Chapter 1 1.2 Descriptive techniques summarize data. Inferential techniques draw inferences about a population based on sample data.
1.3 a The population is the 25,000 registered voters. b The sample is the 200 registered voters. c The 48% figure is the statistic
1.4 a The population is the complete production run. b The sample is comprised of the 1,000 chips. c The parameter is the proportion of defective chips in the production run. d The statistic is the proportion of defective chips in the sample. e The 10% figure refers to the parameter. f The 7.5% figure refers to the statistic. g We can estimate the population proportion is 7.5%. Statistical inference methods will allow us to determine whether we have enough statistical evidence to reject the claim.as the sample proportion.
1.5 Draw a random sample from the population of graduates who have majored in your subject and a random sample of graduates of other majors and record their highest salary offers.
1.6 a Flip the coin (say 100 times) and record the number of heads (assuming that you are interested in the number of heads). b The population is composed of the theoretical result of flipping the coin an infinite number of times and recording either “heads” or “tails”. c The sample is comprised of the “heads” and “tails” in the sample. d The parameter is the proportion of heads (again assuming that your interest is the number of heads rather than tails) in the population. e The statistic is the proportion of heads (or tails depending on the choice made in part d). f The sample statistic can be used to judge whether the coin is actually fair.
1.7 a We would conclude that the coin is not fair. b We may conclude that there is some evidence that the coin is not fair.
1.8 a The population is made up of the propane mileage of all the cars in the fleet. b The parameter is the mean propane mileage of all the cars in the fleet. c The sample is composed of the propane mileage of the 50 cars.
d The statistic is the mean propane mileage of the 50 cars in the sample. e We can use the sample statistic to estimate the population parameter.
Chapter 2 2.1 Nominal: Occupation, undergraduate major. Ordinal: Rating of university professor, Taste test ratings. Interval: age, income 2.2 a Interval b Interval c Nominal d Ordinal
2.3 a Interval b Nominal c Ordinal d Interval e Interval
2.4 a Nominal b Interval c Nominal d Interval e Ordinal
2.5 a Interval b Interval c Nominal d Interval e Nominal
2.6 a Interval b Interval c Nominal d Ordinal e Interval
2.7 a Interval b Nominal c. Nominal
5
d Interval e Interval f Ordinal 2.8 a Interval b Ordinal c Nominal d Ordinal 2.9 a Interval b Nominal c Nominal 2.10 a Ordinal b Ordinal c Ordinal
2.11 a Nominal b Interval c Ordinal
2.12a Nominal b Interval c Interval d Interval
2.13
6
350 000 000 000 300 000 000 000 250 000 000 000 200 000 000 000 150 000 000 000 100 000 000 000 50 000 000 000 0
2.14
Percentage Brazil; 1,0%
United States; 2,3% United Arab Emirates; 6,3%
Venezuela; 19,1%
Canada; 11,0%
China; 1,6%
Iran; 10,1% Iraq; 9,2% Saudi Arabia; 17,2% Kazakhstan; 1,9% Libya; 3,1% Russia; 6,6%
Kuwait; 6,7%
Nigeria; 2,4%
Qatar; 1,6%
2.15
7
2.16 Residual fuel oil 3%
Marketable coke 5% Still gas 5%
Liquified Lubricants refinery 1% gas 3%
Asphalt and road oil 2%
Jet fuel 13%
Gasoline 51%
Distillate fuel oil 15%
2.17
8
Other 2%
United States
United Kingdom
Spain
Thailand
South Korea
Singapore
Saudi Arabia
Russia
Japan
Mexico
Italy
Iran
Indonesia
India
France
Germany
China
Brazil
Canada
Australia
20 000 000 18 000 000 16 000 000 14 000 000 12 000 000 10 000 000 8 000 000 6 000 000 4 000 000 2 000 000 0
6 000 000 000 000 5 000 000 000 000 4 000 000 000 000 3 000 000 000 000 2 000 000 000 000 1 000 000 000 000
Australia Brazil Canada China European Union France Germany India Italy Japan Korea, South Mexico Russia Saudi Arabia South Africa Spain Taiwan Turkey United Kingdom United States
0
2.18 9000,0 7706,8
8000,0 7000,0
5424,5
6000,0 5000,0 4000,0 3000,0 1591,1
2000,0
1000,0
417,7 541,0
765,6
1556,7
1098,0 528,6 407,9
528,1 443,6
0,0
2.19
9
438,2 451,2 519,9
Steel production 900,0 800,0 700,0 600,0 500,0 400,0 300,0 200,0 100,0 0,0
2.20 1 200 000 000 000 1 000 000 000 000 800 000 000 000
600 000 000 000 400 000 000 000 200 000 000 000 0
2.21
10
Other; 17,5% Metal; 4,1% Glass; 5,1%
Organic; 45,8%
Plastic; 10,3%
Paper; 17,2%
2.22 3500 3000 2500 2000 1500 1000 500 0
2.23
11
2.25
12 Tangerines
Strawberries
Plums and Sloes
Plantains
Pineapples
Persimmons
Pears
Peaches & Nectarines
Papayas
Oranges
Mangoes
Lemons & Limes
Kiwi Fruit
Grapes
Grapefruits
Dates
Cherries
Bananas
Avocados
Apricots
Apples
60 000
50 000
40 000
30 000
20 000
10 000
0
2.24
120
100
80
60
40
20
0
Minimum wage 11,40 11,20 11,00 10,80 10,60 10,40 10,20 10,00 9,80 9,60
11,25 10,70 10,20
10,60
10,55
10,45
10,50
10,50
10,30
10,20
Percent Earning Minimum Wage 14,00% 12,00% 10,00% 8,00% 6,00% 4,00% 2,00% 0,00%
11,70%
5,60% 2,20%
4,90%
5,90%
7,00%
3,30%
2.26
13
5,60% 6,00% 5,90%
Number of students 8%
Community 5%
Career focus 16%
Location 39%
Academic reputation 10%
Majors 22%
2.27 Internet 8%
Word of mouth 12%
Consumer guide 52% Dealership 28%
2.28
14
Living/dining room 9%
Basement 32%
Kitchen 27%
Bathroom 23%
Bedroom 9%
The basement is the top choice followed by kitchen, bathroom, bedroom, and living/dining room.
2.29 a
Newspaper
Frequency
Relative Frequency
Daily News
141
39.2%
Post
128
35.6%
Times
32
8.9%
WSJ
59
16.4%
b
New York Times 9%
Wall Street Journal 16% New York Daily News 39%
New York Post 36%
The Daily News and the Post dominate the market
15
2.30a
Degree
Frequency
BA
88
BBA
37
B Eng
51
B Sc
24
Other
30
b. 100
90
88
80 70 60
51
50 37
40
30 30
24
20 10 0 BA
BBA
Beng
BSc
Other
c Other 13%
B.Sc. 11%
B.A. 38%
B.Eng 22% B.B.A. 16%
d. About 4 applicants in 10 have the BA degree, about one-fifth have a BEng. and one-sixth have a BBA.
16
2.31a 45 39
40 35
30 25 25
21
20 13
15 10
5 0 HP
Lenovo
Dell
Other
b
HP; 21 Other; 25
Lenovo; 13
Dell; 39
c Dell is most popular with 40% proportion, followed by other, 26%, HP, 21% and Lenovo, 13%.
2.32 a
Software
Frequency
Excel
34
Minitab
17
SAS
3
SPSS
4
Other
12
b
17
Other 17%
SPSS 6% SAS 4%
Excel 49%
Minitab 24%
c Excel is the choice of about half the sample, one-quarter have opted for Minitab, and a small fraction chose SAS and SPSS.
2.33
Natural Light 9%
Other 6%
Bud Light 31% Miller Lite 21%
Michelob Light 4%
Busch Light 7% Coors Light 22%
2.34
18
Do not know 20% Many share 41%
Some share 39%
2.35 No opinion 3%
Fair share 20%
Too much 15% Too little 62%
2.36 a
19
Republicans Favor Poor 2%
Middle clas 29%
Rich 69%
b
Democrats Favor
Rich 29%
Poor 35%
Middle clas 36%
According to the survey Republicans favor the rich and Democrats are split among the middle class, poor, and rich.
2.37 a Category
`
Frequency
Relative Frequency
Mom: Full time, Dad: Full time
403
46.0%
Mom: Part time, Dad: Full time
149
17.0%
Mom: Not employed, Dad: Full time
228
26.0%
Mom: Full time, Dad: Part time or not employed
53
6.0%
Mom: Not employed, Dad: Not employed
18
2.1%
Other
26
3.0% 20
b Mom FT, Dad PT/Not 6%
Mom Not, Dad Not 2%
Mom Not, Dad FT 26%
Other 3%
Mom FT, Dad FT 46%
Mom PT, Dad FT 17%
c 450 Mom FT, Dad FT; 403 400 350 300
Mom Not, Dad FT; 228
250
Mom PT, Dad FT; 149
200 150
Mom FT, Dad PT/Not; 53 Mom Not, Dad Other; 26 Not; 18
100 50 0 Mom FT, Dad Mom PT, Dad FT FT
Mom Not, Dad FT
Mom FT, Dad PT/Not
Mom Not, Dad Not
Other
d In most households Dad is working full time. There are very few households where neither Mom nor Dad are working.
2.38
21
No opinion 2%
Favor 45% Oppose 53%
A small majority oppose the Affordable Care Act.
2.39a Views on social issues
Frequency
Relative Frequency
Liberal
322
31.4%
Moderate
328
32.0%
Conservative
375
36.6%
b
The country is split among the three views on social issues with a small plurality of conservatives.
22
2.40 a Views on economic issues
Frequency
Relative Frequency
Liberal
208
20.3%
Moderate
354
34.5%
Conservative
463
45.2%
b
Liberal 20%
Conservative 45%
Moderate 35%
Economically the country is conservative.
2.41 80000 70000 60000 50000
40000 30000 20000 10000 0 Education
Less than high school
Datenreihen1
High school
Datenreihen2
Some college College graduate
Datenreihen3
23
Datenreihen4
There is decreasing numbers of Americans who did not finish high school and increasing numbers of those that go to college.
2.42 180 000 160 000 140 000 120 000 100 000 80 000 60 000 40 000 20 000 0
Year 1995
Year 2000
Year 2005
Year 2008
Spending is increasing in all seven areas.
2.43 350 300 250 200 150 100 50 0
In general crime was decreasing until 2014 when it started increasing.
2.44
24
60 50
40 B.A. 30
B.Eng B.B.A.
20
Other 10 0 University 1
University 2
University 3
University 4
Universities 1 and 2 are similar and quite dissimilar from universities 3 and 4, which also differ. The two nominal variables appear to be related.
2.45
The column proportions are similar; the two nominal variables appear to be unrelated. There does not appear to be any brand loyalty.
2.46
25
The two variables are related.
2.47 700 600 500 400 Men 300
Women
200 100 0 Lost job
Left job
Reentrants
New entrants
There are large differences between men and women in terms of the reason for unemployment.
2.48
26
200 180 160 140 120 100 80 60 40 20 0
Year 1995 Year 2000
Year 2005 Year 2010
The number of prescriptions filled by all stores except independent drug stores has increased substantially.
2.49 40% 35%
30% 25%
20% 15%
Male
10%
Female
5% 0%
There appears to be differences between female and male students in their choice of light beer.
2.50
27
120 100
98 83
80
70 68
64
59 60 46
40
51
50
39
40 25
23
13
20 6 0 C. conservative M conservative Many share
Mixed
M liberal
Some share Don't know
C liberal
3
There are differences among the five groups.
2.51 300 259 236
250 187
200
Fair share 150 100
Too much 122
toolittle 81
No opnion
70 34 41
39
50 18
7
7
0 Conservative
Moderate
Liberal
All three groups say that upper-income people pay too little. However Conservatives are more likely to say fair share than Moderates or Liberals
2.52
28
600 481
500 401 400 300 200 96
94
100 10
12
Democrat
Republican
0
Favor
Oppose
No opinion
Democrats support and Republicans oppose the Affordable Care Act.
2.53 250
200
226
173 139
150 114 100
50
108
99
90
41
35
0 Liberal
Moderate Democrat
Independent
Conservative Republican
No surprise-on social issues Democrats are liberal and Republicans are conservative.
2.54
29
300 264 250 200 159 150
125
133 113
100 66
82
69
50 14 0 Democrat
Independent Liberal
Moderate
Republican Conservative
On economic issues Republicans are very conservative whereas Democrats and Moderates are mixed.
2.55 7 000,0 6 000,0
6 542,6 5 699,4
5 000,0 4 000,0 3 000,0
2 897,7
2 648,2
2 000,0 1 000,0
863,6
556,5
0,0 U.S. U.S. Social U.S. Federal U.S. Civil U.S. Military Individuals Security Trust Reserve Service Retirement and Fund Retirement Fund Institutions Fund
2.56
30
Foreign Nations
1 400,0
1 254,8
1 200,0
1 149,2
1 000,0 800,0 600,0 322,0
400,0
291,4
255,0
232,9
4 214
3 895
225,6
210,6
197,0
188,2
200,0 0,0
2.57 16 000
14 732
14 000 12 000
10 043
10 000 8 000 6 000
7 013
4 000
2 397 1 225
2 000 0
2.58
31
1 648
40 35 30 25 20 15 10 5 0 Married 0 children
Food
Married Couple w children
Housing
One Parent, At Least 1 < 18
Transportation
Healthcare
Insurance & pensions
Other
The pattern is about the same for the three households.
2.59
Don't know/refused 4%
Other reasons 12%
Unemployed/wor k doesn't offer/not eligible at work 11%
Too expensive 47%
Told they were ineligible 7% Immigration status Don't know 7% Don't need it how to get it 6% 3%
Opposed to the ECA/prefer to pay penalty 3%
2.60
32
0,250
0,200 0,150 0,100 0,050 0,000
2013 Uninsured Rate
2014 Uninsured Rate
There are decreases in almost every state. However, there are many Americans without health insurance.
2.61 Strongly agree 4%
Strongly disagree 15%
Agree 23%
Disagree 20%
Neither agree nor disagree 38%
More students disagree than agree.
2.62
33
Very good 10%
Excellent 3%
Poor 15%
Fair 27%
Good 45%
More than 40% rate the food as less than good.
2.63
Manual 18%
Computer 44%
Computer and manual 38%
34
2.64 45% 40%
35% 30% 25% Children
20%
No children 15% 10% 5% 0% Poor
Fair
Good
Very good Excellent
Customers with children rated the restaurant more highly than did customers with no children.
2.65 45 40 35 30 25 20 15 10 5 0
Female Male
Males and females differ in their areas of employment. Females tend to choose accounting marketing/sales and males opt for finance.
b
35
40 35 30 25 20 15 10 5 0
Very satisfied Quite satisfied Little satisfied
Not satisfied
Area and job satisfaction are related. Graduates who work in finance and general management appear to be more satisfied than those in accounting, marketing/sales, and others.
2.66
Males 45% Females 55%
The survey oversampled women slightly.
2.67
36
Others 10% Blacks 15%
Whites 75%
2.68a
Married Widowed Divorced Separated Never married
1158 209 411 81 675
b. Pie chart c.
Never married 27%
Married 46%
Separated 3% Divorced 16%
Widowed 8%
37
2.69 Cpmpleted graduate degree 11% Left high school 13% Completed Bachelor's degree 19%
Completed junior college 7%
Graduated high school 50%
2.70 800 700 600 Left high school
500
High schoo; 400
Junior college
300
Bachelor's degree Graduate
200 100 0 Male
Female
The patterns are similar.
2.71
38
Government 19%
Private sector 81%
2.72 1600
1467
1400 1200 1000 800 600 400
White; 340
273 199
200
Black; 94
Other; 34
0 White
Black
Other
The patterns are similar.
2.73
39
1400 1196 1200 1000
949
800 Self-employed 600
Work for someone else
400 200
0 Male
Female
Males are slightly more likely to be self-employed than females.
2.74 Category 1 10% Category 5 30%
Category 2 13%
Category 3 14%
Category 4 33%
The ”married” categories (4 and 5) make up more than 60% of the households. 2.75
40
2500
2000
No high school
1500
High school Some college
1000
College degree 500
0 Male
Female
There are large differences between male and female heads of households.
2.76 Other 5%
Hispanic 9% Black 12%
White 74%
Whites make up three quarters of the survey.
2.77
41
1800 1600 1400 1200 1000 800 600 400 200 0 White
Black
Hispanic 1
2
Other 3
4
5
There are large differences between the four races in terms of family structure.
2.78 College degree; 2227
2500
2000
1500 High school; 953
1000
646
Some college; 567 463
High school
Some college
No high school; 500 252 294
613
0 No high school
Own
Otherwise
College degree holders are much more likely to own their homes.
42
College degree
Chapter 3 3.1 7 to 9
3.2 11 to 13
3.3 a 9 to 10 (or 11) b Upper limits: 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250
3.4 a 5 (or 6) b Upper limits: 5.2, 5.4, 5.6, 5.8, 6.0, 6.2
Frequency
3.5 14 12 10 8 6 4 2 0 5
8
11 Calls
14
17
3.6
Frequency
20 15 10 5 0 24
28
32
36
40
Ages
3.7
5
44
12 Frequency
10 8 6 4 2 0 70
80
90
100
110
20
25
Balls
3.8
Frequency
15 10 5 0 5
10
15 Calls
The number of calls is bimodal.
3.9
Frequency
30 20 10 0 31
32
33
34
35
36
37
38
39 More
Times
3.10
6
120
Frequency
25 20 15 10 5 0
1
2.5
4
5.5
7
8.5
10
More
Stores
b. The number of stores is bimodal and positively skewed.
3.11
Frequency
20 15 10 5
0 8
12
16
20
24
28
Games
The histogram is positively skewed.
3.12
Frequency
20 15 10
5 0 5
10
15
20
25
30
35
40
Rounds
The histogram is symmetric (approximately) and bimodal.
3.13
7
45
Frequency
80 60 40 20 0
The histogram is symmetric, unimodal, and bell shaped.
Frequency
3.14 70 60 50 40 30 20 10 0 2
4
6
8
10 12 Days
14
16
18
50
60
65
70
Most orders arrive within 12 days.
3.15a.
Frequency
80 60 40
20 0 25
30
35
40
45
55
Customers
b. The histogram is unimodal and positively skewed.
3.16 a.
8
More
25 Frequency
20 15 10 5 0 50
100
150
200
250 300 350 Downloads
400
450 More
b. The histogram is somewhat bell shaped.
Frequency
3.17 100 80 60 40 20 0 37
44
51
58
65
72
79
86
93 100
Marks
The histogram is negatively skewed, bimodal, and not bell shaped.
3.18
Frequency
60
40 20 0 18
19
20
21
22
23
24
Lengths
The histogram is unimodal, bell-shaped and roughly symmetric. Most of the lengths lie between 18 and 23 inches.
3.19
9
Frequency
30
20 10 0 200
400
600
800
1000
1200
1400
Copies
The histogram is unimodal and positively skewed. On most days the number of copies made is between 200 and 1000. On a small percentage of days more than 1000 copies are made.
Frequency
3.20 100 80 60 40 20 0
1
2
3
4
5
6
7
8
Weights
The histogram is unimodal, symmetric and bell-shaped. Most tomatoes weigh between 2 and 7 ounces with a small fraction weighing less than 2 ounces or more than 7 ounces.
3.21
Frequency
150 100 50 0 15
20
25
30
35
40
45
50
55
Gallons
The histogram is positively skewed and unimodal. Most households use between 20 and 45 gallons per day. The center of the distributions appears to be around 25 to 30 gallons. 3.22
10
Frequency
30
20 10 0 12000 15000 18000 21000 24000 27000 30000 Books
The histogram of the number of books shipped daily is negatively skewed. It appears that there is a maximum number that the company can ship.
3.23 600 Frequency
500 400 300 200 100 0 20
30
40
50
60
70
80
90
AGE The histogram is slightly positively skewed with few respondents under 20 or over 80.
3.24 1000 Frequency
800 600 400 200 0 2
4
6
8
10 12 EDUC
14
16
18
20
The histogram is bimodal. The modes represent high school completion (12 years) and university completion (16 years).
11
3.25
Frequency
500 400
300 200 100 0
RINCOME As expected the histogram is positively skewed.
3.26 1000 Frequency
800 600 400 200 0 2
4
6
8
10
12 14 16 TVHOURS
18
20
22
24
Many respondents watched less than 2 hours per day and almost all watched for less than 6 hours.
3.27 600 Frequency
500 400 300 200 100 0 16
20
24
28
32 36 40 44 AGEKDBRN
48
52
Most respondents had their first child born before the age of 24.
12
56
60
3.28
Frequency
1500 1000 500 0 20
30
40
50
60 AGE
70
80
90
100
The histogram is bell shaped with the 50-60 interval was the modal class.
3.29 5000 Frequency
4000 3000 2000 1000 0 1
2
3
4
5
6
7
8
KIDS The maximum number in the sample is 8.
3.30 a. Almost all the values are in the first interval. b. Most of the value are greater than 200,000.
2500 2000 1500 1000 500 0 100000 200000 300000 400000 500000 600000 700000 800000 900000 1000000 1100000 1200000 1300000 1400000 1500000 1600000 1700000 1800000 1900000 2000000 More
Frequency
c.
ASSET
13
This last histogram is best because it allows us to see the assets below $2,000,000.
3.31
Frequency
250 200 150 100 50 0
CDS We ignored the responses with 0 (most respondents reported no CDs.)
3.32 40 35 30 25 20 15 10 5 0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 Canada
United States
Both countries are winning an increasing number of medals.
3.33a.
14
0,920 0,910 0,900 0,890 0,880 0,870 0,860 0,850 1 2 3 4 5 6 7 8 9 1011121314151617181920212223242526272829303132
b. 4,50 4,00 3,50
3,00 2,50 2,00 1,50 1,00 0,50 0,00 1 2 3 4 5 6 7 8 9 10 11 1213 1415 1617 18 1920 2122 2324 25 2627 2829 3031 32
The save percentage is increasing and goals against are decreasing.
3.34a.
15
2 500 000
2 000 000
1 500 000
1 000 000
500 000
0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
b. 2 500 000
2 000 000
1 500 000
1 000 000
500 000
0
Both types of crime have been decreasing.
3.35a.
16
Health Care Total ($billions) 1600,000 1400,000 1200,000 1000,000 800,000 600,000 400,000 200,000 0,000 1 3 5 7 9 11131517192123252729313335373941434547495153555759616365
b.
Per capita 5 000 4 500 4 000 3 500 3 000 2 500 2 000 1 500 1 000 500 1 3 5 7 9 1113151719212325272931333537394143454749515355575961636567
c.
17
Per Capita Constant $ 2 000 1 800 1 600 1 400 1 200 1 000 800 600 400 200 1 3 5 7 9 1113151719212325272931333537394143454749515355575961636567
3.36a.
Part A (Hospital Insurance) 250
200
150
100
50
0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
b.
18
Per Capita 700 600 500 400 300 200 100 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
c.
Per Capita Constant $ 300 250 200 150 100 50 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
3.37a.
19
Part B (Suppl. Med. Ins.) -$billions 160 140 120 100 80 60 40 20 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
b.
Per Capita 450 400 350 300 250 200 150 100 50 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
c.
20
Per Capita Constant $ 200 180 160 140 120 100 80 60 40 20 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
3.38a.
Medicaid ($billions) 500,00 450,00 400,00 350,00 300,00 250,00 200,00 150,00 100,00 50,00 0,00 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53
b.
21
Per Capita 1600 1400 1200 1000 800 600 400 200 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53
c.
Per Capita Constant $ 700 600 500 400 300 200 100 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53
3.39a.
22
Disability Ins (DI) -fed 160,00 140,00 120,00 100,00 80,00 60,00 40,00 20,00 0,00 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57
b.
Per Capita 500 450 400 350 300 250 200 150 100 50 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57
c.
23
Per Capita Constant $ 250
200
150
100
50
0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57
3.40a.
Old Age Survivor Ins (OASI)-fed $ billion 800 700 600 500 400 300 200 100 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57
b.
24
Per Capita 2500
2000
1500
1000
500
0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57
c.
Per Capita Constant $ 1200 1000 800 600 400 200 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57
3.41a.
25
Defense ($billions) 1000 900 800 700 600 500 400 300 200 100 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
b.
Per Capita 3000 2500 2000 1500 1000 500 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
c.
26
Per Capita Constant $ 4000 3500 3000 2500 2000 1500 1000 500 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
3.42a.
Total spending ($billions) 7000 6000 5000 4000 3000 2000
1000 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
b.
27
Per Capita 25000
20000
15000
10000
5000
0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
c.
Per Capita Constant $ 10000 9000 8000 7000 6000 5000 4000 3000 2000 1000 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
3.43a.
28
Welfare ($Billions) 600 500 400 300 200 100 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51
b.
Per Capita 1800 1600 1400 1200 1000 800 600 400 200 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51
c.
29
Per Capita Constant $ 800 700 600 500 400 300 200 100 0 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
3.44a.
Education ($billions) 1200 1000 800 600 400 200 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
b.
30
Per Capita 3500 3000 2500 2000 1500 1000 500 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
c.
Per Capita Constant $ 1600 1400 1200 1000 800 600 400 200 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
3.45a.
31
Interest Payments ($billions) 400 350 300 250 200 150 100 50 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
b.
Per Capita 1400 1200 1000 800 600 400 200 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
c.
32
Per Capita Constant $ 800 700 600 500 400 300 200 100 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
3.46b.
Violent Crime Per 100,000 800,0 700,0 600,0 500,0 400,0 300,0 200,0 100,0 0,0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
c.
33
Property Crime per 100,000 5 000 4 500 4 000 3 500 3 000 2 500 2 000 1 500 1 000 500 0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
d. In the period 1993 to 2012 crime was decreasing no matter how it is measured.
3.47 25 000
20 000
15 000
10 000
5 000
0 1
3
5
7
9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 Consumption
Production
The gap is closing.
3.48a.
34
GDP ($billions) 20 000,0 18 000,0 16 000,0 14 000,0 12 000,0 10 000,0 8 000,0 6 000,0 4 000,0 2 000,0 0,0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
b.
GDP Constant $billions 8000 7000 6000 5000 4000 3000 2000 1000 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
c. GDP has been steadily increasing.
3.49a.
35
1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
U.S. Exports to Canada
30000
25000
20000
15000
10000
5000
0
b.
U.S. Imports from Canada
35000
30000
25000
20000
15000
10000
5000
0
c.
36
Trade Balamce 2000
1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
0 -2000 -4000 -6000 -8000 -10000 -12000
d. In recent years the balance has become equal.
3.50a.
U.S. Exports to Japan 7000 6000 5000 4000 3000 2000 1000
1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
0
b.
37
U.S. Imports from Japan 16000 14000 12000 10000
8000 6000 4000 2000 1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
0
c.
Trade Balance -1000
1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
0
-2000 -3000 -4000 -5000 -6000 -7000 -8000 -9000
d. The U.S. imports more from Japan than Japan imports from the U.S.
3.51a.
38
1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
U.S. Exports to China
14000
12000
10000
8000
6000
4000
2000
0
b.
U.S. Imports from China
50000
45000
40000
35000
30000
25000
20000
15000
10000
5000
0
c.
39
Trade Balamce 5000
-5000
1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
0
-10000 -15000 -20000
-25000 -30000 -35000 -40000
d. The U.S. imports far more from China than China imports from the U.S.
3.52
Canadian Dollars to One U.S. Dollar 1,8000 1,6000 1,4000 1,2000 1,0000 0,8000 0,6000 0,4000 0,2000 1 19 37 55 73 91 109 127 145 163 181 199 217 235 253 271 289 307 325 343 361 379 397 415 433 451 469 487 505 523 541
0,0000
The recent trend reveals that the Canadian dollar has been losing value relative to the U.S. dollar.
3.53
40
Japanese Yen to One U.S. Dollar 400,00 350,00 300,00 250,00
200,00 150,00 100,00 50,00 1 19 37 55 73 91 109 127 145 163 181 199 217 235 253 271 289 307 325 343 361 379 397 415 433 451 469 487 505 523 541
0,00
The value of the Japanese yen has been increasing relative to the U.S. dollar.
3.54
DJ Industrial Average Close 20 000,00
18 000,00 16 000,00 14 000,00 12 000,00 10 000,00 8 000,00 6 000,00 4 000,00 2 000,00 1 29 57 85 113 141 169 197 225 253 281 309 337 365 393 421 449 477 505 533 561 589 617 645 673 701 729 757 785
0,00
3.55
41
DJA Constant $ 9000 8000 7000 6000 5000 4000 3000 2000 1000 1 27 53 79 105 131 157 183 209 235 261 287 313 339 365 391 417 443 469 495 521 547 573 599 625 651 677 703 729 755 781
0
The stock index has increased whether measured in actual or constant dollars.
3.56 450 400 350 Savings
300 250 200 150 100 50 0 0,0
1,0
2,0
3,0
4,0 Time
5,0
There is a weak positive linear relationship.
3.57a
42
6,0
7,0
8,0
40 30
Returns
20 10 0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
80
90
100
-10 -20
Inflation
b. There is very weak linear relationship.
3.58a 90 85
Statistics
80 75 70 65 60 40
50
60
70 Calculus
b. There is a positive linear relationship between calculus and statistics marks.
3.59
43
2500
2000
Cost
1500 1000 500 0 0
5
10
15
20
25
Speed
There is a strong positive linear relationship.
3.60 25
Publications
20 15
10 5 0 0
10
20
30
40
50
Age
There is a negative linear relationship.
3.61
44
60
70
80
100 95
90
Mark
85 80 75 70 65 60 55 50 60
65
70
75
80
85
90
95
Time
There is no linear relationship
3.62a 1600
1400
Electricity
1200 1000 800 600
400 200 0 0
1
2
3
4
5
6
Occupants
b. There is a moderately strong positive linear relationship.
3.63
45
7
8
9
700 600 500
Tickets
400 300 200 100 0
-40
-30
-20
-10
0
10
20
30
40
Temperature
There is a moderately strong linear relationship. More lift tickets are sold during warmer days.
3.64a 90 80
70 Income
60 50
40 30 20
10 0 60
65
70
75
Height
b. There is a very weak positive linear relationship.
3.65
46
80
2000000
Compensation
1500000
1000000
500000
0 0
5000
10000
-500000
15000
20000
25000
30000
20000
25000
30000
Profit
500 450
Compensation
400 350 300 250 200 150 100 50 0 0
5000
10000
15000
3-Yr share return
There does not appear to be a linear relationship between compensation of profit and between compensation and 3year share return.
3.66
47
600 500
Tenure
400 300 200 100 0 0
20
40
60
80
100
Age
There is a moderately strong positive linear relationship.
3.67 3000 2500
Sales
2000 1500 1000 500 0 0
5
10
15
20 Time
There is no linear relationship.
3.68
48
25
30
35
400
Livestock sub-index
350 300 250 200 150
100 50 0 0
100
200
300
400
500
600
8
10
12
Grains sub-index
There is moderately strong positive linear relationship.
3.69
Bank prime loan rate
25
20 15 10 5 0 0
2
4
6 Unemployment rate
There is a weak positive linear relationship.
3.70
49
6000 5000
Price
4000 3000
2000 1000 0 0
10
20
30
40
50
60
70
Grade
There is a strong nonlinear positive relationship.
3.71
222 220 218 216 214 212 210 208
0
20
40
60
80
There is a positive linear relationship.
3.72
50
100
120
25
SPEDUC
20
15
10
5
0 0
5
10
15
20
25
15
20
25
EDUC There is a positive linear relationship.
3.73 25
EDUC
20
15
10
5
0 0
5
10
PAEDUC
There is a weak positive linear relationship.
3.74
51
25
EDUC
20
15
10
5
0 0
5
10
15
20
25
MAEDUC
There is a weak positive linear relationship.
3.75 100 90 80 70 60 50 40 30 20 10 0 0
10
20
30
40
50
60
70
There is no relationship.
3.76
52
80
90
100
30
25 20 15
10 5 0 0
10
20
30
40
50
60
70
80
90
100
There is no relationship.
3.77 30 25 20 15 10 5 0 0
5
10
15
20
There is a weak positive linear relationship.
3.78
53
25
60
50 40 30
20 10 0 0
5
10
15
20
25
There is a weak positive linear relationship.
3.79 1 200 000,00 1 000 000,00 800 000,00 600 000,00 400 000,00 200 000,00 0,00 0
10
20
30
40
50
60
There is a negative linear relationship.
3.80
54
70
80
90
100
18 16 14 12 10 8 6 4 2 0 0
10
20
30
40
50
60
70
80
90
100
80
90
100
-2
There is a negative linear relationship.
3.81 900 000,00 800 000,00 700 000,00 600 000,00 500 000,00 400 000,00 300 000,00 200 000,00 100 000,00 0,00 0
10
20
30
40
50
60
70
There is a negative linear relationship.
3.82 a We convert the numbers to accident rate and fatal accident rate.
55
1 2 3 4 5 6 7 8 9
A Age group Under 20 20-24 25-34 35-44 45-54 55-64 65-74 0ver 74
B Accident rate per driver 0.373 0.173 0.209 0.162 0.133 0.108 0.095 0.093
C Fatal accidient rate (per 1,000 drivers) 0.643 0.352 0.305 0.251 0.214 0.208 0.177 0.304
b 0.7
0.6 0.5
0.4
Accidents/driver
0.3 Fatal accidents/1000 drivers
0.2
0.1 0
Under 20-24 25-34 35-44 45-54 55-64 65-74 0ver 20 74
c. The accident rate generally decreases as the ages increase. The fatal accident rate decreases until the over 64 age category where there is an increase.
3.83
56
2.500 2.000 1.500 1.000
Injury rate Fatal injury rate per 100
0.500 0.000
c Older drivers who are in accidents are more likely to be killed or injured. d Exercise 3.9 addressed the issue of accident rates, whereas in this exercise we consider the severity of the accidents.
3.84a 600 550 500 450 400
Verbal All
350
Math All
300 250 200 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
57
b 550 540
530 520
Verbal All Math All
510
500 490 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
c 560 540 520 500 480
Verbal Male
460
Verbal Female
440 420 400 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
58
600 500
400 300
Math Male Math Female
200
100 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
d 550
540
530
520
Verbal Male Verbal Female
510
500
490 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
59
550 540 530 520 510 Math Male 500
Math Female
490 480 470 460 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
3.85a 7.8 7.6
7.4 7.2 7.0 6.8 6.6 6.4 6.2 6.0 1
2
3
4
5
6
7
8
b
60
9
10
11
12
8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 1
2
3
4
5
6
7
8
9
10
11
12
c Caption a: Unemployment rate falling rapidly. Caption b: Unemployment rate virtually unchanged. d The chart in (a) is more honest.
3.86 40%
Return on sub-index
30% 20% 10% 0% -10%
-20% -30% -40% -1.00%
-0.50%
0.00%
0.50%
1.00%
1.50%
2.00%
Inflation rate
There is no linear relationship between the inflation rate and the return on the precious metals sub-index.
61
3.87 1,6 1,4 1,2
1 0,8 0,6 0,4 0,2
1 19 37 55 73 91 109 127 145 163 181 199 217 235 253 271 289 307 325 343 361 379 397 415 433 451 469 487 505 523 541
0
There has been a long-term decline in the value of the Australian dollar.
3.88 160
140
IQ Twin 2
120 100 80 60
40 20 0 0
20
40
60
80
100
IQ Twin 1
There is a strong positive linear relationship.
62
120
140
3.89 180 160
Currencies index
140 120 100
80 60 40
20 0 60
70
80
90
100
110
120
130
Interest rate index
There is a very strong positive linear relationship.
3.90
Frequency
80 60 40
20 0 2
4
6
8 10 12 Spam Emails
14
The histogram is symmetric and bell shaped.
3.91a
63
16
18
140
74 72
Son's height
70 68 66 64 62 60 62
64
66
68
70
72
74
76
Father's height
b. The slope is positive c. There is a moderately strong linear relationship.
3.92 12 10 8 6 4 2 0 0,00 2 000,004 000,006 000,008 000,0010 000,00 12 000,00 14 000,00 16 000,00 18 000,00 20 000,00
There is no linear relationship between the Dow Jones Industrial average and the unemployment rate.
3.93
64
3 2,5 2
1,5 1 0,5
1 19 37 55 73 91 109 127 145 163 181 199 217 235 253 271 289 307 325 343 361 379 397 415 433 451 469 487 505 523 541
0
The value of the British pound has fluctuated quite a bit has stayed at current exchange rate for many months.
3.94 350 330
310 290 Time
270 250 230 210 190 170 150 300
320
340
360
380
400
420
Score
There is a strong positive linear relationship. Poorer players take longer to complete their rounds.
3.95a.
65
1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
U.S. Exports to Mexico
25 000,0
20 000,0
15 000,0
10 000,0
5 000,0
0,0
b.
U.S. Imports from Mexico
30 000,0
25 000,0
20 000,0
15 000,0
10 000,0
5 000,0
0,0
c.
66
Trade Balance 2 000,0 1 000,0
-1 000,0
1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 339 352 365
0,0
-2 000,0 -3 000,0 -4 000,0 -5 000,0 -6 000,0 -7 000,0 -8 000,0
Mexico exports more to the U.S. than the U.S. exports to Mexico.
3.96 1400 1200
Credit
1000 800 600 400 200 0 0
100
200
300 Debit
There is a moderately strong negative linear relationship.
3.97
67
400
500
Frequency
40 30 20 10
0 25000
40000
55000
70000
85000 100000
Pay
Frequency
3.98 50 40 30 20 10 0
7
10
13
16
19
22
25
Meetings
The histogram of the number of meetings is positively skewed.
3.99 45 40 35 30 25 20 15 10 5 0 1 3 5 7 9 1113 1517 1921 2325 2729 3133 3537 3941 4345 4749 51 535557 59 6163 65
The number of fatal accidents has been steadily decreasing.
3.100 68
Frequency
200 150 100
50 0 10
20
30
40
50
60
70
80
90 100 110
Miles
The histogram is positively skewed and bimodal.
3.101
Frequency
200 150 100
50 0 30
45
60
75
90 105 120 135 150 165 Times
The histogram tells us that about 70% of gallery visitors stay for 60 minutes or less and most of the remainder leave within 120 minutes. Although there are other plans, the gallery director proposed the following plan. Admit 200 visitors every hour. We expect that about 140 will leave within 1 hour and about 60 will stay for an additional hour. During the next 1-hour period, 200 new visitors will be admitted. If 60 of the previous hour’s admittances remain, there will be a total of 260 people in the gallery. If this pattern persists during the day, there will be a maximum of 260 visitors at any time. This plan should permit as many people as possible to see the exhibit and yet maintain comfort and safety.
3.102 Business Statistics course (Example 3.3)
69
100 90
Statistics
80 70 60 50 40 25
50
75
100
75
100
Calculus
Mathematical Statistics course (Example 3.4) 100 90
Statistics
80 70 60 50 40 30 25
50 Calculus
There appears to be a stronger linear relationship between marks in the mathematical statistics course and calculus than the relationship between the marks in the business statistics course and the marks in calculus.
3.103 Business Statistics course (Example 3.3)
70
100 90
Final
80 70 60 50 40 50
60
70
80
90
100
80
90
100
Midterm
Mathematical Statistics course (Example 3.4) 100 90
Final
80 70 60 50 40 30 50
60
70 Midterm
The relationship between midterm marks and final marks appear to be similar for both statistics courses. That is, there is weak positive linear relationship.
3.104a.
71
Frequency
80 60 40
20 0 30
34
38
42
46
50
54
58
62
66
Salary
The histogram is approximately bell shaped and symmetric.
b 70 60
Salary
50 40 30 20 10 0 0
1
2
3
4
5
6
7
Search
There is no linear relationship between the amount of time needed to land a job and salary.
3.105
72
Federal Government Debt 20 000 000 000 000 18 000 000 000 000 16 000 000 000 000 14 000 000 000 000 12 000 000 000 000
10 000 000 000 000 8 000 000 000 000 6 000 000 000 000 4 000 000 000 000 2 000 000 000 000 1 10 19 28 37 46 55 64 73 82 91 100 109 118 127 136 145 154 163 172 181 190 199 208 217 226
0
The debt was quite manageable until 100 years ago. It is quite unmanageable now.
3.106
Per Capita Debt 60000 50000 40000 30000 20000 10000 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
Per capita debt is still unmanageable.
3.107
73
Per Capita Debt Constant $ 30000 25000 20000 15000 10000 5000 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
This doesn’t look as bad.
3.108
National Debt 700 000 000 000 600 000 000 000 500 000 000 000 400 000 000 000 300 000 000 000 200 000 000 000 100 000 000 000 0 1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89
3.109
74
Per Capita Debt 20000 18000 16000 14000 12000
10000 8000 6000 4000 2000 1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 141 146
0
3.110
Per Capita Debt Constant $ 25000 20000 15000 10000 5000
1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97 101
0
75
Case 3.1 Line Chart of Temperature Anomalies 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2.5 1
121
241
361
481
601
721
841
961
1081
1201
1321
1441
There is a clear upward trend of about 1 degree Celsius over the 130 years.
Scatter Diagram of National Climate Data Center Land and Sea Temperature Anomalies and CO2 Levels (19582009) 2
Temperature anomalies
1.5 1 0.5 0 -0.5 -1 -1.5 300
320
340
360
380
400
CO2
There is a moderately strong positive linear relationship between carbon dioxide levels and temperature anomalies.
76
Case 3.4 90,000 80,000
70,000
GDP
60,000 50,000
40,000 30,000 20,000
10,000 0 0
20
40
60
80
100
Freedom scores
There is a moderately strong positive linear relationship between freedom scores and gross domestic product.
77
Chapter 4 4.1 a x
x 52 25 15 0 104 44 60 30 33 81 40 5 = 489 = 40.75 i
n
12
12
Ordered data: 0, 5, 15, 25, 30, 33, 40, 44, 52, 60, 81, 104; Median = (33 + 40)/2 = 36.5 Mode = all
4.2 x
x 5 7 0 3 15 6 5 9 3 8 10 5 2 0 12 = 90 = 6.0 i
n
15
15
Ordered data: 0, 0, 2, 3, 3, 5, 5, 5, 6, 7, 8, 9, 10, 12, 15; Median = 5 Mode = 5
4.3 a x
x 5.5 7.2 1.6 22. 0 8.7 2.8 5.3 3.4 12.5 18.6 8.3 6.6 = 102 .5 = 8.54 i
n
12
12
Ordered data: 1.6, 2.8, 3.4, 5.3, 5.5, 6.6, 7.2, 8.3, 8.7, 12.5, 18.6, 22.0; Median = 6.9 Mode = all b The mean number of miles jogged is 8.54. Half the sample jogged more than 6.9 miles and half jogged less.
4.4 a x
x 33 29 45 60 42 19 52 38 36 = 354 = 39.33 i
n
9
9
Ordered data: 19, 29, 33, 36, 38, 42, 45, 52, 60; Median = 38 Mode: all b The mean amount of time is 39.33 minutes. Half the group took less than 38 minutes.
4.5 a x
x 14 8 3 2 6 4 9 13 10 12 7 4 9 13 15 8 11 12 4 0
=
i
n
20
164 = 8.2 20
Ordered data: 0, 2, 3, 4, 4, 4, 6, 7, 8, 8, 9, 9, 10, 11, 12, 12, 13, 13, 14, 15; Median = 8.5 Mode = 4 b The mean number of days to submit grades is 8.2, the median is 8.5, and the mode is 4.
4.6 R g 3 (1 R1)(1 R 2 )(1 R3 ) 1 = 3 (1 .25 )(1 .10 )(1 .50 ) 1 = .19 4.7 R g 4 (1 R1)(1 R 2 )(1 R3 )(1 R 4 ) 1 = 4 (1 .50 )(1 .30 )(1 .50 )(1 .25) 1 = –.075
97
4.8 a x
x .10 .22 .06 .05 .20 = .53 = .106 i
n
5
5
Ordered data: –.05, .06, .10, .20, .22; Median = .10 b R g 5 (1 R1)(1 R 2 )(1 R3 )(1 R 4 )(1 R5 ) 1 = 5 (1 .10 )(1 .22 )(1 .06 )(1 .05)(1 .20 ) 1 = .102 c The geometric mean is best.
4.9 a x
x - .15 - .20 .15 .08 .50 = .22 = .044 i
n
5
5
Ordered data: –.20, –.15, –.08, .15, .50; Median = –.08 b R g 5 (1 R1)(1 R 2 )(1 R3 )(1 R 4 )(1 R5 ) 1 = 5 (1 .15)(1 .20 )(1 .15)(1 .08 )(1 .50 ) 1 = .015 c The geometric mean is best.
4.10 a Year 1 rate of return =
1200 1000 = .20 1000
Year 2 rate of return =
1200 1200 =0 1200
Year 3 rate of return =
1500 1200 = .25 1200
Year 4 rate of return =
2000 1500 = .33 1500
b x
x .20 0 .25 .33 = .78 = .195 i
n
4
4
Ordered data: 0, .20, .25, .33; Median = .225 c R g 4 (1 R1)(1 R 2 )(1 R3 )(1 R 4 ) 1 = 4 (1 .20 )(1 0)(1 .25 )(1 .33) 1 = .188 d The geometric mean is best because 1000(1.188) 4 = 2000.
4.11 a Year 1 rate of return =
10 12 = –.167 12
Year 2 rate of return =
14 10 = .40 10
Year 3 rate of return =
15 14 = .071 14
Year 4 rate of return =
22 15 = .467 15
98
Year 5 rate of return =
30 22 = .364 22
Year 6 rate of return =
25 30 = –.167 30
b x
x - .167 .40 .071 .467 .364 .167 = .968 = .161 i
n
6
6
Ordered data: –.167, –.167, .071, .364, .40, .467; Median = .218 c R g 6 (1 R 1 )(1 R 2 )(1 R 3 )(1 R 4 )(1 R 5 )(1 R 6 ) 1 = 6 (1 .167 )(1 .40 )(1 .071)(1 .467 )(1 .364 )(1 .167 ) 1 = .130 d The geometric mean is best because 12(1.130)6 = 25.
4.12 x = 54.91; median = 55. The mean number of bidders 54.91. Half of the auctions had 55 or fewer bidders. 4.13 x = 65,784; median = 65,805. The mean starting salary is $65,684 and half the sample’s starting salaries were less than or equal to $65,805.
4.14a x = 45.60; median = 45 b The mean commuting time is 45.60 minutes and half the sample took 45 minutes or less.
4.15 x = 26.56; median = 27. The mean commute is 26.56 minutes and half the sample commutes for 27 or less minutes.
4.16a x = .81; median = .84 b The mean percentage is .81. Half the sample paid less than .84. 4.17a x = 32.91; median = 32; mode = 32 b The mean speed is 32.91 mph. Half the sample traveled slower than 32 mph and half traveled faster. The mode is 32.
4.18a x = 122.76; median = 124 b The mean age is 122.76 months. Half the sample was 124 months or less. 4.19 x = 49.01; median = 49 4.20 x = 13.7; median = 14
99
4.21 x = 2.98; median = 2 4.22 x = 45,247, median = 32,500 4.23 x = 51.75; median = 52 4.24 x = 853,135; median = 60,872 4.25 x = 9,282,946; median = 249,000 4.26 x = 264,572; median = 23,690
4.27 x
x 9 3 7 4 1 7 5 4 = 40 = 5 i
n
8
(x x) [(9 5) (3 5) ... (4 5) = 46 = 6.57 2
s2
i
2
2
2
8 1
n 1
4.28 x
7
x 4 5 3 6 5 6 5 6 = 40 = 5 i
n
8
8
(x x) [(4 5) (5 5) ... (6 5) = 8 = 1.14 2
s2
8
4.29 x
i
2
2
2
8 1
n 1
7
x 12 6 22 31 23 13 15 17 21 = 160 = 17.78 i
n
9
9
(x x) [(12 17.78) (6 17.78) ... (21 17.78) = 433 .56 = 54.19 s 2
2
i
n 1
2
2
2
9 1
8
s s 2 = 54.19 = 7.36
100
4.30 x
x 0 (5) (3) 6 4 (4) 1 (5) 0 3 = 3 = –.30 i
n
(x x) [(0 (.3)) ((5) (.3)) ... (3 (.3)) = 136 .1 = 15.12 2
s2
10
10
i
2
n 1
2
2
10 1
9
s s 2 = 15.12 = 3.89
4.31 The data in (b) appear to be most similar to one another.
4.32 a: s 2 = 51.5 b: s 2 = 6.5 c: s 2 = 174.5 4.33 Variance cannot be negative because it is the sum of squared differences.
4.34 6, 6, 6, 6, 6
4.35 a. About 68% b. About 95% c. About 99.7%
4.36 a From the empirical rule we know that approximately 68% of the observations fall between 46 and 54. Thus 16% are less than 46 (the other 16% are above 54). b. Approximately 95% of the observations are between 42 and 58. Thus, only 2.5% are above 58 and all the rest, 97.5% are below 58. c. See (a) above; 16% are above 54.
4.37 a at least 75% b at least 88.9%
4.38 a Nothing b At least 75% lie between 60 and 180. c At least 88.9% lie between 30 and 210.
101
4.39 Range = 25.85, s 2 29.46, and s = 5.43; there is considerable variation between prices; at least 75% of the prices lie within 10.86 of the mean; at least 88.9% of the prices lie within 16.29 of the mean.
4.40 s 2 40.73 mph 2 and s = 6.38 mph; at least 75% of the speeds lie within 12.76 mph of the mean; at least 88.9% of the speeds lie within 19.14 mph of the mean
4.41 a Punter
Variance
Standard deviation
1
40.22
6.34
2
14.81
3.85
3
3.63
1.91
b Punter 3 is the most consistent.
4.42 s 2 .0858 cm 2 , and s = .2929cm; at least 75% of the lengths lie within .5858 of the mean; at least 88.9% of the rods will lie within .8787 cm of the mean. 4.43 x 175.73 and s = 62.1; At least 75% of the withdrawals lie within $124.20 of the mean; at least 88.9% of the withdrawals lie within $186.30 of the mean..
4.44a s = 15.01 b In approximately 68% of the days the number of arrivals falls within 15.01 of the mean; in approximately 95% of the hours the number of arrivals falls within 30.02 of the mean; in approximately 99.7% of the hours the number of arrivals falls within 45.03 of the mean 4.45 x 26.02 and s = 11.81 4.46 x 95.09 and s = 7.51 4.47 x 1937 and s = 950.0 4.48 x 749.7 and s = 215.9 4.49 x 49.01, s2 = 303.2, and s = 17.41 4.50 x 13.70 and s = 3.07 4.51 x 2.98 and s = 2.59
102
4.52 x 45,247 and s = 39,885 4.53 x 51.75 and s = 16.17 4.54 x 853,135 and s = 5,952,380 4.55 x 9,282,946 and s = 54,480,433 4.56 x 264,572 and s = 3,793,046
4.57 First quartile: L25 (10 1) Second quartile: L50 (10 1) Third quartile: L75 (10 1)
25 = (11)(.25) = 2.75; the first quartile is 5.5. 100
50 = (11)(.5) = 5.5; the second quartile is 11. 100
75 = (11)(.75) = 8.25; the third quartile is 16.5. 100
4.58 First quartile: L25 (15 1) Second quartile: L50 (15 1) Third quartile: L75 (15 1)
25 = (16)(.25) = 4; the fourth number is 3. 100
50 = (16)(.5) = 8; the eighth number is 5. 100
75 = (16)(.75) = 12; the twelfth number is 7. 100
4.59 30th percentile: L30 (10 1) 80th percentile: L80 (10 1)
80 = (11)(.80) = 8.8; the 80th percentile 30.8. 100
4.60 20th percentile: L 20 (10 1) 40th percentile: L 40 (10 1)
30 = (11)(.30) = 3.3; the 30th percentile is 22.3. 100
20 = (11)(.20) = 2.2; the 20th percentile is 43 + .2(51–43) = 44.6. 100
40 = (11)(.40) = 4.4; the 40th percentile is 52 +.4(60–52) = 55.2. 100
4.61 First quartile: L25 (13 1) Second quartile: L50 (13 1)
25 = (14)(.25) = 3.5; the first quartile is 13.05. 100
50 = (14)(.5) = 7; the second quartile is 14.7. 100 103
Third quartile: L75 (13 1)
75 = (14)(.75) = 10.5; the third quartile is 15.6. 100
4.62 Third decile: L 30 (15 1) Sixth decile: L60 (15 1)
30 = (16)(.30) = 4.8; the third decile is 5 + .8(7 – 5) = 6.6. 100
60 = (16)(.60) = 9.6; the sixth decile is 17 + .6(18 – 17) = 17.6. 100
4.63 Interquartile range = 15.6 –13.05 = 2.55
4.64 First quartile: L25 (15 1) Third quartile: L75 (15 1)
25 = (16)(.25) = 4; the fourth number is 13. 100
75 = (16)(.75) = 12; the twelfth number is 24. 100
Interquartile range = 24 – 13 = 11 4.65 First quartile = 5.75, third quartile = 15; interquartile range = 15 – 5.75 = 9.25
4.66 First quartile: L25 (15 1) Third quartile: L75 (15 1)
25 = (16)(.25) = 4; the fourth number is 3. 100
75 = (16)(.75) = 12; the twelfth number is 21. 100
Interquartile range = 21 –3 = 18 4.67 L85 = 75; The speed limit should be set at 75 mph. 4.68 a First quartile = 2, second quartile = 4, and third quartile = 8. b Most executives spend little time reading resumes. Keep it short.
4.69 Dogs: First quartile = 1097.5, second quartile = 1204, and third quartile = 1337. Cats: First quartile = 743, second quartile = 856, and third quartile = 988. Dogs cost more money than cats. Both sets of expenses are positively skewed.
4.70 First quartile = 50, second quartile = 125, and third quartile = 260. The amounts are positively skewed.
4.71 BA First quartile = 25,730, second quartile = 27,765, and third quartile = 29836 BSc First quartile = 29,927, second quartile = 33,397, and third quartile = 36,745
104
BBA First quartile = 31,316, second quartile = 34,284, and third quartile = 39,551 Other First quartile = 28,254, second quartile = 29,951, and third quartile = 32,905 The starting salaries of BA and other are the lowest and least variable. Starting salaries for BBA and BSc are higher.
4.72 The quartiles are 145.51, 164.17, and 174.64. The data are positively skewed. One-quarter of the times are below 145.51 and one-quarter are above 174.64.
4.73 Private course: The quartiles are 228, 237, and 245 Public course: The quartiles are 279, 296, and 307 The amount of time taken to complete rounds on the public course are larger and more variable than those played on private courses.
4.74a. The quartiles are 26, 28.5, and 32 b. The times are positively skewed.
4.75 The quartiles are 8081.81, 9890.48, and 11,692.92. One-quarter of mortgage payments are less than $607.19 and one quarter exceed $909.38. 4.76 The quartiles are 2377, 2765, and 3214. One-quarter of the amounts are less than $2377 and one quarter exceed $3214. 4.77 The quartiles are 16,250, 32,500, and 55,000. One-quarter of the amounts are less than $16,500 and one quarter exceed $55,000.
4.78 The quartiles are 12, 14, and 16. One-quarter of the amounts are less than or equal to 12 and one quarter are greater than or equal to 16. 4.79 The quartiles are 1, 2, and 4. One-quarter of the amounts are less than or equal to 1 and one quarter are greater than or equal to 4.
4.80 The quartiles are 28,407, 60,872, and 144,063. One-quarter of the amounts are less than $28,407 and one quarter are greater $144,063.
4.81 The quartiles are 32,100, 246,920, and 1,139,010. One-quarter of the amounts are less than $32,100 and one quarter are greater than $1,139,010.
4.82 The quartiles are 0, 23,690, and 156,100. One-quarter of the amounts are less than or equal to 0 and one quarter are greater than $156,100.
105
4.83 There is a negative linear relationship. The strength is unknown.
4.84 a. r
s xy sxsy
150 .7813 (16 )(12 )
There is a moderately strong negative linear relationship. b. R2 = r2 = (− .7813)2 = .6104 61.04% of the variation in y is explained by the variation in x.
4.85a.
Total
xi
yi
xi2
yi2
x i yi
20
14
400
196
280
40
16
1600
256
640
60
18
3600
324
1080
50
17
2500
289
850
50
18
2500
324
900
55
18
3025
324
990
60
18
3600
324
1080
70
20
4900
400
1400
405
139
22,125
2,437
7,220
n
n
x i = 405
i 1
i 1
n
n
n
y i = 139
x i2 = 22,125
i 1
y i2 = 2,437
i 1
x y = 7,220 i
i 1
n n xi yi n (405 )(139 ) 1 1 i 1 7,220 s xy x i y i i 1 = 26 .16 8 1 8 n 1 i 1 n
2 n xi n 2 1 x i2 i 1 = 1 22,125 (405 ) 231 .7 s 2x n 1 i 1 n 8 1 8
106
i
s x s 2x 231 .7 15 .22 2 n yi n 2 1 y i2 i 1 = 1 2,437 (139 ) 3.13 s 2y n n 1 i 1 8 1 8
s y s 2y 3.13 1.77 r
s xy sxsy
26 .16
..9711
(15 .22 )(1.77 )
R2 = r2 = .97112 = .9430 The covariance is 26.16, the coefficient of correlation is .9711 and the coefficient of determination is .9430. 94.30% of the variation in expenses is explained by the variation in total sales. b.
b1
s xy
=
s 2x
26 .16 .113 231 .7
x
x 405 50.63
y
y 139 17.38
i
n
8
i
n
8
b 0 y b1x = 17.38 – (.113)(50.63) = 11.66 The least squares line is
ŷ = 11.66 + .113x The estimated variable cost is .113 and the estimated fixed cost is 11.66.
4.86
xi
yi
xi2
yi2
x i yi
40 42 37 47 25 44 41 48 35 28
77 63 79 86 51 78 83 90 65 47
1,600 1,764 1,369 2,209 625 1,936 1,681 2,304 1,225 784
5,929 3,969 6,241 7,396 2,601 6,084 6,889 8,100 4,225 2,209
3,080 2,646 2,923 4,041 1,276 3,432 3,403 4,320 2,275 1,316
107
Total
387
719
n
n
n
i 1
28,712
n
n
2 i
2 i
i
i
53,643
x = 15,497 y = 53,643 x y = 28,712
y = 719
x = 387 a
15,497
i 1
i 1
i 1
i
i
i 1
n n xi yi n (387 )( 719 ) 1 1 i 1 28,712 s xy x i y i i 1 = 98 .52 10 1 10 n n 1 i 1
2 n xi n 1 (387 ) 2 1 i 1 2 2 xi sx = 15,497 57 .79 n 1 i 1 n 10 1 10
2 n yi n 2 1 y i2 i 1 = 1 53,643 (719 ) 216 .32 s 2y n 1 i 1 n 10 1 10
b
r
s xy sxsy
98 .52
.8811
(57 .79 )( 216 .32 )
c
R2 = r2 = .88112 = .7763
d
b1
s xy
=
s 2x
98 .52 1.705 57 .79
x
x 387 38.7
y
y 719 71.9
i
n
10
i
n
10
b 0 y b1x = 71.9 – (1.705)(38.7) = 5.917 The least squares line is
ŷ = 5.917 + 1.705x e. There is a strong positive linear relationship between marks and study time. For each additional hour of study time marks increased on average by 1.705.
4.87
xi
yi
xi2
yi2
x i yi
599 689
9.6 8.8
358,801 474,721
92.16 77.44
5750.4 6063.2
108
Total
584 631 594 643 656 594 710 611 593 683 7,587
7.4 10.0 7.8 9.2 9.6 8.4 11.2 7.6 8.8 8.0 106.4
n
n
x i =7,587
341,056 398,161 352,836 413,449 430,336 352,836 504,100 373,321 351,649 466,489 4,817,755
x i2 = 4,817,755
i 1
4321.6 6310.0 4632.2 5915.6 6297.6 4989.6 7952.0 4643.6 5218.4 5464.0 67,559.2 n
n
n
y i = 106.4
i 1
i 1
54.76 100.00 60.84 84.64 92.16 70.56 125.44 57.76 77.44 64.00 957.2 y i2 = 957.2
i 1
x y = 67,559.2 i
i
i 1
n n xi yi n (7,587 )(106 .4) 1 1 i 1 67 ,559 .2 s xy x i y i i 1 = 26 .16 12 1 12 n n 1 i 1
2 n xi (7,587 ) 2 1 n 4 , 817 , 755 1 1,897 .7 x i2 i 1 = 12 1 s 2x 12 n 1 i 1 n
s x s 2x 1,897 .7 43 .56 2 n yi (106 .4) 2 1 n 957 . 2 1 1.25 i 1 y i2 s 2y = 12 1 12 n 1 i 1 n
s Y s 2Y 1.25 1.12 r
s xy sxsy
26 .16
.5362
(43 .56 )(1.12 )
R2 = r2 = .53622 = .2875 The covariance is 26.16, the coefficient of correlation is .5362, and the coefficient of determination is .2875. The coefficient of determination tells us that 28.75% of the variation in MBA GPAs is explained by the variation in GMAT scores.
4.88
109
R2 = r2 = (-.1913)2 = .0366
4.89 a
R2 = r2 = (.2543)2 = .0647. b There is a weak linear relationship between age and medical expenses. Only 6.47% of the variation in average medical bills is explained by the variation in age. c 80.00 70.00
60.00
Expense
50.00
y = 0.2257x - 5.9662 R² = 0.0647
40.00 30.00 20.00 10.00
0.00 -10.00 0
20
40
60
80
100
Age
The least squares line is ŷ 5.966 .2257 x d For each additional year of age mean medical expenses increase on average by $.2257 or 23 cents. e Charge 25 cents per day per year of age.
4.90
R2= (−.0886)2 = .0078 110
Only 0.78% of the variation in the number of houses sold is explained by the variation in interest rates.
4.91 300 250 y = 0,049x + 74,204 R² = 0,0004
200 150 100 50 0 0
20
40
60
80
100
120
140
160
Only 0.04% of the variation in the number of wells drilled is explained by the variation in the price of oil. The relationship is too weak to interpret the value of the slope coefficient.
4.92
R2 = (.0830)2 = .0069. There is a very weak positive relationship between the two variables.
111
4.93 480 460
440 Labor cost
420 400
y = 3.3x + 315.5 R² = 0.5925
380 360 340
320 300 0
10
20
30
40
50
Batch size
ŷ = 315.5 + 3.3x; Fixed costs = $315.50, variable costs = $3.30
4.94 1800 1600 y = 71.654x + 263.4 R² = 0.5437
1400
Cost
1200 1000
800 600 400
200 0 0
2
4
6
8
10
12
14
16
Time
ŷ = 263.4 + 71.65x; Estimated fixed costs = $263.40, estimated variable costs = $71.65
112
4.95
4.96
4.97
4.98
4.99
4.100
113
All five commodities are negatively linearly related to the exchange rate. The relationships are moderately strong.
4.101 y = 366,89x + 771,69 R² = 0,2766
50000 45000 40000 35000 30000
25000 20000 15000 10000 5000 0 0
20
40
60
80
100
a. The slope coefficient is 366.89 and the coefficient of determination is .2766 b. For each additional win home attendance increases on average by 366.89.
4.102
114
120
y = 20,901x + 28811 R² = 0,0124
40000 35000 30000 25000 20000 15000 10000 5000
0 0
20
40
60
80
100
120
a. The coefficient of determination is .0124, which means that there is very little correlation between wins and away attendance. b. For each additional win away attendance increases on average by 20.90.
4.103a. y = 14771x + 1E+06 R² = 0,0753
4 000 000 3 500 000 3 000 000 2 500 000 2 000 000 1 500 000 1 000 000 500 000 0 0
20
40
60
80
100
b. For each additional win the home attendance increases on average by 14,771.
4.104a.
115
120
y = 2278,1x + 2E+06 R² = 0,0371
3 000 000 2 500 000 2 000 000 1 500 000 1 000 000 500 000
0 0
20
40
60
80
100
120
b. For each additional win the home attendance increases on average by 2278.1.
4.105 80
y = 0,641x - 9,6418 R² = 0,271
70 60 50 40 30 20 10 0 0,000
20,000
40,000
60,000
80,000
100,000
a. The marginal cost of one more win is 1 million/.641 = $1,560,000. b.
116
120,000
y = 71,69x + 14910 R² = 0,2625
25 000
20 000
15 000
10 000
5 000
0 0
10
20
30
40
50
60
70
80
b. For each additional win the home attendance increases on average by 71.69. c. y = 8,7705x + 17489 R² = 0,0533
19 500 19 000 18 500 18 000 17 500 17 000 16 500 0
10
20
30
40
50
60
70
b. For each additional win the away attendance increases on average by 8.77.
4.106a.
117
80
y = 0,6812x - 4,4414 R² = 0,3335
70 60 50 40 30 20 10
0 0,000
20,000
40,000
60,000
80,000
100,000
120,000
a. The marginal cost of one more win is 1 million/.6812 = $1,467,998. b.
y = 3328,2x + 574331 R² = 0,2605
1 000 000 900 000 800 000 700 000 600 000 500 000 400 000 300 000 200 000 100 000 0 0
10
20
30
40
50
60
For each additional win the home attendance increases on average by 3328.2. c.
118
70
800 000 y = 1432,3x + 652547 R² = 0,4136
780 000 760 000 740 000 720 000 700 000 680 000 660 000
640 000 0
10
20
30
40
50
60
70
c. For each additional win the away attendance increases on average by 1432.3.
4.107a. y = 0,025x + 4,9339 R² = 0,0206
16 14 12 10 8 6 4 2 0 0,000
20,000
40,000
60,000
80,000 100,000 120,000 140,000 160,000 180,000
The marginal cost of one more win is 1 million/.025 = $40,000,000.
119
b. 100 000 y = 109,98x + 67394 R² = 0,0016
90 000 80 000 70 000 60 000 50 000 40 000 30 000 20 000 10 000 0
0
2
4
6
8
10
12
14
16
For each additional win the home attendance increases on average by 109.98. c. y = 85,184x + 67718 R² = 0,007
76 000 74 000 72 000 70 000 68 000
66 000 64 000 62 000 0
2
4
6
8
10
12
14
c. For each additional win the away attendance increases on average by 85.184.
4.108a.
120
16
y = -0,0329x + 10,511 R² = 0,0135
14 12 10 8 6 4 2 0 0,000
20,000
40,000
60,000
80,000
100,000
120,000
This is a strange result-a negative slope means that higher payrolls lead to fewer wins. b. y = 5796,6x + 490457 R² = 0,0697
800 000 700 000 600 000 500 000 400 000 300 000 200 000 100 000 0 0
2
4
6
8
10
12
For each additional win the home attendance increases on average by 5796.6. c.
121
14
y = 855,4x + 529987 R² = 0,0068
700 000 600 000 500 000 400 000 300 000 200 000 100 000
0 0
2
4
6
8
10
12
14
For each additional win the away attendance increases on average by 855.4.
4.109a. y = 0,394x + 13,304 R² = 0,2955
60 50 40 30 20 10 0 0,000
20,000
40,000
60,000
80,000
a. The marginal cost of one more win is 1 million/.394 = $2,538,071. b.
122
100,000
y = 42,729x + 16886 R² = 0,027
25 000
20 000
15 000
10 000
5 000
0 0
10
20
30
40
50
60
For each additional win the home attendance increases on average by 42.7294. c. 19 500 y = -1,5092x + 17571 R² = 0,0006
19 000 18 500 18 000 17 500 17 000 16 500 0
10
20
30
40
There is a negative relationship-more wins smaller attendance.
4.110a.
123
50
60
y = 0,3745x + 2,8509 R² = 0,1609
40 35 30 25 20 15 10 5
0 0,000
10,000
20,000
30,000
40,000
50,000
60,000
70,000
The marginal cost of one more win is 1 million/.3745 = $2,670,227. b. y = 3131,3x + 350143 R² = 0,1154
600 000 500 000 400 000 300 000 200 000 100 000 0 0
5
10
15
20
25
30
35
For each additional win the home attendance increases on average by 3131.3. c.
124
40
y = 458,6x + 414278 R² = 0,0533
450 000 445 000 440 000 435 000 430 000 425 000 420 000
415 000 410 000 405 000 400 000
395 000 0
5
10
15
20
25
30
35
For each additional win the away attendance increases on average by 458.6.
4.111 .4646; .1946 4.112 1.1193; .4355 4.113 1.1907; .5485 4.114 .8844; .4169 4.115 .6197; .1247 4.116 .8266; .3427 4.117 x
1.1057 1.6057 .8823 3.5937 1.1979 3 3
4.118 x
1.1665 1.1186 .46463 2.7497 .9166 3 3
4.119 x
.6558 .9031 .5841 2.143 .7143 3 3
4.120 1.5876; .1236 4.121 .2312; .0185 4.122 .4031; .0829 4.123 x
.7588 .8089 .8572 2.4249 .8083 3 3
125
40
4.124 x
1.7742 .2312 1.6824 3.6878 1.2293 3 3
4.125 x
.3805 .4031 .7836 .3918 2 2
4.126 1.0815; .2846 4.127 .6446; .0629 4.128 1.7154; .1069 4.129 x
.9556 1.173 .9528 3.0814 1.0271 3 3
4.130 x
1.173 .2818 1.1056 2.5604 .8535 3 3
4.131 x
1.1397 .8228 .8076 2.7701 .9234 3 3
4.132
R2 = .67842 = .4603; 46.03% of the variation in statistics marks is explained by the variation in calculus marks. The coefficient of determination provides a more precise indication of the strength of the linear relationship.
4.133 2500 y = 116.53x - 369.93 R² = 0.9334
2000
Cost
1500 1000 500 0 0
5
10
15 Speed
126
20
25
The least squares line is ŷ = 369.93 + 116.53x. On average for each addition mph the cost of repair increases by $116.53.
4.134 y = 0.6041x + 17.933 R² = 0.0505
90 80
70 Income
60 50
40 30 20
10 0 60
65
70
75
80
Height
a ŷ = 17.933 + .6041x b The coefficient of determination is .0505, which indicates that only 5.05% of the variation in incomes is explained by the variation in heights.
4.135 3000.00
y = 19.059x + 1087.7 R² = 0.0779
2500.00
Sales
2000.00 1500.00 1000.00 500.00 0.00 0
5
10
15
20 Time
127
25
30
35
The coefficient of determination is .0779, which indicates that only 7.79% of the variation in sales is explained by the time between movies.
4.136 400
350
y = 0.07x + 103.44 R² = 0.5201
300
Price
250 200 150
100 50 0 0
500
1000
1500
2000
2500
3000
3500
4000
Size
a ŷ = 103.44 + .07x b. For each additional square foot the price increases on average by $.07 thousand. More simply for each additional square foot the price increases on average by$70. c. From the least squares line we can more precisely measure the relationship between the two variables.
Frequency
4.137 Private course 50 40 30 20 10 0
220
230
240
250
260
Time
Public course
128
Frequency
50 40 30 20 10 0
250
275
300
325
350
375
Time
The information obtained here is more detailed than the information provided by the box plots.
4.138
Frequency
80 60 40 20 0 24
28
32
36
40
44
48
52
Times The times are positively skewed.
4.139
a x 35.01, median = 36
129
56
b s = 7.68 c Half of the bone density losses lie below 36. At least 75% of the numbers lie between 19.64 and 50.38, at least 88.9% of the numbers lie between 11.96 and 58.06.
4.140
a x = 29,913, median = 30,660 b s 2 = 148,213,791; s = 12,174 c The number of coffees sold varies considerably.
4.141
R2 = r2 = .57422 = .3297; 32.97% of the variation in bone loss is explained by the variation in age.
4.142 a & b
130
70000 y = -553,7x + 49337 R² = 0,5489
60000 50000 40000 30000 20000 10000
0 -20
-10
0
10
20
30
40
50
60
70
R2 = .5489 and the least squares line is ŷ = 49,337 – 553.7x c 54.8% of the variation in the number of coffees sold is explained by the variation in temperature. For each additional degree of temperature the number of coffees sold decreases on average by 554 cups. Alternatively for each 1-degree drop in temperature the number of coffees increases on average, by 553.7 cups. d We can measure the strength of the linear relationship accurately and the slope coefficient gives information about how temperature and the number of coffees sold are related.
4.143a mean, median, and standard deviation b
x = 93.90, s = 7.72
131
c We hope Chris is better at statistics than he is golf.
4.144
Mean, median, mode: 34,656, 34,636, 35,149
4.145
There is a moderately strong positive linear relationship.
4.146
132
25 y = 0,3295x + 10,153 R² = 0,2036 20
15
10
5
0 0
5
10
15
20
25
a. The coefficient of determination is .2036. b. The slope coefficient is .3295; for each additional year of education of the father the son’s years of education increase on average by .3295.
4.147 25
y = 0,3452x + 9,8579 R² = 0,1945
20
15
10
5
0 0
5
10
15
20
25
a. The coefficient of determination is .1945. b. The slope coefficient is .3452; for each additional year of education of the father the son’s years of education increase on average by .3452.
4.148
133
30 y = 0,0205x + 1,9801 R² = 0,019
25 20 15 10 5
0 0
20
40
60
80
100
The relationship is very weak. For each additional year of age the number of hours of TV increases on average by .0205.
4.149 y = 0,7744x + 13,903 R² = 0,1514
60 50 40 30 20 10 0 0
5
10
15
20
a. The coefficient of determination is .1514. b. For each additional year of education the age increases by .7744.
4.150
134
25
30 y = -0,1757x + 5,3998 R² = 0,0424
25 20 15 10 5
0 0
5
10
15
20
25
a. The coefficient of determination is .0424. b. For each additional year of education the number of hours of TV decreases by .1757.
4.151
The coefficient of correlation is almost 1.
4.152 y = 6,0186x + 4E+06 R² = 0,4324
1 400 000 000,00 1 200 000 000,00 1 000 000 000,00 800 000 000,00
600 000 000,00 400 000 000,00 200 000 000,00 0,00 0,00
50 000 000,00 100 000 000,00 150 000 000,00 200 000 000,00
-200 000 000,00
The coefficient of determination is .4324. 135
For each additional dollar of income assets increase on average by 6.0186
4.153 y = 17,799x + 7E+06 R² = 0,047
1 400 000 000,00 1 200 000 000,00 1 000 000 000,00 800 000 000,00 600 000 000,00
400 000 000,00 200 000 000,00 0,00 0,00
5 000 000,0010 000 000,0015 000 000,0020 000 000,0025 000 000,00
-200 000 000,00
The coefficient of determination is .047 indicating a weak linear relationship. For each additional dollar of wage income assets increase on average by 17.799.
4.154 9 y = -0,0249x + 2,1289 R² = 0,1176
8 7 6 5 4 3 2 1 0 -1
0
20
40
60
80
100
For each additional year of age the number of children decreases on average by .0249.
136
Case 4.1 a Scatter diagrams with time as the independent variable and temperature anomalies as the dependent variable
3 y = 0,0008x - 0,6381 R² = 0,5404
2,5
Temperature anomalies
2 1,5 1 0,5 0 -0,5
0
200
400
600
800
1000
1200
1400
1600
1800
-1 -1,5 -2 -2,5
Month
Monthly average increase is .0008. For the 1637 month period the increase was 1637(.0008) = 1.31o Celsius.
Scatter diagram with carbon dioxide levels as the independent variable and temperature anomalies as the dependent variable
3 y = 0,0156x - 5,0641 R² = 0,6123
Temperature anomalies
2,5 2 1,5 1 0,5 0 -0,5
300
320
340
360
380
-1 -1,5
Month
137
400
420
The coefficient of determination is .6123, which means that 61.23% of the variation in temperature anomalies is explained by the variation in CO2levels. There is a moderately strong linear relationship.
Case 4.2 1880 to 1940 1,5
y = 0,0007x - 0,5429 R² = 0,1793
Temperature anomalies
1 0,5 0 0
100
200
300
400
500
600
700
800
-0,5 -1 -1,5 -2 -2,5
Month
From 1880 to 1940 the earth warned at an average monthly rate of .0007 o Celsius.
1941 to 1975 y = -0,0001x - 0,0064 R² = 0,0023
1
Temperature anomalies
0,5
0 0
50
100
150
200
250
300
350
400
-0,5
-1
-1,5
Month
From 1941 to 1975 the earth cooled at an average monthly rate of .0001o Celsius
138
450
1976 to 1997 y2 = 0,0021x + 0,0388 R² = 0,2026
Temperature anomalies
1,5 1 0,5 0 0
50
100
150
200
250
300
-0,5 -1
Month
From 1976 to 1997 the earth warmed at an average monthly rate of .0021 o Celsius.
1998 to 2016 y = 0,0018x + 0,7418 R² = 0,1015
Temperature anomalies
2,5
2
1,5
1
0,5
0 0
50
100
150
200
Month
From 1998 to 2012 the earth warmed at an average monthly rate of .001 o Celsius
Over different periods of time the earth has warmed and cooled.
139
250
Case 4.3 2003-04 Season 60 y = 0.1526x + 28.559 R² = 0.0876
50
Wins
40 30 20 10 0 $0
$20
$40
$60
$80
$100
Payroll ($millions)
The cost of winning one additional game is 1million/.1526 = $6.553 million. However, the coefficient of determination is only .0876, which tells us that there are many other variables that determine how well a team will do.
2005-06 Season 70 y = 0.7795x + 14.256 R² = 0.3072
60
Wins
50 40 30 20 10 0 $0
$10
$20
$30
$40
$50
Payroll ($millions)
The cost of winning one additional game is 1million/.7795 = $1.283 million. The coefficient of determination is .3072.
140
The small coefficient of determination in the year before the strike seems to indicate that team owners were spending large amounts of money and getting little in return. The results are markedly different in the year after the strike. There is a much stronger linear relationship between payroll and the number of wins and the cost of winning one additional game is considerably smaller.
Case 4.4
The coefficient of determination is (−.1787)2 = .0319. There is a weak negative linear relationship between percentage of rejected ballots and Percentage of “yes” votes.
The coefficient of determination is (.3600)2 = .1296. There is a moderate positive linear relationship between percentage of rejected ballots and Percentage of Allophones.
The coefficient of determination is (.0678)2 = .0046. There is a very weak positive linear relationship between percentage of rejected ballots and Percentage of Allophones.
The statistics provide some evidence that electoral fraud has taken place.
141
Chapter 5 5.1 In an observational study, there is no attempt to control factors that might influence the variable of interest. In an experimental study, a factor (such as regular use of a fitness center) is controlled by randomly selecting who is exposed to that factor, thereby reducing the influence of other factors on the variable of interest.
5.2a The study is observational. The statistics practitioner did not randomly assign stores to buy cans or bottles. b Randomly assign some stores to receive only cans and others to receive only bottles.
5.3 Randomly sample smokers and nonsmokers and compute the proportion of each group that has lung cancer. b The study is observational. Experimental data would require the statistics practitioner to randomly assign some people to smoke and others not to smoke.
5.4a A survey can be conducted by means of a personal interview, a telephone interview, or a selfadministered questionnaire. b A personal interview has a high response rate relative to other survey methods, but is expensive because of the need to hire well-trained interviewers and possibly pay travel-related costs if the survey is conducted over a large geographical area. A personal interview also will likely result in fewer incorrect responses that arise when respondents misunderstand some questions. A telephone interview is less expensive, but will likely result in a lower response rate. A self-administered questionnaire is least expensive, but suffers from lower response rates and accuracy than interviews.
5.5 Five important points to consider when designing a questionnaire are as follows: (1) The questionnaire should be short. (2) Questions should be short, clearly worded, and unambiguous. (3) Consider using dichotomous or multiple-choice questions, but take care that respondents needn’t make unspecified assumptions before answering the questions. (4) Avoid using leading questions. (5) When preparing the questions, think about how you intend to tabulate and analyze the responses.
5.6a The sampled population will exclude those who avoid large department stores in favor or smaller shops, as well as those who consider their time too valuable to spend participating in a 153
survey. The sampled population will therefore differ from the target population of all customers who regularly shop at the mall. b The sampled population will contain a disproportionate number of thick books, because of the manner in which the sample is selected. c The sampled population consists of those eligible voters who are at home in the afternoon, thereby excluding most of those with full-time jobs (or at school).
5.7a The Literary Digest was a popular magazine in the 1920s and 1930s which had correctly predicted the outcome of many presidential elections. To help predict the outcome of the 1936 presidential election, the Literary Digest mailed sample ballots to 10 million prospective voters. Based on the results of the ballots returned, the magazine predicted that the Republican candidate, Alfred Landon, would defeat the Democratic incumbent, Franklin D. Roosevelt, by a 3 to 2 margin. In fact, Roosevelt won a landslide victory, capturing 62% of the votes. b The main reason for the poll being so wrong was nonresponse bias resulting from a self-selected sample, causing the sample to be unrepresentative of the target population. (Only 2.3 million ballots were returned.) The second reason was selection bias, resulting from poor sampling design, causing the sampled population and the target population to differ. Most of those to whom a ballot was sent were selected from the Literary Digest’s subscription list and from telephone directories. These people tended to be wealthier than average and tended to vote Republican.
5.8a A self-selected sample is a sample formed primarily on the basis of voluntary inclusion, with little control by the designer of the survey. b Choose any recent radio or television poll based on responses of listeners who phone in on a volunteer basis. c Self-selected samples are usually biased, because those who participate are more interested in the issue than those who don’t, and therefore probably have a different opinion.
5.9 We should ignore the results because this is an example of a self-selected sample.
5.10 No, because the sampled population consists of the responses about the professor’s course. We cannot make draw inferences about all courses.
5.11 We used Excel to generate 40 three-digit random numbers. Because we will ignore all randomly generated numbers over 800, we can expect to ignore about 20% (or about 8 to 10) of the randomly generated numbers. We will also ignore any duplication. We therefore chose to generate 40 three-digit random numbers, and will use the first 25 unique random numbers less than 801 to select our sample. The 40 numbers generated are shown below, with a stroke through
154
those to be ignored.
6
357
456
449
862
154
55
412
475
430
999
912
60
207
717
651
10
294
327
165
576
871
990
354
390
540
893
181
496
870
738
820
32
963
160
32
231
86
970
46
5.12 We used Excel to generate 30 six-digit random numbers. Because we will ignore any duplicate numbers generated, we generated 30 six-digit random numbers and will use the first 20 unique random numbers to select our sample. The 30 numbers generated are shown below.
169,470
744,530
22,554
918,730
320,262
503,129
318,858
698,203
822,383
938,262
800,806
56,643
836,116
123,936
80,539
154,211
391,278
940,154
110,630
856,380
222,145
692,313
949,828
561,511
909,269
811,274
288,553
749,627
858,944
39,308
5.13 Stratified random sampling is recommended. The strata are the school of business, the faculty of arts, the graduate school and the all the other schools and faculties would be the fourth stratum. The data can be used to acquire information about the entire campus but also compare the four strata.
5.14 A stratified random sampling plan accomplishes the president’s goals. The strata are the four areas enabling the statistics practitioner to learn about the entire population but also compare the four areas.
5.15 The operations manager can select stratified random samples where the strata are the four departments. Simple random sampling can be conducted in each department.
5.16 Use cluster sampling, letting each city block represent a cluster.
5.17a Sampling error refers to an inaccuracy in a statement about a population that arises because the statement is based only on sample data. We expect this type of error to occur because we are making a statement based on incomplete information. Nonsampling error refers to mistakes made in the acquisition of data or due to the sample observations being selected improperly. b Nonsampling error is more serious because, unlike sampling error, it cannot be diminished by taking a larger sample.
155
5.18 Three types of nonsampling errors: (1) Error due to incorrect responses (2)Nonresponse error, which refers to error introduced when responses are not obtained from some members of the sample. This may result in the sample being unrepresentative of the target population. (3)Error due to selection bias, which arises when the sampling plan is such that some members of the target population cannot possibly be selected for inclusion in the sample.
5.19 Yes. A census will likely contain significantly more nonsampling errors than a carefully conducted sample survey.
156
Chapter 6 6.1 a Relative frequency approach b If the conditions today repeat themselves an infinite number of days rain will fall on 10% of the next days. 6.2 a Subjective approach b If all the teams in major league baseball have exactly the same players the New York Yankees will win 25% of all World Series. 6.3 a {a is correct, b is correct, c is correct, d is correct, e is correct} b P(a is correct) = P(b is correct) = P(c is correct) = P(d is correct) = P(e is correct) = .2 c Classical approach d In the long run all answers are equally likely to be correct. 6.4 a Subjective approach b The Dow Jones Industrial Index will increase on 60% of the days if economic conditions remain unchanged. 6.5 a P(even number) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2 b P(number less than or equal to 4) = P(1) + P(2) + P(3) + P(4) = 1/6 + 1/6 + 1/6 +1/6 = 4/6 = 2/3 c P(number greater than or equal to 5) = P(5) + P(6) = 1/6 + 1/6 = 2/6 = 1/3 6.6 {Adams wins. Brown wins, Collins wins, Dalton wins} 6.7a P(Adams loses) = P(Brown wins) + P(Collins wins) + P(Dalton wins) = .09 + .27 + .22 = .58 b P(either Brown or Dalton wins) = P(Brown wins) + P(Dalton wins) = .09 + .22 = .31 c P(either Adams, Brown, or Collins wins) = P(Adams wins) + P(Brown wins) + P(Collins wins) = .42 + .09 + .27 = .78 6.8 a {0, 1, 2, 3, 4, 5} b {4, 5} c P(5) = .10 d P(2, 3, or 4) = P(2) + P(3) + P(4) = .26 + .21 + .18 = .65 e P(6) = 0 6.9 {Contractor 1 wins, Contractor 2 wins, Contractor 3 wins} 6.10 P(Contractor 1 wins) = 2/6, P(Contractor 2 wins) = 3/6, P(Contractor 3 wins) = 1/6
6.11 a {Shopper pays cash, shopper pays by credit card, shopper pays by debit card} b P(Shopper pays cash) = .30, P(Shopper pays by credit card) = .60, P(Shopper pays by debit card) = .10 c Relative frequency approach 6.12 a P(shopper does not use credit card) = P(shopper pays cash) + P(shopper pays by debit card) = .30 + .10 = .40 b P(shopper pays cash or uses a credit card) = P(shopper pays cash) + P(shopper pays by credit card) = .30 + .60 = .90 6.13 {single, divorced, widowed} 6.14 a P(single) = .15, P(married) = .50, P(divorced) = .25, P(widowed) = .10 b Relative frequency approach 6.15 a P(single) = .15 b P(adult is not divorced) = P(single) + P(married) + P(widowed) = .15+ .50 + .10 = .75 c P(adult is either widowed or divorced) = P(divorced) + P(widowed) = .25 + .10 = .35 6.16 {Spanish, Chinese, Tagalog, Vietnamese, French, Korean, others} 6.17 a .619 b .381 c .0435 d .245 6.18 {Very safe, Somewhat safe, Somewhat unsafe, Very unsafe, Not sure} 6.19a .17 b .45 c .12
6.20 P( A1 ) = .1 + .2 = .3, P( A 2 ) = .3 + .1 = .4, P( A 3 ) = .2 + .1 = .3. P( B1 ) = .1 + .3 + .2 = .6, P( B 2 ) = .2 + .1 + .1 = .4.
6.21 P( A1 ) = .4 + .2 = .6, P( A 2 ) = .3 + .1 = .4. P( B1 ) = .4 + .3 = .7, P( B 2 ) = .2 + .1 = .3.
P(A1 and B1 )
6.22 a P(A1 | B1 )
P(B1 )
.4 .57 .7
P(A 2 and B1 ) .3 .43 P(B1 ) .7
b P(A 2 | B1 )
c Yes. It is not a coincidence. Given B1 the events A1 and A 2 constitute the entire sample space.
6.23 a P(A1 | B 2 ) b P( B 2 | A 1 )
P(A1 and B 2 ) P( B 2 )
.2 .67 .3
P(A1 and B2 ) .2 .33 P(A1 ) .6
c One of the conditional probabilities would be greater than 1, which is not possible. 6.24 The events are not independent because P(A1 | B 2 ) P(A1 ) . 6.25 a P( A 1 or B1 ) = P(A1 ) P(B1 ) P(A1 and B1 ) = .6 + .7 - .4 = .9 b P( A 1 or B 2 ) = P(A1 ) P(B 2 ) P(A1 and B 2 ) = .6 + .3 - .2 = .7 c P( A 1 or A 2 ) = P(A1 ) P(A 2 ) = .6 + .4 = 1
6.26 P(A 1 | B1 )
P(A 1 and B1 ) .20 .25 ; P(A1 ) .20 .05 .25 ; the events are independent. P(B1 ) .20 .60
6.27 P(A1 | B1 )
P(A1 and B1 ) .20 .571 ; P(A1 ) .20 .60 .80 ; the events are dependent. P(B1 ) .20 .15
6.28 P( A1 ) = .15 + .25 = .40, P( A 2 ) = .20 + .25 = .45, P( A 3 ) = .10 + .05 = .15. P( B1 ) = .15 + 20 + .10 = .45, P( B 2 ) = .25 + .25 + .05 = .55.
6.29 a P( A 2 | B 2 ) =
P(A 2 and B 2 ) .25 .455 P( B 2 ) .55
b P( B 2 | A 2 ) =
P(A 2 and B 2 ) .25 .556 P( A 2 ) .45
c P( B1 | A 2 ) =
P(A 2 and B1 ) .20 .444 P( A 2 ) .45
6.30 a P( A 1 or A 2 ) = P( A 1 ) + P( A 2 ) = .40 + .45 = .85 b P( A 2 or B 2 ) = P( A 2 ) + P( B 2 ) – P( A 2 and B 2 ) = .45 + .55 - .25 = .75
c P( A 3 or B1 ) =P( A 3 ) + P( B1 ) – P( A 3 and B1 ) = .15 + .45 - .10 = .50
P(promoted and female ) .03 .20 P(female ) .03 .12
6.31 a P(promoted | female) =
b P(promoted | male) =
P(promoted and male ) .17 .20 P(male ) .17 .68
c No, because promotion and gender are independent events. 6.32 a P(debit card) = .04 + .18 + .14 = .36 b P(over $100 | credit card) =
P(credit card and over $100 .23 .49 P(credit card) .03 .21 .23
c P(credit card or debit card) = P(credit card) + P(debit card) = .47 + .36 = .83 6.33 a P(Less than high school) = .057 + .104 = .161 b P(college/university | female) =
c P(high school | male) =
P(college / university and female ) .095 .226 P(female ) .057 .136 .132 .095
P( high school and male ) .224 .622 P( male ) .136 .224
6.34 a P(He is a smoker) = .12 + .19 = .31 b P(He does not have lung disease) = .19 + .66 = .85 c P(He has lung disease | he is a smoker) =
P(he has lung disease and he is a smo ker) .12 .387 P(he is a smo ker) .31
d P(He has lung disease | he does not smoke) =
P(he has lung disease and he does not smoke ) .03 .043 P(he does not smoke ) .69
6.35 The events are dependent because P(he has lung disease) = .15, P(he has lung disease | he is a smoker) = .387
P(manual and math stats) .23 .390 P(math stats) .23 .36
6.36a. P(manual | math-stats) = b. P(computer) = .36 + .30 = .66
c. No, because P(manual) = .23 + .11 = .34, which is not equal to P(manual | math-stats). 6.37 a P(customer will return and good rating) =.35 b P(good rating | will return) =
P(good rating and will return ) .35 .35 .538 P( will return ) .02 .08 .35 .20 .65
c P(will return| good rating)
P(good rating and will return ) .35 .35 .714 P(good rating ) .35 .14 .49
d (a) is the joint probability and (b) and (c) are conditional probabilities 6.38 a P(ulcer) = .01 + .03 + .03 + .04 = .11 b P(ulcer | none) =
P(ulcer and none) .01 .01 .043 P(none) .01 .22 .23
c P(none | ulcer) =
P(ulcer and none) .01 .01 .091 P(ulcer) .01 .03 .03 .04 .11
d P(One, two, or more than two | ulcer) = 1
6.39 a P(Insufficient work | 25-54) =
P(ulcer and none) 1 .091 .909 P(ulcer)
P(Insufficie nt work and 25 54 ) .180 .252 P(25 54 ) .320 .180 .214
b P(65 and over) = .029 + .011 + .016 = .056 c P(65 and over |plant or company closed or moved) =
P(65 and over and plant or company closed or moved ) .029 .064 P(plant or company closed or moved ) .015 .320 ..089 .029
6.40 a P(remember) = .15 + .18 = .33 b P(remember | violent) =
P(remember and violent ) .15 .15 .30 P( violent ) .15 .35 .50
c Yes, the events are dependent.
6.41 a P(above average | murderer) =
P(above average and murderer ) .27 .27 .563 P(murderer ) .27 .21 .48
b No, because P(above average) = .27 + .24 = .51, which is not equal to P(above average testosterone | murderer). 6.42a P(Health insurance) = .167 +.209 +.225 +.177 = .778 b. P(Person 55-64 | No health insurance) =
P(Person 55 64 and No health insurance) .026 .026 .128 P(Person 55 64 ) .177 .026 .203
c. P(Person 25-34|No health insurance) =
P(Person 25 34 and No health insurance) .085 .085 .385 P( No health insurance) .085 .061 .049 .026 .221
6.43a
P(Violent crime and primary school) .393 .393 .673 P(Pr imary school) .393 .191 .584
b. P(No violent crime) = .191 + .010 + .007 + .015 = .223
6.44a
b.
