

b) α = 0.10
Reject H0 if z0 < −zα /2 or z0 > zα / 2 where z0.05 = 1.65 z0 = 0.8814

Because −1.65 < 0.8814 < 1.65 fail to reject H0. There is not sufficient evidence that there is a
difference in seat belt usage at α = 0.10.

c) The conclusions are the same, but with different levels of significance.
d) n1 = 400, n2 = 500

α = 0.05

10-75
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Reject H0 if z0 < −zα /2 or z0 > zα /2 where z0.025 = 1.96

0.825 − 0.792
z0 = = 1.246
⎛ 1 1 ⎞
0.807(1 − 0.807) ⎜ + ⎟
⎝ 400 500 ⎠

Because −1.96 < 1.246 < 1.96 fail to reject H0. There is not sufficient evidence that there is a
difference in seat belt usage at α = 0.05.
α = 0.10
Reject H0 if z0 < −zα /2 or z0 > zα /2 where z0.05 = 1.65 z0 = 1.246

Because −1.65 < 1.246 < 1.65 fail to reject H0. There is not sufficient evidence that there is a
difference in seat belt usage at α = 0.10.

As the sample size increased, the test statistic also increased (because the denominator of z0
decreased). However, the sample size increase was not enough to change our conclusion.

10-83 a) Yes, there could be some bias in the results due to the telephone survey.
b) If it could be shown that these populations are similar to the respondents, the results may be
extended.

10-84 The parameter of interest is μ1 − 2μ2

H 0 : μ1 = 2μ 2 H 0 : μ1 − 2μ 2 = 0
→
H1 : μ1 > 2μ 2 H1 : μ1 − 2 μ2 > 0

Let n1 = size of sample 1 X1 estimate for μ1

Let n2 = size of sample 2 X2 estimate for μ2

X1 − 2 X2 is an estimate for μ1 − 2μ 2

σ12 4σ 22
The variance is V( X1 − 2 X2 ) = V( X1 ) + V(2 X2 ) = +
n1 n2

The test statistic for this hypothesis is:

( X1 − 2 X 2 ) − 0
Z0 =
σ 12 4σ 22
+
n1 n2

We reject the null hypothesis if z0 > zα/2 for a given level of significance. P-value = P(Z ≥ z0 ).

10-85 x1 = 910 x2 = 905

σ1 = 3 σ 2 = 4.5
n1 = 12 n2 = 10

a) 90% two-sided confidence interval:

10-76
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

σ 12 σ 22 σ 12 σ 22
( x1 − x2 ) − zα / 2 + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + zα / 2 +
n1 n2 n1 n2

32 4.52 32 4.52
(910 − 905) − 1.645 + ≤ μ1 − μ 2 ≤ (910 − 905) + 1.645 +
12 10 12 10
2.259 ≤ μ1 − μ 2 ≤ 7.741

We are 90% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between
2.259 and 7.741 ml.

b) 95% two-sided confidence interval:

σ 12 σ 22 σ 12 σ 22
( x1 − x2 ) − zα / 2 + ≤ μ1 − μ2 ≤ ( x1 − x2 ) + zα / 2 +
n1 n2 n1 n2

32 4.52 32 4.52
(910 − 905) − 1.96 + ≤ μ1 − μ2 ≤ (910 − 905) + 1.96 +
12 10 12 10
1.735 ≤ μ1 − μ 2 ≤ 8.265

We are 95% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between
1.735 and 8.265 ml.
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:

σ 12 σ 22
μ1 − μ2 ≤ ( x1 − x2 ) + zα +
n1 n2

32 4.52
μ1 − μ2 ≤ (910 − 905) + 1.645 +
12 10
μ1 − μ2 ≤ 7.741
With 95% confidence, the fill volume for machine 1 exceeds the fill volume of machine 2 by no
more than 7.741 ml.

d) 1) The parameter of interest is the difference in mean fill volume μ1 − μ2

2) H0 : μ1 − μ 2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ 2 ≠ 0 or μ1 ≠ μ2

4) The test statistic is

( x1 − x2 ) − Δ 0
z0 =
σ 12 σ 22
+
n1 n2

5) Reject H0 if z0 < −zα/2 = −1.96 or z0 > zα/2 = 1.96 for α = 0.05

6) x1 = 910 x2 = 905

σ1 = 3 σ 2 = 4.5

10-77
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

n1 = 12 n2 = 10
(910 − 905)
z0 = =3
32 4.52
+
12 10
7) Because 3 > 1.96 reject the null hypothesis and conclude the mean fill volumes of machine 1 and
machine 2 differ significantly at α = 0.05.
P-value = 2[1 − Φ(3)] = 2(1 − 0.998650) = 0.0027

e) Assume the sample sizes are to be equal, use α = 0.05, β = 0.10, and Δ = 5

( zα + zβ ) (σ 12 + σ 22 ) (1.96 + 1.28) (3 + 4.52 )

