Solutions Manual for
Introduction to Statistics and Data
Analysis 7th Edition By Roxy Peck,
Chris Olsen
(All Chapters 1-16, 100% Original
Verified, A+ Grade)
All Chapters Arranged Reverse: 16-1
This is the Original Solutions Manual
for 7th Edition, All Other Files in the
Market are Wrong/Old Questions.
, Solution and Answer Guide: Peck, Introduction to Statistics and Data Analysis, 7e, 9798214000008; Chapter 16:
Nonparametric (Distribution-Free) Statistical Methods
Solution and Answer Guide
PECK, INTRODUCTION TO STATISTICS AND DATA ANALYSIS, 7E, 9798214000008; CHAPTER 16:
Nonparametric (Distribution-Free) Statistical Methods
TABLE OF CONTENTS
End of Section Exercise Solutions ......................................................................................1
END OF SECTION EXERCISE SOLUTIONS
1. Urinary fluoride concentration (in parts per million) was measured for both a sample of
livestock that had been grazing in an area previously exposed to fluoride pollution and
a similar sample of livestock that had grazed in an unpolluted region.
Do the accompanying data provide evidence that the mean fluoride concentration
for livestock grazing in the polluted region is greater than that for livestock grazing in
the unpolluted region? Assume that the distributions of urinary fluoride concentration
for both grazing areas have the same shape and variability, and use a level 0.05 rank-
sum test.
Polluted 21.3 18.7 23.0 17.1 16.8 20.9 19.7
Unpolluted 14.2 18.3 17.2 18.4 20.0
Solution:
Let 1 denote the mean fluoride concentration for livestock grazing in the polluted
region and 2 denote the mean fluoride concentration for livestock grazing in the
unpolluted regions.
H 0 : 1 2 = 0 H a : 1 2 > 0
= 0.05
The test statistic is: rank sum for polluted area (sample 1).
Sample Ordered Data Rank
2 14.2 1
1 16.8 2
1 17.1 3
2 17.2 4
2 18.3 5
2 18.4 6
© 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 1
accessible website, in whole or in part.
, Solution and Answer Guide: Peck, Introduction to Statistics and Data Analysis, 7e, 9798214000008; Chapter 16:
Nonparametric (Distribution-Free) Statistical Methods
1 18.7 7
1 19.7 8
2 20.0 9
1 20.9 10
1 21.3 11
1 23.0 12
Rank sum = (2 + 3 + 7 + 8 + 10 + 11 + 12) = 53
P-value: This is an upper-tail test. With n1 = 7 and n 2 = 5, Chapter 16 Appendix Table 1
tells us that the P-value > 0.05.
Since the P-value exceeds , Ho is not rejected. The data do not support the
conclusion that there is a larger average fluoride concentration for the polluted area
than for the unpolluted area.
2. A modification has been made to the process for producing a certain type of film.
Because the modification involves extra cost, it will be incorporated only if sample
data provide evidence that the modification decreases mean developing time by more
than 1 second. Assuming that the developing time distributions have the same shape
and variability, use the rank-sum test, a significance level of 0.05, and the following
data to test the appropriate hypotheses.
Original process 8.6 5.1 4.5 5.4 6.3 6.6 5.7 8.5
Modified process 5.5 4.0 3.8 6.0 5.8 4.9 7.0 5.7
Solution:
Let 1 denote the true average developing time without modification and 2 the true
average developing time with modification.
H 0 : 1 2 = 1 H a : 1 2 > 1
= 0.05
The test statistic is: rank sum for (unmodified times - 1).
(A) original process 1 7.6 4.1 3.5 4.4 5.3 5.6 4.7 7.5
(B) modified process 5.5 4.0 3.8 6.0 5.8 4.9 7.0 5.7
Sample Ordered Data Rank
A 3.5 1
B 3.8 2
B 4.0 3
A 4.1 4
© 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 2
accessible website, in whole or in part.
, Solution and Answer Guide: Peck, Introduction to Statistics and Data Analysis, 7e, 9798214000008; Chapter 16:
Nonparametric (Distribution-Free) Statistical Methods
A 4.4 5
A 4.7 6
B 4.9 7
A 5.3 8
B 5.5 9
A 5.6 10
B 5.7 11
B 5.8 12
B 6.0 13
B 7.0 14
A 7.5 15
A 7.6 16
Rank sum = 1 + 4 + 5 + 6 + 8 + 10 + 15 + 16 = 65
P-value: This is an upper-tail test. With n1 = 8 and n2 = 8, Chapter 16 Appendix Table 1
tells us that the P-value is greater than 0.05.
Since the P-value exceeds , H 0 is not rejected. There is not enough evidence
suggesting that the mean reduction in development time resulting from the
modification exceeds 1 second.
3. The study described in “Gait Patterns During Free Choice Ladder Ascents” (Human
Movement Science [1983]: 187–195) was motivated by publicity concerning the
increased accident rate for individuals climbing ladders. A number of different gait
patterns were used by subjects climbing a portable straight ladder according to
specified instructions. The following data consist of climbing times for seven subjects
who used a lateral gait and six subjects who used a four-beat diagonal gait:
Lateral gait 0.86 1.31 1.64 1.51 1.53 1.39 1.09
Diagonal gait 1.27 1.82 1.66 0.85 1.45 1.24
a. Use the rank-sum test to decide whether the data suggest a difference in the
mean climbing times for the two gaits.
b. Interpret the 95% confidence interval for the difference between the mean
climbing times given in the following Minitab output:
lateral N=7 Median = 1.3900
diagonal N=6 Median = 1.3600
Point estimate for ETA1 – ETA2 is 0.0400
96.2 percent C.I. for ETA1 – ETA2 is (–0.4300, 0.3697)
© 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 3
accessible website, in whole or in part.