P(Violent crime and enrollment less than 300 ) .159 .159 .636 P(Enrollment less than 300 ) .159 .091 .250
P(Violent crime and enrollment less than 300 ) .159 .159 .205 P(Violent crime ) .159 .221 .289 .108 .777
6.45 a P(new | overdue) =
b P(overdue | new) =
P(new and overdue) .06 .06 .103 P(overdue) .06 .52 .58
P(new and overdue) .06 .06 .316 P(new) .06 .13 .19
c Yes, because P(new) = .19 P(new | overdue) 6.46 a P(under 20) = .464 + .147 + .237 = .848 b P(retail) = .237 + .035 + .005 = .277 c P(20 to 99 | construction) =
P(20 to 99 and construction) .039 .039 .077 P(construction) .464 .039 .005 .508
6.47 a P(fully repaid) = .19 + .64 = .83 b P(fully repaid | under 400) =
P(fully repaid and under 400 ) .19 .19 .594 P(under 400 ) .19 .13 .32
c P(fully repaid | 400 or more) =
P(fully repaid and 400 or more ) .64 .64 .941 P(400 or more ) .64 .04 .68
d No, because P(fully repaid) P(fully repaid | under 400)
6.48 P(purchase | see ad) =
P(purchase and see ad) .18 .18 .30; P(purchase) = .18 + .12 = .30. The P(see ad) .18 .42 .60
events are independent and thus, the ads are not effective. 6.49 a P(unemployed | high school graduate) =
P( unemployed and high school graduate) .035 .035 .120 P( high school graduate) .257 .035 .292 b P(employed) = .075 + .2572 + .155 + .096 + .211 + .118 = .912 c P(advanced degree | unemployed) = P(advanced deg ree and unemployed) .004 .004 .044 P( unemployed) .015 .035 .016 .008 .012 .004 .090
d P(not a high school graduate) = .075 + .015 = .090
6.50 a P(bachelor’s degree | west) =
P( bachelor' s deg ree and west) .050 .050 .216 P( west) .032 .058 .044 .022 .050 .025 .231
b P(northeast | high school graduate) =
P( northeast and high school graduate) .062 .062 .198 P( high school graduate) .062 .075 .118 .058 .313
c P(south) = .053 + .118 + .062 + .032 + .067 + .036 = .368 d P(not south) = 1 – P(south) = 1−.368 = .632 6.51a. P(Some | White) = .356/.851 = .418 b. P(Very little or no | Black) = .06/.149 = .403 c. P(White | Some) = .356/.404 = .881 6.52a. P(Arthritis | Over 80) = .105/.140 = .750 b. P(No arthritis | 50-60) = .360/.400 = .900 c. P(60-70 | Arthritis) = .075/.292) = .257 6.53a. P(New Democrats | Men) = .044/.490 = .0898 b. P(Women | Liberal) = .224/.415 = .540 c. P(Conservative) = .255 + .215 = .470 6.54a. P(Married | Millennial) = .089/.331) = .269 b. P(Single, never married | Baby boomer) = .030/.310 = .0968 c. P(Married) = .089 + .223 + .201 = .513 d. P(Generation X | Living with not married) = .025/.064 =.391 6.55a. P(Trust) = .0896 + .1386 + .1944 + .0629 + .0144 = .4999 b. P(Distrust | Consistent conservative) = .0558/.0900 = .6200 c. P(Neither | Consistent liberal) = .0576/.1600 = .3600 d. P(Consistent liberal) = .0896 + .0096 + .0576 + .0032 = .1600 6.56a. P(Distrust | Mostly conservative) = .0680/.1700 = .4000 b. P(Neither | Mixed) = .1120/.3600 = .3111 c. P(Trust) = .0832 + .1056 + .1400 + .0442 + .0063 = .3797 d. P(Mostly conservative) = .0442 + .0680 + .0442 + .0136 = .1700 6.57a. P(Distrust | Consistent liberal) = .1296/.1600 = .8100
b. P(Trust | Mostly conservative) = .1224/.1700 = .7200 c. P(Neither | Consistent conservative) = .0045/.0900 = .0500 d. P(Consistent conservative) = .0792 + .0027 + .0045 + .0036 = .0900 6.58a. P(Distrust) = .0192 + .0242 + .0504 + .0561 + .0549 = .2048 b. P(Trust | Consistent conservative) = .0126/.0900 = .1400 c. P(Neither | Mostly liberal) = .0396/.2200 = .1800 d. P(Mixed) = .2196 + .0504 + .0612 + .0288 = .3600 6.59
6.60
6.61
6.62
6.63
6.64
a P(R and R) = .81 b P(L and L) = .01 c P(R and L) + P(L and R) = .09 + .09 = .18 d P(Rand L) + P(L and R) + P(R and R) = .09 + .09 + .81 = .99 6.65 a & b
c
0 right-handers
1
1 right-hander
3
2 right-handers
3
3 right-handers
1
d P(0 right-handers) = .001
P(1 right-hander) = 3(.009) = .027 P(2 right-handers) = 3(.081) = .243 P(3 right-handers) = .729 6.66a
b P(RR) = .8091 c P(LL) = .0091 d P(RL) + P(LR) = .0909 + .0909 = .1818 e P(RL) + P(LR) + P(RR) = .0909 + .0909 + .8091 = .9909 6.67a
P(0 right-handers) = (10/100)(9/99)(8/98) = .0007
P(1 right-hander) = 3(90/100)(10/99)(9/98) = .0249 P)2 right-handers) = 3(90/100)(89/99)(10/98) = .2478 P(3 right-handers) = (90/100)(89/99)(88/98) = .7265
6.68
a P(win both) = .28 b P(lose both) = .30 c P(win only one) = .12 + .30 = .42 6.69
P(sale) = .04
6.70
P(D) = .02 + .018 = .038 6.71
P(Same party affiliation) = P(DD) + P(RR) + P(OO) = .1936 + .1369 + .0361.3666
6.72
Diversity index = .12 + .04 + .12 + .0075 + .04 + .0075 = .335 6.73
P(heart attack) = .0504 + .0792 = .1296
6.74
P(pass) = .228 + .243 + .227 = .698 6.75
P(good ) = .3132 + .0416 = .3548
6.76
P(myopic) = .1008 + .1512 = .2520
6.77
P(does not have to be discarded) = .1848 + .78 = .9648 6.78 Let A = mutual fund outperforms the market in the first year B = mutual outperforms the market in the second year P(A and B) = P(A)P(B | A) = (.15)(.22) = .033 6.79 Let A = DJIA increase and B = NASDAQ increase P(A) = .60 and P(B | A) = .77 P(A and B) = P(A)P(B | A) = (.60)(.77) = .462
6.80 Define the events: M: The main control will fail. B1: The first backup will fail. B2: The second backup will fail
The probability that the plane will crash is P(M and B1 and B2) = [P(M)][ P(B1)][ P(B2)] = (.0001) (.01) (.01) = .00000001 We have assumed that the 3 systems will fail independently of one another. 6.81 P( wireless Web user uses it primarily for e-mail) = .69 P(3 wireless Web users use it primarily for e-mail) = (.69)(.69)(.69) = .3285 6.82
P(Increase) = .05 + .5625 = .6125 6.83 Number saying leaving is a bad thing = 630(.62) + 590(.74) + 480(.57) = 1100.8 P(bad thing) = 1100.8/(630 + 590 + 480) = .6475 6.84 Number disapproving = 385(.94) + 420(.70) + 475(.66) = 820.5 P(Disapprove) = 820.5/(385 + 420 + 475) = .6408 6.85 P(Student debt) = (.32)(.45) + (.15)(.39) + (.53)(.27) = .3456 6.86 P(pass) = (.38)(.79) + (.41)(.74) + (.13)(.68) + (.08)(.57) = .7376 6.87 P(A and B) = .36, P(B) = .36 + .07 = .43 P(A | B) =
P(A and B) .36 .837 P(B) .43
6.88 P(A and B) = .32, P(AC and B) = .14, P(B) = .46, P(B C) = .54
a P(A | B) =
P(A and B) .32 .696 P(B) .46
b P(AC | B) =
P(A C and B) .14 .304 P(B) .46
c P(A and BC) = .48; P(A | BC ) =
P(A and B C )
d P(AC and BC) = .06; P(AC | BC) =
C
P( B )
.48 .889 .54
P(A C and B C ) P( B C )
.06 .111 .54
6.89
P(B) = .4940 + .0115 = .5055 P(A | B) =
P(A and B) .4940 .9773 P(B) .5055
6.90 P(F | D) =
P(F and D) .020 .526 P(D) .038
6.91 Define events: A = crash with fatality, B = BAC is greater than .09) P(A) = .01, P(B | A) = .084, P(B) = .12 P(A and B) = (.01)(.084) = .00084 P(A | B) =
P(A and B) .00084 .007 P(B) .12
6.92 P(CFA I | passed) =
P(CFA I and passed) .228 .327 P(passed) .698
6.93 Define events: A = heart attack, B = periodontal disease P(A) = .10, P(B | A) = .85, P(B | AC ) = .29
P(B ) = .085 + .261 = .346 P(A | B) =
P(A and B) .085 .246 P(B) .346
6.94 P(A) = .40, P(B | A) = .85, P(B | AC ) = .29
P(B ) = .34 + .174 = .514 P(A | B) =
P(A and B) .34 .661 P(B) .514
6.95 Define events: A = smoke, B1 = did not finish high school, B 2 = high school graduate, B 3 = some college, no degree, B 4 = completed a degree
P(A | B1 ) = .40, P(A | B 2 ) = .34, P(A | B 3 ) = .24, P(A | B 4 ) = .14 From Exercise 6.45: P( B1 ) = .1055, P( B 2 ) = .3236, P( B 3 ) = .1847, P( B 4 ) = .3862
P(A) = .0422 + .1100 + .0443 + .0541 = .2506 P( B 4 | A) = .0541/.2506 = .2159
6.96 Define events: A, B, C = airlines A, B, and C, D = on time P(A) = .50, P(B) = .30, P(C) = .20, P(D | A) = .80, P(D | B) = .65, P(D | C) = .40
P(D) = .40 + .195 + .08 = .675
P(A | D) =
P(A and D) .40 .593 P(D) .675
6.97 Define events: A = win series, B = win first game P(A) = .60, P(B | A) = .70, P(B | AC ) =.25
P(BC ) = .18 + .30 = .48 P(A | BC ) =
P(A and B C ) C
P( B )
.18 .375 .48
6.98
P(PT) = .28 + .052 = .332 P(R | PT) =
P(R and PT ) .28 .843 P(PT ) .332
6.99
P(PT) = .0046 + .0269 = .0315 P(H | PT) =
P(H and PT ) .0046 .1460 P(PT ) .0315
6.100 Sensitivity = P(PT | H) = .920 Specificity = P(NT | H C ) = .973 Positive predictive value = P(H | PT) = .1460 Negative predictive value = P H C | NT) =
6.101
P(H C and NT ) .9681 .9681 .9996 P( NT ) .0004 .9681 .9685
P(PT) = .0164 + .6233 = .6397 P(NT) = .0036 + .3567 = .3603 P(C | PT) =
P(C and PT ) .0164 .0256 P(PT ) .6397
P(C | NT) =
P(C and NT ) .0036 .0010 P( NT ) .3603
6.102 P(Light| Myopic) =
P( Light and Myopic ) .1008 .4000 .2520 P( Myopic )
6.103 P(Italy | Bad thing) = Number of Italy and Bad thing / Number of bad thing) = 273.6/1100.8 = .2485 6.104 P(Greece | Disapprove) = Number of Greece and Disapprove / Number Disapprove) = 238.7/820.25 = .2910 6.105 P(Managerial/Professional | Student debt) = P(Managerial/Professional and Student debt) / P(Student Debt) = .1440/.3456 = .4167 6.106 P(BBA | Pass) = P(BBA and Pass) / P(Pass) = .3034/.7376 = .4113 6.107a P(Good Day) = Number of Good Day / Number of Respondents = (525(.30) + 650(.17) + 390(.41)) / 1565 = .2734 b P(United States | Typical Day) = Number of United States and Typical Day) / Number Typical Day = 191.1/1042.1 = .1834 6.108 a P(Marketing A) = .053 + .237 = .290 b P(Marketing A | Statistics not A) =
P(Marketing A and Statistics not A) .23 .237 .290 P(Statistics not A) .237 .580 .817
c Yes, the probabilities in Parts a and b are the same. 6.109 Define events: A = win contract A and B = win contract B
a P(A and B) = .12 b P(A and BC) + P(AC and B ) = .18 + .14 = .32 c P(A and B) + P(A and BC ) + P(AC and B ) = .12 + .18 + .14 = .44 6.110 a P(second) = .05 + .14 = .19 b P(successful | –8 or less) =
P(successfuland 8 or less) .15 .15 .517 P(8 or less) .15 .14 .29
c No, because P(successful) = .66 + .15 = .81, which is not equal to P(successful | –8 or less) . 6.111 Define events: A = woman, B = drug is effective
P(B) = .528 + .221 = .749
6.112 P(AC | B) =
P(A C and B) .221 .295 P(B) .749
6.113 P(Idle roughly) = P(at least one spark plug malfunctions) = 1– P(all function) = 1 – (.90 4 ) = 1-.6561 = .3439
6.114
P(no sale) = .65 + .175 = .825 6.115 a P(pass) = .86 + .03 = .89 b P(pass | miss 5 or more classes) =
P(pass and miss 5 or more classes) .03 .03 .250 P(miss 5 or more classes) .09 .03 .12
c P(pass | miss less than 5 classes) =
P(pass and miss less than 5 classes) .86 .86 .977 . P(miss less than 5 classes) .86 .02 .88
d No since P(pass) P(pass | miss 5 or more classes) 6.116 Define events: R = reoffend, D = detained
a P(D) = P(R and D) + P(R C and D) = .1107 + .2263 = .3370 P(R| D) =
P(R and D) .1107 .3285 P ( D) .3370
b P(D C ) = P(R and D C ) + P(R C and D C ) = .1593 + .5037 = .6630 P(R| D C ) =
P(R and D C ) P( D C )
.1593 .2403 .6630
6.117 a P(excellent) = .27 + .22 = .49 b P(excellent | man) =
P(man and excellent ) .22 .22 .44 P(man ) .22 .10 .12 .06 .50
c P(man | excellent) =
P(man and excellent ) .22 .22 .449 P(excellent ) .27 .22 .49
d No, since P(excellent) P(excellent | man)
6.118
P(R) = .0176 + .5888 = .6064 P(S | R) =
P(S and R ) .5888 .9710 P( R ) .6064
6.119 Define events: A1 = Low-income earner, A 2 = medium-income earner, A 3 = high-income earner, B = die of a heart attack, BC survive a heart attack
P(BC ) = .1848 + .4459 + .2790 = .9097 P( A1 | BC ) =
P(A1 and B C ) C
P( B )
.1848 .2031 .9097
6.120 Define the events: A1 = envelope containing two Maui brochures is selected, A 2 = envelope containing two Oahu brochures is selected, A 3 = envelope containing one Maui and one Oahu brochures is selected. B = a Maui brochure is removed from the selected envelope.
P(B) = 1/3 + 0 + 1/6 = 1/2 P( A1 | B) =
P(A 1 and B) 1 / 3 2/3 P(B) 1/ 2
6.121 Define events: A = purchase extended warranty, B = regular price a P(A | B) =
P(A and B) .21 .21 .2692 P(B) .21 .57 .78
b P(A) = .21 + .14 = .35 c No, because P(A) P(A | B) 6.122 Define events: A = company fail, B = predict bankruptcy
P(B) = .068 + .2392 = .3072 P(A | B) =
P(A and B) .068 .2214 P(B) .3072
6.123 Define events: A = job security is an important issue, B = pension benefits is an important issue P(A) = .74, P(B) = .65, P(A | B) = .60 a P(A and B) = P(B)P(A | B) = (.65)(.60) = .39 b P(A or B) = .74 + .65 – .39 = 1 6.124 Probabilities of outcomes: P(HH) = .25, P(HT) = .25, P(TH) = .25, P(TT) = .25 P(TT | HH is not possible) = .25/(.25 + .25 + .25) = .333 6.125 P(T) = .5 Case 6.1 1. P(Curtain A) = 1/3, P(Curtain B) = 1/3 2. P(Curtain A) = 1/3, P(Curtain B) = 2/3 Switch to Curtain B and double your probability of winning the car.
Case 6.2 Probability
Bases
Probability
Joint
of outcome
Occupied
Outs
of scoring
Probability
1
.75
2nd
1
.42
.3150
2
.10
1st
1
.26
.0260
3
.10
none
2
.07
.0070
4
.05
1st and 2nd
0
.59
.0295
Outcome
P(scoring) = .3775 Because the probability of scoring with a runner on first base with no outs (.39) is greater than the probability of scoring after bunting (.3775) you should not bunt. Case 6.3 0 outs: Probability of scoring any runs from first base = .39 Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (.68)(.57) = .3876 Decision: Do not attempt to steal. 1 out: Probability of scoring any runs from first base = .26 Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (.68) (.42) = .2856 Decision: Attempt to steal. 2 outs: Probability of scoring any runs from first base = .13 Probability of scoring from second base = probability of successful steal probability of scoring any runs from second base = (.68) (.24) = .1632 Decision: Attempt to steal.
Case 6.4
Age 25: P(D) = 1/1,300 P(D and PT) = (1/1,300)(.624) = .00048 P(D and NT) = (1/1,300)(.376) = .00029 P( D C and PT) = (1,299/1,300)(.04) = .03997 P( D C and NT) = (1,299/1,300)(.96) = .95926 P(PT) = .00048 + .03997 = .04045 P(NT) = .00029 + .95926 = .95955 P(D | PT) = .00048/.04045 = .01187 P(D | NT) = .00029/.95955 = .00030 Age 30: P(D) = 1/900 P(D and PT) = (1/900)(.710) = .00079 P(D and NT) = (1/900)(.290) = .00032 P( D C and PT) = (899/900)(.082) = .08190 P( D C and NT) = (899/900)(.918) = .91698 P(PT) = .00079 + .08190 = .08269 P(NT) = .00032 + .91698 = .91730 P(D | PT) = .00079/.08269 = .00955 P(D | NT) = .00032/.91730 = .00035 Age 35: P(D) = 1/350 P(D and PT) = (1/350)(.731) = .00209
P(D and NT) = (1/350)(.269) = .00077 P( D C and PT) = (349/350)(.178) = .17749 P( D C and NT) = (349/350)(.822) = .81965 P(PT) = .00209 + .17749 = .17958 P(NT) = .00077 + .81965 = .82042 P(D | PT) = .00209/.17958 = .01163 P(D | NT) = .00077/.82042 = .00094 Age 40: P(D) = 1/100 P(D and PT) = (1/100)(.971) = .00971 P(D and NT) = (1/100)(.029) = .00029 P( D C and PT) = (99/100)(.343) = .33957 P( D C and NT) = (99/100)(.657) = .65043 P(PT) = .00971 + .33957 = .34928 P(NT) = .00029 + .65043 = .65072 P(D | PT) = .00971/.34928 = .02780 P(D | NT) = .00029/.65072 = .00045 Age 45: P(D) = 1/25 P(D and PT) = (1/25)(.971) = .03884 P(D and NT) = (1/25)(.029) = .00116 P( D C and PT) = (24/25)(.343) = .32928 P( D C and NT) = (24/25)(.657) = .63072 P(PT) = .03884 + .32928 = .36812 P(NT) = .00116 + .63072 = .63188 P(D | PT) = .03884/.36812 = .10551 P(D | NT) = .00116/.63188 = .00184 Age 49: P(D) = 1/12 P(D and PT) = (1/12)(.971) = .08092 P(D and NT) = (1/12)(.029) = .00242 P( D C and PT) = (11/12)(.343) = .31442 P( D C and NT) = (11/12)(.657) = .60255 P(PT) = .08092 + .31442 = .39533 P(NT) = .00242 + .60255 = .60467 P(D | PT) = .08092/.39533 = .20468 P(D | NT) = .00242/.60467 = .00400
Case 6.5 The probability that 23 people have different birthdays is .4927. The probability that at least two people have the same birthday is 1 − .4927 = .5073.
Chapter 7 7.1 a 0, 1, 2, … b Yes, we can identify the first value (0), the second (1), and so on. c It is finite, because the number of cars is finite. d The variable is discrete because it is countable. 7.2 a any value between 0 and several hundred miles b No, because we cannot identify the second value or any other value larger than 0. c No, uncountable means infinite. d The variable is continuous. 7.3 a The values in cents are 0 ,1 ,2, … b Yes, because we can identify the first ,second, etc. c Yes, it is finite because students cannot earn an infinite amount of money. d Technically, the variable is discrete. 7.4 a 0, 1, 2, …, 100 b Yes. c Yes, there are 101 values. d The variable is discrete because it is countable. 7.5 a No the sum of probabilities is not equal to 1. b Yes, because the probabilities lie between 0 and 1 and sum to 1. c No, because the probabilities do not sum to 1. 7.6 P(x) = 1/6 for x = 1, 2, 3, 4, 5, and 6 7.7 a
x 0 1 2 3 4 5
P(x) 1218/101,501 = .012 32,379/101,501 = .319 37,961/101,501 = .374 19,387/101,501 = .191 7714/101,501 = .076 2842/101,501 = .028
b (i) P(X 2) = P(0) + P(1) + P(2) = .012 + .319 + .374 = .705 (ii) P(X > 2) = P(3) + P(4) + P(5) = .191 + .076 + .028 = .295 (iii) P(X 4) = P(4) + P(5) = .076 + .028 = .104
187
7.8 a P(2 X 5) = P(2) + P(3) + P(4) + P(5) = .310 + .340 + .220 + .080 = .950 P(X > 5) = P(6) + P(7) = .019 + .001 = .020 P(X < 4) = P(0) + P(1) + P(2) + P(3) = .005 + .025 + .310 + .340 = .680 b. b. E(X) =
xP (x) = 0(.005) + 1(.025) + 2(.310) + 4(.340) +5(.080) + 6(.019) + 7(.001) =
3.066 c. 2 = V(X) =
(x ) P(x) = (0–3.066) (.005) + (1–3.066) (.025) + (2–3.066) (.310) 2
2
2
2
2
2
2
2
+ (3–3.066) (.340) + (4–3.066) (.220) + (5–3.066) (.080) + (6–3.066) (.019) 2
+ (7–3.066) (.001) = 1.178
=
2 1.178 = 1.085
7.9 P(0) = P(1) = P(2) = . . . = P(10) = 1/11 = .091 7.10 a P(X > 0) = P(2) + P(6) + P(8) = .3 + .4 + .1 = .8 b P(X 1) = P(2) + P(6) + P(8) = .3 + .4 + .1 = .8 c P(X 2) = P(2) + P(6) + P(8) = .3 + .4 + .1 = .8 d P(2 X 5) = P(2) = .3 7.11a P(3 X 6) = P(3) + P(4) + P(5) + P(6) = .04 + .28+ .42 + .21 = .95 b. P(X > 6) = P(X 7) = P(7) + P(8) = .02 + .02 = .04 c. P(X < 3) = P(X 2) = P(0) + P(1) + P(2) = 0 + 0 + .01 = .01
7.12 P(Losing 6 in a row) = .5 6 = .0156
7.13 a P(X < 2) = P(0) + P(1) = .05 + .43 = .48 b P(X > 1) = P(2) + P(3) = .31 + .21 = .52
188
7.14
a P(HH) = .25 b P(HT) = .25 c P(TH) = .25 d P(TT) = .25 7.15 a P(0 heads) = P(TT) = .25 b P(1 head) = P(HT) + P(TH) = .25 + .25 = .50 c P(2 heads) = P(HH) = .25 d P(at least 1 head) = P(1 head) + P(2 heads) = .50 + .25 = .75 7.16
189
7.17 a P(2 heads) = P(HHT) + P(HTH) + P(THH) = .125 + .125 + .125 = .375 b P(1 heads) = P(HTT) + P(THT) = P(TTH) = .125 + .125 + .125 = .375 c P(at least 1 head) = P(1 head) + P(2 heads) + P(3 heads) = .375 + .375 + .125 = .875 d P(at least 2 heads) = P(2 heads) + P(3 heads) = .375 + .125 = .500
7.18a. = E(X) =
2 = V(X) =
xP (x) = –2(.59) +5(.15) + 7(.25) +8(.01) = 1.40
(x ) P(x) = (–2–1.4) (.59) + (5–1.4) (.15) + (7–1.4) (.25) + (8–1.4) (.01) 2
2
2
2
2
= 17.04 b.
x
–2
5
7
8
y
–10
25
35
40
P(y)
.59
.15
.25
.01
yP( y) = –10(.59) + 25(.15) + 35(.25) + 40(.01) = 7.00 V(Y) = ( y ) P( y) = (–10–7.00) (.59) + (25–7.00) (.15) + (35–7.00) (.25)
c. E(Y) =
2
2
2
2
2
+ (40–7.00) (.01) = 426.00 d. E(Y) = E(5X) = 5E(X) = 5(1.4) = 7.00 V(Y) = V(5X) = 5 2 V(X) = 25(17.04) = 426.00.
7.19a = E(X) =
2 = V(X) =
xP (x) = 0(.4) + 1(.3) + 2(.2) + 3(.1) = 1.0
(x ) P(x) = (0–1.0) (.4) + (1–1.0) (.3) + (2–1.0) (.2) + (3–1.0) (.1) 2
2
2
2
2
= 1.0
= b.
2 1.0 = 1.0
x
0
1
2
3
y
2
5
8
11
P(y)
.4
.3
.2
.1
yP( y) = 2(.4) + 5(.3) + 8(.2) + 11(.1) = 5.0 =V(Y) = ( y ) P( y) = (2 – 5) (.4) + (5 – 5) (.3) + (8 – 5) (.2) + (11 – 5) (.1) = 9.0
c. E(Y) = 2
=
2
2
2
2 9.0 = 3.0
d. E(Y) = E(3X + 2) = 3E(X) + 2 = 3(1) + 2 = 5.0
2 = V(Y) = V(3X + 2) = V(3X) = 3 2 V(X) = 9(1) = 9.0.
=
2 9.0 = 3.0
The parameters are identical. 190
2
2
7.20a. P(X 2) = P(2) + P(3) = .4 + .2 = .6
xP (x) = 0(.1) + 1(.3) + 2(.4) + 3(.2) = 1.7 = V(X) = ( x ) P( x ) = (0–1.7) (.1) + (1–1.7) (.3) + (2–1.7) (.4) + (3–1.7) (.2) = .81 b. = E(X) =
2
2
2
2
2
2
7.21 E(Profit) = E(5X) = 3E(X) = 3(1.7) = 5.1 V(Profit) = V(3X) = 3 2 V(X) = 9(.81) = 7.29 7.22 a P(X > 4) = P(5) + P(6) + P(7) = .20 + .10 + .10 = .40 b P(X 2) = 1– P(X 1) = 1 – P(1) = 1 – .05 = .95
7.23 = E(X) =
2 = V(X) =
xP (x) = 1(.05) + 2(.15) + 3(.15) + 4(.25) + 5(.20) + 6(.10) + 7(.10) = 4.1
(x ) P(x) = (1–4.1) (.05) + (2–4.1) (.15) + (3–4.1) (.15) + (4–4.1) (.25) 2
2
2
2
2
2
2
2
+ (5–4.1) (.20) + (6–4.1) (.10) + (7–4.1) (.10) = 2.69 7.24 Y = .25X; E(Y) = .25E(X) = .25(4.1) = 1.025 2
V(Y) = V(.25X) = (.25) (2.69) = .168 7.25 a. x
1
2
3
4
5
6
7
y
.25
.50
.75
1.00
1.25
1.50
1.75
P(y)
.05
.15
.15
.25
.20
.10
.10
b. E(Y) =
yP( y) = .25(.05) + .50(.15) + .75(.15) +1.00(.25) + 1.25(.20) + 1.50(.10) + 1.75(.10)
= 1.025 V(Y) =
(y ) P(y) = (.25–1.025) (.05) + (.50–1.025) (.15) + (.75–1.025) (.15) 2
2
2
2
2
2
2
+ (1.00–1.025) (.25) + (1.25–1.025) (.20) + (1.50–1.025) (.10) + (1.75–1.1025) (.10) = .168 c. The answers are identical. 7.26 a P(4) = .06 b P(8) = 0 c P(0) = .35 d P(X 1) = 1 – P(0) = 1 – .35 = .65 7.27 a P(X 20) = P(20) + P(25) + P(30) + P(40) + P(50) + P(75) + P(100) = .08 + .05 + .04 + .04 + .03 + .03 + .01 = .28 191
2
b P(X = 60) = 0 c P(X > 50) = P(75) + P(100) = .03 + .01 = .04 d P(X > 100) = 0 7.28 a P(X = 3) = P(3) = .21 b P(X 5) = P(5) + P(6) + P(7) + P(8) = .12 + .08 + .06 + .05 = .31 c P(5 X 7) = P(5) + P(6) + P(7) = .12 + .08 + .06 = .26 7.29 a P(X > 1) = P(2) + P(3) + P(4) = .17 + .06 + .01 = .24 b P(X = 0) = .45 c P(1 X 3) = P(1) + P(2) + P(3) = .31 + .17 + .06 = .54
7.30 = E(X) =
2 = V(X) =
xP (x) = 0(.04) + 1(.19) + 2(.22) + 3(.28) + 4(.12) + 5(.09) + 6(.06) = 2.76
(x ) P(x) = (1–2.76) (.04) + (2–2.76) (.19) + (3–2.76) (.28) 2
2
2
2
2
2
2
+ (4–2.76) (.12) + (5–2.76) (.09) + (6–2.76) (.06) = 2.302
=
2 2.302 = 1.517
7.31 Y = 10X; E(Y) = E(10X) = 10E(X) = 10(2.76) = 27.6 2
V(Y) = V(10X) = 10 V(X) =100(2.302) = 230.2
=
2 230 .2 = 15.17
7.32 = E(X) =
xP (x) = 1(.24) + 2(.18) + 3(.13) + 4(.10) + 5(.07) + 6(.04) + 7(.04) + 8(.20) =
3.86
2 = V(X) =
(x ) P(x) = (1–3.86) (.24) + (2–3.86) (.18) + (3–3.86) (.13) + (4–3.86) 2
2
2
2
(.10) 2
2
2
2
+ (5–3.86) (.07) +(6–3.86) (.04) + (7–3.86) (.04) + (8–3.86) (.20) = 6.78
=
2 6.78 = 2.60
7.33 Revenue = 2.50X; E(Revenue) = E(2.50X) = 2.50E(X) = 2.50(3.86) = 9.65 V(Revenue) = V(2.50X) = 2.50 2 (V(X) = 6.25(6.78) = 42.38
=
2 42 .38 = 6.51
7.34 E(Value of coin) = 400(.40) + 900(.30) + 100(.30) = 460. Take the $500. 192
2
7.35 = E(X) =
2 = V(X) =
xP (x) = 0(.10) + 1(.20) + 2(.25) + 3(.25) + 4(.20) = 2.25
(x ) P(x) = (0–2.25) (.10) + (1–2.25) (.20) + (2–2.25) (.25) + (3–2.25) 2
2
2
2
2
(.13) 2
+ (4–2.25) (.20) = 1.59
=
2 1.59 = 1.26
7.36 E(damage costs) = .01(400) + .02(200) + .10(100) + .87(0) = 18. The owner should pay up to $18 for the device.
7.37 E(X) =
xP (x) = 1,000,000(1/10,000,000) + 200,000(1/1,000,000) + 50,000(1/500,000)
+ 10,000(1/50,000) + 1,000(1/10,000) = .1 + .2 + .1 + .2 + .1 = .7 Expected payoff = 70 cents.
7.38 = E(X) =
2 = V(X) =
xP (x) = 1(.05) + 2(.12) + 3(.20) + 4(.30) + 5(.15) + 6(.10) + 7 (.08) = 4.00
(x ) P(x) = (1–4.0) (.05) + (2–4.0) (.12) + (3–4.0) (.20) + (4–4.0) (.30) 2
2
2
2
2
2
2
2
+ (5–4.0) (.15) +(6–4.0) (.10) + (7–4.0) (.08) = 2.40 7.39 Y = .25X; E(Y) = E(.25X) = .25E(X) = .25(4.0) = 1.0 2
V(Y) = V(.25X) = (.25) V(X) =.0625(2.40) = .15
7.40 = E(X) =
xP (x) = 0(.10) + 1(.25) + 2(.40) + 3(.20) + 4(.05) = 1.85
7.41 Profit = 4X; Expected profit = E(4X) = 4E(X) = 4(1.85) = $7.40 7.42a P(X > 4) = P(5) + P(6) + P(7) + P(8) = .04 + .15 + .03 + .10 = .32 b P(X < 5) = P(1) + P(2) + P(3) + P(4) = .03 + .32 + .05 + .28 = .68 P(4≤ X ≤ 6) = P(4) + P(5) + P(6) = .28 + .04 + .15 = .47
7.43 = E(X) =
xP (x) = 1(.03) + 2(.32) + 3(.05) + 4(.28) + 5(.04) + 6 (.15) + 7 (.03) + 8(.10)
= 4.05
193
2 = V(X) =
(x ) P(x) = (1–4.05) (.03) + (2–4.05) (.32) + (3–4.05) (.05) + (4–4.05) 2
2
2
2
2
2
2
(.28) + (5–4.05) (.04) +(6–4.05) (.15) + (7–4.05) (.03) + (8–4.05)2 (.10) = 4.11
=
2 4.11 = 2.03
7.44a P(X > 3) = P(4) + P(5) + P(6) = .02 + .01 + .01 = .04 b P(0) = .78 c P(3 ≤ X ≤ 5) = P(3) + P(4) + P(5) = .03 + .02 + .01 = .06 7.45 a
b
x
P(x)
1
.6
2
.4
y
P(y)
1
.6
2
.4
xP (x) = 1(.6) + 2(.4) = 1.4 = V(X) = ( x ) P( x ) = (1–1.4) (.6) + (2–1.4) (.4) = .24
c = E(X) =
2
2
2
2
d = 1.4, 2 = .24
7.46 a
xyP (x, y) = (1)(1)(.5) + (1)(2)(.1) + (2)(1)(.1) + (2)(2)(.3) = 2.1 all x all y
COV(X, Y) =
xyP (x, y) – = 2.1 – (1.4)(1.4) = .14 x
y
all x all y
x 2x .24 = .49, y 2y .24 = .49
.14 COV (X, Y) = = .58 (.49 )(. 49 ) xy
7.47 E(X + Y) = E(X) + E(Y) = 1.4 + 1.4 = 2.8 V(X + Y) = V(X) + V(Y) + 2COV(X, Y) = .24 + .24 + 2(.14) = .76 7.48 a
x+y
P(x + y)
2
.5
194
2
3
.2
4
.3
(x y)P(x y) = 2(.5) + 3(.2) + 4(.3) = 2.8 ] P( x y) = (2–2.8) (.5) + (3–2.8) (.2) + (4–2.8) (.3) = = V(X+Y) = [( x y)
b x y = E(X+Y) = 2x y
xy
2
2
2
.76 c Yes 7.49 a
b
x
P(x)
1
.4
2
.6
y
P(y)
1
.7
2
.3
xP (x) = 1(.4) + 2(.6) = 1.6 = V(X) = ( x ) P( x ) = (1–1.6) (.4) + (2–1.6) (.6) = .24 d = E(Y) = yP( y) = 1(.7) + 2(.3) = 1.3 = V(Y) = ( y ) P( y) = (1–1.3) (.7) + (2–1.3) (.3) = .21 c = E(X) = 2
2
2
2
2
2
2
2
7.50 a
xyP (x, y) = (1)(1)(.28) + (1)(2)(.12) + (2)(1)(.42) + (2)(2)(.18) = 2.08 all x all y
COV(X, Y) =
xyP (x, y) – = 2.08 – (1.6)(1.3) = 0 x
y
all x all y
x 2x .24 = .49, y 2y .21 = .46
0 COV (X, Y) = =0 (.49 )(. 46 ) xy
7.51 E(X + Y) = E(X) + E(Y) = 1.6 + 1.3 = 2.9 V(X + Y) = V(X) + V(Y) + 2COV(X, Y) = .24 + .21 + 2(0) = .45 7.52 a x + y
P(x + y)
2
.28
3
.54
4
.18
b x y = E(X+Y) =
(x y)P(x y) = 2(.28) + 3(.54) + 4(.18) = 2.9 195
2
2x y = V(X+Y) =
[( x y)
xy ]
2
2
2
= .45 c Yes 7.53 a x
P(x)
y
P(y)
1
.7
1
.6
2
.2
2
.4
3
.1
xP (x) = 1(.7) + 2(.2) + 3(.1) = 1.4 = V(X) = ( x ) P( x ) = (1–1.4) (.7) + (2–1.4) (.2) + (3–1.4) (.1) = .44 = E(Y) = yP( y) = 1(.6) + 2(.4) = 1.4 = V(Y) = ( y ) P( y) = (1–1.4) (.6) + (2–1.4) (.4) = .24 xyP (x, y) = (1)(1)(.42) + (1)(2)(.28) + (2)(1)(.12) + (2)(2)(.08) + (3)(1)(.06) +
b x = E(X) = 2
2
2
2
2
2
2
y
2
all x all y
(3)(2)(.04) = 1.96 COV(X, Y) =
xyP (x, y) – = 1.94 – (1.4)(1.4) = 0 x
y
all x all y
x 2x .44 = .66, y 2y .24 = .49
c
0 COV (X, Y) = =0 xy (.66 )(. 49 )
x+y
P(x + y)
2
.42
3
.40
4
.14
5
.04
7.54
7.55
x y
0
1
2
1
.42
.21
.07
2
.18
.09
.03
2
P( x y) = (2–2.9) (.28) + (3–2.9) (.54) + (4–2.9) (.18)
x
196
2
y
0
1
1
.04
.16
2
.08
.32
3
.08
.32
7.56 a
b
Refrigerators, x
P(x)
0
.22
1
.49
2
.29
Stoves, y
P(y)
0
.34
1
.39
2
.27
xP (x) = 0(.22) + 1(.49) + 2(.29) = 1.07 = V(X) = ( x ) P( x ) = (0–1.07) (.22) + (1–1.07) (.49) + (2–1.07) (.29) = .505 d = E(Y) = yP( y) = 0(.34) + 1(.39) + 2(.27) = .93 = V(Y) = ( y ) P( y) = (0–.93) (.34) + (1–.93) (.39) + (2–.93) (.27) = .605 e xyP ( x, y) = (0)(0)(.08) + (0)(1)(.09) + (0)(2)(.05) + (1)(0)(.14) + (1)(1)(.17) c x = E(X) =
2
2
2
2
2
y
2
2
2
2
all x all y
+ (1)(2)(.18) + (2)(0)(.12) + (2)(1)(.13) + (2)(2)(.04) = .95 COV(X, Y) =
xyP (x, y) – = .95 – (1.07)(.93) = –.045 x
y
all x all y
x 2x .505 = .711, y 2y .605 = .778
.045 COV (X, Y) = = –.081 xy (.711)(. 778 )
7.57 a
Bottles, x
P(x)
0
.72
1
.28
Cartons, y
P(y)
0
.81
1
.19
b
xP (x) = 0(.72) + 1(.28) = .28 = V(X) = ( x ) P( x ) = (0–.28) (.72) + (1–.28) (.28) = .202
c x = E(X) = 2
2
2
2
197
2
yP( y) = 0(.81) + 1(.19) = .19 = V(Y) = ( y ) P( y) = (0–.19) (.81) + (1–.19) (.19) = .154 e xyP ( x, y) = (0)(0)(.63) + (0)(1)(.09) + (1)(0)(.18) + (1)(1)(.10) = .100
d y = E(Y) =
2
2
2
2
all x all y
COV(X, Y) =
xyP (x, y) – = .100 – (.28)(.19) = .0468 x
y
all _ x all _ y
x 2x .202 = .449, y 2y .154 = .392
.0468 COV (X, Y) = = .266 (.449 )(. 392 ) xy
7.58 a P(X = 1 | Y = 0) = P(X =1 and Y = 0)/P(Y = 0) = .14/.34 = .412 b P(Y = 0 | X = 1) = P(X =1 and Y = 0)/P(X = 1) = .14/.49 = .286 c P(X = 2 | Y = 2) = P(X =2 and Y = 2)/P(Y = 2) = .04/.27 = .148 7.59 a P(CMD = 0 and SD = 2) = .06 b P(CMD = 2 and SD = 0) = 0 c P(CMD ≥ 1 and SD ≥ 1) = .07 + .01 + .10 + .05 + .04 + .02 = .29 7.60 a P(CMD = 1 and SD = 2) = .10 b. P(SD = 2 | CMD = 1) = .10/.24 = .4167 c P(CMD = 1 | SD = 2) = .10/.21 = .4762 7.61 a P(CMD = 0) = .68, P(CMD = 1) = .24, P(CMD = 2) = .08
xP (x) = 0(.68) + 1(.24) + 2(.08) = .40 = V(X) = ( x ) P( x ) = (0–.40) (.68) + (1–.40) (.24) + (2-.40) (.08) = .40
b = E(X) =
2
2
=
2
2
2
2 .40 = .632
7.62a. P(SD = 0) = .45, P(SD=1) = .23, P(SD = 2) = .21, P(SD = 3) = .11
xP (x) = 0(.45) + 1(.23) + 2(.21) + 3(.11) = .98 = V(X) = ( x ) P( x ) = (0–.98) (.45) + (1–.98) (.23) + (2-.98) (.21) + (3-.98) (.11)
b. = E(X) =
2
2
2
2
= 1.100
=
2 1.100 = 1.049
198
2
2
7.63a P(Home Team > Visiting Team) = .11 + .09 + .10 + .05 + .02 + .01 = .38 b P(Tie) = .14 + .10 + .04 + 0 = .28 c P(Home Team < Visiting Team) = .12 + .09 + .03 + .07 + .02 + .01 = .34 7.64a. P(HT = 0) = .38, P(HT = 1) =.30, P(HT = 2) = .19, P(HT = 3) = .13
xP (x) = 0(.38) + 1(.30) + 2(.19) + 3(.12) = 1.07 = V(X) = ( x ) P( x ) = (0–1.07) (.38) + (1–1.07) (.30) + (2-1.07) (.19) + (3-1.07) (.12)
b. = E(X) =
2
2
2
2
2
2
= 1.085
2 1.085 = 1.042
=
7.65a. P(VT = 0) = .44, P(VT = 1) = .29, P(VT = 2) = .21, P(VT = 3) = .06
xP (x) = 0(.44) + 1(.29) + 2(.21) + 3(.06) = .89 = V(X) = ( x ) P( x ) = (0–.89) (.44) + (1–.89) (.29) + (2-.89) (.21) + (3-.89) (.06)
b. = E(X) =
2
2
2
2
2
2
= .878
=
2 .878 = .937
7.66 a. P(T = 0) = .14, P(T = 1) = .23, P(T = 2) = .28, P(T = 3) = .25, P(T = 4) = .08, P(T = 5) = .02, P(T = 6) = 0
xP (x) = 0(.14) + 1(.23) + 2(.28) + 3(.25) + 4(.08) + 5(.02) + 6(0) = 1.96 = V(X) = ( x ) P( x ) = (0–1.96) (.14) + (1–1.96) (.23) + (2-1.96) (.28) + (3-1.96) (25) +
b. = E(X) =
2
2
2
2
2
2
(4-1.96)2(.08) + (5-1.96)2(.02) + (6-1.96)2(0) = 1.538
= c.
2 1.538 = 1.240
xyP (x, y) = (0)(0)(.14) + (0)(1)(.12) + (0)(2)(.09) + (0)(3)(.03) + (1)(0)(.11) + all x all y
(1)(1)(.10) + (1)(2)(.07) + (1)(3)(.02) + (2)(0)(.09) + (2)(1)(.05) + (2)(2)(.04) + (2)(3)(.01) + (3)(0)(.10) + (3)(1)(.02) + (3)(2)(.01) + (3)(3)(0) = .74
COV(X, Y) =
xyP (x, y) – = .74 – (1.07)(.89) = -.212 x
y
all x all y
COV ( X , Y )
7.67 E
x y
=
.212 = –.217 (1.085 )(. 937 )
X E(X ) = 18 + 12 + 27 + 8 = 65 i
i
199
V
X V(X ) = 8 + 5 + 6 + 2 = 21 i
i
X E(X ) = 35 + 20 + 20 + 50 + 20 = 145 V X V(X ) = 8 + 5 + 4 + 12 + 2 = 31
7.68 E
i
i
i
i
X E(X ) = 8 + 14 + 5 + 3 + 30 + 30 + 10 = 100 V X V(X ) = 2 + 5 + 1 + 1 + 8 +10 + 3 = 30
7.69 E
i
i
i
i
X E(X ) = 10 + 3 + 30 + 5 + 100 + 20 = 168 V X V(X ) = 9 + 0 + 100 + 1 + 400 + 64 = 574
7.70 E
i
i
i
i
7.71 The expected value does not change. The standard deviation decreases. 7.2 E(Rp) = w1E(R1) + w2E(R2) = (.30)(.12) + (.70)(.25) = .2110 a. V(Rp) =
w12 12 + w22 22 + 2 w1 w2 1 2
= (.30)2 (.02)2 (.70)2 (.152 ) 2(.30)(.70)(.5)(.02)(.15) = .0117 R p .0117 = .1081
b. V(Rp) =
w12 12 + w22 22 + 2 w1 w2 1 2
= (.30) 2 (.02) 2 (.70) 2 (.152 ) 2(.30)(.70)(.2)(.02)(.15) = .0113 R p .0113 = .1064
c. V(Rp) =
w12 12 + w22 22 + 2 w1 w2 1 2
= (.30) 2 (.02) 2 (.70) 2 (.152 ) 2(.30)(.70)(0)(.02)(.15) = .0111 R p .0111 = .1052
7.73 a She should choose stock 2 because its expected value is higher. b. She should choose stock 1 because its standard deviation is smaller. 7.74 E(Rp) = w1E(R1) + w2E(R2) = (.60)(.09) + (.40)(.13) = .1060 V(Rp) =
w12 12 + w22 22 + 2 w1 w2 1 2 = (.60) 2 (.15) 2 (.40) 2 (.212 ) 2(.60)(.40)(.4)(.15)(.21) = .0212
200
R p .0212 = .1456
7.75 E(Rp) = w1E(R1) + w2E(R2) = (.30)(.09) + (.70)(.13) = .1180 V(Rp) =
w12 12 + w22 22 + 2 w1 w2 1 2 = (.30) 2 (.15) 2 (.70) 2 (.212 ) 2(.30)(.70)(.4)(.15)(.21) = .0289
R p .0289 = .1700
The statistics used in Exercises 7.66 to 7.83 were computed by Excel. The variances were taken from the variance-covariance matrix. As a result they are the population parameters. To convert to statistics multiply the variance of the portfolio returns by n/(n–1).
7.76 a .0074 .0667 b .0056 .0675 c .0064 .0590 d (a) e (b)
7.77 a .0158 .0425 b .0153 .0461 c .0143 .0439 d (a) has the largest mean and the smallest standard deviation. 7.78 a .0080 .0394 b .0084 .0397 c .0059 .0437 d (c) has the smallest mean and the largest standard deviation. 7.79 a .0016 .0327 b .0120 .0307 c .0121 .0315 d (c) e (b) 7.80 a .0101 .0338 b .0097 .0339 201
c .0110 .0360 d (c) e (a) 7.81 a .0080 .0448 b .0062 .0472 c .0098 .0428 d (c) has the largest mean and the smallest standard deviation. 7.82 a .0103 .0276 b .0085 .0282 c .0120 .0315 d (c) e (a) 7.84 .0100 .0267 7.85 Because all the stocks are bank stocks there is very little diversification. 7.86
7.87 a .0131 .0364 b .0118 .0424 c .0127 .0531 d (a) has the largest mean and the smallest standard deviation. 7.88 a .0119 .0350 b .0135 .0326 c .0142 .0372 d (c) e (b)
202
7.89 a .0096 .0350 b .0115 .0536 c .0093 .0357 d (b) e (a) 7.90 a .0065 .0277 b .0096 .0303 c .0047 .0371 d (b) e (a) 7.91 a .0077 .0362 b .0102 .0341 c .0086 .0324 d (b) e (c) 7.93 .0100 .0315 7.94 a .0128 .0456 b .0161 .0491 c .0117 .0437 d (b) e (c) 7.95 a .0154 .0464 b .0172 .0465 c .0165 .0519 d (b) e (a) 7.96 a .0186 .0468 b .0185 .0592 c .0213 .0752 d (c) e (a)
203
7.97 a .0147 .0454 b .0212 .0463 c .0119 .0482 d (b) e (a) 7.99 .0200 .0444
7.100 P(X = x) =
n! p x (1 p) n x x! (n x )!
a P(X = 3) =
10! (.3) 3 (1 .3)103 = .2668 3! (10 3)!
b P(X = 5) =
10! (.3) 5 (1 .3)105 = .1029 5! (10 5)!
c P(X = 8) =
10! (.3) 8 (1 .3)108 = .0014 8!(10 8)!
7.101 a P(X = 3) = P(X 3) – P(X 2) = .6496 – .3828 = .2668 b P(X = 5) = P(X 5) – P(X 4) = .9527 – .8497 = .1030 c P(X = 8) = P(X 8) – P(X 7) = .9999 – .9984 = .0015 7.102 a .26683 b .10292 c .00145
7.103 P(X = x) =
n! p x (1 p) n x x! (n x )!
a. P(X = 2) =
6! (.2) 2 (1 .2) 62 = .2458 2! (6 2)!
b. P(X = 3) =
6! (.2) 3 (1 .2) 63 = .0819 3! (6 3)!
c. P(X = 5) =
6! (.2) 5 (1 .2)5 = .0015 5! (6 5)!
7.104 a P(X = 2) = P(X 2) – P(X 1) = .9011 – .6554 = .2457 b P(X = 3) = P(X 3) – P(X 2) = .9830 −.9011 = .0819 204
c P(X = 5) = P(X 5) – P(X 4) = .9999 – 9984 = .0015 7.105 a .24576 b .08192 c .00154 7.106 a P(X = 18) = P(X 18) – P(X 17) = .6593 – .4882 = .1711 b P(X = 15) = P(X 15) – P(X 14) =.1894 – .0978 = .0916 c P(X 20) = .9095 d P(X 16) = 1 – P(X 15) = 1 – .1894 = .8106 7.107 a .17119 b .09164 c .90953 d .81056 7.108 Binomial distribution with p = .25 a P(X = 1) =
4! (.25)1(1 .25)41 = .4219 1! (4 1)!
b Table 1 with n = 8: p(2) = P(X 2) – P(X 1) = .6785 – .3671 = .3114 c Excel: p(3) = .25810 7.109 Table 1 with n = 25 and p = .3: P(X 10) = .9022 7.110 Table 1 with n = 25 and p = .90 a P(X = 20) = P(X 20) – P(X 19) = .0980 – .0334 = .0646 b P(X 20) = 1 – P(X 19) = 1 – .0334 = .9666 c P(X 24) = .9282 d E(X) = np = 25(.90) = 22.5 7.111 Table 1 with n = 25 and p = .75: P(X 15) = 1 – P(X 14) = 1 – .0297 = .9703
7.112 P(X = 0) = P(X = 1) =
4! (.7) 0 (1 .7) 40 = .0081 0!( 4 0)!
4! (.7)1 (1 .7) 41 = .0756 1!( 4 1)!
205
P(X = 2) =
4! (.7) 2 (1 .7) 42 = .2646 2!( 4 2)!
P(X = 3) =
4! (.7)3 (1 .7) 43 = .4116 3!(4 3)!
P(X = 4) =
4! (.7) 4 (1 .7) 44 = .2401 4!( 4 4)!
7.113 Excel with n = 5 and p = .27: a. P(X ≤ 2) = .8743 b P(X = 0) = .2073 c P(X ≥ 3) = 1 – P(≤ 2) = 1-.8743 = .1257 7.114 Excel with n = 10 and p = .30 a P(X = 3) = .2668 b .P(X ≥ 3) = 1 – P(X ≤ 2) = 1 - .3828 = .6172 7.115 Excel with n = 10 and p = .25 a. P(X ≤ 3) = .7759 b. E(X) = np = 100(.25) = 25
7.116 P(X = 20) =
20! (.75) 20 (1 .75) 2020 = .00317 20! (20 20 )!
7.117 Excel with n = 4 and p = .72 a. P(X = 4) = .2687 b. P(X = 2) = .2439 c E(X) = np = 4(.72) = 2.88 7.118 Excel with n = 10 and p = .29 a. P(X ≤ 3) = .6761 b. P(X = 0) = .0326 c. E(X) = np = 10(.29) = 2.9 7.119a Excel with n = 10 and p = 244/495: P(X 5) = 1 – P(X 4) = 1 – .39447 = .60553 b E(X) = np =100(244/495) = 49.29
206
7.120 a P(X = 2) =
5! (.45) 2 (1 .45) 52 = .3369 2! (5 2)!
b Excel with n = 25 and p = .45: P(X 10) = 1 – P(X 9) = 1 – .24237 = .75763 7.121 a Table 1 with n = 5 and p = .5: P(X = 2) = P(X 2) – P(X 1) = .5 – .1875 = .3125 b: Table 1 with n = 25 and p = .5: P(X 10) = 1 – P(X 9) = 1 – .1148 = .8852
7.122 a P(X = 2) =
5! (.52)2 (1 .52)52 = .2990 2!(5 2)!
b Excel with n = 25 and p = .52: P(X 10) = 1 – P(X 9) = 1 – .08033 = .91967 7.123 a Excel with n = 25 and p = 2/38: P(X 2) = 1 – P(X 1) = 1 – .61826 = .38174 b Excel with n = 25 and p = 2/38: P(X = 0)) = .25880 c Excel with n = 25 and p = 18/38: P(X 15) = 1 – P(X 14) = 1 – .85645 = .14355 d Excel with n = 25 and p = 18/38: P(X 10) = .29680 7.124 a Excel with n = 100 and p = .52: P(X 50) = 1 – P(X 49) = 1 – .30815 = .69185 b Excel with n = 100 and p = .36: P(X 30) = .12519 c Excel with n = 100 and p = .06: P(X 5) = .44069 7.125 Excel with n = 20 and p = .38: P(X 10) = 1 – P(X 9) = 1 – .81032 = .18968 7.126a. Excel with n = 10 and p = .23: P(X 5) = 1 – P(X 4) = 1 – .94308 = .05692 b. Excel with n = 25 and p = .23: P(X 5) = .47015 7.127 Excel with n = 100 and p = .53 a P(X > 50) = 1 - P(X ≤ 50) = 1 - .3078 = .6922 b p(X > 60) = 1 – P(X ≤ 60) = 1 - .9341 = .0659 cE(X) = np = 100(.53) = 53
7.128 a P(X = 0) =
e 2 2 0 e x = = .1353 0! x!
b P(X = 3) =
e 2 2 3 e x = = .1804 3! x!
c P(X = 5) =
e 2 2 5 e x = = .0361 5! x! 207
7.129a P(X = 0) =
e x e .5 .5 0 = = .6065 x! 0!
b P(X = 1) =
e .5 .51 e x = = .3033 1! x!
c P(X = 2) =
e .5 .5 2 e x = = .0758 2! x!
7.130 a Table 2 with = 3.5: P(X = 0) = P(X 0) = .0302 b Table 2 with = 3.5: P(X 5) = 1 – P(X 4) = 1 – .7254 = .2746 c Table 2 with = 3.5/7: P(X = 1) = P(X 1) – P(X 0) = .9098 – .6065 = .3033
7.131 a P(X = 5 with = 14/3) = b. P(X = 1 with = 14/3) =
e e x = x!
7.132 a P(X = 0 with = 2) = b P(X = 10 with = 14) =
e e x = x! 1 / 3
14 / 3
(14 / 3) 5 = .1734 5!
(1 / 3)1 = .2388 1!
e x e 2 (2) 0 = = .1353 0! x!
e x e 14 (14)10 = = .0663 10! x!
7.133 a Table 2 with = 5: P(X 10) = 1 – P(X 9) = 1 – .9682 = .0318 b Table 2 with = 10: P(X 20) = 1 – P(X 19) = 1 – .9965 = .0035 7.134 a Excel with = 30: P(X 35) = 1 – P(X 34) = 1 – .79731 = .20269 b Excel with = 15:P(X 12 = .26761 7.135 a Excel with = 1.8: P(X 3) = 1 – P(X 2) = 1 – .73062 = .26938 b Table 2 with = 9: P(10 X 15) = P(X 15) – P(X 9) = .9780 – .5874 = .3906
7.136 P(X = 0 with = 80/200) =
e x e .4 (.4) 0 = =.6703 0! x!
7.137 Poisson with µ = 2.5 a P(X ≥ 5) = 1 – P(X ≤ 4) = 1 - .8912 = .1088 b P(0) = .0821 208
c P(X ≤ 3) =.7576 7.138 Poisson with µ = 4 a P(X ≥ 5) = 1 – P(X ≤ 4) = 1 - .6288 = .3712 b P(X ≤ 3) = .4335 c P(X = 4) = .1954 7.139 Poisson with µ = 5 a P(X ≤ 3) = .2650 b P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – .6160 = .3840 c P(X ≤ 8) = .9319 7.140 Poisson with µ = 1 a P(X = 0) = .3679 b P(X ≥ 5) = 1 – P(X ≤ 4) = 1 - .9963 = .0037 7.141 Poisson with µ = 2 a P(X = 0) = .1353 b P(X ≥ 4) = 1 – P(X ≤ 3) = 1 – .8571 = . 1429 c P(X ≤ 2) = .6767 7.142 a Table 2 with = 1.5: P(X 2) = 1 – P(X 1) = 1 – .5578 = .4422 b Table 2 = 6: P(X < 4) = P(X 3) = .1512
7.143 a P(X = 1 with = 5) =
e x e 5 (5)1 = = .0337 1! x!
b Table 2 with = 15: P(X > 20) = P(X 21) = 1 – P(X 20) = 1 – .9170 = .0830
7.144 a P(X = 0 with = 1.5) =
e x e 1.5 (1.5) 0 = = .2231 0! x!
b Table 2 with = 4.5: P(X 5) = .7029 c Table 2 with = 3.0: P(X 3) = 1 – P(X 2 = 1 – .4232 = .5768
7.145 Binomial with n = 25 and p = .20 a P(X ≤ 5) = .6167 b P(X ≥ 7) = 1 – P(X ≤ 6) = 1 – .7800 = .2200 c P(X = 5) = .1960 209
7.146 Binomial with n = 20 and p = .15 a P(X = 3) = .2428 b P(X ≤ 5) = .9327 c P(X ≥ 3) = 1 – P(X ≤ 2) = 1 - .4049 = .5951
7.147 P(X = 5) =
5! (.774) 5 (1 .774) 55 = .2778 5! (5 5)!
7.148 a E(X) = np = 40(.02) = .8 b P(X = 0) =
40! (.02)0 (1 .02)400 = .4457 0! (40 0)!
xP (x) = 0(.48) + 1(.35) + 2(.08) + 3(.05) + 4(.04) = .82
7.149 a = E(X) =
2 = V(X) =
(x ) P(x) = (0–.82) (.48) + (1–.82) (.35) + (2–.82) (.08) 2
2
2
2
2
2
+ (3–.82) (.05) + (4–.82) (.04) = 1.0876
=
2 1.0876 = 1.0429
7.150 a P(X = 10 with = 8) =
e x e 8 (8)10 = = .0993 10! x!
b Table 2 with = 8: P(X > 5) = P(X 6) = 1 – P(X 5) = 1 – .1912 = .8088 c Table 2 with = 8: P(X < 12) = P(X 11) = .8881
7.151 Binomial with n = 10 and p = .80 a P(X ≥ 9) = 1 – P(X ≤ 8) = 1 – .6242 = .3758 b P(X = 8) = .3020 c P(X ≤ 7) = .3222 7.152 Table 1 with n = 10 and p = .3: P(X > 5) = P(X 6) = 1 – P(X 5) = 1 – .9527 = .0473
7.153 a = E(X) =
2 = V(X) =
xP (x) = 0(.05) + 1(.16) + 2(.41) + 3(.27) + 4(.07) + 5(.04) = 2.27
(x ) P(x) = (0–2.27) (.05) + (1–2.27) (.16) + (2–2.27) (.41) 2
2
2
2
2
2
+ (3–2.27) (.27) + (4–2.27) (.07) + (5–2.27) (.04) = 1.1971
=
2 1.1971 = 1.0941
210
2
7.154 Poisson with µ = 5 a P(X = 1) = .0337 b P(X ≤ 5) = .6160 c P(X ≥ 8) = 1 – P(X ≤ 7) = 1 - .8666 = .1334 7.155 Binomial with n = 25 and p = .10 a P(X ≥ 3) = 1 – P(X ≤ 2) = 1 - .5371 = .4629 b P(X = 2) = .2659 c P(X ≤ 3) = .7636 7.156 Binomial with n = 10 and p = .50 a P(X = 6) = .2051 bP(X ≥ 8) = 1 – P(X ≤ 7) = 1 – .9453 = .0547 c P(X ≤ 4) = .3770 7.157 Poisson with µ = 1.5 a P(X = 1) = .3347 b P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – .8088 = .1912 c P(X ≤ 4) = .9814 7.158 Table 1 with n = 10 and p = .20: P(X 6) = 1 – P(X 5) = 1 – .9936 = .0064
7.159 a P(X = 2) =
10! (.05) 2 (1 .05)10 2 = .0746 2!(10 2)!
b Excel with n = 400 and p = .05: P(X = 25) = .04455 c .05 7.160 a Excel with n = 80 and p = .70: P(X > 65) = P(X 66) = 1 – P(X 65) = 1 – .99207 = .00793 b E(X) = np = 80(.70) = 56 c =
np(1 p) 80 (.70 )(1 .70 ) = 4.10
7.161 a Excel with = 9.6: P(X ≥ 10) = 1 – P(X ≤ 9) = 1 − .5089 = .4911 b. Excel with µ = 6: P(X ≤ 5) = .4457 c Excel with = 2.3: P(X ≥ 3) = 1 – P(X ≤ 2) = 1 − .5960 = .4040
211
7.162 Table 1 with n = 25 and p = .40: a P(X = 10) = P(X 10) – P(X 9) = .5858 – .4246 = .1612 b P(X < 5) = P(X 4) = .0095 c P(X > 15) = P(X 16) = 1 – P(X 15) = 1 – .9868 = .0132 7.163 Excel with n = 100 and p = .45: a P(X > 50) = P(X 49) = 1 – P(X 50) = 1 – .86542 = .13458 b P(X < 44) = P(X 43) = .38277 c P(X = 45) = .07999
xP (x) = 0(.36) + 1(.22) + 2(.20) + 3(.09) + 4(.08) + 5(.05) = 1.46
7.164 a. = E(X) =
2 = V(X) =
(x ) P(x) = (0–1.46) (.36) + (1–1.46) (.22) + (2–1.46) (.20) 2
2
2
2
2
2
2
+ (3–1.46) (.09) + (4–1.46) (.08) + (5–1.46) (.05) = 2.23
=
2 2.23 = 1.49
xP (x) = 0(.15) + 1(.18) + 2(.23) + 3(.26) + 4(.10) + 5(.08) = 2.22 = V(X) = ( x ) P( x ) = (0–2.22) (.15) + (1–2.22) (.18) + (2–2.22) (.23)
b. = E(X) =
2
2
2
2
2
2
2
2
+ (3–2.22) (.26) + (4–2.22) (.10) + (5–2.22) (.08) = 2.11
=
2 2.11 = 1.45
7.165a Binomial with n = 5 and p = .350: P(X ≥ 1) = 1 – P(0) = 1 - .1160 = .8840 b Probability of at least 1 hit in 56 consecutive games = (.8840) 56 = .00100 7.166 Binomial with n = 10 and p = .80 a P(X = 10) = .1074 b P(X ≥ 8) = 1 – P(X ≤ 7) = 1 - .3222 = .6778 c P(X ≤ 8) = .6242 7.167 Binomial with n = 20 and p = .60 a P(X ≥ 15) = 1 – P(X ≤ 14) = 1 – .8744 = .1256 b P(X ≤ 12) = .5841 c P(X = 12) = .1797
7.168 Binomial with n = 5 and p = .01. (using Excel)
212
x
p(x)
0
.95099
1
.04803
2
.00097
3
.00001
4
0
5
0
7.169 Binomial with n = 100 and p = .01 a P(X = 0) = .3660 b P(X = 1) = .3697 c P(X = 2) = .1849 7.170 p = .08755 because P(X 1) = 1– P(X = 0 with n = 10 and p = .08755) = 1– .40 = .60 Case 7.1 Expected number of runs without bunting = .85. If batter bunts: Bases Outcome
Expected Number
Probability
Occupied
Outs
of Runs
1
.75
2nd
1
.69
.5175
2
.10
1st
1
.52
.0520
3
.10
none
2
.10
.0100
4
.05
1st and 2nd
0
1.46
.0730
Expected number of runs = .6255 Decision: Don’t bunt.
213
Chapter 8 8.1a. P(X > 45)
(60 45) 2 (75 60 ) 2 = .0800 50 15 50 15
b. P(10 < X < 40) c. P(X < 25)
(15 [30 ]) 6 (0 [15]) 10 (15 0) 17 (25 15) 7 = .7533 50 15 50 15 50 15 50 15
d. P(35 < X < 65)
8.2a. P(X > 45)
(45 35) 6 (60 45 ) 2 (65 60 ) 2 = .1333 50 15 50 15 50 15
(60 45 ) 3 (75 60 ) 3 = .1200 50 15 50 15
b. P(10 < X < 40) c. P(X < 25)
(15 10 ) 17 (30 15 ) 7 (40 30 ) 6 = .3333 50 15 50 15 50 15
(15 10 ) 16 (30 15 ) 8 (40 30 ) 8 = .4800 50 15 50 15 50 15
(30 [45 ]) 5 (15 [30 ]) 5 (0 [15 ]) 2 (15 0) 16 (25 15 ) 8 50 15 50 15 50 15 50 15 50 15
= .6667 d. P(35 < X < 65)
(45 35 ) 8 (60 45 ) 3 (65 60 ) 3 = .1867 50 15 50 15 50 15
8.3a. P(55 < X < 80)
(60 55 ) 16 (70 60 ) 5 (80 70 ) 24 = .6167 60 10 60 10 60 10
b. P(X > 65)
(70 65 ) 5 (80 70 ) 24 (90 80 ) 7 (100 90 ) 1 = .5750 60 10 60 10 60 10 60 10
c. P(X < 85)
(50 40 ) 7 (60 50 ) 16 (70 60 ) 5 (80 70 ) 24 (85 80 ) 7 60 10 60 10 60 10 60 10 60 10
= .9250 d. P(75 < X < 85)
(80 75 ) 24 (85 80 ) 7 = .2583 60 10 60 10
215
8.4 a.
b. P(X > 25) = 0 c. P(10 < X < 15) = (15 10 )
1 = .25 20
d. P(5.0 < X < 5.1) = (5.1 5)
8.5a. f(x) =
1 = .005 20
1 1 = (60 20 ) 40
b. P(35 < X < 45) = (45–35)
20 < x < 60
1 = .25 40
216
c.
8.6 f(x) =
1 1 30 < x < 60 (60 30 ) 30
a. P(X > 55) = (60 55 )
1 = .1667 30
b. P(30 < X < 40) = (40 30 )
1 = .3333 30
c. P(X = 37.23) = 0
8.7
1 (60 30 ) 7.5 ; The first quartile = 30 + 7.5 = 37.5 minutes 4
8.8 .10 (60 30) 3 ; The top decile = 60–3 = 57 minutes
8.9 f(x) =
1 1 (175 110 ) 65
a. P(X > 150) = (175 150 )
110 < x < 175
1 = .3846 65
b. P(120 < X < 160) = (160 120 )
1 = .6154 65
8.10 .20(175–110) = 13. Bottom 20% lie below (110 + 13) = 123
Note: The area in a triangle is height X base /2
8.11a.
217
b. P(0 < X < 2) = (.5)(2–0)(1) = 1.0 c. P(X > 1) = (.5)(2 – 1)(.5) = .25 d. P(X < .5) = 1 – P(X > .5) = 1 – (.5)(.75)(2–.5) = 1 – .5625 = .4375 e. P(X = 1.5) = 0
8.12 a
b. P(2 < X < 4) = P(X < 4) – P(X < 2) = (.5)(3/8)(4–1) – (.5)(1/8)(2–1) = .5625 – .0625 = .5 c. P(X < 3) = (.5)(2/8)(3–1) = .25
8.13a. 218
b. P(1 < X < 3) = P(X < 3) – P(X < 1) =
1 1 1 3 (3 0) (1 0) = .18 – .02 = .16 2 25 2 25
c. P(4 < X < 8) = P(4 < X < 5) + P(5 < X < 8) P(4 < X < 5)= P(X < 5) – P(X <4) =
1 5 1 4 (5 0) (4 0) = .5 – .32 = .18 2 25 2 25
P(5 < X < 8) = P(X > 5) – P(X > 8) =
1 2 1 5 (10 5) (10 8) = .5 – .08 = .42 2 25 2 25
P(4 < X < 8) = .18 + .42 = .60 d. P(X < 7) = 1 – P(X > 7) P(X > 7) =
1 3 (10 7) = .18 2 25
P(X < 7) = 1 – .18 = .82 e. P(X > 3) = 1 – P(X < 3) P(X < 3) =
1 3 (3 0) = .18 2 25
P(X > 3) = 1 – .18 = .82 8.14 a. f(x) = .10 – .005x 0 x 20 b. P(X > 10) = (.5)(.05)(20–10) = .25 c. P(6 < X < 12) = P(X > 6) – PX > 12) = (.5)(.07)(20–6) – (.5)(.04)(20–12) = .49 – .16 = .33 8.15 b P(X < 8) = P(0 < X < 1) + P(1 < X < 8) = .40(1) + .05(8-1) = .40 + .35 = .75 c P(.4 < X < 10) = P(.4 < X < 1) + P(1 < X < 10) = .4(1-.4) + .05(10-1) = .24 + .45 = .69
8.16 b P(X < 5.5) = P(0 < X < 2) + P(2 < X < 5) + P(5 < X < 5.5) = .10(2-0) + .20(5 – 2) + .15(5.5 – 5) = .20 + .60 + .075 = .875 c P(X > 3.5) = P(3.5 < X < 5) +P(5 < X < 6) + P(6 < X < 7) = .20(5-3.5) + .15(6-5) + .05(7-6) = .30 + .15 + .05 = .50 d P(1 < X < 6.5) = P(1 < X < 2) = P(2 < X < 5) + P(5 < X < 6) + P(6 < X < 6.5)
219
= .10(2-1) + .20(5-2) + .15(6-5) + .05(6.5-6) = .10 + .60 + .15 + .025 = .875
8.17 c P(X < 2) = .4(2-0)/2 = .40 d P(X < 3) = P(0 < X < 2) + P(2 < X < 3) = .40 + .40(3-2) = .80 e P(1 < X < 2) = P(0 < X < 2) – P(0 < X < 1) = .4(2-0)/2 - .2(1-0)/2 = .30 P(1 < X < 2.5) = P(1 < X < 2) + P(2 < X < 2.5) = .30 + .4(2.5-2) = .50 8.18 b P(X < 2) = P(0 < X < 4) – P(2 < X < 4) = .4(4-0)/2 - .20(4-2)/2 = .8 - .20 = .60 c P(X > 5) = P(4 < X < 6) – P(4 < X < 5) = .20(6-4)/2 - .10(5-4)/2 = .20 - .05 = .15 d P(2.5 < X < 5.5) = (2.5 < X < 4) + P(4 < X < 5.5) = .15(4-2.5)/2 + .15(5.5-4)/2 = .1125 + .1125 = .2250
8.19 P( Z < 1.60) = .9452
8.20 P(Z < 1.61) = .9463
8.21 P(Z < 1.65) = .9505 8.22 P(Z < −1.39) = .0823 8.23 P(Z < −1.80) = .0359 8.24 P(Z < − 2.16) = .0154 8.25 P(–1.30 < Z < 0.70) = P( Z < 0.70) − P(Z < −1.30) = .7580− .0968 = .6612 8.26 P(Z > –1.24) = 1 – P(Z < −1.24) = 1 − .1075 = .8925 8.27 P(Z < 2.23) = .9871 8.28 P(Z > 1.87) = 1 – P(Z < 1.87) = 1 – .9693 = .0307 8.29 P(Z < 2.57) = ..9949 8.30 P(1.04 < Z < 2.03) = P(Z < 2.03) – P(Z < 1.04) = .9788 – .8508 = .1280 8.31 P(–0.71 < Z < –0.33) = P(Z < −.33) – P(Z < −.71) = .3707 – .2389 = .1318
220
8.32 P(Z > 3.09) = 1 – P(Z < 3.09) = 1 – .9990 = .0010 8.33 P(Z > 0) = 1 – P(Z < 0) = 1 − .5 = .5 8.34 P(Z > 4.0) = 0 8.35 P(Z < z.03) = 1 – .03 = .9700; z.03 = 1.88 8.36 P(Z < z.065) = 1 – .065 = .9350; z.065 = 1.51 8.37 P(Z < z.28 ) = 1 – .28 = .7200; z.28 = .58
X 145 100 8.38 P(X > 145) = P = P(Z > 2.25) = 1 – P(Z < 2.25) = 1 – .9878 = .0122 20
8.39 P(Z < z.15 ) = 1 – .15 = .8500; z .15 = 1.04; z.15
x 250 x ; 1.04 ; x = 291.6 40
800 1,000 X 1,100 1,000 8.40 P(800 < X < 1100) = P = P(–.8 < Z < .4) 250 250 = P(Z < .4) − P(Z < −.8) = .6554 − .2119 = .4435
8.41 P(Z < z .08 ) = .0800; z .08 1.41; z .08
x x 50 ; 1.41 ; x = 38.72 8
5 6.3 X 10 6.3 8.42 a P(5 < X < 10) = P = P(–.59 < Z > 1.68) 2.2 2.2 = P(Z < 1.68) − P(Z < −.59) = .9535 − .2776 = .6759
X 7 6.3 b P(X > 7) = P = P(Z > .32) = 1 – P(Z < .32) = 1 – .6255 = .3745 2.2 X 4 6.3 c P(X < 4) = P = P(Z < –1.05) = .1469 2.2
8.43 P(Z < z .10 ) = 1 – .10 = .9000; z .10 = 1.28; z .10 Calls last at least 9.116 minutes. 221
x x 6 .3 ; 1.28 ; x = 9.116 2 .2
X 5,000 5,100 8.44 P(X > 5,000) = P = P(Z > –.5) = 1 −P(Z < −.5) = 1 − .3085 = .6915 200
8.45 P(Z < z .02 ) = .02; z .02 2.05; z .02
x x 5100 ; 2.05 ; x = 4690; 200
8.46 z.75 = -.67: Q1 = (-.67)(300) + 1000 = 799, z.50 = 0: Q2 = 1000, z.25 = .67: Q3 = (.67)(300) + 1000 = 1201
X 30 ,000 25,000 = P(Z > 1.00) = 1 – P(Z < 1.00) = 1 – .8413 8.47 a P(X > 30,000) = P 5,000 = .1587
X 22 ,500 25,000 = P(Z < –0.50) = .3085 b P(X < 22,500) = P 5,000
20 ,000 25,000 X 32 ,000 25,000 c P(20,000 < X < 32,000) = P 5,000 5,000 = P(–1.00 < Z < 1.40) = P(Z < 1.40) − P(Z < −1.00) = .9192 − 1587 = .7605
X 210 200 8.48 P(X > 210) = P = P(Z > .5) = 1 −P(Z < .5) = 1 − .6915 = .3085 20 8.49 85th percentile = (z.15)(5) + 68 = 1.04(5) + 68 = 73.2 8.50 First quintile = (z.80)σ + µ = (-.84)(10,000) + 50,000 = 41,600 Second quintile = (z.60)σ + µ = (-.25)(10,000) + 50,000 = 47,500 Third quintile = (z.40)σ + µ = (.25)(10,000) + 50,000 = 52,500 Fourth quintile = (z.20)σ + µ = (.84)(10,000) + 50,000 = 58,400
X 70,000 82,000 = P(Z < –1.88) = .0301 8.51 a P(X < 70,000) = P 6,400 X 100 ,000 82 ,000 b P(X > 100,000) = P = P(Z > 2.81) = 1 – P(Z < 2.81) = 1 – .9975 = 6,400 .0025
222
8.52 Top 5%: P(Z < z .05 ) = 1 – .05 = .9500; z .05 = 1.645; z .05
x x 32 ; 1.645 ;x= 1 .5
34.4675 Bottom 5%: P(Z < z .05 ) = .0500; z .05 1.645 ; z .05
x x 32 ; 1.645 ; 1.5
x = 29.5325
X 36 32 8.53 a P(X > 36) = P = P(Z > 2.67) = 1 – P(Z < 2.67) = 1 – .9962 = .0038 1.5
X 34 32 b P(X < 34) = P = P(Z < 1.33) = .9082 1.5 30 32 X 33 32 c P(30 < X < 33) = P = P(–1.33 < Z < .67) 1.5 1.5 = P(Z < .67) − P(Z < −1.33) = .7486 − .0918 = .6568
X 8 7.2 8.54 P(X > 8) = P = P (Z > 1.20) = 1 – P(Z < 1.20) = 1 – .8849 = .1151 .667
8.55 P(Z < z .25 ) = .7500; z .25 = .67; z .25
x 7.2 x ; .67 ; x = 7.65 hours .667
X 10 7.5 8.56 a P(X > 10) = P = P Z > 1.19) = 1 – P(Z < 1.19) = 1 – .8830 = .1170 2.1 7 7. 5 X 9 7 . 5 b P(7 < X < 9) = P = P(–.24 < Z < .71) 2.1 2.1 = P(Z < .71) − P(0 < Z < −.24) = .7611 − .4052 = .3559
X 3 7.5 c P(X < 3) = P = P Z < –2.14) = .0162 2.1 d P(Z < – z .05 ) = .0500; z .05 = –1.645; z .05
x x 7. 5 ; 1.645 ; x = 4.05 hours 2 .1
X 12 ,000 11,500 8.57 a P(X > 12,000) = P = P Z > .63) = 1– P(Z < .63) = 1 – .7357 = 800 .2643
X 10 ,000 11,500 b P(X < 10,000) = P = P(Z < –1.88) = .0301 800
223
8.58 P(Z < – z .01 ) = .0100; z .05 = –2.33; z .01
x x 11,500 ; 2.33 ; x = 9,636 800
X 4900 6000 8.59 a P(X > 4900) = P = P Z > -.92) = 1 – P(Z < -.92) = 1 – .1788 = 1200 .8212
3800 6000 X 5700 6000 b P(3800 < X < 5700) = P = P(–1.83 < Z < -.25) 1200 1200 = P(Z < -.25) − P(Z < −1.83) = .4013 − .0336 = .3677
X 6500 6000 c P(X < 6500) = P = P Z < .42) = .6628 1200 X 145 155 8.60 P(X < 145) = P = P Z < -1.11) = .1335 9 24 26 X 28 26 8.61 a P(24 < X < 28) = P = P(–.80 < Z < .80) 2. 5 2. 5 = P(Z < .80) – P(Z < −.80) = .7881 − .2119 = .5762
X 28 26 b P(X > 28) = P = P(Z > .80) = 1 – P(Z < .80) = 1 – .7881 = .2119 2.5 X 24 26 c P(X < 24) = P = P(Z < –.80) =.2119 2.5
X 30 27 8.62 a P(X > 30) = P = P(Z > .43) = 1 – P(Z < .43) = 1 – .6664 = .3336 7
X 40 27 b P(X > 40) = P = P(Z > 1.86) = 1 – P(Z < 1.86) = 1 – .9686 = .0314 7 X 15 27 c P(X < 15) = P = P(Z < –1.71) = .0436 7 d P(Z < z .20 ) = 1 – .20 = .8000; z .20 = .84; z .20
x x 27 ; .84 ; x = 32.88 7
X 4 7.5 8.63 a P(X < 4) = P = P(Z < –2.92) = .0018 1.2
224
7 7.5 X 10 7.5 b P(7 < X < 10) = P = P(–.42 < Z < 2.08) 1.2 1.2 = P(Z < 2.08) − P(Z < −.42) = .9812 − .3372 = .6440
X 10 16 .40 8.64a P(X < 10) = P = P(Z < –2.33) = .0099 2.75 b P(Z < – z .10 ) = .1000; – z .10 = –1.28; z .10
x x 16 .40 ; 1.28 ; x = 12.88 2.75
8.65 A: P(Z < z .10 ) = 1– .10 = .9000; z .10 = 1.28; z .10 B: P(Z < z .40 ) = 1 –.40 = .6000; z .40 = .25; z .40 C: P(Z < – z .20 ) = .2000; z .20 .84; z .20
x 70 x ; 1.28 ; x = 82.8 10
x 70 x ; .25 ; x = 72.5 10
x x 70 ; .84 ; x = 61.6; 10
D: P(Z < – z .05 ) = .0500; z .05 1.645 ; z .05
x x 70 ; 1.645 ; x = 53.55 10
8.66 P(Z < z .02 ) = 1 – .02 = .9800; z .02 = 2.05; z .02
x 100 x ; 2.05 ; x = 132.80 16
(rounded to 133)
X 64,000 50,000 8.67 P(x > 64,000) = P = P(Z > 1.75) = 1 – P(Z < 1.75) = 1 – .9599 8000 = .0401
X 500 ,000 325 ,600 8.68 P(x > 500,000) = P = P(Z > 1.74) = 1 – P(Z < 1.74) = 1 – .9591 100 ,000 = .0409
8.69 P(Z < z .06 ) = 1 – .06 = .9400; z .06 = 1.55; z .06
ROP ROP 200 ; 1.55 ; 30
ROP = 246.5 (rounded to 247)
8.70 P(Z < z.20 ) = 1 – .20 = .8000; z .20 = .84; z .20
225
x 150 x ; .84 ; x = 171 25
8.71 P(Z < z .30 ) = 1 – .30 = .7000; z .30 = .52; z .30
x 850 x ; .52 ; x = 896.8 90
(rounded to 897)
8.72 P( Z < z .40 ) = 1– .40 = .6000; z .40 = .25; z .40
x 850 x ; .25 ; .x = 872.5 90
(rounded to 873)
X 25 14 8.73 a P(X > 25 = P = P(Z > .61) = 1 – P(Z < .61) = 1 - .7291 = .2709 18
X 0 14 b P(X < 0) = P = P(Z < –.78) = .2177 18 8.74 First quintile = (z.80)σ + µ = (-.84)(30,000) + 250,000 = 224,800 Second quintile = (z.60)σ + µ = (-.25)(30,000) + 250,000 = 242,500 Third quintile = (z.40)σ + µ = (.25)(30,000) + 250,000 = 257,500 Fourth quintile = (z.20)σ + µ = (.84)(30,000) + 250,000 = 275,200 8.75 From Exercise 7.57: = 65, 2 = 21, and = 4.58
X 60 65 P(X > 60) = P = P(Z > –1.09) = 1 − P(Z < −1.09) = 1 −.1379 = .8621 4.58
X 150 145 8.76 P(X < 150) = P = P(Z < .90) = .8159 5.57
X 105 100 8.77 a P(X > 105) = P = P(Z > .90) = 1- P(Z < .90) = 1 - .8159 = .1841 5.48 X 92 100 b P(X > 92) = P = P(Z > -1.46) = 1- P(Z < -1.46) = 1 - .0721 = .9279 5.48
95 100 X 112 100 c P(95 < X < 112) = P = P(–.91 < Z < 2.19) 5.48 5.48 = P(Z < 2.19) − P(Z < −.91) = .9857 − .1814 = .8043 8.78 z.75 = -.67: Q1 = (-.67)(23.96) + 168 = 152.95, z.50 = 0: Q2 = 168, z.25 = .67: Q3 = (.67)(23.96) + 168 = 185.05
226
8.79
Exponential Distribution (Lambda = 3)
f(x)
3,5 3 2,5 2 1,5 1 0,5 0 0
0,5
1 X
1,5
2
8.80
Exponential Distribution (Lambda = .25) 0,3 0,25
f(x)
0,2 0,15 0,1 0,05 0 0
5
10 X
8.81 a P(X 1) e ..5(1) e ..5 = .6065 a P(X .4) e ..5(.4) e ..2 = .8187 c P(X .5) 1 e ..5(..5) 1 e ..25 = 1 – .7788 = .2212 d P(X 2) 1 e .5(2) 1 e 1 = 1 – .3679 = .6321
8.82 a P(X 2) e ..3(2) e .6 = .5488 b P(X 4) 1 e ..3(4) 1 e 1.2 = 1 – .3012 = .6988
227
15
20
c P(1 X 2) e ..3(1) e ..3(2) e ..3 e .6 = .7408 – .5488 = .1920 d P(X = 3) = 0 8.83 = 6 kilograms/hour = .1 kilogram/minute P(X 15) e .1(15) e 1.5 = .2231
8.84 1 / 25 hours; = .04 breakdowns/hour P(X 50) e .04(50) e 2 = .1353
8.85 = 10 trucks/hour = .167 truck/minute P(X 15) e .167(15) e 2.5 = .0821
8.86 1 / = 5 minutes; = .2 customer/minute P(X 10) 1 e .2(10) 1 e 2 = 1– .1353 = .8647
8.87 1 / = 2.7 minutes; = .37 service/minute P(X 3) 1 e .37(3) 1 e 1.11 = 1– .3296 = .6704
8.88 1 / = 7.5 minutes; = .133 service/minute P(X 5) 1 e .133(5) 1 e .665 = 1– .5143 = .4857
8.89 1 / = 125 seconds; = .008 transactions/second = .48 transactions/minute P(X 3) e .48(3) e 1.44 = .2369
8.90 1 / = 6 minutes; = .167 customers/minute P(X 10) e .167(10) e 1.67 = .1889
8.91 x = 30 and P(X > x) = .01. λ=
ln[ P( X x)] ln[. 01] (4.605 ) = = = .1535 x 30 30
The service rate should be .1535 trucks per minute or 9.21 trucks per hour.
8.92 x = 30 and P(X > x) = .10.
228
λ=
ln[ P( X x)] ln[. 10 ] (2.303 ) = = = .0768 x 30 30
The service rate should be .0768 cars per minute or 4.61 cars per hour.
8.93
a 1.341
b 1.319
c 1.988
d 1.653
8.94
a 2.750
b 1.282
c 2.132
d 2.528
8.95
a 1.3406
b 1.3195
c 1.9890
d 1.6527
8.96
a 1.6556
b 2.6810
c 1.9600
d 1.6602
8.97
a .0189
b .0341
c .0927
d .0324
8.98
a .1744
b .0231
c .0251
d .0267
8.99
a 9.24
b 136
c 9.39
d 37.5
8.100
a 17.3
b 50.9
c 2.71
d 53.5
8.101
a 73.3441
b 102.946
c 16.3382
d 24.7690
8.102
a 33.5705
b 866.911
c 24.3976
d 261.058
8.103
a .2688
b 1.0
c .9903
d 1.0
8.104
a .4881
b .9158
c .9988
d .9077
8.105
a 4.35
b 8.89
c 3.29
d 2.50
8.106
a 2.84
b 1.93
c 3.60
d 3.37
8.107
a 1.4857
b 1.7633
c 1.8200
d 1.1587
8.108
a 1.5204
b 1.5943
c 2.8397
d 1.1670
8.109
a .0510
b .1634
c .0222
d .2133
229
8.110
a .1050
b .1576
c .0001
230
d .0044
Chapter 9 9.91a. 1/6 b. 1/6 9.10 a P( X 1) =P(1,1)= 1/36 b P( X 6) = P(6,6) = 1/36
9.11a P( X = 1) = (1/6) 5 = .0001286 b P( X = 6) = (1/6) 5 = .0001286
9.12 The variance of X is smaller than the variance of X.
9.13 The sampling distribution of the mean is normal with a mean of 40 and a standard deviation of 12/ 100 = 1.2.
9.14 No, because the sample mean is approximately normally distributed.
X 1050 1000 = P(Z > 1.00) = 1 – P(Z < 1.00) = 1 – .8413 9.15 a P( X 1050 ) = P 200 / 16 / n = .1587
X 960 1000 = P(Z < –.80) = .2119 b P( X 960 ) P 200 / 16 / n X 1100 1000 = P(Z > 2.00) = 1 – P(Z < 2.00) = 1 – .9772 = .0228 c P( X 1100 ) P 200 / 16 / n
X 1050 1000 = P(Z > 1.25) = 1 – P(Z < 1.25) = 1 – .8944 9.16 a P( X 1050 ) = P 200 / 25 / n = .1056
X 960 1000 = P(Z < –1.00) = .1587 b P( X 960 ) P 200 / 25 / n X 1100 1000 = P(Z > 2.50) = 1 – P(Z < 2.50) = 1 – .9938 = .0062 c P( X 1100 ) P 200 / 25 / n
229
X 1050 1000 = P(Z > 2.50) = 1 – P(Z < 2.50) = 1 – .9938 9.17 a P( X 1050 ) = P 200 / 100 / n = .0062
X 960 1000 = P(Z < –2.00) = .0228 b P( X 960 ) P / n 200 / 100
X 1100 1000 = P(Z > 5.00) = 0 c P( X 1100 ) P 200 / 100 / n
49 50 X 52 50 = P(–.40 < Z < .80) 9.18 a P(49 X 52 ) P / n 5/ 4 5/ 4 = P(Z < .80) − P(Z < −.40) = .7881 −.3446 = .4435
49 50 X 52 50 = P(–.80 < Z < 1.60) b P(49 X 52 ) P 5 / 16 / n 5 / 16 = P(Z < 1.60) − P(Z < −.80) = .9452 −.2119 = .7333
49 50 X 52 50 = P(–1.00 < Z < 2.00) c P(49 X 52 ) P 5 / 25 / n 5 / 25 = P(Z < 2.00) − P(Z < −1.00) = .9772 −.1587 = .8185
49 50 X 52 50 = P(–.20 < Z < .40) 9.19 a P(49 X 52 ) P 10 / 4 / n 10 / 4 = P(Z < .40) − P(Z < −.20) = .6554 −.4207= .2347
49 50 X 52 50 = P(–.40 < Z < .80) b P(49 X 52 ) P 10 / 16 / n 10 / 16 = P(Z < .80) − P(Z < −.40) = .7881 −.3446= .4435
49 50 X 52 50 = P(–.50 < Z < 1.00) c P(49 X 52 ) P 10 / 25 / n 10 / 25 = P(Z < 1.00) − P(Z < −.50) = .8413 −.3085= .5328
49 50 X 52 50 = P(–.10 < Z < .20) 9.20 a P(49 X 52 ) P 20 / 4 / n 20 / 4 = P(Z < .20) − P(Z < −.10) = .5793 −.4602= .1191
49 50 X 52 50 = P(–.20 < Z < .40) b P(49 X 52 ) P 20 / 16 / n 20 / 16 = P(Z < .40) − P(Z < −.20) = .6554 −.4207= .2347 230
49 50 X 52 50 = P(–.25 < Z < .50) c P(49 X 52 ) P 20 / 25 / n 20 / 25 = P(Z < .50) − P(Z < −.25) = .6915 −.4013= .2902
9.21 a
Nn = N 1
1,000 100 = .9492 1,000 1
b
Nn = N 1
3,000 100 = .9834 3,000 1
c
Nn = N 1
5,000 100 = .9900 5,000 1
d. The finite population correction factor is approximately 1.
9.22 a x =
b x = c x =
n n
n
Nn = N 1
500 1,000
10 ,000 1,000 = 15.00 10 ,000 1
500 Nn = N 1 500
10,000 500 = 21.80 10,000 1
500 Nn = N 1 100
10,000 100 = 49.75 10,000 1
X 66 64 9.23 a P(X > 66) = P = P(Z > 1.00) = 1 – P(Z < 1.00) = 1 – .8413 = .1587 2
X 66 64 = P(Z > 2.00) = 1 – P(Z < 2.00) = 1 – .9772 = .0228 b P( X 66 ) P 2/ 4 / n
X 66 64 = P(Z > 10.00) = 0 c P( X 66 ) P / n 2 / 100
9.24 We can answer part (c) and possibly part (b) depending on how nonnormal the population is. X 120 117 9.25 a P(X > 120) = P = P(Z > 0.58) = 1 – P(Z < .58) = 1 – .7190 = .2810 5 .2
X 120 117 = P(Z > 1.15) = 1 – P(Z < 1.15) = 1 – .8749 = .1251 b P( X 120 ) P 5.2 / 4 / n 4
c [P(X >120)] =[.2810]
4
= .00623
231
X 262 ,000 250 ,000 = P(Z > 2.4) = 1 – P(Z < 2.4) 9.26 P( X 262 ,000 ) P 50,000 / 100 / n = 1 - .9918 = .0082
9.27 No, unless the population is extremely nonnormal.
X 60 52 9.28 a P(X > 60) = P = P(Z > 1.33) = 1 – P(Z < 1.33) = 1 – .9082 = .0918 6
X 60 52 = P(Z > 2.31) = 1 – P(Z < 2.31) = 1 – .9896 = .0104 b P( X 60 ) P 6/ 3 / n 3
3
c [P(X >60)] =[.0918] = .00077 X 12 10 9.29 a P(X > 12) = P = P(Z > .67) = 1 – P(Z < .67) = 1 – .7486 = .2514 3
X 11 10 = P(Z > 1.67) = 1 – P(Z < 1.67) b P( X 275 / 25 ) P( X 11) P / n 3 / 25 = 1 – .9525 = 0475 X 75 78 9.30 a P(X < 75) = P = P(Z < –.50) = .3085 6
X 75 78 = P(Z < –3.54) = 1 – P(Z < 3.54) = 1 – 1 = 0 b P( X 75) P / n 6 / 50
X 76 9.31 a P(X > 7) = P = P(Z > .67) = 1 – P(Z < .67) = 1 – .7486 = .2514 1 .5
X 76 = P(Z > 1.49) = 1 – P(Z < 1.49) = 1 – .9319 = .0681 b P( X 7) P / n 1.5 / 5 5
5
c [P(X >7)] =[.2514] = .00100
X 5.97 6.05 = P(Z < –2.67) =.0038 9.32 a P( X 5.97 ) P .18 / 36 / n b It appears to be false.
232
X 625 600 = P(Z > .50) = 1 – P(Z < .50) 9.33 P( X 10,000 / 16 ) P( X 625 ) P / n 200 / 16 = 1 – .6915 = .3085 9.34 The professor needs to know the mean and standard deviation of the population of the weights of elevator users and that the distribution is not extremely nonnormal.
X 71 .25 75 = P(Z > –1.50) 9.35 P( X 1,140 / 16 ) P( X 71 .25 ) P 10 / 16 / n = 1 − P(Z < −1.50) = 1 − 0668 = .9332
X 5 4.8 = P(Z > 1.19) 9.36 P(Total time > 300) = P( X 300 / 60 ) P(X 5) P / n 1.3 / 60 = 1 – P(Z < 1.19) = 1 – .8830 = .1170 9.37 No because the central limit theorem says that the sample mean is approximately normally distributed.
X 1.92 2.0 9.38 P(Total number of cups > 240) = P( X 240 / 125 ) P( X 1.92 ) P / n .6 / 125 = P(Z > –1.49) = 1 − P(Z < −1.49) = 1 − .0681 = .9319
X 300 275 9.39 P(Total number of faxes > 1500) = P( X 1500 / 5) P( X 300 ) P 75 / 5 / n = P(Z > .75) = 1 – P(Z < .75) = 1 – .7734 = .2266
9.40 a The sample mean is normally distributed with a mean of 2800 and a standard deviation of 200.
2500 2800 X 2900 2800 = P(-1.5 < Z < .5) b P(2500 X 2900 ) P / n 400 / 4 400 / 4 = P(Z < .5) – P(Z < -1.5) = .6915 – .0668 = .6247 9.41 We could not answer the question because the sample size is small.
233
P̂ p 9.42a P( P̂ > .60) = P p(1 p) / n
= P(Z > 3.46) = 0 (.5)(1 .5) / 300
P̂ p b. P( P̂ > .60) = P p(1 p) / n
.60 .55
.60 .5
= P(Z > 1.74) = 1 – P(Z < 1.74) (.55 )(1 .55 ) / 300
= 1 – .9591 = .0409
= P(Z > 0) = 1 – P(Z < 0) = 1 − .5 = .5 (.6)(1 .6) / 300
P̂ p c. P( P̂ > .60) = P p(1 p) / n
.60 .6
P̂ p 9.43a P( P̂ < .22) = P p(1 p) / n
= P(Z < –1.55) = .0606 (.25 )(1 .25 ) / 500 .22 .25
P̂ p b. P( P̂ < .22) = P p(1 p) / n
= P(Z < –1.96) = .0250 (.25 )(1 .25 ) / 800
P̂ p c. P( P̂ < .22) = P p(1 p) / n
= P(Z < –2.19) = .0143 (.25 )(1 .25) / 1000
.22 .25
.22 .25
P̂ p 9.44 P( P̂ < .75) = P p(1 p) / n
= P(Z < –1.25) = .1056 (.80 )(1 .80 ) / 100
P̂ p 9.45 P( P̂ > .35)= P p(1 p) / n
= P(Z > –.79) = .1 − P(Z < −.79) (.40 )(1 .40 ) / 60
.75 .80
.35 .40
1 − .2148= .7852
P̂ p 9.46 P( P̂ < .49) = P p(1 p) / n
= P(Z < –2.70) = .0035 (.55 )(1 .55 ) / 500
P̂ p 9.47 P( P̂ > .04)= P p(1 p) / n
= P(Z > 4.04) = 1 – P(Z < 4.04) (.02 )(1 .02 ) / 800
.49 .55
.04 .02
= 1 – 1= 0; The defective rate appears to be larger than 2%.
