Appendix D: Self-Test Solutions and Answers

to Even-Numbered Exercises

Chapter 1 14. a. Graph with a time series line for each manufacturer
b. Toyota surpasses General Motors in 2006 to become
2. a. 10 the leading auto manufacturer
b. 5 c. A bar chart would show cross-sectional data for 2007;
c. Categorical variables: Size and Fuel bar heights would be GM 8.8, Ford 7.9, DC 4.6, and
Quantitiative variables: Cylinders, City MPG, and Toyota 9.6
Highway MPG
18. a. 36%
d.
b. 189
Variable Measurement Scale c. Categorical
Size Ordinal 20. a. 43% of managers were bullish or very bullish, and 21%
Cylinders Ratio of managers expected health care to be the leading
City MPG Ratio industry over the next 12 months
Highway MPG Ratio b. The average 12-month return estimate is 11.2% for the
Fuel Nominal population of investment managers
3. a. Average for city driving 5 182/10 5 18.2 mpg c. The sample average of 2.5 years is an estimate of how
b. Average for highway driving 5 261/10 5 26.1 mpg long the population of investment managers think it
On average, the miles per gallon for highway driving will take to resume sustainable growth
is 7.9 mpg greater than for city driving 22. a. The population consists of all customers of the chain
c. 3 of 10 or 30% have four-cyclinder engines stores in Charlotte, North Carolina
d. 6 of 10 or 60% use regular fuel b. Some of the ways the grocery store chain could use to
4. a. 7 collect the data are
b. 5 • Customers entering or leaving the store could be
c. Categorical variables: State, Campus Setting, and surveyed
NCAA Division • A survey could be mailed to customers who have a
d. Quantitiative variables: Endowment and Applicants shopper’s club card
Admitted • Customers could be given a printed survey when
6. a. Quantitative they check out
b. Categorical • Customers could be given a coupon that asks them
c. Categorical to complete a brief online survey; if they do, they will
d. Quantitative receive a 5% discount on their next shopping trip
e. Categorical 24. a. Correct
8. a. 1015 b. Incorrect
b. Categorical c. Correct
c. Percentages d. Incorrect
d. .10(1015) 5 101.5; 101 or 102 respondents e. Incorrect

10. a. Quantitative; ratio

b. Categorical; nominal Chapter 2
c. Categorical; ordinal 2. a. .20
d. Quantitative; ratio b. 40
e. Categorical; nominal c/d.
12. a. All visitors to Hawaii Percent
b. Yes Class Frequency Frequency
c. First and fourth questions provide quantitative data
A 44 22
Second and third questions provide categorical data B 36 18
13. a. Federal spending ($ trillions) C 80 40
b. Quantitative D 40 20
c. Time series Total 200 100
d. Federal spending has increased over time
1008 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

3. a. 360° 3 58/120 5 174° Management should be pleased with these results: 64%
b. 360° 3 42/120 5 126° of the ratings are very good to outstanding, and 84% of
c. 48.3% the ratings are good or better; comparing these ratings to
previous results will show whether the restaurant is
making improvements in its customers’ ratings of food
Yes No 16.7%
Opinion
quality

8. a.
No
Relative
Position Frequency Frequency
35% P 17 .309
H 4 .073
d. 1 5 .091
60 2 4 .073
3 2 .036
S 5 .091
40 L 6 .109
C 5 .091
R 7 .127
20
Totals 55 1.000

Yes No No Opinion
b. Pitcher
c. 3rd base
4. a. Categorical d. Right field
b. e. Infielders 16 to outfielders 18
Percent
TV Show Frequency Frequency 10. a/b.
Law & Order 10 20% Percent
CSL 18 36% Rating Frequency Frequency
Without a Trace 9 18% Excellent 20 2
Desp Housewives 13 26% Good 101 10
Total: 50 100% Fair 528 52
Bad 244 24
Terrible 122 12
d. CSI had the largest viewing audience; Desperate House- Total 1015 100
wives was in second place

6. a. c.
Percent 60
Network Frequency Frequency
ABC 15 30 50
Percent Frequency

CBS 17 34
40
FOX 1 2
NBC 17 34 30

20
b. CBS and NBC tied for first; ABC is close with 15
10

7. 0
Terrible Bad Fair Good Excellent
Relative
Rating Frequency Frequency Rating

Outstanding 19 .38
Very good 13 .26 d. 36% a bad or a terrible job
Good 10 .20 12% a good or excellent job
Average 6 .12 e. 50% a bad or a terrible job
Poor 2 .04 4% a good or excellent job
More pessimism in Spain
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1009

12.
Cumulative Salary Percent Frequency
Cumulative Relative 170–179 35
Class Frequency Frequency
180–189 25
#19 10 .20 190–199 5
#29 24 .48 200–209 10
#39 41 .82 210–219 5
#49 48 .96
#59 50 1.00 Total 100

c.
14. b/c.
Cumulative
Percent Percent
Class Frequency Frequency Salary Frequency
6.0 –7.9 4 20 Less than or equal to 159 5
8.0 –9.9 2 10 Less than or equal to 169 20
10.0 –11.9 8 40 Less than or equal to 179 55
12.0 –13.9 3 15 Less than or equal to 189 80
14.0 –15.9 3 15 Less than or equal to 199 85
Totals 20 100 Less than or equal to 209 95
Less than or equal to 219 100
Total 100
15. a/ b.
Waiting Relative
Time Frequency Frequency e. There is skewness to the right
f. 15%
0–4 4 .20
5–9 8 .40 18. a. Lowest $180; highest $2050
10–14 5 .25 b.
15–19 2 .10
20–24 1 .05 Percent
Spending Frequency Frequency
Totals 20 1.00
$0–249 3 12
250–499 6 24
c/d. 500 –749 5 20
Cumulative 750–999 5 20
Waiting Cumulative Relative 1000–1249 3 12
Time Frequency Frequency 1250–1499 1 4
1500–1749 0 0
#4 4 .20 1750–1999 1 4
#9 12 .60 2000–2249 1 4
#14 17 .85
Total 25 100
#19 19 .95
#24 20 1.00
c. The distribution shows a positive skewness
d. Majority (64%) of consumers spend between $250 and
e. 12/20 5 .60 $1000; the middle value is about $750; and two high
16. a. spenders are above $1750
Salary Frequency 20. a.
150–159 1 Off-Course Percent
160–169 3 Income ($1000s) Frequency Frequency
170–179 7 0–4,999 30 60
180–189 5 5,000–9,999 9 18
190–199 1 10,000–14,999 4 8
200–209 2 15,000–19,999 0 0
210–219 1
20,000–24,999 3 6
Total 20 25,000–29,999 2 4
30,000–34,999 0 0
b. 35,000–39,999 0 0
40,000–44,999 1 2
Salary Percent Frequency 45,000–49,999 0 0
150–159 5 Over 50,000 1 2
160–169 15 Total 50 100
1010 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

c. Off-course income is skewed to the right; only Tiger 28. a. 2 14

Woods earns over $50 million 2 67
d. The majority (60%) earn less that $5 million; 78% earn 3 011123
less than $10 million; five golfers (10%) earn between
3 5677
$20 million and $30 million; only Tiger Woods and
Phil Mickelson earn more than $40 million 4 003333344
4 6679
22. 5 7 8 5 00022
6 4 5 8 5 5679
7 0 2 2 5 5 6 8 6 14
8 0 2 3 5 6 6
23. Leaf unit 5 .1 7 2
6 3 b. 40– 44 with 9
7 5 5 7 c. 43 with 5
d. 10%; relatively small participation in the race
8 1 3 4 8
9 3 6
10 0 4 5 29. a.
11 3 y
1 2 Total
24. Leaf unit 5 10
A 5 0 5
11 6 x B 11 2 13
12 0 2 C 2 10 12
13 0 6 7 Total 18 12 30
14 2 2 7
15 5
16 0 2 8 b.
17 0 2 3 y
1 2 Total
25. 9 8 9
A 100.0 0.0 100.0
10 2 4 6 6 x B 84.6 15.4 100.0
11 4 5 7 8 8 9 C 16.7 83.3 100.0
12 2 4 5 7
13 1 2
14 4
c.
15 1
y
26. a. 1 0 3 7 7 b. 0 5 7 1 2
2 4 5 5 1 0 1 1 3 4 A 27.8 0.0
3 0 0 5 5 9 1 5 5 5 8 x B 61.1 16.7
4 0 0 0 5 5 8 2 0 0 0 0 0 0 C 11.1 83.3
5 0 0 0 4 5 5 2 5 5 Total 100.0 100.0
3 0 0 0
3 6
4 d. A values are always in y 5 1
4 B values are most often in y 5 1
5 C values are most often in y 5 2
5
6 3
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1011

30. a. c.
56 Fund Type Frequency
DE 27
40 FI 10
IE 8
24
Total 45
8
y

d. The margin of the crosstabulation shows these fre-

–8 quency distributions
–24 e. Higher returns—International Equity funds
Lower returns—Fixed Income funds
–40
–40 –30 –20 –10 0 10 20 30 40 36. b. Higher 5-year returns are associated with higher net
x asset values.
b. A negative relationship between x and y; y decreases as 38. a.
x increases
Highway MPG
32. a.
Displace 15–19 20–24 25–29 30–34 35–39 Total

1.0–2.9 0 6 72 46 4 128
Household Income ($1000s) 3.0–4.9 3 56 86 0 0 145
Under 25.0– 50.0– 75.0– 100 or 5.0–6.9 23 14 1 0 0 38
Education Level 25 49.9 74.9 99.9 more Total
Total 26 76 159 46 4 311
Not H.S. Graduate 32.10 18.71 9.13 5.26 2.20 13.51
H.S. Graduate 37.52 37.05 33.04 25.73 16.00 29.97
Some College 21.42 28.44 30.74 31.71 24.43 27.21
b. Higher fuel efficiencies are associated with smaller
Bachelor’s Degree 6.75 11.33 18.72 25.19 32.26 18.70 displacement engines
Beyond Bach. Deg. 2.21 4.48 8.37 12.11 25.11 10.61 Lower fuel efficiencies are associated with larger dis-
placement engines
Total 100.00 100.00 100.00 100.00 100.00 100.00
d. Lower fuel efficiencies are associated with larger dis-
placement engines
13.51% of the heads of households did not graduate e. Scatter diagram
from high school
40. a.
b. 25.11%, 53.54%
c. Positive relationship between income and education Division Frequency Percent
level Buick 10 5
Cadillac 10 5
Chevrolet 122 61
34. a. GMC 24 12
Hummer 2 1
5-Year Average Return
Pontiac 18 9
Saab 2 1
0– 10– 20– 30– 40– 50– Saturn 12 6
Fund Type 9.99 19.99 29.99 39.99 49.99 59.99 Total
Total 200 100
DE 1 25 1 0 0 0 27
FI 9 1 0 0 0 0 10
IE 0 2 3 2 0 1 8
b. Chevrolet, 61%
c. Hummer and Saab, both only 1%
Total 10 28 4 2 0 1 45
Maintain Chevrolet and GMC
42. a.
b. SAT Score Frequency
5-Year Average Return Frequency 800–999 1
0–9.99 10 1000–1199 3
10–19.99 28 1200–1399 6
20–29.99 4 1400–1599 10
30–39.99 2 1600–1799 7
40–49.99 0 1800–1999 2
50–59.99 1 2000–2199 1
Total 45 Total 30
1012 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

b. Nearly symmetrical d.
c. 33% of the scores fall between 1400 and 1599 High Low
A score below 800 or above 2200 is unusual Temp Frequency Temp Frequency
The average is near or slightly above 1500 10–19 0 10–19 1
20–29 0 20–29 5
44. a. 30–39 1 30–39 5
Percent 40–49 4 40–49 5
Population Frequency Frequency 50–59 3 50–59 3
0.0–2.4 17 34 60–69 9 60–69 1
2.5–4.9 12 24 70–79 2 70–79 0
5.0–7.4 9 18 80–89 1 80–89 0
7.5–9.9 4 8 Total 20 Total 20
10.0–12.4 3 6
12.5–14.9 1 2
15.0–17.4 1 2 48. a.
17.5–19.9 1 2
20.0–22.4 0 0 Level of Support Percent Frequency
22.5–24.9 1 2 Strongly favor 30.10
25.0–27.4 0 0 Favor more than oppose 34.83
27.5–29.9 0 0 Oppose more than favor 21.13
30.0–32.4 0 0 Strongly oppose 13.94
32.5–34.9 0 0
Total 100.00
35.0–37.4 1 2
Total 50 100
Overall favor higher tax 5 30.10% 1 34.83%
5 64.93%
c. High positive skewness b. 20.2, 19.5, 20.6, 20.7, 19.0
d. 17 (34%) with population less than 2.5 million Roughly 20% per country
29 (58%) with population less than 5 million c. The crosstabulation with column percentages:
8 (16%) with population greater than 10 million
Largest 35.9 million (California)
Smallest .5 million (Wyoming) Country
Great United
46. a. High Temperatures Support Britain Italy Spain Germany States

1 Strongly favor 31.00 31.96 45.99 19.98 20.98

Favor more than oppose 34.04 39.04 32.01 36.99 32.06
2 Oppose more than favor 23.00 17.99 13.98 24.03 26.96
3 0 Strongly oppose 11.96 11.01 8.03 18.99 20.00
Total 100.00 100.00 100.00 100.00 100.00
4 1 2 2 5
5 2 4 5
6 0 0 0 1 2 2 5 6 8 Considering the percentage of respondents who favor
7 0 7 the higher tax by either saying “strongly favor” or “fa-
8 4 vor more than oppose,” 65.04%, 71.00%, 78.00%,
b. Low Temperatures 56.97%, and 53.04% for the five countries; all show
more than 50% support, but all European countries
1 1
show more support for the tax than the United States;
2 1 2 6 7 9
Italy and Spain show the highest level of support.
3 1 5 6 8 9
4 0 3 3 6 7 50. a. Row totals: 247; 54; 82; 121
5 0 0 4 Column totals: 149; 317; 17; 7; 14
b.
6 5
7 Year Freq. Fuel Freq.
8 1973 or before 247 Elect. 149
1974–79 54 Nat. Gas 317
c. The most frequent range for high is in 60s (9 of 20) 1980–86 82 Oil 17
with only one low temperature above 54 1987–91 121 Propane 7
High temperatures range mostly from 41 to 68, while Total 504 Other 14
low temperatures range mostly from 21 to 47 Total 504
Low was 11; high was 84
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1013

c. Crosstabulation of column percentages 25

i5 (8) 5 2; use positions 2 and 3
100
Fuel Type 20 1 25
Year 25th percentile 5 5 22.5
Constructed Elect. Nat. Gas Oil Propane Other 2
1973 or before 26.9 57.7 70.5 71.4 50.0
65
i5 (8) 5 5.2; round up to position 6
1974–1979 16.1 8.2 11.8 28.6 0.0 100
1980–1986 24.8 12.0 5.9 0.0 42.9
1987–1991 32.2 22.1 11.8 0.0 7.1 65th percentile 5 28
Total 100.0 100.0 100.0 100.0 100.0 75
i5 (8) 5 6; use positions 6 and 7
100
28 1 30
d. Crosstabulation of row percentages 75th percentile 5 5 29
2
Fuel Type 4. 59.73, 57, 53
Year Nat. 6. a. 18.42
Constructed Elect. Gas Oil Propane Other Total
b. 6.32
1973 or before 16.2 74.1 4.9 2.0 2.8 100.0
1974–1979 44.5 48.1 3.7 3.7 0.0 100.0
c. 34.3%
1980–1986 45.1 46.4 1.2 0.0 7.3 100.0 d. Reductions of only .65 shots and .9% made shots per game
1987–1991 39.7 57.8 1.7 0.0 0.8 100.0 Yes, agree but not dramatically
o xi 3200
8. a. x̄ 5 5 5 160
52. a. Crosstabulation of market value and profit n 20
Order the data from low 100 to high 360

1100220 5 10; use 10th and

Profit ($1000s) 50
Median: i 5
Market Value 0– 300– 600– 900–
($1000s) 300 600 900 1200 Total 11th positions

1 2
0–8000 23 4 27 130 1 140
8000–16,000 4 4 2 2 12 Median 5 5 135
16,000–24,000 2 1 1 4
2
24,000–32,000 1 2 1 4 Mode 5 120 (occur 3 times)
32,000– 40,000 2 1 3
11002 20 5 5; use 5th and 6th positions
25
Total 27 13 6 4 50 b. i5

Q 51 2 5 115
115 1 115
1
b. Crosstabulation of row percentages 2
i51
100 2
75
20 5 15; use 15th and 16th positions
Profit ($1000s)

Q 51 2 5 187.5
Market Value 0– 300 – 600– 900 – 180 1 195
($1000s) 300 600 900 1200 Total 3
2
c. i 5 1
100 2
0–8000 85.19 14.81 0.00 0.00 100 90
8000–16,000 33.33 33.33 16.67 16.67 100 20 5 18; use 18th and 19th positions
16,000–24,000 0.00 50.00 25.00 25.00 100
90th percentile 5 1 2 5 245
235 1 255
24,000–32,000 0.00 25.00 50.00 25.00 100
32,000– 40,000 0.00 66.67 33.33 0.00 100 2
90% of the tax returns cost $245 or less
c. A positive relationship is indicated between profit and 10. a. .4%, 3.5%
market value; as profit goes up, market value goes up b. 2.3%, 2.5%, 2.7%
c. 2.0%, 2.8%
54. b. A positive relationship is demonstrated between mar- d. optimistic
ket value and stockholders’ equity
12. Disney: 3321, 255.5, 253, 169, 325
Pixar: 3231, 538.5, 505, 363, 631
Pixar films generate approximately twice as much box
Chapter 3 office revenue per film
2. 16, 16.5 14. 16, 4
3. Arrange data in order: 15, 20, 25, 25, 27, 28, 30, 34 15. Range 5 34 2 15 5 19
20 Arrange data in order: 15, 20, 25, 25, 27, 28, 30, 34
i5 (8) 5 1.6; round up to position 2
100 25 20 1 25
20th percentile 5 20 i5 (8) 5 2; Q1 5 5 22.5
100 2
1014 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

75 28 1 30 38 2 30 1
i5 (8) 5 6; Q3 5 5 29 c. z 5 5 1.6; 1 2 5 .61
100 2 5 (1.6)2
IQR 5 Q3 2 Q1 5 29 2 22.5 5 6.5 42 2 30 1
d. z 5 5 2.4; 1 2 5 .83
o xi 204 5 (2.4)2
x̄ 5 5 5 25.5
n 8 48 2 30 1
e. z 5 5 3.6; 1 2 5 .92
5 (3.6)2
xi (xi 2 x̄) (xi 2 x̄)2 28. a. 95%
27 1.5 2.25 b. Almost all
25 2.5 .25 c. 68%
20 25.5 30.25
15 210.5 110.25 29. a. z 5 2 standard deviations
30 4.5 20.25 1 1 3
34 8.5 72.25 1 2 2 5 1 2 2 5 ; at least 75%
z 2 4
28 2.5 6.25
25 2.5 .25 b. z 5 2.5 standard deviations
242.00 1 1
12 2512 5 .84; at least 84%
z 2.52
o(xi 2 x̄)2 242
s2 5 5 5 34.57 c. z 5 2 standard deviations
n21 821
Empirical rule: 95%
s 5 Ï34.57 5 5.88
30. a. 68%
b. 81.5%
c. 2.5%
16. a. Range 5 190 2 168 5 22
o xi 1068 32. a. 2.67
b. x̄ 5 5 5 178 b. 1.50
n 6
o(xi 2 x̄)2 c. Neither an outlier
s2 5 d. Yes; z 5 8.25
n21
42 1 (210)2 1 62 1 122 1 (28)2 1 (24)2 34. a. 76.5, 7
5 b. 16%, 2.5%
621
c. 12.2, 7.89; no
376
5 5 75.2 36. 15, 22.5, 26, 29, 34
5
38. Arrange data in order: 5, 6, 8, 10, 10, 12, 15, 16, 18
c. s 5 Ï75.2 5 8.67
25
s 8.67 i5 (9) 5 2.25; round up to position 3
d. (100) 5 (100%) 5 4.87% 100
x̄ 178
Q1 5 8
18. a. 38, 97, 9.85 Median (5th position) 5 10
b. Eastern shows more variation 75
i5 (9) 5 6.75; round up to position 7
20. Dawson: range 5 2, s 5 .67 100
Clark: range 5 8, s 5 2.58 Q3 5 15
5-number summary: 5, 8, 10, 15, 18
22. a. 1285, 433
Freshmen spend more
b. 1720, 352
c. 404, 131.5 5 10 15 20
d. 367.04, 96.96
e. Freshmen have more variability 40. a. Men’s 1st place 43.73 minutes faster
b. Medians: 109.64, 131.67
24. Quarter-milers: s 5 .0564, Coef. of Var. 5 5.8% Men’s median time 22.03 minutes faster
Milers: s 5 .1295, Coef. of Var. 5 2.9% c. 65.30, 87.18, 109.64, 128.40, 148.70
26. .20, 1.50, 0, 2.50, 22.20 109.03, 122.08, 131.67, 147.18, 189.28
d. Men’s Limits: 25.35 to 190.23; no outliers
27. Chebyshev’s theorem: at least (1 2 1/z2) Women’s Limits: 84.43 to 184.83; 2 outliers
40 2 30 1 e. Women runners show less variation
a. z 5 5 2; 1 2 5 .75
5 (2)2 41. a. Arrange data in order low to high
45 2 30 1 25
b. z 5 5 3; 1 2 5 .89 i5 (21) 5 5.25; round up to 6th position
5 (3)2 100
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1015