2 2 2
/2
n≅ = = 12.3, n = 12, use n1 = n2 = 12
( Δ − Δ0 ) (−5) 2
2

10-86 H0 : μ1 = μ 2

H1 : μ1 ≠ μ2

n1 = n 2 = n
β = 0.10
α = 0.05
Assume normal distribution and σ 12 = σ 22 = σ 2

μ1 = μ2 + σ
| μ − μ2 | σ 1
d= 1 = =
2σ 2σ 2

n∗ + 1 50 + 1
From Chart VII (e) n∗ = 50 and n = = = 25.5 and n1 = n2 = 26
2 2

10-87 a) 1) The parameters of interest are: the proportion of lenses that are unsatisfactory after tumble-polishing,
p1, p2
2) H0 : p1 = p2
3) H1 : p1 ≠ p2
4) The test statistic is
pˆ1 − pˆ 2
z0 =
⎛1 1⎞
pˆ (1 − pˆ ) ⎜ + ⎟
⎝ n1 n2 ⎠
5) Reject H0 if z0 < − zα /2 or z0 > zα / 2 where zα /2 = 2.58 for α = 0.01.

6) x1 = number of defective lenses

x1 47 x1 + x2
pˆ1 = = = 0.1567 pˆ = = 0.2517
n1 300 n1 + n2

10-78
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

x2 104
pˆ 2 = = = 0.3467
n2 300

0.1567 − 0.3467
z0 = = −5.36
⎛ 1 1 ⎞
0.2517(1 − 0.2517) ⎜ + ⎟
⎝ 300 300 ⎠

7) Conclusion: Because −5.36 < −2.58 reject H0 and conclude there is strong evidence to support the
claim that the two polishing fluids are different.

b) The conclusions are the same whether we analyze the data using the proportion unsatisfactory or
proportion satisfactory.
2
⎛ (0.9 + 0.6)(0.1 + 0.4) ⎞
⎜⎜ 2.575 + 1.28 0.9(0.1) + 0.6(0.4) ⎟⎟
2
c) n = ⎝ ⎠
(0.9 − 0.6) 2

5.346
= = 59.4
0.09
n = 60

10-88 a) α = 0.05, β = 0.05, Δ= 1.5. Use sp = 0.7071 to approximate σ in equation 10-19.

Δ 1.5
d= = = 1.06 ≅ 1
2( s p ) 2(.7071)

n∗ + 1 20 + 1
From Chart VII (e), n∗ = 20 n = = = 10.5; n = 11 is needed to reject the null hypothesis
2 2
that the two agents differ by 0.5 with probability of at least 0.95.

b) The original size of n = 5 in Exercise 10-18 was not appropriate to detect the difference because a
sample size of 11 is needed to reject the null hypothesis that the two agents differ by 1.5 with probability
of at least 0.95.

10-89 a) No

10-79
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

b) The normal probability plots indicate that the data follow normal distributions because the data appear
to fall along a straight line. The plots also indicate that the variances appear to be equal because the
slopes appear to be the same.

c) By correcting the data points, it is more apparent the data follow normal distributions. Note that one
unusual observation can cause an analyst to reject the normality assumption.
d) 95% confidence interval on the ratio of the variances, σ 2V /σ M
2

10-80
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

sV2 = 0.27 f9,9,0.025 = 4.03

1 1
sM2 = 0.0037 f 9,9,0.975 = = = 0.248
f 9,9,0.025 4.03

⎛ sV2 ⎞ σ V2 ⎛ sV2 ⎞
⎜ 2 ⎟ f 9,9,0.975 < 2 < ⎜ 2 ⎟ f 9,9,0.025
⎝ sM ⎠ σ M ⎝ sM ⎠
⎛ 0.27 ⎞ σ V2 ⎛ 0.27 ⎞
⎜ ⎟ 0.248 < 2 < ⎜
σ M ⎝ 0.0037 ⎟⎠
4.03
⎝ 0.0037 ⎠
σ V2
18.097 < < 294.08
σ M2
Because the interval does not include the value one, we reject the hypothesis that variability in mileage
performance is the same for the two types of vehicles. There is evidence that the variability is greater
for a Volkswagen than for a Mercedes.

e) 1) The parameters of interest are the variances in mileage performance, σ 12 , σ 22

2) H0 : σ 12 = σ 22 Where Volkswagen is represented by variance 1, Mercedes by variance 2.

