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Solution Manual For Statistics For Engineers And Scientists, 4th Edition

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    Solutions Manual to accompany STATISTICS FOR ENGINEERS AND SCIENTISTS, 4th ed. Prepared by William Navidi

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    Table of Contents Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467

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    SECTION 1.1 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. 2. (iii). It is very unlikely that students whose names happen to fall at the top of a page in the phone book will differ systematically in height from the population of students as a whole. It is somewhat more likely that engineering majors will differ, and very likely that students involved with basketball intramurals will differ. 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 approx- imation. 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. Page 1

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    2 CHAPTER 1 6. (a) False (b) True (c) True 7. A good knowledge of the process that generated the data. 8. (a) An observational study (b) It is not well-justified. Because the study is observational, there could be differences between the groups other than the level of exercise. These other differences (confounders) could cause the difference in blood pressure. 9. (a) A controlled experiment (b) It is well-justified, because it is based on a controlled experiment rather than an observational study. Section 1.2 1. False 2. No. In the sample 1, 2, 4 the mean is 7/3, which does not appear at all. 3. No. In the sample 1, 2, 4 the mean is 7/3, which does not appear at all. 4. No. The median of the sample 1, 2, 4, 5 is 3. 5. The sample size can be any odd number. Page 2

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    SECTION 1.2 3 6. Yes. For example, the list 1, 2, 12 has an average of 5 and a standard deviation of 6.08. 7. Yes. If all the numbers in the list are the same, the standard deviation will equal 0. 8. The mean increases by $50; the standard deviation is unchanged. 9. The mean and standard deviation both increase by 5%. 10. (a) Let X 1 , ..., X 100 denote the 100 numbers of children. 100 ∑ i = 1 X i = 27 ( 0 ) + 22 ( 1 ) + 30 ( 2 ) + 12 ( 3 ) + 7 ( 4 ) + 2 ( 5 ) = 156 X = ∑ 100 i = 1 X i 100 = 156 100 = 1 . 56 (b) The sample variance is s 2 = 1 99 ( 100 ∑ i = 1 X 2 i − 100 X 2 ) = 1 99 [( 27 ) 0 2 + ( 22 ) 1 2 + ( 30 ) 2 2 + ( 12 ) 3 2 + ( 7 ) 4 2 + ( 2 ) 5 2 − 100 ( 1 . 56 2 )] = 1 . 7034 The standard deviation is s = √ s 2 = 1 . 3052. Alternatively, the sample variance can be computed as s 2 = 1 99 100 ∑ i = 1 ( X i − X ) 2 = 1 99 [ 27 ( 0 − 1 . 56 ) 2 + 22 ( 1 − 1 . 56 ) 2 + 30 ( 2 − 1 . 56 ) 2 + 12 ( 3 − 1 . 56 ) 2 + 7 ( 4 − 1 . 56 ) 2 + 2 ( 5 − 1 . 56 ) 2 ] = 1 . 7034 (c) The sample median is the average of the 50th and 51st value when arranged in order. Both these values are equal to 2, so the median is 2. Page 3

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    4 CHAPTER 1 (d) The first quartile is the average of the 25th and 26th value when arranged in order. Both these values are equal to 0, so the first quartile is 0. (e) Of the 100 women, 30 + 12 + 7 + 2 = 51 had more than the mean of 1.56 children, so the proportion is 51/100 = 0.51. (f) The quantity that is one standard deviation greater than the mean is 1.56 + 1.3052 = 2.8652. Of the 100 women, 12 + 7 + 2 = 21 had more than 2.8652 children, so the proportion is 21/100 = 0.21. (g) The region within one standard deviation of the mean is 1 . 56 ± 1 . 3052 = ( 0 . 2548 , 2 . 8652 ) . Of the 100 women, 22 + 30 = 52 are in this range, so the proportion is 52/100 = 0.52. 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. (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 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. Page 4

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    SECTION 1.2 5 (c) 0 . 20 ( 16 ) = 3 . 2 ≈ 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 first 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: Q 1 = 21 . 0, Q 3 = 25 . 0; B: Q 1 = 19 . 85, Q 3 = 22 . 0; C: Q 1 = 20 . 75, Q 3 = 21 . 5; D: Q 1 = 20 . 1, Q 3 = 21 . 3 (e) The variance for A is s 2 = 1 15 [ 18 . 0 2 + 18 . 0 2 + 18 . 0 2 + 20 . 0 2 + 22 . 0 2 + 22 . 0 2 + 22 . 5 2 + 23 . 0 2 + 24 . 0 2 + 24 . 0 2 + 25 . 0 2 + 25 . 0 2 + 25 . 0 2 + 25 . 0 2 + 26 . 0 2 + 26 . 4 2 − 16 ( 22 . 744 2 )] = 8 . 2506 The standard deviation for A is s = √ 8 . 2506 = 2 . 8724. The variance for B is s 2 = 1 15 [ 18 . 8 2 + 18 . 9 2 + 18 . 9 2 + 19 . 6 2 + 20 . 1 2 + 20 . 4 2 + 20 . 4 2 + 20 . 4 2 + 20 . 4 2 + 20 . 5 2 + 21 . 2 2 + 22 . 0 2 + 22 . 0 2 + 22 . 0 2 + 22 . 0 2 + 23 . 6 2 − 16 ( 20 . 700 2 )] = 1 . 8320 The standard deviation for B is s = √ 1 . 8320 = 1 . 3535. The variance for C is s 2 = 1 15 [ 20 . 2 2 + 20 . 5 2 + 20 . 5 2 + 20 . 7 2 + 20 . 8 2 + 20 . 9 2 + 21 . 0 2 + 21 . 0 2 + 21 . 0 2 + 21 . 0 2 + 21 . 0 2 + 21 . 5 2 + 21 . 5 2 + 21 . 5 2 + 21 . 5 2 + 21 . 6 2 − 16 ( 20 . 013 2 )] = 0 . 17583 The standard deviation for C is s = √ 0 . 17583 = 0 . 4193. The variance for D is s 2 = 1 15 [ 20 . 0 2 + 20 . 0 2 + 20 . 0 2 + 20 . 0 2 + 20 . 2 2 + 20 . 5 2 + 20 . 5 2 + 20 . 7 2 + 20 . 7 2 + 20 . 7 2 + 21 . 0 2 + 21 . 1 2 + 21 . 5 2 + 21 . 6 2 + 22 . 1 2 + 22 . 3 2 − 16 ( 20 . 806 2 )] = 0 . 55529 The standard deviation for D is s = √ 0 . 55529 = 0 . 7542. Page 5

