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CONTENTS
Chapter 1 ....................................................................................................................... 1
Chapter 2 ....................................................................................................................... 9
Chapter 3 ..................................................................................................................... 29
Chapter 4 ..................................................................................................................... 45
Chapter 5 ..................................................................................................................... 59
Chapter 6 ..................................................................................................................... 73
Chapter 7 ................................................................................................................... 101
Chapter 8 ................................................................................................................... 117
Chapter 9 ................................................................................................................... 139
Chapter 10 ................................................................................................................. 159
Chapter 11 ................................................................................................................. 199
Chapter 12 ................................................................................................................. 211
Chapter 13 ................................................................................................................. 219
Chapter 14 ................................................................................................................. 235
Chapter 1: Introduction to Statistics 1
Chapter 1: Introduction to Statistics
Section 1-2
1.
Statistical significance is indicated when methods of statistics are used to reach a conclusion that some
treatment or finding is effective, but common sense might suggest that the treatment or finding does not
make enough of a difference to justify its use or to be practical. Yes, it is possible for a study to have
statistical significance but not a practical significance.
2.
If the source of the data can benefit from the results of the study, it is possible that an element of bias is
introduced so that the results are favorable to the source.
3.
A voluntary response sample is a sample in which the subjects themselves decide whether to be included in
the study. A voluntary response sample is generally not suitable for a statistical study because the sample
may have a bias resulting from participation by those with a special interest in the topic being studied.
4.
Even if we conduct a study and find that there is a correlation, or association, between two variables, we
cannot conclude that one of the variables is the cause of the other.
5.
There does appear to be a potential to create a bias.
6.
There does not appear to be a potential to create a bias.
7.
There does not appear to be a potential to create a bias.
8.
There does appear a potential to create a bias.
9.
The sample is a voluntary response sample and is therefore flawed.
10. The sample is a voluntary response sample and is therefore flawed.
11. The sampling method appears to be sound.
12. The sampling method appears to be sound.
13. Because there is a 30% chance of getting such results with a diet that has no effect, it does not appear to
have statistical significance, but the average loss of 45 pounds does appear to have practical significance.
14. Because there is only a 1% chance of getting the results by chance, the method appears to have a statistical
significance. The result of 540 boys in 1000 births is above the approximately 50% rate expected by
chance, but it does not appear to be high enough to have practical significance. Not many couples would
bother with a procedure that raises the likelihood of a boy from 50% to 54%.
15. Because there is a 23% chance of getting such results with a program that has no effect, the program does
not appear to have statistical significance. Because the success rate of 23% is not much better than the 20%
rate that is typically expected with random guessing, the program does not appear to have practical
significance.
16. Because there is a 25% chance of getting such results with a program that has no effect, the program does
not appear to have statistical significance. Because the average increase is only 3 IQ point, the program
does not appear to have practical significance.
17. The male and female pulse rates in the same column are not matched in any meaningful way. It does not
make sense to use the difference between any of the pulse rates that are in the same column.
18. Yes, the source of the data is likely to be unbiased.
19. The data can be used to address the issue of whether males and females have pulse rates with the same
average (mean) value.
20. The results do not prove that the populations of males and females have the same average (mean) pulse
rate. The results are based on a particular sample of five males and five females, and analyzing other
samples might lead to a different conclusion. Better results would be obtained with larger samples.
Copyright © 2014 Pearson Education, Inc.
2 Chapter 1: Introduction to Statistics
21. Yes, each IQ score is matched with the brain volume in the same column, because they are measurements
obtained from the same person. It does not make sense to use the difference between each IQ score and the
brain volume in the same column, because IQ scores and brain volumes use different units of measurement.
For example, it would make no sense to find the difference between an IQ score of 87 and a brain volume
of 1035 cm3.
22. The issue that can be addressed is whether there is a correlation, or association, between IQ score and brain
volume.
23. Given that the researchers do not appear to benefit from the results, they are professionals at prestigious
institutions, and funding is from a U.S. government agency, the source of the data appears to be unbiased.
24. No. Correlation does not imply causation, so a statistical correlation between IQ score and brain volume
should not be used to conclude that larger brain volumes cause higher IQ scores.
25. It is questionable that the sponsor is the Idaho Potato Commission and the favorite vegetable is potatoes.
26. The sample is a voluntary response sample, so there is a good chance that the results are not valid.
27. The correlation, or association, between two variables does not mean that one of the variables is the cause
of the other. Correlation does not imply causation.
28. The correlation, or association, between two variables does not mean that one of the variables is the cause
of the other. Correlation does not imply causation.
29. a.
b.
The number of people is (0.39)(1018) = 397.02
c.
No. Because the result is a count of people among 1018 who were surveyed, the result must be a whole
number.
The actual number is 397 people
d.
The percentage is
30. a.
b.
255
= 0.25049 = 25.049%
1018
The number of women is (0.38)(427) = 162.26
b.
No. Because the result is a count of women among 427 who were surveyed, the result must be a whole
number.
The actual number is 162 women.
d.
The percentage is
31. a.
b.
30
= 0.07026 = 7.026%
427
The number of adults is (0.14)(2302) = 322.28
c.
No. Because the result is a count of adults among 2302 who were surveyed, the result must be a whole
number.
The actual number is 322 adults.
d.
The percentage is
32. a.
b.
46
= 0.01998 = 1.998%
2302
The number of adults is (0.76)(2513) = 1909.88
b.
No. Because the result is a count of adults among 2513 who were surveyed, the result must be a whole
number.
The actual number is 1910 adults.
d.
The percentage is
327
= 0.13012 = 13.012%
2513
33. Because a reduction of 100% would eliminate all of the size, it is not possible to reduce the size by 100%
or more.
Copyright © 2014 Pearson Education, Inc.
Chapter 1: Introduction to Statistics 3
34. If the Club eliminated all car thefts, it would reduce the odds of car theft by 100%, so the 400% figure is
impossible.
35. If foreign investment fell by 100% it would be totally eliminated, so it is not possible for it to fall by more
than 100%.
36. Because a reduction of 100% would eliminate all plague, it is not possible to reduce it by more than 100%.
37. Without our knowing anything about the number of ATVs in use, or the number of ATV drivers, or the
amount of ATV usage, the number of 740 fatal accidents has no context. Some information should be
given so that the reader can understand the rate of ATV fatalities.
38. All percentages of success should be multiples of 5. The given percentage cannot be correct.
39. The wording of the question is biased and tends to encourage negative response. The sample size of 20 is
too small. Survey respondents are self-selected instead of being selected by the newspaper. If 20 readers
respond, the percentages should be multiples of 5, so 87% and 13% are not possible results.
Section 1-3
1.
A parameter is a numerical measurement describing some characteristic of a population, whereas a statistic
is a numerical measurement describing some characteristic of a sample.
2.
Quantitative data consist of numbers representing counts or measurements, whereas categorical data can be
separated into different categories that are distinguished by some characteristic that is not numerical.
3.
Parts (a) and (c) describe discrete data.
4.
The values of 1010 and 55% are both statistics because they are based on the sample. The population
consists of all adults in the United States.
5.
Statistic
17. Discrete
6.
Parameter
18. Discrete
7.
Parameter
19. Continuous
8.
Statistic
20. Continuous
9.
Parameter
21. Nominal
10. Parameter
22. Ratio
11. Statistic
23. Interval
12. Statistic
24. Ordinal
13. Continuous
25. Ratio
14. Discrete
26. Nominal
15. Discrete
27. Ordinal
16. Continuous
28. Interval
29. The numbers are not counts or measures of anything, so they are at the nominal level of measurement, and
it makes no sense to compute the average (mean) of them.
30. The flight numbers do not count or measure anything. They are at the nominal level of measurement, and it
does not make sense to compute the average (mean) of them.
31. The numbers are used as substitutes for the categories of low, medium, and high, so the numbers are at the
ordinal level of measurement. It does not make sense to compute the average (mean) of such numbers.
32. The numbers are substitutes for names and are not counts or measures of anything. They are at the nominal
level of measurement, and it makes no sense to compute the average (mean) of them.
Copyright © 2014 Pearson Education, Inc.
4 Chapter 1: Introduction to Statistics
33. a.
b.
c.
d.
Continuous, because the number of possible values is infinite and not countable.
Discrete, because the number of possible values is finite.
Discrete, because the number of possible values is finite.
Discrete, because the number of possible values is infinite and countable.
34. Either ordinal or interval is a reasonable answer, but ordinal makes more sense because differences
between values are not likely to be meaningful. For example, the difference between a food rated 1 and a
food rated 2 is not necessarily the same as a difference between a food rated 9 and a food rated 10.
35. With no natural starting point, temperatures are at the interval level of measurement, so ratios such as
“twice” are meaningless.
Section 1-4
1.
No. Not every sample of the same size has the same chance of being selected. For example, the sample
with the first two names has no chance of being selected. A simple random sample of (n) items is selected
in such a way that every sample of same size has the same chance of being selected.
2.
In an observational study, you would examine subjects who consume fruit and those who do not. In the
observational study, you run a greater risk of having a lurking variable that affects weight. For example,
people who consume more fruit might be more likely to maintain generally better eating habits, and they
might be more likely to exercise, so their lower weights might be due to these better eating and exercise
habits, and perhaps fruit consumption does not explain lower weights. An experiment would be better,
because you can randomly assign subjects to the fruit treatment group and the group that does not get the
fruit treatment, so lurking variables are less likely to affect the results.
3.
The population consists of the adult friends on the list. The simple random sample is selected from the
population of adult friends on the list , so the results are not likely to be representative of the much larger
general population of adults in the United States.
4.
Because there is nothing about left-handedness or right-handedness that would affect being in the author’s
classes, the results are likely to be typical of the population. The results are likely to be good, but
convenience samples in general are not likely to be so good.
5.
Because the subjects are subjected to anger and confrontation, they are given a form or treatment, so this is
an experiment, not an observational study.
6.
Because the subjects were given a treatment consisting of Lipitor, this is an experiment.
7.
This is an observational study because the therapists were not given any treatment. Their responses were
observed.
8.
This is an observational study because the survey subjects were not given any treatment. Their responses
were observed.
9.
Cluster
15. Systematic
10. Convenience
16. Cluster
11. Random
17. Random
12. Systematic
18. Cluster
13. Convenience
19. Convenience
14. Random
20. Systematic
21. The sample is not a simple random sample. Because every 1000th pill is selected, some samples have no
chance of being selected. For example, a sample consisting of two consecutive pills has no chance of being
selected, and this violates the requirement of a simple random sample.
22. The sample is not a simple random sample. Not every sample of 1500 adults has the same chance of being
selected. For example, a sample of 1500 women has no chance of being selected.
23. The sample is a simple random sample. Every sample of size 500 has the same chance of being selected.
Copyright © 2014 Pearson Education, Inc.
Chapter 1: Introduction to Statistics 5
24. The sample is a simple random sample. Every sample of the same size has the same chance of being
selected.
25. The sample is not a simple random sample. Not every sample has the same chance of being selected. For
example, a sample that includes people who do not appear to be approachable has no chance of being
selected.
26. The sample is not a simple random sample. Not all samples of the same size have the same chance of
being selected. For example, a sample would not be selected which included people who do not appear to
be approachable.
27. Prospective study
31. Matched pairs design
28. Retrospective study
32. Randomized block design
29. Cross-sectional study
33. Completely randomized design
30. Prospective study
34. Matched pairs design
35. Blinding is a method whereby a subject (or a person who evaluates results) in an experiment does not know
whether the subject is treated with the DNA vaccine or the adenoviral vector vaccine. It is important to use
blinding so that results are not somehow distorted by knowledge of the particular treatment used.
36. Prospective: The experiment was begun and results were followed forward in time. Randomized:
Subjects were assigned to the different groups through the process of random selection, and whereby they
had the same chance of belonging to each group. Double-blind: The subjects did not know which of the
three groups they were in, and the people who evaluated results did not know either. Placebo-controlled:
There was a group of subjects who were given a placebo, by comparing the placebo group to the two
treatment groups, the effect of the treatments might be better understood.
Chapter Quick Quiz
1.
No. The numbers do not measure or count anything.
2.
Nominal
7.
No
3.
Continuous
8.
Statistic
4.
Quantitative data
9.
Observational study
5.
Ratio
10 False
6.
False
Review Exercises
1.
a.
b.
c.
d.
e.
Discrete
Ratio
Stratified
Cluster
The mailed responses would be a voluntary response sample, so those with strong opinions are more
likely to respond. It is very possible that the results do not reflect the true opinions of the population of
all costumers.
2.
The survey was sponsored by the American Laser Centers, and 24% said that the favorite body part is the
face, which happens to be a body part often chosen for some type of laser treatment. The source is
therefore questionable.
3.
The sample is a voluntary response sample, so the results are questionable.
Copyright © 2014 Pearson Education, Inc.
6 Chapter 1: Introduction to Statistics
4.
a.
b.
c.
5.
a.
It uses a voluntary response sample, and those with special interests are more likely to respond, so it is
very possible that the sample is not representative of the population.
Because the statement refers to 72% of all Americans, it is a parameter (but it is probably based on a
72% rate from the sample, and the sample percentage is a statistic).
Observational study.
If they have no fat at all, they have 100% less than any other amount with fat, so the 125% figure
cannot be correct.
b. The exact number is (0.58)(1182) = 685.56 . The actual number is 686.
c.
331
= 0.28003 = 28.003%
1182
6.
The Gallop poll used randomly selected respondents, but the AOL poll used a voluntary response sample.
Respondents in the AOL poll are more likely to participate if they have strong feelings about the
candidates, and this group is not necessarily representative of the population. The results from the Gallop
poll were more likely to reflect the true opinions of American voters.
7.
Because there is only a 4% chance of getting the results by chance, the method appears to have statistical
significance. The results of 112 girls in 200 births is above the approximately 50% rate expected by
chance, but it does not appear to be high enough to have practical significance. Not many couples would
bother with a procedure that raises the likelihood of a girl from 50% to 56%.
8.
a.
b.
c.
d.
e.
Random
Stratified
Nominal
Statistic, because it is based on a sample.
The mailed responses would be a voluntary response sample. Those with strong opinions about the
topic would be more likely to respond, so it is very possible that the results would not reflect the true
opinions of the population of all adults.
9.
a.
b.
c.
d.
e.
f.
Systematic
Random
Cluster
Stratified
Convenience
No, although this is a subjective
judgment.
10. a.
0.52 (1500) = 780 adults
b.
345
= 0.23 = 23%
1500
c.
Men:
727
= 0.485 = 48.5% ;
1500
773
Women:
= 0.515 = 51.5%
1500
Cumulative Review Exercises
1.
The mean is 11. Because the flight numbers are not measures or counts of anything, the result does not
have meaning.
2.
The mean is 101, and it is reasonably close to the population mean of 100.
3.
4.
(247 −176)
6
= 11.83 is an unusually high value.
(175 −172)
= 0.46
⎛ 29 ⎞⎟
⎜⎜
⎟⎟
⎝⎜ 20 ⎠
(88 − 88.57)
5.
(1.962 × 0.25)
0.032
2
6.
88.57
= 0.0037
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= 1067
Chapter 1: Introduction to Statistics 7
((96 −100) + (106 −100) + (98 −100) ) = 28.0
2
7.
2
(3 −1)
((96 −100)
2
8.
9.
2
+ (106 −100) + (98 −100)
2
(3 −1)
0.614 = 0.00078364164
10. 812 = 68719476736
2
)=
28 = 5.3
11. 714 = 678223072849
12. 0.310 = 0.0000059049
Copyright © 2014 Pearson Education, Inc.
Chapter 2: Summarizing and Graphing Data
Chapter 2: Summarizing and Graphing Data
Section 2-2
1.
No. For each class, the frequency tells us how many values fall within the given range of values, but there
is no way to determine the exact IQ scores represented in the class.
2.
If percentages are used, the sum should be 100%. If proportions are used, the sum should be 1.
3.
No. The sum of the percentages is 199% not 100%, so each respondent could answer “yes” to more than
one category. The table does not show the distribution of a data set among all of several different
categories. Instead, it shows responses to five separate questions.
4.
The gap in the frequencies suggests that the table includes heights of two different populations: students
and faculty/staff.
5.
Class width: 10.
Class midpoints: 24.5, 34.5, 44.5, 54.5, 64.5, 74.5, 84.5.
Class boundaries: 19.5, 29.5, 39.5, 49.5, 59.5, 69.5, 79.5, 89.5.
6.
Class width: 10.
Class midpoints: 24.5, 34.5, 44.5, 54.5, 64.5, 74.5.
Class boundaries: 19.5, 29.5, 39.5, 49.5, 59.5, 69.5, 79.5.
7.
Class width: 10.
Class midpoints: 54.5, 64.5, 74.5, 84.5, 94.5, 104.5, 114.5, 124.5.
Class boundaries: 49.5, 59.5, 69.5, 79.5, 89.5, 99.5, 109.5, 119.5, 129.5.
8.
Class width: 5.
Class midpoints: 2, 7, 12, 17, 22, 27, 32, 37.
Class boundaries: –0.5, 4.5, 9.5, 14.5, 19.5, 24.5, 29.5, 34.5, 39.5.
9.
Class width: 2.
Class midpoints: 3.95, 5.95, 7.95, 9.95, 11.95.
Class boundaries: 2.95, 4.95, 6.95, 8.95, 10.95, 12.95.
10. Class width: 2.
Class midpoints: 3.95, 5.95, 7.95, 9.95, 11.95.
Class boundaries: 2.95, 4.95, 6.95, 8.95, 10.95, 12.95, 14.95.
11. No. The frequencies do not satisfy the requirement of being roughly symmetric about the maximum
frequency of 34.
12. Yes. The frequencies start low, increase to the maximum frequency of 43, and then decrease. Also, the
frequencies are approximately symmetric about the maximum frequency of 43.
13. 18, 7, 4
14. 12, 12, 6, 2
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Chapter 2: Summarizing and Graphing Data
15. On average, the actresses appear to be younger than the actors.
Age When Oscar Was Won
20 – 29
Relative Frequency
(Actresses)
32.9%
Relative Frequency
(Actors)
1.2%
30 – 39
41.5%
31.7%
40 – 49
15.9%
42.7%
50 – 59
2.4%
15.9%
60 – 69
4.9%
7.3%
70 – 79
1.2%
1.2%
80 – 89
1.2%
0.0%
16. The differences are not substantial. Based on the given data, males and females appear to have about the
same distribution of white blood cell counts.
White Blood Cell Counts
3.0 – 4.9
Relative Frequency
(Males)
20.0%
Relative Frequency
(Females)
15.0%
5.0 – 6.9
37.5%
40.0%
7.0 – 8.9
27.5%
22.5%
9.0 – 10.9
12.5%
17.5%
11.0 – 12.9
2.5%
0.0%
13.0 – 14.9
0.0%
5.0%
17. The cumulative frequency table is
Age (years) of Best Actress When Oscar Was Won
Less than 30
Cumulative Frequency
27
Less than 40
61
Less than 50
74
Less than 60
76
Less than 70
80
Less than 80
81
Less than 90
82
18. The cumulative frequency table is
Age (years) of Best Actor When Oscar Was Won
Less than 30
Cumulative Frequency
1
Less than 40
27
Less than 50
62
Less than 60
75
Less than 70
81
Less than 80
82
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Chapter 2: Summarizing and Graphing Data
19. Because there are disproportionately more 0s and 5s, it appears that the heights were reported instead of
measured. Consequently, it is likely that the results are not very accurate.
x
0
Frequency
9
1
2
2
1
3
3
4
1
5
15
6
2
7
0
8
3
9
1
20. Because there are disproportionately more 0s and 5s, it appears that the heights were reported instead of
measured. Consequently, it is likely that the results are not very accurate.
x
0
Frequency
26
1
1
2
1
3
2
4
2
5
12
6
1
7
0
8
4
9
1
21. Yes, the distribution appears to be a normal distribution.
Pulse Rate (Male)
40 – 49
Frequency
1
50 – 59
7
60 – 69
17
70 – 79
9
80 – 89
5
90 – 99
1
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Chapter 2: Summarizing and Graphing Data
22. Yes. The pulse rates of males appear to be generally lower than the pulse rates of females.
Pulse Rate (Females)
50 – 59
Frequency
1
60 – 69
8
70 – 79
18
80 – 89
5
90 – 99
6
100 – 109
2
23. No, the distribution does not appear to be a normal distribution.
Magnitude
Frequency
0.00 – 0.49
5
0.50 – 0.99
15
1.00 – 1.49
19
1.50 – 1.99
7
2.00 – 2.49
2
2.50 – 2.99
2
24. No, the distribution does not appear to be a normal distribution.
Depth (km)
1.00 – 4.99
Frequency
7
5.00 – 8.99
21
9.00 – 12.99
4
13.00 – 16.99
12
17.00 – 20.99
6
25. Yes, the distribution appears to be roughly a normal distribution.
Red Blood Cell Count
4.00 – 4.39
Frequency
2
4.40 – 4.79
7
4.80 – 5.19
15
5.20 – 5.59
13
5.60 – 5.99
3
26. Yes, the distribution appears to be roughly a normal distribution.
Red Blood Cell Count
3.60 – 3.99
Frequency
2
4.00 – 4.39
13
4.40 – 4.79
15
4.80 – 5.19
7
5.20 – 5.59
2
5.60 – 5.99
1
Copyright © 2014 Pearson Education, Inc.
Chapter 2: Summarizing and Graphing Data
13
27. Yes. Among the 48 flights, 36 arrived on time or early, and 45 of the flights arrived no more than 30
minutes late.
Arrival Delay (min)
(–60) – (–31)
Frequency
11
(–30) – (–1)
25
0 – 29
9
30 – 59
1
60 – 89
0
90 – 119
2
28. No. The times vary from a low of 12 minutes to a high of 49 minutes. It appears that many flights taxi out
quickly, but many other flights require much longer times, so it would be difficult to predict the taxi-out
time with reasonable accuracy.
Taxi-Out Time (min)
10 – 14
Frequency
10
15 – 19
20
20 – 24
9
25 – 29
1
30 – 34
2
35 – 39
2
40 – 44
2
45 – 49
2
29.
Category
Male Survivors
Relative Frequency
16.2%
Males Who Died
62.8%
Female Survivors
15.5%
Females Who Died
5.5%
Cause
Bad Track
Relative Frequency
46%
Faulty Equipment
18%
Human Error
24%
Other
12%
30.
31. Pilot error is the most serious threat to aviation safety. Better training and stricter pilot requirements can
improve aviation safety.
Cause
Pilot Error
Relative Frequency
50.5%
Other Human Error
6.1%
Weather
12.1%
Mechanical
22.2%
Sabotage
9.1%
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Chapter 2: Summarizing and Graphing Data
32. The digit 0 appears to have occurred with a higher frequency than expected, but in general the differences
are not very substantial, so the selection process appears to be functioning correctly. The digits are
qualitative data because they do not represent measures or counts of anything. The digits could be replaced
by the first 10 letters of the alphabet, and the lottery would be essentially the same.
Digit
0
Relative Frequency
16.7%
1
8.3%
2
10.0%
3
10.0%
4
6.7%
5
9.2%
6
7.5%
7
8.3%
8
7.5%
9
15.8%
33. An outlier can dramatically affect the frequency table.
Weight (lb)
200 – 219
With Outlier
6
Without Outlier
6
229 – 239
5
5
240 – 259
12
12
260 – 279
36
36
280 – 299
87
87
300 – 319
28
28
320 – 339
0
340 – 359
0
360 – 379
0
380 – 399
0
400 – 419
0
420 – 439
0
440 – 459
0
460 – 479
0
480 – 499
0
500 – 519
1
34.
Number of Data Values
16 – 22
Ideal Number of Classes
5
23 – 45
6
46 – 90
7
91 – 181
8
182 – 362
9
363 – 724
10
725 – 1448
11
1449 – 2896
12
Copyright © 2014 Pearson Education, Inc.
Chapter 2: Summarizing and Graphing Data
15
Section 2-3
1.
It is easier to see the distribution of the data by examining the graph of the histogram than by the numbers
in the frequency distribution.
2.
Not necessarily. Because those with special interests are more likely to respond, and the voluntary
response sample is likely to consist of a group having characteristics that are fundamentally different than
those of the population.
3.
With a data set that is so small, the true nature of the distribution cannot be seen with a histogram. The
data set has an outlier of 1 minute. That duration time corresponds to the last flight, which ended in an
explosion that killed seven crew members.
4.
When referring to a normal distribution, the term normal has a meaning that is different from its meaning in
ordinary language. A normal distribution is characterized by a histogram that is approximately bell-shaped.
Determination of whether a histogram is approximately bell-shaped does require subjective judgment.
5.
Identifying the exact value is not easy, but answers not too far from 200 are good answers.
6.
Class width of 2 inches. Approximate lower limit of first class of 43 inches. Approximate upper limit of
first class of 45 inches.
7.
The tallest person is about 108 inches, or about 9 feet tall. That tallest height is depicted in the bar that is
farthest to the right in the histogram. That height is an outlier because it is very far from all of the other
heights. The height of 9 feet must be an error, because the height of the tallest human ever recorded was 8
feet 11 inches.
8.
The first group appears to be adults. Knowing that the people entered a museum on a Friday morning, we
can reasonably assume that there were many school children on a field trip and that they were accompanied
by a smaller group of teachers and adult chaperones and other adults visiting the museum by themselves.
9.
The digits 0 and 5 seem to occur much more than the other digits, so it appears that the heights were
reported and not actually measured. This suggests that the results might not be very accurate.
10. The digits 0 and 5 seem to occur much more often than the other digits, so it appears that the heights were
reported and not measured. This suggests that the results might not be very accurate.
11. The histogram does appear to depict a normal distribution. The frequencies increase to a maximum and
then tend to decrease, and the histogram is symmetric with the left half being roughly a mirror image of the
right half.
Copyright © 2014 Pearson Education, Inc.
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Chapter 2: Summarizing and Graphing Data
11. (continued)
12. The histogram appears to roughly approximate a normal distribution. The frequencies generally increase to
a maximum and then tend to decrease, and the histogram is symmetric with the left half being roughly a
mirror image of the right half.
13. The histogram appears to roughly approximate a normal distribution. The frequencies increase to a
maximum and then tend to decrease, and the histogram is symmetric with the left half being roughly a
mirror image of the right half.
14. No, the histogram does not appear to approximate a normal distribution. The frequencies do not increase to
a maximum and then decrease, and the histogram is not symmetric with the left half being a mirror image
of the right half.
Copyright © 2014 Pearson Education, Inc.
Chapter 2: Summarizing and Graphing Data
14. (continued)
15. The histogram appears to roughly approximate a normal distribution. The frequencies increase to a
maximum and then tend to decrease, and the histogram is symmetric with the left half being roughly a
mirror image of the right half.
16. The histogram appears to roughly approximate a normal distribution. The frequencies increase to a
maximum and then tend to decrease, and the histogram is symmetric with the left half being roughly a
mirror image of the right half.
Copyright © 2014 Pearson Education, Inc.
17
18
Chapter 2: Summarizing and Graphing Data
17. The two leftmost bars depict flights that arrived early, and the other bars to the right depict flights that
arrived late.
18. Yes, the entire distribution would be more concentrated with less spread.
19. The ages of actresses are lower than those of actors.
20. a.
b.
107 inches to 109 inches; 8 feet 11 inches to 9 feet 1 inch.
The heights of the bars represent numbers of people, not heights. Because there are many more people
between 43 inches tall and 55 inches tall, they have the tallest bars in the histogram, but they have the
lowest actual heights. They have the tallest bars because there are more of them.
Section 2-4
1.
In a Pareto chart, the bars are arranged in descending order according to frequencies. The Pareto chart
helps us understand data by drawing attention to the more important categories, which have the highest
frequencies.
Copyright © 2014 Pearson Education, Inc.
Chapter 2: Summarizing and Graphing Data
19
2.
A scatter plot is a plot of paired quantitative data, and each pair of data is plotted as a single point. The
scatterplot requires paired quantitative data. The configuration of the plotted points can help us determine
whether there is some relationship between two variables.
3.
The data set is too small for a graph to reveal important characteristics of the data. With such a small data
set, it would be better to simply list the data or place them in a table.
4.
The sample is a voluntary response sample since the students report their scores to the website. Because
the sample is a voluntary response sample , it is very possible that it is not representative of the population,
even if the sample is very large. Any graph based on the voluntary response sample would have a high
chance of showing characteristics that are not actual characteristics of the population.
5.
Because the points are scattered throughout with no obvious pattern, there does not appear to be a
correlation.
6.
The configuration of the points does not support the hypothesis that people with larger brains have larger
IQ scores.
Copyright © 2014 Pearson Education, Inc.
20
Chapter 2: Summarizing and Graphing Data
7.
Yes. There is a very distinct pattern showing that bears with larger chest sizes tend to weigh more.
8.
Yes. There is a very distinct pattern showing that cans of Coke with larger volumes tend to weigh more.
Another notable feature of the scatterplot is that there are five groups of points that are stacked above each
other. This is due to the fact that the measured volumes were rounded to one decimal place, so the different
volume amounts are often duplicated, with the result that points are stacked vertically.
9.
The first amount is highest for the opening day, when many Harry Potter fans are most eager to see the
movie; the third and fourth values are from the first Friday and the first Saturday, which are the popular
weekend days when movie attendance tends to spike.
Copyright © 2014 Pearson Education, Inc.
Chapter 2: Summarizing and Graphing Data
21
10. The numbers of home runs rose from 1990 to 2000, but after 2000 there was a very gradual decline.
11. Yes, because the configuration of the points is roughly a bell shape, the volumes appear to be from a
normally distributed population. The volume of 11.8 oz. appears to be an outlier.
12. No, because the configuration of points is not at all a bell shape, the amounts do not appear to be from a
normally distributed population.
13. No. The distribution is not dramatically far from being a normal distribution with a bell shape, so there is
not strong evidence against a normal distribution.
4|5
5|3335579
6|11167
7|11115568
8|4
14. There are no outliers. The distribution is not dramatically far from being a normally distribution with a bell
shape, so there is not strong evidence against a normal distribution.
12 | 6 8
13 | 1 2 3 4 5 5 6 6 6 7 7 8 9 4
14 | 0 0 0 3 3 5
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Chapter 2: Summarizing and Graphing Data
15.
16. To remain competitive in the world, the United States should require more weekly instruction time.
17.
18. Because there is not a single total number of hours of instruction time that is partitioned among the five
countries, it does not make sense to use a pie chart for the given data.
Copyright © 2014 Pearson Education, Inc.
Chapter 2: Summarizing and Graphing Data
23
19. The frequency polygon appears to roughly approximate a normal distribution. The frequencies increase to
a maximum and then tend to decease, and the graph is symmetric with the left half being roughly a mirror
image of the right half.
20. No, the frequency polygon does not appear to approximate a normal distribution. The frequencies do not
increase to a maximum and then decrease, and the graph is not symmetric with the left half being a mirror
image of the right half.
21. The vertical scale does not start at 0, so the difference is exaggerated. The graphs make it appear that
Obama got about twice as many votes as McCain, but Obama actually got about 69 million votes compared
to 60 million to McCain.
22. The fare doubled from $1 to $2, but when the $2 bill is shown with twice the width and twice the height of
the $1 bill, the $2 bill has an area that is four times that of the $1 bill, so the illustration greatly exaggerates
the increase in fare.
23. China’s oil consumption is 2.7 times (or roughly 3 times) that of the United States, but by using a larger
barrel that is three times as wide and three times as tall (and also three times as deep) as the smaller barrel,
the illustration has made it appear that the larger barrel has a volume that is 27 times that of the smaller
barrel. The actual ratio of US consumption to China’s consumption is roughly 3 to 1, but the illustration
makes it appear to be 27 to 1.
24. The actual braking distances are 133 ft., 136 ft., and 143 ft., so the differences are relatively small, but the
illustration has a scale that begins at 130 ft., so the differences are grossly exaggerated.
Copyright © 2014 Pearson Education, Inc.
24
Chapter 2: Summarizing and Graphing Data
25. The ages of actresses are lower than those of actors.
26. a.
b.
96 | 5 9
97 | 0 0 0 1 1 1 2 3 3 3 4 4 4
97 | 5 5 6 6 6 6 6 6 7 8 8 8 8 8 9 9 9
98 | 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 3 3 4 4 4 4 4 4 4 4 4 4 4 4
98 | 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 8 9 9
99 | 0 0 1 2 4
99 | 5 6
The condensed stemplot reduces the number of rows so that the stemplot is not too large to be
understandable.
6 – 7 | 79 * 778
8 – 9 | 45678 * 049
10 – 11 | 348 * 234477
12 – 13 | 01234 * 5
14 – 15 | 05 * 4569
16 – 17 | * 049
18 – 19 | * 6
20 – 21 | 1 * 3
Chapter Quick Quiz
1.
The class width is 1.00
6.
Bar graph
2.
The class boundaries are –0.005 and 0.995
7.
Scatterplot
3.
No
8.
Pareto Chart
4.
61 min., 62 min., 62 min., 62 min., 62 min.,
67 min., and 69 min.
9.
The distribution of the data set
5.
No
10. The bars of the histogram start relatively low, increase to a maximum value and then decrease. Also, the
histogram is symmetric with the left half being roughly a mirror image of the right half.
Review Exercises
1.
Volume (cm3)
900 – 999
Frequency
1
1000 – 1099
10
1100 – 1199
4
1200 – 1299
3
1300 – 1399
1
1400 – 1499
1
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Chapter 2: Summarizing and Graphing Data
2.
No, the distribution does not appear to be normal because the graph is not symmetric.
3.
Although there are differences among the frequencies of the digits, the differences are not too extreme
given the relatively small sample size, so the lottery appears to be fair.
4.
The sample size is not large enough to reveal the true nature of the distribution of IQ scores for the
population from which the sample is obtained.
25
8 |779
9 |66
10 | 1 3 3
5.
A time-series graph is best. It suggests that the amounts of carbon monoxide emissions in the United States
are increasing.
Copyright © 2014 Pearson Education, Inc.
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Chapter 2: Summarizing and Graphing Data
6.
A scatterplot is best. The scatterplot does not suggest that there is a relationship.
7.
A Pareto chart is best.
Cumulative Review Exercises
1.
Pareto chart.
2.
Nominal, because the responses consist of names only. The responses do not measure or count anything,
and they cannot be arranged in order according to some quantitative scale.
3.
Voluntary response sample. The voluntary response sample is not likely to be representative of the
population, because those with special interests or strong feelings about the topic are more likely than
others to respond and their views might be very different from those of the general population.
4.
By using a vertical scale that does not begin at 0, the graph exaggerates the differences in the numbers of
responses. The graph could be modified by starting the vertical scale at 0 instead of 50.
5.
The percentage is
241
= 0.376 = 37.6% . Because the percentage is based on a sample and not a population
641
that percentage is a statistic.
6.
Grooming Time (min.) Frequency
0–9
2
10 – 19
3
20 – 29
9
30 – 39
4
40 – 49
2
Copyright © 2014 Pearson Education, Inc.
Chapter 2: Summarizing and Graphing Data 27
7.
Because the frequencies increase to a maximum and then decrease and the left half of the histogram is
roughly a mirror image of the right half, the data appear to be from a population with a normal distribution.
8.
Stemplot
0|05
1|255
2|024555778
3|0055
4|05
Copyright © 2014 Pearson Education, Inc.
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
29
Chapter 3: Statistics for Describing, Exploring, and Comparing
Data
Section 3-2
1.
No. The numbers do not measure or count anything, so the mean would be a meaningless statistic.
2.
The term average is not used in statistic. The term mean should be used for the value obtained when data
values are added, then the sum is divided by the number of data values.
3.
No. The price exactly in between the highest and lowest is the midrange, not the median.
4.
They use different approaches for providing a value (or values) of the center or middle of a set of data
values.
5.
The mean is
6.
The mean is
7.
The mean is
8.
The mean is
332 + 302 + 235 + 225 + 100 + 90 + 88 + 84 + 75 + 67
= 159.8 million.
10
90 + 100
The median is
= $95 million.
2
There is no mode.
332 + 67
The midrange is
= $199.5 million.
2
Apart from the obvious and trivial fact that the mean annual earnings of all celebrities is less than $332
million, nothing meaningful can be known about the mean of the population.
54410 + 51991 + 51730 + 51300 + 51196 + 51190 + 51122 + 51115 + 51037 + 50875
10
= $51,596.6 .
51190 + 51196
The median is
= $51,193 .
2
There is no mode.
50875 + 54410
The midrange is
= $52, 642.5 .
2
Apart from the obvious and trivial fact that all other colleges have tuition amounts less than those listed,
nothing meaningful can be known about the mean of the population.
371 + 356 + 393 + 544 + 326 + 520 + 501
= 430.1 hic.
7
The median is 393 hic.
There is no mode.
326 + 544
The midrange is
= 435 hic.
2
The safest of these cars appears to be the Hyundai Elantra. Because the measurements appear to vary
substantially from a low of 326 hic to a high of 544 hic, it appears that some small cars are considerably
safer than others.
774 + 649 + 1210 + 546 + 431 + 612
= 703.7 hic.
6
612 + 649
The median is
= 630.5 hic.
2
There is no mode.
1210 + 431
The midrange is
= 820.5 hic.
2
Copyright © 2014 Pearson Education, Inc.
30
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
8.
(continued)
All of the measures of center are less than 1000 hic, but that does not indicate that all of the individual
booster seats satisfy the requirement. One of the booster seats has a measurement of 1210 hic, which does
not satisfy the specified requirement of being less than 1000 hic.
9.
58 + 22 + 27 + 29 + 21 + 10 + 10 + 8 + 7 + 9 + 11 + 9 + 4 + 4
= $16.4 million.
14
10 + 10
The median is
= 10 million.
2
The modes are $4 million, $9 million, and $10 million.
4 + 58
The midrange is
= $31 million.
2
The measures of center do not reveal anything about the pattern of the data over time, and that pattern is a
key component of a movie’s success. The first amount is highest for the opening day when many Harry
Potter fans are most eager to see the movie, the third and fourth values are from the first Friday and the first
Saturday, which are the popular weekend days when movie attendance tends to spike.
The mean is
78 + 81 + 95 + 73 + 69 + 79 + 92 + 73 + 90 + 97
= 82.7 manatees.
10
79 + 81
The median is
= 80 manatees.
2
The mode is 73 manatees.
69 + 97
The midrange is
= 83 manatees.
2
The measures of center do not reveal anything about the pattern of the data over time, and it is important to
monitor the number of manatee deaths caused by collisions with watercraft, so that corrective action might
be taken.
10. The mean is
55.99 + 69.99 + 48.95 + 48.92 + 71.77 + 59.68
= $59.22 .
6
55.99 + 59.68
The median is
= $57.84 .
2
There is no mode.
48.92 + 71.77
The midrange is
= $60.35 .
2
None of the measures of center are most important here. The most relevant statistic in this case is the
minimum value of $48.92, because that is the lowest price for the software. Here, we generally care about
the lowest price not the mean price or median price.
11. The mean is
17, 688, 241 + 1 + 19, 628,585 + 12, 407,800 + 14, 765, 410
= $12,898, 007.40 .
5
The median is $14,765,410.
There is no mode.
1 + 19628585
The midrange is
= $9,814, 293 .
2
The compensation amount of $1 for Jobs is an outlier because it is very far from all the other values.
12. The mean is
3 + 6.5 + 6 + 5.5 + 20.5 + 7.5 + 12 + 11.5 + 17.5
= 11.05 μg/g .
10
7.5 + 11.5
The median is
= 9.5 μg/g .
2
The mode is 20.5 μg/g .
13. The mean is
Copyright © 2014 Pearson Education, Inc.
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
31
13. (continued)
3 + 20.5
= 11.75 μg/g .
2
There is not enough information given here to assess the true danger of these drugs, but ingestion of any
lead is generally detrimental to good health. All of the decimal values are either 0 or 5, so it appears that
the lead concentrations were rounded to the nearest one-half unit of measurement.
The midrange is
0.56 + 0.75 + 0.10 + 0.95 + 1.25 + 0.54 + 0.88
= 0.719 ppm.
7
The median is 0.75 ppm.
There is no mode.
0.1 + 1.25
The midrange is
= 0.675 ppm.
2
Fairway has the tuna with the lowest level of mercury, so it has the healthiest tuna. Because of the large
range of values, it does not appear that the different stores are getting their tuna from the same supplier.
14. The mean is
15. The mean is
4 + 4 + 4 + 4 + 4 + 4 + 4.5 + 4.5 + 4.5 + 4.5 + 4.5 + 4.5 + 6 + 6 + 8 + 9 + 9 + 13 + 13 + 15
=
20
6.5 years.
4.5 + 4.5
= 4.5 years.
2
The modes are 4 years and 4.5 years.
4 + 15
The midrange is
= 9.5 years.
2
It is common to earn a bachelor’s degree in four years, but the typical college student requires more than
four years.
The median is
0.38 + 0.55 + 1.54 + 1.55 + 0.5 + 0.6 + 0.92 + 0.96 + 1.00 + 0.86 + 1.46
= 0.938 W/kg.
11
The median is 0.92 W/kg.
There is no mode.
0.38 + 1.55
The midrange is
= 0.965 W/kg.
2
If purchasing a cell phone with concern about radiation emissions, you might be more interested in the fact
that the maximum emission is 1.55 W/kg, which is less than the FCC standard of 1.6 W/kg. You might
also be interested in the radiation emission for the particular cell phone you are considering.
16. The mean is
17. The mean is
(−15) + (−18) + (−32) + (−21) + (−9) + (−32) + 11 + 2
The median is
8
(−15) + (−18)
2
= −14.3 min.
= −16.5 .
The mode is –32 min.
(−32) + 11
The midrange is
= −10.5 .
2
Because the measures of center are all negative values, it appears that the flights tend to arrive early before
the scheduled arrival times, so the on-time performance appears to be very good.
Copyright © 2014 Pearson Education, Inc.
32 Chapter 3: Statistics for Describing, Exploring, and Comparing Data
18. The mean is
11 + 3 + 0 + (−2) + 3 + (−2) + (−2) + 5 + (−2) + 7 + 2 + 4 + 1 + 8 + 1 + 0 + (−5) + 2
18
.
= 1.9 kg.
1+ 2
= 1.5 kg.
2
The mode is –2 kg.
(−5) + 11
The midrange is
= 3 kg.
2
No, because the mean weight gain is only 1.9 kg, which is below the 6.8 kg weight gain given in the
legend.
The median is
9 + 23 + 25 + 88 + 12 + 19 + 74 + 77 + 76 + 73 + 78
= 50.4 .
11
The median is 73.
There is no mode.
9 + 78
= 48.5 .
The midrange is
2
The numbers do not measure or count anything; they are simply replacements for names. The data are at
the nominal level of measurement, and it makes no sense to compute the measures of center for these data.
19. The mean is
20. The mean is
2 +1 +1 +1 +1 + 1 + 1 + 4 + 1 + 2 + 2 + 1+ 2 + 3 + 3 + 2 + 3 + 1+ 3 + 1+ 3 + 1 + 3 + 2 + 2
25
= 1.9.
The median is 2.
The mode is 1.
1+ 4
= 2.5 .
2
The mode of 1 correctly indicates that the smooth-yellow peas occur more than any other phenotype, but
the other measures of center do not make sense with these data at the nominal level of measurement.
The midrange is
21. White drivers’ mean is 73 mi/h.
White drivers’ median is 73 mi/h.
African American drivers’ mean is 74 mi/h.
African American drivers’ median is 74 mi/h.
Although the African American drivers have a mean speed greater than the white drivers, the difference is
very small, so it appears that drivers of both races appear to speed about the same amount.
22. Collection contractor was Brinks had a mean of $1.55 million, and a median of $1.55 million.
Collection contractor was not Brinks had a mean of $1.73 million and a median of $1.65 million.
The data do suggest that collections were considerably lower when Brinks was the collection contractor.
23. Obama had a mean of $653.9 and a median of $452.
McCain had a mean of $458.5 and a median of $350.
The contributions appear to favor Obama because his mean and median are substantially higher. With 66
contributions to Obama and 20 to McCain, Obama collected substantially more in total contributions.
24. Jefferson Valley had a mean of 7.15 min. and a median of 7.2 min.
Providence had the same results as Jefferson Valley. Although the measures of center are the same, the
Providence times are much more varied than the Jefferson Valley times.
25. The mean is 1.184 the median is 1.235. Yes, it is an outlier because it is a value that is very far away from
all the other sample values.
26. The mean is 21 min. and the median is 18.5 min. The mean taxi-out time is important for calculating and
scheduling the arrival times.
Copyright © 2014 Pearson Education, Inc.
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
33
27. The mean is 15 years and the median is 16 years. Presidents receive Secret Service protection after they
leave office, so the mean is helpful in planning for the cost and resources used for that protection.
28. The mean is 101 and the median is 96.5. The mean of 101 does not differ from the population mean of 100
by an amount that is substantial, so it appears that the sample is consistent with the population.
29.
30.
31.
32.
27 ( 24.5) + 34 (34.5) + 13(44.5) + 2 (54.5) + 4 (64.5) + 1( 74.5) + 1(84.5)
27 + 34 + 13 + 2 + 4 + 1 + 1
to the mean of 35.9 years found by using the original list of data values.
= 35.8 . This result is quite close
24.5(1) + 34.5(26) + 44.5(35) + 54.5(13) + 64.5(6) + 74.5(1)
= 44.5 years. This result is not substantially
1 + 26 + 35 + 13 + 6 + 1
different from the mean of 44.1 found by using the original list of data values.
4 (54.5) + 10 (64.5) + 25 (74.5) + 43(84.5) + 26 (94.5) + 8(104.5) + 3(114.5) + 2(124.5)
4 + 10 + 25 + 43 + 26 + 8 + 3 + 2
is close to the mean of 84.4 found using the original list of data values.
= 84.7 . This result
2 (8) + 7 (2) + 12 (5) + 17 (7) + 22 ( 4) + 27 (6) + 32( 0) + 37 (1)
= 15 years. When rounded, this result is the
8 + 2 + 5 + 7 + 4 + 6 + 0 +1
same mean of 15 years found using the original list of data values.
33. a.
b.
x = 5 (0.62) − 0.3 − 0.4 −1.1− 0.7 = 0.6 parts per million
n–1
34. The mean ignoring the presidents who are still alive is 15 years. The mean including the presidents who
are still alive is at least 15.2 years. The results do not differ by much.
35. The mean is 39.07, the 10% trimmed mean is 27.677, and the 20% trimmed mean is 27.176. By deleting
the outlier of 472.2, the trimmed means are substantially different from the untrimmed mean.
36. The mean of 47 mi/h is not the actual average speed, because more time was spent at the lower speed. The
harmonic mean is 45.3 mi/h, and it does represent the true “average” value.
37. The geometric mean is 5 1.017 ⋅1.037 ⋅1.052 ⋅1.051⋅1.027 = 1.036711036 , or 1.0367 when rounded. Single
percentage growth rate is 3.67%. The result is not exactly the same as the mean which is 3.68%.
38. The root mean square (RMS) is 114.8 volts, which is very different from the mean of 0 volts.
⎛ 27 + 34 + 13 + 2 + 4 + 1 + 1 + 1
⎞
− ( 27 + 1)⎟⎟⎟
⎜⎜⎜
2
⎟⎟ = 33.970588 years, which is rounded
39. The median is 30 + (10)⎜⎜
⎟⎟
⎜⎜
34
⎟
⎜⎝⎜
⎠⎟⎟
to 34 years. The value of 33 years is better because it is based on the original data and does not involve
interpolation.
Section 3-3
1.
The IQ scores of a class of statistics students should have less variation, because those students are a much
more homogeneous group with IQ scores that are likely to be closer together.
2.
Parts (a), (b), and (d) are true.
3.
Variation is a general descriptive term that refers to the amount of dispersion or spread among the data
values, but the variance refers specifically to the square of the standard deviation.
4.
s, σ , s2, σ 2
Copyright © 2014 Pearson Education, Inc.
34 Chapter 3: Statistics for Describing, Exploring, and Comparing Data
5.
The range is $332 − $67 = $265 million.
10 (350, 292) − (2,553, 604)
2
The variance is s =
2
10 (9)
= 10548 square of million dollars.
The standard deviation is s = 10,548 = $102.703 million.
Because the data values are 10 highest from the population, nothing meaningful can be known about the
standard deviation of the population.
6.
The range is $54, 410 − $50,875 = $3535 .
10 (26,631,884,700) − (515,966)
2
The variance is s 2 =
10 (9)
= 1,088,153.8 square dollars.
The standard deviation is s = 1,088,153.8 = $1043.10 .
Because the data values are the 10 highest from the population, nothing meaningful can be known about the
standard deviation of the population.
7.
The range is 544 − 326 = 218 hic.
7 (1,342, 439) − (9, 066,121)
The variance is
= 7879.8 hic squared.
7 (6 )
The standard deviation is 7879.8 = 88.8 hic.
Although all of the cars are small, the range from 326 hic to 544hic appears to be relatively large, so the
head injury measurements are not about the same.
8.
The range is 1210 − 431 = 779 hic.
6 (3,342,798) − ( 4222)
2
The variance is s 2 =
6 (5)
= 74,383.5 hic squared.
The standard deviation is s = 74,383.5 = 272.7 hic.
Because the data values are the 10 highest from the population, nothing meaningful can be known about the
standard deviation of the population.
9.
The range is 58 − 4 = $54 million.
14 (6487) − 52441
The variance is
= 210.9 square of million dollars.
14 (13)
The standard deviation is 210.9 = $14.5 .
An investor would care about the gross from opening day and the rate of decline after that, but the
measures of center and variation are less important.
10. The range is 97 − 69 = 28 manatees.
10 (69,303) − (827)
2
The variance is s =
2
10 (9)
= 101.1 manatees squared.
The standard deviation is s = 101.1 = 10.1 manatees.
The measures of variation reveal nothing about the pattern over time.
11. The range is $71.77 − $48.92 = $22.85 .
6 ( 21,535.3844) −126, 238
The variance is
= 99.141 dollars squared.
6 (5)
The standard deviation is 99.141 = $9.957 .
The measures of variation are not very helpful in trying to find the best deal.
Copyright © 2014 Pearson Education, Inc.
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
35
12. The range is $19,628,584 − $1 = $19,628,584 .
10 (1,070,126,052,084,410) − (64, 490, 037)
2
The variance is s 2 =
10 (9)
= 59,583,269,405,325.10 dollars
squared.
The standard deviation is 59,583,269,405,325.1 = $7,719,020 .
The amount of $1 for Jobs is an outlier, and it has a great effect on the measures of deviation.
13. The range is 20.5 − 3 = 17.5 μg/g .
The variance is
10 (1596.75) −12, 210.25
10 (9)
= 41.75( μg/g ) .
2
41.75 = 6.46 μg/g .
The standard deviation is
If the medicines contained no lead, all of the measures would be 0 μg / g , and the measures of variation
would all be 0 as well.
14. The range is 1.25 − 0.10 = 1.15 ppm.
7 ( 4.42) − (5.03)
2
The variance is s 2 =
= 0.134 ppm squared.
7 (6 )
The standard deviation is 0.134 = 0.366 ppm.
If the tuna sushi contained no mercury, all of the measures would be 0 ppm, and the measures of variation
would all be 0 as well.
15. The range is 15 − 4 = 11 years.
20 (1078.5) −16,900
The variance is
= 12.3 years2.
20 (19)
The standard deviation is 12.3 = 3.5 years.
No, because 12 years is within 2 standard deviations of the mean.
16. The range is 1.55 − 0.38 = 1.17 W/kg.
11(11.41) − (10.32)
2
The variance is s 2 =
11(10)
= 0.179 (W/kg)2.
The standard deviation is 0.179 = 0.423 W/kg.
No. Same models of cell phones are sold much more than others, so the measures from the different models
should be weighted according to their size in the population.
17. The range is 11− (−32) = 43 min.
The variance is
8(3244) −12,996
8 (7 )
= 231.4 min. squared.
The standard deviation is 231.4 = 15.2 min.
The standard deviation can never be negative.
18. The range is 11− (−5) = 16 kg.
18 (340) − (32)
2
The variance is s 2 =
18(17)
= 16.5 kg2.
The standard deviation is 16.5 = 4.1 kg.
The weight gain of 6.8 kg is not unusual because it is within 2 standard deviations of the mean. Although a
gain of 6.8 kg is not unusual, the mean weight gain of 1.9 kg is not close to the legendary 6.8 kg, so an
individual weight gain of 6.8 kg does not support the legend.
Copyright © 2014 Pearson Education, Inc.
36
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
19. The range is 88 − 9 = 79 .
11(38, 078) − 306,916
The variance is
= 1017.7 .
11(10)
The standard deviation is 1017.7 = 31.9 .
The data are at the nominal level of measurement and it makes no sense to compute the measures of
variation for these data.
20. The range is 4 −1 = 3 .
25 (72) − ( 47)
2
The variance is s =
2
25( 24)
= 0.9 .
The standard deviation is 0.9 = 0.95 .
Because the data are at the nominal level of measurement, these results make no sense.
21. The mean of the White drivers is 73 and the standard deviation is 2.906 the coefficient of variation for the
2.906
White drivers is
⋅100% = 4% . The mean for the African American 74 and the standard deviation is
73
2.749
2.749 the coefficient of variation for the African American drivers is
⋅100% = 3.7% . The variation is
74
about the same.
22. The mean of the collection contractor was Brinks is 1.55 and the standard deviation is 0.178 the coefficient
0.178
⋅100% = 11.5% . The mean of the collection contractor was not Brinks is 1.73 and the
of variation is
1.55
0.2214
standard deviation is 0.2214 the coefficient of variation is
⋅100% = 12.8% . The variation is about
1.73
the same.
23. The mean of Obama contributors is $654 and the standard deviation is $523 the coefficient of variation is
$523
⋅100% = 80% . The mean of McCain contributors is $459 and the standard deviation is $418 the
$654
$418
coefficient of variation is
⋅100% = 90% . The variation among Obama contributors is a little less than
$459
the variation among the McCain contributors.
24. The mean of Jefferson Valley is 7.15 and the standard deviation is 0.477 the coefficient of variation
0.477
⋅100% = 6.7% . The mean of Providence is 7.15 and the standard deviation is 1.822 the coefficient
is
7.16
1.822
of variation is
⋅100% = 25.5% . The variation among Jefferson Valley waiting times is much less
7.15
than among the Providence waiting times.
25. The range is 2.95, the variance is 0.345, and the standard deviation is 0.587.
26. The range is 37 min., the variance is 85.5 min. squared, and the standard deviation is 9.2 min.
27. The range is 36 years, the variance is 94.5 years squared, and the standard deviation is 9.7 years.
28. The range is 42, the variance is 174.5, and the standard deviation is 13.2
29. The standard deviation
2.95
= 0.738 , which is not substantially different from 0.587
4
30. The standard deviation
37
= 9.3 min., which is very close to 9.2 min.
4
Copyright © 2014 Pearson Education, Inc.
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
31. The standard deviation
36
= 9 years, this is reasonably close to 9.7 years.
4
32. The standard deviation
42
= 10.5 , which is not substantially different from 13.2.
4
37
33. No. The pulse rate of 99 beats per minute is between the minimum usual value of 54.3 beats per minute
and the maximum usual value of 100.7 beats per minute.
34. Yes. The pulse rate of 45 beats per minute is not between the minimum usual value of 46.7 beats per
minutes and the maximum usual value of 87.9 beats per minute.
35. Yes. The volume of 11.9 oz. is not between the minimum usual value of 11.97 oz. and the maximum usual
value of 12.41 oz.
36. No. The weight of 0.8133 lb. is between the minimum usual value of 0.8127 and the maximum usual value
of 0.8355 lb.
37. s =
82 (84, 408.5) − 8, 637, 721
82 (81)
= 12.3 years. This result is not substantially different from the standard
deviation of 11.1 years found from the original list of data values.
38. s =
82 (169,980.5) −13,315, 201
82 (81)
= 9.7 years. The result is not substantially different from the standard
deviation of 9 years found from the original list of sample values.
39. s =
121(889,106.69) −104,941,584.81
121(120)
= 13.5 . The result is very close to the standard deviation of 13.4
found from the original list of sample values.
40. s =
33(10,552) − 246, 016
33(32)
= 9.8 years. The result is very close to the standard deviation of 9.7 years
found from the original list of sample values.
41. a.
b.
95%
68%
42. a.
b.
68%
99.7%
43. At least 75% of women have platelet counts within 2 standard deviations of the mean. The minimum is
150 and the maximum is 410.
44. At least 89% of healthy adults have body temperatures within 3 standard deviations of the mean. The
minimum is 96.34◦F and the maximum is 100.06◦F.
(2 − 4.33) + (3 − 4.33) + (8 − 4.33)
2
45. a.
σ =
2
2
3
2
= 6.9 min2
b. The nine possible samples of two values are the following: [(2 min, 2 min), (2 min, 3 min), (2 min, 8
min), (3 min, 2 min), (3 min, 3 min), (3 min, 8 min), (8 min, 2 min), (8 min, 3 min), (8 min, 8 min)]
and they have the following corresponding variances: [0, 0.707, 18, 0.707, 0, 12.5, 18, 12.5, 0] which
have the mean of 6.934.
c. The population variances of the nine samples above are [0, 0.3535, 9, 0.3535, 0, 6.25, 9, 6.25, 0]
d. Part (b), because repeated samples result in variances that target the same value (6.9 min.2) as the
population variance. Use division by n −1 .
e.
No. The mean of the sample variances (6.9 min.2) equals the population variance, but the mean of the
sample standard deviations (1.9 min.) does not equal the mean of the population standard deviation
(2.6 min.)
Copyright © 2014 Pearson Education, Inc.
38
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
46. The mean absolute deviation of the population is 2.4 minutes. With repeated samplings of size 2, the nine
different possible samples have mean absolute deviations of 0, 0, 0, 0.5, 0.5, 2.5, 2.5, 3, and 3. With many
such samples, the mean of those nine results is 1.3 minutes, showing that the sample mean absolute
deviations tend to center about the value of 1.3 minutes instead if the mean absolute deviation of the
population, which is 2.4 minutes. The sample mean deviations do not target the mean deviation of the
population. This is not good. This indicates that a sample mean absolute deviation is not a good estimator
of the mean absolute deviation of a population.
Section 3-4
1.
Madison’s height is below the mean. It is 2.28 standard deviations below the mean.
2.
2.00 should be preferred, because it is 2.00 standard deviations above the mean and would correspond to
the highest of the five different possible scores.
3.
The lowest amount is $5 million, the first quartile Q1 is $47 million, the second quartile Q2 (or median) is
$104 million, the third quartile Q3 is $121 million, and the highest gross amount is $380 million.
4.
All three values are the same.
5.
a.
The difference is $3, 670,505 − $4,939, 455 = −$1, 268,950
b.
$1, 268,950
= 0.16 standard deviations
$7, 775,948
6.
7.
8.
9.
z = −0.16
c.
d.
Usual
a.
The difference is 1.766
b.
1.766
= 3.01 standard deviations
0.587
c.
z = 3.01
d.
Unusual
a.
The difference is $1 − $1, 449, 779 = −$1, 449, 778
b.
$1, 449, 778
= 2.75 standard deviation
$527, 651
c.
z = −2.75
d.
Unusual
a.
The difference is 15.3 beats per minute
b.
15.3
= 1.49 standard deviations
10.3
c.
z = −1.49
d.
Usual
Z scores of –2 and 2. A z score of –2 means a score of x = −2 ⋅15 + 100 = 70 . A z score of 2 means a
score of x = 2 ⋅15 + 100 = 130
10. Z scores of –2 and 2. A z score of –2 means a hip breadth of x = −2 ⋅ 2.5 + 36.6 = 31.6 cm. A z score of 2
means a hip breadth of x = 2 ⋅ 2.5 + 36.6 = 41.6 cm
11. Two standard deviations from the mean: 1.240 − 2 ⋅ 0.578 = 0.084 and 1.240 + 2 ⋅ 0.578 = 2.396
12. Two standard deviations from the mean: 16215 − 2 ⋅ 7301 = 1613 words and 16215 + 2 ⋅ 7301 = 30817
words
Copyright © 2014 Pearson Education, Inc.
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
39
247 −175
236 −162
= 10.29 the tallest women z score is z =
= 12.33 . De7
6
Fen Yao is relatively taller, because her z score of 12.33, which is greater than the z score of 10.29 for
Sultan Kosen. De-Fen Yao is more standard deviations above the mean than Sultan Kosen.
13. The tallest man z score is z =
45 − 35.9
= 0.82 , Sandra Bullock was relatively younger than Jeff Bridges, who has
11.1
60 − 44.1
a z score of z =
= 1.77 .
9.0
14. With a z score of z =
15. The SAT score of 1490 has a z score of z =
1490 −1518
= −0.09 , and the ACT score of 17 has a z score
325
17 − 21.1
= −0.85 . The z score of –0.09 is a larger number than the z score of –0.85, so the SAT
4.8
score of 1490 is relatively better.
of z =
16. The male has a higher count because his z score is z =
than the z score of z =
4.91− 5.072
= −0.41 , which is a higher number
0.395
4.32 − 4.577
= −0.67 for the female.
0.382
17. The percentile for 213 sec. is
3
⋅100 = 13 , so the 13th percentile
24
18. The percentile for 240 sec. is
8
⋅100 = 33 , so the 33rd percentile
24
19. The percentile for 250 sec. is
12
⋅100 = 50 , so the 50th percentile
24
20. The percentile for 260 sec. is
20
⋅100 = 83 , so the 83rd percentile
24
21. P60 =
60 ⋅ 24
= 14.4 , pick 15th entry which is 251 sec.
100
22. Q1 =
234 + 235
= 234.5 sec.
2
23. Q3 =
255 + 255
= 255 sec.
2
24. P40 =
40 ⋅ 24
= 9.6 , pick 10th entry which is 243 sec.
100
25. P50 =
245 + 250
= 247.5 sec.
2
26. P75 =
75 ⋅ 24
= 18 , which is entry 255 sec.
100
27. P25 = Q1 = 234.5 sec.
28. P85 =
85 ⋅ 24
= 20.4 , pick the 21st entry which is 260 sec.
100
Copyright © 2014 Pearson Education, Inc.
40
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
29. The five number summary: 1 sec, 8709 sec, 10,074.5 sec, 11,445 sec, 11,844 sec
30. The five number summary: 81 min, 88 min, 94.5 min, 98 min, 106 min
31. The five number summary : 4 min, 14 min, 18 min, 32 min, 63 min
32. The five number summary: 70 mi/h, 72 mi/h, 74 mi/h, 78 mi/h, 79 mi/h
33. It appears that males have lower pulse rates than females
Male Pulse
Female Pulse
34. Although actresses include the oldest age of 80 years, the boxplot for actresses shows that they have ages
that are generally lower than those of actors.
Actresses
Actors
35. The weights of regular Coke appear to be generally greater than those of diet Coke, probably due to the
sugar in cans of regular Coke.
CKREGWT
CKDIETWT
Copyright © 2014 Pearson Education, Inc.
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
41
36. The low lead level group has much more variation and the IQ scores tend to be higher than the IQ scores
from the high lead level group.
Low Lead
High Lead
37. Outliers for actresses 60 years, 61 years, 63 years, 70 years, and 80 years. Outliers for actors: 76 years.
The modified boxplots show that only one actress has an age that is greater than any actor.
38. Using interpolation, P17 = 21.6 . Using figure 3-5, P17 = 22 . In this case, the results are close, but in some
other cases the results might be quite different.
Chapter Quick Quiz
1.
The mean is 14 minutes
2.
The median is 12 minutes
3.
The mode is 12 minutes
4.
The variance is (5 min) 2 = 25 min 2
5.
z=
6.
Standard deviation, variance, range, mean absolute deviation
7.
Sample mean x , population mean μ
8.
s, σ , s 2 , σ 2
9.
75%
6 −11.4
= −0.77
7
10. Minimum, first quartile Q1, second quartile Q2 (or median), third quartile Q3, maximum
Review Exercises
1.
1550 + 1642 + 1538 + 1497 + 1571
= 1559.6 mm
5
a.
x=
b.
c.
The median is 1550 mm
There is no mode
Copyright © 2014 Pearson Education, Inc.
42
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
1.
(continued)
e.
1497 + 1642
= 1569.5 mm
2
The range is 1642 −1497 = 145 mm
f.
s=
g.
s = 53.37 2 = 2849.3 mm2
h.
Q1 =
25 ⋅ 5
= 1.25 , pick second entry (in ordered list) which is 1538 mm
100
i.
Q3 =
75 ⋅ 5
= 3.75 , pick the fourth entry (in the ordered list) which is 1571 mm
100
d.
The midrange is
(1550 − 1559.6) + (1642 −1559.6) + (1538 − 1559.6) + (1497 − 1559.6) + (1571 − 1559.6)
2
2
2
2
2
5 −1
= 53.37 mm
2.
3.
2
1642 −1559.6
= 1.54 . The eye height is not unusual because its z score is between –2 and 2, so it is
53.37
within two standard deviations of the mean.
z=
The five number summary: 1497, 1538, 1550, 1571, 1642
Because the boxplot shows a distribution of data that is roughly symmetric, the data could be from a
population with a normal distribution, but the data are not necessarily from a population with a normal
distribution, because there is no way to determine whether a histogram is roughly a bell shape.
4.
The mean is 10053.5. The ZIP codes do not measure or count anything. They are at the nominal level of
measurement, so the mean is a meaningless statistic.
5.
The male z score is z =
6.
a.
b.
7.
Based on a minimum age of 23 years and a maximum age of 70 years an estimate of the age standard
70 − 23
= 11.75 years.
deviation would be
4
8.
A minimum usual sitting height of 914 − 2 ⋅ 36 = 842 mm and a maximum sitting height of
914 + 2 ⋅ 36 = 986 mm. The maximum usual height of 986 mm is more relevant for designing overhead bin
storage.
9.
The minimum value is 963 cm3, the first quartile is 1034.5 cm3, the second quartile (or median) is 1079
cm3, the third quartile is 1188.5 cm3, and the maximum value is 1439 cm3.
28 − 26.601
29 − 28.441
= 0.26 . The female z score is z =
= 0.08 . The male has
5.359
7.394
a larger relative BMI because the male has the larger z score.
The answers may vary but a mean around $8 or $9 is reasonable.
