Solution Manual for Interactive Statistics, 3rd Edition
Solution Manual for Interactive Statistics, 3rd Edition provides expert-verified solutions to help you study smarter.
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Solutions to All Exercises 275
Section III. Solutions to all
Chapter Exercises
Section III. Solutions to all
Chapter Exercises
276 Chapter 1
Interactive Statistics 3rd Edition: Chapter 1 Full
Solutions
1.1
In hypothesis testing, the purpose is to determine whether there is sufficient evidence with which to reject the null
hypothesis (H0), which generally reflects the prevailing viewpoint. The alternative hypothesis (H1) is often what
someone is hopeful that the data will support.
1.2
(a) True.
(b) True.
(c) False.
(d) False.
1.3
H0: The 5-year survival rate for all those using the vaccine is equal to 10%.
H1: The 5-year survival rate for all those using the vaccine is greater than 10%.
1.4
H0: Using the new technique, the percentage of all whales leaving the area is 40%
H1: Using the new technique, the percentage of all whales leaving the area is more than 40%
1.5
(a) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the gun is not loaded when it is
loaded. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the gun is loaded
when it is not. A Type I error may be more serious as one might accidentally shoot a loaded gun.
(b) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the dog does not bite when it does
bite. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the dog bites when it
does not. A Type I error may be more serious as we might approach a dog that could bite.
(c) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the mall is closed when it is open.
Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the mall is open when it is
closed. A Type II error may be more serious as we might waste the time and gas to drive to the mall expecting
it open when it is closed.
(d) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the watch is waterproof when it is
not. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the watch is not
waterproof when it is waterproof. A Type I error may be more serious as we might ruin our watch if it gets wet.
1.6
(a) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the electricity is not turned on
when it is. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the electricity is
turned on when it is not. A Type I error may be more serious as one might accidentally be electrocuted.
(b) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the brakes are operational when
they are not. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the brakes are
not operational when they are operational. A Type I error may be more serious as it may result in an accident.
(c) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the snake is not poisonous when it
is poisonous. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the snake is
poisonous when it is not poisonous. A Type I error may be more serious as it may result in getting bit by a
poisonous snake.
(d) Since a Type I error is rejecting H0 when H0 is true, we would be concluding it is not safe to cross the street
when it is safe. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking it is safe to
cross the street when it is not safe. A Type II error may be more serious as it may result in an accident.
Interactive Statistics 3rd Edition: Chapter 1 Full
Solutions
1.1
In hypothesis testing, the purpose is to determine whether there is sufficient evidence with which to reject the null
hypothesis (H0), which generally reflects the prevailing viewpoint. The alternative hypothesis (H1) is often what
someone is hopeful that the data will support.
1.2
(a) True.
(b) True.
(c) False.
(d) False.
1.3
H0: The 5-year survival rate for all those using the vaccine is equal to 10%.
H1: The 5-year survival rate for all those using the vaccine is greater than 10%.
1.4
H0: Using the new technique, the percentage of all whales leaving the area is 40%
H1: Using the new technique, the percentage of all whales leaving the area is more than 40%
1.5
(a) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the gun is not loaded when it is
loaded. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the gun is loaded
when it is not. A Type I error may be more serious as one might accidentally shoot a loaded gun.
(b) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the dog does not bite when it does
bite. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the dog bites when it
does not. A Type I error may be more serious as we might approach a dog that could bite.
(c) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the mall is closed when it is open.
Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the mall is open when it is
closed. A Type II error may be more serious as we might waste the time and gas to drive to the mall expecting
it open when it is closed.
(d) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the watch is waterproof when it is
not. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the watch is not
waterproof when it is waterproof. A Type I error may be more serious as we might ruin our watch if it gets wet.
1.6
(a) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the electricity is not turned on
when it is. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the electricity is
turned on when it is not. A Type I error may be more serious as one might accidentally be electrocuted.
(b) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the brakes are operational when
they are not. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the brakes are
not operational when they are operational. A Type I error may be more serious as it may result in an accident.
(c) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the snake is not poisonous when it
is poisonous. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the snake is
poisonous when it is not poisonous. A Type I error may be more serious as it may result in getting bit by a
poisonous snake.
(d) Since a Type I error is rejecting H0 when H0 is true, we would be concluding it is not safe to cross the street
when it is safe. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking it is safe to
cross the street when it is not safe. A Type II error may be more serious as it may result in an accident.
276 Chapter 1
Interactive Statistics 3rd Edition: Chapter 1 Full
Solutions
1.1
In hypothesis testing, the purpose is to determine whether there is sufficient evidence with which to reject the null
hypothesis (H0), which generally reflects the prevailing viewpoint. The alternative hypothesis (H1) is often what
someone is hopeful that the data will support.
1.2
(a) True.
(b) True.
(c) False.
(d) False.
1.3
H0: The 5-year survival rate for all those using the vaccine is equal to 10%.
H1: The 5-year survival rate for all those using the vaccine is greater than 10%.
1.4
H0: Using the new technique, the percentage of all whales leaving the area is 40%
H1: Using the new technique, the percentage of all whales leaving the area is more than 40%
1.5
(a) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the gun is not loaded when it is
loaded. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the gun is loaded
when it is not. A Type I error may be more serious as one might accidentally shoot a loaded gun.
(b) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the dog does not bite when it does
bite. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the dog bites when it
does not. A Type I error may be more serious as we might approach a dog that could bite.
(c) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the mall is closed when it is open.
Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the mall is open when it is
closed. A Type II error may be more serious as we might waste the time and gas to drive to the mall expecting
it open when it is closed.
(d) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the watch is waterproof when it is
not. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the watch is not
waterproof when it is waterproof. A Type I error may be more serious as we might ruin our watch if it gets wet.
1.6
(a) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the electricity is not turned on
when it is. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the electricity is
turned on when it is not. A Type I error may be more serious as one might accidentally be electrocuted.
(b) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the brakes are operational when
they are not. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the brakes are
not operational when they are operational. A Type I error may be more serious as it may result in an accident.
(c) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the snake is not poisonous when it
is poisonous. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the snake is
poisonous when it is not poisonous. A Type I error may be more serious as it may result in getting bit by a
poisonous snake.
(d) Since a Type I error is rejecting H0 when H0 is true, we would be concluding it is not safe to cross the street
when it is safe. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking it is safe to
cross the street when it is not safe. A Type II error may be more serious as it may result in an accident.
Interactive Statistics 3rd Edition: Chapter 1 Full
Solutions
1.1
In hypothesis testing, the purpose is to determine whether there is sufficient evidence with which to reject the null
hypothesis (H0), which generally reflects the prevailing viewpoint. The alternative hypothesis (H1) is often what
someone is hopeful that the data will support.
1.2
(a) True.
(b) True.
(c) False.
(d) False.
1.3
H0: The 5-year survival rate for all those using the vaccine is equal to 10%.
H1: The 5-year survival rate for all those using the vaccine is greater than 10%.
1.4
H0: Using the new technique, the percentage of all whales leaving the area is 40%
H1: Using the new technique, the percentage of all whales leaving the area is more than 40%
1.5
(a) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the gun is not loaded when it is
loaded. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the gun is loaded
when it is not. A Type I error may be more serious as one might accidentally shoot a loaded gun.
(b) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the dog does not bite when it does
bite. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the dog bites when it
does not. A Type I error may be more serious as we might approach a dog that could bite.
(c) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the mall is closed when it is open.
Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the mall is open when it is
closed. A Type II error may be more serious as we might waste the time and gas to drive to the mall expecting
it open when it is closed.
(d) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the watch is waterproof when it is
not. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the watch is not
waterproof when it is waterproof. A Type I error may be more serious as we might ruin our watch if it gets wet.
1.6
(a) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the electricity is not turned on
when it is. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the electricity is
turned on when it is not. A Type I error may be more serious as one might accidentally be electrocuted.
(b) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the brakes are operational when
they are not. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the brakes are
not operational when they are operational. A Type I error may be more serious as it may result in an accident.
(c) Since a Type I error is rejecting H0 when H0 is true, we would be concluding the snake is not poisonous when it
is poisonous. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking the snake is
poisonous when it is not poisonous. A Type I error may be more serious as it may result in getting bit by a
poisonous snake.
(d) Since a Type I error is rejecting H0 when H0 is true, we would be concluding it is not safe to cross the street
when it is safe. Since a Type II error is failing to reject H0 when H1 is true, we would be thinking it is safe to
cross the street when it is not safe. A Type II error may be more serious as it may result in an accident.
Solutions to All Exercises 277
1.7
H0: The average tomato yield for Brand A fertilizer is the same as the average tomato yield for the more expensive
Brand B fertilizer.
H1: The average tomato yield for the more expensive Brand B fertilizer is greater than the average tomato yield for
the Brand A fertilizer.
Type I Error: Spend more money on the Brand B fertilizer when it really is not better than the Brand A fertilizer
regarding the average tomato yield. Type II Error: Continue to use the Brand A fertilizer when the Brand B
fertilizer results in a higher tomato yield on average.
1.8
(a) Type I error: to classify a person as a drug user when he/she is not. Type II error: to classify a person as a non-
drug user when he/she actually is.
(b) Since the test classifies the person as a drug user (i.e. rejects H0) 4% of the time when the person is not a drug
user (i.e. H0 is true), the chance of a Type I error is 4%.
1.9
A Type I error is rejecting the null hypothesis when it is true. So the owner would conclude the patrons are older
and the owner would spend the time and money to remodel, when the crowd is actually not older. The owner would
have spent money unnecessarily and the remodeling may not appeal to some of the patrons, but in general, it is not a
serious error.
1.10
Based on the article "U.S. Health Improves but Rural Areas Lag," from CNN, September 10, 2001. Summary of the
article appears below. Contains many possible hypotheses.
H0: People in Rural areas of the US are as healthy as their urban counterparts.
H1: People in Rural areas of the US are less healthy than their urban counterparts.
Type I error: conclude that rural people are less healthy than urban people when in fact they are not.
Type II error: conclude that rural people are as healthy as urban people when in fact it isn't the case.
Summary of article: " Americans overall are healthier today than they were 25 years ago. A new government report
offers some reasons: longer life expectancy, better infant survival, fewer smokers, less hypertension and lower
cholesterol levels. But in small-town America, the news on health is far from good, said an annual report released
Monday by the Centers for Disease Control and Prevention. Rural residents tend to smoke more, lose more teeth as
they age and die sooner than suburban and many big-city counterparts, the government snapshot of the country's
health shows. For instance: --10.6 percent of the wealthiest residents in rural areas and 10 percent of urban residents
lacked health insurance in 1997 and 1998, compared with about 6.6 percent of suburbanites. --37.6 percent of rural
residents over 65 had edentulism, a total loss of teeth, in 1997 and 1998, compared with about 25.7 percent in the
suburbs and 26.8 percent in cities. --18.9 percent of children age 12 to 17 in the most rural areas were regular
smokers in 1999, compared with 11 percent in urban areas and 15.9 percent in the suburbs. Rural adults also smoked
at higher rates than urban or suburban adults. --46.5 of men and women in the most rural areas did not exercise, play
sports or pursue active hobbies in 1997 and 1998, compared with 40.9 percent of urban dwellers and 31.1 percent of
suburbanites who were not fitness-minded. The youth death rate from all causes was higher in rural areas from 1996
to 1998, as was the adult death rate.
1.11
(a) The null hypothesis could not be rejected.
(b) No, a complaint was not registered.
(c) Yes, a Type II error may have been made. The cans are thought to contain the stated sodium content when
actually they contain higher amounts of sodium on average.
1.12
(a) H0: Septaphine is not better than Cephaline for reducing blood pressure.
H1: Septaphine is better than Cephaline for reducing blood pressure.
(b) (i) The null hypothesis was rejected and the alternative hypothesis was supported.
(ii) A Type I error could have been made, namely, concluding Septaphine is better when it really is not better at
reducing blood pressure.
1.7
H0: The average tomato yield for Brand A fertilizer is the same as the average tomato yield for the more expensive
Brand B fertilizer.
H1: The average tomato yield for the more expensive Brand B fertilizer is greater than the average tomato yield for
the Brand A fertilizer.
Type I Error: Spend more money on the Brand B fertilizer when it really is not better than the Brand A fertilizer
regarding the average tomato yield. Type II Error: Continue to use the Brand A fertilizer when the Brand B
fertilizer results in a higher tomato yield on average.
1.8
(a) Type I error: to classify a person as a drug user when he/she is not. Type II error: to classify a person as a non-
drug user when he/she actually is.
(b) Since the test classifies the person as a drug user (i.e. rejects H0) 4% of the time when the person is not a drug
user (i.e. H0 is true), the chance of a Type I error is 4%.
1.9
A Type I error is rejecting the null hypothesis when it is true. So the owner would conclude the patrons are older
and the owner would spend the time and money to remodel, when the crowd is actually not older. The owner would
have spent money unnecessarily and the remodeling may not appeal to some of the patrons, but in general, it is not a
serious error.
1.10
Based on the article "U.S. Health Improves but Rural Areas Lag," from CNN, September 10, 2001. Summary of the
article appears below. Contains many possible hypotheses.
H0: People in Rural areas of the US are as healthy as their urban counterparts.
H1: People in Rural areas of the US are less healthy than their urban counterparts.
Type I error: conclude that rural people are less healthy than urban people when in fact they are not.
Type II error: conclude that rural people are as healthy as urban people when in fact it isn't the case.
Summary of article: " Americans overall are healthier today than they were 25 years ago. A new government report
offers some reasons: longer life expectancy, better infant survival, fewer smokers, less hypertension and lower
cholesterol levels. But in small-town America, the news on health is far from good, said an annual report released
Monday by the Centers for Disease Control and Prevention. Rural residents tend to smoke more, lose more teeth as
they age and die sooner than suburban and many big-city counterparts, the government snapshot of the country's
health shows. For instance: --10.6 percent of the wealthiest residents in rural areas and 10 percent of urban residents
lacked health insurance in 1997 and 1998, compared with about 6.6 percent of suburbanites. --37.6 percent of rural
residents over 65 had edentulism, a total loss of teeth, in 1997 and 1998, compared with about 25.7 percent in the
suburbs and 26.8 percent in cities. --18.9 percent of children age 12 to 17 in the most rural areas were regular
smokers in 1999, compared with 11 percent in urban areas and 15.9 percent in the suburbs. Rural adults also smoked
at higher rates than urban or suburban adults. --46.5 of men and women in the most rural areas did not exercise, play
sports or pursue active hobbies in 1997 and 1998, compared with 40.9 percent of urban dwellers and 31.1 percent of
suburbanites who were not fitness-minded. The youth death rate from all causes was higher in rural areas from 1996
to 1998, as was the adult death rate.
1.11
(a) The null hypothesis could not be rejected.
(b) No, a complaint was not registered.
(c) Yes, a Type II error may have been made. The cans are thought to contain the stated sodium content when
actually they contain higher amounts of sodium on average.
1.12
(a) H0: Septaphine is not better than Cephaline for reducing blood pressure.
H1: Septaphine is better than Cephaline for reducing blood pressure.
(b) (i) The null hypothesis was rejected and the alternative hypothesis was supported.
(ii) A Type I error could have been made, namely, concluding Septaphine is better when it really is not better at
reducing blood pressure.
Loading page 4...
278 Chapter 1
1.13
If
is decreased then
will increase, so the possible value is 0.30.
1.14
If you do not reject the null hypothesis, you may be making either a Type II error or a correct decision, so the
answer is (e).
1.15
(a) The significance level is 2/30=0.067 and the level of is 20/30=0.667.
(b) Decision Rule #2: Reject H0 if the selected voucher is $2 or is $9. The significance level
is 6/30=0.20
and the level of is 12/30=0.40. Enlarging the rejection region resulted in increasing the level of
from 0.067
to 0.20 while decreasing the level of from 0.667 to 0.40.
