SOLUTIONS MANUAL For Statistics for Engineers and Scientists, 6th Edition by William Navidi by ACADEMIAMILL

Solutions Manual to accompany

STATISTICS FOR ENGINEERS AND SCIENTISTS, 6th ed. Prepared by

William Navidi

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Table of Contents

Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508

SECTION 1.1

Chapter 1 Section 1.1 1. (a) The population consists of all the times the process could be run. It is conceptual. (b) The population consist of all the registered voters in the state. It is tangible. (c) The population consist of all people with high cholesterol levels. It is tangible. (d) The population consist of all concrete specimens that could be made from the new formulation. It is conceptual. (e) The population consist of all bolts manufactured that day. It is tangible.

(iii). It is very unlikely that students whose names happen to fall at the top of a page in the phone book will di˙er systematically in height from the population of students as a whole. It is somewhat more likely that engineering majors will di˙er, and very likely that students involved with basketball intramurals will di˙er.

3. (a) False (b) True

4. (a) False (b) True

5. (a) No. What is important is the population proportion of defectives; the sample proportion is only an approximation. The population proportion for the new process may in fact be greater or less than that of the old process. (b) No. The population proportion for the new process may be 0.12 or more, even though the sample proportion was only 0.11. (c) Finding 2 defective circuits in the sample.

6. (a) False

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CHAPTER 1

(b) True (c) True

A good knowledge of the process that generated the data.

8. (a) An observational study (b) It is not well-justifed. Because the study is observational, there could be di˙erences between the groups other than the level of exercise. These other di˙erences (confounders) could cause the di˙erence in blood pressure. 9. (a) A controlled experiment (b) It is well-justifed, because it is based on a controlled experiment rather than an observational study.

Section 1.2 1.

False

No. In the sample 1, 2, 4, the mean is 7/3, which does not appear at all.

No. The median of the sample 1, 2, 4, 5 is 3.

The sample size can be any odd number.

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SECTION 1.2

Yes. For example, the list 1, 2, 12 has an average of 5 and a standard deviation of 6.08.

Yes. If all the numbers in the list are the same, the standard deviation will equal 0.

The mean increases by $50; the standard deviation is unchanged.

The mean and standard deviation both increase by 5%.

10. (a) Let X1 , ..., X100 denote the 100 numbers of occupants. 100 É

Xi = 70(1) + 15(2) + 10(3) + 3(4) + 2(5) = 152

i=1

∑100 X=

i=1 Xi

100

152 = 1.52 100

(b) The sample variance is s2

= = =

H 100 I 2 1 É 2 X − 100X 99 i=1 i 1 [(70)12 + (15)22 + (10)32 + (3)42 + (2)52 − 100(1.522 )] 99 0.87838

The standard deviation is s = s2 = 0.9372. Alternatively, the sample variance can be computed as 1 É (X − X)2 99 i=1 i 1 = [70(1 − 1.52)2 + 15(2 − 1.52)2 + 10(3 − 1.52)2 + 3(4 − 1.52)2 + 2(5 − 1.52)2 ] 99 = 0.87838 100

(c) The sample median is the average of the 50th and 51st value when arranged in order. Both these values are equal to 1, so the median is 1.

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CHAPTER 1

(d) The frst quartile is the average of the 25th and 26th value when arranged in order. Both these values are equal to 1, so the frst quartile is 1. The third quartile is the average of the 75th and 76th value when arranged in order. Both these values are equal to 2, so the frst quartile is 2.

(e) Of the 100 cars, 15 + 10 + 3 + 2 = 30 had more than the mean of 1.52 occupants, so the proportion is 30/100 = 0.3.

(f) The quantity that is one standard deviation greater than the mean is 1.52 + 0.9372 = 2.5472. Of the 100 cars, 10 + 3 + 2 = 15 had more than 2.8652 children, so the proportion is 15/100 = 0.15.

(g) The region within one standard deviation of the mean is 1.52 ± 0.9372 = (0.5828, 2.4572). Of the 100 cars, 70 + 15 = 85 are in this range, so the proportion is 85/100 = 0.85.

11.

The total height of the 20 men is 20 × 178 = 3560. The total height of the 30 women is 30 × 164 = 4920. The total height of all 50 people is 3560 + 4920 = 8480. There are 20 + 30 = 50 people in total. Therefore the mean height for both groups put together is 8480∕50 = 169.6 cm.

12.

The total weight of the 5 basketball players is 5 × 77.6 = 388. The total weight of all 16 athletes is 16 × 74.5 = 1192. The total weight of the 11 football players is 1192 − 388 = 804. Therefore the mean weight for the football players is 804∕11 = 73.09.

13. (a) The total number of accidents is 80 × 0 + 15 × 1 + 4 × 2 + 1 × 3 = 26. The mean number of accidents is 26∕100 = 0.26.

(b) The median is the average of the 50th and 51st value when arranged in order. Both these values are equal to 0, so the median is 0.

(c) The mean is more useful, because it can be used to compute the total number of claims fled. The median indicates that more than half of policy holders fle no claims, but gives no information about the total number of claims.

14. (a) The mean for A is (18.0+18.0+18.0+20.0+22.0+22.0+22.5+23.0+24.0+24.0+25.0+25.0+25.0+25.0+26.0+26.4)∕16 = 22.744

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SECTION 1.2

The mean for B is (18.8+18.9+18.9+19.6+20.1+20.4+20.4+20.4+20.4+20.5+21.2+22.0+22.0+22.0+22.0+23.6)∕16 = 20.700 The mean for C is (20.2+20.5+20.5+20.7+20.8+20.9+21.0+21.0+21.0+21.0+21.0+21.5+21.5+21.5+21.5+21.6)∕16 = 20.013 The mean for D is (20.0+20.0+20.0+20.0+20.2+20.5+20.5+20.7+20.7+20.7+21.0+21.1+21.5+21.6+22.1+22.3)∕16 = 20.806

(b) The median for A is (23.0 + 24.0)/2 = 23.5. The median for B is (20.4 + 20.4)/2 = 20.4. The median for C is (21.0 + 21.0)/2 = 21.0. The median for D is (20.7 + 20.7)/2 = 20.7. (c) 0.20(16) = 3.2 3. ≈ 3. Trim the 3 highest and 3 lowest observations. The 20% trimmed mean for A is (20.0 + 22.0 + 22.0 + 22.5 + 23.0 + 24.0 + 24.0 + 25.0 + 25.0 + 25.0)∕10 = 23.25 The 20% trimmed mean for B is (19.6 + 20.1 + 20.4 + 20.4 + 20.4 + 20.4 + 20.5 + 21.2 + 22.0 + 22.0)∕10 = 20.70 The 20% trimmed mean for C is (20.7 + 20.8 + 20.9 + 21.0 + 21.0 + 21.0 + 21.0 + 21.0 + 21.5 + 21.5)∕10 = 21.04 The 20% trimmed mean for D is (20.0 + 20.2 + 20.5 + 20.5 + 20.7 + 20.7 + 20.7 + 21.0 + 21.1 + 21.5)∕10 = 20.69

(d) 0.25(17) = 4.25. Therefore the frst quartile is the average of the numbers in positions 4 and 5. 0.75(17) = 12.75. Therefore the third quartile is the average of the numbers in positions 12 and 13. A: Q1 = 21.0, Q3 = 25.0; B: Q1 = 19.85, Q3 = 22.0; C: Q1 = 20.75, Q3 = 21.5; D: Q1 = 20.1, Q3 = 21.3 (e) The variance for A is s2

1 [18.02 + 18.02 + 18.02 + 20.02 + 22.02 + 22.02 + 22.52 + 23.02 + 24.02 15 + 24.02 + 25.02 + 25.02 + 25.02 + 25.02 + 26.02 + 26.42 − 16(22.7442 )] = 8.2506

The standard deviation for A is s = The variance for B is s2

8.2506 = 2.8724.

1 [18.82 + 18.92 + 18.92 + 19.62 + 20.12 + 20.42 + 20.42 + 20.42 + 20.42 15 + 20.52 + 21.22 + 22.02 + 22.02 + 22.02 + 22.02 + 23.62 − 16(20.7002 )] = 1.8320

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CHAPTER 1

The standard deviation for B is s The variance for C is s2

1 [20.22 + 20.52 + 20.52 + 20.72 + 20.82 + 20.92 + 21.02 + 21.02 + 21.02 15 + 21.02 + 21.02 + 21.52 + 21.52 + 21.52 + 21.52 + 21.62 − 16(20.0132 )] = 0.17583

The standard deviation for C is s = The variance for D is s2

.8320 = 1.3535.

0.17583 = 0.4193.

1 [20.02 + 20.02 + 20.02 + 20.02 + 20.22 + 20.52 + 20.52 + 20.72 + 20.72 15 + 20.72 + 21.02 + 21.12 + 21.52 + 21.62 + 22.12 + 22.32 − 16(20.8062 )] = 0.55529

The standard deviation for D is s =

0.55529 = 0.7542.

(f) Method A has the largest standard deviation. This could be expected, because of the four methods, this is the crudest. Therefore we could expect to see more variation in the way in which this method is carried out, resulting in more spread in the results.

(g) Other things being equal, a smaller standard deviation is better. With any measurement method, the result is somewhat di˙erent each time a measurement is made. When the standard deviation is small, a single measurement is more valuable, since we know that subsequent measurements would probably not be much di˙erent.

15. (a) All would be divided by 2.54.

(b) Not exactly the same, because the measurements would be a little di˙erent the second time.

16. (a) We will work in units of $1000. Let S0 be the sum of the original 10 numbers and let S1 be the sum after the change. Then S0 ∕10 = 70, so S0 = 700. Now S1 = S0 − 100 + 1000 = 1600, so the new mean is S1 ∕10 = 160. (b) The median is unchanged at 55. ∑ 2 2 (c) Let X1 , ..., X10 be the original 10 numbers. Let T0 = 10 i=1 Xi . Then the variance is (1∕9)[T0 − 10(70 )] = 2 2 20 = 400, so T0 = 52,600. Let T1 be the sum of the squares after the change. Then T1 = T0 − 100 + 10002 = 1, 042, 600. The new standard deviation is (1∕9)[T1 − 10(1602 )] = 295.63.

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SECTION 1.3

17. (a) The sample size is n = 16. The tertiles have cutpoints (1∕3)(17) = 5.67 and (2∕3)(17) = 11.33. The frst tertile is therefore the average of the sample values in positions 5 and 6, which is (44 + 46)∕2 = 45. The second tertile is the average of the sample values in positions 11 and 12, which is (76 + 79)∕2 = 77.5.

(b) The sample size is n = 16. The quintiles have cutpoints (i∕5)(17) for i = 1, 2, 3, 4. The quintiles are therefore the averages of the sample values in positions 3 and 4, in positions 6 and 7, in positions 10 and 11, and in positions 13 and 14. The quintiles are therefore (23 + 41)∕2 = 32, (46 + 49)∕2 = 47.5, (74 + 76)∕2 = 75, and (82 + 89)∕2 = 85.5.

18. (a) Seems certain to be an error. (b) Could be correct.

Section 1.3 Stem 0 1 1. (a) 2 3 4 5 6 7 8 9 10 11 12 13

Leaf 011112235677 235579 468 11257 14699 5 16 9 0099

0 7 7

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CHAPTER 1 0.45 0.4

(b) Here is one histogram. Other choices for the endpoints are possible.

Relative Frequency

0.35 0.3 0.25 0.2 0.15 0.1 0.05 0

(c)

6 8 Rainfall (inches)

6 8 10 Rainfall (inches)

Rainfall (inches)

(d)

The boxplot shows one outlier. 5

2. (a) Stem 14 15 16 17 18 19 20 21 22

Leaf 12333444467788889 0023344567799 124456 2222338

9 4

(b) Here is one histogram. Other choices for the endpoints are possible.

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SECTION 1.3

(c)

(d)

The boxplot shows 2 outliers.

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CHAPTER 1

Stem Leaf 1 1588 2 00003468 3 0234588 4 0346 5 2235666689 6 00233459 7 113558 8 568 9 1225 10 1 11 12 2 13 06 14 15 16 17 1 18 6 19 9 20 21 22 23 3 There are 23 stems in this plot. An advantage of this plot over the one in Figure 1.6 is that the values are given to the tenths digit instead of to the ones digit. A disadvantage is that there are too many stems, and many of them are empty.

4. (a) Here are histograms for each group. Other choices for the endpoints are possible. 0.6

Relative Frequency

0.4 0.3 0.2 0.1 0

0.5 0.4 0.3 0.2 0.1

150 250 350 450 550 Concentration (mg/kg)

50 150 250350450550650750850950

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SECTION 1.3

1000 Concentration (mg/kg)

(b)

800 600 400 200 0 Nickel

Chromium

(c) The concentrations of nickel are on the whole lower than the concentrations of chromium. The nickel concentrations are highly skewed to the right, which can be seen from the median being much closer to the frst quartile than the third. The chromium concentrations are somewhat less skewed. Finally, the nickel concentrations include an outlier.

5. (a) Here are histograms for each group. Other choices for the endpoints are possible. Catalyst A

0.5 0.4

0.4

Relative Frequency

0.5

0.3 0.2 0.1 0

0.3 0.2 0.1

Yield

4 Yield

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CHAPTER 1

(b)

(c) The yields for catalyst B are considerably more spread out than those for catalyst A.The median yield for catalyst A is greater than the median for catalyst B. The median yield for B is closer to the frst quartile than the third, but the lower whisker is longer than the upper one, so the median is approximately equidistant from the extremes of the data. Thus the yields for catalyst B are approximately symmetric. The largest yield for catalyst A is an outlier; the remaining yields for catalyst A are approximately symmetric.

Yield

5 4 3 2 1

Catalyst A

Catalyst B

6. (a) The histogram should be skewed to the right. Here is an example.

(b) The histogram should be skewed to the left. Here is an example.

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SECTION 1.3

7. (a) The proportion is the sum of the relative frequencies (heights) of the rectangles above 240. This sum is approximately 0.14 + 0.10 + 0.05 + 0.01 + 0.02. This is closest to 30%.

(b) The height of the rectangle over the interval 240–260 is greater than the sum of the heights of the rectangles over the interval 280–340. Therefore there are more men in the interval 240–260 mg/dL.

The relative frequencies of the rectangles shown are 0.05, 0.1, 0.15, 0.25, 0.2, and 0.1. The sum of these relative frequencies is 0.85. Since the sum of all the relative frequencies must be 1, the missing rectangle has a height of 0.15.

18 15 12

Frequency

9. (a)

9 6 3 0

9 11 13 15 17 19 21 23 25 Emissions (g/gal)

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CHAPTER 1

0.15

(b) Density

0.1

0.05

9 11 13 15 17 19 21 23 25 Emissions (g/gal)

0.3

Relative Frequency

0.25

10. (a)

0.2

0.15 0.1

0.05 0

9 11 15 Emissions (g/gal)

(b) No

(d) The classes 11–<15 and 15–<25

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SECTION 1.3

100

11. (a) Number Absent

90 80 70

(b) Yes. The value 100 is an outlier.

60 50 40

12.

The mean. The median, and the frst and third quartiles are indicated directly on a boxplot, and the interquartile range can be computed as the di˙erence between the frst and third quartiles.

13.

The fgure on the left is a sketch of separate histograms for each group. The histogram on the right is a sketch of a histogram for the two groups combined. There is more spread in the combined histogram than in either of the separate ones. Therefore the standard deviation of all 200 resistances is greater than 5Ω. The answer is (ii).

14. (a) True (b) False

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CHAPTER 1

15. (a) IQR = 3rd quartile − 1st quartile. A: IQR = 6.02 − 1.42 = 4.60, B: IQR = 9.13 − 5.27 = 3.86

12 10 8

(b) Yes, since the minimum is within 1.5 IQR of the frst quartile and the maximum is within 1.5 IQR of the third quartile, there are no outliers, and the given numbers specify the boundaries of the box and the ends of the whiskers.

6 4 2 0

(c) No. The minimum value of −2.235 is an “outlier,” since it is more than 1.5 times the interquartile range below the frst quartile. The lower whisker should extend to the smallest point that is not an outlier, but the value of this point is not given.

16. (a) (4) (b) (2) (c) (1) (d) (3)

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SECTION 1.3 500

17. (a) Fracture Stress (MPa)

400 300 200 100 0

(b) The boxplot indicates that the value 470 is an outlier.

((c) (c c)

100

200 300 Fracture Strength (MPa)

400

500

(d) The dotplot indicates that the value 384 is detached from the bulk of the data, and thus could be considered to be an outlier.

18.

Catalyst B 61 965 887541 3 98520 743

Stem 1 2 3 4 5 6

Catalyst A 6669 14689 011344556789 27 7

The distribution of yields for Catalyst A is unimodal, while the distribution of yields for Catalyst B is bimodal.

19. (a) The mean is 0.7619; the median is 1. (b) Skewed to the left.

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CHAPTER 1 0.5 0.4

Density

0.3 0.2 0.1 0 0

20. (a) iii (b) i (c) iv (d) ii

21. (a)

60 50 40 y 30

The relationship is non-linear.

20 10 0 0

(b)

x ln y

1.4 0.83

2.4 1.31

4.0 1.74

4.9 2.29

5.7 1.93

6.3 2.76

7.8 2.73

9.0 3.61

9.3 3.54

11.0 3.97

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SECTION 1.3 4 3.5 3 2.5 ln y 2

The relationship is approximately linear.

1.5 1 0.5 0

0.35

22. (a)

Relative Frequency

0.3 0.25 0.2 0.15 0.1 0.05 0 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000

FEV (mL)

(b) Skewed to the right

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CHAPTER 1

6000

(c)

5500 5000

FEV (mL)

4500

4000 3500 3000 2500 2000 1500 1000

(d) Yes, unusually large values.

0.25

Relative Frequency

23. (a)

0.2 0.15 0.1 0.05 0 −160 −140 −120 −100 −80

−60

−40

−20

100

120

140

160

Acceleration Due To Gravity (deviations from base value in units of 10−8 m/s2)

180

(b) Unimodal

200

Acceleration Due To Gravity (deviations from base value in units of 10−8 m/s2)

(c)

150 100 50 0

−50 −100 −150 −200

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SUPPLEMENTARY EXERCISES FOR CHAPTER 1

(d) One

24. (a)

(b) The depths at 75% humidity are more spread out than those at 40% humidity. In particular, the median and frst quartile are smaller, but the largest depths are bigger.

Supplementary Exercises for Chapter 1 1. (a) The mean will be divided by 2.2.

(b) The standard deviation will be divided by 2.2.

2. (a) The mean will increase by 50 g.

(b) The standard deviation will be unchanged.

3. (a) False. The true percentage could be greater than 5%, with the observation of 4 out of 100 due to sampling variation.

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CHAPTER 1

(b) True

(d) True. If the result di˙ers greatly from 5%, it is unlikely to be due to sampling variation.

4. (a) No. This could well be sampling variation.

(b) Yes. It is virtually impossible for sampling variation to be this large.

5. (a) It is not possible to tell by how much the mean changes, because the sample size is not known.

(b) If there are more than two numbers on the list, the median is unchanged. If there are only two numbers on the list, the median is changed, but we cannot tell by how much.

(c) It is not possible to tell by how much the standard deviation changes, both because the sample size is unknown and because the original standard deviation is unknown.

6. (a) The sum of the numbers decreases by 12.9 − 1.29 = 11.61, so the mean decreases by 11.61/15 = 0.774.

(b) No, it is not possible to determine the value of the mean after the change, since the original mean is unknown.

(d) It is not possible to tell by how much the standard deviation changes, because the original standard deviation is unknown.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 1

7. (a) The mean decreases by 0.774.

(b) The value of the mean after the change is 25 − 0.774 = 24.226.

(d) It is not possible to tell by how much the standard deviation changes, because the original standard deviation is unknown.

8. (a) The sum of the numbers 284.34, 173.01, 229.55, 312.95, 215.34, 188.72, 144.39, 172.79, 139.38, 197.81, 303.28, 256.02, 658.38, 105.14, 295.24, 170.41 is 3846.75. The mean is therefore 3846.75/16 = 240.4219.

(b) The 16 values arranged in increasing order are: 105.14, 139.38, 144.39, 170.41, 172.79, 173.01, 188.72, 197.81, 215.34, 229.55, 256.02, 284.34, 295.24, 303.28, 312.95, 658.38 The median is the average of the 8th and 9th numbers, which is (197.81 + 215.34)∕2 = 206.575.

(c) 0.25(17) = 4.25, so the frst quartile is the average of the 4th and 5th numbers, which is (170.41 + 172.79)∕2 = 171.60.

(d) 0.75(17) = 12.75, so the third quartile is the average of the 12th and 13th numbers, which is (284.34 + 295.24)∕2 = 289.79.

(e)

The median is closer to the frst quartile than to the third quartile, which indicates that the sample is skewed a bit to the right. In addition, the sample contains an outlier.

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24 9.

CHAPTER 1

Statement (i) is true. The sample is skewed to the right.

10. (a) False. The length of the whiskers is at most 1.5 IQR.

(b) False. The length of the whiskers is at most 1.5 IQR.

(d) True. A whisker extends to the most extreme data point that is within 1.5 IQR of the frst or third quartile.

11. (a) Incorrect, the total area is greater than 1.

(b) Correct. The total area is equal to 1.

(d) Correct. The total area is equal to 1.

12.

(i) It would be skewed to the right. The mean is greater than the median. Also note that half the values are between 0 and 0.10, so the left-hand tail is very short.

13. (a) Skewed to the left. The 85th percentile is much closer to the median (50th percentile) than the 15th percentile is. Therefore the histogram is likely to have a longer left-hand tail than right-hand tail.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 1

(b) Skewed to the right. The 15th percentile is much closer to the median (50th percentile) than the 85th percentile is. Therefore the histogram is likely to have a longer right-hand tail than left-hand tail.

14. Class Interval 0–< 1 1–< 2 2–< 3 3–< 4 4–< 5 5–< 6 6–< 7 7–< 8 8–< 9 9–< 10 10–< 11 11–< 12 12–< 13 13–< 14

15. (a)

Relative Frequency 0.2857 0.1429 0.0714 0.1190 0.1190 0.0238 0.0476 0.0238 0.0952 0.0000 0.0000 0.0238 0.0238 0.0238

Frequency 12 6 3 5 5 1 2 1 4 0 0 1 1 1

Cumulative Frequency 12 18 21 26 31 32 34 35 39 39 39 40 41 42

Cumulative Relative Frequency 0.2857 0.4286 0.5000 0.6190 0.7381 0.7619 0.8095 0.8333 0.9286 0.9286 0.9286 0.9524 0.9762 1.0000

0.25

Density

0.2 0.15 0.1 0.05 0

1213 141516 1718 Log population

(b) 0.14

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CHAPTER 1

(d)

0.25

Density

0.2 0.15 0.1 0.05 0

4 6 Population (millions)

The data on the raw scale are skewed so much to the right that it is impossible to see the features of the histogram.

16. (a) The mean is 1 (2099 + 528 + 2030 + 1350 + 1018 + 384 + 1499 + 1265 + 375 + 424 + 789 + 810 23 + 522 + 513 + 488 + 200 + 215 + 486 + 257 + 557 + 260 + 461 + 500) = 740.43

(b) The variance is s2

1 [20992 + 5282 + 20302 + 13502 + 10182 + 3842 + 14992 + 12652 + 3752 + 4242 + 7892 + 8102 22 + 5222 + 5132 + 4882 + 2002 + 2152 + 4862 + 2572 + 5572 + 2602 + 4612 + 5002 − 23(740.432 )] = 302320.26

The standard deviation is s =

302320.26 = 549.84.

(c) The 23 values, arranged in increasing order, are: 200, 215, 257, 260, 375, 384, 424, 461, 486, 488, 500, 513, 522, 528, 557, 789, 810, 1018, 1265, 1350, 1499, 2030, 2099 The median is the 12th value, which is 513.

(d)

500

1000

1500

2000

Electrical Conductivity (µS/cm)

2500

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SUPPLEMENTARY EXERCISES FOR CHAPTER 1

(e) Since (0.1)(23) 2, the 10% trimmed mean is computed by deleting the two highest and two lowest values, and averaging the rest. The 10% trimmed mean is 1 (257 + 260 + 375 + 384 + 424 + 461 + 486 + 488 + 500 + 513 19 + 522 + 528 + 557 + 789 + 810 + 1018 + 1265 + 1350 + 1499) = 657.16

(f) 0.25(24) = 6. Therefore, when the numbers are arranged in increasing order, the frst quartile is the number in position 6, which is 384.

(g) 0.75(24) = 18. Therefore, when the numbers are arranged in increasing order, the third quartile is the number in position 18, which is 1018.

(h) IQR = 3rd quartile − 1st quartile = 1018 − 384 = 634. 2500 Electrical Conductivity (µS/cm)

(i)

2000

1500

1000

500

(j) The points 2030 and 2099 are outliers.

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CHAPTER 1

17. (a)

0.25

Density

0.2 0.15 0.1 0.05 0

024

10 15 20 25 30 Number of shares owned

(b) The sample size is 651, so the median is approximated by the point at which the area to the left is 0.5 = 325.5/651. The area to the left of 3 is 295/651, and the area to the left of 4 is 382/651. The point at which the area to the left is 325.5/651 is 3 + (325.5 − 295)∕(382 − 295) = 3.35. (c) The sample size is 651, so the frst quartile is approximated by the point at which the area to the left is 0.25 = 162.75/651. The area to the left of 1 is 18/651, and the area to the left of 2 is 183/651. The point at which the area to the left is 162.75/651 is 1 + (162.75 − 18)∕(183 − 18) = 1.88. (d) The sample size is 651, so the third quartile is approximated by the point at which the area to the left is 0.75 = 488.25/651. The area to the left of 5 is 425/651, and the area to the left of 10 is 542/651. The point at which the area to the left is 488.25/651 is 5 + (10 − 5)(488.25 − 425)∕(542 − 425) = 7.70.

18. (a)

Frequency

40 30 20 10 0

3 4 5 6 Time (months)

(b) The sample size is 171, so the median is the value in position (171 + 1)∕2 = 86 when the values are arranged in order. There are 45 + 17 + 18 = 80 values less than or equal to 3, and 80 + 19 = 99 values less than or equal to 4. Therefore the class interval 3 – < 4 contains the median.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 1

(c) The sample size is 171, so the frst quartile is the value in position 0.25(171 + 1) = 43 when the values are arranged in order. There are 45 values in the frst class interval 0 – < 1. Therefore the class interval 0 – < 1 contains the frst quartile.

(d) The sample size is 171, so the third quartile is the value in position 0.75(171 + 1) = 129 when the values are arranged in order. There are 45+17+18+19+12+14 = 125 values less than or equal to 6, and 125+13 = 138 values less than or equal to 7. Therefore the class interval 6 – < 7 contains the third quartile.

19. (a)

70 60

Load (kg)

50 40 30 20 10 0

Sacaton

Gila Plain

Casa Grande

(b) Each sample contains one outlier.

(c) In the Sacaton boxplot, the median is about midway between the frst and third quartiles, suggesting that the data between these quartiles are fairly symmetric. The upper whisker of the box is much longer than the lower whisker, and there is an outlier on the upper side. This indicates that the data as a whole are skewed to the right. In the Gila Plain boxplot data, the median is about midway between the frst and third quartiles, suggesting that the data between these quartiles are fairly symmetric. The upper whisker is slightly longer than the lower whisker, and there is an outlier on the upper side. This suggest that the data as a whole are somewhat skewed to the right. In the Casa Grande boxplot, the median is very close to the frst quartile. This suggests that there are several values very close to each other about one-fourth of the way through the data. The two whiskers are of about equal length, which suggests that the tails are about equal, except for the outlier on the upper side.

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CHAPTER 2

Chapter 2 Section 2.1 1.

P (does not fail) = 1 − P (fails) = 1 − 0.87 = 0.13

2. (a) {1, 2, 3} (b) P (odd number) = P (1) + P (3) = 3∕6 + 1∕6 = 2∕3 (c) No, the set of possible outcomes is still {1, 2, 3}. (d) Yes, a list of equally likely outcomes is then {1, 1, 1, 2, 2, 3, 3}, so P (odd) = P (1) + P (3) = 3∕7 + 2∕7 = 5∕7.

3. (a) The outcomes are the 16 di˙erent strings of 4 true-false answers. These are {TTTT, TTTF, TTFT, TTFF, TFTT, TFTF, TFFT, TFFF, FTTT, FTTF, FTFT, FTFF, FFTT, FFTF, FFFT, FFFF}. (b) There are 16 equally likely outcomes. The answers are all the same in two of them, TTTT and FFFF. Therefore the probability is 2∕16 or 1∕8. (c) There are 16 equally likely outcomes. There are four of them, TFFF, FTFF, FFTF, and FFFT, for which exactly one answer is “True.” Therefore the probability is 4∕16 or 1∕4. (d) There are 16 equally likely outcomes. There are fve of them, TFFF, FTFF, FFTF, FFFT, and FFFF, for which at most one answer is “True.” Therefore the probability is 5∕16.

4. (a) The outcomes are the 27 di˙erent strings of 3 chosen from conforming (C), downgraded (D), and scrap (S). These are {CCC, CCD, CCS, CDC, CDD, CDS, CSC, CSD, CSS, DCC, DCD, DCS, DDC, DDD, DDS, DSC, DSD, DSS, SCC, SCD, SCS, SDC, SDD, SDS, SSC, SSD, SSS}. (b) A = {CCC, DDD, SSS} (c) B = {CDS, CSD, DCS, DSC, SCD, SDC} (d) C = {CCD, CCS, CDC, CSC, DCC, SCC, CCC} (e) The only outcome common to A and C is CCC. Therefore A ∩ C = {CCC}.

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SECTION 2.1

(f) The set A ∪ B contains the outcomes that are either in A, in B, or in both. Therefore A ∪ B = {CCC, DDD, SSS, CDS, CSD, DCS, DSC, SCD, SDC}. (g) C c contains the outcomes that are not in C. A ∩ C c contains the outcomes that are in A but not in C. Therefore A ∩ C c = {DDD, SSS}. (h) Ac contains the outcomes that are not in A. Ac ∩ C contains the outcomes that are in C but not in A. Therefore Ac ∩ C = {CCD, CCS, CDC, CSC, DCC, SCC}. (i) No. They both contain the outcome CCC. (j) Yes. They have no outcomes in common.

5. (a) The outcomes are the sequences of candidates that end with either #1 or #2. These are {1, 2, 31, 32, 41, 42, 341, 342, 431, 432}. (b) A = {1, 2} (c) B = {341, 342, 431, 432} (d) C = {31, 32, 341, 342, 431, 432} (e) D = {1, 31, 41, 341, 431} (f) A and E are mutually exclusive because they have no outcomes in commom. B and E are not mutually exclusive because they both contain the outcomes 341, 342, 431, and 432. C and E are not mutually exclusive because they both contain the outcomes 341, 342, 431, and 432. D and E are not mutually exclusive because they both contain the outcomes 41, 341, and 431.

6. (a) The equally likely outcomes are the sequences of two distinct candidates. These are {12, 13, 14, 21, 23, 24, 31, 32, 34, 41, 42, 43}. (b) Of the 12 equally likely outcomes, there are 2 (12 and 21) for which both candidates are qualifed. The probability is therefore 2∕12 or 1∕6. (c) Of the 12 equally likely outcomes, there are 8 (13, 14, 23, 24, 31, 32, 41, and 42) for which exactly one candidate is qualifed. The probability is therefore 8∕12 or 2∕3.

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CHAPTER 2

7. (a)

P (between 18 and 64) = = =

P (between 18 and 44) + P (between 45 and64) 0.19 + 0.30 0.49

(b) P (less than 85) = 1 − P (85 and up) = 1 − 0.04 = 0.96

8. (a) 0.71 (b) P (paid late) = 1 − P (paid on time) = 1 − 0.71 = 0.29

9. (a) The events of having a major faw and of having only minor faws are mutually exclusive. Therefore P (major faw or minor faw) = P (major faw) + P (only minor faws) = 0.15 + 0.05 = 0.20. (b) P (no major faw) = 1 − P (major faw) = 1 − 0.05 = 0.95.

10.

It is not possible, because we don’t know the proportion of people who have both a freplace and a garage.

11.

The proportion of people who have both one and two children is 0. Therefore the probability that a randomly chosen family has either one or two children is 0.16 + 0.21 = 0.37.

12. (a) False (b) True (c) True. This is the defnition of probability.

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SECTION 2.1

13. (a) False (b) True

14. (a)

P (V ∩ W )

= P (V ) + P (W ) − P (V ∪ W ) = 0.15 + 0.05 − 0.17 = 0.03

(b) P (V c ∩ W c ) = 1 − P (V ∪ W ) = 1 − 0.17 = 0.83. (c) We need to fnd P (V ∩ W c ). Now P (V ) = P (V ∩ W ) + P (V ∩ W c ) (this can be seen from a Venn diagram). We know that P (V ) = 0.15, and from part (a) we know that P (V ∩ W ) = 0.03. Therefore P (V ∩ W c ) = 0.12.

15. (a)

P (S ∪ C) =

P (S) + P (C) − P (S ∩ C)

0.4 + 0.3 − 0.2

0.5

(b) P (S c ∩ C c ) = 1 − P (S ∪ C) = 1 − 0.5 = 0.5. (c) We need to fnd P (S ∩ C c ). Now P (S) = P (S ∩ C) + P (S ∩ C c ) (this can be seen from a Venn diagram). Now P (S ∩ C) = P (S) + P (C) − P (S ∪ C) = 0.4 + 0.3 − 0.5 = 0.2 Since P (S) = 0.4 and P (S ∩ C) = 0.2, P (S ∩ C c ) = 0.2.

16. (a) Since 562 stones were neither cracked nor discolored, 38 stones were cracked, discolored, or both. The probability is therefore 38∕600 = 0.0633. (b) Let A be the event that the stone is cracked and let B be the event that the stone is discolored. We need to fnd P (A ∩ B). We know that P (A) = 15∕600 = 0.025 and P (B) = 27∕600 = 0.045. From part (a) we know that P (A ∪ B) = 38∕600. Now P (A ∪ B) = P (A) + P (B) − P (A ∩ B). Substituting, we fnd that 38∕600 = 15∕600 + 27∕600 − P (A ∩ B). It follows that P (A ∩ B) = 4∕600 = 0.0067.

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CHAPTER 2 (c) We need to fnd P (A ∩ B c ). Now P (A) = P (A ∩ B) + P (A ∩ B c ) (this can be seen from a Venn diagram). We know that P (A) = 15∕600 and P (A ∩ B) = 4∕600. Therefore P (A ∩ B c ) = 11∕600 = 0.0183.

17. (a) Let R be the event that a student is profcient in reading, and let M be the event that a student is profcient in mathematics. We need to fnd P (Rc ∩ M). Now P (M) = P (R ∩ M) + P (Rc ∩ M) (this can be seen from a Venn diagram). We know that P (M) = 0.78 and P (R ∩ M) = 0.65. Therefore P (Rc ∩ M) = 0.13. (b) We need to fnd P (R ∩ M c ). Now P (R) = P (R ∩ M) + P (R ∩ M c ) (this can be seen from a Venn diagram). We know that P (R) = 0.85 and P (R ∩ M) = 0.65. Therefore P (R ∩ M c ) = 0.20. (c) First we compute P (R ∪ M): P (R ∪ M) = P (R) + P (M) − P (R ∩ M) = 0.85 + 0.78 − 0.65 = 0.98. Now P (Rc ∩ M c ) = 1 − P (R ∪ M) = 1 − 0.98 = 0.02.

18.

P (A ∪ B)

= P (A) + P (B) − P (A ∩ B) = 0.95 + 0.90 − 0.88 = 0.97

17.

P (A ∩ B)

= P (A) + P (B) − P (A ∪ B) = 0.98 + 0.95 − 0.99 = 0.94

18. (a) P (O) =

1 − P (not O)

= 1 − [P (A) + P (B) + P (AB)] = 1 − [0.35 + 0.10 + 0.05] = 0.50

(b) P (does not contain B) =

1 − P (contains B)

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SECTION 2.2

= 1 − [P (B) + P (AB)] = 1 − [0.10 + 0.05] = 0.85

21. (a) False (b) True (c) False (d) True

22. (a) A and B are mutually exclusive, since it is impossible for both events to occur. (b) If bolts #5 and #8 are torqued correctly, but bolt #3 is not torqued correctly, then events B and D both occur. Therefore B and D are not mutually exclusive. (c) If bolts #5 and #8 are torqued correctly, but exactly one of the other bolts is not torqued correctly, then events C and D both occur. Therefore C and D are not mutually exclusive. (d) If the #3 bolt is the only one not torqued correctly, then events B and C both occur. Therefore B and C are not mutually exclusive.

Section 2.2 1. (a) (4)(4)(4) = 64 (b) (2)(2)(2) = 8 (c) (4)(3)(2) = 24

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CHAPTER 2

(4)(2)(3) = 24

0 1 8 8! = = 70 4 4!4!

0 4.

18 9

1 =

18! = 48,620 9!9!

5. (a) (8)(7)(6) = 336 (b)

0 1 8 8! = = 56 3 3!5!

(10)(9)(8) = 720

(210 )(45 ) = 1, 048, 576

8. (a) (263 )(103 ) = 17, 576, 000 (b) (253 )(93 ) = 11, 390, 625

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SECTION 2.3

(c)

11, 390, 625 = 0.6481 17, 576, 000

9. (a) 368 = 2.8211 × 1012 (b) 368 − 268 = 2.6123 × 1012

(c)

368 − 268 = 0.9260 368

10.

18 6, 6, 6

11.

P (match)

18! = 17, 153, 136 6!6!6!

P (BB) + P (W W )

= (8∕14)(4∕6) + (6∕14)(2∕6) = 44∕84 = 0.5238

12.

P (match)

P (RR) + P (GG) + P (BB)

= =

(6∕12)(5∕11) + (4∕12)(3∕11) + (2∕12)(1∕11) 1∕3

Section 2.3 1.

A and B are independent if P (A ∩ B) = P (A)P (B). Therefore P (B) = 0.25.

A and B are independent if P (A ∩ B) = P (A)P (B). Now P (A) = P (A ∩ B) + P (A ∩ B c ). Since P (A) = 0.5 and P (A ∩ B c ) = 0.4, P (A ∩ B) = 0.1. Therefore 0.1 = 0.5P (B), so P (B) = 0.2.

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CHAPTER 2

3. (a) 5∕15

(b) Given that the frst resistor is 50Ω, there are 14 resistors remaining of which 5 are 100Ω. P (2nd is 100Ω|1st is 50Ω) = 5∕14.

Therefore

(c) Given that the frst resistor is 100Ω, there are 14 resistors remaining of which 4 are 100Ω. Therefore P (2nd is 100Ω|1st is 100Ω) = 4∕14.

4. (a) (10∕15)(9∕14) = 3∕7

(b) P (2 resistors selected)

= P (1st is 50Ω and 2nd is 100Ω) = (10∕15)(5∕14) =

5∕21

= P (1st 3 resistors are all 50Ω) = (10∕15)(9∕14)(8∕13) = 24∕91

Given that a student is an engineering major, it is almost certain that the student took a calculus course. Therefore P (B|A) is close to 1. Given that a student took a calculus course, it is much less certain that the student is an engineering major, since many non-engineering majors take calculus. Therefore P (A|B) is much less than 1, so P (B|A) > P (A|B).

(0.056)(0.027) = 0.001512

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SECTION 2.3

Let A represent the event that the bolt meets specifcations, and let B represent the event that the nail meet specifcations. Then P (A) = 0.9 and P (B) = 0.85. (a) P (A ∩ B) = P (A)P (B) = (0.9)(0.85) = 0.765. (b) P (Ac ∩ B c ) = P (Ac )P (B c ) = (1 − 0.9)(1 − 0.85) = 0.015.

= = = =

P (A) + P (B) − P (A ∩ B) P (A) + P (B) − P (A)P (B) 0.9 + 0.85 − (0.9)(0.85) 0.985

Let M denote the event that the main parachute deploys, and let B denote the event that backup parachute deploys. Then P (M) = 0.99 and P (B | M c ) = 0.98.

(a) P (M ∪ B) =

1 − P (M c ∩ B c )

= =

1 − P (B c | M c )P (M c ) 1 − (1 − P (B| M c ))(1 − P (M))

1 − (1 − 0.98)(1 − 0.99)

0.9998

(b) The backup parachute does not deploy if the main parachute deploys. Therefore P (B) = P (B ∩ M c ) = P (B | M c )P (M c ) = (0.98)(0.01) = 0.0098

Let S denote the event that a person buys an SUV, and let B denote the event that a person buys a black SUV. Then P (B | S)

= =

P (B ∩ S) P (S) P (B) P (S)

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CHAPTER 2 0.03 0.20 = 0.15 =

10.

Let E denote the event that the student is an engineering major, and let S denote the event that the student plays club sports. Then P (E) = 0.3, P (S) = 0.2, and P (E ∩ S) = 0.1.

(a) P (E) = 0.3

(b) P (S) = 0.2

= = =

(d) P (E|S)

= = =

(e) P (S c |E)

= = = =

P (E ∩ S) P (E) 0.1 0.3 1∕3

P (E ∩ S) P (S) 0.1 0.2 1∕2

P (E ∩ S c ) P (E) P (E) − P (E ∩ S) P (E) 0.3 − 0.1 0.3 2∕3

Equivalently, one can compute P (S c |E) = 1 − P (S|E) = 1 − 1∕3 = 2∕3

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SECTION 2.3

(f) P (E c |S)

P (E c ∩ S) P (S) P (S) − P (E ∩ S) = P (S) 0.2 − 0.1 = 0.2 = 1∕2 =

Equivalently, one can compute P (E c |S) = 1 − P (E|S) = 1 − 1∕2 = 1∕2

11.

Let OK denote the event that a valve meets the specifcation, let R denote the event that a valve is reground, and let S denote the event that a valve is scrapped. Then P (OK ∩ Rc ) = 0.7, P (R) = 0.2, P (S ∩ Rc ) = 0.1, P (OK|R) = 0.9, P (S|R) = 0.1. (a) P (Rc ) = 1 − P (R) = 1 − 0.2 = 0.8

(b) P (S|Rc ) =

P (S ∩ Rc ) 0.1 = = 0.125 P (Rc ) 0.8

P (S ∩ Rc ) + P (S ∩ R)

= P (S ∩ Rc ) + P (S|R)P (R) = 0.1 + (0.1)(0.2) =

(d) P (R|S)

0.12

P (S ∩ R) P (S) P (S|R)P (R) = P (S) (0.1)(0.2) = 0.12 = 0.167 =

(e) P (OK) = P (OK ∩ Rc ) + P (OK ∩ R) =

P (OK ∩ Rc ) + P (OK|R)P (R)

= 0.7 + (0.9)(0.2) = 0.88

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P (R ∩ OK) P (OK) P (OK|R)P (R) = P (OK) (0.9)(0.2) = 0.88 = 0.205

(f) P (R|OK) =

(g) P (Rc |OK)

12.

P (Rc ∩ OK) P (OK) 0.7 = 0.88 = 0.795

Let S denote Sarah’s score, and let T denote Thomas’s score.

(a) P (S > 175 ∩ T > 175) = P (S > 175)P (T > 175) = (0.4)(0.2) = 0.08. (b) P (T > 175 ∩ S > T ) = P (T > 175)P (S > T | T > 175) = (0.2)(0.3) = 0.06.

13.

Let T 1 denote the event that the frst device is triggered, and let T 2 denote the event that the second device is triggered. Then P (T 1) = 0.9 and P (T 2) = 0.8.

(a) P (T 1 ∪ T 2) =

P (T 1) + P (T 2) − P (T 1 ∩ T 2)

= P (T 1) + P (T 2) − P (T 1)P (T 2) = 0.9 + 0.8 − (0.9)(0.8) = 0.98

(b) P (T 1c ∩ T 2c ) = P (T 1c )P (T 2c ) = (1 − 0.9)(1 − 0.8) = 0.02

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SECTION 2.3

Let L denote the event that Laura hits the target, and let P ℎ be the event that Philip hits the target. Then P (L) = 0.5 and P (P ℎ) = 0.3.

14.

(a) P (L ∪ P ℎ) = =

P (L) + P (P ℎ) − P (L ∩ P ℎ) P (L) + P (P ℎ) − P (L)P (P ℎ)

= 0.5 + 0.3 − (0.5)(0.3) = 0.65

(b) P (exactly one hit) = P (L ∩ P ℎc ) + P (Lc ∩ P ℎ) = P (L)P (P ℎc ) + P (Lc )P (P ℎ) = (0.5)(1 − 0.3) − (1 − 0.5)(0.3) = 0.5

15. (a)

P (L ∩ exactly one hit) P (exactly one hit) P (L ∩ P ℎc ) P (exactly one hit) P (L)P (P ℎc ) P (exactly one hit) (0.5)(1 − 0.3) 0.5 0.7

88 = 0.88 88 + 12

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CHAPTER 2

(b)

88 = 0.1715 88 + 165 + 260

(c)

88 + 165 = 0.4932 88 + 65 + 260

(d)

88 + 165 = 0.8433 88 + 12 + 165 + 35

513 88 + 165 + 260 = = 0.855. From Problem 15(a), P (E2 | E1 ) = 0.88. Since P (E2 | E1 ) ≠ 600 600 P (E1 ), E1 and E2 are not independent.

16.

P (E2 ) =

17. (a)

56 + 24 = 0.80 100

(b)

56 + 14 = 0.70 100

P (Gene 1 dominant ∩ Gene 2 dominant) P (Gene 1 dominant) 56∕100 0.8 0.7

(d) Yes. P (Gene 2 dominant | Gene 1 dominant) = P (Gene 2 dominant)

18. (a)

(b)

71 71 = 102 + 71 + 33 + 134 340 86 43 = 102 + 86 + 26 107

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SECTION 2.3

19.

(c)

22 11 = 26 + 32 + 22 + 40 60

(d)

22 22 = 33 + 36 + 22 91

(e)

86 + 63 + 36 + 26 + 32 + 22 53 = 86 + 63 + 36 + 105 + 26 + 32 + 22 + 40 82

Let R, D, and I denote the events that the senator is a Republian, Democrat, or Independent, respectively, and let M and F denote the events that the senator is male or female, respectively. (a) P (R ∩ M) = 0.42

(b) P (D ∪ F ) = P (D) + P (F ) − P (D ∩ F ) =

(0.32 + 0.16) + (0.16 + 0.08) − 0.16

= 0.56

P (R ∩ M) + P (R ∩ F ) 0.42 + 0.08 0.50

(d) P (Rc ) = 1 − P (R) = 1 − 0.50 = 0.50

(e) P (D) = = =

(f) P (I) = = =

P (D ∩ M) + P (D ∩ F ) 0.32 + 0.16 0.48

P (I ∩ M) + P (I ∩ F ) 0.02 + 0 0.02

(g) P (D ∪ I) = P (D) + P (I) = 0.48 + 0.02 = 0.50

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CHAPTER 2

20.

Let G denote the event that a customer is a good risk, let M denote the event that a customer is a medium risk, let P denote the event that a customer is a poor risk, and let C be the event that a customer has fled a claim. Then P (G) = 0.7, P (M) = 0.2, P (P ) = 0.1, P (C | G) = 0.005, P (C | M) = 0.01, and P (C | P ) = 0.025. (a) P (G ∩ C) = P (G)P (C | G) = (0.70)(0.005) = 0.0035

(b) P (C) = P (G ∩ C) + P (M ∩ C) + P (P ∩ C) = P (G)P (C | G) + P (M)P (C | M) + P (P )P (C | P ) = (0.7)(0.005) + (0.2)(0.01) + (0.1)(0.025) = 0.008

P (C | G)P (G) (0.005)(0.7) = = 0.4375. P (C) 0.008

21. (a) That the gauges fail independently.

(b) One cause of failure, a fre, will cause both gauges to fail. Therefore, they do not fail independently.

(c) Too low. The correct calculation would use P (second gauge fails|frst gauge fails) in place of P (second gauge fails). Because there is a chance that both gauges fail together in a fre, the condition that the frst gauge fails makes it more likely that the second gauge fails as well. Therefore P (second gauge fails|frst gauge fails) > P (second gauge fails).

22.

No. P (both gauges fail) = P (frst gauge fails)P (second gauge fails|frst gauge fails).

Since P (second gauge fails|frst gauge fails) ≤ 1, P (both gauges fail) ≤ P (frst gauge fails) = 0.01.

23. (a) P (A) = 3∕10

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SECTION 2.3

(b) Given that A occurs, there are 9 components remaining, of which 2 are defective. Therefore P (B|A) = 2∕9. (c) P (A ∩ B) = P (A)P (B|A) = (3∕10)(2∕9) = 1∕15 (d) Given that Ac occurs, there are 9 components remaining, of which 3 are defective. Therefore P (B|A) = 3∕9. Now P (Ac ∩ B) = P (Ac )P (B|Ac ) = (7∕10)(3∕9) = 7∕30. (e) P (B) = P (A ∩ B) + P (Ac ∩ B) = 1∕15 + 7∕30 = 3∕10 (f) No. P (B) ≠ P (B|A) [or equivalently, P (A ∩ B) ≠ P (A)P (B)].

24. (a) P (A) = 300∕1000 = 3∕10 (b) Given that A occurs, there are 999 components remaining, of which 299 are defective. Therefore P (B|A) = 299∕999. (c) P (A ∩ B) = P (A)P (B|A) = (3∕10)(299∕999) = 299∕3330 (d) Given that Ac occurs, there are 999 components remaining, of which 300 are defective. Therefore P (B|A) = 300∕999. Now P (Ac ∩ B) = P (Ac )P (B|Ac ) = (7∕10)(300∕999) = 70∕333. (e) P (B) = P (A ∩ B) + P (Ac ∩ B) = 299∕3330 + 70∕333 = 3∕10

(f) P (A|B) =

P (A ∩ B) 299∕3330 299 = = P (B) 3∕10 999

(g) A and B are not independent, but they are very nearly independent. To see this note that P (B) = 0.3, while P (B|A) = 0.2993. So P (B) is very nearly equal to P (B|A), but not exactly equal. Alternatively, note that P (A ∩ B) = 0.0898, while P (A)P (B) = 0.09. Therefore in most situations it would be reasonable to treat A and B as though they were independent.

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CHAPTER 2

25.

n = 10, 000. The two components are a simple random sample from the population. When the population is large, the items in a simple random sample are nearly independent.

26.

Let E denote the event that a parcel is sent express (so E c denotes the event that a parcel is sent standard), and let N denote the event that a parcel arrives the next day. Then P (E) = 0.25, P (N|E) = 0.95, and P (N|E c ) = 0.80). (a) P (E ∩ N) = P (E)P (N|E) = (0.25)(0.95) = 0.2375.

(b) P (N) = P (N|E)P (E) + P (N|E c )P (E c ) =

(0.95)(0.25) + (0.80)(1 − 0.25)

= 0.8375

P (N|E)P (E) P (N|E)P (E) + P (N|E c )P (E c ) (0.95)(0.25) = (0.95)(0.25) + (0.80)(1 − 0.25) = 0.2836

27.

Let R denote the event of a rainy day, and let C denote the event that the forecast is correct. Then P (R) = 0.1, P (C|R) = 0.8, and P (C|Rc ) = 0.9.

(a) P (C) = P (C|R)P (R) + P (C|Rc )P (Rc ) =

(0.8)(0.1) + (0.9)(1 − 0.1)

= 0.89 (b) A forecast of no rain will be correct on every non-rainy day. Therefore the probability is 0.9.

28.

Let A denote the event that the faw is found by the frst inspector, and let B denote the event that the faw is found by the second inspector. (a) P (A ∩ B) = P (A)P (B) = (0.9)(0.7) = 0.63 (b) P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.9 + 0.7 − 0.63 = 0.97

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SECTION 2.3 (c) P (Ac ∩ B) = P (Ac )P (B) = (1 − 0.9)(0.7) = 0.07

29.

Let F denote the event that an item has a faw. Let A denote the event that a faw is detected by the frst inspector, and let B denote the event that the faw is detected by the second inspector.

(a) P (F |Ac ) =

P (Ac |F )P (F ) P (Ac |F )P (F ) + P (Ac |F c )P (F c )

(0.1)(0.1) (0.1)(0.1) + (1)(0.9) = 0.011 =

(b) P (F |Ac ∩ B c )

30.

Let D denote the event that a person has the disease, and let + denote the event that the test is positive. Then P (D) = 0.05, P (+|D) = 0.99, and P (+|Dc ) = 0.01. P (+|D)P (D) P (+|D)P (D) + P (+|Dc )P (Dc ) (0.99)(0.05) = (0.99)(0.05) + (0.01)(0.95) = 0.8390

(a) P (D|+) =

(b) P (Dc |−)

P (−|Dc )P (Dc ) P (−|Dc )P (Dc ) + P (−|D)P (D)

(0.99)(0.95) (0.99)(0.95) + (0.01)(0.05) = 0.9995

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CHAPTER 2

31. (a) Each child has probability 0.25 of having the disease. Since the children are independent, the probability that both are disease-free is 0.752 = 0.5625. (b) Each child has probability 0.5 of being a carrier. Since the children are independent, the probability that both are carriers is 0.52 = 0.25. (c) Let D denote the event that a child has the disease, and let C denote the event that the child is a carrier. Then P (D) = 0.25, P (C) = 0.5, and P (Dc ) = 0.75. We frst compute P (C | Dc ), the probability that a child who does not have the disease is a carrier. First, P (C ∩ Dc ) = P (C) = 0.5. Now P (C | Dc ) =

P (C ∩ Dc ) 0.5 2 = = P (Dc ) 0.75 3

Since children are independent, the probability that both children are carriers given that neither has the disease is (2∕3)2 = 4∕9. (d) Let WD denote the event that the woman has the disease, Let WC denote the event that the woman is a carrier, and let WF denote the event that the woman does not have the disease and is not a carrier. Then P (WD ) = 0.25, P (WC ) = 0.5, and P (WF ) = 0.25, Let CD denote the event that the child has the disease. P (CD )

= P (CD ∩ WD ) + P (CD ∩ WC ) + P (CD ∩ WF ) = P (CD | WD )P (WD ) + P (CD | WC )P (WC ) + P (CD | WF )P (WF ) = (0.5)(0.25) + (0.25)(0.5) + (0)(0.25) = 0.25

32.

(a) P (F l|F ) =

(b) i. Given that a bottle failed inspection, the probability that it had a faw is only 0.01952.

P (F c |F lc )P (F lc ) P (F c |F lc )P (F lc ) + P (F c |F l)P (F l)

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SECTION 2.3

P (F c |F lc )P (F lc ) P (F c |F lc )P (F lc ) + [1 − P (F |F l)]P (F l)

(0.99)(0.9998) (0.99)(0.9998) + (1 − 0.995)(0.0002) = 0.999999

(d) ii. Given that a bottle passes inspection, the probability that is has no faw is 0.999999.

(e) The small probability in part (a) indicates that some good bottles will be scrapped. This is not so serious. The important thing is that of the bottles that pass inspection, very few should have faws. The large probability in part (c) indicates that this is the case.

33.

Let D represent the event that the man actually has the disease, and let + represent the event that the test gives a positive signal. We are given that P (D) = 0.005, P (+|D) = 0.99, and P (+|Dc ) = 0.01. It follows that P (Dc ) = 0.995, P (−|D) = 0.01, and P (−|Dc ) = 0.99. (a) P (+ + |D) = 0.992 = 0.9801 (b) P (+ + |Dc ) = 0.012 = 0.0001

34.

P (+ + |D)P (D) P (+ + |D)P (D) + P (+ + |Dc )P (Dc ) (0.9801)(0.005) = (0.9801)(0.005) + (0.0001)(0.995) = 0.8312

P (system functions) = P [(A ∩ B) ∪ (C ∪ D)]. Now P (A ∩ B) = P (A)P (B) = (1 − 0.10)(1 − 0.05) = 0.855, and P (C ∪ D) = P (C) + P (D) − P (C ∩ D) = (1 − 0.10) + (1 − 0.20) − (1 − 0.10)(1 − 0.20) = 0.98. Therefore P [(A ∩ B) ∪ (C ∪ D)] = =

P (A ∩ B) + P (C ∪ D) − P [(A ∩ B) ∩ (C ∪ D)] P (A ∩ B) + P (C ∪ D) − P (A ∩ B)P (C ∪ D)

= =

0.855 + 0.98 − (0.855)(0.98) 0.9971

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35.

P (system functions) = P [(A ∩ B) ∩ (C ∪ D)]. Now P (A ∩ B) = P (A)P (B) = (1 − 0.05)(1 − 0.03) = 0.9215, and P (C ∪ D) = P (C) + P (D) − P (C ∩ D) = (1 − 0.07) + (1 − 0.14) − (1 − 0.07)(1 − 0.14) = 0.9902. Therefore P [(A ∩ B) ∩ (C ∪ D)] = = =

P (A ∩ B)P (C ∪ D) (0.9215)(0.9902) 0.9125

36. (a) P (A ∩ B) = P (A)P (B) = (1 − 0.05)(1 − 0.03) = 0.9215 (b) P (A ∩ B) = (1 − p)2 = 0.9. Therefore p = 1 −

√

0.9 = 0.0513.

37.

Let C denote the event that component C functions, and let D denote the event that component D functions.

(a) P (system functions)

= P (C ∪ D) = P (C) + P (D) − P (C ∩ D) = (1 − 0.08) + (1 − 0.12) − (1 − 0.08)(1 − 0.12) = 0.9904

Alternatively, P (system functions) =

1 − P (system fails) = 1 − P (C c ∩ Dc ) = 1 − P (C c )P (Dc ) = 1 − (0.08)(0.12) = 0.9904

(b) P (system functions) = 1 − P (C c ∩ Dc ) = 1 − p2 = 0.99. Therefore p =

√

1 − 0.99 = 0.1.

(c) P (system functions) = 1 − p3 = 0.99. Therefore p = (1 − 0.99)1∕3 = 0.2154. (d) Let n be the required number of components. Then n is the smallest integer such that 1 − 0.5n ≥ 0.99. It follows that n ln(0.5) ≤ ln 0.01, so n ≥ (ln 0.01)(ln 0.5) = 6.64. Since n must be an integer, n = 7.

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SECTION 2.4

38.

To show that Ac and B are independent, we show that P (Ac ∩ B) = P (Ac )P (B). Now B = (Ac ∩ B) ∪ (A ∩ B), and (Ac ∩ B) and (A ∩ B) are mutually exclusive. Therefore P (B) = P (Ac ∩ B) + P (A ∩ B), from which it follows that P (Ac ∩ B) = P (B) − P (A ∩ B). Since A and B are independent, P (A ∩ B) = P (A)P (B). Therefore P (Ac ∩ B) = P (B) − P (A)P (B) = P (B)[1 − P (A)] = P (Ac )P (B). To show that A and B c are independent, it suÿces to interchange A and B in the argument above. To show that Ac and B c are independent, replace B with B c in the argument above, and use the fact that A and B c are independent.

39. (a) P (A) = 1∕2, P (B) = 1∕2, P (A ∩ B) = 1∕4 = P (A)P (B), so A and B are independent. P (C) = 1∕6, P (A ∩ C) = 1∕12 = P (A)P (C), so A and C are independent. P (B ∩ C) = 1∕12 = P (B)P (C), so B and C are independent. (b) P (A ∩ B ∩ C) = 0 ≠ P (A)P (B)P (C), so A, B, and C are not independent.

Section 2.4 1. (a) Discrete (b) Continuous (c) Discrete (d) Continuous (e) Discrete

2. (a) P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2) = 0.4 + 0.3 + 0.15 = 0.85. (b) P (X > 1) = P (X = 2) + P (X = 3) + P (X = 4) = 0.15 + 0.1 + 0.05 = 0.3. (c) X = 0(0.4) + 1(0.3) + 2(0.15) + 3(0.1) + 4(0.05) = 1.1 2 = (0 − 1.1)2 (0.4) + (1 − 1.1)2 (0.3) + (2 − 1.1)2 (0.15) + (3 − 1.1)2 (0.1) + (4 − 1.1)2 (0.05) = 1.39 (d) X 2 = 02 (0.4) + 12 (0.3) + 22 (0.15) + 32 (0.1) + 42 (0.05) − 1.12 = 1.39 Alternatively, X

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CHAPTER 2

3. (a) X = 1(0.4) + 2(0.2) + 3(0.2) + 4(0.1) + 5(0.1) = 2.3 2 = (1 − 2.3)2 (0.4) + (2 − 2.3)2 (0.2) + (3 − 2.3)2 (0.2) + (4 − 2.3)2 (0.1) + (5 − 2.3)2 (0.1) = 1.81 (b) X 2 = 12 (0.4) + 22 (0.2) + 32 (0.2) + 42 (0.1) + 52 (0.1) − 2.32 = 1.81 Alternatively, X

√ 1.81 = 1.345

(d) Y = 10X. Therefore the probability density function is as follows. y p(y)

10 0.4

20 0.2

30 0.2

40 0.1

50 0.1

(e) Y = 10(0.4) + 20(0.2) + 30(0.2) + 40(0.1) + 50(0.1) = 23 (f) Y2 = (10 − 23)2 (0.4) + (20 − 23)2 (0.2) + (30 − 23)2 (0.2) + (40 − 23)2 (0.1) + (50 − 23)2 (0.1) = 181 Alternatively, Y2 = 102 (0.4) + 202 (0.2) + 302 (0.2) + 402 (0.1) + 502 (0.1) − 232 = 181

(g) Y =

√ 181 = 13.45

4. (a) p3 (x) is the only probability mass function, because it is the only one whose probabilities sum to 1. (b) X = 0(0.1) + 1(0.2) + 2(0.4) + 3(0.2) + 4(0.1) = 2.0, 2 = (0 − 2)2 (0.1) + (1 − 2)2 (0.2) + (2 − 2)2 (0.4) + (3 − 2)2 (0.2) + (4 − 2)2 (0.1) = 1.2 X

5. (a)

x p(x)

1 0.70

2 0.15

3 0.10

4 0.03

5 0.02

(b) P (X ≤ 2) = P (X = 1) + P (X = 2) = 0.70 + 0.15 = 0.85 (c) P (X > 3) = P (X = 4) + P (X = 5) = 0.03 + 0.02 = 0.05 (d) X = 1(0.70) + 2(0.15) + 3(0.10) + 4(0.03) + 5(0.02) = 1.52

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SECTION 2.4

(e) X =

√

12 (0.70) + 22 (0.15) + 32 (0.10) + 42 (0.03) + 52 (0.02) − 1.522 = 0.9325

6. (a) X = 24(0.0825) + 25(0.0744) + 26(0.7372) + 27(0.0541) + 28(0.0518) = 25.9183 2 = (24−25.9183)2 (0.0825)+(25−25.9183)2 (0.0744)+(26−25.9183)2 (0.7372)+(27−25.9183)2 (0.0541)+ (b) X (28 − 25.9183)2 (0.0518) = 0.659025 2 = 242 (0.0825) + 252 (0.0744) + 262 (0.7372) + 272 (0.0541) + 282 (0.0518) − 25.91832 = Alternatively, X 0.659025 √ X = 0.659025 = 0.8118.

7. (a)

∑5

x=1 cx = 1, so c(1 + 2 + 3 + 4 + 5) = 1, so c = 1∕15.

(b) P (X = 2) = c(2) = 2∕15 = 0.2

∑5

x=1 xP (X = x) =

∑5

2 2 2 2 2 2 x=1 x ∕15 = (1 + 2 + 3 + 4 + 5 )∕15 = 11∕3

2 = ∑5 (x − )2 P (X = x) = ∑5 2 (d) X X x=1 x=1 x(x − 11∕3) ∕15 = (64∕135) + 2(25∕135) + 3(4∕135) + 4(1∕135) + 5(16∕135) = 14∕9 ∑5 2 = ∑5 2 2 3 2 3 3 3 3 3 2 Alternatively, X x=1 x P (X = x)− X = x=1 x ∕15−(11∕3) = (1 +2 +3 +4 +5 )∕15−(11∕3) = 14∕9

(e) X =

√

14∕9 = 1.2472

8. (a) P (X ≤ 2) = F (2) = 0.83 (b) P (X > 3) = 1 − P (X ≤ 3) = 1 − F (3) = 1 − 0.95 = 0.05 (c) P (X = 1) = P (X ≤ 1) − P (X ≤ 0) = F (1) − F (0) = 0.72 − 0.41 = 0.31 (d) P (X = 0) = P (X ≤ 0) = F (0) = 0.41

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CHAPTER 2

(e) For any integer x, P (X = x) = F (x) − F (x − 1) [note that F (−1) = 0]. The value of x for which this quantity is greatest is x = 0.

x 0 1 9. (a) 2 3 4 5

p1 (x) 0.2 0.16 0.128 0.1024 0.0819 0.0655

x 0 1 (b) 2 3 4 5

p2 (x) 0.4 0.24 0.144 0.0864 0.0518 0.0311

(c) p2 (x) appears to be the better model. Its probabilities are all fairly close to the proportions of days observed in the data. In contrast, the probabilities of 0 and 1 for p1 (x) are much smaller than the observed proportions. (d) No, this is not right. The data are a simple random sample, and the model represents the population. Simple random samples generally do not refect the population exactly.

10.

Let Q denote a qualifed candidate, and U an unqualifed one. (a) P (Q) = 0.3

(b) P (U Q) = P (U )P (Q) = (0.7)(0.3) = 0.21

(0.3)(0.7)x−1 0

x = 1, 2, 3, ... otherwise

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SECTION 2.4

11.

Let Q denote a qualifed candidate, and U an unqualifed one. (a) If the frst two candidates are both qualifed, then Y = 2. This is the smallest possible value. (b) P (Y = 2) = P (QQ) = (0.3)2 = 0.09 P (Y = 4 and X = 1) . P (X = 1) Now P (Y = 4 and X = 1) = P (QU U Q) = (0.3)(0.7)(0.7)(0.3) = 0.0441, and P (X = 1) = P (Q) = 0.3.

Therefore P (Y = 4|X = 1) = 0.0441∕0.3 = 0.147. P (Y = 4 and X = 2) . P (X = 2) Now P (Y = 4 and X = 2) = P (U QU Q) = (0.7)(0.3)(0.7)(0.3) = 0.0441, and

(d) P (Y = 4|X = 2) =

P (X = 2) = P (U Q) = (0.7)(0.3) = 0.21. Therefore P (Y = 3|X = 2) = 0.0441∕0.21 = 0.21.

(e) P (Y = 4) = =

P (QU U Q) + P (U QU Q) + P (U U QQ) (0.3)(0.7)(0.7)(0.3) + (0.7)(0.3)(0.7)(0.3) + (0.7)(0.7)(0.3)(0.3)

= 0.0441 + 0; 0441 + 0.0441 = 0.1323

12. (a) 0, 1, 2, 3 (b) P (X = 3) = P (SSS) = (0.8)3 = 0.512 (c) P (F SS) = (0.2)(0.8)2 = 0.128 (d) P (SF S) = P (SSF ) = (0.8)2 (0.2) = 0.128

(e) P (X = 2) = P (F SS) + P (SF S) + P (SSF ) = 0.384

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CHAPTER 2

(f) P (X = 1) =

P (SF F ) + P (F SF ) + P (F F S)

= (0.8)(0.2)2 + (0.8)(0.2)2 + (0.8)(0.2)2 = 0.096

(g) P (X = 0) = P (F F F ) = (0.2)3 = 0.008

(h) X = 0(0.08) + 1(0.096) + 2(0.384) + 3(0.512) = 2.4 2 = (0 − 2.4)2 (0.08) + (1 − 2.4)2 (0.096) + (2 − 2.4)2 (0.384) + (3 − 2.4)2 (0.512) = 0.48 (i) X 2 = 02 (0.08) + 12 (0.096) + 22 (0.384) + 32 (0.512) − 2.42 = 0.48 Alternatively, X

(j) P (Y = 3) = P (SSSF ) + P (SSF S) + P (SF SS) + P (F SSS) = (0.8)3 (0.2) + (0.8)3 (0.2) + (0.8)3 (0.2) + (0.8)3 (0.2) = 0.4096

13. (a)

Ê80

x − 80 x2 − 160x dx = 800 1600

120

(b)

Ê80

2 = (c) X Ê

X =

= 0.0625 80

x − 80 x3 − 120x dx = 800 2400

120

√

= 320∕3 = 106.67 80

x3 x4 dx − (320∕3) = − 800 3200 30

2 x − 80

x 80

120

− (320∕3)2 = 800∕9

800∕9 = 9.428 x

(d) F (x) =

Ê−∞

f (t) dt 80

If x < 80, F (x) =

Ê−∞

0 dt = 0

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SECTION 2.4 If 80 ≤ x < 120, F (x) = If x ≥ 120, F (x) =

14. (a)

Ê25

(b) =

(d) X =

30 20

Ê80

120

0 dt +

Ê80

t − 80 dt = x2 ∕1600 − x∕10 + 4. 800 x

t − 80 dt + 0 dt = 1. Ê120 800

= 0.55 25 30

x x3 x dx = 250 750

Ê20

2 = (c) X Ê

Ê−∞

0 dt +

Ê−∞

x x2 dx = 250 500

= 76∕3 = 25.33 20

x x4 x dx − (76∕3)2 = 250 1000

− (76∕3)2 = 74∕9

t √ 2 = X 74∕9 = 2.867 x

(e) F (x) =

Ê−∞

f (t) dt 20

If x < 20, F (x) =

Ê−∞

0 dt = 0

If 20 ≤ x < 30, F (x) = If x ≥ 30, F (x) =

(f)

Ê28

15. (a)

Ê0

∞

Ê−∞

0 dt +

Ê−∞

x x2 dx = 250 500

0 dt +

Ê20

t dt = x2 ∕500 − 4∕5. Ê20 250 x

t dt + 0 dt = 1. Ê30 250

= 0.232 28

0.1te−0.1t dt ∞

= −te−0.1t

− 0

Ê0

∞

−e−0.1t dt

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CHAPTER 2 ∞

0 − 10e−0.1t

(b) 2

∞

Ê0

0.1t2 e−0.1t dt − 2 ∞

2 −0.1t

= −t e

− 0 ∞

Ê0

∞

−2te−0.1t dt − 100

0.1te−0.1t dt − 100 Ê0 0 + 20(10) − 100

100

= 0 + 20

X =

√ 100 = 10

Ê−∞

f (t) dt.

If x ≤ 0, F (x) =

Ê−∞

0 dt = 0.

If x > 0, F (x) =

Ê−∞

0 dt +

Ê0

0.1e−0.1t dt = 1 − e−0.1x .

(d) Let T represent the lifetime. P (T < 12) = P (T ≤ 12) = F (12) = 1 − e−1.2 = 0.6988.

10.25

16. (a) =

Ê9.75

10.25

3x[1 − 16(x − 10)2 ] dx = −12x4 + 320x3 − 2398.5x2

= 10 9.75

10.25

Ê9.75 √ = 0.0125 = 0.1118

(b) 2 =

3(x − 10)2 [1 − 16(x − 10)2 ] dx = (x − 10)3 − 9.6(x − 10)5

= 0.0125. 9.75

Ê−∞

f (t) dt.

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SECTION 2.4 If x ≤ 9.75, F (x) =

Ê−∞ If 9.75 < x < 10.25, x

F (x) =

Ê−∞

0 dt = 0. x

0 dt +

Ê9.75

3[1 − 16(t − 10)2 ] dt = 3t − 16(t − 10)3

= 3x − 16(x − 10)3 − 29.5

If x ≥ 10.25, F (x) = 1.

9.75

(d) None of them. F (9.75) = 0.

(e) All of them. F (10.25) = 1, so all of the rings have diameters less than or equal to 10.25 cm. Since none of the rings have diameters less than 9.75 cm, all of them have diameters between 9.75 and 10.25 cm.

17.

With this process, the probability that a ring meets the specifcation is 10.1

Ê9.9

15[1 − 25(x − 10.05)2 ]∕4 dx =

0.05

Ê−0.15

0.05

15[1 − 25x2 ]∕4 dx = 0.25(15x − 125x3 )

= 0.641. −0.15

With the process in Exercise 16, the probability is 10.1

Ê9.9

3[1 − 16(x − 10) ] dx =

0.1

Ê−0.1

3[1 − 16x ] dx = 3x − 16x

= 0.568. −0.1

Therefore this process is better than the one in Exercise 16.

18. (a) P (X > 2) =

Ê2

∞

(b) P (2 < X < 4) =

(d) 2

Ê0

∞

64 16 dx = − (x + 2)5 (x + 2)4

Ê2

∞

= 1∕16 2

64 16 dx = − 5 (x + 2) (x + 2)4

= 65∕1296 2

∞ ∞ 64 64 1 1 dx = (u − 2) du = 64 (u−4 − 2u−5 ) du = 64 − u−3 + u−4 Ê2 Ê2 3 2 (x + 2)5 u5

Ê0

∞

= 2∕3 2

64 dx − 2 (x + 2)5

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CHAPTER 2 Ê2

∞

(u − 2)2

64 du − (2∕3)2 u5

∞

(u−3 − 4u−4 + 4u−5 ) du − 4∕9 Ê2 ∞ 1 −2 4 −3 −4 64 − u + u − u − 4∕9 2 3

8∕9

Ê−∞

(e) F (x) =

f (t) dt. x

If x < 0, F (x) = If x ≥ 0, F (x) =

Ê−∞

0 dt = 0.

Ê0

64 16 dt = − (t + 2)5 (t + 2)4

16 (x + 2)4

=1− 0

(f) The median xm solves F (xm ) = 0.5. Therefore 1 −

16 = 0.5, so xm = 0.3784. (xm + 2)4

(g) The 60th percentile x60 solves F (x60 ) = 0.6. Therefore 1 −

0 3 1 x 3x4 − (3∕64)x (4 − x) dx = 19. (a) P (X > 3) = Ê3 16 256 4

= 67∕256 3

0 3 1 x 3x4 − (3∕64)x (4 − x) dx = (b) P (2 < X < 3) = Ê2 16 256 3

Ê0

(d) 2 =

Ê0

0 (3∕64)x3 (4 − x) dx =

16 = 0.6, so x60 = 0.5149. (x60 + 2)4

3x4 3x5 − 64 320 0

(3∕64)x4 (4 − x) dx − 2 =

= 109∕256 2

= 2.4 0

3x5 x6 − 80 128

− 2.42 = 0.64 0

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SECTION 2.4 x

(e) F (x) =

Ê−∞

f (t) dt

If x ≤ 0, F (x) =

Ê−∞

If 0 < x < 4, F (x) = If x ≥ 3, F (x) =

20. (a) P (X < 0.02) =

(b) =

Ê0

0.04

Ê0

0 dt = 0

Ê0

(3∕64)t2 (4 − t) dt = (16x3 − 3x4 )∕256

(3∕64)t2 (4 − t) dt = 1

0.02

625x2 625x dx = 2

0.08

= 0.125 0

625x3 (50x − 625x ) dx = 625x dx + Ê0.04 3 2

0.04

625x3 + 25x − 3

0.08

= 0.04 0.04

0.04

Ê0

625x4 4

625x3 dx + 0

0.04

+ 0

0.08

Ê0.04

(50x2 − 625x3 ) dx − 2

50x3 625x4 − 3 4

0.08

− 0.042 0.04

= 0.0002667 The standard deviation is =

√

0.0002667 = 0.01633.

(d) F (x) =

Ê−∞

f (t) dt

If x ≤ 0, F (x) =

Ê−∞

0 dt = 0

If 0 < x ≤ 0.04, F (x) =

Ê0

If 0.04 < x ≤ 0.08, F (x) = If x > 0.08, F (x) =

Ê0

0.04

625t dt = 625x2 ∕2 Ê0

0.04

625t dt +

0.08

Ê0.04

(50 − 625t) dt = 50x −

625 2 x −1 2

(50 − 625t) dt = 1

(e) The median xm solves F (xm ) = 0.5. Since F (x) is described by di˙erent expressions for x ≤ 0.04 and x > 0.04, we compute F (0.04). F (0.04) = 625(0.04)2 ∕2 = 0.5 so xm = 0.04.

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CHAPTER 2 0.04

(f) P (0.015 < X < 0.063) =

Ê0.015

0.063

625x dx +

Ê0.04

625x2 (50 − 625x) dx = 2

0.9097

21. (a) P (X < 0.2) =

(b) =

Ê0

0.04

625x2 + 50x − 2

0.063

= 0.04

0.2

12(x − x ) dx = 4x − 3x

= 0.0272 0

12 12x(x − x ) dx = 3x − x5 5 2

= 0.6 0

Ê0

12x2 (x2 + x3 ) dx − 2

12 5 x + 2x6 5

− 0.62 0

= 0.04

(d) F (x) =

Ê−∞

f (t) dt

If x ≤ 0, F (x) =

Ê−∞

If 0 < x ≤ 1, F (x) = If x > 1, F (x) =

Ê0

(e) P (0 < X < 0.8) =

22. (a) P (X > 0.5) =

0 dt = 0

Ê0

12(t2 − t3 ) dt = 4t3 − 3t4

= 4x3 − 3x4 0

12(t2 − t3 ) dt = 1

Ê0

0.8

�

12(x − x ) dx = 4x − 3x

2e−2x dx = Ê0.5 1 − e−2

0.8

= 0.81922 0

1 1 − e−2

(−e−2x )

= 0.2689 0.5

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SECTION 2.4

(b)

= = = =

2xe−2x dx Ê0 1 − e−2 0 2 1 1 e 2xe−2x dx 2 e − 1 Ê0 0 1 2 e2 ue−u du 2e2 − 2 Ê0 I 0 1H 2 2 e2 −u −u −ue e du + Ê0 2e2 − 2 0 0 1H 2I e2 −2 −u −2e − e 2e2 − 2 0

e2 − 3 2e2 − 2 = 0.34348 =

P ( − 0.1 < X < + 0.1)

2e−2x dx Ê0.24348 1 − e−2

−e−2x 1 − e−2 0.24348

0.44348

e−0.48696 − e−0.88696 1 − e−2 = 0.23429 =

(d) The variance is 2

= = = =

2x2 e−2x dx − 2 1 − e−2 1 1 e2 2x2 e−2x dx − 2 e 2 − 1 Ê0 0 1 2 e2 u2 e−u du − 2 4e2 − 4 Ê0 I 0 0 1H 12 2 2 e2 e2 − 3 2 −u −u + −u e 2ue du − Ê0 4e2 − 4 2e2 − 2 0 H IM 0 0 1L 12 2 2 e2 e2 − 3 −2 −u −u e du + − −4e + 2 −ue Ê0 4e2 − 4 2e2 − 2 0 Ê0 0

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CHAPTER 2 0

2 IM

e2 − 3 − = −4e + 2 −2e − e 2e2 − 2 0 0 1 0 1 2 � e2 e2 − 3 = −4e−2 + 2 1 − 3e−2 − 4e2 − 4 2e2 − 2 0 1 2 e2 − 5 e2 − 3 = − 2e2 − 2 2e2 − 2 = 0.0689845 √ The standard deviation is = 0.0689845 = 0.26265. e2 4e2 − 4

−2

−u

(e) Let X denote the concentration. The mean is X = 0.34348. The standard deviation is = 0.26265. 0.60613

P ( − < X < + )

2e−2x dx 1 − e−2

Ê0.08083

−e−2x 1 − e−2 0.08083

0.60613

e−0.16167 − e−1.21226 1 − e−2 = 0.63979

(f) F (x) =

Ê−∞

f (t) dt

If x ≤ 0, F (x) =

Ê−∞

If 0 < x < 1, F (x) = If x ≥ 1, F (x) =

23. (a) P (X < 2.5) =

Ê0

Ê2

0 dt = 0

Ê0

2e−2t 1 − e−2x dt = 2 1−e 1 − e−2

e−2t dt = 1 1 − e−2

2.5

(3∕52)x(6 − x) dx = (9x − x )∕52

3.5

(b) P (2.5 < X < 3.5) =

= 0.2428 2

Ê2.5

(3∕52)x(6 − x) dx =

9x2 − x3 52

3.5

= 0.5144 2.5

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SECTION 2.4 4

24x3 − 3x4 (3∕52)x (6 − x) dx = (c) = Ê2 208

=3 2

(d) The variance is 2

Ê2

9x4 3x5 − 104 260

(3∕52)x3 (6 − x) dx − 2 4

− 32 2

= 0.3230769 The standard deviation is =

√

0.3230769 = 0.5684.

(e) Let X represent the thickness. Then X is within ± of the mean if 2.4316 < X < 3.5684. 3.5684

P (2.4316 < X < 3.5684) =

Ê2.4316

(3∕52)x(6 − x) dx =

9x2 − x3 52

3.5684

= 0.5832 2.4316

(f) F (x) =

Ê−∞

f (t) dt

If x ≤ 2, F (x) =

Ê−∞

0 dt = 0. 2

If 2 < x < 4, F (x) = If x ≥ 4, F (x) =

Ê−∞

24. (a) c solves the equation

(b) X =

Ê1

∞

Ê−∞

0 dt +

Ê1

cx∕x dx =

∞

Ê2

(3∕52)t(6 − t) dt =

(3∕52)t(6 − t) dt +

Ê4

9x2 − x3 − 28 . 52

0 dt = 1.

∞

c∕x3 dx = 1. Therefore −0.5c∕x2

= 1, so c = 2. 1

Ê1

∞

2 2∕x dx = − x

∞

=2 1

Ê−∞

f (t) dt x

If x < 1, F (x) =

Ê−∞

0 dt = 0.

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CHAPTER 2 If x ≥ 1, F (x) =

Ê−∞

0 dt +

Ê1

2∕t3 dt = −

1 = 1 − 1∕x2 . t2 1

(d) The median xm solves F (xm ) = 0.5. Therefore 1 − 1∕x2m = 0.5, so xm = 1.414. (e) P (X ≤ 10) = F (10) = 1 − 1∕102 = 0.99 (f) P (X ≤ 2.5) = F (2.5) = 1 − 1∕2.52 = 0.84

P (X ≤ 2.5 and X ≤ 10) P (X ≤ 2.5) 0.84 = = = 0.85 P (X ≤ 10) P (X ≤ 10) 0.99

(g) P (X ≤ 2.5|X ≤ 10) =

25. (a) P (X < 2) =

Ê0

−x

dx =

−xe

−x

+ 0

(b) P (1.5 < X < 3) =

Ê1.5

Ê0

−x

dx =

−x

−2e

−2

−e

−x

= 1 − 3e−2 = 0.5940 0

−xe

−x

+ 1.5

Ê1.5

−x

H =

−3e

−3

+ 1.5e

−1.5

−e

−x 1.5

= 2.5e−1.5 − 4e−3 = 0.3587 (c) =

Ê0

∞

x2 e−x dx = −x2 e−x

+ 0

Ê0

∞

2xe−x dx = 0 + 2xe−x

=2 0

Ê−∞

(d) F (x) =

f (t) dt x

If x < 0, F (x) = If x > 0, F (x) =

Ê−∞ Ê0

0 dt = 0

te−t dt = 1 − (x + 1)e−x

12.5

26. (a) P (X < 12.5) =

Ê12

6(x − 12)(13 − x) dx = −2x + 75x − 936x

(b) =

6x(x − 12)(13 − x) dx = −

= 0.5 12

3x4 + 50x3 − 468x2 2

= 12.5 12

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SECTION 2.5 13

6x5 75x4 6x (x − 12)(13 − x) dx − = − + − 312x3 − 12.52 = 0.05 Ê12 5 2 12 √ The standard deviation is = 0.05 = 0.2236. 2

(d) F (x) =

Ê−∞

f (t) dt

If x ≤ 12, F (x) =

Ê−∞

If 12 < x < 13, F (x) = If x ≥ 13, F (x) =

Ê12

0 dt = 0 x

Ê12

6(t − 12)(13 − t) dt = −2x3 + 75x2 − 936x + 3888

6(t − 12)(13 − t) dt = 1

12.7

(e) P (12.3 < X < 12.7) =

Ê12.3

12.7

6(x − 12)(13 − x) dx = −2x3 + 75x2 − 936x

= 0.568 12.3

Section 2.5 1. (a) 3X = 3 X = 3(9.5) = 28.5 3X = 3 X = 3(0.4) = 1.2 (b) Y −X = Y − X = 6.8 − 9.5 = −2.7 t √ 2 = 0.12 + 0.42 = 0.412 Y −X = Y2 + X (c) X+4Y = X + 4 Y = 9.5 + 4(6.8) = 36.7 t √ 2 + 42 2 = X+4Y = X 0.42 + 16(0.12 ) = 0.566 Y

2. (a) Let H denote the height. Then V = 10H = 10 H = 10(5) = 50 (b) V = 10H = 10 H = 10(0.1) = 1

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70 3.

CHAPTER 2

Let X1 , ..., X4 be the lifetimes of the four transistors. Let S = X1 + ⋯ + X4 be the total lifetime. ∑ S = Xi = 4(900) = 3600 t∑ √ 2 = S = X 4(302 ) = 60 i

4. (a) V = V1 + V2 = 12 + 6 = 18

(b) V =

Let X1 , ..., X5 be the thicknesses of the fve layers. Let S = X1 + ⋯ + X5 be the total thickness. (a) S =

(b) S =

t √ V2 + V2 = 12 + 0.52 = 1.118

∑

Xi = 5(1.2) = 6.0

t∑

2 = X i

√

5(0.042 ) = 0.0894

Let X1 and X2 be the two measurements. The average is X. √ 7 × 10−15 X = Xi ∕ 2 = = 4.95 × 10−15 . √ 2

7. (a) M = X+1.5Y = X + 1.5 Y = 0.125 + 1.5(0.350) = 0.650 t (b) M = X+1.5Y =

2 + 1.52 2 = X Y

√

0.052 + 1.52 (0.12 ) = 0.158

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SECTION 2.5

Let X1 , ..., X24 be the volumes of the 24 bottles. Let S = X1 + ⋯ + X24 be the total weight. The average volume per bottle is X. (a) S =

(b) S =

∑

Xi = 24(20.01) = 480.24

t∑

2 = X

√

24(0.022 ) = 0.098

(c) X = Xi = 20.01 √ √ (d) X = Xi ∕ 24 = 0.02∕ 24 = 0.00408 √ (e) Let n be the required number of bottles. Then 0.02∕ n = 0.0025, so n = 64.

Let X1 and X2 denote the lengths of the pieces chosen from the population with mean 30 and standard deviation 0.1, and let Y1 and Y2 denote the lengths of the pieces chosen from the population with mean 45 and standard deviation 0.3. (a) X1 +X2 +Y1 +Y2 = X1 + X2 + Y1 + Y2 = 30 + 30 + 45 + 45 = 150

(b) X1 +X2 +Y1 +Y2 =

10.

t √ 2 + 2 + 2 + 2 = 0.12 + 0.12 + 0.32 + 0.32 = 0.447 X X Y Y 1

The daily revenue is R = 2.60X1 + 2.75X2 + 2.90X3 . (a) R = 2.60 1 + 2.75 2 + 2.90 3 = 2.60(1500) + 2.75(500) + 2.90(300) = 6145

(b) R =

t √ 2.602 12 + 2.752 22 + 2.902 32 = 2.602 (1802 ) + 2.752 (902 ) + 2.902 (402 ) = 541.97

11. (a) The number of passenger-miles is 8000(210) = 1,680,000. Let X be the number of gallons of fuel used. Then X = 1,680,000(0.15) = 252,000 gallons.

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CHAPTER 2

(b) X =

√ (1,680,000)(0.01) = 12.9615

(c) X∕1,680,000 = (1∕1,680,000) X = (1∕1,680,000)(252,000) = 0.15. (d) X∕1,680,000 = (1∕1,680,000) X = (1∕1,680,000)(12.9615) = 7.7152 × 10−6

12.

Let X1 , X2 , X3 , X4 denote the lengths of Laura’s sticks. Then X1 , X2 , X3 , and X4 are independent with mean 8 and standard deviation 15. Let Y denote the length of Philip’s sticks. Then Y has mean 8 and standard deviation 0.15. Let A denote the total length of Laura’s sticks, let B denote the total length of Philip’s sticks, and let C denote the length of two sticks from Laura plus two sticks from Philip. (a) A = X1 +X2 +X3 +X4 = X1 + X2 + X3 + X4 = 8 + 8 + 8 + 8 = 32. √ A = X1 +X2 +X3 +X4 = 0.152 + 0.152 + 0.152 + 0.152 = 0.3 (b) B = 4Y = 4 Y = 32 B = 4Y = 4 Y = 0.6. (c) C = X1 +X2 +2Y = 8 + 8 + 2(8) = 32 √ C = X1 +X2 +2Y = 0.152 + 0.152 + 4(0.152 ) = 0.3674.

13. (a) = 0.0695 +

1.0477 0.8649 0.7356 0.2171 2.8146 0.5913 0.0079 + + + + + + + 5(0.0006) = 0.2993 20 20 20 30 60 15 10

t (b) =

0.00182 + (

0.0269 2 ) + ( 0.0225 )2 + ( 0.0113 )2 + ( 0.0185 )2 + ( 0.0284 )2 + ( 0.0031 )2 + ( 0.0006 )2 + 52 (0.0002)2 = 0.00288 20 20 20 30 60 15 10

14. (a) X = O+2N+2C∕3 = O + 2 N + (2∕3) C = 0.1668 + 2(0.0255) + (2∕3)(0.0247) = 0.2343 (b) X = O+2N+2C∕3 =

t √ 2 + (2∕3)2 2 = 0.03402 + 4(0.0194)2 + (2∕3)2 (0.0131)2 = 0.05232 O2 + 4 N C

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SECTION 2.5 9.98

9.98

Ê9.95

15. (a) P (X < 9.98) =

10 dx = 10x

5.1

(b) P (Y > 5.01) =

Ê5.01

= 0.3 9.95

5 dy = 5y

= 0.45 5.01

10.05

(d) X =

Ê9.95

5.1

(e) Y =

16. (a) X =

Ê4.9

Ê4

2 = (b) X Ê

= 10 9.95

5.1

5y dy = 2.5y2

60 4

10.05

10x dx = 5x2

=5 4.9

3x(x − 5)2 3 x− 4 4

0 1 5x3 3x4 2 dx = −9x + − 2 16

3(x − 5)2 3(x − 5)4 − 4 4

0 dx =

=5 4

(x − 5)3 3(x − 5)5 − 4 20

= 0.2 4

(d) Let X1 , X2 , and X3 be the three thicknesses, in millimeters. Then S = X1 + X2 + X3 is the total thickness. S = X1 + X2 + X3 = 3(5) = 15. 2 + 2 + 2 = 3(0.2) = 0.6. S2 = X X X 1

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CHAPTER 2

17. (a) Let = 40.25 be the mean SiO2 content, and let = 0.36 be the standard deviation of the SiO2 content, in a randomly chosen rock. Let X be the average content in a random sample of 10 rocks. 0.36 Then X = = 40.25, and X = √ = √ = 0.11. 10 10 0.36 (b) Let n be the required number of rocks. Then √ = √ = 0.05. n n Solving for n yields n = 51.84. Since n must be an integer, take n = 52.

18. (a) Yes. Let Xi denote the number of bytes downloaded in the ith second. The total number of bytes is X1 + X2 + X3 + X4 + X5 . The mean of the total number is the sum of the fve means, each of which is equal to 105 . (b) No. X1 , ..., X5 are not independent, so the standard deviation of the sum depends on the covariances between the Xi .

Section 2.6 1. (a) 0.17 (b) P (X ≥ 1 and Y < 2) = P (1, 0) + P (1, 1) + P (2, 0) + P (2, 1) = 0.17 + 0.23 + 0.06 + 0.14 = 0.60 (c) P (X < 1) = P (X = 0) = P (0, 0) + P (0, 1) + P (0, 2) = 0.10 + 0.11 + 0.05 = 0.26 (d) P (Y ≥ 1) = 1 − P (Y = 0) = 1 − P (0, 0) − P (1, 0) − P (2, 0) = 1 − 0.10 − 0.17 − 0.06 = 0.67 (e) P (X ≥ 1) = 1 − P (X = 0) = 1 − P (0, 0) − P (0, 1) − P (0, 2) = 1 − 0.10 − 0.11 − 0.05 = 0.74 (f) P (Y = 0) = P (0, 0) + P (1, 0) + P (2, 0) = 0.10 + 0.17 + 0.06 = 0.33 (g) P (X = 0 and Y = 0) = 0.10

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SECTION 2.6

2. (a) The marginal probability mass function pX (x) is found by summing along the rows of the joint probability mass function. x 0 1 2 pY (y)

0 0.10 0.17 0.06 0.33

y 1 0.11 0.23 0.14 0.48

2 0.05 0.08 0.06 0.19

pX (x) 0.26 0.48 0.26

pX (0) = 0.26, pX (1) = 0.48, pX (2) = 0.26, pX (x) = 0 if x ≠ 0, 1, or 2 (b) The marginal probability mass function pY (y) is found by summing down the columns of the joint probability mass function. So pY (0) = 0.33, pY (1) = 0.48, pY (2) = 0.19, pY (y) = 0 if y ≠ 0, 1, or 2 (c) X = 0pX (0) + 1pX (1) + 2pX (2) = 0(0.26) + 1(0.48) + 2(0.26) = 1.00 (d) Y = 0pY (0) + 1pY (1) + 2pY (2) = 0(0.33) + 1(0.48) + 2(0.19) = 0.86 2 = 02 p (0) + 12 p (1) + 22 p (2) − 2 = 02 (0.26) + 12 (0.48) + 22 (0.26) − 1.002 = 0.5200 (e) X X X X X √ X = 0.5200 = 0.7211

(f) Y2 = 02 pY (0) + 12 pY (1) + 22 pY (2) − Y2 = 02 (0.33) + 12 (0.48) + 22 (0.19) − 0.862 = 0.5004 √ Y = 0.5004 = 0.7074 (g) Cov(X, Y ) = XY − X Y XY

= (0)(0)pX,Y (0, 0) + (0)(1)pX,Y (0, 1) + (0)(2)pX,Y (0, 2) + (1)(0)pX,Y (1, 0) + (1)(1)pX,Y (1, 1) + (1)(2)pX,Y (1, 2) + (2)(0)pX,Y (2, 0) + (2)(1)pX,Y (2, 1) + (2)(2)pX,Y (2, 2) =

(0)(0)(0.10) + (0)(1)(0.11) + (0)(2)(0.05) + (1)(0)(0.17) + (1)(1)(0.23) + (1)(2)(0.08) + (2)(0)(0.06) + (2)(1)(0.14) + (2)(2)(0.06)

= 0.91 X = 1.00, Y = 0.86 Cov(X, Y ) = 0.91 − (1.00)(0.86) = 0.0500

(h) X,Y =

Cov(X, Y ) 0.0500 = = 0.0980 X Y (0.7211)(0.7074)

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CHAPTER 2

(i) No. The joint probability density function is not equal to the product of the marginals. For example, P (X = 0 and Y = 0) = 0.10, but P (X = 0)P (Y = 0) = (0.26)(0.33) = 0.0858.

3. (a) pY |X (0 | 0) = pY |X (1 | 0) = pY |X (2 | 0) =

(b) pX|Y (0 | 1) = pX|Y (1 | 1) = pX|Y (2 | 1) =

pX,Y (0, 0) pX (0) pX,Y (0, 1) pX (0) pX,Y (0, 2) pX (0) pX,Y (0, 1) pY (1) pX,Y (1, 1) pY (1) pX,Y (2, 1) pY (1)

0.10 = 0.3846 0.26

0.11 = 0.4231 0.26

0.05 = 0.1923 0.26

0.11 = 0.2292 0.48

0.23 = 0.4792 0.48

0.14 = 0.2917 0.48

(c) E(Y | X = 0) = 0pY |X (0 | 0) + 1pY |X (1 | 0) + 2pY |X (2 | 0) = 0.8077 (d) E(X | Y = 1) = 0pX|Y (0 | 1) + 1pX|Y (1 | 1) + 2pX|Y (2 | 1) = 1.0625

4. (a) The marginal probability mass function pX (x) is found by summing along the rows of the joint probability mass function. x 0 1 2 3 pY (y)

0 0.15 0.09 0.06 0.04 0.34

y 1 0.12 0.07 0.05 0.03 0.27

2 0.11 0.05 0.04 0.02 0.22

3 0.10 0.04 0.02 0.01 0.17

pX (x) 0.48 0.25 0.17 0.10

pX (0) = 0.48, pX (1) = 0.25, pX (2) = 0.17, pX (3) = 0.10, pX (x) = 0 if x ≠ 0, 1, 2, or 3

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SECTION 2.6

(b) The marginal probability mass function pY (y) is found by summing down the columns of the joint probability mass function. So pY (0) = 0.34, pY (1) = 0.27, pY (2) = 0.22, pY (3) = 0.17, pY (y) = 0 if y ≠ 0, 1, 2, or 3 (c) No, X and Y are not independent. For example, P (X = 0 and Y = 0) = 0.15, but P (X = 0)P (Y = 0) = (0.48)(0.34) = 0.1632 ≠ 0.15. (d) X = 0pX (0) + 1pX (1) + 2pX (2) + 3pX (3) = 0(0.48) + 1(0.25) + 2(0.17) + 3(0.10) = 0.89 Y = 0pY (0) + 1pY (1) + 2pY (2) + 3pY (3) = 0(0.34) + 1(0.27) + 2(0.22) + 3(0.17) = 1.22

2 = 02 p (0)+12 p (1)+22 p (2)+32 p (3)− 2 = 02 (0.48)+12 (0.25)+22 (0.17)+32 (0.10)−0.892 = 1.0379 (e) X X X X X X √ X = 1.0379 = 1.0188

Y2 = 02 pY (0)+12 pY (1)+22 pY (2)+32 pY (3)− Y2 = 02 (0.34)+12 (0.27)+22 (0.22)+32 (0.17)−1.222 = 1.1916 √ Y = 1.1916 = 1.0916 (f) Cov(X, Y ) = XY − X Y XY

(0)(0)pX,Y (0, 0) + (0)(1)pX,Y (0, 1) + (0)(2)pX,Y (0, 2) + (0)(3)pX,Y (0, 3) + (1)(0)pX,Y (1, 0) + (1)(1)pX,Y (1, 1) + (1)(2)pX,Y (1, 2) + (1)(3)pX,Y (1, 2) + (2)(0)pX,Y (2, 0) + (2)(1)pX,Y (2, 1) + (2)(2)pX,Y (2, 2) + (2)(3)pX,Y (2, 2) + (3)(0)pX,Y (3, 0) + (3)(1)pX,Y (3, 1) + (3)(2)pX,Y (3, 3) + (3)(3)pX,Y (3, 3)

(0)(0)(0.15) + (0)(1)(0.12) + (0)(2)(0.11) + (0)(3)(0.10)

(1)(0)(0.09) + (1)(1)(0.07) + (1)(2)(0.05) + (1)(3)(0.04)

(2)(0)(0.06) + (2)(1)(0.05) + (2)(2)(0.04) + (2)(3)(0.02)

(3)(0)(0.04) + (3)(1)(0.03) + (3)(2)(0.02) + (4)(3)(0.01)

= 0.97 X = 0.89, Y = 1.22 Cov(X, Y ) = 0.97 − (0.89)(1.22) = −0.1158

(g) X,Y =

Cov(X, Y ) −0.1158 = = −0.1041 X Y (1.0188)(1.0196)

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CHAPTER 2

5. (a) X+Y = X + Y = 0.89 + 1.22 = 2.11

(b) X+Y =

t √ 2 + 2 + 2Cov(X, Y ) = 1.0379 + 1.1916 + 2(−0.1158) = 1.4135 X Y

6. (a) pY |X (0 | 1) = pY |X (1 | 1) = pY |X (2 | 1) = pY |X (3 | 1) =

(b) pX|Y (0 | 2) = pX|Y (1 | 2) = pX|Y (2 | 2) = pX|Y (3 | 2) =

pX,Y (1, 0) pX (1) pX,Y (1, 1) pX (1) pX,Y (1, 2) pX (1) pX,Y (1, 3) pX (1) pX,Y (0, 2) pY (2) pX,Y (1, 2) pY (2) pX,Y (2, 2) pY (2) pX,Y (3, 2) pY (2)

0.09 = 0.36 0.25

0.07 = 0.28 0.25

0.05 = 0.20 0.25

0.04 = 0.16 0.25

0.11 = 1∕2 0.22

0.05 = 5∕22 0.22

0.04 = 2∕11 0.22

0.02 = 1∕11 0.22

(c) E(Y | X = 1) = 0pY |X (0 | 1) + 1pY |X (1 | 1) + 2pY |X (2 | 1) + 3pY |X (3 | 1) = 0(0.36) + 1(0.28) + 2(0.20) + 3(0.16) = 1.16 (d) E(X | Y = 2) = 0pX|Y (0 | 2) + 1pX|Y (1 | 2) + 2pX|Y (2 | 2) + 3pX|Y (3 | 2) = 0(1∕2) + 1(5∕22) + 2(2∕11) + 3(1∕11) = 19∕22

7. (a) 2X + 3Y

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SECTION 2.6

(b) 2X+3Y = 2 X + 3 Y = 2(0.89) + 3(1.22) = 5.44

t (c) 2X+3Y

2 + 32 2 + 2(2)(3)Cov(X, Y ) 22 X Y

√ = 22 (1.0379) + 32 (1.1916) − 2(2)(3)(−0.1158) = 3.6724

8. (a) P (X = 2 and Y = 2) = P (X = 2)P (Y = 2 | X = 2). Now P (X = 2) = 0.30. Given that X = 2, Y = 2 if and only if each of the two customers purchases one item. Therefore P (Y = 2 | X = 2) = 0.052 = 0.0025, so P (X = 2 and Y = 2) = (0.30)(0.0025) = 0.00075. (b) P (X = 2 and Y = 6) = P (X = 2)P (Y = 6 | X = 2). Now P (X = 2) = 0.30. Given that X = 2, Let Y1 be the number of items purchased by the frst customer, and let Y2 be the number of items purchased by the second customer. Then the event Y = 6 is equivalent to the event {Y1 = 5 and Y2 = 1} or {Y1 = 4 and Y2 = 2} or {Y1 = 3 and Y2 = 3} or {Y1 = 2 and Y2 = 4} or {Y1 = 1 and Y2 = 5}. Therefore P (Y = 6 | X = 2) =

P (Y1 = 5 and Y2 = 1) + P (Y1 = 4 and Y2 = 2) + P (Y1 = 3 and Y2 = 3) + P (Y1 = 2 and Y2 = 4) + P (Y1 = 1 and Y2 = 5) = (0.15)(0.05) + (0.30)(0.15) + (0.25)(0.25) + (0.15)(0.30) + (0.05)(0.15) =

0.1675

P (X = 2 and Y = 6) = (0.30)(0.1675) = 0.05025

(c) P (Y = 2) = P (X = 1 and Y = 2) + P (X = 2 and Y = 2). P (X = 1 and Y = 2) = P (X = 1)P (Y = 2 | X = 1) = (0.25)(0.15) = 0.0375. From part (a), P (X = 2 and Y = 2) = 0.00075. P (Y = 2) = 0.0375 + 0.00075 = 0.03825.

9. (a) The marginal probability mass function pX (x) is found by summing along the rows of the joint probability mass function.

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CHAPTER 2

x 0 1 2 3 4 pY (y)

0 0.03 0.09 0.09 0.06 0.04 0.31

1 0.01 0.05 0.05 0.04 0.03 0.18

y 2 0.02 0.08 0.07 0.07 0.04 0.28

3 0.01 0.03 0.04 0.04 0.03 0.15

4 0.01 0.01 0.02 0.02 0.02 0.08

pX (x) 0.08 0.26 0.27 0.23 0.16

pX (0) = 0.08, pX (1) = 0.26, pX (2) = 0.27, pX (3) = 0.23, pX (4) = 0.16, pX (x) = 0 if x ≠ 0, 1, 2, 3, or 4. (b) The marginal probability mass function pY (y) is found by summing down the columns of the joint probability mass function. So pY (0) = 0.31, pY (1) = 0.18, pY (2) = 0.28, pY (3) = 0.15, pY (4) = 0.08, pY (y) = 0 if y ≠ 0, 1, 2, 3, or 4. (c) No. The joint probability mass function is not equal to the product of the marginals. For example, pX,Y (0, 0) = 0.03 ≠ pX (0)pY (0). (d) X = 0pX (0) + 1pX (1) + 2pX (2) + 3pX (3) + 4pX (4) = 0(0.08) + 1(0.26) + 2(0.27) + 3(0.23) + 4(0.16) = 2.13 Y = 0pY (0) + 1pY (1) + 2pY (2) + 3pY (3) + 4pY (4) = 0(0.31) + 1(0.18) + 2(0.28) + 3(0.15) + 4(0.08) = 1.51

2 (e) X

2 = 02 pX (0) + 12 pX (1) + 22 pX (2) + 32 pX (3) + 42 pX (4) − X

= 02 (0.08) + 12 (0.26) + 22 (0.27) + 32 (0.23) + 42 (0.16) − 2.132 = 1.4331 √ X = 1.4331 = 1.1971

02 pY (0) + 12 pY (1) + 22 pY (2) + 32 pY (3) + 42 pY (4) − Y2

= 02 (0.31) + 12 (0.18) + 22 (0.28) + 32 (0.15) + 42 (0.08) − 1.512 = 1.6499 √ Y = 1.6499 = 1.2845 (f) Cov(X, Y ) = XY − X Y . XY

(0)(0)pX,Y (0, 0) + (0)(1)pX,Y (0, 1) + (0)(2)pX,Y (0, 2) + (0)(3)pX,Y (0, 3) + (0)(4)pX,Y (0, 4) + (1)(0)pX,Y (1, 0) + (1)(1)pX,Y (1, 1) + (1)(2)pX,Y (1, 2) + (1)(3)pX,Y (1, 3) + (1)(4)pX,Y (1, 4) + (2)(0)pX,Y (2, 0) + (2)(1)pX,Y (2, 1) + (2)(2)pX,Y (2, 2) + (2)(3)pX,Y (2, 3) + (2)(4)pX,Y (2, 4) + (3)(0)pX,Y (3, 0) + (3)(1)pX,Y (3, 1) + (3)(2)pX,Y (3, 2) + (3)(3)pX,Y (3, 3) + (3)(4)pX,Y (3, 4) + (4)(0)pX,Y (4, 0) + (4)(1)pX,Y (4, 1) + (4)(2)pX,Y (4, 2) + (4)(3)pX,Y (4, 3) + (4)(4)pX,Y (4, 4)

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SECTION 2.6

= (0)(0)(0.03) + (0)(1)(0.01) + (0)(2)(0.02) + (0)(3)(0.01) + (0)(4)(0.01) + (1)(0)(0.09) + (1)(1)(0.05) + (1)(2)(0.08) + (1)(3)(0.03) + (1)(4)(0.01) + (2)(0)(0.09) + (2)(1)(0.05) + (2)(2)(0.07) + (2)(3)(0.04) + (2)(4)(0.02) + (3)(0)(0.06) + (3)(1)(0.04) + (3)(2)(0.07) + (3)(3)(0.04) + (3)(4)(0.02) + (4)(0)(0.04) + (4)(1)(0.03) + (4)(2)(0.04) + (4)(3)(0.03) + (4)(4)(0.02) = 3.38 X = 2.13, Y = 1.51 Cov(X, (X, Y ) = 3.38 − (2.13)(1.51) = 0.1637

(g) X,Y =

Cov(X, Y ) 0.1637 = 0.1065 = (1.1971)(1.2845) X Y

10. (a) X+Y = X + Y = 2.13 + 1.51 = 3.64 2 2 + 2 + 2Cov(X, Y ) = 1.4331 + 1.6499 + 2(0.1637) = 3.4104 (b) X+Y = X Y √ X+Y = 3.4104 = 1.85

11. (a) pY |X (0 | 4) = pY |X (1 | 4) = pY |X (2 | 4) = pY |X (3 | 4) = pY |X (4 | 4) =

(b) pX|Y (0 | 3) =

pX,Y (4, 0) pX (4) pX,Y (4, 1) pX (4) pX,Y (4, 2) pX (4) pX,Y (4, 3) pX (4) pX,Y (4, 4) pX (4) pX,Y (0, 3) pY (3)

0.04 = 0.25 0.16

0.03 = 0.1875 0.16

0.04 = 0.25 0.16

0.03 = 0.1875 0.16

0.02 = 0.125 0.16

0.01 = 1∕15 0.15

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pX|Y (1 | 3) = pX|Y (2 | 3) = pX|Y (3 | 3) = pX|Y (4 | 3) =

pX,Y (1, 3) pY (3) pX,Y (2, 3) pY (3) pX,Y (3, 3) pY (3) pX,Y (4, 3) pY (3)

0.03 = 0.2 0.15

0.04 = 4∕15 0.15

0.03 = 0.2 0.15

(c) E(Y | X = 4) = 0pY |X (0 | 4) + 1pY |X (1 | 4) + 2pY |X (2 | 4) + 3pY |X (3 | 4) + 4pY |X (4 | 4) = 1.75 (d) E(X | Y = 3) = 0pX|Y (0 | 3) + 1pX|Y (1 | 3) + 2pX|Y (2 | 3) + 3pX|Y (3 | 3) + 4pX|Y (4 | 3) = 2.33

12. (a) The marginal probability mass function pX (x) is found by summing along the rows of the joint probability mass function. y x 0 1 2 3 pY (y)

0 0.13 0.12 0.02 0.01 0.28

1 0.10 0.16 0.06 0.02 0.34

2 0.07 0.08 0.08 0.02 0.25

3 0.03 0.04 0.04 0.02 0.13

pX (x) 0.33 0.40 0.20 0.07

pX (0) = 0.33, pX (1) = 0.40, pX (2) = 0.20, pX (3) = 0.07, pX (x) = 0 if x ≠ 0, 1, 2, or 3 (b) The marginal probability mass function pY (y) is found by summing down the columns of the joint probability mass function. So pY (0) = 0.28, pY (1) = 0.34, pY (2) = 0.25, pY (3) = 0.13, pY (y) = 0 if y ≠ 0, 1, 2, or 3. (c) X = 0pX (0) + 1pX (1) + 2pX (2) + 3pX (3) = 0(0.33) + 1(0.40) + 2(0.20) + 3(0.07) = 1.01 (d) Y = 0pY (0) + 1pY (1) + 2pY (2) = 0(0.28) + 1(0.34) + 2(0.25) + 3(0.13) = 1.23

2 (e) X

2 = 02 pX (0) + 12 pX (1) + 22 pX (2) + 32 pX (3) − X

= 02 (0.33) + 12 (0.40) + 22 (0.20) + 32 (0.07) − 1.012 = 0.8099

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SECTION 2.6 X =

√ 0.8099 = 0.8999

(f) Y2 = 02 pY (0) + 12 pY (1) + 22 pY (2) − Y2 = 02 (0.28) + 12 (0.34) + 22 (0.25) + 32 (0.13) − 1.232 = 0.9971. √ Y = 0.9971 = 0.9985 (g) Cov(X, Y ) = XY − X Y XY

(0)(0)pX,Y (0, 0) + (0)(1)pX,Y (0, 1) + (0)(2)pX,Y (0, 2) + (0)(3)pX,Y (0, 3) + (1)(0)pX,Y (1, 0) + (1)(1)pX,Y (1, 1) + (1)(2)pX,Y (1, 2) + (1)(3)pX,Y (1, 3) + (2)(0)pX,Y (2, 0) + (2)(1)pX,Y (2, 1) + (2)(2)pX,Y (2, 2) + (2)(3)pX,Y (2, 3) + (3)(0)pX,Y (3, 0) + (3)(1)pX,Y (3, 1) + (3)(2)pX,Y (3, 2) + (3)(3)pX,Y (3, 3)

(0)(0)(0.13) + (0)(1)(0.10) + (0)(2)(0.07) + (0)(3)(0.03) + (1)(0)(0.12) + (1)(1)(0.16) + (1)(2)(0.08) + (1)(3)(0.04)

+ (2)(0)(0.02) + (2)(1)(0.06) + (2)(2)(0.08) + (2)(3)(0.04) + (3)(0)(0.01) + (3)(1)(0.02) + (3)(2)(0.02) + (3)(3)(0.02) = 1.48 X = 1.01, Y = 1.23 Cov(X, Y ) = 1.48 − (1.01)(1.23) = 0.2377

(h) X,Y =

Cov(X, Y ) 0.2377 = = 0.2645 X Y (0.8999)(0.9985)

13. (a) Z = X+Y = X + Y = 1.01 + 1.23 = 2.24

(b) Z = X+Y =

= = = =

t √ 2 + 2 + 2Cov(X, Y ) = X 0.8099 + 0.9971 + 2(0.2377) = 1.511 Y

P (X + Y = 2) P (X = 0 and Y = 2) + P (X = 1 and Y = 1) + P (X = 2 and Y = 0) 0.07 + 0.16 + 0.02 0.25

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CHAPTER 2

14. (a) T = 50X + 100Y , so T = 50X+100Y = 50 X + 100 Y = 50(1.01) + 100(1.23) = 173.50.