3) H1 : σ 12 ≠ σ 22

4) The test statistic is

s12
f0 =
s22
1 1
5) Reject H0 if f0 < f0.975,9 ,9 where f0.975,9,9 = = = 0.248 for α = 0.05 or f0 > f0.025,9 ,9 where
f0.025,9,9 4.03

f0.025,9 ,9 = 4.03 for α = 0.05

6) s1 = 0.5226 s2 = 0.061
n1 = 10 n2 = 10
(0.5226) 2
f0 = = 73.4
(0.061)2

7) Conclusion: Because 72.78 > 4.03 reject H0 and conclude that there is a significant difference
between Volkswagen and Mercedes in terms of mileage variability. The same conclusions are
reached in part (d).

10-90 a) Underlying distributions appear to be normally distributed because the data fall along a straight line
on the normal probability plots. The slopes appear to be similar so it is reasonable to assume that
σ 12 = σ 22 .

10-81
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Normal Probability Plot Normal Probability Plot

.999 .999
.99 .99
.95 .95

Probability

Probability
.80 .80
.50 .50
.20 .20
.05 .05
.01 .01
.001 .001

751 752 753 754 755 754 755 756 757

ridgecre valleyvi
Average: 752.7 Anderson-Darling Normality Test Average: 755.6 Anderson-Darling Normality Test
StDev: 1.25167 A-Squared: 0.384 StDev: 0.843274 A-Squared: 0.682
N: 10 P-Value: 0.323 N: 10 P-Value: 0.051

b) 1) The parameter of interest is the difference in mean volumes, μ1 − μ2

2) H0 : μ1 − μ2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ2 ≠ 0 or μ1 ≠ μ2

4) The test statistic is

( x1 − x2 ) − δ
t0 =
1 1
sp +
n1 n2

5) Reject H0 if t0 < − t α / 2, ν or z0 > t α / 2 , ν where t α / 2 , ν = t 0.025,18 = 2.101 for α = 0.05

9(1.252) 2 + 9(0.843) 2
6) x1 = 752.7 x2 = 755.6 sp = = 1.07
18
s1 = 1.252 s2 = 0.843

n1 = 10 n2 = 10

(752.7 − 755.6)
t0 = = −6.06
1 1
1.07 +
10 10
7) Conclusion: Because −6.06 < −2.101 reject H0 and conclude there is a significant difference between
the two wineries with respect to mean fill volumes at a 5% significance level.

c) From Section 10-3.3, d = 2/2(1.07) = 0.93, giving a power of just under 80%. Because the power is
relatively low, an increase in the sample size would improve the power of the test.

10-91 a) The assumption of normality appears to be reasonable. This is evident by the fact that the data lie along
a straight line in the normal probability plot.

10-82
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Normal Probability Plot

.999
.99
.95

Probability
.80
.50
.20
.05
.01
.001

-2 -1 0 1 2
diff
Average: -0.222222 Anderson-Darling Normality Test
StDev: 1.30171 A-Squared: 0.526
N: 9 P-Value: 0.128

b) 1) The parameter of interest is the mean difference in tip hardness, μd

2) H0 : μd = 0

3) H1 : μd ≠ 0

4) The test statistic is

d
t0 =
sd / n

5) Since no significance level is given, we will calculate P-value. Reject H0 if the P-value is
significantly small.

6) d = −0.222
sd = 1.30
n=9
−0.222
t0 = = −0.512
1.30 / 9
P-value = 2P(T < −0.512) = 2P(T > 0.512) and 2(0.25) < P-value < 2(0.40). Thus, 0.50 < P-value <
0.80

7) Conclusion: Because the P-value is larger than common levels of significance, fail to reject H0 and
conclude there is no significant difference in mean tip hardness.

c) β = 0.10
μd = 1
1 1
d= = = 0.769
σd 1.3

From Chart VII (f) with α = 0.01, n = 30

10-92 a) From the normal probability plot the data fall along a line and consequently they appear to follow a
normal distribution.

10-83
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

b) 1) The parameter of interest is the mean difference in depth using the two gauges, μd
2) H0 : μd = 0

3) H1 : μd ≠ 0

4) The test statistic is

d
t0 =
sd / n

5) Since no significance level is given, we will calculate P-value. Reject H0 if the P-value is
significantly small.

