SOLUTION MANUAL FOR BASIC STATISTICS FOR BUSINESS AND ECONOMICS 7TH EDITION BY DOUGLAS A. LIND, WILLIAM G. MARCHAL, SAMUEL A. W ATHEN, CAROL ANN W AITE, KEVIN MURPHY
Chapter 1-17 CHAPTER 1 WHAT IS STATISTICS? 1.
a. b. c. d. e. f. g. h. i. j.
Interval Ratio Ratio Nominal Ordinal Ratio Nominal Ordinal Nominal Ratio (LO1-5)
2.
a. b. c. d. e. f.
Ratio Ratio Ratio Ratio Ratio Ratio
(LO1-5)
3.
Answers will vary
(LO1-6)
4.
a. b. c. d.
5.
Qualitative data are not numerical, whereas quantitative data are numerical. Examples will vary by student. (LO1-4)
6.
A population is the entire group which you are studying. A sample is a subset taken from a population. (LO1-3)
7.
Discrete variables can assume only certain values, whereas continuous variables can assume any value within a specific range. Examples will vary. (LO1-4)
8.
The cell phone provider is nominal level data. The minutes used are ratio level. Satisfaction is ordinal level. (LO1-5)
9.
If you were using one store as typical of all of the stores selling electronic book readers in the mall then it would be sample data. However, if you were considering all of the stores
Sample Population Population Sample (LO1-3)
-1-
selling electronic book readers in the mall, then the data would be population data. (LO1-3) 10.
Based on these findings, we can infer that 270/300 or 90 percent of the executives would move. (LO1-3)
11.
If you were using this store as typical of all Best Buy stores, then the daily number sold last month would be a sample. However, if you considered the store as the only store of interest, then the daily number sold last month would be a population. (LO1-3)
12.
The clear majority of customers surveyed (400/500, or 80%) believe the take-out service is excellent. Based on this finding, we can expect a similar proportion of all customers to feel the same way. (LO1-3)
13.
a b.
This year total sales = 1 000 772; last year total sales = 942 973; total sales increased about 6% from last year to this year. Increases: Hockey Men’s Finals by 19.9% and Hockey Women’s by 23.5%. It appears that there has been a significant shift within the market from last year to this year. (LO1-3)
14.
a. b.
qualitative nominal
(LO1-4) (LO1-5)
15.
a. b. c.
quantitative discrete interval
(LO1-4) (LO1-4) (LO1-5)
16.
a. b. c.
quantitative discrete ratio
(LO1-4) (LO1-4) (LO1-5)
17.
a. b. c.
quantitative discrete ratio
(LO1-4) (LO1-4) (LO1-5)
18.
sample
19.
a. b. c.
20.
population
(LO1-3)
21.
sample
(LO1-3)
22.
a. b.
qualitative nominal
(LO1-4) (LO1-5)
23.
a. b.
qualitative nominal
(LO1-4) (LO1-5)
(LO1-3) quantitative continuous ratio
(LO1-4) (LO1-4) (LO1-5)
-2-
24.
a. b. c. d.
A sample is used because it is difficult to locate every student. A population is employed because the information is easy to find. A population is used because the information is easy to find. A sample works because it is difficult to locate every musical. (LO1-3)
25.
a. b. c. d. e. f. g. h. i.
discrete, quantitative, ratio discrete, qualitative, nominal discrete, quantitative, ratio discrete, qualitative, nominal continuous, quantitative, interval continuous, quantitative, interval discrete, qualitative, ordinal discrete, qualitative, ordinal discrete, quantitative, ratio (LO1-4 & 1-5)
26.
a.
Qualitative: Company, Industry, Established, Public, Parent, Physical rating, Cafeteria, and on site fitness Quantitative: Full time, Part time, % Contract, and Starting vacation (LO1-4) Nominal: Company, Industry, Established, Public, Parent, Cafeteria, on-site fitness Ordinal: Physical rating Ratio: Full time, Part time, % Contract, and Starting vacation (LO1-5)
b.
27
a. b.
28.
a.
b.
Qualitative: region Quantitative: average house prices (LO1-4) Nominal: region Ratio: average house prices (LO1-5) Qualitative: Team name, Country, Star power. Quantitative: FMV worth, Revenue, Temperature, Total Snow, Average Attendance, Post season, Appearance in the cup, Cup wins, 2019/2020 season points (LO1-4) Nominal: Team name, Country Ordinal: Star power Interval: Temperature Ratio: FMV worth, Revenue Total Snow, Average attendance, Post Season, Appearances in the Cup, Cup wins, 2019/2020 Points. (LO1-5)
CHAPTER 2 DESCRIBING DATA: FREQUENCY DISTRIBUTIONS AND GRAPHIC PRESENTATION
-3-
1.
Maxwell Heating & Air Conditioning far exceeds the other corporations in sales. Mancell Electric & Plumbing and Mizelle Roofing & Sheet Metals are the two corporations with the least amount of fourth quarter sales.
Maxwell has the highest sales, and Mizelle the lowest. (LO2-2) 2.
Three classes are needed, one for each player. (LO2-1)
3.
There are four classes: winter, spring, summer, and fall. The relative frequencies are 0.1, 0.3, 0.4, and 0.2, respectively.
4.
(LO2-1)
a.
(LO2-2) b.
-4-
Aboriginal Identity
Number
Relative Frequency
Percent
First Nations
851 560
0.608
60.8
Métis
451 795
0.323
32.3
Inuit
59 445
0.042
4.2
Other
26 475
0.019
1.9
Multiple identities
11 415
0.008
0.8
(LO2-1) c.
Aboriginal Identities in Canada 4.2% 1.9%
0.8%
First Nations Métis 32.3%
Inuit 60.8%
Others Multiple Identities
(LO22) 5.
a.
-5-
(LO2-2) b. Colour Bright white Metallic black Magnetic lime Tangerine orange Fusion red Total
Frequency 130 104 325 455 286 1300
Relative Frequency 0.10 0.08 0.25 0.35 0.22 1.00 (LO21)
c.
Pie Chart for Frequency vs Colour 10% 22%
Bright white
8%
Metallic black Magnetic lime 25%
Tangerine orange Fusion red
35%
(LO2-2)
-6-
6.
Student loan amounts for the current year is shown in the bar chart. From the chart, Susan Chan’s loan is the largest and Enrique Lopes’ is the lowest.
(LO2-2) 7.
2 5 32, 2 6 64 ; therefore 6 classes are recommended. (LO2-3)
8.
25 = 32, 26 = 64 suggests 6 classes. i
9.
27 = 128, 28 = 256 suggests 8 classes. i
10.
11.
12.
$29 $0 $4.83 . Use interval of $5. (LO2-3) 6 $567 $235 $41.5 . Use interval of $42 or 8
$45.
(LO2-3)
a.
25 = 32, 26 = 64 suggests 6 classes.
b.
i
a. b. c. d.
e.
24 =16, 25 = 32, suggests 5 classes (31-25)/5 = 6/5 = 1.2 Use interval of 1.5 24 Units f Relative frequency 24 to under 25.5 2 0.125 25.5 to under 27 4 0.250 27 to under28.5 8 0.500 28.5 to under 30 0 0.000 30 to under 31.5 2 0.125 Total 16 1.000 The largest concentration is in the 27 up to 28.5 class (8).
a.
24 = 16, 25 = 32, suggests 5 classes
129 42 14.5 . Use interval of 15 and start first class at 40. (LO2-3) 6
-7-
(LO2-3)
b. c. d.
e.
13.
a.
b.
c.
14.
a.
b. c.
d.
47/5 = 9.4 Use interval of 10. 50 Oil changes f Relative frequency 50 to under 60 4 0.20 60 to under 70 5 0.25 70 to under 80 6 0.30 80 to under 90 2 0.10 90 to under 100 3 0.15 Total 20 1.00 The fewest number is about 50, the highest about 100. The greatest concentration is in classes 60 up to 70 and 70 up to 80. (LO2-3) Number of Shoppers f 0 to under 3 9 3 to under 6 21 6 to under 9 13 9 to under 12 4 12 to under 15 3 15 to under 18 1 Total 51 The largest group of shoppers (21) shop at BiLo Supermarket 3, 4 or 5 times during a month and only one customer visits the store as many as 15 times in a month. Number of Relative Shoppers Frequency 0 to under 3 0.1765 3 to under 6 0.4118 6 to under 9 0.2549 9 to under 12 0.0784 12 to under 15 0.0588 15 to under 18 0.1960 Total 1.0000 (LO2-3) An interval of 10 is more convenient to work with. The distribution using 10 is: $ Spent f 15 to under 25 1 25 to under 35 2 35 to under 45 5 45 to under 55 10 55 to under 65 15 65 to under 75 4 75 to under 85 3 Total 40 Data tends to cluster in classes 45 up to 55 and 55 up to 65. Based on the distribution, the least spent is $15 (actually $18 from the raw data). The most spent was less than $85. The largest concentration of spending is between $45 up to $65. $ Spent Relative Frequency 15 to under 25 0.025
-8-
25 to under 35 35 to under 45 45 to under 55 55 to under 65 65 to under 75 75 to under 85 Total 15.
16.
17.
0.050 0.125 0.250 0.375 0.100 0.075 1.000
(LO2-3)
a. b. c. d. e. f. g.
Histogram 100 5 28 0.28 12.5 13
(LO2-4)
a. b. c. d. e.
3 approximately 27 84 2 frequency polygon
(LO2-4)
a. b. c.
50 1.5 (thousands) or 1500 miles Using lower limits on the X-axis
d. e.
1.5, 5
-9-
f. The frequent flier miles for most of the employees are between 6000-9000 and they are evenly spread on both sides of the graphs. (LO2-4) 18.
a. b. c. d.
40 2.5 2.5, 6 (always draw a frequency polygon using the midpoints)
e.
- 10 -
19.
20.
21.
f.
Most orders take around 10-15 days.
a. b. c. d. e. f.
40 5 11 or 12 about $20 per hour about $9 per hour about 75%
a. b. c. d. e. f.
200 50 or $50,000 approximately $190,000 about $230,000 about 90 homes about 140 homes
a. b.
5 Frequent Flier Miles 0 to under 3 3 to under 6 6 to under 9 9 to under 12 12 to under 15
(LO2-4)
(LO2-5)
(LO2-5)
Frequency 5 12 23 8 2
c.
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Less than Cumulative Frequency 5 17 40 48 50
22.
d.
about 8500 miles
(LO2-5)
a.
Frequent Flier Miles 0 to under 3 3 to under 6 6 to under 9 9 to under 12 12 to under 15
Frequency 5 12 23 8 2
c.
about 7500 miles
(LO2-5)
a. b.
13, 25 Lead Time
More than Cumulative Frequency 50 45 33 10 2
b.
23.
Frequency
Less than Cumulative Frequency
- 12 -
0 to under 5 5 to under 10 10 to under 15 15 to under 20 20 to under 25
6 7 12 8 7
6 13 25 33 40
c.
d.
About 14 days
(LO2-5)
- 13 -
24.
a.
Lead Time Frequency 0 to under 5 6 5 to under 10 7 10 to under 15 12 15 to under 20 8 20 to under 25 7
More than Cumulative Frequency 40 34 27 15 7
b.
c.
About 18 days
a. b. c.
621 to 629 5 621, 623, 623, 627, 629
a. b. c.
210 to 219 6 210, 211, 213, 215, 217, 219
27.
a. b. c. d. e. f. g. h. i.
25 1 38, 106 60, 61, 63, 63, 65, 65, 69 No values 9 9 76 16 (LO2-8)
28.
a. b. c. d. e.
50 1 270, 126 155, 158, 159 No values
25.
26.
(LO2-5)
(LO2-8)
(LO2-8)
- 14 -
29.
30.
31.
32.
f. g. h. i.
13 12 193.5 19
Stem 0 1 2 3 4 5
Leaf 5 28
Stem 3 4 5 6 7 8 9
Leaf 6 7 22499 0113458 035678 0344447 055 The daily usage ranged from 36 to 95. In a typical day the ATM is typically used between 52-87 times, the smallest was 36, the largest was 95, and clustered between 52-87 times. (LO2-8)
(LO2-8)
0024789 12366 2 There were a total of 16 calls studied. The number of calls received ranged from 5 to 52. Typical was 30-39 calls, smallest was 5, and largest was 52. (LO2-8)
a. Qualitative variables are ordinarily nominal level of measurement, but some are ordinal. Quantitative variables are commonly of interval or ratio level of measurement. b. Yes, both types depict samples and populations. (LO2-1&3) a.
(LO2-2)
- 15 -
b.
Pie Chart of preference vs Activity 8% 21%
Planned 26%
Non-planned Unsure No Answer 45%
(LO 2-2) c.
Both are readable, but the pie chart may be easier to comprehend. 2)
33.
2 6 64 and 2 7 128 suggest 7 classes. (LO2-3)
34.
27 = 128, 28 = 256 suggests 8 classes. i
(LO2-
490 56 54.25 . Use the interval of 55. 8
(LO2-3) 35.
A pie chart is also acceptable. From the graph we can see that insurance and license fees are the highest expense at close to $1500 per year. (LO2-2)
- 16 -
36.
(LO2-2) 37.
Business School Expenses Professional development
8.6%
Equipment 10.9%
32.3%
Mileage
12.1%
PAC meetings Office supplies
12.6% 23.5%
Other
Professional development is the largest expense. (LO2-2)
- 17 -
38.
(LO 2-2) 39.
(LO2-2) 40.
Canada won the most medals in 2018 at 29, but 2014 and 2010 are a very close second with 25 and 26 medals won.
- 18 -
(LO 2-2) 41.
Grades 7.1% 7.1%
Over 90% 19.0% 28.6%
80-89% 70-79% 60-69%
Below 60%
38.1%
The largest group had grades between 70 and 79% (38.1%). Three students (7.1%) had grades at 90% or more and 3 students (7.1%) had grades less than 60%. (LO 2-2) 42. Class Frequency 0 to under 200 19 200 to under 400 1 400 to under 600 4 600 to under 800 1 800 to under 1000 2
- 19 -
This distribution is skewed with a large number of observations in the first class. Notice that the top 19 tunes account for 1318 plays out of a total of 5958 or about 70% of all plays. (LO2-3) 43. City Vancouver Calgary Edmonton Saskatoon
Frequency Relative Frequency 100 0.050 450 0.225 1300 0.650 150 0.075 2000 The preference among frequent business travelers is definitely Edmonton (65%). The least preferred is Vancouver (5%). (LO 2-3) 44.
Newcastle Inc. Sales Revenue 3%
8%
5%
Operating Expenses Dividends
11%
Interest Profit 73%
Sinking Fund
By far the largest part of sales revenue goes towards operating expenses. (LO22) 45.
a.
4 5 5 because 2 16 25 and 2 32 25
b.
i
c. d.
15 Class 15 to under 22 22 to under 29 29 to under 36 36 to under 43 43 to under 50
48 16 6.4 use interval of 7. 5
Frequency 3 8 7 5 2 25 e. The distribution is skewed with most of the values are clustered between 22 and 36. (LO2-3)
- 20 -
46.
There will be many answers. The following pie chart shows the breakdown of the six colours. About 77% of the candies are either brown, yellow, or red. Each of these colours represents about 25% of the total. The percent of orange and blue is less than 10% each. About 4% of the candies are green.
M&M Colours in a 500g Bag 9.7%
Brown
24.8%
9.2%
Yellow 4.0%
Red Green Blue
23.5%
Orange
28.7%
(LO22) 47.
48.
49.
a. b. c. d. e. f. g. h.
70 1 0, 145 30, 30, 32, 39 24 21 77.5 25
a.
6 because 2 = 32 < 45 and 2 = 64 > 45
b.
90, found by
c. d.
40 Class 40 to under 130 130 to under 220 220 to under 310 310 to under 400 400 to under 490 490 to under 580
a. b. c.
(LO2-6)
5
6
570 41 88.17 6 Frequency 6 10 17 8 3 1 45
$36.60, (found by 265 – 82)/5 $40 Class Frequency
- 21 -
(LO2-4)
d.
50.
a.
80 to under 120 8 120 to under 160 19 160 to under 200 10 200 to under 240 6 240 to under 280 1 Total 44 The purchases ranged from a low of about $80 to a high of about $280. The concentration is in the $120 to under $160 class. (LO2-3) Student Chequing Accounts End of Month Balances
Balance f CF 0 to under 100 9 9 100 to under 200 6 15 200 to under 300 6 21 300 to under 400 6 27 400 to under 500 5 32 500 to under 600 2 34 600 to under 700 1 35 700 to under 800 3 38 800 to under 900 1 39 900 to under 1000 1 40 Total 40 Probably a class interval of $200 would be better. b.
c. d. 51.
a.
About 67% have less than a $400 balance. Therefore, about 33% would be considered “preferred.” Approximately $50 would be a convenient cutoff point. (LO2-3) Since 2 64 70 128 2 , 7 classes are recommended. The interval should be at least (1002.2 3.3)/7 = 142.7; use 150 as a convenient value. 6
7
- 22 -
Investment Portfolios lower 0 150 300 450 600 750 900
< < < < < < <
cumulative upper midpoint width frequency percent 150 75 150 28 40.0 300 225 150 19 27.1 450 375 150 15 21.4 600 525 150 2 2.9 750 675 150 4 5.7 900 825 150 1 1.4 1,050 975 150 1 1.4 70 100.0
frequency percent 28 40.0 47 67.1 62 88.6 64 91.4 68 97.1 69 98.6 70 100.0
b.
There will be many answers for the interpretation.
(LO2-3&4)
52.
a. b. c. d. e.
less-than cumulative frequency diagram or ogive 250 50 (found by 100 – 50) approximately $240,000 approximately $230,000 (LO2-5)
53.
a. b.
2^5 = 32 <36 < 64 = 2^6. Thus 6 classes are recommended. The interval width should be at least 2, found by (15-3) /6. Use 2.2 for convenience and to ensure there are only 6 classes. 2.2
c. d.
Class 2.2 to under 4.4 4.4 to under 6.6 6.6 to under 8.8
Frequency 2 7 11
- 23 -
8.8 to under 11 11 to under 13.2 13.2 to under 15.4
54.
7 7 2
e.
The distribution is slightly right-skewed with the largest concentration in the class of 6.6 up to 8.8. (LO2-3)
a. b.
Ordinal scale and qualitative variable is used. Performance Relative Frequency Early 0.22 On-time 0.67 Late 0.09 Lost 0.02
c.
d.
Delivery Performance 2% 9% 22%
Early On-time
Late Lost 67%
- 24 -
e. 89% of the packages are either early or on-time and 2% of the packages are lost. So they are missing both of their objectives. They must eliminate all lost packages and reduce the late percentage to below 1%. (LO2-1&2) 55.
a. b. c. d.
25 = 32 < 33 < 64 = 26. Thus 6 classes are recommended. The interval width should be at least 1253, found by (7829-312) /6. Use 1500 for simplicity. 0 Class 0 to under 1500 1500 to under 3000 3000 to under 4500 4500 to under 6000 6000 to under 7500 7500 to under 9000
56.
Frequency 1 2 0 7 20 3
e.
This distribution is skewed with a few very small values which likely correspond to the “start up” phase of this publication. Most observations fall in the 6000 up to 7500 class which contains 20 of the 33 (60.6%) months. (LO 2-3)
a.
Since 25 = 32, 26 = 64, and 60 < 64, 6 classes are recommended. The interval should be at least (10.1 0.4)/6 = 1.6. So we will use 2 as a convenient value.
Personal Computer Usage (Hours) Lower 0 2 4 6 8 10
< < < < < <
cumulative
upper midpoint width frequency percent Frequency percent 2 1 2 7 11.7 7 11.7 4 3 2 11 18.3 18 30.0 6 5 2 19 31.7 37 61.7 8 7 2 12 20.0 49 81.7 10 9 2 10 16.7 59 98.3 12 11 2 1 1.7 60 100.0 60
- 25 -
100.0
b.
Interpretations will vary. The “typical” person used the computer about 5 hours per week and most of them are within about five hours. (LO2-3&4) 57.
There are 50 observations, so the recommended number of classes is 6.
Twenty-three of the 50 days, or 46%, have fewer than 35 calls waiting. There are two days that have more than 105 calls waiting. (LO2-3&4) 58.
a. b. c. d. e.
55 2 91, 237 141, 143, 145 8
- 26 -
59.
f. g. h.
12 3 180
a. b. c. d.
56 10 (found by 60 – 50) 55 17 (LO2-4)
(LO2-6)
60. a. Class Cumulative Frequency Less than 15 1 Less than 30 6 Less than 45 15 Less than 60 26 Less than 75 30 b.
c. d. e.
6 days saw fewer than 30 and 11 days saw fewer than 40. The highest 80% of the days had at least 30 families.
Class Cumulative Frequency More than 0 30 More than 15 29 More than 30 24 More than 45 15 More than 60 4 More than 75 0
- 27 -
f.
61.
g.
About 27; about 12
a. b. c. d.
2^5 = 32 <45 < 64 = 2^6. Thus 6 classes are recommended. The interval width should be at least 1.5, found by (10-1) /6. Use 2 for simplicity. 0 Class 0 to under 2 2 to under 4 4 to under 6 6 to under 8 8 to under 10 10 to under 12
(LO 2-3&5)
Frequency 1 5 12 17 8 2
The distribution is fairly symmetric or “bell-shaped” with a large peak in the middle two classes of 4 up to 8. e. Class Frequency Less than 2 1 Less than 4 6 Less than 6 18 Less than 8 35 Less than 10 43 Less than 12 45
- 28 -
f.
g. h.
About 7 songs
Class Frequency More than 0 45 More than 2 44 More than 4 39 More than 6 27 More than 8 10 More than 10 2 More than 12 0 i.
- 29 -
j. 62.
63.
About 5 songs
(LO 2-3&5)
a. Approximately 180 b. 400 c. 23/180 = 0.128 d. (32+19)/180 = .283 = 28.3% e. 2000, 45 (LO2-4) a.
Class Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 b.
cumulative frequency 4 19 46 64 70
c. About 30; about 57 d.
Class More than 0 More than 5 More than 10 More than 15 More than 20 More than 25
cumulative frequency 70 66 51 24 6 0
- 30 -
e.
f.
About 25%
(LO 2-5)
a.
Lowest Value = 228 600, Highest Value = 1 548 100, Range = 1 319 500 with 6 classes the class width is 1 319 500 ÷ 6 = 219 917 round to 220 000
64.
List Price$ 228 600 448 600 668 600 888 600 1 108 600 1 328 600
< < < < < <
448 600 668 600 888 600 1 108 600 1 328 600 1 548 600 Total
Frequency 15 17 4 2 2 2 42
List prices gather around the 1st two classes. The largest list price is $1 548 100 and the smallest is $228 600. (LO2-3)
- 31 -
b.
The prices cluster around the $558 600 point.
(LO2-4)
c.
Approximately 12 homes listed for $400 000 or less. 0.75 × 42 = 31.5; Around 75% of homes listed for $668 600 or less. 5) 65. a.
Answers may vary. Highest = 21,441, Lowest = 12,618, Range = 8,823, n = 31, 5 classes, Width = 8,823/5 = 1,764.6 ~ 1800, Starting point 12,600
- 32 -
(LO2-
Attendance 12,600 to under 14,400 14,400 to under 16,200 16,200 to under 18,000 18,000 to under 19,800 19,800 to under 21,600
Frequency 2 4 10 12 3 31 Most teams have an average attendance between 16,200 and 19,800. Range is 8,823. Only 2 teams have an average attendance less than 14,400. (LO2-3) The majority of attendance is to the right. The data is not bell shaped but rather pulled from the left or negatively skewed.
b.
(LO2-4) Average Attendance Less than 12,600 Less than 14,400 Less than 16,200 Less than 18,000 Less than 19,800 Less than 21,600
- 33 -
Cumulative Frequency 0 2 6 16 28 31
Average Attendance More than 12,600 More than 14,400 More than 16,200 More than 18,000 More than 19,800 More than 21,600
Cumulative Frequency 31 29 25 15 3 0
0.40 × 31 = 12 teams. 12 teams or 40% of the teams have less than 17,000 in attendance. Around 23 teams have less than 19,000 in attendance. About 90% of the teams have over 15,000 in attendance. (LO2-5)
CHAPTER 3
- 34 -
DESCRIBING DATA: NUMERICAL MEASURES 1.
5.4 , found by 27/5
(LO3-1)
2.
55 . , found by 33/6
(LO3-1)
3.
a. b.
Mean = 7.0, found by 28/4 (5 7) (9 7) (4 7) (10 7) 0
4.
a. b.
4.2, found by 21/5 (13 . 4.2) (7.0 4.2) (3.6 4.2) (41 . 4.2) (5.0 4.2) 0 (LO3-1)
5.
14.58, found by 43.74/3
(LO3-1)
6.
$20.95, found by $125.68/6
(LO3-1)
7.
a. b.
15.4, found by 154/10 (LO3-1) Population parameter since it includes all the salespersons at Midtown Ford.
8.
a. b.
23.9, found by 167/7 (LO3-1) Population parameter since it includes all the calls during a seven-day period.
9.
(LO3-1)
a. b. customers.
$54.55, found by $1091/20 (LO3-1) A sample statistic, assuming that the power company serves more than 20
10.
a. b.
10.73, found by 161/15 (LO3-1) A sample statistic since a sample of 15 workers was studied.
11.
Yes, $162,900, found by 30($5430)
12.
Veteran sales people likely are above average and new recruits are below average. So, the sales goal is impractical and leads to more exits. (LO3-1)
13.
$22.91, found by
14.
$1.50, found by ($40 + $35)/50
15.
$23.00, found by ($800 + $1000 + $2800)/200
(LO3-2)
16.
$143.75, found by ($1000 +$750 + $4000)/40
(LO3-2)
17.
a. b. c.
(LO3-1)
18.
(LO3-1)
300($20) 400($25) 400($23) 300 400 400
$25, 200 1100
(LO3-2)
No mode The given value would be the mode. 3 and 4; bimodal
Median = 33
Mode = 15
(LO3-1)
- 35 -
(LO3-2)
19.
Median = 5
Mode = 5
(LO3-1)
20.
Median = 10.5 Mode = 8
(LO3-1)
21.
a. b.
22.
Median = 9.2; Modes are 8.2, 8.5, and 10.3.
Median = ($24.51 + $25.19)/2 = $24.85 Mode = no mode (LO3-1) (LO3-1)
23. Mean = 58.82; Median = 58.00; Mode = 58.00. All three measures are nearly identical. (LO3-1) 24. Mean = 94.5; Median = 99.5; Applicants are not better than the general population. (LO3-1) 25.
a. 6.72, found by 80.6/12 b. 6.6 is both the median and the mode c. positively skewed; the mean is the highest value
(LO3-1)
26.
Treat Wind Direction as nominal, Temperature as interval, and Pavement as ordinal data. That would lead to a mode (Southwest) for the first column, a mean (1.375) for the second column, and a median (Trace) for a third. (LO3-1)
27.
12.78% increase, found by 5 (1.08)(1.12)(1.14)(1.26)(1.05) 1
28. 3)
5.97% increase, found by 8 (1.02)(1.08)(1.06)(1.04)(1.10)(1.06)(1.08)(1.04) 1
29.
12.28% increase, found by 5 (1.094)(1.138)(1.117)(1.119)(1.147) 1
30.
1.67% increase, found by
31.
1.14% increase, found by 48
37 797 496 1 21 961 999
(LO3-3)
32.
1.90% increase, found by 29
121 .9 1 70.7
(LO3-3)
33.
10.76%, found by 5
34.
2.5% increase, found by 20
12
136 .0 1 111 .5
70 1 42
(LO3-3)
(LO3-3) $12 892 1 $7863
- 36 -
(LO3-3)
(LO3-3) (LO3-
(LO3-3)
35.
a. b. c. d.
7, found by 10 – 3 (LO3-1&4) 6, found by 30/5 2.4, found by 12/5 The difference between the highest number sold (10) and the smallest number sold (3) is 7. On an average the number of HDTVs these service reps sold deviates by 2.4 from the mean of 6.
36.
a. b. c. d.
24, found by 52 – 28 (LO3-1&4) 38, found by 304/8 6.25, found by 50/8 The difference between 28 and 52 is 24. On an average the number of students enrolled deviates 6.25 from the mean of 38.
37.
a. b. c. d.
30, found by 54 – 24 (LO3-1&4) 38, found by 380/10 7.2, found by 72/10 The difference between 54 and 24 is 30. On an average the number of minutes required to install a door deviates 7.2 minutes from the mean of 38 minutes.
38.
a. b. c. d.
7.6%, found by 18.2% – 10.6% (LO3-1&4) 13.85%, found by 110.8%/8 2%, found by 16%/8 The difference between 18.2% and 10.6% is 7.6%. On an average the return on investment deviates two percent from the mean of 13.85%.
39.
British Columbia: Median = 34, Mean = 33.1 Mode = 34, and Range = 32 Manitoba: Median = 25, Mean = 24.5, Mode = 25, and Range = 19 In BC, there was a greater average preference for the pizza than in Manitoba, however BC also had a greater dispersion in preference. (LO3-1&4)
40.
Sales employees: Mean = 3.5, Median = 2.5, Mode = 2, and Range = 10 Warehouse employees: Mean = 1.125, Median = 0.5, Mode = 0, and Range = 5 Sales employees are more likely on average to experience days off due to illness than warehouse employees, however, there is greater dispersion amongst days lost with sales employees. (LO3-1&4)
41.
a.
$5
b.
4.4$2, found by
a.
8
42.
(8 5)2 (3 5)2 (7 5)2 (3 5)2 (4 5)2 (LO3-1&4) 5
(LO3-1&4) (13 8) (3 8) (8 8) (10 8) (8 8) (6 8)2 9.67, found by 6 2
b. 43.
44.
2
2
2
2
a.
$2.77
b.
(2.68 2.77)2 ... (4.30 2.77)2 (3.58 2.77) 2 2 1.26$2 5
(LO3-1&4)
a.
11.76%, found by 58.8%/5
- 37 -
b.
16.89, found by
(13.2 11.76)2 (5 11.76)2 5
(12.9 11.76)2
(LO3-1&4) 45.
a.
b.
46.
47.
48.
49.
50.
51.
52.
Range = 7.3, found by 11.6 – 4.3 (LO3-1&4) Arithmetic mean = 6.94, found by 34.7/5 (4.3 6.94)2 (4.9 6.94)2 (11.6 6.94)2 Variance = 6.5944, found by 5 Standard Deviation = 2.568 Dennis has a higher mean return (11.76 > 6.94). However, Dennis has greater spread in their returns on equity (16.89 > 6.59).
d.
$18 000, found by $140 000 – $122 000 (LO3-1&4) $129 600, found by $648 000/5 Variance = 40 240 000$2, found by $2201 200 000/5 Standard Deviation = $6343.50 Means are about the same, but less dispersion in salary for TMV vice presidents.
a.
Mean = 4
b.
s = 2.3452
a.
Mean = 8
b.
s = 2.3452
a.
Mean = 38
b.
s = 9.0921
a. b. c.
a.
Mean = 13.85
b.
s = 2.4512
s2
( 7 4) 2 ...(3 4) 2 5.5 5 1
(LO3-4) s2
(11 8) 2 ...(7 8) 2 5.5 5 1
(LO3-4)
s2
(28 38)2 ... (42 38)2 82.67 10 1 (LO3-4)
(10.6 13.85)2 ... (15.6 13.85)2 s 6.01 8 1
Mean = $95.1 (101 95.1)2 (97 95.1)2 s2 10 1 b. s = $11.1201
2
(LO3-4)
a.
(88 95.1)2
Mean = $104.1 (110 104.1)2 (126 104.1)2 s2 10 1
123.66 $2 (LO3-4)
a.
b.
(100 104.1)2
s = $10.99
53.
1 – 1/(1.8)2 = .69136 = 69.14%
54.
($500 – $400)/$40 = 2.5
120.77 $2 (LO3-4) (LO3-5)
- 38 -
1 – 1/(2.5)2 = .84 = 84%
(LO3-5)
a. b.
About 95% 47.5%, 2.5%
(LO3-5)
56.
a. b. c. d.
85, halfway between the endpoints of 140 and 30 About 18, found by (140 – 30)/6 103 and 67, found by 85 (1)18 121 and 49, found by 85 (2)18 (LO3-5)
57.
8.06%, found by (0.25/3.10)(100)
58.
Domestic: 23.81%, found by (5/21)(100) Overseas: 20%, found by (7/35)(100) There is slightly more relative dispersion in the weights of luggage for domestic passengers. (LO3-6)
59.
a. b.
Because the two series are in different units of measurement. P.E. ratio = 36.73%; ROI 52%, less spread in the P.E. ratios
60.
a. b.
The data are in the same units but the means, relatively speaking, are far apart. The relative dispersion in stocks under $10 is 28.95%. For stocks over $60, 5.71%. Less relative dispersion in stocks over $60 (LO3-6)
61.
a.
The mean is $30.8, found by $154/5. The median is $31.0, and the standard
55.
(LO3-6)
(LO3-
6)
4806 deviation is $3.96, found as
62.
4
154 2 5 .
3($30.8 $31.0) $3.96
b.
0.151, found by
a.
The mean is $542, found by $8130/15. The median is $546, and the standard
(LO3-1,2&6)
4 415 268 deviation is $25.08, found as
63.
14
81302 15 .
b.
(LO3-1,2&6) 3($542 $546) –0.478, found by $25.08
a.
The mean is $21.93, found by $328.9/15. The median is $15.8, and the standard
13 494.67 deviation is $21.18, found as b.
$0.868, found by
14
3($21.93 $15.8) . $21.18
- 39 -
328.92 15 . (LO3-1,2&6)
64.
a.
The mean is $5961.82, found by $166 931/28. The median is $1375 and the
2 375 037 105 standard deviation is $7148.75, found as b.
27
3($5961.82 $1375) $1.925, found by $7148.75
166 9312 28 . (LO3-1,2&6)
65.
Median = 53, found by (11 + 1)(1/2); therefore 6th value in from lowest. Q1 = 49, found by (11 + 1)(1/4); therefore 3rd value in from lowest Q3 = 55, found by (11 + 1)(3/4); therefore 9th value in from lowest (LO3-7)
66.
Median = 9.53, found by (9.45 + 9.61)/2 Q1 = 7.69, found by 7.59 + (7.99 – 7.59) ¼ Q3 = 12.59, found by 12.22 + (12.71 – 12.22)3/4
(LO3-7)
a. b. c.
Q1 = 33.25 D2 = 27.8 P67 = 47
(LO3-7)
a. b. c. d.
Median = 58 Q1 = 51.25 D1 = 45.3 P33 = 53.53
a. b. c. d. e. f.
350 Q1 = 200 Q3 = 950 750, found by 950 – 200 Less than zero, or more than about 2075 There are no outliers The distribution is positively skewed.
(LO3-7&8)
a. b. c. d. e. f.
450 Q1 = 300 Q3 = 700 400, found by 700 – 300 Less than zero or more than 1300 One outlier at about 1500 Distribution is positively skewed
(LO3-7&8)
67.
68.
69.
70.
71.
Q3 = 50.25 D8 = 52.6
Q3 = 66.0 (Megastat shows Q1 as 51.75) D9 = 76.4 (LO3-7)
The distribution is somewhat positively skewed. Note that the line above 15.5 is longer than below 7.8. (LO3-7&8)
- 40 -
BoxPlot
0
72.
5
10
15
20
25
The median is $253, the smallest value is $116 and the largest is $353. About 25% of the suites are less than $224 and 25% above $298.75. The distribution is negatively skewed.
One-Bedroom Suite Daily Charges
$100
$150
$200
$250
$300
$350
$400
Amount
(LO3-7&8) 73.
Because the exact values in a frequency distribution are not known, the midpoint of the class is used for every member of that class. (LO3-9)
74.
Class f 0 to under 5 2 5 to under 10 7 10 to under 15 12 15 to under 20 6 20 to under 25 3 Total 30
M 2.5 7.5 12.5 17.5 22.5
Mean = 380/30 = 12.67 s
fM2 12.50 393.75 1875.00 1837.50 1518.75 5637.50
fM 5.0 52.5 150.0 105.0 67.5 380.0
5637.5 (380)2 / 30 5.33 29
30 9 2 (5) = 12.5 Median = 10 + 12 75.
Class f 20 to under 30 7 30 to under 40 12 40 to under 50 21 50 to under 60 18 60 to under 70 12 Total 70
M 25 35 45 55 65
(LO3-9)
fM 175 420 945 990 780 3 310
- 41 -
fM2 4 375 14 700 42 525 54 450 50 700 166 750
166 750 (3 310)2 / 70 3310 12.179 47.2857 s 69 70 70 19 2 (10) = 47.6 Median = 40 + (LO3-9) 21
x
76.
Age 10 to under 20 20 to under 30 30 to under 40 40 to under 50 50 to under 60 Total
f 3 7 18 20 12 60
M 15 25 35 45 55
s
Mean = 2410/60 = 40.17
fM2 675 4 375 22 050 40 500 36 300 103 900
fM 45 175 630 900 660 2410
103 900 (2 410) 2 / 60 10.97 60 1
60 28 (10) = 41.0 Median = 40 + 2 20 77.
Amount f 20 to under 30 1 30 to under 40 15 40 to under 50 22 50 to under 60 8 60 to under 70 4 Total 50
M 25 35 45 55 65
fM 25 525 990 440 260 2240
Mean = $2240/50 = $44.8
s
(LO3-9) fM2 625 18 375 44 550 24 200 16 900 104 650
104 650 (2240)2 / 50 $9.37 50 1
50 16 2 (10) = $44.1 Median = 40 + 22 78.
Expenditure 25 to under 35 35 to under 45 45 to under 55 55 to under 65 65 to under 75 Total
f 5 10 21 16 8 60
Mean = $3120/60 = $52
M 30 40 50 60 70
(LO3-9)
fM fM2 150 4500 400 16 000 1050 52 500 960 57 600 560 39 200 3120 169 800
s
169 800 (3120)2 / 60 $11.32 60 1
60 15 2 (10) = $52.14 Median = 45 + 21
- 42 -
(LO3-9)
79.
a.
Mean = 5, found by (6 + 4 + 3 + 7 + 5)/5 Median is 5, found by ordering the values and selecting the middle value. Population because all partners were included. ( x ) = (6 – 5) + (4 – 5) + (3 – 5) + (7 – 5) + (5 – 5) = 0 (LO3-1)
b. c. 80.
a. b.
Mean = 21.71, Median = 22.00 (LO3-1) (23 – 21.71) + (19 – 21.71) + (26 – 21.71) + (17 – 21.71) + (21 – 21.71) + (24 – 21.71) + (22 – 21.71) = 0
81.
Mean = 545/16 = 34.06, Median = 37.50
82. 1)
Mean = 2116/30 = 70.533
83.
The Communications industry has older workers than the Retail Trade. Production workers have the most age difference. (LO31)
84.
4.84, found by 121/25 Median = 4.0 On an average she made 4.84 appointments per hour. Fifty percent of her appointments were below 4 and fifty percent were above 4. (LO3-1)
85.
xw
(LO3-1) (LO3-
270($5.00) 300($6.50) 100($8.00) = $6.12 270 300 100
(LO3-
2) 86.
xw
3(4) 3(4) 5(3) 2(3) 1(4) = 3.50 3 3 5 2 1
87.
xw
15, 300(4.5) 10, 400(3.0) 150, 600(10.2) = 9.28 176, 300
88.
a. b. c.
89.
(LO3-2)
(LO3-2)
Arithmetic mean = 6.49 km Median = 6.3 km Mode = 5.3 km and 4.6 km (bimodal) Wage (x) 13.00 15.50 18.00
Totals:
Freq (f) 20 12 8
fx2
fx 260 186 144
40
(LO3-1)
3380 2883 2592 590
- 43 -
8855
2 590 8855
Mean = $590/40 = $14.75
40
Variance =
40
= 3.8125$2
Standard deviation = $1.95 90.
91.
92.
(LO3-2&4)
GM = 10
$33 598 1 1.03 1 0.03 or 3.0 percent $25 000
GM = 10
$44 771 1 1.06 1 0.06 or 6.0 percent $25 000
a. b. c.
Population 183.47 94.92%; a lot of variability compared to the mean
(LO3-3) (LO3-3)
(LO3-4&6)
Range = 30, found by 30 – 0
(2094)2 150 149
34 758 Standard deviation = 6.09, found by
(LO3-
9) 93. 8)
Comments may vary. The following results were found using statistical software. (LO3-
Q1 = 44.25 Q3 = 68.50 Median = 55.50 The distribution is approximately symmetric. The box plot is as follows. Distance Flown
0
20
40
60
Thousands of Kilometres
94.
a.
- 44 -
80
100
National Muffler Company Muffler Changes
0
b. c.
10
20 30 Time in Minutes
40
50
No outliers Summaries may vary. The distribution is positively skewed. The median time to change a muffler is 23.50 minutes. The first quartile is 15.75 minutes and the third quartile is 29.25 minutes. The range of time is 10 minutes to 44 minutes. (LO3-8)
95.
The distribution is positively skewed. The first quartile is approximately $20 and the third quartile is approximately $90. There is one outlier located at approximately $255. The median is about $50. (LO3-8)
96.
The distribution is positively skewed. The first quartile is equal to 10 and the third quartile is equal to 40. There are four outliers located at approximately 85, 90, 95 and 100. The median is about 20. (LO3-8)
97.
s
Mean is 13 kms, found by 910/70
70 19 2 (5) = 12.96 kms Median = 10 + 27 98.
99.
100.
13 637.50 69
(910)2 70 5.118 kms
(LO3-9)
a. The wait times are a sample because only 1 night was selected. b. The mean is 40.84, found by 1021/25. The median is 39. c. The range is 44, found by 67 – 23. The standard deviation is 14.85, found by the square root of 5291.36/24. (LO3-1&4) Q3 + 1.5(IQR) = 100 + 1.5(20) = 130; Q1 − 1.5(IQR) = 80 − 1.5(20) = 50; Low outliers: 35 and 48; not an outlier: 52, 66, 105, and 108; (LO3-8)
Mean = $96.55, found by $10 620/110
- 45 -
s
1 029 937.5 109
(10 620)2 110 $6.514
110 41 2 (5) = $97 Median = 95 + 35 101.
a.
(LO3-9)
The mean is 173.77 hours, found by 2259/13. The median is 195 hours.
526 391 s = 105.61 hours, found by
12
22592 13 .
b.
CV = 60.78%, found by (105.61/173.77) × 100 Coefficient of skewness is –0.697; slight negative skewness
c.
L45 = (14)(.45) = 6.3. So the 45th percentile is 192 + 0.3(195 192) = 192.9. L82 = (14)(.82) = 11.48. So the 82nd percentile is 260 + 0.48(295 260) = 276.8.
d. Apollo Space Missions
0
50
100
150
200
250
300
350
Duration in Hours
There is a slight negative skewness visible, but no outliers. (LO3-1.4.6.7&8) 102.
a. Mean = $2706.67; median = $2235
(LO3-1&8)
b. Cost of Computer Systems
0
1000
2000
3000
4000
5000
6000
Amount ($)
There is one extreme outlier. This observation will distort the value of the mean. c.
The median will be more representative of the data than the mean due to the extreme outlier.
103. Grade
Number of
fx
- 46 -
7000
Students 4 4 6 20 6 1 41
3 4 5 7 8 10 Total
12 16 30 140 48 10 256
Mean = 256/41 = 6.24 Mode = 7 Median = 7
256 2 41 = 1.70; 40
1714 s=
IQR = 7 − 5 = 2 104.
(LO3-9)
Mean = 76.83; Mode = 80 ; Median = 80; Standard deviation = 11.91; IQR = 80 − 70 = 10 (LO3-9)
105. Class
f
x
fx
fx2
0 to under 200 200 to under 400 400 to under 600 600 to under 800 800 to under 1000
18 5 4 4 2
100 300 500 700 900
1800 1500 2000 2800 1800 9900
180 000 450 000 1 000 000 1 960 000 1 620 000 5 210 000
33
Mean = 9900/33 = 300 Median observation = (33 + 1)*50/100 = 17th observation; Median class = 1st class Median = 0 + (33/2 – 0)/18 (200) = 183.33 2
Standard deviation =
(9900 ) 5 210 000 33 = 264.575 33 1
The median is a smaller value than the mean, so positively skewed. 9) 106. Class
f
x
fx
fx2
0 to under 2 2 to under 4 4 to under 6 6 to under 8
2 4 5 8
1 3 5 7
2 12 25 56
2 36 125 392
- 47 -
(LO 3-
8 to under 10 10 to under 12
14 12
9 11
126 132 353
45
1134 1452 3141
Mean = 353/45 = 7.84 Median observation = (45 + 1)*50/100 = 23rd observation; Median class = 5th class Median = 8 + (45/2 – 19)/14 (2) = 8.5 2
(353) 45 = 2.907 45 1
3141 Standard deviation =
The median is a larger value than the mean, so negatively skewed. 9) 107.
(LO 3-
CV(Calgary) = 7.12/112.50(100) = 6.33% CV(Vaughan) = 7.84/66.33(100) = 11.82% The variability in production is higher at the Vaughan plant. A manager may want to investigate why this is the case. (LO3-6)
108. Class
f
x
fx
fx2
50 to under 60 60 to under 70 70 to under 80 80 to under 90 90 to under 100
3 8 16 12 3
55 65 75 85 95
165 520 1200 1020 285
9075 33 800 90 000 86 700 27 075
3190
246 650
42
Mean = 3190/42 = 75.95% Median observation = (42 + 1)*50/100 = 21.5th observation; Median class = 3rd class Median = 70 + (42/2 – 11)/16 (10) = 76.25% 2
(3190 ) 42 = 10.314% 42 1
246 650 Standard deviation =
The median is a larger value than the mean, so slightly negatively skewed. 9) 109.
a. b. c. d. the mean. e. values. 110.