(b) T

50X+100Y t 2 + 1002 2 + 2(50)(100)Cov(X, Y ) = 502 X Y √ 2 = 50 (0.8099) + 1002 (0.9971) + 2(50)(100)(0.2377) = 119.9 =

15. (a) pY |X (0 | 3) = pY |X (1 | 3) = pY |X (2 | 3) = pY |X (3 | 3) =

(b) pX|Y (0 | 1) = pX|Y (1 | 1) = pX|Y (2 | 1) = pX|Y (3 | 1) =

pX,Y (3, 0) pX (3) pX,Y (3, 1) pX (3) pX,Y (3, 2) pX (3) pX,Y (3, 3) pX (3) pX,Y (0, 1) pY (1) pX,Y (1, 1) pY (1) pX,Y (2, 1) pY (1) pX,Y (3, 1) pY (1)

0.01 = 0.1429 0.07

0.02 = 0.2858 0.07

0.10 = 0.2941 0.34

0.16 = 0.4706 0.34

0.06 = 0.1765 0.34

0.02 = 0.0588 0.34

(c) E(Y | X = 3) = 0pY |X (0 | 3) + 1pY |X (1 | 3) + 2pY |X (2 | 3) + 3pY |X (3 | 3) = 1.71. (d) E(X | Y = 1) = 0pX|Y (0 | 1) + 1pX|Y (1 | 1) + 2pX|Y (2 | 1) + 3pX|Y (3 | 1) = 1

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SECTION 2.6

16. (a) P (X > 1 and Y > 1)

= = =

Ê1

∞

Ê1

∞

Ê1

∞

Ê H1

∞

xe−(x+xy) dy dx ∞I

−e−(x+xy)

dx 1

e−2x dx ∞ −2x

−0.5e

0.0676676

∞

f (x, y) dy. Ê0 If x ≤ 0 then f (x, y) = 0 for all y so fX (x) = 0.

(b) fX (x) =

If x > 0 then fX (x) =

Ê0

∞

xe−(x+xy) dy = −e−(x+xy)

= e−x . 0

∞

f (x, y) dx. Ê0 If y ≤ 0 then f (x, y) = 0 for all x so fY (y) = 0.

fY (y) =

If y > 0 then fY (y) =

Ê0

∞

xe−(x+xy) dx =

1 . (1 + y2 )

17. (a) Cov(X, Y ) = XY − X Y 2

1 xy(x + y) dy dx Ê1 Ê4 6 1 5 2 0 2 2 xy3 1 x y = + dx Ê1 6 2 3 4 1 2 0 1 9x2 61x = + dx Ê1 6 2 3 0 1 2 1 3x3 61x2 = + 6 2 6 =

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CHAPTER 2

41 6

1 9 x+ for 1 ≤ x ≤ 2 (see Example 2.55). 6 2 0 1 2 2 1 9 1 x3 9x2 109 x x+ dx = + = . X = Ê1 6 2 6 3 4 72 1 1 3 y+ for 4 ≤ y ≤ 5 (see Example 2.55). fY (y) = 6 2 0 1 5 5 1 3 1 y3 3y2 325 y y+ dy = + = . Y = Ê4 6 2 6 3 4 72 4 41 109 325 − = −0.000193. Cov(X, Y ) = XY − X Y = 6 72 72 fX (x) =

Cov(X, Y ) . X Y 0 1 2 2 1 2 9 1 x4 3x3 109 2 2 2 x x+ dx − X = + − = 0.08314. X = Ê1 6 2 6 4 2 72 1 0 1 5 5 325 2 1 2 3 1 y4 y3 2 2 Y = y y+ dx − Y = + − = 0.08314. Ê4 6 2 6 4 2 72

(b) X,Y =

−0.000193

X,Y = √ = −0.00232. (0.08314)(0.08314)

18. (a) P (X > 0.5 and Y > 0.5)

3(x2 + y2 ) dy dx Ê0.5 Ê0.5 2 1 1 10 3x2 y y3 = + dx Ê0.5 2 2 0.5 1 10 3x2 21 = + dx Ê0.5 4 48 =

x3 21x + 4 48

21 48

0.5

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SECTION 2.6

(b) For 0 < x < 1, fX (x) = For 0 < y < 1, fY (y) =

Ê0 Ê0

3(x2 + y2 ) dy = 0.5 + 1.5x2 . For x ≤ 0 and x ≥ 1, fX (x) = 0. 2

3(x2 + y2 ) dx = 0.5 + 1.5y2 . For y ≤ 0 and y ≥ 1, fY (y) = 0. 2

19. (a) Cov(X, Y ) = XY − X Y . 1

3(x2 + y2 ) dx dy Ê0 Ê0 2 1 1 10 3x4 y 3x2 y3 = + dy Ê0 8 4 0 1 10 3y 3y3 = + dy Ê0 8 4 0 2 1 1 3y 3y4 = + dy 16 16 xy

3 8

0 2 1 1 x 3x4 5 + = . X = Ê0 4 8 8 0 0 2 1 1 1 1 + 3y2 y 3y4 5 y dy = + = . Y = Ê0 2 4 8 8 0 2 3 5 1 Cov(X, Y ) = − =− . 8 8 64 1

(b) X,Y =

1 + 3x2 x dx = 2

Cov(X, Y ) . X Y

0 3 1 1 2 5 73 x 3x5 + − = . x 2 6 10 8 960 0 0 0 3 1 1 2 1 2 y 3y5 5 73 2 1 + 3y 2 2 y dy − Y = + − = . Y = Ê0 2 6 10 8 960

2 = X Ê

2 1 + 3x

2 = dx − X

−1∕64 15 X,Y = √ =− . √ 73 73∕960 73∕960

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fX,Y (0.5, y) fX (0.5)

3 + 12y2 7 , fX (0.5) = . 8 8 3 + 12y2 So for 0 < y < 1, fY | X (y | 0.5) = 7 For 0 < y < 1, fX,Y (0.5, y) =

(d) E(Y | X = 0.5) =

Ê0

yfY | X (y | 0.5) dy =

20. (a) P (X > 1 and Y > 1)

Ê1

= =

−

∞

Ê1

∞

Ê1 H

Ê1

6e−3

3y2 + 6y4 3 + 12y2 y dy = 7 14

= 0

9 14

4xye−(2x+y) dy dx ∞I

∞

4x(1 + y)e

−(2x+y)

dx 1 ∞

∞

Ê0

8xe

−(2x+1)

dx = −2(1 + 2x)e

−(2x+1) 1

(b) For x ≤ 0, fX (x) = 0. For x > 0, fX (x) =

Ê0

∞

4xye−(2x+y) dy = −4x(1 + y)e−(2x+y) 0

⎧ 4xe−2x ⎪ Therefore fX (x) = ⎨ ⎪ 0 ⎩ For y ≤ 0, fY (y) = 0. For y > 0, fY (y) = Therefore fY (y) =

Ê0 <

= 4xe−2x .

x>0 x≤0 ∞

∞

4xye

−(2x+y)

dx = −y(1 + 2x)e

= ye−y .

−(2x+y) 0

ye−y 0

y>0 y≤0

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SECTION 2.6

21. (a) The probability mass function of Y is the same as that of X, so fY (y) = e−y if y > 0 and fY (y) = 0 if y ≤ 0. Since X and Y are independent, f (x, y) = fX (x)fY (y). < −x−y e x > 0 and y > 0 Therefore f (x, y) = 0 otherwise (b) P (X ≤ 1 and Y > 1)

P (X ≤ 1)P (Y > 1) H I0

= =

H =

Ê0

−e

−x

Ê1

e−x dx 1I H

∞

1 e−y dy

∞I

−y

−e 0

1 −1

−1

(1 − e )(e )

= e−1 − e−2 = 0.2325

Ê0

∞

xe−x dx = −xe−x

− 0

Ê0

∞

e−x dx = 0 − (−e−x )

=0+1=1 0

(d) Since X and Y have the same probability mass function, Y = X = 1. Therefore X+Y = X + Y = 1 + 1 = 2. (e) P (X + Y ≤ 2)

∞

2−x

Ê−∞ Ê−∞ 2

= = = =

Ê0 Ê0 Ê0

Ê0

f (x, y) dy dx

2−x

e−x−y dy dx

−x

2−x I −y

−e

e−x (1 − ex−2 ) dx (e−x − e−2 ) dx 2

−x

(−e

−2

− xe ) 0

= 1 − 3e−2 = 0.5940

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CHAPTER 2

22. (a) P (X = 1) = 1∕3, P (Y = 1) = 2∕3, P (X = 1 and Y = 1) = 1∕3 ≠ P (X = 1)P (Y = 1). (b) XY = (−1)(1∕3) + (0)(1∕3) + (1)(1∕3) = 0. Now X = (−1)(1∕3) + (0)(1∕3) + (1)(1∕3) = 0, and Y = (0)(1∕3) + (1)(2∕3) = 2∕3. So Cov(X, Y ) = XY − X Y = 0 − (0)(2∕3) = 0. Since Cov(X, Y ) = 0, X,Y = 0.

23. (a) R = 0.3X + 0.7Y

(b) R = 0.3X+0.7Y = 0.3 X + 0.7 Y = (0.3)(6) + (0.7)(6) = 6. t 2 + 0.72 2 + 2(0.3)(0.7)Cov(X, Y ). The risk is R = 0.3X+0.7Y = 0.32 X Y Cov(X, Y ) = X,Y X Y = (0.3)(3)(3) = 2.7. √ Therefore R = 0.32 (32 ) + 0.72 (32 ) + 2(0.3)(0.7)(2.7) = 2.52. (c) R = (0.01K)X+(1−0.01K)Y = (0.01K) X + (1 − 0.01K) Y = (0.01K)(6) + (1 − 0.01K)(6) = 6. t 2 + (1 − 0.01K)2 2 + 2(0.01K)(1 − 0.01K)Cov(X, Y ). R = (0.01K)2 X Y √ √ Therefore R = (0.01K)2 (32 ) + (1 − 0.01K)2 (32 ) + 2(0.01K)(1 − 0.01K)(2.7) = 0.03 1.4K 2 − 140K + 10, 000. (d) R is minimized when 1.4K 2 − 140K + 10000 is minimized. d d Now (1.4K 2 − 140K + 10000) = 2.8K − 140, so (1.4K 2 − 140K + 10000) = 0 if K = 50. dK dK R is minimized when K = 50. √ (e) For any correlation , the risk is 0.03 K 2 + (100 − K)2 + 2 K(100 − K). If ≠ 1 this quantity is minimized when K = 50.

24.

Ê19 Ê5 21

= 3

Ê19

3 r2 ℎ(ℎ − 20)2 (r − 5) dr dℎ

ℎ(ℎ − 20)2 dℎ

Ê5

r2 (r − 5) dr

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SECTION 2.6

0 4 1 21 0 4 1 ℎ r 5r3 40ℎ3 2 − (3 ) − + 200ℎ 4 3 4 3

2021.0913cm3

t 25. (a) M1 =

2 = M1

t t √ 2 2 + 2 = = R R+E 22 + 12 = 2.2361. Similarly, M2 = 2.2361. E1 1

(b) M1 M2 = R2 +E1 R+E2 R+E1 E2 = R2 + E1 R + E2 R + E1 E2 = R2 2 (c) M1 M2 = R+E1 R+E2 = ( R + E1 )( R + E2 ) = R R = R 2 = 2 (d) Cov(M1 , M2 ) = M1 M2 − M1 M2 = R2 − R R

(e) M1 ,M2 =

26.

2 R Cov(M1 , M2 ) 4 = = 0.8 = M1 M2 M1 M2 (2.2361)(2.2361)

2. Cov(X, X) = X⋅X − X X = X 2 − ( X )2 = X

27. (a) Cov(aX, bY ) = aX⋅bY − aX bY = abXY − a X b Y = ab XY − ab X Y = ab( XY − X Y ) = abCov(X, Y ). (b) aX,bY = Cov(aX, bY )∕( aX bY ) = abCov(X, Y )∕(ab X Y ) = Cov(X, Y )∕( X Y ) = X,Y .

28.

Cov(X + Y , Z)

= = =

(X+Y )Z − X+Y Z XZ+Y Z − ( X + Y ) Z XZ + Y Z − X Z − Y Z

= =

XZ − X Z + Y Z − Y Z Cov(X, Z) + Cov(Y , Z)

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CHAPTER 2

29. (a) V (X − ( X ∕ Y )Y )

2 = X + ( X ∕ Y )2 Y2 − 2( X ∕ Y )Cov(X, Y ) 2 = 2 X − 2( X ∕ Y )Cov(X, Y )

(b)

V (X − ( X ∕ Y )Y ) 2 2 X − 2( X ∕ Y )Cov(X, Y ) 2 2 X − 2( X ∕ Y ) X,Y X Y 2 2 2 X − 2 X,Y X 1 − X,Y X,Y

(c)

V (X + ( X ∕ Y )Y ) 2 2 X + 2( X ∕ Y )Cov(X, Y ) 2 2 X + 2( X ∕ Y ) X,Y X Y 2 2 2 X + 2 X,Y X 1 + X,Y X,Y

30. (a) X

≥ 0 ≥ 0 ≥

≥

≥ ≤

≥ ≥

≥

≥ ≥

0 0 0 1

0 0 0 0 0 −1

= 1.12C+2.69N+O−0.21F e = 1.12 C + 2.69 N + O − 0.21 F e = 1.12(0.0247) + 2.69(0.0255) + 0.1668 − 0.21(0.0597) = 0.2505

(b) Cov(C, N) = C,N C N = −0.44(0.0131)(0.0194) = −1.118 × 10−4 Cov(C, O) = C,O C O = 0.58(0.0131)(0.0340) = 2.583 × 10−4 Cov(C, F e) = C,F e C F e = 0.39(0.0131)(0.0413) = 2.110 × 10−4 Cov(N, O) = N,O N O = −0.32(0.0194)(0.0340) = −2.111 × 10−4 Cov(N, F e) = N,F e N F e = 0.09(0.0194)(0.0413) = 7.211 × 10−5 Cov(O, F e) = O,F e O F e = −0.35(0.0340)(0.0413) = −4.915 × 10−4

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SECTION 2.6

2 (c) X

2 1.12C+2.69N+O−0.21F e

2 = 1.122 C2 + 2.692 N + O2 + 0.212 F2 e + 2(1.12)(2.69)Cov(C, N) + 2(1.12)Cov(C, O) − 2(1.12)(0.21)Cov(C, F e) + 2(2.69)Cov(N, O) − 2(2.69)(0.21)Cov(N, F e) − 2(0.21)Cov(O, F e)

= 1.122 (0.0131)2 + 2.692 (0.0194)2 + 0.03402 + 0.212 (0.0413)2 + 2(1.12)(2.69)(−0.0001118) + 2(1.12)(0.0002583) − 2(1.12)(0.21)(0.0002110) + 2(2.69)(−0.0002111) − 2(2.69)(0.21)(0.00007211) − 2(0.21)(0.0004915) = 0.0029648 =

31.

√ 0.0029648 = 0.05445

= 7.84C+11.44N+O−1.58F e = 7.84 C + 11.44 N + O − 1.58 F e = 7.84(0.0247) + 11.44(0.0255) + 0.1668 − 1.58(0.0597) = 0.5578

2 7.84C+11.44N+O−1.58F e

2 7.842 C2 + 11.442 N + O2 + 1.582 F2 e + 2(7.84)(11.44)Cov(C, N) + 2(7.84)Cov(C, O) − 2(7.84)(1.58)Cov(C, F e)

+2(11.44)Cov(N, O) − 2(11.44)(1.58)Cov(N, F e) − 2(1.58)Cov(O, F e) 7.842 (0.0131)2 + 11.442 (0.0194)2 + 0.03402 + 1.582 (0.0413)2 + 2(7.84)(11.44)(−0.0001118) +2(7.84)(0.0002583) − 2(7.84)(1.58)(0.0002110) + 2(11.44)(−0.0002111) − 2(11.44)(1.58)(0.00007211) −2(1.58)(0.0004915)

0.038100

√ = 0.038100 = 0.1952

32. (a) Let c = ∫−∞ ℎ(y) dy. ∞

Now fX (x) = ∫−∞ f (x, y) dy = ∫−∞ g(x)ℎ(y) dy = g(x) ∫−∞ ℎ(y) dy = cg(x). ∞

∞

fY (y) = ∫−∞ f (x, y) dx = ∫−∞ g(x)ℎ(y) dx = ℎ(y) ∫−∞ g(x) dx = (1∕c)ℎ(y) ∫−∞ cg(x) dx ∞

∞

= (1∕c)ℎ(y) ∫−∞ fX (x) dx = (1∕c)ℎ(y)(1) = (1∕c)ℎ(y). ∞

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CHAPTER 2

(b) f (x, y) = g(x)ℎ(y) = cg(x)(1∕c)ℎ(y) = fX (x)gY (y). Therefore X and Y are independent.

33. (a) ∫−∞ ∫−∞ f (x, y) dx dy = ∫c ∫a k dx dy = k ∫c ∫a dx dy = k(d − c)(b − a) = 1. ∞

∞

Therefore k =

1 . (b − a)(d − c)

(b) fX (x) = ∫c k dy =

d−c 1 = (b − a)(d − c) b − a

b−a 1 = (b − a)(d − c) d − c

(d) f (x, y) =

1 1 1 = = fX (x)fY (y) (b − a)(d − c) b−a d−c

Supplementary Exercises for Chapter 2 1.

The system functions if at least one of the four components A, B, C, D, function. Therefore P(system functions) = 1 − P(all components fail). Let A be the event that component A functions, let B be the event that component B functions, let C be the event that component C functions, and let D be the event that component D functions. Then P (system functions) = 1 − P (Ac ∩ B c ∩ C c ∩ Dc ) = 1 − P (Ac )P (B c )P (C c )P (Dc ) = 1 − (0.1)(0.2)(0.05)(0.3) = 0.9997.

P (more than 3 tosses necessary) = P (frst 3 tosses are tails) =

Let A denote the event that the resistance is above specifcation, and let B denote the event that the resistance is below specifcation. Then A and B are mutually exclusive.

3 1 1 = . 2 8

(a) P (doesn’t meet specifcation) = P (A ∪ B) = P (A) + P (B) = 0.05 + 0.10 = 0.15

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SUPPLEMENTARY EXERCISES FOR CHAPTER 2

(b) P [B | (A ∪ B)] =

P [(B ∩ (A ∪ B)] P (B) 0.10 = = = 0.6667 P (A ∪ B) P (A ∪ B) 0.15

Let D denote the event that the bag is defective, let L1 denote the event that the bag came from line 1, and let L2 denote them event that the bag came from line 2. Then P (D | L1 ) = 1∕100 and P (D | L2 ) = 3∕100. (a) P (L1 ) =

2 3

(b) P (D) = P (D | L1 )P (L1 ) + P (D | L2 )P (L2 ) = (1∕100)(2∕3) + (3∕100)(1∕3) =

P (D | L1 )P (L1 ) (1∕100)(2∕3) 2 = = P (D) 1∕60 5

(d) P (L1 | Dc ) =

1 60

P (Dc | L1 )P (L1 ) [1 − P (D | L1 )]P (L1 ) (1 − 1∕100)(2∕3) 198 = = = = 0.6712 P (Dc ) P (Dc ) 59∕60 295

Let R be the event that the shipment is returned. Let B1 be the event that the frst brick chosen meets the specifcation, let B2 be the event that the second brick chosen meets the specifcation, let B3 be the event that the third brick chosen meets the specifcation, and let B4 be the event that the fourth brick chosen meets the specifcation. Since the sample size of 4 is a small proportion of the population, it is reasonable to treat these events as independent, each with probability 0.9. P (R) = 1 − P (Rc ) = 1 − P (B1 ∩ B2 ∩ B3 ∩ B4 ) = 1 − (0.9)4 = 0.3439.

6. (a) (0.99)10 = 0.904 (b) Let p be the required probability. Then p10 = 0.95. Solving for p, p = 0.9949.

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CHAPTER 2

Let A be the event that the bit is reversed at the frst relay, and let B be the event that the bit is reversed at the second relay. Then P (bit received is the same as the bit sent) = P (Ac ∩ B c ) + P (A ∩ B) = P (Ac )P (B c ) + P (A)P (B) = 0.92 + 0.12 = 0.82.

8. (a)

(b)

Ê−1

Ê0

k(1 − x ) dx = k

Ê−1

(1 − x ) dx = 1. Since

0 1 x3 0.75(1 − x2 ) dx = 0.75 x − 3

(f) 2

Ê−1

= −1

4 3 , k = = 0.75. 3 4

0.25

= 0.3672 −0.25

(e) The median xm solves

(1 − x ) dx =

x3 x− 3

= 0.5

0 2 1 x x4 − 0.75x(1 − x ) dx = 0.75 (d) = Ê−1 2 4 1

Ê−1

0 1 x3 0.75(1 − x2 ) dx = 0.75 x − Ê−0.25 3 0.25

(c)

=0 −1

0 1 x3 0.75(1 − x2 ) dx = 0.5. Therefore 0.75 x − 3

= 0.5, so xm = 0 −1

0.75x2 (1 − x2 ) dx − 2

= 0.75

0 3 1 x x5 − 3 5

− 02 −1

= 0.2 √ = 0.2 = 0.4472

Let A be the event that two di˙erent numbers come up, and let B be the event that one of the dice comes up 6. Then A contains 30 equally likely outcomes (6 ways to choose the number for the frst die times 5 ways to choose the number for the second die). Of these 30 outcomes, 10 belong to B, specifcally (1,6), (2,6), (3,6), (4,6), (5,6), (6,1), (6,2), (6,3), (6,4), and (6,5). Therefore P (B | A) = 10∕30 = 1∕3.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 2

10.

Let A be the event that the frst component is defective and let B be the event that the second component is defective. (a) P (X = 0) = P (Ac ∩ B c ) = P (Ac )P (B c |Ac ) =

8 10

7 = 0.6222 9

(b) P (X = 1) = P (A ∩ B c ) + P (Ac ∩ B) = P (A)P (B c |A) + P (Ac )P (B|Ac ) 2 8 8 2 = + 10 9 10 9 = 0.3556 (c) P (X = 2) = P (A ∩ B) = P (A)P (B|A) =

2 10

1 = 0.0222 9

(d) pX (0) = 0.6222, pX (1) = 0.3556, pX (2) = 0.0222, pX (x) = 0 if x ≠ 0, 1, or 2. (e) X = 0pX(0) + 1pX (1) + 2pX (2) = 0(0.6222) + 1(0.3556) + 2(0.0222) = 0.4 t (f) X =

2 = 02 pX(0) + 12 pX (1) + 22 pX (2) − X

11. (a) P (X ≤ 2 and Y ≤ 3)

= =

√ 02 (0.6222) + 12 (0.3556) + 22 (0.0222) − 0.42 = 0.5333

1 −x∕2−y∕3 e dy dx Ê0 Ê0 6 H 3I 2 1 −x∕2 −e−y∕3 dx e Ê0 2 0

Ê0

1 −x∕2 (1 − e−1 ) dx e 2 2

−1

− 1)e

= =

(1 − e−1 )2 0.3996

−x∕2 0

(b) P (X ≥ 3 and Y ≥ 3)

Ê3

∞

Ê3

∞

1 −x∕2−y∕3 dy dx e 6

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CHAPTER 2

= =

Ê3

∞

Ê3

∞

H 1 −x∕2 e 2

∞I −y∕3

−e

1 −x∕2 −1 e e dx 2 ∞

−1 −x∕2

= −e e

e−5∕2

= 0.0821

If x > 0, fX (x) =

Ê0 T

Therefore fX (x) =

1 1 −x∕2−y∕3 dy = e−x∕2 e 6 2 1 −x∕2 e 2 0

If y > 0, fY (y) =

Ê3 T

Therefore fY (y) =

1 −x∕2 e . 2

1 −y∕3 e . 3

x≤0

1 −x∕2−y∕3 1 e dx = e−y∕3 6 3 1 −y∕3 e 3 0

−e

x>0

(d) If y ≤ 0, f (x, y) = 0 for all x so fY (y) = 0. ∞

∞I −y∕3

∞I

−e

−x∕2 0

y>0

y≤0

(e) Yes, f (x, y) = fX (x)fY (y).

12. (a) A and B are mutually exclusive if P (A ∩ B) = 0, or equivalently, if P (A ∪ B) = P (A) + P (B). So if P (B) = P (A ∪ B) − P (A) = 0.7 − 0.3 = 0.4, then A and B are mutually exclusive.

(b) A and B are independent if P (A ∩ B) = P (A)P (B). Now P (A ∪ B) = P (A) + P (B) − P (A ∩ B). So A and B are independent if P (A ∪ B) = P (A) + P (B) − P (A)P (B), that is, if 0.7 = 0.3 + P (B) − 0.3P (B). This equation is satisfed if P (B) = 4∕7.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 2

13.

Let D denote the event that a snowboard is defective, let E denote the event that a snowboard is made in the eastern United States, let W denote the event that a snowboard is made in the western United States, and let C denote the event that a snowboard is made in Canada. Then P (E) = P (W ) = 10∕28, P (C) = 8∕28, P (D|E) = 3∕100, P (D|W ) = 6∕100, and P (D|C) = 4∕100.

(a) P (D) = P (D|E)P (E) + P (D|W )P (W ) + P (D|C)P (C) 10 3 10 6 8 4 = + + 28 100 28 100 28 100 122 = = 0.0436 2800

(b) P (D ∩ C) = P (D|C)P (C) =

8 28

4 100

32 = 0.0114 2800

(c) Let U be the event that a snowboard was made in the United States. 122 32 90 Then P (D ∩ U ) = P (D) − P (D ∩ C) = − = . 2800 2800 2800 90∕2800 P (D ∩ U ) 90 P (U |D) = = = = 0.7377. P (D) 122∕2800 122

14. (a) Discrete. The possible values are 10, 60, and 80.

(b) X = 10pX (10) + 60pX (60) + 80pX (80) = 10(0.40) + 60(0.50) + 80(0.10) = 42 2 = 102 p (10) + 602 p (60) + 802 p (80) − 2 = 102 (0.40) + 602 (0.50) + 802 (0.10) − 422 = 716 (c) X X X X X √ X = 716 = 26.76

(d) P (X > 50) = P (X = 60) + P (X = 80) = 0.5 + 0.1 = 0.6

15.

� 6! = 15. Each is equally likely to be chosen. Of these pairs, The total number of pairs of cubicles is 62 = 2!4! fve are adjacent (1 and 2, 2 and 3, 3 and 4, 4 and 5, 5 and 6). Therefore the probability that an adjacent pair of cubicles is selected is 5∕15, or 1∕3.

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100 16.

CHAPTER 2 � 8! The total number of combinations of four shoes that can be selected from eight is 84 = 4!4! = 70. The four shoes will contain no pair if exactly one shoe is selected from each pair. Since each pair contains two shoes, the number of ways to select exactly one shoe from each pair is 24 = 16. Therefore the probability that the four shoes contain no pair is 16∕70, or 8∕35.

17. (a) 3X = 3 X = 3(2) = 6,

2 = 32 2 = (32 )(12 ) = 9 3X X

(b) X+Y = X + Y = 2 + 2 = 4,

2 2 + 2 = 12 + 32 = 10 X+Y = X Y

2 2 + 2 = 12 + 32 = 10 X−Y = X Y

(d) 2X+6Y = 2 X + 6 Y = 2(2) + 6(2) = 16,

18.

2 2 + 62 2 = (22 )(12 ) + (62 )(32 ) = 328 2X+6Y = 22 X Y

Cov(X, Y ) = X,Y X Y = (0.5)(2)(1) = 1 (a) X +Y = X + Y = 1 + 3 = 4,

(b) X−Y = X − Y = 1 − 3 = −2,

2 2 + 2 + 2Cov(X, Y ) = 22 + 12 + 2(1) = 7 X+Y = X Y

2 2 + 2 − 2Cov(X, Y ) = 22 + 12 − 2(1) = 3 X−Y = X Y

(c) 3X+2Y = 3 X + 2 Y = 3(1) + 2(3) = 9, 2 2 + 22 2 + 2(3)(2)Cov(X, Y ) = (32 )(22 ) + (22 )(12 ) + 2(3)(2)(1) = 52 = 32 X 3X+2Y Y

(d) 5Y −2X = 5 Y − 2 X = 5(3) − 2(1) = 13, 2 2 + 2(5)(−2)Cov(X, Y ) = 52 (12 ) + (−2)2 (22 ) + 2(5)(−2)(1) = 21 5Y = 52 Y2 + (−2)2 X −2X

19.

The marginal probability mass function pX (x) is found by summing along the rows of the joint probability mass function.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 2

x 0.02 0.04 0.06 0.08 pY (y)

100 0.05 0.01 0.04 0.04 0.14

y 150 0.06 0.08 0.08 0.14 0.36

200 0.11 0.10 0.17 0.12 0.50

pX (x) 0.22 0.19 0.29 0.30

(a) For additive concentration (X): pX (0.02) = 0.22, pX (0.04) = 0.19, pX (0.06) = 0.29, pX (0.08) = 0.30, and pX (x) = 0 for x ≠ 0.02, 0.04, 0.06, or 0.08. For tensile strength (Y ): The marginal probability mass function pY (y) is found by summing down the columns of the joint probability mass function. Therefore pY (100) = 0.14, pY (150) = 0.36, pY (200) = 0.50, and pY (y) = 0 for y ≠ 100, 150, or 200. (b) No, X and Y are not independent. For example P (X = 0.02 ∩ Y = 100) = 0.05, but P (X = 0.02)P (Y = 100) = (0.22)(0.14) = 0.0308. P (Y ≥ 150 and X = 0.04) P (X = 0.04) P (Y = 150 and X = 0.04) + P (Y = 200 and X = 0.04) = P (X = 0.04) 0.08 + 0.10 = 0.19 = 0.947

(d) P (Y > 125 | X = 0.08)

P (Y > 125 and X = 0.08) P (X = 0.08) P (Y = 150 and X = 0.08) + P (Y = 200 and X = 0.08) = P (X = 0.08) 0.14 + 0.12 = 0.30 = 0.867

(e) The tensile strength is greater than 175 if Y = 200. Now P (Y = 200 and X = 0.02) 0.11 = = 0.500, P (Y = 200 | X = 0.02) = P (X = 0.02) 0.22 P (Y = 200 and X = 0.04) 0.10 = = 0.526, P (Y = 200 | X = 0.04) = P (X = 0.04) 0.19 P (Y = 200 and X = 0.06) 0.17 P (Y = 200 | X = 0.06) = = = 0.586, P (X = 0.06) 0.29

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102

CHAPTER 2 P (Y = 200 and X = 0.08) 0.12 = = 0.400. P (X = 0.08) 0.30 The additive concentration should be 0.06. P (Y = 200 | X = 0.08) =

20. (a) X

= 0.02pX (0.02) + 0.04pX (0.04) + 0.06pX (0.06) + 0.08pX (0.08) = 0.02(0.22) + 0.04(0.19) + 0.06(0.29) + 0.08(0.30) = 0.0534

(b) Y = 100pY (100) + 150pY (150) + 200pY (200) = 100(0.14) + 150(0.36) + 200(0.50) = 168

2 (c) X

2 = 0.022 pX (0.02) + 0.042 pX (0.04) + 0.062 pX (0.06) + 0.082 pX (0.08) − X

= 0.022 (0.22) + 0.042 (0.19) + 0.062 (0.29) + 0.082 (0.30) − 0.05342 = 0.00050444 √ X = 0.00050444 = 0.02246

(d) Y2

1002 pY (100) + 1502 pY (150) + 2002 pY (200) − Y2

= 1002 (0.14) + 1502 (0.36) + 2002 (0.50) − 1682 = 1276 √ Y = 1276 = 35.721 (e) Cov(X, Y ) = XY − X Y . XY

(0.02)(100)P (X = 0.02 and Y = 100) + (0.02)(150)P (X = 0.02 and Y = 150) + (0.02)(200)P (X = 0.02 and Y = 200) + (0.04)(100)P (X = 0.04 and Y = 100) + (0.04)(150)P (X = 0.04 and Y = 150) + (0.04)(200)P (X = 0.04 and Y = 200) + (0.06)(100)P (X = 0.06 and Y = 100) + (0.06)(150)P (X = 0.06 and Y = 150) + (0.06)(200)P (X = 0.06 and Y = 200) + (0.08)(100)P (X = 0.08 and Y = 100) + (0.08)(150)P (X = 0.08 and Y = 150) + (0.08)(200)P (X = 0.08 and Y = 200)

(0.02)(100)(0.05) + (0.02)(150)(0.06) + (0.02)(200)(0.11) + (0.04)(100)(0.01) + (0.04)(150)(0.08) + (0.04)(200)(0.10) + (0.06)(100)(0.04) + (0.06)(150)(0.08) + (0.06)(200)(0.17) + (0.08)(100)(0.04) + (0.08)(150)(0.14) + (0.08)(200)(0.12) = 8.96

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103

Cov(X, Y ) = 8.96 − (0.0534)(168) = −0.0112

(f) X,Y =

Cov(X, Y ) −0.0112 = = −0.01396 X Y (0.02246)(35.721)

21. (a) pY |X (100 | 0.06) =

p(0.06, 100) 0.04 4 = = = 0.138 pX (0.06) 0.29 29

pY |X (150 | 0.06) =

p(0.06, 150) 0.08 8 = = = 0.276 pX (0.06) 0.29 29

pY |X (200 | 0.06) =

p(0.06, 200) 0.17 17 = = = 0.586 pX (0.06) 0.29 29

(b) pX|Y (0.02 | 100) =

p(0.02, 100) 0.05 5 = = = 0.357 pY (100) 0.14 14

pX|Y (0.04 | 100) =

p(0.04, 100) 0.01 1 = = = 0.071 pY (100) 0.14 14

pX|Y (0.06 | 100) =

p(0.06, 100) 0.04 4 = = = 0.286 pY (100) 0.14 14

pX|Y (0.08 | 100) =

p(0.08, 100) 0.04 4 = = = 0.286 pY (100) 0.14 14

= 100pY |X (100 | 0.06) + 150pY |X (150 | 0.06) + 200pY |X (200 | 0.06) = 100(4∕29) + 150(8∕29) + 200(17∕29) =

(d) E(X | Y = 100)

172.4

= 0.02pX|Y (0.02 | 100) + 0.04pX|Y (0.04 | 100) + 0.06pX|Y (0.06 | 100) + 0.08pX|Y (0.08 | 100) = 0.02(5∕14) + 0.04(1∕14) + 0.06(4∕14) + 0.08(4∕14) = 0.0500

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104

CHAPTER 2

22.

Let D denote the event that an item is defective, let S1 denote the event that an item is produced on the frst shift, let S2 denote the event that an item is produced on the second shift, and let S3 denote the event that an item is produced on the third shift. Then P (S1 ) = 0.50, P (S2 ) = 0.30, P (S3 ) = 0.20, P (D|S1 ) = 0.01, P (D|S2 ) = 0.02, and P (D|S3 ) = 0.03.

(a) P (S1 |D)

(b) P (S3 |Dc )

P (D|S1 )P (S1 ) P (D|S1 )P (S1 ) + P (D|S2 )P (S2 ) + P (D|S3 )P (S3 ) (0.01)(0.50) = (0.01)(0.50) + (0.02)(0.30) + (0.03)(0.20) = 0.294 =

23. (a) Under scenario A: = 0(0.65) + 5(0.2) + 15(0.1) + 25(0.05) = 3.75 √ = 02 (0.65) + 52 (0.2) + 152 (0.1) + 252 (0.05) − 3.752 = 6.68

(b) Under scenario B: = 0(0.65) + 5(0.24) + 15(0.1) + 20(0.01) = 2.90 √ = 02 (0.65) + 52 (0.24) + 152 (0.1) + 202 (0.01) − 2.902 = 4.91

(c) Under scenario C: = 0(0.65) + 2(0.24) + 5(0.1) + 10(0.01) = 1.08 √ = 02 (0.65) + 22 (0.24) + 52 (0.1) + 102 (0.01) − 1.082 = 1.81 (d) Let L denote the loss. Under scenario A, P (L < 10) = P (L = 0) + P (L = 5) = 0.65 + 0.2 = 0.85. Under scenario B, P (L < 10) = P (L = 0) + P (L = 5) = 0.65 + 0.24 = 0.89.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 2

Under scenario C, P (L < 10) = P (L = 0) + P (L = 2) + P (L = 5) = 0.65 + 0.24 + 0.1 = 0.99.

24.

Let L denote the loss. (a) P (A ∩ L = 5) = P (L = 5|A)P (A) = (0.20)(0.20) = 0.040

(b) P (L = 5) =

P (L = 5|A)P (A) + P (L = 5|B)P (B) + P (L = 5|C)P (C)

= (0.20)(0.20) + (0.30)(0.24) + (0.50)(0.1) = 0.162 (c) P (A|L = 5) =

P (A ∩ L = 5) 0.040 = = 0.247 P (L = 5) 0.162

1 2 = = 0.0667 15 9 3 4 4 3 4 + = = 0.2667 p(1, 0) = P (X = 1 and Y = 0) = 10 9 15 10 9 2 3 4 = = 0.1333 p(2, 0) = P (X = 2 and Y = 0) = 15 10 9 3 3 3 3 3 + = = 0.2000 p(0, 1) = P (X = 0 and Y = 1) = 10 9 15 10 9 3 4 4 3 4 + = = 0.2667 p(1, 1) = P (X = 1 and Y = 1) = 10 9 15 10 9 1 3 2 = = 0.0667 p(0, 2) = P (X = 0 and Y = 2) = 15 10 9

25. (a) p(0, 0) = P (X = 0 and Y = 0) =

3 10

p(x, y) = 0 for all other pairs (x, y).

The joint probability mass function is

x 0 1 2

0 0.0667 0.2667 0.1333

y 1 0.2000 0.2667 0

2 0.0667 0 0

(b) The marginal probability density function of X is: 1 3 1 1 pX (0) = p(0, 0) + p(0, 1) + p(0, 2) = + + = 15 15 15 3

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CHAPTER 2

pX (1) = p(1, 0) + p(1, 1) = pX (2) = p(2, 0) =

4 4 8 + = . 15 15 15

2 15

X = 0pX (0) + 1pX (1) + 2pX (2) = 0

1 8 2 12 +1 +2 = = 0.8 3 15 15 15

(c) The marginal probability density function of Y is: 1 4 2 7 pY (0) = p(0, 0) + p(1, 0) + p(2, 0) = + + = 15 15 15 15 3 4 7 pY (1) = p(0, 1) + p(1, 1) = + = . 15 15 15 1 pY (2) = p(0, 2) = 15 7 7 1 9 Y = 0pY (0) + 1pY (1) + 2pY (2) = 0 +1 +2 = = 0.6 15 15 15 15 t (d) X

= = =

(e) Y

= = =

2 02 pX (0) + 12 pX (1) + 22 pX (2) − X

√

02 (1∕3) + 12 (8∕15) + 22 (2∕15) − (12∕15)2 96∕225 = 0.6532

t 02 pY (0) + 12 pY (1) + 22 pY (2) − Y2 t 02 (7∕15) + 12 (7∕15) + 22 (1∕15) − (9∕15)2 √ 84∕225 = 0.6110

(f) Cov(X, Y ) = XY − X Y .

(0)(0)p(0, 0) + (1)(0)p(1, 0) + (2)(0)p(2, 0) + (0)(1)p(0, 1) + (1)(1)p(1, 1) + (0)(2)p(0, 2) 4 4 = (1)(1) = 15 15 4 12 9 48 − =− = −0.2133 Cov(X, Y ) = 15 15 15 225 =

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SUPPLEMENTARY EXERCISES FOR CHAPTER 2

(g) X,Y =

−48∕225 Cov(X, Y ) =√ = −0.5345 √ X Y 96∕225 84∕225

26. (a) The constant c solves 1

Since

Ê0 Ê0

c(x + y)2 dx dy = 1.

(x + y)2 dx dy =

(b) For 0 < x < 1, fX (x) =

Ê0

7 6 ,c = . 6 7

2(x + y)3 6 (x + y)2 dy = 7 7

For x ≤ 0 or x ≥ 1 f (x, y) = 0 for all y, so fX (x) = 0. ⎧ 6x2 + 6x + 2 ⎪ Therefore fX (x) = ⎨ 7 ⎪ 0 ⎩

= 0

6x2 + 6x + 2 . 7

0<x<1 otherwise

f (x, y) . If x ≤ 0 or x ≥ 1 fX (x) = 0, so fY |X (y | x) is undefned. fX (x) Now assume 0 < x < 1. If y ≤ 0 or y ≥ 1 then f (x, y) = 0 for all x so fY |X (y | x) = 0.

If 0 < y < 1 then fY |X (y | x) =

(6∕7)(x + y)2 3(x + y)2 = . (6x2 + 6x + 2)∕7 3x2 + 3x + 1

⎧ 3(x + y)2 ⎪ Therefore fY |X (y | x) = ⎨ 3x2 + 3x + 1 ⎪ 0 ⎩ (d) E(Y | X = 0.4)

∞

Ê−∞ 1

0<y<1 otherwise

yfY |X (y | 0.4) dy 3y(0.4 + y)2 dy 3(0.4)2 + 3(0.4) + 1

Ê0

0.24y2 + 0.8y3 + 0.75y4 2.68

= 0.6679

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CHAPTER 2

(e) No, fY |X (y | x) ≠ fY (y).

∞

27. (a) X =

Ê−∞

2 = (b) X Ê

xfX (x) dx =

Ê0

2 2 6x + 6x + 2

6x2 + 6x + 2 1 x dx = (3x4 + 4x3 + 2x2 ) 7 14

1 2 dx − X = 7

6x5 3x4 2x3 + + 5 2 3

− 0

9 14

= 0

9 = 0.6429 14

2 =

199 = 0.06769 2940

Ê0 Ê0

6 (x + y)2 dx dy 7

H 6 x4 2x3 y x2 y2 = y + + Ê0 7 4 3 2 1 1 0 3 2y2 y 6 y = + + dy Ê0 7 2 3 4 0 1 1 6 y2 2y3 y4 = + + 7 8 9 8 1

dy 0

17 42

9 , computed in part (a). To compute Y , note that the joint density is symmetric in x and y, so the 14 9 marginal density of Y is the same as that of X. It follows that Y = X = . 14 17 9 9 −5 − = = −0.008503. Cov(X, Y ) = 42 14 14 588 X =

2 = (d) Since the marginal density of Y is the same as that of X, Y2 = X

Therefore X,Y =

199 . 2940

−5∕588 Cov(X, Y ) −25 =√ = = −0.1256. √ X Y 199 199∕2940 199∕2940

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28.

109

Since the coin is fair, P (H) = P (T ) = 1∕2. The tosses are independent. So P (HT T HH) = P (H)P (T )P (T )P (H)P (H) = (1∕2)5 = 1∕32, and P (HHHHH) = P (H)P (H)P (H )P (H)P (H) = (1∕2)5 = 1∕32 as well.

29. (a) pX (0) = 0.6, pX (1) = 0.4, pX (x) = 0 if x ≠ 0 or 1. (b) pY (0) = 0.4, pY (1) = 0.6, pY (y) = 0 if y ≠ 0 or 1. (c) Yes. It is reasonable to assume that knowledge of the outcome of one coin will not help predict the outcome of the other.

(d) p(0, 0) = pX (0)pY (0) = (0.6)(0.4) = 0.24, p(0, 1) = pX (0)pY (1) = (0.6)(0.6) = 0.36, p(1, 0) = pX (1)pY (0) = (0.4)(0.4) = 0.16, p(1, 1) = pX (1)pY (1) = (0.4)(0.6) = 0.24, p(x, y) = 0 for other values of (x, y).

30.

The probability mass function of X is pX (x) = 1∕6 for x = 1, 2, 3, 4, 5, or 6, and pX (x) = 0 for other values of x. Therefore X = 1(1∕6)+2(1∕6)+3(1∕6)+4(1∕6)+5(1∕6)+6(1∕6) = 3.5. The probability mass function of Y is the same as that of X, so Y = X = 3.5. Since X and Y are independent, XY = X Y = (3.5)(3.5) = 12.25.

31. (a) The possible values of the pair (X, Y ) are the ordered pairs (x, y) where each of x and y is equal to 1, 2, or 3. There are nine such ordered pairs, and each is equally likely. Therefore pX,Y (x, y) = 1∕9 for x = 1, 2, 3 and y = 1, 2, 3, and pX,Y (x, y) = 0 for other values of (x, y). (b) Both X and Y are sampled from the numbers {1, 2, 3}, with each number being equally likely. Therefore pX (1) = pX (2) = pX (3) = 1∕3, and pX (x) = 0 for other values of x. pY is the same. (c) X = Y = 1(1∕3) + 2(1∕3) + 3(1∕3) = 2

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CHAPTER 2

(d) XY =

∑3

1 ÉÉ 1 xy = (1 + 2 + 3)(1 + 2 + 3) = 4. y=1 xypX,Y (x, y) = 9 9 x=1 y=1 3

∑3

x=1

Another way to compute XY is to note that X and Y are independent, so XY = X Y = (2)(2) = 4. (e) Cov(X, Y ) = XY − X Y = 4 − (2)(2) = 0

32. (a) The values of X and Y must both be integers between 1 and 3 inclusive, and may not be equal. There are six possible values for the ordered pair (X, Y ), specifcally (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2). Each of these six ordered pairs is equally likely. Therefore pX,Y (x, y) = 1∕6 for (x, y) = (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2), and pX,Y (x, y) = 0 for other values of (x, y). (b) The value of X is chosen from the integers 1, 2, 3 with each integer being equally likely. Therefore pX (1) = pX (2) = pX (3) = 1∕3. The marginal probability mass function pY is the same. To see this, compute pY (1) = pX,Y (2, 1) + pX,Y (3, 1) = 1∕6 + 1∕6 = 1∕3 pY (2) = pX,Y (1, 2) + pX,Y (3, 2) = 1∕6 + 1∕6 = 1∕3 pY (3) = pX,Y (1, 3) + pX,Y (2, 3) = 1∕6 + 1∕6 = 1∕3 (c) X = Y = 1(1∕3) + 2(1∕3) + 3(1∕3) = 2

(d) XY

(1)(2)pX,Y (1, 2) + (1)(3)pX,Y (1, 3) + (2)(1)pX,Y (2, 1) + (2)(3)pX,Y (2, 3) + (3)(1)pX,Y (3, 1) + (3)(2)pX,Y (3, 2)

[(1)(2) + (1)(3) + (2)(1) + (2)(3) + (3)(1) + (3)(2)](1∕6) 11 = 3

(e) Cov(X, Y ) = XY − X Y =

11 1 − (2)(2) = − 3 3

33. (a) X = ∫−∞ xf (x) dx. Since f (x) = 0 for x ≤ 0, X = ∫0 xf (x) dx. ∞

∞

(b) X = ∫0 xf (x) dx ≥ ∫k xf (x) dx ≥ ∫k kf (x) dx = kP (X ≥ k) ∞

∞

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SUPPLEMENTARY EXERCISES FOR CHAPTER 2

111

(c) X ∕k ≥ kP (X ≥ k)∕k = P (X ≥ k) (d) X = (Y − Y )2 = Y2 (e) P (|Y − Y | ≥ k Y ) = P ((Y − Y )2 ≥ k2 Y2 ) = P (X ≥ k2 Y2 ) (f) P (|Y − Y | ≥ k Y ) = P (X ≥ k2 Y2 ) ≤ X ∕(k2 Y2 ) = Y2 ∕(k2 Y2 ) = 1∕k2

34.

2. A = R2 . We now fnd R 2 = 2 2 R R2 − R . Substituting, we obtain 1 = R2 − 10 . Therefore R2 = 101, and A = 101 .

35. (a) If the pooled test is negative, it is the only test performed, so X = 1. If the pooled test is positive, then n additional tests are carried out, one for each individual, so X = n + 1. The possible values of X are therefore 1 and n + 1. (b) The possible values of X are 1 and 5. Now X = 1 if the pooled test is negative. This occurs if none of the individuals has the disease. The probability that this occurs is (1 − 0.1)4 = 0.6561. Therefore P (X = 1) = 0.6561. It follows that P (X = 5) = 0.3439. So X = 1(0.6561) + 5(0.3439) = 2.3756. (c) The possible values of X are 1 and 7. Now X = 1 if the pooled test is negative. This occurs if none of the individuals has the disease. The probability that this occurs is (1 − 0.2)6 = 0.262144. Therefore P (X = 1) = 0.262144. It follows that P (X = 7) = 0.737856. So X = 1(0.262144) + 7(0.737856) = 5.4271. (d) The possible values of X are 1 and n + 1. Now X = 1 if the pooled test is negative. This occurs if none of the individuals has the disease. The probability that this occurs is (1 − p)n . Therefore P (X = 1) = (1 − p)n . It follows that P (X = n + 1) = 1 − (1 − p)n . So X = 1(1 − p)n + (n + 1)(1 − (1 − p)n ) = n + 1 − n(1 − p)n . (e) The pooled method is more economical if 11 − 10(1 − p)10 < 10. Solving for p yields p < 0.2057.

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Chapter 3 Section 3.1 1.

(ii). The measurements are close together, and thus precise, but they are far from the true value of 100◦ C, and thus not accurate.

2. (a) No. Both measurements di˙ered from the true value by 0.2◦ C , so we cannot tell which one is on the average closer to the true value. (b) Yes, the second one is more precise because its uncertainty is smaller.

3. (a) True (b) False (c) False (d) True

4. (a) The uncertainty is 0.02(100) = 2 g (b) The uncertainty is 0.02(50) = 1 g

5. (a) No, we cannot determine the standard deviation of the process from a single measurement. (b) Yes, the bias can be estimated to be 2 pounds, because the reading is 2 pounds when the true weight is 0.

6. (a) Yes, the uncertainty can be estimated with the standard deviation of the four measurements, which is 1.71.

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SECTION 3.2

(b) Yes, the bias can be estimated to be 2 pounds, because the reading is 2 pounds when the true weight is 0.

7. (a) Yes, the uncertainty can be estimated with the standard deviation of the fve measurements, which is 21.3 g. (b) No, the bias cannot be estimated, since we do not know the true value.

8. (a) Yes, the uncertainty can be estimated with the standard deviation of the fve measurements, which is 0.0109 ppm. (b) Yes, the bias can be estimated to be the average measurement minus the true value. This di˙erence is −0.0624 ppm.

9. (a) No, they are in increasing order, which would be highly unusual for a simple random sample. (b) No, since they are not a simple random sample from a population of possible measurements, we cannot estimate the uncertainty.

Section 3.2 1. (a) 4X = 4 X = 4(0.3) = 1.2 (b) X−2Y =

t √ 2 + 22 2 = 0.32 + (22 )(0.22 ) = 0.5 X Y

t √ 2 + 32 2 = 22 X (22 )(0.32 ) + (32 )(0.22 ) = 0.8485 Y

√ Let n be the number of measurements required. Then 1.5∕ n = 0.5, so n = 9.

√ Let n be the necessary number of measurements. Then X = 3∕ n = 1. Solving for n yields n = 9.

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CHAPTER 3

r = 5, ℎ = 6, ℎ = 0.02. V = r2 ℎ∕3 = 26.1799ℎ = 26.1799(6) = 157.1 cm3 . V = 26.1799ℎ = 26.1799 ℎ = 26.1799(0.02) = 0.52 cm3 .

Let X represent the estimated mean annual level of land uplift for the years 1774–1884, and let Y represent the estimated mean annual level of land uplift for the years 1885–1984. Then X = 4.93, Y = 3.92, X = 0.23, and Y = 0.19. The di˙erence is X − Y = 1.01 mm. t √ 2 + 2 = 0.232 + 0.192 = 0.30 mm. The uncertainty is X−Y = X Y

Let X represent the measured diameter of the hole, and let Y represent the measured diameter of the piston. Then X = 20.00, Y = 19.90, X = 0.01, and Y = 0.02. The clearance is 0.5X − 0.5Y = 0.05 cm. t √ 2 + (−0.5)2 2 = The uncertainty is 0.5X−0.5Y = 0.52 X (0.52 )(0.012 ) + (0.52 )(0.022 ) = 0.011 cm. Y

d = 3, F = 2.2, F = 0.1. W = F d = (2.2)(3) = 6.6 Nm. W = 3F = 3 F = 3(0.1) = 0.3 Nm.

T = 1.5, L = 0.559, L = 0.005 g = (4 2 ∕T 2 )L = 17.546L = 17.546(0.559) = 9.81 m/s2 g = 17.546L = 17.546 L = 17.546(0.005) = 0.088 m/s2

DW = 1000, DS = 500, DS = 5. G = DS ∕DW = 500∕1000 = 0.5 G = 0.001DS = 0.001 DS = 0.001(5) = 0.005 kg/m3 .

10. (a) = 1.49, ℎ = 10, = 30.0, = 0.1 V = ℎ∕ = 6.7114 = 6.7114(30) = 201.34 mm/s V = 6.7114 = 6.7114 = 6.7114(0.1) = 0.67 mm/s (b) V = 6.7114 = 0.2, so = 0.0298 mm/s

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SECTION 3.2

11. (a) Ta = 36, k = 0.025, t = 10, T0 = 72, T0 = 0.5 T = Ta + (T0 − Ta )e−kt = 36 + 0.7788(T0 − 36) = 36 + 0.7788(72 − 36) = 64.04◦ F T = Ta +(T0 −Ta )e−kt = 36+0.7788(T0 −36) = 0.7788(T0 −36) = 0.7788 (T0 −36) = 0.7788 T0 = 0.7788(0.5) = 0.39◦ F (b) T0 = 72, k = 0.025, t = 10, Ta = 36, Ta = 0.5 T = Ta + (T0 − Ta )e−kt = Ta + 0.7788(72 − Ta ) = 56.074 + 0.2212Ta = 56.074 + 0.2212(36) = 64.04◦ F T = 56.074+0.2212Ta = 0.2212Ta = 0.2212 Ta = 0.2212(0.5) = 0.11◦ F

12.

0 = 50, w = 1.0, k = 0.29, k = 0.05 = 0 (1 − kw) = 50(1 − k) = 50 − 50k = 50 − 50(0.29) = 35.5 MPa = 50−50k = −50k = 50k = 50 k = 50(0.05) = 2.5 MPa

13. (a) The √uncertainty in the average of 9 measurements is approximately equal to s∕ 9 = 0.081∕3 = 0.027 cm. (b) The uncertainty in a single measurement is approximately equal to s, which is 0.081 cm.

14. (a) The bias is 2 g and the uncertainty is = 3 g. √ (b) The bias is 2 g and the uncertainty is ∕ 4 = 1.5 g. √ (c) The bias is 2 g and the uncertainty is ∕ 400 = 0.15 g. √ (d) The uncertainty gets smaller; it is proportional to 1∕ n where n is the number of measurements. (e) The bias stays the same. The mean of the sample average is the same no matter what the sample size, so the bias is the same as well.

15. (a) Let X = 87.0 denote the average of the eight measurements, let s = 2.0 denote the sample standard deviation, and let de√ note the unk standard deviation. The volume is estimated with X. The uncertainty √ nown population √ is X = ∕ 8 ≈ s∕ 8 = 2.0∕ 8 = 0.7 mL.

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√ √ (b) X ≈ s∕ 16 = 2.0∕ 16 = 0.5 mL √ (c) Let n be the required number of measurements. Then ∕ n = 0.4. √ Approximating with s = 2.0 yields 2.0∕ n ≈ 0.4, so n ≈ 25.

16.

The average of the fve measurements is X = 37.0, and the sample standard deviation is s = 1.28841. √ √ (a) k = X ± s∕ 5 = 37.0 ± 1.28841∕ 5 = 37.0 ± 0.6 N/m. √ (b) The uncertainty is approximately s∕ 10 = 0.41 N/m. √ (c) Let n be the required number of measurements. Then ∕ n = 0.3. √ Approximating with s = 1.28841 yields 1.28841∕ n = 0.3. Solving for n yields n = 18.44. Since n must be an integer, n = 18 or 19. (d) Since the second spring is similar to the frst, it is reasonable to assume that the uncertainty in a measurement of its spring constant will be similar to that of the frst. Therefore the uncertainty in a single measurement is approximated with s = 1.29 N/m.

17.

Let X denote the average of the 10 yields at 65◦ C, and let Y denote the average of the 10 yields at 80◦ C. Let sX denote the sample standard deviation of the 10 yields at 65◦ C, and let sY denote the sample standard deviation of the 10 yields at 80◦ C. Then X = 70.14, sX = 0.897156, Y = 90.50, sY = 0.794425. √ √ (a) At 65◦ C, the yield is X ± sX ∕ 10 = 70.14 ± 0.897156∕ 10 = 70.14 ± 0.28. √ √ At 80◦ C, the yield is Y ± sY ∕ 10 = 90.50 ± 0.794425∕ 10 = 90.50 ± 0.25. (b) The di˙erence is estimated with Y − X = 90.50 − 70.14 = 20.36. t √ The uncertainty is Y −X = 2 + 2 = 0.252 + 0.282 = 0.38. Y

18.

No, the 8 measurements are not a simple random sample from a single population. Instead, they consist of two random samples, each of size 4, from two di˙erent populations.

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SECTION 3.2 √ √ 19. (a) Let X = 0.05 denote the uncertainty in the instrument. Then X = X ∕ 10 = 0.05∕ 10 = 0.016. √ √ (b) Let Y = 0.02 denote the uncertainty in the instrument. Then Y = Y ∕ 5 = 0.02∕ 5 = 0.0089. (c) The uncertainty in 12 X + 12 Y is t √ 0.5X+0.5Y = 0.52 2 + 0.52 2 = 0.52 (0.052 ∕10) + 0.52 (0.022 ∕5) = 0.0091. X

5 The uncertainty in 10 X + 15 Y is 15

(2∕3)X+(1∕3)Y =

(2∕3)2 2 + (1∕3)2 2 = X

√ (2∕3)2 (0.052 ∕10) + (1∕3)2 (0.022 ∕5) = 0.011.

The uncertainty in 12 X + 21 Y is smaller. 2

0.022 ∕5 = 0.24. 0.052 ∕10 + 0.022 ∕5 2 + 2 X Y t 2 2 + (1 − c )2 2 = 0.0078. The minimum uncertainty is cbest best Y

(d) cbest =

20. (a) X+Y =

(b) X+Y

2 + 2 = X

t √ 12 ∕4 + 22 ∕12 = 0.022 ∕4 + 0.082 ∕12 = 0.025.

t √ √ 12 ∕n + 22 ∕(16 − n) = 0.022 ∕n + 0.082 ∕(16 − n) = 0.0004∕n + 0.0064∕(16 − n)

(c) The integer value of n than minimizes X+Y is n = 3. Therefore measure the frst component 3 times and the second component 13 times.

Section 3.3 X = 2.0, X = 0.3.

1. (a)

dY = 3X 2 = 12, dX

dY = 3.6 dX X

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CHAPTER 3

(b)

dY 1 =√ = 0.5, dX 2X

dY = 0.15 dX X

(c)

dY −3 = = −0.75, dX X2

Y =

dY = 0.225 dX X

(d)

dY 1 = = 0.5, dX X

(e)

dY = eX = e2 = 7.389, dX

(f)

dY = − sin X = − sin 2 = 0.909297, dX

Y =

dY = 2.217 dX X

Y =

dY = 0.273 dX X

X = 3.0, X = 0.1

(a) Y =

1 = 0.33, X

(b) Y = 2X = 6,

9 = 3, X

dY −1 dY = 0.0111, = = −0.111, Y = dX X dX X2 dY dY = 2, Y = = 0.2, dX dX X

dY 9 =− = −1, dX X2

4 (d) Y = √ = 2.309, X

dY = 0.15 dX X

Y =

Y = 0.33 ± 0.01

Y = 6.0 ± 0.2.

dY = 0.1, dX X

Y = 3.0 ± 0.1.

dY dY = −2X −3∕2 = −0.3849, Y = = 0.038, dX dX X

Y = 2.309 ± 0.038.

ℎ = 6, r = 5, r = 0.02, V = r2 ℎ∕3 = 157.0796. dV dV = 2 rℎ∕3 = 62.8319 V = = 62.8319(0.02) = 1.3 cm3 . dr dr r

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SECTION 3.3

√ T = 300, T = 0.4, V = 20.04 T = 347.10 dV dV = 0.23 = 10.02T −1∕2 = 0.578505 V = dT dT T V = 347.10 ± 0.23 m/s

√ √ 5. (a) g = 9.80, L = 0.742, L = 0.005, T = 2 L∕g = 2.00709 L = 1.7289 dT = 1.003545L−1∕2 = 1.165024 dL T = 1.7289 ± 0.0058 s

T =

dT dL

L = 0.0058

(b) L = 0.742, T = 1.73, T = 0.01, g = 4 2 LT −2 = 29.292986T −2 = 9.79 dg dg = −58.5860T −3 = −11.315 T = = 0.11 dT dT T g = 9.79 ± 0.11 m/s2

c = 448, ΔQ = 1210, m = 0.54, m = 0.01, ΔT = ΔQ∕mc = 1210∕[(0.54)(448)] = 5.0017 dΔT dΔT = −ΔQm−2 c −1 = −9.2623 ΔT = m = 9.2623(0.01) = 0.093◦ C. dm dm

7. (a) g = 9.80, d = 0.15, l = 30.0, ℎ = 5.33, ℎ = 0.02, F = dF = 0.055340ℎ−1∕2 = 0.023970 dℎ F = 0.2555 ± 0.0005 m/s

F =

√ gdℎ∕4l = 0.110680 ℎ = 0.2555

dF = 0.0005 dℎ ℎ

(b) g = 9.80, l = 30.0, ℎ = 5.33, d = 0.15, d = 0.03, F = dF = 0.329880d −1∕2 = 0.851747 dℎ F = 0.256 ± 0.026 m/s

√

√ √ gdℎ∕4l = 0.659760 d = 0.2555

dF = 0.026 dd d

Note that F is given to only three signifcant digits in the fnal answer, in order to keep the uncertainty to no more than two signifcant digits.

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F =

√

gdℎ∕4l = 1.399562l−1∕2 = 0.2555

dF = 0.0002 dl l

= 0.70, = 0.02, n = 1∕ sin = 1.55227 dn cos dn =− = 1.84292 n = = 0.037 2 d d sin n = 1.55 ± 0.037

m = 750, V0 = 500.0, V1 = 813.2, V0 = V1 = 0.1, D =

m 750 = = 2.3946. V 1 − V0 V1 − V0

Let V = V1 − V0 . Then V = 813.2 − 500.0 = 313.2, and t √ V = V1 −V0 = V2 + V2 = 0.12 + 0.12 = 0.141421. 1

dD 750 =− = −0.007646, dV V2 D = 2.3946 ± 0.0011 g/mL Now

D =

dD = 0.0011. dV V

10. (a) C0 = 0.1, t = 45, C = 0.0811, C = 0.0005, k = (1∕t)(1∕C − 1∕C0 ) = (1∕45)(1∕C) − 2∕9 = 0.051788 dk 1 dk = 0.0017 =− = −3.37867 k = dC C dC 45C 2 k = 0.0518 ± 0.0017 L/mol⋅min

(b) C0 = 0.1, C = 0.0750, k = 0.0518, k = 0.0017, t = (1∕k)(1∕C − 1∕C0 ) = (10∕3)(1∕k) = 64.350064 dt 10 dt = 2.1 = − 2 = −1242.28 t = dk dk k 3k t = 64.4 ± 2.1 minutes

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121

SECTION 3.3

11. (a) The relative uncertainty is 0.4/20.9 = 1.9% (b) The relative uncertainty is 0.8/15.1 = 5.3% (c) The relative uncertainty is 23/388 = 5.9% (d) The relative uncertainty is 0.009/2.465 = 0.37%

12. (a) The absolute uncertainty is 0.003(48.41) = 0.15 (b) The absolute uncertainty is 0.006(991.7) = 6.0 (c) The absolute uncertainty is 0.09(0.011) = 0.001 (d) The absolute uncertainty is 0.01(7.86) = 0.08

13.

14.

s = 2.2, t = 0.67, t = 0.02, g = 2s∕t2 = 4.4∕t2 = 9.802 d ln g d ln g ln g = ln 4.4 − 2 ln t, = −2∕t = −2.985, ln g = t = 0.060 dt dt g = 9.802 m/s2 ± 6.0%

√ T = 298.4, T = 0.2, V = 20.04 T = 346.18 d ln V ln V = ln 20.04 + 0.5 ln T , = 0.5∕T = 0.001676, dT V = 346.18 m/s ± 0.034%

ln V =

d ln V T = 0.0003351 dT

√ √ 15. (a) g = 9.80, L = 0.855, L = 0.005, T = 2 L∕g = 2.00709 L = 1.85588

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ln T = ln 2.00709 + 0.5 ln L,

d ln T = 0.5∕L = 0.584795, dL

ln T =

d ln T L = 0.0029 dL

T = 1.856 s ± 0.29% (b) L = 0.855, T = 1.856, T = 0.005, g = 4 2 LT −2 = 33.754047T −2 = 9.799 d ln g d ln g ln g = ln 33.754047 − 2 ln T , = −2∕T = −1.0776, ln g = T = 0.0054 dT dT g = 9.799 m/s2 ± 0.54%

16.

c = 448, ΔQ = 1210, m = 0.75, m = 0.01, ΔT = ΔQ∕mc = 1210∕[(0.75)(448)] = 3.6012 d ln ΔT d ln ΔT ln ΔT = ln ΔQ − ln m − ln c, = −1∕m = −1.3333 ln ΔT = m = 1.3333(0.01) = 0.013. dm dm ΔT = 3.6012 ± 1.3%

17. (a) g = 9.80, d = 0.20, l = 35.0, ℎ = 4.51, ℎ = 0.03, F = ln F = ln 0.118332 + 0.5 ln ℎ,

√

√ gdℎ∕4l = 0.118332 ℎ = 0.2513

d ln F d ln F = 0.5∕ℎ = 0.110865, ln F = ℎ = 0.0033 dℎ dℎ

F = 0.2513 m/s ± 0.33%

(b) g = 9.80, l = 35.0, ℎ = 4.51, d = 0.20, d = 0.008, F = ln F = ln 0.561872 + 0.5 ln d,

d ln F = 0.5∕d = 2.5, dd

√ √ gdℎ∕4l = 0.561872 d = 0.2513 ln F =

d ln F d = 0.02 dd

F = 0.2513 m/s ± 2.0%

√

gdℎ∕4l = 1.486573l−1∕2 = 0.2513

d ln F d ln F = −0.5∕l = −0.014286, ln F = l = 0.0057 dl dl

F = 0.2513 m/s ± 0.57%

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SECTION 3.4

18.

= 0.90, = 0.01, n = 1∕ sin = 1.276606 d ln n d ln n cos = 0.0079 ln n = − ln(sin ), =− = −0.793551 ln n = d d sin n = 1.28 ± 0.79%

19.

m = 288.2, V0 = 400.0, V1 = 516.0, V0 = 0.1, V1 = 0.2, D =

m 288.2 = = 2.484 V1 − V 0 V1 − V 0

Let V = V1 − V0 . Then V = 516.0 − 400.0 = 116.0, and t √ V = V1 −V0 = V2 + V2 = 0.12 + 0.22 = 0.223607 1

d ln D = −1∕V = −0.008621, dV D = 2.484 g/mL ± 0.19% ln D = ln 288.2 − ln V ,

ln D =

d ln D V = 0.0019. dV

20. (a) C0 = 0.04, t = 30, C = 0.0038, C = 0.0002, k = (1∕t)(1∕C − 1∕C0 ) = 1∕(30C) − 5∕6 = 7.94 d ln k d ln k 1 ln k = ln(1∕(30C) − 5∕6), =− = −290.782, ln k = C = 0.058 2 dC dC C − 25C k = 7.94 L/mol⋅s ± 5.8%

(b) C0 = 0.04, t = 50, C = 0.0024, C = 0.0002, k = (1∕t)(1∕C − 1∕C0 ) = 1∕(50C) − 1∕2 = 7.83 d ln k 1 d ln k C = 0.089 ln k = ln(1∕(50C) − 1∕2), =− = −443.262, ln k = 2 dC dC C − 25C k = 7.83 L/mol⋅s ± 8.9% (c) ln k = 0.5 ln k1 + 0.5 ln k2 , u √ ln k = 0.52 2 + 0.52 2 = 0.52 (0.058)2 + 0.52 (0.089)2 = 0.053. ln k1

ln k2

The relative uncertainty is 5.3%.

Section 3.4

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1. (a) X = 10.0, X = 0.5, Y = 5.0, Y = 0.1, U = XY 2 = 250 u )U )U )U 2 2 )U 2 2 2 = 2XY = 100.0, U = X + Y = 16.0 = Y = 25.0, )X )Y )X )Y U = 250 ± 16 (b) X = 10.0, X = 0.5, Y = 5.0, Y = 0.1, U = X 2 + Y 2 = 125 u )U )U )U 2 2 )U 2 2 X + Y = 10.0 = 2X = 20.0, = 2Y = 10.0, U = )X )Y )X )Y U = 125 ± 10 (c) X = 10.0, X = 0.5, Y = 5.0, Y = 0.1, U = (X + Y 2 )∕2 = 17.50 u )U )U )U 2 2 )U 2 2 X + Y = 0.56 = 1∕2, = Y = 5.0, U = )X )Y )X )Y U = 17.50 ± 0.56

2. (a) r = 5.00, ℎ = 6.00, r = 0.02, ℎ = 0.01, V = r2 ℎ∕3 = 157.0796 u )V )V )V 2 2 )V 2 2 2 r + ℎ = 1.3 = 2 rℎ∕3 = 62.8319, = r ∕3 = 26.1799, V = )r )ℎ )r )ℎ V = 157.08 ± 1.3 cm3 u

)V 2 2 )V 2 2 )V r + ℎ , = 62.8319, )r )ℎ )r If r = 0.02 and ℎ = 0.005, then V = 1.3.

(b) V =

)V = 26.1799 )ℎ

If r = 0.01 and ℎ = 0.01, then V = 0.68. Reducing the uncertainty in r to 0.01 mm provides the greater reduction.

3. (a) s = 55.2, = 0.50, s = 0.1, = 0.02, ℎ = s tan = 30.1559 )ℎ )ℎ = tan = 0.5463, = s sec2 = 71.6742 )s ) u 2 2 )ℎ )ℎ s2 + 2 = 1.4345 ℎ = )s ) ℎ = 30.2 ± 1.4 u (b) ℎ =

2 )ℎ 2 2 )ℎ s + 2 , )s )

)ℎ = 0.5463, )s

)ℎ = 71.6742 )

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SECTION 3.4

If s = 0.05 and = 0.02, then ℎ = 1.43. If s = 0.1 and = 0.01, then ℎ = 0.72. Reducing the uncertainty in to 0.01 radians provides the greater reduction.

4. (a) = 1.49, = 30.0, = 0.1, ℎ = 10.0, ℎ = 0.2, V = ℎ∕ = ℎ∕1.49 = 201.3 u )V )V )V 2 2 )V 2 2 ℎ = 4.1 = ℎ∕1.49 = 6.71141, = ∕1.49 = 20.1342, V = + ) )ℎ ) )ℎ V = 201.3 ± 4.1 mm/s u

)V 2 2 )V 2 2 )V + ℎ , = 6.71141, ) )ℎ ) If = 0.01 and ℎ = 0.2, then V = 4.0.

(b) V =

)V = 20.1342 )ℎ

If = 0.1 and ℎ = 0.1, then V = 2.1. Reducing the uncertainty in ℎ to 0.1 mm provides the greater reduction.

5. (a) P1 = 10.1, P1 = 0.3, P2 = 20.1, P2 = 0.4, P3 = √ )P3 = 0.5 P2 ∕P1 = 0.705354 )P1 √ )P3 = 0.5 P1 ∕P2 = 0.354432 )P2 v 0 1 0 1 )P3 2 2 )P3 2 2 P3 = P + P = 0.25 1 2 )P1 )P2

√ P1 P2 = 14.25

P3 = 14.25 ± 0.25 MPa v 0 (b) P3 =

)P3 )P1

12 0 1 )P3 2 2 2 P + P , 1 2 )P2

)P3 = 0.705354, )P1

)P3 = 0.354432 )P2

If P1 = 0.2 and P2 = 0.4, then P3 = 0.20. If P1 = 0.3 and P2 = 0.2, then P3 = 0.22. Reducing the uncertainty in P1 to 0.2 MPa provides the greater reduction.

6. (a) M1 = 1.32, M1 = 0.01, M2 = 1.04, M2 = 0.01, W = (M1 − M2 )∕M1 = 0.2121 )W = M2 ∕M12 = 0.596878 )M1

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CHAPTER 3 )W = −1∕M1 = −0.757576 )M2 v 0 12 0 12 )W )W 2 2 = 0.0096 M + M W = 1 2 )M1 )M2 W = 0.2121 ± 0.0096 v

(b) W =

)W )M1

12 0 12 )W 2 + 2 , M M 1 2 )M2

)W = 0.596878, )M1

)W = −0.757576 )M2

If M1 = 0.005 and M2 = 0.01, then W = 0.0081. If M1 = 0.01 and M2 = 0.005, then W = 0.0071. Reducing the uncertainty in M2 to 0.005 kg provides the greater reduction.

7. (a) p = 2.3, p = 0.2, q = 3.1, q = 0.2, f = pq∕(p + q) = 1.320 )f = q 2 ∕(p + q)2 = 0.329561 )p )f = p2 ∕(p + q)2 = 0.181413 )q v0 1 0 12 2 )f )f f = q2 = 0.075 p2 + )p )q f = 1.320 ± 0.075 cm v (b) f =

)f )p

12 0 12 )f p2 + q2 , )q

)f = 0.329561, )p

)f = 0.181413 )q

If p = 0.1 and q = 0.2, then f = 0.049. If p = 0.2 and q = 0.1, then f = 0.068. Reducing the uncertainty in p to 0.1 cm provides the greater reduction.

8. (a) P = 242.52, P = 0.03, V = 10.103, V = 0.002, T = P V ∕8.31 = 294.847 )T = V ∕8.31 = 1.21576 )P )T = P ∕8.31 = 29.1841 )V u 2 )T )T 2 2 T = V = 0.069 P2 + )P )V T = 294.847 ± 0.069 K

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SECTION 3.4

(b) P = 242.52, P = 0.03, T = 290.11, T = 0.02, V = 8.31T ∕P = 9.9407 )V = −8.31T ∕P 2 = −0.0409891 )P )V = 8.31∕P = 0.0342652 )T u )V 2 2 )V 2 2 V = T = 0.0014 P + )P )T V = 9.9407 ± 0.0014 L

(c) V = 10.103, V = 0.002, T = 290.11, T = 0.02, P = 8.31T ∕V = 238.624 )P = −8.31T ∕V 2 = −23.6191 )V )P = 8.31∕V = 0.822528 )T u 2 )P 2 2 )P P = T2 = 0.050 V + )V )T P = 238.624 ± 0.050 kPa

9. (a) C = 1.25, C = 0.03, L = 1.2, L = 0.1, A = 1.30, A = 0.05, M = A∕LC = 0.8667 )M = −A∕(LC 2 ) = −0.6933 )C )M = −A∕(L2 C) = −0.7222 )L )M = 1∕(LC) = 0.6667 )A u )M 2 2 )M 2 2 )M 2 2 C + L + A = 0.082 M = )C )L )A M = 0.867 ± 0.082 u

2 )M 2 2 )M 2 2 C2 + L + A , )L )A )M )M = −A∕(LC 2 ) = −0.6933 = −A∕(L2 C) = −0.7222 )C )L If C = 0.01, L = 0.1, and A = 0.05, then M = 0.080.

(b) M =

)M )C

)M = 1∕(LC) = 0.6667 )A

If C = 0.03, L = 0.05, and A = 0.05, then M = 0.053. If C = 0.03, L = 0.1, and A = 0.01, then M = 0.076. Reducing the uncertainty in L to 0.05 cm provides the greatest reduction.

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CHAPTER 3

10.

Let I1 be the estimated index at a temperature of 166 K, let I2 be the estimated index at a temperature of 196 K, and let R = I1 ∕I2 be the ratio. Then I1 = 0.00116, I1 = 0.0001, I2 = 0.00129, I2 = 0.0001. )R = 1∕I2 = 775.194 )I1 )R = −I1 ∕I22 = −697.073 )I2 v 0 12 0 12 )R )R I2 = 0.10 I2 + R = 1 2 )I1 )I2

11. (a) 0 = 50, 0 = 1, w = 1.2, w = 0.1, k = 0.29, k = 0.05, = 0 (1 − kw) = 32.6 ) = 1 − kw = 0.652 ) 0 ) = − 0 w = −60 )k ) = − 0 k = −14.5 )w v0 12 2 ) ) ) 2 2 2 + w = 3.4 k2 + = 0 ) 0 )k )w = 32.6 ± 3.4 MPa v

(b) =

) ) 0

12 2 ) ) 2 2 w , k2 + 2 + 0 )k )w

) ) ) = 1 − kw = 0.652, = − 0 w = −60, = − 0 k = −14.5 ) 0 )k )w If 0 = 0.1, k = 0.05, and w = 0.1, then = 3.3. If 0 = 1.0, k = 0.025, and w = 0.1, then = 2.2. If 0 = 1.0, k = 0.05, and w = 0.01, then = 3.1. Reducing the uncertainty in k to 0.025 mm−1 provides the greatest reduction. (c) If 0 = 0, k = 0.05, and w = 0, = 3.0. Thus implementing the procedure would reduce the uncertainty in only to 3.0 MPa. It is probably not worthwhile to implement the new procedure for a reduction this small.

12.

1 = 0.3672, 1 = 0.005, 2 = 0.2943, 2 = 0.004, n =

sin 1 = 1.238 sin 2

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SECTION 3.4 cos 1 )n = = 3.21762 ) 1 sin 2 sin 1 cos 2 )n =− = −4.08326 ) 2 sin2 2 v 0 12 0 12 )n )n 2 = 0.023 n = 2 + 1 2 ) 1 ) 2 n = 1.238 ± 0.023

13. (a) g = 3867.4, g = 0.3, b = 1037.0, b = 0.2, m = 2650.4, m = 0.1, y = mb∕g = 710.68. )y = −mb∕g 2 = −0.18376 )g )y = m∕g = 0.685318 )b )y = b∕g = 0.268139 )m v0 1 0 12 0 12 2 )y )y )y m2 + b2 = 0.15 y = g2 + )g )m )b y = 710.68 ± 0.15 g v0 (b) y =

)y )g

12 0 12 0 12 )y )y 2 2 m + b2 , g + )m )b

)y )y = −mb∕g 2 = −0.18376, = m∕g = 0.685318, )g )b If g = 0.1, b = 0.2, and m = 0.1, y = 0.14.

)y = b∕g = 0.268139 )m

If g = 0.3, b = 0.1, and m = 0.1, y = 0.09. If g = 0.3, b = 0.2, and m = 0, y = 0.15. Reducing the uncertainty in b to 0.1 g provides the greatest reduction.

14. (a) l = 14.0, l = 0.1, d = 4.4, d = 0.1, R = kl∕d 2 = 0.723k )R = k∕d 2 = 0.0516529k )l )R = −2kl∕d 3 = −0.3287k )d u 2 2 )R )R d2 = 0.033k l2 + R = )l )d

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CHAPTER 3

R = 0.723k ± 0.033k Ω u

2 )R 2 2 )R )R l + d2 , = k∕d 2 = 0.0516529k, )l )d )l If l = 0.05 and d = 0.1, R = 0.033k.

(b) R =

)R = −2kl∕d 3 = −0.3287k )d

If l = 0.1 and d = 0.05, R = 0.017k. Reducing the uncertainty in d to 0.05 cm provides the greater reduction.

15. (a) F = 800, F = 1, R = 0.75, R = 0.1, L0 = 25.0, L0 = 0.1, L1 = 30.0, L1 = 0.1, F L0 = 2264 R2 (L1 − L0 ) L0 )Y = = 2.82942 2 )F R (L1 − L0 ) −2F L0 )Y = = −6036.1 3 )R R (L1 − L0 ) F L1 )Y = = 543.249 2 )L0 R (L1 − L0 )2 −F L0 )Y = = −452.707 )L1 R2 (L1 − L0 )2 v 0 12 0 12 2 )Y 2 2 )Y )Y )Y 2 + R L2 + L2 = 608 Y = F + 0 1 )F )R )L0 )L1

Y =

Y = 2264 ± 608 N/mm2 v (b) Y =

)Y )F

0 12 0 12 2 2 )Y )Y )Y 2 + R L2 + F2 + L2 , 0 1 )R )L0 )L1

−2F L0 L0 )Y )Y = = 2.82942, = = −6036.1, 2 3 )F )R R (L1 − L0 ) R (L1 − L0 ) −F L0 F L1 )Y )Y = = 543.249, = = −452.707 2 2 2 )L0 )L1 R (L1 − L0 ) R (L1 − L0 )2 If F = 0, R = 0.1, L0 = 0.1, and L1 = 0.1, then Y = 608. If F = 1, R = 0, L0 = 0.1, and L1 = 0.1, then Y = 71. If F = 1, R = 0.1, L0 = 0, and L1 = 0.1, then Y = 605. If F = 1, R = 0.1, L0 = 0.1, and L1 = 0, then Y = 606. R is the only variable that substantially a˙ects the uncertainty in Y .

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131

SECTION 3.4

16.

T = 50, k = 0.025, T0 = 70.1, T0 = 0.2, Ta = 35.7, Ta = 0.1, t = ln(T0 − Ta )∕k − ln(T − Ta )∕k = 40 ln(T0 − Ta ) − 40 ln(50 − Ta ) = 35.11 )t = 40∕(T0 − Ta ) = 1.16279 )T0 )t = 40∕(50 − Ta ) − 40∕(T0 − Ta ) = 1.63441 )Ta v0 12 0 12 )t )t 2 T + T2 = 0.28 t = 0 a )T0 )Ta t = 35.11 ± 0.28 min

17.

t = 10, T = 54.1, T = 0.2, T0 = 70.1, T0 = 0.2, Ta = 35.7, Ta = 0.1, k = ln(T0 − Ta )∕t − ln(T − Ta )∕t = 0.1 ln(T0 − Ta ) − 0.1 ln(T − Ta ) = 0.0626 )k −1 = = −0.00543478 )T 10(T − Ta ) )k 1 = = 0.00290698 )T0 10(T0 − Ta ) )k 1 1 = − = 0.00252781 )Ta 10(T − Ta ) 10(T0 − Ta ) v 0 12 0 12 2 )k )k )k 2 2 T + T + T2 = 0.0013 k = 0 a )T )T0 )Ta k = 0.0626 ± 0.0013 min−1

18. (a) a = 2.5, a = 0.1, b = 0.05, b = 0.01, w = 1.2, w = 0.1, v = a + bw = 2.56 )v =1 )a )v = w = 1.2 )b )v = b = 0.05 )w u 2 2 )v )v )v 2 2 b2 + w = 0.10 v = a2 + )a )b )w v = 2.56 ± 0.10 mm u

2 )v 2 2 )v )v 2 2 )v b2 + w , = 1, a + )a )b )w )a If a = 0, b = 0.01, and w = 0.1, then v = 0.013.

(b) v =

)v = w = 1.2 )b

)v = b = 0.05 )w

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132

CHAPTER 3

If a = 0.1, b = 0, and w = 0.1, then v = 0.10. If a = 0.1, b = 0.01, and w = 0, then v = 0.10. a is the only variable that substantially a˙ects the uncertainty in v.

19. (a) No, they both involve the quantities ℎ and r.

(b) r = 0.9, r = 0.1, ℎ = 1.7, ℎ = 0.1, R = c

2 r(ℎ + 2r) 2ℎ + 4r =c = 2.68c ℎr + 4r2 ∕3 r2 (ℎ + 4r∕3)

2ℎ2 + 16rℎ∕3 + 16r2 ∕3 )R = −c = −2.68052c )r (4r2 ∕3 + rℎ)2 4r2 ∕3 )R = −c = −0.158541c )ℎ (4r2 ∕3 + rℎ)2 u 2 2 )R )R r2 + ℎ2 = 0.27c R = )r )ℎ R = 2.68c ± 0.27c

20.

X = 5.0, X = 0.2, Y = 10.0, Y = 0.5 √ (a) U = X Y = 15.8, ln U = ln X + 0.5 ln Y ) ln U 1 = = 0.2 )X X ) ln U 1 = = 0.05 )Y 2Y u ) ln U 2 2 ) ln U 2 2 X + Y = 0.047 ln U = )X )Y U = 15.8 ± 4.7% √ (b) U = 2Y ∕ X = 8.94, ln U = ln 2 − 0.5 ln X + ln Y −1 ) ln U = = −0.1 )X 2X ) ln U 1 = = 0.1 )Y Y u ) ln U 2 2 ) ln U 2 2 X + Y = 0.054 ln U = )Y )X U = 8.94 ± 5.4% (c) U = X 2 + Y 2 = 125, ln U = ln(X 2 + Y 2 )

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133

SECTION 3.4 ) ln U 2X = = 0.08 )X X2 + Y 2 ) ln U 2Y = = 0.16 )Y X2 + Y 2 u ) ln U 2 2 ) ln U 2 2 X + Y = 0.082 ln U = )Y )X U = 125 ± 8.2% Alternatively (and more easily), U = X 2 + Y 2 = 125, )U = 2X = 10 )X )U = 2Y = 20 )Y u )U 2 2 )U 2 2 X + Y = 10.198 U = )X )Y The relative uncertainty is U ∕U = 10.198∕125 = 0.082. U = 125 ± 8.2%

21.

= 1.49, = 35.2, = 0.1, ℎ = 12.0, ℎ = 0.3, V = ℎ∕ = ℎ∕1.49 = 283.49, ln V = ln + ln ℎ − ln 1.49 ) ln V = 1∕ = 0.028409 ) ) ln V = 1∕ℎ = 0.083333 )ℎ u ) ln V 2 2 ) ln V 2 2 ln V = + ℎ = 0.025 ) )ℎ V = 283.49 mm/s ± 2.5%

22.

P1 = 15.3, P1 = 0.2, P2 = 25.8, P2 = 0.1, P3 =

√

P1 P2 = 19.87, ln P3 = 0.5 ln P1 + 0.5 ln P2

) ln P3 1 = = 0.0326797 )P1 2P1 ) ln P3 1 = = 0.0193798 )P2 2P2 v 0 1 1 0 ) ln P3 2 2 ) ln P3 2 2 ln P3 = P + P = 0.0068 1 2 )P1 )P2 P3 = 19.87 MPa ± 0.68%

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134 23.

CHAPTER 3

p = 4.3, p = 0.1, q = 2.1, q = 0.2, f = pq∕(p + q) = 1.41, ln f = ln p + ln q − ln(p + q) ) ln f = 1∕p − 1∕(p + q) = 0.0763081 )p ) ln f = 1∕q − 1∕(p + q) = 0.31994 )q v 0 12 0 12 ) ln f ) ln f p2 + q2 = 0.064 ln f = )p )q f = 1.41 cm ± 6.4%

24. (a) P = 224.51, P = 0.04, V = 11.237, V = 0.002, T = P V ∕8.31 = 303.59, ln T = ln P + ln V − ln 8.31 ) ln T = 1∕P = 0.00445414 )P ) ln T = 1∕V = 0.0889917 )V u ) ln T 2 2 ) ln T 2 2 P + V = 0.00025 ln T = )V )P T = 303.59 K ± 0.025% (b) P = 224.51, P = 0.04, T = 289.33, T = 0.02, V = 8.31T ∕P = 10.71, ln V = ln 8.31 + ln T − ln P ) ln V = −1∕P = −0.00445414 )P ) ln V = 1∕T = 0.00345626 )T u ) ln V 2 2 ) ln V 2 2 ln V = P + T = 0.00019 )P )T V = 10.71 L ± 0.019% (c) V = 11.203, V = 0.002, T = 289.33, T = 0.02, P = 8.31T ∕V = 214.62, ln V = ln 8.31 + ln T − ln V ) ln P = −1∕V = −0.0892618 )V ) ln P = 1∕T = 0.00345626 )T u ) ln P 2 2 ) ln P 2 2 T = 0.00019 ln P = V + )V )T P = 214.62 kPa ± 0.019%

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135

SECTION 3.4

25.

1 = 0.216, 1 = 0.003, 2 = 0.456, 2 = 0.005, n =

sin 1 = 0.487, ln n = ln(sin 1 ) − ln(sin 2 ) sin 2

) ln n = cot 1 = 4.5574 ) 1 ) ln n = − cot 2 = −2.03883 ) 2 v 0 12 0 12 ) ln n ) ln n ln n = 2 + 2 = 0.017 1 2 ) 1 ) 2 n = 0.487 ± 1.7%

26. (a) l = 10.0, l ∕l = 0.005, d = 10.4, d ∕d = 0.005, R = kl∕d 2 = 0.0925k, ln R = ln k + ln l − 2 ln d ) ln R ) ln R = 1∕l, = −2∕d )l )d u u 2 l 2 ) ln R 2 2 ) ln R 2 2 l + d = +4 d = 0.011 ln R = )l )d l d R = 0.0925k Ω ± 1.1% The relative uncertainty does not depend on k. u

2 l 2 +4 d l d If l ∕l = 0.002 and d ∕d = 0.005, then ln R = 1.0%.

(b) ln R =

If l ∕l = 0.005 and d ∕d = 0.002, then ln R = 0.64%. d should be remeasured so as to provide the greater improvement in the relative uncertainty of the resistance.

27.

F = 750, F = 1, R = 0.65, R = 0.09, L0 = 23.7, L0 = 0.2, L1 = 27.7, L1 = 0.2, Y =

F L0 R2 (L

1 − L0 )

= 3347.9, ln Y = ln F + ln L0 − ln − 2 ln R − ln(L1 − L0 )

) ln Y = 1∕F = 0.001333 )F ) ln Y = 1∕L0 + 1∕(L1 − L0 ) = 0.292194 )L0 ) ln Y = −2∕R = −3.07692 )R ) ln Y = −1∕(L1 − L0 ) = −0.25 ) L1

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CHAPTER 3

ln Y =

) ln Y )F

0 0 12 12 2 ) ln Y ) ln Y 2 2 ) ln Y 2 2 F + L2 = 0.29 L + R + 0 1 )L0 )R )L1

Y = 3347.9 N/mm2 ± 29%

28.

T = 50, k = 0.032, T0 = 73.1, T0 = 0.1, Ta = 37.5, Ta = 0.2, t = ln(T0 − Ta )∕k − ln(T − Ta )∕k = 31.25 ln(T0 − Ta ) − 31.25 ln(50 − Ta ) = 32.71 It is easier to compute t and then divide by t than to compute ln t . )t 31.25 = = 0.877809 )T0 T 0 − Ta )t 31.25 31.25 = − = 1.62219 )Ta 50 − Ta T0 − Ta v0 12 0 12 )t )t 2 T + T2 = 0.3361 t = 0 a )T0 )Ta t = 0.3361∕32.71 = 0.010 t t = 32.7 min ± 1.0%

29.

r = 0.8, r = 0.1, ℎ = 1.9, ℎ = 0.1 (a) S = 2 r(ℎ + 2r) = 17.59, ln S = ln(2 ) + ln r + ln(ℎ + 2r) ) ln S 1 2 = + = 1.82143 )r r ℎ + 2r ) ln S 1 = = 0.285714 )ℎ ℎ + 2r u ) ln S 2 2 ) ln S 2 2 ln S = r + ℎ = 0.18 )r )ℎ S = 17.59 m ± 18% (b) V = r2 (ℎ + 4r∕3) = 5.965, ln V = ln( ) + 2 ln r + ln(ℎ + 4r∕3) ) ln V 2 4 = + = 2.94944 )r r 3ℎ + 4r ) ln V 3 = = 0.337079 )ℎ 3ℎ + 4r u ) ln V 2 2 ) ln V 2 2 r + ℎ = 0.30 ln V = )r )ℎ V = 5.965 m3 ± 30%

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SECTION 3.4 2 r(ℎ + 2r) 2ℎ + 4r =c = 2.95c, ln R = ln c + ln(2ℎ + 4r) − ln(rℎ + 4r2 ∕3) rℎ + 4r2 ∕3 r2 (ℎ + 4r∕3) ) ln R 2 8r + 3ℎ = − = −1.12801 )r 2r + ℎ 4r2 + 3rℎ ) ln R 1 3 = − = −0.0513644 )ℎ 2r + ℎ 4r + 3ℎ u ) ln R 2 2 ) ln R 2 2 r + ℎ = 0.11 ln R = )r )ℎ R = 2.95c ± 11%

Note that the uncertainty in R cannot be determined directly from the uncertainties in S and V , because S and V are not independent.

(d) No

30.

P3 = P10.5 P20.5 . The relative uncertainties are P3 P3

P1 P1

= 0.05 and

P2 P2

= 0.02.

0 1 2 0 12 √ P P 0.5 1 + 0.5 2 = 0.0252 + 0.012 = 0.027. P1 P2

The relative uncertainty is 2.7%. 31.

R = kld −2 . The relative uncertainties are k = 0 (since k is a constant), l = 0.03, and d = 0.02. l d k u √ k 2 2 2 R = + l + −2 d = 02 + 0.032 + (−0.04)2 = 0.05. R k l d The relative uncertainty is 5.0%.

Supplementary Exercises for Chapter 3 1. (a) X = 25.0, X = 1, Y = 5.0, Y = 0.3, Z = 3.5, Z = 0.2. Let U = XY + Z. )U )U )U = Y = 5.0, = X = 25.0, =1 )X )Y )Z u )U 2 2 )U 2 2 )U 2 2 U = X + Y + Z = 9.0 )X )Y )Z

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(b) X = 25.0, X = 1, Y = 5.0, Y = 0.3, Z = 3.5, Z = 0.2. Let U = Z∕(X + Y ). )U )U )U = −Z∕(X + Y )2 = −0.003889, = −Z∕(X + Y )2 = −0.003889, = 1∕(X + Y ) = 0.03333 )X )Y )Z u )U 2 2 )U 2 2 )U 2 2 Y + Z = 0.0078 U = X + )X )Y )Z √ (c) X = 25.0, X = 1, Y = 5.0, Y = 0.3, Z = 3.5, Z = 0.2. Let U = X(ln Y + Z). u u u )U 1 ln Y + Z )U 1 X )U 1 X = = 0.226041, = = 0.221199, = = 1.105996, )X 2 X )Y 2Y ln Y + Z )Z 2 ln Y + Z u )U 2 2 )U 2 2 )U 2 2 Y + Z = 0.32 U = X + )X )Y )Z

(d) X = 25.0, X = 1, Y = 5.0, Y = 0.3, Z = 3.5, Z = 0.2. Let U = XeZ −2Y . 2 2 2 )U )U )U = 2XZeZ −2Y = −1660.353771 = eZ −2Y = 9.487736, = −2XeZ −2Y = −474.38679, )X )Y )Z u )U 2 2 )U 2 2 )U 2 2 U = X + Y + Z = 361.41 )X )Y )Z

The relative uncertainty in X is ln X = 0.05, the relative uncertainty in Y is ln Y = 0.10, and the relative uncertainty in Z is ln Z = 0.15. (a) Let U = XY ∕Z, so ln U = ln X + ln Y − ln Z. ln U

t 2 2 + 2 + ln ln X Y ln Z √ = 0.052 + 0.102 + 0.152 = 0.19

The relative uncertainty in U is 19%.

(b) Let U =

√ XY 2 Z 3 , so ln U = 0.5 ln X + ln Y + 1.5 ln Z. t

ln U

2 2 + 1.52 2 + ln 0.52 ln X Y ln Z

√ = (0.5)2 (0.05)2 + 0.102 + (1.5)2 (0.15)2 = 0.245

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SUPPLEMENTARY EXERCISES FOR CHAPTER 3

The relative uncertainty in U is 24.5%.

XY 1∕3 , so ln U = (1∕3) ln X + (1∕3) ln Y − (1∕3) ln Z. Z t 2 2 + (1∕3)2 2 + (1∕3)2 ln = (1∕3)2 ln X Y ln Z t = (1∕3)2 (0.05)2 + (1∕3)2 (0.10)2 + (1∕3)2 (0.15)2

= 0.062 The relative uncertainty in U is 6.2%.

3. (a) Let X, Y , and Z represent the measured lengths of the components, and let T = X + Y + Z be the combined length. Then X = Y = Z = 1.2. t 2 + 2 + 2 = 2.1 mm T = X Y Z

(b) Let be the required uncertainty in X and Y . Then T = = 0.029 mm.

√

2 + 2 + 2 = 0.05. Solving for yields

t1 = 20, t2 = 40, m1 = 17.7, m1 = 1.7, m2 = 35.9, m2 = 5.8,

r = (m2 − m1 )∕(t2 − t1 ) = 0.05m2 − 0.05m1 = 0.91 (a) r =

t 0.052 m2 + 0.052 m2 = 0.30 1

r = 0.91 ± 0.30

(b)

r = 0.30∕0.91 = 33% r

5. (a) 훾 = 9800, = 0.85, = 0.02, Q = 60, Q = 1, H = 3.71, H = 0.10 P = 훾 QH = 9800 QH = 1.854 × 106 )P = 9800QH = 2.18148 × 106 )

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CHAPTER 3 )P = 9800 H = 30904.3 )Q )P = 9800 Q = 499800 )H v 0 12 0 12 2 )P )P 2 + )P 2 = 73188.9 P = Q H 2 + ) )Q )H P = (1.854 ± 0.073) × 106 W

(b) The relative uncertainty is

)P )

P 0.073 × 106 = = 0.039 = 3.9%. P 1.854 × 106

12 0 12 2 )P 2 + )P 2 , 2 + Q H )Q )H

)P = 9800QH = 2.18148 × 106 , )

)P = 9800 H = 30904.3, )Q

)P = 9800 Q = 499800 )H

If = 0.01, Q = 1, and H = 0.1, then P = 6.3 × 104 . If = 0.02, Q = 0.5, and H = 0.1, then P = 6.8 × 104 . If = 0.02, Q = 1, and H = 0.05, then P = 5.9 × 104 . Reducing the uncertainty in H to 0.05 m provides the greatest reduction.

6. (a) p = 0.360, p = 0.048, q = 0.250, q = 0.043, PAB = pq = 0.090 )PAB = q = 0.25 )p )PAB = p = 0.36 )q v0 1 0 1 )PAB 2 2 )PAB 2 2 PAB = p + q = 0.020 )p )q PAB = 0.090 ± 0.020

PAB

0.020 = 0.22 = 22%. 0.090 ) ln PAB ) ln PAB = 1∕q = 4, = 1∕p = 2.77778, Alternatively, ln PAB = ln p + ln q, so )q )p

(b) The relative uncertainty is

PAB

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SUPPLEMENTARY EXERCISES FOR CHAPTER 3 v

PAB

= ln PAB =

PAB

)PAB )p

) ln PAB )p

0 1 12 ) ln PAB 2 2 2 p + q = 0.22 = 22% )q

12 0 1 )PAB 2 2 p2 + q , )q

)PAB = q = 0.25 )p

)PAB = p = 0.36 )q

If p = 0.02 and q = 0.043, then PAB = 0.016. If p = 0.048 and q = 0.02, then PAB = 0.014. Reducing the uncertainty in the type B proportion to 0.02 provides the greater reduction.

7. (a) H = 4, c = 0.2390, ΔT = 2.75, ΔT = 0.02, m = 0.40, m = 0.01 C=

cH(ΔT ) 0.956(ΔT ) = = 6.57 m m

)C 0.956 = = 2.39 m )ΔT −0.956(ΔT ) )C = = −16.4312 )m m2 u )C 2 2 )C 2 2 ΔT + m = 0.17 C = ) ΔT )m C = 6.57 ± 0.17 kcal.

C 0.17 = = 0.026 = 2.6%. C 6.57 ) ln C 1 ) ln C −1 Alternatively, ln C = ln 0.956 + ln(ΔT ) − ln m, so = = 0.363636, = = −2.5, )ΔT ΔT )m m v C ) ln C 2 2 ) ln C 2 2 = ln C = m = 0.026 = 2.6% ΔT + C )ΔT )m

(b) The relative uncertainty is

)C 2 2 )C 2 2 )C 0.956 ΔT + m , = = 2.39 )ΔT )m )ΔT m If ΔT = 0.01 and m = 0.01, then C = 0.17.

−0.956(ΔT ) )C = = −16.4312 )m m2

If ΔT = 0.02 and m = 0.005, then C = 0.10. Reducing the uncertainty in the mass to 0.005 g provides the greater reduction.

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CHAPTER 3

8. (a) The estimate of the resistance is the average of the 16 measurements, which is X = 52.37. The uncertainty is √ √ X = ∕ n = 0.12∕ 16 = 0.03 Ω. (b) The uncertainty in a single measurement is = 0.12 Ω.

9. (a) Let X be the measured northerly component of velocity and let √ Y be the measured easterly component of velocity. The velocity of the earth’s crust is estimated with V = X 2 + Y 2 . √ Now X = 22.10, X = 0.34 and Y = 14.3, Y = 0.32, so V = 22.102 + 14.32 = 26.32, and √ )V = X∕ X 2 + Y 2 = 0.83957 )X √ )V = Y ∕ X 2 + Y 2 = 0.543251 )Y u )V 2 2 )V 2 2 Y = 0.33 X + V = )X )Y V = 26.32 ± 0.33 mm/year

(b) Let T be the estimated number of years it will take for the earth’s crust to move 100 mm. By part (a), V = 26.3230, V = 0.334222. We are using extra precision in V and V to get good precision for T and T . dT Now T = 100∕V = 3.799, = −100∕V 2 = −0.144321 dV dT T = = 0.144321(0.334222) = 0.048 dV V T = 3.799 ± 0.048 years

10. (a) Let M1 represent the estimated molar mass of the unknown gas, and let M2 represent the molar mass of carbon dioxide. Then M2 = 44, and R = 1.66, R = 0.03. dM1 dM1 Now M1 = M2 R2 = 44R2 = 121.2, and = 88R = 146.08, so M = = 146.08(0.03) = 4.4. dR dR R M1 = 121.2 ± 4.4 g/mol

(b) The relative uncertainty is

M1 M1

= 4.4∕121.2 = 0.036 = 3.6%.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 3

Alternatively, ln M1 = ln 44 + 2 ln R, so M1 M1

11.

= ln M1 =

d ln M1 = 2∕R = 2∕1.66 = 1.204819. dR

d ln M1 R = 1.204819(0.03) = 0.036 = 3.6% dR

Let X1 and X2 represent the estimated thicknesses of the outer layers, and let Y1 , Y2 , Y3 represent the estimated thicknesses of the inner layers. Then X1 = X2 = 0.50, Y1 = Y2 = Y3 = 6.25, Xi = 0.02 for i = 1, 2, and Yj = 0.05 for j = 1, 2, 3. The estimated thickness of the item is T = X1 + X2 + Y1 + Y2 + Y3 = 19.75. t 2 + 2 + 2 + 2 + 2 = 0.09. The uncertainty in the estimate is T = X X Y Y Y 1

The thickness is 19.75 ± 0.09 mm.

12.

Let X be the estimated concentration in the frst experiment and let Y be the estimated concentration in the second experiment. Then X = 43.94, X = 1.18, and Y = 48.66, Y = 1.76. The estimated di˙erence is Y − X = 48.66 − 43.94 = 4.72. t √ 2 = The uncertainty in the estimate is Y −X = Y2 + X 1.762 + 1.182 = 2.12. The di˙erence is 4.72 ± 2.12 g/L.

13. (a) The relative uncertainty in is ∕ = ln . The given relative uncertainties are ln V = 0.0001, ln I = 0.0001, ln A = 0.001, ln l = 0.01, ln a = 0.01. ln = ln V + ln I + ln A − ln − ln l − ln a. t √ 2 2 + 2 + 2 + 2 = ln = ln + 0.00012 + 0.00012 + 0.0012 + 0.012 + 0.012 = 0.014 = 1.4% V ln I ln A ln l ln a (b) If ln V = 0.0001, ln I = 0.0001, ln A = 0.001, ln l = 0.005, and ln a = 0.01, then ln V = 0.011. If ln V = 0, ln I = 0, ln A = 0, ln l = 0.01, and ln a = 0.01, then ln V = 0.014. Reducing the uncertainty in l to 0.5% provides the greater reduction.

