Introductory statistics 9th edition solution manual pdf

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Solution Manual for Introductory Statistics, 9th Edition, Prem S. Mann,,

Table of Contents

1. Introduction

2. Organizing and Graphing Data

3. Numerical Descriptive Measures

4. Probability

5. Discrete Random Variables and Their Probability Distribution

6. Continuous Random Variables and the Normal Distribution

7. Sampling Distributions

8. Estimation of the Mean and Proportion

9. Hypothesis Tests About the Mean and Proportion

10. Estimation and Hypothesis Testing: Two Populations

11. Chi-Square Tests

12. Analysis of Variance

13. Simple Linear Regression

[Test Bank for ch24 ~ 17]

14. Multiple Regression (online only)

15. Nonparametric Methods (online only)

16. Appendix A: Explanation of Data Sets

17. Appendix B: Statistical Tables

16
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 2 ANSWERS
Exercises 2.1
2.1 (a) Hair color, model of car, and brand of popcorn are qualitative variables.
(b) Number of eggs in a nest, number of cases of flu, and number of employees are discrete, quantitative variables.
(c) Temperature, weight, and time are quantitative continuous variables.
2.2 (a) A qualitative variable is a nonnumerically valued variable. Its
(b) A discrete, quantitative variable is one whose possible values can be listed. It is usually obtained by counting rather than by measuring.
(c) A continuous, quantitative variable is one whose possible values form some interval of numbers. It usually results from measuring.
2.3 (a) Qualitative data result from observing and recording values of a qualitative variable, such as, color or shape.
(b) Discrete, quantitative data are values of a discrete quantitative variable. Values usually result from counting something.
(c) Continuous, quantitative data are values of a continuous variable. Values are usually the result of measuring something such as temperature that can take on any value in a given interval.
2.4 The classification of data is important because it will help you choose the correct statistical method for analyzing the data.
2.5 Of qualitative and quantitative (discrete and continuous) types of data, only qualitative yields nonnumerical data.
2.6 (a) The first column lists states. Thus, it consists of qualitative data.
(b) The second column gives the number of serious doctor disciplinary actions in each state in 2005-2007. These data are integers and therefore are quantitative, discrete data.
(c) The third column gives ratios of actions per 1,000 doctors for the years 2005-2007. The hint tells us that the possible ratios of positive whole numbers can be listed. For example, 8.33 out of 1,000 could also be listed as 833 out of 100,000. Ratios of whole numbers cannot be irrational. Therefore these data are quantitative, discrete.
2.7 (a) The second column consists of quantitative, discrete data. This column provides the ranks of the cities with the highest temperatures.
(b) The third column consists of quantitative, continuous data since temperatures can take on any value from the interval of numbers found on the temperature scale. This column provides the highest temperature in each of the listed cities.
(c) The information that Phoenix is in Arizona is qualitative data since it is nonnumeric.
2.8 (a) The first column consists of quantitative, discrete data. This column provides the ranks of the deceased celebrities with the top 5 earnings during the period from October 2004 to October 2005.
(b) The third column consists of quantitative, discrete data, the earnings of the celebrities. Since money involves discrete units, such as dollars and cents, the data is discrete, although, for all practical purposes, this data might be considered quantitative continuous data.
2.9 (a) The first column consists of quantitative, discrete data. This column
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Section2.2,OrganizingQualitativeData 17
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
provides the ranks of the top ten countries with the highest number of Wi-Fi locations, as of October 28, 2009. These are whole numbers.
(b) The countries listed in the second column are qualitative data since they are nonnumerical.

(c) The third column consists of quantitative, discrete data. This column provides the number of Wi-Fi locations in each of the countries. These are whole numbers.
2.10 (a) The first column contains types of products. They are qualitative data since they are nonnumerical.
(b) The second column contains number of units shipped in the millions. These are whole numbers and are quantitative, discrete.
(c) The third column contains money values. Technically, these are quantitative, discrete data since there are gaps between possible values at the cent level. For all practical purposes, however, these are quantitative, continuous data.
2.11 The first column contains quantitative, discrete data in the form of ranks. These are whole numbers. The second and third columns contain qualitative data in the form of names. The last column contains the number of viewers of the programs. Total number of viewers is a whole number and therefore quantitative, discrete data.
2.12 Duration is a measure of time and is therefore quantitative, continuous. One might argue that workshops are frequently done in whole numbers of weeks, which would be quantitative, discrete. The number of students, the number of each gender, and the number of each ethnicity are whole numbers and are therefore quantitative, discrete. The genders and ethnicities themselves are nonnumerical and are therefore qualitative data. The number of web reports is a whole number and is quantitative, discrete data.
2.13 The first column contains quantitative, discrete data in the form of ranks. These are whole numbers. The second and fourth columns are nonnumerical and are therefore qualitative data. The third and fifth columns are measures of time and weight, both of which are quantitative, continuous data.
2.14 Of the eight items presented, only high school class rank involves ordinal data. The rank is ordinal data.
Exercises 2.2
2.15 A frequency distribution of qualitative data is a table that lists the distinct values of data and their frequencies. It is useful to organize the data and make it easier to understand.
2.16 (a) The frequency of a class is the number of observations in the class, whereas, the relative frequency of a class is the ratio of the class frequency to the total number of observations.
(b) The percentage of a class is 100 times the relative frequency of the class. Equivalently, the relative frequency of a class is the percentage of the class expressed as a decimal.
2.17 (a) True. Having identical frequency distributions implies that the total number of observations and the numbers of observations in each class are identical. Thus, the relative frequencies will also be identical.
(b) False. Having identical relative frequency distributions means that the ratio of the count in each class to the total is the same for both frequency distributions. However, one distribution may have twice (or some other multiple) the total number of observations as the other. For example, two distributions with counts of 5, 4, 1 and 10, 8, 2
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Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
would be different, but would have the same relative frequency distribution.
(c) If the two data sets have the same number of observations, either a frequency distribution or a relative-frequency distribution is suitable. If, however, the two data sets have different numbers of observations, using relative-frequency distributions is more appropriate because the total of each set of relative frequencies is 1, putting both distributions on the same basis for comparison.
2.18 (a)-(b)
The classes are the days of the week and are presented in column 1. The frequency distribution of the networks is presented in column 2. Dividing each frequency by the total number of shows, which is 20, results in each class's relative frequency. The relative frequency distribution is presented in column 3.
Network Frequency Relative Frequency
ABC 5 0.25 CBS 9 0.45 Fox 6 0.30
20 1.00
(c) We multiply each of the relative frequencies by 360 degrees to obtain the portion of the pie represented by each network. The result is
CBSFoxABC
Category
ABC25.0%
Fox30.0%
CBS45.0%
NETWORK
(d) We use the bar chart to show the relative frequency with which each
network occurs. The result is
FoxCBSABC
50
40
30
20
10
0
NETWORK
Perc
ent
NETWORK
Percent within all data.
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Section2.2,OrganizingQualitativeData 19
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
2.19 (a)-(b)
The classes are the NCAA wrestling champions and are presented in column 1. The frequency distribution of the champions is presented in column 2. Dividing each frequency by the total number of champions, which is 25, results in each class's relative frequency. The relative frequency distribution is presented in column 3.
Champion Frequency Relative Frequency
Iowa 13 0.52 Iowa St. 1 0.04
Minnesota 3 0.12 Arizona St. 1 0.04 Oklahoma St. 7 0.28
25 1.00
(c) We multiply each of the relative frequencies by 360 degrees to obtain the portion of the pie represented by each team. The result is
IowaOklahoma St.MinnesotaArizona St.Iowa St.
CategoryIowa St.
4.0%Arizona St.4.0%
Minnesota12.0%
Oklahoma St.28.0%
Iowa52.0%
CHAMPION
(d) We use the bar chart to show the relative frequency with which each TEAM occurs. The result is
Iowa St.Arizona St.MinnesotaOklahoma St.Iowa
50
40
30
20
10
0
CHAMPION
Perc
ent
CHAMPION