6) d = 0.2
sd = 5.401
n = 15
0.2
t0 = = 0.14
5.401/ 15
P-value = 2P(T > 0.14), 2(0.44) < P-value, 0.88 < P-value
7) Conclusion: Because the P-value is larger than common levels of significance, fail to reject H0 and
conclude there is no significant difference in mean depth measurements for the two gauges.

c) Power = 0.8. Because Power = 1 – β , β = 0.20

μd = 4.2
4.2 4.2
d= = = 0.778
σd (5.401)

From Chart VII (f) with α = 0.01 and β = 0.20, we find n = 30.

10-93 a) The data from both depths appear to be normally distributed, but the slopes do not appear to be equal.
Therefore, it is not reasonable to assume that σ 12 = σ 22 .

10-84
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Normal Probability Plot for surface...bottom

ML Estimates

surface
99
bottom

95
90

80
70

Percent
60
50
40
30
20

10
5

4 5 6 7 8
Data

b) 1) The parameter of interest is the difference in mean HCB concentration, μ1 − μ2 , with Δ0 = 0

2) H0 : μ1 − μ2 = 0 or μ1 = μ 2

3) H1 : μ1 − μ2 ≠ 0 or μ1 ≠ μ2

4) The test statistic is

( x1 − x2 ) − Δ 0
t0 =
s12 s22
+
n1 n2

5) Reject the null hypothesis if t0 < − t 0.025,15 or t0 > t 0.025,15 where t 0.025,15 = 2.131 for α = 0.05. Also

2
⎛ s12 s22 ⎞
⎜ + ⎟
ν = ⎝ 1 2 2 ⎠ = 15.06
n n
⎛ s1 ⎞
2
⎛ s22 ⎞
⎜ ⎟ ⎜ ⎟
⎝ n1 ⎠ + ⎝ n2 ⎠
n1 − 1 n2 − 1
ν ≅ 15
(truncated)
6) x1 = 4.804 x2 = 5.839 s1 = 0.631 s2 = 1.014

n1 = 10 n2 = 10
(4.804 − 5.839)
t0 = = −2.74
(0.631) 2 (1.014) 2
+
10 10

7) Conclusion: Because –2.74 < –2.131 reject the null hypothesis. Conclude that the mean HCB
concentration is different at the two depths sampled at the 0.05 level of significance.

10-85
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

c) Assume the variances are equal. Then Δ = 2, α = 0.05, n = n1 = n2 = 10, n* = 2n – 1 = 19, sp = 0.84 and
2
d= = 1.2 . From Chart VII (e) we find β ≈ 0.05, and then calculate power = 1 – β = 0.95
2(0.84)

d) Assume the variances are equal. Then Δ = 1, α = 0.05, n = n1 = n2 , n* = 2n – 1, β = 0.1, sp = 0.84 ≈ 1

1 50 + 1
and d = = 0.6 . From Chart VII (e) we find n* = 50 and n = = 25.5 , so n = 26.
2(0.84) 2

10-86
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Mind-Expanding Exercises

10-94 The estimate of μ is given by X . Therefore X =

1
2
( X1 + X 2 ) − X 3 . The variance of X can be shown
to be V ( X ) = 1 ⎛⎜ σ 1 + σ 2 ⎞⎟ + σ 3 . Use s1, s2, and s3 as estimates for σ1 σ2 and σ3 respectively.
2 2 2

4 ⎜⎝ n1 n2 ⎟⎠ n3

a) A 100(1 – α)% confidence interval on μ is then:

1 ⎛ s12 s22 ⎞ s32 1 ⎛ s12 s22 ⎞ s32

X − Zα / 2 ⎜ + ⎟+ ≤ μ ≤ X + Zα / 2 ⎜ + ⎟+
4 ⎝ n1 n2 ⎠ n3 4 ⎝ n1 n2 ⎠ n3

b) A 95% confidence interval for μ is

⎛1 ⎞ 1 ⎛ 0.7 2 0.62 ⎞ 0.82 ⎛1 ⎞ 1 ⎛ 0.7 2 0.62 ⎞ 0.82

⎜ ( 4.6 + 5.2 ) − 6.1⎟ − 1.96 ⎜ + ⎟+ ≤ μ ≤ ⎜ ( 4.6 + 5.2 ) − 6.1⎟ + 1.96 ⎜ + ⎟+
⎝2 ⎠ 4 ⎝ 100 120 ⎠ 130 ⎝2 ⎠ 4 ⎝ 100 120 ⎠ 130
−1.2 − 0.163 ≤ μ ≤ −1.2 + 0.163
−1.363 ≤ μ ≤ −1.037

Because zero is not contained in this interval, and because the possible mean difference (–1.363, –1.037)
is negative, we can conclude that there is sufficient evidence to indicate that pesticide three is more
effective.