(LO3-
55, found by 72 – 17 14.4, found by 144/10 where X = 43.2 17.62 1 – 1/k2 = 1 – ¼ = .75 = 75%; all the values are within 2 standard deviations of 43.2 ± 2(17.6245) = 7.95 to 78.45; all the values are occurring between these two (LO3-4&5)
CV(Stock A) = $0.4720/$0.6517(100) = 72.43% CV(Stock B) = $4.0355/$69.1820(100) = 5.833%
- 48 -
The variability in Stock A is high. A risk averse investor would probably not want to invest in Stock A, but would pick Stock B instead. (LO3-6) 111. a. 1,4,5,6,7&8)
Mean = $857.90/50 = $17.158 Median = $16.35
c.
(857.90)2 50 s $10.582 50 1 $17.158 (2)($10.582) $4.006 and $38.322
d.
CV
(LO3-
20, 206.73
b.
$10.582 (100) 61.67 % $17.158
3($17.158 $16.35) 0.23 $10.582 25 75 L p (50 1) 12.75 L p (50 1) 38.25 100 100 Excel will give you $8.075 & $27.025 sk
e. f.
Q1 = $7.825
Q3 = $27.4 Auto Parts Company Market Values
0
5
10
15
20
25
30
35
40
Millions of Dollars
g.
The distribution is nearly symmetrical. The mean is $17.158, the median is $16.35, and the standard deviation is $10.582. About 75 percent of the companies have a value less than $27.4 and 25 percent have a value less than $7.825.
112.
a. 9, found by 12 – 3 b. 2.72, found by 13.6/5 where mean = 7.6 c. 3.51 d. 1 – 1/k2 = 1 – 1/6.25 = 0.84 = 84%; all the values are within 2.5 standard deviation of the mean. d. 7.6 ± 3.5071 = 4.09 to 11.11; About 68 e. % of the values are within these two values. (LO34&5)
113.
a. 68% b. 95% c. 99.7%
- 49 -
114.
115.
d. 2.5% e. 0.15% f. 2.5% + 0.15% = 2.65%
(LO3-5)
a. 16% b. 2.5% c. 95% d. 81.5% e. 47.5% + 49.85% = 97.35% f. 16% + 2.5% = 18.5%
(LO3-6,7&8)
a. sort the data: 11 12 14
15
26
25 L25 (13 1) 3 .5 100 L75 (13 1)
33
34
36
37
39
41
42
55
Q1 = 14 + 0.5 (15 − 14) = 14.5; Median = 34
75 10.5 100
Q3 = 39 + 0.5 (41 − 39) = 40
b. BoxPlot
0
10
20
30
40
50
60
c. 14.5 – 1.5(40 − 14.5) = −23.75; 40 + 1.5(40 − 14.5) = 78.25; an outlier would be less than -23.75 or more than 78.25 d. There are no outliers.
20 2.8 100 60 L60 (13 1) 8.4 100
e. L20 (13 1)
P20 = 12 + 0.8 (14 − 12) = 13.6 P60 = 36 + 0.4 (37 − 36) = 36.4
(LO3-
6,7&8) 116.
a. sort the data: 110 114 115
165
265
333
344
362
390
410
412
537
25 3.75 100
Q1 = 115 + 0.75 (125 − 115) = 122.5; Median = 338.5
75 11.25 100
Q3 = 410 + 0.25 (412 − 410) = 410.5
L25 (14 1)
L75 (14 1)
125
- 50 -
855
b. BoxPlot
0
200
400
600
800
1000
1200
1400
#1
c.
Q3 + 1.5(IQR) = 410.5 + 1.5(410.5 − 122.5) = 410.5 + 432 = 842.5 Q1 − 1.5(IQR) = 122.5 − 1.5(410.5 − 122.5) = 122.5 – 432 = −309.5
117.
f. g.
A low outlier is below −309.5 and a high outlier is more than 842.5. There is one high, mild outlier at 855. The data is positively skewed as the outlier will pull up the value of the mean.
h.
L15 (14 1)
15 2.25 100 80 L80 (14 1) 12 100
P15 = 114 + 0.25 (115 − 114) = 114.25 P80 = 12th value = 412
(LO3-6,7&8)
(LO3-1,4,5)
a.
The Histogram above shows that most employers have between 49 and 9,249 full time employees. The mean full time employee level is 7416.44. The median full-time employee level is 2359. There are only 5 employers that employ over 40,000 full time employees which skews the data positively. The median can be considered more representative of the data than the mean.
- 51 -
b. Range in full time employees is 57 193 and the population standard deviation is 12 196.88. 95% of employers employ between −16 977.32 and 31 810.2 full time employees, found by 7416.44 ± 2(12 196.88). The lower limit should be 0 since you cannot employ a negative number of employees. The dispersion of the data as measured by the range and standard deviation is large. c.
The Histogram above shows that most employers have between 0 and 5 400 part time employees. The mean part time employee level is 1929.63. The median part time level is only 33. There are only 5 employers that employ over 10,000 part time employees which skews the data positively. The median can be considered more representative of the data than the mean. Range in part time employees is 60 252 and the population standard deviation is 7074.85. 95% of employers employ between −12 220.1 and 16 079.3 part time employees, found by 1929.63 ± 2(7074.85). The lower limit should be 0 since you can’t employ a negative number of employees. The dispersion of the data as measured by the range and standard deviation is large. 118.
(LO3-1,3,4,5)
- 52 -
a. As observed by the above histogram, most teams revenues cluster in the two bars between $102 million and $196 million. The mean team revenue is $164.06 million, and the median is $164 million. With these two numbers being so close it shows very little skewness in the data and therefore both would be equally representative to summarize the data. b. The range in revenues is $168 million, found by ($270 − $102). The standard deviation of the population is $41.8 million. 95% of all teams’ revenues are between $80.5 million and $247.6 million, found by $164.06 ± 2($41.78) c. The rate of increase in revenues from 2010 to 2020 is 4.76%, found by 1] × 100 119.
(LO3-1,3,4,5)
- 53 -
√
–
a. As observed by the above histogram, most listings cluster in the first bar between $228 600 and $538 600. The mean region listing is $580 743 and the median is $501 700. The mean is larger than the median because a few average listing prices were above $1M. This results in the data being positively skewed and would make the median a more representative value of the combined average listing. b. The range in listing price is $1 319 500, found by ($1 548 100 – $228 600). The standard deviation of the population is $301 352 rounded to the nearest dollar. 95% of all home listings are between $−21 961 and $1 183 447, found by $580 743 ± 2($301 352). A lower boundary of $0 is more appropriate. c. The mean list price in November 2005 is $232 288. The rate of increase in revenues from November 2005 to November 2020 is 6.3%, found by √ – 1] × 100
CHAPTER 4 A SURVEY OF PROBABILITY CONCEPTS 1.
2.
Outcome 1 2 3 4
1 A A F F
2 A F A F
(LO 4-1)
Outcome 1 2 3 4 5 6 7 8 9
1 A A A R R R S S S
2 A R S A R S A R S
(LO 4-1)
3.
a. b.
0.176, found by 6/34 Empirical (LO 4-2)
4.
a. b.
0.40, found by 2/5 Classical (LO 4-2)
5.
a. b. c. d.
Empirical Classical Classical Subjective
(LO 4-2)
- 54 -
6.
a.
b.
Outcome 1 2 3 4 Classical
1st 2nd M M M F F F F M (LO 4-1&2)
7.
a. The survey of 40 people about environmental issues b. For example, 26 or more respond “yes” c. 0.25, found by 10/40 d. Empirical e. The events are probably not equally likely (we don’t know for sure) but they are mutually exclusive. (LO 4-1&2)
8.
a. b. c. d.
Recording the number of violations For example, at least one violation 0.009, found by 18/2000 Empirical (LO 4-1&2)
9.
a.
Answers will vary. Here are some possibilities: 1234, 1248, 1251, 9999
b.
1 10
c.
Classical
10.
a. b. c.
Answers will vary. Value goes up 1/16, goes up 1/8, goes down 1 0.001, 0.001, 0.01 Subjective (LO 4-1&2)
11.
a. b. c.
78 960 960 840, found by (7)(6)(5)(4). That is 7!/3! 10, found by 5!/3!2! (LO 4-3)
12.
a. b. c.
6 840 504, found by (9)(8)(7). That is 9!/6! 21, found by 7!/2!5! (LO 4-3)
13.
210, found by (10)(9)(8)(7)/(4)(3)(2)
14.
10 000, found by (10)4 (LO 4-3)
15.
120, found by 5!
(LO 4-3)
16.
3003, found by 15 C10
15 14 13 12 11 5 432
17.
10 897 286 400, found by 15 P10 15 14 13 12 11 10 9 8 7 6 (LO4-3)
18.
210
4
(LO 4-1&2)
(LO 4-3)
(LO 4-3)
- 55 -
(LO 4-3)
19
P(A or B) P(A) + P(B)
(LO 4-4)
0.30 0.20 0.50 P(neither) 1 0.50 0.50 20.
P(X or Y ) P(X ) + P(Y )
(LO 4-4)
0.05 0.02 0.07 P(neither) 1 0.07 0.93 21.
a. b.
0.51, found by 102/200 0.49, found by (1 − 0.51) or by 61/200 + 37/200 = 0.305 + 0.185 Special rule of addition was applied. (LO 4-4)
22.
a. b.
80%, found by 50% + 30% 80%, found by 100% – 20%
(LO 4-4)
23.
0.75, found by 0.25 + 0.50
(LO 4-4)
24.
Yes, A and B are mutually exclusive. No, A and B are not complements. (LO 4-4)
25.
P(A or B) = P(A) + P(B) P(A and B)
(LO 4-4)
0.20 0.30 0.15 0.35 26.
P(X or Y ) P(X ) + P(Y ) P(X and Y )
(LO 4-4)
0.55 0.35 0.20 0.70 27.
When two events are mutually exclusive it means that if one occurs the other event cannot occur. Therefore, the probability of their joint occurrence is zero. (LO 4-5).
28.
P(H or M) P(H) + P(M) P(H and M) (LO 4-4) 0.60 0.70 0.50 0.80
29.
a. b. c. d. e.
0.20 0.30 No, because a store could have both. Joint probability 0.90, found by 1.0 – 0.10 (LO 4-4)
30.
a. b. c.
0.55, found by 0.50 + 0.40 – 0.35 Joint probability No, a vacationer can visit both attractions.
- 56 -
(LO 4-4)
31.
P(A and B) P(A) P(B|A)
(LO 4-5)
0.40 0.30. 0.12 32.
P(X1 and Y2 ) = P(X1 ) P(Y2|X1 ) (LO 4-5) 0.75 0.40 0.30
33.
0.90, found by (0.80 + 0.60) – 0.50 0.10, found by (1 – 0.90) (LO 4-4)
34.
0.05, found by (1 – 0.95)
35.
a. b. c.
P(Red) = 3/10 = 0.30 P(Clear | Blue) = 1/3 = 0.33 P(Solid and Green) = 1/10 = 0.10
a. b.
6/380 or 0.0158, found by 3/20 × 2/19 272/380 or 0.7158, found by 17/20 × 16/19
a. b. c.
A contingency table 0.27, found by 135/500 A tree diagram would appear as:
36.
37.
38.
Outcome 1 2 3 4 5
A U U U U U
B U U U D D
C U D S U D
(LO 4-4)
Outcome 10 11 12 13 14
(LO 4-6)
(LO 4-5)
(LO 4-6&7)
A D D D D D
- 57 -
B U U U D D
C U D S U D
Outcome 19 20 21 22 23
A S S S S S
B U U U D D
C U D S U D
6 U D S 15 D D S 24 S 7 U S U 16 D S U 25 S 8 U S D 17 D S D 26 S 9 U S S 18 D S S 27 S U = Stock is up D = Stock is down S = Stock is same indicates the criteria that two stocks went up is met. At least 2 of the stocks went up on seven of the 27 outcomes P(At least two of the stock increase in value) = 7/27 = 0.259 (LO4- 5) 39.
No. Probability the 1st presentation wins is 3/5 = 0.60 Probability the 2nd presentation wins is 2/5(3/4) = 0.30 Probability the 3rd presentation wins is (2/5)(1/4)(3/3) = 0.10
D S S S
S U D S
(LO 4-5)
3
40.
a. b.
1 0.0029 7 7 6 5 0.612 7 7 7 3
c.
6 0.630 7
41.
a. b. c. d.
Nominal (LO 4-6) 32/200 = 16% 85/200 = 42.5% Yes, as 32% of men ordered dessert compared to 15% of women
42.
a. b. c. d.
19/50 = 38% 8/50 = 16% 15/50 + 26/50 – 8/50 = 33/50 = 66% 6/26 = 23.1% (LO 4-6)
43.
a. b. c. d. e.
106/659 = 16.1% 143/659 = 21.7% 12/659 = 1.8% 233/659 + 233/659 – 87/659 = 57.5% 40/659 = 6.1%
a. b. c.
55/195 = 28.2% 25/195 = 12.8% 60/90 = 66.7%
45.
a. b.
Asking teenagers their reactions to a newly developed soft drink Answers will vary. One possibility is more than half of the respondents like it. (LO4- 1)
46
Empirical
(LO 4-1)
47
Subjective
(LO 4-1)
44.
(LO 4-5)
(LO 4-6)
(LO 4-6)
- 58 -
48.
a. b. c.
Venn diagram Complement rule 1
(LO 4-4)
49. a. The likelihood an event will occur, assuming that another event has already occurred. b. The collection of one or more outcomes of an experiment. c. A measure of the likelihood that two or more events will happen concurrently.(LO 4- 1) 50.
26 = 64
51.
456 976, found by 264 (LO 4-3)
52.
4 4
53.
52
54.
6 × 6 × 6 = 216 ways (numbers repeated)
55.
15 6
56.
57.
58.
59
(LO 4-3)
P = 24 ways (LO 4-3) C 7 = 133 784 560 ways
(LO 4-3) (LO 4-3)
P = 3 603 600 ways (LO 4-3)
a. b. c.
4/52, or 0.0769 3/51, or 0.0588 0.0045, found by (4/52)(3/51)
a. b. c.
0.8145, found by (0.95)4 Special rule of multiplication P(A and B and C and D) = P(A) × P(B) × P(C) × P(D)
a. b. c. d. e.
0.10, found by 50/500 Yes, mutually exclusive, because a given tip cannot fall in more than one category. 1.00 0.60, found by 300/500 0.90, found by 450/500 or 1 – (50/500) (LO 4-6)
a. b.
0.08, found by 0.80 × 0.10 The text answer is labeled
(LO 4-5)
0.90 0.80 0.10
(LO 4-5)
0.80 × 0.90 = 0.72 0.80 × 0.10 = 0.08 0.20 × 0.78 = 0.156
0.78 0.20
0.20 × 0.22 = 0.044 0.22
- 59 -
c.
Yes, because all the possible outcomes are shown on the tree diagram.
a. b. c.
0.6561, found by (0.9)(0.9)(0.9)(0.9) 0.0001, found by (0.1)(0.1)(0.1)(0.1) 0.3439, found by 1 0.6561
61.
a. b. c. d.
0.57, found by 57/100 0.97, found by (57/100) + (40/100) Yes, because an employee cannot be both. 0.03, found by 1 0.97 (LO 4-4)
62.
a. b.
Empirical 3 0.046, found by 0.359
4&7) 60.
(LO 4-5)
c. d.
0.263, found by 1 0.359 0.737, found by 1 0.263
(LO4-1&5)
a. b. c.
0.5, found by (2/3)(3/4) 0.0833, found by (1/3)(1/4) 0.9167, found by 1 – 1/12
(LO 4-5)
a. b. c.
1/27, found by (1/3)(1/3)(1/3) 1/27, found by (1/3)(1/3)(1/3) 2/9, found by (3/3)(2/3)(1/3)
a. b.
0.9039, found by (0.98)5 0.0961, found by 1 – 0.9039
66.
a. b. c. d.
0.064, found by (0.4)3 0.216, found by (0.6)3 0.784, found by 1 – 0.216 Independent (LO 4-5)
67.
a. b. c. d.
0.0333, found by (4/10)(3/9)(2/8) 0.1667, found by (6/10)(5/9)(4/8) 0.8333, found by 1 – 0.1667 Dependent (LO 4-5)
68.
0.0889, found by (2/10)(4/9)
69.
a. b.
63.
64.
65.
3
(LO 4-5)
(LO4-5)
(LO 4-5)
0.3818, found by (9/12)(8/11)(7/10) 0.6182, found by 1 – 0.3818
(LO 4-5)
70.
7 C 2 50 C 3 = (21)(19 600) = 411 600 ways
(LO 4-3)
71.
20 C 4 15 C 3 = (4845)(455) = 2 204 475 ways
- 60 -
(LO 4-3)
(LO4-
72.
10 C 3 12 C 3 + 10 C 2 12 C 4 + 10 C 1 12 C 5 + 10 C 0 12 C 6 = 57 519 ways
(LO4-
3) 73.
30 C 4 20 C 4 + 30 C 5 20 C 3 + 30 C 6 20 C 2 + 30 C 7 20 C 1 + 30 C 8 20 C 0 = 454 620 240 ways
74.
d. e.
0.40, found by 200/500 0.60, found by 100/500 + 200/500 0.60, found by (200/500) + (200/500) – (100/500) General rule of addition 0.33, found by 100/300 0.1595, found by (200/500)(199/499) (LO4-6)
a. b. c. d.
0.30, found by 6/20 0.45, found by (6 + 7 − 4)/20 0.5714, found by 4/7 0.0789, found by (6/20) × (5/19)
76.
a. b.
0.143, found by 1 – (857/1000) 0.2168, found by 31/143 (LO 4-3)
77.
0.2373, found by (¾)5 (LO 4-5)
78.
a. Independence requires that P(A|B) = P(A).
75.
a. b. c.
(LO4-3)
(LO 4-6)
One possibility is: P(visit often | yes enclosed mall) = P(visit often) 160/300 ≠ 280/705; therefore, the two variables are not independent. So, any joint probability in the table must be computed by using the general rule of multiplication. b.
- 61 -
(LO 4-6&7)
79.
Yes, 256 is found by 28
80.
0.70, found by = P(A) + P(B) P(A and B)
(LO 4-3) (LO 4-4)
0.60 0.40 0.30 0.70
81.
a. b. c. d.
82.
15, found by 5 × 3
83.
a. b.
0.50, found by 10/20 0.2368, found by (10/20)(9/19) 0.1053, found by (10/20)(9/18)(8/18) 0.7158, found by (17/20)(16/19) (LO 4-5) (LO 4-3)
7! 35 3! 4! Seven colours would not be adequate to colour code the 42 different lines. Yes. There are 45 combinations, found by: 7
C3
10 C 2
10! 45 2!(10 2)!
(LO 4-3)
84.
a. b.
17 576 000, found by (26)(26)(26)(10)(10)(10) (LO 4-3) 264 103 = 456 976 000
85.
a. b. c.
P(F and > 60) = 0.25, found by 1/4 0 0.3333, found by 1/3 (LO 4-5)
- 62 -
86.
a. b. c.
87.
26 × 10 × 26 × 10 × 26 × 10 = 17 576 000 ways
88.
a. b.
0.333, found by (6/10)(5/9) 0.9286, found by 1 – [(6/10)(5/9)(4/8)(3/7)] Dependent (LO4-5)
24! 3!(24 3)! 0.125, found by 1 – [(23/24)(22/23)(21/22)]
(LO4-3)
2024, found by 24 C3
(LO4-3&5)
89. a. 60/377 = 0.159 b. (47 + 105)/377 = 0.403 c. 1 – 133/377 = 0.647 d. 23/377 = 0.061 e. (15 + 55)/377 = 0.186 f. 105/377 + 121/377 – 55/377 = 0.454 (LO 4-6) 90.
For the system to operate both components in the series must work. The probability they both work is 0.81, found by (0.90)(0.90) (LO 4-5)
91.
0.512, found by (0.8)3
92.
0.999875, found by 1 – (0.05)3 (LO 4-5)
93.
0.525, found by 1 – (0.78)3 (LO 4-5)
94.
a. b.
51%, found by 0.60(0.85) × 100 9%, found by 0.60(0.15) × 100 (LO4-5)
95.
a. b. c. d.
P(P or D) =(1/50) + (1/10) = 0.10 + 0.02 = 0.12 P(No) = 1 – 0.12 = 0.88 P(No on 3) = (0.88)3 = 0.681 P(at least one prize) = 1 – 0.681 = 0.319 (LO 4-4&5)
96.
a. b. c.
0.42 0.70(0.40) = 0.28 0.88 (LO 4-5)
97.
0.9744, found by 1 – (0.40)4
98.
a. b.
99.
a. b.
(LO 4-5)
(LO 4-5)
4 22 , found by 9 33 3 32 1 , because (LO 4-5) 4 43 2 0.185, found by (0.15)(0.95) + (0.05)(0.85) 0.0075, found by (0.15)(0.05) (LO 4-5)
- 63 -
100.
a. b. c.
(10)(9)(8) = 720 0.0014, found by 1/720 0.9958, found by 1 – 3/720
(LO4-3&5)
101. 3 628 800 matches are possible. So the probability is 1/3 628 800. (LO 4-3) 102. Café Public Yes No Total a. b. c. d.
Yes 32 32 64
No 15 21 36
Total 47 53 100
47/100 = 0.47 32/100 = 0.32 (47 + 36 − 15)/100 = 0.68 32/64 = 0.50
(LO 4-6)
103.
Worth < $500M ≥ $500M Total a. b. c. d.
Country Canada USA 2 9 5 15 7 24
Total 11 20 31
20/31 = 0.6452 5/31 = 0.1613 (7 + 11 − 2)/31 = 0.5161 9/24 = 0.375
(LO 4-6)
104.
<$500K ≥$500K Total a. b. c. d.
Region West East/Atlantic 5 16 7 14 12 30
12/42 = 0.286 7/42 = 0.167 (30 + 21 − 16)/42 = 0.833 7/21 = 0.333
Total 21 21 42
(LO 4-6)
CHAPTER 5 DISCRETE PROBABILITY DISTRIBUTIONS - 64 -
1.
Mean = 1.3 Standard deviation = 0.9, found by: xP( x) 0(0.2) 1(0.4) 2(0.3) 3(0.1) 1.3
(LO5-3)
( x ) P( x) 2
2
= (0 1.3) (0.2) (1 1.3) (0.4) (2 1.3) (0.3) (3 1.3) (0.1) = 0.81 2
2
2
2
0.81 0.9 2.
Mean = 5.4 Variance = 12.04, found by: xP( x) 2(0.5) 8(0.3) 10(0.2) 5.4
(LO5-3)
2 ( x )2 P( x)
= (2 − 5.4)2 (0.5) + (8 − 5.4)2 (0.3) + (10 − 5.4)2 (0.2) = 12.04 3.
a. b.
c.
4.
5.
a. b. c. d. e.
The middle one is a probability distribution. (1) 0.3 = 30% (2) 0.3 = 30% (3) 0.9 = 90% µ = 5(0.1) + 10(0.2) + 15(0.3) + 20(0.4) = 15 σ2 = (5 – 15)2 (0.1) + (10 – 15)2 (0.2) + (15 – 15)2 (0.3) + (20 – 15)2 (0.4) = 25 σ=5 (LO5-3) Discrete Continuous Discrete Continuous Continuous
(LO5-2)
a.
(LO5-2&3) Number of Calls Probability 0 0.16 1 0.20 2 0.44 3 0.18 4 0.02
b. c. d.
Discrete 1.7, found by 0(0.16) + 1(0.2) + 2(0.44) + 3(0.18) + 4(0.02) 1.005, found by
0.16(0 1.7)2 + 0.20(1 1.7)2 + 0.44(2 1.7)2 + 0.18(3 1.7)2 + 0.02(4 1.7)2 6.
1000(0.6) 1200(0.3) 1500(0.1) 1110 2 (1000 1110) 2 (0.6) (1200 1110) 2 (0.3) (1500 1110) 2 (0.1) 24 900 = 157.8 (LO5-3)
- 65 -
7.
a. b. c. d.
0.20 0.55 0.95 $0(0.45) $10(0.3) $100(0.2) $500(0.05) $48
2 ($0 $48) 2 (0.45) ($10 $48) 2 (0.3) ($100 $48) 2 (0.2) ($500 $48) 2 (0.05) $212 226
= $110.57 8.
a. b. c. d.
(LO5-3)
0.5 0.667 0
0(0.333) 1(0.5) 2(0) 3(0.167) 1.001
σ2 = (0 − 1.001)2 (0.333) + (1 − 1.001)2 (0.5) + (2 − 1.001)2 (0) + (3 − 1.001)2 (0.167) = 1.001 σ = 1.000 9. 3)
(LO5-3)
a.
$21, found by $10(0.5) + $25(0.4) + $50(0.08) + $100(0.02)
b.
$16.09, found by
(LO5-
0.50(10 21) 2 + 0.40(25 21) 2 + 0.08(50 21) 2 + 0.02(100 21) 2 10.
a.
(LO5-3) Hours Probability 1 0.080 2 0.152 3 0.212 4 0.180 5 0.160 6 0.052 7 0.020 8 0.144
b.
c.
This is a discrete probability distribution. The mean is 4.144, found by 1(0.080) + 2(0.152) + 3(0.212) + 4(0.180) + 5(0.160) + 6(0.052) + 7(0.020) + 8(0.144) The standard deviation is 2.0908, found by the square root of 0.08(1 − 4.144)2 + 0.152(2 − 4.144)2 + 0.212(3 − 4.144)2 + 0.180(4 − 4.144)2 + 0.160(5 − 4.144)2 +0.052(6 − 4.144)2 + 0.020(7 − 4.144)2 + 0.144(8 − 4.144)2 The “typical” customer is parked for 4.144 hours. The mean is $11.532, found by $3(0.08) + $6(0.152) + $9(0.212) + $12(0.180) + $14(0.160) + $16(0.052) + $18(0.020) + $20(0.144) The standard deviation is $5.0049, found by the square root of 0.08($3 − $11.532)2 + 0.152($6 − $11.532)2 + 0.212($9 − $11.532)2 + 0.180($12 − $11.532)2 + 0.160($14 − $11.532)2 + 0.052($16 − $11.532)2 + 0.020($18 − $11.532)2 +0.144($20 − $11.532)2
- 66 -
11.
c. d.
4! (0.25)2 (0.75)4 2 0.2109 2!(4 2)! 4! P(3) (0.25)3 (0.75)4 3 0.0469 3!(4 3)! P(2) + P(3) + P(4) = 0.2109 + 0.0469 + 0.0039 = 0.2617 P(0) + P(1) + P(2) = 0.3164 + 0.4219 + 0.2109 = 0.9492
a.
P (1)
a. b.
12.
b.
13.
P(2)
(LO5-4)
5! (0.4)1 (0.6)51 0.2592 1!(5 1)! 5! P (2) (0.4) 2 (0.6)5 2 0.3456 2!(5 2)!
c. d.
P(3) + P(4) + P(5) = 0.2304 + 0.0768 + 0.0102 = 0.3174 1 – P(5) = 1 – 0.0102 = 0.9898 (LO5-4)
a.
x P(x) 0 0.064 1 0.288 2 0.432 3 0.216 0(0.064) ... 3(0.216) 1.8
b.
2 (0 18 . ) 2 0.064...(3 18 . ) 2 0.216 0.72 0.72 0.8485 14.
a.
b.
(LO5-4)
x P(x) 0 0.168 1 0.360 2 0.309 3 0.132 4 0.028 5 0.002 0(0.168) ... 5(0.002) 1.496
2 (0 1.496)2 0.168 ... (5 1.496)2 0.002 1.0417 1.0417 1.0207 15.
a. b. c.
16.
a.
(LO5-4)
9! (0.3)2 (0.7)7 2!(9 2)! 9! (0.3)4 (0.7)5 0.1715, found by P(4) 4!(9 4)! 9! (0.3)0 (0.7)9 0.0404, found by P(0) 0!(9 0)!
0.2668, found by P(2)
0.7351, found by P(6)
6! (0.95)6 (0.05)0 6!(6 6)!
- 67 -
(LO5-4)
b. c. d.
6! (0.95)5 (0.05)1 5!(6 5)! = 5.7, found by 6(0.95) 2 0.285 , found by 6(0.95)(0.05)
0.2321, found by P(5)
0.285 0.5339
(LO5-
4) 17.
a. b. c. d. e.
12! (0.1)0 (0.9)12 0!(12 0)! 12! 0.3766, found by P(1) (0.1)1 (0.9)11 1!(12 1)! 12! 0.2301, found by P(2) (0.1)2 (0.9)10 2!(12 2)! P(0) + P(1) + P(2) = 0.2824 + 0.3766 + 0.2301 = 0.8891 = 1.2, found by 12(0.1) 2 1.08 , found by 12(0.1)(0.9)
0.2824, found by P(0)
1.08 1.04 18.
19.
20.
(LO5-4) 4
8
b.
0.2311, found by 12 C4 (0.3) (0.7) 0 12 0.0138, found by 12 C0 (0.3) (0.7)
c. d.
2 10 0.1678, found by 12 C2 (0.3) (0.7) 3.6, found by 12(0.3)
a.
15! 2 13 0.1858, found by 2!13! (0.23) (0.77)
b.
0.1416, found by
a.
(LO5-4)
c.
15! (0.23)5 (0.77)10 5!10! 3.45 found by (0.23)(15)
a.
0.2479, found by
b. c.
0.7521, found by 1 – 0.2479 0.0038, found by 0.0035 + 0.0003 + 0.0000 + 0.0000, where P(5) = 0.0035,
(LO5-4)
8! 0 8 0.16 0.84 0!8!
8! 8! 5 3 6 2 0.16 0.84 , P(6) = 0.0003, found by 0.16 0.84 , P(7) = 5!3! 6!2! 8! 7 1 (LO5-4) 0.16 0.84 and so forth. 0.0000, found by 7!1! found by
21. 4)
a.
0.296, found by using Appendix B.3 with n = 8, p = 0.30 and x = 2.
b. c.
P(x 2) 0.058 0.198 0.296 0.552 0.448, found by P(x 3) 1 P(x 2) 1 0.552
- 68 -
(LO5-
(LO5-4)
22.
a. b. c.
0.101, found from Appendix B.3, n = 12, p = 0.60, x = 5
23. 4)
a.
0.387, found from Appendix B.3 with n = 9, p = 0.90, and an x = 9
b. c. d.
P(x < 5) = 0.001 0.992, found by 1 – 0.008 0.947, found by 1 – 0.053
a.
0.3585, found by
b. c.
0.6415, found by 1 – 0.3585 0.0755, found by 1 – [0.3585 + 0.3774 + 0.1887]
24.
P(x 5) 0.0003 0.0025 0.0125 0.0420 0.1009 0.1582 P(x 6) 1 P(x 5) 1 0.1582 0.8418
20! 0 20 0.05 0.95 0!20!
(LO5-
(LO5-
4)
25.
a. b.
= 10.5, found by 15(0.7) and 15(0.7)(0.3) 1.7748 15! (0.7)10 (0.3)5 0.2061, found by 10!5!
c. d.
0.4247, found by 0.2061 + 0.2186 0.5154, found by 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047
a.
6, found by (0.3)(20) 20! 5 15 0.1789, found by 5!15! (0.3) (0.7) 0.0479, found by 0.0308 + 0.0120 + 0.0039 + 0.0010 + 0.0002 + 0.0000 (LO5Yes, probability is 0.9992, found by 1 0.0008
(LO5-
4) 26.
b. c. d. 4) 27.
P(2)
( 6 C2 )( 4 C1 ) 15(4) 0.5 120 10 C3
(LO5-5)
28.
( C )( C ) 120(5) P(3) 10 3 5 1 0.4396 C 1365 15 4
(LO5-5)
29.
P(0)
30.
a. b.
( 3 C0 )( 7 C2 ) 1(21) 0.4667 45 10 C2
(LO5-5)
( 6 C3 )( 2 C0 ) 20(1) 0.3571 56 8 C3 P(x 2) 1 P(x 3) 1 0.3571 0.6429 P(3)
- 69 -
(LO5-5)
31.
P(2)
( 6 C2 )( 9 C3 ) 15(84) 0.4196 3003 15 C5
32.
P(0)
( 4 C0 )( 11 C5 ) (1)(462) 0.1538 3003 15 C5
P(x 1) 1 P(0) 1 0.1538 0.8462
33.
34.
35.
36.
(LO5-5)
(LO5-5)
a. b.
0.6703 0.3297
(LO5-6)
a. b. c. a. b.
0.1465 0.2381 0.7619 0.0613 0.0803
(LO5-6)
a. b.
0.1353 0.8647
(LO5-6)
(LO5-6)
37.
6, P(x 5) 0.7149 , found by 1 − (0.0025 + 0.0149 + 0.0446 + 0.0892 + 0.1339) (LO5-6)
38.
0.8088
39.
A random variable is a quantitative or qualitative outcome, which results from a chance experiment. A probability distribution also includes the likelihood of each possible outcome. (LO5-1)
40.
A discrete probability distribution can take on only values that are clearly separated from each other. On the other hand, a continuous probability distribution can take on any value in a range. a. Continuous b. Discrete c. Discrete d. Discrete e. Continuous f. Discrete (LO5-2) g. Continuous
41.
The binomial distribution is a discrete probability distribution where there are only two possible outcomes. A second important part is that data collected is a result of counts. Additionally, one trial is independent from the next and the chance for success remains (LO5-4) the same from one trial to the next.
42.
When n is large and p is small the Poisson and the binomial distribution will yield (LO5-6) approximately the same results.
(LO5-6)
- 70 -
43.
0(0.1) 1(0.2) 2(0.3) 3(0.4) 2
2 (0 2)2 (0.1) ... (3 2)2 (0.40) 1 1 44.
(LO5-4)
$1000(0.25) $2000(0.60) $5000(0.15) $2200
2 ($1000 $2200)2 (0.25) ... ($5000 $2200)2 (0.15) $21 560 000 45.
(LO5-4)
0(0.4) 1(0.2) 2(0.2) 3(0.1) 4(0.1) 1.3
2 (0 1.3)2 (0.4) ... (4 1.3)2 (0.1) 1.81 13454 .
46.
(LO5-4)
3 4 e 3 810.049787 (LO 5-6) 0.168031, found by 4! 24
47.
= 13.2, found by 12(0.25) + 13(0.4) + 14(0.25) + 15(0.1) = 3.0 + 5.2 + 3.5 + 1.5 2 = 0.86, found by 0.36 + 0.016 + 0.16 + 0.324 0.86 0.9274 (LO5-4)
48.
a. b. c.
3.5, found by (0.35)(10) 0.2377, found by 210(0.01501)(0.07542) 0.4862, found by 0.2377 + 0.1536 + 0.0689 + 0.0212 + 0.0043 + 0.0005 (LO5-
a. b. c. d. e.
Discrete Continuous Discrete Discrete Continuous
a. b.
The 3rd table 1. 0.1 2. 0.8 3. 0.5
c.
µ = 25(0.5) + 50(0.3) + 75(0.1) + 100(0.1) = 45 σ2 = (25 – 45)2 (0.5) + (50 – 45)2 (0.3) + (75 – 45)2 (0.1) + (100 – 45)2 (0.1) = 600 σ = 24.5 (LO5-4)
a.
6, found by (0.4)(15) 15! 10 5 0.0245, found by 10!5! (0.4) (0.6) 0.0338, found by 0.0245 + 0.0074 + 0.0016 + 0.0003 + 0.0000 0.4032, found by 0.0005 + 0.0047 + 0.0219 + 0.0634 + 0.1268 + 0.1859 (LO5-4)
4) 49.
50.
51.
b. c. d.
(LO5-2)
- 71 -
52.
53.
13! 39! 2!11! 2! 37! 0.2135, found by 52! 4! 48! a. b.
54.
0.7897, found by 1 0.2103
a.
=12(0.07) = 0.84 and 12(0.07)(0.93) 0.8839 12! (0.07)0 (0.93)12 0.4186, found by 0!12!
c.
0.5814, found by 1 0.4186
a.
0.1311, found by
b. c. 56.
a.
b. c. d. 57
= 20(0.075) = 1.5 and 20(0.075)(0.925) 1.1779 20! (0.075)0 (0.925)20 0.2103, found by 0!20!
c.
b.
55.
(LO5-5)
a.
(LO5-4)
(LO5-4)
16! (0.15) 4 (0.85)12 4!12! 2.4, found by (0.15)(16) 0.2100, found by 1 0.0743 0.2097 0.2775 0.2285
(LO5-4)
0 0.0025 1 0.0207 2 0.0763 3 0.1665 4 0.2384 5 0.2340 6 0.1596 7 0.0746 8 0.0229 9 0.0042 10 0.0003 = 10(0.45) = 4.5 and 10(0.45)(0.55) 1.5732 0.2384 0.5044, found by 0.0025 + 0.0207 + 0.0763 + 0.1665 + 0.2384
(LO5-4)
0 1 2 3 4 5 6 7 8 9 10
0.0001 0.0019 0.0116 0.0418 0.1020 0.1768 0.2234 0.2075 0.1405 0.0676 0.0220
- 72 -
11 12
0.0043 0.0004
b. c. d.
=12(0.52) = 6.24 and 12(0.52)(0.48) 1.7307
a.
0.002(1000) = 2
b. c.
20 e 2 0.1353, found by 0! 0.8647, found by 1 – 0.1353
59.
a. b.
0.0498 0.7746, found by (1 – 0.0498)5 (LO5-6)
60.
a. b.
= 3, probability = 0.0498
0.1768 0.3342, found by 0.0001 + 0.0019 + 0.0116 + 0.0418 + 0.1020 + 0.1768 (LO5-
4) 58.
61.
a. b. c.
63.
a. b.
64.
(LO5-6)
0.5768, found by 1 – (0.0498 + 0.1494 + 0.2240)
= 4.0; from Appendix B.4 a. b. c. d.
62.
(LO5-6)
0.0183 0.1954 0.6289 0.5665
(LO5-6)
21 e 2 0.2707, found by 1! 0.0527, found by 0.0361 + 0.0120 + 0.0034 + 0.0009 + 0.0002 + 0.0000 20 e2 (LO5-6) 0.1353, found by 0!
(18.4)4 e 18.4 4! (18.4)0 e 18.4 Almost 0, found by 0! 0.00005, found by
c.
0.38489, found by 1 0.61511
a.
P( x 5 | 2) 0.94735
(LO5-6)
They are very close, but not quite over the goal. (LO5-6) b.
- 73 -
Unfilled orders 0.3 0.25 0.2 0.15 0.1 0.05 0 0
1
2
3
4
5
65.
P(2) = 0.0474 and P(0) = 0.6859
66.
a.
6
7
8
9
10 11 12
(LO5-6)
( 7 C2 )( 3 C1 ) 21(3) 0.5250 C 120 10 3 ( 7 C3 )( 3 C0 ) 35(1) P(0) 0.2917 120 10 C3
P(1)
b.
P(x 1) = 1 – P(0) = 1 – 0.2917 = 0.7083 (LO5-5) 67.
Let n = 34, and p = 0.5
P(29) 34 C29 (0.5)29 (0.5)5 0.00002 (LO5-4) 68.
a. x 0 1 2 3 4 5
b. c. (LO5-3)
Days 4 15 5 3 2 1 30
P(x Sold) 0.1333 0.5000 0.1667 0.1000 0.0667 0.0333 1.0000
µ = 1.5667 per day; this represents the average number of big-screen TVs sold 1.5667($3575)(0.15) = $840.14 per day, so for a 30-day month, multiply by 30.
69.
Using the Binomial Probability Distribution Tables with n = 8 and a probability of success equal to 0.40: a. P(x = 3) = 0.279 or P(x = 3) = 8C3(0.4)3(0.6)5 = 0.2787 b. P(x ≥ 1 ) = 0.983 or P(x ≥ 1 ) = 1 − 8C0(0.4)0(0.6)8 = 0.9832 (LO5-4)
70.
Using the Binomial Probability Distribution Tables with n = 10 and a probability of success equal to 0.25: a. Very close to 0 (0.00000095) b. 0.056 5 5 c. 0.25 0.75 = 0.000232
- 74 -
d. 0.07813 71.
(LO5-4)
6! 6 0 0.8 0.2 6!0!
a.
0.262144, found by
b.
0.98304, found by 0.08192 + 0.24576 + 0.393216 + 0.262144 where P(3) = 0.08192 found by
6! 3 3 0.8 0.2 ,……, P(6) = 0.262144 found by 3!3!
6! 6 0 0.8 0.2 6!0! c.
0.000064, found by (0.2)6 (LO 5-4) (LO5-4)
72.
¼ ($0.75) + ½ ($0.50) + ¼ (−$1.00) = $0.1875
73.
a. b. c.
74.
0.2784, found by P(6 or more) = P(6) +P(7) + P(8) +…+ P(15) = 0.1472 + 0.0811 + 0.0348 + 0.0116 + 0.0030 + 0.0006 + 0.0001 + 0.0000 (LO 5-4)
75.
a. b. c.
2.2059, found by 300(1/136). 0.2680, found by ((2.2059)2 e–2.2059)/2! 0.8898, found by 1 – ((2.2059)0 e–2.2059)/0! (LO 5-6)
76.
a. b.
0.5987, found by (0.95)10 0.4013, found by 1 – 0.5987 (LO 5-4)
77.
a.
Mean is 10.05, found by 15(0.67)
b.
Standard deviation is 1.8211, found by 15(0.67)(0.33) 8 7 0.1114, found by 15 C8 (0.67) (0.33)
c. (LO 5-4)
0.9831, found by 0.8179 + 0.1652 0.9401, found by 0.6676 + 0.2725 0.7359, found by 0.3585 + 0.3774 (LO 5-4)
0.9163, found by 1 – [0.0549 + 0.0210 + 0.0062 + 0.0014 + 0.0002 + 0.0000]
14
78. 79.
18! 2 1 0.1294, found by 14!4! 3 3
Let =155(1/3709) = 0.042 0.0424 e0.042 P(4) 0.0000001 4!
4
(LO5-4)
Very Unlikely! (LO5-6)
80. Divisions 5 + PB Match 5
Payout(x) $50 000 000 $200 000
Odds Probability(x) x.P(X) 146 107 962 0.000000006844 $0.34220 3 563 609 0.000000280614 $0.05612
- 75 -
4 + PB Match 4 3 + PB Match 3 2 + PB 1 + PB 0 + PB
$10 000 $100 $100 $7 $7 $4 $3
Divisions 5 + PB Match 5 4 + PB Match 4 3 + PB Match 3 2 + PB 1 + PB 0 + PB
x - mean $49 999 999.46 $199 999.46 $9 999.46 $99.46 $99.46 $6.46 $6.46 $3.46 $2.46
584 432 14 255 11 927 291 745 127 69
0.000001711060 0.000070145903 0.000083836351 0.003424657534 0.001340482574 0.007812500000 0.014285714286
(x - mean)
2
$0.01711 $0.00701 $0.00838 $0.02397 $0.00938 $0.03125 $0.04286
(x - mean)2 × P(x)
2 499 999 946 170 530.0000 17 109 999.6316 39 999 784 682.3902 11 224.4996 99 989 234.3948 171.0876 9892.6308 0.6939 9892.6308 0.8294 41.7536 0.1430 41.7536 0.0560 11.9834 0.0936 6.0600 0.0866
The mean payout is $0.53829, and the standard deviation of the payout is 4137.8010. If the cost of a ticket is taken into account, then the mean is a loss of $0.46171 and the standard deviation is still $4137.8010. (LO 5-3) 81. a. Wins (x) 0 1 2 3 4 5 6 11 13 23
Frequency 11 6 4 1 2 2 2 1 1 1 31
P(x) 0.3548 0.1935 0.1290 0.0323 0.0645 0.0645 0.0645 0.0323 0.0323 0.0323 1
Mean Variance Standard
3.03 23.1925
(x)(P(x) 0 0.1935 0.2581 0.0968 0.2581 0.3226 0.3871 0.3548 0.4194 0.7419 3.0323
- 76 -
(x − µ)2 9.1946 4.1301 1.0656 0.0010 0.9365 3.8720 8.8075 63.4849 99.3559 398.7107
(x − µ)2 P(x) 3.2626 0.7994 0.1375 0.0000 0.0604 0.2498 0.5682 2.0479 3.2050 12.8616 23.1925
Deviation
4.8
b. 0 wins with a probability of 35.48%
(L05-3)
a. # of Weeks (x) 2 3 3.2 3.6 3.7 3.8 4
(x − µ)2 1.1068 0.0027 0.0219 0.3003 0.4199 0.5594 0.8986
82.
b.
Frequency 8 75 1 1 1 2 10 98
P(x) 0.0816 0.7653 0.0102 0.0102 0.0102 0.0204 0.1020 1.0
Mean Variance Standard Deviation
3.0520 0.2031
(x)(P(x) 0.1633 2.2959 0.0327 0.0367 0.0378 0.0776 0.4082 3.0520
(x − µ)2 P(x) 0.0904 0.0021 0.0002 0.0031 0.0043 0.0114 0.0917 0.2031
0.451
3 weeks has a probability of 76.53% (L05-3)
CHAPTER 6 CONTINUOUS PROBABILITY DISTRIBUTIONS 1.
2.
a. b.
a = 6 b = 10 8 minutes, found by (6 + 10)/2
(10 6) 2 12
c.
1.15 minutes, found by
d. e. f.
[1/(10 – 6)](10 – 6) = 1 0.75, found by [1/(10 – 6)](10 – 7) 0.5, found by [1/(10 – 6)](9 – 7)
a. b.
a=2 b=5 3.5 minutes, found by (2 + 5)/2
(LO6-1)
(5 2) 2 12
c.
0.87 minutes, found by
d. e. f.
[1/(5 – 2)](5 – 2) = 1 0.8, found by [1/(5 – 2)](5 – 2.6) 0.2667, found by [1/(5 – 2)](3.7 – 2.9)
- 77 -
(LO6-1)
3.
4.
5.
a. b.
0.3, found by [1/($30 – $20)]($30 – $27) 0.4, found by [1/($30 – $20)]($24 – $20)
a.
Mean is $2100, found by ($400 + $3800)/2
b.
$981.50, found by
c. d.