1.16
(a)
= chance of a Type I error = chance of rejecting H0 when H0 is true
= chance of observing $1 or $10 from bag E
= 2/30 = 1/15 = 0.0667
(b) p-value = chance of observing $3 or more extreme under H0
= chance of observing $3 or $8 from bag E
= 12/30 = 6/15 = 0.40
(c) (i)
= chance of a Type II error = chance of failing to reject H0 when H1 is true
= chance of observing $2 - $9 from bag G
= 23/30 ~ 0.767
(ii)
= chance of a Type II error = chance of failing to reject H0 when H1 is true
= chance of observing $2 - $9 from bag H
= 23/30 ~ 0.767
1.17
No, we need a decision rule that states when we reject or fail to reject H0.
1.18
(a) H0: The proportion of newborns that are girls equals 0.50.
H1: The proportion of newborns that are girls does not equal 0.50.
(b) Two-sided.
1.19
(a) False:
+
does not need to equal 1. The value of
is calculated under H0 while the value of
is calculated
under H1.
(b) False: Type II error is the chance of failing to reject H0 when H1 is true.
(c) True.
(d) False: H0 is rejected if the sample shows evidence against it.
(e) False: The sample size does not influence the alternative hypothesis. The alternative hypothesis can be one-
sided no matter what the sample size.
1.20
The 0.05 indicates: (b) if H0 is true, the chance of falsely rejecting it is 0.05.
1.21
(a) H0: The shown box is Box A. H1: The shown box is Box B.
(b) The direction of extreme is one-sided to the left.
(c) Reject H0 if the selected token is $5 or less.
(d) The significance level
= 2/25 = 0.08 which is less than 0.10.
(e) The chance of a Type II error is
= 11/25 = 0.44.
(f) Our decision is to reject H0.
1.13
If
is decreased then
will increase, so the possible value is 0.30.
1.14
If you do not reject the null hypothesis, you may be making either a Type II error or a correct decision, so the
answer is (e).
1.15
(a) The significance level is 2/30=0.067 and the level of is 20/30=0.667.
(b) Decision Rule #2: Reject H0 if the selected voucher is $2 or is $9. The significance level
is 6/30=0.20
and the level of is 12/30=0.40. Enlarging the rejection region resulted in increasing the level of
from 0.067
to 0.20 while decreasing the level of from 0.667 to 0.40.
1.16
(a)
= chance of a Type I error = chance of rejecting H0 when H0 is true
= chance of observing $1 or $10 from bag E
= 2/30 = 1/15 = 0.0667
(b) p-value = chance of observing $3 or more extreme under H0
= chance of observing $3 or $8 from bag E
= 12/30 = 6/15 = 0.40
(c) (i)
= chance of a Type II error = chance of failing to reject H0 when H1 is true
= chance of observing $2 - $9 from bag G
= 23/30 ~ 0.767
(ii)
= chance of a Type II error = chance of failing to reject H0 when H1 is true
= chance of observing $2 - $9 from bag H
= 23/30 ~ 0.767
1.17
No, we need a decision rule that states when we reject or fail to reject H0.
1.18
(a) H0: The proportion of newborns that are girls equals 0.50.
H1: The proportion of newborns that are girls does not equal 0.50.
(b) Two-sided.
1.19
(a) False:
+
does not need to equal 1. The value of
is calculated under H0 while the value of
is calculated
under H1.
(b) False: Type II error is the chance of failing to reject H0 when H1 is true.
(c) True.
(d) False: H0 is rejected if the sample shows evidence against it.
(e) False: The sample size does not influence the alternative hypothesis. The alternative hypothesis can be one-
sided no matter what the sample size.
1.20
The 0.05 indicates: (b) if H0 is true, the chance of falsely rejecting it is 0.05.
1.21
(a) H0: The shown box is Box A. H1: The shown box is Box B.
(b) The direction of extreme is one-sided to the left.
(c) Reject H0 if the selected token is $5 or less.
(d) The significance level
= 2/25 = 0.08 which is less than 0.10.
(e) The chance of a Type II error is
= 11/25 = 0.44.
(f) Our decision is to reject H0.
Loading page 5...
Solutions to All Exercises 279
1.22
(a) The frequency plots are given below. The direction of extreme is two-sided.
Distribution in Bag A
X
X
X
X
X
X
X X X
X X X
X X X
X X X
X X X
X X X
X X X X X
X X X X X
X X X X X X X
X X X X X X X X X
2 4 6 8 10 12 14 16 18
Token Value (in $)
Distribution in Bag B
X X
X X
X X
X X X X
X X X X
X X X X X X
X X X X X X
X X X X X X
X X X X X X X X X
X X X X X X X X X
2 4 6 8 10 12 14 16 18
Token Value (in $)
(b) H0: The shown bag is Bag A. H1: The shown bag is Bag B.
(c) Answers may vary. One reasonable rule is to reject H0 is the selected token is $4 or $16.
(d) The significance level
= 6/50 = 0.12.
(e) The chance of a Type II error is
= 16/50 = 0.32.
(f) (i) Fail to reject H0.
(ii) A Type II error.
(g) (i) Reject H0.
(ii) A Type I error.
1.23
(a) False.
+
does not need to equal 1. The value of
is calculated under H0 while the value of
is calculated
under H1.
(b) False. A Type II error occurs if H1 is true and we fail to reject H0. Here we don’t know if H1 is true.
(c) True.
1.22
(a) The frequency plots are given below. The direction of extreme is two-sided.
Distribution in Bag A
X
X
X
X
X
X
X X X
X X X
X X X
X X X
X X X
X X X
X X X X X
X X X X X
X X X X X X X
X X X X X X X X X
2 4 6 8 10 12 14 16 18
Token Value (in $)
Distribution in Bag B
X X
X X
X X
X X X X
X X X X
X X X X X X
X X X X X X
X X X X X X
X X X X X X X X X
X X X X X X X X X
2 4 6 8 10 12 14 16 18
Token Value (in $)
(b) H0: The shown bag is Bag A. H1: The shown bag is Bag B.
(c) Answers may vary. One reasonable rule is to reject H0 is the selected token is $4 or $16.
(d) The significance level
= 6/50 = 0.12.
(e) The chance of a Type II error is
= 16/50 = 0.32.
(f) (i) Fail to reject H0.
(ii) A Type II error.
(g) (i) Reject H0.
(ii) A Type I error.
1.23
(a) False.
+
does not need to equal 1. The value of
is calculated under H0 while the value of
is calculated
under H1.
(b) False. A Type II error occurs if H1 is true and we fail to reject H0. Here we don’t know if H1 is true.
(c) True.
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280 Chapter 1
1.24
(a) H0: The shown jar is Jar A. H1: The shown jar is Jar B.
(b) No, the coins are of different sizes and weights. If we mixed up the contents of the shown jar and reached in to
pick one coin, the coins that we are more likely to pick might be a quarter as it has a larger size as compared to
say a dime. The larger coins may be more likely to be selected over the smaller coins. This will be referred to
later in Chapter 2 as a length-biased sampling method.
(c) The responses are currently listed as P, N, D, and Q, but there is an ordering to them with respect to their value.
So a better way to record the response would be using their worth of 1, 5, 10, or 25 cents. Correspondingly, a
better picture to depict the two models for the shown jar is given below. The direction of extreme would be to
the right, to the larger coin values.
Distribution in Jar A
X X
X X
X X
X X
X X
X X X
X X X
X X X X
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Coin Value (in cents)
Distribution in Jar B
X X
X X
X X
X X
X X
X X X
X X X
X X X X
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Coin Value (in cents)
1.25
(a) The frequency plots are provided below:
Bag X Bag Y
X X X
X X X
X X X X X
X X X X X
X X X X X
X X X X X
X X X X X X X
X X X X X X X
X X X X X X X X X X
Blue Brown Yellow Green Red Blue Brown Yellow Green Red
(b) No, the response being recorded is the color, which has no particular ordering for the outcomes. If the colors
had been listed in the order of Blue, Red, Brown, Green, and Yellow, then the apparent direction of extreme
would be one-sided to the left. So it is not appropriate to discuss a direction of extreme in this case.
1.24
(a) H0: The shown jar is Jar A. H1: The shown jar is Jar B.
(b) No, the coins are of different sizes and weights. If we mixed up the contents of the shown jar and reached in to
pick one coin, the coins that we are more likely to pick might be a quarter as it has a larger size as compared to
say a dime. The larger coins may be more likely to be selected over the smaller coins. This will be referred to
later in Chapter 2 as a length-biased sampling method.
(c) The responses are currently listed as P, N, D, and Q, but there is an ordering to them with respect to their value.
So a better way to record the response would be using their worth of 1, 5, 10, or 25 cents. Correspondingly, a
better picture to depict the two models for the shown jar is given below. The direction of extreme would be to
the right, to the larger coin values.
Distribution in Jar A
X X
X X
X X
X X
X X
X X X
X X X
X X X X
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Coin Value (in cents)
Distribution in Jar B
X X
X X
X X
X X
X X
X X X
X X X
X X X X
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Coin Value (in cents)
1.25
(a) The frequency plots are provided below:
Bag X Bag Y
X X X
X X X
X X X X X
X X X X X
X X X X X
X X X X X
X X X X X X X
X X X X X X X
X X X X X X X X X X
Blue Brown Yellow Green Red Blue Brown Yellow Green Red
(b) No, the response being recorded is the color, which has no particular ordering for the outcomes. If the colors
had been listed in the order of Blue, Red, Brown, Green, and Yellow, then the apparent direction of extreme
would be one-sided to the left. So it is not appropriate to discuss a direction of extreme in this case.
Loading page 7...
Solutions to All Exercises 281
1.26
The p-value is a number between 0 and 1 which measures how likely the observed result is, or a result that is even
more extreme (in the direction of H0), assuming the null hypothesis H0 is true.
1.27
The p-value should be small in order to reject the null hypothesis H0. A small p-value indicates that the observed
data or data even more extreme is very unlikely or unusual if the null hypothesis is true. In general, we reject H0 if
p-value is less than or equal to , the significance level.
1.28
(a) It is stated there was no significant difference in the caesarean delivery rates between the two groups, so the
data supported the null hypothesis H0.
(b) The p-value would have been "large". Since the null hypothesis was not rejected, the observed data were not
considered unlikely under the null hypothesis.
1.29
(a) Frequency plots for the two completing hypotheses.
(b)
= chance of a Type I error = chance of rejecting H0 when H0 is true
= chance of observing $1 from the winning bag
= 2/10 = 0.20
(c)
= chance of a Type II error = chance of failing to reject H0 when H1 is true
= chance of observing $10 or $100 from the losing bag
= 3/8 =0.375
(d) No, we did not actually observe a voucher.
1.30
(a) The direction of extreme is one-sided to the left, since the smaller values are more likely under H1 and less
likely under H0.
(b) The chance of a Type I error is
= 3/15 = 0.20 (found under the Machine A model for 2 or fewer flaws). The
chance of a Type II error is
= 6/15 = 0.40 (found under the Machine B model for more than 2 flaws).
(c) The p-value is the chance of getting the observed 4 flaws or something even more extreme (less than 4 flaws),
assuming the null hypothesis is true and the machine is Machine A. The p-value = 10/15 = 0.667.
(d) Since the data were not usually assuming the null hypothesis is true, that is, the p-value is larger than
the data
are not statistically significant at the level
1.26
The p-value is a number between 0 and 1 which measures how likely the observed result is, or a result that is even
more extreme (in the direction of H0), assuming the null hypothesis H0 is true.
1.27
The p-value should be small in order to reject the null hypothesis H0. A small p-value indicates that the observed
data or data even more extreme is very unlikely or unusual if the null hypothesis is true. In general, we reject H0 if
p-value is less than or equal to , the significance level.
1.28
(a) It is stated there was no significant difference in the caesarean delivery rates between the two groups, so the
data supported the null hypothesis H0.
(b) The p-value would have been "large". Since the null hypothesis was not rejected, the observed data were not
considered unlikely under the null hypothesis.
1.29
(a) Frequency plots for the two completing hypotheses.
(b)
= chance of a Type I error = chance of rejecting H0 when H0 is true
= chance of observing $1 from the winning bag
= 2/10 = 0.20
(c)
= chance of a Type II error = chance of failing to reject H0 when H1 is true
= chance of observing $10 or $100 from the losing bag
= 3/8 =0.375
(d) No, we did not actually observe a voucher.
1.30
(a) The direction of extreme is one-sided to the left, since the smaller values are more likely under H1 and less
likely under H0.
(b) The chance of a Type I error is
= 3/15 = 0.20 (found under the Machine A model for 2 or fewer flaws). The
chance of a Type II error is
= 6/15 = 0.40 (found under the Machine B model for more than 2 flaws).
(c) The p-value is the chance of getting the observed 4 flaws or something even more extreme (less than 4 flaws),
assuming the null hypothesis is true and the machine is Machine A. The p-value = 10/15 = 0.667.
(d) Since the data were not usually assuming the null hypothesis is true, that is, the p-value is larger than
the data
are not statistically significant at the level
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282 Chapter 1
1.31
(a) The direction of extreme is one-sided to the left (to the smaller values).
(b) The chance of a Type I error is
= 1/6 = 0.1667 (found under the Die A model for 1 or less). The chance of a
Type II error is
= 7/10 = 0.70 (found under the Die B model for 2 or more).
(c) The p-value is the chance of getting the observed value of 2 or less, assuming the die is Die A, so the p-value =
2/6 = 0.333. Since this p-value is greater than the significance level of 0.1667, we fail to reject H0 and
conclude that the selected die appears to be Die A.
1.32
(a) The direction of extreme is one-sided to the left. Note that the smaller
values show the most support for H1 and the least support for H0.
(b) See pictures at the right for the circling and labeling.
i.
= 3/35 = 0.0857.
ii.
= 14/35 = 0.40.
(c) p-value = 7/35 = 0.20
(d) The observed result is not statistically significant because the p-value
is more than
.
(e) A new decision rule is reject H0 if the observed voucher value is $6 or
more extreme, that is less than or equal to $6. Using a cut-off value of
$6 or more will yield a larger rejection region and thus a larger value
for
. The new
= 7/35 = 0.20 is larger than
= 3/35 = 0.0857.
1.33
(a) H0: The shown bag is Bag A. H1: The shown bag is Bag B.
(b) The direction of extreme is two-sided.
(c) (i) The p-value is10.040
4 = .
(ii) Yes, since the p-value is
.
(iii) No, since the p-value is >
.
(d) (i) The p-value is 1.
(ii) No, since the p-value is >
.
(iii) No, since the p-value is >
.
1.34
(a) Two possible p-values that are statistically significant at 0.01are 0.002 and 0.004. Note that any two values 0 ≤
p-value ≤ 0.01 will work.
(b) Two possible p-values that are statistically significant at 0.05, but not statistically significant at 0.01 are
0.03 and 0.04. Note that any two values 0.01 < p-value ≤ 0.05 will work.
(c) Two possible p-values that are not statistically significant at 0.10 are 0.20 and 0.30. Note that any two values
p-value > 0.10 but not >1 will work.
1.35
(a) All new model 100-watt light bulbs produced at Claude's plant.
(b) H0: The population of all new model 100-watt light bulbs (produced at Claude's plant) has an average lifetime
equal to 40 hours. H1: The population of all new model 100-watt light bulbs (produced at Claude’s plant) has
an average lifetime greater than 40 hours.
(c) 10%
(d) The p-value can be any value between 0 and 0.10.