Q1 5 1872 50. b. .910

Median (11th position) 5 4019 c. Strong positive linear relationship; no
75
i5 (21) 5 15.75; round up to 16th position 52. a. 3.69
100 b. 3.175
Q3 5 8305
5-number summary: 608, 1872, 4019, 8305, 14,138 53. a.
b. IQR 5 Q3 2 Q1 5 8305 2 1872 5 6433 fi Mi fi Mi
Lower limit: 1872 2 1.5(6433) 5 27777.5 4 5 20
Upper limit: 8305 1 1.5(6433) 5 17,955 7 10 70
c. No; data are within limits 9 15 135
d. 41,138 . 27,604; 41,138 would be an outlier; data 5 20 100
value would be reviewed and corrected 25 325
e. ofi Mi 325
x̄ 5 5 5 13
n 25

0 3,000 6,000 9,000 12,000 15,000

b.
42. a. 73.5
b. 68, 71.5, 73.5, 74.5, 77 fi Mi (Mi 2 x̄) (Mi 2 x̄)2 fi (Mi 2 x̄)2
c. Limits: 67 and 79; no outliers 4 5 28 64 256
d. 66, 68, 71, 73, 75; 60.5 and 80.5 7 10 23 9 63
63, 65, 66, 67.6, 69; 61.25 and 71.25 9 15 2 4 36
5 20 7 49 245
75, 77, 78.5, 79.5, 81; 73.25 and 83.25
No outliers for any of the services 25 600
e. Verizon is highest rated o f (M 2 x̄)2 600
s2 5 i i 5 5 25
Sprint is lowest rated n21 25 2 1
44. a. 18.2, 15.35 s 5 Ï25 5 5
b. 11.7, 23.5
c. 3.4, 11.7, 15.35, 23.5, 41.3 54. a.
d. Yes; Alger Small Cap 41.3 Grade xi Weight wi
45. b. There appears to be a negative linear relationship 4 (A) 9
between x and y 3 (B) 15
2 (C) 33
c. 1 (D) 3
xi yi xi 2 x̄ yi 2 ȳ (xi 2 x̄)( yi 2 ȳ) 0 (F) 0
4 50 24 4 216 60 credit hours
6 50 22 4 28
11 40 3 26 218 owi xi 9(4) 1 15(3) 1 33(2) 1 3(1)
x̄ 5 5
3 60 25 14 270 owi 9 1 15 1 33 1 3
16 30 8 216 2128 150
40 230 0 0 2240 5 5 2.5
60
x̄ 5 8; ȳ 5 46
o(xi 2 x̄)( yi 2 ȳ) 2240 b. Yes
sxy 5 5 5 260
n21 4
56. 3.8, 3.7

The sample covariance indicates a negative linear 58. a. 1800, 1351

association between x and y b. 387, 1710
sxy 260 c. 7280, 1323
d. rxy 5 5 5 2.969 d. 3,675,303, 1917
sx sy (5.43)(11.40)
e. High positive skewness
The sample correlation coefficient of 2.969 is indica- f. Using a box plot: 4135 and 7450 are outliers
tive of a strong negative linear relationship
60. a. 2.3, 1.85
46. b. There appears to be a positive linear relationship b. 1.90, 1.38
between x and y c. Altria Group 5%
c. sxy 5 26.5 d. 2.51, below mean
d. rxy 5 .693 e. 1.02, above mean
48. 2.91; negative relationship f. No
1016 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

62. a. $670 12. a. 3,478,761

b. $456 b. 1/3,478,761
c. z 5 3; yes c. 1/146,107,962
d. Save time and prevent a penalty cost
14. a. ¹⁄₄
64. a. 215.9 b. ¹⁄₂
b. 55% c. ³⁄₄
c. 175.0, 628.3 15. a. S 5 {ace of clubs, ace of diamonds, ace of hearts, ace
d. 48.8, 175.0, 215.9, 628.3, 2325.0 of spades}
e. Yes, any price over 1308.25 b. S 5 {2 of clubs, 3 of clubs, . . . , 10 of clubs, J of clubs,
f. 482.1; prefer median Q of clubs, K of clubs, A of clubs}
66. a. 364 rooms c. There are 12; jack, queen, or king in each of the four suits
b. $457 d. For (a): 4/52 5 1/13 5 .08
c. 2.293; slight negative correlation For (b): 13/52 5 1/4 5 .25
Higher cost per night tends to be associated with For (c): 12/52 5 .23
smaller hotels 16. a. 36
68. a. .268, low or weak positive correlation c. ¹⁄₆
b. Very poor predictor; spring training is practice and d. ⁵⁄₁₈
does not count toward standings or playoffs e. No; P(odd) 5 P(even) 5 12
f. Classical
70. a. 60.68
b. s 2 5 31.23; s 5 5.59 17. a. (4, 6), (4, 7), (4, 8)
b. .05 1 .10 1 .15 5 .30
c. (2, 8), (3, 8), (4, 8)
d. .05 1 .05 1 .15 5 .25
Chapter 4 e. .15
6 .5 .4 .3 .2 .1
132 5 3!3! 5 (3 .2 .1)(3 .2 .1) 5 20
6 6! 18. a. .0222
2.
b. .8226
ABC ACE BCD BEF c. .1048
ABD ACF BCE CDE 20. a. .108
ABE ADE BCF CDF b. .096
ABF ADF BDE CEF c. .434
ACD AEF BDF DEF
22. a. .40, .40, .60
4. b. (H,H,H), (H,H,T), (H,T,H), (H,T,T), b. .80, yes
(T,H,H), (T,H,T), (T,T,H), (T,T,T) c. Ac 5 {E3, E4, E5}; C c 5 {E1, E4 };
c. ¹⁄₈ P(Ac ) 5 .60; P(C c ) 5 .40
6. P(E1) 5 .40, P(E2) 5 .26, P(E3) 5 .34 d. (E1, E2, E5); .60
The relative frequency method was used e. .80

8. a. 4: Commission Positive—Council Approves 23. a. P(A) 5 P(E1) 1 P(E4 ) 1 P(E6 )

Commission Positive—Council Disapproves 5 .05 1 .25 1 .10 5 .40
Commission Negative—Council Approves P(B) 5 P(E2) 1 P(E4 ) 1 P(E7)
Commission Negative—Council Disapproves 5 .20 1 .25 1 .05 5 .50
P(C ) 5 P(E2) 1 P(E3) 1 P(E5) 1 P(E7)
50 .49 .48 .47
1 2
50 50! 5 .20 1 .20 1 .15 1 .05 5 .60
9. 5 5 5 230,300
4 4!46! 4 .3 .2 .1 b. A < B 5 {E1, E2, E4, E6, E7};
P(A < B) 5 P(E1) 1 P(E2) 1 P(E4 ) 1 P(E6 ) 1 P(E7)
10. a. Using the table, P(Debt) 5 .94 5.05 1 .20 1 .25 1 .10 1 .05
b. Five of the 8 institutions, P(over 60%) 5 5/8 5 .625 5.65
c. Two of the 8 institutions, P(more than $30,000) c. A ù B 5 {E4}; P(A ù B) 5 P(E4) 5 .25
5 2/8 5 .25 d. Yes, they are mutually exclusive
d. P(No debt) 5 1 2 P(debt) 5 1 2 .72 5 .28 e. B c 5 {E1, E3, E5, E6};
e. A weighted average with 72% having average debt of P(B c ) 5 P(E1) 1 P(E3) 1 P(E5) 1 P(E6 )
$32,980 and 28% having no debt 5.05 1 .20 1 .15 1 .10
.72($32,980) 1 .28($0) 5.50
Average debt per graduate 5
.72 1 .28 24. a. .05
5 $23,746 b. .70
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1017

26. a. .64 b. Southwest (.40)

b. .48 c. .7718
c. .36 d. US Airways (.3817); Southwest (.2910)
d. .76
36. a. .7921
28. Let B 5 rented a car for business reasons
b. .9879
P 5 rented a car for personal reasons
c. .0121
a. P(B < P) 5 P(B) 1 P(P) 2 P(B ù P)
d. .3364, .8236, .1764
5.540 1 .458 2 .300
Don’t foul Jerry Stackhouse
5.698
b. P(Neither) 5 1 2 .698 5 .302 38. a. .70
P(A > B) .40 b. .30
30. a. P(A | B) 5 5 5 .6667 c. .67, .33
P(B) .60
d. .20, .10
P(A > B) .40
b. P(B | A) 5 5 5 .80 e. .40
P(A) .50 f. .20
c. No, because P(A | B) Þ P(A) g. No; P(S | M)ÞP(S)
32. a.
39. a. Yes, because P(A1 ù A2 ) 5 0
Car Light Truck Total
b. P(A1 ù B) 5 P(A1)P(B | A1) 5 .40(.20) 5 .08
U.S. .1330 .2939 .4269 P(A2 ù B) 5 P(A2 )P(B | A2 ) 5 .60(.05) 5 .03
Non-U.S. .3478 .2253 .5731 c. P(B) 5 P(A1 ù B) 1 P(A2 ù B) 5 .08 1 .03 5 .11
Total .4808 .5192 1.0000 .08
d. P(A1 | B) 5 5 .7273
.11
b. .4269, .5731 Non-U.S. higher .03
.4808, .5192 Light Truck slightly higher P(A2 | B) 5 5 .2727
.11
c. .3115, .6885 Light Truck higher
d. .6909, .3931 Car higher 40. a. .10, .20, .09
e. .5661, U.S. higher for Light Trucks b. .51
33. a. c. .26, .51, .23

42. M 5 missed payment

Reason for Applying D1 5 customer defaults
Cost/ D2 5 customer does not default
Quality Convenience Other Total P(D1) 5 .05, P(D2 ) 5 .95, P(M | D2 ) 5 .2, P(M | D1) 5 1
Full-time .218 .204 .039 .461 P(D1)P(M| D1)
Part-time .208 .307 .024 .539 a. P(D1 | M) 5
P(D1)P(M|D1) 1 P(D2 )P(M| D2 )
Total .426 .511 .063 1.000 (.05)(1)
5
(.05)(1) 1 (.95)(.2)
b. A student is most likely to cite cost or convenience as .05
the first reason (probability 5 .511); school quality is 5 5 .21
.24
the reason cited by the second largest number of stu-
b. Yes, the probability of default is greater than .20
dents (probability 5 .426)
c. P(quality | full-time) 5 .218/.461 5 .473 44. a. .47, .53, .50, .45
d. P(quality | part-time) 5 .208/.539 5 .386 b. .4963
e. For independence, we must have P(A)P(B) 5 P(A ù B); c. .4463
from the table d. 47%, 53%
P(A ù B) 5 .218, P(A) 5 .461, P(B) 5 .426
46. a. .60
P(A)P(B) 5 (.461)(.426) 5 .196
b. .26
Because P(A)P(B) Þ P(A ù B), the events are not c. .40
independent d. .74
34. a.
On Time Late Total 48. a. 315
b. .29
Southwest .3336 .0664 .40 c. No
US Airways .2629 .0871 .35
d. Republicans
JetBlue .1753 .0747 .25
Total .7718 .2282 1.00 50. a. .76
b. .24
1018 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

52. b. .2022 4. x 5 0, 1, 2, . . . , 9
c. .4618
d. .4005 6. a. 0, 1, 2, . . . , 20; discrete
54. a. .49 b. 0, 1, 2, . . . ; discrete
b. .44 c. 0, 1, 2, . . . , 50; discrete
c. .54 d. 0 # x # 8; continuous
d. No e. x . 0; continuous
e. Yes
7. a. f (x) $ 0 for all values of x
56. a. .25 of (x) 5 1; therefore, it is a valid probability
b. .125 distribution
c. .0125 b. Probability x 5 30 is f (30) 5 .25
d. .10 c. Probability x # 25 is f (20) 1 f (25) 5 .20 1 .15 5 .35
e. No d. Probability x . 30 is f (35) 5 .40
58. a.
Young Older 8. a.
Adult Adult Total x f(x)
Blogger .0432 .0368 .08 1 3/20 5 .15
Nonblogger .2208 .6992 .92 2 5/20 5 .25
Total .2640 .7360 1.00 3 8/20 5 .40
4 4/20 5 .20
b. .2640 Total 1.00
c. .0432
d. .1636
b. f(x)
60. a. .40
b. .67 .4

.3
Chapter 5 .2
1. a. Head, Head (H, H)
Head, Tail (H, T ) .1
Tail, Head (T, H)
x
Tail, Tail (T, T ) 1 2 3 4
b. x 5 number of heads on two coin tosses
c.
c. f (x) $ 0 for x 5 1, 2, 3, 4
Outcome Values of x of (x) 5 1
(H, H ) 2
(H, T ) 1 10. a. x 1 2 3 4 5
(T, H ) 1
(T, T ) 0 f (x) .05 .09 .03 .42 .41

b. x 1 2 3 4 5
d. Discrete; 0, 1, and 2
2. a. x 5 time in minutes to assemble product f (x) .04 .10 .12 .46 .28
b. Any positive value: x . 0
c. .83
c. Continuous
d. .28
3. Let Y 5 position is offered e. Senior executives are more satisfied
N 5 position is not offered
a. S 5 {(Y, Y, Y ), (Y, Y, N ), (Y, N, Y ), (Y, N, N ), (N, Y, Y ), 12. a. Yes
(N, Y, N ), (N, N, Y ), (N, N, N )} b. .15
b. Let N 5 number of offers made; N is a discrete random c. .10
variable
c. Experimental (Y, Y, (Y, Y, (Y, N, (Y, N, (N, Y, (N, Y, (N, N, (N, N, 14. a. .05
Outcome Y) N) Y) N) Y) N) Y) N)
b. .70
Value of N 3 2 2 1 2 1 1 0 c. .40
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1019

16. a. 24. a. Medium: 145; large: 140

y f( y) yf ( y) b. Medium: 2725; large: 12,400
2 .20 .4 25. a. S
4 .30 1.2
7 .40 2.8
8 .10 .8
S
Totals 1.00 5.2 F
E( y) 5 µ 5 5.2

S
b. F

y y2µ ( y 2 µ)2 f( y) ( y 2 µ)2f ( y)

2 23.20 10.24 .20 2.048 F
4 21.20 1.44 .30 .432
112(.4) (.6)
2 1 1 2!
7 1.80 3.24 .40 1.296 b. f (1) 5 5 (.4)(.6) 5 .48
8 2.80 7.84 .10 .784 1!1!
f (0) 5 1 2 (.4) (.6)
2 0 2 2!
Total 4.560 c. 5 (1)(.36) 5 .36
0 0!2!
Var( y) 5 4.56
f (2) 5 1 2 (.4) (.6)
2 2 0 2!
σ 5 Ï4.56 5 2.14 d. 5 (.16)(.1) 5 .16
2 2!0!
e. P(x $ 1) 5 f (1) 1 f (2) 5 .48 1 .16 5 .64
18. a/ b. f. E(x) 5 np 5 2(.4) 5 .8
Var(x) 5 np(1 2 p) 5 2(.4)(.6) 5 .48
x f(x) xf(x) x 2 µ (x 2 µ)2 (x 2 µ)2f (x) σ 5 Ï.48 5 .6928
0 0.04 0.00 21.84 3.39 0.12 26. a. .3487
1 0.34 0.34 20.84 0.71 0.24 b. .1937
2 0.41 0.82 0.16 0.02 0.01 c. .9298
3 0.18 0.53 1.16 1.34 0.24 d. .6513
4 0.04 0.15 2.16 4.66 0.17 e. 1
Total 1.00 1.84 0.79 f. .9, .95
↑ ↑ 28. a. .2789
E(x) Var(x) b. .4181
c. .0733
c/d.
30. a. Probability of a defective part being produced must
be .03 for each part selected; parts must be selected
y f ( y) yf ( y) y2µ ( y 2 µ)2 y 2 µ2f ( y) independently
0 0.00 0.00 22.93 8.58 0.01 b. Let D 5 defective
1 0.03 0.03 21.93 3.72 0.12 G 5 not defective
2 0.23 0.45 20.93 0.86 0.20
3 0.52 1.55 0.07 0.01 0.00 Experimental
Outcome Number
4 0.22 0.90 1.07 1.15 0.26 1st part 2nd part Defective
Total 1.00 2.93 0.59
↑ ↑ (D, D) 2
D
E( y) Var( y)

e. The number of bedrooms in owner-occupied houses is D

greater than in renter-occupied houses; the expected num- G (D, G) 1
ber of bedrooms is 2.93 2 1.84 5 1.09 greater, and the
variability in the number of bedrooms is less for the (G, D) 1
D
owner-occupied houses G
20. a. 430
b. 290; concern is to protect against the expense of a
G (G, G) 0
large loss
22. a. 445
b. $1250 loss c. Two outcomes result in exactly one defect
1020 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

1121 4 2 1 2 11!2!213!4!2
d. P(no defects) 5 (.97)(.97) 5 .9409 3 10 2 3 3! 7!
P(1 defect) 5 2(.03)(.97) 5 .0582
P(2 defects) 5 (.03)(.03) 5 .0009 46. a. f (1) 5 5
142
10 10!
32. a. .90 4!6!
b. .99 (3)(35)
5 5 .50
c. .999 210

1 21 2
d. Yes 3 10 2 3
34. a. .2262 2 222 (3)(1)
b. f (2) 5 5 5 .067
1 2
b. .8355 10 45
2
36. a. .1897
1 21 2
3 10 2 3
b. .9757
0 220 (1)(21)
c. f (12) 5 .0008; yes c. f (0) 5 5 5 .4667
1 2
d. 5 10 45
2
3xe23
1 21 2
3 10 2 3
38. a. f (x) 5
x! 2 422 (3)(21)
d. f (2) 5 5 5 .30
b. .2241
1 2
10 210
c. .1494 4
d. .8008 e. x 5 4 is greater than r 5 3; thus, f(4) 5 0
x 22
2e 48. a. .5250
39. a. f (x) 5
x! b. .8167
b. µ 5 6 for 3 time periods 50. N 5 60, n 5 10
6xe26 a. r 5 20, x 5 0
c. f (x) 5
1 21 2 1 2
x! 20 40 40!
(1)
22e22 4(.1353) 0 10 10!30!
d. f (2) 5 5 5 .2706 f (0) 5 5
2! 2
1 2
60 60!
6 26
6e 10 10!50!
e. f (6) 5 5 .1606
6!
1 21 2
40! 10!50!
45e24 5
f. f (5) 5 5 .1563 10!30! 60!
5! 40 .39 .38 .37 .36 .35 .34 .33 .32 .31
5 . . . . . . . . .
40. a. .1952 60 59 58 57 56 55 54 53 52 51
b. .1048 5 .0112
c. .0183 b. r 5 20, x 5 1

1 21 2
d. .0907 20 40

1 21 2
1 9 40! 10!50!
7 0e27 f (1) 5 5 20
5 e27 5 .0009
1 2
42. a. f (0) 5 60 9!31! 60!
0!
10
b. probability 5 1 2 [f (0) 1 f (1)]
5 .0725
71e27
f (1) 5 5 7e27 5 .0064 c. 1 2 f(0) 2 f (1) 5 1 2 .0112 2 .0725 5 .9163
1!
d. Same as the probability one will be from Hawaii; .0725
probability 5 1 2 [.0009 1 .0064]5 .9927
c. µ 5 3.5 52. a. .2917
3.5 0e23.5 b. .0083
f (0) 5 5 e23.5 5 .0302 c. .5250, .1750; 1 bank
0!
probability 5 1 2 f (0) 5 1 2 .0302 5 .9698 d. .7083
d. e. .90, .49, .70
probability 5 1 2 [f (0) 1 f (1) 1 f (2) 1 f (3) 1 f (4)] 54. a. x 1 2 3 4 5
5 1 2 [.0009 1 .0064 1 .0223 1 .0521 1 .0912] f (x) .24 .21 .10 .21 .24
5 .8271 b. 3.00, 2.34
c. Bonds: E(x) 5 1.36, Var(x) 5 .23
44. a. µ 5 1.25
Stocks: E(x) 5 4, Var(x) 5 1
b. .2865
c. .3581 56. a. .0596
d. .3554 b. .3585
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1021

c. 100 12. a. .2967

d. 95, 9.75 b. .4418
58. a. .9510 c. .3300
b. .0480 d. .5910
c. .0490 e. .8849
f. .2389
60. a. 240
b. 12.96 13. a. P(21.98 # z # .49) 5 P(z # .49) 2 P(z , 21.98)
c. 12.96 5 .6879 2 .0239 5 .6640
b. P(.52 # z # 1.22) 5 P(z # 1.22) 2 P(z , .52)
62. .1912 5 .8888 2 .6985 5 .1903
64. a. .2240 c. P(21.75 # z # 21.04) 5 P(z # 21.04) 2 P(z ,
b. .5767 21.75) 5 .1492 2 .0401 5 .1091
66. a. .4667 14. a. z 5 1.96
b. .4667 b. z 5 1.96
c. .0667 c. z 5 .61
d. z 5 1.12
e. z 5 .44
Chapter 6 f. z 5 .44
1. a. 15. a. The z value corresponding to a cumulative probability
f(x) of .2119 is z 5 2.80
3 b. Compute .9030/2 5 .4515; the cumulative probability
2 of .5000 1 .4515 5 .9515 corresponds to z 5 1.66
c. Compute .2052/2 5 .1026; z corresponds to a cumula-
1 tive probability of .5000 1 .1026 5 .6026, so z 5 .26
x d. The z value corresponding to a cumulative probability
.50 1.0 1.5 2.0 of .9948 is z 5 2.56
e. The area to the left of z is 1 2 .6915 5 .3085,
b. P(x 5 1.25) 5 0; the probability of any single point is
so z 5 2.50
zero because the area under the curve above any single
point is zero 16. a. z 5 2.33
c. P(1.0 # x # 1.25) 5 2(.25) 5 .50 b. z 5 1.96
d. P(1.20 , x , 1.5) 5 2(.30) 5 .60 c. z 5 1.645
d. z 5 1.28
2. b. .50
18. µ 5 30 and σ 5 8.2
c. .60
d. 15 40 2 30
a. At x 5 40, z 5 5 1.22
e. 8.33 8.2
P(z # 1.22) 5 .8888
4. a.
P(x $ 40) 5 1.000 2 .8888 5 .1112
f(x)
20 2 30
1.5 b. At x 5 20, z 5 5 21.22
8.2
1.0 P(z # 21.22) 5 .1112
.5 P(x # 20) 5 .1112
c. A z value of 1.28 cuts off an area of approximately 10%
x
0 1 2 3 in the upper tail
x 5 30 1 8.2(1.28)
b. P(.25 , x , .75) 5 1(.50) 5 .50 5 40.50
c. P(x # .30) 5 1(.30) 5 .30 A stock price of $40.50 or higher will put a company
d. P(x . .60) 5 1(.40) 5 .40 in the top 10%
20. a. .0885
6. a. .125 b. 12.51%
b. .50 c. 93.8 hours or more
c. .25
22. a. .7193
10. a. .9332 b. $35.59
b. .8413 c. .0233
c. .0919 24. a. 200, 26.04
d. .4938 b. .2206
1022 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