0.4706, found by [1/($3800 - $400)] * ($2000 – $400) 0.2353, found by [1/($3800 - $400)] * ($3800 – $3000)
a. b.
a = 0.5 b = 8.0 Mean is 4.25 minutes, found by (0.5 + 8.0)/2
($3800 $400) 2 12
Standard deviation is 2.17 minutes, found by
6.
c. d. e.
0.2, found by [1/(8.0 – 0.5)](2.0 – 0.5) 0.0, found by [1/(8.0 – 0.5)](3.0 – 3.0) 0.4, found by [1/(8.0 – 0.5)](8.0 – 5.0)
a. b.
a = 0.5 b = 10.0 (using minutes as the units) Mean is 5.25 minutes, found by (0.5 + 10)/2
(LO6-1)
(8 0.5) 2 12 (LO6-1)
Standard deviation is 2.74 minutes, found by c. d.
(LO6-1)
(10 0.5) 2 12
0.5263, found by [1/(10 – 0.5)]*(10 – 5) 2.875 minutes, found from [1/(10 – 0.5)]*(x – .5) = 0.25 and 7.625 minutes, found from [1/(10 – 0.5)]*(10 – x) = 0.25
(LO6-
1) 7.
The actual shape of a normal distribution depends on its mean and standard deviation. Thus, there is a normal distribution, and an accompanying normal curve, for a mean of 7 and a standard deviation of 2. There is another normal curve for a mean of $25,000 and a standard deviation of $1742, and so on. (LO6-2)
8.
It is bell shaped and symmetrical about its mean. It is asymptotic. There is a family of normal curves. The mean, median, and the mode are equal. (LO6-2)
9.
a. b. c.
490 and 510, found by 500 1(10) 480 and 520, found by 500 2(10) 470 and 530, found by 500 3(10)
a. b. c.
68.26% 95.44% 99.74%
a. b. c.
68.26% 95.44% 99.74%
a. b.
275 and 425, found by 350 1(75) 200 and 500, found by 350 2(75)
10.
11.
12.
(LO6-3)
(LO6-3)
(LO6-3)
- 78 -
c. 13.
125 and 575, found by 350 3(75)
(LO6-3)
$50 000 – $60 000 $50 000 – $35 000 –2.00 zRachel 1.875 $5000 $8000 Adjusting for their industries, Rob is well below average and Rachel well above. (LO6zRob
3) 14.
$75 $90 $100 $90 0.68 z2 0.45 $22 $22 The first is slightly less expensive than average and the second is slightly more. (LO6z1
3) 15.
a. b. c. d.
0.8413; 0.1587 0.1056; 0.8944 0.9977; 0.0023 0.0094; 0.9906
(LO6-3)
16. a.
The area under the curve for a z of 1 is 0.3413 (from Appendix B.1). The area under the curve for a z of –1 is 0.3413. Thus, the area under the normal curve between z values of 1 and –1 is 0.3413 + 0.3413 = 0.6826. b. The area under the curve for a z of –1.25 is 0.3944 (from Appendix B.1). The area under the curve for a z of –2 is 0.4772. Thus, the area under the normal curve between z values of –1.25 and –2 is 0.4772 – 0.3944 = 0.0828. c. The area under the curve for a z of 2.83 is 0.4977 (from Appendix B.1). The area under the curve for a z of 1.75 is 0.4599. Thus, the area under the normal curve between z values of 2.83 and 1.75 is 0.4977 – 0.4599 = 0.0378.0.0378 d. The area under the curve for a z of –2.35 is 0.4906 (from Appendix B.1). The area under the curve for a z of 1.5 is 0.4332. Thus, the area under the normal curve between z values of –2.35 and 1.5 is 0.4906 + 0.4332 = 0.9238. (LO63)
17.
a. b.
25 20 125 . 4.0 0.3944 = 39.44%, found in Appendix B.1
c.
0.3085 = 30.85%, found by z =
1.25, found by z
18 20 = –0.5 4.0
Find 0.1915 in Appendix B.1 for z = –0.5; then 0.5000 – 0.1915 = 0.3085 (LO63) 18.
b.
14.3 12.2 0.84 2.5 0.2995 = 29.95%, found in Appendix B.1
c.
0.1894 = 18.94%, found by z
a.
z = 0.84, found by z
10 12.2 0.88 2.5
Find 0.3106 in Appendix B.3 for z = –0.88, then 0.5000 – 0.3106 = 0.1894 (LO6-3)
- 79 -
19.
$20.00 $16.50 1.00 $3.50
a.
0.3413, found by z
b.
Then find 0.3413 in Appendix B.3 for z = 1 0.1587, found by 0.5000 – 0.3413 = 0.1587
c.
0.3336, found by z
$15.00 $16.50 0.43 $3.50
Find 0.1664 in Appendix B.3, for a z = –0.43, then 0.5000 – 0.1664 = 0.3336 (LO6-3) 20.
a. b. c.
About 0.4332 from Appendix B.1, where z = 1.50 About 0.1915, where z = –0.50 About 0.3085, found by 0.5000 – 0.1915 (LO6-3)
21.
a.
0.8276, first find z = –1.5, found by (($44 – $50)/$4) and z = 1.25 = ($55 – $50)/$4). The area between –1.5 and 0 is 0.4332 and the area between 0 and 1.25 is 0.3944, both from Appendix B.3. Then adding the two areas, we find that 0.4332 + 0.3944 = 0.8276. 0.1056, found by 0.5000 – 0.3994, where z = 1.25 0.2029, recall that the area for z = 1.25 is 0.3944, and the area for z = 0.5, found by (($52 – $50)/$4) is 0.1915. Then subtract 0.3944 – 0.1915 and find 0.2029. (LO6-3)
b. c.
22.
a.
b. c.
23.
a. b. c.
24.
a. b.
c.
0.4017, first find z = –0.36, found by ((75 – 80)/14) and z = 0.71, found by ((90 – 80)/14). The area between –0.36 and 0 is 0.1406 and the area between 0 and 0.71 is 0.2611, both from Appendix B.1, then adding the two area we find 0.1406 + 0.2611 = 0.4017. 0.3594, found by 0.5000 – 0.1406, where z = –0.36 0.2022 found by z = (55 – 80)/14 = –1.79, for which the area is 0.4633. The z value for 70 is –0.71 and the corresponding area is 0.2611. So 0.4633 – 0.2611 = 0.2022(LO6-3) 0.1525, found by subtracting 0.4938 – 0.3413, which are the areas associated with z values of 2.5 and 1, respectively. 0.0062, found by 0.5000 – 0.4938 0.9710, found by recalling that the area of the z value of 2.5 is 0.4938. Then find z= –2.00 found by ((205 – 225)/10). Thus, 0.4938 + 0.4772 = 0.9710.(LO6-3) 0.3085, found by z = ($80 000 – $70 000)/$20 000 = 0.50. The area is 0.1915. Then 0.5000 – 0.1915 = 0.3085 0.2902, found by z = ((80 000 – 70 000)/20 000) = 0.50, the area is 0.1915 z = ((65 000 – 70 000)/20 000) = –0.25, the area is 0.0987 Adding these values together: 0.1915 + 0.0987 = 0.2902 0.5987, found by the area under the curve with a z = –0.25, 0.0987 + 0.5000 =
0.5987 (LO6-3) 25.
a. b.
0.0764, found by z = (20 – 15)/3.5 = 1.43, then 0.5000 – 0.4236 = 0.0764 0.9236, found by 0.5000 + 0.4236, where z = 1.43 - 80 -
26.
c.
0.1185, found by z = (12 – 15)/3.5 = –0.86. The area under the curve is 0.3051, then z = ((10 – 15)/3.5) = –1.43. The area is 0.4236, finally, 0.4236 – 0.3051 = 0.1185(LO6-3)
a. b. c.
0.7348, found by 0.3186 + 0.4162 0.0059, found by 0.5000 0.4941 0.0779, found by 0.4941 0.4162
(LO6-3)
27.
x = $56.58, found by adding 0.5000(the area left of the mean) and then finding a z value that forces 45% of the data to fall inside the curve. Solving for x: 1.645 = (x – $50)/$4 = $56.58.(LO6-4)
28.
0.84
x 80 14
x = 68.24 hours (LO6-4) 29.
200.7; find a z value where 0.4900 of area is between 0 and z. That value is z = 2.33, then solve for x: 2.33 = (x – 200)/0.3 so x = 200.7. (LO6-4)
30.
a. b.
$107600, found by $70000 + 1.88(20000) $44400, found by $70000 1.28(20000)
(LO6-4)
31.
1630, found by 2100 1.88(250)
(LO6-4)
32.
1570, found by 1200 + 1.645(225)
(LO6-4)
33.
1026, found by 900 + 0.84(150)
(LO6-4)
34.
10289; found by 12200 – 2.33(820)
(LO6-4)
35.
a.
= np = 50(0.25) = 12.5 2 = np(1 – p) = 12.5(1 – 0.25) = 9.375 9.375 3.06
b.
0.2578, found by (14.5 – 12.5)/3.06 = 0.65.The area is 0.2422. Then 0.5000 – 0.2422 = 0.2578. 0.2578, found by (10.5 – 12.5)/3.06 = –0.65. The area is 0.2422. Then 0.5000 – 0.2422 = 0.2578. (LO6-5)
c.
36.
a. b. c. d.
37.
a. b. c.
= (40)(0.55) = 22
2 = 9.9 = 3.15
0.2148, found by (24.5 – 22)/3.15 = 0.79. The area is 0.2852. Then 0.5000 – 0.2852 = 0.2148. 0.0197, found by (15.5 – 22)/3.15 = –2.06. The area is 0.4803. Then 0.5000 – 0.4803 = 0.0197. z = (14.5 – 22.0)/3.15 = –2.38 and (25.5 – 22.0)/3.15 = 1.11, so 0.8578, found by 0.4913 + 0.3665 = 0.8578 (LO6-5) 0.0192, found by 0.5000 0.4808 0.0694, found by 0.5000 0.4306 0.0502, found by 0.0694 0.0192
- 81 -
(LO6-5)
38.
a. b. c. d.
39.
a.
b.
10, which is the same as 0.1894, found by ((7.5 – 10)/2.83) = –0.88. The area is 0.3106. Then 0.5000 – 0.3106 = 0.1894. 0.2981, found by ((8.5 – 10)/2.83) = –0.53. The area is 0.2019. Then 0.5000 – 0.2019 = 0.2981. 0.1087, found by 0.3106 – 0.2019 (LO6-5) Yes. (1) There are two mutually exclusive outcomes-overweight and not overweight. (2) It is the result of counting the number of successes (overweight members). (3) Each trial is independent. (4) The probability of 0.30 remains the same for each trial. 0.0084, found by = 500(0.30) = 150 2 = 500(0.30)(0.70) =105
105 10.25
z
x
174.5 150 2.39 10.24695
The area under the curve for 2.39 is 0.4916. Then 0.5000 – 0.4916 = 0.0084. c.
0.8461, found by z
139.5 150 1.02 10.25
The area between 139.5 and 150 is 0.3461. Adding 0.3461 + 0.5000 = 0.8461. (LO6-5) 40.
c.
About 0.9599, found by = 100(0.38) = 38 2 = 100(0.38)(1 – 0.38) = 23.56 2356 . 4.85 Then z = (29.5 – 38)/4.85 = –1.75. Area under the curve for – 1.75 is 0.4599. Adding 0.4599 + 0.5000 = 0.9599 0.6985, found by (40.5 – 38)/4.85= 0.52, for which the area is 0.1985. Then 0.5000 + 0.1985 = 0.6985 0.6584, found by 0.4599 + 0.1985 (LO6-5)
a.
12.005 ounces, found by (11.96 + 12.05)/2
b.
0.03 ounces, found by
(12.05 11.96)2 12
c. d. e.
0.4444, found by [1/(12.05 – 11.96)](12 – 11.96) 0.7778, found by [1/(12.05 – 11.96)](12.05 – 11.98) 11.6667, found by [1/(12.05 – 11.96)](12.05 – 11.00) (LO6-1)
a.
2.1 ounces, found by (0 + 4.2)/2
a.
b.
41.
42.
43.
(4.2 0) 2 12
b.
1.21 ounces, found by
c. d.
0.7143, found by [1/(4.2 – 0)](3 – 0) 0.6429, found by [1/(4.2 – 0)](4.2 – 1.5) (LO6-1)
a.
7 minutes, found by (4 + 10)/2
b.
1.73 minutes, found by
c. d.
0.3333, found by [1/(10 – 4)](6 – 4) 0.8333, found by [1/(10 – 4)](10 – 5) (LO6-1)
(10 4) 2 12
- 82 -
a. b.
1 = [1/(3.5 - 0)](3.5 – 0) 1.75 minutes, found by (0 + 3.5)/2
c.
1.01 minutes, found by
d. e.
0.2857, found by [1/(3.5 – 0)](1 – 0) 0.4286, found by [1/(3.5 – 0)](3.5 – 2) (LO6-1)
a. b. c. d.
0.9406 and 0.0594 0.9664 and 0.0336 0.2177 and 0.7823 0.0071 and 0.9929
a. b. c. d.
0.8931, found by 0.4332 + 0.4599 0.1059, found by 0.4966 – 0.3907 0.1420, found by 0.4951 – 0.3531 0.2077, found by 0.4929 – 0.2852
47.
a. b. c. d. e.
0.3413 + 0.3413 = 0.6826 0.4772 + 0.4772 = 0.9544 0.4987 + 0.4987 = 0.9974 0.3413 + 0.4772 = 0.8185 0.0228 + 0.1587 = 0.1815(LO 6-3)
48.
a. b. c. d. e. f.
0.0475, found by 0.5 – 0.4525 0.0150, found by 0.5 – 0.4850 0.5934, found by 2(0.2967) 0.8664, found by 2(0.4332) –0.84 = (x – $45 000)/$6000; x = $39 960 1.28 = (x – $45 000)/$6000; x = $52 680
44.
45.
46.
(3.5 0) 2 12
(LO6-3)
(LO6-3)
(LO6-3 & 4)
49.
a. –0.4 for net sales, found by ($170 – $180)/$25 and 2.92 for employees, found by (1850 – 1500)/120 b. Net sales are 0.4 standard deviations below the mean. Employees is 2.92 standard deviations above the mean. c. 65.54% of the aluminum fabricators have greater net sales compared with Clarion, found by 0.1554 + 0.5000. Only 0.18% have more employees than Clarion, found by 0.5000 – 0.4982. (LO6-3)
50.
a.
51.
b. c. d.
29 32 34 32 1.5 1.0 2 2 0.7745, found by 0.4332 + 0.3413 0.0495, found by 0.5000 0.4505 35.29 hours, found by 32 + 1.645(2)
a.
almost 0.5000, because z
b.
0.3413
$30 $490 5.11 $90 0.2514, found by 0.5000 0.2486
- 83 -
(LO6-3 & 4)
52.
53.
54.
55.
c. d.
0.6374, found by 0.2486 + 0.3888 0.3450, found by 0.3888 0.0438
a.
0.4082 = 40.82%, because z
(LO6-3)
b. c. d. e.
5 4.2 1.33 0.6 0.0918 = 9.18%, found by 0.5000 0.4082 0.0905 = 9.05%, found by 0.4987 0.4082 0.6280 = 62.80%, found by 0.4987 + 0.1293 5.25 minutes, found by 4.2 + 1.75(0.6)
a. b. c. d.
0.3015 = 30.15%, found by 0.5000 0.1985 0.2579 = 25.79%, found by 0.4564 0.1985 0.0011 = 0.11%, found by 0.5000 0.4989 $1817.6, found by $1280 + 1.28($420)
(LO6-3 & 4)
a. b. c. d.
0.3446, found by 0.5000 – 0.1554 0.6006, found by 0.1554 + 0.4452 0.1039, found by 0.4452 – 0.3413 $44200, found by $40000 + 0.84($5000)
(LO6-3 & 4)
a. b. c.
90.82%, found by z = (40 – 34)/4.5 = 1.33, then 0.5000 + 0.4082 78.23%, found by 0.5000 + 0.2823, where z = (25 – 29)/5.1 = –0.78 P(z > x) = 0.01 implies P(0 < z < x) = 0.49 and x = 2.33 Women: 34 + 2.33(4.5) = 44.485 hours Men: 29 + 2.33(5.1) = 40.883 hours (LO6-3 & 4)
56. a. 0.3686 b. 2.24 c. 0.4864
(LO6-3 & 4)
0.1314 or 13.14%, found by z = ($2500 – $1994)/$450 = 1.12, then 0.5000 – 0.1189 or 11.89%, found by 0.4875 – 0.3686, when z = ($3000 – $1994)/$450 = 0.0136 or 1.36%, found by z = ($1000 – $1994)/$450 = –2.21, then 0.5000 – (LO6-3)
57.
About 4099 units, found by solving for x. 1.645 = (x – 4000)/60
58.
a. b. c. d.
59.
a. b.
(LO6-5)
0.0044, found by 0.5000 – 0.4956 0.0578, found by 0.4247 – 0.3669 0.8413, found by 0.5000 + 0.3413 No, because 112 700 – 96 600 z 5.03 (LO6-3) 3200
c.
15.39%, found by (8 – 10.3)/2.25 = –1.02, then 0.5000 – 0.3461 = 0.1539 17.31%, found by: z = (12 – 10.3)/2.25 = 0.76. Area is 0.2764 z = (14 – 10.3)/2.25 = 1.64. Area is 0.4495 The area between 12 and 14 is 0.1731, found by 0.4495 – 0.2764. Yes, but it is rather remote. Reasoning: On 99.73% of the days, returns are between 3.55 and 17.05, found by 10.3 3(2.25). Thus, the chance of less than 3.55 returns is rather remote. (LO6-3)
- 84 -
60.
a. b. c. d.
0.0262, found by (4.5 – 10)/2.83 = –1.94, for which the area is 0.4738. Then 0.5000 – 0.4738 = 0.0262 0.9441, found by (5.5 – 10)/2.83 = –1.59, for which the area is 0.4441. Then 0.5000 + 0.4441 = 0.9441 0.0297, found by (4.5 – 10)/2.83 = –1.94 and (5.5 – 10)/2.83 = –1.59. Then 0.4738 – 0.4441 = 0.0297 0.8882, found by adding the area between z = –1.59 and z = 1.59. Then 2(0.4441) = 0.8882 (LO6-5)
61.
a. b. c. d.
(37 – 39.5)/1.5 = –1.67. Then 0.4525 + 0.5 = 0.9525 (41.5 – 39.5)/1.5 = 1.33. Then 0.4082 + 0.5 = 0.9082 (36 – 39.5)/1.5 = –2.33. Then 0.4901– 0.4525 = 0.0376 (0.5000–0.4901) + 0.9525 = 0.9624 (LO6-3)
62.
a. b. c.
0.0091, found by z = (2500 – 4200)/720 = –2.36, then 0.5000 – 0.4909 0.0062, found by z = (6000 – 4200)/720 = 2.5, then 0.5000 – 0.4938 0.9847, found by 0.4909 + 0.4938 (LO6-3)
63.
a.
c. d.
0.9678, found by: = 60(0.64) = 38.4 2 = 60(0.64)(0.36) = 13.824 13824 . 3.72 Then (31.5 – 38.4)/3.72 = –1.85, for which the area is 0.4678. Then 0.5000 + 0.4678 = 0.9678 0.0853, found by (43.5 – 38.4)/3.72 = 1.37, for which the area is 0.4147. Then 0.5000 – 0.4147 = 0.0853 0.8084, found by 0.4441 + 0.3643 0.0348, found by 0.4495 – 0.4147 (LO6-5)
a.
= 60(0.1) = 6 and 60(0.1)(0.9) 2.32
b. c.
0.0393, found by 0.4738 0.4345 0.9738, found by 0.5000 + 0.4738
b.
64.
65.
0.0968, found by: = 50(0.40) = 20
= 50(0.40)(0.60) = 12 2
(LO6-5)
(LO6-5)
12 3.46
z = (24.5 – 20)/3.46 = 1.30. The area is 0.4032. Then for 25 or more, 0.5000 – 0.4032 = 0.0968. 66.
67.
= 800(0.80) = 640 664.5 640 z 2.17 11.31
800(0.80)(0.20) 11.31
(LO6-5)
Probability is 0.5000 – 0.4850 = 0.0150
a. b. c.
1.645 = (45 – )/5 = 36.78 minutes 1.645 = (45 – )/10 = 28.55 minutes z = (30 – 28.55)/10 = 0.15, then 0.5000 + 0.0596 = 0.5596
a.
2 3.1 3.67 0.3 almost 0
(LO6-
4) 68.
0.1293 b.
3 3.1 0.33 0.3
- 85 -
0.3707, found by 0.5000
c. 000(0.0228).
0.0228, found by 0.5000 0.4772; leads to 228 students, found by 10 d. 3.484, found by 3.1 + 1.28(0.3) (LO6-3 & 4)
69.
a. b.
215 hours, found by 195 + 2.33(8.5) 270 hours, found by 290 – 2.33(8.5) (LO 6-4)
70.
a. b.
z = ($30 000 – $25 000)/$4500 = 1.11; p = 0.3665; 0.3665(2) = 0.7330 $32403; found by $25000 + 1.645($4500) (LO6-3 & 4)
71.
a. b. c.
z = (100 000 – 85 000)/8000 = 1.88, so 0.5000 – 0.4699 = 0.0301 Let z = 0.67, so 0.67 = (x– 85 000)/8000 and x = 90360, set mileage at 90 360 km z = (72 000 – 85 000)/8000 = – 1.63, so 0.5000 – 0.4484 = 0.0516 (LO6-
a.
z = ($45.00 – $42.000)/$2.25 = 1.33, p(z > 1.33) = 0.5000 – 0.4082 = 0.0918. It is over $45 about 22 days, found by 240(0.0918). z = ($38.00 – $42.00)/$2.25 = –1.78 and z = ($40.00 – $42.00)/$2.25 = –0.89. So 0.4625 – 0.3133 = 0.1492 or 14.92% of the days. $45.44 (z =1.53) (LO6-3 & 4)
3) 72.
b. c. 73.
a
b. c.
0.2620; the z value for 30 is 0.2, found by ((30 – 29)/5, with a corresponding area of 0.0793, and the z value for 34 is 1.0, found by (34 – 29)/5, with a corresponding area of 0.3413, which leads to 0.3413 – 0.0793 = 0.2620. 0.8849, found by 0.3849 + 0.5000. The z value for 23 is –1.2, found by (23 – 29)/5. 0.0139, found by 0.5000 – 0.4861. The z value for 40 is 2.2, found by (40 – 29)/5. (LO 6-3)
74.
32.56 months, found by –1.28 = (x – 36.84)/3.34
75.
= 150(0.15) = 22.5 z = (30.5 – 22.5)/4.37 = 1.83
(LO6-4)
150(0.15)(0.85) 4.37 P(z > 1.83) = 0.5000 – 0.4664 = 0.0336(LO6-5)
76.
z = (1900 – 1500)/378.5 = 1.06. This person drinks more water than the average as stated in the survey. (LO 6-3)
77.
a. b. c. d.
78.
a. b. c.
z = (154 – 150)/5 = 0.8; p = 0.2881 z = (164 – 150)/5 = 2.8; p = 0.4974; (0.5000 – 0.4974) = 0.0026 = 0.26% z = (146 – 150)/5 = –0.8; p = 0.2881; z = (156 – 150)/5 = 1.2; p = 0.3849 0.2881 + 0.3849 = 0.6730 z = (156 – 150)/5 = 1.2; p = 0.3849; z = (162 – 150)/5 = 2.4; p = 0.4918 0.4918 – 0.3849 = 0.1069 (LO 6-3) 0.4052, found by 0.5000 – 0.0948, where z = 0.24, found by (($3100 – $3000)/$410). 0.2940; the z value for $3500 is 1.22, found by (($3500 – $3000)/$410, and the corresponding area is 0.3888, which leads to 0.3888 – 0.0948 = 0.2940. 0.8552; the z value for $2250 is –1.83, found by (($2250 – $3000)/$410, and the corresponding area is 0.4664, which leads to 0.4664 + 0.3888 = 0.8552. (LO 6-3)
- 86 -
470 000 –
79.
0.25
500 000 –
1.28
= 29 126 and = 462 718
(LO6-3) 80.
81.
a. b. c. d.
0.1335, found by 0.5000 0.3665 0.1965, found by 0.3665 0.1700 0.4575, found by 0.3665 + 0.0910 0.0019, found by 0.5000 0.4981
a. b. c. d. e. f.
–0.71 0.2611+ 0.4686 = 0.7297 = 72.97% 0.2611 + 0.5000 = 0.7611 = 76.11% 0.4236 + 0.5000 = 0.9236 = 92.36% 0.0764 + 0.2389 = 0.3153 = 31.53% 0.84 = (x – 50)/7; x = 55.88
(LO 6-3)
(LO6-3 & 4)
82.
1884, found by 1500 + 1.28(300) (LO 6-4)
83.
a. b.
84.
0.5000 – 0.3333 = 0.1667, so z = 0.43
0.43
0.5000 – 0.2000 = 0.3000, so z = 0.84
0.84
0.2119, found by z = (3 – 3.1)/0.125 = –0.80; so 0.5000 – 0.2881 = 0.2119 Increase the mean. z = (3 – 3.15)/0.125 = –1.2; probability is 0.5000 – 0.3849 = 0.1151 Reduce the standard deviation. z = (3 – 3.1)/0.1 = 1.0; the probability = 0.5000 – 0.3413 = 0.1587 Increasing the mean is better because a smaller percent of the hams will be below the limit. (LO6-3)
40 43.9
49 43.9
= 9.07 = 6.07
There is about a 50 percent difference between the two standard deviations. The distribution is not normal. Something may be wrong with the report. (LO6-3 & 4) 85.
42, found by 12 + 1.645(18)
(LO 6-4)
86.
Population Mean = 42.2831 years, Population Standard Deviation = 13.2445 years, z = (30 42.2831)/13.2445, z = 0.93, Area = 0.3238 + 0.5000 = 0.8238 Actual number of employers with Longest service over 30 years is 82/100 = 0.82. The normal Distribution is a good fit. (LO6-3)
87.
Population Mean = 17 589.23 fans, Population Standard Deviation = 1995.4775 fans z = (18 500 17 589.23)/1995.4775 = 0.46 Area = 0.1772 + 0.5000 = 0.6772 Actual number is 19/31, which is 0.6129. The normal distribution is a roughly close approximation. (LO6-3)
88.
Population Mean = 580 742.9, Population Standard Deviation = 301 352.4 z = (500 000 580 742.9)/301 352.4 = 0.27 Area = 0.1064 z = (600 000 580 742.9)/301 352.4 = 0.06 Area = 0.0239
- 87 -
Area = 0.1064 + 0.0239 = 0.1303 Actual number is 7/42, which is 0.1667. The normal distribution is a roughly close approximation. (LO6-3)
CHAPTER 7 SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 1.
a. b. c. d.
2.
Cluster sampling
3.
Systematic random sampling
4.
Stratified random sampling
5.
a.
b.
303 Louisiana Av, 5155 S Main, 3501 Monroe St, 2652 W Central Av Answers will vary 630 Dixie Hwy, 835 S McCord Rd, 4624 Woodville Rd Answers will vary (LO7-1)
Sample 1 2 3 4 5 6
(LO7-1)
Values 12, 12 12, 14 12, 16 12, 14 12, 16 14, 16
(LO7-1) (LO7-1) Sum 24 26 28 26 28 30
Mean 12 13 14 13 14 15
x (12 13 14 13 14 15) / 6 13.5
= (12 + 12 + 14 + 16)/4 = 13.5 c.
6.
a.
b.
c.
The means are equal. More dispersion with population compared to the sample means. The sample means vary from 12 to 15 whereas the population varies from 12 to 16. (LO7-2) Sample Values Sum 1 2,2 4 2 2,4 6 3 2,4 6 4 2,8 10 5 2,4 6 6 2,4 6 7 2,8 10 8 4,4 8 9 4,8 12 10 4,8 12 = (2 + 2 + 4 + 4 + 8)/5 = 4
Mean 2 3 3 5 3 3 5 4 6 6
x (2 3 3 5 3 3 5 4 6 6) /10 4
The means are equal. The dispersion for the population is greater than that for the sample means. The population varies from 2 to 8, whereas the sample means only vary from 2 to 6. (LO7-2) - 88 -
7.
a.
b.
Sample Values Sum Mean 1 12,12,14 38 12.67 2 12,12,15 39 13.00 3 12,12,20 44 14.67 4 14,15,20 49 16.33 5 12,14,15 41 13.67 6 12,14,15 41 13.67 7 12,15,20 47 15.67 8 12,15,20 47 15.67 9 12,14,20 46 15.33 10 12,14,20 46 15.33 x (12.66 13.00 .... 15.33 15.33) / 10 14.6
= (12 +12 + 14 + 15 + 20)/5 = 14.6 c.
8.
a.
b.
The means are equal. The dispersion of the population is greater than that of the sample means. The sample means vary from 12.67 to 16.33 whereas the population varies from 12 to 20. (LO7-2) Sample 1 2 3 4 5 6 7 8 9 10
Values 0,0,1 0,0,3 0,0,6 0,1,3 0,3,6 0,1,3 0,3,6 1,3,6 0,1,6 0,1,6
Sum 1 3 6 4 9 4 9 10 7 7
Mean 0.33 1.00 2.00 1.33 3.00 1.33 3.00 3.33 2.33 2.33
x (0.33 1.00 .... 2.33 2.33) /10 2
= (0 + 0 + 1 + 3 + 6)/5 = 2
9.
c.
The dispersion of the population is greater than the sample means. The sample means vary from 0.33 to 3.33, the population varies from 0 to 6. (LO72)
a. b.
20, found by 6 C3 Sample Cases Ruud, Austin, Sass 3,6,3 Ruud, Sass, Palmer 3,3,3 Ruud, Palmer, Wilhelms 3,3,0 Ruud, Wilhelms, Schueller 3,0,1 Austin, Sass, Palmer 6,3,3 Austin, Palmer, Wilhelms 6,3,0 Austin, Wilhelms, Schueller 6,0,1 Sass, Palmer, Wilhelms 3,3,0 Sass, Wilhelms, Schueller 3,0,1 Palmer, Wilhelms, Schueller 3,0,1 Austin, Sass, Wilhelms 6,3,0
- 89 -
Sum 12 9 6 4 12 9 7 6 4 4 9
Mean 4.0 3.0 2.0 1.33 4.0 3.0 2.33 2.0 1.33 1.33 3.00
Ruud, Austin, Palmer Ruud, Austin, Wilhelms Ruud, Austin, Schueller Ruud, Sass, Wilhelms Ruud, Sass, Schueller Ruud, Palmer, Schueller Austin, Sass, Schueller Austin, Palmer, Schueller Sass, Palmer, Schueller c.
x
3,6,3 3,6,0 3,6,1 3,3,0 3,3,1 3,3,1 6,3,1 6,3,1 3,3,1
12 9 10 6 7 7 10 10 7
4.0 3.0 3.33 2.0 2.33 2.33 3.33 3.33 2.33
53.33 = 2.67 20
= (3 + 6 + 3 + 3 + 0 + 1)/6 = 2.67 They are equal. d.
Sample Mean 1.33 2.00 2.33 3.00 3.33 4.00
Number of Means 3 3 4 4 3 3 20
Probability 0.15 0.15 0.20 0.20 0.15 0.15 1.00
More of a dispersion in population compared to sample means. The sample means vary from 1.33 to 4.0. The population varies from 0 to 6.
- 90 -
(LO7-2) 10.
a. b.
c. identical. d.
10, found by (5!)/3!2! Cars Sample Cars Sample sold mean sold mean 8,6 7 6,10 8 8,4 6 6,6 6 8,10 9 4,10 7 8,6 7 4,6 5 6,4 5 10,6 8 The population mean is 6.8, and the mean of sample means is 6.8. They are
- 91 -
(LO7-2) 11.
a.
0 1...9 4.5 10
0
1
0.12 0.1 0.08 0.06 0.04 0.02 0
b.
Sample 1 2 3 4 5 6 7 8 9 10
2
3
4
Sum 11 31 21 24 21 20 23 29 35 27
x (2.2 6.2
5
6
7
8
9
x 2.2 6.2 4.2 4.8 4.2 4.0 4.6 5.8 7.0 5.4
7.0 5.4) / 10 = 4.84
- 92 -
2.5
Frequency
2 1.5 1 0.5 0 2.2 4.0 4.2 4.6 4.8 5.4 5.8 6.2 7.0 Value s
The mean of the 10 sample means is 4.84, which is close to the population mean of 4.5. The sample means range from 2.2 to 7.0, whereas the population values range from 0 to 9. From the above graph, the sample means tend to cluster between 4 and 5. (LO7-3 &4) a. 10 Frequency
12.
8 6 4 2 0 2
3
4
5
6
7
8
Units Sold
b. c.
23 5 3.3 20
Answers will vary, below is one sample x Sample Sample Values Sum 1 2,3,2,3,3 13 2.6 2 3,3,4,2,4 16 3.2 3 3,3,4,4,2 16 3.2 4 3,2,5,5,3 18 3.6 5 3,4,4,2,7 20 4.0
x d.
2.6 3.2 3.2 3.6 4.0 = 3.32 5
Sample mean is very close to the population mean. It is not to be expected that they are exact.
e.
- 93 -
There is less dispersion in the sample means than the population. 3&4) 13.
a. Answers will vary. b. Answers will vary. c. The sample distribution should be more bell-shaped.
14.
Answers will vary.
15.
a.
z
(LO7-3&4)
(LO7-3&4)
63 60 0.75 12 / 9
So, the probability is 0.2266, found by 0.5000 0.2734.
16.
a.
56 60 1 12 / 9
b.
z
c.
So, the probability is 0.1587, found by 0.5000 0.3413 0.6147, found by 0.3413 + 0.2734. (LO7-5)
z
74 75 1.26 5 / 40
So, probability is 0.1038, found by 0.5000 0.3962. b.
z
76 75 1.26 5 / 40
So, probability is 0.7924, found by 2(0.3962).
17.
77 75 2.53 5 / 40
c.
z
d.
So, probability is 0.0981, found by 0.4943 0.3962 0.0057, found by 0.5000 0.4943 (LO7-5)
z
$950 $1200 7.07 $250 / 50
So, probability is very close to 1 or virtually certain.
- 94 -
(LO7-5)
(LO7-
18.
a. b.
80 12.649 40 320 330 z 0.79 80 / 40
sx
So, probability is 0.7852, found by 0.2852 + 0.5000. c.
d. e
z
350 330 1.58 80 / 40
So, probability is 0.7281, found by 0.2852 + 0.4429. 0.0571, found by 0.5000 0.4429. No assumptions are necessary as n > 30. (LO7-5)
19.
0.45(1 0.45) = 0.035178 200
(LO7-6)
20.
0.09(1 0.09) = 0.04047 50
(LO7-6)
21.
z = (0.06 − 0.02)/
0.02(1 0.02) = 2.02; then 0.5 − 0.4783 = 0.0217 50
(LO7-
z = (0.35 − 0.30)/
0.30(1 0.30) = 2.78; then 0.5 − 0.4973 = 0.0027 650
(LO7-
z = (0.80 − 0.75)/
0.75(1 0.75) = 2; then 0.5 − 0.4772 = 0.0228 300
(LO7-
0.25(1 0.25) = −2.58; then 0.5 − 0.4951 = 0.0049 500
(LO7-
6) 22. 6) 23.
6) 24.
z = (0.20 − 0.25)/ 6)
25.
a. b. c.
Formal Man, Summit Stationers, Bootleggers, Leather Ltd., Petries Answers will vary. Gap, Frederick’s of Hollywood, Summit Stationers, M Studios, Leather Ltd., Things Remembered, County Seat, Coach House Gifts, Regis Hairstylists (LO7-1)
26.
Answers will vary. One possible answer is to sample every 50th family. This would be a systematic random sample. (LO7-1)
- 95 -
27.
The difference between a sample statistic and the population parameter. Yes, the difference could be zero. The sample mean and the population parameter are equal. (LO7-2)
28.
1. 2. 3. 4. 5.
Results adequate Destructive nature of some tests Physically impossible to check all items Costly to check all items Time consuming to check all items (LO7-1)
29.
a.
The standard error of the mean declines as the sample size grows because the sample size is in the denominator and as the denominator increases the proportion decreases. If the sample size is increased, the Central Limit theorem guarantees the distribution of the sample means becomes more normal. The shape of the distribution becomes narrower since the dispersion is less and estimates of the mean are more precise. (LO7-3)
b. c.
30.
a.
b. c. d.
31.
Samples Mean 1,1 1.0 1,2 1.5 1,3 2.0 2,1 1.5 2,2 2.0 2,3 2.5 3,1 2.0 3,2 2.5 3,3 3.0 Mean of sample means is (1.0 + 1.5 + 2.0 + … +3.0)/ 9 = 18/9 = 2.0 The population mean is (1 + 2 +3)/3 = 6/3= 2.0 They are the same value. The dispersion for the population is equal that for the sample means. The population varies from 1 to 3, whereas the sample means also vary from 1 to 3. The shape of the sampling distribution of the sample mean is approximately bell shaped peaking at 2. The population is uniform between 1 and 3. (LO7-3)
Use of either a proportional or nonproportional stratified random sample would be appropriate. For example, suppose the number of banks in the financial district were as follows: Assets Number Percent of Total $500 million and more 20 2.0 $100-499 million 324 32.4 less than $100 million 656 65.6 1000 100 For a proportional stratified sample, if the sample size is 100, then two banks worth assets of $500 million and more, 32 medium-sized banks, and 66 small banks would be selected. For a nonproportional sample, 10 or even all-20 large banks could be selected, and fewer medium and small size banks could be selected and the sample results weighted by the appropriate percentage of the total. (LO7-1)
- 96 -
32.
A simple random sample would be appropriate, but this means each pipe would have to be numbered 1, 2, 3, …, 720. A faster method would be to (1) select a pipe from the first say, 20 pipes produced, and (2) select every 20th pipe produced thereafter and measure its inside diameter. Thus, the sample would include about 36 PVC pipes. (LO7-1)
33.
a. b. c.
We selected 60, 104, 75, 72, and 48. Answers will vary. We selected the third observation. So the sample consists of 75, 72, 68, 82, 48. Answers will vary. Use a stratified random sample. (LO7-1)
34.
a. b. c. d.
Answers will vary. Answers will vary. Answers will vary. For example, stratified random sampling by gender can be used. (LO7-1)
35.
a. b.
15 found by 6 C2 Sample Value 1 79,64 2 79,84 3 79,82 4 79,92 5 79,77 6 64,84 7 64,82 8 64,92 9 64,77 10 84,82 11 84,92 12 84,77 13 82,92 14 82,77 15 92,77
c.
x
Sum 143 163 161 171 156 148 146 156 141 166 176 161 174 159 169
Mean 71.5 81.5 80.5 85.5 78.0 74.0 73.0 78.0 70.5 83.0 88.0 80.5 87.0 79.5 84.5 1195.0
1195 = 79.67 15
= 478/6 = 79.67 d.
36.
a. b.
They are equal. No, the student is not graded on all available information. He/she is as likely to get a lower grade based on the sample as a higher grade. By dropping a lower grade, the average is 82.8. This is preferable. (LO7-3) 10, found by 5 C2 Sample Value 1 2,3 2 2,5 3 2,3 4 2,5 5 3,5 6 3,3 7 3,5
Sum 5 7 5 7 8 6 8
Mean 2.5 3.5 2.5 3.5 4.0 3.0 4.0
- 97 -
8 9 10
5,3 5,5 3,5
8 10 8
4.0 5.0 4.0 36.0
x = 36/10 = 3.6
c.
=18/5 = 3.6
They are equal. 37.
a. b.
(LO7-3)
10, found by 5 C2 Shutdowns Mean 4,3 3.5 4,5 4.5 4,3 3.5 4,2 3.0 3,5 4.0
Shutdowns 3,3 3,2 5,3 5,2 3,2
Mean 3.0 2.5 4.0 3.5 2.5
x = (3.5 + 4.5 + … + 2.5)/10 = 3.4
c.
= (4 + 3 + 5 + 3 + 2)/5 = 3.4 The two means are equal. The population values are uniform in shape. The distribution of the sample means tends toward normality. See the following charts. (LO73)
d.
Histogram
Histogram
45
35
40
30
35
25 Percent
25 20 15
20 15
Data
38.
a. b.
c.
5. 0
4. 5
4. 0
3. 5
3. 0
6. 0
5. 0
4. 0
0 3. 0
5
0 2. 0
5
2. 5
10
10
2. 0
Percent
30
Mean
15, found by 6 C2 # Sold Mean # Sold Mean 54,50 52 50,52 51 54,52 53 52,48 50 54,48 51 52,50 51 54,50 52 52,52 52 54,52 53 48,50 49 50,52 51 48,52 50 50,48 49 50,52 51 50,50 50 Sample Means Frequency 49 2 50 3 51 5 52 3
- 98 -
Probability 0.1333 0.2000 0.3333 0.2000
53 x = 51
2
0.1333
d.
=51
e.
The shape of the population distribution is tending toward normal. Histogram 35 30
Percent
25 20 15 10 5
55 .0
54 .0
53 .0
52 .0
51 .0
50 .0
49 .0
48 .0
0
Data
f.
The shape of the distribution of the sample mean is somewhat normal. Histogram 35 30
Percent
25 20 15 10 5
54 .0
53 .0
52 .0
51 .0
50 .0
49 .0
0
Mean
(LO7-3) 39.
a.
The distribution of sample mean will be normal.
b.
x
c.
5.5 1.1 25 36 35 z 0.91 5.5 25
So, probability is 0.1814, found by 0.5000 − 0.3186.
40.
34.5 35 0.45 5.5 25
d.
z
e.
So, probability is 0.6736, found by 0.5000 + 0.1736. 0.4922, found by 0.3186 + 0.1736
a.
The distribution of sample mean will be normal.
b.
x
c.
z
8 2 16 140 135
8 16
2.5
So, probability is 0.0062, found by 0.5000 0.4938.
- 99 -
(LO7-5)
41.
128 135 3.5 8 16
d.
z
e.
So, probability is 0.9998, found by 0.5000 + 0.4998. 0.9936, found by 0.4998 + 0.4938. (LO7-5)
z
$335 $350 2.11 $45 40
So, probability is 0.9826, found by 0.5000 + 0.4826. 42.
(LO7-5)
$40 000 5656.85 50
a.
sx
b.
The distribution will be normal.
c.
z
$112 000 $110 000 0.35 $40 000 50
So, probability is 0.3632, found by 0.5000 − 0.1368.
43.
$100 000 $110 000 1.77 $40 000 50
d.
z
e.
So, probability is 0.9616, found by 0.5000 + 0.4616. 0.5984, found by 0.4616 + 0.1368. (LO7-5)
z
31.5 30.6 2 .5
60
= 2.79
So, probability is 0.9974, found by 0.5000 + 0.4974. 44.
(LO7-5)
$20 $21 $23 $21 1.11 and z2 2.21 $3.5 15 $3.5 15
a.
z1
b.
So probability is 0.8529, found by 0.3665 + 0.4864. Since the sample size is small, you assume the population is normally distributed. (LO7-5)
45.
Between 2679 and 2721, found by 2700 ± 1.96(68/ 40 ).
46.
a.
z
(LO7-5)
$25 $23.5 2.12 $5 50
So, probability is 0.0170, found by 0.5000 0.4830. b.
z
$22.5 $23.5 1.41 $5 50
So, probability is 0.9037, found by 0.4207 + 0.4830. c.
$5 . 50
Between $22.33 and $24.67, found by $23.50 1.65
5)
- 100 -
(LO7-
47.
z
900 947 1.78 205 60
So, probability is 0.0375, found by 0.5000 − 0.4625. 48.
a. b.
36 Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
1st roll 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6
2nd roll 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6
Mean 1 1.5 2 2.5 3 3.5 1.5 2 2.5 3 3.5 4 2 2.5 3 3.5 4 4.5 2.5 3 3.5 4 4.5 5 3 3.5 4 4.5 5 5.5 3.5 4 4.5 5 5.5 6
c.
- 101 -
(LO7-5)
6 5
Frequency
4 3 2 1 0 1
2
3
4
5
6
1st roll 6
Frequency
5 4 3 2 1 0 1.0
d.
49.
z
1.5
2.0
2.5
3.0
3.5 4.0
73 69 12.5
50
5.0
5.5
6.0
= 2.26
So, probability is 0.0119, found by 0.5 – 0.4881.
50.
4.5
Mean Both means are 3.5. The standard deviation of individual rolls is 1.708, while the standard deviation of sample means is 1.225 (LO7-3)
(LO7-56
z = (0.5 − 0.35)/
0.35(1 0.35) = 2.04; then 0.5 − 0.4793 = 0.0207 42
(LO7-
z = (0.3 − 0.25)/
0.25(1 0.25) = 1.63; then 0.5 + 0.4484 = 0.9484 200
(LO7-
6) 51. 6)
- 102 -
52.
53.
54.
a.
z = (0.10 − 0.05)/
0.05(1 0.05) = 1.62; then 0.5 − 0.4474 = 0.0526 50
b.
z = (0.01 − 0.05)/
0.05(1 0.05) = −1.30; then 0.5 − 0.4032 = 0.0968 50
c.
0.0526 + 0.0968 = 0.1494
a.
z = (0.85 − 0.90)/
0.90(1 0.90) = −2.89; then 0.5 + 0.4981 = 0.9981 300
b.
z = (0.92 − 0.90)/
0.90(1 0.90) = 1.15; then 0.5 − 0.3749 = 0.1251 300
c.
0.4981 + 0.3749 = 0.8730
a.
x
b.
(LO7-6)
(LO7-6)
2 .1
0.2333 81 6 6.5 7 6.5 z1 2.14 and z 2 2.14 2.1 2.1 81 81
Probability is 0.9676, found by 0.4838 + 0.4838. c.
z1
6.25 6.5 6.75 6.5 1.07 and z 2 1.07 2.1 2.1 81 81
Probability is 0.7154, found by 0.3577 + 0.3577.
