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Exam
Name___________________________________
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) A college alumni fund appeals for donations by phoning or emailing recent graduates. A random
sample of 300 alumni shows that 40% of the 150 who were contacted by telephone actually made
contributions compared to only 30% of the 150 who received email requests. Which formula
calculates the 98% confidence interval for the difference in the proportions of alumni who may
make donations if contacted by phone or by email?
A) (0.40 – 0.30) ± 2.33 (0.35)(0.65)
300
B) (0.40 – 0.30) ± 2.33 (0.40)(0.60)
150 + (0.30)(0.70)
150
C) (0.40 – 0.30) ± 2.33 (0.40)(0.60)
300 + (0.30)(0.70)
300
D) (0.40 – 0.30) ± 2.33 (0.35)(0.65)
150
E) (0.40 – 0.30) ± 2.33 (0.35)(0.65)
150 + (0.35)(0.65)
150
Answer: B
Explanation: A)
B)
C)
D)
E)
1)
12) An online catalog company wants on–time delivery for at least 90% of the orders they ship. They
have been shipping orders via UPS and FedEx but will switch to a more expensive service
(ShipFast) if there is evidence that this service can exceed the 90% on–time goal. As a test the
company sends a random sample of orders via ShipFast, and then makes follow–up phone calls to
see if these orders arrived on time. Which hypotheses should they test?
A) H0: p = 0.90
HA: p < 0.90
B) H0: p > 0.90
HA: p = 0.90
C) H0: p = 0.90
HA: p 0.90
D) H0: p < 0.90
HA: p = 0.90
E) H0: p = 0.90
HA: p > 0.90
Answer: E
Explanation: A)
B)
C)
D)
E)
3) A researcher investigating whether joggers are less likely to get colds than people who do not jog
found a P–value of 3%. This means that:
A) There’s a 3% chance that joggers get fewer colds.
B) There’s a 3% chance that joggers don’t get fewer colds.
C) None of these.
D) Joggers get 3% fewer colds than non–joggers.
E) 3% of joggers get colds.
Answer: C
Explanation: A)
B)
C)
D)
E)
2
2)
3)4) To plan the course offerings for the next year a university department dean needs to estimate what
impact the “No Child Left Behind” legislation might have on the teacher credentialing program.
Historically, 40% of this university’s pre–service teachers have qualified for paid internship
positions each year. The Dean of Education looks at a random sample of internship applications to
see what proportion indicate the applicant has achieved the content–mastery that is required for
the internship. Based on these data he creates a 90% confidence interval of (33%, 41%). Could this
confidence interval be used to test the hypothesis H0: p = 0.40 versus HA: p < 0.40 at the = 0.05
level of significance?
A) No, because the dean only reviewed a sample of the applicants instead of all of them.
B) Yes, since 40% is not the center of the confidence interval he rejects the null hypothesis,
concluding that the percentage of qualified applicants will decrease.
C) Yes, since 40% is in the confidence interval he accepts the null hypothesis, concluding that the
percentage of applicants qualified for paid internship positions will stay the same.
D) Yes, since 40% is in the confidence interval he fails to reject the null hypothesis, concluding
that there is not strong enough evidence of any change in the percent of qualified applicants.
E) No, because the dean should have used a 95% confidence interval.
Answer: D
Explanation: A)
B)
C)
D)
E)
5) To plan the budget for next year a college needs to estimate what impact the current economic
downturn might have on student requests for financial aid. Historically, this college has provided
aid to 35% of its students. Officials look at a random sample of this year’s applications to see what
proportion indicate a need for financial aid. Based on these data they create a 90% confidence
interval of (32%, 40%). Could this interval be used to test the hypothesis H0: p = 0.35 versus HA: p
0.35 at the = 0.10 level of significance?
A) Yes; since 35% is in the confidence interval they fail to reject the null hypothesis, concluding
that there is not strong evidence of any change in financial aid requests.
