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Solutions to Even numbered problems in Chapters 5-
CHAPTER 5 (Sections 5.1-5.8)
5-2. 1) The parameter of interest is the difference in breaking strengths, μ 1 − μ 2 and ∆ 0 = 10
- H 0 : μ 1 − μ 2 = 10 or μ 1 =μ 2
- H 1 : μ 1 − μ 2 > 10 or μ 1 >μ 2
- α = 0.
- The test statistic is z x^ x
n n
0 1 2 0 12 1
22 2
= −^ −
σ σ
- Reject H 0 if z 0 > zα = 1.
- x 1 = 162.5 x 2 = 155.0 δ = 10 σ 1 = 1.0 σ 2 = 1. n 1 = 10 n 2 = 12 z 0 162 5^2 155 0 210 10 10
= −^ − 5 84
- Since -5.84 < 1.645 do not reject the null hypothesis and conclude there is insufficient evidence to support the use of plastic 1 at α = 0.05.
5-4. x 1 = 30.87 x 2 = 30. σ 1 = 0.10 σ 2 = 0. n 1 = 12 n 2 = 10
a) 90% two-sided confidence interval:
( x^1 x^2 ) z^2 n 1 n ( x^ x^ ) z^ n n
2 1
22 2
1 2 1 2 2 1
2 1
22 2
− − (^) α / σ^ + σ^ ≤ μ − μ ≤ − + (^) α/ σ^ +σ
30 87 30 68 1645 (.^010 )^ (.^ )^... (.^ )^ (.^ )
2 2 1 2
2 2 − − + ≤ μ − μ ≤ − + +
0 0987. ≤ μ 1 − μ 2 ≤0 2813.
We are 90% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between 0.0987 and 0.2813 fl. oz.
b) 95% two-sided confidence interval:
( x^1 x^2 ) z^2 n 1 n ( x^ x^ ) z^ n n
2 1
22 2
1 2 1 2 2 1
2 1
22 2
− − (^) α / σ^ + σ^ ≤ μ − μ ≤ − + (^) α/ σ^ +σ
30 87 30 68 196 ( .0 10^ )^ (.^ )^... (.^ )^ (.^ )
2 2 1 2
2 2 − − + ≤ μ − μ ≤ − + +
0 081. ≤ μ 1 − μ 2 ≤0 299.
We are 95% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between 0.081 and 0.299 fl. oz.
Comparison of parts a and b:
As the level of confidence increases, the interval width also increases (with all other values held constant).
c) 95% upper-sided confidence interval:
σ σ 1 2 1 2 α^1
2
1
2
2
2
− ≤ x − x + z + n n
2 2 30 87 30 68 1645 010 12
− ≤. −. +. (.^ )^ +(.^ )
μ 1 − μ 2 ≤ 0 2813.
With 95% confidence, we bellieve the fill volumne for machine 1 exceeds the fill volume of machine 2 by no more than 0.2813 fl. oz.
5-6. x 1 = 89.6 x 2 = 92.
σ 12 = 1.5 σ 22 = 1. n 1 = 15 n 2 = 20
a) 95% confidence interval:
( x^1 x^2 ) z^2 n 1 n ( x^ x^ ) z^ n n
2 1
22 2
1 2 1 2 2 1
2 1
22 2
− − (^) α / σ^ + σ^ ≤ μ − μ ≤ − + (^) α/ σ^ +σ
89 6 92 5 196 15.^.^....^.
− − + ≤ μ − μ ≤ − + +
− 3 684. ≤ μ 1 − μ 2 ≤ −2 116.
With 95% confidence, we believe the mean road octane number for formulation 2 exceeds that of formulation 1 by between 2.116 and 3.684.
b) 1) The parameter of interest is the difference in mean road octane number, μ 1 − μ 2 and ∆ 0 = 0
- H 0 : μ 1 − μ 2 = 0 or μ 1 =μ 2
- H 1 : μ 1 − μ 2 < 0 or μ 1 <μ 2
- α = 0.