55.
d. a. b. c.
56.
57.
a. b. c.
1.5/ 100 = 0.15 z = (6.25 – 6)/0.15 = 1.67; p = 0.4525 0.5 − 0.4082 = 0.0918; found by z = (5.8 – 6)/0.15 = −1.33; p = 0.4082 5)
a.
d.
0.5 − 0.4525 = 0.0475 = 4.75%; found by z = ($675 – $650)/($120/ 64 ) = 1.67; p = 0.4525 0.5 + 0.2486 = 0.7486 = 74.86%; found by z = ($660 – $650)/($120/ 64 ) = 0.67; p = 0.2486 0.4082 − 0.2486 = 0.1596 = 15.96%; found by z = ($670 – $650)/($120/ 64 ) = 1.33; p = 0.4082 0.5 + 0.0475 = 0.5475 = 54.75% (LO 7-5)
a.
x
b. c.
58.
Probability is 0.0162, found by 0.5000 − 0.4838. (LO7-5) 20/ 36 = 3.33 z = (490 – 500)/3.33 = −3.00 and z = (510 – 500)/3.33 = 3.00; p = 0.4987(2) = 0.9974 0.4987 + 0.5 = 0.9987 (LO 7-5)
$100 $12.91 60
- 103 -
(LO 7-
b.
c.
d.
59.
a. b.
c.
d.
60.
$550 $502 3.72 $100 60 So, probability is virtually 0. z
(LO 7-5)
$0.33 $0.052 40 $3.46 $3.50 $3.54 $3.50 z1 0.77 and z2 0.77 $0.052 $0.052 Probability is 0.5588, found by 0.2794 + 0.2794. $0.01 $0.01 z1 0.19 and z2 0.19 $0.052 $0.052 Probability is 0.1506, found by 0.0753 + 0.0753. $3.60 $3.50 z 1.92 $0.052 Probability is 0.0274, found by 0.5000 – 0.4726. (LO 7-5)
x
c.
600 510 19.93 14.28 10 So, probability is virtually 0. 500 510 z 2.21 14.28 10 Probability is 0.9864, found by 0.4864 + 0.5000. 0.0136, found by 0.5000 – 0.4864. (LO 7-5)
a.
p = 420/1400 = 0.3
b.
0.0122, found by
c.
0.5 − 0.4599 = 0.0401, found by
a.
b.
61.
$477 $502 $527 $502 1.94 and z2 1.94 $100 $100 60 60 So, probability is 0.9476, found by 0.4738 + 0.4738. $492 $502 $512 $502 z1 0.77 and z2 0.77 $100 $100 60 60 So, probability is 0.5588, found by 0.2794 + 0.2794. z1
z
0.3(1 0.3) 1400 390 1400 0.3 1.75 ; p = 0.4599 0.0122
6) 62.
a.
p = (600 + 150)/2000 = 0.375
- 104 -
(LO 7-
0.375(1 0.375) 2000
b.
0.0108, found by
c.
0.4896 − 0.3770 = 0.1126; found by
700 2000 0.375 2.31 ; p = 0.4896; 0.0108
725 2000 0.375 1.16 ; p = 0.3770; 0.0108 63.
0.02 0.05 = −0.92; p = 0.3212 0.05(1 0.05) 45 0.10 0.05 = 1.54; p = .4382 0.05(1 0.05) 45 0.0 0.05 = −1.54; p = 0.4382 (LO 7-6) 0.05(1 0.05) 45
a. 0.5 − 0.3212 = 0.1788, found by
b. 0.5 − 0.4382 = 0.0618, found by
c. 0.5 − 0.4382 = 0.0618, found by
64.
a. 0.4738, found by
(LO 7-6)
115 250 0.40 = 1.94 0.4(1 0.4) 250
b. 0.5 – 0.4951 = 0.0049, found by
120 250 0.40 = 2.58; p = 0.4951 0.4(1 0.4) 250 90 250 0.40
c. 0.0985 + 0.0049 = 0.1034, found by
0.4(1 0.4) 250
p = 0.4015; 0.5 − 0.4015 = 0.0985 6)
= −1.29.
(LO 7-
65. a. µ = 41.5, σ = 3.23
- 105 -
Average Age of Employees 50
Frequency
40 30 20 10 0 28-31
31-34
34-37
37-40
40-43
43-46
46-49
Average Age of Employees
Most of the values occur towards the right, leaving a few categories at the beginning. The histogram appears to not follow a normal distribution. It resembles a negatively skewed distribution. b. Sample mean of 7 employers = 43. z = (43 − 41.5)/(3.23/√ ) = 1.23 The answer is 0.5 − 0.3907 = 0.1093
(LO7-5)
66. µ = $164.1 M, σ = $41.8 M
a.
Team Revenue 12
Frequency
9
6
3
0 100-135
135-170
170-205
205-240
240-275
Team Revenue in $ Millions
Most revenues happen in the first two categories. The population does not follow a normal distribution.
- 106 -
The Distribution is positively skewed also known as skewed to the right. b.
Sample mean of last ten teams is 173. z = (173 − 164.1)/(41.8/√ ) = 0.67. The answer is 0.5 − 0.2486 = 0.2514 (LO7-5)
CHAPTER 8 ESTIMATION AND CONFIDENCE INTERVALS 1. 2)
$51 314.286 and $58 685.714, found by $55 000 2.58($10 000 / 49)
2.
38.911 and 41.089, found by 40 1.96 5/ 81
3.
a. b. c.
1.581, found by X 5 / 10 The population is normally distributed and the population variance is known. 16.901 and 23.099, found by 20 3.099 (LO8-2)
4.
a. b. c.
3.953, found by 25 / 40 68.498 and 81.502, found by 75 1.645(3.953) Increase the confidence level (LO8-2)
5.
a.
0.95, found by 4.75/ 25
b. c.
14.637 to 19.064, found by 16.85 ± 2.33(4.75/ 25 ) Decrease the confidence level (LO8-2)
(LO8-
(LO8-2)
6.
1.44
7.
a. b.
$20. It is our best estimate of the population mean. . ($5 / 49 ) . About 95 percent of the $18.60 and $21.40, found by $20 196 intervals similarly constructed will include the population mean. (LO81&2)
8.
a.
. ($5 / 64 ) $18.775 and $21.225, found by $20 196
b.
The confidence interval is based on the standard error computed by s / n . As n, the sample size, increases (in this case from 49 to 64) the standard error decreases and the confidence interval becomes smaller. (LO8-2)
a.
40 litres
b. c.
36.669 and 43.331litres, found by 40 ± 2.58(10 / 60 ) If 100 such intervals were determined, the population mean would be included in about 99 intervals. (LO8-1&2)
9.
10.
(LO8-2)
. (2.44 / 40 ) 5.294 and 6.806 errors, found by 6.05 196
- 107 -
(LO8-2)
11.
a. b. c.
2.201 1.729 3.499
(LO8-3)
- 108 -
12.
a. b. c.
2.145 2.500 1.796
13.
a. b. c.
5.870, found by 26.25/ 20 64.851 to 85.149, found by 75 ± 1.729(5.870) Increase the confidence level (LO8-3)
14.
a. b. c.
0.950, found by 4.75/ 25 14.483 to19.217, found by 16.85 ± 2.492(0.950) Decrease the confidence level (LO8-3)
15.
a. b. c.
The population mean is unknown, but the best estimate is 20, the sample mean. Use the t distribution as the population standard deviation is unknown. However, we must assume that the population is normally distributed. 2.093
d.
Between 19.064 and 20.936, found by 20 2.093
e.
Neither value is reasonable because they are not inside the interval.
a. b. c.
The population mean is unknown, but the best estimate is 27, the sample mean. Use the t distribution as the standard deviation is unknown and the population is normally distributed. 1.753
d. e.
Between 23.056 and 30.944 kg, found by 27 ± 1.753(9 / 16 ). That value is reasonable because it is inside the interval. (LO8-3)
(LO8-3)
2 . 20 (LO8-
3) 16.
5.54 . 10
(LO8-3)
6.02 . 15
(LO8-3)
17.
Between 95.389 and 101.811, found by 98.6 1.833
18.
Between 30.991 and 39.149, found by 35.07 2.624
19.
a.
0.375, found by 75/200
b.
0.0342, found by
c. d.
0.319 to 0.431, found by 0.375 ± 1.645(0.0342) If 200 such intervals were determined, the population proportion would be included in about 180 intervals. (LO8-4)
a.
73% = 0.73
b.
0.0199, found by
c.
0.691 to 0.769, found by 0.73 ± 1.96(0.0199)
20.
(0.375)(1 0.375) = 0.0342 200
(0.73)(1 0.73) = 0.0199 500
- 109 -
21.
22.
23.
24.
d.
If 500 such intervals were determined, the population proportion would be included in about 475 intervals. (LO8-4)
a.
0.8, found by 80/100.
b.
0.04, found by
c.
Between 0.722 and 0.878, found by 0.8 1.96
d.
We are reasonably sure the population proportion is between 72% and 88%.
a.
0.75, found by 300/400.
b.
0.0217, found by
c.
Between 0.694 and 0.806, found by 0.75 2.58
d.
We are reasonably sure the population proportion is between 69% and 81%.
a.
0.625, found by 250/400.
b.
0.0242, found by
c.
Between 0.563 and 0.687, found by 0.625 2.58
d.
We are reasonably sure the population proportion is between 56% and 69%.
a.
0.05, found by 15/300.
b.
Between 0.025 and 0.075, found by 0.05 1.96
c.
No, because 0.10 is not in the interval.
(LO8-4)
0.8(1 0.8) . 100 0.8(1 0.8) . 100
(LO8-4)
0.75(1 0.75) . 400 0.75(1 0.75) . 400 (LO8-4)
0.625(1 0.625) . 400 0.625(1 0.625) . 400
0.05(1 0.05) . 300
5 300 36 36 300 1
25.
33.410 and 36.590, found by 35 ± 2.030
26.
$36.722 and $43.278, found by $40 ± 2.682
(LO8-4)
(LO8-6)
9 500 49 49 500 1
The advertisement is likely true because $50 is more than the upper limit of the confidence interval. (LO8-6)
27.
0.50 400 50 50 400 1
1.683 up to 2.037, found by 1.86 ± 2.680
- 110 -
(LO8-6)
0.2(1 0.2) 900 60 60 900 1
28.
0.102 to 0.298, found by 0.2 ± 1.96
29.
0.023 to 0.110, found by 0.067 ± 1.645
(LO8-6)
0.067(1 0.067) 500 75 75 500 1
10/75 = 0.13; therefore, it would not be reasonable as 0.13 is not in the CI.
(LO8-
6) 0.433 and 0.767, found by 0.60 1.96
(0.60)(0.40) 300 30 30 300 1
31.
1.96 30 55, found by = 54.0225 8
(LO8-5)
32.
2.58 15 60, found by = 59.9076 5
(LO8-5)
33.
196, found by n = 0.15(0.85)
34.
2.58 165, found by n = 0.45(0.55) = 164.7459 0.10
35.
1.96 3 554, found by = 553.1904 0.25
(LO8-5)
36.
26, found by
1.96 0.23 = 25.0889 0.09
(LO8-5)
30.
(LO8-
6) 2
2
2
1.96 = 195.9216 0.05
(LO8-5)
2
(LO8-5)
2
2
2
37.
a.
1.96 577, found by 0.60(0.40) = 576.24 0.04
b.
1.96 601, found by 0.50(0.50) = 600.25 0.04
a.
5683, found by 0.30(0.70)
2
(LO8-5)
2
38.
1.645 = 5682.6525 0.01
- 111 -
(LO8-5)
b.
Increase the allowable error from 0.01 to 0.05 (and/or reduce the confidence level). Thus, the sample size would be reduced to 228, found by 0.30(0.70) 2
1.645 = 227.3061 0.05 39.
6.133 years to 6.867 years, found by 6.5 1.989(1.7 / 85 )
40.
a. b.
1.401 kg (LO8-3) 1.398 to 1.404 kg, found by 1.401 2.030(0.01/ 36) About 95 percent of similarly constructed intervals would include the population
a. b. 3)
Between $1.273 to $1.327, found by 1.30 ±2.680(0.07/ 50 ) $1.40 is not reasonable because it is outside of the confidence interval.
(LO8-3)
mean. 41.
42.
6.2 . 50
Between 24.238 and 27.762, found by 26 2.01
28 is not reasonable because it is outside of the confidence interval. 3) 43.
3 . 40
Between 7.201 and 8.799, found by 8 1.685
b.
9 is not reasonable because it is outside of the confidence interval. 3)
2.545 up to 2.975 meals, found by 2.76 2.224(0.75/ 60)
45.
a. b. c.
(LO8-
(LO8-3)
65.492 up to 71.708 hours, found by 68.6 ± 2.680(8.2/ 50 ) The value suggested by the manager is included in the confidence interval. Therefore, it is reasonable. Changing the confidence level to 95% would reduce the width of the interval. The value of 2.608 would change to 2.010. (LO8-3)
a.
From $1621.75 to $2018.25, found by $1820 ± 2.015($660/ 45 )
b.
The population mean could be $1700 because it is in the interval constructed
above.
1.96 16 = 61.5 4 2
47.
(LO8-
a.
44.
46. 3)
(LO8-
62, found by n =
(LO8-5)
- 112 -
(LO8-
48.
$125 . 20
Between $231.499 and $348.501, found by $290 2.093
(LO8-
3) 49.
$1892 . 25
Between $13 733.558 up to $15 028.442, found by $14 381 1.711 $15 000 is reasonable because it is inside of the confidence interval. 3)
(LO8-
5 . 15
50.
Between 20.612 and 27.388, found by 24 2.624
51.
a.
$62.583, found by $751/12.
b.
Between $60.541 and $64.626, found by $62.583 1.796
c.
$60 is not reasonable because it is outside of the confidence interval.
a.
2408.8, found by 24 088/10.
b.
Between 2191.039 and 2626.561, found by 2408 .8 2.262
a.
89.467, found by 1342/15.
b.
Between 84.992 and 93.942, found by 89.4667 2.145
c. 3)
Yes, because even the lower limit of the confidence interval is above 80. (LO8-
(LO8-3)
$3.94 . 12 (LO8-
3) 52.
304.4 . 10
(LO8-
3) 53.
8.08 . 15
0.69(1 0.69) . 600
54.
Between 0.641 and 0.739, found by 0.69 2.58
55.
Do not use PCF as n < 5% N. Between 0.647 and 0.753, found by 0
(LO8-4)
0.7(1 0.7) . .7 2.58 500 Yes, because even the lower limit of the confidence interval is above 0.500.
(LO8-4
& 6) 56.
a. b.
0.56(0.44) 1000 The lower point of the interval is greater than 0.50. So we can conclude the majority feel the President is doing a good job. (LO8-4) . p = 560/1000 = 0.560, from 0.529 up to 0.591, found by 0.56 196
- 113 -
57.
$52.508 and $55.492, found by $54.00 ± 2.032
$4.50 (500 35) 500 1 35
(LO8-
6)
(0.52)(0.48) (650 50) 50 650 1
58.
0.345 and 0.695, found by 0.52 2.58
59.
369, found by n = 0.60(1 – 0.60)[1.96/0.05]2 = 368.794
60.
133, found by {(1.645 14)/2}2 = 132.595
61.
1.96 500 97, found by = 96.04 100
62.
1.96 865, found by 0.10(0.90) = 864.36 0.02
(LO8-6) (LO8-5)
(LO8-5)
2
(LO8-5)
2
63.
(LO8-5)
a. b.
708.135, rounded up to 709, found by: 0.21(1 – 0.21)[1.96/0.03]2 1068, found by 0.50(1 – 0.50)[1.96/0.03]2
a.
25%, found by 25/100
b.
. 0.172 to 0.328, found by 0.25 196
(LO 8-
5) 64.
(0.25)(0.75) 605 100 The interval is for 100 604
those who do not use. Since 0.40 is not in the interval, then at least 0.40 will use the ATM. c. 1.65, found by 165/100 14659 . 505 . 196 . d. 1.387 to 1.913, found by 165 Note: s = 1.4659 100 604 e. No, because 0 is not in the interval between 1.387 and 1.913. (LO86)
65.
0.613(1 0.613) . Yes, because 1000
Between 0.573 and 0.653, found by 0.613 2.58
even the lower limit of the confidence interval is above 0.500.
(LO8-4)
66.
Between $29 688.967 and $34 311.033, found by $32 000 ± 1.691($8200/ 36 ) (LO83)
67.
Between 12.69 and 14.11 hours, found by 13.4 1.96
6.8 . 352
- 114 -
(LO8-3)
2
1.96 . 0.02
68.
2185, found by 0.35 1 0.35
69.
a.
$30, found by $ 180
b.
Between $355.10 and $476.90, found by $416 2.030
c.
1245, found by
a.
Between 7849.05 and 8150.95 feet, found by 8000 2.756
b.
554, found by {(1.96 300)/25}2 = 553.19
a.
Between $313.41 and $332.59, found by $323 2.426
b.
$350 is not reasonable because it is outside of the confidence interval. (LO 8-3)
a.
The population mean is unknown, but close to 8.32 years. 3.07 Between 7.502 and 9.138 years, found by 8.32 1.685 . 40 No, it is not reasonable because 10 is above the upper limit of 9.14 years. (LO 83)
70.
71.
72.
b. c.
73.
a. b. c.
74.
75.
76.
(LO8-5)
36
$180 36
1.96($180) $10
2
(LO8-3&5)
300 . 30
(LO8-3&5)
$25 . 40
65.492 up to 71.708 hours, found by 68.6 2.68(8.2 / 50) The value suggested by the NCAA is included in the confidence interval, therefore it is reasonable. Changing the confidence interval to 95% would reduce the width of the interval. The value of 2.680 would change to 2.010.(LO 8-3)
a.
$65 is the best estimate of the unknown population mean.
b
Between $61.504 and $68.496, found by 65 2.797
c.
151, found by {(1.96 6.25)/1}2 = 150.063
a.
Between 75.442 and 80.558 hours, found by 78 2.01
b.
220, found by {(1.645 9)/1}2 = 219.188
6.25 . 25
(LO 8-3&5)
9 . 50 (LO 8-3&5)
The confidence interval is between 0.011 and 0.059, found by 0.035(1 0.035) 0.035 2.58 . It would be reasonable to conclude that less than 5% 400
- 115 -
of the employees are now failing the test because 0.05 is inside the confidence interval. (LO 8-4) 77.
The confidence interval is between 0.021 and 0.106, found by 0.064(1 0.064) 0.064 2.58 . 220 It would be reasonable to conclude that more than 10% the applicants are now failing the test because 0.10 is inside the confidence interval. (LO 8-4)
78.
Between 0.608 and 0.692, found by 0.65 1.96
0.65(0.35) . 500
The lower point of the interval is greater than 0.50. So we can conclude the majority do not approve.(LO 8-4)
$1000 . 20
a.
Between $10 592.384 and $11 365.616, found by $10 979 1.729
b.
107, found by {(2.58 1000)/250}2 = 106.502
80.
a. b.
708.13, rounded up to 709, found by: 0.21(1 – 0.21)[1.96/0.03]2 1068, found by 0.50(1 – 0.50)[1.96/0.03]2 (LO 8-5)
81.
a.
n = 40, first 40 companies, sample mean = 9452.425, s = 14 023.5606, standard error = 2217.3196, t = 2.023. 95% CI: 4966.787 ≤ µ ≤ 13 938.063 The population mean = 7416.44 which is in the interval. (LO8-3) n = 40, first 40 companies, sample mean = 2067.85, s = 5602.96, standard error = 885.9058, t = 2.023. 95% CI: 275.663 ≤ µ ≤ 3860.037 The population mean = 1929.63 which is in the interval. (LO8-3) n = 40, x = 19 are public, p = 0.475, standard error = √
79.
b.
c. 8,
(LO 8-3 & 8-5)
95% CI for P: 0.32 ≤ P ≤ 0.63 found by: 0.475 ± 1.96(0.0790) The population proportion = 47/100, 47% which is in the interval.
(LO8-
4) 82.
a.
n = 7, sample mean of Canadian teams = 76.29, s = 7.34. 95% CI for the Population Mean: 70.89 to 81.68 points found by 76.29 ±
1.943(7.34/√ ) b. Yes, it is reasonable that the True Population Mean is 81 points since 81 falls within the confidence interval found in part (a). The True population mean is 77.9 which is also in the interval. (LO8-3) 83.
a.
Considering the selected 42 cities as a sample; the sample mean is $580 742.86. The sample standard deviation is $305 005.28 and the standard error = $47
063.34
b.
The 80% confidence interval lower bound is $519 419.33 and upper bound is $642 066.38 found by $580 742.86 ± 1.303($47 063.34) (LO8-3) $500 000 does not fall in the confidence interval found above. Based on an 80% confidence interval it is not reasonable to say that the population mean is $500 000.
- 116 -
CHAPTER 9 ONE-SAMPLE TESTS OF HYPOTHESIS 1.
a. b. c. d.
2.
a. b. c. d.
3.
a. b c.
d. from 50. e.
4.
a. b.
5.
a. b.
H0: µ = 2 H1: µ ≠ 2 H0: µ ≥ 40 H1: µ < 40 H0: µ ≤ $1750 H1: µ > $1750 H0: µ = $3.25 H1: µ ≠ $3.25
(LO9-5)
H0: µ ≥ 1.5 H1: µ < 1.5 H0: µ ≥ 1500 H1: µ < 1500 H0: µ = $2000 H1: µ ≠ $2000 H0: µ = 500 mL H1: µ ≠ 500 mL
(LO9-5)
Two-tailed Reject H0 and accept H1 when z does not fall in the region from –1.96 and 1.96 49 50 1.2, found by z (5 / 36) There is not enough evidence to reject H0 and conclude the mean is not different p = 0.2302, found by 2(0.5000 – 0.3849). A 23.02% chance of finding a z value this large when H0 is true. (LO9-6 & 9-7)
One-tailed (LO9-6 & 9-7) Reject H0 and accept H1 when z > 2.33 12 10 c. 4, found by z (3 / 36) d. Reject H0 and conclude that the mean delivery time is greater than the advertised 10 days. e. p-value is very close to 0, given a z value of 4.00. Very little chance H0 is true (LO 9-6 & 9-7)
c. d.
One-tailed Reject H0 and accept H1 when z > 1.645 21 20 1.2, found by z (5 / 36) There is not enough evidence to reject H0 at the 0.05 significance level so we can conclude that wait times have not increased.
- 117 -
e. p = 0.1151, found by 0.5000 – 0.3849. An 11.51% chance of finding a z-value this large or larger. (LO9-6 & 9-7) 6.
a. b. c. d. e. & 9-7)
7.
8.
b.
H0: = 96 600 H1: 96 600 Reject H0 if z < 1.96 or z > 1.96
c.
0.69, found by z
d. e.
There is not enough evidence to reject H0. p = 0.4902, found by 2(0.5000 – 0.2549). Crosset’s experience is not different from that claimed by the manufacturer. If the H0 is true, the probability of finding a value more extreme than this is 0.4902. p-value from computer is 0.4884. (LO9-6 & 9-7)
a.
H0: 3 H1: < 3; choose α = 0.05 Reject H0 if z < 1.645 2.75 3.0 1.77, found by z (1 / 50 ) Reject H0 p = 0.0384, found by (0.5000 – 0.4616). We conclude that the mean waiting time is less than three minutes. When H0 is true, the probability of obtaining a value smaller than 1.77 is 0.0384. Computer p-value is 0.0385. (LO9-6 & 9-7)
a.
b. c. d. e.
9.
a. b. c. d. e.
10.
One-tailed Reject H0 and accept H1 when z < 1.88 215 220 2.67, found by z (15 / 64 ) Reject H0 and conclude that the population mean is less than 220 at the 0.03 significance level. p = 0.0038, found by 0.5000 – 0.4962. Less than 0.5% chance H0 is true (LO9-6
a. b.
95 795 96 600 8050 48
H0: 6.8 H1: < 6.8 Reject H0 if z < 1.645 6.2 6.8 7.2, found by z (0.5 / 36 ) Reject H0 at a significance level of 0.05. p-value is almost zero; the mean number of movies watched is less than 6.8 per month. If H0 is true, there is virtually no chance of getting a statistic this small. (LO9-6 & 9-7) H0: 80 H1: > 80 Reject H0 if z > 2.33
- 118 -
$84.65 $80
c.
8.49, found by z
d. e.
Reject H0 at a significance level of 0.01. p-value is almost zero; the mean amount of tips per day is larger than $80. If
$3.24 / 35
H0 is true, the probability of obtaining a sample mean this far above 80 is virtually zero. (LO9-6 & 9-7) 11.
a.
c.
Reject H0 when t > 1.833 12 10 t 2.108 (3 / 10 ) Reject H0; the mean delivery time is greater than 10 days.
a.
Reject H0 if t < 3.106 or t > 3.106
b.
t
c.
Reject H0, the mean does not equal 400.
b.
(LO9-
8) 12.
13.
407 400 4.041 (6 / 12 )
(LO9-8)
Choose α = 0.05 H0: 40 H1: > 40 Reject H0 if t > 1.703
t
42 40 5.04 (2.1 / 28)
Reject H0 and conclude that the mean number of calls is greater than 40 per week. (LO98) 14.
H0: 42.3 H1: < 42.3 Reject H0 if t < 1.319
t
40.6 42.3 3.085 (2.7 / 24 )
Reject H0. The mean assembly time is less than 42.3 minutes. 15.
(LO9-8)
Ho: 35 600 H1: > 35 600 Reject H0 if t > 1.740
t
37 675 35 600 = 3.645 2415 / 18
Reject H0 and conclude that the mean life of the spark plugs is greater than 35 600 km, the claim is true. 16.
a. b.
(LO9-8)
The population of complaints should follow a normal probability distribution. The assumption of normality appears reasonable.
- 119 -
Histogram of Complaints, with Normal Curve
5
Frequency
4
3
2
1
0 10
11
12
13
14
15
16
Complaints
c.
H0: 15 H1: < 15 Reject H0 if t < –1.729
t
(13.5 15) 4.46 (1.504 / 20)
Reject the null hypothesis. The mean number of complaints is less than 15. (LO9-8) 17.
a.
b.
Reject H0 if t < 3.747
x 17 and s t
c.
(85) 2 5 3536 . 51
1495
17 20
190 . 3536 . / 5 Do not reject H0. We cannot conclude that the population mean wait time is less
than 20. d. Between 0.05 and 0.10, about 0.065; by computer, the p-value = 0.0653. (LO9-7 & 9-8) 18.
a. b. c. d. e.
Reject H0 if t < 2.571 or t > 2.571 111667 . 100 t 4.72 6.055 / 6 Reject H0. The population mean is not equal to 100 Less than 0.01 (between 0.001 and 0.01); using a computer, the p-value = .0052 105.31 to 118.02; the hypothesized mean does not fall in the confidence interval,
so H0 is rejected.
(LO9-7 & 9-8)
- 120 -
19.
H0: 4.35 H1: > 4.35 Reject H0 if t > 2.821 4.368 4.35 t 168 . (0.0339 / 10 ) Do not reject H0. The additive did not increase the mean weight of the puppies. The pvalue is between 0.10 and 0.05. Using a computer, the p-value is 0.0639. (LO9-7 & 9-8)
20.
H0: 2160 H1: > 2160 Reject H0 if t > 2.306 2172.44 2160 t 3.98 (9.3823 / 9 ) Reject H0. The mean chlorine shelf life has increased. The p-value is less than 0.005.
Using a computer, the p-value is 0.002.
(LO9-7 & 9-8)
21.
H0: 4.0 H1: > 4.0 Reject H0 if t > 1.796 4.50 4.0 t 0.65 (2.68 / 12 ) Do not reject H0. The mean number of kilometres travelled has not been shown to be greater than 4.0. The p-value is greater than 0.10. Using a computer, the p-value is 0.2657. (LO9-7 & 9-8)
22.
H0: 53 H1: > 53 Reject H0 if t > 1.761
t
56.4 53 3.52 (3.7378 / 15)
Reject H0. The mean number of surveys conducted is greater than 53. The p-value is less than 0.005. Using a computer, the p-value is 0.0017. (LO9-7 & 9-8) 23.
a. b.
c. 24.
a. b.
H0 is rejected if z > 1.645 0.75 0.70 1.09, found by z 0.70(0.30) 100 H0 is not rejected. (LO9-9) H0 is rejected if z < 1.96 or z > 1.96 0.30 0.40 2.24, found by z 0.40(0.60) 120
- 121 -
c.
H0: p 0.52 H1: p > 0.52 b. The level of significance selected is 0.01. c. H0 is rejected if z > 2.33 0.5667 0.52 d. 1.62, found by z 0.52(0.48) 300 e. H0 is not rejected. We cannot conclude that the proportion of men driving on Highway 400 is larger than 0.52. (LO9-9)
25.
a.
26.
a. b. c. d.
e. (LO9-9) 27.
H0 is rejected; the proportion of customers who prefer black teas is not 40%. (LO9-9)
a. b. c. d.
e.
H0: p 0.33 H1: p > 0.33 The level of significance selected is 0.02. H0 is rejected if z > 2.05 0.40 0.3333 2.00, found by z 0.3333(0.6667) 200 H0 is not rejected. The proportion of students with jobs is not larger than 1 in 3. H0: p 0.90 H1: p < 0.90 The level of significance selected is 0.01. H0 is rejected if z < 2.33 0.82 0.90 2.67, found by z 0.90(010 . ) 100 H0 is rejected. Less than 90% of the customers receive their orders in less than
10 minutes.
(LO9-9)
H0: p 0.50 H1: p < 0.50 b. Choose α = 0.05 c. H0 is rejected if z < 1.645 0.48 0.50 d. 0.40, found by z 0.50(0.50) 100 e. H0 is not rejected. The proportion of student changing their major has not decreased. (LO9-9)
28.
a.
29.
H0: = $60 000 H1: $60 000 Reject H0 if z < −1.645 or z > 1.645
- 122 -
z
$62 500 $60 000 $6000
120
= 4.56
Reject H0. We can conclude that the mean income is not $60 000. The p-value is very close to zero. The confidence interval is from $61 599 to $63 401. As $60 000 falls outside the confidence interval, the null hypothesis is rejected. (LO9-6 & 9-7) 30.
H0: 50 H1: < 50 Reject H0 if z < 2.33
a.
x 48.18 48.18 50 z 1.92 3 10
b.
c. & 9-7) 31.
H0 is not rejected. The mean weight is not less than 50 kg. Mr. Rutter can use the z distribution as the test statistic because the population follows a normal distribution and the population standard deviation ( = 3) is known. p-value = 0.5000 – 0.4726 = 0.0274 (by computer p-value = 0.0275) (LO9-6
H0: 10 H1: < 10 Reject H0 if z < 1.645
z
9 10 2.53 2.8 / 50
Reject H0. The mean weight loss is less than 10 pounds. The p-value = 0.5000 – 0.4943 = 0.0057. (LO9-6 & 9-7) 32.
H0: 450 H1: > 450 Reject H0 if z > 1.645
z
451.4 450 11.65 0.85 50
Reject H0. The cans are being overfilled. The p-value is very close to 0. & 9-7) 33.
(LO9-6
H0: 7 H1: < 7 Reject H0 if t < −1.677
t
(6.8 7) 1.57 0.9 / 50
The p-value is 0.0613, so at a significance level of 5%, do not reject the null hypothesis. University students sleep no less than the typical adult male. (LO9-7 & 9-8) 34.
H0: 90 H1: > 90 Reject H0 if t > 1.290
- 123 -
t
94 90 = 1.82 22 / 100
Reject H0. At the 0.10 level, we can conclude that the vacation property selling time has increased. The p-value is 0.0360. (LO9-8) 35.
H0: $1.25 H1: > $1.25 Reject H0 if t > 1.691
t
$1.27 $1.25 2.37 $0.05 / 35
Reject H0. The mean price of gasoline is greater than $1.25. The p-value is 0.0119 (by computer). (LO9-6 & 9-7) 36.
H0: = 40 H1: ≠ 40 Reject H0 if t < 2.001 or t > 2.001
t
37.8 40 = −1.40 12.2 / 60
H0 is not rejected. The p-value is 0.1677 (by computer). We cannot conclude that the mean leisure time is untrue. (LO9-6 & 9-7) 37.
H0: 14% H1: > 14% Reject H0 if t > 2.821
t
15.64 14 = 3.32 1.561 10
Reject H0. The mean rate charged is greater than 14 percent. 38.
(LO9-8)
H0: 6 H1: < 6 Reject H0 if t < –2.998 [Assuming the population is normally distributed]
t
5.6375 6 1.62 0.6346 / 8
Do not reject H0. The mean rate could be 6.0 percent. The p-value is 0.0751. & 9-8) 39.
(LO9-7
H0: = 1.92 H1: 1.92 Reject H0 if t < 2.201 or t > 2.201 x = 2.08667 s = 0.40484
t
2.08667 1.92 = 1.43 0.40484 12
Do not reject H0. There is no difference in the mean amount of water consumed at the college surveyed and the national average.
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The confidence interval is from 1.8294 to 2.3439. The hypothesized mean falls in the confidence interval, so H0 is not rejected. (LO9-6) 40.
H0: 25 H1: > 25 Reject H0 if t > 2.624 x = 26.07 s = 1.5337
t
26.07 25 2.69 1.5337 / 15
Reject H0. The mean number of patients per day is more than 25. The p-value is less than 0.01. Using a computer, the p-value is 0.0087. This is lower than the significance level, so, the null hypothesis is rejected. (LO9-7&8) 41.
H0: 6.5 H1: < 6.5 Reject H0 if t < 2.718 x = 5.1667 s = 3.1575
t
(LO9-8)
5.1667 6.5 1.46 3.1575 / 12
Do not reject H0. The mean is not less than 6.5. 42.
H0: 3.5 H1: < 3.5 Reject H0 if t < 1.746
t
2.9553 3.5 4.014 0.5596 / 17
Reject H0. The mean time to complete a game is less than 3.5 hours. The p-value (by computer) is 0.0005. (LO9-7 & 9-8) 43.
H0: = 0 H1: 0 Reject H0 if t < 2.110 or t > 2.110 x = 0.2322 s = 0.3120 0.2322 0 t 3158 . 0.3120 / 18 Reject H0. The mean gain or loss does not equal 0. The p-value is less than 0.01, but greater than 0.001, so the probability of no time gain or loss is very small. Using a computer, the p-value is 0.0057, and as this is less than the significance level, the null hypothesis is rejected. (LO9-7 & 9-8)
44.
H0: 4.50% H1: > 4.50% Reject H0 if t > 1.796 x = 4.5717 s = 0.2405
t
4.5717 4.50 1.03 0.2405 / 12
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Do not reject H0. The mean rate of return is not more than 4.50%. 8) 45.
H0: ≤ 100 H1: > 100 Reject H0 if t > 1.761. [Assuming the population is normally distributed]
t
109.4 100 3.65 9.96 / 15
Reject H0. The mean is greater than 100. 46.
(LO9-
(LO9-8)
Ho: $267 H1: > $267 Reject H0 if t > 2.681
t
$288.31 $267 3.421 $22.46 13
Reject the H0. The average amount spent on textbooks per semester has increased. pvalue < 0.005; using a computer, the p-value is 0.0025, and as this is less than the significance level, the null hypothesis is rejected. (LO9-7 & 9-8) 47.
H0: p 0.60 H1: p > 0.60 H0 is rejected if z > 2.33 0.70 0.60 z 2.89 0.60(0.40) 200 H0 is rejected. Ms. Dennis is correct. She can conclude that more than 60% of the accounts are in arrears for more than three months. (LO9-9)
48.
H0: p 0.55 H1: p > 0.55 H0 is rejected if z > 1.645 0.60 0.55 0.05 z 0.841 0.55(0.45) 0.0595 70 H0 is not rejected. We cannot conclude that more than 55% of the commuters would use the route. (LO9-9)
49.
H0: p 0.44 H1: p > 0.44 H0 is rejected if z > 1.645
z
0.48 0.44 2.55 0.44(0.56) 1000
- 126 -
H0 is rejected. We conclude that there has been an increase in the proportion of people wanting to go to Europe. (LO9-9) 50.
H0: p 0.10 H1: p > 0.10 H0 is rejected if z > 1.645 9 0.10 (LO9-7 & 9-9) z 50 189 . 0.10( 0.90) 50 H0 is rejected. More than 10% of the sets need repair. p-value = 0.5000 – 0.4706 = 0.0294
51.
H0: p 0.20 H1: p > 0.20 H0 is rejected if z > 2.33 56 0.20 200 z 2.83 0.20(0.80) 200 H0 is rejected. More than 20% of the owners move during a particular year. The p-value = 0.5000 – 0.4977 = 0.0023 (LO9-7 & 9-9)
52.
H0: $10 H1: < $10 Reject Ho if t < 1.895
x t
$78.3 = $9.7875 8
s
$5.889 $0.9172 8 1
$9.7875 $10
0.655 $0.9172 / 8 Do not reject H0 at a significance level of 0.05. We cannot conclude that the cost is less than $10,000. (LO9-8)
53.
H0: 42 H1: > 42 Reject H0 if t > 1.796 51 42 t 3.90 8 / 12 Reject H0. The mean time for delivery is more than 42 days. The p-value is less than
0.005; using a computer, the p-value is 0.0012. 54.
H0: 9 H1: > 9 Reject H0 if t > 2.998 x 9.488 s = 0.467
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(LO9-7 & 9-8)
9.488 9 2.95 0.467 / 8
t
Do not reject Ho. The mean waiting time does not exceed 9 minutes. The p-value is more than 0.01; using a computer, the p-value is 0.0107. (LO9-7 & 9-8) 55.
Ho: 8 H1: < 8 Reject H0 if t < 1.714
7.5 8 0.77 3.2 24
t
Do not reject H0. The waiting time is not less than 8 minutes. 56.
(LO9-8)
H0: p = 0.50 H1: p 0.50 Reject H0 if z is not between –1.96 and 1.96
0.482 0.5 1.14 (0.5)(0.5) 1002
z
Do not reject H0. There is not enough evidence to indicate that the survey results have changed. (LO9-9) 57.
H0: 50 H1: > 50 Reject H0 if t > 1.796 82.5 50 t 1.89 59.5 / 12
Reject H0 and conclude that the mean number of text messages is greater than 50. The pvalue (0.0425) is less than 0.05. There is a slight probability (less than one chance in 20) this could happen by chance. (LO 9-7 & 9-8) 58.
H0: p ≥ 0.0008 H1: p < 0.0008 H0 is rejected if z < –1.645
z
0.0006 0.0008 0.71 0.0008(0.9992) 10 000
H0 is not rejected. This data does not prove there is a reduced fatality rate. (LO 9-9) 59.
H0: = 1.5 H1: ≠1.5 Reject H0 if t is not between −3.25 and 3.25
t
1.3 1.5 0.703 0.9 10
Do not reject H0. This data does not contradict the publisher. (LO 9-8)
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60.
a.
This is a binomial situation with both the mean number of successes (6, found by 60(0.10)) and failures (54, found by 60(0.90)) greater than five. b. H0: p = 0.10 H1: p ≠ 0.10 c. Reject H0 if z is not between –1.960 and 1.960. 3 0.1 z 60 1.29 0.1(1 0.1) 60 We fail to reject the null hypothesis. This data does not show a lower failure rate at Short’s slots. d. The p-value is 0.1970, found by 2(0.5000 – 0.4015). A value this extreme will happen about once out of five times with that failure rate. (LO 9-7 & 9-9)
61.
H0: p ≤ 0.40 H1: p > 0.40 Reject H0 if z is greater than 2.33
z
(16 / 30) 0.40 1.49 (0.40(1 0.40)) / 30
Do not reject the null hypothesis. This data does not show college students are more likely to skip breakfast. (LO 9-9) 62.
H0: ≤ 2.5 H1: > 2.5 Reject H0 if t > 1.703 t=
2.86 2.5 1.59 1.20 28
Do not reject H0 and conclude that the mean number residents is not more than 2.5. (LO 9-8) 63.
H0: $370 H1: > $370 Reject H0 if t > 2.681 $388.31 $370 t 2.939 $22.46 13 Reject the null hypothesis. The fare is increased. p-value = 0.0062 < 0.01 (LO 9-7 & 9-8)
64.
H0: p ≥ 0.56 H1: p < 0.56 H0 is rejected if z < −1.28
180 0.56 300 The test statistic is 1.396, found by z 0.560.44 300
- 129 -
There is not enough evidence to reject H0. This indicates the proportion is at least 56% which supports the journal’s claim. The p-value is 0.9186 which is not < 0.10 confirming the decision to not reject H0. (LO 9-9) 65.
H0: p ≥ 0.5 H1: p < 0.5 H0 is rejected if z < –1.645
z
0.4 0.5
0.5(0.5) 200
2.828
H0 is rejected. The proportion of students who supports the coach is less than 50%. (LO 9-9) 66. equal
a. This is a binomial situation with both the mean number of successes and failures to 22.5, found by 0.5(45). b. H0: p = 0.50 H1: p ≠ 0.50 c. Distribution Plot Normal, Mean=0, StDev=1 0.4
Density
0.3
0.2
0.1
0.0
0.005
0.005 -2.576
0 z value
2.576
Reject Ho if z is not between –2.58 and 2.58. d.
31 0.5 45 z 2.534 0.5(1 0.5) / 45 Do not reject the null hypothesis. This data does not prove the coin flip is biased.
e.
67.
The p-value is 0.0114, found by 2(0.5000 – 0.4943). A value this extreme will happen about once out of ninety times with a fair coin. (LO 9-7 & 9-9)
a.
Sample mean = 7.397%, sample standard deviation = 9.861%. 95% confidence interval: 5.1% ≤ µ ≤ 9.7%, found by: 7.397% ± 1.984(9.861%/√
- 130 -
b.
c.
d.
68.
a.
b.
c.
69.
a.
b.
, t-stat = (7.397 − 10)/(9.861/√ ,t critical: (2 tail, α = 0.02, df = 99) = ±2.365. t-stat falls in the reject H0 area, we can conclude that the average amount of % contract employees does not equal 10%. %, t-stat = (7.397 − 6)/(9.861/√ t critical: (2 tail, α = 0.05, df = 99) = ±1.984. t-stat falls in the do not reject H0 area, we can conclude that the average amount of % contract employees is not different from 6%. The confidence interval supports this conclusion because 6% falls in the interval of: 5.4% ≤ µ ≤ 9.4% The p-value for part c, where the t-stat = 1.42, df = 99, 2 tails is 0.1597. Since the p-value is greater than alpha = 0.05, we do not reject H0 which supports our previous decision. (LO-9-7&9-8) Sample mean = $667.742, sample standard deviation = $335.78. H0: µ ≤ $500, H1: µ > $500 t-stat = ($667.742 − $500)/($335.78/√ t-critical: (0.01, df = 30, 1 tail) = 2.457. 2.78 > 2.457, therefore reject H0 and conclude that team worth is more than $500 million. The p-value is 0.0046 which is less than 0.01 which confirms our decision to reject H0. Sample mean = $164.065, sample standard deviation = $42.468 H0: µ ≤ $150 H1: µ > $150 t-stat = ($164.065 − $150)/($42.468/√ t-critical: (0.01, df = 30, 1 tail) = 2.457. 1.84 < 2.457, therefore do not reject H0 and conclude that the average team revenues are not more than $500 million. The p-value is 0.0375 which is not less than 0.01 which confirms our decision not to reject H0. H0: p ≤ 0.60 H1: p > 0.60 p = 20/31 = 0.645, z-stat = (0.645 − 0.60)/√ = 0.51 z-critical = 1.28. Since z-stat < z critical, we do not reject H0 and conclude we can not confirm that over 60% of the team have won the cup. The p-value of 0.3039 is greater than the significance level of 0.10 which confirms our decision not to reject the null hypothesis. (LO9-7, 9-8 & 9-9) Sample mean = $580 742.9, sample standard deviation = $305 005.3 H0: µ = $500 000 H1: µ ≠ $500 000 t-stat = ($580 742.9 – $500 000)/($305 005.3/√ t-crit (0.10, df = 41, 2 tail) = ±1.683. 1.716 > 1.683, therefore reject H0 and conclude that the mean city listing value does not equal $500 000. 95% confidence interval: $501 541.004 ≤ µ ≤ $659 944.71, found by $580 742.9 ± 1.683($305 005.3/√ . Note $500 000 is not in the 90% confidence interval. The p-value is 0.0938 which is less than the significance level of 0.10 which confirms our decision to reject H0. H0: p ≤ 0.10 H1: p > 0.10 p = 6/42 = 0.14286, z-stat = (0.14286 − 0.10)/√ = 0.93.
- 131 -
z-critical = 1.645. Since z-stat < z critical, we do not reject H0 and conclude we can not confirm that over 10% of home listings are over $1 000 000. The p-value of 0.1773 is greater than the significance level of 0.05 which confirms our decision not to reject the null hypothesis. (LO97, 9-8 & 9-9)
CHAPTER 10 TWO-SAMPLE TESTS OF HYPOTHESIS 1.
a. b.
d. e.
Two-tailed test Reject H0 if z < 2.05 or z > 2.05 102 99 2.59, found by z 52 62 40 50 Reject H0 p = 0.0096, found by 2(0.5000 – 0.4952)
a. b.
One-tailed test Reject H0 if z > 1.41
c.
0.61, found by z
c.
2.
d. e.
3.
(LO10-1)
2.67 2.59
(0.75)2 (0.66)2 65 50 Not enough evidence to reject H0 p = 0.2709, found by 0.5000 – 0.2291 p = 0.2719 by computer
(LO10-1)
Step 1: H0: 1 ≥ 2 H1: 1 < 2 Step 2: The 0.05 significance level was chosen Step 3: Reject H0 if z < 1.645 Step 4: 0.83, found by z
3.5 3.7
1.052 1.32 40
55
Step 5: There is not enough evidence to reject H0. Babies using the Gibbs brand did not gain less weight. p = 0.2033 found by 0.5000 – 0.2967, which is > than the significance level, and so, there is not enough evidence to reject H0. Using a computer, the p-value = 0.2037.(LO10-1) 4.