P̂ p 9.48 a P( P̂ < .50) = P p(1 p) / n
= P(Z < –1.20) =.1151; the claim may (.53)(1 .53) / 400 .50 .53
be true
234
P̂ p b P( P̂ < .50) = P p(1 p) / n
= P(Z < –1.90) = .0287; the claim (.53)(1 .53) / 1,000 .50 .53
appears to be false
P̂ p 9.49 P( P̂ > .10) = P p(1 p) / n
= P(Z > –1.15) = 1 – P(Z < – 1.15) (.14 )(1 .14 ) / 100 .10 .14
= 1 – .1251 = .8749
P̂ p 9.50 P( P̂ > .05)= P p(1 p) / n
= P(Z > 2.34) = 1 – P(Z < 2.34) (.03)(1 .03) / 400 .05 .03
= 1 – .9904 = .0096; The commercial appears to be dishonest
P̂ p 9.51 P( P̂ > .32) = P p(1 p) / n
= P(Z > 1.38) = 1 – P(Z < 1.38) (.30 )(1 .30 ) / 1,000 .32 .30
= 1 – .9162 = .0838
P̂ p 9.52 a P( P̂ < .45) = P p(1 p) / n
= P(Z < –2.45) = .0071 (.50 )(1 .50 ) / 600 .45 .50
b The claim appears to be false.
P̂ p 9.53 P( P̂ < .75) = P p(1 p) / n
= P(Z < –2.34) = .0096 (.80 )(1 .80 ) / 350
P̂ p 9.54 P( P̂ < .70) = P p(1 p) / n
= P(Z < –2.48) = .0066 (.75)(1 .75 ) / 460
P̂ p 9.55 P( P̂ > .28) = P p(1 p) / n
= P(Z > 2.40) = 1 – P(Z < 2.40) (.25 )(1 .25 ) / 1200
.75 .80
.70 .75
.28 .25
= 1 – .9918 = .0082 9.56 The claim appears to be false.
235
9.57 P( P̂ > .15) = P
Pˆ p p(1 p) / n
= P(Z > 1.19) = 1 – P(Z < 1.19) (.13)(1 .13) / 400 .15 .13
= 1 – .8830 = .1170
9.58 P( P̂ > .22) = P
Pˆ p p (1 p) / n
= P(Z > 1.12) = 1 – P(Z < 1.12) (.20 )(1 .20 ) / 500 .22 .20
= 1 – .8686 = .1314
9.59 P( P̂ < .08) = P
9.60 P( P̂ < .08) = P
Pˆ p p(1 p) / n
Pˆ p p(1 p) / n
= P(Z < -.82) = .2061 (.10 )(1 .10 ) / 150
= P(Z < -2.40) = .0082 (.15)(1 .15) / 150
.08 .10
.08 .15
( X X ) ( ) 25 (280 270 ) 1 2 1 2 = P(Z > 1.21) 9.61 P( X 1 X 2 25 ) P 25 2 30 2 12 22 10 10 n1 n 2
= 1 – P(Z < 1.21) = 1 – .8869 = .1131 ( X X ) ( ) 25 (280 270 ) 2 1 2 = P(Z > 2.72) 9.62 P( X 1 X 2 25 ) P 1 2 2 1 2 25 2 30 2 50 50 n1 n 2
= 1 – P(Z < 2.72) = 1 – .9967 = .0033 ( X X ) ( ) 25 (280 270 ) 2 1 2 = P(Z > 3.84) 9.63 P( X 1 X 2 25 ) P 1 2 2 25 2 30 2 1 2 100 100 n1 n 2
= 1 – P(Z < 3.84) = 1 – 1= 0
236
( X X ) ( ) 0 (40 38 ) 2 1 2 = P(Z > –1.00) 9.64 P( X 1 X 2 0) P 1 2 2 2 2 6 8 1 2 25 25 n n 2 1
= 1 – P(Z < –1.00) = 1 – .1587 = .8413 ( X X ) ( ) 0 (40 38 ) 2 1 2 = P(Z > –.50) = 1 – P(Z < –.50) 9.65 P( X 1 X 2 0) P 1 2 2 12 2 16 2 1 2 25 25 n1 n 2
= 1 – .3085 = .6915 ( X X ) ( ) 0 (140 138 ) 2 1 2 = P(Z > –1.00) 9.66 P( X 1 X 2 0) P 1 12 22 62 82 25 25 n1 n 2
= 1 – P(Z < –1.00) = 1 – .1587 = .8413 ( X X ) ( ) 0 (75 65 ) 2 1 2 = P(Z > –.77) = 1 – P(Z < −.77) 9.67 P( X 1 X 2 0) P 1 2 2 1 2 20 2 21 2 5 5 n1 n 2
= 1 – .2206 = .7794 ( X X ) ( ) 0 (73 77 ) 1 2 1 2 = P(Z > .51) = 1 – P(Z < .51) 9.68 P( X 1 X 2 0) P 12 22 12 2 10 2 4 4 n1 n 2
= 1 – .6950 = .3050 ( X X ) ( ) 0 (18 15 ) 2 1 2 = P(Z > –2.24) = 1– P(Z < –2.24) 9.69 P( X 1 X 2 0) P 1 2 2 32 32 1 2 10 10 n1 n 2
= 1 – .0125 = .9875
237
( X X ) ( ) 0 (10 15 ) 2 1 2 = P(Z < 5.89) = 1 9.70 P( X 1 X 2 0) P 1 2 2 2 2 3 3 1 2 25 25 n n 2 1
238
Chapter 10 10.1 A point estimator is a single value; an interval estimator is a range of values.
10.2 An unbiased estimator of a parameter is an estimator whose expected value equals the parameter.
10.3
10.4
10.5 An unbiased estimator is consistent if the difference between the estimator and the parameter grows smaller as the sample size grows.
10.6
239
10.7 If there are two unbiased estimators of a parameter, the one whose variance is smaller is relatively efficient.
10.8
10.13 a. x z / 2 / n = 100 1.645(25/ 50 ) = 100 5.82; LCL = 94.18, UCL = 105.82 b. x z / 2 / n = 100 1.96(25/ 50 ) = 100 6.93; LCL = 93.07, UCL = 106.93 c. x z / 2 / n = 100 2.575(25/ 50 ) = 100 9.11; LCL = 90.89, UCL = 109.11 d. The interval widens.
10.14 a. x z / 2 / n = 200 1.96(50/ 25 ) = 200 19.60; LCL = 180.40, UCL = 219.60 b. x z / 2 / n = 200 1.96(25/ 25 ) = 200 9.80; LCL = 190.20, UCL = 209.80 c. x z / 2 / n = 200 1.96(10/ 25 ) = 200 3.92; LCL = 196.08, UCL = 203.92 d. The interval narrows.
10.15 a. x z / 2 / n = 80 1.96(5/ 25 ) = 80 1.96; LCL = 78.04, UCL = 81.96 b. x z / 2 / n = 80 1.96(5/ 100 ) = 80 .98; LCL = 79.02, UCL = 80.98 c. x z / 2 / n = 80 1.96(5/ 400 ) = 80 .49; LCL = 79.51, UCL = 80.49 d. The interval narrows.
10.16 a. x z / 2 / n = 500 2.33(12/ 50 ) = 500 3.95; LCL = 496.05, UCL = 503.95 b. x z / 2 / n = 500 1.96(12/ 50 ) = 500 3.33; LCL = 496.67, UCL = 503.33 c. x z / 2 / n = 500 1.645(12/ 50 ) = 500 2.79; LCL = 497.21, UCL = 502.79 d. The interval narrows.
240
10.17 a. x z / 2 / n = 500 2.575(15/ 25 ) = 500 7.73; LCL = 492.27, UCL = 507.73 b. x z / 2 / n = 500 2.575(30/ 25 ) = 500 15.45; LCL = 484.55, UCL = 515.45 c. x z / 2 / n = 500 2.575(60/ 25 ) = 500 30.91; LCL = 469.09, UCL = 530.91 d. The interval widens.
10.18 a. x z / 2 / n = 10 1.645(5/ 100 ) = 10 .82; LCL = 9.18, UCL = 10.82 b. x z / 2 / n = 10 1.645(5/ 25 ) = 10 1.64; LCL = 8.36, UCL = 11.64 c. x z / 2 / n = 10 1.645(5/ 10 ) = 10 2.60; LCL = 7.40, UCL = 12.60 d. The interval widens.
10.19 a. x z / 2 / n = 100 1.96(20/ 25 ) = 100 7.84; LCL = 92.16, UCL = 107.84 b. x z / 2 / n = 200 1.96(20/ 25 ) = 200 7.84; LCL = 192.16, UCL = 207.84 c. x z / 2 / n = 500 1.96(20/ 25 ) = 500 7.84; LCL = 492.16, UCL = 507.84 d. The width of the interval is unchanged.
10.20 a. x z / 2 / n = 400 2.575(5/ 100 ) = 400 1.29; LCL = 398.71, UCL = 401.29 b. x z / 2 / n = 200 2.575(5/ 100 ) = 200 1.29; LCL = 198.71, UCL = 201.29 c. x z / 2 / n = 100 2.575(5/ 100 ) = 100 1.29; LCL = 98.71, UCL = 101.29 d. The width of the interval is unchanged.
10.21 Yes, because the expected value of the sample median is equal to the population mean.
10.22 The variance decreases as the sample size increases, which means that the difference between the estimator and the parameter grows smaller as the sample size grows larger.
10.23 Because the variance of the sample mean is less than the variance of the sample median, the sample mean is relatively more efficient than the sample median.
10.24 a sample median z / 2
1.2533
= 500 1.645
1.2533 (12 ) 50
n
241
= 500 3.50
b. The 90% confidence interval estimate of the population mean using the sample mean is 500 2.79. The 90% confidence interval of the population mean using the sample median is wider than that using the sample mean because the variance of the sample median is larger. The median is calculated by placing all the observations in order. Thus, the median loses the potential information contained in the actual values in the sample. This results in a wider interval estimate.
10.25 x z / 2 / n = 6.89 1.645(2/ 9 ) = 6.89 1.10; LCL = 5.79, UCL = 7.99
10.26 x z / 2 / n = 43.75 1.96(10/ 8 ) = 43.75 6.93; LCL = 36.82, UCL = 50.68 We estimate that the mean age of men who frequent bars lies between 36.82 and 50.68. This type of estimate is correct 95% of the time.
10.27 x z / 2 / n = 22.83 1.96(12/ 12 ) = 22.83 6.79; LCL = 16.04, UCL = 29.62
10.28 x z / 2 / n = 9.85 1.645(8/ 20 ) = 9.85 2.94; LCL = 6.91, UCL = 12.79
10.29 x z / 2 / n = 68.6 1.96(15/ 15 ) = 68.6 7.59; LCL = 61.01, UCL = 76.19 We estimate that the mean number of cars sold annually by all used car salespersons lies between 61.01 and 76.19. This type of estimate is correct 95% of the time.
10.30 x z / 2 / n = 16.9 2.575(5/ 10 ) = 16.9 4.07; LCL = 12.83, UCL = 20.97
10.31 x z / 2 / n = 147.33 1.96(40/ 15 ) = 147.33 20.24; LCL = 127.09, UCL = 167.57
10.32 x z / 2 / n = 13.15 1.645(6/ 13 ) = 13.15 2.74; LCL = 10.41, UCL = 15.89
10.33 x z / 2 / n = 75.625 2.575(15/ 16 ) = 75.625 9.656; LCL = 65.969, UCL = 85.281
10.34 x z / 2 / n = 252.38 1.96(30/ 400 ) = 252.38 2.94; LCL = 249.44, UCL = 255.32
242
10.35 x z / 2 / n = 1,810.16 1.96(400/ 64 ) = 1,810.16 98.00; LCL = 1,712.16, UCL = 1,908.16
10.36 x z / 2 / n = 12.10 1.645(2.1/ 200 ) = 12.10 .24; LCL = 11.86, UCL = 12.34. We estimate that the mean rate of return on all real estate investments lies between 11.86% and 12.34%. This type of estimate is correct 90% of the time.
10.37 x z / 2 / n = 10.21 2.575(2.2/ 100 ) = 10.21 .57; LCL = 9.64, UCL = 10.78
10.38 x z / 2 / n = .510 2.575(.1/ 250 ) = .510 .016; LCL = .494, UCL = .526. We estimate that the mean growth rate of this type of grass lies between .494 and .526 inch . This type of estimate is correct 99% of the time.
10.39 x z / 2 / n = 26.81 1.96(1.3/ 50 ) = 26.81 .36; LCL = 26.45, UCL = 27.17. We estimate that the mean time to assemble a cell phone lies between 26.45 and 27.17 minutes. This type of estimate is correct 95% of the time.
10.40 x z / 2 / n = 19.28 1.645(6/ 250 ) = 19.28 .62; LCL = 18.66, UCL = 19.90. We estimate that the mean leisure time per week of Japanese middle managers lies between 18.66 and 19.90 hours. This type of estimate is correct 90% of the time.
10.41 x z / 2 / n = 15.00 2.575(2.3/ 100 ) = 15.00 .59; LCL = 14.41, UCL = 15.59. We estimate that the mean pulse-recovery time lies between 14.41 and 15.59 minutes. This type of estimate is correct 99% of the time.
10.42 x z / 2 / n = 585,063 1.645(30,000/ 80 ) = 585,063 5,518; LCL = 579,545, UCL = 590,581. We estimate that the mean annual income of all company presidents lies between $579,545 and $590,581. This type of estimate is correct 90% of the time.
10.43 x z / 2 / n = 109.6 1.96(20/ 200 ) = 109.60 2.77; LCL = 106.8, UCL = 112.4
243
10.44 x z / 2 / n = 227.48 2.575(50/ 300 ) = 227.48 7.43; LCL = 220.05, UCL = 234.91
10.45 x z / 2 / n = 314,245 1.645(75,000/ 150 ) = 314,245 10,074; LCL = 304,171, UCL = 324,319
10.46 x z / 2 / n = 27.19 1.96(8/ 100 ) = 27.19 1.57; LCL = 25.62, UCL = 28.76
2
2
z 1.645 50 10.47 a. n / 2 = = 68 B 10 2
2
z 1.645 100 b. n / 2 = = 271 B 10 2
2
z 1.96 50 c. n / 2 = = 97 10 B 2
2
z 1.645 50 d. n / 2 = = 17 20 B
10.48 a The sample size increases. b The sample size increases. c The sample size decreases.
2
2
z 2.575 250 10.49 a. n / 2 = = 166 B 50 2
2
2
2
z 2.575 50 b. n / 2 = =7 B 50
z 1.96 250 c. n / 2 = = 97 B 50 2
2
z 2.575 250 d. n / 2 = = 4,145 10 B
10.50 a The sample size decreases. b The sample size decreases. c The sample size increases.
244
2
2
z 1.645 10 10.51 a. n / 2 = = 271 1 B
b. 150 1
10.52 a. x z / 2 b. x z / 2
150 1.645
5
150 .5
271
n
150 1.645
n
20
150 2
271
10.53 a. The width of the confidence interval estimate is equal to what was specified. b. The width of the confidence interval estimate is smaller than what was specified. c. The width of the confidence interval estimate is larger than what was specified.
2
2
z 1.96 200 10.54 a. n / 2 = = 1,537 10 W
b. 500 10
10.55 a. x z / 2 b. x z / 2
500 1.96
100
500 5
1537
n
500 1.96
n
400
500 20
1537
10.56 a The width of the confidence interval estimate is equal to what was specified. b The width of the confidence interval estimate is smaller than what was specified. c The width of the confidence interval estimate is larger than what was specified.
2
2
10.57
z 1.645 10 n / 2 = = 68 2 B
10.58
z 2.575 360 n / 2 = = 2,149 20 B
10.59
z 1.96 12 n / 2 = = 139 2 B
2
2
2
2
245
2
2
10.60
z 1.645 20 n / 2 = = 1,083 1 B
10.61
z 1.96 25 n / 2 = = 97 5 B
10.62
z 1.96 15 n / 2 = = 217 2 B
2
2
2
2
246
Chapter 11 11.1
H 0 : The drug is not safe and effective H 1 : The drug is safe and effective
11.2
H 0 : I will complete the Ph.D. H 1 : I will not be able to complete the Ph.D.
11.3
H 0 : The batter will hit one deep H 1 : The batter will not hit one deep
11.4
H 0 : Risky investment is more successful H 1 : Risky investment is not more successful
11.5
H 1 : The plane is on fire H 1 : The plane is not on fire
11.6 The defendant in both cases was O. J. Simpson. The verdicts were logical because in the criminal trial the amount of evidence to convict is greater than the amount of evidence required in a civil trial. The two juries concluded that there was enough (preponderance of) evidence in the civil trial, but not enough evidence (beyond a reasonable doubt) in the criminal trial.
All p-values and probabilities of Type II errors were calculated manually using Table 3 in Appendix B.
11.7 Rejection region: z < z .005 2.575 or z > z.005 = 2.575
z
x / n
980 1000
1.00
200 / 100
p-value = 2P(Z < –1.00) = 2(.1587) = .3174 There is not enough evidence to infer that 1000.
247
11.8 Rejection region: z > z.03 = 1.88
z
x / n
51 50
.60
5/ 9
p-value = P(Z > .60) = 1 – .7257 = .2743 There is not enough evidence to infer that > 50.
11.9 Rejection region: z < z.10 1.28 z
x / n
14 .3 15 2 / 25
1.75
p-value = P(Z < –1.75) = .0401 There is enough evidence to infer that < 15.
11.10 Rejection region: z < z .025 1.96 or z > z.025 = 1.96 z
x / n
100 100 10 / 100
0
p-value = 2P(Z > 0) = 2(.5) = 1.00
248
There is not enough evidence to infer that 100.
11.11 Rejection region: z > z.01 = 2.33 z
x / n
80 70 20 / 100
5.00
p-value = p(z > 5.00) = 0 There is enough evidence to infer that > 70.
11.12 Rejection region: z < z.05 1.645 z
x / n
48 50 15 / 100
1.33
p-value = P(Z < –1.33) = .0918 There is not enough evidence to infer that < 50.
11.13 z
x
/ n
240 250 40 / 70
2.09
p-value = P(Z < -2.09) = .0182
249
11.14 z
x
/ n
1525 1500 220 / 125
1.27
p-value = 2P(Z > 1.27) = 2(1-P(Z < 1.27) = 2(1 - .8980) = 2(.1020) = .2040
11.15 z
x
/ n
8.5 7.5
3.65
1.5 / 30
p-value = P(Z > 3.65) = 0.
11.16 z
x
1.5 0
/ n
1.50
10 / 100
p-value = P(Z > 1.50) = 1 – P(Z < 1.5) = 1 - .9332 = .0668
11.17 z
x
/ n
2.3 0 25 / 400
1.84
p-value = P(Z < -1.84) = .0329
11.18 z
x
/ n
5.5 0 50 / 90
1.04
p-value = 2P(Z < -1.04) = 2(.1492) = .2984
11.19 z
x
/ n
4 (5) 5 / 25
1.00
p-value =2P(Z > 1.00) = 2(1 – P(Z < 1.00) = 2(1 - .8413) = .3174
11.20 a. No because the test statistic will be negative. b The p-value will be larger than .5.
11.21a. z
x / n
52 50
1.20
5/ 9
p-value = P(Z > 1.20) = 1 – .8849 = .1151 b. z
x / n
52 50 5 / 25
2.00
p-value = P(Z > 2.00) = .5 – .4772 = .0228. c. z
x / n
52 50 5 / 100
4.00
p-value = P(Z > 4.00) = 0.
250
d. The value of the test statistic increases and the p-value decreases.
11.22a. z
x / n
190 200
.60
50 / 9
p-value = P(Z < –.60) = .5 – .2257 = .2743 b. z
x
190 200
/ n
1.00
30 / 9
p-value = P(Z < –1.00) = .1587 c z
x
/ n
190 200
3.00
10 / 9
p-value = P(Z < –3.00) = .0013 d. The value of the test statistic decreases and the p-value decreases. x
11.23 a. z
/ n
21 20 5 / 25
1.00
p-value = 2P(Z > 1.00) = 2(1 – .8413) = .3174 b. z
x / n
22 20
2.00
5 / 25
p-value = 2P(Z > 2.00) = 2(1 – .9772) = .0456 c. z
x
/ n
23 20
3.00
5 / 25
p-value = 2P(Z > 3.00) = 2(1 – .9987) = .0026 d. The value of the test statistic increases and the p-value decreases. x
11.24 a. z
/ n
99 100 8 / 100
1.25
p-value = 2P(Z < –1.25) = 2(.1056) = .2112 b. z
x / n
99 100
.88
8 / 50
p-value = 2P(Z < –.88) = 2(.1894) = .3788 c. z
x / n
99 100
.56
8 / 20
p-value = 2P(Z < –.56) = 2(.2877) = .5754 d. The value of the test statistic increases and the p-value increases.
11.25 a. z
x / n
990 1000
4.00
25 / 100 251
p-value = P(Z < –4.00) = 0 b. z
x / n
990 1000
2.00
50 / 100
p-value = P(Z < –2.00) = .0228 c. z
x
/ n
990 1000
1.00
100 / 100
p-value = P(Z < –1.00) = .1587 d. d. The value of the test statistic increases and the p-value increases.
x
11.26 a. z
/ n
72 60
3.00
20 / 25
p-value = P(Z > 3.00) = 1 – .9987 = .0013 b. z
x / n
68 60
2.00
20 / 25
p-value = P(Z > 2.00) = 1 – .9772 = .0228 c. z
x / n
64 60
1.00
20 / 25
p-value = P(Z > 1.00) = 1 – .8413 = .1587 d. The value of the test statistic decreases and the p-value increases.
11.27 a z
x / n
178 170 65 / 200
1.74
p-valu.e = P(Z > 1.74) = 1 – .9591 = .0409 b. z
x / n
178 170 65 / 100
1.23
p-value = P(Z > 1.23) = 1 – .8907 = .1093 c. The value of the test statistic increases and the p-value decreases.
11.28 a z
x / n
178 170 35 / 400
4.57
p-value = P(Z > 4.57) = 0. b z
x / n
178 170 100 / 400
1.60
p-value = P(Z > 1.60) = 1 – .9452 = .0548 The value of the test statistic decreases and the p-value increases.
11.29 a Yes, because the test statistic will be positive.
252
b The p-value will be larger than .5. x
11.30 a z
/ n
21.63 22
.62
6 / 100
p-value = P(Z < –.62) = .2676 bz
x / n
21 .63 22 6 / 500
1.38
p-value = P(Z < –1.38) = .0838 The value of the test statistic decreases and the p-value decreases.
11.31 a z
x / n
21 .63 22
3 / 220
1.83
p-value = P(Z < –1.83) = .0336 bz
x / n
21.63 22 12 / 220
.46
p-value = P(Z < –.46) = .3228 The value of the test statistic increases and the p-value increases.
11.32
x 22
z
x
p-value
6 / 220 22.0 21.8 21.6 21.4 21.2 21.0 20.8 20.6 20.4
0 –.49 –.99 –1.48 –1.98 –2.47 –2.97 –3.46 –3.96
x
11.33 a z
/ n
.5 .3121 .1611 .0694 .0239 .0068 .0015 0 0
17 .55 17 .09
.84
3.87 / 50
p-value = 2P(Z > .84) = 2(1 – .7995) = 2(.2005) = .4010 bz
x / n
17.55 17.09 3.87 / 400
2.38
p-value = 2P(Z > 2.38) = 2(1 – .9913) = 2(.0087) = .0174 The value of the test statistic increases and the p-value decreases.
11.34 a z
x / n
17 .55 17 .09
2.30
2 / 100
p-value = 2P(Z > 2.30) = 2(1 – .9893) = 2(.0107) = .0214 253
x
bz
/ n
17 .55 17 .09
.46
10 / 100
p-value = 2P(Z > .46) = 2(1 – .6772) = 2(.3228) = .6456 The value of the test statistic decreases and the p-value increases.
11.35a
11.36
x 17 .09
x
z
15.0 15.5 16.0 16.5 17.0 17.5 18.0 18.5 19.0
3.87 / 100 –5.40 –4.11 –2.82 –1.52 –.23 1.06 2.35 3.64 4.94
p-value 0 0 .0048 .1286 .8180 .2892 .0188 0 0
H0 : = 5
H1 : > 5 z
x / n
65 1.5 / 10
2.11
p-value = P(Z > 2.11) = 1 – .9826 = .0174 There is enough evidence to infer that the mean is greater than 5 cases.
11.37
H 0 : = 50
H1 : > 50 z
x / n
59 .17 50 10 / 18
3.89
p-value = P(Z > 3.89) = 0 There is enough evidence to infer that the mean is greater than 50 minutes.
11.38
H 0 : = 12
H1 : < 12 z
x / n
11.00 12 3 / 15
1.29
p-value = P(Z < –1.29) = .0985 There is enough evidence to infer that the average number of golf balls lost is less than 12.
11.39
H 0 : = 36
254
H1 : < 36 z
x / n
34 .25 36 8 / 12
.76
p-value = P(Z < –.76) = .2236 There is not enough evidence to infer that the average student spent less time than recommended.
11.40
H0 : = 6
H1 : > 6 z
x / n
6.60 6
.95
2 / 10
p-value = P(Z > .95) = 1 – .8289 = .1711 There is not enough evidence to infer that the mean time spent putting on the 18 th green is greater than 6 minutes.
11.41
H 0 : = .50
H1 : .50 z
x / n
.493 .50 .05 / 10
.44
p-value = 2P(Z < –.44) = 2(.3300) = .6600 There is not enough evidence to infer that the mean diameter is not .50 inch.
11.42
H 0 : = 25
H1 : > 25 z
x / n
30 .22 25 12 / 18
1.85
p-value = P(Z > 1.85) = 1 – .9678 =.0322 There is not enough evidence to conclude that the manager is correct.
11.43
H 0 : = 5,000
H1 : > 5,000 z
x / n
5,065 5,000
1.62
400 / 100
p-value = P(Z > 1.62) = 1 – .9474 =.0526 There is not enough evidence to conclude that the claim is true.
255
11.44
H 0 : = 30,000
H1 : < 30,000
z
x / n
29 ,120 30 ,000
2.06
8,000 / 350
p-value = P(Z < –2.06) = .0197 There is enough evidence to infer that the president is correct
11.45
H 0 : = 560
H1 : > 560 z
x / n
569 .0 560
.80
50 / 20
p-value = P(Z > .80) = 1 – .7881 = .2119 There is not enough evidence to conclude that the dean’s claim is true.
11.46a
H 0 : = 17.85
H1 : > 17.85 z
x / n
19 .13 17 .85
1.65
3.87 / 25
p-value = P(Z > 1.65) = 1 – .9505 = .0495 There is enough evidence to infer that the campaign was successful. b We must assume that the population standard deviation is unchanged.
11.47
H0 : = 0
H1 : < 0 z
x / n
1.20 0
1.41
6 / 50
p-value = P(Z < –1.41) = .0793 There is not enough evidence to conclude that the safety equipment is effective.
11.48
H 0 : = 55
H1 : > 55 z
x / n
55 .80 55
2.26
5 / 200
p-value = P(Z > 2.26) = 1 – .9881 = .0119 There is not enough evidence to support the officer’s belief. 256
11.49
H0 : = 4
H1 : > 4 z
x / n
5.04 4
4.90
1.5 / 50
p-value = P(Z > 4.90) = 0 There is enough evidence to infer that the expert is correct.
11.50
H 0 : = 20
H1 : < 20 z
x / n
19 .39 20
1.22
3 / 36
p-value = P(Z < –1.22) = .1112 There is not enough evidence to infer that the manager is correct.
11.51
H 0 : = 100
H1 : > 100 z
x / n
105 .7 100
2.25
16 / 40
p-value = P(Z > 2.25) = 1 – .9878 = .0122 There is not enough evidence to infer that the site is acceptable.
11.52
H0 : = 4
H1 : 4 z
x / n
4.84 4
3.33
2 / 63
p-value = 2P(Z > 3.33) = 0 There is enough evidence to infer that the average Alpine skier does not ski 4 times per year.
11.53
H0 : = 5
H1 : > 5 z
x / n
5.64 5 2 / 25
1.60
p-value = P(Z > 1.60) = 1 – .9452 = .0548 There is enough evidence to infer that the golf professional’s claim is true.
257
11.54
H 0 : = 32
H1 : < 32 z
x / n
29 .92 32
2.73
8 / 110
p-value = P(Z < –2.73) = 1– .9968 = .0032 There is enough evidence to infer that there has been a decrease in the mean time away from desks. A type I error occurs when we conclude that the plan decreases the mean time away from desks when it actually does not. This error is quite expensive. Consequently we demand a low pvalue. The p-value is small enough to infer that there has been a decrease.
11.55
H 0 : = 230
H1 : > 230 z
x / n
231 .56 230 10 / 100
1.56
p-value = P(Z > 1.56) = 1 – .9406 = .0594 There is not enough evidence to infer that Nike is correct. 11.56
H 0 : = 10
H1 : > 10 z
x
/ n
10.44 10 3 / 174
1.93
p-value = P(Z > 1.93) = 1 – .9732 = .0268 There is enough evidence to infer that heating costs increased faster than inflation.
11.57
H 0 : = 30
H1 : 30 z
x
/ n
29 .51 30
1.63
5 / 277
p-value = 2P(Z < -1.63) = 2(.0516) = .1032 There is not enough evidence to infer that the average mean monthly expenditures on bakery products is not equal to $30.
11.58
H 0 : = 125,000
H1 : > 125,000
258
z
x
/ n
126 ,837 125 ,000 25,000 / 410
1.49
p-value = P(Z > 1.49) = 1 – .9319 = .0681 There is not enough evidence to infer that mean value of 401k accounts is greater than $125,000.
11.59
H 0 : = 7500
H1 : > 7500 z
x
/ n
7625 7500 1200 / 163
1.33
p-value = P(Z > 1.33) = 1 – .9082 = .0918 There is not enough evidence to infer that the mean income exceeds $7500.
11.60 Rejection region:
x 200
x
> z / 2 or
/ n
> z .025 1.96 or
10 / 100 x > 201.96 or x < 198.04
x 200 10 / 100
x / n
< z / 2
< –1.96
= P(198.04 < x < 201.96 given = 203)
198 .04 203 x 201 .96 203 = P( –4.96 < z < –1.04) = .1492 – 0 = .1492 = P / n 10 / 100 10 / 100
11.61 Rejection region:
x / n
> z
x 1000
> z.01 2.33 50 / 25 x > 1023.3
x 1023 .3 1050 = P(z < –2.67) = .0038 = P( x < 1023.3 given = 1050) = P 50 / 25 / n
11.62 Rejection region:
x 50 10 / 40 x < 47.40
x / n
< z
< z.05 1.645
x 47 .40 48 = P(z > –.38) = 1 − .3520 = .6480 = P( x > 47.40 given = 48) = P 10 / 40 / n
11.63 Exercise 11.48 259
Exercise 11.49
Exercise 11.50
260
11.64 a. Rejection region:
x 100 20 / 100 x > 102.56
x / n
> z
> z .10 1.28
x 102 .56 102 = P(z < .28) = .6103 = P( x < 102.56 given = 102) = P 20 / 100 / n x b. Rejection region: > z / n x 100 > z .02 2.55 20 / 100 x > 104.11 x 104 .11 102 = P(z < 1.06) = .8554 = P( x < 104.11 given = 102) = P 20 / 100 / n c. increases.
11.65 a. Rejection region:
x / n
< z
x 40
< z .05 1.645 5 / 25 x < 38.36
x 38 .36 37 = P(z > 1.36) = 1 – .9131 = .0869 = P( x > 38.36 given = 37) = P 5 / 25 / n x < z b. Rejection region: / n x 40 < z .15 1.04 5 / 25 x < 38.96 x 38 .96 37 = P(z > 1.96) = 1 – .9750 = .0250 = P( x > 38.96 given = 37) = P 5 / 25 / n c. decreases.
11.66 Exercise 11.52 a
261
Exercise 11.52 b
Exercise 11.53 a
262
Exercise 11.53 b
11.67 a. Rejection region:
x / n
< z
x 200
< z .10 1.28 30 / 25 x < 192.31 x 192 .31 196 = P(z > –.62) = 1 − .2676 = .7324 = P( x > 192.31 given = 196) = P 30 / 25 / n x b. Rejection region: < z / n x 200 < z .10 1.28 30 / 100 x < 196.16 x 196 .16 196 = P(z > .05) = 1 – .5199 = .4801 = P( x > 196.16 given = 196) = P 30 / 100 / n
263
c. decreases.
11.68 a. Rejection region:
x / n
> z
x 300
> z .05 1.645 50 / 81 x > 309.14 x 309 .14 310 = P(z < –.15) = .4404 = P( x < 309.14 given = 310) = P 50 / 81 / n x > z b. Rejection region: / n x 300 > z .05 1.645 50 / 36 x > 313.71 x 313 .71 310 = P(z < .45) = .6736 = P( x < 313.71 given = 310) = P 50 / 36 / n c. increases.
11.69 Exercise 11.55 a
Exercise 11.55 b
264
Exercise 11.56 a
Exercise 11.56 b
265
11.70
11.71 266
11.72
H 0 : = 170
H1 : < 170 A Type I error occurs when we conclude that the new system is not cost effective when it actually is. A Type II error occurs when we conclude that the new system is cost effective when it actually is not.
The test statistic is the same. However, the p-value equals 1 minus the p-value calculated Example 11.1. That is, p-value = 1 – .0069 = .9931 We conclude that there is no evidence to infer that the mean is less than 170. That is, there is no evidence to infer that the new system will not be cost effective.
11.73 Rejection region:
x0 6 / 50
x / n
< z
< z .10 1.28
x < –1.09
x 1.09 (2) = P(z > 1.07) = 1 – .8577 = .1423 = P( x > –1.09 given = –2) = P 6 / 50 / n can be decreased by increasing and/or increasing the sample size.
267
11.74 Rejection region:
x 22
x / n
< z
< z .10 1.28
6 / 220 x < 21.48
x 21 .48 21 = P(z > 1.19) =1 – .8830 = .1170 = P( x > 21.48 given = 21) = P 6 / 220 / n
The company can decide whether the sample size and significance level are appropriate.
11.75 Rejection region:
x / n
> z
x 100
> z .01 2.33 16 / 40 x > 105.89 x 105 .89 104 = P(z < .75) = .7734 = P( x < 105.89 given = 104) = P 16 / 40 / n
11.76 Rejection region:
x 32
x / n
< z
< z .05 1.645
8 / 110 x < 30.75
x 30 .75 30 ) = P(z > .98) = 1 – .8365 = .1635 = P( x > 30.75 given = 30) = P 8 / 110 / n can be decreased by increasing and/or increasing the sample size.
11.77 i Rejection region:
x 10 3 / 100
x / n
< z
< z.01 2.33
x < 9.30
x 9.30 9 = P(z > 1) = 1 – .8413 = .1587 = P( x > 9.30 given = 9) = P / n 3 / 100 ii Rejection region:
x 10 3 / 75
x / n
< z
< z.05 1.645
268
x < 9.43
x 9.43 9 = P(z > 1.24) = 1 – .8925 = .1075 = P( x > 9.43 given = 9) = P / n 3 / 75
iii Rejection region:
x 10 3 / 50
x / n
< z
< z.10 1.28
x < 9.46
x 9.46 9 = P(z > 1.08) = 1 – .8599 = .1401 = P( x > 9.46 given = 9) = P / n 3 / 50 Plan ii has the lowest probability of a type II error.
11.78 A Type I error occurs when we conclude that the site is feasible when it is not. The consequence of this decision is to conduct further testing. A Type II error occurs when we do not conclude that a site is feasible when it actually is. We will do no further testing on this site, and as a result we will not build on a good site. If there are few other possible sits, this could be an expensive mistake.
11.79
H 0 : 20 H1 : 25
Rejection region:
x / n
> z
x 20
> z .01 2.33 8 / 25 x > 23.72
x 23 .72 25 = P(z < –.80) = .2119 = P( x < 23.72 given = 25) = P 8 / 25 / n The process can be improved by increasing the sample size.
269
Chapter 12 12.1a x t / 2 s / n = 70 2.004(12/ 56 ) = 70 ± 3.21; LCL = 66.79, UCL = 73.21 b x t / 2 s / n = 35 ± 2.004(12/ 56 ) = 30 ± 3.21; LCL = 26.79, UCL = 33.21 c. The interval width does not change.
12.2a x t / 2 s / n = 50 ± 1.711(10/ 25 ) = 50 ± 3.42; LCL = 46.58, UCL = 53.42 b x t / 2 s / n = 100 ± 1.711(10/ 25 ) = 100 ± 3.42; LCL = 96.58, UCL = 103.42 c. The interval width does not change.
12.3 a x t / 2 s / n = 510 ± 2.064(125/ 25 ) = 510 ± 51.60; LCL = 458.40, UCL = 561.60 b x t / 2 s / n = 510 ± 2.009(125/ 50 ) = 510 ± 35.51; LCL = 474.49, UCL = 545.51 c x t / 2 s / n = 510 ± 1.984(125/ 100 ) = 510 ± 24.80; LCL = 485.20, UCL = 534.80 d. The interval narrows.
12.4 a x t / 2 s / n = 1,500 ± 1.984(300/ 100 ) = 1,500 ± 59.52; LCL = 1,440.48, UCL = 1,559.52 b x t / 2 s / n = 1,500 ± 1.984(200/ 100 ) = 1,500 ± 39.68; LCL = 1,460.32, UCL = 1,539.68 c x t / 2 s / n = 1,500 ± 1.984(100/ 100 ) = 1,500 ± 19.84; LCL = 1,480.16, UCL = 1,519.84 d. The interval narrows.
12.5 a x t / 2 s / n = 700 ± 1.645(100/ 400 ) = 700 ± 8.23; LCL = 691.77, UCL = 708.23 b x t / 2 s / n = 700 ± 1.96(100/ 400 ) = 700 ± 9.80; LCL = 690.20, UCL = 709.80 a x t / 2 s / n = 700 ± 2.576(100/ 400 ) = 700 ± 12.88; LCL = 687.12, UCL = 712.88 d. The interval widens.
12.6 a x t / 2 s / n = 10 ± 1.984(1/ 100 ) = 10 ± .20; LCL = 9.80, UCL = 10.20 b x t / 2 s / n = 10 ± 1.984(4/ 100 ) = 10 ± .79; LCL = 9.21, UCL = 10.79 c x t / 2 s / n = 10 ± 1.984(10/ 100 ) = 10 ± 1.98; LCL = 8.02, UCL = 11.98 d The interval widens.
251
12.7 a x t / 2 s / n = 120 ± 2.009(15/ 51 ) = 120 ± 4.22; LCL = 115.78, UCL = 124.22 b x t / 2 s / n = 120 ± 1.676(15/ 51 ) = 120 ± 3.52; LCL = 116.48, UCL = 123.52 c x t / 2 s / n = 120 ± 1.299(15/ 51 ) = 120 ± 2.73; LCL = 117.27, UCL = 122.73 d The interval narrows.
12.8 a x t / 2 s / n = 63 ± 1.990(8/ 81 ) = 63 ± 1.77; LCL = 61.23, UCL = 64.77 b x t / 2 s / n = 63 ± 2.000(8/ 64 ) = 63 ± 2.00; LCL = 61.00, UCL = 65.00 c x t / 2 s / n = 63 ± 2.030(8/ 36 ) = 63 ± 2.71; LCL = 60.29, UCL = 65.71 d The interval widens.
H 0 : = 20
12.9
H1 : > 20 a Rejection region: t t , n 1 t.05,9 = 1.833
x
t
s/ n
23 20
1.05, p-value = .1597. There is not enough evidence to infer that the population mean is
9 / 10
greater than 20. b Rejection region: t t , n 1 t.05, 29 = 1.699
x
t
s/ n
23 20
1.83, p-value = .0391. There is enough evidence to infer that the population mean is greater
9 / 30
than 20. c Rejection region: t t , n 1 t .05, 49 1.676
x
t
s/ n
23 20
2.36, p-value = .0112. There is enough evidence to infer that the population mean is greater
9 / 50
than 20. d As the sample size increases the test statistic increases [and the p-value decreases].
12.10
H 0 : = 180
H1 : 180 Rejection region: t t / 2, n 1 t .025,199 1.972 or t t / 2,n 1 t .025,199 = 1.972 at
x
s/ n
175 180
3.21, p-value = .0015. There is enough evidence to infer that the population mean is not
22 / 200
equal to 180.
252
x
b t
s/ n
175 180
1.57 , p-value = .1177. There is not enough evidence to infer that the population mean is
45 / 200
not equal to 180.
x
c t
s/ n
175 180
1.18, p-value = .2400. There is not enough evidence to infer that the population mean is
60 / 200
not equal to 180. d. As s increases, the test statistic increases and the p-value increases. 12.11 Rejection region: t t ,n 1 t .05,99 1.660
x
at
145 150
s/ n
1.00, p-value = .1599. There is not enough evidence to infer that the population mean is
50 / 100
less than 150.
x
bt
s/ n
140 150
2.00, p-value = .0241. There is enough evidence to infer that the population mean is less
50 / 100
than 150.
x
ct
s/ n
135 150
3.00, p-value = .0017. There is enough evidence to infer that the population mean is less
50 / 100
than 150 d The test statistics decreases and the p-value decreases.
12.12
H0: µ = 50 H1: µ ≠ 50
a Rejection region: t t / 2,n 1 t .05, 24 1.711 or t t / 2,n 1 t .05, 24 1.711
t
x
s/ n
52 50
.67 , p-value = .5113. There is not enough evidence to infer that the population mean is not
15 / 25
equal to 50. b Rejection region: t t / 2,n 1 t .05,14 1.761 or t t / 2, n 1 t .05,14 1.761
t
x
s/ n
52 50
.52, p-value = .6136. There is not enough evidence to infer that the population mean is not
15 / 15
equal to 50. c Rejection region: t t / 2,n 1 t .05, 4 2.132 or t t / 2,n 1 t .05, 4 2.132
t
x
s/ n
52 50
.30, p-value = .7804. There is not enough evidence to infer that the population mean is not
15 / 5
equal to 50. d The test statistic decreases and the p-value increases.
253
12.13 Rejection region: t t , n 1 t .10, 49 1.299 at
x
s/ n
585 600
2.36, p-value = .0112. There is enough evidence to infer that the population mean is less
45 / 50
than 600. bt
x
s/ n
590 600
1.57 , p-value = .0613. There is enough evidence to infer that the population mean is less
45 / 50
than 600. ct
x
s/ n
595 600
.79, p-value = .2179. There is not enough evidence to infer that the population mean is
45 / 50
less than 600. d The test statistic increases and the p-value increases. 12.14 Rejection region: t t , n 1 t .01,99 2.364 at
x
s/ n
106 100
1.71, p-value = .0448. There is not enough evidence to infer that the population mean is
35 / 100
greater than 100. bt
x
s/ n
106 100
2.40, p-value = .0091. There is enough evidence to infer that the population mean is
25 / 100
greater than 100. ct
x s/ n
106 100
4.00, p-value = .0001. There is enough evidence to infer that the population mean is
15 / 100
greater than 100 d The test statistic increases and the p-value decreases.
12.15 a x t / 2 s / n = 40 2.365(10/ 8 ) = 40 ± 8.36; LCL = 31.64, UCL = 48.36 b x z / 2 / n = 40 1.96(10/ 8 ) = 40 ± 6.93; LCL = 33.07, UCL = 46.93 c The student t distribution is more widely dispersed than the standard normal; thus, z / 2 is smaller than t / 2 .
12.16 a x t / 2 s / n = 175 2.132(30/ 5 ) = 175± 28.60; LCL = 146.40, UCL = 203.60 b x z / 2 / n = 175 1.645(30/ 5 ) = 175± 22.07; LCL = 152.93, UCL = 197.07 c The student t distribution is more widely dispersed than the standard normal; thus, z / 2 is smaller than t / 2 .
12.17 a x t / 2 s / n = 15,500 1.645(9,950/ 1,000 ) = 15,500 ± 517.59; LCL = 14,982.41, UCL = 16,017.59 b x z / 2 / n = 15,500 1.645(9,950/ 1,000 ) = 15,500 ± 517.59; LCL = 14,982.41, UCL = 16,017.59 254
c With n = 1,000 the student t distribution with 999 degrees of freedom is almost identical to the standard normal distribution.
12.18 a x t / 2 s / n = 350 2.576(100/ 500 ) = 350± 11.52; LCL = 338.48, UCL = 361.52 b x z / 2 / n = 350 2.575(100/ 500 ) = 350 ± 11.52; LCL = 338.48, UCL = 361.52 c With n = 500 the student t distribution with 999 degrees of freedom is almost identical to the standard normal distribution.
12.19
H0: µ = 70 H1: µ > 70
a Rejection region: t t , n 1 t .05,10 = 1.812
t
x
74 .5 70
s/ n
1.66, p-value = .0641. There is not enough evidence to infer that the population mean is
9 / 11
greater than 70. b Rejection region: z z z .05 = 1.645
z
x / n
74 .5 70
9 / 11
1.66, p-value = P(Z > 1.66) = 1 – P(Z< 1.66) = 1 – .9515 = .0485. There is enough
evidence to infer that the population mean is greater than 70. c The Student t distribution is more dispersed than the standard normal.
12.20
H0: µ = 110 H1: µ < 110
a Rejection region: t t , n 1 t .10,9 = –1.383
t
x
103 110
s/ n
1.30, p-value = .1126. There is not enough evidence to infer that the population mean is
17 / 10
less than 110. b Rejection region: z z z .10 = –1.28
z
x / n
103 110 17 / 10
1.30, p-value = P(Z < –1.30) = .0968. There is enough evidence to infer that the
population mean is less than 110. c The Student t distribution is more dispersed than the standard normal.
12.21
H0: µ = 15 H1: µ < 15
a Rejection region: t t ,n 1 t .05,1499 = –1.645
255
t
x
14 15
s/ n
1.55, p-value = .0608. There is not enough evidence to infer that the population mean is
25 / 1,500
less than 15. b Rejection region: z z z .05 = –1.645
z
x / n
14 15
25 / 1,500
1.55, p-value = P(Z < –1.55) = .0606. There is not enough evidence to infer that the
population mean is less than 15. c With n = 1,500 the student t distribution with 1,499 degrees of freedom is almost identical to the standard normal distribution. 12.22 a. Rejection region: t t , n 1 t .05,999 = 1.645
t
x
405 400
s/ n
1.58, p-value = .0569. There is not enough evidence to infer that the population mean is
100 / 1,000
less than 15. b Rejection region: z z z .05 = 1.645
z
x
/ n
405 400 100 / 1,000
1.58, p-value = P( Z > 1.58) = 1 – .9429 = .0571. There is not enough evidence to infer
that the population mean is less than 15. c With n = 1,000 the student t distribution with 999 degrees of freedom is almost identical to the standard normal distribution.
12.23
H0: µ = 6 H1: µ < 6
a Rejection region: t t , n 1 t .05,11 = –1.796
t
x
s/ n
5.69 6
.68, p-value = .2554. There is not enough evidence to support the courier’s
1.58 / 12
advertisement.
12.24 x t / 2 s / n = 24,051 2.145(17,386/ 15 ) = 24,051 9,628; LCL = 14,422, UCL = 33,680
12.25
H0: µ = 20 H1: µ > 20
Rejection region: t t , n 1 t .05,19 1.729
t
x s/ n
20 .85 20
.56, p-value = .2902. There is not enough evidence to support the doctor’s claim.
6.76 / 20 256
12.26
H 0: µ = 8 H 1: µ < 8
Rejection region: t t ,n 1 t.01,17 2.567
t
x
7.91 8 .085 / 18
s/ n
–4.49, p-value = .0002. There is enough evidence to conclude that the average container is
mislabeled.
12.27 x t / 2 s / n = 18.13 2.145(9.75/ 15 ) = 18.13± 5.40; LCL = 12.73, UCL =23.53
12.28 x t / 2 s / n = 26.67 1.796(16.52/ 12 ) = 26.67± 8.56; LCL = 18.11, UCL = 35.23
12.29 x t / 2 s / n = 17.70 2.262(9.08/ 10 ) = 17.70± 6.49; LCL = 11.21, UCL =24.19
12.30
H0: µ = 10 H1: µ < 10
Rejection region: t t ,n 1 t .10,9 1.383
t
x s/ n
7.10 10
2.45, p-value = .0185. There is enough evidence to infer that the mean proportion of
3.75 / 10
returns is less than 10%.
12.31 x t / 2 s / n = 7.15 1.972(1.65/ 200 ) = 7.15 ± .23; LCL = 6.92, UCL = 7.38
12.32 x t / 2 s / n = 4.66 2.576(2.37/ 240 ) = 4.66 ± .39; LCL = 4.27, UCL = 5.05 Total number: LCL = 100 million (4.27) = 427 million, UCL = 100 million (5.05) = 505 million
12.33 x t / 2 s / n =17.00 1.975(4.31/ 162 = 17.00 ±.67; LCL = 16.33, UCL = 17.67 Total number: LCL = 50 million (16.33) = 816.5 million, UCL = 50 million (17.67) = 883.5 million
12.34 x t / 2 s / n = 15,137 1.96(5,263/ 306 = 15,137 ±590; LCL = 14,547, UCL = 15,727 Total credit card debt: LCL = 50 million (14,547) = $727,350 million, UCL = 50 million (15,727) = $786,350 million
12.35a. x t / 2 s / n = 59.04 1.980(20.62/ 122 ) = 59.04 ± 3.70; LCL = 55.34, UCL = 62.74 257
Total spent on other products: LCL = 2800(55.34 = $154,952, UCL = 2800(62.74) = $175,672
12.36 x t / 2 s / n = 2.67 1.973(2.50/ 188 ) = 2.67 ± .36; LCL = 2.31, UCL = 3.03
12.37 x t / 2 s / n = 29.69 1.96(7.88/ 475 ) = 44.14 ± .71; LCL = 43.43, UCL = 44.85
12.38 x t / 2 s / n = 591.87 1.972(125.06/ 205 ) = 591.87 ± 17.22; LCL = 574.65, UCL = 609.09 Total cost of congestion: LCL = 171 million (574.65) = $98,265 million, UCL = 171 million (609.09) = $104,154 million
12.39 x t / 2 s / n = 3.79 1.960(4.25/ 564 ) = 3.79 ± .35; LCL = 3.44, UCL = 4.14 Package of 10: LCL = 13.65(10) = 136.5 days, UCL = 14.23(10) = 142.3 days.
12.40
H0: µ = 15 H1: µ > 15
Rejection region: t t ,n 1 t.05,115 1.658
t
x
s/ n
15 .27 15 5.72 / 116
.51, p-value = .3061. There is not enough evidence to infer that the mean number of
commercials is greater than 15.
12.41 x t / 2 s / n = 4.34 1.960(4.22/ 950 ) = 4.34 ± .27; LCL = 4.07, UCL = 4.61 Total number of visits: LCL = 307,439,000(4.07) = 1,251,276,730
12.42
UCL = 307,439,000 (4.61) = 1,417,293,790
H0: µ = 85 H1: µ > 85
Rejection region: t t , n 1 t .05,84 1.663
t
x s/ n
89 .27 85 17 .30 / 85
2.28, p-value = .0127. There is enough evidence to infer that an e-grocery will be
successful.
12.43 x t / 2 s / n = 15.02 1.990(8.31/ 83 ) = 15.02 1.82; LCL = 13.20, UCL = 16.84
12.44 x t / 2 s / n = 96,100 1.960(34,468/ 473 ) = 96,100 ± 3106; LCL = 92,994, UCL = 99,206 Total amount of debt: LCL = 7 million(92,994) = 650,958 million 258
UCL = 7 million(99,206) = 694,442 million
12.45 x t / 2 s / n = 1.507 1.645(.640/ 473 ) = 1.507 ± .048; LCL = 1.459, UCL = 1.555
12.46 x t / 2 s / n = 27,852 1.96(9252/ 347 ) = 27,852 ± 977; LCL = 26,875, UCL = 28,829 Total: LCL = 43.3million (26,875) = $1,163,687,500,000, UCL = 43.3 million (28,829) = $1,248,295,700,000
12.47 x t / 2 s / n = 354.55 2.576(90.32/ 681 ) = 354.55 ± 8.94; LCL = 345.61, UCL = 363.49
12.48a x t / 2 s / n = 25,228 1.96(5544/ 184 ) = 25,228 ± 806; LCL = 24,422, UCL = 26,034 b x t / 2 s / n = 27,751 1.96(6098/ 184 ) = 27,751 ± 887; LCL = 26,864, UCL = 28,638
12.49 x t / 2 s / n = 366,203 1.96(122,277/ 452 ) = 366,203 ± 11,303; LCL = 354,900, UCL = 377,506 Total: 2,500,000(354,900) = $887,250,000,000, UCL = 2,500,000(377,506) = $943,765,000,000.
12.50 x t / 2 s / n = 46,699 1.96(9032/ 608 ) = 46,699 ± 719; LCL = 45,980, UCL = 47,418 Total: LCL = 112,236(45,980) = $5,160,611,280, UCL = 112,236(47,418) = 5,322,006,648.
12.51 x t / 2 s / n = 7.31 1.973(5.58/ 178 ) = 7.31 ± .83; LCL = 6.48, UCL = 8.14
12.52 x t / 2 s / n = 1157.77 1.645(396.51/ 325 ) = 1157.77 ± 36.28; LCL = 1121.49, UCL = 1194.05
12.53 x t / 2 s / n = 530.69 1.96(97.17/ 485 ) = 530.69 ± 8.67; LCL = 522.02, UCL = 539.36 Total: LCL = 24,000,000(522.02) = $12,528,480,000, UCL = 24,000,000(539.36) = $12,944,640,000
12.54a.
b The required condition is that the variable is normally distributed. c The histogram is somewhat bell shaped. 259
12.55a H0: µ = 12 H1: µ > 12
p-value = 0. There is enough evidence to infer that the average American completed high school. b The required condition is that the variable is normally distributed. c The histogram is bell shaped.
12.56
12.57a H0: µ = 40 H1: µ > 40
p-value = 0. There is enough evidence to infer that the average American is working more than 40 hours per week. b The required condition is that the variable is normally distributed. c The histogram is bell shaped.
12.58
260
12.59
H0: µ = 50.4 H1: µ > 50.4
p-value = 0. There is enough evidence to support the expectation. The required condition is that the variable is normally distributed. The histogram is bell shaped.
12.60a.
b The required condition is that the variable is normally distributed. c The histogram is positively skewed.
12.61a.
b The required condition is that the variable is normally distributed. c The histogram is positively skewed.
12.62a.
261
b The required condition is that the variable is normally distributed. The histogram is positively skewed.
12.63
12.64
12.65
12.66
H0: µ = 2625 H1: µ < 2625
262
p-value = 0. There is enough evidence to infer that the average middle class household spends less than $2625.
12.67
12.68
H0: µ = 12.9 H1: µ > 12.9
p-value = 0. There is enough evidence to infer that the average middle class head of household has more education that 12.9 years,
12.69
12.70
H0: σ2 = 300 H1: σ2 ≠ 300
263
a Rejection region: 2 12 / 2,n 1 .2975,99 74 .2 or 2 2 / 2,n 1 .2025,99 130
2
(n 1)s 2
2
=
(100 1)( 220 ) = 72.60, p-value = .0427. There is enough evidence to infer that the population 300
variance differs from 300. b Rejection region: 2 12 / 2,n 1 .2975,49 32.4 or 2 2 / 2,n 1 .2025,49 71 .4
2
(n 1)s 2
2
=
(50 1)( 220 ) = 35.93, p-value = .1643. There is not enough evidence to infer that the population 300
variance differs from 300. c Decreasing the sample size decreases the test statistic and increases the p-value of the test. H0: σ2 = 100
12.71
H1: σ2 < 100 a Rejection region: 2 12,n 1 .299,49 29 .7
2
(n 1)s 2
2
=
(50 1)(80 ) = 39.20, p-value = .1596. There is not enough evidence to infer that the population 100
variance is less than 100. b Rejection region: 2 12,n 1 .299,99 70 .1
2
(n 1)s 2 2
=
(100 1)(80 ) = 79.20, p-value = .0714. There is not enough evidence to infer that the population 100
variance is less than 100. c Increasing the sample size increases the test statistic and decreases the p-value.
12.72 a LCL =
UCL =
2 / 2,n 1
(n 1)s 2 12 / 2, n 1
b LCL =
UCL =
(n 1)s 2
(n 1)s 2 2 / 2,n 1
(n 1)s 2 12 / 2, n 1
=
=
=
=
(n 1)s 2 .205,14
=
(15 1)(12 ) = 7.09 23 .7
(n 1)s 2 (15 1)(12 ) = 25.57 6.57 .295,14
(n 1)s 2 .205, 29 (n 1)s 2 .295, 29
=
(30 1)(12 ) = 8.17 42 .6
=
(30 1)(12 ) = 19.66 17 .7
c Increasing the sample size narrows the interval.
264
12.73 LCL =
UCL =
(n 1)s 2
=
2 / 2,n 1
(n 1)s 2 12 / 2, n 1
=
(n 1)s 2 .205,7
(n 1)s 2 .295,7
=
=
(8 1)(. 00093 ) = .00046, 14 .1
(8 1)(. 00093 ) = .00300 2.17
H0: σ2 = 250
12.74
H1: σ2 < 250 Rejection region: 2 12,n 1 .290,9 4.17
2
(n 1)s 2
2
=
(10 1)( 210 .22 ) 7.57 , p-value = .4218. There is not enough evidence to infer that the population 250
variance has decreased. H0: σ2 = 23
12.75
H1: σ2 ≠ 23 Rejection region: 2 12 / 2,n 1 .295,7 2.17 or 2 2 / 2,n 1 .205,7 14 .1
2
(n 1)s 2
2
=
(8 1)(16 .50 ) 5.02 , p-value = .6854. There is not enough evidence to infer that the population 23
variance has changed.
12.76 LCL =
UCL =
(n 1)s 2 2 / 2,n 1
(n 1)s 2 12 / 2, n 1
=
=
(n 1)s 2 .2025,9
(n 1)s 2 .2975,9
=
=
(10 1)(15 .43) 7.31 19 .0
(10 1)(15 .43) 51 .43 2.70
12.77 a H0: σ2 = 250 H1: σ2 ≠ 250 Rejection region: 2 12 / 2,n 1 .2975,24 12.4 or 2 2 / 2,n 1 .2025,24 39.4
2
(n 1)s 2
2
=
(25 1)( 270 .58 ) = 25.98, p-value = .7088. There is not enough evidence to infer that the 250
population variance is not equal to 250. b Demand is required to be normally distributed. c The histogram (not shown here) is approximately bell shaped.
265
H0: σ2 = 18
12.78
H1: σ2 > 18 Rejection region: 2 2 ,n 1 .210,244 272.704 (from Excel)
2
(n 1)s 2
2
=
(245 1)( 22 .56 ) = 305.81; p-value = .0044. There is enough evidence to infer that the population 18
variance is greater than 18.
12.79 LCL =
UCL =
(n 1)s 2
=
2 / 2,n 1
(n 1)s 2 12 / 2, n 1
=
(n 1)s 2 .205,89
(n 1)s 2 .295,89
=
=
(90 1)( 4.72 ) 3.72 113
(90 1)( 4.72 ) 6.08 69 .1
H0: σ2 = 200
12.80
H1: σ2 < 200 Rejection region: 2 12,n 1 .295,99 77.9
2
(n 1)s 2
2
=
(100 1)(174 .47 ) = 86.36; p-value = .1863. There is not enough evidence to infer that the 200
population variance is less than 200. Replace the bulbs as they burn out.
12.81 LCL =
UCL =
(n 1)s 2 2 / 2,n 1
(n 1)s 2 12 / 2, n 1
=
=
(n 1)s 2 .2025, 24
(n 1)s 2 .2975, 24
=
=
(25 1)(19 .68) 11 .99 39 .4
(25 1)(19 .68 ) 38 .09 12 .4
12.82 a p̂ z / 2 p̂(1 p̂) / n = .48 ±1.96 .48 (1 .48) / 200 = .48 .0692 b p̂ z / 2 p̂(1 p̂) / n = .48 ±1.96 .48 (1 .48 ) / 500 = .48 .0438 c p̂ z / 2 p̂(1 p̂) / n = .48 ±1.96 .48 (1 .48 ) / 1000 = .48 .0310 d The interval narrows. 12.83 a p̂ z / 2 p̂(1 p̂) / n = .50 ±1.96 .50 (1 .50 ) / 400 = .50 .0490 b p̂ z / 2 p̂(1 p̂) / n = .33 ±1.96 .33(1 .33) / 400 = .33 .0461
266
c p̂ z / 2 p̂(1 p̂) / n = .10 ±1.96 .10 (1 .10 ) / 400 = .10 .0294 d The interval narrows.
12.84
H0: p = .60 H1: p > .60
a z
b z
c z
p̂ p p(1 p) / n p̂ p p(1 p) / n p̂ p p(1 p) / n
=
=
=
.63 .60 .60 (1 .60 ) / 100 .63 .60 .60 (1 .60 ) / 200 .63 .60 .60 (1 .60 ) / 400
= .61, p-value = P(Z > .61) = 1 – .7291 =.2709
= .87, p-value = P(Z > .87) = 1 – .8078 = .1922 = 1.22, p-value = P(Z > 1.22) = 1 – .8888 = .1112
d The p-value decreases.
12.85 a z
b z
c z
p̂ p p(1 p) / n
p̂ p p(1 p) / n p̂ p p(1 p) / n
=
=
.73 .70
=
.70 (1 .70 ) / 100
.72 .70 .70 (1 .70 ) / 100 .71 .70 .70 (1 .70 ) / 100
= .65, p-value = P(Z > .65) = 1 – .7422 =.2578
= .44, p-value = P(Z > .44) = 1 – .6700 =.3300
= .22, p-value = P(Z > .22) = 1 – .5871 =.4129
d. The z statistic decreases and the p-value increases.
2
2
z 1.645 .5(1 .5) p̂(1 p̂) = 752 12.86 n = / 2 = .03 B
12.87a. .5 .03 b. Yes, because the sample size was chosen to produce this interval.
12.88a. p̂ z / 2 p̂(1 p̂) / n = .75 ±1.645 .75 (1 .75) / 752 = .75 .0260 b. The interval is narrower. c. Yes, because the interval estimate is better than specified. 2
2
z 1.645 .75 (1 .75 ) p̂(1 p̂) = 564 12.89 n = / 2 = B .03
267
12.90a. .75 .03 b. Yes, because the sample size was chosen to produce this interval.
12.91a. p̂ z / 2 p̂(1 p̂) / n = .92 ±1.645 .92 (1 .92 ) / 564 = .92 .0188 b. The interval is narrower. c. Yes, because the interval estimate is better than specified.
12.92a. p̂ z / 2 p̂(1 p̂) / n = .5 ±1.645 .5(1 .5) / 564 = .5 .0346 b. The interval is wider. c. No because the interval estimate is wider (worse) than specified.
12.93 p̂ = 259/373 = .69
p̂ z / 2 p̂(1 p̂) / n = .69 1.96 .69 (1 .69 ) / 373 = .69 ± .0469; LCL = .6431, UCL = .7369
12.94
H0: p = .25 H1: p < .25
p̂ = 41/200 = .205
z
p̂ p p(1 p) / n
=
.205 .25 .25(1 .25) / 200
1.47 , p-value = P(Z < –1.47) = .0708. There is enough evidence to
support the officer’s belief.
12.95 p̂ = 204/314 = .65
p̂ z / 2 p̂(1 p̂) / n = .65 1.645 .65 (1 .65 ) / 314 = .65 .0443; LCL = .6057, UCL = .6943
12.96
H0: p = .92 H1: p > .92
p̂ = 153/165 = .927
z
p̂ p p(1 p) / n
=
.927 .92 .92 (1 .92 ) / 165
.33, p-value = P(Z > .33) = 1 – .6293 =.3707. There is not enough evidence
to conclude that the airline’s on-time performance has improved.
268
12.97 p̂ = 97/344 = .28
p̂ z / 2 p̂(1 p̂) / n = .28 ±1.96 .28 (1 .28 ) / 344 = .28 ± .0474; LCL = .2326, UCL = .3274
12.98 p̂ = 68/400 = .17
p̂ z / 2 p̂(1 p̂) / n = .17 ±1.96 .17 (1 .17 ) / 400 = .17 ± .0368; LCL = .1332, UCL = .2068
12.99 LCL = .1332(1,000,000)(3.00) = $399,600, UCL = .2068(1,000,000)(3.00) = $620,400
p 12.100 ~
x2 1 2 .0147 n 4 200 4
~ .0147 (1 .0147 ) p (1 ~ p) ~ = .0147 1.96 = .0147 ± .0165; LCL = 0 (increased from –.0018), UCL p z / 2 200 4 n4 = .0312
p 12.101 ~
x2 3 2 .0132 n 4 374 4
~ .0132 (1 .0132 ) p (1 ~ p) ~ = .0132 1.645 = .0132 ± .0097; LCL = .0035, UCL = .0229 p z / 2 374 4 n4
p 12.102 ~
x2 1 2 .0077 n 4 385 4
~ .0077 (1 .0077 ) p (1 ~ p) ~ = .0077 2.575 = .0077 ± .0114; LCL = 0 (increased from –.0037), UCL p z / 2 385 4 n4 = .0191 12.103 p̂ z / 2 p̂(1 p̂) / n = .2680 1.96 .2680 (1 .2680 ) / 974 = .2680 ± .0278; LCL = .2402, UCL = .2958 Number: LCL = 234,564,000(.2402) = 56,342,273, UCL = 234,564,000(.2958) = 69,384,031
12.104 p̂ z / 2 p̂(1 p̂) / n = .1192 ±1.96 .1192 (1 .1192 ) / 562 = .1192 ± .0268; LCL = .0924, UCL = .1460
12.105 p̂ z / 2 p̂(1 p̂) / n = .1056 ±1.96 .1056 (1 .1056 ) / 521 = .1056 ± .0264; LCL = .0792, UCL = .1320
12.106 LCL = 75,000(.0792) =5,940, UCL = 75,000(.1320) = 9,900
269
12.107 p̂ z / 2 p̂(1 p̂) / n = .1202 ±1.96 .1202 (1 .1202 ) / 391 = .1202 ± .0322; LCL = .0880, UCL = .1524
12.108 H0: p = .90 H1: p < .90
z
p̂ p p(1 p) / n
=
.8644 .90 .90 (1 .90 ) / 177
= –1.58, p-value = P(Z < –1.58) = .0571. There is not enough evidence to
infer that the satisfaction rate is less than 90%.
12.109 p̂ z / 2 p̂(1 p̂) / n = .2333 ±1.96 .2333 (1 .2333 ) / 120 = .2333 ± .0757; LCL = .1576, UCL = .3090
12.110 p̂ z / 2 p̂(1 p̂) / n = .600 ±1.96 .600 (1 .600 ) / 1508 = .600 ± .025; LCL = .575, UCL = .625 Total number of Canadians who prefer artificial Christmas trees: LCL = 6 million (.575) = 3.452 million, UCL = 6 million (.625) = 3.749 million
12.111a. p̂ z / 2 p̂(1 p̂) / n = .7840 ±1.96 .7840 (1 .7840 ) / 426 = .7840 ± .0391; LCL = .7449, UCL = .8231
12.112 H0: p = .50 H1: p > .50
z
p̂ p p(1 p) / n
=
.57 .50 .50 (1 .50 ) / 100
= 1.40, p-value = P(Z > 1.40) = 1 – .9192 =.0808. There is not enough
evidence to conclude that more than 50% of all business students would rate the book as excellent.
12.113 Codes 1, 2, and 3 have been recoded to 5. H0: p = .90 H1: p > .90
z
p̂ p p(1 p) / n
=
.96 .90 .90 (1 .90 ) / 100
= 2.00, p-value = P(Z > 2.00) = 1 – .9772 =.0228. There is enough evidence
to conclude that more than 90% of all business students would rate the book as at least adequate.
12.114 p̂ z / 2 p̂(1 p̂) / n = .0360 ±1.96 .0360 (1 .0360 ) / 5000 = .0360 ± .0052; LCL = .0308, UCL = .0412
270
Proportion: LCL = .0308, UCL = .0412. Number: LCL = 126.54million (.0308) = 3.89 million , UCL = 126.54 million (.0412) = 5.21 million 12.115 H0: p = .2155 H1: p ≠ .2155
z
p̂ p p(1 p) / n
=
.2442 .2155 .2155 (1 .2155 ) / 1040
= 2.25, p-value = 2P(Z > 2.25) = 2(1 – .9878) =.0244. There is enough
evidence to conclude that the proportion of 4-4-3-2 hands differs from the theoretical probability. This deviation is likely caused by insufficient shuffling.
12.116a. p̂ z / 2 p̂(1 p̂) / n = .2031 ±1.96 .2031 (1 .2031 ) / 650 = .2031 ± .0309; LCL = .1722, UCL = .2340 Number: LCL = 5 million (.1722) = .861 million, UCL = 5 million (.2340) = 1.17 million
12.117 H0: p = .795 H1: p < .795
z = -6.28, p-value = 0. There is enough evidence to infer that the proportion of White Americans has decreased.
12.118 H0: p = .269 H1: p ≠ .269
z = -.30, p-value = .7659. There is not enough evidence to conclude that the proportion has changed.
12.119 H0: p = .658 H1: p ≠ .658
271
p-value = .0137. There is enough evidence to infer that the proportion of home ownership has changed.
12.120 H0: p = .129 H1: p ≠ .129
p-value = .8777. There is not enough evidence to conclude that the proportion of Americans who did not finish high school has changed.
12.121 H0: p = .104 H1: p > .104
p-value = 0. There is enough evidence to conclude that the proportion has increased.
12.122 H0: p = .102 H1: p > .102
272
p-value = .6904. There is not enough evidence to conclude that the proportion is larger than 10.2%.
12.123 H0: p = .14 H1: p > .14
p-value = .3088. There is not enough evidence to conclude that the proportion has increased.
12.124 H0: p = .129 H1: p < .129
p-value = .1329. There is not enough evidence
12.125 H0: p = .658 H1: p > .658
273
p-value = .1318. There is not enough evidence
12.126 H0: p = .168 H1: p > .168
p-value = .2508. There is not enough evidence
12.127 H0: p = .124 H1: p < .124
p-value = .1583. There is not enough evidence
12.128 H0: p = .158 H1: p < .158
274
p-value = .0223. There is enough evidence
12.129 H0: p = .105 H1: p > .105
p-value = 0. There is enough evidence
12.130 Codes 3 and 4 were changed to 5
p̂ z / 2 p̂(1 p̂) / n = .7305 1.96 .7305 (1 .7305 ) / 475 = .7305 .0399; LCL = .6906, UCL = .7704; Market segment size: LCL = 41,580,000 (.6906) = 28,715,148, UCL = 41,580,000 (.7704) = 32,033,232
12.131 Code 2 was changed to 3.
p̂ z / 2 p̂(1 p̂) / n = .5313 ±1.96 .5313 (1 .5313 ) / 320 = .5313 ± .0547; LCL = .4766, UCL = .5860; Market segment size: LCL = 16,015,493 (.4766) = 7,632,984 , UCL = 16,015,493 (.5860) = 9,385,079
12.132a. p̂ z / 2 p̂(1 p̂) / n = .2919 ±1.96 .2919 (1 .2919 ) / 1836 = .2919 ± .0208; LCL = .2711, UCL = .3127 b LCL = 124,723,003 (.2711) = 33,812,406, UCL = 124,723,003 (.3127) = 39,000,883
12.133 p̂ z / 2 p̂(1 p̂) / n = .1077 ±1.96 .1077 (1 .1077 ) / 325 = .1077 ± .0337; LCL = .0740, UCL = .1414; Market segment size: LCL = 44,679,192(.0740) = 3,306,260, UCL = 44,679,192(.1414) = 6,317,638
275
12.134 p̂ z / 2 p̂(1 p̂) / n = .1748 ±1.645 .1748 (1 .1748 ) / 412 = .1748 ± .0308; LCL = .1440, UCL = .2056; Number: LCL = 211,306,936(.1440) = 30,428.199, UCL = 211,306,936(.2056) = 43,444,706
12.135 p̂ z / 2 p̂(1 p̂) / n = .1500 ±1.96 .1500 (1 .1500 ) / 340 = .1500 ± .0380; LCL = .1120, UCL = .1880; Number: LCL = 211,306,936(.1120) = 23,666,377, UCL = 211,306,936(.1880) = 39,725,704
12.136
Proportion: LCL = .0861, UCL = .1605. Amount: LCL =$8,354, UCL = $15,573
12.137
LCL = 701.26, UCL = 760.02
12.138a. H0: µ = 30 H1: µ > 30
276
t = 3.04, p-value = .0015; there is enough evidence to infer that the candidate is correct. b
LCL = $30.68, UCL = $33.23 c The costs are required to be normally distributed.
12.139a.
LCL = .0892, UCL = .1461 b.
277
LCL = .1105, UCL = .1483
12.40
H0: σ2 = 17 H1: σ2 > 17
2 = 30.71, p-value = .0435. There is enough evidence to infer that problems are likely.
12.141
LCL = 1193.39, UCL = 1258.43 Total: 354,520(1193.39) = 423,080,623, UCL = 354,520(1258.43) = 446,138,604
12.142 278
LCL = .3718, UCL = .4428
12.143
LCL = 10.51, UCL = 11.33
12.144
LCL = 9.92, UCL = 10.66
12.145
LCL = .3044, UCL = .4119 279
12.146a
LCL = 69.03, UCL = 74.73
b
H0: µ = 68 H1: µ > 68
t = 2.74, p-value = .0043; there is enough evidence to infer that students with a calculus background would perform better in statistics than students with no calculus.
12.147
LCL = .0945, UCL = .1492
280
12.148 H0: p = .75 H1: p > .75
z = .64, p-value = .2611; there is not enough evidence to infer that more than 75% of workers drove alone to work.
12.149
LCL = .0765, UCL = .1271
12.150
LCL = 26.44, UCL = 27.55
281
12.151a.
LCL = 6.84, UCL = 6.98 b. The histogram is bell shaped. c.
H 0: µ = 7 H 1: µ < 7
t = –3.48, p-value = .0004; there is enough evidence to infer that postal workers are spending less than seven hours doing their jobs.
12.152
LCL = .5818, UCL = .6822
282
12.153
LCL = 5.11, UCL = 6.47
12.154
Proportion: LCL = .2977, UCL = .3680. Number: LCL = 69,833,339, UCL = 86,316,357 12.155 H0: σ2 = 4 H1: σ2 > 4
2 = 161.25, p-value = .0001; there is enough evidence to conclude that the number of springs requiring reworking is unacceptably large.
283
12.156 H0: p = .90 H1: p < .90
z = –1.33, p-value = .0912; there is enough evidence to infer that less than 90% of the springs are the correct length. 12.157
LCL = .937, UCL = 1.263
12.158a. H0: µ = 9.8 H1: µ < 9.8
284
t = –2.97, p-value = .0018; there is enough evidence to infer that enclosure of preaddressed envelopes improves the average speed of payments. b.
H0: σ2 = (3.2)2 = 10.24 H1: σ2 < 10.24
2 = 101.58, p-value = .0011; there is enough evidence to infer that the variability in payment speeds decreases when a preaddressed envelope is sent.
2
2
z 2.575 .5(1 .5) p̂(1 p̂) = 4144 12.159 n = / 2 = . 02 B
285
12.160
Proportion: LCL = .1245, UCL = .1822. Total: LCL = 400,000(.1245) = 49,800 UCL = 400,000(.1822) = 72,880
12.161 Number of cars: H0: µ = 125 H1: µ > 125
t = .459, p-value = .3351; there is not enough evidence to infer that the employee is stealing by lying about the number of cars.
Amount of time H0: µ = 3.5 H1: µ > 3.5
286
t = 7.00, p-value = 0; there is enough evidence to infer that the employee is stealing by lying about the amount of time.
12.162a.
LCL = -5.54 UCL = 29.61 b.
H0: µ = 16 H1: µ < 16
287
t = -.473, p-value = .3210; there is not enough evidence to infer that Mr. Cramer does less well than the S&P 500 index.
12.163
LCL = .0508, UCL = .0740
12.164 H0: µ = 14.35 H1: µ > 14.35
t = .908, p-value = .1823; There is not enough evidence to that the debt ratio increased since 2006.
12.165 H0: µ = 17.62 H1: µ > 17.62
288
t = 4.23, p-value = 0; there is enough evidence to infer that financial obligations increased since 2006.
12.166 H0: µ = 25.97 H1: µ > 25.97
t = .959, p-value = .1693; there is not enough evidence to infer that financial obligations for renters increased since 2006.
12.167
Proportion: LCL = .7267, UCL = .7768. Number: LCL = 57,289,123, UCL = 61,236,017
289
12.168 H0: µ = 65.8 H1: µ > 65.8
t = 2.44, p-value = .0083; there is enough evidence to infer that the percentage of total compensation increased between 2007 and 2008.
12.169
Proportion: LCL = .1439, UCL = .1868. Number: LCL = 33,763,728, UCL = 43,815,015
12.170a.
LCL = .6135, UCL = .7107 b.
290
LCL = 26,823, UCL = 37,144 c.
LCL = 4.67, UCL = 5.79
d.
LCL = 1.41, UCL = 1.84
291
12.171
Total: LCL = 244,137,873(.1260) = 30,761,372, UCL = 244,137,873(.1530) = 37,353,095
12.172
Total: LCL = 244,137,873(.1786) = 43,603,024, UCL = 244,137,873(.2102) = 51,317,781
12.173
Total: LCL = 244,137,873(.1080) = 26,366,891, UCL = 244,137,873(.1338) = 32,665,647
12.174a. H0: µ = 3.0 H1: µ < 3.0
p-value = .3762. There is not enough evidence to support the claim. b. 292
12.175 a H0:p = .5 H1: p >.5
p-value = 0. There is enough evidence to infer that there are more Democrats than Republicans.
12.176 H0: p = .5 H1: p >.5
p-value = .0002. There is enough evidence to infer that there are more conservatives than liberals.
12.177 There are more Democrats than Republicans but there are more conservatives than liberals.
12.178 a H0: µ = 2.08 H1: µ < 2.08
p-value = 0. There is enough evidence to infer that the mean number of children is less than 2.08.
12.179 H0: p = .5 H1: p ≠.5
293
p-value = 0. There is enough evidence to infer that there is no gender equality.
12.180
Total: LCL = 242,823,652(.7475) = 181,510,680, UCL = 242,823,652(.7691) = 186,755,671
12.181
Total: LCL = 123,460,000(.7147) = 88,236,862, UCL = 123,460,000(.7373) = 91,027,058
12.182
Total: LCL = 123,460,000(.1256) = 15,506,576, UCL = 123,460,000(.1400) = 17,284,400
12.183
294
Total: LCL = 123,460,000(.2919) = 36,037,974, UCL = 123,460,000(.3226) = 39,828,196
12.184
12.185a. H0: µ = 60 H1: µ > 60
p-value = .1049. There is not enough evidence to infer that the mean age is greater than 60. b. AGE is required to be normally distributed. The histogram is approximately bell shaped.
12.186
12.187
295
12.188 a.
b. DEBT is required to be normally distributed. The histogram is extremely positively skewed.
12.189a.
b. The required condition is normality. The histogram is extremely positively skewed.
12.190 H0: µ = 5250 H1: µ > 5250
p-value = 0. There is enough evidence to infer that wealthy households spend more than $5250.
296
Case 12.1 95% confidence interval estimate of mean weekly consumption per student:
1 2 3 4 5 6 7
A B t-Estimate: Mean
Mean Standard Deviation LCL UCL
C
D
Cans 1.316 1.115 1.218 1.414
Estimated Mean Number of Cans per Student
Profit
Current Profit
Net
LCL = 1.218
$652,600
$616,000
$ 36,600
UCL = 1.414
$789,800
$616,050
$173,800
Pepsi should sign the exclusivity agreement. They would increase profit by between $36,600 and $173,800.
297
Case 12.2 Estimated total number of soft drink sales per year LCL = 1.218 × 40 × 50,000 = 2,436,000. Pepsi sells 22,000 × 40 = 880,000 Total sales by Coke = 2,436,000- 880,000 = 1,556,000 cans per year Coke’s current estimated profit = 1,556,000 × .70 = $1,089,200 Profit with exclusivity agreement: $652,600
UCL = 1.414 × 40 × 50,000 = 2,828,000. Pepsi sells 22,000 × 40 = 880,000 Total sales by Coke = 2,828,000- 880,000 = 1,948,000 cans per year Coke’s current estimated profit = 1,948,000 × .70 = $1,363,600 Profit with exclusivity agreement: $789,800
Coke would not sign the exclusivity agreement. Coke is expected to lose from the exclusivity agreement because they currently have a much larger share of the market and would not gain by paying for exclusivity.
Case 12.3 a 95% confidence interval estimate of the mean medical costs for each of the four age categories:
1 2 3 4 5 6 7
1 2 3 4 5 6 7
A B t-Estimate: Mean
C
Age:45-64 1830 749 1784 1877
Mean Standard Deviation LCL UCL
A B t-Estimate: Mean
Mean Standard Deviation LCL UCL
D
C
D
Age:65-74 4494 1820 4381 4607
298
1 2 3 4 5 6 7
1 2 3 4 5 6 7
A B t-Estimate: Mean
C
Age:75-84 8074 3186 7876 8272
Mean Standard Deviation LCL UCL
A B t-Estimate: Mean
D
C
D
Age:85+ 15957 6207 15572 16342
Mean Standard Deviation LCL UCL
Estimated annual costs for 2016 Estimate of mean
Estimate of total (1,000s)
Age category Number (1,000s)
LCL
UCL
LCL
UCL
45 to 64
10,094
1,784
1,877
18,007,696
18,946,438
65 to 74
3,375
4,381
4,607
14,785,875
15,548,625
75 to 84
1,745
7,876
8,272
13,743,620
14,434,640
15,572 16,342
12,333,024
12,942,864
58,870,215
61,872,567
85 and over Total
792 16,006
Estimated annual costs for 2021 Estimate of mean
Estimate of total (1,000s)
Age category Number (1,000s)
LCL
UCL
LCL
UCL
45 to 64
10,054
1,784
1,877
17,936,336
18,871,358
65 to 74
4,023
4,381
4,607
17,624,763
18,533,961
75 to 84
2,087
7,876
8,272
16,437,212
17,263,664
15,572 16,342
13,578,784
14,250,224
65,577,095
68,919,207
85 and over Total
872 17,036
Estimated annual costs for 2026 Estimate of mean
Estimate of total (1,000s)
Age category Number (1,000s)
LCL
UCL
LCL
UCL
45 to 64
9,892
1,784
1,877
17,647,238
18,567,284
65 to 74
4,538
4,381
4,607
19,880,978
20,906,566
299
75 to 84 85 and over Total
2,666 973
7,876
8,272
20,997,416
22,053,152
15,572 16,342
15,151,556
15,900,766
73,677,2785
77,427,768
18,069
Estimated annual costs for 2031 Estimate of mean
Estimate of total (1,000s)
Age category Number (1,000s)
LCL
UCL
LCL
UCL
45 to 64
9,823
1,784
1,877
17,524,232
18,437,771
65 to 74
4,874
4,381
4,607
21,352,994
22,454,518
75 to 84
3,209
7,876
8,272
25,274,084
26,544,848
85 and over
1,193
15,572 16,342
18,577,396
19,496,006
Total
19,099
82, 728,706
86, 933,143
Estimated annual costs for 2036 Estimate of mean
Estimate of total (1,000s)
Age category Number (1,000s)
LCL
UCL
LCL
UCL
45 to 64
10,150
1,784
1,877
18,107,600
19,051,550
65 to 74
4,675
4,381
4,607
20,481,175
21,537,725
75 to 84
3,669
7,876
8,272
28,897,044
30,349,968
85 and over
1,543
15,572 16,342
24,027,596
25,215,706
Total
20,037
91,513,415
96,154,949
Case 12.4 a 95% confidence interval estimate of the proportion of Americans with Alzheimer’s disease for each of the three age categories:
300
Estimated number of Americans with Alzheimer’s disease for 2015 Estimate of proportion
Estimate of total (1,000s)
Age category Number (1,000s)
LCL
UCL
LCL
UCL
65 to 74
26,967
.0009
.0025
24.270
67.418
75 to 84
13,578
.0077
.0115
104.551
156.147
6,292
.0074
.0112
46.561
70.470
175.382
294.035
85 and over Total
46,837
Estimated number of Americans with Alzheimer’s disease for 2020 Estimate of proportion
Estimate of total (1,000s)
Age category Number (1,000s)
LCL
UCL
LCL
UCL
65 to 74
32,312
.0009
.0025
29.081
80.780
75 to 84
15,895
.0077
.0115
122.392
182.793
6,597
.0074
.0112
48.818
73.886
200.290
337.459
85 and over Total
54,804
Estimated number of Americans with Alzheimer’s disease for 2025 Estimate of proportion
Estimate of total (1,000s)
Age category Number (1,000s)
LCL
UCL
LCL
UCL
65 to 74
36,356
.0009
.0025
32.720
90.890
75 to 84
20,312
.0077
.0115
156.402
233.588
301
85 and over Total
7,239
.0074
.0112
63,907
53.569
81.077
242.691
405.555
Case 12.5 a. If there is no middle bias the proportion of bets on 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, and 35 is 12/36, or .3333. b.
H0: p = .3333 H1: p > .3333
The number of bets on 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, and 35 are 450, 636, 633, 783, 649, 1079, 983, 746, 703, 878, 925, and 627, respectively. The total number of bets on the middle numbers is 9092. Thus the sample proportion is p̂ = 9092/21,731 = .4184. The test statistic is
z
p̂ p p(1 p) / n
=
.4184 ..3333 .3333 (1 .3333 ) / 21,731
26 .61, p-value = P(Z > 26.61) = 0. There is overwhelming
evidence of middle bias. c. If there is no middle bias the proportion of bets on 17 and 20 is 2/36 = .0556. d.
H0: p = .0556 H1: p > .0556
The number of bets on 17 and 20 are 1079 and 983, respectively. The total number of bets on the middle numbers is 2062. Thus the sample proportion is p̂ = 2062/21,731 = .0949. The test statistic is
z
p̂ p p(1 p) / n
=
.0949 ..0556 .0556 (1 .0556 ) / 21,731
25 .28, p-value = P(Z > 25.28) = 0. There is overwhelming
evidence of the middle of the middle bias.
302
Chapter 13 13.1a Equal-variances estimator
1 1 ( x1 x 2 ) t / 2 s 2p = (524 – 469) 2.009 n1 n 2
(25 1)129 2 (25 1)131 2 1 1 25 25 25 25 2
= 55 73.87 b Equal-variances estimator
1 1 ( x1 x 2 ) t / 2 s 2p = (524 – 469) 2.009 n1 n 2
(25 1)255 2 (25 1)260 2 1 1 25 25 25 25 2
= 55 146.33 c The interval widens. d Equal-variances estimator
1 1 ( x1 x 2 ) t / 2 s 2p = (524 – 469) 1.972 n n 2 1
(100 1)129 2 (100 1)131 2 1 1 100 100 2 100 100
= 55 36.26 e The interval narrows.
13.2
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
a Equal-variances test statistic Rejection region: t t / 2, t .025, 22 –2.074 or t t / 2, t .025, 22 2.074 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(74 71) 0 (12 1)18 2 (12 1)16 2 1 1 12 12 12 12 2
= .43, p-value = .6703. There is not enough
evidence to infer that the population means differ.
b Equal-variances test statistic Rejection region: t t / 2, t .025, 22 –2.074 or t t / 2, t .025, 22 2.074 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(74 71) 0 (12 1)210 2 (12 1)198 2 1 1 12 12 12 12 2
evidence to infer that the population means differ. c The value of the test statistic decreases and the p-value increases. 281
= .04, p-value = .9716. There is not enough
d Equal-variances test statistic Rejection region: t t / 2, t .025, 298 –1.960 or t t / 2, t .025, 298 1.960 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
(74 71) 0
=
(150 1)18 2 (150 1)16 2 1 1 150 150 2 150 150
= 1.53, p-value = .1282. There is not
enough evidence to infer that the population means differ. e The value of the test statistic increases and the p-value decreases. f Rejection region: t t / 2, t .025, 22 –2.074 or t t / 2, t .025, 22 2.074 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
(76 71) 0
=
(12 1)18 2 (12 1)16 2 1 1 12 12 12 12 2
= .72, p-value = .4796. There is not enough
evidence to infer that the population means differ. g The value of the test statistic increases and the p-value decreases.
13.3 a Unequal-variances estimator
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 64.8 (rounded to 65, approximated by 70 )
s2 s2 (x1 x 2 ) t / 2 1 2 = (63 – 60) 1.667 n n 2 1
18 2 7 2 = 3 4.59 50 45
b Unequal-variances estimator
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 63.1 (rounded to 63, approximated by 60 )
s2 s2 (x1 x 2 ) t / 2 1 2 = (63 – 60) 1.671 n n 2 1
41 2 15 2 = 3 10.38 50 45
c The interval widens. d Unequal-variances estimator
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 131 (approximated by 140 )
s2 s2 (x1 x 2 ) t / 2 1 2 = (63 – 60) 1.656 n n 2 1
18 2 7 2 = 3 3.22 100 90
282
e The interval narrows.
13.4
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
a Unequal-variances test statistic
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 200.4 (rounded to 200)
Rejection region: t t , t.05, 200 1.653
t
( x 1 x 2 ) ( 1 2 ) s 12
n1
s 22
=
n 2
(412 405 ) 0 128 2 54 2 150 150
= .62, p-value = .2689. There is not enough evidence to infer that 1
is greater than 2 . b Unequal-variances test statistic
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 223.1 (rounded to 223)
Rejection region: t t , t .05, 223 1.645 t
( x 1 x 2 ) ( 1 2 ) s 12 s 22 n1 n 2
=
(412 405 ) 0 31 2 16 2 150 150
= 2.46, p-value = .0074. There is enough evidence to infer that 1 is
greater than 2 . c The value of the test statistic increases and the p-value decreases.
d Unequal-variances test statistic
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 25.6 (rounded to 26)
Rejection region: t t , t.05,26 1.706
t
( x 1 x 2 ) ( 1 2 ) s 12 s 22 n1 n 2
=
(412 405 ) 0 128 2 54 2 20 20
= .23, p-value = .4118. There is not enough evidence to infer that 1
is greater than 2 . e The value of the test statistic decreases and the p-value increases.
283
f Unequal-variances test statistic Rejection region: t t , t.05, 200 1.653
t
( x 1 x 2 ) ( 1 2 ) s 12 s 22 n1 n 2
=
(409 405 ) 0 128 2 54 2 150 150
= .35, p-value = .3624. There is not enough evidence to infer that 1
is greater than 2 . g The value of the test statistic decreases and the p-value increases.
13.5 a Equal-variances t-test degrees of freedom = 28, Unequal-variances t-test degrees of freedom =26.4 b Equal-variances t-test degrees of freedom = 24, Unequal-variances t-test degrees of freedom = 10.7 c Equal-variances t-test degrees of freedom = 98 , Unequal-variances t-test degrees of freedom = 91.2 d Equal-variances t-test degrees of freedom = 103, Unequal-variances t-test degrees of freedom = 78.5
13.6 a In all cases the equal-variances t-test degrees of freedom is greater than the unequal-variances t-test degrees of freedom.
13.7
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
Two-tail F test: F = 1.85, p-value = .4352; use equal-variances test statistic Rejection region: t < -tα/2,υ = -t.05,14 = -1.761 or t > tα/2,υ = t.05,14 = 1.761 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(6.25 5.88) 0 (8 1)18 .21 (8 1)9.84 1 1 882 8 8
.20, p-value = .8442. There is not enough
evidence of a difference.
13.8
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
Two-tail F test: F = 2.87, p-value = .2608; use equal-variances test statistic Rejection region: t < -tα/2,υ = -t.05,13 = -1.771 or t > tα/2,υ = t.05,13 = 1.771 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(8.44 6.50 ) 0 (9 1)15 .78 (6 1)5.50 1 1 962 9 6
evidence of a difference.
13.9
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0 284
1.07 , p-value = .3028. There is not enough
Two-tail F test: F = .08, p-value = .0694; use unequal-variances test statistic
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n 2 1 n1 1
= 3.49 (rounded to 3)
Rejection region: : t < -tα/2,υ = -t.05,3 = -2.353 or t > tα/2,υ = t.05,3 = 2.353 t
( x 1 x 2 ) ( 1 2 ) s 12
n1
s 22
=
(107 .75 116 .75) 0 7.58 92 .25 4 4
n 2
= -1.80, p-value = .1694. There is not enough evidence to infer
that the batteries differ.
13.10
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = 2.55, p-value = .3272; use equal-variances test statistic Rejection region: t > tα,υ = t.10,10 = 1.372 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(36 .67 34 .50 ) 0 (6 1)5.87 (6 1)2.30 1 1 662 6 6
1.86, p-value = .0465. There is enough evidence
to infer that new putter is better.
13.11
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
Two-tail F test: F = 1.02, p-value = .9856; use equal-variances test statistic Rejection region: t t , t .10,10 1.372 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(361 .50 381 .83) 0 (6 1)6767 .5 (6 1)6653 .4 1 1 662 6 6
.43, p-value = .3382. The manager should
choose to use cameras.
13.12
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
Two-tail F test: F = 1.02, p-value = .9823; use equal-variances test statistic Rejection region: t t , t .10,18 1.330
285
t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(5.10 7.30 ) 0 (10 1)5.88 (10 1)5.79 1 1 10 10 2 10 10
2.04, p-value = .0283. There is enough
evidence to infer that there are fewer errors when the yellow ball is used.
13.13
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
Two-tail F test: F = 1.13, p-value = .8430; use equal-variances test statistic Rejection region: t t , t .10,19 1.328 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(12 .62 14 .20 ) 0 1 (10 1)2.95 (11 1)2.61 1 10 11 2 10 11
2.17 , p-value = .0213. There is enough
evidence to infer that users of the new device leave larger tips.
13.14
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) 0
Two-tail F test: F = 1.04, p-value = .9873; use equal-variances test statistic Rejection region: t t / 2, t .05,13 1.771 or t t / 2, t .05,13 1.771 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(3,372 4,093 ) 0 (9 1)755 ,196 (6 1)725 ,778 1 1 962 9 6
1.59, p-value = .1368. There is not
enough evidence to infer a difference between the two types of vacation expenses.
13.15
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
Two-tail F test: F = .69, p-value = .5486; use equal-variances test statistic Rejection region: t t , t .10,22 1.321 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(172 .0 176 .8) 0 1 (12 1)157 .1 (12 1)227 .7 1 12 12 2 12 12
.84 , p-value = .2053. There is not
enough evidence to conclude that games take longer to complete than 5 years ago.
13.16
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) 0
Two-tail F test: F = 1.67, p-value = .4060; use equal-variances test statistic
286
Rejection region: t t / 2, t .05,22 1.717 or t t / 2, t .05,22 1.717 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(33 .33 31 .50 ) 0 1 (12 1)20 .24 (12 1)12 .09 1 12 12 2 12 12
1.12 , p-value = .2761. There is not
enough evidence to infer a difference in speeds.
13.17a H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = 1.59, p-value = .3050; use equal-variances test statistic Rejection region: t t , t.05,38 1.684 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(36 .93 31 .36 ) 0 (15 1)4.23 2 (25 1)3.35 2 1 1 15 25 15 25 2
= 4.61, p-value = 0. There is enough
evidence to infer that Tastee is superior.
1 1 = (36.93 – 31.36) 2.021 b ( x1 x 2 ) t / 2 s 2p n1 n 2
(15 1)4.23 2 (25 1)3.35 2 1 1 15 25 15 25 2
= 5.57 2.43; LCL = 3.14, UCL = 8.00 c The histograms are somewhat bell shaped. The weight gains may be normally distributed.
13.18
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
Two-tail F test: F = .99, p-value = .9571; use equal-variances test statistic Rejection region: t t / 2, t .025, 238 1.960 or t t / 2, t .025, 238 1.960 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(10 .01 9.12 ) 0 (120 1)4.43 2 (120 1)4.45 2 1 1 120 120 2 120 120
1.55, p-value = .1204. There is not
enough evidence to infer that oat bran is different from other cereals in terms of cholesterol reduction?