- The test statistic is z x^ x
n n
0 1 2 0 12 1
22 2
= −^ −
σ σ
- Reject H 0 if z 0 < −zα = −1.
- x 1 = 89.6 x 2 = 92. σ 12 = 1.5 σ 22 = 1. n 1 = 15 n 2 = 20
z 0 89 6^2 92 5^20 15 15
= −^ − 7 254
- Since −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.
c) P-value = P z( ≤ − 7 25. ) = 1 − P z( ≤ 7 25. ) = 1 − 1 = 0
5-8. 95% level of confidence, E = 1, and z0.025 =1.
μ 1 = mean foam expansion for AFCC μ 2 = mean foam expansion for ATC x 1 = 4.7 x 2 = 6. s 1 = 0.6 s 2 = 0. n 1 = 5 n 2 = 5
95% confidence interval: t0.025,8 = 2.306 sp = 5 0 60^ +^ 5 0 80 = 8
( x^1 x^2 ) t^2 n n 2 s^ p n n ( x^ x^ ) t^ n n s^ p n n
1 2
1 2 1 2 2 2 (^1 2 1 21 )
− − 1 + 1 ≤ − ≤ − + 1 +^1
α / , + − (^ )^ μ^ μ α/ , + −(^ )
− − + ≤ μ − μ ≤ − + +
− 3 23. ≤ μ 1 − μ 2 ≤ − 117.
Yes, with 95% confidence, we believe the mean foam expansion for ATC exceeds that of AFCC by between 1.17 and 2.32.
5-16. a) According to the normal probability plots, the assumption of normality appears to be met since the data 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.
P-Value:A-Squared: 0.463 0.
Anderson-Darling Normality Test N: 15StDev: 10.
Average: 196.
180 190 200 210
. . . . . . . . .
Probability
type
Normal Probability Plot
P-Value:A-Squared: 0.295 0.
Anderson-Darling Normality Test N: 15StDev: 9.
Average: 192.
175 185 195 205
. . . . . . . . .
Probability
type
Normal Probability Plot
175 185 195 205 type
180 190 200 210 type
b) 1) The parameter of interest is the difference in deflection temperature under load, μ 1 −μ 2
s 1 = 12 s 2 = 22 n 1 = 10 n 2 = 16 t 0 290 2 321 20 12 10
= −^ − 4 64
- Since −4.64 < −1.708 reject the null hypothesis and conclude that supplier 2 provides gears with higher mean impact strength at the 0.05 level of significance.
b) P-value = P(t < −4.64): P-value < 0.
c) 1) The parameter of interest is the difference in mean impact strength, μ 2 −μ 1
- H 0 : μ 2 − μ 1 = 25
- H 1 : μ 2 − μ 1 > 25 or μ 2 > μ 1 + 25
- α = 0.
- The test statistic is t x^ x s n
s n
0 2 1 12 1
22 2
= −^ −
( ) δ
- Reject the null hypothesis if t 0 > tα ν, = 1.708 where
ν
ν
s n
s n s n n
s n n
12 1
22 2
2
12 1
2
1
22 2 1 2 1
7) x 1 = 290 x 2 = 321 ∆ 0 =
s 1 = 12 s 2 = 22 n 1 = 10 n 2 = 16 t (^0 321 2290 ) 12 10
= −^ − 0 898
- Since 0.898 < 1.708, do not reject the null hypothesis and conclude that the mean impact strength from supplier 2 is not at least 25 ft-lb higher that supplier 1 using α = 0.05.
5-20. 1) The parameter of interest is the difference in mean melting point, μ 1 −μ 2
- H 0 : μ 1 − μ 2 = 0 or μ 1 =μ 2
- H 1 : μ 1 − μ 2 ≠ 0 or μ 1 ≠μ 2
- α = 0.