Step 1: H0: d = m H1: d ≠ m Step 2: The 0.05 significance level was chosen Step 3: Reject H0 if z less than 1.96 or greater than 1.96 Step 4: 1.43, found by z
595 610
482 42 2 35
40
- 132 -
Step 5: Not enough evidence to reject H0. There is no difference in the mean number of kms traveled per month. The p-value is 0.1528 found by 2(0.5000 – 0.4236), which is > than the significance level, and so, there is not enough evidence to reject H0. Using a computer, the p-value = 0.1525. (LO10-1)
- 133 -
5. Step 1: Two-tailed test. Because we are trying to show a difference exists between two means. H0: 1 = 2 H1: 1 ≠ 2 Step 2: The 0.05 significance level was chosen Step 3: Reject H0 if z < 1.96 or z > 1.96 Step 4:1.59, found by √
Step 5:Do not reject H0 at the 0.05 level. There is no difference in the mean amount of time spent on Facebook. The p-value = 2(0.5 0.4441) = 0.1118. Since this is > than the significance level we do not reject H0. (LO10-1) 6.
Step 1: H0: 1 ≤ 2 H1: 1 >2 Step 2: The 0.02 significance level was chosen Step 3: If z > 2.05, reject H0. 20.75 19.80 Step 4: z 2.09 (2.25)2 (1.90)2 40 45 Step 5: Reject H0. It is reasonable to conclude union nurses earn more. The p-value is 0.0183, which is < than the significance level, and therefore, H0 is rejected. (LO10-1)
7.
a. b. c.
d. e. (LO10-2) 8.
a.
2)
b. c.
d. e. level;
Reject H0 if t > 2.120 or t < 2.120 df = 10 + 8 – 2 = 16 2 2 (10 1)(4) (8 1)(5) s 2p 19.9375 10 8 2 23 26 t 1.416 1 1 19.9375 10 8 There is not enough evidence to reject H0. p-value is greater than 0.10 and less than 0.20. The actual p-value = 0.1759. Reject H0 if t > 1.697 or t < 1.697 df = 17 + 15 – 2 = 30 (LO102 2 (15 1)(12) (17 1)(15) s 2p 187.20 15 17 2 350 342 t 1.651 1 1 187.2 15 17 Do not reject H0. p-value is greater than 0.10 and less than 0.20, and so is > than the significance therefore, there is not enough evidence to rejectH0. The actual p-value = 0.1093 (LO10-
2)
- 134 -
9.
H0: w ≤ m H1: w > m df = 9 + 7 – 2 = 14 2 2 (7 1)(6.88) (9 1)(9.49) s 2p 71.749 792
10.
H0: s ≤ d H1: s > d df = 15 + 12 – 2 = 25 2 (15 1)(15.5) (12 1)(18.1)2 s 2p 278.69 15 12 2
11.
H0: s ≤ a H1: s > a df = 6 + 7 – 2 = 11 Reject H0 if t > 1.363 2 2 (6 1)(12.2) (7 1)(15.8) 142.5 130.3 t 1.538 s 2p 203.82 672 1 1 203.82 6 7 Reject H0. The mean daily expenses are greater for the sales staff. The p-value is between 0.05 and 0.10, which is < than the significance level, and so H0 is rejected. The actual pvalue = 0.0763. (LO10-2)
12.
H0: n ≤ t H1: n > t df = 8 + 12 – 2 = 18 2 2 (8 1)(34.4) (12 1)(22.8) s 2p 777.88 8 12 2
Reject H0 if t > 2.624 79 78 t 0.234 1 1 71.749 7 9 There is not enough evidence to reject H0. The mean grade of women is not higher than that of men. The p-value = 0.4090, which is > than the significance level and supports the decision. The actual p-value = 0.4091. (LO10-2) Reject H0 if t > 2.485 61 48.4 t 1.949 1 1 278.69 15 12 There is not enough evidence to reject H0. There is no difference in the mean amount of time spent watching television. The p-value is between 0.025 and 0.050, which confirms that there is not enough evidence to reject H0. The actual p-value = 0.0313. (LO10-2)
t
835.8 826.8
1 1 777.88 8 12
Reject H0 if t > 2.552
0.707
There is not enough evidence to reject H0. We cannot conclude the mean salary of nurses is higher. The p-value is greater than 0.10, which is > than the significance level, and so there is not enough evidence to reject H0. The actual p-value = 0.2443. (LO102) 13.
Reject H0 if t > 2.353
d
4 1 4
t
1 3.367
sd
4
38 4 2 4 = 3.367 3
= 0.594
There is not enough evidence to reject the H0. We cannot conclude there are more defects on shift I. (The actual p-value = 0.2971) (LO 10-3)
- 135 -
14.
Reject H0 if t < 2.776 or t > 2.776 (23)2 5 1.52 4
115
23 4.6 sd 5 4.6 t 6.767 1.52 / 5 Reject the H0. There is a difference in the mean number of citations given by the two officers. p-value is less than 0.01, but greater than 0.001. (The actual p-value = 0.0025) (LO 10-3) d
15.
H0: d 0 H1: d > 0 Reject H0 if t > 1.796 d = 25.917 sd = 40.791 25.917 t 2.20 Reject H0. The incentive plan resulted in an increase in daily 40.791/ 12 income. The p-value is 0.0250,which is < than the significance level, and so reject H0. (LO 10-3)
16.
H0: d ≥ 0 d = 3.625
H1: d < 0 sd = 4.8385 √
Reject H0 if t < 2.998
2.12
Do not reject H0. The p-value is about 0.0359.
(LO
10-3) 17.
a. b. c.
H0 is rejected if z > 1.645 70 90 0.64, found by pc 100 150 1.61, found by √
18.
d.
There is not enough evidence to reject H0.
e.
p-value = 0.5 – 0.4463 = 0.0537. The p-value is 0.0537 (by computer 0.0533), which is > than the significance level, and therefore, there is not enough evidence to reject H0. (LO10-5)
a.
H0 is rejected if z < 1.96 or z > 1.96 170 110 0.80, found by pc 200 150
b. c.
2.70, found by √
d.
H0 is rejected
- 136 -
e.
19.
p-value = 2(0.5 – 0.4965) = 0.0070. The p-value is < than the significance level, and therefore, H0 is rejected. (LO105)
H0: p1 = p2 H1: p1 ≠ p2 H0 is rejected if z < 1.96 or z > 1.96 24 40 pc 0.08 400 400 2.09, found by √
H0 is rejected. The proportion infested is not the same in the two fields. The p-value = 2(0.5 – 0.4817) = 0.0366. The p-value is < than the significance level, and therefore, H0 is rejected. (Computer p-value = 0.0371) (LO10-5) 20.
H0: p1 ≥ p2 H1: p1 < p2 0.05 is chosen as the significance level. H0 is rejected if z < 1.645 1530 2010 pc 0.59 3000 3000 12.60, found by √
H0 is rejected. The proportion of women who think men are thoughtful has declined. the p-value is very close to 0, which is much smaller than the significance level, and therefore, H0 is rejected. (LO10-5) 21.
H0: pc ≤ pd
H1: pc > pd pc
H0 is rejected if z > 2.05
168 200 0.2044 800 1000
√
There is not enough evidence to reject H0. There is no difference in the proportion of Conservatives and Liberals who favor lowering the standards. The p-value = 0.5 – 0.1985 = 0.3015. The p-value is > than the significance level, and therefore, there is not enough evidence to reject H0. (LO10-5) 22.
H0: ps = pm
H1: ps ≠ pm
H0 is rejected if z < 1.96 or z > 1.96
pc
120 150 0.27 400 600
√
There is not enough evidence to reject H0. There is no difference in the proportion of married and single drivers who have accidents. The p-value = 2(0.5 – 0.4591) = 0.0818. The p-value is > than the significance level, and therefore, there is not enough evidence to reject H0. (LO10-5) 23.
a) H0: 1 = 2
H1: 1 ≠ 2
- 137 -
b) z
24.51 22.69
4.482 3.86 2 35
= 1.87
40
c)The p-value is 0.0614 from 2(0.50 0.4693) d) Reject H0 if z < 2.58 or z > 2.58. z statistic is in the Do not RejectH0 area. There is not enough evidence to reject the null hypothesis. e) There is no difference in the means. Also, the p value of 0.0614 is >than the significance level, and confirms that there is not enough evidence to reject H0. (LO10-1) 24.
25.
26.
a) H0: 1 ≥ 2 H1:1 < 2 4.35 5.84 b) z 5.44 (1.20) 2 (1.36) 2 50 40 c) The p-value is almost zero d)Reject H0 if z < 2.33. H0 is rejected. e) The mean number of cups for regular coffee drinkers is less. a) H0: 1= 2 H1: 1 ≠ 2 36.2 37.0 b) z 2.84 (1.14) 2 (1.30) 2 35 40 c)The p-value is 0.0046, found by 2(0.5000 – 0.4977). d)Reject H0 if z < 2.58 or z > 2.58. Reject H0, z statistic is 2.84 e) There is a difference in the useful life of the two brands of paint. Since the p-value is < than the significance level, H0 is rejected. 1) a) H0: 1 ≤ 2
(LO10-1)
(LO10-
H1: 1 > 2
= 1.30
b) √
c) The p-value is 0.50 0.4032 = 0.0968 (by computer 0.0964) d) Reject H0 if z > 1.645. Z=1.30 which is less than 1.645. There is not enough evidence to reject H0. e) There is not enough evidence to conclude that more units are produced on the nightshift. (LO101) 27.
a) H0: 1 = 2 H1: 1 ≠ 2 4.77 5.02 b) z 1.04 (1.05)2 (1.23)2 40 50 c) The p-value = 2(0.5000 – 0.3508) = 0.2984 d) Reject H0 if z < 1.96 or z > 1.96. z statistic is –1.04 therefore there is not enough evidence to reject H0. e) There is no difference in the mean number of calls. LO10-1) - 138 -
28.
H0:1 = 2
H1: 1 ≠ 2
25 15 (25 1)(3) = 17.0 2
s 2p
t
Reject H0 if t < 2.011 or t > 2.011
2
25 25 2 30 28
1 1 17.0 25 25
= 1.715
There is not enough evidence to reject the null hypothesis. The mean of the 2 programs is not different. The p-value (0.0928) is > than the significance level (0.05) which supports the decision to not reject. (LO10-2) 29.
a) H0:1 = 2
H1: 1 ≠ 2
(15 -1)(40 000) 2 (25 -1)(30 000) 2 1 157 894 737 15 25 - 2 150 000 180 000 t 2.70 1 1 1 157 894 737 15 25
2 b) s p
c) The p-value is 0.0103. d) If t is not between 2.024 and 2.024. Reject H0 the null hypothesis. e) The population means are different. (LO10-2) 30.
a)H0: 12 b)
H1: 1>2 (
)
= 29.96
= 0.86 √
c) p-value by computer = 0.2008. d) Reject H0 if t > 1.323. There is not enough evidence to reject H0. e) There is not a significant difference in the two exchanges. 31.
H0: p1 p2 H1: p1 > p2 180 261 pc 0.882 200 300
(LO10-2)
Reject H0 if z > 1.645 √
There is not enough evidence to reject H0. There is no difference in the proportions that found relief in the new and the old drugs. The p-value is 0.5 – 0.3461 = 0.1539, which is > than the significance level, and confirms that there is not enough evidence to reject H0. (Computer p-value = 0.1542) (LO10-5) 32.
H0: p2 ≤ p1 H1: p2 > p1 160 170 pc 0.56 300 290
If z > 1.645, reject H0. √
There is not enough evidence to reject the null. We cannot conclude an increased proportion believe the economy is expanding. The p-value is 0.5 – 0.4015 = 0.0985, which is > than the significance level, and confirms that there is not enough evidence to
- 139 -
reject H0. (Computer p-value = 0.0980) (LO10-5) 33. 5)
H0: p2 p1
pc
H1: p2 > p1
7 13 0.081 67 180
If z >1.645, reject H0.
(LO10-
√
Do not reject the null. The white board markers from the new supplier do not have a higher defective rate to those in stock. 34.
H0: p1 ≤ p2 H1: p1 > p2 39 11 pc 0.07194 388 307
If z > 1.645, reject H0. √
Therefore, reject the null hypothesis. The Banking Services division has significantly more female top executives. (LO10-5) 35.
H0: p1 = p2 H1: p1 ≠ p2 H0 is rejected if z < 1.96 or z > 1.96 68 45 pc 0.617 98 85
√ H0 is rejected. The proportions are not the same. 36.
(LO10-5)
H0: p1 = p2 H1: p1 ≠ p2 H0 is rejected if z < 1.96 or z > 1.96
pc
56 52 0.22833 270 203
√ H0 is not rejected. The proportions are the same. 37.
H0: p1 = p2
(LO10-5)
If z < 1.96 or z > 1.96, reject H0.
H1: p1 ≠ p2
√ Reject the null. We can conclude the proportion of responses is different between women and men. The p-value is virtually zero. (Computer p-value = 0.0002) (LO105) 38.
H0: p2 ≤ p1
H1: p2 > p1
If z > 2.33, reject H0.
- 140 -
√
There is not enough evidence to reject the null. We cannot conclude the proportion of passengers staying on the ship is greater in the Mediterranean. The p-value = (0.5 – 0.3749) = 0.1251 and confirms that there is not enough evidence to reject H0. (Computer p-value = 0.125) (LO10-5) 39.
a) H0: n= s H1: n ≠ s (10 1)(10.5)2 (12 1)(14.25)2 b) s 2p 161.2969 10 12 2 83.55 78.8 t 0.873 1 1 161.2969 10 12 c) The p-value is > than 0.20, which is > than the significance level, (Computer p-value = 0.3928). d) Reject H0 if t < 2.086 or t > 2.086. Since t-stat = 0.873 is in the DNR area there is not enough evidence to reject H0, there is no difference in the mean number of hamburgers sold at the two locations. (LO10-2)
40.
a) H0: 1 = 2
H1: 1 ≠ 2
.
b) √
c) P-value = 0.2617 which is > than the significance level d) If t is not between 2.086 and 2.086, reject H0.t falls in the Do not reject area. e) There is not enough evidence to reject H0. The mean waiting times are not different. (LO102) 41.
a) H0: 1 2 H1: 1 > 2 (8 1)(2.2638)2 (11 1)(2.4606) 2 b) s 2p 5.672 8 11 2
t
10.375 5.636 1 1 5.672 8 11
4.28
c) p-value = 0.0003 which is < than α = 0.01 d) Reject H0 if t > 2.567. Reject H0 since t = 4.28 is > 2.567 e) The mean number of transaction by the young adults is more than for the senior citizens. (LO10-2) 42.
a) H0: 1 = 2 b) x1 = 83.8
H1: 1 ≠ 2 s1 = 13.6772
x2 = 79.29
s2 = 6.7074 √
c) p-value by computer = 0.2946 which is > α = 0.10
- 141 -
d) Reject H0 if t > 1.717 or t < 1.717. Therefore Do not reject H0. e) There is no difference in the mean exam scores. 2) 43.
a) H0: 1 2 H1: 1 > 2 b) x1 =125.125 s1 = 15.094
s 2p
x2 = 117.714
s2 = 19.914
(8 1)(15.094) (7 1)(19.914) 305.708 872 2
(LO10-
2
t
125.125 117.714 1 1 305.708 8 7
0.819
c) p-value = 0.2138 which is > α = 0.01, therefore DNR H0 d) Reject H0 if t > 2.65. There is not enough evidence to reject H0. There is no increase in the mean number sold at the regular price and the mean number sold at reduced price.(LO10-2) 44.
a) H0: d 0 b) d = 2.5
H1: d < 0 sd = 2.928
t
2.5 = 2.415 2.928 / 8
c) p-value = 0.0232 which is not < the significance level. Based on the p-value approach DNR H0. d) Reject H0 if t < 2.998, t = 2.415 therefore do not reject H0 e) There is not enough evidence to reject H0. The mean number of accidents has not been reduced. (LO103) 45.
a) H0: d 0;
H1: d > 0
Reject H0 if t > 1.440
b) d = 2 (post-training – pre-training)
sd = 2.082
t
2 2.082
7
= 2.542
c) p-value = 0.022 d) Reject H0 if t > 1.440. Therefore Reject H0. e) The survey results show the training program has been effective in improving customer service. (LO 10-3). 46.
H0: d = 0 d = 246
H1: d 0 sd = 547
Reject H0 if t < 1.761 or t > 1.761 ⁄√
p-value = 0.103 There is not enough evidence to reject H0. There is no difference in the mean insurance price. (LO10-3) 47.
a) H0: B ≤ A H1: B > B (40 – 1)(9200)2 (30 – 1)(7100)2 b) s 2p 70 041 912 40 30 – 2
= 1.98 √
c) Computer p-value = 0.0259 d) Reject H0 if t > 1.668.t value is larger than 1.668 therefore reject H0
- 142 -
e) The mean income of those selecting Plan B is larger. The p-value is 0.0259, which is smaller than the significance level, and confirms that the null hypothesis should be rejected. (LO10-2) 48.
a) H0: 1 = 2, H1: 1 ≠ 2 (
b)
) √
c) p-value = 0.1636 d) If t is not between 2.228 and 2.228, reject H0. Since t = 1.50 do not reject H0. e) There is not enough evidence to reject H0. We conclude that there is not a difference in the taste ratings. (LO10-2) 49.
H0: d ≥ 0 d = 3.11
H1: d < 0
Reject H0 if t < 1.895
sd = 2.91
√
Reject H0. The new soap reduces the amount of bacteria. p-value = 0.0096 3) 50.
(LO10-
H0: 1 2 H1: 1 > 2 Reject H0 if t > 1.665 2 2 (35 1)(4.2) (45 1)(3.9) s 2p 16.27 35 45 2
t
18 15.5
1 1 16.27 35 45
= 2.75
Reject H0. Software issues take longer on average. The p-value is 0.0037, which is smaller than the significance level, and confirms that the null hypothesis should be rejected. (Computer p-value = 0.0037) (LO10-2) 51.
H0: p1 ≤ p2;
H1: p1 > p2 = 0.174
H0 is rejected if z > 1.28.
√ H0 is rejected. There is a higher response rate of those with pictures. The p-value is 0.0765 which is < the significance level confirming the decision to reject H0. (LO10-5) 52.
H0: d 0;
H1: d > 0
Reject H0 if t >1.833
d = 0.10 (open-ended – multiple choice)
sd = 2.601
t
0.1 = 0.12 2.601 10
There is not enough evidence to reject H0. Students do not score higher on open-ended questions. p-value = 0.4530 (LO 10-3)
- 143 -
53.
a) H0: d ≥ 0 b) d = 1.75
H1: d < 0 sd = 2.9155
√
c) p-value = 0.0667.The p-value is greater than 0.05 d) Reject H0 if t > 1.895. There is not enough evidence to reject H0. e) There is no difference in the mean number of absences. 54.
H0: p1 p2 H1: p1 < p2 400 450 Pc 0.425 1000 1000
(LO10-3)
If z < 1.645, reject H0. √
Reject the null hypothesis. There is enough evidence to show that sales in January are less than sales in June. The p-value is 0.5 – 0.4881 = 0.0119. (LO10-5) 55.
H0: 1 ≥ 2 H1: 1 < 2 Reject H0 if t < 1.860 1: new web design 2: old web design x1 =18.0 x 2 = 32.0 s1 = 5.70 s2 = 7.71
(5 1)(7.71) 2 (5 1)(5.70) 2 s 45.967 552 2 p
t
18.0 32.0 3.26 1 1 45.967 ( ) 5 5
Reject H0. The new web design allows users to access web sites faster than the old design. (LO10-2) 56.
H0: 1 = 2 x1 = 6.44
H1: 1 ≠ 2 s1 = 3.32
If tis not between 1.740 and 1.740, reject H0. s2 = 1.72
x2 = 9.50
(9 1)(3.32) 2 (10 1)(1.72) 2 6.75 9 10 2 6.44 9.50 t 2.56 1 1 6.75( ) 9 10 s 2p
Reject the null hypothesis. The population means are different. The p-value = 0.0203. (LO10-2) 57.
H0: p1 ≤ p2 H1: p1 > p2 88.5 85.7 pc 0.871 100 100
H0 is rejected if z > 1.645 √
H0 is rejected. Conclude there is an increase in the proportion calling conditions “good”. The p-value is 0.0041.(LO 10-5) 58.
a) H0:1 = 2
s 2p
H1:1 2
If t is not between 2.131 and 2.131, reject H0
(8 1)( 27.61) (9 1)( 25.45) 2 701 .206 892 2
- 144 -
t
87.63 88.44 0.0637 1 1 701.206( ) 8 9
For a 5% significance level the decision rule is: Do not reject the null hypothesis. It is reasonable to conclude there is no difference in the mean number of cars in the two lots. (LO 10-2) b) H0: 1 = 2 H1: 1 2 Reject H0 if t < 2.008 or t > 2.008. 2 (25 1)(23.43) (28 1)(24.12)2 s 2p 566 25 28 2 86.2 92.0 t 0.886 1 1 566 25 28 Do not reject the null hypothesis. It is reasonable to conclude there is no difference in the means. The p-value is 0.38. 59.
H0: 2 ≤ 1
s 2p
(LO 10-3)
H1: 2 > 1
Reject H0 if t > 2.453
(15 1)( 3.17 ) (18 1)( 4.38) 2 15.06 15 18 2 2
√
Reject H0. The mean allowance for children in 11 to 14 year age group is more that for children 8 to 10. The p-value is between 0.005 and 0.010. The computer calculated pvalue = 0.007 (LO10-2) 60.
H0: d ≥ 0
H1: d < 0
Reject H0 if t < 1.729
√
Reject the null hypothesis. It is reasonable to conclude the percent is less now.The pvalue is 0.018. (LO 10-3) 61.
(LO10-2,LO10-5) a. H0: PPublic ≤ PNon public
H1: PPublic > PNon public
= 2.2 √
p-value = 0.014 z critical = 1.645 The z-statistic = 2.2 > 1.645, therefore the null hypothesis is rejected. Conclude that Public employers have a higher proportion of fitness facilities. The p-value equal to 0.0132 < α = 0.05 therefore this confirms the decision to reject the null hypothesis. b. H0: µPublic = µNon public
H0: µPublic ≠ µNon public
- 145 -
(
Pooled Variance =
)
√
(
)
t critical (α = 0.05, two tail, df = 98) = (1.984, 1.984) The t stat is falls in the do not reject the null hypothesis. Conclude that there is no difference in the average amount of full time employees in Public employers as compared to Non public employers. The p-value = 0.2033 which is > 0.05 which confirms not to reject the null hypothesis. 62.
(LO10-2) a. H0: µOrignal ≤ µRemainder
H0: µOriginal >µRemainder (
)
Pooled Variance = (
) √
t critical (α = 0.05, one tail, df = 29) = 1.699 Since the t-stat = 2.38442 is > t critical = 1.699 the null hypothesis is rejected. The conclusion is that the original 6 teams have a higher average attendance than the average attendance of the remaining teams. b.
63. (LO10-2) a.
Using a software t stat = 2.38442, for one tail and df = 29 the computed p-value = 0.011934. The p-value is less than α = 0.05. This confirms that the null hypothesis should be rejected and concluded that the mean attendance is higher for original 6 teams.
The sample mean and standard deviation for Ottawa (Jan 2010-Dec 2015) is $353 681.94 and $12 966.36. The sample mean and standard deviation for Barrie and District (Jan 2010-Dec 2015) is $304 169.44 and $30 979.04. Pooled Variance =
– )(
(
√
(
– )(
)
–
–
(
)
)– (
)
With a t stat = 12.51, reject the null hypothesis and conclude during the period from January 2010 to December 2015 the average house price in Ottawa was higher than in Barrie and District.
- 146 -
b.
The sample mean and standard deviation for Barrie (Jan 2016-Nov 2020) is $500 423.73 and $49 257.33. The sample mean and standard deviation for Ottawa (Jan 2016-Nov 2020) is $441 305.08 and $61 443.54. Pooled Variance =
(
)
√ With a t stat = 5.77, reject the null hypothesis and conclude during the period from January 2016 to November 2020 the average house price in Barrie and District was higher than in Ottawa. c.
An explanation that could be made is that during the period of high increases in list prices throughout Canada, the proximity of Barrie and District to Toronto compared to Ottawa fueled higher growth in the more recent period.
CHAPTER 11 ANALYSIS OF VARIANCE 1.
9.01 from Appendix B.7
(LO11-1)
2.
9.78
(LO11-1)
3.
Reject H0 if F > 10.46, where df in numerator are 7 and 5 in the denominator. s 2 (10) 2 2.04 F = 2.04, found by: F 12 s2 (7) 2 There is not enough evidence to reject H0. There is no difference in the variations of the
two populations. 4.
(LO11-1)
Reject H0 if F > 9.15, where df in numerator are 4 and 6 in the denominator. s 2 (12) 2 2.94 F = 2.94, found by: F 12 s2 (7) 2 There is not enough evidence to reject H0. There is no difference in the variations of the two populations. (LO11-1)
5.
H0: H1: Reject H0 when F > 3.10 (3.10 is about halfway between 3.14 and 3.07), where df in numerator are 11 and 9 in the denominator. (12) 2 1.44 F = 1.44, found by F (10) 2
- 147 -
There is not enough evidence to reject H0. There is no difference in the variations of the two populations. (LO11-1) 6.
H0: H1: Reject H0 when F > 3.68, where df in numerator are 9 and 7 in the denominator. (3.9) 2 1.24 F = 1.24, found by F (3.5) 2 There is not enough evidence to reject H0. There is less than or the same variation in the stocks. (LO11-1) 7.
a. b. c. d.
e.
8.
a. b. c. d.
e. 9.
H0: 1 = 2 = 3 H1: Treatment means are not all the same. Reject H0 if F > 4.26 62.17, 12.75, 74.92 Source SS df MS F Treatments 62.17 2 31.08 21.94 Error 12.75 9 1.42 Total 74.92 11 Since 21.94 is more than 4.26, there is enough evidence to reject H0. The treatment means are not all equal. (LO11-2) H0: 1 = 2 = 3 H1: Treatment means are not all the same. Reject H0 if F > 3.89 70.4, 82.53, 152.93 Source SS df MS F Treatments 70.40 2 35.20 5.12 Error 82.53 12 6.88 Total 152.93 14 Reject H0. The treatment means are not all the same.
H0: 1 = 2 = 3 H1: Treatment means are not all the same. Reject H0 if F > 4.26 Source SS df MS F Treatments 276.50 2 138.25 14.18 Error 87.75 9 9.75 Total 364.25 11 Reject H0. The treatment means are not all the same.
(LO11-2)
(LO11-2)
H0: 1 = 2 = 3 H1: Treatment means are not all the same. Reject H0 if F > 3.89 Source SS df MS F Treatments 22.93 2 11.47 5.73 Error 24.00 12 2.00 Total 46.93 14 Reject H0 as 5.73 > 3.89. The mean number of hours spent on a terminal are not equal. (LO11-2) 10.
- 148 -
11.
e.
H0: 1 = 2 = 3 H1: Treatment means are not all the same. Reject H0 if F > 4.26 SST = 107.20 SSE = 9.47 SS total = 116.67 Source SS df MS F Treatments 107.20 2 53.60 50.96 Error 9.47 9 1.05 Total 116.67 11 Since 50.96 > 4.26, H0 is rejected. At least one of the means differs.
f.
( x1 x2 ) t MSE(1 n1 1 n2 )
a. b. c. d.
= (9.667 − 2.20) 2.262 1.05(1 / 3 1 / 5) = 7.467 ± 1.694 = [5.773, 9.161] Yes, we can conclude that the treatments 1 and 2 have different means. 3) 12.
e.
H0: 1 = 2 = 3 H1: Treatment means are not all the same. Reject H0 if F > 3.47 SST = 46.96 SSE = 53.00 SS total = 99.96 Source SS df MS F Treatments 46.96 2 23.48 9.30 Error 53.00 21 2.52 Total 99.96 23 Since 9.30 > 3.47, H0 is rejected. At least one of the means differ.
f.
( x1 x2 ) t MSE(1 n1 1 n2 )
a. b. c. d.
(LO11-
= (6.0 − 4.25) 2.080 2.52(1 / 10 1 / 8) = 1.75 ± 1.57 = [0.18, 3.32] Yes, we can conclude that the treatments 2 and 3 are different. (LO11-3) 13
H0: 1 = 2 = 3 = 4 H1: Treatment means are not all equal. Reject H0 if F > 3.71 Source SS df MS F P Factor 32.33 3 10.78 2.36 0.133 Error 45.67 10 4.57 Total 78.00 13 Since 2.36 is less than 3.71, there is not enough evidence to reject H0. The p-value is 0.133 which is greater than 0.05 which confirms our decision not to reject H0. There is no difference in the mean number of months. (LO11-2)
14.
a.
H0: 1 = 2 = 3 H1: At least one mean differs. Reject H0 if F > 3.81 Source SS df
- 149 -
MS
F
P
Factor 86.49 2 43.25 13.09 0.001 Error 42.95 13 3.30 Total 129.44 15 Since 13.09 > 3.81, H0 is rejected. The p-value is 0.001 which is less than 0.05 which b.
15.
confirms our decision to reject H0. At least one mean rate of return differs. No, you would need more information, such as confidence intervals to make that decision. (LO112)
H0: 12 22 H1: 12 22 df1 = 21 – 1 = 20 df2 = 18 – 1 = 17 H0 is rejected if F > 3.16 (45 600) 2 F 4.57 (21 330) 2 Reject H0. There is more variation in selling price of waterfront homes. (LO11-1)
16.
H0: n2 o2 H1: n2 o2 dfn = 16 – 1 = 15 dfo = 16 – 1 = 15 H0 is rejected if F > 2.40 (22) 2 F 3.36 (12) 2 Reject H0. There is more variation in load time of the new page.
17.
Sharkey: n = 7 White: n = 8 H0: w2 s2
(LO11-1)
s = 14.79 s = 22.95
H1: w2 s2 dfs = 7 – 1 = 6 dfw = 8 – 1 = 7 H0 is rejected if F > 8.26 (22.95) 2 F 2.41 (14.79) 2 There is not enough evidence to reject H0. There is no difference in the variation of the weekly sales. (LO11-1) 18.
a. b. c.
H0: 1 = 2 = 3 H1: Treatment means are not all equal. = 0.05 Reject H0 if F > 3.89 Source SS df MS Treatments 40 2 20 Error 60 12 5
- 150 -
F 4
d. 19.
a. b. c.
d. 20.
Total 100 14 Reject H0. The treatment means are not all equal.
(LO11-2)
H0: 1 = 2 = 3 = 4 H1: Treatment means are not all equal. = 0.05 Reject if F > 3.10 Source SS df MS F Treatments 50 3 16.67 1.67 Error 200 20 10.00 Total 250 23 Do not reject H0. There is no difference in the treatment means. (LO11-2)
Source SS df MS F Treatments 320 2 160 8.00 Error 180 9 20 Total 500 11 a. 3 b. 12 c. 4.26 d. H0: 1 = 2 = 3 H1: Not all means are equal. e. H0 is rejected. The treatment means differ.
(LO11-2)
21.
H0: 1 = 2 = 3 H1: Not all means are equal. H0 is rejected if F > 3.89 Source SS df MS F Treatments 63.33 2 31.67 13.38 Error 28.40 12 2.37 Total 91.73 14 H0 is rejected. There is a difference in the price of the particular toy at three different types of stores. (LO11-2)
22.
H0: 1 = 2 = 3 H1: Not all means are equal. H0 is rejected if F > 3.89 Source SS df MS F Treatments 26.13 2 13.07 13.52 Error 11.60 12 0.97 Total 37.73 14 H0 is rejected since 13.52 > 3.89. There is a difference in the mean amount of weight loss among the three diets. (LO11-2)
23.
H0: 1 = 2 = 3 = 4 H1: Not all means are equal. H0 is rejected if F > 3.10 Source SS df Factor 87.79 3 Error 64.17 20
MS 29.26 3.21
- 151 -
F 9.12
Total 151.96 23 Since computed F of 9.12 > 3.10, the null hypothesis of no difference is rejected at the 0.05 level. There is a difference in the mean number of crimes in the four districts. (LO11-2) 24.
a. One-way ANOVA: Clothes, Food, Toys H0: 1 = 2 = 3 H1: Not all means are equal. Source DF Factor 2 Error 29 Total 31 S = 6.689 Level +-Clothes Food Toys
SS MS 3182.0 1591.0 1297.5 44.7 4479.5 R-Sq = 71.04%
N
Mean
StDev
9 12 11
28.000 46.417 52.636
8.139 6.186 5.887
F 35.56
P 0.000
R-Sq(adj) = 69.04% Individual 95% CIs For Mean Based on Pooled StDev -------+---------+---------+--------(----*----) (---*---) (---*---) -------+---------+---------+---------
+-30
40
50
60 Pooled StDev = 6.689
The hypothesis of identical means can definitely be rejected since the p-value is virtually zero. b. As seen from the non-overlapping confidence intervals, clothes have a mean attention span at least ten minutes below the other groups. (LO11-2 & 11-3) H0: 1 = 2 H1: 1 2 Critical value of F = 4.75 Source SS df MS F Treatments 219.43 1 219.43 23.10 Error 114.00 12 9.5 Total 333.43 13 19 27 b. t 4.81 1 1 9.5 6 8 2 Since t = F. That is (4.81)2 23.10 (actually 23.14, difference due to rounding). c. H0 is rejected. There is a difference in the mean scores. (LO11-1)
25.
a.
26.
H0: a2 d2 H1: a2 d2 dfa = 10 – 1 = 9 dfd = 15 – 1 = 14
- 152 -
H0 is rejected if F > 2.65 F = 1.47, found by: F
(23.8) 2 1.47 (19.6) 2
There is not enough evidence to reject H0. There is no difference in the variations of the two shifts. (LO11-1) 27.
H0: 1 = 2 = 3 = 4 H1: Treatment means are not all equal. Reject Ho if F > 3.49. The computed value of F is 9.61. The null hypothesis of equal means is rejected because the F statistic (9.61) is greater than the critical value (3.49). The p-value (0.0016) is also less than the significance level (0.05). The mean waiting times are different. (LO112)
28.
H0: 1 = 2 = 3 H1: Treatment means are not all equal. Reject H0 if F > 5.61. The computed value of F is 8.26. The null hypothesis of equal means is rejected because the F statistic (8.26) is greater than the critical value (5.61) using the 0.01 significance level. The p-value (0.0019) is also less than the significance level (0.01). The mean fuel efficiencies are different. (LO11-2)
29.
H0: H1: Reject H0 when F > 2.85 (df: 10, 11) F = 2.25, found by F
30.
(12) 2 2.25 (8) 2
There is not enough evidence to reject H0. We conclude there is no difference in the variation at a 10% level of significance. (LO11-1) H0: 1 = 2 = 3 = 4 H1: Not all means are equal. H0 is rejected if F > 3.10 ANOVA table Source SS df MS F P-value Treatments 31 533.00 3 10 511.000 2.32 0.1057 Error 90 436.33 20 4 521.817 Total 121 969.33 23 There is not enough evidence to reject H0. There is no difference in the sales at the four locations. (LO11-2)
31.
H0: 1 = 2 = 3 = 4 = 5 = 6 H1: The treatment means are not equal. Reject H0 if F > 2.37 ANOVA table Source Treatments
SS 0.03478
df 5
- 153 -
MS 0.006956
F 3.86
Error Total
0.10439 0.13917
58 63
0.001800
H0 is rejected. There is a difference in the mean weight of the colours.
(LO11-
2) 32.
a.
Recall that x x / n so x (n) x For the first treatment
x (n) 51.43(10), so
SST
x 514.3 SST = 306.934, found by
(514 .3) 2 (446 .4) 2 (472 .0) 2 (508 .5) 2 (1941 .2) 2 10 10 10 10 40
d.
650.75 – 306.934 = 343.816 Source SS df MS F Treatments 306.934 3 102.311 10.713 Error 343.816 36 9.550 Total 650.750 39 10.713 > 2.87, so reject H0. There is a difference in the treatment means.
e.
The 95% CI: (51.43 – 50.85) ± 2.028 9.550(1/10 1/10) = 0.58 ± 2.830
b. c.
[2.223, 3.883] Zero is in the interval, so we cannot conclude that the number of minutes of music differ between station 1 and station 4. (LO11-2 & 113) 33.
a.
H0: 1 = 2 = 3 H1: Not all means are equal. H0 is rejected if F > 3.29 Source SS df MS F Treatments 127.1 2 63.6 3.04 Error 669.9 32 20.9 Total 797.0 34 H0 is not rejected since 3.04 < 3.29. There is no difference in the mean flight
times. b.
1 1 (52.07 55.71) 2.037 20.9 14 7
This reduces to −3.64 4.31. So there is no difference as 0 is present in the interval between −7.95 and 0.67 minutes. By computer: -8.15 to 0.86. (LO 11-2 & 11-3) 34.
H0: 1 = 2 = 3 = 4 H1: At least one mean is different. Reject Ho if F > 2.74. Since 2.72 is less than 2.74, H0 is not rejected. The p-value is 0.051, which is greater than 0.05. There is no difference in the means for the different types of mail. (LO 11-2 & 113)
35.
H0: 1 = 2 = 3 = 4 = 5 = 6 = 7 H1: At least one mean is different.
- 154 -
The critical value for a F statistic with 6 and 48 degrees of freedom in the numerator and the denominator, respectively, at a 0.05 significance level is 2.295. Since 5.72 is greater than 2.295 the null is rejected. At least one mean is different. Also, note that the p-value is very close to 0. (LO 11-2 & 11-3) 36.
H0: 1 = 2 = 3 = 4 H1: Not all means are equal. H0 is rejected if F > 2.76 Source SS df MS F Treatments 1 552 3 517 1.84 Error 16 846 60 281 Total 18 399 63 Since computed F of 1.84 < 2.76, the null hypothesis is not rejected at the 0.05 level. The mean percentage of stock investment could be the same for all age groups. (LO 11-2)
37.
a.
b.
H0: 1 = 2 = 3 H1: Treatment means are not all equal. Reject H0 if F > 6.36. The computed value of F is 11.33. The null hypothesis of equal means is rejected because the F statistic (11.33) is greater than the critical value (6.36). The p-value (0.0010) is also less than the significance level (0.01). The mean production rates are different. 1 1 (43.33 41.5) 2.947 0.5444 6 6 This reduces to 1.83 1.26. So the difference is between 0.57 and 3.09.
(LO112 & 11-3) 38.
H0: 1 = 2 = 3 H1: The means are not all equal. Critical value of F = 3.44
68.8 68.8 68.8
ANOVA table Source Treatments Error Total
Mean 49.0 74.7 78.3 68.8
SS 3 872 4 182 8 054
n 7 9 9 25
Std. Dev 7.81 15.31 15.58 18.32
df 2 22 24
High School Undergraduate Master’s Total
MS 1 936.000 190.091
F 10.18
P-value 0.0007
Since the computed value of 10.18 exceeds the critical value of 3.44, the null hypothesis is rejected. The mean salary for those with high school or less is $49 000, it is $74 667 for those with an undergraduate degree, and $78 333 for those with a master’s degree or more. The salary for those with only high school differs from both the other groups. The
- 155 -
salaries for those with college work do not differ. The confidence interval for the difference between high school and undergraduate is computed as follows: 1 1 (49.00 74.67) 2.074 190 25.67 14.41 : (-40,-11) 7 9 Zero is not in this interval so there is a difference between these 2 means
The confidence interval for the difference between high school and master’s is computed as follows:
1 1 x1 and x3 : ( x1 x3 ) 2.074 190 = −29.333 ± 14.41: (-43,-15) 7 9 Zero is not in this interval so there is a difference between these two means The confidence interval for the difference between undergraduate and master’s is computed as follows:
1 1 x2 and x3 : ( x2 x3 ) 2.074 190 = −3.667 ± 13.479: (-17,9.8) 9 9 Note that zero is in this interval so there is no difference between these means. (LO11-2 & 113) 39.
H0: 1 = 2 = 3 H1: Not all means are equal. H0 is rejected if F > 3.89 ANOVA table Source SS df MS F P-value Treatments 62.53 2 31.267 4.86 0.0284 Error 77.20 12 6.433 Total 139.73 14 Computed F > 3.89. Reject the null hypothesis that the production levels are the same on all shifts. (LO 11-2)
40.
a.
b.
41.
There is not enough evidence to reject the null hypothesis of equal means because the F statistic (3.41) is less than the critical value (5.49). The p-value (0.0478) is also greater than the significance level (0.01). The mean amounts of money withdrawn are not different. 1 1 (82.5 38.2) 1.703 1477.633 10 10 This reduces to 44.3 29.3. So, the difference is between 15.0 and 73.6. (LO11-2 & 11-3)
a. H0: 1 = 2 = 3 = 4 H1: Not all means are equal. H0 is rejected if F > 4.07 ANOVA table Source SS
df
MS - 156 -
F
P-value
Treatments Error Total
44.25 22.67 66.92
3 8 11
14.750 2.833
5.21
0.0276
Since the p-value is less than the significance level, reject H0. b. SUMMARY Groups Red Blue Orange Green
Count 3 3 3 3
Sum 11 26 17 23
Average 3.66667 8.66667 5.66667 7.66667
(ⴟ-ⴟ)
t df=8
(MSE)0.5
(1/3+1/3)0.5
Red-Blue
-5
2.306
1.68315
0.8165
Red-Orange
-2
2.306
1.68315
0.8165
Red-Green
-4
2.306
1.68315
0.8165
Blue-Orange
3
2.306
1.68315
0.8165
Blue-Green
1
2.306
1.68315
0.8165
Orange-Green
-2
2.306
1.68315
0.8165
Low 8.169 5.169 7.169 0.169 2.169 5.169
High 1.831
*
1.169 0.831
*
6.169 4.169 1.169
*Note Red-Blue and Red-Green do not contain zero in the interval therefore these colours have different means. (LO11-2) 42.
a) H0: 1 = 2 = 3 H1: Not all means are equal. ANOVA table
Source Treatments Error Total
SS 173.6599 432.1269 605.7869
df 2 24 26
MS 86.82996 18.00529
Since the p-value is less than the significance level, reject H0. b) SUMMARY Groups Count Sum Average Super$ 9 130.4 14.489 Ralph's 9 176.65 19.628 LowName 9 180.73 20.081
- 157 -
F 4.82
P-value 0.0174
(ⴟ-ⴟ) t df=24 (18.00x(1/9+1/9))0.5 Low High Super$-Ralph's 5.139 2.064 2 -9.267 -1.011 Super$LowName 5.592 2.064 2 -9.720 -1.464 Ralph'sLowName 0.453 2.064 2 -4.581 3.675 *These pairs do not contain 0 in the interval. These means are not the same. (LO11-2) 43.
(LO11-1 & 11-2) a. H0: p2 n2 H1: p2 n2 dfp = 47 – 1 = 46 dfn = 53 – 1 = 52 H0 is rejected if F > 1.60 by computer Reject H0. There is more variation in the number of part-time employees in public employers. b. H0: 1 = 2 = 3 H1: Treatment means are not all the same. Reject H0 if F > 3.09 Source SS df MS F Treatments 842 378 149.25 2 421 189 074.62 2.91 Error 14 034 005 897.39 97 144 680 473.17 Total 14 876 384 046.64 99 Since 2.91 is less than 3.09, there is not enough evidence to reject H0. There is no difference in the mean number of full-time employees between these three groups.
44.
Testing whether the mean attendance per division is the same H0: 1 = 2 = 3 = 4 H1: Treatment means are not all the same. Reject H0 if F > 2.96 Source SS df MS F Treatments 16 591 639.69 3 5 530 546.56 1.40 Error 106 848 199.73 27 3 957 340.73 Total 123 439 839.42 30 Since 1.40 is less than 2.96, there is not enough evidence to reject H0. There is no difference in the mean attendance between divisions. Testing whether the mean team worth per division is the same H0: 1 = 2 = 3 = 4
- 158 -
H1: Treatment means are not all the same. Reject H0 if F > 2.96 Source SS df MS F Treatments 479 155.32 3 159 718.44 1.49 Error 2 903 286.61 27 107 529.13 Total 3 382 441.94 30 Since 1.49 is less than 2.96, there is not enough evidence to reject H0. There is no difference in the mean team worth between divisions. (LO11-1) 45.
(LO11-2) a. H0: 1 = 2 = 3 H1: Treatment means are not all the same. Reject H0 if F > 3.05 Source SS df MS F Treatments 41 728 471 864.41 2 20 864 235 932.20 4.69 Error 773 844 029 830.51 174 4 447 379 481.78 Total 815 572 501 694.92 176 Since 4.69 is greater than 3.05, we reject H0. There is a difference in the average single family home list price. b. H0: n2 s2 H1: n2 s2 dfn = 59 – 1 = 58 dfs = 59 – 1 = 58 H0 is rejected if F > 1.40 F= Since the F statistic is > the critical value, there is enough evidence to reject H0. There is more variation of list prices of St John’s than in Newfoundland and Labrador.
CHAPTER 12 LINEAR REGRESSION AND CORRELATION 1.
a.
- 159 -
Sales ($)
Number of Customers vs Sales ($) 8 7 6 5 4 3 2 1 0 0
2
4
6
8
10
12
Number of Customers
y = 3.767 + 0.363x 5(173) (28)(29) b 0.363 5(186) (28)2 6.308, found by y = 3.767 + 0.363(7)
b.
c. 2.
a
29 28 (0.363) 3.7672 5 5 (LO12-1)
a.
Scatter Diagram 16 14
x-values
12 10 8 6 4 2 0 0
2
4
6
8
10
y-values
b.
c. 3.
y = 19.1197 – 1.7425x 8(378) (39)(85) b 1.7425 8(211) (39)2 6.9222, found by 19.1197 – 1.7425(7)
a.