(e) Yes, if the p-value is less than or equal to 0.10, then the p-value is also less than or equal to 0.15 so it is
significant at the 0.15 level. However, if we only know that the p-value is less than or equal to 0.10, we cannot
be sure whether the p-value is also less than or equal to 0.05. Without further information about the value of the
p-value we cannot determine if the data would also be significant at the 0.05 level.Box I
X
X X
X X
X X
X X X
X X X
X X X
X X X X
X X X X
X X X X X
X X X X X X
2 4 6 8 10 12 $
Box II
X
X X
X X
X X
X X X
X X X
X X X
X X X X
X X X X
X X X X X
X X X X X X
2 4 6 8 10 12 $
p-value
1.31
(a) The direction of extreme is one-sided to the left (to the smaller values).
(b) The chance of a Type I error is
= 1/6 = 0.1667 (found under the Die A model for 1 or less). The chance of a
Type II error is
= 7/10 = 0.70 (found under the Die B model for 2 or more).
(c) The p-value is the chance of getting the observed value of 2 or less, assuming the die is Die A, so the p-value =
2/6 = 0.333. Since this p-value is greater than the significance level of 0.1667, we fail to reject H0 and
conclude that the selected die appears to be Die A.
1.32
(a) The direction of extreme is one-sided to the left. Note that the smaller
values show the most support for H1 and the least support for H0.
(b) See pictures at the right for the circling and labeling.
i.
= 3/35 = 0.0857.
ii.
= 14/35 = 0.40.
(c) p-value = 7/35 = 0.20
(d) The observed result is not statistically significant because the p-value
is more than
.
(e) A new decision rule is reject H0 if the observed voucher value is $6 or
more extreme, that is less than or equal to $6. Using a cut-off value of
$6 or more will yield a larger rejection region and thus a larger value
for
. The new
= 7/35 = 0.20 is larger than
= 3/35 = 0.0857.
1.33
(a) H0: The shown bag is Bag A. H1: The shown bag is Bag B.
(b) The direction of extreme is two-sided.
(c) (i) The p-value is10.040
4 = .
(ii) Yes, since the p-value is
.
(iii) No, since the p-value is >
.
(d) (i) The p-value is 1.
(ii) No, since the p-value is >
.
(iii) No, since the p-value is >
.
1.34
(a) Two possible p-values that are statistically significant at 0.01are 0.002 and 0.004. Note that any two values 0 ≤
p-value ≤ 0.01 will work.
(b) Two possible p-values that are statistically significant at 0.05, but not statistically significant at 0.01 are
0.03 and 0.04. Note that any two values 0.01 < p-value ≤ 0.05 will work.
(c) Two possible p-values that are not statistically significant at 0.10 are 0.20 and 0.30. Note that any two values
p-value > 0.10 but not >1 will work.
1.35
(a) All new model 100-watt light bulbs produced at Claude's plant.
(b) H0: The population of all new model 100-watt light bulbs (produced at Claude's plant) has an average lifetime
equal to 40 hours. H1: The population of all new model 100-watt light bulbs (produced at Claude’s plant) has
an average lifetime greater than 40 hours.
(c) 10%
(d) The p-value can be any value between 0 and 0.10.
(e) Yes, if the p-value is less than or equal to 0.10, then the p-value is also less than or equal to 0.15 so it is
significant at the 0.15 level. However, if we only know that the p-value is less than or equal to 0.10, we cannot
be sure whether the p-value is also less than or equal to 0.05. Without further information about the value of the
p-value we cannot determine if the data would also be significant at the 0.05 level.Box I
X
X X
X X
X X
X X X
X X X
X X X
X X X X
X X X X
X X X X X
X X X X X X
2 4 6 8 10 12 $
Box II
X
X X
X X
X X
X X X
X X X
X X X
X X X X
X X X X
X X X X X
X X X X X X
2 4 6 8 10 12 $
p-value
Loading page 9...
Solutions to All Exercises 283
1.36
(a) H0: The shown bag is Bag A. H1: The shown bag is Bag B.
(b) The direction of extreme is one-sided to the right, to the larger values.
(c) The p-value is06.050
3 = .
(d) Yes, the $14 voucher is statistically significant at the 10% level as the p-value is less than or equal to 0.10.
(e) No, the $14 voucher is not statistically significant at the 5% level as the p-value is more than 0.05.
1.37
(a) H0
(b) p-value > 0.10
(c) We failed to reject H0, so we could have made a Type II error.
(d) One-sided to the right. We want to see if the numbers have increased.
1.38
(a) Since the data were statistically significant, the null hypothesis was rejected and the alternative hypothesis was
supported.
(b) The direction of extreme is two-sided.
(c) Answers will vary. Two possible values are 0.02 and 0.03. Note that any two values between 0 and 0.05 will
work.
1.39
(a) We rejected H0, so we could have made a Type I error.
(b) We decide that the average cost is higher than $350 while it is really not. Maybe you decide that the cost is too
high and decide to attend a different college, while in reality you could have attended this college after all.
(c) The p-value is 0.10.
(d) Yes.
1.40
(a) H0: The population of all goats born to mother goats that were trained to walk on a treadmill has a mean birth
weight of 1600 grams. H1: The population of all goats born to mother goats that were trained to walk on a
treadmill has a mean birth weight different from 1600 grams.
(b) The null hypothesis was supported at the 1% level.
(c) The p-value was more than 0.01 (but of course less than 1).
(d) Answers will vary. One possible p-value is 0.08.
(e) Yes, as now the p-value must be larger than 0.01 but less than or equal to 0.05. A value that will satisfy this
statement on significance is 0.04.
(f) A Type II error could have been made.
1.41
(a) For study A, a possible p-value is 0.001; for study B, a possible p-value is 0.11; and for study C, a possible p-
value is 0.03.
(b) Reject H0 if the p-value is small, so support for H0 is shown if the p-value is large, the largest p-value is for
Study B.
(c) We rejected H0, but it was true, so a Type I error was made.
(d) For Study A: one-sided to the right, Study B: two-sided, Study C: one-sided to the left.
1.42
(a) One-sided to the right.
(b) H0
(c) The p-value is 0.05.
(d) Type I error.
1.36
(a) H0: The shown bag is Bag A. H1: The shown bag is Bag B.
(b) The direction of extreme is one-sided to the right, to the larger values.
(c) The p-value is06.050
3 = .
(d) Yes, the $14 voucher is statistically significant at the 10% level as the p-value is less than or equal to 0.10.
(e) No, the $14 voucher is not statistically significant at the 5% level as the p-value is more than 0.05.
1.37
(a) H0
(b) p-value > 0.10
(c) We failed to reject H0, so we could have made a Type II error.
(d) One-sided to the right. We want to see if the numbers have increased.
1.38
(a) Since the data were statistically significant, the null hypothesis was rejected and the alternative hypothesis was
supported.
(b) The direction of extreme is two-sided.
(c) Answers will vary. Two possible values are 0.02 and 0.03. Note that any two values between 0 and 0.05 will
work.
1.39
(a) We rejected H0, so we could have made a Type I error.
(b) We decide that the average cost is higher than $350 while it is really not. Maybe you decide that the cost is too
high and decide to attend a different college, while in reality you could have attended this college after all.
(c) The p-value is 0.10.
(d) Yes.
1.40
(a) H0: The population of all goats born to mother goats that were trained to walk on a treadmill has a mean birth
weight of 1600 grams. H1: The population of all goats born to mother goats that were trained to walk on a
treadmill has a mean birth weight different from 1600 grams.
(b) The null hypothesis was supported at the 1% level.
(c) The p-value was more than 0.01 (but of course less than 1).
(d) Answers will vary. One possible p-value is 0.08.
(e) Yes, as now the p-value must be larger than 0.01 but less than or equal to 0.05. A value that will satisfy this
statement on significance is 0.04.
(f) A Type II error could have been made.
1.41
(a) For study A, a possible p-value is 0.001; for study B, a possible p-value is 0.11; and for study C, a possible p-
value is 0.03.
(b) Reject H0 if the p-value is small, so support for H0 is shown if the p-value is large, the largest p-value is for
Study B.
(c) We rejected H0, but it was true, so a Type I error was made.
(d) For Study A: one-sided to the right, Study B: two-sided, Study C: one-sided to the left.
1.42
(a) One-sided to the right.
(b) H0
(c) The p-value is 0.05.
(d) Type I error.
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284 Chapter 1
1.43
(a) See the chart below for the alternative hypotheses.
(b) See the chart below for the possible p-values.
(c) The results for Study C had the most support for the null hypothesis since the p-value was the largest.
(d) This would be called a Type I error.
1.44
(a) The alternative hypothesis could be stated in words as H1: The proportion of all convicted persons who have
been through the program and later reconvicted is lower than the national proportion of convicted persons who
are released and later reconvicted.
(b) The p-value must be greater than 0.05 but less than or equal to 0.10. Thus there is one possible value of 0.06.
1.45
(a) True.
(b) False.
1.46
(a) False. The p-value is the chance of getting the observed value or values more extreme assuming that H0 is true.
(b) True. The p-value of 0.04 was less than the significance level of 0.05, so the decision is to reject H0.
(c) True. When we reject H0, we say that the results are statistically significant.
(d) True. If the results are statistically significant at the 0.05 level (i.e. the p-value was less than 0.05), then the
results will also be significant at the 0.10 level (as the p-value will also be less than 0.10).
1.47
(c) to be statistically significant at the 5% level means the p-value is less than or equal to 0.05. However, we do not
know if the p-value is less than or equal to 0.01 or if it is between 0.01 and 0.05, so the answer is "sometimes yes"
(if the p-value is also 0.01) and "sometimes no" (if the p-value is > 0.01).
1.43
(a) See the chart below for the alternative hypotheses.
(b) See the chart below for the possible p-values.
(c) The results for Study C had the most support for the null hypothesis since the p-value was the largest.
(d) This would be called a Type I error.
1.44
(a) The alternative hypothesis could be stated in words as H1: The proportion of all convicted persons who have
been through the program and later reconvicted is lower than the national proportion of convicted persons who
are released and later reconvicted.
(b) The p-value must be greater than 0.05 but less than or equal to 0.10. Thus there is one possible value of 0.06.
1.45
(a) True.
(b) False.
1.46
(a) False. The p-value is the chance of getting the observed value or values more extreme assuming that H0 is true.
(b) True. The p-value of 0.04 was less than the significance level of 0.05, so the decision is to reject H0.
(c) True. When we reject H0, we say that the results are statistically significant.
(d) True. If the results are statistically significant at the 0.05 level (i.e. the p-value was less than 0.05), then the
results will also be significant at the 0.10 level (as the p-value will also be less than 0.10).
1.47
(c) to be statistically significant at the 5% level means the p-value is less than or equal to 0.05. However, we do not
know if the p-value is less than or equal to 0.01 or if it is between 0.01 and 0.05, so the answer is "sometimes yes"
(if the p-value is also 0.01) and "sometimes no" (if the p-value is > 0.01).
Loading page 11...
Solutions to All Exercises 285
1.48
(a) Type I error: The subject does not have ESP but you conclude he does.
(b) Type II error: The subject has ESP but you conclude he does not.
(c) Level of significance is the chance of rejecting Ho when Ho is true. If the subject does not have ESP (i.e., H0 is
true) then the chance that he correctly identifies the card (and we reject H0) is 1 in 52, 1/52 = 0.019.
(d) The chance of a Type II error is the chance we fail to reject H0 when H0 is true. If the subject has perfect ESP,
then the chance he correctly identifies the card is 1 or 100%. He will correctly identify it, so the chance of a
Type II error is 0.
(e) The p-value calculates the chance of getting the observed result (here 0 correct answers) or something more
extreme, in the direction of H0 (namely, 1 correct answer), under the assumption that H0 is true (the person does
not have ESP). If the subject does not have ESP, the chance of incorrectly identifying the card is 51/52 = 0.981
and the chance of correctly identifying it is 1/52 = 0.019. So the p-value is 51/52 + 1/52 = 1.
(f) No, the null hypothesis is either true or false (correct or not). The individual either has ESP or does not. The
p-value depends on the sample.
1.49
(a) You observed a yellow ball, which is the most extreme result that you could get. With only one observation
there is no more extreme than observing a yellow. So the p-value is the chance of observing a yellow ball under
the null hypothesis, which is 1/5 = 0.20. Since the p-value is larger than 10%, the result is not statistically
significant.
(b) The data consists of selecting two balls with replacement (and the order is not important). The possible
outcomes are shown in the picture below:
We observed one yellow and one blue ball. Results that are even more extreme would be observing two yellow
balls. So the p-value is (1+8)/25 = 0.36. Since the p-value is larger than 10%, the result is not statistically
significant.
1.50
(a) H0: The percentage of Republicans who are in favor of the death penalty is equal to the percentage of
Democrats who are in favor of the death penalty.
H1: The percentage of Republicans who are in favor of the death penalty is larger than the percentage of
Democrats who are in favor of the death penalty.
(b) It would be a one-sided rejection region since we are looking for a particular direction (larger than), and not just
any difference.
(c) The result could be statistically significant if the p-value were
.
(d) The difference of 2% would probably not be practically significant.
1.48
(a) Type I error: The subject does not have ESP but you conclude he does.
(b) Type II error: The subject has ESP but you conclude he does not.
(c) Level of significance is the chance of rejecting Ho when Ho is true. If the subject does not have ESP (i.e., H0 is
true) then the chance that he correctly identifies the card (and we reject H0) is 1 in 52, 1/52 = 0.019.
(d) The chance of a Type II error is the chance we fail to reject H0 when H0 is true. If the subject has perfect ESP,
then the chance he correctly identifies the card is 1 or 100%. He will correctly identify it, so the chance of a
Type II error is 0.
(e) The p-value calculates the chance of getting the observed result (here 0 correct answers) or something more
extreme, in the direction of H0 (namely, 1 correct answer), under the assumption that H0 is true (the person does
not have ESP). If the subject does not have ESP, the chance of incorrectly identifying the card is 51/52 = 0.981
and the chance of correctly identifying it is 1/52 = 0.019. So the p-value is 51/52 + 1/52 = 1.
(f) No, the null hypothesis is either true or false (correct or not). The individual either has ESP or does not. The
p-value depends on the sample.
1.49
(a) You observed a yellow ball, which is the most extreme result that you could get. With only one observation
there is no more extreme than observing a yellow. So the p-value is the chance of observing a yellow ball under
the null hypothesis, which is 1/5 = 0.20. Since the p-value is larger than 10%, the result is not statistically
significant.
(b) The data consists of selecting two balls with replacement (and the order is not important). The possible
outcomes are shown in the picture below:
We observed one yellow and one blue ball. Results that are even more extreme would be observing two yellow
balls. So the p-value is (1+8)/25 = 0.36. Since the p-value is larger than 10%, the result is not statistically
significant.
1.50
(a) H0: The percentage of Republicans who are in favor of the death penalty is equal to the percentage of
Democrats who are in favor of the death penalty.
H1: The percentage of Republicans who are in favor of the death penalty is larger than the percentage of
Democrats who are in favor of the death penalty.
(b) It would be a one-sided rejection region since we are looking for a particular direction (larger than), and not just
any difference.
(c) The result could be statistically significant if the p-value were
.
(d) The difference of 2% would probably not be practically significant.
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286 Chapter 1
1.51
Statement (a) since the effect is small so we would need a larger sample size to detect it.
1.52
The significance level is set at 0.05. For n=1: the decision rule is reject H0 if the result is $60 and the
corresponding
is 12/20 = 0.60. For n=2: the decision rule is reject H0 if the result is $45 and the corresponding
is 53/190 = 0.28. As the sample size is increased, the level of
decreased from 0.60 to 0.28.
1.53
(a) (i)
= chance of observing an average 45 from bag A = 4+2+1/190 = 0.0368
(ii) 45
(iii)
= 0.2789
(iv) The chance that we decide that the bag shown is bag A, while it really is bag B is 0.28 (or 28%).