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CHAPTER 3

14.

Consider the cable to be made of four strands of each of three types of wire. Let X1 = X2 = X3 = X4 be the estimated strengths of the wires of the frst type, so Xi = 6000 and Xi = 20 for i = 1, 2, 3, 4. Let Y1 = Y2 = Y3 = Y4 be the estimated strengths of the wires of the second type, so Yi = 5700 and Yi = 30 for i = 1, 2, 3, 4. Let Z1 = Z2 = Z3 = Z4 be the estimated strengths of the wires of the third type, so Zi = 6200 and Zi = 40 for i = 1, 2, 3, 4. (a) Let D denote the strength of the cable estimated by the ductile wire method. Then D = X1 + X2 + X3 + X4 + Y1 + Y2 + Y3 + Y4 + Z1 + Z2 + Z3 + Z4 = 71, 600. The uncertainty is t∑ √ 4 2 + ∑ 4 2 + ∑4 2 = D = 4(202 ) + 4(302 ) + 4(402 ) = 108. i=1 Y i=1 Z i=1 X i

The strength of the cable is 71, 600 ± 108 pounds.

(b) The weakest wire has strength Yi = 5700 ± 30 lbs. The strength of the cable estimated by the brittle wire method, is B = 12Yi = 68, 400. The uncertainty is B = 12 Y = 12(30) = 360. So B = 68, 400 ± 360 pounds.

15. (a) Yes, since the strengths of the wires are all estimated to be the same, the sum of the strengths is the same as the strength of one of them multiplied by the number of wires. Therefore the estimated strength is 80,000 pounds in both cases.

(b) √ No, for the ductile wire method the squares of the uncertainties of the 16 wires are added, to obtain = 16 × 202 = 80. For the brittle wire method, the uncertainty in the strength of the weakest wire is multiplied by the number of wires, to obtain = 16 × 20 = 320.

16.

A = 80, A = 5, B = 90, B = 3. The relative increase is R = (B − A)∕A = B∕A − 1. )R = −B∕A2 = −0.0140625 )A )R = 1∕A = 0.0125 )B u 2 2 )R )R B2 = 0.080 A2 + R = )A )B The uncertainty is 0.080.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 3

145

17. (a) r = 3.00, r = 0.03, v = 4.0, v = 0.2. Q = r2 v = 113.1. )Q = 2 rv = 75.3982 )r )Q = r2 = 28.2743 )v v 0 12 0 12 )Q )Q 2 r + v2 = 6.1 Q = )r )v Q = 113.1 ± 6.1 m3 /s

(b) r = 4.00, r = 0.04, v = 2.0, v = 0.1. Q = r2 v = 100.5. )Q = 2 rv = 50.2655 )r )Q = r2 = 50.2655 )v v 0 12 0 12 )Q )Q r2 + v2 = 5.4 Q = )r )v Q = 100.5 ± 5.4 m3 /s

Q Q

= ln Q .

) ln Q 2 ) ln Q 1 ln Q = ln + 2 ln r + ln v, ln r = r = 0.01, ln v = v = 0.05, = , = r v )r r )v v v u 0 12 0 12 2 √ 2 ) ln Q ) ln Q r2 + ln Q = v2 = 22 r + v = 22 (0.012 ) + 0.052 = 0.054 )r )v r v Yes, the relative uncertainty in Q can be determined from the relative uncertainties in r and v, and it is equal to 5.4%.

18. (a) C0 = 0.2, t = 300, C = 0.174, C = 0.005 k = (1∕t)(ln C0 − ln C) = −0.005365 − (1∕300) ln C = 0.000464 dk −1 = = −0.019157 dC 300C dk k = = (0.019157)(0.005) = 0.000096 dC C k = (46.4 ± 9.6) × 10−5 s−1

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CHAPTER 3 k 9.6 × 10−5 = = 0.21 = 21% k 46.4 × 10−5

(c) From part (a), k = 46.4207 × 10−5 and k = 9.5785 × 10−5 . We are using extra precision in k and k in order to get good precision in t1∕2 and t1∕2 . dt1∕2 − ln 2 ln 2 , so = = −3.216639 × 106 k dk k2 dt1∕2 k = (3.216639 × 106 )(9.5785 × 10−5 ) = 308 t1∕2 = dk t1∕2 =

t1∕2 = 308 ≈ 310 s

(d) ln(t1∕2 ) = ln(ln 2) − ln k. Therefore ln t1∕2 = ln k =

k = 0.21 = 21% from part (b). k

19. (a) C0 = 0.03, t = 40, C = 0.0023, C = 0.0002. k = (1∕t)(1∕C − 1∕C0 ) = (1∕40)(1∕C) − 5∕6 = 10.04 dk dk −1 = = −4725.90, k = = 4725.90(0.0002) = 0.95 dC dC C 40C 2 k = 10.04 ± 0.95 s−1

(b) C0 = 0.03, t = 50, C = 0.0018, C = 0.0002. k = (1∕t)(1∕C − 1∕C0 ) = (1∕50)(1∕C) − 2∕3 = 10.4 dk −1 dk = 6172.84(0.0002) = 1.2 = = −6172.84, k = 2 dC C dC 50C k = 10.4 ± 1.2 s−1

k2 )∕2 = 0.5 k1 + 0.5 k2 . From parts (a) and (b), k = 0.945180 and k = 1.234568. We are (c) Let k = ( k1 + 1 2 using extra precision in k and k in order to get good precision in k . 1 2 u √ k = 0.52 2 + 0.52 2 = 0.52 (0.9451802 ) + 0.52 (1.2345682 ) = 0.78 k1

(d) The value of c is cbest =

2 + 2 k1

1.2345682 = 0.63. 0.9451802 + 1.2345682

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SUPPLEMENTARY EXERCISES FOR CHAPTER 3

20. (a) s = 1, v = 3.2, v = 0.1, a1 = v2 ∕2s = v2 ∕2 = 5.12 da1 da1 = v∕s = 3.2, a1 = = (3.2)(0.1) = 0.32 dv dv v a1 = 5.12 ± 0.32 m/s2

(b) s = 1, t = 0.63, t = 0.01, a2 = 2s∕t2 = 2∕t2 = 5.04 da2 da2 = −4s∕t3 = −15.997, a2 = = (15.997)(0.01) = 0.16 dt dt t a2 = 5.04 ± 0.16 m/s2

a2 + a2 1

0.162 = 0.20. 0.322 + 0.162 The weighted average with the smallest uncertainty is 0.20a1 + 0.80a2 . t The uncertainty is 0.202 a2 + 0.802 a2 = 0.14 m/s2 . From part (a), a1 = 0.32, and from part (b), a2 = 0.16. So cbest =

21. (a) Let s be the measured side of the square. Then s = 181.2, s = 0.1. The estimated area is S = s2 = 32, 833. dS dS = 2s = 362.4, S = = (362.4)(0.1) = 36 ds ds s S = 32, 833 ± 36 m2

(b) The estimated area of a semicircle is C = ( s2 )∕8 = 12, 894. dC dC = s∕4 = 142.314, C = = (142.314)(0.1) = 14 ds ds s C = 12, 894 ± 14 m2

(c) This is not correct. Let s denote the length of a side of the square. Since S and C are both computed in terms of s, they are not independent. In order to compute A correctly, we must express A directly in terms of s: dA A = s2 + 2 s2 ∕8 = s2 (1 + ∕4). So A = = 2s(1 + ∕4) s = 65 m2 . ds s

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22. (a) r = 3.0, r = 0.1, A = r2 = 28.27433,

dA = 2 r = 18.85 dr

dA = 18.85(0.1) = 1.9 dr r A = 28.27433 ± 1.9 cm2 A =

(b)

2 1 d A 2 d2A = 2 . The bias corrected estimate is A − = 28.27433 − (0.12 ) = 28.2429 cm2 . 2 dr2 r dr2

23. (a) P1 = 8.1, P1 = 0.1, P2 = 15.4, P2 = 0.2, P =

√

P1 P2 = 11.16871

)P3 P = √2 = 0.689426 )P1 2 P 1 P2 )P3 P = √1 = 0.36262 )P2 2 P 1 P2 v 0 1 0 1 )P3 2 2 )P3 2 2 P3 = P + P = 0.10 1 2 )P1 )P2 P3 = 11.16871 ± 0.10 MPa

(b)

) 2 P3

−3∕2

1∕2

) 2 P3

−3∕2

1∕2

= −(0.25)P2 P1 = −0.011773 )P12 )P22 The bias corrected estimate is LH I H I M ) 2 P3 ) 2 P3 1 2 2 P3 − P + P = 11.16871−(0.5)[−0.042557(0.12 )−0.011773(0.22 )] = 11.16916. 1 2 2 )P22 )P12 = −(0.25)P1

= −0.042557,

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Chapter 4 Section 4.1 2 = p(1 − p) = 0.24. 1. (a) X ∼ Bernoulli(p), where p = 0.4. X = p = 0.4, X

(b) No. A Bernoulli random variable has possible values 0 and 1. The possible values of Y are 0 and 2. (c) Y = 2X, where X ∼ Bernoulli(p). 2 = 4p(1 − p) = 0.96. Therefore Y = 2 X = 2p = 0.80, and Y2 = 22 X

2. (a) pX = 0.2 (b) pY = 0.45 (c) pZ = 0.65 (d) No. If the order is for a red set, it cannot also be for a white set. (e) Yes. pZ = 0.65 = 0.2 + 0.45 = pX + pY . (f) Yes. If the set is red, then X = 1, Y = 0, and Z = 1, so Z = X + Y . If the set is white, then X = 0, Y = 1, and Z = 1, so Z = X + Y . If the set is blue, then X = 0, Y = 0, and Z = 0, so Z = X + Y .

3. (a) pX = 0.05 (b) pY = 0.20 (c) pZ = 0.23 (d) Yes, it is possible for there to be both discoloration and a crack. (e) No. pZ = P (X = 1 or Y = 1) ≠ P (X = 1) + P (Y = 1) because P (X = 1 and Y = 1) ≠ 0.

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(f) No. If the surface has both discoloration and a crack, then X = 1, Y = 1, and Z = 1, but X + Y = 2.

4. (a) If X and Y cannot both be equal to 1, the possible values for Z are 0 and 1, so Z is a Bernoulli random variable. (b) pZ = P (Z = 1) = P (X = 1 or Y = 1) = P (X = 1) + P (Y = 1) − P (X = 1 and Y = 1) = P (X = 1) + P (Y = 1) − 0 = pX + pY (c) P (Z = 2) = P (X = 1 and Y = 1) ≠ 0. Since P (Z = 2) ≠ 0, Z is not a Bernoulli random variable.

5. (a) pX = 1∕2 (b) pY = 1∕2 (c) pZ = 1∕4 (d) Yes. P (X = x and Y = y) = P (X = x)P (Y = y) for all values of x and y. (e) Yes. pZ = 1∕4 = (1∕2)(1∕2) = pX pY . (f) Yes. If both coins come up heads, then X = 1, Y = 1, and Z = 1, so Z = XY . If not, then Z = 0, and either X, Y , or both are equal to 0 as well, so again Z = XY .

6. (a) pX = 1∕6 (b) pY = 5∕36 (c) pZ = 1∕36 (d) No. P (X = 1 and Y = 1) = P (double 3) = 1∕36, but P (X = 1)P (Y = 1) = (1∕6)(5∕36) ≠ 1∕36. (e) No. pZ = P (Z = 1) = P (X = 1 and Y = 1) ≠ P (X = 1)P (Y = 1). (f) Yes. If the dice come up double 3, then X = 1, Y = 1, and Z = 1, so Z = XY . If not, then Z = 0, and either X, Y , or both are equal to 0 as well, so again Z = XY .

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7. (a) Since the possible values of X and Y are 0 and 1, the possible values of the product Z = XY are also 0 and 1. Therefore Z is a Bernoulli random variable. (b) pZ = P (Z = 1) = P (XY = 1) = P (X = 1 and Y = 1) = P (X = 1)P (Y = 1) = pX pY .

The possible values of p are the solutions to p(1 − p) = 0.21. The solutions are p = 0.3 and p = 0.7.

Section 4.2 1. (a) P (X = 1) =

7! (0.3)1 (1 − 0.3)7−1 = 0.2471 1!(7 − 1)!

(b) P (X = 2) =

7! (0.3)2 (1 − 0.3)7−2 = 0.3177 2!(7 − 2)!

(d) P (X > 4)

7! (0.3)0 (1 − 0.3)7−0 = 0.0824 0!(7 − 0)!

P (X = 5) + P (X = 6) + P (X = 7) 7! 7! 7! = (0.3)5 (1 − 0.3)7−5 + (0.3)6 (1 − 0.3)7−6 + (0.3)7 (1 − 0.3)7−7 5!(7 − 5)! 6!(7 − 6)! 7!(7 − 7)! = 0.002500 + 0.003572 + 0.0002187 = 0.0288 =

(e) X = (7)(0.3) = 2.1 2 = (7)(0.3)(1 − 0.3) = 1.47 (f) X

2. (a) P (X > 6)

= =

P (X = 7) + P (X = 8) + P (X = 9) 9! 9! 9! (0.4)8 (1 − 0.4)9−8 + (0.4)9 (1 − 0.4)9−9 (0.4)7 (1 − 0.4)9−7 + 7!(9 − 7)! 8!(9 − 8)! 9!(9 − 9)!

= 0.0250

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(b) P (X ≥ 2) = = =

1 − P (X < 2) 1 − P (X = 0) − P (X = 1) 9! 9! (0.4)1 (1 − 0.4)9−1 1− (0.4)0 (1 − 0.4)9−0 − 0!(9 − 0)! 1!(9 − 1)!

= 0.9295

= =

P (X = 2) + P (X = 3) + P (X = 4) 9! 9! 9! (0.4)3 (1 − 0.4)9−3 + (0.4)4 (1 − 0.4)9−4 (0.4)2 (1 − 0.4)9−2 + 2!(9 − 2)! 3!(9 − 3)! 4!(9 − 4)!

= 0.6629

(d) P (2 < X ≤ 5)

= =

P (X = 3) + P (X = 4) + P (X = 5) 9! 9! 9! (0.4)4 (1 − 0.4)9−4 + (0.4)5 (1 − 0.4)9−5 (0.4)3 (1 − 0.4)9−3 + 3!(9 − 3)! 4!(9 − 4)! 5!(9 − 5)!

= 0.6687

(e) P (X = 0) =

9! (0.4)0 (1 − 0.4)9−0 = 0.0101 0!(9 − 0)!

(f) P (X = 7) =

9! (0.4)7 (1 − 0.4)9−7 = 0.0212 7!(9 − 7)!

(g) X = (9)(0.4) = 3.6 2 = (9)(0.4)(0.6) = 2.16 (h) X

3. (a) P (X = 2) =

(b) P (X > 2)

4! (0.6)2 (1 − 0.6)4−2 = 0.3456 2!(4 − 2)! = 1 − P (X ≤ 2) = 1 − P (X = 0) − P (X = 1) − P (X = 2) 8! 8! 8! (0.2)1 (1 − 0.2)8−1 − (0.2)2 (1 − 0.2)8−2 = 1− (0.2)0 (1 − 0.2)8−0 − 0!(8 − 0)! 1!(8 − 1)! 2!(8 − 2)! = 0.2031

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SECTION 4.2

P (X = 0) + P (X = 1) + P (X = 2) 5! 5! 5! (0.4)1 (1 − 0.4)5−1 + (0.4)2 (1 − 0.4)5−2 = (0.4)0 (1 − 0.4)5−0 + 0!(5 − 0)! 1!(5 − 1)! 2!(5 − 2)! = 0.6826 =

(d) P (3 ≤ X ≤ 5)

P (X = 3) + P (X = 4) + P (X = 5) 6! 6! 6! (0.7)4 (1 − 0.7)6−4 + (0.7)5 (1 − 0.7)6−5 = (0.7)3 (1 − 0.7)6−3 + 3!(6 − 3)! 4!(6 − 4)! 5!(6 − 5)! = 0.8119 =

Let X be the number of fights in the sample that are on time. Then X ∼ Bin(10, 0.75). (a) P (X = 10) =

(b) P (X = 8) =

10! (0.75)10 (1 − 0.75)10−10 = 0.0563 10!(10 − 10)!

10! (0.75)8 (1 − 0.75)10−8 = 0.2816 8!(10 − 8)!

10! 10! (0.75)9 (1 − (0.75)8 (1 − 0.75)10−8 + 8!(10 − 8)! 9!(10 − 9)!

10! (0.75)10 (1 − 0.75)10−10 = 0.2816 + 0.1877 + 0.0563 = 0.5256 10!(10 − 10)!

Let X be the number of automobiles that violate the standard. Then X ∼ Bin(12, 0.1). (a) P (X = 3) =

12! (0.1)3 (1 − 0.1)12−3 = 0.0852 3!(12 − 3)!

(b) P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2) = 0.1)10−1 +

10! 10! (0.1)0 (1 − 0.1)10−0 + (0.1)1 (1 − 0!(10 − 0)! 1!(10 − 1)!

10! (0.1)2 (1 − 0.1)10−2 = 0.2824 + 0.3766 + 0.2301 = 0.8891 2!(10 − 2)!

12! (0.1)0 (1 − 0.1)12−0 = 0.2824 0!(12 − 0)!

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Let X be the number of times the die comes up 6, and let Y be the number of times the die comes up an odd number. Then X ∼ Bin(8, 1∕6) and Y ∼ Bin(8, 1∕2). (a) P (X = 2) =

8! (1∕6)2 (1 − 1∕6)8−2 = 0.2605 2!(8 − 2)!

(b) P (Y = 5) =

8! (1∕2)5 (1 − 1∕2)8−5 = 0.2188 5!(8 − 5)!

(e) X =

(8)(1∕6)(5∕6) = 1.0541

(f) Y =

(8)(1∕2)(1∕2) = 1.4142

Let X be the number of failures that occur in the base metal. Then X ∼ Bin(20, 0.15). (a) P (X = 5) =

20! (0.15)5 (1 − 0.15)20−5 = 0.1028 5!(20 − 5)!

(b) P (X < 4)

= =

P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) 20! 20! 20! (0.15)1 (1 − 0.15)20−1 + (0.15)2 (1 − 0.15)20−2 (0.15)0 (1 − 0.15)20−0 + 0!(20 − 0)! 1!(20 − 1)! 2!(20 − 2)! +

20! (0.15)3 (1 − 0.15)20−3 3!(20 − 3)!

= 0.6477

20! (0.15)0 (1 − 0.15)20−0 = 0.0388 0!(20 − 0)!

(d) X = (20)(0.15) = 3 (e) X =

(20)(0.15)(0.85) = 1.5969

Let X be the number of contracts, out of the 20 chosen, that have a cost overrun. Then X ∼ Bin(20, 0.2).

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SECTION 4.2

(a) P (X = 4) =

20! (0.2)4 (1 − 0.2)20−4 = 0.2182 4!(20 − 4)!

(b) P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2) =

20! (0.2)0 (1 − 0.2)20−0 0!(20 − 0)!

20! 20! (0.2)1 (1 − 0.2)20−1 + (0.2)2 (1 − 0.2)20−2 1!(20 − 1)! 2!(20 − 2)! = 0.01153 + 0.05765 + 0.13691 = 0.2061

20! (0.2)0 (1 − 0.2)20−0 = 0.0115 0!(20 − 0)!

(d) = (20)(0.2) = 4 (e) =

(20)(0.2)(1 − 0.2) = 1.7889

Let X be the number of women among the fve winners. Then X ∼ Bin(5, 0.6). (a) P (X ≤ 3) = 1−P (X > 3) = 1−P (X = 4)−P (X = 5) = 1−

5! 5! (0.6)4 (1−0.6)5−4 − (0.6)5 (1− 4!(5 − 4)! 5!(5 − 5)!

0.6)5−5 = 1 − 0.25920 − 0.07776 = 0.6630 (b) P (3 of one gender and 2 of another) = P (X = 2)+P (X = 3) =

5! 5! (0.6)3 (1− (0.6)2 (1−0.6)5−2 + 2!(5 − 2)! 3!(5 − 3)!

0.6)5−3 = 0.2304 + 0.3456 = 0.5760

10.

Let X be the number of people in the sample who have made a purchase on the internet in the last month. and let p be the population proportion of people who have made a purchase on the internet in the last month. Then X ∼ Bin(150, p). The observed value of X is 87. u (a) p = X∕150 = 87∕150 = 0.58,

p =

p(1 − p) n u

Replacing p with p = 0.58 and n with 150, p =

0.58(1 − 0.58) = 0.0403. 150

p(1 − p) = 0.03. n u 0.58(1 − 0.58) Substitute p = 0.58 for p to obtain = 0.03. Solving for n yields n = 271. n

(b) We must fnd n so that

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11.

Let X be the number of rods that meet specifcations in the sample from mill A, and let Y be the number of rods that meet specifcations in the sample from mill B. Let pA be the probability that a rod from mill A meets specifcations and let pB be the probability that a rod from mill B meets specifcations. Then X ∼ Bin(100, pA ) and Y ∼ Bin(150, pB ). The observed value of X is 88 and the observed value of Y is 135. u (a) p A = X∕100 = 88∕100 = 0.88,

p A =

pA (1 − pA ) 100 u

Replacing pA with p A = 0.88 and n with 100, p A =

0.88(1 − 0.88) = 0.0325. 100

pB (1 − pB ) 150 u 0.9(1 − 0.9) Replacing pB with p B = 0.9 and n with 150, p B = = 0.0245. 150

(b) p B = Y ∕150 = 135∕150 = 0.9, p B =

12. (a) Let D denote the event that the item is defective and let R denote the even that the item can be repaired. Then P (D) = 0.2 and P (R ߀ D) = 0.6. Then P (D ∩ Rc ) = P (Rc ߀ D)P (D) = [1 − P (R ߀ D)]P (D) = (1 − 0.6)(0.2) = 0.08. (b) Let X be the number of items out of 20 that are defective and cannot be repaired. Then X ∼ Bin(20, 0.08). 20! P (X = 2) = (0.08)2 (1 − 0.08)20−2 = 0.2711 2!(20 − 2)!

13. (a) P (can be used) = P (meets spec) + P (long) = 0.90 + 0.06 = 0.96. (b) Let X be the number of bolts out of 10 that can be used. Then X ∼ Bin(10, 0.96). P (X < 9) =

1 − P (X ≥ 9)

1 − P (X = 9) − P (X = 10) 10! 10! (0.96)10 (1 − 0.96)10−10 = 1− (0.96)9 (1 − 0.96)10−9 − 9!(10 − 9)! 10!(10 − 10)! = 1 − 0.27701 − 0.66483 = 0.0582 =

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SECTION 4.2

14. (a) Let X be the number of the ten gears that are scrap. Then X ∼ Bin(10, 0.05). 10! P (X ≥ 1) = 1 − P (X = 0) = 1 − (0.05)0 (1 − 0.05)10−0 = 1 − 0.5987 = 0.4013 0!(10 − 0)! (b) Eight or more are not scrap if and only if two or fewer are scrap. 10! 10! P (X ≤ 2) = (0.05)0 (1 − 0.05)10−0 + (0.05)1 (1 − 0.05)10−1 0!(10 − 0)! 1!(10 − 1)! 10! + (0.05)2 (1 − 0.05)10−2 = 0.9885 2!(10 − 2)! (c) The probability that a gear is either degraded or scrap is 0.15 + 0.05 = 0.20. Let Y be the number of gears that are either degraded or scrap. Then Y ∼ Bin(10, 0.2). P (Y > 2)

= = =

1 − P (Y ≤ 2) 1 − P (Y = 0) − P (Y = 1) − P (Y = 2) 10! 10! 10! (0.2)1 (1 − 0.2)10−1 − (0.2)2 (1 − 0.2)10−2 1− (0.2)0 (1 − 0.2)10−0 − 0!(10 − 0)! 1!(10 − 1)! 2!(10 − 2)!

= 0.3222 (d) The probability that a gear is either conforming or degraded is 0.80 + 0.15 = 0.95. Let Y be the number of gears that are either conforming or degraded. Then Y ∼ Bin(10, 0.95). 10! (0.95)9 (1 − 0.95)10−9 = 0.3151 P (Y = 9) = 9!(10 − 9)!

15. (a) Let Y be the number of lights that are green. Then Y ∼ Bin(3, 0.6). 3! P (Y = 3) = (0.6)3 (1 − 0.6)3−3 = 0.216 3!(3 − 3)! (b) X ∼ Bin(5, 0.216) (c) P (X = 3) =

5! (0.216)3 (1 − 0.216)5−3 = 0.0619 3!(5 − 3)!

16. (a) Let X be the number of components out of 10 that are defective. Then X ∼ Bin(10, 0.1). P (return)

= P (X > 1) = 1 − P (X ≤ 1) 1 − P (X = 0) − P (X = 1) 10! 10! (0.1)1 (1 − 0.1)10−1 = 1− (0.1)0 (1 − 0.1)10−0 − 0!(10 − 0)! 1!(10 − 1)! = 1 − 0.34868 − 0.38742 = 0.2639 =

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(b) Let X be the number of components out of 10 that are defective. Then X ∼ Bin(10, 0.2). P (return)

= P (X > 1) =

1 − P (X ≤ 1)

1 − P (X = 0) − P (X = 1) 10! 10! (0.2)1 (1 − 0.2)10−1 = 1− (0.2)0 (1 − 0.2)10−0 − 0!(10 − 0)! 1!(10 − 1)! = 1 − 0.10737 − 0.26844 =

= 0.6242

= P (X > 1) = 1 − P (X ≤ 1) = 1 − P (X = 0) − P (X = 1) 10! 10! (0.02)1 (1 − 0.02)10−1 = 1− (0.02)0 (1 − 0.02)10−0 − 0!(10 − 0)! 1!(10 − 1)! = 1 − 0.81707 − 0.16675 = 0.0162

(d) Let n be the required sample size. Let X be the number of sampled components that are defective. Then X ∼ Bin(n, 0.2). P (accept) = P (X = 0) = (0.8)n ≤ 0.01. So n ln 0.8 ≤ ln 0.01, −0.2231n ≤ −4.605, n ≥ 20.64.

The minimum sample size is 21.

17. (a) Let X be the number of components that function. Then X ∼ Bin(5, 0.9). 5! 5! (0.9)3 (1 − 0.9)5−3 + P (X ≥ 3) = (0.9)4 (1 − 0.9)5−4 3!(5 − 3)! 4!(5 − 4)! 5! + (0.9)5 (1 − 0.9)5−5 = 0.9914 5!(5 − 5)! (b) We need to fnd the smallest value of n so that P (X ≤ 2) < 0.10 when X ∼ Bin(n, 0.9). Consulting Table A.1, we fnd that if n = 3, P (X ≤ 2) = 0.271, and if n = 4, P (X ≤ 2) = 0.052. The smallest value of n is therefore n = 4.

18. (a) On a rainy day, the number X of components that function is distributed Bin(6, 0.7).

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SECTION 4.2 P (X ≥ 4) =

6! 6! (0.7)4 (1 − 0.7)6−4 + (0.7)5 (1 − 0.7)6−5 4!(6 − 4)! 5!(6 − 5)! 6! + (0.7)6 (1 − 0.7)6−6 = 0.7443 6!(6 − 6)!

(b) On a nonrainy day, the number X of components that function is distributed Bin(6, 0.9). 6! 6! P (X ≥ 4) = (0.9)4 (1 − 0.9)6−4 + (0.9)5 (1 − 0.9)6−5 4!(6 − 4)! 5!(6 − 5)! 6! + (0.9)6 (1 − 0.9)6−6 = 0.9842 6!(6 − 6)! (c) Let R denote the event of a rainy day, and let F denote the event that the system functions. From part (a) we know that P (F ߀R) = 0.7443, and from part (b) we know that P (F ߀Rc ) = 0.9842. P (F ) = P (F ߀R)P (R) + P (F ߀Rc )P (Rc ) = (0.7443)(0.2) + (0.9842)(0.8) = 0.9362

19. (a) X ∼ Bin(10, 0.15). P (X ≥ 7)

P (X = 7) + P (X = 8) + P (X = 9) + P (X = 10) 10! 10! (0.15)8 (1 − 0.15)10−8 = (0.15)7 (1 − 0.15)10−7 + 7!(10 − 7)! 8!(10 − 8)! 10! 10! (0.15)10 (1 − 0.15)10−10 + (0.15)9 (1 − 0.15)10−9 + 9!(10 − 9)! 10!(10 − 10)! = 1.2591 × 10−4 + 8.3326 × 10−6 + 3.2677 × 10−7 + 5.7665 × 10−9 = 1.346 × 10−4 =

(b) Yes, only about 13 or 14 out of every 100,000 samples of size 10 would have 7 or more defective items.

= 1 − P (X < 2) = 1 − P (X = 0) − P (X = 1) 10! 10! (0.15)1 (1 − 0.15)10−1 = 1− (0.15)0 (1 − 0.15)10−0 − 0!(10 − 0)! 1!(10 − 1)! = 1 − 0.19687 − 0.34743 = 0.4557

(e) No, in about 45% of the samples of size 10, 2 or more items would be defective.

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(f) No, because 2 defectives in a sample of size 10 is not an unusually large number for a good shipment.

20. (a) X ∼ Bin(8, 0.80). P (X ≤ 1) =

P (X = 0) + P (X = 1) 8! 8! (0.8)1 (1 − 0.8)8−1 = (0.8)0 (1 − 0.8)8−0 + 0!(8 − 0)! 1!(8 − 1)! = 2.56 × 10−6 + 8.192 × 10−5 = 8.448 × 10−5

(b) Yes, if the claim is true, then only about 8 or 9 out of every 100,000 samples would have 1 or fewer policy holders with smoke detectors.

(c) Yes, because 1 out of 8 is an unusually small number of policy holders with smoke detectors if the claim is true. (d) P (X ≤ 6)

= 1 − P (X > 6) = 1 − P (X = 7) + P (X = 8) 8! 8! (0.80)8 (1 − 0.80)8−8 = 1− (0.80)7 (1 − 0.80)8−7 − 7!(8 − 7)! 8!(8 − 8)! = 1 − 0.33554 − 0.16777 = 0.4967

(e) No, for nearly half the samples of size 8, 6 or fewer policy holders would have smoke detectors.

(f) No, 6 out of 8 is not an unusually small number if the claim is true. 21. (a) Let X be the number of bits that are reversed. Then X ∼ Bin(5, 0.3). The correct value is assigned if X ≤ 2. 5! 5! (0.3)1 (1 − 0.3)5−1 P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2) = (0.3)0 (1 − 0.3)5−0 + 0!(5 − 0)! 1!(5 − 1)! 5! + (0.3)2 (1 − 0.3)5−2 = 0.8369 2!(5 − 2)! (b) We need to fnd the smallest odd value of n so that P (X ≤ (n − 1)∕2) ≥ 0.90 when X ∼ Bin(n, 0.3). Consulting Table A.1, we fnd that if n = 3, P (X ≤ 1) = 0.784, if n = 5, P (X ≤ 2) = 0.837, if n = 7, P (X ≤ 3) = 0.874, and if n = 9, P (X ≤ 4) = 0.901. The smallest value of n is therefore n = 9.

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SECTION 4.2

22.

P (Y = y) = P (n − X = y) = P (X = n − y) =

n! n! (1 − p)y [1 − (1 − p)]n−y pn−y (1 − p)y = (n − y)!y! y!(n − y)!

23. (a) Y = 10X + 3(100 − X) = 7X + 300

(b) X ∼ Bin(100, 0.9), so X = 100(0.9) = 90. Y = 7 X + 300 = 7(90) + 300 = $930 (c) X =

24.

100(0.9)(0.1) = 3. Y = 7 X = $21

Let X be the number of components of the two in the frst design that work. Let Y be the number of components of the four in the second design that work. Then X ∼ Bin(2, 0.9) and Y ∼ Bin(4, 0.9). The probability that the design with two components works is 2! P (X ≥ 1) = 1 − P (X = 0) = 1 − (0.9)0 (1 − 0.9)2−0 = 1 − 0.01 = 0.99. 0!(2 − 0)! The probability that the design with four components works is P (Y ≥ 2)

1 − P (Y < 2)

1 − P (Y = 0) − P (Y = 1) 4! 4! (0.9)1 (1 − 0.9)4−1 = 1− (0.9)0 (1 − 0.9)4−0 − 0!(4 − 0)! 1!(4 − 1)! = 1 − 0.0001 − 0.0036 = 0.9963 =

The design with four components has the greater probability of functioning.

25.

Let p be the probability that a tire has no faw. The results of Example 4.14 show that the sample proportion is p = 93∕100 = 0.93 and p = 0.0255. Let X be the number of tires out of four that have no faw. Then X ∼ Bin(4, p). Let q be the probability that exactly one of four tires has a faw. 4! (p)3 (1 − p)4−3 = 4p3 (1 − p) = 4p3 − 4p4 . Then q = P (X = 3) = 3!(4 − 3)! The estimate of q is q = 4p 3 − 4p 4 = 4(0.93)3 − 4(0.93)4 = 0.225. d q = 12p 2 − 16p 3 = 12(0.93)2 − 16(0.93)3 = −2.4909 d p

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CHAPTER 4 d q p = (2.4909)(0.0255) = 0.064 d p The probability is 0.225 ± 0.064.

q =

26. (a) The sample size is n = p =

p (1 − p )∕n =

and p = 88∕612 = 0.143791. .143791)(0.856209)∕612 = 0.014.

p = 0.144 ± 0.014

(b) Let O =

88∕612 p p 88 be the odds. Then O is estimated with = = = 0.167939. 1−p 1 − p 1 − 88∕612 524

dO = (1 − p) −2 = (1 − 0.143791)−2 = 1.364081. d p dO O = p = (1.364081)(0.014) = 0.019. O = 0.168 ± 0.019

Section 4.3 1. (a) P (X = 1) = e−4

41 = 0.0733 1!

(b) P (X = 0) = e−4

40 = 0.0183 0!

40 41 + e−4 = 0.0183 + 0.0733 = 0.0916 0! 1!

(d) P (X > 1) = 1 − P (X ≤ 1) = 1 − P (X = 0) − P (X = 1) = 1 − 0.0183 − 0.0733 = 0.9084 (e) Since X ∼ Poisson(4), X = 4. (f) Since X ∼ Poisson(4), X =

4 = 2.

2. (a) Since X ∼ Poisson(6), P (X = 8) = e−6

68 = 0.1033. 8!

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SECTION 4.3

(b) P (X = 2) = e−6

62 = 0.0446. 2!

P (X = 0) + P (X = 1) + P (X = 2) 60 61 62 = e−6 + e−6 + e−6 0! 1! 2! = 0.00248 + 0.01487 + 0.04462 = 0.0620

(d) P (X > 1)

1 − P (X = 0) − P (X = 1) 60 61 = 1 − e−6 − e−6 0! 1! = 1 − 0.002479 − 0.014873 = 0.9826 =

(e) Since X ∼ Poisson(6), X = 6. (f) Since X ∼ Poisson(6), X =

6 = 2.4495.

3. (a) Since X ∼ Poisson(5), P (X = 6) = e−5 (b) P (X ≤ 2)

56 = 0.1462. 6!

= P (X = 0) + P (X = 1) + P (X = 2) 60 61 62 = e−6 + e−6 + e−6 0! 1! 2! = 0.0067379 + 0.03369 + 0.084224 = 0.1247

P (X = 6) + P (X = 7) 57 56 = e−5 + e−5 6! 7! = 0.14622 + 0.10444 = 0.2507 =

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CHAPTER 4

(d) Since X ∼ Poisson(5), X = 5. (e) Since X ∼ Poisson(5), X =

5 = 2.236.

4. (a) X ∼ Poisson(6), so P (X = 7) = e−6 (b) P (X ≥ 3)

67 = 0.1377 7!

1 − P (X < 3)

= 1 − P (X = 0) − P (X = 1) − P (X = 2) 60 61 62 = 1 − e−6 − e−6 − e−6 0! 1! 2! = 1 − 0.00248 − 0.01487 − 0.04462 = 0.9380

P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) 63 64 65 66 = e−6 + e−6 + e−6 + e−6 3! 4! 5! 6! = 0.08924 + 0.13385 + 0.16062 + 0.16062 = 0.5443 =

(d) Since X ∼ Poisson(6), X = 6. (e) Since X ∼ Poisson(6), X =

6 = 2.4495.

Let X be the number of servers that fail. Then X is the number of successes in n = 1000 Bernoulli trials, each of which has success probability p = 0.003. The mean of X is np = (1000)(0.003) = 3. Since n is large and p is small, X ∼ Poisson(3) to a very close approximation. (a) P (X = 2) = e−3

32 = 0.2240 2!

(b) The event that fewer than 998 servers function is the same as the event that more than 2 servers fail, or equivalently, that X > 2. P (X > 2) = 1 − P (X ≤ 2) = 1 − e−3

31 32 30 − e−3 − e−3 0! 1! 2! = 1 − 0.04979 − 0.14936 − 0.22404

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SECTION 4.3

= 0.5768 (c) X = 3 (d) X =

3 = 1.732

Let X be the number of individuals in the sample that carry the gene. Then X is the number of successes in n = 1000 Bernoulli trials, each of which has success probability p = 0.0002. The mean of X is np = (1000)(0.0002) = 0.2. Since n is large and p is small, X ∼ Poisson(0.2) to a very close approximation. (a) P (X = 1) = e−0.2

0.21 = 0.1637 1!

(b) P (X = 0) = e−0.2

0.20 = 0.8187 0!

= 1 − P (X ≤ 2) = 1 − P (X = 0) − P (X = 1) − P (X = 2) 0.21 0.22 0.20 = 1 − e−0.2 − e−0.2 − e−0.2 0! 1! 2! = 1 − 0.81873 − 0.16375 − 0.016375 = 0.0011

(d) Since X ∼ Poisson(0.2), X = 0.2. (e) Since X ∼ Poisson(0.2), X =

0.2 = 0.45.

7. (a) Let X be the number of hits in one minute. Since the mean rate is 4 messages per minute, X ∼ Poisson(4). P (X = 5) = e−4

45 = 0.1563 5!

(b) Let X be the number of hits in 1.5 minutes. Since the mean rate is 4 messages per minute, X ∼ Poisson(6). P (X = 9) = e−6

69 = 0.0688 9!

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P (X = 0) + P (X = 1) + P (X = 2) 21 22 20 = e−2 + e−2 + e−2 0! 1! 2! = 0.13534 + 0.27067 + 0.27067

= 0.6767

8. (a) Let X be the number of cars that arrive in a given second. Since the mean rate is 4 per second, X ∼ Poisson(4). P (X = 3) = e−4

43 = 0.1954. 3!

(b) Since the mean rate is 4 per second, X ∼ Poisson(12) P (X = 8) = e−12

128 = 0.0655. 8!

1 − P (X = 0) − P (X = 1) − P (X = 2) − P (X = 3) 81 82 83 80 = 1 − e−8 − e−8 − e−8 − e−8 0! 1! 2! 3! = 1 − 0.000355 − 0.002684 − 0.010735 − 0.028626 = 0.9576

2 = np(1 − p), which is less than 3 because 1 − p < 1. Now (ii). Let X ∼ Bin(n, p) where X = np = 3. Then X let Y have a Poisson distribution with mean 3. The variance of Y is also equal to 3, because the variance of a Poisson random variable is always equal to its mean. Therefore Y has a larger variance than X.

10.

Let X represent the number of particles observed in 3 mL. Let represent the true concentration in particles per mL. Then X ∼ Poisson(3 ). The observed value of X is 48. The estimated concentration is = 48∕3 = 16. The uncertainty is = 16∕3 = 2.3. = 16.0 ± 2.3.

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167

SECTION 4.3

11.

Let X represent the number of bacteria observed in 0.5 mL. Let represent the true concentration in bacteria per mL. Then X ∼ Poisson(0.5 ). The observed value of X is 39. The estimated concentration is = 39∕0.5 = 78. The uncertainty is = 78∕0.5 = 12.49. = 78 ± 12.

12. (a) Let X be the number of plants in a two-acre region. Then X ∼ Poisson(20). P (X = 18) = e−20

2018 = 0.08439 18!

(b) The area of the circle is 10, 000 = 31, 415.9 ft2 . In units of acres, the area is 31, 415.9∕43,560 = 0.72121 acres. Let X be the number of plants in the circle. Then X ∼ Poisson(7.2121). P (X = 12) = e−7.2121

7.212112 = 0.0305 12!

(c) Let Y represent the number of plants observed in 0.1 acres. Let represent the true concentration in plants per acre. Then Y ∼ Poisson(0.1 ). The observed value of Y is 5. The estimated concentration is = 5∕0.1 = 50. The uncertainty is = 50∕0.1 = 22.36. = 50 ± 22.

13. (a) Let N be the number of defective components produced. Then N ∼ Poisson(20). P (N = 15) = e−20

2015 = 0.0516 15!

(b) Let X be the number that are repairable. Then X ∼ Bin(15, 0.6). 15! P (X = 10) = (0.6)10 (1 − 0.6)15−10 = 0.1859 10!(15 − 10)! (c) Given N, X ∼ Bin(N, 0.6). (d) Let N be the number of defective components, and let X be the number that are repairable. P (N = 15 ∩ X = 10)

P (N = 15)P (X = 10߀N = 15) 0 10 1 2015 15! = e−20 (0.6)10 (1 − 0.6)15−10 15! 10!(15 − 10)!

14.

0.00960

Let be the mean number of particles emitted in one minute. Let X be the number of particles actually emitted during a given minute.

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CHAPTER 4 0 = e− . 0! Now e− = 0.1353, so = − ln 0.1353 = 2.000.

Then X ∼ Poisson( ), so P (X = 0) = e−

15. (a) Let X be the number of faws in a 3-meter board. Since the mean number of faws is 0.45 per meter, X ∼ Poisson(1.35). P (X = 0) = e−1.35

1.350 , so P (X = 0) = 0.2592. 0!

(b) Let n be the length in meters needed so that the probability of no faw is 0.5. Then X ∼ Poisson(0.45n), and 0.45n0 P (X = 0) = e−0.45n = 0.5. Solving for n yields n = 1.54. 0!

16. (a) Let X be the number of raisins in a randomly chosen slice. Since the mean number of raisins per slice is 100∕60 = 5∕3, X ∼ Poisson(5∕3). P (X = 0) = e−(5∕3)

(5∕3)0 = 0.1889 0!

(b) Let X be the number of raisins in a randomly chosen slice. Since the mean number of raisins per slice is 200∕60 = 10∕3, X ∼ Poisson(10∕3). P (X = 5) = e−(10∕3)

(10∕3)5 = 0.1223 5!

(c) Let n be the required number of raisins. Let X be the number of raisins in a randomly chosen slice. Then X ∼ Poisson(n∕60). (n∕60)0 = e−(n∕60) = 0.01 0! −n∕60 = ln 0.01 = −4.605, so n = 276.3

P (X = 0) = e−(n∕60)

The smallest value of n for which P (X = 0) ≤ 0.01 is n = 277.

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SECTION 4.3

17.

Let 1 be the mean number of chips per cookie in one of Mom’s cookies, and let 2 be the mean number of chips per cookie in one of Grandma’s cookies. Let X1 and X2 be the numbers of chips in two of Mom’s cookies, and let Y1 and Y2 be the numbers of chips in two of Grandma’s cookies. Then Xi ∼ Poisson( 1 ) and Yi ∼ Poisson( 2 ). The observed values are X1 = 14, X2 = 11, Y1 = 6, and Y2 = 8. (a) The estimate is 1 = X = (14 + 11)∕2 = 12.5. (b) The estimate is 2 = Y = (6 + 8)∕2 = 7.0. (c) X1 = X2 =

1 . The uncertainty is X =

1 ∕2. Replacing 1 with 1 , X =

12.5∕2 = 2.5.

(d) Y1 = Y2 =

2 . The uncertainty is Y =

2 ∕2. Replacing 2 with 2 , Y =

7.0∕2 = 1.9.

(e) The estimate is 1 − 2 = 12.5 − 7.0 = 5.5. u The uncertainty is − = 2 + 2 = 6.25 + 3.5 = 3.1. 1

1 − 2 = 5.5 ± 3.1

18.

If the mean decay rate is exactly 1 per second, then the mean number of events in 10 seconds is 10, so X ∼ Poisson(10). (a) P (X ≤ 1) = P (X = 0) + P (X = 1) = e−10

101 100 + e−10 = 4.994 × 10−4 0! 1!

(b) Yes. If the mean rate is 1 event per second, then there will be 1 or fewer events in 10 seconds only about 5 times in 10,000. (c) Yes, because 1 event in 10 seconds is an unusually small number if the mean rate is 1 event per second. (d) P (X ≤ 8) =

∑8

x=0 P (X = x) =

∑8

x=0 e

x −10 10 = 0.3328

(e) No. If the mean rate is 1 event per second, then there will be 8 or fewer events in 10 seconds about one-third of the time. (f) No, because 8 events in 10 seconds is not an unusually small number if the mean rate is 1 event per second.

19.

If the mean number of particles is exactly 7 per mL, then X ∼ Poisson(7). (a) P (X ≤ 1) = P (X = 0) + P (X = 1) = e−7

70 71 + e−7 = 7.295 × 10−3 0! 1!

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CHAPTER 4

(b) Yes. If the mean concentration is 7 particles per mL, then only about 7 in every thousand 1 mL samples will contain 1 or fewer particles. (c) Yes, because 1 particle in a 1 mL sample is an unusually small number if the mean concentration is 7 particles per mL. (d) P (X ≤ 6) =

∑6

x=0 P (X = x) =

∑6

x=0 e

x −7 7 = 0.4497

(e) No. If the mean concentration is 7 particles per mL, then about 45% of all 1 mL samples will contain 6 or fewer particles. (f) No, because 6 particles in a 1 mL sample is not an unusually small number if the mean concentration is 7 particles per mL.

20.

Let B represent the background rate, let S represent the source rate, and let T = B + S represent the total rate, all in particles per second. (a) Let X be the number of background events counted in 100 seconds. Then X ∼ Poisson(100 B ). The observed value of X is 36. The estimated background rate is B = X∕100 = 36∕100 = 0.36. B ∕100. Replacing B with B yields =

The uncertainty is = B

0.36∕100 = 0.06.

B = 0.36 ± 0.06 (b) Let Y be the number of events, both background and source, counted in 100 seconds. Then Y ∼ Poisson(100 T ). The observed value of Y is 324. The estimated total rate is T = Y ∕100 = 324∕100 = 3.24. The uncertainty is = T ∕100. Replacing T with T yields = 3.24∕100 = 0.18. T

T = 3.24 ± 0.18 (c) S = T − B , so S = T − B = 3.24 − 0.36 = 2.88. Now T and B are independent, because they are based on independent measurements (X and Y ). u Therefore = 2 + 2 = (0.06)2 + (0.18)2 = 0.0036 + 0.0324 = 0.19. S

S = 2.88 ± 0.19 (d) If background and source plus background are each counted for 150 seconds, then 2 = B ∕150 and 2 = T ∕150. B

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SECTION 4.4 Approximating B with B = 0.36, 2 = 0.36∕150 = 0.0024. B

Approximating T with T = 3.24, 2 = 3.24∕150 = 0.0216. T u = 2 + 2 = 0.0024 + 0.0216 = 0.15 T

If background is counted for 100 seconds and source plus background is counted for 200 seconds, then 2 = B ∕100 and 2 = T ∕200. B

Approximating B with B = 0.36, 2 = 0.36∕100 = 0.0036. B

Approximating T with T = 3.24, 2 = 3.24∕200 = 0.0162. T u = 2 + 2 = 0.0036 + 0.0162 = 0.14. Counting background for 100 seconds and source plus T

background for 200 seconds produces the smaller uncertainty.

(e) It is not possible. Assume that the background emissions are counted for 100 seconds, and the source plus background is counted for N seconds. Then the uncertainty in the estimate of the source rate is 0.36∕100 + 3.24∕N. No matter how large N is, this quantity is always greater than 0.36∕100 = 0.06, and thus greater than 0.03.

21.

Let X be the number of faws in a one-square-meter sheet of aluminum. Let be the mean number of faws per square meter. Then X ∼ Poisson( ). Let p be the probability that a one-square-meter sheet of aluminum has exactly one faw. 1 = e− . From Example 4.27, = 2, and 2 = 0.02. 1! − = 2e−2 = 0.271. Therefore p is estimated with p = e

Then p = P (X = 1) = e−

dp − = e−2 − 2e−2 = −0.135335 = e− − e d p =

dp = 0.135335 0.02 = 0.019 d

p = 0.271 ± 0.019

Section 4.4

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CHAPTER 4

Let X be the number of automobiles that need brakes among the nine chosen. Then X ∼ H(25, 10, 9). 0 10 1 10 25 − 10 3 9−3 = 0.2940 P (X = 3) = 0 1 25 9

Let X be the number of restaurants with code violations among the 10 chosen. Then X ∼ H(30, 4, 10). 0 10 1 4 30 − 4 2 10 − 2 (a) P (X = 2) = = 0.3120 0 1 30 10 1 0 10 4 30 − 4 0 10 − 0 = 0.1768 (b) P (X = 0) = 0 1 30 10

Let X be the number of the day on which the computer crashes. Then X ∼ Geom(0.1). P (X = 12) = (0.1)(0.9)11 = 0.0314

Let X be the number days that pass up to and including the frst time the car encounters a red light. Then X ∼ Geom(0.4). (a) P (X = 3) = (0.4)(0.6)2 = 0.144 (b) P (X ≤ 3) =

∑3

x=0 P (X = x) =

∑3

x=0 (0.4)(0.6)

x = 0.784

Alternatively, P (X ≤ 3) = 1 − P (X > 3) = 1 − P (no red light for frst 3 days) = 1 − (0.6)3 = 0.784.

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SECTION 4.4

0 5. (a) Y ∼ NB(3, 0.4). P (Y = 7) =

1 7−1 (0.4)3 (1 − 0.4)7−3 = 0.1244 3−1

(b) Y = 3∕0.4 = 7.5 (c) Y2 = 3(1 − 0.4)∕(0.42 ) = 11.25

Let X1 be the number of green lights, X2 be the number of yellow lights, and X3 be the number of red lights encountered. Then (X1 , X2 , X3 ) ∼ MN(10, 0.5, 0.1, 0.4). 10! (0.5)4 (0.1)1 (0.4)5 = 0.0806 P (X1 = 4, X2 = 1, X3 = 5) = 4!1!5!

(iv). P (X = x) = p(1 − p)x−1 for x ≥ 1. Since 1 − p < 1, this quantity is maximized when x = 1.

Let X be then number of packages that will be flled before the process is stopped. Then X ∼ Geom(0.01). (a) X = 1∕0.01 = 100 2 = (1 − 0.01)∕(0.012 ) = 9900 (b) X

(c) Let Y be the number of packages up to and including the fourth one whose weight falls outside the specifcation. Then Y ∼ NB(4, 0.01). Y = 4∕0.01 = 400,

2 = 4(1 − 0.01)∕(0.012 ) = 39, 600

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174 9.

CHAPTER 4

Let X be the number of hours that have elapsed when the fault is detected. Then X ∼ Geom(0.8). (a) P (X ≤ 3) = P (X = 1) + P (X = 2) + P (X = 3) = (0.8)(1 − 0.8)0 + (0.8)(1 − 0.8)1 + (0.8)(1 − 0.8)2 = 0.992 (b) P (X = 3߀X > 2) =

P (X = 3 ∩ X > 2) P (X = 3) = P (X > 2) P (X > 2)

Now P (X = 3) = 0.8(1 − 0.8)2 = 0.032, and P (X > 2) = 1 − P (X ≤ 2) = 1 − P (X = 1) − P (X = 2) = 1 − 0.8 − 0.8(1 − 0.8) = 0.04. 0.032 Therefore P (X = 3߀X > 2) = = 0.8. 0.04 (c) X = 1∕0.8 = 1.25

10.

Let X be the number of runs up to and including the ffth failure. Then X ∼ NB(5, 0.001). (a) X = 5∕0.001 = 5000 u (b) X =

5(1 − 0.001) = 2234.95 0.0012

0 11. (a) X ∼ H(15, 12, 4). P (X = 3) =

12 3

1 15 − 12 4−3 = 0.4835 0 1 15 4

(b) X = 4(12∕15) = 3.2

u 4

12 15

3 15

15 − 4 15 − 1

= 0.7091

0 12. (a) X ∼ H(500, 100, 20), so P (X = 5) =

100 5

10 1 500 − 100 100! 400! 20 − 5 5!95! 15!385! = . 0 1 500! 500 20!480! 20

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(b) The binomial approximation is X ∼ Bin(20, 0.2). So P (X

20! (0.2)5 (1 − 0.2)20−5 = 0.1746. 5!(20 − 5)!

13. (a) X ∼ H(60, 5, 10), so

0 10 1 0 10 1 5 60 − 5 5 60 − 5 0 10 − 0 1 10 − 1 P (X > 1) = 1 − P (X = 0) − P (X = 1) = 1 − − = 0.1904 0 1 0 1 60 60 10 10

(b) X ∼ H(60, 10, 10), so

0 10 1 0 10 1 10 60 − 10 10 60 − 10 0 10 − 0 1 10 − 1 P (X > 1) = 1 − P (X = 0) − P (X = 1) = 1 − − = 0.5314 0 1 0 1 60 60 10 10

0 10 1 0 10 1 20 60 − 20 20 60 − 20 0 10 − 0 1 10 − 1 P (X > 1) = 1 − P (X = 0) − P (X = 1) = 1 − − = 0.9162 0 1 0 1 60 60 10 10

14. (a) (X1 , X2 , X3 , X4 ) ∼ MN(20, 0.40, 0.10, 0.45, 0.05) 20! P (X1 = 7, X2 = 3, X3 = 8, X4 = 2) = (0.40)7 (0.10)3 (0.45)8 (0.05)2 = 0.006872 7!3!8!2! (b) X1 ∼ Bin(20, 0.4). P (X1 = 7) =

15.

20! (0.4)7 (1 − 0.4)20−7 = 0.1659 7!(20 − 7)!

Let X1 denote the number of orders for a small drink, let X2 denote the number of orders for a medium drink, let X3 denote the number of orders for a large drink. (a) (X1 , X2 , X3 ) ∼ MN(20, 0.25, 0.35, 0.40) 20! (0.25)5 (0.35)7 (0.4)8 = 0.0411 P (X1 = 5, X2 = 7, X3 = 8) = 5!7!8!

(b) X3 ∼ Bin(20, 0.4). P (X3 > 10) =

∑20

20! x 20−x = 0.1275 x=11 x!(20 − x)! (0.4) (1 − 0.4)

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CHAPTER 4 Alternatively, use Table A.1 to fnd P (X3 ≤ 10) = 0.872. Then P (X3 > 10) = 1 − P (X ≤ 10) = 0.128

16. (a) Let X1 be the number of readings that are within 0.1 C of the true temperature, let X2 be the number of readings that are more than 0.1 C above the true temperature, let X3 be the number of readings that are more than 0.1 C below the true temperature. Then (X1 , X2 , X3 ) ∼ MN(10, 0.7, 0.1, 0.2). 10! (0.7)5 (0.1)2 (0.2)3 = 0.0339 P (X1 = 5, X2 = 2, X3 = 3) = 5!2!3! (b) X1 ∼ Bin(10, 0.7) P (X1 > 8)

P (X = 9) + P (X = 10) 10! 10! = (0.7)9 (1 − 0.7)10−9 + (0.7)10 (1 − 0.7)10−10 9!(10 − 9)! 10!(10 − 10)! = 0.1493

0 1 n 1 p (1 − p)n−1 = np(1 − p)n−1 . So P (X = n) = (1∕n)P (Y = 1). 1

17.

P (X = n) = p(1 − p)n−1 . P (Y = 1) =

18.

Let Y ∼ Bin(10, 0.3). From Table A.1, P (Y = 1) = 0.121, so P (X = 10) = (1∕10)(0.121) = 0.0121.

Section 4.5 1. (a) Using Table A.2: 1 − 0.1977 = 0.8023 (b) Using Table A.2: 0.9032 − 0.6554 = 0.2478 (c) Using Table A.2: 0.8159 − 0.3821 = 0.4338 (d) Using Table A.2: 0.0668 + (1 − 0.3264) = 0.7404

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2. (a) Using Table A.2: 0.7123 (b) Using Table A.2: 0.0197 − 0.0017 = 0.0180 (c) Using Table A.2: 0.7580 − 0.1401 = 0.6179 (d) Using Table A.2: 0.8315 + (1 − 0.9474) = 0.8841

3. (a) Using Table A.2: c = 1 (b) Using Table A.2: c = −2.00 (c) Using Table A.2: c = 1.50 (d) Using Table A.2: c = 0.83 (e) Using Table A.2: c = 1.45

4. (a) z = (2 − 2)∕ 9 = 0. The area to the right of z = 0 is 0.5000. (b) For 1, z = (1 − 2)∕ 9 = −0.33. For 7, z = (7 − 2)∕ 9 = 1.67. The area between z = −0.33 and z = 1.67 is 0.9525 − 0.3707 = 0.5818. (c) For −2.5, z = (−2.5 − 2)∕ 9 = −1.50. For −1, z = (−1 − 2)∕ 9 = −1.00. The area between z = −1.50 and z = −1.00 is 0.1587 − 0.0668 = 0.0919. (d) P (−3 ≤ X − 2 < 3) = P (−1 ≤ X < 5). For −1, z = (−1 − 2)∕ 9 = −1.00. For 5, z = (5 − 2)∕ 9 = 1.00. The area between z = −1.00 and z = 1.00 is 0.8413 − 0.1587 = 0.6826.

5. (a) z = (25.0 − 25.1)∕0.08 = −1.25. The area to the right of z = −1.25 is 0.1056.

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(b) The z-score of the 10th percentile is The 10th percentile is therefore

−1.28.

25.1 − 1.28(0.08) = 24.9976.

(c) z = (25.2 − 25.1)∕0.08 = 1.25. The area to the left of z = 1.25 is 0.8944. Therefore a diameter of 25.2 is on the 89th percentile, approximately. (d) For 25.0, z = (25.0 − 25.1)∕0.08 = −1.25. For 25.3, z = (25.3 − 25.1)∕0.08 = 2.50. The area between z = −1.25 and z = 2.50 is 0.9938 − 0.1056 = 0.8882.

6. (a) The z-score of the 10th percentile is The 10th percentile is therefore

−1.28.

822 − 1.28(29) = 784.88.

(b) z = (780 − 822)∕29 = −1.45. The area to the left of z = −1.45 is 0.0735. Therefore a depth of 780 is on the 7th percentile, approximately. (c) For 800, z = (800 − 822)∕29 = −0.76. For 830, z = (830 − 822)∕29 = 0.28. The area between z = −0.76 and z = 0.28 is 0.6103 − 0.2236 = 0.3867.

7. (a) z = (70 − 80.5)∕9.9 = −1.06. The area to the right of z = −1.06 is 0.1446.

(b) The z-score of the 80th percentile is The 80th percentile is therefore

.84.

80.5 + 0.84(9.9) = 88.8.

(c) z = (84 − 80.5)∕9.9 = 0.35. The area to the left of z = 0.35 is 0.6368. Therefore a score of 84 is on the 64th percentile, approximately.

(d) z = (90 − 80.5)∕9.9 = 0.96. The area to the left of z = 0.96 is 0.8315.

8. (a) For 3.7, z = (3.7 − 4.1)∕0.6 = −0.67. For 4.4, z = (4.4 − 4.1)∕0.6 = 0.50. The area between z = −0.67 and z = 0.50 is 0.6915 − 0.2514 = 0.4401.

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SECTION 4.5

(b) The area to the right of z = 0.50 is 0.3085. Therefore approximately 30.85% of cats are heavier than this one. (c) The z-score of the 80th percentile is The 80th percentile is therefore

.84.

.1 + 0.84(0.6) = 4.604.

(d) z = (4.5 − 4.1)∕0.6 = 0.67. The area to the right of z = 0.67 is 0.2514. (e) Let X be the number of cats that weigh more than 4.5 kg. Using part (d), the probability that a cat weighs more than 4.5 kg is 0.2514. Therefore X ∼ Bin(6, 0.2514). 6! P (X = 1) = (0.2514)1 (1 − 0.2514)6−1 = 0.3546. 1!(6 − 1)!

9. (a) z = (1800 − 1400)∕200 = 2.00. The area to the right of z = 2.00 is 0.0228. (b) The z-score of the 10th percentile is The 10th percentile is therefore

−1.28.

1400 − 1.28(200) = 1144.

(c) z = (1645 − 1400)∕200 = 1.23. The area to the left of z = 1.23 is 0.8907. Therefore a lifetime of 1645 is on the 89th percentile, approximately. (d) For 1350, z = (1350 − 1400)∕200 = −0.25. For 1550, z = (1550 − 1400)∕200 = 0.75. The area between z = −0.25 and z = 0.75 is 0.7734 − 0.4013 = 0.3721. (e) Let X be the number of bulbs with lifetimes between 1350 and 1550 hours. Using part (d), the probability that a bulb has a lifetime between 1350 and 1550 hours is 0.3721. Therefore X ∼ Bin(8, 0.2721). 8! P (X = 2) = (0.3721)2 (1 − 0.3721)8−2 = 0.2376. 2!(8 − 2)!

10.

No. The maximum possible score of 800 is 1.5 standard deviations above the mean of 650. If the population were normal, about 7% of the scores would therefore be above 800. In fact, no scores are above 800. Therefore the population cannot be normal.

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11. (a) z = (6 − 4.9)∕0.6 = 1.83. The area to the right of z = 1.83 is 1 − 0.9664 = 0.0336. The process will be shut down on 3.36% of days.

(b) z = (6 − 5.2)∕0.4 = 2.00. The area to the right of z = 2.00 is 1 − 0.9772 = 0.0228. Since a process with this broth will be shut down on 2.28% of days, this broth will result in fewer days of production lost.

12.

Since 20% of the cars are traveling between 55 and 65 mph, 30% are traveling less than 55 mph. The normal distribution is symmetric around the mean of 65. Therefore 30% of the cars are traveling faster than 75 mph.

13.

Let X be the diameter of a hole, and let Y be the diameter of a piston. Then X ∼ N(15, 0.0252 ), and Y ∼ N(14.88, 0.0152 ). The clearance is given by C = 0.5X − 0.5Y .

(a) C = 0.5X−0.5Y = 0.5 X − 0.5 Y = 0.5(15) − 0.5(14.88) = 0.06 cm

(b) C =

t 2 + (−0.5)2 2 = 0.52 X Y

0.52 (0.0252 ) + 0.52 (0.0152 ) = 0.01458 cm

(c) Since C is a linear combination of normal random variables, it is normally distributed, with C and C as given in parts (a) and (b). The z-score of 0.05 is (0.05 − 0.06)∕0.01458 = −0.69. The area to the left of z = −0.69 is 0.2451. Therefore P (C < 0.05) = 0.2451.

(d) The z-score of the 25th percentile is The 25th percentile is therefore

−0.67.

.06 − 0.67(0.01458) = 0.0502 cm.

(e) The z-score of 0.05 is z = (0.05−0.06)∕0.01458 = −0.69. The z-score of 0.09 is z = (0.09−0.06)∕0.01458 = 2.06. The area between z = −0.69 and z = 2.06 is 0.9803 − 0.2451 = 0.7352. Therefore P (0.05 < C < 0.09) = 0.7352.

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SECTION 4.5

(f) The probability is maximized when C = 0.07 cm, the midpoint between 0.05 and 0.09 cm. Now Y = 14.88, and X must be adjusted so that C = 0.5 X − 0.5 Y = 0.07. Substituting C = 0.07 and Y = 14.88, and solving for X yields X = 15.02 cm. To fnd the probability that the clearance will be between 0.05 and 0.09: The z-score of 0.05 is z = (0.05 − 0.07)∕0.01458 = −1.37. The z-score of 0.09 is z = (0.09 − 0.07)∕0.01458 = 1.37. The area between z = −1.37 and z = 1.37 is 0.9147 − 0.0853 = 0.8294. Therefore P (0.05 < C < 0.09) = 0.8294.

14. (a) The specifcation interval is 0.645 to 0.655 cm. The z-score of 0.645 is (0.645 − 0.652)∕0.003 = −2.33. The z-score of 0.655 is (0.655 − 0.652)∕0.003 = 1.00. The area between z = −2.33 and z = 1.00 is 0.8413 − 0.0099 = 0.8314. Therefore 83.14% of shafts will meet the specifcation.

(b) The z-score of 0.645 is (0.645 − 0.650)∕0.003 = −1.67. The z-score of 0.655 is (0.655 − 0.650)∕0.003 = 1.67. The area between z = −1.67 and z = 1.67 is 0.9525 − 0.0475 = 0.9050. Therefore 90.50% of shafts will meet the specifcation.

(c) Let be the required standard deviation. Since the mean of 0.650 is in the center of the specifcation interval, the proportion of diameters that are too thin will be equal to the proportion of diameters that are too thick. Therefore must be chosen so that the proportion of diameters that are greater than 0.655 is 0.005. The z-score for which 0.005 of the area is to the right is approximately z = 2.58. Therefore we fnd so that the z-score of 0.655 is 2.58 by solving 2.58 = (0.655 − 0.650)∕ , so = 0.00194 cm. Note that if z = 2.57 is used instead of z = 2.58, then = 0.00195 cm.

15. (a) The z-score of 12 is (12 − 12.05)∕0.03 = −1.67. The area to the left of z = −1.67 is 0.0475. The proportion is 0.0475.

(b) Let be the required value of the mean. This value must be chosen so that the 1st percentile of the distribution is 12. The z-score of the 1st percentile is approximately z = −2.33. Therefore −2.33 = (12 − )∕0.03. Solving for yields = 12.07 ounces.

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(c) Let be the required standard deviation. The value of must be chosen so that the 1st percentile of the distribution is 12. The z-score of the 1st percentile is approximately z = −2.33. Therefore −2.33 = (12 − 12.05)∕ . Solving for yields = 0.0215 ounces.

16. (a) The z-score of 6.2 is (6.2 − 6)∕0.3 = 0.67. The area to the left of z = 0.67 is 0.7486. The probability is 0.7486. (b) The amount of paint needed is the 90th percentile of the distribution. Let p be the 90th percentile. The z-score of the 90th percentile is approximately z = 1.28. Therefore 1.28 = (p − 6)∕0.3. Solving for p yields p = 6.384 L. (c) Let be the required standard deviation. The value of must be chosen so that the 90th percentile of the distribution is 6.2. The z-score of the 90th percentile is approximately z = 1.28. Therefore 1.28 = (6.2 − 6)∕ . Solving for yields = 0.156 L.

17. (a) The proportion of strengths that are less than 65 is 0.10. Therefore 65 is the 10th percentile of strength. The z-score of the 10th percentile is approximately z = −1.28. Let be the required standard deviation. The value of must be chosen so that the z-score for 65 is −1.28. Therefore −1.28 = (65 − 75)∕ . Solving for yields = 7.8125 N/m2 . (b) Let be the required standard deviation. The value of must be chosen so that the 1st percentile of the distribution is 65. The z-score of the 1st percentile is approximately z = −2.33. Therefore −2.33 = (65−75)∕ . Solving for yields = 4.292 N/m2 . (c) Let be the required value of the mean. This value must be chosen so that the 1st percentile of the distribution is 65. The z-score of the 1st percentile is approximately z = −2.33. Therefore −2.33 = (65 − )∕5. Solving for yields = 76.65 N/m2 .

18. (a) The z-score of 10.3 is (10.3 − 10)∕0.2 = 1.5. The area to the right of z = 1.5 is 1 − 0.9332 = 0.0668. The probability is 0.0668. (b) The mean area covered by 2 L is 2(10) = 20, and the standard deviation is 0.2 2 = 0.28284. The z score of 19.9 is (19.9 − 20)∕0.28284 = −0.35. The area to the right of z = −0.35 is 1 − 0.3632 = 0.6368. The probability is 0.6368.

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SECTION 4.5

19.

A score of 64 is on the 10th percentile, so the z-score of 64 is −1.28. A score of 81 is on the 80th percentile, so the z-score of 81 is 0.84. We therefore have the following equations in and : 64 − 81 − −1.28 = 0.84 = Solving for and yielda = 74.26, = 8.019.

20. (a) Since 10% of the bolts have strengths less than 18.3 kN, the z-score of 18.3 is −z.10 = −1.28. Since 5% of the bolts have strengths greater than 19.76 kN, the z-score of 19.76 is z.05 = 1.645. Therefore (18.3 − )∕ = −1.28, and (19.76 − )∕ = 1.645. Solving for and yields = 18.9389 and = 0.4991. (b) z = (18 − 18.9389)∕0.4991 = −1.88. The area to the right of z = −1.88 is 0.9699.

21.

Let a = 1∕ and let b = − ∕ . Then Z = aX +b. Now a +b = (1∕ ) − ∕ = 0, and a2 2 = (1∕ )2 2 = 1. Equation (4.25) now shows that Z ∼ N(0, 1).

22. (a) The z-score of 160 is (160 − 200)∕10 = −4.00. The area to the left of z = −4.00 is negligible. Therefore P (X ≤ 160)

(b) Yes. If the procedure is functioning correctly, values of 160 N or less would almost never occur.

(d) The z-score of 203 is (203 − 200)∕10 = 0.30. The area to the right of z = 0.30 is 1 − 0.6179 = 0.3821. Therefore P (X ≥ 203) = 0.3821.

(e) No. If the procedure is functioning correctly, values of 203 N or more would occur about 38% of the time.

(f) No, because 203 N is not unusually large if the process is functioning correctly.

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(g) The z-score of 195 is (195 − 200)∕10 = −0.50. The area to the left of z = −0.50 is 0.3085. Therefore P (X ≤ 195) = 0.3085.

(h) No. If the procedure is functioning correctly, values of 195 N or less would occur about 31% of the time.

(i) No, because 195 N is not unusually small if the process is functioning correctly.

23. (a) Let D = R2 − R1 . The event that R2 > R1 is the event that D > 0. D = R2 − R1 = 120 − 100 = 20. Since R1 and R2 are independent, D =

2 + 2 = R R 1

52 + 102 = 11.180.

Since D is a linear combination of independent normal random variables, D is normally distributed. The z-score of 0 is (0 − 20)∕11.180 = −1.79. The area to the right of z = −1.79 is 1 − 0.0367 = 0.9633. Therefore P (D > 0) = 0.9633.

(b) Let D = R2 − R1 . The event that R2 exceeds R1 by more than 30 is the event that D > 30. D = R2 − R1 = 120 − 100 = 20. Since R1 and R2 are independent, D =

2 + 2 = R R 1

52 + 102 = 11.180.

Since D is a linear combination of independent normal random variables, D is normally distributed. The z-score of 30 is (30 − 20)∕11.180 = 0.89. The area to the right of z = 0.89 is 1 − 0.8133 = 0.1867. Therefore P (D > 30) = 0.1867.

t 2 + 2 = R R 1

52 + 102 = 11.180.

Since S is a linear combination of independent normal random variables, S is normally distributed. The z-score of 200 is (200 − 220)∕11.180 = −1.79. The area to the left of z = −1.79 is 0.0367. Therefore P (S < 200) = 0.0367.

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24. (a) Since M is a linear combination of independent normal random variables, M is normally distributed. M = 0.5 X + Y = 0.5(0.450) + 0.250 = 0.475. 2 = 0.52 2 + 2 = 0.52 (0.0502 ) + 0.0252 = 0.00125. M X Y

Therefore M ∼ N(0.475, 0.00125).

(b) The z-score of 0.5 is (0.5 − 0.475)∕ 0.00125 = 0.71. The area to the right of z = 0.71 is 1 − 0.7611 = 0.2389. Therefore P (M > 0.5) = 0.2389.

25. (a) If m = 0, then X = E, so X ∼ N(0, 0.25). P (error) = P (X > 0.5). The z-score of 0.5 is (0.5 − 0)∕ 0.25 = 1.00. The area to the right of z = 1.00 is 1 − 0.8413 = 0.1587. Therefore P (error) = 0.1587. (b) If m = 0, then X = E, so X ∼ N(0, 2 ). P (error) = P (X > 0.5) = 0.01. The z-score of 0.5 is (0.5 − 0)∕ . Since P (X > 0.5) = 0.01, the z-score of 0.5 is 2.33. Therefore (0.5 − 0)∕ = 2.33. Solving for yields = 0.2146, and 2 = 0.04605.

26. (a) If m = 0, then X = −1.5 + E, so X ∼ N(−1.5, 0.64). P (error) = P (X > 0.5). The z-score of 0.5 is [0.5 − (−1.5)])∕ 0.64 = 2.50. The area to the right of z = 2.50 is 1 − 0.9938 = 0.0062. Therefore P (error) = 0.0062. (b) If m = 1, then X = 1.5 + E, so X ∼ N(1.5, 0.64).

P (error) = P (X ≤ 0.5). The z-score of 0.5 is (0.5 − 1.5)∕ 0.64 = −1.25.

The area to the left of z = −1.25 is 0.1056. Therefore P (error) = 0.1056. (c) Let S be the value from the sent string (0 or 1). Let C be the event that the value is received correctly. From part (a) we know that P (C߀S = 0) = 1 − 0.0062 = 0.9938. From part (b) we know that P (C߀S = 1) = 1 − 0.1056 = 0.8944. Now P (C) = P (C߀S = 0)P (S = 0) + P (C߀S = 1)P (S = 1) = (0.9938)(0.4) + (0.8944)(0.6) = 0.9342.

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(d) Let S be the value from the sent string (0 or 1), and let R be the value received. We must fnd P (S = 0߀R = 1). P (R = 1߀S = 0)P (S = 0) P (S = 0߀R = 1) = P (R = 1߀S = 0)P (S = 0) + P (R = 1߀S = 1)P (S = 1) From part (a) we know that P (R = 1߀S = 0) = 0.0062. From part (b) we know that P (R = 1߀S = 1) = 1 − 0.1056 = 0.8944. (0.0062)(0.4) Therefore P (S = 0߀R = 1) = = 0.00460. (0.0062)(0.4) + (0.8944)(0.6) (e) Let S be the value from the sent string (0 or 1), and let R be the value received. We must fnd P (S = 1߀R = 0). P (R = 0߀S = 1)P (S = 1) P (S = 1߀R = 0) = P (R = 0߀S = 1)P (S = 1) + P (R = 0߀S = 0)P (S = 0) From part (a) we know that P (R = 0߀S = 0) = 1 − 0.0062 = 0.9938. From part (b) we know that P (R = 0߀S = 1) = 0.1056. (0.1056)(0.6) Therefore P (S = 1߀R = 0) = = 0.1375. (0.1056)(0.6) + (0.9938)(0.4)

27. (a) The sample mean is 114.8 J and the sample standard deviation is 5.006 J.

(b) The z-score of 100 is (100 − 114.8)∕5.006 = −2.96. The area to the left of z = −2.96 is 0.0015. Therefore only 0.15% of bolts would have breaking torques less than 100 J, so the shipment would be accepted.

(c) The sample mean is 117.08 J; the sample standard deviation is 8.295 J. The z-score of 100 is (100−117.08)∕8.295 = −2.06. The area to the left of z = −2.06 is 0.0197. Therefore about 2% of the bolts would have breaking torques less than 100 J, so the shipment would not be accepted.

(d) The bolts in part (c) are stronger. In fact, the weakest bolt in part (c) is stronger than the weakest bolt in part (a), the second-weakest bolt in part (c) is stronger than the second-weakest bolt in part (a), and so on.

(e) The method is certainly not valid for the bolts in part (c). This sample contains an outlier (140), so the normal distribution should not be used.

28.

k 1 2 3

P (߀X − X ߀ ≥ k X ) 0.3174 0.0456 0.0026

1∕k2 1 0.25 0.111

The actual probabilities are much smaller than the Chebyshev bounds.

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Section 4.6 1.

Let Y be the lifetime of the component. 2

(a) E(Y ) = e + ∕2 = e1.2+(0.4) ∕2 = 3.5966 (b) P (3 < Y < 6) = P (ln 3 < ln Y < ln 6) = P (1.0986 < ln Y < 1.7918). ln Y ∼ N(1.2, 0.42 ). The z-score of 1.0986 is (1.0986 − 1.2)∕0.4 = −0.25. The z-score of 1.7918 is (1.7918 − 1.2)∕0.4 = 1.48. The area between z = −0.25 and z = 1.48 is 0.9306 − 0.4013 = 0.5293. Therefore P (3 < Y < 6) = 0.5293. (c) Let m be the median of Y . Then P (Y ≤ m) = P (ln Y ≤ ln m) = 0.5. Since ln Y ∼ N(1.2, 0.42 ), P (ln Y < 1.2) = 0.5. Therefore ln m = 1.2, so m = e1.2 = 3.3201. (d) Let y90 be the 90th percentile of Y . Then P (Y ≤ y90 ) = P (ln Y ≤ ln y90 ) = 0.90. The z-score of the 90th percentile is approximately z = 1.28. Therefore the z-score of ln y90 must be 1.28, so ln y90 satisfes the equation 1.28 = (ln y90 − 1.2)∕0.4. ln y90 = 1.712, so y90 = e1.712 = 5.540.

Let Y be the amount of active ingredient in the skin. 2

(a) E(Y ) = e + ∕2 = e2.2+(2.1) ∕2 = 81.8591 (b) Let m be the median of Y . Then P (Y ≤ m) = P (ln Y ≤ ln m) = 0.5. Since ln Y ∼ N(2.2, 2.12 ), P (ln Y < 2.2) = 0.5. Therefore ln m = 2.2, so m = e2.2 = 9.0250. (c) P (Y > 100) = P (ln Y > ln 100) = P (ln Y > 4.6052). ln Y ∼ N(2.2, 2.12 ). The z-score of 4.6052 is (4.6052 − 2.2)∕2.1 = 1.15. The area to the right of z = 1.15 is 1 − 0.8749 = 0.1251. Therefore P (Y > 100) = 0.1251. (d) P (Y < 50) = P (ln Y < ln 50) = P (ln Y < 3.9120). ln Y ∼ N(2.2, 2.12 ). The z-score of 3.9120 is (3.9120 − 2.2)∕2.1 = 0.82.

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The area to the left of z = 0.82 is 0.7939. Therefore P (Y < 50) = 0.7939. (e) Let y80 be the 80th percentile of Y . Then P (Y ≤ y80 ) = P (ln Y ≤ ln y80 ) = 0.80. The z-score of the 80th percentile is approximately z = 0.84. Therefore the z-score of ln y80 must be 0.84, so ln y80 satisfes the equation 0.84 = (ln y80 − 2.2)∕2.1. ln y80 = 3.964, so y80 = e3.964 = 52.668. 2

(f) V (Y ) = e2 +2 − e2 + = e2(2.2)+2(2.1) − e2(2.2)+(2.1) = 544580.1. The standard deviation is

V (Y ) =

e2(2.2)+2(2.1)2 − e2(2.2)+(2.1)2 = 737.96.

Let Y represent the BMI for a randomly chosen man aged 25–34. 2

(a) E(Y ) = e + ∕2 = e3.215+(0.157) ∕2 = 25.212 2

(b) V (Y ) = e2 +2 − e2 + = e2(3.215)+2(0.157) − e2(3.215)+(0.157) = 15.86285. The standard deviation is

V (Y ) =

e2(3.215)+2(0.157)2 − e2(3.215)+(0.157)2 =

15.86285 = 3.9828.

(c) Let m be the median of Y . Then P (Y ≤ m) = P (ln Y ≤ ln m) = 0.5. Since ln Y ∼ N(3.215, 0.1572 ), P (ln Y < 3.215) = 0.5. Therefore ln m = 3.215, so m = e3.215 = 24.903.

(d) P (Y < 22) = P (ln Y < ln 22) = P (ln Y < 3.0910). The z-score of 3.0910 is (3.0910 − 3.215)∕0.157 = −0.79. The area to the left of z = −0.79 is 0.2148. Therefore P (Y < 22) = 0.2148. (e) Let y75 be the 75th percentile of Y . Then P (Y ≤ y75 ) = P (ln Y ≤ ln y75 ) = 0.75. The z-score of the 75th percentile is approximately z = 0.67. Therefore the z-score of ln y75 must be 0.67, so ln y75 satisfes the equation 0.67 = (ln y75 − 3.215)∕0.157. ln y75 = 3.3202, so y75 = e3.3202 = 27.666.

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SECTION 4.6

Let Y be the increase in the risk. 2

(a) E(Y ) = e + ∕2 = e−15.65+(0.79) ∕2 = 2.1818 × 10−7 (b) Let m be the median of Y . Then P (Y ≤ m) = P (ln Y ≤ ln m) = 0.5. Since ln Y ∼ N(−15.65, 0.792 ), P (ln Y < −15.65) = 0.5. Therefore ln m = −15.65, so m = e−15.65 = 1.5970 × 10−7 . 2

V (Y ) =

e2(−15.65)+2(0.79)2 − e2(−15.65)+(0.79)2 =

4.1250 × 10−14 = 2.0310 × 10−7 .

(d) Let y5 be the 5th percentile of Y . Then P (Y ≤ y5 ) = P (ln Y ≤ ln y5 ) = 0.05. The z-score of the 5th percentile is approximately z = −1.645. Therefore the z-score of ln y5 must be −1.645, so ln y5 satisfes the equation −1.645 = [ ln y5 − (−15.65)]∕0.79. ln y5 = −16.9496, so y5 = e−16.9496 = 4.354 × 10−8 . (e) Let y95 be the 95th percentile of Y . Then P (Y ≤ y95 ) = P (ln Y ≤ ln y95 ) = 0.95. The z-score of the 95th percentile is approximately z = 1.645. Therefore the z-score of ln y95 must be 1.645, so ln y95 satisfes the equation 1.645 = [ ln y95 − (−15.65)]∕0.79. ln y95 = −14.35045, so y95 = e−14.35045 = 5.857 × 10−7 .

5. (a) ln L ∼ N(0.7, 0.05), ln 2 = 1.83788, and ln g = 2.28238. Therefore ln T = ln 2 + 0.5 ln L − 0.5 ln g = 0.5 ln L + 1.83788 − 0.5 ∗ 2.28238. It follows that ln T ∼ N(1.04669, 0.0125). Since ln T is normal, T is lognormal, with T = 1.04669 and V2 = 0.0125. (b) P (T > 3) = P (ln T > ln 3) = P (ln T > 1.0986). Now ln T ∼ N(1.04669, 0.0125), so the z-score of 1.09896 is (1.09896 − 1.04669)∕ 0.0125 = 0.46. The area to the right of z = 0.46 is 0.3228. Therefore P (T > 3) = 0.3228. (c) P (1.5 < T < 3) = P (ln 1.5 < ln T < ln 3) = P (0.405465 < ln T < 1.098612). Now ln T ∼ N(1.04669, 0.0125), so the z-score of 0.405645 is (0.405645 − 1.04669)∕ 0.0125 = −5.74, and

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the z-score of 1.09896 is (1.09896 − 1.04669)∕ 0.0125 = 0.46. The area between z = 0.46 and z = −5.74 is 1.0000 − 0.3228 = 0.6772. Therefore P (1.5 < T < 3) = 0.6772. 2

(d) The mean of T is E(T ) = e + ∕2 = e1.04669+0.05∕2 = 2.866. (e) Since ln T ∼ N(1.04669, 0.05), the median of ln T is 1.04669. Therefore the median of T is e1.04669 = 2.848. e2 +2 2 − e2 + 2 =

(f) The standard deviation of T is

e2(1.04669)+2(0.0125) − e2(1.04669)+0.0125 = 0.3214.

(g) Let t15 be the 15th percentile of T . Then P (T ≤ t15 ) = P (ln T ≤ ln t15 ) = 0.15. The z-score of the 15th percentile is approximately z = −1.04. Therefore the z-score of ln t15 must be −1.04, so ln t15 satisfes the equation −1.04 = ( ln t15 −1.04669)∕ 0.0125. ln t15 = 0.930414, so t15 = e0.930414 = 2.5356. (h) Let t85 be the 85th percentile of T . Then P (T ≤ t85 ) = P (ln T ≤ ln t85 ) = 0.85. The z-score of the 85th percentile is approximately z = 1.04. Therefore the z-score of ln t85 must be 1.04, so ln t85 satisfes the equation 1.04 = ( ln t85 − 1.04669)∕ 0.0125. ln t85 = 1.1629655, so t85 = e1.1629655 = 3.1994.

6. (a) ln r ∼ N(1.6, 0.04), ln ℎ ∼ N(1.9, 0.05), and r and ℎ are independent. Since ln V = ln + 2 ln r + ln ℎ, ln V ∼ N(6.24473, 0.21), V = 6.24473 and V2 = 0.21. (b) P (V < 600) = P (ln V < ln 600) = P (ln V < 6.3969). Now ln V ∼ N(6.24473, 0.21), so the z-score of 6.3969 is (6.3969 − 6.24473)∕ 0.21 = 0.33. The area to the left of z = 0.33 is 0.6293. Therefore P (V < 600) = 0.6293. (c) P (500 < V < 800) = P (ln 500 < ln V < ln 800) = P (6.21461 < ln V < 6.68461). Now ln V ∼ N(6.2447, 0.21), so the z-score of 6.21461 is (6.21461 − 6.24473)∕ 0.21 = −0.07, and the z-score of 6.68461 is (6.68461 − 6.24473)∕ 0.21 = 0.96. The area between z = 0.96 and z = −0.07 is 0.8315 − 0.4721 = 0.3594. Therefore P (500 < V < 800) = 0.3594. 2

(d) The mean of V is E(V ) = e + ∕2 = e6.24473+0.21∕2 = 572.3381.

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SECTION 4.6

(e) Since ln V ∼ N(6.24473, 0.21), the median of ln V is 6.24473. Therefore the median of V is e6.24473 = 515.2900. (f) The standard deviation of V is

e2 +2 2 − e2 + 2 =

e2(6.24473)+2(0.21) − e2(6.24473)+0.21 = 276.6697.

(g) Let v05 be the 5th percentile of V . Then P (V ≤ v05 ) = P (ln V ≤ ln v05 ) = 0.05. The z-score of the 5th percentile is approximately z = −1.645. Therefore the z-score of ln v05 must be −1.645, so ln v05 satisfes the equation −1.645 = ( ln v05 −6.24473)∕ 0.21. ln v05 = 5.49090, so v05 = e5.49090 = 242.4744. (h) Let v95 be the 95th percentile of V . Then P (V ≤ v95 ) = P (ln V ≤ ln v95 ) = 0.95. The z-score of the 95th percentile is approximately z = 1.645. Therefore the z-score of ln v95 must be 1.645, so ln v95 satisfes the equation 1.645 = ( ln v05 −6.24473)∕ 0.21. ln v05 = 6.998564, so v05 = e6.998564 = 1095.0591.

From Exercise 5b), the probability that a pendulum has a period greater than 3 seconds is 0.3228. Let X be the number of pendulums out of 10 whose period is greater than 3 seconds. Then X ∼ Bin(10, 0.3228). P (X ≥ 4)

= 1 − P (X ≤ 3) = 1 − [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)] 10! 10! (0.3228)1 (1 − 0.3228)10−1 = 1− (0.3228)0 (1 − 0.3228)10−0 − 0!(10 − 0)! 1!(10 − 1)! 10! 10! (0.3228)3 (1 − 0.3228)10−3 − (0.3228)2 (1 − 0.3228)10−2 − 2!(10 − 2)! 3!(10 − 3)! = 1 − 0.020285 − 0.096691 − 0.2074 − 0.26363 = 0.4120

From Exercise 6c), the probability that a cylinder has a volume between 500 and 800 cm2 is 0.3594. Let X be the number of cylinders out of 8 whose volumes are between 500 and 800 cm2 . Then X ∼ Bin(8, 0.3594). P (X < 5)

= 1 − P (X ≥ 5) = 1 − [P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8)] 8! 8! (0.3594)6 (1 − 0.3594)8−6 = 1− (0.3594)5 (1 − 0.3594)8−5 − 5!(8 − 5)! 6!(8 − 6)! 8! 8! (0.3594)8 (1 − 0.3594)8−8 − (0.3594)7 (1 − 0.3594)8−7 − 7!(8 − 7)! 8!(8 − 8)! = 1 − 0.088275 − 0.024763 − 0.0039694 − 0.00027837 = 0.8827

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Let X represent the withdrawal strength for a randomly chosen annularly threaded nail, and let Y represent the withdrawal strength for a randomly chosen helically threaded nail. 2

(a) E(X) = e3.82+(0.219) ∕2 = 46.711 N/mm 2

(b) E(Y ) = e3.47+(0.272) ∕2 = 33.348 N/mm

(c) First fnd the probability for annularly threaded nails. P (X > 50) = P (ln X > ln 50) = P (ln X > 3.9120). The z-score of 3.9120 is (3.9120 − 3.82)∕0.219 = 0.42. The area to the right of z = 0.42 is 1 − 0.6628 = 0.3372. Therefore the probability for annularly threaded nails is 0.3372. Now fnd the probability for helically threaded nails. P (Y > 50) = P (ln Y > ln 50) = P (ln Y > 3.9120). The z-score of 3.9120 is (3.9120 − 3.47)∕0.272 = 1.63. The area to the right of z = 1.63 is 1 − 0.9484 = 0.0516. Therefore the probability for helically threaded nails is 0.0516. Annularly threaded nails have the greater probability. The probability is 0.3372 vs. 0.0516 for helically threaded nails.

(d) First fnd the median strength for annularly threaded nails.

Let m be the median of X. Then P (X ≤ m) = P (ln X ≤ ln m) = 0.5. Since ln X ∼ N(3.82, 0.2192 ), P (ln Y < 3.82) = 0.5. Therefore ln m = 3.82, so m = e3.82 . Now P (Y > e3.82 ) = P (ln Y > ln e3.82 ) = P (ln Y > 3.82). The z-score of 3.82 is (3.82 − 3.47)∕0.272 = 1.29. The area to the right of z = 1.29 is 1 − 0.9015 = 0.0985. Therefore the probability is 0.0985.

(e) The log of the withdrawal strength for this nail is ln 20 = 2.996. For annularly threaded nails, the z-score of 2.996 is (2.996 − 3.82)∕0.219 = −3.76. The area to the left of z = −3.76 is less than 0.0001. Therefore a strength of 20 is extremely small for an annularly threaded nail; less than one in ten thousand such nails have strengths this low. For helically threaded nails, the z-score of 2.996 is (2.996 − 3.47)∕0.272 = −1.74. The area to the left of z = −1.74 is 0.0409. Therefore about 4.09% of helically threaded nails have strengths of 20 or below.

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SECTION 4.6

We can be pretty sure that it was a helically threaded nail. Only about 0.01% of annularly threaded nails have strengths as small as 20, while about 4.09% of helically threaded nails do.

10.

i. Since the lognormal population is skewed to the right, the mean will always be larger than the median. More 2 precisely, the mean is e + ∕2 , while the median is e . Since 2 ∕2 > 0, the mean is greater than the median.

11.

Let X represent the price of a share of company A one year from now. Let Y represent the price of a share of company B one year from now. 2

(a) E(X) = e0.05+(0.1) ∕2 = $1.0565 (b) P (X > 1.20) = P (ln X > ln 1.20) = P (ln X > 0.1823). The z-score of 0.1823 is (0.1823 − 0.05)∕0.1 = 1.32. The area to the right of z = 1.32 is 1 − 0.9066 = 0.0934. Therefore P (X > 1.20) = 0.0934. 2

(c) E(Y ) = e0.02+(0.2) ∕2 = $1.0408 (d) P (Y > 1.20) = P (ln Y > ln 1.20) = P (ln Y > 0.1823). The z-score of 0.1823 is (0.1823 − 0.02)∕0.2 = 0.81. The area to the right of z = 0.81 is 1 − 0.7910 = 0.2090. Therefore P (Y > 1.20) = 0.2090.

12. (a) P (X < 20) = P (ln X < ln 20) = P (ln X < 2.996). The z-score of 2.996 is (2.996 − 5)∕0.5 = −4.01. The area to the left of z = −4.01 is negligible. Therefore P (X < 20)

(b) Yes. Almost none of the specimens will have tensile strengths less than 20 MPa if the claim is true. (c) Yes, because 20 MPa is unusually small if the claim is true.

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(d) P (X < 130) = P (ln X < ln 130) = P (ln X < 4.868). The z-score of 4.868 is (4.868 − 5)∕0.5 = −0.26. The area to the left of z = −0.26 is 0.3974. Therefore P (X < 130) = 0.3974. (e) No. Nearly 40% of the specimens will have tensile strengths less than 130 MPa if the claim is true. (f) No, because 130 MPa is not unusually small if the claim is true.

13.

ln X1 , ..., ln Xn are independent normal random variables, so ln P = a1 ln X1 + ⋯ + an ln Xn is a normal random variable. It follows that P is lognormal.

Section 4.7 1. (a) T = 1∕0.45 = 2.2222 (b) T2 = 1∕(0.452 ) = 4.9383 (c) P (T > 3) = 1 − P (T ≤ 3) = 1 − (1 − e−0.45(3) ) = 0.2592 (d) Let m be the median. Then P (T ≤ m) = 0.5.

P (T ≤ m) = 1 − e−0.45m = 0.5, so e−0.45m = 0.5.

Solving for m yields m = 1.5403.

2. (a) = 1∕ = 1∕0.5 = 2 seconds−1 . (b) Let T be the time between requests, and let m be the median of T . Then P (T ≤ m) = 0.5. P (T ≤ m) = 1 − e−2m = 0.5, so e−2m = 0.5. Solving for m yields m = 0.3466 seconds.

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SECTION 4.7

(c) = 1∕ = 1∕2 = 0.5 seconds (d) Let T be the lifetime of a fuse, and let t80 be the 80th percentile of T . Then P (T ≤ t80 ) = 0.80. P (T ≤ t80 ) = 1 − e−2t80 = 0.8, so e−2t80 = 0.2. Solving for t80 yields t80 = 0.8047 seconds. (e) P (T > 1) = 1 − P (T ≤ 1) = 1 − (1 − e−2(1) ) = 0.1353 (f) The probability is P (T > 3 ߀ T > 2). By the lack of memory property, P (T > 3 ߀ T > 2) = P (T > 1). P (T > 1) = 1 − P (T ≤ 1) = 1 − (1 − e−2(1) ) = 0.1353

Let X be the diameter in microns.

(a) X = 1∕ = 1∕0.25 = 4 microns (b) X = 1∕ = 1∕0.25 = 4 microns (c) P (X < 3) = 1 − e−0.25(3) = 0.5276 (d) P (X > 11) = 1 − (1 − e−0.25(11) ) = 0.0639 (e) Let m be the median. Then P (X ≤ m) = 0.5.

P (X ≤ m) = 1 − e−0.25m = 0.5, so e−0.25m = 0.5. Solving for m yields m = 2.7726 microns.

(f) Let x75 be the 75th percentile, which is also the third quartile. Then P (X ≤ x75 ) = 0.75. P (X ≤ x75 ) = 1 − e−0.25x75 = 0.75, so e−0.25x75 = 0.25. Solving for x75 yields x75 = 5.5452 microns. (g) Let x99 be the 99th percentile. Then P (X ≤ x99 ) = 0.99. P (X ≤ x99 ) = 1 − e−0.25x99 = 0.99, so e−0.25x99 = 0.01. Solving for x99 yields x99 = 18.4207 microns.

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CHAPTER 4

Let T be the distance between faws, in meters. (a) The mean is 1∕ , so = 1∕12. (b) The median tm solves the equation FT (tm ) = 0.5. Therefore 1 − e−(1∕12)tm = 0.5. Solving for tm yields tm = 8.3178 (c) T = 1∕ = 12 meters. (d) The 65th percentile t65 solves the equation FT (t65 ) = 0.65. Therefore 1 − e−(1∕12)t65 = 0.65. Solving for t65 yields t65 = 12.598 months.

5. (a) Let X be the number of pores with diameters less than 3 microns. The probability that a diameter is less than 3 microns is 0.5276. Therefore X ∼ Bin(10, 0.5276). P (X = 8) + P (X = 9) + P (X = 10) 10! 10! (0.5276)9 (1 − 0.5276)10−9 = (0.5276)8 (1 − 0.5276)10−8 + 8!(10 − 8)! 9!(10 − 9)! 10! = + (0.5276)10 (1 − 0.5276)10−10 10!(10 − 10)!

P (X > 7) =

= 0.077 (b) Let X be the number of pores with diameters greater than 11 microns. The probability that a diameter is greater than 11 microns is 0.0639. Therefore X ∼ Bin(10, 0.0639). P (X = 1) =

10! (0.0639)1 (1 − 0.0639)10−1 = 0.353 1!(10 − 1)!

6. (a) The mean time between requests is 0.5 s, so the mean number of requests is two per second. Let X be the number of requests in two seconds. It follows that X ∼ Poisson(4). P (X = 5) = e−4

45 = 0.1563 5!

(b) Let X be the number of requests in a 1.5 second time interval. Then X ∼ Poisson(3). P (X > 1) = 1 − P (X = 0) − P (X = 1) = 1 − e−3

31 30 − e−3 = 0.80085 0! 1!

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SECTION 4.7

20 = 0.1353 0!

(d) Let T be the time between requests. Then P (T > 1) = 1 − P (T ≤ 1) = 1 − (1 − e−2(1) ) = 0.1353 (e) Let T be the time between requests. Then P (1 < T < 2)

= P (T < 2) − P (T ≤ 1)

P (T ≤ 2) − P (T ≤ 1)

= 1 − (1 − e−2(2) ) − (1 − e−2(1) ) = 0.1170 (f) P (T > 3 ߀ T > 2) = P (T > 1) = 0.1353.

7. (a) The mean distance between faws is 12 m, so the mean number of faws is 1∕12 per meter. Let X be the number of faws in a 50 m length. It follows that X ∼ Poisson(25∕6). P (X = 5) = e−(25∕6)

(25∕6)5 = 0.1623 5!

(b) Let X be the number of faws in a 20 m length. Then X ∼ Poisson(5∕3). P (X > 2)

1 − P (X = 0) − P (X = 1) − P (X = 2) (5∕3)0 (5∕3)1 (5∕3)2 = 1 − e−(5∕3) − e−(5∕3) − e−(5∕3) 0! 1! 2! = 1 − 0.18888 − 0.31479 − 0.26233 = 0.2340 =

(5∕4)0 = 0.2865 0!

(d) Let D be the distance between two faws.

P (D > 15) = 1 − P (D ≤ 15) = 1 − (1 − e−15∕12 ) = 0.2865.

(e) Let D be the distance between two faws. P (8 < D < 20)

P (D < 20) − P (D ≤ 8)

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CHAPTER 4 = P (D ≤ 20) − P (D ≤ 8) = (1 − e−20∕12 ) − (1 − e−8∕12 ) = 0.3245

8. (a) P (X ≥ 5) = 1 − P (X < 5) = 1 − P (X ≤ 5) = 1 − (1 − e−1(5) ) = 0.0067 (b) Yes, it would happen less than 1% of the time. (c) No, because 5 minutes is unusually long if the claim is true.

No. If the lifetimes were exponentially distributed, the proportion of used components lasting longer than 5 years would be the same as the proportion of new components lasting longer than 5 years, because of the lack of memory property.

10. (a) T ∼ Exp(2), so T = 1∕2 = 0.5 seconds. (b) Let m be the median. Then P (T ≤ m) = 0.5. P (T ≤ m) = 1 − e−2m = 0.5, so e−2m = 0.5.

Solving for m yields m = 0.3466 seconds. (c) P (T > 2) = 1 − P (T ≤ 2) = 1 − (1 − e−2(2) ) = 0.0183 (d) P (T < 0.1) = P (T ≤ 0.1) = 1 − e−2(0.1) = 0.1813

(e) P (0.3 < T < 1.5)

= P (T < 1.5) − P (T ≤ 0.3) = P (T ≤ 1.5) − P (T ≤ 0.3) = (1 − e−2(1.5) ) − (1 − e−2(0.3) ) = 0.4990

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SECTION 4.7 (f) The probability is P (T ≤ 4 ߀ T > 3) = 1 − P (T > 4 ߀ T > 3) By the lack of memory property, P (T > 4 ߀ T > 3) = P (T > 1),

so P (T ≤ 4 ߀ T > 3) = 1 − P (T > 1) = P (T ≤ 1) = 1 − e−2(1) = 0.8647.

11.

Let T be the waiting time between accidents. Then T ∼ Exp(3). (a) T = 1∕3 year (b) T = 1∕3 year (c) P (T > 1) = 1 − P (T ≤ 1) = 1 − (1 − e−3(1) ) = 0.0498 (d) P (T < 1∕12) = 1 − e−3(1∕12) = 0.2212 (e) The probability is P (T ≤ 1.5 ߀ T > 0.5) = 1 − P (T > 1.5 ߀ T > 0.5) By the lack of memory property, P (T > 1.5 ߀ T > 0.5) = P (T > 1),

so P (T ≤ 1.5 ߀ T > 0.5) = 1 − P (T > 1) = P (T ≤ 1) = 1 − e−3(1) = 0.9502.

12. (a) = 1∕ X = 1∕3 (b) Let Y be the number of faws in a fve-meter length of aluminum. Then Y ∼ Poisson(5 ), so Y ∼ Poisson(5∕3). P (Y = 2) = e−(5∕3)

13.

(5∕3)2 = 0.2623 2!

Let T be the time that the frst particle is emitted, and let be the rate parameter. Then P (T > 2) = e− (2) = 0.5, so = − ln(0.5)∕2 = 0.34657. (a) P (T > 4) = e− (4) = e−0.34657(4) = 0.25.

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CHAPTER 4

(b) P (T > 1) = e− (1) = e−0.34657(1) = 0.7071. (c) Let X be the number of particles emitted in one second. Then E(X) = = 0.34657.

14.

Let T be the lifetime of a transistor, and let be the rate parameter. Then P (T > 5) = e− (5) = 0.8, so = − ln(0.8)∕5 = 0.044629. (a) P (T > 15) = e− (15) = e−0.044629(15) = 0.512. (b) P (T < 2.5) = 1 − e− (2.5) = 1 − e−0.044629(2.5) = 0.1056. (c) E(T ) = 1∕ = 1∕0.044629 = 22.407.

15.

X1 , ..., X5 are each exponentially distributed with = 1∕200 = 0.005. (a) P (X1 > 100) = 1 − P (X1 ≤ 100) = 1 − (1 − e−0.005(100) ) = e−0.5 = 0.6065

(b) P (X1 > 100 and X2 > 100 and X3 > 100 and X4 > 100 and X5 > 100)

5 Ç

P (Xi > 100)

i=1 −0.5 5

= (e ) = e−2.5

= 0.0821

(c) The time of the frst replacement will be greater than 100 hours if and only if each of the bulbs lasts longer than 100 hours. (d) Using parts (b) and (c), P (T ≤ 100) = 1 − P (T > 100) = 1 − 0.0821 = 0.9179. (e) P (T ≤ t) = 1 − P (T > t) = 1 −

5 Ç

P (Xi > t) = 1 − (e−0.005t )5 = 1 − e−0.025t

i=1

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201

SECTION 4.8

(f) Yes, T ∼ Exp(0.025) (g) T = 1∕0.025 = 40 hours (h) T ∼ Exp(n )

Section 4.8 1.

Let D be the distance. (a) T = (30 + 50)∕2 = 40 m u (b) T =

(50 − 30)2 = 5.7735 m. 12

42 − 35 = 0.35 50 − 30

(d) The probability that the distance advanced is greater than 45 m on any given day is (50 − 45)∕(50 − 30) = 0.25. Let X be the number of days on which the distance advanced is greater than 45 m. Then X ∼ Bin(10, 0.25). 10! (0.25)4 (1 − 0.25)10−4 = 0.1460. P (X = 4) = 4!(10 − 4)!

Let X be the resistance of a resistor. (a) X = (95 + 103)∕2 = 99 mm u (b) X =

(103 − 95)2 = 2.3094 mm 12

102 − 98 = 0.5 103 − 95

(d) The probability that any given resistor has a resistance greater than 100 Ω is 103 − 100 = 0.375. P (X > 100) = 103 − 95 Let Y be the number of resistors out of 6 whose resistance is greater than 100 Ω. Then Y ∼ Bin(6, 0.375).

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CHAPTER 4

P (Y = 3) =

6! (0.375)3 (1 − 0.375)6−3 = 0.2575. 3!(6 − 3)!

3. (a) T = 4∕0.5 = 8

(b) T =

4∕0.52 = 4

1−

4−1 É

e−(0.5)(1)

j=0

[(0.5)(1)]j j!

[(0.5)(1)]0 [(0.5)(1)]1 [(0.5)(1)]2 [(0.5)(1)]3 − e−(0.5)(1) − e−(0.5)(1) − e−(0.5)(1) 0! 1! 2! 3! = 1 − 0.60653 − 0.30327 − 0.075816 − 0.012636 = 0.00175 =

(d) P (T ≥ 4)

= =

1 − e−(0.5)(1)

1 − P (T ≤ 4) H I 4−1 j É −(0.5)(4) [(0.5)(4)] 1− 1− e j! j=0

[(0.5)(4)]1 [(0.5)(4)]2 [(0.5)(4)]3 [(0.5)(4)]0 + e−(0.5)(4) + e−(0.5)(4) + e−(0.5)(4) 0! 1! 2! 3! = 0.13534 + 0.27067 + 0.27067 + 0.18045 = 0.8571

e−(0.5)(4)

T = r∕ and T2 = r∕ 2 . It follows that T = T2 . Since T = 8 and T2 = 16, = 0.5. Now 8 = r∕0.5, so r = 4.

T = r∕ and T2 = r∕ 2 . It follows that = r∕ T . Since T = 8 and r = 16, = 2. 2 = 16∕(22 ) = 4.

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203

SECTION 4.8

Let X1 , ..., X6 be the lifetimes of the frst 6 motors. Let T = X1 + ⋯ + X6 be the time that the sixth motor fails. We need to fnd P (T > 1). T is distributed Γ(6, 3.6). P (T > 1) =

6−1 É

e−3.6(1)

j=0

= 0.5, (a) T =

(b) T2 =

5 [3.6(1)]j É −3.6 3.6j = e = 0.8441. j! j! j=0

= 3, 1∕ = 2.

[(1∕ )!] = 0.6667

1 2

[(2∕ )! − [(1∕ )!]2 ] = 20∕9. T =

20∕9 = 1.4907.

P (T < 1) = 0.8231.

P (T > 5) = 0.0208.

(e) P (2 < T < 4) = [1 − e−(4 ) ] − [1 − e−(2 ) ] = e−(2 ) − e−(4 ) = e− 6 − e− 12 = 0.0550.

8. (a) f (t) =

t −1 e−( t) = e−( t)

t −1 e−( t) , and F (t) = 1 − e−( t) . So ℎ(t) =

(b) ℎ (t) = ( − 1) t −2 . Now > 0, > 0, and t > 0, so ℎ (t) > 0 if Therefore ℎ(t) is increasing in t for > 1 and decreasing in t for < 1.

− 1 < 0.

= 1. When

= 1, ℎ (t) = 0 for

− 1 > 0 and ℎ (t) < 0 if

(c) If T has an exponential distribution, then T has a Weibull distribution with all t, so the hazard function is constant.

t −1 .

Let T be the lifetime in hours of a bearing. (a) P (T > 1000) = 1 − P (T ≤ 1000) = 1 − (1 − e−[(0.0004474)(1000)]

2.25

(b) P (T < 2000) = P (T ≤ 2000) = 1 − e−[(0.0004474)(2000)]

2.25

) = e−[(0.0004474)(1000)]

= 0.8490

= 0.5410

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CHAPTER 4

Then P (T ≤ m) = 0.5, so 1 − e−[(0.0004474)(m)]

2.25

= 0.5, and e−[(0.0004474)(m)]

= 0.5.

(0.0004474m)2.25 = − ln 0.5 = 0.693147 0.0004474m = (0.693147)1∕2.25 = 0.849681 m = 0.849681∕0.0004474 = 1899.2 hours

(d) ℎ(t) =

10.

−1 = 2.25(0.00044742.25 )(20002.25−1 ) = 8.761 × 10−4

Let T be the lifetime of a battery. (a) P (T > 10) = 1 − P (T ≤ 10) = 1 − (1 − e−[(0.1)(10)] ) = e−1 = 0.3679 2

(b) P (T < 5) = P (T ≤ 5) = 1 − e−[(0.1)(5)] = 0.2212 2

(d) ℎ(10) =

11.