A reasonable standard deviation would be around $1 or $2.
10. The median would be better because it is not affected much by the one very large income.
Cumulative Review Exercises
1.
a.
b.
Continuous
Ratio
Copyright © 2014 Pearson Education, Inc.
Chapter 3: Statistics for Describing, Exploring, and Comparing Data
43
2.
3.
Hand Length (mm)
150 – 159
Frequency
1
160 – 169
0
170 – 179
2
180 – 189
0
190 – 199
3
200 – 209
1
210 – 219
1
Hand length histogram
4.
15 | 8
16 |
17 |3 9
18 |
19 | 5 6 9
20 | 7
21 | 4
5.
173 + 179 + 207 + 158 + 196 + 195 + 214 + 199
= 190.1 mm
8
a.
x=
b.
The median is 195.5 mm
c.
(173 −190.1)2 + (179 −190.1)2 + (207 −190.1)2 + (158 −190.1)2 + (196 −190.1)2
+ (195 −190.1)2 + ( 214 −190.1)2 + (199 −190.1)2 = 2440.88
s=
2440.88
= 18.7 mm mm
7
d.
s 2 = 18.7 2 = 348.7 mm2
e.
The range is 214 −158 = 56 mm
6.
Yes. The frequencies increase to a maximum, and then they decrease. Also, the frequencies preceding the
maximum are roughly a mirror image of those that follow the maximum.
7.
No. Even though the sample is large, it is a voluntary response sample, so the responses cannot be
considered to be representative of the population of the United States.
8.
The vertical scale does not begin at 0, so the differences among different outcomes are exaggerated.
Copyright © 2014 Pearson Education, Inc.
Chapter 4: Probability 45
Chapter 4: Probability
Section 4-2
1
1
9999
= 0.0001 , P( A) = 1−
=
= 0.9999
10, 000
10, 000 10, 000
1.
P( A) =
2.
The probability of a baby being born a boy is
3.
Part (c).
4.
The answers vary, but an answer in the neighborhood of 0.99 is reasonable.
5.
5:2,
6.
1
or 0.25
4
7.
7
456
, –0.9,
3
123
1
or 0.5
2
13.
14. 0.2
1
or 0.2
5
8.
0
9.
Unlikely, neither unusually low nor
unusually high
15.
1
or 0.5
2
16.
1
or 0.5
2
17.
1
or 0.2
5
18.
1
or 0.0278
36
10. Unlikely, unusually high
11. Unlikely, unusually low
12. Unlikely, neither unusually low nor
unusually high
21.
22.
1
or 0.25
4
19. 0
20. 1
6
or 0.006. The employer would suffer because it would be at a risk by hiring someone who uses
1000
drugs.
90
or 0.09. The person tested would suffer because he or she would be suspected of using drugs when
1000
in reality he or she does not use drugs.
23.
50
or 0.05. This result is not close to the probability of 0.134 for a positive test result.
1000
24.
950
or 0.95. This result is not very close to the probability of 0.866 for a negative test result.
1000
25.
879
or 0.93. Yes, the technique appears to be effective.
945
26.
239
or 0.821. Yes, the technique appears to be effective.
291
27.
304
or 0.00000101. No, the probability of being struck is much greater on an open golf course
300,000,000
during a thunder storm. The golfer should seek shelter.
Copyright © 2014 Pearson Education, Inc.
46
Chapter 4: Probability
28.
428
= 0.738; yes
580
29. a.
1
365
b.
Yes
c.
He already knew
d.
0
30.
31.
32.
33.
35.
2834
= 0.67 . No, it is not unlikely, because the responses are from a voluntary response
169 + 1227 + 2834
survey; the results are not likely to be very good.
10,427,000
or 0.0767. No, a crash is not unlikely. Given that car crashes are so common, we should take
135,933,000
precautions such as not driving after drinking and not using a cell phone or texting.
117
or 0.000000117. Yes, it is unlikely. The air travel fatality rate is much higher than that of
1,000,000,000
cars. The comparison isn’t fair because car trips involve much shorter distances than trips by air.
8
= 0.00985 . It is unlikely
8 + 804
34.
141
= 0.175 . It is unlikely
141 + 663
8
= 0.00993 . Yes, it is unlikely. The middle seat lacks an outside view, easy access to the
492 + 8 + 306
aisle, and a passenger in the middle seat has passengers on both sides instead of on one side only.
36.
19
= 0.0202 . Yes, it is unlikely.
19 + 441 + 235 + 103 + 66 + 75
37.
3
or 0.375
8
38.
3
or 0.375
8
39. {bb, bg, gb, gg};
1
or 0.5
2
40. {bbbb, bbbg, bbgb, bbgg, bgbb, bgbg, bggb, bggg, gbbg, gbbb, gbgb, gbgg, ggbb, ggbg, gggb, gggg};
4
16
or 0.25
41. a.
brown /brown, brown/blue, blue/brown,
blue/blue
b.
1
4
c.
3
4
42. a.
b.
c.
d.
0
0
0.5
0
43. a.
b.
c.
44. a.
b.
c.
d.
999 : 1
499 : 1
The description is not accurate. The odds
against winning are 999:1 and the odds in
favor are 1:999, not 1:1000
18
or 0.474
38
10 : 9
$18
$ 20
Copyright © 2014 Pearson Education, Inc.
Chapter 4: Probability 47
45. a.
b.
c.
d.
$16
8:1
About 9.75 : 1, which becomes 39 : 4
$21.50
46.
1
or 0.0270
37
26
2103
26
26
1−
2103 = 0.938
47. Relative risk: 2103 = 0.939
Odds ratio:
22
22
1671
1671
22
1−
1671
The probability of a headache with Nasonex (0.0124) is slightly less than the probability of a headache with
the placebo (0.0132), so Nasonex does not appear to pose a risk of headache.
48. 1
49.
1
4
Section 4-3
1.
Based on the rule of the complements, the sum of P(A) and its complement must always be 1, so the sum
cannot be 0.5
2.
A is the event of betting on the pass line and not winning (or losing). P( A) =
3.
Because it is possible to select someone who is male and a Republican, events M and R are not disjoint.
Both events can occur at the same time when someone is randomly selected.
4.
It is certain that an event occurs or does not occur.
5.
Disjoint
10. Disjoint
6.
Not disjoint
11. Not disjoint
7.
Not disjoint
12. Disjoint
8.
Not disjoint
13. 1 − 0.47 = 0.53
9.
Disjoint
14. 1 − 0.198 = 0.802
251
= 0.507
495
15. P( D) = 0.45 , where P( D) is the probability of randomly selecting someone who does not choose a direct
in-person encounter as the most fun way to flirt.
16. P( I ) denotes the probability of screening a driver and finding that he or she is not intoxicated, and
P( I ) = 0.99112
17. 1
20.
44 + 6 + 860
= 0.91
1000
18.
44 + 90 + 860
= 0.994
1000
19.
90 + 860 + 6
= 0.956
1000
21.
13
or 0.464. That probability is not as high as it should be.
28
Copyright © 2014 Pearson Education, Inc.
48 Chapter 4: Probability
22.
15
or 0.536. That probability is not as low as it should be.
28
23.
16
or 0.571
28
24.
25. a.
11
= 0.786 or 78.6%
14
b.
2
= 0.143 or 14.3%
14
c.
The physicians given the labels with concentrations appear to have done much better. The results
suggest that labels described as concentrations are much better than labels described as ratios.
26. a.
3
= 0.214 or 21.4%
14
b.
12
= 0.857 or 85.7%
14
c.
17
or 0.607
28
The physicians given the labels with ratios appear to have done much worse. The results suggest that
label described as ratios are much worse than labels described as concentrations
Use the following table for Exercises 27–32
Age
27.
165
251
1205
22–29
30–39
40–49
50–59
Responded
Refused
73
11
255
20
245
33
136
16
Total
84
275
278
152
1049
156
156
= 0.129. Yes. A high refusal rate results in a sample that is not necessarily representative of the
1205
population, because those who refuse may well constitute a particular group with opinions different from
others.
28.
202
= 0.168
1205
29.
1049
84
73
1060
+
−
=
= 0.88
1205 1205 1205 1205
30.
Total
138
27
60 and
over
202
49
18–21
31.
156
251
49
358
+
−
=
= 0.297
1205 1205 1205 1205
32.
1049 275 + 278 255 + 245 1102
+
−
=
1205
1205
1205
1205
= 0.915
156 84 + 251 11 + 49
431
+
−
=
= 0.358
1205
1205
1205
1205
33. 300
Subject Used
Marijuana
Subject Did not Use
Marijuana
Total
Positive Test Result
119
Negative Test Result
3
Total
24
154
178
143
157
300
Copyright © 2014 Pearson Education, Inc.
122
Chapter 4: Probability 49
34.
119 + 3 + 24
= 0.487
300
36.
122
=0.407. No, the general population probably has a marijuana usage rate less than 0.407, or 40.7%.
300
37.
27
= 0.09. With an error rate of 0.09 or 9%, the test does not appear to be highly accurate.
300
38.
39.
35.
3 + 154 + 24
= 0.603
300
273
= 0.91. Exercise 37 results in the probability of a wrong result and this exercise result in the
300
probability of a correct result, so these exercises deal with events that are complements.
3
or 0.75
4
40. No. Here is one example: A = event of selecting a male under 30 years of age, B = selecting a female, C =
selecting a male over 18 years of age.
41. P( A or B) = P( A) + P ( B ) − 2 P( A and B )
42. P( A or B or C )
= P( A) + P ( B ) + P (C ) − P( A and B) − P( A and C ) − P ( B and C ) + P ( A and B and C )
43. a.
1− P ( A) − P( B) + P( A and B )
b.
1− P( A and B )
c.
No
Section 4-4
1.
The probability that the second selected senator is a Democrat given that the first selected senator was a
Republican.
2.
R and D are dependent events, because the probability of a Democrat on the second selection is affected by
the outcome of the first selection. Because it was stipulated that the second selection must be a different
senator, the sampling is done without replacement, so only 99 senators are available for the second
selection.
3.
False. The events are dependent because the radio and air conditioner are both powered by the same
electrical system. If you find that your car’s radio does not work, there is a greater probability that the air
conditioner will also not work.
4.
Because the selections are based on different numbers, the sampling is done without replacement and the
events are dependent. Because the sample size of 1068 is less than 5% of the population size of 28,741,346,
the events can be treated as being independent (based on the 5% guideline for cumbersome calculations).
5.
a.
The events are dependent
b.
1
or 0.00758
132
a.
Independent
b.
1
or 0.25
4
6.
7.
8.
a.
Independent
b.
1
or 0.0833
12
a.
Dependent
b.
1
or 0.0238
42
Copyright © 2014 Pearson Education, Inc.
50 Chapter 4: Probability
9.
a.
Independent
b.
5
5
⋅
= 0.000507
222 222
11. a.
b.
10. a.
Independent
12. a.
b.
1
or 0.01
100
b.
13. a.
90
90
⋅
= 0.0081 . Yes, it is unlikely
1000 1000
b.
90 89
⋅
= 0.00802 . Yes, it is unlikely
1000 999
14. a.
b.
15. a.
b.
16. a.
b.
17.
18.
Dependent
58 1
⋅ = 0.00586
100 99
Dependent
8 7
⋅ = 0.00566
100 99
6
6
6
⋅
⋅
= 0.000000216 . Yes, it is unlikely
1000 1000 1000
6
5
4
⋅
⋅
= 0.00000012 . Yes, it is unlikely
1000 999 998
904 904 904
⋅
⋅
= 0.739 . No, it is not unlikely
1000 1000 1000
904 903 902
⋅
⋅
= 0.739 . No , it is not unlikely
1000 999 998
860 860 860 860
⋅
⋅
⋅
= 0.547 . No, it is not unlikely
1000 1000 1000 1000
860 859 858 857
⋅
⋅
⋅
= 0.546 . No, it is not unlikely
1000 999 998 997
8330 8329 8328
⋅
⋅
= 0.838 . No, the entire batch consists of malfunctioning pacemakers.
8834 8833 8832
708 707 706 705
⋅
⋅
⋅
= 0.583 . The scheme is not likely to detect the large number if defects. With a
810 809 808 807
probability of 0.583, it is more likely that the entire batch will be accepted.
19. a.
2
= 0.02
100
b.
2 2
⋅
= 0.0004
100 100
c.
2 2
2
⋅
⋅
= 0.000008
100 100 100
d.
By using one backup drive, the probability of failure is 0.02, and with three independent disk drives,
the probability drops to 0.000008. By changing from one drive to three, the likelihood of failure drops
from 1 chance in 50 to only 1 chance in 125,000, and that is a very substantial improvement in
reliability. BACK UP YOUR DATA.
20. 0.0035 ⋅ 0.0035 = 0.0000123 . With one radio there is a 0.0035 probability of a serious problem, but with
two independent radios, the probability of a serious problem drops to 0.0000123, which is substantially
lower. The flight becomes much safer with two independent radios.
Copyright © 2014 Pearson Education, Inc.
Chapter 4: Probability 51
21. a.
b.
22. a.
c.
1
or 0.00274
365
c.
1
or 0.00274
365
b.
1
or 0.2
5
1
1
⋅
= 0.00000751
365 365
1 1
⋅ = 0.04
5 5
1 1 1 1 1 1 1 1 1
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 0.000000512 . Yes, it is unlikely, but perhaps there was a strong need to staff
5 5 5 5 5 5 5 5 5
the department so that the hirings were more likely to occur on the same day.
23.
Negative Test Result
False Negative
3
True Negative
154
Total
Subject did not use
marijuana
Positive Test Result
True Positive
119
False Positive
24
Total
143
157
300
Subject used marijuana
119 118 154 153 154 119 119 154
⋅
+
⋅
+
⋅
+
⋅
= 0.828 . No, it is not unlikely
300 299 300 299 300 299 300 299
24.
24 23
3
2
24 3
3 24
⋅
+
⋅
+
⋅
+
⋅
= 0.00783 . Yes, it is unlikely
300 299 300 299 300 299 300 299
25.
24 23 22
⋅
⋅
= 0.000454 . Yes, it is unlikely
300 299 298
26.
154 153 152
⋅
⋅
= 0.134 . No it is not unlikely
300 299 298
27. a.
2518 − 252
= 0.9
2518
⎛ 2518 − 252 ⎞⎟
⎟ = 0.00513 . Using the 5% guideline for cumbersome calculations
⎜⎜⎜
⎝ 2518 ⎠⎟
50
b.
28. a.
1021− 61
= 0.94
1021
⎛1021− 61⎞⎟
⎟ = 0.0851 . Using the 5% guideline for cumbersome calculations
⎜⎜⎜
⎝ 1021 ⎠⎟
40
b.
29. a.
162 161
⋅
= 0.143
427 426
⎛ 427 −162 ⎞⎟
⎜⎜
= 0.00848 . Using the 5% guideline for cumbersome calculations
⎜⎝ 427 ⎠⎟⎟
10
b.
Copyright © 2014 Pearson Education, Inc.
122
178
52
Chapter 4: Probability
30. a.
492 + 306
= 0.99
492 + 8 + 306
b.
798 797
⋅
= 0.98
806 805
c.
⎛ 798 ⎞⎟
⎟ = 0.779
⎜⎜⎜
⎝ 806 ⎠⎟
25
31. a.
32.
0.99 ⋅ 0.99 + 0.99 ⋅ 0.01 + 0.01⋅ 0.99 = 0.9999
b.
0.99 ⋅ 0.99 = 0.9801
c.
The series arrangement provides better protection.
365 364 363
341
⋅
⋅
⋅ …⋅
= 0.431
365 365 365
365
Section 4-5
1.
a.
b.
Answers vary, but 0.98 is a reasonable estimate.
Answers vary, but 0.999 is a reasonable estimate.
2.
A conditional probability is a probability of an event calculated with the knowledge that some other event
has occurred.
3.
The probability that the polygraph indicates lying given that the subject is actually telling the truth.
4.
Confusion of the inverse is to think that the following two probabilities are the same: (1) the probability of
a polygraph indication of lying when the subject is telling the truth; (2) the probability of a subject telling
the truth when the polygraph indicates lying. Confusion of the inverse is to think that
P( A | B) = P( B | A) or to use one of those probabilities in place of the other.
5.
At least one of the five children is a boy.
6.
At least one of the five children is a girl.
7.
⎛9⎞
None of the digits is 0. ⎜⎜ ⎟⎟⎟ = 0.656
⎜⎝10 ⎠
8.
⎛9⎞
At least one of the digits is 7. 1− ⎜⎜ ⎟⎟⎟ = 0.344
⎜⎝10 ⎠
9.
⎛ 4⎞
1− ⎜⎜ ⎟⎟⎟ = 0.893 . The chance of passing is reasonably good
⎜⎝ 5 ⎠
31
or 0.969
32
31
or 0.969
32
4
4
10
10. 0.92 ⋅ 0.92 + 0.92 ⋅ 0.08 + 0.08 ⋅ 0.92 = 0.9936 . The probability of having to complete the exam without a
working calculator drops from 0.08 to 0.0064 (or 64 chances in 10000), so she does gain a substantial
increase in reliability.
11. 0.5 or 50%
12.
13. 1− (0.512) = 0.965
5
1
or 0.2
5
14. 1− (0.545) = 0.952 . The system cannot continue indefinitely because eventually there would be no
women to give birth.
5
Copyright © 2014 Pearson Education, Inc.
Chapter 4: Probability 53
15. 1− (1− 0.0423) = 0.122 . Given that the three cars are in the same family, they are not randomly selected
3
and there is a good chance that the family members have similar driving habits, so the probability might not
be accurate.
16. a.
1− (1− 0.124) = 0.484
5
b.
(0.124) = 0.0000293
c.
The detective is much better than average, or the detective was given five easy cases.
5
17. 1− (1− 0.67) = 0.988 . It is very possible that the result is not valid because it is based on data from a
4
voluntary response survey.
18. 1− (1− 0.41) = 0.998 . It is very possible that the result is not valid because it is based on data from a
12
voluntary response survey.
19.
20.
21.
90
or 0.0947. This is the probability of the test making it appear that the subject uses drugs when the
950
subject is not a drug user.
6
or 0.12. The employer would suffer by hiring a job applicant who appears to not use drugs, but the
50
applicant actually does use drugs.
6
or 0.00693. This result is substantially different from the result found in Exercise 20. The
866
probabilities P(subject uses drugs | negative test result) and P(negative test result | subject uses drugs) are
not equal.
22. a.
860
or 0.905
950
26. a.
b.
860
or 0.993
866
b.
c.
The results are different
23.
44
or 0.328
134
860
24.
or 0.993
866
25. a.
b.
10
or 1
10
27.
10
or 0.5
20
28.
1
or 0.25
4
29. a.
1
or 0.333
3
2
or 0.667
3
b.
1− (0.02) = 0.9996
2
1− (0.02) = 0.999992
3
5
or 0.5
10
30. 1− (0.0035) = 0.99998775 . Rounding the result to three significant digits would yield a probability of
1.00, but that would be misleading because it would suggest that it is certain that both radios will work.
Yes, the probability is high enough to ensure flight safety.
2
31. 1− (1− 0.134) = 0.684 . The probability is not low, so further testing of the individual samples will be
necessary for about 68% of the combined samples.
8
Copyright © 2014 Pearson Education, Inc.
54
Chapter 4: Probability
32. 1− (1− 0.005) = 0.0248 . The probability is quite low, indicating that further testing of the individual
5
samples will be necessary for about 2% of the combined samples.
33. a.
b.
35. a.
b.
c.
365 364 363
341
⋅
⋅
⋅ …⋅
= 0.431
365 365 365
365
34.
1
or 0.333
3
1− 0.431 = 0.569
0.8 ⋅ 0.01
= 0.0748
0.8 ⋅ 0.01 + 0.1⋅ 0.99
0.8
The estimate of 75% is dramatically greater than the actual rate of 7.48%. They exhibited confusion of
the inverse. A consequence is that they would unnecessarily alarm patients who are benign, and they
might start treatments that are not necessary.
Section 4-6
1.
The symbol ! is the factorial symbol that represents the product of decreasing whole numbers, as in
4! = 4 ⋅ 3 ⋅ 2 ⋅1 = 24 . Four people can stand in line 24 different ways.
2.
Combinations, because order does not count and five numbers are selected (from 1 to 39) without
replacement.
3.
Because repetition is allowed, numbers are selected with replacement, so neither of the two permutation
rules applies. The fundamental counting rule can be used to show that the number of possible outcomes is
1
10 ⋅10 ⋅10 ⋅10 = 10, 000 , so the probability of winning is
.
10, 000
4.
Only the fundamental counting rule applies.
5.
1 1 1 1
1
⋅ ⋅ ⋅ =
10 10 10 10 10, 000
6.
1 1 1 1 1 1 1 1 1
1
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =
10 10 10 10 10 10 10 10 10 1, 000, 000, 000
7.
1 1 1 1 1 1 1 1 1 1
1
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = =
9 8 7 6 5 4 3 2 1 9! 362,880
8.
1
1
=
7! 5040
9.
The number of combinations is
10.
1 1
1 1
1
⋅ + ⋅ =
52 51 52 51 1326
11.
1 1 1 1
1
⋅ ⋅ ⋅ =
. No 5,527,200 is too many possibilities to list.
50 49 48 47 5,527, 200
12.
8!
= 336
(8 − 3)!
13.
11!
= 34, 650
4!4!2!
27!
= 17,383,860 . Because that number is so large, it is not
(27 −12)!12!
practical to make a different CD for each possible combination.
14.
10!
= 50, 400
3!3!2!
Copyright © 2014 Pearson Education, Inc.
Chapter 4: Probability 55
15.
44!
1
= 7, 059, 052 . The probability is
(44 − 6)!6!
7, 059, 052
16.
53!
1
= 22,957, 480 . The probability is
(53 − 6)!6!
22,957, 480
17.
1
1
=
4! 24
19. a.
1
1
=
7! 5040
41!
1
= 749,398 . The probability is
(41− 5)!5!
749,398
b.
1
1
=
104 10,000
c.
$10,000
20. a.
18.
38!
1
= 575, 757 . The probability is
575, 757
(39 − 5)!5!
b.
1
1
=
103 1000
c.
$1000
21. a.
12!
= 11,880
12
( − 4)!
22. a.
b.
12!
= 495
(12 − 4)!4!
b.
16!
= 10, 461,394,944, 000
16
( −14)!
16!
= 120
(16 −14)!14!
1
1
c.
495
120
23. The number of possible “combinations” is 50 ⋅ 50 ⋅ 50 = 125, 000 . The fundamental counting rule can be
used. The different possible codes are ordered sequences of numbers, not combinations, so the name of
“combination lock” is not appropriate. Given that “fundamental counting rule lock” is a bit awkward, a
better name would be something like “number lock”.
c.
24.
1
1
=
. No, there are too many different possibilities
100 ⋅100 ⋅100 ⋅100 100, 000, 000
25. 5! = 120 ; AMITY;
26.
1
120
27.
5!
5!
5!
5!
+
+
+
= 26
5!0! 4!1! 3!2! 2!3!
6!
1
= 360 ; HARROW;
2
360
28. a.
c.
1
1
=
16
10
10, 000,000, 000, 000, 000
b.
1
1
=
12
10
1, 000, 000, 000, 000
1
1
=
. The number of possibilities (100,000,000) is still quite large, so there is no
108 100, 000, 000
1
1
reason to worry. 8 =
10
100, 000, 000
Copyright © 2014 Pearson Education, Inc.
56
Chapter 4: Probability
29.
5!
= 10
(5 − 2)!2!
30. a.
c.
3
1
or 0.188
b.
or 0.25
4
16
Trick question. There is no finite number of attempts, because you could continue to get the wrong
position every time.
31. 4 ⋅ 4 ⋅ 4 = 64
32. a.
31
⎛ 1 ⎞⎟
1
⎜⎜ ⎟ =
= 0.000000000466
⎜⎝ 2 ⎠⎟
2,147, 483, 648
31
b.
33.
35.
1
1
=
⋅
195,
249,
054
C
C
59 5 39 1
34.
1
1
=
⋅
175,711,536
C
C
56 5 46 15
2
2
=
. Yes, if everyone treated is of one gender while everyone in the placebo group is of the
252
C
10 5
opposite gender, you would not know if different reactions are due to the treatment or gender.
36. 8 ⋅ 2 ⋅ 9 = 144
37. 26 + 26 ⋅ 36 + 26 ⋅ 362 + 26 ⋅ 363 + 26 ⋅ 364 + 26 ⋅ 365 + 26 ⋅ 366 + 26 ⋅ 367 = 2, 095, 681, 645,538
38. a.
5
C2 = 10
c.
4! = 24
d. (n −1)!
n(n −1)
2
39. 12 ways: {25p, 1n 20p, 2n 15p, 3n 10p, 4n 5p, 5n, 1d 15p, 1d 1n 10p, 1d 2n 5p, 1d 3n, 2d 5p,2d 1n}
(Note: 25p represents 25 pennies, etc.)
b.
n
C2 =
40. The probability is 0. If 9 of the letters are in the correct envelopes, the 10th letter must also be in the correct
envelope, so it is impossible for the 10th letter to go into the wrong envelope.
Chapter Quick Quiz
1.
0 (not an option)
2.
10 − 3
= 0.7
10
3.
1 (all days contain the letter y)
4.
0.2 ⋅ 0.2 = 0.04
5.
Answers vary, but an answer such as 0.01 or
lower is reasonable
6.
288 + 224
512
=
= 0.61
201 + 126 + 288 + 224 839
7.
224 + 288 + 201 713
=
= 0.85
839
839
8.
126
= 0.15
839
9.
126 125
⋅
= 0.0224
839 839
10.
126
126
=
= 0.36
126 + 224 350
3.
58
= 0.806
58 + 14
Review Exercises
1.
392 + 58
= 0.438
1028
2.
392
= 0.41
392 + 564
Copyright © 2014 Pearson Education, Inc.
Chapter 4: Probability 57
4.
It appears that you have a substantially
better chance of avoiding prison if you enter
a guilty plea.
8.
72
578
14
+
−
= 0.619
1028 1028 1028
5.
392 + 58 392 + 564 392
+
−
= 0.986
1028
1028
1028
9.
392
= 0.381
1028
6.
450 449
⋅
= 0.191
1028 1027
10.
14
= 0.0136
1028
7.
72
71
⋅
= 0.00484
1028 1027
11. Answers vary, but DuPont data show that about 8% of cars are red, so any estimate between 0.01 and 0.2
would be reasonable.
12. a.
1− 0.35 = .65
b.
(0.35) = 0.015
c.
Yes, because the probability is so small
4
13. a.
1
365
c.
b.
31
365
d.
⎛
213 ⎞⎟
14. 1− ⎜⎜1−
⎟ = 0.0211 . No
⎜⎝ 100, 000 ⎠⎟
10
15.
1
1
=
5, 245, 786
42 C6
16.
1
1
=
575, 757
39 C5
17.
18.
Answers vary, but it is probably small, such
as 0.02
Yes
1 1 1
1
⋅ ⋅ =
10 10 10 1000
P = 1320 . The probability is
12 3
1
1320
Cumulative Review Exercises
1.
a.
b.
The mean of –8.9 years is not close to the value of 0 years that would be expected with no gender
discrepancy.
The median of –13.5 years is not close to the value of 0 years that would be expected with no gender
discrepancy.
(−20 −(−8.9))
2
+ (−15 − (−8.9)) + … + (−15 − (−8.9))
2
2
c.
s=
d.
s 2 = (10.6) = 113.2 years2
e.
Q1 = −15 years
f.
Q3 = −5 years
g.
The boxplot suggests that the data have a distribution that is skewed.
11
= 10.6 years.
2
Copyright © 2014 Pearson Education, Inc.
58 Chapter 4: Probability
2.
a.
b.
3.
100 − 77.5
= 1.94 . No, the pulse rate of 100 beats per minute is within 2 standard deviations away
11.6
from the mean, so it is not unusual.
z=
50 − 77.5
= −2.37 . Yes, the pulse rate of 50 beats per minutes is more than 2 standard deviations
11.6
away from the mean so it is unusual.
z=
c.
Yes, because the probability of
1
(or 0.0039) is so small.
256
d.
No, because the probability of
1
(or 0.125) is not very small.
8
a.
2346
= .46 = 46%
5100
b.
0.46 = 46%
c.
Stratified sample
4.
The graph is misleading because the vertical scale does not start at 0. The vertical scale starts at the
frequency of 500 instead of 0, so the difference between the two response rates is exaggerated. The graph
incorrectly makes it appear that “no” responses occurred 60 times more often than the number of “yes”
responses, but comparisons of the actual frequencies shows that the “no” responses occurred about four
times more often than the number of “yes” responses.
5.
a.
b.
c.
A convenience sample
If the students at the college are mostly from a surrounding region that includes a large proportion of
one ethnic group, the results will not reflect the general population of the United States.
0.35 + 0.4 = 0.75
d.
1− (0.6) = 0.64
2
6.
The straight-line pattern of the points suggests that there is a correlation between chest size and weight.
7.
a.
1
1
=
575, 757
39 C5
b.
1
19
c.
1
1
=
⋅
10,939,383
C
C
39 5 19 1
Copyright © 2014 Pearson Education, Inc.
Chapter 5: Discrete Probability Distributions 59
Chapter 5: Discrete Probability Distributions
Section 5-2
1.
The random variable is x, which is the number of girls in three births. The possible values of x are 0, 1, 2,
and 3. The values of the random variable x are numerical.
2.
The random variable is discrete because the number of possible values is 4, and 4 is a finite number. The
random variable is discrete if it has a finite number of values or a countable number of values.
3.
Table 5-7 does describe a probability distribution because the three requirements are satisfied. First, the
variable x is a numerical random variable and its values are associated with probabilities. Second,
ΣP( x) = 0.125 + 0.375 + 0.375 + 0.125 = 1 as required. Third, each of the probabilities is between 0 and 1
inclusive, as required.
4.
a.
Yes, because 0.0208 ≤ 0.05
b.
No, because 0.089 > 0.05
5.
7.
6. a. Not a random variable
a. Continuous random variable
b. Continuous random variable
b. Discrete random variable
c. Discrete random variable
c. Not a random variable
d. Discrete random variable
d. Discrete random variable
e. Not a random variable
e. Continuous random variable
f. Discrete random variable
f. Discrete random variable
Probability distribution with
μ = (0 ⋅ 0.0625) + (1⋅ 0.25) + (2 ⋅ 0.375) + (3⋅ 0.25) + (4 ⋅ 0.0625) = 2
σ = (0 − 2) 2 ⋅ 0.0625 + (1− 2) 2 ⋅ 0.25 + (2 − 2) 2 ⋅ 0.375 + (3 − 2)2 ⋅ 0.25 + (4 − 2)2 ⋅ 0.0625 = 1
8.
Probability distribution with
μ = (0 ⋅ 0.659) + (1⋅ 0.287) + (2 ⋅ 0.05) + (3⋅ 0.004) + (4 ⋅ 0.001) + (5 ⋅ 0) = 0.4
σ = (0 − 0.4) 2 ⋅ 0.659 + (1− 0.4) 2 ⋅ 0.287 + (2 − 0.4) 2 ⋅ 0.05 + ... + (4 − 0.4) 2 ⋅ 0.001 + (5 − 0.4) 2 ⋅ 0
= 0.6
9.
Not a probability distribution because the sum of the probabilities is 0.601, which is not 1 as required.
Also, Ted clearly needs a new approach.
10. Not a probability distribution because the responses are not values of a numerical random variable.
11. Probability distribution with
μ = (0 ⋅ 0.041) + (1⋅ 0.2) + (2 ⋅ 0.367) + (3⋅ 0.299) + (4 ⋅ 0.092) = 2.2
σ = (0 − 2.2)2 ⋅ 0.041 + (1− 2.2) 2 ⋅ 0.2 + (2 − 2.2) 2 ⋅ 0.367 + (3 − 2.2)2 ⋅ 0.299 + (4 − 2.2) 2 ⋅ 0.092 = 1
12. Probability distribution with
μ = (0 ⋅ 0.0) + (1⋅ 0.003) + (2 ⋅ 0.025) + (3 ⋅ 0.111) + (4 ⋅ 0.279) + (5 ⋅ 0.373) + (6 ⋅ 0.208) = 4.6
σ = (0 − 4.6) 2 ⋅ 0.0 + (1− 4.6) 2 ⋅ 0.003 + (2 − 4.6)2 ⋅ 0.025 + (3 − 4.6)2 ⋅ 0.111 + ... + (6 − 4.6) 2 ⋅ 0.208
=1
13. Not a probability distribution because the responses are not values of a numerical random variable. Also,
sum of the probabilities is 1.18 instead of 1 as required.
14. Not a probability distribution because the sum of the probabilities is 0.967 instead of 1 as required. The
discrepancy between 0.967 and 1 is too large to attribute to rounding errors.
15. μ = (0 ⋅ 0.001) + (1⋅ 0.01) + (2 ⋅ 0.044) + … + (9 ⋅ 0.01) + (10 ⋅ 0.001) = 5
σ=
(0 − 5) 2 ⋅ 0.001 + (1 − 5) 2 ⋅ 0.01 + (2 − 5) 2 ⋅ 0.044 + … + (9 − 5) 2 ⋅ 0.01 + (10 − 5) 2 ⋅ 0.001 = 1.6
Copyright © 2014 Pearson Education, Inc.
60
Chapter 5: Discrete Probability Distributions
16. Lower limit: μ − 2σ = 5 − 2(1.6) = 1.8 girls; Upper limit: μ + 2σ = 5 + 2(1.6) = 8.2 girls
Yes, 1 girl is an unusually low number of girls, because 1 girl is outside the range of usual values.
17. a.
b.
P ( X = 8) = 0.044
P ( X ≥ 8) = 0.044 + 0.01 + 0.001 = 0.055
c. The result from part (b)
d. No, because the probability of 8 or more girls is 0.055, which is not very low (less than or equal to
0.05)
18. a.
P ( X = 1) = 0.01
b.
P ( X ≤ 1) = 0.001 + 0.01 = 0.011
c.
d.
The result from part (b)
Yes, because the probability of 0.011 is very low (less than or equal to 0.05)
19. μ = (0 ⋅ 0.377) + (1⋅ 0.399) + (2 ⋅ 0.176) + (3 ⋅ 0.041) + (4 ⋅ 0.005) + (5 ⋅ 0) + (6 ⋅ 0) = 0.9
σ = (0 − 0.9) 2 ⋅ 0.377 + (1− 0.9) 2 ⋅ 0.399 + ... + (4 − 0.9)2 ⋅ 0.005 + (5 − 0.9)2 ⋅ 0 + (6 − 0.9)2 ⋅ 0
= 0.9
20. Lower limit: μ − 2σ = 0.9 − 2(0.9) = −0.9 ;Upper limit: μ + 2σ = 0.9 + 2(0.9) = 2.7
Yes; 3 is above the range of usual values, so 3 is an unusually high number of failures among 6 cars tested.
21. a.
b.
P( X = 3) = 0.041
P( X ≥ 3) = 0.041 + 0.005 + 0 + 0 = 0.046
c. The probability from part (b)
d. Yes, because the probability of three or more failures is 0.046 which is very low (less than or equal to
0.05)
22. a.
P( X = 1) = 0.399
b.
P( X ≤ 1) = 0.377 + 0.399 = 0.776
c.
d.
The result from part (b)
No, because the probability of 0.776 is not very low (less than or equal to 0.05)
23. a.
10 ⋅10 ⋅10 = 1000
b.
1
1000
c.
$500 − $1 = $499
d.
−$1⋅1 + $500 ⋅
e.
The $1 bet on the pass line in craps is better because its expected value of –1.4 cents is much greater
than the expected value of –50 cents for the Texas Pick 3 lottery.
24. a.
1
= −$0.50 = −50 cents
1000
10 ⋅10 ⋅10 ⋅ 0 = 10, 000
b.
1
10, 000
c.
$5000 − $1 = $4999
d.
−$1⋅1 + $5000 ⋅
e.
Because both bets have the same expected value of –50 cents, neither bet is better than the other.
1
= −$0.50 = −50 cents
10, 000
Copyright © 2014 Pearson Education, Inc.
Chapter 5: Discrete Probability Distributions 61
25. a.
b.
26. a.
b.
−$0.26 + $30 ⋅
5
33
− $5 ⋅ = −0.39
38
38
The bet on the number 27 is better because its expected value of –26 cents is greater than the expected
value of –39 cents for the other bet.
0.01 1
5 10
750, 000 1, 000, 000
+ + + +…+
+
= $131, 477.54
26
26 26 26
26
26
(0.01 − 131, 477.54) 2 ⋅
1
26
+ (1 − 131, 477.54) 2 ⋅
1
26
+ … + (1, 000, 000 − 131, 477.54) 2 ⋅
c.
Lower limit: $131, 477.54 − 2 ⋅ $253,584.47 = −$375, 691.40
Upper limit: $131, 477.54 + 2 ⋅ $253,584.47 = $638, 646.48
d.
Yes, because the values are above the range of usual values given in part (c)
1
26
= $253, 584.47
Section 5-3
1.
The given calculation assumes that the first two adults include Wal-Mart and the last three adults do not
include Wal-Mart, but there are other arrangements consisting of two adults who include Wal-Mart and
three who do not. The probabilities corresponding to those other arrangements should also be included in
the result.
2.
The format of Formula 5-5 requires that the probability p and the variable x refer to the same outcome. If p
is the probability of an adult including Wal-Mart, then x should count the number of people who include
Wal-Mart.
3.
Because the 30 selections are made without replacement, they are dependent, not independent. Based on
the 5% guideline for cumbersome calculations, the 30 selections can be treated as being independent. (The
30 selections constitute 3% of the population of 1000 responses, and 3% is not more than 5% of the
population.) The probability can be found by using the binomial probability formula.
4.
The 0+ indicates that the probability is a very small positive value. (The actual value is 0.00000296.) The
notation of 0+ does not indicate that the event is impossible; it indicates that the event is possible, but very
unlikely.
5.
Not binomial. Each of the weights has more than two possible outcomes.
6.
Binomial
7.
Binomial
8.
Not binomial. Each of the responses has more than two possible outcomes.
9.
Not binomial. Because the senators are selected without replacement, the selections are not independent.
(The 5% guideline for cumbersome calculations cannot be applied because the 40 selected senators
constitute 40% of the population of 100 senators, and that exceeds 5%.)
10. Not binomial. Because the senators are selected without replacement, they are not independent. . (The 5%
guideline for cumbersome calculations cannot be applied because the 10 selected senators constitute 10%
of the population of 100 senators, and that exceeds 5%.). Also, the numbers of terms have more than two
possible outcomes.
11. Binomial. Although the events are not independent, they can be treated as being independent by applying
the 5% guideline. The sample size of 380 is no more than 5% of the population of all smartphone users.
12. Binomial. Although the events are not independent, they can be treated as being independent by applying
the 5% guideline. The sample size of 427 is not more than 5% of the population of all women.
Copyright © 2014 Pearson Education, Inc.
62 Chapter 5: Discrete Probability Distributions
13. a.
b.
c.
14. a.
b.
c.
4 4 1
⋅ ⋅ = 0.128
5 5 5
{WWC, WCW, CWW}; 0.128 for each
0.128 ⋅ 3 = 0.384
1 1 9 9
⋅ ⋅ ⋅ = 0.0081
10 10 10 10
{MMXX, MXMX, MXXM, XXMM, XMXM, XMMX}; each has a probability of 0.0081
0.0081⋅ 6 = 0.0468
15.
5
C3 ⋅ 0.23 ⋅ 0.82 = 0.051
16.
5
C3 ⋅ 0.23 ⋅ 0.82 + 5 C4 ⋅ 0.2 4 ⋅ 0.81 + 5 C5 ⋅ 0.25 ⋅ 0.80 = 0.057
17.
5
C3 ⋅ 0.23 ⋅ 0.82 + 5 C4 ⋅ 0.2 4 ⋅ 0.81 + 5 C5 ⋅ 0.25 ⋅ 0.80 = 0.057
18.
5
C0 ⋅ 0.20 ⋅ 0.85 + 5 C1 ⋅ 0.21 ⋅ 0.84 + 5 C2 ⋅ 0.22 ⋅ 0.83 = 0.943
19.
5
C5 ⋅ 0.20 ⋅ 0.85 = 0.328
22.
16
C6 ⋅ 0.456 ⋅ 0.5510 = 0.168
20.
5
C5 ⋅ 0.25 ⋅ 0.80 = 0.00032 = 0+
23.
20
C16 ⋅ 0.4516 ⋅ 0.554 = 0.00125
21.
8
C3 ⋅ 0.453 ⋅ 0.555 = 0.257
24.
11
C9 ⋅ 0.459 ⋅ 0.552 = 0.0126
25. P( X ≥ 2) = 0.033 + 0.132 + 0.297 + 0.356 + 0.178 = 0.996 ; yes
26. P( X ≤ 5) = 0.000 + 0.004 + 0.033 + 0.132 + 0.297 + 0.356 = 0.822
27. P( X ≤ 2) = 0.000 + 0.004 + 0.033 = 0.037 ; yes, because the probability of 2 or fewer peas with green
pods is small (less than or equal to 0.05).
28. P( X ≥ 5) = 0.356 + 0.178 = 0.534 ; no, because the probability of 0.534 is not small (less than or equal to
0.05)
29. a.
6
C5 ⋅ 0.205 ⋅ 0.801 = 0.002 (Tech: 0.00154)
b.
6
C6 ⋅ 0.206 ⋅ 0.800 = 0+ 0+ (Tech: 0.000064)
c.
0.002 + 0 = 0.002 (Tech: 0.00160)
d.
Yes, the small probability from part (c) suggests that 5 is an unusually high number.
30. a.
7
C2 ⋅ 0.802 ⋅ 0.205 = 0.004 (Tech: 0.00430)
b.
7
C1 ⋅ 0.801 ⋅ 0.206 = 0+ 0+ (Tech: 0.000358)
c.
0.004 + 0 = 0.004 (Tech: 0.00467)
d.
Yes, the small probability from part (c) suggests that 2 is unusually low
Copyright © 2014 Pearson Education, Inc.
Chapter 5: Discrete Probability Distributions 63
31. a.
5
C0 ⋅ 0.200 ⋅ 0.805 = 0.328
b.
5
C1 ⋅ 0.201 ⋅ 0.804 = 0.410 0.410
c.
0.328 + 0.410 = 0.738 (Tech: 0.737)
d.
No, the probability from part (c) is not small, so 1 is not an unusually low number
32. a.
8
C8 ⋅ 0.908 ⋅ 0.100 = 0.430
b.
8
C7 ⋅ 0.907 ⋅ 0.101 = 0.383
c.
0.43 + 0.383 = 0.813
d.
No, the probability from part (c) is not small, so 7 is not unusually high
33.
34.
C12 ⋅ 0.4812 ⋅ 0.528 = 0.101 . No, because the probability of exactly 12 is 0.101, the probability of 12 or
more is greater than 0.101, so the probability of getting 12 or more is not very small, so 12 us not unusually
high
20
24
C6 ⋅ 0.256 ⋅ 0.7518 = 0.185 . The probability is not very small, so it is not unlikely
35.
C10 ⋅ 0.80510 ⋅ 0.1952 = 0.287 . No, because the flights all originate from New York, they are not randomly
selected flights, so the 80.5% on-time rate might not apply
36.
C20 ⋅ 0.72 20 ⋅ 0.2810 = 0.125 . No, because the probability of exactly 20 is 0.125, the probability of 20 or
fewer having those concerns is greater than 0.125, so the probability of getting 20 or fewer is not very
small, so 20 is not unusually low.
12
30
37. a.
38.
12
C0 ⋅ 0.450 ⋅ 0.5512 = 0.000766
b.
1− 0.000766 = 0.999
c.
12
d.
Yes, the very low probability of 0.00829 would suggest that the 45 share value is wrong
C0 ⋅ 0.450 ⋅ 0.5512 + 12 C1 ⋅ 0.451 ⋅ 0.5511 = 0.00829
C20 ⋅ (1− 0.0995) 20 ⋅ 0.09955 + 21 C21 ⋅ (1− 0.0995)21 ⋅ 0.09950 = 0.368 . With 21 booked passengers, there is
a probability of 0.368 that more than 19 passengers will show up, and that the flight will be overbooked. It
does not seem wise to schedule in such a way that the flights will be overbooked about 37% of the time.
21
39. a.
14
C13 ⋅ 0.513 ⋅ 0.51 = 0.000854
b.
14
C14 ⋅ 0.514 ⋅ 0.50 = 0.000061
c.
14
C14 ⋅ 0.514 ⋅ 0.50 + 14 C13 ⋅ 0.513 ⋅ 0.51 = 0.000916
d.
40. a.
b.
Yes. The probability of getting 13 girls or a result of 14 girls is 0.000916, so chance does not appear to
be a reasonable explanation for the 13 girls. Because 13 is an unusually high number of girls, it
appears that the probability of a girl is higher with the XSORT method, and it appears that the XSORT
method is effective.
20
C8 ⋅ 0.58 ⋅ 0.512 = 0.12
No. If the success rate is equal to 50% , it is likely (with probability 0.252) that we get 8 successes or a
result that is more extreme(fewer than 8 successes). This indicates that with a 50% success rate, the
occurrence of 8 successes in 20 challenges could be reasonably explained by chance.
41. 1− 24 C0 ⋅ 0.0060 ⋅ 0.99424 = 0.134 . It is not unlikely for such a combined sample to test positive.
42. 1− 16 C0 ⋅ 0.001140 ⋅ 0.9988624 = 0.0181 . It is unlikely for such a combined sample to test positive.
Copyright © 2014 Pearson Education, Inc.
64
Chapter 5: Discrete Probability Distributions
43.
C1 ⋅ 0.031 ⋅ 0.9739 + 40 C0 ⋅ 0.030 ⋅ 0.97 40 = 0.662 . The probability shows that about 2/3 of all shipments
will be accepted. With about 1/3 of the shipments rejected, the supplier would be wise to improve quality.
44.
40
C2 ⋅ 0.022 ⋅ 0.9828 + 30 C1 ⋅ 0.021 ⋅ 0.9829 + 30 C0 ⋅ 0.020 ⋅ 0.9830 = 0.978 . About 98% of all shipments will be
accepted. Almost all shipments will be accepted, and only 2% of the shipments will be rejected.
30
45. P( X = 5) = 0.06 (1− 0.06) = 0.0468
4
46.
12! 18 18 2
⋅ ⋅ ⋅ = 0.00485
5!4!3! 38 38 38
47. a.
P(4) =
6!
43!
(6 + 43)!
⋅
÷
= 0.000969
(6 − 4)!4! (43 − 6 + 4)!(6 − 4)! (6 + 43 − 6)!6!
b.
P(6) =
6!
43!
(6 + 43)!
⋅
÷
= 0.0000000715
(6 − 6)!6! (43 − 6 + 6)!(6 − 6)! (6 + 43 − 6)!6!
c.
P(0) =
6!
43!
(6 + 43)!
⋅
÷
= 0.436
(6 − 0)!0! (43 − 6 + 0)!(6 − 0)! (6 + 43 − 6)!6!
Section 5-4
1.
n = 270, p = 0.07, q = 0.93
2.
270 ⋅ 0.07 ⋅ 0.93 = 4.2 people. Yes, both expressions will yield the same result because they are
equivalent. They are equivalent because q = 1− p
3.
Variance is 150 ⋅ 0.933⋅ 0.067 = 9.4 executives2
4.
The mean of 140.0 executives is expressed with μ . The mean is calculated for the population of all groups
of 150 executives, not just one sample group. Because the mean is calculated for a population, it is a
parameter.
5.
μ = np = 60 ⋅ 0.2 = 12 correct guesses and σ = np(1− p) = 60 ⋅ 0.2 ⋅ 0.8 = 3.1 correct guesses.
Minimum: 12 − 2(3.1) = 5.8 correct guesses, maximum: 12 + 2(3.1) = 18.2 correct guesses.
6.
μ = np = 14 ⋅ 0.5 = 7 girls and σ = np(1− p) = 14 ⋅ 0.5 ⋅ 0.5 = 1.9 girls. Minimum: 7 − 2(1.9) = 3.2 ,
maximum 7 + 2(1.9) = 10.8 girls
7.
μ = np = 1013 ⋅ 0.66 = 668.6 worriers and σ = np(1− p ) = 1013⋅ 0.66 ⋅ 0.34 = 15.1 worriers. Minimum:
668.6 − 2(15.1) = 638.4 worriers, maximum: 668.6 + 2(15.1) = 698.8 worriers
8.
μ = np = 94 ⋅ 0.064 = 6 subjects with headaches and σ = np(1− p) = 94 ⋅ 0.064 ⋅ 0.936 = 2.4 subjects
with headaches. Minimum: 6 − 2(2.4) = 1.2 subjects with headaches, maximum: 6 + 2(2.4) = 10.8
subjects with headaches.
9.
a.
μ = np = 291⋅ 0.5 = 145.5 boys and σ = np(1− p) = 291⋅ 0.5 ⋅ 0.5 = 8.5 boys
b.
Yes. Using the range rule of thumb, the minimum value is 145.5 − 2(8.5) = 128.5 boys and the
maximum value is 145.5 + 2(8.5) = 162.5 boys. Because 239 boys is above the range of usual values,
it is unusually high. Because 239 boys is unusually high, it does appear that the YSORT method of
gender selection is effective.
Copyright © 2014 Pearson Education, Inc.
Chapter 5: Discrete Probability Distributions 65
10. a.
b.
11. a.
b.
12. a.
b.
13. a.
μ = np = 580 ⋅ 0.25 = 145 and σ = np(1− p) = 580 ⋅ 0.25 ⋅ 0.75 = 10.4
No, it is within the range of usual values of 145 − 2(10.4) = 124.2 and145 + 2(10.4) = 165.8 . It does
not provide strong evidence against Mendel’s theory.
μ = np = 100 ⋅ 0.20 = 20 and σ = np(1− p) = 100 ⋅ 0.2 ⋅ 0.8 = 4
No, because 25 orange M&Ms is within the range of usual values of
20 − 2(4) = 12 and 20 + 2(4) = 28 . The claimed rate of 20% does not necessarily appear to be wrong,
because that rate will usually result in 12 to 28 orange M&Ms (among 100), and the observed number
of orange M&Ms is within that range.
μ = np = 100 ⋅ 0.14 = 14 and σ = np(1− p) = 100 ⋅ 0.14 ⋅ 0.86 = 3.5
No, because 8 yellow M&Ms is within the range of usual values of 14 − 2(3.5) = 7 and
14 + 2(3.5) = 21 . The claimed rate of 14% does not necessarily appear to be wrong, because that rate
will usually result in 7 to 21 yellow M&Ms (among 100), and the observed number of yellow M&Ms
is within that range.
μ = np = 420, 095 ⋅ 0.00034 = 142.8 and σ = np(1− p) = 420, 095 ⋅ 0.00034 ⋅ 0.999666 = 11.9
b.
No, 135 is not unusually low or high because it is within the range of usual values
142.8 − 2(11.9) = 119 and 142.8 + 2(11.9) = 166.6
c.
Based on the given results, cell phones do not pose a health hazard that increases the likelihood of
cancer of the brain or nervous system.
14. a.
b.
15. a.
b.
16. a.
b.
17. a.
b.
18. a.
b.
μ = np = 280 ⋅ 0.5 = 140 and σ = np(1− p ) = 280 ⋅ 0.5 ⋅ 0.5 = 8.4
The result of 123 correct identifications is just outside the range of usual values of
140 − 2(8.4) = 123.2 and140 + 2(8.4) = 156.8 , but this indicates that 123 is unusually low. If the
touch therapists really had an ability to select the correct hand, they would have made more than 156.8
correct identifications. Therefore, they do not appear to have that ability.
μ = np = 2600 ⋅ 0.06 = 156 and σ = np(1− p) = 2600 ⋅ 0.06 ⋅ 0.94 = 12.1
The minimum usual frequency is 156 − 2(12.1) = 131.8 and the maximum is 156 + 2(12.1) = 180.2 .
The occurrence of r 178 times is not unusually low or high because it is within the range of usual
values.
μ = np = 2600 ⋅ 0.127 = 330.2 and σ = np(1− p) = 2600 ⋅ 0.127 ⋅ 0.813 = 17
The minimum usual frequency is 330.2 − 2(17) = 296.2 and the maximum is 330.2 + 2(17) = 364.2 .
The occurrence of e 290 times is unusually because it is below the range of usual values.
μ = np = 370 ⋅ 0.2 = 74 and σ = np(1− p ) = 370 ⋅ 0.2 ⋅ 0.8 = 7.7
The minimum usual number is 74 − 2(7.7) = 58.6 and the maximum is 74 + 2(7.7) = 89.4 . The value
of 90 is unusually high because it is above the range of usual values.
μ = np = 50 ⋅
1
1 37
= 1.3 and σ = np(1− p) = 50 ⋅ ⋅
= 1.1
38
38 38
The minimum usual value is 1.3 − 2(1.1) = −0.9 and the maximum is 1.3 + 2(1.1) = 3.5
The result of 0 wins is not unusually low because 0 wins is within the range of usual values.
Copyright © 2014 Pearson Education, Inc.
66 Chapter 5: Discrete Probability Distributions
μ = np = 30 ⋅
19. a.
1
1 364
= 0.0821918 and σ = np(1− p) = 30 ⋅
⋅
= 0.2862981
365
365 365
The minimum usual value is 0.0821918 − 2(0.2862981) = −0.4904044 and the maximum is
0.0821918 + 2(0.2862981) = 0.654788 . The result of 2 students born on the 4th of July would be
unusually high, because 2 is above the range of usual values.
b.
μ = np = 2600 ⋅
20. a.
1
= 0.000013 and
195, 249, 054
σ = np(1− p) = 2600 ⋅
1
195, 249, 053
⋅
= 0.003649
195, 249, 054 195, 249, 054
The minimum usual values is 0.000013 − 2(0.003649) = −0.007285 and the maximum is
0.000013 + 2(0.003649) = 0.007311 . It is unusual to buy a ticket each week for 50 years and win at
least once, because 1 win (or more) is outside the range of usual values.
b.
21. From the range of usual values we get μ = 60 and σ = 6 . Using the formulas for the mean and the
standard deviation we get p = 1−
σ2
μ
= 0.4 which leads to n = = 150 and q = 0.6
p
μ
22. 170.
23. The probability of selecting a girl out of 40 is given by P( X = n) =
10
Cn ⋅ 30 C12−n
the following table list
40 C12
the probabilities of selecting the number of girls from 0 to 10
Number of girls (X = n)
0
Probability P(X = n)
0.0154815
1
2
3
4
5
6
7
8
9
10
0.0977782
0.2420012
0.3073032
0.2200011
0.0918266
0.0223190
0.0030608
0.0002207
0.0000073
0.0000001
The mean is μ = ∑ [ x ⋅ P ( x)]
μ = 0 ⋅ 0.0154815 + 1⋅ 0.0977782 + 2 ⋅ 0.2420012 + 3 ⋅ 0.3073032 + 4 ⋅ 0.2200011 + 5 ⋅ 0.0918266 +
6 ⋅ 0.0223190 + 7 ⋅ 0.0030608 + 8 ⋅ 0.0002207 + 9 ⋅ 0.0000073 + 10 ⋅ 0.0000001 = 3
The standard deviation is σ =
σ=
∑[x
2
⋅ P ( x)] − μ 2
02 ⋅ 0.0154815 + 12 ⋅ 0.0977782 + 22 ⋅ 0.2420012 + 32 ⋅ 0.3073032 + 42 ⋅ 0.2200011 + 52 ⋅ 0.0918266 +
62 ⋅ 0.0223190 + 7 2 ⋅ 0.0030608 + 82 ⋅ 0.0002207 + 92 ⋅ 0.0000073 + 102 ⋅ 0.0000001− 33
= 1.3
Copyright © 2014 Pearson Education, Inc.
Chapter 5: Discrete Probability Distributions 67
Section 5-5
1.
535
= 0.929 , which is the mean number of hits per region. x = 2 , because we want the probability
576
that a randomly selected region had exactly 2 hits, and e = 2.71828 which is a constant used in all
applications of Formula 5 – 9.
μ=
194
= 4.2 , the standard deviation is σ = 4.2 = 2.1 and the variance is σ 2 = 4.2
46
2.
The mean is μ =
3.
With n = 50 , the first requirement of n ≥ 100 is not satisfied. With n = 50 and p = 0.001 the second
requirement of np ≤ 10 is satisfied. Because both requirements are not satisfied, we should not use the
Poisson distribution as an approximation to the binomial.
4.
With n = 100 and p = 0.001 the two requirements are satisfied. For 101 wins, the Poisson approximation
gives a small positive probability, but the actual probability of 101 wins is 0 since it is impossible to get
101 wins in 100 tries.
5.
P(0) =
8.50 ⋅ e−8.5
= 0.000203 ; Yes it is unlikely.
0!
6.
P(6) =
8.56 ⋅ e−8.5
= 0.107 ; No it is not unlikely
6!
7.
P(10) =
8.510 ⋅ e−8.5
= 0.11 ; No it is not unlikely
10!
8.
P(12) =
8.512 ⋅ e−8.5
= 0.0604 ; No it is not unlikely
12!
9.
a.
μ=
b.
1−
c.
Yes. Based on the result in part (b), we are quite sure (with probability 0.998) that there is at least one
earthquake measuring 6.0 or higher on the Richter scale, so there is a very low probability (0.002) that
there will be no such earthquake in a year.
10. a.
268
= 6.5
41
6.50 ⋅ e−6.5
= 0.998
0!
μ=
5469
= 133.4
41
133.4133 ⋅ e−133.4
= 0.0346
133!
b.
P(133) =
c.
No. Although the probability of exactly 133 earthquakes measured at 6.0 or higher on the Richter scale
is quite small (0.0346), the number 133 is so close to the mean of 133.4 that this year would be quite
ordinary, and it would not be unusual.
11. a.
b.
μ=
22713
= 62.2
365
P(50) =
62.250 ⋅ e62.2
= 0.0155
50!
Copyright © 2014 Pearson Education, Inc.
68
Chapter 5: Discrete Probability Distributions
12. μ =
196
= 9.8
20
a.
P(0) =
9.80 ⋅ e−9.8
= 0.497
0!
c.
P(2) =
9.82 ⋅ e−9.8
= 0.122
2!
b.
P(1) =
9.81 ⋅ e−9.8
= 0.348
1!
d.
P(3) =
9.83 ⋅ e−9.8
= 0.0284
3!
e.
13. a.
b.
c.
14. a.
b.
9.84 ⋅ e−9.8
= 0.00497 . The expected frequencies of 139, 97, 34, 8 , and 1.4 compare
4!
reasonably well to the actual frequencies, so the Poisson distribution does provide good results.
P(4) =
P(2) =
0.9292 ⋅ e−0.929
= 0.17
2!
The expected number of regions with exactly 2 hits is 98.2
The expected number of regions with 2 hits is close to 93, which is the actual number of regions with 2
hits.
μ = 12429 ⋅ 0.000011 = 0.1367
0.1370 ⋅ e−0.137
0.1371 ⋅ e−0.137
= 0.872 and P(1) =
= 0.119 . So the probability of 0 or 1 is
0!
1!
0.872 + 0.119 = 0.991
P(0) =
c.
1− 0.991 = 0.009
d.
No, the probability of more than one case is extremely small, so the probability of getting as many as
four cases is even smaller.
15. a.
b.
16. a.
b.
17. a.
b.
30.426 ⋅ e−30.4
= 0.0558 . The expected value is 34 ⋅ 0.0558 = 1.9 cookies. The expected
26!
number of cookies is very close to the actual number of cookies with 26 chocolate chips which is 2.
P(26) =
30.430 ⋅ e−30.4
= 0.0724 . The expected value is 34 ⋅ 0.0724 = 2.5 cookies. The expected
30!
number of cookies is very different from the actual number of cookies with 26 chocolate chips which
is 6.
P(30) =
19.618 ⋅ e−19.6
= 0.0875 . The expected value is 40 ⋅ 0.0875 = 3.5 cookies. The expected
18!
number of cookies is not very close to the actual number of cookies with 18 chocolate chips which is 5.
P(18) =
19.621 ⋅ e−19.6
= 0.0826 . The expected value is 40 ⋅ 0.0826 = 3.3 cookies. The expected
21!
number of cookies is very close to the actual number of cookies with 21 chocolate chips which is 3.
P(21) =
No. With n = 12 and p = 1
the requirement of n ≥ 100 is not satisfied, so the Poisson distribution
6
is not a good approximation to the binomial distribution.
No. The Poisson distribution approximation to the binomial distribution yields
3
9
⎛ 1 ⎞⎟ ⎛ 5 ⎞⎟
23 ⋅ e−2
⎜
⎜
= 0.18 and the binomial distribution yields P(3) = 12 C3 ⋅ ⎜ ⎟⎟ ⋅ ⎜ ⎟⎟ = 0.197 . The
P(3) =
⎝⎜ 6 ⎠ ⎝⎜ 6 ⎠
3!
Poisson approximation of 0.18 is too far from the correct result of 0.197.
Copyright © 2014 Pearson Education, Inc.
Chapter 5: Discrete Probability Distributions 69
Chapter Quick Quiz
1.
Yes
2.
1
100 ⋅ = 20
5
3.
σ = 100 ⋅ 0.2 ⋅ 0.8 = 4
4.
The range of usual values has a minimum value of 200 − 2 ⋅10 = 180 and a maximum value
of 200 + 2 ⋅10 = 220 . Therefore, 232 girls in 400 is an unusually high number of girls since it is outside the
range of usual values.
5.
The range of usual values has a minimum value of 200 − 2 ⋅10 = 180 and a maximum value
of 200 + 2 ⋅10 = 220 . Therefore, 185 girls in 400 is not an unusually high number of girls since it is inside
the range of usual values.
6.
Yes. The sum of the probabilities is 0.999 and it can be considered to be 1.
7.
0+ indicates that the probability is a very small positive number. It does not indicate that it is impossible
for none of the five flights to arrive on time.
8.
P( x ≥ 3) = 0.198 + 0.409 + 0.338 = 0.945
9.
μ = 0 ⋅ 0 + 1⋅ 0.006 + 2 ⋅ 0.048 + 3 ⋅ 0.198 + 4 ⋅ 0.409 + 5 ⋅ 0.338 = 4.022 and
σ = 02 ⋅ 0 + 12 ⋅ 0.006 + 22 ⋅ 0.048 + 32 ⋅ 0.198 + 42 ⋅ 0.409 + 52 ⋅ 0.338 − 4.0222 = 0.893
The range of usual values is from 2.236 to 5.808. Since zero is outside the range of usual values it is an
unusually low number.
10. μ = 0 ⋅ 0 + 1⋅ 0.006 + 2 ⋅ 0.048 + 3 ⋅ 0.198 + 4 ⋅ 0.409 + 5 ⋅ 0.338 = 4.022 and
σ = 02 ⋅ 0 + 12 ⋅ 0.006 + 22 ⋅ 0.048 + 32 ⋅ 0.198 + 42 ⋅ 0.409 + 52 ⋅ 0.338 − 4.0222 = 0.893
The range of usual values is from 2.236 to 5.808. Since 5 is inside the range of usual values it is not an
unusually high number.
Review Exercise
1.
P( X = 0) = 6 C0 ⋅ 0.40 ⋅ 0.66 = 0.0467
3.
μ = 600 ⋅ 0.4 = 240 and σ = 600 ⋅ 0.4 ⋅ 0.6 = 12 . The range of usual values has a minimum of
240 − 2 ⋅12 = 216 and a maximum value of 240 + 2 ⋅12 = 264 . The result of 200 with brown eyes is
unusually low.
4.
The probability of 239 or fewer (0.484) is relevant for determining whether 239 is an unusually low
number. Because that probability is not very small, it appears that 239 is not an unusually low number of
people with brown eyes.
5.
Yes. The three requirements are satisfied. There is a numerical random variable x and its values are
associated with corresponding probabilities. The sum of the probabilities is 1.001, so the sum is 1 when we
allow for a small discrepancy due to rounding. Also, each of the probability values is between 0 and 1
inclusive.
6.
μ = 0 ⋅ 0.674 + 1⋅ 0.28 + 2 ⋅ 0.044 + 3 ⋅ 0.003 + 4 ⋅ 0 = 0.4 and
2.
P( X = 4) = 6 C4 ⋅ 0.44 ⋅ 0.62 = 0.138
σ = 02 ⋅ 0.674 + 12 ⋅ 0.28 + 22 ⋅ 0.044 + 32 ⋅ 0.003 + 42 ⋅ 0 − 0.42 = 0.6
The range of usual values has a minimum of 0.4 − 2 ⋅ 0.6 = −0.8 and a maximum value
of 0.4 + 2 ⋅ 0.6 = 1.6 . Yes, 3 is an unusually high number of males with tinnitus among four randomly
selected males.
7.
The sum of the probabilities is 0.902 which is not 1as required. Because the three requirements are not
satisfied, the given information does not describe a probability distribution.
Copyright © 2014 Pearson Education, Inc.
70
8.
9.
Chapter 5: Discrete Probability Distributions
1
1
1
1
1
$75 ⋅ + $300 ⋅ + $75, 000 ⋅ + $500, 000 ⋅ + $1, 000, 000 ⋅ = $315, 075 . Because the offer is well
5
5
5
5
5
below her expected value, she should continue the game (although the guaranteed prize of $193000 had
considerable appeal).
1
1
1
+ $100, 000 ⋅
+ $25, 000 ⋅
900, 000, 000
110, 000, 000
110, 000, 000
1
1
+$5000 ⋅
+ $2500 ⋅
= $0.012
36, 667, 000
27,500,000
a.
$1, 000, 000 ⋅
b.
$0.012 minus the cost of the postage stamp. Since the expected value of winning is much smaller than
the cost of a postage stamp, it is not worth entering the contest.
10. a.
μ=
18
= 0.6
30
0.60 ⋅ e−0.6
= 0.549
0!
b.
P(0) =
c.
30 ⋅ 0.549 = 16.5 days
d.
The expected number of days is 16.5, and that is reasonably close to the actual number of days which
is 18.
Cumulative Review Exercises
1.
The mean is x =
b.
c.
The median is 24.2 hours
The range is 26.9 − 22.2 = 4.7 hours
d.
The standard deviation is
s=
2.