The V ( X − X ) = σ 1 + σ 2 and suppose this is to equal a constant k. Then, we are to minimize

2 2
10-95
1 2
n1 n 2

σ 12 σ 22
C1 n1 + C 2 n 2 subject to + = k . Using a Lagrange multiplier, we minimize by setting the partial
n1 n2

⎛ ⎞
derivatives of f (n1 , n2 , λ ) = C1n1 + C2 n2 + λ ⎜ σ 1 + σ 2 − k ⎟ with respect to n1, n2 and λ equal to zero.
2 2

⎜n ⎟
⎝
n 1 2 ⎠

These equations are

∂ λσ 2
f (n1 , n2 , λ ) = C1 − 21 = 0 (1)
∂n1 n1

∂ λσ 2 (2)
f (n1 , n2 , λ ) = C2 − 22 = 0
∂n2 n2

∂ σ2 σ2
f (n1 , n2 , λ ) = 1 + 2 = k (3)
∂λ n1 n2

Upon adding equations (1) and (2), we obtain C + C − λ ⎛⎜ σ 1 + σ 2 ⎞⎟ = 0

2 2

1 2 ⎜n ⎟
⎝ 1 n2 ⎠
C1 + C2
Substituting from equation (3) enables us to solve for λ to obtain =λ
k
Then, equations (1) and (2) are solved for n1 and n2 to obtain

10-87
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

σ 12 (C1 + C2 ) σ 22 (C1 + C2 )
n1 = n2 =
kC1 kC2

It can be verified that this is a minimum and that with these choices for n1 and n2.
σ 12 σ 22
V ( X1 − X 2 ) = + .
n1 n2

10-96 Maximizing the probability of rejecting H0 is equivalent to minimizing

⎛ ⎞ ⎛ ⎞
⎜ x −x ⎟ ⎜ δ δ ⎟
P ⎜ − zα / 2 < 1 2 22 < zα / 2 | μ1 − μ 2 = δ ⎟ = P⎜ − z α / 2 − < Z < zα / 2 − ⎟
σ1 σ 2
⎜ + ⎟ ⎜ σ 12 σ 22 σ 12 σ 22
⎟
⎝ n1 n2 ⎠ ⎝ n1
+
n2 n1
+
n2 ⎠
δ
where z is a standard normal random variable. This probability is minimized by maximizing .
σ 12 σ 22
+
n1 n2

σ 12 σ 22
Therefore, we are to minimize + subject to n1 + n2 = N.
n1 n2

σ 12 σ 22
From the constraint, n2 = N − n1, and we are to minimize f ( n1 ) = + . Taking the derivative
n1 N − n1

of f(n1) with respect to n1 and setting it equal to zero results in the equation − σ 1 + σ 22
2
= 0.
2
n1 ( N − n1 ) 2

Upon solving for n1, we obtain n = σ 1 N and n 2 = σ 2 N

1
σ1 + σ 2 σ1 + σ 2
Also, it can be verified that the solution minimizes f(n1).

10-97 a) α = P ( Z > z ε or Z < − z α −ε ) where Z has a standard normal distribution. Then,

α = P( Z > zε ) + P( Z < − zα −ε ) = ε + α − ε = α
b) β = P(− zα −ε < Z 0 < zε | μ1 = μ0 + δ )

(
β = P − zα − ε < < z | μ = μ +δ)
x − μ0
σ2 / n ε 1 0

= P (− z − α −ε<Z<z − δ
) ε
δ
σ2 / n σ2 / n

= Φ(z −ε ) − Φ (−z − )
δ
α −ε
δ
σ2 / n σ2 / n

10-98 The requested result can be obtained from data in which the pairs are very different. Example:
pair 1 2 3 4 5
sample 1 100 10 50 20 70
sample 2 110 20 59 31 80

x1 = 50 x2 = 60

s1 = 36.74 s2 = 36.54 spooled = 36.64

10-88
Applied Statistics and Probability for Engineers, 5th edition March 15, 2010