Percent within all data. 2.20 (a)-(b) The classes are the colleges and are presented in column 1. The
frequency distribution of the colleges is presented in column 2. Dividing each frequency by the total number of students in the section of Introduction to Computer Science, which is 25, results in each
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class's relative frequency. The relative frequency distribution is presented in column 3.
College Frequency Relative Frequency
BUS 9 0.36 ENG 12 0.48
LIB 4 0.16
25 1.00
(c) We multiply each of the relative frequencies by 360 degrees to obtain the portion of the pie represented by each college. The result is
ENGBUSLIB
Category
LIB16.0%
BUS36.0%
ENG48.0%
COLLEGE
(d) We use the bar chart to show the relative frequency with which each COLLEGE occurs. The result is
LIBENGBUS
50
40
30
20
10
0
COLLEGE
Perc
ent
COLLEGE
Percent within all data. 2.21 (a)-(b)
The classes are the class levels and are presented in column 1. The frequency distribution of the class levels is presented in column 2. Dividing each frequency by the total number of students in the introductory statistics class, which is 40, results in each class's relative frequency. The relative frequency distribution is presented in column 3.
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Section2.2,OrganizingQualitativeData 21
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
Class Level Frequency Relative Frequency
Fr 6 0.150 So 15 0.375
Jr 12 0.300 Sr 7 0.175
40 1.000
(c) We multiply each of the relative frequencies by 360 degrees to obtain the portion of the pie represented by each class level. The result is
SoJrSrFr
Category
Fr15.0%
Sr17.5%
Jr30.0%
So37.5%
CLASS
(d) We use the bar chart to show the relative frequency with which each CLASS level occurs. The result is
SrJrSoFr
40
30
20
10
0
CLASS
Perc
ent
CLASS
Percent within all data. 2.22 (a)-(b)
The classes are the regions and are presented in column 1. The frequency distribution of the regions is presented in column 2. Dividing each frequency by the total number of states, which is 50, results in each class's relative frequency. The relative frequency distribution is presented in column 3.
Class Level Frequency Relative Frequency
NE 9 0.18 MW 12 0.24
SO 16 0.32 WE 13 0.26
50 1.00
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(c) We multiply each of the relative frequencies by 360 degrees to obtain the portion of the pie represented by each region. The result is
NEMWWESO
Category
SO32.0%
WE26.0%
MW24.0%
NE18.0%
REGION
(d) We use the bar chart to show the relative frequency with which each REGION occurs. The result is
SOWEMWNE
35
30
25
20
15
10
5
0
REGION
Perc
ent
REGION
Percent within all data. 2.23 (a)-(b)
The classes are the days and are presented in column 1. The frequency distribution of the days is presented in column 2. Dividing each frequency by the total number road rage incidents, which is 69, results in each class's relative frequency. The relative frequency distribution is presented in column 3.
Class Level Frequency Relative Frequency
Su 5 0.0725 M 5 0.0725
Tu 11 0.1594 W 12 0.1739
Th 11 0.1594
F 18 0.2609
Sa 7 0.1014
69 1.0000
(c) We multiply each of the relative frequencies by 360 degrees to obtain the portion of the pie represented by each day. The result is
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Section2.2,OrganizingQualitativeData 23
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
FWTuThSaSuM
CategoryM
7.2%Su
7.2%
Sa10.1%
Th25.9%
Tu15.9%
W17.4%
F26.1%
DAY
(d) We use the bar chart to show the relative frequency with which each DAY occurs. The result is
SaFThWTuMSu
25
20
15
10
5
0
DAY
Perc
ent
DAY
Percent within all data.
2.24 (a) We first find each of the relative frequencies by dividing each of the frequencies by the total frequency of 413,403
Robbery Type Frequency Relative
Frequency Street/highway 179,296 0.4337 Commercial house 60,493 0.1463
Gas or service station 11,362 0.0275 Convenience store 25,774 0.0623
Residence 56,641 0.1370
Bank 9,504 0.0230
Miscellaneous 70,333 0.1701
413,403 1.0000
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each robbery type. The result is
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Street/highwayMiscellaneousCommercial houseResidenceConvenience storeGas or service stationBank
CategoryBank2.3%
Gas or service station2.7%Convenience store
6.2%
Residence13.7%
Commercial house14.6%
Miscellaneous17.0%
Street/highway43.4%
TYPE

(c) We use the bar chart to show the relative frequency with which each
robbery type occurs. The result is
2.25 (a) We first find the relative frequencies by dividing each of the
frequencies by the total sample size of 509.
Color Frequency Relative Frequency
Brown 152 0.2986 Yellow 114 0.2240
Red 106 0.2083 Orange 51 0.1002
Green 43 0.0845
Blue 43 0.0845
509 1.0000
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each color of M&M. The result is
TYPE
REL
ATI
VE
FREQ
UENC
Y
Misce
llane
ous
Bank
Resid
ence
Coinv
enien
ce st
ore
Gas o
r serv
ice st
ation
Comm
ercial
hous
e
Stree
t/High
way
0.4
0.3
0.2
0.1
0.0
Chart of RELATIVE FREQUENCY vs TYPE
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Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
(c) We use the bar chart to show the relative frequency with which each
color occurs. The result is

2.26 (a) We first find the relative frequencies by dividing each of the frequencies by the total sample size of 500.
Political View Frequency Relative
Frequency Liberal 160 0.320 Moderate 246 0.492
Conservative 94 0.188
500 1.000
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each political view. The result is
0.0844794, 8.4%BLUE
0.0844794, 8.4%GREEN
0.100196, 10.0%ORANGE

0.208251, 20.8%RED
0.223969, 22.4%YELLOW
0.298625, 29.9%BROWN
Pie Chart of RELATIVE FREQUENCY vs COLOR
COLOR
REL
ATI
VE
FREQ
UENC
Y
BLUEGREENORANGEREDYELLOWBROWN
0.30
0.25
0.20
0.15
0.10
0.05
0.00
Chart of RELATIVE FREQUENCY vs COLOR
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ModerateLiberalConservative
Category
Conservative18.8%
Liberal32.0%
Moderate49.2%
VIEW
(c) We use the bar chart to show the relative frequency with which each
political view occurs. The result is
ConservativeModerateLiberal
50
40
30
20
10
0
VIEW
Perc
ent
of F
REQ
UENC
Y
VIEW
Percent within all data.
2.27 (a) We first find the relative frequencies by dividing each of the frequencies by the total sample size of 98,993.
Rank Frequency Relative Frequency
Professor 24,418 0.2467
Associate professor 21,732 0.2195
Assistant professor 40,379 0.4079
Instructor 10,960 0.1107
Other 1,504 0.0152
98,993 1.0000
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each rank. The result is
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Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.
Assistant professorProfessorAssociate professorInstructorother
Categoryother1.5%Instructor
11.1%
Associate professor22.0%
Professor24.7%
Assistant professor40.8%
RANK
(c) We use the bar chart to show the relative frequency with which each
rank occurs. The result is
OtherInstructorAssociate professorProfessorAssistant professor
40
30
20
10
0
RANK
Perc
ent
of F
REQ
UENC
Y
RANK
Percent within all data.
2.28 (a) We first find the relative frequencies by dividing each of the frequencies by the total sample size of 52,389.
Payer Frequency Relative Frequency
Medicare 9,983 0.1906
Medicaid 8,142 0.1554
Private insurance 26,825 0.5120
Other government 1,777 0.0339
Self pay/charity 5,512 0.1052
Other 150 0.0029
52,389 1.0000
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each payer. The result is
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Private insuranceMedicareMedicaidSelf pay/charityOther governmentother
Categoryother0.3%
Other government3.4%
Self pay/charity10.5%
Medicaid15.5%
Medicare19.1%
Private insurance51.2%
PAYER
(c) We use the bar chart to show the relative frequency with which each
payer occurs. The result is
Othe
r
Othe
r gov
ernme
nt
Self p
ay/ch
arity
Medic
aid
Medic
are
Priva
te ins
uran
ce
50
40
30
20
10
0
PAYER
Perc
ent
of F
REQ
UENC
Y
PAYER
Percent within all data.
2.29 (a) We first find the relative frequencies by dividing each of the frequencies by the total sample size of 200.
Color Frequency Relative Frequency
Red 88 0.44
Black 102 0.51
Green 10 0.05
200 1.00
(b) We multiply each of the relative frequencies by 360 degrees to obtain
the portion of the pie represented by each color. The result is
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BlackRedGreen
CategoryGreen5.0%
Red44.0%
Black51.0%
COLOR
(c) We use the bar chart to show the relative frequency with which each
color occurs. The result is
GreenBlackRed
50
40
30
20
10
0
NUMBER
Perc
ent
of F
REQ
UENC
Y