13.19 a H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0 Two-tail F test: F = .50, p-value = 0; use unequal-variances test statistic
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 449
Rejection region: t t / 2, t .025, 449 1.960 or t t / 2, t .025, 449 1.960 287
t
( x 1 x 2 ) ( 1 2 ) s 12 s 22 n1 n 2
=
(58 .99 52 .96 ) 0 30 .77 2 43 .32 2 250 250
= 1.79, p-value = .0734. There is not enough evidence to
conclude that a difference in mean listening times exist between the two populations.
s2 s2 b ( x 1 x 2 ) t / 2 1 2 = (58.99 –52.96) 1.960 n1 n 2
30 .77 2 43 .32 2 = 6.03 6.59; 250 250
LCL = –.56, UCL = 12.62 c The histograms are bell shaped.
13.20 a H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = 1.01, p-value = .9619; use equal-variances test statistic Rejection region: t t , t .05, 282 1.645 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(59 .81 57 .40 ) 0 (125 1)7.02 2 (159 1)6.99 2 1 1 125 159 125 159 2
2.88, p-value = .0021. There is
enough evidence to infer that the cruise ships are attracting younger customers.
1 1 = (59.81 – 57.40) b ( x1 x 2 ) t / 2 s 2p n1 n 2
(125 1)7.02 2 (159 1)6.99 2 1 1 2.576 125 159 2 125 159
= 2.41 2.16; LCL = .25, UCL = 4.57
13.21a H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0 Two-tail F test: F = .98, p-value = .9254; use equal-variances test statistic Rejection region: t t / 2, t.025,198 1.972 or t t / 2, t.025,198 1.972
t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(10 .23 9.66 ) 0 (100 1)2.87 2 (100 1)2.90 2 1 1 100 100 100 100 2
= 1.40, p-value = .1640. There is not
enough evidence to infer that the distance males and females drive differs.
1 1 = (10.23 – 9.66) 1.972 b ( x1 x 2 ) t / 2 s 2p n n 2 1
(100 1)2.87 2 (100 1)2.90 2 1 1 100 100 100 100 2
= .57 .80; LCL = –.23, UCL = 1.37 288
c The histograms are bell shaped.
13.22
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = .98, p-value = .9520; use equal-variances test statistic Rejection region: t t , t .05,58 1.671 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(115 .50 110 .20 ) 0 (30 1)21 .69 2 (30 1)21 .93 2 1 1 30 30 30 30 2
.94 , p-value = .1753. There is not
enough evidence to retain supplier A - switch to supplier B.
13.23
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
Two-tail F test: F = .92, p-value = .5000; use equal-variances test statistic Rejection region: t t / 2, t .025,594 1.960 or t t / 2, t .025,594 1.960 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(5.56 5.49 ) 0 (306 1)5.36 2 (290 1)5.58 2 1 1 306 290 306 290 2
.16 , p-value = .8759. There is no
evidence of a difference in job tenures between men and women.
13.24a H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0 Two-tail F test: F = 5.18, p-value = .0019; use unequal-variances test statistic
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 33.9 (rounded to 34)
Rejection region: t t / 2, t .005,34 2.724 or t t / 2, t .005,34 2.724 t
( x 1 x 2 ) ( 1 2 ) s 12 s 22 n1 n 2
=
(70 .42 56 .44 ) 0 20 .54 2 9.03 2 24 16
2.94 , p-value = .0060. There is enough evidence to conclude
that the two packages differ in the amount of time needed to learn how to use them.
s2 s2 b ( x1 x 2 ) t / 2 1 2 = (70.42 –56.44) 2.030 n n 2 1
20 .54 2 9.03 2 = 13.98 9.67; 24 16
LCL = 4.31, UCL = 23.65 c The amount of time is required to be normally distributed. 289
d The histograms are somewhat bell shaped.
13.25a H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0 Two-tail F test: F = 5.18, p-value = .0019; use unequal-variances test statistic
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 276.5 (rounded to 277)
Rejection region: t t , t .01, 277 2.326
t
( x 1 x 2 ) ( 1 2 ) s 12
n1
s 22
=
n 2
(5.02 7.80 ) 0 1.39 2 3.09 2 200 200
= –11.60, p-value = 0. There is enough evidence to infer that the
amount of time wasted in unsuccessful firms exceeds that of successful firms.
s2 s2 b ( x1 x 2 ) t / 2 1 2 = (5.02 – 7.80) 1.960 n n 2 1
1.39 2 3.09 2 = –2.78 .47; 200 200
LCL = –3.25, UCL = –2.31. Workers in unsuccessful companies waste on average between 2.31 and 3.25 hours per week more than workers in successful companies.
13.26
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = .73, p-value = .0699; use equal-variances test statistic Rejection region: t t , t .05, 268 1.645 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(.646 .601) 0 (125 1).045 2 (145 1).053 2 1 1 125 145 2 125 145
= 7.54, p-value = 0. There is enough
evidence to conclude that the reaction time of drivers using cell phones is slower that for non-cell phone users.
13.27
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
Two-tail F test: F = 1.10, p-value = .6406; use equal-variances test statistic Rejection region: t t / 2, t .025,183 1.973 or t t / 2, t .025,183 1.973 t
( x1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(.654 .662 ) 0 (95 1).048 2 (90 1).045 2 1 1 95 90 95 90 2
= –1.17, p-value = .2444. There is not
enough evidence to infer that the type of discussion affects reaction times. 290
13.28
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = .97, p-value = .9054; use equal-variances test statistic Rejection region: t t , t.05,143 1.656 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(6.18 5.94 ) 0 (64 1)1.59 2 (81 1)1.61 2 1 1 64 81 64 81 2
= .90, p-value = .1858. There is not enough
evidence to infer that people spend more time researching for a financial planner than they do for a stock broker.
13.29
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
Two-tail F test: F = .74, p-value = .0446; use unequal-variances test statistic
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 373
Rejection region: t t , t.05,373 1.645
t
( x 1 x 2 ) ( 1 2 ) s 12 s 22 n1 n 2
=
(63 .71 66 .80 ) 0 5.90 2 6.85 2 173 202
= –4.69, p-value = 0. There is enough evidence to infer that
students without textbooks outperform those with textbooks.
13.30
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
Two-tail F test: F = .85, p-value = .2494; use equal-variances test statistic Rejection region: t t / 2, t .025, 413 1.960 or t t / 2, t .025, 413 1.960 t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(149 .85 154 .43) 0 (213 1)21 .82 2 (202 1)23 .64 2 1 1 213 202 2 213 202
= –2.05, p-value = .0412. There is
enough evidence to conclude that there are differences in service times between the two chains.
13.31
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
Two-tail F test: F = .80, p-value = .1818; use equal-variances test statistic Rejection region: t t / 2, t .025,309 1.960 or t t / 2, t .025,309 1.960 291
t
( x 1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(488 .4 498 .1) 0 (124 1)19 .62 2 (187 1)21 .93 2 1 1 124 187 124 187 2
= –3.985, p-value = 0. There is
enough evidence to conclude that there are differences in the amount of sleep between men and women.
13.32 a H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0 Two-tail F test: F = 1.51, p-value = .0402; use unequal-variances test statistic
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 190
Rejection region: t t / 2, t.025,190 1.973 or t t / 2, t.025,190 1.973
t
( x1 x 2 ) (1 2 ) s12
n 1
s 22
=
n 2
(130 .93 126 .14 ) 0 31 .99 2 26 .00 2 100 100
= 1.16, p-value = .2467. There is not enough evidence to infer
that differences exist between the two types of customers.
13.33
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = 1.07, p-value = .8792; use equal-variances test statistic Rejection region: t t , t.05,38 1.684
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(73 .60 69 .20 ) 0 (20 1)15 .60 2 (20 1)15 .06 2 1 1 20 20 20 20 2
= .91, p-value = .1849. There is not
enough evidence to infer that the new design tire lasts longer than the existing design tire.
13.34
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
Two-tail F test: F = .95, p-value = .8252; use equal-variances test statistic Rejection region: t t , t .05,178 1.653
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
60,245 63,563 ) 0 (90 1)10,506 2 (90 1)10,755 2 1 1 90 90 2 90 90
= −2.09, p-value = .0189. There is
enough evidence to conclude that commission salespeople outperform fixed-salary salespersons
292
13.35
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
Two-tail F test: F = .88, p-value = .4709; use equal-variances test statistic Rejection region: t t / 2, t.025,429 1.645 and t t / 2, t .025,429 1.645
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
633 .97 661 .86 ) 0 (93 1)49 .45 2 (338 1)52 .69 2 1 1 93 338 93 338 2
= −4.58, p-value = 0. There is enough
evidence to conclude there is a difference in scores between those who have and those who do not have accidents in a three-year period.
13.36
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = .41, p-value = 0; use unequal-variances test statistic
(s12 / n 1 s 22 / n 2 ) 2 (s12 / n 1 ) 2 (s 22 / n 2 ) 2 n1 1 n 2 1
= 222
Rejection region: t t , t .05,222 1.645
t
( x1 x 2 ) (1 2 )
(14 .20 11 .27 ) 0
= 6.28, p-value = 0. There is enough evidence to conclude that s12 s 22 2.84 2 4.42 2 n n 130 130 2 1 bottles of wine with metal caps are perceived to be cheaper.
13.37
=
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
Two-tail F test: F = 1.14, p-value = .2430; use equal-variances test statistic Rejection region: t t , t .05,641 1.645
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(496 .9 511 .3) 0 (355 1)173 .78 2 (288 1)69 .07 2 1 1 355 288 2 355 288
= −2.54, p-value = .0057. There is
enough evidence to conclude that SAT scores increased after the change in school start time.
13.38
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = 3.14, p-value = 0; use unequal-variances test statistic
293
(s12 / n1 s 22 / n 2 ) 2 (s12 / n1 ) 2 (s 22 / n 2 ) 2 n1 1 n2 1
74
Rejection region: t t , t .05,74 1.665
t
( x1 x 2 ) (1 2 ) s12
n 1
s 22
=
n 2
(93 .81 61 .25) 0 15 .89 2 8.96 2 48 48
= 12.37, p-value = 0. There is enough evidence to conclude that the
larger the bucket the more people will eat?
13.39a H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = .48, p-value = .0856; use equal-variances test statistic Rejection region: t t , t .05,46 1.679
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(97 .71 94 .58 ) 0 (24 1)5.10 2 (24 1)7.36 2 1 1 24 24 2 24 24
= 1.71, p-value = .0471. There is enough
evidence to conclude that diners who believe they are drinking a fine wine (California wine) eat more than diners who believe they are drinking an inferior wine?
b
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = 1.07, p-value = .8796; use equal-variances test statistic Rejection region: t t , t .05,46 1.679
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(64 .00 57 .33) 0 (24 1)9.44 2 (24 1)9.14 2 1 1 24 24 24 24 2
= 2.49, p-value = .0083. There is enough
evidence to conclude that diners who believe they are drinking a fine wine (California wine) spend more time in the restaurant than diners who believe they are drinking an inferior wine?
13.40
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = 1.57, p-value = .2486; use equal-variances test statistic Rejection region: t t , t .05,54 1.673
294
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(10 .04 8.64 ) 0 (28 1)2.32 2 (28 1)1.85 2 1 1 28 28 28 28 2
= 2.49, p-value = .0080. There is enough
evidence to conclude that people eat more when they are not aware of how much they have already eaten?
3.41
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = 5.99, p-value = .0001; use unequal-variances test statistic
(s12 / n1 s 22 / n 2 ) 2 (s12 / n1 ) 2 (s 22 / n 2 ) 2 n1 1 n2 1
25
Rejection region: t t / t .05,25 1.708
t
( x1 x 2 ) (1 2 ) s12
n 1
s 22
=
n 2
(136 .80 72 .80 ) 0 124 .16 2 9.87 2 20 20
= 10.97, p-value = 0. There is enough evidence to conclude that
people will eat more M&M’s when given a full pound.
13.42
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
Two-tail F test: F = .92, p-value = .1565; use equal-variances test statistic Rejection region: t < -tα,υ = -t.05,∞ = -1.645
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(4.69 4.86 ) 0 1 (756 1)8.58 (7643 1)9.30 1 756 7643 2 756 7643
enough to conclude that the adage is true.
13.43
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = 2.12, p-value = 0; use unequal-variances test statistic
( s12 / n1 s 22 / n 2 ) 2 ( s12 / n1 ) 2 ( s 22 / n 2 ) 2 n1 1 n2 1
= 926
Rejection region: t t / t.05,926 1.645
295
= -1.48, p-value = .0692. There is not
t
( x1 x 2 ) (1 2 ) s12 s 22 n n 2 1
=
34 .97 31 .63) 0 97 .02 45 .83 483 521
= 6.31, p-value = 0. There is enough evidence to conclude that
guesses this year are higher than they were 5 years ago.
13.44
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = 1.27, p-value = .2120 use equal-variances test statistic Rejection region: t t / t.05, 234 1.645
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(57 ,030 56,055 ) 0 1 (129 1)24,910 ,081 (97 1)19,545 ,241 1 129 97 2 129 97
= 1.53, p-value = .0643.There is
not enough evidence to conclude that electrical engineers receive higher starting salaries than do mechanical engineers.
13.45
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = 2.39, p-value = .0220; use unequal-variances test statistic
( s12 / n1 s 22 / n 2 ) 2 ( s12 / n1 ) 2 ( s 22 / n 2 ) 2 n1 1 n2 1
= 50
Rejection region: t t / t.05,50 1.676
t
( x1 x 2 ) (1 2 ) s12
n 1
s 22
=
n 2
166 .8 116 .7) 0 522 .6 218 .7 30 30
= 10.08, p-value = 0. There is enough evidence to conclude that
chocolate helps improve cognitive memory.
13.46
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = .67, p-value = .0081; use unequal-variances test statistic
( s12 / n1 s 22 / n 2 ) 2 ( s12 / n1 ) 2 ( s 22 / n 2 ) 2 n1 1 n2 1
= 339
Rejection region: t t / t.05,339 1.645
296
t
( x1 x 2 ) (1 2 ) s12 s 22 n n 2 1
=
7.31 6.97 ) 0 31 .18 46 .53 177 178
= .51, p-value = .3053. There is not enough evidence to conclude
that Americans have more missing teeth.
13.47
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = 1.04, p-value = .9378 use equal-variances test statistic Rejection region: t t / t.05,162 1.654
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(20 .15 14 .82 ) 0 1 (134 1)14 .67 (30 1)14 .06 1 134 30 2 134 30
= 6.92, p-value = 0.There is enough
evidence that White American adults get more slow-wave sleep than Black American adults.
13.48
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = 1.02, p-value = .9310 use equal-variances test statistic Rejection region: t t / t.05,182 1.653
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(74 .85 67 .81) 0 1 (86 1)113 .85 (98 1)111 .72 1 86 98 2 86 98
= 4.49, p-value = 0.There is enough
evidence to conclude that the exercise improves cognitive impairment.
13.49
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = .97, p-value = .6983 use equal-variances test statistic Rejection region: t t / t .05, 1.645
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n n 2 1
=
(1452 .56 1246 .61) 0 1 (438 1)130 ,588 (571 1)135 ,277 1 438 571 2 438 571
= 8.88, p-value = 0.There is enough
evidence to infer that low-income smokers spend more than high-income smokers.
13.50
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = .2.70, p-value = 0; use unequal-variances test statistic
297
( s12 / n1 s 22 / n 2 ) 2 ( s12 / n1 ) 2 ( s 22 / n 2 ) 2 n1 1 n2 1
= 498
Rejection region: t t / t.05, 498 1.645
t
( x1 x 2 ) (1 2 ) s12
n 1
s 22
=
n 2
1157 .77 1091 .71) 0 157 ,220 58,327 441 325
= 2.66, p-value = .0040. There is enough evidence to conclude
that low-income alcoholic drinkers spend more than middle-income.
13.51
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
Two-tail F test: F = 1.00, p-value = .9644 use equal-variances test statistic Rejection region: t < -tα/2,υ = -t.025,∞ = -1.96 or t > tα/2,υ = t.025,∞ = 1.96
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
448 .28 443 .03) 0 1 (552 1)9799 (577 1)9837 1 552 577 2 552 577
= .89, p-value = .3741.There is not
enough to infer a difference between the two sexes.
13.52
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
Two-tail F test: F = .91, p-value = .2915 use equal-variances test statistic Rejection region: t t / t.05, 1.645
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(6.44 10 .53) 0 1 (524 1)15 .15 (409 1)16 .71 1 524 409 2 524 409
= -15.57, p-value = 0.There is enough
evidence to infer that white-collar public servants take more sick days than do white-collar private sector workers.
13.53
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = .93, p-value = .4742 use equal-variances test statistic Rejection region: t t / t .05, 1.645
t
( x1 x 2 ) (1 2 ) 1 1 s 2p n1 n 2
=
(10 .53 10 .32 ) 0 1 (409 1)16 .71 (397 1)17 .94 1 409 397 2 409 397
= .68, p-value = .2469.There is not
enough evidence to conclude that white-collar public sector workers take more sick days than they did 5 years ago.
298
13.54a. Two-tail F test: F = 2.24, p-value = 0. Use unequal-variances estimator.
LCL = 15,826, UCL = 23,645 b. Incomes are required to be normally distributed. c. The required condition is not satisfied. However, the sample sizes are large enough to overcome this deficiency.
13.55 Two-tail F test: F = 1.16, p-value = .1318. Use the equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = 7.36, p-value = 0. There is enough evidence to conclude that men work longer hours. b The numbers of hours appear to be bell shaped.
13.56a. Two tail F test: F = .415, p-value = 0. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = 4.99, p-value = 0. There is enough evidence to infer that American born residents are more educated than people born outside the United States. b. The required condition of normality is satisfied. c. A nonparametric test, the Wilcoxon rank sum test can be used.
13.57a Two tail F test: F = 1.03, p-value = .7934. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 299
t = 1.04, p-value = .1490. There is not enough evidence to infer that American born residents have higher incomes than people born outside the United States. b. Incomes are required to be normally distributed. c. The required condition is not satisfied. However, the sample sizes are large enough to overcome this deficiency.
13.58a. Two tail F test: F = .999, p-value = .9879. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -1.73, p-value = .0420. There is enough evidence to infer that Republicans are older than Democrats. b. The required condition of normality is satisfied. c. Wilcoxon rank sum test.
13.59a. Two tail F test: F = .756, p-value = .0054. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -3.37, p-value = .0004. There is enough evidence to infer that Republicans earn more income than Democrats. b.
300
LCL = -16,041, UCL = -4,223 c. Incomes are required to be normally distributed. d The required condition is not satisfied. However, the sample sizes are large enough to overcome this deficiency.
13.60a. Two tail F test: F = .813, p-value = .0417. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -1.46, p-value = .0730. There is not enough evidence to infer that Republicans do not work harder than do Democrats. b. The required condition is satisfied. c. Wilcoxon rank sum test
13.61 Two tail F test: F = 1.30, p-value = .0010. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -1.32, p-value = .0931. There is not enough evidence to infer that Republicans are more educated than Democrats. b.
LCL = -,537, UCL = .104
301
13.62 Two tail F test: F = 1.41, p-value = .0002. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -1.77, p-value = .0389. There is enough evidence to infer that Republicans are older than Democrats when their first child is born. b. Ages are required to be normally distributed. c. We can use the Wilcoxon rank sum test.
13.63a. Two tail F test: F = .961, p-value = .6748. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -2.08, p-value = .0189. There is enough evidence to infer that conservatives have higher incomes than liberals. b. Incomes are required to be normally distributed. This requirement is not satisfied. c. Wilcoxon rank sum test.
13.64 Two tail F test: F = .955, p-value = .5403. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
302
t = 4.65, p-value = 1.00. The p-value tells us that in fact there is enough evidence that Liberals are more educated than conservatives.
13.65a. Two tail F test: F = .865, p-value = .1316. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -1.65, p-value = .0493. There is enough evidence to infer that conservatives work longer hours than do liberals. b. Hours of work are required to be normally distributed. c. Wilcoxon rank sum test.
13.66a. Two tail F test: F = .942, p-value = .4214. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = 9.25, p-value = 0. There is enough evidence to conclude that government workers have more education than do private sector employees. b. The required condition is satisfied. c. Wilcoxon rank sum test
13.67a. Two tail F test: F = .534, p-value = 0. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
303
t = 1.45, p-value = .0744. There is not enough evidence to infer that government workers have higher incomes than do private sector workers. b. Incomes are not normally distributed. c. Wilcoxon rank sum test
13.68a. Two tail F test: F = 2.36, p-value = 0. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = 3.09, p-value = .0011. There is enough evidence to infer that self-employed individuals have higher incomes. b. Incomes are not normally distributed. c. Wilcoxon rank sum test
13.69 Two tail F test: F = .900, p-value = .4602. Use equal variances estimator.
LCL = -35,341, UCL = 19,248.
13.70a. Two tail F test: F = .541, p-value = 0. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
304
t = 1.02, p-value = .1541. There is not enough evidence to infer that people who work for someone else have higher incomes than self-employed people. b
LCL = -6,513, UCL = 20,433.
13.71 Two tail F test: F = .548, p-value = 0. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -3.66, p-value = .0001. There is enough evidence to infer that heads of households who completed high school have more debt than those who did not finish high school.
13.72 Two tail F test: F = .502, p-value = 0. Use unequal variances estimator.
LCL = -23,715, UCL = -10,707.
13.73 Two tail F test: F = 1.031, p-value = .8148. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
305
t = 1.20, p-value = .2321. There is not enough evidence to conclude that a difference exists in unrealized capital gains between those who finished high school and those who did not.
13.74 Two tail F test: F = .927, p-value = .5336. Use equal variances estimator.
LCL = -24,385, UCL = -3,092.
13.75 Two tail F test: F = .534, p-value = 0. Use unequal variances estimator
LCL = -31,633, UCL = -16,571.
13.76 Two tail F test: F = .680, p-value = .0015. Use unequal variances test statistic H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = .34, p-value = .3680. There is not enough evidence to infer that heads of households with college graduates have smaller unrealized capital gains than do heads of households with only some college.
13.77a. Two tail F test: F = 2.07, p-value = 0. Use unequal variances test statistic H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
306
t = 10.67, p-value = 0. There is sufficient evidence to infer that male heads of households have higher incomes than female heads of households. b.
LCL = 23,059, UCL = 33,463. c. Incomes are not normally distributed.
13.78 Two tail F test: F = 1.009, p-value = .9274. Use equal variances test statistic H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -.085, p-value = .4660. The is not enough evidence to infer that heads of households who finished high school have greater net worth than those who did not finish high school.
13.79 Two tail F test: F = .519, p-value = 0. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
307
t = -6.24, p-value = 0. here is enough evidence to infer that heads of households with college degrees have more assets than those with only some college.
13.80 Two tail F test: F = .404, p-value = 0. Use unequal variances test statistic. H0: (µ 4- µ 3) = 0 H1: (µ 4 - µ 3) < 0
t = -6.03, p-value = 0. There is enough evidence to infer that heads of households with some college have less debts than those with college degrees.
t =6.03, p-value = 0. There is enough evidence to conclude that households whose heads have some college have less debt than households whose heads completed a college degree.
13.81 The data are observational. Experimental data could be produced by randomly assigning babies to either Tastee or the competitor’s product.
13.82 Assuming that the volunteers were randomly assigned to eat either oat bran or another grain cereal the data are experimental.
13.83a. Since 40 students were selected randomly, but were given the choice as to which software package to use, the data must be observational. b. Experimental data could have been derived by selecting the 40 students at random and assigning either of the software packages at random.
308
c. Students may choose the software package to which they have prior experience and greater proficiency. The differences in the amount of time needed to learn how to use each software package may be a function of the popularity of one software package over the other.
13.84a Let students select the section they wish to attend and compare test results. b Randomly assign students to either section and compare test results.
13.85 Randomly assign patients with the disease to receive either the new drug or a placebo.
13.86a Randomly select finance and marketing MBA graduates and determine their starting salaries. b Randomly assign some MBA students to major in finance and others to major in marketing. Compare starting salaries after they graduate. c Better students may be attracted to finance and better students draw higher starting salaries.
13.87a The data are observational because to obtain experimental data would entail randomly assigning some people to smoke and others not to smoke. b It is possible that some people smoke because of a genetic defect (Genetics have been associated with alcoholism.), which may also be linked to lung cancer. c In our society the experiment described in part a is impossible.
H0 : D = 0
13.88
H1 : D < 0 Rejection region: t t , t .05,7 1.895 t
x D D sD / n D
4.75 0 4.17 / 8
3.22 , p-value = .0073. There is enough evidence to infer that the Brand A is better than
Brand B.
13.89
H0: µ D = 0 H1: µ D < 0
Rejection region: t t , t .05,7 1.895
t
xD D sD / nD
13.90
.175 0 .225 / 8
2.20, p-value = .0320. There is enough evidence to infer that ABS is better.
H0: µ D = 0 H1: µ D > 0
309
Rejection region: t t , t .05,6 1.943
t
xD D
1.86 0
sD / nD
1.98, p-value = .0473. There is enough evidence to infer that the camera reduces the
2.48 / 7
number of red-light runners.
13.91a H0: µ D = 0 H1: µ D < 0 Rejection region: t t , t.05,11 1.796
t
x D D
1.00 0
sD / nD
3.02 / 12
–1.15, p-value = .1375. There is not enough evidence to infer that the new fertilizer is
better. b xD t / 2
sD nD
= 1.00 2.201
3.02 12
1.00 1.92 ; LCL = –2.92, UCL = .92
c The differences are required to be normally distributed d No, the histogram is bimodal. e The data are experimental. f The experimental design should be independent samples.
13.92 a H0: µ D = 0 H 1: µ D > 0 Rejection region: t t , t.05,11 1.796
t
x D D
3.08 0
sD / nD
1.82, p-value = .0484. There is enough evidence to infer that companies with exercise
5.88 / 12
programs have lower medical expenses. b xD t / 2
sD nD
= 3.08 2.201
5.88 12
3.08 3.74 ; LCL = –.66, UCL = 6.82
c Yes because medical expenses will vary by the month of the year.
13.93
H 0: µ D = 0 H 1: µ D > 0
Rejection region: t t , t.05,149 1.656 t
x D D sD / n D
12 .4 0 99 .1 / 150
1.53, p-value = .0638. There is not enough evidence to infer that mortgage payments
have increases in the past 5 years.
310
13.94
H 0: µ D = 0 H 1: µ D ≠ 0
Rejection region: t t / 2, t .025, 49 2.009 or t t / 2, t .025, 49 2.009 t
x D D sD / n D
1.16 0
2.22 / 50
–3.70, p-value = .0005. There is enough evidence to infer that waiters and waitresses
earn different amounts in tips.
13.95 a x D t / 2 b
sD nD
= 19 .75 1.684
30 .63 40
19 .75 8.16 ; LCL = 11.59, UCL = 27.91
H 0: µ D = 0 H 1: µ D > 0
Rejection region: t t , t.05,39 1.684 t
x D D sD / n D
19 .75 0 30 .63 / 40
4.08, p-value = .0001. There is enough evidence to conclude that companies that
advertise in the Yellow Pages have higher sales than companies that do not. c The histogram of the differences is bell shaped. d No, because we expect a great deal of variation between stores.
13.96a H0: µ D = 0 H1: µ D < 0 Rejection region: t t , t .10,14 1.345 t
xD D sD / n D
57 .40 0 13 .14 / 15
16 .92 , p-value = 0. There is enough evidence to conclude that heating costs for
insulated homes is less than that for uninsulated homes. b xD t / 2
sD nD
= 57 .40 2.145
13 .14 15
57 .40 7.28 ; LCL = -64.68, UCL = -50.12; mean savings lies
between $50.12 and $64.80 c Differences are required to be normally distributed.
13.97
H 0: µ D = 0 H 1: µ D ≠ 0
Rejection region: t t / 2, t .025, 44 2.014 or t t / 2, t .025, 44 2.014
311
t
x D D sD / n D
42 .94 0
–.91, p-value = .3687. There is not enough evidence to infer men and women spend
317 .16 / 45
different amounts on health care.
13.98
H0: µ D = 0 H1: µ D < 0
Rejection region: t t , t.05,169 1.654 t
x D D sD / n D
183 .35 0 1568 .94 / 170
–1.52, p-value = .0647. There is not enough to infer stock holdings have
decreased.
13.99
H0: µ D = 0 H1: µ D > 0
Rejection region: t t , t.05,37 1.690
t
xD D
sD / nD
.0422 0
1.59, p-value = .0597. There is not enough evidence to conclude that ratios are higher
.1634 / 38
this year.
13.100 H0: µ D = 0 H1: µ D > 0 Rejection region: t t , t.05,54 1.676
t
x D D
sD / nD
520 .85 0
2.08, p-value = .0210. There is enough evidence to infer that company 1’s
1854 .92 / 55
calculated tax payable is higher than company 2’s.
13.101 H0: µ D = 0 H1: µ D > 0 Rejection region: t t , t.05,19 1.729 t
x D D sD / n D
4.55 0 7.22 / 20
2.82, p-value = .0055. There is enough evidence to that the new design tire lasts longer
than the existing design.
13.102 The matched pairs experiment reduced the variation caused by different drivers.
312
13.103 H0: µ D = 0 H 1: µ D > 0 Rejection region: t t , t.05,24 1.711 t
x D D sD / n D
4587 0 22 ,851 / 25
1.00, p-value = .1628. There is not enough evidence to infer that finance majors
attract higher salary offers than do marketing majors.
13.104 Salary offers and undergraduate GPA are not as strongly linked as are salary offers and MBA GPA.
13.105 a H0: µ D = 0 H1: µ D < 0 Rejection region: t t , t.05, 41 1.684
t
x D D
sD / nD
.10 0 1.95 / 42
–.33, p-value = .3704. There is not enough evidence to infer that for companies where
an offspring takes the helm there is a decrease in operating income. b
H0: µ D = 0 H1: µ D > 0
Rejection region: t t , t .05,97 1.660
t
x D D sD / nD
1.24 0
4.34, p-value = 0. There is enough evidence to infer that when an outsider becomes
2.83 / 98
CEO the operating income increases.
13.106 H0: µ D = 0 H1: µ D > 0
t = 24.93, p-value = 0. There is enough evidence to conclude that Americans are more educated than their fathers.
13.107 H0: µ D = 0 H1: µ D > 0
313
t = 28.38, p-value = 0. There is enough evidence to conclude that Americans are more educated than their mothers.
13.108
LCL = 23,515, UCL = 26,929.
13.109 H0: µ D = 0 H 1: µ D ≠ 0
t = .64, p-value = 5200. There is not enough evidence of a difference between spouses’ hours or work.
13.110 H0: µ D = 0 H1: µ D < 0
t = -3.28, p-value = .0005. There is enough evidence to infer that this year is worse than normal.
314
13.111
LCL = 3581, UCL = 6767.
13.112 a H 0 : 12 / 22 1
H1 : 12 / 22 1 Rejection region: F F / 2,1 ,2 F.05,29,29 1.88 or F F1 / 2,1 ,2 1 / F / 2,2 ,1 1 / F.05,29,29 1 / 1.88 .53 F = s12 / s 22 = 350/700 =.50, p-value = .0669. There is enough evidence to conclude that the population variances differ. b Rejection region: F F / 2,1 ,2 F.025,14,14 2.98 or F F1 / 2,1 ,2 1 / F / 2,2 ,1 1 / F.025,14,14 1 / 2.98 .34 F = s12 / s 22 = 350/700 =.50, p-value = .2071. There is not enough evidence to conclude that the population variances differ. c The value of the test statistic is unchanged and in this exercise the conclusion changed as well. s2 s2 1 28 28 1 13.113 a LCL = 1 = = .366, UCL = 1 F / 2, 2 ,1 = 4.03 = 5.939 2 s 2 F / 2, , 19 4 . 03 19 s2 2 1 2 s2 s2 1 28 28 1 = = .649, UCL = 1 F / 2, 2 ,1 = 2.27 = 3.345 b LCL = 1 2 s 2 F / 2, , 19 19 2.27 s2 2 1 2
c The interval narrows.
13.114
H 0 : 12 / 22 1 H1 : 12 / 22 1
Rejection region: F F / 2,1 ,2 F.025,9,10 3.78 or F F1 / 2,1 ,2 1 / F / 2,2 ,1 1 / F.025,10,9 1 / 3.96 .25 F = s12 / s 22 = .0000057/.0000114 =.50, p-value = .3179. There is not enough evidence to conclude that the two machines differ in their consistency of fills.
13.115
H 0 : 12 / 22 1
H1 : 12 / 22 1 Rejection region: F F1,1 ,2 1 / F,2 ,1 1 / F.05,9,9 1 / 3.18 .314
315
F = s12 / s 22 = .1854/.1893 =.98, p-value = .4879. There is not enough evidence to infer that the second method is more consistent than the first method.
H 0 : 12 / 22 1
13.1116
H1 : 12 / 22 1 Rejection region: F F / 2,1 ,2 F.025,10,10 3.72 or F F1 / 2,1 ,2 1 / F / 2,2 ,1 1 / F.025,10,10 1 / 3.72 .269 F = s12 / s 22 = 193.67/60.00 = 3.23, p-value = .0784. There is not enough evidence to infer that the variances of the marks differ between the two sections.
13.117
H 0 : 12 / 22 1
H1 : 12 / 22 1 Rejection region: F F,1 ,2 F.05,99,99 1.39 F = s12 / s 22 = 19.38/12.70 = 1.53, p-value = .0183. There is enough evidence to infer that limiting the minimum and maximum speeds reduces the variation in speeds.
13.118 H 0 : 12 / 22 1
H1 : 12 / 22 1 Rejection region: F F / 2,1 ,2 F.025,99,99 1.48 or F F1 / 2,1 ,2 1 / F / 2,2 ,1 1 / F.025,99,99 1 / 1.48 .68 F = s12 / s 22 = 41,309/19,850 = 2.08, p-value = .0003. There is enough evidence to conclude that the variances differ.
13.119
H 0 : 12 / 22 = 1 H1 : 12 / 22 < 1
Rejection region: F F1,1 ,2 1 / F,2 ,1 1 / F.05,51,51 1 / 1.60 .63 F = s12 / s 22 = .0261/.0875 = .298, p-value = 0. There is enough evidence to infer that portfolio 2 is riskier than portfolio 1.
13.120
H 0 : 12 / 22 = 1 H1 : 12 / 22 1
Rejection region: F F / 2,1 ,2 F.05,99,99 1.39 or F F1,1 ,2 1 / F,2 ,1 1 / F.05,99,99 1 / 1.39 .72
316
F = s12 / s 22 = 3.35/10.95 = .31, p-value = 0. There is enough evidence to conclude that the variance in service times differ.
13.121
H 0 : 12 / 22 = 1
H1 : 12 / 22 > 1
F = 2.36, p-value = 0. There is enough evidence to infer that the variance in income is greater for self-employed individuals. Note; That the version of XLSTAT that the author had showed that the p-value is 1.000, which is clearly incorrect. The author changed the printout to show the correct result.
13.122
H 0 : 12 / 22 = 1
H1 : 12 / 22 > 1
F = 2.12, p-value = 0. There is enough evidence to conclude that the hours worked by the self-employed vary more than do the hours of employees who work for someone else. Note; That the version of XLSTAT that the author had showed that the p-value is 1.000, which is clearly incorrect. The author changed the printout to show the correct result.
13.123
H 0 : 12 / 22 = 1
H1 : 12 / 22 < 1 317
F = .54, p-value = 0. There is enough evidence to infer that the variation in income is greater for self-employed individuals.
13.124
H 0 : 12 / 22 = 1
H1 : 12 / 22 < 1
F = .904, p-value = .2390. There is not enough evidence to infer that the variation in net worth is greater for selfemployed individuals.
13.125
H 0 : 12 / 22 = 1 H1 : 12 / 22 < 1
318
F = .90, p-value = .2301. There is not enough evidence to infer that the variation in debt is greater for self-employed individuals.
13.126
H 0 : 12 / 22 = 1
H1 : 12 / 22 < 1
F = .652, p-value = .0014. There is enough evidence to infer that the variation in total capital gains is greater for self-employed individuals. 13.127 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
a z
b z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2 (p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
=
(.45 .40 ) 1 1 .425 (1 .425 ) 100 100 (.45 .40 ) 1 1 .425 (1 .425 ) 400 400
= .72, p-value = 2P(Z > .72) = 2(1 – .7642) = .4716.
= 1.43, p-value = 2P(Z > 1.43) = 2(1 – .9236) = .1528.
c The p-value decreases. 13.128 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0 a z
bz
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2 (p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
=
(.60 .55) 1 1 .575 (1 .575 ) 225 225 (.95 .90 ) 1 1 .925 (1 .925 ) 225 225
1.07 , p-value = 2P(Z > 1.07) = 2(1 – .8577) = .2846
2.01, p-value = 2P(Z > 2.01) = 2(1 – .9778) = .0444.
c. The p-value decreases. 319
d z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n n 2 1
=
(.10 .05) 1 1 .075 (1 .075 ) 225 225
2.01, p-value = 2P(Z > 2.01) = 2(1 – .94778) = .0444.
e. The p-value decreases.
13.129 a (p̂1 p̂ 2 ) z / 2
p̂1 (1 p̂ 1 ) p̂ 2 (1 p̂ 2 ) .18(1 .18) .22 (1 .22 ) = (.18–.22) 1.645 n2 n1 100 100
= –.040 .0929 b (p̂1 p̂ 2 ) z / 2
p̂1 (1 p̂1 ) p̂ 2 (1 p̂ 2 ) .48(1 .48) .52 (1 .52 ) = (.48–.52) 1.645 n1 n2 100 100
= –.040 .1162 c The interval widens. 13.130 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(.205 .140 )
1.70, p-value = P(Z > 1.70) = 1 – .9554 = .0446.
1 1 .177 (1 .177 ) 229 178
There is enough evidence to conclude that those who paid the regular price are more likely to buy an extended warranty. 13.131 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n n 2 1
=
(.145 .234 ) 1 1 .209 (1 .209 ) 83 209
1.69, p-value = 2P(Z < −1.69) = 2(.0455) = .0910.
There is not enough evidence to conclude that new and old accounts are different with respect to overdue accounts. 13.132 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n n 2 1
=
(. 26 .07 ) 1 1 .165 (1 .165 100 100
3.62 , p-value = P(Z > 3.62) = 0.
There is enough evidence to conclude that obese Chinese food eaters are less likely to use chop sticks?
320
13.133a H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .05 = 1.645 z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(.56 .46 ) 1 1 .518 (1 .518 ) 1100 800
= 4.31, p-value = 0. There is enough evidence to infer that
the leader’s popularity has decreased. b
H0: (p1 – p2) = .05 H1: (p1 – p2) > .05
Rejection region: z z z .05 = 1.645
z
(p̂1 p̂ 2 ) (p1 p 2 ) p̂1 (1 p̂1 ) p̂ 2 (1 p̂ 2 ) n2 n1
=
(.56 .46 ) .05
= 2.16, p-value = P(Z > 2.16) = 1 – .9846 = .0154.
.56 (1 .56 ) .46 (1 .46 ) 1100 800
There is enough evidence to infer that the leader’s popularity has decreased by more than 5%. c (p̂1 p̂ 2 ) z / 2
p̂1 (1 p̂1 ) p̂ 2 (1 p̂ 2 ) .56 (1 .56 ) .46 (1 .46 ) = (.56 .46 ) 1.96 n1 n2 1100 800
= .10 .045 13.134 H0: (p1 – p2) = -.08 H1: (p1 – p2) < -.08 Rejection region: z z z.01 = –2.33
z
(p̂1 p̂ 2 ) (p1 p 2 ) p̂1 (1 p̂1 ) p̂ 2 (1 p̂ 2 ) n1 n2
=
(.11 .28) (.08) .11(1 .11) .28(1 .28) 300 300
= –2.85, p-value =P(Z < –2.85) = 1 – .9978 = .0022.
There is enough evidence to conclude that management should adopt process 1. 13.135 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .05 = 1.645 z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(.071 .064 ) 1 1 .068 (1 .068 ) 1604 1109
= .71, p-value = P(Z > .71) = 1 – .7611= .2389.
There is not enough evidence to infer that the claim is false. 13.136 a H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 321
Rejection region: z z z .05 = –1.645
z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(.093 .115 ) 1 1 .104 (1 .104 ) 6281 6281
= –4.04, p-value = 0. There is enough evidence to infer
that Plavix is effective. 13.137 a H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 Rejection region: z z z .01 = –2.33 p̂1
z
104 189 104 189 .0095 p̂ 2 .0172 p̂ .01335 11,000 11,000 22,000
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n n 2 1
=
(.0095 .0172 ) 1 1 .01335 (1 .01335 ) 11,000 11,000
= –4.98, p-value = 0. There is enough evidence
to infer that aspirin is effective in reducing the incidence of heart attacks. 13.138 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .05 1.645 p̂1
z
1,084 1,084 997 997 .0906 p̂ .0946 .0985 p̂ 2 11,000 22,000 11,000
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n n 2 1
=
(.0985 .0906 ) 1 1 .0946 (1 .0946 ) 11,000 11,000
2.00 , p-value = P(Z > 2.00) = 1 – .9772 = .0228.
There is enough evidence to infer that aspirin leads to more cataracts. 13.139 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 Rejection region: z z z .01 = –2.33 p̂1
z
132 75 75 132 .0399 .0289 p̂ 2 .0509 p̂ 2,594 2,594 2,594 2,594
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(.0289 .0509 ) 1 1 .0399 (1 .0399 ) 2,594 2,594
There is enough evidence to infer that Letrozole works.
322
4.04 , p-value = 0.
13.140 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 Rejection region: z z z .05 1.645
p̂1 z
88 105 88 105 .2409 .2228 p̂ 2 .2586 p̂ 395 406 395 406 (p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(.2228 .2586 )
1.19, p-value = P(Z < –1.19) = .1170. There is not
1 1 .2409 (1 .2409 ) 395 406
enough evidence to infer that exercise training reduces mortality. 13.141 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .05 1.645
p̂1 z
27 18 27 18 .1585 .1895 p̂ 2 .1429 p̂ 95 189 95 189 (p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(. 1895 .1429 ) 1 1 .1585 (1 .1585 ) 95 189
1.02 , p-value = P(Z > 1.02) = 1-.8461 = .1539.
There is not enough evidence to infer that women drivers in a following car are more likely to honk when the lead car driver is using a cell phone. 13.142 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .05 1.645
p̂1 z
35 34 34 35 .2536 p̂ 2 .2018 .1667 p̂ 138 138 204 204 (p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(. 2536 .1667 ) 1 1 .2018 (1 .2018 ) 138 204
1.97 , p-value = P(Z > 1.97) = 1-.9756 = .0244.
There is enough evidence to infer that men drivers in a following car are more likely to honk when the lead car driver is using a cell phone. 13.143 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 Rejection region: z z z .05 1.645
323
p̂1 z
33 49 49 33 .2129 p̂ 2 .2547 .2934 p̂ 155 167 155 167 (p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n n 2 1
=
(. 2129 .2934 ) 1 1 .2547 (1 .2547 ) 155 167
1.66, p-value = P(Z < –1.66) = .0485
There is enough evidence to allow us to conclude that drivers behind expensive cars are less likely to honk. 13.144 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .05 1.645
p̂1 z
64 47 47 64 .3000 .3422 p̂ 2 .2568 p̂ 187 183 187 183 (p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n n 2 1
=
(. 3422 .2568 ) 1 1 .3000 (1 .3000 ) 187 183
1.79 , p-value = P(Z > 1.79) = 1- .9633 = .0367
There is enough evidence to allow us to conclude that drivers behind a male driver are more likely to honk. 13.145 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 Rejection region: z z z .05 1.645
p̂1 z
68 105 105 68 .3223 p̂ 2 .3836 .4375 p̂ 211 211 240 240 (p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(. 3223 .4375 ) 1 1 .3836 (1 .3836 ) 211 240
2.51, p-value = P(Z < –2.51) = .0060
There is enough evidence to allow us to conclude that women are less likely to honk. 13.146 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 Rejection region: z z z .05 1.645
p̂1 z
86 21 86 21 .2616 .1858 p̂ 2 .2905 p̂ 113 296 113 296 (p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(. 1858 .2905 ) 1 1 .2616 (1 .2616 ) 113 296
2.15, p-value = P(Z < –2.15) = .0158
There is enough evidence to allow us to draw the inference that drivers of convertibles are less likely to honk.
324
13.147 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0 (p̂1 p̂ 2 )
a z
1 1 p̂(1 p̂) n1 n 2
=
(.9057 .8878 1 1 .8975 (1 .8975 ) 350 294
.75,
p-value = 2P(Z > .75) = 2(1 – .7734)= .4532. There is not enough evidence to infer that the two populations of car owners differ in their satisfaction levels. 13.148a H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .10 = 1.28 z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(.2632 .0741 ) 1 1 .11(1 .11) 38 162
= 3.35, p-value = 0. There is enough evidence to conclude that
smokers have a higher incidence of heart diseases than nonsmokers. b (p̂1 p̂ 2 ) z / 2
p̂1 (1 p̂1 ) p̂ 2 (1 p̂ 2 ) n2 n1
= (.2632–0741) 1.645
.2632 (1 .2632 ) .0741 (1 .0741 ) =.1891 .1223; LCL = .0668, UCL = .3114 38 162
13.149 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .05 = 1.645 z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(.3585 .3420 ) 1 1 .3504 (1 .3504 ) 477 462
= .53, p-value = P(Z > .53) = 1 – .7019 = .2981.
There is not enough evidence to infer that the use of illicit drugs in the United States has increased in the past decade. 13.150 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0 Rejection region: z z / 2 1.96, z z / 2 , z .025 1.96 1 = Success
z
( p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(.5169 .375 ) 1 1 .4463 (1 .4463 ) 445 440 325
= 4.24, p-value = 2P(Z > 4.24) = 0
There is enough evidence to infer that Canadians and Americans differ in their responses to the survey question. 13.151 H0: (p1 – p2) = -.02 H1: (p1 – p2) < -.02 Rejection region: z z z .05 = –1.645
z
(p̂1 p̂ 2 ) (p1 p 2 ) p̂1 (1 p̂1 ) p̂ 2 (1 p̂ 2 ) n1 n2
=
(.055 .11) (.02 ) .055 (1 .055 ) .11(1 .11) 200 200
= –1.28,
p-value = P(Z < –1.28) = .1003. There is not enough evidence to choose machine A. 13.152 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .05 = 1.645 2 = Success
z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(. 8608 .7875 ) 1 1 .8394 (1 .8394 ) 80 194
= 1.50, p-value = .0668.
There is not enough evidence to infer that those with more education are less likely to work 11 hours or more per day. 13.153 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 Rejection region: z z z .05 = -1.645 z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n n 2 1
=
(. 6058 .6480 ) 1 1 .6288 (1 .6288 ) 104 125
= -0.66, p-value = .2551
There is not enough evidence to infer that Americans are more dissatisfied with their jobs in 2011 than they were in 2008. 13.154 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .05 = 1.645
326
z = 3.64, p-value = .0001. There is enough evidence to infer that the proportion of American adults who believe that global warming is a very serious problem decreased between 2012 and 2013. 13.155 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0 Rejection region: z z / 2 1.96, z z / 2 , z .025 1.96
z = 1.56, p-value = .1198. There is not enough evidence to infer that men and women differ in their support of the pipeline. 13.156 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
327
z = 1.47, p-value = .0702. There is not enough evidence to infer that including a reference to religious activity reduces the probability of a callback. 13.157 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
z = -.65, p-value = .2588. There is not enough evidence to infer that the Wallonians had a higher callback frequency than did main stream religions. 13.158 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = 6.21, p-value = 0. There is enough evidence to infer that females are more likely to be in ideal cardiovascular health than males. 13.159 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
z = -3.97, p-value = .0001. There is sufficient evidence that there is a difference in the error rate between the two umpires.
328
13.160 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
z = -.29, p-value = .7708. There is not enough evidence that there is a difference in the error rate between the two umpires. 13.16a. H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
z = 1.90, p-value = .0570. There is not enough evidence of a difference.
b.
H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
z = 4.74, p-value = 0. There is enough evidence of a difference. 13.162 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
329
z = 3.84, p-value = .0001. There is enough evidence to infer that government workers are thriving compared to other workers. 13.163 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
z = -4.81, p-value = .0702. There is enough evidence to infer that American adults are less satisfied than five years ago. 13.164 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .05 = 1.645
z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n n 2 1
=
(.1385 .0905 ) 1 1 .1035 (1 .1035 ) 231 619
z = 2.04, p-value = P(Z > 2.04) = 1 – .9793 = .0207.
There is enough evidence to conclude that health conscious adults are more likely to buy Special X. 13.165a H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 Rejection region: z z z .05 = 1.645
p̂1
68 29 68 29 .3579 .4172 p̂ 2 .2685 p̂ 163 108 163 108
330
z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n n 2 1
=
(.4172 .2685 ) 1 1 .3579 (1 .3579 ) 163 108
2.50,
p-value = P(Z > 2.50) = 1 – .9938 = .0062. There is enough evidence to conclude that members of segment 1 are more likely to use the service than members of segment 4. b
H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
Rejection region: z z / 2 z .05 = –1.96 or z z / 2 z .05 = 1.96
p̂ 1 z
20 10 20 10 .3896 .3704 p̂ 2 .4348 p̂ 54 23 54 23 (p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n n 2 1
=
(.3704 .4348 1 1 .3896 (1 .3896 ) 54 23
.53,
p-value = 2P(Z < –.53) = 2(.2981) = .5962. There is not enough evidence to infer that retired persons and spouses that work in the home differ in their use of services such as Quik Lube. 13.166 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0 Rejection region: z z / 2 z .025 = –1.96 or z z / 2 z .025 = 1.96
z
(p̂1 p̂ 2 ) 1 1 p̂(1 p̂) n1 n 2
=
(.0995 .1297 ) 1 1 .1132 (1 .1132 ) 382 316
= –1.25, p-value = 2P(Z < –1.25) = 2(.1056) = .2112.
There is not enough evidence to infer differences between the two sources.
13.167 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
z = .98, p-value = .2393. There is not enough evidence to infer that a difference exists.
331
13.168 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
z = 3.56, p-value = .0004. There is enough evidence to conclude that a difference exists. 13.169 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
z = -2.54, p-value = .0055. There is enough evidence to infer that foreign-born people are more likely to have a graduate degree than native-born Americans. 13.170 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
z = -2.76, p-value = .0029. There is enough evidence to conclude that women are more likely to work for the government. 13.171 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
332
z = 1.53, p-value = .0624. There is not enough evidence to conclude that Democrats are more likely to work for the government than Republicans. 13.172 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
z = -1.02, p-value = .1528. There is not enough evidence to conclude that Republicans are more likely than Democrats to be working full time. 13.173 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
z = -3.92, p-value = 0. There is enough evidence to conclude that Republicans are more likely than Democrats to be working for themselves. 13.174 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
333
z = .95, p-value = .3439. There is not enough evidence to infer that married people are more likely to be working full time than never married people. 13.175 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = 4.89, p-value = 0. There is enough evidence to infer that married people are more likely to be working for themselves than never married people. 13.176 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
z = 7.31, p-value = 0. There is enough evidence to infer that there is a difference in the proportion of men and women working full time. 13.177 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
334
z = 5.85, p-value = 0. There is enough evidence to infer that there is a difference between married and never married people with respect to completion of a graduate degree. 13.178 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = 1.64, p-value = .0502. There is not enough evidence to infer that married people are more likely to be working for the government than never married people. 13.179 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = 3.25, p-value = .0006. There is enough evidence to conclude that native-born Americans are more likely to work for the government. 13.180 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
335
z = -.33, p-value = .3693. There is not enough evidence to infer that foreign-born people are more likely than nativeborn Americans to work for themselves. 13.181 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
z = .49, p-value = .6260. There is not enough evidence to infer that the proportion of men and women who answer correctly differ. 13.182 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
z = .42, p-value = .6773. There is not enough evidence to infer that the proportion of men and women who answer correctly differ. 13.183 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
336
z = 4.34, p-value = 0. There is enough evidence to infer that the proportion of men and women who answer correctly differ. 13.184 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
z = 6.21, p-value = 0. There is enough evidence to infer that the proportion of men and women who answer correctly differ. 13.185 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = 10.45, p-value = 0. There is enough evidence to infer that male heads of households are more likely to have a college degree than female heads of households. 13.186 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
337
z = 11.34, p-value = 0. There is enough evidence to infer that male heads of households are more likely to have a higher employment rate than female heads of households. 13.187 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = 16.14, p-value = 0. There is enough evidence to infer that male heads of households are more likely to own the home they live in than female heads of households. 13.188 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = 53.36, p-value = 0. There is overwhelming evidence to infer that male heads of households are more likely to be married or living with partner than female heads of households.
13.89 It appears that female heads of households are likely to be never married or divorced and have less education, rent their homes, and less likely to be working. 13.190 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 338
z = 11.82, p-value = 0. There is enough evidence to infer that married people are more likely to have a college degree. 13.191 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
z = -12.02, p-value = 0. There is enough evidence to infer that married people are less likely to be unemployed. 13.192 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = 14.67, p-value = 0. There is enough evidence to infer that married people are more likely to be self-employed. 13.193 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
339
z = -.10, p-value = .4589. There is not enough evidence to infer that married people are less likely to have declared bankruptcy in the last five years. 13.194a. H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = 2.49, p-value = .0065. There is enough evidence to conclude that there has been a decrease in participation among boys over the past 10 years. b.
H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = .893, p-value = .1859. There is not enough evidence to conclude that there has been a decrease in participation among girls over the past 10 years. c.
H0: (p1 – p2) = 0 340
H1: (p1 – p2) > 0
z = 2.61, p-value = .0045. There is enough evidence to infer girls are less likely to participate in sports than boys in 2011.
13.195 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = .859, p-value = .7850; use equal-variances t-test
t = 2.65, p-value = .0059. There is enough evidence to conclude that the campaign is successful.
13.196 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = 2.57, p-value = 0; use unequal-variances t-test
341
t = 15.82, p-value = 0. There is sufficient evidence to infer that government employees work less than private-sector workers.
13.197 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0 Two-tail F test: F = .551, p-value = .0054; use unequal-variances t-test
t = –8.73, p-value = 0. There is enough evidence to conclude that men and women who suffer heart attacks vacation less than those who do not suffer heart attacks.
13.198
H0 : D = 0 H1 : D 0
342
t = –6.09, p-value = 0. There is enough evidence to infer that the new drug is effective. 13.199 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0 Two-tail F test: F = .958, p-value = .7726; use equal-variances t-test
t = –1.29, p-value = .1993. There is not enough evidence of a difference in reading time between the two groups. 13.200 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
343
z = –2.30, p-value = .0106. There is enough evidence to infer an increase in seatbelt use. 13.201a H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0 Two-tail F test: F = .811, p-value = .1404; use equal-variances t-test
t = –2.02, p-value = .0218. There is enough evidence to infer that housing cost a percentage of total income has increased. b The histograms are be bell shaped.
13.202a. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0 Two-tail F test: F = 1.10, p-value = .8430; use equal-variances t-test
344
A B C 1 t-Test: Two-Sample Assuming Equal Variances 2 3 Male Female 4 Mean 39.75 49.00 5 Variance 803.88 733.16 6 Observations 20 20 7 Pooled Variance 768.52 8 Hypothesized Mean Difference 0 9 df 38 10 t Stat -1.06 11 P(T<=t) one-tail 0.1490 12 t Critical one-tail 1.3042 13 P(T<=t) two-tail 0.2980 14 t Critical two-tail 1.6860 t = –1.06, p-value = .2980. There is not enough evidence to conclude that men and women differ in the amount of time spent reading magazines. b
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
Two-tail F test: F = .266, p-value = .0053; use unequal-variances t-test
A B C 1 t-Test: Two-Sample Assuming Unequal Variances 2 Low High 3 33.10 56.84 4 Mean 278.69 1047.81 5 Variance 6 Observations 21.00 19 0.00 7 Hypothesized Mean Difference 26.00 8 df -2.87 9 t Stat 0.0040 10 P(T<=t) one-tail 1.3150 11 t Critical one-tail 0.0080 12 P(T<=t) two-tail 13 t Critical two-tail 1.7056 t = –2.87, p-value = .0040. There is enough evidence to conclude that high-income individuals devote more time to reading magazines than do low-income individuals. 13.203a H 0 : D = 0
H1 : D 0
345
A B 1 t-Test: Paired Two Sample for Means 2 3 Female 4 Mean 55.68 5 Variance 105.64 6 Observations 25 7 Pearson Correlation 0.96 8 Hypothesized Mean Difference 0 9 df 24 10 t Stat -1.13 11 P(T<=t) one-tail 0.1355 12 t Critical one-tail 1.3178 13 P(T<=t) two-tail 0.2710 14 t Critical two-tail 1.7109
C
Male 56.40 116.75 25
t = –1.13, p-value = .2710. There is no evidence to infer that gender is a factor. b A large variation within each gender group was expected. c The histogram of the differences is somewhat bell shaped. 13.204 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
A 1 z-Test: Two Proportions 2 3 4 Sample Proportions 5 Observations 6 Hypothesized Difference 7 z Stat 8 P(Z<=z) one tail 9 z Critical one-tail 10 P(Z<=z) two-tail 11 z Critical two-tail
B
C
D
This Year 3 Years Ago 0.4351 0.3558 393 385 0 2.26 0.0119 1.2816 0.0238 1.6449
z = 2.26, p-value = .0119. There is enough evidence to infer that Americans have become more distrustful of television and newspaper reporting this year than they were three years ago.
13.205 H0: (µ 1 - µ 2) = 25 H1: (µ 1 - µ 2) > 25 Two-tail F test: F = 1.35, p-value = .4809; use equal-variances t-test
346
t = 1.75, p-value = .0433. There is enough evidence to conclude that machine A should be purchased. 13.206 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0 The totals in columns A through D are 5788, 265, 5154, and 332, respectively.
1 2 3 4 5 6 7
A B C D z-Test of the Difference Between Two Proportions (Case 1)
Sample proportion Sample size Alpha
Sample 1 Sample 2 z Stat 0.045800 0.064400 P(Z<=z) one-tail 5788 5154 z Critical one-tail 0.05 P(Z<=z) two-tail z Critical two-tail
E
-4.28 0.0000 1.6449 0.0000 1.9600
z = –4.28, p-value = 0. There is enough evidence to infer that the defective rate differs between the two machines.
13.207
H0 : D = 0 H1 : D 0
Dry Cleaner
347
A B C 1 t-Test: Paired Two Sample for Means 2 3 Dry C Before Dry C After 4 Mean 168.00 165.50 5 Variance 351.38 321.96 6 Observations 14 14 7 Pearson Correlation 0.86 8 Hypothesized Mean Difference 0 9 df 13 10 t Stat 0.96 11 P(T<=t) one-tail 0.1780 12 t Critical one-tail 1.7709 13 P(T<=t) two-tail 0.3559 14 t Critical two-tail 2.1604 t = .96, p-value = .1780. There is not enough evidence to conclude that the dry cleaner sales have decreased. Doughnut shop
A B 1 t-Test: Paired Two Sample for Means 2 3 Donut Before 4 Mean 308.14 5 Variance 809.67 6 Observations 14 7 Pearson Correlation 0.86 8 Hypothesized Mean Difference 0 9 df 13 10 t Stat 3.24 11 P(T<=t) one-tail 0.0032 12 t Critical one-tail 1.7709 13 P(T<=t) two-tail 0.0065 14 t Critical two-tail 2.1604
C
Donut After 295.29 812.07 14
t = 3.24, p-value = .0032. There is enough evidence to conclude that the doughnut shop sales have decreased. Convenience store
348
A B C 1 t-Test: Paired Two Sample for Means 2 3 Convenience Before Convenience After 4 Mean 374.64 348.14 5 Variance 2270.40 2941.82 6 Observations 14 14 7 Pearson Correlation 0.97 8 Hypothesized Mean Difference 0 9 df 13 10 t Stat 7.34 11 P(T<=t) one-tail 0.0000 12 t Critical one-tail 1.7709 13 P(T<=t) two-tail 0.0000 14 t Critical two-tail 2.1604 t = 7.34, p-value = 0. There is enough evidence to conclude that the convenience store sales have decreased.
13.208 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = 1.77, p-value = .0084; use unequal-variances test statistic
z = -4.53, p-value = 0. There is enough evidence to conclude that Facebook users maintain lower GPA scores. 13.209a H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
349
A 1 z-Test: Two Proportions 2 3 4 Sample Proportions 5 Observations 6 Hypothesized Difference 7 z Stat 8 P(Z<=z) one tail 9 z Critical one-tail 10 P(Z<=z) two-tail 11 z Critical two-tail
B
C
D
Depressed Not Depressed 0.2879 0.2004 132 1058 0 2.33 0.0100 2.3263 0.0200 2.5758
z = 2.33, p-value = .0100. There is enough evidence to infer that men who are clinically depressed are more likely to die from heart diseases. b No, we cannot establish a causal relationship.
13.210 a H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = .438, p-value = .0482; use unequal-variances t-test
A B C 1 t-Test: Two-Sample Assuming Unequal Variances 2 Exercise Drug 3 13.52 9.92 4 Mean 5.76 13.16 5 Variance 25 25 6 Observations 0 7 Hypothesized Mean Difference 42 8 df 4.14 9 t Stat 0.0001 10 P(T<=t) one-tail 2.4185 11 t Critical one-tail 12 P(T<=t) two-tail 0.0002 13 t Critical two-tail 2.6981 t = 4.14, p-value = .0001. There is enough evidence that exercise is more effective than medication in reducing hypertension. b 1 2 3 4 5 6 7 8
A B C D t-Estimate of the Difference Between Two Means (Unequal-Variances)
Mean Variance Sample size Degrees of freedom Confidence level
E
Sample 1 Sample 2 Confidence Interval Estimate 13.52 9.92 3.60 5.76 13.16 Lower confidence limit 25 25 Upper confidence limit 41.63 0.95
LCL = 1.84, UCL = 5.36 350
F
1.76 1.84 5.36
c The histograms are bell shaped.
13.211 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0 Two-tail F test: F = 1.93, p-value = .0232; use unequal-variances t-test
A B C 1 t-Test: Two-Sample Assuming Unequal Variances 2 3 Group 1 Group 2 4 Mean 7.46 8.46 5 Variance 25.06 12.98 6 Observations 50 50 7 Hypothesized Mean Difference 0 8 df 89 9 t Stat -1.14 10 P(T<=t) one-tail 0.1288 11 t Critical one-tail 1.6622 12 P(T<=t) two-tail 0.2575 13 t Critical two-tail 1.9870 t = –1.14, p-value = .1288. There is not enough evidence to conclude that people who exercise moderately more frequently lose weight faster
13.212
H0 : D = 0 H1 : D 0
A B C 1 t-Test: Paired Two Sample for Means 2 3 Group 1 Group 2 4 Mean 7.53 8.57 5 Variance 29.77 43.37 6 Observations 50 50 7 Pearson Correlation 0.89 8 Hypothesized Mean Difference 0 9 df 49 10 t Stat -2.40 11 P(T<=t) one-tail 0.0100 12 t Critical one-tail 1.6766 13 P(T<=t) two-tail 0.0201 14 t Critical two-tail 2.0096 t = –2.40, p-value = .0100. There is enough evidence to conclude that people who exercise moderately more frequently lose weight faster
13.213 Two tail F test: F = 1.20, p-value = .0375. Use unequal variances t test
351
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = 7.89, p-value = 0. There is enough evidence to infer consumers are optimistic about the economy.
13.214 Two tail F test: F = 1.69, p-value = .0156. Use unequal variances t test H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = 6.08, p-value = 0. There is enough evidence to conclude that kidneys from living donors last longer than kidneys from deceased donors.
13.215 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0 352
Two-tail F test: F = .363, p-value = .0161; use unequal-variances t-test
A B C 1 t-Test: Two-Sample Assuming Unequal Variances 2 3 Small space Large space 4 Mean 1245.7 1915.8 5 Variance 23812 65566 6 Observations 25 25 7 Hypothesized Mean Difference 0 8 df 39 9 t Stat -11.21 10 P(T<=t) one-tail 0.0000 11 t Critical one-tail 1.6849 12 P(T<=t) two-tail 0.0000 13 t Critical two-tail 2.0227 t = –11.21, p-value = 0. There is enough evidence to infer that students write in such a way as to fill the allotted space.
13.216 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = .624, p-value = .0431; use unequal-variances t-test
A B C 1 t-Test: Two-Sample Assuming Unequal Variances 2 3 Computer No Computer 4 Mean 69,933 48,246 5 Variance 63,359,040 101,588,525 6 Observations 89 61 7 Hypothesized Mean Difference 0 8 df 109 9 t Stat 14.07 10 P(T<=t) one-tail 0.0000 11 t Critical one-tail 1.2894 12 P(T<=t) two-tail 0.0000 13 t Critical two-tail 1.6590 t = 14.07, p-value = 0. There is enough evidence to conclude that single-person businesses that use a PC earn more. 13.217 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
353
A 1 z-Test: Two Proportions 2 3 4 Sample Proportions 5 Observations 6 Hypothesized Difference 7 z Stat 8 P(Z<=z) one tail 9 z Critical one-tail 10 P(Z<=z) two-tail 11 z Critical two-tail
B
C
D
New 0.948 250 0 1.26 0.1037 1.2816 0.2074 1.6449
Older 0.920 250
z = 1.26, p-value = .1037. There is not enough evidence to conclude that the new company is better.
13.218 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0 Two-tail F test: F = 1.62, p-value = .1008; use equal-variances t-test
A B C 1 t-Test: Two-Sample Assuming Equal Variances 2 3 Supplement Placebo 4 Mean 19.02 21.85 5 Variance 41.34 25.49 6 Observations 48 48 7 Pooled Variance 33.41 8 Hypothesized Mean Difference 0 9 df 94 10 t Stat -2.40 11 P(T<=t) one-tail 0.0092 12 t Critical one-tail 1.6612 13 P(T<=t) two-tail 0.0183 14 t Critical two-tail 1.9855 t = –2.40, p-value = .0092. There is enough evidence to infer that taking vitamin and mineral supplements daily increases the body's immune system? 13.219 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
354
A 1 z-Test: Two Proportions 2 3 4 Sample Proportions 5 Observations 6 Hypothesized Difference 7 z Stat 8 P(Z<=z) one tail 9 z Critical one-tail 10 P(Z<=z) two-tail 11 z Critical two-tail
B
C
D
$100-Limit $3000-Limit 0.5234 0.5551 491 490 0 -1.00 0.1598 1.2816 0.3196 1.6449
z = –1.00, p-value = .1598. There is not enough evidence to infer that the dealer at the more expensive table is cheating.
13.220 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0 Two-tail F test: F = 1.43, p-value = 0; use unequal-variances test statistic
t = 0.71, p-value = .4763. There is not enough evidence to conclude that math test scores differ between males and females. 13.221 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = 1.04, p-value = .6531; use equal-variances test statistic
355
t = 2.59, p-value = .0049. There is sufficient evidence to conclude that American cars this year are on average older than cars last year. 13.222 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = 1.57, p-value = .0578. There is not enough evidence to infer that Americans are more generous.
13.223 Time:
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
Two-tail F test: F = .257, p-value = .0009; use unequal-variances test statistic
356
t = 4.96, p-value = 0. There is sufficient evidence to conclude that when the soft relaxing music was played diners spent more time in the restaurant.
Bar bill: H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = 2.05, p-value = .0723; use equal-variances test statistic
t = 2.68, p-value = .0049. There is sufficient evidence to conclude that when the soft relaxing music was played diners spent more money on drinks.
13.224 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = 1.61, p-value = .0429; use unequal-variances test statistic
357
t = 12.67, p-value = 0. There is sufficient evidence to infer that when the restaurant features bright lights and loud music customers spend less time in the restaurant. 13.225 H0: (p1 – p2) = 0 H1: (p1 – p2) > 0
z = 3.32, p-value = .0005. There is enough evidence to conclude that when the restaurant features bright lights and loud music customers are less likely to return.
13.226 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = .869, p-value = .6699; use equal-variances test statistic
358
t = 16.47, p-value = 0. There is enough evidence to conclude that the amounts in the short wide glass are greater than the amounts in the tall skinny glasses.
13.227 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0 Two-tail F test: F = .901, p-value = .6052; use equal-variances t-test
A B C 1 t-Test: Two-Sample Assuming Equal Variances 2 3 Female Male 4 Mean 7.27 7.02 5 Variance 2.57 2.85 6 Observations 103 97 7 Pooled Variance 2.71 8 Hypothesized Mean Difference 0 9 df 198 10 t Stat 1.08 11 P(T<=t) one-tail 0.1410 12 t Critical one-tail 1.6526 13 P(T<=t) two-tail 0.2820 14 t Critical two-tail 1.9720 t = 1.08, p-value = .2820. There is not enough evidence to conclude that female and male high school students differ in the amount of time spent at part-time jobs.
13.228 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = 1.41, p-value = .1433; use equal-variances t-test
359
A B C 1 t-Test: Two-Sample Assuming Equal Variances 2 City Suburb 3 4 Mean 2.42 1.97 5 Variance 1.08 0.77 6 Observations 70 78 7 Pooled Variance 0.92 0 8 Hypothesized Mean Difference 146 9 df 2.85 10 t Stat 11 P(T<=t) one-tail 0.0025 12 t Critical one-tail 1.6554 13 P(T<=t) two-tail 0.0050 14 t Critical two-tail 1.9763
t = 2.85, p-value = .0025. There is enough evidence to infer that city households discard more newspaper than do suburban households.
13.229 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = 2.73, p-value = 0; use unequal-variances t-test
A B C 1 t-Test: Two-Sample Assuming Unequal Variances 2 3 Teenagers 20-to-30 4 Mean 18.18 14.30 5 Variance 357.32 130.79 6 Observations 176 154 7 Hypothesized Mean Difference 0 8 df 293 9 t Stat 2.28 10 P(T<=t) one-tail 0.0115 11 t Critical one-tail 1.6501 12 P(T<=t) two-tail 0.0230 13 t Critical two-tail 1.9681 t = 2.28, p-value = .0115.There is enough evidence to infer that teenagers see more movies than do twenty to thirty year olds. 13.230 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0
360
A 1 z-Test: Two Proportions 2 3 4 Sample Proportions 5 Observations 6 Hypothesized Difference 7 z Stat 8 P(Z<=z) one tail 9 z Critical one-tail 10 P(Z<=z) two-tail 11 z Critical two-tail
B
C
D
No HS HS 0.127 0.358 63 257 0 -3.54 0.0002 1.6449 0.0004 1.96
z = –3.54, p-value = .0002. There is enough evidence to conclude that Californians who did not complete high school are less likely to take a course in the university’s evening program.
13.231 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0 Two-tail F test: F = .433, p-value = 0; use unequal-variances t-test
A B C 1 t-Test: Two-Sample Assuming Unequal Variances 2 3 Group 1 Groups 2-4 4 Mean 11.58 10.60 5 Variance 9.28 21.41 6 Observations 269 981 7 Hypothesized Mean Difference 0 8 df 644 9 t Stat 4.15 10 P(T<=t) one-tail 0.0000 11 t Critical one-tail 1.6472 12 P(T<=t) two-tail 0.0000 13 t Critical two-tail 1.9637 t = 4.15, p-value = 0. There is enough evidence to conclude that on average the market segment concerned about eating healthy food (group 1) outspends the other market segments.
13.232 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0 Two-tail F test: F = .501, p-value = 0; use unequal-variances t-test
361
A B C 1 t-Test: Two-Sample Assuming Unequal Variances 2 3 Sale-CDs Sale-fax 4 Mean 59.04 65.57 5 Variance 425.4 849.7 6 Observations 122 144 7 Hypothesized Mean Difference 0 8 df 256 9 t Stat -2.13 10 P(T<=t) one-tail 0.0171 11 t Critical one-tail 1.6508 12 P(T<=t) two-tail 0.0341 13 t Critical two-tail 1.9693 t = –2.13, p-value = .0171. There is enough evidence to conclude that those who buy the fax/copier outspend those who buy the package of CD-ROMS.
13.233 H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0 Two-tail F test: F = 1.15, p-value = .5373; use equal-variances t-test
A B C 1 t-Test: Two-Sample Assuming Equal Variances 2 3 No Yes 4 Mean 91,467 97,836 5 Variance 461,917,705 401,930,840 6 Observations 466 55 7 Pooled Variance 455,676,297 8 Hypothesized Mean Difference 0 9 df 519 10 t Stat -2.09 11 P(T<=t) one-tail 0.0184 12 t Critical one-tail 1.6478 13 P(T<=t) two-tail 0.0369 14 t Critical two-tail 1.9645 t = –2.09, p-value = .0184. There is enough evidence to infer that professors aged 55 to 64 who plan to retire early have higher salaries than those who don’t plan to retire early. 13.234 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0 The data must first be unstacked. Success = 2
362
A 1 z-Test: Two Proportions 2 3 4 Sample Proportions 5 Observations 6 Hypothesized Difference 7 z Stat 8 P(Z<=z) one tail 9 z Critical one-tail 10 P(Z<=z) two-tail 11 z Critical two-tail
B
C
D
Female Male 0.5945 0.6059 762 746 0 -0.45 0.3256 1.6449 0.6512 1.96
z = −.45, p-value = .6512. There is not enough evidence to conclude that men and women differ in their choices of Christmas trees.
13.235a. Two tail F test: F = .978, p-value = .6929. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
t = .29, p-value = .7735. There is not enough evidence of a difference in years of education between men and women. b. Years of education are required to be normally distributed. c. The histograms are bell shaped. d. The Wilcoxon rank sum test. 13.236 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 363
z = -1.29, p-value = .1000. There is not enough evidence to conclude that the fraction of Americans working for the government increased between 2008 and 2014.
13.237 Two tail F test: F = .974, p-value = .7060. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = .71, p-value = .2378. There is not enough evidence to conclude that men watch more television than women.
13.238 Two tail F test: F = 1.036, p-value = .3994. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
364
t = -1.84, p-value = .0326. There is enough evidence to infer that Americans are more educated in 2014 than in 2012.
13.239 Two tail F test: F = 1.22, p-value = 0. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = 1.06, p-value = .1473. There is not enough evidence to infer that Americans are watching less televison on 2014 than in 2012. 13.240 H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0
365
z = -.74, p-value = .4604. There is not enough evidence to conclude that the percentage of Americans with graduate degrees changed between 2012 and 2014.
13.241 Two tail F test: F = 1.01, p-value = .9109. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -.96, p-value = .1683. There is not enough evidence to conclude that Americans were working longer in 2014 than in 2010.
13.242 Two tail F test: F = 1.02, p-value = .7914. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
366
t = -1.55, p-value = .0612. There is not enough evidence to infer that immigrants have more children than nativeborn Americans.
13.243 Two tail F test: F = .873, p-value = .0147. Use unequal variances estimator.
LCL = -7843, UCL = -1952.
13.244 Two tail F test: F = 1.09, p-value = .4279. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
367
t = .78, p-value = .2170. There is not enough evidence to infer that native-born Americans work longer hours than do immigrants.
13.245 Two tail F test: F = 2.02, p-value = 0. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = 3.67, p-value = .0001. There is enough evidence to infer that Democrats watch more television than do Republicans.
13.246 Two tail F test: F = 1.01, p-value = .8610. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 368
H1: (µ 1 - µ 2) < 0
t = -4.97, p-value = 0. There is enough evidence to infer that conservatives are older than liberals.
13.247
LCL = -.0003, UCL = .0373.
13.248 Two tail F test: F = .621, p-value = 0. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
369
t = .006, p-value = .5023. There is no evidence that government workers work fewer hours than do private sector workers.
13.249 Two tail F test: F = 1.026, p-value = .7124. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = 4.24, p-value = 0. There is enough evidence to conclude that government workers are older than private sector workers. 370
13.250a. Two tail F test: F = 2.12, p-value = 0. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = -1.48, p-value = .9296. The p-value indicates that there is some evidence that self-employed people work less hours than people who work for someone else. b. The required condition of normality is satisfied. c. The Wilcoxon rank sum test.
13.251a. Two tail F test: F = .869, p-value = .1233. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
371
t = 2.67, p-value = .0038. There is enough evidence to infer that self-employed people have more education than other workers. b. The required condition is satisfied.
13.252a. Two tail F test: F = .931, p-value = .0666. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -6.50, p-value = 0. There is enough evidence to infer that the United States has grown older between 2004 and 2014.
372
13.253 Two tail F test: F = .743, p-value = 0. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = 15.64, p-value = 0. There is enough evidence to conclude that the number of children per family decreased between 2004 and 2014.
13.254 Two tail F test: F = .789, p-value = 0. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -1.82, p-value = .0345. There is not enough evidence to conclude that the age at which families have their first child has increased between 2004 and 2014.
373
13.255 Two tail F test: F = .885, p-value = .0016. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = 0, p-value = .4986. There is no evidence to infer that the United States is more educated in 2014 than in 2004.
13.256 Two tail F test: F = 1.096, p-value = .0665. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) > 0
t = .69, p-value = .2452. There is not enough evidence to infer that Americans are working less in 2014 than in 2004.
13.257 Two tail F test: F = 1.02, p-value = .6926. Use equal variances test statistic.
374
H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) ≠ 0
t = -1.09, p-value = .2780. There is not enough evidence to conclude that the amount of television differs between 2004 and 2014.
13.258 Two tail F test: F = .904, p-value = .4779. Use equal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
375
t = -1.61, p-value = .0537. There is not enough evidence to conclude that self-employed heads of households have less net worth than heads of households who work for someone else.
13.259 Two tail F test: F = .652, p-value = .0028. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
t = -2.15, p-value = .0168. There is enough evidence to conclude that heads of households who work for someone else have larger unrealized capital gains than heads of households who are self-employed.
13.260a. Two tail F test: F = .706, p-value = .0251. Use unequal variances test statistic. H0: (µ 1 - µ 2) = 0 H1: (µ 1 - µ 2) < 0
376
t = -2.70, p-value = .0038. There is enough evidence to infer that heads of households with a high school diploma have more household assets than do heads of households who did not finish high school.
Case 13.1 2000 to 2008 Two tail F test of RINCOME: F = .4779, p-value = 0.
377
Chapter 14 14.1a x
5(10 ) 5(15 ) 5(20 ) = 15 555
n (x x) 5(10 – 15) + 5(15 – 15) + 5(20 – 15) = 250 SSE = (n 1)s (5 –1)(50) + (5 – 1)(50) + (5 – 1)(50) = 600 SST =
2
j
j
2
2
2
2 j
j
ANOVA Table Source
Degrees of Freedom Sum of Squares
Treatments
k 1 2
SST = 250
Error
n k 12
SSE = 600
Mean Squares SST 250 = 125 k 1 2 SSE 600 = 50 n k 12
F MST 125 = 2.50 MSE 50
__________________________________________ SS(Total) = 850 Total n k 14 bx
10 (10 ) 10 (15) 10 (20 ) = 15 555
n (x x) 10(10 – 15) + 10(15 – 15) + 10(20 – 15) = 500 SSE = (n 1)s (10 –1)(50) + (10 – 1)(50) + (10 – 1)(50) = 1350 SST =
2
j
j
2
2
2
2 j
j
ANOVA Table Source
Degrees of Freedom Sum of Squares
Treatments
k 1 2
SST = 500
Error
n k 27
SSE = 1350
Mean Squares
F
SST 500 = 250 k 1 2 SSE 1350 50 nk 27
MST 250 = 5.00 MSE 50
__________________________________________ n k 29 Total SS(Total) = 1850 c The F statistic increases.
14.2 a x
4(20 ) 4(22 ) 4(25) = 22.33 444
n (x x) 4(20 – 22.33) + 4(22 – 22.33) + 4(25 – 22.33) = 50.67 SSE = (n 1)s (4 –1)(10) + (4 – 1)(10) + (4 – 1)(10) = 90 SST =
2
j
j
j
2
2
2 j
349
2
.
ANOVA Table Source
Degrees of Freedom
Sum of Squares
Mean Squares
F
Treatments
k 1 2
SST = 50.67
SST 50 .67 25 .33 k 1 2
MST 25 .33 = 2.53 MSE 10
Error
nk 9
SSE = 90
SSE 90 = 10 n k1 9
b SSE =
(n 1)s (4 –1)(25) + (4 – 1)(25) + (4 – 1)(25) = 225 2 j
j
ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 25 .33 = 1.01 MSE 25 .0
Treatments
k 1 2
SST = 50.67
SST 50 .67 25 .33 k 1 2
Error
nk 9
SSE = 225
SSE 225 25 .0 nk 9
c The F statistic decreases.
14.3 a x
10 (30 ) 14 (35) 11(33) 18(40 ) 35 .34 10 14 11 18
n (x x) 10(30– 35.34) +14(35 – 35.34) + 11(33 –35.34) + 18(40 –35.34) = 737.9 SSE = (n 1)s (10 –1)(10) + (14 – 1)(10) + (11 – 1)(10) + (18−1)(10) = 490.0 2
SST =
j
j
2
2
2
2
2 j
j
ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 246 .0 24 .60 MSE 10 .00
Treatments
k 1 3
SST = 737.9
SST 737 .9 246 .0 k 1 3
Error
n k 49
SSE = 490.0
SSE 490 .0 10 .00 nk 49
b x
10 (130 ) 14 (135 ) 11(133 ) 18(140 ) 135 .34 10 14 11 18
SST =
n (x x) j
2
j
= 10(130– 135.34) 2 +14(135 – 135.34) 2 + 11(133 –135.34) 2 + 18(140 –135.34) 2 = 737.9 SSE =
(n 1)s (10 –1)(10) + (14 – 1)(10) + (11 – 1)(10) + (18−1)(10) = 610.0 j
2 j
350
ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 246 .0 24 .60 MSE 10 .00
Treatments
k 1 3
SST = 737.9
SST 737 .9 246 .0 k 1 3
Error
n k 49
SSE = 490.0
SSE 490 .0 10 .00 nk 49
c No change 14.4
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
Rejection region: F F,k 1,n k F.05, 2,9 4.26 Finance
Marketing
Management
Mean
2.25
3.25
5.75
Variance
2.25
2.92
2.92
Grand mean = 3.75
n (x x) = 4(2.25–3.75) + 4(3.25 – 3.75) + 4(5.75 –3.75) = 26.00 SSE = (n 1)s (4 –1)(2.25) + (4 – 1)(2.92) + (4 – 1)(2.92) = 24.25 2
SST =
j
2
j
2
2
2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 13 .00 4.82 MSE 2.69
Treatments
k 1 2
SST = 26.00
SST 26 .00 13 .00 k 1 2
Error
nk 9
SSE = 24.25
SSE 24 .25 2.69 nk 9
.
F = 4.82, p-value = .0377. There is enough evidence to conclude that there are differences in the number of job offers between the three MBA majors. 14.5
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
Rejection region: F F, k 1,n k F.01, 2,15 6.36 Brand 1
Brand 2
Brand 3
Mean
1.33
2.50
2.67
Variance
1.87
2.30
1.47
Grand mean = 2.17
351
n (x x) = 6(1.33 – 2.17) + 6(2.50 – 2.17) + 6(2.67 – 2.17) = 6.33 SSE = (n 1)s (6 –1)(1.87) + (6 – 1)(2.30) + (6 – 1)(1.47) = 28.17 2
SST =
j
2
j
2
2
2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 3.17 1.69 MSE 1.88
Treatments
k 1 2
SST = 6.33
SST 6.33 3.17 k 1 2
Error
n k 15
SSE = 28.17
SSE 28 .17 1.88 nk 15
.
F = 1.69, p-value = .2185. There is not enough evidence to conclude that differences exist between the three brands. 14.6
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
Rejection region: F F, k 1,n k F.05, 2,12 3.89 BA
BSc
BBA
Mean
3.94
4.78
5.76
Variance
1.26
.92
1.00
Grand mean = 4.83
n (x x) = 5(3.94 – 4.83) + 5(4.78 – 4.83) + 5(5.76 – 4.83) = 8.30 SSE = (n 1)s (5 –1)(1.26) + (5 – 1)(.92) + (5 – 1)(1.00) = 12.73 2
SST =
j
2
j
2
2
2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 4.15 3.91 MSE 1.06
Treatments
k 1 2
SST = 8.30
SST 8.30 4.15 k 1 2
Error
n k 12
SSE = 12.73
SSE 12 .73 1.06 nk 12
.
F = 3.91, p-value = .0493. There is enough evidence to conclude that students in different degree program differ in their summer earnings. 14.7
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
. Rejection region: F F,k 1,n k F.025, 2,12 5.10
Mean
Professors
Administrators
Students
6.4
10.4
10.8 352
Variance
48.3
16.3
37.7
Grand mean = 9.2
n (x x) = 5(6.4 – 9.2) + 5(10.4 – 9.2) + 5(10.8 – 9.2) = 59.2 SSE = (n 1)s (5 –1)(48.3) + (5 – 1)(16.3) + (5 – 1)(37.7) = 409.2 2
SST =
j
2
j
2
2
2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 29 .6 .87 MSE 34 .1
Treatments
k 1 2
SST = 59.2
SST 59 .2 29 .6 k 1 2
Error
n k 12
SSE = 409.2
SSE 409 .2 34 .1 nk 12
.
F = .87, p-value = .4445. There is not enough evidence to conclude that the differing university communities differ in the amount of spam they receive in their emails. 14.8
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
Rejection region: F F, k 1,n k F.05,3,8 4.07 IBM
Dell
HP
Other
Mean
13.33
11.00
9.67
17.00
Variance
12.33
79.00
22.33
39.00
Grand mean = 12.75
n (x x) = 3(13.33 – 12.75) + 3(11.00 – 12.75) + 3(9.67 – 12.75) + 3(17.00 – 12.75) = 92.92 SSE = (n 1)s (3 –1)(12.33) + (3 – 1)(79.00) + (3 – 1)(22.33) + (3 – 1)(39.00) = 305.33 2
SST =
j
2
j
2
2
2
2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 30 .97 .81 MSE 38 .17
Treatments
k 1 3
SST = 92.92
SST 92 .92 30 .97 k 1 3
Error
nk 8
SSE = 305.33
SSE 305 .33 38 .17 nk 8
.
F = .81, p-value = .5224. There is not enough evidence to conclude that there are differences in age between the computer brands. 14.9a
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
Rejection region: F F,k 1,n k F.05,3,77 2.76 Grand mean = 65.30 353
SST =
n (x x) j
2
j
= 20(68.83 – 65.30) 2 + 26(65.08 – 65.30) 2 + 16(62.01 – 65.30) 2 + 19(64.64 – 65.30) 2 = 430.95 SSE =
(n 1)s (20 –1)(52.28) + (26 – 1)(37.38) + (16 – 1)(63.46) + (19 – 1)(56.88) = 3903.57 2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
Treatments
k 1 3
SST = 430.95
SST 430 .95 143 .65 k 1 3
Error
n k 77
SSE = 3903.57
SSE 3903 .57 50 .70 nk 77
F
.
MST 143 .65 2.83 MSE 50 .70
F = 2.83, p-value = .0437. There is enough evidence to infer that there are differences in grading standards between the four high schools. b The grades of the students at each high school are required to be normally distributed with the same variance. c The histograms are approximately bell-shaped and the sample variances are similar. 14.10a H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ Rejection region: F F,k 1, n k F.05,3,116 2.68 Grand mean = 101.0 SST =
n (x x) j
2
j
= 30(90.17 – 101.0) 2 + 30(95.77 – 101.0) 2 + 30(106.8 – 101.0) 2 + 30(111.17 – 101.0) 2 = 8,464 SSE =
(n 1)s j
2 j
= (30 –1)(991.52) + (30 – 1)(900.87) + (30 – 1)(928.70) + (30 – 1)(1,023.04) = 111,480 ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 2,821 2.94 MSE 961 .0
Treatments
k 1 3
SST = 8,464
SST 8,464 2,821 k 1 3
Error
n k 116
SSE = 111,480
SSE 111,480 961 .0 nk 116
.
F = 2.94, p-value = .0363. There is enough evidence to infer that there are differences between the completion times of the four income tax forms. b The times for each form must be normally distributed with the same variance. c The histograms are approximately bell-shaped and the sample variances are similar.
354
14.11
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
Rejection region: F F,k 1,n k F.05,3, 275 2.61 Grand mean = 218.0 SST =
n (x x) = 41(196.83 – 218.0) + 73(207.78 – 218.0) + 86(223.38 – 218.0) 2
j
2
j
2
2
+ 79(232.67 – 218.0) 2 = 45,496 SSE =
(n 1)s = (41 –1)(914.05) + (73 – 1)(861.12) + (86 – 1)(1,195.44) 2 j
j
+ (79 – 1)(1,079.81) = 284,400 ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
Treatments
k 1 3
SST = 45,496
SST 45,496 15,165 k 1 3
Error
n k 275
SSE = 284,400
SSE 284 ,400 1,034 nk 275
F
.
MST 15,165 14 .66 MSE 1,034
F = 14.66, p-value = 0. There is enough evidence to infer that there are differences in test scores between children whose parents have different educational levels. 14.12
H0:µ 1= µ 2 = µ 3 = µ 4= µ 5 H1: At least two means differ
Rejection region: F F, k 1, n k F.01, 4,120 3.48 Grand mean = 173.3 SST =
n (x x) = 25(164.6 – 173.3) + 25(185.6 – 173.3) + 25(154.8 – 173.3) 2
j
2
j
2
2
+ 25(182.6 – 173.3) 2 + 25(178.9 – 173.3) 2 = 17,251 SSE =
(n 1)s = (25 –1)(1,164) + (25 – 1)(1,720) + (25 – 1)(1,114) + (25 – 1)(1,658) j
2 j
+ (25 – 1)(841.8) = 155,941 ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
.
Treatments
k 1 4
SST = 17,251
SST 17 ,251 MST 4,312 .6 4,312 .6 3.32 k 1 4 MSE 1,299 .5
Error
n k 120
SSE = 155,941
SSE 155 ,941 1,299 .5 nk 120
F = 3.32, p-value = .0129. There is not enough evidence to allow the manufacturer to conclude that differences exist between the five lacquers. b The times until first sign of corrosion for each lacquer must be normally distributed with a common variance. c The histograms are approximately bell-shaped with similar sample variances. 355
14.13
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
Rejection region: F F,k 1,n k F.05,3, 297 2.61 Grand mean = 16.11 SST =
n (x x) = 39(22.21 – 16.11) + 114(18.46 – 16.11) + 81(15.49 – 16.11) 2
j
2
j
2
2
+ 67(9.31 – 16.11) 2 = 5,202 SSE =
(n 1)s = (39 –1)(121.64) + (114 – 1)(90.39) + (81 – 1)(85.25) + (67 – 1)(65.40) = 25,973 2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
Treatments
k 1 3
SST = 5,202
SST 5,202 1,734 k 1 3
Error
n k 297
SSE = 25,973
SSE 25,973 87 .45 nk 297
F
.
MST 1,734 19 .83 MSE 87 .45
F = 19.83, p-value = 0. There is enough evidence to infer that there are differences exist between the four groups. 14.14
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
Rejection region: F F, k 1, n k F.05, 2,57 3.15 Grand mean = 562.6
n (x x) = 20(551.50 – 562.6) + 20(576.75 – 562.6) + 20(559.45 – 562.6) = 6,667 SSE = (n 1)s = (20 –1)(2,741.95) + (20 – 1)(2,641.14) + (20 – 1)(3,129.31) = 161,736 2
SST =
j
2
j
2
2
2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
Treatments
k 1 2
SST = 6,667
SST 6,667 3,334 k 1 2
Error
n k 57
SSE = 161,736
SSE 161,736 2,837 nk 57
F
.
MST 3,334 1.17 MSE 2,837
F = 1.17, p-value = .3162. There is not enough evidence of a difference between fertilizers in terms of crop yields. 14.15
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
Rejection region: F F, k 1, n k F.05, 2, 297 3.07 Grand mean = 5.48 356
n (x x) = 100(5.81 – 5.48) + 100(5.30 – 5.48) + 100(5.33 – 5.48) = 16.38 SSE = (n 1)s = (100 –1)(6.22) + (100 – 1)(4.05) + (100 – 1)(3.90) = 1,402.5 2
SST =
j
2
j
2
2
2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 8.19 1.73 MSE 4.72
Treatments
k 1 2
SST = 16.38
SST 16 .38 8.19 k 1 2
Error
n k 297
SSE = 1,402.5
SSE 1,402 .5 4.72 nk 297
.
F = 1.73, p-value = .1783. There is not enough evidence of a difference between the three departments. 14.16
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
Rejection region: F F, k 1, n k F.05,3,116 2.68 Grand mean = 77.39 SST =
n (x x) = 30(74.10 – 77.39) + 30(75.67 – 77.39) + 30(78.50 – 77.39) 2
j
2
j
2
2
+ 30(81.30 – 77.39) 2 = 909.42 SSE =
(n 1)s = (30 –1)(249.96) + (30 – 1)(184.23) + (30 – 1)(233.36) + (30 – 1)(242.91) = 26,403 2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 303 .1 1.33 MSE 227 .6
Treatments
k 1 3
SST = 909.4
SST 909 .4 303 .1 k 1 3
Error
n k 116
SSE = 26,403
SSE 26 ,403 227 .6 nk 116
.
F = 1.33, p-value = .2675. There is not enough evidence of a difference between the four groups of companies. 14.17
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
a Leaf size: Rejection region: F F,k 1,n k F.05, 2,147 3.06 Grand mean = 21.49
n (x x) = 50(24.97 – 21.49) + 50(21.65 – 21.49) + 50(17.84 – 21.49) = 1,270 SSE = (n 1)s = (50 –1)(48.23) + (50 – 1)(54.54) + (50 – 1)(33.85) = 6,695
SST =
2
j
j
j
2
2
2
2 j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares 357
F
.
Treatments
k 1 2
SST = 1,270
SST 1,270 635 .0 k 1 2
Error
n k 147
SSE = 6,695
SSE 6,695 45 .54 nk 147
MST 635 .0 13 .95 MSE 45 .54
F = 13.95, p-value = 0. There is sufficient evidence to conclude that the leaf sizes differ between the 3 groups. b Nicotine: Rejection region: F F,k 1,n k F.05, 2,147 3.06 Grand mean = 13.00
n (x x) = 50(15.52 – 13.00) + 50(13.39 – 13.00) + 50(10.08 – 13.00) = 753.2 SSE = (n 1)s = (50 –1)(3.72) + (50 – 1)(3.59) + (50 – 1)(3.83) = 545.54 2
SST =
j
2
j
2
2
2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
k 1 2
SST = 753.2
MST 376 .6 SST 753 .2 101 .47 376 .6 MSE 3.71 k 1 2
Error
n k 147
SSE = 545.54
SSE 545 .54 3.71 nk 147
.
F = 101.47, p-value = 0. There is sufficient evidence to infer that the amounts of nicotine differ between the 3 groups. 14.18
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
a Ages: Rejection region: F F,k 1, n k F.05,3, 291 2.61 Grand mean = 36.23 SST =
n (x x) = 63(31.30 – 36.23) + 81(34.42 – 36.23) + 40(37.38 – 36.23) 2
j
2
j
2
2
+ 111(39.93 – 36.23) 2 = 3,366 SSE =
(n 1)s = (63 –1)(28.34) + (81 – 1)(23.20) + (40 – 1)(31.16) + (111 – 1)(72.03) = 12,752 j
2 j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 1,122 25 .60 MSE 43 .82
Treatments
k 1 3
SST =3,366
SST 3,366 1,122 k 1 3
Error
n k 291
SSE = 12,752
SSE 12 ,752 43 .82 nk 291
.