(b) (i) First note that the observed average voucher value is $35, so we have: p-value = chance of observing an
average of $35 or more extreme under H0 = 17+9+4+2+1/190 = 0.1737
(ii) p-value
so we reject H0
(iii) 20+17+9+4+2+1/190 = 0.2789
(iv) p-value >
so we fail to reject H0
(v) The cut-off value is the average of $35, since we rejected H0 for $35 and we failed to reject H0 for $30.
1.54
(a) The first table shows the possible samples, possible averages, and their frequencies of occurring.
Possible Samples Average Frequency if Bag C Frequency if Bag D
1,1 1 0 10
1,2 1.5 2 20
1,3 2 3 15
1,4 2.5 4 10
1,5 3 5 5
2,2 2 1 6
2,3 2.5 6 12
2,4 3 8 8
2,5 3.5 10 4
3,3 3 3 3
3,4 3.5 12 6
3,5 4 15 3
4,4 4 6 1
4,5 4.5 20 2
5,5 5 10 0
The next table combines the entries in the first table according to the different possible averages.
Average Frequency if Bag C Frequency if Bag D
1 0 10
1.5 2 20
2 4 21
2.5 10 22
3 16 16
3.5 22 10
4 21 4
4.5 20 2
5 10 0
Total 105 105
The corresponding frequency plots are given next.
1.51
Statement (a) since the effect is small so we would need a larger sample size to detect it.
1.52
The significance level is set at 0.05. For n=1: the decision rule is reject H0 if the result is $60 and the
corresponding
is 12/20 = 0.60. For n=2: the decision rule is reject H0 if the result is $45 and the corresponding
is 53/190 = 0.28. As the sample size is increased, the level of
decreased from 0.60 to 0.28.
1.53
(a) (i)
= chance of observing an average 45 from bag A = 4+2+1/190 = 0.0368
(ii) 45
(iii)
= 0.2789
(iv) The chance that we decide that the bag shown is bag A, while it really is bag B is 0.28 (or 28%).
(b) (i) First note that the observed average voucher value is $35, so we have: p-value = chance of observing an
average of $35 or more extreme under H0 = 17+9+4+2+1/190 = 0.1737
(ii) p-value
so we reject H0
(iii) 20+17+9+4+2+1/190 = 0.2789
(iv) p-value >
so we fail to reject H0
(v) The cut-off value is the average of $35, since we rejected H0 for $35 and we failed to reject H0 for $30.
1.54
(a) The first table shows the possible samples, possible averages, and their frequencies of occurring.
Possible Samples Average Frequency if Bag C Frequency if Bag D
1,1 1 0 10
1,2 1.5 2 20
1,3 2 3 15
1,4 2.5 4 10
1,5 3 5 5
2,2 2 1 6
2,3 2.5 6 12
2,4 3 8 8
2,5 3.5 10 4
3,3 3 3 3
3,4 3.5 12 6
3,5 4 15 3
4,4 4 6 1
4,5 4.5 20 2
5,5 5 10 0
The next table combines the entries in the first table according to the different possible averages.
Average Frequency if Bag C Frequency if Bag D
1 0 10
1.5 2 20
2 4 21
2.5 10 22
3 16 16
3.5 22 10
4 21 4
4.5 20 2
5 10 0
Total 105 105
The corresponding frequency plots are given next.
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Solutions to All Exercises 287x
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1.5 4.53.52.52 543
Possible Averages
Bag A
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1.5 4.53.52.52 543
Possible Averages
Bag B
1 1
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(b) For the decision rule to reject the null hypothesis if the average is $2, we have the significance level
is
6/105=0.057 and the level of
is 54/105=0.514.
(c) If you observe an average of $3, the p-value would be the chance of observing $3 or less, which is p-value =
32/105 = 0.305.
1.55
(a) 20 nCr 2 = 190
(b) 20 nCr 3 = 1140
1.56
(a) The height must be 1/(base) = 1/7 or 0.1429.
(b)
(i)
= (1)(1/7) = 1/7 = 0.1429, as represented by the light grey shaded rectangle from 1 and to the left under
the H0 density.
(ii)
= 0.25 + 0.15 + 0.08 + 0.05 + 0.04 + 0.03 = 0.60 (or from 1 – 0.40), as represented by the area to the right
of 1 under the H1 density.
(iii) Power = 1 –
= 1 – 0.60 = 0.40, as represented by the area to the left of 1 under the H0 density.
(c) The p-value is (0.5)(1/7) = 0.0714, as represented by the black striped rectangle from 0.5 and to the left under
the H0 density.
(d) (ii) Increasing the sample size would lead to a reduction in the chance of committing a Type II error.
(e) False.
−
= power
p-value
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1.5 4.53.52.52 543
Possible Averages
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1.5 4.53.52.52 543
Possible Averages
Bag B
1 1
Bag C Bag D
(b) For the decision rule to reject the null hypothesis if the average is $2, we have the significance level
is
6/105=0.057 and the level of
is 54/105=0.514.
(c) If you observe an average of $3, the p-value would be the chance of observing $3 or less, which is p-value =
32/105 = 0.305.
1.55
(a) 20 nCr 2 = 190
(b) 20 nCr 3 = 1140
1.56
(a) The height must be 1/(base) = 1/7 or 0.1429.
(b)
(i)
= (1)(1/7) = 1/7 = 0.1429, as represented by the light grey shaded rectangle from 1 and to the left under
the H0 density.
(ii)
= 0.25 + 0.15 + 0.08 + 0.05 + 0.04 + 0.03 = 0.60 (or from 1 – 0.40), as represented by the area to the right
of 1 under the H1 density.
(iii) Power = 1 –
= 1 – 0.60 = 0.40, as represented by the area to the left of 1 under the H0 density.
(c) The p-value is (0.5)(1/7) = 0.0714, as represented by the black striped rectangle from 0.5 and to the left under
the H0 density.
(d) (ii) Increasing the sample size would lead to a reduction in the chance of committing a Type II error.
(e) False.
−
= power
p-value
Loading page 14...
288 Chapter 1
1.57
(a) The sketches are provided below.1 _
4
1 _
4
p-value = area to the left of 4.1
(b) (i) See the shaded area marked as
above.
(ii)
= (0.4)(1/4) = 0.1
(iii) See the shaded area marked as
above.
(iv) Power = 1 –
= 1 – (0.6)(1/4) = 1 – (6/10)(1/4) = 1 - 0.15 = 0.85.
(c) (i) See the shaded area marked as the p-value above.
(ii) p-value = (1.1)(1/4) = 0.275.
(d) No, the p-value > 0.1 =
.
1 _
4
1 _
4
1.57
(a) The sketches are provided below.1 _
4
1 _
4
p-value = area to the left of 4.1
(b) (i) See the shaded area marked as
above.
(ii)
= (0.4)(1/4) = 0.1
(iii) See the shaded area marked as
above.
(iv) Power = 1 –
= 1 – (0.6)(1/4) = 1 – (6/10)(1/4) = 1 - 0.15 = 0.85.
(c) (i) See the shaded area marked as the p-value above.
(ii) p-value = (1.1)(1/4) = 0.275.
(d) No, the p-value > 0.1 =
.
1 _
4
1 _
4
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Solutions to All Exercises 289
Interactive Statistics 3rd Edition: Chapter 2 Full
Solutions
2.1
35% is a parameter – it is a numerical summary of the population.
28% is a statistic – it is a numerical summary of a sample from the population.
2.2
(a) The value of 9% is a parameter – it is a numerical summary of the population.
(b) The value of 12.5% is a statistics – it is a numerical summary of a sample from the population.
2.3
The proportion of register voters in Ann Arbor who would vote “Yes” on Proposal A is an example of (c) a
parameter.
2.4
(a) The population consists of (the planned vote for) the 100 U.S. Senators.
(b) N = 100.
(c) The sample consists of (the planned vote for) the ten selected U.S. Senators.
(d) n = 10.
2.5
For a simple random sample of n = 200 and the number of items that were defective = 5. All we can say is that (c)
the percent of defective items in the sample is 5/200 = 2.5%.p = 5/200 = 0.025 or 2.5%.
2.6
(a) 22.4.
(b) 19. No.
(c) 22.5. No. No.
(d) 18 and 21. The sample mean age is 19.5.
18 and 26. The sample mean age is 22.
20 and 27. The sample mean age is 23.5.
20 and 21. The sample mean age is 20.5.
20 and 26. The sample mean age is 23.
27 and 21. The sample mean age is 24.
27 and 26. The sample mean age is 26.5.
2.7
(a) The value of 34 is a statistic.
(b) Response bias.
2.8
(a) Statistic since the 54% is a sample percentage, not of the population.
(b) Nonresponse bias.
2.9
Response bias.
2.10
Nonresponse bias.
2.11
Nonresponse bias.
Interactive Statistics 3rd Edition: Chapter 2 Full
Solutions
2.1
35% is a parameter – it is a numerical summary of the population.
28% is a statistic – it is a numerical summary of a sample from the population.
2.2
(a) The value of 9% is a parameter – it is a numerical summary of the population.
(b) The value of 12.5% is a statistics – it is a numerical summary of a sample from the population.
2.3
The proportion of register voters in Ann Arbor who would vote “Yes” on Proposal A is an example of (c) a
parameter.
2.4
(a) The population consists of (the planned vote for) the 100 U.S. Senators.
(b) N = 100.
(c) The sample consists of (the planned vote for) the ten selected U.S. Senators.
(d) n = 10.
2.5
For a simple random sample of n = 200 and the number of items that were defective = 5. All we can say is that (c)
the percent of defective items in the sample is 5/200 = 2.5%.p = 5/200 = 0.025 or 2.5%.
2.6
(a) 22.4.
(b) 19. No.
(c) 22.5. No. No.
(d) 18 and 21. The sample mean age is 19.5.
18 and 26. The sample mean age is 22.
20 and 27. The sample mean age is 23.5.
20 and 21. The sample mean age is 20.5.
20 and 26. The sample mean age is 23.
27 and 21. The sample mean age is 24.
27 and 26. The sample mean age is 26.5.
2.7
(a) The value of 34 is a statistic.
(b) Response bias.
2.8
(a) Statistic since the 54% is a sample percentage, not of the population.
(b) Nonresponse bias.
2.9
Response bias.
2.10
Nonresponse bias.
2.11
Nonresponse bias.
Loading page 16...
290 Chapter 2
2.12
False. When only 272 out of 1000 people respond, for a 27.2% response rate, this results in nonresponse bias.
2.13
The best answer is (c).
2.14
Results will vary.
(a) Using the calculator with a seed value of 291, the selected persons are 39 (79 years old), person 3 (75 years old)
and person 24 (70 years old). The average age is 74.67 years.
(b) 22 (36 years old), 34 (89 years old) and 29 (89 years old). The average age is 71.33 years, different from the
mean calculated in part (a).
(c) 69.19 years.
(d) 74.67 and 71.33 are statistics, while the mean in part (c) is a parameter.
2.15
(a) Yes, each sample of size 20 has the same chance as any other sample of size 20 to be selected (assuming all tags
are exactly the same and the box is thoroughly mixed).
(b) It is drawn without replacement.
2.16
(a) 12/100.
(b) With the graphing calculator the selected employees are: 77, 51, 72, 40, 71, 42, 17, 34, 62, 23, 35, and 12.
From the random table the selected employees are: 7, 5, 69, 76, 28, 33, 78, 70, 99, 98, 42, and 80.
2.17
(a) H0: The population proportion of dissatisfied customers equals 0.10.
H1: The population proportion of dissatisfied customers is less than 0.10.
(b) With the calculator the selected customers are: 34318, 15553, 8461, and 614. With the random table the
selected customers are: 15409, 23336, 29490, and 30414.
(c) Type II error.
(d) 0.21.
(e) No, the p-value is > 0.05.
(f) (ii) Statistic.
2.18
Using the TI: Label the sites from 1 to 80. With a seed value of 29 the five labels, and thus sites, selected at random
are as follows: Sites #50, #66, #43, #49, and #74. Using the Random Number Table: Since there are more than 10,
but less than 100, sites, we can use double-digit labels, such as 01, 02, 03, through 80. With Row 10 of the table,
starting at Column 1, the simple random sample of sites consists of sites #47, #53, #68, #57, and #34.
2.19
Stratified random sampling. You are dividing up your population by class rank then selecting 100 students from
within each stratum at random.
2.20
(a) Stratifying by declared major (field of study) would be a good stratification variable. The cost of textbooks is
likely to vary from one field of study to the next, but not vary as much within a particular field of study (many
students have to take the same basic classes for their field of study and hence may be paying similar textbook
costs.) You would also learn about the costs for each field of major as well as costs overall.
(b) Stratifying by gender may not be as useful as a stratification variable. The cost of textbooks may vary a lot for
females and vary a lot for males, and there may not be much variation between males and females. However, if
you wanted to learn about the costs for each gender as well as overall, you might wish to stratify by gender.
(c) Stratifying by class rank may be useful as a stratification variable. The cost of textbooks may vary from one
class-rank to another. However, there may be quite a lot of variation within each class-rank as well. If you
wanted to learn about the costs for each class-rank as well as overall, you might wish to stratify by class-rank.
2.21
2.12
False. When only 272 out of 1000 people respond, for a 27.2% response rate, this results in nonresponse bias.
2.13
The best answer is (c).
2.14
Results will vary.
(a) Using the calculator with a seed value of 291, the selected persons are 39 (79 years old), person 3 (75 years old)
and person 24 (70 years old). The average age is 74.67 years.
(b) 22 (36 years old), 34 (89 years old) and 29 (89 years old). The average age is 71.33 years, different from the
mean calculated in part (a).
(c) 69.19 years.
(d) 74.67 and 71.33 are statistics, while the mean in part (c) is a parameter.
2.15
(a) Yes, each sample of size 20 has the same chance as any other sample of size 20 to be selected (assuming all tags
are exactly the same and the box is thoroughly mixed).
(b) It is drawn without replacement.
2.16
(a) 12/100.
(b) With the graphing calculator the selected employees are: 77, 51, 72, 40, 71, 42, 17, 34, 62, 23, 35, and 12.
From the random table the selected employees are: 7, 5, 69, 76, 28, 33, 78, 70, 99, 98, 42, and 80.
2.17
(a) H0: The population proportion of dissatisfied customers equals 0.10.
H1: The population proportion of dissatisfied customers is less than 0.10.
(b) With the calculator the selected customers are: 34318, 15553, 8461, and 614. With the random table the
selected customers are: 15409, 23336, 29490, and 30414.
(c) Type II error.
(d) 0.21.
(e) No, the p-value is > 0.05.
(f) (ii) Statistic.
2.18
Using the TI: Label the sites from 1 to 80. With a seed value of 29 the five labels, and thus sites, selected at random
are as follows: Sites #50, #66, #43, #49, and #74. Using the Random Number Table: Since there are more than 10,
but less than 100, sites, we can use double-digit labels, such as 01, 02, 03, through 80. With Row 10 of the table,
starting at Column 1, the simple random sample of sites consists of sites #47, #53, #68, #57, and #34.
2.19
Stratified random sampling. You are dividing up your population by class rank then selecting 100 students from
within each stratum at random.
2.20
(a) Stratifying by declared major (field of study) would be a good stratification variable. The cost of textbooks is
likely to vary from one field of study to the next, but not vary as much within a particular field of study (many
students have to take the same basic classes for their field of study and hence may be paying similar textbook
costs.) You would also learn about the costs for each field of major as well as costs overall.
(b) Stratifying by gender may not be as useful as a stratification variable. The cost of textbooks may vary a lot for
females and vary a lot for males, and there may not be much variation between males and females. However, if
you wanted to learn about the costs for each gender as well as overall, you might wish to stratify by gender.