- 160 -
a
85 39 (1.7425) 19.1197 8 8 (LO12-1)
Bradford Electric Illuminating Company
Kilowatt-Hours (thousands)
12 10 8 6 4 2 0 0
5
10
15
Number of Rooms
y = 1.333 + 0.6667x 74 91 10(718) (91)(74) 446 a (0.6667) 1.333 b 0.6667 2 10 10 10(895) (91) 669 y = 1.333 + 0.6667(6) = 5.3332 kilowatt-hours (thousands) (LO12-
b.
c. 1) 4.
a.
Number of Contacts vs Sales 140
Sales ($ thousands)
120 100 80 60 40 20 0 0
10
20
30
40
50
60
Number of Contacts
b.
c.
y = 12.1996 + 2.1946x 10(26 584) (334)(611) b 2.1946 10(13 970) (334)2 611 334 a (2.1946) 12.1996 10 10 y = 12.1996 + 2.1946(40) = 75.5844 ($ thousands)
1)
- 161 -
(LO12-
5.
a.
Earnings
15 10 5 0 0
25
50
75
100
Sale s
y = 1.8507 + 0.0836x 12(3306.35) (501.1)(64.1) b 0.0836 5(28 459) (501.1)2 64.1 501.10 a (0.0836) 1.8507 12 12 c. y = 1.8507 + 0.0836(50) = 6.0307 ($ million) Note: calculator or computer values may be slightly different due to rounding. (LO12-1) b.
6.
a.
Return
15 10 5 0 0
200
400
600
800
Assets
y = 9.9224 0.0004x 9(34 111) (3504.5)(87.9) b 0.0004 9(1 659 866) (3504.5)2 87.9 3504.5 a (0.0004) 9.9224 9 9 y = 9.9224 0.0004(400.0) = 9.7624 ($ millions).
b.
c. a. b.
Police is the independent variable and crime is the dependent variable
Crimes
7.
(LO12-1)
30 20 10 0 10
15
20
25
30
Police
- 162 -
b
c.
y´ = 29.3877 0.9596Police y´ = 29.3877 0.9596(20) = 10.1957 Inverse relationship. As the number of police increase, crime decreases. (LO12-
d. e. 1) a. b.
Age is the independent variable and price is the dependent variable.
Price
8.
8(1502) (146)(95) 95 146 0.9596 a (0.9596) 29.3877 2 8(2906) (146) 8 8
12.5 10 7.5 5 2.5 0 0
2
4
6
8
10 12 14 16
Age
c.
n = 12
x 107 y 82.9 x 1009 xy 712.9 y 615.29 2
2
12(712.9) 107(82.9) 0.4788 12(1009) (107) 2 82.9 107 a (0.4788)( ) 11.1776 12 12 y = 11.1776 0.4788x y = 11.1776 0.4788(10) = 6.3896 ($ thousands)
b
d. e. Inverse relationship. As the age of the car increases, the selling price decreases. (LO12-1) 9.
10
11.
12.
a.
0.992, found by
b.
y 0.992
a.
1.658, found by
b.
y 3.316
a.
0.913, found by
b.
y 1.826
a.
9.31, found by
b.
y 2(9.31)
175 3.7672(29) 0.363(173) 52 (LO12-2)
983 19.1197(85) (1.7425)(378) 82 (LO12-2)
584 1.333(74) 0.6667(718) 10 2 (LO12-2)
51 581 (12.1996)(611) 2.1946(26 584) 10 2 (LO12-2)
- 163 -
13.
3.378, found by
1419 29.3877(95) (0.9596)(1502) 82
(LO12-2)
14.
1.732, found by
615.29 11.1776(82.9) (0.4788)(712.9) 12 2
(LO12-2)
15.
a.
b.
1 (7 5.6) 2 6.3082 ± 3.182(0.992) = 6.3082 ± 1.6314 5 29.2 = [4.6768, 7.9396] 6.3082 (3.182)(0.992) 1 1 / 5 0.0671 [2.755 9.8614]
(LO12-
3)
16.
a.
1 (7 4.875) 2 6.9222 ± 2.447(1.658) = 6.9222 ± 2.3703 8 20.875
(LO12-
3) = [4.5519, 9.2925] b.
6.9222 ± 2.447(1.658)
1 (7 4.875) 2 1 = 6.9222 ± 4.6988 8 20.875
= [2.2234, 11.621] 17.
18.
19.
a. b.
[4.495, 6.1714] [3.4349, 7.227]
a. b.
[66.4223, 84.7465] [47.1084, 104.0604]
x = 28
y = 29
(LO12-3)
(LO12-3)
x = 186 2
∑xy = 173
y = 175 2
5(173) (28)(29)
r
0.752 [5(186) (28) 2 ][5(175) (29) 2 ] The 0.752 coefficient indicates a rather strong positive correlation between x and y. The coefficient of determination is 0.5655, found by (0.752)2. Therefore, x accounts for more than 56 percent of the variation in y. (LO12-4)
20.
x = 39 r
y = 85
x = 211
8(378) (39)(85) [8(211) (39) 2 ][8(983) (85) 2 ]
2
∑xy = 378
y = 983 2
0.891
The 0.891 indicates a very strong negative relationship between x and y. The coefficient of determination is 0.7939, found by (0.891)2. Therefore, x accounts for close to 80% of the variation in y. (LO12-4) 21.
a.
n=5
∑x = 20 ∑y = 85 ∑x2 = 90 ∑xy = 376 ∑y2 = 1595
- 164 -
r b. c.
22.
a.
b.
23.
24.
25.
26.
27.
5(376) (20)(85) [5(90) (20)2 ][5(1595) (85)2 ]
0.93
r 2 (0.93)2 0.864 The 0.93 indicates a very strong positive relationship between x and y. The coefficient of determination is 0.8649. Therefore, x accounts for about 86.49 percent of the variation in y. (LO12-4) ∑x = 15 ∑y = 120 ∑x2 = 55 ∑xy =430 ∑y2 = 3450 5(430) (15)(120) r 0.927 [5(55) (15) 2 ][5(3450) (120) 2 ] The r2 is 0.8593, so about 86 percent of the variation in production is explained by the variation in the number of assemblers. (LO12-4) n=5
∑x = 146 ∑y = 95 ∑x2 = 2906 ∑xy = 1502 8(1502) (146)(95) r 0.874 [8(2906) (146) 2 ][8(1419) (95) 2 ]
∑y2 = 1419
a.
n=8
b. c.
0.7639, found by (0.874)2 0.874 indicates a strong inverse relationship. As the number of police increase, the crime decreases. The coefficient of determination is 0.7639. Therefore, x accounts for about 76.39 percent of the variation in y. (LO12-4)
a.
r
b. c.
0.2959, found by (0.544)2 Weak negative (inverse) correlation between age of car and selling price. So, 29.59 percent of the variation in the selling price is explained by the variation in the age of the car. (LO12-4)
12(712.9) (107)(82.9) [12(1009) (107) 2 ][12(615.29) (82.9) 2 ]
Reject H0 if t > 1.812 0.32 12 2 t 1.068 1 (0.32)2 Reject H0if t <1.771 0.46 15 2 t 1.868 1 (0.46)2
Do not reject H0.
Reject H0.
0.544
(LO12-5)
(LO12-5)
H0: 0 H1: < 0 Reject H0 if t < 2.552 df = 18 0.78 20 2 t 5.288 1 (0.78)2 Reject H0. There is a negative correlation between litres sold and the pump price. (LO125)
- 165 -
28.
H0: 0 H1: > 0 Reject H0 if t > 1.734 df = 18 0.86 20 2 t 7.150 1 (0.86)2 Reject H0. There is a positive correlation between assets and pretax profit.
(LO12-
5) 29.
H0: 0
t
H1: > 0
0.667 15 2 1 (0.667) 2
Reject H0 if t > 2.650
df = 13
3.228
Reject H0. There is a positive correlation between passengers and cost.
(LO12-
5) a. Scatterplot of BAC vs Beers 0.12
0.10
0.08 BAC
30.
0.06
0.04
0.02 1
2
3
4 Beers
5
6
7
b.
Beers BAC x x y y x x y y x x y y 6 0.1 1.722 0.0211 2.966 0.00045 0.0364 7 0.09 2.722 0.0111 7.411 0.00012 0.0302 7 0.09 2.722 0.0111 7.411 0.00012 0.0302 4 0.1 −0.278 0.0211 0.077 0.00045 −0.0059 5 0.1 0.722 0.0211 0.522 0.00045 0.0152 3 0.07 −1.278 −0.0089 1.633 0.0000 0.0114 3 0.1 −1.278 0.0211 1.633 0.00045 −0.0270 6 0.12 1.722 0.0411 2.966 0.00169 0.0708 6 0.09 1.722 0.0111 2.966 0.00012 0.0191 3 0.07 −1.278 −0.0089 1.633 0.0001 0.0114 3 0.05 −1.278 −0.0289 1.633 0.00083 0.0369 7 0.08 2.722 0.0011 7.411 0.00000 0.0030 1 0.04 −3.278 −0.0389 10.744 0.00151 0.1275 4 0.07 −0.278 −0.0089 0.077 0.0001 0.0025 2 0.06 −2.278 −0.0189 5.188 0.00036 0.0430 7 0.12 2.722 0.0411 7.411 0.00169 0.1119 2 0.05 −2.278 −0.0289 5.188 0.00083 0.0658 1 0.02 −3.278 −0.0589 10.744 0.00347 0.1930 2
- 166 -
2
77
1.42
77.611 0.01278
0.7756
77 1.42 4.278 y 0.0789 18 18 77.61 0.01278 sx 2.1367 sy 0.0274 17 17 0.7756 r 0.779 (18 1)( 2.1367 )(0.0274 ) x
c.
0.607, found by (0.779)2
d.
Ho: 0 H1: > 0 Reject H0 if t > 2.583 df = 16 0.779 18 2 t 4.97 1 (0.779 ) 2 Reject H0. There is a positive correlation between beers consumed and
BAC. p-value = 0.0001, which is very close to zero, and confirms that H0 should be rejected. (LO12-2, 12-4& 12-5) 31.
Coefficient of correlation r = 0.8944, found by
32.
a.
(5)(340) (50)(30)
[(5)(600) (50) 2 ][(5)(200) (30) 2 ] Then (0.8944)2 = 0.80, the coefficient of determination. (LO12-4)
4.5 4
Price ($)
3.5 3 2.5 2 1.5 1 0.5 0 0
2
4
6
Turnover
b.
0.8167, found by (6 – 1.1)/6 also, r
( y y ') = 1.1 and ( y y) = 6 2
(5)(37) (15)(10) [(5)(55) (15) ][(5)(26) (10) ] 2
2
- 167 -
2
0.9037 and r2 = 0.8167
33.
c. 4)
Turnover accounts for 81.67% of the variation in price.
a. b.
r2 = 1000/1500 = 0.6667 0.8165, found by 0.6667
c.
6.2017, found by s e
500 15 2
(LO12-1& 12-
(LO12-4& 12-6)
34. Source Regression Residual Total
SS 7200 1800 9000
df 1 18 19
MS 7200 100
(LO12-4& 12-6) 35. 350 300 250
Y
200 150 100 50 0 -20
-10
0
10
20
X
a. The relationship does not appear to be linear. b. The correlation of X and Y is 0.2975. There is a weak positive correlation between the variables. c. 151.388
(LO12-1& 12-
4) 36. a. inverse
The correlation of Demand and Price is –0.865 which indicates a fairly strong linear relationship.
- 168 -
Price
120
70
20 0
10
20
30
40
50
60
Demand
b.
17.428
(LO12-1& 12-4)
37.
H0: 0 H1: > 0 Reject H0 if t > 1.714 0.94 25 2 t 13.213 1 (0.94)2 Reject H0. There is a positive correlation between passengers and weight of luggage. (LO12-5)
38.
H0: 0 H1: > 0 Reject H0 if t > 2.552 0.40 20 2 t 1.852 1 (0.40)2 Do not reject H0. We cannot conclude that there is a positive correlation between GPA and family income. (LO12-5)
39.
H0: 0 H1: > 0 Reject H0 if t > 2.764 0.47 12 2 t 1.684 1 (0.47)2 Do not reject H0. There is not a positive correlation between engine size and performance. The p-value is greater than 0.05, but less than 0.10. (LO12-5)
40.
H0: 0 H1: > 0 Reject H0 if t > 1.734 0.21 20 2 t 0.911 1 (0.21)2 Do not reject H0. There is not a positive correlation between shots attempted and shots scored. The p-value is greater than 0.10. (LO12-5)
41.
H0: 0
H1: < 0
Reject H0 if t < 1.701
- 169 -
df = 28
t
0.45 30 2
2.666 1 0.2025 Reject H0. There is a negative correlation between the selling price and the number of kilometres driven. (LO125)
42.
a. b. c.
43.
a.
Yes, because the correlation coefficient is positive. 0.119 or 11.9%, found by (0.345)2. H0: 0 H1: > 0 Reject H0 if t > 1.771 0.345 15 2 t 1.325 1 (0.345)2 Do not reject H0. There is not a positive correlation between fat grams consumed and cholesterol level. The p-value is greater than 0.10. (LO12-4 & 125)
Scatterplot of Income vs Occupied 1550
Income
1500
1450
1400
1350 10
20
30
40 Occupied
50
60
Revenue increases slightly as the number of occupied rooms increases. b. Pearson correlation of Income and Occupied = 0.423 c. H0: 0 H1: > 0 Reject H0 if t > 1.319
t
d.
44.
a. b. c.
0.423 25 2 1 (0.423) 2
df = 23
2.239
Reject H0. There is a positive correlation between revenue and occupied rooms. 17.9%, found by (0.423)2, of the variation in revenue is explained by variation in occupied rooms. (LO12-1, 12-4 & 12-5) Decrease, because the correlation coefficient is negative. 0.287 or 28.7%, found by (–0.536)2 H0: 0 H1: < 0 Reject H0 if t < 1.714 0.536 25 2 t 3.045 1 (0.536)2 Reject H0. There is a negative correlation between job satisfaction and stress. (LO12-4 & 12-5)
- 170 -
45
No, the coefficient is 0.5170 which indicates a negative relationship between
a. the b.
variables. SSR = 3119.4256/3960 = 78.77%
c. d.
r = 0.78773 = 0.8875; strong, negative y´=45.85; 46 units; yes, this is reasonable.
a. b.
Yes, the coefficient of Hours of Study is positive. SSR = 1569.0299/2733 = 57.41%
c. d.
r 0.5741 = 0.7577; strong, positive relationship y' 5(4.3284 ) 60.1418 81.78; yes, it is reasonable. (LO12-4& 12-5)
47.
a. b. c.
r = 0.589 r2 = (0.589)2 = 0.3469 H0: 0 H1: > 0 Reject H0 if t > 1.860 0.589 10 2 t 2.061 1 (0.589)2 H0 is rejected. There is a positive association between family size and the amount spent on food. (LO12-4& 12-45)
48.
a. b. c.
r = 0.348 r2 = (0.348)2 = 0.1211 H0: 0 H1: > 0
46.
t
0.348 12 2 1 (0.348) 2
(LO12-4& 12-5)
(LO12-4& 12-5) Reject H0 if t > 1.812
1.174
H0 is not rejected. We have not shown this to be a relationship between the variables. a.
It looks to be an inverse relationship between the variables. As the months owned increases the number of hours exercised decreases. Months Ow ned versus Hours Exercised 15
Hours
49.
10 5 0 0
5
10
15
Months
10(313) (65)(58)
b.
r
c.
H0: 0
[10(523) (65) 2 ][10(396) (58) 2 ]
H1: < 0
0.827
Reject H0 if t < 2.896
- 171 -
t
0.827 10 2
4.652 1 (0.827) 2 Reject H0. We can conclude that there is a negative association between months owned and hours exercised. (LO12-1, 124& 12-5)
50.
Source Regression Error Total a.
SS 300 100 400
se
df 1 18 19
r2 = 300/400 = 0.75 r 0.75 0.866
51.
Source Regression Error Total n = 25
b.
52.
F 54.0
(LO12-2& 12-4)
100 2.3570 18
b. c. negative. a.
MS 300 5.556
The sign of r is negative because the sign of b is
SS 50 450 500
df 1 23 24
MS 50 19.5652
F 2.5556
se 19.5652 4.4233
c. d.
r2 = 50/500 = 0.10
a.
y = 17.08 + 0.16(50) = 25.08
b.
25.08 ± 3.182(4.05)
(LO12-2& 12-4)
1
1 (50 42) 2 = 25.08 ± 14.48 5 1030
= [10.602, 39.558]
(LO12-1 & 12-
3) 53.
a.
n = 15 ∑y2 = 969.92
b
∑x = 107 ∑x2 = 837 ∑y = 118.6 s y x = 1.114 ∑xy = 811.60
15(811.60) (107)(118.6) 516.2 0.4667 15(837.0) (107)2 1106.0
b.
118.6 107 (0.4667) 11.2358 15 15 More bidders decrease winning bid. y = 11.2358 – 0.4667(7.0) = 7.9689 ($ millions)
c.
7.9689 (2.160)(1.114) 1
d.
[5.4835, 10.4543] r2 = 0.499. The number of bidders explains nearly 50 percent of the variation in the amount of the bid. (LO12-1, 12-3& 12-4)
a
1 (7 7.1333) 2 7.9689 2.4854 (107) 2 15 837 15
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54.
a.
b
15(13 114.64) – (1193.8)(163.60) 0.0030 15(126 252.04) – (1193) 2
163.6 1193.8 (0.003) 10.6679 15 15 y = 10.6679+0.003x a
b.
r2 = (0.466)2 = 0.2172; no it is not a strong predictor
a.
b
(LO12-1& 12-
4) 55.
30 18 924 – 320.331575.6 30 4292.5 – 320.33
2
2.41
1575.6 320.33 2.41 26.8 30 30 The regression equation is: Price = 26.8 + 2.41 dividend. For each additional dollar of dividend, the price increases by $2.41. 5057.6 r2 0.658 65.8% of the variation in price is explained by 7682.7 the dividend. r 0.658 0.811 H0: 0 H1: > 0 a
b. c.
At the 5% level, reject H0 when t > 1.701.
t
0.811 30 2 1 0.811
Thus H0 is rejected. The population correlation is positive. 4& 12-5) 56.
57.
2
7.34
(LO12-1, 12-
a.
20, found by one more than the total degrees of freedom (19 + 1)
b.
25.88, found by the square root of mean square error
c.
0.46, found by SSR/SStotal =
d. e.
0.68, found by 0.46 Yes, because the tvalue (3.935) is greater than the critical value (1.734) and the pvalue (0.0005) is less than the significance level (0.05). (LO121, 12-2,12-4& 12-5)
a.
35, found by one more than the total degrees of freedom (34 + 1)
b.
5457, found by the square root of MSE (mean square error)
c.
0.93, found by SSR/SStotal
d. e.
0.97, found by 0.93 H0: 0 H1: > 0 Reject H0 if t > 1.692 Yes, because the tvalue (21.33) is greater than the critical value (1.692) and the pvalue (0.000) is less than the significance level (0.05). Reject the null
670 .
10 354 . 22 408
29 778 406 .
13 548 662 082 . 14 531 349 474
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hypothesis. There is a positive association between market value and size of the home. (LO12-1, 12-2,12-4 & 12-5) 58.
a.
There is a lot of scatter in the graph but there appears to be a positive relationship between the two variables. As the distance increases, so does the shipping time. Distance vs. Time
16
Time (days)
14 12 10 8 6 4 2 0 600
650
700
750
800
850
900
Distance (km)
b.
r
20(125 051) – (14 517)(168) 20(10 682 471) – (14517) 2 20(1550) – (168) 2
H0: 0 t
c. d. 59.
H1: > 0
= 0.692
Reject H0 if t > 1.734
0.692 20 2
4.067 1 (0.692)2 Reject H0. There is positive association between shipping distance and shipping time. r2 = 0.479; only 48% of the variation in shipping time is explained by shipping distance. (LO12-1, 12-2, 12-4& 12-5) S e = 2.0040
a.
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1300 1200 1100
Spent
1000 900 800 700 600 500 400 4000
5000
6000
7000
8000
9000
10000
Income
b.
b
40(245 792 400) (273 384)(33 625) 0.1339 40(1 987 816 326) (273 384) 2
33 625 273 387 0.1339 74.4 40 40 The regression equation is Spent = 74.4 + 0.1339 Income. For each additional dollar of income, 13.4 cents more is spent on groceries. 40 245 792 400 273 384 33 625 r 0.945 2 2 [40 1 987 816 326 273 384 ][40 30 662 885 33 625 ] a
c.
H0: 0 H1: > 0 0.945 40 2 t 17.811 2 1 0.945
At the 5% level, reject H0when t > 1.686.
Thus H0 is rejected. The population correlation is positive. d.
60.
We know is that there is a strong positive association between income and groceries; however, other factors such as location and growth in the area need to be considered. (LO12-1, 12-2, 12-4& 12-5)
a.
The correlation of Weight and Consumption is 0.987. H0: 0 H1: > 0 Reject H0 if t > 1.746 0.987 18 2 t 24.564 1 (0.987)2 Reject H0. It is quite reasonable to say the population correlation is positive! b. The regression equation is Weight = –29.7 + 22.9 Consumption. Each additional
cup increases the estimated weight by 22.9 kms. 61.
a.
y’ = 12.3601 + 4.7956 x
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(LO12-1,12-4& 12-5)
b. c. d. e.
The relationship between distance and damage is direct. $36338 found by 12.3601 + 4.7956(5) 0.581, found by 1865/3209 58.1% of the variation in damage is explained by variation in distance. 0.762 which is 0.581 . It is positive because the slope is positive. There is a fairly solid direct link between the variables. H0: = 0 H1: ≠ 0 Reject H0 if t < 2.763 or t > 2.763
t
0.762 30 2 1 (0.762) 2
6.231
Reject H0.
There is a connection between distance and fire damage. (LO12-1, 12-4& 12-5) 62.
a.
b.
The correlation of Return and Growth is 0.059. H0: 0 H1: > 0 Reject H0 if t < 1.943 0.059 8 2 t 0.145 1 (0.059) 2 No, it is not reasonable to say the population correlation is positive. The regression equation is Return = 17.9 – 0.032 Growth. The slope is negative indicating that return goes down when growth goes up. (LO12-1, 12-4& 12-5)
63.
a. b.
The regression equation is Price = –773 + 1408 Speed. The correlation of Speed and Price is 0.835. H0: 0 H1: > 0 Reject H0 if t > 1.812 0.835 12 2 t 4.799 1 (0.835)2 Reject H0. It is reasonable to say the population correlation is positive. (LO12-1, 12-4& 12-5)
64.
a. b.
The correlation of Wattage and Area is 0.939. The link is direct. H0: 0 H1: > 0 Reject H0 if t > 1.734 0.939 20 2 t 11.584 1 (0.939) 2 Reject H0. It is quite reasonable to say the population correlation is positive! The regression equation is Area = –22.6 + 0.149 Wattage. (LO12-1, 12-4& 12-5)
c.
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65.
a. Distance and Fares 400 350
Fare
300 250 200 150 100 0
500
1000
1500 Distance
2000
2500
3000
The relationship is direct. Fares increase for longer flights. b. The correlation of Distance and Fare is 0.606. H0: ≤ 0 H1: > 0 Reject H0 if t >1.812 df = 10
t
0.606 12 2 1 (0.606) 2
2.409
Reject H0. There is a significant positive correlation between fares and distances. 36.7%, found by (0.606)2, of the variation in fares is explained by the variation in distance. d. The regression equation is Fare = 154.07 + 0.0456 Distance. Each additional mileadds$0.0456 to the fare. A 1500 mile flight would cost $222.47, found by $154.07 +0.0456(1500). e. A flight of 4218 miles is outside the range of the sampled data. So the regression equation may not be useful. (LO12-1, 12-4 & 12-5) c.
66.
H0: 0 H1: < 0 Reject H0 if t < 1.697 0.363 32 2 t 2.134 1 0.1318 Reject H0. There is a negative correlation between square feet and rental rate. (LO12-5)
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67.
a. Store Sales by Square Footge 12.00
Sales (millions $)
10.00 8.00 6.00 4.00 2.00 0.00 0.0
1.0
2.0
3.0 4.0 5.0 Square Feet (000)
6.0
7.0
There is a strong positive connection between the two variables. Larger stores have higher sales. b. The correlation of Square Feet and Sales is 0.824 and the coefficient of determination is 0.679. There exists a strong relationship between the two variables. c. H0: ≤ 0 H1: > 0 Reject H0 if t >1.796 df = 11
t
0.824 13 2 1 (0.824) 2
4.817
Reject H0. There is a significant positive correlation between store size and sales. (LO12-1, 12-2, 12-6& 12-7)
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68.
a. Median Age and Population 34.5 34.0
Median Age
33.5 33.0 32.5 32.0 31.5 31.0 30.5 30.0 0.000
b. c.
a. Emily Smith 20000
Estimated Cost
15000
10000
5000
0 1
2
3
4
5
6
7
Age
b.
10.000
r = 0.452 The slope of 0.272 indicates that for each increase of 100 000 in the population, the median age increases on average by 0.272 years. The median age is 32.1 years, found by 31.4 + 0.272 (2.5). The p-value (0.190) for the population variable is greater than, say 0.05. A test for significance of that coefficient would fail to be rejected. In other words, it is possible the population coefficient is zero. (LO12-1, 12-4& 12-5)
d. e.
69.
2.000 4.000 6.000 8.000 Population (in hundred thousands)
r = –0.822
- 179 -
8
9
10
c. The slope of –1534 indicates that for each increase of 1 year in the age of the car that the estimated cost decreases on average by $1534. d. $10688, found by 18 358 – 1534(5) e. The p-value (0.000) for the age variable is less than, say 0.05. A test for significance of that coefficient would be rejected. In other words, the coefficient is much different from zero. (LO12-1,12-4& 12-5) 70. (LO12-1,12-3,12-4) a. Victoria Regression Statistics Multiple R 0.9164 R Square 0.8397 Adjusted R Square 0.8389 Standard Error 46399.8103 Observations 191 ANOVA df Regression Residual Total
Intercept Greater Vancouver
SS MS 1 2131726806542.910 2131726806542.910 189 406906112828.818 2152942395.920 190 2538632919371.730
Coefficients 234181.27 0.3161
Standard Error 10969.0606 0.0100
F Significance F 990.146 0.000
t Stat 21.3493 31.4666
P-value 0.0000 0.0000
Lower 95% Upper 95% Lower 95.0% Upper 95.0% 212543.7550 255818.7849 212543.7550 255818.7849 0.2963 0.3360 0.2963 0.3360
Predicted values for: Victoria 95% Confidence Interval
Greater Vancouver
Predicted
10 00 000
lower
95% Prediction Interval
upper
5 50 321.907 5 43 653.319
lower
upper
5 56 990.495 4 58 551.260 6 42 092.555
Victoria = 234 181.3 + 0.3161(x). When Greater Vancouver’s list price is $1 000 000 a point estimate for a home in Victoria is $550 321.9; A 95% confidence interval is between $543 653.32 and $556 990.50; The 95% prediction interval is between $458 551.26 and $642 092.55 b. Lower Mainland Regression Statistics Multiple R 0.9955 R Square 0.9910 Adjusted R Square 0.9909 Standard Error 26324.9986 Observations 191 ANOVA df Regression Residual Total
Intercept Greater Vancouver
SS MS 1 14367410460839.400 14367410460839.400 189 130978049003.508 693005550.283 190 14498388509842.900
Coefficients 13621.547 0.821
Standard Error 6223.312 0.006
t Stat 2.189 143.986
- 180 -
F Significance F 20732.028 0.000
P-value 0.030 0.000
Lower 95% Upper 95% Lower 95.0% Upper 95.0% 1345.472 25897.622 1345.472 25897.622 0.809 0.832 0.809 0.832
Predicted values for: Lower Mainland 95% Confidence Interval
Greater Vancouver
Predicted
10 00 000
95% Prediction Interval
lower
upper
8 34 358.147 8 30 574.714
8 38 141.580
lower
upper
7 82 291.940 8 86 424.355
Lower Mainland = 13 621.55 + 0.821(x). When Greater Vancouver’s list price is $1 000 000 a point estimate for a home in the Lower Mainland is $834 358.2; A 95% confidence interval is between $830 574.71 and $838 141.58; The 95% prediction interval is between $782 291.94 and $886 424.35. c. For Victoria the coefficient of determination is 84% and the standard error is $46 400. The Lower Mainland coefficient of determination is 99% with a standard error of $26 325. Therefore, the independent variable Greater Vancouver area has a higher coefficient of determination and a lower standard error when predicting the Lower Mainland’s list price. One reason Greater Vancouver predicts the Lower Mainland’s price better could be attributed to the fact that both house markets are located on the mainland. 71. (LO12-1,12-3,12-4) a. Worth Regression Statistics Multiple R 0.9417 R Square 0.8869 Adjusted R Square 0.8830 Standard Error 14.5290 Observations 31 ANOVA df Regression Residual Total
Intercept Worth
SS MS F Significance F 47984.2227 47984.2227 227.3150 0.0000 6121.6483 211.0913 54105.8710
1 29 30
Coefficients Standard Error 84.5324 5.8852 0.1191 0.0079
t Stat P-value 14.3635 0.0000 15.0770 0.0000
Lower 95% Upper 95% Lower 95.0% Upper 95.0% 72.4957 96.5690 72.4957 96.5690 0.1029 0.1353 0.1029 0.1353
Predicted values for: Revenue* 95% Confidence Interval
95% Prediction Interval
Worth*
Predicted
lower
upper
lower
upper
750
173.862
168.362
179.362
143.642
204.082
Revenue = 84.5324 + 0.1191(x). When Team Worth is $750 $M a point estimate for a Team Revenue is $173.86 M; A 95% confidence interval is between $168.36 M and $179.36 M; The 95% prediction interval is between $143.64 M and $204.08 M b. Post-Season
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Regression Statistics Multiple R 0.8151 R Square 0.6643 Adjusted R Square 0.6528 Standard Error 25.0253 Observations 31 ANOVA df
SS 35944.1369 18161.7341 54105.8710
MS 35944.1369 626.2667
Coefficients Standard Error 114.9610 7.8875 1.6194 0.2138
t Stat 14.5751 7.5759
Regression Residual Total
Intercept Post Season
1 29 30
F Significance F 57.3943 0.0000
P-value 0.0000 0.0000
Lower 95% 98.8293 1.1822
Upper 95% Lower 95.0% Upper 95.0% 131.0927 98.8293 131.0927 2.0565 1.1822 2.0565
Predicted values for: Revenue* 95% Confidence Interval
95% Prediction Interval
Post Season
Predicted
lower
upper
lower
upper
Leverage
30
163.542
154.348
172.736
111.540
215.544
0.032
Revenue = 114.961 + 1.6194(x). When the number of Post Season appearances a team has had is 30 a point estimate for the Team Revenue is $163.54 M; A 95% confidence interval is between $154.35 M and $172.74 M; The 95% prediction interval is between $111.54 M and $215.54 M. c. For Team Worth the coefficient of determination is 88.7% and the standard error is $14.5 M. A Teams number of Post Season appearances has a coefficient of determination of 66.4% with a standard error of $25.0M. Therefore, the independent variable Team Worth has a higher coefficient of determination and a lower standard error when predicting the Team’s Revenues. One reason greater Team Worth predicts the Revenues better could be attributed to newer teams in the league may have not as many post season appearances.
CHAPTER 13 MULTIPLE REGRESSION AND CORRELATION ANALYSIS 1.
a. Multiple regression equation b. 64 100 is the y-intercept, the value of y′ when all the x-values are zero. c. $374 748, found by y = 64 100 + 0.394(796 000) + 9.6(6940) 11 600(6) (LO13-1)
2.
a. Multiple regression equation b. One dependent, four independent c. A regression coefficient d. 0.002 e. 105.014, found by y = 11.6 + 0.4(6) + 0.286(280) + 0.112(97) + 0.002(35) (LO13-1)
- 182 -
3.
a. b.
497.736, found by y = 16.24 + 0.017(18) + 0.0028(26 500) + 42(3) + 0.0012(156 000) + 0.19(141) + 26.8(2.5) Two more social activities. Additional income of $10 000 a year will increase the satisfaction index by only 28; two more social activities a week will increase the satisfaction index by 53.6. (LO131)
4.
a. b. c.
21.72 cubic feet Consumption of natural gas would reduce by 0.582 cubic feet. It is logical, the more insulation the less heat needed, the higher the temperature the less heat needed. (LO13-1)
5.
a. b. c.
65 2 1
d.
se.12
SSE 583.693 9.414 3.068 n (k 1) 65 (2 1)
95% of the residuals will be between ±6.136, found by 2(3.068) e.
R2
SSR 77.907 0.118 SStotal 661.6
The independent variables explain 11.8% of the variation. 3) 6.
a. b. c.
52 5 1
d.
se.12345
(LO13-2 & 13-
SSE 2647.38 57.55 7.59 n (k 1) 52 (5 1)
95% of the residuals will be between ±15.17, found by 2(7.59) e.
R2
SSR 3710 0.584 SStotal 6357.38
The independent variables explain 58.4% of the variation. 3) 7.
a. b. c. d.
e.
y′ = 20 − x1 + 12x2 − 15x3 y′ = 20 − (4) + 12(6) – 15(8) = −32 n = 22; 3 independent variables Source SS df MS F Regression 7 500 3 2 500 18 Residual 2 500 18 138.89 Total 10 000 21 H0: 1 = 2 = 3 = 0 H1: Not all ’s are 0. Reject H0 if F > 3.16 The computed value of F is 18, reject H0. Not all regression coefficients equal
zero. f.
(LO13-2 & 13-
For x1
for x2
for x3
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H0: 1 = 0 H0: 2 = 0 H0: 3 = 0 H1: 1 0 H1: 2 0 H1: 3 0 t = 4.00 t = 1.50 t = 3.00 Reject H0 if t > 2.101 or t < 2.101 Delete variable 2, keep 1 and 3
(LO13-1, 13-2
& 13-4) 8.
a. b. c. d.
e.
f.
y′ = −150 + 2000x1 − 25x2 + 5x3 − 300x4 + 0.6x5 y′ = −150 + 2000(4) – 25(6) + 5(8) − 300(6) + 0.60(8) = 5944.8 n = 21; 5 independent variables Source SS df MS F Regression 1500 5 300 9.00 Residual 500 15 33.33 Total 2000 20 H0: 1 = 2 = 3 = 4 = 5 = 0 H1: Not all ’s are 0. Reject H0 if F > 2.90 Reject H0. Not all of the regression coefficients are zero. For x1 for x2 for x3 for x4 H0: 1 = 0 H0: 2 = 0 H0: 3 = 0 H0: 4 = 0 H1: 1 0 H1: 2 0 H1: 3 0 H1: 4 0 t = 4.00 t = 0.833 t = 1.00 t = 3.00 Reject H0 if t > 2.131 or t < 2.131 Delete variable 2 then rerun, perhaps delete variable 3 also.
for x5 H0: 5 = 0 H1: 5 0 t = 4.00 (LO13-1, 13-2
& 13-4) 9.
a. b. variable. c. 7)
x4 has the strongest correlation with the dependent variable. x2, x3, and x4 because they have the strongest correlation with the dependent
10.
Income has the strongest correlation with the dependent variable. Income (0.806) and Population (0.510) because they have the strongest correlation with the dependent variable. No; there is no evidence of multicollinearity. (LO137)
a. b. c.
11.
Yes, between x3 and x4.
(LO13-
a. b. c. (LO13-7)
Horsepower has the strongest correlation with speed at 0.83. It is reasonable as the weight of a car would probably slow it down. No, none of the independent variables is highly correlated with each other.
12.
a. b.
Location has the strongest correlation with the selling price. Yes, the number of bedrooms and the number of bathrooms are highly correlated. If both are left in the equation, incorrect conclusions about statistical significance may be reached, so one variable should be dropped. (LO13-7)
13.
a.
n = 40
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14.
15.
b. c.
4 R2 = 750/1250 = 0.60
d.
Se1234 500 / 35 3.7796
e.
H0: 1 = 2 = 3 = 4 = 0 H1: Not all ’s equal 0. H0 is rejected if F > 2.65 750 / 4 F 13125 . 500 / 35 H0 is rejected. At least one of the regression coefficients is not equal to zero. (LO13-1, 13-3 & 13-4)
H0: 1 = 0 H0: 2 = 0 H1: 1 0 H1: 2 0 H0 is rejected if t < 2.074 or t > 2.074 0.88 2.676 t 4.779 t 1.239 0.56 0.71 The second variable can be deleted. a. b. c. d.
e.
16.
n = 26 R2 = 100/140 = 0.7143 1.4142, found by 2 H0: 1 = 2 = 3 = 4 = 5 = 0 H1: Not all ’s equal 0. Reject H0 if F > 2.71 Computed F is 10.0. Reject H0. At least one regression coefficient is not zero. H0 is rejected in each case if t < 2.086 or t > 2.086. x1 and x5 should be dropped. (x1 = 1.33, do not reject; x2 = 15, reject; x3 = 4, reject; x4 = −2.5, reject; x5 = 0.75, do not reject) (LO13-1, 13-3 & 13-4)
H0: 1 = 2 = 3 = 4 = 5 = 0 H1: Not all ’s equal zero. df1 = 5 df2 = 20 – (5 + 1) = 14 H0 is rejected if F > 2.96 Source SS df MSE F Regression 448.28 5 89.656 17.58 Error 71.40 14 5.10 Total 519.68 19 So, H0 is rejected. Not all the regression coefficients equal zero.
4) 17.
(LO13-4)
a.
$28 000
b.
0.5810, found by R 2
c. d.
SSR 3050 SStotal 5250 9.199, found by 84.62 H0 is rejected if F > 2.98 (approximately)
- 185 -
(LO13-2 & 13-
e.
18. with
a.
Computed F = 1016.67/84.62 = 12.01 H0 is rejected. At least one regression coefficient is not zero. If computed t is to the left of 2.056 or to the right of 2.056, the null hypothesis in each of these cases is rejected. Computed t for x2 and x3 exceed the critical value. Thus, “population” and “advertising expenses” should be retained and “number of competitors,” should be dropped. (LO132, 13-3 & 13-4) The strongest correlation is between GPA and HS Marks. There is no problem multicollinearity.
R2
c.
H0 is rejected if F > 5.41 F = 1.2989/0.2333 = 5.57. At least one coefficient is not zero. Any H0 is rejected if t < 2.571 or t > 2.571. It appears that only HS Marks is significant. Verbal and Math could be eliminated. Also, Verbal and Math could be eliminated as their p-values are greater than 0.05. R2 = 3.7159/5.0631 = 0.7339, which is not too different from the model in b where Math and Verbal were included. The normality assumption seems reasonable since the graph is fairly linear. The plot does not appear to violate any assumptions of regression. (LO13-2, 13-3, 13-4, 13-6, 13-7 & 13-5)
d.
e. f. g.
19.
a.
b.
c.
d. e.
f. g.
20.
3.8967 = 0.7696 5.0631
b.
a.
The strongest relationship is between sales and income (0.964). A problem could occur if both “outlets” and “income” (0.825) and “automobiles” and “outlets” (0.775) are part of the final solution. This is called multicollinearity. y΄ = −19.6715 − 0.0006(outlets) + 1.7399(automobiles) + 0.4099(income) + 2.0357(age) − 0.0344(supervisors) 1593.81 R2 0.9943 1602.89 H0 is rejected. At least one regression coefficient is not zero. The computed value of F is 140.36. Critical value is 6.26; therefore, reject the null hypothesis (Note that the p-value is 0.0001, which is less than the significance level 0.05, and supports the decision to reject the null. Delete “outlets” and “supervisors”. Critical values are 2.776 and 2.776. Note that “age” is also insignificant. 1593.66 R2 0.9942 1602.89 There was little change in the coefficient of determination. Note that age is now a significant variable. The normality assumption seems reasonable since the graph is fairly linear. The plot does not appear to violate any assumptions of regression. (LO13-2, 13-4, 13-5, 13-6, & 13-7) The correlation matrix is: Salary Experience Rating Salary 1.000
- 186 -
Master's
Experience 0.868 1.000 Rating 0.547 0.187 1.000 Master's 0.311 0.208 0.458 1.000 Experience has the strongest correlation with salary. There does not appear to be a problem with multicollinearity. b. Regression output variables coefficients Intercept 43.9152 Experience 0.8994 Rating 0.1539 Master’s -0.6673
std. error 1.9163 0.0877 0.0314 1.2139
t (df=16) 22.917 10.258 4.895 -0.550
p-value 1.16E-13 1.93E-08 .0002 .5901
The regression equation is Salary = 43.9152 + 0.8994(Experience) + 0.1539(Rating) 0.6673(Master’s) Years 5
c. d.
e.
Rating 60
Masters 0
Predicted 57.6470484
y = 57.647 or $57 647 H0 is rejected if F > 3.24; Computed F = 301.06/5.71 = 52.72 H0 is rejected. At least one regression coefficient is not zero. A regression coefficient is dropped if computed t is to the left of 2.120 or right of 2.120. Keep “Experience” and “Rating”; drop “Master’s.” (See the regression output above). Dropping “Master’s”, we have: Salary = 44.1157 + 0.8926(Experience) + 0.1464(Rating) Regression output variables Intercept Experience Rating
coefficients 44.1157 0.8926 0.1464
std. error 1.8422 0.0850 0.0277
f. Observation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
y 55.10 57.60 53.30 67.00 62.60 69.00 66.00 60.80 72.60 55.70 49.70 54.60 75.80 70.70 62.40 57.60 65.80 54.70
Predicted 56.38 54.87 53.37 66.29 64.62 68.32 63.27 58.27 71.80 59.97 49.40 55.02 76.94 70.42 64.78 60.27 64.43 56.76
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Residual -1.28 2.73 -0.07 0.71 -2.02 0.68 2.73 2.53 0.80 -4.27 0.30 -0.42 -1.14 0.28 -2.38 -2.67 1.37 -2.06
t (df=17) p-value 23.947 1.55E-14 10.499 7.54E-09 5.283 0.0001
19 20
56.80 66.80
57.61 61.80
-0.81 5.00
Residuals
Residual (gridlines = std. error)
6.00 4.00 2.00 0.00 -2.00 -4.00 -6.00 0
5
10 15 Observation
20
25
Normal Probability Plot of Residuals 6.00 4.00
Residual
2.00 0.00 -2.00 -4.00 -6.00 -3.0
-2.0
-1.0
0.0
1.0
Normal Score
The distribution of the residuals is approximately normal.
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2.0
3.0
g. Residuals by Predicted
Residual (gridlines = std. error)
7.02032 4.68022 2.34011 0.00000 -2.34011 -4.68022 -7.02032 10
15
20
25
30
35
40
45
Predicted
The plot does not appear to violate any assumptions of regression. (LO13-1, 134, 13-5, 13-6, & 13-7) 21.
a.
The correlation matrix is: Cars Adv Cars 1.000 Adv 0.808 1.000 Sales 0.872 0.537 City 0.639 0.713
Sales
City
1.000 0.389
1.000
Size of sales force (0.872) has strongest correlation with cars sold. Fairly strong relationship between location of dealership and advertising (0.713). Could be a problem. b. The regression equation is Cars = 31.1328 + 2.1516(Adv) + 5.0140(Sales) + 5.6651(City) y = 31.1328 + 2.1516(15) + 5.0140(20) + 5.6651(1) = 169.352 c. H0: 1 = 2 = 3 = 0 H1: Not all ’s are 0. Reject H0 if computed F > 4.07 Analysis of Variance Source SS df MS Regression 5504.4 3 1834.8 Residual 420.2 8 52.5 Total 5924.7 11 F = 1834.8/52.5 = 34.95. Reject H0. At least one regression coefficient is not 0.
- 189 -
d.
e.
H0 is rejected in all cases if t < 2.306 or if t > 2.306. Advertising and sales force should be retained, city should be dropped. Note that dropping city removes the problem with multicollinearity.) Predictor Coef std. error t (df=8) P Constant 31.1328 13.3961 2.32 0.049 Adv 2.1516 0.8049 2.67 0.028 Sales 5.0140 0.9105 5.51 0.000 City 5.6651 6.3324 0.89 0.397 The new output is Cars = 25.30 + 2.6187adv + 5.0233sales Predictor Coef std. error t (df=9) Constant 25.2952 11.5689 2.19 Adv 2.6187 0.6057 4.32 Sales 5.0233 0.9003 5.58 Analysis of Variance Source SS df MS Regression 5462.4 2 2731.2 Residual 462.3 9 51.4 Total 5924.7 11
f. Observation 1 2 3 4 5 6 7 8 9 10 11 12
Cars 127.0 138.0 159.0 144.0 139.0 128.0 161.0 180.0 102.0 163.0 106.0 149.0
Predicted 122.7 139.9 153.2 145.8 130.1 127.5 161.1 178.8 99.7 168.5 122.7 146.0
- 190 -
Residual 4.3 -1.9 5.8 -1.8 8.9 0.5 -0.1 1.2 2.3 -5.5 -16.7 3.0
Residuals
Residual (gridlines = std. error)
14.3 7.2 0.0 -7.2 -14.3 -21.5 0
2
4
6
8
10
12
14
Observation
Normal Probability Plot of Residuals 15 10 Residual
5 0 -5 -10 -15 -20 -2.0
-1.0
0.0 Normal Score
1.0
2.0
The normality assumption is reasonable. g. The critical value could be a problem. However, with a small sample the residual plot is acceptable.
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Residual (gridlines = std. error)
Residuals by Predicted 14.3 7.2 0.0 -7.2 -14.3 -21.5 50
100
150
200
Predicted
(LO13-1, 13-4, 13-5 & 13-6) 22.
a. b.
The regression equation is y = 1480.7446 + 0.7315x1 + 9.991x2 2.308x3 R2 = 83.5% or 0.835. 83.5% of the variation in the sales is explained by the store area, parking spaces,
and c.
income. H0: 1 = 2 = 3 = 0 H1: Not all ’s = 0 Reject H0 if F > 3.59
F d.