- The test statistic is t x^ x s p n n
0 1 2 0
1 2
= −^ −
Reject the null hypothesis if t 0 < − t (^) α / 2 , n 1 + n 2 − 2 where − t0 01 40. , = −2.423 or t 0 > t (^) α/ 2 , n 1 + n 2 − 2 where t0 01 40. , = 2.
x 1 = 421 x 2 = 426 ∆ 0 = 0 s n^ s^ n^ s p (^) n n
= −^ +^ −
1 2
s 1 = 4 s 2 = 3 = 20 4^ +^ 20 3 = 40
( ) 2 ( )^2
n 1 = 21 n 2 = 21 t 0 421 426 0 2 915 1 20
= −^ − 5 424
- Since −5.424 < −2.423 reject the null hypothesis and conclude that the data do not support the claim that both alloys have the same melting point at α = 0.
P-value = 2P ( t < −5 424. ) P-value < 0.
5-22. a) 1) The parameter of interest is the difference in mean wear amount, μ 1 − μ 2.
- H 0 : μ 1 − μ 2 = 0 or μ 1 =μ 2
- H 1 : μ 1 − μ 2 ≠ 0 or μ 1 ≠μ 2
- α = 0.
- The test statistic is t x^ x s n
s n
0 1 2 0 1
2 1
2
2 2
= −^ −
6) Reject the null hypothesis if t 0 < −t0 025 27. , where −t 0 025 27. , = −2.052 or t 0 > t0 025 27. , where
t 0 025 27. , = 2.052 since
ν
ν
s n
s n s n n
s n n
12 1
22 2
2
1
2 1
2
1
2
2 2 1 2 1
(truncated)
- x 1 = 20 x 2 = 15 ∆ 0 = 0 s 1 = 2 s 2 = 8 n 1 = 25 n 2 = 25 t (^0 20 2 15 ) 2 25
= −^ − 3 03
- Since 3.03 > 2.052 reject the null hypothesis and conclude that the data support the claim that the two companies produce material with significantly different wear at the 0.05 level of significance.
b) P-value = 2P(t > 3.03), 2(0.0025) < P-value < 2(0.005)
0.005 < P-value < 0.
c) 1) The parameter of interest is the difference in mean wear amount, μ 1 −μ 2
- H 0 : μ 1 − μ 2 = 0
- H 1 : μ 1 − μ 2 > 0
- α = 0.
- The test statistic is
- H 0 : μd = 0
- H 1 : μd ≠ 0
- α = 0.
- The test statistic is t d (^0) s (^) d n
6) Reject the null hpothesis if t 0 < −t 0 05 13. , where −t 0 05 13. , = −1.771 or t 0 > t 0 05 13. , where t 0 05 13. , = 1.
- d = 1. sd = 12.
n = 14
t 0 121 12 68 14
- Since −1.771 < 0.357 < 1.771 do not reject the null and conclude the data do not support the claim that the two cars have different mean parking times at the 0.10 level of significance. The result is consistent with the confidence interval constructed since 0 is included in the 90% confidence interval.
5-30. d = 868.375 sd = 1290, n = 8 where di = brand 1 - brand 2 99% confidence interval: d t s n
d t s n (^) n − d^ d n d
α / 2 , − 1 μ α/ 2 , − 1
868 375 3 499^1290
868 375 3 499^1290
μd
−727.46 ≤ μd ≤ 2464.
Since zero is contained within this interval, we are 99% confident there is no significant difference between the two brands of tire. 5-32. 1) The parameter of interest is the difference in blood cholesterol level, μd where di = Before − After.
- H 0 : μd = 0
- H 1 : μd > 0
- α = 0.
- The test statistic is t d (^0) s (^) d n
6) Reject the null hpothesis if t 0 > t 0 05 14. , where t 0 05 14. , = 1.
- d = 26. sd = 19.
n = 15
t 0 26 867 19 04 15
- Since 5.465 > 1.761 reject the null and conclude the data support the claim that low the mean difference in cholesterol levels is significanlty less after fat diet and aerobic exercise program at the 0. level of significance. 5-34. 1) The parameter of interest is the difference in mean weight, μd where di =Weight Before − Weight After.
