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Contents
- 1 Introduction to Statistics and Data Analysis
- 2 Probability
- 3 Random Variables and Probability Distributions
- 4 Mathematical Expectation
- 5 Some Discrete Probability Distributions
- 6 Some Continuous Probability Distributions
- 7 Functions of Random Variables
- 8 Fundamental Sampling Distributions and Data Descriptions
- 9 One- and Two-Sample Estimation Problems
- 10 One- and Two-Sample Tests of Hypotheses
- 11 Simple Linear Regression and Correlation
- 12 Multiple Linear Regression and Certain Nonlinear Regression Models
- 13 One-Factor Experiments: General
- 14 Factorial Experiments (Two or More Factors)
- 15 2k Factorial Experiments and Fractions
- 16 Nonparametric Statistics
- 17 Statistical Quality Control
- 18 Bayesian Statistics
2 Chapter 1 Introduction to Statistics and Data Analysis
1.3 (a) A dot plot is shown below.
200 205 210 215 220 225 230 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) X¯A = 7.950 and X˜A = 8.250; X¯B = 10.260 and X˜B = 10.150. (b) A dot plot is shown below.
6.5 7.5 8.5 9.5 10.5 11. 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.
−10 0 10 20 30 40
In the figure, “×” represents the control group and “◦” represents the treatment group. (b) X¯Control = 5.60, X˜Control = 5.00, and X¯tr(10);Control = 5.13; X¯Treatment = 7.60, X˜Treatment = 4.50, and X¯tr(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. In the figure, “×” represents the 20◦C group and “◦” represents the 45◦C group. (b) X¯ 20 ◦C = 2.1075, and X¯ 45 ◦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. (d) It also seems that the variation of the tensile strength gets larger when the cure temper- ature is increased.
1.7 s^2 = (^151) − 1 [(3. 4 − 3 .787)^2 + (2. 5 − 3 .787)^2 + (4. 8 − 3 .787)^2 + · · · + (4. 8 − 3 .787)^2 ] = 0.94284; s =
s^2 =
Solutions for Exercises in Chapter 1 3
1.8 s^2 = (^201) − 1 [(18. 71 − 20 .7675)^2 + (21. 41 − 20 .7675)^2 + · · · + (21. 12 − 20 .7675)^2 ] = 2.5329; s =
1.9 (a) s^2 No Aging = (^101) − 1 [(227 − 222 .10)^2 + (222 − 222 .10)^2 + · · · + (221 − 222 .10)^2 ] = 23.66; sNo Aging =
s^2 Aging = (^101) − 1 [(219 − 209 .90)^2 + (214 − 209 .90)^2 + · · · + (205 − 209 .90)^2 ] = 42.10; sAging =
(b) Based on the numbers in (a), the variation in “Aging” is smaller that the variation in “No Aging” although the difference is not so apparent in the plot.
1.10 For company A: s^2 A = 1.2078 and sA =
For company B: s^2 B = 0.3249 and sB =
1.11 For the control group: s^2 Control = 69.38 and sControl = 8.33. For the treatment group: s^2 Treatment = 128.04 and sTreatment = 11.32.
1.12 For the cure temperature at 20◦C: s^220 ◦C = 0.005 and s 20 ◦C = 0.071. For the cure temperature at 45◦C: s^245 ◦C = 0.0413 and s 45 ◦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 = s^2 = 10; standard deviation= s = 3.162; range=10; (c) Variation of the diameters seems too big so the quality is questionable.
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
∑^ n
i=
xi − nx¯ =
∑^ n
i=
xi −
∑^ n
i=
xi = 0,
which is the term on the right side.
1.17 (a) X¯smokers = 43.70 and X¯nonsmokers = 30.32; (b) ssmokers = 16.93 and snonsmokers = 7.13; (c) A dot plot is shown below.
10 20 30 40 50 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.