22.2 + 24.8 + 24.2 + 26.9 + 23.8
= 24.4 hours
5
a.
(22.2 − 24.4)2 + (24.8 − 24.4)2 + (24.2 − 24.4) 2 + (26.9 − 24.4) 2 + (23.8 − 24.4) 2
= 1.7
5
e.
f.
The variance is 2.9 hours2
The minimum is 24.4 − 2 ⋅1.7 = 21 hours and the maximum is 24.4 + 2 ⋅1.7 = 27.8 hours.
g.
h.
i.
j.
No, because none of the times are outside the range of the usual values
Ratio
Continuous
The given times come from countries with very different population sizes, so it does not make sense to
treat the given times equally. Calculations of statistics should take the different population sizes into
account. Also, the sample is very small, and there is no indication that the sample is random.
a.
1 1 1 1
1
⋅ ⋅ ⋅ =
= 0.0001
10 10 10 10 10, 000
b.
e.
x
–$1
P(x)
0.9999
$4999
0.0001
c.
365 ⋅ 0.0001 = 0.0365
d.
P(1) =
−$1⋅ 0.9999 + $4999 ⋅ 0.0001 = −$0.50 or –50 cents.
Copyright © 2014 Pearson Education, Inc.
0.03651 ⋅ e−0.0365
= 0.0352
1!
Chapter 5: Discrete Probability Distributions 71
3.
a.
121 + 51
= 0.282
611
e.
121 + 51 121 + 51
⋅
= 0.0792
611
611
b.
121
= 0.303
121 + 279
f.
121 + 279 121 + 51 121
+
−
= 0.738
611
611
611
c.
51
= 0.242
51 + 160
g.
d.
51
= 0.297
51 + 121
⎛ 121 ⎞⎟
⎜⎜
⎜⎝ 611⎠⎟⎟
= 0.703
⎛172 ⎞⎟
⎜⎜
⎟
⎜⎝ 611⎠⎟
4.
Because the vertical scale begins at 60 instead of 0, the difference between the two amounts is exaggerated.
The graph makes it appear that men’s earnings are roughly twice those of women, but men earn roughly 1.2
times the earnings of women.
5.
a.
b.
Frequency distribution or frequency table
Probability distribution
c.
x=
d.
e.
6.
0 ⋅ 9 + 1⋅ 7 + 2 ⋅12 + 3⋅10 + 4 ⋅10 + 5 ⋅11 + 6 ⋅ 8 + 7 ⋅ 8 + 8 ⋅14 + 9 ⋅11
= 4.7
9 + 7 + 12 + 10 + 10 + 11 + 8 + 8 + 14 + 11
This value is a statistic
μ = 0 ⋅ 0.1 + 1⋅ 0.1 + 2 ⋅ 0.1 + 3 ⋅ 0.1 + 4 ⋅ 0.1 + 5 ⋅ 0.1 + 6 ⋅ 0.1 + 7 ⋅ 0.1 + 8 ⋅ 0.1 + 9 ⋅ 0.1 = 4.5 . This value is
a parameter
The random generation of 1000 digits should have a mean close to 4.5 from part (d). The mean of 4.5
is the mean for the population of all random digits; so samples will have means that tend to center
about 4.5
C4 ⋅ 0.14 ⋅ 0.912 = 0.0514
a.
16
b.
1− 16 C0 ⋅ 0.10 ⋅ 0.916 = 0.815
c.
This is a voluntary response sample. This suggests that the results might not be valid, because those
with a strong interest in the topic are more likely to respond.
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 73
Chapter 6: Normal Probability Distributions
Section 6-2
1.
The word “normal” has a special meaning in statistics. It refers to a specific bell-shaped distribution that
can be described by Formula 6-1.
2.
3.
The mean and standard deviation have values of μ = 0 and σ = 1
4.
The notation zα represents the z score that has an area of α to its right.
5.
0.2 (5 −1.25) = 0.75
8.
0.2 ( 4.5 −1.5) = 0.60
6.
0.2 (0.75 − 0) = 0.15
9.
P ( z < 0.44) = 0.6700
7.
0.2 (3 −1) = 0.40
10. P ( z > −1.04) = 0.8508
11. P (−0.84 < z < 1.28) = P ( z < 1.28) − P ( z < −0.84) = 0.8997 − 0.2005 = 0.6992 (Tech: 0.6993)
12. P (−1.07 < z < 0.67) = P ( z < 0.67) − P ( z < −1.07) = 0.7486 − 0.1423 = 0.6063
13. z = 1.23
14. z = −0.51
15. z = −1.45
16. z = 0.82
20. P ( z < 1.96) = 0.9750
21. P ( z > 0.82) = 1− 0.7939 = 0.2061
22. P ( z > 1.82) = 1− 0.9656 = 0.0344
17. P ( z < −2.04) = 0.0207
23. P ( z > −1.50) = 1− 0.0668 = 0.9332
18. P ( z < −0.19) = 0.4247
24. P ( z > −0.84) = 1− 0.2005 = 0.7995
19. P ( z < 2.33) = 0.9901
25. P (0.25 < z < 1.25) = P ( z < 1.25) − P ( z < 0.25) = 0.8944 − 0.5987 = 0.2957 (Tech: 0.2956)
26. P (1.23 < z < 2.37) = P ( z < 2.37) − P ( z < 1.23) = 0.9911− 0.8907 = 0.1004 0.1004 (Tech: 0.1005)
27. P (−2.75 < z < −2.00) = P ( z < −2.00) − P ( z < −2.75) = 0.0228 − 0.0030 = 0.0198
28. P (−1.93 < z < −0.45) = P ( z < −0.45) − P ( z < −1.93) = 0.3264 − 0.0268 = 0.2996
29. P (−2.20 < z < 2.50) = P ( z < 2.50) − P ( z < −2.20) = 0.9938 − 0.0139 = 0.9799
30. P (−0.62 < z < 1.78) = P ( z < 1.78) − P ( z < −0.62) = 0.9625 − 0.2676 = 0.6949 (Tech: 0.6948)
Copyright © 2014 Pearson Education, Inc.
74 Chapter 6: Normal Probability Distributions
31. P (−2.11 < z < 4.00) = P ( z < 4.00) − P ( z < −2.11) = 0.9999 − 0.0174 = 0.9825 (Tech: 0.9827)
32. P (−3.90 < z < 2.00) = P ( z < 2.00) − P ( z < −3.90) = 0.9772 − 0.0001 = 0.9771 0. (Tech: 0.0772)
33. P ( z < 3.65) = 0.9999
39. P2.5 = −1.96 and P97.5 = 1.96
34. P ( z > −3.80) = 0.9999 0.9999
40. P0.5 = −2.575 and P99.5 = 2.575
35. P ( z < 0) = 0.5000
41. z0.025 = 1.96
42. z0.01 = 2.33
36. P ( z > 0) = 0.5000
43. z0.05 = 1.645
37. P90 = 1.28
44. z0.03 = 1.88
38. P5 = −1.645
45. P (−1 < z < 1) = P ( z < 1) − P ( z < −1) = 0.8413 − 0.1587 = 0.6826 = 68.26% (Tech: 68.27%)
46. P (−2 < z < 2) = P ( z < 2) − P ( z < −2) = 0.9772 − 0.0228 = 0.9544 = 95.44% (Tech: 95.45%)
47. P (−3 < z < 3) = P ( z < 3) − P ( z < −3) = 0.9987 − 0.0013 = 0.9974 = 99.74% (Tech: 99.73%)
48. P (−3.5 < z < 3.5) = P ( z < 3.5) − P ( z < −3.5) = 0.9999 − 0.0001 = 0.9998 = 99.98% (Tech: 99.95%)
49. a.
P (−1 < z < 1) = P ( z < 1) − P ( z < −1) = 0.8413 − 0.1587 = 0.6826 = 68.26% (Tech: 68.27%)
b.
P ( z < −2 or z > 2) = P ( z < −2) + P ( z > 2) = 0.0228 + 0.0228 = 0.0456 = 4.56%
c.
P (−1.96 < z < 1.96) = P ( z < 1.96) − P ( z < −1.96) = 0.975 − 0.020 = 0.9500 = 95%
d.
P (−2 < z < 2) = P ( z < 2) − P ( z < −2) = 0.9772 − 0.0228 = 0.9544 = 95.44% (Tech: 95.45%)
e.
P ( z > 3) = 1− P ( z < 3) = 1− 0.9987 = 0.0013 = 0.13%
50. a.
b.
μ = 2.5 min. and σ =
The probability is
1
3
5
12
= 1.4 min.
or 0.5774, and it is very different from the probability of 0.6826 that would be
obtained by incorrectly using the standard normal distribution. The distribution does affect the results
very much.
Section 6-3
a.
μ = 0 and σ = 1
b.
The z scores are numbers without units of measurements
2.
a.
b.
c.
d.
The area equals the maximum probability value of 1.
The median is the middle value and for normally distributed scores that is also the mean, which is 100.
The mode is also 100.
The variance is the square of the standard deviation which is 225.
3.
The standard normal distribution has a mean of 0 and a standard deviation of 1, but a nonstandard normal
distribution has a different value for one or both of those parameters.
4.
No. Randomly generated digits have a uniform distribution, but not a normal distribution. The probability
3
= 0.3
of a digit less than 3 is
10
1.
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 75
5.
z x =118 =
118 −100
= 1.2 which has an area of 0.8849 to the left of it
15
6.
z x =90 =
91−100
= −0.6 which has an area of 0.7257 to the right of it
15
7.
8.
9.
133 −100
110 −100
= 2.2 which has an area of 0.9861 to the left of it. z x =79 =
= −1.4 which has
15
15
an area of 0.0808 to the left of it. The area between the two scores is 0.9861− 0.0808 = 0.9053 .
z x =133 =
124 −100
112 −100
= 1.6 which has an area of 0.9452 to the left of it. z x =112 =
= 0.8 which has
15
15
an area of 0.7881 to the left of it. The area between the two scores is 0.9452 − 0.7881 = 0.1571 .
z x =124 =
z = 2.44 , which means x = 2.44 ⋅15 + 100 = 136
10. z = 1 , which mean x = 1⋅15 + 100 = 115
11. z = −2.07 , which means x = −2.07 ⋅15 + 100 = 69
12. z = 1.33 , which means x = 1.33 ⋅15 + 100 = 120
13. z x =85 =
85 −100
= −1 , which has an area of 0.1587 to the left of it
15
14. z x =70 =
70 −100
= −2 , which has an area of 0.9772 to the right of it
15
90 −100
110 −100
= −0.67 which has an area of 0.2514 to the left of it. z x =110 =
= 0.67 which
15
15
has an area of 0.7486 to the left of it. The area between the two scores is 0.7486 − 0.2514 = 0.4972 .
(Tech: 0.4950)
15. z x =90 =
16. z x =120 = 1.33 which has an area of 0.9082 to the left of it.
110 −100
= 0.67 which has an area of 0.7486 to the left of it. The area between the two scores is
15
0.9082 − 0.7486 = 0.1596 (Tech: 0.1613)
z x =110 =
17. z = 1.27 which means the score is x = 1.27 ⋅15 + 100 = 119
18. z = −0.67 which means the score is x = −0.67 ⋅15 + 100 = 90
19. z = 0.67 which means the score is x = 0.67 ⋅15 + 100 = 110
20. z = 2.07 which means the minimum score is x = 2.07 ⋅15 + 100 = 131
21. a.
b.
78 − 63.8
= 5.46 which has an area of 0.9999 to the left of it.
2.6
62 − 63.8
= −0.69 which has an area of 0.2451 to the left of it. Therefore, the percentage of
z x =62 =
2.6
qualified women is 0.9999 − 0.2451 = 0.7548 or 75.48%. (Tech 95.56%.) Yes, about 25% of women
are not qualified because of their heights.
z x =78 =
78 − 69.5
62 − 69.5
= 3.54 which has an area of 0.9999 to the left of it. z x =62 =
= −3.13
2.4
2.4
which has an area of 0.0009 to the left of it. Therefore, the percentage of men is
0.9999 − 0.0009 = 0.9990 or 99.90%. (Tech: 99.89%.) No, only about 0.1% of men are not qualified
because of their heights.
z x =78 =
Copyright © 2014 Pearson Education, Inc.
76 Chapter 6: Normal Probability Distributions
21. (continued)
c.
The z score with 2% to the left of it is –2.04 which corresponds to a height
of x = −2.04 ⋅ 2.6 + 63.8 = 58.5 in. The z score with 2% to the right of it is 2.04 which corresponds to
a height of x = 2.04 ⋅ 2.6 + 63.8 = 69.1 in.
d.
The z score with 1% to the left of it is –2.33 which corresponds to a height of
x = −2.33 ⋅ 2.4 + 69.5 = 63.9 in. The z score with 1% to the right of it is 2.33 which corresponds to a
height of x = 2.33 ⋅ 2.4 + 69.5 = 75.1 in.
22. a.
b.
c.
23. a.
b.
64 − 63.8
77 − 63.8
= 0.08 and z x =77 =
= 5.08 . The area between the two z scores is
2.6
2.6
0.9999 − 0.5319 = 0.4680 or 46.80%. (Tech: 46.93%.)
z x =64 =
64 − 69.5
77 − 69.5
= −2.29 and z x =77 =
= 3.13 . The area between the two z scores is
2.4
2.4
0.9991− 0.0110 = 0.9881 or 98.81%.
z x =64 =
The z score with 3% to the left of it for women is –1.88 which corresponds to a height of
−1.88 ⋅ 2.6 + 63.8 = 58.9 in. The z score with 3% to the right of it for men is 1.875 or 1.88 which
corresponds to a height of 1.88 ⋅ 2.4 + 69.5 = 74 in
The height minimum is 4 ⋅12 + 8 = 56 in. and the height maximum is 6 ⋅12 + 3 = 75 in. The z score for
56 − 63.8
= −3 , and the z score for women for the maximum is
women for the minimum is
2.6
75 − 63.8
= 4.31 . The area between the z scores is 0.9999 − 0.0013 = 0.9986 or 99.86%
2.6
The z score for men for the minimum is
56 − 69.5
= −5.63 , and the z score for men for the maximum
2.4
56 − 69.5
= 2.29 . The area between the z scores is 0.9890 − 0.0001 = 0.9898 or 98.98%. (Tech:
2.4
98.90%.)
The z score with 5% for women to the left of it is –1.65 which corresponds to a height of
−1.65 ⋅ 2.6 + 63.8 = 59.5 in. the z score with 5% of men to the right of it is 1.625 which corresponds to
a height of 1.625 ⋅ 2.4 + 69.5 = 73.4 in.
is
c.
24. a.
b.
c.
d.
25. a.
51.6 − 69.5
= −7.46 which has an area of 0.0001 or 0.01% to
2.4
the left of it meaning that practically no man can fit without bending. (Tech: 0.00%.)
The z score for the minimum height is
51.6 − 63.8
= −4.69 which has an area of 0.0001 or 0.01% to
2.6
the left of it meaning that practically no women can fit without bending. (Tech: 0.00%.)
The door design is very inadequate, but the jet is relatively small and seats only six people. A much
higher door would require such major changes in the design and cost of the jet, that the greater height
is not practical.
The z score for 60% is 0.25 which corresponds to a height 0.25 ⋅ 2.4 + 69.5 = 70.1 in for men.
The z score for the minimum height is
The z score for 174 lb. is
174 −182.9
= −0.22 which has an area of 0.4129 to the left of it. (Tech:
40.8
0.4137.)
b.
3500
= 25 people
140
c.
3500
= 19.14 , so 19 people
182.9
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 77
25. (continued)
d.
26. a.
b.
The mean weight is increasing over time, so safety limits must be periodically updated to avoid an
unsafe condition
The z score that has 95% of the area to the left of it is 1.67 which corresponds to a height of
1.67 ⋅1.2 + 21.4 = 23.4 in. If there is clearance for 95% of males, there will certainly be clearance for
all women in the bottom 5%
Men’s z score is
23.5 − 21.4
= 1.75 and that has an area of 0.9599 or 99.95%. Women’s z score is
1.2
23.5 −19.6
= 3.55 and that has an area of 0.9999 or 99.99%. (Tech 99.98%.) The table will fit almost
1.1
everyone except about 4% of the men with the largest sitting knee heights
27. a.
b.
28. a.
b.
29. a.
308 − 268
= 2.67 which corresponds to a probability of 0.0038
15
or 0.38%. Either a very rare event occurred or the husband is not the father.
The z score corresponding to 3% is –1.87 which corresponds to a pregnancy of
−1.87 ⋅15 + 268 = 240 days
The z score for a 308 day pregnancy is
100.6 − 98.2
= 3.87 which corresponds to a an area of
0.62
1− 0.9999 = 0.0001 = 0.01% to the right of it.; yes
The z score for a probability of 5% is 1.65 which corresponds to a temperature of
1.65 ⋅ 0.62 + 98.2 = 99.22 degrees.
The z score for a temperature of 100.6 is
The z score for an earthquake of magnitude 2 is
2 −1.184
= 1.39 which is 0.9177 or 91.77% of
0.587
earthquakes. (Tech: 99.78%.)
4 −1.184
= 4.80 which is 0.0001 or 0.01% of earthquakes. (Tech: 0.00%.)
0.587
b.
The z score is
c.
The z score for 95% of earthquakes is 1.645 or 1.65 which corresponds to an earthquake magnitude of
1.645 ⋅ 0.587 + 1.184 = 2.15 , so not all earthquakes about the 95th percentile will cause items to shake.
30. The z score for a probability of 99% is 2.33 which corresponds to a hip breadth of
2.59 ⋅ 0.00436 + 0.78386 = 0.7951 in.
31. The z score for P1 is –2.33 which corresponds to a count of −2.33 ⋅ 2.6 + 24 = 17.9 chocolate chips. (Tech:
18 chocolate chips.) The z score for P99 is 2.33 which corresponds to a count of 2.33⋅ 2.6 + 24 = 30.1
chocolate chips. (Tech: 10.0 chocolate chips.) The values can be used to identify cookies with an unusually
low number of chocolate chips or an unusually high number of chocolate chips, so those numbers can be
used to monitor the production process to ensure that the numbers of chocolate chips stay within reasonable
limits.
32. a.
b.
5.64 − 5.67
= −0.5 which has a corresponding probability of
0.06
5.7 − 5.67
= 0.5 which has a corresponding
0.3085 and the maximum weight has a z score of
0.06
probability of 0.6915. Therefore, the percentage of quarters rejected is
1− (0.6915 − 0.3085) = 0.6170 . (Tech: 61.71%.) That percentage is too high because most quarters
will be rejected.
The minimum weight has a z score of
The z score for a probability of the top 2.5% and the bottom 2.5% is 2 and –2 respectively. Therefore
the weight minimum is −2 ⋅ 0.06 + 5.67 = 5.5 g and the weight maximum is 2 ⋅ 0.06 + 5.67 = 5.79 g
Copyright © 2014 Pearson Education, Inc.
78 Chapter 6: Normal Probability Distributions
33. a.
The mean is 67.25 beats per minute and the standard deviation is 10.335 beats per minute. The
histogram for the data confirms that the distribution is roughly normal.
9
8
Frequency
7
6
5
4
3
2
1
0
45
50
55
60
65
70
75
80
85
90
PULSE
b.
34. a.
The z score for the bottom 2.5% is –1.95 which corresponds to a pulse of −1.95 ⋅10.335 + 67.25 = 47
beats per minute, and the z score for the top 2.5% is 1.95 which corresponds to a pulse of
1.95 ⋅10.335 + 67.25 = 87.5 beats per minute.
The mean is 0.78386 lb. and the standard deviation is 0.00436 lb. The histogram confirms that the
distribution of weights is roughly normal.
9
8
Frequency
7
6
5
4
3
2
1
0
0.774
0.778
0.782
0.786
0.790
0.794
Diet Pepsi Weights
b.
35. a.
b.
c.
d.
36. a.
b.
c.
The z score for the bottom 0.5% is –2.59 which has a corresponding weight of,
−2.59 ⋅ 0.00436 + 0.78386 = 0.7726 lb., the z score for the top 0.5% is 2.59 which has a corresponding
weight of 2.59 ⋅ 0.00436 + 0.78386 = 0.7951 lb.
The new mean is equal to the old one plus the new points which is 75. The standard deviation is
unchanged at 10 (since we added the same amount to each student.)
No, the conversion should also account for variation.
The z score for the bottom 70% is 0.52 which has a corresponding score of 0.52 ⋅10 + 40 = 45.2 , and
the z score for the top 10% is 1.28 which has a corresponding score of 1.28 ⋅10 + 40 = 52.8
Using a scheme like the one in part (c), because variation is included in the curving process.
30 − 24
20 − 24
= 2.31 which has a percentage of 0.9896, z x = 20 =
= −1.54 which has a
2.6
2.6
percentage of 0.0618. Therefore, the percentage between 20 and 30 chocolate chips is
0.9896 − 0.0618 = 0.9278 or 92.78%. (Tech: 92.75%)
z x =30 =
30.5 − 24
19.5 − 24
= 2.5 which has a percentage of 0.9938 and z x=19.5 =
= −1.73 which has
2.6
2.6
a percentage of 0.0418. Therefore, the percentage between 19.5 and 30.5 is
0.9938 − 0.0418 = 0.9520 or 95.20%.
z x=30.5 =
The use of the continuity correction changes the result by a relatively small but not insignificant
amount.
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 79
37. The z score for Q1 is –0.67, and the z score for Q3 is 0.67. The IQR is 0.67 − (−0.67) = 1.34 .
1.5 ⋅ IQR = 2.01 , so Q1 −1.5 ⋅ IQR = −0.67 − 2.01 = −2.68 and Q3 + 1.5 ⋅ IQR = 0.67 + 2.01 = 2.68
The percentage to the left of –2.68 is 0.0037 and the percentage to the right of 2.68 is 0.0037. Therefore,
the percentage of an outlier is 0.0074. (Tech: 0.0070)
38. a.
b.
The z score for the 95th percentile is 1.645. This corresponds to an SAT score of
1.645 ⋅ 312 + 1511 = 2024.24 or 2024. The z score of 1.645 corresponds to an ACT score of
1.645 ⋅ 5.1 + 21.1 = 29.4895 or 29.5.
The score of 2100 corresponds to a z score of
2100 −1511
= 1.89 which corresponds to an ACT score
312
of 1.89 ⋅ 5.1 + 21.1 = 30.73 or 30.7.
Section 6-4
1.
a.
b.
c.
2.
a.
b.
3.
The sample mean will tend to center about the population parameter of 5.67 g.
The sample mean will tend to have a distribution that is approximately normal.
The sample proportions will tend to have a distribution that is approximately normal.
Without replacement
(1) When selecting a relatively small sample from a large population, it makes no significant difference
whether we sample with replacement or without replacement. (2) Sampling with replacement results in
independent events that are unaffected by previous outcomes, and independent events are easier to
analyze and they result in simpler calculations and formulas.
Sample mean, sample variance, sample proportion
4.
No. The data set is only one sample, but the sampling distribution of the mean is the distribution of the
means from all samples, not the one sample mean obtained from this single sample.
5.
No. The sample is not a simple random sample from the population of all college Statistics students. It is
very possible that the students at Broward College do not accurately reflect the behavior of all college
Statistics students.
6.
a.
Normal
b.
0 +1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
= 4.5
10
c.
7.
a.
5
= 0.5 which is the proportion of the five odd numbers {1, 3, 5, 7, 9} to the ten digits {0, 1, 2, 3, 4,
10
5, 6, 7, 8, 9}
The mean of the population is μ =
σ2 =
b.
4+5+9
= 6 , and the variance is
3
(4 − 6) 2 + (5 − 6)2 + (9 − 6)2
= 4.7
3
The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)} which
have the following variances {0, 0.5, 12.5, 0.5, 0, 8, 12.5, 8, 0} respectively.
Sample Variance
0
Probability
3/9
0.5
8
12.5
2/9
2/9
2/9
Copyright © 2014 Pearson Education, Inc.
80 Chapter 6: Normal Probability Distributions
7.
8.
9.
(continued)
3 ⋅ 0 + 2 ⋅ 0.5 + 2 ⋅ 8 + 2 ⋅12.5
= 4.7
9
c.
The sample variances’ mean is
d.
Yes. The mean of the sampling distribution of the sample variances (4.7) is equal to the value of the
population variance (4.7) so the sample variances target the value of the population variance.
a.
The population standard deviation (using the result from the previous problem) is σ = 4.7 = 2.160
b.
By taking the square root of the sample variances from the previous problem we get
Sample Standard Deviation
0.000
Probability
3/9
0.707
2/9
2.828
2/9
3.536
2/9
3 ⋅ 0 + 2 ⋅ 0.707 + 2 ⋅ 2.828 + 2 ⋅ 3.536
= 1.571
9
c.
The mean of the sample standard deviations is
d.
No. The mean of the sampling distribution of the sample standard deviations is 1.571, and it is not
equal to the value of the population standard deviation (2.160), so the sample standard deviations do
not target the value of the population standard deviation.
a.
b.
The population median is 5
The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)} which
have the following medians {4, 4.5, 6.5, 4.5, 5, 7, 6.5, 7, 9}
Sample Median
4
Probability
1/9
4.5
2/9
5
1/9
6.5
2/9
7
2/9
9
1/9
c.
The mean of the sampling distribution of the sampling median is
4 + 4.5 + 4.5 + 5 + 6.5 + 6.5 + 7 + 7 + 9
=6
9
d.
No. The mean of the sampling distribution of the sample medians is 6, and it is not equal to the value
of the population median of 5, so the sample medians do not target the value of the population median.
10. a.
b.
The proportion of odd numbers is 2/3 (there are two odd numbers from the population of 4, 5, and 9)
The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)} which
have the following proportion of odd numbers {0, 0.5, 0.5, 0.5, 1, 1, 0.5, 1, 1}
Sample Proportion
0
c.
Probability
1/9
0.5
4/9
1
4/9
The mean of the sampling distribution of sample proportions is
0 + 0.5 + 0.5 + 0.5 + 0.5 + 1 + 1 + 1 + 1 2
= = 0.67
9
3
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 81
10. (continued)
d.
11. a.
Yes. The mean of the sampling distribution of the sample proportion of odd numbers is 2/3, and it is
equal to the value of the population proportion of odd numbers of 2/3, so the sample proportions target
the value of the population proportion
The possible samples of size 2 are {(56, 56), (56, 49), (56, 58), (56, 46), (49, 56), (49, 49), (49, 58),
(49, 46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)}
Sample Mean Age
46
Probability
1/16
47.5
2/16
49
1/16
51
2/16
52
2/16
52.5
2/16
53.5
2/16
56
1/16
57
2/16
58
1/16
56 + 49 + 58 + 46
= 52.25 and the mean of the sample means is
4
46 + 47.5 + 47.5 + 49 + 51 + 51 + 52 + 52 + 52.5 + 52.5 + 53.5 + 53.5 + 56 + 57 + 57 + 58
= 52.25
16
b.
The mean of the population is
c.
The sample means target the population mean. Sample means make good estimators of population
means because they target the value of the population mean instead of systematically underestimating
or overestimating it.
12. a.
The possible samples of size 2 are {(56, 56), (56, 49), (56, 58), (56, 46), (49, 56), (49, 49), (49, 58),
(49, 46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)}
Sample Median Age
46
b.
c.
The median of the population is
Probability
1/16
47.5
2/16
49
1/16
51
2/16
52
2/16
52.5
2/16
53.5
2/16
56
1/16
57
2/16
58
1/16
49 + 56
= 52.5 and the median of the sample medians is
2
52 + 52.5
= 52.25 . The two values are not equal.
2
The sample medians do not target the population median of 52.5, so the sample medians do not make
good estimators of the population medians
Copyright © 2014 Pearson Education, Inc.
82
Chapter 6: Normal Probability Distributions
13. a.
The possible samples of size 2 are {(56, 56), (56, 49), (56, 58), (56, 46), (49, 56), (49, 49), (49, 58),
(49, 46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)} which have the
following ranges and associated probabilities
Sample Range
0
Probability
4/16
2
2/16
3
2/16
7
2/16
9
2/16
10
2/16
12
2/16
b.
The range of the population is 58 − 46 = 12 , the mean of the sample ranges is
4 ⋅ 0 + 2 ⋅ 2 + 2 ⋅ 3 + 2 ⋅ 7 + 2 ⋅ 9 + 2 ⋅10 + 2 ⋅12
= 5.375 . The values are not equal.
16
c.
The sample ranges do not target the population range of 12, so sample ranges do not make good
estimators of the population range.
14. a.
b.
c.
The possible samples of size 2 are {(56, 56), (56, 49), (56, 58), (56, 46), (49, 56), (49, 49), (49, 58),
(49, 46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)} which have the
following variances and associated probabilities
Sample Variance
0
Probability
4/16
2
2/16
4.5
2/16
24.5
2/16
40.5
2/16
50
2/16
72
2/16
The variance of the population is
(56 − 52.25) 2 + (49 − 52.25) 2 + (58 − 52.25) 2 + (46 − 52.25)2
= 24.1875
4
4 ⋅ 0 + 2 ⋅ 2 + 2 ⋅ 4.5 + 2 ⋅ 24.5 + 2 ⋅ 40.5 + 2 ⋅ 50 + 2 ⋅ 72
The mean of the sample variances is
= 24.1875
16
The two values are equal
The sample variances do target the population variance , so sample variances do make good estimators
of the population variance.
15. The possible birth samples are {(b, b), (b, g), (g, b), (g, g)}
Proportion of Girls
0
Probability
0.25
1/2
0.5
2/2
0.25
Yes. The proportion of girls in 2 births is 0.5, and the mean of the sample proportions is 0.5. The result
suggests that a sample proportion is an unbiased estimator of the population proportion.
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 83
16. The possible birth samples are {bbb, bbg, bgb, gbb, ggg, ggb, gbg, bgg}
Proportion of Girls
0
Probability
1/3
1/3
3/8
2/3
3/8
3/3
1/8
Yes. The proportion of girls in 3 births is 0.5 and the mean of the sample proportions is 0.5. The result
suggests that a sample proportion is an unbiased estimator of the population proportion.
17. The possibilities are: both questions incorrect, one question correct (two choices), both questions correct.
a.
Proportion Correct
Probability
0
2
4 4 16
⋅ =
5 5 25
1
2
⎛1 4⎞ 8
2 ⋅ ⎜⎜ ⋅ ⎟⎟⎟ =
⎜⎝ 5 5 ⎠ 25
2
2
⎛ 1 1 ⎞⎟ 1
⎜⎜ ⋅ ⎟ =
⎜⎝ 5 5 ⎠⎟ 25
16 ⋅ 0 + 8 ⋅ 0.5 + 1⋅1
= 0.2
25
b.
The mean is
c.
Yes. The sampling distribution of the sample proportions has a mean of 0.2 and the population
proportion is also 0.2 (because there is 1 correct answer among 5 choices.) Yes, the mean of the
sampling distribution of the sample proportions is always equal to the population proportion.
18. a.
The proportions of 0, 0.5, and 1 have the following probabilities
Proportion of Defective
Probability
0
9 / 25
0.5
12 / 25
1
4 / 25
9 ⋅ 0 + 12 ⋅ 0.5 + 4 ⋅1
= 0.4
25
b.
The mean is
c.
Yes. The population proportion is 0.4 (2 out of 5) and the mean of the sampling proportions is also
0.4. Yes, the mean of the sampling distribution of proportions is always equal to the population
proportion.
19. P(0) =
1
1
1
= = 0.25 , P(0.5) =
= 0.5 ,
2(2 − 2 ⋅ 0)!(2 ⋅ 0)! 4
2(2 − 2 ⋅ 0.5)!(2 ⋅ 0.5)!
1
= 0.25 . The formula yields values which describes the sampling distribution of
2(2 − 2 ⋅1)!(2 ⋅1)!
the sample proportions. The formula is just a different way of presenting the same information in the table
that describes the sampling distribution.
P(1) =
Copyright © 2014 Pearson Education, Inc.
84
Chapter 6: Normal Probability Distributions
20. Sample values of the mean absolute deviation (MAD) do not usually target the value of the population
MAD, so a MAD statistic is not good for estimating a population MAD. If the population of {4, 5, 9} from
Example 5 is used, the sample MAD values of 0, 0.5, 2, and 2.5 have corresponding probabilities of 3/9,
2/9, 2/9, and 2/9. For these values, the population MAD is 2, but the sample MAD values have a mean of
1.1, so the mean of the sample MAD values is not equal to the population MAD.
Section 6-5
1.
Because the sample size is greater than 30, the sampling distribution of the mean ages can be approximated
σ
by a normal distribution with mean μ and standard deviation
.
40
2.
No. Because the original population is normally distributed, the sample means will be normally distributed
for any sample size, not just those greater than 30.
3.
μx = 60.5 cm and it represents the mean of the population consisting of all sample means.
σx =
6.6
36
= 1.1 cm, and it represents the standard deviation of the population consisting of all sample
means.
4.
Because the digits are equally likely to occur, they have a uniform distribution. Because the sample means
are based on samples of size 3 drawn from a population that does not have a normal distribution, we should
not treat the sample means as having a normal distribution.
5.
a.
z x = 222.7 =
b.
z x = 207 =
a.
z x =196.9 =
b.
z x = 205 =
a.
z x = 218.4 =
b.
z x = 204 =
c.
0.6996.)
Because the original population has a normal distribution, the distribution of sample means is normal
for any sample size.
a.
z x =195 =
6.
7.
8.
222.7 − 205.5
= 2 , which has a probability of 0.9772.
8.6
207 − 205.5
= 1.22 , which has a probability of 0.8888. (Tech: 0.8889.)
8.6
49
196.9 − 205.5
= −1 , which has a probability of 0.1587.
8.6
205 − 205.5
= −0.35 , which has a probability of 0.3632. (Tech: 0. 3636.)
8.6
36
218.4 − 205.5
= 1.5 , which has a probability of 1− 0.9332 = 0.0668 to the right of it
8.6
204 − 205.5
= −0.52 which has a probability of 1− 0.3015 = 0.6985 to the right of it. (Tech:
8.6
9
195 − 205.5
= −1.22 which has an area of 1− 0.1112 = 0.8888 to the right of it. (Tech:
8.6
0.8889.)
203 − 205.5
= −1.45 which has an area of 1− 0.0735 = 0.9265 to the right of it. (Tech: 0.9270.)
8.6
25
b.
z=
c.
Because the original population has a normal distribution, the distribution of sample means is normal
for any sample size
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 85
9.
a.
b.
10. a.
b.
231.5 − 205.5
179.7 − 205.5
= 3.02 and z x =179.7 =
= −3 which has a probability of
8.6
8.6
0.9987 − 0.0013 = 0.9974 between them. (Tech: 0.9973.)
z x = 231.5 =
206 − 205.5
204 − 205.5
= 0.37 and z x = 204 =
= −1.1 which has a probability of
8.6
8.6
40
40
0.6443 − 0.1357 = 0.5086 between them. (Tech: 0.5085.)
z x = 206 =
200 − 205.5
180 − 205.5
= −0.64 and z x =180 =
= −2.97 which have a probability of
8.6
8.6
0.2611− 0.0015 = 0.2596 between them. (Tech: 0.2597.)
z x = 200 =
206 − 205.5
198 − 205.5
= 0.41 and z x =198 =
= −6.17 which has a probability of
8.6
8.6
50
50
0.6951− 0.0001 = 0.6950 between them. (Tech: 0.6955.)
z x = 206 =
195.3 −182.9
= 1.22 which has a probability of 1− 0.8888 = 0.1112 to the right of it. The
40.8
16
elevator appears to be relatively safe because there is a very small chance that it will be overloaded with 16
male passengers. (Tech: 0.1121.)
11. z x =195.3 =
195.3 −174
= 2.09 which has a probability of 1− 0.9817 = 0.0183 to the right of it. The elevator
40.8
16
appears to be relatively safe because there is a very small chance of overloading. Using the outdated mean
that is too low has the effect of making the elevator appear to be much safer than it actually is. (Tech:
0.0183.)
12. z x =195.3 =
13. a.
b.
25 − 22.65
21− 22.65
= 2.94 and z x = 21 =
= −2.06 , so 0.9984 − 0.0197 = 0.9787 = 97.87%
0.8
0.8
of women can fit into the hats. (Tech: 0.9788.)
The z scores for the smallest 2.5% and the largest 2.5% head circumferences are –1.96 and 1.96
respectively. This corresponds to head circumferences of 0.8 ⋅ (−1.96) + 22.65 = 21.08 and
z x = 25 =
0.8 ⋅1.96 + 22.65 = 24.22
c.
14. a.
b.
23 − 22.65
22 − 22.65
= 3.5 and z x = 22 =
= −6.5 which have a probability of
0.8
0.8
64
64
0.9998 − 0.0000 = 0.9998 = 99.98% between them. No, the hats must fit individual women, not the
mean from 64 women. If all hats are made to fit head circumferences between 22 in. and 23 in., the
hats will not fit about half those women.
z x = 23 =
z x = 22 =
22 −18.2
= 3.8 which has a probability of 0.9999. So the percentage is 99.99%
1
18.5 −18.2
= 1.8 which has a probability of 0.9641. No, when considering the diameters of
1
36
manholes, we should use a design based on individual men, not samples of 36 men.
z x =18.5 =
Copyright © 2014 Pearson Education, Inc.
86 Chapter 6: Normal Probability Distributions
15. a.
b.
The mean weight of passengers is
3500
= 140 lb.
25
140 −182.9
= −5.26 which has a probability of 0.99999 (or 1.0000) to the right of it. (Tech:
40.8
25
0.0.99999993.)
z x =140 =
175 −182.9
= −0.87 which has a probability of 0.8078 to the right of it. (Tech: 0.8067.)
40.8
20
c.
z x =175 =
d.
Given that there is a 0.8078 probability of exceeding the 3500 lb. limit when the water taxi is loaded
with 20 random men, the new capacity of 20 passengers does not appear to be safe enough because the
probability of overloading is too high.
16. a.
b.
17. a.
0.8535 − 0.8565
= −0.06 which has a probability of 1− 0.4761 = 0.5239 to the right of it.
0.0518
(Tech: 0.5231.)
z x =0.8535 =
0.8535 − 0.8565
= −1.25 which has a probability of 1− 0.1056 = 0.8944 to the right of it.
0.0518
465
(Tech: 0.8941.) Instead of filling each bag with exactly 465 M&Ms, the company probably fills the
bags so that the weight is as stated. In any event, the company appears to be doing a good job of filling
the bags.
z x =0.8535 =
z x =167 =
167 −182.9
= −0.39 which has a probability of 1− 0.3483 = 0.6517 to the right of it. (Tech:
40.8
0.6516.)
167 −182.9
= −1.35 which has a probability of 1− 0.0885 = 0.9115 to the right of it
40.8
12
b.
z x =167 =
c.
There is a high probability that the gondola will be overloaded if it is occupied by 12 more people, so it
appears that the number of allowed passengers should be reduced.
18. a.
b.
c.
The z score for 1% is –2.33 which corresponds to a pulse rate of −2.33 ⋅11.6 + 77.5 = 50.5 beats per
minute. The z score for 99% is 2.33 which corresponds to a pulse rate of
2.33⋅11.6 + 77.5 = 104.5 beats per minute.
85 − 77.5
70 − 77.5
= 3.23 and z x =70 =
= −3.23 which have a probability of
11.6
11.6
25
25
0.9994 − 0.0006 = 0.9988 between them
Instead of the mean pulse rate from the patients in a day , the cutoff values should be based on
individual patients, so it would be better to use the pulse rates of 50.5 beats per minute and 104.5 beats
per minute.
z x =85 =
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 87
19. a.
b.
c.
20. a.
b.
211−165
140 −165
= 1.01 and z x =140 =
= −0.55 which have a probability of
45.6
45.6
0.8438 − 0.2912 = 0.5526 between them. (Tech: 0.5517.)
z x = 211 =
211−165
140 −165
= 6.05 and z x =140 =
= −3.29 which have a probability of
45.6
45.6
36
36
0.9999 − 0.0005 = 0.9994 between them. (Tech: 0.9995.)
z x = 211 =
Part (a) because the ejection seats will be occupied by individual women, not groups of women.
140 −182.9
= −7.44 which has a probability of 1− 0.0001 = 0.9999 to the right of it. (Tech:
40.8
50
1.0000 when rounded to four decimal places.)
z x =140 =
z x =140 =
174 −182.9
= −0.82 which has a probability of 1− 0.2061 = 0.7939 to the right of it. (Tech:
40.8
14
0.7928.)
21. a.
b.
c.
d.
72 − 69.5
= 1.04 which has a probability of 0.8508. (Tech: 0.8512.)
2.4
72 − 69.5
= 14.76 which has a probability of 0.9999. (Tech: 1.0000 when rounded to four
z x=72 =
2.4
100
decimal places.)
The probability of Part (a) is more relevant because it shows that 85.08% of male passengers will not
need to bend. The result from part (b) gives us useful information about the comfort and safety of
individual male passengers.
Because men are generally taller than women, a design that accommodates a suitable proportion of
men will necessarily accommodate a greater proportion of women.
z x=72 =
167.6 −182.9
= −2.28 which has a probability of 1− 0.0113 = 0.9887 to the right of it. There is
40.8
37
a 0.9887 probability that the aircraft is overloaded. Because that probability is so high, the pilot should
take action, such as removing excess fuel and/or requiring that some passengers disembark and take a later
flight.
22. z x =167.6 =
23. a.
Yes. The sampling is without replacement and the sample size of 50 is greater than 5% of the finite
population size of 275. σ x =
b.
24. a.
50
275 − 50
= 2.0504584
275 −1
105 − 95.5
95 − 95.5
= 4.63 and z x =95.5 =
= −0.24 which have a probability of
2.0504584
2.0504584
1− 0.4053 = 0.5947 . (Tech: 0.5963.)
z x =105 =
Yes. The sampling is without replacement (because each sample of 16 elevator passengers consists of
16 different people) and the sample size of 16 is greater than 5% of the finite population size of 300.
σx =
b.
16
40
16
300 −16
= 9.7459365
300 −1
3000
= 187.5 lb.
16
Copyright © 2014 Pearson Education, Inc.
88
Chapter 6: Normal Probability Distributions
24. (continued)
c.
d.
187.5 −177
= 1.08 which has a probability of 1− 0.8599 = 0.1401 . The probability is not as
9.7459365
low as it should be, since it would be overloaded 14% of the time. (Tech: 0.1407.)
The z score for 0.999 is 3.1 which means that we need to solve for n in the equation
z x =187.5 =
3.1 =
25. a.
μ=
40 16 − n
which has a solution of n = 14 passengers.
n 16 −1
4+5+9
(4 − 6)2 + (5 − 6) 2 + (9 − 6)2
= 6, σ =
= 2.160246899
3
3
b.
The possible samples of size 2 are:{ (4, 5), (4, 9), (5, 4), (5, 9), (9, 4), (9, 5)} which have the following
means {4.5, 6.5, 4.5, 7, 6.5, 7} respectively.
c.
μx =
σx =
d.
4.5 + 6.5 + 4.5 + 7 + 6.5 + 7
= 6 and
6
(4.5 − 6)2 + (6.5 − 6) 2 + (4.5 − 6) 2 + (7 − 6)2 + (6.5 − 6) 2 + (7 − 6) 2
= 1.08012345
6
It is clear that μ = μx = 6 . σ x =
2.160246899 3 − 2
= 1.08012345 = σ
3 −1
2
Section 6-6
1.
The histogram should be approximately bell-shaped, and the normal quantile plot should have points that
approximate a straight line pattern.
2.
Either the points are not reasonably close to a straight line pattern, or there is some systematic pattern that
is not a straight line pattern.
3.
We must verify that the sample is from a population having a normal distribution. We can check for
normality using a histogram, identifying the number of outliers, and constructing a normal quantile plot.
4.
Because the histogram is roughly bell-shaped, conclude that the data are from a population having a normal
distribution.
5.
Not normal. The points show a systematic pattern that is not a straight line pattern.
6.
Normal. The points are reasonably close to a straight line pattern, and there is no other pattern that is not a
straight line pattern.
7.
Normal. The points are reasonably close to a straight line pattern, and there is no other pattern that is not a
straight line pattern.
8.
Not normal. The points are not reasonably close to a straight line pattern, and there appears to be a pattern
that is not a straight line pattern.
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 89
9.
Not normal
14
12
Frequency
10
8
6
4
2
0
-60
-30
0
30
60
90
120
Arrival Delay
10. Normal
12
10
Frequency
8
6
4
2
0
160
168
176
184
192
Height
11. Normal
7
Frequency
6
5
4
3
2
1
0
100
110
120
130
140
150
Systolic
12. Not normal
40
Frequency
30
20
10
0
0
80
160
240
No Exposure
Copyright © 2014 Pearson Education, Inc.
320
90
Chapter 6: Normal Probability Distributions
13. Not normal
14. Normal
15. Normal
16. Not normal
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 91
17. Normal. The points have coordinates (–131, –1.28), (134, –0.52), (139, 0), (143, 0.52), (145, 1.28)
18. Not normal. The points have coordinates (13, –1.38), (14, –0.67), (15, –0.21), (15, 0.21), (31, 0.67), (37,
1.38)
19. Not normal. The points have coordinates (1034, –1.53), (1051, –0.89), (1067, –0.49), (1070, –0.16), (1079,
0.16), (1079, 0.49), (1173, 0.89), (1272, 1.53)
Copyright © 2014 Pearson Education, Inc.
92
Chapter 6: Normal Probability Distributions
20. Normal. The points have coordinates (0.825, –1.59), (0.825, –0.97), (0.855, –0.59), (0.864, –0.28), (0.869,
0), (0.886, 0.28), (0.887, 0.59), (0.912, 0.97), (0.942, 1.59)
21. a.
b.
c.
Yes.
Yes.
No.
22. a.
The magnitudes are from a normally distributed
b.
The original measurements have a lognormal distribution
c.
We can reverse the process of taking values be an exponent of 10. The normal quantile plot indicates
that the original values are not from a population with a normal distribution.
23. The original values are not from a normally distributed population.
0.99
0.95
Probability
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
0.01
-40000
-20000
0
20000
40000
60000
Net Worth
Copyright © 2014 Pearson Education, Inc.
80000
Chapter 6: Normal Probability Distributions 93
23. (continued)
After taking the logarithm of each value, the values appear to be from a normally distributed population.
0.99
0.95
Probability
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
0.01
1
2
3
4
5
Log(Net Worth)
The original values are from a population with a lognormal distribution.
Section 6-7
1.
The Minitab display shows that the region representing 235 wins is a rectangle. The result of 0.0068 is an
approximation, but the result of 0.0066 is better because it is based on an exact calculation. The
approximation differs from the exact result by a very small amount.
2.
The continuity correction is used to compensate for the fact that a continuous distribution (normal) is used
to approximate a discrete distribution (binomial). The discrete number of 13 is represented by the interval
from 12.5 to 13.5.
3.
1
4
= 0.2 , q = = 0.8 , μ = 25 ⋅ 0.2 = 5 , σ = 25 ⋅ 0.2 ⋅ 0.8 = 2 . The value of 5 for the mean shows
5
5
that for many people who make random guesses for the 25 questions, the mean number of correct answers
is 5. For many people who make random guesses, the standard deviation of 2 is a measure of how much
the numbers of correct responses vary.
p=
4.
Yes. The circumstances correspond to 25 independent trials of a binomial experiment in which the
probability of success is 0.2. Also, with n = 25 , p = 0.2 , q = 0.8 , the requirements of np ≥ 5 and
nq ≥ 5 are both satisfied.
5.
The requirements are satisfied with a mean of 13 ⋅ 0.4 = 5.2 and the standard deviation of
2.5 − 5.2
= −1.53 which has a probability of 0.0630. (Tech:
13 ⋅ 0.4 ⋅ 0.6 = 1.766 . Therefore, z x=2.5 =
13 ⋅ 0.4 ⋅ 0.6
0.0632.)
6.
The requirement of nq ≥ 5 is not satisfied. Normal approximation should not be used.
7.
The requirement of nq ≥ 5 is not satisfied. Normal approximation should not be used.
8.
The requirements are satisfied with a mean of 10 and a standard deviation of 25 ⋅ 0.4 ⋅ 0.6 = 2.449 .
9.5 −10
= −0.20 which has a probability of 1− 0.4207 = 0.5793 to the right of it.
Therefore, z x=9.5 =
25 ⋅ 0.4 ⋅ 0.6
(Tech: 0.5809.)
9.
μ = 100 ⋅ 0.22 = 22, σ = 100 ⋅ 0.22 ⋅ 0.78 = 4.1425
19.5 − 22
z x=19.5 =
= −0.60 which has a probability of 0.2743. (Tech: 0.2731.)
100 ⋅ 0.22 ⋅ 0.78
Copyright © 2014 Pearson Education, Inc.
94 Chapter 6: Normal Probability Distributions
10. z x=24.5 =
24.5 − 22
= 0.60 which has a probability of 1− 0.7257 = 0.2743 to the right of it. (Tech:
100 ⋅ 0.22 ⋅ 0.78
0.2731.)
22.5 − 22
23.5 − 22
= 0.12 and z x = 23.5 =
= 0.36 which have a probability of
100 ⋅ 0.22 ⋅ 0.78
100 ⋅ 0.22 ⋅ 0.78
0.6406 − 0.5478 = 0.0928 between them. (Tech: 0.0933.)
11. z x = 22.5 =
18.5 − 22
19.5 − 22
= −0.84 and z x=19.5 =
= −0.60 which have a probability of
100 ⋅ 0.22 ⋅ 0.78
100 ⋅ 0.22 ⋅ 0.78
0.2743 − 0.2005 = 0.0738 . (Tech: 0.0738.)
12. z x=18.5 =
13. μ = 611⋅ 0.3 = 183.3, σ = 611⋅ 0.3 ⋅ 0.7 = 11.3274
a.
b.
c.
d.
172.5 −183.3
171.5 −183.3
= −0.95 and z x=171.5 =
= −1.04 which have a probability of
611⋅ 0.3⋅ 0.7
611⋅ 0.3⋅ 0.7
0.1711− 0.1492 = 0.0219 between them. (Tech using normal approximation: 0.0214; Tech using
binomial: 0.0217)
z x=172.5 =
172.5 −183.3
= −0.95 which has a probability of 0.1711. The result of 172 overturned calls
11.3274
is not unusually low. (Tech using normal approximation: 0.1702; Tech using binomial: 0.1703.)
The result from part (b) is useful. We want the probability of getting a result that is at least as extreme
as the one obtained.
If the 30% rate is correct, there is a good chance (17.11%) of getting 172 or fewer calls overturned, so
there is not strong evidence against the 30% rate.
z x=172.5 =
14. μ = 611⋅ 0.33 = 201.63, σ = 611⋅ 0.33 ⋅ 0.67 = 11.6229
a.
b.
c.
d.
172.5 − 201.63
171.5 − 201.63
= −2.51 and z x=171.5 =
= −2.59 which have a probability of
611⋅ 0.33 ⋅ 0.67
611⋅ 0.33 ⋅ 0.67
0.0060 − 0.0048 = 0.0012 between them. (Tech using normal approximation: 0.0013; Tech using
binomial: 0.0013)
z x=172.5 =
172.5 − 201.63
= −2.51 which has a probability of 0.006. The result of 172 overturned
611⋅ 0.33 ⋅ 0.67
calls is unusually low. (Tech using normal approximation: 0.0061; Tech using binomial: 0.0056.)
The result from part (b) is useful. We want the probability of getting a result that is at least as extreme
as the one obtained.
If the 33% rate is correct, there is a very small chance (0.6%) of getting 172 or fewer calls overturned,
so there is not strong evidence against the 33% rate.
z x=172.5 =
15. μ = 580 ⋅ 0.75 = 435, σ = 580 ⋅ 0.75 ⋅ 0.25 = 10.4283
a.
b.
c.
428.5 − 435
427.5 − 435
= −0.62 and z x=427.5 =
= −0.72 which have a probability
580 ⋅ 0.75 ⋅ 0.25
580 ⋅ 0.75 ⋅ 0.25
of 0.2676 − 0.2358 = 0.0318 between them . (Tech using normal approximation: 0.0305; Tech using
binomial: 0.0301.)
z x=428.5 =
428.5 − 435
= −0.62 which has a probability of 0.2676. The result of 428 peas with
580 ⋅ 0.75 ⋅ 0.25
green pods is not unusually low. (Tech using normal approximation: 0.2665; Tech using binomial:
0.2650.)
The result from part (b) is useful. We want the probability of getting a result that is at least as extreme
as the one obtained.
z x=428.5 =
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 95
15. (continued)
d.
No. Assuming that Mendel’s probability of 3/4 is correct, there is a good chance (26.76%) of getting
the results that were obtained. The obtained results do not provide strong evidence against the claim
that the probability of a pea having a green pod is 3/4
16. μ = 1004 ⋅ 0.25 = 251, σ = 1004 ⋅ 0.25 ⋅ 0.75 = 13.7204
a.
b.
c.
290.5 − 251
= 2.88 which has a probability of 1.0000 − 0.9980 = 0.0020 to the right of
1004 ⋅ 0.25 ⋅ 0.75
it. (Tech using normal approximation: 0.0020; Tech using binomial: 0.0023.)
Because the probability of getting 291 or more with the value of 25% is so small, the result of 291 is
unusually high.
The results do suggest that the rate is greater than 25%.
z x=290.5 =
17. μ = 945 ⋅ 0.5 = 472.5, σ = 945 ⋅ 0.5 ⋅ 0.5 = 15.3704
a.
b.
c.
d.
879.5 − 472.5
878.5 − 472.5
= 26.48 and z x=878.5 =
= 26.41 which have a probability of
945 ⋅ 0.5 ⋅ 0.5
945 ⋅ 0.5 ⋅ 0.5
0.0000 or 0+ (a very small positive probability that is extremely close to 0) between them.
z x=879.5 =
878.5 − 472.5
= 26.41 which has a probability of 0.0001. (Tech: 0.0000 or 0+, which is a
945 ⋅ 0.5 ⋅ 0.5
very small positive probability that is extremely close to 0). If boys and girls are equally likely, 879
girls in 945 births is unusually high.
The result from part (b) is more relevant, because we want the probability of a result that is at least as
extreme as the one obtained.
Yes. It is very highly unlikely that we would get 879 girls in 945 births by chance. Given that the 945
couples were treated with the XSORT method, it appears that this method is effective in increasing the
likelihood that a baby will be a girl.
z x=878.5 =
18. μ = 523 ⋅ 0.85 = 444.55, σ = 523 ⋅ 0.85 ⋅ 0.15 = 8.1659
517.5 − 444.55
= 8.93 which has a probability of 0.0001 to the right of it. (Tech using normal
z x=517.5 =
523 ⋅ 0.85 ⋅ 0.15
approximation: 0.0000 or 0+; Tech using binomial: 0.0000 or 0+.) It appears that many adult males say
that they wash their hands in a public restroom when they actually do not.
19. μ = 1002 ⋅ 0.61 = 611.22, σ = 1002 ⋅ 0.61⋅ 0.39 = 15.4394
700.5 − 611.22
= 5.78 which has a probability of 0.0001 to the right of it. (Tech 0.0000.) The
z x=700.5 =
1002 ⋅ 0.61⋅ 0.39
result suggests that the surveyed people did not respond accurately.
20. μ = 420, 095 ⋅ 0.00034 = 142.83, σ = 420, 095 ⋅ 0.000344 ⋅ 0.999656 = 11.9492
135.5 −142.83
= −0.61 which has a probability of 0.2709. (Tech using normal
z x=135.5 =
420,095 ⋅ 0.000344 ⋅ 0.999656
approximation: 0.2697; Tech using binomial: 0.2726.) Media reports appear to be wrong.
21. The probability of six or fewer should be computed. μ = 50 ⋅ 0.2 = 10, σ = 50 ⋅ 0.2 ⋅ 0.8 = 2.8284
6.5 −10
= −1.24 which has a probability of 0.1075. (Tech using normal approximation:
z x=6.5 =
50 ⋅ 0.2 ⋅ 0.8
0.1080; Tech using binomial: 0.1034.) Because that probability is not very small, the evidence against the
rate of 20% is not very strong.
Copyright © 2014 Pearson Education, Inc.
96 Chapter 6: Normal Probability Distributions
22. The probability of three or fewer should be computed. μ = 50 ⋅ 0.2 = 10, σ = 50 ⋅ 0.2 ⋅ 0.8 = 2.8284
3.5 −10
= −2.30 which has a probability of 0.0107. (Tech using normal approximation: 0.0108;
z x=3.5 =
2.8284
Tech using binomial: 0.0057.) Because that probability is very small, the evidence against the rate of 20%
is very strong. It appears that the rate of smoking among statistics students is lower than the 20% rate for
the general population.
23. The probability of 170 or fewer should be computed. μ = 1000 ⋅ 0.20 = 200, σ = 1000 ⋅ 0.2 ⋅ 0.8 = 12.6491
170.5 − 200
= −2.33 which has a probability of 0.0099. (Tech using normal approximation:
z x=170.5 =
1000 ⋅ 0.2 ⋅ 0.8
0.0098; Tech using binomial: 0.0089.) Because the probability of 170 or fewer is so small with the
assumed 20% rate, it appears that the rate is actually less than 20%.
24. The probability of 175 or more should be computed.
μ = 250 ⋅ 0.67 = 167.5, σ = 250 ⋅ 0.67 ⋅ 0.33 = 7.4347
174.5 −167.5
z x=174.5 =
= 0.94 which has a probability of 1.0000 − 0.8264 = 0.1736 to the right of it.
250 ⋅ 0.67 ⋅ 0.33
(Tech using normal approximation: 0.1732; Tech using binomial: 0.1734.) If the internet access rate is
67%, there is a relatively high probability of 17.32% of getting 175 or more households with internet access
when 250 households are surveyed. It does not appear that the 67% rate is too low.
25. a.
b.
In order to make a profit Marc will need to win over $1000. With 35:1 odds a $5 bet wins $175.
Therefore, Marc needs 6 winning bets in order to make a profit.
1
1 37
μ = 200 ⋅ = 5.2632, σ = 200 ⋅ ⋅
= 2.2638
38
38 38
5.5 − 5.2632
= 0.10 which has a probability of 1.0000 − 0.5398 = 0.4602 to the right of it.
z x=5.5 =
2.2638
(Tech using normal approximation: 0.4583; tech using binomial: 0.4307)
Since the odds of winning are 1:1 Marc would need 101 wins or more to make a
244
244 251
= 98.5859, σ = 200 ⋅
⋅
= 7.0704
495
495 495
100.5 − 98.5859
= 0.27 which has a probability of 1.0000 − 0.6064 = 0.3936
z x=100.5 =
7.0704
(Tech using normal approximation: 0.3933; tech using binomial: 0.3932)
The roulette game provides a better likelihood of making a profit.
profit. μ = 200 ⋅
c.
26. The z score that corresponds to a 0.95 probability is 1.645. This means that we have to solve the equation
1.645 ⋅ n ⋅ 0.9005 ⋅ 0.0995 + n ⋅ 0.9005 = 213 for n. This has a solution of 229 reservations. (Tech:
230.)
Chapter Quick Quiz
1.
μ = 0 and σ = 1
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 97
2.
3.
P98 = 2.05 (Tech: 2.05375)
4.
P ( z > −1) = 1− P ( z < −1) = 1− 0.1587 = 0.8413
5.
P (1.37 < z < 2.42) = P ( z < 2.42) − P ( z < 1.37) = 0.9922 − 0.9147 = 0.0775 (Tech: 0.0076)
6.
z x=4.2 =
4.2 − 4.577
= −0.99 which have a probability of 0.1611. (Tech: 0.1618.)
0.382
7.
z x=5.4 =
5.4 − 4.577
= 2.15 which has a probability of 1.0000 − 0.9842 = 0.0158 . (Tech: 0.0156.)
0.382
8.
The z score for P80 = 0.84 which corresponds to a red blood count of 0.84 ⋅ 0.382 + 4.577 = 4.898
9.
z=
4.444 − 4.577
= −1.74 which has a probability of 0.0409
0.382
25
4.2 − 4.577
5.4 − 4.577
= −0.99 and z x=5.4 =
= 2.15 which have a probability of
0.382
0.382
0.9842 − 0.1611 = 0.8231 or 82.31%. (Tech: 82.26%)
10. z x=4.2 =
Review Exercises
1.
2.
a.
b.
The probability to the left of a z score of 2.93 is 0.9983
The probability to the right of a z score of –1.53 is 1.0000 − 0.0630 = 0.9370
c.
The probability between z scores –1.07 and 2.07 is 0.9808 − 0.1423 = 0.8385
d.
The z score for P30 = −0.52
e.
z x=0.27 =
a.
b.
3.
a.
b.
0.27 − 0
= 1.08 which has a probability of 1− 0.8599 = 0.1401
1
16
1605 −1516
= 1.41 which has a probability of 1− 0.9207 = 0.0793 or 7.93%. (Tech: 7.89%.)
63
The z score for the lowest 1% is –2.33 which corresponds to a standing eye height of
−2.33⋅ 63 + 1516 = 1369.2 mm. (Tech: 1369.4.)
z x=1605 =
1500 −1634
= −2.03 which has a probability of 1− 0.0212 = 0.9788 or 97.88%
66
The z score for the lowest 95% is 1.645 which corresponds to a standing eye height of
1.645 ⋅ 66 + 1634 = 1742.6 mm
z x=1500 =
Copyright © 2014 Pearson Education, Inc.
98
Chapter 6: Normal Probability Distributions
4.
a.
Normal distribution
b.
μx = 21.1
c.
σx =
a.
An unbiased estimator is a statistic that targets the value of the population parameter in the sense that
the sampling distribution of the statistic has a mean that is equal to the mean of the corresponding
parameter.
Mean, variance and proportion
True
5.
b.
c.
6.
7.
80
= 0.57
b.
72 − 69.5
= 1.04 which has a probability of 0.8508 or 85.08%. (Tech: 85.12.) With about
2.4
15% of all men needing to bend, the design does not appear to be adequate, but the Mark VI monorail
appears to be working quite well in practice.
The z score for 99% is 2.33 which corresponds to a doorway height of 2.33 ⋅ 2.4 + 69.5 = 75.1
a.
z x=175 =
a.
b.
8.
5.1
a.
z x=72 =
175 −182.9
= −0.19 which has a probability of 1− 0.4247 = 0.5753 . (Tech: 0.5766.)
40.9
175 −182.9
= −2.82 which has a probability of 1− 0.0024 = 0.9976 . Yes, if the plane is full
40.9
213
of male passengers, it is highly likely that it is overweight.
z x=175 =
No. A histogram is far from bell shaped.
12
Frequency
10
8
6
4
2
0
0
8000
16000
24000
32000
Salary
b.
9.
No. The sample has a size of 26 which does not satisfy the condition at least 30, and the values do not
appear to be from a population having a normal distribution.
3
3 1
μ = 1064 ⋅ = 798, σ = 1064 ⋅ ⋅ = 14.1244
4
4 4
787.5 − 798
= −0.74 which has a probability of 0.2296. (Tech using normal approximation:
z z=787.5 =
3 1
1064 ⋅ ⋅
4 4
0.2286; Tech using binomial: 0.2278.) The occurrence of 787 offspring plants with long stem is not
unusually low because its probability is not small. The results are consistent with Mendel’s claimed
proportion of 3/4
Copyright © 2014 Pearson Education, Inc.
Chapter 6: Normal Probability Distributions 99
10. μ = 64 ⋅ 0.8 = 51.2, σ = 64 ⋅ 0.8 ⋅ 0.2 = 3.2
a.
b.
49.5 − 51.2
= −0.53 which has a probability of 1− 0.2981 = 0.7019 to the right of it. (Tech
3.2
using normal approximation: 0.7024; Tech using binomial: 0.7100)
z x=49.5 =
50.5 − 51.2
= −0.22 which have a probability of
3.2
0.4129 − 0.2981 = 0.1148 between them. (Tech using normal approximation: 0.1158; Tech using
binomial: 0.1190)
z x=49.5 = −0.53 and z x=50.5 =
Cumulative Review Exercises
1.
14,500, 000 + 145, 000, 000 + 14, 000, 000 + 5, 000, 000 + 3,500, 000
= $10,300,000
5
a.
x=
b.
The median is $14,000,000
(14,500 −10,300) 2 + (14,500 −10,300) 2 + ... + (5000 −10,300) 2 + (3500 −10,300) 2
5 −1
= $5552.027 (in thousands of dollars) which is $5,552,027.
c. s =
2.
3.
d.
s 2 = 30,825, 003,810, 000 square dollars
e.
z x=14,500,000 =
f.
g.
h.
Ratio
Discrete
No, the starting players are likely to be the best players who receive the highest salaries.
a.
A is the event of selecting someone who does not have the belief that college is not a good
investment. NOTE: This is not the same as selecting someone who believes that college is a good
investment.
b.
P( A) = 1− 0.1 = 0.9
14,500,000 −10,300,000
= 0.76
5,552,027
c.
P = 0.1⋅ 0.1⋅ 0.1 = 0.001
d.
The sample is a voluntary response sample. This suggests that the 10% rate might not be very
accurate, because people with strong feelings or interest about the topic are more likely to respond.
a.
b.
c.
d.
2500 − 3369
= −1.53 which has a probability of 0.0630. (Tech: 0.0627)
567
The z score for the bottom 10% is –1.28, which correspond to the weight
−1.28 ⋅ 567 + 3369 = 2642.24 g. (Tech: 2642 g.)
z x=2500 =
1500 − 3369
= −3.3 which has a probability of 0.0005
567
3400 − 3369
= 0.27 which has a probability of 1− 0.6064 = 0.3936 . (Tech: 0.3923) to the
z x=3400 =
567
25
right of it.
z x=1500 =
Copyright © 2014 Pearson Education, Inc.
100 Chapter 6: Normal Probability Distributions
4.
c.
5.
a.
The vertical scale does not start at 0, so differences are somewhat distorted. By using a scale ranging
from 1 to 29 for frequencies that range 2 to 14, the graph is flattened, so differences are not shown as
they should be.
b. The graph depicts a distribution that is not exactly normal, but it is approximately normal because it is
roughly bell shaped.
Minimum: 42 years; maximum: 70 years. Using the range rule of thumb, the standard deviation is
70 − 42
= 7 years. The estimate of 7 years is very close to the actual standard
estimated to be
4
deviation of 6.6 years, so the range rule of thumb works quite well here.
a.