10 −1 = 2(0.12 )(102−1 ) = 0.2000

Let T be the lifetime of a fan. 1.5

1.5

(a) P (T > 10, 000) = 1 − (1 − e−[(0.0001)(10,000)] ) = e−[(0.0001)(10,000)] (b) P (T < 5000) = P (T ≤ 5000) = 1 − e−[(0.0001)(5000)]

1.5

= 0.3679

= 0.2978

P (T ≤ 9000) − P (T ≤ 3000) 1.5

1.5

= (1 − e−[(0.0001)(9000)] ) − (1 − e−[(0.0001)(3000)] ) = 0.4227

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SECTION 4.8 12. (a) P (T ≤ 1) = 1 − e−[(0.01)(1)] = 1.000 × 10−6 3

(b) Yes. Only one component in 106 will have a lifetime of 1 day or less if the model is correct. (c) No, because a lifetime of 1 day is unusually short if the model is correct. (d) P (T ≤ 90) = 1 − e−[(0.01)(90)] = 0.5176 3

(e) No. About half the components will have lifetimes of 90 days or less if the model is correct. (f) Yes, since 90 days is neither an unusually short nor an unusually long lifetime.

13. (a) P (X1 > 5) = 1 − P (X1 ≤ 5) = 1 − (1 − e−[(0.2)(5)] ) = e−1 = 0.3679 2

(b) Since X2 has the same distribution as X1 , P (X2 > 5) = P (X1 > 5) = e−1 . Since X1 and X2 are independent, P (X1 > 5 and X2 > 5) = P (X1 > 5)P (X2 > 5) = (e−1 )2 = e−2 = 0.1353. (c) The lifetime of the system will be greater than 5 hours if and only if the lifetimes of both components are greater than 5 hours. (d) P (T ≤ 5) = 1 − P (T > 5) = 1 − e−2 = 0.8647 (e) P (T ≤ t) = 1 − P (T > t) = 1 − P (X1 > t)P (X2 > t) = 1 − (e−[(0.2)(t)] )2 = 1 − e−0.08t = 1 − e−( 0.08t) 2

(f) Yes, T ∼ Weibull(2,

14.

X =

Êa

0.08) = Weibull(2, 0.2828).

1 a+b dx = b−a 2

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15.

CHAPTER 4

X 2 =

Êa

Now X =

1 a2 + ab + b2 dx = b−a 3

2 (b − a)2 2 2 a+b 2 = 2 = a + ab + b − a + b − , therefore X = . 2 X X 2 3 2 12

16. (a) Since P (X ≤ a) = 0, P (X ≤ x) = 0 if x ≤ a. (b) Since P (X ≤ b) = 1, P (X ≤ x) = 1 if x > b. (c) If a < x ≤ b, P (X ≤ x) = P (X ≤ a) + P (a < X ≤ x) = 0 +

1 x−a dx = . Êa b − a b−a Combining the results of (a), (b), and (c) shows that the cumulative distribution function of X is

⎧ 0 ⎪ x−a F (x) = ⎨ b−a ⎪ 1 ⎩

x≤a a<x≤b

x>b

17. (a) The cumulative distribution function of U is ⎧ 0 ⎪ FU (x) = ⎨ x ⎪ 1 ⎩

x≤0 0<x≤1 x>1

⎧ 0 ⎪ x− a x−a x−a = FU (b) FX (x) = P (X ≤ x) = P ((b − a)U + a ≤ x) = P U ≤ =⎨ b − a b−a b−a ⎪ 1 ⎩

x≤a a<x≤b x>b

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207

SECTION 4.9

Section 4.9 1.

(iii) By defnition, an estimator is unbiased if its mean is equal to the true value.

(ii) The variance measures the spread in repeated values of an estimator, that is, how close repeated values are to each other.

3. (a) We denote the mean of 1 by E( 1 ) and the variance of 1 by V ( 1 ). X1 + X2 + = = . E( 1 ) = 2 2 The bias of 1 is E( 1 ) − = − = 0. The variance of 1 is V ( 1 ) =

2 1 2 + 2 = = . 4 2 2

The mean squared error of 1 is the sum of the variance and the square of the bias, so 1 1 MSE( 1 ) = + 02 = . 2 2 (b) We denote the mean of 2 by E( 2 ) and the variance of 2 by V ( 2 ). X1 + 2 X2 + 2 = = . E( 2 ) = 3 3 The bias of 2 is E( 2 ) − = − = 0. The variance of 2 is V ( 2 ) =

2 + 4 2 5 2 5 = = . 9 9 9

The mean squared error of 2 is the sum of the variance and the square of the bias, so 5 5 MSE( 2 ) = + 02 = . 9 9 (c) We denote the mean of 3 by E( 3 ) and the variance of 3 by V ( 3 ). X1 + X2 + = = . E( 3 ) = 4 4 2 The bias of 3 is E( 3 ) − = − = − . 2 2 The variance of 3 is V ( 3 ) =

2 + 2 2 1 = = . 16 8 8

The mean squared error of 3 is the sum of the variance and the square of the bias, so 2 2 2 + 1 2 + − = . MSE( 3 ) = 8 2 8 (d) 3 has smaller mean squared error than 1 whenever

2 2 + 1 1 < . 8 2

Solving for yields −1.2247 < < 1.2247.

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CHAPTER 4

(e) 3 has smaller mean squared error than 2 whenever

2 2 + 1 5 < . 8 9

Solving for yields −1.3123 < < 1.3123.

4. (a) We denote the mean of k2 by E( k2 ). 0 1 (n − 1) s2 (n − 1)s2 (n − 1) 2 E( k2 ) = E = = k k k The bias of k2 is E( k2 ) − 2 =

(n − 1 − k) 2 (n − 1) 2 . − 2 = k k

(b) We denote the variance of k2 by V ( k2 ). 0 1 2(n − 1) 4 (n − 1)s2 (n − 1)2 2 (n − 1)2 2 4 V ( k2 ) = V = s2 = = k k2 k2 n − 1 k2 (c) The mean squared error of k2 is the sum of the variance and the square of the bias, so MSE( k2 ) =

2(n − 1) 4 (n − 1 − k)2 4 + . k2 k2

2(n − 1) + (n − 1)2 4 2(n − 1) 4 2(n − 1) 4 (n − 1 − k)2 4 + = − . k k2 k2 k2 We di˙erentiate MSE( k2 ) with respect to k and set the derivative equal to 0. 0 1 −4(n − 1) − 2(n − 1)2 4 2(n − 1) 4 d d 2(n − 1) + (n − 1)2 4 2(n − 1) 4 2 MSE( k ) = =0 − = + dk dk k k2 k2 k3 It follows that −4 − 2(n − 1) + 2k = 0. Solving for k yields k = n + 1.

k2 ) = (d) MSE(

The probability mass function of X is f (x; p) = p(1 − p)x−1 . The MLE is the value of p that maximizes f (x; p), or equivalently, ln f (x; p). d d 1 x−1 ln f (x; p) = [ln p + (x − 1) ln(1 − p)] = − = 0. dp dp p 1−p 1 1 Solving for p yields p = . The MLE is p = . x X

The joint probability mass function of X1 , ..., Xn is

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SECTION 4.9

f (x1 , ..., xn ; ) =

n Ç

−

i=1

∑n xi −n i=1 = e ∏n . xi ! i=1 xi !

The MLE is the value of that maximizes f (x1 , ..., xn ; ), or equivalently, ln f (x1 , ..., xn ; ). ∑n n n É É xi d d ln xi !) = −n + i=1 = 0. ln f (x1 , ..., xn ; ) = (−n + xi ln − d d i=1 i=1 Now we solve for : ∑n xi −n + i=1 = 0. ∑n xi = i=1 = x. n The MLE is = X.

n! (p)x (1 − p)n−x . x!(n − x)! The MLE of p is the value of p that maximizes f (x; p), or equivalently, ln f (x; p). d d x n−x ln f (x; p) = [ln n! − ln x! − ln(n − x)! + x ln p + (n − x) ln(1 − p)] = − = 0. dp dp p 1−p x X Solving for p yields p = . The MLE of p is p = . n n p p X The MLE of is therefore = . 1−p 1 − p n − X

7. (a) The probability mass function of X is f (x; p) =

(b) The MLE of p is p =

p p 1 1 . The MLE of is therefore = . X 1−p 1 − p X − 1

The joint probability density function of X1 , ..., Xn is f (x1 , ..., xn ; ) =

n Ç

e−(xi − ) ∕2 = (2 )−n∕2 e−

∑n

2 i=1 (xi − ) ∕2

. 2 The MLE is the value of that maximizes f (x1 , ..., xn ; ), or equivalently, ln f (x1 , ..., xn ; ). M L n n É É (xi − )2 d d −(n∕2) ln 2 − = (xi − ) = 0. ln f (x1 , ..., xn ; ) = d d 2 i=1 i=1 i=1

Now we solve for :

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CHAPTER 4 ∑n

i=1 (xi − ) =

∑n

i=1 xi

i=1 xi − n = 0, so =

= x.

The MLE of is = X.

The joint probability density function of X1 , ..., Xn is f (x1 , ..., xn ; ) =

n Ç

e−xi ∕2 = (2 )−n∕2 −n e−

∑n

2 2 i=1 xi ∕2

. 2 The MLE is the value of that maximizes f (x1 , ..., xn ; ), or equivalently, ln f (x1 , ..., xn ; ). L M ∑n n 2 É xi2 n d d i=1 xi ln f (x1 , ..., xn ; ) = −(n∕2) ln 2 − n ln − =− + = 0. 2 d d 3 i=1 2 i=1

Now we solve for : ∑n 2 n i=1 xi − + =0 3 ∑n −n 2 + i=1 x2i = 0 ∑n 2 i=1 xi 2 = n ∑n 2 i=1 xi = n

∑n

= The MLE of is

10.

2 i=1 Xi

The joint probability density function of X1 , ..., Xn is f (x1 , ..., xn ; , ) =

n Ç

e−(xi − ) ∕2 = (2 )−n∕2 −n e−

∑n

2 2 i=1 (xi − ) ∕2

. 2 The MLEs are the values of and that maximizes f (x1 , ..., xn ; , ), or equivalently, ln f (x1 , ..., xn ; , ). L M ∑n n É (xi − ) (xi − )2 ) ) ln f (x1 , ..., xn ; , ) = −(n∕2) ln 2 − n ln − = i=1 = 0. ) ) 2 2 2 i=1 i=1

Now we solve for : ∑n i=1 (xi − ) = 0. 2 ∑n i=1 (xi − ) = 0. ∑n i=1 xi − n = 0

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SECTION 4.10 ∑n =

i=1 xi

= x. n The MLE of is = X. Now we substitute for in ln f (x, ..., xn ; , ) and maximize over : M L ∑n n É (xi − ) 2 (xi − ) 2 ) ) n ln f (x1 , ..., xn ; , ) = −(n∕2) ln 2 − n ln − = − + i=1 = 0. ) ) 2 2 3 i=1

Now we solve for : ∑n (xi − )2 n − + i=1 =0 3 ∑ −n 2 + ni=1 (xi − ) 2=0 ∑n (xi − )2 2 = i=1 n v ∑n )2 i=1 (xi − = n = Now = X, so the MLE of is

∑n

i=1 (Xi − X)

Section 4.10 1. (a) No (b) No (c) Yes

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CHAPTER 4

These data appear to come from an approximately normal distribution.

These data do not appear to come from an approximately normal distribution.

The PM data do not appear to come from an approximately normal distribution.

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SECTION 4.10

0.999 0.99 0.95 0.9 0.75 0.5 0.25 0.1 0.05 0.01 0.001 0

0.5

1.5

2.5

The logs of the PM data do appear to come from an approximately normal distribution.

Yes. If the logs of the PM data come from a normal population, then the PM data come from a lognormal population, and vice versa.

0.999 0.99 0.95 0.9 0.75 0.5 0.25 0.1 0.05 0.01 0.001 3

3.5

4.5

5.5

The plot is a probability plot of the logs of the data. The points follow a line reasonably well, with the exception of the smallest value. The distribution is approximately normal except perhaps for the left tail, which may be somewhat longer than that of a normal distribution.

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214 9. (a)

CHAPTER 4

0.999 0.99 0.95 0.9 0.75 0.5 0.25 0.1 0.05 0.01 0.001 0

100

150

The data do not come from an approximately normal distribution. (b)

0.999 0.99 0.95 0.9 0.75 0.5 0.25 0.1 0.05 0.01 0.001 0

The square roots come from an approximately normal distribution.

10. (a)

0.999 0.99 0.95 0.9 0.75 0.5 0.25 0.1 0.05 0.01 0.001 −2

The data do not come from an approximately normal distribution.

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215

SECTION 4.11

(b)

0.999 0.99 0.95 0.9 0.75 0.5 0.25 0.1 0.05 0.01 0.001 −4

−2

The reciprocals come from an approximately normal distribution.

Section 4.11 1. (a) Let X1 , ..., X144 be the volumes in the 144 bottles. Then X is approximately normally distributed with mean X = 12.01 and X = 0.2∕ 144 = 0.01667. The z-score of 12.00 is (12.00 − 12.01)∕0.01667 = −0.60. The area to the left of z = −0.60 is 0.2743. P (X < 12) = 0.2743. (b) X is approximately normally distributed with mean X = 12.03 and X = 0.2∕ 144 = 0.01667. The z-score of 12.00 is (12.00 − 12.03)∕0.01667 = −1.80. The area to the left of z = −1.80 is 0.0359. P (X < 12) = 0.0359.

2. (a) Let X1 , ..., X250 denote the thicknesses of the 250 sheets of paper. Let S = X1 + ⋯ + X250 be the total thickness of the 250 pages. Then S is approximately normally distributed with mean S = 250(0.08) = 20 and X = 0.01 250 = 0.158114. The z-score of 20.2 is (20.2 − 20)∕0.158114 = 1.26. The area to the right of z = 1.26 is 1 − 0.8962 = 0.1038. P (S > 20.2) = 0.1038.

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(b) Let s10 denote the 10th percentile The z-score of the 10th percentile is approximately z = −1.28. Therefore s10 satisfes the equation −1.28 = (s10 − 20)∕0.158114. s10 = 19.798 mm. (c) No, because we don’t know the distribution of page thicknesses. In particular, we do not know whether the page thicknesses are normally distributed.

The mean number of red lights encountered per day is X = 0(0.1) + 1(0.3) + 2(0.3) + 3(0.2) + 4(0.1) = 1.9. The standard deviation is X = (0 − 1.9)2 (0.1) + (1 − 1.9)2 (0.3) + (2 − 1.9)2 (0.3) + (3 − 1.9)2 (0.2) + (4 − 1.9)2 (0.1) = 1.1358. Let X1 , ..., X100 be the numbers of red lights encountered on 100 days. Then X is approximately normally distributed with mean X = 1.9 and X = 1.1358∕ 100 = 0.11358. The z-score of 2 is z = (2 − 1.9)∕0.11358 = 0.88. The area to the right of z = 0.88 is 1 − 0.8106 = 0.1894. P (X > 2) = 0.1894.

4. (a) Let X1 , ..., X900 be the amounts on the 625 bills. Then X is approximately normally distributed with mean X = 485 and X = 300∕ 900 = 10. The z-score of 500 is (500 − 485)∕10 = 1.50. The area to the right of z = 1.50 is 1 − 0.9332 = 0.0668. P (X > 500) = 0.0668. (b) Let Y be the number of bills whose amount is greater than $1000. Then Y ∼ Bin(900, 0.15), so Y is approximately normal with mean 900(0.15) = 135 and standard deviation 900(0.15)(0.85) = 10.712. To fnd P (Y > 150), use the continuity correction and fnd the z-score of 150.5. The z-score of 150.5 is (150.5 − 135)∕10.712 = 1.45. The area to the right of z = 1.45 is 1 − 0.9265 = 0.0735. P (Y > 150) = 0.0735.

5. (a) Let X1 , ..., X60 be the weights of the 60 bags. Then X is approximately normally distributed with mean X = 15 and X = 5∕ 60 = 0.6455.

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SECTION 4.11

The z-score of 14 is (14 − 15)∕0.6455 = −1.55. The area to the left of z = −1.55 is 0.0606. P (X < 14) = 0.0606. (b) Let x70 denote the 70th percentile The z-score of the 70th percentile is approximately z = 0.52. Therefore x70 satisfes the equation 0.52 = (x70 − 15)∕0.6455. x70 = 15.34 kg. (c) Let n be the necessary sample size. Then X is approximately normally distributed with mean X = 15 and X = 5∕ n. Since P (X < 14) = 0.01, 14 is the 1st percentile of the distribution of X. The z-score of the 1st percentile is approximately z = −2.33. Therefore 14 = 15 − 2.33(5∕ n). Solving for n yields n

136.

6. (a) Let X1 , ..., X200 be the amounts of warpage in the 200 wafers. Then X is approximately normally distributed with mean X = 1.3 and X = 0.1∕ 200 = 0.00707107. The z-score of 1.305 is (1.305 − 1.3)∕0.00707107 = 0.71. The area to the right of z = 0.71 is 1 − 0.7611 = 0.2389. P (X > 1.305) = 0.2389. (b) Let x25 denote the 25th percentile. The z-score of the 25th percentile is approximately z = −0.67. Therefore x25 satisfes the equation −0.67 = (x25 − 1.3)∕0.00707107. x25 = 1.2953 mm. (c) Let n be the necessary sample size. Then X is approximately normally distributed with mean X = 1.3 and X = 0.1∕ n. Since P (X > 1.305) = 0.05, 1.305 is the 95th percentile of the distribution of X. The z-score of the 95th percentile is approximately z = 1.645. Therefore 1.305 = 1.3 + 1.645(0.1∕ n). Solving for n yields n

1083.

7. (a) Let X1 , ..., X50 be the times taken by each of the 50 customers. Let S = X1 + ⋯ + X50 . Then S is approximately normally distributed with mean S = 50(4) = 200 and standard deviation S = 2 50 = 14.142. The z-score of 180 is (180 − 200)∕14.142 = −1.41.

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CHAPTER 4

The area to the left of z = −1.41 is 0.0793. P (S < 180) = 0.0793.

(b) Let X1 , ..., X50 be the times taken by each of the 50 customers. Let S = X1 + ⋯ + X50 . Then S is approximately normally distributed with mean S = 50(4) = 200 and standard deviation S = 2 50 = 14.142. Let x30 denote the 30th percentile of the total time taken by 50 customers. The z-score of the 30th percentile is approximately z = −0.52. Therefore x30 = 200 − 0.52(14.142) = 192.65.

8. (a) Let X1 , ..., X50 be the times to complete 50 oil changes. Then Xi = 29.5 and Xi = 3. Let S = X1 + ⋯ + X50 be the total time to complete 50 oil changes. Then S is approximately normally distributed with S = 50(29.5) = 1475, S = 3 50 = 21.213. The z-score of 1500 is (1500 − 1475)∕21.213 = 1.18. The area to the right of z = 1.18 is 1 − 0.8810 = 0.1190. P (S > 1500) = 0.1190.

(b) Let X1 , ..., X80 be the times to complete 80 oil changes. Then Xi = 29.5 and Xi = 3. Let T = X1 + ⋯ + X80 be the total time to complete 80 oil changes. Then T is approximately normally distributed with T = 80(29.5) = 2360, T = 3 80 = 26.833.

The probability that 80 or more oil changes can be completed in 40 hours (2400 minutes) is P (T ≤ 2400). The z-score of 2400 is (2400 − 2360)∕26.833 = 1.49. The area to the left of z = 1.49 is 0.9319. P (T ≤ 2400) = 0.9319.

(c) Let m be the required mean time. Then T is approximately normally distributed with mean 80m and standard deviation 3 80 = 26.833. 2400 − 80m P (T < 2400) = 0.99, so the z-score for 2400 is z = 2.33. Therefore 2.33 = . Solving for m yields 26.833 m = 29.22.

Let n be the required number of measurements. Let X be the average of the n measurements. Then the true value is X , and the standard deviation is X = ∕ n = 0.5∕ n. Now P ( X − 0.10 < X < X + 0.10) = 0.90. In any normal population, 90% of the population is within 1.645

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219

SECTION 4.11

standard deviations of the mean. Therefore 1.645 X = 0.1. Since X = 0.5∕ n, n = 67.65. The smallest value of n is therefore n = 68.

10.

Let X represent the number of adults who have a college degree. Then X ∼ Bin(100, 0.30), so X is approximately normal with mean X = 100(0.30) = 30 and standard deviation X = 100(0.3)(0.7) = 4.5826. To fnd P (X > 35), use the continuity correction and fnd the z-score of 35.5. The z-score of 35.5 is (35.5 − 30)∕4.5826 = 1.20. The area to the right of z = 1.20 is 1 − 0.8849 = 0.1151. P (X > 35) = 0.1151.

11. (a) Let X represent the number of nondefective bearings in a shipment. Then X ∼ Bin(500, 0.90), so X is approximately normal with mean X = 500(0.90) = 450 and standard deviation X = 500(0.1)(0.9) = 6.7082. To fnd P (X ≥ 440), use the continuity correction and fnd the z-score of 439.5. The z-score of 439.5 is (439.5 − 450)∕6.7802 = −1.57. The area to the right of z = −1.57 is 1 − 0.0582 = 0.9418. P (X ≥ 440) = 0.9418.

(b) Let Y represent the number of shipments out of 300 that are acceptable. From part (a) the probability that a shipment is acceptable is 0.9418, so Y ∼ Bin(300, 0.9418). It follows that Y is approximately normal with mean Y = 300(0.9418) = 282.54 and standard deviation Y = 300(0.9418)(0.0582) = 4.0551. To fnd P (Y > 285), use the continuity correction and fnd the z-score of 285.5. The z-score of 285.5 is (285.5 − 282.54)∕4.0551 = 0.73. The area to the right of z = 0.73 is 1 − 0.7673 = 0.2327. P (X > 285) = 0.2327. (c) Let p be the required proportion of defective bearings, and let X represent the number of defective bearings in a shipment. Then X ∼ Bin(500, p), so X is approximately normal with mean X = 500p and standard deviation X = 500p(1 − p).

The probability that a shipment is acceptable is P (X ≥ 440) = 0.99.

Using the continuity correction, 439.5 is the 1st percentile of the distribution of X. The z-score of the 1st percentile is approximately z = −2.33.

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The z-score can be expressed in terms of p by −2.33 = (439.5 − 500p)∕ 500p(1 − p). This equation can be rewritten as 252, 714.45p2 − 442, 214.45p + 193160.25 = 0. Solving for p yields p = 0.909. (0.841 is a spurious root.)

12. (a) Let X be the number of blocks that meet the specifcation. Then X ∼ Bin(1000, 0.85), so X is approximately normal with mean X = 2000(0.85) = 1700 and standard deviation X = 2000(0.85)(0.15) = 15.969. To fnd P (X < 1690), use the continuity correction and fnd the z-score of 1689.5. The z-score of 1689.5 is (1689.5 − 1700)∕15.969 = −0.66. The area to the left of z = −0.66 is 0.2546. P (X < 1690) = 0.2546. (b) Let x70 denote the 60th percentile The z-score of the 70th percentile is approximately z = 0.52. Therefore x70 satisfes the equation 0.52 = (x70 − 1700)∕15.969. x70 = 1708.3 or 1708. (c) From part (a), the probability that fewer than 1690 blocks in any given lot meet the specifcation is 0.2546. Let Y be the number of lots for which fewer than 1690 blocks meet the specifcation. Then Y ∼ Bin(6, 0.2546). P (Y ≥ 3)

1 − P (Y = 0) − P (Y = 1) − P (Y = 2) 5! 5! (0.2546)1 (1 − 0.2546)5−1 = 1− (0.2546)0 (1 − 0.2546)5−0 − 0!(5 − 0)! 1!(5 − 1)! 5! − (0.2546)2 (1 − 0.2546)5−2 2!(5 − 2)! = 0.1768

13. (a) Let X be the number of particles emitted by mass A and let Y be the number of particles emitted by mass B in a fve-minute time period. Then X ∼ Poisson(100) and Y ∼ Poisson(125). Now X is approximately normal with mean 100 and variance 100, and Y is approximately normal with mean 125 and variance 125. The number of particles emitted by both masses together is S = X + Y . Since X and Y are independent, and both approximately normal, S is approximately normal as well, with mean S = 100 + 125 = 225 and standard deviation S = 100 + 125 = 15. The z-score of 200 is (200 − 225)∕15 = −1.67. The area to the left of z = −1.67 is 0.0475. P (S < 200) = 0.0475. (b) Let X be the number of particles emitted by mass A and let Y be the number of particles emitted by mass B in a two-minute time period. Then X ∼ Poisson(40) and Y ∼ Poisson(50).

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SECTION 4.11

Now X is approximately normal with mean 40 and variance 40, and Y is approximately normal with mean 50 and variance 50. The di˙erence between the numbers of particles emitted by the two masses is D = Y − X. Mass B emits more particles than mass A if D > 0. Since X and Y are independent, and both approximately normal, D is approximately normal as well, with mean D = 50 − 40 = 10 and standard deviation D = 50 + 40 = 9.486833. The z-score of 0 is (0 − 10)∕9.486833 = −1.05. The area to the right of z = −1.05 is 1 − 0.1469 = 0.8531. P (D > 0) = 0.8531.

14. (a) Let X be the number of particles contained in a 2 ml sample. Then X ∼ Poisson(60), so X is approximately normal with mean X = 60 and standard deviation X =

60 = 7.745967.

The z-score of 50 is (50 − 60)∕7.7459674 = −1.29. The area to the right of z = −1.29 is 1 − 0.0985 = 0.9015. P (X > 50) = 0.9015.

(b) Let Y be the number of samples that contain more than 50 particles. From part (a), the probability that a sample contains more than 50 particles is 0.9015. Therefore Y ∼ Bin(10, 0.9015). P (Y ≥ 9)

= =

P (Y = 9) + P (Y = 10) 10! 10! (0.9015)10 (1 − 0.9015)10−10 (0.9015)9 (1 − 0.9015)10−9 + 9!(10 − 9)! 10!(10 − 10)!

= 0.7419 (c) Let W be he number of samples that contain more than 50 particles. From part (a), the probability that a sample contains more than 50 particles is 0.9015. Therefore W ∼ Bin(100, 0.9015), so W is approximately normal with mean W = 100(0.9015) = 90.15 and standard deviation 100(0.9015)(0.0985) = 2.979895. To fnd P (W ≥ 90), use the continuity correction and fnd the z-score of 89.5. The z-score of 89.5 is (89.5 − 90.15)∕2.979895 = −0.22. The area to the right of z = −0.22 is 1 − 0.4129 = 0.5871.

P (W ≥ 90) = 0.5871.

15. (a) Let X be the number of particles withdrawn in a 5 mL volume. Then the mean of X is 50(5) = 250, so X ∼ Poisson(250), and X is approximately normal with mean X = 250 and standard deviation X = 250 = 15.8114.

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The z-score of 235 is (235 − 250)∕15.8114 = −0.95, and the z-score of 265 is (265 − 250)∕15.8114 = 0.95. The area between z = −0.95 and z = 0.95 is 0.8289 − 0.1711 = 0.6578. The probability is 0.6578. (b) Since the withdrawn sample contains 5 mL, the average number of particles per mL will be between 48 and 52 if the total number of particles is between 5(48) = 240 and 5(52) = 260. From part (a), X is approximately normal with mean X = 250 and standard deviation X =

250 = 15.8114.

The z-score of 240 is (240 − 250)∕15.8114 = −0.63, and the z-score of 260 is (260 − 250)∕15.8114 = 0.63. The area between z = −0.63 and z = 0.63 is 0.7357 − 0.2643 = 0.4714. The probability is 0.4714. (c) Let X be the number of particles withdrawn in a 10 mL volume. Then the mean of X is 50(10) = 500, so X ∼ Poisson(500), and X is approximately normal with mean X = 500 and standard deviation X = 500 = 22.3607. The average number of particles per mL will be between 48 and 52 if the total number of particles is between 10(48) = 480 and 10(52) = 520. The z-score of 480 is (480 − 500)∕22.3607 = −0.89, and the z-score of 520 is (520 − 500)∕22.3607 = 0.89. The area between z = −0.89 and z = 0.89 is 0.8133 − 0.1867 = 0.6266. The probability is 0.6266. (d) Let v be the required volume. Let X be the number of particles withdrawn in a volume of v mL. Then X ∼ Poisson(50v), so X is approximately normal with mean X = 50v and standard deviation X = 50v. The average number of particles per mL will be between 48 and 52 if the total number of particles X is between 48v and 52v. Since X = 50v, P (48v < X < 52v) = 0.95 if the z-score of 48v is −1.96 and the z-score of 52v is 1.96. The z-score of 1.96 therefore satisfes the equation 1.96 = (52v−50v)∕ 50v, or equivalently,

v = 1.96 50∕2.

Solving for v yields v = 48.02 ml.

16. (a) If the claim is true, then X is approximately normal with mean X = 40 and X = 5∕ 100 = 0.5. The z-score of 36.7 is (36.7 − 40)∕0.5 = −6.6. The area to the left of z = −6.6 is negligible. P (X ≤ 36.7)

(b) Yes. Almost none of the samples will have mean lifetimes 36.7 hours or less if the claim is true.

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SECTION 4.11

(d) If the claim is true, then X is approximately normal with mean X = 40 and X = 5∕ 100 = 0.5. The z-score of 39.8 is (39.8 − 40)∕0.5 = −0.4. The area to the left of z = −0.4 is 0.3446.

P (X ≤ 39.8) = 0.3446.

(e) No. More than 1/3 of the samples will have mean lifetimes 39.8 hours or less if the claim is true. (f) Yes, because a sample mean lifetime of 39.8 hours or less is not unusual if the claim is true.

17. (a) If the claim is true, then X ∼ Bin(1000, 0.05), so X is approximately normal with mean X = 1000(0.05) = 50 and X = 1000(0.05)(0.95) = 6.89202. To fnd P (X ≥ 75), use the continuity correction and fnd the z-score of 74.5.

The z-score of 74.5 is (74.5 − 50)∕6.89202 = 3.55. The area to the right of z = 3.55 is 1 − 0.9998 = 0.0002. P (X ≥ 75) = 0.0002.

(b) Yes. Only about 2 in 10,000 samples of size 1000 will have 75 or more nonconforming tiles if the goal has been reached. (c) No, because 75 nonconforming tiles in a sample of 1000 is an unusually large number if the goal has been reached. (d) If the claim is true, then X ∼ Bin(1000, 0.05), so X is approximately normal with mean X = 1000(0.05) = 50 and X = 1000(0.05)(0.95) = 6.89202. To fnd P (X ≥ 53), use the continuity correction and fnd the z-score of 52.5. The z-score of 52.5 is (52.5 − 50)∕6.89202 = 0.36. The area to the right of z = 0.36 is 1 − 0.6406 = 0.3594. P (X ≥ 53) = 0.3594.

(e) No. More than 1/3 of the samples of size 1000 will have 53 or more nonconforming tiles if the goal has been reached. (f) Yes, because 53 nonconforming tiles in a sample of 1000 is not an unusually large number if the goal has been reached.

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18.

Let X1 , ..., X100 be the times taken on machine 1 by the 100 parts. Let Y1 , ..., Y100 be the times taken on machine 2 by the 100 parts. Let SX = X1 + ⋯ + X100 be the total time on machine 1, and let SY = Y1 + ⋯ + Y100 be the total time on machine 2. Then SX is approximately normal with mean SX = 100(0.5) = 50 and standard deviation SX = 0.4 100 = 4, and SY is approximately normal with mean SY = 100(0.6) = 60 and standard deviation SY = 0.5 100 = 5. (a) The z-score of 55 is (55 − 50)∕4 = 1.25. The area to the right of z = 1.25 is 1 − 0.8944 = 0.1056. P (SX > 55) = 0.1056. (b) The z-score of 55 is (55 − 60)∕5 = −1.00. The area to the left of z = −1.00 is 0.1587. P (SY < 55) = 0.1587. (c) Let T = SX + SY be the total time used by both machines. Since SX and SY are independent and approximately normal, T istapproximately normal with mean

T = SX + SY = 50 + 60 = 110, and standard deviation T =

S2 + S2 = X

42 + 52 = 6.403124.

The z-score of 115 is (115 − 110)∕6.403124 = 0.78. The area to the right of z = 0.78 is 1 − 0.7823 = 0.2177. P (T > 115) = 0.2177. (d) Let D = SX − SY be the di˙erence between the time on machine 1 and the time on machine 2. Since SX and SY are independent and approximately normal, D ist approximately normal with mean

D = SX − SY = 50 − 60 = −10, and standard deviation T =

S2 + S2 = X

42 + 52 = 6.403124.

The time on machine 1 is greater than the time on machine 2 if D > 0. The z-score of 0 is [0−(−10)]∕6.403124 = 1.56. The area to the right of z = 1.56 is 1 − 0.9406 = 0.0594. P (D > 0) = 0.0594.

19.

Let X be the number of rivets from vendor A that meet the specifcation, and let Y be the number of rivets from vendor B that meet the specifcation. Then X ∼ Bin(500, 0.7), and Y ∼ Bin(500, 0.8). 2 = 500(0.7)(0.3) = It follows that X is approximately normal with mean X = 500(0.7) = 350 and variance X 2 105, and Y is approximately normal with mean Y = 500(0.8) = 400 and variance X = 500(0.8)(0.2) = 80.

Let T = X + Y be the total number of rivets that meet the specifcation. Then t T is approximately normal with mean T = X + Y = 350 + 400 = 750, and standard deviation T =

2 + 2 = X Y

105 + 80 = 13.601471.

To fnd P (T > 775), use the continuity correction and fnd the z-score of 775.5.

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SECTION 4.11

The z-score of 775.5 is (775.5 − 750)∕13.601471 = 1.87. The area to the right of z = 1.87 is 1 − 0.9693 = 0.0307. P (T > 775) = 0.0307.

20. (a) solves the equation 11, 000 = (ln 15.3 − ln )∕0.0001210. Therefore = 4.0425 emissions per minute.

(b) From part (a), = 4.0425 emissions per minute, and X ∼ Poisson(25 ). The mean of X is X = 25 = 101.06, and the standard deviation is X =

25 = 10.053.

(c) = X ∕25 = 4.0425, = X ∕25 = 0.4021 (d) The required value of solves the equation 10,000 = (ln 15.3 − ln )∕0.0001210. Therefore = 4.5624 emissions per minute. (e) The required value of solves the equation 12,000 = (ln 15.3 − ln )∕0.0001210. Therefore = 3.5818 emissions per minute.

(f) The age estimate is correct to within ±1000 years if it is between 10,000 and 12,000, or equivalently, if 3.5818 < < 4.5624. Since = X∕25, P (3.5818 < < 4.5624) = P (3.5818 < X∕25 < 4.5624) = P (89.545 < X < 114.06). The z-score of 89.545 is (89.545 − 101.06)∕10.053 = −1.15. The z-score of 114.06 is (114.06 − 101.06)∕10.053 = 1.29. The area between z = −1.15 and z = 1.29 is 0.9015 − 0.1251 = 0.7764. Note that one may compute P (90 ≤ X ≤ 114) since X must be an integer. In this case the answer obtained is 0.7658.

Section 4.12 1. (a) X ∼ Bin(100, 0.03), Y ∼ Bin(100, 0.05)

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CHAPTER 4

(b) Answers will vary. (c)

.72

(d)

.18

(e) The distribution deviates somewhat from the normal.

2. (a)

.84

(b)

.77

3. (a) A = 6 exactly (simulation results will be approximate), A2 (b)

.25.

.16

4. (a)

(b)

(c)

(d) Yes, the cable is acceptable. The probability that the cable breaks under a load of 28 kN is approximately 0.005.

5. (a)

.25

(b)

.25

(c)

.61

6. (a–d) Answers will vary.

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SECTION 4.12

7. (a–c) Answers will vary. (d)

.025

8. (a,b) Answers will vary. (c) Replacing B results in a longer mean system lifetime (

.25 compared with

(d) Replacing A results in a smaller probability that the system fails within a month ( (e) Replacing A results in a greater 10th percentile (

.76 compared with

.0). .15 compared with

.18).

.68).

9. (a) Answers will vary. (b)

(c)

.34

(d)

(e) System lifetime is not approximately normally distributed. (f) Skewed to the right.

10. (a) Answers will vary. (b)

.75

(c)

.75

(d)

.28

(e)

(f)

.46

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CHAPTER 4

11. (a) Answers will vary. (b)

10, 090

(c)

1250

(d)

.58

(e)

.095

(f) The distribution di˙ers somewhat from normal.

12. (a,b) Answers will vary. (c) median = 0 exactly (simulation results will be approximate), median

.536.

(d) Answers will vary. (e) X = 0 exactly, X = 1∕ 5 exactly (simulation results will be approximate). (f) Bias is exactly 0 in both cases (simulation results will be approximate). median (simulation results will be approximate).

.536, X = 1∕ 5 exactly

(g) Bias estimate should be close to 0 because X = so the true bias is 0. The uncertainty should be close to 1∕ 5 because X = ∕ n = 1∕ 5.

13. (a) = 0.25616 (b–e) Answers will vary.

14. (a–e) Answers will vary.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 4

229

Supplementary Exercises for Chapter 4 1.

Let X be the number of people out of 105 who appear for the fight. Then X ∼ Bin(105, 0.9), so X is approximately normal with mean X = 105(0.9) = 94.5 and standard deviation X = 105(0.9)(0.1) = 3.0741. To fnd P (X ≤ 100), use the continuity correction and fnd the z-score for 100.5.

The z-score of 100.5 is (100.5 − 94.5)∕3.0741 = 1.95. The area to the left of z = 1.95 is 0.9744. P (X ≤ 100) = 0.9744.

2. (a) Let X be the number of cracks in a 500 meter length of pavement. Then the mean of X is 5, so X ∼ Poisson(5). P (X = 8) = e−5

58 = 0.06528. 8!

(b) Let Y be the number of cracks in a 100 meter length of pavement. Then the mean of Y is 1, so Y ∼ Poisson(1). P (Y = 0) = e−1

10 = 0.3679. 0!

(c) Since T is the distance between two consecutive events in a Poisson process, T has an exponential distribution. Since the rate at which events occur is 0.01 per meter, T ∼ Exp(0.01). The probability density function is fT (t) = 0.01e−0.01t for t > 0, fT (t) = 0 for t ≤ 0.

(d) P (T > 50) = 1 − P (T ≤ 50) = 1 − (1 − e−0.01(50) ) = e−0.5 = 0.6065

3. (a) Let X be the number of plants out of 10 that have green seeds. Then X ∼ Bin(10, 0.25). 10! P (X = 3) = (0.25)3 (1 − 0.25)10−3 = 0.2503. 3!(10 − 3)!

(b) P (X > 2) =

1 − P (X ≤ 2)

1 − P (X = 0) − P (X = 1) − P (X = 2) 10! 10! (0.25)1 (1 − 0.25)10−1 = 1− (0.25)0 (1 − 0.25)10−0 − 0!(10 − 0)! 1!(10 − 1)! 10! − (0.25)2 (1 − 0.25)10−2 2!(10 − 2)! = 0.4744 =

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(c) Let Y be the number of plants out of 100 that have green seeds. Then Y ∼ Bin(100, 0.25) so Y is approximately normal with mean Y = 100(0.25) = 25 and standard deviation Y = 100(0.25)(0.75) = 4.3301. To fnd P (Y > 30), use the continuity correction and fnd the z-score for 30.5. The z-score of 30.5 is (30.5 − 25)∕4.3301 = 1.27. The area to the right of z = 1.27 is 1 − 0.8980 = 0.1020. P (Y > 30) = 0.1020. (d) To fnd P (30 ≤ Y ≤ 35), use the continuity correction and fnd the z-scores for 29.5 and 35.5. The z-score of 29.5 is (29.5 − 25)∕4.3301 = 1.04. The z-score of 35.5 is (35.5 − 25)∕4.3301 = 2.42. The area to between z = 1.04 and z = 2.42 is 0.9922 − 0.8508 = 0.1414.

P (30 ≤ Y ≤ 35) = 0.1414.

(e) Fewer than 80 have yellow seeds if more than 20 have green seeds. To fnd P (Y > 20), use the continuity correction and fnd the z-score for 20.5. The z-score of 20.5 is (20.5 − 25)∕4.3301 = −1.04. The area to the right of z = −1.04 is 1 − 0.1492 = 0.8508. P (Y > 20) = 0.8508.

4. (a) True. If ln X1 , ..., ln Xn come from an approximately normal population, then X1 , ..., Xn come from an approximately lognormal population. (b) False. X1 , ..., Xn come from an approximately lognormal population. (c) False. ln X1 , ..., ln Xn come from an approximately normal population. (d) True. ln X1 , ..., ln Xn come from an approximately normal population.

5. Let X denote the number of devices that fail. Then X ∼ Bin(10, 0.01). (a) P (X = 0) = (b) P (X ≥ 2)

10! (0.01)0 (1 − 0.01)10−0 = 0.9910 = 0.9044 0!(10 − 0)! = 1 − P (X ≤ 1) = 1 − P (X = 0) − P (X = 1) 10! 10! (0.01)1 (1 − 0.01)10−1 = 1− (0.01)0 (1 − 0.01)10−0 − 0!(10 − 0)! 1!(10 − 1)!

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SUPPLEMENTARY EXERCISES FOR CHAPTER 4

231

= 0.00427

(c) Let p be the required probability. Then X ∼ Bin(10, p). 10! P (X = 0) = p0 (1 − p)10−0 = (1 − p)10 = 0.95. 0!(10 − 0)! Solving for p yields p = 0.00512.

The concentrations are not approximately normally distributed. There are no concentrations lower than 0, and 0 is only 0.71/1.09 = 0.65 standard deviations below the mean. In a normal distribution, approximately 26% of the values would be farther below the mean than this. In addition, the mean is much larger than the median, which is uncharacteristic of a sample from a normal distribution.

7. (a) The probability that a normal random variable is within one standard deviation of its mean is the area under the normal curve between z = −1 and z = 1. This area is 0.8413 − 0.1587 = 0.6826.

(b) The quantity + z is the 90th percentile of the distribution of X. The 90th percentile of a normal distribution is 1.28 standard deviations above the mean. Therefore z = 1.28.

(c) The z-score of 15 is (15 − 10)∕ 2.6 = 3.10. The area to the right of z = 3.10 is 1 − 0.9990 = 0.0010. P (X > 15) = 0.0010.

8. Let X1 , ..., X5 be the thicknesses of the fve layers. The thickness of the plywood is S = X1 + ⋯ + X5 . (a) S = 5(5) = 25.

(b) S = 0.2 5 = 0.4472. (c) S is normally distributed with mean 25 and standard deviation 0.4472. The z-score of 24 is z = (24 − 25)∕0.4472 = −2.24. The area to the left of z = −2.24 is 0.0125. P (S < 24) = 0.0125.

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9. (a) The z-score of 215 is (215 − 200)∕10 = 1.5. The area to the right of z = 1.5 is 1 − 0.9332 = 0.0668. The probability that the clearance is greater than 215 m is 0.0668.

(b) The z-score of 180 is (180 − 200)∕10 = −2.00. The z-score of 205 is (205 − 200)∕10 = 0.50. The area between z = −2.00 and z = 0.50 is 0.6915 − 0.0228 = 0.6687. The probability that the clearance is between 180 and 205 m is 0.6687.

(c) Let X be the number of valves whose clearances are greater than 215 m. From part (a), the probability that a valve has a clearance greater than 215 X ∼ Bin(6, 0.0668). 6! P (X = 2) = (0.0668)2 (1 − 0.0668)6−2 = 0.0508. 2!(6 − 2)!

m is 0.0668, so

10. (a) No, we do not know the distribution of the population of beam sti˙nesses. In particular, we do not know whether the population is normal.

(b) Yes, the mean of a large sample is normally distributed, so the Central Limit Theorem can be used. Let X be the sample mean sti˙ness. The X is approximately normally distributed with mean X = 30 and standard deviation X = 2∕ 100 = 0.2. The z-score of 30.2 is (30.2 − 30)∕0.2 = 1.00. The area to the right of z = 1.00 is 1 − 0.8413 = 0.1587. P (X > 30.2) = 0.1587.

11. (a) Let X be the number of components in a sample of 250 that are defective. Then X ∼ Bin(250, 0.07), so X is approximately normal with mean X = 250(0.07) = 17.5 and standard deviation X = 250(0.07)(0.93) = 4.0342. To fnd P (X < 20), use the continuity correction and fnd the z-score of 19.5. The z-score of 19.5 is (19.5 − 17.5)∕4.0342 = 0.50. The area to the left of z = 0.50 is 0.6915. P (X < 20) = 0.6915.

(b) Let Y be the number of components in a sample of 10 that are defective. Then X ∼ Bin(10, 0.07). 10! P (X ≥ 1) = 1 − P (X = 0) = 1 − (0.07)0 (1 − 0.07)10−0 = 0.5160. 0!(10 − 0)!

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SUPPLEMENTARY EXERCISES FOR CHAPTER 4

(c) Let p be the required probability, and let X represent the number of components in a sample of 250 that are defective. Then X ∼ Bin(250, p), so X is approximately normal with mean X = 250p and standard deviation X = 250p(1 − p). P (X ≥ 20) = 0.01.

Using the continuity correction, 19.5 is the 1st percentile of the distribution of X. The z-score of the 1st percentile is approximately z = −2.33. The z-score can be expressed in terms of p by −2.33 = (19.5 − 250p)∕ 250p(1 − p). This equation can be rewritten as 63, 857.225p2 − 11, 107.225p + 380.25 = 0. Solving for p yields p = 0.04686. (0.1271 is a spurious root.)

12. (a) The mean concentration of faws is 0.4 per m2 , so the mean number of faws in a 15 m2 area is 15(0.4) = 6.0. Let X be the number of faws in a 15 m2 area. Then X ∼ Poisson(6). P (X > 5)

1 − P (X ≤ 5)

1 − P (X = 0) − P (X = 1) − P (X = 2) − P (X = 3) − P (X = 4) − P (X = 5) 60 61 62 63 64 65 = 1 − e−6 − e−6 − e−6 − e−6 − e−6 − e−6 0! 1! 2! 3! 4! 5! = 1 − 0.0024788 − 0.014873 − 0.044618 − 0.089235 − 0.13385 − 0.16062 = 0.5543

(b) The mean number of faws in a 6 m2 area is 6(0.4) = 2.4. Let Y be the number of faws in a 6 m2 area. Then Y ∼ Poisson(2.4). P (Y = 0) = e−2.4

2.40 = e−2.4 = 0.09072. 0!

(c) The total area of the 50 surfaces is 50(18) = 900 m2 . The mean number of faws in a 900 m2 area is 900(0.4) = 360. Let W be the number of faws in a 900 m2 area. Then W ∼ Poisson(360), so W is approximately normal with mean X = 360 and standard deviation X = 360 = 18.974. The z-score of 350 is (350 − 360)∕18.974 = −0.53. The area to the right of z = −0.53 is 1 − 0.2981 = 0.7019. P (W > 350) = 0.7019.

13. (a) Let be the true concentration. Then = 56∕2 = 28. The uncertainty is = ∕2. Substituting for , =

28∕2 = 3.7.

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CHAPTER 4

= 28.0 ± 3.7. ∕v = 1.

(b) Let v be the required volume, in mL. Then =

Substituting = 28 for and solving for v yields v = 28 mL.

14. (a) X ∼ NB(10, 0.9). 0 1 12 − 1 P (X = 12) = (0.9)10 (1 − 0.9)12−10 = 0.1918 10 − 1 (b) X = 10∕0.9 = 11.11 (c) X =

10(1 − 0.9)∕0.92 = 1.1111

15. (a) Let X1 , X2 , X3 be the three thicknesses. Let S = X1 + X2 + X3 be the thickness of the stack. Then S is normally distributed with mean S = 3(1.5) = 4.5 and standard deviation 0.2 3 = 0.34641. The z-score of 5 is (5 − 4.5)∕0.34641 = 1.44. The area to the right of z = 1.44 is 1 − 0.9251 = 0.0749. P (S > 5) = 0.0749. (b) Let x80 denote the 80th percentile. The z-score of the 80th percentile is approximately z = 0.84. Therefore x80 satisfes the equation 0.84 = (x80 − 4.5)∕0.34641. Solving for x80 yields x80 = 4.7910 cm. (c) Let n be the required number. Then S = 1.5n and S = 0.2 n.

P (S > 5) ≥ 0.99, so 5 is less than or equal to the 1st percentile of the distribution of S.

The z-score of the 1st percentile is z = −2.33.

Therefore n satisfes the inequality −2.33 ≥ (5 − 1.5n)∕(0.2 n). The smallest value of n −2.33 = (5 − 1.5n)∕(0.2 n).

satisfying

this

inequality

the

solution

the

equation

This equation can be rewritten 1.5n − 0.466 n − 5 = 0. Solving for n with the quadratic formula yields therefore n = 4.

n = 1.9877, so n = 3.95. The smallest integer value of n is

16. Let T be the lifetime of a microprocessor.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 4 (a) T ∼ Exp(1∕3000), so P (T ≤ 300) = 1 − e−300∕3000 = 0.0952 (b) P (T > 6000) = 1 − P (T ≤ 6000) = 1 − (1 − e−6000∕3000 ) = 0.1353

(c) Let S represent the time to failure of the new microprocessor, and let T represent the additional time, past the frst 1000 hours, to failure of the old microprocessor. By the lack of memory property, both S and T have the same exponential distribution. Since S and T are independent, P (S < T ) = 1∕2.

17. (a) Let T represent the lifetime of a bearing.

P (T > 1) = 1 − P (T ≤ 1) = 1 − (1 − e−[(0.8)(1)] ) = 0.4889 1.5

(b) P (T ≤ 2) = 1 − e−[(0.8)(2)]

1.5

18.

= 0.8679

Let X1 , ..., X16 be the times needed to complete 16 oil changes. Let T = X1 + ⋯ + X16 be the total time needed to complete all 16 oil changes. Then T is normally distributed with mean T = 16(29.5) = 472 and standard deviation T = 3 16 = 12. The mechanic completes 16 oil changes in an eight-hour day if T < 480. The z-score of 480 is (480 − 472)∕12 = 0.67. The area to the left of z = 0.67 is 0.7486. P (T < 480) = 0.7486.

19. (a) S is approximately normal with mean S = 75(12.2) = 915 and S = 0.1 75 = 0.86603. The z-score of 914.8 is (914.8 − 915)∕0.86603 = −0.23. The area to the left of z = −0.23 is 0.4090. P (S < 914.8) = 0.4090. (b) No. More than 40% of the samples will have a total weight of 914.8 ounces or less if the claim is true. (c) No, because a total weight of 914.8 ounces is not unusually small if the claim is true. (d) S is approximately normal with mean S = 75(12.2) = 915 and S = 0.1 75 = 0.86603. The z-score of 910.3 is (910.3 − 915)∕0.86603 = −5.43. The area to the left of z = −5.43 is negligible. P (S < 910.3)

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CHAPTER 4

(e) Yes. Almost none of the samples will have a total weight of 910.3 ounces or less if the claim is true. (f) Yes, because a total weight of 910.3 ounces is unusually small if the claim is true.

20. (a) The mean number of hits in fve hours is 100, so X ∼ Poisson(100). Therefore X is approximately normal with mean X = 100 and X = 100 = 10. The z-score of 95 is (95 − 100)∕10 = −0.50. The area to the left of z = −0.50 is 0.3085. P (X ≤ 95) = 0.3085.

(b) No. More than 30% of fve-hour periods will have 95 or fewer hits if the claim is true. (c) No, because a total of 95 hits is not unusually small if the claim is true. (d) The mean number of hits in fve hours is 100, so X ∼ Poisson(100). Therefore X is approximately normal with mean X = 100 and X = 100 = 10. The z-score of 65 is (65 − 100)∕10 = −3.50. The area to the left of z = −3.50 is 0.0002. P (X ≤ 65) = 0.0002.

(e) Yes. Only about 2 in 10,000 fve-hour periods will have 65 or fewer hits if the claim is true. (f) Yes, because a total of 65 hits is unusually small if the claim is true.

21. (a) P (X ≤ 0) = F (0) = e−e

−0

= e−1

(b) P (X > ln 2) = 1 − P (X ≤ ln 2) = 1 − F (ln 2) = 1 − e−e

− ln 2

−xm

= 1 − e−1∕2

= 0.5.

Solving for xm yields e−xm = − ln 0.5 = ln 2, so xm = − ln(ln 2) = 0.3665.

22. (a) The cumulative distribution function of X is F (x) =

Ê−∞

f (t) dt.

If x < , F (x) =

Ê−∞

0 dt = 0.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 4 If x ≥ , F (x) = T So F (x) =

(b) X =

2 (c) X

∞

r r dt = r r t−r−1 dt = r r Ê tr+1

t−r −r

= r r

−x−r + −r r

=1−

r x

0 r 1− x

x≥

r r x dx = r r Ê xr+1 ∞

∞

−r

dx = r

1−r rx

1−r

∞

= r r

1−r r = r−1 r−1

r r 2 dx − X Ê xr+1 ∞ r 2 = r r x−r+1 dx − Ê r−1 ∞ 2 2−r r x = r r − 2−r r−1 2−r r 2 − = r r r−2 r−1 r 2 = (r − 1)2 (r − 2) =

(d) If r ≤ 1, then the integral

∞

r r x dx = r r Ê xr+1

∞

−r

dx = r

1−r rx

∞. (e) If r ≤ 2, then the integral

∞.

23. (a) fX (x) = F ¨ (x) =

∞

1−r

∞

diverges, because limx→∞

x1−r = 1−r

∞ ∞ r 2−r x2−r 2 r r −r+1 rx = x dx = r dx = r diverges, because limx→∞ x Ê 2−r 2−r xr+1

e−(x− )∕ [1 + e−(x− )∕ ]2

(b) fX ( − x) = fX ( + x) =

ex∕ [1 + ex∕ ]2

24. (a) X

Ê0

∞

x[p 1 e− 1 x + (1 − p) 2 e− 2 x ] dx

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CHAPTER 4 ∞

= −pxe− 1 x

+p 0

Ê0

∞

e− 1 x dx − (1 − p)xe− 2 x

+ (1 − p) 0

Ê0

∞

e− 2 x dx

1 1 = −p(0) + p − (1 − p)(0) + (1 − p) 1 2 p 1−p = + 1 2 x

(b) The cumulative distribution function of X is F (x) =

Ê−∞

f (t) dt.

If x < 0, then F (x) = If x ≥ 0, then F (x) =

Ê0

Ê−∞

0 dt = 0.

p 1 e− 1 t + (1 − p) 2 e− 2 t dt = p(1 − e− 1 x ) + (1 − p)(1 − e− 2 x ) = 1 − pe− 1 x − (1 − p)e− 2 x

(c) P (X ≤ 2) = F (2) = 1 − 0.5e−2(2) − (1 − 0.5)e−1(2) = 1 − 0.5e−4 − 0.5e−2 = 0.9232 (d) Let M1 be the event that mass 1 was selected. P (X ≤ 2 ∩ M1) . P (M1߀X ≤ 2) = P (X ≤ 2)

Now P (X ≤ 2 ∩ M1) = P (X ≤ 2߀M1)P (M1) = (1 − e−2(2) )(0.5). From part (c), P (X ≤ 2) = 1 − 0.5e−4 − 0.5e−2 . (1 − e−2(2) )(0.5) = 0.5317. Therefore P (M1߀X ≤ 2) = 1 − 0.5e−4 − 0.5e−2

25. (a) P (X > s) = P (First s trials are failures) = (1 − p)s

(b) P (X > s + t ߀ X > s)

= P (X > s + t and X > s)∕P (X > s) = P (X > s + t)∕P (X > s) = (1 − p)s+t ∕(1 − p)s = =

(1 − p)t P (X > t)

Note that if X > s + t, it must be the case that X > s, which is the reason that P (X > s + t and X > s) = P (X > s + t). (c) Let X be the number of tosses of the penny needed to obtain the frst head. Then P (X > 5 ߀ X > 3) = P (X > 2) = 1∕4. The probability that the nickel comes up tails twice is also 1/4.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 4

Let U be the length of the part of the stick used as the length of the rectangle. Then the width is 1 − U , so the area is U (1 − U ).

26.

Since U ∼ U (0, 1), the probability density function of U is f (u) = 1 for 0 < u < 1 and f (u) = 0 otherwise. The mean area is therefore

Ê0

u(1 − u)(1) du =

1 . 6

27. (a) FY (y) = P (Y ≤ y) = P (7X ≤ y) = P (X ≤ y∕7). Since X ∼ Exp( ), P (X ≤ y∕7) = 1 − e− y∕7 .

(b) fY (y) = FY¨ (y) = ( ∕7)e− y∕7

P (X = x) 28. (a) P (X = x − 1)

n! px (1 − p)n−x x!(n − x)! = n! px−1 (1 − p)n−x+1 (x − 1)!(n − x + 1)! 4 50 x 14 5 p (1 − p)n−x n!(x − 1)!(n − x + 1)! = n!x!(n − x)! px−1 (1 − p)n−x+1 0 p 1 n−x+1 = x 1−p

(b) Let x be an integer such that P (X = x) ≥ P (X = x − 1). Then

0 p 1 n−x+1 x 1−p p(n − x + 1) np − xp + p np + p

≥

≥ x(1 − p) ≥ x − xp ≥ x

Reversing the above steps shows that if x ≤ np + p then P (X = x) ≥ P (X = x − 1).

29. (a)

e− x ∕x! P (X = x) e− x (x − 1)! = = = P (X = x − 1) e− x−1 ∕(x − 1)! x e− x−1 x!

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CHAPTER 4

(b) P (X = x) ≥ P (X = x − 1) if and only if

≥ 1 if and only if x ≤ . x

30. (a) FX (x) = P (X ≤ x) = P ( Z + ≤ x) = P (Z ≤ x− ) = Φ( x− ) ) = (1∕ ) ( x− )= (b) FX¨ (x) = (1∕ )Φ¨ ( x−

− (x− ) 2 2

) = 1 − Φ( x− ), (c) FX (x) = P (X ≤ x) = P (− Z + ≤ x) = P (Z ≥ x− − − so FX¨ (x) = (1∕ )Φ¨ ( x− ) = (1∕ ) ( x− )= − −

− (x− ) 2 2

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241

SECTION 5.1

Chapter 5 Section 5.1 1. (a) 1.96 (b) 2.33 (c) 2.57 or 2.58 (d) 1.28

2. (a) 95% (b) 97% (c) 80% (d) 99.9%

The level is the proportion of samples for which the confdence interval will cover the true value. Therefore as the level goes up, the reliability goes up. This increase in reliability is obtained by increasing the width of the confdence interval. Therefore as the level goes up the precision goes down.

The wider the confdence interval, the greater the confdence level. Therefore the 95% confdence interval is (6.8, 11.9), the 98% confdence interval is (6.4, 12.3), and the 99.9% confdence interval is (5.1, 13.6).

5. (a) The sample mean is midway between the endpoints of the confdence interval. So the sample mean is (0.88 + 1.02)∕2 = 0.95.

(b) The width of the confdence interval is ±1.96s∕ 114, where s is the sample standard deviation. Therefore 1.02 − 0.88 = 3.92s∕ 114. Solving for s yields s = 0.3813.

6. (a) X = 348.2, s = 5.1, n = 67, z.05 = 1.645. The confdence interval is 348.2 ± 1.645(5.1∕ 67), or (347.175, 349.225).

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CHAPTER 5

(b) X = 348.2, s = 5.1, n = 67, z.025 = 1.96. The confdence interval is 348.2 ± 1.96(5.1∕ 67), or (346.979, 349.421).

(c) X = 348.2, s = 5.1, n = 67, so the upper confdence bound 348.9 satisfes 348.9 = 348.2 + z ∕2 (5.1∕ 67). Solving for z ∕2 yields z ∕2 = 1.12. The area to the right of z = 1.12 is 1 − 0.8686 = 0.1314, so ∕2 = 0.1314. The level is 1 −

= 1 − 2(0.1314) = 0.7372, or 73.72%.

7. (a) X = 1217, s = 52, n = 80, z.025 = 1.96. The confdence interval is 1217 ± 1.96(52∕ 80), or (1205.6, 1228.4).

(b) X = 1217, s = 52, n = 80, z.01 = 2.33. The confdence interval is 1217 ± 2.33(52∕ 80), or (1203.5, 1230.5).

(c) X = 1217, s = 52, n = 80, so the upper confdence bound 1226 satisfes 1226 = 1217 + z ∕2 (52∕ 80). Solving for z ∕2 yields z ∕2 = 1.55. The area to the right of z = 1.55 is 1 − 0.9394 = 0.0606, so ∕2 = 0.0606. The level is 1 −

= 1 − 2(0.0606) = 0.8788, or 87.88%.

8. (a) X = 85, s = 2, n = 60, z.025 = 1.96. The confdence interval is 85 ± 1.96(2∕ 60), or (84.494, 85.506).

(b) X = 85, s = 2, n = 60, z.0025 = 2.81. The confdence interval is 85 ± 2.81(2∕ 60), or (84.274, 85.726).

(c) X = 85, s = 2, n = 60, so the upper confdence bound 85.37 satisfes 85.37 = 85 + z ∕2 (2∕ 60). Solving for z ∕2 yields z ∕2 = 1.43. The area to the right of z = 1.43 is 1 − 0.9236 = 0.0764, so ∕2 = 0.0764. The level is 1 −

= 1 − 2(0.0764) = 0.8472, or 84.72%.

9. (a) X = 11.9, s = 1.1, n = 140, z.025 = 1.96. The confdence interval is 11.9 ± 1.96(1.1∕ 140), or (11.718, 12.082).

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SECTION 5.1

(b) X = 11.9, s = 1.1, n = 140, z.005 = 2.58. The confdence interval is 11.9 ± 2.58(1.1∕ 140), or (11.660, 12.140).

(c) X = 11.9, s = 1.1, n = 140, so the upper confdence bound 11.99 satisfes 11.99 = 11.9 + z ∕2 (1.1∕ 140). Solving for z ∕2 yields z ∕2 = 0.97. The area to the right of z = 0.97 is 1 − 0.8340 = 0.1660, so ∕2 = 0.1660. The level is 1 −

= 1 − 2(0.1660) = 0.6680, or 66.80%.

10. (a) X = 1217, s = 52, n = 80, z.10 = 1.28. The upper confdence bound is 1217 + 1.28(52∕ 80) = 1224.4.

(b) The upper confdence bound 1220 satisfes 1220 = 1217 + z (52∕ 80). Solving for z yields z = 0.52. The area to the left of z = 0.52 is 0.6985. The level is 0.6985, or 69.85%.

11. (a) X = 85, s = 2, n = 60, z.02 = 2.05. The lower confdence bound is 85 − 2.05(2∕ 60) = 84.471.

(b) The lower confdence bound 11.7 satisfes 11.7 = 85 − z (2∕ 60). Solving for z yields z = 1.55. The area to the left of z = 1.55 is 1 −

= 0.9394.

The level is 0.9394, or 93.94%.

12. (a) X = 11.9, s = 1.1, n = 140, z.05 = 1.645. The upper confdence bound is 11.9 + 1.645(1.1∕ 140) = 12.053.

(b) The lower confdence bound 11.7 satisfes 11.7 = 11.9 − z (1.1∕ 140). Solving for z yields z = 2.15. The area to the left of z = 2.15 is 1 −

= 0.9842.

The level is 0.9842, or 98.42%.

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CHAPTER 5

13.

The sample mean X is the midpoint of the interval, so X = 0.227. The upper confdence bound 0.241 satisfes 0.241 = 0.227 + 1.96(s∕ n). Therefore s∕ n = 0.00714286. A 90% confdence interval is 0.227±1.645(s∕ n) = 0.227±1.645(0.00714286), or (0.21525, 0.23875).

14. (a) True. This results from the expression X ± 1.96(s∕ n), which is a 95% confdence interval. (b) False. This is a specifc interval, so it does not make sense to talk about the probability that it covers the true value. (c) False. A confdence interval says something about where the true value is likely to be. It does not say anything about the proportion of the sample that is in the interval. (d) False. A confdence interval says something about where the true value is likely to be. It does not say where the next measurement is likely to be.

15. (a) False. The confdence interval is for the population mean, not the sample mean. The sample mean is known, so there is no need to construct a confdence interval for it. (b) True. This results from the expression X ± 1.96(s∕ n), which is a 95% confdence interval for the population mean. (c) False. A confdence interval says something about where the true value is likely to be. It does not say anything about the proportion of the sample that is in the interval. (d) False. The standard deviation of the mean involves the square root of the sample size, not of the population size.

16.

Let X be the number of 90% confdence intervals that fail to cover the true mean. Then X ∼ Bin(200, 0.10), so X is approximately normal with mean X = 200(0.10) = 20 and standard deviation 200(0.10)(0.90) = 4.242641. To fnd P (X > 15), use the continuity correction and fnd the z-score of 15.5. The z-score of 15.5 is (15.5 − 20)∕4.242641 = −1.06. The area to the right of z = −1.06 is 1 − 0.1446 = 0.8554. P (X > 15) = 0.8554.

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SECTION 5.2

17.

The supervisor is underestimating the confdence. The statement that the mean cost is less than $160 is a onesided upper confdence bound with confdence level 97.5%.

18.

It is not a valid confdence interval because the data upon which it is based are not a simple random sample.

Section 5.2 1. (a) 1.796 (b) 2.447 (c) 63.657 (d) 2.048

2. (a) 1.363 (b) 1.943 (c) 31.821 (d) 1.701

3. (a) 95% (b) 98% (c) 99% (d) 80% (e) 90%

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CHAPTER 5

False. The population must be approximately normal.

X = 2.938, s = 0.47199, n = 5, t5−1,.025 = 2.776. The confdence interval is 2.938 ± 2.776(0.47199∕ 5), or (2.352, 3.524).

No. The maximum value is 39.920, yet the third quartile is only 8.103, and the interquartile range is 6.136. The distance between the third quartile and the maximum is thus more than 5 times the interquartile range. The maximum value is an outlier, so it is not appropriate to assume that this is a sample from a normal population.

Yes it is appropriate, since there are no outliers. X = 205.1267, s = 1.7174, n = 9, t9−1,.025 = 2.306. The confdence interval is 205.1267 ± 2.306(1.7174∕ 9), or (203.81, 206.45).

8. (a) X = 196.64, s = 0.6, n = 6, t6−1,.025 = 2.571. The confdence interval is 196.64 ± 2.571(0.6∕ 6), or (195.93, 197.35). (b) X = 196.64, s = 0.6, n = 6, t6−1,.01 = 3.365. The confdence interval is 196.64 ± 3.365(0.6∕ 6), or (195.71, 197.57). (c) No, the value 198.02 is an outlier.

9. (a)

1.3

1.305

1.31

1.315

1.32

1.325

(b) Yes, there are no outliers. X = 1.3115, s = 0.00628490, n = 6, t6−1,.005 = 4.032. The confdence interval is 1.3115 ± 4.032(0.00628490∕ 6), or (1.3012, 1.3218).

1.31

1.32

1.33

1.34

1.35

1.36

1.37

1.38

1.39

(d) No, the data set contains an outlier.

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SECTION 5.2

10.

X = 13, s = 2, n = 15, t15−1,.005 = 2.977. The confdence interval is 13 ± 2.977(2∕ 15), or (11.463, 14.537).

11.

X = 2.03, s = 0.090, n = 6, t6−1,.05 = 2.015. The confdence interval is 2.03 ± 2.015(0.090∕ 6), or (1.956, 2.104).

12.

X = 2.64, s = 1.02, n = 15, t15−1,.025 = 2.145. The confdence interval is 2.64 ± 2.145(1.02∕ 15), or (2.075, 3.205).

13.

X = 0.242, s = 0.031, n = 10, t10−1,.025 = 2.262. The confdence interval is 0.242 ± 2.262(0.031∕ 10), or (0.2198, 0.2642).

14. (a) X = 654.1, s = 311.7, n = 50, t50−1,.025 = 2.009575. The confdence interval is 654.1 ± 2.009575(311.7∕ 50), or (565.52, 742.68).

(b) X = 654.1, s = 311.7, n = 50, t50−1,.01 = 2.404892. The confdence interval is 654.1 ± 2.404892(311.7∕ 50), or (548.09, 760.11). (c) t50−1,.025 = 2.009575. n = (2.0095752 )(311.72 )∕(502 ) = 156.94, round up to 157. (d) t50−1,.01 = 2.404892. n = (2.4048922 )(311.72 )∕(502 ) = 224.76, round up to 225.

15. (a) X = 8.1, s = 0.5, n = 100, t100−1,.025 = 1.984217. The confdence interval is 8.1 ± 1.984217(8.1∕ 100), or (8.001, 8.199).

(b) X = 8.1, s = 0.5, n = 100, t100−1,.005 = 2.626405. The confdence interval is 8.1 ± 2.626405(8.1∕ 100), or (7.969, 8.231).

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(c) t100−1,.025 = 1.984217. n = (1.984272 )(0.52 )∕(0.082 ) = 153.794, round up to 154. (d) t100−1,.005 = 2.626405. n = (2.6264052 )(0.52 )∕(0.082 ) = 269.453, round up to 270.

16. (a) X = 136.9, s = 22.6, n = 123, t123−1,.025 = 1.9796. The confdence interval is 136.9 ± 1.9796(22.6∕ 123), or (132.866, 140.934).

(b) X = 136.9, s = 22.6, n = 123, t123−1,.0025 = 2.858984. The confdence interval is 136.9 ± 2.858984(22.6∕ 123), or (131.074, 142.726). (c) t123−1,.025 = 1.9796. n = (1.97962 )(22.62 )∕(32 ) = 222.397, round up to 223. (d) t123−1,.0025 = 2.858984. n = (2.8589842 )(22.62 )∕(32 ) = 463.872, round up to 464.

17. (a) X = 38.3, s = 0.6, n = 75, t75−1,.025 = 1.992543. The confdence interval is 38.3 ± 1.992543(0.6∕ 75), or (38.162, 38.438).

(b) X = 38.3, s = 0.6, n = 75, t75−1,.005 = 2.643913. The confdence interval is 38.3 ± 2.643913(0.6∕ 75), or (38.117, 38.483). (c) t75−1,.025 = 1.992543. n = (1.9925432 )(0.62 )∕(0.12 ) = 142.928, round up to 143. (d) t75−1,.005 = 2.643913. n = (2.6439132 )(0.62 )∕(0.12 ) = 251.650, round up to 252.

18. (a) X = 72, s = 10, n = 150, t150−1,.025 = 1.976013. The confdence interval is 72 ± 1.976013(10∕ 150), or (70.387, 73.613). (b) X = 72, s = 10, n = 150, t150−1,.025 = 1.655145. The confdence interval is 72 ± 1.655145(10∕ 150), or (70.649, 73.351). (c) t150−1,.025 = 1.976013. n = (1.9760132 )(102 )∕(12 ) = 390.463, round up to 391.

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(d) t150−1,.05 = 1.655145. n = (1.6551452 )(102 )∕(12 ) = 273.950, round up to 274.

19. (a) X = 1.56, s = 0.1, n = 80, t80−1,.025 = 1.99045. The confdence interval is 1.56 ± 1.99045(0.1∕ 80), or (1.538, 1.582). (b) X = 1.56, s = 0.1, n = 80, t80−1,.01 = 2.374482. The confdence interval is 1.56 ± 2.374482(0.1∕ 80), or (1.5335, 1.5865). (c) t80−1,.025 = 1.99045. n = (1.990452 )(0.12 )∕(0.012 ) = 396.189, round up to 397. (d) t80−1,.01 = 2.374482. n = (2.3744822 )(0.12 )∕(0.012 ) = 563.816, round up to 564.

20. (a) X = 654.1, s = 311.7, n = 50, t50−1,.05 = 1.676551. The upper confdence bound is 654.1 + 1.676551(311.7∕ 50), or 728.004. (b) X = 654.1, s = 311.7, n = 50, t50−1,.02 = 2.109873. The lower confdence bound is 654.1 − 2.109873(311.7∕ 50), or 561.095.

21. (a) X = 8.1, s = 0.5, n = 100, t100−1,.05 = 1.660391. The lower confdence bound is 8.1 − 1.660391(0.5∕ 100), or 8.017. (b) X = 8.1, s = 0.5, n = 100, t100−1,.01 = 2.364606. The upper confdence bound is 8.1 + 2.364606(0.5∕ 100), or 8.218.

22. (a) X = 136.9, s = 22.6, n = 123, t123−1,.05 = 1.6574395. The lower confdence bound is 136.9 − 1.6574395(22.6∕ 123), or 133.523. (b) X = 136.9, s = 22.6, n = 123, t123−1,.10 = 1.288529. The upper confdence bound is 136.9 + 1.288529(22.6∕ 123), or 139.526.

23. (a) X = 38.3, s = 0.6, n = 75, t75−1,.05 = 1.665707. The upper confdence bound is 38.3 + 1.665707(0.6∕ 75), or 38.415.

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(b) X = 38.3, s = 0.6, n = 75, t75−1,.005 = 2.643913. The lower confdence bound is 38.3 − 2.643913(0.6∕ 75), or 38.117.

24. (a) X = 72, s = 10, n = 150, t150−1,.05 = 1.665145. The upper confdence bound is 72 + 1.665145(10∕ 150), or 73.351. (b) X = 72, s = 10, n = 150, t150−1,.02 = 2.071884. The lower confdence bound is 72 − 2.071884(10∕ 150), or 70.308.

25. (a) X = 1.56, s = 0.1, n = 80, t80−1,.05 = 1.664371. The lower confdence bound is 1.56 − 1.664371(0.1∕ 80), or 1.541. (b) X = 1.56, s = 0.1, n = 80, t80−1,.10 = 1.29236. The lower confdence bound is 1.56 − 1.29236(0.1∕ 80), or 1.574.

26.

(iii) one-half as wide. The ratio of the widths is approximately equal to the ratio of the standard deviations of ∕ 400 the sample mean, which is = 100∕400 = 1∕2. ∕ 100

27.

With a sample size of 70, the standard deviation of X is ∕ 70. The interval will be approximately half as wide if the standard deviation of X is ∕(2 70) = ∕ 280. The sample size needs to be approximately 280.

28. (a) N − 1 = 10 − 1 = 9. (b) X = 6.59635, s = 0.11213, n = 10, t10−1,.005 = 3.250. The confdence interval is 6.59635 ± 3.250(0.11213∕ 10), or (6.481, 6.712). Alternatively, one may compute 6.59635 ± 3.250(0.03546).

29. (a) SE Mean is StDev/ N, so 0.52640 = StDev/ 20, so StDev = 2.3541.

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(b) X = 2.39374, s = 2.3541, n = 20, t20−1,.005 = 2.861. The lower limit of the 99% confdence interval is 2.39374 − 2.861(2.3541∕ 20) = 0.888. Alternatively, one may compute 2.39374 − 2.861(0.52640). (c) X = 2.39374, s = 2.3541, n = 20, t20−1,.005 = 2.861. The upper limit of the 99% confdence interval is 2.39374 + 2.861(2.3541∕ 20) = 3.900. Alternatively, one may compute 2.39374 + 2.861(0.52640).

The value 85 comes from a normal population with standard deviation = 8. Since is known, a 95% confdence interval is based on z.025 = 1.96. The confdence interval is 85 ± 1.96(8), or (69.32, 100.68).

30.

31. (a) X = 0.32, s = 0.05, n = 8, t8−1,.025 = 2.365. The confdence interval is 0.32 ± 2.365(0.05∕ 8), or (0.2782, 0.3618). (b) No. The minimum possible value is 0, which is less than two sample standard deviations below the sample mean. Therefore it is impossible to observe a value that is two or more sample standard deviations below the sample mean. This suggests that the sample may not come from a normal population.

32. (a) X = 18.322, s = 3.8237, n = 306, t306−1,.025 = 1.9678. The confdence interval is (17.892, 18.752).

(b)

X = 27.611, s = 6.62, n = 15, t15−1,.025 = 2.1448. The confdence interval is (23.945, 31.277)

(d) X = 26.713, s = 11.219, n = 100, t100−1,.05 = 1.6604. The confdence interval is (24.85, 28.576).

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(e) X = 29.012, s = 10.575, n = 1267, t1267−1,.05 = 2.5797. The confdence interval is (28.245, 29.778). (f) X = 27.561, s = 12.646, n = 166, t166−1,.05 = 2.606. The confdence interval is (25.003, 30.119).

33.

(iv) There are more pits at 40% humidity than at 75% and there is less spread in their depths.

34. (a) X = 49.925, s = 20.535, n = 656, t656−1,.05 = 1.9636. The confdence interval is (48.35, 51.499). (b) X = 60.917, s = 11.473, n = 328, t328−1,.05 = 2.3378. The confdence interval is (59.436, 62.398). (c) X = 75.485, s = 11.56, n = 278, t278−1,.05 = 2.5937. The confdence interval is (77.687, 77.283). (d) X = 30.385, s = 11.367, n = 858, t858−1,.05 = 1.6466. The confdence interval is (29.746, 31.024).

35. (a) X = 1735.1, s = 456.95, n = 656, t656−1,.05 = 1.9636. The confdence interval is (1700.1, 1770.2). (b) X = 1223.5, s = 269.09, n = 328, t328−1,.05 = 2.3378. The confdence interval is (1188.8, 1258.2). (c) X = 984.32, s = 215.68, n = 278, t278−1,.05 = 2.5937. The confdence interval is (950.77, 1017.9). (d) X = 2469.3, s = 294.48, n = 858, t858−1,.05 = 1.6466. The confdence interval is (2452.7, 2485.8).

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36. (a) Two

(b) The sample size is large (200).

Section 5.3 1. (a) 28∕70 = 0.4, or 40%. (b) X = 28, n = 70, p̃ = (28 + 2)∕(70 + 4) = 0.405405, z.025 = 1.96. The confdence interval is 0.405405 ± 1.96 0.405405(1 − 0.405405)∕(70 + 4), or (0.294, 0.517).

(c) X = 28, n = 70, p̃ = (28 + 2)∕(70 + 4) = 0.405405, z.01 = 2.33. The confdence interval is 0.405405 ± 2.33 0.405405(1 − 0.405405)∕(70 + 4), or (0.272, 0.538).

(d) Let n be the required sample size. Then n satisfes the equation 0.1 = 1.96 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.405405 and solving for n yields n = 89.

(e) Let n be the required sample size. Then n satisfes the equation 0.10 = 2.33 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.405405 and solving for n yields n = 127.

(f) The upper confdence bound 0.5 satisfes the equation 0.5 = 0.405405 + z Solving for z yields z = 1.66. The area to the left of z = 1.66 is 1 −

0.405405(1 − 0.405405)∕(70 + 4) = 0.9515.

The level is 0.9515, or 95.15%.

2. (a) X = 73, n = 100, p̃ = (73 + 2)∕(100 + 4) = 0.72115, z.025 = 1.96. The confdence interval is 0.72115 ± 1.96 0.72115(1 − 0.72115)∕(100 + 4), or (0.635, 0.807).

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(b) X = 73, n = 100, p̃ = (73 + 2)∕(100 + 4) = 0.72115, z.005 = 2.58. The confdence interval is 0.72115 ± 2.58 0.72115(1 − 0.72115)∕(100 + 4), or (0.608, 0.835).

(c) Let n be the required sample size. Then n satisfes the equation 0.05 = 1.96 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.72115 and solving for n yields n = 305.

(d) Let n be the required sample size. Then n satisfes the equation 0.05 = 2.58 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.72115 and solving for n yields n = 532.

(e) The lower confdence bound 0.7 satisfes the equation 0.7 = 0.72115 − z Solving for z yields z = 0.48. The area to the left of z = 0.48 is 1 −

0.72115(1 − 0.72115)∕(100 + 4) = 0.6844.

The level is 0.6844, or 68.44%.

(f) Let X be the number of 95% confdence intervals that cover the true proportion. Then X ∼ Bin(200, 0.95), so X is approximately normal with mean X = 200(0.95) = 190 and standard deviation 200(0.95)(0.05) = 3.082207. To fnd P (X > 192), use the continuity correction and fnd the z-score of 192.5. The z-score of 192.5 is (192.5 − 190)∕3.082207 = 0.81. The area to the right of z = 0.81 is 1 − 0.7910 = 0.2090. P (X > 192) = 0.2090.

3. (a) X = 52, n = 70, p̃ = (52 + 2)∕(70 + 4) = 0.72973, z.025 = 1.96. The confdence interval is 0.72973 ± 1.96 0.72973(1 − 0.72973)∕(70 + 4), or (0.629, 0.831).

(b) X = 52, n = 70, p̃ = (52 + 2)∕(70 + 4) = 0.72973, z.05 = 1.645. The confdence interval is 0.72973 ± 1.645 0.72973(1 − 0.72973)∕(70 + 4), or (0.645, 0.815).

(c) Let n be the required sample size. Then n satisfes the equation 0.05 = 1.96 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.72973 and solving for n yields n = 300.

(d) Let n be the required sample size.

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Then n satisfes the equation 0.05 = 1.645 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.72973 and solving for n yields n = 210.

(e) Let X be the number of 90% confdence intervals that cover the true proportion. Then X ∼ Bin(300, 0.90), so X is approximately normal with mean X = 300(0.90) = 270 and standard deviation 300(0.90)(0.10) = 5.196152. To fnd P (X > 280), use the continuity correction and fnd the z-score of 280.5. The z-score of 280.5 is (280.5 − 270)∕5.196152 = 2.02. The area to the right of z = 2.02 is 1 − 0.9783 = 0.0217. P (X > 280) = 0.0217.

4. (a) X = 2636, n = 5409, p̃ = (2636 + 2)∕(5409 + 4) = 0.48735, z.025 = 1.96. The confdence interval is 0.48735 ± 1.96

.48735(1 − 0.48735)∕(5409 + 4), or (0.474, 0.501).

(b) X = 2636, n = 5409, p̃ = (2636 + 2)∕(5409 + 4) = 0.48735, z.005 = 2.58. The confdence interval is 0.48735 ± 2.58 0.48735(1 − 0.48735)∕(5409 + 4), or (0.470, 0.505).

0.48735(1 − 0.48735)∕(5409 + 4)

Solving for z yields z = 1.86. The area to the left of z = 1.86 is 0.9686. The level is 0.9686, or 96.86%.

(d) Let n be the required sample size. Then n satisfes the equation 0.01 = 1.96 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.48735 and solving for n yields n = 9594.

(e) Let n be the required sample size. Then n satisfes the equation 0.01 = 2.58 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.48735 and solving for n yields n = 16627.

5. (a) X = 47, n = 155, p̃ = (47 + 2)∕(155 + 4) = 0.30818, z.025 = 1.96. The confdence interval is 0.30818 ± 1.96 0.30818(1 − 0.30818)∕(155 + 4), or (0.236, 0.380).

(b) X = 47, n = 155, p̃ = (47 + 2)∕(155 + 4) = 0.30818, z.005 = 2.58.

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The confdence interval is 0.30818 ± 2.58 0.30818(1 − 0.30818)∕(155 + 4), or (0.214, 0.403).

(c) The upper confdence bound 0.35 satisfes the equation 0.35 = 0.30818 + z Solving for z yields z = 1.14. The area to the left of z = 1.14 is 1 −

0.30818(1 − 0.30818)∕(155 + 4)

= 0.8729.

The level is 0.8729, or 87.29%.

X = 28, n = 70, p̃ = (28 + 2)∕(70 + 4) = 0.405045, z.05 = 1.645. The lower confdence bound is 0.405405 − 1.645 0.405405(1 − 0.405405)∕(70 + 4), or 0.312.

X = 73, n = 100, p̃ = (73 + 2)∕(100 + 4) = 0.72115, z.005 = 2.58. The upper confdence bound is 0.72115 + 2.05 0.72115(1 − 0.72115)∕(100 + 4), or 0.811.

X = 2636, n = 5409, p̃ = (2636 + 2)∕(5409 + 4) = 0.48735, z.01 = 2.33. The lower confdence bound is 0.48735 − 2.33 0.48735(1 − 0.48735)∕(5409 + 4), or 0.472.

9. (a) X = 30, n = 400, p̃ = (30 + 2)∕(400 + 4) = 0.079208, z.025 = 1.96. The confdence interval is 0.079208 ± 1.96 0.079208(1 − 0.079208)∕(400 + 4), or (0.0529, 0.1055). (b) Let n be the required sample size. Then n satisfes the equation 0.02 = 1.96 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.079208 and solving for n yields n = 697. (c) Let X be the number of defective components in a lot of 200. Let p be the population proportion of components that are defective. Then X ∼ Bin(200, p), so X is approximately normally distributed with mean X = 200p and X = 200p(1 − p). Let r represent the proportion of lots that are returned. Using the continuity correction, r = P (X > 20.5). To fnd a 95% confdence interval for r, express the z-score of P (X > 20.5) as a function of p and substitute the upper and lower confdence limits for p. The z-score of 20.5 is (20.5 − 200p)∕ 200p(1 − p). Now fnd a 95% confdence interval for z by substituting the upper and lower confdence limits for p. From part (a), the 95% confdence interval for p is (0.052873, 0.10554). Extra precision is used for this confdence interval to get good precision in the fnal answer.

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SECTION 5.3

Substituting 0.052873 for p yields z = 3.14. Substituting 0.10554 for p yields z = −0.14. Since we are 95% confdent that 0.052873 < p < 0.10554, we are 95% confdent that −0.14 < z < 3.14. The area to the right of z = −0.14 is 1 − 0.4443 = 0.5557. The area to the right of z = 3.14 is 1 − 0.9992 = 0.0008. Therefore we are 95% confdent that 0.0008 < r < 0.5557. The confdence interval is (0.0008, 0.5557).

10.

q = (1 − p)2 . From part (a), we are 95% confdent that 0.0529 < p < 0.1055. We are therefore 95% confdent that (1 − 0.1055)2 < q < (1 − 0.0529)2 , or equivalently, that 0.800 < q < 0.897. The confdence interval is (0.800, 0.897).

11. (a) X = 89, n = 710, p̃ = (89 + 2)∕(710 + 4) = 0.12745, z.05 = 1.645. The confdence interval is 0.12745 ± 1.645 0.12745(1 − 0.12745)∕(710 + 4), or (0.107, 0.148).

(b) X = 89, n = 710, p̃ = (89 + 2)∕(710 + 4) = 0.12745, z.025 = 1.96. The confdence interval is 0.12745 ± 1.96 0.12745(1 − 0.12745)∕(710 + 4), or (0.103, 0.152).

(c) X = 89, n = 710, p̃ = (89 + 2)∕(710 + 4) = 0.12745, z.005 = 2.58. The confdence interval is 0.12745 ± 2.58 0.12745(1 − 0.12745)∕(710 + 4), or (0.095, 0.160).

12. (a) X = 104, n = 427, p̃ = (104 + 2)∕(427 + 4) = 0.24594, z.025 = 1.96. The confdence interval is 0.24594 ± 1.96 0.24594(1 − 0.24594)∕(427 + 4), or (0.205, 0.287).

(b) X = 104, n = 427, p̃ = (104 + 2)∕(427 + 4) = 0.24594, z.005 = 2.58. The confdence interval is 0.24594 ± 2.58 0.24594(1 − 0.24594)∕(427 + 4), or (0.192, 0.299).

(c) Let n be the required sample size. Then n satisfes the equation 0.03 = 1.96 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.24594 and solving for n yields n = 788.

(d) Let n be the required sample size. Then n satisfes the equation 0.03 = 2.58 p̃(1 − p̃)∕(n + 4).

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Replacing p̃ with 0.24594 and solving for n yields n = 1368.

13. (a) Let n be the required sample size. Then n satisfes the equation 0.05 = 1.96 p̃(1 − p̃)∕(n + 4). Since there is no preliminary estimate of p̃ available, replace p̃ with 0.5. Solving for n yields n = 381. (b) X = 20, n = 100, p̃ = (20 + 2)∕(100 + 4) = 0.21154, z.025 = 1.96. The confdence interval is 0.21154 ± 1.96 0.21154(1 − 0.21154)∕(100 + 4), or (0.133, 0.290). (c) Let n be the required sample size. Then n satisfes the equation 0.05 = 1.96 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.21154 and solving for n yields n = 253.

14. (a) Let n be the required sample size. Then n satisfes the equation 0.05 = 2.33 p̃(1 − p̃)∕(n + 4). Since there is no preliminary estimate of p̃ available, replace p̃ with 0.5. The equation becomes 0.05 = 2.33 0.5(1 − 0.5)∕(n + 4). Solving for n yields n = 539. (b) X = 30, n = 200, p̃ = (30 + 2)∕(200 + 4) = 0.15686, z.01 = 2.33. The confdence interval is 0.15686 ± 2.33 0.15686(1 − 0.15686)∕(200 + 4), or (0.0975, 0.2162).

(c) Let n be the required sample size. Then n satisfes the equation 0.05 = 2.33 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.15686 and solving for n yields n = 284.

15. (a) X = 293, n = 335, p̃ = (293 + 2)∕(335 + 4) = 0.87021, z.05 = 1.645. The confdence interval is 0.87021 ± 1.645 0.87021(1 − 0.87021)∕(293 + 4), or (0.840, 0.900).

(b) Let n be the required sample size. Then n satisfes the equation 0.025 = 1.645 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.87021 and solving for n yields n = 486. (c) Let n be the required sample size. Then n satisfes the equation 0.03 = 1.645 p̃(1 − p̃)∕(n + 4). Since there is no preliminary estimate of p̃ available, replace p̃ with 0.5. The equation becomes 0.03 = 1.645 0.5(1 − 0.5)∕(n + 4). Solving for n yields n = 748.

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SECTION 5.4

16.

No, the data are not a simple random sample.

17. (a) X = 4, n = 306, p̃ = 0.019355, z.025 = 1.96. The confdence interval is 0.019355 ± 1.96 0.019355(1 − 0.019355)∕(306 + 4), or (0.004018, 0.03469). (b) X = 5, n = 15, p̃ = 0.36842, z.025 = 1.96. The confdence interval is 0.36842 ± 1.96 0.36842(1 − 0.36842)∕(15 + 4), or (0.15152, 0.58532). (c) X = 281, n = 1001, p̃ = 0.28159, z.05 = 1.645. The confdence interval is 0.28159 ± 1.645 0.28159(1 − 0.28159)∕(1001 + 4), or (0.25825, 0.30493). (d) X = 33, n = 100, p̃ = 0.33654, z.05 = 1.645. The confdence interval is 0.33654 ± 1.645 0.33654(1 − 0.33654)∕(100 + 4), or (0.26032, 0.41276). (e) X = 518, n = 1267, p̃ = 0.40913, z.005 = 2.58. The confdence interval is 0.40913 ± 2.58 0.40913(1 − 0.40913)∕(1267 + 4), or (0.37355, 0.44471). (f) X = 63, n = 166, p̃ = 0.38235, z.005 = 2.58. The confdence interval is 0.38235 ± 2.58 0.38235(1 − 0.38235)∕(166 + 4), or (0.28619, 0.47851).

18.

(i) The number of pits is less at 75% humidity than at 40%, but the proportion that are deep is about the same or greater.

19.

X = 48, n = 200, p̃ = 0.245098, z.025 = 1.96. The confdence interval is 0.245098 ± 1.96 0.245098(1 − 0.245098)∕(200 + 4), or (0.18607, 0.30413).

Section 5.4 1.

X = 620, sX = 20, nX = 80, Y = 750, sY = 30, nY = 95, z.025 = 1.96. The confdence interval is 750 − 620 ± 1.96 202 ∕80 + 302 ∕95, or (122.54, 137.46).

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CHAPTER 5

X = 169.9, sX = 24.8, nX = 110, Y = 163.3, sY = 25.8, nY = 225, z.025 = 1.96. The confdence interval is 169.9 − 163.3 ± 1.96 24.82 ∕110 + 25.82 ∕225, or (0.8690, 12.331).

X = 313, sX = 25, nX = 210, Y = 233, sY = 19, nY = 210. The number of degrees of freedom is 4 2 52 192 25 + 210 210 = = 390, rounded down to the nearest integer. (252 ∕210)2 (192 ∕210)2 + 210 − 1 210 − 1 u 252 192 + , t390,.005 = 2.5885, so the confdence interval is 313 − 233 ± 2.5885 210 210 or (74.391, 85.609).

X = 72.7, sX = 4.4, nX = 296, Y = 54.1, sY = 4.7, nY = 296. The number of degrees of freedom is 4 2 52 4.4 4.72 + 296 296 = = 587, rounded down to the nearest integer. 2 2 (4.72 ∕296)2 (4.4 ∕296) + 296 − 1 296 − 1 u 4.42 4.72 t587,.025 = 1.964, so the confdence interval is 72.7 − 54.1 ± 1.964 + , 296 296 or (17.865, 19.335).

X = 3.64, sX = 0.23, nX = 40, Y = 3.40, sY = 0.28, nY = 43. The number of degrees of freedom is 4 52 0.232 0.282 + 40 43 = = 79, rounded down to the nearest integer. (0.232 ∕40)2 (0.282 ∕43)2 + 40 − 1 43 − 1 u 0.232 0.282 t79,.025 = 1.9905, so the confdence interval is 3.64 − 3.40 ± 1.9905 + , 40 43 or (0.1284, 0.3516).

X = 25, sX = 9, nX = 78, Y = 14, sY = 7, nY = 43.

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SECTION 5.4

The number of degrees of freedom is 4 2 52 72 9 + 78 43 = = 105, rounded down to the nearest integer. (92 ∕78)2 (72 ∕43)2 + 78 − 1 43 − 1 u 92 72 + , t105,.025 = 1.9828, so the confdence interval is 25 − 14 ± 1.9828 78 43 or (8.0738, 13.926).

X = 117, sX = 15, nX = 35, Y = 105, sY = 20, nY = 49. The number of degrees of freedom is 4 2 52 15 202 + 35 49 = = 81, rounded down to the nearest integer. 2 2 (202 ∕49)2 (15 ∕35) + 35 − 1 49 − 1 u 152 202 t81,.01 = 2.3733, so the confdence interval is 117 − 105 ± 2.3733 + , 35 49 or (2.9343, 21.066).

X = 80, sX = 2.7, nX = 85, Y = 70, sY = 2.5, nY = 90. The number of degrees of freedom is 4 2 52 2.52 2.7 + 85 90 = = 169, rounded down to the nearest integer. 2 2 (2.52 ∕90)2 (2.7 ∕85) + 85 − 1 90 − 1 u 2.72 2.52 t169,.01 = 2.3497, so the confdence interval is 80 − 70 ± 2.3497 + , 85 90 or (9.0747, 10.925).

X = 242, sX = 20, nX = 47, Y = 220, sY = 31, nY = 42. The number of degrees of freedom is 4 2 52 312 20 + 47 42 = = 68, rounded down to the nearest integer. 2 2 (312 ∕42)2 (20 ∕47) + 47 − 1 42 − 1

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CHAPTER 5 u t68,.025 = 1.9955, so the confdence interval is 242 − 220 ± 1.9955 or (10.820, 33.180).

10.

202 312 + , 47 42

X = 1827, sX = 168, nX = 60, Y = 1658, sY = 225, nY = 180. The number of degrees of freedom is 4 52 1682 2252 + 60 180 = = 134, rounded down to the nearest integer. 2 2 (2252 ∕180)2 (168 ∕60) + 60 − 1 180 − 1 u 1682 2252 + , t134,.025 = 1.9778, so the confdence interval is 1827 − 1658 ± 1.9778 60 180 or (114.775, 223.225).

11. (a) X = 91.1, sX = 6.23, nX = 50, Y = 90.7, sY = 4.34, nY = 40. The number of degrees of freedom is 4 52 6.232 4.342 + 50 40 = = 86, rounded down to the nearest integer. 2 2 (4.342 ∕40)2 (6.23 ∕50) + 50 − 1 40 − 1 u 6.232 4.342 + , t86,.025 = 1.9879, so the confdence interval is 91.1 − 90.7 ± 1.9879 50 40 or (−1.820, 2.620).

(b) No. Since 0 is in the confdence interval, it may be regarded as being a plausible value for the mean di˙erence in hardness.

t 12.

The precision of the confdence interval is ±t ,.025

s2X ∕nX + sY2 ∕nY . sX = 6.23 and sY = 4.34.

If all 10 new welds are cooled at 10 C, then nX = 60 and nY = 40. The number of degrees of freedom is = 97. t97,.025 = 1.9847, so the precision of the confdence interval is ±1.9847 6.232 ∕60 + 4.34∕40 = ±2.0983. If all 10 new welds are cooled at 30 C, then nX = 50 and nY = 50. The number of degrees of freedom is = 87. t87,.025 = 1.9876, so the precision of the confdence interval is ±1.9876 6.232 ∕50 + 4.34∕50 = ±2.1342. If 5 of the new welds are cooled at 10 C and 5 at 30 C, then nX = 55 and nY = 45. The number of degrees of freedom is = 95. t95,.025 = 1.9853, so the precision of the confdence interval is ±1.9853 6.232 ∕55 + 4.34∕45 = ±2.105.

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SECTION 5.4

Therefore the confdence interval would be most precise if all 10 new welds were cooled at 10 C.

13.

It is not possible. The amounts of time spent in bed and spent asleep in bed are not independent.

14.

X = 348, sX = 80, nX = 48, Y = 309, sY = 89, nY = 46. The number of degrees of freedom is 4 2 52 80 892 + 48 46 = = 90, rounded down to the nearest integer. 2 2 (892 ∕46)2 (80 ∕48) + 48 − 1 46 − 1 u 802 892 t90,.025 = 1.9867, so the confdence interval is 348 − 309 ± 1.9867 + , 48 46 or (4.2741, 73.726).

15.

X = 457, sX = 38, nX = 16, Y = 394, sY = 23, nY = 12. The number of degrees of freedom is 4 2 52 232 38 + 16 12 = = 25, rounded down to the nearest integer. 2 2 (232 ∕12)2 (38 ∕16) + 16 − 1 12 − 1 u 382 232 t25,.005 = 2.787, so the confdence interval is 457 − 394 ± 2.787 + , 16 12 or (30.693, 95.307).

16.

X = 6.83, sX = 1.72, nX = 5, Y = 3.32, sY = 1.17, nY = 7. The number of degrees of freedom is 4 52 1.722 1.172 + 5 7 = = 6, rounded down to the nearest integer. 2 2 (1.172 ∕7)2 (1.72 ∕5) + 5−1 7−1 u 1.722 1.172 t6,.025 = 2.447, so the confdence interval is 6.83 − 3.32 ± 2.447 + , 5 7 or (1.3389, 5.6811).

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264 17.

CHAPTER 5

X = 59.6, sX = 5.295701, nX = 10, Y = 50.9, sY = 5.321863, nY = 10. The number of degrees of freedom is 4 52 5.2957012 5.3218632 + 10 10 = = 17, rounded down to the nearest integer. 2 2 (5.3218632 ∕10)2 (5.295701 ∕10) + 10 − 1 10 − 1 u 5.2957012 5.3218632 t17,.005 = 2.898, so the confdence interval is 59.6 − 50.9 ± 2.898 + , 10 10 or (1.8197, 15.5803).

18.

X = 28.714286, sX = 7.387248, nX = 7, Y = 17.714286, sY = 4.4614753, nY = 7. The number of degrees of freedom is 4 52 7.3872482 4.46147532 + 7 7 = = 9, rounded down to the nearest integer. 2 2 (4.46147532 ∕7)2 (7.387248 ∕7) + 7−1 7−1 u 7.3872482 4.46147532 t9,.01 = 2.821, so the confdence interval is 28.714286 − 17.714286 ± 2.821 + , 7 7 or (1.7984, 20.2016).

19.

X = 73.1, sX = 9.1, nX = 10, Y = 53.9, sY = 10.7, nY = 10. The number of degrees of freedom is 4 2 52 9.1 10.72 + 10 10 = = 17, rounded down to the nearest integer. 2 2 (10.72 ∕10)2 (9.1 ∕10) + 10 − 1 10 − 1 u 9.12 10.72 t17,.01 = 2.567, so the confdence interval is 73.1 − 53.9 ± 2.567 + , or (7.798, 30.602). 10 10

20.

X = 18.7, sX = 3.3, nX = 11, Y = 11.2, sY = 2.4, nY = 9. The number of degrees of freedom is 4 2 52 3.3 2.42 + 11 9 = = 17, rounded down to the nearest integer. 2 2 (2.42 ∕9)2 (3.3 ∕11) + 11 − 1 9−1

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265

SECTION 5.4 u t17,.025 = 2.110, so the confdence interval is 18.7 − 11.2 ± 2.110 or (4.8061, 10.1939).

21.

3.32 2.42 + , 11 9

X = 33.8, sX = 0.5, nX = 4, Y = 10.7, sY = 3.3, nY = 8. The number of degrees of freedom is 4 2 52 0.5 3.32 + 4 8 = = 7, rounded down to the nearest integer. 2 2 (3.32 ∕8)2 (0.5 ∕4) + 4−1 8−1 u 0.52 3.32 t7,.025 = 2.365, so the confdence interval is 33.8 − 10.7 ± 2.365 + , or (20.278, 25.922). 4 8

22.

X = 7.9, sX = 0.6, nX = 8, Y = 3.4, sY = 0.6, nY = 10. The number of degrees of freedom is 4 2 52 0.6 0.62 + 8 10 = = 15, rounded down to the nearest integer. 2 2 (0.62 ∕10)2 (0.6 ∕8) + 8−1 10 − 1 u 0.62 0.62 t15,.01 = 2.602, so the confdence interval is 7.9 − 3.4 ± 2.602 + , or (3.579, 5.241). 8 10

23.

X = 4.8, sX = 1.9, nX = 24, Y = 2.8, sY = 1.0, nY = 24. The number of degrees of freedom is 4 2 52 1.9 1.02 + 24 24 = = 34, rounded down to the nearest integer. (1.92 ∕24)2 (1.02 ∕24)2 + 24 − 1 24 − 1 u 1.92 1.02 t34,.025 = 2.032, so the confdence interval is 4.8 − 2.8 ± 2.032 + , or (1.1093, 2.8907). 24 24

24.

X = 648.833, sX = 11.805, nX = 6, Y = 400.625, sY = 5.8049, nY = 8. The number of degrees of freedom is

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CHAPTER 5 4 =

11.8052 5.80492 + 6 8

(11.8052 ∕6)2 (5.80492 ∕8)2 + 6−1 8−1

= 6, rounded down to the nearest integer. u

t6,.01 = 3.143, so the confdence interval is 648.833 − 400.625 ± 3.143 or (231.75, 264.67).

25.

11.8052 5.80492 + , 6 8

X = 36.893077, sX = 3.221054, nX = 13, Y = 33.000000, sY = 2.544882, nY = 9. The number of degrees of freedom is 4 52 3.2210542 2.5448822 + 13 9 = = 19, rounded down to the nearest integer. 2 2 (2.5448822 ∕9)2 (3.221054 ∕13) + 13 − 1 9−1 u 3.2210542 2.5448822 t19,.01 = 2.539, so the confdence interval is 36.893077 − 33.000000 ± 2.539 + , 13 9 or (0.7646, 7.0216).

26.

X = 62.71429, sX = 3.8607, nX = 7, Y = 60.40000, sY = 5.46097, nY = 10. The number of degrees of freedom is 4 52 3.86072 5.460972 + 7 10 = = 14, rounded down to the nearest integer. 2 2 (5.460972 ∕10)2 (3.8607 ∕7) + 7−1 10 − 1 u 3.86072 5.460972 t14,.025 = 2.145, so the confdence interval is 62.71429 − 60.40000 ± 2.145 + , 7 10 or (−2.535, 7.164).

27.

X = 229.54286, sX = 14.16885, nX = 7, Y = 143.95556, sY = 59.75699, nY = 9. The number of degrees of freedom is 4 52 14.168852 59.756992 + 7 9 = = 9, rounded down to the nearest integer. (14.168852 ∕7)2 (59.756992 ∕9)2 + 7−1 9−1 u 14.168852 59.756992 t9,.025 = 2.262, so the confdence interval is 229.54286 − 143.95556 ± 2.262 + , 7 9

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267

SECTION 5.4

or (38.931, 132.244).

28.

X = 95, sX = 1, nX = 5, Y = 70, sY = 6, nY = 5. The number of degrees of freedom is 4 2 52 1 62 + 5 5 = = 4, rounded down to the nearest integer. (12 ∕5)2 (62 ∕5)2 + 5−1 5−1 u 12 62 t4,.025 = 2.776, so the confdence interval is 95 − 70 ± 2.776 + , or (17.448, 32.552). 5 5

29.

X = 4500.8, sX = 271.6, nX = 5, Y = 1299.8, sY = 329.8, nY = 5. The number of degrees of freedom is 4 52 271.62 329.82 + 5 5 = = 7, rounded down to the nearest integer. 2 2 (329.82 ∕5)2 (271.6 ∕5) + 5−1 5−1 u t7,.01 = 2.998, so the confdence interval is 4500.8 − 1299.8 ± 2.998 3773.821).

271.62 329.82 + , or (2628.179, 5 5

30. (a) Median torque is 1877.85, median speed is 50.85.

(b) Median torque is 1229.75, median speed is 64.25.

(d) Median torque is 2515.3, median speed is 29.6.

31.

(ii) For soils where the median torque is greater, the median speed is less.

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CHAPTER 5

32. (a) X = 56.185, sX = 17.078, nX = 328, Y = 43.664, sY = 21.781, nY = 328. The number of degrees of freedom is 52 4 17.0782 43.6642 + 328 328 = 618, rounded down to the nearest integer. = (17.0872 ∕328)2 (21.7812 ∕328)2 + 328 − 1 328 − 1 u 17.0782 43.6642 t618,.025 = 1.9638, so the confdence interval is 56.185−43.664±1.9638 + , or (9.5195, 15.522). 328 328 (b) X = 62.443, sX = 9.8553, nX = 164, Y = 59.391, sY = 12.737, nY = 164. The number of degrees of freedom is 4 52 9.85532 12.7572 + 164 164 = = 306, rounded down to the nearest integer. (9.85532 ∕164)2 (12.7572 ∕164)2 + 164 − 1 164 − 1 u 9.85332 12.7372 t306,.025 = 1.9677, so the confdence interval is 62.443−59.931±1.9677 + , or (0.57662, 5.5258). 16 164 (c) X = 79.016, sX = 10.436, nX = 139, Y = 71.954, sY = 11.585, nY = 139. The number of degrees of freedom is 4 52 10.4362 11.5852 + 139 139 = = 273, rounded down to the nearest integer. (10.4362 ∕139)2 (11.5852 ∕139)2 + 139 − 1 139 − 1 u 10.4362 11.5852 t273,.025 = 1.9687, so the confdence interval is 79.016−71.954±1.9687 + , or (4.4583, 9.6655). 139 139 (d) X = 31.678, sX = 11.127, nX = 429, Y = 29.093, sY = 11.469, nY = 429. The number of degrees of freedom is 4 52 11.1272 11.4692 + 429 429 = = 855, rounded down to the nearest integer. 2 2 (11.4692 ∕429)2 (11.127 ∕429) + 429 − 1 429 − 1 u 11.1272 11.4692 t855,.025 = 1.9627, so the confdence interval is 31.678−29.093±1.9627 + , or (1.0708, 4.0994). 429 429

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SECTION 5.4

33.

(ii) For each of the four soils, we are 95% confdent that the mean speed is greater when the torque is greater.

34.

It’s the opposite. Within soils, higher torques are associated with higher speeds. Between soils, higher torques are associated with lower speeds.

35.

X = 36.533, sX = 16.375, nX = 1060, Y = 57.606, sY = 21.272, nY = 1060. The number of degrees of freedom is 4 52 16.3752 21.2722 + 1060 1060 = = 1987, rounded down to the nearest integer. 2 2 (21.2722 ∕1060)2 (16.375 ∕429) + 1060 − 1 1060 − 1 u 16.3752 21.2722 + , or (−22.69, −19.456). t1987,.025 = 1.9612, so the confdence interval is 36.533−57.606±1.9612 1060 1060

36.

(i) For observations over all soils, we are 95% confdent that the mean speed is greater when the torque is lower.

37.

It’s more like the between soils relationship.