Solutions for Exercises in Chapter 1 5
(b) The following is the relative frequency distribution table. Relative Frequency Distribution of Years Class Interval Class Midpoint Frequency, f Relative Frequency
- 0 − 0. 9
- 0 − 1. 9
- 0 − 2. 9
- 0 − 3. 9
- 0 − 4. 9
- 0 − 5. 9
- 0 − 6. 9
(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 56667777777889999 17 1* 0000001223333344 16 1 5566788899 10 2* 034 3 2 7 1 3* 2 1
(b) The relative frequency distribution table is shown next. 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
(c) A histogram plot is shown next.
2 7 12 17 22 27 32 Fruit fly lives (seconds)
Relative Frequency
(d) X˜ = 10.50.
6 Chapter 1 Introduction to Statistics and Data Analysis
1.21 (a) X¯ = 74.02 and X˜ = 78; (b) s = 39.26.
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. Relative Frequency Histogram for Diameter (c) The data appear to be skewed to the left.
1.23 (a) A dot plot is shown next.
0 100 200 300 400 500 600 700 800 900 1000
160.15 395.
(b) X¯ 1980 = 395.1 and X¯ 1990 = 160.2. (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. Salaries
Relative Frequency
8 Chapter 1 Introduction to Statistics and Data Analysis
(c) A plot of wears is shown next.
700 800 900 1000 1100 1200 1300
100
300
500
700
load
wear
(d) The relationship between load and wear in (c) is not as strong as the case in (a), especially for the load at 1300. One reason is that there is an extreme value (750) which influence the mean value at the load 1300.
1.28 (a) A dot plot is shown next.
71.45 71.65 71.85 72.05 72.25 72.45 72.65 72.85 73.
High Low
In the figure, “×” represents the low-injection-velocity group and “◦” represents the high-injection-velocity group. (b) It appears that shrinkage values for the low-injection-velocity group is higher than those for the high-injection-velocity group. Also, the variation of the shrinkage is a little larger for the low injection velocity than that for the high injection velocity.
1.29 A box plot is shown next.
Solutions for Exercises in Chapter 1 9
1.30 A box plot plot is shown next.
700
800
900
1000
1100
1200
1300
1.31 (a) A dot plot is shown next.
76 79 82 85 88 91 94
Low High
In the figure, “×” represents the low-injection-velocity group and “◦” represents the high-injection-velocity group. (b) In this time, the shrinkage values are much higher for the high-injection-velocity group than those for the low-injection-velocity group. Also, the variation for the former group is much higher as well. (c) Since the shrinkage effects change in different direction between low mode temperature and high mold temperature, the apparent interactions between the mold temperature and injection velocity are significant.
1.32 An interaction plot is shown next.
Low high injection velocity
low mold temp
high mold temp
mean shrinkage value
It is quite obvious to find the interaction between the two variables. Since in this experimental data, those two variables can be controlled each at two levels, the interaction can be inves-
Chapter 2
Probability
2.1 (a) S = { 8 , 16 , 24 , 32 , 40 , 48 }. (b) For x^2 + 4x − 5 = (x + 5)(x − 1) = 0, the only solutions are x = −5 and x = 1. S = {− 5 , 1 }. (c) S = {T, HT, HHT, HHH}. (d) S = {N. America, S. America, Europe, Asia, Africa, Australia, Antarctica}. (e) Solving 2x − 4 ≥ 0 gives x ≥ 2. Since we must also have x < 1, it follows that S = φ.
2.2 S = {(x, y) | x^2 + y^2 < 9; x ≥ 0 , y ≥ 0 }.
2.3 (a) A = { 1 , 3 }. (b) B = { 1 , 2 , 3 , 4 , 5 , 6 }. (c) C = {x | x^2 − 4 x + 3 = 0} = {x | (x − 1)(x − 3) = 0} = { 1 , 3 }. (d) D = { 0 , 1 , 2 , 3 , 4 , 5 , 6 }. Clearly, A = C.
2.4 (a) S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}. (b) S = {(x, y) | 1 ≤ x, y ≤ 6 }.