c. .1251
d. P(x # 5) 5 1 2 e25/3 5 1 2 .1889 5 .8111
d. 242.84 million
e. P(2 # x # 5) 5 P(x # 5) 2 P(x # 2)
26. a. µ 5 np 5 100(.20) 5 20 5 .8111 2 .4866 5 .3245
σ 2 5 np(1 2 p) 5 100(.20)(.80) 5 16 34. a. .5624
σ 5 Ï16 5 4 b. .1915
b. Yes, because np 5 20 and n(1 2 p) 5 80 c. .2461
c. P(23.5 # x # 24.5) d. .2259
24.5 2 20 35. a. f(x)
z5 5 1.13 P(z # 1.13) 5 .8708
4
23.5 2 20 .09
z5 5 .88 P(z # .88) 5 .8106
4 .08
P(23.5 # x # 24.5) 5 P(.88 # z # 1.13) .07
5 .8708 2 .8106 5 .0602 .06
d. P(17.5 # x # 22.5) .05
22.5 2 20
z5 5 .63 P(z # .63) 5 .7357 .04
4
.03
17.5 2 20
z5 5 2.63 P(z # 2.63) 5 .2643 .02
4
P(17.5 # x # 22.5) 5 P(2.63 # z # .63) .01
5 .7357 2 .2643 5 .4714 x
0 6 12 18 24
e. P(x # 15.5)
15.5 2 20 b. P(x # 12) 5 1 2 e212@12 5 1 2 .3679 5 .6321
z5
4
5 21.13 P(z # 21.13) 5 .1292 c. P(x # 6) 5 1 2 e26 @12 5 1 2 .6065 5 .3935
P(x # 15.5) 5 P(z # 21.13) 5 .1292 d. P(x $ 30) 5 1 2 P(x , 30)
5 1 2 (1 2 e230@12 )
28. a. µ 5 np 5 250(.20) 5 50 5 .0821
b. σ 2 5 np(1 2 p) 5 250(.20)(1 2 20) 5 40 36. a. .3935
σ 5 Ï40 5 6.3246 b. .2386
P(x , 40) 5 P(x # 39.5) c. .1353
x2µ 39.5 2 50 38. a. f (x) 5 5.5e25.5x
z5 5 5 21.66 Area 5 .0485
σ 6.3246 b. .2528
P(x # 39.5) 5 .0485 c. .6002
c. P(55 # x # 60) 5 P(54.5 # x # 60.5) 40. a. $3780 or less
x2µ 54.5 2 50 b. 19.22%
z5 5 5 .71 Area 5 .7611
σ 6.3246 c. $8167.50
x2µ 60.5 2 50
z5 5 5 1.66 Area 5 .9515 42. a. 3229
σ 6.3246
b. .2244
P(54.5 # x # 60.5) 5 .9515 2 .7611 5 .1904
c. $12,383 or more
d. P(x $ 70) 5 P(x $ 69.5)
x2µ 69.5 2 50 44. a. .0228
z5 5 5 3.08 Area 5 .9990
σ 6.3246 b. $50
P(x $ 69.5) 5 1 2 .9990 5 .0010 46. a. 38.3%
30. a. 220 b. 3.59% better, 96.41% worse
b. .0392 c. 38.21%
c. .8962 48. µ 5 19.23 ounces
32. a. .5276 50. a. Lose $240
b. .3935 b. .1788
c. .4724 c. .3557
d. .1341 d. .0594
33. a. P(x # x0 ) 5 1 2 e2x0y3 52. a. ¹⁄₇ minute
b. P(x # 2) 5 1 2 e22/3 5 1 2 .5134 5 .4866 b. 7e27x
c. P(x $ 3) 5 1 2 P(x # 3) 5 1 2 (1 2 e23/3 ) c. .0009
5 e21 5 .3679 d. .2466
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1023

54. a. 2 minutes 16. a. .10

b. .2212 b. 20
c. .3935 c. .72
d. .0821
18. a. 200
b. 5
Chapter 7 c. Normal with E(x̄) 5 200 and σx̄ 5 5
1. a. AB, AC, AD, AE, BC, BD, BE, CD, CE, DE d. The probability distribution of x̄
b. With 10 samples, each has a ¹⁄₁₀ probability
c. E and C because 8 and 0 do not apply; 5 identifies E; 19. a. The sampling distribution is normal with
7 does not apply; 5 is skipped because E is already in E(x̄) 5 µ 5 200
the sample; 3 identifies C; 2 is not needed because the σx̄ 5 σ/ Ïn 5 50/ Ï100 5 5
sample of size 2 is complete For 65, 195 # x̄ # 205
Using the standard normal probability table:
2. 22, 147, 229, 289
x̄ 2 µ 5
3. 459, 147, 385, 113, 340, 401, 215, 2, 33, 348 At x̄ 5 205, z 5 5 51
σx̄ 5
4. a. Bell South, LSI Logic, General Electric
P(z # 1) 5 .8413
b. 120
x̄ 2 µ 25
6. 2782, 493, 825, 1807, 289 At x̄ 5 195, z 5 5 5 21
σx̄ 5
8. ExxonMobil, Chevron, Travelers, Microsoft, Pfizer, and
P(z , 21) 5 .1587
Intel
P(195 # x̄ # 205) 5 .8413 2 .1587 5 .6826
10. a. finite; b. infinite; c. infinite; d. finite; e. infinite
b. For 610, 190 # x̄ # 210
oxi 54 Using the standard normal probability table:
11. a. x̄ 5 5 59
n 6 x̄ 2 µ 10

Î
At x̄ 5 210, z 5 5 52
o(xi 2 x̄)2 σx̄ 5
b. s 5
n21 P(z # 2) 5 .9772
2 2 2 2
o(xi 2 x̄ ) 5 (24) 1 (21) 1 1 1 (22) 1 1 1 5 2 2 2 x̄ 2 µ 210
At x̄ 5 190, z 5 5 5 22
5 48 σx̄ 5

s5 Î621
48
5 3.1
P(z , 22) 5 .0228
P(190 # x̄ # 210) 5 .9722 2 .0228 5 .9544

12. a. .50 20. 3.54, 2.50, 2.04, 1.77

b. .3667 σx̄ decreases as n increases
oxi 465 22. a. Normal with E(x̄) 5 51,800 and σx̄ 5 516.40
13. a. x̄ 5 5 5 93
n 5 b. σx̄ decreases to 365.15
b. c. σx̄ decreases as n increases

23. a.

xi (xi 2 x̄) (xi 2 x̄)2

94 11 1
100 17 49
85 28 64
94 11 1
92 21 1
Totals 465 0 116

s5 Î o(xi 2 x̄)2
n21
5 Î 116
4
5 5.39
51,300 51,800 52,300
x

σ 4000
σx̄ 5 5 5 516.40
Ïn Ï60
14. a. .45
b. .15 52,300 2 51,800
At x̄ 5 52,300, z 5 5 .97
c. .45 516.40
1024 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

P(x̄ # 52,300) 5 P(z # .97) 5 .8340 35. a.

51,300 2 51,800
At x̄ 5 51,300, z 5 5 2.97
516.40
P(x̄ , 51,300) 5 P(z , 2.97) 5 .1660
P(51,300 # x̄ # 52,300) 5 .8340 2 .1660 5 .6680
σ 4000
b. σx̄ 5 5 5 365.15
Ïn Ï120
52,300 2 51,800
At x̄ 5 52,300, z 5 5 1.37
365.15
p
P(x̄ # 52,300) 5 P(z # 1.37) 5 .9147 .30

Î Î
51,300 2 51,800
At x̄ 5 51,300, z 5 5 21.37 p(1 2 p) .30(.70)
365.15 σp̄ 5 5 5 .0458
n 100
P(x̄ , 51,300) 5 P(z , 21.37) 5 .0853
P(51,300 # x̄ # 52,300) 5 .9147 2 .0853 5 .8294 The normal distribution is appropriate because np 5
100(.30) 5 30 and n(1 2 p) 5 100(.70) 5 70 are
24. a. Normal with E(x̄) 5 17.5 and σx̄ 5 .57 both greater than 5
b. .9198 b. P(.20 # p̄ # .40) 5 ?
c. .6212 .40 2 .30
z5 5 2.18
.0458
26. a. .4246, .5284, .6922, .9586
P(.20 # p̄ # .40) 5 P(22.18 # z # 2.18)
b. Higher probability the sample mean will be close to
5 .9854 2 .0146
population mean
5 .9708
28. a. Normal with E(x̄) 5 95 and σx̄ 5 2.56 c. P(.25 # p̄ # .35) 5 ?
b. .7580 .35 2 .30
c. .8502 z5 5 1.09
.0458
d. Part (c), larger sample size
P(.25 # p̄ # .35) 5 P(21.09 # z # 1.09)
30. a. n/N 5 .01; no 5 .8621 2 .1379
b. 1.29, 1.30; little difference 5 .7242
c. .8764 36. a. Normal with E( p̄) 5 .66 and σp̄ 5 .0273
b. .8584
32. a. E( p̄) 5 .40 c. .9606
σp̄ 5 Î
p(1 2 p)
n
5 Î
(.40)(.60)
200
5 .0346
d.
e.
Yes, standard error is smaller in part (c)
.9616, the probability is larger because the increased
Within 6.03 means .37 # p̄ # .43 sample size reduces the standard error
38. a. Normal with E( p̄) 5 .56 and σp̄ 5 .0248
p̄ 2 p .03
z5 5 5 .87 b. .5820
σp̄ .0346
c. .8926
P(.37 # p̄ # .43) 5 P(2.87 # z # .87) 40. a. Normal with E( p̄) 5 .76 and σp̄ 5 .0214
5 .8078 2 .1922 b. .8384
5 .6156 c. .9452
p̄ 2 p .05 42. 122, 99, 25, 55, 115, 102, 61
b. z 5 5 5 1.44
σp̄ .0346
44. a. Normal with E(x̄) 5 115.50 and σx̄ 5 5.53
P(.35 # p̄ # .45) 5 P(21.44 # z # 1.44) b. .9298
5 .9251 2 .0749 c. z 5 22.80, .0026
5 .8502 46. a. 955
34. a. .6156 b. .50
b. .7814 c. .7062
c. .9488 d. .8230
d. .9942 48. a. 625
e. Higher probability with larger n b. .7888
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1025

1Ï652
50. a. Normal with E( p̄) 5 .28 and σp̄ 5 .0290 5.2
19.5 6 1.998
b. .8324
c. .5098 19.5 6 1.29 or (18.21 to 20.79)
52. a. .8882 16. a. 1.69
b. .0233 b. 47.31 to 50.69
54. a. 48 c. Fewer hours and higher cost for United
b. Normal, E( p̄) 5 .25, σp̄ 5 .0625 18. a. 22 weeks
c. .2119 b. 3.8020
c. 18.20 to 25.80
Chapter 8 d. Larger n next time
2. Use x̄ 6 zα/2(σyÏn ) 20. x̄ 5 22; 21.48 to 22.52
a. 32 6 1.645(6yÏ50) 22. a. $9,269 to $12,541
32 6 1.4; 30.6 to 33.4 b. 1523
b. 32 6 1.96(6yÏ50) c. 4,748,714, $34 million
32 6 1.66; 30.34 to 33.66 Range 36
24. a. Planning value of σ 5 5 59
c. 32 6 2.576(6yÏ50) 4 4
32 6 2.19; 29.81 to 34.19 2 2
z .025 σ (1.96)2(9)2
b. n 5 2 5 5 34.57; use n 5 35
4. 54 E (3)2
5. a. 1.96σ yÏn 5 1.96(5yÏ49 ) 5 1.40 (1.96)2(9)2
c. n 5 5 77.79; use n 5 78
b. 24.80 6 1.40; 23.40 to 26.20 (2)2
6. 8.1 to 8.9 z2α/2 σ 2
25. a. Use n 5
8. a. Population is at least approximately normal E2
b. 3.1 (1.96)2(6.84)2
c. 4.1 n5 5 79.88; use n 5 80
(1.5)2
10. a. $113,638 to $124,672 (1.645)2(6.84)2
b. $112,581 to $125,729 b. n5 5 31.65; use n 5 32
(2)2
c. $110,515 to $127,795
26. a. 18
d. Width increases as confidence level increases
b. 35
12. a. 2.179 c. 97
b. 21.676
28. a. 328
c. 2.457
b. 465
d. 21.708 and 1.708
c. 803
e. 22.014 and 2.014
d. n gets larger; no to 99% confidence
oxi 80
13. a. x̄ 5 5 5 10 30. 81
n 8

Î Î
100
o(xi 2 x̄)2 84 31. a. p̄ 5 5 .25
b. s 5 5 5 3.464 400

Î Î
n21 7
p̄(1 2 p̄) .25(.75)
1 2 1 2
s 3.46 b. 5 5 .0217
c. t.025 5 2.365 5 2.9 n 400

Î
Ïn Ï8
p̄(1 2 p̄)
1 2
s c. p̄ 6 z.025
d. x̄ 6 t.025 n
Ïn
10 6 2.9 (7.1 to 12.9) .25 6 1.96(.0217)
14. a. 21.5 to 23.5 .25 6 .0424; .2076 to .2924
b. 21.3 to 23.7 32. a. .6733 to .7267
c. 20.9 to 24.1 b. .6682 to .7318
d. A larger margin of error and a wider interval
34. 1068
15. x̄ 6 tα/2(syÏn )
1760
90% confidence: df 5 64 and t.05 5 1.669 35. a. p̄ 5 5 .88
2000
1 2
5.2
19.5 6 1.669 b. Margin of error
Ï65
19.5 6 1.08 or (18.42 to 20.58)
95% confidence: df 5 64 and t.025 5 1.998
z.05 5 Î p̄(1 2 p̄)
n
Î
5 1.645
.88(1 2 .88)
2000
5 .0120
1026 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

c. Confidence interval 4. a. H0: µ $ 220

.88 6 .0120 or .868 to .892 Ha: µ , 220
d. Margin of error 5. a. Rejecting H0: µ # 56.2 when it is true

z.05 5Îp̄(1 2 p̄)

n
5 1.96 Î.88(1 2 .88)
2000
5 .0142
b. Accepting H0: µ # 56.2 when it is false
6. a. H0: µ # 1
95% confidence interval Ha: µ . 1
.88 6. 0142 or .8658 to .8942 b. Claiming µ . 1 when it is not true
c. Claiming µ # 1 when it is not true
36. a. .23
b. .1716 to .2884 8. a. H0: µ $ 220
Ha: µ , 220
38. a. .1790 b. Claiming µ , 220 when it is not true
b. .0738, .5682 to .7158 c. Claiming µ $ 220 when it is not true
c. 354 x̄ 2 µ0 26.4 2 25
10. a. z 5 5 5 1.48
z2 p*(1 2 p*) (1.96)2(.156)(1 2 .156) σyÏn 6y Ï40
39. a. n 5 .025 5
E2 (.03) 2 b. Using normal table with z 5 1.48: p-value 5
5 562 1.0000 2 .9306 5 .0694
z2.005 p*(1 2 p*) (2.576)2(.156)(1 2 .156) c. p-value . .01, do not reject H0
b. n 5 2 5 d. Reject H0 if z $ 2.33
E (.03) 2
5 970.77; use 971 1.48 , 2.33, do not reject H0

40. .0346 (.4854 to .5546) x̄ 2 µ0 14.15 2 15

11. a. z 5 5 5 22.00
σyÏn 3y Ï50
42. a. .0442
b. 601, 1068, 2401, 9604 b. p-value 5 2(.0228) 5 .0456
c. p-value # .05, reject H0
44. a. 4.00 d. Reject H0 if z # 21.96 or z $ 1.96
b. $29.77 to $37.77 22.00 # 21.96, reject H0
46. a. 122
12. a. .1056; do not reject H0
b. $1751 to $1995
b. .0062; reject H0
c. $172, 316 million
c. < 0; reject H0
d. Less than $1873
d. .7967; do not reject H0
48. a. 14 minutes
b. 13.38 to 14.62 14. a. .3844; do not reject H0
c. 32 per day b. .0074; reject H0
d. Staff reduction c. .0836; do not reject H0
50. 37 15. a. H0: µ $ 1056
52. 176 Ha: µ , 1056
x̄ 2 µ0 910 2 1056
54. a. .5420 b. z 5 5 5 21.83
b. .0508 σyÏn 1600y Ï400
c. .4912 to .5928 p-value 5 .0336
56. a. .8273 c. p-value # .05, reject H0; the mean refund of
b. .7957 to .8589 “last-minute” filers is less than $1056
d. Reject H0 if z # 21.645
58. a. 1267 21.83 # 21.645; reject H0
b. 1509
16. a. H0: µ # 3173
60. a. .3101
Ha: µ . 3173
b. .2898 to .3304
b. .0207
c. 8219; no, this sample size is unnecessarily
c. Reject H0, conclude mean credit card balance for
large
undergraduate student has increased
18. a. H0: µ 5 4.1
Chapter 9 Ha: µ Þ 4.1
2. a. H0: µ # 14 b. 22.21, .0272
Ha: µ . 14 c. Reject H0; return for Mid-Cap Growth Funds differs
b. No evidence that the new plan increases sales from that for U.S. Diversified Funds
c. The research hypothesis µ . 14 is supported; the new 20. a. H0: µ $ 32.79
plan increases sales Ha: µ , 32.79
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1027

b. 22.73 34. a. H0: µ 5 2

c. .0032 Ha: µ Þ 2
d. Reject H0; conclude the mean monthly Internet bill is b. 2.2
less in the southern state c. .52
22. a. H0: µ 5 8 d. Between .20 and .40
Ha: µ Þ 8 Exact p-value 5 .2535
b. .1706 e. Do not reject H0; no reason to change from 2 hours for
c. Do not reject H0; we cannot conclude the mean waiting cost estimating
time differs from 8 minutes
p̄ 2 p0 .68 2 .75

Î Î
d. 7.83 to 8.97; yes 36. a. z 5 5 5 22.80
p0(1 2 p0) .75(1 2 .75)
x̄ 2 µ0 17 2 18
24. a. t5 5 5 21.54 n 300
syÏn 4.5y Ï48 p-value 5 .0026
b. Degrees of freedom 5 n 2 1 5 47 p-value # .05; reject H0
Area in lower tail is between .05 and .10
.72 2 .75

Î
p-value (two-tail) is between .10 and .20 b. z 5 5 21.20
Exact p-value 5 .1303 .75(1 2 .75)
c. p-value . .05; do not reject H0 300
d. With df 5 47, t.025 5 2.012 p-value 5 .1151
Reject H0 if t # 22.012 or t $ 2.012 p-value . .05; do not reject H0
t 5 21.54; do not reject H0 .70 2 .75

Î
c. z 5 5 22.00
26. a. Between .02 and .05; exact p-value 5 .0397; reject H0 .75(1 2 .75)
b. Between .01 and .02; exact p-value 5 .0125; reject H0 300
c. Between .10 and .20; exact p-value 5 .1285; do not p-value 5 .0228
reject H0 p-value # .05; reject H0
27. a. H0: µ $ 238 .77 2 .75

Î
Ha: µ , 238 d. z 5 5 .80
.75(1 2 .75)
x̄ 2 µ0 231 2 238
b. t 5 5 5 2.88 300
syÏn 80y Ï100 p-value 5 .7881
Degrees of freedom 5 n 2 1 5 99 p-value . .05; do not reject H0
p-value is between .10 and .20
Exact p-value 5 .1905 38. a. H0: p 5 .64
c. p-value . .05; do not reject H0 Ha: p Þ .64
Cannot conclude mean weekly benefit in Virginia is less b. p̄ 5 52/100 5 .52
than the national mean p̄ 2 p0 .52 2 .64

