Exam (elaborations) TEST BANK FOR Probability & Statistics for Engineers & Scientists, 8th Edition By Sharon Myers , Keying Ye and Walpole (Instructors Solution Manual ONLY)

Exam (elaborations) TEST BANK FOR Probability & Statistics for Engineers & Scientists, 8th Edition By Sharon Myers , Keying Ye and Walpole (Instructors Solution Manual ONLY) INSTRUCTOR’S SOLUTION MANUAL KEYING YE AND SHARON MYERS for PROBABILITY & STATISTICS FOR ENGINEERS & SCIENTISTS EIGHTH EDITION WALPOLE, MYERS, MYERS, YE Contents 1 Introduction to Statistics and Data Analysis 1 2 Probability 11 3 Random Variables and Probability Distributions 29 4 Mathematical Expectation 45 5 Some Discrete Probability Distributions 59 6 Some Continuous Probability Distributions 71 7 Functions of Random Variables 85 8 Fundamental Sampling Distributions and Data Descriptions 91 9 One- and Two-Sample Estimation Problems 103 10 One- and Two-Sample Tests of Hypotheses 121 11 Simple Linear Regression and Correlation 149 12 Multiple Linear Regression and Certain Nonlinear Regression Models 171 13 One-Factor Experiments: General 185 14 Factorial Experiments (Two or More Factors) 213 15 2k Factorial Experiments and Fractions 237 16 Nonparametric Statistics 257 iii iv CONTENTS 17 Statistical Quality Control 273 18 Bayesian Statistics 277 Chapter 1 Introduction to Statistics and Data Analysis 1.1 (a) 15. (b) ¯x = 1 15 (3.4 + 2.5 + 4.8 + · · · + 4.8) = 3.787. (c) Sample median is the 8th value, after the data is sorted from smallest to largest: 3.6. (d) A dot plot is shown below. 2.5 3.0 3.5 4.0 4.5 5.0 5.5 (e) After trimming total 40% of the data (20% highest and 20% lowest), the data becomes: 2.9 3.0 3.3 3.4 3.6 3.7 4.0 4.4 4.8 So. the trimmed mean is ¯xtr20 = 1 9 (2.9 + 3.0 + · · · + 4.8) = 3.678. 1.2 (a) Mean=20.768 and Median=20.610. (b) ¯xtr10 = 20.743. (c) A dot plot is shown below. 23 1 2 Chapter 1 Introduction to Statistics and Data Analysis 1.3 (a) A dot plot is shown below. 225 230 In the figure, “×” represents the “No aging” group and “◦” represents the “Aging” group. (b) Yes; tensile strength is greatly reduced due to the aging process. (c) MeanAging = 209.90, and MeanNo aging = 222.10. (d) MedianAging = 210.00, and MedianNo aging = 221.50. The means and medians for each group are similar to each other. 1.4 (a) ¯XA = 7.950 and ˜XA = 8.250; ¯X B = 10.260 and ˜XB = 10.150. (b) A dot plot is shown below. 6.5 7.5 8.5 9.5 10.5 11.5 In the figure, “×” represents company A and “◦” represents company B. The steel rods made by company B show more flexibility. 1.5 (a) A dot plot is shown below. − 40 In the figure, “×” represents the control group and “◦” represents the treatment group. (b) ¯XControl = 5.60, ˜XControl = 5.00, and ¯Xtr(10);Control = 5.13; ¯X Treatment = 7.60, ˜XTreatment = 4.50, and ¯Xtr(10);Treatment = 5.63. (c) The difference of the means is 2.0 and the differences of the medians and the trimmed means are 0.5, which are much smaller. The possible cause of this might be due to the extreme values (outliers) in the samples, especially the value of 37. 1.6 (a) A dot plot is shown below. 1.95 2.05 2.15 2.25 2.35 2.45 2.55 In the figure, “×” represents the 20◦C group and “◦” represents the 45◦C group. (b) ¯X20◦C = 2.1075, and ¯X45◦C = 2.2350. (c) Based on the plot, it seems that high temperature yields more high values of tensile strength, along with a few low values of tensile strength. Overall, the temperature does have an influence on the tensile strength. Solutions for Exercises in Chapter 1 3 (d) It also seems that the variation of the tensile strength gets larger when the cure temperature is increased. 1.7 s2 = 1 15−1 [(3.4−3.787)2+(2.5−3.787)2+(4.8−3.787)2+· · ·+(4.8−3.787)2] = 0.94284; s = √s2 = √0.9428 = 0.971. 1.8 s2 = 1 20−1 [(18.71 − 20.768)2 + (21.41 − 20.768)2 + · · · + (21.12 − 20.768)2] = 2.5345; s = √2.5345 = 1.592. 1.9 s2 No Aging = 1 10−1 [(227 − 222.10)2 + (222 − 222.10)2 + · · · + (221 − 222.10)2] = 42.12; sNo Aging = √42.12 = 6.49. s2 Aging = 1 10−1 [(219 − 209.90)2 + (214 − 209.90)2 + · · · + (205 − 209.90)2] = 23.62; sAging = √23.62 = 4.86. 1.10 For company A: s2 A = 1.2078 and sA = √1.2078 = 1.099. For company B: s2 B = 0.3249 and sB = √0.3249 = 0.570. 1.11 For the control group: s2 Control = 69.39 and sControl = 8.33. For the treatment group: s2 Treatment = 128.14 and sTreatment = 11.32. 1.12 For the cure temperature at 20◦C: s2 20◦C = 0.005 and s20◦C = 0.071. For the cure temperature at 45◦C: s2 45◦C = 0.0413 and s45◦C = 0.2032. The variation of the tensile strength is influenced by the increase of cure temperature. 1.13 (a) Mean = ¯X = 124.3 and median = ˜X = 120; (b) 175 is an extreme observation. 