(c) Stratifying by class rank may be useful as a stratification variable. The cost of textbooks may vary from one
class-rank to another. However, there may be quite a lot of variation within each class-rank as well. If you
wanted to learn about the costs for each class-rank as well as overall, you might wish to stratify by class-rank.
2.21
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Solutions to All Exercises 291
(a) 45 8th-grade students; 25 10th-grade students; 48 12th-grade students. For a total in the sample of 118 students.
(b) With the calculator the selected students are: 193, 127, 430, 100, and 427. With the random number table the
selected students are: 241, 304, 22, 364, and 151.
2.22
(a) 100/200 = 1/2 = 0.50 or 50%.
(b) 100/1000 = 1/10 = 0.10 or 10%.
(c) The chance of being chosen is 0.10. This is NOT a simple random sample. For this stratified random sampling
plan each possible sample would contain exactly 100 males and 20 females. All samples of size 120 are not
equally likely (as it should be for simple random sample). Some samples of size 120 are not even possible, for
example, having 120 males.
2.23
We should use a weighted average:40
100
70( ) + 60
100
63( ) = 65.8 inches
2.24
(a) 20/200 = 0.10.
(b) With the calculator the first five females selected in the sample are: 132, 195, 147, 171, and 85. With the
random number table: To each woman we assigned 5 numbers, for example 1, 201, 401, 601 and 801. The
numbers from the table were 521, 625, 391, 646, and 369. So the first five selected females are: 121, 25, 191,
46, and 169.
(c) The weighted average is: (0.75)8 + (0.25)5 = 7.25.
2.25
(a) Stratified random sampling.
(b) With the calculator the first five selected homes are: 386, 81, 379, 211, and 156. With the random number table
the first five selected homes are: 94, 299, 396, 378, and 363.
(c) (0.60)(2100) + (0.40)(2600) = 2300 square feet.
2.26
(0.20)(16) + (0.50)(43) + (0.30)(71) = 46 years.
2.27
(a) Stratified random sampling.
(b) With the calculator the first six homes selected from the 1000 homes in County I are: 918, 193, 902, 502, 370,
and 5. With the random number table the first six homes selected from the 1000 homes in County I are: 963, 19,
197, 705, 463, and 79 where the homes are labeled 0 to 999.
(c) (0.50)(175) + (0.30)(200) + (0.20)(195) = 186.5 thousands of dollars or $186,500.
2.28
(a) With the calculator, the label of the first student selected is 3. With the random number table, the label of the
first student selected is 3.
(b) The students in the sample are those with ID numbers 3, (3 + 4 =) 7, (7 + 4 =) 11, and (11 + 4=) 15, for a total
of 4 students.
2.29
(a) With the calculator with the first 100 addresses labeled 1 through 100, the sample is: Addresses = #79, #179,
#279, #379, #479. Using the random number table with the first 100 addresses labeled 01, 02, ... , 98, 99, 00,
the sample is: Addresses = #30, #130, #230, #330, #430.
(b) 1/100 = 0.01 or 1% There are 100 possible systematic samples of size 5, each equally likely.
(a) 45 8th-grade students; 25 10th-grade students; 48 12th-grade students. For a total in the sample of 118 students.
(b) With the calculator the selected students are: 193, 127, 430, 100, and 427. With the random number table the
selected students are: 241, 304, 22, 364, and 151.
2.22
(a) 100/200 = 1/2 = 0.50 or 50%.
(b) 100/1000 = 1/10 = 0.10 or 10%.
(c) The chance of being chosen is 0.10. This is NOT a simple random sample. For this stratified random sampling
plan each possible sample would contain exactly 100 males and 20 females. All samples of size 120 are not
equally likely (as it should be for simple random sample). Some samples of size 120 are not even possible, for
example, having 120 males.
2.23
We should use a weighted average:40
100
70( ) + 60
100
63( ) = 65.8 inches
2.24
(a) 20/200 = 0.10.
(b) With the calculator the first five females selected in the sample are: 132, 195, 147, 171, and 85. With the
random number table: To each woman we assigned 5 numbers, for example 1, 201, 401, 601 and 801. The
numbers from the table were 521, 625, 391, 646, and 369. So the first five selected females are: 121, 25, 191,
46, and 169.
(c) The weighted average is: (0.75)8 + (0.25)5 = 7.25.
2.25
(a) Stratified random sampling.
(b) With the calculator the first five selected homes are: 386, 81, 379, 211, and 156. With the random number table
the first five selected homes are: 94, 299, 396, 378, and 363.
(c) (0.60)(2100) + (0.40)(2600) = 2300 square feet.
2.26
(0.20)(16) + (0.50)(43) + (0.30)(71) = 46 years.
2.27
(a) Stratified random sampling.
(b) With the calculator the first six homes selected from the 1000 homes in County I are: 918, 193, 902, 502, 370,
and 5. With the random number table the first six homes selected from the 1000 homes in County I are: 963, 19,
197, 705, 463, and 79 where the homes are labeled 0 to 999.
(c) (0.50)(175) + (0.30)(200) + (0.20)(195) = 186.5 thousands of dollars or $186,500.
2.28
(a) With the calculator, the label of the first student selected is 3. With the random number table, the label of the
first student selected is 3.
(b) The students in the sample are those with ID numbers 3, (3 + 4 =) 7, (7 + 4 =) 11, and (11 + 4=) 15, for a total
of 4 students.
2.29
(a) With the calculator with the first 100 addresses labeled 1 through 100, the sample is: Addresses = #79, #179,
#279, #379, #479. Using the random number table with the first 100 addresses labeled 01, 02, ... , 98, 99, 00,
the sample is: Addresses = #30, #130, #230, #330, #430.
(b) 1/100 = 0.01 or 1% There are 100 possible systematic samples of size 5, each equally likely.
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292 Chapter 2
2.30
(a) Stratified random sampling.
(b) No, if there are more people whose family name begins with say A and fewer people whose family name begins
with Z, then those with Z will have a higher chance of being selected.
(c) Selection bias, a systematic tendency to exclude those with unlisted phone numbers.
2.31
Yes, the chance is 1/5 or 0.20 or 20%. There are 5 possible systematic samples, each equally likely.
2.32
(a) 1/14 = 0.0714.
(b) Note that there are 555 members so 555/14 → 39.64 or 39 groups of 14 and one last group of 9. For the 17th
member to be selected, the starting point must have been a 3, that is, the 3rd member in each group of 14 is
selected. This will result in a sample of 40 members.
2.33
(a) 1/8 = 0.125.
(b) Note that there are 350 members so 350/8 → 43.75 or 43 groups of 8 and one last group of 6. For the 300th
member to be selected, the starting point must have been a 4, that is, the 4th member in each group of 8 is
selected. This will result in a sample of 44 members.
2.34
(a) 1/40 = 0.025.
(b) Using the TI calculator with seed of 19, and with the first 40 members labeled 1 through 40, the first five
members are 7, 47, 87, 127, and 167. Using row 15, column 1, and labeling the first 40 members as 01, 02, ... ,
39, 40, the first 5 members are 7, 47, 87, 127, and 167.
(c) Note that there are 2220 members so 2220/40 → 55.5 or 55 groups of 40 and one last group of 20. Since the
starting point was a 7, that is, the 7th member in each group of 40 was selected. This would result in a complete
sample of 56 members.
2.35
False, the chance depends on the number of clusters.
2.36
(a) With the calculator, the two departments selected were 2 and 4; Chemistry and Mathematics. Using the random
number table, the two department selected were 2 and 3; Chemistry and Biology.
(b) With calculator, the sample size is 80. With random number table, the sample size is 90.
2.37
(a) Cluster sampling. The faculty are grouped into departments which serve as clusters. Six of the clusters are
selected at random. All of the units in the cluster are in the sample.
(b) Label the list of the 60 departments 1, 2, 3, ..., 60. The numbers generated and thus the departments
selected using the TI calculator with seed = 79 are: 54, 37, 49, 5, 15, and 43. If you are using the random
number table, you might label the list of the 60 departments 01, 02, 03, ... , 60. The numbers generated and thus
the departments selected starting at row 60, column 1 are: 23, 22, 47, 40, 25, and 37.
(c) Yes we can determine the chance that any specific professor will be selected. The chance of selecting a
professor is the same chance that his/her department or cluster will be selected. The reason being if a cluster is
selected, every element in the cluster is selected. Therefore the probability that a cluster is selected is 6/60 =
0.10, the chance that any specific professor will be selected.
2.30
(a) Stratified random sampling.
(b) No, if there are more people whose family name begins with say A and fewer people whose family name begins
with Z, then those with Z will have a higher chance of being selected.
(c) Selection bias, a systematic tendency to exclude those with unlisted phone numbers.
2.31
Yes, the chance is 1/5 or 0.20 or 20%. There are 5 possible systematic samples, each equally likely.
2.32
(a) 1/14 = 0.0714.
(b) Note that there are 555 members so 555/14 → 39.64 or 39 groups of 14 and one last group of 9. For the 17th
member to be selected, the starting point must have been a 3, that is, the 3rd member in each group of 14 is
selected. This will result in a sample of 40 members.
2.33
(a) 1/8 = 0.125.
(b) Note that there are 350 members so 350/8 → 43.75 or 43 groups of 8 and one last group of 6. For the 300th
member to be selected, the starting point must have been a 4, that is, the 4th member in each group of 8 is
selected. This will result in a sample of 44 members.
2.34
(a) 1/40 = 0.025.
(b) Using the TI calculator with seed of 19, and with the first 40 members labeled 1 through 40, the first five
members are 7, 47, 87, 127, and 167. Using row 15, column 1, and labeling the first 40 members as 01, 02, ... ,
39, 40, the first 5 members are 7, 47, 87, 127, and 167.
(c) Note that there are 2220 members so 2220/40 → 55.5 or 55 groups of 40 and one last group of 20. Since the
starting point was a 7, that is, the 7th member in each group of 40 was selected. This would result in a complete
sample of 56 members.
2.35
False, the chance depends on the number of clusters.
2.36
(a) With the calculator, the two departments selected were 2 and 4; Chemistry and Mathematics. Using the random
number table, the two department selected were 2 and 3; Chemistry and Biology.
(b) With calculator, the sample size is 80. With random number table, the sample size is 90.
2.37
(a) Cluster sampling. The faculty are grouped into departments which serve as clusters. Six of the clusters are
selected at random. All of the units in the cluster are in the sample.
(b) Label the list of the 60 departments 1, 2, 3, ..., 60. The numbers generated and thus the departments
selected using the TI calculator with seed = 79 are: 54, 37, 49, 5, 15, and 43. If you are using the random
number table, you might label the list of the 60 departments 01, 02, 03, ... , 60. The numbers generated and thus
the departments selected starting at row 60, column 1 are: 23, 22, 47, 40, 25, and 37.
(c) Yes we can determine the chance that any specific professor will be selected. The chance of selecting a
professor is the same chance that his/her department or cluster will be selected. The reason being if a cluster is
selected, every element in the cluster is selected. Therefore the probability that a cluster is selected is 6/60 =
0.10, the chance that any specific professor will be selected.
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Solutions to All Exercises 293
2.38
(a) This is cluster sampling since the students are first divided into clusters (undergraduate classes). A class/cluster
is then selected using a simple random sample and all students from that class/cluster are sampled.
(b) No, since not all students take the same number of classes. Students who attend more classes have a greater
chance of being selected.
(c) No it is not biased since the cluster was selected at random. It is a case of poor design together with bad luck.
When clustering the variability between clusters should not be more important than the variability within
clusters. Here we have poor design because there might be more variability between a class with many students
on athletic scholarship and a class without any, than variability within each of those classes.
2.39
(a) The type of sampling performed in each dorm is cluster sampling, with the rooms forming the clusters and 3
clusters were selected at random from each of the four dorms.
(b) If each cluster selected has one student, we would have 3 students from each of the four dorms for a minimum
sample size of 12 students.
(c) If each cluster selected has three students, we would have 9 students from each of the four dorms for a
maximum sample size of 36 students.
(d) We do not know, it will depend on how many clusters selected are rooms with women. The number of women
could be as low as 0 and as high as 36.
(e) No, there will be anywhere from 3 to 9 freshmen students sampled from the freshmen dorm, as well as from 3
to 9 sophomores, from 3 and 9 juniors, and from 3 to 9 seniors.
2.40
(a) Cluster sampling.
(b) No, if the number of adults per city block is unknown. Yes, if you have the seed value and you know how
many adults are in each block.
2.41
(a) With the calculator or the random number table, the selected region is 3 = Southwest.
(b) Stratified random sampling.
(c) (i) 1-in-10 systematic sampling.
(ii) 0.10.
(iii) With the calculator, the first five selected cans are 7, 17, 27, 37, and 47. With the random number table we
might label the first can 1, the second can 2, …, and the 10th can 0. Then the first five selected cans are 7,
17, 27, 37, and 47.
(iv) Note that 125/10 = 12.5 → 12 or 13 cans. However, there will not be a 7th can to select in last group. Thus
the total number of cans in the sample will be 12.
(d) (i) No.
(ii) Two possible values are 0.12 and 0.15.
(iii) Yes, a Type II error.
2.42
(a) (i) Convenience sampling.
(ii) Yes, a selection bias.
(iii) The calculated average is expected to be higher than the true average, as all of the books in the sample have
already been checked out at least once, and may include some of the more popular books.
(b) Cluster sampling.
(c) (i) Stratified random sampling.
(ii) Overall estimate: (40/1200)(20) + (200/1200)(15) + (600/1200)(10) = 14.2 times checked out.
(d) For each of the three categories of books the following stages are followed.
Stage 1: Divide the books into clusters according to the last digit of the call number (0 through 9). Take a
simple random sample of 3 digits from the list of 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
The clusters of books (in that category) with call numbers ending with those selected digits are selected.
Stage 2: Within each of the selected clusters of books from Stage 2, select a simple random sample of 7 books.
Note that with this multistage sampling plan, we will have a total of 3 categories x 3 clusters x 7 books = 63
books.
2.43
2.38
(a) This is cluster sampling since the students are first divided into clusters (undergraduate classes). A class/cluster
is then selected using a simple random sample and all students from that class/cluster are sampled.
(b) No, since not all students take the same number of classes. Students who attend more classes have a greater
chance of being selected.
(c) No it is not biased since the cluster was selected at random. It is a case of poor design together with bad luck.
When clustering the variability between clusters should not be more important than the variability within
clusters. Here we have poor design because there might be more variability between a class with many students
on athletic scholarship and a class without any, than variability within each of those classes.
2.39
(a) The type of sampling performed in each dorm is cluster sampling, with the rooms forming the clusters and 3
clusters were selected at random from each of the four dorms.
(b) If each cluster selected has one student, we would have 3 students from each of the four dorms for a minimum
sample size of 12 students.
(c) If each cluster selected has three students, we would have 9 students from each of the four dorms for a
maximum sample size of 36 students.
(d) We do not know, it will depend on how many clusters selected are rooms with women. The number of women
could be as low as 0 and as high as 36.
(e) No, there will be anywhere from 3 to 9 freshmen students sampled from the freshmen dorm, as well as from 3
to 9 sophomores, from 3 and 9 juniors, and from 3 to 9 seniors.
2.40
(a) Cluster sampling.
(b) No, if the number of adults per city block is unknown. Yes, if you have the seed value and you know how
many adults are in each block.
2.41
(a) With the calculator or the random number table, the selected region is 3 = Southwest.
(b) Stratified random sampling.
(c) (i) 1-in-10 systematic sampling.
(ii) 0.10.
(iii) With the calculator, the first five selected cans are 7, 17, 27, 37, and 47. With the random number table we
might label the first can 1, the second can 2, …, and the 10th can 0. Then the first five selected cans are 7,
17, 27, 37, and 47.