F = 25.60, p-value = 0. There is sufficient evidence to infer that the ages of the four groups of cereal buyers differ. b Incomes: Rejection region: F F,k 1, n k F.05,3, 291 2.61 Grand mean = 39.97
358
SST =
n (x x) = 63(37.22 – 39.97) + 81(38.91 – 39.97) + 40(41.48 – 39.97) 2
j
2
j
2
2
+ 111(41.75 – 39.97 ) 2 = 1,008 SSE =
(n 1)s = (63 –1)(39.82) + (81 – 1)(40.85) + (40 – 1)(61.38) + (111 – 1)(46.59) = 13,256 2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 336 .0 7.37 MSE 45 .55
Treatments
k 1 3
SST =1,008
SST 1,008 336 .0 k 1 3
Error
n k 291
SSE = 13,256
SSE 13,256 45 .55 nk 291
.
F = 7.37, p-value = .0001. There is sufficient evidence to conclude that incomes differ between the four groups of cereal buyers. c Education: Rejection region: F F,k 1, n k F.05,3, 291 2.61 Grand mean = 11.98 SST =
n (x x) = 63(11.75 – 11.98) + 81(12.41 – 11.98) + 40(11.73 – 11.98) 2
j
2
j
2
2
+ 111(11.89 – 11.98) 2 = 21.71 SSE =
(n 1)s = (63 –1)(3.93) + (81 – 1)(3.39) + (40 – 1)(4.26) + (111 – 1)(4.30) = 1,154 2 j
j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 7.24 1.82 MSE 3.97
Treatments
k 1 3
SST =21.71
SST 21 .71 7.24 k 1 3
Error
n k 291
SSE = 1,154
SSE 1,154 3.97 nk 291
.
F = 1.82, p-value = .1428. There is not enough evidence to infer that education differs between the four groups of cereal buyers. d Using the F-tests and the descriptive statistics we see that the mean ages and mean household incomes are in ascending order. For example, Sugar Smacks buyers are younger and earn less than the buyers of the other three cereals. Cheerio purchasers are older and earn the most. 14.19
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
Rejection region: F F, k 1, n k F.05, 2,57 3.15 Grand mean = 146.1
n (x x) = 20(153.60 – 146.1) + 20(151.50 – 146.1) + 20(133.25 – 146.1) = 5,011 SSE = (n 1)s = (20 –1)(654.25) + (20 – 1)(924.05) + (20 – 1)(626.83) = 41,898 SST =
2
j
j
j
2
2
2 j
359
2
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 2,506 3.41 MSE 735 .0
Treatments
k 1 2
SST = 5,011
SST 5,011 2,506 k 1 2
Error
n k 57
SSE = 41,898
SSE 41,898 735 .0 nk 57
.
F = 3.41, p-value = .0400. There is enough evidence to infer that sales will vary according to price. 14.20
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
Rejection region: F F, k 1, n k F.05, 2, 232 3.07 Grand mean = 19.50
n (x x) = 61(18.54 – 19.50) + 83(19.34 – 19.50) + 91(20.29 – 19.50) = 114.5 SSE = (n 1)s = (61 –1)(177.95) + (83 – 1)(171.42) + (91 – 1)(297.50) = 51,508 2
SST =
j
2
j
j
2
2
2 j
ANOVA table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 57 .3 .26 MSE 222 .0
Treatments
k 1 2
SST = 114.5
SST 114 .5 57 .3 k 1 2
Error
n k 232
SSE = 51,508
SSE 51,508 222 .0 nk 232
F = .26, p-value = .7730. There is not enough evidence of a difference between the three segments. 14.21
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
360
.
F = 36.7, p-value = 0. There is enough evidence to conclude that at least two means differ. 14.22 Reading H0:µ 1= µ 2 = µ 3 H1: At least two means differ
F = 182.43, p-value = 0. There is enough evidence to conclude that at least two means differ. Mathematics H0:µ 1= µ 2 = µ 3 H1: At least two means differ
361
F = 353.45, p-value = 0. There is enough evidence to conclude that at least two means differ. Science H0:µ 1= µ 2 = µ 3 H1: At least two means differ
F = 121.0, p-value = 0. There is enough evidence to conclude that at least two means differ. 14.23
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 = µ 8 H1: At least two means differ
362
F = 9.19, p-value = 0. 14.24
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 = µ 8 H1: At least two means differ
F = 6.09, p-value = 0. 14.25
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 = µ 8 H1: At least two means differ
F = 2.38, p-value = .0201. 14.26
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 = µ 8 H1: At least two means differ
F = 56.97, p-value = 0. 14.27
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 = µ 8 H1: At least two means differ
363
F = 44.02, p-value = 0. 14.28
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 = µ 8 H1: At least two means differ
F = 52.41, p-value = 0. 14.29
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 = µ 8 H1: At least two means differ
F = 62.35, p-value = 0. 14.30
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 H1: At least two means differ
F = 12.84, p-value = 0. 14.31
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 H1: At least two means differ
364
F = 4.68, p-value = 0. 14.32
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 H1: At least two means differ
F = .87, p-value = .5189. 14.33
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 H1: At least two means differ
F = 7.22, p-value = 0. 14.34
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 H1: At least two means differ
F = 41.85, p-value = 0. 14.35
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 H1: At least two means differ
365
F = 27.61, p-value = 0. 14.36
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 H1: At least two means differ
F = 41.91, p-value = 0. 14.37
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 H1: At least two means differ
F = 40.96, p-value = 0. 14.38
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 H1: At least two means differ
F = 83.86, p-value = 0. 14.39
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
366
F = 23.72, p-value = 0. 14.40
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 H1: At least two means differ
F = 2.50, p-value = .0406. 14.41
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
F = 8.57, p-value = 0. 14.42
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
F = 18.23, p-value = 0. 14.43
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
367
F = 8.67, p-value = 0. 14.44
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
F = 8.29, p-value = 0. 14.45
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 H1: At least two means differ
F = 7.07, p-value = 0. 14.46
H0:µ 1= µ 2 = µ 3 = µ 4 = µ 5 H1: At least two means differ
F = .745, p-value = .5610. 14.47
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
368
F = 32.45, p-value = 0. 14.48
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
F = 39.25, p-value = 0. 14.49
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
F = 20.24, p-value = 0. 14.50
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
F = 2.52, p-value = .0565. 14.51
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
F = 110.0, p-value = 0.
369
14.52
H0:µ 1= µ 2 = µ 3 = µ 4 H1: At least two means differ
F = 152.9, p-value = 0. 14.53
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
F = 12.25, p-value = 0. 14.54
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
F = 14.27, p-value = 0. 14.55
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
F = 3.07, p-value = .0466. 14.56
H0:µ 1= µ 2 = µ 3 H1: At least two means differ
370
F = 9.73, p-value = 0. 14.57 a = .05: t / 2, n k t .025, 27 2.052 1 1 1 1 = 2.052 700 = 24.28 LSD = t / 2,n k MSE ni n j 10 10
Treatment
Means
Difference
i = 1, j = 2
128.7
101.4
27.3
i = 1, j = 3
128.7
133.7
−5.0
i = 2, j = 3
101.4
133.7
−32.3
Conclusion: 2 differs from 1 and 3 . b C = 3(2)/2 = 3, E = .05, E / C .0167: t / 2,n k t .0083, 27 2.552 (from Excel) 1 1 1 1 LSD = t / 2,n k MSE = 2.552 700 = 30.20 ni n j 10 10
Treatment
Means
Difference
i = 1, j = 2
128.7
101.4
27.3
i = 1, j = 3
128.7
133.7
−5.0
i = 2, j = 3
101.4
133.7
−32.3
Conclusion: 2 and 3 differ. c q (k, ) q .05 (3,27 ) 3.53 q (k , ) Treatment
Means
MSE 700 = 3.53 = 29.53 ng 10
Difference
i = 1, j = 2
128.7
101.4
27.3
i = 1, j = 3
128.7
133.7
−5.0
i = 2, j = 3
101.4
133.7
−32.3
Conclusion: 2 and 3 differ. 14.58 a = .05: t / 2, n k t.025, 20 2.086 1 1 1 1 LSD = t / 2,n k MSE = 2.086 125 = 14.75 ni n j 5 5 Treatment Means Difference
i = 1, j = 2
227
205
22
i = 1, j = 3
227
219
8
i = 1, j = 4
227
248
−21
i = 1, j = 5
227
202
25
371
i = 2, j = 3
205
219
−14
i = 2, j = 4
205
248
−43
i = 2, j = 5
205
202
3
i = 3, j = 4
219
248
−29
i = 3, j = 5
219
202
17
i = 4, j = 5
248
202
46
Conclusion: The following pairs of means differ. 1 and 2 , 1 and 4 , 1 and 5 , 2 and 4 , 3 and 4 , 3 and 5 , and 4 and 5 . b. C = 5(4)/2 = 10, E = .05, E / C .005 t / 2,n k t .0025, 20 3.153 (from Excel) 1 1 1 1 LSD = t / 2,n k MSE = 3.153 125 = 22.30 ni n j 5 5
Treatment
Means
Difference
i = 1, j = 2
227
205
22
i = 1, j = 3
227
219
8
i = 1, j = 4
227
248
−21
i = 1, j = 5
227
202
25
i = 2, j = 3
205
219
−14
i = 2, j = 4
205
248
−43
i = 2, j = 5
205
202
3
i = 3, j = 4
219
248
−29
i = 3, j = 5
219
202
17
i = 4, j = 5
248
202
46
Conclusion: The following pairs of means differ. 1 and 5 , 2 and 4 , 3 and 4 , and 4 and 5 . c q (k, ) q .05 (5, 20) = 4.23 q (k , ) Treatment
Means
MSE 125 = 4.23 = 21.15 ng 5
Difference
i = 1, j = 2
227
205
22
i = 1, j = 3
227
219
8
i = 1, j = 4
227
248
−21
i = 1, j = 5
227
202
25
i = 2, j = 3
205
219
−14
i = 2, j = 4
205
248
−43
i = 2, j = 5
205
202
3
i = 3, j = 4
219
248
−29
i = 3, j = 5
219
202
17
372
i = 4, j = 5
248
202
46
Conclusion: The following pairs of means differ. 1 and 2 , 1 and 5 , 2 and 4 , 3 and 4 , and 4 and 5 .
14.59 q (k, ) q .05 (3, 15) = 3.67 q (k , ) Treatment
Means
MSE 1.88 = 3.67 = 2.05 ng 6
Difference
i = 1, j = 2
1.33
2.50
−1.17
i = 1, j = 3
1.33
2.67
−1.34
i = 2, j = 3
2.50
2.67
−.17
There are no differences.
1 1 1 1 = 1.782 1.06 = 1.16 14.60 a. LSD = t / 2,n k MSE ni n j 5 5
Treatment
Means
Difference
i = 1, j = 2
3.94
4.78
−.84
i = 1, j = 3
3.94
5.76
−1.82
i = 2, j = 3
4.78
5.76
−1.02
Means of BAs and BBAs differ. b. C = 3(2)/2 = 3, E = .10, E / C .0333: t / 2,n k t .0167,12 2.404 (from Excel) 1 1 1 1 LSD = t / 2, n k MSE = 2.404 1.06 = 1.57 ni n j 5 5 Treatment Means Difference
i = 1, j = 2
3.94
4.78
−.84
i = 1, j = 3
3.94
5.76
−1.82
i = 2, j = 3
4.78
5.76
−1.02
Means of BAs and BBAs differ. 14.61 LSD method: C = 4(3)/2 = 6, E = .05, E / C .0083 1 1 LSD = t / 2, n k MSE (LSD must be calculated for each pair of treatments.) ni n j
Tukey’s method: n g
4 k 19 .6 (rounded to 20) = 1 1 1 1 1 1 1 ... 20 26 16 19 n1 n 2 nk
q (k, ) q .05 (4, 77) 3.74 = 3.74
50 .70 = 5.95 20
373
Treatment
Means
Difference
LSD
i = 1, j = 2
68.83
65.08
3.75
5.74
i = 1, j = 3
68.83
62.01
6.82
6.47
i = 1, j = 4
68.83
64.64
4.19
6.18
i = 2, j = 3
65.08
62.01
3.07
6.13
i = 2, j = 4
65.08
64.64
.44
5.82
i = 3, j = 4
62.01
64.64
−2.63
6.55
a The mean grades from high schools A and C differ. b The mean grades from high schools A and C differ.
14.62 Tukey’s method: q (k, ) q .05 (4, 116) 3.68 = 3.68
961 .0 = 20.83 30
LSD method with the Bonferroni adjustment: C = 4(3)/2 = 6, E = .05, E / C .0083 1 1 1 1 LSD = t / 2,n k MSE = 2.64 961 .0 = 21.13 ni n j 30 30
Treatment
Means
Difference
Tukey
LSD
i = 1, j = 2
90.17
95.77
−5.60
20.83
21.13
i = 1, j = 3
90.17
106.8
−16.67
20.83
21.13
i = 1, j = 4
90.17
111.17
−21.00
20.83
21.13
i = 2, j = 3
95.77
106.8
−11.07
20.83
21.13
i = 2, j = 4
95.77
111.17
−15.40
20.83
21.13
i = 3, j = 4
106.8
111.17
−4.33
20.83
21.13
a The means for Forms 1 and 4 differ. b No means differ.
14.63a
H 0 : 1 2 3 = 4 H 1 : At least two means differ.
x
10 (61 .60 ) 10 (57 .30 ) 10 (61 .80 ) 10 (51 .80 ) 58 .13 10 10 10 10
n (x x) = 10(61.6–58.13) +10(57.3–58.13) + 10(61.8–58.13) + 10(51.8–58.13) = 662.7 SSE = (n 1)s (10 –1)(80.49) + (10 – 1)(70.46) + (10 – 1)(22.18) + (10−1)(75.29) = 2,236
SST =
2
j
j
j
2
2
2 j
374
2
2
ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 220 .9 3.56 MSE 62 .11
Treatments
k 1 3
SST = 662.7
SST 662 .7 220 .9 k 1 3
Error
n k 36
SSE = 2,236
SSE 2,236 62 .11 nk 36
F = 3.56, p-value = .0236. There is enough evidence to infer that differences exist between the flares with respect to burning times. LSD method: C = 4(3)/2 = 6, E = .05, E / C .0083 t / 2, n k t .0042,36 2.794 (from Excel) 1 1 1 1 LSD = t / 2, n k MSE = 2.794 62 .11 9.85 ni n j 10 10
Tukey’s method: q (k, ) q .05 (4, 136) 3.79 = 3.79 Treatment
Means
62 .11 = 9.45 10
Difference
i = 1, j = 2
61.6
57.3
4.3
i = 1, j = 3
61.6
61.8
−.2
i = 1, j = 4
61.6
51.8
9.8
i = 2, j = 3
57.3
61.8
−4.5
i = 2, j = 4
57.3
51.8
5.5
i = 3, j = 4
61.8
51.8
10.0
b. The means of flares C and D differ. c The means of flares A and D, and C and D differ. 14.64 a LSD method: C = 5(4)/2 = 10, E = .05, E / C .005 t / 2,n k t .0025,120 2.860 (from Excel) 1 1 1 1 LSD = t / 2, n k MSE = 2.860 1300 29 .17 ni n j 25 25
b Tukey’s method: q (k, ) q .05 (5,120) = 3.92 = 3.92 Treatment
Means
Difference
i = 1, j = 2
164.6
185.6
−21.0
i = 1, j = 3
164.6
154.8
9.8
i = 1, j = 4
164.6
182.6
−18.0
i = 1, j = 5
164.6
178.9
−14.3
i = 2, j = 3
185.6
154.8
30.8
i = 2, j = 4
185.6
182.6
3.0
i = 2, j = 5
185.6
178.9
6.7
375
1300 = 28.27 25
i = 3, j = 4
154.8
182.6
−27.8
i = 3, j = 5
154.8
178.9
−24.1
i = 4, j = 5
182.6
178.9
3.7
a The means of lacquers 2 and 3 differ b The means of lacquers 2 and 3 differ.
14.65a
H 0 : 1 2 3 H1 : At least two means differ.
x
30 (53 .17 ) 30 (49 .37 ) 30 (44 .33) 48 .96 30 30 30
n (x x) = 30(53.17–48.96) +30(49.37–48.96) + 30(44.33–48.96) = 1,178 SSE = (n 1)s (30 –1)(194.6) + (30 – 1)(152.6) + (30 – 1)(129.9) = 13,836 SST =
2
j
2
j
2
2
2 j
j
ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
MST 589 .0 3.70 MSE 159 .0
Treatments
k 1 2
SST = 1,178
SST 1,178 589 .0 k 1 2
Error
n k 87
SSE = 13,836
SSE 13,836 159 .0 nk 87
F = 3.70, p-value = .0286. There is enough evidence to infer that speed of promotion varies between the three sizes of engineering firms. b q (k, ) q .05 (3,87) 3.40 = 3.40
159 .0 7.83 30
Treatment
Difference
Means
i = 1, j = 2
53.17
49.37
3.80
i = 1, j = 3
53.17
44.44
8.83
i = 2, j = 3
49.37
44.33
5.04
The means of small and large firms differ. Answer (v) is correct. 14.66 Tukey’s method: q (k, ) q .05 (3,57) 3.40 = 3.40
2,838 40 .50 20
LSD method: C = 3(2)/2 = 3, E = .05, E / C .0167 t / 2, n k t .0083,57 2.466 (from Excel) 1 1 1 1 LSD = t / 2, n k MSE = 2.466 2,838 41 .54 ni n j 20 20
376
Treatment
Means
Difference
i = 1, j = 2
551.5
576.8
−25.3
i = 1, j = 3
551.5
559.5
−8.0
i = 2, j = 3
576.8
559.5
17.3
a There are no differences. b There are no differences. 14.67a XLSTAT
PARTYID 3 differs from all the others. b. Excel Workbook PARTYID
Means
| Difference |
LSD
i = 0, j = 1
13.986 13.670
.316
.507
i = 0, j = 2
13.986 13.846
.140
.532
i = 1, j = 2
13.670 13.846
.176
.536
None of the pairs differ. c. Excel Workbook
377
PARTYID
Means
| Difference |
LSD
i = 4, j = 5
13.928 14.055
.127
.652
i = 4, j = 6
13.928 14.037
.109
.655
i = 5, j = 6
14.055 14.037
.018
.630
None of the three pairs differ. 14.68
PARTYID
Means
| Difference |
LSD
i = 2, j = 3
45,501 34,423
11,078
8599
i = 2, j = 4
45,501 49,374
3873
9912
i= 3, j = 4
34,423 49,374
14,951
8973
PARTYID 3 differs from 2 and 4. 14.69a. XLSTAT
PARTYID 4 differs from 0 and 3 only. At least two means differ. b. The analysis of variance F test produced the same result; at least two means differ. 14.70a. XLSTAT
378
b. Excel Workbook PARTYID
Means
| Difference |
LSD
i = 2, j = 3
3.196
3.433
.237
.377
i = 2, j = 4
3.196
4.768
1.572
.450
i= 3, j = 4
3.433
4.768
1.355
.417
PARTYID 4 differs from 2 and 3. 4.71a. XLSTAT
379
b. Excel Workbook PARTYID
Means
| Difference |
LSD
i = 2, j = 3
2.793
2.696
.097
.230
i = 2, j = 4
2.793
3.549
.756
.276
i= 3, j = 4
2.696
3.549
.853
.255
PARTYID 4 differs from 2 and 3 4.72a XLSTAT
F = 14.23, p-value = 0. b.
380
Whites differ from Blacks and Others. 14.73 a XLSTAT
F = .772, p-value = .4624. b.
There are no differences. c. Tukey and ANOVA produced the same result. 14.74a XLSTAT
b.
Independents differ from Democrats and Republicans. 14.75 Excel Workbook PARTYID
Means
| Difference |
LSD
i = 1, j = 3
2.265
2.490
.225
.229
i = 1, j = 5
2.265
3.259
.994
.264
i= 3, j = 5
2.490
3.259
.769
.252
PARTYID 5 differs from 1 and 3.
381
14.76 a. Bonferroni with C = 3. b. Excel Workbook POLVIEWS
Means
| Difference |
LSD
i = 1, j = 2
14.681 14.635
.046
.842
i = 1, j = 3
14.681 14.297
.384
.858
i= 2, j = 3
14.635 14.297
.338
.601
There are no differences. 14.77 XLSTAT
The mean of 4 differs from the means of 5, 6, and 7. 14.78 a. XLSTAT
382
F = 33.00, p-value = 0. b.
All three pairs differ. 14.79 XLSTAT
Blacks differ from Whites and others. 14.80 Excel Workbook POLVIEWS
Means
| Difference |
LSD
i = 1, j = 2
2.094
2.786
.692
.662
i = 1, j = 3
2.094
3.288
1.194
.671
i= 2, j = 3
2.786
3.288
.502
.485
All three differ. 14.81 XLSTAT
383
The mean of 4 differs from the means of 1, 2, and 3. 14.82 XLSTAT
All pairs of means differ except the means of 1 and 2. 14.83 ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
2
100
50.00
24.04
Blocks
6
50
8.33
4.00
Error
12
25
2.08
Total
20
175
.
a Rejection region: F F,k 1, n k b 1 F.05, 2,12 = 3.89 Conclusion: F = 24.04, p-value = .0001. There is enough evidence to conclude that the treatment means differ. b Rejection region: F F , b 1, n k b 1 F.05,6,12 = 3.00 F = 4.00, p-value = .0197. There is enough evidence to conclude that the block means differ. 14.84 ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
4
1,500
375.0
16.50
Blocks
11
1,000
90.91
4.00
Error
44
1,000
22.73
Total
59
3,500
a Rejection region: F F,k 1, n k b 1 F.01, 4, 44 3.83 F = 16.50, p-value = 0. There is enough evidence to conclude that the treatment means differ. b Rejection region: F F,b 1,n k b 1 F.01,11, 44 2.80 Conclusion: F = 4.00, p-value = .0005. There is enough evidence to conclude that the block means differ.
384
14.85 ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
3
275
91.67
7.99
Blocks
9
625
69.44
6.05
Error
27
310
11.48
Total
39
1,210
a Rejection region: F F,k 1, n k b 1 F.01,3, 27 = 4.60 Conclusion: F = 7.99, p-value = .0006. There is enough evidence to conclude that the treatment means differ. b Rejection region: F F , b 1, n k b 1 F.01,9, 27 = 3.15 Conclusion: F = 6.05, p-value = .0001. There is enough evidence to conclude that the block means differ. 14.86 Rejection region: F F,k 1, n k b 1 F.05, 2,14 = 3.74 a. ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
2
1,500
750.0
7.00
Blocks
7
500
71.43
.67
Error
14
1,500
107.1
Total
23
3,500
Conclusion: F = 7.00, p-value = .0078. There is enough evidence to conclude that the treatment means differ. b. ANOVA Table Source
Degrees of Freedom
Sum of Squares
Mean Squares
F
Treatments
2
1,500
750.0
10.50
Blocks
7
1,000
142.86
2.00
Error
14
1,000
71.43
Total
23
3,500
Conclusion: F = 10.50, p-value = .0016. There is enough evidence to conclude that the treatment means differ c. ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
2
1,500
750
21.00
Blocks
7
1,500
214.3
6.00
Error
14
500
35.71
Total
23
3,500
Conclusion: F = 21.00, p-value = .0001. There is enough evidence to conclude that the treatment means differ d. The test statistic increases.
385
14.87 a. k = 3, b = 5, Grand mean = 10.4 b
k
(x x) (7 10.4) (10 10.4) (12 10.4) (9 10.4) (12 10.4) 2
2
SS(Total) =
2
2
2
2
ij
j1
i 1
(12 10.4) 2 (8 10.4) 2 (16 10.4) 2 (13 10.4) 2 (10 10.4) 2 (8 10.4) 2 (9 10.4) 2 (13 10.4) 2 (6 10.4) 2 (1110.4) 2 = 99.6 k
SST =
b(x[T] x) 5[(10 10.4) (11.8 10.4) (9.4 10.4) ] 15.6 2
2
2
2
j
j1 b
SSB =
k(x[B] x) 3[(9 10.4) (9 10.4) (13.7 10.4) (9.3 10.4) (11 10.4) ] 48.3 2
2
2
2
2
2
i
i 1
SSE = SS(Total) – SST – SSB = 99.6− 15.6 – 48.3 = 35.7 k
b SS(Total) =
b
(x x) (7 10.4) (10 10.4) (12 10.4) (9 10.4) (12 10.4) 2
2
2
2
2
2
ij
j1
i 1
(12 10.4) 2 (8 10.4) 2 (16 10.4) 2 (13 10.4) 2 (10 10.4) 2
(8 10.4) 2 (9 10.4) 2 (13 10.4) 2 (6 10.4) 2 (1110.4) 2 = 99.6 k
SST =
n (x x) 5(10 10.4) 5(11.8 10.4) 5(9.4 10.4) 15.6 2
2
j
2
2
j
j1
SSE = SS(Total) – SST = 99.6 – 15.6 = 84.0 c The variation between all the data is the same for both designs. d The variation between treatments is the same for both designs. e Because the randomized block design divides the sum of squares for error in the one-way analysis of variance into two parts. 14.88 a k = 4, b = 3, Grand mean = 5.6 k
SS(Total) =
b
(x x) (6 5.6) (8 5.6) (7 5.6) (5 5.6) (5 5.6) (6 5.6) 2
2
2
2
2
2
ij
j1
i 1
(4 5.6) 2 (5 5.6) 2 (5 5.6) 2 (4 5.6) 2 (6 5.6) 2 (6 5.6) 2 = 14.9 k
SST =
b(x[T] x) 3[(7 5.6) (5.3 5.6) (4.7 5.6) (5.3 5.6) ] 8.9 2
2
2
2
j
j1 b
SSB =
k(x[B] x) 4[(4.8 5.6) (6 5.6) (6 5.6) 4.2 2
2
2
2
i
i 1
SSE = SS(Total) – SST – SSB = 14.9− 8.9 – 4.2 = 1.8 386
2
2
b
k
b SS(Total) =
(x x) (6 5.6) (8 5.6) (7 5.6) (5 5.6) (5 5.6) (6 5.6) 2
2
2
2
2
2
2
ij
j1
i 1
(4 5.6) 2 (5 5.6) 2 (5 5.6) 2 (4 5.6) 2 (6 5.6) 2 (6 5.6) 2 = 14.9 k
SST =
b(x[T] x) 3[(7 5.6) (5.3 5.6) (4.7 5.6) (5.3 5.6) ] 8.9 2
2
2
2
2
j
j1
SSE = SS(Total) – SST = 14.9 – 8.9 = 6.0 c The variation between all the data is the same for both designs. d The variation between treatments is the same for both designs. e Because the randomized block design divides the sum of squares for error in the one-way analysis of variance into two parts.
H0 : 1 2 3
14.89
H 1 : At least two means differ. Rejection region: F F,k 1, n k b 1 F.05, 2,6 = 5.14 k = 3, b = 4, Grand mean = 2.38 b
k
SS(Total) =
(x x) (1.4 2.38) (3.1 2.38) (2.8 2.38) (3.4 2.38) 2
2
2
2
2
ij
j1
i 1
(1.5 2.38) 2 (2.6 2.38) 2 (2.1 2.38) 2 (3.6 2.38) 2
(1.3 2.38) 2 (2.4 2.38) 2 (1.5 2.38) 2 (2.9 2.38) 2 7.30 k
SST =
b(x[T] x) 4[(2.68 2.38) (2.45 2.38) (2.03 2.38) ] .87 2
2
2
2
j
j1 b
SSB =
k(x[B] x) 3[(1.4 2.38) (2.7 2.38) (2.1 2.38) (3.3 2.38) ] 5.91 2
2
2
2
2
i
i 1
SSE = SS(Total) – SST – SSB = 7.30 − .87 – 5.91 = .52 ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
2
.87
.44
5.06
Blocks
3
5.91
1.97
22.64
Error
6
.52
.087
Total
11
7.30
F = 5.08, p-value = .0512. There is not enough evidence to conclude that there are differences between the subjects being measured.
387
H 0 : 1 2 3 4
14.90
H1 : At least two means differ. Rejection region: F F,k 1, n k b 1 F.01,3,12 = 5.95 k = 4, b = 5, Grand mean = 8.3 k
SS(Total) =
b
(x x)
2
ij
j1
i 1
(5 8.3)2 (4 8.3)2 (6 8.3)2 (7 8.3)2 (9 8.3)2
(2 8.3) 2 (7 8.3) 2 (12 8.3) 2 (11 8.3) 2 (8 8.3) 2 (6 8.3) 2 (8 8.3) 2 (9 8.3) 2 (16 8.3) 2 (15 8.3) 2 (8 8.3) 2 (10 8.3) 2 (2 8.3) 2 (7 8.3) 2 (14 8.3) 2 = 286.2 k
SST =
b(x[T] x) 5[(6.2 8.3) (8.0 8.3) (10.8 8.3) (8.2 8.3) ] 53.8 2
2
2
2
2
j
j1
SSB
b
k(x[B] x) i
2
4[(5.25 8.3)2 (7.25 8.3)2 (7.25 8.3)2 (10.25 8.3)2 (11.5 8.3)2 ] 102.2
i 1
SSE = SS(Total) – SST – SSB = 286.2 – 53.8 – 102.2 = 130.2 ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
3
53.8
17.93
1.65
Blocks
4
102.2
25.55
2.35
Error
12
130.2
10.85
Total
19
286.2
F = 1.65, p-value = .2296. There is not enough evidence to conclude there are differences between the four diets. 14.91 ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
2
204.2
102.11
4.54
Blocks
11
1150.2
104.57
4.65
Error
22
495.1
22.51
a
H 0 : 1 2 3 H1 : At least two means differ.
F = 4.54, p-value = .0224. There is enough evidence to conclude that there are differences between the three couriers. 388
b
H 0 : 1 2 … 12 H1 : At least two means differ.
F = 4.65, p-value = .0011. The block means differ; the practitioner used the correct design. 14.92 ANOVA Table Source
Degrees of Freedom
Sum of Squares
Mean Squares
F
Treatments
2
7,131
3,566
123.36
Blocks
19
177,465
9,340
323.16
Error
38
1,098
28.90
a
H 0 : 1 2 3 H1 : At least two means differ.
Rejection region: F F,k 1, n k b 1 F.05, 2,38 3.23 F = 123.36, p-value = 0. There is sufficient evidence to conclude that the three fertilizers differ with respect to crop yield. b F = 323.16, p-value = 0. There is sufficient evidence to indicate that there are differences between the plots. 14.93 ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
2
10.26
5.13
.86
Blocks
19
3,020
159.0
26.64
Error
38
226.7
5.97
a
H 0 : 1 2 3 H1 : At least two means differ.
Rejection region: F F,k 1, n k b 1 F.05, 2,38 3.23 F = .86, p-value = .4313. There is not enough evidence to conclude that there are differences in sales ability between the holders of the three degrees. b
H 0 : 1 2 … 20 H1 : At least two means differ.
F = 26.64, p-value = 0. There is sufficient evidence to indicate that there are differences between the blocks of students. The independent samples design would not be recommended. c The commissions for each type of degree are required to be normally distributed with the same variance. d The histograms are bell shaped and the sample variances are similar.
389
14.94 ANOVA Table Source
Degrees of Freedom
Sum of Squares
Mean Squares
F
Treatments
3
4,206
1,402
21.16
Blocks
29
126,843
4,374
66.02
Error
87
5,764
66.25
a
H 0 : 1 2 3 = 4 H1 : At least two means differ.
Rejection region: F F,k 1, n k b 1 F.01,3,87 4.01 F = 21.16, p-value = 0. There is sufficient evidence to conclude that differences in completion times exist between the four forms. b
H 0 : 1 2 … 30 H1 : At least two means differ.
F = 66.02, p-value = 0. There is sufficient evidence to indicate that there are differences between the taxpayers, which tells us that this experimental design is recommended.
14.95
H 0 : 1 2 3 = 4 = 5 = 6 = 7 H1 : At least two means differ.
ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
6
28,674
4,779
11.91
Blocks
199
209,835
1,054
2.63
Error
1194
479,125
401.3
F = 11.91, p-value = 0. There is enough evidence to conclude that there are differences in time spent listening to music between the days of the week 14.96
ANOVA Table
Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Treatments
4
1,406.4
351.6
10.72
Blocks
35
7,309.7
208.9
6.36
Error
140
4,593.9
32.81
a
H 0 : 1 2 3 = 4 = 5 H1 : At least two means differ.
F = 10.72, p-value = 0. There is enough evidence to infer differences between medical specialties. b
H 0 : 1 2 … 36 H1 : At least two means differ.
390
F = 6.36, p-value = 0. There is sufficient evidence to indicate that there are differences between the physicians’ ages, which tells us that this experimental design is recommended.
14.97
H 0 : 1 2 3 = 4 H 1 : At least two means differ.
ANOVA Table Source
Degrees of Freedom
Sum of Squares
Mean Squares
F
Treatments
3
563.82
187.9
15.06
Blocks
20
1,327.33
66.37
5.32
Error
60
748.70
12.48
F = 15.06, p-value = 0. There is enough evidence to infer differences in grading standards between the four high schools.
14.98
H 0 : 1 2 3 H 1 : At least two means differ.
F = 476.7, p-value = 0.
14.99
H 0 : 1 2 3 H 1 : At least two means differ.
F = 355.0, p-value = 0. 14.100 ANOVA Table
391
Source
Degrees of Freedom
Sum of Squares
Mean Squares
F
Factor A
2
1,560
780
5.86
Factor B
3
2,880
960
7.18
Interaction
6
7,605
1268
9.53
Error
228
30,405
133
Total
239
42,450
Test for interaction: Rejection region: F F,(a 1)( b 1),n ab F.01,6, 228 2.80 F = 9.53. There is enough evidence to conclude that factors A and B interact. The F-tests in Parts b and c are irrelevant. 14.101 ANOVA Table Source
Degrees of Freedom Sum of Squares
Mean Squares
F
Factor A
3
203
67.67
.72
Factor B
2
859
429.5
4.60
Interaction
6
513
85.5
.92
Error
84
7845
93.39
Total
95
9420
a Rejection region: F F,(a 1)( b 1),n ab F.05,6,84 2.25 F = .92. There is not enough evidence to conclude that factors A and B interact. b Rejection region: F F,a 1,n ab F.05,3,84 2.76 F = .72. There is not enough evidence to conclude that differences exist between the levels of factor A. c Rejection region: F F , b 1, n ab F.05, 2,84 3.15 F = 4.60. There is enough evidence to conclude that differences exist between the levels of factor B. 14.102 ANOVA Table
A 23 ANOVA 24 Source of Variation 25 Sample 26 Columns 27 Interaction 28 Within 29 30 Total
B
C
SS 5.33 56.33 1.33 34.67
df
97.67
1 1 1 8
D
E
F
MS 5.33 56.33 1.33 4.33
F P-value 1.23 0.2995 13.00 0.0069 0.31 0.5943
G F crit 5.32 5.32 5.32
11
a F = .31, p-value = .5943. There is not enough evidence to conclude that factors A and B interact. b F = 1.23, p-value = .2995. There is not enough evidence to conclude that differences exist between the levels of factor A.
392
c F = 13.00, p-value = .0069. There is enough evidence to conclude that differences exist between the levels of factor B. 14.103 ANOVA Table
A 29 ANOVA 30 Source of Variation 31 Sample 32 Columns 33 Interaction 34 Within 35 36 Total
B
C
SS 177.25 0.38 9.25 159.75
df
346.63
2 1 2 18
D
E
MS 88.63 0.38 4.63 8.88
F P-value 9.99 0.0012 0.04 0.8394 0.52 0.6025
F
G F crit 3.55 4.41 3.55
23
a F = .52, p-value = .6025. There is not enough evidence to conclude that factors A and B interact. b F = 9.99, p-value = .0012. There is enough evidence to conclude that differences exist between the levels of factor A. c F = .04, p-value = .8394. There is not enough evidence to conclude that differences exist between the levels of factor B. 14.104 ANOVA Table A 35 ANOVA 36 Source of Variation 37 Sample 38 Columns 39 Interaction 40 Within 41 42 Total
B
C
SS 135.85 151.25 6.25 726.20
df 3 1 3 72
1019.55
D
E
F
G
MS 45.28 151.25 2.08 10.09
F
P-value 0.0060 0.0002 0.8915
F crit 2.7318 3.9739 2.7318
4.49 15.00 0.21
79
The test for interaction yields (F = .21, p-value = .8915) and the test for the differences between educational levels (F = 4.49, p-value = .0060) is the same as in Example 14.4. However, in this exercise there is evidence of a difference between men and women (F = 15.00, p-value = .0002). 14.105 ANOVA Table A 35 ANOVA 36 Source of Variation 37 Sample 38 Columns 39 Interaction 40 Within 41 42 Total
B
C
SS 345.85 61.25 72.25 726.20
df
1205.55
3 1 3 72
D
E
MS 115.28 61.25 24.08 10.09
F 11.43 6.07 2.39
79
393
F
G
P-value 3.25E-06 0.0161 0.0760
F crit 2.7318 3.9739 2.7318
Compared to Example 14.4, the test for interaction has the same conclusion, although the value of F is larger and the p-value is smaller. Moreover, the mean number of jobs differs between the educational levels (F = 11.43, p-value = 0) and between men and women (F = 6.07, p-value = .0161). 14.106a. There are 12 treatments. b. There are two factors, tax form and income group. c. There are a = 4 forms and b = 3 income groups.
A 28 29 ANOVA 30 Source of Variation 31 Sample 32 Columns 33 Interaction 34 Within 35 36 Total
B
C
D
SS 6719 6280 5102 88217
df
MS 3359.4 2093.3 850.3 816.8
2 3 6 108
106317
E
F
F P-value 4.11 0.0190 2.56 0.0586 1.04 0.4030
G
F crit 3.08 2.69 2.18
119
d. F = 1.04, p-value = .4030. There is not enough evidence to conclude that forms and income groups interact e. F = 2.56, p-value = .0586. There is not enough evidence to conclude that differences exist between the forms. f. F = 4.11, p-value = .0190. There is enough evidence to conclude that differences exist between the three income groups. 14.107a. Detergents and temperatures b. The response variable is the whiteness score. c. There are a = 5 detergents and b = 3 temperatures.
A 29 ANOVA 30 Source of Variation 31 Sample 32 Columns 33 Interaction 34 Within 35 36 Total
B
C
D
SS 3937 2967 2452 14910
df
MS 1968.5 741.9 306.5 110.4
24267
2 4 8 135
E
F
F P-value 17.82 0.0000 6.72 0.0001 2.78 0.0071
G F crit 3.06 2.44 2.01
149
d. Test for interaction: F = 2.78, p-value = .0071. There is sufficient evidence to conclude that detergents and temperatures interact. The F-tests in Parts e and f are irrelevant. 14.108a. Factor A is the drug mixture and factor B is the schedule. b. The response variable is the improvement index. c. There are a = 4 drug mixtures and b = 2 schedules.
394
A 23 ANOVA 24 Source of Variation 25 Sample 26 Columns 27 Interaction 28 Within 29 30 Total
B
C
SS 14.40 581.80 548.60 804.80
df
1949.60
D MS 1 14.40 3 193.93 3 182.87 32 25.15
E
F
G
F P-value 0.57 0.4548 7.71 0.0005 7.27 0.0007
F crit 4.15 2.90 2.90
39
d Test for interaction: F = 7.27, p-value = .0007. There is sufficient evidence to conclude that the schedules and drug mixtures interact. 14.109a. There are 2 factors--class configuration and time period. b. The response variable is the number of times students ask and answer questions. c. There are 2 levels of class configuration and 3 levels of time period.
ANOVA Source of Variation Sample Columns Interaction Within
SS 13.33 46.67 206.67 202.00
Total
468.67
df 1 2 2 24
MS 13.33 23.33 103.33 8.42
F
P-value 1.58 0.2203 2.77 0.0826 12.28 0.0002
F crit 4.26 3.40 3.40
29
d. Interaction: F = 12.28, p-value = .0002. There is sufficient evidence to conclude that the class configuration and time interact. The other two F-tests are invalid. 14.110
A 23 ANOVA 24 Source of Variation 25 Sample 26 Columns 27 Interaction 28 Within 29 30 Total
B
C
SS 16.04 6.77 0.025 39.17
df
62.00
1 1 1 36
D
E
F
MS 16.04 6.77 0.025 1.09
F P-value 14.74 0.0005 6.22 0.0173 0.023 0.8814
G F crit 4.11 4.11 4.11
39
The p-values for interaction, machines, and alloys are .8814, .0173, .0005, and, respectively. Both machines and alloys are sources of variation. 14.111
395
A 35 ANOVA 36 Source of Variation 37 Sample 38 Columns 39 Interaction 40 Within 41 42 Total
B
C
SS 0.000309 0.000515 0.000183 0.004953
df
0.005959
D
3 1 3 32
MS 0.000103 0.000515 0.000061 0.000155
E
F
F P-value 0.66 0.5798 3.33 0.0775 0.39 0.7584
G F crit 2.90 4.15 2.90
39
The p-values for interaction, devices, and alloys are .7584, .0775, and .5798, respectively. There are no sources of variation. 14.112
A 29 ANOVA 30 Source of Variation 31 Sample 32 Columns 33 Interaction 34 Within 35 36 Total
B
C
SS 211.78 0.59 0.13 211.42
df
D
2 1 2 42
423.91
MS 105.89 0.59 0.0640 5.03
E
F
F P-value 21.04 0.0000 0.12 0.7348 0.0127 0.9874
G F crit 3.22 4.07 3.22
47
The p-values for interaction, methods, and skills are .9874, .7348, and 0, respectively. The only source of variation is skill level. 14.113a. The factors are mental outlook (2 levels) and physical condition (3 levels).
A 29 ANOVA 30 Source of Variation 31 Sample 32 Columns 33 Interaction 34 Within 35 36 Total
B
C
SS 2118.4 166.7 20.0 2336.2
df
4641.3
D MS 2 1059.22 1 166.67 2 10.02 54 43.26
E
F
F P-value 24.48 0.0000 3.85 0.0548 0.23 0.7941
G F crit 3.17 4.02 3.17
59
Test for interaction: F = .23, p-value = .7941. There is not enough evidence to infer interaction. b. F = 3.85, p-value = .0548. There is not enough evidence to conclude that differences exist between optimists and pessimists. c. F = 24.48, p-value = 0. There is sufficient evidence to conclude that differences exist between the three levels of physical condition.
14.114
H 0 : 1 2 3 = 4 H1 : At least two means differ. 396
F = 7.63, p-value = 0. There is enough evidence to infer that there are differences between the four degrees. 14.115a H 0 : 1 2 3 = 4 = 5
H1 : At least two means differ.
A 12 ANOVA 13 Source of Variation 14 Between Groups 15 Within Groups 16 17 Total
B
C
D
SS 1747.4 23983.7
df
MS 436.86 97.89
25731.1
4 245
E
F
F P-value 4.46 0.0017
G F crit 2.41
249
F = 4.46, p-value = .0017. There is enough evidence to infer that there are differences in the effect of the new assessment system between the five boroughs. b
397
Multiple Comparisons LSD Omega Treatment Treatment Difference Alpha = 0.05 Alpha = 0.05 Borough A Borough B -7.42 3.90 5.40 Borough C -6.18 3.90 5.40 Borough D -2.42 3.90 5.40 Borough E -3.94 3.90 5.40 1.24 3.90 5.40 Borough B Borough C Borough D 5.00 3.90 5.40 Borough E 3.48 3.90 5.40 Borough C Borough D 3.76 3.90 5.40 Borough E 2.24 3.90 5.40 Borough D Borough E -1.52 3.90 5.40 The mean assessments in borough A differs from the means in boroughs B and C. c The assessments for each borough are required to be normally distributed with equal variances. d The histograms are approximately bell-shaped and the sample variances are similar.
14.116
H 0 : 1 2 3 = 4 H1 : At least two means differ.
A 31 ANOVA 32 Source of Variation 33 Rows 34 Columns 35 Error 36 37 Total
B
C
SS 43980 4438 6113
df
D MS 19 2314.72 3 1479.21 57 107.25
54530
E
F
F P-value 21.58 0.0000 13.79 0.0000
G F crit 1.77 2.77
79
F = 13.79, p-value = 0. There is sufficient evidence to conclude that the reading speeds differ between the four typefaces. The typeface that was read the fastest should be used.
14.117
H 0 : 1 2 3 H1 : At least two means differ.
A 10 ANOVA 11 Source of Variation 12 Between Groups 13 Within Groups 14 15 Total
B
C
D
SS 406.5 16445.8
df
MS 203.25 111.88
16852.3
2 147
E
F
F P-value 1.82 0.1662
G F crit 3.06
149
F = 1.82, p-value = .1662. There is not enough evidence to infer that differences in attention span exist between the three products.
398
14.118
H 0 : 1 2 3 H1 : At least two means differ
A 17 ANOVA 18 Source of Variation 19 Rows 20 Columns 21 Error 22 23 Total
B
C
SS 195.33 43.52 33.81
df 6 2 12
272.67
D
E
MS 32.56 21.76 2.82
F P-value 11.55 0.0002 7.72 0.0070
F
G F crit 3.00 3.89
20
F = 7.72, p-value = .0070. There is enough evidence to infer that differences in attention span exist between the three products. 14.119
A 23 ANOVA 24 Source of Variation 25 Sample 26 Columns 27 Interaction 28 Within 29 30 Total
B
C
SS 123553 3965110 30006 4856578
df
8975248
D
MS 1 123553 2 1982555 2 15003 144 33726
E
F
F P-value 3.66 0.0576 58.78 0.0000 0.44 0.6418
G F crit 3.91 3.06 3.06
149
Interaction: F = .44, p-value = .6418. There is not enough evidence to conclude that age and gender interact. Age: F = 58.78, p-value = 0. There is sufficient evidence to conclude that differences in offers exist between the three age groups. Gender: F = 3.66, p-value = .0576. There is not enough evidence to conclude that differences in offers exist between males and females 14.120a H 0 : 1 2 3
H1 : At least two means differ.
A 10 ANOVA 11 Source of Variation 12 Between Groups 13 Within Groups 14 15 Total
B
C
D
SS 1769.5 2409.8
df
MS 884.74 6.48
4179.3
2 372
E
F
F P-value 136.58 0.0000
G F crit 3.02
374
F = 136.58, p-value = 0. There is sufficient evidence to infer that differences exist between the effects of the three teaching approaches. b
399
A B C D E 1 Multiple Comparisons 2 LSD Omega 3 4 Treatment Treatment Difference Alpha = 0.0167 Alpha = 0.05 Whole LanguagEmbedded -0.856 0.774 0.754 5 Pure -4.976 0.774 0.754 6 7 Embedded Pure -4.120 0.774 0.754 All three means differ from one another. From the sample means we may infer that the pure method is best, followed by embedded, and by whole-language. 14.121a H 0 : 1 2 3
H1 : At least two means differ.
A 10 ANOVA 11 Source of Variation 12 Between Groups 13 Within Groups 14 15 Total
B
C
SS 1913 8727
df
10640
D
2 87
E
MS 956.70 100.31
F
G
F P-value 9.54 0.0002
F crit 3.10
89
F = 9.54, p-value = .0002. There is sufficient evidence to infer that there are differences between the three groups. b
A B C D E 1 Multiple Comparisons 2 LSD Omega 3 Treatment Difference Alpha = 0.0167 Alpha = 0.05 4 Treatment Mozart White noise -9.30 6.31 6.14 5 Glass -10.20 6.31 6.14 6 Glass -0.90 6.31 6.14 7 White noise The mean time of the Mozart group differs from the mean times of white noise and the Glass groups.
14.122
H 0 : 1 2 3 = 4 H1 : At least two means differ.
A 11 ANOVA 12 Source of Variation 13 Between Groups 14 Within Groups 15 16 Total
B
C
SS 5990284 40024172
df
46014456
D
MS 3 1996761 290 138014
E
F
F P-value 14.47 0.0000
G F crit 2.64
293
F = 14.47, p-value = 0. There is enough evidence to infer differences in debt levels between the four types of degrees. 400
14.123
H 0 : 1 2 3 = 4 H1 : At least two means differ.
A 11 ANOVA 12 Source of Variation 13 Between Groups 14 Within Groups 15 16 Total
B
C
D
SS 3263 29685
df
MS 1087.8 106.0
32948
3 280
E
F
F P-value 10.26 0.0000
G F crit 2.64
283
F = 10.26, p-value = 0. There is enough evidence of differences between the four groups of investors.
14.124
H 0 : 1 2 3 = 4 H1 : At least two means differ.
A 11 ANOVA 12 Source of Variation 13 Between Groups 14 Within Groups 15 16 Total
B
C
D
SS 3007 10576
df
MS 1002.3 72.4
13583
3 146
E
F
F P-value 13.84 0.0000
G F crit 2.67
149
F = 13.84, p-value = 0. There is enough evidence to infer that the length of time depends on the size of the party 14.125 H0: µ 1 = µ 2 = µ 3 = µ 4 H1: At least two means differ.
A 11 ANOVA 12 Source of Variation 13 Between Groups 14 Within Groups 15 16 Total
B
C
D
SS 2.12 7.99
df
MS 0.705 0.0769
10.11
3 104
E
F
F P-value 9.17 0.0000
G F crit 2.69
107
F = 9.17, p-value = 0. There is sufficient evidence to infer that there are differences in changes to the TSE depending on the loss the previous day. 14.126 H0: µ 1 = µ 2 = µ 3 H1: At least two means differ.
401
A 10 ANOVA 11 Source of Variation 12 Between Groups 13 Within Groups 14 15 Total
B
C
SS 1.57 46.98
df
48.55
2 97
D
E
F
MS 0.787 0.484
F P-value 1.62 0.202233
G F crit 3.09
99
F = 1.62, p-value = .2022. There is no evidence to infer that at least one buy indicator is useful. 14.127 H0: µ 1 = µ 2 = µ 3 H1: At least two means differ.
F = 25.98, p-value = 0. There is sufficient evidence to infer that the amount of sleep differs between commuting categories?
14.128 H0: µ 1 = µ 2 = µ 3 H1: At least two means differ.
402
F = 2.70, p-value = .0684. There is not enough evidence to infer a difference between the three groups of teenagers. 14.129 a.H0: µ 1 = µ 2 = µ 3 = µ 4 H1: At least two means differ.
F = 111.4, p-value = 0. b. The incomes are required to be normally distributed. c. The histograms are bell shaped. 14.130 H0: µ 1 = µ 2 = µ 3 H1: At least two means differ.
403
F = 21.63, p-value = 0. 14.131 H0: µ 1 = µ 2 = µ 3 = µ 4 H1: At least two means differ.
F = 113.4, p-value = 0. 14.132a. H0: µ 1 = µ 2 = µ 3 = µ 4 = µ 5 = µ 6 = µ 7 H1: At least two means differ.
404
F = 9.42, p-value = 0. b.
405
All six days differ from Thursday. 14.133 H0: µ 1 = µ 2 = µ 3 = µ 4 = µ 5 H1: At least two means differ.
406
F = 445.5, p-value = 0. b.
All pairs differ except for 2011 and 2016. 14.134 H0: µ 1 = µ 2 = µ 3 = µ 4 = µ 5 H1: At least two means differ.
407
F = 23.19, p-value = 0. b.
All pairs differ except Electrical and Mechanical and Chemical and Computer. 14.135 H0: µ 1 = µ 2 = µ 3 = µ 4 H1: At least two means differ.
408
F = 45.48, p-value = 0. 14.136 H0: µ 1 = µ 2 = µ 3 H1: At least two means differ.
F = 35.90, p-value = 0.
409
b. Excel Workbook Groups
Means
| Difference |
LSD
i = 1, j = 2
4958
5956
998
294.6
i = 1, j = 3
4958
5708
750
297.6
i= 2, j = 3
5956
5708
248
300.5
Group 1 differs from groups 2 and 3. 14.137 H0: µ 1 = µ 2 = µ 3 = µ 4 = µ 5 = µ 6 H1: At least two means differ.
F = 6.77, p-value = 0. b.
410
The only pairs that differ are Guelph and Windsor and Brantford and Windsor. 14.138 Two-factor ANOVA
Interaction: F = 15.63, p-value = 0. 14.139 H0: µ 1 = µ 2 = µ 3 = µ 4 = µ 5 H1: At least two means differ.
411
F = 14.76, p-value = 0. b.
Pairs that differ: 1&2, 1&3, 1&4, 1&5, 2&5, 3&4, 3&5 14.140a. H0: µ 1 = µ 2 = µ 3 = µ 4 H1: At least two means differ.
412
F = .451, p-value = .7166. There is not enough evidence to infer that differences exist. 14.141 H0: Factors A and B do not interact H1: Factors A and B do interact F = 3.82, p-value = .0543. There is not enough evidence of interaction H0: The means of the bowls are equal H1: The means of the bowls differ F = 26.51, p-value = 0. There is enough evidence to conclude that the mean amount of ice cream differs between the two sizes of bowls. H0: The means of the scoops are equal H1: The means of the scoops differ F = 9.51, p-value = .0028. There is enough evidence to conclude that the mean amount of ice cream differs between the two sizes of scoop.
413
14.142
H 0 : 1 2 3 H1 : At least two means differ.
F = 39.93, p-value = 0.There is enough evidence to conclude that the number of kisses differ by location.
414
14.143
H 0 : 1 2 3 H1 : At least two means differ.
F = 127.43, p-value = 0. There Is sufficient evidence to conclude that the prices differed according to the way the brownies were offered.
14.144
H 0 : 1 2 3 H1 : At least two means differ.
1 2 3 4 5 6
A ANOVA Source of Variation Between Groups Within Groups
B
C
D
E
SS 11374 229170
df
MS 5687 125.0
F P-value 45.49 0.0000
Total
240544
1835
2 1833
F
G F crit 3.00
F = 45.49, p-value = 0. There is enough evidence to infer that family incomes differ between the three market segments.
14.145
H 0 : 1 2 3 H1 : At least two means differ.
1 2 3 4 5 6
A ANOVA Source of Variation Between Groups Within Groups
B
C
SS 52.82 1607.82
df
Total
1660.6
2 197
D
E
MS 26.41 8.16
F
199
415
3.24
F
G
P-value 0.0414
F crit 3.04
F = 3.24, p-value = .0414. There is enough evidence to infer that the distances driven differ between drivers who have had 0, 1, or 2 accidents.
H 0 : 1 2 3 = 4
14.146
H1 : At least two means differ. 1 2 3 4 5 6
A ANOVA Source of Variation Between Groups Within Groups
B
C
SS 6636.1 3595.9
df
Total
10232.0
3 344
D
E
F
MS 2212.03 10.45
F P-value 211.61 0.0000
G F crit 2.63
347
F = 211.61, p-value = 0. There is enough evidence to conclude that there are differences in the age of the car between the four market segments. Case 14.1 One-way analysis of variance H0: All 91 means are equal H1: At least two means differ
F = .845, p-value = .8489. There is not enough evidence to infer that a difference exists. Case 14.2 Episodes A 1 Anova: Single Factor 2 3 SUMMARY 4 Groups 5 Surgery 6 Drug 7 Placebo 8 9 10 ANOVA 11 Source of Variation 12 Between Groups 13 Within Groups 14 15 Total
B
C
D
E
Count 60 60 60
Sum Average Variance 198 3.30 1.74 178 2.97 1.19 207 3.45 1.68
SS 7.34 271.4
df
278.7
2 177
MS 3.67 1.53
179
416
F 2.40
F
G
P-value 0.0941
F crit 3.0470
F = 2.40, p-value = .0941. There is not enough evidence to conclude that there are differences in the number of episodes between the three types of treatments. Visits A 1 Anova: Single Factor 2 3 SUMMARY 4 Groups 5 Surgery 6 Drug 7 Placebo 8 9 10 ANOVA 11 Source of Variation 12 Between Groups 13 Within Groups 14 15 Total
B
C
D
E
Count 60 60 60
Sum Average Variance 130 2.17 2.51 114 1.90 1.38 147 2.45 2.83
SS 9.08 396.6
df 2 177
405.7
MS 4.54 2.24
F 2.03
F
G
P-value 0.1349
F crit 3.0470
179
F = 2.03, p-value = .1349. There is not enough evidence to conclude that there are differences in the number of physician visits between the three types of treatments. Prescriptions A 1 Anova: Single Factor 2 3 SUMMARY 4 Groups 5 Surgery 6 Drug 7 Placebo 8 9 10 ANOVA 11 Source of Variation 12 Between Groups 13 Within Groups 14 15 Total
B
C
D
E
Count 60 60 60
Sum Average Variance 201 3.35 2.20 178 2.97 3.05 205 3.42 3.43
SS 7.08 512.2
df
519.2
2 177
MS 3.54 2.89
F 1.22
F
G
P-value 0.2968
F crit 3.0470
179
F = 1.22, p-value = .2968. There is not enough evidence to conclude that there are differences in the number of prescriptions between the three types of treatments. Days
417
A 1 Anova: Single Factor 2 3 SUMMARY 4 Groups 5 Surgery 6 Drug 7 Placebo 8 9 10 ANOVA 11 Source of Variation 12 Between Groups 13 Within Groups 14 15 Total
B
C
D
E
Count 60 60 60
Sum Average Variance 689 11.48 22.76 663 11.05 19.71 779 12.98 28.22
SS 123.5 4170.8
df
4294.3
2 177
MS 61.76 23.56
F 2.62
F
G
P-value 0.0756
F crit 3.0470
179
F = 2.62, p-value = .0756. There is not enough evidence to conclude that there are differences in the number of days with respiratory infections between the three types of treatments.
418
Chapter 15 15.1
H0: p1 = .1, p2 = .2, p3 = .3, p4 = .2, p5 = .2 H1: At least one pi is not equal to its specified value. Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
28
0.2
20
8
3.20
2
17
0.2
20
-3
0.45
3
19
0.2
20
-1
0.05
4
17
0.2
20
-3
0.45
5
19
0.2
20
-1
0.05
Total
100
1
100
0
4.20
p-value
0.3796
Rejection region: 2 > 2 ,k 1 .201,4 13.3
2 = 4.53, p-value = .3386. There is not enough evidence to infer that at least one p i is not equal to its specified value.
15.2
H0: p1 = .1, p2 = .2, p3 = .3, p4 = .2, p5 = .2 H1: At least one pi is not equal to its specified value. Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
12
0.1
15
-3
0.60
2
32
0.2
30
2
0.13
3
42
0.3
45
-3
0.20
4
36
0.2
30
6
1.20
5
28
0.2
30
-2
0.13
Total
150
1
150
0
2.27
p-value
0.6868
Rejection region: 2 > 2 ,k 1 .201,4 13.3
2 = 2.27, p-value = .6868. There is not enough evidence to infer that at least one p i is not equal to its specified value.
15.3
H0: p1 = .1, p2 = .2, p3 = .3, p4 = .2, p5 = .2 H1: At least one pi is not equal to its specified value. Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
6
0.1
7.5
-1.5
0.30
2
16
0.2
15
1
0.07
3
21
0.3
22.5
-1.5
0.10
4
18
0.2
15
3
0.60
5
14
0.2
15
-1
0.07
Total
75
1
75
0
1.13
457
p-value
0.8889
Rejection region: 2 > 2 ,k 1 .201,4 13.3
2 = 1.13, p-value = .8889. There is not enough evidence to infer that at least one p i is not equal to its specified value.
15.4 The 2 statistic decreases.
15.5
H0: p1 = .3, p2 = .3, p3 = .2, p4 = .2 H1: At least one pi is not equal to its specified value. Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
38
0.3
45
-7
1.09
2
50
0.3
45
5
0.56
3
38
0.2
30
8
2.13
4
24
0.2
30
-6
1.20
Total
150
1
150
0
4.98
p-value
0.1734
Rejection region:
> 2 ,k 1 .205,3 7.81
2
2 = 4.98, p-value = .1734. There is not enough evidence to infer that at least one p i is not equal to its specified value.
15.6
H0: p1 = .3, p2 = .3, p3 = .2, p4 = .2 H1: At least one pi is not equal to its specified value. Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
76
0.3
90
-14
2.18
2
100
0.3
90
10
1.11
3
76
0.2
60
16
4.27
4
48
0.2
60
-12
2.40
Total
300
1
300
0
9.96
p-value
0.0189
Rejection region: 2 > 2 ,k 1 .205,3 7.81
2 = 9.96, p-value = .0189. There is enough evidence to infer that at least one p i is not equal to its specified value.
15.7
H0: p1 = .2, p2 = .2, p3 = .2, p4 = .2, p5 = .2 H1: At least one pi is not equal to its specified value.
458
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
28
0.2
20
8
3.20
2
17
0.2
20
-3
0.45
3
19
0.2
20
-1
0.05
4
17
0.2
20
-3
0.45
5
19
0.2
20
-1
0.05
Total
100
1
100
0
4.20
p-value
0.3796
Rejection region: 2 > 2 ,k 1 .210,4 7.78
2 = 4.20, p-value = .3796. There is not enough evidence to infer that at least one p i is not equal to its specified value.
15.8
H0: p1 = .15, p2 = .40, p3 = .35, p4 = .10 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
41
0.15
34.95
6.05
1.05
2
107
0.4
93.2
13.8
2.04
3
66
0.35
81.55
-15.55
2.97
4
19
0.1
23.3
-4.3
0.79
Total
233
1
233
0
6.85
p-value
0.0769
Rejection region: 2 > 2 ,k 1 .205,3 7.81
2 = 6.85, p-value = .0769. There is not enough evidence to infer that at least one p i is not equal to its specified value.
15.9
H0: p1 = 1/6, p2 = 1/6, p3 = 1/6, p4 = 1/6, p5 = 1/6, p6 = 1/6 H1: At least one pi is not equal to its specified value. Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
114
0.167
100
14
1.96
2
92
0.167
100
-8
0.64
3
84
0.167
100
-16
2.56
4
101
0.167
100
1
0.01
5
107
0.167
100
7
0.49
6
102
0.167
100
2
0.04
Total
600
1
600
459
0
5.70
p-value
0.3365
Rejection region: 2 > 2 ,k 1 .205,5 11.07
2 = 5.70, p-value = .3365. There is not enough evidence to infer that the die is not fair.
15.10
H0: p1 = .05, p2 = .25, p3 = .40, p4 = .25, p5 = .05 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
1
11
0.05
2
32
0.25
3
62
0.4
4
29
0.25
5
16
Total
150
Expected value E
F-E
((F-E)^2)/E
7.5
3.5
1.63
37.5
-5.5
0.81
60
2
0.07
37.5
-8.5
1.93
0.05
7.5
8.5
9.63
1
150
0
14.07
p-value
0.0071
Rejection region: 2 > 2 ,k 1 .210,4 7.78
2 = 14.07, p-value = .0071. There is enough evidence to infer that grades are distributed differently from grades in the past.
15.11
H0: p1 = .2, p2 = .2, p3 = .2, p4 = .2, p5 = .2 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
8
0.2
5
3
1.80
2
4
0.2
5
-1
0.20
3
8
0.2
5
3
1.80
4
3
0.2
5
-2
0.80
5
2
0.2
5
-3
1.80
Total
25
1
25
0
6.40
p-value
0.1712
Rejection region: 2 > 2 ,k 1 .205,4 9.49
2 = 6.40, p-value = .1712. There is not enough evidence to infer that the professor does not randomly distribute the correct answer over the five choices.
15.12
H0: p1 = .72, p2 = .15, p3 = .10, p4 = .03 H1: At least one pi is not equal to its specified value.
460
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
159
0.72
180
-21
2.45
2
28
0.15
37.5
-9.5
2.41
3
47
0.1
25
22
19.36
4
16
0.03
7.5
8.5
9.63
Total
250
1
250
0
33.85
p-value
0.0000
Rejection region: 2 > 2 ,k 1 .205,3 7.81
2 =33.85, p-value = 0. There is enough evidence to infer that the aging schedule has changed.
15.13
H0: p1 = .15, p2 = .25, p3 = .40, p4 = .20 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
36
0.15
29.55
6.45
1.41
2
58
0.25
49.25
8.75
1.55
3
74
0.4
78.8
-4.8
0.29
4
29
0.2
39.4
-10.4
2.75
Total
197
1
197
0.00
6.00
p-value
0.1116
Rejection region: 2 > 2 ,k 1 .205,3 7.81
2 = 6.00, p-value = .1116. There is not enough evidence to infer that certain sizes of cars are involved in a higher than expected percentage of accidents.
15.14
H0: p1 = .31, p2 = .51, p3 = .18 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
408
0.31
372
36
3.48
2
571
0.51
612
-41
2.75
3
221
0.18
216
5
0.12
Total
1200
1
1200
0
6.35
p-value
0.0419
Rejection region:
2
> 2 ,k 1 .210,2 4.61
2 = 6.35, p-value = .0419. There is enough evidence to infer that voter support has changed since the election.
15.15
H0: p1 = .0528, p2 = .3472, p3 = .51, p4 = .2, p5 = .09 H1: At least one pi is not equal to its specified value. 461
Cell
Frequency F
Hypothesized P
1
9
0.0528
2
123
0.3472
3
149
0.51
4
39
Total
320
Expected value E
F-E
((F-E)^2)/E
16.896
-7.896
3.69
111.104
11.896
1.27
163.2
-14.2
1.24
0.09
28.8
10.2
3.61
1
320
0.00
9.81
p-value
0.0202
Rejection region: 2 > 2 ,k 1 .205,3 7.81
2 = 9.81, p-value = .0202. There is enough evidence to infer that this umpire differs.
15.16
H0: p1 = .40, p2 = .144, p3 = .178, p4 = .095, p5 = .183
H1 : At least one p i is not equal to its specified value. Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
36
0.4
46
-10
2.17
2
26
0.144
16.56
9.44
5.38
3
24
0.178
20.47
3.53
0.61
4
14
0.095
10.925
3.075
0.87
5
15
0.183
21.045
-6.045
1.74
Total
115
1
115
0
10.77
p-value
0.0293
Rejection region: 2 > 2 ,k 1 .205,4 9.49
2 = 10.77, p-value = .0293. There is enough evidence to infer that this rookie differs.
15.17
H0: p1 = .52, p2 = .19, p3 = .15, p4 = .14 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
248
0.52
266.240
-18.24
1.25
2
108
0.19
97.280
10.72
1.18
3
47
0.15
76.800
-29.80
11.56
4
109
0.14
71.680
37.32
19.43
Total
512
1
512
0.00
33.42
p-value
0.0000
Rejection region: 2 > 2 ,k 1 .205,3 7.81
462
2 = 33.42, p-value = 0. There is enough evidence to infer that the pitching distribution has changed.
15.18
H0: p1 = .17, p2 = .18, p3 = .17, p4 = .18, p5 = .21, P6 = .09 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
137
0.170
139.06
-2.06
0.03
2
133
0.180
147.24
-14.24
1.38
3
142
0.170
139.06
2.94
0.06
4
136
0.180
147.24
-11.24
0.86
5
171
0.210
171.78
-0.78
0.00
6
99
0.090
73.62
25.38
8.75
Total
818
1
818
0.00
11.08
p-value
0.0498
Rejection region: 2 > 2 ,k 1 .205,5 11.07
2 = 11.08, p-value = .0498. There is enough evidence to infer that the distribution has changed.
15.19
H0: p1 = .024, p2 = .030, p3 = .067, p4 = .169, p5 = .212, P6 = .174, p7 = .080, p8 = .057, p9 = .059, p10 = .049, p11 = .049, p12 = .028 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
26
0.024
30.22
-4.22
0.59
2
2
0.030
37.77
-35.77
33.88
3
13
0.067
84.35
-71.35
60.36
4
185
0.169
212.77
-27.77
3.62
5
414
0.212
266.91
147.09
81.06
6
184
0.174
219.07
-35.07
5.61
7
122
0.080
100.72
21.28
4.50
8
44
0.057
71.76
-27.76
10.74
9
24
0.059
74.28
-50.28
34.04
10
57
0.049
61.69
-4.69
0.36
11
95
0.049
61.69
33.31
17.98
12
93
0.028
35.25
57.75
94.60
Total
1259
0.998
1256.482
2.52
347.33
p-value
0.0000
Rejection region: 2 > 2 ,k 1 .205,11 19.68
2 = 347.33, p-value = 0. There is enough evidence to infer that the percentages have changed. 463
H 0 : p1 .23, p 2 .40, p 3 .15, p 4 .22
15.20
H1: At least one pi is not equal to its specified value. Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
63
0.23
73.600
-10.60
1.53
2
125
0.40
128.000
-3.00
0.07
3
45
0.15
48.000
-3.00
0.19
4
87
0.22
70.400
16.60
3.91
Total
320
1
320
0.00
5.70
p-value
0.1272
Rejection region:
2
> 2 ,k 1 .205,3 7.81
2 = 5.70, p-value = .1272. There is not enough evidence to infer that there has been a change in proportions.
15.21
H0: p1 = .773, p2= .132, p3= .095 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
1890
0.773
1961.874
-71.874
2.63
2
386
0.132
335.016
50.984
7.76
3
262
0.095
241.11
20.89
1.81
Total
2538
1
2538
0.00
12.20
p-value
0.0022
2 = 12.20, p-value = .0022. There is sufficient evidence to conclude that the survey overrepresented at least one race.
15.22
H0: p1+4 = .525, p2 = .057, p3 = .100, p5 = .317 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1+4
1239
0.525
1330.350
-91.35
6.27
2
209
0.057
144.438
64.56
28.86
3
411
0.100
253.400
157.60
98.02
5
675
0.317
803.278
-128.28
20.49
Total
2534
0.999
2531.466
2.53
153.63
p-value
0.0000
2 = 153.63, p-value = 0. There is sufficient evidence to conclude that the survey overrepresented at least one category.
464
15.23
H0: p0 = .123, p1 = .296, p2 = .194, p3+4 = .386 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
0
330
0.123
312.174
17.83
1.02
1
1269
0.296
751.248
517.75
356.83
2
186
0.194
492.372
-306.37
190.64
3+4
753
0.386
979.668
-226.67
52.44
Total
2538
0.999
2535.462
2.54
600.93
p-value
0.0000
2 = 600.93, p-value = 0. There is sufficient evidence to conclude that the survey overrepresented at least one category.
15.24
H0: p1 = .616, p2 = .132, p3 = .176, p5 = .075 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
4425
0.616
3705.240
719.76
139.82
2
747
0.132
793.980
-46.98
2.78
3
556
0.176
1058.640
-502.64
238.65
5
287
0.075
451.125
-164.13
59.71
Total
6015
0.999
6008.985
6.02
440.96
p-value
0.0000
2 = 440.96, p-value = 0. There is sufficient evidence to conclude that the survey overrepresented at least one race.
15.25
H0: p1 = .126, p2 = .296, p3 = .196, p5 = .383 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
546
0.126
757.890
-211.89
59.24
2
1599
0.296
1780.440
-181.44
18.49
3
1030
0.196
1178.940
-148.94
18.82
4
2840
0.383
2303.745
536.26
124.83
Total
6015
1.001
6021.015
-6.01
221.37
p-value
0.0000
2 = 221.37, p-value = 0. There is sufficient evidence to conclude that the survey overrepresented at least one category.
465
15.26
H0: The two variables are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)( c 1 .205,1 3.84
2 = 19.10, p-value = 0. There is enough evidence to infer that the two variables are dependent. 15.27
H0: The two variables are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)(c 1 .205,1 3.84
2 = 9.56, p-value = .0020. There is enough evidence to infer that the two classifications L and M are dependent.
15.28
H0: The two variables are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)(c 1 .205,1 3.84
2 = 4.78, p-value = .0289. There is enough evidence to infer that the two classifications L and M are dependent.
15.29 The 2 statistic decreases.
466
15.30
H0: The two variables are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)( c 1 .210, 2 4.61
2 = 4.40, p-value = .1110. There is not enough evidence to infer that the two classifications R and C are dependent.
15.31
H0: The two variables (responses and employee group) are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)( c 1 .205, 2 5.99
2 = 2.27, p-value = .3221. There is not enough evidence to infer that responses differ among the three groups of employees.
15.32
H0: The two variables (shirt condition and shift) are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)( c 1 .205, 2 5.99
2 = 2.35, p-value = .3087. There is not enough evidence to infer that there are differences in quality among the three shifts.
467
15.33
H0: The two variables (economic option and political affiliation) are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)(c 1 .201,6 16.8
2 = 70.67, p-value = 0. There is sufficient evidence to infer that political affiliation affects support for economic options.
15.34
H0: The two variables (inducement and return) are independent H1: The two variables are dependent
Rejection region: > 2 ,( r 1)( c1 .205, 2 5.99 2
2 = 19.68, p-value = 0. There is sufficient evidence to infer that the return rates differ among the different inducements.
15.35
H0: The two variables (newspaper and occupation) are independent H1: The two variables are dependent
468
Rejection region: > 2 ,( r 1)( c 1 .205,6 12.59 2
2 = 32.57, p-value = 0. There is sufficient evidence to infer that occupation and newspaper are related.
15.36
H0: The two variables (remedy/placebo and side effects) are independent H1: The two variables are dependent
Rejection region: > 2 ,( r 1)(c 1 .205,3 7.81 2
2 = .918, p-value = .8211. There is not enough evidence to infer that the reported side effects differ.
15.37
H0: The two variables (last purchase and second-last purchase) are independent H1: The two variables are dependent
Rejection region: > 2 ,( r 1)( c 1 .205,9 16.9 2
469
2 = .67, p-value = .9999. There is no evidence of a relationship.
15.38
H0: The two variables (education and smoker) are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)( c 1 .205,3 7.81
2 = 41.76, p-value = 0. There is sufficient evidence to infer that the amount of education is a factor in determining whether a smoker will quit.
15.39
H0: The two variables (education and smoker) are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)( c 1 .205,15 25.0
2 = 22.36, p-value = .0988. There is not enough evidence to infer that there is a relationship between an adult’s source of news and his or her heartburn condition.
15.40
H0: The two variables (university and degree) are independent
470
H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)( c 1 .205,9 16.92
2 = 43.36, p-value = 0. There is enough evidence to infer that undergraduate degree and the university applied to are related.
15.41
H0: The two variables (results and financial ties) are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)( c 1 .205, 2 5.99
2 = 20.99, p-value = 0. There is sufficient evidence to infer that the research findings are related to whether drug companies fund the research.
15.42
H0: The two variables (degree and approach) are independent H1: The two variables are dependent
471
Rejection region: > 2 ,( r 1)( c 1 .205,6 12.59 2
2 = 20.89, p-value = .0019. There is sufficient evidence to infer that there are differences in teaching approach among the four types of degree. The editor can design books and sales campaigns based on the distribution of degrees.
15.43
H0: The two variables (year and weapon) are independent H1: The two variables are dependent
2 = 4.76, p-value = .5751. There is not sufficient evidence to infer that the frequency of the use of weapons in robberies differed over the three years.
15.44
H0: The two variables are independent H1: The two variables are dependent
472
2 = 1.22, p-value = .9761. There is not enough evidence to conclude that the deteriorating condition of American bridges has changed over the years.
15.45
H0: The two variables are independent H1: The two variables are dependent
2 = 26.69, p-value = .0008. There is sufficient evidence to infer that there are differences in household types between the three countries.
473
15.46
H0: The two variables are independent H1: The two variables are dependent
2 = 311.4, p-value = 0. There is sufficient evidence to infer that there are differences in household types between the four countries.
15.47
H0: The two variables are independent H1: The two variables are dependent
2 = .395, p-value = .9413. There is not enough evidence to conclude that there are differences in obesity rates between the four Commonwealth nations
474
15.48
H0: The two variables are independent H1: The two variables are dependent
2 = 11.82, p-value = .0080. There is enough evidence to conclude that there are differences in smoking between the four Scandinavian countries.
15.49
H0: The two variables are independent H1: The two variables are dependent
2 = 2.32, p-value = .5084.
15.50
H0: The two variables are independent H1: The two variables are dependent
475
2 = 22.49, p-value = .0002. There is enough evidence to infer that there are differences in adult cigarette smoking between the four Commonwealth countries.
15.51
H0: The two variables are independent H1: The two variables are dependent
2 = 4.53, p-value = .1038. There is not enough evidence to infer that there are differences.
15.52
476
2 = 8.52, p-value = .0365. There is enough evidence to conclude that differences exist.
15.53
H0: The two variables are independent H1: The two variables are dependent
2 = 12.57, p-value = .0019.
15.54
H0: The two variables are independent H1: The two variables are dependent
477
2 = 29.70, p-value = 0.
15.55
H0: The two variables are independent H1: The two variables are dependent
2 = 7.16, p-value = .0279.
15.56
H0: The two variables are independent H1: The two variables are dependent
2 = 15.80, p-value = .0004.
15.57
H0: The two variables are independent H1: The two variables are dependent
478
2 = 2.02, p-value = .3646
15.58
H0: The two variables are independent H1: The two variables are dependent
2 = 40.19, p-value = 0.
15.59
H0: The two variables are independent H1: The two variables are dependent
2 = 159.79, p-value = 0. 479
15.60
H0: The two variables are independent H1: The two variables are dependent
2 = 146.54, p-value = 0.
15.61
H0: The two variables are independent H1: The two variables are dependent
2 = 80.36, p-value = 0.
15.62
H0: The two variables are independent H1: The two variables are dependent
480
2 = 29.42, p-value = 0.
15.63
H0: The two variables are independent H1: The two variables are dependent
2 = 2.95, p-value = .4000.
15.64
H0: The two variables are independent H1: The two variables are dependent
481
2 = .271, p-value = .9653.
15.65
H0: The two variables are independent H1: The two variables are dependent
2 = 5.13, p-value = .1625.
15.66
H0: The two variables are independent H1: The two variables are dependent
482
2 = 50.56, p-value = 0.
15.67
H0: The two variables are independent H1: The two variables are dependent
2 = 9.04, p-value = .0287.
15.68
H0: The two variables are independent H1: The two variables are dependent
483
2 = 5.99, p-value = .1121.
15.69
H0: The two variables are independent H1: The two variables are dependent
2 = 2.90, p-value = .4077.
15.70
H0: The two variables are independent H1: The two variables are dependent
484
2 = 4.41, p-value = .2204.
15.71
H0: The two variables are independent H1: The two variables are dependent
2 = 10.13, p-value = .0175.
15.72
H0: The two variables are independent H1: The two variables are dependent
485
2 = 1.97, p-value = .5796.
15.73
H0: The two variables are independent H1: The two variables are dependent
2 = 46.71, p-value = 0.
15.74
H0: The two variables are independent H1: The two variables are dependent
486
2 = .684, p-value = .7102.
15.75
H0: The two variables are independent H1: The two variables are dependent
2 = 5.00, p-value = .0254.
15.76
H0: The two variables are independent H1: The two variables are dependent
487
2 = 1.79, p-value = .1810.
15.77
H0: The two variables are independent H1: The two variables are dependent
2 = 20.03, p-value = 0.
15.78
H0: The two variables are independent H1: The two variables are dependent
488
2 = 4.51, p-value = .0337.
15.79
H 0 : The data are normally distributed H1 : The data are not normally distributed Expected
Observed
Value e i 6.68
Value f i 10
f i ei 3.32
(f i ei ) 2 / ei 1.65
-1.5 < Z -0.5 .2417
24.17
18
-6.17
1.58
-0.5 < Z 0.5
.3829
38.29
48
9.71
2.46
0.5 < Z 1.5
.2417
24.17
16
-8.17
Z > 1.5
.0668
6.68
8
1.32
Total
1
100
100
Interval Z -1.5
Probability .0668
2.76 0.26
2 = 8.71
Rejection region: 2 > 2 ,k 3 .205,2 5.99
2 = 8.71, p-value = .0128. There is enough evidence to infer that the data are not normally distributed.
15.80
H 0 : The data are normally distributed H1 : The data are not normally distributed Expected
Observed
Value e i 7.94
Value f i 6
f i ei -1.94
(f i ei ) 2 / ei 0.47
Interval Z -1
Probability .1587
-1 < Z 0
.3413
17.07
27
9.93
5.78
0<Z 1
.3413
17.07
14
-3.07
0.55
Z>1
.1587
7.94
3
-4.94
3.07
Total
1
50
50
489
2 = 9.87
Rejection region: 2 > 2 ,k 3 .210,1 2.71
2 = 9.87, p-value = .0017. There is sufficient evidence to infer that the data are not normally distributed.
15.81
H 0 : Times are normally distributed H1 : Times are not normally distributed.
A B C D 1 Chi-Squared Test of Normality 2 Hours 3 7.15 4 Mean 1.65 5 Standard deviation 200 6 Observations 7 Probability Expected Observed Intervals 8 (z <= -1.5) 0.0668 13.36 11 9 0.2417 48.35 55 10 (-1.5 < z <= -0.5) (-0.5 < z <= 0.5) 0.3829 76.59 52 11 (0.5 < z <= 1.5) 0.2417 48.35 67 12 (z > 1.5) 0.0668 13.36 15 13 14 15 16.62 16 chi-squared Stat 2 17 df 0.0002 18 p-value 5.9915 19 chi-squared Critical
2 = 16.62, p-value = .0002. There is sufficient evidence to infer that the amount of time at parttime jobs is not normally distributed.
15.82
H 0 : Costs are normally distributed H1 : Costs are not normally distributed
490
A B 1 Chi-Squared Test of Normality 2 Drug Cost 3 29.69 4 Mean 27.53 5 Standard deviation 900 6 Observations 7 Intervals 8 Probability 9 (z <= -2) 0.0228 (-2 < z <= -1) 10 0.1359 (-1 < z <= 0) 11 0.3413 (0 < z <= 1) 12 0.3413 (1 < z <= 2) 13 0.1359 14 (z > 2) 0.0228 15 16 chi-squared Stat 506.76 17 df 3 18 p-value 0 19 chi-squared Critical 7.8147
C
D
Expected Observed 20.48 0 122.31 9 307.21 599 307.21 248 122.31 17 20.48 27
2 = 506.76, p-value = 0. There is sufficient evidence to infer that drug costs are not normally distributed. 15.83
Successful firms:
H 0 : Productivity in successful firms is normally distributed H1 : Productivity in successful firms is not normally distributed
A B C D 1 Chi-Squared Test of Normality 2 3 Successful 4 Mean 5.02 5 Standard deviation 1.39 6 Observations 200 7 8 Probability Expected Observed Intervals 9 (z <= -1.5) 0.0668 13.36 12 10 (-1.5 < z <= -0.5) 0.2417 48.35 52 11 (-0.5 < z <= 0.5) 0.3829 76.59 72 12 (0.5 < z <= 1.5) 0.2417 48.35 55 13 (z > 1.5) 0.0668 13.36 9 14 15 3.0288 16 chi-squared Stat 2 17 df 18 p-value 0.2199 19 chi-squared Critical 5.9915
2 =3.03, p-value = .2199. There is not enough evidence to infer that productivity in successful firms is not normally distributed. 491
Unsuccessful firms:
H 0 : Productivity in unsuccessful firms is normally distributed H1 : Productivity in unsuccessful firms is not normally distributed
A B 1 Chi-Squared Test of Normality 2 Unsuccessful 3 7.80 4 Mean 3.09 5 Standard deviation 200 6 Observations 7 Probability Intervals 8 (z <= -1.5) 0.0668 9 0.2417 10 (-1.5 < z <= -0.5) (-0.5 < z <= 0.5) 0.3829 11 (0.5 < z <= 1.5) 0.2417 12 (z > 1.5) 0.0668 13 14 15 1.1347 16 chi-squared Stat 2 17 df 0.567 18 p-value 5.9915 19 chi-squared Critical
C
D
Expected Observed 13.36 12 48.35 47 76.59 83 48.35 44 13.36 14
2 = 1.13, p-value = .5670. There is not enough evidence to infer that productivity in unsuccessful firms is not normally distributed.
15.84
H 0 : Reaction times are normally distributed H1 : Reaction times re not normally distributed
Phone
492
A B C D 1 Chi-Squared Test of Normality 2 3 Phone 4 Mean 0.646 5 Standard deviation 0.045 6 Observations 125 7 8 Intervals Probability Expected Observed 9 (z <= -1.5) 0.0668 8.35 8 10 (-1.5 < z <= -0.5) 0.2417 30.22 32 11 (-0.5 < z <= 0.5) 0.3829 47.87 47 12 (0.5 < z <= 1.5) 0.2417 30.22 29 13 (z > 1.5) 0.0668 8.35 9 14 15 16 chi-squared Stat 0.2351 17 df 2 18 p-value 0.8891 19 chi-squared Critical 5.9915
2 = .235, p-value = .8891. There is not enough evidence to infer that reaction times of those using the cell phone are not normally distributed.
Not on phone
A B 1 Chi-Squared Test of Normality 2 Not 3 0.601 4 Mean 0.053 5 Standard deviation 145 6 Observations 7 Intervals Probability 8 9 (z <= -1.5) 0.0668 10 (-1.5 < z <= -0.5) 0.2417 11 (-0.5 < z <= 0.5) 0.3829 12 (0.5 < z <= 1.5) 0.2417 13 (z > 1.5) 0.0668 14 15 16 chi-squared Stat 3.1752 17 df 2 18 p-value 0.2044 19 chi-squared Critical 5.9915
C
D
Expected Observed 9.69 8 35.05 40 55.52 55 35.05 29 9.69 13
2 = 3.18, p-value = .2044. There is not enough evidence to infer that reaction times of those not using the cell phone are not normally distributed.
493
15.85
H 0 : Matched pairs differences of sales are normally distributed H1 : Matched pairs differences of sales are not normally distributed
A B C D 1 Chi-Squared Test of Normality 2 Difference 3 19.75 4 Mean 30.63 5 Standard deviation 40 6 Observations 7 Probability Expected Observed Intervals 8 (z <= -1) 0.1587 6.35 6 9 (-1 < z <= 0) 0.3413 13.65 14 10 (0 < z <= 1) 0.3413 13.65 14 11 (z > 1) 0.1587 6.35 6 12 13 14 15 0.0553 16 chi-squared Stat 1 17 df 0.8140 18 p-value 2.7055 19 chi-squared Critical
2 = .055, p-value = .8140. There is not enough evidence to infer that matched pairs difference of sales is not normally distributed.
15.86
H0: p1= 1/3. P2 = 1/3, p3 = 1/3 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
14
0.333
10
4
1.60
2
10
0.333
10
0
0.00
3
6
0.333
10
-4
1.60
Total
30
1
30
0.00
3.20
p-value
0.2019
Rejection region: 2 > 2 ,k 1 .210,2 4.61
2 = 3.20, p-value = .2019. There is not enough evidence to infer that the game is unfair.
15.87
H0: p1= .2. P2 = .2, p3 = .2, p4 = .2, p5 = .2 H1: At least one pi is not equal to its specified value.
Cell
Frequency F
Hypothesized P
Expected value E
F-E
((F-E)^2)/E
1
87
0.2
72.4
14.6
2.94
2
62
0.2
72.4
-10.4
1.49
494
3
71
0.2
72.4
-1.4
0.03
4
68
0.2
72.4
-4.4
0.27
5
74
0.2
72.4
1.6
0.04
Total
362
1
362
0.00
4.77
p-value
0.3119
Rejection region: 2 > 2 ,k 1 .205, 4 9.49
2 = 4.77, p-value = .3119. There is not enough evidence to infer that absenteeism is higher on some days of the week.
15.88
H0: The two variables (shift and day) are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)( c 1 .210, 4 7.78
2 = 5.42, p-value = .2465. There is not enough evidence to infer that there is a relationship between the days an employee is absent and the shift on which the employee works.
15.89
H0: The two variables (satisfaction and relationship) are independent H1: The two variables are dependent
Rejection region: 2 > 2 ,( r 1)( c 1 .205,6 12.59
2 = 74.47, p-value = 0. There is sufficient evidence to infer that the level of job satisfaction depends on boss/employee gender relationship.
15.90
H0: The two variables (Country and stress) are independent H1: The two variables are dependent
495
A B C 1 Contingency Table 2 3 Stress 4 Country 1 5 1 266 6 2 347 7 3 153 8 4 164 9 5 92 10 TOTAL 1022 11 12 13 chi-squared Stat 14 df 15 p-value 16 chi-squared Critical
D
E
2 315 276 187 128 79 985
TOTAL 581 623 340 292 171 2007
20.3755 4 0.0004 9.4877
2 = 20.38, p-value = .0004. There is enough evidence to infer that Americans and Canadians differ in their sources of stress.
15.91
H0: The two variables (method and quit) are independent H1: The two variables are dependent
A B C 1 Contingency Table 2 Quit 3 1 4 Method 1 104 5 2 14 6 TOTAL 118 7 8 9 chi-squared Stat 10 df 11 p-value 12 chi-squared Critical 13
D
E
F
G
2 125 17 142
3 32 5 37
4 49 9 58
TOTAL 310 45 355
0.5803 3 0.9009 7.8147
2 = .580, p-value = .9009. There is not enough evidence to infer that the four methods differ in their success rates.
15.92
H0: The two variables (education and section) are independent H1: The two variables are dependent
496
A B C 1 Contingency Table 2 3 Section 4 Education 1 5 1 4 6 2 27 7 3 1 8 4 10 9 TOTAL 42 10 11 12 chi-squared Stat 13 df 14 p-value 15 chi-squared Critical
D
E
F
G
2 21 32 20 44 117
3 31 18 42 22 113
4 14 2 22 3 41
TOTAL 70 79 85 79 313
86.6154 9 0 16.919
2 = 86.62, p-value = 0. There is sufficient evidence to infer that educational level affects the way adults read the newspaper.
15.93a The expected frequency is 1/49.
H 0 : p1 = 1/49, p2 = 1/49, . . . , p49 = 1/49
b
H1:At least one pi is not equal to its specified value. Number i f i
ei
(f i e i )
(f i ei ) 2 / ei
1
5
312(1/49) = 6.37
-1.38
0.29
2
6
312(1/49) = 6.37
-0.38
0.02
3
7
312(1/49) = 6.37
0.63
0.06
.
.
.
.
.
.
47
6
312(1/49) = 6.37
-0.37
0.02
48
10
312(1/49) = 6.37
3.63
2.07
49
6
312(1/49) = 6.37
-0.37
0.02
Total
312
2 = 38.22
312
2 = 38.22, p-value = .8427. There is not enough evidence to infer that the numbers were not generated randomly.
15.94
H0: The two variables are independent H1: The two variables are dependent
497
2 = 15.39, p-value = .0005.
15.95
H0: The two variables are independent H1: The two variables are dependent
2 = 11.36, p-value = .0034.
15.96
H0: The two variables are independent H1: The two variables are dependent
498
2 = 166.8, p-value = 0.
15.97
H0: The two variables are independent H1: The two variables are dependent
2 = 51.16, p-value = 0.
15.98 Binomial probabilities with n = 5 and p = .5: P(X = 0) = .0313, P(X = 1) = .1563, P(X = 2) = .3125, P(X = 3) = .3125, P(X = 4) = .1563, P(X = 5) = .0313
H0: p0 = .0313, p1 = .1563, p2 = .3125, p3 = .3125, p4 = .1563, p5 = .0313 H1: At least one pi is not equal to its specified value. Cell i
fi
ei
(f i e i )
(f i ei ) 2 / ei
0
8
200(.0313) = 6.26
1.74
0.48
1
35
200(.1563) = 31.26
3.74
0.45
2
57
200(.3125) = 62.50
-5.50
0.48
3
69
200(.3125) = 62.50
6.50
0.68
499
4
28
200(.1563) = 31.26
-3.26
0.34
5
3
200(.0313) = 6.26
-3.26
1.70
Total
200
2 = 4.13
200
Rejection region: 2 > 2 ,61 .205,5 11.1
2 = 4.13, p-value = .5310. There is not enough evidence to infer that at the number of boys in families with 5 children is not a binomial random variable with p =.5.
15.99
H0: The two variables (cold and group) are independent H1: The two variables are dependent
A B C 1 Contingency Table 2 3 Group 4 Cold 1 5 1 17 6 2 12 7 3 9 8 4 16 9 TOTAL 54 10 11 12 chi-squared Stat 13 df 14 p-value 15 chi-squared Critical
D
E
2 11 13 18 18 60
TOTAL 28 25 27 34 114
4.139 3 0.2468 7.8147
2 = 4.14, p-value = .2468. There is not enough evidence to infer there are differences between the four groups.
15.100 H0: The two variables (faculty and retire) are independent H1: The two variables are dependent
500
A B C 1 Contingency Table 2 Retire 3 1 4 Faculty 1 174 5 2 13 6 TOTAL 187 7 8 9 chi-squared Stat 10 df 11 p-value 12 chi-squared Critical 13
D
E
F
G
H
2 51 7 58
3 113 22 135
4 42 7 49
5 86 6 92
TOTAL 466 55 521
9.732 4 0.0452 9.4877
2 = 9.73, p-value = .0452. There is enough evidence to infer that whether a professor wishes to retire is related to the faculty.
15.101 H0: The two variables (tree choice and age category) are independent H1: The two variables are dependent A B C 1 Contingency Table 2 Tree choice 3 4 Age category 1 5 1 136 6 2 196 TOTAL 332 7 8 9 10 chi-squared Stat df 11 p-value 12 13 chi-squared Critical
D
E
F
2 309 339 648
3 158 370 528
TOTAL 603 905 1508
38.41 2 0 5.9915
2 = 38.41, p-value = 0. There is enough evidence to infer that there are differences in the choice of Christmas tree between the three age categories.
15.102 H0: The two variables (network and ask) are independent H1: The two variables are dependent
501
A B C 1 Contingency Table 2 Ask 3 1 4 Network 1 19 5 2 104 6 TOTAL 123 7 8 9 chi-squared Stat 10 df 11 p-value 12 chi-squared Critical 13
D
E
F
2 30 107 137
3 43 123 166
TOTAL 92 334 426
4.573 2 0.1016 5.9915
2 = 4.57, p-value = .1016. There is not enough evidence to conclude that there are differences in responses between the three network news shows.
15.103 a H0: The two variables (education and group) are independent H1: The two variables are dependent
A B C 1 Contingency Table 2 3 Group 4 Education 1 5 1 5 6 2 70 7 TOTAL 75 8 9 10 chi-squared Stat 11 df 12 p-value 13 chi-squared Critical
D
E
F
G
2 113 305 418
3 73 189 262
4 40 55 95
TOTAL 231 619 850
26.7059 3 0 7.8147
2 = 26.71, p-value = 0. There is enough evidence to infer that there are differences in educational attainment between those who belong and those who do not belong to the health conscious group. b
H0: The two variables (education and buy Special X) are independent H1: The two variables are dependent
502
A B C 1 Contingency Table 2 Buy Sp X 3 1 4 Education 1 70 5 2 5 6 TOTAL 75 7 8 9 chi-squared Stat 10 df 11 p-value 12 chi-squared Critical 13
D
E
F
G
2 376 42 418
3 229 33 262
4 87 8 95
TOTAL 762 88 850
2.9416 3 0.4007 7.8147
2 = 2.94, p-value = .4007. There is not enough evidence to infer that there is a relationship between the four education groups and whether a person buys Special X.
15.104 a H0: The two variables (gender and vote) are independent H1: The two variables are dependent
A B C 1 Contingency Table 2 3 Votes 4 Gender 1 5 1 189 6 2 203 7 TOTAL 392 8 9 10 chi-squared Stat 11 df 12 p-value 13 chi-squared Critical
D
E
2 169 204 373
TOTAL 358 407 765
0.6483 1 0.4207 3.8415
2 = .648, p-value = .4207. There is not enough evidence to infer that voting and gender are related. b
H0: The two variables (education and vote) are independent H1: The two variables are dependent
503
A B C 1 Contingency Table 2 Votes 3 1 4 Educ 1 48 5 2 34 6 TOTAL 82 7 8 9 chi-squared Stat 10 df 11 p-value 12 chi-squared Critical 13
D
E
F
G
2 164 178 342
3 107 134 241
4 39 61 100
TOTAL 358 407 765
7.7214 3 0.0521 7.8147
2 = 7.72, p-value = .0521. There is not enough evidence to infer that voting and educational level are related. c
H0: The two variables (income category and vote) are independent H1: The two variables are dependent
A B C 1 Contingency Table 2 3 Votes 4 Income 1 5 1 38 6 2 21 7 TOTAL 59 8 9 10 chi-squared Stat 11 df 12 p-value 13 chi-squared Critical
D
E
F
G
2 186 185 371
3 105 128 233
4 29 73 102
TOTAL 358 407 765
23.108 3 0 7.8147
2 = 23.11, p-value = 0. There is enough evidence to infer that voting and income are related.