38. (a) X = 18.322, sX = 3.8237, nX = 306, Y = 27.611, sY = 6.62, nY = 15. The number of degrees of freedom is 4 52 3.82372 6.622 + 306 15 = = 14, rounded down to the nearest integer. 2 2 (6.622 ∕15)2 (3.8237 ∕306) + 306 − 1 15 − 1 u 3.82372 6.622 + , or (−12.985, −5.5934). t14,.025 = 2.1448, so the confdence interval is 18.322−27.611±2.1448 306 14 (b) X = 25.827, sX = 7.8274, nX = 1001, Y = 26.713, sY = 11.219, nY = 100. The number of degrees of freedom is 4 52 7.82742 11.2192 + 1001 100 = = 108, rounded down to the nearest integer. (7.82742 ∕1001)2 (11.2192 ∕100)2 + 1001 − 1 100 − 1 u 7.82742 11.2192 + , or (−2.7924, 1.0197). t108,.05 = 1.6591, so the confdence interval is 25.827−26.713±1.6591 1001 100

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(c) X = 29.012, sX = 10.575, nX = 1267, Y = 27.561, sY = 12.646, nY = 166. The number of degrees of freedom is 4 52 10.5752 12.6462 + 1267 166 = = 196, rounded down to the nearest integer. (10.5752 ∕1267)2 (12.6462 ∕166)2 + 1267 − 1 166 − 1 u 10.5752 12.6462 + , or (−1.2166, 4.1183). t196,.005 = 2.6011, so the confdence interval is 29.012−27.561±2.6011 1267 166

39. (a) X = 18.322, sX = 3.8237, nX = 306, Y = 25.827, sY = 7.8274, nY = 1001. The number of degrees of freedom is 4 52 3.82372 7.82742 + 306 1001 = = 1057, rounded down to the nearest integer. 2 2 (7.82742 ∕1001)2 (3.8237 ∕306) + 306 − 1 1001 − 1 u 3.82372 7.82742 + , or (−8.1523, −6.8567). t1057,.025 = 2.6011, so the confdence interval is 18.322−25.827±2.6011 306 1001 (b) X = 25.827, sX = 7.8274, nX = 1001, Y = 29.012, sY = 10.575, nY = 1267. The number of degrees of freedom is 4 52 7.82742 10.5752 + 1001 1267 = = 2256, rounded down to the nearest integer. 2 2 (10.5752 ∕1267)2 (7.8274 ∕1001) + 1001 − 1 1267 − 1 u 7.82742 10.5752 + , or (−3.8216, −2.5492). t2256,.05 = 1.6455, so the confdence interval is 25.827−29.012±1.6455 1001 1267 (c) X = 27.611, sX = 6.62, nX = 15, Y = 26.713, sY = 11.219, nY = 100. The number of degrees of freedom is 4 52 6.622 11.2192 + 15 100 = = 27, rounded down to the nearest integer. (6.622 ∕15)2 (11.2192 ∕100)2 + 15 − 1 100 − 1 u 6.622 11.2192 + , or (−4.1571, 5.9540). t27,.01 = 2.4727, so the confdence interval is 27.611−26.713 ± 2.4727 15 100

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(d) X = 26.713, sX = 11.219, nX = 100, Y = 27.561, sY = 12.646, nY = 166. The number of degrees of freedom is 4 52 11.2192 12.6462 + 100 166 = = 228, rounded down to the nearest integer. 2 2 (12.6462 ∕166)2 (11.219 ∕100) + 100 − 1 166 − 1 u 11.2192 12.6462 t228,.005 = 2.5976, so the confdence interval is 26.713−27.561 ± 2.5976 + , or (−4.7202, 3.0238). 100 166

Section 5.5 1.

X = 125, nX = 600, p̃X = (125 + 1)∕(600 + 2) = 0.2093, Y = 130, nY = 800, p̃Y = (130 + 1)∕(800 + 2) = 0.16334, z.05 = 1.645. u 0.2093(1 − 0.2093) 0.16334(1 − 0.16334) The confdence interval is 0.2093 − 0.16334 ± 1.645 + , 600 + 2 800 + 2 or (0.1125, 0.0807).

X = 68, nX = 85, p̃X = (68 + 1)∕(85 + 2) = 0.793103, Y = 105, nY = 120, p̃Y = (105 + 1)∕(120 + 2) = 0.868852, z.025 = 1.96. u 0.793103(1 − 0.793103) 0.868852(1 − 0.868852) The confdence interval is 0.793103−0.868852 ± 1.96 + , 85 + 2 120 + 2 or (−0.180, 0.0283).

3. (a) X = 841, nX = 5320, p̃X = (841 + 1)∕(5320 + 2) = 0.158211, Y = 134, nY = 1120, p̃Y = (134 + 1)∕(1120 + 2) = 0.120321, z.01 = 2.33. u 0.158211(1 − 0.158211) 0.120321(1 − 0.120321) The confdence interval is 0.158211−0.120321 ± 2.33 + , 5320 + 2 1120 + 2 or (0.0124, 0.0633). (b) The standard deviation of the di˙erence between the proportions is Estimate pX

p̃X = 0.120321 and pY

pX (1 − pX )∕(nX + 2) + pY (1 − pY )∕(nY + 2).

p̃Y = 0.158211.

If 1000 additional patients are treated with bare metal stents, the standard deviation of the di˙erence between the proportions is then 0.120321(1 − 0.120321)∕(1120 + 2) + 0.158211(1 − 0.158211)∕(6320 + 2) = 0.01074. The width of the 98% confdence interval will be ±2.33(0.01074) = 0.0250.

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If 500 additional patients are treated with drug-coated stents, the standard deviation of the di˙erence between the proportions is then 0.120321(1 − 0.120321)∕(1620 + 2) + 0.158211(1 − 0.158211)∕(5320 + 2) = 0.009502. The width of the 98% confdence interval will be ±2.33(0.009502) = 0.0221. If 500 additional patients are treated with bare metal stents and 250 additional patients are treated with drugcoated stents, the standard deviation of the di˙erence between the proportions is then 0.120321(1 − 0.120321)∕(1370 + 2) + 0.158211(1 − 0.158211)∕(5820 + 2) = 0.01000. The width of the 98% confdence interval will be ±2.33(0.01000) = 0.0233. Therefore the confdence interval would be most precise if 500 new patients are treated with drug-coated stents.

X = 43, nX = 50, p̃X = (43 + 1)∕(50 + 2) = 0.846154, Y = 25, nY = 40, p̃Y = (25 + 1)∕(40 + 2) = 0.619048, z.005 = 2.58. u 0.846154(1 − 0.846154) 0.619048(1 − 0.619048) The confdence interval is 0.846154−0.619048 ± 2.58 + , 50 + 2 40 + 2 or (−0.00536, 0.4596).

X = 8, nX = 12, p̃X = (8 + 1)∕(12 + 2) = 0.642857, Y = 5, nY = 15, p̃Y = (5 + 1)∕(15 + 2) = 0.352941, z.025 = 1.96. u 0.642857(1 − 0.642857) 0.352941(1 − 0.352941) The confdence interval is 0.642857−0.352941 ± 1.96 + , 12 + 2 15 + 2 or (−0.04862, 0.6285).

X = 2527, nX = 11805, p̃X = (2527 + 1)∕(11805 + 2) = 0.21411, Y = 2503, nY = 14985, p̃Y = (2503 + 1)∕(14985 + 2) = 0.16708, z.025 = 1.96. u 0.21411(1 − 0.21411) 0.16708(1 − 0.16708) The confdence interval is 0.21411 − 0.16708 ± 1.96 + , 11805 + 2 14985 + 2 or (0.0375, 0.0565).

No. The sample proportions come from the same sample rather than from two independent samples.

X = 85, nX = 150, p̃X = (85 + 1)∕(150 + 2) = 0.57237, Y = 74, nY = 150, p̃Y = (74 + 1)∕(150 + 2) = 0.49342, z.025 = 1.96. u 0.57237(1 − 0.57237) 0.49342(1 − 0.49342) The confdence interval is 0.57237 − 0.49342 ± 1.96 + , 150 + 2 150 + 2 or (−0.0329, 0.191).

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SECTION 5.5

X = 30, nX = 164, p̃X = (30 + 1)∕(164 + 2) = 0.186747, Y = 19, nY = 185, p̃Y = (19 + 1)∕(185 + 2) = 0.106952, z.005 = 2.58. u 0.186747(1 − 0.186747) 0.106952(1 − 0.106952) + , The confdence interval is 0.186747−0.106952 ± 2.58 164 + 2 185 + 2 or (−0.0176, 0.1772).

10.

X = 32, nX = 91, p̃X = (32 + 1)∕(91 + 2) = 0.35484, Y = 13, nY = 67, p̃Y = (13 + 1)∕(67 + 2) = 0.20290, z.05 = 1.645. u 0.35484(1 − 0.35484) 0.20290(1 − 0.20290) + , The confdence interval is 0.35484 − 0.20290 ± 1.645 91 + 2 67 + 2 or (0.0379, 0.266).

11.

No, these are not simple random samples.

12. (a) X = 5, nX = 15, p̃X = (5 + 1)∕(15 + 2) = 0.35294, Y = 4, nY = 306, p̃Y = (4 + 1)∕(306 + 2) = 0.01623, z.025 = 1.96. u 0.35294(1 − 0.35294) 0.01623(1 − 0.01623) The confdence interval is 0.35294 − 0.01623 ± 1.96 + , 15 + 2 306 + 2 or (0.1091, 0.56432). (b) X = 33, nX = 100, p̃X = (33 + 1)∕(100 + 2) = 0.33333, Y = 281, nY = 1001, p̃Y = (281 + 1)∕(1001 + 2) = 0.28116, z.05 = 1.645. u 0.33333(1 − 0.33333) 0.28116(1 − 0.28116) + , The confdence interval is 0.33333 − 0.28116 ± 1.645 100 + 2 1001 + 2 or (−0.028077, 0.13243) (c) X = 518, nX = 1267, p̃X = (518 + 1)∕(1267 + 2) = 0.40898, Y = 63, nY = 166, p̃Y = (63 + 1)∕(166 + 2) = 0.38095, z.005 = 2.58. u 0.40898(1 − 0.40898) 0.38095(1 − 0.38095) + , The confdence interval is 0.40898 − 0.38095 ± 2.58 1267 + 2 166 + 2 or (−0.074982, 0.13104)

13. (a) X = 281, nX = 1001, p̃X = (281 + 1)∕(1001 + 2) = 0.28116, Y = 4, nY = 306, p̃Y = (4 + 1)∕(306 + 2) = 0.01623, z.025 = 1.96. u 0.28116(1 − 0.28116) 0.01623(1 − 0.01623) + , The confdence interval is 0.28116 − 0.01623 ± 1.96 1001 + 2 306 + 2 or (0.23373, 0.29612).

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(b) b. X = 518, nX = 1267, p̃X = (518 + 1)∕(1267 + 2) = 0.40898, Y = 281, nY = 1001, p̃Y = (281 + 1)∕(1001 + 2) = 0.28116, z0.005 = 2.58. u 0.40898(1 − 0.40898) 0.28116(1 − 0.28116) The confdence interval is 0.40898 − 0.28116 ± 2.58 + , 1267 + 2 1001 + 2 or (0.076747, 0.17891)

(c) c. X = 5, nX = 15, p̃X = (5 + 1)∕(15 + 2) = 0.35294, Y = 33, nY = 100, p̃Y = (33 + 1)∕(100 + 2) = 0.33333, z.05 = 1.645. u 0.35294(1 − 0.35294) 0.33333(1 − 0.33333) The confdence interval is 0.35294 − 0.33333 ± 1.645 + , 15 + 2 100 + 2 or (−0.18593, 0.22515)

(d) d. X = 63, nX = 166, p̃X = (63 + 1)∕(166 + 2) = 0.38095, Y = 33, nY = 100, p̃Y = (33 + 1)∕(100 + 2) = 0.33333, z.01 = 2.33. u 0.38095(1 − 0.38095) 0.33333(1 − 0.33333) The confdence interval is 0.38095 − 0.33333 ± 2.33 + , 166 + 2 100 + 2 or (−0.091839, 0.18708)

Section 5.6 1.

D = 6.736667, sD = 6.045556, n = 9, t9−1,.025 = 2.306. The confdence interval is 6.736667 ± 2.306(6.045556∕ 9), or (2.090, 11.384).

D = 0.19375, sD = 0.120230, n = 8, t8−1,.025 = 2.365. The confdence interval is 0.19375 ± 2.365(0.120230∕ 8), or (0.0932, 0.2943).

The di˙erences are: 9, 7, 5, 9, 10, 9, 2, 8, 8, 10. D = 7.7, sD = 2.496664, n = 10, t10−1,.01 = 2.821. The confdence interval is 7.7 ± 2.821(2.496664∕ 10), or (5.473, 9.927).

The di˙erences are: 15, 21, 18, 20, 22, 24, 15, 19, 20, 20. D = 19.4, sD = 2.836273, n = 10, t10−1,.025 = 2.262. The confdence interval is 19.4 ± 2.262(2.836273∕ 10), or (17.3712, 21.4288).

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SECTION 5.6

The di˙erences are: 25, 32, 39, 23, 38. D = 31.4, sD = 7.300685, n = 5, t5−1,.05 = 2.132. The confdence interval is 31.4 ± 2.132(7.300685∕ 5), or (24.4391, 38.3609).

The di˙erences are: 0.30, 0.38, 0.66, 0.41, 0.38, 0.58, 0.20, 0.16, 0.41, 0.50. D = 0.398, sD = 0.155835, n = 10, t10−1,.01 = 2.821. The confdence interval is 0.398 ± 2.821(0.155835∕ 10), or (0.259, 0.537).

The di˙erences are: 8.4, 8.6, 10.5, 9.6, 10.7, 10.8, 10.7, 11.3, 10.7. D = 10.144444, sD = 1.033333, n = 9, t9−1,.025 = 2.306. The confdence interval is 10.144444 ± 2.306(1.033333∕ 9), or (9.350, 10.939).

8. (a) Yes. This design involves two independent samples, so a confdence interval for the di˙erence between the two means can be constructed by using expression (5.16) in Section 5.4.

(b) It is likely to be less precise, because there is likely to be a lot of variation in brake pad life from one car to another.

9. (a) The di˙erences are: 3.8, 2.6, 2.0, 2.9, 2.2, −0.2, 0.5, 1.3, 1.3, 2.1, 4.8, 1.5, 3.4, 1.4, 1.1, 1.9, −0.9, −0.3. D = 1.74444, sD = 1.46095, n = 18, t18−1,.005 = 2.898. The confdence interval is 1.74444 ± 2.898(1.46095∕ 18), or (0.747, 2.742).

(b) The level 100(1 − )% can be determined from the equation t17, ∕2 (1.46095∕ 18) = 0.5. From this equation, t17, ∕2 = 1.452. The t table indicates that the value of ∕2 is between 0.05 and 0.10, and closer to 0.10. Therefore the level 100(1 − )% is closest to 80%.

10. (a) Expression (5.16) in Section 5.4 should be used, because the design involves two independent samples.

(b) X = 23.93, sX = 1.79, nX = 18, Y = 22.19, sY = 1.95, nY = 18. The number of degrees of freedom is

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CHAPTER 5 4 =

1.792 1.952 + 18 18

(1.792 ∕18)2 (1.952 ∕18)2 + 18 − 1 18 − 1

= 33, rounded down to the nearest integer. u

t33,.005 = 2.733, so the confdence interval is 23.93 − 22.19 ± 2.733

1.792 1.952 + , or (0.0349, 3.445). 18 18

(c) The confdence interval based on the two-sample design is wider because there is considerable variation in mileage between the cars.

11. (a) Paired samples (b) Independent samples (c) Independent samples (d) Paired samples

12. (a) Independent samples (b) Paired samples (c) Paired samples (d) Independent samples

13.

D = 44.06, sD = 52.249, n = 366, t366−1,.025 = 1.9665. The confdence interval is 44.06 ± 1.9665(52.249)∕ 366, or (38.689, 49.431). Here is the CI for every seventh day starting January 1. D = 48.868, sD = 58.499, n = 53, t53−1,.025 = 2.0066. The confdence interval is 48.868 ± 2.0066(58.499)∕ 53, or (32.744, 64.992).

14.

Yes, because the confdence interval suggests that record high temperatures have been occuring more recently than record lows.

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SECTION 5.7

Section 5.7 2 1. (a) 12,.025 = 23.337 2 (b) 12,.975 = 4.404 2 (c) 5,.005 = 16.750 2 (d) 5,.995 = 0.412 2 (e) 22,.1 = 30.813 2 (f) 22,.9 = 14.041

2 2 = 0.05. 24,.025 = 39.364, 24,.975 = 12.401. 0 1 2 24(15 ) 24(152 ) 2 , , or (137.18, 435.45). A 95% confdence interval for is 39.364 12.401

s = 7.5, n = 8,

2 2 = 0.02. 19,.01 = 36.191, 19,.99 = 7.633. Hu I u 19(872 ) 19(872 ) A 95% confdence interval for is , , or (63.04, 137.26). 36.191 7.633

2 2 = 0.05. 17,.025 = 30.191, 17,.975 = 7.564. 0 1 2 2 17(8 ) 17(8 ) 2 A 95% confdence interval for is , , or (36.04, 143.84). 30.191 7.564

s = 15, n = 25,

2 2 = 0.01. 7,.005 = 20.278, 7,.995 = 0.989. I Hu u 7(7.52 ) 7(7.52 ) A 99% confdence interval for is , , or (4.41, 19.95). 20.278 0.989

s = 87, n = 20,

s = 8, n = 18,

6. (a) s = 1.798. 2 2 = 0.05. 11,.025 = 21.920, 11,.975 = 3.816. Hu I u 11(1.7982 ) 11(1.7982 ) , , or (1.274, 3.053). A 95% confdence interval for is 21.920 3.816

(b) s = 1.798, n = 12,

7. (a) s = 0.0614

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CHAPTER 5 2 2 = 24.725, 11,.99 = 3.053. = 0.08. 11,.01 Hu I u 11(0.06412 ) 11(0.06412 ) A 95% confdence interval for is , , or (0.041, 0.117). 24.725 3.053

(b) s = 0.0614, n = 12,

2 2 = 0.01. 5,.005 = 16.750, 5,.995 = 0.412. 0 1 5(0.83772 ) 5(0.83772 ) A 95% confdence interval for 2 is , , or (0.21, 8.52). 16.750 0.412

s = 0.18053, n = 12,

10.

s = 40, n = 101,

11.

s = 40 and the number of degrees of freedom is k = 100.

s = 0.8377, n = 6,

2 2 = 0.05. 11,.025 = 21.920, 11,.975 = 3.816. 0 1 11(0.180532 ) 11(0.180532 ) A 95% confdence interval for 2 is , , or (0.016355, 0.093947). 21.920 3.816 2 2 = 0.05. 100,.025 = 129.561, 100,.975 = 74.222. Hu I u 100(402 ) 100(402 ) A 95% confdence interval for is , , or (35.142, 46.429). 129.561 74.222

2 distribution with the N(100, 200) distribution. We therefore approximate the 100

The z score for the 2.5 percentile of the normal distribution is z = −1.96. 2 We therefore estimate 100,.025 = 100 − 1.96 (200) = 72.281.

The z score for the 97.5 percentile of the normal distribution is z = 1.96. 2 = 100 + 1.96 (200) = 127.719. We therefore estimate 100,.975 Hu I u 100(402 ) 100(402 ) A 95% confdence interval for is , , or (35.394, 47.049). 127.719 72.281

12.

s = 40 and the number of degrees of freedom is k = 100. 2 2 = 0.5 −1.96 + 2(100) − 1 = 73.772, We approximate 100,.025 2 2 100,.975 = 0.5 1.96 + 2(100) − 1 = 129.070. Hu I u 100(402 ) 100(402 ) A 95% confdence interval for is , , or (35.208, 46.571). 129.070 73.772

Section 5.8 1. (a) X = 101.4, s = 2.3, n = 25, t25−1,.025 = 2.064.

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SECTION 5.9

The prediction interval is 101.4 ± 2.064(2.3 1 + 1∕25), or (96.559, 106.241). (b) X = 101.4, s = 2.3, n = 25, k25,.05,.10 = 2.2083 The tolerance interval is 101.4 ± 2.2083(2.3), or (96.321, 106.479).

2. (a) X = 89.7, s = 8.2, n = 20, t20−1,.005 = 2.861. The prediction interval is 89.7 ± 2.861(8.2 1 + 1∕20), or (65.660, 113.740). (b) X = 89.7, s = 8.2, n = 20, k20,.01,.05 = 3.1681 The tolerance interval is 89.7 ± 3.1681(8.2), or (63.722, 115.678).

3. (a) X = 5.9, s = 0.56921, n = 6, t6−1,.01 = 3.365. The prediction interval is 5.9 ± 3.365(0.56921 1 + 1∕6), or (3.8311, 7.9689). (b) X = 5.9, s = 0.56921, n = 6, k6,.05,.05 = 4.4140 The tolerance interval is 5.9 ± 4.4140(0.56921), or (3.3875, 8.4125).

4. (a) X = 19.35, s = 0.577, n = 6, t6−1,.05 = 2.015. The prediction interval is 19.35 ± 2.015(0.577 1 + 1∕6), or (18.094, 20.606). (b) X = 19.35, s = 0.577, n = 6, k6,.05,.01 = 5.7746 The tolerance interval is 19.35 ± 5.7746(0.577), or (16.018, 22.682).

5. (a) X = 86.56, s = 1.02127, n = 5, t5−1,.025 = 2.776. The prediction interval is 86.56 ± 2.776(1.02127 1 + 1∕5), or (83.454, 89.666). (b) X = 86.56, s = 1.02127, n = 5, k5,.01,.10 = 6.6118 The tolerance interval is 86.56 ± 6.6118(1.02127), or (79.808, 93.312).

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Section 5.9 1. (a) X ∗ ∼ N(8.5, 0.22 ), Y ∗ ∼ N(21.2, 0.32 )

(b) Answers will vary.

(d) Yes, P is approximately normally distributed.

(e) Answers will vary.

2. (a) X ∗ ∼ N(1.18, 0.022 ), Y ∗ ∼ N(0.85, 0.022 )

(b) Answers will vary.

(d) Yes, W is approximately normally distributed.

(e) Answers will vary.

3. (a) Yes, A is approximately normally distributed.

(b) Answers will vary.

4. (a) Yes, T is approximately normally distributed.

(b) Answers will vary.

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281

SECTION 5.9

5. (a) N(0.27, 0.402 ∕349) and N(1.62, 1.702 ∕143). Since the values 0.27 and 1.62 are sample means, their variances are equal to the population variances divided by the sample sizes. (b) No, R is not approximately normally distributed. (c) Answers will vary. (d) It is not appropriate, since R is not approximately normally distributed.

6. (a) Using Method 1, the limits of the 95% confdence interval are the 2.5 and 97.5 percentiles of the bootstrap means, X .025 and X .975 . The 2.5 percentile is (Y25 + Y26 )∕2 = 38.3818, and the 97.5 percentile is (Y975 + Y976 )∕2 = 38.53865. The confdence interval is (38.3818, 38.53865). (b) Using Method 2, the 95% confdence interval is (2X − X .975 , 2X − X .025 ). X = 38.44727, X .025 = (Y25 + Y26 )∕2 = 38.3818, and X .975 = (Y975 + Y976 )∕2 = 38.53865. The confdence interval is (38.3559, 38.5127). (c) Using Method 1, the limits of the 90% confdence interval are the 5th and 95th percentiles of the bootstrap means, X .05 and X .95 . The 5th percentile is (Y50 + Y51 )∕2 = 38.39135, and the 95th percentile is (Y950 + Y951 )∕2 = 38.5227. The confdence interval is (38.39135, 38.5227). (d) Using Method 2, the 90% confdence interval is (2X − X .95 , 2X − X .05 ). X = 38.44727, X .05 = (Y50 + Y51 )∕2 = 38.39135, and X .95 = (Y950 + Y951 )∕2 = 38.5227. The confdence interval is (38.3718, 38.5032).

7. (a,b,c) Answers will vary.

8. (a) 95% exactly (simulation results will be approximate). (b)

91.5%

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282 (c)

CHAPTER 5

82.5%

9. (a) Coverage probability for traditional interval is

.89, mean length is

Coverage probability for the Agresti–Coull interval is (b) Coverage probability for traditional interval is

.98, mean length is

.95, mean length is

Coverage probability for the Agresti–Coull interval is

.51.

.46.

.95, mean length is

.92, mean length is

Coverage probability for the Agresti–Coull interval is

.585.

.42.

.305.

.96, mean length is

.29.

(d) The traditional method has coverage probability close to 0.95 for n = 17, but less than 0.95 for both n = 10 and n = 40. (e) The Agresti–Coull interval has greater coverage probability for sample sizes 10 and 40, and nearly equal for 17. (f) The Agresti–Coull method does.

10. (a) General coverage is

.95, pooled coverage is exactly 0.95 (simulation results will be approximate).

(b) General coverage is

.95, pooled coverage is

.94.

.95, pooled coverage is

.999.

(d) General coverage is

.95, pooled coverage is

.66.

(e) No. (f) (iii)

11. (a) Answers will vary. (b) Answers will vary. (c) Answers will vary.

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283

12. (a) Answers will vary. (b) Answers will vary. (c) Answers will vary.

Supplementary Exercises for Chapter 5 1.

The di˙erences are 21, 18, 5, 13, −2, 10. D = 10.833, sD = 8.471521, n = 6, t6−1,.025 = 2.571. The confdence interval is 10.833 ± 2.571(8.471521∕ 6), or (1.942, 19.725).

X = 27.833333, sX = 8.328665, nX = 6, Y = 17.500000, sY = 4.370355, nY = 6. The number of degrees of freedom is 4 52 8.3286652 4.3703552 + 6 6 = = 7, rounded down to the nearest integer. (8.3286652 ∕6)2 (4.3703552 ∕6)2 + 6−1 6−1 u 8.3286652 4.3703552 + , t7,.025 = 2.365, so the confdence interval is 27.833333 − 17.500000 ± 2.365 6 6 or (1.252, 19.415).

X = 1919, nX = 1985, p̃X = (1919 + 1)∕(1985 + 2) = 0.966281, Y = 4561, nY = 4988, p̃Y = (4561 + 1)∕(4988 + 2) = 0.914228, z.005 = 2.58. u 0.966281(1 − 0.966281) 0.914228(1 − 0.914228) + , The confdence interval is 0.966281−0.914228 ± 2.58 1985 + 2 4988 + 2 or (0.0374, 0.0667).

4. (a) X = 4.2, s = 0.1, n = 87, t87−1,.01 = 2.3705. The confdence interval is (4.1746, 4.2254). (b) t75−1,.01 = 2.3705. n = (2.37052 )(0.12 )∕(0.0152 ) = 249.75, round up to 250.

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284 5.

CHAPTER 5

X = 6.1, sX = 0.7, nX = 125, Y = 5.8, sY = 1, nY = 75. The number of degrees of freedom is 4 2 52 12 0.7 + 125 75 = = 117, rounded down to the nearest integer. (0.72 ∕125)2 (12 ∕75)2 + 125 − 1 75 − 1 u 0.72 12 t117,.05 = 1.658, so the confdence interval is 6.1 − 5.8 ± 1.658 + , 125 75 or (0.082221, 0.51778).

t 6.

The precision of the confdence interval is ±t ,.05

s2X ∕nX + sY2 ∕nY . sX = 0.7 and sY = 1.0.

If 50 pieces of yarn are sampled from the old batch, then nX = 175 and nY = 75. The number of degrees of freedom is = 106. t106,.05 = 1.6594, so the precision of the confdence interval is ±1.6594 0.72 ∕175 + 1.02 ∕75 = 0.2108. If 50 pieces of yarn are sampled from the new batch, then nX = 125 and nY = 125. The number of degrees of freedom is = 221. t221,.05 = 1.6518, so the precision of the confdence interval is ±1.6518 0.72 ∕125 + 1.02 ∕125 = 0.1803. If 25 pieces of yarn are sampled from each batch, then nX = 150 and nY = 100. The number of degrees of freedom is = 149. t149,.05 = 1.6551, so the precision of the confdence interval is ±1.6551 0.72 ∕150 + 1.02 ∕100 = 0.1847. Therefore sampling 50 additional pieces from the new batch increases the precision the most.

7. (a) X = 13, n = 87, p̃ = (13 + 2)∕(87 + 4) = 0.16484, z.025 = 1.96. The confdence interval is 0.16484 ± 1.96 0.16484(1 − 0.16484)∕(87 + 4), or (0.0886, 0.241).

(b) Let n be the required sample size. Then n satisfes the equation 0.03 = 1.96 p̃(1 − p̃)∕(n + 4). Replacing p̃ with 0.16484 and solving for n yields n = 584.

X = 72.5, s = 5.8, n = 10, t10−1,.005 = 3.250. The confdence interval is 72.5 ± 3.250(5.8∕ 10), or (66.539, 78.461).

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SUPPLEMENTARY EXERCISES FOR CHAPTER 5

The higher the level, the wider the confdence interval. Therefore the narrowest interval, (4.20, 5.83), is the 90% confdence interval, the widest interval, (3.57, 6.46), is the 99% confdence interval, and (4.01, 6.02) is the 95% confdence interval.

10.

Let n be the required sample size. Then n satisfes the equation 0.04 = 1.96 p̃(1 − p̃)∕(n + 4). Since there is no preliminary value of p̃ we replace p̃ with 0.5. Solving for n yields n = 597.

11.

X = 7.909091, sX = 0.359039, nX = 11, Y = 8.00000, sY = 0.154919, nY = 6. The number of degrees of freedom is 4 52 0.3590392 0.1549192 + 11 6 = = 14, rounded down to the nearest integer. 2 2 (0.1549192 ∕6)2 (0.359039 ∕11) + 11 − 1 6−1 u 0.3590392 0.1549192 + , t14,.01 = 2.624, so the confdence interval is 7.909091 − 8.00000 ± 2.624 11 6 or (−0.420, 0.238).

12. (a) X = 1.69, s = 0.25, n = 80, t80−1,.01 = 1.9905. The confdence interval is (1.6344, 1.7456). (b) t85−1,.01 = 1.9905. n = (1.99052 )(0.252 )∕(0.022 ) = 619.08, round up to 610.

13.

Let n be the required sample size. t64−1,.005 = 2.6561, s = 3.6. The required width is 1.0. n = (1.99052 )(0.252 )∕(0.022 ) = 91.434, round up to 92.

14.

(ii) ±0.03. The original confdence has width ±t100−1,.025 s 100. The new confdence interval has width ±t400−1,.025 s 400. Now t100−1,.025 and t400−1,.025 are similar, so the width of the new confdence interval divided by the width of the old is approximately

100∕ 400, or 1∕2.

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CHAPTER 5

15. (a) False. This a specifc confdence interval that has already been computed. The notion of probability does not apply. (b) False. The confdence interval specifes the location of the population mean. It does not specify the location of a sample mean. (c) True. This says that the method used to compute a 95% confdence interval succeeds in covering the true mean 95% of the time. (d) False. The confdence interval specifes the location of the population mean. It does not specify the location of a future measurement.

16. (a) True. t400−1,.025 = 1.9659, so a 95% confdence interval is found by adding and subtracting 1.9659(650∕ 400) from the mean of 370. (b) False. The confdence interval specifes the location of the population mean. It does not specify the proportion of the sample items that fall into any given interval. (c) True. t400−1,.16 = 0.9957, so the level of the confdence interval X ± s∕ 400 is approximately 68%.Therefore this method produces an interval that covers the true value approximately 68% of the times that it is used. (d) False. So long as the sample mean is approximately normal, which will be the case when the sample is large, the method can be used. (e) False. The confdence interval specifes the location of the population mean. It does not specify the proportion of the population that falls into any given interval.

17. (a) X = 37, and the uncertainty is sX = s∕ n = 0.1. Now t100−1,.025 = 1.9842, so a 95% confdence interval is 37 ± 1.9842(0.1), or (36.802, 37.198). (b) The measurements come from a normal population. (c) t10−1,.025 = 2.262. A 95% confdence interval is therefore 37 ± 2.262(s∕ n) = 37 ± 2.262(0.1), or (36.774, 37.226).

18. (a) X = 1500, s = 5, n = 10, t10−1,.025 = 2.262. The confdence interval is 1500 ± 2.262(5∕ 10), or (1496.4, 1503.6).

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287

SUPPLEMENTARY EXERCISES FOR CHAPTER 5 (b) nail = box 100

(c) From part (a), we are 95% confdent that the mean weight of a box of 100 nails is in the interval (1496.4, 1503.6). Therefore we are 95% confdent that the mean weight of a single nail is in the interval (14.964, 15.036).

19. (a) Since X is normally distributed with mean n , it follows that for a proportion 1 − −z ∕2 X < X − n < z ∕2 X .

of all possible samples,

Multiplying by −1 and adding X across the inequality yields X − z ∕2 X < n < X + z ∕2 X , which is the desired result.

(b) Since n is a constant, X∕n = X ∕n =

n ∕n =

∕n.

Therefore = X ∕n. (c) Divide the inequality in part (a) by n. t for in part (c) to show that for a proportion 1 − of all possible samples, (d) Substitute ∕n t t − z ∕2 ∕n < < + z ∕2 ∕n. t The interval ± z ∕2 ∕n is therefore a level 1 − confdence interval for . (e) n = 5, = 300∕5 = 60, = or (53.210, 66.790).

60∕5 = 3.4641. A 95% confdence interval for is therefore 60±1.96(3.4641),

20. (a) n = 1, = 64∕1 = 64, = or (48.32, 79.68).

64∕1 = 8. A 95% confdence interval for is therefore 64 ± 1.96(8),

(b) n = 4, = 256∕4 = 64, = or (56.16, 71.84).

64∕4 = 4. A 95% confdence interval for is therefore 64 ± 1.96(4),

(c) Let n be the required number of minutes. Then n satisfes the equation 1 = 1.96 ∕n. Substituting = 64 for and solving for n yields n = 245.86 minutes.

21. (a) = 1.6, = 0.05, ℎ = 15, ℎ = 1.0, = 25, = 1.0. V = ℎ∕ = 234.375

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CHAPTER 5 )V )V )V = − ℎ∕ 2 = −146.484375, = ∕ = 15.625, = ℎ∕ = 9.375 ) )ℎ ) v 0 12 )V )V 2 2 )V 2 2 2 + ℎ + = 19.63862 V = ) )ℎ )

(b) If the estimate of V is normally distributed, then a 95% confdence interval for V is 234.375 ± 1.96(19.63862) or (195.883, 272.867). (c) The confdence interval is valid. The estimate of V is approximately normal. 22. (a) X = 2.3367, s = 0.093737, n = 6, t6−1,.025 = 2.571. The prediction interval is 2.3367 ± 2.571(0.093737 1 + 1∕6), or (2.076, 2.597). (b) X = 2.3367, s = 0.093737, n = 6, k6,.05,.10 = 3.7123. The tolerance interval is 2.3367 ± 3.7123(0.093737), or (1.989, 2.685). 23. (a) X = 8.95, s = 0.5682, n = 8, t8−1,.01 = 2.998. The prediction interval is 8.95 ± 2.998(0.5682 1 + 1∕8), or (7.143, 10.757). (b) X = 8.95, s = 0.5682, n = 8, k8,.05,.01 = 4.8907. The tolerance interval is 8.95 ± 4.8907(0.5682), or (6.171, 11.729). 2 2 = 0.05. 7,.025 = 16.013, 7,.975 = 1.690. Hu I u 7(6.08862 ) 7(6.08862 ) A 95% confdence interval for is , , or (4.026, 12.391). 16.013 1.690

24.

s = 6.0886, n = 8,

25.

It is not appropriate. The sample contains an outlier, which suggests that the population is not normal.

26. (a) Using Method 1, the limits of the 95% confdence interval are the 2.5 and 97.5 percentiles of the bootstrap means, X .025 and X .975 . The 2.5 percentile is (Y250 + Y251 )∕2 = 1305.5, and the 97.5 percentile is (Y9750 + Y9751 )∕2 = 1462.1. The confdence interval is (1305.5, 1462.1). (b) Using Method 2, the 95% confdence interval is (2X − X .975 , 2X − X .025 ). X = 1385.1, X .025 = (Y250 + Y251 )∕2 = 1305.5, and X .975 = (Y9750 + Y9751 )∕2 = 1462.1. The confdence interval is (1308.1, 1464.7).

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SUPPLEMENTARY EXERCISES FOR CHAPTER 5

289

(c) Using Method 1, the limits of the 99% confdence interval are the 0.5 and 99.5 percentiles of the bootstrap means, X .005 and X .995 . The 0.5 percentile is (Y50 + Y51 )∕2 = 1283.4, and the 99.5 percentile is (Y9950 + Y9951 )∕2 = 1483.8. The confdence interval is (1283.4, 1483.8).

(d) Using Method 2, the 99% confdence interval is (2X − X .995 , 2X − X .005 ). X = 1385.1, X .005 = (Y50 + Y51 )∕2 = 1283.4, and X .995 = (Y9950 + Y9951 )∕2 = 1483.8. The confdence interval is (1286.4, 1486.8).

27. (a) Answers may vary somewhat from the 0.5 and 99.5 percentiles in Exercise 26 parts (c) and (d). (b) Answers may vary somewhat from the answer to Exercise 26 part (c). (c) Answers may vary somewhat from the answer to Exercise 26 part (d).

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CHAPTER 6

Chapter 6 Section 6.1 1. (a) X = 1.03, s = 0.1, n = 50. The null and alternative hypotheses are H0 ∶ ≤ 1 versus H1 ∶ > 1. z = (1.03 − 1)∕(0.1∕ 50) = 2.12. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of z = 2.12. Thus P = 0.0170.

(b) The P -value is 0.0170, so if H0 is true then the sample is in the most extreme 1.70% of its distribution. 2. (a) X = 0.595, s = 0.05, n = 65. The null and alternative hypotheses are H0 ∶ ≥ 0.6 versus H1 ∶ < 0.6. z = (0.595 − 0.6)∕(0.05∕ 65) = −0.81. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of z = −0.81. Thus P = 0.2090.

(b) The P -value is 0.2090, so if H0 is true then the sample is in the most extreme 20.90% of its distribution. 3. (a) X = 231.7, = 2.19, n = 66. The null and alternative hypotheses are H0 ∶ = 232 versus H1 ∶ ≠ 232.

z = (231.7 − 232)∕(2.19∕ 66) = −1.11. Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of z = 1.11 and to the left of z = −1.11.

Thus P = 0.1335 + 0.1335 = 0.2670.

(b) The P -value is 0.2670, so if H0 is true then the sample is in the most extreme 26.70% of its distribution. 4. (a) X = 2.6, = 0.3, n = 50. The null and alternative hypotheses are H0 ∶ ≤ 2.5 versus H1 ∶ > 2.5. z = (2.6 − 2.5)∕(0.3∕ 50) = 2.36. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of z = 2.36. Thus P = 0.0091. (b) The P -value is 0.0091, so if H0 is true then the sample is in the most extreme 0.91% of its distribution.

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SECTION 6.1 5. (a) X = 4.5, = 2.7, n = 80. The null and alternative hypotheses are H0 ∶ ≥ 5.4 versus H1 ∶ < 5.4.

z = (4.5 − 5.4)∕(2.7∕ 80) = −2.98. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of z = −2.98. Thus P = 0.0014. (b) If the mean number of sick days were 5.4, the probability of observing a sample mean less than or equal to the observed value of 4.5 would be 0.0014. Since 0.0014 is a small probability, we are convinced that the mean number of sick days is less than 5.4. 6. (a) X = 15.2, = 1.8, n = 87. The null and alternative hypotheses are H0 ∶ = 15 versus H1 ∶ ≠ 15.

z = (15.2 − 15)∕(1.8∕ 87) = 1.04. Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of z = 1.04 and to the left of z = −1.04.

Thus P = 0.1492 + 0.1492 = 0.2984. (b) If the mean particle diameter were 15 m, the probability of observing a sample mean as far from 15 as the value of 15.2 that was actually observed would be 0.2984. Since 0.2984 is not a small probability, it is plausible that the mean is 15 m. 7. (a) X = 850, = 153, n = 60. The null and alternative hypotheses are H0 ∶ ≥ 900 versus H1 ∶ < 900. z = (850 − 900)∕(153∕ 60) = −2.53. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of z = −2.53. Thus P = 0.0057. (b) If the mean depth were 900 m or more, the probability of observing a sample mean as small as the value of 850 that was actually observed would be 0.0057. Since this is a small probability, we are convinced that the mean daily output is not 900 m or more, but is instead less than 900 m tons. 8. (a) X = 5.02, = 0.06, n = 100. The null and alternative hypotheses are H0 ∶ = 5 versus H1 ∶ ≠ 5. z = (5.02 − 5)∕(0.06∕ 100) = 3.33. Since the alternative hypothesis is of the form = 0 , the P -value is the sum of the areas to the right of z = 3.33 and to the left of z = −3.33. Thus P = 0.0008. (b) If the mean length were 5, the probability of observing a sample mean as far from 5 as the value of 5.02 that was actually observed would be 0.0008. Since this is a small probability, we are convinced that the mean error is not equal to 5.

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CHAPTER 6

9. (a) X = 8.24, = 16.33, n = 126. The null and alternative hypotheses are H0 ∶ ≥ 10 versus H1 ∶ < 10. z = (8.24 − 10)∕(16.33∕ 126) = −1.21. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of z = −1.21. Thus P = 0.1131.

(b) If the mean proft margin were 10%, the probability of observing a sample mean less than or equal to the observed value of 8.24 would be 0.1131. Since 0.1131 is not a small probability, it is plausible that the mean is 10% or more. 10. (a) X = 1356, = 70, n = 100. The null and alternative hypotheses are H0 ∶ ≤ 1350 versus H1 ∶ > 1350. z = (1356 − 1350)∕(70∕ 100) = 0.86. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of z = 0.86. Thus P = 0.1949.

(b) If the mean compressive strength were 1350 kPa, the probability of observing a sample mean greater than or equal to the observed value of 1356 would be 0.195. Since 0.195 is not a small probability, it is plausible that the mean is 1350 kPa or less, and that the blocks do not meet the specifcation.

11.

(ii) 5. The null distribution specifes that the population mean, which is also the mean of X, is the value on the boundary between the null and alternative hypotheses.

12.

(iii) 4. The P -value is the probability, under the assumption that H0 is true, of observing a result as extreme as or more extreme than that actually observed. Since P = 0.04, we expect to obtain a value of X less than or equal to 8 approximately 4 times in 100.

13.

(iii) 78 The P -value is the probability, when H0 is true, of observing a value of the test statistic whose disagreement with H0 is as great or greater than that actually observed. The P -value is 0.22. Therefore, the probability is 0.22 of observing a value of X less than 4 or greater than 6. The probability of observing a value between 4 and 6 is therefore 1 − 0.22 = 0.78.

14.

X = 7.4 and X = ∕ n = 0.2. The null and alternative hypotheses are H0 ∶ = 7.0 versus H1 ∶ ≠ 7.0.

z = (7.4 − 7.0)∕0.2 = 2.00. Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of z = 2.00 and to the left of z = −2.00.

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SECTION 6.1

Thus P = 0.0228 + 0.0228 = 0.0456.

15.

X = 5.2 and X = ∕ n = 0.1. The null and alternative hypotheses are H0 ∶ = 5.0 versus H1 ∶ ≠ 5.0.

z = (5.2 − 5.0)∕0.1 = 2.00. Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of z = 2.00 and to the left of z = −2.00. Thus P = 0.0228 + 0.0228 = 0.0456.

16. (a) Two-tailed. The alternative hypothesis is of the form ≠ 0 . (b) H0 ∶ = 73.5 (c) P = 0.196 (d) X = 73.2461 and s∕ n = 0.1963. The null and alternative hypotheses are H0 ∶ ≥ 73.6 versus H1 ∶ < 73.6. z = (73.2461 − 73.6)∕0.1963 = −1.80. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of z = −1.80. Thus P = 0.0359.

(e) X = 73.2461, s∕ n = 0.1963, z.005 = 2.58. or (72.74, 73.75).

The confdence interval is 73.2461 ± 2.58(0.1963),

17. (a) One-tailed. The alternative hypothesis is of the form > 45. (b) H0 ∶ ≤ 45. (c) P = 0.003. (d) X = 46.2598 and s∕ n = 0.4610. The null and alternative hypotheses are H0 ∶ ≤ 46 versus H1 ∶ > 46. z = (46.2598 − 46)∕0.1963 = 0.56. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of z = 0.56. Thus P = 0.2877.

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(e) X = 46.2598 and s∕ n = 0.4610. z.01 = 2.33. The confdence interval is 46.2598 ± 2.33(0.4610), or (45.19, 47.33).

18. (a) SE Mean = s∕ n, so 0.2767 = 2.2136∕ n. Solving for n yields n = 64. (b) z = (X − 0 )∕(SE Mean), so −1.03 = (X − 10.5)∕0.2767. Solving for X yields X = 10.215. (c) Since the alternative hypothesis is of the form H1 ∶ < 0 < the P -value is the area to the left of z = −1.03. Therefore P = 0.1515. 19. (a) SE Mean = s∕ n = 2.00819∕ 87 = 0.2153.

(b) X = 4.07114. From part (a), s∕ n = 0.2153. The null and alternative hypotheses are H0 ∶ = 3.5 versus H1 ∶ ≠ 3.5. z = (4.07114 − 3.5)∕0.2153 = 2.65. (c) Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of area to the right of z = 2.65 and the area to the left of z = −2.65. Thus P = 0.0080.

20. (a) One-tailed. The alternative hypothesis is of the form > 0 . (b) H0 ∶ ≤ 20 (c) P = 0.02471 (d) X = 21.537, = 5.5311, and n = 50, so ∕ n = 0.7822. The null and alternative hypotheses are H0 ∶ ≤ 20.2 versus H1 ∶ > 20.2. z = (21.537 − 20.2)∕0.7822 = 1.71. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of z = 1.71. Thus P = 0.0436. 21. (a) Two-tailed. The alternative hypothesis is of the form ≠ 0 .

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(b) H0 ∶ = 50 (c) P = 0.2625 (d) X = 52.17, = 11.6189, and n = 36, so ∕ n = 1.9635. The null and alternative hypotheses are H0 ∶ ≤ 48 versus H1 ∶ > 48. z = (52.17 − 48)∕1.9635 = 2.15. Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of z = 2.15 and the left of z = −2.15.

Thus P = 0.0316.

22. (a) One-tailed. The alternative hypothesis is of the form < 0 . (b) H0 ∶ = 69.5 (c) P = 0.007354 (d) X = 69.1, = 1.6397, and n = 100, so ∕ n = 0.16397. The null and alternative hypotheses are H0 ∶ ≥ 69.3 versus H1 ∶ < 69.3. z = (69.1 − 69.3)∕0.16397 = −1.22. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of z = −1.22. Thus P = 0.1112.

Section 6.2 1.

P = 0.5. The larger the P -value, the more plausible the null hypothesis.

2. (a) True. (b) False. We conclude that H0 is plausible. (c) True. (d) False. We conclude that H0 is plausible, and therefore that H1 is plausible as well.

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(iv). A P -value of 0.01 means that if H0 is true, then the observed value of the test statistic was in the most extreme 1% of its distribution. This is unlikely, but not impossible.

(v). A P -value of 0.50 means that if H0 is true, then the observed value of the test statistic was in the most extreme 50% of its distribution. This is not at all unusual, so H0 is plausible. We can never conclude that H1 is false, so therefore we conclude that H1 is plausible as well.

5. (a) True. The result is statistically signifcant at any level greater than or equal to 2%. (b) False. P > 0.01, so the result is not statistically signifcant at the 1% level. (c) True. The null hypothesis is rejected at any level greater than or equal to 2%. (d) False. P > 0.01, so the null hypothesis is not rejected at the 1% level. 6. (a) (ii) If we reject H0 , we conclude that ≠ 10, which means the scale is not in calibration. (b) (iii) If we do not we reject H0 , we conlude that might be equal to 10. We therefore conclude that the scale might be in calibration.

7. (a) (i) If we reject H0 , we conclude that > 100, which means that the classes are successful. (b) (iii) If we do not reject H0 , we conlude that might be less than or equal to 100. We therefore conclude that the classes might not be successsful.

8. (a) No. If the P -value is 0.20, then the result is not statistically signifcant at the 5% level. (b) Yes. If the P -value is 0.02, then the result is statistically signifcant at the 5% level.

10.

(iii) Since the decrease is statistically signifcant, it is reasonable to conclude that the homicide rate did actually decrease. However, we cannot determine what caused the decrease. H0 ∶ ≥ 5 is the best. If H0 is rejected, we can be convinced that < 5, and that the workplace is safe. If H0 is not rejected, then the workplace might not be safe.

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SECTION 6.2 11. (a) H0 ∶ ≤ 8. If H0 is rejected, we can conclude that > 8, and that the new battery should be used.

(b) H0 ∶ ≤ 60, 000. If H0 is rejected, we can conclude that > 60, 000, and that the new material should be used. (c) H0 ∶ = 10. If H0 is rejected, we can conclude that ≠ 10, and that the fow meter should be recalibrated.

12. (a) There is uncertainty in the mean measurement. It is possible to get a mean measurement less than 4 even when the true value is 4 or more. (b) H0 ∶ ≥ 4 versus H1 ∶ < 4. If H0 is not rejected, then it is plausible that the mean radon concentration may be 4 pCi/L or more. (c) X = 3.72, = 1.93, n = 75. The null and alternative hypotheses are H0 ∶ ≥ 4 versus H1 ∶ < 4. z = (3.72 − 4)∕(1.93∕ 75) = −1.26. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of z = −1.26. Thus P = 0.1038. The null hypothesis has some plausibility. Radon abatement is recommended. 13. (a) (ii) The scale is out of calibration. If H0 is rejected, we conclude that H0 is false, so ≠ 10. (b) (iii) The scale might be in calibration. If H0 is not rejected, we conclude that H0 is plausible, so might be equal to 10. (c) No. The scale is in calibration only if = 10. The strongest evidence in favor of this hypothesis would occur if X = 10. But since there is uncertainty in X, we cannot be sure even then that = 10.

14. (a) No. P = 0.30 is not small. Both the null and alternative hypotheses are therefore plausible. (b) No, we cannot conclude that the null hypothesis is true. The alternative hypothesis is also plausible.

15.

No, she cannot conclude that the null hypothesis is true, only that it is plausible.

16. (a) z = (21 − 20)∕(10∕ 100) = 1. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of z = 1. Therefore P = 0.1587. Since P is not small, we do not reject H0 .

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(b) z = (21 − 20)∕(10∕ 1000) = 3.16. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of z = 3.16. Therefore P = 0.0008. Since P is small, we reject H0 . (c) No. The di˙erence is of practical sigifcance only if > 25, which we cannot conclude. (d) When the sample size is larger, the standard error of the mean is smaller, so the magnitude of z is larger, and the P -value is smaller.

17. (a) False. We cannot conclude that the reduction is important. (b) True. We can conclude that there is a reduction.

18.

If a 95% confdence interval is (94.6, 98.3), then the value 100 lies outside the 95% confdence interval, so the P -value for the test of H0 ∶ = 100 versus H1 ∶ ≠ 100 is less than 0.05.

19.

(ii) less than 0.01. Since the value 6 is outside the 95% confdence interval, the P -value for testing H0 ∶ = 6 versus H1 ∶ ≠ 6 will be less than 0.01.

20.

(i) greater than 0.05. Since the value 1.4 is inside the 95% confdence interval, the P -value for testing H0 ∶ = 1.4 versus H1 ∶ ≠ 1.4 will be greater than 0.05.

21.

(i) H0 ∶ = 1.2. For either of the other two hypotheses, the P -value would be 0.025.

22.

(iii) 0.01 < P < 0.05. Since the value 4 is not inside the 95% confdence interval, P < 0.05. Since it is inside the 99% confdence interval, P > 0.01.

23. (a) Yes. The value 3.5 is greater than the upper confdence bound of 3.45. Quantities greater than the upper confdence bound will have P -values less than 0.05. Therefore P < 0.05. (b) No, we would need to know the 99% upper confdence bound to determine whether P < 0.01.

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24. (a) No, we would need to know the 99% lower confdence bound in order to determine whether P < 0.01. (b) Yes, since 50 is less than the 98% lower confdence bound, we know that P < 0.02. Therefore P < 0.05.

25.

Yes, we can compute the P -value exactly. Since the 95% upper confdence bound is 3.45, we know that 3.40 + 1.645 ∕ n = 3.45. Therefore ∕ n = 0.0304. The z-score is (3.40 − 3.50)∕0.0304 = −3.29. The P -value is 0.0005, which is less than 0.01.

26.

No, the confdence bound itself tells us only that P < 0.02. To determine the P -value more precisely, we would need to know the standard deviation of the sample mean. This is equal to ∕ n where is the population standard deviation and n is the sample size. We know that s = 5, but we do not know n.

27. (a) No. The P -value is 0.196, which is greater than 0.05. (b) The sample mean is 73.2461. The sample mean is therefore closer to 73 than to 73.5. The P -value for the null hypothesis = 73 will therefore be larger than the P -value for the null hypothesis = 73.5, which is 0.196. Therefore the null hypothesis = 73 cannot be rejected at the 5% level.

Section 6.3 1. (a) X = 100.01, s = 0.0264575, n = 3. There are 3 − 1 = 2 degrees of freedom. The null and alternative hypotheses are H0 ∶ = 100 versus H1 ∶ ≠ 100.

t = (100.01 − 100)∕(0.0264575∕ 3) = 0.6547.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of t = 0.6547 and to the left of t = −0.6547. From the t table, 0.50 < P < 0.80. A computer package gives P = 0.580. We conclude that the scale may well be calibrated correctly. (b) The t-test cannot be performed, because the sample standard deviation cannot be computed from a sample of size 1.

2. (a) Reject the assumptions: the value 221.03 is an outlier. (b) The assumptions appear to be met.

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(c) Reject the assumptions: the values are in increasing order, so this does not appear to be a simple random sample. 3. (a) H0 ∶ ≤ 5 versus H1 ∶ > 5 (b) X = 6.5, s = 1.9, n = 8. There are 8 − 1 = 7 degrees of freedom.

From part (a), the null and alternative hypotheses are H0 ∶ ≤ 5 versus H1 ∶ > 5. t = (6.5 − 5)∕(1.9∕ 8) = 2.233. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 2.233. From the t table, 0.025 < P < 0.050. A computer package gives P = 0.030.

(c) Yes, the P -value is small, so we conclude that > 5. 4. (a) The null and alternative hypotheses are H0 ∶ = 23 versus H1 ∶ ≠ 23. (b) X = 23.2, s = 0.2, n = 10. There are 10 − 1 = 9 degrees of freedom. t = (23.2 − 23)∕(0.2∕ 10) = 3.162.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of t = 3.162 and to the left of t = −3.162. From the t table, 0.01 < P < 0.02. A computer package gives P = 0.0115.

(c) Yes. Because the P -value is small, we can conclude that the mean percentage di˙ers from 23, so the machine should be recalibrated.

5. (a) X = 6.7, s = 3.9, n = 20. There are 20 − 1 = 19 degrees of freedom.

The null and alternative hypotheses are H0 ∶ ≥ 10 versus H1 ∶ < 10.

t = (6.7 − 10)∕(3.9∕ 20) = −3.784. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −3.784. From the t table, 0.0005 < P < 0.001. A computer package gives P = 0.0006. We can conclude that the system meet this guideline. (b) X = 6.7, s = 3.9, n = 20. There are 20 − 1 = 19 degrees of freedom.

The null and alternative hypotheses are H0 ∶ ≥ 7.5 versus H1 ∶ < 7.5.

t = (6.7 − 7.5)∕(3.9∕ 19) = −0.917. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −0.917. From the t table, 0.10 < P < 0.25. A computer package gives P = 0.1852. We cannot conclude that the system meet this guideline.

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X = 3.506667, s = 0.08406, n = 6. There are 6 − 1 = 5 degrees of freedom.

The null and alternative hypotheses are H0 ∶ ≤ 3.5 versus H1 ∶ > 3.5. t = (3.506667 − 3.5)∕(0.08406∕ 6) = 0.1943.

Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 0.1943. From the t table, P > 0.40. A computer package gives P = 0.4268. We cannot conclude that the mean cycle time is greater than 3.5 hours.

7. (a) 3.8

4.2

(b) Yes, the sample contains no outliers. X = 4.032857, s = 0.061244, n = 7. There are 7 − 1 = 6 degrees of freedom.

The null and alternative hypotheses are H0 ∶ = 4 versus H1 ∶ ≠ 4. t = (4.032857 − 4)∕(0.061244∕ 7) = 1.419.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of t = 1.419 and to the left of t = −1.419. From the t table, 0.20 < P < 0.50. A computer package gives P = 0.2056. It cannot be concluded that the mean thickness di˙ers from 4 mils. (c) 3.9

4.1

4.2

4.3

(d) No, the sample contains an outlier. 8. (a) H0 ∶ ≥ 15 versus H1 ∶ < 15 (b) X = 12.383, s = 2.9322, n = 12. There are 12 − 1 = 11 degrees of freedom. The null and alternative hypotheses are H0 ∶ ≥ 15 versus H1 ∶ < 15. t = (12.383 − 15)∕(2.9322∕ 12) = −3.091. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −3.091. From the t table, 0.005 < P < 0.01. A computer package gives P = 0.0051. (c) Yes, the P -value is small, so we conclude that < 15.

X = 457.8, s = 317.7, n = 18. There are 18 − 1 = 17 degrees of freedom.

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CHAPTER 6 The null and alternative hypotheses are H0 ∶ = 0 versus H1 ∶ ≠ 0. t = (457.8 − 0)∕(317.7∕ 18) = 6.114.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of t = 6.114 and to the left of t = −6.114. From the t table, P < 0.001. A computer package gives P = 1.15 × 10−5 .

Yes, we can conclude that the mean error is not equal to 0.

10.

X = 0.32, s = 0.05, n = 8. There are 8 − 1 = 7 degrees of freedom.

The null and alternative hypotheses are H0 ∶ ≤ 0.3 versus H1 ∶ > 0.3. t = (0.32 − 0.3)∕(0.05∕ 8) = 1.1314. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 1.1314. From the t table, 0.10 < P < 0.25. A computer package gives P = 0.148. We cannot conclude that the mean arsenic concentration is greater than 0.3 mg/kg.

11.

X = 27.125, s = 10.092, n = 8. There are 8 − 1 = 7 degrees of freedom.

The null and alternative hypotheses are H0 ∶ ≥ 40 versus H1 ∶ < 40. t = (27.125 − 40)∕(10.092∕ 8) = −3.6086. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −3.6086. From the t table, 0.001 < P < 0.005. A computer package gives P = 0.004320. We can conclude that the mean number of pits is less than 40.

12.

X = 1.102, s = 0.02, n = 65. There are 65 − 1 = 64 degrees of freedom.

The null and alternative hypotheses are H0 ∶ ≤ 1.1 versus H1 ∶ > 1.1. t64 = (1.102 − 1.1)∕(0.02∕ 65) = 0.80623. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the right of t = 0.80623. P = 0.21155. We cannot conclude that the mean extension is greater than 1.1 mm.

13.

X = 1.9, s = 21.2, n = 160. There are 160 − 1 = 159 degrees of freedom. The null and alternative hypotheses are H0 ∶ = 0 versus H1 ∶ ≠ 0.

t159 = (1.9 − 0)∕(21.2∕ 160) = 1.1336.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of t = 1.1336 and to the left of t = −1.1336. P = 0.25865. It is plausible that the mean error measurement is equal to 0.

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14.

X = 11.98, s = 0.19, n = 100. There are 100 − 1 = 99 degrees of freedom.

The null and alternative hypotheses are H0 ∶ = 12 versus H1 ∶ ≠ 12. t99 = (11.98 − 12)∕(0.19∕ 100) = −1.0526.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of t = 1.0526 and to the left of t = −1.0526. P = 0.29507. It is plausible that the mean fll volume is 12 ounces.

15.

X = 715, s = 24, n = 60. There are 60 − 1 = 59 degrees of freedom.

The null and alternative hypotheses are H0 ∶ ≥ 740 versus H1 ∶ < 740. t59 = (715 − 12)∕(24∕ 60) = −8.0867. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −8.0867. P = 2.0939 × 10−11 . We can conclude that the plant is not operating properly.

16.

X = 84.8, s = 0.5, n = 50. There are 50 − 1 = 49 degrees of freedom.

The null and alternative hypotheses are H0 ∶ = 85 versus H1 ∶ ≠ 85. t49 = (84.8 − 85)∕(0.5∕ 50) = −2.8284.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of t = 2.8284 and to the left of t = −2.8284. P = 0.00676. It is not plausible that the mean is on target.

17.

X = 39.6, s = 7, n = 58. There are 58 − 1 = 57 degrees of freedom.

The null and alternative hypotheses are H0 ∶ ≥ 40 versus H1 ∶ < 40. t57 = (39.6 − 40)∕(7∕ 58) = −0.4352. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −0.4352. P = 0.33254. We cannot conclude that the mean is less than 40 cm/sec.

18. (a) Two-tailed. The alternative hypothesis is of the form ≠ 0 . (b) H0 ∶ = 1 (c) No, the P -value is 0.08552, which is greater than 0.05.

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(d) z = 1.7303. The P -value is the sum of the areas under the standard normal curve to the right of 1.7303 and to the left of −1.7303. P = 0.083577. The result is similar to that obtained with the t-test. (e) t159 = (3.9 − 0)∕(21.2∕ 160) = 2.327. P = 0.02123.

19. (a) One-tailed. The alternative hypothesis is of the form > 0 . (b) H0 ∶ ≤ 90 (c) No, the P -value is 0.3153, which is greater than 0.05. (d) X = 90.3, s = 1.29. There are 4 degrees of freedom, so the sample size is n = 5. t4 = (90.3 − 87.5)∕1.29∕ 5) = 4.8535. The P -value is the area to the right of t = 4.8535. P = 0.00416. (e) It’s not a good idea because the sample size is small (n = 5).

20. (a) One-tailed. The alternative hypothesis is of the form > 0 . (b) H0 ∶ ≤ 5.5 (c) Yes, the P -value is 0.002, which is less than 0.01. (d) X = 5.92563, n = 5, s∕ n = 0.07046. There are 5 − 1 = 4 degrees of freedom. The null and alternative hypotheses are H0 ∶ ≥ 6.5 versus H1 ∶ < 6.5. t = (5.92563 − 6.5)∕0.07046 = −8.152. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −8.152. From the t table, 0.0005 < P < 0.001. A computer package gives P = 0.00062. (e) X = 5.92563, n = 5, s∕ n = 0.07046. There are 5 − 1 = 4 degrees of freedom. t4,.005 = 4.604. The confdence interval is 5.92563 ± 4.604(0.07046), or (5.6012, 6.2500).

21. (a) StDev = (SE Mean) N = 1.8389 11 = 6.0989. (b) t10,.025 = 2.228. The lower 95% confdence bound is 13.2874 − 2.228(1.8389) = 9.190. (c) t10,.025 = 2.228. The upper 95% confdence bound is 13.2874 + 2.228(1.8389) = 17.384.

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(d) t = (13.2874 − 16)∕1.8389 = −1.475.

22.

X = 1961.9, s = 38.8784. n = 366. There are 366 − 1 = 365 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≤ 1945.5 versus H1 ∶ X > 1945.5. t = (1961.9 − 1945.5)∕(38.784∕ 366) = 8.0702. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 8.0702. P = 5.107 × 10−15 . We conclude that the mean year of the record high is greater than 1945.5.

23.

Consecutive observations are not independent.

24.

The most recent occurence would on the average be greater than 1945.5, even if records are equally likely to occur in any year.

25.

X = 1960.9, s = 36.91. n = 47. There are 47 − 1 = 46 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≤ 1945.5 versus H1 ∶ X > 1945.5. t = (1960.9 − 1945.5)∕(36.91∕ 46) = 2.8592. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 2.8592. P = 0.0031814. We conclude that the mean year of the record high is greater than 1945.5.

26.

X = 1908.7, s = 31.345. n = 43. There are 43 − 1 = 42 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≥ 1945.5 versus H1 ∶ X < 1945.5. t = (1908.7 − 1945.5)∕(31.345∕ 43) = −7.6943. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −7.6943. P = 7.509 × 10−10 . We conclude that the mean year of the record low is less than 1945.5.

27.

No. For the data starting on January 4 and January 7, mean of year of record high is not signifcantly greater than 1945.5 (P = 0.2053, P = 0.1198). Otherwise everything is still signifcant. So the conclusions don’t change. Starting January 2 High X = 1964, s = 35.073, n = 45. There are 45 − 1 = 44 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≤ 1945.5 versus H1 ∶ X > 1945.5. t = (1964 − 1945.5)∕(35.073∕ 45) = 3.5384. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 3.5384. P = 0.00048179. We conclude that the mean year of the record high is greater than 1945.5. Starting January 2 Low X = 1913.8, s = 35.732, n = 43. There are 43 − 1 = 42 degrees of freedom.

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CHAPTER 6 The null and alternative hypotheses are H0 ∶ X ≥ 1945.5 versus H1 ∶ X < 1945.5. t = (1913.8 − 1945.5)∕(35.732∕ 43) = −5.8193. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the right of t = −5.8193. P = 3.6125 × 10−7 . We conclude that the mean year of the record low is less than 1945.5. Starting January 3 High X = 1962.8, s = 40.223, n = 39. There are 39 − 1 = 38 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≤ 1945.5 versus H1 ∶ X > 1945.5. t = (1962.8 − 1945.5)∕(40.228∕ 39) = 2.6892. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the right of t = 2.6892. P = 0.0052893. We conclude that the mean year of the record high is greater than 1945.5. Starting January 3 Low X = 1915.2, s = 35.278, n = 39. There are 39 − 1 = 38 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≥ 1945.5 versus H1 ∶ X < 1945.5. t = (1915.2 − 1945.5)∕(35.278∕ 39) = −5.3674. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −5.3674. P = 2.0962 × 10−5 . We conclude that the mean year of the record low is less than 1945.5. Starting January 4 High X = 1951.3, s = 44.917, n = 42. There are 42 − 1 = 41 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≤ 1945.5 versus H1 ∶ X > 1945.5. t = (1951.3 − 1945.5)∕(44.917∕ 42) = 0.83134. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 0.83134. P = 0.2053. We cannot conclude that the mean year of the record high is greater than 1945.5. Starting January 4 Low X = 1909, s = 29.426, n = 40. There are 39 − 1 = 38 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≥ 1945.5 versus H1 ∶ X < 1945.5. t = (1909 − 1945.5)∕(29.426∕ 40) = −7.8449. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −7.8449. P = 7.6311 × 10−10 . We conclude that the mean year of the record low is less than 1945.5. Starting January 5 High X = 1957.6, s = 39.153, n = 44. There are 44 − 1 = 43 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≤ 1945.5 versus H1 ∶ X > 1945.5. t = (1957.6 − 1945.5)∕(39.153∕ 44) = 2.0523. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 2.0523. P = 0.023132. We conclude that the mean year of the record high is greater than 1945.5.

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Starting January 5 Low X = 1916.3, s = 32.478, n = 41. There are 41 − 1 = 40 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≥ 1945.5 versus H1 ∶ X < 1945.5. t = (1916.3 − 1945.5)∕(32.478∕ 41) = −5.7632. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −5.7632. P = 5.0864 × 10−7 . We conclude that the mean year of the record low is less than 1945.5. Starting January 6 High X = 1959.5, s = 40.304, n = 44. There are 44 − 1 = 43 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≤ 1945.5 versus H1 ∶ X > 1945.5. t = (1959.5 − 1945.5)∕(40.304∕ 44) = 2.3078. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 2.3078. P = 0.012942. We conclude that the mean year of the record high is greater than 1945.5. Starting January 6 Low X = 1912.5, s = 30.365, n = 42. There are 41 − 1 = 40 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≥ 1945.5 versus H1 ∶ X < 1945.5. t = (1916.3 − 1945.5)∕(30.365∕ 41) = −7.043. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −7.043. P = 7.1652 × 10−9 . We conclude that the mean year of the record low is less than 1945.5. Starting January 7 High X = 1952.6, s = 36.647, n = 38. There are 38 − 1 = 37 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≤ 1945.5 versus H1 ∶ X > 1945.5. t = (1952.6 − 1945.5)∕(36.647∕ 38) = 1.1952. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 1.1952. P = 0.11981. We cannot conclude that the mean year of the record high is greater than 1945.5. Starting January 7 Low X = 1915.5, s = 30.927, n = 43. There are 43 − 1 = 42 degrees of freedom. The null and alternative hypotheses are H0 ∶ X ≥ 1945.5 versus H1 ∶ X < 1945.5. t = (1915.5 − 1945.5)∕(30.927∕ 43) = −6.3584. Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −7.043. P = 6.0566 × 10−8 . We conclude that the mean year of the record low is less than 1945.5.

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CHAPTER 6

X = 35, n = 1000, p = 35∕1000 = 0.035.

The null and alternative hypotheses are H0 ∶ p ≥ 0.05 versus H1 ∶ p < 0.05. z = (0.035 − 0.05)∕ 0.05(1 − 0.05)∕1000 = −2.18. Since the alternative hypothesis is of the form p < p0 , the P -value is the area to the left of z = −2.18, so P = 0.0146. We can conclude that the goal has been met.

X = 281, n = 444, p = 281∕444 = 0.632883.

The null and alternative hypotheses are H0 ∶ p ≤ 0.60 versus H1 ∶ p > 0.60. z = (0.632883 − 0.60)∕ 0.60(1 − 0.60)∕444 = 1.41. Since the alternative hypothesis is of the form p > p0 , the P -value is the area to the right of z = 1.41, so P = 0.0793. The null hypothesis is somewhat suspect, but we would probably not consider the evidence strong enough to frmly conclude that more than 60% of HIV-positive smokers are male.

X = 29, n = 50, p = 29∕50 = 0.58.

The null and alternative hypotheses are H0 ∶ p ≤ 0.50 versus H1 ∶ p > 0.50. z = (0.58 − 0.50)∕ 0.50(1 − 0.50)∕50 = 1.13. Since the alternative hypothesis is of the form p > p0 , the P -value is the area to the right of z = 1.13, so P = 0.1292. We cannot conclude that more than half of bathroom scales underestimate weight.

X = 49, n = 73, p = 0.671233.

The null and alternative hypotheses are H0 ∶ p ≤ 0.6 versus H1 ∶ p > 0.6. z = (0.671233 − 0.6)∕ 0.6(1 − 0.6)∕73 = 1.24. Since the alternative hypothesis is of the form p > p0 , the P -value is the area to the right of z = 1.24, so P = 0.1075. We cannot conclude that more than half of the vehicles exceed the speed limit.

X = 274, n = 500, p = 274∕500 = 0.548.

The null and alternative hypotheses are H0 ∶ p ≤ 0.50 versus H1 ∶ p > 0.50. z = (0.548 − 0.50)∕ 0.50(1 − 0.50)∕500 = 2.15. Since the alternative hypothesis is of the form p > p0 , the P -value is the area to the right of z = 2.15, so P = 0.0158. We can conclude that more than half of residents are opposed to building a new shopping mall.

X = 51, n = 85, p = 51∕85 = 0.6.

The null and alternative hypotheses are H0 ∶ p ≤ 0.50 versus H1 ∶ p > 0.50.

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z = (0.6 − 0.50)∕ 0.50(1 − 0.50)∕85 = 1.84. Since the alternative hypothesis is of the form p > p0 , the P -value is the area to the right of z = 1.84, so P = 0.0329. We can conclude that more than half of ACL injuries are to the left knee.

X = 54, n = 124, p = 54∕124 = 0.43548.

The null and alternative hypotheses are H0 ∶ p ≥ 0.45 versus H1 ∶ p < 0.45. z = (0.43548 − 0.45)∕ 0.45(1 − 0.45)∕124 = −0.32. Since the alternative hypothesis is of the form p < p0 , the P -value is the area to the left of z = −0.32, so P = 0.3745. We cannot conclude that less than 45% of nurses feel that a busy workload has a positive e˙ect on their enjoyment of their job.

X = 12, n = 300, p = 12∕300 = 0.04.

The null and alternative hypotheses are H0 ∶ p ≥ 0.08 versus H1 ∶ p < 0.08. z = (0.04 − 0.08)∕ 0.08(1 − 0.08)∕300 = −2.55. Since the alternative hypothesis is of the form p < p0 , the P -value is the area to the left of z = −2.55, so P = 0.0054. The machine can be qualifed.

X = 42, n = 300, p = 42∕300 = 0.14.

The null and alternative hypotheses are H0 ∶ p = 0.12 versus H1 ∶ p ≠ 0.12. z = (0.14 − 0.12)∕ 0.12(1 − 0.12)∕300 = 1.07. Since the alternative hypothesis is of the form p = p0 , the P -value is the sum of the areas to the right of z = 1.07 and to the left of z = −1.07, so P = 0.2846. We cannot conclude that the locus is not in Hardy–Weinberg equilibrium.

10.

X = 42, n = 200, p = 42∕200 = 0.21.

The null and alternative hypotheses are H0 ∶ p ≤ 0.15 versus H1 ∶ p > 0.15. z = (0.21 − 0.15)∕ 0.15(1 − 0.15)∕200 = 2.38. Since the alternative hypothesis is of the form p > p0 , the P -value is the area to the right of z = 2.38, so P = 0.0087. We can conclude that more than 15% of computer-related workers have changed jobs in the past six months.

11.

X = 517, n = 1294, p = 517∕1294 = 0.39954.

The null and alternative hypotheses are H0 ∶ p ≤ 0.35 versus H1 ∶ p > 0.35.

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z = (0.39954 − 0.35)∕ 0.35(1 − 0.35)∕1294 = 3.74. Since the alternative hypothesis is of the form p > p0 , the P -value is the area to the right of z = 3.74, so P = 0.0001. We can conclude that more than 35% of of adults performed volunteer work during the past year.

12. (a) One-tailed. The alternative hypothesis is of the form p < p0 . (b) H0 ∶ p ≥ 0.4 (c) Yes, P = 0.001, which is less than 0.02. (d) The sample proportion is 0.304167. Therefore the P -value for the null hypothesis p ≥ 0.45 will be less than the P -value for the null hypothesis p ≥ 0.40, which is 0.001. It follows that the null hypothesis p ≥ 0.45 can be rejected at the 2% level.

(e) X = 73, n = 240, p = 73∕240 = 0.304167.

The null and alternative hypotheses are H0 ∶ p ≤ 0.25 versus H1 ∶ p > 0.25. z = (0.304167 − 0.25)∕

.25(1 − 0.25)∕240 = 1.94.

Since the alternative hypothesis is of the form p > p0 , the P -value is the area to the right of z = 1.94. Thus P = 0.0262.

(f) X = 73, n = 240, p̃ = (73 + 2)∕(240 + 4) = 0.307377, z.05 = 1.645. The confdence interval is 0.307377 ± 1.645 0.307377(1 − 0.307377)∕(240 + 4), or (0.2588, 0.3560).

13. (a) Sample p = p = 345∕500 = 0.690. (b) The null and alternative hypotheses are H0 ∶ p ≥ 0.7 versus H1 ∶ p < 0.7. n = 500. From part (a), p = 0.690. z = (0.690 − 0.700)∕ 0.7(1 − 0.7)∕500 = −0.49.

(c) Since the alternative hypothesis is of the form p < p0 , the P -value is the area to the left of z = −0.49. Thus P = 0.3121.

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SECTION 6.4

14.

Consecutive observations are not independent.

15.

X = 42, n = 53, p = 42∕53 = 0.79245.

The null and alternative hypotheses are H0 ∶ p ≤ 0.5 versus H1 ∶ p > 0.5. z = (0.79245 − 0.5)∕ 0.5(1 − 0.5)∕53 = 4.2582. P = 1.0305 × 10−5 . We conclude that it is more likely for the record high to be more recent.

16.

The most recent occurence would on the average be greater than 1945.5, even if records are equally likely to occur in any year.

17.

X = 30, n = 47, p = 30∕47 = 0.63830.

The null and alternative hypotheses are H0 ∶ p ≤ 0.5 versus H1 ∶ p > 0.5. z = (0.63830 − 0.5)∕ 0.5(1 − 0.5)∕47 = 1.8962. P = 0.02896. We conclude that it is more likely for the record high to be more recent.

18.

X = 5, n = 43, p = 5∕43 = 0.11628.

The null and alternative hypotheses are H0 ∶ p ≥ 0.5 versus H1 ∶ p < 0.5. z = (0.11628 − 0.5)∕ 0.5(1 − 0.5)∕43 = −5.0325. P = 2.421 × 10−7 . We conclude that it is more likely for the record high to be more recent.

19.

For Exercise 17: Starting January 2: X = 30, n = 45, P = 0.01267366 Starting January 3: X = 25, n = 39, P = 0.03908454 Starting January 4: X = 23, n = 42, P = 0.26854699 Starting January 5: X = 29, n = 44, P = 0.01740424 Starting January 6: X = 27, n = 44, P = 0.06583401 Starting January 7: X = 21, n = 38, P = 0.25820613 For Exercise 18: Starting January 2: X = 9, n = 43, P = 6.8974 × 10−5 Starting January 3: X = 8, n = 39, P = 1.1528 × 10−4 Starting January 4: X = 5, n = 40, P = 1.0507 × 10−6 Starting January 5: X = 6, n = 41, P = 2.9626 × 10−6 Starting January 6: X = 5, n = 42, P = 3.9523 × 10−7 Starting January 7: X = 7, n = 43, P = 4.8792 × 10−6

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For January 4, 6, and 7, P -values for record high are 0.2685, 0.0658, 0.2582. All other results for the record high are between 0.01 and 0.05. All results for the record low are less than 0.001. Conclusions do not change.

Section 6.5 1.

X = 8.5, X = 1.9, nX = 58, Y = 11.9, Y = 3.6, nY = 58.

The null and alternative hypotheses are H0 ∶ X − Y ≥ 0 versus H1 ∶ X − Y < 0. z = (8.5−11.9−0)∕ 1.92 ∕58 + 3.62 ∕58 = −6.36. Since the alternative hypothesis is of the form X − Y < Δ, the P -value is the area to the left of z = −6.36. Thus P

We can conclude that the mean hospital stay is shorter for patients receiving C4A-rich plasma.

X = 20.95, X = 14.5, nX = 482, Y = 22.79, Y = 15.6, nY = 614.

The null and alternative hypotheses are H0 ∶ X − Y ≥ 0 versus H1 ∶ X − Y < 0.

z = (20.95 − 22.79 − 0)∕ 14.52 ∕482 + 15.62 ∕614 = −2.02. Since the alternative hypothesis is of the form X − Y < Δ, the P -value is the area to the left of z = −2.02. Thus P = 0.0217. We can conclude that the mean weight of male spotted founder is greater than that of females.

X = 169.9, sX = 24.8, nX = 110, Y = 163.3, sY = 25.8, nY = 225. The number of degrees of freedom is 4 52 24.82 25.82 + 110 225 = = 224, rounded down to the nearest integer. 2 2 (25.82 ∕225)2 (24.8 ∕110) + 110 − 1 225 − 1 t224 = (169.9 − 163.3 − 0)∕ 24.82 ∕110 + 25.82 ∕225 = 2.2572.

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is the sum of the areas to the right of t = 2.2572 and to the left of t = −2.2572. Thus P = 0.02496. We can conclude that the mean peak systolic blood pressure di˙ers between these two groups of women.

X = 2.49, sX = 4.87, nX = 413, Y = 1.22, sY = 3.24, nY = 382.

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The number of degrees of freedom is 4 52 4.872 3.242 + 413 382 = = 721, rounded down to the nearest integer. (4.872 ∕413)2 (3.242 ∕382)2 + 413 − 1 382 − 1 t721 = (2.49 − 1.22 − 0)∕ 4.872 ∕413 + 3.242 ∕382 = 4.3585.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0.

Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 4.3585. Thus P = 7.501 × 10−6 . We can conclude that the mean number of energy drinks is greater for male students than for female students.

X = 68.4, sX = 12.1, nX = 50, Y = 71.3, sY = 14.2, nY = 50. The number of degrees of freedom is 4 52 12.12 14.22 + 50 50 = = 95, rounded down to the nearest integer. 2 2 (14.22 ∕50)2 (12.1 ∕50) + 50 − 1 50 − 1 t95 = (68.4 − 71.3 − 0)∕ 12.12 ∕50 + 14.22 ∕50 = −1.0992.

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is the sum of the areas to the right of t = 1.0992 and to the left of t = −1.0992.

Thus P = 0.27447. We cannot conclude that the mean scored di˙er between online and paper tests.

6. (a) The 60 counts may not be independent if they are taken in consecutive time periods.

(b) X = 73.8, sX = 5.2, nX = 60, Y = 76.1, sY = 4.1, nY = 60. The number of degrees of freedom is 4 2 52 5.2 4.12 + 60 60 = = 111, rounded down to the nearest integer. (5.22 ∕60)2 (4.12 ∕60)2 + 60 − 1 60 − 1 t111 = (73.8 − 76.1 − 0)∕ 5.22 ∕60 + 4.12 ∕60 = −2.6904.

The null and alternative hypotheses are H0 ∶ X − Y ≥ 0 versus H1 ∶ X − Y < 0.

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Since the alternative hypothesis is of the form X − Y < Δ, the P -value is the area to the left of t = −2.6904. Thus P = 0.0041196. We can conclude that machine 2 is faster. 7. (a) X = 7.79, sX = 1.06, nX = 80, Y = 7.64, sY = 1.31, nY = 80. The number of degrees of freedom is 4 52 1.062 1.312 + 80 80 = = 151, rounded down to the nearest integer. 2 2 (1.312 ∕80)2 (1.06 ∕80) + 80 − 1 80 − 1 t151 = (7.79 − 7.64 − 0)∕ 1.062 ∕80 + 1.312 ∕80 = 0.79616.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 0.79616. Thus P = 0.21359. We cannot conclude that the mean score on one-tailed questions is greater.

(b) X = 7.79, sX = 1.06, nX = 80, Y = 7.64, sY = 1.31, nY = 80. The number of degrees of freedom is 4 52 1.062 1.312 + 80 80 = = 151, rounded down to the nearest integer. 2 2 (1.312 ∕80)2 (1.06 ∕80) + 80 − 1 80 − 1 t151 = (7.79 − 7.64 − 0)∕ 1.062 ∕80 + 1.312 ∕80 = 0.79616.

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is the sum of the areas to the right of t = 0.79616 and to the left of t = −0.79616. Thus P = 0.42719. We cannot conclude that the mean score on one-tailed questions di˙ers from the mean score on two-tailed questions.

8. (a) X = 495.6, sX = 19.4, nX = 50, Y = 481.2, sY = 14.3, nY = 50. The number of degrees of freedom is 4 52 19.42 14.32 + 50 50 = = 90, rounded down to the nearest integer. 2 2 (14.32 ∕50)2 (19.4 ∕50) + 50 − 1 50 − 1

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t90 = (495.6 − 481.2 − 0)∕ 19.42 ∕50 + 14.32 ∕50 = 4.2249.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 4.2249.

Thus P = 2.8576 × 10−5 . We can conclude that the mean speed for the new chips is greater.

. (b) X = 495.6, sX = 19.4, nX = 50, Y = 391.2, sY = 17.2, nY = 60. The number of degrees of freedom is 4 52 19.42 17.22 + 50 60 = = 98, rounded down to the nearest integer. 2 2 (17.22 ∕60)2 (19.4 ∕50) + 50 − 1 60 − 1 t98 = (495.6 − 391.2 − 0)∕ 19.42 ∕50 + 17.22 ∕60 = 1.2466.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 1.2466.

Thus P = 0.10776. The data do not provide convincing evidence for this claim.

9. (a) X = 4.7, sX = 7.2, nX = 77, Y = 2.6, sY = 5.9, nY = 79. The number of degrees of freedom is 4 2 52 7.2 5.92 + 77 79 = = 146, rounded down to the nearest integer. 2 2 (5.92 ∕79)2 (7.2 ∕77) + 77 − 1 79 − 1 t146 = (4.7 − 2.6 − 0)∕ 7.22 ∕77 + 5.92 ∕79 = 1.9898.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 1.9898.

Thus P = 0.024243. We can conclude that the mean weight loss is greater for those on the low-carbohydrate diet.

(b) X = 4.7, sX = 7.2, nX = 77, Y = 2.6, sY = 5.9, nY = 79. The number of degrees of freedom is

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CHAPTER 6 4 =

7.22 5.92 + 77 79

(7.22 ∕77)2 (5.92 ∕79)2 + 77 − 1 79 − 1

= 146, rounded down to the nearest integer.

t146 = (4.7 − 2.6 − 1)∕ 7.22 ∕77 + 5.92 ∕79 = 1.0423.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 1 versus H1 ∶ X − Y > 1. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 1.0423.

Thus P = 0.14951. We cannot conclude that the mean weight loss on the low-carbohydrate diet is more than 1 kg greater than that of the low-fat diet.

10. (a) X = 5.2, sX = 3.1, nX = 60, Y = 6.1, sY = 2.8, nY = 72. The number of degrees of freedom is 4 2 52 3.1 2.82 + 60 72 = = 120, rounded down to the nearest integer. 2 2 (2.82 ∕72)2 (3.1 ∕60) + 60 − 1 72 − 1 t120 = (5.2 − 6.1 − 0)∕ 3.12 ∕60 + 2.82 ∕72 = −1.7351.

The null and alternative hypotheses are H0 ∶ X − Y ≥ 0 versus H1 ∶ X − Y < 0. Since the alternative hypothesis is of the form X − Y < Δ, the P -value is the area to the left of t = −1.7351.

Thus P = 0.042646. We can conclude that the mean checkout time is less for people who use the self-service lane.

(b) We cannot conclude that the mean checkout time would decrease if everyone used the self-service checkout lane. One reason is that those currently using the cashier may have more items than those currently using the self-service lane.

11.

X = 348, sX = 80, nX = 48, Y = 309, sY = 89, nY = 46. The number of degrees of freedom is 4 2 52 892 80 + 48 46 = = 90, rounded down to the nearest integer. 2 2 (892 ∕46)2 (80 ∕48) + 48 − 1 46 − 1 t90 = (348 − 309 − 0)∕ 802 ∕48 + 892 ∕46 = 2.2312.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0.

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SECTION 6.5

Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 2.2312. Thus P = 0.014076. We can conclude that the mean distance walked for patients using a treadmill is greater than the mean for the controls.

12.

X = 27.2, sX = 1.2, nX = 40, Y = 28.1, sY = 2, nY = 40. The number of degrees of freedom is 4 2 52 22 1.2 + 40 40 = = 63, rounded down to the nearest integer. 2 2 (22 ∕40)2 (1.2 ∕40) + 40 − 1 40 − 1 t63 = (27.2 − 28.1 − 0)∕ 1.22 ∕40 + 22 ∕40 = −2.4405.

The null and alternative hypotheses are H0 ∶ X − Y ≥ 0 versus H1 ∶ X − Y < 0. Since the alternative hypothesis is of the form X − Y < Δ, the P -value is the area to the left of t = −2.4405.

Thus P = 0.0087462. We can conclude that this type of car gets better mileage with premium fuel.

13. (a) X = 3.05, sX = 0.34157, nX = 4, Y = 1.8, sY = 0.90921, nY = 4. The number of degrees of freedom is 4 52 0.341572 0.909212 + 4 4 = = 3, rounded down to the nearest integer. 2 2 (0.909212 ∕4)2 (0.34157 ∕4) + 4−1 4−1 t3 = (3.05 − 1.8 − 0)∕ 0.341572 ∕4 + 0.909212 ∕4 = 2.574.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 2.574. From the t table, 0.025 < P < 0.050. A computer package gives P = 0.041. We can conclude that the mean strength of crayons made with dye B is greater than that made with dye A. (b) X = 3.05, sX = 0.34157, nX = 4, Y = 1.8, sY = 0.90921, nY = 4. The number of degrees of freedom is 4 52 0.341572 0.909212 + 4 4 = = 3, rounded down to the nearest integer. 2 2 (0.909212 ∕4)2 (0.34157 ∕4) + 4−1 4−1

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CHAPTER 6

t3 = (3.05 − 1.8 − 1)∕ 0.341572 ∕4 + 0.909212 ∕4 = 0.5148.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 1 versus H1 ∶ X − Y > 1. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 0.5148. From the t table, 0.25 < P < 0.40. A computer package gives P = 0.321. We cannot conclude that the mean strength of crayons made with dye B exceeds that of those made with dye A by more than 1 J.

14.

X = 255.8, sX = 8.216325, nX = 6, Y = 270.7375, sY = 11.902933, nY = 8. The number of degrees of freedom is 4 52 8.2163252 11.9029332 + 6 8 = = 11, rounded down to the nearest integer. 2 2 (11.9029332 ∕8)2 (8.216325 ∕6) + 6−1 8−1 t11 = (255.8 − 270.7375 − 0)∕ 8.2163252 ∕6 + 11.9029332 ∕8 = −2.776.

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is the sum of the areas to the right of t = 2.776 and to the left of t = −2.776. From the t table, 0.01 < P < 0.02. A computer package gives P = 0.018. We can conclude that the dissolve times di˙er between the two shapes.

15.

X = 2.31, sX = 0.89, nX = 15, Y = 2.8, sY = 1.1, nY = 15. The number of degrees of freedom is 4 52 0.892 1.12 + 15 15 = = 26, rounded down to the nearest integer. 2 2 (1.12 ∕15)2 (0.89 ∕15) + 15 − 1 15 − 1 t26 = (2.31 − 2.8 − 0)∕ 0.892 ∕15 + 1.12 ∕15 = −1.341.

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is the sum of the areas to the right of t = 1.341 and to the left of t = −1.341. From the t table, 0.10 < P < 0.20. A computer package gives P = 0.191. We cannot conclude that the penetration resistance di˙ers between the two depths.

16. (a) X = 47.79, sX = 2.19, nX = 14, Y = 47.15, sY = 2.65, nY = 40.

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SECTION 6.5

The number of degrees of freedom is 4 52 2.192 2.652 + 14 40 = = 27, rounded down to the nearest integer. (2.192 ∕14)2 (2.652 ∕40)2 + 14 − 1 40 − 1 t27 = (47.79 − 47.15 − 0)∕ 2.192 ∕14 + 2.652 ∕40 = 0.8891.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 0.8891.

From the t table, 0.10 < P < 0.25. A computer package gives P = 0.191. We cannot conclude that the time to perform a lift of low diÿculty is less with the video system than with the tagman system.

(b) X = 69.33, sX = 6.26, nX = 12, Y = 58.5, sY = 5.59, nY = 24. The number of degrees of freedom is 4 52 6.262 5.592 + 12 24 = = 19, rounded down to the nearest integer. 2 2 (5.592 ∕24)2 (6.26 ∕12) + 12 − 1 24 − 1 t19 = (69.33 − 58.5 − 0)∕ 6.262 ∕12 + 5.592 ∕24 = 5.067.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 5.067.

From the t table, P < 0.0005. A computer package gives P = 3.4 × 10−5 . We can conclude that the time to perform a lift of moderate diÿculty is less with the video system than with the tagman system.

(c) X = 109.71, sX = 17.02, nX = 17, Y = 84.52, sY = 13.51, nY = 29. The number of degrees of freedom is 4 52 17.022 13.512 + 17 29 = = 27, rounded down to the nearest integer. 2 2 (13.512 ∕29)2 (17.02 ∕17) + 17 − 1 29 − 1 t27 = (109.71 − 84.52 − 0)∕ 17.022 ∕17 + 13.512 ∕29 = 5.215.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 5.215. From the t table, P < 0.0005. A computer package gives P = 8.6 × 10−6 . We can conclude that the time to perform a lift of high diÿculty is less with the video system than with the

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CHAPTER 6

tagman system.

17.

X = 24.8, sX = 0.91043, nX = 10, Y = 19.5, sY = 1.3106, nY = 10. The number of degrees of freedom is 4 52 0.910432 1.31062 + 10 10 = = 16, rounded down to the nearest integer. 2 2 (1.31062 ∕10)2 (0.91043 ∕10) + 10 − 1 10 − 1 t16 = (24.8 − 19.5 − 0)∕ 0.910432 ∕10 + 1.31062 ∕10 = 10.502.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 10.502. From the t table, P < 0.0005. A computer package gives P = 6.914 × 10−9 . We can conclude that the mean height of plants grown with slow-release fertilizer is greater.

18.

X = 66.1667, sX = 19.630758, nX = 6, Y = 28.2, sY = 17.767949, nY = 5. Since the population standard deviations are assumed to be equal, estimate their common value with the pooled standard deviation v (nX − 1)s2X + (nY − 1)s2Y sp = = 18.825613. nX + nY − 2 The number of degrees of freedom is = nX + nY − 2 = 6 + 5 − 2 = 9. t9 =

66.1667 − 28.2

= 3.331.

18.825613 1∕6 + 1∕5

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is the sum of the areas to the right of t = 3.331 and to the left of t = −3.331. From the t table, 0.002 < P < 0.01. A computer package gives P = 0.0088. We can conclude that the weights di˙er.

19.

X = 53.0, sX = 1.41421, nX = 6, Y = 54.5, sY = 3.88587, nY = 6. The number of degrees of freedom is 4 52 1.414212 3.885872 + 6 6 = = 6, rounded down to the nearest integer. 2 2 (3.885872 ∕6)2 (1.41421 ∕6) + 6−1 6−1

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SECTION 6.5

t6 = (53.0 − 54.5 − 0)∕ 1.414212 ∕6 + 3.885872 ∕6 = −0.889.

The null and alternative hypotheses are H0 ∶ X − Y ≥ 0 versus H1 ∶ X − Y < 0. Since the alternative hypothesis is of the form X − Y < Δ, the P -value is the area to the left of t = −0.889. From the t table, 0.10 < P < 0.25. A computer package gives P = 0.204. We cannot conclude that the mean cost of the new method is less than that of the old method.

20.

X = 70, sX = 12.4, nX = 25, Y = 75, sY = 13.6, nY = 11. The number of degrees of freedom is 4 52 12.42 13.62 + 25 11 = = 17, rounded down to the nearest integer. 2 2 (13.62 ∕11)2 (12.4 ∕25) + 25 − 1 11 − 1 t17 = (70 − 75 − 0)∕ 12.42 ∕25 + 13.62 ∕11 = −1.043.

The null and alternative hypotheses are H0 ∶ X − Y ≥ 0 versus H1 ∶ X − Y ≤ 0.

Since the alternative hypothesis is of the form X − Y ≤ Δ, the P -value is the sum of the areas to the left of t = −1.043. From the t table, 0.010 < P < 0.25. A computer package gives P = 0.1557. We cannot conclude that the mean life span of those exposed to the mummy’s curse is less than the mean of those not exposed.

21.

X = 8.38, sX = 0.96, nX = 15, Y = 9.83, sY = 1.02, nY = 15. The number of degrees of freedom is 4 52 0.962 1.022 + 15 15 = = 27, rounded down to the nearest integer. 2 2 (1.022 ∕15)2 (0.96 ∕15) + 15 − 1 15 − 1 t27 = (8.38 − 9.83 − 0)∕ 0.962 ∕15 + 1.022 ∕15 = −4.009.

The null and alternative hypotheses are H0 ∶ X − Y ≥ 0 versus H1 ∶ X − Y < 0. Since the alternative hypothesis is of the form X − Y < Δ, the P -value is the area to the left of t = −4.009. From the t table, P < 0.0005. A computer package gives P = 0.000216. We can conclude that 6" spacing provides a higher mean failure pressure.

22.

X = 9.30, sX = 2.71, nX = 14, Y = 9.47, sY = 2.22, nY = 7. The number of degrees of freedom is

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CHAPTER 6 4 =

2.712 2.222 + 14 7

(2.712 ∕14)2 (2.222 ∕7)2 + 14 − 1 7−1

= 14, rounded down to the nearest integer.

t14 = (9.30 − 9.47 − 0)∕ 2.712 ∕14 + 2.222 ∕7 = −0.1534.

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is the sum of the areas to the right and to the left of t = −0.1534. From the t table, P > 0.80. A computer package gives P = 0.8803. We cannot conclude that the mean iron oxide content di˙ers between limenite grain and limenite lamella.

23.

X = 8.27, sX = 0.62, nX = 15, Y = 6.11, sY = 1.31, nY = 15. The number of degrees of freedom is 4 52 0.622 1.312 + 15 15 = = 12, rounded down to the nearest integer. 2 2 (1.312 ∕15)2 (0.62 ∕15) + 15 − 1 15 − 1 t12 = (8.27 − 6.11 − 0)∕ 0.622 ∕15 + 1.312 ∕15 = 4.7129.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0.

Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area to the right of t = 3.1601. From the t table, P < 0.0005. A computer package gives P = 0.0002515. We can conclude that the mean load at failure is greater for rounded connections than for double-rounded connections.

24.

X = 112.5, sX = 13.9, nX = 16, Y = 110.34, sY = 8.65, nY = 29. The number of degrees of freedom is 52 4 13.92 8.652 + 16 29 = = 21, rounded down to the nearest integer. 2 2 (8.652 ∕29)2 (13.9 ∕16) + 16 − 1 29 − 1 t21 = (112.5 − 110.34 − 0)∕ 13.92 ∕16 + 8.652 ∕29 = 0.5642.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y ≥ 0.

Since the alternative hypothesis is of the form X − Y ≥ Δ, the P -value is the area to the right of t = 0.5642.

From the t table, 0.25 < P < 0.40. A computer package gives P = 0.2893. We cannot conclude that the mean systolic blood pressure is higher in those su˙ering from depression.

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SECTION 6.5

25. (a) X = 27.3, sX = 5.2, nX = 12, Y = 32.7, sY = 4.1, nY = 14. The number of degrees of freedom is 4 2 52 5.2 4.12 + 12 14 = = 20, rounded down to the nearest integer. 2 2 (4.12 ∕14)2 (5.2 ∕12) + 12 − 1 14 − 1 t20 = (27.3 − 32.7 − 0)∕ 5.22 ∕12 + 4.12 ∕14 = −2.9056.

The null and alternative hypotheses are H0 ∶ X − Y ≥ 0 versus H1 ∶ X − Y < 0.

Since the alternative hypothesis is of the form X − Y < Δ, the P -value is the area to the left of t = −2.9056. From the t table, 0.001 < P < 0.005. A computer package gives P = 0.0043724. We can conclude that the mean time to sleep is less for the new drug.

26.

X = 17.714, sX = 4.4615, nX = 7, Y = 28.714, sY = 7.3872, nY = 7. The number of degrees of freedom is 4 52 4.46152 7.38722 + 7 7 = = 9, rounded down to the nearest integer. 2 2 (7.38722 ∕7)2 (4.4615 ∕7) + 7−1 7−1 t9 = (17.714 − 28.714 − 0)∕ 4.46152 ∕7 + 7.38722 ∕7 = −3.372.

The null and alternative hypotheses are H0 ∶ X − Y ≥ 0 versus H1 ∶ X − Y < 0. Since the alternative hypothesis is of the form X − Y < Δ, the P -value is the area to the left of t = −3.372. From the t table, 0.001 < P < 0.005. A computer package gives P = 0.004113. We can conclude that the mean recovery time is less for patients receiving the new treatment.

27.

X = 80.1, sX = 9.3, nX = 14, Y = 76.8, sY = 8.6, nY = 12. The number of degrees of freedom is 4 2 52 9.3 8.62 + 14 12 = = 23, rounded down to the nearest integer. 2 2 (8.62 ∕12)2 (9.3 ∕14) + 14 − 1 12 − 1 t23 = (80.1 − 76.8 − 0)∕ 9.32 ∕14 + 8.62 ∕12 = 0.9394.

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y ≥ 0.

Since the alternative hypothesis is of the form X − Y ≥ Δ, the P -value is the area to the right of t = 0.9394. From the t table, 0.10 < P < 0.25. A computer package gives P = 0.1787.

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We cannot conclude that the traditional classroom instruction was more e˙ective.

28. (a) One-tailed. The alternative hypothesis is of the form X − Y > Δ. (b) H0 ∶ X − Y ≤ 0 (c) Yes. P = 0.003, which is less than 0.01.

29. (a) SE Mean = StDev/ N = 0.482∕ 6 = 0.197. (b) StDev = (SE Mean) N = 0.094 13 = 0.339. (c) X − Y = 1.755 − 3.239 = −1.484. (d) t =

1.755 − 3.239 0.4822 ∕6 + 0.0942

= −6.805.

30. (a) One-tailed. The alternative hypothesis is of the form X − Y < Δ. (b) H0 ∶ X − Y ≥ 0 (c) Yes. P = 0.04438, which is less than 0.05. (d) Since P > 0.01, the 99% upper confdence bound is greater than 0, and is therefore positive. (e) z = −1.7082, so P = 0.0438. The results are similar to those of the t-test. 31. (a) Two-tailed. The alternative hypothesis is of the form X − Y ≠ Δ. (b) H0 ∶ X − Y = 0 (c) No. P = 0.01466, which is greater than 0.01. (d) Since P < 0.02, the 98% confdence interval would not contain 0. (e) z = 2.4883, so P = 0.012836. The results are similar to those of the t-test.

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SECTION 6.5

32.

X = 2480.2, sX = 767.78, nX = 90, Y = 2674.9, sY = 867.23, nY = 803. The number of degrees of freedom is 4 52 767.782 867.232 + 90 803 = = 116, rounded down to the nearest integer. 2 2 (867.232 ∕803)2 (767.78 ∕90) + 90 − 1 803 − 1 t116 = (2480.2 − 2674.9 − 0)∕ 767.782 ∕90 + 867.232 ∕803 = −2.2499. P = 0.013172. We conclude that the mean FEV is lower for those whose ozone exposure is in the top 10%.

33.

X = 2651.3, sX = 804.20, nX = 90, Y = 2655.7, sY = 865.77, nY = 803. The number of degrees of freedom is 4 52 804.202 865.772 + 90 803 = = 113, rounded down to the nearest integer. (804.202 ∕90)2 (865.772 ∕803)2 + 90 − 1 803 − 1 t113 = (2651.3 − 2655.7 − 0)∕ 804.202 ∕90 + 865.772 ∕803 = −0.04816. P = 0.48084. We cannot conclude that the mean FEV is lower for those whose PM10 exposure is in the top 10%.

34.

X = 2703.8, sX = 901.29, nX = 90, Y = 2649.8, sY = 854.92, nY = 803. The number of degrees of freedom is 4 52 901.292 854.922 + 90 803 = = 107, rounded down to the nearest integer. 2 2 (854.922 ∕803)2 (901.29 ∕90) + 90 − 1 803 − 1 t107 = (2703.8 − 2649.8 − 0)∕ 901.292 ∕90 + 854.922 ∕803 = 0.54194. P = 0.70551. We cannot conclude that the mean FEV is lower for those whose NO2 exposure is in the top 10%.

35.

Mean with asthma is 2824.7; mean without asthma is 2637.7. It is counterintuitive that those with asthma would have greater lung capacity, on the average, than those without.

36.

X = 2824.7, sX = 829.77, nX = 84, Y = 2637.7, sY = 860.93, nY = 809. The number of degrees of freedom is

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CHAPTER 6 4 =

829.772 860.932 + 84 809

(829.772 ∕84)2 (860.932 ∕809)2 + 84 − 1 809 − 1

= 102, rounded down to the nearest integer.

t102 = (2824.7 − 2637.7 − 0)∕ 829.772 ∕84 + 860.932 ∕809 = 1.9595. P = 0.026392. Taken at face value, we would conclude that the mean FEV is higher for those with asthma than those without.

37.

Of those older than 10.82 years, 56 out of 446, or 12.556%, have asthma. Of those 10.82 years or younger, 28 out of 447, or 6.264%, have asthma. Since those with asthma are more likely to be older, they are more likely to have larger lungs, and thus greater lung capacity even though they have asthma.

38.

X = 3197.1, sX = 768.26, nX = 56, Y = 3283.6, sY = 797.94, nY = 390. The number of degrees of freedom is 4 52 768.262 797.942 + 56 390 = = 73, rounded down to the nearest integer. 2 2 (797.942 ∕390)2 (768.26 ∕56) + 56 − 1 390 − 1 t73 = (3197.1 − 3283.6 − 0)∕ 768.262 ∕56 + 797.942 ∕390 = −0.78391. P = 0.43563. Among those older than 10.82 years, we cannot conclude that the mean FEV for those with asthma di˙ers from the mean FEV for those without.

39.

X = 2079.9, sX = 226.39, nX = 28, Y = 2036.4, sY = 297.55, nY = 419. The number of degrees of freedom is 4 52 226.392 297.552 + 28 419 = = 33, rounded down to the nearest integer. 2 2 (297.552 ∕419)2 (226.39 ∕28) + 28 − 1 419 − 1 t33 = (2079.9 − 2036.4 − 0)∕ 226.392 ∕28 + 297.552 ∕419 = 0.96204. P = 0.34303. Among those 10.82 years and younger, we cannot conclude that the mean FEV for those with asthma di˙ers from the mean FEV for those without.

40.

No. When comparisons are made within age groups, there is no signifcant di˙erence. The di˙erence observed when all individuals are compared is likely to be due to the fact that those with asthma tend to be older.

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SECTION 6.6

Section 6.6 1. (a) H0 ∶ pX − pY ≥ 0 versus H1 ∶ pX − pY < 0 (b) X = 960, nX = 1000, p X = 960∕1000 = 0.960, Y = 582, nY = 600, p Y = 582∕600 = 0.970, p = (960 + 582)∕(1000 + 600) = 0.96375.

The null and alternative hypotheses are H0 ∶ pX − pY ≥ 0 versus H1 ∶ pX − pY < 0. 0.960 − 0.970

= −1.04.

0.96375(1 − 0.96375)(1∕1000 + 1∕600) Since the alternative hypothesis is of the form pX − pY < 0, the P -value is the area to the left of z = −1.04. Thus P = 0.1492. (c) Since P = 0.1492, we cannot conclude that machine 2 is better. Therefore machine 1 should be used. 2. (a) H0 ∶ pX − pY ≥ 0 versus H1 ∶ pX − pY < 0 (b) X = 150, nX = 180, p X = 150∕180 = 0.83333, Y = 233, nY = 270, p Y = 233∕270 = 0.86296, p = (150 + 233)∕(180 + 270) = 0.85111.

The null and alternative hypotheses are H0 ∶ pX − pY ≥ 0 versus H1 ∶ pX − pY < 0.

0.83333 − 0.86296

= −0.86.

0.85111(1 − 0.85111)(1∕180 + 1∕270) Since the alternative hypothesis is of the form pX − pY < 0, the P -value is the area to the left of z = −0.86. Thus P = 0.1949. (c) No. Since P = 0.1949, we cannot conclude that the proportion from vendor B is greater than that from vendor A.

X = 245, nX = 307, p X = 245∕307 = 0.7980, Y = 304, nY = 347, p Y = 304∕347 = 0.8761, p = (245 + 304)∕(307 + 347) = 0.8394.

The null and alternative hypotheses are H0 ∶ pX − pY ≥ 0 versus H1 ∶ pX − pY < 0.

0.7980 − 0.8761

= −2.71.

0.8394(1 − 0.8394)(1∕307 + 1∕347) Since the alternative hypothesis is of the form pX − pY < 0, the P -value is the area to the left of z = −2.71. Thus P = 0.0034. We can conclude that the method is more accurate for new age songs.

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CHAPTER 6

X = 21, nX = 154, p X = 21∕154 = 0.13636, Y = 20, nY = 183, p Y = 20∕183 = 0.10929, p = (21 + 20)∕(154 + 183) = 0.12166.

The null and alternative hypotheses are H0 ∶ pX − pY = 0 versus H1 ∶ pX − pY ≠ 0.

0.13636 − 0.10929

= 0.76.

0.12166(1 − 0.12166)(1∕154 + 1∕183)

Since the alternative hypothesis is of the form pX − pY ≠ 0, the P -value is the sum of the areas to the right of z = 0.76 and to the left of z = −0.76. Thus P = 0.4472. We cannot conclude that the proportion of red-light runners di˙ers between the two intersections.

X = 102, nX = 230, p X = 102∕230 = 0.443478, Y = 20, nY = 72, p Y = 20∕72 = 0.277778, p = (102 + 20)∕(230 + 72) = 0.403974.

The null and alternative hypotheses are H0 ∶ pX − pY ≤ 0 versus H1 ∶ pX − pY > 0.

0.443478 − 0.277778

= 2.50. Since the alternative hypothesis is of the form pX − 0.403974(1 − 0.403974)(1∕230 + 1∕72) pY > 0, the P -value is the area to the right of z = 2.50. Thus P = 0.0062. We can conclude that the proportion of patch users is greater among European-Americans.

X = 77, nX = 365, p X = 77∕365 = 0.210959, Y = 23, nY = 179, p Y = 23∕179 = 0.128492, p = (77 + 23)∕(365 + 179) = 0.1838235.

The null and alternative hypotheses are H0 ∶ pX − pY ≤ 0 versus H1 ∶ pX − pY > 0.

z =

0.210959 − 0.128492

= 2.33. Since the alternative hypothesis is of the form 0.1838235(1 − 0.1838235)(1∕365 + 1∕179) pX − pY > 0, the P -value is the area to the right of z = 2.33. Thus P = 0.0099. We can conclude that the frequency of wheezing symptoms is greater among those residents who participated in the cleaning of food-damaged homes.