- H 0 : μd = 0
- H 1 : μd > 0
- α = 0.
- The test statistic is
t d (^0) s (^) d n
6) Reject the null hpothesis if t 0 > t 0 05 9. , where t 0 05 9. , = 1.
- d = 17 sd = 6.
n = 10
t 0 17 6 41 10
- Since 8.387 > 1.833 reject the null and conclude there is evidence to conclude that the mean weight loss is significantly greater than 0; that is, the data support the claim that this particular diet modification program is significantly effective in reducing weight at the 0.05 level of significance.
5-36. 1) The parameter of interest is the difference in mean weight loss, μd where di = Before − After.
- H 0 : μd = 10
- H 1 : μd > 10
- α = 0.
- The test statistic is t d (^0) s (^) d n
= − ∆^0
6) Reject the null hpothesis if t 0 > t 0 05 9. , where t 0 05 9. , = 1.
- d = 17 sd = 6.
n = 10
t 0 17 10 6 41 10
= −^ =3 45
- Since 3.45 > 1.833 reject the null and conclude there is evidence to support the claim that this particular diet modification program is effective in producing a mean weight loss of at least 10 lbs at the 0.05 level of significance.
5-38. a) f0.25,5,10 = 1.59 d) f0.75,5,10 = 1 1 189
f 0 25 10 5 (^). , ,.
b) f0.10,24,9 = 2.28 e) f0.90,24,9 = 1 1 191
f 0 10 9 24 (^). , ,.
c) f0.05,8,15 = 2.64 f) f0.95,8,15 = 1 1 3 22
f 0 05 15 8 (^). , ,.
5-40. 1) The parameters of interest are the variances of concentration, σ 12 ,σ 22
- H 0 : σ 12 =σ 22
- H 1 : σ 12 ≠σ 22
- α = 0.
- The test statistic is
f s (^0) s 12 22
6) Reject the null hypothesis if f 0 < f0 975 9 15. , , where f0 975 9 15. , , = 0.265 or f 0 > f0 025 9 15. , , where
f0 025 9 15. , , =3.
- n 1 = 10 n 2 = 16
s 1 = 0.11 s 2 = 0.
f 0
2 2
= (.^ ) = 149
- Since 0.143 < 1.232 < 6.99 do not reject the null hypothesis and conclude the thickness variances do not significantly differ at the 0.02 level of significance.
5-46. 1) The parameters of interest are the melting variances, σ 12 ,σ 22
- H 0 : σ 12 =σ 22
- H 1 : σ 12 ≠σ 22
- α = 0.
- The test statistic is
f s (^0) s 12 22
- Reject the null hypothesis if f 0 < f0 975 20 20. , , where f0 975 20 20. , , =0.4065 or f 0 > f0 025 20 20. , , where f0 025 20 20. , , =2.
- n 1 = 21 n 2 = 21 s 1 = 4 s 2 = 3
f 0
2 2
- Since 0.4065 < 1.78 < 2.46 do not reject the null hypothesis and conclude the population variances do not significantly differ at the 0.05 level of significance.
5-48. 1) The parameters of interest are the time to assemble standard deviations, σ 1 ,σ 2
- H 0 : σ 12 =σ 22
- H 1 : σ 12 ≠σ 22
- α = 0.
- The test statistic is
f s (^0) s 1
2
22
- Reject the null hypothesis if f 0 < f (^1) − α / , 2 n 1 − 1 , n 2 − 1 =0..365 or f 0 > f (^) α/ 2 , n 1 − 1 , n 2 − 1 = 2.
- n 1 = 25 n 2 = 21 s 1 = 0.98 s 2 = 1.
f 0
2 2
= (.^ ) =0 923
- Since 0.365 < 0.923 < 2.86 do not reject the null hypothesis and conclude there is no evidence to support the claim that men and women differ significantly in repeatability for this assembly task at the 0.02 level of significance.