15.105 H0: The two variables are independent H1: The two variables are dependent
504
2 = 5.49, p-value = .2409.
15.106 H0: The two variables are independent H1: The two variables are dependent
2 = 57.45, p-value = 0.
15.107 H0: The two variables are independent H1: The two variables are dependent
2 = 12.93, p-value = .0441. 505
15.108 H0: The two variables are independent H1: The two variables are dependent
2 = 28.97, p-value = 0.
15.109 H0: The two variables are (type of work and segment) independent H1: The two variables are dependent
A B C 1 Contingency Table 2 Segment 3 1 4 Work 1 157 5 2 219 6 3 256 7 TOTAL 632 8 9 10 11 chi-squared Stat 12 df 13 p-value 14 chi-squared Critical
D
E
F
2 44 53 102 199
3 217 264 524 1005
TOTAL 418 536 882 1836
23.0946 4 0.0001 9.4877
2 = 23.0946, p-value = .0001. There is enough evidence to infer that there are differences in employment between the three market segments.
15.110 H0: The two variables (value and segment) are independent H1: The two variables are dependent
506
A B C 1 Contingency Table 2 Segment 3 1 4 Value 1 147 5 2 221 6 3 339 7 TOTAL 707 8 9 10 chi-squared Stat 11 df 12 p-value 13 chi-squared Critical 14
D
E
F
2 135 155 254 544
3 136 160 289 585
TOTAL 418 536 882 1836
4.5122 4 0.3411 9.4877
2 = 4.51, p-value = .3411. There is not enough evidence to infer that there are differences in the definition of value between the three market segments.
15.111 H0: The two variables (breakfast and group) are independent H1: The two variables are dependent
A B C 1 Contingency Table 2 3 Group 4 Breakfast 1 5 1 3 6 2 44 7 3 17 8 4 25 9 TOTAL 89 10 11 12 chi-squared Stat 13 df 14 p-value 15 chi-squared Critical
D
E
F
G
2 15 53 10 77 155
3 29 95 79 77 280
4 222 292 135 77 726
TOTAL 269 484 241 256 1250
206.4984 9 0 16.919
2 = 206.5, p-value = 0. There is enough evidence to infer that there are differences in frequency of healthy breakfasts between the three four segments.
507
Chapter 16 16.1 a The slope coefficient tells us that for additional inch of father’s height the son’s height increases on average by .516. The y-intercept is meaningless. b On average the son will be shorter than his father. c On average the son will be taller than his father.
16.2 a Scatter Diagram 20
Sales
15 10 5 0 0
20
40
60
80
100
120
Advertising
b
Total
xi 23 46 60 54 28 33 25 31 36 88 90 99 613 n
xi2 529 2,116 3,600 2,916 784 1,089 625 961 1,296 7,744 8,100 9,801 39,561
yi 9.6 11.3 12.8 9.8 8.9 12.5 12.0 11.4 12.6 13.7 14.4 15.9 144.9 n
x i = 613
i 1
i 1
x i yi 220.8 519.8 768.0 529.2 249.2 412.5 300.0 353.4 453.6 1205.6 1296.0 1,574.1 7,882.2
n
n
y i = 144.9
yi2 92.16 127.69 163.84 96.04 79.21 156.25 144.00 129.96 158.76 187.69 207.36 252.81 1,795.77
x i2 = 39,561
i 1
x y = 7,882.2 i
i
i 1
n n xi yi n 1 (613 )(144 .9 1 i 1 s xy x i y i i 1 = 12 1 7,882 .2 43.66 n 1 i 1 n 12
509
2 n xi n 2 1 x i2 i 1 = 1 39,561 (613 ) 749 .7 s 2x n 1 i 1 n 12 1 12
b1
s xy
=
s 2x
43 .66 .0582 749 .7
x
x 613 51.08
y
y 144 .9 12.08
i
n
12
i
n
12
b 0 y b1x = 12.08 – (.0582)(51.08) = 9.107 The sample regression line is
ŷ = 9.107 + .0582x The slope tells us that for each additional thousand dollars of advertising sales increase on average by .0582 million. The y-intercept has no practical meaning.
16.3 a
Total
xi 8.5 7.8 7.6 7.5 8.0 8.4 8.8 8.9 8.5 8.0 82.0
x i2 72.25 60.84 57.76 56.25 64.00 70.56 77.44 79.21 72.25 64.00 674.56
yi 115 111 185 201 206 167 155 117 133 150 1,540
n
x = 82.0 i
i 1
n
n
yi2 13,225 12,321 34,225 40,401 42,436 27,889 24,025 13,689 17,689 22,500 248,400
x i yi 977.5 865.8 1,406.0 1,507.5 1,648.0 1,402.8 1,364.0 1,041.3 1,130.5 1,200.0 12,543.4
n
y = 1,540 x = 674.56 x y = 12,543.4 2 i
i
i 1
i 1
i
i
i 1
n n xi yi n (82 .0)(1,540 1 1 i 1 12 ,543 .4 = s xy x i y i i 1 9.40 10 1 10 n 1 i 1 n
2 n xi n 2 1 x i2 i 1 = 1 674 .56 (82 .0) .24 s 2x n 1 i 1 n 10 1 10
510
b1
s xy
=
s 2x
9.40 39 .17 .24
x
x 82.0 8.20
y
y 1,540 154 .0
i
n
10
i
n
10
b 0 y b1x = 154.0 – (–39.17)(8.20) = 475.2 The sample regression line is
ŷ = 475.2 – 39.17x b. The slope coefficient tells us that for each additional 1 percentage point increase in mortgage rates, the number of housing starts decreases on average by 39.17. The y-intercept has no meaning.
16.4a
Scatter Diagram 20 15
Overweight
10 5 0 -5
0
10
20
30
40
50
-10 -15 Television
b
xi 42 34 25 35 37 38 31 33 19 29 38 28
yi 18 6 0 –1 13 14 7 7 –9 8 8 5
x i2 1,764 1,156 625 1,225 1,369 1,444 961 1,089 361 841 1,444 784
511
yi2 324 36 0 1 169 196 49 49 81 64 64 25
x i yi 756 204 0 –35 481 532 217 231 –171 232 304 140
Total
29 36 18 472
3 14 –7 86
n
841 1,296 324 15,524
n
x i = 472
i 1
x i2 = 15,524
i 1
i 1
87 504 –126 3,356
n
n
y i = 86
9 196 49 1,312
x y = 3,356 i
i
i 1
n n xi yi n (472 )(86 1 1 i 1 x i y i i 1 s xy = 15 1 3,356 15 46 .42 n 1 i 1 n
2 n xi n 2 1 x i2 i 1 = 1 15,524 (472 ) 47 .98 s 2x n 1 i 1 n 15 1 15
b1
s xy
=
s 2x
46 .42 .9675 47 .98
x
x 472 31.47
y
y 86 5.73
i
n
15
i
n
15
b 0 y b1x = 5.73 – (.9675)(31.47) = –24.72 The sample regression line is
ŷ = –24.72 + .9675x The slope coefficient indicates that for each additional hour of television weight increases on average by .9675 pounds. The y-intercept is the point at which the regression line hits the y–axis; it has no practical meaning.
16.5a
Total
xi 80 68 78 79 87 74 86 92 77 84 805
yi 20,533 1,439 13,829 21,286 30,985 17,187 30,240 37,596 9,610 28,742 211,447
x i2 6,400 4,624 6,084 6,241 7,569 5,476 7,396 8,464 5,929 7,056 65,239
512
yi2 421,604,089 2,070,721 191,241,241 453,093,796 960,070,225 295,392,969 914,457,600 413,459,100 92,352,100 826,102,564 5,569,844,521
x i yi 1,642,640 97,852 1,078,662 1,681,594 2,695,695 1,271,838 2,600,640 3,458,832 739,970 2,414,328 17,682,051
n
n
n
n
y = 211,447 x = 65,239 x y = 17,682,051
x = 805
2 i
i
i
i 1
i 1
i 1
i
i
i 1
n n xi yi n (805 )( 211,447 1 1 i 1 17 ,682 ,051 = s xy x i y i i 1 73,396 10 10 1 n n 1 i 1
2 n xi n 2 1 x i2 i 1 = 1 65,239 (805 ) 48 .50 s 2x n 1 i 1 n 10 1 10
b1
s xy
=
s 2x
73,396 1,513 48 .50
x
x 805 80.5
y
y 211,447 21,145
i
n
10
i
n
10
b 0 y b1x = 21,145 – (1,513)(80.5) = –100,652 The sample regression line is
ŷ = –100,652 + 1,513x b. For each additional one degree increase in temperature the number of beers sold increases on average by 1,513. The y-intercept is the point at which the regression line hits the y–axis; it has no practical meaning.
513
16.6 a Scatter Diagram 30
Test scores
25 20 15 10 5 0 0
20
40
60
80
Lengths
b b1
s xy s 2x
51 .86 = .2675, b 0 y b1x = 13.80 – .2675(38.00) = 3.635 193 .9
=
Regression line: ŷ = 3.635 + .2675x (Excel: ŷ = 3.636 + .2675x) c b1 = .2675; for each additional second of commercial, the memory test score increases on average by .2675. b0 = 3.64 is the y-intercept.
16.7a b1
s xy
=
s 2x
86 .93 1.465 , b 0 y b1x = 210.4 – 1.465(13.68) =190.4. 59 .32
Regression line: ŷ = 190.4 + 1.465x (Excel: ŷ = 190.4 + 1.465x)
b For each additional floor prices increase on average by $1.465 thousand ($1,465). The yintercept has no practical meaning.
16.8 b1
s xy s 2x
=
9.67 .0899 , b 0 y b1x = 11.55 – .0899(45.49) =7.460. 107 .51
Regression line: ŷ =7.460 + .0899x (Excel: ŷ = 7.462 + .0900x) The slope coefficient tells us that for each additional year of age time increases on average by .0899 minutes. The y-intercept has no meaning.
16.9 a b1
s 2xy s 2x
=
6.44 .1169 , b 0 y b1x = 26.28 – (–.11697)(37.29) = 30.64. 55 .11
514
Regression line: ŷ = 30.64 – .1169x (Excel: ŷ = 30.63 – .1169x) b The slope coefficient indicates that for each additional year of age, the employment period decreases on average by .1169. b0 = 30.63 is the y-intercept.
16.10a b1
s xy s 2x
=
20 .55 .1898 , b 0 y b1x = 14.43 – .1898(37.64) = 7.286. 108 .3
Regression line: ŷ = 7.286 +.1898x (Excel: ŷ = 7.287 +.1897x) b For each additional cigarette the number of days absent from work increases on average by .1898. The y-intercept has no meaning.
16.11 b1
s xy s 2x
=
22 .83 5.347 , b 0 y b1x = 49.22 – 5.347(4.885) = 23.10. 4.270
Regression line: ŷ = 23.10 + 5.347x (Excel: ŷ = 23.11 + 5.347x) For each addition kilometer a house is away from its nearest fire station the percentage damage increases on average by 5.347.
16.12a b1
s xy s 2x
=
30 ,945 44 .97 , b 0 y b1x = 6,465 – 44.97(53.93) = 4040. 688 .2
Regression line: ŷ = 4040 + 44.97x (Excel: ŷ = 4040 + 44.97x) b. For each additional thousand square feet the price increases on average by $44.97 thousand.
16.13 b1
s xy s 2x
=
81.78 .00138 , b 0 y b1 x = 27.73 – (–.00138)(1199) = 29.39. 59,153
Regression line: ŷ = 29.39–.00138x (Excel: 29.39–.00138x) For each additional hour the price decreases on average by .00138 thousand dollars or $1.38.
16.14 b1
s xy s 2x
=
310 .0 64 .05, b 0 y b1x = 762.6 –64.05(4.75) = 458.4. 4.84
Regression line: ŷ = 458.4 + 64.05x (Excel: ŷ = 458.9 + 64.00x) For each additional occupant the electrical use increases on average by 64.05.
16.15 b1
s xy s 2x
=
225 .7 1.959 , b 0 y b1x = 270.3 –1.959(59.42) = 153.9. 115 .2
Regression line: ŷ = 153.9 + 1.959x (Excel: : ŷ = 153.9 + 1.958x) For each additional $1,000 of income the weekly food budget increases on average by $1.96. 515
16.16 a b1
s xy s 2x
=
10 .78 .3039 , b 0 y b1x = 17.20 – (–.3039)(11.33) = 20.64. 35 .47
Regression line: ŷ = 20.64 – .3039x (Excel: ŷ = 20.64 – .3038x) b The slope indicates that for each additional one percentage point increase in the vacancy rate rents on average decrease by $.3039.
16.17 a b1
s xy s 2x
=
6.020 .604 , b 0 y b1x = 59.59 –.604(68.95) =17.94. 9.966
Regression line: ŷ = 17.94 + .604x (Excel: ŷ = 17.93 + .604x) b For each additional inch of height income increases on average by $.604 thousand or $604.
16.18 b1
s xy s 2x
=
.8258 .0514 , b 0 y b1x = 93.89 –.0514(79.47) = 89.81. 16 .07
Regression line: ŷ = 89.81 + .0514x (Excel: ŷ = 89.81 + .0514x) For each additional mark on the test the number of non-defective products increases on average by .0514.
16.19 For each commercial length, the memory test scores are normally distributed with constant variance and a mean that is a linear function of the commercial lengths.
16.20 For each number of years of education incomes are normally distributed with constant variance and a mean that is a linear function of the number of years of education.
16.21 For each number of hours prices are normally distributed with constant variance and a mean that is a linear function of the number of hour.
16.22 b x i
yi
x i2
yi2
x i yi
1 3 4 6 9 8 10 41
1 8 15 33 75 70 95 297
1 9 16 36 81 64 100 307
1 64 225 1089 5625 4900 9025 20,929
1 24 60 198 675 560 950 2,468
Total
n
i 1
n
x i = 41
i 1
n
y i = 297
n
n
x i2 = 307
i 1
516
i 1
y i2 = 20,929
x y = 2,468 i
i 1
i
n n xi yi n (41)( 297 1 1 i 1 2,468 = s xy x i y i i 1 121 .4 7 7 1 n 1 i 1 n
2 n xi n 1 (41) 2 1 i 1 x i2 s 2x = 307 11 .14 n 1 i 1 n 7 1 7
2 n yi n 2 1 y i2 i 1 = 1 20,929 (297 ) 1,388 .0 s 2y n 1 i 1 n 7 1 7
b1
s xy s 2x
=
121 .4 10 .90 11 .14
s 2xy (121 .4) 2 SSE (n 1) s 2y 2 = (7 1)1,388 .0 390 .1 11 .14 s x s
SSE = n2
390 .1 8.83 (Excel: s = 8.85) 72
H 0 : 1 0 H 1 : 1 0 Rejection region: t t / 2, n 2 t .025,5 2.571 or t t / 2, n 2 t .025,5 2.571 s b1
t
s (n 1)s 2x
=
8.83 (7 1)(11 .14 )
1.08
b1 1 10 .90 0 = 10 .09 (Excel: t = 10.07, p–value = 0. There is enough evidence to infer a 1.08 s b1
linear relationship.
517
Scatter Diagram 120
100 80 60 40 20
0 -20
0
2
4
6
8
10
12
There does appear to be a linear relationship.
16.23a
Scatter Diagram 300
250 200 150 100 50
0 -50
0
1
2
3
4
5
6
7
There does appear to be a linear relationship.
b
xi
yi
x i2
yi2
x i yi
Total
3 5 2 6 1 4 21
25 110 9 250 3 71 468
9 25 4 36 1 16 91
625 12100 81 62500 9 5041 80,356
75 550 18 1500 3 284 2,430
518
n
n
x = 21
y = 468 i
i
i 1
i 1
n
x = 91 2 i
i 1
n
n
y = 80,356 x y = 2,430 2 i
i 1
i
i
i 1
n n xi yi n (21)( 468 ) 1 1 i 1 2,430 = s xy x i y i i 1 158 .4 6 6 1 n n 1 i 1
2 n xi n 2 1 x i2 i 1 = 1 91 (21) 3.50 s 2x n 1 i 1 n 6 1 6
2 n yi n 2 1 y i2 i 1 = 1 80,356 (468 ) 8,770 s 2y n 1 i 1 n 6 1 6
b1
s xy s 2x
=
158 .4 45 .26 3.5
s 2xy (158 .4) 2 SSE (n 1) s 2y 2 = (6 1) 8,770 8,006 3.50 s x s
8006 44 .74 (Excel: s = 44.75) 62
SSE = n2
H 0 : 1 0 H 1 : 1 0 Rejection region: t t / 2,n 2 t .025, 4 2.776 or t t / 2,n 2 t .025, 4 2.776 s b1
t
s (n 1)s 2x
=
44 .74 (6 1)(3.50 )
10 .69
45 .26 0 b1 1 4.23 (Excel: t = 4.23, p–value = .0134. There is enough evidence to = 10 .69 s b1
infer a linear relationship.
519
2 n yi n 2 1 y i2 i 1 = 1 1,795 .77 (144 .9) 4.191 16.24 a s 2y n 1 i 1 n 12 1 12
s 2xy (43 .66 ) 2 SSE (n 1) s 2y 2 = (12 1) 4.191 18 .13 749 .7 s x SSE = n2
s
18 .13 1.347 (Excel: s = 1.347) 12 2
H 0 : 1 0
b
H 1 : 1 0 Rejection region: t t / 2, n 2 t.025,10 2.228 or t t / 2, n 2 t.025,10 2.228 s
s b1
t
(n 1)s 2x
=
1.347 (12 1)(749 .7)
.0148
b1 1 .0582 0 3.93 (Excel: t = 3.93, p–value = .0028. There is enough evidence to = .0148 s b1
infer a linear relationship between advertising and sales. c b1 t / 2,n 2 s b1 .0582 2.228(.0148) .0582 .0330 LCL = .0252, UCL = .0912 d R2
s 2xy
(43 .66 ) 2 = .6067 (Excel: R 2 = .6066). 60.67% of the variation in sales is s 2x s 2y (749 .7)( 4.191)
explained by the variation in advertising. e There is evidence of a linear relationship. For each additional dollar of advertising sales increase, on average by .0582.
2 n yi n 2 1 y i2 i 1 = 1 248 ,400 (1,540 ) 1249 16.25 s 2y n 1 i 1 n 10 1 10
R 2
s 2xy
(9.40 ) 2 = .2948 (Excel: R 2 = .2948). 2 2 (. 24 )( 1 , 249 ) s xs y
s 2xy (9.49 ) 2 SSE (n 1) s 2y 2 = (10 1)1,249 7,864 .24 s x
520
s
SSE = n2
7,864 31 .35 (Excel: s = 31.48) 10 2
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2,n 2 t .025,8 2.306 or t t / 2, n 2 t .025,8 2.306 s b1
t
s (n 1)s 2x
=
31 .35 (10 1)(. 24 )
21 .33
39 .17 0 b1 1 1.84 (Excel: t = 1.83, p–value = .1048. There is not enough evidence = .21 .33 s b1
to infer a linear relationship between interest rates and housing starts.
2 n yi n 2 1 y i2 i 1 = 1 1,312 (86 ) 58 .50 16.26 s 2y n 1 i 1 n 15 1 15
2 s 2xy (46 .42 ) 2 SSE (n 1) s y 2 = (15 1) 58 .50 190 .2 47 .98 s x s
SSE = n2
190 .2 3.825 15 2
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2, n 2 t .025,13 2.160 or t t / 2, n 2 t .025,13 2.160 s b1
t
s (n 1)s 2x
=
3.825 (15 1)( 47.98)
.1476
b1 1 .9675 0 6.55 (Excel: t = 6.55, p–value = 0.) There is enough evidence to = .1476 s b1
conclude that there is a linear relationship between hours of television viewing and how overweight the child is.
521
2 n yi n 2 1 y i2 i 1 = 1 5,569 ,844 ,521 (211,447 ) 122 ,095 ,682 16.27 s 2y n 1 i 1 n 10 1 10
s 2xy (73,396 ) 2 SSE (n 1) s 2y 2 = (10 1)122 ,095 ,682 99,216 ,698 48 .50 s x s
99 ,216 ,698 3,522 10 2
SSE = n2
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2,n 2 t .025,8 2.306 or t t / 2, n 2 t .025,8 2.306 s
s b1
t
(n 1)s 2x
=
3522 (10 1)( 48 .50 )
168.6
b1 1 1,513 0 8.97 (Excel: t = 8.98, p–value = 0.) There is evidence of a linear = 168 .6 s b1
relationship between temperature and the number of beers sold.
s 2xy (51 .86 ) 2 16.28 a SSE (n 1) s 2y 2 = (60 1) 47 .96 2,011 193 . 9 s x s
2,011 5.888 (Excel: s = 5.888). Relative to the values of the dependent 60 2
SSE = n2
variable the standard error of estimate appears to be large indicating a weak linear relationship. b R2
s 2xy s 2x s 2y
=
(51 .86 ) 2 .2892 (Excel: R 2 = .2893). (193 .9)( 47 .96 )
H 0 : 1 0
c
H1 : 1 0 Rejection region: t t / 2, n 2 t .025,58 2.000 or t t / 2, n 2 t .025,58 2.000 s b1
t
s (n 1)s 2x
=
5.888 (60 1)(193 .9)
.0550
b1 1 .2675 0 4.86 (Excel: t = 4.86, p–value = 0). There is enough evidence to infer a = .0550 s b1
linear relationship between memory test scores and length of commercial.
522
d b1 t / 2,n 2 s b1 .2675 1.671(.0550) .2675 .0919 LCL = .1756, UCL = .3594
s 2xy (86 .93) 2 16.29 SSE (n 1) s 2y 2 = (50 1) 496 .4 18,081 59 .32 s x s
SSE = n2
18,081 19 .41 (Excel: s = 19.41). Relative to the values of the dependent 50 2
variable the standard error of estimate appears to be large indicating a weak linear relationship. R2
s 2xy
(86.93) 2 = .2566 (Excel: R 2 = .2566). 2 2 s x s y (59.32)( 496 .4)
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2, n 2 t .025, 48 2.009 or t t / 2, n 2 t .025, 48 2.009 s b1
t
s (n 1)s 2x
=
19 .41 (50 1)(59 .32 )
.3600
b1 1 1.465 0 4.07 (Excel: t = 4.07, p–value = .0002). There is evidence of a linear = .3600 s b1
relationship. The relationship however, is weak.
s 2xy (9.67 ) 2 9500 .8 16.30 SSE (n 1) s 2y 2 = (229 1) 42 .54 107 .51 s x s
SSE = n2
9500 .8 6.47 229 2
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2,n 2 t .025,227 1.96 or t t / 2,n 2 t .025,227 1.96 s b1
t
s (n 1)s 2x
=
6.47 ( 229 1)(107 .51)
.0413
b1 1 .0899 0 = 2.17 (Excel: t =2.18, p–value = .0305.) There is evidence of a linear .0413 s b1
relationship between age and time to complete census.
s 2xy (6.44 ) 2 16.31 SSE (n 1) s 2y 2 = (80 1) 4.00 256 .5 55 .11 s x 523
s
SSE = n2
256 .5 1.813 (Excel: s = 1.813). Relative to the values of the dependent 80 2
variable the standard error of estimate appears to be large indicating a weak linear relationship. R2
s 2xy
(6.44 ) 2 = .1881 (Excel: R 2 = .1884). There is a weak linear relationship s 2x s 2y (55 .11)( 4.00 )
between age and number of weeks of employment.
s 2xy (20 .55) 2 16.32 SSE (n 1) s 2y 2 = (231 1)19 .80 3657 108 .3 s x s
SSE = n2
3657 3.996 231 2
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2, n 2 t .025, 229 1.960 or t t / 2,n 2 t .025, 229 1.960 s b1
t
s (n 1)s 2x
=
3.996 (231 1)(108 .3)
.02532
b1 1 .1898 0 7.50 (Excel: t =7.49, p–value = 0.) There is evidence of a positive linear = .02532 s b1
relationship between cigarettes smoked and the number of sick days.
s 2xy (22 .83) 2 16.33 SSE (n 1) s 2y 2 = (85 1) 243 .9 10,234 4.270 s x s
SSE = n2
10 ,234 11 .10 (Excel: s = 11.11). 85 2
H 0 : 1 0
a
H1 : 1 0 Rejection region: t t / 2, n 2 t .025,83 1.990 or t t / 2, n 2 t .025,83 1.990 s b1
t
s (n 1)s 2x
=
11.10 (85 1)( 4.270 )
.5861
b1 1 5.347 0 9.12 (Excel: t =9.12, p–value = 0.) There is evidence of a linear = .5861 s b1
relationship between distance and fire damage. b b1 t / 2,n 2 s b1 5.347 1.988(.5861) 5.347 1.166 LCL = 4.18, UCL = 6.51 524
c R2
s 2xy
(22 .83) 2 = .5005 (Excel: R 2 = .5004). There is a moderately strong linear 2 2 ( 4 . 270 )( 243 . 9 ) s xs y
relationship between distance and fire damage. 2 2 s 2xy s = (40 1)11,918 ,489 (30,945 ) 410 ,554 ,683 SSE ( n 1 ) 16.34 y 688 .2 s 2x
410 ,554 ,683 3,287 (Excel: s = 3,287). There is a weak linear relationship. 40 2
a s
SSE = n2
b
H 0 : 1 0 H1 : 1 0
Rejection region: t t / 2, n 2 t .025,38 2.021 or t t / 2, n 2 t .025,38 2.021 s b1
t
s (n 1)s 2x
=
3,287 (40 1)(688 .2)
20 .06
44 .97 0 b1 1 2.24 (Excel: t = 2.24, p–value = .0309.) There is enough evidence of a = 20 .06 s b1
linear relationship. c R2
s 2xy
(30,945 ) 2 = .1167 (Excel: R 2 = .1168) 11.67% of the variation in s 2x s 2y (688 .2)(11,918 ,489 )
percent damage is explained by the variation in distance to the fire station.
s 2xy (81 .78) 2 16.35 SSE (n 1) s 2y 2 = (60 1) 3.623 207 .1 59,153 s x s
SSE = n2
207 .1 1.890 (Excel: s =1.889 ). 60 2
H 0 : 1 0 H1 : 1 0 Rejection region: t t ,n 2 t .05,58 1.671 s b1
t
s (n 1)s 2x
=
1.890 (60 1)(59,153 )
.001012
.00138 0 b1 1 1.364 (Excel: t = –1.367, p–value = .1769/2 = .0885.) There is not = .001012 s b1
enough evidence to infer that as hours of engine use increase the price decreases.
525
s 2xy (310 .0) 2 16.36 SSE (n 1) s 2y 2 = (200 1) 56,725 7,337 ,056 4.84 s x
s R2
7,337 ,056 191 .1 (Excel: s = 192.5). 200 2
SSE = n2 s 2xy
(310 .0) 2 = .3500 (Excel: R 2 = .3496) 35.00% of the variation in the s 2x s 2y (4.84 )(56,725 )
electricity use is explained by the variation in the number of occupants.
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2, n 2 t .025,198 1.972 or t t / 2, n 2 t .025,198 1.972 s b1
t
s (n 1)s 2x
191 .1
=
(200 1)( 4.84)
6.16
b1 1 64 .05 0 10 .39 (Excel: t =10.32, p–value = 0.) There is enough evidence of a = 6.16 s b1
linear relationship.
16.37 a R 2
s 2xy s 2x s 2y
=
(225 .7) 2 .2461 (Excel: R 2 = .2459) 24.61% of the variation in food (115 .2)(1,797 )
budgets is explained by the variation in household income.
s 2xy (225 .7) 2 b SSE (n 1) s 2y 2 = (150 1)1,797 201,866 115 .2 s x s
SSE = n2
201,866 36 .93 (Excel: s = 36.94 ). 150 2
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2, n 2 t .025,148 1.977 or t t / 2, n 2 t .025,148 1.977 s b1
t
s (n 1)s 2x
=
36 .93 (150 1)(115 .2)
.2819
b1 1 1.959 0 6.949 (Excel: t = 6.95, p–value = 0.) There is evidence of a linear = .2819 s b1
relationship between food budget and household income.
526
s 2xy (10 .78) 2 16.38 SSE (n 1) s 2y 2 = (30 1)11 .24 230 .9 35 .47 s x s
SSE = n2
230 .9 2.872 (Excel: s = 2.873 ). 30 2
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2, n 2 t .025, 28 2.048 or t t / 2, n 2 t .025, 28 2.048 s b1
t
s (n 1)s 2x
=
2.872 (30 1)(35.47 )
.08955
.3039 0 b1 1 3.39 (Excel: t = –3.39, p–value = .0021.) There is sufficient evidence = .08955 s b1
to conclude that office rents and vacancy rates are linearly related.
s 2xy (6.020 ) 2 16.39 SSE (n 1) s 2y 2 = (250 1) 71 .95 17 ,010 9.966 s x s
SSE = n2
17 ,010 8.28 (Excel: s = 8.28). 250 2
H 0 : 1 0 H1 : 1 0 Rejection region: t t , n 2 t .05, 248 1.645 s b1
t
s (n 1)s 2x
=
8.28 (250 1)(9.966
.166
b1 1 .604 0 3.64 (Excel: t = 3.63, p–value = .00034/2 = .00017) There is enough = .166 s b1
evidence to conclude that height and income are positively linearly related.
16.40 a R 2
s 2xy
(.8258 ) 2 = .0331 (Excel: R 2 = .0331) 3.31% of the variation in s 2x s 2y (16 .07 )(1.283 )
percentage of defectives is explained by the variation in aptitude test scores.
2 s 2xy (.8258 ) 2 b. SSE (n 1) s y 2 = (45 1)1.283 54 .58 16 .07 s x s
SSE = n2
54 .58 1.127 (Excel: s = 1.127). 45 2 527
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2, n 2 t .025, 43 2.014 or t t / 2, n 2 t .025, 43 2.014 s
s b1
t
(n 1)s 2x
=
1.127 (45 1)(16 .07 )
.04238
b1 1 .0516 0 1.22 (Excel: t = 1.21, p–value = .2319) There is not enough evidence to = .04238 s b1
conclude that aptitude test scores and percentage of defectives are linearly related.
H0 : 0
16.41
H1 : 0 Rejection region: t t 2, n 2 t .05,58 1.671
r
s xy sxsy
.1767
(59,153 )(3.623 )
n2
tr
81 .78
1 r 2
(.1767 )
60 2 1 (.1767 ) 2
1.367 (Excel: t = –1.367, p–value = .0885) There is not
enough evidence to infer a negative linear relationship.
H0 : 0
16.42
H1 : 0 Rejection region: t t / 2, n 2 t .025,58 2.000 or t t / 2, n 2 t .025,58 2.000
r
s xy sxsy
tr
51 .86
.5378
(193 .9)( 47 .96 )
n2 1 r
2
(.5378 )
60 2 1 (.5378 ) 2
4.86 (Excel: t = 4.86, p–value = 0) This result is identical
to the one produced in Exercise 16.6.
16.43
H0 : 0 H1 : 0
Rejection region: t t / 2, n 2 t .025,148 1.977 or t t / 2, n 2 t .025,148 1.977
528
r
s xy sxsy
.4961
(115 .2)(1,797 )
n2
tr
225 .7
1 r
2
(.4961 )
150 2 1 (.4961 ) 2
6.95 (Excel: t = 6.95, p–value = 0.) There is evidence of a
linear relationship between food budget and household income.
H0 : 0
16.44
H1 : 0 Rejection region: t t , n 2 t .05, 229 1.645
r
s xy sxsy
tr
20 .55
.4438
(108 .3)(19 .80 )
n2 1 r
2
(.4438 )
231 2 1 (.4438 ) 2
7.49 (Excel: t = 7.49, p–value = 0.) There is evidence of a
positive linear relationship between cigarettes smoked and the number of sick days.
16.45
a.
H0:β1 = 0 H1:β1 ≠ 0
529
t = 11.30, p-value = 0. There is enough evidence to conclude that there is a linear relationship between the two variables. b. LCL = 4.59, UCL = 6.53
16.46
a.
H0:β1 = 0 H1:β1 ≠ 0
t = 7.64, p-value = 0. There is enough evidence to infer that there is a linear relationship between grade and price. b. R2 = .4932; 49.32% of the variation in price is explained by the variation in grade.
16.47
530
a.
H0:β1 = 0 H1:β1 ≠ 0
t = 21.19, p-value = 0. There is enough evidence to infer that there is a linear relationship between age and number of days watching national news on television. b. R2 = .5854; 58.54% of the variation in number of days is explained by the variation in age.
16.48
531
H0:β1 = 0 H1:β1 ≠ 0 t = 10.81, p-value = 0. There is enough evidence to infer that age and intention to vote are linearly related.
16.49
a.
H0:β1 = 0 H1:β1 ≠ 0
t = 5.89, p-value = 0. There is not enough evidence to infer that there is a positive linear relationship between temperature and distance. b. R2 = .3166; 31.66% of the variation in distance is explained by the variation in temperature.
16.50
H0:β1 = 0 H1:β1 ≠ 0
t = 5.04, p-value = 0. There is enough evidence to infer that there is a linear relationship between income and position on the question should the government reduce income differences between rich and poor.
532
16.51
LCL = 554.3, UCL = 837.1
16.52
H0:β1 = 0 H1:β1 < 0
t = -8.58, p-value = 0. There is enough evidence to conclude that more educated people watch less television.
16.53
H0:β1 = 0 H1:β1 > 0
a. t = 13.81, p-value = 0. There is enough evidence to infer that more hours of work leads to higher income. b. LCL = 870.4, UCL = 1159.
16.54
H0:β1 = 0 H1:β1 ≠ 0
t = -.239, p-value = .8115. There is not enough evidence to infer a linear relationship between age and hours of work per week.
16.55
H0:β1 = 0 H1:β1 > 0
533
t = 5.67, p-value = 0. There is enough evidence to infer a positive linear relationship between age and hours of watching television per day.
16.56
H0:β1 = 0 H1:β1 > 0
t = 20.26, p-value = 0. There is enough evidence of a positive linear relationship between total family income and the number of earners is the family. LCL = 18,260 UCL = 22,173
16.57
H0:β1 = 0 H1:β1 > 0
t = 3.45, p-value = .00058/2 = .00029. There is enough evidence to conclude that more educated people are more likely to support government action to reduce income differences across the country differences.
16.58
H0:β1 = 0 H1:β1 > 0
t = 27.08, p-value = 0. There is sufficient evidence to conclude that a married person’s years of education are positively linearly related to his or her spouse’s level of education.
534
16.59
H0:β1 = 0 H1:β1 > 0
t = 1.17, p-value = .2439. There is not enough evidence to infer that as people become richer they tend to have more children.
16.60
H0:β1 = 0 H1:β1 > 0
t = 2.60, p-value = .0096. There is enough evidence to conclude that if one spouse works longer hours so does the spouse.
16.61
H0:β1 = 0 H1:β1 > 0
t = 17.99, p-value = 0. There is sufficient evidence to infer a positive linear relationship between years of education and the age when one has his or her first child.
16.62
H0:β1 = 0 H1:β1 < 0
t = -10.04, p-value = 0. There is enough evidence to conclude that more educated people have fewer children.
16.63
H0:β1 = 0 H1:β1 > 0
535
t = 21.94, p-value = 0. There is enough evidence to infer that the amount of education one completes is positively linearly related to his or her father. 16.64
H0:β1 = 0 H1:β1 > 0
t = 23.41, p-value = 0. There is enough evidence to conclude that there is a positive linear relationship between the years of education and the years of education of one’s mother.
16.65
H0:β1 = 0 H1:β1 > 0
t = 2.87, p-value = .0041/2 = .0021. There is sufficient evidence to conclude that harder working Americans are more likely to urge to want government not to reduce income differences.
16.66
H0:β1 = 0 H1:β1 > 0
t = 12.40, p-value = 0. There is enough evidence to conclude that more education leads to higher incomes. b. LCL = 5330, UCL = 7334
16.67
H0:β1 = 0 H1:β1 > 0
536
t= 10.83, p-value = 0. There is enough evidence to conclude that more education increases financial assets. b. LCL = 6181, UCL = 8917 c.
The coefficient of determination is .0889, which means that 8.89% of the variation in financial assets is explained by the variation in education.
16.68
H0:β1 = 0 H1:β1 > 0
t = 11.10, p-value = 0. There is enough evidence to conclude that education and debt are positively linearly related. b. LCL = 11,833, UCL = 16,913 c,
The coefficient of determination is .0931, which means that 9.31% of the variation in debt is explained by the variation in education.
16.69
H0:β1 = 0 H1:β1 < 0
t = -12.22, p-value = 0. There is evidence of a negative linear relationship.
16.70
H0:β1 = 0 H1:β1 > 0
a.t = 6.67, p-value = 0. There is enough evidence to infer a positive linear relationship. b. LCL = 815.6, UCL = 1495.
537
16.71
H0:β1 = 0 H1:β1 < 0
t = -9.12, p-value = 0. There is enough evidence to infer that age and amount spent on food away from home are negatively related.
16.72
H0:β1 = 0 H1:β1 < 0
t = -9.08, p-value = 0. There is enough evidence of a negative linear relationship.
16.73
H0:β1 = 0 H1:β1 > 0
t = 8.90, p-value = 0. There is enough evidence to conclude that as one grows older one increases unrealized capital gains.
16.74
a.
H0:β1 = 0 H1:β1 > 0
t = 18.01, p-value = 0. There is enough evidence of a positive linear relationship. b. LCL = 1324, UCL = 1648
16.75
538
a.
H0:β1 = 0 H1:β1 > 0
t = 2.89, p-value = .0039/2 = .0040. There is enough evidence of a positive linear relationship. b.LCL = 58.52, UCL = 306.6.
16.76 The prediction interval provides a prediction for a value of y. The confidence interval estimator of the expected value of y is an estimator of the population mean for a given x.
16.77 Yes, because
1
2 1 (x g x) n ( n 1)s 2x
2 1 (x g x) n ( n 1)s 2x
16.78 ŷ b 0 b1x g 9.107 .0582 (80 ) 13 .76 Prediction interval: ŷ t / 2,n 2 s 1
= 13 .76 1.812 (1.347 ) 1
2 1 (x g x) (where t / 2,n 2 t .05,10 1.812 ) n (n 1)s 2x
1 (80 51 .08 ) 2 13 .76 2.657 12 (12 1)( 749 .7)
Lower prediction limit = 11.10, Upper prediction limit = 16.42 (Excel: 11.10, 16.41) 16.79 ŷ b 0 b1x g 475 .2 39 .17 (7) 201 .0 Confidence interval estimate: ŷ t / 2,n 2 s
201 .0 1.860 (31 .35)
2 1 (x g x) (where t / 2, n 2 t .05,8 1.860 ) n (n 1)s 2x
1 (7 8.20 ) 2 201 .0 51 .06 10 (10 1)(. 24 )
LCL = 149.94, UCL = 252.06 (Excel: 149.75, 252.25) 16.80 ŷ b 0 b1x g 24 .72 .9675 (35) 9.143 a Prediction interval: ŷ t / 2,n 2 s 1
= 9.143 1.771(3.825 ) 1
2 1 (x g x) (where t / 2,n 2 t .05,13 1.771) n (n 1)s 2x
1 (35 31 .47 ) 2 9.143 7.057 15 (15 1)( 47 .98 )
Lower prediction limit =2.086, Upper prediction limit = 16.200 (Excel: 2.095, 16.209) b Confidence interval estimate: ŷ t / 2,n 2 s
2 1 (x g x) n (n 1)s 2x
539
9.143 1.771(3.825 )
1 (35 31 .47 ) 2 9.143 1.977 15 (15 1)( 47 .98 )
LCL = 7.166, UCL = 11.120 (Excel: 7.174, 11.130) 16.81 ŷ b 0 b1x g 100 ,652 1,513 (75) 12,823 Prediction interval: ŷ t / 2,n 2 s 1
= 12 ,823 1.860 (3,522 ) 1
2 1 (x g x) (where t / 2, n 2 t .05,8 1.860 ) n (n 1)s 2x
1 (75 80 .5) 2 12 ,823 7,084 10 (10 1)( 48 .50 )
Lower prediction limit = 5,739, Upper prediction limit = 19,907 (Excel: 5,740, 19,902) 16.82 ŷ b 0 b1 x g 3.636 + .2675(30) = 11.61 a Prediction interval: ŷ t / 2,n 2 s 1
= 11 .61 2.000 (5.888 ) 1
2 1 (x g x) (where t / 2, n 2 t .025,58 2.000 ) n (n 1)s 2x
1 (30 38 ) 2 11 .61 11 .91 60 (60 1)(193 .9)
Lower prediction limit = -.3 (changed to 0), Upper prediction limit = 23.52 (Excel: -.26, 23.58) b Confidence interval estimate: ŷ t / 2,n 2 s
11 .61 2.000 (5.888 )
2 1 (x g x) n (n 1)s 2x
1 (30 38 ) 2 11 .61 1.757 60 (60 1)(193 .9)
LCL = 9.85, UCL = 13.37 (Excel: 9.90, 13.42) 16.83 ŷ b 0 b1 x g 190.4 + 1.465(20) = 219.7 a Prediction interval: ŷ t / 2,n 2 s
= 219 .7 2.009 (19 .41) 1
2 1 (x g x) (where t / 2, n 2 t .025, 48 2.009 ) 1 n (n 1)s 2x
1 ( 20 13 .68 ) 2 219 .7 39 .65 50 (50 1)( 59 .32 )
Lower prediction limit = 180.1, Upper prediction limit = 259.4 (Excel: 180.0, 259.4) b ŷ b 0 b1 x g 190.4 + 1.465(15) = 212.4
540
Confidence interval estimate: ŷ t / 2,n 2 s
212 .4 2.678 (19 .41)
2 1 (x g x) (where t / 2, n 2 t .005, 48 2.678 ) n (n 1)s 2x
1 (15 13 .68 ) 2 212 .4 7.460 50 (50 1)( 59 .32 )
LCL = 204.9, UCL = 219.9 (Excel: 204.9, 219.8) 16.84 ŷ b 0 b1 x g 7.46 + .0899(40) = 11.06 Confidence interval estimate: ŷ t / 2,n 2 s
11 .06 1.645 (6.47 )
2 1 (x g x) (where t / 2,n 2 t .05,227 1.645 ) n (n 1)s 2x
1 ( 40 45 .49 ) 2 11 .06 .80 229 ( 229 1)(107 .51)
LCL = 10.26, UCL = 11.86 (Excel: 10.26, 11.86) 16.85 ŷ b 0 b1 x g 30.64 – .1169(22) = 28.07 Prediction interval: ŷ t / 2,n 2 s 1
= 28 .07 1.990 (1.813 ) 1
2 1 (x g x) (where t / 2,n 2 t .025,78 1.990 ) n (n 1)s 2x
1 ( 22 37 .29 ) 2 28 .07 3.73 80 (80 1)( 55 .11)
Lower prediction limit = 24.34, Upper prediction limit = 31.80 (Excel: 24.33, 31.79) 16.86 ŷ b 0 b1 x g 7.286 + .1898(40) = 14.88 Prediction interval: ŷ t / 2,n 2 s 1
= 14 .88 1.960 (3.996 ) 1
2 1 (x g x) (where t / 2,n 2 t .025, 229 1.960 ) n (n 1)s 2x
1 ( 40 37 .64 ) 2 14 .88 7.85 231 ( 231 1)(108 .3)
Lower prediction limit = 7.03, Upper prediction limit = 22.73 (Excel: 6.98, 22.77) 16.87 ŷ b 0 b1 x g 23.10 + 5.347(8) = 65.88 a Prediction interval: ŷ t / 2,n 2 s
= 65 .88 1.990 (11 .10 ) 1
2 1 (x g x) (where t / 2, n 2 t .025,83 1.990 ) 1 n (n 1)s 2x
1 (8 4.885 ) 2 65 .88 22 .51 85 (85 1)( 4.270 )
541
Lower prediction limit = 43.37, Upper prediction limit = 88.39 (Excel:43.37, 88.39) b ŷ b 0 b1 x g 23.10 + 5.347(5) = 49.84 Confidence interval estimate: ŷ t / 2,n 2 s
2 1 (x g x) n (n 1)s 2x
1 (5 4.885 ) 2 49 .84 2.40 85 (85 1)( 4.270 )
49 .84 1.990 (11 .10 )
LCL = 47.44, UCL = 52.24 (Excel: 47.44, 52.24) 16.88 ŷ b 0 b1 x g 4,040 + 44.97(60) = 6,738 Confidence interval estimate: ŷ t / 2,n 2 s
6,738 2.021(3,287 )
2 1 (x g x) (where t / 2,n 2 t .025,38 2.021) n (n 1)s 2x
1 (60 53 .93) 2 6,738 1,079 40 ( 40 1)( 688 .2)
LCL = 5,659, UCL = 7,817 (Excel: LCL = 5,657, UCL = 7,818) 16.89 ŷ b 0 b1 x g 29.39 – .00138(400) = 28.84 Prediction interval: ŷ t / 2,n 2 s
2 1 (x g x) (where t / 2, n 2 t .005,58 2.660 ) 1 n (n 1)s 2x
= 28 .84 2.660 (1.889 ) 1
1 ( 400 1,199 ) 2 28 .84 5.50 60 (60 1)( 59 ,153 )
Lower prediction limit = 23.34, Upper prediction limit = 34.34 (Excel: 23.33, 34.35) 16.90 ŷ b 0 b1 x g 458.4 + 64.05(4) = 714.6 Confidence interval estimate: ŷ t / 2,n 2 s
714 .6 1.653 (191 .1)
2 1 (x g x) (where t / 2, n 2 t .05,198 1.653 ) n (n 1)s 2x
1 ( 4 4.75) 2 714 .6 23 .6 200 (200 1)( 4.84 )
LCL = 691.0, UCL = 738.2 (Excel: 691.2, 738.7) 16.91 ŷ b 0 b1 x g 153.9 + 1.959(60) = 271.4 Prediction interval: ŷ t / 2,n 2 s 1
2 1 (x g x) (where t / 2, n 2 t .05,148 1.656 ) n (n 1)s 2x
542
= 271 .4 1.656 (36 .93) 1
1 (60 59 .42 ) 2 271 .4 61 .4 150 (150 1)(115 .2)
Lower prediction limit = 210.0, Upper prediction limit = 332.8 (Excel: 210.0, 332.7) 16.92 ŷ b 0 b1 x g 20.64 – .3039(8) = 18.21 Prediction interval: ŷ t / 2,n 2 s 1
= 18 .21 2.048 (2.872 ) 1
2 1 (x g x) (where t / 2, n 2 t .025, 28 2.048 ) n (n 1)s 2x
1 (8 11 .33) 2 18 .21 6.01 30 (30 1)( 35 .47 )
Lower prediction limit = 12.20, Upper prediction limit = 24.22 (Excel: 12.19, 24.22) 16.93 a ŷ b 0 b1 x g 17.94 + .604(74) = 62.64 Confidence interval: ŷ t / 2,n 2 s
= 62 .64 1.96 (8.28
2 1 (x g x) (where t / 2, n 2 t .025, 248 1.96 ) n (n 1)s 2x
1 (74 68 .95) 2 62 .64 1.94 250 ( 250 1)( 9.966 )
Lower confidence limit = 60.70, Upper confidence limit = 64.58 (Excel: 60.70, 64.59) b ŷ b 0 b1 x g 17.94 + .604(68) = 59.01 Prediction interval: ŷ t / 2,n 2 s 1
= 59 .01 1.96 (8.28 1
2 1 (x g x) (where t / 2, n 2 t .025, 248 1.96 ) n (n 1)s 2x
1 (68 68 .95) 2 59 .01 16 .26 250 ( 250 1)( 9.966 )
Lower prediction limit = 42.75, Upper prediction limit = 75.27 (Excel: 42.67, 75.36) 16.94 ŷ b 0 b1 x g 89.81 + .0514(80) = 93.92 Confidence interval estimate: ŷ t / 2,n 2 s
93 .92 2.014 (1.127 )
2 1 (x g x) where t / 2,n 2 t .025,43 2.014 n (n 1)s 2x
1 (80 79 .47 ) 2 93 .92 .34 45 (45 1)(16 .07 )
LCL = 93.58, UCL = 94.26 (Excel: 93.57, 94.26)
16.95 543
LCL = 74.163, UCL = 80.057.
16.96
Lower prediction interval = -235.426, upper predication interval = 2070.219
16.97
LCL = 1.892, UCL = 2.103
16.98
544
Lower prediction interval = 2.468, upper predication interval = 8.604
16.99
Lower prediction interval = 211.350, upper predication interval = 219.702.
16.100 In all cases the linear relationship is far too weak to produce accurate predictions.
16.101
Lower prediction interval = 7.679, upper predication interval = 19.723. LCL = 13.578, UCL = 13.823
16.102
545
Lower prediction interval = -1.680, upper predication interval = 8.262. LCL = 3.150, UCL = 3.432.
16.103
Lower prediction interval = -28,401, upper predication interval = 119,175. LCL = 43,322, UCL = 47,387.
16.104
Lower prediction interval = 13.64, upper predication interval = 69.99 LCL = 40.69, UCL = 42.93
16.105
546
Lower prediction interval = -1.722, upper predication interval = 8.346. LCL = 3.144, UCL = 3.480.
16.106
Lower prediction interval = 26,400, upper predication interval = 205,033. LCL = 110,196, UCL = 121,237.
16.107
Lower prediction interval = -33,469, upper predication interval = 123,084. LCL = 42,645, UCL = 46,970.
16.108
Lower prediction interval = -1.021, upper predication interval = 5.224. LCL = 2.019, UCL = 2.184.
547
16.109
Lower prediction interval = 9.72, upper predication interval = 20.47. LCL = 14.94, UCL = 15.25.
16.110
Lower prediction interval = 11.39, upper predication interval = 22.13. LCL = 16.50, UCL = 17.03.
16.111
Lower prediction interval = -.06, upper predication interval = 7.84. LCL = 3.75, UCL = 4.03.
16.112 The relationship between the dependent and independent variables is too weak to provide accurate predictions.
16.113 a x i –5 –2 0 3
yi 15 9 7 6
x i2 25 4 0 9
548
yi2 225 81 49 36
x i yi –75 –18 0 18
Total
4 7 7
4 1 42 n
16 49 103 n
n
xi = 7
i 1
16 1 408
y i = 42
n
x y = –52
x i2 = 103
i
i 1
i 1
16 7 –52 i
i 1
n n x yi n i (7)( 42 1 1 i 1 x i y i i 1 s xy = 6 1 52 6 20 .20 n n 1 i 1
2 n xi n 2 1 x i2 i 1 = 1 103 (7) 18 .97 s 2x n 1 i 1 n 6 1 6
b1
s xy s 2x
20 .2 1.065 18 .97
=
x
x 7 1.167
y
y 42 7.000
i
n
6
i
n
6
b 0 y b1x = 7.000 – (–1.065)(1.167) = 8.253 The sample regression line is
ŷ = 8.253 – 1.065x 2 n yi n 1 (42 ) 2 1 i 1 2 2 yi = b, c, &d s y 22 .80 408 n 1 i 1 n 6 1 6
s 2xy (20 .20 ) 2 SSE (n 1) s 2y 2 = (6 1) 22 .80 6.451 18 .97 s x s
xi –5 –2 0 3
SSE = n2
6.451 1.270 (Excel: s = 1.268) 62
yi 15 9 7 6
e i y i ŷ i 1.42 –1.38 –1.253 .942
ŷ i 8.253 1.065 x 13.58 10.38 8.253 5.058
549
ei / s 1.118 –1.087 –.987 .742
4
4
7 1 There are no outliers.
3.993
.007
.0055
.798
.202
.159
16.114
xi 23 46 60 54 28 33 25 31 36 88 90 99
yi 9.6 11.3 12.8 9.8 8.9 12.5 12.0 11.4 12.6 13.7 14.4 15.9
ŷ i 9.107 .0582 x 10.45 11.78 12.60 12.25 10.74 11.03 10.56 10.91 11.20 14.23 14.35 14.87
16.115
xi 8.5 7.8 7.6 7.5 8.0 8.4 8.8 8.9 8.5 8.0
yi 115 111 185 201 206 167 155 117 133 150
ŷ = 475.2 – 39.17x 142.3 169.7 177.5 181.4 161.8 146.2 130.5 126.6 142.3 161.8
16.116 a & b
xi 42 34 25 35 37 38 31 33 19 29 38 28 29 36 18
e i y i ŷ i –.85 –.48 .20 –2.45 –1.84 1.47 1.44 .49 1.40 –.53 .06 1.03 e i y i ŷ i –27.3 –58.7 7.5 19.6 44.2 20.8 24.5 –9.6 –9.3 –11.8
ŷ = – 24.72 + .9675x 15.92 8.18 –.53 9.14 11.08 12.05 5.27 7.21 –6.34 3.34 12.05 2.37 3.34 10.11 –7.31
yi 18 6 0 –1 13 14 7 7 –9 8 8 5 3 14 –7
c
550
e i y i ŷ i 2.09 –2.18 .53 –10.14 1.92 1.96 1.73 –.21 –2.66 4.66 –4.05 2.63 –.34 3.89 .31
Plot of Residuals vs Predicted 6 4
Residuals
2 0 -10
-5
-2 0
5
10
15
20
-4 -6 -8 -10 -12 Predicted
16.117
xi 80 68 78 79 87 74 86 92 77 84
ŷ = –100,652 + 1,513x 20,388 2,232 17,362 18,875 30,979 11,310 29,466 38,544 15,849 26,440
yi 20,533 1,439 13,829 21,286 30,985 17,187 30,240 37,596 9,610 28,742
e i y i ŷ i 145 –793 –3,533 2,411 6 5,877 774 –948 –6,239 2,302
The histograms drawn below are of the standardized residuals, which make it easier to see whether the shape is extremely nonnormal. It also makes it easier to identify outliers. The shape of the resulting histogram is identical to the histogram of the residuals using the equivalent class limits.
16.118 b & c
Frequency
Histogram 40 20 0 -3
-2
-1
0
1
2
3
Standardized Residuals
Because the histogram is approximately bell shaped the errors appear to be normally distributed. There are two residuals whose absolute value exceeds 2.0.
551
d Plot of Residuals vs Predicted 20 15
Residuals
10 5 0 -5 0
5
10
15
20
25
-10 -15 -20 Predicted
There is no indication of heteroscedasticity.
16.119 a
Frequency
Histogram 40 20 0 -3
-2
0
1
1
2
3
Standardized Residuals
The histogram is not bell shaped. However, residuals do not appear to be extremely nonnormal. c
552
Residuals
Plot of Residuals vs Predicted 60 50 40 30 20 10 0 -10180 -20 -30 -40 -50
190
200
210
220
230
240
3
4
Predicted
There is no clear indication of heteroscedasticity.
16.120
Frequency
Histogram 100 50 0 -3
-2
-1
0
1
2
Standard Residuals The error variable appears to be normally distributed.
Plot of Residuals versus Predicted 25 20 15
Residuals
10
5 0 -5 8
9
10
11
12
-10 -15 -20
Predicted
553
13
14
15
The variance of the error variable is constant.
16.121 b & c
Frequency
Histogram 40 20 0 -3
-2
-1
0
1
2
3
Standard Residuals
The error variable appears to be normally distributed. There are three observations whose standardized residuals are greater than 2.0.
d Plot of Residuals vs Predicted 6
Residuals
4 2 0 -2
24
25
26
27
28
-4 -6 Predicted
554
29
The variance of the error variable is constant.
16.122
Frequency
Histogram 100 50 0 -3
-2
-1
0
1
2
More
Standard Residuals
The error variable appears to be normally distributed. Plot of Residuals vs Predicted 15 10
Residuals
5 0 -5
8
12
16
20
-10 -15 Predicted
The variance of the error variable is constant.
16.123b
Frequency
Histogram 40 20 0 -3
-2
-1
0
1
Standard Residuals
The errors appear to be normally distributed. c There are no outliers. d
555
2
3
Plot of Residuals vs Predicted 40 30 Residuals
20 10 0 -10
20
30
40
50
60
70
80
-20 -30 Predicted
The variance of the error variable appears to decrease somewhat as the predicted values increase. However, the effect is not large enough to be a problem.
16.124
Frequency
Histogram 20 10 0 -3
-2
-1
0
1
2
3
8000
9000
Standard Residuals
The error variable appears to be normally distributed. Plot of Residuals vs Predicted 8000 6000
Residuals
4000 2000 0 -20004000
5000
6000
7000
-4000 -6000 -8000 Predicted
There is no clear sign of heteroscedasticity. 556
16.125
Frequency
Histogram 40 20 0 -3
-2
-1
0
1
2
3
Standard Residuals
The error variable appears to be normally distributed. Plot of Residuals vs Predicted 8 6
Residuals
4 2 0 -2
27
27.5
28
28.5
-4 -6 Predicted
The variance of the error variable is constant.
16.126
Frequency
Histogram 100 0 -3
-2
-1
0
1
Standard Residuals
The error variable appears to be normal.
557
2
3
Plot of Residuals vs Predicted 600
400 200 0 -200
400
500
600
700
800
900
-400
-600 -800
The variance of the error variable is constant.
16.127
Frequency
Histogram 100 50 0 -3
-2
-1
0
1
Standard Residuals
The error variable appears to be normal.
558
2
3
1000
Plot of Residuals vs Predicted 150
Residuals
100 50 0 200 -50
250
300
350
-100 -150 Predicted
The variance of the error variable is constant.
16.128
Frequency
Histogram 20 10 0 -3
-2
-1
0
1
2
3
Standard Residuals
The error variable appears to be normal.
Plot of Residuals vs Predicted 6
Residuals
4 2 0 -2
14
18
22
-4 -6 Predicted
The variance of the error variable is constant.
559
16.129
Frequency
Histogram 100 0 -3
-2
-1
0
1
2
3
Standard Residuals
The error variable appears to be normal. 25
Residuals vs Predicted
20 15 10 5 0 -5 54
56
58
60
62
64
-10 -15 -20 -25
The variance of the error variable is constant.
16.130
Frequency
Histogram 20 10 0 -3
-2
-1
0
1
Standard Residuals
The error variable appears to be normal.
560
2
3
66
Plot of Residuals vs Predicted 3 2
Residuals
1 0 93.2 -1
93.4
93.6
93.8
94
94.2
94.4
-2 -3 -4 Predicted
The variance of the error variable is constant.
16.131 a b1
s xy s 2x
=
74 .02 21 .33, b 0 y b1x = 384.81 – 21.33(4.12) = 296.93 3.47
Regression line: ŷ = 296.93 + 21.33x (Excel: ŷ = 296.92 + 21.36x) b On average each additional ad generates 21.33 customers.
2 s 2xy (74 .02 ) 2 c SSE (n 1) s y 2 = (26 1)18,552 424 ,326 3.47 s x s
SSE = n2
424 ,326 132 .97 (Excel: s = 132.96). 26 2
H 0 : 1 0 H1 : 1 0 Rejection region: t t , n 2 t .05, 24 1.711 s b1
t
s (n 1)s 2x
=
132 .97 (26 1)(3.47 )
14 .28
21 .33 0 b1 1 1.49 (Excel: t = 1.50, p–value = .1479/2 = .0740.) There is not enough = 14 .28 s b1
evidence to conclude that the larger the number of ads the larger the number of customers. d R2
s 2xy
(74 .02 ) 2 = .0851 (Excel: R 2 = .0852). There is a weak linear relationship s 2x s 2y (3.47 )(18,552 )
between the number of ads and the number of customers. e The linear relationship is too weak for the model to produce predictions. 561
16.132 a b1
s xy s 2x
=
936 .82 2.47 b 0 y b1x = 395.21 – 2.47(113.35) = 115.24. 378 .77
Regression line: ŷ = 115.24 + 2.47x (Excel: ŷ = 114.85 + 2.47x) b b1 = 2.47; for each additional month of age, repair costs increase on average by $2.47.
b0 = 114.85 is the y-intercept. c R2
s 2xy
(936 .82)2 = .5659 (Excel: R 2 = .5659) 56.59% of the variation in repair s 2x s 2y (378 .77 )( 4,094 .79)
costs s explained by the variation in ages.
s 2xy (936 .82) 2 d SSE (n 1) s 2y 2 = (20 1) 4,094 .79 33,777 378 .77 s x s
33,777 43 .32 (Excel: s = 43.32). 20 2
SSE = n2
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2, n 2 t .025,18 2.101 or t t / 2, n 2 t .025,18 2.101 s b1
t
s (n 1)s 2x
=
43 .32
.511
(20 1)(378 .77 )
2.47 0 b1 1 4.84 (Excel: t = 4.84, p–value = .0001. There is enough evidence to infer = .511 s b1
that repair costs and age are linearly related. e ŷ b0 b1xg 115.24 2.47(120) 411.64 Prediction interval: ŷ t / 2,n 2 s 1
= 411 .64 2.101(43 .32) 1
2 1 (x g x) (where t / 2, n 2 t.025,18 2.101) n (n 1)s 2x
1 (120 113 .35)2 411 .64 93 .54 20 (20 1)(378 .77 )
Lower prediction limit = 318.1, upper prediction limit = 505.2 (Excel: 318.1, 505.2)
16.133 a b1
s xy s 2x
=
2538 .123 b 0 y b1x = 318.60 –.123(300) = 281.7. 20,690
Regression line: ŷ = 281.7 + .123x (Excel: ŷ = 281.8 + .123x)
562
The slope is .123, which tells us that for each additional unit of fertilizer, corn yield increases on average by .123. The y-intercept is 281.7, which has no real meaning.
s 2xy (2,538 ) 2 b SSE (n 1) s 2y 2 = (30 1) 5,230 142 ,641 20,690 s x s
142 ,641 71 .37 (Excel: s = 71.38). 30 2
SSE = n2
H 0 : 1 0 H1 : 1 0 Rejection region: t t / 2, n 2 t.025, 28 2.048 or t t / 2, n 2 t.025, 28 2.048 s b1
t
s (n 1)s 2x
=
71 .37 (30 1)( 20,690 )
.0921
b1 1 .123 0 1.34 (Excel: t = 1.33, p–value = .1938. There is not enough evidence to = .0921 s b1
infer a linear relationship between amount of fertilizer and corn yield. c R2
s 2xy s 2x s 2y
=
(2,538 ) 2 .0595 ( Excel: R 2 = .0595) 5.95% of the variation in corn yield (20,690 )(5,230 )
is explained by the variation in amount of fertilizer. d The model is too poor to be used to predict. 16.134a H 0 : 0
H1 : 0
A 1 Correlation 2 3 Tar and Nicotine 4 Pearson Coefficient of Correlation 5 t Stat 6 df 7 P(T<=t) one tail 8 t Critical one tail 9 P(T<=t) two tail 10 t Critical two tail
B
0.9766 21.78 23 0 1.7139 0 2.0687
r = .9766, t = 21.78, p–value = 0. There is sufficient evidence to infer that levels of tar and nicotine are linearly related.
b
H0 : 0
563
H1 : 0
A 1 Correlation 2 3 Nicotine and CO 4 Pearson Coefficient of Correlation 5 t Stat 6 df 7 P(T<=t) one tail 8 t Critical one tail 9 P(T<=t) two tail 10 t Critical two tail
B
0.9259 11.76 23 0 1.7139 0 2.0687
r = .9259, t = 11.76, p–value = 0. There is sufficient evidence to infer that levels of nicotine and carbon monoxide are linearly related.
H0 : 0
16.135
H1 : 0 Rejection region: t t , n 2 t .05, 428 1.645 or
r
s xy s xs y
tr
255 ,877
(99 .11)( 2,152 ,602 ,614 )
n2 1 r
2
(.5540 )
430 2 1 (.5540 ) 2
.5540 (Excel: .5540)
13 .77 (Excel: t = 13.77, p–value = 0). There is enough
evidence of a positive linear relationship. The theory appears to be valid.
H0 : 0
16.136
H1 : 0 Rejection region: t t / 2, n 2 t .025, 48 2.009 or t t / 2, n 2 t .025, 48 2.009
r
s xy sxsy
tr
13 .08 (90 .97 )(11 .84 )
n2 1 r
2
(.3985 )
.3985 (Excel: .3984)
50 2 1 (.3985 ) 2
3.01 (Excel: t = 3.01, p–value = .0042). There is enough
evidence of a linear relationship. The theory appears to be valid.
16.137
H0 : 0 H1 : 0
564
A B C 1 Correlation 2 3 Fund and Gold 4 Pearson Coefficient of Correlation 5 t Stat 6 df 7 P(T<=t) one tail 8 t Critical one tail 9 P(T<=t) two tail 10 t Critical two tail
D
0.7929 6.63 26 0 1.7056 0 2.0555
t = 6.63, p-value = 0. There is enough evidence to conclude that there is a positive linear relationship between the value of the fund and the price of gold.
16.138
H0 : 0 H1 : 0
A B C 1 Correlation 2 3 Time and Sales 4 Pearson Coefficient of Correlation 5 t Stat 6 df 7 P(T<=t) one tail 8 t Critical one tail 9 P(T<=t) two tail 10 t Critical two tail
D
0.2791 1.67 33 0.0522 1.6924 0.1044 2.0345
t = 1.67, p-value = .0522. There is not enough evidence to infer that when the times between movies increase so do sales.
565
16.139 a 195,0 190,0 185,0
Height
180,0 175,0 170,0 165,0 160,0 155,0 150,0 4,0
5,0
6,0
7,0
8,0
9,0
10,0
Index Finger Length
b.
H0 : 0 H1 : 0
t = 7.60, p-value = 0. There is enough evidence to infer that index finger length and height are linearly related.
566
c.
Lower prediction limit = 150.5, Upper prediction limit = 182.5
16.140
H0 : 0 H1 : 0
A 1 Correlation 2 3 Hours and GPA 4 Pearson Coefficient of Correlation 5 t Stat 6 df 7 P(T<=t) one tail 8 t Critical one tail 9 P(T<=t) two tail 10 t Critical two tail
B
-0.5748 -9.88 198 0 1.6526 0 1.9720
t = –9.88, p–value = 0. There is enough evidence of a linear relationship between time spent at part time jobs and grade point average.
16.141
H0 : 0 H1 : 0
567
A 1 Correlation 2 3 Times and Amount 4 Pearson Coefficient of Correlation 5 t Stat 6 df 7 P(T<=t) one tail 8 t Critical one tail 9 P(T<=t) two tail 10 t Critical two tail
B
0.7976 29.51 498 0 1.6479 0 1.9647
t = 29.51, p–value = 0. There is overwhelming evidence of a linear relationship between listening times and amounts spent on music.
16.142 The intervals between values must be constant, which is arguable.
16.143 H0:β1 = 0 H1:β1 > 0
t = 5.34, p-value = 0. There is enough evidence to infer that older people are more likely to be conservative. 16.144 H0:β1 = 0 H1:β1 > 0
t = 2.30, p-value = .0218/2 = .0109. There is enough evidence to infer that as income increases people are more likely to be conservative. 16.145 H0:β1 = 0 H1:β1 < 0
t = -5.05, p-value = 0. There is enough evidence to infer that as education increases people are more likely to be liberal.
568
16.146 H0:β1 = 0 H1:β1 > 0
t = 15.67, p-value = 0. 16.147 H0:β1 = 0 H1:β1 > 0
t = 12.43, p-value = 0. 16.148 H0:β1 = 0 H1:β1 > 0
t = 15.51, p-value = 0. 16.149 H0:β1 = 0 H1:β1 > 0
t = 15.60, p-value = 0.
16.150 In general liberals want government to act and conservatives want no government action. However, the coefficients of determination indicate weak relationships.
Case 16.1 a
569
A B C D E F 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.9896 5 R Square 0.9794 6 Adjusted R Square 0.9787 7 Standard Error 355.6 8 Observations 32 9 10 ANOVA 11 df SS MS F Significance F 12 Regression 1 180,066,717 180,066,717 1424 0.0000 13 Residual 30 3,793,289 126,443 14 Total 31 183,860,006 15 16 Coefficients Standard Error t Stat P-value 17 Intercept 16.23 114.70 0.14 0.8884 18 A-Park 0.693 0.0184 37.74 0.0000
The regression equation is ŷ = 16.23 + .693x. This equation was used to predict museum attendance when it was closed (observations 33 to 179). The sum of the predictions is 785,009.
b
570
A B C D E F 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.9909 4 Multiple R 0.9819 5 R Square 0.9811 6 Adjusted R Square 572.9 7 Standard Error 8 Observations 26 9 10 ANOVA 11 df SS MS F Significance F 12 Regression 1 426,295,375 426,295,375 1299 0.0000 13 Residual 24 7,875,875 328,161 14 Total 25 434,171,250 15 16 Coefficients Standard Error t Stat P-value 17 Intercept 459.5 295.4 1.56 0.1330 18 A-Park 0.970 0.0269 36.04 0.0000
The regression equation is ŷ = 459.5 + .970x. This equation was used to predict museum attendance when it was closed (observations 33 to 179). The sum of the predictions is 1,162,994. c The predicted lost revenue should be based on the regression using the first 32 weeks. Multiply 785,009 by the price of tickets and subtract fixed costs to produce the amount the insurance company should pay the museum.
Case 16.2 Regression using the best 6 OACs:
A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.4883 5 R Square 0.2385 6 Adjusted R Square 0.2363 7 Standard Error 0.8295 8 Observations 363 9 10 ANOVA 11 df SS 12 Regression 1 77.78 13 Residual 361 248.39 14 Total 362 326.17 15 16 Coefficients Standard Error 17 Intercept -5.35 1.31 18 Best-6 0.15 0.01
571
D
E
MS 77.78 0.69
F Significance F 113.04 0.0000
t Stat P-value -4.08 0.0001 10.63 0.0000
F
t = 10.63, p–value = 0. There is evidence of a linear relationship between the average of the best 6 OACs and university GPA. R 2 = .2385, s = .8295
Regression using the best 4 OACs plus English and calculus: A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.5924 5 R Square 0.3509 6 Adjusted R Square 0.3491 7 Standard Error 0.7658 8 Observations 363 9 10 ANOVA 11 df SS 12 Regression 1 114.46 13 Residual 361 211.71 14 Total 362 326.17 15 16 Coefficients Standard Error 17 Intercept -3.32 0.85 18 B4+E+C 0.14 0.01
D
E
MS 114.46 0.59
F Significance F 195.17 0.0000
t Stat -3.89 13.97
P-value 0.0001 0.0000
t = 13.97, p–value = 0; there is evidence of a linear relationship between the average of the best 4 OACs plus English and calculus and university GPA. R 2 = .3509, s = .7658.
The second model fits better (higher coefficient of determination and lower standard error of estimate) and as such is likely to be a better predictor of university GPA.
572
F
Chapter 17 17.1
a ŷ 51.39 .700 x1 .679 x 2 .378 x3 b The standard error of estimate is s = 40.24. It is an estimate of the standard deviation of the error variable. c The coefficient of determination is R 2 = .2425; 24.25% of the variation in prices is explained by the model. d The coefficient of determination adjusted for degrees of freedom is .2019. It differs from R 2 because it includes an adjustment for the number of independent variables. e
H0: β1 = β2 = β3 = 0 H1: At least one i is not equal to zero
F = 5.97, p-value = .0013. There is enough evidence to conclude that the model is valid. f b1 = .700; for each addition thousand square feet the price on average increases by .700 thousand dollars provided that the other variables remain constant.
b2 = .679; for each addition tree the price on average increases by .679 thousand dollars provided that the other variables remain constant.
b3 = –.378; for each addition foot from the lake the price on average decreases by .378 thousand dollars provided that the other variables remain constant. g
H0: βi = 0 H1: βi ≠ 0 593
Lot size: t = 1.25, p-value = .2156 Trees: t = 2.96, p-value = .0045 Distance: t = –1.94, p-value = .0577 Only for the number of trees is there enough evidence to infer a linear relationship with price. h
We predict that the lot in question will sell for between $35,500 and $172,240 i
Estimated average price lies between $39,290 and $90,300.
594
17.2
a ŷ 13.01 .194 x1 1.11x 2 b The standard error of estimate is s = 3.75. It is an estimate of the standard deviation of the error variable. c The coefficient of determination is R 2 = .7629; 76.29% of the variation in final exam marks is explained by the model. d
H0: β1 = β2 = 0 H1: At least one i is not equal to zero
F = 43.43, p-value = 0. There is enough evidence to conclude that the model is valid. e b1 = .194; for each addition mark on assignments the final exam mark on average increases by .194 provided that the other variable remains constant.
b2 = 1.112; for each addition midterm mark the final exam mark on average increases by 1.112 provided that the other variable remains constant. f
H 0 : 1 0 H1 : 1 0
t = .97, p-value = .3417. There is not enough evidence to infer that assignment marks and final exam marks are linearly related. g
H0 : 2 0 H1 : 2 0
t = 9.12, p-value = 0. There is sufficient evidence to infer that midterm marks and final exam marks are linearly related. 595
h.
Pat’s final exam mark is predicted to lie between 23 and 39 i. Pat’s predicted final grade: LCL = 12 + 14 + 23 = 49, UCL = 12 + 14 + 39 = 65
17.3a
b The standard error of estimate is s = 40.13. It is an estimate of the standard deviation of the error variable. c The coefficient of determination is R 2 = .8935; 89.35% of the variation in monthly sales of drywall is explained by the model.
596
d
H0: β1 = β2 = β3 = β4 = 0 H1: At least one i is not equal to zero
F = 39.86, p-value = 0. There is enough evidence to conclude that the model is valid. e b1 = 4.76; for each addition building permit monthly sales on average increase by 4.76 hundred sheets provided that the other variables remain constant.
b2 = 16.99; for each addition one point increase in mortgage rates monthly sales on average increase by 16.99 hundred sheets provided that the other variables remain constant.
b3 = –10.53; for each one percentage point increase in the apartment vacancy rate monthly sales decrease on average by 10.53 hundred sheets provided that the other variables remain constant.
b4 = 1.31; for each one percentage point increase in the office vacancy rate monthly sales increase on average by 1.31 hundred sheets provided that the other variables remain constant. f
H 0 : i 0 H1 : i 0
Permits: t = 12.06, p-value = 0 Mortgage: t = 1.12, p-value = .2764 A Vacancy: t = –1.65, p-value = .1161 O Vacancy: t = .47, p-value = .6446 Only the number of building permits is linearly related to monthly sales. g
Next month’s drywall sales are predicted to lie between 16,710 and 35,290 sheets.
597
17.4a
b b1 = .666; for each additional minor league home run the number of major league home runs increases on average by .666 provided that the other variables remain constant.
b2 = .136; for each additional year of age the number of major league home runs increases on average by .14 provided that the other variables remain constant.
b3 = 1.18; for each additional year as a professional the number of major league home runs increases on average by 1.18 provided that the other variables remain constant. c s = 6.99 and R 2 = .3511; the model's fit is not very good. d
H0: β1 = β2 = β3 = 0 H1: At least one i is not equal to zero
F = 22.01, p-value = 0. There is enough evidence to conclude that the model is valid. e
H 0 : i 0 H1 : i 0
Minor league home runs: t = 7.64, p-value = 0 Age: t = .26, p-value = .7961 Years professional: t = 1.75, p-value = .0819 At the 5% significance level only the number of minor league home runs is linearly related to the number of major league home runs.
598
f
We predict that the player will hit between 9.86 (rounded to 10) and 38.76 (rounded to 39) home runs. g
It is estimated that the average player will hit between 14.66 and 24.47 home runs.
599
17.5
a The regression equation is ŷ = 6.06 –.00781x 1 + .603x 2 –.0702x 3 b s = 1.92, R 2 .7020, F = 36.12, p-value = 0. The model is valid and the fit is reasonably good. c
H 0 : i 0 H1 : i 0
Age: t = –.118, p-value = .9069 Years: t = 6.25, p-value = 0 Pay: t = –1.34, p-value = .1864 Only the number of years with the company is linearly related to severance pay d.
600
95% prediction interval: Lower prediction limit = 5.64, upper prediction limit = 13.50. The offer of 5 weeks severance pay falls below the prediction interval and thus Bill is correct.
17.6
b The coefficient of determination is R 2 = .2882; 28.82% of the variation in university GPAs is explained by the model. c
H0: β1 = β2 = β3 = 0 H1: At least one i is not equal to zero
F = 12.96, p-value = 0. There is enough evidence to conclude that the model is valid. d
H 0 : i 0 H1 : i 0
High school GPA: t = 6.06,, p-value = 0 SAT: t = .942, p-value = .3485 Activities: t = .722, p-value = .4720 At the 5% significance level only the high school GPA is linearly related to the university GPA
601
e
We predict that the student's GPA will fall between 4.45 and 12.00 (12 is the maximum). f
The mean GPA is estimated to lie between 6.90 and 8.22.
602
17.7
a The regression equation is ŷ = 12.31 + .570x 1 + 3.32x 2 + .732x 3 b The coefficient of determination is R 2 = .1953; 19.53% of the variation in sales is explained by the model. The coefficient of determination adjusted for degrees of freedom is .0803. The model fits poorly. c The standard error of estimate is s = 2.59. It is an estimate of the standard deviation of the error variable d
H0: β1 = β2 = β3 = 0 H1: At least one i is not equal to zero
F = 1.70, p-value = .1979. There is not enough evidence to conclude that the model is valid. e
H 0 : i 0 H1 : i 0
Direct: t = .331, p-value = .7437 Newspaper: t = 2.16, p-value = .0427 Television: t = .374, p-value = .7123 Only expenditures on newspaper advertising is linearly related to sales.
603
f&g
f We predict that sales will fall between $12,270 and $24,150. g We estimate that mean sales will fall between $15,700 and $20,730. h The interval in part f predicts one week’s gross sales, whereas the interval in part h estimates the mean weekly gross sales.
17.8 a
b
H0: β1 = β2 = β3 = 0 H1: At least one i is not equal to zero
F = 29.80, p-value = 0. There is enough evidence to conclude that the model is valid.
604
c b1= .0019; for each additional square foot the amount of garbage increases on average by .0019 pounds holding the other variables constant. b2 = 1.10; for each additional child in the home the amount of garbage increases on average by 1.10 pounds holding the other variables constant. b3= 1.04; for each additional adult at home during the day the amount of garbage increases on average by 1.04 holding the other variables constant. d
H 0 : i 0 H1 : i 0
House size : t = 3.21, p-value = .0014 Number of children: t = 7.84 p-value = 0 Number of adults at home: t = 4.48, p-value = 0 All three independent variable are linearly related to the amount of garbage.
17.9a
a ŷ = 35.68 + .247x 1 + .245x 2 + .133x 3 b
H0: β1 = β2 = β3 = 0 H1: At least one i is not equal to zero
F = 6.66, p-value = .0011. There is enough evidence to conclude that the model is valid. c b1 = .247; for each one percentage point increase in the proportion of teachers with mathematics degrees the test score increases on average by .247 provided the other variables are constant.
605
b2 = .245; for each one year increase in mean age test score increases on average by .245 provided the other variables are constant .
b3 = .135; for each one thousand dollar increase in salary test score increases on average by .135 provided the other variables are constant.
H 0 : i 0 H1 : i 0 Proportion of teachers with at least one mathematics degree: t = 3.54, p-value = .0011 Age: t = 1.32, p-value = .1945 Income: t = .87, p-value = .3889. The proportion of teachers with at least one mathematics degree is linearly related to test scores. The other two variables appear to be unrelated.
d The school's test score is predicted to fall between 49.02 and 81.02.
606
17.10a
b
H0: β1 = β2 = β3 = β4 = 0 H1: At least one i is not equal to zero
F = 67.97, p-value = 0. There is enough evidence to conclude that the model is valid. c b1 = .451; for each one year increase in the mother's age the customer's age increases on average by .451 provided the other variables are constant (which may not be possible because of the multicollinearity).
b2 = .411; for each one year increase in the father's age the customer's age increases on average by .411 provided the other variables are constant.
b3 = .0166; for each one year increase in the grandmothers' mean age the customer's age increases on average by .0166 provided the other variables are constant.
b 4 = .0869; for each one year increase in the grandfathers' mean age the customer's age increases on average by .0869 provided the other variables are constant.
H 0 : i 0 H1 : i 0 Mothers: t = 8.27, p-value = 0 Fathers: t = 8.26, p-value = 0 Grandmothers: t = .250, p-value .8028 Grandfathers: t = 1.32, p-value = .1890
607
The ages of mothers and fathers are linearly related to the ages of their children. The other two variables are not. d
The man is predicted to live to an age between 65.54 and 77.31 e
The mean longevity is estimated to fall between 68.75 and 74.66.
608
17.11
Assessing the Model:
s = 7.01 and R 2 = .7209; the model fits well. Testing the validity of the model: H0: β1 = β2 = 0 H1: At least one i is not equal to zero F = 60.70, p-value = 0. There is enough evidence to conclude that the model is valid. Drawing inferences about the independent variables:
H 0 : i 0 H1 : i 0 Evaluations: t = .598, p-value = .5529 Articles: t = 8.08, p-value = 0. The number of articles a professor publishes is linearly related to salary. Teaching evaluations are not.
609
17.12
a ŷ = –28.43 + .604x 1 + .374x 2 b s = 7.07 and R 2 = .8072; the model fits well. c b1 = .604; for each one additional box, the amount of time to unload increases on average by .604 minutes provided the weight is constant.
b2 = .374; for each additional hundred pounds the amount of time to unload increases on average by .374 minutes provided the number of boxes is constant.
H 0 : i 0 H1 : i 0 Boxes: t = 10.85, p-value = 0 Weight: t = 4.42, p-value = .0001 Both variables are linearly related to time to unload.
610
d&e
d It is predicted that the truck will be unloaded in a time between 35.16 and 66.24 minutes. e The mean time to unload the trucks is estimated to lie between 44.43 and 56.96 minutes.
17.13
b
H0: β1 = β2 = β3 = β4 = 0 H1: At least one i is not equal to zero
F = 18.17, p-value = 0. There is enough evidence to conclude that the model is valid.
611
e
H 0 : i 0 H1 : i < 0 (for beliefs 1 and 4) H1 : i > 0 for beliefs 2 and 3)
Belief 1: t = –3.26, p-value = .0016/2 = .0008 Belief 2: t = 1.16, p-value = .2501/2 = .1251 Belief 3: t = .417, p-value = .6780/2 = .3390 Belief 4: t = –2.69, p-value = .0085/2 = .0043 There is enough evidence to support beliefs 1 and 4. There is not enough evidence to support beliefs 2 and 3.
17.14 a
b
H0: β1 = β2 = β3 = 0 H1: At least one i is not equal to zero
F = 24.48, p-value = 1.64E-11 .≈ 0. There is sufficient evidence to conclude that the model is valid. c
H 0 : i 0 H1 : i 0
Undergraduate GPA: t = .524, p-value = .6017 GMAT: t = 8.16, p-value = 0
612
Work experience: t = 3.00, p-value = .0036 Both the GMAT and work experience are linearly related to MBA GPA
17.15
ŷ = 3.02 + .0289 AGE + .2461 EDUC + 3.75X10-6 INCOME b.
H0: β1 = β2 = β3 = 0 H1: At least one i is not equal to zero
F = 41.92, p-value = 0. There is sufficient evidence to conclude that the model is valid. c.
H0: βi = 0 H1: βi ≠ 0
Variable
t-statistic
p-value
AGE
6.93
5.99E-12 ≈ 0
EDUC
8.07
1.36E-15 ≈ 0
INCOME
1.72
.0858
AGE and EDUC are linearly related to DEFINITE.
613
17.16
ŷ = 6.365 + .135 DAYS1 + .036 DAYS2 + .060 DAYS3 + .107 DAYS4 + .142 DAYS5 + .134 DAYS6 b.
H0: β1 = β2 = β3 = β4 = β5 = β6 = 0 H1: At least one i is not equal to zero
F = 11.72, p-value = 0. There is enough evidence to infer that the model is valid. c.
H0: βi = 0 H1: βi ≠ 0
Variable
t-statistic
p-value
DAYS1
3.33
.0009
DAYS2
.81
.4183
DAYS3
1.41
.1582
DAYS4
3.00
.0027
DAYS5
3.05
.0024
DAYS6
3.71
.0002
Only DAYS2 and DAYS3 are not linearly related to DEFINITE.
614
7.17
b.
H0: β1 = β2 = β3 = β4 = β5 = β6 = 0 H1: At least one i is not equal to zero
F = 17.14, p-value = 3.03E-13 ≈ 0. There is enough evidence to conclude that the model is valid. c.
H 0 : i 0 H1 : i 0
Variable
t-statistic
p-value
Number
-6.07
0
Nearest
2.60
.0108
Office space
5.80
0
Enrollment
1.59
.1159
Income
2.96
.0039
Distance
-1.26
.2107
The intercept is b 0 38.14. This is the average operating margin when all the independent variables are zero.
615
The relationship between operating margin and the number of motel and hotel rooms within 3 miles is described by b1 −.0076. Changing the units we can interpret b1 to say that for each additional 1,000 rooms, the margin decreases by 7.6%. The coefficient b 2 1.65 specifies that for each additional mile that the nearest competitor is to a La Quinta Inn, the average operating margin increases by 1.65%, assuming the constancy of the other independent variables. The relationship between office space and operating margin is expressed by b 3 .020. For every extra 100,000 square feet of office space, the operating margin increases on average by 2.0%.
The relationship between operating margin and college and university enrollment is described by
b 4 .21, which we interpret to mean that for each additional thousand students the average operating margin increases by .21% when the other variables are constant. The relationship between operating margin and median household income is described by b 5 .41. For each additional thousand dollar increase in median household income, the average operating margin increases by .41%, holding all other variables constant. This statistic suggests that motels in more affluent communities have higher operating margins.
The last variable in the model is distance to the downtown core. Its relationship with operating margin is described by b 6 −.23. This tells us that for each additional mile to the downtown center, the operating margin decreases on average by .23%, keeping the other independent variables constant. It may be that people prefer to stay at motels that are closer to town centers. e. & f.
616
e. Lower prediction limit n= 25.40, Upper prediction limit = 48.79 f. LCL = 32.97, UCL = 41.21
17.18
b.
H0: β1 = β2 = 0 H1: At least one i is not equal to zero
F = 284.0, p-value = 0. There is sufficient evidence to conclude that the model is valid.
c
H 0 : i 0 H1 : i 0
PAEDUC: t = 10.16, p-value = 0 MAEDUC: t = 8.49, p-value = 0 Both variables are linearly related to EDUC. d. b1: For each additional year of the father’s education the son or daughter’s years of education increases on average by .212 years holding the other variable constant. b2: For each additional year of the mother’s education the son or daughter’s years of education increases on average by .194 years holding the other variable constant.
17.19
617
b.
H0: β1 = β2 = β3 = β4 = 0 H1: At least one i is not equal to zero
F = 11.51, p-value = 0. There is enough evidence to conclude that the model is valid. c.
H 0 : i 0 H1 : i 0
AGE: t = 3.62, p-value = 0 RINCOME: t = 3.33, p-value = .0001 EDUC: t = .49, p-value = .6214 HRS1: t = 1.44, p-value = .1515 Only AGE and RINCOME are linearly related to EQWLTH d. R2 = .0503; 5.03% of the variation in EQWLTH is explained by the model.
618
17.20
b.
H0: β1 = β2 = β3 = β4 = 0 H1: At least one i is not equal to zero
F = 16.41, p-value = 0. There is enough evidence to conclude that the model is valid. c.
H 0 : i 0 H1 : i 0
AGE: t = 2.74, p-value = .0062 EDUC: t = -6.14, p-value = 0 HRS1: t = -3.92, p-value = 0 CHILDS: t = .11, p-value = .9108 AGE, EDUC, and HRS1 are linearly related to TVHOURS d. R2 = .0618; 6.18% of the variation in TVHOURS is explained by the model.
619
17.21
b.
H0: β1 = β2 = β3 = β4 = 0 H1: At least one i is not equal to zero
F = 2.27, p-value = .0060. There is not enough evidence to conclude that the model is valid. c.
H 0 : i 0 H1 : i 0
AGE: t = 2.15, p-value = .0316 EDUC: t = .43, p-value = .6703 RINCOME: t = .89, p-value = .3716 HRS1: t = .65, p-value = .5189 Only AGE is linearly related to HELPPOOR d. R2 = .0105; 1.05% of the variation in HELPPOOR is explained by the model.
620
17.22
ŷ = 56,326 - 883.7 AGE - 4503 EDUC b.
H0: β1 = β2 = 0 H1: At least one i is not equal to zero
F = 140.30, p-value = 0. There is enough evidence to infer that the model is valid. c.
H0: β1 = 0 H1: β1 > 0
t = 8.69, p-value = 0. There is enough evidence to infer that more education leads to more income. d. R2 = .1895; 18.95% of the variation in income is explained by the model.
17.23
ŷ = 7612 -37.88 AGE – 72.39 EDUC + .0190 INCOME 621
b.
H0: β1 = β2 = β3 = 0 H1: At least one i is not equal to zero
F = 54.73, p-value = 0. There is enough evidence to infer that the model is valid. c.
AGE: t = -5.77, p-value = 0. EDUC: t = -.180, p-value = .0722 INCOME: t = 8.74, p-value = 0.
AGE and INCOME are linearly related to FOODHOME.
17.24
ŷ = 2206 -21.95 AGE + 4.52 EDUC + .0140 INCOME b.
H0: β1 = β2 = β3 = 0 H1: At least one i is not equal to zero
F = 62.20, p-value = 0. There is enough evidence to infer that the model is valid. c.
AGE: t = -4.95, p-value = 0. EDUC: t = .17, p-value = .8680 INCOME: t = 9.50, p-value = 0.
AGE and INCOME are linearly related to FOODAWAY.
622
17.25 a
Frequency
Histogram 40 20 0 -2
-1
0
1
2
3
-3
The normality requirement is satisfied. b Plot of Residuals vs predicted
Residuals
100 50 0 40
60
80
100
120
140
-50 -100 Predicted
The variance of the error variable appears to be constant.
17.26
A 1 2 Lot size 3 Trees 4 Distance
B Lot size
C Trees
1 0.2857 -0.1895
1 0.0794
D Distance
1
There does not appear to be a multicollinearity problem. The t–tests are valid.
623
17.27b
Frequency
Histogram 20 10 0 -3
-2
-1
0
1
2
3
Standard Residuals
The normality requirement has not been violated. c
Residuals
Plot of Residuals vs Predicted 10 8 6 4 2 0 -2 20 -4 -6 -8 -10
30
40
50
Predicted
The variance of the error variable appears to be constant. d
A 1 2 Assignment 3 Midterm
B C Assignment Midterm 1 0.1037 1
The lack of multicollinearity means that the t–tests were valid.
624
60
17.28a
Frequency
Histogram 20 10 0 -3
-2
-1
0
1
2
3
Standard Residuals
The error variable may be normally distributed. b Plot of Residuals vs Predicted 100
Residuals
50 0 0
100
200
300
400
500
-50 -100 Predicted
The variance of the error variable is constant. c
1 2 3 4 5
A
B Permits
Permits Mortgage A Vacancy O Vacancy
1 0.0047 -0.1505 -0.1027
C D Mortgage A Vacancy 1 -0.0399 -0.0332
1 0.0652
Multicollinearity is not a problem.
17.29a
A B 1 Minor HR 2 Minor HR 1 3 Age 0.0354 4 Years Pro -0.0392
C Age
D Years Pro
1 0.7355
1
625
E O Vacancy
1
Age and years as a professional are highly correlated. The correlations of the other combinations are small. b The t–tests may not be valid.
17.30 a
Frequency
Histogram 40 20 0 -3
-2
-1
0
1
2
3
Standard Residuals
The error variable is approximately normally distributed. b Plot of Residuals vs Predicted 6
Residuals
4 2 0 -2
0
5
10
15
20
-4 -6 Predicted
The variance of the error variable is constant. c
A 1 2 Age 3 Years 4 Pay
B Age
C Years
1 0.8080 0.1725
1 0.2610
D Pay
1
The correlation between age and years is high indicating that multicollinearity is a problem.
626
17.31
Frequency
Histogram 50 0 -3
-2
-1
0
1
2
Standard Residuals
Residuals
Plot of Residuals vs Predicted 5 4 3 2 1 0 -1 3 -2 -3 -4 -5
5
7
9
Predicted
The error variable is approximately normally distributed and the variance is constant.
17.32a
Frequency
Histogram 20 10 0 -3
-2
-1
0
1
Standard Residuals
627
2
3
Plot of Residuals vs Predicted 6
Residuals
4 2 0 -2 15
17
19
21
23
-4 -6 Predicted
The error variable is approximately normal. However, the variance is not constant. b
A 1 2 Direct 3 Newspaper 4 Television
B Direct
C D Newspaper Television
1 -0.1376 -0.1246
1 0.1468
1
Multicollinearity is not a problem. c There is one observation whose standardized residual exceeds 2.0 that should be checked
17.33
Frequency
Histogram 200 100
0 -3
-2
-1
0
1
Standard Residuals
The error appears to be normally distributed.
628
2
3
15
Plot of Residuals vs Predicted
10
5 0 8
10
12
14
16
18
20
22
-5
-10 -15
There are indications that the error grows smaller as the predicted value increases. However, overall the requirement of constant variance may be valid.
17.34 a
Frequency
Histogram 20 10 0 -3
-2
-1
0
1
2
3
Standard Residuals
The normality requirement is satisfied. Plot of Residuals vs Predicted
Residuals
20 10 0 -10 50
55
60
65
70
-20 -30 Predicted
The variance of the error variable is constant.
629
75
c
A B 1 Math Degree 2 Math Degree 1 3 Age 0.0766 4 Income 0.0994
C Age
D Income
1 0.5698
1
The correlation between age and income is high. Multicollinearity is a problem.
17.35a
Frequency
Histogram 40 20 0 -3
-2
-1
0
1
2
3
More
Standard Residuals
The normality requirement is satisfied. b Plot of Residuals vs Predicted 10 8 6
Residuals
4 2 0 -2 60
` 65
70
75
80
-4 -6 -8 -10 Predicted
The variance of the error variable is constant. c
630
85
A 1 2 3 4 5
B Mother
Mother Father Gmothers Gfathers
C Father
1 0.2766 0.4343 0.3910
1 0.2409 0.3752
D E Gmothers Gfathers
1 -0.0077
1
The correlations are large enough to cause problems with the t–tests.
17.36
Frequency
Histogram 40 20 0 -3
-2
-1
0
1
2
3
Standard Residuals
The error variable appears to be normal. The error variable's variance appears to be constant. The required conditions are satisfied.
17.37
Frequency
Histogram 20 10 0 -3
-2
-1
0
1
Standard Residuals
631
2
3
Plot of Residuals vs Predicted
Residuals
20 10 0 20
40
60
80
100
-10 -20 Predicted
c The error variable appears to be normally distributed. The variance of the errors appears to be constant.
17.38a
Frequency
Histogram 40 20 0 -3
-2
-1
0
1
2
3
Standard Residuals
Plot of Residuals vs Predicted
Residuals
10 5 0 -5
-5
0
5
10
-10 Predicted
The errors appear to be normally distributed. The variance of the errors is not constant.
632
b
Lottery
Education
Age
1 -0.6202 0.1767 -0.0230 -0.5891
1 -0.1782 0.1073 0.7339
1 0.1072 -0.0418
Lottery Education Age Children Income
Children
Income
1 0.0801
1
There is a strong correlation between income and education. The t–tests of these two coefficients may be distorted.
Frequency
17.39
25 20 15 10 5 0 -2
-1,5
-1
-0,5
0
0,5
1
1,5
2
Residuals The histogram of the residuals is bell shaped.
2
1,5 1
Residuals
0,5 0 -0,5
6
7
8
9
-1 -1,5
-2 -2,5
Predicted
There is no evidence of heteroscedasticity. The requirements are satisfied.