X = 20, nX = 1200, p X = 20∕1200 = 0.016667, Y = 15, nY = 1500, p Y = 15∕1500 = 0.01, p = (20 + 15)∕(1200 + 1500) = 0.012963.

The null and alternative hypotheses are H0 ∶ pX − pY ≤ 0 versus H1 ∶ pX − pY > 0.

0.016667 − 0.01

= 1.52.

0.012963(1 − 0.012963)(1∕1200 + 1∕1500) Since the alternative hypothesis is of the form pX − pY > 0, the P -value is the area to the right of z = 1.52.

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SECTION 6.6

Thus P = 0.0643. The evidence suggests that heavy packaging reduces the proportion of damaged shipments, but may not be conclusive.

p X = 0.070, nX = 10292, p Y = 0.098, nY = 51460. p = (nX p X + nY p Y )∕(nX + nY ) = 0.0933333.

The null and alternative hypotheses are H0 ∶ pX − pY ≥ 0 versus H1 ∶ pX − pY < 0. 0.070 − 0.098

z =

= −8.91. Since the alternative hypothesis is of the 0.0933333(1 − 0.0933333)(1∕10292 + 1∕51460) form pX − pY < 0, the P -value is the area to the left of z = −8.91.

Thus P

. A computer package gives 2.55 × 10−19 .

We can conclude that the percentage of people who have had colonoscopies is greater in those without colorectal cancer.

X = 76, nX = 164, p X = 76∕164 = 0.46341, Y = 48, nY = 96, p Y = 48∕96 = 0.5, p = (76 + 48)∕(164 + 96) = 0.47692.

The null and alternative hypotheses are H0 ∶ pX − pY ≥ 0 versus H1 ∶ pX − pY < 0.

0.46341 − 0.5

= −0.57.

0.47692(1 − 0.47692)(1∕164 + 1∕96) Since the alternative hypothesis is of the form pX − pY < 0, the P -value is the area to the left of z = −0.57. Thus P = 0.2843. We cannot conclude that the proportion of patients who fall is less for those who exercise.

10.

X = 74, nX = 150, p X = 74∕150 = 0.49333, Y = 86, nY = 150, p Y = 86∕150 = 0.57333, p = (74 + 86)∕(150 + 150) = 0.53333.

The null and alternative hypotheses are H0 ∶ pX − pY = 0 versus H1 ∶ pX − pY ≠ 0.

0.49333 − 0.57333

= −1.39.

0.53333(1 − 0.53333)(1∕150 + 1∕150)

Since the alternative hypothesis is of the form pX − pY ≠ 0, the P -value is the sum of the area to the right of z = 1.39 and the the area to the left of z = −1.39. Thus P = 0.1646. We cannot conclude that the proportion of seeds that germinate di˙ers with the percent of mushroom compost in the soil.

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CHAPTER 6

X = 53, nX = 118, p X = 53∕118 = 0.44915, Y = 30, nY = 70, p Y = 30∕70 = 0.42857, p = (53 + 30)∕(118 + 70) = 0.44149.

The null and alternative hypotheses are H0 ∶ pX − pY ≤ 0 versus H1 ∶ pX − pY > 0.

0.44915 − 0.42857

= 0.27.

0.44149(1 − 0.44149)(1∕118 + 1∕70)

Since the alternative hypothesis is of the form pX − pY ≤ 0, the P -value is the area to the right of z = 0.27. Thus P = 0.3936. We cannot conclude that children who live in high ozone areas are more likely to miss a day of school.

12.

X = 87, nX = 546, p X = 87∕546 = 0.159341, Y = 74, nY = 508, p Y = 74∕508 = 0.145669, p = (87 + 74)∕(546 + 508) = 0.15275.

The null and alternative hypotheses are H0 ∶ pX − pY = 0 versus H1 ∶ pX − pY ≠ 0.

0.159341 − 0.145669

= 0.62.

0.15275(1 − 0.15275)(1∕546 + 1∕508)

Since the alternative hypothesis is of the form pX − pY ≠ 0, the P -value is the sum of the areas to the right of z = 0.62 and to the left of z = −0.62. Thus P = 0.2676 + 0.2676 = 0.5352. We cannot conclude that the proportion of boys who are overweight di˙ers from the proportion of girls who are overweight.

13.

No, because the two samples are not independent.

14. (a) Two-tailed. The alternative hypothesis is of the form p1 − p2 ≠ 0. (b) H0 ∶ p1 − p2 = 0 (c) Yes, P = 0.024, which is less than 0.05.

15. (a) 101∕153 = 0.660131. (b) 90(0.544444) = 49. (c) X1 = 101, n1 = 153, p 1 = 101∕153 = 0.660131, X2 = 49, n2 = 90, p 2 = 49∕90 = 0.544444, p = (101 + 49)∕(153 + 90) = 0.617284. z=

0.660131 − 0.544444

= 1.79.

0.617284(1 − 0.617284)(1∕153 + 1∕90)

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SECTION 6.7

(d) Since the alternative hypothesis is of the form p1 − p2 ≠ 0, the P -value is the sum of the areas to the right of z = 1.79 and to the left of z = −1.79. Thus P = 0.0367 + 0.0367 = 0.0734.

16.

X = 49, nX = 446, Y = 35, nY = 447, z = 1.61565, P = 0.05308. We cannot reject H0 at the 5% level. It is barely plausible that those whose exposure is greater than the median are no more likely to have asthma than those whose exposure is less than or equal to the median.

17.

. X = 237, nX = 366, Y = 69, nY = 366, z = 12.58923, P We conclude that the record maximum is more likely than the record minimum to have occurred after 1945.

Section 6.7 1.

D = 0.17214, sD = 0.44136, n = 14. There are 14 − 1 = 13 degrees of freedom.

The null and alternative hypotheses are H0 ∶ D = 0 versus H1 ∶ D ≠ 0. t = (0.17214 − 0)∕(0.44136∕ 14) = 1.459.

Since the alternative hypothesis is of the form D ≠ Δ, the P -value is the sum of the areas to the right of t = 1.459 and to the left of t = −1.459. From the t table, 0.10 < P < 0.20. A computer package gives P = 0.16821. We cannot conclude that the mean amount absorbed di˙ers between the brand name and the generic drug.

D = −102.1667, sD = 324.93771, n = 12. There are 12 − 1 = 11 degrees of freedom.

The null and alternative hypotheses are H0 ∶ D = 0 versus H1 ∶ D ≠ 0. t = (−102.16667 − 0)∕(324.93771∕ 12) = −1.089.

Since the alternative hypothesis is of the form D ≠ Δ, the P -value is the sum of the areas to the right of t = 1.089. and to the left of t = −1.089. From the t table, 0.20 < P < 0.50. A computer package gives P = 0.2994. We cannot conclude that the monthly means di˙er between the two types of trucks.

The di˙erences are 0.04, 0.05, 0.02, 0.02, −0.01, 0.02, 0.05, −0.01, 0.05, −0.03 D = 0.02, sD = 0.028674, n = 10. There are 10 − 1 = 9 degrees of freedom. The null and alternative hypotheses are H0 ∶ D ≤ 0 versus H1 ∶ D > 0.

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CHAPTER 6

t = (0.02 − 0)∕(0.028674∕ 10) = 2.206. Since the alternative hypothesis is of the form D > Δ, the P -value is the area to the right of t = 2.206. From the t table, 0.025 < P < 0.05. A computer package gives P = 0.02742. We can conclude that the mean weight on scale 2 is greater than the mean weight on scale 1.

The di˙erences are 0.0195, −0.0120, 0.0131, 0.1041, 0.0211, 0.0434, 0.0991, 0.0458, −0.0249, 0.0539. D = 0.03631, sD = 0.042323, n = 10. There are 10 − 1 = 9 degrees of freedom. The null and alternative hypotheses are H0 ∶ D = 0 versus H1 ∶ D ≠ 0. t = (0.03631 − 0)∕(0.042323∕ 10) = 2.713.

Since the alternative hypothesis is of the form D ≠ Δ, the P -value is the sum of the areas to the right of t = 2.713 and to the left of t = −2.713. From the t table, 0.02 < P < 0.05. A computer package gives P = 0.0239. There is fairly strong evidence that the mean emissions di˙er between the two temperatures.

The di˙erences are −7, −21, 4, −16, 2, −9, −20, −13. D = −10, sD = 9.3808, n = 8. There are 8 − 1 = 7 degrees of freedom.

The null and alternative hypotheses are H0 ∶ D = 0 versus H1 ∶ D ≠ 0. t = (−10 − 0)∕(9.3808∕ 8) = −3.015.

Since the alternative hypothesis is of the form D ≠ Δ, the P -value is the sum of the areas to the right of t = 3.015 and to the left of t = −3.015. From the t table, 0.01 < P < 0.02. A computer package gives P = 0.0195. We can conclude that the mean amount of corrosion di˙ers between the two formulations.

The di˙erences are 3.1, −1.2, 0.4, −0.4, −3.1, −7.7. D = −1.48333, sD = 3.6625, n = 6. There are 6 − 1 = 5 degrees of freedom.

The null and alternative hypotheses are H0 ∶ D = 0 versus H1 ∶ D ≠ 0. t = (−1.48333 − 0)∕(3.6625∕ 6) = −0.9921.

Since the alternative hypothesis is of the form D ≠ Δ, the P -value is the sum of the areas to the right of t = 0.9921 and to the left of t = −0.9921. From the t table, 0.20 < P < 0.50. A computer package gives P = 0.367. We cannot conclude that the mean speeds of the two processors di˙er.

The di˙erences are 35, 57, 14, 29, 31. D = 33.2, sD = 15.498, n = 5.

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SECTION 6.7

There are 5 − 1 = 4 degrees of freedom.

The null and alternative hypotheses are H0 ∶ D ≤ 0 versus H1 ∶ D > 0.

t = (33.2 − 0)∕(15.498∕ 5) = 4.790. Since the alternative hypothesis is of the form D > Δ, the P -value is the area to the right of t = 4.790. From the t table, 0.001 < P < 0.005. A computer package gives P = 0.0044. We can conclude that the mean strength after 6 days is greater than the mean strength after 3 days.

The di˙erences are 0.56, 0.61, 0.64, 0.67. D = 0.62, sD = 0.046904, n = 4. There are 4 − 1 = 3 degrees of freedom.

The null and alternative hypotheses are H0 ∶ D ≤ 0 versus H1 ∶ D > 0.

t = (0.62 − 0)∕(0.046904∕ 5) = 26.437. Since the alternative hypothesis is of the form D > Δ, the P -value is the area to the right of t = 26.437. From the t table, P < 0.0005. A computer package gives P = 5.93 × 10−5 . We can conclude that the mean defection is greater for the Boeing 777.

The di˙erences are 1, 2, −1, 0, 2, −1, 2. D = 0.714286, sD = 1.38013, n = 7. There are 7 − 1 = 6 degrees of freedom. The null and alternative hypotheses are H0 ∶ D = 0 versus H1 ∶ D ≠ 0. t = (0.714286 − 0)∕(1.38013∕ 7) = 1.3693.

Since the alternative hypothesis is of the form D ≠ Δ, the P -value is the sum of the areas to the right of t = 1.3693 and to the left of t = −1.3693. From the t table, 0.20 < P < 0.50. A computer package gives P = 0.220. We cannot conclude that the mean response di˙ers between the two drugs.

10. (a) The di˙erences are 68, 93, 87, 72, 70, 63, 101, 81. D = 79.375, sD = 13.383759, n = 8. There are 8 − 1 = 7 degrees of freedom.

The null and alternative hypotheses are H0 ∶ D ≤ 0 versus H1 ∶ D > 0. t = (79.375 − 0)∕(13.383759∕ 8) = 16.775.

Since the alternative hypothesis is of the form D > Δ, the P -value is the area to the right of t = 16.775. From the t table, P < 0.0005. A computer package gives P = 3.27 × 10−7 . We can conclude that the mean cholesterol level after treatment is less than the mean before treatment. (b) The di˙erences are 68, 93, 87, 72, 70, 63, 101, 81. D = 79.375, sD = 13.383759, n = 8.

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CHAPTER 6

There are 8 − 1 = 7 degrees of freedom.

The null and alternative hypotheses are H0 ∶ D ≤ 75 versus H1 ∶ D > 75.

t = (79.375 − 75)∕(13.383759∕ 8) = 0.9246. Since the alternative hypothesis is of the form D > Δ, the P -value is the area to the right of t = 0.9246. From the t table, 0.10 < P < 0.25. A computer package gives P = 0.193. We cannot conclude that the reduction in mean cholesterol level after treatment is greater than 75 mg/dL.

11. (a) The di˙erences are 5.0, 4.6, 1.9, 2.6, 4.4, 3.2, 3.2, 2.8, 1.6, 2.8. Let R be the mean number of miles per gallon for taxis using radial tires, and let B be the mean number of miles per gallon for taxis using bias tires. The appropriate null and alternative hypotheses are H0 ∶ R − B ≤ 0 versus H1 ∶ R − B > 0. D = 3.21, sD = 1.1338, n = 10. There are 10 − 1 = 9 degrees of freedom. t = (3.21 − 0)∕(1.1338∕ 10) = 8.953. Since the alternative hypothesis is of the form D > Δ, the P -value is the area to the right of t = 8.953. From the t table, P < 0.0050. A computer package gives P = 4.5 × 10−6 . We can conclude that the mean number of miles per gallon is higher with radial tires. (b) The appropriate null and alternative hypotheses are H0 ∶ R − B ≤ 2 vs. H1 ∶ R − B > 2. D = 3.21, sD = 1.1338, n = 10. There are 10 − 1 = 9 degrees of freedom. t = (3.21 − 2)∕(1.1338∕ 10) = 3.375. Since the alternative hypothesis is of the form D > Δ, the P -value is the area to the right of t = 3.375. From the t table, 0.001 < P < 0.005. A computer package gives P = 0.0041. We can conclude that the mean mileage with radial tires is more than 2 miles per gallon higher than with bias tires.

12. (a) One-tailed. The alternative hypothesis is of the form D > 0 . (b) H0 ∶ X − Y ≤ 0 (c) Yes. P = 0.038058, which is less than 0.05. (d) X = 33.632, 0 = 0, n = 12, t = 1.9576, so 1.9576 = (33.632 − 0)∕(s∕ 12). Solving for s we obtain s = 59.514.

13. (a) SE Mean = StDev/ N = 2.9235∕ 7 = 1.1050.

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SECTION 6.8

(b) StDev = (SE Mean) N = 1.0764 7 = 2.8479. (c) D = X − Y = 12.4141 − 8.3476 = 4.0665. (d) t = (4.0665 − 0)∕1.19723 = 3.40.

Section 6.8 1. (a) The signed ranks are x x − 14 20 6 21 7 5 −9 24 10 25 11 32 18 43 29

Signed Rank 1 2 −3 4 5 6 7

The null and alternative hypotheses are H0 ∶ ≤ 14 versus H1 ∶ > 14. The sum of the positive signed ranks is S+ = 25. n = 7. Since the alternative hypothesis is of the form > 0 , the P -value is the area under the signed rank probability mass function corresponding to S+ ≥ 25. From the signed-rank table, P = 0.0391. We can conclude that the mean concentration is greater than 14 g/L.

(b) The signed ranks are x x − 30 32 2 25 −5 24 −6 21 −9 20 −10 43 13 5 −25

Signed Rank 1 −2 −3 −4 −5 6 −7

The null and alternative hypotheses are H0 ∶ ≥ 14 versus H1 ∶ < 30. The sum of the positive signed ranks is S+ = 7. n = 7. Since the alternative hypothesis is of the form < 0 , the P -value is the area under the signed rank probability mass function corresponding to S+ ≤ 7. From the signed-rank table, P > 0.1094.

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CHAPTER 6

We cannot conclude that the mean concentration is less than 30 g/L.

x − 18 2 3 6 7 −13 14 25

Signed Rank 1 2 3 4 −5 6 7

The null and alternative hypotheses are H0 ∶ = 18 versus H1 ∶ ≠ 18. The sum of the positive signed ranks is S+ = 23. n = 7.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is twice the area under the signed rank probability mass function corresponding to S+ ≥ 23. From the signed-rank table, P = 2(0.0781) = 0.1562. We cannot conclude that the mean concentration di˙ers from 18 g/L.

2. (a) The signed ranks are x x − 41 41.01 0.01 41.37 0.37 41.42 0.42 40.50 −0.50 41.70 0.70 41.83 0.83 41.83 0.83 42.68 1.68

Signed Rank 1 2 3 −4 5 6.5 6.5 8

The null and alternative hypotheses are H0 ∶ ≤ 41 versus H1 ∶ > 41. The sum of the positive signed ranks is S+ = 32. n = 8. Since the alternative hypothesis is of the form > 0 , the P -value is the area under the signed rank probability mass function corresponding to S+ ≥ 32. From the signed-rank table, P = 0.0273. We can conclude that the mean thickness is greater than 41 mm.

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SECTION 6.8

(b) The signed ranks are

Signed Rank 1.5 1.5 −3 −4 −5 −6 7 −8

x x − 41.8 41.83 0.03 41.83 0.03 41.70 −0.10 41.42 −0.38 41.37 −0.43 41.01 −0.79 42.68 0.88 40.50 −1.30

The null and alternative hypotheses are H0 ∶ ≥ 41.8 versus H1 ∶ < 41.8. The sum of the positive signed ranks is S+ = 10. n = 8. Since the alternative hypothesis is of the form < 0 , the P -value is the area under the signed rank probability mass function corresponding to S+ ≤ 10. From the signed-rank table, P > 0.1250. We cannot conclude that the mean thickness is less than 41.8 mm.

(c) The signed ranks are x x − 42 41.83 −0.17 41.83 −0.17 41.70 −0.30 41.42 −0.58 41.37 −0.63 42.68 0.68 41.01 −0.99 40.50 −1.50

Signed Rank −1.5 −1.5 −3 −4 −5 6 −7 −8

The null and alternative hypotheses are H0 ∶ = 42 versus H1 ∶ ≠ 42. The sum of the positive signed ranks is S+ = 6. n = 8.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is twice the area under the signed rank probability mass function corresponding to S+ ≤ 6. From the signed-rank table, P = 2(0.0547) = 0.1094. We cannot conclude that the mean thickness di˙ers from 42 mm.

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3. (a) The signed ranks are x x − 45 41.1 −3.9 41.0 −4.0 40.0 −5.0 38.1 −6.9 52.3 7.3 37.3 −7.7 36.2 −8.8 34.6 −10.4 33.6 −11.4 33.4 −11.6 59.4 14.4 30.3 −14.7 30.1 −14.9 28.8 −16.2 63.0 18.0 26.8 −18.2 67.8 22.8 19.0 −26.0 72.3 27.3 14.3 −30.7 78.6 33.6 80.8 35.8 9.1 −35.9 81.2 36.2

Signed Rank −1 −2 −3 −4 5 −6 −7 −8 −9 −10 11 −12 −13 −14 15 −16 17 −18 19 −20 21 22 −23 24

The null and alternative hypotheses are H0 ∶ ≥ 45 versus H1 ∶ < 45. The sum of the positive signed ranks is S+ = 134. n = 24. Since n > 20, compute the z-score of S+ and use the z table. z=

S+ − n(n + 1)∕4 n(n + 1)(2n + 1)∕24

= −0.46.

Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of z = −0.46. From the z table, P = 0.3228. We cannot conclude that the mean conversion is less than 45.

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SECTION 6.8

(b) The signed ranks are x 30.1 30.3 28.8 26.8 33.4 33.6 34.6 36.2 37.3 38.1 40.0 19.0 41.0 41.1 14.3 9.1 52.3 59.4 63.0 67.8 72.3 78.6 80.8 81.2

x − 30 0.1 0.3 −1.2 −3.2 3.4 3.6 4.6 6.2 7.3 8.1 10.0 −11.0 11.0 11.1 −15.7 −20.9 22.3 29.4 33.0 37.8 42.3 48.6 50.8 51.2

Signed Rank 1 2 −3 −4 5 6 7 8 9 10 11 −12.5 12.5 14 −15 −16 17 18 19 20 21 22 23 24

The null and alternative hypotheses are H0 ∶ ≤ 30 versus H1 ∶ > 30. The sum of the positive signed ranks is S+ = 249.5. n = 24. Since n > 20, compute the z-score of S+ and use the z table. z=

S+ − n(n + 1)∕4 n(n + 1)(2n + 1)∕24

= 2.84.

Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of z = 2.84. From the z table, P = 0.0023. We can conclude that the mean conversion is greater than 30.

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(c) The signed ranks are x x − 55 52.3 −2.7 59.4 4.4 63.0 8.0 67.8 12.8 41.1 −13.9 41.0 −14.0 40.0 −15.0 38.1 −16.9 72.3 17.3 37.3 −17.7 36.2 −18.8 34.6 −20.4 33.6 −21.4 33.4 −21.6 78.6 23.6 30.3 −24.7 30.1 −24.9 80.8 25.8 28.8 −26.2 81.2 26.2 26.8 −28.2 19.0 −36.0 14.3 −40.7 9.1 −45.9

Signed Rank −1 2 3 4 −5 −6 −7 −8 9 −10 −11 −12 −13 −14 15 −16 −17 18 −19.5 19.5 −21 −22 −23 −24

The null and alternative hypotheses are H0 ∶ = 55 versus H1 ∶ ≠ 55. The sum of the positive signed ranks is S+ = 70.5. n = 24. Since n > 20, compute the z-score of S+ and use the z table. z=

S+ − n(n + 1)∕4 n(n + 1)(2n + 1)∕24

= −2.27.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is the sum of the areas to the right of z = 2.27 and to the left of z = −2.27. From the z table, P = 0.0116 + 0.0116 = 0.0232. We can conclude that the mean conversion di˙ers from 55.

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SECTION 6.8

4. (a) The signed ranks are x x − 41 40.76 −0.24 40.57 −0.43 41.48 0.48 39.68 −1.32 43.06 2.06 43.53 2.53 43.68 2.68 43.76 2.76 44.86 3.86 46.14 5.14

Signed Rank −1 −2 3 −4 5 6 7 8 9 10

The null and alternative hypotheses are H0 ∶ ≤ 41 versus H1 ∶ > 41. The sum of the positive signed ranks is S+ = 48. n = 10. Since the alternative hypothesis is of the form > 0 , the P -value is the area under the signed rank probability mass function corresponding to S+ ≥ 48. From the signed-rank table, 0.0137 < P < 0.0244. We can conclude that the mean thickness is greater than 41 mm.

(b) The signed ranks are x x − 43 0.06 43.06 43.53 0.53 43.68 0.68 43.76 0.76 41.48 −1.52 44.86 1.86 40.76 −2.24 40.57 −2.43 46.14 3.14 39.68 −3.32

Signed Rank 1 2 3 4 −5 6 −7 −8 9 −10

The null and alternative hypotheses are H0 ∶ ≥ 43 versus H1 ∶ < 43. The sum of the positive signed ranks is S+ = 25. n = 10. Since the alternative hypothesis is of the form < 0 , the P -value is the area under the signed rank probability mass function corresponding to S+ ≤ 25. From the signed-rank table, P > 0.1162. We cannot conclude that the mean thickness is less than 43 mm.

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(c) The signed ranks are x x − 44 43.76 −0.24 43.68 −0.32 43.53 −0.47 44.86 0.86 43.06 −0.94 46.14 2.14 41.48 −2.52 40.76 −3.24 40.57 −3.43 39.68 −4.32

Signed Rank −1 −2 −3 4 −5 6 −7 −8 −9 −10

The null and alternative hypotheses are H0 ∶ = 44 versus H1 ∶ ≠ 44. The sum of the positive signed ranks is S+ = 10. n = 10.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is twice the area under the signed rank probability mass function corresponding to S+ ≤ 10. From the signed-rank table, P = 2(0.0420) = 0.0840. We cannot conclude that the mean thickness di˙ers from 44 mm.

The signed ranks are x 0.01 0.01 −0.01 0.03 0.05 −0.05 −0.07 −0.11 −0.13 0.15

x−0 0.01 0.01 −0.01 0.03 0.05 −0.05 −0.07 −0.11 −0.13 0.15

Signed Rank 2 2 −2 4 5.5 −5.5 −7 −8 −9 10

The null and alternative hypotheses are H0 ∶ = 0 versus H1 ∶ ≠ 0. The sum of the positive signed ranks is S+ = 23.5. n = 10.

Since the alternative hypothesis is of the form ≠ 0 , the P -value is twice the area under the signed rank probability mass function corresponding to S+ ≤ 23.5. From the signed-rank table, P > 2(0.1162) = 0.2324. We cannot conclude that the gauges di˙er.

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SECTION 6.8

The ranks of the combined samples are

Value Rank 0.43 1 0.73 2 0.93 3 1.12 4 1.24 5 1.91 6 2.56 7 2.93 8 3.72 9 6.19 10 11.00 11

Sample Y X Y X X Y Y X Y Y Y

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0. The test statistic W is the sum of the ranks corresponding to the X sample. W = 19. The sample sizes are m = 4 and n = 7.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is twice the area under the rank-sum probability mass function corresponding to W ≤ 19. From the rank-sum table, P > 2(0.0545) = 0.1090. We cannot conclude that the mean rates di˙er.

The ranks of the combined samples are

Value 12 13 15 18 19 20 21 23 24 27 30 32 35 40

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Sample X X X Y X X X Y Y X Y Y Y Y

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0. The test statistic W is the sum of the ranks corresponding to the X sample. W = 34. The sample sizes are m = 7 and n = 7.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is twice the area under the rank-sum probability mass function corresponding to W ≤ 34. From the rank-sum table, P = 2(0.0087) = 0.0174.

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We can conclude that the mean recovery times di˙er.

The ranks of the combined samples are

Value 1255 1255 1270 1280 1287 1291 1296 1301 1302 1306 1310 1314 1318 1321 1326 1328 1329 1332 1341 1343 1349 1364 1372 1374 1376 1378 1387 1396 1397 1399

Rank 1.5 1.5 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Sample X X X X X X X Y X X X X X Y X X X Y Y Y Y Y Y Y Y Y Y Y Y Y

The null and alternative hypotheses are H0 ∶ X − Y ≥ 0 versus H1 ∶ X − Y < 0. The test statistic W is the sum of the ranks corresponding to the X sample. W = 131. The sample sizes are m = 15 and n = 15. Since n and m are both greater than 8, compute the z-score of W and use the z table. z=

W − m(m + n + 1)∕2 mn(m + n + 1)∕12

= −4.21.

Since the alternative hypothesis is of the form X − Y < Δ, the P -value is the area to the left of z = −4.21. . From the z table, P We can conclude that the mean strength is greater for blocks cured 6 days.

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SECTION 6.8

The ranks of the combined samples are

Value 51 59 64 64 64 66 69 72 74 74 75 75 77 77 77 80 80 81 81 83 83 85 85 86 88 91

Rank 1 2 4 4 4 6 7 8 9.5 9.5 11.5 11.5 14 14 14 16.5 16.5 18.5 18.5 20.5 20.5 22.5 22.5 24 25 26

Sample Y Y X Y Y X X X X Y X Y X X Y Y Y Y Y X Y X Y Y X X

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0. The test statistic W is the sum of the ranks corresponding to the X sample. W = 168. The sample sizes are m = 12 and n = 14. Since n and m are both greater than 8, compute the z-score of W and use the z table. z=

W − m(m + n + 1)∕2 mn(m + n + 1)∕12

= 0.31.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is the sum of the areas to the right of z = 0.31 and to the left of z = −0.31. From the z table, P = 0.3783 + 0.3783 = 0.7566. We cannot conclude that the mean scores di˙er.

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CHAPTER 6

The ranks of the combined samples are

Value Rank Sample 15.7 1 Y 15.9 2 Y 16 3 Y 16.1 4 X 16.2 5 Y 16.3 6 Y 16.4 7 Y 16.7 8 X 16.9 9 X 17.2 10 X 19.8 11 X

The null and alternative hypotheses are H0 ∶ X − Y ≤ 0 versus H1 ∶ X − Y > 0. The test statistic W is the sum of the ranks corresponding to the X sample. W = 42. The sample sizes are m = 5 and n = 6. Since the alternative hypothesis is of the form X − Y > Δ, the P -value is the area under the rank-sum probability mass function corresponding to W ≥ 42. From the rank-sum table, P = 0.0152. We can conclude that the mean time is less for route A.

Section 6.9 1. (a) Let p1 represent the probability that a randomly chosen fastener is conforming, let p2 represent the probability that it is downgraded, and let p3 represent the probability that it is scrap. Then the null hypothesis is H0 ∶ p1 = 0.85, p2 = 0.10, p3 = 0.05. (b) The total number of observation is n = 500. The expected values are np1 , np2 , and np3 , or 425, 50, and 25. (c) The observed values are 405, 55, and 40. 2 = (405 − 425)2 ∕425 + (55 − 50)2 ∕50 + (40 − 25)2 ∕25 = 10.4412. (d) There are 3 − 1 = 2 degrees of freedom. From the 2 table, 0.005 < P < 0.01. A computer package gives P = 0.00540. The true percentages di˙er from 85%, 10%, and 5%.

2. (a) Let pij denote the probability that an observation is in column j, given that it is in row i. Then the null hypothesis is H0 ∶ For each column j, p1j = p2j = p3j .

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SECTION 6.9

(b) The row totals are O1. = 700, O2. = 650, O3. = 400. The column totals are O.1 = 1166, O.2 = 491, O.3 = 93. The grand total is O.. = 1750. The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table. 466.400 433.086 266.514

196.400 182.371 112.229

37.200 34.543 21.257

∑ 3 ∑3 i=1

2 j=1 (Oij − Eij ) ∕Eij = 5.760.

(d) There are (3 − 1)(3 − 1) = 4 degrees of freedom. From the 2 table, P > 0.10. A computer package gives P = 0.218. There is no evidence that the quality of welds di˙ers among the shifts.

The row totals are O.1 = 1603 and O.2 = 2036. The column totals are O1. = 504, O2. = 229, O3. = 686, O4. = 1105, O5. = 1115. The grand total is O.. = 3639. The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table.

Men Women

Poor 222.01 281.99

Near Poor 100.88 128.12

Low Income 302.19 383.81

Middle Income 486.76 618.24

High Income 491.16 623.84

There are (2 − 1)(5 − 1) = 4 degrees of freedom. ∑ ∑ 2 = 2i=1 5j=1 (Oij − Eij )2 ∕Eij = 108.35. From the 2 table, P < 0.005. A computer package gives P

We can conclude that the proportions in the various income categories di˙er between men and women.

2 =

∑9

2 i=1 (Oi − Ei ) ∕Ei = 21.475.

There are 9 − 1 = 8 degrees of freedom. From the 2 table, 0.005 < P < 0.01. A computer package gives P = 0.00599. The theoretical model does not explain the observed values well.

The row totals are O1. = 41, O2. = 39, and O3. = 412. The column totals are O.1 = 89, O.2 = 163, O.3 = 240. The grand total is O.. = 492. The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table.

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Diseased Sensitized Normal

<1 7.4167 7.0549 74.528

1 to < 5 13.583 12.921 136.50

≥5 20.000 19.024 200.98

There are (3 − 1)(3 − 1) = 4 degrees of freedom. ∑ ∑ 2 = 3i=1 3j=1 (Oij − Eij )2 ∕Eij = 10.829. From the 2 table, 0.025 < P < 0.05. A computer package gives P = 0.0286. There is some evidence that the proportions of workers in the various disease categories di˙er among exposure levels.

The row totals are O1. = 3557, O2. = 306, O3. = 887. The column totals are O.1 = 1454, O.2 = 1670, O.3 = 1626. The grand total is O.. = 4750. The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table. 1088.8 93.668 271.52

1250.6 107.58 311.85

1217.6 104.75 303.63

There are (3 − 1)(3 − 1) = 4 degrees of freedom. ∑ ∑ 2 = 3i=1 3j=1 (Oij − Eij )2 ∕Eij = 122.81. From the 2 table, P < 0.005. A computer package gives P

We can conclude that the proportions of infants using the various types of restraints varies with age.

7. (a) The row totals are O1. = 37, O2. = 25, and O3. = 35. The column totals are O.1 = 27, O.2 = 35, O.3 = 35. The grand total is O.. = 97. The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table. 10.2990 6.9588 9.7423

13.3505 9.0206 12.6289

(b) The 2 test is appropriate here, since all the expected values are greater than 5. There are (3 − 1)(3 − 1) = 4 degrees of freedom. ∑ ∑ 2 = 3i=1 3j=1 (Oij − Eij )2 ∕Eij = 6.4808. From the 2 table, P > 0.10. A computer package gives P = 0.166. There is no evidence that the rows and columns are not independent.

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SECTION 6.9

8. (a) The row totals are O1. = 40, O2. = 10, O3. = 50. The column totals are O.1 = 70, O.2 = 10, O.3 = 20. The grand total is O.. = 100. The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table. 28 7 35

4 1 5

8 2 10

(b) The 2 test should not be performed, because some of the expected values are less than 5.

(iii) Both row totals and column totals in the observed table must be the same as the row and column totals, respectively, in the expected table.

10.

It is possible to compute the expected values. The grand total is given to be 150. Therefore the missing row and column totals must be the values that make the grand total ,O.. , come out to 150. Therefore the row totals are O1. = 25, O2. = 10, O3. = 40, O4. = 75. The column totals are O.1 = 50, O.2 = 20, O.3 = 80. The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table. 8.33 3.33 13.33 25.00

11.

3.33 1.33 5.33 10.00

13.33 5.33 21.33 40.00

Let p1 represent the probability that a randomly chosen plate is classifed as premium, let p2 represent the probability that it is conforming, let p3 represent the probability that it is downgraded, and let p4 represent the probability that it is unacceptable. Then the null hypothesis is H0 ∶ p1 = 0.10, p2 = 0.70, p3 = 0.15, p4 = 0.05. The total number of observation is n = 200. The expected values are np1 , np2 , np3 , and np4 , or 20, 140, 30, and 10. The observed values are 19, 133, 35, and 13. 2 = (19 − 20)2 ∕20 + (133 − 140)2 ∕140 + (35 − 30)2 ∕30 + (13 − 10)2 ∕10 = 2.133. There are 4 − 1 = 3 degrees of freedom. From the 2 table, P > 0.10. A computer package gives P = 0.545. We cannot conclude that the engineer’s claim is incorrect.

12.

The row totals are O1. = 47, O2. = 63, O3. = 32, O4. = 25. The column totals are O.1 = 58, O.2 = 71, O.3 = 38. The grand total is O.. = 167.

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The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table. Infuenza Headache Weakness Shortness of Breath

Morning 16.323 21.880 11.114 8.6826

Evening 19.982 26.784 13.605 10.629

Night 10.695 14.335 7.281 5.689

There are (4 − 1)(3 − 1) = 6 degrees of freedom. ∑ ∑ 2 = 4i=1 3j=1 (Oij − Eij )2 ∕Eij = 17.572. From the 2 table, 0.005 < P < 0.01. A computer package gives P = 0.00740. We can conclude that the proportions of workers with the various symptoms di˙er among shifts.

13.

The row totals are O1. = 217 and O2. = 210. The column totals are O.1 = 32, O.2 = 15, O.3 = 37, O.4 = 38, O.5 = 45, O.6 = 48, O.7 = 46, O.8 = 42, O.9 = 34, O.10 = 36, O.11 = 28, O.12 = 26. The grand total is O.. = 427. The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table.

Known Unknown

1 16.26 15.74

2 7.62 7.38

3 18.80 18.20

4 19.31 18.69

5 22.87 22.13

Month 6 7 24.39 23.38 23.61 22.62

8 21.34 20.66

9 17.28 16.72

10 18.30 17.70

11 14.23 13.77

12 13.21 12.79

There are (2 − 1)(12 − 1) = 11 degrees of freedom. ∑ ∑ 2 2 = 2i=1 12 j=1 (Oij − Eij ) ∕Eij = 41.33. From the 2 table, P < 0.005. A computer package gives P = 2.1 × 10−5 . We can conclude that the proportion of false alarms whose cause is known di˙ers from month to month.

14. (a) 22 (b) 39.86 (c) C2 (d) C3 (e) No. The P -value is 0.301, so the null hypothesis of independence cannot be rejected. (f) No. It cannot be concluded that the null hypothesis is true.

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SECTION 6.9

15.

The observed values are: 4 weeks 4 302

Deep Not deep

26 weeks 281 720

52 weeks 518 749

The row totals are O.1 = 803 and O.2 = 1771. The grand total is O.. = 2574. The expected values are Eij − Oi. O.j ∕O.. , as shown in the following table: > 30 <= 30

4 weeks 95.462 210.538

26 weeks 312.278 688.722

52 weeks 395.261 871.739

There are (3 − 1)(2 − 1) = 2 degrees of freedom. ∑ ∑ 2 = 2i=1 3j=1 (Oij − Eij )2 ∕Eij = 187.31026. From the 2 table, P < 0.005. A computer package gives P . We can conclude that the proportion of deep pits varies with duration of exposure.

16.

The observed values are:

> 50 <= 50

GSG 338 318

CCS 295 33

GSF 270 8

GSGU 42 816

The row totals are O.1 = 941 and O.2 = 1179. The grand total is O.. = 2120. The expected values are Eij − Oi. O.j ∕O.. , as shown in the following table: > 50 <= 50

GSG 292.415 363.585

CCS 146.208 181.792

GSF 123.920 154.080

GSGU 382.458 475.542

There are (4 − 1)(2 − 1) = 3 degrees of freedom. ∑ ∑ 2 = 2i=1 4j=1 (Oij − Eij )2 ∕Eij = 1143.54177. From the 2 table, P < 0.005. A computer package gives P . We can conclude that the proportion of times the speed exceeds 50 mm/min di˙ers among soil types.

Section 6.10 1.

s2 = 96, n = 11. The null and alternative hypotheses are H0 ∶ 2 ≤ 50, H1 ∶ 2 > 50. The test statistic is (11 − 1)(96)∕50 = 19.2. There are 11 − 1 = 10 degrees of freedom. 2 = 19.2. Since the alternative hypothesis is of the form 2 > 02 , the P -value is the area to the right of 10

From the 2 table, 0.025 < P < 0.05. A computer package gives P = 0.0378.

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There is reasonably strong evidence that 2 > 50. 2.

s2 = 24, n = 29. The null and alternative hypotheses are H0 ∶ 2 ≥ 30, H1 ∶ 2 < 30. The test statistic is (29 − 1)(24)∕30 = 22.4. There are 29 − 1 = 28 degrees of freedom. 2 = 22.4. Since the alternative hypothesis is of the form 2 < 02 , the P -value is the area to the left of 28

From the 2 table, P > 0.10. A computer package gives P = 0.2376. We cannot conclude that 2 < 30. 3.

s2 = 64, n = 25. The null and alternative hypotheses are H0 ∶ 2 = 225, H1 ∶ 2 ≠ 225. The test statistic is (25 − 1)(64)∕225 = 6.8267. There are 25 − 1 = 24 degrees of freedom.

2 = 6.8267. Since the alternative hypothesis is of the form 2 ≠ 02 , the P -value is twice the area to the left of 24

From the 2 table, P < 0.01. A computer package gives P = 0.000463. There is suÿcient evidence to contradict the claim. 4.

s2 = 0.03628, n = 10. The null and alternative hypotheses are H0 ∶ 2 = 0.01, H1 ∶ 2 ≠ 0.01. The test statistic is (10 − 1)(0.03628)∕0.01 = 32.652. There are 10 − 1 = 9 degrees of freedom.

Since the alternative hypothesis is of the form 2 ≠ 02 , the P -value is twice the area to the left of 92 = 32.652. From the 2 table, P < 0.01. A computer package gives P = 0.000307. We can conclude that the variance di˙ers from 0.01. 5.

s2 = 18.49, n = 25. The null and alternative hypotheses are H0 ∶ 2 ≥ 25, H1 ∶ 2 < 25. The test statistic is (25 − 1)(18.49)∕25 = 17.750. There are 25 − 1 = 24 degrees of freedom. 2 = 17.750. Since the alternative hypothesis is of the form 2 < 02 , the P -value is the area to the left of 24

From the 2 table, P > 0.10. A computer package gives P = 0.185. We cannot conclude that the pediatrician’s claim is true. 6.

s2 = 3.61, n = 40. The null and alternative hypotheses are H0 ∶ 2 ≥ 7.0756, H1 ∶ 2 < 7.0756. The test statistic is (40 − 1)(3.61)∕7.0756 = 19.898. There are 40 − 1 = 39 degrees of freedom. 2 = 19.898. Since the alternative hypothesis is of the form 2 < 02 , the P -value is the area to the left of 39

From the 2 table, P < 0.005. A computer package gives P = 0.00475. We can conclude that the population standard deviation is less than 2.66. 7.

s2 = 7569, n = 20. The null and alternative hypotheses are H0 ∶ 2 = 10,000, H1 ∶ 2 ≠ 10, 000. The test statistic is (20 − 1)(7569)∕10,000 = 14.381. There are 20 − 1 = 19 degrees of freedom.

2 = 14.381. Since the alternative hypothesis is of the form 2 ≠ 02 , the P -value is twice the area to the left of 19

From the 2 table, P > 0.20. A computer package gives P = 0.478. There is not suÿcient evidence to contradict the claim. 8.

s2 = 2.25, n = 25. The null and alternative hypotheses are H0 ∶ 2 ≤ 2.0449, H1 ∶ 2 > 2.0449.

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The test statistic is (25 − 1)(2.25)∕2.0449 = 26.407. There are 25 − 1 = 24 degrees of freedom. 2 = 26.407. Since the alternative hypothesis is of the form 2 > 02 , the P -value is the area to the right of 19

From the 2 table, P > 0.10. A computer package gives P = 0.333. We cannot conclude that the true standard deviation is greater than the value reported by the National Institutes of Health. 9.

1 = 7, 2 = 20. From the F table, the upper 5% point is 2.51.

10.

1 = 2, 2 = 5. From the F table, the upper 1% point is 13.27.

11. (a) The upper 1% point of the F5, 7 distribution is 7.46. Therefore the P -value is 0.01. (b) The P -value for a two-tailed test is twice the value for the corresponding one-tailed test. Therefore P = 0.02.

12. (a) The sample variance for day 1 is s21 = 0.0192308. The sample variance for day 2 is s22 = 0.148077. The sample variance for day 3 is s23 = 0.268590. The sample sizes are n1 = n2 = n3 = 13.

The null and alternative hypotheses are H0 ∶ 22 ∕ 12 ≤ 1 versus H1 ∶ 22 ∕ 12 > 1. The test statistic is F = s22 ∕s12 = 7.7000. The numbers of degrees of freedom are 12 and 12. The P -value is the area to the right of 7.7000 under the F12, 12 probability density function. From the F table, P < 0.001. A computer package gives P = 0.00064. We can conclude that the variability of the process on the second day is greater than on the frst day. (b) The null and alternative hypotheses are H0 ∶ 32 ∕ 22 ≤ 1 versus H1 ∶ 32 ∕ 22 > 1. The test statistic is F = s23 ∕s22 = 1.8139. The numbers of degrees of freedom are 12 and 12. The P -value is the area to the right of 1.8139 under the F12, 12 probability density function. From the F table, P > 0.1. A computer package gives P = 0.158. We cannot conclude that the variability of the process on the third day is greater than on the second day.

13.

The sample variance of the sodium contents for the brand A is s21 = 6.4764. The sample size is n1 = 9. The sample variance of the sodium contents for brand B is s22 = 10.3572. The sample size is n2 = 13.

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CHAPTER 6 The null and alternative hypotheses are H0 ∶ 12 ∕ 22 = 1 versus H1 ∶ 12 ∕ 22 ≠ 1. The test statistic is F = s22 ∕s12 = 1.6020. The numbers of degrees of freedom are 12 and 8. Since this is a two-tailed test, the P -value is twice the area to the right of 1.6020 under the F12, 8 probability density function. From the F table, P > 0.2. A computer package gives P = 0.513. We cannot conclude that the variance of the sodium content di˙ers between the brands.

14.

The sample variance of the times for the frst month is s21 = 200.7562. The sample size is n1 = 7. The sample variance of the times for the seventh month is s22 = 3570.8978. The sample size is n2 = 9. The null and alternative hypotheses are H0 ∶ 22 ∕ 12 ≤ 1 versus H1 ∶ 22 ∕ 12 > 1.

The test statistic is F = s22 ∕s12 = 17.7872. The numbers of degrees of freedom are 8 and 6. The P -value is the area to the right of 17.7872 under the F8, 6 probability density function. From the F table, 0.001 < P < 0.01. A computer package gives P = 0.0012. We can conclude that the time to freeze-up is more variable in the seventh month than in the frst month.

Section 6.11 1. (a) True. H0 is rejected at any level greater than or equal to 0.03. (b) False. 2% is less than 0.03. (c) False. 10% is greater than 0.03.

2. (a) Let X be the sample mean of the 80 specimens. Under H0 , the population mean is = 2, and the population standard deviation is = 0.6. The null distribution of X is therefore normal with mean 2 and standard deviation 0.6∕ 80 = 0.0670820. Since the alternative hypothesis is of the form < 0 , the rejection region for a 5% level test consists of the lower 5% of the null distribution. The z-score corresponding to the lower 5% of the normal distribution is z = −1.645. Therefore the rejection region consists of all values of X less than or equal to 2 − 1.645(0.0670820) = 1.890.

H0 will be rejected if X ≤ 1.890.

(b) Since the alternative hypothesis is of the form < 0 , the rejection region for a 10% level test consists of the lower 10% of the null distribution.

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The z-score corresponding to the lower 10% of the normal distribution is z = −1.28. Therefore the rejection region consists of all values of X less than or equal to 2 − 1.28(0.0670820) = 1.9141. Since 1.85 < 1.9141, H0 will be rejected at the 10% level. (c) Since the alternative hypothesis is of the form < 0 , the rejection region for a 1% level test consists of the lower 1% of the null distribution. The z-score corresponding to the lower 1% of the normal distribution is z = −2.33. Therefore the rejection region consists of all values of X less than or equal to 2 − 2.33(0.0670820) = 1.8437. Since 1.85 > 1.8437, H0 will not be rejected at the 1% level. (d) Under H0 , the z-score of 1.9 is (1.9 − 2)∕0.0670820 = −1.49. Since the alternative hypothesis is of the form < 0 , the level is the area to the left of z = −1.49. Therefore the level is

= 0.0681.

3. (a) H0 ∶ ≥ 90 versus H1 ∶ < 90 (b) Let X be the sample mean of the 150 times. Under H0 , the population mean is = 90, and the population standard deviation is = 5. The null distribution of X is therefore normal with mean 90 and standard deviation 5∕ 150 = 0.408248. Since the alternative hypothesis is of the form < 0 , the rejection region for a 5% level test consists of the lower 5% of the null distribution. The z-score corresponding to the lower 5% of the normal distribution is z = −1.645. Therefore the rejection region consists of all values of X less than or equal to 90 − 1.645(0.408248) = 89.3284. H0 will be rejected if X < 89.3284. (c) This is not an appropriate rejection region. The rejection region should consist of values for X that will make the P -value of the test less than or equal to a chosen threshold level. Therefore the rejection region must be of the form X ≤ x0 . This rejection region is of the form X ≥ x0 , and so it consists of values for which the P -value will be greater than some level.

(d) This is an appropriate rejection region. Under H0 , the z-score of 89.4 is (89.4 − 90)∕0.408248 = −1.47. Since the alternative hypothesis is of the form < 0 , the level is the area to the left of z = −1.47. Therefore the level is

= 0.0708.

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(e) This is not an appropriate rejection region. The rejection region should consist of values for X that will make the P -value of the test less than a chosen threshold level. This rejection region contains values of X greater than 90.6, for which the P -value will be large.

4. (a) Type I error. H0 is true and was rejected. (b) Correct decision. H0 is true and was not rejected. (c) Type II error. H0 is false and was not rejected. (d) Correct decision. H0 is false and was rejected.

5. (a) H0 says that the claim is true. Therefore H0 is true, so rejecting it is a type I error. (b) Correct decision. H0 is false and was rejected. (c) Correct decision. H0 is true and was not rejected. (d) Type II error. H0 is false and was not rejected.

6. (a) When a type I error is made, H0 is true but rejected. The conclusion is that the mean breaking strength is greater than 50 kN. (b) When a type II error is made, H0 is false but not rejected. The conclusion is that the mean breaking strength may be less than or equal to 50 kN.

7. (a) A type I error occurs when H0 is true but rejected. The conclusion is that the mean thickness is not equal to 2 mm. (b) A type II error occurs when H0 is false but not rejected. The conclusion is that the mean thickness might be equal to 2 mm.

The maximum probability of rejecting H0 when true is the level

= 0.10.

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SECTION 6.12

The 1% level. The error in question is rejecting H0 when true, which is a type I error. When the level is smaller, it is less likely to reject H0 , and thus less likely to make a type I error.

10.

The smaller the level, the greater the disagreement with H0 must be in order to reject. Therefore X > 258.4 has level 0.05, X > 261.9 has level 0.01, and X > 256.5 has level 0.10.

11. (a) H0 is true but rejected. This is a type I error. (b) A type I error occurs when H0 is true but rejected. H0 is in fact false, so a type I error cannot be made. (c) A type II error occurs when H0 is false but not rejected. H0 is in fact false, so a type II error will be made if H0 is not rejected.

12. (a) H0 is false but not rejected. This is a type II error. (b) A type I error occurs when H0 is true but rejected. H0 is in fact true, so a type I error will be made if H0 is rejected. (c) A type II error occurs when H0 is false but not rejected. H0 is in fact true, so a type II error cannot be made.

13.

The rejection region for a 5% level test is 0 − 1.645 X , so 10 − 1.645 X = 7.9, and X = 1.2766. The rejection region for a 1% level test is 0 − 2.33 X . Therefore the rejection region for a 1% level test is X < 7.0255.

14. (a) A type I error occurs if H0 is rejected, which will happen if the coin comes up heads. The probability of this is p = 0.1. (b) A type II error occurs if H0 is not rejected, which will happen if the coin comes up tails. The probability of this is 1 − p = 1 − 0.4 = 0.6.

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Section 6.12 1. (a) True. This is the defnition of power. (b) True. When H0 is false, making a correct decision means rejecting H0 . (c) False. The power is 0.90, not 0.10. (d) False. The power is not the probability that H0 is true.

2. (a) True. This is the defnition of power. (b) False. The power is the probability of not making a type II error, not the probability of making a type I error. (c) False. The power does not specify the probability of a type I error. (d) False. The power is the probability of not making a type II error, not the probability of making a type II error. (e) True. The probability of making a type II error is equal to 1 minus the power. (f) False. The power is not the probability that H0 is false.

increase. If the level increases, the probability of rejecting H0 increases, so in particular, the probability of rejecting H0 when it is false increases.

increase. If the sample size increases, the probability of rejecting H0 when it is false increases.

5. (a) H0 ∶ ≥ 50, 000 versus H1 ∶ < 50, 000. H1 is true, since the true value of is 49,500. (b) The level is the probability of rejecting H0 when it is true. Under H0 , X is approximately normally distributed with mean 50,000 and standard deviation X = 5000∕ 100 = 500. The probability of rejecting H0 is P (X ≤ 49, 400).

Under H0 , the z-score of 49,400 is z = (49, 400 − 50, 000)∕500 = −1.20. The level of the test is the area under the normal curve to the left of z = −1.20.

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SECTION 6.12

Therefore the level is 0.1151. The power is the probability of rejecting H0 when = 49, 500. X is approximately normally distributed with mean 49,500 and standard deviation X = 5000∕ 100 = 500. The probability of rejecting H0 is P (X ≤ 49, 400).

The z-score of 49,400 is z = (49, 400 − 49, 500)∕500 = −0.20. The power of the test is thus the area under the normal curve to the left of z = −0.20. Therefore the power is 0.4207. (c) Since the alternative hypothesis is of the form < 0 , the 5% rejection region will be the region X ≤ x5 , where x5 is the 5th percentile of the null distribution. The z-score corresponding to the 5th percentile is z = −1.645. Therefore x5 = 50, 000 − 1.645(500) = 49, 177.5. The rejection region is X ≤ 49, 177.5.

The power is therefore P (X ≤ 49, 177.5) when = 49, 500. The z-score of 49,177.5 is z = (49, 177.5 − 49, 500)∕500 = −0.645. We will use z = −0.65. The power is therefore the area to the left of z = −0.65. Thus the power is 0.2578. (d) For the power to be 0.80, the rejection region must be X ≤ x0 where P (X ≤ x0 ) = 0.80 when = 49, 500. Therefore x0 is the 80th percentile of the normal curve when = 49, 500. The z-score corresponding to the 80th percentile is z = 0.84. Therefore x0 = 49, 500 + 0.84(500) = 49, 920.

Now compute the level of the test whose rejection region is X ≤ 49, 920.

The level is P (X ≤ 49, 920) when = 50, 000.

The z-score of 49,920 is z = (49, 920 − 50, 000)∕500 = −0.16. The level is the area under the normal curve to the left of z = −0.16. Therefore the level is 0.4364.

(e) Let n be the required number of tires. The null distribution is normal with = 50, 000 and X = 5000∕ n. The alternative distribution is normal with = 49, 500 and X = 5000∕ n. Let x0 denote the boundary of the rejection region. Since the level is 5%, the z-score of x0 is z = −1.645 under the null distribution. Therefore x0 = 50, 000 − 1.645(5000∕ n). Since the power is 0.80, the z-score of x0 is z = 0.84 under the alternative distribution.

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Therefore x0 = 49, 500 + 0.84(5000∕ n). It follows that 50, 000 − 1.645(5000∕ n) = 49, 500 + 0.84(5000∕ n). Solving for n yields n = 618.

6. (a) The null distribution of X is normal with mean = 1000 and standard deviation X = 100∕ 75 = 11.5470. The level is P (X ≤ 980), computed under the null distribution. The z-score of 980 is z = (980 − 1000)∕11.5470 = −1.73. The level is the area to the left of z = −1.73. Thus the level is 0.0418.

(b) The alternative distribution of X is normal with mean = 965 and standard deviation X = 100∕ 75 = 11.5470. The power is P (X ≤ 980), computed under the alternative distribution. The z-score of 980 is z = (980 − 965)∕11.5470 = 1.30. The power is the area to the left of z = 1.30. Thus the power is 0.9032.

(c) To make the power 0.95, the rejection region must be X ≤ x0 where P (X ≤ x0 ) = 0.95 under the alternative distribution. The z-score corresponding to the 95th percentile is z = 1.645. Therefore x0 = 965 + 1.645(11.5470) = 983.99. The rejection region should be X ≤ 983.99.

(d) To make the level 5%, the rejection region must be X ≤ x0 where P (X ≤ x0 ) = 0.05 under the null distribution. The z-score corresponding to the 5th percentile is z = −1.645. Therefore x0 = 1000 − 1.645(11.5470) = 981.01.

The rejection region should be X ≤ 981.01.

(e) The rejection region for the 5% level test is X ≤ 981.01.

The power is then P (X ≤ 981.01), computed under the alternative distribution.

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The z-score of 981.01 is z = (981.01 − 965)∕11.547 = 1.39. The power is the area to the left of z = 1.39. Thus the power is 0.9177.

(f) Let n be the required sample size. The null distribution is normal with = 1000 and X = 100∕ n. The alternative distribution is normal with = 965 and X = 100∕ n. Let x0 denote the boundary of the rejection region. Since the level is 5%, the z-score of x0 is z = −1.645 under the null distribution. Therefore x0 = 1000 − 1.645(100∕ n). Since the power is 0.95, the z-score of x0 is z = 1.645 under the alternative distribution. Therefore x0 = 965 + 1.645(100∕ n). It follows that 1000 − 1.645(100∕ n) = 965 + 1.645(100∕ n). Solving for n yields n = 89.

(ii). Since 7 is farther from the null mean of 10 than 8 is, the power against the alternative = 7 will be greater than the power against the alternative = 8.

8. (a) The null distribution of X is Bin(250, 0.10). Thus the null distribution is approximately normal with mean = 250(0.10) = 25 and standard deviation = 250(0.1)(1 − 0.1) = 4.743416. The level is P (X ≤ 18), computed under the null distribution.

To fnd P (X ≤ 18), use the continuity correction, fnd the z-score of 18.5. The z-score of 18.5 is z = (18.5 − 25)∕4.743416 = −1.37. The level is the area to the left of z = −1.37. Thus the level is 0.0853.

(b) The alternative distribution of X is Bin(250, 0.06). Thus the alternative distribution is approximately normal with mean = 250(0.06) = 15 and standard deviation = 250(0.06)(1 − 0.06) = 3.75500. The power is P (X ≤ 18), computed under the alternative distribution.

To fnd P (X ≤ 18), use the continuity correction, fnd the z-score of 18.5. The z-score of 18.5 is z = (18.5 − 15)∕3.75500 = 0.93. The power is the area to the left of z = 0.93.

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Thus the power is 0.8238.

(c) No. To compute the level, use the standard deviation assumed by the null hypothesis, which is 250(0.1)(0.9) = 4.7434. To compute the power, use the standard deviation assumed by the alternative hypothesis, which is 250(0.06)(0.94) = 3.7550. (d) Let n be the required sample size. The null distribution is approximately normal with = 0.1n and = n(0.1)(0.9). The alternative distribution is normal with = 0.06n and = n(0.06)(0.94). Let x0 denote the boundary of the rejection region. Since the level is 5%, the z-score of x0 is z = −1.645 under the null distribution. Therefore x0 = 0.1n − 1.645 n(0.1)(0.9). Since the power is 0.90, the z-score of x0 is z = 1.28 under the alternative distribution. Therefore x0 = 0.06n + 1.28 n(0.06)(0.94). It follows that 0.1n − 1.645 n(0.1)(0.9) = 0.06n + 1.28 n(0.06)(0.94). Dividing by

n yields the equation 0.1 n − 1.645 (0.1)(0.9) = 0.06 n + 1.28 (0.06)(0.94).

Solving for n yields n = 398. 9. (a) Two-tailed. The alternative hypothesis is of the form p ≠ p0 . (b) p = 0.5 (c) p = 0.4 (d) Less than 0.7. The power for a sample size of 150 is 0.691332, and the power for a smaller sample size of 100 would be less than this. (e) Greater than 0.6. The power for a sample size of 150 is 0.691332, and the power for a larger sample size of 200 would be greater than this. (f) Greater than 0.65. The power against the alternative p = 0.4 is 0.691332, and the alternative p = 0.3 is farther from the null than p = 0.4. So the power against the alternative p = 0.3 is greater than 0.691332. (g) It’s impossible to tell from the output. The power against the alternative p = 0.45 will be less than the power against p = 0.4, which is 0.691332. But we cannot tell without calculating whether the power will be less than 0.65.

10. (a) One-tailed. The alternative hypothesis is of the form > 0 . (b) 4. The alternative mean is equal to the null plus the di˙erence, and the di˙erence is 1, so, if the null mean is 3, the alternative is 4.

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(c) Greater than 0.85. The power is 0.857299 when the sample size is 18, so it will be greater than this when the sample size is 25. (d) Less than 0.90. The power against a di˙erence of 1 is 0.857299. The power against a smaller di˙erence of 0.5 will be less than this. (e) Less than 0.85. The sample size of 18 is the smallest that will produce power greater than or equal to the target power of 0.85.

11. (a) Two-tailed. The output indicates that the alternative is two-sided. (b) Less than 0.9. A sample size of 60 is the smallest that will produce power greater than or equal to the target power of 0.9. (c) Greater than 0.9. The power is equal to 0.9 against a di˙erence of 3, so it will be greater than 0.9 against any di˙erence greater than 3.

Section 6.13 1.

Several tests have been performed, so we cannot interpret the P -values in the way that we do when only one test is performed.

The Bonferroni-adjusted P -value is 5(0.03) = 0.15, which is not small enough to reject H0 .

3. (a) There are six tests, so the Bonferroni-adjusted P -values are found by multiplying the original P -values by 6. For the setting whose original P -value is 0.002, the Bonferroni-adjusted P -value is therefore 0.012. Since this value is small, we can conclude that this setting reduces the proportion of defective parts. (b) The Bonferroni-adjusted P -value is 6(0.03) = 0.18. Since this value is not so small, we cannot conclude that this setting reduces the proportion of defective parts.

(iii). The Bonferroni-adjusted P -value is 5(0.03) = 0.15, which is not small enough to conclude that this process is better than the current process. However, it might be. The best thing to do is to run it again, and

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see if its P -value remains small. It is of less interest to rerun processes whose P -values were big the frst time around, since there is no evidence that any of them are better than the current process.

The original P -value must be 0.05∕20 = 0.0025.

6. (a) For additive A: X = 12.938, s = 1.1168, n = 10. t = (12.938 − 12)∕(1.1168∕ 10) = 2.655. The P -value is the area to the left of t = 2.655. There are 10 − 1 = 9 degrees of freedom. From the t table, the area to the right is between 0.01 and 0.025. Therefore the P -value is between 0.975 and 0.99. A computer package gives P = 0.987. For additive B: X = 10.890, s = 1.2395, n = 10. t = (10.890 − 12)∕(1.2395∕ 10) = −2.832. The P -value is the area to the left of t = −2.832. There are 10 − 1 = 9 degrees of freedom. From the t table, 0.005 < P < 0.01. A computer package gives P = 0.0098. For additive C: X = 11.721, s = 2.3433, n = 10. t = (11.721 − 12)∕(2.3433∕ 10) = −0.3768. The P -value is the area to the left of t = −0.3768. There are 10 − 1 = 9 degrees of freedom. From the t table, 0.25 < P < 0.40. A computer package gives P = 0.358. For additive D: X = 10.473, s = 1.3996, n = 10. t = (10.473 − 12)∕(1.3996∕ 10) = −3.4503. The P -value is the area to the left of t = −3.4503. There are 10 − 1 = 9 degrees of freedom. From the t table, 0.001 < P < 0.005. A computer package gives P = 0.0036. For additive E: X = 11.524, s = 1.0892, n = 10. t = (11.524 − 12)∕(1.0892∕ 10) = −1.3822. The P -value is the area to the left of t = −1.3822. There are 10 − 1 = 9 degrees of freedom. From the t table, P

.10. A computer package gives P = 0.100.

(b) (i) At least one of the new additives results in an improvement. The Bonferroni-adjusted P -value for method D is 0.018. This is obtained by multiplying the unadjusted P -value, which is 0.0036, by 5. Since the Bonferroniadjusted P -value is small, we may conclude that additive D results in an improvement.

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Alternatively, the t table shows that P < 0.005. Therefore the Bonferroni-adjusted P -value is less than 5(0.005) = 0.025. Again the adjusted P -value is small, so we may conclude that additive D results in an improvement.

7. (a) No. Let X represent the number of times in 200 days that H0 is rejected. If the mean burn-out amperage is equal to 15 A every day, the probability of rejecting H0 is 0.05 each day, so X ∼ Bin(200, 0.05). The probability of rejecting H0 10 or more times in 200 days is then P (X ≥ 10), which is approximately equal to 0.5636. So it would not be unusual to reject H0 10 or more times in 200 trials if H0 is always true.

Alternatively, note that if the probability of rejecting H0 is 0.05 each day, the mean number of times that H0 will be rejected in 200 days is (200)(0.05) = 10. Therefore observing 10 rejections in 200 days is consistent with the hypothesis that the mean burn-out amperage is equal to 15 A every day. (b) Yes. Let X represent the number of times in 200 days that H0 is rejected. If the mean burn-out amperage is equal to 15 A every day, the probability of rejecting H0 is 0.05 each day, so X ∼ Bin(200, 0.05). The probability of rejecting H0 20 or more times in 200 days is then P (X ≥ 20) which is approximately equal to 0.0010.

So it would be quite unusual to reject H0 20 times in 200 trials if H0 is always true. We can conclude that the mean burn-out amperage di˙ered from 15 A on at least some of the days.

Section 6.14 1. (a) Using Method 1, the limits of the 95% confdence interval are the 2.5 and 97.5 percentiles of the bootstrap means, X .025 and X .975 . The 2.5 percentile is (Y25 + Y26 )∕2 = 38.3818, and the 97.5 percentile is (Y975 + Y976 )∕2 = 38.53865. The confdence interval is (38.3818, 38.53865). Values outside this interval can be rejected at the 5% level. Therefore null hypotheses (ii) and (iv) can be rejected. (b) Using Method 1, the limits of the 90% confdence interval are the 5th and 95th percentiles of the bootstrap means, X .05 and X .95 . The 5th percentile is (Y50 + Y51 )∕2 = 38.39135, and the 95th percentile is (Y950 + Y951 )∕2 = 38.5227. The confdence interval is (38.39135, 38.5227). Values outside this interval can be rejected at the 10% level. Therefore null hypotheses (i), (ii), and (iv) can be rejected.

2. (a) Answers may vary.

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(b) Answers may vary. (c) No, because the simulated samples di˙er somewhat each time the bootstrap is performed.

No, the value 103 is an outlier.

4. (a) There are 16 outcomes in both groups together. Under H0 , any group of eight is equally likely to comprise the 16 values for regular gasoline. Therefore the total number of possible outcomes is = 12, 870. 8 (b) R = 25.8500, P = 31.1375. (c) P

.013.

(d) R = 25.85, sR = 5.7249, nR = 8, S = 31.1375, sS = 3.0156, nS = 8. The number of degrees of freedom is 4 52 5.72492 3.01562 + 8 8 = = 10, rounded down to the nearest integer. 2 2 (3.01562 ∕8)2 (5.7249 ∕8) + 8−1 8−1 t10 = (25.85 − 31.1375)∕ 5.72492 ∕8 + 3.01562 ∕8 = −2.3113.

The null and alternative hypotheses are H0 ∶ R − S ≥ 0 versus H1 ∶ R − S < 0. Since the alternative hypothesis is of the form R − S < Δ, the P -value is the area to the left of t = −2.3113. From the t table, 0.01 < P < 0.025. A computer package gives P = 0.022. This result is not reliable, because the data contain outliers.

5. (a) s2A = 200.28, s2B = 39.833, s2A ∕s2B = 5.02. (b) No, the F -test requires the assumption that the data are normally distributed. These data contain an outlier (103), so the F -test should not be used. (c) P

.37

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.19

7. (a) The test statistic is t =

X−7 s∕ 7

. H0 will be rejected if |t| > 2.447.

(b) The power is approximately 0.60.

8. (a) The power of the two-sample test is approximately 0.46, and the power of the paired test is approximately 0.41. The two-sample test has slightly greater power. (b) The power of the two-sample test is approximately 0.46, and the power of the paired test is approximately 0.97. The paired test has much greater power.

9. (a) X = 22.10, X = 0.34, Y = 14.30, Y = 0.32, V = X 2 + Y 2 = 26.323. )V )V = X∕ X 2 + Y 2 = 0.83957, = Y ∕ X 2 + Y 2 = 0.54325, )X )Y u )V 2 2 )V 2 2 V = Y = 0.3342. X + )X )Y V = 26.323, V = 0.3342 (b) V = 26.323, V = 0.3342. The null and alternative hypotheses are H0 ∶ V ≤ 25 versus H1 ∶ V > 25. z = (26.323 − 25)∕0.3342 = 3.96. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of z = 3.96. Thus P

10. (a) p 1 = 0.42, p 1 = 0.049, p 2 = 0.23, p 2 = 0.043, p = p 1 p 2 = 0.0966. v0 0 12 12 ) p ) p ) p )V = p 1 = 0.42, p = p 2 + p 2 = 0.0213. = p 2 = 0.23, 1 2 ) p2 ) p1 ) p2 ) p1 p = 0.0966, p = 0.0213. (b) p = 0.0966, p = 0.0213. The null and alternative hypotheses are H0 ∶ p ≥ 25 versus H1 ∶ p < 25. z = (0.0966 − 0.10)∕0.0213 = −0.16. Since the alternative hypothesis is of the form p < p0 , the P -value is the area to the left of z = −0.16. Thus P = 0.4364.

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11. (a) Answers may vary.

(b) Answers may vary.

12.

Answers may vary.

Supplementary Exercises for Chapter 6 1.

This requires a test for the di˙erence between two means. The data are unpaired. Let 1 represent the population mean annual cost for cars using regular fuel, and let 2 represent the population mean annual cost for cars using premium fuel. Then the appropriate null and alternative hypotheses are H0 ∶ 1 − 2 ≥ 0 versus H1 ∶ 1 − 2 < 0. The test statistic is the di˙erence between the sample mean costs between the two groups. The t-distribution should be used to fnd the P -value.

The data are paired, so this requires a paired test for the mean di˙erence. Let 1 represent the population mean pulse recovery rate when using the old breathing style, and let 2 represent the population mean pulse recovery rate when using the new breathing style. The appropriate null and alternative hypotheses are H0 ∶ 1 − 2 ≤ 0 versus H1 ∶ 1 − 2 > 0. For each swimmer, let Di denote the di˙erence in recovery rates between the old and new styles, and let D and s denote the sample mean and standard deviation, respectively, of these di˙erences. The test statistic is D∕(s∕ 15). The t-distribution should be used to fnd the P -value. There are 14 degrees of freedom.

This requires a test for a population proportion. Let p represent the population proportion of defective parts under the new program. The appropriate null and alternative hypotheses are H0 ∶ p ≥ 0.10 versus H1 ∶ p < 0.10. The test statistic is the sample proportion of defective parts. The normal distribution should be used to fnd the P -value.

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This requires an F -test. Let 12 represent the population variance for the crushing strength of the old material, and let 22 represent the population variance for the crushing strength of the new material. Then the appropriate null and alternative hypotheses are H0 ∶ 12 ∕ 22 ≤ 1 versus H1 ∶ 12 ∕ 22 > 1. The test statistic is the ratio s12 ∕s22 of the two sample variances. The F -distribution should be used to fnd the P -value. There are 15 and 19 degrees of freedom.

5. (a) H0 ∶ ≥ 16 versus H1 ∶ < 16 (b) X = 15.887, s = 0.13047, n = 10. There are 10 − 1 = 9 degrees of freedom. The null and alternative hypotheses are H0 ∶ ≥ 16 versus H1 ∶ < 16. t = (15.887 − 16)∕(0.13047∕ 10) = −2.739.

(c) Since the alternative hypothesis is of the form < 0 , the P -value is the area to the left of t = −2.739. From the t table, 0.01 < P < 0.025. A computer package gives P = 0.011. We conclude that the mean fll weight is less than 16 oz. 6. (a) H0 ∶ p = 0.20 versus H1 ∶ p ≠ 0.20. (b) X = 192, n = 1280, p = 192∕1280 = 0.15. z = (0.15 − 0.20)∕ 0.20(1 − 0.20)∕1280 = −4.47. (c) Since the alternative hypothesis is of the form p ≠ p0 , the P -value is the sum of the areas to the right of z = 4.47 and to the left of z = −4.47. Thus P

= 0.

We conclude that choices for the SAT are not generated at random. 7. (a) H0 ∶ 1 − 2 = 0 versus H1 ∶ 1 − 2 ≠ 0 (b) D = 1608.143, sD = 2008.147, n = 7. There are 7 − 1 = 6 degrees of freedom. The null and alternative hypotheses are H0 ∶ D = 0 versus H1 ∶ D ≠ 0. t = (1608.143 − 0)∕(2008.147∕ 7) = 2.119. (c) Since the alternative hypothesis is of the form D ≠ 0 , the P -value is the sum of the areas to the right of t = 2.119 and to the left of t = −2.119. From the t table, 0.05 < P < 0.010. A computer package gives P = 0.078.

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The null hypothesis is suspect, but one would most likely not frmly conclude that it is false.

s1 = 3.06, s2 = 1.41. The sample sizes are n1 = 21 and n2 = 21.

The null and alternative hypotheses are H0 ∶ 12 ∕ 22 ≤ 1 versus H1 ∶ 12 ∕ 22 > 1. The test statistic is F = 22 ∕ 12 = 3.062 ∕1.412 = 4.7098. The numbers of degrees of freedom are 20 and 20. The P -value is the area to the right of 4.7098 under the F20, 20 probability density function. From the F table, P < 0.001. A computer package gives P = 0.00053. We can conclude that the measurements made on the base metal are more variable than those made on the weld metal.

X = 5.6, s = 1.2, n = 85. There are 85 − 1 = 84 degrees of freedom. The null and alternative hypotheses are H0 ∶ ≤ 5 versus H1 ∶ > 5. t = (5.6 − 5)∕(1.2∕ 85) = 4.61. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 4.6098. Thus P = 7.1319 × 10−6 . We can conclude that that the silicon content of the water is greater today than it was 10 years ago.

10.

D = 0.17, sD = 1.0122, n = 10. There are 10 − 1 = 9 degrees of freedom. The null and alternative hypotheses are H0 ∶ D = 0 versus H1 ∶ D ≠ 0.

t = (0.17 − 0)∕(1.0122∕ 10) = 0.5311.

Since the alternative hypothesis is of the form D ≠ 0 , the P -value is the sum of the areas to the right of t = 0.5311 and to the left of t = −0.5311. From the t table, 0.50 < P < 0.80. A computer package gives P = 0.608. We cannot conclude that there is a di˙erence in the mean noise levels between acceleration and deceleration lanes.

11. (a) The null distribution of X is normal with mean = 100 and standard deviation X = 0.1∕ 100 = 0.01.

Since the alternative hypothesis is of the form ≠ 0 , the rejection region will consist of both the upper and lower 2.5% of the null distribution. The z-scores corresponding to the boundaries of upper and lower 2.5% are z = 1.96 and z = −1.96, respectively.

Therefore the boundaries are 100 + 1.96(0.01) = 100.0196 and 100 − 1.96(0.01) = 99.9804. Reject H0 if X ≥ 100.0196 or if X ≤ 99.9804.

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(b) The null distribution of X is normal with mean = 100 and standard deviation X = 0.1∕ 100 = 0.01.

Since the alternative hypothesis is of the form ≠ 0 , the rejection region will consist of both the upper and lower 5% of the null distribution. The z-scores corresponding to the boundaries of upper and lower 5% are z = 1.645 and z = −1.645, respectively. Therefore the boundaries are 100 + 1.645(0.01) = 100.01645 and 100 − 1.645(0.01) = 99.98355. Reject H0 if X ≥ 100.01645 or if X ≤ 99.98355.

(c) Yes (d) No (e) Since this is a two-tailed test, there are two critical points, equidistant from the null mean of 100. Since one critical point is 100.015, the other is 99.985. The level of the test is the sum P (X ≤ 99.985) + P (X ≥ 100.015), computed under the null distribution.

The null distribution is normal with mean = 100 and standard deviation X = 0.01. The z-score of 100.015 is (100.015 − 100)∕0.01 = 1.5. The z-score of 99.985 is (99.985 − 100)∕0.01 = −1.5. The level of the test is therefore 0.0668 + 0.0668 = 0.1336, or 13.36%.

12. (a) First fnd the rejection region. The null distribution of X is normal with mean = 100 and standard deviation X = 5∕ 100 = 0.5. Since the level is 5%, and the alternative hypothesis is of the form > 0 , the rejection region must be the upper 5% of the null distribution. The z-score of the boundary of the upper 5% of the normal distribution is z = 1.645. The boundary is therefore 100 + 1.645(0.5) = 100.8225. The rejection region is therefore X ≥ 100.8225.

The power is P (X ≥ 100.8225), computed under the alternative distribution. The alternative distribution is normal with mean = 101 and standard deviation X = 0.5. The z-score of 100.8225 is (100.8225 − 101)∕0.5 = −0.355. We will use z = −0.36. The power is the area to the right of z = −0.36, which is 0.6406. (b) Let n be the required sample size. The null distribution is normal with = 100 and X = 5∕ n. The alternative distribution is normal with = 101 and X = 5∕ n. Let x0 denote the boundary of the rejection region. Since the level is 5%, the z-score of x0 is z = 1.645 under the null distribution. Therefore x0 = 100 + 1.645(5∕ n). Since the power is 0.95, the z-score of x0 is z = −1.645 under the alternative distribution.

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Therefore x0 = 101 − 1.645(5∕ n). It follows that 100 + 1.645(5∕ n) = 101 − 1.645(5∕ n). Solving for n yields n = 271. (c) Let x0 be the boundary of the rejection region.

Since the power is 0.90, P (X ≥ x0 ) = 0.90, computed under the alternative distribution. Therefore x0 is the 10th percentile of the alternative distribution.

The alternative distribution is normal with = 101 and X = 5∕ 100 = 0.5. The z-score corresponding to the 10th percentile is z = −1.28. Therefore x0 = 101 − 1.28(0.5) = 100.36.

The level is P (X ≥ 100.36), computed under the null distribution. The z-score of 100.36 is (100.36 − 100)∕0.5 = 0.72. The level is the area to the right of z = 0.72, which is 0.2358, or 23.58%.

(d) The power is P (X ≥ 100.5), computed under the alternative hypothesis. The alternative distribution is normal with mean = 101 and standard deviation X = 0.5. The z-score of 100.5 is (100.5 − 101)∕0.5 = −1.00. The power is the area to the right of z = −1.00, which is 0.8413.

13. (a) The null hypothesis specifes a single value for the mean: = 3. The level, which is 5%, is therefore the probability that the null hypothesis will be rejected when = 3. The machine is shut down if H0 is rejected at the 5% level. Therefore the probability that the machine will be shut down when = 3 is 0.05. (b) First fnd the rejection region. The null distribution of X is normal with mean = 3 and standard deviation X = 0.10∕ 50 = 0.014142.

Since the alternative hypothesis is of the form ≠ 0 , the rejection region will consist of both the upper and lower 2.5% of the null distribution.

The z-scores corresponding to the boundaries of upper and lower 2.5% are z = 1.96 and z = −1.96, respectively. Therefore the boundaries are 3 + 1.96(0.014142) = 3.0277 and 3 − 1.96(0.014142) = 2.9723. H0 will be rejected if X ≥ 3.0277 or if X ≤ 2.9723.

The probability that the equipment will be recalibrated is therefore equal to P (X P (X ≤ 2.9723), computed under the assumption that = 3.01.

≥

3.0277) +

The z-score of 3.0277 is (3.0277 − 3.01)∕0.014142 = 1.25. The z-score of 2.9723 is (2.9723 − 3.01)∕0.014142 = −2.67.

Therefore P (X ≥ 3.0277) = 0.1056, and P (X ≤ 2.9723) = 0.0038. The probability that the equipment will be recalibrated is equal to 0.1056 + 0.0038 = 0.1094.

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14.

373

The row totals are O1. = 1811 and O2. = 200. The column totals are O.1 = 539, O.2 = 526, O.3 = 495, O.4 = 451. The grand total is O.. = 2011. The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table. Line Pass Fail

1 485.395 53.605

2 473.688 52.312

3 445.771 49.229

4 406.147 44.853

There are (2 − 1)(4 − 1) = 3 degrees of freedom. ∑2 ∑4 2 2 = i=1 j=1 (Oij − Eij ) ∕Eij = 4.676. From the 2 table, P > 0.10. A computer package gives P = 0.197. We cannot conclude that the failure rates di˙er.

15.

X = 37, n = 37 + 458 = 495, p = 37∕495 = 0.074747.

The null and alternative hypotheses are H0 ∶ p ≥ 0.10 versus H1 ∶ p < 0.10. z = (0.074747 − 0.10)∕ 0.10(1 − 0.10)∕495 = −1.87. Since the alternative hypothesis is of the form p < p0 , the P -value is the area to the left of z = −1.87, so P = 0.0307. Since there are four samples altogether, the Bonferroni-adjusted P -value is 4(0.0307) = 0.1228. We cannot conclude that the failure rate on line 3 is less than 0.10.

16. (a) Both samples have a mean of 131 and a variance of 107,210.

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(b) The ranks of the combined samples are

Value −738 0 2 3 4 10 20 40 100 162 222 242 252 258 259 260 262 1000

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Sample Y X X X X X X X X Y Y Y Y Y Y Y Y X

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0. The test statistic W is the sum of the ranks corresponding to the X sample. W = 62. The sample sizes are m = 9 and n = 9. Since n and m are both greater than 8, compute the z-score of W and use the z table. z=

W − m(m + n + 1)∕2 mn(m + n + 1)∕12

= −2.08.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is the sum of the areas to the right of z = 2.08 and to the left of z = −2.08. From the z table, P = 0.0188 + 0.0188 = 0.0376. It seems reasonable to conclude that the population means are unequal. (c) No, the X sample is skewed to the right, while the Y sample is skewed to the left. It does not seem reasonable to assume that these samples came from populations of the same shape.

17. (a) Both samples have a median of 20.

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(b) The ranks of the combined samples are

Value −10 −9 −8 −7 −6 −5 −4 1 2 3 4 5 6 7 20 20 21 22 23 24 25 26 27 40 50 60 70 80 90 100

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15.5 15.5 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Sample Y Y Y Y Y Y Y X X X X X X X X Y Y Y Y Y Y Y Y X X X X X X X

The null and alternative hypotheses are H0 ∶ median of X − median of Y = 0 versus H1 ∶ median of X − median of Y ≠ 0. The test statistic W is the sum of the ranks corresponding to the X sample. W = 281.5. The sample sizes are m = 15 and n = 15. Since n and m are both greater than 8, compute the z-score of W and use the z table. z=

W − m(m + n + 1)∕2 mn(m + n + 1)∕12

= 2.03.

Since the alternative hypothesis is of the form median of X − median of Y ≠ Δ, the P -value is the sum of the areas to the right of z = 2.03 and to the left of z = −2.03. From the z table, P = 0.0212 + 0.0212 = 0.0424. The P -value is fairly small, and so it provides reasonably strong evidence that the population medians are different. (c) No, the X sample is heavily skewed to the right, while the Y sample is strongly bimodal. It does not seem

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CHAPTER 6

reasonable to assume that these samples came from populations of the same shape.

18. (a) X = 15.95, sX = 0.20976, nX = 16, Y = 16.00, sY = 0.18974, nY = 16. The number of degrees of freedom is 4 52 0.209762 0.189742 + 16 16 = = 29, rounded down to the nearest integer. 2 2 (0.189742 ∕16)2 (0.20976 ∕16) + 16 − 1 16 − 1 t29 = (15.95 − 16.00 − 0)∕ 0.209762 ∕16 + 0.189742 ∕16 = −0.7071.

The null and alternative hypotheses are H0 ∶ X − Y = 0 versus H1 ∶ X − Y ≠ 0.

Since the alternative hypothesis is of the form X − Y ≠ Δ, the P -value is the sum of the areas to the right of t = 0.7071 and to the left of t = −0.7071. From the t table, 0.20 < P < 0.50. A computer package gives P = 0.485.

Since P > 0.05, we cannot conclude at the 5% level that the means are di˙erent. (b) The sample variance for the old process is 12 = 0.044. The sample variance for the new process is 22 = 0.036. The sample sizes are n1 = n2 = 16. The null and alternative hypotheses are H0 ∶ 12 ∕ 22 ≤ 1 versus H1 ∶ 12 ∕ 22 > 1.

The test statistic is F = 12 ∕ 22 = 1.2222. The numbers of degrees of freedom are 15 and 15. The P -value is the area to the right of 1.2222 under the F15, 15 probability density function. From the F table, P > 0.10. A computer package gives P = 0.351. Since P > 0.05, we cannot conclude at the 5% level that the variance of the new process is less than that of the old process.

19. (a) Let A be the mean thrust/weight ratio for Fuel A, and let B be the mean thrust/weight ratio for Fuel B. The appropriate null and alternative hypotheses are H0 ∶ A − B ≤ 0 versus H1 ∶ A − B > 0. (b) A = 54.919, sA = 2.5522, nA = 16, B = 53.019, sB = 2.7294, nB = 16. The number of degrees of freedom is 4 52 2.55222 2.72942 + 16 16 = = 29, rounded down to the nearest integer. 2 2 (2.72942 ∕16)2 (2.5522 ∕16) + 16 − 1 16 − 1 t29 = (54.919 − 53.019 − 0)∕ 2.55222 ∕16 + 2.72942 ∕16 = 2.0339.

The null and alternative hypotheses are H0 ∶ A − B ≤ 0 versus H1 ∶ A − B > 0.

Since the alternative hypothesis is of the form A − B > Δ, the P -value is the area to the right of t = 2.0339.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 6

377

From the t table, 0.025 < P < 0.05. A computer package gives P = 0.026.

Since P ≤ 0.05, we can conclude at the 5% level that the means are di˙erent.

20. (a) X = 1.59, s = 0.25, n = 80. There are 80 − 1 = 79 degrees of freedom. The null and alternative hypotheses are H0 ∶ ≤ 1.50 versus H1 ∶ > 1.50. t = (1.59 − 1.50)∕(0.25∕ 80) = 3.2199. Since the alternative hypothesis is of the form > 0 , the P -value is the area to the right of t = 3.2199. Thus P = 0.00093. Since P ≤ 0.05, H0 is rejected. (b) First fnd the rejection region. The null distribution of X is normal with mean = 1.5 and standard deviation X = 0.33∕ 80 = 0.036895. Since the level is 5%, and the alternative hypothesis is of the form > 0 , the rejection region must be the upper 5% of the null distribution. The z-score of the boundary of the upper 5% of the normal distribution is z = 1.645. The boundary is therefore 1.5 + 1.645(0.036895) = 1.5607. The rejection region is therefore X ≥ 1.5607.

The power is P (X ≥ 1.5607), computed under the alternative distribution. The alternative distribution is normal with mean = 1.6 and standard deviation X = 0.036895. The z-score of 1.5607 is (1.5607 − 1.6)∕0.036895 = −1.07. The power is the area to the right of z = −1.07, which is 0.8577. (c) Let n be the required sample size. The null distribution of X is normal with mean = 1.5 and standard deviation X = 0.33∕ n. The alternative distribution is normal with mean = 1.6 and standard deviation X = 0.33∕ n. Let x0 denote the boundary of the rejection region. Since the level is 5%, the z-score of x0 is z = 1.645 under the null distribution. Therefore x0 = 1.5 + 1.645(0.33∕ n). Since the power is 0.99, the z-score of x0 is z = −2.33 under the alternative distribution. Therefore x0 = 1.6 − 2.33(0.33∕ n). It follows that 1.5 + 1.645(0.33∕ n) = 1.6 − 2.33(0.33∕ n). Solving for n yields n = 173.

21. (a) Yes. (b) The conclusion is not justifed. The engineer is concluding that H0 is true because the test failed to reject.

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378 22.

CHAPTER 6

X = 347, n = 510, p = 347∕510 = 0.68039.

The null and alternative hypotheses are H0 ∶ p ≤ 0.48 versus H1 ∶ p > 0.48. z = (0.68039 − 0.48)∕ 0.48(1 − 0.48)∕510 = 9.06. Since the alternative hypothesis is of the form p > p0 , the P -value is the area to the right of z = 9.06, so P

We can conclude that the survey method tends to oversample males.

23.

The row totals are O1. = 214 and O2. = 216. The column totals are O.1 = 65, O.2 = 121, O.3 = 244. The grand total is O.. = 430. The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table. Site Casa da Moura Wandersleben

0–4 years 32.349 32.651

Numbers of Skeletons 5–19 years 20 years or more 60.219 121.43 60.781 122.57

There are (2 − 1)(3 − 1) = 2 degrees of freedom. ∑ ∑ 2 = 2i=1 3j=1 (Oij − Eij )2 ∕Eij = 2.1228. From the 2 table, P > 0.10. A computer package gives P = 0.346. We cannot conclude that the age distributions di˙er between the two sites.

24.

The row totals are O1. = 127, O2. = 123, O3. = 135. The column totals are O.1 = 102, O.2 = 39, O.3 = 48, O.4 = 88, O.5 = 57, O.6 = 51. The grand total is O.. = 385. The expected values are Eij = Oi. O.j ∕O.. , as shown in the following table. State Haryana Bihar Uttar Pradesh

0 33.647 32.587 35.766

1–4 12.865 12.460 13.675

Years of Education 5–6 7–9 10–11 15.834 29.029 18.803 15.335 28.114 18.210 16.831 30.857 19.987

12 or more 16.823 16.294 17.883

There are (3 − 1)(6 − 1) = 10 degrees of freedom. ∑ ∑ 2 = 3i=1 6j=1 (Oij − Eij )2 ∕Eij = 63.301. From the 2 table, P < 0.005. A computer package gives P = 8.6 × 10−10 . We can conclude that the educational levels di˙er among the three states.

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379

SECTION 7.1

Chapter 7 Section 7.1 1.

x = y = 4,

∑n ∑n

r= t

∑n

i=1 (xi − x)

2 = 28,

∑n

i=1 (yi − y)

i=1 (xi − x)(yi − y)

2 i=1 (xi − x)

t∑

n 2 i=1 (yi − y)

2 = 28,

∑n

i=1 (xi − x)(yi − y) = 23.

= 0.8214.

2. (a) y has been replaced with y + 3, x is unchanged. This involves only adding a constant, which does not change the correlation coefficient. (b) x has been replaced with 10x + 1, y has been replaced with y + 3. This involves only adding a constant and multiplying by a constant, neither of which changes the correlation coefficient. (c) x has been replaced with 10y + 33, y has been replaced with 2x + 2. This involves only adding a constant, multiplying by a constant, and interchanging x and y, none of which changes the correlation coefficient.

3. (a) The correlation coefficient is appropriate. The points are approximately clustered around a line. (b) The correlation coefficient is not appropriate. The relationship is curved, not linear. (c) The correlation coefficient is not appropriate. The plot contains outliers.

4. (a) True. This is a result of the plot being clustered around a line with positive slope. (b) False. Since the plot is clustered around a line with negative slope, below average values of one variable will tend to be associated with above average values of the other. (c) False. The correlation coefficient describes the shape of the plot, not its location relative to the axes.

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380

CHAPTER 7

5. The heights and weights for the men (dots) are on the whole greater than those for the women (xs). Therefore the scatterplot for the men is shifted up and to the right. The overall plot exhibits a higher correlation than either plot separately. The correlation between heights and weights for men and women taken together will be more than 0.6.

6. (a) Let x represent velocity and let y represent acceleration. ∑n ∑n 2 2 x = 1.662, y = 6.668, i=1 (xi − x) = 2.23928, i=1 (yi − y) = 16.48108, ∑n i=1 (xi − x)(yi − y) = −5.37248. ∑n i=1 (xi − x)(yi − y) r= t = −0.88436. t∑ ∑n n 2 2 (x − x) (y − y) i=1 i i=1 i

(b)

Acceleration

7 6 5 4 3 0.5

1.5

2 2.5 Velocity

3.5

(c) No, the point (2.92, 3.25) is an outlier. (d) No effect. Converting units from meters to centimeters and from seconds to minutes involves multiplying by a constant, which does not change the correlation coefficient.

7. (a) Let x represent temperature, y represent stirring rate, and z represent yield.

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381

SECTION 7.1 ∑n 2 Then x = 119.875, y = 45, z = 75.590625, i=1 (xi − x) = 1845.75, ∑ ∑ ∑n n n 2 2 i=1 (yi − y) = 1360, i=1 (zi − z) = 234.349694, i=1 (xi − x)(yi − y) = 1436, ∑n ∑n i=1 (xi − x)(zi − z) = 481.63125, i=1 (yi − y)(zi − z) = 424.15. The correlation coefficient between temperature and yield is ∑n i=1 (xi − x)(zi − z) r= t = 0.7323. t∑ ∑n n 2 2 (x − x) (z − z) i=1 i i=1 i The correlation coefficient between stirring rate and yield is ∑n i=1 (yi − y)(zi − z) r= t = 0.7513. t∑ ∑n n 2 2 (y − y) (z − z) i i i=1 i=1 The correlation coefficient between temperature and stirring rate is ∑n i=1 (xi − x)(yi − y) r= t = 0.9064. t∑ ∑n n 2 2 (x − x) (y − y) i=1 i i=1 i

(b) No, the result might be due to confounding, since the correlation between temperature and stirring rate is far from 0. (c) No, the result might be due to confounding, since the correlation between temperature and stirring rate is far from 0.

8. (a) Let x represent temperature, y represent stirring rate, and z represent yield. ∑n ∑n 2 2 Then x = 126.5, y = 45, z = 75.59, i=1 (xi − x) = 2420, i=1 (yi − y) = 2000, ∑ ∑ ∑n n n 2 i=1 (zi − z) = 225.986, i=1 (xi − x)(yi − y) = 0, i=1 (xi − x)(zi − z) = 0.66, ∑n i=1 (yi − y)(zi − z) = 518.1. The correlation coefficient between temperature and yield is ∑n i=1 (xi − x)(zi − z) r= t = 0.0009. t∑ ∑n n 2 2 (x − x) (z − z) i=1 i i=1 i The correlation coefficient between stirring rate and yield is ∑n i=1 (yi − y)(zi − z) r= t = 0.7707. t∑ ∑n n 2 2 (y − y) (z − z) i i i=1 i=1 The correlation coefficient between temperature and stirring rate is

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382

CHAPTER 7 ∑n r= t

∑n

i=1 (xi − x)(yi − y)

2 i=1 (xi − x)

t∑

n 2 i=1 (yi − y)

= 0.

(b) Yes, the correlation between temperature and yield is nearly 0, and there is no confounding of temperature with stirring rate since their correlation is 0. (c) Yes, the correlation between stirring rate and yield is fairly large and positive, and there is no confounding of temperature with stirring rate since their correlation is 0. (d) This design is better because temperature and stirring rate are not confounded.

∑n ∑n 2 2 9. (a) x = 117.36, y = 76.643, i=1 (xi − x) = 1173.2, i=1 (yi − y) = 799.21, ∑n i=1 (xi − x)(yi − y) = 667.79, n = 14. ∑n i=1 (xi − x)(yi − y) r= t = 0.68963. t∑ ∑n n 2 2 i=1 (xi − x) i=1 (yi − y) 1 1+r ln = 0.84725, σW = 1∕(14 − 3) = 0.30151. 2 1−r A 95% confidence interval for μW is W ± 1.96σW , or (0.25629, 1.4382). 0 2(0.25629) 1 e − 1 e2(1.4382) − 1 , , or (0.25082, 0.89334). A 95% confidence interval for ρ is e2(0.25629) + 1 e2(1.4382) + 1 W =

(b) The null and alternative hypotheses are H0 ρ ≤ 0.5 versus H1 ρ > 0.5. 1 1+r = 0.84725, σW = 1∕(14 − 3) = 0.30151. r = 0.68963, W = ln 2 1−r 1 1 + 0.5 = 0.54931. Under H0 , take ρ = 0.5, so μW = ln 2 1 − 0.5 The null distribution of W is therefore normal with mean 0.54931 and standard deviation 0.30151. The z-score of 0.84725 is (0.84725 − 0.54931)∕0.30151 = 0.99. Since the alternative hypothesis is of the form ρ > ρ0 , the P -value is the area to the right of z = 0.99. Thus P = 0.1611. We cannot conclude that ρ > 0.5.

(c) r = 0.68963, n = 14, U = r n − 2∕ 1 − r2 = 3.2989. Under H0 , U has a Student’s t distribution with 14 − 2 = 12 degrees of freedom. Since the alternative hypothesis is of the form ρ > ρ0 , the P -value is the area to the right of t = 3.2989. From the t table, 0.001 < P < 0.005. A computer package gives P = 0.00318. We conclude that ρ > 0.

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383

SECTION 7.1

10. (a) r = 0.15, n = 300, U = r n − 2∕ 1 − r2 = 2.61903. Under H0 , U has a Student’s t distribution with 300−2 = 298 degrees of freedom. Since the number of degrees of freedom is large, use the z table. Since the alternative hypothesis is H1 ρ > 0, the P -value is the area to the right of z = 2.62. Thus P = 0.0044. Alternatively, t298 = 2.619033, P = 0.00463. We conclude that ρ > 0. (b) No, we can only conclude that ρ > 0. We cannot conclude that ρ is large.

1 1+r ln = −1.847395, σW = 1∕(30000 − 3) = 0.00577379. 2 1−r A 95% confidence interval for μW is W ± 1.96σW , or (−1.85871, −1.83608). 0 2(−1.85871) 1 e − 1 e2(−1.83608) − 1 , , or (−0.9526, −0.9504). A 95% confidence interval for ρ is e2(−1.85871) + 1 e2(−1.83608) + 1

11.

r = −0.9515, W =

12.

r = 0.25, n = 134, U = r n − 2∕ 1 − r2 = 2.96648. Under H0 , U has a Student’s t distribution with 134−2 = 132 degrees of freedom. Since the number of degrees of freedom is large, use the z table. Since the alternative hypothesis is H1 ρ > 0, the P -value is the area to the right of z = 2.97. Thus P = 0.0015. Alternatively, t132 = 2.96648, P = 0.0018. We conclude that ρ > 0.

13.

r = −0.509, n = 23, U = r n − 2∕ 1 − r2 = −2.7098. Under H0 , U has a Student’s t distribution with 23 − 2 = 21 degrees of freedom.

Since the alternative hypothesis is H1 ρ ≠ 0, the P -value is the sum of the areas to the right of t = 2.7098 and to the left of t = −2.7098. From the t table, 0.01 < P < 0.02. A computer package gives P = 0.0131. We conclude that ρ ≠ 0.

14.

x = 0 and

∑n

i=1 (xi − x)

2 = 10.

y = [−2 + (−1) + 0 + 1 + y]∕5 = (y − 2)∕5, so y = 5y + 2. ∑ ∑ Express ni=1 (xi − x)(yi − y) , ni=1 (yi − y)2 , and r in terms of y: ∑n i=1 (xi − x)(yi − y) = −2(−2 − y) + (−1)(−1 − y) + 0(0 − y) + 1(1 − y) + 2(5y + 2 − y) = 10y + 10.

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CHAPTER 7 ∑n

i=1 (yi − y)

Now r =

2 = (−2 − y)2 + (−1 − y)2 + (0 − y)2 + (1 − y)2 + (5y + 2 − y)2 = 20y2 + 20y + 10.

10y + 10 y+1 =t , so (1 − 2r2 )y2 + (2 − 2r2 )y + (1 − r2 ) = 0. t y2 + 2y + 1 10 20y2 + 20y + 10

For any given value of r, substitute r into (1 − 2r2 )y2 + (2 − 2r2 )y + (1 − r2 ) = 0 and solve for y, then compute y = 5y + 2. (a) Substitute r = 1 to obtain −y2 = 0, so y = 0, and y = 5(0) + 2 = 2. (b) Substitute r = 0 to obtain y2 + 2y + 1 = 0, so y = −1, and y = 5(−1) + 2 = −3. (c) Substitute r = 0.5 to obtain (1∕2)y2 + (3∕2)y + (3∕4) = 0. There are two roots, y = −0.633975 and y = −2.366025. The two corresponding values of y are y = −1.169873 and y = −9.830127. If y = −1.169873 then r = 0.5, If y = −9.830127 then r = −0.5.

(d) From part (c), y = −9.830127.

(e) For the correlation to be equal to −1, the points would have to lie on a straight line with negative slope. There is no value for y for which this is the case.

Section 7.2 1. (a) 245.82 + 1.13(65) = 319.27 pounds

(b) The difference in y predicted from a one-unit change in x is the slope 1 = 1.13. Therefore the change in the number of lbs of steam predicted from a change of 5 C is 1.13(5) = 5.65 pounds.

2. (a) −196.32 + 2.42(102.7) = 52.21 ksi.

(b) The difference in y predicted from a one-unit change in x is the slope 1 = 2.42. Therefore the change in tensile strength predicted from a change of 3 in the Brinell hardness is 2.42(3) = 7.26 ksi.

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385

SECTION 7.2 ∑n 3.

r2 = 1 −

^i )2 1450 i=1 (yi − y =1− = 0.8492. 2 9615 (y − y) i=1 i

∑n

^i )2 33.9 i=1 (yi − y =1− = 0.8129. 2 181.2 i=1 (yi − y)

∑n

5. (a) −0.2967 + 0.2738(70) = 18.869 in. (b) Let x be the required height. Then 19 = −0.2967 + 0.2738x, so x = 70.477 in.

∑n ∑n 2 2 n = 40, i=1 (xi − x) = 98,775, i=1 (yi − y) = 19.10, x = 26.36, y = 0.5188, ∑n i=1 (xi − x)(yi − y) = 826.94. ∑n (a) r = t

∑n

i=1 (xi − x)(yi − y)

2 i=1 (xi − x)

t∑

n 2 i=1 (yi − y)

= 0.602052.

∑n

∑n ^i )2 = (1 − r2 ) i=1 (yi − y)2 = 12.177. i=1 (yi − y ∑ ∑n The regression sum of squares is i=1 (yi − y)2 − ni=1 (yi − y^i )2 = 19.10 − 12.177 = 6.923. ∑ The total sum of squares is ni=1 (yi − y)2 = 19.10.

(b) The error sum of squares is

∑n (c) β^1 =

i=1 (xi − x)(yi − y) = 0.008371956 and ^0 = y − ^1 x = 0.298115. ∑n 2 (x − x) i=1 i

The equation of the least-squares line is y = 0.298115 + 0.008371956x

(d) 0.298115 + 0.008371956(40) = 0.633 mm (e) Let x be the required temperature. Then 0.5 = 0.298115 + 0.008371956x, so x = 24.1 C.

(f) No. If the actual amount of warping happens to come out above the value predicted by the least-squares line, it could exceed 0.5 mm.

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CHAPTER 7

1 = rsy ∕sx = (0.85)(1.9)∕1.2 = 1.3458.

0 = y − 1 x = 30.4 − 1.3458(8.1) = 19.499.

The equation of the least-squares line is y = 19.499 + 1.3458x.

^1 = rsy ∕sx = 0.7(100∕2) = 35. ^0 = y − ^1 x = 1350 − 35(5) = 1175.

The equation of the least-squares line is y = 1175 + 35x.

9. (a) Right foot temperature

The linear model is appropriate.

75 75

Left foot temperature

∑n ∑n 2 2 (b) x = 83.66667, y = 83.38889, i=1 (xi − x) = 466.00000, i=1 (yi − y) = 248.277778, ∑n i=1 (xi − x)(yi − y) = 276.33333. ∑n (xi − x)(yi − y) ^1 = i=1 = 0.592990 and ^0 = y − ^1 x = 33.775393. ∑n 2 − x) (x i=1 i The equation of the least-squares line is y = 33.775393 + 0.592990x. (c) By 0.592990(2) = 1.1860 F. (d) 33.775393 + 0.592990(81) = 81.805 F.

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387

10. (a)

Diastolic pressure (mmHg)

SECTION 7.2 110 100

The linear model is appropriate.

90 80 70 60 100

110

120

130

140

150

Systolic blood presure (mmHg)

160

∑n ∑n 2 2 (b) x = 122.125, y = 79.375, i=1 (xi − x) = 3723.75, i=1 (yi − y) = 1675.75, ∑n i=1 (xi − x)(yi − y) = 2140.25. ∑n (xi − x)(yi − y) ^1 = i=1 = 9.1828466 and ^0 = y − ^1 x = 0.574757. ∑n 2 i=1 (xi − x) The equation of the least-squares line is y = 9.1828466 + 0.574757x. (c) By 0.574757(10) = 5.74757 mmHg. (d) 9.1828466 + 0.574757(125) = 81.03 mmHg.

Frequency (Hz)

11. (a)

The linear model is appropriate.

0 0.1

0.2

0.3

0.4

0.5

0.6

Damping Ratio

0.7

0.8

0.9

∑n ∑n 2 2 (b) x = 0.527778, y = 1.752778, i=1 (xi − x) = 0.476111, i=1 (yi − y) = 13.672761, ∑n i=1 (xi − x)(yi − y) = −2.335389. ∑n (xi − x)(yi − y) ^1 = i=1 = −4.905134 and 0 = y − ^1 x = 4.341599. ∑n 2 (x − x) i=1 i The equation of the least-squares line is y = 4.341599 − 4.905134x.

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(c) By 4.905134(0.2) = 0.9810 Hz. (d) 4.341599 − 4.905134(0.75) = 0.6627 (e) No, values greater than 1 are outside the range of the data. (f) Let x be the required damping ratio. Then 2.0 = 4.341599 − 4.905134x, so x = 0.47738.

2 Corrosion (mm/yr)

12. (a)

1.5

The linear model is appropriate.

1 0.5 0 5

15 20 Temperature (°C)

∑n ∑n 2 2 (b) x = 17.75556, y = 1.02444, i=1 (xi − x) = 485.5022, i=1 (yi − y) = 0.88302, ∑n i=1 (xi − x)(yi − y) = 17.2398. ∑n (xi − x)(yi − y) ^1 = i=1 = 0.035509 and ^0 = y − ^1 x = 0.393960. ∑n 2 (x − x) i i=1 The equation of the least-squares line is y = 0.393960 + 0.035509x.

(d) 0.393960 + 0.035509(20) = 1.1041 mm/yr (e) The fitted values are the values y^i = ^0 + ^1 xi , for each value xi . They are shown in the following table.

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SECTION 7.2

x 26.6 26.0 27.4 21.7 14.9 11.3 15.0 8.7 8.2

Fitted Value y^ = ^0 + β^1 x 1.3385 1.3172 1.3669 1.1645 0.92304 0.79521 0.92659 0.70289 0.68513

y 1.58 1.45 1.13 0.96 0.99 1.05 0.82 0.68 0.56

(f) The residuals are the values ei = yi − y^i for each i. They are shown in the following table. x 26.6 26.0 27.4 21.7 14.9 11.3 15.0 8.7 8.2

Fitted Value Residual y y^ = ^0 + ^1 x e = y − y^ 1.58 1.3385 0.24150 1.45 1.3172 0.13281 1.13 1.3669 −0.23691 0.96 1.1645 −0.20451 0.99 0.92304 0.06696 1.05 0.79521 0.25479 0.82 0.92659 −0.10659 0.68 0.70289 −0.02289 0.56 0.68513 −0.12513

The point whose residual has the largest magnitude is (11.3, 1.05). ∑n (g) r = t

∑n

i=1 (xi − x)(yi − y)

2 i=1 (xi − x)

t∑

n 2 i=1 (yi − y)

= 0.832627.

∑n

2 i=1 (yi − y) = 0.8830. ∑ ∑ The error sum of squares is ni=1 (yi − y^i )2 = (1 − r2 ) ni=1 (yi − y)2 = 0.2709. ∑ ∑ The regression sum of squares is ni=1 (yi − y)2 − ni=1 (yi − y^i )2 = 0.6121.

(h) The total sum of squares is

(i) The quotient of the regression sum of squares divided by the total sum of squares is 0.6121∕0.8830 = 0.6932. This is the square of the correlation coefficient. ∑n ∑n ∑n 2 ^i )2 ^i )2 i=1 (yi − y) − i=1 (yi − y i=1 (yi − y To see this, note that = 1 − = 1 − (1 − r2 ) = r2 . ∑n ∑n 2 2 (y − y) (y − y) i=1 i i=1 i

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13. (a) Vertical Expansion

75 70 65 60 55 50 45

Horizontal Expansion

∑n ∑n ∑n 2 2 (b) x = 21.857143, y = 64.285714, i=1 (xi − x) = 994.857143, i=1 (yi − y) = 463.428571, i=1 (xi − x)(yi − y) = −557.714286. ∑n (xi − x)(yi − y) ^1 = i=1 = 76.538771 and ^0 = y − ^1 x = −0.560597. ∑n 2 i=1 (xi − x) The equation of the least-squares line is y = 76.538771 − 0.560597x. (c) The fitted values are the values y^i = ^0 + ^1 xi , and the residuals are the values ei = yi − y^i , for each value xi . They are shown in the following table. x 43 5 18 24 32 10 21

Fitted Value Residual y y^ = ^0 + β^1 x e = y − y^ 55 52.433 2.5669 80 73.736 6.2642 58 66.448 −8.448 68 63.084 4.9156 57 58.600 −1.5997 69 70.933 −1.9328 63 64.766 −1.7662

(d) −0.560597(10) = −5.60597. Decrease by 5.60597 parts per hundred thousand. (e) 76.538771 − 0.560597(30) = 59.7209 parts per hundred thousand. (f) No, because 60 is outside the range of the horizontal expansion values present in the data. (g) Let x be the required horizontal expansion. Then 65 = 76.538771 − 0.560597x, so x = 20.583.

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SECTION 7.2

14. (a)

12 10 y

8 6 4 2 0 0

(b) No, because x and y are not linearly related. 3

(c)

1 0

−1 −2 0

∑n ∑n 2 2 (d) x = 5.5, z = 0.41, i=1 (xi − x) = 82.5, i=1 (zi − z) = 18.2494, ∑n (xi − x)(zi − z) ^1 = i=1 = 0.4672727 and ^0 = z − ^1 x = −2.16. ∑n 2 (x − x) i i=1

∑n

i=1 (xi − x)(zi − z) = 38.55.