COLOR
Percent within all data. 2.30 (a) Using Minitab, retrieve the data from the Weiss-Stats-CD. Column 1
contains the type of the car. From the tool bar, select Stat Tables
Tally Individual Variables, double-click on TYPE in the first box so
that TYPE appears in the Variables box, put a check mark next to Counts and Percents under Display, and click OK. The result is
TYPE Count Percent Large 47 9.40 Luxury 71 14.20
Midsize 249 49.80 Small 133 26.60 N= 500 (b) The relative frequencies were calculated in part(a) by putting a check
mark next to Percents. 9.4% of the cars were Large, 14.2% were Luxury, 49.8% were Midsize, and 26.6% were Small.
(c) Using Minitab, select Graph Pie Chart, check Chart counts of unique values, double-click on TYPE in the first box so that TYPE appears in the Categorical Variables box. Click Pie Options, check decreasing volume, click OK. Click Labels, enter TYPE in for the title, click Slice Labels, check Category Name, Percent, and Draw a line from label to slice, Click OK twice. The result is
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MidsizeSmallLuxuryLarge
CategoryLarge9.4%
Luxury14.2%
Small26.6%
Midsize49.8%
TYPE
(d) Using Minitab, select Graph Bar Chart, select Counts of unique values, select Simple option, click OK. Double-click on TYPE in the first box so that TYPE appears in the Categorical Variables box. Select Chart Options, check decreasing Y, check show Y as a percent, click OK. Select Labels, enter in TYPE as the title. Click OK twice. The result is
LargeLuxurySmallMidsize
50
40
30
20
10
0
TYPE
Perc
ent
TYPE
Percent within all data. 2.31 (a) Using Minitab, retrieve the data from the Weiss-Stats-CD. Column 1
contains the type of the hospital. From the tool bar, select Stat
Tables Tally Individual Variables, double-click on TYPE in the first
box so that TYPE appears in the Variables box, put a check mark next to Counts and Percents under Display, and click OK. The result is
TYPE Count Percent NPC 2919 50.79
IOC 889 15.47 SLC 1119 19.47 FGH 221 3.85 NLT 129 2.24 NFP 451 7.85 HUI 19 0.33 N= 5747
(b) The relative frequencies were calculated in part(a) by putting a check
mark next to Percents. 50.79% of the hospitals were Nongovernment not-for-profit community hospitals, 15.47% were Investor-owned (for-profit) community hospitals, 19.47% were State and local government community hospitals, 3.85% were Federal government hospitals, 2.24% were Nonfederal long term care hospitals, 7.85% were Nonfederal psychiatric hospitals, and 0.33% were Hospital units of institutions.
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(c) Using Minitab, select Graph Pie Chart, check Chart counts of unique values, double-click on TYPE in the first box so that TYPE appears in the Categorical Variables box. Click Pie Options, check decreasing volume, click OK. Click Labels, enter TYPE in for the title, click Slice Labels, check Category Name, Percent, and Draw a line from label to slice, Click OK twice. The result is
NPCSLCIOCNFPFGHNLTHUI
CategoryHUI