10 057.7 / 3 18.60 1982.3 /11
H0 is rejected. At least one of the net regression coefficients do not equal zero. H0: 1 = 0 H0: 2 = 0 H0: 3 = 0 H1: 1 0 H1: 2 0 H1: 3 0 Reject H0 if t < 2.201 or t > 2.201 Reject H0 for store area and parking space, do not reject for income. Delete
income. e. R2 = 0.804, y = 1342.4902 + 0.7727x1 + 11.6338x2 & 13-4) 23.
a. b.
c.
(LO13-1, 13-3
The regression equation is y = 965.281 + 2.865x1 + 6.754x2 + 0.2873x3 y = 965.281 + 2.865(220) + 6.754(64) + 0.2873(1500) = 2458.775 or $2 458 775 Analysis of Variance Source SS DF MS Regression 45 510 101 3 15 170 034 Error 12 215 892 12 1 017 991 Total 57 725 994 15 F = 15 170 034/1 017 991 = 14.902 H0 is rejected because computed F of 14.9 is greater than the critical value of 3.49. At least one of the regression coefficients is not zero. H0: 1 = 0 H0: 2 = 0 H0: 3 = 0 H1: 1 0 H1: 2 0 H1: 3 0
- 192 -
Reject H0 if t < 2.179 or t > 2.179 variables Intercept x1 x2 x3
d.
t (df=12)
p-value
1.810 0.657 2.586
0.0954 0.5236 0.0238
x1 = 1.810, do not reject; x2 = 0.657, do not reject; x3 = 2.586, reject. Both workers and dividends are not significant variables. Inventory is significant. Delete dividends and rerun the regression equation. The regression equation (if we used x1 and x3) is y = 1134.7613 + 3.258x1 + 0.3099x3 Predictor Coef std. error t (df=13) Constant 1134.7613 418.574 2.71 Workers 3.2575 1.434 2.27 Inventory 0.3099 0.103 3.00 Analysis of Variance Source SS df MS F Regression 45 070 638 2 22 535 319 23.15 Error 12 655 356 13 973 489 Total 57 725 994 15
e. Observation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Profit 2 800 1 300 1 230 1 600 4 500 5 700 3 150 640 3 400 6 700 3 700 6 440 1 280 4 160 3 870 980
Predicted 2 148.6 1 445.7 1 812.3 1 532.9 3 520.8 5 920.3 1 960.0 1 534.7 4 402.7 6 833.4 3 539.9 5 885.9 2 939.9 2 322.7 3 624.0 2 026.2
- 193 -
Residual 651.4 −145.7 -582.3 67.1 979.2 −220.3 1 190.0 −894.7 −1 002.7 −133.4 160.1 554.1 −1 659.9 1 837.3 246.0 −1 046.2
Residuals
Residual (gridlines = std. error)
2,960.0 1,973.3 986.7 0.0 -986.7 -1,973.3 0
5
10
15
20
1.0
2.0
Observation
Normal Probability Plot of Residuals 2,500.0 2,000.0 1,500.0 Residual
1,000.0 500.0 0.0 -500.0 -1,000.0 -1,500.0 -2,000.0 -2.0
-1.0
0.0 Normal Score
The normality assumption is reasonable.
- 194 -
f. Residuals by Predicted
Residual (gridlines = std. error)
2,960.0 1,973.3 986.7 0.0 -986.7 -1,973.3 0
2000
4000
6000
8000
Predicted
The plot does not appear to violate any assumptions of regression. (LO13-1, 13-4, 13-5 & 13-6) 24.
y = 28.2425 + 0.0287x1 + 0.6497x2 0.0490x3 0.0004x4 + 0.7227x5 R2 = 0.750 or 75% 75% of variations in the dependent variable is explained by the independent variables. c & d. Predictor Coef std. error t (df=19) P Constant 28.2425 2.9857 9.46 0.000 Value 0.0287 0.0050 5.77 0.000 Years 0.6497 0.2412 2.69 0.014 Age 0.0490 0.0313 1.57 0.134 Mortgage 0.0004 0.0013 0.32 0.753 Gender 0.7227 0.2491 2.90 0.009 s = 0.5911 R-sq = 75.0% R-sq(adj) = 68.4% Analysis of Variance Source df SS MS F P Regression 5 19.8914 3.9783 11.39 0.000 Error 19 6.6390 0.3494 Total 24 26.5304 Since the p-value of ANOVA is virtually zero, we reject the null hypothesis and conclude that at least one of regression coefficient is not equal to zero. Also, since the p-value for age and mortgage is greater than 0.05, we drop the variables age and mortgage. e. Drop mortgage first. a. b.
R² Adjusted R² R Std. Error
0.748 0.698 0.865 0.578
n k Dep. Var.
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25 4 Income
ANOVA table Source Regression Residual Total
SS 19.8559 6.6745 26.5304
Regression output variables coefficients Intercept 28.0638 Value 0.0281 Years 0.6587 Gender 0.7385 Age −0.0490
df 4 20 24
std. error 2.8661 0.0046 0.2342 0.2386 0.0305
MS 4.9640 0.3337
F 14.87
t (df=20) 9.791 6.173 2.813 3.096 −1.602
p-value 4.50E-09 4.96E-06 0.0107 0.0057 0.1247
P-value 8.62E-06
confidence interval 95% lower 95% upper 22.0851 34.0424 0.0186 0.0376 0.1702 1.1471 0.2409 1.2362 -0.1127 0.0148
The new regression equation is y = 28.0638 + 0.0281(Value) + 0.6587(Years) + 0.7385(Gender) − 0.0490(Age). The R-square value is 0.748. Age is still an insignificant variable. Drop Age. R² Adjusted R² R Std. Error ANOVA table Source Regression Residual Total
0.717 0.661 0.847 0.612
n k Dep. Var.
25 4 Income
SS 19.0344 7.4960 26.5304
df 4 20 24
MS 4.7586 0.3748
F 12.70
Regression output variables coefficients Intercept 29.9896 Value 0.0259 Years 0.3974 Gender 0.6920 Mortgage −0.0004
std. error 2.8683 0.0048 0.1860 0.2572 0.0013
t (df=20) 10.456 5.385 2.137 2.690 −0.308
p-value 1.49E-09 2.86E-05 0.0452 0.0141 0.7612
P-value 2.65E-05
confidence interval 95% lower 95% upper 24.0064 35.9728 0.0159 0.0359 0.0094 0.7853 0.1554 1.2285 −0.0031 0.0023
The new regression equation is y = 29.9896 + 0.0259(Value) + 0.3974(Years) + 0.6920(Gender) − 0.0004(Mortgage). The R-square value is 0.717. Mortgage is still an insignificant variable. Drop age and mortgage. R² Adjusted R²
0.716 0.676
n
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25
R Std. Error ANOVA table Source Regression Residual Total
SS 18.9989 7.5315 26.5304
0.846 0.599
k Dep. Var.
df 3 21 24
3 Income
MS 6.3330 0.3586
F 17.66
Regression output Variables Intercept Value Years Gender
coefficients 29.8109 0.0253 0.4063 0.7079
std. error 2.7479 0.0044 0.1797 0.2465
t (df=21) 10.849 5.803 2.262 2.872
p-value 4.57E-10 9.25E-06 .0344 .0091
P-value 5.90E-06
confidence interval 95% 95% lower upper 24.0964 35.5254 0.0163 0.0344 0.0327 0.7800 0.1952 1.2205
The new regression equation is y = 29.8109 + 0.0253(Value) + 0.4063(Years) + 0.7079(Gender). The R-square value is 0.716. All variables in the equation are significant. (LO13-1, 13-3, 13-4 & 13-5) 25.
a. be a b. c. d.
e.
The correlation of median income and median age is 0.721. So there does appear to linear relationship between the two. Median income is the “dependent” variable. The regression equation is Income = 22 805 + 362(Median Age). For each year increase in age, the income increases $362 on an average. Using an indicator variable for Population > 400 000, the regression equation is Income = 24 865 + 251(Median Age) + 6888(Population). That changes the estimated effect of an additional year of age to $251 and the effect of the population > 400 000 as adding $6888 to income. Notice the p-values are not more than 5%. This indicates that the independent variables are significant influences on income. Predictor Constant med age pop
Coef 24865 250.5 6888
SE Coef 4552 102.4 2501
T 5.46 2.45 2.75
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P 0.002 0.050 0.033
f.
The normality assumption is reasonable. Normal Probability Plot of Residuals 4,000.0 2,000.0
Residual
0.0 -2,000.0 -4,000.0 -6,000.0 -8,000.0 -2.0
g
-1.5
-1.0
-0.5 0.0 0.5 Normal Score
1.0
1.5
2.0
The variation is about the same across the different fitted values. Residuals by Predicted
Residual (gridlines = std. error)
6,853.7
3,426.8
0.0
-3,426.8
-6,853.7
-10,280.5 30000
35000
(LO13-1, 13-4, 13-5, 13-6, & 13-8) 26.
a.
The correlation matrix is: Food Income Food 1.000 Income 0.156 1.000 Size 0.876 −0.098
Size
1.000 - 198 -
40000 Predicted
45000
50000
b.
c.
d.
e.
Income and size are not related. There is no multicollinearity. The regression equation is Food = 2.84 + 0.0061(Income) + 0.4248(Size). Another thousand dollar of income leads to an increased food bill of $6.1. Another member of the family adds $424.8 to the food bill.
R2
18.4095 0.826 22.2935
To conduct the global test: H0: β1 = β 2 = 0 versus H1: Not all ’s = 0 At the 0.05 significance level, H0 is rejected if F > 3.44. Source DF SS MS F P Regression 2 18.4095 9.2048 52.14 0.000 Error 22 3.8840 0.1765 Total 24 22.2935 The computed value of F is 52.14, so H0 is rejected. Some of the regression coefficients and R2 are not zero. Since the p-values are less than 0.05, there is no need to delete variables. Predictor Coef SE Coef t P Constant 2.8436 0.2618 10.86 0.000 Income 0.0061 0.0023 2.72 0.012 Size 0.4248 0.0422 10.06 0.000 The residuals appear normally distributed. Observation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Food 5.04 4.08 5.76 3.48 4.20 4.80 4.32 5.04 6.12 3.24 4.80 3.24 6.60 4.92 6.60 5.40 6.00 5.40 3.36 4.68 4.32 5.52 4.56 5.40 4.80
Predicted 4.996 4.030 5.120 3.587 4.096 4.871 4.607 4.963 5.982 3.665 4.966 4.039 6.292 4.743 7.169 4.984 5.194 4.891 3.709 4.544 4.026 5.505 4.352 6.085 5.264
- 199 -
Residual 0.044 0.050 0.640 −0.107 0.104 −0.071 −0.287 0.077 0.138 −0.425 −0.166 −0.799 0.308 0.177 −0.569 0.416 0.806 0.509 −0.349 0.136 0.294 0.015 0.208 −0.685 −0.464
Normal Probability Plot of Residuals 1.000 0.800 0.600 Residual
0.400 0.200 0.000 -0.200 -0.400 -0.600 -0.800 -1.000 -3.0
-2.0
-1.0
0.0
1.0
2.0
3.0
20
25
30
Normal Score
Residuals
Residual (gridlines = std. error)
1.260 0.840 0.420 0.000 -0.420 -0.840 -1.260 0
5
10
15 Observation
f.
The variance is the same as we move from small values to large. So, there is no homoscedasticity problem.
- 200 -
Residuals Versus the Fitted Values (response is Food)
Residual
1
0
-1 4
5
6
7
Fitted Value
(LO13-1, 13-3, 13-4, 13-5, 13-6, & 13-7) 27.
a. Salary GPA Business
b.
c.
d.
Salary 1.000 0.902 0.911
GPA
Business
1.000 0.851
1.000
Yes, multicollinearity occurs between GPA and Business (.851) Salary = 23.4474 + 2.7748(GPA) + 1.3071(Business) As GPA increases by one point salary increases by $277.48. The average business school graduate makes $1307 more than a corresponding non-business graduate. Estimated salary is $33 079, found by 23.4474 + 2.7748(3.00) + 1.3071(1).
R2
21.182 0.888 23.857
So, 88.8% of the variation in the dependent variable is explained or accounted by the independent variables. Since the p-values are less than 0.05, there is no need to delete variables. Predictor Coef SE Coef T P Constant 23.4474 3.4901 6.72 0.000 GPA 2.7748 1.1068 2.51 0.028 Business 1.3071 0.4660 2.80 0.016
- 201 -
e.
The residuals appear normally distributed. Normal Probability Plot of Residuals 0.80 0.60 0.40
Residual
0.20 0.00 -0.20 -0.40 -0.60 -0.80 -1.00 -1.20 -2.0
f.
-1.0
0.0 Normal Score
1.0
2.0
The variance is the same as we move from small values to large. So, there is no homoscedasticity problem.
RESI1
0.5
0.0
-0.5
-1.0 32
33
34
35
FITS2
(LO13-1, 13-4, 13-5, 13-6, & 13-8) 28.
a. b.
c.
The regression equation is P/E = 29.913 – 5.324(EPS) + 1.449(Yield). Here is part of the software output: Predictor Coef SE Coef t P Constant 29.913 5.767 5.19 0.000 EPS –5.324 1.634 –3.26 0.005 Yield 1.449 1.798 0.81 0.431 Thus, EPS has a significant relationship with P/E while Yield does not. As EPS increases by one, P/E decreases by 5.324 and when Yield increases by one, P/E increases by 1.449.
- 202 -
The second stock has a residual of −18.43, signifying that it’s fitted P/E is around 21.
d. e.
Normal Probability Plot of Residuals 15.000 10.000
Residual
5.000 0.000 -5.000 -10.000 -15.000 -20.000 -3.0
-2.0
-1.0 0.0 1.0 Normal Score
2.0
3.0
The distribution of residuals may show a negative skewness. f. Residuals Versus the Fitted Values (response is P/E) 10
Residual
0
-10
-20 10
20
30
Fitted Value
g.
There does not appear to be any problem with homoscedasticity. The correlation matrix is P/E EPS Yield P/E 1.000 EPS −0.602 1.000 Yield 0.054 0.162 1.000
- 203 -
40
The correlation between the independent variables is shown in the correlation matrix. There is no multicollinearity. (LO13-1, 13-5, 13-6, 13-7, & 13-8) 29.
a.
b.
The regression equation is Sales = 1.0188 + 0.0829(Infomercials) Predictor Coef SE Coef t P Constant 1.0188 0.3105 3.28 0.006 Infomercials 0.0829 0.0168 4.94 0.000 S = 0.3087 R-sq = 65.2% R-sq(adj) = 62.5% Analysis of Variance Source DF SS MS F P Regression 1 2.3214 2.3214 24.36 0.000 Residual Error 13 1.2386 0.0953 Total 14 3.5600 The global test on F demonstrates there is a substantial connection between sales and the number of commercials.
c. 0.60
Normal Probability Plot of Residuals
0.40
Residuals
0.20 0.00 -0.20 -0.40 -0.60 -2.0
30.
-1.0
0.0 Normal Score
1.0
2.0
d. 13-6)
The residuals appear to be fairly normally distributed.
a. b.
Sub = 5.7329 + 0.0075(Popul) + 0.0509(Adv) + 1.0974(Income) H0: 1 = 2 = 3 = 0 H1: Not all ’s = 0 Reject H0 if F > 3.07 11.3437 / 3 F 35.38 2.2446 / 21 All coefficients are significant. Do not delete any. The residuals appear to be random. No problem.
c. d.
- 204 -
(LO13-1, 13-3, 13-5 &
Residual (gridlines = std.error)
Residuals by Predicted 0.7 0.5 0.3 0.1 -0.1 -0.3 -0.5 -0.7
37
38
39 Predicted
40
41
e. The residuals appear normally distributed. No problem. Normal Probability Plot of Residuals 0.8 0.6
Residual
0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -3.0
-2.0
-1.0
0.0 1.0 Normal Score
2.0
3.0
(LO13-1, 13-4, 13-5 & 13-6) 31.
a.
The correlation of Screen and Price is 0.893. So, there does appear to be a linear relationship between the two. b. Price is the “dependent” variable. c. The regression equation is Price = –2484 + 101(Screen). For each inch increase in screen size, the price increases $101 on an average. d. Using “dummy” indicator variables for Sharp and Sony, the regression equation is Price = –2308.2 + 94.1(Screen) + 15.1(Manufacturer Sharp) + 381.4(Manufacturer Sony) Sharp can obtain on an average $15.1 more than Samsung and Sony can collect an additional benefit of $381.4. e. Here is some of the output. Predictor Coef SE Coef t P
- 205 -
Constant
-2308.2
492.0
-4.69
94.1
10.83
8.69
Manufacturer_Sharp 15.1
171.6
0.09
Manufacturer_Sony 381.4
168.8
2.26
0.000 Screen 0.000 0.931 0.036
f.
The p-value for Sharp is relatively large. A test of their coefficient would not be rejected. That means they may not have any real advantage over Samsung. On the other hand, the p-value for the Sony coefficient is quite small. That indicates that it did not happen by chance and there is some real advantage to Sony over Samsung. A histogram of the residuals indicates they follow a normal distribution. Residual plot Normal
4 -5.931
Frequency
3
2
1
0
-400
-200
0 RESI1
200
400
600
The residual variation may be increasing for larger fitted values. Residuals vs. Fits for LCD's 500
250
RESI1
g.
-600
0
-250
-500
1000
1500
2000 FITS1
(LO13-1, 13-4, 13-5, 13-6, & 13-8)
- 206 -
2500
3000
a. Scatterplot of Sales vs Advertising, Accounts, Competitors, Potential Advertising
Accounts 300 200 100 0
Sales
32.
2
4
6 Competitors
8
30
40
50 Potential
60
70
300 200 100 0 4
6
8
10
12
5
10
15
20
Sales seem to fall with the number of competitors and rise with the number of accounts and potential. b. Pearson correlations Sales Advertising Accounts Competitors Advertising 0.159 Accounts 0.783 0.173 Competitors -0.833 -0.038 -0.324 Potential 0.407 -0.071 0.468 -0.202 The number of accounts and the market potential are moderately correlated. c. The regression equation is Sales = 178 + 1.81 Advertising + 3.32 Accounts - 21.2 Competitors + 0.325 Potential Predictor Coef SE Coef T P Constant 178.32 12.96 13.76 0.000 Advertising 1.807 1.081 1.67 0.109 Accounts 3.3178 0.1629 20.37 0.000 Competitors -21.1850 0.7879 -26.89 0.000 Potential 0.3245 0.4678 0.69 0.495 S = 9.60441 R-Sq = 98.9% R-Sq(adj) = 98.7% Analysis of Variance Source DF SS MS F P Regression 4 176777 44194 479.10 0.000 Residual Error 21 1937 92 Total 25 178714 The computed F value is quite large. So we can reject the null hypothesis that all of the regression coefficients are zero. We conclude that some of the independent variables are effective in explaining sales. d. Market potential and advertising have large p-values (0.495 and 0.109, respectively). You would probably drop them. e. If you omit potential, the regression equation is Sales = 180 + 1.68 Advertising + 3.37 Accounts - 21.2 Competitors Predictor Coef SE Coef T P
- 207 -
Constant 179.84 12.62 14.25 0.000 Advertising 1.677 1.052 1.59 0.125 Accounts 3.3694 0.1432 23.52 0.000 Competitors -21.2165 0.7773 -27.30 0.000 Now advertising is not significant. That would also lead you to cut out the advertising variable and report the polished regression equation is: Sales = 187 + 3.41 Accounts - 21.2 Competitors. Predictor Coef SE Coef T P Constant 186.69 12.26 15.23 0.000 Accounts 3.4081 0.1458 23.37 0.000 Competitors -21.1930 0.8028 -26.40 0.000 f. Histogram of the Residuals (response is Sales)
9 8 7
Frequency
6 5 4 3 2 1 0
-20
-10
0 Residual
10
20
Normal Probability Plot of Residuals 30.00
Residual
20.00 10.00 0.00 -10.00 -20.00 -30.00 -3.0
-2.0
-1.0 0.0 1.0 Normal Score
2.0
3.0
The histogram and normal probability plot looks to be normal. There are no problems shown in this plot. (LO13-1, 13-4, 13-5, 13-6, & 13-7)
- 208 -
33. (LO13-1, 13-3, 13-4, 13-7, 13-8) a.
Regression Analysis R² Adjusted R² R Std. Error ANOVA table Source Regression Residual Total
0.905 0.886 0.951 113.547
n k Dep. Var.
31 5 Worth
SS 3,060,119.4907 322,322.4448 3,382,441.9355
df 5 25 30
MS 612,023.8981 12,892.8978
F 47.47
coefficients -516.225 7.546 5.811 -133.311 0.064 -0.003
std. error 211.421 0.581 5.942 65.497 0.444 0.012
t (df=25) -2.442 12.994 0.978 -2.035 0.143 -0.283
p-value 0.022 0.000 0.337 0.053 0.887 0.779
Regression output variables Intercept Revenue Temp C0 Country cm Attendance
p-value 5.78E-12
b0 = −516.225: If all the independent variables (x’s) were equal to zero, the net worth of the team would be $-516.225 M. b1 = 7.546: As a team’s Revenue increases by $1M, the team Worth also increases by $7.546 M, all other things being equal. b2 = 5.811: As the mean annual temperature in a team’s city increases by 1 0 Celsius the team Worth also increases by $5.811 M, all other things being equal. b3 = −133.311: As you change the Country from Canada to USA the team Worth will decrease by $133.311 M, all other things being equal. b4 = 0.064: As the total amount of Snow a team’s annual snowfall increases by 1 cm the team Worth also increases by $0.06 M, all other things being equal. b5 = −0.003: As the team’s average attendance increases by 1 person the team Worth decreases by $0.003 M, all other things being equal. b. R2 = 0.905. 90.5% of the variation in the team’s Worth can be explained by these 5 independent variables. c. Correlation Matrix Worth
Revenue
Temperature C0
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Country
Snowfall cm
Attendance
Worth Revenue Temperature 0 C Country Snowfall cm Attendance
1.00 0.94
1.00
−0.26 −0.20 0.29 0.41
−0.25 −0.09 0.26 0.45
1.00 0.64 −0.66 0.01
1.00 −0.53 −0.02
1.00 0.01
Dependent Variable = Worth The only variable that has strong correlation with the teams Worth is teams Revenue with a positive correlation of 0.94 Marked in Yellow, these variables flipped their sign from the regression equation. This is a result of too strong correlation between the independent variables (Temperature and Revenue r = −0.25) and between the independent variables (Attendance and Revenue r = 0.45). These two relationships will lead to multicollinearity. Other cautionary independent relations are shown in orange. d. H0: H1: At least one of the regression coefficient is not equal to zero. At a significance level of 0.05, the (Global) F-Stat is 47.47. The Critical F(df N = 5, df D = 25) = 2.60 Our conclusion is to reject H0 in favour of H1 indicating at least one of the variables is useful. e. H0: for k = 1, 2, 3, 4, 5. The only p-value < 0.05 is for Revenues at 0.000. Therefore we can reject Ho for the X1 variable Revenue and do not reject Ho for the remaining varibles. This means we should keep only Revenue to predict team Worth. Regression Statistics Multiple R 0.951 R Square 0.904 Adjusted R Square 0.894 Standard Error 109.492 Observations 31.000 ANOVA Regression Residual Total
Intercept Revenue Tempearture 0C Country
df 3 27 30 Coefficients −547.4933 7.4805 5.1669 −133.4304
SS 3 058 754.47 323 687.47 3 382 441.94 Standard Error 119.774 0.4874 5.0441 61.5796
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MS 1 019 585 11 988
F 85.045
t Stat −4.57 15.35 1.02 −2.17
P-value 9.64E-05 7.38E-15 0.314763 0.039242
Significance F 7.1E-14
1
f. Regression Analysis -- Stepwise Selection displaying the best model of each size 31 Worth
Nvar 1 2 3 4 5
Revenue .0000 .0000 .0000 .0000 .0000
observations is the dependent variable p-values for the coefficients Temp C Country cm .0593 .0392 .0412 .0525
.3148 .3066 .3374
.8874
Attendance
.7682 .7794
s 114.876 109.588 109.492 111.387 113.547
Adj R² .883 .893 .894 .890 .886
R² .887 .901 .904 .905 .905
Cp 2.683 1.082 2.106 4.020 6.000
p-value 2.94E-15 9.21E-15 7.10E-14 6.89E-13 5.78E-12
Using Stepwise regression the 3-variable model which includes: Revenue, Temperature and Country produces the lowest standard error (s) and the highest adjusted R squared. 34 (LO13-1, 13-3, 13-4, 13-7, 13-8) a. Regression Statistics Multiple R 0.955 R Square 0.911 Adjusted R Square 0.898 Standard Error 1.567 Observations 31 ANOVA
Regression Residual Total
df 4 26 30
SS 655.14 63.82 718.97
Intercept Post In the Cup Points 19/20 Star Power
Coefficients −1.86412 −0.07402 0.79033 0.03598 −0.02836
Standard Error 2.384792 0.031671 0.088101 0.031389 0.016062
MS 163.79 2.45
F 66.72
t Stat −0.78167 −2.33724 8.97077 1.14624 −1.76559
P-value 0.44147 0.02740 1.93E-09 0.26214 0.08921
Significance F 0.0000
b0 = 1.86: If all the independent variables (x’s) were equal to zero, the number of wins for a team is 1.86.
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b1 = −0.07: As the number of times the teams is in the post season increases by 1, the number of times they are expected to win the cup decreases by 0.07 ceterus paribus. This is counter intuitive. b2 = 0.79: As the number of times a team is playing for the cup increases by 1, the number of times they win the cup also increases by nearly 1(0.79) ceterus paribus. b3 = 0.036: As the number of points a team achieves in the 2019/2020 hockey season increases by 1, the number of win increases by 0.036 ceterus paribus. b4 = −0.029: As the total number of Star Power a team has increases by 1, the number of wins will actually decreases by 0.028 ceterus paribus. b. R-Squared = 0.911. 91.1% of the variation in the number of times a team wins can be determined by the variation in these 4 independent variables. c. Correlation Matrix
Wins Post In the Cup Points 19/20 Star Power
Wins
Post
1 0.787 0.939 −0.184 0.161
1.000 0.902 −0.071 0.294
In the Cup
Points 2019/2020
Star Power
1.000 −0.173 0.269
1.000 0.535
1.000
Both number of times in the Post Season and number of times playing in the Cup have high positive correlations. In yellow Post Season, Points 2019/2020, and Star Power all had slopes that flipped their signs compared to the type of correlation in the matrix. This is a warning of multicollinearity being present in the model. The orange colouring highlights where too much correlation is occurring between pairs of independent variables. For instance, 0.535 shows too much correlation between the two independent variables Star Power and Points 2019/2020 when compared to their correlations with the dependent variable Wins. d. H0: H1: At least one of β’s not equal to zero. At a significance level of 0.05, the (Global) F-Stat is 66.72. The Critical F(df N = 4, df D = 26) = 2.74 Our conclusion is to reject H0 in favour of H1 indicating at least one of the variables is useful. e H0: for k = 1, 2, 3, 4. The p-values less than 0.05 are Post Season (0.027) and In the Cup (0.0000). However, when only including number of appearances in the cup the standard error is low and the adjusted R-squared value is high. The p-value for the overall model is 0.000 and the pvalue for just appearances in the cup is 0.000. We have deleted the other variables because of multicollinearity (flipped sign) or the variable was not significant. SUMMARY OUTPUT Regression Statistics Multiple R 0.939 R Square 0.881
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Adjusted R Square Standard Error Observations
0.877 1.715 31
ANOVA
Regression Residual Total
df 1 29 30
SS 633.703 85.265 718.968
MS 633.703 2.940
F 215.53
Intercept In the Cup
Coefficients −0.4776 0.5788
Standard Error 0.3899 0.0394
t Stat −1.2251 14.6810
P-value 0.230401 5.84E-15
Significance F 5.84033E-15
f. Using stepwise regression from megastat, identify the best fit model. Regression Analysis -- Stepwise Selection displaying the best model of each size 31 Wins
Nvar 1 2 3 4
Seasons .0277 .0389 .0274
observations is the dependent variable p-values for the coefficients In the Cup Points Star Power .0000 .0000 .0000 .1918 .0000 .2621 .0892
s 1.715 1.598 1.576 1.567
Adj R² .877 .893 .896 .898
R² .881 .901 .907 .911
Cp 7.735 4.127 4.314 5.000
p-value 5.84E-15 9.25E-15 5.02E-14 2.73E-13
The model with all 4 variables produces the lowest standard error (s) and the highest adjusted R2
CHAPTER 14 CHI-SQUARE APPLICATIONS FOR NOMINAL DATA 1.
2.
3.
a. b.
3 7.815
(LO14-1)
a. b.
5 15.086
(LO14-1)
a.
Reject H0 if 2 > 7.815 (10 20) 2 (20 20) 2 (30 20) 2 (20 20) 2 2 10.0 20 20 20 20
b.
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4.
c.
Reject H0. The frequencies are not equal.
a.
Reject H0 if 2 > 5.991 (10 20) 2 (20 20) 2 (30 20) 2 2 10.0 20 20 20 Reject H0. The frequencies are not equal.
b. c. 5.
(LO14-1)
(LO14-1)
H0: The outcomes are the same H1: The outcomes are not the same Reject H0 if 2 > 9.236 (3 5) 2 (7 5) 2 2 ... 7.60 5 5 Do not reject H0. Cannot reject H0 that outcomes are the same. Die is fair.
(LO14-
1) 6.
H0: Rounds played is the same for each day H1: Rounds played is not the same 2 Reject H0 if > 9.488 (124 104) 2 (120 104) 2 2 ... 15.308 104 104 Reject H0. The number of rounds played is not the same for each day. (LO14-
1) 7. H0: There is no difference in the proportions H1: There is a difference in the proportions Reject H0 if 2 > 15.086 (47 40) 2 (34 40) 2 2 ... 3.400 40 40 Do not reject H0. There is no difference in the proportions. (LO14-1) 8.
H0: The proportions are the same H1: The proportions are not the same Reject H0 if 2 > 18.475 (6 10) 2 (19 10) 2 (6 10) 2 2 ... 24.60 10 10 10 Reject H0. The accidents are not evenly distributed throughout the day.
(LO14-
1) 9.
a. b. c.
Reject H0 if 2 > 9.210 (30 24) 2 (20 24) 2 (10 12) 2 2 2.50 24 24 12 Do not reject H0.
(LO14-
3) 10.
H0: The proportions are as stated H1: The proportions are not as stated Reject H0 if 2 > 7.824 (60 50) 2 (30 25) 2 (10 25) 2 2 12.00 50 25 25 Reject H0. Proportions are not as stated.
3)
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(LO14-
11.
H0: The proportions are as stated H1: The proportions are not as stated 2 Reject H0 if > 11.345 (50 25) 2 (100 75) 2 (190 125) 2 (160 275) 2 2 115.22 25 75 125 275 Reject H0. Proportions are not as stated.
(LO14-
3) 12.
H0: The proportions are as stated Reject H0 if 2 > 7.815
2
H1: The proportions are not as stated
(28 42) 2 (32 27.6) 2 (30 25.2) 2 (30 25.2) 2 7.197 42 27.6 25.2 25.2
=7.197 Do not reject H0. Proportion of viewers is as stated.
(LO14-
3) 13.
H0: There is no relationship between the person resides and section read H1: There is a relationship Reject H0 if 2 > 9.488 (170 157.50) 2 (88 83.62) 2 2 ... 7.340 157.50 83.62 Do not reject H0. There is no relationship between where the person resides and section
read. (LO14-4) 14.
H0: There is no relationship between quality and manufacturer H1: There is a relationship Reject H0 if 2 > 7.815 (12 9) 2 (8 9) 2 (89 91) 2 2 ... 3.663 9 9 91 Do not reject H0. There is no relationship between quality and manufacturer.
(LO14-
H0: No relationship between error rates and item type H1: There is a relationship between error rates and item type Reject H0 if 2 > 9.21 (20 14.1) 2 (10 15.9) 2 (200 199.75) 2 (225 225.25) 2 2 ... 8.033 14.1 15.9 199.75 225.25 Do not reject H0. There is no relationship between error rates and item type. (Computer value = 8.036)
(LO14-
4) 15.
4) 16.
H0: No relationship between phone use and accidents H1: There is a relationship between phone use and accidents Reject H0 if 2 > 3.841
(25 31.45) 2 (300 293.55)2 (50 43.55)2 (400 406.45)2 2.523 31.45 293.55 43.55 406.45 2
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Do not reject H0. There is no relationship between phone use and accidents.
(LO14-
H0: ps = 0.50, pr = pl = 0.25 H1: Distribution is not as given above. df = 2 Reject H0 if 2 > 4.605. Turn fo fe fo − fe (fo − fe)2/fe Straight 112 100 12 1.44 Right 48 50 2 0.08 Left 40 50 10 2.00 Total 200 200 3.52 H0 is not rejected. The proportions are as given in the null hypothesis.
(LO14-
H0: ps = pc = pe H1: The proportions are not equal. df = 2 Reject H0 if 2 > 5.991. Gift fo fe fo − fe (fo − fe)2/fe Sweatshirt 183 166.67 16.33 1.6000 Coffee Cup 175 166.67 8.33 0.4163 Earrings 142 166.67 24.67 3.6516 Total 500 500.00 0 5.6679 H0 is not rejected. There is not a preference for the gifts.
(LO14-
H0: There is no preference with respect to TV stations. H1: There is a preference with respect to TV stations. df = 3 1 = 2 H0 is rejected if 2 > 5.991 TV Station fo fe fo − fe (fo − fe)2 (fo − fe)2/fe CTV 53 50 3 9 0.18 Global 64 50 14 196 3.92 Citytv 33 50 17 289 5.78 Total 150 150 9.88 H0 is rejected. There is a preference for TV stations.
(LO14-
H0: p1 = p2 = p3 = p4 H1: The proportions are not equal. 2 df = 3 Reject H0 if > 11.345. Entrance fo fe fo − fe (fo − fe)2/fe Main 140 100 40 16.00 Broad 120 100 20 4.00 Cherry 90 100 10 1.00 Walnut 50 100 50 25.00 Total 400 400 0 46.00 H0 is rejected. The entrances are not equally likely.
(LO14-
4) 17.
1) 18.
1) 19.
3) 20.
1) 21.
H0: pne = 0.21, pm = 0.24, ps= 0.35 and pw = 0.20 H1: The distribution is not as given. Reject H0 if 2 > 11.345.
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Area fo fe fo − fe (fo − fe)2/fe New York State 68 84 16 3.0476 Ont(exc GTA) 104 96 8 0.6667 GTA 155 140 15 1.6071 Other 73 80 7 0.6125 Total 400 400 0 5.9339 There is not enough evidence to reject H0. The geographical distribution of her club members has not changed. (LO14-3) 22. 2
H0: Distribution with Poisson with = 2
H1: Distribution is not Poisson with =
23.
H0: p0 0.4, p1 0.3, p2 0.2, p3 0.1 H1: The proportions are not as given Reject H0 if 2 > 7.815 Applications fo fe (fofe)2/fe 0 46 48 0.083 1 40 36 0.444 2 22 24 0.167 3 12 12 0.000 Total 120 120 0.694 Do not reject H0. Evidence does not show a change in the accident distribution.
(LO14-
H0: Store size and advertising amount are not related H1: Store size and advertising amount are related Reject H0 if 2 > 5.991 (40 56.97) 2 (32 37.70) 2 2 ... 18177 . 56.97 37.70 Reject H0. Store size and advertising amount are related.
(LO14-
Decision rule: If 2 > 11.070 reject H0. Applications fo fe (fo − fe)2/fe 0 50 40.59 2.1815 1 77 81.21 0.2183 2 81 81.21 0.0005 3 48 54.12 0.6921 4 31 27.06 0.5737 5 or more 13 15.78 0.4898 Total 300 300 4.1558 Do not reject H0. It is reasonable to conclude the distribution is Poisson with = 2. (LO14-1)
1) 24.
3) 25.
H0: Level of management and concern regarding the environment are not related H1: Level of management and concern regarding the environment are related Reject H0 if 2 > 16.812
2
(15 14) 2 (31 28) 2 ... 1.550 14 28
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Do not reject H0. Levels of management and environmental concerns are not related. (LO14-3) 26.
H0: Age and pressure are not related H1: Age and pressure are related 2 Reject H0 if > 16.812 (20 19.44) 2 (43 40.32) 2 2 ... 2.191 19.44 40.32 Do not reject H0. Age and pressure are not related. (LO14-
3) 27.
H0: Whether a claim are filed and age is not related H1: Whether a claim are filed and age is related Reject H0 if 2 > 7.815 (170 203.33) 2 (74 40.67) 2 (24 35.67) 2 2 ... 53.639 203.33 40.67 35.67 Reject H0. Age is related to whether a claim is filed.
(LO14-
3) 28.
H0: Pension plan preference and job class are not related H1: Pension plan preference and job class are related Reject H0 if 2 > 13.277 (10 17.33) 2 (22 33.94) 2 2 ... 84.04 17.33 33.94 Reject H0. There is a relationship between pension plan preference and job class. (LO14-
3) 29.
H0: p0 0.55, p1 0.28 p2 0.17 H1: The proportions are not as given Reject H0 if 2 >5.991 Applications fo fe (fo − fe)2/fe 0 220 247.5 3.056 1 158 126 8.127 2 72 76.5 0.265 Total 450 450 11.448 Reject H0.Young adults differ from the general population.
(LO14-
H0: The distributions are the same for human resource and technical personnel. H1: The distributions are different. df = 3 Reject H0 if 2 > 11.345 (35 59.1111)2 (16 25.5556) 2 2 ... 35.944 59.1111 25.5556 Reject H0. The distributions are different.
(LO14-
1) 30.
3) 31.
H0: The proportions are the same Reject H0 if 2 > 16.919
H1: The proportions are not the same
- 218 -
fo 44 32 23 27 23 24 31 27 28 21
fe 28 28 28 28 28 28 28 28 28 28
280
280
fo – fe 16 4 5 1 5 4 3 1 0 7
(fo − fe)2 256 16 25 1 25 16 9 1 0 49
(fo − fe)2/fe 9.143 0.571 0.893 0.036 0.893 0.571 0.321 0.036 0.000 1.750 14.214
Do not reject H0. The digits are evenly distributed.
(LO14-
1) 32.
H0: There is no difference from the survey distribution. H1: There is a difference from the survey distribution. Reject H0 if 2 > 11.345 Visit Type f0 fe (fo − fe)2/fe Primary care Medical specialist Surgical specialist Emergency
29 31.8 0.247 11 11.4 0.014 16 10.2 3.298 4 6.6 1.024 60 60 4.583 Since 4.583 < 11.345there is not enough evidence to reject the null hypothesis. Conclude there is no difference from the survey distribution. (LO14-1) 33.
H0: Gender and attitude toward the deficit are not related H1: Gender and attitude toward the deficit are related Reject H0 if 2 > 5.991
244 292.41 194 164.05 68 49.53 305 256.59 114 143.95 25 43.47 43.578 2
2
292.41
2
164.05
2
49.53
2
256.59
2
143.95
Since 43.578 > 5.991 you reject H0. A person’s position on the deficit is influenced by his or her gender. (LO14-4) 34.
H0: pBL 0.23, pB R 0.12, pY 0.15 p R 0.12, pO 0.23 pG 0.15 H1: The proportions are not as given. df = 5 Reject H0 if 2 >11.070 Color fo fe (fo – fe)2/fe Blue 12 16.6 1.256 Brown 14 8.6 3.325 Yellow 13 10.8 0.448 Red 14 8.6 3.325 Orange 7 16.6 5.519 Green 12 10.8 0.133
- 219 -
2
43.47
Total 72 72 14.006 Reject H0. The colour distribution does not agree with the manufacturer’s information. (LO 14-3) 35.
H0: Gender and type of movie are not related H1: Gender and type of movie are related Reject H0 if 2 > 11.345 Action Gender Male Female
fo 75 50
fe 61.25 63.75
Documentary
Romantic
Comedy
fo 50 50
fo 60 95
fo 60 60
fe 49 51
fe 75.95 79.05
fe 58.8 61.2
2 = 3.087+2.966+ 0.020+ 0.020+3.350+3.218+0.024+ 0.024 = 12.708 The computed value lies to the right of 11.345, so the null hypothesis is rejected. Gender and type of movie are related. The p-value is 0.0053 which is smaller than the significance level of 0.01. This confirms the null hypothesis should be rejected. (LO144) 36.
H0: Coffee consumption and age are not related H1: Coffee consumption and age are related Reject H0 if 2 > 9 .488 low under 30 30 to 50 50 or more
fo 36 28 26 90
moderate
fe 27.6 38.7 23.7 90
fo 32 54 24 110
fe 33.7 47.3 29.0 110
High fo 24 47 29 100
fe 30.7 43.0 26.3 100
Total 92 129 79 300
2 = 2.557 + 2.958 + 0.223 + 0.089 + 0.949 + 0.852 + 1.449 + 0.372 + 0.270 = 9.719 The computed value lies to the right of 9.488, so the null hypothesis is rejected. Coffee consumption and age are related. The p-value is 0.0454 which is less than the significance level of 0.05. This confirms the null hypothesis should be rejected. (LO144) 37.
H0: Loyalty and years of service are not related H1: Loyalty and years of service are related Reject H0 if 2 > 7.815 Less 1 year 1-5 years Would remain Would not remain
6-10 years
More than 10 years
fo
fe
fo
fe
fo
fe
fo
fe
10
21
30
27
5
9
75
63
25 35
14
15 45
18
10 15
6
30 105
42
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2 = 5.762 + 8.643 + 0.333 + 0.500 + 1.778 + 2.667 + 2.286 + 3.429 = 25.397 The computed value lies to the right of 7.815, so the null hypothesis is rejected. Loyalty and years of service are related. The p-value is 0.00001 which is smaller than the significance level of 0.05. This confirms the null hypothesis should be rejected. (LO144) 38. H0: There is no difference in the proportion of adults who prefer one flavour over another. H1: There is a difference in the proportion of adults who prefer one flavour over another. Reject H0 if 2 > 6.251
Carrot cake Coffee walnut Apple-taffy Coconut cream
fo 30 115 50 105
fe 75 75 75 75
(fo − fe)2/fe 27.00 21.33 8.33 12.00 68.67
The computed value lies to the right of 6.251, so the null hypothesis is rejected. There is a difference in the preference of ice cream flavours. (LO14-1) 39.
H0: There is no difference in the proportion of students who are absent on Fridays. H1: There is a difference in the proportion of students who are absent on Fridays. Reject H0 if 2 > 9.488 Days of the Week Monday Tuesday Wednesday Thursday Friday
fo 11 6 8 7 12
fe 8.8 8.8 8.8 8.8 8.8
(fo – fe)2/fe 0.550 0.891 0.073 0.368 1.164 3.045
The computed value lies to the left of 9.488, so there is not enough evidence to reject the null hypothesis. There is no difference in the proportion of students who are absent on Fridays. (LO14-1) 40.
H0: p0 0.08; p12 0.56; p34 0.28; p5 0.08 H1: The proportions are not as stated Reject H0 if 2 > 6.251 fo 7 55 28 10
fe 8 56 28 8
(fofe)2/fe 0.125 0.018 0 0.50
- 221 -
0.643 There is not enough evidence to reject H0. Proportions are as stated in the report by the Cable and Telecommunications Association. (LO143) 41.
H0: The proportions are as stated H1: The proportions are not as stated Reject H0 if 2 > 9.488 fo 7 11 16 3 3
fe 3 8 21 5 3
(fofe)2/fe 5.333 1.125 1.190 0.800 0.000
8.449 There is not enough evidence to reject H0. Proportions are as stated and follow the distribution as seen in Professor Brown’s classes in the past. (LO14-3) 42.
H0: The variables: Type of Employer and On-site café are not related (independent) H1: The variables: Type of Employer and On-site café are related (not independent) Public Yes Café Yes Café No Subtotal
32 15 47
Public No 32 21 53
Subtotal 64 36 100
Observed Expected (O E)2/E 32 30.08 0.122553 32 33.92 0.108679 15 16.92 0.217872 21 19.08 0.193208 2-STAT = 0.642312 Decision Rule: If the 2-STAT is > than the 2-Critical reject H0. 2-Critical (df = (21)(21) = 1, α = 0.05) = 3.841. Since 2-STAT is not > 3.841 we fail to reject H0 and conclude the Type of Employer being Public or Non Public is not related to whether they have an on-site cafeteria or not. (LO14-4) 43. H0: The variables: Original Six and All other teams and team Worth (above or below $1B) are not related (independent) H1: The variables:(Original Six and All other teams and team Worth (below or above $1B) are related (dependent)
- 222 -
Note: Caution should be observed with this table as two cell values are less than 5. ≤$1B >$1B Subtotal
Original Six Others Subtotal 1 25 26 5 0 5 6 25 31
Observed Expected (O E)2/E 5.032258 3.230976 1 20.96774 0.775434 25 0.967742 16.80108 5 4.032258 4.032258 0 2 -STAT = 24.83974 Decision: If the 2-STAT is > than the 2-Critical reject H0. 2-Critical (df = (2 1)(2 1) = 1, α = 0.05) = 3.841. Since 2-STAT is > 3.841, we reject H0 and conclude that being an Original Six team is related to the team’s Worth. (LO14-4) 44. H0: The Variables: Regions: (BC, Prairies, Ont and combined Quebec/Atlantic Canada) and List Price (below or above $600K) are not related (independent) H1: The Variables: Regions: (BC, Prairies, Ont and combined Quebec/Atlantic Canada) and List Price (below or above $600K) are related (dependent) Note: Caution should be observed with this table as some cell values are less than 5. <$600 BC Prairies ONT QC&Atl Subtotal
1 5 17 5 28
>$600 6 0 8 0 14
Subtotal 7 5 25 5 42
Observed Expected (O E)2/E 1 4.67 2.88 6 2.33 5.76 5 3.33 0.83 0 1.67 1.67 17 16.67 0.007 8 8.33 0.013 5 3.33 0.83 0 1.67 1.67 2 -STAT = 13.66 Decision: If the 2-STAT is > than the 2-Critical reject H0. 2-Critical (df = (41)(21) = 3, α = 0.05) = 7.815. Since 2-STAT is > 7.815 we reject H0 and conclude there is
- 223 -
relationship between the region a home was listed and the listing price. (LO14-4)
CHAPTER 15 INDEX NUMBERS 1.