5-50. 1) the parameters of interest are the proportion of defective parts, p 1 and p 2
- H 0 : p 1 =p 2
- H 1 : p 1 ≠p 2
- α = 0.
- Test statistic is z p^ p p p n n
0 1 2
1 2
where
p^ $ x^ x n n
1 2 1 2
- Reject the null hypothesis if z 0 < −z0 025. where −z0 025. = −1.96 or z 0 > z0 025. where z0 025. = 1.
- n 1 = 300 n 2 = 300 x 1 = 15 x 2 = 8 $p 1 = 0.05 p$ 2 = 0.0267 p$ (^) = +.
z 0 0 05^ 0 0267 0 0383 1 0 0383 1 300
- Since −1.96 < 1.49 < 1.96 do not reject the null hypothesis and conclude that yes the evidence indicates that there is not a 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 < 1.49)) = 0.
5-52. a) β =
z pq n n
p p z pq n n
p p
p p p p
α α
σ σ
/
$ $
/ $ $ $ $
2 1 2
1 2 2 1 2
1 2
1 2 1 2
− −
p = 300 0 05^ 300 0 02 300 300
= 0.035 q = 0.
σ^ $^ p$ (^1) − $p 2 = 0 05 1^ 0 05 300
β =
Power = 1 − 0.48401 = 0.
b)
n
z p^ p^ q^ q^ z p q p q
p p
α /2 1 2 1 2 β 1 1 2 2
2
1 2
2
196 0 05^ 0 02^ 0 95^ 0 98
2
2
..^.^.^.^.. (. ). (. )
n = 791
5-54. 95% confidence interval on the difference:
( p$^ p$^ ) z (^) / $p^ (^ p$^ )^ $^ (^ $^ )^ ( $^ $^ ) (^) / $^ (^ $^ )^ $^ (^ $^ ) n
p p n
p p p p z p^ p n
p p (^1 2 2) n 1 1 1
2 2 2
1 2 1 2 2 1 1 1
2 2 2
− − 1 −^ + 1 −^ ≤ − ≤ − + 1 −^ + 1 −
α α
Means plot: LSD Confidence level: 95 Range test: LSD
Analysis of variance
Source of variation Sum of Squares d.f. Mean square F-ratio Sig. level
Between groups .1391104 3 .0463701 2.616. Within groups .3190714 18.
Total (corrected) .4581818 21
0 missing value(s) have been excluded.
Do not reject H 0. There is insignificant evidence to indicate the four firing temperatures affect the density of the brick.
b) P-value = 0.
5-60. One-Way Analysis of Variance
Data: Conductivity
Level codes: Coating
Labels:
Means plot: LSD Confidence level: 95 Range test: LSD
Analysis of variance
Source of variation Sum of Squares d.f. Mean square F-ratio Sig. level
Between groups 1060.5000 4 265.12500 16.349. Within groups 243.2500 15 16.
Total (corrected) 1303.7500 19
0 missing value(s) have been excluded.
Reject H 0. There appears to be a significant difference among the five coating types in their effect on conductivity.
Chapter 6 (Section 6.1-6.4)
6-2. a) y 0 = β 0 +β 1 x 1
Sxx = 1432158. − 1478202 = 339916. Sxy = 1083 67. − (^1478 20 )(12 75^.^ )= 141445. $. .
β
β
1
0 12 75 20 147820
S
S
xy xx
y $ = 0 3299892. +0 0041612. x
0
0.
0.
0.
0.
0.
0.
0.
0.
-50 0 50 100
x
y
b) $y^ = 0 3299892. + 0 0041612 85. ( ) =0 683689.
c) y$ = 0 3299892. + 0 0041612 90. ( ) =0 7044949.
d) β$^1 =0 00416.
6-4. a) Regression Analysis - Linear model: Y = a+bX
Dependent variable: Games Independent variable: Yards
Standard T Prob. Parameter Estimate Error Value Level
Intercept 21.7883 2.69623 8.081. Slope -7.0251E-3 1.25965E-3 -5.57703.