2.5 S = { 1 HH, 1 HT, 1 T H, 1 T T, 2 H, 2 T, 3 HH, 3 HT, 3 T H, 3 T T, 4 H, 4 T, 5 HH, 5 HT, 5 T H, 5 T T, 6 H, 6 T }.
2.6 S = {A 1 A 2 , A 1 A 3 , A 1 A 4 , A 2 A 3 , A 2 A 4 , A 3 A 4 }.
2.7 S 1 = {M M M M, M M M F, M M F M, M F M M, F M M M, M M F F, M F M F, M F F M, F M F M, F F M M, F M M F, M F F F, F M F F, F F M F, F F F M, F F F F }. S 2 = { 0 , 1 , 2 , 3 , 4 }.
2.8 (a) A = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}. (b) B = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6)}. Copyright ©c2012 Pearson Education, Inc. Publishing as Prentice Hall.
11
12 Chapter 2 Probability
(c) C = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}. (d) A ∩ C = {(5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}. (e) A ∩ B = φ. (f) B ∩ C = {(5, 2), (6, 2)}. (g) A Venn diagram is shown next.
A
A C
B
B C
C
S
∩
∩
2.9 (a) A = { 1 HH, 1 HT, 1 T H, 1 T T, 2 H, 2 T }. (b) B = { 1 T T, 3 T T, 5 T T }. (c) A′^ = { 3 HH, 3 HT, 3 T H, 3 T T, 4 H, 4 T, 5 HH, 5 HT, 5 T H, 5 T T, 6 H, 6 T }. (d) A′^ ∩ B = { 3 T T, 5 T T }. (e) A ∪ B = { 1 HH, 1 HT, 1 T H, 1 T T, 2 H, 2 T, 3 T T, 5 T T }. 2.10 (a) S = {F F F, F F N, F N F, N F F, F N N, N F N, N N F, N N N }. (b) E = {F F F, F F N, F N F, N F F }. (c) The second river was safe for fishing. 2.11 (a) S = {M 1 M 2 , M 1 F 1 , M 1 F 2 , M 2 M 1 , M 2 F 1 , M 2 F 2 , F 1 M 1 , F 1 M 2 , F 1 F 2 , F 2 M 1 , F 2 M 2 , F 2 F 1 }. (b) A = {M 1 M 2 , M 1 F 1 , M 1 F 2 , M 2 M 1 , M 2 F 1 , M 2 F 2 }. (c) B = {M 1 F 1 , M 1 F 2 , M 2 F 1 , M 2 F 2 , F 1 M 1 , F 1 M 2 , F 2 M 1 , F 2 M 2 }. (d) C = {F 1 F 2 , F 2 F 1 }. (e) A ∩ B = {M 1 F 1 , M 1 F 2 , M 2 F 1 , M 2 F 2 }. (f) A ∪ C = {M 1 M 2 , M 1 F 1 , M 1 F 2 , M 2 M 1 , M 2 F 1 , M 2 F 2 , F 1 F 2 , F 2 F 1 }.
(g)
A
A B
B
C
S
∩
14 Chapter 2 Probability
A B
S
1 2 3 4
(a) From the above Venn diagram, (A ∩ B)′^ contains the regions of 1, 2 and 4. (b) (A ∪ B)′^ contains region 1. (c) A Venn diagram is shown next.
A B
C
S
1
2 3
4
5
6
7
8
(A ∩ C) ∪ B contains the regions of 3, 4, 5, 7 and 8.
2.18 (a) Not mutually exclusive. (b) Mutually exclusive. (c) Not mutually exclusive. (d) Mutually exclusive. 2.19 (a) The family will experience mechanical problems but will receive no ticket for traffic violation and will not arrive at a campsite that has no vacancies. (b) The family will receive a traffic ticket and arrive at a campsite that has no vacancies but will not experience mechanical problems. (c) The family will experience mechanical problems and will arrive at a campsite that has no vacancies. (d) The family will receive a traffic ticket but will not arrive at a campsite that has no vacancies. (e) The family will not experience mechanical problems.