P( X = 3) = 0.1⋅ 0.1⋅ 0.1 = 0.001
b.
P( X ≥ 1) = 1− P ( X = 0) = 1− (0.9 ⋅ 0.9 ⋅ 0.9) = 0.271
c.
The requirement that np ≥ 5 is not satisfied, indicating that the normal approximation would result in
errors that are too large.
μ = 50 ⋅ 0.1 = 5
d.
e.
f.
σ = 50 ⋅ 0.1⋅ 0.9 = 2.1213
No, 8 is within two standard deviations of the mean and is within the range of values that could easily
occur by chance.
Copyright © 2014 Pearson Education, Inc.
Chapter 7: Estimates and Sample Sizes 101
Chapter 7: Estimates and Sample Sizes
Section 7-2
1.
The confidence level (such as 95%) was not provided.
2.
When using 26% to estimate the value of the population percentage, the maximum likely difference
between 26% and the true population percentage is three percentage points, so the interval from 23% to
29% is likely to contain the true population percentage.
3.
p̂ = 0.26 is the sample proportion; q̂ = 0.74 (found from evaluating 1 – p̂ ); n = 1910 is the sample size;
E = 0.03 is the margin of error; p is the population proportion, which is unknown. The value of α is 0.05.
4.
The 95% confidence interval will be wider than the 80% confidence interval. A confidence interval must be
wider in order for us to be more confident that it captures the true value of the population proportion.
(Think of estimating the age of a classmate. You might be 90% confident that she is between 20 and 30, but
you might be 99.9% confident that she is between 10 and 40.)
5.
1.28
7.
1.645
6.
2.575 (Tech: 2.576)
8.
2.05
9.
E=
10. E =
0.186 − 0.0641
= 0.061, so 0.125 ± 0.061
2
0.335 − 0.165
= 0.085, so 0.250 ± 0.085
2
11. 0.0268 < p < 0.133
13. a.
pˆ =
12. 0.183 < p < 0.357
531
= 0.530
1002
ˆˆ
pq
= 1.96
n
531
471
( 1002
)( 1002
)
b.
E = zα / 2
c.
pˆ − E < p < pˆ − E ⇒ 0.530 − 0.0309 < p < 0.530 − 0.0309 ⇒ 0.499 < p < 0.561
d.
14. a.
b.
1002
= 0.0309
We have 95% confidence that the interval from 0.499 to 0.561 actually does contain the true value of
the population proportion.
pˆ =
490
= 0.610
806
E = zα / 2
ˆˆ
pq
= 2.58
n
490
316
( 806
)( 806
)
806
= 0.0443
pˆ − E < p < pˆ − E
c.
0.610 − 0.0443 < p < 0.610 − 0.0443
0.566 < p < 0.654
d.
We have 99% confidence that the interval from 0.566 to 0.654 actually does contain the true value of
the population proportion.
Copyright © 2014 Pearson Education, Inc.
102 Chapter 7: Estimates and Sample Sizes
15. a.
pˆ =
1083
= 0.430
2518
ˆˆ
pq
= 1.65
n
1435
( 1083
2518 )( 2518 )
b.
E = zα / 2
c.
pˆ − E < p < pˆ − E
0.430 − 0.0162 < p < 0.430 − 0.0162
2518
= 0.0162
0.414 < p < 0.446
d.
16. a.
b.
We have 90% confidence that the interval from 0.414 to 0.446 actually does contain the true value of
the population proportion.
pˆ =
543
= 0.540
1005
ˆˆ
pq
= 1.28
n
E = zα / 2
543
462
( 1005
)( 1005
)
1005
= 0.0201
pˆ − E < p < pˆ − E
c.
0.540 − 0.0201 < p < 0.540 − 0.0201
0.520 < p < 0.560
d.
17. a.
b.
c.
18. a.
b.
c.
19. a.
We have 80% confidence that the interval from 0.520 to 0.560 actually does contain the true value of
the population proportion.
pˆ =
879
= 0.930
945
pˆ ± zα / 2
( 879 )( 66 )
ˆ ˆ 879
pq
=
± 1.96 945 945
n
945
945
0.914 < p < 0.946
Yes. The true proportion of girls with the XSORT method is substantially greater than the proportion
of (about) 0.5 that is expected when no method of gender selection is used.
pˆ =
239
= 0.821
291
pˆ ± zα / 2
( 239 )( 52 )
ˆ ˆ 239
pq
=
± 2.56 291 291
n
291
945
0.763 < p < 0.879
Yes. The true proportion of boys with the YSORT method is substantially greater than the proportion
of (about) 0.5 that is expected when no method of gender selection is used.
0.5
123
= 0.439
280
b.
pˆ =
c.
( 123 )( 157 )
ˆ ˆ 123
pq
=
± 2.56 280 280
n
280
280
0.363 < p < 0.515 or 36.3% < p < 51.5%
d.
pˆ ± zα / 2
If the touch therapists really had an ability to select the correct hand by sensing an energy field, their
success rate would be significantly greater than 0.5, but the sample success rate of 0.439 and the
confidence interval suggest that they do not have the ability to select the correct hand by sensing an
energy field.
Copyright © 2014 Pearson Education, Inc.
Chapter 7: Estimates and Sample Sizes 103
20. a.
b.
21. a.
b.
c.
d.
e.
( 152 )( 428 )
ˆ ˆ 152
pq
=
± z0.025 580 580
580
580
n
0.226 < p < 0.298 or 22.6% < p < 29.8%
pˆ ± zα / 2
No, the confidence interval includes 0.25, so the true percentage could easily equal 25%.
427 (0.29) = 124
(0.29)(0.71)
ˆˆ
pq
= 0.29 ± z0.025
427
n
0.247 < p < 0.333 or 24.7% < p < 33.3%
pˆ ± zα / 2
Yes. Because all values of the confidence interval are less than 0.5, the confidence interval shows that
the percentage of women who purchase books online is very likely less than 50%.
No. The confidence interval shows that it is possible that the percentage of women who purchase
books online could be less than 25%.
Nothing.
22. If the subjects chose to respond to the posted question, the sample is a voluntary response sample, so the
confidence interval could be very misleading.
(0.208)(0.792)
ˆˆ
pq
= 0.208 ± z0.025
(using x = 30: 14.2% < p < 27.5%).
144
n
0.142 < p < 0.274 or 14.2% < p < 27.4%
pˆ ± zα / 2
23. a.
514 (0.459) = 236
b.
pˆ ± zα / 2
c.
pˆ ± zα / 2
d.
24. a.
(0.459)(0.541)
ˆˆ
pq
= 0.459 ± z0.10
n
514
0.431 < p < 0.487
The 95% confidence interval is wider than the 80% confidence interval. A confidence interval must be
wider in order to be more confident that it captures the true value of the population proportion. (See
Exercise 4.)
514 (0.90) = 463
b.
pˆ ± zα / 2
c.
pˆ ± zα / 2
d.
(0.459)(0.541)
ˆˆ
pq
= 0.459 ± z0.025
(using x = 236: 0.403 < p < 0.516).
n
514
0.402 < p < 0.516
(0.90)(0.10)
ˆˆ
pq
= 0.90 ± z0.005
(using x = 463: 0.867 < p < 0.935).
514
n
0.866 < p < 0.934
(0.90)(0.10)
ˆˆ
pq
= 0.90 ± z0.10
(using x = 463: 0.884 < p < 0.918).
514
n
0.883 < p < 0.917
The 95% confidence interval is wider than the 80% confidence interval. A confidence interval must be
wider in order to be more confident that it captures the true value of the population proportion. (See
Exercise 4.)
Copyright © 2014 Pearson Education, Inc.
104 Chapter 7: Estimates and Sample Sizes
25. No, the confidence interval limits contain the value of 0.13, so the claimed rate of 13% could be the true
percentage for the population of brown M&Ms.
(0.08)(0.92)
ˆˆ
pq
= 0.08 ± z0.01
. (Tech: 0.0169 < p < 0.143)
n
100
0.0168 < p < 0.143
pˆ ± zα / 2
26. a.
b.
27. a.
(0.70)(0.30)
ˆˆ
pq
= 0.70 ± z0.01
(Tech: 0.666 < p < 0.733)
n
1002
0.666 < p < 0.734
pˆ ± zα / 2
No. Because 0.61 is not included in the confidence interval, it does not appear that the responses are
consistent with the actual voter turnout.
(0.000321)(0.999679)
ˆˆ
pq
= 0.000321 ± z0.05
(using x = 135: 0.0276% < p < 0.0367%).
n
420, 095
pˆ ± zα / 2
0.0276% < p < 0.0366%
b.
28. a.
No, because 0.0340% is included in the confidence interval.
3005(0.817) = 2455
(0.817)(0.183)
ˆˆ
pq
= 0.817 ± z0.005
n
3005
0.805 < p < 0.829 or 80.5% < p < 82.9%
b.
pˆ ± zα / 2
c.
Nothing.
ˆˆ
[ zα / 2 ] pq
2
29. n =
E2
ˆˆ
[ zα / 2 ] pq
2
30. n =
E2
ˆˆ
[ zα / 2 ] pq
E2
ˆˆ
[ zα / 2 ] pq
E2
[1.28] (0.25)
0.042
n=
b.
n=
= 256 (Tech: 257)
[ 2.575] (0.15)(0.85)
0.052
[ 2.33] (0.15)(0.85)
= 339
2
=
ˆˆ
[ zα / 2 ] pq
0.032
2
33. a.
= 752
2
=
2
32. n =
0.032
2
=
2
31. n =
[1.645] (0.25)
2
=
E2
ˆˆ
[ zα / 2 ] pq
2
=
2
E2
ˆˆ
[ zα / 2 ] pq
[1.96] (0.25)
= 770 (Tech: 767)
0.0252
= 1537
[1.96] (0.38)(0.62)
2
=
2
0.0252
[ 2.575] (0.25)
= 1449
2
34. a.
n=
b.
n=
c.
Yes. Using the additional survey information from part (b) dramatically reduces the sample size.
E2
ˆˆ
[ zα / 2 ] pq
=
2
E2
0.012
= 16,577 (Tech: 16,588)
[ 2.575] (0.90)(0.10)
2
=
0.012
= 5968 (Tech: 5972)
Copyright © 2014 Pearson Education, Inc.
Chapter 7: Estimates and Sample Sizes 105
ˆˆ
[ zα / 2 ] pq
2
[1.645] (0.25)
2
35. a.
n=
b.
n=
c.
No. A sample of students at the nearest college is a convenience sample, not a simple random sample,
so it is very possible that the results would not be representative of the population of adults.
E2
ˆˆ
[ zα / 2 ] pq
=
2
E2
ˆˆ
[ zα / 2 ] pq
0.052
= 271
[1.645] (0.85)(0.15)
2
=
2
0.052
[1.28] (0.25)
= 139 (Tech: 138)
2
36. a.
n=
b.
n=
c.
No. Flights between New York and San Francisco might not be representative of the population of all
Southwest flights.
E2
ˆˆ
[ zα / 2 ] pq
=
2
E2
0.032
= 456 (Tech: 457)
[1.28] (0.84)(0.16)
2
=
0.032
= 245 (Tech: 246)
37. Greater height does not appear to be an advantage for presidential candidates. If greater height is an
advantage, then taller candidates should win substantially more than 50% of the elections, but the
confidence interval shows that the percentage of elections won by taller candidates is likely to be anywhere
between 36.2% and 69.7%.
( 18 )( 16 )
ˆ ˆ 18
pq
= ± 1.96 34 34
n
34
34
0.362 < p < 0.697 or 36.2% < p < 69.7%.
pˆ ± zα / 2
18
pˆ =
= 0.529 .
34
38. No, the confidence interval is based on sample data consisting of flights from New York (JFK) to Los
Angeles, and arrival delays for that route might be very different from arrival delays for the population that
includes all routes.
( 44 )( 4 )
ˆ ˆ 44
pq
=
± 1.645 48 48
n
48
48
0.851 < p < 0.980 or 85.1% < p < 98.0%.
pˆ ± zα / 2
44
pˆ =
= 0.917 .
48
ˆ ˆ [ zα / 2 ]
Npq
200 (0.5)(0.5)[1.96]
2
39. a.
n=
b.
n=
ˆ ˆ [ zα / 2 ] + ( N −1) E 2
pq
2
ˆ ˆ [ zα / 2 ]
Npq
2
=
(0.5)(0.5)[1.96] + (200 −1) 0.0252
2
200 (0.38)(0.62)[1.96]
2
40.
ˆ ˆ [ zα / 2 ] + ( N −1) E 2
pq
2
= 178
2
=
(0.38)(0.62)[1.96] + ( 200 −1) 0.0252
2
= 176
( 3 )( 5 )
ˆˆ 3
pq
= ± 1.96 8 8
no
8
8
n
0.0395 < p < 0.710;
pˆ ± zα / 2
41. The upper confidence interval limit is greater than 100%. Given that the percentage cannot exceed 100%,
change the upper limit to 100%.
( 44 )( 4 )
ˆ ˆ 44
pq
= ± 2.575 48 48
48
48
n
0.814 < p < 1.019 or 81.4% < p < 101.9%.
pˆ ± zα / 2
Copyright © 2014 Pearson Education, Inc.
106 Chapter 7: Estimates and Sample Sizes
42. a.
The requirement of at least 5 successes and at least 5 failures is not satisfied, so the normal distribution
cannot be used.
3
= 0.075
40
b.
43. Because we have 95% confidence that p is greater than 0.831, we can safely conclude that more than 75%
of adults know what Twitter is.
pˆ + zα
(0.85)(0.15)
ˆ ˆ 44
pq
=
+ 1.645
(Tech: p > 0.832).
n
48
1007
p > 0.831
Section 7-3
1.
a.
sec 233.4 sec < μ < 256.65 sec
b.
The best point estimate of μ is x =
E=
256.65 + 233.4
= 245.025 sec . The margin of error is
2
256.65 − 233.4
= 11.625 sec.
2
2.
a.
b.
c.
df = 39
2.023
In general, the number of degrees of freedom for a collection of sample data is the number of sample
values that can vary after certain restrictions have been imposed on all data values.
3.
We have 95% confidence that the limits of 233.4 sec and 256.65 sec contain the true value of the mean of
the population of all duration times.
4.
When we say that the confidence interval methods of this section are robust against departures from
normality, we mean that these methods work reasonably well with distributions that are not normal,
provided that departures from normality are not too extreme. The given dotplot does appear to satisfy the
loose normality requirement. Also, there are 40 dots, so the sample size of 40 satisfies the condition of n >
30.
5.
Neither the normal nor the Student t
distribution applies.
6.
tα / 2 = 1.729
9.
Because the sample size is greater than 30, the confidence interval yields a reasonable estimate of μ , even
though the data appear to be from a population that is not normally distributed.
x ± tα / 2
s
7.
tα / 2 = 2.708
8.
zα / 2 = 2.575 (Tech: 2.576)
= 9.808 ± 2.403⋅
5.013
n
50
8.104 km < μ < 11.512 km
(Tech: 8.103 km < μ < 513 km)
10. x ± tα / 2
s
= 0.719 ± 1.943 ⋅
0.366
n
7
0.450 ppm < μ < 0.988 ppm
(If the original values are used, the upper limit is 0.987 ppm.)
Copyright © 2014 Pearson Education, Inc.
Chapter 7: Estimates and Sample Sizes 107
11. The $1 salary of Jobs is an outlier that is very far away from the other values, and that outlier has a
dramatic effect on the confidence interval.
x ± tα / 2
s
= 12898 ± 2.776 ⋅
7719.05
n
5
3315.1 thousand dollars < μ < 22480.9 thousand dollars
(Tech: 3313.5 thousand dollars < μ < 6 22,482.5 thousand dollars)
12. The confidence interval is an estimate of the population mean and it does not apply to individual sample
values.
2.55
= 23.95 ± 2.71⋅
n
40
22.86 chocolate chips < μ < 25.04 chocolate chips
x ± tα / 2
s
13. Because the confidence interval does not contain 98.6°F, it appears that the mean body temperature is not
98.6°F, as is commonly believed.
x ± tα / 2
s
= 98.2 ± 1.98 ⋅
0.62
n
106
D
D
98.08 F < μ < 98.32 F
14. Because the confidence interval does not include 0 or negative values, it does appear that the weight loss
program is effective with a positive loss of weight. Because the amount of weight lost is relatively small,
the weight loss program does not appear to be very practical.
x ± tα / 2
s
= 2.1 ± 1.68 ⋅
n
0.8 lb < μ < 3.4 lb
4.8
40
15. Because the confidence interval includes the value of 0, it is very possible that the mean of the changes in
LDL cholesterol is equal to 0, suggesting that the garlic treatment did not affect LDL cholesterol levels. It
does not appear that garlic is effective in reducing LDL cholesterol.
x ± tα / 2
s
= 0.4 ± 2.4 ⋅
21
n
49
−6.8 mg/dL < μ < 7.6 mg/dL
16. The confidence interval includes the mean of 102.8 min that was measured before the treatment, so the
mean could be the same after the treatment. This result suggests that the zoplicone treatment has no effect.
x ± tα / 2
s
= 98.9 ± 2.6 ⋅
42.3
n
16
71.4 min < μ < 126.4 min
17. The data appear to have a distribution that is far from normal, so the confidence interval might not be a
good estimate of the population mean. The population is likely to be the list of box office receipts for each
day of the movie’s release. Because the values are from the first 14 days of release, the sample values are
not a simple random sample, and they are likely to be the largest of all such values, so the confidence
interval is not a good estimate of the population mean.
x ± tα / 2
s
= 16.4 ± 3.01⋅
14.5
n
14
4.7 million dollars < μ < 28.1 million dollars
Copyright © 2014 Pearson Education, Inc.
108 Chapter 7: Estimates and Sample Sizes
18. The confidence interval does not contain the value of 4 years. The data appear to have a distribution that is
far from normal, so the confidence interval might not be a good estimate of the population mean.
x ± tα / 2
s
6.5 ± 1.73 ⋅
3.51
n
20
5.1 years < μ < 7.9 years
19. The sample data meet the loose requirement of having a normal distribution. Because the confidence
interval is entirely below the standard of 1.6 W/kg, it appears that the mean amount of cell phone radiation
is less than the FCC standard, but there could be individual cell phones that exceed the standard.
s
0.423
= 0.938 ± 1.81⋅
n
11
0.707 W/kg < μ < 1.169 W/kg
x ± tα / 2
20. The sample data meet the loose requirement of having a normal distribution
x ± tα / 2
s
= 33.6 ± 2.62 ⋅
7.66998
n
15
28.4 years < μ < 38.8 years
21. The sample data meet the loose requirement of having a normal distribution. We cannot conclude that the
population mean is less than 7 μg/g , because the confidence interval shows that the mean might be greater
than that level.
s
x ± tα / 2
= 11.05 ± 2.26 ⋅
6.46
n
10
6.43 μg/g < μ < 15.67 μg/g
22. The sample data meet the loose requirement of having a normal distribution. The values are typical because
they are between 950 cm3 and 1800 cm3.
s
x ± tα / 2
= 1130.2 ± 3.25 ⋅
117.44
n
10
3
1009.5 cm < μ < 1250.9 cm3
23. Although final conclusions about means of populations should not be based on the overlapping of
confidence intervals, the confidence intervals do overlap, so it appears that both populations could have the
same mean, and there is not clear evidence of discrimination based on age.
CI for ages of unsuccessful applicants
x ± tα / 2
s
= 46.96 ± 2.07 ⋅
7.2
n
23
43.9 years < μ < 50.1 years
CI for ages of successful applicants
x ± tα / 2
s
= 44.5 ± 2.05 ⋅
5.03
n
30
42.6 years < μ < 46.4 years
24. Although final conclusions about means of populations should not be based on the overlapping of
confidence intervals, the confidence intervals do overlap, so it appears that both populations could have the
same mean, and there is not clear evidence that skull breadths changed from 4000 b.c. to 150 a.d.
CI for 4000 b.c.
x ± tα / 2
s
CI for 150 a.d.
= 128.7 ± 2.201⋅
4.6
n
12
125.8 mm < μ < 131.6 mm
133.33 ± 2.201⋅
5.0155
12
130.1 mm < μ < 136.5 mm
Copyright © 2014 Pearson Education, Inc.
Chapter 7: Estimates and Sample Sizes 109
2
⎡ zα / 2 σ ⎤
⎡1.645 ⋅15 ⎤
⎢
⎥
⎢
⎥
=
= 68 , and it does appear to be very reasonable.
25. The sample size is n =
⎢⎣ E ⎥⎦
⎢⎣
⎥⎦
3
2
2
⎡z σ⎤
⎡ 2.58 ⋅ 0.79 ⎤
⎥ = 104 . Limiting the sample to students at your
26. The required sample size is n = ⎢ α / 2 ⎥ = ⎢
⎢⎣ E ⎥⎦
⎢⎣ 0.2 ⎥⎦
college would result in a convenience sample that might not be representative of the population of all
college students, so it does not make sense to collect the entire sample at your college.
2
2
⎡z σ⎤
⎡ 2.33 ⋅ 2157 ⎤
⎥ = 405 (Tech: 403). It is not likely that you would
27. The required sample size is n = ⎢ α / 2 ⎥ = ⎢
⎢⎣ E ⎥⎦
⎢⎣ 250 ⎥⎦
find that many two-year-old used Corvettes in your region.
2
2
⎡z σ⎤
⎡1.96 ⋅ 210 ⎤
⎥ = 753 . A major obstacle to getting a good estimate
28. The required sample size is n = ⎢ α / 2 ⎥ = ⎢
⎢⎣ E ⎥⎦
⎢⎣ 15 ⎥⎦
of the population mean is that it would be very difficult to actually measure times spent on Facebook, so
you must rely on reported times that can be very inaccurate.
2
⎡z σ⎤
⎡ 2.33 ⋅ 450 ⎤
2400 − 600
⎥ = 110 . The margin of error
= 450 to get a sample size of n = ⎢ α / 2 ⎥ = ⎢
⎢
⎥
4
⎣⎢ 100 ⎦⎥
⎣ E ⎦
of 100 points seems too high to provide a good estimate of the mean SAT score.
2
2
29. Use σ =
⎡z σ⎤
⎡ 2.575 ⋅11250 ⎤
45, 000 − 0
⎥ = 83,919
30. Use σ =
= 11, 250 to get a sample size of n = ⎢ α / 2 ⎥ = ⎢
⎢⎣ E ⎥⎦
4
100
⎣⎢
⎦⎥
(Tech: 83,973). The sample size seems too large to be practical.
2
31. With the range rule of thumb, use σ =
2
90 − 46
= 11 to get a required sample size of
4
2
⎡z σ⎤
⎡1.96 ⋅11⎤
⎥ = 117 .
n = ⎢ α/2 ⎥ = ⎢
⎢⎣ E ⎥⎦
⎢⎣ 2 ⎦⎥
2
2
⎡z σ⎤
⎡1.96 ⋅10.3 ⎤
⎥ = 102 . The better estimate of s is the
With s = 10.3, the required sample size is n = ⎢ α / 2 ⎥ = ⎢
⎢⎣ E ⎥⎦
2
⎣⎢
⎦⎥
standard deviation of the sample, so the correct sample size is likely to be closer to 102 than 117.
2
32. With the range rule of thumb, use σ =
2.95 − 0
= 0.7375 to get a required sample size of
4
2
⎡ zα / 2 σ ⎤
⎡1.96 ⋅ 0.7375 ⎤
⎢
⎥
⎢
⎥
=
= 53 . With s = 0.587, the required sample size is 34. The better estimate of s
n=
⎢⎣ E ⎥⎦
⎢⎣
⎥⎦
0.2
is the standard deviation of the sample, so the sample size of 34 is the better result.
2
0.5873
= 1.1842 ± 2.8 ⋅
n
50
0.963 < μ < 1.407
(Tech: 0.962 < μ < 1.407)
33. x ± tα / 2
34. x ± zα / 2
s
σ
= 172.5 ± 2.023⋅
σ
5.013
= 9.808 ± 2.33 ⋅
n
50
8.156 km < μ < 11.46 km
35. x ± zα / 2
(Tech: 8.159 km < μ < 11.457 km)
119.5
n
40
134.3 ng/mL < μ < 210.7 ng/mL
Copyright © 2014 Pearson Education, Inc.
110 Chapter 7: Estimates and Sample Sizes
σ
36. x ± zα / 2
= 0.719 ± 1.645 ⋅
0.366
n
7
0.491 ppm < μ < 0.947 ppm
37.
38.
x ± zα / 2
σ
(If the original values are used, the upper limit is 0.946 ppm.)
= 12898 ± 1.96 ⋅
7717.8
n
5
6133.05 thousand dollars < μ < 19662.95 thousand dollars
(Tech: 6131.9 thousand dollars < μ < 19,663.3 thousand dollars)
x ± zα / 2
σ
= 23.95 ± 2.576 ⋅
2.55
40
n
22.91 chocolate chips < μ < 24.99 chocolate chips
39. The sample data do not appear to meet the loose requirement of having a normal distribution. The effect of
the outlier on the confidence interval is very substantial. Outliers should be discarded if they are known to
be errors. If an outlier is a correct value, it might be very helpful to see its effects by constructing the
confidence interval with and without the outlier included.
x ± tα / 2
s
= 11.375 ± 2.26 ⋅
7.1851
10
n
−24.54 m < μ < 106.04 m
(Tech: − 24.55 m < μ < 106.05 m)
40. The second confidence interval is narrower, indicating that we have a more accurate estimate when the
relatively large sample is from a relatively small finite population.
Large population:
x ± tα / 2
s
= 0.8565 ± 2.26 ⋅
0.0518
n
100
0.8462 g < μ < 0.8668 g
N −n
0.0518 465 −100
= 0.8565 ± 2.26 ⋅
Finite population:
100 −1
n n −1
100
0.8474 g < μ < 0.8656 g
41. The confidence interval based on the first sample value is much wider than the confidence interval based on
all 10 sample values.
x ± tα / 2
s
x ± 9.68 3.0
−26.0 m < μ < 32.0 m
Section 7-4
1.
916.591 (mg/dL) < σ 2 <
2
2252.1149 (mg/dL) ⇒ 30.3 mg/dL < σ < 47.5 mg/dL . We have 95%
2
confidence that the limits of 30.3 mg/dL and 47.5 mg/dL contain the true value of the standard deviation of
the LDL cholesterol levels of all women.
2.
The format implies that s = 15.7, but s is given as 14.3. In general, a confidence interval for σ does not
have s at the center.
3.
The original sample values can be identified, but the dotplot shows that the sample appears to be from a
population having a uniform distribution, not a normal distribution as required. Because the normality
requirement is not satisfied, the confidence interval estimate of s should not be constructed using the
methods of this section.
Copyright © 2014 Pearson Education, Inc.
Chapter 7: Estimates and Sample Sizes 111
4.
The normality requirement for a confidence interval estimate of σ has a much stricter normality
requirement than the loose normality requirement for a confidence interval estimate of μ . Departures from
normality have a much greater effect on confidence interval estimates of σ than on confidence interval
estimates of μ .
5.
df = 24. χ L2 = 9.886 and χ R2 = 45.559.
(n −1) s 2
χ
2
R
(25 −1) 0.242
<σ <
<σ <
6.
(n −1) s 2
(n −1) s 2
χ
χ
2
L
(25 −1) 0.242
χ
(40 −1) 65.22
(n −1) s 2
<σ <
2
R
χ L2
(40 −1) 65.22
;df = 40
59.342
24.433
52.9 < σ < 82.4 (Tech: 53.4 < σ < 83.7)
8.
<σ <
df = 49. χ L2 = 32.357 (Tech: 31.555) and χ R2 = 71.420 (Tech: 70.222).
(n −1) s 2
χ
(50 −1) 0.587 2
<σ <
2
R
(n −1) s 2
χ L2
(50 −1) 0.587 2
;df = 50
71.420
32.357
0.486 < σ < 0.722 (Tech: 0.490 < σ < 0.731)
9.
<σ <
(n −1) s 2
χ
(106 −1) 0.622
2
R
<σ <
(n −1) s 2
χ L2
(106 −1) 0.622
;df = 100
124.342
77.929
0.579D F < σ < 0.720D F (Tech: 0.557D F 6 s 6 0.700D F)
10.
<σ <
(n −1) s 2
χ R2
(40 −1) 2.552
<σ <
<σ <
<σ <
<σ <
(n −1) s 2
χ L2
(20 −1) 0.041112
38.582
6.844
0.02885 g < σ < 0.06850 g
df = 39. χ L2 = 24.433 (Tech: 23.654) and χ R2 = 59.342 (Tech: 58.120).
(n −1) s 2
2
R
(20 −1) 0.041112
45.559
9.886
0.17 mg < σ < 0.37 mg
7.
df = 19. χ L2 = 6.844 and χ R2 = 38.582.
(n −1) s 2
χ L2
(40 −1) 2.552
;df = 40
55.758
26.509
2.13 chocolate chips < σ < 3.09 chocolate chips
(Tech: 2.16 chocolate chips < σ < 3.14 chocolate chips)
Copyright © 2014 Pearson Education, Inc.
112 Chapter 7: Estimates and Sample Sizes
11. The confidence interval shows that the standard deviation is not likely to be less than 30 mL, so the
variation is too high instead of being at an acceptable level below 30 mL. (Such one-sided claims should be
tested using the formal methods presented in Chapter 8.)
(n −1) s 2
(n −1) s 2
<σ <
χ R2
(24 −1) 42.82
χ L2
(24 −1) 42.82
<σ <
44.181
9.260
30.9 mL < σ < 67.45 mL
(n −1) s 2
12. a.
χ R2
(40 −1)10.32
<σ <
<σ <
(n −1) s 2
χ L2
(40 −1)10.32
;df = 40
53.672
13.787
7.9 beats per minute < σ < 14.1 beats per minute
(Tech: 7.9 beats per minute < σ < 14.4 beats per minute)
(n −1) s 2
b.
χ R2
(40 −1)11.62
<σ <
<σ <
(n −1) s 2
χ L2
(40 −1)11.62
53.672
13.787
8.9 beats per minute < σ < 15.9 beats per minute
(Tech: 9.0 beats per minute < σ < 16.2 beats per minute)
c.
13.
The confidence intervals are not dramatically different, so it appears that the populations of pulse rates
of men and women have about the same standard deviation.
(n −1) s 2
χ R2
(7 −1) 0.365762
<σ <
<σ <
(n −1) s 2
χ L2
(7 −1) 0.365762
12.592
1.635
0.252 ppm < σ < 0.701 ppm
14.
0
Because traffic conditions vary considerably at different times during the day, the confidence interval
is an estimate of the standard deviation of the population of speeds at 3:30 on a weekday, not other times.
(n −1) s 2
χ R2
(12 −1) 4.0752
<σ <
<σ <
(n −1) s 2
χ L2
(12 −1) 4.0752
19.675
4.575
2.9 mi/h < σ < 6.9 mi/h
Copyright © 2014 Pearson Education, Inc.
Chapter 7: Estimates and Sample Sizes 113
15. CI for ages of unsuccessful applicants:
(n −1) s 2
χ R2
(25 −1) 7.222
(n −1) s 2
<σ <
χ L2
(25 −1) 7.222
<σ <
45.559
9.886
5.2 years < σ < 11.5 years
CI for ages of successful applicants:
(n −1) s 2
χ
2
R
(29 −1)5.0262
<σ <
<σ <
(n −1) s 2
χ L2
(29 −1) 5.0262
50.993
12.461
3.7 years < σ < 7.5 years
Although final conclusions about means of populations should not be based on the overlapping of
confidence intervals, the confidence intervals do overlap, so it appears that the two populations have
standard deviations that are not dramatically different.
16. a.
(n −1) s 2
χ
2
R
(10 −1) 0.4767 2
<σ <
<σ <
(n −1) s 2
χ L2
(10 −1) 0.4767 2
19.023
2.700
0.33 min < σ < 0.87 min
b.
(n −1) s 2
χ R2
(10 −1)1.82162
c.
17.
<σ <
<σ <
(n −1) s 2
χ L2
(10 −1)1.82162
19.023
2.700
1.25 min < σ < 3.33 min
The variation appears to be significantly lower with a single line. The single line appears to be better.
(n −1) s 2
χ R2
(37 −1) 0.01652
<σ <
<σ <
(n −1) s 2
χ L2
(37 −1) 0.01652
63.691
22.164
0.01239 g < σ < 0.02111 g
; df=40
(Tech: 0.01291 g < σ < 0.02255 g)
18.
(n −1) s 2
χ R2
(37 −1) 6.56132
<σ <
<σ <
(n −1) s 2
χ L2
(37 −1) 6.56132
63.691
22.164
5.2 years < σ < 8.9 years
; df=40
Copyright © 2014 Pearson Education, Inc.
114 Chapter 7: Estimates and Sample Sizes
19. 33,218 is too large. There aren’t 33,218 statistics professors in the population, and even if there were, that
sample size is too large to be practical.
20. The sample size of 48 is very practical, although the sample should be selected from the population of all
McDonald’s restaurants with drive-up windows.
21. The sample size is 768. Because the population does not have a normal distribution, the computed
minimum sample size is not likely to be correct.
22. The sample size is 1336. The population of incomes does not have a normal distribution, so the computed
sample size is not likely to be correct.
2
2
1⎡
1
−zα / 2 + 2k −1⎤⎥ = ⎡⎢−1.645 + 2 ⋅105 −1⎤⎥ = 82.072 and
⎢
⎣
⎦
⎣
⎦
2
2
2
2
1⎡
1
2
χ R = ⎢ zα / 2 + 2k −1⎤⎥ = ⎡⎢1.645 + 2 ⋅105 −1⎤⎥ = 129.635
⎦
⎦
2⎣
2⎣
23. χ L2 =
(Tech using zα / 2 = 1.644853626 : χ L2 = 82.073 and χ R2 = 129.632). The approximate values are quite close
to the actual critical values.
Chapter Quick Quiz
1.
40% − 3.1% < p < 40% + 3.1%
36.9% < p < 43.1%
2.
pˆ =
0.511 + 0.449
= 0.480
2
3.
We have 95% confidence that the limits of 0.449 and 0.511 contain the true value of the proportion of
females in the population of medical school students.
4.
z = 1.645
5.
n=
6.
2
⎡z σ⎤
⎡ 2.575 ⋅15 ⎤
⎥ = 373 (Tech: 374)
n = ⎢ α/2 ⎥ = ⎢
⎢⎣ E ⎥⎦
⎢⎣
⎥⎦
2
2
ˆˆ
[ zα / 2 ] pq
E2
[1.645] (0.25)
2
=
0.032
= 752
2
7.
The sample must be a simple random sample and there is a loose requirement that the sample values appear
to be from a normally distributed population.
8.
The degrees of freedom is the number of sample values that can vary after restrictions have been imposed
on all of the values. For the sample data in Exercise 7, df = 5.
9.
t = 2.571
10. χ L2 = 0.831 and χ R2 = 12.833
Review Exercises
1.
284
= 0.510 = 51.0%
557
a.
pˆ =
b.
pˆ ± zα / 2
c.
( 284 )( 273 )
ˆ ˆ 284
pq
=
± 1.96 557 557
n
557
557
46.8% < p < 55.1%
No, the confidence interval shows that the population percentage might be 50% or less, so we cannot
safely conclude that the majority of adults say that they are underpaid.
Copyright © 2014 Pearson Education, Inc.
Chapter 7: Estimates and Sample Sizes 115
ˆˆ
[ zα / 2 ] pq
2
[ 2.575] (0.25)
2
2.
n=
3.
2
⎡z σ⎤
⎡ 2.33⋅16 ⎤
⎥ = 155 (Tech: 154)
n = ⎢ α/2 ⎥ = ⎢
⎢⎣ E ⎥⎦
⎢⎣ 3 ⎥⎦
E2
=
0.022
= 4145 (Tech: 4147)
2
4.
a.
b.
c.
Student t distribution
Normal distribution
The distribution is not normal, Student t, or chi-square.
d.
χ 2 (chi-square distribution)
e.
Normal distribution
a.
n=
ˆˆ
[ zα / 2 ] pq
2
5.
E2
[ 2.33] (0.25)
2
=
0.052
= 543 (Tech: 542)
2
⎡z σ⎤
⎡ 2.33 ⋅ 337 ⎤
⎥ = 247 (Tech: 246)
n = ⎢ α/2 ⎥ = ⎢
⎢⎣ E ⎥⎦
⎣⎢ 50 ⎦⎥
543
2
b.
c.
6.
Because the entire confidence interval is above 50%, we can safely conclude that the majority of adults
consume alcoholic beverages.
(0.64)(0.36)
ˆˆ
pq
= 0.64 ± 1.65
n
1011
61.5% < p < 66.5%
pˆ ± zα / 2
s
= 143 ± 2.201⋅
259.7754
7.
x ± tα / 2
8.
Because women and men have some notable physiological differences, the confidence interval does not
necessarily serve as an estimate of the mean white blood cell count of men.
n
12
−22.1 sec < μ < 308.1 sec
x ± tα / 2
9.
s
= 7.15 ± 1.685 ⋅
n
6.54 < μ < 7.76
40
There is 95% confidence that the limits of 37.5 g and 47.9 g contain the true mean deceleration
measurement for all small cars.
x ± tα / 2
s
= 42.7 ± 2.447 ⋅
n
37.5 g < μ < 47.9 g
10.
2.28
(n −1) s 2
χ R2
(7 −1) 5.62
<σ <
<σ <
5.6
7
(n −1) s 2
χ L2
(7 −1) 5.62
14.449
1.237
3.6 g < σ < 12.3 g
Cumulative Review Exercises
1.
x = 5.5 ; median = 5.0; s = 3.8
Copyright © 2014 Pearson Education, Inc.
116 Chapter 7: Estimates and Sample Sizes
2.
The range of usual values is from 5.5 − 2 (3.8) = −2.1 to 5.5 + 2 (3.8) = 3.8 (or from 0 to 13.1).
2
⎡z σ⎤
⎡1.96 ⋅ 5.8 ⎤
⎥ = 33 campuses
n = ⎢ α/2 ⎥ = ⎢
⎢⎣ E ⎥⎦
⎢⎣
2 ⎥⎦
The population should include only colleges of the same type as the sample, so the population consists of
all large urban campuses with residence halls.
2
3.
5.
4.
Ratio level of measurement; discrete data.
x ± tα / 2
s
5.8
= 5.5 ± 2.02 ⋅
n
40
3.6 < μ < 7.4
6.
The graphs suggest that the population has a distribution that is skewed (to the right) instead of being
normal. The histogram shows that some taxi-out times can be very long, and that can occur with heavy
traffic, but little or no traffic cannot make the taxi-out time very low. There is a minimum time required,
regardless of traffic conditions. Construction of a confidence interval estimate of a population standard
deviation has a strict requirement that the sample data are from a normally distributed population, and the
graphs show that this strict normality requirement is not satisfied.
7.
a.
(0.59)(0.31)
ˆˆ
pq
= 0.59 ± 1.96
(or 0.560 < p < 0.621 if using x = 592)
n
1003
0.560 < p < 0.620
pˆ ± zα / 2
b.
Because the survey was about shaking hands and because it was sponsored by a supplier of hand
sanitizer products, the sponsor could potentially benefit from the results, so there might be some
pressure to obtain results favorable to the sponsor.
c.
n=
ˆˆ
[ zα / 2 ] pq
2
E2
[1.96] (0.25)
2
=
0.0252
= 1083
8.
There does not appear to be a correlation between HDL and LDL cholesterol levels.
9.
a.
b.
10. a.
b.
185 −175
= 1.11 and P ( z > 1.11) = 13.35% (Tech: 13.32%).
9
Yes, losing about 13% of the market would be a big loss.
5th percentile: x = μ + z ⋅ σ = 175 −1.645 ⋅ 9 = 160.2 mm
95th percentile: x = μ + z ⋅ σ = 175 + 1.645 ⋅ 9 = 189.8 mm
z=
There are 103 possible tickets so the probability of winning by purchasing one ticket is
1−
1
999
=
.
1000 1000
⎛ 999 ⎞⎟
⎜⎜
= 0.990.
⎜⎝1000 ⎠⎟⎟
10
c.
Copyright © 2014 Pearson Education, Inc.
1
.
1000
Chapter 8: Hypothesis Testing 117
Chapter 8: Hypothesis Testing
Section 8-2
1.
Rejection of the aspirin claim is more serious because the aspirin is a drug treatment. The wrong aspirin
dosage can cause adverse reactions. M&Ms do not have those same adverse reactions. It would be wise to
use a smaller significance level for testing the aspirin claim.
2.
Estimates and hypothesis tests are both methods of inferential statistics, but they have different objectives.
We could use the sample weights to construct a confidence interval estimate of the mean weight of all
M&Ms, but hypothesis testing is used to test some claim made about the mean weight of all M&Ms.
3.
a.
H0: μ = 98.6°F
b.
H1: μ ≠ 98.6°F
c.
d.
Reject the null hypothesis or fail to reject the null hypothesis.
No. In this case, the original claim becomes the null hypothesis. For the claim that the mean body
temperature is equal to 98.6°F, we can either reject that claim or fail to reject it, but we cannot state
that there is sufficient evidence to support that claim.
4.
The P-value of 0.001 is preferred because it corresponds to the sample evidence that most strongly supports
the alternative hypothesis that the XSORT method is effective.
5.
a.
b.
6.
7.
8.
9.
a.
p = 0.20
H0: p = 0.20 and H1: p ≠ 0.20
p > 0.5
b.
H0: p = 0.5 and H1: p > 0.5
a.
μ ≤ 76
b.
H0: μ = 76 and H1: μ < 76
a.
σ ≥ 50
b.
H0: σ = 50 and H1: σ > 50
There is not sufficient evidence to warrant rejection of the claim that 20% of adults smoke.
10. There is sufficient evidence to support the claim that when parents use the XSORT method of gender
selection, the proportion of baby girls is greater than 0.5.
11. There is not sufficient evidence to warrant rejection of the claim that the mean pulse rate of adult females is
76 or lower.
12. There is sufficient evidence to reject the claim that pulse rates of adult females have a standard deviation of
at least 50.
ˆ
13. z = p − p = 0.89 − 0.75 = 10.33 (or z = 10.35 if using x = 909)
(0.75)(0.25)
pq
n
1021
ˆ
14. z = p − p = 0.48 − 0.50 = −1.27 (or z = –1.26 if using x = 481)
(0.48)(0.52)
pq
n
15. χ 2 =
1002
(n −1) s 2
σ2
=
(40 −1) 2.282
52
= 8.110
16. t = x − μ = 7.15 − 8 = −2.358
s n 2.28 40
Copyright © 2014 Pearson Education, Inc.
118
Chapter 8: Hypothesis Testing
17. P-value = P ( z > 2) = 0.0228. Critical value: z = 1.645 .
18. P-value = P ( z < −2) = 0.0228. Critical value: z = −1.645 .
19. P-value = 2 ⋅ P ( z < −1.75) = 0.0802. (Tech: 0.0801). Critical values: z = −1.96, z = 1.96 .
20. P-value = 2 ⋅ P ( z > 1.50) = 0.1336. . Critical values: z = −1.96, z = 1.96 .
21. P-value = 2 ⋅ P ( z < −1.23) = 0.2186. (Tech: 0.2187). Critical values: z = −1.96, z = 1.96 .
22. P-value = 2 ⋅ P ( z > 2.50) = 0.0124. Critical values: z = −1.96, z = 1.96 .
23. P-value = P ( z < −3.00) = 0.0013. Critical value: z = −1.645 .
24. P-value = P ( z > 2.88) = 0.0020. Critical value: z = 1.645 .
25. a.
b.
Reject H0.
There is sufficient evidence to support the claim that the percentage of blue M&Ms is greater than 5%.
26. a.
b.
Fail to reject H0.
There is not sufficient evidence to support the claim that fewer than 20% of M&M candies are green.
27. a.
b.
Fail to reject H0.
There is not sufficient evidence to warrant rejection of the claim that women have heights with a mean
equal to 160.00 cm.
28. a.
b.
Reject H0.
There is sufficient evidence to warrant rejection of the claim that women have heights with a standard
deviation equal to 5.00 cm.
29. a.
H0: p = 0.5 and H1: p > 0.5
30. a.
b.
α = 0.01
b.
c.
d.
e.
Normal distribution.
Right-tailed.
z = 1.00
c.
d.
e.
f.
P-value = P ( z > 1.00) = 0.1587.
f.
g.
z = 2.33
h.
0.01
H0: p = 0.5 and H1: p ≠ 0.5
α = 0.05
Normal distribution.
Two-tailed.
z = 1.00
P-value = 2 ⋅ P ( z > 1.00) = 0.3174.
(Tech: 0.3173)
g.
h.
z = −1.96, z = 1.96
0.05
31. Type I error: In reality p = 0.1 , but we reject the claim that p = 0.1 . Type II error: In reality p ≠ 0.1 , but
we fail to reject the claim that p = 0.1 .
32. Type I error: In reality p = 0.001 , but we reject the claim that p = 0.001 . Type II error: In reality
p ≠ 0.001 , but we fail to reject the claim that p = 0.001 .
33. Type I error: In reality p = 0.5 , but we support the claim that p > 0.5 . Type II error: In reality p > 0.5 ,
but we fail to support that conclusion.
34. Type I error: In reality p = 0.9 , but we support the claim that p < 0.9 . Type II error: In reality p < 0.9 ,
but we fail to support that conclusion.
Copyright © 2014 Pearson Education, Inc.
Chapter 8: Hypothesis Testing 119
35. The power of 0.96 shows that there is a 96% chance of rejecting the null hypothesis of p = 0.08 when the
true proportion is actually 0.18. That is, if the proportion of Chantix users who experience abdominal pain
is actually 0.18, then there is a 96% chance of supporting the claim that the proportion of Chantix users
who experience abdominal pain is greater than 0.08.
36. a.
From p = 0.5 , pˆ = 0.5 + 1.645
From p = 0.65 , z =
(0.5)(0.5)
64
0.6028125 − .65
(0.65)(0.35)
= 0.6028125
= −0.791 ; Power = P ( z > −0.791) = 0.7852. (Tech: 0.7857)
64
b.
Assuming that p = 0.5 , as in the null hypothesis, the critical value of z = 1.645 corresponds to
pˆ = 0.6028125 , so any sample proportion greater than 0.6028125 causes us to reject the null
hypothesis, as shown in the shaded critical region of the top graph. If p is actually 0.65, then the null
hypothesis of p = 0.5 is false, and the actual probability of rejecting the null hypothesis is found by
finding the area greater than pˆ = 0.6028125 in the bottom graph, which is the shaded area. That is, the
shaded area in the bottom graph represents the probability of rejecting the false null hypothesis.
37. From p = 0.5 , pˆ = 0.5 + 1.645
(0.5)(0.5)
n
, from p = 0.55 , pˆ = 0.55 − 0.842
(0.55)(0.45)
n
; Since
( P ( z > −0.842) = 0.8000) , so:
0.5 + 1.645
(0.5)(0.5)
n
= 0.55 − 0.842
(0.55)(0.45)
n
0.5 n + 1.645 0.25 = 0.55 n − 0.842 0.2475
0.05 n = 1.645 0.25 + 0.842 0.2475
⎛1.645 0.25 + 0.842 0.2475 ⎞⎟
n = ⎜⎜⎜
⎟⎟ = 617
⎜⎝
0.05
⎠⎟
2
Section 8-3
1.
2.
The P-value method and the critical value method always yield the same conclusion. The confidence
interval method might or might not yield the same conclusion obtained by using the other two methods.
pˆ =
411
= 0.410. The symbol p̂ is used to represent a sample proportion.
1003
3.
P-value = 0.00000000550. Because the P-value is so low, we have sufficient evidence to support the claim
that p < 0.5 .
4.
a.
b.
c.
5.
The symbol p represents the population proportion, but the P-value is a probability of getting sample
results that are at least as extreme as those obtained (assuming that the null hypothesis is true).
If the P-value is very low (such as less than or equal to 0.05), “the null must go” means that we should
reject the null hypothesis.
The statement that “if the P is high, the null will fly” suggests that with a high P-value, the null
hypothesis has been proved or is supported, but we should never make such a conclusion.
a.
b.
Left-tailed.
z = −1.94
c.
d.
P-value = 0.0260 (rounded)
H0: p = 0.1 . Reject the null hypothesis.
e.
There is sufficient evidence to support the claim that less than 10% of treated subjects experience
headaches.
Copyright © 2014 Pearson Education, Inc.
120
6.
7.
8.
9.
Chapter 8: Hypothesis Testing
a.
b.
Two-tailed.
z = 1.45
c.
d.
P-value = 0.146
H0: p = 0.35 . Fail to reject the null hypothesis.
e.
There is not sufficient evidence to warrant rejection of the claim that 35% of homes have guns in them.
a.
b.
Two-tailed.
z = −0.82
c.
d.
P-value = 0.4106
H0: p = 0.35 . Fail to reject the null hypothesis.
e.
There is not sufficient evidence to warrant rejection of the claim that 35% of adults have heard of the
Sony Reader.
a.
b.
Left-tailed.
z = −2.53
c.
d.
P-value = 0.0057
H0: p = 0.5 . Reject the null hypothesis.
e.
There is sufficient evidence to support the claim that fewer than half of adults say that public speaking
is the activity that they dread most.
H0: p = 0.25 . H1: p ≠ 0.25 . Test statistic: z =
152
580
− 0.25
(0.25)(0.75)
= 0.67 . Critical values: z = ±2.575
580
(Tech: ±2.576 ). P-value = 2 ⋅ P ( z > 0.67) = 0.5028 (Tech: 0.5021). Fail to reject H0. There is not
sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow.
MINITAB
Test of p = 0.25 vs p not = 0.25
Sample
X
N Sample p
1
152
580 0.262069
95% CI
(0.226280, 0.297858)
10. H0: p = 0.13 . H1: p ≠ 0.13 . Test statistic: z =
0.08 − 0.13
(0.13)(0.87)
Z-Value
0.67
P-Value
0.502
= −1.49 . Critical values: z = ±1.96 .
100
P-value = 2 ⋅ P ( z > 1.49) = 0.1362 (Tech: 0.1371). Fail to reject H0. There is not sufficient evidence to
warrant rejection of the claim that 13% of M&Ms are brown.
MINITAB
Test of p = 0.13 vs p not = 0.13
Sample
X
N Sample p
1
8
100 0.080000
(0.026828, 0.133172)
11. H0: p = 0.5 . H1: p > 0.5 . Test statistic: z =
531
1002
− 0.5
(0.5)(0.5)
95% CI
-1.49
Z-Value
0.137
P-Value
= 1.90 . Critical value: z = 1.645 . P-value
1002
= P ( z > 1.90) = 0.0287 (Tech: 0.0290). Reject H0. There is sufficient evidence to support the claim that
the majority of adults feel vulnerable to identify theft.
MINITAB
Test of p = 0.5 vs p > 0.5
Sample
X
N Sample p
1
531 1002 0.529940
Z-Value
1.90
P-Value
0.029
Copyright © 2014 Pearson Education, Inc.
Chapter 8: Hypothesis Testing 121
12. H0: p = 0.5 . H1: p > 0.5 . Test statistic: z =
492
806
− 0.5
(0.5)(0.5)
= 6.27 . Critical value: z = 2.33 . P-value
806
= P ( z > 6.27) = 0.0001 (Tech: 0.000000000182). Reject H0. There is sufficient evidence to support the
claim that the majority of adults prefer window seats when they fly.
MINITAB
Test of p = 0.5 vs p > 0.5
Sample
X
N Sample p
1
492
806 0.610422
Z-Value
6.27
13. H0: p = 0.5 . H1: p > 0.5 . Test statistic: z =
879
945
P-Value
0.000
− 0.5
(0.5)(0.5)
= 26.45 . Critical value: z = 2.33 . P-value
945
= P ( z > 26.45) = 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that
the XSORT method is effective in increasing the likelihood that a baby will be a girl.
MINITAB
Test of p = 0.5 vs p > 0.5
Sample
X
N Sample p
1
879
945 0.930159
Z-Value
26.45
14. H0: p = 0.5 . H1: p > 0.5 . Test statistic: z =
239
291
P-Value
0.000
− 0.5
(0.5)(0.5)
= 10.96 . Critical value: z = 2.33 . P-value
291
= P ( z > 10.96) = 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that
the YSORT method is effective in increasing the likelihood that a baby will be a boy.
MINITAB
Test of p = 0.5 vs p > 0.5
Sample
X
N Sample p
1
239
291 0.821306
Z-Value
10.96
15. H0: p = 0.5 . H1: p ≠ 0.5 . Test statistic: z =
123
280
P-Value
0.000
− 0.5
(0.5)(0.5)
= −2.03 . Critical values: z = ±1.645 . P-value
280
= 2 ⋅ P ( z < −2.03) = 0.0424 (Tech: 0.0422). Reject H0. There is sufficient evidence to warrant rejection of
the claim that touch therapists use a method equivalent to random guesses. However, their success rate of
123/280 (or 43.9%) indicates that they performed worse than random guesses, so they do not appear to be
effective.
MINITAB
Test of p = 0.5 vs p not = 0.5
Sample
X
N Sample p
1
123
280 0.439286
95% CI
Z-Value
(0.381154, 0.497417)
-2.03
16. H0: p = 0.5 . H1: p ≠ 0.5 . Test statistic: z =
123
280
− 0.5
(0.5)(0.5)
P-Value
0.042
= −2.03 . Critical values: z = ±2.575 (Tech:
280
±2.576 ). P-value = 2 ⋅ P ( z < −2.03) = 0.0424 (Tech: 0.0422). Reject H0. There is not sufficient evidence
to warrant rejection of the claim that touch therapists use a method equivalent to random guesses. However,
their success rate of 123/280 (or 43.9%) indicates that they performed worse than random guesses, so they
do not appear to be effective.
MINITAB
Test of p = 0.5 vs p not = 0.5
Sample
X
N Sample p
1
123
280 0.439286
95% CI
Z-Value
(0.381154, 0.497417)
-2.03
Copyright © 2014 Pearson Education, Inc.
P-Value
0.042
122
Chapter 8: Hypothesis Testing
17. H0: p = 13 . H1: p < 13 . Test statistic: z =
172
611
− 13
= −2.72 . Critical value: z = −2.33 . P-value
( 13 )( 23 )
611
= P ( z < −2.72) = 0.0033 . Reject H0. There is sufficient evidence to support the claim that fewer than 1/3
of the challenges are successful. Players don’t appear to be very good at recognizing referee errors.
MINITAB
Test of p = 0.3333 vs p < 0.3333
Sample
X
N Sample p
1
172
611 0.281506
Z-Value
-2.72
18. H0: p = 0.43 . H1: p ≠ 0.43 . Test statistic: z =
P-Value
0.003
308
611
− 0.43
(0.43)(0.57)
= 3.70 . Critical values: z = ±1.645 .
601
P-value = 2 ⋅ P ( z > 3.70) = 0.0002 . Reject H0. There is sufficient evidence to warrant rejection of the
claim that the percentage who believe that they voted for the winning candidate is equal to 43%. There
appears to be a substantial discrepancy between how people said that they voted and how they actually did
vote.
MINITAB
Test of p = 0.43 vs p not = 0.43
Sample
X
N Sample p
1
308 611 0.504092
95% CI
Z-Value
(0.464447, 0.543736)
3.70
19. H0: p = 0.000340 . p ≠ 0.000340 . Test statistic: z =
135
420,095
− 0.000340
(0.000340)(0.99966)
P-Value
0.000
= −0.66 . Critical values:
420,095
z = ±2.81 . P-value = 2 ⋅ P ( z < −0.66) = 0.5092 (Tech: 0.5122). Fail to reject H0. There is not sufficient
evidence to support the claim that the rate is different from 0.0340%. Cell phone users should not be
concerned about cancer of the brain or nervous system.
MINITAB
Test of p = 0.00034 vs p not = 0.00034
Sample
X
N Sample p
95% CI
Z-Value
1
135 420095 0.000321 (0.000267, 0.000376)
-0.66
20. H0: p = 0.75 . H1: p > 0.75 . Test statistic: z =
856
1007
− 0.75
(0.75)(0.25)
P-Value
0.512
= 7.33 . Critical value: z = 2.33 . P-value
1007
= P ( z > 7.33) = 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that
more than 75% of adults know what Twitter is.
MINITAB
Test of p = 0.75 vs p > 0.75
Sample
X
N Sample p
1
856 1007 0.850050
Z-Value
7.33
21. H0: p = 0.5 . H1: p ≠ 0.5 . Test statistic: z =
235
414
P-Value
0.000
− 0.5
(0.5)(0.5)
414
= 2.75 . Critical values: z = ±1.96 . P-value
= 2 ⋅ P ( z > 2.75) = 0.0060 (Tech: 0.0059). Reject H0. There is sufficient evidence to warrant rejection of
the claim that the coin toss is fair in the sense that neither team has an advantage by winning it. The coin
toss rule does not appear to be fair.
MINITAB
Test of p = 0.5 vs p not = 0.5
Sample
X
N Sample p
1
235
414 0.567633
95% CI
Z-Value
(0.519912, 0.615354)
2.75
Copyright © 2014 Pearson Education, Inc.
P-Value
0.006
Chapter 8: Hypothesis Testing 123
22. H0: p = 0.5 . H1: p > 0.5 . Test statistic: z =
39
71
− 0.5
= 0.83 . Critical value: z = 1.645 . P-value
(0.5)(0.5)
71
= 2 ⋅ P ( z > 0.83) = 0.2033 (Tech: 0.2031). Fail to reject H0. There is not sufficient evidence to support the
claim that among smokers who try to quit with nicotine patch therapy, the majority are smoking a year after
the treatment. The results show that about half of those who use nicotine patch therapy are successful in
quitting smoking.
MINITAB
Test of p = 0.5 vs p > 0.5
Sample
X
N Sample p
1
39
71 0.549296
Z-Value
0.83
23. H0: p = 0.5 . H1: p < 0.5 . Test statistic: z =
152
380
P-Value
0.203
− 0.5
(0.5)(0.5)
380
= −3.90 . Critical value: z = −2.33 . P-value
= P ( z < −3.90) = 0.0001 (Tech: 0.0000484). Reject H0. There is sufficient evidence to support the claim
that fewer than half of smartphone users identify the smartphone as the only thing they could not live
without. Because only smartphone users were surveyed, the results do not apply to the general population.
MINITAB
Test of p = 0.5 vs p < 0.5
Sample
X
N Sample p
1
152
380 0.400000
Z-Value
-3.90
24. H0: p = 0.5 . H1: p < 0.5 . Test statistic: z =
P-Value
0.000
6062
12000
− 0.5
(0.5)(0.5)
= 1.13 . Critical value: z = −1.645 . P-value
12000
= P ( z < 1.13) = 0.8708 (Tech: 0.8712). Fail to reject H0. There is not sufficient evidence to support the
claim that less than 0.5 of the deaths occur the week before Thanksgiving. Based on these results, there is
no indication that people can temporarily postpone their death to survive Thanksgiving.
MINITAB
Test of p = 0.5 vs p < 0.5
Sample
X
N Sample p
1
6062 12000 0.505167
Z-Value
1.13
25. H0: p = 0.25 . H1: p > 0.25 . Test statistic: z =
P-Value
0.871
0.29 − 0.25
(0.25)(0.75)
= 1.91 or z = 1.93 (using x = 124). Critical
427
value: z = 1.645 (assuming a 0.05 significance level). P-value = P ( z > 1.92) = 0.0281 (using pˆ = 0.29 )
or 0.0268 (using x = 124) (Tech P-value = 0.0269). Reject H0. There is sufficient evidence to support the
claim that more than 25% of women purchase books online.
MINITAB
Test of p = 0.25 vs p > 0.25
Sample
X
N Sample p
1
124
427 0.290398
Z-Value
1.93
26. H0: p = 0.5 . H1: p < 0.5 . Test statistic: z =
P-Value
0.027
0.459 − 0.5
(0.5)(0.5)
= −1.86 or z = –1.85 (using x = 236). Critical
514
value: z = −2.33 . P-value = P ( z < −1.86) = 0.0314 (using pˆ = 0.459 ) or 0.0322 (using x = 236) (Tech
P-value = 0.0320). Fail to reject H0. There is not sufficient evidence to support the claim that less than half
of all human resource professionals say that body piercings are big grooming red flags.
MINITAB
Test of p = 0.5 vs p < 0.5
Sample
X
N Sample p
1
236
514 0.459144
Z-Value
-1.85
P-Value
0.032
Copyright © 2014 Pearson Education, Inc.
124 Chapter 8: Hypothesis Testing
27. H0: p = 0.75 . H1: p > 0.75 . Test statistic: z =
0.90 − 0.75
(0.75)(0.25)
= 7.85 or z = 7.89 (using x = 463). Critical
514
value: z = 2.33 . P-value = P ( z > 7.85) = 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence
to support the claim that more than 3/4 of all human resource professionals say that the appearance of a job
applicant is most important for a good first impression.
MINITAB
Test of p = 0.75 vs p > 0.75
Sample
X
N Sample p
1
463
514 0.900778
Z-Value
7.89
28. H0: p = 0.61 . H1: p ≠ 0.61 . Test statistic: z =
P-Value
0.000
0.70 − 0.61
(0.61)(0.39)
= 5.84 or z = 5.81 (using x = 701). Critical
1002
values: z = ±1.96 (assuming a 0.05 significance level). P-value = 2 ⋅ P ( z > 5.81) = 0.0002 (Tech:
0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the percentage of all
voters who say that they voted is equal to 61%. The results suggest that either survey respondents are not
being truthful or they have an incorrect perception of reality.
MINITAB
Test of p = 0.61 vs p not = 0.61
Sample
X
N Sample p
1
701 1002 0.699601
95% CI
Z-Value
(0.671216, 0.727986)
5.81
29. H0: p = 0.791 . H1: p < 0.791 . Test statistic: z =
0.39 − 0.791
(0.791)(0.209)
P-Value
0.000
= −29.09 or z = –29.11 (using x = 339).
870
Critical value: z = −2.33 . P-value = 2 ⋅ P ( z < −29.09) = 0.0001 (Tech: 0.0000). Reject H0. There is
sufficient evidence to support the claim that the percentage of selected Americans of Mexican ancestry is
less than 79.1%, so the jury selection process appears to be unfair.
MINITAB
Test of p = 0.791 vs p < 0.791
Sample
X
N Sample p
1
339
870 0.389655
Z-Value
-29.11
30. H0: p = 0.5 . H1: p > 0.5 . Test statistic: z =
P-Value
0.000
0.61− 0.5
(0.5)(0.5)
= 5.83 or z = 5.85 (using x = 429). Critical value:
703
z = 1.645 . P-value = P ( z > 5.83) = 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to
support the claim that most workers get their jobs through networking.
MINITAB
Test of p = 0.5 vs p > 0.5
Sample
X
N Sample p
1
429
703 0.610242
Z-Value
5.85
31. H0: p = 0.75 . H1: p > 0.75 . Test statistic: z =
P-Value
0.000
0.77 − 0.75
(0.75)(0.25)
= 7.30 . Critical value: z = 2.33 . P-value
25,000
= P ( z > 7.30) = 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that
more than 75% of television sets in use were tuned to the Super Bowl.
MINITAB
Test of p = 0.75 vs p > 0.75
Sample
X
N Sample p
1
19250 25000 0.770000
Z-Value
7.30
P-Value
0.000
Copyright © 2014 Pearson Education, Inc.
Chapter 8: Hypothesis Testing 125
32. H0: p = 0.5 . H1: p < 0.5 . Test statistic: z =
0.47 − 0.5
(0.5)(0.5)
= −2.32 . Critical value: z = −2.33 . P-value
1500
= P ( z < −2.32) = 0.0102 (Tech: 0.0101). Fail to reject H0. There is not sufficient evidence to support the
claim that fewer than half of all households have a high-definition television. Because the use of highdefinition televisions is growing rapidly, these results are not likely to be valid today.
MINITAB
Test of p = 0.5 vs p < 0.5
Sample
X
N Sample p
1
705 1500 0.470000
Z-Value
-2.32
P-Value
0.010
33. Among 100 M&Ms, 19 are green. H0: p = 0.16 . H1: p ≠ 0.16 . Test statistic: z =
0.19 − 0.16
(0.16)(0.84)
= 0.82 .
100
Critical values: z = ±1.96 . P-value = 2 ⋅ P ( z > 0.82) = 0.4122 (Tech: 0.4132). Fail to reject H0. There is
not sufficient evidence to warrant rejection of the claim that 16% of plain M&M candies are green.
MINITAB
Test of p = 0.16 vs p not = 0.16
Sample
X
N Sample p
1
19
100 0.190000
95% CI
Z-Value
(0.113110, 0.266890)
0.82
P-Value
0.413
34. Among 48 flights, 44 are on time. H0: p = 0.795 . H1: p ≠ 0.795 . Test statistic: z =
44
48
− 0.795
(0.795)(0.205)
= 2.09 .
48
Critical values: z = ±1.96 . P-value = 2 ⋅ P ( z > 2.09) = 0.0366 (Tech: 0.0368). Reject H0. There is
sufficient evidence to warrant rejection of the claim that 79.5% of flights are on time. With 91.7% of the 48
flights arriving on time, American Airlines appears to have a better on-time performance.
MINITAB
Test of p = 0.795 vs p not = 0.795
Sample
X
N Sample p
1
44
48 0.916667
95% CI
Z-Value
(0.838478, 0.994855)
2.09
P-Value
0.037
35. H0: p = 0.5 . H1: p > 0.5 . Using the binomial probability distribution with an assumed proportion of
p = 0.5 , the probability of 7 or more heads is 0.0352, so the P-value is 0.0352. Reject H0. There is
sufficient evidence to support the claim that the coin favors heads.
36. a.
H0: p = 0.10 . H1: p ≠ 0.1 . Test statistic: z =
0.119 − 0.1
(0.1)(0.9)
= 2.00 . Critical values: z = ±1.96 . Reject
1000
H0. There is sufficient evidence to warrant rejection of the claim that the proportion of zeros is 0.1.
b.
H0: p = 0.10 . H1: p ≠ 0.1 . Test statistic: z =
0.119 − 0.1
(0.1)(0.9)
= 2.00 . P-value
1000
c.