2.
(LO15-1) Year Loan($M) 2013 55177 2014 65694 2015 83040 2016 88378 2017 97420 2018 98608 2019 101364 2020 110527 2021 116364 (LO15-1) Year EPS 2013 1.57 2014 2.01 2015 2.47 2016 3.76 2017 4.71 2018 5.46 2019 6.45 2020 7.29 2021 9.73
Index(base = 2013) 100.0 119.1 150.5 160.2 176.6 178.7 183.7 200.3 210.9
Index(base = 2013) 100.0 128.0 157.3 239.5 300.0 347.8 410.8 464.3 619.7
3.
The mean sales for the earliest 3 years is $(486.6 + 506.8 + 522.2)/3 or $505.2. 2020: 90.4, found by ($456.6/$505.2) (100) 2021: 85.8, found by ($433.3/$505.2) (100) Net sales decreased by 9.6% and 14.2% from the average of the 2012–2014 period to 2020 and 2021, respectively. (LO15-1)
4.
(1.873/1.753) × 100 = 106.8. Prices have increased by 6.8 %
5.
a. b.
Pt = (3.35/2.49) × 100 = 134.5 Ps = (4.49/3.29) × 100 = 136.5 Pc = (4.19/1.59) × 100 = 263.52 Pa = (2.49/1.79) × 100 = 139.11 Simple Aggregate = (14.52/9.16) × 100 = 158.5
c.
Laspeyres =
d.
(LO15-1)
(3.35 6) (4.49 4) (4.19 2) (2.49 3) 100 147.1 (2.49 6) (3.29 4) (1.59 2) (1.79 3) (3.35 6) (4.49 5) (4.19 3) (2.49 4) 100 150.2 Paasche = (2.49 6) (3.29 5) (1.59 3) (1.79 4)
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6.
e.
Fisher’s = 147.1150.2 148.64
a.
b. c
Pp = (0.69/0.23) × 100 = 300 Pg = (1.0/0.29) × 100 = 344.8 Pa = (1.89/0.35)× 100=540 Ps = (3.79/1.02)× 100 = 371.6 Po = (2.99/0.89) × 100 = 336 Simple Aggregate = (10.39/2.78) × 100 = 372.7 LPI=
d.
PPI=
e.
Fisher’s = 406.1 397.6 401.8
a. b. c.
Pw = 0.38/0.34 × 100 = 142.9 Pc = 0.03/0.04 × 100 = 75.0 Ps = 0.15/0.15 × 100 = 100 Ph = 0.10/0.08 × 100 = 125.0 Simple Aggregate=0.38/0.34×100=111.8 LPI=
d.
PPI=
e.
Fisher’s = 102.92 103.31 102.11 (LO15-3)
(LO15-3)
(0.69 100) (1.00 50) (1.89 85) (3.79 8) (2.99 6) 100 406.1 (0.23 100) (0.29 50) (0.35 85) (1.02 8) (0.89 6) (0.69 120) (1.00 55) (1.89 85) (3.79 10) (2.99 8) 100 397.6 (0.23 120) (0.29 55) (0.35 85) (1.02 10) (0.89 8)
7.
8.
(LO15-3)
(0.117000) (0.03 125000) (0.15 40000) (0.1 62000) 100 102.92 (0.07 17000) (0.04 125000) (0.15 40000) (0.08 62000)
(0.1 20000) (0.03 130000) (0.15 42000) (0.1 65000) 100 103.31 (0.07 20000) (0.04 130000) (0.15 42000) (0.08 65000)
a. Ps = (6.83/6.1) × 100 = 112 Pt = (9.35/8.10) × 100 = 115.4 Pa = (4.62/4.00) × 100 = 115.5 Pp = (6.85/6.00) × 100 = 114.2 Pc = (13.65/12) × 100 = 113.8 b. Simple Aggregate= (41.3/36.2) × 100 = 114.1 c. LPI=
(6.83 1500) (9.35 10) (4.62 250) (6.85 1000) (13.65 30) 100 113 (6.11500) (8.110) (4 250) (6 1000) (12 30)
9.
d.
PPI=
e.
Fisher’s = 113 112.8 112.9
(6.83 2000) (9.35 12) (4.62 250) (6.85 900) (13.65 40) 100 112.8 (6.1 2000) (8.1 12) (4 250) (6 900) (12 40)
(LO15-3)
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(LO15-3)
Grain Oats Wheat Corn Barley Value=
2018 P 2018 Q P0 × Q0 2021 P 2021 Q Pt × Qt 2.4878 1298 3229 2.5712 815 2096 5.0936 26 117 133 030 4.9757 51 287 255 189 3.7829 345 506 1 307 015 3.7043 366 287 1 356 837 5.0836 4750 24 147 4.9757 3333 16 584 1 467 420 1 630 705 111.1
10. (LO15-3) Product 2010 P 2019 P 2010 Q 2019 Q Pt × Qt P0 × Q0 Small $23.6 $28.8 1760 4259 $122 659.2 $41 536 Scrubbin 2.96 3.08 86 450 62 949 193 882.9 255 892 Nails 0.4 0.48 9460 22 370 10 737.6 3784 $327279.7 $301 212 Value = 108.6543 11.
a. (6.8/5.3) × 0.20 + (362.26/265.88) × 0.40 + (125.0/109.6) × 0.25 + 622 864/529 917 × 0.15 = 1.263 Index is 126.3 b. Business activity increased by 26.3% from 2000 to 2018 (LO15-4)
12. a. 0.10+
b.
(1162/1088) × 0.2 + (520/627) × 0.25) + (162.6/191.7) × 0.4 + (29841/24119) × (39.9/41.6) × 0.05 = 0.9319 Index is 93.19 Business activity has decreased by 6.81%
(LO15-4)
13. (LO15-5) Year Salary CPI Real Income 2002 44 800/100 × 100 44800 2021 86829/163.1 × 100 53237 % Change = (53 237 – 44 800)/44 800 × 100 = 18.8% increase 14. (LO15-5) Plumbers Electricians Year Divide by 133.8 × 100 Divide by 126.0 × 100 2014 100 100 2021 119.1 126.0 Conclude that electrical workers wages have increased by nearly 7% more than plumbers from 2014-2021 15.(LO15-5) Year 2000 2010 2021
Lindsay Wages Wages/56660 56660 100 69950 123.5 78500 138.5
National Index 100 122.5 136.9
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Overall, the Lindsay increases in wages were slightly higher from 2000 to 2010 and also for the period from 2000-2021 Lindsay salaries grew slightly higher than the national increases. 16 (LO15-5) Year Income ($000) Index Real Income ($000) 2015 134.8 divided by 160.6 × 100 = 83.9 2016 145.2 divided by 173.6 × 100 = 83.6 2017 156.6 divided by 187.9 × 100 = 83.3 2018 168.8 divided by 203.3 × 100 = 83.0 Note how Sam’s income was going up but his real income was decreasing. Sam’s increases did not keep up to the industries increases. 17.
(LO15-5) Year 2010=100 2010 100.0 2011 43.8 2012 101.3 2013 108.4 2014 118.2 2015 121.2 2016 128.4 2017 135.4 2018 142.3 Domestic Sales are increasing year over year except for 2011.
18.
(LO15-5) Average (29 437 + 12 907 + 29 830)/3 = 24 058. Divide each sale by 24 058 Year 2010-2012=100 2013 132.6 2014 144.6 2015 148.3 2016 157.2 2017 165.7 2018 174.1 Domestic sales are increasing year over year to a larger amount with the lower base because of 2011
19.
(LO15-5) Year 2010 2011 2012 2013 2014 2015 2016
2010=100 100.0 112.4 116.4 122.7 123.1 107.0 106.1
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2017 113.9 2018 123.6 International sales have a steady increase until 2015 where they dip for 3 years until recovering in 2018 20.
(LO15-5) Average (32 124+36 107+37 394)/3 = 35 208.33. Divide each year by 35 208.33 Year 2010-2012 = 100 2013 111.9 2014 112.3 2015 97.7 2016 96.8 2017 103.9 2018 112.7 International Sales have increased for 2013 and 2014. There is a strong fall starting in 2015 however slowly the sales recover back to the 2014 levels by 2018.
21.
(LO15-5) Year 2010-100 2010 100.0 2011 103.4 2012 111.9 2013 112.4 2014 111.0 2015 111.5 2016 110.9 2017 117.5 2018 118.5 The employee index shows that from 2010 to 2018 the employee has increased by 18.5%
22.
(LO15-5) Average (114.0 + 117.9 + 127.6)/3 = 119.8333 Year 2010-2012 = 100 2013 106.9 2014 105.6 2015 106.1 2016 105.5 2017 111.8 2018 112.7 With an average index we find employee increase is only 12.7% from 2013 to 2018.
23.
(LO15-5) Year 2013 = 100 2013 100.0 2014 103.5 2015 103.7
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2016 109.2 2017 107.8 2018 110.9 GE sales have increased by 10.9% since 2013 24.
(LO15-5) Average (113 245 + 117 184)/2 = 115 214.5 Year 2013-2014 = 100 2015 101.9 2016 107.4 2017 106.0 2018 109.0 With an averaged base year, the Sales have increased by only 9%
25.
(LO15-5) Year 2013=100 2013 100.0 2014 99.3 2015 108.5 2016 96.1 2017 102.0 2018 92.2 GE’s employee group has decreased by 7.8%
26.
(LO15-5) Average = 306 Year 2013-2014 = 100 2015 108.8 2016 96.4 2017 102.3 2018 92.5 Using an average index employees decreased by 7.5%
27.
(LO15-1) SPI Ma (2.00/0.81) × 100 SPI S (1.88/0.84) × 100 SPI Mi (2.89/1.44) × 100 SPI P (3.99/2.91) × 100
246.9 223.8 200.7 137.1
28. 2)
(2.00 + 1.88 + 2.89 + 3.99)/(0.81 + 0.84 + 1.44 + 2.91) × 100 = 179.3
29.
(LO15-3) Item Margarine Shortening Milk
P0 0.81 0.84 1.44
Q0 18 5 70
Pt 2 1.88 2.89
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Qt 27 9 65
Pt × Q0 36 9.4 202.3
(LO15-
P0 × Q0 14.58 4.2 100.8
Potato chips
30.
2.91
27
3.99
33
107.73 355.43
78.57 198.15
LPI = (355.43/198.15) × 100 =
179.4
(LO15-3) Item Margarine Shortening Milk Potato chips
P0 0.81 0.84 1.44
Q0 18 5 70
Pt 2 1.88 2.89
Qt 27 9 65
Pt × Qt 54 16.92 187.85
P0 × Qt 21.87 7.56 93.6
2.91
27
3.99
33
131.67 390.44
96.03 219.06
PPI = (390.44/219.06) × 100 =
178.2
31.
Fisher's =Sqrt(179.4x178.2)=178.8
(LO15-3)
32.
(LO15-3) Item Margarine Shortening Milk Potato chips
P0 0.81 0.84 1.44
Q0 18 5 70
Pt 2 1.88 2.89
Qt 27 9 65
Pt × Qt 54 16.92 187.85
P0 × Q0 14.58 4.2 100.8
2.91
27
3.99
33
131.67 390.44
78.57 198.15
VI = (390.44/198.15) × 100 = 33.
34.
(LO15-1) Part RC-33 SM-14 WC50
P13 0.5 1.2 0.85
(LO15-2) Part RC-33 SM-14 WC50 2.5/2.55 × 100 =
35.
(LO15-3) Part RC-33 SM-14 WC50
197.0
P21 0.6 0.9 1
P 13 0.5 1.2 0.85 2.55 98.0
P 13 0.5 1.2 0.85
Q13 320 110 230
Q21 340 130 250
(P21/P13) × 100 120.0 75.0 117.6
P21 0.6 0.9 1 2.5
Q13 320 110 230
Q21 340 130 250
P21 0.6 0.9 1
Q13 320 110 230
Q21 340 130 250
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P21 × Q13 192.0 99.0 230.0
P13 × Q13 160 132 195.5
521.0
487.5
P21 × Q21 204.0 117.0 250.0 571.0
P13 × Q21 170 156 212.5 538.5
LPI = (521.0/487.5) × 100 = 106.9 36.
(LO15-3) Part RC-33 SM-14 WC50
P 13 0.5 1.2 0.85
P21 0.6 0.9 1
Q13 320 110 230
Q21 340 130 250
PPI = (571.0/538.5) × 100 =106 37.
Fisher’s = 106.9 106 106.5
38.
(LO15-3) Part RC-33 SM-14 WC50
P 13 0.5 1.2 0.85
(LO15-3)
P21 0.6 0.9 1
Q13 320 110 230
Q21 340 130 250
P21 × Q21 204.0 117.0 250.0 571.0
P13 × Q13 160 132 195.5 487.5
VI = (571/487.5) × 100 = 117.1 39.
(LO15-1) Item Cabbage Carrots Peas Endive
40.
P13 0.06 0.1 0.2 0.15
Q13 2000 200 400 100
Q21 1500 200 500 200
Q21 1500 200 500 200
SPI = P21/P13 × 100 83.3 120.0 90.0 100.0
SPI= (P21/P13) × 100 83.3 120.0 90.0 100.0
(LO15-1) Item Cabbage Carrots Peas Endive
P13 P21 Q13 0.06 0.05 2000 0.1 0.12 200 0.2 0.18 400 0.15 0.15 100 0.51 0.5 Aggregate = (0.50/0.51) × 100 = 98.0 41.
P21 0.05 0.12 0.18 0.15
(LO15-3) Item P13 Cabbage 0.06 Carrots 0.1 Peas 0.2 Endive 0.15
P21 0.05 0.12 0.18 0.15
Q13 2000 200 400 100
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Q21 1500 200 500 200
P21 × Q13 100.0 24.0 72.0 15.0 211.0
P13 × Q13 120 20 80 15 235.0
LPI = (211/235) × 100 = 89.8 42.
(LO15-3) Item P13 Cabbage 0.06 Carrots 0.1 Peas 0.2 Endive 0.15
P21 0.05 0.12 0.18 0.15
Q13 2000 200 400 100
Q21 1500 200 500 200
P21 × Q21 75.0 24.0 90.0 30.0 219.0
P13 × Q21 90 20 100 30 240.0
P21 × Q21 75.0 24.0 90.0 30.0 219.0
P13 × Q13 120 20 80 15 235.0
PPI = (219/240) × 100 = 91.3 43. Fisher’s = 89.8 91.3 90.5 44.
(LO15-3) Item P13 Cabbage 0.06 Carrots 0.1 Peas 0.2 Endive 0.15
(LO15-3)
P21 0.05 0.12 0.18 0.15
Q13 2000 200 400 100
Q21 1500 200 500 200
VI = (219.0/235.0) × 100 = 93.2 45.
(LO15-1) Item P13 P21 Q13 Q21 P21/P13 × 100 Aluminum 0.82 0.86 1000 1200 104.9 Nat Gas 4.37 2.99 5000 4000 68.4 Petroleum 71.21 58.15 60 000 60 000 81.7 Platinum 1743.6 1292.53 500 600 74.1
46.
(LO15-2) Item P13 P21 Q13 Q21 Aluminum 0.82 0.86 1000 1200 Nat Gas 4.37 2.99 5000 4000 Petroleum 71.21 58.15 60 000 60 000 Platinum 1743.6 1292.53 500 600 1820 1354.53 Aggregate = (1354.53/1820) × 100 = 74.4
47.
(LO15-3) Item Aluminum Nat Gas Petroleum Platinum
P13 0.82 4.37 71.21 1743.6
P21
Q13 0.86 2.99 58.15 1292.53
1000 5000 60 000 500
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Q21 P21 × Q13 P13 × Q13 1200 860 820 4000 14 950 21 850 60 000 3 489 000 4 272 600 600 646 265 871 800 4 151 075 5 167 070
LPI = (5 167 070/4 151 075) × 100 = 80.3 48.
(LO15-3) Item Aluminum Nat Gas Petroleum Platinum
P13 0.82 4.37 71.21 1743.6
P21 0.86 2.99 58.15 1292.53
Q13 Q21 P21 × Q21 P13 × Q21 1000 1200 1032 984 5000 4000 11 960 17 480 60 000 60 000 3 489 000 4 272 600 500 600 775 518 1 046 160 4 277 510 5 337 224
PPI = (4 277 510/5 337 224) ×100 = 80.1 49.
Fisher’s= 80.3 80.1 80.2 (LO15-3)
50.
(LO15-3) Item P13 P21 Q13 Q21 P21 × Q21 P13 × Q13 Aluminum 0.82 0.86 1000 1200 1032 820 Nat Gas 4.37 2.99 5000 4000 11 960 21 850 Petroleum 71.21 58.15 60 000 60 000 3 489 000 4 272 600 Platinum 1743.6 1292.53 500 600 775 518 871 800 4 277 510 5 167 070 VI = (4 277 510/5 167 070) × 100 = 82.8
51.
Special-purpose index = (1971/1159) × 0.20 + (91/87) × 0.10 + (114.7/110.6) × 0.40 + (251 000/214 000 × 0.3) × 100 = 121.1 The overall economy has expanded by 21.1% from 2015 to 2021. (LO15-4)
52.
(LO15-5) Year CPI/72.1 × 100 Labour/6.4 × 100 Production/64.9 × 100 GNP/28.6 × 100 1950 100.0 100.0 100.0 100.0 1967 138.7 126.6 154.1 274.8 1971 168.2 135.9 170.0 371.7 1975 223.6 148.4 177.0 530.1 1980 342.3 167.2 225.9 568.5 All facets of the Canadian economy have expanded with the following increases from smallest to largest: Labour 67.2%, Production 125.9%, CPI 242.3%, and GNP 468.5%
53.
January = 100 February = (6.8/8) × 40 + (23/20) × 35 + (303/300) × 25 = 99.5 March = (6.4/8) × 40 + (21/20) × 35 + (297/300) × 25 = 93.5 The index reveals the economy is contracting hence inventories should be kept at a low level. (LO15-4)
54.
a. b.
PP$ = (1/251.1) × 100 = $0.398 Real Income = $(7000/251.1) × 100 = $2787.73 (LO15-5)
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55.
56.
(2 400 000/160.5) × 100 = $1 495 327 (3 500 000/204.0) × 10 = $1 715 686 a. City/228 600 × 100 228 600 261 900 272 700 284 000 289 100 292 700 297 800 323 300 323 400 377 800 378 100 402 900 412 900 417 800 439 000 463 500 471 800 471 900 481 000 492 100 492 500 510 900 517 200 519 500 520 100 522 200 540 300 593 300 619 600 627 300 639 900 644 600 674 200 684 900 785 400 813 700 1 028 600 1 059 200
City Greater Moncton North Bay St John's NL Newfoundland and Labrador Regina Quevec-CMA Winnipeg Bancroft and Area Saskatoon Grey Bruce Owen Sound Edmonton Quinte and District Tilsonburg District Huron Perth Simcoe and District Lakelands Calgary Montreal-CMA Woodstock Ingersol Kawartha Lakes Peterborough and Kawarthas London St. Thomas Southern Georgian Bay Northumberland Hills Brantford Region Niagra Region Vancouver Is. Ottawa Cambridge Barrie and District Okanagan Valley Chilliwack and District Kitchener Waterloo Guelph and District Hamilton Burlington Victoria Toronto-GTA Fraser Valley
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(LO15-5)
100.0 114.6 119.3 124.2 126.5 128.0 130.3 141.4 141.5 165.3 165.4 176.2 180.6 182.8 192.0 202.8 206.4 206.4 210.4 215.3 215.4 223.5 226.2 227.3 227.5 228.4 236.4 259.5 271.0 274.4 279.9 282.0 294.9 299.6 343.6 355.9 450.0 463.3
1 126 000 1 200 000 1 341 400 1 548 100
Mississauga Oakville Milton Lower Mainland Greater Vancouver
492.6 524.9 586.8 677.2
Increase of 677.2 – 100 = 577.2% b. Base Listing = $543 000 November 2020 Index = 1 548 100/543 000×100=285.1 or 185% increase since 2005 (LO15-5) CHAPTER 16 FORECASTING WITH TIME SERIES ANALYSIS
a. The graph shows a stationary pattern.
Demand Demand
1.
200 150 100 50 0 1
2
3
4
5
6
7
8
9 10
Time Period
b.
c. The forecast demand for period 11 is 119.
- 235 -
d. The MAD of 12.48 is the error associated with the eleventh-period forecast. (LO16-2) 2.
a. The graph shows a stationary pattern.
b. The forecast demand for period 11 is 124.5. c. The MAD of 8.3 is the error associated with the eleventh-period forecast. (LO162) 3. a. The graph shows a stationary pattern.
- 236 -
b. and c.
d. The six-period simple moving average model has less error. (LO16-2) 4.
a. The graph shows a stationary pattern.
- 237 -
b. and c.
d. The four-period simple moving average model has less error. a. Sales 1200 1000 800
Sales
5.
(LO16-2)
600 400 200 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Time Period
b. The graph shows a negative time series trend in sales. c. A trend model is appropriate because we want to estimate the decreasing change in sales per time period. d. Based on the regression analysis; the trend equation is: Sales = 1062.86 – 36.24 (time period);
- 238 -
sales are declining at a historical rate of 36.24 for each increment of 1 time period. e. Sales are declining at a historical rate of $36.24 for each increment of 1 time period. f. Sales (19) = 1062.86 – 36.24 (19) = $374.37 Sales (20) = 1062.86 – 36.24 (20) = $338.13 Sales (21) = 1062.86 – 36.24 (21) = $301.89 a.
Sales 1800 1600 1400 1200
Sales
6.
(LO16-3 & LO16-4)
1000 800 600 400 200 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Time Period
b. The graph shows a positive time series trend in sales.
- 239 -
c. A trend model is appropriate because we want to estimate the increasing change in sales per time period.
d. Sales = 484.405 + 64.180 (time period); Sales are increasing at a historical rate of $64.180 for each increment of 1 time period. e. Sales (19) = 484.405 + 64.180 (19) = $1703.817 Sales (20) = 484.405 + 64.180 (20) = $1767.997 Sales (21) = 484.405 + 64.180 (21) = $1832.176 a.
Net Sales $100,000
Millions $
7.
(LO16-3 & LO16-4)
$80,000 $60,000 $40,000 $20,000 $0 0
5
10
15
20
25
Time Period
b. The graph shows a positive time series trend in sales. c. A trend model is appropriate because we want to estimate the increasing change in sales per time period.
- 240 -
d. Net Sales = 11392.18 + 3892.578 (Time Period)
e. Sales are increasing at 3 892.578 (in millions of dollars) for each increment of 1 time period. f. 2021: 11 392.18 + 3892.578(21) = 93 136 (in millions of dollars) 2022: 11 392.18 + 3892.578(22) = 97 029 (in millions of dollars) 2023: 11 392.18 + 3892.578(23) = 100 921 (in millions of dollars) (LO16-3 & LO16-4) a.
Scrap 8.0
Scrap (tonnes)
8.
6.0 4.0 2.0 0.0 0
1
2
3
4
5
6
7
8
Time Period
b. The graph shows a positive time series trend in output. c. A trend model is appropriate because we want to estimate the increasing change in output per time period. d. Scrap = 1.714286 + 0.75 (Time Period) SUMMARY OUTPUT
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Regression Statistics Multiple R 0.900368 R Square 0.810662 Adjusted R Square 0.772794 Standard Error 0.857738 Observations 7 ANOVA df Regression Residual Total
1 5 6
Intercept Time Period
Coefficient s 1.714286 0.75
SS MS F 15.75 15.75 21.40777 3.678571 0.735714 19.428571 Standard Error 0.724921 0.162097
t Stat 2.36479 4.62685
Significance F 0.0057
P-value 0.0644 0.0057
e. Output is increasing at 0.75 tonnes for each increment of 1 time period. f. 2021: 1.714286 + 0.75(8) = 7.714286 2022: 1.714286 + 0.75(9) = 8.464286 2023: 1.714286 + 0.75(10) = 9.214286 (LO16-3 & LO16-4) 9.
a.
Sales ($M)
Millions ($)
30
20
10
0 1
2
3
4
5
6
7
Time Period
b. The graph shows a positive time series trend in sales. c. A trend model is appropriate because we want to estimate the increasing change in sales per time period. d. Sales = 1.57714 + 1.725 (Time Period) SUMMARY OUTPUT
- 242 -
Regression Statistics Multiple R 0.9958 R Square 0.9917 Adjusted R Square 0.9900 Standard Error 0.3733 Observations 7 ANOVA
Regression Residual Total
df 1 5 6
SS 83.3175 0.6968 84.0143
MS 83.3175 0.1394
F 597.8703
Intercept Year
Coefficients 15.77142857 1.725
Standard Error 0.3155 0.0705
t Stat 49.9885 24.4514
P-value 0.0000 0.0000
e. Sales are increasing at 1.725 millions of dollars for each increment of 1 time period. f. 2021: 15.7714 + 1.725(8) = 29.57 ($M) 2022: 15.7714 + 1.725(9) = 31.30 ($M) 2023: 15.7714 + 1.725(10) = 33.02 ($M) (LO16-3 & LO16-4) a.
Quarterly Sales 1400 1200 1000
Sales
10.
800 600 400 200 0 1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16
Time Period
- 243 -
Significance F 0.0000
b. The quarterly time series shows two patterns, negative trend and seasonality. The seasonality indicated with quarters 3 and 4 is always with high sales, and quarters 1 and 2 is always with low sales. c. A seasonality adjusted trend model is appropriate because we want to estimate the decreasing change in sales per quarter and we want to quantify the seasonal effects for each quarter. d. e.
QTR 1 2 3 4
Index 0.866 0.838 1.079 1.216
f. Sales = [1041.875 – 10.081 (time period)] [quarterly index for the time period] Period 17 Sales = [1041.875 – 10.081 (17)] [0.866] = [870.498] [0.866] = 754.121 Period 18 Sales = [1041.875 – 10.081 (18)] [0.838] = [860.417] [0.838] = 720.940 Period 19 Sales = [1041.875 – 10.081 (19)] [1.079] = [850.336] [1.079] = 917.928
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Period 20 Sales = [1041.875 – 10.081 (20)] [1.216] = [840.255] [1.216] = 1021.852 (LO16-4) 11.
a.
Quarterly Sales 1400 1200 1000 800 600 400 200 0 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Time Period
b. The quarterly time series shows two patterns, positive trend and seasonality. The seasonality is indicated with quarter 4 has the lowest sales among quarters 1, 2, 3, and 4 in years 1, 3, and 4. In years 2, 3, and 4, quarter 1 is the highest sales among quarters 1, 2, 3, and 4. In the first year, Q1 is not the highest, but close. c. A seasonality adjusted trend model is appropriate because we want to estimate the decreasing change in sales per quarter and we want to quantify the seasonal effects for each quarter.
- 245 -
d. and e.
Quarter 1 2 3 4
Index 1.206 1.030 0.948 0.817
f. Sales = [774.950 + 17.249 (time period)] [quarterly index for the time period] Period 17 Sales = [774.950 + 17.249 (17)] [1.206] = [1068.175] [1.206] = 1287.78 Period 18 Sales = [774.950 + 17.249 (18)] [1.030] = [1085.424] [1.030] = 1118.29 Period 19 Sales = [774.950 + 17.249 (19)] [0.948] = [1102.672] [0.948] = 1044.91 Period 20 Sales = [774.950 + 17.249 (20)] [0.817] = [1119.921] [0.817] = 914.70 (LO16-4) 12. Average SI Seasonal Quarter Component 1 0.6522 2 1.6522 3 1.2174 4 0.4783
Index 0.6929 1.6742 1.1792 0.4511
(LO16-4 & LO 16-5)
- 246 -
13. Average SI Component 0.9281 0.7564 1.1369 1.1787
Quarter 1 2 3 4 14. t 21 22 23 24
estimated pairs (millions) 40.05 41.80 43.55 45.30
Seasonal Index 0.9339 0.7576 1.1354 1.1731
(LO16-5) Quarterly forecast (millions) 44.055 50.160 34.840 40.770 (LO16-5)
Seasonal index 110.0 120.0 80.0 90.0
15.
Sales for each quarter are 500, found by 2000/4. The estimated sales for the second quarter are 725, found by 500(1.45). (LO16-5)
16.
The three possible patterns are: 1. a positive, increasing trend, 2. a negative decreasing trend, 3. no trend where there is no evidence of increases or decreases over time. For positive or negative trends, regression analysis can be applied to create a forecasting model. When no trend is apparent, a simple moving average should be applied. (LO16-1)
17.
a. Four. One for each quarter. b. Twelve. One for each month. c. Two. One for each six-month period.
18.
(LO16-1)
a. The graph does not show any trend or seasonality. The graph shows a stationary pattern. Simple moving average models would be a good choice to compute forecasts.
Demand 150 140 130 120 110 100 90 80 1
2
3
4
5
6
7
8
9
10
11
Time Period
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12
13
14
15
16
b.
c. The forecast demand for period 17 is 112.6 units. (LO16-1 & LO 16-2) 19. a. The graph does not show any trend or seasonality. The graph shows a stationary pattern. Simple moving average models would be a good choice to compute forecasts. Demand 450 400 350 300 250 200 150 100 50 0 1
2
3
4
5
6
7
8
9
10
11
12
Time Period
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13
14
15
16
b. and c.
d. The period 17 forecast for the 3 period moving average is 192.33. The period 17 forecast for the 6 period moving average is 238.00. e. The 6 period moving average appears to be slightly better. Practically, either model is a valid choice. (LO16-1 & LO 16-2) 20.
a. Sales 1200 1000 800 600 400 200 0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18
Time Period
b. The graph shows a downward, negative trend in sales. c. Forecasting with a trend model will reveal the average, per period, change in sales.
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d. The Trend forecast model is: Sales = 930.954 – 27.457 (time period).
e. Based on the slope from the regression analysis, sales are decreasing at a rate of 27.457 units per period. f. Forecasts for periods 19, 20, and 21 are: Sales = 930.954 – 27.457 (19) = 409.271 Sales = 930.954 – 27.457 (20) = 381.814 Sales = 930.954 – 27.457 (21) = 354.357 g. Forecasts are based on historical data and assume no changes. Given that, forecasts can be reliably used along with other factors such as economic indexes etc. in order to be more accurate. (LO 16-3 & 16-4) 21.
a.
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Asset Turnover
Turnover
2.00 1.50 1.00 0.50 0
2
4
6
8
10
12
Time Period
b. The graph shows an increasing trend in asset trunover. c. Forecasting with a trend model will reveal the average, per period, change in asset turnover.
d. The trend forecast model is: Asset Turnover = 1.0045 + 0.0441 (time period). e. Based on the slope from the regression analysis, asset turnovers are increasing at a rate of 0.0441 units per period. f. Forecasts for periods 12, 13, and 14 are: Sales = 1.0045 + 0.04401 (12) = 1.534 Sales = 1.0045 + 0.04401 (13) = 1.578 - 251 -
Sales = 1.0045 + 0.04401 (14) = 1.622 g. Forecasts are based on historical data and assume no changes. Given that, forecasts can be reliably used along with other factors such as economic indexes etc. in order to be more accurate. (LO 16-3 & 16-4) 22. a.
Sales per Year 8.20
$B
8.00 7.80 7.60 7.40 0
2
4
6
8
10
Time Period
b.
Sales ($B) = 7.6336 + 0.0522 (Time Period) c. The predicted annual change per year is an increase of 0.0522 ($B). d. Forecasts for next 2 years, the time periods are 10 and 11. (LO 16-3 & 16-4)
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Time Period 10 11
Sales ($B) 8.156 8.208
23. a.
Employees (thousands)
Keller Overhead Door Inc. 50.0 45.0 40.0 35.0 30.0 25.0 20.0 15.0 10.0 5.0 1
2
3
4
5
6
7
8
Time period
b. The graph shows a downward trend in the number of employees. c.
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9
d. 2021: 25.275 = 25 275 2022: 23.15 = 23 150 e. The number of employees are decreasing on average by 2.125 (thousands) per year. (LO 16-3 & 16-4) 24. a.
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b.
c.
d.
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e. As we move from a three-year to a seven-year moving average, the fluctuations in the data are smoothed out. (LO16-2) 25.
The trend shows that there is a regular increase in the number of tickets sold. 2)
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(LO16-
26.
Rentals = 8 842.1273 – 88.1273 (time period)
Time Period Predicted 12 7 784.6 13 7 696.5 14 7 608.3 15 7 520.2 16 7 432.1 (LO16-3 & 16-4) a. XYZ Corp 130,000 120,000 Salary of Accountants
27.
110,000 100,000 90,000 80,000 70,000 60,000 50,000 40,000 0
5
10 15 Time Period
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20
25
b. Salary = 53 081.364 + 2 879.34 (time period) c. Time period = 24; Predicted = $122 185.52 d. The average salary has increased by $2 879.34 per year over the time period. (LO16-3 & 16-4) 28.
a. Banner Rocker Company 175
Sales (000)
170 165 160 155 150 145 0
2
4
6 Time Period
8
10
12
b. Sales = 147.564 + 2.3 (time period) c. Time Period Predicted 12 175.2 13 177.5 14 179.8 15 182.1 16 184.4 d. The average sales have increased by 2.3 ($2 300) per year over the time period. (LO16-3 & 16-4) 29.
30.
If sales are the same each month, monthly sales = (825 000/12) = $68 750. For Feb: 68 750(126.9/100) = $87 244; for Mar: 68 750(86.3/100) = $59 331. 5)
(LO16-
Listing = 302 463.11 + 4 059.95 (time period). The y-intercept is the point when t = 0 (December 2014) . The value is $302 463. Telling us an estimated value of the average listing on Vancouver Island for that period. The slope is $4 059.95 which shows an estimated increase of $4 059.95 every successive month. (LO16-4)
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31.
The 12 monthly seasonal index values have a low in December of 0.980 and a high of 1.018 in July. This tells us that Listings are 2.0% lower on average in December and in July the Listing values typically rise by 1.8%. (LO16-5)
32.
(LO16-4 & LO16-5) Month Jan-20 Mar 20 Jun-20 Sep-20
t 61 63 66 69
Trend Forecast 550 120.1 558 240.0 570 419.8 582 599.7
Seasonal Index 0.980 0.990 1.013 1.009
Forecast 539 117.7 552 657.6 577 835.3 587 843.1.
Actual 516 700 520 900 534 800 527 300
Difference 22 416.6 31 757.6 43 035.3 60 543.1
CI= Y/YhatxS 0.958 0.943 0.926 0.898
The forecasts were closest for the first forecast but became less accurate as the year continued. The CI or Cyclical-Irregular index values continue to decrease as the year moves forward. As we know, this was the year of the pandemic which was a huge irregular event that could not be predicted This caused listing prices to fall below the forecasts. The (CI) cyclical-irregual activity illustrates how the listing economy is also shrinking.
CHAPTER 17 AN INTRODUCTION TO DECISION MAKING 1.
2.
EMV ( A1 ) 0.30($50) 0.50($70) 0.20($100) $70 EMV ( A2 ) 0.30($90) 0.50($40) 0.20($80) $63 EMV ( A3 ) 0.30($70) 0.50($60) 0.20($90) $69 Decision: Choose alternative 1
Choose returnables because EMV (returnables) is higher EMV (returnables) = $80(0.70) + $40(0.30) = $68 (in thousands) EMV (nonreturnables) = $25(0.70) + $60(0.30) = $35.5 (in thousands) Use returnable bottles (LO17-2)
3. A1 A2 A3 4.
5.
6.
(LO17-2)
Opportunity loss S1 S2 S3 $40 $0 $0 0 30 20 20 10 10
(LO17-3)
Opportunity loss ($000) Type of bottle Passed Not passed Returnable 0 $20 Nonreturnable $55 0
(LO17-3)
EOL(A1) = 0.30($40) + 0.50($0) + 0.20($0) = $12 EOL(A2) = 0.30($0) + 0.50($30) + 0.20($20) = $19 EOL(A3) = 0.30($20) + 0.50($10) + 0.20($10) = $13
(LO17-3)
Answers in $000
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EOL (Returnables) = $0(0.70) + $20(0.30) = $6 (in thousands) EOL (Nonreturnables) = $55(0.70) + $0(0.30) = $38.5 (in thousands)
(LO17-3)
7.
Expected value under conditions of certainty is $82, found by 0.30($90) + 0.50($70) + 0.20($100) EVPI = $82 $70 = $12 (LO17-5)
8.
Condition of Certainty = $80(0.70) + $60(0.30) = $74 (in thousands) Then EVPI = $74 $68 = $6 thousand (LO17-5)
9.
Yes, it changes the decision. Choose alternative 2 EMV(A1) = 0.50($50) + 0.20($70) + 0.30($100) = $69 EMV(A2) = 0.50($90) + 0.20($40) + 0.30($80) = $77 EMV(A3) = 0.50($70) + 0.20($60) + 0.30($90) = $74
10.
11.
Choose returnables. Does not alter decision. EMV (returnables) = $80(0.30) + $40(0.70) = $52 (in thousands) EMV (nonreturnables) = $25(0.30) + $60(0.70) = $49.5 (in thousands) a.
b. c.
d.
e.
(LO17-6)
Answers in ($000) EMV (neither) = 0.30($0) + 0.50($0) + 0.20($0) = $0 EMV (1) = 0.30($125) + 0.50($65) + 0.20($30) = $76 EMV (2) = 0.30($105) + 0.50($60) + 0.20($30) = $67.50 EMV (both) = 0.30($220) + 0.50($110) + 0.20($40) = $129 Choose both Opportunity loss S1 S2 S3 Neither $220 $110 $40 1 95 45 10 2 115 50 10 Both 0 0 0 EOL(neither) = 0.30($220) + 0.50($110) + 0.20($40) = $129.00 EOL(1) = 0.30($95) + 0.50($45) + 0.20($10) = $53.00 EOL(2) = 0.30($115) + 0.50($50) + 0.20($10) = $61.50 EOL(both) = 0.30($0) + 0.50($0) + 0.20($0) = $0 EVPI = $0, found by $129 $129 Certainty = 0.30($220) + 0.50($110) + 0.20($40) = $129 (LO17-2, 17-3
& 17-5) 12.
(LO17-6)
Weather Transportation Good Bad Plane $366.25 $372.50 Train 490 502.50 Car 370 445 EMV(plane) = $366.25(0.40) + $372.50(0.60) = $370.00 EMV(train) = $490(0.40) + $502.5(0.60) = $497.50 EMV(car) = $370(0.40) + $445(0.60) = $415.00 Choose the plane because $370 is the least cost. Certainty = $366.25(0.40) + $372.50(0.60) = $370.00 EPVI = $370 $370 = $0 (LO17-5)
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13.
The payoff table is as follows in $000 Recession No Recession S1 S2 Production $10.0 $15.0 Stock 5.0 12.0 CD 6.0 6.0 a. Purchase CD b. Increase production c. (Answers in $000) EMV(Prod.) = 0.20(10) + 0.80(15.0) = 10.0 EMV(Stock) = 0.20(5) + 0.80(12.0) = 8.6 EMV(CD) = 0.20(6) + 0.80(6) = 6.0 Expand Production d. EVPI = [0.20(6) + 0.80(15)] [10.0] = 13.2 – 10.0 = 3.2 & 17-5)
(LO17-2, 17-4
14.
Payoff Table a. S1 S2 S3 No Inspection $7.2 $14.4 $21.60 Inspection 10.0 10.0 10.00 b. EMV(No inspect) = 0.70(7.2) + 0.20(14.4) + 0.1(21.60) = $10.08 Inspect = $10.0 c. EVPI = [0.70(7.2) + 0.20(10) + 0.10(10)] – 10.0 = 1.96 (LO17-2 & 17-5)
15.
a.
b.
c.
d.
Event Act 10 11 12 13 14 10 $500 $500 $500 $500 $500 11 200 550 550 550 550 12 100 250 600 600 600 13 400 50 300 650 650 14 700 350 0 350 700 Act Expected profit 10 $500.00 11 504.50 12 421.50 13 233.50 14 31.50 Order 11 mobile homes because expected profit of $504.50 is the highest. Opportunity Loss Supply 10 11 12 13 14 10 $0 $50 $100 $150 $200 11 300 0 50 100 150 12 600 300 0 50 100 13 900 600 300 0 50 14 1200 900 600 300 0 Act 10 11 12 13 14 Expect. Opp. Loss $95.50 $91 $174 $362 $627 Decision: Order 11 homes because the opportunity loss of $91 is the smallest.
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16.
e.
$91, found by $595.50 – 504.50 = $91.00 value of perfect information (LO17-2, 17-3 & 17-5)
a.
Event Act 7 8 9 10 7 $35 $35 $35 $35 8 15 40 40 40 9 5 20 45 45 10 25 0 25 50 Expected profits are: Demand Expected Payoff 7 $35.00 8 37.50 9 33.75 10 18.75 The computation for 9 snowmobiles is: Event Probability Payoff Expected Profit X P(X) Y P(X)Y 7 0.10 $5 $0.50 8 0.25 20 5.00 9 0.45 45 20.25 10 0.20 45 9.00 Total 33.75 Lease 8 snowmobiles because the expected profit of $37.50 is the highest. Opportunity loss table is: Act Opportunity loss Number available 7 8 9 10 7 0 5 10 15 8 20 0 5 10 9 40 20 0 5 10 60 40 20 0 Computations for 8 snowmobiles: Demand Probability Opp.Loss Expected Loss X P(X) OL P(X)OL 7 0.10 $20 $2.00 8 0.25 0 0.00 9 0.45 5 2.25 10 0.20 10 2.00 Total $6.25 The expected opportunity losses are: Number available 7 8 9 10 Expected Opp. Loss $8.75 $6.25 $10.00 $25.00 Lease 8 snowmobiles The expected value of perfect information is $6.25. Profit under certainty is $43.75, found by: Total Expected Event Profit Profit Probability Profit 7 $5 $35 0.10 $3.50 8 5 40 0.25 10.00 9 5 45 0.45 20.25
b.
c. d.
e.
f. g.
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h.
17.
10 5 50 0.20 10.00 Total $43.75 Then, $43.75 $37.50 = $6.25 All evidence indicates that leasing 8 snowmobiles would be the most profitable. (LO17-2, 17-3 & 17-5)
a.
b.
c. d.
e. f.
Event Act 41 42 43 44 45 46 41 $410 $410 $410 $410 $410 $410 42 405 420 420 420 420 420 43 400 415 430 430 430 430 44 395 410 425 440 440 440 45 390 405 420 435 450 450 46 385 400 415 430 445 460 Expected profits are: Act Expected Payoff 41 $410.00 42 419.10 43 426.70 44 432.20 45 431.70 46 427.45 Order 44 because $432.20 is the largest expected profit. Expected opportunity loss: 41 42 43 44 45 46 $28.30 $19.20 $11.60 $6.10 $6.60 $10.85 Order 44 because the opportunity loss of $6.10 is the smallest. Yes, it agrees. $6.10, found by $438.30 $432.20 = $6.10 value of perfect information The maximum we should pay for perfect information is $6.10. (LO17-2, 17-3
& 17-5) 18.
a. The cost per car is $4000, found by $6000 $2000. If Tim purchased 20 cars and he can rent 20 cars the payoff is $12 500. It is computed as follows: (20 cars)(5 days)(50 weeks)($20 $1.50) (20 cars)($4000) = $92 500 $80 000 = $12 500. The other payoffs are computed in a similar fashion. Payoff (in$000) States of Nature 20 21 22 23 EMV(A1) A1 20 12.5 12.500 12.500 12.500 12.5000 A2 21 8.5 13.125 13.125 13.125 12.6625 A3 22 4.5 9.125 13.750 13.750 11.9000 A4 23 0.5 5.125 9.750 14.375 8.8250 EMV(A1) = 1(12.5) = 12.50 EMV(A2) = 0.10(8.5) + 0.90(13.125) = 12.6625 EMV(A3) = 0.10(0.45) + 0.20(9.125) + 0.70(13.75) = 11.90 EMV(A4) = 0.10(0.5) + 0.20(5.125) + 0.50(9.75) + 0.20(14.375)) = 8.825 b. EVCP = 0.1(12.5) + 0.20(13.125) + 0.50(13.75) + 0.20(14.375) = 13.625 EVPI – 13.625 – 12.6625 = 0.9625 $962.50 (LO17-5)
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19.
a.
Event Option 100 300 500 700 1 $29.99 $39.99 $59.99 $79.99 2 34.99 34.99 44.99 64.99 3 59.99 59.99 59.99 59.99 b. Expected costs are: Option Expected Cost 1 $52.49 found by 0.25(29.99) + 0.25(39.99) + 0.25(59.99) + 0.25(79.99) 2 44.99 found by 0.25(34.99) + 0.25(34.99) + 0.25(44.99) + 0.25(64.99) 3 59.99 found by 0.25(59.99) + 0.25(59.99) + 0.25(59.99) + 0.25(59.99) Option 2 is best. c. Option 1, because 29.99 is lower than 34.99 or 59.99. d. Option 3, because 59.99 is lower than 79.99 or 64.99. e. Event Option 100 300 500 700 1 $0 $5 $15 $20 2 5 0 0 5 3 30 25 15 0 f. Option 2, because 5 is lower than 20 or 30. g. EVPI = 44.99 – [0.25(29.99) + 0.25(34.99) + 0.25(44.99) + 0.25(59.99)] = 44.99 – 42.49 = 2.50 (LO17-2, 17-4 & 17-5)
20. Test Yes No Probability
Engine Condition Out of tune Okay $80 $20 100 0 0.30 0.70
EMV(Test) = 0.30($80) + 0.70($20) = $38 EMV(No Test) = 0.30($100) + 0.70($0) = $30 It seems the test is not advisable because $30 is less than $38. (LO17-3)
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