0.3%NLT
2.2%FGH3.8%NFP
7.8%
IOC15.5%
SLC19.5%
NPC50.8%
TYPE
(d) Using Minitab, select Graph Bar Chart, select Counts of unique values, select Simple option, click OK. Double-click on TYPE in the first box so that TYPE appears in the Categorical Variables box. Select Chart Options, check decreasing Y, check show Y as a percent, click OK. Select Labels, enter in TYPE as the title. Click OK twice. The result is
HUINLTFGHNFPIOCSLCNPC
50
40
30
20
10
0
TYPE
Perc
ent
TYPE
Percent within all data. 2.32 (a) Using Minitab, retrieve the data from the Weiss-Stats-CD. Column 2
contains the marital status and column 3 contains the number of drinks
per month. From the tool bar, select Stat Tables Tally Individual
Variables, double-click on STATUS and DRINKS in the first box so that both STATUS and DRINKS appear in the Variables box, put a check mark next to Counts and Percents under Display, and click OK. The results are
STATUS Count Percent DRINKS Count Percent Single 354 19.98 Abstain 590 33.30 Married 1173 66.20 1-60 957 54.01 Widowed 143 8.07 Over 60 225 12.70 Divorced 102 5.76 N= 1772 N= 1772
(b) The relative frequencies were calculated in part(a) by putting a check
mark next to Percents. For the STATUS variable; 19.98% of the US Adults are single, 66.20% are married, 8.07% are widowed, and 5.76% are
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divorced. For the DRINKS variable, 33.30% of US Adults abstain from drinking, 54.01% have 1-60 drinks per month, and 12.70% have over 60 drinks per month.
(c) Using Minitab, select Graph Pie Chart, check Chart counts of unique values, double-click on STATUS and DRINKS in the first box so that STATUS and DRINKS appear in the Categorical Variables box. Click Pie Options, check decreasing volume, click OK. Click Multiple Graphs, check On the Same Graphs, Click OK. Click Labels, click Slice Labels, check Category Name, Percent, and Draw a line from label to slice, Click OK twice. The results are
STATUS DRINKSMarriedSingleWidowedDivorcedAbstain1-60Over 60
Category
Divorced5.8%Widowed
8.1%
Single20.0%
Married66.2%
Over 6012.7%
Abstain33.3%
1-6054.0%
STATUS and DRINKS
(d) Using Minitab, select Graph Bar Chart, select Counts of unique values, select Simple option, click OK. Double-click on STATUS and DRINKS in the first box so that STATUS and DRINKS appear in the Categorical Variables box. Select Chart Options, check decreasing Y, check show Y as a percent, click OK. Click OK twice. The results are
DivorcedWidowedSingleMarried
70
60
50
40
30
20
10
0
STATUS
Perc
ent
Chart of STATUS
Percent within all data.
Over 60Abstain1-60
60
50
40
30
20
10
0
DRINKS
Perc
ent
Chart of DRINKS
Percent within all data. 2.33 (a) Using Minitab, retrieve the data from the Weiss-Stats-CD. Column 2
contains the preference for how the members want to receive the ballots and column 3 contains the highest degree obtained by the members. From
the tool bar, select Stat Tables Tally Individual Variables,
double-click on PREFERENCE and DEGREE in the first box so that both PREFERENCE and DEGREE appear in the Variables box, put a check mark next to Counts and Percents under Display, and click OK. The results are
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PREFERENCE Count Percent DEGREE Count Percent Both 112 19.79 MA 167 29.51 Email 239 42.23 Other 11 1.94 Mail 86 15.19 PhD 388 68.55 N/A 129 22.79 N= 566
N= 566 (b) The relative frequencies were calculated in part(a) by putting a check
mark next to Percents. For the PREFERENCE variable; 19.79% of the members prefer to receive the ballot by both e-mail and mail, 42.23% prefer e-
obtained a PhD, and 1.94% received a different degree.
(c) Using Minitab, select Graph Pie Chart, check Chart counts of unique values, double-click on PREFERENCE and DEGREE in the first box so that PREFERENCE and DEGREE appear in the Categorical Variables box. Click Pie Options, check decreasing volume, click OK. Click Multiple Graphs, check On the Same Graphs, Click OK. Click Labels, click Slice Labels, check Category Name, Percent, and Draw a line from label to slice, Click OK twice. The results are
PREFERENCE DEGREEEmailN/ABothMailMAotherPhD
Category
Mail15.2%
Both29.8%
N/A22.8%
Email42.2%
other1.9%
MA29.5%
PhD68.6%
Pie Chart of PREFERENCE, DEGREE
(d) Using Minitab, select Graph Bar Chart, select Counts of unique values, select Simple option, click OK. Double-click on PREFERENCE and DEGREE in the first box so that PREFERENCE and DEGREE appear in the Categorical Variables box. Select Chart Options, check decreasing Y, check show Y as a percent, click OK. Click OK twice. The results are
MailBothN/AEmail
40
30
20
10
0
PREFERENCE
Perc
ent
Chart of PREFERENCE
Percent within all data.
OtherMAPhD
70
60
50
40
30
20
10
0
DEGREE
Perc
ent
Chart of DEGREE
Percent within all data.
Exercises 2.3
2.34 One important reason for grouping data is that grouping often makes a large and complicated set of data more compact and easier to understand.
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2.35 For class limits, marks, cutpoints and midpoints to make sense, data must be numerical. They do not make sense for qualitative data classes because such data are nonnumerical.
2.36 The most important guidelines in choosing the classes for grouping a data set are: (1) the number of classes should be small enough to provide an effective summary, but large enough to display the relevant characteristics of the data; (2) each observation must belong to one, and only one, class; and (3) whenever feasible, all classes should have the same width.
2.37 In the first method for depicting classes called cutpoint grouping, we used the notation a under b to mean values that are greater than or equal to a and up to, but not including b, such as 30 under 40 to mean a range of values greater than or equal to 30, but strictly less than 40. In the alternate method called limit grouping, we used the notation a-b to indicate a class that extends from a to b, including both. For example, 30-39 is a class that includes both 30 and 39. The alternate method is especially appropriate when all of the data values are integers. If the data include values like 39.7 or 39.93, the first method is more advantageous since the cutpoints remain integers; whereas, in the alternate method, the upper limits for each class would have to be expressed in decimal form such as 39.9 or 39.99.
2.38 (a) For continuous data displayed to one or more decimal places, using the cutpoint grouping is best since the description of the classes is simpler, regardless of the number of decimal places displayed.
(b) For discrete data with relatively few distinct observations, the single value grouping is best since either of the other two methods would result in combining some of those distinct values into single classes, resulting in too few classes, possibly less than 5.
2.39 For limit grouping, we find the class mark, which is the average of the lower and upper class limit. For cutpoint grouping, we find the class midpoint, which is the average of the two cutpoints.
2.40 A frequency histogram shows the actual frequencies on the vertical axis; whereas, the relative frequency histogram always shows proportions (between 0 and 1) or percentages (between 0 and 100) on the vertical axis.
2.41 An advantage of the frequency histogram over a frequency distribution is that it is possible to get an overall view of the data more easily. A disadvantage of the frequency histogram is that it may not be possible to determine exact frequencies for the classes when the number of observations is large.
2.42 By showing the lower class limits (or cutpoints) on the horizontal axis, the range of possible data values in each class is immediately known and the class mark (or midpoint) can be quickly determined. This is particularly helpful if it is not convenient to make all classes the same width. The use of the class mark (or midpoint) is appropriate when each class consists of a single value (which is, of course, also the midpoint). Use of the class marks (or midpoints) is not appropriate in other situations since it may be difficult to determine the location of the class limits (or cutpoints) from the values of the class marks (or midpoints), particularly if the class marks (or midpoints) are not evenly spaced. Class Marks (or midpoints) cannot be used if there is an open class.
2.43 If the classes consist of single values, stem-and-leaf diagrams and frequency histograms are equally useful. If only one diagram is needed and the classes consist of more than one value, the stem-and-leaf diagram allows one to retrieve all of the original data values whereas the frequency histogram does not. If two or more sets of data of different sizes are to
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be compared, the relative frequency histogram is advantageous because all of the diagrams to be compared will have the same total relative frequency of 1.00. Finally, stem-and-leaf diagrams are not very useful with very large data sets and may present problems with data having many digits in each number.
2.44 The histogram (especially one using relative frequencies) is generally preferable. Data sets with a large number of observations may result in a stem of the stem-and-leaf diagram having more leaves than will fit on the line. In that case, the histogram would be preferable.
2.45 You can reconstruct the stem-and-leaf diagram using two lines per stem. For
es from 10 to 14 and on the second, the values from 15 to 19. If there are still two few stems, you can reconstruct the diagram using five lines per stem, recording 10 and 11 on the first line, 12 and 13 on the second, and so on.
2.46 For the number of bedrooms per single-family dwelling, single-value grouping is probably the best because the data is discrete with relatively few distinct observations.
2.47 For the ages of householders, given as a whole number, limit grouping is probably the best because the data are given as whole numbers and there are probably too many distinct observations to list them as single-value grouping.
2.48 For additional sleep obtained by a sample of 100 patients by using a particular brand of sleeping pill, cutpoint grouping is probably the best because the data is continuous and the data was recorded to the nearest tenth of an hour.
2.49 For the number of automobiles per family, single-value grouping is probably the best because the data is discrete with relatively few distinct observations.
2.50 For gas mileages, rounded to the nearest number of miles per gallon, limit grouping is probably the best because the data are given as whole numbers and there are probably too many distinct observations to list them as single-value grouping.
2.51 For carapace length for a sample of giant tarantulas, cutpoint grouping is probably the best because the data is continuous and the data was recorded to the nearest hundredth of a millimeter.
2.52 (a) Since the data values range from 0 to 4, we construct a table with classes based on a single value. The resulting table follows.
Number of Siblings Frequency 0 8 1 17 2 11 3 3 4 1 40
(b) To get the relative frequencies, divide each frequency by the sample size of 40.
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SIBLINGS
Freq
uenc
y
43210
18
16
14
12
10
8
6
4
2
0
Histogram of SIBLINGS
SIBLINGS
Perc
ent
43210
40
30
20
10
0
Histogram of SIBLINGS
Number of Siblings Relative Frequency 0 0.200 1 0.425 2 0.275 3 0.075 4 0.025 1.000
(c) The frequency histogram in Figure (a) is constructed using the frequency distribution presented in part (a) of this exercise. Column 1 demonstrates that the data are grouped using classes based on a single value. These single values in column 1 are used to label the horizontal axis of the frequency histogram. Suitable candidates for vertical axis units in the frequency histogram are the integers within the range 0 through 17, since these are representative of the magnitude and spread of the frequencies presented in column 2. When classes are based on a single value, the middle of each histogram bar is placed directly over the single numerical value represented by the class. Also, the height of each bar in the frequency histogram matches the respective frequency in column 2.
Figure (a) Figure (b)
(d) The relative-frequency histogram in Figure (b) is constructed using the relative-frequency distribution presented in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 2 range in size from 0.025 to 0.425. Thus, suitable candidates for vertical-axis units in the relative-frequency histogram are increments of 0.05 (or 5%), from zero to 0.45 (or 45%). The middle of each histogram bar is placed directly over the single numerical value represented by the class. Also, the height of each bar in the relative-frequency histogram matches the respective relative frequency in column 2.
2.53 (a) Since the data values range from 1 to 7, we construct a table with classes based on a single value. The resulting table follows.
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0
5
10
15
20
25
30
35
1 2 3 4 5 6 7
Number of People
Perc
ent
Number of Persons Frequency
1 7 2 13 3 9 4 5 5 4 6 1 7 1
40
(b) To get the relative frequencies, divide each frequency by the sample size of 40.
Number of Persons Relative Frequency
1 0.175 2 0.325 3 0.225 4 0.125 5 0.100 6 0.025 7 0.025
1.000
(c) The frequency histogram in Figure (a) is constructed using the frequency distribution presented in part (a) of this exercise. Column 1 demonstrates that the data are grouped using classes based on a single value. These single values in column 1 are used to label the horizontal axis of the frequency histogram. Suitable candidates for vertical axis units in the frequency histogram are the integers within the range 0 through 13, since these are representative of the magnitude and spread of the frequencies presented in column 2. When classes are based on a single value, the middle of each histogram bar is placed directly over the single numerical value represented by the class. Also, the height of each bar in the frequency histogram matches the respective frequency in column 2.
Figure (a) Figure (b)
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(d) The relative-frequency histogram in Figure (b) is constructed using the relative-frequency distribution presented in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 2 range in size from 0.025 to 0.325. Thus, suitable candidates for vertical-axis units in the relative-frequency histogram are increments of 0.05 (or 5%), from zero to 0.35 (or 35%). The middle of each histogram bar is placed directly over the single numerical value represented by the class. Also, the height of each bar in the relative-frequency histogram matches the respective relative frequency in column 2.
2.54 (a) Since the data values range from 1 to 8, we construct a table with classes based on a single value. The resulting table follows.
Litter Size Frequency
1 1 2 0 3 1 4 3 5 7 6 6 7 4 8 2
24
(b) To get the relative frequencies, divide each frequency by the sample size of 24.
Litter Size Relative Frequency
1 0.042 2 0.000 3 0.042 4 0.125 5 0.292 6 0.250 7 0.167 8 0.083
1.000
(c) The frequency histogram in Figure (a) is constructed using the frequency distribution presented in part (a) of this exercise. Column 1 demonstrates that the data are grouped using classes based on a single value. These single values in column 1 are used to label the horizontal axis of the frequency histogram. Suitable candidates for vertical axis units in the frequency histogram are the integers within the range 0 through 7, since these are representative of the magnitude and spread of the frequencies presented in column 2. When classes are based on a single value, the middle of each histogram bar is placed directly over the single numerical value represented by the class. Also, the height of each bar in the frequency histogram matches the respective frequency in column 2.
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Figure (a) Figure (b)
87654321
7
6