Introduction to Statistics and Data
Analysis 7th Edition By Roxy Peck,
Chris Olsen
(All Chapters 1-16, 100% Original
Verified, A+ Grade)
All Chapters Arranged Reverse: 16-1
This is the Original Solutions Manual
for 7th Edition, All Other Files in the
Market are Wrong/Old Questions.
, Solution and Answer Guide: Peck, Introduction to Statistics and Data Analysis, 7e, 9798214000008; Chapter 16:
Nonparametric (Distribution-Free) Statistical Methods
Solution and Answer Guide
PECK, INTRODUCTION TO STATISTICS AND DATA ANALYSIS, 7E, 9798214000008; CHAPTER 16:
Nonparametric (Distribution-Free) Statistical Methods
TABLE OF CONTENTS
End of Section Exercise Solutions ......................................................................................1
END OF SECTION EXERCISE SOLUTIONS
1. Urinary fluoride concentration (in parts per million) was measured for both a sample of
livestock that had been grazing in an area previously exposed to fluoride pollution and
a similar sample of livestock that had grazed in an unpolluted region.
Do the accompanying data provide evidence that the mean fluoride concentration
for livestock grazing in the polluted region is greater than that for livestock grazing in
the unpolluted region? Assume that the distributions of urinary fluoride concentration
for both grazing areas have the same shape and variability, and use a level 0.05 rank-
sum test.
Polluted 21.3 18.7 23.0 17.1 16.8 20.9 19.7
Unpolluted 14.2 18.3 17.2 18.4 20.0
Solution:
Let 1 denote the mean fluoride concentration for livestock grazing in the polluted
region and 2 denote the mean fluoride concentration for livestock grazing in the
unpolluted regions.
H 0 : 1 2 = 0 H a : 1 2 > 0
= 0.05
The test statistic is: rank sum for polluted area (sample 1).
Sample Ordered Data Rank
2 14.2 1
1 16.8 2
1 17.1 3
2 17.2 4
2 18.3 5
2 18.4 6
© 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 1
accessible website, in whole or in part.
, Solution and Answer Guide: Peck, Introduction to Statistics and Data Analysis, 7e, 9798214000008; Chapter 16:
Nonparametric (Distribution-Free) Statistical Methods
1 18.7 7
1 19.7 8
2 20.0 9
1 20.9 10
1 21.3 11
1 23.0 12
Rank sum = (2 + 3 + 7 + 8 + 10 + 11 + 12) = 53
P-value: This is an upper-tail test. With n1 = 7 and n 2 = 5, Chapter 16 Appendix Table 1
tells us that the P-value > 0.05.
Since the P-value exceeds , Ho is not rejected. The data do not support the
conclusion that there is a larger average fluoride concentration for the polluted area
than for the unpolluted area.
2. A modification has been made to the process for producing a certain type of film.
Because the modification involves extra cost, it will be incorporated only if sample
data provide evidence that the modification decreases mean developing time by more
than 1 second. Assuming that the developing time distributions have the same shape
and variability, use the rank-sum test, a significance level of 0.05, and the following
data to test the appropriate hypotheses.
Original process 8.6 5.1 4.5 5.4 6.3 6.6 5.7 8.5
Modified process 5.5 4.0 3.8 6.0 5.8 4.9 7.0 5.7
Solution:
Let 1 denote the true average developing time without modification and 2 the true
average developing time with modification.
H 0 : 1 2 = 1 H a : 1 2 > 1
= 0.05
The test statistic is: rank sum for (unmodified times - 1).
(A) original process 1 7.6 4.1 3.5 4.4 5.3 5.6 4.7 7.5
(B) modified process 5.5 4.0 3.8 6.0 5.8 4.9 7.0 5.7
Sample Ordered Data Rank
A 3.5 1
B 3.8 2
B 4.0 3
A 4.1 4
© 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 2
accessible website, in whole or in part.
, Solution and Answer Guide: Peck, Introduction to Statistics and Data Analysis, 7e, 9798214000008; Chapter 16:
Nonparametric (Distribution-Free) Statistical Methods
A 4.4 5
A 4.7 6
B 4.9 7
A 5.3 8
B 5.5 9
A 5.6 10
B 5.7 11
B 5.8 12
B 6.0 13
B 7.0 14
A 7.5 15
A 7.6 16
Rank sum = 1 + 4 + 5 + 6 + 8 + 10 + 15 + 16 = 65
P-value: This is an upper-tail test. With n1 = 8 and n2 = 8, Chapter 16 Appendix Table 1
tells us that the P-value is greater than 0.05.
Since the P-value exceeds , H 0 is not rejected. There is not enough evidence
suggesting that the mean reduction in development time resulting from the
modification exceeds 1 second.
3. The study described in “Gait Patterns During Free Choice Ladder Ascents” (Human
Movement Science [1983]: 187–195) was motivated by publicity concerning the
increased accident rate for individuals climbing ladders. A number of different gait
patterns were used by subjects climbing a portable straight ladder according to
specified instructions. The following data consist of climbing times for seven subjects
who used a lateral gait and six subjects who used a four-beat diagonal gait:
Lateral gait 0.86 1.31 1.64 1.51 1.53 1.39 1.09
Diagonal gait 1.27 1.82 1.66 0.85 1.45 1.24
a. Use the rank-sum test to decide whether the data suggest a difference in the
mean climbing times for the two gaits.
b. Interpret the 95% confidence interval for the difference between the mean
climbing times given in the following Minitab output:
lateral N=7 Median = 1.3900
diagonal N=6 Median = 1.3600
Point estimate for ETA1 – ETA2 is 0.0400
96.2 percent C.I. for ETA1 – ETA2 is (–0.4300, 0.3697)
© 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 3
accessible website, in whole or in part.