= 2 ⋅ P ( z > 2.00) = 0.0456 (Tech: 0.0452). There is sufficient evidence to warrant rejection of the
claim that the proportion of zeros is 0.1.
0.0989 < p < 0.139 ; because 0.1 is contained within the confidence interval, fail to reject
H0: p = 0.10 . There is not sufficient evidence to warrant rejection of the claim that the proportion of
zeros is 0.1.
MINITAB
Sample
X
1
119
d.
N
1000
Sample p
0.119000
95% CI
(0.098932, 0.139068)
The traditional and P-value methods both lead to rejection of the claim, but the confidence interval
method does not lead to rejection of the claim.
Copyright © 2014 Pearson Education, Inc.
126 Chapter 8: Hypothesis Testing
From p = 0.40 , pˆ = 0.4 −1.645
37. a.
From p = 0.25 , z =
0.286 − 0.25
(0.25)(0.75)
(0.4)(0.6)
50
= 0.286
= 0.588 ; Power = P ( z < 0.588) = 0.7224. (Tech: 0.7219)
50
b.
1 – 0.7224 = 0.2776 (Tech: 0.2781)
c.
The power of 0.7224 shows that there is a reasonably good chance of making the correct decision of
rejecting the false null hypothesis. It would be better if the power were even higher, such as greater
than 0.8 or 0.9.
Section 8-4
1.
The requirements are (1) the sample must be a simple random sample, and (2) either or both of these
conditions must be satisfied: The population is normally distributed or n > 30 . There is not enough
information given to determine whether the sample is a simple random sample. Because the sample size is
not greater than 30, we must check for normality, but the value of 583 sec appears to be an outlier, and a
normal quantile plot or histogram suggests that the sample does not appear to be from a normally
distributed population.
Probability Plot
5
0.99
0.95
4
Probability
Frequency
0.9
3
2
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
1
0.05
0
0
100
200
300
400
500
600
0.01
-300
-200
-100
0
100
200
300
400
500
600
2.
df denotes the number of degrees of freedom. For the sample of 12 times, df = 12 −1 = 11 .
3.
A t test is a hypothesis test that uses the Student t distribution, such as the method of testing a claim about a
population mean as presented in this section. The t test methods are much more likely to be used than the z
test methods because the t test does not require a known value of σ , and realistic hypothesis tests of claims
about μ typically involve a population with an unknown value of σ .
4.
Use a 90% confidence level. The given confidence interval does contain the value of 90 sec, so it is
possible that the value of μ is equal to 90 sec or some lower value, so there is not sufficient evidence to
support the claim that the mean is greater than 90 sec.
5.
P-value < 0.005 (Tech: 0.0013).
7.
0.02 < P-value < 0.05 (Tech: 0.0365).
6.
0.025 < P-value < 0.05 (Tech: 0.0480).
8.
0.01 < P-value < 0.02 (Tech: 0.0183)
9.
H0: μ = 24 . H1: μ < 24 . Test statistic: t = −7.323 . Critical value: t = −1.685 . P-value < 0.005. (The
display shows that the P-value is 0.00000000387325.) Reject H0. There is sufficient evidence to support the
claim that Chips Ahoy reduced-fat cookies have a mean number of chocolate chips that is less than 24 (but
this does not provide conclusive evidence of reduced fat).
10. H0: μ = 10 km. H1: μ ≠ 10 km. Test statistic: t = −0.27 . Critical values: t = ±2.678 (approximately). Pvalue > 0.20. (Minitab shows a P-value of 0.790.) Fail to reject H0. There is not sufficient evidence to
warrant rejection of the claim that the earthquakes are from a population with a mean depth equal to 10 km.
Copyright © 2014 Pearson Education, Inc.
Chapter 8: Hypothesis Testing 127
35.9 − 33
= 2.367 . Critical values t = ±2.639
11.1 82
(approximately). P-value > 0.02 (Tech: 0.0204). Fail to reject H0. There is not sufficient evidence to
warrant rejection of the claim that the mean age of actresses when they win Oscars is 33 years.
11. H0: μ = 33 years. H1: μ ≠ 33 years. Test statistic: t =
MINITAB
Test of mu = 33 vs not = 33
N
Mean StDev SE Mean
82 35.90 11.10
1.23
95% CI
T
(33.46, 38.34) 2.37
P
0.020
1.911−1.800
= 0.821 . Critical value: t = 1.671
1.065 62
(approximately). P-value > 0.10 (Tech: 0.2075). Fail to reject H0. There is not sufficient evidence to
support the claim that the mean weight of discarded plastic from the population of households is greater
than 1.800 lb.
12. H0: μ = 1.800 lb. H1: μ > 1.800 lb. Test statistic: t =
MINITAB
Test of mu = 1.8 vs > 1.8
N
Mean StDev SE Mean
62 1.911 1.065
0.135
T
0.82
P
0.208
13. H0: μ = 0.8535 g. H1: μ ≠ 0.8535 g. Test statistic: t =
0.8635 − 0.8535
= 0.765 . Critical values:
0.0570 19
t = ±2.101 . P-value > 0.20 (Tech: 0.4543). Fail to reject H0. There is not sufficient evidence to warrant
rejection of the claim that the mean weight of all green M&Ms is equal to 0.8535 g. The green M&Ms do
appear to have weights consistent with the package label.
MINITAB
Test of mu = 0.8535 vs not = 0.8535
N
Mean StDev SE Mean
95% CI
T
19 0.8635 0.0570
0.0131 (0.8360, 0.8910) 0.76
P
0.454
98.2 − 98.6
= −6.642 . Critical values: t = ±1.984
0.62 106
(approximately). P-value < 0.01 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection
of the claim that the mean body temperature of the population is equal to 98.6°F. There is sufficient
evidence to conclude that the common belief is wrong.
14. H0: μ ≠ 98.6 °F. H1: μ ≠ 98.6 °F. Test statistic: t =
MINITAB
Test of mu = 98.6 vs not = 98.6
N
Mean StDev SE Mean
95% CI
T
P
106 98.2000 0.6200
0.0602 (98.0806, 98.3194) -6.64 0.000
3− 0
= 3.872 . Critical value: t = 2.426 .
4.9 40
P-value < 0.005 (Tech: 0.0002). Reject H0. There is sufficient evidence to support the claim that the mean
weight loss is greater than 0. Although the diet appears to have statistical significance, it does not appear to
have practical significance, because the mean weight loss of only 3.0 lb does not seem to be worth the
effort and cost.
15. H0: μ = 0 lb. H1: μ > 0 lb. Test statistic: t =
MINITAB
Test of mu = 0 vs > 0
N
Mean StDev
40 3.000 4.900
SE Mean
0.775
T
3.87
P
0.000
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128
Chapter 8: Hypothesis Testing
10.5 −12.0
= −0.337 . Critical value: t = −2.412
430.8 48
(approximately). P-value > 0.10 (Tech: 0.3687). Fail to reject H0. There is not sufficient evidence to
support the claim that the mean departure delay time for all such flights is less than 12.0 min. A flight
operations manager is not justified in reporting that the mean departure time is less than 12.0 min.
16. H0: μ = 12.0 min. H1: μ < 12.0 min. Test statistic: t =
MINITAB
Test of mu = 12 vs < 12
N
Mean StDev SE Mean
48 10.50 30.80
4.45
T
-0.34
P
0.369
0.4 − 0
= 0.133 . Critical value: t = 1.676 (approximately,
21.0 49
assuming a 0.05 significance level). P-value > 0.10 (Tech: 0.4472). Fail to reject H0. There is not sufficient
evidence to support the claim that with garlic treatment, the mean change in LDL cholesterol is greater than
0. The results suggest that the garlic treatment is not effective in reducing LDL cholesterol levels.
17. H0: μ = 0 . H1: μ > 0 . Test statistic: t =
MINITAB
Test of mu = 0 vs > 0
N
Mean StDev
49
0.40 21.00
SE Mean
3.00
T
0.13
P
0.447
18. H0: μ = 102.8 min. H1: μ < 102.8 min. Test statistic: t =
98.9 −102.8
= −0.369 . Critical value:
42.3 16
t = −1.753 (assuming a 0.05 significance level). P-value > 0.10 (Tech: 0.3587). Fail to reject H0. There is
not sufficient evidence to support the claim that after treatment with Zopiclone, subjects have a mean wake
time less than 102.8 min. This result suggests that the Zoplicone treatment is not effective.
MINITAB
Test of mu = 102.8 vs < 102.8
N
Mean StDev SE Mean
16
98.9
42.3
10.6
T
-0.37
P
0.359
19. H0: μ = 4 years. H1: μ > 4 years. Test statistic: t = 3.189 . Critical value: t = 2.539 .
P-value < 0.005 (Tech: 0.0024). Reject H0. There is sufficient evidence to support the claim that the mean
time required to earn a bachelor’s degree is greater than 4.0 years. Because n ≤ 30 and the data do not
appear to be from a normally distributed population, the requirement that “the population is normally
distributed or n > 30 ” is not satisfied, so the conclusion from the hypothesis test might not be valid.
However, some of the sample values are equal to 4 years and others are greater than 4 years, so the claim
does appear to be justified.
12
Frequency
10
8
6
4
2
0
4
6
8
10
12
Years of College
MINITAB
Test of mu = 4 vs > 4
Variable
N Mean
CollegeYears 20 6.500
StDev
3.506
SE Mean
0.784
T
3.19
P
0.002
Copyright © 2014 Pearson Education, Inc.
14
Chapter 8: Hypothesis Testing 129
20. The sample data meet the loose requirement of having a normal distribution. H0: μ = 30 years.
H1: μ > 30 years. Test statistic: t = 1.818 . Critical value: t = 1.761 . P-value < 0.05 (Tech: 0.0453). Reject
H0. There is sufficient evidence to support the claim that the mean age of all race car drivers is greater than
30 years.
MINITAB
Test of mu = 30 vs > 30
Variable
N
Mean
Ages
15 33.60
StDev
7.67
SE Mean
1.98
T
1.82
P
0.045
21. The sample data meet the loose requirement of having a normal distribution. H0: μ = 14 mg/g.
H1: μ < 14 mg/g. Test statistic: t = −1.444 . Critical value: t = −1.833 . P-value > 0.05 (Tech: 0.0913).
Fail to reject H0. There is not sufficient evidence to support the claim that the mean lead concentration for
all such medicines is less than 14 mg/g.
MINITAB
Test of mu = 14 vs < 14
Variable
N
Mean
Lead
10 11.05
StDev
6.46
SE Mean
2.04
T
-1.44
P
0.091
22. The sample data meet the loose requirement of having a normal distribution. H0: μ = 1100 cm3.
H1: μ ≠ 1100 cm3. Test statistic : t = 0.813 . Critical values: t = ±3.250 . P-value > 0.20 (Tech: 0.4371).
Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the population of
brain volumes has a mean equal to 1100.0 cm3.
MINITAB
Test of mu = 1100 vs not = 1100
Variable
N
Mean
StDev
Volume 10 1130.2
117.4
SE Mean
37.1
95% CI
(1046.2, 1214.2)
T
0.81
P
0.437
23. The sample data meet the loose requirement of having a normal distribution. H0: μ = 63.8 in.
H1: μ > 63.8 in. Test statistic: t = 23.824 . Critical value: t = 2.821 . P-value < 0.005 (Tech: 0.0000).
Reject H0. There is sufficient evidence to support the claim that supermodels have heights with a mean that
is greater than the mean height of 63.8 in. for women in the general population. We can conclude that
supermodels are taller than typical women.
MINITAB
Test of mu = 63.8 vs > 63.8
Variable
N
Mean
StDev
Heights 10 69.825
0.800
SE Mean
0.253
T
23.82
P
0.000
24. The sample data meet the loose requirement of having a normal distribution. H0: μ = 65 mi/h. H1:
μ < 65 mi/h. Test statistic: t = −3.684 . Critical value: t = −1.796 . P-value < 0.005 (Tech: 0.0018).
Reject H0. There is sufficient evidence to support the claim that the sample is from a population with a
mean that is less than the speed limit of 65 mi/h.
MINITAB
Test of mu = 65 vs < 65
Variable
N
Mean
HWY
12 60.67
StDev
4.08
SE Mean
1.18
T
-3.68
P
0.002
25. The sample data meet the loose requirement of having a normal distribution. H0: μ = 1.00 . H1: μ > 1.00 .
Test statistic: t = 2.218 . Critical value: t = 1.676 (approximately). P-value < 0.025 (Tech: 0.0156). Reject
H0. There is sufficient evidence to support the claim that the population of earthquakes has a mean
magnitude greater than 1.00.
MINITAB
Test of mu = 1 vs > 1
Variable
N
Mean
MAG
5 0 1.1842
StDev
0.5873
SE Mean
0.0831
Bound
1.0449
T
2.22
Copyright © 2014 Pearson Education, Inc.
P
0.016
130
Chapter 8: Hypothesis Testing
26. The sample data meet the loose requirement of having a normal distribution. H0: μ = 120 mm Hg. H1:
μ < 120 mm Hg. Test statistic: t = −0.424 . Critical value: t = −1.685 . P-value > 0.10 (Tech: 0.3370).
Fail to reject H0. There is not sufficient evidence to support the claim that the female population has a mean
systolic blood pressure level less than 120.0 mm Hg.
MINITAB
Test of mu = 120 vs < 120
Variable
N
Mean
StDev
SYS
40 118.50
22.39
SE Mean
3.54
T
-0.42
P
0.337
27. The sample data meet the loose requirement of having a normal distribution. H0: μ = 83 kg. H1: μ < 83 kg.
Test statistic: t = −5.524 . Critical value: t = −2.453 . P-value < 0.005 (Tech: 0.0000). Reject H0. There is
sufficient evidence to support the claim that male college students have a mean weight that is less than the
83 kg mean weight of males in the general population.
MINITAB
Test of mu = 83 vs < 83
Variable
N
Mean
WTSEP 32 72.72
StDev
10.53
SE Mean
1.86
T
-5.52
P
0.000
28. The sample data meet the loose requirement of having a normal distribution. H0: μ = 120 V. H1:
μ ≠ 120 V. Test statistic: t = 96.358 . Critical values: t = ±2.708 . P-value < 0.01 (Tech: 0.0000). Reject
H0. There is sufficient evidence to warrant rejection of the claim that the mean voltage amount is 120 volts.
MINITAB
Test of mu = 120 vs not = 120
Variable
N
Mean
StDev
Home
40 123.663
0.240
SE Mean
0.038
29. H0: μ = 24 . H1: μ < 24 . Test statistic: z =
95% CI
(123.586, 123.739)
19.6 − 24
3.8
40
T
96.36
P
0.000
= −7.32 . Critical value: z = −1.645 .
P-value = P ( z < −7.32) = 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the
claim that Chips Ahoy reduced-fat cookies have a mean number of chocolate chips that is less than 24 (but
this does not provide conclusive evidence of reduced fat).
30. H0: μ = 10 km. H1: μ ≠ 10 km. Test statistic: z =
9.810 −10
5.01 50
= −0.27 . Critical values: z = ±2.575 . P-
value = 2 ⋅ P ( z < −0.27) = 0.7872 (Tech: 0.7886). Fail to reject H0. There is not sufficient evidence to
warrant rejection of the claim that the earthquakes are from a population with a mean depth equal to 10 km.
31. H0: μ = 33 years. H1: μ ≠ 33 years. Test statistic: z =
35.9 − 33
11.1 82
= 2.37 . Critical values: z = ±2.575 . P-
value = 2 ⋅ P ( z > 2.37) = 0.0178 (Tech: 0.0180). Fail to reject H0. There is not sufficient evidence to
warrant rejection of the claim that the mean age of actresses when they win Oscars is 33 years.
32. H0: μ = 1.800 lb. H1: μ > 1.800 lb. Test statistic: z =
1.911−1.800
1.065
62
= 0.82 . Critical value: z = 1.645 . P-
value = P ( z > 0.82) = 0.2061 (Tech: 0.2059). Fail to reject H0. There is not sufficient evidence to support
the claim that the mean weight of discarded plastic from the population of households is greater than 1.800
lb.
33. A =
1.645(8 ⋅149 + 3)
8 ⋅149 −1
(
t = 149 e1.6505247
2
/149
= 1.6505247 . The approximation yields a critical value of
)
−1 = 1.655 , which is the same as the result from STATDISK or a TI-83/84 Plus
calculator.
Copyright © 2014 Pearson Education, Inc.
Chapter 8: Hypothesis Testing 131
34. Using the normal distribution makes you more likely to reject the null hypothesis because the critical z
values are not as extreme as the corresponding critical t values.
35. a.
b.
The power of 0.4274 shows that there is a 42.74% chance of supporting the claim that μ < 1 W/kg
when the true mean is actually 0.80 W/kg. This value of power is not very high, and it shows that the
hypothesis test is not very effective in recognizing that the mean is less than 1.00 W/kg when the
actual mean is 0.80 W/kg.
β = 0.5726. The probability of a type II error is 0.5726. That is, there is a 0.5726 probability of
making the mistake of not supporting the claim that μ < 1 W/kg when in reality the population mean is
0.80 W/kg.
Section 8-5
1.
a.
b.
c.
d.
2.
a.
b.
The mean waiting time remains the same.
The variation among waiting times is lowered.
Because customers all have waiting times that are roughly the same, they experience less stress and are
generally more satisfied. Customer satisfaction is improved.
The single line is better because it results in lower variation among waiting times, so a hypothesis test
of a claim of a lower standard deviation is a good way to verify that the variation is lower with a single
waiting line.
The normality requirement for a hypothesis test of a claim about a standard deviation is much more
strict, meaning that the distribution of the population must be much closer to a normal distribution.
With only 10 sample values, a histogram doesn’t really give us a good picture of the distribution, so a
normal quantile plot would be better. Also, we should determine that there are no outliers.
3.
Use a 90% confidence interval. The conclusion based on the 90% confidence interval will be the same as
the conclusion from a hypothesis test using the P-value method or the critical value method.
4.
a.
b.
c.
d.
e.
5.
H0: σ = 1.8 min. H1: σ < 1.8 min.
χ2 =
(n −1) s 2
σ2
=
(10 −1) 0.52
= 0.694
Reject H0, the null hypothesis.
There is sufficient evidence to support the claim that the standard deviation of waiting times of all
customers is less than 1.8 min.
The change to a single waiting line is effective because the variation among waiting times is less than
it was with multiple lines.
H0: σ = 0.15 oz. H1: σ < 0.15 oz. Test statistic: χ 2 =
(36 −1) 0.112
= 18.822 . Critical value of χ 2 is
0.152
between 18.493 and 26.509, so it is estimated to be 22.501 (Tech: 22.465). P-value < 0.05 (Tech: 0.0116).
Reject H0. There is sufficient evidence to support the claim that the population of volumes has a standard
deviation less than 0.15 oz.
MINITAB
Method Chi-Square
Standard
18.82
6.
1.82
DF
35
P-Value
0.012
H0: σ = 0.15 oz. H1: σ < 0.15 oz. Test statistic: χ 2 =
(36 −1) 0.092
= 12.600 . Critical value of χ 2 is
0.152
between 18.493 and 26.509, so it is estimated to be 22.501 (Tech: 22.465). P-value < 0.05 (Tech: 0.0002).
Reject H0. There is sufficient evidence to support the claim that the population of volumes has a standard
deviation less than 0.15 oz.
MINITAB
Method Chi-Square
Standard
12.60
DF
35
P-Value
0.000
Copyright © 2014 Pearson Education, Inc.
132
7.
Chapter 8: Hypothesis Testing
H0: σ = 0.0230 g. H1: σ < 0.0230 g. Test statistic: χ 2 =
= 18.483 . Critical value of χ 2
0.02302
is between 18.493 and 26.509, so it is estimated to be 22.501 (Tech: 23.269). P-value < 0.05 (Tech:
0.0069). Reject H0. There is sufficient evidence to support the claim that the population of weights has a
standard deviation less than the specification of 0.0230 g.
MINITAB
Method Chi-Square
Standard
18.48
8.
DF
36
P-Value
0.007
H0: σ = 0.0230 g. H1: σ > 0.0230 g. Test statistic: χ 2 =
(35 −1) 0.039102
= 98.260 . Critical value of χ 2
0.02302
is between 43.773 and 55.758. P-value < 0.005 (Tech: 0.0000). Reject H0. There is sufficient evidence to
support the claim that pre-1983 pennies have a standard deviation greater than 0.0230 g. Weights of pre1983 pennies appear to vary more than those of post-1983 pennies.
MINITAB
Method Chi-Square
Standard
98.26
9.
(37 −1) 0.016482
DF
34
P-Value
0.000
The data appear to be from a normally distributed population. H0: σ = 10 bpm. H1: σ ≠ 10 bpm. Test
(40 −1)10.32
= 41.375 . Critical value of χ 2 = 24.433 and χ 2 = 59.342 (approximately).
statistic: χ 2 =
2
10
P-value > 0.20 (Tech: 0.7347). Fail to reject H0. There is not sufficient evidence to warrant rejection of the
claim that pulse rates of men have a standard deviation equal to 10 beats per minute.
MINITAB
Method Chi-Square
Standard
41.38
DF
39
P-Value
0.735
10. The data appear to be from a normally distributed population. H0: σ = 10 bpm. H1: σ ≠ 10 bpm. Test
(40 −1)11.62
= 52.478 . Critical values of χ 2 = 24.433 and χ 2 = 59.342 (approximately).
statistic: χ 2 =
102
P-value > 0.10 (Tech: 0.1463). Fail to reject H0. There is not sufficient evidence to warrant rejection of the
claim that pulse rates of women have a standard deviation equal to 10 beats per minute.
MINITAB
Method Chi-Square
Standard
52.48
DF
39
P-Value
0.146
11. H0: σ = 3.2 mg. H1: σ ≠ 3.2 mg. Test statistic: χ 2 =
(25 −1) 3.7 2
= 32.086 . Critical values: χ 2 = 12.401
3.22
and χ 2 = 39.364. P-value > 0.20 (Tech: 0.2498). Fail to reject H0. There is not sufficient evidence to
support the claim that filtered 100-mm cigarettes have tar amounts with a standard deviation different from
3.2 mg. There is not enough evidence to conclude that filters have an effect.
MINITAB
Method Chi-Square
Standard
32.09
DF
24
P-Value
0.250
12. H0: σ = 28.866 cents. H1: σ ≠ 28.866 cents. Test statistic: χ 2 =
(100 −1) 33.52
= 133.337 . Critical values:
28.8662
χ 2 = 67.328 and χ 2 = 140.169 (approximately). P-value > 0.02 (Tech: 0.0244). Fail to reject H0. There is
not sufficient evidence to warrant rejection of the claim that the standard deviation is 28.866 cents. Because
the amounts from 0 cents to 99 cents are all equally likely, the requirement of a normal distribution is
violated, so the results are highly questionable.
Copyright © 2014 Pearson Education, Inc.
Chapter 8: Hypothesis Testing 133
12. (continued)
MINITAB
Method Chi-Square
Standard
133.34
DF
99
P-Value
0.024
13. The data appear to be from a normally distributed population. H0: σ = 22.5 years. H1: σ < 22.5 years. Test
(15 −1) 7.67 2
statistic: χ 2 =
= 1.627 . Critical value: χ 2 = 4.660. P-value < 0.005 (Tech: 0.0000). Reject
22.52
H0. There is sufficient evidence to support the claim that the standard deviation of ages of all race car
drivers is less than 22.5 years.
MINITAB
Variable
Ages
Method
Standard
Chi-Square
1.63
DF
14.00
P-Value
0.000
14. The data appear to be from a normally distributed population. H0: σ = 5 mi/h. H1: σ ≠ 5 mi/h. Test
(12 −1) 4.082
= 7.307 . Critical values of χ 2 = 3.816 and χ 2 = 21.920. P-value > 0.20
statistic: χ 2 =
5.02
(Tech: 0.4525). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the
standard deviation of speeds is equal to 5.0 mi/h.
MINITAB
Variable
HWY
Method
Standard
Chi-Square
7.31
DF
11.00
P-Value
0.453
15. The data appear to be from a normally distributed population. H0: σ = 32.2 ft. H1: σ > 32.2 ft. Test
(12 −1)52.42
= 29.176 . Critical value: χ 2 = 19.675. P-value = 0.0021. Reject H0. There is
statistic: χ 2 =
32.22
sufficient evidence to support the claim that the new production method has errors with a standard
deviation greater than 32.2 ft. The variation appears to be greater than in the past, so the new method
appears to be worse, because there will be more altimeters that have larger errors. The company should take
immediate action to reduce the variation.
MINITAB
Variable
Errors
Method
Standard
Chi-Square
29.18
DF
11.00
P-Value
0.002
16. The data appear to be from a normally distributed population. H0: σ = 15 . H1: σ < 15 . Test statistic:
(12 −1)9.502
χ2 =
= 4.416 . Critical value: χ 2 = 4.575. P-value < 0.05 (Tech: 0.0439). Reject H0. There
152
is sufficient evidence to support the claim that IQ scores of professional pilots have a standard deviation
less than 15.
MINITAB
Variable
IQ
Method
Standard
Chi-Square DF
4.42
11.00
P-Value
0.044
17. The data appear to be from a normally distributed population. H0: σ = 0.15 oz. H1: σ < 0.15 oz. Test
(36 −1) 0.08092
= 10.173 . Critical value of χ 2 is between 18.493 and 26.509, so it is
statistic: χ 2 =
0.152
estimated to be 22.501 (Tech: 22.465). P-value < 0.01 (Tech: 0.0000). Reject H0. There is sufficient
evidence to support the claim that the population of volumes has a standard deviation less than 0.15 oz.
MINITAB
Variable
Method Chi-Square DF
CKDTVOL Standard
10.17 35.00
P-Value
0.000
Copyright © 2014 Pearson Education, Inc.
134
Chapter 8: Hypothesis Testing
18. The data appear to be from a normally distributed population. H0: σ = 0.0230 g. H1: σ ≠ 0.0230 g. Test
(35 −1) 0.03912
= 156.155 . Critical values of χ 2 = 13.787 and χ 2 = 53.672
statistic: χ 2 =
0.02302
(approximately). P-value < 0.01 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection
of the claim that the population of weights has a standard deviation equal to 0.0230 g.
MINITAB
Variable
Wheat
Method
Standard
Chi-Square
156.16
(
DF
34.00
P-Value
0.000
)
2
1
−1.645 + 2 ⋅ 35 −1 = 22.189 , which is reasonably close to the value of 22.465 obtained
2
from STATDISK and Minitab.
19. Critical χ 2 =
⎛
⎛
2
2 ⎞⎟⎟⎞⎟⎟
⎜
20. Critical χ 2 = 35⎜⎜1−
+ ⎜⎜⎜−1.645
⎟⎟ = 22.642 , which is very close to the value of 22.465
⎜⎝ 9 ⋅ 35 ⎜⎝
9 ⋅ 35 ⎟⎠⎟⎠⎟
3
obtained from STATDISK and Minitab.
Chapter Quick Quiz
1.
H0: μ = 0 sec. H1: μ ≠ 0 sec.
2.
a.
b.
3.
a. Fail to reject H0.
b. There is not sufficient evidence to warrant rejection of the claim that the sample is from a population
with a mean equal to 0 sec.
4.
There is a loose requirement of a normally distributed population in the sense that the test works reasonably
well if the departure from normality is not too extreme.
5.
a.
H0: p = 0.5 . H1: p > 0.5 .
b.
z=
Two-tailed.
Student t.
0.64 − 0.5
(0.5)(0.5)
= 6.33
511
c.
P-value = 0.0000000001263996. There is sufficient evidence to support the claim that the majority of
adults are in favor of the death penalty for a person convicted of murder.
6.
= 2 ⋅ P ( z < −2.00) = 0.0456 (Tech: 0.0455)
7.
The only true statement is the one given in part (a).
8.
No. All critical values of x2 are greater than zero.
9.
True.
10. False.
Review Exercises
1.
a.
b.
c.
False.
True.
False.
d.
e.
False.
False.
Copyright © 2014 Pearson Education, Inc.
Chapter 8: Hypothesis Testing 135
2.
H0: p = 23 . H1: p ≠ 23 . Test statistic: z =
657
1010
− 32
( 23 )( 13 )
= −1.09 . Critical values: z = ±2.575 (Tech: ±2.576 ).
1010
P-value = 2 ⋅ P ( z < −1.09) = 0.2758 (Tech: 0.2756). Fail to reject H0. There is not sufficient evidence to
warrant rejection of the claim that 2/3 of adults are satisfied with the amount of leisure time that they have.
MINITAB
Test of p = 0.666667 vs p not = 0.666667
Sample
X
N Sample p
95% CI
Z-Value P-Value
1
657
1010 0.650495 (0.621089, 0.679901) -1.09 0.27
3.
H0: p = 0.75 . H1: p > 0.75 . Test statistic: z =
678
737
− 0.75
(0.75)(0.25)
= 10.65 or z = 10.66 (if using pˆ = 0.92 ).
737
Critical value: z = ±2.33 . P-value = 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to
support the claim that more than 75% of us do not open unfamiliar e-mail and instant-message links. Given
that the results are based on a voluntary response sample, the results are not necessarily valid.
MINITAB
Test of p = 0.75 vs p > 0.75
Sample
X
N Sample p
1
678
737 0.919946
Z-Value P-Value
10.65
0.000
3245 − 567
4.
= −19.962 . Critical value: t = −2.328
446 81
(approximately). P-value < 0.005 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the
claim that the mean birth weight of Chinese babies is less than the mean birth weight of 3369 g for
Caucasian babies.
5.
H0: σ = 567 g. H1: σ ≠ 567 g. Test statistic: χ 2 =
H0: μ = 3369 g. H1: μ < 3369 g. Test statistic: t =
= 54.038 . Critical values of χ 2 = 51.172
567 2
and χ 2 = 116.321. P-value is between 0.02 and 0.05 (Tech: 0.0229). Fail to reject H0. There is not
sufficient evidence to warrant rejection of the claim that the standard deviation of birth weights of Chinese
babies is equal to 567 g.
MINITAB
Method Chi-Square
Standard
54.04
DF
80
P-Value
0.023
H0: μ = 1.5 mg/m3. H1: μ > 1.5 mg/m3. Test statistic : t = 0.049 . Critical value: t = 2.015 .
P-value > 0.10 (Tech: 0.4814). Fail to reject H0. There is not sufficient evidence to support the claim that
the sample is from a population with a mean greater than the EPA standard of 1.5 mg/m3. Because the
sample value of 5.40 mg/m3 appears to be an outlier and because a normal quantile plot suggests that the
sample data are not from a normally distributed population, the requirements of the hypothesis test are not
satisfied, and the results of the hypothesis test are therefore questionable.
Probability Plot of Air Lead
0.99
0.95
0.9
Probability
6.
(81−1) 4662
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
0.01
-4
-2
0
2
4
Air Lead
Copyright © 2014 Pearson Education, Inc.
6
136
6.
Chapter 8: Hypothesis Testing
(continued)
MINITAB
Test of mu = 1.5 vs > 1.5
Variable
N Mean
Air Lead
6 1.538
7.
StDev SE Mean
1.914 0.781
a.
b.
9.
P
0.481
24.2 − 25
= −0.567 . Critical values: t = ±1.984
14.1 100
(approximately). P-value > 0.20 (Tech: 0.5717). Fail to reject H0. There is not sufficient evidence to
warrant rejection of the claim that the sample is selected from a population with a mean equal to 25.
H0: μ = 25 . H1: μ ≠ 25 . Test statistic: t =
MINITAB
Test of mu = 25 vs not = 25
N
Mean StDev SE Mean
100 24.20 14.10
1.41
8.
T
0.05
95% CI
(21.40, 27.00)
T
-0.57
P
0.572
A type I error is the mistake of rejecting a null hypothesis when it is actually true. A type II error is the
mistake of failing to reject a null hypothesis when in reality it is false.
Type I error: Reject the null hypothesis that the mean of the population is equal to 25 when in reality,
the mean is actually equal to 25. Type II error: Fail to reject the null hypothesis that the population
mean is equal to 25 when in reality, the mean is actually different from 25.
The χ 2 test has a reasonably strict requirement that the sample data must be randomly selected from a
population with a normal distribution, but the numbers are selected in such a way that they are all equally
likely, so the population has a uniform distribution instead of the required normal distribution. Because the
requirements are not all satisfied, the χ 2 2 test should not be used.
10. The sample data meet the loose requirement of having a normal distribution. H0: μ = 1000 HIC.
H1: μ < 1000 HIC. Test statistic: t = −10.177 . Critical value: t = −3.747 . P-value < 0.005 (Tech:
0.0003). Reject H0. There is sufficient evidence to support the claim that the population mean is less than
1000 HIC. The results suggest that the population mean is less than 1000 HIC, so they appear to satisfy the
specified requirement.
MINITAB
Test of mu = 1000 vs < 1000
Variable
N Mean
StDev SE Mean
Booster
5 653.8
76.1
34.0
T
-10.18
P
0.000
Cumulative Review Exercises
s2 = 245.1 words2
Range = 45 words
a.
x = 53.3 words
b.
c.
Median = 52.0 words
s = 15.7 words
2.
a.
c.
Ratio.
b. Discrete.
The sample is a simple random sample if it was selected in such a way that all possible samples of the
same size have the same chance of being selected.
3.
42.1 words < μ < 64.5 words
1.
d.
e.
MINITAB
Variable N Mean StDev SE Mean
X
10 53.30 15.66
4.95
95% CI
(42.10, 64.50)
Copyright © 2014 Pearson Education, Inc.
Chapter 8: Hypothesis Testing 137
4.
MINITAB
Test of mu = 48 vs > 48
N Mean StDev SE Mean
10 53.30 15.70
4.96
5.
6.
53.3 − 48.0
= 1.070 . Critical value: t = 1.833 .
15.7 10
P-value > 0.10 (Tech: 0.1561). Fail to reject H0. There is not sufficient evidence to support the claim that
the mean number of words on a page is greater than 48.0. There is not enough evidence to support the
claim that there are more than 70,000 words in the dictionary.
H0: μ = 48.0 words. H1: μ > 48.0 words. Test statistic: t =
T
P
1.07 0.157
38.8 − 36.0
= 2; P ( z > 2) = 2.28%.
1.4
a.
z=
b.
98th percentile: x = μ + z ⋅ σ = 36.0 + 2.054 ⋅1.4 = 38.9 in.
c.
z=
a.
(0.125) = 0.00195 . It is unlikely because the probability of the event occurring is so small.
b.
(0.097)(0.125) = 0.0121
37.0 − 36.0
1.4
4
= 1.43; P ( z < 1.43) = 92.36%. (Tech: 0.9234)
3
c.
1− (0.875) = 0.487
5
7.
No. The distribution is very skewed. A normal distribution would be approximately bell-shaped, but the
displayed distribution is very far from being bell-shaped.
8.
Because the vertical scale starts at 7000 and not at 0, the difference between the number of males and the
number of females is exaggerated, so the graph is deceptive by creating the wrong impression that there are
many more male graduates than female graduates.
9.
a.
0.372 (1003) = 373
b.
34.2% < p < 40.2%
MINITAB
Sample X
N Sample p
1
373 1003 0.371884
c.
Yes. With test statistic z = −8.11 and with a P-value close to 0, there is sufficient evidence to support
the claim that less than 50% of adults answer “yes.”
MINITAB
Test of p = 0.5 vs p < 0.5
Sample X
N Sample p
1
373 1003 0.371884
d.
95% CI
(0.341974, 0.401795)
Z-Value P-Value
-8.11
0.000
The required sample size depends on the confidence level and the sample proportion, not the
population size.
10. H0: p = 0.5 . H1: p < 0.5 . Test statistic: z =
0.372 − 0.5
(0.5)(0.5)
= −8.11 . Critical value: z = −2.33 . P-value
1003
= P ( z < −8.11) = 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that
fewer than 50% of Americans say that they have a gun in their home.
Copyright © 2014 Pearson Education, Inc.
Chapter 9: Inferences from Two Samples 139
Chapter 9: Inferences from Two Samples
Section 9-2
1.
The samples are simple random samples that are independent. For each of the two groups, the number of
successes is at least 5 and the number of failures is at least 5. (Depending on what we call a success, the
four numbers are 33, 115, 201,229 and 200,745 and all of those numbers are at least 5.) The requirements
are satisfied.
2.
n1 = 201,229 , pˆ1 =
33
= 0.000163992 , qˆ1 = 1− 0.000163992 = 0.999836 , n2 = 200,745,
201,299
115
= 0.000572866 , qˆ2 = 1− 0.000572866 = 0.999427 ,
200, 745
33 + 115
p=
= 0.000368183 , and q = 1− 0.000368183 = 0.999632 .
201, 299 + 200, 745
pˆ 2 =
3.
4.
a.
H0: p1 = p2 . H1: p1 < p2 .
b.
If the P-value is less than 0.001 we should reject the null hypothesis and conclude that there is
sufficient evidence to support the claim that the rate of polio is less for children given the Salk vaccine
than it is for children given a placebo.
a.
b.
0.90, or 90%
Because the confidence interval limits do not contain 0, there appears to be a significant difference
between the two proportions. Because the confidence interval consists of negative values only, it
appears that the first proportion is less than the second proportion. There is sufficient evidence to
support the claim that the rate of polio is less for children given the Salk vaccine than it is for children
given a placebo.
The P-value method and the critical value method are equivalent in the sense that they will always lead
to the same conclusion, but the confidence interval method is not equivalent to them.
c.
5.
Test statistic: z = −12.39 (rounded). The P-value of 3.137085E–35 is 0.0000 when rounded to four
decimal places. There is sufficient evidence to warrant rejection of the claim that the vaccine has no effect.
6.
Test statistic: z = 2.17 . P-value: 0.030. Because the P-value is greater than the significance level of 0.01,
conclude that there is not sufficient evidence to warrant rejection of the claim that for those saying that
monitoring e-mail is seriously unethical, the proportion of workers is the same as the proportion of
managers.
For Exercises 7 – 18, assume that the data fit the requirements for the statistical methods for two proportions
unless otherwise indicated.
7.
a.
H0: p1 = p2 . H1: p1 > p2 . Test statistic: z = 6.44 . Critical value: z = 2.33 . P-value: 0.0001 (Tech:
0.0000). Reject H0. There is sufficient evidence to support the claim that the proportion of people over
55 who dream in black and white is greater than the proportion for those under 25.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs > 0): Z = 6.44 P-Value = 0.000
b.
98% CI: 0.117 < p1 − p2 < 0.240. Because the confidence interval limits do not include 0, it appears
that the two proportions are not equal. Because the confidence interval limits include only positive
values, it appears that the proportion of people over 55 who dream in black and white is greater than
the proportion for those under 25.
MINITAB
Difference = p (1) - p (2)
98% CI for difference: (0.116836, 0.240360)
Copyright © 2014 Pearson Education, Inc.
140 Chapter 9: Inferences from Two Samples
7.
8.
(continued)
c.
The results suggest that the proportion of people over 55 who dream in black and white is greater than
the proportion for those under 25, but the results cannot be used to verify the cause of that difference.
a.
H0: p1 = p2 . H1: p1 < p2 . Test statistic: z = −1.66 . Critical value: z = −2.33 . P-value: 0.0485
(Tech: 0.0484). Fail to reject H0. There is not sufficient evidence to support the claim that the rate of
dementia among those who use ginkgo is less than the rate of dementia among those who use a
placebo. There is not sufficient evidence to support the claim that ginkgo is effective in preventing
dementia.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs < 0): Z = -1.66 P-Value = 0.048
b.
98% CI: –0.0542 < p1 − p2 < 0.00909 (Tech: –0.0541 < p1 − p2 < 0.00904). Because the confidence
interval limits include 0, there does not appear to be a significant difference between dementia rates for
those treated with ginkgo and those given a placebo. There is not sufficient evidence to support the
claim that the rate of dementia among those who use ginkgo is less than the rate of dementia among
those who use a placebo. There is not sufficient evidence to support the claim that ginkgo is effective
in preventing dementia.
MINITAB
Difference = p (1) - p (2)
98% CI for difference: (-0.0541115, 0.00904103)
9.
c.
The sample results suggest that ginkgo is not effective in preventing dementia.
a.
H0: p1 = p2 . H1: p1 > p2 . Test statistic: z = 6.11 . Critical value: z = 1.64 5. P-value: 0.0001 (Tech:
0.0000). Reject H0. There is sufficient evidence to support the claim that the fatality rate is higher for
those not wearing seat belts.
MINITAB
Test for difference = 0 (vs > 0): Z = 6.11 P-Value = 0.000
b.
90% CI: 0.00556 < p1 − p2 < 0.0122. Because the confidence interval limits do not include 0, it
appears that the two fatality rates are not equal. Because the confidence interval limits include only
positive values, it appears that the fatality rate is higher for those not wearing seat belts.
MINITAB
Difference = p (1) - p (2)
90% CI for difference: (0.00558525, 0.0122561)
c.
10. a.
The results suggest that the use of seat belts is associated with lower fatality rates than not using seat
belts.
H0: p1 = p2 . H1: p1 ≠ p2 . Test statistic: z = 18.26 . Critical values: z = ±2.575 (Tech: ±2.576 ). Pvalue: 0.0002 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim
that the survival rates are the same for day and night.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 18.26 P-Value = 0.000
b.
99% CI: 0.0441 < p1 − p2 < 0.0579. Because the confidence interval limits do not contain 0, there
appears to be a significant difference between the two proportions. There is sufficient evidence to
warrant rejection of the claim that the survival rates are the same for day and night.
MINITAB
Difference = p (1) - p (2)
99% CI for difference: (0.0441419, 0.0579311)
c.
The data suggest that for in-hospital patients who suffer cardiac arrest, the survival rate is not the same
for day and night. It appears that the survival rate is higher for in-hospital patients who suffer cardiac
arrest during the day.
Copyright © 2014 Pearson Education, Inc.
Chapter 9: Inferences from Two Samples 141
11. a.
H0: p1 = p2 . H1: p1 ≠ p2 . Test statistic: z = 0.57 . Critical values: z = ±1.96 . P-value: 0.5686
(Tech: 0.5720). Fail to reject H0. There is not sufficient evidence to support the claim that echinacea
treatment has an effect.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 0.57 P-Value = 0.572
b.
95% CI: –0.0798 < p1 − p2 < 0.149. Because the confidence interval limits do contain 0, there is not
a significant difference between the two proportions. There is not sufficient evidence to support the
claim that echinacea treatment has an effect.
MINITAB
Difference = p (1) - p (2)
95% CI for difference: (-0.0798112, 0.148851)
c.
12. a.
Echinacea does not appear to have a significant effect on the infection rate. Because it does not appear
to have an effect, it should not be recommended.
H0: p1 = p2 . H1: p1 < p2 . Test statistic: z = −2.44 . Critical value: z = −2.33 . P-value: 0.0074.
Reject H0. There is sufficient evidence to support the claim that the incidence of malaria is lower for
infants who use the bednets.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs < 0): Z = -2.44 P-Value = 0.007
b.
98% CI: 0.0950 < p1 − p2 < –0.00118 (Tech: –0.0950 < p1 − p2 < –0.00125). Because the
confidence interval does not include 0 and it includes only negative values, it appears that the rate of
malaria is lower for infants who use the bednets.
MINITAB
Difference = p (1) - p (2)
98% CI for difference: (-0.0949568, -0.00125315)
c.
13. a.
The bednets appear to be effective.
H0: p1 = p2 . H1: p1 ≠ p2 . Test statistic: z = 0.40 . Critical values: z = ±1.96 . P-value: 0.6892
(Tech: 0.6859). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that
men and women have equal success in challenging calls.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 0.40 P-Value = 0.686
b.
95% CI: –0.0318 < p1 − p2 < 0.0484. Because the confidence interval limits contain 0, there is not a
significant difference between the two proportions. There is not sufficient evidence to warrant rejection
of the claim that men and women have equal success in challenging calls.
MINITAB
Difference = p (1) - p (2)
95% CI for difference: (-0.0318350, 0.0484421)
c.
14. a.
It appears that men and women have equal success in challenging calls.
H0: p1 = p2 . H1: p1 ≠ p2 . Test statistic: z = 1.91 . Critical values: z = ±1.96 . P-value: 0.0562 (Tech:
0.0567). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that New
York City police and Los Angeles police have the same proportion of hits.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 1.91 P-Value = 0.057
b.
95% CI:–0.000455 < p1 − p2 < 0.130 (Tech: –0.000454 < p1 − p2 < 0.130). Because the confidence
interval limits contain 0, there does not appear to be a significant difference between the two
proportions. There is not sufficient evidence to warrant rejection of the claim that New York City
police and Los Angeles police have the same proportion of hits.
Copyright © 2014 Pearson Education, Inc.
142 Chapter 9: Inferences from Two Samples
14. (continued)
MINITAB
Difference = p (1) - p (2)
95% CI for difference: (-0.000453716, 0.130358)
c.
15. a.
There does not appear to be a difference between the hit rates of New York City police and Los
Angeles police.
H0: p1 = p2 . H1: p1 > p2 . Test statistic: z = 9.97 . Critical value: z = 2.33 . P-value: 0.0001 (Tech:
0.0000). Reject H0. There is sufficient evidence to support the claim that the cure rate with oxygen
treatment is higher than the cure rate for those given a placebo. It appears that the oxygen treatment is
effective.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs > 0): Z = 9.97 P-Value = 0.000
b.
98% CI: 0.467 < p1 − p2 < 0.687. Because the confidence interval limits do not include 0, it appears
that the two cure rates are not equal. Because the confidence interval limits include only positive
values, it appears that the cure rate with oxygen treatment is higher than the cure rate for those given a
placebo. It appears that the oxygen treatment is effective.
MINITAB
Difference = p (1) - p (2)
98% CI for difference: (0.467454, 0.687321)
c.
16. a.
The results suggest that the oxygen treatment is effective in curing cluster headaches.
H0: p1 = p2 . H1: p1 < p2 . Test statistic: z = −1.85 . Critical value: z = −1.64 5. P-value: 0.0322
(Tech: 0.0324). Reject H0. There is sufficient evidence to support the claim that when given a single
large bill, a smaller proportion of women in China spend some or all of the money when compared to
the proportion of women in China given the same amount in smaller bills.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs < 0): Z = -1.85 P-Value = 0.032
b.
90% CI: –0.201 < p1 − p2 < –0.0127. Because the confidence interval does not include 0 and it
includes only negative values, it appears that the first proportion is less than the second proportion.
There is sufficient evidence to support the claim that when given a single large bill, a smaller
proportion of women in China spend some or all of the money when compared to the proportion of
women in China given the same amount in smaller bills.
MINITAB
Difference = p (1) - p (2)
90% CI for difference: (-0.200605, -0.0127280)
c.
17. a.
Because the P-value is 0.0322 (Tech: 0.0324), the difference is significant at the 0.05 significance
level, but not at the 0.01 significance level. The conclusion does change.
H0: p1 = p2 . H1: p1 < p2 . Test statistic: z = −1.17 . Critical value: z = −2.33 . P-value: 0.1210
(Tech: 0.1214). Fail to reject H0. There is not sufficient evidence to support the claim that the rate of
left-handedness among males is less than that among females.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs < 0): Z = -1.17 P-Value = 0.121
b.
98% CI: –0.0849 < p1 − p2 < 0.0265 (Tech: –0.0848 < p1 − p2 < 0.0264). Because the confidence
interval limits include 0, there does not appear to be a significant difference between the rate of lefthandedness among males and the rate among females. There is not sufficient evidence to support the
claim that the rate of left-handedness among males is less than that among females.
Copyright © 2014 Pearson Education, Inc.
Chapter 9: Inferences from Two Samples 143
17. (continued)
MINITAB
Difference = p (1) - p (2)
98% CI for difference: (-0.0847744, 0.0264411)
c.
18. a.
The rate of left-handedness among males does not appear to be less than the rate of left-handedness
among females.
H0: p1 = p2 . H1: p1 ≠ p2 . Test statistic: z = 2.30 . Critical values: z = ±2.575 (Tech: ±2.576 ). Pvalue: 0.0214 (Tech: 0.0213). Fail to reject H0. There is not sufficient evidence to warrant rejection of
the claim that the rate of those who finish is the same for men and women.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 2.30 P-Value = 0.021
b.
99% CI: –0.000409 < p1 − p2 < 0.00553 (Tech: –0.000409 < p1 − p2 < 0.00554). Because the
confidence interval limits contain 0, there does not appear to be a significant difference between the
two proportions. There is not sufficient evidence to warrant rejection of the claim that the rate of those
who finish is the same for men and women.
MINITAB
Difference = p (1) - p (2)
99% CI for difference: (-0.000386669, 0.00562869)
c.
19. a.
It appears that men and women finish the New York City marathon at the same rate.
0.0227 < p1 − p2 < 0.217; because the confidence interval limits do not contain 0, it appears that
p1 = p2 can be rejected.
MINITAB
Difference = p (1) - p (2)
95% CI for difference: (0.0227099, 0.217290)
b.
0.491 < p1 < 0.629; 0.371 < p2 < 0.509; because the confidence intervals do overlap, it appears that
p1 = p2 cannot be rejected.
MINITAB
Sample X
1
112
2
88
c.
N
200
200
Sample p
0.560000
0.440000
95% CI
(0.488250, 0.629944)
(0.370056, 0.511750
H0: p1 = p2 . H1: p1 ≠ p2 . Test statistic: z = 2.40 . P-value: 0.0164. Critical values: z = ±1.96 .
Reject H0. There is sufficient evidence to reject p1 = p2 .
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 2.40 P-Value = 0.016
d.
Reject p1 = p2 . Least effective: Using the overlap between the individual confidence intervals.
20. Hypothesis test: With a test statistic of z = −1.96 15, P-value = 0.05 (Tech: 0.0498), reject p1 = p2 .
Confidence interval: –0.422 < p1 − p2 < 0.0180, which suggests that we should not reject p1 = p2
(because 0 is included). The hypothesis test and confidence interval lead to different conclusions about the
equality of p1 = p2 .
MINITAB
Difference = p (1) - p (2)
95% CI for difference: (-0.422046, 0.0180456)
Test for difference = 0 (vs not = 0): Z = -1.96 P-Value = 0.050
21. n =
zα2 / 2
1.6452
=
= 3383 (Tech: 3382)
2 E 2 2 ⋅ 0.022
Copyright © 2014 Pearson Education, Inc.
144 Chapter 9: Inferences from Two Samples
Section 9-3
1.
Independent: b, d, e
2.
–17.32 cm < μ1 − μ2 < –11.61 cm
3.
Because the confidence interval does not contain 0, it appears that there is a significant difference between
the mean height of women and the mean height of men. Based on the confidence interval, it appears that
the mean height of men is greater than the mean height of women.
4.
a.
b.
Yes.
90%
5.
a.
H0: μ1 = μ2 . H1: μ1 ≠ μ2 . Test statistic: t = −2.979 . Critical values: t = ±2.032 (Tech: ±2.002 ). Pvalue < 0.01 (Tech: 0.0042). Reject H0. There is sufficient evidence to warrant rejection of the claim
that the samples are from populations with the same mean. Color does appear to have an effect on
creativity scores. Blue appears to be associated with higher creativity scores.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = -2.98 P-Value = 0.004 DF = 58
b.
95% CI: –0.98 < μ1 − μ2 < –0.18 (Tech: –0.97 < μ1 − μ2 < –0.19)
MINITAB
Difference = mu (1) - mu (2)
95% CI for difference: (-0.970, -0.190)
6.
a.
H0: μ1 = μ2 . H1: μ1 ≠ μ2 . Test statistic: t = 2.647 . Critical values: t = ±2.032 (Tech: ±1.995 ). Pvalue < 0.02 (Tech: 0.0101). Reject H0. There is sufficient evidence to warrant rejection of the claim
that the samples are from populations with the same mean. Color does appear to have an effect on
word recall scores. Red appears to be associated with higher word memory recall scores.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = 2.65 P-Value = 0.010 DF = 68
b.
95% CI: 0.83 < μ1 − μ2 < 6.33 (Tech: 0.88 < μ1 − μ2 < 6.28)
MINITAB
Difference = mu (1) - mu (2)
95% CI for difference: (0.88, 6.28)
7.
a.
H0: μ1 = μ2 . H1: μ1 > μ2 . Test statistic: t = 0.132 . Critical value: t = 1.729 . P-value > 0.10 (Tech:
0.4480). Fail to reject H0. There is not sufficient evidence to support the claim that the magnets are
effective in reducing pain. It is valid to argue that the magnets might appear to be effective if the
sample sizes are larger.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs >): T-Value = 0.13 P-Value = 0.448 DF = 33
b.
90% CI: –0.61 < μ1 − μ2 < 0.71 (Tech: –0.59 < μ1 − μ2 < 0.69)
MINITAB
Difference = mu (1) - mu (2)
90% CI for difference: (-0.592, 0.692)
8.
a.
H0: μ1 = μ2 . H1: μ1 < μ2 . Test statistic: t = −0.676 . Critical value: t = −2.345 (Tech: –2.337). Pvalue > 0.10 (Tech: 0.2499). Fail to reject H0. There is not sufficient evidence to support the claim that
the mean number of words spoken in a day by men is less than that for women.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs <): T-Value = -0.68 P-Value = 0.250 DF = 364
Copyright © 2014 Pearson Education, Inc.
Chapter 9: Inferences from Two Samples 145
8.
(continued)
b.
98% CI: –2443.6 words < μ1 − μ2 < 1350.6 words (Tech: –2436.8 words < μ1 − μ2 < 1343.8 words)
MINITAB
Difference = mu (1) - mu (2)
98% CI for difference: (-2437, 1344)
9.
a.
The sample data meet the loose requirement of having a normal distribution. H0: μ1 = μ2 . H1: μ1 > μ2 .
Test statistic: t = 0.852 . Critical value: t = 2.426 (Tech: 2.676). P-value > 0.10 (Tech: 0.2054). Fail
to reject H0. There is not sufficient evidence to support the claim that men have a higher mean body
temperature than women.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs >): T-Value = 0.85 P-Value = 0.206 DF = 12
b.
98% CI: –0.54 D F < μ1 − μ2 < 1.02 D F (Tech: –0.51 D F < μ1 − μ2 < 0.99 D F)
MINITAB
Difference = mu (1) - mu (2)
98% CI for difference: (-0.515, 0.995)
10. a.
H0: μ1 = μ2 . H1: μ1 ≠ μ2 . Test statistic: t = 1.559 . Critical values: t = ±2.977 (Tech: ±2.789 ). Pvalue > 0.10 (Tech: 0.1316). Fail to reject H0. There is not sufficient evidence to support the claim that
men and women have different mean body temperatures.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = 1.56 P-Value = 0.132 DF = 24
b.
99% CI: –0.19°F < μ1 − μ2 < 0.61°F (Tech: –0.17°F < μ1 − μ2 < 0.59°F)
MINITAB
Difference = mu (1) - mu (2)
99% CI for difference: (-0.167, 0.587)
11. a.
H0: μ1 = μ2 . H1: μ1 < μ2 . Test statistic: t = −3.547 . Critical value: t = −2.462 (Tech: –2.392). Pvalue < 0.005 (Tech: 0.0004). Reject H0. There is sufficient evidence to support the claim that the
mean maximal skull breadth in 4000 b.c. is less than the mean in a.d. 150.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs <): T-Value = -3.55 P-Value = 0.000 DF = 57
b.
98% CI: –8.13 mm < μ1 − μ2 < –1.47 mm (Tech: –8.04 mm < μ1 − μ2 < –1.56 mm)
MINITAB
Difference = mu (1) - mu (2)
98% CI for difference: (-8.04, -1.56)
12. a.
H0: μ1 = μ2 . H1: μ1 ≠ μ2 . Test statistic: t = −0.941 . Critical value: t = ±2.201 (Tech: 2.080).
P-value > 0.20 (Tech: 0.3573). Fail to reject H0. There is not sufficient evidence to warrant rejection of
the claim that Flight 1 and Flight 3 have the same mean arrival delay time.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = -0.94 P-Value = 0.358 DF = 20
b.
95% CI: –18.1 min < μ1 − μ2 < 7.3 min (Tech: –17.4 min < μ1 − μ2 < 6.6 min)
MINITAB
Difference = mu (1) - mu (2)
95% CI for difference: (-17.42, 6.59)
Copyright © 2014 Pearson Education, Inc.
146 Chapter 9: Inferences from Two Samples
13. a.
H0: μ1 = μ2 . H1: μ1 < μ2 . Test statistic: t = −3.142 . Critical value: t = −2.462 (Tech: –2.403). Pvalue < 0.005 (Tech: 0.0014). Reject H0. There is sufficient evidence to support the claim that students
taking the nonproctored test get a higher mean than those taking the proctored test.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs <): T-Value = -3.17 P-Value = 0.001 DF = 49
b.
98% CI: –25.54 < μ1 − μ2 < –3.10 (Tech: –25.27 < μ1 − μ2 < –3.37)
MINITAB
Difference = mu (1) - mu (2)
98% CI for difference: (-25.27, -3.37)
14. a.
H0: μ1 = μ2 . H1: μ1 ≠ μ2 . Test statistic: t = −0.770 . Critical values: t = ±2.756 (Tech: ±2.666 ). Pvalue > 0.20 (Tech: 0.4443). Fail to reject H0. There is not sufficient evidence to warrant rejection of
the claim that the samples are from populations with the same mean.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = -0.77 P-Value = 0.444 DF = 56
b.
99% CI: –18.17 < μ1 − μ2 < 10.23 (Tech: –17.71 < μ1 − μ2 < 9.77)
MINITAB
Difference = mu (1) - mu (2)
99% CI for difference: (-17.71, 9.77)
15. a.
H0: μ1 = μ2 . H1: μ1 ≠ μ2 . Test statistic: t = 1.274 . Critical values: t = ±2.023 (Tech: ±1.994 ). Pvalue > 0.20 (Tech: 0.2066). Fail to reject H0. There is not sufficient evidence to warrant rejection of
the claim that males and females have the same mean BMI.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = 1.27 P-Value = 0.207 DF = 71
b.
95% CI: –1.08 < μ1 − μ2 < 4.76 (Tech: –1.04 < μ1 − μ2 < 4.72)
MINITAB
Difference = mu (1) - mu (2)
95% CI for difference: (-1.04, 4.72)
16. a.
H0: μ1 = μ2 . H1: μ1 > μ2 . Test statistic: t = 2.282 . Critical value: t = 1.725 (Tech: 2.004).
P-value < 0.05 (Tech: 0.0132). Reject H0. There is sufficient evidence to support the claim that the
mean IQ score of people with low lead levels is higher than the mean IQ score of people with high lead
levels.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs >): T-Value = 2.28 P-Value = 0.013 DF = 54
b.
90% CI: 1.5 < μ1 − μ2 < 10.5 (Tech: 1.6 < μ1 − μ2 < 10.4)
MINITAB
Difference = mu (1) - mu (2)
90% CI for difference: (1.59, 10.37)
17. a.
H0: μ1 = μ2 . H1: μ1 > μ2 . Test statistic: t = 0.089 . Critical value: t = 1.725 (Tech: 2.029).
P-value > 0.10 (Tech: 0.4648.) Fail to reject H0. There is not sufficient evidence to support the claim
that the mean IQ score of people with medium lead levels is higher than the mean IQ score of people
with high lead levels.
MINITAB
Variable
N
Mean StDev
LOW LEAD
22 87.23 14.29
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs >): T-Value = 0.09 P-Value = 0.464 DF = 35
Copyright © 2014 Pearson Education, Inc.
Chapter 9: Inferences from Two Samples 147
17. (continued)
b.
90% CI: –5.9 < μ1 − μ2 < 6.6 (Tech: –5.8 < μ1 − μ2 < 6.4)
MINITAB
Difference = mu (1) - mu (2)
Estimate for difference: 0.33
90% CI for difference: (-5.80, 6.45)
18. a.
The sample data meet the loose requirement of having a normal distribution. H0: μ1 = μ2 . H1: μ1 > μ2 .
Test statistic: t = 12.533 . Critical value: t = 2.821 (Tech: 2.411). P-value < 0.005 (Tech: 0.0000).
Reject H0. There is sufficient evidence to support the claim that supermodels have heights with a mean
that is greater than the mean height of women in the general population. We can conclude that
supermodels are taller than typical women.
MINITAB
Variable
N
Mean StDev
Model Height 10 69.825 0.800
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs >): T-Value = 12.53 P-Value = 0.000 DF = 45
b.
98% CI: 4.7 in < μ1 − μ2 < 7.4 in. (Tech: 4.9 in. < μ1 − μ2 < 7.2 in.)
MINITAB
Difference = mu (1) - mu (2)
98% CI for difference: (4.880, 7.207)
19. a.
H0: μ1 = μ2 . H1: μ1 < μ2 . Test statistic: t = −1.810 . Critical value: t = −2.650 (Tech: –2.574). Pvalue > 0.025 (Tech: 0.0442). Fail to reject H0. There is not sufficient evidence to support the claim
that the mean longevity for popes is less than the mean for British monarchs after coronation.
MINITAB
Difference = mu (Popes) - mu (Kings and Queens)
T-Test of difference = 0 (vs <): T-Value = -1.81 P-Value = 0.045 DF = 16
b.
98% CI: –23.6 years < μ1 − μ2 < 4.4 years (Tech: –23.2 years < μ1 − μ2 < 4.0 years)
MINITAB
Difference = mu (Popes) - mu (Kings and Queens)
98% CI for difference: (-23.28, 4.10)
20. a.
H0: μ1 = μ2 . H1: μ1 > μ2 . Test statistic: t = 3.265 . Critical value: t = 1.796 (Tech: t = 1.746 ). Pvalue < 0.005 (Tech: 0.0024). Reject H0. There is sufficient evidence to support the claim that the
mean amount of strontium-90 from Pennsylvania residents is greater than the mean from New York
residents.
MINITAB
Difference = mu (Pennsylvania) - mu (New York)
T-Test of difference = 0 (vs >): T-Value = 3.27 P-Value = 0.003 DF = 15
b.
90% CI: 5.0 mBq < μ1 − μ2 < 17.3 mBq (Tech: 5.2 mBq < μ1 − μ2 < 17.1 mBq)
MINITAB
Difference = mu (Pennsylvania) - mu (New York)
90% CI for difference: (5.17, 17.16)
21. H0: μ1 = μ2 . H1: μ1 ≠ μ2 . Test statistic: t = 32.773 . Critical values: t = ±2.023 (Tech: ±1.994 ). P-value
< 0.01 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the two
populations have equal means. The difference is highly significant, even though the samples are relatively
small.
MINITAB
Difference = mu (Pre-1964 Quarters) - mu (Post-1964 Quarters)
T-Test of difference = 0 (vs not =): T-Value = 32.77 P-Value = 0.000 DF = 70
Copyright © 2014 Pearson Education, Inc.
148 Chapter 9: Inferences from Two Samples
22. –9.1 years < μ1 − μ2 < 5.4 years (Tech: –9.0 years < μ1 − μ2 < 5.3 years). Because the confidence
interval includes 0, there is not a significant difference between the two population means. It appears that
the sample of men and the sample of women are from populations with the same mean.
MINITAB
N
Mean StDev
MALE AGE
40 36.4 1
6.5
FEMALE AGE 40
38.3
15.6
Difference = mu (MALE AGE) - mu (FEMALE AGE)
95% CI for difference: (-9.00, 5.30)
23. 0.03795 lb < μ1 − μ2 < 0.04254 lb (Tech: 0.03786 lb < μ1 − μ2 < 0.04263 lb). Because the confidence
interval does not include 0, there appears to be a significant difference between the two population means.
It appears that the cola in cans of regular Pepsi weighs more than the cola in cans of Diet Pepsi, and that is
probably due to the sugar in regular Pepsi that is not in Diet Pepsi.
MINITAB
N
Mean StDev
PPREGWT
36 0.82410 0.00570
PPDIETWT
36 0.78386 0.00436
Difference = mu (PPREGWT) - mu (PPDIETWT)
95% CI for difference: (0.03786, 0.04263)
24. H0: μ1 = μ2 . H1: μ1 ≠ μ2 . Test statistic: t = 22.095 . Critical values: t = ±2.023 (Tech: ±2.003 ).
P-value < 0.01 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that
the two populations have equal means. The difference is due to the sugar in regular Coke that is not in diet
Coke.
MINITAB
N
Mean StDev
CKREGWT
36 0.81682 0.00751
CKDIETWT
36 0.78479 0.00439
Difference = mu (CKREGWT) - mu (CKDIETWT)
T-Test of difference = 0 (vs not =): T-Value = 22.10 P-Value = 0.000 DF = 56
25. a.
The sample data meet the loose requirement of having a normal distribution. H0: μ1 = μ2 . H1: μ1 > μ2 .
Test statistic: t = 1.046 . Critical value: t = 2.381 (Tech: 2.382). P-value > 0.10 (Tech: 0.1496). Fail
to reject H0. There is not sufficient evidence to support the claim that men have a higher mean body
temperature than women.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs >): T-Value = 1.05 P-Value = 0.150 DF = 68
Both use Pooled StDev = 0.6986
b.
98% CI: –0.31°F < μ1 − μ2 < 0.79°F. The test statistic became larger, the P-value became smaller,
and the confidence interval became narrower, so pooling had the effect of attributing more significance
to the results.
MINITAB
Difference = mu (1) - mu (2)
98% CI for difference: (-0.307, 0.787)
Both use Pooled StDev = 0.6986
26. a.
H0: μ1 = μ2 . H1: μ1 < μ2 . Test statistic: t = −0.682 . Critical value: t = −2.336 . P-value > 0.10
(Tech: 0.2477). Fail to reject H0. There is not sufficient evidence to support the claim that the mean
number of words spoken in a day by men is less than that for women.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs <): T-Value = -0.68 P-Value = 0.248 DF = 394
Both use Pooled StDev = 7954.1009
Copyright © 2014 Pearson Education, Inc.
Chapter 9: Inferences from Two Samples 149
26. (continued)
b.
98% CI: –2417.4 words < μ1 − μ2 < 1324.4 words (Tech: –2417.2 words < μ1 − μ2 < 1324.3 words).