5
4
3
2
1
0
SIZE
Freq
uenc
y
Litter Size
87654321
30
25
20
15
10
5
0
SIZE
Perc
ent
Litter Size
(d) The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 2 range in size from 0.000 to 0.292. Thus, suitable candidates for vertical-axis units in the relative-frequency histogram are increments of 0.05 (or 5%), from zero to 0.30 (or 30%). The middle of each histogram bar is placed directly over the single numerical value represented by the class. Also, the height of each bar in the relative-frequency histogram matches the respective relative frequency in column 2.
2.55 (a) Since the data values range from 1 to 10, we construct a table with classes based on a single value. The resulting table follows.
Number of Radios Frequency
1 1 2 1 3 3 4 12 5 6 6 4 7 5 8 4
9 6
10 3
45
(b) To get the relative frequencies, divide each frequency by the sample size of 45.
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Number of Radios Relative Frequency
1 0.022 2 0.022 3 0.067 4 0.267 5 0.133 6 0.089 7 0.111 8 0.089
9 0.133
10 0.067
1.000
(c) The frequency histogram in Figure (a) is constructed using the frequency distribution presented in part (a) of this exercise. Column 1 demonstrates that the data are grouped using classes based on a single value. These single values in column 1 are used to label the horizontal axis of the frequency histogram. Suitable candidates for vertical axis units in the frequency histogram are the integers within the range 1 through 12, since these are representative of the magnitude and spread of the frequencies presented in column 2. When classes are based on a single value, the middle of each histogram bar is placed directly over the single numerical value represented by the class. Also, the height of each bar in the frequency histogram matches the respective frequency in column 2.
Figure (a) Figure (b)
108642
12
10
8
6
4
2
0
RADIOS
Freq

uenc
y
Radios per Household
108642
30
25
20
15
10
5
0
RADIOS
Perc
ent
Radios per Household
(d) The relative-frequency histogram in Figure (b) is constructed using
the relative-frequency distribution presented in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 2 range in size from 0.022 to 0.267. Thus, suitable candidates for vertical-axis units in the relative-frequency histogram are increments of 0.05 (or 5%), from zero to 0.30 (or 30%). The middle of each histogram bar is placed directly over the single numerical value represented by the class. Also, the height of each bar in the relative-frequency histogram matches the respective relative frequency in column 2.
2.56 (a) The first class to construct is 40-49. Since all classes are to be of equal width, and the second class begins with 50, we know that the width of all classes is 50 - 40 = 10. All of the classes are presented
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in column 1. The last class to construct is 150-159, since the largest single data value is 155. Having established the classes, we tally the energy consumption figures into their respective classes. These results are presented in column 2, which lists the frequencies.
(b) Dividing each frequency by the total number of observations, which is 50, results in each class's relative frequency. The relative frequencies for all classes are presented in column 2. The resulting table follows.
(c) The frequency histogram in Figure (a) is constructed using the frequency distribution presented in part (a) of this exercise. The lower class limits of column 1 are used to label the horizontal axis of the frequency histogram. Suitable candidates for vertical-axis units in the frequency histogram are the even integers 0 through 10, since these are representative of the magnitude and spread of the frequency presented in column 2. The height of each bar in the frequency histogram matches the respective frequency in column 2.
(d) The relative-frequency histogram in Figure (b) is constructed using the relative-frequency distribution presented in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 2 vary in size from 0.00 to 0.20. Thus, suitable candidates for vertical axis units in the relative-frequency histogram are increments of 0.05 (or 5%), from zero to 0.20 (or 20%). The height of each bar in the relative-frequency histogram matches the respective relative frequency in column 2.
Consumption (mil. BTU) Frequency 40-49 1 50-59 7 60-69 7 70-79 3 80-89 6 90-99 10 100-109 5 110-119 4 120-129 2 130-139 3 140-149 0 150-159 2 50
Consumption (mil. BTU) Relative Frequency 40-49 0.02 50-59 0.14 60-69 0.14 70-79 0.06 80-89 0.12 90-99 0.20 100-109 0.10 110-119 0.08 120-129 0.04 130-139 0.06 140-149 0.00 150-159 0.04 1.00
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Figure (a) Figure (b)
160150140130120110100908070605040
10
8

6
4
2
0
ENERGY
Freq
uenc
y
Histogram of ENERGY
160150140130120110100908070605040
20
15
10
5
0
ENERGY
Perc
ent
Histogram of ENERGY
2.57 (a) The first class to construct is 40-44. Since all classes are to be of
equal width, and the second class begins with 45, we know that the width of all classes is 45 - 40 = 5. All of the classes are presented in column 1. The last class to construct is 60-64, since the largest single data value is 61. Having established the classes, we tally the age figures into their respective classes. These results are presented in column 2, which lists the frequencies.
(b) Dividing each frequency by the total number of observations, which is 21, results in each class's relative frequency. The relative frequencies for all classes are presented in column 2. The resulting table follows.
(c) The frequency histogram in Figure (a) is constructed using the frequency distribution presented in part (a) of this exercise. The lower class limits of column 1 are used to label the horizontal axis of the frequency histogram. Suitable candidates for vertical-axis units in the frequency histogram are the even integers 2 through 8, since these are representative of the magnitude and spread of the frequency presented in column 2. The height of each bar in the frequency histogram matches the respective frequency in column 2.
(d) The relative-frequency histogram in Figure (b) is constructed using the relative-frequency distribution presented in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 2 vary in size from 0.095 to 0.381. Thus, suitable candidates for vertical axis units in the relative-frequency histogram are increments of 0.10 (or
Age Frequency 40-44 4 45-49 3 50-54 4 55-59 8 60-64 2 21
Age Relative Frequency 40-44 0.190 45-49 0.143 50-54 0.190 55-59 0.381 60-64 0.095 1.000
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10%), from zero to 0.40 (or 40%). The height of each bar in the relative-frequency histogram matches the respective relative frequency in column 2.
Figure (a) Figure (b)
656055504540
9
8
7
6
5
4
3
2
1
0
AGE
Freq
uenc
y
Early-Onset Dementia
656055504540
40
30
20
10
0
AGE
Perc
ent
Early-Onset Dementia
2.58 (a) The first class to construct is 20-22. Since all classes are to be of
equal width, and the second class begins with 23, we know that the width of all classes is 23 - 20 = 3. All of the classes are presented in column 1. The last class to construct is 44-46, since the largest single data value is 45. Having established the classes, we tally the cheese consumption figures into their respective classes. These results are presented in column 2, which lists the frequencies.
(b) Dividing each frequency by the total number of observations, which is 35, results in each class's relative frequency. The relative frequencies for all classes are presented in column 2. The resulting table follows.
(c) The frequency histogram in Figure (a) is constructed using the frequency distribution presented in part (a) of this exercise. The
Cheese Consumption Frequency 20-22 2 23-25 3 26-28 4 29-31 7 32-34 6 35-37 5 38-40 3 41-43 3 44-46 2 35
Cheese Consumption Relative Frequency 20-22 0.057 23-25 0.086 26-28 0.114 29-31 0.200 32-34 0.171 35-37 0.143 38-40 0.086 41-43 0.086 44-46 0.057 1.000
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lower class limits of column 1 are used to label the horizontal axis of the frequency histogram. Suitable candidates for vertical-axis units in the frequency histogram are the even integers 2 through 7, since these are representative of the magnitude and spread of the frequency presented in column 2. The height of each bar in the frequency histogram matches the respective frequency in column 2.
(d) The relative-frequency histogram in Figure (b) is constructed using the relative-frequency distribution presented in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 2 vary in size from 0.057 to 0.200. Thus, suitable candidates for vertical axis units in the relative-frequency histogram are increments of 0.05 (or 5%), from zero to 0.20 (or 20%). The height of each bar in the relative-frequency histogram matches the respective relative frequency in column 2.
Figure (a) Figure (b)
47444138353229262320
7
6
5
4
3
2
1
0
CHEESE
Freq
uenc
y
Cheese Consumption
47444138353229262320
20
15
10
5
0
CHEESE
Perc
ent
Cheese Consumption
2.59 (a) The first class to construct is 12-17. Since all classes are to be of
equal width, and the second class begins with 18, we know that the width of all classes is 18 - 12 = 6. All of the classes are presented in column 1. The last class to construct is 60-65, since the largest single data value is 61. Having established the classes, we tally the anxiety questionnaire score figures into their respective classes. These results are presented in column 2, which lists the frequencies.
(b) Dividing each frequency by the total number of observations, which is 31, results in each class's relative frequency. The relative frequencies for all classes are presented in column 2. The resulting table follows.
Anxiety Frequency 12-17 2 18-23 3 24-29 6 30-35 5 36-41 10 42-47 4 48-53 0 54-59 0 60-65 1 31
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(c) The frequency histogram in Figure (a) is constructed using the frequency distribution presented in part (a) of this exercise. The lower class limits of column 1 are used to label the horizontal axis of the frequency histogram. Suitable candidates for vertical-axis units in the frequency histogram are the even integers 0 through 10, since these are representative of the magnitude and spread of the frequency presented in column 2. The height of each bar in the frequency histogram matches the respective frequency in column 2.
(d) The relative-frequency histogram in Figure (b) is constructed using the relative-frequency distribution presented in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 2 vary in size from 0.000 to 0.323. Thus, suitable candidates for vertical axis units in the relative-frequency histogram are increments of 0.05 (or 5%), from zero to 0.35 (or 35%). The height of each bar in the relative-frequency histogram matches the respective relative frequency in column 2.
Figure (a) Figure (b)
66605448423630241812
10
8