The equation of the least-squares line is z = −2.16 + 0.4672727x. The predicted value of z when x = 4.5 is z ^ = −2.16 + 0.4672727(4.5) = −0.057273. This is an appropriate method of prediction, because x and z are linearly related. (e) y ^ = ez ^ = e−0.057273 = 0.944. Since z ^ is a reasonable predictor of z, and since y = ez , ez ^ is a reasonable predictor of y.

15.

(iii) equal to $34,900. Since 70 inches is equal to x, the predicted y value, y ^ will be equal to y = 34, 900. To see this, note that y ^ = ^0 + ^1 x = (y − ^1 x) + ^1 x = y.

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16. (a) Let x represent time and y = T1 represent temperature. ∑n ∑n 2 2 x = 10, y = 64.428571, i=1 (xi − x) = 770, i=1 (yi − y) = 17111.14286, ∑n i=1 (xi − x)(yi − y) = 3597. ∑n (xi − x)(yi − y) ^1 = i=1 = 4.671428 and ^0 = y − ^1 x = 17.714286. ∑n 2 i=1 (xi − x) The equation of the least-squares line is y = 17.714286 + 4.671428x.

(b) Let x represent time and y = T2 represent temperature. ∑n ∑n 2 2 x = 10, y = 63.761905, i=1 (xi − x) = 770, i=1 (yi − y) = 16999.80952, ∑n i=1 (xi − x)(yi − y) = 3576. ∑n (xi − x)(yi − y) ^1 = i=1 = 4.644156 and ^0 = y − ^1 x = 17.320346. ∑n 2 (x − x) i i=1 The equation of the least-squares line is y = 17.320346 + 4.644156x.

(c) Let x represent time and y = T3 represent temperature. ∑n ∑n 2 2 x = 10, y = 63.857143, i=1 (xi − x) = 770, i=1 (yi − y) = 18480.57143, ∑n i=1 (xi − x)(yi − y) = 3710. ∑n (xi − x)(yi − y) ^1 = i=1 = 4.818182 and ^0 = y − ^1 x = 15.675325. ∑n 2 − x) (x i=1 i The equation of the least-squares line is y = 15.675325 + 4.818182x.

(d) ^0 = (17.714286 + 17.320346 + 15.675325)/3 = 16.9033. ^1 = (4.671428 + 4.644156 + 4.818182)∕3 = 4.7113. The equation of the least-squares line is y = 16.9033 + 4.7113x.

(e) Let x represent time and y = (T1 + T2 + T3 )∕3 represent temperature. ∑n ∑n 2 2 x = 10, y = 64.015873, i=1 (xi − x) = 770, i=1 (yi − y) = 17321.21693, ∑n i=1 (xi − x)(yi − y) = 3267.666667. ∑n (xi − x)(yi − y) ^1 = i=1 = 4.7113 and ^0 = y − ^1 x = 16.9033. ∑n 2 − x) (x i=1 i

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SECTION 7.3

(f) The results in parts (d) and (e) are the same.

17.

The slope of the least-squares line is r −1 and 1, so Carrie is right.

sy sx

. Since sx = sy , the slope is r. Therefore the slope must be between

∑ ∑ ∑ 18. (a) x = 32.796, y = 2655.247, ni=1 (xi −x)2 = 54739.832, ni=1 (yi −y)2 = 658702850.473, ni=1 (xi −x)(yi −y) = −497992.007. ∑n (xi − x)(yi − y) ^1 = i=1 = −9.0974 and ^0 = y − ^1 x = 2953.6064. ∑n 2 i=1 (xi − x) The equation of the least-squares line is y = 2953.606 − 9.0974x. (b) The null and alternative hypotheses are H0 : 1 ≥ 0 versus H1 : ∑n [ i=1 (xi − x)(yi − y)]2 2 r = ∑n = 0.0068778. ∑n (xi − x)2 (xi − x)2 i=1 i=1 v ∑n (1 − r2 )( i=1 (yi − y)2 s= = 856.85845. n−2 s ^1 = −9.0974, s ^ = t = 3.6623, 1 ∑n 2 (x − x) i=1 i

1 < 0.

t = (−9.0974 − 0)∕3.6623 = −2.48407. There are 893 − 2 = 891 degrees of freedom. Since the alternative hypothesis is of the form P = 0.006586. We conclude that 1 < 0.

1 < b, the P -value is the area to the right of t = −2.48407.

∑n (c) r = t

∑n

19.

i=1 (xi − x)(yi − y)

2 i=1 (xi − x)

∑n

= −0.0829327.

2 i=1 (xi − x)

(i) Higher exposure to ozone is associated with lower values of FEV1, but the value of the correlation shows that the association is weak.

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Section 7.3 ∑n ∑n ∑n 2 2 1. (a) x = 65.0, y = 29.05, i=1 (xi − x) = 6032.0, i=1 (yi − y) = 835.42, i=1 (xi − x)(yi − y) = 1988.4, n = 12. ∑n (xi − x)(yi − y) ^1 = i=1 = 0.329642 and ^0 = y − ^1 x = 7.623276. ∑n 2 i=1 (xi − x)

(b) r2 =

∑ ∑ [ ni=1 (xi − x)(yi − y)]2 (1 − r2 ) ni=1 (yi − y)2 2 = 0.784587. s = = 17.996003. ∑n ∑n 2 2 n−2 i=1 (xi − x) i=1 (yi − y) v

17.996003 = 4.242170. s ^ = s 0

s ^ = t 1 ∑n

1 x2 + ∑n = 3.755613. 2 n i=1 (xi − x)

= 0.0546207.

2 i=1 (xi − x)

There are n − 2 = 10 degrees of freedom. t10,.025 = 2.228. Therefore a 95% confdence interval for The 95% confdence interval for

0 is 7.623276 ± 2.228(3.755613), or (−0.744, 15.991).

1 is 0.329642 ± 2.228(0.0546207), or (0.208, 0.451).

(d) ^1 = 0.329642, s ^ = 0.0546207, n = 12. There are 12 − 2 = 10 degrees of freedom. 1 ≥ 0.5 versus H1

The null and alternative hypotheses are H0

1 < 0.5.

t = (0.329642 − 0.5)∕0.0546207 = −3.119. Since the alternative hypothesis is of the form

1 < b, the P -value is the area to the left of t = −3.119.

From the t table, 0.005 < P < 0.01. A computer package gives P = 0.00545. We can conclude that the claim is false. (e) x = 40, y ^ = 7.623276 + 0.329642(40) = 20.808952. v (x − x)2 1 = 1.834204. There are 10 degrees of freedom. t10,.025 = 2.228. sy ^ = s + ∑n 2 n i=1 (xi − x) Therefore a 95% confdence interval for the mean response is 20.808952 ± 2.228(1.834204), or (16.722, 24.896). (f) x = 40, y ^ = 7.623276 + 0.329642(40) = 20.808952. v (x − x)2 1 = 4.621721. There are 10 degrees of freedom. t10,.025 = 2.228. spred = s 1 + + ∑n 2 n i=1 (xi − x) Therefore a 95% prediction interval is 20.808952 ± 2.228(4.621721), or (10.512, 31.106).

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SECTION 7.3

2. (a) n − 2 = 7 − 2 = 5 (b) ^1 = −0.27985, s ^ = 0.079258, t5,.01 = 3.365, so a 98% confidence interval is −0.27985 ± 3.365(0.079258), 1 or (−0.5466, −0.0131). (c) ^0 = 0.77289, s ^ = 0.14534, t5,.01 = 3.365, so a 98% confidence interval is 0.77289 ± 3.365(0.14534), 0 or (0.2838, 1.2620). (d) The null and alternative hypotheses are H0 1 = −0.6 versus H1 1 ≠ −0.6. ^1 = −0.27985, s ^ = 0.079258, t = (−0.27985 − (−0.6))∕0.079258 = 4.039.

Since the alternative hypothesis is of the form 1 ≠ b, the P -value is the sum of the areas to the right of t = 4.039 and to the left of t = −4.039. 1

From the t table, 0.002 < P < 0.01. A computer package gives P = 0.00993. The claim is not plausible.

3. (a) The slope is −0.7524; the intercept is 88.761. (b) Yes, the P -value for the slope is

, so ozone level is related to humidity.

(c) 88.761 − 0.7524(50) = 51.141 ppb. (d) Since ^1 < 0, r < 0. So r = − r2 = − 0.220 = −0.469. (e) Since n = 120, there are 120 − 2 = 118 degrees of freedom. t118,.05 = 1.6579, so a 90% confidence interval is 43.62 ± 1.6579(1.20), or (41.63, 45.61). (f) No. A reasonable range of predicted values is given by the 95% prediction interval, which is (20.86, 66.37).

4. (a) The slope is −0.1347. (b) ^1 = −0.1347, s ^ = 0.0378. Since n = 120, there are 120 − 2 = 118 degrees of freedom. 1

t118,.025 = 1.9803. A 95% confidence interval is −0.1347 ± 1.9803(0.0378), or (−0.2096, −0.0598).

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CHAPTER 7 1 ≥ −0.1 versus H1 :

1 < −0.1.

Since n = 120, there are 120 − 2 = 118 degrees of freedom. ^1 = −0.1347, s ^ = 0.0378. t = (−0.1347 − (−0.1))∕0.0378 = −0.918. 1

Since the alternative hypothesis is of the form

1 < b, the P -value is the area to the left of t = −0.918.

P = 0.18025. It is plausible that the slope is greater than or equal to −0.1.

5. (a) H0 :

A− B =0

(b) ^A and ^B are independent and normally distributed with means A and B , respectively, and estimated standard deviations s ^ = 0.13024 and s ^ = 0.03798. A

Since n = 120 is large, the estimated standard deviations can be treated as good approximations to the true standard deviations. ^A = −0.7524 and ^B = −0.13468. u The test statistic is z = ( ^A − ^B )∕ s2 ^ + s2 ^ = −4.55. A

Since the alternative hypothesis is of the form z = 4.55 and to the left of z = −4.55. Thus P

A − B ≠ 0, the P -value is the sum of the areas to the right of

= 0.

We can conclude that the effect of humidity differs between the two cities.

∑n ∑n 2 2 6. (a) x = 1.237222, y = 1.274074, i=1 (xi − x) = 4.757483, i=1 (yi − y) = 2.244304, ∑n i=1 (xi − x)(yi − y) = 1.896211, n = 54. ∑n (xi − x)(yi − y) ^1 = i=1 = 0.398574 and ^0 = y − ^1 x = 0.780949. ∑n 2 i=1 (xi − x) The least-squares line is y = 0.780949 + 0.398574x. ∑n

(xi − x)(yi − y)]2 = 0.336755. s = (b) r2 = ∑n i=1 ∑n 2 2 i=1 (xi − x) i=1 (yi − y) [

∑n

(1 − r2 )

∑n

i=1 (yi − y)

n−2

= 0.169191.

= 0.0775689.

2 i=1 (xi − x)

There are n − 2 = 52 degrees of freedom. t52,.025 = 2.0066. Therefore a 95% confidence interval for

1 is 0.398574 ± 2.0066(0.0775689), or (0.2429, 0.5543).

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SECTION 7.3 v (d) x = 1.2, y ^ = 0.780949 + 0.398574(1.2) = 1.259238. sy ^ = s

(x − x)2 1 + ∑n = 0.0232042. 2 n i=1 (xi − x)

t52,.025 = 2.0066. A 95% confidence interval for the mean response is 1.259238 ± 2.0066(0.0232042) or (1.213, 1.306).

(e) Let y1.2 =

0 + 1 (1.2) denote the mean long-term measurement for those with short-term measurements of 1.2.

The null and alternative hypotheses are H0 y1.2 ≤ 1.2 versus H1 y1.2 > 1.2. v (x − x)2 1 = 0.0232042. y ^ = ^0 + ^1 (1.2) = 1.259238, sy ^ = s + ∑n 2 n i=1 (xi − x)

t = (1.259238 − 1.2)∕0.0232042 = 2.553. There are 54 − 2 = 52 degrees of freedom. Since the alternative hypothesis is y1.2 > 1.2, the P -value is the area to the right of t = 2.553. From the t table (using 40 or 60 degrees of freedom), 0.005 < P < 0.01. A computer package (using 52 degrees of freedom) gives P = 0.00683. We may conclude that the mean long-term measurement for those with short-term measurements of 1.2 is greater than 1.2. v ^1 (1.2) = 1.259238, spred = s (f) y ^ = ^0 + ^

(x − x)2 1 + ∑n = 0.170774. 2 n i=1 (xi − x)

t52,.025 = 2.0066. A 95% prediction interval is 1.259238 ± 2.0066(0.170774) or (0.9166, 1.6019). (g) The prediction interval, because the short-term measurement is used to estimate a particular long-term measurement.

∑n ∑n 2 2 7. (a) x = 31.87625, y = 79.0625, i=1 (xi − x) = 101.7619875, i=1 (yi − y) = 6.358750, ∑n i=1 (xi − x)(yi − y) = 24.345785, n = 8. ∑n (xi − x)(yi − y) ^1 x = 71.436320. ^0 = y − ^ ^1 = i=1 = 0.239243 and ^ ∑n 2 − x) (x i=1 i The least-squares line is y = 71.436230 + 0.23924x v

∑n

(xi − x)(yi − y)]2 = 0.915996, (b) r2 = ∑n i=1 ∑n 2 2 i=1 (xi − x) i=1 (yi − y) [

(1 − r2 )

∑n

i=1 (yi − y)

n−2

= 0.298374,

= 0.0295780.

2 i=1 (xi − x)

There are 8 − 2 = 6 degrees of freedom. t6,.025 = 2.447.

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CHAPTER 7

Therefore a 95% confidence interval for v (d) y ^ = 78.613620, sy ^ = s

1 is 0.239243 ± 2.447(0.0295780), or (0.167, 0.312).

(x − x)2 1 = 0.119198. + ∑n 2 n i=1 (xi − x)

t6,.025 = 2.447. A 95% confidence interval is 78.613620 ± 2.447(0.119198) or (78.322, 78.905). (e) Let y30 =

0 + 1 (30) denote the mean noise level for streets whose mean speed is 30 kilometers per hour.

The null and alternative hypotheses are H0 y30 ≤ 78 versus H1 y30 > 78. v 2 ^0 + ^ ^1 (30) = 78.613620, sy ^ = s 1 + ∑ (x − x) y ^ = ^ = 0.119198. n 2 n i=1 (xi − x)

t = (78.613620 − 78)∕0.119198 = 5.148. There are 8 − 2 = 6 degrees of freedom. Since the alternative hypothesis is y30 < 78, the P -value is the area to the right of t = 5.148. From the t table, 0.001 < P < 0.005. A computer package gives P = 0.00106. We may conclude that the mean noise level for streets whose mean speed is 30 kilometers per hour is greater than 78 decibels. v (f) y ^ = 78.613620, spred = s

(x − x)2 1 + ∑n = 0.321302. 2 n i=1 (xi − x)

t6,.025 = 2.447. A 95% prediction interval is 78.613620 ± 2.447(0.321302) or (77.827, 79.400).

∑n ∑n 2 2 8. (a) x = 0.840370, y = 0.826667, i=1 (xi − x) = 2.967896, i=1 (yi − y) = 2.9446, ∑n i=1 (xi − x)(yi − y) = 2.897233, n = 27. ∑n i=1 (xi − x)(yi − y) ^1 x = 0.006304765. ^0 = y − ^ = 0.976191 and ^ ∑n 1 = 2 (x − x) i i=1 The least-squares line is y = 0.006304765 + 0.0.976191x (b) The null and alternative hypotheses are H0 0 = 0 versus H1 0 ≠ 0. v ∑ (1 − r2 ) ni=1 (yi − y)2 = 0.0682194, s= n−2 v 2 ^0 = 0.006304765, ^ ^ = s 1 + ∑ x ^ = 0.03577397 n 2 0 n i=1 (xi − x) t = (0.006304765 − 0)∕0.03577397 = 0.1762. There are 27 − 2 = 25 degrees of freedom. Since the alternative hypothesis is of the form −0.1762 and to the right of t = 0.1762.

0 ≠ b, the P -value is the sum of the areas to the left of t =

From the t table, P > 0.80. A computer package gives P = 0.8615.

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399

SECTION 7.3

It is plausible that the intercept is equal to 0. (c) The null and alternative hypotheses are H0 1 = 1 versus H1 1 ≠ 1. v ∑n ∑ 2 − y)] (1 − r2 ) ni=1 (yi − y)2 [ (x − x)(y i i=1 i 2 = 0.9604879, s = = 0.0682194, r = ∑n ∑n 2 2 n−2 i=1 (xi − x) i=1 (yi − y) ^1 = 0.976191, ^

=t ∑n

= 0.0395989.

2 i=1 (xi − x)

t = (0.976191 − 1)∕0.0395989 = −0.6013. There are 27 − 2 = 25 degrees of freedom. Since the alternative hypothesis is of the form −0.6013 and to the right of t = 0.6013.

1 ≠ b, the P -value is the sum of the areas to the left of t =

From the t table, 0.50 < P < 0.80. A computer package gives P = 0.553. It is plausible that the slope is equal to 1.

(d) No. It is plausible that v (e) y ^ = 0.787257, sy ^ = s

0 = 0 and β1 = 1, so we cannot conclude that the prediction method is not accurate.

(x − x)2 1 + ∑n = 0.0132258. 2 n i=1 (xi − x)

t25,.025 = 2.060. A 95% confidence interval is 0.787257 ± 2.060(0.0132258) or (0.760, 0.814). (f) Let y0.8 =

0 + 1 (0.8) denote the mean predicted roughness for steel specimens whose true roughness is 0.8 μm.

The null and alternative hypotheses are H0 y0.8 = 0.75 versus H1 y0.8 ≠ 0.75. v (x − x)2 1 ^ ^ ^ ^ y ^ = 0 + 1 (0.8) = 0.787257, sy ^ = s + ∑n = 0.0132258. 2 n i=1 (xi − x)

t = (0.787257 − 0.75)∕0.0132258 = 2.817. There are 27 − 2 = 25 degrees of freedom.

Since the alternative hypothesis is y0.8 ≠ 0.75, the P -value is the sum of the areas to the left of t = −2.817 and to the right of t = 2.817. From the t table, 0.002 < P < 0.01. A computer package gives P = 0.00933. We can conclude that the claim is false.

∑n ∑n 2 2 9. (a) x = 21.5075, y = 4.48, i=1 (xi − x) = 1072.52775, i=1 (yi − y) = 112.624, ∑n i=1 (xi − x)(yi − y) = 239.656, n = 40. ∑n (xi − x)(yi − y) ^1 x = −0.325844. ^0 = y − ^ ^1 = i=1 = 0.223450 and ^ ∑n 2 − x) (x i=1 i The least-squares line is y = −0.325844 + 0.223450

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v ∑ 2 − y)] (1 − r2 ) ni=1 (yi − y)2 (x − x)(y i i=1 i 2 (b) r = ∑n = 0.475485, s = = 1.246816, ∑n 2 2 n−2 i=1 (xi − x) i=1 (yi − y) v 1 x2 s + ∑n = 0.842217, s ^ = t = 0.0380713. s ^ ^ = s 2 0 1 ∑ n n i=1 (xi − x) (x − x)2 [

∑n

i=1

There are 40 − 2 = 38 degrees of freedom. t38,.025 Therefore a 95% confidence interval for The 95% confidence interval for

.024.

0 is −0.325844 ± 2.024(0.842217), or (−2.031, 1.379).

1 is 0.223450 ± 2.024(0.0380713), or (0.146, 0.301).

^1 (20) = −0.325844 + 0.223450(20) = 4.143150. ^0 + ^ (c) The prediction is ^ v (d) y ^ = 4.14350, sy ^ = s t38,.025

(x − x)2 1 + ∑n = 0.205323. 2 n i=1 (xi − x)

.024. A 95% confidence interval is 4.14350 ± 2.024(0.205323) or (3.727, 4.559). v

(e) y ^ = 4.14350, spred = s t38,.025

(x − x)2 1 = 1.263609. + ∑n 2 n i=1 (xi − x)

.024. A 95% prediction interval is 4.14350 ± 2.024(1.263609) or (1.585, 6.701).

t∑ n 2 10. (a) If the standard deviation s is held constant, the width of a confidence interval is proportional to 1∕ i=1 (xi − x) . ∑n Let SA , SB , and SC denote the values of i=1 (xi − x)2 for Engineers A, B, and C, respectively. Then SA = 22.8, SB = 2SA = 45.6, and SC = 4SA = 91.2. Let WA , WB , and WC denote the widths of the confidence intervals for Engineers A, B, and C, respectively. Then WA ∕WB

SB ∕SA =

The answer is approximate because the values of s will in fact vary somewhat among the engineers.

(b) WA ∕WC

SC ∕SA = 2.

The answer is approximate because the values of s will in fact vary somewhat among the engineers. v (c) If the standard deviation s is held constant, the width of a confidence interval is proportional to

(x − x)2 1 . + ∑n 2 n i=1 (xi − x)

The value of x is 2.5. The values of x are xA = 2.8, xB = 2.8, xC = 5.6.

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SECTION 7.3

The values of n are nA = 5, nB = 10, nC = 5. Let WA , WB , and WC denote the widths of the confidence intervals for Engineers A, B, and C, respectively. u u (2.5 − 2.8)2 1 (2.5 − 2.8)2 1 + = 0.45, + = 0.32, Then WA , WB , WC are proportional to 5 22.8 10 45.6 u 1 (2.5 − 5.6)2 + = 0.55 respectively. 5 91.2 These results are approximate because the values of s will in fact vary somewhat among the engineers. B is most likely to be the shortest; C is most likely to be the longest.

v 11.

The width of a confidence interval is proportional to sy ^ = s

(x − x)2 1 + ∑n . 2 n i=1 (xi − x)

∑ Since s, n, x, and ni=1 (xi − x)2 are the same for each confidence interval, the width of the confidence interval is an increasing function of the difference x − x. x = 21.7548. The value 20 is closest to x and the value 15 is the farthest from x. Therefore the confidence interval at 20 would be the shortest, and the confidence interval at 15 would be the longest.

12.

The mean temperature is (x) = 65.0. Since 60 C is closest to (x), its confidence interval would be the shortest. Since 45 C is furthest from (x), its confidence interval would be the longest.

13.

s2 = (1 − r2 )

14.

r2 satisfies the equation s2 = (1 − r2 )

∑n

2 i=1 (yi − y) ∕(n − 2) = (1 − 0.9111)(234.19)∕(17 − 2) = 1.388.

∑n

2 i=1 (yi − y) ∕(n − 2). Therefore 0.0670 = (1 − r2 )(57.313)∕(25 − 2), so r2 = 0.9731.

15. (a) t = 1.71348∕6.69327 = 0.256.

(b) n = 25, so there are n − 2 = 23 degrees of freedom. The P -value is for a two-tailed test, so it is equal to the sum of the areas to the right of t = 0.256 and to the left of t = −0.256. Thus P = 0.40 + 0.40 = 0.80.

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CHAPTER 7

satisfies the equation 3.768 = 4.27473∕s ^ , so s

= 1.13448.

(d) n = 25, so there are n − 2 = 23 degrees of freedom. The P -value is for a two-tailed test, so it is equal to the sum of the areas to the right of t = 3.768 and to the left of t = −3.768. Thus P = 0.0005 + 0.0005 = 0.001.

16. (a)

0 satisfies the equation 0.688 =

^ ^

0 ∕0.43309, so β0 = 0.29797.

(b) n = 20, so there are n − 2 = 18 degrees of freedom. The P -value is for a two-tailed test, so it is equal to the sum of the areas to the right of t = 0.688 and to the left of t = −0.688. Thus P = 0.25 + 0.25 = 0.50.

(d) n = 20, so there are n − 2 = 18 degrees of freedom. The P -value is for a two-tailed test, so it is equal to the sum of the areas to the right of t = 2.878 and to the left of t = −2.878. Thus P = 0.005 + 0.005 = 0.010.

17. (a) y ^ = 106.11 + 0.1119(4000) = 553.71. (b) y ^ = 106.11 + 0.1119(500) = 162.06. (c) Below. For values of x near 500, there are more points below the least squares estimate than above it. (d) There is a greater amount of vertical spread on the right side of the plot than on the left.

Section 7.4 1. (a) ln y = −0.4442 + 0.79833 ln x (b) y ^ = eln y ^ = e−0.4442+0.79833(ln 2500) = 330.95.

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SECTION 7.4 (c) y ^ = eln y ^ = e−0.4442+0.79833(ln 1600) = 231.76. (d) The 95% prediction interval for ln y is given as (3.9738, 6.9176). The 95% prediction interval for y is therefore (e3.9738 , e6.9176 ), or (53.19, 1009.89).

2. (a)

The least-squares line is y = 5.576 + 0.0142x. The linear model is not appropriate. The point (276.02, 9.36) is highly influential.

Residual

1 0

−1 −2 6

(b)

7 8 Fitted Value

0.4

The least-squares line is y = 6.148 + 0.629 ln x. The linear model appears to be appropriate.

Residual

0.2

−0.2

−0.4

6 7 Fitted Value

(c)

1.5

Residual

The least-squares line is y = 5.172 + 0.286x. The linear model is not appropriate. The point (276.02, 9.36) is highly influential.

0.5 0

−0.5 −1 −1.5 −2 5

7 8 Fitted Value

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CHAPTER 7

(d) The most appropriate model is y = 6.148 + 0.629 ln x. The prediction is y ^ = 6.148 + 0.629 ln 50 = 8.609.

(e) The 95% confidence interval is (8.025, 9.192).

3. (a) y = 20.162 + 1.269x

(b)

Residual

There is no apparent pattern to the residual plot. The linear model looks fine.

5 0 −5

−10 50

65 Fitted Value

The residuals increase over time. The linear model is not appropriate as is. Time, or other variables related to time, must be included in the model.

5 0 −5 −10 −15 0

10 15 Order of Observations

∑n ∑n 2 2 4. (a) x = 2000.0, y = 1782.5, i=1 (xi − x) = 10500000, i=1 (yi − y) = 1417370, ∑n i=1 (xi − x)(yi − y) = 3780500. ∑n (xi − x)(yi − y) ^1 = i=1 = 0.360048 and ^0 = y − ^1 x = 1062.404762. ∑n 2 i=1 (xi − x) ∑n

(b) r2 =

2 i=1 (xi − x)(yi − y) = 0.9603. ∑ n 2 2 i=1 (xi − x) i=1 (yi − y)

∑n

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SECTION 7.4

(c)

100

Residual

The relationship does not appear to be linear.

0 −50

−100

1000

1500

2000

2500

Fitted value

(d) False. The coefficient of determination can be large even when the relationship is not linear.

∑n ∑n 2 2 5. (a) x = 15.68, y = 36.56, i=1 (xi − x) = 430.644, i=1 (yi − y) = 63.136, ∑n i=1 (xi − x)(yi − y) = 160.538. ∑n (xi − x)(yi − y) ^1 = i=1 = 0.372786 and 0 = y − ^1 x = 30.714717. ∑n 2 i=1 (xi − x) The equation of the least-squares line is y = 30.714717 + 0.372786x. ∑n

(b) r2 =

2 i=1 (xi − x)(yi − y) = 0.9479. ∑n ∑n 2 2 i=1 (xi − x) i=1 (yi − y) 1

(c)

0.8 0.6

Residual

0.4 0.2

The relationship does not appear to be linear.

0 −0.2 −0.4 −0.6 −0.8 −1 34

Fitted value

(d) False. The coefficient of determination can be large even when the relationship is not linear.

∑n ∑n 2 2 6. (a) x = 6.0, y = 8.176190, i=1 (xi − x) = 150.0, i=1 (yi − y) = 230.418095, ∑n i=1 (xi − x)(yi − y) = 131.6.

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CHAPTER 7 ∑n

i=1 (xi − x)(yi − y) ^1 x = 2.91219. = 0.87733 and ^0 = y − ^ ∑n 2 (x − x) i=1 i

^1 =

The equation of the least-squares line is y = 2.91219 + 0.87733x. 20

(b)

Depth

The point (14, 16.8) has an unusually large x-value.

Age

∑n ∑n 2 2 x = 5.6, y = 7.745, i=1 (xi − x) = 82.8, i=1 (yi − y) = 152.3295, ∑n i=1 (xi − x)(yi − y) = 59.16. ∑n (xi − x)(yi − y) ^1 = i=1 = 0.71449 and ^0 = y − ^1 x = 3.74384. ∑n 2 i=1 (xi − x) The equation of the least-squares line is y = 3.74384 + 0.71449x.

(c) The point (6, 15.5) can be classified as an outlier. ∑n ∑n 2 2 x = 6.0, y = 7.81, i=1 (xi − x) = 150.0, i=1 (yi − y) = 174.098, ∑n (x − x)(y − y) = 131.6. i i=1 i ∑n (xi − x)(yi − y) ^1 = i=1 = 0.87733 and 0 = y − ^1 x = 2.546. ∑n 2 i=1 (xi − x) The equation of the least-squares line is y = 2.546 + 0.87733x. (d) The point (14, 16.8) is more influential. Both the slope and intercept change more when this point is removed.

7. (a) y = −235.32 + 0.695x.

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SECTION 7.4 1500

(b)

1000

Residual

500

The residual plot shows a pattern, with positive residuals at the higher and lower fitted values, and negative residuals in the middle. The model is not appropriate.

0 −500 −1000 −1500

1000

2000 3000 Fitted Value

4000

5000

0.6

(d)

0.4

Residual

0.2 0

The residual plot shows no obvious pattern. The model is appropriate.

−0.2 −0.4 −0.6 −0.8 4.5

5.5

6 6.5 7 Fitted Value

7.5

(e) The log model is more appropriate. The 95% prediction interval is (197.26, 1559.76).

8. (a) y = −0.9853 + 0.5483x

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CHAPTER 7

(b)

−3

x 10

Residual

4 2 0 −2 −4 −6 −8 −0.98

−0.975

−0.97

−0.965 −0.96 Fitted Value

−0.955

−0.95

One residual is somewhat large. The model seems appropriate. (c) y = −0.9757 + 0.07038x2

(d)

−3

x 10

Residual

4 2 0 −2 −4 −6 −8 −0.975

−0.97

−0.965 −0.96 Fitted Value

−0.955

−0.95

One residual is somewhat large. The model seems appropriate.

(e) For the model y = −0.9853 + 0.05483x, the 95% confidence interval is (−0.9716, −0.9639). For the model y = −0.9757 + 0.07038x2 , the 95% confidence interval is (−0.9724, −0.9647). The confidence intervals are quite similar.

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409

SECTION 7.4

9. (a)

80 75

California

70 65 60

(b) y = 30.122 + 0.596x.

(d) Yes, the outlier is influential. The equation of the least-squares line changes substantially when the outlier is removed.

10. (a) The least-squares line is y = 36.632 − 3.2927x.

(b)

Residual

10 0 −10 −20 −20

−10

20 10 Fitted Value

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CHAPTER 7

(d)

Residual

5 0 −5 −10 −20

20 Fitted Value

(e) The line y = −5.3657 + 85.439(1∕x) fits better. The confidence interval is (8.7434, 14.701).

0.6

11. (a)

0.4

Residual

0.2

The least-squares line is y = 0.833 + 0.235x.

−0.2 −0.4 −0.6 −0.8 −1 1

(b)

1.5

2.5 Fitted Value

3.5

0.4

Residual

0.2

The least-squares line is y = 0.199 + 1.207 ln x.

−0.2 −0.4 −0.6 0.5

1.5 2 Fitted Value

2.5

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SECTION 7.4 0.4

(c)

0.2 0 −0.2

The least-squares line is ln y = −0.0679 + 0.137x.

Residual

−0.4 −0.6 −0.8 −1 −1.2 −1.4 0

0.5

1 Fitted Value

1.5

0.2

(d)

0 Residual

The least-squares line is

y = 0.956 + 0.0874x.

−0.2

−0.4

−0.6 1

1.2

1.4 1.6 Fitted Value

1.8

(e) The model y = 0.199 + 1.207 ln x fits best. Its residual plot shows the least pattern.

(f)

0.5

The residuals show no pattern with time. 0

−0.5

−1 0

(g) The model y = 0.199 + 1.207 ln x fits best. The predicted value is y ^ = 0.199 + 1.207 ln 5.0 = 2.14.

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CHAPTER 7

(h) The 95% prediction interval is (1.689, 2.594).

12. (a) R2 = 0.464 + 0.571R1

0.5

Residual

(b)

The linear model is not appropriate. The residual plot has a strong pattern. 0

−0.5 1

1.5

2.5 3 3.5 Fitted Value

4.5

0.3

0.2

0 Residual

Residual

0.1

−0.1

−0.2

−0.2 −0.3

−0.3 1.6

1.8 2 Fitted Value

2.2

2.5

For R1 < 4, the least squares line is R2 = 1.233 + 0.264R1 . There is a pattern to the residual plot, so the linear model does not seem appropriate.

3.5 4 Fitted Value

4.5

For R1 ≥ 4, the least squares line is R2 = −0.190 + 0.710R1 . The residual plot does not have much pattern, so the linear model seems appropriate.

(d) The linear model seems to work well when R1 ≥ 4, but less well when R1 < 4. It might make sense to use a linear model to predict R2 when R1 ≥ 4, but to fit a non-linear model to predict R2 from R1 when R1 < 4.

13. (a) The model is log10 y = 0 + 1 log10 x + ε. Note that the natural log (ln) could be used in place of log10 , but common logs are more convenient since partial pressures are expressed as powers of 10.

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SECTION 7.4 0.4

(b)

0.3 0.2

Residual

0.1

The least-squares line is log10 y = −3.277 − 0.225 log10 x. The linear model appears to fit quite well.

0 −0.1 −0.2 −0.3 −0.4 −0.5 −2

−1.5

Fitted Value

−1

−0.5

(c) The theory says that the coefficient 1 of log10 x in the linear model is equal to −0.25. The estimated value is ^ = −0.225. We determine whether the data are consistent with the theory by testing the hypotheses H0 1 = −0.25 vs. H1 1 ≠ −0.25. The value of the test statistic is t = 0.821. There are 21 degrees of freedom, so 0.20 < P < 0.50 (a computer package gives P = 0.210). We do not reject H0 , so the data are consistent with the theory.

450

14. (a)

Acid Level (mg/l)

400 350 300 250 Outlier

200 150 100

200

300 400 500 Ester Level (mg/l)

600

700

(b) ^0 = 352.32, s ^ = 52.774, ^1 = −0.2495, s ^ = 0.1951. 0

(c) t14 = −1.28, 0.20 < P < 0.50. A computer package gives P = 0.221. (d) ^0 = 359.23, s ^ = 91.076, ^1 = −0.2834, s ^ = 0.4108. 0

(e) t13 = −0.690, P

.50. A computer package gives P = 0.502.

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CHAPTER 7

(f) Not useful. The P -values for the slope are not small, whether or not the outlier is included. Therefore it is plausible that the slope is equal to 0, so a linear model is not helpful for prediction.

15. (a) y = 2049.87 − 4.270x

(b) (12, 2046) and (13, 1954) are outliers. The least-squares line with (12, 2046) deleted is y = 2021.85 − 2.861x. The least-squares line with (13, 1954) deleted is y = 2069.30−5.236x. The least-squares line with both outliers deleted is y = 2040.88 − 3.809x.

16. (a) y = 457.270059 + 87.555863x

Residual

5000

−5000 0

2000

4000 6000 Fitted Value

8000

6000 5000 4000 Residual

(b) y = −504.001756 + 1165.271152 ln x

3000 2000 1000 0 −1000 −2000 −2000 −1000

1000 2000 3000 4000 5000 Fitted Value

Residual

0.5 0 −0.5 −1 4

6 7 Fitted Value

(d) The residual plot shows that the model ln y = 5.258 + 0.799 ln x fits best. (e) Let y ^ be the predicted flow when the discharge is 50 km3 /yr. Then ln y ^ = 5.258333 + 0.799002 ln 50 = 8.38405.

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415

SUPPLEMENTARY EXERCISES FOR CHAPTER 7

So y ^ = e8.38405 = 4376.7 m3 /s

(f) The 95% prediction interval for ln y is (7.282961, 9.485138). The 95% prediction interval for y is (e7.282961 , e9.485138 ), or (1455.3, 13162.6).

17.

The equation becomes linear upon taking the log of both sides: ln W = 0 + β1 ln L, where 0 = ln a and 1 = b.

18.

ln y =

0 + 1 ln x1 + 2 ln x2 , where 0 = ln a, 1 = −b, and 2 = −c.

19. (a) A physical law.

(b) It would be better to redo the experiment. If the results of an experiment violate a physical law, then something was wrong with the experiment, and you can’t fix it by transforming variables.

Supplementary Exercises for Chapter 7 ∑n ∑n 2 2 1. (a) x = 1.48, y = 1.466, i=1 (xi − x) = 0.628, i=1 (yi − y) = 0.65612, ∑n i=1 (xi − x)(yi − y) = 0.6386, n = 5. ∑n i=1 (xi − x)(yi − y) ^1 x = −0.038981. ^0 = y − ^ = 1.016879 and ^ ∑n 1 = 2 i=1 (xi − x) (b) 0

(xi − x)(yi − y)]2 (d) r2 = ∑n i=1 = 0.989726, s = ∑n 2 2 i=1 (xi − x) i=1 (yi − y) [

(1 − r2 )

∑n

i=1 (yi − y)

n−2

= 0.0474028,

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416

CHAPTER 7 v s

1 x2 + ∑n = 0.0910319. 2 n i=1 (xi − x)

The null and alternative hypotheses are H0

0 = 0 versus H1

0 ≠ 0.

There are n − 2 = 3 degrees of freedom. t = (−0.038981 − 0)∕0.0910319 = −0.428.

Since the alternative hypothesis is of the form 0 ≠ b, the P -value is the sum of the areas to the right of t = 0.428 and to the left of t = −0.428. From the t table, 0.50 < P < 0.80. A computer package gives P = 0.698. The data are consistent with the Beer–Lambert law.

2. (a) y = 4.9x2 , so the relationship between x and y is not linear. The correlation coefficient is not appropriate.

(b) y = 4.9w, so the relationship between w and y is linear. The correlation coefficient is appropriate.

4.9x, so the relationship between v and x is linear. The correlation coefficient is appropriate.

(d) v =

4.9 w, so the relationship between v and w is not linear. The correlation coefficient is not appropriate.

(e) ln y = ln 4.9+2 ln x, so the relationship between ln y and ln x is linear. The correlation coefficient is appropriate.

3. (a)

100 90 80 70 60 50 40 40

100

∑n ∑n 2 2 (b) x = 71.101695, y = 70.711864, i=1 (xi − x) = 10505.389831, i=1 (yi − y) = 10616.101695, ∑n i=1 (xi − x)(yi − y) = −7308.271186, n = 59. ∑n i=1 (xi − x)(yi − y) ^1 x = 120.175090. = −0.695669 and ^0 = y − ^ ∑n 1 = 2 (x − x) i i=1

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SUPPLEMENTARY EXERCISES FOR CHAPTER 7

Ti+1 = 120.175090 − 0.695669Ti . ∑n

(xi − x)(yi − y)]2 (c) r2 = ∑n i=1 = 0.478908, s = ∑n 2 2 i=1 (xi − x) i=1 (yi − y) [

s ^ ^ = t

∑n

(1 − r2 )

∑n

2 i=1 (yi − y)

n−2

= 9.851499,

= 0.096116.

2 i=1 (xi − x)

There are n − 2 = 57 degrees of freedom. t57,.025 Therefore a 95% confidence interval for

.002.

1 is −0.695669 ± 2.002(0.096116), or (−0.888, −0.503).

(d) y ^ = 120.175090 − 0.695669(70) = 71.4783 minutes. v (e) y ^ = 71.4783, sy ^ = s

(x − x)2 1 + ∑n = 1.286920. 2 n i=1 (xi − x)

There are 59 − 2 = 57 degrees of freedom. t57,.01

.394.

Therefore a 98% confidence interval is 71.4783 ± 2.394(1.286920), or (68.40, 74.56). v (f) y ^ = 71.4783, spred = s

(x − x)2 1 + ∑n = 9.935200. 2 n i=1 (xi − x)

.6649. A 95% prediction interval is 71.4783 ± 2.6649(9.935200) or (45.00, 97.95).

t57,.005 30

4. (a)

Residual

20 10

There appears to be some heteroscedasticity, with a larger spread in residuals for smaller fitted values.

−10 −20 −30

70 75 80 Fitted Value

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418

CHAPTER 7 30

(b)

Residual

20 10

No violations of the standard assumptions are indicated.

−10 −20 −30 0

20 30 40 Order of Observations

∑n ∑n 2 2 5. (a) x = 50, y = 47.909091, i=1 (xi − x) = 11000, i=1 (yi − y) = 9768.909091, ∑n i=1 (xi − x)(yi − y) = 10360, n = 11. ∑n i=1 (xi − x)(yi − y) ^1 x = 0.818182. ^0 = y − ^ = 0.941818 and ^ ∑n 1 = 2 − x) (x i=1 i v ∑n ∑n 2 y)] (1 − r2 ) i=1 (yi − y)2 [ (x − x)(y − i i=1 i 2 (b) r = ∑n = 0.998805, s = = 1.138846. ∑ n 2 2 n−2 i=1 (xi − x) i=1 (yi − y) v 2 ^0 = 0.818182, s ^^ = s 1 + ∑ x ^ = 0.642396. n 2 0 n i=1 (xi − x) The null and alternative hypotheses are H0

0 = 0 versus H1

0 ≠ 0.

There are 11 − 2 = 9 degrees of freedom. t = (0.818182 − 0)∕0.642396 = 1.274.

Since the alternative hypothesis is of the form 0 ≠ b, the P -value is the sum of the areas to the right of t = 1.274 and to the left of t = −1.274. From the t table, 0.20 < P < 0.50. A computer package gives P = 0.235. It is plausible that

0 = 0.

1 = 1 versus H1

1 ≠ 1.

s ^1 = 0.941818, s ^^ = t ^ = 0.010858. 1 ∑n 2 (x − x) i=1 i There are 11 − 2 = 9 degrees of freedom. t = (0.941818 − 1)∕0.010858 = −5.358.

Since the alternative hypothesis is of the form 1 ≠ b, the P -value is the sum of the areas to the right of t = 5.358 and to the left of t = −5.358. From the t table, P < 0.001. A computer package gives P = 0.00046.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 7

We can conclude that

1 ≠ 1.

(d) Yes, since we can conclude that

1 ≠ 1, we can conclude that the machine is out of calibration.

Since two coefficients were tested, some may wish to apply the Bonferroni correction, and multiply the P -value for 1 by 2. The evidence that 1 ≠ 1 is still conclusive. ^1 (20) = 19.65455. ^0 + ^ (e) x = 20, y ^ = ^ v (x − x)2 1 sy ^ = s + ∑n = 0.47331. There are 9 degrees of freedom. t9,.025 = 2.262. 2 n i=1 (xi − x) Therefore a 95% confidence interval for the mean response is 19.65455 ± 2.262(0.47331), or (18.58, 20.73). ^1 (80) = 76.163636. (f) x = 80, y ^ = 0 + ^ v (x − x)2 1 sy ^ = s + ∑n = 0.47331. There are 9 degrees of freedom. t9,.025 = 2.262. 2 n i=1 (xi − x) Therefore a 95% confidence interval for the mean response is 76.163636 ± 2.262(0.47331), or (75.09, 77.23).

(g) No, when the true value is 20, the result of part (e) shows that a 95% confidence interval for the mean of the measured values is (18.58, 20.73). Therefore it is plausible that the mean measurement will be 20, so that the machine is in calibration.

∑n ∑n 2 2 6. (a) x = 4.036667, y = 51.3333, i=1 (xi − x) = 21.150533, i=1 (yi − y) = 3411.333333, ∑n i=1 (xi − x)(yi − y) = 268.166667, n = 6. ∑n i=1 (xi − x)(yi − y) ^1 x = 0.152617. = 12.678955 and 0 = y − ^ ∑n 1 = 2 (x − x) i i=1 y = 0.152617 + 12.678955x

(b) y ^ =

^ ^

0 + 1 (3.00) = 38.189.

7. (a) x = 92.0, y = 0.907407,

∑n

i=1 (xi − x)

2 = 514.0,

∑n

i=1 (yi − y)

2 = 4.738519,

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420

CHAPTER 7 ∑n

i=1 (xi − x)(yi − y) = 20.1,

n = 27.

∑n

1 =

i=1 (xi − x)(yi − y) ^1 x = −2.690258 ^0 = y − ^ = 0.0391051 and ^ ∑n 2 − x) (x i i=1

y = −2.690258 + 0.0391051x v ∑ 2 − y)] (1 − r2 ) ni=1 (yi − y)2 (x − x)(y i i = 0.165877, s = = 0.397618. (b) r2 = ∑n i=1 ∑n 2 2 n−2 i=1 (xi − x) i=1 (yi − y) v x2 1 = −2.690258, = s + = 1.615327. ∑ 0 n 2 0 n i=1 (xi − x) [

∑n

1 = 0.0391051,

∑n

= 0.0175382.

2 i=1 (xi − x)

There are 27 − 2 = 25 degrees of freedom. t25,.025 = 2.060. Therefore a 95% confidence interval for 0 is −2.690258 ± 2.060(1.615327), or (−6.0171, 0.6366). A 95% confidence interval for 1 is 0.0391051 ± 2.060(0.0175382), or (0.00298, 0.0752). ^1 (93) = 0.94651. ^0 + ^ (c) x = 93, y ^ = ^

^1 (93) = 0.94651. ^0 + ^ (d) x = 93, y ^ = ^ v (x − x)2 1 + ∑n = 0.078506. sy ^ = s 2 n i=1 (xi − x) There are 25 degrees of freedom. t25,.025 = 2.060. Therefore a 95% confidence interval for the mean response is 0.94651 ± 2.060(0.078506), or (0.785, 1.108).

(e) x = 93, y ^ = v spred = s

^ ^

0 + 1 (93) = 0.94651.

(x − x)2 1 + ∑n = 0.405294. 2 n i=1 (xi − x)

There are 25 degrees of freedom. t25,.025 = 2.060. Therefore a 95% prediction interval is 0.94651 ± 2.060(0.405294), or (0.112, 1.781).

8. (a) x = 4.43, y = 3.451875,

∑n

i=1 (xi − x)

2 = 11.2828,

∑n

i=1 (yi − y)

2 = 37.7994875,

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SUPPLEMENTARY EXERCISES FOR CHAPTER 7 ∑n

i=1 (xi − x)(yi − y) = 14.3206,

421

n = 32.

∑n

^1 =

i=1 (xi − x)(yi − y) = 1.269242 and ^0 = y − ^1 x = −2.170866. ∑n 2 − x) (x i i=1

y = −2.170866 + 1.269242x v ∑ 2 (x − x)(y − y)] (1 − r2 ) ni=1 (yi − y)2 i i i=1 2 = 0.480861, s = = 0.808768. (b) r = ∑n ∑n 2 2 n−2 i=1 (xi − x) i=1 (yi − y) v 2 ^0 = −2.170866, s ^ = s 1 + ∑ x = 1.076183. n 2 0 n i=1 (xi − x) [

∑n

s ^1 = 1.269242, s ^ = t = 0.240777. 1 ∑n 2 (x − x) i=1 i There are 32 − 2 = 30 degrees of freedom. t30,.025 = 2.042. Therefore a 95% confidence interval for 0 is −2.170866 ± 2.042(1.076183), or (−4.368, 0.027). A 95% confidence interval for 1 is 1.269242 ± 2.042(0.240777), or (0.778, 1.761). (c) By ^1 (1.5) = 1.269242(1.5) = 1.904. (d) x = 5.0, y ^ = ^0 + ^1 (5.0) = 4.175343. v (x − x)2 1 s =s + ∑n = 0.198183. 2 n i=1 (xi − x) There are 30 degrees of freedom. t30,.025 = 2.042. Therefore a 95% confidence interval for the mean response is 4.175343 ± 2.042(0.198183), or (3.771, 4.580). (e) Let y5.0 =

0 + 1 (5.0) denote the mean lipase production when the cell mass is 5.0 g/L.

The null and alternative hypotheses are H0 y5.0 ≥ 4.4 versus H1 y5.0 < 4.4. v (x − x)2 1 = 0.198183. + ∑n y ^ = ^0 + ^1 (5.0) = 4.175342, sy ^ = s 2 n i=1 (xi − x)

t = (4.175342 − 4.4)∕0.198183 = −1.1336 There are 27 − 2 = 25 degrees of freedom. Since the alternative hypothesis is y5.0 < 4.4, the P -value is the area to the left of t = −1.134. From the t table, 0.10 < P < 0.25. A computer package gives P = 0.134. We cannot conclude that the mean lipase production is less than 4.4 when the cell mass is 5.0.

9. (a) ln y =

0 + 1 ln x, where 0 = ln k and 1 = r.

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422

CHAPTER 7

(b) Let ui = ln xi and let vi = ln yi .

∑n ∑ 2 u = 1.755803, v = −0.563989, ni=1 (ui − u)2 = 0.376685, i=1 (vi − v) = 0.160487, ∑n i=1 (ui − u)(vi − v) = 0.244969, n = 5. ∑n (ui − u)(vi − v) ^1 u = −1.705838. ^0 = v − β^ ^1 = i=1 = 0.650328 and β ^ ∑n 2 i=1 (ui − u) The least-squares line is ln y = −1.705838 + 0.650328 ln x. ^ Therefore = 0.650328 and ^ k = e−1.705838 = 0.18162.

(c) The null and alternative hypotheses are H0 r = 0.5 versus H1 r ≠ 0.5. s = 0.0322616. = 0.650328, s = 0.0198005, s ^^r = t ∑n 2 (u − u) i=1 i There are 5 − 2 = 3 degrees of freedom. t = (0.650328 − 0.5)∕0.0322616 = 4.660.

Since the alternative hypothesis is of the form r ≠ r0 , the P -value is the sum of the areas to the right of t = 4.660 and to the left of t = −4.660.

From the t table, 0.01 < P < 0.02. A computer package gives P = 0.019. We can conclude that ≠ 0.5.

∑n ∑ 2 10. (a) x = 28.078261, y = 29.221739, ni=1 (xi − x)2 = 287.999130, i=1 (yi − y) = 1692.179130, ∑n i=1 (xi − x)(yi − y) = −347.219130, n = 23. ∑n (xi − x)(yi − y) ^1 = i=1 = −1.205626 and β ^0 = y − β ^1 x = 63.073610, ∑n 2 i=1 (xi − x) The least-squares line is y = 63.073610 − 1.205626x. v ∑ 2 (x − x)(y − y)] (1 − r2 ) ni=1 (yi − y)2 i i i=1 2 = 0.247383, s = = 7.787545. (b) r = ∑n ∑n 2 2 n−2 i=1 (xi − x) i=1 (yi − y) v 2 ^0 = 63.073610, s ^ = s 1 + ∑ x β ^ = 12.986644. n β0 2 n i=1 (xi − x) [

∑n

There are 23 − 2 = 21 degrees of freedom. t21,.025 = 2.080. Therefore a 95% confidence interval for ^ ^1 = −1.205626, s ^ = t β 1

∑n

0 is 63.073610 ± 2.080(12.986644), or (36.1, 90.1).

= 0.458886.

2 i=1 (xi − x)

There are 23 − 2 = 21 degrees of freedom. t21,.025 = 2.080. Therefore a 95% confidence interval for

1 is −1.205626 ± 2.080(0.458886), or (−2.160, −0.251).

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423

SUPPLEMENTARY EXERCISES FOR CHAPTER 7 (d) The null and alternative hypotheses are H0 β1 ≥ 0 versus H1 s ^ ^1 = −1.205626, s ^ = t = 0.458886. β1 ∑n 2 (x − x) i=1 i

1 < 0.

There are 23 − 2 = 21 degrees of freedom. t = (−1.205626 − 0)∕0.458886 = −2.627. Since the alternative hypothesis is of the form

1 < b, the P -value is the area to the left of t = −2.627.

From the t table, 0.005 < P < 0.01. A computer package gives P = 0.00787. We can conclude that butterflies with larger wingspans have shorter flight periods on the average. (e) Let y30 = β0 + β1 (30) denote the mean flight period for butterflies whose wingspan is 30 mm. The null and alternative hypotheses are H0 y30 ≥ 28 versus H1 y30 < 28. v 2 ^1 (30) = 26.904841, sy ^ = s 1 + ∑ (x − x) ^0 + β^ y ^ = β ^ = 1.847824. n 2 n i=1 (xi − x)

t = (26.904841 − 28)∕1.847824 = −0.593. There are 23 − 2 = 21 degrees of freedom. Since the alternative hypothesis is y30 < 0.93, the P -value is the area to the left of t = −0.593. From the t table, 0.25 < P < 0.40. A computer package gives P = 0.280. We cannot conclude that the mean flight period for butterflies with wingspan 30 mm is less than 28 days. ^1 (28.5) = 28.713280. ^0 + ^ (f) x = 28.5, y ^ = β^ v (x − x)2 1 = 7.957392. spred = s 1 + + ∑n 2 n i=1 (xi − x) There are 21 degrees of freedom. t21,.025 = 2.080. Therefore a 95% prediction interval is 28.713280 ± 2.080(7.957392), or (12.16, 45.26). ∑n ∑ 2 11. (a) x = 2812.130435, y = 612.739130, ni=1 (xi − x)2 = 335069724.6, i=1 (yi − y) = 18441216.4, ∑n i=1 (xi − x)(yi − y) = 32838847.8. ∑n (xi − x)(yi − y) ^0 = y − β ^1 x = 337.133439. ^1 = i=1 = 0.098006 and β ^ ∑n 2 − x) (x i=1 i The least-squares line is y = 337.133 + 0.098x. 3000

(b) Residual

2000 1000 0

−1000 −2000 0

500

1000 1500 Fitted Value

2000

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CHAPTER 7

(c) Let ui = ln xi and let vi = ln yi . ∑n ∑ 2 u = 6.776864, v = 5.089504, ni=1 (ui − u)2 = 76.576108, i=1 (vi − v) = 78.642836, ∑n i=1 (ui − u)(vi − v) = 62.773323. ∑n (ui − u)(vi − v) ^1 u = −0.465835. ^0 = v − β^ ^ = 0.819751 and β ^ β1 = i=1 ∑n 2 i=1 (ui − u) The least-squares line is ln y = −0.466 + 0.820 ln x. 3

(d) Residual

2 1 0

−1 −2 −3 0

4 Fitted Value

(e) Let ui = ln xi and let vi = ln yi . ^1 (ln 3000) = 6.09739080. ^0 + β^ x = 3000, so u = ln x = ln 3000, and v ^ = β ^ v (u − u)2 1 spred = s 1 + + ∑n = 1.17317089. 2 n i=1 (ui − u) There are 21 degrees of freedom. t21,.025 = 2.080. Therefore a 95% prediction interval for v = ln y is 6.09739080 ± 2.080(1.17317089), or (3.657195, 8.537586). A 95% prediction interval for y is (e3.657195 , e8.537586 ), or (38.75, 5103.01).

12.

(iv). Since Newton’s law of cooling is a physical law, the plot of ln temperature versus time should be linear. The non-linearity of the plot suggests that something went wrong with the experiment.

13. (a)

∑n

i=1 (xi − x)

2 = 0.712761,

∑n

r= t

∑n

i=1 (yi − y)

i=1 (xi − x)(yi − y)

2 i=1 (xi − x)

∑n

2 = 19338.195122, ∑n (x − x)(y − y) = −102.158537. i i=1 i

= −0.870151

2 i=1 (yi − y)

∑n ∑n 2 2 (b) x = 0.449024, y = 750.536585, i=1 (xi − x) = 0.712761, i=1 (yi − y) = 19338.195122, ∑n i=1 (xi − x)(yi − y) = −102.158537.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 7

425

∑n

i=1 (xi − x)(yi − y) ^1 x = 814.894310. ^0 = y − ^ = −143.327904 and ^ ∑n 2 (x − x) i=1 i

^1 =

The equation of the least-squares line is y = 814.894310 − 143.327904x. 40

(c)

30 20

The residuals corresponding to x ≥ 0.5 are the ones with the smaller fitted values. They are more spread out than the residuals corresponding to x < 0.5.

Residual

10 0 −10 −20 −30 −40 700

720

740

760

Fitted Value

780

800

∑n ∑n 2 2 (d) x = 0.351304, y = 766.782609, i=1 (xi − x) = 0.121061, i=1 (yi − y) = 2627.913043, ∑n i=1 (xi − x)(yi − y) = −13.453478. ∑n (xi − x)(yi − y) ^1 x = 805.823014. ^0 = y − β^ ^1 = i=1 = −111.129866 and β ^ ∑n 2 − x) (x i i=1 The equation of the least-squares line is y = 805.823014 − 111.129866x. ∑n ∑n 2 2 (e) x = 0.573889, y = 729.777778, i=1 (xi − x) = 0.091428, i=1 (yi − y) = 2883.111111, ∑n i=1 (xi − x)(yi − y) = −5.534444. ∑n (xi − x)(yi − y) ^0 = y − 1 x = 764.517287. ^1 = i=1 = −60.533512 and ^ ∑n 2 i=1 (xi − x) The equation of the least-squares line is y = 764.517287 − 60.533512x. ∑n

2 i=1 (xi − x)(yi − y) = 0.5689. ∑n 2 2 i=1 (xi − x) i=1 (yi − y)

(f) r2 = ∑n

∑n

(g) r2 =

2 i=1 (xi − x)(yi − y) = 0.1162. ∑ n 2 2 i=1 (xi − x) i=1 (yi − y)

∑n

(h) The coefficient of determination is greater when x < 0.5, so the average number of late payments is more strongly related to average credit score when the average number of late payments is less than 0.5.

14. (a) y ^ = −0.23429 + 0.72868(2.5) = 1.5874. (b) The null and alternative hypotheses are H0 β1 = 1 versus H1 β1 ≠ 1. ^1 = 0.72868, s ^ = 0.06353 β1

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426

CHAPTER 7

There are 13 − 2 = 11 degrees of freedom. t = (0.72868 − 1)∕0.06353 = −4.271.

Since the alternative hypothesis is of the form 1 ≠ b, the P -value is the sum of the areas to the right of t = 4.271 and to the left of t = −4.271. From the t table, 0.001 < P < 0.002. A computer package gives P = 0.0013. We can conclude that β1 ≠ 1. The claim is not plausible.

0 ≠ 0.

From the output, P = 0.350. The data are consistent with the fact that there is no runoff when there is no rainfall.

15.

(ii). The standard deviation sy ^ is not given in the output. To compute sy ^, the quantity known. v

16.

The width of the confidence interval is proportional to sy ^ = s

∑n

i=1 (xi − x)

2 must be

(x − x)2 1 + ∑n . 2 n i=1 (xi − x)

∑ Since s, n, and ni=1 (xi − x)2 are the same for each confidence interval, the width of the confidence interval is an increasing function of (x − x). x = 53. Therefore the confidence interval for 54 MPa would be the shortest, and the one for 11 MPa would be the longest.

17. (a) If f = 1∕2 then 1∕f = 2. The estimate is ^t = 145.736 − 0.05180(2) = 145.63.

(b) Yes. r = − R-Sq = −0.988. Note that r is negative because the slope of the least-squares line is negative. (c) If f = 1 then 1∕f = 1. The estimate is ^t = 145.736 − 0.05180(1) = 145.68.

18. (a) If y is unbiased, then y = true value + ε. If the calculated value x is equal to the true value, then y = x + ε. ∑n ∑n 2 2 (b) x = 0.775714, y = 0.767857, i=1 (xi − x) = 0.955743, i=1 (yi − y) = 0.968036, ∑n i=1 (xi − x)(yi − y) = 0.935871, n = 14.

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427

SUPPLEMENTARY EXERCISES FOR CHAPTER 7 ∑n β ^1 =

i=1 (xi − x)(yi − y) = 0.979208 and ^0 = y − ^1 x = 0.00827120. ∑n 2 (x − x) i=1 i

y = 0.00827120 + 0.979208x.

(c) If the calculated value is accurate, then y = 0 + 1x + ε. v ∑n ∑ 2 [ (x − x)(y − y)] (1 − r2 ) ni=1 (yi − y)2 i i=1 i 2 (d) r = ∑n = 0.946673, s = = 0.0655887. ∑n 2 2 n−2 i=1 (xi − x) i=1 (yi − y) v 2 ^0 = 0.00827120, s ^ = s 1 + ∑ x = 0.0549156. n 2 0 n i=1 (xi − x) The null and alternative hypotheses are H0 β0 = 0 versus H1 β0 ≠ 0.

There are 14 − 2 = 12 degrees of freedom. t = (0.00827120 − 0)∕0.0549156 = 0.151.

Since the alternative hypothesis is of the form β0 ≠ b, the P -value is the sum of the areas to the right of t = 0.151 and to the left of t = −0.151. From the t table, P > 0.80. A computer package gives P = 0.88. It is plausible that β0 = 0. The null and alternative hypotheses are H0 β1 = 1 versus H1 β1 ≠ 1. s = 0.0670901. β ^1 = 0.979208, s ^ = t 1 ∑n 2 (x − x) i=1 i There are 14 − 2 = 12 degrees of freedom. t = (0.979208 − 1)∕0.0670901 = −0.310. Since the alternative hypothesis is of the form t = −0.310 and to the left of t = −0.310.

≠ b, the P -value is the sum of the areas to the right of

From the t table, 0.50 < P < 0.80. A computer package gives P = 0.762. It is plausible that

1 = 1.

(e) It is plausible that the calculated value is accurate. Neither of the hypotheses β0 = 0 nor β1 = 1 can be rejected.

∑ 19. (a) We need to minimize the sum of squares S = (yi − β ^xi )2 . ∑ We take the derivative with respect to β ^ and set it equal to 0, obtaining −2 xi (yi − β ^xi ) = 0. ∑ ∑ ∑ ∑ Then xi yi − β ^ x2i = 0, so β ^ = xi yi ∕ x2i . (b) Let ci = xi ∕

∑

x2i . Then β ^ =

∑

ci yi , so σ 2 ^ = β

∑ 2 2 ∑ ∑ ∑ ci σ = σ 2 x2i ∕( xi2 )2 = σ 2 ∕ x2i .

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428 20.

CHAPTER 7 ∑n

∑n i=1 (xi − x) = i=1 xi − nx = 0. ∑n ∑ Since x is constant, i=1 x(xi − x) = x ni=1 (xi − x) = 0 as well. ∑n ∑ We now show that ni=1 xi (xi − x) = i=1 (xi − x)2 . This follows since ∑n ∑n ∑n ∑n i=1 xi (xi − x) = i=1 xi (xi − x) − i=1 x(xi − x) = i=1 [xi (xi − x) − x(xi − x)] ∑n ∑n 2 = i=1 (xi − x)(xi − x) = i=1 (xi − x) . First note that

Now μβ ^

n É

(xi − x) μ ∑n 2 yi i=1 (xi − x) i=1

n É

(xi − x) ( β − β 1 xi ) ∑n 2 0 i=1 (xi − x) i=1 M L n É 1 = ∑n (xi − x)( 0 + 1 xi ) 2 i=1 (xi − x) i=1 L M n n É É 1 β0 (xi − x)xi (xi − x) + 1 = ∑n 2 i=1 (xi − x) i=1 i=1 L M n É 1 2 = ∑n (xi − x) 0 (0) + 1 2 i=1 (xi − x) i=1 L M n É 1 β1 = ∑n (xi − x)2 2 (x − x) i i=1 i=1 =

21.

β1

From the answer to Exercise 20, we know that ∑n ∑n 2 i=1 xi (xi − x) = i=1 (xi − x) . Now μ ^

= = =

∑n

i=1 (xi − x) = 0,

∑n

i=1 x(xi − x) = 0, and

L M n É x(xi − x) 1 − ∑n μy i 2 n i=1 (xi − x) i=1 L M n É x(xi − x) 1 ( β 0 + β 1 xi ) − ∑n 2 n i=1 (xi − x) i=1 ∑n ∑n n n É É xi 1 i=1 x(xi − x) i=1 xi x(xi − x) + β1 − β0 ∑ n − β1 ∑ β0 n 2 2 n n i=1 (xi − x) i=1 (xi − x) i=1 i=1 ∑n i=1 xi (xi − x) 0 + 1 x − 0 − 1 x ∑n 2 i=1 (xi − x)

0 + 1x − 0 − 1x

β0

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SUPPLEMENTARY EXERCISES FOR CHAPTER 7

22.

σ 2 ^ β1

∑n

σ 2 ^

[

(xi − x)

∑n

i=1

23.

n É

2 i=1 (xi − x)

429

M2 σ2

2 i=1 (xi − x) σ2 2 ]2 (x − x) i i=1 σ2

∑n

i=1 (xi − x)

L M2 n É x(xi − x) 1 − ∑n σ2 2 n − x) (x i i=1 i=1 L M ∑n n 2 É 1 2x (xi − x) 2 i=1 (xi − x) − + x ∑n σ2 ∑n 2 2 2 ]2 n n [ (x − x) − x) (x i=1 i i=1 i i=1 M L n ∑n ∑n 2 É 1 x i=1 (xi − x) 2 i=1 (xi − x) − 2 ∑n + x ∑n σ2 2 2 2 ]2 n n [ (x − x) − x) (x i=1 i i=1 i i=1 L M 2 1 x x σ2 − 2 (0) + ∑n 2 n n − x) (x i i=1 L M 2 1 x σ2 + ∑n 2 n − x) (x i=1 i

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430

CHAPTER 8

Chapter 8 Section 8.1 1. (a) The predicted strength is 26.641 + 3.3201(8.2) − 0.4249(10) = 49.62 kg/mm2 . (b) By 3.3201(10) = 33.201 kg/mm2 . (c) By 0.4249(5) = 2.1245 kg/mm2 .

2. (a)

1 = 3.3201, s 1 = 0.33198, n = 20.

There are 20 − 3 = 17 degrees of freedom. t17,.025 = 2.110. A 95% confdence interval is 3.3201 ± 2.110(0.33198), or (2.620, 4.021). (b) 2 = −0.4249, s = 0.12606, n = 20. 2

There are 20 − 3 = 17 degrees of freedom. t17,.005 = 2.898. A 99% confdence interval is −0.4249 ± 2.898(0.12606), or (−0.790, −0.060). (c) 1 = 3.3201, s = 0.33198, n = 20.

The null and alternative hypotheses are H0 ∶ 1 ≤ 3 versus H1 ∶ 1 > 3. There are 20 − 3 = 17 degrees of freedom. t = (3.3201 − 3)∕0.33198 = 0.964. Since the alternative hypothesis is 1 > 3, the P -value is the area to the right of t = 0.964. 1

From the t table, 0.10 < P < 0.25. A computer package gives P = 0.174. We cannot conclude that

1 > 3.

̂2 = −0.4249, s = 0.12606, n = 20. (d)

The null and alternative hypotheses are H0 ∶ 2 ≥ −0.1 versus H1 ∶ 1 < −0.1. There are 20 − 3 = 17 degrees of freedom. t = [−0.4249 − [−0.1])∕0.12606 = −2.577. Since the alternative hypothesis is 2 < −0.1, the P -value is the area to the left of t = −2.577. 2

From the t table, 0.005 < P < 0.01. A computer package gives P = 0.00979. We can conclude that

2 < −0.1.

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431

SECTION 8.1

Residuals

The linear model is reasonable. There is no obvious pattern to the plot.

0 −1 −2 44

48 50 Fitted Values

4. (a) y = −0.83 + 0.017(20) + 0.0895(17) + 42.771(0.006) + 0.027(10) − 0.0043(17)(10) = 0.8271 seconds. (b) Yes, the predicted change is 0.017(2) = 0.034 seconds. (c) No, the predicted change depends on the value of x4 , because of the interaction term x2 x4 . (d) R2 = SSR∕SST = SSR∕(SSR + SSE) = 1.08613∕(1.08613 + 0.036310) = 0.967651. (e) There are 5 degrees of freedom for regression and 44 degrees of freedom for error. SSR∕5 = 263.23. From the F table, P < 0.001. A computer package gives P F5,44 = SSE∕44 The null hypothesis can be rejected.

5. (a) y = 56.145−9.046(3)−33.421(1.5)+0.243(20)−0.5963(3)(1.5)−0.0394(3)(20)+0.6022(1.5)(20)+0.6901(32 )+ 11.7244(1.52 ) − 0.0097(202 ) = 25.465.

(b) No, the predicted change depends on the values of the other independent variables, because of the interaction terms. (c) R2 = SSR∕SST = (SST − SSE)∕SST = (6777.5 − 209.55)∕6777.5 = 0.9691.

(d) There are 9 degrees of freedom for regression and 27 − 9 − 1 = 17 degrees of freedom for error. SSR∕9 (SST − SSE)∕9 F9,17 = = = 59.204. SSE∕17 SSE∕17 From the F table, P < 0.001. A computer package gives P = 4.6 × 10−11 .

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432

CHAPTER 8

The null hypothesis can be rejected.

6. (a) Yes, the predicted productivity would increase by 151.680 m3 .

(b) No, the number of trucks, the match factor, and the truck travel time must also be known, because these variables appear in interactions with haul length.

7. (a) y = −0.219471 + 0.779023(2.113) − 0.108270(0) + 1.353613(1.4) − 0.001343(730) = 2.3413 liters

(b) By 1.353613(0.05) = 0.06768 liters

(c) Nothing is wrong. In theory, the constant estimates FEV1 for an individual whose values for the other variables are all equal to zero. Since these values are outside the range of the data (e.g., no one has zero height), the constant need not represent a realistic value for an actual person.

8. (a) 1 = 0.779023, s

= 0.0490921. Since n = 165, there are 160 degrees of freedom for error.

t160,.025 = 1.9749. A 95% confdence interval is 0.779023 ± 1.9749(0.0490921), or (0.6821, 0.8760). (b) 3 = 1.353613, s = 0.2880034. Since n = 165, there are 160 degrees of freedom for error. 3

t160,.01 = 2.3499. A 95% confdence interval is 1.353613 ± 2.3499(0.2880034), or (0.6768, 2.0304). (c) 2 = −0.108270, s = 0.0352002.

2 ≥ −0.08 versus H1 ∶

The null and alternative hypotheses are H0 ∶

2 < −0.08.

Since n = 164, there are 160 degrees of freedom for error. t = [−0.108270 − (−0.08)]∕0.0352002 = −0.80312. Since the alternative hypothesis is of the form P = 0.21155. We cannot conclude that

2 < b, the P -value is the area to the left of t = −0.80312.

2 < −0.08.

(d) 3 = 1.353613, s = 0.2880034.

3 ≤ 0.5 versus H1 ∶

The null and alternative hypotheses are H0 ∶

3 > 0.5.

Since n = 164, there are 160 degrees of freedom for error. t = (1.353613 − 0.5)∕0.2880034 = 2.9639. Since the alternative hypothesis is of the form

3 > b, the P -value is the area to the right of t = 2.9639.

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433

SECTION 8.1

P = 0.00175. We can conclude that

3 > 0.5.

9. (a) y = −1.7914 + 0.00026626(1500) + 9.8184(1.04) − 0.29982(17.5) = 3.572. (b) By 9.8184(0.01) = 0.098184. (c) Nothing is wrong. The constant estimates the pH for a pulp whose values for the other variables are all equal to zero. Since these values are outside the range of the data (e.g., no pulp has zero density), the constant need not represent a realistic value for an actual pulp. (d) From the output, the confdence interval is (3.4207, 4.0496). (e) From the output, the prediction interval is (2.2333, 3.9416). (f) Pulp B. The standard deviation of its predicted pH (SE Fit) is smaller than that of Pulp A (0.1351 vs. 0.2510).

10.

ln y =

0 + 1 x1 + 2 x2 + ", where 0 = ln

and " = ln .

11. (a) t = −0.5876∕0.2873 = −2.045. (b) s satisfes the equation 4.301 = 1.5102∕s , so s = 0.3511. 1

(c) satisfes the equation −0.622 = 2 ∕0.3944, so 2 = −0.2453. (d) t = 1.8233∕0.3867 = 4.715. (e) MSR = 41.76∕3 = 13.92. (f) F = MSR∕MSE = 13.92∕0.76 = 18.316. (g) SSE = 0.760 ∗ 6 = 4.56.

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434

CHAPTER 8

12. (a) 0 satisfes the equation 5.913 = 0 ∕1.4553, so 0 = 8.5934. (b) s satisfes the equation 1.710 = 1.2127∕s , so s = 0.7092. 1

(c) t = 7.8369∕3.2109 = 2.441. (d) 3 satisfes the equation −3.561 = 3 ∕0.8943, so 3 = −3.1846 (e) There are three variables in the model other than the intercept, so there are 3 degrees of freedom for the model. (f) SSR = 8.1291(3) = 24.3876. (g) MSE = 6.8784∕10 = 0.68784.

13. (a) y = 267.53 − 1.5926(30) − 1.3897(35) − 1.0934(30) − 0.002658(30)(30) = 135.92 F (b) No. The change in the predicted fash point due to a change in acetic acid concentration depends on the butyric acid concentration as well, because of the interaction between these two variables. (c) Yes. The predicted fash point will change by −1.3897(10) = −13.897 F.

14.

ln p =

0 + 1 t + 2 (1∕t) + 3 ln t + ", where " = ln .

15. (a) The residuals are the values ei = yi − y i for each i. They are shown in the following table. x y 150 10.4 175 12.4 200 14.9 225 15 250 13.9 275 11.9

(b) SSE =

∑6

Fitted Value y 10.17143 12.97429 14.54858 14.89429 14.01143 11.90001

2 i=1 ei = 0.52914,

Residual e = y − y 0.22857 −0.57429 0.35142 0.10571 −0.11144 −0.00001

SST =

∑n

i=1 (yi − y)

2 = 16.70833.

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435

SECTION 8.1

(e) F =

SSR∕2 (SST − SSE)∕2 = = 45.864. There are 2 and 3 degrees of freedom. SSE∕3 SSE∕3

(f) Yes. From the F table, 0.001 < P < 0.01. A computer package gives P = 0.0056. Since P ≤ 0.05, the hypothesis H0 ∶ 1 = 2 = 0 can be rejected at the 5% level.

16. (a) The residuals are the values ei = yi − y i for each i. They are shown in the following table. Fitted Value Residual y e = y − y 99.62857 −0.62857 94.88572 1.11428 87.57143 0.42857 77.68572 −1.68572 65.22857 0.77143

x y 1 99 1.5 96 2 88 2.5 76 3 66

(b) SSE =

∑6

2 i=1 ei = 5.25714,

SST =

∑n

i=1 (yi − y)

2 = 768.

(e) F =

SSR∕2 (SST − SSE)∕2 = = 145.09. There are 2 and 2 degrees of freedom. SSE∕2 SSE∕2

(f) Yes. From the F table, 0.001 < P < 0.01. A computer package gives P = 0.0068. Since P ≤ 0.05, the hypothesis H0 ∶ 1 = 2 = 0 can be rejected at the 5% level.

17. (a) y = 1.18957 + 0.17326(0.5) + 0.17918(5.7) + 0.17591(3.0) − 0.18393(4.1) = 2.0711.

(b) 0.17918

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436

CHAPTER 8

(d) The percent change in GDP would be expected to be larger in Sweden, because the coeÿcient of PP is negative.

18. (a)

Predictor Constant Days Ice Cover Days Below 8 Avg Snow Depth

Coef −372.98 3.5368 3.7345 −2.1661

T −1.5389 2.9602 3.3738 −2.5430

StDev 242.37 1.1948 1.1069 0.85177

P 0.158 0.016 0.008 0.032

(b) By 3.5368(2) = 7.0736. (c) Less thickness, because the coeÿcient is negative.

19. (a)

Predictor Constant Time Time2

Coef −0.012167 0.043258 2.9205

StDev 0.01034 0.043186 0.038261

T −1.1766 1.0017 76.33

P 0.278 0.350 0.000

y = −0.012167 + 0.043258t + 2.9205t2 2 = 2.9205, s (b) ̂

= 0.038261. There are n − 3 = 7 degrees of freedom.

t7,.025 = 2.365. A 95% confdence interval is therefore 2.9205 ± 2.365(0.038261), or (2.830, 3.011). (c) Since a = 2 2 , the confdence limits for a 95% confdence interval for a are twice the limits of the confdence interval for 2 . Therefore a 95% confdence interval for a is (5.660, 6.022). (d)

0 : t7 = −1.1766, P = 0.278, 1 : t7 = 1.0017, P = 0.350, 2 : t7 = 76.33, P = 0.000.

(e) No, the P -value of 0.278 is not small enough to reject the null hypothesis that

0 = 0.

(f) No, the P -value of 0.350 is not small enough to reject the null hypothesis that

1 = 0.

Section 8.2

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437

SECTION 8.2

1. (a)

Predictor Constant x1

Coef 6.3347 1.2915

StDev 2.1740 0.1392

T 2.9138 9.2776

P 0.009 0.000

0 di˙ers from 0 (P = 0.009), 1 di˙ers from 0 (P = 0.000).

(b)

Predictor Constant x2

Coef 53.964 −0.9192

StDev 8.7737 0.2821

T 6.1506 −3.2580

P 0.000 0.004

0 di˙ers from 0 (P = 0.000), 1 di˙ers from 0 (P = 0.004).

(c)

Predictor Constant x1 x2

Coef 12.844 1.2029 −0.1682

StDev 7.5139 0.1707 0.1858

T 1.7094 7.0479 −0.90537

P 0.104 0.000 0.377

0 may not di˙er from 0 (P = 0.104), 1 di˙ers from 0 (P = 0.000), 2 may not di˙er from 0 (P = 0.377).

(d) The model in part (a) is the best. When both x1 and x2 are in the model, only the coeÿcient of x1 is signifcantly di˙erent from 0. In addition, the value of R2 is only slightly greater (0.819 vs. 0.811) for the model containing both x1 and x2 than for the model containing x1 alone.

Torque and HP are most nearly collinear. We can tell because their coeÿcients are signifcantly di˙erent from 0 when either one is in the model without the other, but not signifcantly di˙erent from 0 when both are in the model together.

3. (a) Plot (i) came from Engineer B, and plot (ii) came from Engineer A. We know this because the variables x1 and x2 are both signifcantly di˙erent from 0 for Engineer A but not for Engineer B. Therefore Engineer B is the one who designed the experiment to have the dependent variables nearly collinear. (b) Engineer A’s experiment produced the more reliable results. In Engineer B’s experiment, the two dependent variables are nearly collinear.

4. (a)

Predictor Constant x1

Coef −1.6415 3.1415

StDev 0.54133 0.61125

T −3.0323 5.1394

P 0.013 0.000

0 di˙ers from 0 (P = 0.013), 1 di˙ers from 0 (P = 0.000).

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438 (b)

CHAPTER 8

Predictor Constant x2

Coef −1.3701 2.8892

StDev 0.57649 0.6634

T −2.3765 4.3552

P 0.039 0.001

0 di˙ers from 0 (P = 0.039), 1 di˙ers from 0 (P = 0.001).

(c)

Predictor Constant x1 x2

Coef −1.8339 2.0828 1.3006

StDev 0.53319 0.94564 0.91539

T −3.4395 2.2026 1.4209

P 0.007 0.055 0.189

0 di˙ers from 0 (P = 0.007), the evidence regarding (P = 0.189).

1 is borderline (P = 0.055),

2 may not di˙er from 0

(d) The model y = 0 + 1 x1 + ". When both x1 and x2 are in the model, the coeÿcient of x2 is not signifcantly di˙erent from 0. In addition, the value of R2 is only slightly greater (0.776 vs. 0.725) for the model containing both x1 and x2 than for the model containing x1 alone. 5. (a) For R1 < 4, the least squares line is R2 = 1.233 + 0.264R1 . For R1 ≥ 4, the least squares line is R2 = −0.190 + 0.710R1 . (b) The relationship is clearly non-linear when R1 < 4. Quadratic Model Predictor Constant R1 R21

Coef 1.2840 0.21661 0.0090189

Cubic Model Predictor Constant R1 R21 R31

Coef −1.8396 4.4987 −1.7709 0.22904

Quartic Model Predictor Constant R1 R21 R31 R41

Coef −2.6714 6.0208 −2.7520 0.49423 −0.02558

T 4.8536 0.91947 0.20049

StDev 0.26454 0.23558 0.044984

StDev 0.56292 0.75218 0.30789 0.039454

StDev 2.0117 3.6106 2.2957 0.61599 0.05930

T −3.2680 5.9809 −5.7518 5.8053

T −1.3279 1.6675 −1.1988 0.80234 −0.43143

P 0.000 0.368 0.843

P 0.004 0.000 0.000 0.000

P 0.200 0.112 0.245 0.432 0.671

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SECTION 8.2

−0.5 1.5

2 Fitted Value

Residual

.. ....

Cubic Model

0.5

. ..

−0.5 1

2.5

0.5

Residual

Quadratic Model

0.5

(c)

1.5 2 Fitted Value

2.5

Quartic Model

−0.5 1

1.5 2 Fitted Value

250

(d)

200

R4 1

150

The correlation coeÿcient between R31 and R41 is 0.997.

100 50 0 0

(e) R31 and R41 are nearly collinear. (f) The cubic model is best. The quadratic is inappropriate because the residual plot exhibits a pattern. The residual plots for both the cubic and quartic models look good, however, there is no reason to include R41 in the model since it merely confounds the e˙ect of R31 .

6. (a)

(b)

T 18.839 −5.1614

Predictor Coef StDev Constant 95.9120 5.0911 −7.3597 1.4259 x1 Yes, there is a linear relationship.

Predictor Coef StDev Constant 119.14 6.8619 −0.68333 0.095153 x2 Yes, there is a linear relationship.

P 0.000 0.000

T 17.363 −7.1814

P 0.000 0.000

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440 (c)

CHAPTER 8

Predictor Constant x1 x2

Coef 120.15 0.89597 −0.74088

StDev 7.4121 2.3684 0.17982

T 16.210 0.37831 −4.1202

P 0.000 0.707 0.000

(d) x2 . If x2 were known, it would not be much additional help to know x1 . The value of the coeÿcient of x1 is not signifcantly di˙erent from 0 when x2 is in the model. Equivalently, the value of R2 does not increase signifcantly when x1 is added to the model containing x2 . The value of R2 for the model in part (b) is 0.47067, and the value of R2 for the model in part (c) is 0.47199.

Section 8.3 1. (a) False. There are usually several models that are about equally good. (b) True. (c) False. Model selection methods can suggest models that ft the data well. (d) True.

(iii). x1 x2 , x1 x3 , and x2 x3 all have large P -values and thus may not contribute signifcantly to the ft.

(v). x22 , x1 x2 , and x1 x3 all have large P -values and thus may not contribute signifcantly to the ft.

(iv). x1 and x3 both have large P -values and thus may not contribute signifcantly to the ft.

(iv). Carbon and Silicon both have large P -values and thus may not contribute signifcantly to the ft.

6. (a) X1 , X3 , and X4 (b) X1 , X2 , X3 , and X4

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SECTION 8.3

The four-variable model with the highest value of R2 has a lower R2 than the three-variable model with the highest value of R2 . This is impossible.

T 10.259 −6.6174

8. (a)

Predictor Constant Distance

Coef 2.7782 −1.6303

StDev 0.27081 0.24637

(b)

Predictor Constant Distance Distance2

Coef 3.2574 −3.5151 1.0300

StDev 0.05916 0.15138 0.079477

(c)

Predictor Constant Distance Distance2 Distance3

Coef 3.2739 −3.6951 1.2955 −0.096716

P 0.000 0.001

T 55.061 −23.221 12.960

P 0.000 0.000 0.000

T 45.472 −9.9129 2.5934 −0.53981

StDev 0.071998 0.37275 0.49955 0.17917

P 0.000 0.002 0.081 0.627

(d) The model in part (b) is preferred. All the variables in this model have coeÿcients that are signifcantly di˙erent from 0. In model (c), the coeÿcient of the variable Distance3 may not be di˙erent from 0. It is better to leave this variable out of the model. (e) The estimated defection is y = 3.2574 − 3.5151(1) + 1.0300(12 ) = 0.7723 mm.

9. (a) SSEfull = 7.730, SSEreduced = 7.772, n = 165, p = 7, k = 4. F =

(SSEreduced − SSEfull )∕(p − k) = 0.28455. SSEfull ∕(n − p − 1)

(b) 3 degrees of freedom in the numerator and 157 in the denominator. (c) P > 0.10 (a computer package gives P = 0.8365). The reduced model is plausible.

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CHAPTER 8

(d) This is not correct. It is possible for a group of variables to be fairly strongly related to an independent variable, even though none of the variables individually is strongly related.

SSEfull = 62.068, SSEreduced = 66.984, n = 10, p = 5, k = 2. The value of the F statistic for testing the plausibility of the reduced model is

10.

F =

(SSEreduced − SSEfull )∕(p − k) = 0.1056. SSEfull ∕(n − p − 1)

There are p − k = 3 and n − p − 1 = 4 degrees of freedom. From the F table, P > 0.10. A computer package gives P = 0.952. The reduced model is plausible.

SSEfull = 9.37, SSEreduced = 27.49, n = 24, p = 6, k = 3. The value of the F statistic for testing the plausibility of the reduced model is

11.

F =

(SSEreduced − SSEfull )∕(p − k) = 10.958. SSEfull ∕(n − p − 1)

There are p − k = 3 and n − p − 1 = 17 degrees of freedom. From the F table, P < 0.001. A computer package gives P = 0.000304. The reduced model is not plausible.

12. (a)

Predictor Constant ln Sc. Dist. ln Charge

Coef 6.6320 −1.6778 −0.19583

StDev 1.6806 0.36395 0.20846

T 3.9462 −4.6101 −0.9394

P 0.002 0.001 0.366

(b)

Predictor Constant ln Sc. Dist.

Coef 5.2782 −1.4652

StDev 0.86076 0.28366

T 6.1320 −5.1653

P 0.000 0.000

(c) The model in part (b) is preferred. In model (a), the coeÿcient of the variable ln Charge may not be di˙erent from 0. It is better to leave this variable out of the model.

13. (a)

Predictor Constant x

Coef 37.989 1.0774

StDev 53.502 0.041608

T 0.71004 25.894

P 0.487 0.000

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SECTION 8.3

(b)

Predictor Constant x x2

Coef −253.45 1.592 −0.00020052

StDev 132.93 0.22215 0.000085328

T −1.9067 7.1665 −2.3499

P 0.074 0.000 0.031

Linear Model

(c)

200 150 Residual

100 50 0 −50 −100 −150 500

(d)

1000

1500 Fitted Value

2000

Quadratic Model

150

Residual

100 50 0 −50

−100 −150 500

1000

1500 2000 Fitted Value

2500

(e) The quadratic model seems more appropriate. The P -value for the quadratic term is fairly small (0.031), and the residual plot for the quadratic model exhibits less pattern. (There are a couple of points somewhat detached from the rest of the plot, however.) (f) 1683.5 (g) (1634.7, 1732.2)

14. (a)

Predictor Constant x1 x2

Coef 0.046975 −0.00084581 0.00041568

StDev 0.021292 0.00028086 0.00017662

T 2.2062 −3.0115 2.3536

P 0.041 0.008 0.031

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CHAPTER 8

y = 0.046975 − 0.00084581x1 + 0.00041568x2 .

Residual

0.2

The residual plot shows a defnite pattern. The model is not appropriate.

−0.2 −0.1

(b)

Predictor Constant lnx1 lnx2

0 Fitted Value

0.1

Coef −1.8898 −3.0315 2.06

StDev 1.761 0.45905 0.46648

T −1.0732 −6.6039 4.416

P 0.298 0.000 0.000

ln y = −1.8898 − 3.0315x1 + 2.0600x2 . 2

Residual

The residual plot shows no defnite pattern. The model appears to be appropriate.

−1 −2 −10

−8

−6 −4 Fitted Value

−2

Predictor Constant lnx1 lnx2 lnx1 lnx2

Coef −5.6416 −2.0919 2.9432 −0.2174

StDev 10.491 2.6307 2.4793 0.5988

T −0.53777 −0.79518 1.1871 −0.36306

P 0.598 0.438 0.253 0.721

No. The P -value for testing the null hypothesis that the coeÿcient of the interaction term ln x1 ln x2 is zero is 0.721. Therefore the interaction term should not be included.

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SECTION 8.3

15. (a)

Predictor Constant x1 x2

Coef 25.613 0.18387 −0.015878

(b)

Predictor Constant x1

Coef 14.444 0.17334

(c)

Predictor Constant x2

Coef 40.370 −0.015747

StDev 10.424 0.12353 0.0040542

StDev 16.754 0.20637

T 2.4572 1.4885 −3.9164

T 0.86215 0.83993

StDev 3.4545 0.0043503

P 0.044 0.180 0.006

P 0.414 0.425

T 11.686 −3.6197

P 0.000 0.007

(d) The model containing x2 as the only independent variable is best. There is no evidence that the coeÿcient of x1 di˙ers from 0.

16.

Answers will vary. One good model is y = 73.228 + 0.17511x2 . Predictor Constant x2

17.

Coef 73.228 0.17511

StDev 2.0671 0.063783

T 35.425 2.7454

P 0.000 0.025

The model y = 0 + 1 x2 + " is a good one. One way to see this is to compare the ft of this model to the full quadratic model. The ANOVA table for the full model is Source Regression Residual Error Total

DF 5 9 14

SS 4.1007 3.9241 8.0248

DF 1 13 14

F 1.881

P 0.193

F 6.8285

P 0.021

0 + 1 x2 + " is

The ANOVA table for the model y = Source Regression Residual Error Total

MS 0.82013 0.43601

SS 2.7636 5.2612 8.0248

MS 2.7636 0.40471

From these two tables, the F statistic for testing the plausibility of the reduced model is (5.2612 − 3.9241)∕(5 − 1) = 0.7667. 3.9241∕9 The null distribution is F4,9 , so P > 0.10 (a computer package gives P = 0.573). The large P -value indicates that the reduced model is plausible.

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CHAPTER 8

18. (a) False (b) True (c) False (d) True

Supplementary Exercises for Chapter 8 1. (a) y = 46.802 − 130.11(0.15) − 807.10(0.01) + 3580.5(0.15)(0.01) = 24.6%. (b) By 130.11(0.05) − 3580.5(0.006)(0.05) = 5.43%. (c) No, we need to know the oxygen content, because of the interaction term.

̂1 = −130.11, s = 20.467. 2. (a) 1

There are 35 degrees of freedom for error. t35,.025 = 2.030. A 95% confdence interval is −130.11 ± 2.030(20.467), or (−171.66, −88.56).

(b)

2 = −807.10, s 2 = 158.03.

There are 35 degrees of freedom for error. t35,.005 = 2.724. A 99% confdence interval is −807.10 ± 2.724(158.03), or (−1237.6, −376.63).

(c)

3 = 3580.5, s 3 = 958.05.

There are 35 degrees of freedom for error. t35,.01 = 2.438. A 98% confdence interval is 3580.5 ± 2.438(958.05), or (1244.8, 5916.2). (d) 1 = −130.11, s

= 20.467.

The null and alternative hypotheses are H0 ∶

1 ≥ −75 versus H1 ∶

1 < −75.

There are 35 degrees of freedom for error. t = [−130.11 − (−75)]∕20.467 = −2.693. Since the alternative hypothesis is of the form 1 < b, the P -value is the area to the left of t = −2.693. From the t table, 0.005 < P < 0.01. A computer package gives P = 0.0054. We can conclude that 1 < −75.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 8 (e) 2 = −807.10, s = 158.03. 2

The null and alternative hypotheses are H0 ∶

2 ≤ −1000 versus H1 ∶

2 > −1000.

There are 35 degrees of freedom for error. t = [−807.10 − (−1000)]∕158.03 = 1.221. Since the alternative hypothesis is of the form 2 > b, the P -value is the area to the right of t = 1.221. From the t table, 0.10 < P < 0.25. A computer package gives P = 0.115. We cannot conclude that 2 > −1000.

0 = 0.207. 3. (a) 0 satisfes the equation 0.59 = 0 ∕0.3501, so ̂ (b) s satisfes the equation 2.31 = 1.8515∕s , so s = 0.8015. 1

MSE =

1.44 = 1.200.

(e) There are 2 independent variables in the model, so there are 2 degrees of freedom for regression. (f) SSR = SST − SSE = 104.09 − 17.28 = 86.81. (g) MSR = 86.81∕2 = 43.405. (h) F = MSR∕MSE = 43.405∕1.44 = 30.14. (i) 2 + 12 = 14.

5. (a)

The colleague is right. The models with the second or third best R2 or Mallows’ Cp are often just as good in practice as the model with the best values.

Predictor Constant Speed Pause Speed2 Pause2 Speed⋅Pause

Coef 10.84 −0.073851 −0.12743 0.0011098 0.0016736 −0.00024272

StDev 0.2749 0.023379 0.013934 0.00048887 0.00024304 0.00027719

T 39.432 −3.1589 −9.1456 2.2702 6.8861 −0.87563

P 0.000 0.004 0.000 0.032 0.000 0.390

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CHAPTER 8

S = 0.33205

R-sq = 92.2%

Analysis of Variance Source DF Regression 5 Residual Error 24 Total 29

R-sq(adj) = 90.6% MS 6.2608 0.11026

F 56.783

P 0.000

StDev 0.23213 0.022223 0.01260 0.00048658 0.0002419

T 47.246 −3.5961 −10.518 2.2809 6.9185

P 0.000 0.001 0.000 0.031 0.000

SS 31.304 2.6462 33.95

(b) We drop the interaction term Speed⋅Pause. Predictor Constant Speed Pause Speed2 Pause2

Coef 10.967 −0.079919 −0.13253 0.0011098 0.0016736

S = 0.33050

R-sq = 92.0%

Analysis of Variance Source DF Regression 4 Residual Error 25 Total 29

R-sq(adj) = 90.7%

SS 31.22 2.7307 33.95

MS 7.8049 0.10923

F 71.454

P 0.000

(2.7307 − 2.6462)∕(5 − 4) = 0.77, P > 0.10. A 2.6462∕24 computer package gives P = 0.390 (the same as the P -value for the dropped variable). Comparing this model with the one in part (a), F1,24 =

(c)

Residual

0.5

There is a some suggestion of heteroscedasticity, but it is hard to be sure without more data.

−0.5 −1 6

(d)

8 9 Fitted Value

Predictor Constant Pause Pause2 S = 0.53888

Coef 9.9601 -0.13253 0.0016736

StDev 0.21842 0.020545 0.00039442

R-sq = 76.9%

T 45.601 -6.4507 4.2431

P 0.000 0.000 0.000

R-sq(adj) = 75.2%

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SUPPLEMENTARY EXERCISES FOR CHAPTER 8

Analysis of Variance Source DF Regression 2 Residual Error 27 Total 29

SS 26.11 7.8405 33.95

F 44.957

MS 13.055 0.29039

Comparing this model with the one in part (a), F3,24 =

P 0.000

(7.8405 − 2.6462)∕(5 − 2) = 15.70, P < 0.001. A 2.6462∕24

computer package gives P = 7.3 × 10−6 .

(e)

Vars 1 1 2 2 3 3 4 4 5

R-Sq 61.5 60.0 76.9 74.9 90.3 87.8 92.0 90.5 92.2

R-Sq(adj) 60.1 58.6 75.2 73.0 89.2 86.4 90.7 89.0 90.6

Cp 92.5 97.0 47.1 53.3 7.9 15.5 4.8 9.2 6.0

S 0.68318 0.69600 0.53888 0.56198 0.35621 0.39903 0.33050 0.35858 0.33205

Speed

Pause X

Speed2

Pause2

Speed * Pause X

X X X X X

X X X X X X X

X X

X X X X X

X X X

(f) The model containing the dependent variables Speed, Pause, Speed2 and Pause2 has both the lowest value of Cp and the largest value of adjusted R2 .

6. (a)

Predictor Constant x1 x2 x3

Coef 10.552 −0.033435 0.54648 1.8305

StDev 3.5390 0.097034 0.20168 0.24774

T 2.9818 −0.34457 2.7097 7.3887

P 0.004 0.731 0.008 0.000

StDev 3.0544 0.10546 0.24639

T 3.2562 4.6212 7.4392

P 0.002 0.000 0.000

(b) We drop x1 : Predictor Constant x2 x3 (c)

Predictor Constant x2 x3 x 2 x3

Coef 9.9458 0.48736 1.8329

Coef 17.956 0.16285 0.54885 0.051789

StDev 9.1033 0.36306 1.3965 0.055437

T 1.9725 0.44854 0.39302 0.93418

P 0.051766 0.65489 0.69528 0.35282

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CHAPTER 8

(d) y = 9.9458 + 0.48736x2 + 1.8329x3 (e) Yes

The residual plot shows an obvious curved pattern, so the linear model is not appropriate.

Quadratic Model

Residual

There is no obvious pattern to the residual plot, so the quadratic model appears to ft well.

−10 −20 0

100

200 300 Fitted Value

400

Cubic Model

Residual

10 5

There is no obvious pattern to the residual plot, so the cubic model appears to ft well.

−5

−10 −15 0

100

200 300 Fitted Value

400

(i) The linear model is best. The plot shows that all three models make nearly identical predictions, so the best model is the one that contains the fewest independent variables.

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451

SUPPLEMENTARY EXERCISES FOR CHAPTER 8

9. (a) Under model 1, the prediction is −320.59 + 0.37820(1500) − 0.16047(400) = 182.52. Under model 2, the prediction is −380.1 + 0.41641(1500) − 0.5198(150) = 166.55. Under model 3, the prediction is −301.8 + 0.3660(1500) − 0.2106(400) + 0.164(150) = 187.56.

(b) Under model 1, the prediction is −320.59 + 0.37820(1600) − 0.16047(300) = 236.39. Under model 2, the prediction is −380.1 + 0.41641(1600) − 0.5198(100) = 234.18. Under model 3, the prediction is −301.8 + 0.3660(1600) − 0.2106(300) + 0.164(100) = 237.02.

(c) Under model 1, the prediction is −320.59 + 0.37820(1400) − 0.16047(200) = 176.80. Under model 2, the prediction is −380.1 + 0.41641(1400) − 0.5198(75) = 163.89. Under model 3, the prediction is −301.8 + 0.3660(1400) − 0.2106(200) + 0.164(75) = 180.78.

(d) (iv). The output does not provide much to choose from between the two-variable models. In the three-variable model, none of the coeÿcients are signifcantly di˙erent from 0, even though they were signifcant in the twovariable models. This suggest collinearity.

10. (a) Yes, the new model is y =

(b) Yes, the new model is y =

∗ + ∗ F + ", where ∗ = −57.6 ∗ 1 and 1 = 1.8 1 . 0 1 0

∗ − ∗ F + ∗ F 2 + ", where ∗ = 0 1 2 0

∗ ∗ 2 0 + 57.6 2 , 1 = −207.36 2 , and 2 = 3.24 2 .

11. (a)

2 2 z then y =

0 + 2 (a + bx)

Linear Model Predictor Constant Hardwood

Coef 40.751 0.54013

S = 12.308

R-sq = 4.2%

Analysis of Variance Source DF Regression 1 Residual Error 18 Total 19

2 ≠

StDev 5.4533 0.61141

SS 118.21 2726.5 2844.8

2 0 + 2x .

T 7.4728 0.88341

P 0.000 0.389

R-sq(adj) = −1.2% MS 118.21 151.47

F 0.78041

P 0.38866

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Quadratic Model Predictor Constant Hardwood Hardwood2 S = 5.3242

Coef 12.683 10.067 −0.56928

StDev 3.9388 1.1028 0.063974

R-sq = 83.1%

Analysis of Variance Source DF 2 Regression Residual Error 17 Total 19

Coef 27.937 0.48749 0.85104 −0.057254

S = 2.6836

R-sq = 95.9%

Analysis of Variance Source DF Regression 3 Residual Error 16 Total 19

Coef 30.368 −1.7962 1.4211 −0.10878 0.0015256

S = 2.7299

R-sq = 96.1%

Analysis of Variance Source DF Regression 4 Residual Error 15 Total 19

SS 2733 111.78 2844.8

F 41.678

P 0.000

T 9.5755 0.3355 4.2204 −7.1354

P 0.000 0.742 0.001 0.000

MS 1181.4 28.347

StDev 2.9175 1.453 0.20165 0.0080239

SS 2729.5 115.23 2844.8

Quartic Model Predictor Constant Hardwood Hardwood2 Hardwood3 Hardwood4

P 0.005 0.000 0.000

R-sq(adj) = 81.1%

SS 2362.9 481.90 2844.8

Cubic Model Predictor Constant Hardwood Hardwood2 Hardwood3

T 3.2199 9.1287 −8.8986

R-sq(adj) = 95.2% MS 909.84 7.2018

F 126.34

P 0.000

StDev 4.6469 3.6697 0.8632 0.076229 0.0022438

T 6.5351 −0.48946 1.6463 −1.4271 0.67989

P 0.000 0.632 0.120 0.174 0.507

R-sq(adj) = 95.0% MS 683.24 7.4522

F 91.683

P 0.00

The values of SSE and their degrees of freedom for models of degrees 1, 2, 3, and 4 are:

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SUPPLEMENTARY EXERCISES FOR CHAPTER 8 Linear Quadratic Cubic Quartic

18 17 16 15

2726.55 481.90 115.23 111.78

(2726.55 − 481.90)∕(18 − 17) = 79.185. 481.90∕17 . A computer package gives P = 8.3 × 10−8 .

To compare quadratic vs. linear, F1,17 = P

(481.90 − 115.23)∕(17 − 16) = 50.913. 115.23∕16 . A computer package gives P = 2.4 × 10−6 .

To compare cubic vs. quadratic, F1,16 = P

(115.23 − 111.78)∕(16 − 15) = 0.463. 111.78∕15 P > 0.10. A computer package gives P = 0.507. To compare quartic vs. cubic, F1,15 =

The cubic model is selected by this procedure. (b) The cubic model is y = 27.937 + 0.48749x + 0.85104x2 − 0.057254x3 . The estimate y is maximized when dy∕dx = 0. dy∕dx = 0.48749 + 1.70208x − 0.171762x2 . Therefore x = 10.188 (x = −0.2786 is a spurious root).

12. (a)

Linear Model Predictor Constant x Quadratic Model Predictor Constant x x2 Cubic Model Predictor Constant x x2 x3

Coef 16.887 −1.2482

StDev 4.1077 0.069411

T 4.111 −17.983

P 0.003 0.000

Coef 3.814 −0.44982 −0.0077691

StDev 1.234 0.057106 0.00053834

T 3.0908 −7.8769 −14.432

P 0.015 0.000 0.000

Coef 5.4977 −0.64732 −0.003024 −0.000030317

StDev 1.6043 0.14185 0.0031997 0.000020192

T 3.4268 −4.5635 −0.94511 −1.5015

P 0.011 0.003 0.376 0.177

(b) The quadratic model is most appropriate. This can be seen by noticing that the coeÿcient of x3 in the cubic model is not signifcantly di˙erent from 0. Another way to see this is to perform successive F tests, comparing each model with the model of degree one less. (c) There are 11 observations, and 2 independent variables, so there are 11 − 2 − 1 = 8 degrees of freedom for error

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CHAPTER 8

in the quadratic model. t8,.005 = 3.355. A 99% confdence interval for

0 is 3.814 ± 3.355(1.234), or (−0.326, 7.95).

A 99% confdence interval for

1 is −0.44982 ± 3.355(0.057106), or (−0.641, −0.258).

A 99% confdence interval for

2 is −0.0077691 ± 3.355(0.00053834), or (−0.009575, −0.005963).

13. (a) Let y1 represent the lifetime of the sponsor’s paint, y2 represent the lifetime of the competitor’s paint, x1 represent January temperature, x2 represent July temperature, and x3 represent Precipitation. Then one good model for y1 is y1 = −4.2342 + 0.79037x1 + 0.20554x2 − 0.082363x3 − 0.0079983x1 x2 − 0.0018349x21 . A good model for y2 is y2 = 6.8544+0.58898x1 +0.054759x2 −0.15058x3 −0.0046519x1 x2 +0.0019029x1 x3 − 0.0035069x21 . (b) Substitute x1 = 26.1, x2 = 68.9, and x3 = 13.3 to obtain y 1 = 13.83 and y 2 = 13.90.

14. (a)

(b)

(c)

Linear Model Predictor Constant x

Quadratic Model Predictor Constant x x2

Cubic Model Predictor Constant x x2 x3

Coef −171.9 209.66

Coef 453.22 −1306.7 919.47

Coef −12071 44265 −54351 22344

StDev 1.9125 2.3188

StDev 53.355 129.42 78.475

StDev 5832 21219 25735 10404

T −89.885 90.417

P 0.000 0.000

T 8.4945 −10.097 11.717

T −2.070 2.086 −2.112 2.148

P 0.000 0.000 0.000

P 0.107 0.105 0.102 0.098

Note: The design matrix for the cubic model is highly ill-conditioned, so the values calculated for the coeÿcients and standard deviations are unstable.

(d) The quadratic model. The coeÿcients are not signifcant in the cubic model.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 8

(e) The log of the fow rate is estimated to be y = 453.22 − 1306.7(0.83) + 919.47(0.832 ) = 2.1, so the fow rate is estimated to be e2.1 = 8.2.

15. (a)

(b)

(c)

(d)

Linear Model Predictor Constant x

Quadratic Model Predictor Constant x x2

Coef 0.25317 −0.041561

StDev 0.0065217 0.040281

T 38.819 −1.0318

P 0.000 0.320

Coef 0.21995 0.58931 −2.2679

StDev 0.0038434 0.06146 0.2155

T 57.23 9.5886 −10.524

P 0.000 0.000 0.000

Cubic Model Predictor Constant x x2 x3

Coef 0.22514 0.41105 −0.74651 −3.6728

StDev 0.0068959 0.20576 1.6887 4.043

T 32.648 1.9977 −0.44206 −0.90843

P 0.000 0.069 0.666 0.382

Quartic Model Predictor Constant x x2 x3 x4

Coef 0.23152 0.10911 3.4544 −26.022 40.157

StDev 0.013498 0.58342 7.7602 40.45 72.293

T 17.152 0.18702 0.44515 −0.64333 0.55548

P 0.000 0.855 0.665 0.533 0.590

(e) The quadratic model. The coeÿcient of x3 in the cubic model is not signifcantly di˙erent from 0. Neither is the coeÿcient of x4 in the quartic model. (f) y = 0.21995 + 0.58931(0.12) − 2.2679(0.122 ) = 0.258

16. (a)

Predictor Constant x1 x2 x1 x2

Coef 0.094806 0.096663 −0.074354 0.0034715

StDev 3.0827 0.17372 0.071246 0.0042389

T 0.030754 0.55643 −1.0436 0.81896

P 0.976 0.585 0.310 0.424

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Analysis of Variance Source DF Regression 3 Residual Error 18 Total 21

(b)

(c)

Predictor Constant x1 x2

SS 28.303 49.06 77.363

Coef −1.9749 0.21634 −0.017619

MS 9.4344 2.7256

StDev 1.7499 0.093111 0.016492

F 3.4615

T −1.1286 2.3235 −1.0683

P 0.273 0.031 0.299

Analysis of Variance Source DF Regression 2 Residual Error 19 Total 21

SS 26.475 50.888 77.363

MS 13.238 2.6783

F 4.9425

Coef −3.0827 0.25439

StDev 1.4146 0.086334

T −2.1792 2.9466

P 0.041 0.008

Analysis of Variance Source DF Regression 1 Residual Error 20 Total 21

SS 23.418 53.945 77.363

MS 23.418 2.6972

F 8.6823

Predictor Constant x1

P 0.038

P 0.019

P 0.008

(50.888 − 49.06)∕(3 − 2) = 0.671. 49.06∕18 There are 1 and 18 degrees of freedom.

(d) For comparing (b) with (a), F =

From the F table, P > 0.10. A computer package gives P = 0.423. (53.945 − 49.06)∕(3 − 1) = 0.896. 49.06∕18 There are 2 and 18 degrees of freedom.