(iv) Note that 125/10 = 12.5 → 12 or 13 cans. However, there will not be a 7th can to select in last group. Thus
the total number of cans in the sample will be 12.
(d) (i) No.
(ii) Two possible values are 0.12 and 0.15.
(iii) Yes, a Type II error.
2.42
(a) (i) Convenience sampling.
(ii) Yes, a selection bias.
(iii) The calculated average is expected to be higher than the true average, as all of the books in the sample have
already been checked out at least once, and may include some of the more popular books.
(b) Cluster sampling.
(c) (i) Stratified random sampling.
(ii) Overall estimate: (40/1200)(20) + (200/1200)(15) + (600/1200)(10) = 14.2 times checked out.
(d) For each of the three categories of books the following stages are followed.
Stage 1: Divide the books into clusters according to the last digit of the call number (0 through 9). Take a
simple random sample of 3 digits from the list of 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
The clusters of books (in that category) with call numbers ending with those selected digits are selected.
Stage 2: Within each of the selected clusters of books from Stage 2, select a simple random sample of 7 books.
Note that with this multistage sampling plan, we will have a total of 3 categories x 3 clusters x 7 books = 63
books.
2.43
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294 Chapter 2
(a) A cluster sampling of blocks.
(b) Yes, the chance is 5/50 = 0.10.
(c) No, we do not know how many households are in each block. Additionally, if the number of households in
each block was not the same across all blocks, then we would also need to know which of the 5 blocks were
selected.
(d) With the calculator, the five selected blocks are 8, 44, 33, 6, and 38. With the random number table we might
label the blocks 01 to 50, then the five selected blocks are 12, 18, 11, 43, and 05.
(e) Response bias because the interviewers are college students. People may not feel comfortable telling these
students they want to forbid loud music at parties in the college dorms.
2.44
Answer is (c) statistic is to a sample.
2.45
The proportion 153/200 is a parameter since the instructor polled her entire class and that was the group that she was
interested in learning about.
2.46
(a) Population: Adults U.S. residents. The sample size n = 1500.
(b) Population: Today’s shipment of 1-gallon milk cartons. The sample size n = 5.
(c) Population: The 740 members of the local women’s business association. The sample size n = 100.
2.47
Nonresponse bias.
2.48
(a) Selection bias, where the sampling frame is either incomplete (such as sampling only one dealership) or
incorrect.
(b) Nonresponse bias, where persons who do not respond to a survey (such as the lazy car owners) may have
different opinions to those who do.
(c) Response bias, where respondents may have a tendency to lie (such as car dealerships that believe in lower
miles per gallon) or refuse to answer.
2.49
This survey may be subject to nonresponse bias. Only those alumni who respond and report their income will be
included. Alumni who perhaps are currently unemployed or in a low paying position may elect not to respond.
Therefore, the reported average income based on such a survey may be biased upwards -- the average may be larger
than the actual average for all alumni.
2.50
(a) The sampling design is a simple random sample of size 227 taken from the cocaine users who called.
(b) The population this sample is drawn from is cocaine users who called the National help line between February
and March.
(c) The sample is not from the population of workers, but only from those who called the hotline, so the survey
results do not generalize to the population of workers. Also, "more people" implies a comparison for which no
data were given.
2.51
(a) With the calculator the selected ID numbers are: 179, 2274, and 3327. With the random number table the
selected ID numbers are: 2398, 2258, and 3540.
(b) A statistic.
2.52
(a) True. If we fail to reject the null hypothesis, them the population consists of all males.
(b) False.
(c) True.
2.53
(a) A cluster sampling of blocks.
(b) Yes, the chance is 5/50 = 0.10.
(c) No, we do not know how many households are in each block. Additionally, if the number of households in
each block was not the same across all blocks, then we would also need to know which of the 5 blocks were
selected.
(d) With the calculator, the five selected blocks are 8, 44, 33, 6, and 38. With the random number table we might
label the blocks 01 to 50, then the five selected blocks are 12, 18, 11, 43, and 05.
(e) Response bias because the interviewers are college students. People may not feel comfortable telling these
students they want to forbid loud music at parties in the college dorms.
2.44
Answer is (c) statistic is to a sample.
2.45
The proportion 153/200 is a parameter since the instructor polled her entire class and that was the group that she was
interested in learning about.
2.46
(a) Population: Adults U.S. residents. The sample size n = 1500.
(b) Population: Today’s shipment of 1-gallon milk cartons. The sample size n = 5.
(c) Population: The 740 members of the local women’s business association. The sample size n = 100.
2.47
Nonresponse bias.
2.48
(a) Selection bias, where the sampling frame is either incomplete (such as sampling only one dealership) or
incorrect.
(b) Nonresponse bias, where persons who do not respond to a survey (such as the lazy car owners) may have
different opinions to those who do.
(c) Response bias, where respondents may have a tendency to lie (such as car dealerships that believe in lower
miles per gallon) or refuse to answer.
2.49
This survey may be subject to nonresponse bias. Only those alumni who respond and report their income will be
included. Alumni who perhaps are currently unemployed or in a low paying position may elect not to respond.
Therefore, the reported average income based on such a survey may be biased upwards -- the average may be larger
than the actual average for all alumni.
2.50
(a) The sampling design is a simple random sample of size 227 taken from the cocaine users who called.
(b) The population this sample is drawn from is cocaine users who called the National help line between February
and March.
(c) The sample is not from the population of workers, but only from those who called the hotline, so the survey
results do not generalize to the population of workers. Also, "more people" implies a comparison for which no
data were given.
2.51
(a) With the calculator the selected ID numbers are: 179, 2274, and 3327. With the random number table the
selected ID numbers are: 2398, 2258, and 3540.
(b) A statistic.
2.52
(a) True. If we fail to reject the null hypothesis, them the population consists of all males.
(b) False.
(c) True.
2.53
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Solutions to All Exercises 295
(a) Based on her decision rule, Jane rejects H0.
(b) Since Jane rejected the null hypothesis, the data are statistically significant.
(c) No, based on a sample of two $1 Jane can not be certain of which purse she has since both purses contain at
least two $1.
(d) Type I error: reject H0 when H0 is true.
(e) Simple random samples of size 2 from null purse:
$11 and $12, $11 and $51, $11 and $52
$12 and $51, $12 and $52, $51 and $52
(f) Simple random samples of size 2 from alternative purse:
$11 and $12, $11 and $13, $11 and $14
$12 and $13, $12 and $14, $13 and $14
(g) The p-value is the chance of getting two $1 bills or more extreme (in the direction of H0, but in this case, there
is no “more extreme”) if H0 is true. The p-value is the chance of getting two $1 bills if H0 is true, i.e., 1/6.
2.54
You first need to label the 4000 signatures. Hopefully no one signed the petition more than one time. If you use a
calculator or computer, the labels can simply be 1 to 4000. If you use the random number table, you could label the
signatures from 0001 to 4000. Using a seed value of say 29 with the calculator or row 120, column 11 of the
random number table, you can proceed to take a simple random sample of 400 signatures.
2.55
Stratified random sampling. You are dividing up your population by gender then selecting your sample within the
strata at random.
2.56
(a) Stratified random sampling.
(b) 75/416.
(c) (0.20)(15/25) + (0.80)(60/75) = 0.76.
(d) Statistic.
2.57
(a) All former university graduate students.
(b) Stratified random sampling.
(c) False.
2.58
(a) Stratified random sampling.
(b) Selection bias.
2.59
(a) Stratified random sampling
(b) The chance is 0, only 1 of the two fiction books (A, B) will be selected.
(c) Two books that could be selected are Book A and Book C.
(d) Another pair of books that could be selected is Book A and Book D.
(e) The chance that the total number of pages exceeds 800 is the same as the chance that the two selected books are
Book B and Book D. There are 4 possible pairs of books that could be selected of which the (B, D) pair is 1, so
the chance is 1/4 = 0.25.
2.60
(a) Stratified random sampling.
(b) High: 20 clients, Moderate: 125 clients, Low: 45 clients.
(c) With the calculator the selected clients are: 163, 2196, 214, 2462, 740.
With the random number table the selected clients are: 1887, 1209, 2294, 954, and 1869.
(a) Based on her decision rule, Jane rejects H0.
(b) Since Jane rejected the null hypothesis, the data are statistically significant.
(c) No, based on a sample of two $1 Jane can not be certain of which purse she has since both purses contain at
least two $1.
(d) Type I error: reject H0 when H0 is true.
(e) Simple random samples of size 2 from null purse:
$11 and $12, $11 and $51, $11 and $52
$12 and $51, $12 and $52, $51 and $52
(f) Simple random samples of size 2 from alternative purse:
$11 and $12, $11 and $13, $11 and $14
$12 and $13, $12 and $14, $13 and $14
(g) The p-value is the chance of getting two $1 bills or more extreme (in the direction of H0, but in this case, there
is no “more extreme”) if H0 is true. The p-value is the chance of getting two $1 bills if H0 is true, i.e., 1/6.
2.54
You first need to label the 4000 signatures. Hopefully no one signed the petition more than one time. If you use a
calculator or computer, the labels can simply be 1 to 4000. If you use the random number table, you could label the
signatures from 0001 to 4000. Using a seed value of say 29 with the calculator or row 120, column 11 of the
random number table, you can proceed to take a simple random sample of 400 signatures.
2.55
Stratified random sampling. You are dividing up your population by gender then selecting your sample within the
strata at random.
2.56
(a) Stratified random sampling.
(b) 75/416.
(c) (0.20)(15/25) + (0.80)(60/75) = 0.76.
(d) Statistic.
2.57
(a) All former university graduate students.
(b) Stratified random sampling.
(c) False.
2.58
(a) Stratified random sampling.
(b) Selection bias.
2.59
(a) Stratified random sampling
(b) The chance is 0, only 1 of the two fiction books (A, B) will be selected.
(c) Two books that could be selected are Book A and Book C.
(d) Another pair of books that could be selected is Book A and Book D.
(e) The chance that the total number of pages exceeds 800 is the same as the chance that the two selected books are
Book B and Book D. There are 4 possible pairs of books that could be selected of which the (B, D) pair is 1, so
the chance is 1/4 = 0.25.
2.60
(a) Stratified random sampling.
(b) High: 20 clients, Moderate: 125 clients, Low: 45 clients.
(c) With the calculator the selected clients are: 163, 2196, 214, 2462, 740.
With the random number table the selected clients are: 1887, 1209, 2294, 954, and 1869.
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296 Chapter 2
2.61
(a) With the calculator the first student selected is 1. With the random number table the first student selected is 1.
(b) The sample size is 6.
2.62
(a) Systematic sampling (1-in-30). Once you select the sample you can determine the number of freshmen you
selected. Also knowing the population list you can calculate the number of freshmen you would select for each
of the 30 possible samples (depending on the starting point.)
(b) Stratified random sampling. Yes, you would have 25 freshmen.
2.63
(a) (i) 0.02(500)+0.03(1200)+0.05(18000) = 10+36+900 = 946.
(ii) Stratified Random Sampling.
(iii) With the calculator the selected labels are 432, 232, 304, 412, and 372. With the random number table, we
might assign the first High category driver the labels 001 and 501. We would assign the second High
category driver 002 and 502. This assignment pattern would continue until the 500th High category driver
who would be assigned the labels 000 and 500. Reading off labels from row 60, column 1, we have: 789,
191, 947, 423, and 632. This would correspond to selecting the 289th, 191st, 447th, 423rd, and 132nd High
category drivers in the list of 500 High category drivers.
(b) (i) With the calculator or the random number table, the first selected label is 15. Thus the selected labels will
be 15, 35, 55, 75, 95, and so on.
(ii) Since the 500 High category drivers divide evenly into groups of 20 (500/20 = 25), there will be a total of
25 High category drivers in the systematic 1-in-20 sample.
2.64
(a) 907.
(b) Cluster sample.
(c) (iii) Selection bias.
2.65
These results are based on a study of 125 aerobic classes in five health clubs, not all aerobic classes in all health
clubs. Thus, the 60% figure is a statistic and the sample size is n = 125.
2.66
(a) It is a systematic 1-in-45 sampling resulting in 50 students (1 from each of the 50 sections, the 33rd in each of
the 50 lists of 45 students).
(b) It is a cluster sample and you cannot know the sample size (number of students selected) because we do not
know how many students are in the various major clusters.
2.67
(a) Cluster sampling.
(b) 1/5.
(c) Response bias.
2.68
(a) False.
(b) False.
(c) True.
2.69
(a) A 1-in-40 systematic sample.
(b) 1/40 is the chance that any specific address is chosen since one of the first 40 addresses is picked at random.
The other addresses are directly linked to that first random pick.
2.61
(a) With the calculator the first student selected is 1. With the random number table the first student selected is 1.
(b) The sample size is 6.
2.62
(a) Systematic sampling (1-in-30). Once you select the sample you can determine the number of freshmen you
selected. Also knowing the population list you can calculate the number of freshmen you would select for each
of the 30 possible samples (depending on the starting point.)
(b) Stratified random sampling. Yes, you would have 25 freshmen.
2.63
(a) (i) 0.02(500)+0.03(1200)+0.05(18000) = 10+36+900 = 946.
(ii) Stratified Random Sampling.
(iii) With the calculator the selected labels are 432, 232, 304, 412, and 372. With the random number table, we
might assign the first High category driver the labels 001 and 501. We would assign the second High
category driver 002 and 502. This assignment pattern would continue until the 500th High category driver
who would be assigned the labels 000 and 500. Reading off labels from row 60, column 1, we have: 789,
191, 947, 423, and 632. This would correspond to selecting the 289th, 191st, 447th, 423rd, and 132nd High
category drivers in the list of 500 High category drivers.
(b) (i) With the calculator or the random number table, the first selected label is 15. Thus the selected labels will
be 15, 35, 55, 75, 95, and so on.
(ii) Since the 500 High category drivers divide evenly into groups of 20 (500/20 = 25), there will be a total of
25 High category drivers in the systematic 1-in-20 sample.
2.64
(a) 907.
(b) Cluster sample.
(c) (iii) Selection bias.
2.65
These results are based on a study of 125 aerobic classes in five health clubs, not all aerobic classes in all health
clubs. Thus, the 60% figure is a statistic and the sample size is n = 125.
2.66
(a) It is a systematic 1-in-45 sampling resulting in 50 students (1 from each of the 50 sections, the 33rd in each of
the 50 lists of 45 students).
(b) It is a cluster sample and you cannot know the sample size (number of students selected) because we do not
know how many students are in the various major clusters.
2.67
(a) Cluster sampling.
(b) 1/5.
(c) Response bias.
2.68
(a) False.
(b) False.
(c) True.
2.69
(a) A 1-in-40 systematic sample.
(b) 1/40 is the chance that any specific address is chosen since one of the first 40 addresses is picked at random.
The other addresses are directly linked to that first random pick.
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Solutions to All Exercises 297
2.70
(a) H0: The population mean increase in the GMAT score is 40 points.
H1: The population mean increase in the GMAT score is less than 40 points.
(d) With the calculator the first four selected students are: 209, 218, 7, and 750.
With the random number table the first four selected students are: 070, 569, 762, and 833.
(c) (i) A possible p-value is 0.03.
(ii) Reject H0.
(iii) Type I error.
(iv) Yes, since the p-value was ≤ 0.05, it would also be ≤ 0.10.
(d) (ii) Statistic.
2.71
(a) Multistage, with the first stage being a cluster sample of 3 lab sections and the second stage being a simple
random sample of 25% of the students in the 3 selected labs.
(b) 0.0625.
(c) The 78% is a statistic since it was based on the sample of students surveyed.
2.72
Nonresponse bias is the distortion that can arise because a large number of units selected for the sample do not
respond or refuse to respond, and these nonresponders have a tendency to be different from the responders. So
nonresponse bias has to do with who responds. Response bias is the distortion that can arise because the wording of
a question and the behavior of the interviewer can affect the responses received. So response bias has to do with
how the responders answer.