Î Î
z5 5 5 22.50
d. df 5 99, t.05 5 21.66 p0(1 2 p0) .64(1 2 .64)
Reject H0 if t # 21.66
n 100
2.88 . 21.66; do not reject H0
p-value 5 2(.0062) 5 .0124
28. a. H0: µ $ 9 c. p-value # .05; reject H0
Ha: µ , 9 Proportion differs from the reported .64
b. Between .005 and .01 d. Yes, because p̄ 5 .52 indicates that fewer believe the
Exact p-value 5 .0072 supermarket brand is as good as the name brand
c. Reject H0; mean tenure of a CEO is less than 9 years
40. a. .2702
30. a. H0: µ 5 600 b. H0: p # .22
Ha: µ Þ 600 Ha: p . .22
b. Between .20 and .40 p-value < 0; reject H0; there is a significant increase
Exact p-value 5 .2491 after viewing commercials
c. Do not reject H0; cannot conclude there has been a c. Helps evaluate the effectiveness of
change in mean CNN viewing audience commercials
d. A larger sample size
42. a. p̄ 5 .15
32. a. H0: µ 5 10,192
b. .0718 to .2282
Ha: µ Þ 10,192
c. The return rate for the Houston store is different than
b. Between .02 and .05
the national average
Exact p-value 5 .0304
c. Reject H0; mean price at dealership differs from 44. a. H0: p # .51
national mean price Ha: p . .51
1028 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

b. p̄ 5 .58, p-value 5 .0026 (zα 1 zβ )2σ 2 (1.645 1 1.28)2(5)2

c. Reject H0; people working the night shift get drowsy 54. n 5 2 5 5 214
( µ0 2 µa ) (10 2 9)2
more often
56. 109
46.
57. At µ0 5 400, α 5 .02; z.02 5 2.05
At µa 5 385, β 5 .10; z.10 5 1.28
With σ 5 30,
(zα 1 zβ)2σ 2 (2.05 1 1.28)2(30)2
c n5 5 44.4 or 45
2 5
Ha: µ < 10 ( µ0 2 µa) (400 2 385)2
58. 324
H0: µ ≥ 10 60. a. H0: µ 5 16
.05 Ha: µ Þ 16
b. .0286; reject H0
10 Readjust line
c. .2186; do not reject H0
c 5 10 2 1.645(5y Ï120 ) 5 9.25 Continue operation
Reject H0 if x̄ # 9.25 d. z 5 2.19; reject H0
a. When µ 5 9, z 5 21.23; do not reject H0
9.25 2 9 Yes, same conclusion
z5 5 .55
5y Ï120 62. a. H0: µ # 119,155
P(Reject H0 ) 5 (1.0000 2 .7088) 5 .2912 Ha: µ . 119,155
b. Type II error b. .0047
c. When µ 5 8, c. Reject H0; mean annual income for theatergoers in Bay
9.25 2 8 Area is higher
z5 5 2.74
5y Ï120 64. t 5 21.05
β 5 (1.0000 2 .9969) 5 .0031 p-value between .20 and .40
Exact p-value 5 .2999
48. a. Concluding µ # 15 when it is not true
Do not reject H0; there is no evidence to conclude that the
b. .2676
age at which women had their first child has changed
c. .0179
66. t 5 2.26
49. a. H0: µ $ 25
p-value between .01 and .025
Ha: µ , 25
Exact p-value 5 .0155
Reject H0 if z # 22.05 Reject H0; mean cost is greater than $125,000
x̄ 2 µ0 x̄ 2 25 68. a. H0: p # .50
z5 5 5 22.05
σyÏn 3y Ï30 Ha: p . .50
Solve for x̄ 5 23.88 b. .64
Decision Rule: Accept H0 if x̄ . 23.88 c. .0026; reject H0; college graduates have a greater stop
Reject H0 if x̄ # 23.88 smoking success rate
b. For µ 5 23, 70. a. H0: p # .80
23.88 2 23 Ha: p . .80
z5 5 1.61 b. .84
3y Ï30
c. .0418
β 5 1.0000 2 .9463 5 .0537
d. Reject H0; more than 80% of customers are satisfied
c. For µ 5 24, with service of home agents
23.88 2 24 72. H0: p $ .90
z5 5 2.22
3y Ï30 Ha: p , .90
β 5 1.0000 2 .4129 5 .5871 p-value 5 .0808
d. The Type II error cannot be made in this case; note that Do not reject H0; claim of at least 90% cannot be rejected
when µ 5 25.5, H0 is true; the Type II error can only be 74. a. H0: µ # 72
made when H0 is false Ha: µ . 72
50. a. Concluding µ 5 28 when it is not true b. .2912
b. .0853, .6179, .6179, .0853 c. .7939
c. .9147 d. 0, because H0 is true
52. .1151, .0015 76. a. 45
Increasing n reduces β b. .0192, .2358, .7291, .7291, .2358, .0192
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1029

Chapter 10 s 21 s 22 2

1n 1 n 2
1 2
1. a. x̄1 2 x̄ 2 5 13.6 2 11.6 5 2 b. df 5 2 2 2 2

n 2 1 1n 2 n 2 1 1n 2
1 s 1 1 s 2
b. zα/2 5 z.05 5 1.645 1

x̄1 2 x̄ 2 6 1.645 Î σ 21
n1 1
σ 22
n2
1
5.2 2

1 35 1 40 2
8.5
1 2
2 2
2

(2.2)2
Î (3)2 5 2 2 5 65.7 2 2

34 1 35 2 39 1 40 2
2 6 1.645 1 1 5.2 1 8.5
50 35 1
2 6 .98 (1.02 to 2.98)
c. zα/2 5 z.05 5 1.96 Use df 5 65

2 6 1.96
(2.2)
50
Î 1
(3)
35
2 2 c. df 5 65, area in tail is between .01 and .025;
two-tailed p-value is between .02 and .05
Exact p-value 5 .0329
2 6 1.17 (.83 to 3.17) d. p-value # .05; reject H0
(x̄1 2 x̄ 2 ) 2 D0 (25.2 2 22.8) 2 0

Î Î
2. a. z 5 5 5 2.03 12. a. x̄1 2 x̄ 2 5 22.5 2 18.6 5 3.9 miles
σ 21 σ2 (5.2)2 (6)2 s 21 s2 2
n1
1 2
n2 40
1
50 1n1
1 2
n2 2
b. df 5
b. p-value 5 1.0000 2 .9788 5 .0212 s 21 2 s 22 2
1 2 1 2
1 1
c. p-value # .05; reject H0 1
n1 2 1 n1 n2 2 1 n2
2 2 2

1 50 1 40 2
4. a. x̄1 2 x̄ 2 5 85.36 2 81.40 5 3.96 8.4 7.4

b. z.025 Î σ 21
1
σ 22
5 1.96 Î (4.55) 2
1
(3.97) 2
5 1.88 5 2 2 2 2 5 87.1

49 1 50 2 39 1 40 2
n1 n2 37 44 1 8.4 1 7.4
1
c. 3.96 6 1.88 (2.08 to 5.84)
Use df 5 87, t.025 5 1.988

Î
6. p-value 5 .0351
8.42 7.42
Reject H0; mean price in Atlanta lower than mean price in 3.9 6 1.988 1
Houston 50 40
3.9 6 3.3 (.6 to 7.2)
8. a. Reject H0; customer service has improved for Rite Aid
b. Do not reject H0; the difference is not statistically 14. a. H0: µ1 2 µ 2 $ 0
significant Ha: µ1 2 µ 2 , 0
c. p-value 5 .0336; reject H0; customer service has b. 22.41
improved for Expedia c. Using t table, p-value is between .005 and .01
d. 1.80 Exact p-value 5 .009
e. The increase for J.C. Penney is not statistically signif- d. Reject H0; nursing salaries are lower in Tampa
icant
16. a. H0: µ1 2 µ 2 # 0
9. a. x̄1 2 x̄ 2 5 22.5 2 20.1 5 2.4 Ha: µ1 2 µ 2 . 0
s 21 s2 2
1 2
b. 38
1 2 c. t 5 1.80, df 5 25
n1 n2
b. df 5 Using t table, p-value is between .025 and .05
s 21 2 s 22 2
1 2 1 2
1 1
1 Exact p-value 5 .0420
n1 2 1 n1 n2 2 1 n2
d. Reject H0; conclude higher mean score if college grad
2.52 4.82 2
20 1 1
30 2 18. a. H0: µ1 2 µ 2 $ 120
5 5 45.8 Ha: µ1 2 µ 2 , 120
1 2.52 2 1 4.82 2
19 20 1 2 1
29 30 1 2 b. 22.10
c. df 5 45, t.025 5 2.014 Using t table, p-value is between .01 and .025

Î Î
Exact p-value 5 .0195
s 21 s2 2.52 4.82 c. 32 to 118
t.025 1 2 5 2.014 1 5 2.1
n1 n2 20 30 d. Larger sample size
d. 2.4 6 2.1 (.3 to 4.5)
19. a. 1, 2, 0, 0, 2
(x̄1 2 x̄ 2 ) 2 0 (13.6 2 10.1) 2 0 b. d̄ 5 odiyn 5 5y5 5 1

Î Î Î Î
10. a. t 5 5 5 2.18
s 21 s2 5.22 8.52 o(di 2 d̄ )2 4
1 2 1 c. sd 5 5 51
n1 n2 35 40 n21 521
1030 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

d̄ 2 µ 120 .22 2 .16

Î
d. t 5 5 5 2.24 5 5 1.70
1y Ï5
sd yÏn
1 2
1 1
.1840(1 2 .1840) 1
df 5 n 2 1 5 4 200 300
Using t table, p-value is between .025 and .05 p-value 5 1.0000 2 .9554 5 .0446
Exact p-value 5 .0443 b. p-value # .05; reject H0
p-value # .05; reject H0 30. p̄1 5 .55, p̄2 5 .48
.07 6 .0691
20. a. 3, 21, 3, 5, 3, 0, 1
b. 2 32. a. H0: pw # pm
c. 2.08 Ha: pw . pm
d. 2 b. p̄w 5 .3699
e. .07 to 3.93 c. p̄m 5 .3400
d. p-value 5 .1093
21. H0: µ d # 0 Do not reject H0; cannot conclude women are more
Ha: µ d . 0 likely to ask directions
d̄ 5 .625
34. a. .64
sd 5 1.30
d̄ 2 µd .625 2 0 b. .45
t5 5 5 1.36 c. .19 6 .0813 (.1087 to .2713)
sd y Ïn 1.30y Ï8
36. a. H0: p1 2 p2 5 0
df 5 n 2 1 5 7
Ha: p1 2 p2 Þ 0
Using t table, p-value is between .10 and .20
b. .13
Exact p-value 5 .1080
c. p-value 5 .0404
p-value . .05; do not reject H0; cannot conclude com-
d. Reject H0; there is a significant difference between
mercial improves mean potential to purchase
the younger and older age groups
22. $.10 to $.32; earnings have increased 38. a. H0: µ 1 2 µ 2 5 0
Ha: µ 1 2 µ 2 Þ 0
24. t 5 1.32
z 5 2.79
Using t table, p-value is greater than .10
p-value 5 .0052
Exact p-value 5 .1142
Reject H0; a significant difference between systems
Do not reject H0; cannot conclude airfares from Dayton
exists
are higher
40. a. H0: µ 1 2 µ 2 # 0
26. a. t 5 21.42 Ha: µ 1 2 µ 2 . 0
Using t table, p-value is between .10 and .20 b. t 5 .60, df 5 57
Exact p-value 5 .1718 Using t table, p-value is greater than .20
Do not reject H0; no difference in mean scores Exact p-value 5 .2754
b. 21.05 Do not reject H0; cannot conclude that funds with loads
c. 1.28; yes have a higher mean rate of return
28. a. p̄1 2 p̄2 5 .48 2 .36 5 .12 42. a. A decline of $2.45

Î
b. 2.45 6 2.15 (.30 to 4.60)
p̄1(1 2 p̄1) p̄ (1 2 p̄2 ) c. 8% decrease
b. p̄1 2 p̄2 6 z.05 1 2
n1 n2 d. $23.93

.12 6 1.645 Î
.48(1 2 .48)
400
1
.36(1 2 .36)
300
44. a. p-value < 0, reject H0
b. .0468 to .1332
.12 6 .0614 (.0586 to .1814) 46. a. .35 and .47
b. .12 6 .1037 (.0163 to .2237)
c. .12 6 1.96 Î .48(1 2 .48)
400
1
.36(1 2 .36)
300
c. Yes, we would expect occupancy rates to be higher

.12 6 .0731 (.0469 to .1931) Chapter 11

2. s 2 5 25
n p̄ 1 n2 p̄2 200(.22) 1 300(.16)
29. a. p̄ 5 1 1 5 5 .1840 a. With 19 degrees of freedom, χ 2.05 5 30.144 and
n1 1 n2 200 1 300 χ 2.95 5 10.117
p̄1 2 p̄2 19(25) 19(25)
Î
z5 # σ2 #
1 2
1 1 30.144 10.117
p̄(1 2 p̄) 1
n1 n2 15.76 # σ 2 # 46.95
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1031

b. With 19 degrees of freedom, χ 2.025 5 32.852 and 17. a. Population 1 is 4-year-old automobiles
χ 2.975 5 8.907 H0: σ 21 # σ 22
19(25) 19(25) Ha: σ 21 . σ 22
# σ2 #
32.852 8.907 s2 170 2
b. F 5 12 5 5 2.89
14.46 # σ 2 # 53.33 s2 100 2
c. 3.8 # σ # 7.3 Degrees of freedom: 25, 24
4. a. .22 to .71 From tables, p-value is less than .01
b. .47 to .84 p-value # .01; reject H0
Conclude that 4-year-old automobiles have a larger
6. a. .2205, 47.95, 6.92 variance in annual repair costs compared to 2-year-old
b. 5.27 to 10.11 automobiles, which is expected because older auto-
8. a. .4748 mobiles are more likely to have more expensive repairs
b. .6891 that lead to greater variance in the annual repair costs
c. .2383 to 1.3687 18. F 5 1.44
.4882 to 1.1699 p-value greater than .20
9. H0: σ 2 # .0004 Do not reject H0; the difference between the variances is
Ha: σ 2 . .0004 not statistically significant
(n 2 1)s 2 (30 2 1)(.0005) 20. F 5 5.29
χ2 5 5 5 36.25 p-value < 0
σ 20 .0004
From table with 29 degrees of freedom, p-value is greater Reject H0; population variances are not equal for seniors
than .10 and managers
p-value . .05; do not reject H0 22. a. F 5 4
The product specification does not appear to be violated p-value less than .01
10. H0: σ 2 # 331.24 Reject H0; greater variability in stopping distance on
Ha: σ 2 . 331.24 wet pavement
χ 2 5 52.07, df 5 35 24. 10.72 to 24.68
p-value between .025 and .05
Reject H0; standard deviation for Vanguard is greater 26. a. χ 2 5 27.44
p-value between .01 and .025
12. a. .8106 Reject H0; variance exceeds maximum requirements
b. χ 2 5 9.49 b. .00012 to .00042
p-value greater than .20
Do not reject H0; cannot conclude the variance for the 28. χ 2 5 31.50
other magazine is different p-value between .05 and .10
Reject H0; conclude that population variance is greater
14. a. F 5 2.4
than 1
p-value between .025 and .05
Reject H0 30. a. n 5 15
b. F.05 5 2.2; reject H0 b. 6.25 to 11.13
15. a. Larger sample variance is s 21 32. F 5 1.39
s2 8.2 Do not reject H0; cannot conclude the variances of grade
F 5 21 5 5 2.05 point averages are different
s2 4
Degrees of freedom: 20, 25 34. F 5 2.08
From table, area in tail is between .025 and .05 p-value between .05 and .10
p-value for two-tailed test is between .05 and .10 Reject H0; conclude the population variances are not equal
p-value . .05; do not reject H0
b. For a two-tailed test:
Fα/2 5 F.025 5 2.30
Reject H0 if F $ 2.30 Chapter 12
2.05 , 2.30; do not reject H0 1. a. Expected frequencies: e1 5 200(.40) 5 80
16. F 5 1.59 e2 5 200(.40) 5 80
p-value less than .05 e3 5 200(.20) 5 40
Reject H0; the Fidelity Fund has greater variance Actual frequencies: f1 5 60, f2 5 120, f3 5 20
1032 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

(60 2 80)2 (120 2 80)2 (20 2 40)2 (20 2 28.5)2 (44 2 39.9)2 (50 2 45.6)2
χ2 5 1 1 χ2 5 1 1
80 80 40 28.5 39.9 45.6
400 1600 400
5 1 1 (30 2 21.5)2 (26 2 30.1)2 (30 2 34.4)2
80 80 40 1 1 1
21.5 30.1 34.4
5 5 1 20 1 10 5 35
Degrees of freedom: k 2 1 5 2 5 7.86
χ 2 5 35 shows p-value is less than .005 Degrees of freedom: (2 2 1)(3 2 1) 5 2
p-value # .01; reject H0; the proportions are not .40, χ 2 5 7.86, p-value between .01 and .025
.40, and .20 Reject H0; column variable and row variable are not
b. Reject H0 if χ 2 $ 9.210 independent
χ 2 5 35; reject H0 10. χ 2 5 19.77, df 5 4
2. χ 2 5 15.33, df 5 3 p-value less than .005
p-value less than .005 Reject H0; column variable and row variable are not
Reject H0; the proportions are not all .25 independent
11. H0: Type of ticket purchased is independent of the type of
3. H0: pABC 5 .29, pCBS 5 .28, pNBC 5 .25, pIND 5 .18
flight
Ha: The proportions are not
pABC 5 .29, pCBS 5 .28, pNBC 5 .25, pIND 5 .18 Ha: Type of ticket purchased is not independent of the type
Expected frequencies: 300(.29) 5 87, 300(.28) 5 84 of flight
300(.25) 5 75, 300(.18) 5 54 Expected frequencies:
e1 5 87, e2 5 84, e3 5 75, e4 5 54 e11 5 35.59 e12 5 15.41
Actual frequencies: f1 5 95, f2 5 70, f3 5 89, f4 5 46 e21 5 150.73 e22 5 65.27
(95 2 87)2 (70 2 84)2 (89 2 75)2 e31 5 455.68 e32 5 197.32
χ2 5 1 1
87 84 75
(46 2 54)2
1 5 6.87 Observed Expected
54
Frequency Frequency
Degrees of freedom: k 2 1 5 3 Ticket Flight ( fi ) (ei ) ( fi 2 ei )2/ei
χ 2 5 6.87, p-value between .05 and .10 First Domestic 29 35.59 1.22
Do not reject H0; cannot conclude that the audience First International 22 15.41 2.82
proportions have changed Business Domestic 95 150.73 20.61
Business International 121 65.27 47.59
4. χ 2 5 29.51, df 5 5 Full-fare Domestic 518 455.68 8.52
p-value is less than .005 Full-fare International 135 197.32 19.68
Reject H0; the percentages differ from those reported by Totals 920 χ 2 5 100.43
the company
Degrees of freedom: (3 2 1)(2 2 1) 5 2
6. a. χ 2 5 12.21, df 5 3
χ 2 5 100.43, p-value is less than .005
p-value is between .005 and .01 Reject H0; type of ticket is not independent of type of
Conclude difference for 2003 flight
b. 21%, 30%, 15%, 34%
Increased use of debit card 12. a. χ 2 5 7.95, df 5 3
c. 51% p-value is between .025 and .05
Reject H0; method of payment is not independent of
8. χ 2 5 16.31, df 5 3 age group
p-value less than .005 b. 18 to 24 use most
Reject H0; ratings differ, with telephone service slightly
better 14. a. χ 2 5 8.47; p-value is between .025 and .05
Reject H0; intent to purchase again is not independent
9. H0: The column variable is independent of the row of the automobile
variable b. Accord 77, Camry 71, Taurus 62, Impala 57
Ha: The column variable is not independent of the row c. Impala and Taurus below, Accord and Camry above;
variable Accord and Camry have greater owner satisfaction,
Expected frequencies: which may help future market share
16. a. 6446
A B C b. χ 2 5 425.4; p-value 5 0
P 28.5 39.9 45.6 Reject H0; attitude toward nuclear power is not
Q 21.5 30.1 34.4 independent of country
c. Italy (58%), Spain (32%)
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1033

18. χ 2 5 3.01, df 5 2 Degrees of freedom: 6 2 2 2 1 5 3

p-value is greater than .10 χ 2 5 3.20, p-value greater than .10
Do not reject H0; couples working is independent of loca- Do not reject H0
tion; 63.3% Assumption of a normal distribution is not rejected
20. First estimate µ from the sample data (sample size 5 120) 22. χ 2 5 4.30, df 5 2
0(39) 1 1(30) 1 2(30) 1 3(18) 1 4(3) p-value greater than .10
µ5 Do not reject H0; assumption of Poisson distribution is not
120
156 rejected
5 5 1.3
120 24. χ 2 5 2.8, df 5 3
Therefore, we use Poisson probabilities with µ 5 1.3 to p-value greater than .10
compute expected frequencies Do not reject H0; assumption of normal distribution is not
rejected

Observed Poisson Expected Difference 26. χ 2 5 8.04, df 5 3

x Frequency Probability Frequency ( fi 2 ei ) p-value between .025 and .05
0 39 .2725 32.70 6.30 Reject H0; potentials are not the same for each sales
1 30 .3543 42.51 212.51 territory
2 30 .2303 27.63 2.37
3 18 .0998 11.98 6.02 28. χ 2 5 4.64, df 5 2
4 or more 3 .0431 5.16 22.17 p-value between .05 and .10
Do not reject H0; cannot conclude market shares have
changed
(6.30)2 (212.51)2 (2.37)2 (6.02)2
χ2 5 1 1 1 30. χ 2 5 42.53, df 5 4
32.70 42.51 27.63 11.98
p-value is less than .005
(22.17)2
1 5 9.04 Reject H0; conclude job satisfaction differs
5.16
Degrees of freedom: 5 2 1 2 1 5 3 32. χ 2 5 23.37, df 5 3
χ 2 5 9.04, p-value is between .025 and .05 p-value is less than .005
Reject H0; not a Poisson distribution Reject H0; employment status is not independent of region

21. With n 5 30 we will use six classes with .1667 of the 34. a. 71%, 22%, slower preferred
probability associated with each class b. χ 2 5 2.99, df 5 2
p-value greater than .10
x̄ 5 22.8, s 5 6.27
Do not reject H0; cannot conclude men and women dif-
The z values that create 6 intervals, each with probability fer in preference
.1667 are 2.98, 2.43, 0, .43, .98
36. χ 2 5 6.17, df 5 6
p-value is greater than .10
z Cutoff Value of x Do not reject H0; assumption that county and day of week
2.98 22.8 2 .98(6.27) 5 16.66 are independent cannot be rejected
2.43 22.8 2 .43(6.27) 5 20.11
0 22.8 1 .00(6.27) 5 22.80 38. χ 2 5 7.75, df 5 3
.43 22.8 1 .43(6.27) 5 25.49 p-value is between .05 and .10
.98 22.8 1 .98(6.27) 5 28.94 Do not reject H0; cannot conclude office vacancies differ
by metropolitan area

Observed Expected Chapter 13

Interval Frequency Frequency Difference
1. a. x̄¯ 5 (156 1 142 1 134)/3 5 144
less than 16.66 3 5 22 k
16.66–20.11
20.11–22.80
7
5
5
5
2
0
SSTR 5 o n (x̄ 2 x̄¯)
j51
j j
2