Two-sample t-test : t0 = −0.43 P-value = 0.68

xd = −10 sd = 0.707

Paired t-test: t 0 = −31.62 P-value ≈ 0

p1 pˆ
10-99 a) θ = and θˆ = 1 and ln(θˆ) ~ N [ln(θ ), (n1 − x1 ) / n1 x1 + (n2 − x2 ) / n2 x2 ]
p2 pˆ 2

ln(θˆ) − ln(θ )
The (1 – α) confidence interval for ln(θ) can use the relationship Z = 1/ 4
⎛ ⎛ n1 − x1 ⎞ ⎛ n2 − x2 ⎞ ⎞
⎜⎜ ⎟+⎜ ⎟⎟
⎝ ⎝ n1 x1 ⎠ ⎝ n2 x2 ⎠ ⎠

1/ 4 1/ 4
⎛ ⎛ n − x ⎞ ⎛ n − x2 ⎞⎞ ⎛⎛ n − x ⎞ ⎛ n − x ⎞⎞
ln(θˆ) − Zα ⎜⎜ ⎜ 1 1 ⎟ + ⎜ 2 ⎟ ⎟⎟ ≤ ln(θ ) ≤ ln(θˆ) + Zα ⎜⎜ ⎜ 1 1 ⎟ + ⎜ 2 2 ⎟ ⎟⎟
2
⎝ ⎝ n1 x1 ⎠ ⎝ n2 x2 ⎠⎠ 2
⎝ ⎝ n1 x1 ⎠ ⎝ n2 x2 ⎠⎠

b) The (1 – α) confidence interval for θ can use the CI developed in part (a) where θ = e^( ln(θ))
1/ 4 1/ 4
⎛⎛ n − x ⎞ ⎛ n − x ⎞⎞ ⎛⎛ n − x ⎞ ⎛ n − x ⎞⎞
− Zα ⎜ ⎜ 1 1 ⎟ + ⎜ 2 2 ⎟ ⎟ Zα ⎜ ⎜ 1 1 ⎟ + ⎜ 2 2 ⎟ ⎟
2 ⎝ ⎝ n1 x1 ⎠ ⎝ n2 x2 ⎠ ⎠ 2 ⎝ ⎝ n1 x1 ⎠ ⎝ n2 x2 ⎠ ⎠
θˆ e ≤ θ ≤ θˆ e

⎛ ⎛ n − x ⎞ ⎛ n − x ⎞ ⎞ .25 ⎛ ⎛ n − x ⎞ ⎛ n − x ⎞ ⎞ .25
− Zα ⎜ ⎜ 1 1 + 2 2
⎟ Zα ⎜ ⎜ 1 1 + 2 2
⎟
2 ⎜⎝ ⎝ n1 x1 ⎟⎠ ⎜⎝ n2 x2 ⎟⎠ ⎟⎠ 2 ⎜⎝ ⎝ n1 x1 ⎟⎠ ⎜⎝ n2 x2 ⎟⎠ ⎟⎠
θˆ e ≤ θ ≤ θˆ e
1/ 4 1/ 4
⎛ ⎛ 100 − 27 ⎞ ⎛ 100 −19 ⎞ ⎞ ⎛ ⎛ 100 − 27 ⎞ ⎛ 100 −19 ⎞ ⎞
−1.96⎜ ⎜ ⎟ +⎜ ⎟⎟ 1.96⎜ ⎜ ⎟ +⎜ ⎟⎟
⎝ ⎝ 2700 ⎠ ⎝ 1900 ⎠ ⎠ ⎝ ⎝ 2700 ⎠ ⎝ 1900 ⎠ ⎠
c) 1.42e ≤ θ ≤ 1.42e
0.519 ≤ θ ≤ 3.887

Because the confidence interval contains the value one, we conclude that there is no significant
difference in the proportions at the 95% level of significance.

10-100 H 0 : σ 12 = σ 22

H1 : σ12 ≠ σ 22

⎛ σ ⎞ 2

β = P ⎜ f1−2α / 2,n1−1,n2 −1 < S < fα2/ 2,n1−1,n2 −1 |

2
1
= δ ≠ 1⎟ 1

⎝ S 2
2 σ ⎠ 2
2

⎛σ2 S2 /σ2 σ
2
σ2 ⎞
= P ⎜ 22 f1−α / 2,n1−1,n2 −1 < 12 12 < 22 f α / 2,n1−1,n2 −1 | 12 = δ ⎟
⎝ σ1 S2 / σ 2 σ1 σ2 ⎠

S12 / σ12
where has an F distribution with n1 − 1 and n2 − 1 degrees of freedom.
S22 / σ 22

10-89