1.14 (a) Mean = ¯X = 570.5 and median = ˜X = 571; (b) Variance = s2 = 10; standard deviation= s = 3.162; range=10; (c) Variation of the diameters seems too big. 1.15 Yes. The value 0.03125 is actually a P-value and a small value of this quantity means that the outcome (i.e., HHHHH) is very unlikely to happen with a fair coin. 1.16 The term on the left side can be manipulated to Xn i=1 xi − n¯x = Xn i=1 xi − Xn i=1 xi = 0, which is the term on the right side. 1.17 (a) ¯Xsmokers = 43.70 and ¯Xnonsmokers = 30.32; (b) ssmokers = 16.93 and snonsmokers = 7.13; 4 Chapter 1 Introduction to Statistics and Data Analysis (c) A dot plot is shown below. 60 70 In the figure, “×” represents the nonsmoker group and “◦” represents the smoker group. (d) Smokers appear to take longer time to fall asleep and the time to fall asleep for smoker group is more variable. 1.18 (a) A stem-and-leaf plot is shown below. Stem Leaf Frequency 1 057 3 2 35 2 3 246 3 4 1138 4 5 22457 5 6 79 11 9 0258 4 (b) The following is the relative frequency distribution table. Relative Frequency Distribution of Grades Class Interval Class Midpoint Frequency, f Relative Frequency 10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69 70 − 79 80 − 89 90 − 99 14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 3 2 3 4 5 11 14 14 4 0.05 0.03 0.05 0.07 0.08 0.18 0.23 0.23 0.07 (c) A histogram plot is given below. 14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 Final Exam Grades Relative Frequency Solutions for Exercises in Chapter 1 5 The distribution skews to the left. (d) ¯X = 65.48, ˜X = 71.50 and s = 21.13. 1.19 (a) A stem-and-leaf plot is shown below. Stem Leaf Frequency 0 8 1 6 2 035 3 3 03 2 4 057 3 5 0569 4 6 0005 4 (b) The following is the relative frequency distribution table. Relative Frequency Distribution of Years Class Interval Class Midpoint Frequency, f Relative Frequency 0.0 − 0.9 1.0 − 1.9 2.0 − 2.9 3.0 − 3.9 4.0 − 4.9 5.0 − 5.9 6.0 − 6.9 0.45 1.45 2.45 3.45 4.45 5.45 6.45 8 6 3 2 3 4 4 0.267 0.200 0.100 0.067 0.100 0.133 0.133 (c) ¯X = 2.797, s = 2.227 and Sample range is 6.5 − 0.2 = 6.3. 1.20 (a) A stem-and-leaf plot is shown next. Stem Leaf Frequency 0* 34 2 0 17 1* 16 1 9 10 2* 034 3 2 7 1 3* 2 1 (b) The relative frequency distribution table is shown next. 6 Chapter 1 Introduction to Statistics and Data Analysis Relative Frequency Distribution of Fruit Fly Lives Class Interval Class Midpoint Frequency, f Relative Frequency 0 − 4 5 − 9 10 − 14 15 − 19 20 − 24 25 − 29 30 − 34 2 7 12 17 22 27 32 2 17 16 10 3 1 1 0.04 0.34 0.32 0.20 0.06 0.02 0.02 (c) A histogram plot is shown next. 27 32 Fruit fly lives (seconds) Relative Frequency (d) ˜X = 10.50. 1.21 (a) ¯X = 1.7743 and ˜X = 1.7700; (b) s = 0.3905. 1.22 (a) ¯X = 6.7261 and ˜X = 0.0536. (b) A histogram plot is shown next. 6.62 6.66 6.7 6.74 6.78 6.82 Relative Frequency Histogram for Diameter (c) The data appear to be skewed to the left. 1.23 (a) A dot plot is shown next. 1000 160.15 395.10 (b) ¯X1980 = 395.1 and ¯X1990 = 160.2. Solutions for Exercises in Chapter 1 7 (c) The sample mean for 1980 is over twice as large as that of 1990. The variability for 1990 decreased also as seen by looking at the picture in (a). The gap represents an increase of over 400 ppm. It appears from the data that hydrocarbon emissions decreased considerably between 1980 and 1990 and that the extreme large emission (over 500 ppm) were no longer in evidence. 1.24 (a) ¯X = 2.8973 and s = 0.5415. (b) A histogram plot is shown next. 1.8 2.1 2.4 2.7 3 3.3 3.6 3.9 Salaries Relative Frequency (c) Use the double-stem-and-leaf plot, we have the following. Stem Leaf Frequency 1 (84) 1 2* (05)(10)(14)(37)(44)(45) 6 2 (52)(52)(67)(68)(71)(75)(77)(83)(89)(91)(99) 11 3* (10)(13)(14)(22)(36)(37) 6 3 (51)(54)(57)(71)(79)(85) 6 1.25 (a) ¯X = 33.31; (b) ˜X = 26.35; (c) A histogram plot is shown next. Percentage of the families Relative Frequency 8 Chapter 1 Introduction to Statistics and Data Analysis (d) ¯Xtr(10) = 30.97. This trimmed mean is in the middle of the mean and median using the full amount of data. Due to the skewness of the data to the right (see plot in (c)), it is common to use trimmed data to have a more robust result. 1.26 If a model using the function of percent of families to predict staff salaries, it is likely that the model would be wrong due to several extreme values of the data. Actually if a scatter plot of these two data sets is made, it is easy to see that some outlier would influence the trend. 1.27 (a) The averages of the wear are plotted here. load wear (b) When the load value increases, the wear value also increases. It does show certain relationship. (c