22.80–25.49 7 5 2
5 6(156 2 144)2 1 6(142 2 144)2 1 6(134 2 144)2
25.49–28.94 3 5 22
28.94 and up 5 5 0 5 1488
SSTR 1488
b. MSTR 5 5 5 744
k21 2
(22)2 (2)2 (0)2 (2)2 (22)2 (0)2
χ2 5 1 1 1 1 1 c. s 1 5 164.4, s 2 5 131.2, s 23 5 110.4
2 2
5 5 5 5 5 5
k

5
16
5
5 3.20 SSE 5 o (n 2 1)s
j51
j
2
j
1034 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

5 5(164.4) 1 5(131.2) 1 5(110.4) MSTR 258

F5 5 5 9.00
5 2030 MSE 28.67
SSE 2030
d. MSE 5 5 5 135.3 Source of Sum of Degrees of Mean
nT 2 k 18 2 3
e. Variation Squares Freedom Square F p-value
Treatments 516 2 258 9.00 .003
Source of Sum of Degrees of Mean Error 430 15 28.67
Variation Squares Freedom Square F p-value Total 946 17
Treatments 1488 2 744 5.50 .0162
Error 2030 15 135.3 Using F table (2 numerator degrees of freedom and 15
Total 3518 17 denominator), p-value is less than .01
Using Excel or Minitab, the p-value corresponding to
MSTR 744 F 5 9.00 is .003
f. F 5 5 5 5.50 Because p-value # α 5 .05, we reject the null hypothesis
MSE 135.3
that the means for the three plants are equal; in other
From the F table (2 numerator degrees of freedom and words, analysis of variance supports the conclusion that
15 denominator), p-value is between .01 and .025 the population mean examination scores at the three NCP
Using Excel or Minitab, the p-value corresponding to plants are not equal
F 5 5.50 is .0162
10. p-value 5 .0000
Because p-value # α 5 .05, we reject the hypothesis Because p-value # α 5 .05, we reject the null hypothesis
that the means for the three treatments are equal that the means for the three groups are equal
2. 12. p-value 5 .0038
Because p-value # α 5 .05, we reject the null hypothesis
Source of Sum of Degrees of Mean that the mean meal prices are the same for the three types
Variation Squares Freedom Square F p-value of restaurants
Treatments 300 4 75 14.07 .0000 13. a. x̄¯ 5 (30 1 45 1 36)/3 5 37
Error 160 30 5.33 k
Total 460 34 SSTR 5 on (x̄ 2 x̄¯)
j51
j j
2
5 5(30 2 37)2 1 5(45 2 37)2

4. 1 5(36 2 37)2 5 570

SSTR 570
MSTR 5 5 5 285
k21 2
Source of Sum of Degrees of Mean k
Variation Squares Freedom Square F p-value SSE 5 o (n 2 1)sj j
2
5 4(6) 1 4(4) 1 4(6.5) 5 66
Treatments 150 2 75 4.80 .0233 j51

Error 250 16 15.63 SSE 66

MSE 5 5 5 5.5
Total 400 18 n T 2 k 15 2 3
MSTR 285
Reject H0 because p-value # α 5 .05 F5 5 5 51.82
MSE 5.5
6. Because p-value 5 .0082 is less than α 5 .05, we reject Using F table (2 numerator degrees of freedom and
the null hypothesis that the means of the three treatments 12 denominator), p-value is less than .01
are equal Using Excel or Minitab, the p-value corresponding to
F 5 51.82 is .0000
8. x̄¯ 5 (79 1 74 1 66)/3 5 73 Because p-value # α 5 .05, we reject the null hypothesis
k
SSTR 5 on (x̄ 2 x̄¯) 2
5 6(79 2 73)2 1 6(74 2 73)2 that the means of the three populations are equal

Î
j j
j51

1n 1 n 2
1 1
1 6(66 2 73)2 5 516 b. LSD 5 tα/2 MSE
i j
SSTR 516
MSTR 5
2
k21
s 1 5 34
5
2
2
5 258

s2 5 20 s23 5 32
Î
5 t.025 5.5 15 1 52
1 1

k
5 2.179Ï2.2 5 3.23
SSE 5 o (n 2 1)s
j51
j j
2
5 5(34) 1 5(20) 1 5(32) 5 430
| x̄1 2 x̄ 2 | 5 | 30 2 45 | 5 15 . LSD; significant difference
SSE 430 | x̄1 2 x̄3 | 5 | 30 2 36 | 5 6 . LSD; significant difference
MSE 5 5 5 28.67
n T 2 k 18 2 3 | x̄ 2 2 x̄3 | 5 | 45 2 36 | 5 9 . LSD; significant difference
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1035

c. x̄1 2 x̄ 2 6 tα/2 MSE Î 1n

1
1
1
1
n2 2
18. a. Significant; p-value 5 .0000
b. Significant; 2.3 . LSD 5 1.19

(30 2 45) 6 2.179 5.5 1 1 1

5 5
Î 1 2
20. a. Significant; p-value 5 .011
b. Comparing North and South
215 6 3.23 5 218.23 to 211.77 )7702 2 5566) 5 2136 . LSD 5 1620.76
significant difference
14. a. Significant; p-value 5 .0106
Comparing North and West
b. LSD 5 15.34
1 and 2; significant )7702 2 8430) 5 728 . LSD 5 1620.76
1 and 3; not significant no significant difference
2 and 3; significant Comparing South and West
15. a. )5566 2 8430) 5 2864 . LSD 5 1775.45
significant difference
Manufacturer Manufacturer Manufacturer 21. Treatment Means
1 2 3
x̄.1 5 13.6, x̄.2 5 11.0, x̄.3 5 10.6
Sample Mean 23 28 21
Sample Variance 6.67 4.67 3.33
Block Means
x̄1. 5 9, x̄2. 5 7.67, x̄3. 5 15.67, x̄4. 5 18.67, x̄5. 5 7.67

x̄¯ 5 (23 1 28 1 21)/3 5 24 Overall Mean

k x̄¯ 5 176/15 5 11.73
SSTR 5 on (x̄ 2 x̄¯)
j51
j j
2
Step 1
2
5 4(23 2 24) 1 4(28 2 24) 1 4(21 2 24) 2 2
SST 5 oo (x
i j
ij 2 x̄¯)2
5 104
5 (10 2 11.73)2 1 (9 2 11.73)2 1 . . . 1 (8 2 11.73)2
SSTR 104
MSTR 5 5 5 52 5 354.93
k21 2
k Step 2
SSE 5 o (n 2 1)s 2

o(x̄ . 2 x̄¯)
j j
2
j51 SSTR 5 b j
j
5 3(6.67) 1 3(4.67) 1 3(3.33) 5 44.01
SSE 44.01 5 5[(13.6 2 11.73)2 1 (11.0 2 11.73)2
MSE 5 5 5 4.89 1 (10.6 2 11.73)2] 5 26.53
n T 2 k 12 2 3
MSTR 52 Step 3
F5 5 5 10.63
MSE 4.89 SSBL 5 k o(x̄ . 2 x̄¯)
j
i
2

Using F table (2 numerator degrees of freedom and

9 denominator), p-value is less than .01 5 3[(9 2 11.73)2 1 (7.67 2 11.73)2
1 (15.67 2 11.73)2 1 (18.67 2 11.73)2
Using Excel or Minitab, the p-value corresponding to
1 (7.67 2 11.73)2] 5 312.32
F 5 10.63 is .0043
Because p-value # α 5 .05, we reject the null hypothesis Step 4
that the mean time needed to mix a batch of material is the SSE 5 SST 2 SSTR 2 SSBL
same for each manufacturer. 5 354.93 2 26.53 2 312.32 5 16.08

b. LSD 5 tα/2 MSE Î 1n

1
1
1
1
n3 2 Source of Sum of Degrees of Mean
Variation Squares Freedom Square F p-value

Î 1
5 t.025 4.89 1
4
1
4 1 2
Treatments 26.53
Blocks 312.32
2
4
13.27 6.60 .0203
78.08
Error 16.08 8 2.01
5 2.262Ï2.45 5 3.54 Total 354.93 14
Since | x̄1 2 x̄3 | 5 | 23 2 21 | 5 2 , 3.54, there does not
appear to be any significant difference between the means From the F table (2 numerator degrees of freedom and
for manufacturer 1 and manufacturer 3 8 denominator), p-value is between .01 and .025
16. x̄1 2 x̄2 6 LSD Actual p-value 5 .0203
23 2 28 6 3.54 Because p-value # α 5 .05, we reject the null hypothe-
25 6 3.54 5 28.54 to 21.46 sis that the means of the three treatments are equal
1036 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

22. Because p-value . α 5 .05, Factor A is not significant

Factor B: F 5 4.06
Source of Sum of Degrees of Mean Using F table (2 numerator degrees of freedom and
Variation Squares Freedom Square F p-value 6 denominator), p-value is between .05 and .10
Treatments 310 4 77.5 17.69 .0005 Using Excel or Minitab, the p-value corresponding to
Blocks 85 2 42.5 F 5 4.06 is .0767
Error 35 8 4.38 Because p-value . α 5 .05, Factor B is not significant
Total 430 14
Interaction: F 5 7.66
Significant; p-value # α 5 .05 Using F table (2 numerator degrees of freedom and
6 denominator), p-value is between .01 and .025
24. p-value 5 .0453 Using Excel or Minitab, the p-value corresponding to
Because p-value # α 5 .05, we reject the null hypothesis F 5 7.66 is .0223
that the mean tune-up times are the same for both analyzers Because p-value # α 5 .05, interaction is significant
26. a. Significant; p-value 5 .0231 30. Design: p-value 5 .0104; significant
Size: p-value 5 .1340; not significant
b. Writing section
Interaction: p-value5 .2519; not significant
28. Step 1 32. Class: p-value 5 .0002; significant
SST 5 ooo (x ijk 2 x̄¯)2 Type: p-value 5 .0006; significant
i j k Interaction: p-value 5 .4229; not significant
5 (135 2 111)2 1 (165 2 111)2
34. Significant; p-value 5 .0134
1 . . . 1 (136 2 111)2 5 9028
36. Significant; p-value 5 .046
Step 2
38. Not significant; p-value 5 .2455
SSA 5 br o (x̄ . 2 x̄¯)
i
j
2

40. a. Significant; p-value 5 .0175

5 3(2)[(104 2 111)2 1 (118 2 111)2] 5 588 42. Significant; p-value 5 .004
Step 3 44. Type of machine ( p-value 5 .0226) is significant; type
SSB 5 ar o (x̄. 2 x̄¯) j
2
of loading system ( p-value 5 .7913) and interaction
j ( p-value 5 .0671) are not significant
5 2(2)[(130 2 111)2 1 (97 2 111)2 1 (106 2 111)2]
5 2328
Step 4 Chapter 14
SSAB 5 r oo i j
(x̄ ij 2 x̄ i. 2 x̄ .j 1 x̄¯)2
1. a. y
5 2[(150 2 104 2 130 1 111)2 14
1 (78 2 104 2 97 1 111)2 12
1 . . . 1 (128 2 118 2 106 1 111)2] 5 4392 10
Step 5 8
SSE 5 SST 2 SSA 2 SSB 2 SSAB 6
5 9028 2 588 2 2328 2 4392 5 1720 4
2
Source of Sum of Degrees of Mean p- 0 x
Variation Squares Freedom Square F value 0 1 2 3 4 5

Factor A 588 1 588 2.05 .2022 b. There appears to be a positive linear relationship between
Factor B 2328 2 1164 4.06 .0767 x and y
Interaction 4392 2 2196 7.66 .0223 c. Many different straight lines can be drawn to provide
Error 1720 6 286.67 a linear approximation of the relationship between x
Total 9028 11 and y; in part (d) we will determine the equation of a
straight line that “best” represents the relationship
Factor A: F 5 2.05 according to the least squares criterion
Using F table (1 numerator degree of freedom and 6 d. Summations needed to compute the slope and y-intercept:
denominator), p-value is greater than .10 ox 15 oy 40
x̄ 5 i 5 5 3, ȳ 5 i 5 5 8,
Using Excel or Minitab, the p-value corresponding to n 5 n 5
F 5 2.05 is .2022 o(xi 2 x̄)( yi 2 ȳ) 5 26, o(xi 2 x̄)2 5 10
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1037

o(xi 2 x̄)( yi 2 ȳ) 26

b1 5 5 5 2.6
o(xi 2 x̄)2 10 xi yi ŷi yi 2 ŷi ( yi 2 ŷi)2 yi 2 ȳ ( yi 2 ȳ)2
b0 5 ȳ 2 b1x̄ 5 8 2 (2.6)(3) 5 0.2 1 3 2.8 .2 .04 25 25
ŷ 5 0.2 2 2.6x 2 7 5.4 1.6 2.56 21 1
e. ŷ 5 .2 1 2.6x 5 .2 1 2.6(4) 5 10.6 3 5 8.0 23.0 9.00 23 9
4 11 10.6 .4 .16 3 9
2. b. There appears to be a negative linear relationship 5 14 13.2 .8 .64 6 36
between x and y SSE 5 12.40 SST 5 80
d. ŷ 5 68 2 3x SSR 5 SST 2 SSE 5 80 2 12.4 5 67.6
e. 38
SSR 67.6
4. a. y b. r 2 5 5 5 .845
SST 80
140 The least squares line provided a good fit; 84.5% of the
130 variability in y has been explained by the least squares line
c. rxy 5 Ï.845 5 1.9192
Weight

120
16. a. SSE 5 230, SST 5 1850, SSR 5 1620
110
b. r 2 5 .876
100 c. rxy 5 2.936
x 18. a. The estimated regression equation and the mean for the
60 62 64 66 68 70
dependent variable:
Height
ŷ 5 1790.5 1 581.1x, ȳ 5 3650
b. There appears to be a positive linear relationship The sum of squares due to error and the total sum of
between x 5 height and y 5 weight squares:
c. Many different straight lines can be drawn to provide SSE 5 o( yi 2 ŷi )2 5 85,135.14
a linear approximation of the relationship between SST 5 o( yi 2 ȳ)2 5 335,000
height and weight; in part (d) we will determine the
Thus, SSR 5 SST 2 SSE
equation of a straight line that “best” represents the
5 335,000 2 85,135.14 5 249,864.86
relationship according to the least squares criterion
d. Summations needed to compute the slope and y-intercept: SSR 249,864.86
b. r 2 5 5 5 .746
ox 325 oy 585 SST 335,000
x̄ 5 i 5 5 65, ȳ 5 i 5 5 117, The least squares line accounted for 74.6% of the total
n 5 n 5
sum of squares
o(xi 2 x̄)( yi 2 ȳ) 5 110, o(xi 2 x̄)2 5 20
c. rxy 5 Ï.746 5 1.8637
o(xi 2 x̄)( yi 2 ȳ) 110 20. a. ŷ 5 12.0169 1 .0127x
b1 5 5 5 5.5
o(xi 2 x̄) 2
20 b. r 2 5 .4503
b0 5 ȳ 2 b1x̄ 5 117 2 (5.5)(65) 5 2240.5 c. 53
ŷ 5 2240.5 1 5.5x 22. a. .77
e. ŷ 5 2240.5 1 5.5(63) 5 106 b. Yes
The estimate of weight is 106 pounds c. rxy 5 1.88, strong
SSE 12.4
6. c. ŷ 5 8.9412 2 .02633x 23. a. s 2 5 MSE 5 5 5 4.133
n22 3
e. 6.3 or approximately $6300
b. s 5 ÏMSE 5 Ï4.133 5 2.033
8. c. ŷ 5 359.2668 2 5.2772x c. o(xi 2 x̄)2 5 10
d. $254 s 2.033
sb1 5 5 5 .643
10. c. ŷ 526,745.44 1 149.29x Ïo(xi 2 x̄)2
Ï10
d. 4003 or $4,003,000 b1 2 β1 2.6 2 0
d. t 5 sb1 5 .643 5 4.044
12. c. ŷ 5 28129.4439 1 22.4443x From the t table (3 degrees of freedom), area in tail is
d. $8704 between .01 and .025
14. c. ŷ 5 37.1217 1 .51758x p-value is between .02 and .05
d. 73 Using Excel or Minitab, the p-value corresponding to
t 5 4.04 is .0272
15. a. ŷi 5 .2 1 2.6xi and ȳ 5 8 Because p-value # α, we reject H0: β1 5 0
1038 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

SSR 30. Significant; p-value 5 .0042

e. MSR 5 5 67.6
1 32. a. s 5 2.033
MSR 67.6 x̄ 5 3, o(xi 2 x̄)2 5 10
F5 5 5 16.36
MSE 4.133
From the F table (1 numerator degree of freedom and sŷp 5 sÎ 1
n
1
(xp 2 x̄)2
o(xi 2 x̄)2

Î
3 denominator), p-value is between .025 and .05
Using Excel or Minitab, the p-value corresponding to 1 (4 2 3)2
5 2.033 1 5 1.11
F 5 16.36 is .0272 5 10
Because p-value # α, we reject H0: β1 5 0 b. ŷ 5 .2 1 2.6x 5 .2 1 2.6(4) 5 10.6
ŷp 6 tα/2sŷp
Source of Sum of Degrees of Mean 10.6 6 3.182(1.11)
Variation Squares Freedom Square F p-value 10.6 6 3.53, or 7.07 to 14.13
Regression
Error
Total
67.6
12.4
80
1
3
4
67.6
4.133
16.36 .0272
Î
c. sind 5 s 1 1
1
n
1
(xp 2 x̄)2
o(xi 2 x̄)2

24. a. 76.6667
5 2.033 1 1 Î 1
5
1
(4 2 3)2
10
5 2.32

b. 8.7560 d. ŷp 6 tα/2 sind

c. .6526 10.6 6 3.182(2.32)
d. Significant; p-value 5 .0193 10.6 6 7.38, or 3.22 to 17.98
e. Significant; p-value 5 .0193 34. Confidence interval: 8.65 to 21.15
SSE 85,135.14 Prediction interval: 24.50 to 41.30
26. a. s2 5 MSE 5 5 5 21,283.79
n22 4 35. a. s 5 145.89, x̄ 5 3.2, o(xi 2 x̄)2 5 .74
s 5 ÏMSE 5 Ï21,283.79 5 145.89 ŷ 5 1790.5 1 581.1x 5 1790.5 1 581.1(3)
o(xi 2 x̄)2 5 .74 5 3533.8

sb1 5
s
Ïo(xi 2 x̄)2
5
145.89
Ï.74
5 169.59 sŷp 5 s Î 1
n
1
(xp 2 x̄)2
o(xi 2 x̄)2

t5 s
b1 2 β1 581.08 2 0
b1
5
169.59
5 3.43 5 145.89 Î
1
6
1
(3 2 3.2)2
.74
5 68.54
From the t table (4 degrees of freedom), area in tail is ŷp 6 tα/2sŷp
between .01 and .025
3533.8 6 2.776(68.54)
p-value is between .02 and .05
3533.8 6 190.27, or $3343.53 to $3724.07

Î
Using Excel or Minitab, the p-value corresponding to
t 5 3.43 is .0266 1 (xp 2 x̄)2
b. sind 5 s 1 1 1
Because p-value # α, we reject H0: β1 5 0 n o(xi 2 x̄)2

b. MSR 5
SSR 249,864.86
1
5
1
5 249,864.86 5 145.89 1 1 Î 1
6
1
(3 2 3.2)2
.74
5 161.19

MSR 249,864.86 ŷp 6 tα/2 sind

F5 5 5 11.74 3533.8 6 2.776(161.19)
MSE 21,283.79
3533.8 6 447.46, or $3086.34 to $3981.26
From the F table (1 numerator degree of freedom and
4 denominator), p-value is between .025 and .05 36. a. $201
b. 167.25 to 234.65
Using Excel or Minitab, the p-value corresponding to
c. 108.75 to 293.15
F 5 11.74 is .0266
38. a. $5046.67
Because p-value # α, we reject H0: β1 5 0
b. $3815.10 to $6278.24
c. c. Not out of line

Source of Sum of Degrees of Mean

40. a. 9
Variation Squares Freedom Square F p-value b. ŷ 5 20.0 1 7.21x
c. 1.3626
Regression 249,864.86 1 249,864.86 11.74 .0266
Error 85,135.14 4 21,283.79 d. SSE 5 SST 2 SSR 5 51,984.1 2 41,587.3 5 10,396.8
Total 335,000 5 MSE 5 10,396.8/7 5 1485.3
MSR 41,587.3
F5 5 5 28.0
28. They are related; p-value 5 .000 MSE 1485.3
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1039