2.20 (a) 6; (b) 2; (c) 2, 5, 6; (d) 4, 5, 7, 8.
2.21 With n 1 = 6 sightseeing tours each available on n 2 = 3 different days, the multiplication rule gives n 1 n 2 = (6)(3) = 18 ways for a person to arrange a tour.
Solutions for Exercises in Chapter 2 15
2.22 With n 1 = 8 blood types and n 2 = 3 classifications of blood pressure, the multiplication rule gives n 1 n 2 = (8)(3) = 24 classifications.
2.23 Since the die can land in n 1 = 6 ways and a letter can be selected in n 2 = 26 ways, the multiplication rule gives n 1 n 2 = (6)(26) = 156 points in S.
2.24 Since a student may be classified according to n 1 = 4 class standing and n 2 = 2 gender classifications, the multiplication rule gives n 1 n 2 = (4)(2) = 8 possible classifications for the students.
2.25 With n 1 = 5 different shoe styles in n 2 = 4 different colors, the multiplication rule gives n 1 n 2 = (5)(4) = 20 different pairs of shoes.
2.26 Using Theorem 2.8, we obtain the followings.
(a) There are
5
= 21 ways. (b) There are
3
= 10 ways.
2.27 Using the generalized multiplication rule, there are n 1 × n 2 × n 3 × n 4 = (4)(3)(2)(2) = 48 different house plans available.
2.28 With n 1 = 5 different manufacturers, n 2 = 3 different preparations, and n 3 = 2 different strengths, the generalized multiplication rule yields n 1 n 2 n 3 = (5)(3)(2) = 30 different ways to prescribe a drug for asthma.
2.29 With n 1 = 3 race cars, n 2 = 5 brands of gasoline, n 3 = 7 test sites, and n 4 = 2 drivers, the generalized multiplication rule yields (3)(5)(7)(2) = 210 test runs.
2.30 With n 1 = 2 choices for the first question, n 2 = 2 choices for the second question, and so forth, the generalized multiplication rule yields n 1 n 2 · · · n 9 = 2^9 = 512 ways to answer the test.
2.31 Since the first digit is a 5, there are n 1 = 9 possibilities for the second digit and then n 2 = 8 possibilities for the third digit. Therefore, by the multiplication rule there are n 1 n 2 = (9)(8) = 72 registrations to be checked.
2.32 (a) By Theorem 2.3, there are 6! = 720 ways. (b) A certain 3 persons can follow each other in a line of 6 people in a specified order is 4 ways or in (4)(3!) = 24 ways with regard to order. The other 3 persons can then be placed in line in 3! = 6 ways. By Theorem 2.1, there are total (24)(6) = 144 ways to line up 6 people with a certain 3 following each other. (c) Similar as in (b), the number of ways that a specified 2 persons can follow each other in a line of 6 people is (5)(2!)(4!) = 240 ways. Therefore, there are 720 − 240 = 480 ways if a certain 2 persons refuse to follow each other.
2.33 (a) With n 1 = 4 possible answers for the first question, n 2 = 4 possible answers for the second question, and so forth, the generalized multiplication rule yields 4^5 = 1024 ways to answer the test.
Solutions for Exercises in Chapter 2 17
2.43 By Theorem 2.5, there are 4! = 24 ways.
2.44 By Theorem 2.5, there are 7! = 5040 arrangements. 2.45 By Theorem 2.6, there are (^) 3!2!8! = 3360.
2.46 By Theorem 2.6, there are (^) 3!4!2!9! = 1260 ways.
2.47 By Theorem 2.8, there are
3
= 56 ways. 2.48 Assume February 29th as March 1st for the leap year. There are total 365 days in a year. The number of ways that all these 60 students will have different birth dates (i.e, arranging 60 from 365) is 365 P 60. This is a very large number. 2.49 (a) Sum of the probabilities exceeds 1. (b) Sum of the probabilities is less than 1. (c) A negative probability. (d) Probability of both a heart and a black card is zero.