6
4
2
0
ANXIETY
Freq
uenc
y
Chronic Hemodialysis and Axiety
66605448423630241812
35
30
25
20
15
10
5
0
ANXIETY
Perc
ent
Chronic Hemodialysis and Axiety
2.60 (a) The first class to construct is 12 under 13. Since all classes are
to be of equal width 1, the second class is 13 under 14. All of the classes are presented in column 1. The last class to construct is 19 under 20, since the largest single data value is 19.492. Having established the classes, we tally the audience sizes into their respective classes. These results are presented in column 2, which lists the frequencies.
Anxiety Relative Frequency 12-17 0.065 18-23 0.097 24-29 0.194 30-35 0.161 36-41 0.323 42-47 0.129 48-53 0.000 54-59 0.000 60-65 0.032 1.000
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Audience (Millions) Frequency 12 under 13 3 13 under 14 5 14 under 15 4 15 under 16 4 16 under 17 1 17 under 18 1 18 under 19 1 19 under 20 1 20
(b) Dividing each frequency by the total number of observations, which is 20, results in each class's relative frequency. The relative frequencies for all classes are presented in column 2
Audience (Millions) Relative Frequency 12 under 13 0.15 13 under 14 0.25 14 under 15 0.20 15 under 16 0.20 16 under 17 0.05 17 under 18 0.05 18 under 19 0.05 19 under 20 0.05 1.00
(c) The frequency histogram in Figure (a) is constructed using the frequency distribution obtained in part (a) of this exercise Column 1 demonstrates that the data are grouped using classes with class widths of 1. Suitable candidates for vertical axis units in the frequency histogram are the integers within the range 1 through 5, since these are representative of the magnitude and spread of the frequencies presented in column 2. Also, the height of each bar in the frequency histogram matches the respective frequency in column 2.
(d) The relative-frequency histogram in Figure (b) is constructed using the relative-frequency distribution obtained in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 3 range in size from 0.05 to 0.20. Thus, suitable candidates for vertical axis units in the relative-frequency histogram are increments of 0.05 (5%), from zero to 0.20 (20%). The height of each bar in the relative-frequency histogram matches the respective relative frequency in column 2.
Figure (a) Figure (b)
201918171615141312
5
4
3
2
1
0
AUDIENCE
Freq
uenc
y
Top Broadcast Shows
201918171615141312
25
20
15
10
5
0
AUDIENCE
Perc
ent
Top Broadcast Shows
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2.61 (a) The first class to construct is 52 under 54. Since all classes are to be of equal width 2, the second class is 54 under 56. All of the classes are presented in column 1. The last class to construct is 74 under 76, since the largest single data value is 75.3. Having established the classes, we tally the cheetah speeds into their respective classes. These results are presented in column 2, which lists the frequencies.
Speed Frequency 52 under 54 2 54 under 56 5 56 under 58 6 58 under 60 8 60 under 62 7 62 under 64 3 64 under 66 2 66 under 68 1 68 under 70 0 70 under 72 0 72 under 74 0 74 under 76 1 35
(b) Dividing each frequency by the total number of observations, which is 35, results in each class's relative frequency. The relative frequencies for all classes are presented in column 2
Speed Relative Frequency 52 under 54 0.057 54 under 56 0.143 56 under 58 0.171 58 under 60 0.229 60 under 62 0.200 62 under 64 0.086 64 under 66 0.057 66 under 68 0.029 68 under 70 0.000 70 under 72 0.000 72 under 74 0.000 74 under 76 0.029 1.000
(c) The frequency histogram in Figure (a) is constructed using the frequency distribution obtained in part (a) of this exercise Column 1 demonstrates that the data are grouped using classes with class widths of 2. Suitable candidates for vertical axis units in the frequency histogram are the integers within the range 0 through 8, since these are representative of the magnitude and spread of the frequencies presented in column 2. Also, the height of each bar in the frequency histogram matches the respective frequency in column 2.
(d) The relative-frequency histogram in Figure (b) is constructed using the relative-frequency distribution obtained in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 3 range in size from 0.000 to 0.229. Thus, suitable candidates for vertical axis units in the relative-frequency histogram are increments of 0.05 (5%), from zero to 0.25 (25%). The height of each bar in the relative-
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frequency histogram matches the respective relative frequency in column 2.
Figure (a) Figure (b)
76747270686664626058565452
9
8
7
6
5
4
3
2
1
0
SPEED
Freq
uenc
y
Speeds of Cheetahs
76747270686664626058565452
25
20
15
10
5
0
SPEED
Perc
ent
Speeds of Cheetahs
2.62 (a) The first class to construct is 12 under 14. Since all classes are
to be of equal width 2, the second class is 14 under 66. All of the classes are presented in column 1. The last class to construct is 26 under 28, since the largest single data value is 26.4. Having established the classes, we tally the fuel tank capacities into their respective classes. These results are presented in column 2, which lists the frequencies.
Fuel Tank Capacity Frequency 12 under 14 2 14 under 16 6 16 under 18 7 18 under 20 6 20 under 22 6 22 under 24 3 24 under 26 3 26 under 28 2 35
(b) Dividing each frequency by the total number of observations, which is 35, results in each class's relative frequency. The relative frequencies for all classes are presented in column 2
Fuel Tank Capacity Relative Frequency 12 under 14 0.057 14 under 16 0.171 16 under 18 0.200 18 under 20 0.171 20 under 22 0.171 22 under 24 0.086 24 under 26 0.086 26 under 28 0.057 1.000
(c) The frequency histogram in Figure (a) is constructed using the frequency distribution obtained in part (a) of this exercise Column 1 demonstrates that the data are grouped using classes with class widths of 2. Suitable candidates for vertical axis units in the frequency histogram are the integers within the range 2 through 7, since these are representative of the magnitude and spread of the frequencies presented in column 2. Also, the height of each bar in the frequency histogram matches the respective frequency in column 2.
(d) The relative-frequency histogram in Figure (b) is constructed using
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the relative-frequency distribution obtained in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 3 range in size from 0.057 to 0.200. Thus, suitable candidates for vertical axis units in the relative-frequency histogram are increments of 0.05 (5%), from zero to 0.20 (20%). The height of each bar in the relative-frequency histogram matches the respective relative frequency in column 2.
Figure (a) Figure (b)
282624222018161412
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Fuel Tank Capacity
2.63 (a) The first class to construct is 0 under 1. Since all classes are to
be of equal width 1, the second class is 1 under 2. All of the classes are presented in column 1. The last class to construct is 7 - under 8, since the largest single data value is 7.6. Having established the classes, we tally the fuel tank capacities into their respective classes. These results are presented in column 2, which lists the frequencies.
Oxygen Distribution Frequency 0 under 1 1 1 under 2 10 2 under 3 5 3 under 4 4 4 under 5 0 5 under 6 0 6 under 7 1 7 under 8 1 22