From the F table (1 numerator degree of freedom and satisfied; the scatter diagram for these data also indi-
7 denominator), p-value is less than .01 cates that the underlying relationship between x and y
Using Excel or Minitab, the p-value corresponding to may be curvilinear
F 5 28.0 is .0011 d. s 2 5 23.78
Because p-value # α 5 .05, we reject H0: β1 5 0 1 (x 2 x̄)2
hi 5 1 i
e. ŷ 5 20.0 1 7.21(50) 5 380.5, or $380,500 n o(xi 2 x̄)2
1 (x 2 14)2
42. a. ŷ 5 80.0 1 50.0x 5 1 i
5 126
b. 30
c. Significant; p-value 5 .000
d. $680,000 Standardized
xi hi syi 2 ŷi yi 2 ŷi Residuals
44. b. Yes 6 .7079 2.64 3.48 1.32
c. ŷ 5 2044.38 2 28.35 weight 11 .2714 4.16 22.47 2.59
d. Significant; p-value 5 .000 15 .2079 4.34 24.83 21.11
e. .774; a good fit 18 .3270 4.00 21.60 2.40
20 .4857 3.50 5.22 1.49
oxi 70 oy 76
45. a. x̄ 5 5 5 14, ȳ 5 i 5 5 15.2,
n 5 n 5 e. The plot of the standardized residuals against ŷ has the
o(xi 2 x̄)( yi 2 ȳ) 5 200, o(xi 2 x̄)2 5 126 same shape as the original residual plot; as stated in
o(xi 2 x̄)( yi 2 ȳ) 200 part (c), the curvature observed indicates that the
b1 5 5 5 1.5873
o(xi 2 x̄)2 126 assumptions regarding the error term may not be
b0 5 ȳ 2 b1x̄ 5 15.2 2 (1.5873)(14) 5 27.0222 satisfied
ŷ 5 27.02 1 1.59x 46. a. ŷ 5 2.32 1 .64x
b. No; the variance appears to increase for larger values
b.
of x
xi yi ŷi yi 2 ŷi
47. a. Let x 5 advertising expenditures and y 5 revenue
6 6 2.52 3.48
11 8 10.47 22.47 ŷ 5 29.4 1 1.55x
15 12 16.83 24.83 b. SST 5 1002, SSE 5 310.28, SSR 5 691.72
18 20 21.60 21.60 SSR
20 30 24.78 5.22 MSR 5 5 691.72
1
SSE 310.28
MSE 5 5 5 62.0554
c. y – ^y n22 5
MSR 691.72
5 F5 5 5 11.15
4 MSE 62.0554
3 From the F table (1 numerator degree of freedom and
2 5 denominator), p-value is between .01 and .025
1 Using Excel or Minitab, p-value 5 .0206
0
Because p-value # α 5 .05, we conclude that the two
–1
–2 variables are related
–3 c.
–4 xi yi ŷi 5 29.40 1 1.55xi yi 2 ŷi
–5
x 1 19 30.95 211.95
5 10 15 20 25 2 32 32.50 2.50
4 44 35.60 8.40
With only five observations, it is difficult to determine 6 40 38.70 1.30
whether the assumptions are satisfied; however, the 10 52 44.90 7.10
plot does suggest curvature in the residuals, which 14 53 51.10 1.90
would indicate that the error term assumptions are not 20 54 60.40 26.40
1040 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

y – ^y c. The scatter diagram is shown:

y
10 150

140
0 130

120
–10
110

^
y 100
30 40 50 60
90 x
100 110 120 130 140 150 160 170 180
d. The residual plot leads us to question the assumption
of a linear relationship between x and y; even though The scatter diagram also indicates that the observation
the relationship is significant at the α 5 .05 level, it x 5 135, y 5 145 may be an outlier; the implication is
would be extremely dangerous to extrapolate beyond that for simple linear regression outliers can be identi-
the range of the data fied by looking at the scatter diagram

48. b. Yes 52. a. Aportion of the Minitab output is shown in Figure D14.52
b. Minitab identifies observation 1 as having a large stan-
50. a. Using Minitab, we obtained the estimated regression dardized residual; thus, we would consider observation
equation ŷ 5 66.1 1 .402x; a portion of the Minitab 1 to be an outlier
output is shown in Figure D14.50; the fitted values and
54. b. Value 5 2252 1 5.83 Revenue
standardized residuals are shown:
c. There are five unusual observations (9, 19, 21, 22, and
32).
Standardized 58. a. ŷ 5 9.26 1 .711x
xi yi ŷi Residuals b. Significant; p-value 5 .001
135 145 120.41 2.11 c. r 2 5 .744; good fit
110 100 110.35 21.08 d. $13.53
130 120 118.40 .14
145 120 124.43 2.38 60. b. GR(%) 5 25.4 1 .285 RR(%)
175 130 136.50 2.78 c. Significant; p-value 5 .000
160 130 130.47 2.04 d. No; r 2 5 .449
120 110 114.38 2.41 e. Yes
f. Yes

b. 62. a. ŷ 5 22.2 2 .148x

b. Significant relationship; p-value 5 .028
Standardized c. Good fit; r2 5 .739
Residuals d. 12.294 to 17.271
2.5
64. a. ŷ 5 220 1 132x
2.0 b. Significant; p-value 5 .000
1.5 c. r 2 5 .873; very good fit
1.0 d. $559.50 to $933.90
0.5
66. a. Market beta 5 .95
0.0
b. Significant; p-value 5 .029
–0.5 c. r 2 5 .470; not a good fit
–1.0 d. Xerox has a higher risk
^
–1.5 y
105 110 115 120 125 130 135 140 68. b. There appears to be a positive linear relationship
between the two variables
The standardized residual plot indicates that the c. ŷ 5 9.37 1 1.2875 Top Five (%)
observation x 5 135, y 5 145 may be an outlier; note d. Significant; p-value 5 .000
that this observation has a standardized residual e. r2 5 .741; good fit
of 2.11 f. rxy 5 .86
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1041

FIGURE D14.50
The regression equation is
Y = 66.1 + 0.402 X

Predictor Coef SE Coef T p

Constant 66.10 32.06 2.06 0.094
X 0.4023 0.2276 1.77 0.137

S = 12.62 R-sq = 38.5% R-sq(adj) = 26.1%

Analysis of Variance

SOURCE DF SS MS F p
Regression 1 497.2 497.2 3.12 0.137
Residual Error 5 795.7 159.1
Total 6 1292.9

Unusual Observations
Obs X Y Fit SE Fit Residual St Resid
1 135 145.00 120.42 4.87 24.58 2.11R

R denotes an observation with a large standardized residual

FIGURE D14.52
The regression equation is
Shipment = 4.09 + 0.196 Media$

Predictor Coef SE Coef T p

Constant 4.089 2.168 1.89 0.096
Media$ 0.19552 0.03635 5.38 0.000

S = 5.044 R-Sq = 78.3% R-Sq(adj) = 75.6%

Analysis of Variance
Source DF SS MS F p
Regression 1 735.84 735.84 28.93 0.000
Residual Error 8 203.51 25.44
Total 9 939.35

Unusual Observations
Obs Media$ Shipment Fit SE Fit Residual St Resid
1 120 36.30 27.55 3.30 8.75 2.30R

R denotes an observation with a large standardized residual

Chapter 15 c. The estimated regression equation is

ŷ 5 218.37 1 2.01x1 1 4.74x2
2. a. The estimated regression equation is An estimate of y when x1 5 45 and x2 5 15 is
ŷ 5 45.06 1 1.94x1 ŷ 5 218.37 1 2.01(45) 1 4.74(15) 5 143.18
An estimate of y when x1 5 45 is
4. a. $255,000
ŷ 5 45.06 1 1.94(45) 5 132.36
b. The estimated regression equation is 5. a. The Minitab output is shown in Figure D15.5a
ŷ 5 85.22 1 4.32x2 b. The Minitab output is shown in Figure D15.5b
An estimate of y when x2 5 15 is c. It is 1.60 in part (a) and 2.29 in part (b); in part (a)
ŷ 5 85.22 1 4.32(15) 5 150.02 the coefficient is an estimate of the change in revenue
1042 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

FIGURE D15.5a
The regression equation is
Revenue = 88.6 + 1.60 TVAdv

Predictor Coef SE Coef T p

Constant 88.638 1.582 56.02 0.000
TVAdv 1.6039 0.4778 3.36 0.015

S = 1.215 R-sq = 65.3% R-sq(adj) = 59.5%

Analysis of Variance

SOURCE DF SS MS F p
Regression 1 16.640 16.640 11.27 0.015
Residual Error 6 8.860 1.477
Total 7 25.500

FIGURE D15.5b
The regression equation is
Revenue = 83.2 + 2.29 TVAdv + 1.30 NewsAdv

Predictor Coef SE Coef T p

Constant 83.230 1.574 52.88 0.000
TVAdv 2.2902 0.3041 7.53 0.001
NewsAdv 1.3010 0.3207 4.06 0.010

S = 0.6426 R-sq = 91.9% R-sq(adj) = 88.7%

Analysis of Variance

SOURCE DF SS MS F p
Regression 2 23.435 11.718 28.38 0.002
Residual Error 5 2.065 0.413
Total 7 25.500

due to a one-unit change in television advertising

c. PCT 5 21.23 1 4.82 FG% 2 2.59 Opp 3 Pt% 1
expenditures; in part (b) it represents an estimate of the
.0344 Opp TO
change in revenue due to a one-unit change in televi-
d. Increase FG%; decrease Opp 3 Pt%; increase Opp TO
sion advertising expenditures when the amount of
e. .638
newspaper advertising is held constant
d. Revenue 5 83.2 1 2.29(3.5) 1 1.30(1.8) 5 93.56 or SSR 14,052.2
$93,560 12. a. R2 5 5 5 .926
SST 15,182.9
6. a. Proportion Won 5 .354 1 .000888 HR n21
b. R2a 5 1 2 (1 2 R2)
b. Proportion Won 5 .865 2 .0837 ERA n2p21
c. Proportion Won 5 .709 1 .00140 HR 2 .103 ERA 10 2 1
5 1 2 (1 2 .926) 5 .905
8. a. ŷ 5 31054 1 1328.7 Reliability 10 2 2 2 1
b. ŷ 5 21313 1 136.69 Score 2 1446.3 Reliability c. Yes; after adjusting for the number of independent vari-
c. $26,643 ables in the model, we see that 90.5% of the variability
10. a. PCT 5 21.22 1 3.96 FG% in y has been accounted for
b. Increase of 1% in FG% will increase PCT by .04 14. a. .75 b. .68
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1043

SSR 23.435 24. a. ŷ 5 2.682 1 .0498 Revenue 1 .0147 % Wins

15. a. R2 5 5 5 .919
SST 25.5 b. Significant; p-value 5 .001
n21 c. Revenue is significant; p-value 5 .001
R2a 5 1 2 (1 2 R2)
n2p21 %Wins is significant; p-value 5 .025
821
5 1 2 (1 2 .919) 5 .887 26. a. Significant; p-value 5 .000
82221 b. All significant; p-values are all , α 5 .05
b. Multiple regression analysis is preferred because both
R2 and R2a show an increased percentage of the variabil- 28. a. Using Minitab, the 95% confidence interval is 132.16 to
ity of y explained when both independent variables are 154.16
used b. Using Minitab, the 95% prediction interval is 111.13 at
16. a. No, R 2 5 .153 175.18
b. Better fit with multiple regression 29. a. See Minitab output in Figure D15.5b.
18. a. R 2 5 .564, R2a 5 .511 ŷ 5 83.23 1 2.29(3.5) 1 1.30(1.8) 5 93.555 or
b. The fit is not very good $93,555
SSR 6216.375 b. Minitab results: 92.840 to 94.335, or $92,840
19. a. MSR 5 5 5 3108.188
p 2 to $94,335
SSE 507.75 c. Minitab results: 91.774 to 95.401, or $91,774 to $95,401
MSE 5 5 5 72.536
n2p21 10 2 2 2 1
30. a. 46.758 to 50.646
MSR 3108.188 b. 44.815 to 52.589
b. F 5 5 5 42.85
MSE 72.536
32. a. E( y) 5 β0 1 β1x1 1 β2x2
From the F table (2 numerator degrees of freedom and
7 denominator), p-value is less than .01
Using Excel or Minitab the p-value corresponding to
where x2 5 H 0 if level 1
1 if level 2
F 5 42.85 is .0001 b. E( y) 5 β0 1 β1x1 1 β2(0) 5 β0 1 β1x1
Because p-value # α, the overall model is significant c. E( y) 5 β0 1 β1x1 1 β2(1) 5 β0 1 β1x1 1 β2
b1 .5906 d. β2 5 E(y | level 2) 2 E(y | level 1)
c. t 5 5 5 7.26
sb1 .0813 β1 is the change in E( y) for a 1-unit change in x1 hold-
ing x2 constant
p-value 5 .0002
Because p-value # α, β1 is significant 34. a. $15,300
b2 .4980 b. ŷ 5 10.1 2 4.2(2) 1 6.8(8) 1 15.3(0) 5 56.1
d. t 5 5 5 8.78 Sales prediction: $56,100
sb2 .0567
c. ŷ 5 10.1 2 4.2(1) 1 6.8(3) 1 15.3(1) 5 41.6
p-value 5 .0001
Sales prediction: $41,600
Because p-value # α, β2 is significant
36. a. ŷ 5 1.86 1 0.291 Months 1 1.10 Type 2 0.609 Person
20. a. Significant; p-value 5 .000
b. Significant; p-value 5 .000 b. Significant; p-value 5 .002
c. Significant; p-value 5 .002 c. Person is not significant; p-value 5 .167
22. a. SSE 5 4000, s 2 5 571.43, 38. a. ŷ 5 291.8 1 1.08 Age 1 .252 Pressure 1 8.74 Smoker
MSR 5 6000 b. Significant; p-value 5 .01
b. Significant; p-value 5 .008 c. 95% prediction interval is 21.35 to 47.18 or a proba-
23. a. F 5 28.38 bility of .2135 to .4718; quit smoking and begin some
p-value 5 .002 type of treatment to reduce his blood pressure
Because p-value # α, there is a significant 39. a. The Minitab output is shown in Figure D15.39
relationship b. Minitab provides the following values:
b. t 5 7.53
p-value 5 .001 Standardized
Because p-value # α, β1 is significant and x1 should xi yi ŷi Residual
not be dropped from the model 1 3 2.8 .16
c. t 5 4.06 2 7 5.4 .94
3 5 8.0 21.65
p-value 5 .010
4 11 10.6 .24
Because p-value # α, β2 is significant and x2 should 5 14 13.2 .62
not be dropped from the model
1044 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

FIGURE D15.39
The regression equation is
Y = 0.20 + 2.60 X

Predictor Coef SE Coef T p

Constant 0.200 2.132 0.09 0.931
X 2.6000 0.6429 4.04 0.027

S = 2.033 R-sq = 84.5% R-sq(adj) = 79.3%

Analysis of Variance
SOURCE DF SS MS F p
Regression 1 67.600 67.600 16.35 0.027
Residual Error 3 12.400 4.133
Total 4 80.000

Standardized 41. a. The Minitab output appears in Figure D15.5b; the esti-
Residuals mated regression equation is
1.0 Revenue 5 83.2 1 2.29 TVAdv 1 1.30 NewsAdv

0.5 b. Minitab provides the following values:

0.0
Standardized Standardized
–0.5 ŷi Residual ŷi Residual
–1.0 96.63 21.62 94.39 1.10
90.41 21.08 94.24 2.40
–1.5 94.34 1.22 94.42 21.12
^ 92.21 2.37 93.35 1.08
–2.0 y
0 3 6 9 12 15

The point (3,5) does not appear to follow the trend of Standardized
the remaining data; however, the value of the standard- Residuals
ized residual for this point, 21.65, is not large enough 1.5
for us to conclude that (3,5) is an outlier
1.0
c. Minitab provides the following values:
0.5
Studentized 0.0
xi yi Deleted Residual
– 0.5
1 3 .13
2 7 .91 –1.0
3 5 24.42
4 11 .19 –1.5
5 14 .54 –2.0
^
y
90 91 92 93 94 95 96 97

t.025 5 4.303 (n 2 p 2 2 5 5 2 1 2 2 5 2 degrees of With relatively few observations, it is difficult to

freedom) determine whether any of the assumptions regarding e
Because the studentized deleted residual for (3,5) is have been violated; for instance, an argument could be
24.42 , 24.303, we conclude that the 3rd observa- made that there does not appear to be any pattern in the
tion is an outlier plot; alternatively, an argument could be made that
40. a. ŷ 5 253.3 1 3.11x there is a curvilinear pattern in the plot
b. 21.94, 2.12, 1.79, .40, 21.90; no c. The values of the standardized residuals are greater
c. .38, .28, .22, .20, .92; no than 22 and less than 12; thus, using this test, there
d. .60, .00, .26, .03, 11.09; yes, the fifth observation are no outliers
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1045

As a further check for outliers, we used Minitab to e β01β1 x

compute the following studentized deleted residuals: 48. a. E( y) 5
1 1 e β01β1 x
b. ĝ(x) 5 22.805 1 1.1492x
Studentized Studentized c. .86
Deleted Deleted d. Estimated odds ratio 5 3.16
Observation Residual Observation Residual
1 22.11 5 1.13 50. b. 67.39
2 21.10 6 2.36
52. a. ŷ 5 21.41 1 .0235x1 1 .00486x2
3 1.31 7 21.16
4 2.33 8 1.10 b. Significant; p-value 5 .0001
c. Both significant
d. R2 5 .937; R2a 5 9.19; good fit
t.025 5 2.776 (n 2 p 2 2 5 8 2 2 2 2 5 4 degrees of
freedom) 54. a. Buy Again 5 27.522 1 1.8151 Steering
Because none of the studentized deleted residuals are b. Yes
less than 22.776 or greater than 2.776, we conclude c. Buy Again 5 25.388 1 .6899 Steering 1 .9113
that there are no outliers in the data Treadwear
d. Minitab provides the following values: d. Significant; p-value 5 .001
56. a. ŷ 5 4.9090 1 10.4658 FundDE 1 21.6823 FundIE
Observation hi Di b. R2 5 .6144; reasonably good fit
1 .63 1.52 c. ŷ 5 1.1899 1 6.8969 FundDE 1 17.6800 FundIE
2 .65 .70 1 0.0265 Net Asset Value ($)
3 .30 .22 1 6.4564 Expense Ratio (%)
4 .23 .01 Net Asset Value ($) is not significant and can be
5 .26 .14
deleted
6 .14 .01
7 .66 .81 d. ŷ 5 24.6074 1 8.1713 FundDE 1 19.5194 FundIE
8 .13 .06 1 5.5197 Expense Ratio (%) 1 5.9237 3StarRank
1 8.2367 4StarRank 1 6.6241 5StarRank
The critical leverage value is e. 15.28%
3( p 1 1) 3(2 1 1)
5 5 1.125
n 8
Because none of the values exceed 1.125, we conclude Chapter 16
that there are no influential observations; however,
using Cook’s distance measure, we see that D1 . 1 1. a. The Minitab output is shown in Figure D16.1a
(rule of thumb critical value); thus, we conclude that b. Because the p-value corresponding to F 5 6.85 is
the first observation is influential .059 . α 5 .05, the relationship is not significant
Final conclusion: observation 1 is an influential c.
observation y
42. b. Unusual trend 40
c. No outliers 35
d. Observation 2 is an influential observation
30
e β01β1 x
44. a. E( y) 5
1 1 e β01β1 x 25
b. Estimate of the probability that a customer who does 20
not have a Simmons credit card will make a purchase
c. ĝ(x) 5 20.9445 1 1.0245x 15
d. .28 for customers who do not have a Simmons credit card 10 x
.52 for customers who have a Simmons credit card 20 25 30 35 40
e. Estimated odds ratio 5 2.79
e β01β1 x The scatter diagram suggests that a curvilinear relation-
46. a. E( y) 5 ship may be appropriate
1 1 e β01β1 x
e22.635510.22018x d. The Minitab output is shown in Figure D16.1d
b. E( y) 5 e. Because the p-value corresponding to F 5 25.68 is
1 1 e22.635510.22018x
.013 , α 5 .05, the relationship is significant
c. Significant; p-value 5 .0002
f. ŷ 5 2168.88 1 12.187(25) 2 .17704(25)2 5 25.145
d. .39
e. $1200 2. a. ŷ 5 9.32 1 .424x; p-value 5 .117 indicates a weak
f. Estimated odds ratio 5 1.25 relationship between x and y
1046 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

FIGURE D16.1a
The regression equation is
Y = - 6.8 + 1.23 X

Predictor Coef SE Coef T p

Constant -6.77 14.17 -0.48 0.658
X 1.2296 0.4697 2.62 0.059

S = 7.269 R-sq = 63.1% R-sq(adj) = 53.9%

Analysis of Variance

SOURCE DF SS MS F p
Regression 1 362.13 362.13 6.85 0.059
Residual Error 4 211.37 52.84
Total 5 573.50

FIGURE D16.1d
The regression equation is
Y = - 169 + 12.2 X - 0.177 XSQ

Predictor Coef SE Coef T p

Constant -168.88 39.79 -4.74 0.024
X 12.187 2.663 4.58 0.020
XSQ -0.17704 0.04290 -4.13 0.026

S = 3.248 R-sq = 94.5% R-sq(adj) = 90.8%

Analysis of Variance

SOURCE DF SS MS F p
Regression 2 541.85 270.92 25.68 0.013
Residual Error 3 31.65 10.55
Total 5 573.50

b. ŷ 5 28.10 1 2.41x 2 .0480x 2 b. Price 5 33829 2 4571 Rating 1 154 RatingSq

R 2a 5 .932; a good fit c. logPrice 5 210.2 1 10.4 logRating
c. 20.965 d. Part (c); higher percentage of variability is explained
4. a. ŷ 5 943 1 8.71x 10. a. Significant; p-value 5 .000
b. Significant; p-value 5 .005 , α 5 .01 b. Significant; p-value 5 .000
5. a. The Minitab output is shown in Figure D16.5a 11. a. SSE 5 1805 2 1760 5 45

1 2
b. Because the p-value corresponding to F 5 73.15 is MSR 1760/4
F5 5 5 244.44
.003 , α 5 .01, the relationship is significant; we MSE 45/25
would reject H0: β1 5 β2 5 0 Because p-value 5 .000 the relationship is
c. See Figure D16.5c significant
6. b. No, the relationship appears to be curvilinear b. SSE(x1, x2, x3, x4) 5 45
c. Several possible models; e.g., c. SSE(x2, x3) 5 1805 2 1705 5 100
ŷ 5 2.90 2 .185x 1 .00351x 2 (100 2 45)/2
d. F 5 5 15.28
8. a. It appears that a simple linear regression model is not 1.8
appropriate Because p-value 5 .000, x1 and x2 are significant
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1047

FIGURE D16.5a
The regression equation is
Y = 433 + 37.4 X -0.383 XSQ

Predictor Coef SE Coef T p

Constant 432.6 141.2 3.06 0.055
X 37.429 7.807 4.79 0.017
XSQ -0.3829 0.1036 -3.70 0.034

S = 15.83 R-sq = 98.0% R-sq(adj) = 96.7%

Analysis of Variance

SOURCE DF SS MS F p
Regression 2 36643 18322 73.15 0.003
Residual Error 3 751 250
Total 5 37395

FIGURE D16.5c
Fit Stdev.Fit 95% C.I. 95% P.I.
1302.01 9.93 (1270.41, 1333.61) (1242.55, 1361.47)

12. a. The Minitab output is shown in Figure D16.12a 20.

b. The Minitab output is shown in Figure D16.12b x1 x2 x3 Treatment
[SSE(reduced) 2 SSE(full)]y(# extra terms) 0 0 0 A
c. F 5 1 0 0 B
MSE(full)
0 1 0 C
(7.2998 2 4.3240)/2 0 0 1 D
5 5 8.95
.1663
E( y) 5 β0 1 β1 x1 1 β2 x2 1 β3 x3
The p-value associated with F 5 8.95 (2 numerator
degrees of freedom and 26 denominator) is .001; with a
p-value , α 5 .05, the addition of the two independent 22. Factor A: x1 5 0 if level 1 and 1 if level 2
variables is significant Factor B:
14. a. ŷ 5 2111 1 1.32 Age 1 .296 Pressure
b. ŷ 5 2123 1 1.51 Age 1 .448 Pressure
x2 x3 Level
1 8.87 Smoker 2 .00276 AgePress
c. Significant; p-value 5 .000 0 0 1
1 0 2
16. a. Weeks 5 28.9 1 1.51 Age 0 1 3
b. Weeks 5 2.07 1 1.73 Age 2 2.7 Manager
E( y) 5 β0 1 β1 x1 1 β2 x2 1 β3 x1x2 1 β4x1x3
2 15.1 Head 2 17.4 Sales
c. Same as part (b)
d. Same as part (b) 24. a. Not significant at the .05 level of significance;
e. Weeks 5 13.1 1 1.64 Age 2 9.76 Married p-value 5 .093
2 19.4 Head 2 29.0 Manager 2 19.0 Sales b. 139
18. a. RPG 5 24.05 1 27.6 OBP
b. A variety of models will provide a good fit; the five- 26. Overall significant; p-value 5 .029
variable model identified using Minitab’s Stepwise Individually, none of the variables are significant at the
Regression procedure with Alpha-to-Enter 5 .10 and .05 level of significance; a larger sample size would be
Alpha-to-Remove 5 .10 follows: helpful
RPG 5 2.0909 1 32.2 OBP 1 .109 HR 2 21.5 AVG
1 .244 3B 2 .0223 BB 28. d 5 1.60; test is inconclusive
1048 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

FIGURE D16.12a
The regression equation is
Scoring Avg. = 46.3 + 14.1 Putting Avg.