633
10
11
17.40
Frequency
30 20 10 0 -10
-7
-4
-1
2
5
8
11
14
Residuals
The histograms is somewhat bimodal. There residuals may not be normally distributed.
15 10
Residuals
5 0
30
35
40
45
-5 -10 -15
Predicted
There is no evidence of heteroscedasticity.
17.41 Histogram of standardized residuals
634
50
55
60
600 Frequency
500 400 300 200 100 0 -3 -2,5 -2 -1,5 -1 -0,5 0
0,5
1
1,5
2
2,5
3
Std Res The error term appears to be negatively skewed.
Plot of residuals vs predicted 6 4
Residuals
2 0 0
2
4
6
8
10
-2 -4 -6 -8 -10
Predicted There is some form of heteroscedasticity.
17.42 Histogram of standardized residuals
Histogram Frequency
400 300 200 100 0 -3 -2,5 -2 -1,5 -1 -0,5 0 0,5 1 1,5 2 2,5 3 3,5 Std Res
635
12
The error term appears to be negatively skewed.
Plot of residuals versus predicted 6 4
Residuals
2 0 0
2
4
6
8
10
12
-2
-4 -6 -8 -10
Predicted
There is some form of heteroscedasticity.
17.43a
b. The correlations between RINCOME and AGE, EDUC, and HRS1 may be large enough to produce a multicollinearity effect.
636
c. AGE (r = .0855, p-value = .0121) and RINCOME (r = .0695, p-value = .0416) are linearly related to HELPPOOR. In the analysis in Exercise 17.21 only AGE was significant.
17.44
The correlation between PAEDUC and MAEDUC is .6934. Multicollinearity is a problem in Exercise 17.18.
17.45a.
The correlations between RINCOME AGE, EDUC and HRS1 are large enough to indicate the presence of multicollinearity. b. All four independent variables are linearly related to EQWLTH. In the analysis in Exercise 17.19 only RINCOME and AGE were linearly related to EQWLTH.
637
17.46a.
The correlations between AGE and CHILDS and EDUC and CHILDS indicate that multicollinearity may be a problem. b. AGE (r = .0883, p-value = .0051), EDUC (r = -.1971, p-value = 0), HRS1 (r = -.1322, p-value = 0) are linearly related to TVHOURS. These are the same independent variables that were shown to be linearly related to TVHOURS in the regression analysis.
17.47 d L = 1.12, d U = 1.66. There is evidence of positive first–order autocorrelation. 17.48 d L = 1.16, d U = 1.59, 4 – d U = 2.41, 4 – d L = 2.84. There is evidence of negative first– order autocorrelation. 17.49 d L = .95, d U = 1.89, 4 – d U = 2.11, 4 – d L = 3.05. There is evidence of first–order autocorrelation.
17.50 d L = 1.46, d U = 1.63. There is evidence of positive first–order autocorrelation. 17.51 d L = 1.41, d U = 1.64, 4 – d U = 2.36, 4 – d L = 2.59. The test is inconclusive 17.52 4 – d U = 4 – 1.73 = 2.27, 4 – d L = 4 – 1.19 = 2.81. There is no evidence of negative first– order autocorrelation.
638
17.53 a The regression equation is ŷ = 303.3 + 14.94 x1 + 10.52 x 2 b Plot of Residuals vs Time 1000
Y
500 0 0
20
40
60
80
100
120
-500 -1000 Tim e
The graph indicates that autocorrelation exists. c
A B C 1 Durbin-Watson Statistic 2 3 d = 0.7749 d L = 1.63, d U = 1.72, 4 – d U = 2.37, 4 – d L = 2.28. There is evidence of autocorrelation. d The model is y 0 + 1 x1 + 2 x 2 + 3t + The regression equation is ŷ = 10.00 + 6.78 x1 + 9.37 x 2 + 9.64t e Plot of Residuals vs Time 600 400 200
Y
0 -200
0
20
40
60
80
-400 -600 -800 Tim e
A B C 1 Durbin-Watson Statistic 2 3 d = 2.1237
639
100
120
There is no evidence of autocorrelation. First model: s = 348.7 and R 2 = .2825. Second model: s = 208.1 and R 2 = .7471 The second model fits better.
17.54 a The regression equation is ŷ = 2260 + .423x b Plot of Residuals vs Time 1500 1000
Y
500 0 -500 0 -1000 -1500
10
20
30
40
50
60
-2000 -2500 Tim e
There appears to be a strong autocorrelation. c
A B C 1 Durbin-Watson Statistic 2 3 d = 0.7859 d L 1.50, d U 1.59, 4 d U 2.41, 4 d L 2.50. There is evidence of first–order autocorrelation. d The model is y 0 + 1 x + 2 t + The regression equation is ŷ = 446.2 + 1.10x + 38.92t e
640
Plot of Residuals vs Time 1000
Sales
500 0 0
10
20
30
40
50
60
-500 -1000 -1500 Tim e
A B C 1 Durbin-Watson Statistic 2 3 d = 2.2631 There is no evidence of autocorrelation. First model: s = 709.7 and R 2 = .0146. Second model: s = 413.7 and R 2 = .6718. The second model fits better.
17.55
A B C 1 Durbin-Watson Statistic 2 3 d = 1.755 d L = 1.01, d U = 1.78. There is no evidence of positive first–order autocorrelation.
17.56
A B C 1 Durbin-Watson Statistic 2 3 d = 2.2003 d = 2.2003; d L = 1.30, d U = 1.46, 4 – d U = 2.70, 4 – d L = 2.54. There is no evidence of first– order autocorrelation.
17.57 a ŷ = 898.0 + 11.33x b
641
Frequency
Histogram 10 5 0 -3
-2
-1
0
1
2
3
Standard Residuals
Plot of Residuals vs Predicted 200
Residuals
150 100 50 0 -50900
950
1000
1050
1100
-100 -150 Predicted
The normality requirement is satisfied. However, the constant variance requirement is not.
A B C 1 Durbin-Watson Statistic 2 3 d = 1.0062 d L = 1.24, d U = 1.43, 4 – d U = 2.76, 4 – d L = 2.57. There is evidence of first–order autocorrelation. c The problem is that the errors are not independent. We add a time variable (week number) to the model. Thus, the new model is y = 0 + 1 x + 2 t + . The regression equation is ŷ = 960.6 + 13.88x – 7.69t
Frequency
Histogram 10 5 0 -3
-2
-1
0
1
Standard Residuals
642
2
3
Plot of Residuals vs Years 100
Tires
50 0 800
900
1000
1100
1200
-50 -100 -150 Years
A B C 1 Durbin-Watson Statistic 2 3 d = 1.9012 d L = 1.15, d U = 1.54, 4 – d U = 2.85, 4 – d L = 2.46. There is no evidence of first–order autocorrelation. d s = 48.55 and R 2 = .7040 Snowfall: b1 13 .88; for each additional inch of snowfall tire sales increase on average by 13.88 (holding the time period constant). t = 5.862, p-value = 0. Week: b2 7.687 ; weekly sales decrease on average by 7.687 (holding snowfall constant). t = –4.579, p-value = .0002
643
17.58a
A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.6894 5 R Square 0.4752 6 Adjusted R Square 0.4363 7 Standard Error 63.08 8 Observations 30 9 10 ANOVA 11 df SS 12 Regression 2 97283 13 Residual 27 107428 14 Total 29 204711 15 16 Coefficients Standard Error 17 Intercept 164.01 35.9 18 Fetilizer 0.140 0.081 19 Water 0.0313 0.0067
D
E
MS 48641 3979
F Significance F 12.23 0.0002
t Stat P-value 4.57 9.60E-05 1.72 0.0974 4.64 8.08E-05
a ŷ 164 .01 .140 x 1 .0313 x 2 For each additional unit of fertilizer crop yield increases on average by .140 (holding the amount of water constant). For each additional unit of water crop yield increases on average by .0313 (holding the fertilizer constant). b
H 0 : 1 0 H1 : 1 0
t = 1.72, p-value = .0974. There is not enough evidence to conclude that there is a linear relationship between crop yield and amount of fertilizer. c
H 0 : 2 0 H1 : 2 0
t = 4.64, p-value = .0001. There is enough evidence to conclude that there is a linear relationship between crop yield and amount of water. d s = 63.08 and R 2 = .4752; the model fits moderately well.
644
F
e
Frequency
Histogram 20 10
0 -3
-2
-1
0
1
2
3
Standard Residuals
Plot of Residuals vs Predicted 150 100
Residuals
50
0 0
100
200
300
400
500
-50 -100 -150 Predicted
The errors appear to be normal, but the plot of residuals vs predicted aeems to indicate a problem. f
A 1 Prediction Interval 2 3 4 5 Predicted value 6 7 Prediction Interval 8 Lower limit 9 Upper limit 10 11 Interval Estimate of Expected Value 12 Lower limit 13 Upper limit 14
B
Yield 209.3
69.2 349.3
155.7 262.8
We predict that the crop yield will fall between 69.2 and 349.3.
645
17.59
A B C D E F 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.6882 4 Multiple R 0.4736 5 R Square 0.4134 6 Adjusted R Square 2,644 7 Standard Error 8 Observations 40 9 10 ANOVA 11 df SS MS F Significance F 12 Regression 4 220,130,124 55,032,531 7.87 0.0001 13 Residual 35 244,690,939 6,991,170 14 Total 39 464,821,063 15 16 Coefficients Standard Error t Stat P-value 17 Intercept 1,433 2,093 0.68 0.4980 18 Size -14.55 20.70 -0.70 0.4866 19 Apartments 113.0 24.01 4.70 0.0000 20 Age -50.10 98.81 -0.51 0.6153 21 Floors -223.8 171.1 -1.31 0.1994 b
H 0 : 1 2 3 4 0 H1 : At least one i is not equal to zero
F = 7.87, p-value = .0001. There is enough evidence to conclude that the model is valid. c The regression equation for Exercise 16.12 is ŷ = 4040 + 44.97x. The addition of the new variables changes the coefficients of the regression line in Exercise 17.12. 17.60a ŷ 29 .60 .309 x 1 1.11x 2
646
A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.7825 4 Multiple R 0.6123 5 R Square 0.5835 6 Adjusted R Square 2.16 7 Standard Error 8 Observations 30 9 10 ANOVA 11 df SS 12 Regression 2 199.65 13 Residual 27 126.44 14 Total 29 326.09 15 16 Coefficients Standard Error 17 Intercept 29.60 2.08 18 Vacancy -0.309 0.067 19 Unemployment -1.11 0.24
D
E
MS 99.82 4.68
F Significance F 21.32 0.0000
t Stat P-value 14.22 0.0000 -4.58 0.0001 -4.73 0.0001
2 b R .6123 ; 61.23 % of the variation in rents is explained by the independent variables.
H 0 : 1 2 0
c
H1 : At least one i is not equal to zero F = 21.32, p-value = 0. There is enough evidence to conclude that the model is valid.
H 0 : i 0
d
H1 : i 0 Vacancy rate: t = –4.58, p-value = .0001 Unemployment rate: t = –4.73, p-value = .0001 Both vacancy and unemployment rates are linearly related to rents. e
Frequency
Histogram 20 10 0 -3
-2
-1
0
1
Standard Residuals
647
2
3
F
Plot of Residuals vs Predicted 6
Residuals
4 2 0 -2
10
12
14
16
18
20
22
-4 Predicted
The error is approximately normally distributed with a constant variance. f
A B C 1 Durbin-Watson Statistic 2 3 d = 2.0687 d L = 1.28, d U = 1.57, 4 – d U = 2.72, 4 – d L = 2.43. There is no evidence of first–order autocorrelation. g
A B C 1 Prediction Interval 2 Rent 3 4 18.72 5 Predicted value 6 7 Prediction Interval 14.18 8 Lower limit 23.27 9 Upper limit 10 11 Interval Estimate of Expected Value 17.76 12 Lower limit 19.68 13 Upper limit
D
The city's office rent is predicted to lie between $14.18 and $23.27.
648
Case 17.1 A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.1228 4 Multiple R 0.0151 5 R Square 0.0131 6 Adjusted R Square 8.30 7 Standard Error 8 Observations 2029 9 10 ANOVA 11 df SS 12 Regression 4 2,137 13 Residual 2024 139,561 14 Total 2028 141,698 15 16 Coefficients Standard Error 17 Intercept -1.15 2.20 18 SAT 0.0051 0.0013 19 MBA 0.674 0.376 20 Age -0.141 0.042 21 Tenure 0.082 0.176
D
E
F
MS 534.29 68.95
F
Significance F 0.0000
7.75
t Stat P-value -0.52 0.6012 3.96 0.0001 1.79 0.0730 -3.31 0.0009 0.47 0.6412
The model is valid (F = 7.75, p-value = 0) but the model does not fit well (R 2 = .0151; only 1.51% of the variation in returns is explained by the model).
Interpreting the coefficients in this sample: For each additional one–point increase in the SAT score, returns increase on average by .0051 provided the other variables remain constant. The returns of mutual funds managed by MBAs are on average .674 larger than the returns of mutual funds managed by people without an MBA For each additional one–year increase in age of the manager , returns decrease on average by .141 provided the other variables remain constant. For each additional one–year increase in the manager’s job tenure, returns increase on average by .082 provided the other variables remain constant.
Testing the coefficients: SAT: t = 3.96, p-value = .0001 MBA: t = 1.79, p-value = .0730 Age: t = –3.31, p-value = .0009 Tenure: t = .47, p-value = .6412
649
There is overwhelming evidence to infer that SAT scores of the undergraduate university and age of the manager are linearly related to returns. There is weak evidence that MBAs and non–MBAs have different mean returns. There is not enough evidence to conclude that job tenure is linearly related to returns.
Case 17.2 Analysis of Betas
A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.3597 5 R Square 0.1294 6 Adjusted R Square 0.1277 7 Standard Error 0.2245 8 Observations 2029 9 10 ANOVA 11 df SS 12 Regression 4 15.15 13 Residual 2024 101.97 14 Total 2028 117.12 15 16 Coefficients Standard Error 17 Intercept 0.152 0.059 18 SAT 0.00050 0.000035 19 MBA 0.0366 0.0102 20 Age 0.0088 0.0011 21 Tenure -0.0352 0.0047
D
E
F
MS 3.79 0.050
F Significance F 75.20 0.0000
t Stat P-value 2.56 0.0107 14.55 0.0000 3.60 0.0003 7.66 0.0000 -7.42 0.0000
The model is valid (F = 75.20, p-value = 0) with R 2 = .1294; only 12.94% of the variation in betas is explained by the model.
Interpreting the coefficients in this sample: For each additional one–point increase in the SAT score, betas increase on average by .00050 provided the other variables remain constant. The betas of mutual funds managed by MBAs are on average .0366 larger than the betas of mutual funds managed by people without an MBA For each additional one–year increase in age of the manager, betas increase on average by .0088 provided the other variables remain constant. For each additional one–year increase in the manager’s job tenure, betas decrease on average by .0352 provided the other variables remain constant.
650
Testing the coefficients: SAT: t = 14.55, p-value = 0 MBA: t = 3.60, p-value = .0003 Age: t = 7.66, p-value = 0 Tenure: t = –7.42, p-value = 0 There is overwhelming evidence to infer that all four independent variables are linearly related to mutual fund betas.
Analysis of MERs
A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.2697 4 Multiple R 0.0728 5 R Square 0.0705 6 Adjusted R Square 0.6847 7 Standard Error 8 Observations 2029 9 10 ANOVA 11 df SS 12 Regression 5 74.42 13 Residual 2023 948.48 14 Total 2028 1022.90 15 16 Coefficients Standard Error 17 Intercept 2.89 0.183 18 SAT -0.00055 0.00011 19 MBA -0.082 0.0310 20 Age 0.0133 0.0035 21 Tenure 0.0375 0.0145 22 Log Assets -0.209 0.0229
D
E
MS 14.88 0.47
F Significance F 31.74 0.0000
t Stat 15.73 -5.21 -2.65 3.80 2.59 -9.13
F
P-value 0.0000 0.0000 0.0081 0.0001 0.0097 0.0000
The model is valid (F = 31.74, p-value = 0) with R 2 = .0728; only 7.28% of the variation in MERs is explained by the model.
Interpreting the coefficients in this sample: For each additional one–point increase in the SAT score, MERs decrease on average by .00055 provided the other variables remain constant. The MERs of mutual funds managed by MBAs are on average .082 smaller than the MERs of mutual funds managed by people without an MBA For each additional one–year increase in age of the manager, MERs increase on average by .0133 provided the other variables remain constant.
651
For each additional one–year increase in the manager’s job tenure, MERs increase on average by .0375 provided the other variables remain constant. For each additional one–point increase in the log of the assets, MERs decrease on average by .209 provided the other variables remain constant.
Testing the coefficients: SAT: t = –5.21, p-value = 0 MBA: t = –2.65, p-value = .0081 Age: t = 3.80, p-value = .0001 Tenure: t = 2.59, p-value = .0097 Log Assets: t = –9.13, p-value = 0 There is overwhelming evidence to infer that all five independent variables are linearly related to mutual fund MERs.
652
Chapter 18 18.1 a
b
18.2 a
671
b
2
18.3 a Sales = 0 + 1 Space + 2 Space + b A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.6378 4 Multiple R 0.4068 5 R Square 0.3528 6 Adjusted R Square 41.15 7 Standard Error 8 Observations 25 9 10 ANOVA 11 df SS 12 Regression 2 25,540 13 Residual 22 37,248 14 Total 24 62,788 15 16 Coefficients Standard Error 17 Intercept -108.99 97.24 18 Space 33.09 8.59 19 Space-sq -0.666 0.177
D
E
F
MS 12,770 1,693
F
Significance F 0.0032
7.54
t Stat P-value -1.12 0.2744 3.85 0.0009 -3.75 0.0011
s = 41.15 and R 2 = .4068. The model's fit is relatively poor. F = 7.54, p-value = .0032. However, there is enough evidence to support the validity of the model.
18.4a
First–order model: a Demand = 0 + 1 Price+ 2
Second–order model: a Demand = 0 + 1 Price + 2 Price +
672
First–order model: A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.9249 5 R Square 0.8553 6 Adjusted R Square 0.8473 7 Standard Error 13.29 8 Observations 20 9 10 ANOVA 11 df SS 12 Regression 1 18,798 13 Residual 18 3,179 14 Total 19 21,977 15 16 Coefficients Standard Error 17 Intercept 453.6 15.18 18 Price -68.91 6.68
D
E
MS 18,798 176.6
F Significance F 106.44 0.0000
F
t Stat P-value 29.87 0.0000 -10.32 0.0000
Second–order model: A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.9862 4 Multiple R 0.9726 5 R Square 0.9693 6 Adjusted R Square 5.96 7 Standard Error 8 Observations 20 9 10 ANOVA 11 df SS 12 Regression 2 21,374 13 Residual 17 603 14 Total 19 21,977 15 16 Coefficients Standard Error 17 Intercept 766.9 37.40 18 Price -359.1 34.19 19 Price-sq 64.55 7.58
D
E
MS 10,687 35.49
F Significance F 301.15 0.0000
t Stat P-value 20.50 0.0000 -10.50 0.0000 8.52 0.0000
c The second order model fits better because its standard error of estimate is 5.96, whereas that of the first–order models is 13.29 d ŷ .= 766.9 –359.1(2.95) + 64.55(2.95) 2 = 269.3
673
F
18.5a
First–order model: a Time = 0 + 1 Day+ 2
Second–order model: a Time = 0 + 1 Day + 2 Day + b First–order model A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.9222 4 Multiple R 0.8504 5 R Square 0.8317 6 Adjusted R Square 1.79 7 Standard Error 8 Observations 10 9 10 ANOVA 11 df SS 12 Regression 1 145.34 13 Residual 8 25.56 14 Total 9 170.90 15 16 Coefficients Standard Error 17 Intercept 41.40 1.22 18 Day -1.33 0.197
D
E
F
MS 145.34 3.20
F Significance F 45.48 0.0001
t Stat P-value 33.90 0.0000 -6.74 0.0001
F = 45.48, p-value = 0. The model is valid. Second–order model A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.9408 4 Multiple R 0.8852 5 R Square 0.8524 6 Adjusted R Square 1.67 7 Standard Error 8 Observations 10 9 10 ANOVA 11 df SS 12 Regression 2 151.28 13 Residual 7 19.62 14 Total 9 170.90 15 16 Coefficients Standard Error 17 Intercept 43.73 1.97 18 Day -2.49 0.822 19 Dat-sq 0.106 0.073
D
E
MS 75.64 2.80
F Significance F 26.98 0.0005
t Stat P-value 22.21 0.0000 -3.03 0.0191 1.46 0.1889
F = 26.98, p-value = .0005. The model is valid. c The second–order model is only slightly better because its standard error of estimate is smaller.
674
F
18.6a MBA GPA= 0 + 1 UnderGPA + 2 GMAT + 3 Work + 4 UnderGPA GMAT + b A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.6836 4 Multiple R 0.4674 5 R Square 0.4420 6 Adjusted R Square 0.790 7 Standard Error 8 Observations 89 9 10 ANOVA 11 df SS 12 Regression 4 45.97 13 Residual 84 52.40 14 Total 88 98.37 15 16 Coefficients Standard Error 17 Intercept -11.11 14.97 18 UnderGPA 1.19 1.46 19 GMAT 0.0311 0.0255 20 Work 0.0956 0.0312 21 UGPA-GMAT -0.0019 0.0025
D
E
F
MS 11.49 0.62
F Significance F 18.43 0.0000
t Stat P-value -0.74 0.4601 0.82 0.4159 1.22 0.2265 3.06 0.0030 -0.78 0.4392
F = 18.43, p-value = 0; s = .790 and R 2 = .4674. The model is valid, but the fit is relatively poor. c MBA example s = .788 and R 2 = .4635. There is little difference between the fits of the two models.
18.7 a (Excel output shown below) b
H 0 : 1 2 3 0 H1 : At least on i is not equal to 0
F = 80.65, p-value = 0. There is enough evidence to infer that the model is valid.
675
A B C D E F 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.9330 5 R Square 0.8705 6 Adjusted R Square 0.8597 7 Standard Error 4745 8 Observations 40 9 10 ANOVA 11 df SS MS F Significance F 12 Regression 3 5,446,857,189 1,815,619,063 80.65 0.0000 13 Residual 36 810,445,659 22,512,379 14 Total 39 6,257,302,848 15 16 Coefficients Standard Error t Stat P-value 17 Intercept -82,044 48,530 -1.69 0.0996 18 Home % 98,443 97,463 1.01 0.3192 19 Visiting % 106,779 98,313 1.09 0.2846 20 Home-Visit 53,204 196,610 0.27 0.7882
18.8a A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.9255 5 R Square 0.8566 6 Adjusted R Square 0.8362 7 Standard Error 5.20 8 Observations 25 9 10 ANOVA 11 df SS 3 3398.7 12 Regression 21 568.8 13 Residual 14 Total 24 3967.4 15 16 Coefficients Standard Error 17 Intercept 260.7 162.3 18 Temperature -3.32 2.09 19 Currency -164.3 667.1 20 Temp-Curr 3.64 8.54
676
D
E
MS 1132.9 27.08
F Significance F 41.83 0.0000
t Stat P-value 1.61 0.1230 -1.59 0.1270 -0.25 0.8078 0.43 0.6741
F
b C B A 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.9312 5 R Square 0.8671 6 Adjusted R Square 0.8322 7 Standard Error 5.27 25 8 Observations 9 10 ANOVA SS 11 df 3440.3 5 12 Regression 527.1 19 13 Residual 3967.4 24 14 Total 15 16 Coefficients Standard Error 283.8 274.8 17 Intercept 6.88 -1.72 18 Temperature 888.5 -828.6 19 Currency 0.0475 -0.0024 20 Temp-sq 1718.5 2054.0 21 Curr-sq 10.57 -0.870 22 Temp-Curr
F
D
E
MS 688.1 27.74
Significance F F 0.0000 24.80
P-value t Stat 0.3449 0.97 0.8053 -0.25 0.3627 -0.93 0.9608 -0.05 0.2467 1.20 0.9353 -0.08
c Both models fit equally well. The standard errors of estimate and coefficients of determination are quite similar.
18.9a A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.3788 4 Multiple R 0.1435 5 R Square 0.1167 6 Adjusted R Square 1.58 7 Standard Error 8 Observations 100 9 10 ANOVA 11 df SS 12 Regression 3 40.38 13 Residual 96 241.06 14 Total 99 281.44 15 16 Coefficients Standard Error 17 Intercept -4.86 1.83 18 Faceoff 0.121 0.0366 19 PM-diff 0.135 0.399 20 Face-PM -0.0009 0.0080
677
D
E
F
MS 13.46 2.51
F
Significance F 0.0019
5.36
t Stat P-value -2.66 0.0092 3.31 0.0013 0.34 0.7360 -0.12 0.9086
b
H 0 : 1 2 3 0 H1 : At least on i is not equal to 0
F = 5.36, p-value = .0019. There is enough evidence to infer that the model is valid. c
H 0 : 3 0 H1 : 3 0
t = –.12, p-value = .9086. There is not enough evidence to infer that there is an interaction effect between face–offs won and penalty minutes differential.
18.10a Yield = 0 + 1 Pressure + 2 Temperature + 3 Pressure
2
2
+ 4 Temperature + 5 Pressure Temperature + b A B C D E F 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.8290 4 Multiple R 0.6872 5 R Square 0.6661 6 Adjusted R Square 512 7 Standard Error 8 Observations 80 9 10 ANOVA 11 df SS MS F Significance F 12 Regression 5 42,657,846 8,531,569 32.52 0.0000 13 Residual 74 19,413,277 262,342 14 Total 79 62,071,123 15 16 Coefficients Standard Error t Stat P-value 17 Intercept 74462 7526 9.89 0.0000 18 Pressure 14.40 5.92 2.43 0.0174 19 Temperature -613.3 59.95 -10.23 0.0000 20 Press-sq -0.0159 0.0032 -5.04 0.0000 21 Temp-sq 1.23 0.12 9.86 0.0000 22 Press-temp 0.0381 0.0174 2.19 0.0316 2
c s = 512 and R = .6872. The model's fit is good. 18.11 The number of indicator variables is m – 1 = 5 – 1 = 4.
678
18.12 a
I 1 = 1 if Catholic I 1 = 0 otherwise
I 2 = 1 if Protestant I 2 = 0 otherwise b
I 1 = 1 if 8:00 A.M. to 4:00 P.M. I 1 = 0 otherwise
I 2 = 1 if 4:00 P.M. to midnight
I 2 = 0 otherwise c
I 1 = 1 if Jack Jones I 1 = 0 otherwise
I 2 = 1 if Mary Brown I 2 = 0 otherwise I 3 = 1 if George Fosse I 3 = 0 otherwise 18.13 a Macintosh
18.14
b IBM
c other
I1 1 if B.A. = 0 otherwise
I 2 1 if B.B.A. = 0 otherwise
I 3 1 if B.Sc. or B.Eng. = 0 otherwise
679
I1: t = -1.54, p-value = .1269 I2: t = 2.93, p-value = .0043 I3: t = .166, p-value = .8684 Only I2 is statistically significant. However, this allows us to answer the question affirmatively. 18.15a
A B C 1 Prediction Interval 2 MBA GPA 3 4 10.11 5 Predicted value 6 7 Prediction Interval 8.55 8 Lower limit 11.67 9 Upper limit 10 11 Interval Estimate of Expected Value 9.53 12 Lower limit 10.68 13 Upper limit
D
Prediction: MBA GPA will lie between 8.55 and 11.67
680
b
A B C 1 Prediction Interval 2 MBA GPA 3 4 9.73 5 Predicted value 6 7 Prediction Interval 8.15 8 Lower limit 11.31 9 Upper limit 10 11 Interval Estimate of Expected Value 9.10 12 Lower limit 10.36 13 Upper limit
D
Prediction: MBA GPA will lie between 8.15 and 11.31
18.16a A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.8973 4 Multiple R 0.8051 5 R Square 0.7947 6 Adjusted R Square 2.32 7 Standard Error 8 Observations 100 9 10 ANOVA 11 df SS 12 Regression 5 2096.3 13 Residual 94 507.5 14 Total 99 2603.8 15 16 Coefficients Standard Error 17 Intercept 23.57 5.98 18 Mother 0.306 0.0542 19 Father 0.303 0.0476 20 Gmothers 0.0316 0.0577 21 Gfathers 0.0778 0.0573 22 Smoker -3.72 0.669
D
E
MS 419.26 5.40
F Significance F 77.66 0.0000
t Stat P-value 3.94 0.0002 5.65 0.0000 6.37 0.0000 0.55 0.5853 1.36 0.1777 -5.56 0.0000
b Exercise 18.10: ŷ = 3.24 + .451Mother + .411Father + .0166Gmothers + .0869Gfathers There are large differences to all the coefficients. c
H 0 : 5 0 H1 : 5 0
t = –5.56, p-value = 0. There is enough evidence to infer that smoking affects longevity. 681
F
18.17a A B C D 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.8368 4 Multiple R 0.7002 5 R Square 0.6659 6 Adjusted R Square 810.8 7 Standard Error 8 Observations 40 9 10 ANOVA 11 df SS MS 12 Regression 4 53,729,535 13,432,384 13 Residual 35 23,007,438 657,355 14 Total 39 76,736,973 15 16 Coefficients Standard Error t Stat 17 Intercept 3490 469.2 7.44 18 Yest Att 0.369 0.078 4.73 19 I1 1623 492.5 3.30 20 I2 733.5 394.4 1.86 21 I3 -765.5 484.7 -1.58
b
E
F Significance F 20.43 0.0000
P-value 0.0000 0.0000 0.0023 0.0713 0.1232
H 0 : 1 2 3 4 0 H1 : At least on i is not equal to 0
F = 20.43, p-value = 0. There is enough evidence to infer that the model is valid. c
H 0 : i 0 H1 : i 0
I 2 : t = 1.86, p-value = .0713 I 3 : t = –1.58, p-value = .1232 Weather is not a factor in attendance. d
H0 : 2 0 H1 : 2 > 0
t = 3.30, p-value = .0023/2 = .0012. There is sufficient evidence to infer that weekend attendance is larger than weekday attendance.
682
F
18.18a
A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.5602 4 Multiple R 0.3138 5 R Square 0.2897 6 Adjusted R Square 5.84 7 Standard Error 8 Observations 60 9 10 ANOVA 11 df SS 12 Regression 2 887.9 13 Residual 57 1941.7 14 Total 59 2829.6 15 16 Coefficients Standard Error 17 Intercept 7.02 3.24 18 Length 0.250 0.056 19 Type -1.35 0.947 b
D
E
MS 443.95 34.06
F Significance F 13.03 0.0000
t Stat P-value 2.17 0.0344 4.46 0.0000 -1.43 0.1589
H0 : 2 0 H1 : 2 0
t = –1.43, p-value = .1589. There is not enough evidence to infer that the type of commercial affects memory test scores. c Let
I1 = 1 if humorous I1 = 0 otherwise I 2 = 1 if musical I 2 = 0 otherwise See Excel output below. d
H 0 : i 0 H1 : i 0
I1: t = 1.61, p-value = .1130 I2: t = 3.01, p-value = .0039 There is enough evidence to infer that there is a difference in memory test scores between watchers of humorous and serious commercials. e The variable type of commercial in parts (a) and (b) is nominal. It is usually meaningless to conduct a regression analysis with such variables without converting them to indicator variables.
683
F
A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.6231 5 R Square 0.3882 6 Adjusted R Square 0.3554 7 Standard Error 5.56 8 Observations 60 9 10 ANOVA 11 df SS 12 Regression 3 1099 13 Residual 56 1731 14 Total 59 2830 15 16 Coefficients Standard Error 17 Intercept 2.53 2.15 18 Length 0.223 0.054 19 I1 2.91 1.81 20 I2 5.50 1.83
D
E
MS 366.17 30.91
F Significance F 11.85 0.0000
F
t Stat P-value 1.18 0.2445 4.10 0.0001 1.61 0.1130 3.01 0.0039
18.19 a A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.9233 5 R Square 0.8525 6 Adjusted R Square 0.8429 7 Standard Error 6.25 8 Observations 50 9 10 ANOVA 11 df SS 12 Regression 3 10384 13 Residual 46 1796 14 Total 49 12181 15 16 Coefficients Standard Error 17 Intercept -41.42 7.00 18 Boxes 0.644 0.050 19 Weight 0.349 0.075 20 Codes 4.54 1.21
b Let
I1 = 1 if morning I1 = 0 otherwise I 2 = 1 if early afternoon I 2 = 0 otherwise 684
D
E
MS 3461.40 39.05
F Significance F 88.64 0.0000
t Stat P-value -5.92 0.0000 12.79 0.0000 4.65 0.0000 3.76 0.0005
F
A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.9727 4 Multiple R 0.9461 5 R Square 0.9414 6 Adjusted R Square 3.82 7 Standard Error 8 Observations 50 9 10 ANOVA 11 df SS 12 Regression 4 11525 13 Residual 45 655.9 14 Total 49 12181 15 16 Coefficients Standard Error 17 Intercept -29.72 3.73 18 Boxes 0.618 0.031 19 Weight 0.346 0.046 20 I1 -6.76 1.50 21 I2 6.48 1.45
D
MS 2881.16 14.58
E
F Significance F 197.66 0.0000
t Stat P-value -7.97 0.0000 19.99 0.0000 7.54 0.0000 -4.51 0.0000 4.47 0.0001
c Model 1: s = 6.25 and R 2 = .8525. Model 2: s = 3.82 and R 2 = .9461. The second model fits better. d
H 0 : i 0 H1 : i 0
I1: t = –4.51, p-value = 0. There is enough evidence to infer that the average time to unload in the morning is different from that in the late afternoon. I2: t = 4.47, p-value = .0001. There is enough evidence to infer that the average time to unload in the early afternoon is different from that in the late afternoon.
18.20a Let
I1 = 1 if no scorecard I1 = 0 otherwise I 2 = 1 if scorecard overturned more than 10% of the time I 2 = 0 otherwise
685
F
b A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.7299 5 R Square 0.5327 6 Adjusted R Square 0.5181 7 Standard Error 4.20 8 Observations 100 9 10 ANOVA 11 df SS 12 Regression 3 1933 13 Residual 96 1696 14 Total 99 3629 15 16 Coefficients Standard Error 17 Intercept 4.65 2.06 18 Loan Size 0.00012 0.00015 19 I1 4.08 1.14 20 I2 10.18 1.01
D
E
F
MS 644.46 17.67
F Significance F 36.48 0.0000
t Stat P-value 2.26 0.0260 0.83 0.4084 3.57 0.0006 10.08 0.0000
c s = 4.20 and R 2 = .5327. The model's fit is mediocre. d
1 2 3 4 5
A
B Pct Bad
C Loan Size
D I1
Pct Bad Loan Size I1 I2
1 0.1099 -0.1653 0.6835
1 -0.0346 0.0737
1 -0.5471
E I2
1
There is a high correlation between I1 and I 2 that may distort the t–tests. e b1 =.00012; in this sample for each additional dollar lent the default rate increases by .00012 provided the other variables remain the same.
b 2 = 4.08; In this sample banks that don't use scorecards on average have default rates 4.08 percentage points higher than banks that overturn their scorecards less than 10% of the time.
b 3 = 10.18; In this sample banks that overturn their scorecards more than 10% of the time on average have default rates 10.18 percentage points higher than banks that overturn their scorecards less than 10% of the time.
686
f
A B C 1 Prediction Interval 2 Pct Bad 3 4 9.94 5 Predicted value 6 7 Prediction Interval 1.39 8 Lower limit 18.49 9 Upper limit 10 11 Interval Estimate of Expected Value 8.08 12 Lower limit 11.81 13 Upper limit
D
We predict that the bank's default rate will fall between 1.39 and 18.49%.
18.21 a Let
I1 = 1 if welding machine I1 = 0 otherwise I 2 = 1 if lathe I 2 = 0 otherwise A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.7706 5 R Square 0.5938 6 Adjusted R Square 0.5720 7 Standard Error 48.59 8 Observations 60 9 10 ANOVA 11 df SS 12 Regression 3 193,271 13 Residual 56 132,223 14 Total 59 325,494 15 16 Coefficients Standard Error 17 Intercept 119.3 35.00 18 Age 2.54 0.402 19 I1 -11.76 19.70 20 I2 -199.4 30.71
687
D
E
MS 64,424 2,361
F Significance F 27.29 0.0000
t Stat P-value 3.41 0.0012 6.31 0.0000 -0.60 0.5531 -6.49 0.0000
F
b b1 = 2.54; in this sample for each additional month repair costs increase on average by $2.54 provided that the other variable remains constant.
b 2 = –11.76; in this sample welding machines cost on average $11.76 less to repair than stamping machines for the same age of machine.
b 3 = –199.4; in this sample lathes cost on average $199.40 less to repair than stamping machines for the same age of machine. c
H 0 : 2 0 H1 : 2 < 0
t = –.60, p-value .5531/2 = .2766. There is no evidence to infer that welding machines cost less to repair than stamping machines.
18.22
a. The coefficient of determination in Exercise 16.107 was .3270. In this model the coefficient of determination is .6385. This model is better.
688
b
Lower prediction limit = 150.5, upper prediction limit = 174.1 c
Lower prediction limit = 162.6, upper prediction limit = 186.3
d No, because the width of the prediction intervals are far too wide.
689
18.23a
A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.7296 5 R Square 0.5323 6 Adjusted R Square 0.5075 7 Standard Error 2.36 8 Observations 100 9 10 ANOVA 11 df SS 12 Regression 5 593.9 13 Residual 94 521.7 14 Total 99 1115.6 15 16 Coefficients Standard Error 17 Intercept 10.26 1.17 18 Wage -0.00020 0.000036 19 Pct PT -0.107 0.029 20 Pct U 0.060 0.012 21 Av Shift 1.56 0.50 22 UM Rel -2.64 0.492 b
D
E
F
MS 118.78 5.55
F Significance F 21.40 3.08E-14
t Stat P-value 8.76 8.12E-14 -5.69 1.43E-07 -3.62 0.0005 4.83 5.38E-06 3.11 0.0025 -5.36 5.99E-07
H0 : 4 0 H1 : 4 0
t = 3.11, p-value = .0025. There is enough evidence to infer that the availability of shiftwork affects absenteeism. c
H 0 : 5 0 H1 : 5 < 0
t = –5.36, p-value = (5.99E-07) /2 = 3.00×10-7= virtually 0. There is enough evidence to infer that in organizations where the union–management relationship is good absenteeism is lower.
690
18.24a Let I1= 1 if respondent works for him or herself (0 if not)
b. b4 = 8429; after removing the effect of age, education, and weekly hours of work people who work for themselves earn on average $8429 more than people who work for someone else. c.
H0:β4 = 0 H1:β4 > 0
t = 3.07, p-value = .0022/2 = .0011. There is enough evidence to infer that people who work for themselves have larger incomes after removing the effect of age, education, and weekly hours of work.
18.25 Let
I1 = 1 if Democrat (PARTYID3 = 1) = 0 if not I2 = 1 if Republican (PARTYID3 - 3 = 0, if not
691
a.
H0:β5 = 0 H1:β5 < 0
t = -6.77, p-value = 0. There is sufficient evidence to infer that Democrats are more likely than Independents to believe that government should reduce income differences after removing the effects of age, income, education, and weekly hours. b.
H0:β6 = 0 H1:β6 > 0
t = 7.06, p-value = 0. There is sufficient evidence to infer that Republicans are more likely than Independents to believe that government should take no action to reduce income differences after removing the effects of age, income, education, and weekly hours.
18.26
I1 = 1 if Liberal (POLVIEWS3 = 1) = 0 if not I2 = 1 if Conservative (POLVIEWS3 = 3 = 0, if not
692
a.
H0:β5 = 0 H1:β5 < 0
t = -5.79, p-value = 0. There is sufficient evidence to infer that liberals are more likely than moderates to believe that government should reduce income differences after removing the effects of age, income, education, and weekly hours. b.
H0:β6 = 0 H1:β6 > 0
t = 6.47, p-value = 0. There is sufficient evidence to infer that conservatives are more likely than moderates to believe that government should take no action to reduce income differences after removing the effects of age, income, education, and weekly hours.
693
18.27
I1 = 1 if Male = 0 if not
H0:β3 = 0 H1:β3 ≠ 0 t = .735, p-value = .4624. There is not enough evidence to infer that men and women differ in the amount of television per day after removing the effects of age and education.
18.28 Let
I1 = 1 if Liberal (POLVIEWS3 = 1) = 0 if not I2 = 1 if Conservative (POLVIEWS3 = 3) = 0, if not
694
a.
H0:β5 = 0 H1:β5 < 0
t = -4.59, p-value = 0. There is sufficient evidence to infer that liberals are more likely than moderates to believe that government help poor people after removing the effects of age, income, education, and weekly hours. b.
H0:β6 = 0 H1:β6 > 0
t = 4.59, p-value = 0. There is sufficient evidence to infer that conservatives are more likely than moderates to believe that people should help themselves after removing the effects of age, income, education, and weekly hours.
18.29
Let I1= 1 if respondent works for government (0 if not)
695
H0:β4 = 0 H1:β4 ≠ 0 t = -2.16, p-value = .0312. There is enough evidence to infer that there are differences in mean income between people who work for the government and people who work for private employers after removing the effects of age, education, and weekly hours of work.
18.30
I1= 1, if White =0, if not I2 =1, if black = 0, if not
696
Compare whites and others H0:β3 = 0 H1:β3 ≠ 0 t = -1.12, p-value = .2639
Compare blacks and others H0:β4 = 0 H1:β4 ≠ 0 t = 4.00, p-value = 0. There is enough evidence to infer that there are differences between the blacks and others after removing the effects of age and education.
18.31
I1 = 1 if Democrat (PARTYID3 = 1) = 0 if not I2 = 1 if Republican (PARTYID3 - 3 = 0, if not
697
a.
H0:β5 = 0 H1:β5 < 0
t = -5.83, p-value = 0. There is sufficient evidence to infer that Democrats are more likely than Independents to believe that government should help poor people after removing the effects of age, income, education, and weekly hours. b.
H0:β6 = 0 H1:β6 > 0
t = 5.94, p-value = 0. There is sufficient evidence to infer that Republicans are more likely than Independents to believe that people should help themselves after removing the effects of age, income, education, and weekly hours.
18.32
I1 = 1 if born in the U.S. (0 if not)
698
H0:β4 = 0 H1:β4 ≠ 0 t = -1.35, p-value = .1768. There is not enough evidence to infer that there are differences in income between Americans born in the United States and those born elsewhere after removing the effects of age, education , and weekly hours of work.
18.33
I1 = 1 if UNION = 1, 2, or 3 I1 = 0 if UNION = 4 (neither belong)
H0:β4 = 0 H1:β4 ≠ 0
699
t = .356, p-value = .7216. There is not enough evidence to infer that union membership affects income after removing the effects of age, education, and hours of work.
18.34 A B C D E F 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.9737 5 R Square 0.9482 6 Adjusted R Square 0.9454 7 Standard Error 3015 8 Observations 100 9 10 ANOVA 11 df SS MS F Significance F 12 Regression 5 15,636,303,318 3,127,260,664 344.04 0.0000 13 Residual 94 854,451,113 9,089,905 14 Total 99 16,490,754,431 15 16 Coefficients Standard Error t Stat P-value 17 Intercept -5916 3141 -1.88 0.0627 18 Years 1022 48.93 20.88 0.0000 19 PhD 725.7 961.5 0.75 0.4523 20 Evaluation 3729 619.8 6.02 0.0000 21 Articles 439.1 80.7 5.44 0.0000 22 Gender 1090 632.0 1.72 0.0879
a
H 0 : 1 2 3 4 5 0 H1 : At least on i is not equal to 0
F = 344.04, p-value = 0. There is enough evidence to infer that the model is valid. b
H 0 : 5 0 H1 : 5 > 0
t = 1.72, p-value = .0879/2 = .0440. There is evidence that male professors are better paid than female professors with the same qualifications.
700
18.35 A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.8311 4 Multiple R 0.6907 5 R Square 0.5670 6 Adjusted R Square 1.86 7 Standard Error 8 Observations 8 9 10 ANOVA 11 df SS 12 Regression 2 38.47 13 Residual 5 17.23 14 Total 7 55.70 15 16 Coefficients Standard Error 17 Intercept 2.01 4.02 18 Score 3.25 1.00 19 Gender -0.039 1.35
D
E
F
MS 19.24 3.45
F
Significance F 0.0532
5.58
t Stat P-value 0.50 0.6385 3.25 0.0227 -0.03 0.9782
In this case male–dominated jobs are paid on average $.039 (3.9 cents) less than female– dominated jobs after adjusting for the value of each job.
18.36 All weights = .2 A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.7623 4 Multiple R 0.5812 5 R Square 0.4136 6 Adjusted R Square 2.16 7 Standard Error 8 Observations 8 9 10 ANOVA 11 df SS 12 Regression 2 32.37 13 Residual 5 23.33 14 Total 7 55.70 15 16 Coefficients Standard Error 17 Intercept 4.70 4.07 18 Score 2.57 1.01 19 Gender 0.26 1.56
701
D
E
F
MS 16.19 4.67
F
Significance F 0.1135
3.47
t Stat P-value 1.15 0.3011 2.55 0.0514 0.16 0.8761
In this case male–dominated jobs are paid on average $.26 (26 cents) more than female–dominated jobs after adjusting for the value of each job.
18.37 The strength of this approach lies in regression analysis. This statistical technique allows us to determine whether gender is a factor in determining salaries. However, the conclusion is very much dependent upon the subjective assignment of weights. Change the value of the weights and a totally different conclusion is achieved.
18.38a.
b. In this regression analysis the variables DAYS2 and DAYS3 are not included in the equation.
18.39a
702
b. The variables Enrollment and Distance are excluded. 18.40
18.41
703
18.42
2
18.43a Mileage = 0 + 1 Speed + 2 Speed + b
704
A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.8428 4 Multiple R 0.7102 5 R Square 0.6979 6 Adjusted R Square 3.86 7 Standard Error 8 Observations 50 9 10 ANOVA 11 df SS 12 Regression 2 1719.9 13 Residual 47 701.64 14 Total 49 2421.5 15 16 Coefficients Standard Error 17 Intercept 9.34 1.71 18 Speed 0.802 0.077 19 Speed-sq -0.0079 0.00073
D
E
MS 859.94 14.93
F Significance F 57.60 0.0000
F
t Stat P-value 5.47 0.0000 10.39 0.0000 -10.73 0.0000
c s = 3.86 and R 2 = .7102. The model fits moderately well.
18.44a Apply a first–order model with interaction. b A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.8623 5 R Square 0.7436 6 Adjusted R Square 0.7299 7 Standard Error 1.27 8 Observations 60 9 10 ANOVA 11 df SS 12 Regression 3 260.2 13 Residual 56 89.72 14 Total 59 349.9 15 16 Coefficients Standard Error 17 Intercept 640.8 53.80 18 Cars -64.17 5.27 19 Speed -10.63 0.897 20 Cars-Speed 1.08 0.088
c:
H 0 : 1 2 3 0 H1 : At least on i is not equal to 0
705
D
E
MS 86.74 1.60
F Significance F 54.14 0.0000
t Stat P-value 11.91 0.0000 -12.19 0.0000 -11.85 0.0000 12.26 0.0000
F
F = 54.14, p-value = 0. There is enough evidence to infer that the model is valid.
18.45a A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.8668 5 R Square 0.7514 6 Adjusted R Square 0.7284 7 Standard Error 1.27 8 Observations 60 9 10 ANOVA 11 df SS 12 Regression 5 262.95 13 Residual 54 86.99 14 Total 59 349.93 15 16 Coefficients Standard Error 17 Intercept 404.5 327.0 18 Cars -66.57 6.54 19 Speed -2.35 10.54 20 Cars-sq 0.107 0.097 21 Speed-sq -0.070 0.085 22 Cars-Speed 1.08 0.096
D
E
MS 52.59 1.61
F Significance F 32.65 0.0000
t Stat P-value 1.24 0.2214 -10.19 0.0000 -0.22 0.8246 1.10 0.2741 -0.82 0.4180 11.21 0.0000
b F = 32.65, p-value = 0. There is enough evidence to infer that the model is valid.
18.46 a Let
I1 = 1 if ad was in newspaper I1 = 0 otherwise I 2 = 1 if ad was on radio I 2 = 0 otherwise
b
706
F
A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.6946 5 R Square 0.4824 6 Adjusted R Square 0.4501 7 Standard Error 44.87 8 Observations 52 9 10 ANOVA 11 df SS 12 Regression 3 90057 13 Residual 48 96627 14 Total 51 186684 15 16 Coefficients Standard Error 17 Intercept 282.6 17.46 18 Ads 25.23 3.98 19 I1 -23.36 15.83 20 I2 -46.59 16.44
b
D
E
MS 30019 2013
F Significance F 14.91 0.0000
t Stat P-value 16.19 0.0000 6.34 0.0000 -1.48 0.1467 -2.83 0.0067
H 0 : 1 2 3 0 H1 : At least on i is not equal to 0
F = 14.91, p-value = 0. There is enough evidence to infer that the model is valid. c
H 0 : i 0 H1 : i 0
I–1: t = –1.48, p-value = .1467 I–2: t = –2.83, p-value = .0067 There is enough evidence to infer that the advertising medium makes a difference.
18.47 (See Excel output below) b
H 0 : 3 0 H1 : 3 < 0
t = –8.61, p-value = 0. There is enough evidence to infer that a team that fires its manager within 12 months wins less frequently than other teams.
707
F
A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.9347 5 R Square 0.8736 6 Adjusted R Square 0.8654 7 Standard Error 0.0183 8 Observations 50 9 10 ANOVA 11 df SS 12 Regression 3 0.107 13 Residual 46 0.0154 14 Total 49 0.122 15 16 Coefficients Standard Error 17 Intercept 0.357 0.0592 18 BA -0.401 0.236 19 ERA 0.0764 0.00478 20 Fired -0.0509 0.00591
D
MS 0.0355 0.00034
E
F
F Significance F 106.01 0.0000
t Stat 6.03 -1.70 15.98 -8.61
P-value 0.0000 0.0964 0.0000 0.0000
D
E
MS 87,646 7,740
F Significance F 11.32 0.0000
2
18.48a Units = 0 + 1 Years + 2 Years + b A B C 1 SUMMARY OUTPUT 2 Regression Statistics 3 0.4351 4 Multiple R 0.1893 5 R Square 0.1726 6 Adjusted R Square 87.98 7 Standard Error 8 Observations 100 9 10 ANOVA 11 df SS 12 Regression 2 175,291 13 Residual 97 750,764 14 Total 99 926,056 15 16 Coefficients Standard Error 17 Intercept 331.2 17.55 18 Years 21.45 5.50 19 Years-sq -0.848 0.325
c s = 87.98 and R 2 = .1893. The model fits poorly.
708
t Stat P-value 18.87 0.0000 3.90 0.0002 -2.61 0.0105
F
2
18.49a Depletion = 0 + 1 Temperature + 2 PH–level + 3 PH–level + 4 I 4 + 5 I 5 + where
I1 = 1 if mainly cloudy I1 = 0 otherwise I 2 = 1 if sunny I 2 = 0 otherwise b A B C 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.8085 5 R Square 0.6537 6 Adjusted R Square 0.6452 7 Standard Error 4.14 8 Observations 210 9 10 ANOVA 11 df SS 12 Regression 5 6596 13 Residual 204 3495 14 Total 209 10091 15 16 Coefficients Standard Error 17 Intercept 1003 55.12 18 Temperature 0.194 0.029 19 PH Level -265.6 14.75 20 PH-sq 17.76 0.983 21 I1 -1.07 0.700 22 I2 1.16 0.700
c
D
E
MS 1319 17.13
F Significance F 77.00 0.0000
t Stat P-value 18.19 0.0000 6.78 0.0000 -18.01 0.0000 18.07 0.0000 -1.53 0.1282 1.65 0.0997
H 0 : 1 2 3 4 5 0 H1 : At least on i is not equal to 0
F = 77.00, p-value = 0. There is enough evidence to infer that the model is valid. d
H 0 : 1 0 H1 : 1 > 0
t = 6.78, p-value = 0. There is enough evidence to infer that higher temperatures deplete chlorine more quickly. e
H 0 : 3 0 H1 : 3 > 0
t = 18.07, p-value = 0. There is enough evidence to infer that there is a quadratic relationship between chlorine depletion and PH level. 709
F
f
H 0 : i 0 H1 : i 0
I1 : t = –1.53, p-value = .1282. There is not enough evidence to infer that chlorine depletion differs between mainly cloudy days and partly sunny days.
I 2 : t = 1.65, p-value = .0997. There is not enough evidence to infer that chlorine depletion differs between sunny days and partly sunny days. Weather is not a factor in chlorine depletion.
710
Chapter 19 19.1
H0: The two population locations are the same H1: The location of population 1 is different from the location of population 2
E (T ) T
n1 (n1 n 2 1) 15(15 15 1) 232 .5 2 2 n1n 2 (n1 n 2 1) (15)(15)(15 15 1) 24 .11 12 12
a z
T E(T) 250 232 .5 = = .73, p-value = 2P(Z > .73) = 2(1 – .7673) = .4654. 24 .11 T
b z
T E(T) 275 232 .5 = = 1.76, p-value 2P(Z > 1.76) = 2(1 – .9608) = .0784. 24 .11 T
c The value of the test statistic increases and the p-value decreases.
19.2
H 0 : The two population locations are the same H1 : The location of population 1 is to the right of the location of population 2
Rejection region: z z z .05 = 1.645
E (T )
n 1 (n 1 n 2 1) 30 (30 40 1) 1,065 2 2
T
n 1 n 2 (n 1 n 2 1) 12
a z
(30 )( 40 )(30 40 1) 84 .26 12
T E(T) 1,205 1,065 = = 1.66, p-value P(Z > 1.66) = 1 – .9515 = .0485. There is enough 84 .26 T
evidence to infer that the location of population 1 is to the right of the location of population 2. b z
T E(T) 1,065 1,065 = = 0, p-value = P(Z > 0) = .5. There is not enough evidence to infer 84 .26 T
that the location of population 1 is to the right of the location of population 2. c The value of the test statistic decreases and the p-value increases.
19.3
H 0 : The two population locations are the same H1 : The location of population 1 is to the left of the location of population 2
Rejection region: T TL = 19
707
Sample 1 75 60 73 66 81
Rank Sample 2 Rank 5 90 9 1 72 3 4 103 10 2 82 8 7 78 6 T1 = 19 T2 = 36 There is enough evidence to infer that the location of population 1 is to the left of the location of population 2.
19.4
H 0 : The two population locations are the same H1 : The location of population 1 is different from the location of population 2
Rejection region: T TU = 127 or T TL = 83 Sample 1 Rank Sample 2 Rank 15 4.0 8 2.0 7 1.0 27 18.0 22 14.0 17 7.0 20 .5 25 16.0 32 20.0 20 11.5 18 9.5 16 5.0 26 17.0 21 13.0 17 7.0 17 7.0 23 15.0 10 3.0 30 19.0 18 9.5 T1 = 118 T2 = 92 There is not enough evidence to infer that the location of population 1 is different from the location of population 2.
19.5
H 0 : The two population locations are the same H1 : The location of population 1 is to the left of the location of population 2
A B C D 1 Wilcoxon Rank Sum Test 2 Rank Sum Observations 3 4 New 623.5 25 651.5 25 5 Leading 6 z Stat -0.2716 7 P(Z<=z) one-tail 0.3929 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.7858 1.96 10 z Critical two-tail
E
a z = –.27, p-value = .3929. There is not enough evidence to infer that the new beer is less highly rated than the leading brand. b The printout is identical to that of part a. c All codes that preserve the order produce the same results. 708
19.6
H 0 : The two population locations are the same H1 : The location of population 1 is different from the location of population 2
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 Business 4004 40 5 Economy 8086 115 6 z Stat 3.6149 7 P(Z<=z) one-tail 0.0002 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.0004 10 z Critical two-tail 1.96
E
a z = 3.61, p-value = .0004. There is enough evidence to infer that the business and economy class differ in their degree of satisfaction. b The printout is identical to that of part a. c All codes that preserve the order produce the same results.
19.7
H 0 : The two population locations are the same H1 : The location of population 1 is to the right of the location of population 2
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 New 276.5 15 5 Aspirin 188.5 15 6 z Stat 1.825 7 P(Z<=z) one-tail 0.034 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.068 10 z Critical two-tail 1.96
E
a z = 1.83, p-value = .0340. There is enough evidence to infer that the new painkiller is more effective than aspirin. b The results are identical because the codes in this exercise and in Example 19.2 are ranked identically.
19.8
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
Rejection region: z z z.05 = 1.645
709
E(T)
T
z
n1 (n1 n 2 1) 82(82 75 1) 6478 2 2 n1n 2 (n1 n 2 1) (82 )(75)(82 75 1) 284 .6 12 12
T E(T) 6,807 6,478 = = 1.16, p-value P(Z > 1.16) = 1 – .8770 = .1230. There is not 284 .6 T
enough evidence to infer that members of the Mathematics department rate nonparametric techniques as more important than do members of other departments.
19.9
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
Rejection region: z z z.05 1.645
E (T ) T
z
n1 (n1 n 2 1) 30 (30 30 1) 915 2 2 n1n 2 (n1 n 2 1) (30 )(30 )(30 30 1) 67 .6 12 12
T E(T) 797 915 = 1.75 , p-value P(Z < –1.75) = .0401. There is enough evidence to 67 .6 T
infer that companies that provide exercise programs should be given discounts.
19.10
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
Rejection region: z z z.05 1.645
E (T )
n1 (n1 n 2 1) 125 (125 125 1) 15,687 .5 2 2
T
n1n 2 (n1 n 2 1) 12
z
(125 )(125 )(125 125 1) 571 .7 12
T E(T) 14 ,873 15,687 .5 1.42 , p-value P(Z < – 1.42) =.0778. There is not enough = 571 .7 T
evidence to infer that women are doing less housework today than last year.
19.11
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
Rejection region: z z z.05 1.645
710
E (T ) T
z
n1 (n1 n 2 1) 100 (100 100 1) 10,050 2 2 n1n 2 (n1 n 2 1) (100 )(100 )(100 100 1) 409 .3 12 12
T E(T) 10 ,691 10 ,050 = 1.57 , p-value P(Z > 1.57) = 1 – .9418 = .0582. There is not 409 .3 T
enough evidence to conclude that public support has decreased between this year and last year.
19.12
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2
Rejection region: z z / 2 z.025 1.96 or z z / 2 z .025 1.96
E (T ) T
z
n1 (n1 n 2 1) 50 (50 50 1) 2525 2 2 n1n 2 (n1 n 2 1) (50 )(50 )(50 50 1) 145 .1 12 12
T E(T) 2810 2525 = 1.964 , p-value = 2P(Z > 1.964), which is slightly less than 2P(Z > 145 .1 T
1.96) = 2(1 – .9750) = .0500. There is enough evidence to infer that men and women experience different levels of stomach upset.
19.13
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
Rejection region: z z z.05 1.645
E (T ) T
z
n1 (n1 n 2 1) 15(15 25 1) 307 .5 2 2 n1n 2 (n1 n 2 1) (15)( 25)(15 25 1) 35 .8 12 12
T E(T) 383 .5 307 .55 2.12, p-value = P(Z > 2.12) = 1 – .9830 = .0170. There is = 35 .8 T
enough evidence to infer that Tastee is superior.
19.14
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
Rejection region: z z z.05 1.645
711
E (T ) T
z
n1 (n1 n 2 1) 20 (20 20 1) 410 , 2 2 n1n 2 (n1 n 2 1) (20 )( 20 )( 20 20 1) 37 .0 12 12
T E(T) 439 .5 410 = .80 , p-value = P(Z > .80) = 1 – .7881 = .2119. There is not enough 37 .0 T
evidence to infer that women perceive another woman wearing a size 6 dress as more professional than one wearing a size 14 dress. 19.15a. H0: The two population locations are the same H1: The location of population 1 is to the left of the location of population 2 Rejection region: z z z.05 1.645
E (T ) T
z
n1 (n1 n 2 1) 125 (125 125 1) 15,687 .5 2 2 n1n 2 (n1 n 2 1) (125 )(125 )(125 125 1) 571 .7 12 12
T E(T) 13,078 15,687 .5 4.56 , p-value = P(Z < –4.56) = 0. There is enough evidence = 571 .7 T
to infer that changing the name of prunes to dried plums will increase the likelihood that shoppers will buy.
19.16
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2
Rejection region: z z / 2 z.025 1.96 or z z / 2 z .025 1.96
E (T ) T
z
n1 (n1 n 2 1) 182 (182 163 1) 31,486 2 2 n1n 2 (n1 n 2 1) (182 )(163 )(182 163 1) 924 .9 12 12
T E(T) 32,225 .5 31,486 .80 , p-value = 2P(Z > .80) = 2(1 – .7881) = .4238. There is = 924 .9 T
not enough evidence to infer that the night and day shifts rate the service differently. 19.17
H0: The two population locations are the same H1: The location of population 1 (males) is to the left of the location of population 2(females)
712
z = -2.53, p-value = .0116. There is enough evidence to infer that men prefer jobs with higher incomes than do women. 19.18
H0: The two population locations are the same H1: The location of population 1(government worker) is to the left of the location of population 2(private-sector worker)
z = -.179, p-value = .4290. There is not enough evidence to infer that government workers show a greater preference for job security than do private-sector workers. 19.19
H0: The two population locations are the same H1: The location of population 1 (self-employed) is to the left of the location of population 2(work for someone else)
713
z = -3.77, p-value = .0001. There is enough evidence to infer that people who work for themselves prefer shorter work hours with lots of free time than do workers who work for someone else. 19.20
H0: The two population locations are the same H1: The location of population 1 is different from the location of population 2
z = 2.90, p-value = .0036. There is enough evidence to infer that men and women differ in their preference for work that is important and gives a feeling of accomplishment. 19.21
H0: The two population locations are the same
714
H1: The location of population 1 is different from the location of population 2
z = -.345, p-value = .7300. There is not enough evidence to conclude that those who would stop working and those who would continue working differ in their preference for high income. 19.22
H0: The two population locations are the same H1: The location of population 1 (continue working) is to the left of the location of population 2(stop working)
z = -2.18, p-value = .0147. There is enough evidence to conclude that those who would continue working have a higher preference for work they consider important and gives them a feeling of accomplishment. 19.23
H0: The two population locations are the same H1: The location of population 1 is different from the location of population 2
715
z = .493, p-value = .6220. There is not enough evidence to conclude that men and women differ in their preference for jobs where there is a chance for advancement? 19.24
H0: The two population locations are the same H1: The location of population 1 is different from the location of population 2
z = 3.73, p-value = .0002. There is enough evidence to conclude that Democrats and Republicans differ in their views about the federal income tax that they have to pay.
19.25
H0: The two population locations are the same H1: The location of population 1 is different from the location of population 2
716
z = 1.56, p-value = .1198. There is not enough evidence to conclude that people who are selfemployed differ from people who work for someone else differ in their views about the federal income tax that they have to pay.
19.26
H0: The two population locations are the same H1: The location of population 1 (men) is to the right of the location of population 2 (women)
z = .707, p-value = .2399. There is not enough evidence to infer that women are more likely than men to lose their jobs in the next 12 months. 19.27
H0: The two population locations are the same H1: The location of population 1 (men) is to the left of the location of population 2 (women)
717
z = -.017, p-value = .4934. There is not enough evidence to infer that men consider themselves to be healthier than women. 19.28
H0: The two population locations are the same H1: The location of population 1 (2012) is to the left of the location of population 2 (2014)
z = -.452, p-value = .3257. There is not enough evidence to infer that Americans were healthier in 2012 than in 2014. 19.29
H0: The two population locations are the same H1: The location of population 1 (2012) is to the left of the location of population 2 (2014)
718
z = -1.23, p-value = .1089. There is not enough evidence to infer that Americans were more worried about their chances of losing their jobs in 2012 than in 2014. 19.30
H0: The two population locations are the same H1: The location of population 1 (2012) is to the right of the location of population 2 (2014)
z = .630, p-value = .2642. There is not enough evidence to conclude that Americans were more optimistic about their children’s standard of living in 2014 than they were in 2012. 19.31a. Assets are required to be normally distributed. b. The histograms are not bell shaped. c.
H0: The two population locations are the same H1: The location of population 1 (Some college) is to the left of the location of population 2 (College)
719
z = -5.21, p-value = 0. There is enough evidence to infer that heads of households with college degrees have more assets than do heads of households who have some college. 19.32a. Income is required to be normally distributed. The histograms are not bell shaped. The Wilcoxon rank sum test should be used. b.
H0: The two population locations are the same H1: The location of population 1(Work for someone else) is to the right of the location of population 2 (Self-employed0
z = 3.07, p-value = .0011. There is enough evidence to conclude that heads of households who work for someone else have higher incomes than self-employed heads of households. 19.33a. Incomes are required to be normally distributed. b. The histograms are not bell shaped. c.
H0: The two population locations are the same H1: The location of population 1 (Male) is to the right of the location of population 2 (Female)
720
z = 9.91, p-value = 0. There is enough evidence to conclude that male heads of households have higher incomes than female heads of households. 19.34a. Debt is required to be normally distributed. The histograms are not bell shaped. The Wilcoxon rank sum test should be used. b.
H0: The two population locations are the same H1: The location of population 1 (Some college) is to the left of the location of population 2 (College)
z = -4.65, p-value = 0. There is evidence to conclude that heads of households with some college have less debt than heads of heads of households who have graduated college.
19.35
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2
Rejection region: z z / 2 z.025 1.96 or z z / 2 z.025 1.96 z=
x .5n .5 n
=
15 .5(45 ) .5 45
= –2.24, p-value = 2P(Z < – 2.24) = 2(.0125) = .0250. There is enough
evidence to infer that the population locations differ.
721
19.36
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
Rejection region: z z z.10 1.28 z=
x .5n .5 n
=
28 .5(69 ) .5 69
= –1.57, p-value = P(Z < –1.57) = .0582. There is enough evidence to
infer that the location of population 1 is to the left of the location of population 2.
19.37
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of right of the location of population 2
Rejection region: z z z.05 1.645 z=
x .5n .5 n
=
18 .5(30 ) .5 30
= 1.10, p-value = P(Z > 1.10) = 1 – .8643 = .1357. There is not enough
evidence to infer that the location of population 1 is to the right of the location of population 2.
19.38
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
Pair 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
A 5 3 4 2 3 4 3 5 4 3 4 5 4 5 3 2
B 3 2 4 3 3 1 3 4 2 5 1 2 2 3 2 2
Sign of Difference + + 0 – 0 + 0 + + – + + + + + 0
Rejection region: z z z.05 1.645 x = 10, n = 12, z
x .5n .5 n
=
10 .5(12 ) .5 12
= 2.31, p-value = P(Z > 2.31) = 1 – .9896 = .0104
There is enough evidence to infer that the population 1 is located to the right of population 2.
722
19.39
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2
Rejection region: z z / 2 z .025 1.96 or z z / 2 z.025 1.96
E (T )
z
n (n 1) 55(56 ) 770 ; T 4 4
n (n 1)( 2n 1) 24
55(56 )(111) 119 .35 24
T E(T) 660 770 = = –.92, p-value = 2P(z < –.92) = 2(.1788) = .3576. There is not enough 119 .35 T
evidence to infer that the population locations differ.
19.40
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
Rejection region: z z z.01 2.33
E (T )
z
n (n 1) 108 (109 ) 2943 ; T 4 4
n (n 1)( 2n 1) 108 (109 )( 217 ) 326 .25 24 24
T E(T) 3457 2943 = = 1.58, p-value = P(Z > 1.58) = 1 – .9429 = .0571. There is not 326 .25 T
enough evidence to conclude that population 1 is located to the right of the location of population 2.
19.41
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2 Rejection region: T TU = 19 or T TL = 2
Pair 1 2 3 4 5 6
|Difference| Ranks 4 5.5 2 3.5 2 3.5 1 1.5 4 5.5 1 1.5 _____________________________ T = 19.5 T = 1.5 T = 19.5. There is enough evidence to infer that the population locations differ.
19.42
Sample 1 9 12 13 8 7 10
Sample 2 5 10 11 9 3 9
Difference 4 2 2 –1 4 1
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2
Pair 1
Rejection region: T TU = 39 or T TL = 6 Sample 1 Sample 2 Difference |Difference| 18.2 18.2 0 0 723
Ranks
2 3 4 5 6 7 8 9 10 11 12
14.1 24.5 11.9 9.5 12.1 10.9 16.7 19.6 8.4 21.7 23.4
14.1 23.6 12.1 9.5 11.3 9.7 17.6 19.4 8.1 21.9 21.6
0 .9 –.2 0 .8 1.2 –.9 .2 .3 –.2 1.8
0 .9 6.5 .2 2 0 .8 5 1.2 8 .9 6.5 .2 2 .3 4 .2 2 1.8 9 _____________________________ T = 34.5 T = 10.5 T = 34.5. There is not enough evidence to conclude that the population locations differ.
19.43
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
New - Leading 46 30 24 1.84 0.0332 1.6449 0.0664 1.96
a z = 1.84, p-value = .0332. There is enough evidence to indicate that the new beer is more highly rated than the leading brand. b The printout is identical to that of part a. c All codes that preserve the order produce the same results.
19.44
H 0 : H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
724
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
Brand A - Brand B 21 15 14 1.00 0.1587 1.6449 0.3174 1.96
a z = 1.00, p-value = .1587. There is no evidence to infer that Brand A is preferred. b The printout is identical to that of part a. c All codes that preserve the order produce the same results.
19.45
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
European - American 17 6 2 2.29 0.0109 1.6449 0.0218 1.96
a z = 2.29, p-value = .0109. There is enough evidence to infer that the European car is perceived to be more comfortable. b The results are identical. All codes that preserve the order produce the same results.
19.46
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2
725
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
Sample 1 - Sample 2 51 74 0 -2.06 0.0198 1.6449 0.0396 1.96
a. z = –2.06, p-value = .0396. There is enough evidence to infer that the population locations differ. b
A B C D 1 Wilcoxon Signed Rank Sum Test 2 3 Difference Sample 1 - Sample 2 4 5 T+ 3726.5 6 T4148.5 125 7 Observations (for test) -0.52 8 z Stat 0.3016 9 P(Z<=z) one-tail 1.6449 10 z Critical one-tail 0.6032 11 P(Z<=z) two-tail 1.96 12 z Critical two-tail
E
z = –.52, p-value = .6032. There is not enough evidence to infer that the population locations differ. c The sign test ignores the magnitudes of the paired differences whereas the Wilcoxon signed rank sum test does not.
19.47
H 0 : The two population locations are the same
H1 : The location of population 1 is different from the location of population 2
726
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
Sample 1 - Sample 2 19 29 22 -1.44 0.0745 1.6449 0.149 1.96
a z = –1.44, p-value .1490. There is not enough evidence to infer that the population locations differ. b
A B C D 1 Wilcoxon Signed Rank Sum Test 2 Sample 1 - Sample 2 3 Difference 4 304 5 T+ 872 6 T48 7 Observations (for test) -2.91 8 z Stat 0.0018 9 P(Z<=z) one-tail 1.6449 10 z Critical one-tail 0.0036 11 P(Z<=z) two-tail 1.96 12 z Critical two-tail z = –2.91, p-value = .0036. There is enough evidence to conclude that the population locations differ. c The sign test ignores the magnitudes of the paired differences whereas the Wilcoxon signed rank sum test does not.
19.48
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
Rejection region: z z z.05 1.645
E (T )
n (n 1) 72 (72 1) 1314 ; T 4 4
n (n 1)( 2n 1) 24
72 (72 1)( 2[72 ] 1) 178 .2 24
T E(T) 378 .5 1314 = = –5.25, p-value = P(Z < –5.25) = 0. There is enough evidence to 178 .2 T infer that the drug is effective. z
727
19.49
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
Rejection region: z z z.01 2.33
E (T )
z
n (n 1) 40 (40 1) 410 ; T 4 4
n (n 1)( 2n 1) 24
40 (40 1)( 2[40 ] 1) 74 .4 24
T E(T) 62 410 = 4.68, p-value = P(z < –4.68) = 0. There is enough evidence to infer 74 .4 T
that women are doing less housework now than last year.
19.50
H 0 : The two population locations are the same
H 1 : The location of population 1 is to the right of the location of population 2 Rejection region: z z z.05 1.645
x .5n
60 .5(98 )
2.22, p-value = P(Z > 2.22) = 1 – .9868 = .0132. There is enough .5 n .5 98 evidence to conclude that concern about a gasoline shortage exceeded concern about an electricity shortage.
z=
19.51
=
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2
Rejection region T ≤ TL = 73 or T ≥ TU = 203 T = 40.5. There is enough evidence of a difference between machines.
19.52
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
Rejection region: T ≤ TL = 110 T = 111. There is not enough evidence to infer that the swimming department has higher gross sales.
19.53
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
Rejection region: z z z.10 1.28
x .5n
30 .5(38)
3.57 , p-value = P(Z > 3.57) = 0. There is enough evidence to conclude .5 38 .5 n that the European brand is preferred. z=
=
728
19.54
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
Rejection region: z z z.01 2.33
x .5n
5 .5(20 )
2.24, p-value = P(Z < –2.24) = .0125. There is not enough evidence to .5 n .5 20 conclude that children feel less pain. z=
19.55
=
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2
Rejection region T ≤ TL = 90 or T ≥ TU = 235 T = 190. There is not enough evidence of a difference in salary offers between men and women
19.56
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
Rejection region: z z z.05 1.645
x .5n
32 .5(53)
1.51, p-value = P(Z > 1.51) = 1 – .9345 = .0655. There is not enough .5 n .5 53 evidence to infer that preference should be given to students for high school 1.
z=
19.57
=
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
Rejection region: z z z.05 1.645
E (T )
z
n (n 1) 39 (39 1) 390 ; T 4 4
n (n 1)( 2n 1) 24
39 (39 1)( 2[39 ] 1) 71 .7 24
T E(T) 48 390 4.77 , p-value = P(Z < –4.77) = 0. There is enough evidence to = 71 .7 T
support the belief. 19.58
H0: The two population locations are the same H1: The location of population 1 PARSOL) is to the right of the location of population 2 (KIDDSOL)
729
z = 5.17, p-value = 0. There is enough evidence to conclude that Americans are more optimistic about their children than themselves. 19.59
H0: The two population locations are the same H1: The location of population 1 (DEGREE) is different from the location of population 2 (SPDEG)
z = 1.27 p-value = .2032. There is not enough evidence to conclude that married couples have different amounts of education.
19.60
H 0 : The locations of all 3 populations are the same. H 1 : At least two population locations differ.
Rejection region: H 2 , k 1 .205,2 = 5.99
730
12 H n (n 1)
12 984 2 1502 2 1430 2 Tj2 3(88 1) = 1.56, p 3(n 1) 36 29 n j 88(88 1) 23
value = .4584. There is not enough evidence to conclude that the population locations differ.
19.61
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
Rejection region: H 2 , k 1 .201,3 = 11.3
12 T j2 1207 2 1088 2 1310 2 1445 2 12 3(100 1) 3(n 1) H 25 nj 100 ( 100 1 ) 25 25 25 n (n 1) = 3.28, p-value = .3504. There is not enough evidence to conclude that the population locations
differ.
19.62
H 0 : The locations of all 3 populations are the same. H1 : At least two population locations differ.
Rejection region: H 2 , k 1 .210, 2 = 4.61 12 H n (n 1)
3741 2 1610 2 4945 2 Tj2 12 3(143 1) = 6.30, p-value 3(n 1) n j 29 67 143 (143 1) 47
= .0429. There is enough evidence to conclude that the population locations differ.
19.63
H 0 : The locations of all 3 populations are the same. H1 : At least two population locations differ.
Rejection region: H 2 , k 1 .205, 2 = 5.99 1 Rank 27 8 33 12.5 18 4.5 29 10 41 15.5 52 17 75 18 _ T 1 = 85.5 12 H n (n 1)
.
2 37 12 17 22 30
Rank 14 1.5 3 7 11
3 19 12 33 41 28 18
T 2 = 36.5 Tj2
Rank 6 1.5 12.5 15.5 9 4.5 T 3 = 49
12 85 .5 36 .5 49 2 3(n 1) 3(18 1) = 3.03, p-value = n j 5 6 18 (18 1) 7 2
2
.2195. There is no evidence to conclude that at least two population locations differ.
19.64
H 0 : The locations of all 3 populations are the same.
731
H1 : At least two population locations differ. Rejection region: H 2 , k 1 .205, 2 = 5.99 1 25 15 20 22 23
Rank 10.5 1 3 6 8.5 T 1 = 29
12 H n (n 1)
.
2 Rank 19 2 21 4 23 8.5 22 6 28 13.5 T 2 = 34
3 27 25 22 29 28
Rank 12 10.5 6 15 13.5 T 3 = 57
12 29 2 34 2 57 2 Tj2 3(15 1) = 4.46, p-value = .1075. 3(n 1) n j 5 5 15(15 1) 5
There is not enough evidence to conclude that at least two population locations differ.
19.65
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
A B C 1 Kruskal-Wallis Test 2 3 Group Rank Sum Observations 4 Printer 1 4889.5 50 5 Printer 2 5350 50 6 Printer 3 4864.5 50 7 Printer 4 4996 50 8 9 H Stat 0.899 10 df 3 11 p-value 0.8257 12 chi-squared Critical 7.8147
D
a H = .899, p-value = .8257. There is not enough evidence to conclude that differences exist between the ratings of the four printings. b The printout is identical to that of part a. c All codes that preserve the order produce the same results.
19.66
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
Rejection region: Fr 2 , k 1 = .210,3 = 6.25 Block 1 2 3 4
1 Rank 10 2 8 2 13 2 9 1.5
2 12 10 14 9
Treatment Rank 3 3 3 1.5
3 15 11 16 12
Rank 4 4 4 3 732
4 9 6 11 13
Rank 1 1 1 4
5
7
1 T 1 = 8.5
8
2 T 2 = 12.5
14
4 T 3 = 19
10 3 T 4 = 10
k 12 12 Fr Tj2 3b(k 1) (8.5 2 12 .5 2 19 2 10 2 3(5)(5) = 7.74, b(k )( k 1) j1 (5)( 4)(5)
p-value = .0517. There is enough evidence to infer that at least two population locations differ.
19.67
H 0 : The locations of all 3 populations are the same. H1 : At least two population locations differ.
Rejection region: Fr 2 , k 1 = .205, 2 = 5.99 Block 1 2 3 4 5
1 7.3 8.2 5.7 6.1 5.9
Rank 2 3 1 1 1 T1 = 8
Treatment 2 Rank 6.9 1 7.0 1 6.0 2 6.5 2 6.1 2 T2 = 8
3 Rank 8.4 3 7.3 2 8.1 3 9.1 3 8.0 3 T 3 = 14
k 12 12 Fr Tj2 3b(k 1) (8 2 8 2 14 2 3(5)( 4) = 4.8, p-value = .0907. b(k )( k 1) j1 (5)(3)( 4)
There is not enough evidence to infer that at least two population locations differ.
19.68
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
A B 1 Friedman Test 2 3 Group 4 Brand A 5 Brand B 6 Brand C 7 Brand D 8 9 Fr Stat 10 df 11 p-value 12 chi-squared Critical
C
Rank Sum 65 65 85 85 8.00 3 0.0460 7.8147
a Fr = 8.00, p-value = .0460. There is enough evidence to infer that differences exist between the ratings of the four brands of coffee. b Printout is identical to that of part a. c Different codes produce identical results provided the codes are in order.
733
19.69
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
A B 1 Friedman Test 2 3 Group 4 Manager 1 5 Manager 2 6 Manager 3 7 Manager 4 8 9 Fr Stat 10 df 11 p-value 12 chi-squared Critical
C
Rank Sum 21 10 24.5 24.5 10.613 3 0.014 7.8147
a Fr = 10.613, p-value = .0140. There is enough evidence to infer that differences exist between the ratings of the four managers. b The results are identical because the codes are in order.
19.70
H 0 : The locations of all 3 populations are the same. H1 : At least two population locations differ.
Rejection region: H 2 , k 1 = .205, 2 = 5.99 k 12 767 .5 2 917 2 1165 2 Tj2 12 3(75 1) = 6.81, 3(n 1) = H 75(75 1) 25 25 25 n (n 1) j1 n j
p-value = .0333. There is enough evidence to infer that there are differences in student satisfaction between the teaching methods.
19.71
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
Rejection region: H 2 , k 1 = .205,3 = 7.81
734
k 12 Tj2 3(n 1) = H n (n 1) j1 n j
17 ,116 .5 2 16,816 .5 2 17 ,277 2 29,391 2 12 3(401 1) = 6.65, p-value = .0838. 401(401 1) 80 90 77 154 There is not enough evidence to infer that there are differences between the four groups of GMAT scores.
19.72
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
Rejection region: Fr 2 , k 1 = .205,3 = 7.81 Judge 1 2 3 4 5 6 7 8 9 10
1 3 2 4 3 2 4 3 2 4 2
Rank 1.5 1 3 2 1 2 1.5 1 2.5 1 T1 = 16.5
Orange Juice Brand 2 Rank 3 Rank 5 4 4 3 3 2 5 4 4 3 3 1 4 3 5 4 4 3.5 4 3.5 5 3.5 5 3.5 3 1.5 4 3.5 3 3 3 3 3 1 5 4 4 3 5 4 T2 = 27.5 T3 = 33.5
4 3 4 4 2 3 3 4 3 4 3
Rank 1.5 3 3 1 2 1 3.5 3 2.5 2 T4 = 22.5
k 12 12 (16 .52 27 .52 33 .52 22 .52 3(10 )( 5) = Tj2 3b(k 1) Fr b(k )( k 1) j1 (10 )( 4)( 5)
9.42, p-value = .0242. There is enough evidence to infer that differences in sensory perception exist between the four brands of orange juice. 19.73a The randomized block experiment of the analysis of variance and the Friedman test should be considered. The analysis of variance requires the number of pedestrians to be normally distributed. b
H 0 : The locations of all 3 populations are the same. H1 : At least two population locations differ.
Rejection region: Fr 2 , k 1 = .205, 2 = 5.99
735
k 12 12 Fr Tj2 3b(k 1) (46 2 72 2 62 2 3(30 )( 4) = 11.47, p-value = b(k )( k 1) j1 (30 )(3)( 4)
.0032. There is enough evidence to infer that there are differences in the number of people passing between the three locations.
19.74
H 0 : The locations of all 3 populations are the same. H1 : At least two population locations differ.
Rejection region: Fr 2 ,k 1 = .205, 2 = 5.99 k 12 12 Fr Tj2 3b(k 1) (28 .5 2 22 .5 2 21 2 3(12 )( 4) = 2.63, ( 12 )( 3 )( 4 ) b ( k )( k 1 ) j1
p-value = .2691. There is not enough evidence to infer that there are differences in delivery times between the three couriers. 19.75a The randomized block experimental design of the analysis of variance and the Friedman test. b
H 0 : The locations of all 3 populations are the same. H1 : At least two population locations differ.
Rejection region: Fr 2 , k 1 = .205, 2 = 5.99 Job Advertised Receptionist Systems analyst Junior secretary Computer programmer Legal secretary Office manager
1 14 8 25 12 7 5
Rank 2 2 3 2 2 2 T1 = 13
Newspaper 2 Rank 17 3 9 3 20 1 15 3 10 3 9 3 T2 = 16
3 12 6 23 10 5 4
Rank 1 1 2 1 1 1 T3 = 7
k 12 12 Fr Tj2 3b(k 1) (13 2 16 2 7 2 3(6)( 4) = 7.00, p-value = ( 6 )( 3 )( 4 ) b(k )( k 1) j1
.0302. There is enough evidence to infer that differences exist between the newspapers.
19.76
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
Rejection region: H 2 , k 1 = .205,3 = 7.81
736
k 12 Tj2 3(n 1) = H n (n 1) j1 n j
2195 2 1650 .5 2 2830 2 2102 .5 2 12 3(132 1) 132 (132 1) 33 34 34 31 = 14.04, p-value = .0029. There is enough evidence to conclude that there are differences in grading standards between the four high schools.
19.77
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
Rejection region: Fr 2 , k 1 = .205,3 = 7.81 k 12 12 Tj2 3b(k 1) Fr (59 .5 2 63 .5 2 64 2 63 2 3(25)(5) = .300, b(k )( k 1) j1 (25)( 4)(5)
p-value = .9600. There is not enough evidence to infer that differences exist between the four drugs.
19.78
H 0 : The locations of all 3 populations are the same. H1 : At least two population locations differ.
Rejection region: Fr 2 ,k 1 = .205, 2 = 5.99 k 12 12 Fr Tj2 3b(k 1) (33 2 39 .5 2 47 .5 2 3(20 )( 4) = 5.28, b(k )( k 1) j1 (20 )(3)( 4)
p-value = .0715. There is not enough evidence to infer that there are differences in the ratings of the three recipes.
19.79 H 0 : The locations of all 3 populations are the same.
H1 : At least two population locations differ. Rejection region: H 2 , k 1 = .205,2 = 5.99 k 12 Tj2 13850 .52 14909 .52 16390 2 12 3(300 1) 3(n 1) = H 100 100 300 (300 1) 100 n (n 1) j1 n j
=4.32, p-value = .1151. There is not enough evidence to infer that differences exist between the three shifts. 19.80a The one-way analysis of variance and the Kruskal-Wallis test should be considered. 737
b
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
Rejection region: H 2 , k 1 = .205,3 = 7.81 k 12 4180 2 5262 2 5653 2 5005 2 Tj2 12 3(200 1) 3(n 1) = H 200 (200 1) 50 50 50 50 n (n 1) j1 n j
= 6.96, p-value = .0733. There is not enough evidence to infer that differences exist between the speeds at which the four brands perform. 19.81
H 0 : The locations of all 3 populations are the same. H1 : At least two population locations differ.
Rejection region: H 2,k 1 = .205, 2 = 5.99 k 12 Tj2 1565 2 1358 .52 1171 .52 12 3(90 1) = 3.78, p 3(n 1) = H 30 30 90 (90 1) 30 n (n 1) j1 n j
value = .1507. There is not enough evidence to infer that Democrat’s ratings of their chances changed over the 3–month period.
19.82
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
Rejection region: H 2 , k 1 = .205,3 = 7.81 k 12 Tj2 3(n 1) = H n (n 1) j1 n j
21,246 2 19,784 2 20,976 18,194 2 12 3(400 1) = 4.34, p-value = .2269. 400 (400 1) 100 100 100 100 There is not enough evidence to infer that differences in believability exist between the four ads.
19.83
H 0 : The locations of all 4 populations are the same. H1 : At least two population locations differ.
Rejection region: H 2 , k 1 = .205,3 = 7.81
738
k 12 Tj2 3(n 1) = H n (n 1) j1 n j
28,304 2 21,285 2 21,796 2 20,421 2 12 3(428 1) = 4.64, p-value = .1999. 428 (428 1) 123 109 102 94 There is not enough evidence to infer that there are differences in support between the four levels of students. 19.84
H0: The locations of all 5 populations are the same. H1: At least two population locations differ.
Rejection region: H 2 , k 1 = .205, 4 = 9.49 k 12 Tj2 3(n 1) H n (n 1) j1 n j
=
638 .5 2 1233 .5 2 1814 .5 2 3159 .5 2 2065 2 12 3(133 1) = 18.73, p-value = 133 (133 1) 18 14 26 42 33
.0009. There is enough evidence to infer that differences in perceived ease of use between the five brands of scanners. 19.85
H0: The locations of all 5 populations are the same. H1: At least two population locations differ.
H = 1637, p-value = 0. There is enough evidence to conclude that there are differences in job satisfaction between the five groups of American adults.
739
19.86
H0: The locations of all 3 populations are the same. H1: At least two population locations differ.
H = 22.27, p-value = 0. There is enough evidence to infer that differences in perception exist between the three races. 19.87
H0: The locations of all 5 populations are the same. H1: At least two population locations differ.
H = 7.11, p-value = .1302. There is not enough evidence to infer that there are differences between the five categories with respect to health. 19.88
H0: The locations of all 5 populations are the same. H1: At least two population locations differ.
740
H = 11.55, p-value = .0210. There is enough evidence to conclude that there are differences in perceived likelihood of losing their jobs between the five degree categories. 19.89
H0: The locations of all 3 populations are the same. H1: At least two population locations differ.
H = 12.04, p-value = .0024. There is enough evidence to infer that there are differences in perceived likelihood of losing their jobs between the three races. 19.90
H0: The locations of all 3 populations are the same. H1: At least two population locations differ.
741
H = 14.39, p-value = .0008. There is enough evidence to conclude that there are differences in their views about the federal income tax they have to pay between Democrats, Independents, and Republicans. 19.91
H0: The locations of all 3 populations are the same. H1: At least two population locations differ.
H = 37.99, p-value = 0. There is enough evidence to conclude that there are differences in their views about the federal income tax they have to pay between liberals, moderates, and conservatives. 19.92
H0: The locations of all 3 populations are the same. H1: At least two population locations differ.
742
H = 1.91, p-value = .3858. There is not enough evidence to infer that the races differ with respect to health. 19.93
H0: The locations of all 5 populations are the same. H1: At least two population locations differ.
H = 122.8, p-value = 0. There is enough evidence to infer that there are differences between the five educational attainment groups with respect to the newspaper readership. 19.94 Rejection region: z z / 2 z .025 1.96 or z z / 2 z .025 1.96 z rS n 1 (.23) 50 1 1.61, p-value = 2P(Z > 1.61) = 2(1 – .9463) = .1074 There is not
enough evidence to reject the null hypothesis.
19.95
H 0 : S 0 H1 : S 0 743
Rejection region: rS .497 rS .15. There is not enough evidence to conclude that there is a positive relationship between the two variables.
19.96
H 0 : S 0 H1 : S 0
Mathematics 4 2 5 4 2 2 1 Totals n
n
a i = 28
i 1
a2 30.25 9 49 30.25 9 9 1 137.5
a Economics b 5.5 5 6.5 3 2 1.5 7 3 4 5.5 5 6.5 3 3 4 3 3 4 1 2 1.5 28 28
n
n
b i = 28
i 1
b2 42.25 2.25 16 42.25 16 16 2.25 137.0
a i2 = 137.5
i 1
ab 35.75 4.5 28 35.75 12 12 1.5 129.5
n
b i2 = 137.0
a b = 129.5 i
i
i 1
i 1
n n ai bi n (28 )( 28 1 1 129 .5 2.917 s ab a i b i i 1 i 1 = 7 7 1 n n 1 i 1
2 n ai n 2 1 a i2 i 1 = 1 137 .5 (28) 4.250 , s a s a2 4.250 2.062 s a2 n n 1 i 1 7 1 7
2 n bi n 2 1 b i2 i 1 = 1 137 .0 (28) 4.167 , s b s 2b 4.167 2.041 s 2b n 1 i 1 n 7 1 7
rS
s ab 2.917 .6931 s b s b (2.062 )( 2.041)
H 0 : S 0 H1 : S 0 Rejection region: rS .786 or rS .786 There is not enough evidence to infer a relationship between the grades in the two courses.
744
19.97
H 0 : S 0 H1 : S 0
Number a 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 Totals 36 n
a i = 36
i 1
Rating 4 5 3 3 3 2 3 1
b 7 8 4.5 4.5 4.5 2 4.5 1 36
n
n
b i = 36
i 1
a2 1 4 9 16 25 36 49 64 204
b2 49 64 20.25 20.25 20.25 4 20.25 1 199
n
a i2 = 204
i 1
i 1
ab 7 16 13.5 18 22.5 12 31.5 8 128.5 n
b i2 = 199
a b = 128.5 i
i
i 1
n n ai bi n (36 )(36 1 1 4.786 128 .5 s ab a i b i i 1 i 1 = 8 1 8 n 1 i 1 n
2 n ai n 2 1 a i2 i 1 = 1 204 (36 ) 6.000 , s a s a2 6.000 2.450 s a2 n n 1 i 1 8 1 8
2 n bi n 2 1 b i2 i 1 = 1 199 (36 ) 5.286 , s b s 2b 5.286 2.299 s 2b n 1 i 1 n 8 1 8
rS
s ab 4.786 .8497 s b s b (2.450 )( 2.299 )
H 0 : S 0 H1 : S 0 Rejection region: rS .643 or rS .643 There is enough evidence to infer a relationship between the number of commercials and the rating.
19.98
H 0 : S 0
745
H1 : S 0 Stock 1 a –7 4.5 –4 6.5 –7 4.5 –3 8 2 10.5 –10 2.5 –10 2.5 5 12 1 9 –4 6.5 2 10.5 6 13 –13 1 Totals 91 n
n
Stock 2 6 6 –4 9 3 –3 7 –3 4 7 9 5 –7
a i = 91
n
b i = 91
i 1
i 1
n
a i2 = 817
i 1
a2 20.25 42.25 20.25 64 110.25 6.25 6.25 144 81 42.25 110.25 169 1 817
b 8.5 8.5 2 12.5 5 3.5 10.5 3.5 6 10.5 12.5 7 1 91
i 1
b2 72.25 72.25 4 156.25 25 12.25 110.25 12.25 36 110.25 156.25 49 1 817
ab 38.25 55.25 9 100 52.5 8.75 26.25 42 54 68.25 131.25 91 1 677.5
n
b i2 = 817
a b = 677.5 i
i
i 1
n n ai bi n (91)(91 1 1 677 .5 3.375 s ab a i b i i 1 i 1 = 13 13 1 n n 1 i 1
2 n ai n 2 1 a i2 i 1 = 1 817 (91) 15 .00, sa sa2 15 .00 3.873 s a2 n n 1 i 1 13 1 13
2 n bi n 2 1 b i2 i 1 = 1 817 (91) 15 .00, s b s 2b 15 .00 3.873 s 2b n 1 i 1 n 13 1 13
rS
sab 3.375 .2250 s bs b (3.873 )(3.873 )
H 0 : S 0 H1 : S 0 Rejection region: rS .566 or rS .566 There is not enough evidence to infer a relationship between the two stock returns.
19.99
H 0 : S 0
746
H1 : S 0 Experience 1 17 20 9 2 13 9 23 7 10 12 24 8 20 21 19 1 22 20 11 18 14 21 21 Totals
a 1.5 13 17 6.5 3 11 6.5 23 4 8 10 24 5 17 20 15 1.5 22 17 9 14 12 20 20 300
i 1
a i = 300
b i = 300
b2 ab 4 3 256 208 256 272 462.25 139.75 36 18 256 176 121 71.5 462.25 494.5 36 24 462.25 172 462.25 215 36 144 36 30 462.25 365.5 121 220 36 90 4 3 121 242 256 272 121 99 462.25 301 256 192 121 220 4 40 4,850.5 4,012.25 n
n
a i2 = 4,895
i 1
i 1
a2 2.25 169 289 42.25 9 121 42.25 529 16 64 100 576 25 289 400 225 2.25 484 289 81 196 144 400 400 4,895
b 2 16 16 21.5 6 16 11 21.5 6 21.5 21.5 6 6 21.5 11 6 2 11 16 11 21.5 16 11 2 300
n
n
n
Rating 1 4 4 5 2 4 3 5 2 5 5 2 2 5 3 2 1 3 4 3 5 4 3 1
b i2 = 4,850.5
a b = 4,012.25 i
i
i 1
i 1
n n a bi n i (300 )(300 ) 1 1 4,012 .25 s ab a i b i i 1 i 1 = 11 .40 24 24 1 n n 1 i 1
2 n ai n 2 1 a i2 i 1 = 1 4,895 (300 ) 49 .78, sa sa2 49 .78 7.056 s a2 n n 1 i 1 24 1 24
2 n bi n 2 1 b i2 i 1 = 1 4,850 .5 (300 ) 47 .85, s b s 2b 47 .85 6.917 s 2b n 1 i 1 n 24 1 24
747
rS
sab 11 .40 .2336 s bs b (7.056 )( 6.917 )
Rejection region: rS .409 or rS −.409 There is not enough evidence to infer a relationship between experience and quality.