The test statistic became larger, the P-value became smaller, and the confidence interval became
narrower, so pooling had the effect of attributin
MINITAB
98% CI for difference: (-2417, 1324)
Both use Pooled StDev = 7954.1009
27. H0: μ1 = μ2 . H1: μ1 ≠ μ2 . Test statistic: t = 15.322 . Critical values: t = ±2.080 . P-value < 0.01 (Tech:
0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the two populations
have the same mean.
t=
(0.049 − 0.000) − ( μ1 − μ2 )
s 2p
22
+
s 2p
; s 2p =
(22 −1) 0.0152 + (22 −1) 02
= 0.000125
(22 −1) + ( 22 −1)
22
28. df = 77.3502249. Using “df = smaller of n1 −1 and n2 −1 ” is a more conservative estimate of the number
of degrees of freedom (than the estimate obtained with Formula 9-1) in the sense that the confidence
interval is wider, so the difference between the sample means needs to be more extreme to be considered a
significant difference.
29. a.
b.
H0: μ1 = μ2 . H1: μ1 < μ2 . Test statistic: t = −3.002 . Critical value based on 68.9927614 degrees of
freedom: t = −2.381 (Tech: –2.382). P-value < 0.005 (Tech: 0.0019). Reject H0. There is sufficient
evidence to support the claim that students taking the nonproctored test get a higher mean than those
taking the proctored test.
–25.68 < μ1 − μ2 < –2.96 (Tech: –25.69 < μ1 − μ2 < –2.95)
Section 9-4
1.
Parts (c) and (e) are true.
2.
d = 2.4 mi/gal and sd = 1.1 mi/gal. μd represents the mean of the differences from the population of
paired data.
3.
The test statistic will remain the same. The confidence interval limits will be expressed in the equivalent
values of km/L.
4.
The first confidence interval shows that we have 95% confidence that the limits of 1.0 mi/gal and 3.8
mi/gal contain the mean of the population of differences, but the second confidence interval shows that we
have 95% confidence that the limits of –7.8 mi/gal and 12.6 mi/gal contain the difference between the two
population means. Because the first confidence interval does not include 0 mi/gal and consists of positive
values only, it appears that the old ratings are higher than the new ratings. Because the second confidence
interval does include 0 mi/gal, there does not appear to be a significant different between the mean of the
old ratings and the mean of the new ratings.
5.
H0: μd = 0 cm. H1: μd > 0 cm. Test statistic: t = 0.036 (rounded). Critical value: t = 1.692 .
P-value > 0.10 (Tech: 0.4859). Fail to reject H0. There is not sufficient evidence to support the claim that
for the population of heights of presidents and their main opponents; the differences have a mean greater
than 0 cm (with presidents tending to be taller than their opponents).
6.
–2.7 cm < μd < 2.8 cm. The confidence interval includes 0 cm, so it is very possible that the mean of the
differences is equal to 0 cm, indicating that there is no significant difference between heights of presidents
and heights of their opponents.
Copyright © 2014 Pearson Education, Inc.
150 Chapter 9: Inferences from Two Samples
7.
8.
9.
a.
d = −11.6 years
c.
Test statistic t =
d.
H0: μd = 0 . H1: μd ≠ 0 . Critical values: t = ±2.776
a.
d = −0.35D F
c.
Test statistic: t =
d.
H0: μd = 0 . H1: μd ≠ 0 . Critical values: t = ±3.182
d − μd
sd
n
d − μd
sd
n
=
−11.6 − 0
=
17.2
5
sd = 17.2 years
b.
sd = 0.30D F
= −1.508
−0.35 − 0
0.30
b.
4
= −2.333
−11.6 − 0
= −1.507 . Critical values: t = ±2.776 .
17.21 5
P-value > 0.20 (Tech: 0.2063). Fail to reject H0. There is not sufficient evidence to support the claim that
there is a difference between the ages of actresses and actors when they win Oscars.
H0: μd = 0 . H1: μd ≠ 0 . Test statistic: t =
MINITAB
Paired T for Actress - Actor
T-Test of mean difference = 0 (vs not = 0): T-Value = -1.51 P-Value = 0.206
10. H0: μd = 0 . H1: μd ≠ 0 . Test statistic: t =
−0.35 − 0
= −2.333 . Critical values: t = ±3.182 .
0.30 4
P-value > 0.10 (Tech: 0.1018). Fail to reject H0. There is not sufficient evidence to warrant rejection of the
claim that there is no difference between body temperatures measured at 8 A.M. and at 12 A.M.
MINITAB
Paired T for 8A.M. - 12A.M.
T-Test of mean difference = 0 (vs not = 0): T-Value = -2.33 P-Value = 0.102
11. 1.0 min < μd < 12.0 min. Because the confidence interval includes only positive values and does not
include 0 min, it appears that the taxi-out times are greater than the corresponding taxi-in times, so there is
sufficient evidence to support the claim of the flight operations manager that for flight delays, more of the
blame is attributable to taxi-out times at JFK than taxi-in times at LAX.
MINITAB
Paired T for Out - In
90% CI for mean difference: (0.99, 12.01)
d = 6.5 min ; df = 12 −1 = 11
s
10.63
E = tα / 2 ⋅ d = 1.796 ⋅
= 5.5 min
12
n
12. –66.8 cm3 < μd < 49.8 cm3 (Tech: –66.7 cm3 < μd < 49.7 cm3). Because the confidence interval includes
0 cm3, the mean of the differences could be equal to 0 cm3, so there does not appear to be a significant
difference.
MINITAB
Paired T for First Born - Second Born
99% CI for mean difference: (-66.7, 49.7)
13. H0: μd = 0 . H1: μd > 0 . Test statistic: t =
d = −8.5 cm3 ; df = 10 −1 = 9
s
56.7
E = tα / 2 ⋅ d = 3.250 ⋅
= 58.3 cm3
10
n
7279 − 0
= 2.579 . Critical value: t = 2.015 .
6913 6
P-value < 0.025 (Tech: 0.0247). Reject H0. There is sufficient evidence to support the claim that among
couples, males speak more words in a day than females.
MINITAB
Paired T for Male - Female
T-Test of mean difference = 0 (vs > 0): T-Value = 2.58 P-Value = 0.025
Copyright © 2014 Pearson Education, Inc.
Chapter 9: Inferences from Two Samples 151
14. H0: μd = 0 . H1: μd ≠ 0 . Test statistic: t =
−72.20 − 0
= −17.339 . Critical values: t = ±4.604 .
9.31 4
P-value > 0.01 (Tech: 0.0001). Reject H0. There is sufficient evidence to support the claim of a difference
in measurements between the two arms. The right and left arms should yield the same measurements, but
the given data show that this is not happening.
MINITAB
Paired T for Right arm - Left arm
T-Test of mean difference = 0 (vs not = 0): T-Value = -17.34 P-Value = 0.000
15. –6.5 < μd < –0.2. Because the confidence interval does not include 0, it appears that there is sufficient
evidence to warrant rejection of the claim that when the 13th day of a month falls on a Friday, the numbers
of hospital admissions from motor vehicle crashes are not affected. Hospital admissions do appear to be
affected.
MINITAB
Paired T for Friday the 6th - Friday the 13th
95% CI for mean difference: (-6.49, -0.17)
d = −3.33 cm3 ; df = 6 −1 = 5
s
3.01
E = tα / 2 ⋅ d = 2.571⋅
= 3.2 cm3
6
n
16. –4.2 in. < μd < 2.2 in. Because the confidence interval limits contain 0, there is not sufficient evidence to
support a claim that there is a difference between self-reported heights and measured heights. We might
believe that males would tend to exaggerate their heights, but the given data do not provide enough
evidence to support that belief.
MINITAB
Paired T for Reported - Measured
99% CI for mean difference: (-4.16, 2.16)
d = −1.0 in. ; df = 12 −1 = 11
s
3.52
E = tα / 2 ⋅ d = 3.106 ⋅
= 3.2 in.
12
n
−1.57 − 0
= −1.080 . Critical value: t = −1.833 .
4.60 10
P-value > 0.10 (Tech: 0.1540). Fail to reject H0. There is not sufficient evidence to support the claim that
Harry Potter and the Half-Blood Prince did better at the box office. After a few years, the gross amounts
from both movies can be identified, and the conclusion can then be judged objectively without using a
hypothesis test.
17. H0: μd = 0 . H1: μd < 0 . Test statistic: t =
MINITAB
Paired T for Phoenix - Prince
T-Test of mean difference = 0 (vs < 0): T-Value = -1.08 P-Value = 0.154
18.58 − 0
= 6.371 . Critical value: t = 2.718 .
10.10 12
P-value < 0.005 (Tech: 0.00003). Reject H0. There is sufficient evidence to support the claim that Captopril
is effective in lowering systolic blood pressure.
18. H0: μd = 0 . H1: μd > 0 . Test statistic: t =
MINITAB
Paired T for Before - After
T-Test of mean difference = 0 (vs < 0): T-Value = 6.37 P-Value = 1.000
19. 0.69 < μd < 5.56. Because the confidence interval limits do not contain 0 and they consist of positive
values only, it appears that the “before” measurements are greater than the “after” measurements, so
hypnotism does appear to be effective in reducing pain.
MINITAB
Paired T for Before - After
95% CI for mean difference: (0.69, 5.56)
d = 3.13 ; df = 8 −1 = 7
s
2.91
E = tα / 2 ⋅ d = 2.365 ⋅
= 2.43
8
n
Copyright © 2014 Pearson Education, Inc.
152 Chapter 9: Inferences from Two Samples
20. –7.3°F < μd < 6.3°F. Because the confidence interval limits do contain 0°F, there is not a significant
difference between the actual high temperatures and those that were forecast five days earlier. This
suggests that the forecast temperatures are reasonably accurate.
MINITAB
Paired T for Actual High - Forecast High
99% CI for mean difference: (-7.28, 6.28)
d = −0.5 °F ; df = 8 −1 = 7
s
5.48
E = tα / 2 d = 3.500
= 6.8 °F
8
n
21. H0: μd = 0 . H1: μd ≠ 0 . Test statistic: t = −5.553 . Critical values: t = ±1.990 . P-value < 0.01 (Tech:
0.0000). Reject H0. There is sufficient evidence to support the claim that there is a difference between the
ages of actresses and actors when they win Oscars.
MINITAB
Paired T for Actresses - Actors
T-Test of mean difference = 0 (vs not = 0): T-Value = -5.55 P-Value = 0.000
22. H0: μd = 0 . H1: μd ≠ 0 . Test statistic: t = 0.124 . Critical values: t = ±2.028 . P-value > 0.20 (Tech:
0.9023). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that there is no
difference between body temperatures measured at 8 a.m. and at 12 a.m.
MINITAB
Paired T for 8 AM - 12 AM
T-Test of mean difference = 0 (vs not = 0): T-Value = 0.12 P-Value = 0.902
23. H0: μd = 0 . H1: μd < 0 . Test statistic: t = −1.560 . Critical value of t is between –1.671 and –1.676
(Tech: –1.673). P-value > 0.05 (Tech: 0.0622). Fail to reject H0. There is not sufficient evidence to support
the claim that among couples, males speak fewer words in a day than females.
MINITAB
Paired T for M1 - F1
95% upper bound for mean difference: 135
T-Test of mean difference = 0 (vs < 0): T-Value = -1.56 P-Value = 0.062
24. H0: μd = 0 sec. H1: μd > 0 sec. Test statistic: t = 0.938 . Critical value: t = 1.694 . P-value > 0.10 (Tech:
0.1776). Fail to reject H0. There is not sufficient evidence to support the claim that the mean of the
differences is greater than 0 sec. There is not sufficient evidence to support the claim that more time is
devoted to showing tobacco than alcohol. For animated children’s movies, no time should be spent showing
the use of tobacco or alcohol.
MINITAB
Paired T for Tobacco Use (sec) - Alcohol Use (sec)
T-Test of mean difference = 0 (vs > 0): T-Value = 0.94 P-Value = 0.176
25. H0: μd = 6.8 kg. H1: μd ≠ 6.8 kg. Test statistic: t = −11.833 . Critical values: t = ±1.994 (Tech:
±1.997 ). P-value < 0.01 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the
claim that μd = 6.8 kg. It appears that the “Freshman 15” is a myth, and college freshman might gain some
weight, but they do not gain as much as 15 pounds.
MINITAB
Paired T for WTAPR - WTSEP
T-Test of mean difference = 6.8 (vs not = 6.8): T-Value = -11.83 P-Value = 0.000
Section 9-5
1.
a.
b.
c.
No.
No.
The two samples have the same standard deviation (or variance).
Copyright © 2014 Pearson Education, Inc.
Chapter 9: Inferences from Two Samples 153
2.
a.
s12 = 6.602 = 43.56 cm2 and s22 = 6.022 = 36.2404 cm2
b.
H0: σ12 = σ 22
c.
F=
d.
There is not sufficient evidence to support the claim that heights of men and heights of women have
different variances.
s12
43.56
=
= 1.2120
s22 36.2404
3.
The F test is very sensitive to departures from normality, which means that it works poorly by leading to
wrong conclusions when either or both of the populations has a distribution that is not normal. The F test is
not robust against sampling methods that do not produce simple random samples. For example, conclusions
based on voluntary response samples could easily be wrong.
4.
No. Unlike some other tests which have a requirement that samples must be from normally distributed
populations or the samples must have more than 30 values, the F test has a requirement that the samples
must be from normally distributed populations, regardless of how large the samples are.
5.
H0: σ1 = σ 2 . H1: σ1 ≠ σ 2 . Test statistic: F = 1.7341 . Upper critical F value is between 1.8752 and 2.0739
(Tech: 1.9611). P-value: 0.1081. Fail to reject H0. There is not sufficient evidence to support the claim that
weights of regular Coke and weights of regular Pepsi have different standard deviations.
6.
H0: σ1 = σ 2 . H1: σ1 > σ 2 . Test statistic: F = 1.0110 . Critical F value is less than 1.3519 (Tech: 1.2848).
P-value: 0.4745. Fail to reject H0. There is not sufficient evidence to support the claim that ages of student
cars vary more than the ages of faculty cars.
7.
H0: σ1 = σ 2 . H1: σ1 ≠ σ 2 . Test statistic: F =
5.902
= 1.1592 . Upper critical F value is between 1.8752
5.482
and 2.0739 (Tech: 1.9678). P-value: 0.6656. Fail to reject H0. There is not sufficient evidence to warrant
rejection of the claim that the samples are from populations with the same standard deviation. The
background color does not appear to have an effect on the variation of word recall scores.
MINITAB
Test for Equal Variances
F-Test (Normal Distribution)
Test statistic = 1.16, p-value = 0.666
8.
8632.52
= 1.3979 . Critical F value is between 1.0000 and
7301.22
1.3519 (Tech: 1.2642). P-value: 0.0094. Reject H0. There is sufficient evidence to support the claim that
the numbers of words spoken in a day by men vary more than the numbers of words spoken in a day by
women.
H0: σ1 = σ 2 . H1: σ1 > σ 2 . Test statistic: F =
MINITAB
Test for Equal Variances
F-Test (Normal Distribution)
Test statistic = 1.40, p-value = 0.164
9.
2.22
= 9.3364 . Critical F value is between 12.0540 and
0.722
2.0960 (Tech: 2.0842). P-value: 0.0000. Reject H0. There is sufficient evidence to support the claim that
the treatment group has errors that vary more than the errors of the placebo group.
H0: σ1 = σ 2 . H1: σ1 > σ 2 . Test statistic: F =
MINITAB
F-Test (Normal Distribution)
Test statistic = 9.34, p-value = 0.000
Copyright © 2014 Pearson Education, Inc.
154
Chapter 9: Inferences from Two Samples
0.892
= 1.8184 . Critical F value is between 1.9926 and
0.662
2.0772 (Tech: 1.9983). P-value: 0.0774. Fail to reject H0. There is not sufficient evidence to support the
claim that men have body temperatures that vary more than the body temperatures of women.
10. H0: σ1 = σ 2 . H1: σ1 > σ 2 . Test statistic: F =
MINITAB
Test for Equal Variances
F-Test (Normal Distribution)
Test statistic = 1.82, p-value = 0.155
1.42
= 2.1267 . Critical F value is between 2.1555 and
0.962
2.2341 (Tech: 2.1682). P-value: 0.0543. Fail to reject H0. There is not sufficient evidence to support the
claim that those given a sham treatment (similar to a placebo) have pain reductions that vary more than the
pain reductions for those treated with magnets.
11. H0: σ1 = σ 2 . H1: σ1 > σ 2 . Test statistic: F =
MINITAB
Test for Equal Variances
F-Test (Normal Distribution)
Test statistic = 2.13, p-value = 0.109
5.352
= 1.0876 . Upper critical F value is between 2.0923
5.132
and 2.1540 (Tech: 2.1010). P-value: 0.8226. Fail to reject H0. There is not sufficient evidence to warrant
rejection of the claim that the variation of maximal skull breadths in 4000 b.c. is the same as the variation
in a.d. 150.
12. H0: σ1 = σ 2 . H1: σ1 ≠ σ 2 . Test statistic: F =
MINITAB
Test for Equal Variances
F-Test (Normal Distribution)
Test statistic = 1.09, p-value = 0.823
10.63832
= 4.1648 . Critical F value is between 2.7876 and
5.21292
2.8536 (Tech: 2.8179). P-value: 0.0130. Reject H0. There is sufficient evidence to support the claim that
amounts of strontium-90 from Pennsylvania residents vary more than amounts from New York residents.
13. H0: σ1 = σ 2 . H1: σ1 > σ 2 . Test statistic: F =
MINITAB
Test for Equal Variances
F-Test (Normal Distribution)
Test statistic = 4.16, p-value = 0.026
18.60282
= 4.3103 . Upper critical F value is between 2.4665
8.96042
and 2.5699 (Tech: 2.5308). P-value: 0.0023. Reject H0. There is sufficient evidence to warrant rejection of
the claim that both populations of longevity times have the same variation.
14. H0: σ1 = σ 2 . H1: σ1 ≠ σ 2 . Test statistic: F =
MINITAB
Test for Equal Variances: Kings/Queens, Popes
F-Test (Normal Distribution)
Test statistic = 4.31, p-value = 0.002
6.064652
= 1.0073 . Upper critical F value: 4.0260. P-value:
6.042642
0.9915. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that females and
males have heights with the same amount of variation.
15. H0: σ1 = σ 2 . H1: σ1 ≠ σ 2 . Test statistic: F =
MINITAB
Test for Equal Variances
F-Test (Normal Distribution)
Test statistic = 0.99, p-value = 0.992
Copyright © 2014 Pearson Education, Inc.
Chapter 9: Inferences from Two Samples 155
22.6627 2
= 1.7619 . Critical F value: 3.1789. P-value:
17.07342
0.2058. Fail to reject H0. There is not sufficient evidence to support the claim that males have weights with
more variation than females.
16. H0: σ1 = σ 2 . H1: σ1 > σ 2 . Test statistic: F =
MINITAB
Test for Equal Variances
F-Test (Normal Distribution)
Test statistic = 1.76, p-value = 0.412
20.68832
= 1.2397 . Critical F value is between 1.6928 and
18.52812
1.8409 (Tech: 1.7045). P-value: 0.2527. Fail to reject H0. There is not sufficient evidence to support the
claim that males have weights with more variation than females.
17. H0: σ1 = σ 2 . H1: σ1 > σ 2 . Test statistic: F =
MINITAB
Test for Equal Variances
F-Test (Normal Distribution)
Test statistic = 1.25, p-value = 0.494
0.05762
= 1.3213 . Upper critical F value: 2.5411 (Tech:
0.05012
2.5412). P-value: 0.5399. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim
that the two samples are from populations with the same amount of variation.
18. H0: σ1 = σ 2 . H1: σ1 ≠ σ 2 . Test statistic: F =
MINITAB
Test for Equal Variances
F-Test (Normal Distribution)
Test statistic = 1.32, p-value = 0.540
19. a.
No solution provided.
b.
c1 = 4 , c2 = 0
c.
Critical value =
d.
Fail to reject σ12 = σ 22 .
log (0.05 / 2)
=5.
⎛ 40 ⎞⎟
⎜
log ⎜
⎜⎝ 40 + 40 ⎠⎟⎟
20. Test statistic: t = 2.055 . Critical values: t = ±2.023 (Tech: ±1.996 ). P-value > 0.05 (Tech: 0.0438).
Using Table A-3 with df = the smaller of n1 −1 and n2 −1 , fail to reject σ12 = σ 22 . Using technology with
df found from Formula 9-1, reject σ12 = σ 22 .
MINITAB
Difference = mu (pre) - mu (post)
T-Test of difference = 0 (vs not =): T-Value = 2.05 P-Value = 0.044 DF = 67
21. FL = 0.2727 , FR = 2.8365
Chapter Quick Quiz
1.
H0: p1 = p2 . H1: p1 ≠ p2 .
347 + 305
= 0.875
386 + 359
2.
p=
3.
2 ⋅ P ( z > 2.04) = 0.0414 0.0414
4.
0.00172 < p1 − p2 < 0.0970
5.
Because the data consist of matched pairs, they
are dependent.
6.
H0: μd = 0 . H1: μd > 0 .
7.
There is not sufficient evidence to support the
claim that front repair costs are greater than the
corresponding rear repair costs.
8.
F distribution
Copyright © 2014 Pearson Education, Inc.
156 Chapter 9: Inferences from Two Samples
9.
False.
10. True.
Review Exercises
1.
H0: p1 = p2 . H1: p1 > p2 . Test statistic: z = 3.12 . Critical value: z = 2.33 . P-value: 0.0009. Reject H0.
There is sufficient evidence to support a claim that the proportion of successes with surgery is greater than
the proportion of successes with splinting. When treating carpal tunnel syndrome, surgery should generally
be recommended instead of splinting.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs > 0): Z = 3.12 P-Value = 0.001
2.
98% CI: 0.0581 < p1 − p2 < 0.332 (Tech: 0.0583 < p1 − p2 < 0.331). The confidence interval limits do
not contain 0; the interval consists of positive values only. This suggests that the success rate with surgery
is greater than the success rate with splints.
MINITAB
Difference = p (1) - p (2)
98% CI for difference: (0.0583369, 0.331496)
3.
H0: p1 = p2 . H1: p1 < p2 . Test statistic: z = −1.91 . Critical value: z = −1.645 . P-value: 0.0281 (Tech:
0.0280). Reject H0. There is sufficient evidence to support the claim that the fatality rate of occupants is
lower for those in cars equipped with airbags.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs < 0): Z = -1.91 P-Value = 0.028
4.
H0: μd = 0 . H1: μd > 0 . Test statistic: t = 4.712 . Critical value: t = 3.143 . P-value < 0.005 (Tech:
0.0016). Reject H0. There is sufficient evidence to support the claim that flights scheduled 1 day in advance
cost more than flights scheduled 30 days in advance. Save money by scheduling flights 30 days in advance.
MINITAB
Paired T for Flight scheduled one day in adv - Flight scheduled 30 days in adv
T-Test of mean difference = 0 (vs > 0): T-Value = 4.71 P-Value = 0.002
5.
H0: μd = 0 . H1: μd ≠ 0 . Test statistic: t = −0.574 . Critical values: t = ±2.365 . P-value > 0.20 (Tech:
0.5840). Fail to reject H0. There is not sufficient evidence to support the claim that there is a difference
between self-reported heights and measured heights of females aged 12–16.
MINITAB
Paired T for Reported Height - Measured Height
T-Test of mean difference = 0 (vs not = 0): T-Value = -0.57 P-Value = 0.584
6.
H0: μ1 = μ2 . H1: μ1 > μ2 . Test statistic: t = 2.879 . Critical value: t = 2.429 (Tech: 2.376). P-value <
0.005 (Tech: 0.0026). Reject H0. There is sufficient evidence to support the claim that “stress decreases the
amount recalled.”
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs >): T-Value = 2.88 P-Value = 0.003 DF = 76
7.
98% CI: 1.3 < μ1 − μ2 < 14.7 (Tech: 1.4 < μ1 − μ2 < 14.6). The confidence interval limits do not contain
0; the interval consists of positive values only. This suggests that the numbers of details recalled are lower
for those in the stress population.
MINITAB
Difference = mu (1) - mu (2)
98% CI for difference: (1.40, 14.60)
Copyright © 2014 Pearson Education, Inc.
Chapter 9: Inferences from Two Samples 157
8.
H0: p1 = p2 . H1: p1 ≠ p2 . Test statistic: z = −4.20 . Critical values: z = ±2.575 . P-value: 0.0002 (Tech:
0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the acceptance rate is
the same with or without blinding. Without blinding, reviewers know the names and institutions of the
abstract authors, and they might be influenced by that knowledge.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = -4.20 P-Value = 0.000
9.
H0: μ1 = μ2 . H1: μ1 ≠ μ2 . Test statistic: t = 0.679 . Critical values: t = ±2.014 approximately (Tech:
±1.985 ). P-value > 0.20 (Tech: 0.4988). Fail to reject H0. There is not sufficient evidence to warrant
rejection of the claim of no difference between the mean LDL cholesterol levels of subjects treated with
raw garlic and subjects given placebos. Both groups appear to be about the same.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = 0.68 P-Value = 0.499 DF = 94
10. H0: σ1 = σ 2 . H1: σ1 ≠ σ 2 . Test statistic: F = 1.1480 . Upper critical F value is between 1.6668 and 1.8752
(Tech: 1.7799). P-value: 0.6372. Fail to reject H0. There is not sufficient evidence to warrant rejection of
the claim that the two populations have LDL levels with the same standard deviation.
MINITAB
F-Test (Normal Distribution)
Test statistic = 1.15, p-value = 0.637
Cumulative Review Exercises
1.
a.
b.
c.
2.
Because the sample data are matched with each column consisting of heights from the same family, the
data are dependent.
Mean: 63.81 in.; median: 63.70 in.; mode: 62.2 in.; range: 8.80 in.; standard deviation: 2.73 in.;
variance: 7.43 in2
Ratio
There does not appear to be a correlation or association between the heights of mothers and the heights of
their daughters.
Heights of Daughters (in.)
69
68
67
66
65
64
63
62
61
60
60
61
62
63
64
65
66
67
68
69
Heights of Mothers (in.)
3.
61.86 in. < μ < 65.76 in. We have 95% confidence that the limits of 61.86 in. and 65.76 in. actually
contain the true value of the mean height of all adult daughters.
MINITAB
Variable
N Mean
Daughters 10 63.810
StDev SE Mean
2.726
0.862
95% CI
(61.860, 65.760)
Copyright © 2014 Pearson Education, Inc.
158
4.
Chapter 9: Inferences from Two Samples
H0: μd = 0 . H1: μd ≠ 0 . Test statistic: t = 0.283 . Critical values: t = ±2.262 . P-value > 0.20 (Tech:
0.7834). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim of no significant
difference between the heights of mothers and the heights of their daughters.
MINITAB
Paired T for Heights of Mothers (in.) - Heights of Daughters (in.)
T-Test of mean difference = 0 (vs not = 0): T-Value = 0.28 P-Value = 0.783
5.
Because the points lie reasonably close to a straight-line pattern and there is no other pattern that is not a
straight-line pattern and there are no outliers, the sample data appear to be from a population with a normal
distribution.
6.
0.109 < p1 < 0.150. Because the entire range of values in the confidence interval lies below 0.20, the
results do justify the statement that “fewer than 20% of Americans choose their computer and/or Internet
access when identifying what they miss most when electrical power is lost.”
MINITAB
Sample X
N
1
134 1032
Sample p
0.129845
95% CI
(0.109337, 0.150353)
7.
No. Because the Internet users chose to respond, we have a voluntary response sample, so the results are
not necessarily valid.
8.
n=
= 2944 . The survey should not be conducted using only local phone
E2
0.022
numbers. Such a convenience sample could easily lead to results that are dramatically different from results
that would be obtained by randomly selecting respondents from the entire population, not just those having
local phone numbers.
9.
a.
z=
b.
z=
c.
80th percentile: x = μ + z ⋅ σ = 162.0 + 0.842 ⋅ 6.6 = 167.6 cm.
ˆˆ
[ zα / 2 ] pq
2
[ 2.17 ] (0.25)
2
=
152.1−162.0
= −1.5; P ( z > −1.5) = 0.9332.
6.6
152.1−162.0
6.6
4
= −3; P ( z > −1.5) = 0.0.9987.
10. No. Because the states have different population sizes, the mean cannot be found by adding the 50 state
means and dividing the total by 50. The mean income for the U.S. population can be found by using a
weighted mean that incorporates the population size of each state.
Copyright © 2014 Pearson Education, Inc.
Chapter 10: Correlation and Regression 159
Chapter 10: Correlation and Regression
Section 10-2
1.
r represents the value of the linear correlation computed by using the paired sample data. ρ represents the
value of the linear correlation coefficient that would be computed by using all of the paired data in the
population. The value of r is estimated to be 0 (because there is no correlation between sunspot numbers
and the Dow Jones Industrial Average).
2.
No. The value of r = 0 suggests that there is no linear relationship, but there might be some other
relationship that is not linear in the sense that the pattern of points in the scatterplot is not a straight-line
pattern.
3.
The headline is not justified because it states that increased salt consumption is the cause of higher blood
pressure levels, but the presence of a correlation between two variables does not necessarily imply that one
is the cause of the other. Correlation does not imply causality. A correct headline would be this: “Study
Shows That Increased Salt Consumption Is Associated with Higher Blood Pressure.”
4.
Table A-6 shows that the critical values of r are ±0.312 (assuming a 0.05 significance level), so there is
sufficient evidence to support a claim of a linear correlation between the before and after weights. The
value of r does not indicate that the diet is effective in reducing weight. While the diet might be effective in
reducing weight, there could be a linear correlation if the diet has no effect so that the before and after
weights are about the same, or there could be a linear correlation if the diet causes people to gain weight.
5.
H0: ρ = 0 . H1: ρ ≠ 0; Yes. With r = 0.687 and critical values of ±0.312 , there is sufficient evidence to
support the claim that there is a linear correlation between the durations of eruptions and the time intervals
to the next eruptions.
6.
H0: ρ = 0 . H1: ρ ≠ 0; No. With r = 0.091 and critical values of ±0.312 , there is not sufficient evidence
to support the claim that there is a linear correlation between the durations of eruptions and the heights of
eruptions.
7.
H0: ρ = 0 . H1: ρ ≠ 0; No. With r = 0.149 and a P-value of 0.681 (or critical values of ±0.632 ), there is
not sufficient evidence to support the claim that there is a linear correlation between the heights of fathers
and the heights of their sons.
8.
H0: ρ = 0 . H1: ρ ≠ 0; Yes. With r = 0.765 and critical values of ±0.497 , there is sufficient evidence to
support the claim that there is a linear correlation between calories and sugar in a gram of cereal.
9.
a.
Copyright © 2014 Pearson Education, Inc.
160 Chapter 10: Correlation and Regression
9.
(continued)
b.
H0: ρ = 0 . H1: ρ ≠ 0; r = 0.816. Critical values: r = ±0.602. P-value = 0.002. There is sufficient
evidence to support the claim of a linear correlation between the two variables.
MINITAB
Pearson correlation of x and y = 0.816
P-Value = 0.002
c.
The scatterplot reveals a distinct pattern that is not a straight line pattern.
10. a.
13
12
11
y
10
9
8
7
6
5
5.0
7.5
10.0
12.5
15.0
x
b.
H0: ρ = 0 . H1: ρ ≠ 0; r = 0.816. Critical values: r = ±0.602. P-value = 0.002. There is sufficient
evidence to support the claim of a linear correlation between the two variables.
MINITAB
Pearson correlation of x and y = 0.816
P-Value = 0.002
c.
11. a.
b.
The scatterplot reveals a perfect straight-line pattern, except for the presence of one outlier.
There appears to be a linear correlation.
H0: ρ = 0 . H1: ρ ≠ 0; r = 0.906. Critical values: r = ±0.632 (for a 0.05 significance level). There is
a linear correlation.
MINITAB
Pearson correlation of x and y = 0.906
P-Value = 0.000
c.
H0: ρ = 0 . H1: ρ ≠ 0; r = 0 . Critical values: r = ±0.666 (for a 0.05 significance level). There does
not appear to be a linear correlation.
MINITAB
Pearson correlation of x and y = 0.000
P-Value = 1.000
d.
12. a.
b.
c.
The effect from a single pair of values can be very substantial, and it can change the conclusion.
There does not appear to be a linear correlation.
There does not appear to be a linear correlation.
H0: ρ = 0 . H1: ρ ≠ 0; r = 0 . Critical values: r = ±0.950 (for a 0.05 significance level). There does
not appear to be a linear correlation. The same results are obtained with the four points in the upper
right corner.
MINITAB
Pearson correlation of x and y = 0.000
P-Value = 1.000
Copyright © 2014 Pearson Education, Inc.
Chapter 10: Correlation and Regression 161
12. (continued)
d.
H0: ρ = 0 . H1: ρ ≠ 0; r = 0.985 . Critical values: r = ±0.707 (for a 0.05 significance level). There is
a linear correlation.
MINITAB
Pearson correlation of x and y = 0.985
P-Value = 0.000
e.
There are two different populations that should be considered separately. It is misleading to use the
combined data from women and men and conclude that there is a relationship between x and y.
13. H0: ρ = 0 . H1: ρ ≠ 0; r = −0.959 . Critical values: r = ±0.878 . P-value = 0.010. There is sufficient
evidence to support the claim that there is a linear correlation between weights of lemon imports from
Mexico and U.S. car fatality rates. The results do not suggest any cause-effect relationship between the two
variables.
MINITAB
Pearson correlation of Lemon Imports and
Fatality Rate = -0.959
P-Value = 0.010
Scatterplot of Fatality Rate vs Lemon Imports
16.0
Fatality Rate
15.8
15.6
15.4
15.2
15.0
200
250
300
350
400
450
500
550
Lemon Imports
14. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.543 . Critical values: r = ±0.707 . P-value = 0.164. There is not sufficient
evidence to support the claim that there is a linear correlation between PSAT scores and SAT scores.
Because the data are from a voluntary response sample, the results are very questionable.
MINITAB
Pearson correlation of PSAT and SAT
= 0.543
P-Value = 0.164
Scatterplot of SAT vs PSAT
2400
2300
SAT
2200
2100
2000
1900
140
150
160
170
180
PSAT
Copyright © 2014 Pearson Education, Inc.
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200
210
162 Chapter 10: Correlation and Regression
15. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.561. Critical values: r = ±0.632 . P-value = 0.091. There is not sufficient
evidence to support the claim that there is a linear correlation between enrollment and burglaries. The
results do not change if the actual enrollments are listed as 32,000, 31,000, 53,000, etc.
MINITAB
Pearson correlation of Enrollment and
Burglaries = 0.561
P-Value = 0.091
Scatterplot of Burglaries vs Enrollment
160
140
Burglaries
120
100
80
60
40
20
0
30
35
40
45
50
55
Enrollment
16. H0: ρ = 0 . H1: ρ ≠ 0; r = −0.997. Critical values: r = ±0.754. P-value = 0.000. There is sufficient
evidence to support the claim that there is a linear correlation between altitude and outside air temperature.
The results do not change if the altitudes are converted to meters and the temperatures are converted to the
Celsius scale.
MINITAB
Pearson correlation of Alt and Temp
= -0.997
P-Value = 0.000
Scatterplot of Temp vs Alt
50
Temp
25
0
-25
-50
0
5
10
15
20
25
30
35
Alt
17. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.864. Critical values: r = ±0.666. P-value = 0.003. There is sufficient
evidence to support the claim that there is a linear correlation between court incomes and justice salaries.
The correlation does not imply that court incomes directly affect justice salaries, but it does appear that
justices might profit by levying larger fines, or perhaps justices with higher salaries impose larger fines.
MINITAB
Pearson correlation of Income and Salary
= 0.864
P-Value = 0.003
Scatterplot of Salary vs Income
100
90
80
Salary
70
60
50
40
30
20
10
0
200
400
600
800
Income
Copyright © 2014 Pearson Education, Inc.
1000
1200
1400
1600
Chapter 10: Correlation and Regression 163
18. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.947. Critical values: r = ±0.878. P-value = 0.015. There is sufficient
evidence to support the claim that there is a linear correlation between the opening bids suggested by the
auctioneer and the final winning bids.
MINITAB
Pearson correlation of Open and Win
= 0.947
P-Value = 0.015
Scatterplot of Win vs Open
700
600
Win
500
400
300
200
100
200
400
600
800
1000
1200
1400
1600
Open
19. H0: ρ = 0 . H1: ρ ≠ 0; r = 1.000. Critical values: r = ±0.811. P-value = 0.000. There is sufficient
evidence to support the claim that there is a linear correlation between amounts of redshift and distances to
clusters of galaxies. Because the linear correlation coefficient is 1.000, it appears that the distances can be
directly computed from the amounts of redshift.
MINITAB
Pearson correlation of Red and Dist
= 1.000
P-Value = 0.000
Scatterplot of Dist vs Red
1.0
0.9
0.8
Dist
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
Red
20. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.968. Critical values: r = ±0.811. P-value = 0.002. There is sufficient
evidence to support the claim that there is a linear correlation between weights and prices. The results do
not necessarily apply to other populations of diamonds, such as those with different color and clarity
ratings.
MINITAB
Pearson correlation of Weight and Price
= 0.968
P-Value = 0.002
Scatterplot of Price vs Weight
6000
5000
Price
4000
3000
2000
1000
0
0.3
0.4
0.5
0.6
0.7
Weight
Copyright © 2014 Pearson Education, Inc.
0.8
0.9
1.0
164 Chapter 10: Correlation and Regression
21. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.948. Critical values: r = ±0.811. P-value = 0.004. There is sufficient
evidence to support the claim of a linear correlation between the overhead width of a seal in a photograph
and the weight of a seal.
MINITAB
Pearson correlation of Width and Weight
= 0.948
P-Value = 0.004
Scatterplot of Weight vs Width
260
240
Wght
220
200
180
160
140
120
100
7.0
7.5
8.0
8.5
9.0
9.5
10.0
Width
22. H0: ρ = 0 . H1: ρ ≠ 0; r = −0.283. Critical values: r = ±0.754. P-value = 0.539. There is not sufficient
evidence to support the claim of a linear correlation between the repair costs from full-front crashes and
full-rear crashes.
MINITAB
Pearson correlation of Front and Rear
= -0.283
P-Value = 0.539
Scatterplot of Rear vs Front
3500
3000
Rear
2500
2000
1500
1000
1000
1500
2000
2500
3000
3500
4000
4500
Front
23. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.867. Critical values: r = ±0.878. P-value = 0.057. There is not sufficient
evidence to support the claim of a linear correlation between the systolic blood pressure measurements of
the right and left arm.
MINITAB
Pearson correlation of Right Arm and Left
Arm = 0.867
P-Value = 0.057
Scatterplot of Left Arm vs Right Arm
180
Left Arm
170
160
150
140
80
85
90
Right Arm
Copyright © 2014 Pearson Education, Inc.
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100
105
Chapter 10: Correlation and Regression 165
24. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.874. Critical values: r = ±0.707. P-value = 0.005. There is sufficient
evidence to support the claim of a linear correlation between number of cricket chirps and temperature.
MINITAB
Pearson correlation of Chirps and Temp
= 0.874
P-Value = 0.005
Scatterplot of Temp(°F) vs Chirps
95
Temp(°F)
90
85
80
75
70
850
900
950
1000
1050
1100
1150
1200
Chirps
25. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.197. Critical values: r = ±0.707. P-value = 0.640. There is not sufficient
evidence to support the claim that there is a linear correlation between prices of regular gas and prices of
premium gas. Because there does not appear to be a linear correlation between prices of regular and
premium gas, knowing the price of regular gas is not very helpful in getting a good sense for the price of
premium gas.
MINITAB
Pearson correlation of Reg and Prem
= 0.197
P-Value = 0.640
Scatterplot of Prem vs Reg
3.09
3.08
3.07
Prem
3.06
3.05
3.04
3.03
3.02
3.01
3.00
2.750
2.775
2.800
2.825
2.850
2.875
Reg
26. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.399. Critical values: r = ±0.707. P-value = 0.327.
There is not sufficient
evidence to support the claim that there is a linear correlation between prices of regular gas and prices of
mid-grade gas. Because there does not appear to be a linear correlation between prices of regular and midgrade gas, knowing the price of regular gas is not very helpful in getting a good sense for the price of midgrade gas.
MINITAB
Pearson correlation of Reg and Mid
= 0.399
P-Value = 0.327
Scatterplot of Mid vs Reg
3.00
Mid
2.95
2.90
2.85
2.80
2.750
2.775
2.800
Reg
Copyright © 2014 Pearson Education, Inc.
2.825
2.850
2.875
166 Chapter 10: Correlation and Regression
27. H0: ρ = 0 . H1: ρ ≠ 0; r = 1.000. Critical values: r = ±0.707. P-value = 0.000. There is sufficient
evidence to support the claim that there is a linear correlation between diameters and circumferences. A
scatterplot confirms that there is a linear association between diameters and volumes.
MINITAB
Pearson correlation of Diam and Circ
= 1.000
P-Value = 0.000
Scatterplot of Circum vs Diam
80
70
Circum
60
50
40
30
20
10
5
10
15
20
25
Diam
28. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.978. Critical values: r = ±0.707. P-value = 0.000. There is sufficient
evidence to support the claim that there is a linear correlation between diameters and volumes. Although
the results suggest that there is a linear correlation between diameters and volumes, the scatterplot suggests
that there is a very strong correlation that is not linear.
MINITAB
Pearson correlation of Diam and Vol = 0.978
P-Value = 0.000
Scatterplot of Vol vs Diam
8000
7000
6000
Vol
5000
4000
3000
2000
1000
0
5
10
15
20
25
Diam
29. H0: ρ = 0 . H1: ρ ≠ 0; r = −0.063. Critical values: r = ±0.444. P-value = 0.791. There is not sufficient
evidence to support the claim of a linear correlation between IQ and brain volume.
MINITAB
Pearson correlation of IQ and VOL = -0.063
P-Value = 0.791
Scatterplot of VOL vs IQ
1500
1400
VOL
1300
1200
1100
1000
900
80
90
100
110
IQ
Copyright © 2014 Pearson Education, Inc.
120
130
Chapter 10: Correlation and Regression 167
30. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.917. Critical values: r = ±0.279 (approximately) (Tech: ±0.285 ). P-value
= 0.000. There is sufficient evidence to support the claim of a linear correlation between departure delay
times and arrival delay times.
MINITAB
Pearson correlation of Dep Delay and Arr
Delay = 0.917
P-Value = 0.000
Scatterplot of Arr Delay vs Dep Delay
125
100
Arr Delay
75
50
25
0
-25
-50
0
25
50
75
100
125
150
Dep Delay
31. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.319. Critical values: r = ±0.254 (approximately) (Tech: ±0.263). P-value =
0.017. There is sufficient evidence to support the claim of a linear correlation between the numbers of
words spoken by men and women who are in couple relationships.
MINITAB
Pearson correlation of M1 and F1 = 0.319
P-Value = 0.017
Scatterplot of M1 vs F1
50000
40000
M1
30000
20000
10000
0
5000
10000
15000
20000
25000
30000
35000
40000
F1
32. H0: ρ = 0 . H1: ρ ≠ 0; r = 0.027. Critical values: r = ±0.279. P-value = 0.852. There is not sufficient
evidence to support the claim of a linear correlation between magnitudes of earthquakes and their depths.
MINITAB
Pearson correlation of MAG and DEPTH
= 0.027
P-Value = 0.852
Scatterplot of MAG vs DEPTH
3.0
2.5
MAG
2.0
1.5
1.0
0.5
0.0
0
5
10
DEPTH
Copyright © 2014 Pearson Education, Inc.
15
20
168 Chapter 10: Correlation and Regression
33. a.
r = 0.911
r = 0.976
d.
MINITAB
Pearson correlation of y and x = 0.911
P-Value = 0.031
b.
MINITAB
Pearson correlation of y and SQRT(x)
= 0.976
P-Value = 0.005
r = 0.787
c.
r = −0.948
e.
MINITAB
Pearson correlation of y and x^2 =
0.787
P-Value = 0.114
MINITAB
Pearson correlation of y and 1/x
= -0.948
P-Value = 0.014
r = 0.9999 (largest)
MINITAB
Pearson correlation of y and LOG(x)
= 1.000
P-Value = 0.000
34. r = ±
t
t + n−2
2
=±
2.485
2.4852 + 27 − 2
= ±0.445
Section 10-3
1.
The symbol ŷ represents the predicted pulse rate. The predictor variable represents height. The response
variable represents pulse rate.
2.
The regression line has the property that the sum of squares of the residuals is the lowest possible sum
(where a residual is the difference between an observed value of y and a predicted value of y).
3.
If r is positive, the regression line has a positive slope and rises from left to right. If r is negative, the slope
of the regression line is negative and it falls from left to right.
4.
The first equation represents the regression line that best fits sample data, whereas the second equation
represents the regression line that best fits all paired data in a population.
5.
The regression line fits the points well, so the best predicted time for an interval after the eruption is
yˆ = 47.4 + 0.180 (120) = 69 min.
6.
The regression line does not fit the points well, so the best predicted height is y = 127.2 ft.
7.
The regression line does not fit the points well, so the best predicted height is y = 68.0 in.
8.
The regression line fits the points well, so the best predicted value is yˆ = 3.46 + 1.01(0.40)
= 3.86 calories.
yˆ = 3.00 + 0.500 x . The data have a pattern that is not a straight line.
MINITAB
Predictor Coef
SE Coef
Constant 3.001 1.125
x
0.5000 0.1180
T
2.67
4.24
P
0.026
0.002
Scatterplot of y vs x
10
9
8
7
y
9.
6
5
4
3
5.0
7.5
10.0
x
Copyright © 2014 Pearson Education, Inc.
12.5
15.0
Chapter 10: Correlation and Regression 169
10. yˆ = 3.00 + 0.500 x . There is an outlier.
MINITAB
Predictor Coef
SE Coef
Constant 3.002 1.124
x
0.4997 0.1179
T
2.67
4.24
P
0.026
0.002
Scatterplot of y vs x
13
12
11
10
y
9
8
7
6
5
4
5.0
7.5
10.0
12.5
15.0
x
11. a.
yˆ = 0.264 + 0.906 x
MINITAB
Predictor
Constant
x
b.
12. a.
Coef
2.0000
-0.0000
Coef
0.0846
0.98462
SE Coef
0.8165
0.3780
T
2.45
-0.00
P
0.044
1.000
SE Coef
0.4864
0.07134
T
0.17
13.80
P
0.868
0.000
SE Coef
1.118
0.7071
T
1.34
0.00
P
0.312
1.000
T
6.727
0.00
P
1.41
1.000
yˆ = 1.5 + 0 x (or yˆ = 1.5 )
Coef
1.500
0.0000
yˆ = 9.5 + 0 x (or yˆ = 9.5 )
MINITAB
Predictor Coef
Constant
x
0.0000
d.
P
0.653
0.000
yˆ = 0.0846 + 0.985 x
MINITAB
Predictor
Constant
x
c.
T
0.47
6.04
The results are very different, indicating that one point can dramatically affect the regression equation.
MINITAB
Predictor
Constant
x
b.
SE Coef
0.5649
0.1499
yˆ = 2 + 0 x (or yˆ = 2 )
MINITAB
Predictor
Constant
x
c.
Coef
0.2642
0.9057
SE Coef
9.500
0.7071
0.293
The results are very different, indicating that combinations of clusters can produce results that differ
dramatically from results within each cluster alone.
Copyright © 2014 Pearson Education, Inc.
170 Chapter 10: Correlation and Regression
13. yˆ = 16.5 − 0.00282 x ; The regression line fits the points well, so the best predicted value is
yˆ = 16.5 − 0.00282 (500) = 15.1 fatalities per 100,000 population.
MINITAB
Predictor Coef
SE Coef T
P
Constant 16.4909 0.1880
87.70 0.000
Lemon
-0.00282 0.0004815 -5.86 0.010
Scatterplot of Fatality Rate vs Lemon Imports
16.0
Fatality Rate
15.8
15.6
15.4
15.2
15.0
200
250
300
350
400
450
500
550
Lemon Imports
Scatterplot of Residuals vs Lemon Imports
Probability Plot of Residuals
0.20
Normal
0.99
0.15
0.95
0.9
Probability
RESI1
0.10
0.05
0.00
-0.05
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-0.10
200
250
300
350
400
450
500
550
0.01
Lemon Imports
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
Residuals
14. yˆ = 1314 + 4.56 x ; The regression line does not fit the points well, so the best predicted value is y = 2153.
The result is not close to the actual reported value of 2400. Because the data are from a voluntary response
sample, the results have questionable validity.
SE Coef
532.5
2.878
T
2.47
1.59
P
0.049
0.164
Scatterplot of SAT vs PSAT
2400
2300
2200
SAT
MINITAB
Predictor Coef
Constant 1313.7
PSAT
4.562
2100
2000
1900
140
150
160
170
180
PSAT
Copyright © 2014 Pearson Education, Inc.
190
200
210
Chapter 10: Correlation and Regression 171
14. (continued)
Scatterplot of Residuals vs PSAT
Probability Plot of Residuals
200
Normal
0.99
0.95
0.9
Probability
Residuals
100
0
-100
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
-200
0.05
140
150
160
170
180
190
200
210
0.01
-400
PSAT
-300
-200
-100
0
100
200
300
400
Residuals
15. yˆ = −36.8 + 3.47 x ; The regression line does not fit the points well, so the best predicted value is y = 87.7
burglaries. The predicted value is not close to the actual value of 329 burglaries.
MINITAB
Predictor Coef
Constant -36.77
Enrollment 3.467
SE Coef
66.50
1.807
T
-0.55
1.92
P
0.595
0.091
Scatterplot of Burglaries vs Enrollment
160
140
Burglaries
120
100
80
60
40
20
0
30
35
40
45
50
55
Enrollment
Scatterplot of Residuals vs Enrollment
Probability Plot of Residuals
50
Normal
0.99
0.95
0.9
0
Probability
Residuals
25
-25
-50
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-75
30
35
40
Enrollment
45
50
55
0.01
-100
-50
0
Residuals
Copyright © 2014 Pearson Education, Inc.
50
100
172 Chapter 10: Correlation and Regression
16. yˆ = 72.5 − 3.68 x ; The best predicted value is yˆ = 72.5 − 3.68(6.327) = 49.2°F. The predicted value is
close to the actual value of 48°F.
MINITAB
Predictor Coef
Constant 72.498
Altitude
-3.6843
SE Coef
3.017
0.1326
T
P
24.03 0.000
-27.78 0.000
Scatterplot of Temp(°F) vs Altitude
75
Temp(°F)
50
25
0
-25
-50
0
5
10
15
20
25
30
35
Altitude
Scatterplot of Residuals vs Altitude
Probability Plot of Residuals
4
Normal
3
0.99
0.95
1
0.9
0
0.8
Probability
Residuals
2
-1
-2
-3
-4
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-5
0
5
10
15
20
25
30
35
0.01
Altitude
-10
-5
0
5
10
Residuals
17. yˆ = 27.7 + 0.0373 x ; The best predicted value is yˆ = 27.7 + 0.0373(83.941) = 30.8, which represents
$30,800. The predicted value is not very close to the actual salary of $26,088. The possible outliers might
explain the inaccuracy.
Coef
SE Coef T
27.701 5.519
5.02
0.03728 0.008201 4.55
P
0.002
0.003
Scatterplot of Justice Salary vs Court Income
100
90
80
Justice Salary
MINITAB
Predictor
Constant
Income
70
60
50
40
30
20
10
0
200
400
600
800
1000
Court Income
Copyright © 2014 Pearson Education, Inc.
1200
1400
1600
Chapter 10: Correlation and Regression 173
17. (continued)
Scatterplot of Residuals vs Court Income
Probability Plot of Residuals
Normal
20
0.99
15
0.95
0.9
10
Probability
Residuals
25
5
0
-5
-10
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-15
0
200
400
600
800
1000
1200
1400
1600
0.01
Court Income
-30
-20
-10
0
10
20
30
Residuals
18. yˆ = −4.62 + 0.429 x ; The best predicted value is yˆ = −4.62 + 0.429 (300) = $124. The predicted value is
not very close to the actual winning bid of $250. The one influential outlier would account for this
inaccuracy.
P
0.948
0.015
Scatterplot of Winning Bid vs Opening Bid
700
600
Winning Bid
MINITAB
Predictor
Coef
SE Coef T
Constant
-4.62
65.17
-0.07
Opening Bid 0.4291 0.0841 5.10
500
400
300
200
100
200
400
600
800
1000
1200
1400
1600
Opening Bid
Scatterplot of Residuals vs Opening Bid
Probability Plot of Residuals
Normal
0.99
0.95
50
0.9
Probability
Residuals
100
0
-50
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-100
200
400
600
800
1000
Opening Bid
1200
1400
1600
0.01
-200
-100
0
Residuals
Copyright © 2014 Pearson Education, Inc.
100
200
174 Chapter 10: Correlation and Regression
19. yˆ = −0.00440 + 14.0 x ; The best predicted value is yˆ = −0.00440 + 14.0 (0.0126) = 0.172 billion light-
years. The predicted value is very close to the actual distance of 0.18 light-years.
MINITAB
Predictor Coef
SE Coef
Constant -0.004396 0.00125
Redshift 13.9999 0.0278
T
P
-3.51 0.025
503.40 0.000
Scatterplot of Distance vs Redshift
1.0
0.9
Distance
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
Redshift
Scatterplot of Residuals vs Redshift
Probability Plot of Residuals
Normal
0.0010
0.99
0.0005
0.95
0.0000
0.8
0.9
Probability
Residuals
0.0015
-0.0005
-0.0010
-0.0015
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-0.0020
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.01
-0.003
Redshift
-0.002
-0.001
0.000
0.001
0.002
0.003
Residuals
20. yˆ = −2010 + 7180 x ; best predicted value is yˆ = −2010 + 7180 (1.5) = $8760. (Tech: $8759). The
predicted value is far from the actual price of $16,097. The weight of 1.50 carats is well beyond the scope
of the available sample weights, so the extrapolation might be off by a considerable amount.
Coef
-2007.0
7177.0
SE Coef
571.8
935.8
T
P
-3.51 0.025
7.67 0.002
Scatterplot of Price vs Weight
6000
5000
4000
Price
MINITAB
Predictor
Constant
Weight
3000
2000
1000
0
0.3
0.4
0.5
0.6
0.7
Weight
Copyright © 2014 Pearson Education, Inc.
0.8
0.9
1.0
Chapter 10: Correlation and Regression 175
20. (continued)
Scatterplot of Residuals vs Weight
Probability Plot of Residuals
500
Normal
0.99
0.95
0.9
0
Probability
Residuals
250
-250
-500
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-750
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.01
-1000
Weight
-500
0
500
1000
Residuals
21. yˆ = −157 + 40.2 x ; The best predicted weight is yˆ = −157 + 40.2 (2) = −76.6 kg. (Tech: –76.5 kg). That
prediction is a negative weight that cannot be correct. The overhead width of 2 cm is well beyond the scope
of the available sample widths, so the extrapolation might be off by a considerable amount.
MINITAB
Predictor
Constant
Width
Coef
-156.88
40.182
SE Coef
57.41
6.712
T
P
-2.73 0.052
5.99 0.004
Scatterplot of Weight vs Width
260
240
Weight
220
200
180
160
140
120
100
7.0
7.5
8.0
8.5
9.0
9.5
10.0
Width
Scatterplot of Residuals vs Width
Probability Plot of Residuals
Normal
10
0.99
5
0.95
0
0.8
0.9
Probability
Residuals
15
-5
-10
-15
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-20
7.0
7.5
8.0
8.5
Width
9.0
9.5
10.0
0.01
-40
-30
-20
-10
0
Residuals
Copyright © 2014 Pearson Education, Inc.
10
20
30
40
176 Chapter 10: Correlation and Regression
22. yˆ = 2060 − 0.186 x ; The regression line does not fit the data well, so the best predicted cost is y = $1615.
The predicted cost of $1615 is very different from the actual cost of $982.
MINITAB
Predictor
Constant
Front
Coef
2062.6
-0.1856
SE Coef
782.0
0.2818
T
P
2.64 0.046
-0.66 0.539
Scatterplot of Rear vs Front
3500
3000
Rear
2500
2000
1500
1000
1000
1500
2000
2500
3000
3500
4000
4500
Front
Scatterplot of Residuals vs Front
Probability Plot of Residuals
1500
Normal
0.99
1000
0.9
Probability
Residuals
0.95
500
0
-500
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-1000
1000
1500
2000
2500
3000
3500
4000
4500
0.01
Front
-2000
-1000
0
1000
2000
Residuals
23. yˆ = 43.6 + 1.31x ; The regression line does not fit the data well, so the best predicted value is
y = 163.2 mm Hg.
Coef
43.56
1.3147
SE Coef
39.93
0.4361
T
1.09
3.01
P
0.355
0.057
Scatterplot of Left Arm vs Right Arm
180
170
Left Arm
MINITAB
Predictor
Constant
Right Arm
160
150
140
80
85
90
Right Arm
Copyright © 2014 Pearson Education, Inc.
95
100
105
Chapter 10: Correlation and Regression 177
23. (continued)
Scatterplot of Residuals vs Right Arm
Probability Plot of Residuals
15
Normal
0.99
0.95
0.9
5
Probability
Residuals
10
0
-5
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-10
80
85
90
95
100
105
0.01
Right Arm
-20
-10
0
10
20
Residuals
24. yˆ = 27.6 + 0.0523x ; best predicted value is yˆ = 27.6 + 0.0523(3000) = 185°F (Tech: 184°F). The value of
3000 chirps in 1 minute is well beyond the scope of the available sample data, so the extrapolation might be
off by a considerable amount.
MINITAB
Predictor
Constant
Chirps
Coef
SE Coef
27.63
12.17
0.05227 0.01188
T
2.27
4.40
P
0.064
0.005
Scatterplot of Temp(°F) vs Chirps
95
Temp(°F)
90
85
80
75
70
850
900
950
1000
1050
1100
1150
1200
Chirps
Scatterplot of Residuals vs Chirps
Probability Plot of Residuals
5.0
Normal
0.99
0.95
0.9
0.0
Probability
Residuals
2.5
-2.5
-5.0
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-7.5
850
900
950
1000
1050
Chirps
1100
1150
1200
0.01
-10
-5
0
Residuals
Copyright © 2014 Pearson Education, Inc.
5
10
178 Chapter 10: Correlation and Regression
25. yˆ = 2.57 + 0.172 x ; The regression line does not fit the data well, so the best predicted value is y = $3.05.
The predicted price is not very close to the actual price of $2.93.
MINITAB
Predictor
Constant
Regular
Coef
2.5662
0.1718
SE Coef
0.9732
0.3491
T
2.64
0.49
P
0.039
0.640
Scatterplot of Premium vs Regular
3.09
3.08
Premium
3.07
3.06
3.05
3.04
3.03
3.02
3.01
3.00
2.750
2.775
2.800
2.825
2.850
2.875
Regular
Scatterplot of Residuals vs Regular
Probability Plot of Residuals
0.050
Normal
0.99
0.95
0.9
Probability
Residuals
0.025
0.000
-0.025
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-0.050
2.750
2.775
2.800
2.825
2.850
2.875
0.01
-0.08
Regular
-0.06
-0.04
-0.02
0.00
0.02
0.04
0.06
0.08
Residuals
26. yˆ = 0.640 + 0.813x ; The regression line does not fit the data well, so the best predicted value is
y = $2.91. The predicted price is not too far from the actual price.
Coef
0.640
0.8129
SE Coef
2.125
0.7621
T
0.30
1.07
P
0.773
0.327
Scatterplot of Mid-Grade vs Regular
3.00
2.95
Mid-Grade
MINITAB
Predictor
Constant
Regular
2.90
2.85
2.80
2.750
2.775
2.800
2.825
Regular
Copyright © 2014 Pearson Education, Inc.
2.850
2.875
Chapter 10: Correlation and Regression 179
26. (continued)
Scatterplot of Residuals vs Regular
Probability Plot of Residuals
Normal
0.10
0.99
0.95
0.9
0.00
Probability
Residuals
0.05
-0.05
-0.10
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-0.15
2.750
2.775
2.800
2.825
2.850
2.875
0.01
Regular
-0.15
-0.10
-0.05
0.00
0.05
0.10
0.15
Residuals
27. yˆ = −0.00396 + 3.14 x ; The best predicted value is yˆ = −0.00396 + 3.14 (1.50) = 4.7 cm. Even though
the diameter of 1.50 cm is beyond the scope of the sample diameters, the predicted value yields the actual
circumference.
MINITAB
Predictor Coef
SE Coef T
P
Constant -0.00396 0.01883 -0.21
0.840
Diameter 3.14274 0.00129 2443.98 0.000
Scatterplot of Circumference vs Diameter
80
Circumference
70
60
50
40
30
20
10
5
10
15
20
25
Diameter
Scatterplot of Residuals vs Diameter
Probability Plot of Residuals
0.04
Normal
0.03
0.99
0.95
0.01
0.9
0.00
Probability
Residuals
0.02
-0.01
-0.02
-0.03
-0.04
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
-0.05
0.05
5
10
15
Diameter
20
25
0.01
-0.075
-0.050
-0.025
0.000
Residuals
Copyright © 2014 Pearson Education, Inc.
0.025
0.050
180 Chapter 10: Correlation and Regression
28. yˆ = −2010 + 347 x ; The best predicted value is yˆ = −2010 + 347 (1.50) = −1489.5 cm3 (Tech: –1489.8
cm3). The predicted value is negative and is far from the actual volume of 1.8 cm3. The diameter of 1.50 cm
is beyond the scope of the sample diameters, and the predicted value is way wrong. The scatterplot and
residual plot suggest that a nonlinear model would yield better results.
MINITAB
Predictor
Constant
Diameter
Coef
-2010.7
347.30
SE Coef
441.0
30.11
T
P
-4.56 0.004
11.53 0.000
Scatterplot of Volume vs Diameter
8000
7000
6000
Volume
5000
4000
3000
2000
1000
0
-1000
5
10
15
20
25
Diameter
Scatterplot of Residuals vs Diameter
Probability Plot of Residuals
1000
Normal
0.99
0.95
0.9
Probability
Residuals
500
0
-500
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-1000
5
10
15
20
25
0.01
Diameter
-1500
-1000
-500
0
500
1000
1500
Residuals
29. yˆ = 109 − 0.00670 x ; The regression line does not fit the data well, so the best predicted IQ score is
y = 101 .
Coef
SE Coef
108.55 28.17
-0.00670 0.02487
T
P
3.85 0.001
-0.27 0.791
Scatterplot of IQ vs VOLUME
130
120
110
IQ
MINITAB
Predictor
Constant
VOLUME
100
90
80
900
1000
1100
1200
VOLUME
Copyright © 2014 Pearson Education, Inc.
1300
1400
1500
Chapter 10: Correlation and Regression 181
29. (continued)
Scatterplot of Residuals vs VOLUME
Probability Plot of Residuals
30
Normal
0.99
20
0.9
Probability
Residuals
0.95
10
0
-10
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-20
900
1000
1100
1200
1300
1400
1500
0.01
-30
VOLUME
-20
-10
0
10
20
30
Residuals
30. yˆ = −18.4 + 904 x ; best predicted arrival delay time is yˆ = −18.4 + 904 (0) = −18.4 minutes. That is, if a
flight has no departure delay, we can predict that the flight will arrive 18.4 minutes early.
MINITAB
Predictor
Constant
Dep Delay
Coef
SE Coef
-18.366 1.870
0.90356 0.05801
T
P
-9.82 0.000
15.58 0.000
Scatterplot of Arr Delay vs Dep Delay
125
100
Arr Delay
75
50
25
0
-25
-50
0
25
50
75
100
125
150
Dep Delay
Scatterplot of Residuals vs Dep Delay
Probability Plot of Residuals
30
Normal
0.99
20
0.9
Probability
Residuals
0.95
10
0
-10
-20
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-30
0
25
50
75
Dep Delay
100
125
150
0.01
-30
-20
-10
0
Residuals
Copyright © 2014 Pearson Education, Inc.
10
20
30
182 Chapter 10: Correlation and Regression
31. yˆ = 13, 400 + 0.302 x ; The best predicted value is yˆ = 13, 400 + 0.302 (10, 000) = 16,400 (Tech: 16,458).
MINITAB
Predictor
Constant
M1
Coef
SE Coef
13439 2239
0.3019 0.1222
T
6.00
2.47
P
0.000
0.017
Scatterplot of F1 vs M1
40000
35000
30000
F1
25000
20000
15000
10000
5000
0
10000
20000
30000
40000
50000
M1
Scatterplot of Residuals vs M1
Probability Plot of Residuals
20000
Normal
0.99
0.95
10000
0.9
5000
Probability
Residuals
15000
0
-5000
-10000
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-15000
0
10000
20000
30000
40000
50000
0.01
-20000
M1
-10000
0
10000
20000
Residuals
32. yˆ = 9.53 + 0.231x ; The regression line does not fit the data well, so the best predicted value is
y = 9.81 km .
MINITAB
Predictor
Coef
SE Coef T
P
Constant
9.535 1.625
5.87 0.000
MAG
0.231 1.232
0.19 0.852
Scatterplot of DEPTH vs MAG
20
DEPTH
15
10
5
0
0.0
0.5
1.0
1.5
MAG
Copyright © 2014 Pearson Education, Inc.
2.0
2.5
3.0
Chapter 10: Correlation and Regression 183
32. (continued)
Scatterplot of Residuals vs MAG
Probability Plot of Residuals
10
Normal
0.99
0.95
0.9
Probability
Residuals
5
0
-5
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-10
0.0
0.5
1.0
1.5
2.0
2.5
3.0
0.01
MAG
-15
-10
-5
0
5
10
Residuals
33. With β1 = 0 , the regression line is horizontal so that different values of x result in the same y value, and
there is no correlation between x and y.
34. a.
b.
61.8
The sum of squares of the residuals is 101.3, which is larger than 61.8.
yˆ = 125.4 + 1.727 x
176.6919
176.6919
179.6278
180.3186
173.0652
( ŷ − y )
2
ŷ − y
1.3919
–1.1081
–5.7722
5.0186
0.3652
Sum
yˆ = 120 + 2.00 x
1.937386
1.227886
33.31829
25.18635
0.133371
61.80328
179.4
179.4
182.8
183.6
175.2
ŷ − y
4.1
1.6
–2.6
8.3
2.5
Sum
( ŷ − y )
2
16.81
2.56
6.76
68.89
6.25
101.27
Section 10-4
1.