Predictor Coef SE Coef T p

Constant 46.277 6.026 7.68 0.000
Putting Avg. 14.103 3.356 4.20 0.000

S = 0.510596 R-Sq = 38.7% R-Sq(adj) = 36.5%

Analysis of Variance

SOURCE DF SS MS F p
Regression 1 4.6036 4.6036 17.66 0.0000
Residual Error 28 7.2998 0.2607
Total 29 11.9035

FIGURE D16.12b
The regression equation is
Scoring Avg. = 59.0 - 10.3 Greens in Reg.
+ 11.4 Putting Avg - 1.81 Sand Saves

Predictor Coef SE Coef T p

Constant 59.022 5.774 10.22 0.000
Greens in Reg. -10.281 2.877 -3.57 0.001
Putting Avg. 11.413 2.760 4.14 0.000
Sand Saves -1.8130 0.9210 -1.97 0.060

S = 0.407808 R-Sq = 63.7% R-Sq(adj) = 59.5%

Analysis of Variance

Source DF SS MS F p
Regression 3 7.5795 2.5265 15.19 0.000
Residual Error 26 4.3240 0.1663
Total 29 11.9035

30. a.
2000
1800
1600
1400
1200
Price ($)

1000
800
600
400
200
0
15 20 25 30 35 40
Weight (lb.)

There appears to be a curvilinear relationship between weight and price

 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1049

b. A portion of the Minitab output follows:

The regression equation is

Price = 11376 - 728 Weight + 12.0 WeightSq

Predictor Coef SE Coef T p

Constant 11376 2565 4.43 0.000
Weight -728.3 193.7 -3.76 0.002
WeightSq 11.974 3.539 3.38 0.004

S = 242.804 R-Sq = 77.0% R-Sq(adj) = 74.1%

Analysis of Variance

SOURCE DF SS MS F p
Regression 2 3161747 1580874 26.82 0.000
Residual Error 16 943263 58954
Total 18 4105011

The results obtained support the conclusion that there is a curvilinear relationship between weight and price
c. A portion of the Minitab output follows:

The regression equation is

Price = 1284 - 572 Type_Fitness - 907 Type_Comfort

Predictor Coef SE Coef T p

Constant 1283.75 95.22 13.48 0.000
Type_Fitness -571.8 153.5 -3.72 0.002
Type_Comfort -907.1 145.5 -6.24 0.000

S = 269.328 R-Sq = 71.7% R-Sq(adj) = 68.2%

Analysis of Variance

SOURCE DF SS MS F p
Regression 2 2944410 1472205 20.30 0.000
Residual Error 16 1160601 72538
Total 18 4105011

Type of bike appears to be a significant factor in predicting price, but the estimated regression equation developed in
part (b) appears to provide a slightly better fit
d. A portion of the Minitab output follows; in this output WxF denotes the interaction between the weight of the bike and the
dummy variable Type_Fitness and WxC denotes the interaction between the weight of the bike and the dummy variable
Type_Comfort

The regression equation is

Price = 5924 - 214 Weight - 6343 Type_Fitness - 7232
Type_Comfort + 261 WxF + 266 WxC

Predictor Coef SE Coef T p

Constant 5924 1547 3.83 0.002
Weight -214.56 71.42 -3.00 0.010
Type_Fitness -6343 2596 -2.44 0.030
1050 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

Type_Comfort -7232 2518 -2.87 0.013

WxF 261.3 111.8 2.34 0.036
WxC 266.41 93.98 2.83 0.014

S = 224.438 R-Sq = 84.0% R-Sq(adj) = 77.9%

Analysis of Variance

SOURCE DF SS MS F p
Regression 5 3450170 690034 13.70 0.000
Residual Error 13 654841 50372
Total 18 4105011

By taking into account the type of bike, the weight, and the 6.
interaction between these two factors, this estimated re-
gression equation provides an excellent fit Base Period
Weighted
Price Price
32. a. Delay 5 63.0 1 11.1 Industry; no significant positive
Item Relative Price Usage Weight Relative
autocorrelation
A 150 22.00 20 440 66,000
34. Significant differences between comfort levels for the B 90 5.00 50 250 22,500
three types of browsers; p-value 5 .034 C 120 14.00 40 560 67,200
Totals 1250 155,700
Chapter 17 I5
155,700
5 125
1250
1. a.
Item Price Relative
7. a. Price relatives for A 5 (3.95/2.50)100 5 158
A 103 5 (7.75/7.50)(100) B 5 (9.90/8.75)100 5 113
B 238 5 (1500/630)(100)
C 5 (.95/.99)100 5 96
b.
7.75 1 1500.00 1507.75
b. I2009 5 (100) 5 (100) 5 237
7.50 1 630.00 637.50 Weighted
7.75(1500) 1 1500.00(2) Price Base Weight Price
c. I2009 5 (100) Item Relative Price Quantity Pi0Qi Relative
7.50(1500) 1 630.00(2)
14,625.00 A 158 2.50 25 62.5 9,875
5 (100) 5 117 B 113 8.75 15 131.3 14,837
12,510.00 C 96 .99 60 59.4 5,702
7.75(1800) 1 1500.00(1) Totals 253.2 30,414
d. I2009 5 (100)
7.50(1800) 1 630.00(1) 30,414
15,450.00 I5 5 120
5 (100) 5 109 253.2
14,130.00

2. a. 32% Cost of raw materials is up 20% for the chemical

b. $8.14
8. I 5 105; portfolio is up 5%
3. a. Price relatives for A 5 (6.00/5.45)100 5 110
$11.86
B 5 (5.95/5.60)100 5 106 10. a. Deflated 1996 wages: (100) 5 $7.66
C 5 (6.20/5.50)100 5 113 154.9
6.00 1 5.95 1 6.20 $18.55
b. I2009 5 (100) 5 110 Deflated 2009 wages: (100) 5 $8.74
5.45 1 5.60 1 5.50 212.2
6.00(150) 1 5.95(200) 1 6.20(120)
c. I2009 5 (100) 18.55
5.45(150) 1 5.60(200) 1 5.50(120) b. (100) 5 156.4; the percentange increase in
5 109 11.86
actual wages is 56.4%
9% increase over the two-year period 8.74
c. (100) 5 114.1; the change in real wages is an
4. I2009 5 114 7.66
increase of 14.1%
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1051

12. a. 2428, 2490, 2451 16. I 5 83

Manufacturing shipments increased slightly in con-
stant dollars
18. a. 151, 197, 143, 178
b. 3043, 3132, 3050
b. I 5 170
c. PPI
300(18.00) 1 400(4.90) 1 850(15.00) 20. IJan 5 73.5, IMar 5 70.1
14. I 5 (100)
350(18.00) 1 220(4.90) 1 730(15.00)
20,110 22. I 5 182.7
5 (100) 5 110
18,328
95(1200) 1 75(1800) 1 50(2000) 1 70(1500) 24. $36,082; $32,528; $27,913; $34,387; $40,551; $42,651;
15. I 5 (100)
120(1200) 1 86(1800) 1 35(2000) 1 60(1500) $46,458; $56,324
5 99
Quantities are down slightly 26. I 5 143; quantity is up 43%

Chapter 18
1. The following table shows the calculations for parts (a), (b), and (c):

Time Absolute Value Squared Absolute Value

Series Forecast of Forecast Forecast Percentage of Percentage
Week Value Forecast Error Error Error Error Error
1 18
2 13 18 25 5 25 238.46 38.46
3 16 13 3 3 9 18.75 18.75
4 11 16 25 5 25 245.45 45.45
5 17 11 6 6 36 35.29 35.29
6 14 17 23 3 9 221.43 21.43
Totals 22 104 251.30 159.38

22
a. MAE 5 5 4.4
5
104
b. MSE 5 5 20.8
5
159.38
c. MAPE 5 5 31.88
5
d. Forecast for week 7 is 14

2. The following table shows the calculations for parts (a), (b), and (c):

Time Absolute Value Squared Absolute Value

Series Forecast of Forecast Forecast Percentage of Percentage
Week Value Forecast Error Error Error Error Error
1 18
2 13 18.00 25.00 5.00 25.00 238.46 38.46
3 16 15.50 0.50 0.50 0.25 3.13 3.13
4 11 15.67 24.67 4.67 21.81 242.45 42.45
5 17 14.50 2.50 2.50 6.25 14.71 14.71
6 14 15.00 21.00 1.00 1.00 27.14 7.14
Totals 13.67 54.31 270.21 105.86
1052 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

13.67 d. The three-week moving average provides a better fore-

a. MAE 5 5 2.73
5 cast since it has a smaller MSE
54.31 e. Smoothing constant 5 .4
b. MSE 5 5 10.86
5
105.89 Time Squared
c. MAPE 5 5 21.18
5 Series Forecast Forecast
Week Value Forecast Error Error
d. Forecast for week 7 is
18 1 13 1 16 1 11 1 17 1 14 1 18
5 14.83 2 13 18.00 25.00 25.00
6 3 16 16.00 0.00 0.00
4 11 16.00 25.00 25.00
363 5 17 14.00 3.00 9.00
4. a. MSE 5 5 60.5
6 6 14 15.20 21.20 1.44
Forecast for month 8 is 15 Total 60.44
216.72 60.44
b. MSE 5 5 36.12 MSE 5 5 12.09
6 5
Forecast for month 8 is 18
c. The average of all the previous values is better because
MSE is smaller The exponential smoothing forecast using α 5 .4 provides a
better forecast than the exponential smoothing forecast us-
5. a. The data appear to follow a horizontal pattern ing α 5 .2 since it has a smaller MSE
b. Three-week moving average. 6. a. The data appear to follow a horizontal pattern
110
b. MSE 5 5 27.5
Time Squared 4
Series Forecast Forecast The forecast for week 8 is 19
Week Value Forecast Error Error 252.87
c. MSE 5 5 42.15
1 18 6
2 13 The forecast for week 7 is 19.12
3 16 d. The three-week moving average provides a better fore-
4 11 15.67 24.67 21.78 cast since it has a smaller MSE
5 17 13.33 3.67 13.44 e. MSE 5 39.79
6 14 14.67 20.67 0.44
The exponential smoothing forecast using α 5 .4 pro-
Total 35.67 vides a better forecast than the exponential smoothing
35.67 forecast using α 5 .2 since it has a smaller MSE
MSE 5 5 11.89
3
8. a.

(11 1 17 1 14) Week 4 5 6 7 8 9 10 11 12

The forecast for week 7 5 5 14 Forecast 19.33 21.33 19.83 17.83 18.33 18.33 20.33 20.33 17.83
3
c. Smoothing constant 5 .2
b. MSE 5 11.49
Prefer the unweighted moving average here; it has a
Time Squared
Series Forecast Forecast smaller MSE
Week Value Forecast Error Error c. You could always find a weighted moving average at
least as good as the unweighted one; actually the un-
1 18
2 13 18.00 25.00 25.00 weighted moving average is a special case of the
3 16 17.00 21.00 1.00 weighted ones where the weights are equal
4 11 16.80 25.80 33.64 10. b. The more recent data receive the greater weight or im-
5 17 15.64 1.36 1.85
portance in determining the forecast; the moving aver-
6 14 15.91 21.91 3.66
ages method weights the last n data values equally in
Total 65.15 determining the forecast
65.15
MSE 5 5 13.03 12. a. The data appear to follow a horizontal pattern
5
b. MSE(3-Month) 5 .12
MSE(4-Month) 5 .14
The forecast for week 7 is .2(14) 1 (1 2 .2)15.91 Use 3-Month moving averages
5 15.53 c. 9.63
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1053

13. a. The data appear to follow a horizontal pattern b. The methods discussed in this section are only
b. applicable for a time series that has a horizontal pat-
tern, so if there is a really a long-term linear trend in
the data, the methods discussed in this section are not
3-
Month appropriate
Time- Moving c. The time series plot for the data for years 2002–2008
Series Average α 5 .2 exhibits a horizontal pattern; it seems reasonable to con-
Month Value Forecast (Error)2 Forecast (Error)2 clude that the extreme values observed in 1997 and
1 240 2001 are more attributable to viewer interest in the per-
2 350 240.00 12100.00 formance of Tiger Woods; basing the forecast on years
3 230 262.00 1024.00
2002–2008 does seem reasonable, but, because of the
4 260 273.33 177.69 255.60 19.36
5 280 280.00 0.00 256.48 553.19 injury that Tiger Woods experienced in the 2008 reason,
6 320 256.67 4010.69 261.18 3459.79 if he is able to play in the 2009 Masters, then the rating
7 220 286.67 4444.89 272.95 2803.70 for 2009 may be significantly higher than suggested by
8 310 273.33 1344.69 262.36 2269.57 the data for years 2002–2008
9 240 283.33 1877.49 271.89 1016.97
10 310 256.67 2844.09 265.51 1979.36 17. a. The time series plot shows a linear trend
11 240 286.67 2178.09 274.41 1184.05 n n
12 230 263.33 1110.89 267.53 1408.50
ot
t51 15 oY
t51
t
55
Totals 17,988.52 27,818.49 b. t 5 5 53 Y5 5 5 11
n 5 n 5
MSE (3-Month) 5 17,988.52/9 5 1998.72
MSE (α 5 .2) 5 27,818.49/11 5 2528.95 o(t 2 t̄ )(Yt 2 Ȳ ) 5 21 o(t 2 t̄ )2 5 10
n

Based on the preceding MSE values, the 3-Month o(t 2 t)(Y 2 Y)

t51
t
21
moving averages appear better; however, exponential b1 5 n 5 5 2.1
10
smoothing was penalized by including month 2, which o(t 2 t)
t51
2

was difficult for any method to forecast; using only the

errors for months 4 to 12, the MSE for exponential b0 5 Ȳ 2 b1t̄ 5 11 2 (2.1)(3) 5 4.7
smoothing is
Tt 5 4.7 1 2.1t
MSE(α 5 .2) 5 14,694.49/9 5 1632.72
c. T6 5 4.7 1 2.1(6) 5 17.3
Thus, exponential smoothing was better considering
months 4 to 12 18. Forecast for week 6 is 21.16
c. Using exponential smoothing, 20. a. The time series plot exhibits a curvilinear trend
F13 5 αY12 1 (1 2 α)F12 b. Tt 5 107.857 2 28.9881t 1 2.65476t2
c. 45.86
5.20(230) 1 .80(267.53) 5 260
21. a. The time series plot shows a linear trend
14. a. The data appear to follow a horizontal pattern n n
b. Values for months 2–12 are as follows: o
t51
t
45 t51
oY t
108
105.00 114.00 115.80 112.56 105.79 110.05 b. t 5 5 55 Y5 5 5 12
n 9 n 9
120.54 126.38 118.46 106.92 104.85
o(t 2 t̄ )(Yt 2 Ȳ ) 5 87.4 o(t 2 t̄ )2 5 60
MSE 5 510.29
n
c. Values for months 2–12 are as follows:
o(t 2 t)(Y 2 Y)
t51
t
87.4
105.00 120.00 120.00 112.50 101.25 110.63 b1 5 n 5 5 1.4567
60
127.81 133.91 116.95 98.48 99.24 o(t 2 t)
t51
2

MSE 5 540.55
b0 5 Ȳ 2 b1t̄ 5 12 2 (1.4567)(5) 5 4.7165
Conclusion: A smoothing constant of .3 is better than a
smoothing constant of .5 since the MSE is less for 0.3 Tt 5 4.7165 1 1.4567t
16. a. The time series plot indicates a possible linear trend in c. T10 5 4.7165 1 1.4567(10) 5 19.28
the data; this could be due to decreasing viewer interest 22. a. The time series plot shows a downward linear trend
in watching the Masters, but closer inspection of the b. Tt 5 13.8 2 .7t
data indicates that the two highest ratings correspond to c. 8.2
years 1997 and 2001, years in which Tiger Woods won d. If SCF can continue to decrease the percentage
the tournament; the pattern observed may be simply due of funds spent on administrative and fund-raising by
to the effect Tiger Woods has on ratings and not neces- .7% per year, the forecast of expenses for 2015 is
sarily on any long-term decrease in viewer interest 4.70%
1054 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

24. a. The time series plot shows a linear trend b. A portion of the Minitab regression output follows:
b. Tt 5 7.5623 2 .07541t
c. 6.7328 The regression equation is
d. Given the uncertainty in global market conditions, Revenue = 70.0 + 10.0 Qtr1 + 105
making a prediction for December using only time is Qtr2 + 245 Qtr3
not recommended
26. a. A linear trend is not appropriate Quarter 1 forecast is 80
b. Tt 5 5.702 1 2.889t 2 1618t2 Quarter 2 forecast is 175
c. 17.90 Quarter 3 forecast is 315
28. a. The time series plot shows a horizontal pattern, but Quarter 4 forecast is 70
there is a seasonal pattern in the data; for instance, in c. A portion of the Minitab regression output follows
each year the lowest value occurs in quarter 2 and the
highest value occurs in quarter 4 The regression equation is
b. A portion of the Minitab regression output is shown; Revenue = -70.1 + 45.0 Qtr1 + 128
Qtr2 + 257 Qtr3 + 11.7 Period
The regression equation is
Value = 77.0 - 10.0 Qtr1 - 30.0
Quarter 1 forecast 5 is 221
Qtr2 - 20.0 Qtr3
Quarter 2 forecast 5 is 315
Quarter 3 forecast 5 is 456
c. The quarterly forecasts for next year are as follows: Quarter 4 forecast 5 is 211
Quarter 1 forecast 5 77.0 2 10.0(1) 2 30.0(0)
34. a. The time series plot shows seasonal and linear trend
2 20.0(0) 5 67
effects
Quarter 2 forecast 5 77.0 2 10.0(0) 2 30.0(1)
b. Note: Jan 5 1 if January, 0 otherwise; Feb 5 1 if
2 20.0(0) 5 47
February, 0 otherwise; and so on
Quarter 3 forecast 5 77.0 2 10.0(0) 2 30.0(0)
2 20.0(1) 5 57 A portion of the Minitab regression output follows:
Quarter 4 forecast 5 77.0 2 10.0(0) 2 30.0(0)
2 20.0(0) 5 77 The regression equation is
30. a. There appears to be a seasonal pattern in the data and Expense = 175 - 18.4 Jan - 3.72 Feb +
perhaps a moderate upward linear trend 12.7 Mar + 45.7 Apr + 57.1
b. A portion of the Minitab regression output follows: May + 135 Jun + 181 Jul + 105
Aug + 47.6 Sep + 50.6 Oct +
The regression equation is 35.3 Nov + 1.96 Period
Value = 2492 - 712 Qtr1 - 1512
Qtr2 + 327 Qtr3 c. Note: The next time period in the time series is
Period 5 37 (January of Year 4); the forecasts for Janu-
c. The quarterly forecasts for next year are as follows: ary–December are 229; 246; 264; 299; 312; 392; 440;
Quarter 1 forecast is 1780 366; 311; 316; 302; 269
Quarter 2 forecast is 980 35. a. The time series plot indicates a linear trend and a sea-
Quarter 3 forecast is 2819 sonal pattern
Quarter 4 forecast is 2492 b.
d. A portion of the Minitab regression output follows:
Time Four-Quarter Centered
The regression equation is Series Moving Moving
Value = 2307 - 642 Qtr1 - 1465 Year Quarter Value Average Average
Qtr2 + 350 Qtr3 + 23.1 t 1 1 4

2 2
The quarterly forecasts for next year are as follows:
3.50
Quarter 1 forecast is 2058 3 3 3.750
Quarter 2 forecast is 1258 4.00
Quarter 3 forecast is 3096 4 5 4.125
Quarter 4 forecast is 2769 4.25
2 1 6 4.500
32. a. The time series plot shows both a linear trend and sea- 4.75
sonal effects
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1055