(b) Dividing each frequency by the total number of observations, which is 22, results in each class's relative frequency. The relative frequencies for all classes are presented in column 2
Oxygen Distribution Relative Frequency 0 under 1 0.045 1 under 2 0.455 2 under 3 0.227 3 under 4 0.182 4 under 5 0.000 5 under 6 0.000 6 under 7 0.045 7 under 8 0.045 1.000
(c) The frequency histogram in Figure (a) is constructed using the frequency distribution obtained in part (a) of this exercise Column 1 demonstrates that the data are grouped using classes with class widths
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of 2. Suitable candidates for vertical axis units in the frequency histogram are the integers within the range 0 through 10, since these are representative of the magnitude and spread of the frequencies presented in column 2. Also, the height of each bar in the frequency histogram matches the respective frequency in column 2.
(d) The relative-frequency histogram in Figure (b) is constructed using the relative-frequency distribution obtained in part (b) of this exercise. It has the same horizontal axis as the frequency histogram. We notice that the relative frequencies presented in column 3 range in size from 0.000 to 0.455. Thus, suitable candidates for vertical axis units in the relative-frequency histogram are increments of 0.05 (5%), from zero to 0.50 (50%). The height of each bar in the relative-frequency histogram matches the respective relative frequency in column 2.
Figure (a) Figure (b)
876543210
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OXYGEN
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Oxygen Distribution
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2.64 The horizontal axis of this dotplot displays a range of possible exam
scores. To complete the dotplot, we go through the data set and record each exam score by placing a dot over the appropriate value on the horizontal axis.
SCORE9990817263544536
Dotplot of SCORE
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2.65 The horizontal axis of this dotplot displays a range of possible ages. To complete the dotplot, we go through the data set and record each age by placing a dot over the appropriate value on the horizontal axis.
2.66 (a) The data values range from 52 to 84, so the scale must accommodate those values. We stack dots above each value on two different lines using the same scale for each line. The result is
(b) The two sets of pulse rates are both centered near 68, but the Intervention data are more concentrated around the center than are the Control data.
2.67 (a) The data values range from 7 to 18, so the scale must accommodate those values. We stack dots above each value on two different lines using the same scale for each line. The result is
(b) The Dynamic system does seem to reduce acute postoperative days in the hospital on the average. The Dynamic data are centered at about 7
AGE181512963
Dotplot of AGE
Data85807570656055
INTERVENTION
CONTROL
Dotplot of INTERVENTION, CONTROL
Data181614121086
DYNAMIC
STATIC
Dotplot of DYNAMIC, STATIC
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days, whereas the Static data are centered at about 11 days and are much more spread out than the Dynamic data.
2.68 Since each data value consists of 3 or 4 digits ranging from 914 to 1060. The last digit becomes the leaf and the remaining digits are the stems, so we have stems of 91 to 106. The resulting stem-and-leaf diagram is
91| 4 92| 93| 94| 6 95| 79 96| 4 97| 4577 98| 46789 99| 015679 100| 1 101| 0478 102| 58 103| 01 104| 105| 106| 0
2.69 Since each data value consists of 2 digits, each beginning with 1, 2, 3, or 4, we will construct the stem-and-leaf diagram with these four values as the stems. The result is
1| 238 2| 1678899 3| 34459 4| 04
2.70 (a) Since each data value consists of a 2 digit number with a one digit decimal, we will make the leaf the decimal digit and the stems the remaining two digit numbers of 28, 29, 30, and 31. The result is
28| 8 29| 368 30| 1247 31| 02
(b) Splitting into two lines per stem, leafs of 0-4 belong in the first stem and leafs of 5-9 belong in the second stem. The result is
28| 8 29| 3 29| 68 30| 124 30| 7 31| 02
(c) The stem-and-leaf diagram in part (b) is more useful because by splitting the stems into two lines per stem, you have created more lines. Part (a) had too few lines.
2.71 (a) Since each data value lies between 2 and 93, we will construct the stem-and-leaf diagram with one line per stem. The result is
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0| 2234799 1| 11145566689 2| 023479 3| 004555 4| 19 5| 5 6| 9 7| 9 8| 9| 3
(b) Using two lines per stem, the same data result in the following diagram: 0| 2234
0| 799 1| 1114 1| 5566689 2| 0234 2| 79 3| 004 3| 555 4| 1 4| 9 5| 5| 5 6| 6| 9 7| 7| 9 8| 8| 9| 3
(c) The stem with one line per stem is more useful. One gets the same impression regarding the shape of the distribution, but the two lines per stem version has numerous lines with no data, making it take up more space than necessary to interpret the data and giving it too many lines.
2.72 (a) Since we have two digit numbers, the last digit becomes the leaf and the first digit becomes the stem. For this data, we have stems of 6 and 7. Splitting the data into five lines per stem, we put the leaves 0-1 in the first stem, 2-3 in the second stem, 4-5 in the third stem, 6-7 in the fourth stem, and 8-9 in the fifth stem. The result is
6| 899 7| 0001 7| 22222333333333 7| 44555555 7| 6666677 7| 88
(b) Using one or two lines per stem would have given us too few lines.
2.73 (a) Since we have two digit numbers, the last digit becomes the leaf and the first digit becomes the stem. For this data, we have stems of 6, 7, and 8. Splitting the data into five lines per stem, we put the leaves 0-1 in the first stem, 2-3 in the second stem, 4-5 in the third stem, 6-7 in the fourth stem, and 8-9 in the fifth stem. The result is
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6| 99 7| 11 7| 222233 7| 444444444445555555555 7| 666667 7| 88 8| 1
(b) Using one or two lines per stem would have given us too few lines.
2.74 The heights of the bars of the relative-frequency histogram indicate that:
(a) About 27.5% of the returns had an adjusted gross income between $10,000 and $19,999, inclusive.
(b) About 37.5% were between $0 and $9,999; 27.5% were between $10,000 and $19,999; and 19% were between $20,000 and $29,999. Thus, about 84% (i.e., 37.5% + 27.5% + 19%) of the returns had an adjusted gross income less than $30,000.
(c) About 11% were between $30,000 and $39,999; and 5% were between $40,000 and $49,999. Thus, about 16% (i.e., 11% + 5%) of the returns had an adjusted gross income between $30,000 and $49,999. With 89,928,000 returns having an adjusted gross income less than $50,000, the number of returns having an adjusted gross income between $30,000 and $49,999 was 14,388,480 (i.e., 0.16 x 89,928,000).
2.75 The graph indicates that:
(a) 20% of the patients have cholesterol levels between 205 and 209, inclusive.
(b) 20% are between 215 and 219; and 5% are between 220 and 224. Thus, 25% (i.e., 20% + 5%) have cholesterol levels of 215 or higher.
(c) 35% of the patients have cholesterol levels between 210 and 214, inclusive. With 20 patients in total, the number having cholesterol levels between 210 and 214 is 7 (i.e., 0.35 x 20).
2.76 (a) Using Minitab, retrieve the data from the Weiss-Stats-CD. Column 1 contains the numbers of pups borne in a lifetime for each of 80 female
Great White Sharks. From the tool bar, select Stat Tables Tally
Individual Variables, double-click on PUPS in the first box so that PUPS appears in the Variables box, put a check mark next to Counts and Percents under Display, and click OK. The result is
PUPS Count Percent
3 2 2.50
4 5 6.25
5 10 12.50
6 11 13.75
7 17 21.25
8 17 21.25
9 11 13.75
10 4 5.00
11 2 2.50
12 1 1.25
(b) After retrieving the data from the WeissStats CD, select Graph