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    6 CHAPTER 1 (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 different each time a measurement is made. When the standard deviation is small, a single measure- ment is more valuable, since we know that subsequent measurements would probably not be much different. 13. (a) All would be divided by 2.54. (b) Not exactly the same, because the measurements would be a little different the second time. 14. (a) We will work in units of $1000. Let S 0 be the sum of the original 10 numbers and let S 1 be the sum after the change. Then S 0 / 10 = 70, so S 0 = 700. Now S 1 = S 0 − 100 + 1000 = 1600, so the new mean is S 1 / 10 = 160. (b) The median is unchanged at 55. (c) Let X 1 , ..., X 10 be the original 10 numbers. Let T 0 = ∑ 10 i = 1 X 2 i . Then the variance is ( 1 / 9 )[ T 0 − 10 ( 70 2 )] = 20 2 = 400, so T 0 = 52 , 600. Let T 1 be the sum of the squares after the change. Then T 1 = T 0 − 100 2 + 1000 2 = 1 , 042 , 600. The new standard deviation is √ ( 1 / 9 )[ T 1 − 10 ( 160 2 )] = 295 . 63. 15. (a) The sample size is n = 16. The tertiles have cutpoints ( 1 / 3 )( 17 ) = 5 . 67 and ( 2 / 3 )( 17 ) = 11 . 33. The first 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. Page 6

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    SECTION 1.3 7 16. (a) Seems certain to be an error. (b) Could be correct. Section 1.3 1. (a) Stem Leaf 0 011112235677 1 235579 2 468 3 11257 4 14699 5 5 6 16 7 9 8 0099 9 10 11 0 12 7 13 7 (b) Here is one histogram. Other choices for the endpoints are possi- ble. 0 2 4 6 8 10 12 14 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 Relative Frequency Rainfall (inches) (c) 0 2 4 6 8 10 12 14 Rainfall (inches) Page 7

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    8 CHAPTER 1 (d) 0 5 10 15 Rainfall (inches) The boxplot shows one outlier. 2. (a) Stem Leaf 14 12333444467788889 15 0023344567799 16 124456 17 2222338 18 19 20 21 9 22 4 (b) Here is one histogram. Other choices for the endpoints are possible. 14 15 16 17 18 19 20 21 22 23 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Relative Frequency Sulfur Trioxide (percent) Page 8

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    SECTION 1.3 9 (c) 14 16 18 20 22 24 (d) 14 16 18 20 22 24 The boxplot shows 2 outliers. 3. 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 Page 9

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    10 CHAPTER 1 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. 50 150 250 350 450 550 0 0.1 0.2 0.3 0.4 Relative Frequency Concentration (mg/kg) 50 150250 350 450550650750850950 0 0.1 0.2 0.3 0.4 0.5 0.6 Relative Frequency (b) 0 200 400 600 800 1000 Concentration (mg/kg) Chromium Nickel (c) The concentrations of nickel are on the whole lower than the concentrations of chromium. The nickel con- centrations are highly skewed to the right, which can be seen from the median being much closer to the first quartile than the third. The chromium concentrations are somewhat less skewed. Finally, the nickel concentra- tions include an outlier. Page 10

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    SECTION 1.3 11 5. (a) Here are histograms for each group. Other choices for the endpoints are possible. 2 3 4 5 6 7 0 0.1 0.2 0.3 0.4 0.5 Relative Frequency Yield Catalyst A 1 2 3 4 5 6 7 0 0.1 0.2 0.3 0.4 0.5 Relative Frequency Yield (b) 1 2 3 4 5 6 7 Catalyst A Catalyst B Yield (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 first 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. 6. (a) The histogram should be skewed to the right. Here is an example. Page 11

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    12 CHAPTER 1 (b) The histogram should be skewed to the left. Here is an example. (c) The histogram should be approximately symmetric. Here is an example. 7. (a) The proportion is the sum of the relative frequencies (heights) of the rectangles above 240. This sum is approx- imately 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. 8. 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. Page 12

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    SECTION 1.3 13 9. (a) 1 3 5 7 9 11 13 15 17 19 21 23 25 0 3 6 9 12 15 18 Frequency Emissions (g/gal) (b) 1 3 5 7 9 11 13 15 17 19 21 23 25 0 0.05 0.1 0.15 Density Emissions (g/gal) (c) Yes, the shapes of the histograms are the same. 10. (a) 1 3 5 7 9 11 15 25 0 0.05 0.1 0.15 0.2 0.25 0.3 Relative Frequency Emissions (g/gal) (b) No (c) The class interval widths are unequal. (d) The classes 11– < 15 and 15– < 25 Page 13

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