For comparing (c) with (a), F =

From the F table, P > 0.10. A computer package gives P = 0.426. The comparison of (c) with (a) shows that it is reasonable to drop both x2 and x1 x2 from the model. Therefore model (c) is preferred.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 8

17. (a)

Predictor Constant x1 x2 x21 x22 x1 x2

Coef −0.093765 0.63318 2.5095 5.318 −0.3214 0.15209

StDev 0.092621 2.2088 0.30151 8.2231 0.17396 1.5778

T −1.0123 0.28666 8.3233 0.64672 −1.8475 0.09639

Analysis of Variance Source DF Regression 5 Residual Error 10 Total 15

SS 20.349 0.045513 20.394

MS 4.0698 0.0045513

P 0.335 0.780 0.000 0.532 0.094 0.925 F 894.19

P 0.000

(b) The model containing the variables x1 , x2 , and x22 is a good one. Here are the coeÿcients along with their standard deviations, followed by the analysis of variance table. Predictor Constant x1 x2 x22

Coef −0.088618 2.1282 2.4079 −0.27994

Analysis of Variance Source DF Regression 3 Residual Error 12 Total 15

StDev 0.068181 0.30057 0.13985 0.059211

T −1.2997 7.0805 17.218 −4.7279

P 0.218 0.000 0.000 0.000

SS 20.346 0.048329 20.394

MS 6.782 0.0040275

F 1683.9

P 0.000

The F statistic for comparing this model to the full quadratic model is F2,10 =

(0.048329 − 0.045513)∕(12 − 10) = 0.309, P > 0.10, 0.045513∕10

so it is reasonable to drop x21 and x1 x2 from the full quadratic model. All the remaining coeÿcients are significantly di˙erent from 0, so it would not be reasonable to reduce the model further.

Response is y

Vars 1 1

R-Sq 98.4 91.8

R-Sq(adj) 98.2 91.2

Mallows C-p 61.6 354.1

S 0.15470 0.34497

x x x 1 2 1 x x ^ ^ x 1 2 2 2 2 X X

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2 2 3 3 4 4 5

99.3 99.2 99.8 99.8 99.8 99.8 99.8

99.2 99.1 99.7 99.7 99.7 99.7 99.7

20.4 25.7 2.2 2.6 4.0 4.1 6.0

0.10316 0.11182 0.062169 0.063462 0.064354 0.064588 0.067463

X X X X X X X X X X X X X X X X X X X X X X X

The model with the best adjusted R2 (0.99716) contains the variables x2 , x21 , and x22 . This model is also the model with the smallest value of Mallows’ Cp (2.2). This is not the best model, since it contains x21 but not x1 . The model containing x1 , x2 , and x22 , suggested in the answer to part (b), is better. Note that the adjusted R2 for the model in part (b) is 0.99704, which di˙ers negligibly from that of the model with the largest adjusted R2 value.

18. (a)

Linear Model Predictor Constant x

Coef 10.408 6.7508

StDev 3.9496 0.58519

T 2.6353 11.536

P 0.027 0.000

Residual

5 0

The least-squares line is y = 10.408 + 6.5708x. The residual plot shows a curved pattern, so the linear model is not appropriate.

−5 −10 −15 0

(b)

40 60 Fitted Value

Log-linear Model Predictor Constant lnx

Coef 1.5037 31.397

100

StDev 1.107 0.64619

T 1.3584 48.589

P 0.207 0.000

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SUPPLEMENTARY EXERCISES FOR CHAPTER 8 3

Residual

2 1

The least-squares line is y = 1.5037 + 31.397 ln(x). The residual plot shows a curved pattern, so the loglinear model is not appropriate.

0 −1 −2 0

(c)

40 Fitted Value

Quadratic Model Predictor Constant x x2

Coef −4.3677 13.733 −0.575

StDev 2.7567 1.0758 0.086308

T −1.5844 12.766 −6.6622

P 0.152 0.000 0.000

Residual

The residual plot shows a curved pattern, so the quadratic model is not appropriate.

−2 −4 0

Cubic Model Predictor Constant x x2 x3

40 Fitted Value

Coef −14.542 22.112 −2.2255 0.089953

StDev 1.3195 0.92605 0.17391 0.0093835

T −11.021 23.878 −12.797 9.5863

P 0.000 0.000 0.000 0.000

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CHAPTER 8 1.5

Residual

1 0.5

The residual plot shows a curved pattern, so the cubic model is not appropriate.

−0.5 −1 0

Quartic Model Predictor Constant x x2 x3 x4

40 60 Fitted Value

100

Coef −19.005 27.258 −3.9152 0.29679 −0.0084092

StDev 0.94669 0.96693 0.29976 0.035807 0.001447

T −20.075 28.19 −13.061 8.2885 −5.8116

P 0.000 0.000 0.000 0.000 0.001

0.4

Residual

0.2 0

The residual plot for the quartic model is acceptable.

−0.2 −0.4 −0.6 0

Quintic Model Predictor Constant x x2 x3 x4 x5

40 60 Fitted Value

100

Coef −18.657 26.743 −3.6711 0.24775 −0.0040467 −0.00014162

StDev 1.8428 2.4948 1.1193 0.21856 0.019197 0.00062108

T −10.124 10.719 −3.2797 1.1335 −0.2108 −0.22802

P 0.000 0.000 0.022 0.308 0.841 0.829

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SUPPLEMENTARY EXERCISES FOR CHAPTER 8 0.4

Residual

0.2 0

The residual plot for the quintic model is acceptable.

−0.2 −0.4 −0.6 0

40 60 Fitted Value

100

(d) Yes. All the coeÿcients of the quartic model are signifcantly di˙erent from 0, while several of the coeÿcients of the quintic model are not.

19. (a)

Predictor Constant t t2

Coef 1.1623 0.059718 −0.00027482

StDev 0.17042 0.0088901 0.000069662

T 6.8201 6.7174 −3.9450

P 0.006 0.007 0.029

(b) Let x be the time at which the reaction rate will be equal to 0.05. Then 0.059718 − 2(0.00027482)x = 0.05, so x = 17.68 minutes. (c) 1 = 0.059718, s

= 0.0088901.

There are 6 observations and 2 dependent variables, so there are 6 − 2 − 1 = 3 degrees of freedom for error. t3,.025 = 3.182. A 95% confdence interval is 0.059718 ± 3.182(0.0088901), or (0.0314, 0.0880). (d) The reaction rate is decreasing with time if

2 < 0. We therefore test H0 ∶

From the output, the test statistic for testing H0 ∶

2 = 0 versus H1 ∶

≥ 0 versus H1 ∶

2 ≠ 0 is is t = −3.945.

2 < 0.

The output gives P = 0.029, but this is the value for a two-tailed test. For the one-tailed test, P = 0.029∕2 = 0.0145. It is reasonable to conclude that the reaction rate decreases with time.

20. (a)

Predictor Constant x1 x2

Coef −5.3862 −0.3545 5.7697

StDev 11.184 1.0714 2.8274

T −0.48160 −0.33087 2.0406

P 0.638 0.746 0.061

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(b)

Predictor Constant x2

Coef −6.7265 5.0425

StDev 10.11 1.7251

T −0.66532 2.923

P 0.516 0.010

(c)

The two uppermost points in the plot could be characterized as outliers. If these are dropped and the model is reft, the least-squares line is y = −2.4236 + 4.0416x2 . These coeÿcients are well inside the 95% confdence intervals computed from the data that includes the outliers. Therefore it is reasonable to conclude that the outliers are not particularly infuential, and to accept the model ft to the entire data set.

Residual

5 0 −5 −10 10

20 25 Fitted Value

0 + 1 x1 + 2 x2 + 3 x1 x2 + ".

21.

22.

There are several good models. The variable y should be transformed to y or perhaps ln y to avoid heteroscedasticity. Models that obey the exceptions to the principle of parsimony are better than ones that have better goodness-of-ft statistics such as Mallows’ Cp or adjusted R2 , but which do not obey these exceptions. One reasonably good model is y = −6.7381 + 1.7014m − 0.016352d + 3.1571s2 − 0.67773ms2 , where m is magnitude and d is distance. There are others. 40

Duration

There is no clear relationship between duration and time. It does not appear to be necessary to include time in the model.

0 0

60 Time

100

120

23. (a) The 17-variable model containing the independent variables x1 , x2 , x3 , x6 , x7 , x8 , x9 , x11 , x13 , x14 , x16 , x18 , x19 , x20 , x21 , x22 , and x23 has adjusted R2 equal to 0.98446. The ftted model is y

−1569.8 − 24.909x1 + 196.95x2 + 8.8669x3 − 2.2359x6

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SUPPLEMENTARY EXERCISES FOR CHAPTER 8

463

− 0.077581x7 + 0.057329x8 − 1.3057x9 − 12.227x11 + 44.143x13 + 4.1883x14 + 0.97071x16 + 74.775x18 + 21.656x19 − 18.253x20 + 82.591x21 − 37.553x22 + 329.8x23

(b) The 8-variable model containing the independent variables x1 , x2 , x5 , x8 , x10 , x11 , x14 , and x21 has Mallows’ Cp equal to 1.7. The ftted model is y = −665.98 − 24.782x1 + 76.499x2 + 121.96x5 + 0.024247x8 + 20.4x10 − 7.1313x11 + 2.4466x14 + 47.85x21

(c) Using a value of 0.15 for both -to-enter and -to-remove, the equation chosen by stepwise regression is y = −927.72 + 142.40x5 + 0.081701x7 + 21.698x10 + 0.41270x16 + 45.672x21 . (d) The 13-variable model below has adjusted R2 equal to 0.95402. (There are also two 12-variable models whose adjusted R2 is only very slightly lower.) z

= 8663.2 − 313.31x3 − 14.46x6 + 0.358x7 − 0.078746x8 +13.998x9 + 230.24x10 − 188.16x13 + 5.4133x14 + 1928.2x15 −8.2533x16 + 294.94x19 + 129.79x22 − 3020.7x23

(e) The 2-variable model z = −1660.9 + 0.67152x7 + 134.28x10 has Mallows’ Cp equal to −4.0. (f) Using a value of 0.15 for both -to-enter and -to-remove, the equation chosen by stepwise regression is z = −1660.9 + 0.67152x7 + 134.28x10 (g) The 17-variable model below has adjusted R2 equal to 0.97783. w

700.56 − 21.701x2 − 20.000x3 + 21.813x4 + 62.599x5 + 0.016156x7 − 0.012689x8 + 1.1315x9 + 15.245x10 + 1.1103x11 − 20.523x13 − 90.189x15 − 0.77442x16 + 7.5559x19 + 5.9163x20 − 7.5497x21 + 12.994x22 − 271.32x23

(h) The 13-variable model below has Mallows’ Cp equal to 8.0. w

567.06 − 23.582x2 − 16.766x3 + 90.482x5 + 0.0082274x7 − 0.011004x8 + 0.89554x9 + 12.131x10 − 11.984x13 − 0.67302x16 + 11.097x19 + 4.6448x20 + 11.108x22 − 217.82x23

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CHAPTER 8

(i) Using a value of 0.15 for both -to-enter and -to-remove, the equation chosen by stepwise regression is w = 130.92 − 28.085x2 + 113.49x5 + 0.16802x9 − 0.20216x16 + 11.417x19 + 12.068x21 − 78.371x23 .

24. (a, b) The predicted values and prediction errors for the new dataset are: y Pred Error 1713 507.82 1205.2 344 296.3 47.695 474 369.33 104.67 1336 1280.9 55.063 1916 668.93 1247.1 1280 557.7 722.3 1683 1512.5 170.52 901 826.94 74.062 460 −88.779 548.78 826 599 227

(c) The ftted values and residuals for the 28 row dataset are: y 1850 932 556 217 199 56 64 397 1926 724 711 1818 619 375

Fit 1900.3 895.21 518.32 256.95 237.73 7.0631 74.546 404.85 1923.3 722.13 716.06 1768.1 674.77 374.13

Resid y −50.307 214 36.793 300 37.676 771 −39.953 189 −38.733 494 48.937 389 −10.546 441 −7.8539 768 2.7333 797 1.8669 261 −5.0634 524 49.891 1262 −55.772 839 0.8659 1003

Fit 266.69 272.96 710.37 181.04 574.73 395.78 451.23 746.81 786.35 238.32 495.5 1253.3 776.76 1072.7

Resid −52.693 27.038 60.632 7.9603 −80.734 −6.7767 −10.225 21.19 10.652 22.684 28.501 8.6612 62.24 −69.667

(d) The prediction errors are on the whole larger than the residuals. The model is chosen to ft the frst data set, in other words, the model is chosen to make the residuals small. The ft to another data set will therefore in general be less good.

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SECTION 9.1

Chapter 9 Section 9.1 1. (a) Source Temperature Error Total

DF 3 16 19

SS 202.44 18.076 220.52

F 59.731

MS 67.481 1.1297

P 0.000

(b) Yes. F3, 16 = 59.731, P < 0.001 (P = 6.55 × 10−9 ).

2. (a) Source Fertilizer Error Total

DF 2 27 29

SS 229.74 439.39 669.13

MS 114.87 16.274

F 7.0587

P 0.003

(b) Yes. F2, 27 = 7.0587, 0.001 < P < 0.01 (P = 0.003).

3. (a) Source Treatment Error Total

DF 4 11 15

SS 19.009 22.147 41.155

F 2.3604

MS 4.7522 2.0133

P 0.117

(b) No. F4, 11 = 2.3604, P > 0.10 (P = 0.117).

4. (a) Source DF SS MS F P Amount 5 3.4778 0.69556 1.8385 0.180 Error 12 4.5400 0.37833 Total 17 8.0178 (b) No, F5,12 = 1.8385, P > 0.10 (P = 0.180).

5. (a) Source DF SS MS F P Age 5 3.8081 0.76161 7.9115 0.000 Error 73 7.0274 0.096266 Total 78 10.835

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(b) Yes, F5,73 = 7.9115, P < 0.01 (P = 5.35 × 10−6 ).

6. (a) Source Age Error Total

DF 3 45 48

SS 20.964 352.95 373.91

MS 6.988 7.8433

F 0.89095

P 0.453

(b) No. F3, 45 = 0.89095, P > 0.10 (P = 0.453).

7. (a) Source Group Error Total

DF 3 62 65

SS 0.19218 2.1133 2.3055

MS 0.064062 0.034085

F 1.8795

P 0.142

F 4.8179

P 0.003

F 8.7186

P 0.001

(b) No. F3, 62 = 1.8795, P > 0.10 (P = 0.142).

8. (a) Source Plant Error Total

DF 3 116 119

SS 12712.7 102027.8 114740.5

MS 4237.6 879.55

(b) Yes. F3, 116 = 4.8179, P < 0.01 (P = 0.003).

9. (a) Source Strength Error Total

DF 2 25 27

SS 481.45 690.26 1171.7

MS 240.73 27.611

(b) Yes. F2, 25 = 8.7186, 0.001 < P < 0.01 (P = 0.001).

10. (a) MSTr = 176.482∕3 = 58.827 (b) 19 − 3 = 16 (c) SSE = SST − SSTr = 235.958 − 176.482 = 59.476

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SECTION 9.1

(d) MSE = 59.476∕16 = 3.717 (e) F = MSTr∕MSE = 15.83 (f) There are 3 and 16 degrees of freedom. From the F table, P < 0.001. A computer package gives P = 4.8 × 10−5 .

11.

No, F3,16 = 15.83, P < 0.001 (P

12. (a) Source Temperature Error Total

DF 2 6 8

SS 9.2505 54.544 63.795

.8 × 10−5 ).

MS 4.6252 9.0907

F 0.50879

P 0.625

F 8.4914

P 0.001

(b) No. F2, 6 = 0.50879, P > 0.10 (P = 0.625).

13. (a) Source Temperature Error Total

DF 3 16 19

SS 58.650 36.837 95.487

MS 19.550 2.3023

(b) Yes, F3, 16 = 8.4914, 0.001 < P < 0.01 (P = 0.0013).

14. (a) From Exercise 12, MSE = 9.0907, so s =

9.0907 = 3.0151.

(b) The MINITAB output for the power calculation is

Power and Sample Size One-way ANOVA Alpha = 0.05 Factors: 1

Assumed standard deviation = 3.0151 Number of levels: 3

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Maximum Difference 5

Sample Size 11

Target Power 0.9

Actual Power 0.921575

The sample size is for each level. (c) The MINITAB output for the power calculation is

Power and Sample Size One-way ANOVA Alpha = 0.05

Assumed standard deviation = 4.522618

Factors: 1

Number of levels: 3

Maximum Difference 5

Sample Size 22

Target Power 0.9

Actual Power 0.903765

The sample size is for each level.

15. (a) From Exercise 13, MSE = 2.3023, so s =

2.3023 = 1.517.

(b) The MINITAB output for the power calculation is

Power and Sample Size One-way ANOVA Alpha = 0.05

Assumed standard deviation = 1.517

Factors: 1

Number of levels: 4

Maximum Difference 2

Sample Size 18

Target Power 0.9

Actual Power 0.912468

The sample size is for each level.

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SECTION 9.1

Power and Sample Size One-way ANOVA Alpha = 0.05

Assumed standard deviation = 2.2755

Factors: 1

Number of levels: 4

Maximum Difference 2

Sample Size 38

Target Power 0.9

Actual Power 0.902703

The sample size is for each level.

16. (a) Source Diameter Error Total

DF 2 8 10

SS 20.565 10.974 31.539

F 7.4959

MS 10.283 1.3718

P 0.015

(b) Yes, F2,8 = 7.4959, 0.01 < P < 0.05 (P = 0.015). 17. (a) Source DF SS MS F P Machine 4 6862 1715.5 7.8825 0.000 Error 30 6529.1 217.64 Total 34 13391 (b) (b) Yes, F4,30 = 7.8825, P 18. (a) Source Nail Type Error Total

DF 2 27 29

(P = 0.00018171).

SS 316.62 1867.6 2184.2

MS 158.31 69.171

F 2.2887

P 0.121

(b) No, F2,27 = 2.2887, P > 0.10 (P = 0.121). 19. (a) Source Soil Error Total

DF 2 23 25

SS 2.1615 4.4309 6.5924

MS 1.0808 0.19265

F 5.6099

P 0.0104

(b) Yes. F2, 23 = 5.6099, 0.01 < P < 0.05 (P = 0.0104).

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20. (a) The target power is 0.85. (b) The are 4 levels, or groups, with a sample size of 14 in each group, for a total of 56 observations in all. (c) The maximum di˙erence is 200. (d) Greater than 0.864138. The larger the di˙erence is, the greater the probability is to detect it.

21. (a)

Source Duration Error Total

DF 2 2571 2573

SS 28913.175 207317.893 236231.069

MS 14456.587 80.637

(b) Yes. F2,2573 = 179.280, P < 0.001 (P

22. (a)

Source Duration Error Total

DF 2 278 280

SS 46.831 39460 39507

MS 23.415 141.94

F 179.280

P 0.000

F 0.16496

P 0.848

(b) No. F2,278 = 0.16496, P > 0.10 (P = 0.848).

23. (a)

Source Soil Type Error Total

DF 3 2116 2119

SS 531544.959 466995.116 998540.075

MS 177181.653 220.697

(b) Yes. F3,2116 = 802.827, P < 0.001 (P

24. (a)

Source Soil Type Error Total

DF 3 2116 2119

SS 674709926.706 247646411.483 922356338.190

(b) Yes. F3,2116 = 1921.673, P < 0.001 (P

F 802.827

P 0.000

0).

MS 224903308.902 117035.166

F 1921.673

P 0.000

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SECTION 9.2

Section 9.2 1. (a) Yes, F5,6 = 46.64, P

(b) q6,6,.05 = 5.63. The value of MSE is 0.00508. The 5% critical value is therefore 5.63 0.00508∕2 = 0.284. Any pair that di˙ers by more than 0.284 can be concluded to be di˙erent. The following pairs meet this criterion: A and B, A and C, A and D, A and E, B and C, B and D, B and E, B and F, D and F. (c) t6,.025∕15 = 4.698 (the value obtained by interpolating is 4.958). The value of MSE is 0.00508. The 5% critical value is therefore 4.698 2(0.00508)∕2 = 0.335. Any pair that di˙ers by more than 0.335 may be concluded to be di˙erent. The following pairs meet this criterion: A and B, A and C, A and D, A and E, B and C, B and D, B and E, B and F, D and F. (d) The Tukey–Kramer method is more powerful, since its critical value is smaller (0.284 vs. 0.335). (e) Either the Bonferroni or the Tukey–Kramer method can be used.

2. (a) Yes, F2,24 = 5.071, P = 0.015. (b) q3,24,.05 = 3.53. The value of MSE is 5023.0. The 5% critical value is therefore 3.53 5023.0∕9 = 83.394. We may conclude that treatments B and C di˙er. (c) t24,.025∕3 = 2.5736 (the value obtained by interpolating is 2.594). The value of MSE is 5023.0. The 5% critical value is therefore 2.5736 2(5023.0)∕9 = 85.984. We may conclude that treatments B and C di˙er. (d) The Tukey–Kramer method is more powerful, because its critical value is smaller.

3. (a) MSE = 2.9659, Ji = 12 for all i. There are 7 comparisons to be made. Now t88,.025∕7 = 2.754, so the 5% critical value is 2.754 2.9659(1∕12 + 1∕12) = 1.936. All the sample means of the non-control formulations di˙er from the sample mean of the control formulation by more than this amount. Therefore we conclude at the 5% level that all the non-control formulations di˙er from the control formulation. (b) There are 7 comparisons to be made. We should use the Studentized range value q7,88,.05 . This value is not in the table, so we will use q7,60,.05 = 4.31, which is only slightly larger. The 5% critical value is 4.31 .9659∕12 = 2.14. All of the non-control formulations di˙er from the sample mean of the control formulation by more than this amount. Therefore we conclude at the 5% level that all the non-control formulations di˙er from the control formulation. (c) The Bonferroni method is more powerful, because it is based on the actual number of comparisons being made, which is 7. The Tukey–Kramer method is based on the largest number of comparisons that could be made, which is (7)(8)∕2 = 28.

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4. (a) t27,.025∕2 = 2.3734 (the value obtained by interpolating is 2.403). The value of MSE is 16.274. The 5% critical value is therefore 2.3734 2(16.274)∕10 = 4.282. We may conclude that the mean for the defcient fertilizer di˙ers from the control mean. (b) We should use the value q3,27,.05 . This value is not in the table, so we will use q3,24,.05 = 3.53, which is only slightly larger. The 5% critical value is therefore 3.53 16.274∕10 = 4.503. Any pair that di˙ers by more than 4.503 can be concluded to be di˙erent. We may conclude that the mean for the defcient fertilizer di˙ers from the control mean. (c) The Bonferroni method is more powerful, because its critical value is smaller.

5. (a) t16,.025∕6 = 3.0083 (the value obtained by interpolating is 3.080). The value of MSE is 2.3023. The 5% critical value is therefore 3.0083 2(2.3023)∕5 = 2.8869. We may conclude that the mean for 750 C di˙ers from the means for 850 C and 900 C, and that the mean for 800 C di˙ers from the mean for 900 C. (b) q4,16,.05 = 4.05. The value of MSE is 2.3023. The 5% critical value is therefore 4.05 2.3023∕5 = 2.75. We may conclude that the mean for 750 C di˙ers from the means for 850 C and 900 C, and that the mean for 800 C di˙ers from the mean for 900 C. (c) The Tukey–Kramer method is more powerful, because its critical value is smaller.

6. (a) t16,.025∕3 = 2.6730 (the value obtained by interpolating is 2.696). The value of MSE is 1.1297. The 5% critical value is therefore 2.6730 2(1.1297)∕5 = 1.797. We may conclude that the mean strengths for temperatures of 0 C and 10 C di˙er from the mean strength for a temperature of 30 C. (b) q4,16,.05 = 4.05. The value of MSE is 1.1297. The 5% critical value is therefore 4.05 1.1297∕5 = 1.9251. We may We may conclude that the mean strengths for temperatures of 0 C and 10 C di˙er from the mean strength for a temperature of 30 C. (c) The Bonferroni method is more powerful, because its critical value is smaller.

7. (a) t16,.025∕3 = 2.6730 (the value obtained by interpolating is 2.696). The value of MSE is 2.3023. The 5% critical value is therefore 2.6730 2(2.3023)∕5 = 2.5651. We may conclude that the mean for 900 C di˙ers from the means for 750 C and 800 C. (b) q4,16,.05 = 4.05. The value of MSE is 2.3023. The 5% critical value is therefore 4.05 2.3023∕5 = 2.75. We may conclude that the mean for 900 C di˙ers from the means for 750 C and 800 C. (c) The Bonferroni method is more powerful, because its critical value is smaller.

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SECTION 9.2

8. (a) t11,.025 = 2.201, MSE = 2.0133, the sample sizes are 4 and 3. The sample means are X B = 5.2525, X D = 6.990. The 95% confidence interval is 6.990 - 5.2525  2.201 2.0133(1/4 + 1/3), or (−0.6477, 4.1227). (b) The sample sizes are J1 = 2, J 2 = 4, J 3 = 3, J 4 = 3, J 5 = 4. MSE = 2.0133. The 0.95 quantile of the Studentized Range distribution is q5,11,.05 = 4.57. The values of q5,11,.05

(MSE/2)(1/Ji + 1/J j ) are presented

in the table on the left, and the values of the differences |X i. - X j. | are presented in the table on the right. 1 2 3 4 5

1 − 3.9709 4.1857 4.1857 3.9709

2 3.9709 − 3.5020 3.5020 3.2422

3 4.1857 3.5020 − 3.7438 3.5020

4 4.1857 3.5020 3.7438 − 3.5020

5 3.9709 3.2422 3.5020 3.5020 −

1 2 3 4 5

1 0 0.3225 1.2667 2.06 1.03

2 0.3225 0 0.94417 1.7375 1.3525

3 1.2667 0.94417 0 0.79333 2.2967

4 2.06 1.7375 0.79333 0 3.09

5 1.03 1.3525 2.2967 3.09 0

None of the differences exceeds its critical value, so we cannot conclude at the 5% level that any of the treatment means differ. 9. (a) t73,.025 = 1.993, MSE = 0.096266, the sample sizes are 12 and 15. The sample means are X 1 = 1.6825, X 6 = 2.0353. The 95% confidence interval is 0.3528 ± 1.993

0.096266(1/12 + 1/15), or (0.1133, 0.5923).

(b) The sample sizes are J1 = 12, J 2 = 12, J 3 = 13, J 4 = 12, J 5 = 15, J 6 = 15. MSE = 0.096266. We should use the Studentized range value q6,73,.05. This value is not in the table, so we will use q6,60,.05 = 4.16, which is only slightly larger. The values of q6,60,.05 (MSE /2)(1/Ji + 1/J j ) and the values of the differences |X i. - X j. | are presented in the following two tables.

q6,60,.05 (MSE /2)(1/Ji + 1/Jj ) 1 1 2 3 4 5 6

--0.37260 0.36536 0.37260 0.35348 0.35348

2 0.37260 --0.36536 0.37260 0.35348 0.35348

1 2 3 4 5 6

1 0 0.0075 0.49904 0.15083 0.5475 0.35283

2 0.0075 0 0.49154 0.14333 0.54 0.34533

3 0.36536 0.36536 --0.36536 0.34584 0.34584

4 0.37260 0.37260 0.36536 --0.35348 0.35348

5 0.35348 0.35348 0.34584 0.35348 --0.33326

6 0.35348 0.35348 0.34584 0.35348 0.33326 ---

|X i . - X j . | 3 4 0.49904 0.15083 0.49154 0.14333 0 0.34821 0.34821 0 0.048462 0.39667 0.14621 0.202

5 0.5475 0.54 0.048462 0.39667 0 0.19467

6 0.35283 0.34533 0.14621 0.202 0.19467 0

The differences that are significant at the 5% level are: mean 1 differs from means 3 and 5; mean 2 differs from means 3 and 5; and mean 4 differs from mean 5.

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10. (a) t25,.025 = 2.060, MSE = 27.611, the sample sizes are 11 and 7. The sample means are X 1 = 47.930909, X 6 = 54.67. The 95% confidence interval is 6.7391 ± 2.060 27.611(1/11 + 1/7), or (1.5055, 11.9727). (b) The sample sizes are J1 = 11, J2 = 10, J3 = 7. MSE = 27.611. We should use the Studentized range value q3,25,.05. This value is not in the table, so we will use q3,24,.05 = 3.53, which is only slightly larger. The values of q3,24,.05 (MSE /2)(1/Ji + 1/Jj ) and the values of the differences |X i. - X j. | are presented in the following two tables.

(MSE /2)(1/Ji + 1/J j )

q3,24,.05 1

1 2 3

− 5.73078 6.34150

1 2 3

1 0 4.06291 6.73909

5.73078 − 6.46363 |X i. - X j. | 2 4.06291 0 10.802

3 6.34150 6.46363 − 3 6.73909 10.802 0

The differences between the mean for plant C and the means for plants A and B are both significant at the 5% level. 11. (a) t8,.025 = 2.306, MSE = 1.3718. The sample means are X 1 = 1.998 and X 3 = 5.300. The sample sizes are J1 = 5 and J3 = 3. The 95% confidence interval is therefore 3.302  2.306 1.3718(1/5 + 1/3), or (1.330, 5.274). (b) The sample means are X 1 = 1.998, X 2 = 3.0000, X 3 = 5.300. The sample sizes are J1 = 5, J2 =

J3 = 3. The upper 5% point of the Studentized range is q3,8,.05 = 4.04. The 5% critical value for |X 1 - X 2 | and for |X 1 - X 3 | is 4.04 (1.3718/2)(1/5 + 1/3) = 2.44, and the 5% critical value for

|X 2 - X 3 | is 4.04 (1.3718/2)(1/3 + 1/3) = 2.73. Therefore means 1 and 3 differ at the 5% level.

12. (a) t27,.025 = 2.052, MSE = 69.171. The sample means are X 1 = 32.019 and X 3 = 24.239. The sample sizes are J1 = J3 = 10. The 95% confidence interval is therefore 7.780  2.052 69.171(1/10 + 1/10), or (0.148, 15.412). (b) The table doesn’t contain q3,27,.05, so we’ll use q3,24,.05 = 3.53. The 5% critical value is 3.53 69.171/10 = 9.28. The three sample means are X 1 = 32.019, X 2 = 26.681, X 3 = 24.239. No pair of means differs at the 5% level. 13. (a) X .. = 88.04, I = 4, J = 5, MSTr = å i =1 J (X i . - X .. )2 /(I - 1) = 19.554. I

F = MSTr∕MSE = 19.554∕3.85 = 5.08. There are 3 and 16 degrees of freedom, so 0.01 < P < 0.05 (a computer package gives P = 0.012). The null hypothesis of no difference is rejected at the 5% level. (b) q4, 16 .05 = 4.05, so catalysts whose means differ by more than 4.05 3.85/5 = 3.55 are significantly different at the 5% level. Catalyst 1 and Catalyst 2 both differ significantly from Catalyst 4.

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SECTION 9.3

14. (a) X .. = 1351.5, I = 4, J = 4, MSTr =

∑I

2 i=1 J (X i. − X .. ) ∕(I − 1) = 5158.67.

F = MSTr∕MSE = 5158.67∕875.2 = 5.89. There are 3 and 12 degrees of freedom, so P (a computer package gives P = 0.010). The null hypothesis of no di˙erence is rejected at the 5% level.

.01

(b) q4, 12 .05 = 4.20, so catalysts whose means di˙er by more than 4.20 875.2∕4 = 62.13 are signifcantly di˙erent at the 5% level. Curing times 1 and 2 both di˙er signifcantly from curing time 4.

15.

The value of the F statistic is F = MSTr∕MSE = 19.554∕MSE. The upper 5% point of the F3,16 distribution is 3.24. Therefore the F test will reject at the 5% level if 19.554∕MSE ≥ 3.24, or, equivalently, if MSE ≤ 6.035. The largest di˙erence between the sample means is 89.88−85.79 = 4.09. The upper 5% point of the Studentized .05. Therefore the Tukey–Kramer test will fail to fnd any di˙erences signifcant range distribution is q4,16,.05 MSE∕5, or equivalently, if MSE > 5.099. at the 5% level if 4.09 < 4. Therefore the F test will reject the null hypothesis that all the means are equal, but the Tukey–Kramer test will not fnd any pair of means to di˙er at the 5% level, for any value of MSE satisfying 5.099 < MSE < 6.035.

16.

The value of the F statistic is F = MSTr∕MSE = 5158.67∕MSE. The upper 5% point of the F3,12 distribution is 3.49. Therefore the F test will reject at the 5% level if 5158.67∕MSE ≥ 3.49, or, equivalently, if MSE ≤ 1478.13. The largest di˙erence between the sample means is 1389 − 1316 = 73. The upper 5% point of the Studentized .20. Therefore the Tukey–Kramer test will fail to fnd any di˙erences signifcant range distribution is q4,12,.05 ∕4, or equivalently, if MSE > 1208.39. at the 5% level if 73 < 4.20 Therefore the F test will reject the null hypothesis that all the means are equal, but the Tukey–Kramer test will not fnd any pair of means to di˙er at the 5% level, for any value of MSE satisfying 1208.39 < MSE < 1478.13.

Section 9.3 1.

Let I be the number of levels of oil type, let J be the number of levels of piston ring type, and let K be the number of replications. Then I = 4, J = 3, and K = 3. (a) The number of degrees of freedom for oil type is I − 1 = 3. (b) The number of degrees of freedom for piston ring type is J − 1 = 2. (c) The number of degrees of freedom for interaction is (I − 1)(J − 1) = 6.

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(d) The number of degrees of freedom for error is IJ (K − 1) = 24. (e) The mean squares are found by dividing the sums of squares by their respective degrees of freedom. The F statistics are found by dividing each mean square by the mean square for error. The number of degrees of freedom for the numerator of an F statistic is the number of degrees of freedom for its e˙ect, and the number of degrees of freedom for the denominator is the number of degrees of freedom for error. P -values may be obtained from the F table, or from a computer software package. Source Oil Ring Interaction Error Total

DF 3 2 6 24 35

SS 1.0926 0.9340 0.2485 1.7034 3.9785

MS 0.36420 0.46700 0.041417 0.070975

F 5.1314 6.5798 0.58354

P 0.007 0.005 0.740

(f) Yes. F6, 24 = 0.58354, P > 0.10 (P = 0.740). (g) No, some of the main e˙ects of oil type are non-zero. F3, 24 = 5.1314, 0.001 < P < 0.01 (P = 0.007). (h) No, some of the main e˙ects of piston ring type are non-zero. F2, 24 = 6.5798, 0.001 < P < 0.01 (P = 0.005).

Let I be the number of levels of operator, let J be the number of levels of machine, and let K be the number of replications. Then I = 3, J = 3, and K = 4. (a) The number of degrees of freedom for operator is I − 1 = 2. (b) The number of degrees of freedom for machine is J − 1 = 2. (c) The number of degrees of freedom for interaction is (I − 1)(J − 1) = 4. (d) The number of degrees of freedom for error is IJ (K − 1) = 27. (e) The mean squares are found by dividing the sums of squares by their respective degrees of freedom. The F statistics are found by dividing each mean square by the mean square for error. The number of degrees of freedom for the numerator of an F statistic is the number of degrees of freedom for its e˙ect, and the number of degrees of freedom for the denominator is the number of degrees of freedom for error. P -values may be obtained from the F table, or from a computer software package.

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SECTION 9.3

Source Operator Machine Interaction Error Total

DF 2 2 4 27 35

SS 3147.0 136.50 411.70 1522.0 5217.2

F 27.914 1.2107 1.8259

MS 1573.5 68.250 102.92 56.370

P 0.000 0.314 0.153

(f) Yes. F4, 27 = 1.8259, P > 0.10 (P = 0.153). (g) No, some of the main e˙ects of operator are non-zero. F2, 27 = 27.914, P < 0.001 (P

(h) Yes. F2, 27 = 1.2107, P > 0.10 (P = 0.314).

3. (a) Let I be the number of levels of mold temperature, let J be the number of levels of alloy, and let K be the number of replications. Then I = 5, J = 3, and K = 4. The number of degrees of freedom for mold temperature is I − 1 = 4. The number of degrees of freedom for alloy is J − 1 = 2. The number of degrees of freedom for interaction is (I − 1)(J − 1) = 8. The number of degrees of freedom for error is IJ (K − 1) = 45. The mean squares are found by dividing the sums of squares by their respective degrees of freedom. The F statistics are found by dividing each mean square by the mean square for error. The number of degrees of freedom for the numerator of an F statistic is the number of degrees of freedom for its e˙ect, and the number of degrees of freedom for the denominator is the number of degrees of freedom for error. P -values may be obtained from the F table, or from a computer software package. Source Mold Temp. Alloy Interaction Error Total

DF 4 2 8 45 59

SS 69738 8958 7275 115845 201816

MS 17434.5 4479.0 909.38 2574.3

F 6.7724 1.7399 0.35325

P 0.000 0.187 0.939

(b) Yes. F8,45 = 0.35325, P > 0.10 (P = 0.939). (c) No, some of the main e˙ects of mold temperature are non-zero. F4, 45 = 6.7724, P < 0.001 (P

(d) Yes. F3, 45 = 1.7399, P > 0.10, (P = 0.187).

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4. (a) Let I be the number of levels of adhesive, let J be the number of levels of curing pressure, and let K be the number of replications. Then I = 4, J = 3, and K = 5. The number of degrees of freedom for adhesive is I − 1 = 3. The number of degrees of freedom for curing pressure is J − 1 = 2. The number of degrees of freedom for interaction is (I − 1)(J − 1) = 6. The number of degrees of freedom for error is IJ (K − 1) = 48. The mean squares are found by dividing the sums of squares by their respective degrees of freedom. The F statistics are found by dividing each mean square by the mean square for error. The number of degrees of freedom for the numerator of an F statistic is the number of degrees of freedom for its e˙ect, and the number of degrees of freedom for the denominator is the number of degrees of freedom for error. P -values may be obtained from the F table, or from a computer software package. Source Adhesive Curing Press. Interaction Error Total

DF 3 2 6 48 59

SS 155.7 287.9 156.7 397 997.3

MS 51.900 143.95 26.117 8.2708

F 6.2751 17.405 3.1577

P 0.001 0.000 0.011

(b) No. F6,48 = 3.1577, 0.001 < P < 0.01 (P = 0.011). (c) No, some of the main e˙ects of curing pressure are non-zero. F2, 48 = 17.405, P < 0.001 (P ). Note that since the additive model does not hold, it is diÿcult to assign a practical interpretation to the main e˙ects. .001. Note that since the additive (d) No, some of the main e˙ects of adhesive are non-zero. F3, 48 = 6.2751, P model does not hold, it is diÿcult to assign a practical interpretation to the main e˙ects.

5. (a)

Main E˙ects of Solution NaCl −9.1148 9.1148 Na2 HPO4

(b) Source Solution Temperature Interaction Error Total

DF 1 1 1 20 23

Main E˙ects of Temperature 1.8101 25 C 37 C −1.8101

SS 1993.9 78.634 5.9960 7671.4 9750.0

MS 1993.9 78.634 5.9960 383.57

Interactions Temperature Solution 25 C 37 C 0.49983 NaCl −0.49983 0.49983 −0.49983 Na2 HPO4 F 5.1983 0.20500 0.015632

P 0.034 0.656 0.902

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SECTION 9.3

(c) Yes, F1,20 = 0.015632, P > 0.10 (P = 0.902). (d) Yes, since the additive model is plausible. The mean yield stress di˙ers between Na2 HPO4 and NaCl: F1,20 = 5.1983, 0.01 < P < 0.05 (P = 0.034). (e) There is no evidence that the temperature a˙ects yield stress: F1,20 = 0.20500, P > 0.10 (P = 0.656).

6. (a)

Main E˙ects of System Serial Line −1.22 Inventory 1.22

Main E˙ects of Scheme A 0.31 B −0.31

Interactions Scheme A B System Serial Line 0.12 −0.12 Inventory −0.12 0.12

(b) Source DF SS MS F P System 1 59.5360 59.5360 17.0238 0.000 Scheme 1 3.8440 3.8440 1.0992 0.301 Interaction 1 0.5760 0.5760 0.1647 0.687 Error 36 125.9000 3.4972 Total 39 189.8560

(c) Yes. F1,36 = 0.1647, P > 0.10 (P = 0.687). (d) Yes, since the additive model is plausible. The mean reduction di˙ers between the two systems: F1,36 = 17.0238, P < 0.001 (P = 0.0002).

(e) Yes, since the additive model is plausible. There is no evidence that the mean reduction di˙ers between the two schemes: F1,36 = 1.0992, P > 0.10 (P = 0.301).

7. (a) Source DF SS MS F P Adhesive 1 17.014 17.014 10.121 0.008 Pressure 2 35.663 17.832 10.607 0.002 Interaction 2 4.3544 2.1772 1.2951 0.310 Error 12 20.173 1.6811 17 77.205 Total

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(b) Yes. F2,12 = 1.2951, P > 0.10 (P = 0.310). (c) Yes, since the additive model is plausible. The mean strength di˙ers between the two adhesives: F1,12 = 10.121, P < 0.01 (P = 0.008). (d) Yes, since the additive model is plausible. The mean strength di˙ers among the pressures: F2,12 = 10.607, P < 0.01 (P = 0.002).

8. (a) Main E˙ects of Content 5 −11.487 10 −9.9267 15 21.413

Main E˙ects of Size Coarse −8.3687 Fine 8.3687

Interactions Size Content Coarse Fine 5 6.1867 −6.1867 10 5.9267 −5.9267 15 −12.113 12.113

(b) Source DF SS MS F P Content 2 6890.1 3445.1 24.26 0.000 Size 1 2110.1 2110.1 14.859 0.001 Interaction 2 2201.3 1100.7 7.7509 0.003 Error 24 3408.1 142.01 Total 29 14610

(c) No, F2,24 = 7.7509, P < 0.01 (P = 0.0025301). (d) No, because the additive model is rejected. (e) No, because the additive model is rejected.

9. (a) Main E˙ects of Speed 80 −13.074 120 −5.7593 150 19.463

Main E˙ects of Time 5 −8.4259 10 −0.2037 15 8.6296

Speed 5 80 5.6481 120 2.3704 150 −8.0185

Interactions Time 10 15 0.75926 −6.4074 −0.018519 −2.3519 −0.74074 8.7593

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SECTION 9.3

(b) Source DF SS MS F P Speed 2 10796 5397.9 63.649 0.000 Time 2 2619.1 1309.6 15.442 0.000 Interaction 4 1357.5 339.38 4.0018 0.007 Error 45 3816.3 84.807 Total 53 18589

(e) No, because the additive model is rejected.

10. (a) Source Feed Rate Speed Interaction Error Total

DF 2 2 4 27 35

SS 1718.4 224.18 48.758 197.51 2188.8

MS 859.19 112.09 12.189 7.3153

F 117.45 15.323 1.6663

P 0.000 0.000 0.187

(b) Yes, F4, 27 = 1.6663, P > 0.10 (P = 0.187). (c) Yes, since the additive model is plausible. The mean lifetime stress varies with feed rate. F2, 27 = 117.45, P < 0.001 (P ).

(d) Yes, since the additive model is plausible. The mean lifetime stress varies with speed. F2, 27 = 15.323, P < 0.001 (P ).

11. (a)

Main E˙ects of Material 0.044367 CPTi-ZrO2 TiAlloy-ZrO2 −0.044367

Main E˙ects of Neck Length Short −0.018533 Medium −0.024833 Long 0.043367

Interactions

CPTi-ZrO2 TiAlloy-ZrO2

Short 0.0063333 0.0063333

Neck Length Long Medium −0.023767 0.017433 0.023767 −0.017433

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482 (b) Source Taper Material Neck Length Interaction Error Total

CHAPTER 9

DF 1 2 2 24 29

SS 0.059052 0.028408 0.0090089 0.059976 0.15652

F 23.630 5.6840 1.8185

MS 0.059052 0.014204 0.0045444 0.002499

P 0.000 0.010 0.184

(c) Yes, the interactions may plausibly be equal to 0. The value of the test statistic is 1.8185, its null distribution is F2,24 , and P > 0.10 (P = 0.184). (d) Yes, since the additive model is plausible. The mean coeÿcient of friction di˙ers between CPTi-ZrO2 and TiAlloy-ZrO2 : F1,24 = 23.630, P < 0.001. (e) Yes, since the additive model is plausible. The mean coeÿcient of friction is not the same for all neck lengths: F2, .6840, P .01. To determine which pairs of e˙ects di˙er, we use q3,24,.05 = 3.53. We compute 3. 0.002499∕10 = 0.056. We conclude that the e˙ect of long neck length di˙ers from both short and medium lengths, but we cannot conclude that the e˙ects of short and medium lengths di˙er from each other.

12. (a) Main E˙ects of Concentration 15 0.72472 40 −0.32528 100 −0.39944

(b) Source Concentration Delivery Ratio Interaction Error Total

Main E˙ects of Delivery Ratio 1∶0 5.2267 1 ∶ 1 −1.0089 1 ∶ 5 −2.0422 1 ∶ 10 −2.1756

DF 2 3 6 24 35

SS 9.487 335.16 20.295 9.7799 374.72

Interactions Concentration 15 40 100

MS 4.7435 111.72 3.3825 0.40749

1∶0 1.6742 −0.77250 −0.90167

F 11.641 274.16 8.3008

Delivery Ratio 1∶1 1∶5 1 ∶ 10 0.10972 −0.86028 −0.92361 0.056389 0.19306 0.52306 −0.16611 0.66722 0.40056

P 0.000 0.000 0.000

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SECTION 9.3

13. (a)

Main E˙ects of Concentration 15 0.16667 40 −0.067778 100 −0.098889

(b) Source Concentration Delivery Ratio Interaction Error Total

Main E˙ects of Delivery Ratio 1∶1 0.73333 1 ∶ 5 −0.30000 1 ∶ 10 −0.43333

DF 2 2 4 18 26

SS 0.37936 7.34 3.4447 0.8814 12.045

Interactions Delivery Ratio Concentration 1∶1 1∶5 1 ∶ 10 15 0.66778 −0.30222 −0.36556 40 −0.20111 −0.064444 0.26556 100 −0.46667 0.36667 0.10000

MS 0.18968 3.67 0.86118 0.048967

F 3.8736 74.949 17.587

P 0.040 0.000 0.000

(d)

Sorption (%)

2.5 2

concentration = 15 concentration = 40

The slopes of the line segments are quite di˙erent from one another, indicating a high degree of interaction.

1.5 concentration = 100

1 0.5 0

1:1

14. (a) Main E˙ects of Travel Speed 10 −61.389 20 18.722 30 42.667

1:5 Delivery Ratio

1:10

Main E˙ects of Accelerating Voltage 10 66.056 25 −2.9444 50 −63.111

Interactions Accelerating Voltage Travel Speed 10 25 50 10 9.0556 −22.944 13.889 20 −16.722 9.6111 7.1111 30 7.6667 13.333 −21.000

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(b) Source Travel Speed Acceleration Interaction Error Total

DF 2 2 4 45 53

SS 106912.1 150390.3 11408.8 275228.0 543939.3

MS 53456.1 75195.2 2852.2 6116.2

F 8.7401 12.294 0.46634

P 0.001 0.000 0.760

(c) Yes, F4,45 = 0.46634, P = 0.760. .001. To determine which e˙ects di˙er at the (d) The travel speed does a˙ect the hardness: F2,45 = 8.7401, P 5% level, we should use q3,45,.05 . This value is not found in Table A.8, so we approximate it with q3,40,.05 = 3.44. We compute 3.44 6116.2∕18 = 63.41. We conclude that the e˙ect of a travel speed of 10 di˙ers from those of both 20 and 30, but we cannot conclude that the e˙ects of 20 and 30 di˙er from each other.

(e) The accelerating voltage does a˙ect the hardness: F2,45 = 12.294, P = 0.000. To determine which e˙ects di˙er at the 5% level, we should use q3,45,.05 . This value is not found in Table A.8, so we approximate it with q3,40,.05 = 3.44. We compute 3.44 6116.2∕18 = 63.41. We conclude that the e˙ect of an accelerating voltage of 10 di˙ers from those of both 25 and 50, but we cannot conclude that the e˙ects of 25 and 50 di˙er from each other.

15. (a)

Main E˙ects of Attachment Nail −1.3832 Adhesive 1.3832

(b) Source Attachment Length Interaction Error Total

DF 1 2 2 54 59

Main E˙ects of Length Quarter −7.1165 Half −2.5665 Full 9.683

SS 114.79 3019.8 10.023 107.29 3251.9

MS 114.79 1509.9 5.0115 1.9869

Interactions

Nail Adhesive

F 57.773 759.94 2.5223

Length Quarter Half Full 0.48317 0.33167 −0.51633 −0.48317 −0.33167 0.51633

P 0.000 0.000 0.090

(c) The additive model is barely plausible: F2,54 = 2.5223, 0.05 < P < 0.10 (P = 0.090). (d) Yes, the attachment method does a˙ect the critical buckling load: F1,54 = 57.773, P

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SECTION 9.3

(e) Yes, the side member length does a˙ect the critical buckling load: F2,54 = 759.94, P . To determine which e˙ects di˙er at the 5% level, we should use q3,54,.05 . This value is not found in Table A.8, so we approximate it with q3,40,.05 = 3.44. We compute 3.44 1.9869∕20 = 1.08. We conclude that the e˙ects of quarter, half and full all di˙er from each other.

16. (a)

Main E˙ects of Attachment Nail −0.043833 Adhesive 0.043833

(b) Source Attachment Length Interaction Error Total

DF 1 2 2 54 59

Interactions

Main E˙ects of Length Quarter −0.11917 Half 0.31933 Full −0.20017

SS 0.11528 3.1248 0.029323 44.391 47.66

MS 0.11528 1.5624 0.014662 0.82206

Nail Adhesive

F 0.14024 1.9006 0.017835

Quarter −0.029167 0.029167

Length Half 0.024333 −0.024333

Full 0.0048333 −0.0048333

P 0.710 0.159 0.982

(c) The additive model is quite plausible: F2,54 = 0.017835, P > 0.10 (P = 0.982). (d) There is no evidence that Young’s modulus di˙ers between attachment methods: F1,54 = 0.14024, P > 0.10 (P = 0.710).

(e) There is no evidence that Young’s modulus di˙ers among side member lengths: F2,54 = 1.9006, P > 0.10 (P = 0.159).

17. (a) Source Wafer Operator Interaction Error Total

DF 2 2 4 9 17

SS 114661.4 136.78 6.5556 45.500 114850.3

MS 57330.7 68.389 1.6389 5.0556

F 11340.1 13.53 0.32

P 0.000 0.002 0.855

(b) There are di˙erences among the operators. F2, 9 = 13.53, 0.01 < P < 0.001 (P = 0.002).

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18. (a) Source Wafer Operator Interaction Error Total

DF 2 2 4 9 17

SS 262627.1 24.778 16.222 49.000 262717.1

MS 131313.6 12.389 4.0556 5.4444

F 24118.8 2.28 0.74

P 0.000 0.159 0.585

(b) With the balance powered up continuously, there are no apparent di˙erences in measurements obtained by different operators. When one of the operators was using the balance shortly after power-up, there were di˙erences. It is reasonable to conclude that the balance should be powered up continuously.

19. (a) Source PVAL DCM Interaction Error Total

DF 2 2 4 18 26

SS 125.41 1647.9 159.96 136.94 2070.2

MS 62.704 823.94 39.990 7.6075

F 8.2424 108.31 5.2567

P 0.003 0.000 0.006

(b) Since the interaction terms are not equal to 0, (F4,18 = 5.2567, P = 0.006), we cannot interpret the main e˙ects. Therefore we compute the cell means. These are

PVAL 0.5 1.0 2.0

DCM (ml) 50 40 30 97.8 92.7 74.2 93.5 80.8 75.4 94.2 88.6 78.8

We conclude that a DCM level of 50 ml produces greater encapsulation eÿciency than either of the other levels. If DCM = 50, the PVAL concentration does not have much e˙ect. Note that for DCM = 50, encapsulation eÿciency is maximized at the lowest PVAL concentration, but for DCM = 30, it is maximized at the highest PVAL concentration. This is the source of the signifcant interaction.

Section 9.4 1. (a) NaOH concentration is the blocking factor, age is the treatment factor.

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487

SECTION 9.4

(b) Source DF SS MS F P Treatment 3 386.33 128.78 211.14 0.000 Blocks 4 13.953 3.4882 5.7192 0.008 Error 12 7.3190 0.6099 Total 19 407.60

(P = 1.202 × 10−10 )

(d) q4, 12 .05 = 4.20, MSAB = 0.6099, and J = 5. The 5% critical value is therefore 4.20 .6099∕5 = 1.4669. The sample means are X 0 = 55.46, X 4 = 45.22, X 8 = 46.26, and X 12 = 44.47. We therefore conclude that age 0 di˙ers from ages 4, 8, and 12, and that age 8 di˙ers from age 12.

2. (a) Source Machine Block Interaction Error Total

DF 3 2 6 12 23

SS 257.678 592.428 30.704 215.836 1096.646

MS 85.893 296.21 5.1173 17.986

F 4.7754 16.469 0.28451

P 0.021 0.000 0.933

(b) Yes, the P -value for interactions is large.

3. (a) Let I be the number of levels for lighting method, let J be the number of levels for blocks, and let K be the number of replications. Then I = 4, J = 3, and K = 3. The number of degrees of freedom for treatments is I − 1 = 3. The number of degrees of freedom for blocks is J − 1 = 2. The number of degrees of freedom for interaction is (I − 1)(J − 1) = 6. The number of degrees of freedom for error is IJ (K − 1) = 24. The mean squares are found by dividing the sums of squares by their respective degrees of freedom. The F statistics are found by dividing each mean square by the mean square for error. The number of degrees of freedom for the numerator of an F statistic is the number of degrees of freedom for its e˙ect, and the number of degrees of freedom for the denominator is the number of degrees of freedom for error. P -values may be obtained from the F table, or from a computer software package.

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CHAPTER 9

Source Lighting Block Interaction Error Total

DF 3 2 6 24 35

SS 9943 11432 6135 23866 51376

F 3.3329 5.7481 1.0282

MS 3314.33 5716.00 1022.50 994.417

P 0.036 0.009 0.431

(b) Yes. The P -value for interactions is large (0.431).

4. (a) Let I be the number of levels for coating, let J be the number of levels for blocks, and let K be the number of replications. Then I = 3, J = 10, and K = 2. The number of degrees of freedom for treatments is I − 1 = 2. The number of degrees of freedom for blocks is J − 1 = 9. The number of degrees of freedom for interaction is (I − 1)(J − 1) = 18. The number of degrees of freedom for error is IJ (K − 1) = 30. The mean squares are found by dividing the sums of squares by their respective degrees of freedom. The F statistics are found by dividing each mean square by the mean square for error. The number of degrees of freedom for the numerator of an F statistic is the number of degrees of freedom for its e˙ect, and the number of degrees of freedom for the denominator is the number of degrees of freedom for error. P -values may be obtained from the F table, or from a computer software package. Source Coating Block Interaction Error Total

DF 2 9 18 30 59

SS 4.8 11.2 18.4 10.3 44.7

MS 2.4000 1.2444 1.0222 0.3433

F 6.99029 3.62460 2.97735

P 0.003 0.004 0.004

(b) No. The P -value for interactions is 0.004, so we cannot conclude that there is no interaction between treatment and block.

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489

SECTION 9.4

5. (a) Source Variety Block Error Total

DF 9 5 45 59

SS 339032 1860838 660198 2860069

F 2.5677 25.367

MS 37670 372168 14671

P 0.018 0.000

(b) Yes, F9,45 = 2.5677, P = 0.018.

6. (a) Day is the blocking factor, Sprinkler is the treatment factor.

(b) Source Sprinkler Day Interaction Error Total

DF 3 3 9 16 31

SS 328.69 30.693 31.553 31.695 422.63

F 55.308 5.1648 1.7698

MS 109.56 10.231 3.5059 1.9809

P 0.000 0.011 0.153

(c) Yes, the P -value for interactions is reasonably large (0.153), so it is plausible that the interactions are equal to 0.

(d) Yes, the P -value for the main e˙ect of Sprinkler is

(e) q4,16,.05 = 4.05, MSE = 1.9809, so the critical value is 4.05 1.9809∕(4 ⋅ 2) = 2.015. The mean runo˙s for the sprinklers are A: 7.975, B: 6.4875, C: 1.4625, D: 0.4375. We can therefore conclude at the 5% level that A di˙ers from both C and D, and B di˙ers from both C and D.

7. (a) Source Waterworks Block Error Total

DF 3 14 42 59

SS 1253.5 1006.1 3585.0 5844.6

MS 417.84 71.864 85.356

F 4.8953 0.84193

P 0.005 0.622

(b) Yes, F3,42 = 4.8953, P = 0.005.

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(c) To determine which e˙ects di˙er at the 5% level, we should use q4,42,.05 This value is not found in Table A.8, so we approximate it with q4,40,.05 = 3.79. The 5% critical value is 3.79 85.356∕15 = 9.04. The sample means are X A = 34.000, X B = 22.933, X C = 24.800, X D = 31.467. We can conclude that A di˙ers from both B and C. (d) The P -value for the blocking factor is large (0.622), suggesting that the blocking factor (time) has only a small e˙ect on the outcome. It might therefore be reasonable to ignore the blocking factor and perform a one-way ANOVA.

8. (a) Blocking factor is job, treatment factor is machine. (b) Source DF SS MS F P Treatment 4 543.71 135.93 10.849 0.000 Blocks 6 323.89 53.981 4.3086 0.004 Error 24 300.69 12.529 Total 34 1168.3 (c) Yes, F4,24 = 10.849, P

(P = 3.63 × 10−5 )

(d) q5,24,.05 = 4.17. The value of MSE is 12.529. The 5% critical value is therefore 4.17 12.529∕7 = 5.58. The sample means are X A = 21.143, X B = 23.143, X C = 15.286, X D = 22.571, X E = 13.571. We conclude that C and E di˙er from A, B, and D.

9. (a) One motor of each type should be tested on each day. The order in which the motors are tested on any given day should be chosen at random. This is a randomized block design, in which the days are the blocks. It is not a completely randomized design, since randomization occurs only within blocks. ∑5 (b) The test statistic is ∑

4 j=1

∑5

i=1 (X i. − X .. )

2 i=1 (Xij − X i. − X .j − X .. ) ∕12

10. (a) Use of the randomized block design assumes that there are no interactions between treatments and blocks. Since di˙erent students often do best in di˙erent courses, there are likely to be interactions between students and courses.

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SECTION 9.5

(b) There are clearly interactions present. For example, student 1 did best in chemistry, while student 2 did best in statistics, and student 4 did best in physics.

Section 9.5 1. 1 ad bd ab cd ac bc abcd

A B − − + − − + + + − − + − − + + +

C − − − − + + + +

D − + + − + − − +

The alias pairs are {A, BCD}, {B, ACD}, {C, ABD}, {D, ABC}, {AB, CD}, {AC, BD}, and {AD, BC}

An experiment in which the e˙ect of the presence or absence of several catalysts on yield is being studied, and in which the ambient temperature also a˙ects yield. If all the runs with catalyst A present are run in the morning, when the ambient temperature is low, and all the runs with catalyst A absent are run in the afternoon, when the ambient temperature is high, then the e˙ect of catalyst A will be incorrectly estimated.

3. (a) Variable A B C AB AC BC ABC Error Total

Sum of Ef fect DF Squares 6.75 1 182.25 9.50 1 361.00 1.00 1 4.00 2.50 1 25.00 0.50 1 1.00 0.75 1 2.25 −2.75 1 30.25 8 122.00 15 727.75

Mean Square F 182.25 11.9508 361.00 23.6721 4.00 0.2623 25.00 1.6393 1.00 0.0656 2.25 0.1475 30.25 1.9836 15.25

P 0.009 0.001 0.622 0.236 0.804 0.711 0.197

(b) Factors A and B (temperature and concentration) seem to have an e˙ect on yield. There is no evidence that pH has an e˙ect. None of the interactions appear to be signifcant. Their P -values are all greater than 0.19.

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(c) Since the e˙ect of temperature is positive and statistically signifcant, we can conclude that the mean yield is higher when temperature is high.

4. (a) Variable A B C AB AC BC ABC Error Total

Sum of Mean Squares Ef fect DF Square −0.04625 1 0.0085562 0.0085562 −0.19875 1 0.15801 0.15801 0.01625 1 0.0010563 0.0010563 −0.02375 1 0.0022563 0.0022563 −0.02375 1 0.0022563 0.0022563 0.00875 1 0.00030625 0.00030625 −0.01125 1 0.00050625 0.00050625 8 0.043550 0.0054437 15 0.21649

F 1.5718 29.025 0.19403 0.41447 0.41447 0.056257 0.092997

P 0.245 0.001 0.671 0.538 0.538 0.818 0.768

(b) The additive model is appropriate, since none of the interactions are signifcant.

5. (a) Variable A B C AB AC BC ABC

Ef fect 3.3750 23.625 1.1250 −2.8750 −1.3750 −1.6250 1.8750

(b) No, since the design is unreplicated, there is no error sum of squares.

(d) If the additive model is known to hold, then the ANOVA table below shows that the main e˙ect of B is not equal to 0, while the main e˙ects of A and C may be equal to 0.

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SECTION 9.5

Variable A B C Error Total

Ef fect 3.3750 23.625 1.1250

Sum of Mean DF Squares Square 1 22.781 22.781 1 1116.3 1116.3 1 2.5312 2.5312 4 32.625 8.1562 7 1174.2

F P 2.7931 0.170 136.86 0.000 0.31034 0.607

E˙ect −0.0325 −0.0525 −0.0125 0.0175 0.0475 −0.0625 −0.0025

Term A B C 6. (a) AB AC BC ABC

(b) No, since the design is unreplicated, there is no error sum of squares. (c)

0.999 0.99 0.95 0.9 0.75

The estimates lie nearly on a straight line, so none of the factors can clearly be said to infuence the thickness.

0.5 0.25 0.1 0.05 0.01 0.001 −0.06 −0.04 −0.02 0 Effect

7. (a) Term A B C AB AC BC ABC

0.02

0.04

E˙ect −119.25 259.25 −82.75 101.75 −6.25 −52.75 −2.25

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(b) No, since the design is unreplicated, there is no error sum of squares.

(d) No, because the AB interaction is large.

8. (a) Let x be the required value. Then x satisfes the equation −x + 353 + 778 − 769 + 544 − 319 − 651 + 625 = 0. Therefore x = 561.

(b) Variable A B C AB AC BC ABC

Ef fect −117 261.5 −80.5 99.5 −8.5 −55 0

9. (a) Variable A B C D AB AC AD BC BD CD ABC ABD ACD BCD ABCD

Ef fect 1.2 3.25 −16.05 −2.55 2 2.9 −1.2 1.05 −1.45 −1.6 −0.8 −1.9 −0.15 0.8 0.65

(b) Factor C is the only one that really stands out.

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SECTION 9.5 10. (a) Term E˙ect A −2251.3 B 72.6 C 3727.5 D 503.0 E 3.8 AB −12.1 AC −972.5 AD −163.0 AE 2.5 BC 105.4 BD 13.6 BE −19.6 CD 199.0 CE 4.3 DE 5.6

Term ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE

E˙ect −16.1 −1.3 −0.7 −48.8 0.3 3.1 16.7 −7.7 4.9 3.7 4.0 −3.9 −12.2 −0.9 7.6 −1.9

(b) A, C, and possibly D. The largest e˙ects are the main e˙ects of A and C, and the interaction of A and C. The main e˙ect of D is also larger than the others.

11. (a) Variable A B C AB AC BC ABC Error Total

Sum of Mean Ef fect DF Squares Square F 1 811.68 811.68 691.2 14.245 8.0275 1 257.76 257.76 219.5 −6.385 1 163.07 163.07 138.87 −1.68 1 11.29 11.29 9.6139 −1.1175 1 4.9952 4.9952 4.2538 −0.535 1 1.1449 1.1449 0.97496 −1.2175 1 5.9292 5.9292 5.0492 8 9.3944 1.1743 15 1265.3

P 0.000 0.000 0.000 0.015 0.073 0.352 0.055

(b) All main e˙ects are signifcant, as is the AB interaction. Only the BC interaction has a P -value that is reasonably large. All three factors appear to be important, and they seem to interact considerably with each other.

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496 12. (a) Variable A B C D E AB AC AD

CHAPTER 9

Ef fect 49.337 51.788 17.812 −8.7125 18.537 40.112 17.188 −3.2375

Variable AE BC BD BE CD CE DE

Ef fect 16.112 15.588 −9.9375 12.862 13.637 −8.0625 12.263

0.999

(b) The probability plot shows that factors A, B and D appear to a˙ect the solubility. The CE interaction is somewhat large as well; this is more diÿcult to interpret.

0.99 0.95 0.9 0.75 0.5 0.25 0.1 0.05 0.01

B A AB

CE D BD

0.001 0

Ef fect 7.50 19.75 1.25 0.00 44.50 5.25 1.25 −4.00

Variable AE BC BD BE CD CE DE

20 Effect

Ef fect 7.00 3.00 −1.75 0.25 2.25 −6.25 3.50

0.999

(d) The probability plot shows that factors B and E appear to a˙ect the yield. None of the interactions appear to be noticeably large.

0.99 0.95 0.9 0.75 0.5 0.25 0.1 0.05 0.01

E B

0.001 −10

20 Effect

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SUPPLEMENTARY EXERCISES FOR CHAPTER 9

13.

(ii) The sum of the main e˙ect of A and the BCDE interaction.

Supplementary Exercises for Chapter 9 1.

Source Gypsum Error Total

DF 3 8 11

SS 0.013092 0.12073 0.13383

MS 0.0043639 0.015092

F 0.28916

P 0.832

The value of the test statistic is F3,8 = 0.28916; P > 0.10 (P = 0.832). There is no evidence that the pH di˙ers with the amount of gypsum added.

Source Gypsum Soil Type Interaction Error Total

DF 2 1 2 6 11

SS 0.096717 0.10268 0.03515 0.30595 0.54049

MS 0.048358 0.10268 0.017575 0.050992

F 0.94836 2.0136 0.34466

P 0.439 0.206 0.722

(a) No. F2,6 = 0.34466, P > 0.10 (P = 0.722). (b) No. F1,6 = 2.0136, P > 0.10 (P = 0.206). (c) No. F2,6 = 0.94836, P > 0.10 (P = 0.439).

Source Day Error Total

DF 2 36 38

SS 1.0908 0.87846 1.9692

MS 0.54538 0.024402

F 22.35

P 0.000

We conclude that the mean sugar content di˙ers among the three days (F2,36 = 22.35, P

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4. (a) 23 − 3 − 2 − 6 = 12 (b) SSAB = 4.200(6) = 25.200 (c) SSE = SST − SSA − SSB − SSAB = 32.253 (d) MSA = 145.375∕3 = 48.4583 (e) MSB = 15.042∕2 = 7.521 (f) MSE = 32.253∕12 = 2.68775 (g) MSA∕MSE = 48.4583∕2.68775 = 18.029 (h) MSB∕MSE = 7.521∕2.68775 = 2.798 (i) MSAB∕MSE = 4.200∕2.68775 = 1.563 (j) P < 0.001 (F3,12 = 18.029 (k) P

.7 × 10−5 )

.10 (F2,12 = 2.798, P = 0.101)

(l) P > 0.10 (F6,12 = 1.563

.240)

5. (a) No. The variances are not constant across groups. In particular, there is an outlier in group 1. (b) No, for the same reasons as in part (a).

DF 4 35 39

SS 5.2029 5.1080 10.311

MS 1.3007 0.14594

F 8.9126

P 0.000

We conclude that the mean dissolve time di˙ers among the groups (F4,35 = 8.9126, P

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SUPPLEMENTARY EXERCISES FOR CHAPTER 9

6. (a)

Main E˙ects of Al2 O3 1.0 0.3244 5.0 −0.01222 10.0 −0.3122

(b) Source Al2 O3 C4 S3 Interaction Error Total

DF 2 2 4 27 35

Main E˙ects of C4 S4 Low 1.0003 Medium −0.11722 High −0.88306

SS 2.4348 21.529 0.075744 2.8566 26.896

MS 1.2174 10.765 0.018936 0.1058

Al2 O3 = 1.0 Al2 O3 = 5.0 Al2 O3 = 10.0 F 11.507 101.75 0.17898

Interactions C4 S3 Low C4 S3 medium 0.021389 0.033889 0.040556 −0.074444 −0.061944 0.040556

C4 S3 High −0.055278 0.033889 0.021389

P 0.000 0.000 0.947

(d) Yes. F2,27 = 11.507, P

(e) Yes. F2,27 = 101.75, P

The recommendation is not a good one. The engineer is trying to interpret the main e˙ects without looking at the interactions. The small P -value for the interactions indicates that they must be taken into account. Looking at the cell means, it is clear that if design 2 is used, then the less expensive material performs just as well as the more expensive material. The best recommendation, therefore, is to use design 2 with the less expensive material.

8. (a) The main e˙ects are: A: (1∕4)(−79.8 + 69.0 − 72.3 + 71.2 − 91.3 + 95.4 − 92.7 + 91.5) = −2.25. B: (1∕4)(−79.8 − 69.0 + 72.3 + 71.2 − 91.3 − 95.4 + 92.7 + 91.5) = −1.95. C: (1∕4)(−79.8 − 69.0 − 72.3 − 71.2 + 91.3 + 95.4 + 92.7 + 91.5) = 19.65. D: (1∕4)(−79.8 + 69.0 + 72.3 − 71.2 + 91.3 − 95.4 − 92.7 + 91.5) = −3.75.

(b) The interaction sums of squares are AB: 2.42, AC: 27.38, BC: 0.98. The pooled sum of squares is 2.42 + 27.38 + 0.98 = 30.78. This is used as an error sum of squares.

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Ef fect −2.25 −1.95 19.65 −3.75

Sum of DF Squares 1 10.125 1 7.605 1 772.25 1 28.125

Interaction

30.78

Total

848.88

Mean Square F P 10.125 0.98684 0.394 7.605 0.74123 0.453 772.25 75.268 0.003 28.125 2.7412 0.196 10.26

The interaction mean square is 30.78∕3 = 10.26. The test statistics for the main e˙ects are found by dividing their respective mean squares by the interaction mean square. Each test statistic has null distribution F1,3 . We can conclude that factor C a˙ects the yield (F1,3 = 75.268, P = 0.003).

9. (a) Source Base Instrument Interaction Error Total

DF 3 2 6 708 719

SS 13495 90990 12050 422912 539447

MS 4498.3 45495 2008.3 597.33

F 7.5307 76.164 3.3622

P 0.000 0.000 0.003

(b) No, it is not appropriate because there are interactions between the row and column e˙ects (F6,708 = 3.3622, P = 0.003).

10. (a)

Main E˙ects of Fertilizer Low −5.2630 Medium 5.8926 High −0.62963

Low Fertilizer Medium Fertilizer High Fertilizer

Main E˙ects of Zone Southern −2.3852 Central 3.8370 Northern −1.4519 Interactions Southern Zone Central Zone −2.3259 −0.74815 1.2852 −0.27037 1.0407 1.0185

Northern Zone 3.0741 −1.0148 −2.0593

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SUPPLEMENTARY EXERCISES FOR CHAPTER 9

(b) Source Fertilizer Zone Interaction Error Total

DF 2 2 4 18 26

SS 565.36 202.68 73.606 150.11 991.76

F 33.896 12.152 2.2065

MS 282.68 101.34 18.401 8.3396

P 0.000 0.000 0.109

(c) It is plausible that there is no interaction. (F4,18 = 2.2065, P = 0.109). (d) There are di˙erences in sap production between the three fertilizer levels (F2,18 = 33.896, P

11. (a) Source Channel Type Error Total

DF 4 15 19

SS 1011.7 435.39 1447.1

MS 252.93 29.026

F 8.7139

P 0.001

Yes. F4,15 = 8.7139, P = 0.001. (b) q5,20,.05 = 4.23, MSE = 29.026, J = 4. The 5% critical value is therefore 4.23 29.026∕4 = 11.39. The sample means for the fve channels are X 1 = 44.000, X 2 = 44.100, X 3 = 30.900, X 4 = 28.575, X 5 = 44.425. We can therefore conclude that channels 3 and 4 di˙er from channels 1, 2, and 5.

12. (a) The estimates are:

A: 1.6875, B: 0.8125, C: 1.1875, AB: 0.6875, AC: 1.8125, BC: −0.3125,

ABC: −0.1875. (b) Variable A B C AB AC BC ABC Error Total

Sum of Mean Ef fect DF Squares Square F 1.6875 1 22.781 22.781 9.3863 0.8125 1 5.2812 5.2812 2.176 1.1875 1 11.281 11.281 4.6481 0.6875 1 3.7812 3.7812 1.5579 1.8125 1 26.281 26.281 10.828 −0.3125 1 0.78125 0.78125 0.32189 −0.1875 1 0.28125 0.28125 0.11588 24 58.25 2.4271 31 128.72

P 0.005 0.153 0.041 0.224 0.003 0.576 0.737

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CHAPTER 9

(c) The sum of squares for interactions is 3.7812 + 26.281 + 0.78125 + 0.28125 = 31.1247. There are four interactions (AB, AC, BC, and ABC), so there are four degrees of freedom for interaction. The mean square for interaction is therefore 31.1247∕4 = 7.781175. The mean square for error is 2.4271, so the F statistic for testing interactions is 7.781175∕2.4271 = 3.206. The null distribution is F4,24 , and 0.01 < P < 0.05 (P = 0.030). The additive model does not seem appropriate. (d) The e˙ects with small P -values are A, C, and the AC interaction. It is reasonable to conclude that factors A and C a˙ect the outcome, while B may have no e˙ect.

13.

Source Well Type Error Total

DF 4 289 293

SS 5.7523 260.18 265.93

F 1.5974

MS 1.4381 0.90028

P 0.175

No. F4,289 = 1.5974, P > 0.10 (P = 0.175).

14. (a) Source Concentration Error Total

DF 3 12 15

No. F3,12 = 345.58,

SS 8.5768 0.099275 8.6761

MS 2.8589 0.0082729

F 345.58

P 0.000

(b) q4,12,.05 = 4.20, MSE = 0.0082729, so the 5% critical point is 4. 0.0082729∕4 = 0.191. The sample means are X 1 = 1.625, X 2 = 2.7, X 3 = 3.545, X 4 = 3.2475. We can conclude that all pairs of concentrations have di˙ering enthalpies.

15. (a) From Exercise 11, MSE = 29.026, so s =

.026 = 5.388.

(b) The MINITAB output for the power calculation is

Power and Sample Size One-way ANOVA

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SUPPLEMENTARY EXERCISES FOR CHAPTER 9

Alpha = 0.05

Assumed standard deviation = 5.388

SS Means 50

Target Power 0.9

Sample Size 10

Actual Power 0.901970

Number of Levels = 5

Maximum Difference 10

The sample size is for each level.

Power and Sample Size One-way ANOVA Alpha = 0.05

Assumed standard deviation = 8.082

SS Means 50

Target Power 0.9

Sample Size 22

Actual Power 0.913650

Number of Levels = 5

Maximum Difference 10

The sample size is for each level.

16. (a) From Exercise 14, MSE = 0.0082729, so s

0.0082729 = 0.0910.

(b) The MINITAB output for the power calculation is

Power and Sample Size One-way ANOVA Alpha = 0.05

Assumed standard deviation = 0.091

SS Means 0.005

Target Power 0.8

Sample Size 20

Actual Power 0.821699

Number of Levels = 4

Maximum Difference 0.1

The sample size is for each level.

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Power and Sample Size One-way ANOVA Alpha = 0.05

Assumed standard deviation = 0.1365

SS Means 0.005

Target Power 0.8

Sample Size 42

Actual Power 0.804081

Number of Levels = 4

Maximum Difference 0.1

The sample size is for each level.

E˙ect Variable E˙ect Variable E˙ect 17. (a) Variable E˙ect Variable A 3.9875 AB −0.1125 BD −0.0875 ACD 0.4875 B 2.0375 AC 0.0125 CD 0.6375 BCD −0.3125 C 1.7125 AD −0.9375 ABC −0.2375 ABCD −0.7125 D 3.7125 BC 0.7125 ABD 0.5125 (b) The main e˙ects are noticeably larger than the interactions, and the main e˙ects for A and D are noticeably larger than those for B and C.

(c)

Sum of Mean Variable Ef fect DF Squares Square F P A 3.9875 1 63.601 63.601 68.415 0.000 B 2.0375 1 16.606 16.606 17.863 0.008 C 1.7125 1 11.731 11.731 12.619 0.016 D 3.7125 1 55.131 55.131 59.304 0.001 AB −0.1125 1 0.050625 0.050625 0.054457 0.825 AC 0.0125 1 0.000625 0.000625 0.00067231 0.980 AD −0.9375 1 3.5156 3.5156 3.7818 0.109 BC 0.7125 1 2.0306 2.0306 2.1843 0.199 BD −0.0875 1 0.030625 0.030625 0.032943 0.863 CD 0.6375 1 1.6256 1.6256 1.7487 0.243 Interaction 5 4.6481 0.92963 Total 15 158.97 We can conclude that each of the factors A, B, C, and D has an e˙ect on the outcome.

(d) The F statistics are computed by dividing the mean square for each e˙ect (equal to its sum of squares) by the

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SUPPLEMENTARY EXERCISES FOR CHAPTER 9

error mean square 1.04. The degrees of freedom for each F statistic are 1 and 4. The results are summarized in the following table. Variable A B C D AB AC AD BC BD CD ABC ABD ACD BCD ABCD

Sum of Mean Ef fect DF Squares Square F 1 63.601 63.601 61.154 3.9875 16.606 2.0375 1 16.606 15.967 1.7125 1 11.731 11.731 11.279 3.7125 1 55.131 55.131 53.01 −0.1125 1 0.050625 0.050625 0.048678 0.0125 1 0.000625 0.000625 0.00060096 −0.9375 1 3.5156 3.5156 3.3804 0.7125 1 2.0306 2.0306 1.9525 −0.0875 1 0.030625 0.030625 0.029447 0.6375 1 1.6256 1.6256 1.5631 −0.2375 1 0.22563 0.22563 0.21695 0.5125 1 1.0506 1.0506 1.0102 0.4875 1 0.95063 0.95063 0.91406 −0.3125 1 0.39062 0.39062 0.3756 −0.7125 1 2.0306 2.0306 1.9525

P 0.001 0.016 0.028 0.002 0.836 0.982 0.140 0.235 0.872 0.279 0.666 0.372 0.393 0.573 0.235

(e) Yes. None of the P -values for the third- or higher-order interactions are small. (f) We can conclude that each of the factors A, B, C, and D has an e˙ect on the outcome.

18. (a) Source Circumference Depth Interaction Error Total

DF 7 3 21 96 127

SS 6589643 2877565 3455204 16911875 29834287

MS 941378 959188 164534 176165.4

F 5.3437 5.4448 0.93397

P 0.000 0.002 0.550

(b) No. F21,96 = 0.93397, P > 0.10 (P = 0.550). (c) Yes. F7,96 = 5.3437, P ≈ 0. (d) Yes. F3,96 = 5.4448, 0.001 < P < 0.01 (P = 0.002).

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CHAPTER 9

Source Location Error Total

DF 2 107 109

SS 3.0287 17.16 20.189

MS 1.5144 0.16037

F 9.4427

P 0.000

Yes, F2,107 = 9.4427, P ≈ 0.

20.

The sample standard deviations vary by a factor of about 9. Since the sample sizes are reasonably large, the sample standard deviations are close to the population standard deviations, so we conclude that the assumption of equal population variances is seriously violated.

21. (a) Source H2 SO4 CaCl2 Interaction Error Total

DF 2 2 4 9 17

SS 457.65 38783 279.78 232.85 39753

MS 228.83 19391 69.946 25.872

F 8.8447 749.53 2.7036

P 0.008 0.000 0.099

(b) The P -value for interactions is 0.099. One cannot rule out the additive model.

22.

Source Area Error Total

DF 4 16 20

SS 245.13 74.147 319.28

MS 61.283 4.6342

F 13.224

P 0.000

Yes, F4,16 = 13.224, P ≈ 0.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 9

23.

Source Angle Error Total

DF 6 39 45

SS 936.4 299.81 1236.2

F 20.302

MS 156.07 7.6874

P 0.000

Yes, F6,39 = 20.302, P ≈ 0.

24. (a) Source Distance Time Interaction Error Total

DF 5 2 10 36 53

SS 3.2226 153.33 3.5999 6.6929 166.84

MS 0.64452 76.665 0.35999 0.18591

F 3.4668 412.37 1.9363

P 0.012 0.000 0.072

(b) F10,36 = 1.9363, 0.05 < P < 0.10 (P = 0.072). The additive model is barely plausible.

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Chapter 10 Section 10.1 1. (a) Count (b) Continuous (c) Binary (d) Continuous

2. (a) True (b) False. If no special causes are operating, then the variability is constant. It is still possible for much of the output to fail to conform to specifcations, for example through incorrect calibration, or because of a large variation in common causes. (c) True. This is part of the defnition of a common cause. (d) False. Rational subgroups should be chosen in such way that the variation within them is due to common causes. (e) False. There will be variation from common causes.

3. (a) is in control (b) has high capability

(i) It must still be monitored continually.

5. (a) False. Being in a state of statistical control means only that no special causes are operating. It is still possible for the process to be calibrated incorrectly, or for the variation due to common causes to be so great that much of the output fails to conform to specifcations.

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SECTION 10.2

(b) False. Being out of control means that some special causes are operating. It is still possible for much of the output to meet specifcations. (c) True. This is the defnition of statistical control. (d) True. This is the defnition of statistical control.

(ii) it is more important to sample frequently than to choose large samples, so that special causes can be detected more quickly.

Section 10.2 1. (a) The sample size is n = 4. The upper and lower limits for the R-chart are D3 R and D4 R, respectively. From the control chart table, D3 = 0 and D4 = 2.282. R = 143.7∕30 = 4.79. Therefore LCL = 0, and UCL = 10.931. (b) The sample size is n = 4. The upper and lower limits for the S-chart are B3 s and B4 s, respectively. From the control chart table, B3 = 0 and B4 = 2.266. s = 62.5∕30 = 2.08333. Therefore LCL = 0 and UCL = 4.721. (c) The upper and lower limits for the X-chart are X − A2 R and X + A2 R, respectively. From the control chart table, A2 = 0.729. R = 143.7∕30 = 4.79 and X = 712.5∕30 = 23.75. Therefore LCL = 20.258 and UCL = 27.242. (d) The upper and lower limits for the X-chart are X − A3 s and X + A3 s, respectively. From the control chart table, A3 = 1.628. s = 62.5∕30 = 2.08333 and X = 712.5∕30 = 23.75. Therefore LCL = 20.358 and UCL = 27.142.

The process is never out of control.

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3. (a) The sample size is n = 5. The upper and lower limits for the R-chart are D3 R and D4 R, respectively. From the control chart table, D3 = 0 and D4 = 2.114. R = 0.1395. Therefore LCL = 0 and UCL = 0.2949. The variance is in control. (b) The upper and lower limits for the X-chart are X − A2 R and X + A2 R, respectively. From the control chart table, A2 = 0.577. R = 0.1395 and X = 2.505. Therefore LCL = 2.4245 and UCL = 2.5855. The process is out of control for the frst time on sample 8. (c) The 1 limits are X − A2 R∕3 = 2.478 and X + A2 R∕3 = 2.5318, respectively. The 2 limits are X − 2A2 R∕3 = 2.4513 and X + 2A2 R∕3 = 2.5587, respectively. The process is out of control for the frst time on sample 7, where 2 out of the last three samples are below the lower 2 control limit.

4. (a) The sample size is n = 5. The upper and lower limits for the S-chart are B3 s and B4 s, respectively. From the control chart table, B3 = 0 and B4 = 2.089. s = 0.057. Therefore LCL = 0 and UCL = 0.119. The variance is in control. (b) The upper and lower limits for the X-chart are X − A3 s and X + A3 s, respectively. From the control chart table, A3 = 1.427. s = 0.057 and X = 2.505. Therefore LCL = 2.4237 and UCL = 2.5863. The process is out of control for the frst time on sample 8. (c) The 1 limits are X − A3 s∕3 = 2.4779 and X + A3 s∕3 = 2.5321, respectively. The 2 limits are X − 2A3 s∕3 = 2.4508 and X + 2A3 s∕3 = 2.5592, respectively. The process is out of control for the frst time on sample 7, where 2 out of the last three samples are below the lower 2 control limit.

5. (a) X has a normal distribution with = 14 and X = 3∕

.341641.

The 3 limits are 12 ± 3(1.341641), or 7.97508 and 16.02492. The probability that a point plots outside the 3 limits is p = P (X < 7.97508) + P (X > 16.02492). The z-score for 7.97508 is (7.97508 − 14)∕1.341641 = −4.49. The z-score for 16.02492 is (16.02492 − 14)∕1.341641 = 1.51. The probability that a point plots outside the 3 limits is the sum of the area to the left of z = −4.49 and the area to the right of z = 1.51. Therefore p = 0.0000 + 0.0655 = 0.0655. The ARL is 1∕p = 1∕0.0655 = 15.27.

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SECTION 10.2

(b) Let m be the required value. Since the shift is upward, m > 12. The probability that a point plots outside the 3 limits is p = P (X < 7.97508) + P (X > 16.02492). Since ARL = 4, p = 1∕4. Since m > 12, P (X > 16.02492) > P (X < 7.97508). Find m so that P (X > 16.02492) = 1∕4, and check that P (X < 7.97508)

The z-score for 16.02492 is (16.02492 − m)∕1.341641. The z-score with an area of 1/4 = 0.25 to the right is approximately z = 0.67. Therefore 0.67 = (16.02492 − m)∕1.341641, so m = 15.126. Now check that P (X < 7.97508)

The z-score for 7.97508 is (7.97508 − 15.126)∕1.341641 = −5.33. So P (X < 7.97508)

Therefore m = 15.126. (c) We will fnd the required value for X . The probability that a point plots outside the 3 limits is p = P (X < 12 − 3 X ) + P (X > 12 + 3 X ). Since ARL = 4, p = 1∕4. Since the process mean is 14, P (X > 12 + 3 X ) > P (X > 12 − 3 X ). Find so that P (X > 12 + 3 X ) = 1∕4, and check that P (X < 12 − 3 X

The z-score for 12 + 3 X is (12 + 3 X − 14)∕ X . The z-score with an area of 1/4 = 0.25 to the right is approximately z = 0.67. Therefore (12 + 3 X − 14)∕ X = 0.67, so X = 0.8584. Now check that P (X < 12 − 3 X

12 − 3 X = 9.425. The z-score for 9.425 is (9.425 − 14)∕0.8584 = −5.33, so P (X < 12 − 3 X

Therefore X = 0. Since n = 5, X =

Therefore = 1.92.

(d) Let n be the required sample size. Then =

. = 0.8584, so n = 12.214.

From part (c), X = 0.8584. Therefore Round up to obtain n = 13.

6. (a) X has a normal distribution with = 9 and X = 2∕

The 3 limits are 8 ± 3(1), or 5 and 11. The probability that a point plots outside the 3 limits is p = P (X < 5) + P (X > 11). The z-score for 5 is (5 − 9)∕1 = −4.00. The z-score for 11 is (11 − 9)∕1 = 2.00. The probability that a point plots outside the 3 limits is the sum of the area to the left of z = −4.00 and the area to the right of z = 2.00.

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Therefore p = 0.0000 + 0.0228 = 0.0228. The ARL is 1∕p = 1∕0.0228 = 43.86. (b) Let m be the required value. Since the shift is upward, m > 8. The probability that a point plots outside the 3 limits is p = P (X < 5) + P (X > 11). Since ARL = 8, p = 1∕8. Since m > 8, P (X > 11) > P (X < 5). Find m so that P (X > 11) = 1∕8, and check that P (X < 5)

The z-score for 11 is (11−m)∕1. The z-score with an area of 1/8 = 0.1250 to the right is approximately z = 1.15. Therefore 1.15 = (11 − m)∕1, so m = 9.85. Now check that P (X < 5)

The z-score for 5 is (5 − 9.85)∕1 = −4.85. So P (X < 5)

Therefore m = 9.85. (c) We will fnd the required value for X . The probability that a point plots outside the 3 limits is p = P (X < 8 − 3 X ) + P (X > 8 + 3 X ). Since ARL = 8, p = 1∕8. Since the process mean is 9, P (X > 8 + 3 X ) > P (X < 8 − 3 X ). Find so that P (X > 8 + 3 X ) = 1∕8, and check that P (X < 8 − 3 X

The z-score for 8 + 3 X is (8 + 3 X − 9)∕ X . The z-score with an area of 1/8 = 0.1250 to the right is approximately z = 1.15. Therefore (8 + 3 X − 9)∕ X = 1.15, so X = 0.5405. Now check that P (X < 8 − 3 X

8 − 3 X = 6.38. The z-score for 6.38 is (6.38 − 9)∕0.5405 = −4.85, so P (X < 8 − 3 X

Since n = 4, X = ∕ 4. Therefore = 1.081.

(d) Let n be the required sample size. Then X = 2∕ n. From part (c), X = 0.5405. Therefore 2∕ n = 0.5405, so n = 13.69. Round up to obtain n = 14.

The probability of a false alarm on any given sample is 0.0027, and the probability that there will not be a false alarm on any given sample is 0.9973.

(a) The probability that there will be no false alarm in the next 50 samples is 0.997350 = 0.874. Therefore the probability that there will be a false alarm within the next 50 samples is 1 − 0.874 = 0.126.

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SECTION 10.2

(b) The probability that there will be no false alarm in the next 100 samples is 0.9973100 = 0.763. Therefore the probability that there will be a false alarm within the next 50 samples is 1 − 0.763 = 0.237. (c) The probability that there will be no false alarm in the next 200 samples is 0.9973200 = 0.582. (d) Let n be the required number. Then 0.9973n = 0.5, so n ln 0.9973 = ln 0.5. Solving for n yields n = 256.37

257.

8. (a) The sample size is n = 8. The upper and lower limits for the R-chart are D3 R and D4 R, respectively. From the control chart table, D3 = 0.136 and D4 = 1.864. R = 0.2628. Therefore LCL = 0.0357 and UCL = 0.4899. The variance is in control. (b) The upper and lower limits for the X-chart are X − A2 R and X + A2 R, respectively. From the control chart table, A2 = 0.373. R = 0.2628 and X = 9.9892. Therefore LCL = 9.891, UCL = 10.087. The process is out of control for the frst time on sample 3. (c) The 1 limits are X − A2 R∕3 = 9.9565 and X + A2 R∕3 = 10.0219, respectively. The 2 limits are X − 2A2 R∕3 = 9.9239 and X + 2A2 R∕3 = 10.0545, respectively. The process is out of control for the frst time on sample 3, where one sample exceeds the upper 3 control limit.

9. (a) The sample size is n = 8. The upper and lower limits for the S-chart are B3 s and B4 s, respectively. From the control chart table, B3 = 0.185 and B4 = 1.815. s = 0.0880. Therefore LCL = 0.01628 and UCL = 0.1597. The variance is in control. (b) The upper and lower limits for the X-chart are X − A3 s and X + A3 s, respectively. From the control chart table, A3 = 1.099. s = 0.0880 and X = 9.9892. Therefore LCL = 9.8925 and UCL = 10.0859. The process is out of control for the frst time on sample 3. (c) The 1 limits are X − A3 s∕3 = 9.9570 and X + A3 s∕3 = 10.0214, respectively. The 2 limits are X − 2A3 s∕3 = 9.9247 and X + 2A3 s∕3 = 10.0537, respectively. The process is out of control for the frst time on sample 3, where one sample exceeds the upper 3 control limit.

10. (a) The sample size is n = 5. The upper and lower limits for the R-chart are D3 R and D4 R, respectively. From the control chart table, D3 = 0 and D4 = 2.114. R = 1.14. Therefore LCL = 0 and UCL = 2.410. The variance is in control.

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(b) The upper and lower limits for the X -chart are X − A2 R and X + A2 R, respectively. From the control chart table, A2 = 0.577. R = 1.14 and X = 9.81. Therefore LCL = 9.152 and UCL = 10.468. The process is in control. (c) The 1 limits are X − A2 R∕3 = 9.591 and X + A2 R∕3 = 10.029, respectively. The 2 limits are X − 2A2 R∕3 = 9.371 and X + 2A2 R∕3 = 10.249, respectively. The process is out of control for the frst time on sample 9, where 2 of the last three sample means are below the lower 2 control limit.

11. (a) The sample size is n = 5. The upper and lower limits for the S-chart are B3 s and B4 s, respectively. From the control chart table, B3 = 0 and B4 = 2.089. s = 0.4647. Therefore LCL = 0 and UCL = 0.971. The variance is in control. (b) The upper and lower limits for the X-chart are X − A3 s and X + A3 s, respectively. From the control chart table, A3 = 1.427. s = 0.4647 and X = 9.81. Therefore LCL = 9.147 and UCL = 10.473. The process is in control. (c) The 1 limits are X − A3 s∕3 = 9.589 and X + A3 s∕3 = 10.031, respectively. The 2 limits are X − 2A3 s∕3 = 9.368 and X + 2A3 s∕3 = 10.252, respectively. The process is out of control for the frst time on sample 9, where 2 of the last three sample means are below the lower 2 control limit.

12. (a) The sample size is n = 4. The upper and lower limits for the R-chart are D3 R and D4 R, respectively. From the control chart table, D3 = 0 and D4 = 2.282. R = 6.97. Therefore LCL = 0 and UCL = 15.906. The variance is out of control on sample 8. After deleting this sample, X = 150.166 and R = 6.538. The new limits for the R-chart are 0 and 14.920. The variance is now in control. (b) The upper and lower limits for the X-chart are X − A2 R and X + A2 R, respectively. From the control chart table, A2 = 0.729. R = 6.538 and X = 150.166. Therefore LCL = 145.400 and UCL = 154.932. The process is in control. (c) The 1 limits are X − A2 R∕3 = 148.577 and X + A2 R∕3 = 151.755, respectively. The 2 limits are X − 2A2 R∕3 = 146.989 and X + 2A2 R∕3 = 153.343, respectively. The process is in control (recall that sample 8 has been deleted).

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SECTION 10.3

13. (a) The sample size is n = 4. The upper and lower limits for the S-chart are B3 s and B4 s, respectively. From the control chart table, B3 = 0 and B4 = 2.266. s = 3.082. Therefore LCL = 0 and UCL = 6.984. The variance is out of control on sample 8. After deleting this sample, X = 150.166 and s = 2.911. The new limits for the S-chart are 0 and 6.596. The variance is now in control. (b) The upper and lower limits for the X-chart are X − A3 s and X + A3 s, respectively. From the control chart table, A3 = 1.628. s = 2.911 and X = 150.166. Therefore LCL = 145.427 and UCL = 154.905. The process is in control. (c) The 1 limits are X − A3 s∕3 = 148.586 and X + A3 s∕3 = 151.746, respectively. The 2 limits are X − 2A3 s∕3 = 147.007 and X + 2A3 s∕3 = 153.325, respectively. The process is in control (recall that sample 8 has been deleted).

Section 10.3 1.

The sample size is n = 200. p = 1.64∕30 = 0.054667. The centerline is

= 0.054667.

The LCL is p

(1 − p)∕200 = 0.00644.

The UCL is p

p(1 − p)∕200 = 0.1029.

2. (a) The sample size is n = 300. The mean number of underweight boxes over the last 25 days is 26.84. Therefore p = 26.84∕300

0894667.

The control limits are p ± 3 p(1 − p)∕n. Therefore LCL = 0.0400 and UCL = 0.1389. (b) The process is out of control for the frst time on sample #22.

Yes, the only information needed to compute the control limits is p and the sample size n. In this case, n = 100, and p = (622∕50)∕100 = 0.1244.

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The control limits are p ± 3

(1 − p)∕n, so

LCL = 0.0254 and UCL = 0.2234.

Yes, the process is out of control on the sample with 24 defectives, since the sample proportion of 0.24 exceeds the upper control limit of 0.2234.

(iv). The sample size must be large enough so the mean number of defectives per sample is at least 10.

c = 876∕70 = 12.51. The centerline is The LCL is c The UCL is c

√ √

12.51. = 1.90. = 23.13.

It was out of control. The UCL is 23.13.

8. (a) The average number of defectives per sample is c = 45.3. The control limits are c ± 3 c Therefore LCL = 25.11 and UCL = 65.49. (b) The process is in control.

Section 10.4 1. (a) No samples need be deleted. (b) The estimate of X is A2 R∕3. The sample size is n = 5. R = 0.1395. From the control chart table, A2 = 0.577. Therefore X = (0.577)(0.1395)∕3 = 0.0268.

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SECTION 10.4

CUSUM Chart 0.4

(c)

Cumulative Sum

0.3 0.2 UCL = 0.107

0.1 0 −0.1

LCL = −0.107

−0.2 −0.3 0

10 15 Sample Number

(d) The process is out of control on sample 8. (e) The Western Electric rules specify that the process is out of control on sample 7.

2. (a) No samples need be deleted.

(b) The estimate of X is A2 R∕3. The sample size is n = 8. R = 0.2628. From the control chart table, A2 = 0.373. Therefore X = (0.373)(0.2628)∕3 = 0.0327.

CUSUM Chart 0.15

UCL = 0.131

0.1 0.05

(c)

0 −0.05 −0.1 LCL = −0.131

−0.15 −0.2 −0.25 0

10 15 Sample Number

(d) The process is out of control on sample 10. (e) The Western Electric rules specify that the process is out of control on sample 3.

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CHAPTER 10

3. (a) No samples need be deleted.

(b) The estimate of X is A2 R∕3. The sample size is n = 5. R = 1.14. From the control chart table, A2 = 0.577. Therefore X = (0.577)(1.14)∕3 = 0.219.

CUSUM Chart 3

(c)

Cumulative Sum

2 1

UCL = 0.877

0 LCL = −0.877

−1 −2 −3 0

15 20 Sample Number

(d) The process is out of control on sample 9.

(e) The Western Electric rules specify that the process is out of control on sample 9.

4. (a) LCL = 0, UCL = 15.906. The variance is out of control on sample 8. After deleting this sample, X = 150.166 and R = 6.538. The new limits for the R-chart are 0 and 14.920. The variance is now in control.

(b) The estimate of X is A2 R∕3. The sample size is n = 4. R = 6.538. From the control chart table, A2 = 0.729. Therefore X = (0.729)(6.538)∕3 = 1.589.

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519

SECTION 10.4

CUSUM Chart 8 UCL = 6.355

(c)

Cumulative Sum

4 2 0 −2 −4 −6

LCL = −6.355

−8 0

15 20 Sample Number

(d) The process is in control.

(e) The Western Electric rules specify that the process is in control.

CUSUM Chart 80 UCL =60

5. (a)

Cumulative Sum

40 20 0 −20 −40 LCL =−60

−60 −80 0

20 Sample Number

(b) The process is in control.

The value of SL can never be positive, and the value of SH can never be negative. Therefore the values of 0.2371 and 0.7104 for SL, and the value of −0.7345 for SH, are incorrect.

Section 10.5

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520

CHAPTER 10

1. (a) = X = 0.248, s = 0.002, LSL = 0.246, U SL = 0.254. The sample size is n = 6. = s∕c4 . From the control chart table, c4 = 0.9515. Therefore = 0.002102. Since is closer to LSL than to U SL, Cpk = ( − LSL)∕(3 ) = 0.3172. (b) No. Since Cpk < 1, the process capability is not acceptable.

2. (a) = X = 15.52, R = 0.05, LSL = 15.40, U SL = 15.60. The sample size is n = 8. = R∕d2 . From the control chart table, d2 = 2.847. Therefore = 0.017562. Since is closer to U SL than to LSL, Cpk = (U SL − )∕(3 ) = 1.5184. (b) Yes. Since Cpk > 1, the process capability is acceptable.

3. (a) The capability is maximized when the process mean is equidistant from the specifcation limits. Therefore the process mean should be set to 15.50.

(b) LSL = 15.40, U SL = 15.60, = 0.017562. If = 15.50, then Cpk = (15.60 − 15.50)∕[3(0.017562)] = 1.898.

4. (a) The capability is maximized when the process mean is equidistant from the specifcation limits. Therefore the process mean should be set to 0.250.

(b) No. LSL = 0.246, U SL = 0.254, = 0.002109. If = 0.250, then Cpk = (0.254 − 0.250)∕[3(0.002109)] = 0.63. Since Cpk < 1, the process capability is not acceptable. (c) Let be the required value. Then Cpk = (0.254 − 0.250)∕(3 ) = 1, so = 0.00133. (d) Let be the required value. Then Cpk = (0.254 − 0.250)∕(3 ) = 2, so = 0.000667.

5. (a) Let be the optimal setting for the process mean.

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SUPPLEMENTARY EXERCISES FOR CHAPTER 10

521

Then Cp = (U SL − )∕(3 ) = ( − LSL)∕(3 ), so 1.2 = (U SL − )∕(3 ) = ( − LSL)∕(3 ). Solving for LSL and U SL yields LSL = − 3.6 and U SL = + 3.6 .

(b) The z-scores for the upper and lower specifcation limits are z = ±3.60. Therefore, using the normal curve, the proportion of units that are non-conforming is the sum of the areas under the normal curve to the right of z = 3.60 and to the left of z = −3.60. The proportion is 0.0002 + 0.0002 = 0.0004.

Supplementary Exercises for Chapter 10 1.

The sample size is n = 300. p = 5.83∕100 = 0.0583. The centerline is

= 0.0583

The LCL is p − 3

− p)∕300 = 0.0177.

The UCL is p + 3

(1 − p)∕300 = 0.0989.

2. (a) The probability, p, that a point plots outside the limits ±2.0 X is the sum of the areas under the normal curve to the right of z = 2.00 and to the left of z = −2.00. Therefore p = 0.0228 + 0.0228 = 0.0456. The ARL is 1∕p = 1∕0.0456 = 21.93.

(b) If = 0.5 X , then the z-scores for the upper and lower control limits are (2.0 X − 0.5 X )∕ X = 1.50 and (−2.0 X − 0.5 X )∕ X = −2.50 respectively. The probability, p, that a point plots outside the limits ±2.0 X is the sum of the areas under the normal curve to the right of z = 1.50 and to the left of z = −2.50. Therefore p = 0.0668 + 0.0062 = 0.0730. The ARL is 1∕p = 1∕0.730 = 13.699.

(c) Let m be the process mean after the shift. The probability that a point plots outside the ±2.0 X limits is p = P (X < −2.0 X ) + P (X > 2.0 X ). Since this is an upward shift, P (X > 2.0 X ) > P (X < −2.0 X ).

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522

CHAPTER 10

Since ARL = 10, p = 0.1. Find m so that P (X > 2.0 X ) = 0.1, then check that P (X < −2.0 X

The z-score for 2.0 X is (2.0 X −m)∕ X . The z-score with an area of 0.1 to the right is approximately z = 1.28. Therefore 1.28 = (2.0 X − m)∕ X , so m = 0.72 X . Now check that P (X < −2.0 X

The z-score for −2.0 X is (−2.0 X − m)∕ X = −2.72. So P (X < −2.0 X

Therefore m = 0.72 X .

3. (a) The sample size is n = 3. The upper and lower limits for the R-chart are D3 R and D4 R, respectively. From the control chart table, D3 = 0 and D4 = 2.575. R = 0.110. Therefore LCL = 0 and UCL = 0.283. The variance is in control.

(b) The upper and lower limits for the X-chart are X − A2 R and X + A2 R, respectively. From the control chart table, A2 = 1.023. R = 0.110 and X = 5.095. Therefore LCL = 4.982 and UCL = 5.208. The process is out of control on sample 3.

(c) The 1 limits are X − A2 R∕3 = 5.057 and X + A2 R∕3 = 5.133, respectively. The 2 limits are X − 2A2 R∕3 = 5.020 and X + 2A2 R∕3 = 5.170, respectively. The process is out of control for the frst time on sample 3, where a sample mean is above the upper 3 control limit.

4. (a) The sample size is n = 3. The upper and lower limits for the S-chart are B3 s and B4 s, respectively. From the control chart table, B3 = 0 and B4 = 2.568. s = 0.058. Therefore LCL = 0, UCL = 0.149. The variance is in control.

(b) The upper and lower limits for the X-chart are X − A3 s and X + A3 s, respectively. From the control chart table, A3 = 1.954. s = 0.058 and X = 5.095. Therefore LCL = 4.982 and UCL = 5.208. The process is out of control on sample 3.

(c) The 1 limits are X − A3 s∕3 = 5.057 and X + A3 s∕3 = 5.133, respectively. The 2 limits are X − 2A3 s∕3 = 5.019 and X + 2A3 s∕3 = 5.171, respectively. The process is out of control for the frst time on sample 3, where a sample mean is above the upper 3 control limit.

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523

SUPPLEMENTARY EXERCISES FOR CHAPTER 10

5. (a) No samples need be deleted.

(b) The estimate of X is A2 R∕3. The sample size is n = 3. R = 0.110. From the control chart table, A2 = 1.023. Therefore X = (1.023)(0.110)∕3 = 0.0375.

CUSUM Chart 1

(c)

Cumulative Sum

0.5 UCL = 0.15 0 LCL = −0.15 −0.5

−1 0

15 20 Sample Number

(d) The process is out of control on sample 4.

(e) The Western Electric rules specify that the process is out of control on sample 3. The CUSUM chart frst signaled an out-of-control condition on sample 4.

6. (a) The average number of defectives per sample is c = 1085∕50 = 21.7. The centerline is c = 21.7. The control limits are c Therefore LCL = 7.725 and UCL = 35.675.

(b) Yes, the process was out of control, because the number of faws was below the lower control limit. However, the number of faws was unusually low, so when the special cause is determined, steps should be taken to preserve it rather than to eliminate it.

7. (a) The sample size is n = 300. The mean number of defectives over the last 20 days is 12.0. Therefore p = 12.0∕300 = 0.04. The control limits are p ± 3 p(1 − p)∕n.

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524

CHAPTER 10

Therefore LCL = 0.00606 and UCL = 0.0739.

(b) Sample 7. The proportion of defective chips is then 1∕300 = 0.0033, which is less than the lower control limit.

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525

APPENDIX B

Appendix B

𝜕𝑣 = 3 + 2𝑦4 , 𝜕𝑥

𝜕𝑣 = 8𝑥𝑦3 𝜕𝑦

2𝑥(𝑥3 + 𝑦3 ) 3𝑥2 𝜕𝑤 = − , 𝜕𝑥 𝑥2 + 𝑦2 (𝑥2 + 𝑦2 )2

𝜕𝑧 = − sin 𝑥 sin 𝑦2 , 𝜕𝑥

𝜕𝑣 = 𝑦𝑒𝑥𝑦 , 𝜕𝑥

𝜕𝑣 = 𝑒𝑥 (cos 𝑦 + sin 𝑧), 𝜕𝑥

𝜕𝑤 𝑥 =√ , 𝜕𝑥 2 𝑥 + 4𝑦2 + 3𝑧2

𝜕𝑧 2𝑥 = , 𝜕𝑥 𝑥2 + 𝑦2

2𝑥𝑦 2 𝜕𝑣 = − 𝑧𝑒𝑦 sin(𝑥𝑧), 2 𝜕𝑥 𝑥 𝑦 + 𝑧

𝜕𝑣 = 𝜕𝑥

10.

𝑥𝑦 cos(𝑥2 𝑦) 𝜕𝑧 , = √ 𝜕𝑥 sin(𝑥2 𝑦)

−

𝜕𝑣 = −𝑒𝑥 sin 𝑦, 𝜕𝑦

𝜕𝑣 = 𝑒𝑥 cos 𝑧 𝜕𝑧

4𝑦 𝜕𝑤 =√ , 𝜕𝑦 2 𝑥 + 4𝑦2 + 3𝑧2

𝜕𝑤 3𝑧 =√ 𝜕𝑧 2 𝑥 + 4𝑦2 + 3𝑧2

2𝑦 𝜕𝑧 = 𝜕𝑦 𝑥2 + 𝑦2

√ 𝑥

𝜕𝑧 = 2𝑦 cos 𝑥 cos 𝑦2 𝜕𝑦

𝜕𝑣 = 𝑥𝑒𝑥𝑦 𝜕𝑦

√ 𝑦5

3𝑦2 2𝑦(𝑥3 + 𝑦3 ) 𝜕𝑤 = − 𝜕𝑦 𝑥2 + 𝑦2 (𝑥2 + 𝑦2 )2

3 2

𝑦3 , 𝑥

2 𝜕𝑣 𝑥2 = + 2𝑦𝑒𝑦 cos(𝑥𝑧), 2 𝜕𝑦 𝑥 𝑦 + 𝑧

2 𝜕𝑣 1 = − 𝑥𝑒𝑦 sin(𝑥𝑧) 2 𝜕𝑧 𝑥 𝑦 + 𝑧

√ 𝜕𝑣 9√ = 5 𝑥𝑦3 − 𝑥𝑦 𝜕𝑦 2 𝑥2 cos(𝑥2 𝑦) 𝜕𝑧 = √ 𝜕𝑦 2 sin(𝑥2 𝑦)

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