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Histogram, choose Simple and click OK. Double click on PUPS in the first box to enter PUPS in the Graphs variables box, and click OK. The frequency histogram is
To change to a relative-frequency histogram, before clicking OK the second time, click on the Scale button and the Y-Scale type tab, and choose Percent and click OK. The graph will look like the frequency histogram, but will have relative frequencies on the vertical scale instead of counts.
The numbers of pups range from 1 to 12 per female with 7 and 8 pups occurring more frequently than any other values.
2.77 (a) Using Minitab, there is not a direct way to get a grouped frequency distribution. However, you can use an option in creating your histogram that will report the frequencies in each of the classes, essentially creating a grouped frequency distribution. Retrieve the data from the Weiss-Stats-CD. Column 2 contains the number of albums sold, in millions, for the top recording artists. From the tool bar, ,
select Graph Histogram, choose Simple and click OK. Double click on
ALBUMS to enter ALBUMS in the Graph variables box. Click Labels, click the Data Labels tab, then check Use y-value labels. Click OK twice. The result is
150120906030
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ALBUMS
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1000111101
7610
17
35
84
74
Histogram of ALBUMS
Above each bar is a label for each of the frequencies. Also, the
labeling on the horizontal axis is the midpoint for class, where each class has a width of 10. The first class would be 5 under 15, the second class would be 15 under 25, etc. To get the relative-
PUPS
Freq
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0
Histogram of PUPS
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frequency distribution, follow the same steps as above, but also Click the Scale button, click the Y-scale Type, check Percent, then click OK twice. The result is
150120906030
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Perc
ent
0.418410000.418410.418410.418410.4184100.41841
2.928872.510464.1841
7.11297
14.6444
35.1464
30.9623
Histogram of ALBUMS
Above each bar is the percentage for that class, essentially creating a
relative-frequency distribution. You could also transfer these results into a table.
(b) After entering the data from the WeissStats CD, in Minitab, select
Graph Histogram, choose Simple and click OK. Double click on ALBUMS
to enter ALBUMS in the Graph variables box and click OK. The result is
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Histogram of ALBUMS
The graph shows that there are only a few artists who sell many units
and many artists who sell relatively few units.
(c) To obtain the dotplot, select Graph Dotplot, select Simple in the
One Y row, and click OK. Double click on ALBUMS to enter ALBUMS in the Graph variables box and click OK. The result is
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175150125100755025ALBUMS
Dotplot of ALBUMS
Each symbol represents up to 2 observations.
(d) The graphs are similar, but not identical. This is because Minitab grouped the data values slightly differently for the two graphs. The overall impression, however, remains the same.
2.78 (a) After entering the data from the WeissStats CD, in Minitab, select
Graph Stem-and-Leaf, double click on PERCENT to enter PERCENT in the
Graph variables box and enter a 10 in the Increment box, and click OK. The result is
Stem-and-leaf of PERCENT N = 51 Leaf Unit = 1.0
1 7 9 (33) 8 001112223333444566777778888889999
17 9 00000001111111223
(b) Repeat part (a), but this time enter a 5 in the Increment box. The result is
Stem-and-leaf of PERCENT N = 51 Leaf Unit = 1.0
1 7 9 16 8 001112223333444 (18) 8 566777778888889999 17 9 00000001111111223
(c) Repeat part (a) again, but this time enter a 2 in the Increment box. The result is
Stem-and-leaf of PERCENT N = 51 Leaf Unit = 1.0
1 7 9 6 8 00111 13 8 2223333 17 8 4445 24 8 6677777 (10) 8 8888889999 17 9 00000001111111
3 9 223 (d) The last graph is the most useful since it gives a better idea of the
shape of the distribution. Typically, we like to have five to fifteen classes and this is the only one of the three graphs that satisfies that condition.
2.79 (a) After entering the data from the WeissStats CD, in Minitab, select
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Graph Stem-and-Leaf, double click on RATE to enter RATE in the Graph
variables box and enter a 10 in the Increment box, and click OK. The result is
Stem-and-leaf of RATE N = 51 Leaf Unit = 1.0
1 1 9 14 2 0145678888999 (13) 3 0014445667788 24 4 012223345566677899 6 5 00124
1 6 2 (b) Repeat part (a), but this time enter a 5 in the Increment box. The
result is
Stem-and-leaf of RATE N = 51 Leaf Unit = 1.0
1 1 9 4 2 014 14 2 5678888999 20 3 001444 (7) 3 5667788 24 4 01222334 16 4 5566677899 6 5 00124 1 5
1 6 2 (c) Repeat part (a) again, but this time enter a 2 in the Increment box.
The result is
Stem-and-leaf of RATE N = 51 Leaf Unit = 1.0
1 1 9 3 2 01 3 2 5 2 45 7 2 67 14 2 8888999 17 3 001 17 3 21 3 4445 25 3 6677 (2) 3 88 24 4 01 22 4 22233 17 4 455 14 4 66677 9 4 899 6 5 001 3 5 2 2 5 4 1 5 1 5 1 6
1 6 2 (d) The second graph is the most useful. The third one has more classes
than necessary to comprehend the shape of the distribution and has a number of empty stems. Typically, we like to have five to fifteen classes and the first and second diagrams satisfy that condition, but the second one provides a better idea of the shape of the distribution.
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TEMP
Freq
uenc
y
99.599.098.598.097.597.0
20
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Histogram of TEMP
TEMP99.298.898.498.097.697.296.8
Dotplot of TEMP
2.80 (a) After entering the data from the WeissStats CD, in Minitab, select
Graph Histogram, select Simple and click OK. double click on TEMP to
enter TEMP in the Graph variables box and click OK. The result is
(b) Now select Graph Dotplot, select Simple in the One Y row, and click
OK. Double click on TEMP to enter TEMP in the Graph variables box and click OK. The result is
(c) Now select Graph Stem-and-Leaf, double click on TEMP to enter TEMP in
the Graph variables box and click OK. Leave the Increment box blank to allow Minitab to choose the number of lines per stem. The result is
Stem-and-leaf of TEMP N = 93 Leaf Unit = 0.10 1 96 7 3 96 89 8 97 00001 13 97 22233 19 97 444444 26 97 6666777 31 97 88889 45 98 00000000000111 (10) 98 2222222233 38 98 4444445555 28 98 66666666677 17 98 8888888 10 99 00001 5 99 2233 1 99 4
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(d) The dotplot shows all of the individual values. The stem-and-leaf diagram used five lines per stem and therefore each line contains leaves with possibly two values. The histogram chose classes of width 0.25. This resulted in, for example, the class with midpoint 97.0 including all of the values 96.9, 97.0, and 97.1, while the class with midpoint 97.25 includes only the two values 97.2 and 97.3. Thus the
stem-and-leaf diagram. Overall, the dotplot gives the truest picture of the data and allows recovery of all of the data values.
2.81 (a) The classes are presented in column 1. With the classes established, we then tally the exam scores into their respective classes. These results are presented in column 2, which lists the frequencies. Dividing each frequency by the total number of exam scores, which is 20, results in each class's relative frequency. The relative frequencies for all classes are presented in column 3. The class mark of each class is the average of the lower and upper limits. The class marks for all classes are presented in column 4.
Score Frequency Relative Frequency Class Mark
30-39 2 0.10 34.5 40-49 0 0.00 44.5 50-59 0 0.00 54.5 60-69 3 0.15 64.5 70-79 3 0.15 74.5 80-89 8 0.40 84.5 90-100 4 0.20 95.0 20 1.00
(b) The first six classes have width 10; the seventh class had width 11.
(c) Answers will vary, but one choice is to keep the first six classes the same and make the next two classes 90-99 and 100-109. Another possibility is 31-40, 41- -100.
2.82 Answers will vary, but by following