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
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exam elaborations
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test bank for probability amp statistics for engineers amp scientists
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8th edition by sharon myers
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keying ye and walpole instructors solution manual only