The value of se = 17.5436 cm is the standard error of estimate, which is a measure of the differences
between the observed weights and the weights predicted from the regression equation. It is a measure of the
variation of the sample points about the regression line.
2.
We have 95% confidence that the limits of 50.7 kg and 123.0 kg contain the value of the weight for a male
with a height of 180 cm. The major advantage of using a prediction interval is that it provides us with a
range of likely weights, so we have a sense of how accurate the predicted weight is likely to be. The
terminology of prediction interval is used for an interval estimate of a variable, whereas the terminology of
confidence interval is used for an interval estimate of a population parameter.
3.
The coefficient of determination is r 2 = (0.356) = 0.127. We know that 12.7% of the variation in weight
is explained by the linear correlation between height and weight, and 87.3% of the variation in weight is
explained by other factors and/or random variation.
4.
For the paired weights, se = 0 because there is an exact conversion formula. For a textbook that weighs
2
4.5 lb, the predicted weight is
4.5
= 2.04 kg , and there is no prediction interval because the conversion
2.205
yields an exact result.
5.
r 2 = (0.933) = 0.870. 87.0% of the variation in waist size is explained by the linear correlation between
2
weight and waist size, and 13.0% of the variation in waist size is explained by other factors and/or random
variation.
Copyright © 2014 Pearson Education, Inc.
184 Chapter 10: Correlation and Regression
6.
r 2 = (0.963) = 0.927. 92.7% of the variation in weight is explained by the linear correlation between
2
chest size and weight, and 7.3% of the variation in weight is explained by other factors and/or random
variation.
7.
r 2 = (−0.793) = 0.629. 62.9% of the variation in highway fuel consumption is explained by the linear
2
correlation between weight and highway fuel consumption, and 37.1% of the variation in highway fuel
consumption is explained by other factors and/or random variation.
8.
r 2 = (0.751) = 0.564. 56.4% of the variation in household size is explained by the linear correlation
2
between weight of discarded plastic and household size, and 43.6% of the variation in household size is
explained by other factors and/or random variation.
9.
r = 0.842. Critical values: r = ±0.312 (assuming a 0.05 significance level). P-value = 0.000. There is
sufficient evidence to support a claim of a linear correlation between foot length and height.
10. r = (0.842) = 0.709. 70.9% of the variation in height is explained by the linear correlation between foot
length and height.
2
11. yˆ = 64.1 + 4.29 ( 29.0) =189 cm
12. 177 cm < y < 200 cm . We have 95% confidence that the limits of 177 cm and 200 cm contain the height
of someone with a foot length of 29.0 cm.
13. 160 cm < y < 183 cm
yˆ = 64.1 + 4.29 ( 25) = 171.35
n ( x0 − x )
40 (25 − 25.68)
1
1
= 2.024 (5.50571) 1 + +
= 11.299
E = tα / 2 se 1 + +
2
2
n n (Σx ) − (Σx )
40 40 (26530.92) − (1027.2)2
2
2
14. 156 cm < y < 186 cm (Tech: 156 cm < y < 187 cm )
yˆ = 64.1 + 4.29 ( 25) = 171.35
n ( x0 − x )
40 (25 − 25.68)
1
1
= 2.712 (5.50571) 1 + +
= 15.139
E = tα / 2 se 1 + +
2
2
n n (Σx ) − (Σx )
40 40 (26530.92) − (1027.2)2
2
2
15. 149 cm < y < 168 cm
yˆ = 64.1 + 4.29 (22) = 158.48
n ( x0 − x )
40 (22 − 25.68)
1
1
= 1.686 (5.50571) 1 + +
= 9.797
E = tα / 2 se 1 + +
n n (Σx 2 ) − (Σx )2
40 40 ( 26530.92) − (1027.2)2
2
2
16. 164 cm < y < 187 cm
yˆ = 64.1 + 4.29 ( 26) = 175.64
n ( x0 − x )
40 (26 − 25.68)
1
1
= 2.204 (5.50571) 1 + +
= 12.289
E = tα / 2 se 1 + +
2
2
n n (Σx ) − (Σx )
40 40 (26530.92) − (1027.2)2
2
2
Copyright © 2014 Pearson Education, Inc.
Chapter 10: Correlation and Regression 185
17. a.
b.
c.
10,626.59
68.83577
38.0°F < y < 60.4°F
MINITAB
Analysis of Variance
Source
DF
SS
Regression
1 10627
Residual Error 5
69
Total
6 10695
MS
F
10627 771.88
14
Predicted Values for New Observations
Obs
Fit
SE Fit 95% CI
1
49.19 2.31
(43.26, 55.12)
18. a.
3210.364
b.
1087.191
c.
$10,400 < y < $105,000
MINITAB
Analysis of Variance
Source
DF
SS
MS
Regression
1 3210.4 3210.4
Residual Error 7 1087.2 155.3
Total
8 4297.6
19. a.
c.
95% PI
(37.96, 60.42)
F
20.67
Predicted Values for New Observations
Obs
Fit
SE Fit 99% CI
1
57.53 5.08
(39.75, 75.31)
P
0.000
P
0.003
99% PI
(10.43, 104.63)
0.466276
b.
0.000007359976
0.168 billion light-years < y < 0.176 billion light-years
MINITAB
Analysis of Variance
Source
DF
Regression
1
Residual Error 4
Total
5
SS
0.46628
0.00001
0.46628
MS
F
0.46628 253411.69
0.00000
Predicted Values for New Observations
Obs
Fit
SE Fit
90% CI
1
0.17200 0.00095 (0.16997, 0.17403)
20. a.
b.
c.
P
0.000
90% PI
(0.16847, 0.17554)
16,139,685
1,097,655
$2051 < y < $5419
MINITAB
Analysis of Variance
Source
DF
Regression
1
Residual Error 4
Total
5
SS
MS
16139685 16139685
1097655
274414
17237340
F
58.82
P
0.002
Predicted Values for New Observations
Obs
1
Fit
3735
SE Fit
306
95% CI
(2886, 4583)
95% PI
(2051, 5419)
Copyright © 2014 Pearson Education, Inc.
186 Chapter 10: Correlation and Regression
21. 58.9 < β0 < 103 ; 2.46 < β1 < 3.98
CI for β0
b0 − E < β0 < b0 + E
2
b0 = 80.93; E = tα / 2 se
x
1
1
29.022
+
= 2.024 (5.94376)
+
= 22.06
2
2
n Σx 2 − (Σx)
40 33933 − 1160.7
40
n
CI for β1
b1 − E < β1 < b1 + E
se
b1 = 3.2186; E = tα / 2 ⋅
(Σx)
2
Σx −
2
= 2.024 ⋅
n
5.94376
1160.7 2
33933 −
40
= 0.757
22. 172 cm < y < 176 cm
yˆ − E < y < yˆ + E
yˆ = 80.93 + 3.2186 (29) = 174.269
40 ( 29 − 29.02)
n ( x0 − x )
1
1
+
= 2.024 (5.94376)
+
= 1.902
2
2
40 40 (33933) − (1160.7)2
n n ( Σx ) − ( Σx )
2
E = tα / 2 se
2
Section 10-5
1.
The response variable is weight and the predictor variables are length and chest size.
2.
No, it is not better to use the regression equation with the three predictor variables of length, chest size, and
neck size. The adjusted R 2 value of 0.925 is just a little less than 0.933, so in this case it is better to use
two predictor variables instead of three.
3.
The unadjusted R 2 increases (or remains the same) as more variables are included, but the adjusted R 2 is
adjusted for the number of variables and sample size. The unadjusted R 2 incorrectly suggests that the best
multiple regression equation is obtained by including all of the available variables, but by taking into
account the sample size and number of predictor variables, the adjusted R 2 is much more helpful in
weeding out variables that should not be included.
4.
92.8% of the variation in weights of bears can be explained by the variables of length and chest size, so
7.2% of the variation in weights can be explained by other factors and/or random variation.
5.
LDL = 47.4 + 0.085 WT + 0.497 SYS.
6.
a.
b.
c.
7.
No. The P-value of 0.149 is not very low, and the values of R 2 (0.098) and adjusted R 2 (0.049) are not
high. Although the multiple regression equation fits the sample data best, it is not a good fit.
8.
Predicted LDL = 47.4 + 0.085(59.3) + 0.497 (122) = 113 mg/dL. This result is not likely to be a good
0.149
9.8%, or 0.098
4.9%, or 0.049
predicted value because the multiple regression equation is not a good model (based on the results from
Exercise 7).
9.
HWY (highway fuel consumption) because it has the best combination of small P-value (0.000) and
highest adjusted R 2 (0.920).
Copyright © 2014 Pearson Education, Inc.
Chapter 10: Correlation and Regression 187
10. WT (weight) and HWY (highway fuel consumption) because they have the best combination of small Pvalue (0.000) and highest adjusted R 2 (0.935).
11. CITY = –3.15 + 0.819 HWY. That equation has a low P-value of 0.000 and its adjusted R 2 value of 0.920
isn’t very much less than the values of 0.928 and 0.935 that use two predictor variables, so in this case it is
better to use the one predictor variable instead of two.
12. Predicted city fuel consumption is CITY = −3.15 + 0.819 (36) = 26.3 mi/gal (based on the result from
Exercise 11). The predicted value is a good estimate, but it might not be very accurate because the sample
consists of only 21 cars.
13. The best regression equation is yˆ = 0.127 + 0.0878 x1 − 0.0250 x2 , where x1 represents tar and x2
represents carbon monoxide. It is best because it has the highest adjusted R 2 value of 0.927 and the lowest
P-value of 0.000. It is a good regression equation for predicting nicotine content because it has a high value
of adjusted R 2 and a low P-value.
MINITAB
Predictor
Coef SE Coef
T
P
Constant
0.08000 0.06611
1.21 0.239
100 Tar
0.063333 0.004832 13.11 0.000
S = 0.0869783 R-Sq = 88.2% R-Sq(adj) = 87.7%
Predictor
Coef SE Coef
T
P
Constant
0.3281
0.1378
2.38 0.026
100 CO
0.039721 0.008967
4.43 0.000
S = 0.185937 R-Sq = 46.0% R-Sq(adj) = 43.7%
Predictor
Coef SE Coef
T
P
Constant
0.12714 0.05230
2.43 0.024
100 Tar
0.087797 0.007062 12.43 0.000
100 CO
-0.025004 0.006130
-4.08 0.000
S = 0.0671065 R-Sq = 93.3% R-Sq(adj) = 92.7%
14. The best regression equation is yˆ = 0.251 + 0.101x1 − 0.0454 x2 , where x1 represents tar and x2 represents
carbon monoxide. It is best because it has the highest adjusted R 2 value of 0.908 and the lowest P-value of
0.000. It is a good regression equation for predicting nicotine content because it has a high value of
adjusted R 2 and a low P-value.
MINITAB
Predictor
Coef SE Coef
T
P
Constant
0.13884 0.08874
1.56 0.131
Menth Tar 0.056746 0.006609
8.59 0.000
S = 0.120760 R-Sq = 76.2% R-Sq(adj) = 75.2%
Predictor
Coef SE Coef
T
P
Constant
0.3851
0.1559
2.47 0.021
Menth CO
0.03255 0.01005
3.24 0.004
S = 0.205218 R-Sq = 31.3% R-Sq(adj) = 28.3%
Predictor
Coef SE Coef
T
P
Constant
0.25073 0.05699
4.40 0.000
Menth Tar 0.100692 0.008053 12.50 0.000
Menth CO -0.045432 0.007206
-6.30 0.000
S = 0.0737007 R-Sq = 91.5% R-Sq(adj) = 90.8%
Copyright © 2014 Pearson Education, Inc.
188 Chapter 10: Correlation and Regression
15. The best regression equation is yˆ = 109 − 0.00670 x1 , where x1 represents volume. It is best because it has
the highest adjusted R 2 value of –0.0513 and the lowest P-value of 0.791. The three regression equations
all have adjusted values of R 2 that are very close to 0, so none of them are good for predicting IQ. It does
not appear that people with larger brains have higher IQ scores.
MINITAB
Predictor
Coef SE Coef
T
P
Constant
108.55
28.17
3.85 0.001
VOL
-0.00670 0.02487
-0.27 0.791
S = 13.5455 R-Sq = 0.4% R-Sq(adj) = 0.0%
Predictor
Coef SE Coef
T
P
Constant
101.14
12.46
8.11 0.000
WT
-0.0018
0.1554
-0.01 0.991
S = 13.5728 R-Sq = 0.0% R-Sq(adj) = 0.0%
Predictor
Coef SE Coef
T
P
Constant
108.26
29.72
3.64 0.002
VOL
-0.00694 0.02616
-0.27 0.794
WT
0.0072
0.1631
0.04 0.965
S = 13.9375 R-Sq = 0.4% R-Sq(adj) = 0.0%
16. The best regression equation is yˆ = −10.0 + 0.567 x1 + 0.532 x2 , where x1 represents verbal IQ score and
x2 represents performance IQ score. It is best because it has the highest adjusted R 2 value of 0.999 and the
lowest P-value of 0.000. Because the adjusted R 2 is so close to 1, it is likely that predicted values will be
very accurate.
MINITAB
Predictor
Coef S
E Coef
T
P
Constant
11.504
4.091
2.81 0.006
IQV
0.94024 0.04790 19.63 0.000
S = 7.03711 R-Sq = 76.4% R-Sq(adj) = 76.2%
Predictor
Coef SE Coef
T
P
Constant
10.465
3.548
2.95 0.004
IQP
0.80620 0.03515 22.94 0.000
S = 6.22165 R-Sq = 81.6% R-Sq(adj) = 81.4%
Predictor
Coef SE Coef
T
P
Constant
-10.0200
0.3285 -30.50 0.000
IQV
0.566561 0.004261 132.95 0.000
IQP
0.532216 0.003537 150.49 0.000
S = 0.508785 R-Sq = 99.9% R-Sq(adj) = 99.9%
0.7072
= 5.486 , the P-value is 0.000, and the critical values are
0.1289
t = ±2.110 , so reject H0 and conclude that the regression coefficient of b1 = 0.707 should be kept. For H0:
17. For H0: β1 = 0, the test statistic is t =
0.1636
= 1.292 , the P-value is 0.213, and the critical values are t = ±2.110 ,
0.1266
so fail to reject H0 and conclude that the regression coefficient of b2 = 0.164 should be omitted. It appears
that the regression equation should include the height of the mother as a predictor variable, but the height of
the father should be omitted.
β2 = 0, the test statistic is t =
Copyright © 2014 Pearson Education, Inc.
Chapter 10: Correlation and Regression 189
18. 0.435 < β1 < 0.979 ; −0.104 < β2 < 0.431 . The confidence interval for β2 includes 0, suggesting that the
father’s height be eliminated as a predictor variable.
CI for β1
b1 − E < β1 < b1 + E
b1 − tα / 2 s1 < β1 < b1 + tα / 2 s1
0.7072 − 2.11⋅ 0.1289 < β1 < 0.7072 + 2.11⋅ 0.1289
0.435 < β1 < 0.979
CI for β2
b2 − E < β2 < b2 + E
b2 − tα / 2 s2 < β2 < b2 + tα / 2 s2
0.1636 − 2.11⋅ 0.1266 < β2 < 0.1636 + 2.11⋅ 0.1266
−0.104 < β2 < 0.431
19. yˆ = 3.06 + 82.4 x1 + 2.91x2 , where x1 represents sex and x2 represents age.
Female: yˆ = 3.06 + 82.4 (0) + 2.91( 20) = 61 lb ; male: yˆ = 3.06 + 82.4 (1) + 2.91( 20) = 144 lb . The sex of
the bear does appear to have an effect on its weight. The regression equation indicates that the predicted
weight of a male bear is about 82 lb more than the predicted weight of a female bear with other
characteristics being the same.
MINITAB
Predictor
Constant
SEX
AGE
Coef
3.06
82.38
2.9053
SE Coef
22.46
20.80
0.2974
T
P
0.14 0.892
3.96 0.000
9.77 0.000
Section 10-6
1.
Since the area of a square is the square of its side, the best model is y = x 2 ; quadratic; R 2 = 1
2.
Quadratic is best because it has the highest R 2 value, but this is not a good model because the value of R 2
is so low. Using the models discussed in this section, it appears that we cannot make accurate predictions of
the numbers of points scored in future Super Bowl games. Common sense suggests that no such model
could be found.
3.
10.3% of the variation in Super Bowl points can be explained by the quadratic model that relates the
variable of year and the variable of points scored. Because such a small percentage of the variation is
explained by the model, the model is not very useful.
4.
Instead of showing a pattern that approximates the graph of the quadratic equation, the points are scattered
about with no obvious pattern. The points do not fit the graph of the quadratic equation well, so the value of
R 2 = 0.103 is very low.
Copyright © 2014 Pearson Education, Inc.
190 Chapter 10: Correlation and Regression
5.
Quadratic: d = −4.88t 2 + 0.0214t + 300
Model
Linear
Quadratic
Logarithmic
Exponential
Power
R2
0.962
1.000
0.831
0.933
0.783
y = -4.8786x2 + 0.0214x + 299.96
R2 = 1
6.
Power: y = 35 x3
Model
Linear
Quadratic
Logarithmic
Exponential
Power
7.
R2
0.929
0.999
0.828
0.974
1.000
y = 35.001x3
R2 = 1
Exponential: y = 100 (1.03x ) The value of R 2 is slightly higher for the exponential model.
Model
Linear
Quadratic
Logarithmic
Exponential
Power
R2
0.999
1.000
0.900
0.999
0.918
y = 100(1.03)x
R2 = 1
8.
Logarithmic: y = 0.00476 + 4.34 ln x
Model
Linear
Quadratic
Logarithmic
Exponential
Power
R2
0.895
0.988
1.000
0.861
0.997
y = 4.3426Ln(x) + 0.0048
R2 = 1
Copyright © 2014 Pearson Education, Inc.
Chapter 10: Correlation and Regression 191
9.
−0.945
Power: y = 65.7 x−0.945 . Prediction for the 22nd day: y = 65.7 (22)
= $3.5 million, which isn’t very
close to the actual amount of $2.2 million. The model does not take into account the fact that movies do
better on weekend days.
Model
Linear
Quadratic
Logarithmic
Exponential
Power
R2
0.562
0.774
0.820
0.792
0.842
y = 65.7x-0.945
R2 = 0.8421
10. Quadratic: y = −323x 2 + 1365 x + 45, 084 . (with 1975 coded as 1). Projected value for 2020:
y = −323(10) + 1365(10) + 45, 084 = 26,434 (Tech: 26,454).
2
Model
Linear
Quadratic
Logarithmic
Exponential
Power
R2
0.555
0.652
0.388
0.549
0.377
y = -322.77x2 + 1364.7x + 45084
R2 = 0.652
11. Logarithmic: y = 3.22 + 0.293ln x
Model
Linear
Quadratic
Logarithmic
Exponential
Power
R2
0.620
0.901
0.997
0.566
0.989
y = 0.2933Ln(x) + 3.2178
R2 = 0.9972
Copyright © 2014 Pearson Education, Inc.
192 Chapter 10: Correlation and Regression
12. Power: y = 0.313x−0.863
Model
Linear
Quadratic
Logarithmic
Exponential
Power
R2
0.746
0.945
0.947
0.931
0.999
y = 0.3133x-0.8631
R2 = 0.9985
13. Exponential: y = 10 ( 2 x )
Model
Linear
Quadratic
Logarithmic
Exponential
Power
R2
0.771
0.975
0.549
1.000
0.927
y = 10(2)x
R2 = 1
14. Exponential: y = 115 (0.938) . (Result is based on 1980 coded as 1.) With R 2 = 0.253 , this best model is
not a good model. There is a cyclical pattern that does not fit any of the five models included in this
section.
x
Model
Linear
Quadratic
Logarithmic
Exponential
Power
R2
0.218
0.218
0.252
0.253
0.233
y = 115.00(0.938)x
R2 = 0.2535
Copyright © 2014 Pearson Education, Inc.
Chapter 10: Correlation and Regression 193
15. Quadratic: y = 125 x 2 − 439 x + 3438 . The projected value for 2010 is y = 125 (21) − 439 (21) + 3438
2
= 49,344 (Tech: 49,312), which is dramatically greater than the actual value of 11,655.
Model
Linear
Quadratic
Logarithmic
Exponential
Power
R2
0.893
0.995
0.657
0.958
0.767
y = 124.93x2 - 438.96x + 3437.7
R2 = 0.995
16. Quadratic: y = −31.4 x 2 + 1233 x + 280 . The projected value for 2010 is
y = −31.4 ( 21) + 1233( 21) + 280 = 12,326 (Tech: 12,345), which is not dramatically different from the
2
actual value of 11,655.
Model
Linear
Quadratic
Logarithmic
Exponential
Power
R2
0.834
0.899
0.822
0.811
0.879
y = -31.378x2 + 1233.4x + 280.2
R2 = 0.8992
17. a.
Exponential: y = 2 3 (
2
x −1)
[or y = 0.629961(1.587401) for an initial value of 1 that doubles every 1.5
x
years].
b.
Exponential: y = 1.36558(1.42774) , where 1971 is coded as 1.
x
Model
Linear
Quadratic
Logarithmic
Exponential
Power
c.
R2
0.380
0.55
0.158
0.990
0.790
y = 1.3656(1.42774)x
R2 = 0.9899
Moore’s law does appear to be working reasonably well. With R 2 = 0.990 , the model appears to be
very good.
Copyright © 2014 Pearson Education, Inc.
194 Chapter 10: Correlation and Regression
18. a.
b.
c.
6641.8
73.2
The quadratic sum of squares of residuals (73.2) is less than the sum of squares of residuals from the
linear model (6641.8).
Chapter Quick Quiz
1.
r = ±0.878
2.
Based on the critical values of ±0.878 (assuming a 0.05 significance level), conclude that there is not
sufficient evidence to support the claim of a linear correlation between systolic and diastolic readings.
3.
The best predicted diastolic reading is 90.6, which is the mean of the five sample diastolic readings.
4.
The best predicted diastolic reading is yˆ = −1.99 + 0.698(125) = 85.3 , which is found by substituting 125
for x in the regression equation.
5.
r 2 = 0.342
6.
False; there could be another relationship.
7.
False, correlation does not imply causation.
8.
r =1
9.
Because r must be between –1 and 1 inclusive, the value of 3.335 is the result of an error in the
calculations.
10. r = −1
Review Exercises
a.
r = 0.926 . Critical values: r = ±0.707 (assuming a 0.05 significance level). P-value = 0.001. There
is sufficient evidence to support the claim that there is a linear correlation between duration and
interval-after time.
MINITAB
Pearson correlation of After and Duration = 0.926
P-Value = 0.001
b.
r 2 = (0.926) = 0.857 = 85.7%
c.
yˆ = 34.8 + 0.234 x
2
MINITAB
Predictor Coef
Constant 34.770
Duration 0.23406
SE Coef
8.732
0.03908
T
3.98
5.99
P
0.007
0.001
Scatterplot of After vs Duration
100
90
After
1.
80
70
60
100
120
140
160
180
200
Duration
Copyright © 2014 Pearson Education, Inc.
220
240
260
280
Chapter 10: Correlation and Regression 195
1.
(continued)
Scatterplot of Residuals vs Duration
Probability Plot of Residuals
Normal
0.99
0.95
2.5
0.9
Probability
Residuals
5.0
0.0
-2.5
-5.0
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
-7.5
100
120
140
160
180
200
220
240
260
280
0.01
Duration
-5
0
5
10
Residuals
d.
yˆ = 34.8 + 0.234 ( 200) = 81.6 min
a.
The scatterplot suggests that there is not sufficient sample evidence to support the claim of a linear
correlation between heights of eruptions and interval-after times.
Scatterplot of After vs Height
100
90
After
2.
-10
80
70
60
110
120
130
140
150
Height
b.
r = 0.269. Critical values: r = ±0.707 (assuming a 0.05 significance level). P-value = 0.519. There
is not sufficient evidence to support the claim that there is a linear correlation between height and
interval-after time.
MINITAB
Pearson correlation of Height and After = 0.269
P-Value = 0.519
c.
yˆ = 54.3 + 0.246 x
MINITAB
Predictor
Constant
Height
d.
Coef
SE Coef T
54.27 46.53 1.17
0.2465 0.3597 0.69
P
0.288
0.519
yˆ = 54.3 + 0.246 (100) = 78.9 min
Copyright © 2014 Pearson Education, Inc.
196 Chapter 10: Correlation and Regression
3.
a.
The scatterplot suggests that there is not sufficient sample evidence to support the claim of a linear
correlation between duration and height.
Scatterplot of Height vs Duration
150
Height
140
130
120
110
100
120
140
160
180
200
220
240
260
280
Duration
b.
r = 0.389. Critical values: r = ±0.707 (assuming a 0.05 significance level). P-value = 0.340. There is
not sufficient evidence to support the claim that there is a linear correlation between duration and
height.
MINITAB
Pearson correlation of Height and Duration = 0.389
P-Value = 0.340
c.
yˆ = 105 + 0.108 x
MINITAB
Predictor
Coef
Constant 105.19
Duration 0.1076
d.
4.
SE Coef
23.22
0.1039
T
P
4.53 0.004
1.04 0.340
The regression line does not fit the points well, so the best predicted height y = 128.8 ft.
r = 0.450. Critical values: r = ±0.632 (assuming a 0.05 significance level). P-value = 0.192. There is not
sufficient evidence to support the claim that there is a linear correlation between time and height. Although
there is no linear correlation between time and height, the scatterplot shows a very distinct pattern revealing
that time and height are associated by some function that is not linear.
MINITAB
Pearson correlation of Height(m) and Time(sec)
= 0.450
P-Value = 0.192
Scatterplot of Height(m) vs Time(sec)
5
Height(m)
4
3
2
1
0
0.0
0.5
1.0
1.5
2.0
Time(sec)
5.
AFTER = 50.1 + 0.242 Duration – 0.178 BEFORE, or yˆ = 50.1 + 0.242 x1 − 0.178 x2 . R 2 = 0.872; adjusted
R 2 = 0.820; P-value = 0.006. With high values of R 2 and adjusted R 2 and a small P-value of 0.006, it
appears that the regression equation can be used to predict the time interval after an eruption given the
duration of the eruption and the time interval before that eruption.
MINITAB
Predictor
Coef SE Coef
T
P
Constant
50.09
22.07
2.27 0.072
Duration
0.24179 0.04177
5.79 0.002
Before 0.1779
0.2336
-0.76 0.481
S = 5.15785 R-Sq = 87.2% R-Sq(adj) = 82.0%
Copyright © 2014 Pearson Education, Inc.
Chapter 10: Correlation and Regression 197
Cumulative Review Exercises
1.
x = 3.3 lb , s = 5.7 lb
2.
The highest weight before the diet is 212 lb, which converts to z =
3.
H 0 : μd = 0 . H 0 : μd > 0 . Test statistic: t = 1.613 . Critical value: t = 1.895 . P-value > 0.05 (Tech:
0.075). Fail to reject H0. There is not sufficient evidence to support the claim that the diet is effective.
212 −179.4
= 1.55 . The highest weight
21.0
is not unusual because its z score of 1.55 shows that it is within 2 standard deviations of the mean.
MINITAB
Paired T for Before - After
95% lower bound for mean difference: -0.57
T-Test of mean difference = 0 (vs > 0): T-Value = 1.61 P-Value = 0.075
4.
161.8 lb < μ < 197.0 lb. We have 95% confidence that the interval limits of 161.8 lb and 197.0 lb contain
the true value of the mean of the population of all subjects before the diet.
MINITAB
Variable N
Before 8
5.
a.
Mean StDev SE Mean
95% CI
179.38 21.04
7.44 (161.79, 196.96)
r = 0.965. Critical values: r = ±0.707 (assuming a 0.05 significance level). P-value = 0.000. There is
sufficient evidence to support the claim that there is a linear correlation between before and after
weights.
MINITAB
Pearson correlation of Before and After = 0.965
P-Value = 0.000
6.
b.
d.
r =1
c.
r =1
The effectiveness of the diet is determined by the amounts of weight lost, but the linear correlation
coefficient is not sensitive to different amounts of weight loss. Correlation is not a suitable tool for
testing the effectiveness of the diet.
a.
z=
b.
10th percentile: x = μ + z ⋅ σ = 3420 −1.28 ⋅ 495 = 2786.4 g (Tech: 2785.6 g)
c.
z=
2450 − 3420
= −1.96; P ( z < −1.96) = 0.0250.
495
z=
4390 − 3420
= 1.96; P ( z > 1.96) = 0.0250.
495
3500 − 3420
= 0.162; P ( z > 0.162) = 43.64%. (Tech: 43.58%)
495
0.0250 + 0.0250 = 0.0500 = 5.00%. Yes, many of the babies do require special treatment.
7.
a.
H0: p = 0.5. H1: p > 0.5. Test statistic: z = 3.84 . Critical value: z = 1.645 . P-value: 0.0001. Reject H0.
There is sufficient evidence to support the claim that the majority of us say that honesty is always the
best policy.
MINITAB
Test of p = 0.5 vs p > 0.5
Sample
1
b.
95% Lower
X
N Sample p
Bound Z-Value P-Value
269 456 0.589912 0.552026
3.84 0.000
The sample is a voluntary response (or self-selected) sample. This type of sample suggests that the
results given in part (a) are not necessarily valid.
Copyright © 2014 Pearson Education, Inc.
198 Chapter 10: Correlation and Regression
8.
a.
b.
c.
Nominal
Ratio
Discrete
a.
⎛ 304 ⎞⎟
⎜⎜
= 0.330
⎜⎝ 529 ⎠⎟⎟
b.
d.
304
= 0.575
529
304 + 156 460
=
= 0.870
529
529
e.
Parameter
c.
514
= 0.972
529
d.
39
= 0.0737 = 7.37%
529
2
9.
10.
300
Number
250
200
150
100
50
0
Protestant
Catholic
Jewish
Mormon
Members of Congress
Copyright © 2014 Pearson Education, Inc.
Other
Chapter 11: Goodness-of-Fit and Contingency Tables 199
Chapter 11: Goodness-of-Fit and Contingency Tables
Section 11-2
1.
The test is to determine whether the observed frequency counts agree with the claimed uniform distribution
so that frequencies for the different days are equally likely.
2.
E=
3.
Because the given frequencies differ substantially from frequencies that are all about the same, the χ 2 test
statistic should be large and the P-value should be small.
4.
df = 6. Critical value: χ 2 = 12.592.
5.
Test statistic: χ 2 = 1934.979. Critical value: χ 2 = 12.592. P-value = 0.000. There is sufficient evidence to
warrant rejection of the claim that the days of the week are selected with a uniform distribution with all
days having the same chance of being selected.
6.
Test statistic: χ 2 = 6.6. Critical value: χ 2 = 16.919. P-value = 0.679. There is not sufficient evidence to
support the claim that the sample is from a population of heights in which the last digits do not occur with
the same frequency.
7.
Critical value: χ 2 = 16.919. P-value > 0.10 (Tech: 0.516). There is not sufficient evidence to warrant
rejection of the claim that the observed outcomes agree with the expected frequencies. The slot machine
appears to be functioning as expected.
8.
Test statistic: χ 2 = 4.600. Critical value: χ 2 = 7.815. P-value > 0.10 (Tech: P-value = 0.204). There is not
sufficient evidence to warrant rejection of the claim that the tires selected by the students are equally likely.
It appears that students do not have the ability to select the same tire.
1005
, or 143.571, for each of the seven days of the week. For Sunday, O = 523 and E = 143.571.
7
Tire
Left Front
Right Front
Left Rear
Right Rear
9.
O
11
15
8
6
E
40 ⋅ 0.25 = 10
40 ⋅ 0.25 = 10
40 ⋅ 0.25 = 10
40 ⋅ 0.25 = 10
O−E
1
5
–2
–4
(O − E )
2
1
25
4
16
Sum
(O − E )
2
E
0.1
2.5
0.4
1.6
4.6
Test statistic: χ 2 = 10.375. Critical value: χ 2 = 19.675. P-value > 0.10 (Tech: 0.497). There is not
sufficient evidence to warrant rejection of the claim that homicides in New York City are equally likely for
each of the 12 months. There is not sufficient evidence to support the police commissioner’s claim that
homicides occur more often in the summer when the weather is better.
MINITAB
N
DF Chi-Sq P-Value
512 11 10.375 0.497
10. Test statistic: χ 2 = 93.072. Critical value: χ 2 = 19.675. P-value < 0.005 (Tech: 0.000). There is sufficient
evidence to warrant rejection of the claim that American born major league baseball players are born in
different months with the same frequency. The sample data appear to support Gladwell’s claim.
MINITAB
N
DF Chi-Sq
4515 11 93.0718
P-Value
0.000
Copyright © 2014 Pearson Education, Inc.
200
Chapter 11: Goodness-of-Fit and Contingency Tables
11. Test statistic: χ 2 = 5.860. Critical value: χ 2 = 11.071. P-value > 0.10 (Tech: P-value = 0.320). There is
not sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes appear
to be equally likely, so the loaded die does not appear to behave differently from a fair die.
MINITAB
N
DF Chi-Sq P-Value
200 5 5.86
0.320
12. Test statistic: χ 2 = 16.895. Critical value: χ 2 = 16.812. P-value < 0.01 (Tech: 0.0097). There is sufficient
evidence to warrant rejection of the claim that births occur on the days of the week with equal frequency.
Because many births are induced or involve Caesarean section, they are scheduled for days other than
Saturday or Sunday, so those two days have smaller numbers of births.
MINITAB
N
DF Chi-Sq
773 6
16.8952
P-Value
0.010
13. Test statistic: χ 2 = 13.483. Critical value: χ 2 = 16.919. P-value > 0.10 (Tech: 0.142). There is not
sufficient evidence to warrant rejection of the claim that the likelihood of winning is the same for the
different post positions. Based on these results, post position should not be considered when betting on the
Kentucky Derby race.
MINITAB
N
DF Chi-Sq
116 9
13.4828
P-Value
0.142
14. Test statistic: χ 2 = 8.021. Critical value: χ 2 = 16.919. P-value > 0.10 (Tech: 0.532). There is not sufficient
evidence to warrant rejection of the claim that the digits are selected in a way that they are equally likely.
MINITAB
N
DF Chi-Sq
2920 9
8.02055
P-Value
0.532
15. Test statistic: χ 2 = 29.814. Critical value: χ 2 = 16.812. P-value < 0.005 (Tech: 0.000). There is sufficient
evidence to warrant rejection of the claim that the different days of the week have the same frequencies of
police calls. The highest numbers of calls appear to fall on Friday and Saturday, and these are weekend
days with disproportionately more partying and drinking.
MINITAB
N
DF Chi-Sq
1095 6
29.8137
P-Value
0.000
16. Test statistic: χ 2 = 31.963. Critical value: χ 2 = 16.812. P-value < 0.005 (Tech: 0.000). There is sufficient
evidence to warrant rejection of the claim that the different days of the week have the same frequencies of
police calls. Because March has 31 days, three of the days of the week occur more often than the other days
of the week, so the comparison does not make sense with the given data.
MINITAB
N
DF Chi-Sq
1451 6
31.9628
P-Value
0.000
17. Test statistic: χ 2 = 7.579. Critical value: χ 2 = 7.815. P-value > 0.05 (Tech: 0.056). There is not sufficient
evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by
the proportions listed in the given table.
Copyright © 2014 Pearson Education, Inc.
Chapter 11: Goodness-of-Fit and Contingency Tables 201
17. (continued)
Games Played
O
103 ⋅ 0.125 = 12.875
103 ⋅ 0.2500 = 25.75
103 ⋅ 0.3125 = 32.1875
103 ⋅ 0.3125 = 32.1875
20
23
23
37
4
5
6
7
(O − E )
2
O−E
E
7.125
–2.75
–9.1875
4.8125
50.76563
7.5625
84.41016
23.16016
Sum
(O − E )
2
E
3.942961
0.293689
2.622451
0.719539
7.578641
18. Test statistic: χ 2 = 5.624. Critical value: χ 2 = 12.592. P-value > 0.10 (Tech: 0.467). There is not sufficient
evidence to warrant rejection of the claim that the actual eliminations agree with the expected numbers. The
leadoff singers do appear to be at a disadvantage because 20 of them were eliminated compared to the
expected value of 12.9 eliminations, but that result is not significant in the context of the available sample
data.
MINITAB
N DF Chi-Sq
69 6
5.62408
P-Value
0.467
19. Test statistic: χ 2 = 6.682. Critical value: χ 2 = 11.071 (assuming a 0.05 significance level). P-value > 0.10
(Tech: 0.245). There is not sufficient evidence to warrant rejection of the claim that the color distribution is
as claimed.
Color
Red
Orange
Yellow
Brown
Blue
Green
O
13
25
8
8
27
19
E
100 ⋅ 0.13 = 13
100 ⋅ 0.20 = 20
100 ⋅ 0.14 = 14
100 ⋅ 0.13 = 13
100 ⋅ 0.24 = 24
100 ⋅ 0.16 = 16
O − E (O − E )
2
0
5
–6
–5
3
3
0
25
36
25
9
9
Sum
(O − E )
2
E
0
1.25
2.571429
1.923077
0.375
0.5625
6.682005
20. Test statistic: χ 2 = 0.976. Critical value: χ 2 = 9.488. P-value > 0.10 (Tech: 0.913). There is not sufficient
evidence to warrant rejection of the claim that the actual frequencies fit a Poisson distribution.
MINITAB
N
DF Chi-Sq
576 4
0.976153
P-Value
0.913
21. Test statistic: χ 2 = 3650.251. Critical value: χ 2 = 20.090. P-value < 0.005 (Tech: 0.000). There is
sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a
distribution that conforms to Benford’s law. It does appear that the checks are the result of fraud (although
the results cannot confirm that fraud is the cause of the discrepancy between the observed results and the
expected results).
MINITAB
N
DF Chi-Sq
784 8
3650.25
P-Value
0.000
22. Test statistic: χ 2 = 14.432. Critical value: χ 2 = 15.507. P-value > 0.05 (Tech: 0.071). There is not
sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a
distribution that conforms to Benford’s law. The author’s check amounts appear to be legitimate.
Copyright © 2014 Pearson Education, Inc.
202
Chapter 11: Goodness-of-Fit and Contingency Tables
22. (continued)
MINITAB
N
DF Chi-Sq
200 8 14.4316
P-Value
0.071
23. Test statistic: χ 2 = 1.762. Critical value: χ 2 = 15.507. P-value > 0.10 (Tech: 0.988). There is not sufficient
evidence to warrant rejection of the claim that the leading digits are from a population with a distribution
that conforms to Benford’s law. The tax entries do appear to be legitimate.
MINITAB
N
DF Chi-Sq
511 8 1.76216
P-Value
0.987
24. Test statistic: χ 2 = 10.299. Critical value: χ 2 = 15.507. P-value > 0.10 (Tech: 0.245). There is not
sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a
distribution that conforms to Benford’s law.
MINITAB
N
DF Chi-Sq
150 8
10.2989
25. a.
b.
c.
P-Value
0.245
6, 13, 15, 6
z=
155.41−162
= −1; P ( z < −1) = 0.1587 ;
6.595
z=
162.005 −162
= 0; P (−1 < z < 0) = 0.5000 − 0.1587 = 0.3413 ;
6.595
z=
168.601−162
= 1; P (0 < z < 1) = 0.8413 − 0.5000 = 0.3413 :
6.595
z=
215 − 280
= −1; P ( z > 1) = 0.1587
65
(Tech: 0.1587, 0.3413, 0.3414, 0.1586)
40 ⋅ 0.1587 = 6.348 , 40 ⋅ 0.3413 = 13.652 , , 40 ⋅ 0.3413 = 13.652 , 40 ⋅ 0.1587 = 6.348
(Tech: 6.348, 13.652, 13.656, 6.344)
d.
Test statistic: χ 2 = 0.202 (Tech: 0.201). Critical value: χ 2 = 11.345. P-value > 0.10 (Tech: 0.977).
There is not sufficient evidence to warrant rejection of the claim that heights were randomly selected
from a normally distributed population. The test suggests that the data are from a normally distributed
population.
Height
Less than 155.410
155.410–162.005
162.005–168.601
Greater than 168.601
MINITAB
N DF Chi-Sq
40 3
0.202395
O
6
13
15
6
E
6.348
13.652
13.652
6.348
O−E
–0.348
–0.652
1.348
–0.348
P-Value
0.977
Copyright © 2014 Pearson Education, Inc.
(O − E )
2
0.121104
0.425104
1.817104
0.121104
Sum
(O − E )
2
E
0.019078
0.031139
0.133102
0.019078
0.202395
Chapter 11: Goodness-of-Fit and Contingency Tables 203
Section 11-3
1.
Because the P-value of 0.216 is not small (such as 0.05 or lower), fail to reject the null hypothesis of
independence between the treatment and whether the subject stops smoking. This suggests that the choice
of treatment doesn’t appear to make much of a difference.
2.
In this context, the word contingency refers to a dependency of one variable on another, and we use a test
of independence between the row variable and the column variable to determine whether one variable
appears to be contingent on the other. We use the terminology of two-way table because the frequency
counts are arranged in a table with two variables: the row variable and the column variable.
3.
df = (3 −1)(2 −1) = 2 and the critical value is χ 2 = 5.991.
4.
The test is right-tailed. The test statistic is based on differences between observed frequencies and the
frequencies expected with the assumption of independence between the row and column variables. Only
large values of the test statistic correspond to substantial differences between the observed and expected
values, and such large values are located in the right tail of the distribution.
5.
Test statistic: χ 2 = 3.409. Critical value: χ 2 = 3.841. P-value > 0.05 (Tech: 0.0648). There is not sufficient
evidence to warrant rejection of the claim that the form of the 100-Yuan gift is independent of whether the
money was spent. There is not sufficient evidence to support the claim of a denomination effect.
6.
Test statistic: χ 2 = 9.750. Critical value: χ 2 = 6.635. P-value < 0.005 (Tech: 0.002). There is sufficient
evidence to warrant rejection of the claim that success is independent of the type of treatment. The results
suggest that the surgery treatment is better.
7.
Test statistic: χ 2 = 25.571. Critical value: χ 2 = 3.841. P-value < 0.005 (Tech: 0.000). There is sufficient
evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test
indication. The results suggest that polygraphs are effective in distinguishing between truths and lies, but
there are many false positives and false negatives, so they are not highly reliable.
MINITAB
Expected counts are printed below observed counts
No (Did
Not Lie)
1
15
27.34
2
32
19.66
Total
47
Yes
(Lied)
42
29.66
9
21.34
51
Total
57
41
98
Chi-Sq = 25.571, DF = 1, P-Value = 0.000
8.
Test statistic: χ 2 = 4.423. Critical value: χ 2 = 6.635. P-value > 0.025 (Tech: 0.0355). There is not
sufficient evidence to support the claim that the results are discriminatory.
MINITAB
Expected counts are printed below observed counts
Passed
17
12.81
2
9
13.19
Total
26
1
Failed
16
20.19
25
20.81
41
Total
33
34
67
Chi-Sq = 4.423, DF = 1, P-Value = 0.035
Copyright © 2014 Pearson Education, Inc.
204
9.
Chapter 11: Goodness-of-Fit and Contingency Tables
Test statistic: χ 2 = 42.557. Critical value: χ 2 = 3.841. P-value < 0.005 (Tech: 0.000). There is sufficient
evidence to warrant rejection of the claim that the sentence is independent of the plea. The results
encourage pleas for guilty defendants.
MINITAB
Expected counts are printed below observed counts
Guilty Not Guilty
Plea
Plea
1
392
58
418.48
31.52
2
564
14
537.52
40.48
Total
956
72
Total
450
578
1028
Chi-Sq = 42.557, DF = 1, P-Value = 0.000
10. Test statistic: χ 2 = 86.481. Critical value: χ 2 = 6.635. P-value < 0.005 (Tech: 0.000). There is sufficient
evidence to warrant rejection of the claim that deaths on shifts are independent of whether Gilbert was
working. The results favor the guilt of Gilbert.
MINITAB
Expected counts are printed below observed counts
With Death
40
11.59
2
34
62.41
Total
74
1
Without Death
217
245.41
1350
1321.59
1567
Total
257
1384
1641
Chi-Sq = 86.481, DF = 1, P-Value = 0.000
11. Test statistic: χ 2 = 0.164. Critical value: χ 2 = 3.841. P-value > 0.10 (Tech: 0.686). There is not sufficient
evidence to warrant rejection of the claim that the gender of the tennis player is independent of whether the
call is overturned.
MINITAB
Expected counts are printed below observed counts
Yes
421
416.90
2
220
224.10
Total
641
1
No
991
995.10
539
534.90
1530
Total
1412
759
2171
Chi-Sq = 0.164, DF = 1, P-Value = 0.686
12. Test statistic: χ 2 = 1.364. Critical value: χ 2 = 3.841. P-value > 0.10 (Tech: 0.243). There is not sufficient
evidence to warrant rejection of the claim that left-handedness is independent of gender.
MINITAB
Expected counts are printed below observed counts
Yes
No
Total
23
217
240
27.79
212.21
2
65
455
520
60.21
459.79
Total
88
672
760
Chi-Sq = 1.364, DF = 1, P-Value = 0.243
1
Copyright © 2014 Pearson Education, Inc.
Chapter 11: Goodness-of-Fit and Contingency Tables 205
13. Test statistic: χ 2 = 14.589. Critical value: χ 2 = 9.488. P-value < 0.01 (Tech: 0.0056). There is sufficient
evidence to warrant rejection of the claim that the direction of the kick is independent of the direction of the
goalkeeper jump. The results do not support the theory that because the kicks are so fast, goalkeepers have
no time to react, so the directions of their jumps are independent of the directions of the kicks.
MINITAB
Chi-Sq = 14.589, DF = 4, P-Value = 0.006
14. Test statistic: χ 2 = 1.358. Critical value: χ 2 = 7.815 (assuming a 0.05 significance level). P-value > 0.10
(Tech: 0.715). There is not sufficient evidence to warrant rejection of the claim that the amount of smoking
is independent of seat belt use. The theory is not supported by the given data.
MINITAB
Chi-Sq = 1.358, DF = 3, P-Value = 0.715
15. Test statistic: χ 2 = 2.925. Critical value: χ 2 = 5.991. P-value > 0.10 (Tech: 0.232). There is not sufficient
evidence to warrant rejection of the claim that getting a cold is independent of the treatment group. The
results suggest that echinacea is not effective for preventing colds.
MINITAB
Chi-Sq = 2.925, DF = 2, P-Value = 0.232
16. Test statistic: χ 2 = 9.971. Critical value: χ 2 = 9.488 (assuming a 0.05 significance level). P-value < 0.05
(Tech: 0.041). There is sufficient evidence to warrant rejection of the claim that injuries are independent of
helmet color. It appears that motorcycle drivers should use yellow or orange helmets.
MINITAB
Chi-Sq = 9.971, DF = 4, P-Value = 0.041
17. Test statistic: χ 2 = 20.271. Critical value: χ 2 = 15.086. P-value < 0.005 (Tech: 0.0011). There is sufficient
evidence to warrant rejection of the claim that cooperation of the subject is independent of the age
category. The age group of 60 and over appears to be particularly uncooperative.
MINITAB
Chi-Sq = 20.271, DF = 5, P-Value = 0.001
18. Test statistic: χ 2 = 20.054. Critical value: χ 2 = 19.675. P-value > 0.05 (Tech: 0.0446). There is sufficient
evidence to warrant rejection of the claim that months of births of baseball players are independent of
whether they are born in America. The data do appear to support Gladwell’s claim.
MINITAB
Chi-Sq = 20.054, DF = 11, P-Value = 0.045
19. Test statistic: χ 2 = 0.773. Critical value: χ 2 = 11.345. P-value > 0.10 (Tech: 0.856). There is not sufficient
evidence to warrant rejection of the claim that getting an infection is independent of the treatment. The
atorvastatin treatment does not appear to have an effect on infections.
MINITAB
Chi-Sq = 0.773, DF = 3, P-Value = 0.856
20. Test statistic: χ 2 = 784.647. Critical value: χ 2 = 11.345. P-value < 0.005 (Tech: 0.000). There is sufficient
evidence to warrant rejection of the claim that left-handedness is independent of parental handedness. It
appears that handedness of the parents has an effect on handedness of the offspring, so left-handedness
appears to be an inherited trait.
Copyright © 2014 Pearson Education, Inc.
206
Chapter 11: Goodness-of-Fit and Contingency Tables
20. (continued)
MINITAB
Expected counts are printed below observed counts
Yes
5360
6067.90
2
767
377.63
3
741
475.19
4
94
41.29
Total 6962
1
No Total
50928 56288
50220.10
2736 3503
3125.37
3667 4408
3932.81
289
383
341.71
57620 64582
Chi-Sq = 784.647, DF = 3, P-Value = 0.000
21. Test statistics: χ 2 = 12.1619258 and z = 3.487395274, so that z 2 = χ 2 . Critical values: χ 2 = 3.841 and
z 2 = ±1.96 , so z 2 = χ 2 (approximately).
MINITAB
Expected counts are printed below observed counts
Purchased
Gum
1
27
18.84
2
12
20.16
Total
39
Kept the
Money
16
24.16
34
25.84
50
MINITAB
Difference = p (1) - p (2)
Estimate for difference: 0.372308
95% CI for difference: (0.178143,
0.566473)
Test for difference = 0 (vs not = 0): Z =
3.49 P-Value = 0.000
Total
43
46
89
Chi-Sq = 12.162, DF = 1, P-Value = 0.000
22. Without Yates’s correction, the test statistic is χ 2 = 12.162. With Yates’s correction, the test statistic is χ 2
= 10.717. Yates’s correction decreases the test statistic so that sample data must be more extreme in order
to reject the null hypothesis of independence.
Without Yates’s correction
(27 −18.84)
2
χ =
2
18.84
(16 − 24.16)
2
+
24.16
(12 − 20.16)
2
+
20.16
(34 − 25.84)
2
+
25.84
= 12.162
With Yates’s Correction
( 27 −18.84 − 0.5)
2
χ =
2
18.84
( 16 − 24.16 − 0.5)
2
+
24.16
( 12 − 20.16 − 0.5)
2
+
20.16
( 34 − 25.84 − 0.5)
2
+
25.84
= 10.717
Chapter Quick Quiz
1.
H0: p1 = p2 = p3 = p4 = p5 . H1: At least one of the probabilities is different from the others.
2.
O = 23 and E =
3.
Right-tailed.
4.
df = 4 and the critical value is χ 2 = 9.488.
5.
There is not sufficient evidence to warrant rejection of the claim that occupation injuries occur with equal
frequency on the different days of the week.
107
= 21.4 .
5
Copyright © 2014 Pearson Education, Inc.
Chapter 11: Goodness-of-Fit and Contingency Tables 207
6.
H0: Response to the question is independent of gender. H1: Response to the question and gender are
dependent.
7.
Chi-square distribution.
8.
Right-tailed.
9.
df = (2 −1)(3 −1) = 2 and the critical value is χ 2 = 5.991.
10. There is not sufficient evidence to warrant rejection of the claim that response is independent of gender.
Review Exercises
1.
Test statistic: χ 2 = 931.347. Critical value: χ 2 = 16.812. P-value: 0.000. There is sufficient evidence to
warrant rejection of the claim that auto fatalities occur on the different days of the week with the same
frequency. Because people generally have more free time on weekends and more drinking occurs on
weekends, the days of Friday, Saturday, and Sunday appear to have disproportionately more fatalities.
2.
Test statistic: χ 2 = 6.500. Critical value: χ 2 = 16.919. P-value > 0.10 (Tech: 0.689). There is not sufficient
evidence to warrant rejection of the claim that the last digits of 0, 1, 2, . . . , 9 occur with the same
frequency. It does appear that the weights were obtained through measurements.
MINITAB
N DF Chi-Sq P-Value
80 9 6.5
0.689
3.
Test statistic: χ 2 = 288.448. Critical value: χ 2 = 24.725. P-value < 0.005 (Tech: 0.000). There is sufficient
evidence to warrant rejection of the claim that weather-related deaths occur in the different months with the
same frequency. The summer months appear to have disproportionately more weather-related deaths, and
that is probably due to the fact that vacations and outdoor activities are much greater during those months.
MINITAB
N
DF Chi-Sq
489 11 288.448
P-Value
0.000
4.
Test statistic: χ 2 = 10.708. Critical value: χ 2 = 3.841. P-value: 0.00107. There is sufficient evidence to
warrant rejection of the claim that wearing a helmet has no effect on whether facial injuries are received. It
does appear that a helmet is helpful in preventing facial injuries in a crash.
5.
Test statistic: χ 2 = 4.955. Critical value: χ 2 = 3.841. P-value < 0.05 (Tech: 0.0260). There is sufficient
evidence to warrant rejection of the claim that when flipping or spinning a penny, the outcome is
independent of whether the penny was flipped or spun. It appears that the outcome is affected by whether
the penny is flipped or spun. If the significance level is changed to 0.01, the critical value changes to 6.635,
and we fail to reject the given claim, so the conclusion does change. All expected counts are greater than 5.
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
Heads
Tails
2048
1992
2007.29 2032.71
2
953
1047
993.71 1006.29
Total 3001 3039
1
Total
4040
2000
6040
Chi-Sq = 4.955, DF = 1, P-Value = 0.026
Copyright © 2014 Pearson Education, Inc.
208
6.
Chapter 11: Goodness-of-Fit and Contingency Tables
Test statistic: χ 2 = 4.737. Critical value: χ 2 = 7.815. P-value > 0.10 (Tech: 0.192). There is not sufficient
evidence to warrant rejection of the claim that home/visitor wins are independent of the sport.
MINITAB
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
Basketball Baseball Hockey Football Total
127
53
50
57 287
115.97
58.57 54.47
57.99
2
71
47
43
42 203
82.03
41.43 38.53
41.01
Total
198
100
93
99 490
1
Chi-Sq = 4.737, DF = 3, P-Value = 0.192
Cumulative Review Exercises
1.
H0: p = 0.5. H1: p ≠ 0.5. Test statistic: z = 7.28. Critical values: z = ±1.96 . P-value: 0.0002 (Tech:
0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that among those who die
in weather-related deaths, the percentage of males is equal to 50%.
MINITAB
Test of p = 0.5 vs p not = 0.5
Sample
1
2.
X
N Sample p
325 489 0.664622
95% CI
(0.620854, 0.706384)
Exact
P-Value
0.000
59.0% < p < 65.0% . Because the confidence interval does not include 50% (or “half”), we should reject
the stated claim.
MINITAB
Sample X
N
Sample p
1
620 1000 0.620000
95% CI
(0.589098, 0.650193)
3.
x = 53.7 years, median = 60.0 years, s = 16.1 years. Because an age of 16 differs from the mean by more
than 2 standard deviations, it is an unusual age.
4.
42.2 years < μ < 65.2 years . Yes, the confidence interval limits do contain the value of 65.0 years that was
found from a sample of 9269 ICU patients.
MINITAB
Variable N Mean StDev
AGES
10 53.70 16.09
5.
a.
SE Mean
5.09
95% CI
(42.19, 65.21)
r = –0.0458. Critical values: r = ±0.632 . P-value = 0.900. There is not sufficient evidence to support
the claim that there is a linear correlation between the numbers of boats and the numbers of manatee
deaths.
MINITAB
Pearson correlation of Boats and Manatee Deaths = -0.046
P-Value = 0.900
b.
yˆ = 96.1− 0.137 x
MINITAB
The regression equation is
Manatee Deaths = 96.1 - 0.14 Boats
Predictor Coef
Constant 96.14
Boats
-0.137
SE Coef
99.89
1.053
T
P
0.96 0.364
-0.13 0.900
Copyright © 2014 Pearson Education, Inc.
Chapter 11: Goodness-of-Fit and Contingency Tables 209
5.
6.
(continued)
c.
yˆ = 96.1− 0.137 (84) = 84.6 manatee deaths (the value of y). The predicted value is not very accurate
because it is not very close to the actual value of 78 manatee deaths.
a.
5th percentile: x = μ + z ⋅ σ = 686 −1.645 ⋅ 34 = 630 mm
b.
650 − 686
= −1.06 and P ( z < −1.06) = 14.46% (Tech: 14.48%). That percentage is too high,
34
because too many women would not be accommodated.
z=
680 − 686
c.
= −0.706 and P ( z > −0.706) = 76.11% (Tech: 0.7599). Groups of 16 women do not
34 16
occupy a cockpit; because individual women occupy the cockpit, this result has no effect on the design.
7.
a.
b.
c.
d.
e.
Statistic.
Quantitative.
Discrete.
The sampling is conducted so that all samples of the same size have the same chance of being selected.
The sample is a voluntary response sample (or self-selected sample), and those with strong feelings
about the topic are more likely to respond, so it is not a valid sampling plan.
8.
a.
(0.6) = 0.1296
b.
1− 0.6 = 0.4
z=
4
Copyright © 2014 Pearson Education, Inc.
Chapter 12: Analysis of Variance 211
Chapter 12: Analysis of Variance
Section 12-2
1.
a.
b.
The chest deceleration measurements are categorized according to the one characteristic of size.
The terminology of analysis of variance refers to the method used to test for equality of the three
population means. That method is based on two different estimates of a common population variance.
2.
As we increase the number of individual tests of significance, we increase the risk of finding a difference
by chance alone (instead of a real difference in the means). The risk of a type I error—finding a difference
in one of the pairs when no such difference actually exists—is too high. The method of analysis of variance
helps us avoid that particular pitfall (rejecting a true null hypothesis) by using one test for equality of
several means, instead of several tests that each compare two means at a time.
3.
The test statistic is F = 3.288, and the F distribution applies.
4.
The P-value is 0.061. Because the P-value is greater than the significance level of 0.05, we fail to reject the
null hypothesis of equal means. There is not sufficient evidence to warrant rejection of the claim that the
three different categories of car sizes have the same mean chest deceleration in the standard car crash test.
5.
Test statistic: F = 0.39. P-value: 0.677. Fail to reject H0: μ1 = μ2 = μ3 . There is not sufficient evidence to
warrant rejection of the claim that the three categories of blood lead level have the same mean verbal IQ
score. Exposure to lead does not appear to have an effect on verbal IQ scores.
6.
Test statistic: F = 2.3034. P-value: 0.1044. Fail to reject H0: μ1 = μ2 = μ3 . There is not sufficient evidence
to warrant rejection of the claim that the three categories of blood lead level have the same mean full IQ
score. Exposure to lead does not appear to have an effect on full IQ scores.
7.
Test statistic: F = 11.6102. P-value: 0.000577. Reject H0: μ1 = μ2 = μ3 . There is sufficient evidence to
warrant rejection of the claim that the three size categories have the same mean highway fuel consumption.
The size of a car does appear to affect highway fuel consumption.
8.
Test statistic: F = 23.9457. P-value: 0.000008. Reject H0: μ1 = μ2 = μ3 . There is sufficient evidence to
warrant rejection of the claim that the three size categories have the same mean city fuel consumption. The
size of a car does appear to affect city fuel consumption.
9.
Test statistic: F = 0.161. P-value: 0.852. Fail to reject H0: μ1 = μ2 = μ3 . There is not sufficient evidence to
warrant rejection of the claim that the three size categories have the same mean head injury measurement.
The size of a car does not appear to affect head injuries.
10. Test statistic: F = 0.3476. P-value: 0.7111. Fail to reject H0: μ1 = μ2 = μ3 . There is not sufficient evidence
to warrant rejection of the claim that the three size categories have the same mean pelvis injury
measurement. The size of a car does not appear to affect pelvis injuries.
11. Test statistic: F = 27.2488. P-value: 0.000. Reject H0: μ1 = μ2 = μ3 . There is sufficient evidence to warrant
rejection of the claim that the three different miles have the same mean time. These data suggest that the
third mile appears to take longer, and a reasonable explanation is that the third lap has a hill.
EXCEL
ANOVA
Source of Variation
Between Groups
Within Groups
SS
0.103444
0.022778
Total
0.126222
df
2
12
MS
0.051722
0.001898
F
27.24878
14
Copyright © 2014 Pearson Education, Inc.
P-value
3.45E-05
F crit
3.885294
212 Chapter 12: Analysis of Variance
12. Test statistic: F = 9.4695. P-value: 0.000562. Reject H0: μ1 = μ2 = μ3 . There is sufficient evidence to
warrant rejection of the claim that the three books have the same mean Flesch Reading Ease score. The data
suggest that the books appear to have mean scores that are not all the same.
EXCEL
ANOVA
Source of Variation
Between Groups
Within Groups
SS
1338.002
2331.387
Total
3669.389
df
2
33
MS
669.0011
70.64808
F
9.469487
P-value
0.000562
F crit
3.284918
35
13. Test statistic: F = 6.1413. P-value: 0.0056. Reject H0: μ1 = μ2 = μ3 = μ4 . There is sufficient evidence to
warrant rejection of the claim that the four treatment categories yield poplar trees with the same mean
weight. Although not justified by the results from analysis of variance, the treatment of fertilizer and
irrigation appears to be most effective.
EXCEL
ANOVA
Source of Variation
Between Groups
Within Groups
SS
3.346455
2.9062
Total
6.252655
df
3
16
MS
1.115485
0.181638
F
6.14127
P-value
0.005566
F crit
3.238872
19
14. Test statistic: F = 0.3801. P-value: 0.769. Fail to reject H0: μ1 = μ2 = μ3 = μ4 . There is not sufficient
evidence to warrant rejection of the claim that the four treatment categories yield poplar trees with the same
mean weight. In the sandy and dry region, there does not appear to be a treatment that is more effective
than the others.
EXCEL
ANOVA
Source of Variation
Between Groups
Within Groups
SS
0.3114
4.36932
Total
4.68072
df
3
16
MS
0.1038
0.273083
F
0.380105
P-value
0.768664
F crit
3.238872
19
15. Test statistic: F = 18.9931. P-value: 0.000. Reject H0: μ1 = μ2 = μ3 . There is sufficient evidence to warrant
rejection of the claim that the three different types of cigarettes have the same mean amount of nicotine.
Given that the king-size cigarettes have the largest mean of 1.26 mg per cigarette, compared to the other
means of 0.87 mg per cigarette and 0.92 mg per cigarette, it appears that the filters do make a difference
(although this conclusion is not justified by the results from analysis of variance).
EXCEL
ANOVA
Source of Variation
Between Groups
Within Groups
SS
2.208267
4.1856
Total
6.393867
df
2
72
MS
1.104133
0.058133
F
18.99312
74
Copyright © 2014 Pearson Education, Inc.
P-value
2.38E-07
F crit
3.123907
Chapter 12: Analysis of Variance 213
16. Test statistic: F = 20.8562. P-value: 0.000. Reject H0: μ1 = μ2 = μ3 = μ4 . There is sufficient evidence to
warrant rejection of the claim that the three samples are from populations with the same mean. It appears
that cotinine levels are greater with more exposure to tobacco smoke. (The samples do not appear to be
from normally distributed populations, but ANOVA is robust against departures from normality.)
EXCEL
ANOVA
df
Source of Variation SS
Between Groups 518033.017 2
Within Groups
1453040.85 117
Total
MS
F
P-value
259016.508 20.8562144 1.7912E-08
12419.1526
1971073.87 119
Dotplot of SMOKER, ETS, NOETS
SMOKER
ETS
NOETS
0
80
160
240
320
400
480
560
Data
Each symbol represents up to 2 observations.
17. The Tukey test results show different P-values, but they are not dramatically different. The Tukey results
suggest the same conclusions as the Bonferroni test.
18. a.
b.
In Exercise 13 we reject the null hypothesis of equal means. The displayed Bonferroni results show
that with a P-value of 0.039, there is a significant difference between the mean of the no treatment
group (group 1) and the mean of the group treated with both fertilizer and irrigation (group 4).
The test statistic is t = –4.007. P-value = 6(0.001018) = 0.00611. Reject the null hypothesis that the
mean weight from the irrigation treatment group is equal to the mean from the group treated with both
fertilizer and irrigation.
Section 12-3
1.
The load values are categorized using two different factors of (1) femur (left or right) and (2) size of car
(small, midsize, large).
2.
No. To use two individual tests of one-way analysis of variance is to completely ignore the very important
feature of the possible effect from an interaction between femur and size. If there is an interaction, it
doesn’t make sense to consider the effects of one factor without the other.
3.
An interaction between two factors or variables occurs if the effect of one of the factors changes for
different categories of the other factor. If there is an interaction effect, we should not proceed with
individual tests for effects from the row factor and column factor. If there is an interaction, we should not
consider the effects of one factor without considering the effects of the other factor.
4.
Yes, the result is a balanced design because each cell has the same number (7) of values.
Copyright © 2014 Pearson Education, Inc.
214 Chapter 12: Analysis of Variance
5.
For interaction, the test statistic is F = 1.72 and the P-value is 0.194, so there is not sufficient evidence to
conclude that there is an interaction effect. For the row variable of femur (right, left), the test statistic is F =
1.39 and the P-value is 0.246, so there is not sufficient evidence to conclude that whether the femur is right
or left has an effect on measured load. For the column variable of size of the car, the test statistic is F =
2.23 and the P-value is 0.122, so there is not sufficient evidence to conclude that the car size category has
an effect on the measured load.
6.
For interaction, the test statistic is F = 0.34 and the P-value is 0.717, so there is not sufficient evidence to
conclude that there is an interaction effect. For the row variable of type (foreign, domestic), the test statistic
is F = 5.44 and the P-value is 0.038, so there is sufficient evidence to conclude that the type of car (foreign,
domestic) has an effect on measured chest deceleration. For the column variable of size of the car, the test
statistic is F = 3.58 and the P-value is 0.060, so there is not sufficient evidence to conclude that the car size
category has an effect on the measured chest deceleration.
7.
For interaction, the test statistic is F = 1.05 and the P-value is 0.365, so there is not sufficient evidence to
conclude that there is an interaction effect. For the row variable of sex, the test statistic is F = 4.58 and the
P-value is 0.043, so there is sufficient evidence to conclude that the sex of the subject has an effect on
verbal IQ score. For the column variable of blood lead level (LEAD), the test statistic is F = 0.14 and the Pvalue is 0.871, so there is not sufficient evidence to conclude that blood lead level has an effect on verbal
IQ score. It appears that only the sex of the subject has an effect on verbal IQ score.
8.
