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

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Solutions Manualto accompanySTATISTICS FOR ENGINEERS ANDSCIENTISTS, 4th ed.Prepared byWilliam Navidi

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

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SECTION 1.11Chapter 1Section 1.11. (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 willdiffer systematically in height from the population of students as a whole.It is somewhat more likely thatengineering majors will differ, and very likely that students involved with basketball intramurals will differ.3. (a) False(b) True4. (a) False(b) True5. (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 oldprocess.(b) No. The population proportion for the new process may be 0.12 or more, even though the sample proportionwas only 0.11.(c) Finding 2 defective circuits in the sample.Page 1

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2CHAPTER 16. (a) False(b) True(c) True7.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 otherthan 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 1.21.False2.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 1.236.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) LetX1, ...,X100denote the 100 numbers of children.100∑i=1Xi=27(0) +22(1) +30(2) +12(3) +7(4) +2(5) =156X=∑100i=1Xi100=156100=1.56(b) The sample variance iss2=199(100∑i=1X2i−100X2)=199[(27)02+ (22)12+ (30)22+ (12)32+ (7)42+ (2)52−100(1.562)]=1.7034The standard deviation iss=√s2=1.3052.Alternatively, the sample variance can be computed ass2=199100∑i=1(Xi−X)2=199[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 areequal to 2, so the median is 2.Page 3

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4CHAPTER 1(d) The first quartile is the average of the 25th and 26th value when arranged in order. Both these values are equalto 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 themean 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=22.744The 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=20.700The 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=20.013The 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 1.25(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=23.25The 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=20.70The 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=21.04The 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: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 iss2=115[18.02+18.02+18.02+20.02+22.02+22.02+22.52+23.02+24.02+24.02+25.02+25.02+25.02+25.02+26.02+26.42−16(22.7442)] =8.2506The standard deviation for A iss=√8.2506=2.8724.The variance for B iss2=115[18.82+18.92+18.92+19.62+20.12+20.42+20.42+20.42+20.42+20.52+21.22+22.02+22.02+22.02+22.02+23.62−16(20.7002)] =1.8320The standard deviation for B iss=√1.8320=1.3535.The variance for C iss2=115[20.22+20.52+20.52+20.72+20.82+20.92+21.02+21.02+21.02+21.02+21.02+21.52+21.52+21.52+21.52+21.62−16(20.0132)] =0.17583The standard deviation for C iss=√0.17583=0.4193.The variance for D iss2=115[20.02+20.02+20.02+20.02+20.22+20.52+20.52+20.72+20.72+20.72+21.02+21.12+21.52+21.62+22.12+22.32−16(20.8062)] =0.55529The standard deviation for D iss=√0.55529=0.7542.Page 5

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6CHAPTER 1(f) Method A has the largest standard deviation.This could be expected, because of the four methods, this isthe 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 issomewhat 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. LetS0be the sum of the original 10 numbers and letS1be the sum after thechange. ThenS0/10=70, soS0=700. NowS1=S0−100+1000=1600, so the new mean isS1/10=160.(b) The median is unchanged at 55.(c) LetX1, ...,X10be the original 10 numbers. LetT0=∑10i=1X2i. Then the variance is(1/9)[T0−10(702)] =202=400, soT0=52,600.LetT1be the sum of the squares after the change.ThenT1=T0−1002+10002=1,042,600. The new standard deviation is√(1/9)[T1−10(1602)] =295.63.15. (a) The sample size isn=16. The tertiles have cutpoints(1/3)(17) =5.67 and(2/3)(17) =11.33. The first tertileis therefore the average of the sample values in positions 5 and 6, which is(44+46)/2=45. The second tertileis the average of the sample values in positions 11 and 12, which is(76+79)/2=77.5.(b) The sample size isn=16. The quintiles have cutpoints(i/5)(17)fori=1,2,3,4. The quintiles are thereforethe averages of the sample values in positions 3 and 4, in positions 6 and 7, in positions 10 and 11, and inpositions 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 1.3716. (a) Seems certain to be an error.(b) Could be correct.Section 1.31. (a)StemLeaf001111223567712355792468311257414699556167980099910110127137(b) Hereisonehistogram.Otherchoices for the endpoints are possi-ble.0246810121400.050.10.150.20.250.30.350.40.45Relative FrequencyRainfall (inches)(c)02468101214Rainfall (inches)Page 7

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8CHAPTER 1(d)051015Rainfall (inches)The boxplot shows one outlier.2. (a)StemLeaf141233344446778888915002334456779916124456172222338181920219224(b) Here is one histogram. Other choices for the endpoints are possible.1415161718192021222300.050.10.150.20.250.30.35Relative FrequencySulfur Trioxide (percent)Page 8

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SECTION 1.39(c)141618202224(d)141618202224The boxplot shows 2 outliers.3.StemLeaf115882000034683023458840346522356666896002334597113558856891225101111221306141516171186199202122233Page 9

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10CHAPTER 1There are 23 stems in this plot. An advantage of this plot over the one in Figure 1.6 is that the values are givento the tenths digit instead of to the ones digit. A disadvantage is that there are too many stems, and many ofthem are empty.4. (a) Here are histograms for each group. Other choices for the endpoints are possible.5015025035045055000.10.20.30.4Relative FrequencyConcentration (mg/kg)5015025035045055065075085095000.10.20.30.40.50.6Relative Frequency(b)02004006008001000Concentration (mg/kg)ChromiumNickel(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 firstquartile 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 1.3115. (a) Here are histograms for each group. Other choices for the endpoints are possible.23456700.10.20.30.40.5Relative FrequencyYieldCatalyst A123456700.10.20.30.40.5Relative FrequencyYield(b)1234567Catalyst ACatalyst BYield(c) The yields for catalyst B are considerably more spread out thanthose for catalyst A.The median yield for catalyst A is greaterthan the median for catalyst B. The median yield for B is closerto the first quartile than the third, but the lower whisker is longerthan the upper one, so the median is approximately equidistantfrom the extremes of the data. Thus the yields for catalyst B areapproximately symmetric. The largest yield for catalyst A is anoutlier; the remaining yields for catalyst A are approximatelysymmetric.6. (a) The histogram should be skewed to the right. Here is an example.Page 11

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12CHAPTER 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 rectanglesover 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 relativefrequencies is 0.85. Since the sum of all the relative frequencies must be 1, the missing rectangle has a heightof 0.15.Page 12

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SECTION 1.3139. (a)1357911131517192123 250369121518FrequencyEmissions (g/gal)(b)13579111315171921232500.050.10.15DensityEmissions (g/gal)(c) Yes, the shapes of the histograms are the same.10. (a)1357911152500.050.10.150.20.250.3Relative FrequencyEmissions (g/gal)(b) No(c) The class interval widths are unequal.(d) The classes 11–<15 and 15–<25Page 13

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