19.100
H 0 : S 0 H1 : S 0
1 2 3 4 5 6 7 8 9
A B C Spearman Rank Correlation Price and Odometer Spearman Rank Correlation z Stat P(Z<=z) one tail z Critical one tail P(Z<=z) two tail z Critical two tail
D
-0.0201 -0.20 0.4206 1.6449 0.8412 1.96
z = –.20, p-value = .8412. There is not enough evidence to infer that odometer reading and price are related.
19.101
H 0 : S 0 H1 : S 0
1 2 3 4 5 6 7 8 9
A B C Spearman Rank Correlation Rating and Grade Spearman Rank Correlation z Stat P(Z<=z) one tail z Critical one tail P(Z<=z) two tail z Critical two tail
D
-0.2628 -2.93 0.0017 1.6449 0.0034 1.96
z = –2.93, p-value = .0034. There is enough evidence to support the theory. 19.102
H 0 : S 0 H1 : S 0
748
1 2 3 4 5 6 7 8 9
A B C Spearman Rank Correlation Age and Heartburn Spearman Rank Correlation z Stat P(Z<=z) one tail z Critical one tail P(Z<=z) two tail z Critical two tail
D
0.0302 0.54 0.2931 1.6449 0.5862 1.96
z = .54, p-value = .2931. There is not enough evidence to conclude that age and severity of heartburn are positively related.
19.103
H 0 : S 0 H1 : S 0
1 2 3 4 5 6 7 8 9
A B C Spearman Rank Correlation Test and Length Spearman Rank Correlation z Stat P(Z<=z) one tail z Critical one tail P(Z<=z) two tail z Critical two tail
D
0.5460 4.19 0 1.6449 0 1.96
z = 4.19, p-value = 0. There is enough evidence to conclude that the longer the commercial the higher the test score will be.
19.104
H 0 : S 0 H1 : S 0
1 2 3 4 5 6 7 8 9
A B C Spearman Rank Correlation Floor and Price Spearman Rank Correlation z Stat P(Z<=z) one tail z Critical one tail P(Z<=z) two tail z Critical two tail
D
0.553 3.87 0.0001 1.6449 0.0002 1.96
749
z = 3.87, p-value = .0001. There is sufficient evidence to conclude that price and floor number are positively related.
19.105
H 0 : S 0 H1 : S 0
1 2 3 4 5 6 7 8 9
A B C Spearman Rank Correlation Cigarettes and Taste Spearman Rank Correlation z Stat P(Z<=z) one tail z Critical one tail P(Z<=z) two tail z Critical two tail
D
-0.3026 -5.05 0 1.6449 0 1.96
z = –5.05, p-value = 0. There is sufficient evidence to conclude that the more a person smokes the less taste sensation he or she has. 19.106 H0: ρs= 0 H1: ρs> 0
1 2 3 4 5 6 7 8 9
A B C Spearman Rank Correlation Wager and Enjoyment Spearman Rank Correlation z Stat P(Z<=z) one tail z Critical one tail P(Z<=z) two tail z Critical two tail
D
0.3912 5.52 0 1.6449 0 1.96
z = 5.52, p-value = 0. There is enough evidence to infer that the greater the wager the more enjoyable the game is. 19.107 H0: ρs= 0 H1: ρs > 0
750
z = 2.45, p-value = .0144. There is enough evidence to conclude that as one gets older the probability of losing one’s job decreases. 19.108 H0: ρs= 0 H1: ρs < 0
z = -10.24, p-value = 0. There is enough evidence to infer that more educated people read newspapers more often.
19.109 H0: ρs= 0 H1: ρs < 0
751
z = -6.18, p-value = 0. There is enough evidence to infer that more satisfying jobs have higher incomes.
19.110 H0: ρs= 0 H1: ρs < 0
z = -6.59, p-value = 0. There is enough evidence to conclude that jobs that are most secure are also most satisfying.
19.111 H0: ρs= 0 H1: ρs < 0
752
z = -13.09, p-value = 0. There is enough evidence to conclude that more educated people perceive themselves as healthier.
19.112 H0: ρs= 0 H1: ρs < 0
z = -.435, p-value = .3318. There is not enough evidence to conclude that more educated people are more likely to believe that their standard of living is better than that of their parents.
19.113 H0: ρs= 0 H1: ρs < 0
753
z = -10.77, p-value = 0. There is enough evidence to conclude that younger people read newspapers less frequently than do older people. 19.114 H0: ρs= 0 H1: ρs > 0
z = 3.04, p-value = .0012. There is enough evidence to conclude that the longer one works, the probability of losing one’s job decreases. 19.115 H0: ρs= 0 H1: ρs ≠ 0
754
z = 1.10, p-value = .2698. There is not enough evidence to conclude that age affects one’s belief concerning the federal income tax one must pay.
19.116 H0: ρs= 0 H1: ρs < 0
z = -4.86, p-value = 0. There is enough evidence to conclude that higher income individuals are healthier.
19.117 H0: ρs= 0 H1: ρs> 0
z = 3.90, p-value = 0. There is enough evidence to conclude that richer Americans are pessimistic about their children’s chances of having a better standard of living.
19.118
H 0 : S 0 H1 : S 0 755
1 2 3 4 5 6 7 8 9
A B C Spearman Rank Correlation Education and Income Spearman Rank Correlation z Stat P(Z<=z) one tail z Critical one tail P(Z<=z) two tail z Critical two tail
D
0.5742 9.64 0 1.6449 0 1.96
z = 9.64, p-value = 0. There is sufficient evidence to conclude that more education and higher incomes are linked.
19.119
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 Section 1 15297.5 113 5 Section 2 14592.5 131 6 z Stat 2.65 7 P(Z<=z) one-tail 0.0041 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.0082 10 z Critical two-tail 1.96
E
z = 2.65, p-value = .0082. There is enough evidence to infer that the two teaching methods differ
19.120
H 0 : The two population locations are the same H 1 : The location of population 1 is left of the location of population 2
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 New 207.5 15 5 Existing 257.5 15 6 z Stat -1.04 7 P(Z<=z) one-tail 0.1499 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.2998 10 z Critical two-tail 1.96
E
z = –1.04, p-value = .1499. There is not enough evidence to infer that the new method is better. 19.121
H 0 : The locations of all 4 populations are the same. 756
H1 : At least two population locations differ.
A B 1 Friedman Test 2 3 Group 4 Typeface 1 5 Typeface 2 6 Typeface 3 7 Typeface 4 8 9 Fr Stat 10 df 11 p-value 12 chi-squared Critical
C
Rank Sum 50.5 38 66 45.5 12.615 3 0.0055 7.8147
Fr = 12.615, p-value = .0055. There is enough evidence to conclude that there are differences between typefaces.
19.122
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
Drug A - Drug B 2 18 10 -3.58 0.0002 1.6449 0.0004 1.96
z = –3.58, p-value = .0002. There is enough evidence to conclude that drug B is more effective. 19.123
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
757
A B C D 1 Wilcoxon Signed Rank Sum Test 2 3 Difference Drug A - Drug B 4 5 T+ 36 6 T342 7 Observations (for test) 27 8 z Stat -3.68 9 P(Z<=z) one-tail 0.0001 10 z Critical one-tail 1.6449 11 P(Z<=z) two-tail 0.0002 12 z Critical two-tail 1.96 Note that the number of non-zero observations is less than 30, invalidating the use of z statistic. T = 36, rejection region: T ≤ 120. There is enough evidence to conclude that drug B is more effective. 19.124a The one-way analysis of variance and the Kruskal-Wallis test should be considered. If the data are normal apply the analysis of variance, otherwise use the Kruskal-Wallis test. b
H 0 : The locations of all 3 populations are the same. H1 : At least two population locations differ.
A B C 1 Kruskal-Wallis Test 2 Rank Sum Observations 3 Group 827 25 4 Binding 1 1110 25 5 Binding 2 913 25 6 Binding 3 7 3.55 8 H Stat 2 9 df 0.1699 10 p-value 5.9915 11 chi-squared Critical
D
H = 3.55, p-value = .1699. There is not enough evidence to infer that there are differences between bindings.
19.125
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
758
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 New Material 2747 50 5 Old Material 2303 50 6 z Stat 1.53 7 P(Z<=z) one-tail 0.063 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.126 10 z Critical two-tail 1.96
E
z = 1.53, p-value = .0630. There is not enough evidence to conclude that the new material takes longer to burst into flames.
19.126
H 0 : The locations of all 7 populations are the same. H1 : At least two population locations differ.
A B C 1 Kruskal-Wallis Test 2 3 Group Rank Sum Observations 4 Sunday 10060 63 5 Monday 2977 26 6 Tuesday 2932.5 29 7 Wednesday 3834.5 31 8 Thursday 4060.5 30 9 Friday 6045 42 10 Saturday 6405.5 48 11 12 H Stat 14.87 13 df 6 14 p-value 0.0213 15 chi-squared Critical 12.5916
D
H = 14.87, p-value = .0213. There is enough evidence to infer that there are differences in the perceptions of speed of service between the days of the week.
19.127
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2
759
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
F
Commercial 1 - Commercial 2 15 21 24 -1.00 0.1587 1.6449 0.3174 1.96
z = –1.00, p-value = .3174. There is not enough evidence to infer differences in believability between the two commercials.
19.128
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
A B C D 1 Wilcoxon Signed Rank Sum Test 2 3 Difference Men - Women 4 5 T+ 324 6 T204 32 7 Observations (for test) 1.12 8 z Stat 0.1309 9 P(Z<=z) one-tail 1.6449 10 z Critical one-tail 0.2618 11 P(Z<=z) two-tail 1.96 12 z Critical two-tail z = 1.12, p-value = .1309. There is not enough evidence to conclude that men lose a greater percentage of their hearing than women.
19.129
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
760
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 This Year 37525.5 200 5 10 Years Ago 42674.5 200 6 z Stat -2.23 7 P(Z<=z) one-tail 0.013 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.026 10 z Critical two-tail 1.96
E
z = –2.23, p-value = .0130. There is enough evidence to infer that people perceive newspapers as doing a better job 10 years ago than today.
19.130
H 0 : S 0 H1 : S 0
1 2 3 4 5 6 7 8 9
A B C Spearman Rank Correlation Reference and GPA Spearman Rank Correlation z Stat P(Z<=z) one tail z Critical one tail P(Z<=z) two tail z Critical two tail
D
0.0775 1.05 0.148 1.6449 0.296 1.96
z = 1.05, p-value = .2960. There is not enough evidence to infer that the letter of reference and MBA GPA are related.
19.131
H 0 : The two population locations are the same H 1 : The location of population 1 is different from the location of population 2
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 Males 10336.5 100 5 Females 9763.5 100 6 z Stat 0.70 7 P(Z<=z) one-tail 0.24 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.484 10 z Critical two-tail 1.96
E
z = .70, p-value = .4840. There is not enough evidence to conclude that businesswomen and business men differ in the number of business trips taken per year. 761
19.132
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
Before - After 19 5 16 2.86 0.0021 1.6449 0.0042 1.96
z = 2.86, p-value = .0021. There is enough evidence to infer that the midterm test negatively influences student opinion.
19.133
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
A B C D 1 Wilcoxon Rank Sum Test 2 Rank Sum Observations 3 4 2 Years Ago 10786.5 100 5 ThisYear 9313.5 100 6 z Stat 1.80 7 P(Z<=z) one-tail 0.036 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.072 10 z Critical two-tail 1.96
E
z = 1.80, p-value = .0360. There is enough evidence to indicate that the citizens of Stratford should be concerned.
19.134
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
762
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 Low 9055 100 5 High 11045 100 6 z Stat -2.43 7 P(Z<=z) one-tail 0.0075 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.015 10 z Critical two-tail 1.96
E
z = –2.43, p-value = .0075. There is enough evidence to conclude that boys with high levels of lead are more aggressive than boys with low levels.
19.135a H 0 : The two population locations are the same
H 1 : The location of population 1 is to the right of the location of population 2
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
F
G
Female Professor - Male Professor 45 7 48 5.27 0 1.6449 0 1.96
z = 5.27, p-value = 0. There is enough evidence to infer that female students rate female professors higher than they rate male professors. b
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
F
Female Professor - Male Professor 21 31 48 -1.39 0.0828 1.6449 0.1656 1.96 763
G
z = –1.39, p-value = .0828. There is not enough evidence to infer that male students rate male professors higher than they rate female professors.
19.136
H 0 : The locations of all 3 populations are the same. H1 : At least two population locations differ.
A B C 1 Kruskal-Wallis Test 2 Rank Sum Observations 3 Group 4 Unattractive 16844.5 134 5 Neutral 13313 68 6 Attractive 26122.5 133 7 8 H Stat 42.59 9 df 2 10 p-value 0 11 chi-squared Critical 5.9915
D
H = 42.59, p-value = 0. There is enough evidence to conclude that incomes of lawyers are affected by physical attractiveness.
19.137
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
A B C D 1 Wilcoxon Rank Sum Test 2 Rank Sum Observations 3 4 Telecommuters 10934.5 100 5 Office 9165.5 100 6 z Stat 2.16 7 P(Z<=z) one-tail 0.0153 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.0306 10 z Critical two-tail 1.96
E
z = 2.16, p-value = .0153. There is enough evidence to conclude that telecommuters are more satisfied with their jobs.
19.138
H 0 : The two population locations are the same H 1 : The location of population 1 is to the left of the location of population 2
764
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 3 Hours Before 22553.5 180 5 Closing 42426.5 180 6 z Stat -10.06 7 P(Z<=z) one-tail 0 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0 10 z Critical two-tail 1.96
E
z = –10.06, p-value = 0. There is enough evidence to conclude that alcohol impairs judgment. 19.139
H 0 : The two population locations are the same H 1 : The location of population 1 is to the right of the location of population 2
Day 1versus Before
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
Before - Day 1 67 0 5 8.19 0 1.6449 0 1.96
z = 8.19, p-value = 0 Day 2 versus Before
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
Before - Day 2 59 0 13 7.68 0 1.6449 0 1.96
z = 7.68, p-value = 0 Day 3 versus Before 765
E
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
Before - Day 3 35 0 37 5.92 0 1.6449 0 1.96
z = 5.92, p-value = 0 There is enough evidence to infer that exercisers who abstain from physical activity are less happy than when they are exercising.
H 0 : The two population locations are the same
b
H 1 : The location of population 1 (Day 2) is to the left of the location of population 2 (Day 3)
A B 1 Sign Test 2 3 Difference 4 5 Positive Differences 6 Negative Differences 7 Zero Differences 8 z Stat 9 P(Z<=z) one-tail 10 z Critical one-tail 11 P(Z<=z) two-tail 12 z Critical two-tail
C
D
E
Day 2 - Day 3 3 37 32 -5.38 0 1.6449 0 1.96
z = –5.38, p-value = 0. There is enough evidence to conclude that by the third day their moods were improving. c 1. Exercisers who abstained were adjusting to their inactivity by the third day. 2. By the third day exercisers realized that they were closer to the end of their inactivity.
Case 19.1 a Wilcoxon rank sum tests
H 0 : The two population locations are the same. H 1 : The location of population 1 is to the left of the location of population 2. 766
Quality of work
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 No 875.5 26 5 Yes 8169.5 108 6 z Stat -4.95 7 P(Z<=z) one-tail 0 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0 10 z Critical two-tail 1.96
E
z = –4.95, p-value = 0. There is overwhelming evidence to infer that customers who say they will return assess quality of work higher than customers who do not plan to return. Fairness of price
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 No 1477 26 5 Yes 7568 108 6 z Stat -1.56 7 P(Z<=z) one-tail 0.0589 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.1178 10 z Critical two-tail 1.96
E
z = = –1.56, p-value = .0589. There is not enough evidence to infer that customers who say they will return assess fairness of price higher than customers who do not plan to return. Explanation of work and guarantee
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 No 1656 26 7389 108 5 Yes -0.56 6 z Stat 7 P(Z<=z) one-tail 0.2888 8 z Critical one-tail 1.6449 0.5776 9 P(Z<=z) two-tail 10 z Critical two-tail 1.96
E
z = –.56, p-value = .2888. There is no evidence to infer that customers who say they will return assess explanation of work and guarantee higher than customers who do not plan to return.
767
Checkout process
A B C D 1 Wilcoxon Rank Sum Test 2 3 Rank Sum Observations 4 No 1461 26 5 Yes 7584 108 6 z Stat -1.65 7 P(Z<=z) one-tail 0.049 8 z Critical one-tail 1.6449 9 P(Z<=z) two-tail 0.098 10 z Critical two-tail 1.96
E
z = –1.65, p-value = .0490. There is evidence to infer that customers who say they will return assess the checkout process higher than customers who do not plan to return. b Kruskal-Wallis tests
H 0 : The location of all 3 populations is the same H 1 : At least two population locations differ Quality of work
A B C 1 Kruskal-Wallis Test 2 Rank Sum Observations 3 Group 2495.5 33 4 Positive 1018 21 5 Negative 80 6 No commen 5531.5 7 6.63 8 H Stat 2 9 df 0.0364 10 p-value 5.9915 11 chi-squared Critical
D
H = 6.63, p-value = .0364. There is evidence to conclude that customers who make positive, negative, and no comment differ in their assessment of quality of work performed.
768
Fairness of price
A B C 1 Kruskal-Wallis Test 2 Rank Sum Observations 3 Group 2381.5 33 4 Positive 1143.5 21 5 Negative 5520 80 6 No commen 7 2.97 8 H Stat 2 9 df 0.2268 10 p-value 5.9915 11 chi-squared Critical
D
H = 2.97, p-value = .2268. There is no evidence to conclude that customers who make positive, negative, and no comment differ in their assessment of fairness of price Explanation of work and guarantee
A B C 1 Kruskal-Wallis Test 2 Rank Sum Observations 3 Group 2291 33 4 Positive 1335 21 5 Negative 5419 80 6 No commen 7 0.299 8 H Stat 2 9 df 0.8611 10 p-value 5.9915 11 chi-squared Critical
D
H = .299, p-value = .8611. There is no evidence to conclude that customers who make positive, negative, and no comment differ in their assessment of explanation of work and guarantee. Checkout process
A B C 1 Kruskal-Wallis Test 2 Rank Sum Observations 3 Group 1933.5 33 4 Positive 1200 21 5 Negative 80 6 No commen 5911.5 7 5.40 8 H Stat 2 9 df 0.0672 10 p-value 5.9915 11 chi-squared Critical
D
769
H = 5.40, p-value = .0672. There is not enough evidence to conclude that customers who make positive, negative, and no comment differ in their assessment of the checkout process.
770
Chapter 20 20.1 Time series
Moving average
48 41 37 32 36 31 43 52 60 48 41 30
(48+41+37)/3 = 42.00 (41+37+32)/3 = 36.67 (37+32+36)/3 = 35.00 (32+36+31)/3 = 33.00 (36+31+43)/3 = 36.67 (31+43+52)/3 = 42.00 (43+52+60)/3 = 51.67 (52+60+48)/3 = 53.33 (60+48+41)/3 = 49.67 (48+41+30)/3 = 39.67
20.2 Time series
Moving average
48 41 37 32 36 31 43 52 60 48 41 30
(48 +41+37+32+36)/5 = 38.8 (41+37+32+36+31)/5 = 35.4 (37+32+36+31+43)/5 = 35.8 (32+36+31+43+52)/5 = 38.8 (36+31+43+52+60)/5 = 44.4 (31+43+52+60+48)/5 = 46.8 (43+52+60+48+41)/5 = 48.8 (52+60+48+41+30)/5 = 46.2
20.3 70
60
Series1
50
Series2 40 30
Series3
20 10 0 1
2
3
4
5
6
7
8
793
9
10
11
12
20.4 Time series
Moving average
16 22 19 24 30 26 24 29 21 23 19 15
(16+22+19)/3 = 19.00 (22+19+24)/3 = 21.67 (19+24+30)/3 = 24.33 (24+30+26)/3 = 26.67 (30+26+24)/3 = 26.67 (26+24+29)/3 = 26.33 (24+29+21)/3 = 24.67 (29+21+23)/3 = 24.33 (21+23+19)/3 = 21.00 (23+19+15)/3 = 19.00
20.5 Time series 16 22 19 24 30 26 24 29 21 23 19 15
Moving average
(16+22+19+24+30)/5 = 22.2 (22+19+24+30+26)/5 = 24.2 (19+24+30+26+24)/5 = 24.6 (24+30+26+24+29)/5 = 26.6 (30+26+24+29+21)/5 = 26.0 (26+24+29+21+23)/5 = 24.6 (24+29+21+23+19)/5 = 23.2 (29+21+23+19+15)/5 = 21.4
20.6 35
30
Series1
25
Series2 20 15
Series3
10 5 0 1
2
3
4
5
6
7
8
794
9
10
11
12
20.7 Time series
Exponentially smoothed time series
12 18 16 24 17 16 25 21 23 14
12 .1(18) +. 9(12) = 12.60 .1(16) +. 9(12.60) = 12.94 .1(24) +. 9(12.94) = 14.05 .1(17) +. 9(14.05) = 14.34 .1(16) +. 9(14.34) = 14.51 .1(25) +. 9(14.51) = 15.56 .1(21) +. 9(15.56) = 16.10 .1(23) + .9(16.10) = 16.79 .1(14) + .9(16.79) = 16.51
20.8 Time series
Exponentially smoothed time series
12 18 16 24 17 16 25 21 23 14
12.00 .8(18) +. 2(12) = 16.80 .8(16) +. 2(16.80) = 16.16 .8(24) +. 2(16.16) = 22.43 .8(17) +. 2(22.43) = 18.09 .8(16) +. 2(18.09) = 16.42 .8(25) +. 2(16.42) = 23.28 .8(21) +. 2(23.28) = 21.46 .8(23) + .2(21.46) = 22.69 .8(14) + .2(22.69) = 15.74
20.9 30 25
Series1
20 Series2 15 Series3 10
5 0 1
2
3
4
5
6
7
There appears to be a gradual upward trend.
795
8
9
10
20.10 Time series
Exponentially smoothed time series
38 43 42 45 46 48 50 49 46 45
38.00 .1(43) +. 9(38) = 38.50 .1(42) +. 9(38.50) = 38.85 .1(45) +. 9(38.85) = 39.47 .1(46) +. 9(39.47) = 40.12 .1(48) +. 9(40.12) = 40.91 .1(50) +. 9(40.91) = 41.82 .1(49) +. 9(41.82) = 42.53 .1(46) + .9(42.53) = 42.88 .1(45) + .9(42.88) = 43.09
There appears to be a gradual upward trend.
20.11 Time series
Exponentially smoothed time series
38 43 42 45 46 48 50 49 46 45
38 .8(43) +. 2(38) = 42.00 .8(42) +. 2(42.00) = 42.00 .8(45) +. 2(42.00) = 44.40 .8(46) +. 2(44.40) = 45.68 .8(48) +. 2(45.68) = 47.54 .8(50) +. 2(47.54) = 49.51 .8(49) +. 2(49.51) = 49.10 .8(46) + .2(49.10) = 46.62 .8(45) + .2(46.62) = 45.32
20.12 55
Series1 50
Series2
45
40
Series3
35
30 1
2
3
4
5
6
7
There is a trend component.
796
8
9
10
20.13 & 20.14 Sales
3-Day moving average
5-Day moving average
43 45 22 25 31 51 41 37 22 25 40 57 30 33 37 64 58 33 38 25
(43+45+22)/3 = 36.67 (45+22+25)/3 = 30.67 (22+25+31)/3 = 26.00 (25+31+51)/3 = 35.67 (31+51+41)/3 = 41.00 (51+41+37)/3 = 43.00 (41+37+22)/3 = 33.33 (37+22+25)/3 = 28.00 (22+25+40)/3 = 29.00 (25+40+57)/3 = 40.67 (40+57+30)/3 = 42.33 (57+30+33)/3 = 40.00 (30+33+37)/3 = 33.33 (33+37+64)/3 = 44.67 (37+64+58)/3 = 53.00 (64+58+33)/3 = 51.67 (58+33+38)/3 = 43.00 (33+38+25)/3 = 32.00
(43+45+22+25+31)/5 = 33.20 (45+22+25+31+51)/5 = 34.80 (22+25+31+51+41)/5 = 34.00 (25+31+51+41+37)/5 = 37.00 (31+51+41+37+22)/5 = 36.40 (51+41+37+22+25)/5 = 35.20 (41+37+22+25+40)/5 = 33.00 (37+22+25+40+57)/5 = 36.20 (22+25+40+57+30)/5 = 34.80 (25+40+57+30+33)/5 = 37.00 (40+57+30+33+37)/5 = 39.40 (57+30+33+37+64)/5 = 44.20 (30+33+37+64+58)/5 = 44.40 (33+37+64+58+33)/5 = 45.00 (37+64+58+33+38)/5 = 46.00 (64+58+33+38+25)/5 = 43.60
70 Series1
60 50
Series2
40 30
Series3
20 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
c There appears to be a seasonal (weekly) pattern.
797
20.15a Sales 18
4-quarter moving average
Centered moving average
22 (18+22+27+31)/4 = 24.50 27
(24.50+28.25)/2 = 26.375 (22+27+31+33)/4 = 28.25
31
(28.25+27.75)/2 = 28.000 (27+31+33+20)/4 = 27.75
33
(27.75+30.50)/2 = 29.125 (31+33+20+38)/4 = 30.50
20
(30.50+29.25)/2 = 29.875 (33+20+38+26)/4 = 29.25
38
(29.25+27.25)/2 = 28.250 (20+38+26+25)/4 = 27.25
26
(27.25+31.25)/2 = 29.250 (38+26+25+36)/4 = 31.25
25
(31.25+32.75)/2 = 32.000 (26+25+36+44)/4 = 32.75
36
(32.75+33.50)/2 = 33.125 (25+36+44+29)/4 = 33.50
44
(33.50+37.50)/2 = 35.500 (36+44+29+41)/4 = 37.50
29
(37.50+36.75)/2 = 37.125 (44+29+41+33)/4 = 36.75
41
(36.75+38.75)/2 = 37.750 (29+41+33+52)/4 = 38.75
33
(38.75+42.75)/2 = 40.750 (41+33+52+25)/4 = 42.75
52 45 b 60 50 Series1
40 30
Series2
20
10 0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16
c There appears to be a gradual trend of increasing sales.
798
20.16 & 20.17 Sales
Exponentially smoothed w = .4
Exponentially smoothed w = .8
18 22 27 31 33 20 38 26 25 36 44 29 41 33 52 45
18.00 .4(22)+.6(18) = 19.60 .4(27)+.6(19.6) = 22.56 .4(31)+.6(22.56) = 25.94 .4(33)+.6(25.94) = 28.76 .4(20)+.6(28.76) = 25.26 .4(38)+.6(25.26) = 30.35 .4(26)+.6(30.35) = 28.61 .4(25)+.6(28.61) = 27.17 .4(36)+.6(27.17) = 30.70 .4(44)+.6(30.70) = 36.02 .4(29)+.6(36.02) = 33.21 .4(41)+.6(33.21) = 36.33 .4(33)+.6(36.33) = 35.00 .4(52)+.6(35.00) = 41.80 .4(45)+.6(41.80) = 43.08
18.00 .8(22)+.2(18) = 21.20 .8(27)+.2(21.2) = 25.84 .8(31)+.2(25.84) = 29.97 .8(33)+.2(29.97) = 32.39 .8(20)+.2(32.39) = 22.48 .8(38)+.2(22.48) = 34.90 .8(26)+.2(34.90) = 27.78 .8(25)+.2(27.78) = 25.56 .8(36)+.2(25.56) = 33.91 .8(44)+.2(33.91) = 41.98 .8(29)+.2(41.98) = 31.60 .8(41)+.2(31.60) = 39.12 .8(33)+.2(39.12) = 34.22 .8(52)+.2(34.22) = 48.44 .8(45)+.2(48.44) = 45.69
20.16 b 60 50
Series1
40 30
Series2
20
10 0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16
799
20.17 b 60 50 Series1
40 30
Series2
20
10 0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16
20.18
Line chart 25
Time series
20 15 10
5 0 1
2
3
4
5 Period
The quadratic model would appear to be the best model.
800
6
7
8
20.19
Line chart 60
Time series
50
40 30 20 10 0 1
2
3
4
5
6
Period
The linear trend model appears to be best.
20.20 ŷ 4.96 2.38 t
(R 2 .81)
ŷ 3.14 2.48 t .54 t 2
(R 2 .98)
The quadratic trend line fits better.
20.21 ŷ 63 .87 3.94 t
(R 2 .94 )
ŷ 57 .2 .61t .30 t 2
(R 2 .98)
The quadratic trend line fits slightly better.
801
7
8
9
10
20.22 Week
Day
Period t
y
ŷ
y / yˆ
1
1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
12 18 16 25 31 11 17 19 24 27 14 16 16 28 25 17 21 20 24 32
17.2 17.5 17.9 18.3 18.6 19.0 19.4 19.7 20.1 20.5 20.8 21.2 21.6 21.9 22.3 22.7 23.0 23.4 23.8 24.1
0.699 1.027 0.894 1.369 1.664 0.579 0.878 0.963 1.194 1.320 0.672 0.755 0.742 1.277 1.122 0.750 0.912 0.855 1.010 1.327
Monday .699 .579 .672 .750 .675
Tuesday 1.027 .878 .755 .912 .893
Day Wednesday .894 .963 .742 .855 .864
Thursday 1.369 1.194 1.277 1.010 1.213
Friday 1.664 1.320 1.122 1.327 1.358
5.003
.675
.892
.864
1.212
1.357
5.000
2
3
4
Week 1 2 3 4 Average Seasonal Index
802
Total
20.23 Year
Quarter
y
ŷ
y / yˆ
1
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
55 44 46 39 41 38 37 30 43 39 39 35 36 32 30 25 50 25 24 22
46.6 45.6 44.5 43.5 42.4 41.3 40.3 39.2 38.2 37.1 36.0 35.0 33.9 32.9 31.8 30.7 29.7 28.6 27.6 26.5
1.179 0.965 1.033 0.897 0.967 0.919 0.919 0.765 1.127 1.051 1.082 1.001 1.061 0.974 0.943 0.813 1.685 0.874 0.871 0.830
2
3
4
5
Year 1 2 3 4 5 Average Seasonal Index
1 1.179 0.967 1.127 1.061 1.685 1.204
2 0.965 0.919 1.051 0.974 0.874 0.957
Quarter 3 1.033 0.919 1.082 0.943 0.871 0.970
1.207
0.959
0.972
803
4 0.897 0.765 1.001 0.813 0.830 0.861
Total
3.991
0.863
4.000
20.24 Year
Quarter
Period t
y
ŷ
y / yˆ
2001
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
52 67 85 54 57 75 90 61 60 77 94 63 66 82 98 67
62.9 64.1 65.3 66.5 67.7 68.8 70.0 71.2 72.4 73.6 74.7 75.9 77.1 78.3 79.5 80.6
0.827 1.045 1.302 0.812 0.842 1.090 1.286 0.857 0.829 1.046 1.258 0.830 0.856 1.047 1.233 0.831
2002
2003
2004
Year 2001 2002 2003 2004 Average Seasonal Index
1 .827 .842 .829 .856 .838
2 1.045 1.090 1.046 1.047 1.057
Quarter 3 1.302 1.286 1.258 1.233 1.270
4 .812 .857 .830 .831 .833
3.998
.839
1.058
1.270
.833
4.000
Total
20.25a
Scatter Diagram 18 000 17 500
y = 253,78x - 491595
17 000 Enrollment
16 500 16 000
15 500
15 000 14 500 14 000 13 500 1992
1994
1996
1998
2000 Year
b ŷ 491,595 253 .78 Year
804
2002
2004
2006
2008
20.26a
Balance of Trade
Scatter Diagram 0 -100 -200 -300 -400 -500 -600 -700 -800 -900 1975
y = -25,039x + 49661
1980
1985
1990
1995
2000
2005
2010
20
25
30
Year b ŷ = 49,661 – 25.039Year
20.27
Scatter Diagram 400
y = 7.416x + 143.0
350
Subscribers
300 250 200 150 100 50 0 0
5
10
15 Quarter
ŷ = 143 + 7.42 t
805
Year
Quarter
Period t
y
ŷ =143+7.42t
y / yˆ
1
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
184 173 160 189 191 185 184 200 205 192 200 229 236 219 211 272 280 261 275 322 331 301 306 351
150.42 157.84 165.26 172.68 180.1 187.52 194.94 202.36 209.78 217.2 224.62 232.04 239.46 246.88 254.3 261.72 269.14 276.56 283.98 291.4 298.82 306.24 313.66 321.08
1.223 1.096 0.968 1.095 1.061 0.987 0.944 0.988 0.977 0.884 0.890 0.987 0.986 0.887 0.830 1.039 1.040 0.944 0.968 1.105 1.108 0.983 0.976 1.093
1 1.223 1.060 .977 .986 1.040 1.108 1.066
2 1.096 .987 .884 .887 .944 .983 .964
Quarter 3 .968 .944 .890 .830 .968 .976 .929
4 1.095 .988 .987 1.039 1.105 1.093 1.051
4.010
1.063
.962
.927
1.048
4.000
2
3
4
5
6
Year 1 2 3 4 5 6 Average Seasonal Index
20.28 Regression line: ŷ = 145 + 1.66 t
806
Total
Week
Day
Period t
y
ŷ =145+1.66t
y / yˆ
1
1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
240 85 93 106 125 188 314 221 80 75 121 110 202 386 235 86 74 100 117 205 402 219 91 102 89 105 192 377
146.66 148.32 149.98 151.64 153.3 154.96 156.62 158.28 159.94 161.6 163.26 164.92 166.58 168.24 169.9 171.56 173.22 174.88 176.54 178.2 179.86 181.52 183.18 184.84 186.5 188.16 189.82 191.48
1.636 0.573 0.620 0.699 0.815 1.213 2.005 1.396 0.500 0.464 0.741 0.667 1.213 2.294 1.383 0.501 0.427 0.572 0.663 1.150 2.235 1.206 0.497 0.552 0.477 0.558 1.011 1.969
2
3
4
Week 1 2 3 4 Average Seasonal Index
Sunday 1.636 1.396 1.383 1.206 1.405
Monday .573 .500 .501 .497 .518
Tuesday .620 .464 .427 .552 .516
Day Wednesday .699 .741 .572 .477 .657
Thursday .815 .667 .663 .558 .676
Friday 1.213 1.213 1.150 1.011 1.187
Saturday 2.005 2.294 2.235 1.969 2.126
7.085
1.404
.517
.515
.621
.675
1.145
2.123
7.000
20.29 Regression line: ŷ = 90.4 + 2.02 t
807
Total
Year
Quarter
t
y
ŷ =90.4 + 2.02t
y / yˆ
1
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
106 92 65 121 115 100 73 135 114 105 79 140 121 111 82 163
92.42 94.44 96.46 98.48 100.5 102.52 104.54 106.56 108.58 110.6 112.62 114.64 116.66 118.68 120.7 122.72
1.147 0.974 0.674 1.229 1.144 0.975 0.698 1.267 1.050 0.949 0.701 1.221 1.037 0.935 0.679 1.328
1 1.147 1.144 1.050 1.037 1.095
2 .974 .975 .949 .935 .959
3 .674 .698 .701 .679 .688
4 1.229 1.267 1.221 1.328 1.261
4.003
1.094
.958
.688
1.260
4.000
2
3
4
Quarter Year 1 2 3 4 Average Seasonal Index
Total
20.30
MAD
166 173 179 186 195 192 214 211 220 223
5 7 7 3 3 3 23 4.60 5 5 SSE (166 173 ) 2 (179 186 ) 2 (195 192 ) 2 ( 214 211 ) 2 ( 220 223 ) 2 49 49 9 9 9 125
20.31
Model 1:
MAD
6.0 7.5 6.6 6.3 7.3 5.4 9.4 8.2 4
1.5 .3 1.9 1.2 4 4.9 1.225 4
808
Model 2:
MAD
6.0 6.3 6.6 6.7 7.3 7.1 9.4 7.5
4 .3 .1 .2 1.9 2.5 .625 4 4
Model 2 is more accurate. Model 1: SSE = (6.0 - 7.5)2 + (6.6 - 6.3)2 + (7.3 - 5.4)2 + (9.4 - 8.2)2 = 2.25 + .09 + 3.61 + 1.44 = 7.39 Model 2: SSE = (6.0 - 6.3)2 + (6.6 - 6.7)2 + 7.3 - 7.1)2 + (9.4 - 7.5)2 = .09 + .01 + .04 + 3.61 = 3.75 Model 2 is more accurate.
20.32
57 63 60 72 70 86 75 71 70 60 MAD
5 6 12 16 4 10 48 9.6 5 5
SSE = (57 - 63)2 + (60 - 72)2 + (70 - 86)2 + (75 - 71)2 +(70-60)2 = 36 + 144 + 256 + 16 + 100 = 552
20.33
Technique 1:
MAD
19 21 24 27 28 29 32 31 38 35 5
2 3 1 1 3 10 2.0 5 5
SSE = (19 - 22)2 + (24 - 27)2 + (32 - 31)2 + (38 - 35)2 = 4 + 9 + 1 + 1 + 9 = 24 Technique 2:
MAD
19 22 24 24 28 26 32 28 38 30 5
3 0 2 4 8 15 3.0 5 5 SSE = (19 - 22)2 + (24 - 24)2 + (28 - 26)2 + (32 - 28)2 + (28 - 30)2 = 9 + 0 + 4 + 16 + 64 = 93
809
Technique 3:
MAD
19 17 24 20 28 25 32 31 38 39 5
2 4 3 1 1 11 2.2 5 5 SSE = (19 - 17)2 + (24 - 20)2 + (28 - 25)2 + (32 - 31)2 + (38 - 39)2 = 4 + 16 + 9 + 1+ 1 = 31 By both measures, technique 1 is the most accurate.
20.34 Quarter
t
1 2 3 4
41 42 43 44
20.35
20.36
yˆ 150 3t 273 276 279 282
SI
Forecast
.7 1.2 1.5 .6
191.1 331.2 418.5 169.2
Day
t
yˆ 120 2.3t
SI
Forecast
Sunday Monday Tuesday Wednesday Thursday Friday Saturday
29 30 31 32 33 34 35
186.7 189.0 191.3 193.6 195.9 198.2 200.5
1.5 .4 .5 .6 .7 1.4 1.9
280.1 75.6 95.7 116.2 137.1 277.5 381.0
yˆ t 625 1.3 yt 1 625 1.3(65) 540.5
20.37 yˆ t 155 21yt 1 155 21(11) 386 20.38 F17 F18 F19 F20 S16 43.08
20.39 Day
t
1 2 3 4 5
21 22 23 24 25
SI yˆ 16 .8 .366 t 24.49 .675 24.85 .892 25.22 .864 25.58 1.212 25.95 1.357
810
Forecast 16.53 22.17 21.79 31.01 35.21
20.40 Quarter
t
1 2 3 4
21 22 23 24
20.41 Quarter 1 2 3 4 1 2 3 4
yˆ 47 .7 1.06t 25.44 24.38 23.32 22.26
yˆ 61 .75 1.18t
t 17 18 19 20 21 22 23 24
81.81 82.99 84.17 85.35 86.53 87.71 88.89 90.07
SI
Forecast
1.207 .959 .972 .863
30.71 23.38 22.67 19.21
SI 0.839 1.058 1.270 0.833 0.839 1.058 1.270 0.833
Forecast 68.64 87.80 106.90 71.10 72.60 92.80 112.89 75.03
20.42a ŷ 2007 1,775 .2 .9054 y 2006 = 1,775.2 + .9054(17,672) = 17,775 b F2007 S2006 17,146. 20.43 a ŷ 2007 4.2245 1.1203 2006 = −4.2245 + 1.1203(−817.3) = −919.8
F2007 S2006 −719.5
20.44 Quarter
t
1 2 3 4
25 26 27 28
20.45 Day 1 2 3 4 5 6 7
t 29 30 31 32 33 34 35
yˆ 143 7.42t 328.50 335.92 343.34 350.76
yˆ 145 1.66t 193.14 194.80 196.46 198.12 199.78 201.44 203.10
811
SI 1.404 0.517 0.515 0.621 0.675 1.145 2.123
SI
Forecast
1.063 .962 .927 1.048
349.20 323.16 318.28 367.60
Forecast 271.17 100.71 101.18 123.03 134.85 230.65 431.18
20.46 Day 1 2 3 4
yˆ 90.4 2.02t
t 17 18 19 20
124.74 126.76 128.78 130.80
SI 1.094 0.958 0.688 1.260
Forecast 136.47 121.44 88.60 164.81
20.47 60
Revenues
50 40 30 20 10 0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 Quarter
20.48 There is a small upward trend and seasonality.
20.49 ŷ = 20.21 + .732t
20.50 A 1 Seasonal Indexes 2 3 Season 4 1 5 2 6 3 7 4
B
Index 0.646 1.045 1.405 0.904
20.51 Quarter 1 2 3 4
t 21 22 23 24
yˆ 20 .21 .732 t 35.58 36.31 37.05 37.78 812
SI 0.646 1.045 1.405 0.904
Forecast 23.00 37.95 52.05 34.14
20.52a
b. Period Month
ŷ = 16.34-.1008t Seasonal Index
Forecasts
Actual
61
January
10.19
.7192
7.33
2.2
62
February
10.09
.6729
6.79
3.6
63
March
9.99
.8936
8.93
5.3
64
April
9.89
1.0500
10.38
4.5
65
May
9.79
1.2148
11.89
5.7
66
June
9.69
1.3998
13.56
8.1
67
July
9.59
1.1908
11.42
6.1
68
August
9.49
1.1651
11.05
6.5
69
September
9.38
1.0246
9.62
5.8
70
October
9.28
1.0288
9.55
4.9
71
November
9.18
.8535
7.84
4.8
72
December
9.08
.7870
7.15
4.4
c. MAD = 4.47 SSE = 254.32
813
814
Chapter 21 21.1Chance variation is caused by a number of randomly occurring events that are part of the production process and that in general cannot be eliminated without changing the process. 21.3 Special variation is caused by specific events or factors that are frequently temporary and that can usually be identified and eliminated 21.4a Chance variation represents the variation in student achievement caused by differences in preparation, motivation, and ability. b Special variation represents variation due to specific event or factors that can be corrected. 21.5 P(| z | 2.5) 2(1 .9938 ) .0124
21.6 ARL =
1 = 81 .0124
21.7 P(| z | 2.0) 2(1 .9772 ) .0456
21.8 ARL =
1 = 22 .0456
21.9 ARL =
1 = 385 .0026
Number of units = Production ARL = 100(385) = 38,500 21.10a From Beta-mean spreadsheet, = .6603 b Probability = .6603 8 = .0361
21.11 P = 1 − = 1 − .6603 = .3397; ARL =
1 1 = 2.94 P .3397
21.12 Number of units = Production ARL = 50(385) = 19,250 21.13a From Beta-mean spreadsheet, = .8133 b Probability = .8133 8 = .1914 815
21.14 P = 1 − = 1 − .8133 = .1867; ARL =
1 1 = 5.36 P .1867
21.15 Sampling 3 units per hour means that on average we will produce 38,500 units before erroneously concluding that the process is out of control when it isn’t. Sampling 2 units per half hour reduces this figure by 50%. Sampling 4 units per hour means that when the process goes out of control, the probability of not detecting a shift of 1.5 standard deviations is .6603 and we will produce on average 2.94 100 = 294 units until the chart indicates a problem. Sampling 2 units per half hour increases the probability of not detecting the shift to .8133 and decreases the average number of units produced when the process is out of control to 50 5.36 = 268. 21.16 Number of units = Production ARL = 2000(385) = 770,000 21.17a From Beta-mean spreadsheet, = .7388 b Probability = .7388 4 = .2979
21.18 P = 1 − = 1 − .7388 = .2612; ARL =
1 1 = 3.83 P .2612
21.19 Number of units = Production ARL = 4000(385) = 1,540,000 21.20a From Beta-mean spreadsheet, = .3659 b Probability = .3659 4 = .0179
21.21 P = 1 − = 1 − .3659 = .6341; ARL =
1 1 = 1.58 P .6341
21.22 Sampling 10 units per half hour means that on average we will produce 770,000 units before erroneously concluding that the process is out of control when it isn’t. Sampling 20 units per hour doubles this figure. Sampling 10 units per half hour means that when the process goes out of control, the probability of not detecting a shift of .75 standard deviations is .7388 and we will produce on average 3.83 2000 = 7660 units until the chart indicates a problem. Sampling 20 units per hour decreases the probability of not detecting the shift to .3659 and decreases the average number of units produced when the process is out of control to 4000 1.58 = 6320.
21.23 Centerline = x = 453.6
816
12 .5 = 434.85 = 453.6 − 3 n 4
Lower control limit = x
3S
Upper control limit = x
3S
12 .5 = 472.35 = 453.6 + 3 n 4
21.24 Centerline = x = 181.1
11 .0 = 170.1 = 181.1 − 3 n 9
Lower control limit = x
3S
Upper control limit = x
3S
11 .0 = 192.1 = 181.1 + 3 n 9
Zone boundaries: 170.10, 173.77, 177.44, 181.10, 184.77, 188.43, 192.10
21.25 a Centerline = x = 13.3
3.8 = 7.6 = 13.3 − 3 n 4
Lower control limit = x
3S
Upper control limit = x
3S
3.8 = 19.0 = 13.3 + 3 n 4
c The process is out of control at points 8, 9, 21, 22, and 25. 21.26a S Chart
1 2 3 4 5 6
A B C Statistical Process Control
Upper control limit Centerline Lower control limit
D
Data 10.0885 4.452 0
817
12
10
8
6
4
2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
x Chart
1 2 3 4 5 6 7 8
A B C Statistical Process Control
D
Data Upper control limit 19.9668 Centerline 12.7386 Lower control limit 5.5103 Pattern Test #2 Failed at Points: 29 Pattern Test #6 Failed at Points: 29, 30
818
25
20
15
10
5
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
c The process is out of control at samples 29 and 30. d A level shift occurred. 21.27a S Chart
1 2 3 4 5 6
A B C Statistical Process Control
Upper control limit Centerline Lower control limit
2by4 lumber 0.1851 0.0721 0
819
0.2 0.18 0.16 0.14 0.12 0.1
0.08 0.06 0.04 0.02 0 1
1 2 3 4 5 6 7 8 9
3
5
7
9
11
13
15
17
A B C Statistical Process Control
D
19
21
23
E
25
27
29
31
F
2by4 lumber Upper control limit 96.0635 Centerline 95.9231 Lower control limit 95.7827 Pattern Test #1 Failed at Points: 1, 4, 5, 28, 35, 36, 37, 39, 40 Pattern Test #5 Failed at Points: 3, 4, 5, 6, 9, 10, 35, 36, 37, 38, 39, 40 Pattern Test #6 Failed at Points: 5, 6, 7, 8, 9, 35, 36, 37, 38, 39, 40
820
33
G
35
37
39
96.2 96.1 96 95.9 95.8 95.7 95.6 95.5 1
3
5
7
9
11
13
15
17
b The process is out of control at the first sample. c Clean out the sawdust more frequently. 21.28
1 2 3 4 5 6
A B C Statistical Process Control
Upper control limit Centerline Lower control limit
AEU 0.0031 0.0015 0
821
19
21
23
25
27
29
31
33
35
37
39
0.0035 0.003 0.0025 0.002 0.0015 0.001 0.0005 0 1
1 2 3 4 5 6
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
A B C Statistical Process Control AEU 0.4408 0.4387 0.4366
Upper control limit Centerline Lower control limit
0.442 0.441 0.44 0.439 0.438
0.437 0.436 0.435 0.434 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
The process is under control. 822
S
21.29
S 5
n
Upper control limit Centerline .4408 .4387 .0007 3 3
.0007 ; S .0016
CPL
x LSL .4387 .4335 1.08 3S 3(.0016 )
CPU
USL x 4435 .4387 1.00 3S 3(.0016 )
C pk Min (CPL, CPU) = 1.00
21.30
1 2 3 4 5 6
A B C Statistical Process Control
Upper control limit Centerline Lower control limit
Volume 1.8995 0.9093 0
2 1.8 1.6 1.4 1.2 1
0.8 0.6 0.4 0.2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
1 2 3 4 5 6
A B C Statistical Process Control
Upper control limit Centerline Lower control limit
Volume 1.3871 0.092 -1.2031 823
2 1.5 1 0.5 0 -0.5 -1 -1.5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
The process is under control.
S
21.31
S 5
n
Upper control limit Centerline 1001 .3871 1000 .092 .4317 3 3
.4317 ; S .9653
CPL
x LSL 1000 .092 995 1.76 3S 3(.9653 )
CPU
USL x 1005 1000 .092 1.69 3S 3(.9653 )
Cpk Min (CPL, CPU) = 1.69
21.32
1 2 3 4 5 6
A B C Statistical Process Control
Upper control limit Centerline Lower control limit
Headrest 2.3313 0.9078 0
824
2.5
2
1.5
1
0.5
0 1
1 2 3 4 5 6 7 8
2
3
4
5
6
7
8
A B C Statistical Process Control
9
10
11
12
13
14
15
16
17
18
19
20
9
10
11
12
13
14
15
16
17
18
19
20
D
Headrest Upper control limit 241.3248 Centerline 239.5617 Lower control limit 237.7986 Pattern Test #1 Failed at Points: 19, 20 Pattern Test #5 Failed at Points: 20
242 241 240 239 238 237 236 235 1
2
3
4
5
6
7
8
825
a The process is out of control. b The process is out of control at sample 19. c The width became too small. 21.33
1 2 3 4 5 6 7
A B C Statistical Process Control
D
Nuts Upper control limit 4.6143 Centerline 2.0363 Lower control limit 0 Pattern Test #1 Failed at Points: 21
6
5
4
3
2
1
0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
The process is out of control at sample 21. It is not necessary to draw the x chart.
21.34
1 2 3 4 5 6 7
A B C Statistical Process Control
D
Seats Upper control limit 11.1809 Centerline 5.3523 Lower control limit 0 Pattern Test #1 Failed at Points: 23, 24
826
14 12 10 8 6 4 2 0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
The process is out of control at sample 23. It is not necessary to draw the x chart.
21.35
1 2 3 4 5 6
A B C Statistical Process Control
Upper control limit Centerline Lower control limit
Headers 0.0045 0.002 0
827
0.005
0.004
0.003
0.002
0.001
0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
A B C 1 Statistical Process Control 2 3 Headers 4 Upper control limit 4.9872 5 Centerline 4.9841 6 Lower control limit 4.9809 4.988
4.986
4.984
4.982
4.98
4.978
4.976 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
The process is under control.
828
S
21.36
S 4
n
Upper control limit Centerline 4.9873 4.9841 .00107 3 3
.00107 ; S .00214
CPL
x LSL 4.9841 4.978 .95 3S 3(.00214 )
CPU
USL x 4.990 4.9841 .92 3S 3(.00214 )
Cpk Min (CPL, CPU) = .92
21.37
1 2 3 4 5 6
A B C Statistical Process Control Bolts 1.2895 0.5021 0
Upper control limit Centerline Lower control limit
1.4 1.2 1 0.8 0.6 0.4 0.2 0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
829
A B C Statistical Process Control
1 2 3 4 5 6 7 8
D
E
Bolts Upper control limit 12.7864 Centerline 11.81 Lower control limit 10.8336 Pattern Test #5 Failed at Points: 25 Pattern Test #6 Failed at Points: 23, 24, 25 13
12.5 12 11.5 11 10.5 10 9.5 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
The process went out of control at sample 23. 21.38
A B C 1 Statistical Process Control 2 3 Bottles 4 Upper control limit 69.6821 5 Centerline 33.3567 6 Lower control limit 0
830
80 70 60 50 40
30 20 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
1 2 3 4 5 6 7 8 9
A B C Statistical Process Control
D
Bottles Upper control limit 803.9376 Centerline 756.4267 Lower control limit 708.9157 Pattern Test #1 Failed at Points: 30 Pattern Test #5 Failed at Points: 29, 30 Pattern Test #6 Failed at Points: 30
850 830 810 790 770 750
730 710 690 670 650 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
831
The process went out of control at sample 29. 21.39a
A B C Statistical Process Control
1 2 3 4 5 6
Upper control limit Centerline Lower control limit
Calls 40.592 17.9131 0
50 45 40 35 30 25
20 15 10 5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
1 2 3 4 5 6 7 8 9
A B C Statistical Process Control
D
Calls Upper control limit 114.1052 Centerline 85.0217 Lower control limit 55.9381 Pattern Test #1 Failed at Points: 25 Pattern Test #5 Failed at Points: 30 Pattern Test #6 Failed at Points: 29, 30
832
140 120 100 80 60 40 20 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
b The process went out of control at sample 25 21.40
A B C 1 Statistical Process Control 2 Pipes 3 0.0956 4 Upper control limit 0.0372 5 Centerline 0 6 Lower control limit
833
0.12
0.1
0.08
0.06
0.04
0.02
0 1
1 2 3 4 5 6
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
8
9
10
11
12
13
14
15
16
17
18
19
20
A B C Statistical Process Control Pipes 3.0615 2.9892 2.9168
Upper control limit Centerline Lower control limit 3.1
3.05
3
2.95
2.9
2.85
2.8 1
2
3
4
5
6
7
The process is under control.
834
S
21.41
S 3
n
Upper control limit Centerline 3.0615 2.9892 .0241 3 3
.0241; S .0417
CPL
x LSL 2.9892 2.9 .71 3S 3(.0417 )
CPU
USL x 3.1 2.9892 .89 3S 3(.0417 )
Cpk Min (CPL, CPU) = .71
21.42
S 5
S n
Upper control limit Centerline 1504 .572 1496 .952 2.54 3 3
2.54; S 5.68
CPL
x LSL 1496 .952 1486 .64 3S 3(5.68 )
CPU
USL x 1506 1496 .952 .53 3S 3(5.68 )
Cpk Min (CPL, CPU) = .53 The value of the index is low because the statistics used to calculate the control limits and centerline were taken when the process was out of control. 21.43 Centerline = p = .035 Lower control limit = p 3
(.035 )(1 .035 ) p (1 p ) = .035 − 3 = .0174 1000 n
Upper control limit = p + 3
(.035 )(1 .035 ) p (1 p ) = .035 + 3 = .0524 1000 n
21.44 Centerline = p = .0324 Lower control limit = p 3
p (1 p ) (.0324 )(1 .0324 ) = .0324 − 3 = −.00516 (= 0) 200 n
Upper control limit = p + 3
p (1 p ) (.0324 )(1 .0324 ) = .0324 + 3 = .06996 200 n
835
1 2 3 4 5 6 7
A B C Statistical Process Control
D
Copiers Upper control limit 0.07 Centerline 0.0324 Lower control limit 0 Pattern Test #1 Failed at Points: 25
0.014 0.012 0.01 0.008 0.006 0.004 0.002 0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
The process is out of control at sample 25. 21.45 Centerline = p = .00352 Lower control limit = p 3
p (1 p ) (.00352 )(1 .00352 ) = .00352 − 3 = −.00443 (=0) 500 n
Upper control limit = p + 3
p (1 p ) (.00352 )(1 .00352 ) = .00352 + 3 = .01147 500 n
1 2 3 4 5 6
A B C Statistical Process Control
Upper control limit Centerline Lower control limit
PCBs 0.0115 0.0035 0
836
0.012
0.01
0.008
0.006
0.004
0.002
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
The process is under control. 21.46 Centerline = p = . 0383 Lower control limit = p 3
(.0383 )(1 .0383 ) p (1 p ) = .0383 − 3 = −.0193 (= 0) 100 n
Upper control limit = p + 3
(.0383 )(1 .0383 ) p (1 p ) = .0383 + 3 = .0959 100 n
1 2 3 4 5 6 7 8
A B C Statistical Process Control
D
Telephones Upper control limit 0.0959 Centerline 0.0383 Lower control limit 0 Pattern Test #1 Failed at Points: 25, 30
837
0.12
0.1
0.08
0.06
0.04
0.02
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
The process is out of control at samples 25 and 30. 21.47 Centerline = p = .0169 Lower control limit = p 3
p (1 p ) (.0169 )(1 .0169 ) = .0169 − 3 = .0047 1000 n
Upper control limit = p + 3
p (1 p ) (.0169 )(1 .0169 ) = .0169 + 3 = .0291 1000 n
1 2 3 4 5 6 7
A B C Statistical Process Control
D
Pages Upper control limit 0.0291 Centerline 0.0169 Lower control limit 0.0047 Pattern Test #2 Failed at Points: 37
838
0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 1
3
5
7
9
11 13 15 17 19 21 23 25 27 29 31 33 35 37 39
The process is out of control at samples 37-40. 21.48
A B C D 1 Statistical Process Control 2 Batteries 3 0.047 4 Upper control limit 0.0257 5 Centerline 0.0045 6 Lower control limit 7 Pattern Test #1 Failed at Points: 28, 29, 30
839
E
0.06
0.05
0.04
0.03
0.02
0.01
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
The process is out of control at sample 28. 21.49
A B C D 1 Statistical Process Control 2 3 Courier 4 Upper control limit 0.0088 5 Centerline 0.0044 6 Lower control limit 0 7 Pattern Test #1 Failed at Points: 23, 26 8 Pattern Test #2 Failed at Points: 9, 29
840
0.01
0.008
0.006
0.004
0.002
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
The process went out of control at sample 9. 21.50
A B C 1 Statistical Process Control 2 Scanners 3 0.0275 4 Upper control limit 0.0126 5 Centerline 0 6 Lower control limit 7 Pattern Test #1 Failed at Points: 24
D
841
0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
The process is out of control at sample 24.
842
Chapter 22 22.1
a1
a2
s1
0
29
s2
0
5
s3
14
0
s4
36
0
22.2
22.3 EMV(a 1 ) = .4(55) + .1(43) + .3(29) + .2(15) = 38.0 EMV(a 2 ) = .4(26) + .1(38) + .3(43) + .2(51) = 37.3
843
22.4
22.5
a1
a2
a3
s1
0
15
21
s2
0
3
4
s3
20
5
0
22.6 EOL(a 1 ) = .2(0) + .6(0) + .2(20) = 4.0 EOL(a 2 ) = .2(15) + .6(3) + .2(5) = 5.8 EOL(a 3 ) = .2(21) + .6(4) + .2(0) = 6.6 The EOL decision is a 1 .
22.7a
Produce
Demand
a0
a1
a2
a3
s0
0
-3.00
-6.00
-9.00
s1
0
5.00
2.00
-1.00
s2
0
5.00
10.00
7.00
s3
0
5.00
10.00
15.00
844
b
Produce
Demand
a0
a1
a2
a3
s0
0
3.00
6.00
9.00
s1
5.00
0
3.00
6.00
s2
10.00
5.00
0
3.00
s3
15.00
10.00
5.00
0
c
22.8a EMV(a 0 ) = 0 EMV(a 1 ) = .25(-3.00) + .25(5.00) + .25(5.00) + .25(5.00) = 3.00 EMV(a 2 ) = . 25(-6.00) + .25(2.00) + .25(10.00) + .25(10.00) = 4.00 EMV(a 3 ) = . 25(-9.00) + .25(-1.00) + .25(7.00) + .25(15.00) = 3.00
845
EMV decision is a 2 (bake 2 cakes) b EOL(a 0 ) = .25(0) + .25(5.00) + .25(10.00) + .25(15.00) = 7.50 EOL(a 1 ) = .25(3.00) + .25(0) + .25(5.00) + .25(10.00) = 4.50 EOL(a 2 ) = . 25(6.00) + .25(3.00) + .25(0) + .25(5.00) = 3.50 EOL(a 2 ) = . 25(9.00) + .25(6.00) + .25(3.00) + .25(0) = 4.50 EOL decision is a 2 (bake 2 cakes)
22.9
a 1 (flat fee)
a 2 Pay per snowfall
s0
-40,000
0
s1
-40,000
-18,000
s2
-40,000
-36,000
s3
-40,000
-54,000
s4
-40,000
-72,000
22.10 EMV(a 1 ) = -40,000 EMV(a 2 ) = . 05(0) + .15(-18,000) + .30(-36,000) + .40(-54,000) +.10(-72,000) = -42,300 EMV decision is a 1
22.11a Payoff Table
s 100
s 150
s 200
s 250
a 100
a 200
a 300
12(100)-10(100)
12(100)-9(200)+6(100)
12(100)-8.50(300)+6(200)
= 200
=0
= -150
12(100)- 10(100)
12(150)-9(200)+6(50)
12(150)-8.50(300)+6(150)
= 200
= 300
= 150
12(100)-10(100)
12(200)-9(200)
12(200)-8.50(300)+6(100)
= 200
= 600
= 450
12(100)-10(100)
12(200)–9(200)
12(250)-8.50(300)+6(50)
= 200
= 600
= 750
846
b
Opportunity Loss Table a 100
a 200
a 300
s 100
0
200
350
s 150
100
0
150
s 200
400
0
150
s 250
550
150
0
c
22.12 EMV(a 100 ) = 200 EMV(a 200 ) = . 20(0) + .25(300) + .40(600) + .15(600) = 405 EMV(a 300 ) = . 20(-150) + .25(150) + .40(450) + .15(750) = 300 EMV decision is order 200 shirts.
847
22.13 P(s 0 ) = .607, P(s 1 ) = .303, P(s 2 ) = .076, P(s 3 ) = .012, P(s 4 ) = .002 Payoff Table a0
a1
a2
a3
s0
0
-6,000
-12,000
-18,000
s1
0
7,000
1,000
-5,000
s2
0
7,000
14,000
8,000
s3
0
7,000
14,000
21,000
Opportunity Loss Table a0
a1
a2
a3
s0
0
6,000
12,000
18,000
s1
7,000
0
6,000
12,000
s2
14,000
7,000
0
6,000
s3
21,000
14,000
7,000
0
22.14a EMV(Small) = .15(-220) + .55(-330) + .30(-440) = -346.5 EMV(Medium) = .15(-300) + .55(-320) + .30(-390) = -338.0 EMV(Large) = .15(-350) + .55(-350) + .30(-350) =-350.0 EMV decision: build a medium size plant; EMV*= -338.0 b
Opportunity Loss Table Small
Medium
Large
Low
0
80
130
Moderate
10
0
30
High
90
40
0
c EOL(Small) = .15(0) + .55(10) + .30(90) = 32.5 EOL(Medium) = .15(80) + .55(0) + .30(40) = 24.0 EOL(Large) = .15(130) + .55(30) + .30(0) = 36.0 EOL decision: build a medium size plant
848
22.15a P(s 10 ) = 9/90 = .10, P(s 11 ) = 18/90 = .20, P(s 12 ) = 36/90 = .40, P(s 13 ) = 27/90 = .30 Payoff Table
s 10
s 11
s 12
s 13
a 10
a 11
a 12
a 13
30
10(5)- 11(2)+2
10(5)-12(2)+3.50
10(5)-13(2)+4.50
= 30
= 29.50
= 28.50
11(5)-11(2)
11(5)-12(2)+2
11(5)-13(2)+3.50
= 33
= 33
= 32.50
11(5)-11(2)
12(5)-12(2)
12(5)- 13(2)+2
= 33
= 36
= 36
11(5)-11(2)
12(5)-12(2)
13(5)-13(2)
= 33
= 36
= 39
30
30
30
b EMV(a 10 ) = 30 EMV(a 11 ) = .10(30) + .20(33) + .40(33) + .30(33) = 32.70 EMV(a 12 ) = .10(29.50) + .20(33) + .40(36) + .30(36) = 34.75 EMV(a 13 ) = .10(28.50) + .20(32.50) + .40(36) + .30(39) = 35.45 EMV decision: buy 13 bushels 22.16
Payoff Table Decision Produce
Don’t produce
5%
-28 million
0
10%
2 million
0
15%
8 million
0
Market share
EMV(produce) = .15(-28 million) + .45(2 million) + .40 (8 million) = -.1 million EMV (don’t produce) = 0 EMV decision: don’t produce 22.17 EPPI = .10(110) + .25(150) + .50(220) + .15(250) = 196 EMV(a 1 ) = .10(60) + .25(40) + .50(220) + .15(250) = 163.5 EMV(a 2 ) = .10(110) + .25(110) + .50(120) + .15(120) = 116.5 EMV(a 3 ) = .10(75) + .25(150) + .50(85) + .15(130) = 107 EVPI = EPPI – EMV* = 196 – 163.5 = 32.5 849
22.18
Opportunity Loss Table a1
a2
a3
s1
50
0
35
s2
110
40
0
s3
0
100
135
s4
0
130
120
EOL(a 1 ) = .10(50) + .25(110) + .50(0) + .15(0) = 32.5 EOL(a 2 ) = .10(0) + .25(40) + .50(100) + .15(130) = 79.5 EOL(a 3 ) = .10(35) + .25(0) + .50(135) + .15(120) = 89 EOL* = 32.5 22.19 EPPI = .5(65) + .5(110) = 87.5 EMV(a 1 ) = .5(65) + .5(70) = 67.5 EMV(a 2 ) = .5(20) + .5(110) = 65.0 EMV(a 3 ) = .5(45) + .5(80) = 62.5 EMV(a 4 ) = .5(30) + .5(95) = 62.5 EVPI = EPPI – EMV* = 87.5 – 67.5 = 20 22.20 a EPPI = .75(65) + .25(110) = 76.25 EMV(a 1 ) = .75(65) + .25(70) = 66.25 EMV(a 2 ) = .75(20) + .25(110) = 42.5 EMV(a 3 ) = .75(45) + .25(80) = 53.75 EMV(a 4 ) = .75(30) + .25(95) = 46.25 EVPI = EPPI – EMV* = 76.25 – 66.25 = 10 b EPPI = .95(65) + .05(110) = 67.25 EMV(a 1 ) = .95(65) + .05(70) = 65.25 EMV(a 2 ) = .95(20) + .05(110) = 24.5 EMV(a 3 ) = .95(45) + .05(80) = 46.75 EMV(a 4 ) = .95(30) + .05(95) = 33.25 EVPI = EPPI – EMV* = 67.25 – 65.25 = 2 22.21 As the difference between the two prior probabilities increases EVPI decreases.
850
22.22 Posterior Probabilities for I 1 sj
P(s j )
P(I 1 |s j )
P(s j and I 1 )
P(s j | I 1 )
__________________________________________________________________________ s1 .25 .40 (.25)(.40) = .10 .12/.20 = .500 s2
.40
.25
(.40)(.25) = .10
.10/.20 = .500
s3
.35
0
(.35)(0) = .0
0/.20 = 0
P(I 1 ) = .20 Posterior Probabilities for I 2 sj
P(s j )
P(I 2 |s j )
P(s j and I 2 )
P(s j | I 2 )
__________________________________________________________________________ s1 .25 .30 (.25)(.30) = .075 .075/.28 = .268 s2
.40
.25
(.40)(.25) = .10
.10/.28 = .357
s3
.35
30
(.35)(.30) = .105
.105/.28 = .375
P(I 2 ) = .28 Posterior Probabilities for I 3 sj
P(s j )
P(I 3 |s j )
P(s j and I 3 )
P(s j | I 3 )
__________________________________________________________________________ s1 .25 .20 (.25)(.20) = .05 .05/.29 = .172 s2
.40
.25
(.40)(.25) = .10
.10/.29 = .345
s3
.35
.40
(.35)(.40) = .14
.14/.29 = .483
P(I 3 ) = .29 Posterior Probabilities for I 4 sj
P(s j )
P(I 4 |s j )
P(s j and I 4 )
P(s j | I 4 )
__________________________________________________________________________ s1 .25 .10 (.25)(.10) = .025 .025/.23 = .109 s2
.40
.25
(.40)(.25) = .10
.10/.23 = .435
s3
.35
.30
(.35)(.30) = .105
.105/.23 = .456
P(I 4 ) = .23 22.23 Posterior Probabilities for I 1 sj
P(s j )
P(I 1 |s j )
P(s j and I 1 )
P(s j | I 1 )
__________________________________________________________________________ s1 .5 .98 (.5)(.98) = .49 .49/.515 = .951 s2
.5
.05
(.5)(.05) = .025 P(I 1 ) = .515
851
.025/.515 = .049
Posterior Probabilities for I 2 sj
P(s j )
P(I 2 |s j )
P(s j and I 2 )
P(s j | I 2 )
__________________________________________________________________________ s1 .5 .02 (.5)(.02) = .01 .01/.485 = .021 s2
.5
.95
(.5)(.95) = .475
.475/.485 = .979
P(I 2 ) = .485 22.24a Prior probabilities: EMV(a 1 ) = .5(10) + .5(22) = 16 EMV(a 2 ) = .5(18) + .5(19) = 18.5 EMV(a 3 ) = .5(23) + .5(15) = 19 EMV* = 19
I 1 : EMV(a 1 ) = .951(10) + .049(22) = 10.588 EMV(a 2 ) = .951(18) + .049(19) = 18.049 EMV(a 3 ) = .951(23) + .049(15) = 22.608 Optimal act: a 3
I 2 : EMV(a 1 ) = .021(10) + .979(22) = 21.748 EMV(a 2 ) = .021(18) + .979(19) = 18.979 EMV(a 3 ) = .021(23) + .979(15) = 15.168 Optimal act: a 1 b EMV` = .515(22.608) + .485(21.748) = 22.191 EVSI = EMV` - EMV* = 22.191 – 19 = 3.191 22.25 Prior probabilities: EMV(a 1 ) = .333(60) + .333(90) + .333(150) = 100 EMV(a 2 ) = 90 EMV* = 100 Posterior Probabilities for I 1 sj
P(s j )
P(I 1 |s j )
P(s j and I 1 )
P(s j | I 1 )
__________________________________________________________________________ s1 .333 .7 (.333)(.7) = .233 .233/.467 = .499 s2
.333
.5
(.333)(.5) = .167
.167/.467 = .358
s3
.333
.2
(.333)(.2) = .067
.067/.467 = .143
P(I 1 ) = .467
852
Posterior Probabilities for I 2 sj
P(s j )
P(I 2 |s j )
P(s j and I 2 )
P(s j | I 2 )
__________________________________________________________________________ s1 .333 .3 (.333)(.3) = .100 .100/.534 = .187 s2
.333
.5
(.333)(.5) = .167
.167/.534 = .313
s3
.333
.8
(.333)(.8) = .267
.267/.534 = .500
P(I 2 ) = .534 I 1 : EMV(a 1 ) = .499(60) + .358(90) + .143(150) = 83.61 EMV(a 2 ) = 90 I 2 : EMV(a 1 ) = .187(60) + .313(90) + .500(150) = 114.39 EMV(a 2 ) = 90 EMV` = .467(90) + .534(114.39) = 103.11 EVSI = EMV` - EMV* = 103.11 – 100 = 3.11 22.26 Prior probabilities: EMV(a 1 ) = .5(60) + .4(90) + .1(150) = 81 EMV(a 2 ) = 90 EMV* = 90
Posterior Probabilities for I 1 sj
P(s j )
P(I 1 |s j )
P(s j and I 1 )
P(s j | I 1 )
__________________________________________________________________________ s1 .5 .7 (.5)(.7) = .35 .35/.57 = .614 s2
.4
.5
(.4)(.5) = .20
.20/.57 = .351
s3
.1
.2
(.1)(.2) = .02
.02/.57 = .035
P(I 1 ) = .57 Posterior Probabilities for I 2 sj
P(s j )
P(I 2 |s j )
P(s j and I 2 )
P(s j | I 2 )
__________________________________________________________________________ .5 .3 (.5)(.3) = .15 .15/.43 = .349 s1 s2
.4
.5
(.4)(.5) = .20
.20/.43 = .465
s3
.1
.8
(.1)(.8) = .08
.08/.43 = .186
P(I 1 ) = .43 I 1 : EMV(a 1 ) = .614(60) + .351(90) + .035(150) = 73.68 EMV(a 2 ) = 90
I2: EMV(a 1 ) = .349(60) + .465(90) + .186(150) = 90.69 EMV(a 2 ) = 90 853
EMV` = .57(90) + .43(90.69) = 90.30 EVSI = EMV` - EMV* = 90.30 – 90 = .30 22.27 Prior probabilities: EMV(a 1 ) = .90(60) + .05(90) + .05(150) = 66 EMV(a 2 ) = 90 EMV* = 90 Posterior Probabilities for I 1 sj
P(s j )
P(I 1 |s j )
P(s j and I 1 )
P(s j | I 1 )
__________________________________________________________________________ .90 .7 (.90)(.7) = .63 .63/.665 = .947 s1 s2
.05
.5
(.05)(.5) = .025
.025/.665 = .038
s3
.05
.2
(.05)(.2) = .01
.01/.665 = .015
P(I 1 ) = .665 Posterior Probabilities for I 2 sj
P(s j )
P(I 2 |s j )
P(s j and I 2 )
P(s j | I 2 )
__________________________________________________________________________ .90 .3 (.90)(.3) = .27 .27/.335 = .806 s1 s2
.05
.5
(.05)(.5) = .025
.025/.335 = .075
s3
.05
.8
(.05)(.8) = .04
.04/.335 = .119
P(I 1 ) = .335 I 1 : EMV(a 1 ) = .947(60) + .038(90) + .015(150) = 62.49 EMV(a 2 ) = 90 I 2 : EMV(a 1 ) = .806(60) + .075(90) + .119(150) = 72.96 EMV(a 2 ) = 90 EMV` = .665(90) + .335(90) = 90 EVSI = EMV` - EMV* = 90 – 90 = 0 22.28 As the prior probabilities become more diverse EVSI decreases. 22.29 Payoff Table Demand
Purchase lot
Don’t purchase lot
10,000
10,000(5)-125,000 = -75,000
0
30,000
30,000(5) – 125,000 = 25,000
0
50,000
50,000(5)-125,000 = 125,000
0
EMV(purchase) = .2(-75,000) + .5(25,000) + .3(125,000) = 35,000 854
EMV(don’t purchase) = 0 EPPI = .2(0) + .5(25,000) + .3(125,000) = 50,000 EVPI = EPPI – EMV* = 50,000 – 35,000 = 15,000 22.30 EMV* = 0 EPPI = .15(0) + .45(2 million) + .40(8 million) = 4.1 million EVPI = EPPI – EMV* = 4.1 million – 0 = 4.1 million 22.31 Likelihood probabilities (binomial probabilities) P(I | s 1 ) = P(x = 3, n= 25 | p = .05) = .0930 P(I | s 2 ) = P(x = 3, n= 25 | p = .10) = .2265 P(I | s 3 ) = P(x = 3, n= 25 | p = .15) = .2174 Posterior Probabilities P(s j ) sj
P(I | s j )
P(s j and I)
P(s j | I)
__________________________________________________________________________ s1 .15 .0930 (.15)(.0930) = .0140 .0140/.2029 = .0690 s2 .45 .2265 (.45)(.2265) = .1019 .1019/.2029 = .5022 s3 .40 .2174 (.40)(.2174) = .0870 .0870/.2029 = .4288 P(I) = .2029 EMV(produce) = .0690(-28 million) + .5022(2 million) + .4288(8 million) = 2.503 million EMV (don’t produce) = 0 EMV decision: produce 22.32a
Payoff Table
Market share
Switch
Don’t switch
5%
5(100,000) – 700,000 = -200,000
285,000
10%
10(100,000) – 700,000 = 300,000
285,000
20%
20(100,000)-700,000 = 1,300,000
285,000
b EMV(switch) = .4(-200,000) + .4(300,000) + .2(1,300,000) = 300,000 EMV(don’t switch) = 285,000 Optimal act: switch (EMV* = 300,000) c EPPI = .4(285,000) + .4(300,000) + .2(1,300,000) = 494,000 EVPI = EPPI – EMV* = 494,000 – 300,000= 194,000
855
22.33
Payoff Table
Participating Households
Proceed
Don’t proceed
50,000
50(500) – 55,000 = -30,000
0
100,000
100(500) – 55,000 = -5,000
0
200,000
200(500) – 55,000 = 45,000
0
300,000
300(500) – 55,000 = 95,000
0
Likelihood probabilities (binomial probabilities) P(I | s 1 ) = P(x = 3, n= 25 | p = .05) = .0930 P(I | s 2 ) = P(x = 3, n= 25 | p = .10) = .2265 P(I | s 3 ) = P(x = 3, n= 25 | p = .20) = .1358 P(I | s 4 ) = P(x = 3, n= 25 | p = .30) = .0243
Posterior Probabilities sj P(s j )
P(I | s j )
P(s j and I)
P(s j | I)
__________________________________________________________________________ s1 .5 .0930 (.5)(.0930) = .0465 .0465/.1305 = .3563 s2
.3
.2265
(.3)(.2265) = .0680
.0680/.1305 = .5211
s3
.1
.1358
(.1)(.1358) = .0136
.0136/.1305 = .1042
s4
.1
.0243
(.1)(.0243) = .0024 .0024/.1305 = .0184 P(I) = .1305 EMV(proceed) = .3563(-30,000) + .5211(-5,000) + .1042(45,000) + .0184(95,000) = -6,858 EMV (don’t proceed = 0 EMV decision: don’t proceed 22.34 Likelihood probabilities (binomial probabilities) P(I | s 1 ) = P(x = 12, n= 100 | p = .05) = .0028 P(I | s 2 ) = P(x = 12, n= 100 | p = .10) = .0988 P(I | s 3 ) = P(x = 12, n= 100 | p = .20) = .0128 P(I | s 4 ) = P(x = 12, n= 100 | p = .30) = .000013
Posterior Probabilities sj P(s j )
P(I | s j )
P(s j and I)
P(s j | I)
__________________________________________________________________________ s1 .5 .0028 (.5)(.0028) = .0014 .0014/.0323 = .0433 s2
.3
.0988
(.3)(.0988) = .0296
.0296/.0323 = .9164
s3
.1
.0128
(.1)(.0128) = .0013
.0013/.0323 = .0402
s4
.1
.000013
(.1)(.000013) = .000001 .000001/.0323 = .000031 P(I) = .0323 856
EMV(proceed) = .0433(-30,000) + .9164(-5,000) + .0402(45,000) + .000031(95,000) = -4,069 EMV (don’t proceed = 0 EMV decision: don’t proceed 22.35 Posterior Probabilities for I 1 sj
P(s j )
P(I 1 |s j )
P(s j and I 1 )
P(s j | I 1 )
__________________________________________________________________________ .15 .5 (.15)(.5) = .075 .075/.30 = .25 s1 s2
.55
.3
(.55)(.3) = .165
.165/.30 = .55
s3
.30
.2
(.30)(.2) = .06
.06/.30 = .20
P(I 1 ) = .30 Posterior Probabilities for I 2 sj
P(s j )
P(I 2 |s j )
P(s j and I 2 )
P(s j | I 2 )
__________________________________________________________________________ s1 .15 .3 (.15)(.3) = .045 .045/.435 = .103 s2
.55
.6
(.55)(.6) = .33
.33/.435 = .759
s3
.30
.2
(.30)(.2) = .06
.06/.435 = .138
P(I 2 ) = .435 Posterior Probabilities for I 3 sj
P(s j )
P(I 3 |s j )
P(s j and I 3 )
P(s j | I 3 )
__________________________________________________________________________ .15 .2 (.15)(.2) = .03 .03/.265 = .113 s1 s2
.55
.1
(.55)(.1) = .055
.055/.265 = .208
s3
.30
.6
(.30)(.6) = .18
.18/.265 = .679
P(I 3 ) = .265 I 1 : EMV(a 1 ) = .25(-220) + .55(-330) + .20(-440) = -324.5 EMV(a 2 ) = .25(-300) + .55(-320) + .20(-390) = -329.0 EMV(a 3 ) = .251(-350) + .55(-350) + .20(-350) = -350 Optimal act: a 1 I 2 : EMV(a 1 ) = .103(-220) + .759(-330) + .138(-440) = -333.85 EMV(a 2 ) = .103(-300) + .759(-320) + .138(-390) = -327.59 EMV(a 3 ) = .103(-350) + .759(-350) + .138(-350) = -350 Optimal act: a 2
857
I 3 : EMV(a 1 ) = .113(-220) + .208(-330) + .679(-440) = -392.26 EMV(a 2 ) = .113(-300) + .208(-320) + .679(-390) = -365.28 EMV(a 3 ) = .113(-350) + .208(-350) + .679(-350) = -350 Optimal act: a 3 EMV` = .30(-324.5) + .435(-327.59) + .265(-350) = -332.60 EVSI = EMV` - EMV* = -332.60 – (-338) = 5.40 22.36 I 0 = neither person supports format change I 1 = one person supports format change I 2 = both people support format change Likelihood probabilities P( I i | s j ) I1 .0950 .18 .32
I0 .9025 .81 .64
5% 10% 20%
I2 .0025 .01 .04
Posterior Probabilities for I 0 sj
P(s j )
P(I 0 |s j )
P(s j and I 0 )
P(s j | I 0 )
__________________________________________________________________________ .4 .9025 (.4)(.9025) = .361 .361/.813 = .444 s1 s2
.4
.81
(.4)(.81) = .324
.324/.813 = .399
s3
.2
.64
(.2)(.64) = .128
.128/.813 = .157
P(I 0 ) = .813 Posterior Probabilities for I 1 sj
P(s j )
P(I 1 |s j )
P(s j and I 1 )
P(s j | I 1 )
__________________________________________________________________________ s1 .4 .0950 (.4)(.0950) = .038 .038/.174 = .218 s2
.4
.18
(.4)(.18) = .072
.072/.174 = .414
s3
.2
.32
(.2)(.32) = .064
.064/.174 = .368
P(I 1 ) = .174 Posterior Probabilities for I 3 sj
P(s j )
P(I 2 |s j )
P(s j and I 2 )
P(s j | I 2 )
__________________________________________________________________________ s1 .4 .0025 (.4)(.0025) = .001 .001/.013 = .077 s2
.4
.01
(.4)(.01) = .004
.004/.013 = .308
s3
.2
.04
(.2)(.04) = .008
.008/.013 = .615
P(I 2 ) = .013
858
I 1 : EMV(switch) = .444(-200,000) + .399(300,000) + .157(1,300,000) = 235,000 EMV(don’t switch) = 285,000 Optimal act: don’t switch I 2 : EMV(switch) = .218(-200,000) + .414(300,000) + .368(1,300,000) = 559,000 EMV(don’t switch) = 285,000 Optimal act: switch I 3 : EMV(switch) = .077(-200,000) + .308(300,000) + .615(1,300,000) = 876,500 EMV(don’t switch) = 285,000 Optimal act: switch EMV` = .813(285,000) + .174(546,000) + .013(876,500) = 338,104 EVSI = EMV` - EMV* = 338,104 – 300,000 = 38,104 22.37 Likelihood probabilities (binomial probabilities) P(I | s 1 ) = P(x = 2, n= 25 | p = .05) = .2305 P(I | s 2 ) = P(x = 2, n= 25 | p = .10) = .2659 P(I | s 3 ) = P(x = 2, n= 25 | p = .20) = . 0708
Posterior Probabilities for I P(s j ) P(I|s j ) sj
P(s j and I)
P(s j | I)
__________________________________________________________________________ .4 .2305 (.4)(.2305) = .0922 .0922/.2127 = .4334 s1 s2
.4
.2659
(.4)(.2659) = .1064
.1064/.2127 = .5000
s3
.2
.0708
(.2)(.0708) = .0142
.0142/.2127 = .0667
P(I) = .2127 EMV(switch) = .4334(-200,000) + .5000(300,000) + .0667(1,300,000) = 149,873 EMV(don’t switch) = 285,000 Optimal act: don’t switch
859
22.38a Payoff Table Demand
Battery 1
Battery 2
Battery 3
50,000
20(50,000)-900,000
23(50,000)-1,150,000
25(50,000)-1,400,000
= 100,000
0
-150,000
20(100,000)-900,000
23(100,000)-1,150,000
25(100,000)-1,400,000
=1,100,000
1,150,000
1,100,000
20(150,000)-900,000
23(150,000)-1,150,000
25(150,000)-1,400,000
=2,100,000
2,300,000
2,350,000
100,000 150,000
b
Opportunity Loss table
Demand
Battery 1
Battery 2
Batter3
50,000
0
100,000
250,000
100,000
50,000
0
50,000
150,000
250,000
50,000
9
c EMV(Battery 1) = .3(100,000) + .3(1,100,000) + .4(2,100,000) = 1,200,000 EMV(Battery 2) = .3(0) + .3(1,150,000) + .4(2,300,000) = 1,265,000 EMV(Battery 3) = .3(-150,000) + .3(1,100,000) + .4(2,350,000) = 1,225,000 EMV decision: Battery 2 d EOL(Battery 2) = .3(100,000) + .3(0) + .4(50,000) = 50,000 EVPI = EOL* = 50,000 22.39
Payoff Table
Percentage change
Change ad
-2
-258,000
0
-1
-158,000
0
0
-58,000
0
1
42,000
0
2
142,000
0
Don’t change
EMV(Change ad) = -1(-258,000) + .1(-158,000) + .2(-58,000) + .3(42,000) + .3(142,000) = 2,000 EMV (don’t change) = 0. Optimal decision: change ad
860
22.40 I 0 = person does not believe the ad I 1 = person believes the ad Likelihood probabilities P( I i | s j ) I0 .70 .69 .68 .67 .66
30% 31% 32% 33% 34%
I1 .30 .31 .32 .33 .34
Posterior Probabilities for I 0 sj
P(s j )
P(I 0 |s j )
P(s j and I 0 )
P(s j | I 0 )
__________________________________________________________________________ s1 .1 .70 (.1)(.70) = .070 .070/.674 = .104 s2
.1
.69
(.1)(.69) = .069
.069/.674 = .102
s3
.2
.68
(.2)(.68) = .136
.136/.674 = .202
s4
.3
.67
(.3)(.67) = .201
.201/.674 = .298
s5
.3
.66
(.3)(.66) = .198
.198/.674 = .294
P(I 0 ) = .674 Posterior Probabilities for I 1 sj
P(s j )
P(I 1 |s j )
P(s j and I 1 )
P(s j | I 1 )
__________________________________________________________________________ s1 .1 .30 (.1)(.30) = .030 .030/.326 = .092 s2
.1
.31
(.1)(.31) = .031
.031/.326 = .095
s3
.2
.32
(.2)(.32) = .064
.064/.326 = .196
s4
.3
.33
(.3)(.33) = .099
.099/.326 = .304
s5
.3
.34
(.3)(.34) = .102
.102/.326 = .313
P(I 1 ) = .326 I 0 : EMV(Change ad) = .104(-258,000) + .102(-158,000) + .202(-58,000) + .298(42,000) + .294(142,000) = -400 EMV (don’t change) = 0. Optimal decision: don’t change ad
I 1 : EMV(Change ad) = .092(-258,000) + .095(-158,000) + .196(-58,000) + .304(42,000) + .313(142,000) = 7,100
861
EMV (don’t change) = 0. Optimal decision: change ad EMV` = .674(0) + .326(7,100) = 2,315 EVSI = EMV` - EMV* = 2,315 – 2,000 = 315 22.41 Likelihood probabilities (binomial probabilities) P(I | s 1 ) = P(x = 1, n = 5 | p = .30) = .3602 P(I | s 2 ) = P(x = 1, n = 5 | p = .31) = .3513 P(I | s 3 ) = P(x = 1, n = 2 | p = .32) = . 3421 P(I | s 4 ) = P(x = 1, n = 2 | p = .33) = . 3325 P(I | s 5 ) = P(x = 1, n = 2 | p = .34) = . 3226 Posterior Probabilities for I sj P(s j ) P(I|s j )
P(s j and I)
P(s j | I)
__________________________________________________________________________ .1 .3602 (.1)(.3602) = .0360 .0360/.3361 = .1072 s1 s2
.1
.3513
(.1)(.3513) = .0351
.0351/.3361 = .1045
s3
.2
.3421
(.2)(.3421) = .0684
.0684/.3361 = .2036
s4
.3
.3325
(.3)(.3325) = .0997
.0997/.3361 = .2968
s5
.3
.3226
(.3)(.3226) = .0968 P(I) = .3361
.0968/.3361 = .2879
EMV(Change ad) = .1072(-258,000) + .1045(-158,000) + .2036(-58,000) + .2968(42,000) + .2879(142,000) = -2,620 EMV (don’t change) = 0. Optimal decision: don’t change ad 22.42 EMV(25 telephones) = 50,000 EMV(50 telephones) = .50(30,000) + .25(60,000) + .25(60,000) = 45,000 EMV(100 telephones) = .50(20,000) + .25(40,000) + .25(80,000) = 40,000 Optimal decision: 25 telephones (EMV* = 50,000) I 1 = small number of calls I 2 = medium number of calls I 3 = large number of calls
862
Likelihood probabilities (Poisson distribution)
=5
I1
I2
I3
P(X < 8 | = 5)
P(8 X < 17 | = 5)
P(X 17 | = 5)
= .8667
= .1334
=0
P(8 X < 17 | = 10)
P(X 17 | = 10)
= .7527
= .0270
P(8 X < 17 | = 15)
P(X 17 | = 15)
= .6461
= .3359
= 10 P(X < 8 | = 10) = .2202
= 15 P(X < 8 | = 15) = .0180
Posterior Probabilities for I 1 sj
P(s j )
P(I 1 |s j )
P(s j and I 1 )
P(s j | I 1 )
__________________________________________________________________________ .50 .8667 (.50)(.8667) = .4333 .4333/.4929 = .8792 s1 s2
.25
.2202
(.25)(.2202) = .0551
.0551/.4929 = .1117
s3
.25
.0180
(.25)(.0180) = .0045
.0045/.4929 = .0091
P(I 1 ) = .4929 Posterior Probabilities for I 2 sj
P(s j )
P(I 2 |s j )
P(s j and I 2 )
P(s j | I 2 )
__________________________________________________________________________ .50 .1334 (.50)(.1334) = .0667 .0667/.4164 = .1601 s1 s2
.25
.7527
(.25)(.7527) = .1882
.1882/.4164 = .4519
s3
.25
.6461
(.25)(.6461) = .1615
.1615/.4164 = .3879
P(I 2 ) = .4164 Posterior Probabilities for I 3 sj
P(s j )
P(I 3 |s j )
P(s j and I 3 )
P(s j | I 3 )
__________________________________________________________________________ s1 .50 .0 (.50)(0) = 0 0/.0907 = 0 s2
.25
.0270
(.25)(.0270) = .0068
.0068/.0907 = .0745
s3
.25
.3359
(.25)(.3359) = .0840
.0840/.0907 = .9254
P(I 3 ) = .0907 I 1 : EMV(25 telephones) = 50,000 EMV(50 telephones) = .8792(30,000) + .1117(60,000) + .0091(60,000) = 33,624 EMV(100 telephones) = .8792(20,000) + .1117(40,000) + .0091(80,000) = 22,780 Optimal act: 25 telephones
863
I 2 : EMV(25 telephones) = 50,000 EMV(50 telephones) = .1601(30,000) + .4519(60,000) + .3879(60,000) = 55,191 EMV(100 telephones) = .1601(20,000) + .4519(40,000) + .38791(80,000) = 52,310 Optimal act: 50 telephones
I 3 : EMV(25 telephones) = 50,000 EMV(50 telephones) = 0(30,000) + .0745(60,000) + .9254(60,000) = 60,000 EMV(100 telephones) = 0(20,000) + .0745(40,000) + .9254(80,000) = 77,012 Optimal act: 100 telephones EMV` = .4929(50,000) + .4164(55,191) + .0907(77,012) = 54,612 EVSI = EMV` - EMV* = 54,612 – 50,000 = 4,612 Because the value is greater than the cost ($4,000) Max should not sample. If he sees a small number of calls install 25 telephones. If there is a medium number install 50 telephones. If there is a large number of calls, install 100 telephones. 22.43a EMV(Model 101) = .2(20 million) + .4(100 million) + .4(210 million) = 128 million EMV (Model 202) = .1(70 million) + .4(100 million) + .5(150 million) = 122 million Optimal decision: Model 101 b Likelihood probabilities (binomial distribution) for Model 101 P(X =1, n = 10| p = .05) = .3151 P(X =1, n = 10| p = .10) = .3874 P(X =1, n = 10| p = .15) = .3474 Posterior Probabilities for Model 101 P(s j ) P(I|s j ) sj
P(s j and I)
P(s j | I)
__________________________________________________________________________ s1 .2 .3151 (.2)(.3151) = .0630 .0630/.3570 = .1766 s2
.4
.3874
(.4)(.3874) = .1550
.1550/.3570 = .4341
s3
.4
.3474
(.4)(.3474) = .1390
.1390/.3570 = .3893
P(I) = .3570 EMV(Model 101) = .1766(20 million) + .4341(100 million) + .3893(210 million) = 128.7 million Likelihood probabilities (binomial distribution) for Model 202 P(X =9, n = 20| p = .30) = .0654 P(X =9, n = 20| p = .40) = .1597 P(X =9, n = 20| p = .50) = .1602 864
Posterior Probabilities for Model 202 sj P(s j ) P(I|s j )
P(s j and I)
P(s j | I)
__________________________________________________________________________ s1 .1 .0654 (.1)(.0654) = .0065 .0065/.1505 = .0434 s2
.4
.1597
(.4)(.1597) = .0639
.0639/.1505 = .4245
s3
.5
.1602
(.5)(.1602) = .0801
.0801/.1505 = .5321
P(I) = .1505 EMV(Model 101) = .0434(70 million) + .4245(100 million) + .5321(150 million) = 125.3 million Optimal decision: Model 101 22.44 EMV( Release in North America) = .5(33 million) + .3(12 million) + .2(-15 million) = 17.1 million EMV(European distributor) = 12 million Optimal decision: Release in North America
Posterior Probabilities for I 1 (Rave review) sj
P(s j )
P(I 1 |s j )
P(s j and I 1 )
P(s j | I 1 )
__________________________________________________________________________ .5 .8 (.5)(.8) = .40 .40/.63 = .635 s1 s2
.3
.5
(.3)(.5) = .15
.15/.63 = .238
s3
.2
.4
(.2)(.4) = .08
.08/.63 = .127
P(I 1 ) = .63 EMV( Release in North America) = .635(33 million) + .238(12 million) + .127(-15 million) =21.9 million EMV(European distributor) = 12 million Optimal decision: Release in North America
Posterior Probabilities for I 2 (lukewarm response) sj
P(s j )
P(I 2 |s j )
P(s j and I 2 )
P(s j | I 2 )
__________________________________________________________________________ s1 .5 .1 (.5)(.1) = .05 .05/.20 = .25 s2
.3
.3
(.3)(.3) = .09
.09/.20 = .45
s3
.2
.3
(.2)(.3) = .06
.06/.20 = .30
P(I 2 ) = .20
865
EMV( Release in North America) = .25(33 million) + .45(12 million) + .30(-15 million) = 9.2 million EMV(European distributor) = 12 million Optimal decision: Sell to European distributor
Posterior Probabilities for I 3 (poor response) sj
P(s j )
P(I 3 |s j )
P(s j and I 3 )
P(s j | I 3 )
__________________________________________________________________________ .5 .1 (.5)(.1) = .05 .05/.17 = .294 s1 s2
.3
.2
(.3)(.2) = .06
.06/.17 = .353
s3
.2
.3
(.2)(.3) = .06
.06/.17 = .353
(I 3 ) = .17 EMV( Release in North America) = .294(33 million) + .353(12 million) + .353(-15 million) = 8.6 million EMV(European distributor) = 12 million Optimal decision: Sell to European distributor. EMV` = .63(21.9 million) + .20(12 million) + .17(12 million) = 18.2 million EVSI = EMV` - EMV* = 18.2 million – 17.1 million = 1.1 million Because EVSI is greater than the sampling cost (100,000) the studio executives should show the movie to a random sample of North Americans. If the response is a rave review release the movie in North America. If not sell it to Europe.
866