For interaction, the test statistic is F = 41.38 and the P-value is 0.000, so there is sufficient evidence to
conclude that there is an interaction effect. The ratings appear to be affected by an interaction between the
use of a supplement and the amount of whey. Because there appears to be an interaction effect, we should
not proceed with individual tests of the row factor (supplement) and the column factor (amount of whey).
EXCEL
ANOVA
9.
Source of Variation
Sample
Columns
Interaction
Within
SS
0.510417
6.824583
3.724583
0.48
Total
11.53958
df
1
3
3
16
MS
0.510417
2.274861
1.241528
0.03
F
P-value
17.01389 0.00079419
75.8287 1.12627E-09
41.38426 9.12956E-08
F crit
4.493998
3.238872
3.238872
23
For interaction, the test statistic is F = 3.7332 and the P-value is 0.0291, so there is sufficient evidence to
conclude that there is an interaction effect. The measures of self-esteem appear to be affected by an
interaction between the self-esteem of the subject and the self-esteem of the target. Because there appears
to be an interaction effect, we should not proceed with individual tests of the row factor (target’s selfesteem) and the column factor (subject’s self-esteem).
EXCEL
ANOVA
Source of Variation
Sample
Columns
Interaction
Within
SS
4.5
2.861111
6.75
59.66667
df
1
2
2
66
MS
Total
73.77778
71
4.5
1.430556
3.375
0.90404
F
P-value
4.977654 0.029079736
1.582402 0.213176735
3.73324 0.029107515
Copyright © 2014 Pearson Education, Inc.
F crit
3.986269
3.135918
3.135918
Chapter 12: Analysis of Variance 215
10. For interaction, the test statistic is F = 3.6872 and the P-value is 0.0628, so there is not sufficient evidence
to conclude that there is an interaction effect. For the row variable of gender (male, female), the test
statistic is F = 10.8629 and the P-value is 0.0022, so there is sufficient evidence to conclude that gender has
an effect on pulse rate. For the column variable of age bracket, the test statistic is
F = 4.8700 and the P-value is 0.0338, so there is sufficient evidence to conclude that the age bracket has an
effect on pulse rate.
EXCEL
ANOVA
Source of Variation
Sample
Columns
Interaction
Within
Total
11. a.
b.
c.
d.
SS
1322.5
592.9
448.9
4382.8
6747.1
df
1
1
1
36
MS
1322.5
592.9
448.9
121.7444
F
P-value
10.86292 0.00221114
4.870037 0.033790054
3.687232 0.062780461
F crit
4.113165
4.113165
4.113165
39
Test statistics and P-values do not change.
Test statistics and P-values do not change.
Test statistics and P-values do not change.
An outlier can dramatically affect and change test statistics and P-values.
Chapter Quick Quiz
1.
H0: μ1 = μ2 = μ3 . Because the displayed P-value of 0.000 is small, reject H0.
2.
No. Because we reject the null hypothesis of equal means, it appears that the three different power sources
do not produce the same mean voltage level, so we cannot expect electrical appliances to behave the same
way when run from the three different power sources.
3.
Right-tailed.
4.
Test statistic: F = 183.01. In general, larger test statistics result in smaller P-values.
5.
The sample voltage measurements are categorized using only one factor: the source of the voltage.
6.
Test a null hypothesis that three or more samples are from populations with equal means.
7.
With one-way analysis of variance, the different samples are categorized using only one factor, but with
two-way analysis of variance, the sample data are categorized into different cells determined by two
different factors.
8.
For interaction, the test statistic is F = 0.19 and the P-value is 0.832. Fail to reject the null hypothesis of no
interaction. There does not appear to be an effect due to an interaction between sex and major.
9.
The test statistic is F = 0.78 and the P-value is 0.395. There is not sufficient evidence to support a claim
that the length estimates are affected by the sex of the subject.
10. The test statistic is F = 0.13 and the P-value is 0.876. There is not sufficient evidence to support a claim
that the length estimates are affected by the subject’s major.
Review Exercises
1.
H0: μ1 = μ2 = μ3 . Test statistic: F = 10.10. P-value: 0.001. Reject the null hypothesis. There is sufficient
evidence to warrant rejection of the claim that 4-cylinder cars, 6-cylinder cars, and 8-cylinder cars have the
same mean highway fuel consumption amount.
Copyright © 2014 Pearson Education, Inc.
216 Chapter 12: Analysis of Variance
2.
For interaction, the test statistic is F = 0.17 and the P-value is 0.915, so there is not sufficient evidence to
conclude that there is an interaction effect. For the row variable of site, the test statistic is F = 0.81 and the
P-value is 0.374, so there is not sufficient evidence to conclude that the site has an effect on weight. For the
column variable of treatment, the test statistic is F = 7.50 and the P-value is 0.001, so there is sufficient
evidence to conclude that the treatment has an effect on weight.
3.
Test statistic: F = 42.9436. P-value: 0.000. Reject H0: μ1 = μ2 = μ3 . There is sufficient evidence to warrant
rejection of the claim that the three different types of cigarettes have the same mean amount of tar. Given
that the king-size cigarettes have the largest mean of 21.1 mg per cigarette, compared to the other means of
12.9 mg per cigarette and 13.2 mg per cigarette, it appears that the filters do make a difference (although
this conclusion is not justified by the results from analysis of variance).
EXCEL
ANOVA
Source of Variation
Between Groups
Within Groups
Total
4.
SS
df
MS
1083.707
2
541.8533
908.48
72
12.61778
1992.187
74
F
42.94364
P-value
F crit
5.29E-13 3.123907449
For interaction, the test statistic is F = 0.8733 and the P-value is 0.3685, so there does not appear to be an
effect from an interaction between gender and whether the subject smokes. For gender, the test statistic is F
= 0.0178 and the P-value is 0.8960, so gender does not appear to have an effect on body temperature. For
smoking, the test statistic is F = 3.0119 and the P-value is 0.1082, so there does not appear to be an effect
from smoking on body temperature.
EXCEL
ANOVA
Source of Variation
Sample(Gender)
Columns(Smokes)
Interaction
Within
SS
0.005625
0.950625
0.275625
3.7875
Total
5.019375
df
1
1
1
12
MS
0.005625
0.950625
0.275625
0.315625
F
0.017822
3.011881
0.873267
P-value
0.896012
0.108238
0.368476
F crit
4.747225
4.747225
4.747225
15
Cumulative Review Exercises
1.
a.
b.
c.
d.
15.5 years, 13.1 years, 22.7 years
9.7 years, 9.0 years, 18.6 years
94.5 years2, 80.3 years2, 346.1 years2
Ratio.
2.
Test statistic: t = –1.383. Critical values t = ±2.160 (assuming a 0.05 significance level). (Tech:
P-value = 0.1860.) Fail to reject H0: μ1 = μ2 . There is not sufficient evidence to support the claim that
there is a difference between the means for the two groups.
MINITAB
Difference = mu (Presidents) - mu (Monarchs)
Estimate for difference: -7.21
95% CI for difference: (-18.33, 3.90)
T-Test of difference = 0 (vs not =): T-Value = -1.38 P-Value = 0.187 DF = 15
Copyright © 2014 Pearson Education, Inc.
Chapter 12: Analysis of Variance 217
3.
Normal, because the histogram is approximately bell-shaped (or the points in a normal quantile plot are
reasonably close to a straight-line pattern with no other pattern that is not a straight-line pattern).
Histogram of Presidents
9
8
Frequency
7
6
5
4
3
2
1
0
0
10
20
30
40
Presidents
4.
12.3 years < μ < 18.7 years
MINITAB
One-Sample T: Presidents
N
Mean StDev
Variable
Presidents 38 15.50 9.72
5.
6.
SE Mean
1.58
95% CI
(12.30, 18.70)
a.
H0: μ1 = μ2 = μ3
b.
Because the P-value of 0.051 is greater than the significance level of 0.05, fail to reject the null
hypothesis of equal means. There is not sufficient evidence to warrant rejection of the claim that the
three means are equal. The three populations do not appear to have means that are significantly
different.
a.
r = 0.918. Critical values: r = ±0.707 . P-value = 0.001. There is sufficient evidence to support the
claim that there is a linear correlation between September weights and the subsequent April weights.
MINITAB
Correlations: September, April
Pearson correlation of September and April = 0.918
P-Value = 0.001
b.
yˆ = 9.28 + 0.823x
MINITAB
Regression Analysis: April versus September
The regression equation is
April = 9.3 + 0.823 September
Predictor Coef
SE Coef T
P
Constant 9.28
11.04
0.84 0.433
September 0.8227 0.1450 5.67 0.001
7.
c.
yˆ = 9.28 + 0.823(94) = 86.6 kg, which is not very close to the actual April weight of 105 kg.
a.
z=
345 − 280
= 1; P ( z > 1) = 0.1587.
65
b.
z=
215 − 280
= −1; P (−1 < z < 1) = 0.8413 − 0.1587 = 0.6826 (Tech: 0.6827)
65
c.
z=
d.
80th percentile: x = μ + z ⋅ σ = 280 + .84 ⋅ 65 = 334.6 (Tech: 334.7)
319 − 280
65 / 25
= 3; P ( z < 3) = 0.9987.
Copyright © 2014 Pearson Education, Inc.
218 Chapter 12: Analysis of Variance
8.
a.
0.20 ⋅1000 = 200 200
b.
0.175 < p < 0.225
MINITAB
Test and CI for One Proportion
Sample X
N
Sample p
1
200 1000 0.200000
9.
95% CI
(0.175621, 0.226159)
c.
Yes. The confidence interval shows us that we have 95% confidence that the true population
proportion is contained within the limits of 0.175 and 0.225, and 1/4 is not included within that range.
a.
The distribution should be uniform, with a flat shape. The given histogram agrees (approximately) with
the uniform distribution that we expect.
No. A normal distribution is approximately bell-shaped, but the given histogram is far from being bellshaped.
b.
10. Test statistic: χ 2 = 10.400. Critical value: χ 2 = 16.919 (assuming a 0.05 significance level). P-value >
0.10 (Tech: 0.319). There is not sufficient evidence to warrant rejection of the claim that the digits are
selected from a population in which the digits are all equally likely. There does not appear to be a problem
with the lottery.
MINITAB
Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: C1
N
DF Chi-Sq P-Value
200 9 10.4
0.319
Copyright © 2014 Pearson Education, Inc.
Chapter 13: Nonparametric Statistics
219
Chapter 13: Nonparametric Statistics
Section 13-2
1.
The only requirement for the matched pairs is that they constitute a simple random sample. There is no
requirement of a normal distribution or any other specific distribution. The sign test is “distribution free” in
the sense that it does not require a normal distribution or any other specific distribution.
2.
There are 2 positive signs, 7 negative signs, 1 tie, n = 9 , and the test statistic is x = 2 (the smaller of 2
and 7).
3.
H0: There is no difference between the populations of September weights and April weights. H1: There is a
difference between the populations of September weights and April weights. The sample data do not
contradict H1 because the numbers of positive signs (2) and negative signs (7) are not exactly the same.
4.
The efficiency of 0.63 indicates that with all other things being equal, the sign test requires 100 sample
observations to achieve the same results as 63 sample observations analyzed through a parametric test.
5.
The test statistic of x = 1 is less than or equal to the critical value of 2 (from Table A-7.) There is sufficient
evidence to warrant rejection of the claim of no difference. There does appear to be a difference.
6.
The test statistic of x = 5 is less than or equal to the critical value of 5 (from Table A-7.) There is
sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.
7.
The test statistic of z =
8.
The test statistic of z =
9.
There are 9 positive signs, 1 negative signs, 0 ties, and n = 10 . The test statistic of x = 1 is less than or
equal to the critical value of 1 (from Table A-7). There is sufficient evidence to warrant rejection of the
claim of no difference. There does appear to be a difference.
(151 + 0.5) − 1007
2
= −22.18 falls in the critical region bounded by z = −1.96 and
1007 2
1.96. There is sufficient evidence to warrant rejection of the claim of no difference. There does appear to be
a difference.
(172 + 0.5)− 6112
= −10.76 falls in the critical region bounded by z = −1.96 and
611 2
1.96. There is sufficient evidence to warrant rejection of the claim of no difference. There does appear to be
a difference.
(16 + 0.5) − 802
= −5.25 falls in the critical region bounded by z = ±1.96 . There is
80 2
sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.
10. The test statistic of z =
(14 + 0.5) − 322
= −0.53 does not fall in the critical region bounded by z = ±1.96 .
32 2
There is not sufficient evidence to warrant rejection of the claim of no difference. There does not appear to
be a difference.
11. The test statistic of z =
MINITAB
Sign Test for Median: Ht - HtOppSign test of median = 0.00000 versus not = 0.00000
N Below Equal Above
P Median
C7
34
14
2
18 0.5966 1.000
12. There are 12 positive signs, 0 negative signs, 0 ties, and n = 12 . The test statistic of x = 0 is less than or
equal to the critical value of 2 (from Table A-7). There is sufficient evidence to warrant rejection of the
claim of no difference. There does appear to be a difference.
Copyright © 2014 Pearson Education, Inc.
220
Chapter 13: Nonparametric Statistics
(52 + 0.5)− 2912
= −10.90 is in the critical region bounded by z = ±2.575 . There
291 2
is sufficient evidence to warrant rejection of the claim of no difference. The YSORT method appears to
have an effect on the gender of the child. (Because so many more boys were born than would be expected
with no effect, it appears that the YSORT method is effective in increasing the likelihood that a baby will
be a boy.)
13. The test statistic of z =
(47 + 0.5) − 1042
= −0.88 is not in the critical region bounded by z = ±1.96 . There
104 2
is not sufficient evidence to warrant rejection of the claim of no difference. It appears that women do not
have the ability to predict the sex of their babies.
14. The test statistic of z =
(123 + 0.5) − 2802
= −1.97 is not in the critical region bounded by z = ±2.575 .
280 2
There is not sufficient evidence to warrant rejection of the claim that the touch therapists make their
selections with a method equivalent to random guesses. The touch therapists do not appear to be effective
in selecting the correct hand.
15. The test statistic of z =
(242 + 0.5) − 6382
= −6.06 is in the critical region bounded by z = ±1.96 . There is
638 2
sufficient evidence to warrant rejection of the claim that respondents did not have a strong opinion one way
or the other. They do appear to favor the opinion that NFL games are not too long. However, the validity of
the results is highly questionable because the sample is a voluntary response sample.
16. The test statistic of z =
(12 + 0.5) − 402
= −2.37 is not in the critical region bounded by z = ±2.575 . There
40 2
is not sufficient evidence to warrant rejection of the claim that the median is equal to 5.670 g. The quarters
appear to be minted according to specifications.
17. The test statistic of z =
MINITAB
Sign Test for Median: Post-1964 Quarters
Sign test of median = 5.670 versus not = 5.670
N Below Equal Above
P Median
Post-1964 Quarters
40
28
0
12 0.0166 5.636
(20 + 0.5) − 472
= −0.88 is not in the critical region bounded by z = ±2.575 .
47 2
There is not sufficient evidence to warrant rejection of the claim that the median is equal to 1.00.
18. The test statistic of z =
MINITAB
Sign Test for Median: MAGNITUDE
Sign test of median = 1.000 versus not = 1.000
N Below Equal Above
P Median
MAG
50
20
3
27 0.3817 1.235
(1 + 0.5) − 342
= −5.32 is in the critical region bounded by z = ±1.96 . There is
34 2
sufficient evidence to warrant rejection of the claim that the median amount of Coke is equal to 12 oz.
Consumers are not being cheated because they are generally getting more than 12 oz of Coke, not less.
19. The test statistic of z =
Copyright © 2014 Pearson Education, Inc.
Chapter 13: Nonparametric Statistics
221
19. (continued)
MINITAB
Sign Test for Median: CKREGVOL
Sign test of median = 12.00 versus not = 12.00
N Below Equal Above
P Median
CKREGVOL 36
1
2
33 0.0000 12.20
(27 + 0.5)− 792
= −2.70 is in the critical region bounded by z = ±1.96 . There is
79 2
sufficient evidence to warrant rejection of the claim that the median age is 30 years.
20. The test statistic of z =
MINITAB
Sign Test for Median: Actresses
Sign test of median = 30.00 versus not = 30.00
N Below Equal Above
P Median
Actresses
82
27
3
52 0.0069 33.00
21. Second approach: The test statistic of z =
(30 + 0.5)− 1052
= −4.29 is in the critical region bounded by
105 2
z = −1.645 , so the conclusions are the same as in Example 4.
(38 + 0.5) − 1062
Third approach: The test statistic of z =
= −2.82 is in the critical region bounded by
106 2
z = −1.645 , so the conclusions are the same as in Example 4. The different approaches can lead to very
different results; as seen in the test statistics of –4.21, –4.29, and –2.82. The conclusions are the same in
this case, but they could be different in other cases.
22. The column entries are *, *, *, *, *, 0, 0, 0.
Section 13-3
1.
The only requirements are that the matched pairs be a simple random sample and the population of
differences be approximately symmetric. There is no requirement of a normal distribution or any other
specific distribution. The Wilcoxon signed-ranks test is “distribution free” in the sense that it does not
require a normal distribution or any other specific distribution.
2.
a.
b.
c.
d.
e.
1, 1, –4, –3, 0, –1, –5, –8, –3, –1
2.5, 2.5, 7, 5.5, 2.5, 8, 9, 5.5, 2.5
2.5, 2.5, –7, –5.5, –2.5, –8, –9, –5.5, –2.5
5 and 40
T =5
f.
Critical value of T is 6.
3.
The sign test uses only the signs of the differences, but the Wilcoxon signed-ranks test uses ranks that are
affected by the magnitudes of the differences.
4.
The efficiency of 0.95 indicates that with all other things being equal, the Wilcoxon signed-ranks test
requires 100 sample observations to achieve the same results as 95 sample observations analyzed through a
parametric test.
5.
Test statistic: T = 6 . Critical value: T = 8 . Reject the null hypothesis that the population of differences
has a median of 0. There is sufficient evidence to warrant rejection of the claim of no difference. There
does appear to be a difference.
MINITAB
Test of median = 0.000000 versus median not = 0.000000
N for
Wilcoxon
Estimated
N
Test
Statistic
P
Median
AGE DIFF 10
10
6.0
0.032
-11.50
Copyright © 2014 Pearson Education, Inc.
222
6.
Chapter 13: Nonparametric Statistics
501.5 −
80 (80 + 1)
4
= −5.36 .
80 (80 + 1)(2 ⋅ 80 + 1)
24
Critical values: z = ±1.96 . (Tech: P-value = 0.000.) Reject the null hypothesis that the population of
differences has a median of 0. There is sufficient evidence to warrant rejection of the claim of no
difference. There does appear to be a difference.
Convert T = 501.5 to the test statistic z =
MINITAB
Test of median = 0.000000 versus median not = 0.000000
N for
Wilcoxon
Estimated
N
Test
Statistic
P
Median
AGE DIFF 82
80
501.5
0.000
-9.000
7.
247 −
32 (32 + 1)
4
= −0.32 .
32 (32 + 1)(2 ⋅ 32 + 1)
24
Critical values: z = ±1.96 . (Tech: P-value = 0.751.) Fail to reject the null hypothesis that the population
of differences has a median of 0. There is not sufficient evidence to warrant rejection of the claim of no
difference. There does not appear to be a difference.
Convert T = 247 to the test statistic z =
MINITAB
Test of median = 0.000000 versus median not = 0.000000
N for
Wilcoxon
Estimated
N
Test
Statistic
P
Median
HtDiff
34
32
247.0
0.758
-0.5000
8.
Test statistic: T = 0 . Critical value: T = 14 . Reject the null hypothesis that the population of differences
has a median of 0. There is sufficient evidence to warrant rejection of the claim of no difference. There
does appear to be a difference.
MINITAB
Test of median = 0.000000 versus median not = 0.000000
N for
Wilcoxon
Estimated
N
Test
Statistic
P
Median
IN-OUT
12
12
0.0 0
.003
-8.500
40 ( 40 + 1)
4
Convert T = 196 to the test statistic z =
= −2.88 .
40 (40 + 1)(2 ⋅ 40 + 1)
24
Critical values: z = ±2.57 5. (Tech: P-value = 0.004.) There is sufficient evidence to warrant rejection of
the claim that the median is equal to 5.670 g. The quarters do not appear to be minted according to
specifications.
196 −
9.
MINITAB
Test of median = 5.670 versus median not = 5.670
N for
Wilcoxon
N
Test
Statistic
Post-1964 Quarters 40
40
196.0
P
0.004
Estimated
Median
5.638
47 (47 + 1)
4
10. Convert T = 376 to the test statistic z =
= −1.99 .
47 ( 47 + 1)( 2 ⋅ 47 + 1)
24
Critical values: z = ±2.57 . (Tech: P-value = 0.047.) There is not sufficient evidence to warrant rejection
of the claim that the median is equal to 1.00.
376 −
Copyright © 2014 Pearson Education, Inc.
Chapter 13: Nonparametric Statistics 223
10. (continued)
MINITAB
Test of median = 1.000000 versus median not = 1.000000
N for
Wilcoxon
Estimated
N
Test
Statistic
P
Median
Mag
50
47
376.0
0.047
-0.1400
34 (34 + 1)
4
11. Convert T = 15.5 to the test statistic z =
= −4.82 .
34 (34 + 1)(2 ⋅ 34 + 1)
24
Critical values: z = ±1.96 . (Tech: P-value = 0.000.) There is sufficient evidence to warrant rejection of
the claim that the median amount of Coke is equal to 12 oz. Consumers are not being cheated because they
are generally getting more than 12 oz of Coke, not less.
15.5 −
MINITAB
Test of median = 12.000000 versus median not = 12.000000
N for
Wilcoxon
Estimated
N
Test
Statistic
P
Median
Volume
36
34
15.5
0.000
-0.2000
79 (79 + 1)
4
12. Convert T = 672.5 to the test statistic z =
= −4.44 .
79 (79 + 1)( 2 ⋅ 79 + 1)
24
Critical values: z = ±1.96 . (Tech: P-value = 0.000.) There is sufficient evidence to warrant rejection of
the claim that the median age is 30 years.
672.5 −
MINITAB
Test of median = 30.000000 versus median not = 30.000000
N for
Wilcoxon
Estimated
N
Test
Statistic
P
Median
AGE
82
79
672.5
0.000
-4.000
13. a.
b.
c.
d.
Min: 0 and Max: 1 + 2 + … +74 + 75 = 2850
2850
= 1425
2
2850 − 850 = 2000
n (n + 1)
2
−k
Section 13-4
1.
Yes. The two samples are independent because the flight data are not matched. The samples are simple
random samples. Each sample has more than 10 values.
2.
R = 137.5
3.
H0: Arrival delay times from Flights 19 and 21 have the same median. There are three different possible
alternative hypotheses: H1: Arrival delay times from Flights 19 and 21 have different medians. H1: Arrival
delay times from Flight 19 have a median greater than the median of arrival delay times from Flight 21. H1:
Arrival delay times from Flight 19 have a median less than the median of arrival delay times from Flight
21.
4.
The efficiency rating of 0.95 indicates that with all other factors being the same, the Wilcoxon rank-sum
test requires 100 sample observations to achieve the same results as 95 observations with the parametric t
test of Section 9-3, assuming that the stricter requirements of the parametric t test are satisfied.
Copyright © 2014 Pearson Education, Inc.
224
5.
Chapter 13: Nonparametric Statistics
R1 = 137.5 , R2 = 162.5 , μR =
12 (12 + 12 + 1)
2
= 150 , σ R =
12 ⋅12 (12 + 12 + 1)
12
= 17.321 , test statistic:
137.5 −150
= −0.72 . Critical values: z = ±1.96 . (Tech: P-value = 0.4705.) Fail to reject the null
17.321
hypothesis that the populations have the same median. There is not sufficient evidence to warrant rejection
of the claim that Flights 19 and 21 have the same median arrival delay time.
z=
6.
R1 = 128 , R2 = 172 R2 = 172, μR =
12 (12 + 12 + 1)
2
= 150 , σ R =
12 ⋅12 (12 + 12 + 1)
12
= 17.321 , test
128 −150
= −1.27 . Critical values: z = ±1.96 . (Tech: P-value = 0.2040.) Fail to reject the
17.321
null hypothesis that the populations have the same median. There is not sufficient evidence to warrant
rejection of the claim that Flights 19 and 21 have the same median taxi-out time.
statistic: z =
7.
R1 = 253.5 , R2 = 124.5 , μR =
13(13 + 14 + 1)
2
= 182 , σ R =
13 ⋅14 (13 + 14 + 1)
12
= 20.607 , test statistic:
253.5 −182
= 3.47 . Critical values: z = ±1.96 . (Tech: P-value = 0.0005.) Reject the null hypothesis
20.607
that the populations have the same median. There is sufficient evidence to reject the claim that for those
treated with 20 mg of atorvastatin and those treated with 80 mg of atorvastatin, changes in LDL cholesterol
have the same median. It appears that the dosage amount does have an effect on the change in LDL
cholesterol.
z=
8.
R1 = 194.5 , R2 = 105.5 , μR =
12 (12 + 12 + 1)
2
= 150 , σ R =
12 ⋅12 (12 + 12 + 1)
12
= 17.321 , test statistic:
194.5 −150
= 2.57 . Critical values: z = ±1.96 . (Tech: P-value = 0.0102.) Reject the null hypothesis
17.321
that the populations have the same median. There is sufficient evidence to reject the claim that the median
amount of strontium-90 from Pennsylvania residents is the same as the median from New York residents.
z=
9.
R1 = 501 , R2 = 445 , μR =
22 ( 22 + 21 + 1)
2
= 484 , σ R =
22 ⋅ 21(22 + 21 + 1)
12
= 41.158 , test statistic:
501− 484
= 0.41 . Critical value: z = 1.645 . (Tech: P-value = 0.3398.) Fail to reject the null
41.158
hypothesis that the populations have the same median. There is not sufficient evidence to support the claim
that subjects with medium lead levels have full IQ scores with a higher median than the median full IQ
score for subjects with high lead levels. It does not appear that lead level affects full IQ scores.
z=
10. R1 = 4178 , R2 = 772 , μR =
78 (78 + 21 + 1)
2
= 3900 , σ R =
78 ⋅ 21(78 + 21 + 1)
12
= 116.833 , test statistic:
4178 − 3900
= 2.38 . Critical value: z = 1.645 . (Tech: P-value = 0.0087.) Reject the null hypothesis
116.833
that the populations have the same median. There is sufficient evidence to support the claim that subjects
with low lead levels have performance IQ scores with a higher median than the median performance IQ
score for subjects with high lead levels. It appears that exposure to lead does have an adverse effect.
z=
Copyright © 2014 Pearson Education, Inc.
Chapter 13: Nonparametric Statistics 225
40 ( 40 + 40 + 1)
11. R1 = 2420 , R2 = 820 , μR =
2
= 1620 , σ R =
40 ⋅ 40 ( 40 + 40 + 1)
12
= 103.923 , test
2420 −1620
= 7.70 . Critical values: z = ±1.96 . (Tech: P-value = 0.0000.) Reject the null
103.923
hypothesis that the populations have the same median. It appears that the design of quarters changed in
1964.
statistic: z =
12. R1 = 1958 , R2 = 670 , μR =
36 (36 + 36 + 1)
2
= 1314 , σ R =
36 ⋅ 36 (36 + 36 + 1)
12
= 88.792 , test statistic:
1958 −1314
= 7.25 . Critical values: z = ±1.96 . (Tech: P-value = 0.0000.) Reject the null hypothesis
88.792
that the populations have the same median. It appears that cans of regular Coke have weights different from
cans of diet Coke. The cans of regular Coke appear to weigh more because they include more sugar.
z=
12 ⋅11
2
13. Using U = 12 ⋅11 +
= 1.26 . The test statistic is
−123.5 = 86.5 , we get z =
2
12 ⋅11⋅ (12 + 11 + 1)
12
the same value with opposite sign.
86.5 −
12 (12 + 1)
14. a.
Rank Sum for
Treatment A
Rank
1
2
3
4
A
A
B
B
3
A
B
A
B
4
A
B
B
A
5
B
B
A
A
7
B
A
A
B
5
B
A
B
A
6
1 1 2 1
1
, , , , and , respectively
6 6 6 6
6
b.
The R values of 3, 4, 5, 6, 7 have probabilities of
c.
No, none of the probabilities for the values of R would be less than 0.10.
Section 13-5
1.
R1 = 1 + 10 + 12.5 + 5 + 8 = 36.5 , R2 = 3 + 14 + 15 + 2 + 12.5 + 6 = 52.5 , R3 = 7 + 16 + 10 + 4 + 10 = 47
Low Lead Level
70 (1)
Medium Lead Level
72 (3)
High Lead Level
82 (7)
85 (10)
86 (12.5)
76 (5)
84 (8)
90 (14)
92 (15)
71 (2)
86 (12.5)
79 (6)
93 (16)
85 (10)
75 (4)
85 (10)
2.
Yes. The samples are independent simple random samples, and each sample has at least five data values.
3.
n1 = 5 , n2 = 6 , n3 = 5 , and N = 5 + 6 + 5 = 16 .
Copyright © 2014 Pearson Education, Inc.
226 Chapter 13: Nonparametric Statistics
4.
5.
6.
The efficiency rating of 0.95 indicates that with all other factors being the same, the Kruskal-Wallis test
requires 100 sample observations to achieve the same results as 95 observations with the parametric oneway analysis of variance test, assuming that the stricter requirements of the parametric test are satisfied.
⎛ 332 222 652 ⎞⎟
12
⎟ − 3(15 + 1) = 9.9800 . Critical value: χ 2 = 5.991 . (Tech:
+
+
⎜⎜
15(15 + 1) ⎜⎝ 5
5
5 ⎠⎟⎟
P-value = 0.0068.) Reject the null hypothesis of equal medians. The data suggest that the different miles
present different levels of difficulty.
Test statistic: H =
Test statistic: H =
⎛ 201.52 337.02 128.52 ⎞⎟
12
⎜⎜
⎟ − 3(36 + 1) = 16.9486 . Critical value:
+
+
36 (36 + 1) ⎜⎝ 12
12
12 ⎠⎟⎟
χ 2 = 5.991 . (Tech: P-value = 0.0002.) Reject the null hypothesis of equal medians. The data suggest that
the books have levels of reading difficulty that are not all the same.
⎛ 862 97 2 482 ⎞⎟
12
⎜⎜
⎟ − 3(21 + 1) = 4.9054 . Critical value: χ 2 = 5.991 . (Tech:
+
+
21( 21 + 1) ⎜⎝ 7
7
7 ⎠⎟⎟
P-value = 0.0861.) Fail to reject the null hypothesis of equal medians. The data do not suggest that larger
cars are safer.
7.
Test statistic: H =
8.
Test statistic: H =
⎛ 33.52 85.52 1122 ⎞⎟
12
⎜⎜
⎟ − 3(21 + 1) = 11.8349 . Critical value: χ 2 = 5.991 .
+
+
21( 21 + 1) ⎜⎝ 7
7
7 ⎠⎟⎟
(Tech: P-value = 0.0027.) Reject the null hypothesis of equal medians. The size of a car does appear to
affect highway fuel consumption.
9.
Test statistic: H =
⎛ 5277.52 11122 991.52 ⎞⎟
12
⎟ − 3(121 + 1) = 8.0115 . Critical value:
+
+
⎜⎜
121(121 + 1) ⎜⎝ 78
22
21 ⎠⎟⎟
χ 2 = 9.210 . (Tech: P-value = 0.0182.) Fail to reject the null hypothesis of equal medians. The data do not
suggest that lead exposure has an adverse effect.
10. Test statistic: H =
⎛ 3629.52 2393.52 1237 2 ⎞⎟
12
⎜⎜
⎟ − 3(120 + 1) = 59.1546 . Critical value:
+
+
120 (120 + 1) ⎜⎝ 40
40
40 ⎠⎟⎟
χ 2 = 9.210 . (Tech: P-value = 0.0000.) Reject the null hypothesis of equal medians. The data suggest that
the amounts of nicotine absorbed by smokers is different from the amounts absorbed by people who don’t
smoke.
11. Test statistic: H =
⎛1413.52 650.52 7862 ⎞⎟
12
⎟ − 3(75 + 1) = 27.9098 . Critical value:
+
+
⎜⎜
75(75 + 1) ⎜⎝ 25
25
25 ⎠⎟⎟
χ 2 = 5.991 . (Tech: P-value: 0.0000.) Reject the null hypothesis of equal medians. There is sufficient
evidence to warrant rejection of the claim that the three different types of cigarettes have the same median
amount of nicotine. It appears that the filters do make a difference.
12. Test statistic: H =
⎛ 822 66.52 82.52 ⎞⎟
12
⎜⎜
⎟ − 3( 21 + 1) = 0.6141 . Critical value: χ 2 = 5.991 .
+
+
21( 21 + 1) ⎜⎝ 7
7
7 ⎠⎟⎟
(Tech: P-value = 0.7356.) Fail to reject the null hypothesis of equal medians. The data do not suggest that
larger cars are safer.
Copyright © 2014 Pearson Education, Inc.
Chapter 13: Nonparametric Statistics
227
13. Using ΣT = 16,836 (see table below) and N = 25 + 25 + 25 = 75 , the corrected value of H is
27.9098
=29.0701 , which is not substantially different from the value found in Exercise 11.
16,836
1− 3
75 − 75
In this case, the large numbers of ties do not appear to have a dramatic effect on the test statistic H.
Nicotine Level
Rank
t
t3 −t
0.2
1.5
2
6
0.6
4.5
2
6
0.7
6.5
2
6
0.8
17.0
19
6840
0.9
28.0
3
24
1.0
33.5
8
504
1.1
48.0
21
9240
1.2
61.0
5
120
1.3
65.5
4
60
1.4
69.0
3
24
1.7
73.5
2
6
SUM
16,836
Section 13-6
1.
The methods of Section 10-3 should not be used for predictions. The regression equation is based on a
linear correlation between the two variables, but the methods of this section do not require a linear
relationship. The methods of this section could suggest that there is a correlation with paired data
associated by some nonlinear relationship, so the regression equation would not be a suitable model for
making predictions.
2.
Data at the nominal level of measurement have no ordering that enables them to be converted to ranks, so
data at the nominal level of measurement cannot be used with the methods of rank correlation.
3.
r represents the linear correlation coefficient computed from sample paired data; ρ represents the
parameter of the linear correlation coefficient computed from a population of paired data; rs denotes the
rank correlation coefficient computed from sample paired data; ρs represents the rank correlation
coefficient computed from a population of paired data. The subscript s is used so that the rank correlation
coefficient can be distinguished from the linear correlation coefficient r. The subscript does not represent
the standard deviation s. It is used in recognition of Charles Spearman, who introduced the rank correlation
method.
4.
The efficiency rating of 0.91 indicates that with all other factors being the same, rank correlation requires
100 pairs of sample observations to achieve the same results as 91 pairs of observations with the parametric
test using linear correlation, assuming that the stricter requirements for using linear correlation are met.
5.
rs = 1 . Critical values are rs = ±0.886 (From Table A-9.) Reject the null hypothesis of ρs = 0 . There is
sufficient evidence to support a claim of a correlation between distance and time.
6.
rs = −1 . Critical values are rs = ±0.648 (From Table A-9.) Reject the null hypothesis of ρs = 0 . There is
sufficient evidence to support a claim of a correlation between altitude and time.
Copyright © 2014 Pearson Education, Inc.
228 Chapter 13: Nonparametric Statistics
7.
rs = 0.821 . Critical values: rs = ±0.786 (From Table A-9.) Reject the null hypothesis of ρs = 0 . There is
sufficient evidence to support the claim of a correlation between the quality scores and prices. These results
do suggest that you get better quality by spending more.
MINITAB
Pearson correlation of Rank Price and Rank Quality = 0.821
8.
rs = 0.467 . Critical values: rs = ±0.738 (From Table A-9.) Fail to reject the null hypothesis of ρs = 0 .
There is not sufficient evidence to support the claim of a correlation between the quality scores and prices.
These results do not suggest that you get better quality by spending more.
MINITAB
Pearson correlation of Rank Quality and Rank Price = 0.467
9.
rs = −0.929 . Critical values: rs = ±0.786 (From Table A-9.) Reject the null hypothesis of ρs = 0 . There
is sufficient evidence to support the claim of a correlation between the two judges. Examination of the
results shows that the first and third judges appear to have opposite rankings.
MINITAB
Pearson correlation of First and Second = -0.929
10. rs = 0.607 . Critical values: rs = ±0.786 (From Table A-9.) Fail to reject the null hypothesis of ρs = 0 .
There is not sufficient evidence to support the claim of a correlation between the two judges. The two
judges appear to rank the bands very differently.
MINITAB
Pearson correlation of First and Third = 0.607
11. rs = 1 . Critical values: rs = ±0.886 (From Table A-9.) Reject the null hypothesis of ρs = 0 . There is
sufficient evidence to conclude that there is a correlation between overhead widths of seals from
photographs and the weights of the seals.
MINITAB
Pearson correlation of Rank Width and Rank Weight = 1.000
12. rs = 0.857 . Critical values: rs = ±0.738 (From Table A-9.) Reject the null hypothesis of ρs = 0 . There is
sufficient evidence to conclude that there is a correlation between the number of chirps in 1 min and the
temperature.
MINITAB
Pearson correlation of Rank Chirps and Rank Temp = 0.857
13. rs = 0.394 . Critical values: rs = ±
1.96
= ±0.314 . Reject the null hypothesis of ρs = 0 . There is
40 −1
sufficient evidence to conclude that there is a correlation between the systolic and diastolic blood pressure
levels in males.
MINITAB
Pearson correlation of RANK SYS and RANK DIAS = 0.394
14. rs = 0.106 . Critical values: rs = ±0.447 (From Table A-9.) Fail to reject the null hypothesis of ρs = 0 .
There is not sufficient evidence to conclude that there is a correlation between brain volumes and IQ
scores.
MINITAB
Pearson correlation of RANK VOL and RANK IQ = 0.106
15. rs = 0.651 . Critical values: rs = ±
1.96
= ±0.286 . Reject the null hypothesis of ρs = 0 . There is
48 −1
sufficient evidence to conclude that there is a correlation between departure delay times and arrival delay
times.
MINITAB
Pearson correlation of RANK DEPART and RANK ARRIVE = 0.651
Copyright © 2014 Pearson Education, Inc.
Chapter 13: Nonparametric Statistics 229
16. rs = 0.0428 . Critical values: rs = ±
1.96
= ±0.280 . Fail to reject the null hypothesis of ρs = 0 . There
50 −1
is not sufficient evidence to conclude that there is a correlation between magnitudes and depths of
earthquakes.
MINITAB
Pearson correlation of RANK MAG and RANK DEPTH = 0.043
17. a.
rs = ±
2.447 2
= ±0.707 is not very close to the values of rs = ±0.738 found in Table A-9.
2.447 2 + 8 − 2
b.
rs = ±
2.7632
= ±0.463 is quite close to the values of rs = ±0.467 found in Table A-9.
2.7632 + 30 − 2
Section 13-7
1.
No. The runs test can be used to determine whether the sequence of World Series wins by American
League teams and National League teams is not random, but the runs test does not show whether the
proportion of wins by the American League is significantly greater than 0.5.
2.
n1 = 12 , n2 = 8 , G = 9
3.
a.
b.
Answers vary, but here is a sequence that leads to rejection of randomness because the number of runs
is 2, which is very low: W W W W W W W W W W W W E E E E E E E E
Answers vary, but here is a sequence that leads to rejection of randomness because the number of runs
is 17, which is very high: W E W E W E W E W E W E W E W E W W W W
4.
No. It is very possible that the sequence of data appears to be random, yet the sampling method (such as
voluntary response sampling) might be very unsuitable for statistical methods.
5.
n1 = 19, n2 = 15, G = 16 , critical values: 11, 24 (From Table A-10.) Fail to reject randomness. There is
not sufficient evidence to support the claim that we elect Democrats and Republicans in a sequence that is
not random. Randomness seems plausible here.
6.
n1 = 17, n2 = 13, G = 15 , critical values: 10, 22 (From Table A-10.) Fail to reject randomness. There is
not sufficient evidence to warrant rejection of the claim that odd and even digits occur in random order.
7.
n1 = 20, n2 = 10, G = 16 , critical values: 9, 20 (From Table A-10.) Fail to reject randomness. There is not
sufficient evidence to reject the claim that the dates before and after July 1 are randomly selected.
8.
n1 = 10, n2 = 10, G = 2 , critical values: 6, 16 (From Table A-10.) Reject randomness. The numbers of
daily newspapers do not appear to be in a random sequence. Because all of the values above the median
occur in the beginning and all of the values below the median occur at the end, there appears to be a
downward trend in the numbers of daily newspapers.
9.
n1 = 24, n2 = 21, G = 17 , μG =
2 ⋅ 24 ⋅ 21(2 ⋅ 24 ⋅ 21− 24 − 21)
2 ⋅ 24 ⋅ 21
+ 1 = 23.4 , σ G =
= 3.3007 .
2
24 + 21
(24 + 21) (24 + 21−1)
17 − 23.4
= −1.94 . Critical values: z = ±1.96 . (Tech: P-value = 0.05252.) Fail to reject
3.3007
randomness. There is not sufficient evidence to reject randomness. The runs test does not test for
disproportionately more occurrences of one of the two categories, so the runs test does not suggest that
either conference is superior.
Test statistic: z =
Copyright © 2014 Pearson Education, Inc.
230 Chapter 13: Nonparametric Statistics
10. n1 = 62, n2 = 44, G = 62 , μG =
2 ⋅ 62 ⋅ 44 ( 2 ⋅ 62 ⋅ 44 − 62 − 44)
2 ⋅ 62 ⋅ 44
+ 1 = 52.4717 , σ G =
= 4.9741 .
2
62 + 44
(62 + 44) (62 + 44 −1)
62 − 52.4717
= 1.92 . Critical values: z = ±1.96 . (Tech: P-value = 0.0554.) Fail to
4.9741
reject randomness. There is not sufficient evidence to reject randomness.
Test statistic: z =
11. The median is 2453, n1 = 23, n2 = 23, G = 4 , μG =
σG =
2 ⋅ 23 ⋅ 23(2 ⋅ 23⋅ 23 − 23 − 23)
(23 + 23) (23 + 23 −1)
2
2 ⋅ 23 ⋅ 23
+ 1 = 24 ,
23 + 23
= 3.3553 . Test statistic: z =
4 − 24
= −5.96 . Critical values:
3.3553
z = ±1.96 . (Tech: P-value = 0.0000.) Reject randomness. The sequence does not appear to be random
when considering values above and below the median. There appears to be an upward trend, so the stock
market appears to be a profitable investment for the long term, but it has been more volatile in recent years.
12. The mean is 14.250°C, n1 = 26, n2 = 24, G = 8 , μG =
σG =
2 ⋅ 26 ⋅ 24 ( 2 ⋅ 26 ⋅ 23 − 26 − 24)
(26 + 24) ( 26 + 24 −1)
2
2 ⋅ 26 ⋅ 24
+ 1 = 25.96 ,
26 + 24
= 3.4936 . Test statistic: z =
8 − 25.96
= −5.14 . Critical values:
3.4935
z = ±1.96 . (Tech: P-value = 0.0000.) Reject randomness. The sequence does not appear to be random
when considering values above and below the mean. There appears to be an upward trend, so global
warming appears to be occurring.
13. a.
b.
c.
d.
No solution provided.
The 84 sequences yield these results: 2 sequences have 2 runs, 7 sequences have 3 runs, 20 sequences
have 4 runs, 25 sequences have 5 runs, 20 sequences have 6 runs, and 10 sequences have 7 runs.
With P(2 runs) = 2/84, P(3 runs) = 7/84, P(4 runs) = 20/84, P(5 runs) = 25/84, P(6 runs) = 20/84, and
P(7 runs) = 10/84, each of the G values of 3, 4, 5, 6, 7 can easily occur by chance, whereas G = 2 is
unlikely because P(2 runs) is less than 0.025. The lower critical value of G is therefore 2, and there is
no upper critical value that can be equaled or exceeded.
Critical value of G = 2 agrees with Table A-10. The table lists 8 as the upper critical value, but it is
impossible to get 8 runs using the given elements.
Chapter Quick Quiz
1.
Distribution-free test
2.
57 has rank
3.
The efficiency rating of 0.91 indicates that with all other factors being the same, rank correlation requires
100 pairs of sample observations to achieve the same results as 91 pairs of observations with the parametric
test for linear correlation, assuming that the stricter requirements for using linear correlation are met.
4.
The Wilcoxon rank-sum test does not require that the samples be from populations having a normal
distribution or any other specific distribution.
5.
G=4
6.
Because there are only two runs, all of the values below the mean occur at the beginning and all of the
values above the mean occur at the end, or vice versa. This indicates an upward (or downward) trend.
7.
Sign test and Wilcoxon signed-ranks test
8.
Rank correlation
1+ 2 + 3
= 2 , 58 has rank 4, and 61 has rank 5.
3
Copyright © 2014 Pearson Education, Inc.
Chapter 13: Nonparametric Statistics 231
9.
Kruskal-Wallis test
10. Test claims involving matched pairs of data; test claims involving nominal data; test claims about the
median of a single population
Review Exercises
The test statistic of z =
(44 + 0.5) − 1062
1.
= −1.65 is not less than or equal to the critical value of
106 2
z = −1.96 . Fail to reject the null hypothesis of p = 0.5 . There is not sufficient evidence to warrant
rejection of the claim that in each World Series, the American League team has a 0.5 probability of
winning.
2.
There are 6 positive signs, 0 negative signs, 0 ties, and n = 7 . The test statistic of x = 0 is less than or
equal to the critical value of 0. There is sufficient evidence to reject the claim of no difference. It appears
that there is a difference in cost between flights scheduled 1 day in advance and those scheduled 30 days in
advance. Because all of the flights scheduled 30 days in advance cost less than those scheduled 1 day in
advance, it is wise to schedule flights 30 days in advance.
3.
The test statistic of T = 0 is less than or equal to the critical value of 0. There is sufficient evidence to
reject the claim that differences between fares for flights scheduled 1 day in advance and those scheduled
30 days in advance have a median equal to 0. Because all of the flights scheduled 1 day in advance have
higher fares than those scheduled 30 days in advance, it appears that it is generally less expensive to
schedule flights 30 days in advance instead of 1 day in advance.
MINITAB
Test of median = 0.000000 versus median not = 0.000000
N for
Wilcoxon
Estimated
N
Test
Statistic
P
Median
One Day – 30 Days 7
7
28.0 0.022
357.8
4.
The sample mean is 54.8 years. n1 = 19 , n2 = 19 , and the number of runs is G = 18 . The critical values
are 13 and 27 (From Table A-10.) Fail to reject the null hypothesis of randomness. There is not sufficient
evidence to warrant rejection of the claim that the sequence of ages is random relative to values above and
below the mean. The results do not suggest that there is an upward trend or a downward trend.
5.
rs = 0.714 . Critical values: rs = ±0.738 (From Table A-9.) Fail to reject the null hypothesis of ρs = 0 .
There is not sufficient evidence to support the claim that there is a correlation between the student ranks
and the magazine ranks. When ranking colleges, students and the magazine do not appear to agree.
MINITAB
Pearson correlation of Student Ranks and USNEWS Ranks = 0.714
6.
7.
(13 + 0.5)− 322
= −0.88 is not in the critical region bounded by z = ±1.96 . There
32 2
is not sufficient evidence to warrant rejection of the claim that the population of differences has a median
of zero. Based on the sample data, it appears that the predictions are reasonably accurate, because there
does not appear to be a difference between the actual high temperatures and the predicted high
temperatures.
The test statistic of z =
230.5 −
32 (32 + 1)
4
= −0.62 .
32 (32 + 1)(2 ⋅ 32 + 1)
24
Critical values: z = ±1.96 . (Tech: P-value = 0.531.) There is not sufficient evidence to warrant rejection
of the claim that the population of differences has a median of zero. Based on the sample data, it appears
that the predictions are reasonably accurate, because there does not appear to be a difference between the
actual high temperatures and the predicted high temperatures.
Convert T = 230.5 to the test statistic z =
Copyright © 2014 Pearson Education, Inc.
232 Chapter 13: Nonparametric Statistics
7.
(continued)
MINITAB
Test of median = 0.000000 versus median not = 0.000000
N for
Wilcoxon
Estimated
N
Test
Statistic
P
Median
TEMP
35
32
230.5
0.537
-0.5000
8.
Test statistic: H =
⎛ 912 112.52 174.52 ⎞⎟
12
⎜⎜
⎟ − 3( 27 + 1) = 6.6305 .
+
+
27 ( 27 + 1) ⎜⎝ 9
9
9 ⎠⎟⎟
Critical value: χ 2 = 5.991 . (Tech: P-value = 0.0363.) Reject the null hypothesis of equal medians.
Interbreeding of cultures is suggested by the data.
9.
R1 = 60 , R2 = 111 , μR =
9 (9 + 9 + 1)
2
9 ⋅ 9 (9 + 9 + 1)
= 85.5 , σ R =
12
= 11.3248 , test statistic:
60 − 85.5
= −2.25 . Critical values: z = ±1.96 . (Tech: P-value = 0.0243.) Reject the null hypothesis
11.3248
that the populations have the same median. Skull breadths from 4000 b.c. appear to have a different median
than those from a.d. 150.
z=
10. rs = 0.473 . Critical values: rs = ±0.587 . Fail to reject the null hypothesis of ρs = 0 . There is not
sufficient evidence to support the claim that there is a correlation between weights of plastic and weights of
food.
Pearson correlation of Rank Plastic and Rank Rood = 0.473
Cumulative Review Exercises
1.
x = 14.6 hours, median = 15.0 hours, s = 1.7 hours, s 2 = 2.9 hour2, range = 6.0 hours
2.
a.
b.
c.
d.
3.
Convenience sample
Because the sample is from one class of statistics students, it is not likely to be representative of the
population of all fulltime college students.
Discrete
Ratio
The data meet the requirement of being from a normal distribution. H0: μ = 14 hours . H1: μ > 14 hours .
Test statistic: t =
14.55 −14
= 1.446 . Critical value: t = 1.729 (assuming a 0.05 significance level). P1.701/ 20
value > 0.05 (Tech: 0.0822). Fail to reject H0. There is not sufficient evidence to support the claim that the
mean is greater than 14 hours.
0.99
0.95
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
0.01
10
11
12
13
14
15
16
17
Copyright © 2014 Pearson Education, Inc.
18
19
Chapter 13: Nonparametric Statistics
233
4.
The test statistic of x = 5 is not less than or equal to the critical value of 4 (from Table A-7.) There is not
sufficient evidence to support the claim that the sample is from a population with a median greater than 14
hours.
5.
13.8 hours < μ < 15.3 hours . We have 95% confidence that the limits of 13.8 hours and 15.3 hours contain
the true value of the population mean.
14.55 − 2.101⋅
6.
1.701
20
< μ < 14.55 + 2.101⋅
1.701
20
r = 0.205 . Critical values: r = ±0.811 . P-value = 0.697. There is not sufficient evidence to support the
claim of a linear correlation between price and quality score. It appears that you don’t get better quality by
paying more.
MINITAB
Pearson correlation of Price and Quality = 0.205
P-Value = 0.697
7.
rs = −0.543 . Critical values: rs = ±0.886 . Fail to reject the null hypothesis of ρs = 0 . There is not
sufficient evidence to support the claim that there is a correlation between price and rank.
MINITAB
Pearson correlation of Rank and Price Rank = -0.543
P-Value = 0.266
8.
0.276 < p < 0.343 . Because the value of 0.25 is not included in the range of values in the confidence
interval, the result suggests that the percentage of all such telephones that are not functioning is different
from 25%.
229
−1.96
740
[ zα / 2 ] (0.25) 1.6452 (0.25)
229
511
( 740
)( 740
)
740
< p<
229
+ 1.96
740
229
511
( 740
)( 740
)
740
2
9.
n=
E2
=
0.022
= 1692 (Tech: 1691)
10. There must be an error, because the rates of 13.7% and 10.6% are not possible with samples of size 100.
Copyright © 2014 Pearson Education, Inc.
Chapter 14: Statistical Process Control 235
Chapter 14: Statistical Process Control
Section 14-2
1.
No. If we know that the manufacture of quarters is within statistical control, we know that the three out-ofcontrol criteria are not violated, but we know nothing about whether the specification of 5.670 g is being
met. It is possible to be within statistical control by manufacturing quarters with weights that are very far
from the desired target of 5.670 g.
2.
An x control chart is a plot of sample means and it includes a centerline as well as a line representing an
upper control limit and a line representing a lower control limit. An R control chart is a plot of sample
ranges and it also includes a centerline as well as a line representing an upper control limit and a line
representing a lower control limit. x denotes the mean of all of the 20 sample means, R denotes the mean
of the 20 ranges, UCL denotes the value used to locate the upper control limit in a control chart, and LCL
denotes the value used to locate the lower control limit in a control chart.
3.
To use an x chart without an R chart is to ignore variation, and amounts of variation that are too large will
result in too many defective goods or services, even though the mean might appear to be acceptable. To use
an R chart without an x chart is to ignore the central tendency, so the goods or services might not vary
much, but the process could be drifting so that daily process data do not vary much, but the daily means are
steadily increasing or decreasing.
4.
The range and mean are both out of statistical control. The control charts show that the elevations are
decreasing substantially, so Lake Mead is becoming shallower. The variation is becoming more stable, so
the elevations are not varying as much as they did earlier.
5.
x = 267.11 lb, R = 54.96 lb, n = 7.
For the R chart: LCL = D3 R = 0.076 ⋅ 54.96 = 4.18 lb and UCL = D4 R = 1.924 ⋅ 54.96 = 105.74 lb .
For the x chart: LCL = x − A2 R = 267.11− 0.419 ⋅ 54.96 = 244.08 lb and
UCL = x + A2 R = 267.11 + 0.419 ⋅ 54.96 = 290.14 lb .
6.
The run chart does not reveal any patterns suggesting problems that need correcting.
Copyright © 2014 Pearson Education, Inc.
236 Chapter 14: Statistical Process Control
7.
The R chart does not violate any of the out-of-control criteria, so the variation of the process appears to be
within statistical control
8.
The x chart does not violate any of the out-of-control criteria, so the mean of the process appears to be
within statistical control.
9.
x = 14.250°C, R = 0.414°C, n = 10.
For the R chart: LCL = D3 R = 0.223 ⋅ 0.414 = 0.092D C and UCL = D4 R = 1.777 ⋅ 0.414 = 0.736D C .
For the x chart: LCL = x − A2 R = 14.250 − 0.308 ⋅ 0.414 = 14.122D C and
UCL = x + A2 R = 14.250 + 0.308 ⋅ 0.414 = 14.377D C .
10. The R chart does not violate any of the out-of-control criteria, so the variation of the process appears to be
within statistical control.
Copyright © 2014 Pearson Education, Inc.
Chapter 14: Statistical Process Control 237
11. Because there is a pattern of an upward trend and there are points lying beyond the control limits, the x
chart shows that the process is out of statistical control.
12. The run chart reveals a very clear pattern of an upward trend, so this graph provides evidence in favor of
the theory that we are undergoing global warming.
13. s = 0.0823 g, n = 5. The R chart and the s chart are very similar in their pattern.
LCL = B3 s = 0 ⋅ 0.0823 = 0 g and UCL = B4 s = 2.089 ⋅ 0.0823 = 0.1719 g .
Copyright © 2014 Pearson Education, Inc.
238 Chapter 14: Statistical Process Control
14. x = 5.6955 g, s = 0.0823 g, n = 5. The two x charts are very similar.
LCL = x − A3 s = 5.6955 −1.427 ⋅ 0.0823 = 5.578 g and
UCL = x + A3 s = 5.6955 + 1.427 ⋅ 0.0823 = 5.813 g .
Section 14-3
1.
No, the process does not appear to be within statistical control. There is a downward trend, there are at least
8 consecutive points all lying above the centerline, and there are at least 8 consecutive points all lying
below the centerline. Because the proportions of defects are decreasing, the manufacturing process is not
deteriorating; it is improving.
2.
p is the pooled estimate of the proportion of defective items. It is obtained by finding the total number of
defects in all samples combined and dividing that total by the total number of items sampled.
3.
LCL denotes the lower control limit. Because the value of –0.000025 is negative and the actual proportion
of defects cannot be less than 0, we should replace that value by 0.
4.
No. It is very possible that the proportions of defects in repeated samplings behave in a way that makes the
process appear to be within statistical control, but the actual proportions of defects could be very high (such
as 90%) so that almost all of the dimes fail to meet the manufacturing specifications. Upper and lower
control limits of a control chart for the proportion of defects are based on the actual behavior of the process,
not the desired behavior.
5.
The process appears to be within statistical control.
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Chapter 14: Statistical Process Control 239
6.
p = 0.282, LCL = p − 3
(0.282)(0.718)
pq
= 0.282 − 3
= 0.1470
100
n
(0.282)(0.718)
pq
= 0.282 + 3
= 0.4170
100
n
The process appears to be within statistical control. The control chart is the same as the control chart from
Exercise 5, except for the values used to locate the centerline and control limits. In this exercise, the
proportions of defects are very high. Even though the process is within statistical control, this
manufacturing process is yielding far too many defective dimes, so corrective action should be taken to
lower the defect rate.
LCL = p + 3
7.
p = 0.01407, LCL = p − 3
UCL = p + 3
(0.01407)(0.98593)
pq
= 0.01407 − 3
= 0.012953
n
100, 000
(0.01407)(0.98593)
pq
= 0.01407 + 3
= 0.015187
n
100, 000
Because there appears to be a pattern of a downward shift and there are at least 8 consecutive points all
lying above the centerline, the process is not within statistical control.
8.
p = 0.005176, LCL = p − 3
UCL = p + 3
(0.005176)(0.994824)
pq
= 0.005176 − 3
= 0.004495
n
100,000
(0.005176)(0.994824)
pq
= 0.005176 + 3
= 0.005857
n
100, 000
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240 Chapter 14: Statistical Process Control
8.
(continued)
There is a pattern of a downward trend and there are at least 8 consecutive points all below the centerline,
so the process is not within statistical control. Because there is a downward trend in the rate of violent
crimes, we are becoming safer.
9.
p = 0.55231, LCL = p − 3
UCL = p + 3
(0.55231)(0.44769)
pq
= 0.55231− 3
= 0.50513
n
1000
(0.55231)(0.44769)
pq
= 0.55231 + 3
= 0.0.59948
n
1000
The process is out of control because there are points lying beyond the control limits and there are at least 8
points all lying below the centerline. The percentage of voters started to increase in recent years, and it
should be much higher than any of the rates shown.
10. p = 0.65361, LCL = p − 3
UCL = p + 3
(0.65361)(0.34639)
pq
= 0.65361− 3
= 0.60847
n
1000
(0.65361)(0.34639)
pq
= 0.65361 + 3
= 0.69875
n
1000
The process appears to be within statistical control. Ideally, there would be an upward trend due to
increasing rates of college enrollments among high school graduates.
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Chapter 14: Statistical Process Control 241
11. p = 0.0268, LCL = p − 3
UCL = p + 3
(0.0268)(0.9732)
pq
= 0.0268 − 3
= 0.00513
n
500
(0.0268)(0.9732)
pq
= 0.0268 + 3
= 0.04847
n
500
There is a pattern of a downward trend and there are at least 8 consecutive points all below the centerline,
so the process does not appear to be within statistical control. Because the rate of defects is decreasing, the
process is actually improving and we should investigate the cause of that improvement so that it can be
continued.
12. p = 0.059335, LCL = p − 3
UCL = p + 3
(0.059335)(0.940665)
pq
= 0.059335 − 3
= 0.009218
n
200
(0.059335)(0.940665)
pq
= 0.059335 + 3
= 0.10945
n
200
There appears to be a pattern of increasing variation, so the process is not within statistical control. The
cause of the increasing variation should be identified and corrected.
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242 Chapter 14: Statistical Process Control
13. np = 10, 000 ⋅ 0.00126 = 12.6 , LCL = np − 3 npq = 12.6 − 3 10, 000 (0.00126)(0.99874) = 1.9578
UCL = np + 3 npq = 12.6 + 3 10, 000 (0.00126)(0.99874) = 23.2422
Except for the vertical scale, the control chart is identical to the one obtained for Example 1.
Chapter Quick Quiz
1.
Process data are data arranged according to some time sequence. They are measurements of a characteristic
of goods or services that result from some combination of equipment, people, materials, methods, and
conditions.
2.
Random variation is due to chance, but assignable variation results from causes that can be identified, such
as defective machinery or untrained employees.
3.
There is a pattern, trend, or cycle that is obviously not random. There is a point lying outside of the region
between the upper and lower control limits. There are at least 8 consecutive points all above or all below
the centerline.
4.
An R chart uses ranges to monitor variation, but an x chart uses sample means to monitor the center
(mean) of a process.
5.
No. The R chart has at least 8 consecutive points all lying below the centerline and there are points lying
beyond the upper control limit. Also, there is a pattern showing that the ranges have jumped in value for the
most recent samples.
6.
R = 52.8 ft. In general, a value of R is found by first finding the range for the values within each
individual subgroup; the mean of those ranges is the value of R .
7.
No. The x chart has a point lying below the lower control limit.
8.
x = 3.95 ft. In general, a value of x is found by first finding the mean of the values within each individual
subgroup; the mean of those subgroup means is the value of x .
9.
A p chart is a control chart of the proportions of some attribute, such as defective items.
10. Because there is a downward trend, the process is not within statistical control, but the rate of defects is
decreasing, so we should investigate and identify the cause of that trend so that it can be continued.
Review Exercises
1.
x = 2781.71 kWh, R = 1729.38 kWh, n = 6.
For the R chart: LCL = D3 R = 0 ⋅1729.38 = 0 kWh and
UCL = D4 R = 2.004 ⋅1729.38 = 3465.678 kWh .
For the x chart: LCL = x − A2 R = 2781.71− 0.483 ⋅1729.38 = 1946.419 kWh and
UCL = x − A2 R = 2781.71− 0.483 ⋅1729.38 = 3617.001 kWh .
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Chapter 14: Statistical Process Control 243
2.
R = 1729.4 kWh, n = 6. The process variation is within statistical control.
LCL = D3 R = 0 ⋅1729.4 = 0.0000 kWh and UCL = D4 R = 2.004 ⋅1729.4 = 3465.7176 kWh .
3.
x = 2781.71 kWh, R = 1729.4 kWh, n = 6. The process mean is within statistical control.
LCL = x − A2 R = 2781.71− 0.483 ⋅1729.4 = 1946.4098 kWh and
UCL = x + A2 R = 2781.71 + 0.483 ⋅1729.4 = 3617.0102 kWh .
4.
There does not appear to be a pattern suggesting that the process is not within statistical control. There is 1
point that appears to be exceptionally low. (The author’s power company made an error in recording and
reporting the energy consumption for that time period.)
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244 Chapter 14: Statistical Process Control
5.
p = 0.056, LCL = p − 3
LCL = p + 3
(0.056)(0.944)
pq
= 0.056 − 3
= −0.01298; use 0
n
100
(0.056)(0.944)
pq
= 0.056 + 3
= 0.12498
n
100
Because there are 8 consecutive points above the centerline and there is an upward trend, the process does
not appear to be within statistical control.
Cumulative Review Exercises
1.
0.519 < p < 0.581 . Because all of the values in the confidence interval estimate of the population
proportion are greater than 0.5, it does appear that the majority of adults believe that it is not appropriate to
wear shorts at work.
MINITAB
Sample X N Sample p
95% CI
1
550 1000 0.550000 (0.518557, 0.581148)
2.
3.
a.
1− 0.55 = 0.45
b.
(0.55) = 0.0503
c.
1− (0.55) = 0.950
5
5
r = 0.820. Critical values: r = ±0.602 . P-value = 0.00202. There is sufficient evidence to support the
claim that there is a linear correlation between yields from regular seed and kiln-dried seed. The purpose of
the experiment was to determine whether there is a difference in yield from regular seed and kiln-dried seed
(or whether kiln-dried seed produces a higher yield), but results from a test of correlation do not provide us
with the information we need to address that issue.
MINITAB
Pearson correlation of Regular and Kiln-dried = 0.820
P-Value = 0.002
4.
H0: μd = 0 . H1: μd < 0 . Test statistic: t = –1.532. Critical value: t = –1.812 (assuming a 0.05 significance
level). P-value < 0.05 (Tech: 0.0783). Fail to reject H0. There is not sufficient evidence to support the claim
that kiln-dried seed is better in the sense that it produces a higher mean yield than regular seed. (The sign
test can be used to arrive at the same conclusion; the test statistic is x = 3 and the critical value is 1. Also,
the Wilcoxon signed-ranks test can be used; the test statistic is T = 13.5 and the critical value is 8.)
Minitab
Paired T for Regular - Kiln-dried
95% upper bound for mean difference: 0.200
T-Test of mean difference = 0 (vs < 0): T-Value = -1.53
P-Value = 0.078
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Chapter 14: Statistical Process Control 245
5.
6.
For the sample of yields from regular seed, x = 20.0 and for the sample of yields from kiln-dried seed, x
= 21.0, so there does not appear to be a significant difference. For the sample of yields from regular seed, s
= 3.4 and for the sample of yields from kiln-dried seed, s = 4.1, so there does not appear to be a significant
difference.
p = 0.122, LCL = p − 3
LCL = p + 3
(0.122)(0.878)
pq
= 0.122 − 3
= −0.01686; use 0
50
n
(0.122)(0.878)
pq
= 0.122 + 3
= 0.26086
n
50
There appears to be a pattern of an upward trend, so the process is not within statistical control.
7.
8.
a.
b.
9.
17 −15.2
= 0.72; P ( z > 0.72) = 23.58%. With 23.58% of males with head breadths greater than
2.5
17 cm, too many males would be excluded.
5th percentile: x = μ + z ⋅ σ = 15.2 −1.645 ⋅ 2.5 = 11.08 cm
95th percentile: x = μ + z ⋅ σ = 15.2 + 1.645 ⋅ 2.5 = 19.3 cm
z=
With a voluntary response sample, the subjects decide themselves whether to be included. With a simple
random sample, subjects are selected through some random process in such a way that all samples of the
same size have the same chance of being selected. A simple random sample is generally better for use with
statistical methods.
10. Sampling method (part c)
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