2.50 Assuming equal weights
(a) P (A) = 185 ; (b) P (C) = 13 ; (c) P (A ∩ C) = 367.
2.51 S = {$10, $25, $100} with weights 275/500 = 11/20, 150/500 = 3/10, and 75/500 = 3/20, respectively. The probability that the first envelope purchased contains less than $100 is equal to 11/20 + 3/10 = 17/20.
2.52 (a) P (S ∩ D′) = 88/500 = 22/125. (b) P (E ∩ D ∩ S′) = 31/500. (c) P (S′^ ∩ E′) = 171/500.
2.53 Consider the events S: industry will locate in Shanghai, B: industry will locate in Beijing.
(a) P (S ∩ B) = P (S) + P (B) − P (S ∪ B) = 0.7 + 0. 4 − 0 .8 = 0.3. (b) P (S′^ ∩ B′) = 1 − P (S ∪ B) = 1 − 0 .8 = 0.2.
2.54 Consider the events B: customer invests in tax-free bonds, M : customer invests in mutual funds.
(a) P (B ∪ M ) = P (B) + P (M ) − P (B ∩ M ) = 0.6 + 0. 3 − 0 .15 = 0.75. (b) P (B′^ ∩ M ′) = 1 − P (B ∪ M ) = 1 − 0 .75 = 0.25.
2.55 By Theorem 2.2, there are N = (26)(25)(24)(9)(8)(7)(6) = 47, 174 , 400 possible ways to code the items of which n = (5)(25)(24)(8)(7)(6)(4) = 4, 032 , 000 begin with a vowel and end with an even digit. Therefore, (^) Nn = 11710.
18 Chapter 2 Probability
2.56 (a) Let A = Defect in brake system; B = Defect in fuel system; P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.25 + 0. 17 − 0 .15 = 0.27. (b) P (No defect) = 1 − P (A ∪ B) = 1 − 0 .27 = 0.73.
2.57 (a) Since 5 of the 26 letters are vowels, we get a probability of 5/26. (b) Since 9 of the 26 letters precede j, we get a probability of 9/26. (c) Since 19 of the 26 letters follow g, we get a probability of 19/26.
2.58 (a) Of the (6)(6) = 36 elements in the sample space, only 5 elements (2,6), (3,5), (4,4), (5,3), and (6,2) add to 8. Hence the probability of obtaining a total of 8 is then 5/36. (b) Ten of the 36 elements total at most 5. Hence the probability of obtaining a total of at most is 10/36=5/18.
2.59 (a) (
4 3 )( 48 2 ) (^525 )
(b) (
13 4 )( 13 1 ) (^525 ) =^
143
2.60 (a) (
(^93 ) =^
1
(b) (
(^93 ) =^
5
2.61 (a) P (M ∪ H) = 88/100 = 22/25; (b) P (M ′^ ∩ H′) = 12/100 = 3/25; (c) P (H ∩ M ′) = 34/100 = 17/50.
2.62 (a) 9; (b) 1/9.
2.63 (a) 0.32; (b) 0.68; (c) office or den.
2.64 (a) 1 − 0 .42 = 0.58; (b) 1 − 0 .04 = 0.96.
2.65 P (A) = 0.2 and P (B) = 0. 35
(a) P (A′) = 1 − 0 .2 = 0.8; (b) P (A′^ ∩ B′) = 1 − P (A ∪ B) = 1 − 0. 2 − 0 .35 = 0.45; (c) P (A ∪ B) = 0.2 + 0.35 = 0.55.
2.66 (a) 0.02 + 0.30 = 0.32 = 32%; (b) 0.32 + 0.25 + 0.30 = 0.87 = 87%; (c) 0.05 + 0.06 + 0.02 = 0.13 = 13%; (d) 1 − 0. 05 − 0 .32 = 0.63 = 63%.