Time Four-Quarter Centered Deseasonalized

Series Moving Moving Year Quarter Value
Year Quarter Value Average Average
2 1 4.979
2 3 5.000 2 4.021
5.25 3 5.834
3 5 5.375 4 5.877
5.50 3 1 5.809
4 7 5.875 2 8.043
6.25 3 7.001
3 1 7 6.375 4 6.717
6.50
2 6 6.625
6.75 b. Let Period 5 1 denote the time series value in Year 1—
3 6 Quarter 1; Period 5 2 denote the time series value in
Year 1—Quarter 2; and so on; a portion of the Minitab
4 8 regression output treating Period as the independent
variable and the Deseasonlized Values as the values of
the dependent variable follows:
c.
The regression equation is
Time Centered Seasonal- Deseasonalized Value = 2.42 + 0.422
Series Moving Irregular Period
Year Quarter Value Average Component
1 1 4
2 2 c. The quarterly deseasonalized trend forecasts for Year 4
3 3 3.750 0.800 (Periods 13, 14, 15, and 16) are as follows:
4 5 4.125 1.212 Forecast for quarter 1 is 7.906
2 1 6 4.500 1.333 Forecast for quarter 2 is 8.328
2 3 5.000 0.600 Forecast for quarter 3 is 8.750
3 5 5.375 0.930
Forecast for quarter 4 is 9.172
4 7 5.875 1.191
3 1 7 6.375 1.098 d. Adjusting the quarterly deseasonalized trend forecasts
2 6 6.625 0.906 provides the following quarterly estimates:
3 6 Forecast for quarter 1 is 9.527
4 8
Forecast for quarter 2 is 6.213
Forecast for quarter 3 is 7.499
Forecast for quarter 4 is 10.924
Seasonal- Adjusted 38. a. The time series plot shows a linear trend and seasonal
Irregular Seasonal Seasonal effects
Quarter Values Index Index
b. 0.71 0.78 0.83 0.97 1.02 1.30 1.50 1.23
1 1.333 1.098 1.216 1.205 0.98 0.99 0.93 0.79
2 0.600 0.906 0.752 0.746 c.
3 0.800 0.930 0.865 0.857
4 1.212 1.191 1.201 1.191
Deaseasonalized
Total 4.036
Month Expense
Adjustment for 4.000
5 5 0.991 1 239.44
seasonal index 4.036 2 230.77
3 246.99
4 237.11
36. a.
5 235.29
6 242.31
Deseasonalized 7 240.00
Year Quarter Value 8 235.77
9 244.90
1 1 3.320 10 242.42
2 2.681 11 247.31
3 3.501
4 4.198 (Continued)
1056 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

time period, and then decreases again. There may also

Deaseasonalized be some linear trend in the data
Month Expense b.
12 246.84
13 253.52 Time Period Adjusted Seasonal Index
14 262.82 12–4 A.M. 0.3256
15 259.04 4–8 A.M. 0.4476
16 252.58 8–12 noon 1.3622
17 259.80 12–4 P.M. 1.6959
18 253.85 4–8 P.M. 1.4578
19 266.67 8–12 midnight 0.7109
20 272.36
21 265.31
22 272.73 c. The following Minitab output shows the results of fitting
23 274.19 a linear trend equation to the deseasonalized time series:
24 278.48
25 274.65
26 269.23 The regression equation is
27 277.11 Deseasonalized Power = 63108 + 1854 t
28 288.66
29 284.31 Deaseasonalized Power (t 5 19) 5 63,108 1 1854(19)
30 300.00
31 280.00 5 98,334
32 268.29 Forecast for 12–4 P.M. 5 1.6959(98,334) 5 166,764.63 or
33 295.92 approximately 166,765 kWh
34 297.98
Deaseasonalized Power (t 5 20) 5 63,108 1 1854(20)
35 301.08
36 316.46 5 100,188
Forecast for 4–8 P.M. 5 1.4578(100,188) 5 146,054.07 or
approximately 146,054 kWh
d. Let Period 5 1 denote the time series value in January Thus, the forecast of power consumption from noon to
-Year 1; Period 5 2 denote the time series value in Feb-
8 P.M. is 166,765 1 146,054 5 312,819 kWh
ruary-Year 2; and so on; a portion of the Minitab re-
gression output treating Period as the independent 42. a. The time series plot indicates a horizontal pattern
variable and the Deseasonlized Values as the values of b. MSE(α 5 .2) 5 1.40
the dependent variable follows: MSE(α 5 .3) 5 1.27
MSE(α 5 .4) 5 1.23
The regression equation is A smoothing constant of α 5 .4 provides the best forecast
Deseasonalized Expense = 228 + 1.96 because it has a smaller MSE
Period c. 31.00
44. a. There appears to be an increasing trend in the data
e. b. A portion the Minitab regression output follows (Note:
t 5 1 corresponds to 2001, t 5 2 corresponds to 2002,
Month Monthly Forecast
and so on)
January 213.37
February 235.93 The regression equation is
March 252.69 Balance($) = 1984 + 146 t
April 297.21
May 314.53 The forecast for 2009 (t 5 9) is Balance($)
June 403.42
5 1984 1 146(9) 5 $3298
July 486.42
August 386.52 c. A portion of the Minitab regression output follows
September 309.88 (Note: t 5 1 corresponds to 2001, t 5 2 corresponds to
October 314.98 2002, and so on)
November 297.71
December 254.44 The regression equation is
Balance($) = 2924 - 419 t + 62.7 tsq
40. a. The time series plot indicates a seasonal effect; power
consumption is lowest in the time period 12–4 A.M., The forecast for 2009 (t 5 9) is Balance ($) 5 2924 2
steadily increases to the highest value in the 12–4 P.M. 419(9) 1 62.7(9)2 5 $4232
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1057

d. The quadratic trend equation provides the best forecast d. Hudson Marine experiences the largest seasonal in-
accuracy for the historical data crease in quarter 2; since this quarter occurs prior to the
e. Linear trend equation peak summer boating season, this result seems reason-
46. a. The forecast for July is 236.97 able, but the largest seasonal effect is the seasonal de-
Forecast for August, using forecast for July as the ac- crease in quarter 4; this is also reasonable because of
tual sales in July, is 236.97 decreased boating in the fall and winter
Exponential smoothing provides the same forecast for
every period in the future; this is why it is not usually
recommended for long-term forecasting Chapter 19
b. Using Minitab’s regression procedure we obtained the 1. n 5 27 cases with a value different than 150
linear trend equation Normal approximation µ 5 .5n 5 .5(27) 5 13.5
Tt 5 149.72 1 18.451t σ 5 Ï.25 n 5 Ï.25(27) 5 2.5981
Forecast for July is 278.88 With the number of plus signs 5 22 in the upper tail,
Forecast for August is 297.33 use continuity correction factor as follows
c. The proposed settlement is not fair since it does not ac-
1 2
21.5 2 13.5
count for the upward trend in sales; based upon trend P(x $ 21.5) 5 P z $ 5 P(z $ 3.08)
2.5981
projection, the settlement should be based on forecasted
lost sales of $278,880 in July and $297,330 in August p-value 5 (1.0000 2 .9990) 5 .0010
p-value # .01; reject H0; conclude population median . 150
48. a. The time series plot shows a linear trend
b. Tt 5 25 1 15t 2. Dropping the no preference, the binomial probabilities
The slope of 15 indicates that the average increase in for n 5 9 and p 5 .50 are as follows
sales is 15 pianos per year
c. 85, 100 x Probability x Probability
50. a. 0 0.0020 5 0.2461
1 0.0176 6 0.1641
2 0.0703 7 0.0703
Quarter Adjusted Seasonal Index
3 0.1641 8 0.0176
1 1.2717 4 0.2461 9 0.0020
2 0.6120
3 0.4978 Number of plus signs 5 7
4 1.6185
P(x $ 7) 5 P(7) 1 P(8) 1 P(9)
5 .0703 1 .0176 1 .0020
4 5 .0899
Note: Adjustment for seasonal index 5 5 1.0260
3.8985 Two-tailed p-value 5 2(.0899) 5 .1798
b. The largest effect is in quarter 4; this seems reasonable p-value . .05, do not reject H0; conclude no indication
since retail sales are generally higher during October, that a difference exists
November, and December
4. a. H0: Median $ 15
52. a. Yes, a linear trend pattern appears to be present Ha: Median , 15
b. A portion of the Minitiab regression output follows: b. n 5 9; number of plus signs 5 1
p-value 5 .0196
The regression equation is
Reject H0; bond mutual funds have lower median
Number Sold = 22.9 + 15.5 Year
6. n 5 48; z 5 1.88
c. Forecast in year 8 is or approximately 147 units p-value 5 .0301
Reject H0; conclude median . $56.2 thousand
54. b. The centered moving average values smooth out the
time series by removing seasonal effects and some of 8. a. n 5 15
the random variability; the centered moving average p-value 5 .0768
time series shows the trend in the data Do not reject H0; no significant difference for the pace
c. b. 25%, 68.8%; recommend larger sample
10. n 5 600; z = 2.41
Quarter Adjusted Seasonal Index p-value 5 .0160
1 0.899 Reject H0; significant difference, American Idol preferred
2 1.362
3 1.118 12. H0: Median for Additive 1 2 Median for Additive 2 5 0
4 0.621 Ha: Median for Additive 1 2 Median for Additive 2 Þ 0
1058 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

Reject H0; conclude significant difference; on-time %

Absolute Signed Ranks better in 2006
Difference Difference Rank Negative Positive
16. n 5 10; T1 5 12.5; z 5 21.48
2.07 2.07 9 9 p-value 5 .1388
1.79 1.79 7 7
20.54 0.54 3 23 Do not reject H0; conclude no difference between median
2.09 2.09 10 10 scores
0.01 0.01 1 1
0.97 0.97 4 4 18. H0: The two populations of additives are identical
21.04 1.04 5 25 Ha: The two populations of additives are not identical
3.57 3.57 12 12
1.84 1.84 8 8
3.08 3.08 11 11 Additive 1 Rank Additive 2 Rank
0.43 0.43 2 2 17.3 2 18.7 8.5
1.32 1.32 6 6 18.4 6 17.8 4
Sum of Positive Signed Ranks T1 5 70 19.1 10 21.3 15
16.7 1 21.0 14
18.2 5 22.1 16
n(n 1 1) 12(13) 18.6 7 18.7 8.5
µT 1 5 5 5 39 17.5 3 19.8 11
4 4

Î Î
20.7 13
n(n 1 1)(2n 1 1) 12(13)(25) 20.2 12
σT 1 5 5 5 12.7475
24 24 W 5 34

1 2
69.5 2 39
P(T 1 $ 70) 5 P z $ 5 P(z $ 2.39)
12.7475 1 1
p-value 5 2(1.0000 2 .9916) 5 .0168 µW 5 n (n 1 n2 1 1) 5 7(7 1 9 1 1) 5 59.5
2 1 1 2
p-value # .05, reject H0; conclude significant difference
between additives σW 5 Î 1
n n (n 1 n2 1 1) 5
12 1 2 1
Î 1
12
7(9)(7 1 9 1 1)

13. H0: Median time without Relaxant 2 Median time with 5 9.4472
Relaxant # 0 With W 5 34 in lower tail, use the continuity correction
Ha: Median time without Relaxant 2 Median time with
1 2
34.5 2 59.5
Relaxant . 0 P(W # 34) 5 P z # 5 P(z # 22.65)
9.4472
p-value 5 2(.0040) 5 .0080
Absolute Signed Ranks
Difference Difference Rank Negative Positive p-value , .05; reject H0; conclude additives are not
5 5 9 9
identical
2 2 3 3 Additive 2 tends to provide higher miles per gallon
10 10 10 10
23 3 6.5 26.5 19. a. H0: The two populations of salaries are identical
1 1 1 1 Ha: The two populations of salaries are not identical
2 2 3 3
22 2 3 23
3 3 6.5 6.5 Public Financial
3 3 6.5 6.5 Accountant Rank Planner Rank
3 3 6.5 6.5 50.2 5 49.0 2
Sum of Positive Signed Ranks T1 5 45.5 58.8 19 49.2 3
56.3 16 53.1 10
58.2 18 55.9 15
n(n 1 1) 10(11) 54.2 13 51.9 8.5
µT 1 5 5 5 27.5 55.0 14 53.6 11
4 4

Î Î
50.9 6 49.7 4
n(n 1 1)(2n 1 1) 10(11)(12) 59.5 20 53.9 12
σT 1 5 5 5 9.8107
24 24 57.0 17 51.8 7
51.9 8.5 48.9 1
1 2
45 2 27.5
P(T 1 $ 45.5) 5 P z $ 5 P(z $ 1.78) W 5 136.5
12.7475
1 1
p-value 5 (1.0000 2 .9925) 5 .0375 µW 5 n (n 1 n2 1 1) 5 10(10 1 10 1 1) 5 105
2 1 1 2

Î Î
p-value # .05; reject H0; conclude without the relaxant
1 1
has a greater median time σW 5 n n (n 1 n2 1 1) 5 10(10)(10 1 10 1 1)
12 1 2 1 12
14. n 5 11; T1 5 61; z 5 2.45 5 13.2288
p-value 5 .0142
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1059

412612 182
315(16) 1 5 24
With W 5 136.5 in upper tail, use the continuity 12
H5 11 2 3(16) 5 9.26
correction 5 5

1 2
136 2 105 χ 2 table with df 5 2, χ 2 5 9.26; the p-value is between
P(W $ 136.5) 5 P z $ 5 P(z $ 2.34)
13.2288 .005 and .01
p-value 5 2(1.0000 2 .9904) 5 .0192 p-value # .05 reject H0; conclude that the populations of
p-value # .05; reject H0; conclude populations are not calories burned are not identical
identical 30. H 5 8.03 with df 5 3
Public accountants tend to have higher salaries p-value is between .025 and .05
(55.0 1 56.3) Reject H0; conclude a difference between quality of courses
b. Public Accountant 5 $55.65 thousand
2
32. a. od 2i 5 52
(51.8 1 51.9)
Financial Planner 5 $51.85 thousand 6od 2i 6(52)
2 rs 5 1 2 512 5 .685
n(n2 2 1) 10(99)
20. a. $54,900, $40,400
b. W 5 69; z 5 2.04
p-value 5 .0414
b. σrs 5 Î 1
n21
5 Î 1
9
5 .3333
Reject H0; conclude a difference between salaries; men
rs 2 0 .685
higher z5 5 5 2.05
σrs .3333
22. W 5 157; z 5 2.74
p-value 5 2(1.0000 2 .9798) 5 .0404
p-value 5 .0062
Reject H0; conclude a difference between ratios; Japan p-value # .05 reject H0; conclude significant positive
tends to be higher rank correlation

24. W 5 116; z 5 2.22 34. od 2i 5 250

p-value 5 .8258 6od 2i 6(250)
Do not reject H0; conclude no evidence prices differ rs 5 1 2 512 5 2.136
n(n2 2 1) 11(120)
26. H0: All populations of product ratings are identical
Ha: Not all populations of product ratings are identical
σrs 5 În21
1
5
1
10
Î
5 .3162

r 20 2.136
z5 s 5 5 2.43
A B C σrs .3162
4 11 7 p-value 5 2(.3336) 5 .6672
8 14 2 p-value . .05 do not reject H0; we cannot conclude that
10 15 1
3 12 6 there is a significant relationship
9 13 5 36. rs 5 2.709, z 5 22.13
Sum of Ranks 34 65 21 p-value 5 .0332
Reject H0; conclude a significant negative rank correlation
342 652 212
315(16) 1 5 24
12 38. Number of plus signs 5 905, z 5 23.15
H5 1 1 2 3(16) 5 10.22
5 5 p-value less than .0020
χ 2 table with df 5 2, χ 2 5 10.22; the p-value is between Reject H0; conclude a significant difference between the
.005 and .01 preferences
p-value # .01; reject H0; conclude the populations of 40. n 5 12; T1 5 6; z 5 22.55
ratings are not identical p-value 5 .0108
28. H0: All populations of calories burned are identical Reject H0; conclude significant difference between prices
Ha: Not all populations calories burned are identical 42. W 5 70; z 5 22.93
p-value 5 .0034
Reject H0; conclude populations of weights are not identical
Swimming Tennis Cycling
44. H 5 12.61 with df 5 2
8 9 5
p-value is less than .005
4 14 1
11 13 3 Reject H0; conclude the populations of ratings are not
6 10 7 identical
12 15 2 46. rs 5 .757, z 5 2.83
Sum of Ranks 41 61 18 p-value 5 .0046
Reject H0; conclude a significant positive rank correlation
1060 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises

Chapter 20 22. a. UCL 5 .0817, LCL 5 2.0017 (use LCL 5 0)

2. a. 5.42
b. UCL 5 6.09, LCL 5 4.75 24. a. .03
4. R chart: b. β 5 .0802
UCL 5 R̄D4 5 1.6(1.864) 5 2.98
LCL 5 R̄D3 5 1.6(.136) 5 .22
x̄ chart:
UCL 5 x̄¯ 1 A2R̄ 5 28.5 1 .373(1.6) 5 29.10 Chapter 21
LCL 5 x̄¯ 2 A2R̄ 5 28.5 2 .373(1.6) 5 27.90
1. a. s1
6. 20.01, .082 250
8. a. .0470
d1 s2
b. UCL 5 .0989, LCL 5 20.0049 (use LCL 5 0) 2 100
c. p̄ 5 .08; in control
d. UCL 5 14.826, LCL 5 20.726 (use LCL 5 0) s3
Process is out of control if more than 14 defective 25
e. In control with 12 defective 1
f. np chart s1
100
n!
10. f (x) 5 p x(1 2 p)n2x d2 s2
x!(n 2 x)! 3 100
When p 5 .02, the probability of accepting the lot is
25! s3
f (0) 5 (.02)0(1 2 .02)25 5 .6035 75
0!(25 2 0)!
When p 5 .06, the probability of accepting the lot is b. EV(d1 ) 5 .65(250) 1 .15(100) 1 .20(25) 5 182.5
25! EV(d2 ) 5 .65(100) 1 .15(100) 1 .20(75) 5 95
f (0) 5 (.06)0(1 2 .06)25 5 .2129
0!(25 2 0)! The optimal decision is d1
12. p0 5 .02; producer’s risk 5 .0599
p0 5 .06; producer’s risk 5 .3396 2. a. d1; EV(d1 ) 5 11.3
Producer’s risk decreases as the acceptance number c is b. d4; EV(d4) 5 9.5
increased

14. n 5 20, c 5 3 3. a. EV(own staff) 5 .2(650) 1 .5(650) 1 .3(600)

5 635
16. a. 95.4 EV(outside vendor) 5 .2(900) 1 .5(600)
b. UCL 5 96.07, LCL 5 94.73 1 .3(300) 5 570
c. No EV(combination) 5 .2(800) 1 .5(650) 1 .3(500)
5 635
18. Optimal decision: hire an outside vendor with an ex-
R Chart x̄ Chart pected cost of $570,000
UCL 4.23 6.57 b. EVwPI 5 .2(650) 1 .5(600) 1 .3(300)
LCL 0 4.27 5 520
EVPI 5 ) 520 2 570 ) 5 50, or $50,000
Estimate of standard deviation 5 .86
4. b. Discount; EV 5 565
20. c. Full Price; EV 5 670
R Chart x̄ Chart
UCL .1121 3.112 6. c. Chardonnay only; EV 5 42.5
LCL 0 3.051 d. Both grapes; EV 5 46.4
e. Both grapes; EV 5 39.6
 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises 1061

8. a. 6: 1150 10: 2000 7: 2000

4: 1870 3: 2000 2: 1560
Profit 1: 1560
Payoff
s1 c. Cost would have to decrease by at least $130,000
100
d1
6
s2 300
12. b. d1, 1250
F c. 1700
3
s1
d. If N, d1
400 If U, d 2; 1666
d2
7 s2
200
Market 14.
Research 2
s1
100
d1 State
8 s2 of Nature P(sj ) P(I | sj ) P(I ù sj ) P(sj | I)
300
s1 .2 .10 .020 .1905
U s2 .5 .05 .025 .2381
4
1 s3 .3 .20 .060 .5714
s1
400 1.0 P(I) 5 .105 1.0000
d2
9 s2
200

16. a. .695, .215, .090

s1 .98, .02
100 .79, .21
d1
10 s2 .00, 1.00
300
c. If C, Expressway
No Market If O, Expressway
5
Research If R, Queen City
s1
400 26.6 minutes
d2
11 s2
200
18. a. The Technology Sector provides the maximum ex-
pected annual return of 16.97%. Using this recommen-
b. EV (node 6) 5 .57(100) 1 .43(300) 5 186
dation, the minimum annual return is 220.1% and the
EV (node 7) 5 .57(400) 1 .43(200) 5 314
maximum annual return is 93.1%
EV (node 8) 5 .18(100) 1 .82(300) 5 264
b. 15.20%; 1.77%
EV (node 9) 5 .18(400) 1 .82(200) 5 236
c. Because the Technology Sector mutual fund shows the
EV (node 10) 5 .40(100) 1 .60(300) 5 220
greater variation in annual return, it is considered to
EV (node 11) 5 .40(400) 1 .60(200) 5 280
have more risk
EV (node 3) 5 Max(186,314) 5 314 d2 d. This is a judgement recommendation and opinions may
EV (node 4) 5 Max(264,236) 5 264 d1 vary, but because the investor is described as being con-
EV (node 5) 5 Max(220,280) 5 280 d2 servative, we recommend the lower risk small-cap
stock mutual fund
EV (node 2) 5 .56(314) 1 .44(264) 5 292
EV (node 1) 5 Max(292,280) 5 292
20. a. Optimal Strategy:
[ Market Research
If Favorable, decision d2 Start the R&D project
If Unfavorable, decision d1 If it is successful, build the facility
Expected value 5 $10M million
10. a. 5000 2 200 2 2000 2 150 5 2650
b. At node 3, payoff for sell rights would have to be
3000 2 200 2 2000 2 150 5 650
$25million or more, in order to recover the $5million
b. Expected values at nodes R&D cost, the selling price would have to be $30mil-
8: 2350 5: 2350 9: 1100 lion or more