2.73
(a) Since 15 patients were selected with a 1-in-9 systematic sample, there were at least 15 groups of 9 patients each
or 135 patients in all.
(b) (ii) Statistic.
2.74
(a) Since there is no prior information, we will sample in proportion to the size of the stratum relative to the size of
the population. So the sample size from Stratum I is 3 (since there are 10/40 large facilities in all, 25% of the
sample could be large facilities, for 25% of 12 = 3), and the sample size from Stratum II is 12 – 3 or 9.
(b) Using the calculator, the stratum I selected facilities and responses are: 7 (Yes), 10 (Yes), 8 (No), and the
stratum II selected facilities and responses are: 7 (No), 17 (No), 9 (Yes), 24 (No), 3 (No), 21 (No), 8 (Yes),
12 (No), and 1 (Yes),
Using the random number table we will label the stratum I facilities as: 1 has label 1, Facility 2 has label 2, ... ,
Facility 10 has label 0. Then the selected facilities and responses are: 2 (Yes), 6 (Yes), 4 (Yes). We will label
the stratum II facilities as: 1 has label 01; Facility 2 has label 02; up to Facility 30 has label 30. Then the
selected facilities and responses are: 05 (No), 36 (skip), 60 (skip), 42 (skip), 13 (Yes), 25 (Yes), 66 (skip), 92
(skip), 64 (skip), 22 (Yes), ..., 04 (Yes), 06 (No), 20 (No), 12 (No), 18 (Yes).
(c) Using the calculator we have:== estimateˆp416.0
9
3
40
30
3
2
40
10 =
+
.
Using the table we have:== estimateˆp667.0
9
5
40
30
3
3
40
10 =
+
.
(d) The true population proportion is p = 20/40 = 0.50.
(e) In general, estimates may not be exactly equal to p. The calculator estimate was slightly too small, while the
random number table estimate was too large.
2.75
(a) Answers will vary. See the web site.
(b) Summaries will vary.
2.70
(a) H0: The population mean increase in the GMAT score is 40 points.
H1: The population mean increase in the GMAT score is less than 40 points.
(d) With the calculator the first four selected students are: 209, 218, 7, and 750.
With the random number table the first four selected students are: 070, 569, 762, and 833.
(c) (i) A possible p-value is 0.03.
(ii) Reject H0.
(iii) Type I error.
(iv) Yes, since the p-value was ≤ 0.05, it would also be ≤ 0.10.
(d) (ii) Statistic.
2.71
(a) Multistage, with the first stage being a cluster sample of 3 lab sections and the second stage being a simple
random sample of 25% of the students in the 3 selected labs.
(b) 0.0625.
(c) The 78% is a statistic since it was based on the sample of students surveyed.
2.72
Nonresponse bias is the distortion that can arise because a large number of units selected for the sample do not
respond or refuse to respond, and these nonresponders have a tendency to be different from the responders. So
nonresponse bias has to do with who responds. Response bias is the distortion that can arise because the wording of
a question and the behavior of the interviewer can affect the responses received. So response bias has to do with
how the responders answer.
2.73
(a) Since 15 patients were selected with a 1-in-9 systematic sample, there were at least 15 groups of 9 patients each
or 135 patients in all.
(b) (ii) Statistic.
2.74
(a) Since there is no prior information, we will sample in proportion to the size of the stratum relative to the size of
the population. So the sample size from Stratum I is 3 (since there are 10/40 large facilities in all, 25% of the
sample could be large facilities, for 25% of 12 = 3), and the sample size from Stratum II is 12 – 3 or 9.
(b) Using the calculator, the stratum I selected facilities and responses are: 7 (Yes), 10 (Yes), 8 (No), and the
stratum II selected facilities and responses are: 7 (No), 17 (No), 9 (Yes), 24 (No), 3 (No), 21 (No), 8 (Yes),
12 (No), and 1 (Yes),
Using the random number table we will label the stratum I facilities as: 1 has label 1, Facility 2 has label 2, ... ,
Facility 10 has label 0. Then the selected facilities and responses are: 2 (Yes), 6 (Yes), 4 (Yes). We will label
the stratum II facilities as: 1 has label 01; Facility 2 has label 02; up to Facility 30 has label 30. Then the
selected facilities and responses are: 05 (No), 36 (skip), 60 (skip), 42 (skip), 13 (Yes), 25 (Yes), 66 (skip), 92
(skip), 64 (skip), 22 (Yes), ..., 04 (Yes), 06 (No), 20 (No), 12 (No), 18 (Yes).
(c) Using the calculator we have:== estimateˆp416.0
9
3
40
30
3
2
40
10 =
+
.
Using the table we have:== estimateˆp667.0
9
5
40
30
3
3
40
10 =
+
.
(d) The true population proportion is p = 20/40 = 0.50.
(e) In general, estimates may not be exactly equal to p. The calculator estimate was slightly too small, while the
random number table estimate was too large.
2.75
(a) Answers will vary. See the web site.
(b) Summaries will vary.
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298 Chapter 2
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Solutions to All Exercises 299
Interactive Statistics 3rd Edition: Chapter 3 Full
Solutions
3.1
(a) Explanatory variable: Amount of coffee consumed.
Response variable: Exam performance.
(b) Explanatory variable: Hours of counseling per week.
Response variable: Grade point average.
(c) Explanatory variable: Number of roommates.
Response variable: Physical health at the end of the semester.
(d) Explanatory variable: Type of driver (levels are good or bad).
Response variable: Reaction time on a driving test.
3.2
(a) A possible confounding variable is type of school (public versus parochial).
(b) A possible confounding variable is type of sport of the subject (track – which are used to running, and
wresting).
(c) A possible confounding variable is type of job (modeling – which may have more fun, and temporary staff).
3.3
(a) Experiment.
(b) The response variable is improvement in blood flow. The explanatory variable is the amount of flavonoid
(High versus Placebo).
(c) p-value ≤ 0.05
(d) The 178.8 is a statistic and a sample mean.
3.4
(a) This was an observational study. The explanatory variable of hormone estrogen level was not actively imposed
on the subjects, but rather simply recorded, along with the various quality of sleep measurements.
(b) The explanatory variable is hormone estrogen level. The general response variable is ‘sleep quality’ which was
measured and compared using different criteria --- total sleep time, time spent awake during the night, and
efficiency of sleep time (based on sleep stages and REM sleep).
(c) The null hypothesis could not be refuted, so the p-value must have been larger than 0.05. Two possible values
could be 0.12 and 0.26.
(d) This design technique is called matching (or pairing) and it was done to help reduce the effect of confounding
due to age.
(e) Some are: oxygen saturation level, muscle movement, hot flash patterns.
3.5
(a) Observational study.
(b) Blood cholesterol level.
(c) Frequency of egg consumption.
(d) Diet, age, amount of exercise, and history of high cholesterol are a few possible confounding variables.
3.6
(a) Observational study.
(b) Heart attack status.
(c) Depression status.
(d) Statistic.
(e) Confounding variable.
Interactive Statistics 3rd Edition: Chapter 3 Full
Solutions
3.1
(a) Explanatory variable: Amount of coffee consumed.
Response variable: Exam performance.
(b) Explanatory variable: Hours of counseling per week.
Response variable: Grade point average.
(c) Explanatory variable: Number of roommates.
Response variable: Physical health at the end of the semester.
(d) Explanatory variable: Type of driver (levels are good or bad).
Response variable: Reaction time on a driving test.
3.2
(a) A possible confounding variable is type of school (public versus parochial).
(b) A possible confounding variable is type of sport of the subject (track – which are used to running, and
wresting).
(c) A possible confounding variable is type of job (modeling – which may have more fun, and temporary staff).
3.3
(a) Experiment.
(b) The response variable is improvement in blood flow. The explanatory variable is the amount of flavonoid
(High versus Placebo).
(c) p-value ≤ 0.05
(d) The 178.8 is a statistic and a sample mean.
3.4
(a) This was an observational study. The explanatory variable of hormone estrogen level was not actively imposed
on the subjects, but rather simply recorded, along with the various quality of sleep measurements.
(b) The explanatory variable is hormone estrogen level. The general response variable is ‘sleep quality’ which was
measured and compared using different criteria --- total sleep time, time spent awake during the night, and
efficiency of sleep time (based on sleep stages and REM sleep).
(c) The null hypothesis could not be refuted, so the p-value must have been larger than 0.05. Two possible values
could be 0.12 and 0.26.
(d) This design technique is called matching (or pairing) and it was done to help reduce the effect of confounding
due to age.
(e) Some are: oxygen saturation level, muscle movement, hot flash patterns.
3.5
(a) Observational study.
(b) Blood cholesterol level.
(c) Frequency of egg consumption.
(d) Diet, age, amount of exercise, and history of high cholesterol are a few possible confounding variables.
3.6
(a) Observational study.
(b) Heart attack status.
(c) Depression status.
(d) Statistic.
(e) Confounding variable.
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300 Chapter 3
3.7
(a) Observational study.
(b) Homelessness.
(c) Some are: separation status (whether or not they were separated from their parents), poverty Status (whether or
not they were raised in poverty), family problem status (whether or not they have family problems), abuse status
(whether or not they were sexually or physically abused).
(d) All homeless people in Los Angeles.
(e) The homeless people in Los Angeles that were surveyed.
(f) The 2.5% figure is a statistic since it is based on a sample of homeless people in Los Angeles.
3.8
(a) The response variable is payment status.
(b) The explanatory variable is source of payment.
(c) The levels of the explanatory variable are: private insurance, government, and other.
(d) This is an observational study because the data were obtained from records.
(e) (i) A statistic since it was computed from a sample of 120 private insurance records.
(ii) Stratified random sampling.
(iii) With the TI calculator we have: 377, 1138, 559, 850, and 1010.
With the random number table we have: 705, 488, 448, 168, and 253.
(f) The population is all hospital discharges in the U.S. or perhaps for hospitals in a certain state.
(g) The sample consists of the 2000 records (120 private insurance records, 60 government records, and 20 “other”
records) from the past two months (for the studied hospital.)
3.9
(a) This is a retrospective observational study.
(b) The response variable is car crash status. The explanatory variable is gender.
(c) We should take into account that men typically drive more than women. In this case, "miles driven" is a
confounding variable that can skew your results. Another confounding variable could be weather conditions at
the time of the crash.
3.10
This may not be valid. It is similar to trying to replace a "nonresponder" in a survey with a "responder". There can
be differences in these two groups, those who survive their suicide attempts perhaps had some uncertainty about
their interest, they may have been trying to get help.
3.11
(a) H1.
(b) p-value was less or equal to 0.05.
(c) (iii) Observational study, Prospective.
3.12
(a) This is an observational study and it is of the prospective type. A group of nurses was identified and followed
up over time to ascertain whether they developed heart disease or not.
(b) The response variable is heart disease status (whether or not the subject gets heart disease). The explanatory
variable (risk factor) is the amount of stick margarine in their diet.
3.13
(a) Observational study (retrospective).
(b) Response variable is kidney stones status (whether or not the subject develops kidney stones); Explanatory
variable is amount of calcium in diet.
(c) A diet high in calcium lowers the risk of developing kidney stones in men and women.
3.7
(a) Observational study.
(b) Homelessness.
(c) Some are: separation status (whether or not they were separated from their parents), poverty Status (whether or
not they were raised in poverty), family problem status (whether or not they have family problems), abuse status
(whether or not they were sexually or physically abused).
(d) All homeless people in Los Angeles.
(e) The homeless people in Los Angeles that were surveyed.
(f) The 2.5% figure is a statistic since it is based on a sample of homeless people in Los Angeles.
3.8
(a) The response variable is payment status.
(b) The explanatory variable is source of payment.
(c) The levels of the explanatory variable are: private insurance, government, and other.
(d) This is an observational study because the data were obtained from records.
(e) (i) A statistic since it was computed from a sample of 120 private insurance records.
(ii) Stratified random sampling.
(iii) With the TI calculator we have: 377, 1138, 559, 850, and 1010.
With the random number table we have: 705, 488, 448, 168, and 253.
(f) The population is all hospital discharges in the U.S. or perhaps for hospitals in a certain state.
(g) The sample consists of the 2000 records (120 private insurance records, 60 government records, and 20 “other”
records) from the past two months (for the studied hospital.)
3.9
(a) This is a retrospective observational study.
(b) The response variable is car crash status. The explanatory variable is gender.
(c) We should take into account that men typically drive more than women. In this case, "miles driven" is a
confounding variable that can skew your results. Another confounding variable could be weather conditions at
the time of the crash.
3.10
This may not be valid. It is similar to trying to replace a "nonresponder" in a survey with a "responder". There can
be differences in these two groups, those who survive their suicide attempts perhaps had some uncertainty about
their interest, they may have been trying to get help.
3.11
(a) H1.
(b) p-value was less or equal to 0.05.
(c) (iii) Observational study, Prospective.
3.12
(a) This is an observational study and it is of the prospective type. A group of nurses was identified and followed
up over time to ascertain whether they developed heart disease or not.
(b) The response variable is heart disease status (whether or not the subject gets heart disease). The explanatory
variable (risk factor) is the amount of stick margarine in their diet.
3.13
(a) Observational study (retrospective).
(b) Response variable is kidney stones status (whether or not the subject develops kidney stones); Explanatory
variable is amount of calcium in diet.
(c) A diet high in calcium lowers the risk of developing kidney stones in men and women.
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Solutions to All Exercises 301
3.14
(a) This is an observational study and it is of the retrospective type. Women with breast disease were identified and
then levels of PCBs measured.
(b) These are called confounding variables.
(c) Older women having been exposed for a longer period of time to PCBs might have a higher level of PCBs in
their breast tissue. It would be hard to tell whether it is age or level of PCBs that is increasing the incidence of
malignant breast cancer.
3.15
(a) Observational study (retrospective).
(b) Response variable is Alzheimer’s disease status, explanatory variable is linguistic ability.
(c) The chance of low linguistic ability given a person has Alzheimer’s.
3.16
(a) This study is an observational study --- observations were collected from the U.S. Census. The sample size was
very large, approximately 3 million.
(b) The response variable is divorce status (whether or not the couple has a divorce) and the explanatory variable is
gender of the child (or the number of girl children versus number of boy children).
(c) With such a large sample size, we would expect the various groups being compared to be somewhat balanced
with respect to any other confounding or lurking variables. With such a large sample size, we are not surprised
to see the difference of 5% being statistically significant. Still, a statistically significant difference does not
imply the explanatory variable caused the response. And in fact, Steven’s presents a possible third factor
involving the concept of importance of inherited wealth that fits with the statistics and doesn’t rely on the
preference for boys as the causal variable.
3.17
(a) It is an experiment because the various treatment combinations were actively imposed on the experimental
units.
(b) The experimental units are the metal clutches.
(c) The response variable is the lifetime of the clutch.
(d) The factors or explanatory variables are type of oven and temperature. We have 4 types of oven and 3 levels of
temperature.
(e) There are 12 treatments.
(f) We would need 24 clutches, 2 at each of the 12 treatments.
(g) A design layout table for this experiment is given as:Factor 2:
Temperature
Factor 1: Type of Oven
2 clutches
gears
2 clutches2 clutches
2 clutches
2 clutches
2 clutches
Type 1 Type 2 Type 3
Temp 1
Temp 2
Type 4
2 clutches
2 clutches
Temp 3 2 clutches 2 clutches 2 clutches 2 clutches
3.18
Six treatment combinations, so (6)(4) = 24 experimental units or pizzas.
3.19
(a) Durability.
(b) Dye color and type of cloth.
(c) 20
(d) 20 x 6 = 120.