Analysis of Variance
Source Sum of Squares Df Mean Square F-Ratio Prob. Level Model 178.09231 1 178.09231 31.1032. Residual 148.87197 26 5.
Total (Corr.) 326.96429 27 Correlation Coefficient = -0.738027 R-squared = 54.47 percent Stnd. Error of Est. = 2.
d) $y^ = − 6 3355. + 9 20836 47. ( ) =426 458. y^ $^ = 426 458. e = y − y$^ = 424 84. − 426 458. = −1 618.
6-8. Model fitting results for: y
Independent variable coefficient std. error t-value sig.level
CONSTANT 350.994271 74.753074 4.
x1 -1.271994 1.16914 -1.
x2 -0.153904 0.08953 -1.
R-SQ. (ADJ.) = 0.7696 SE= 25.497858 MAE= 16.319125 DurbWat=
Previously: 0.0000 0.000000 0.
6 observations fitted, forecast(s) computed for 0 missing val. of dep. var.
If the calculations were to be done by hand use Equation (6-21).
a) $y^ = 350 9943. − 1272. x 1 −0 1539. x 2
b) y$ = 350 9943. − 1272 25. ( ) − 01539 1000. ( ) =165 29.
c) Model fitting results for: y
Independent variable coefficient std. error t-value sig.level
CONSTANT 125.865548 197.957166 0.
x1 7.758641 7.514795 1.
x2 0.094304 0.220657 0.
x1*x2 0.009186 0.007564 -1.
R-SQ. (ADJ.) = 0.8011 SE= 23.691404 MAE= 12.527757 DurbWat=
Previously: 0.7696 25.497858 16.
6 observations fitted, forecast(s) computed for 0 missing val. of dep. var.
If the calculations were to be done by hand, add a cross-product column to the X matrix and use
Equation (6-21).
y^ $^ ′ = 125 8655. − 7 7586. x 1 − 0 0943. x 2 −0 0092. x x 1 2
d) y$ (^) ′ = 125 8655. − 7 7586 25. ( ) − 0 0943 1000. ( ) −0 0092 25. ( )( 1000 ) $y (^) ′ =184 13. The predicted value is larger
Model fitting results for: y
Independent variable coefficient std. error t-value sig.level
CONSTANT 47.173999 49.581476 0.
x1 -9.735202 3.691625 -2.
x2 0.428287 0.223933 1.
x3 18.237455 1.311802 13.
R-SQ. (ADJ.) = 0.9925 SE= 3.479627 MAE= 2.511105 DurbWat=
Previously: 0.0000 0.000000 0.
20 observations fitted, forecast(s) computed for 0 missing val. of dep. var.
If the calculations were to be done by hand, add a cross-product column to the X matrix and use
Equation (6-21).
a) $y^ = 47 174. − 9 7352. x 1 + 0 4283. x 2 +18 2375. x 3
b) y$ = 47 174. − 9 7352 14 5. (. ) + 0 4283 220. ( ) + 18 2375 5. ( ) =91 43.
6-12. a) 1) The parameter of interest is the regressor variable coefficient, β 1
- H 0 :β 1 = 0
- H 1 :β 1 ≠ 0
- α = 0.
- The test statistic is f MS MS
SS
SS n
R E
R E
0
- Reject H 0 if f 0 > fα,1,12 where f0.05,1,12 = 4.
- Using results from Exercise 6- SS S
SS S SS
R xy
E yy R
β 1 2 3298017 59 057143 137 59
f 0 137 59 22 123 12
- Since 74.63 > 4.75 reject H 0 and conclude that compressive strength is a significant in predicting intrinsic permeability of concrete at α = 0.05. We can therefore conclude model specifies a useful linear relationship between these two variables.
P − value≅ 0. 000002
b) σ$^2.^. 2
MS SS = =
E (^) n E
6-14. a) Refer to ANOVA table of Exercise 6-4.
- The parameter of interest is the regressor variable coefficient, β 1.