3.14
(a) This is an observational study and it is of the retrospective type. Women with breast disease were identified and
then levels of PCBs measured.
(b) These are called confounding variables.
(c) Older women having been exposed for a longer period of time to PCBs might have a higher level of PCBs in
their breast tissue. It would be hard to tell whether it is age or level of PCBs that is increasing the incidence of
malignant breast cancer.
3.15
(a) Observational study (retrospective).
(b) Response variable is Alzheimer’s disease status, explanatory variable is linguistic ability.
(c) The chance of low linguistic ability given a person has Alzheimer’s.
3.16
(a) This study is an observational study --- observations were collected from the U.S. Census. The sample size was
very large, approximately 3 million.
(b) The response variable is divorce status (whether or not the couple has a divorce) and the explanatory variable is
gender of the child (or the number of girl children versus number of boy children).
(c) With such a large sample size, we would expect the various groups being compared to be somewhat balanced
with respect to any other confounding or lurking variables. With such a large sample size, we are not surprised
to see the difference of 5% being statistically significant. Still, a statistically significant difference does not
imply the explanatory variable caused the response. And in fact, Steven’s presents a possible third factor
involving the concept of importance of inherited wealth that fits with the statistics and doesn’t rely on the
preference for boys as the causal variable.
3.17
(a) It is an experiment because the various treatment combinations were actively imposed on the experimental
units.
(b) The experimental units are the metal clutches.
(c) The response variable is the lifetime of the clutch.
(d) The factors or explanatory variables are type of oven and temperature. We have 4 types of oven and 3 levels of
temperature.
(e) There are 12 treatments.
(f) We would need 24 clutches, 2 at each of the 12 treatments.
(g) A design layout table for this experiment is given as:Factor 2:
Temperature
Factor 1: Type of Oven
2 clutches
gears
2 clutches2 clutches
2 clutches
2 clutches
2 clutches
Type 1 Type 2 Type 3
Temp 1
Temp 2
Type 4
2 clutches
2 clutches
Temp 3 2 clutches 2 clutches 2 clutches 2 clutches
3.18
Six treatment combinations, so (6)(4) = 24 experimental units or pizzas.
3.19
(a) Durability.
(b) Dye color and type of cloth.
(c) 20
(d) 20 x 6 = 120.
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302 Chapter 3
3.20
(a) Progress status.
(b) Teacher and Method of teaching.
(c) (Grade)(Teacher)(# students) = (5)(4)(6) = 120 students.
3.21
(a) Batches of feed stock.
(b) Yield of the process.
(c) Temperature (2 levels) and Stirring Rate (3 levels).
(d) 6 treatments.
(e) 24 units.
(f) Design Layout Table:
3.22
(a) The experimental units are the subjects who responded to a local questionnaire and who own other advanced
electronic equipment at home.
(b) The response variables are: subject's recall of the ad, attitude toward the camera, and intention to purchase it.
(c) Factor 1 is length (2 levels) and Factor 2 is frequency (3 levels).
(d) The number of treatments is 6, obtained by multiplying the 2 levels for Factor 1 by the 3 levels for Factor 2.
(e) One way to construct the design layout table is given below.Factor 1: Length
30 sec 90 sec
Factor 2:
Frequency
1
3
5
3.23
(a) Weight.
(b) Fiber content and carbohydrate content.
(c) Fiber levels (3) Low, Medium and High. Carbohydrate levels (2) Low and Medium.
(d) 3 x 2 x 20 = 120.
3.24
(a) Number of apple per tree.
(b) Type of Fertilizer and Soil Moisture status.
(c) Type of fertilizer: 3 levels. Soil moisture: two levels.
(d) 3x2x3 = 18.
3.25
Answer is (d).
3.26
The correct answer is (a), treatments are assigned to subjects.
3.27
3.20
(a) Progress status.
(b) Teacher and Method of teaching.
(c) (Grade)(Teacher)(# students) = (5)(4)(6) = 120 students.
3.21
(a) Batches of feed stock.
(b) Yield of the process.
(c) Temperature (2 levels) and Stirring Rate (3 levels).
(d) 6 treatments.
(e) 24 units.
(f) Design Layout Table:
3.22
(a) The experimental units are the subjects who responded to a local questionnaire and who own other advanced
electronic equipment at home.
(b) The response variables are: subject's recall of the ad, attitude toward the camera, and intention to purchase it.
(c) Factor 1 is length (2 levels) and Factor 2 is frequency (3 levels).
(d) The number of treatments is 6, obtained by multiplying the 2 levels for Factor 1 by the 3 levels for Factor 2.
(e) One way to construct the design layout table is given below.Factor 1: Length
30 sec 90 sec
Factor 2:
Frequency
1
3
5
3.23
(a) Weight.
(b) Fiber content and carbohydrate content.
(c) Fiber levels (3) Low, Medium and High. Carbohydrate levels (2) Low and Medium.
(d) 3 x 2 x 20 = 120.
3.24
(a) Number of apple per tree.
(b) Type of Fertilizer and Soil Moisture status.
(c) Type of fertilizer: 3 levels. Soil moisture: two levels.
(d) 3x2x3 = 18.
3.25
Answer is (d).
3.26
The correct answer is (a), treatments are assigned to subjects.
3.27
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Solutions to All Exercises 303
The rats are labeled 1 through 8. Use a calculator with seed= 1209 or a random number table with row=24,
column=6 to select the 4 rats to receive the treatment. The other 4 rats will not receive the treatment. At the end of
the week the effectiveness of the vaccine against the virus will be measured.
3.28
Assign a number to each dog. Rover=1, Spot=2, Jasmin=3, Chelsea=4, Klaus=5, Charlie=6, Maggie=7, Bo=8. With
the calculator use the seed=27 to select the first 4 different numbers at random from 1-8. The dogs with those
numbers will receive the treatment. The first four numbers are: 2, 1, 6, and 5. With the random numbers table, the
four different numbers selected are: 8, 3, 4, and 7.
3.29
(a) Experiment.
(b) Cause of death.
(c) Estrogen therapy with two levels: receive the treatment, do not receive the treatment.
(d) Placebo.
(e) H0: Women who receive Estrogen therapy do not have a lower risk of dying from a heart disease than women
who do not receive the Estrogen therapy
H1: Women who receive Estrogen therapy have a lower risk of dying from a heart disease than women who do
not receive the Estrogen therapy.
(f) The p-value was less than or equal to 0.05.
(g) Yes, Type I error.
3.30
(a) Here is one possible allocation scheme that we will use: The first 10 randomly selected subjects will be
assigned to the aspirin group. The remaining 10, which do not have to be randomly selected because they are all
that are left, will be assigned to the placebo group. Another possible allocation scheme would be to alternate:
The first randomly selected subject will be assigned to the aspirin group, the next randomly selected subject will
be assigned to the placebo group, and so on, until all subjects are assigned to one of the two groups. We will use
the first allocation scheme—the first 10 selected will be allocated to the treatment group.
(b) Using the calculator, the first 10 selected labels are: 4, 19, 16, 17, 2, 13, 11, 10, 9, and 5. So the subjects in
Group 1 (treatment group) are: Steve, James, Eric, Ralph, Kyle, Stan, Ed, Herb, Ryan, and Pablo. Thus, the
remaining subjects are assigned to the placebo group: Lee, Mark, Bill, Matt, Tim, Robb, Phil, Mike, Doug, and
Henry. Using the table, the first 10 selected labels are: 06, 16, 07, 05, 15, 04, 17, 03, 12, and 19. So the
subjects in Group 1 (treatment group) are: Bill, Eric, Matt, Pablo, Mike, Steve, Ralph, Mark, Robb, and James.
Thus the remaining subjects are assigned to the placebo group: Lee, Kyle, Tim, Ryan, Herb, Ed, Stan, Phil,
Doug, and Henry.
3.31
(a) With the calculator the selected rats were: 14, 13, 5, 16, 9, 3, 4, 2, 7, and 10. With the random number table the
selected rats were: 4, 13, 8, 5, 12, 15, 2, 6, 9, and 20.
(b) Have another researcher make and record the measurements.
3.32
(a) Completely randomized block design.
(b) Paired design.
(c) Randomized block design.
3.33
Blocking can help to reduce the bias due to possible confounding variables that you know of and thus build into the
design of the experiment. Randomization can then be used to help reduce the bias due to other possible confounding
variables that you do not know of or have not measured (sometimes these are referred to as lurking variables).
The rats are labeled 1 through 8. Use a calculator with seed= 1209 or a random number table with row=24,
column=6 to select the 4 rats to receive the treatment. The other 4 rats will not receive the treatment. At the end of
the week the effectiveness of the vaccine against the virus will be measured.
3.28
Assign a number to each dog. Rover=1, Spot=2, Jasmin=3, Chelsea=4, Klaus=5, Charlie=6, Maggie=7, Bo=8. With
the calculator use the seed=27 to select the first 4 different numbers at random from 1-8. The dogs with those
numbers will receive the treatment. The first four numbers are: 2, 1, 6, and 5. With the random numbers table, the
four different numbers selected are: 8, 3, 4, and 7.
3.29
(a) Experiment.
(b) Cause of death.
(c) Estrogen therapy with two levels: receive the treatment, do not receive the treatment.
(d) Placebo.
(e) H0: Women who receive Estrogen therapy do not have a lower risk of dying from a heart disease than women
who do not receive the Estrogen therapy
H1: Women who receive Estrogen therapy have a lower risk of dying from a heart disease than women who do
not receive the Estrogen therapy.
(f) The p-value was less than or equal to 0.05.
(g) Yes, Type I error.
3.30
(a) Here is one possible allocation scheme that we will use: The first 10 randomly selected subjects will be
assigned to the aspirin group. The remaining 10, which do not have to be randomly selected because they are all
that are left, will be assigned to the placebo group. Another possible allocation scheme would be to alternate:
The first randomly selected subject will be assigned to the aspirin group, the next randomly selected subject will
be assigned to the placebo group, and so on, until all subjects are assigned to one of the two groups. We will use
the first allocation scheme—the first 10 selected will be allocated to the treatment group.
(b) Using the calculator, the first 10 selected labels are: 4, 19, 16, 17, 2, 13, 11, 10, 9, and 5. So the subjects in
Group 1 (treatment group) are: Steve, James, Eric, Ralph, Kyle, Stan, Ed, Herb, Ryan, and Pablo. Thus, the
remaining subjects are assigned to the placebo group: Lee, Mark, Bill, Matt, Tim, Robb, Phil, Mike, Doug, and
Henry. Using the table, the first 10 selected labels are: 06, 16, 07, 05, 15, 04, 17, 03, 12, and 19. So the
subjects in Group 1 (treatment group) are: Bill, Eric, Matt, Pablo, Mike, Steve, Ralph, Mark, Robb, and James.
Thus the remaining subjects are assigned to the placebo group: Lee, Kyle, Tim, Ryan, Herb, Ed, Stan, Phil,
Doug, and Henry.
3.31
(a) With the calculator the selected rats were: 14, 13, 5, 16, 9, 3, 4, 2, 7, and 10. With the random number table the
selected rats were: 4, 13, 8, 5, 12, 15, 2, 6, 9, and 20.
(b) Have another researcher make and record the measurements.
3.32
(a) Completely randomized block design.
(b) Paired design.
(c) Randomized block design.
3.33
Blocking can help to reduce the bias due to possible confounding variables that you know of and thus build into the
design of the experiment. Randomization can then be used to help reduce the bias due to other possible confounding
variables that you do not know of or have not measured (sometimes these are referred to as lurking variables).
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304 Chapter 3
3.34
(a) A treatment was actively imposed.
(b) With calculator: 2651, 4194, 4510, 6646, 754. With random numbers table: 3744, 6206, 6945, 4690, 405.
(c) 240.
(d) Neither the doctor nor the patients know who receive each treatment.
(e) (1) We don't know whether the result of the study would be published if the data were not favorable to
Celebrex.
(2) Maybe the other two drugs need more time to produce relieve.
3.35
Answers will vary.
3.36
Answers will vary.
3.37
Answers will vary.
3.38
Answers will vary.
3.39
Observational study. No active treatment was imposed.
3.40
For the TI calculator with seed 133, we have:
Treatment 1: (2, 7); Treatment 2: (8, 4); Treatment 3: (3, 6); Treatment 4: (1, 5)
For the random number table with row 18 and column 1, we have:
Treatment 1: (1, 5); Treatment 2: (4, 2); Treatment 3: (3, 6); Treatment 4: (7, 8)
3.41
(a) Temperature (3 levels) and Baking Time (2 levels).
(b) Taste.
(c) 6 treatments.
(d) 36 batches.
3.42
(a) There are 3 explanatory variables: time (3 levels), temperature (3 levels), and cheese (2 levels). So we have
3x3x2 = 18 treatments.
(b) With 5 batches for each treatment, we have 5x18 = 90 total units.
3.43
Lack of blinding of the subjects and of the experimenter(s) or evaluator(s).
3.44
(a) An experiment. A treatment was actively imposed.
(b) Explanatory variable: Whether or not marigolds are planted with broccoli.
Response variable: nematode damage status.
(c) Control for confounding variables.
3.45
(a) Experiment.
(b) The response variable is skin rash status. The explanatory variable is skin rash treatment.
(c) Using the calculator, the selected labels are: 10, 14, 23, 12, 11, 18, 3, 25, 26, 22, 27, 24, 13, 2, and 4. Using the
random number table, the selected labels are: 09, 26, 06, 17, 13, 25, 18, 10, 16, 19, 24, 28, 14, 30, and 02.
(d) The 20% value is a statistic.
(e) Two possible values are 0.08 and 0.10. Any two values larger than 0.05 (but less than 1) would work.
3.34
(a) A treatment was actively imposed.
(b) With calculator: 2651, 4194, 4510, 6646, 754. With random numbers table: 3744, 6206, 6945, 4690, 405.
(c) 240.
(d) Neither the doctor nor the patients know who receive each treatment.
(e) (1) We don't know whether the result of the study would be published if the data were not favorable to
Celebrex.
(2) Maybe the other two drugs need more time to produce relieve.
3.35
Answers will vary.
3.36
Answers will vary.
3.37
Answers will vary.
3.38
Answers will vary.
3.39
Observational study. No active treatment was imposed.
3.40
For the TI calculator with seed 133, we have:
Treatment 1: (2, 7); Treatment 2: (8, 4); Treatment 3: (3, 6); Treatment 4: (1, 5)
For the random number table with row 18 and column 1, we have:
Treatment 1: (1, 5); Treatment 2: (4, 2); Treatment 3: (3, 6); Treatment 4: (7, 8)
3.41
(a) Temperature (3 levels) and Baking Time (2 levels).
(b) Taste.
(c) 6 treatments.
(d) 36 batches.
3.42
(a) There are 3 explanatory variables: time (3 levels), temperature (3 levels), and cheese (2 levels). So we have
3x3x2 = 18 treatments.
(b) With 5 batches for each treatment, we have 5x18 = 90 total units.
3.43
Lack of blinding of the subjects and of the experimenter(s) or evaluator(s).
3.44
(a) An experiment. A treatment was actively imposed.
(b) Explanatory variable: Whether or not marigolds are planted with broccoli.
Response variable: nematode damage status.
(c) Control for confounding variables.
3.45
(a) Experiment.
(b) The response variable is skin rash status. The explanatory variable is skin rash treatment.
(c) Using the calculator, the selected labels are: 10, 14, 23, 12, 11, 18, 3, 25, 26, 22, 27, 24, 13, 2, and 4. Using the
random number table, the selected labels are: 09, 26, 06, 17, 13, 25, 18, 10, 16, 19, 24, 28, 14, 30, and 02.
(d) The 20% value is a statistic.
(e) Two possible values are 0.08 and 0.10. Any two values larger than 0.05 (but less than 1) would work.
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