B) No, because financial aid amounts may not be normally distributed.
C) Yes; since 35% is in the confidence interval they accept the null hypothesis, concluding that
the percentage of students requiring financial aid will stay the same.
D) No, because they only used a sample of the applicants instead of all of them.
E) Yes; since 35% is not at the center of the confidence interval they reject the null hypothesis,
concluding that the percentage of students requiring aid will increase.
Answer: A
Explanation: A)
B)
C)
D)
E)
3
4)
5)6) A pharmaceutical company investigating whether drug stores are less likely than food stores to
remove over–the–counter drugs from the shelves when the drugs are past the expiration date
found a P–value of 2.8%. This means that:
A) There is a 2.8% chance the drug stores remove more expired over–the–counter drugs.
B) 97.2% more drug stores remove over–the–counter drugs from the shelves when the drugs are
past the expiration date than food stores.
C) There is a 97.2% chance the drug stores remove more expired over–the–counter drugs.
D) 2.8% more drug stores remove over–the–counter drugs from the shelves when the drugs are
past the expiration date.
E) None of these.
Answer: E
Explanation: A)
B)
C)
D)
E)
7) A statistics professor wants to see if more than 80% of her students enjoyed taking her class. At the
end of the term, she takes a random sample of students from her large class and asks, in an
anonymous survey, if the students enjoyed taking her class. Which set of hypotheses should she
test?
A) H0: p > 0.80
HA: p = 0.80
B) H0: p < 0.80
HA: p 0.80
C) H0: p < 0.80
HA: p > 0.80
D) H0: p = 0.80
HA: p < 0.80
E) H0: p = 0.80
HA: p > 0.80
Answer: E
Explanation: A)
B)
C)
D)
E)
6)
7)
48) Suppose that a conveyor used to sort packages by size does not work properly. We test the
conveyor on several packages (with H0: incorrect sort) and our data results in a P–value of 0.016.
What probably happens as a result of our testing?
A) We reject H0, making a Type II error.
B) We fail to reject H0, committing a Type II error.
C) We correctly reject H0.
D) We correctly fail to reject H0.
E) We reject H0, making a Type I error.
Answer: E
Explanation: A)
B)
C)
D)
E)
9) A relief fund is set up to collect donations for the families affected by recent storms. A random
sample of 400 people shows that 28% of those 200 who were contacted by telephone actually made
contributions compared to only 18% of the 200 who received first class mail requests. Which
formula calculates the 95% confidence interval for the difference in the proportions of people who
make donations if contacted by telephone or first class mail?
A) (0.28 – 0.18) ± 1.96 (0.28)(0.72)
200 + (0.18)(0.82)
200
B) (0.28 – 0.18) ± 1.96 (0.23)(0.77)
400
C) (0.28 – 0.18) ± 1.96 (0.23)(0.77)
200
D) (0.28 – 0.18) ± 1.96 (0.23)(0.77)
200 + (0.23)(0.77)
200
E) (0.28 – 0.18) ± 1.96 (0.28)(0.72)
400 + (0.18)(0.82)
400
Answer: A
Explanation: A)
B)
C)
D)
E)
8)
9)
510) Suppose that a manufacturer is testing one of its machines to make sure that the machine is
producing more than 97% good parts (H0 : p = 0.97 and HA : p > 0.97) . The test results in a
P–value of 0.122. Unknown to the manufacturer, the machine is actually producing 99% good
parts. What probably happens as a result of the testing?
A) They fail to reject H0, making a Type I error.
B) They correctly fail to reject H0.
C) They correctly reject H0.
D) They fail to reject H0, making a Type II error.
E) They reject H0, making a Type I error.
Answer: D
Explanation: A)
B)
C)
D)
E)
11) We test the hypothesis that p = 35% versus p < 35%. We don’t know it but actually p = 26%. With
which sample size and significance level will our test have the greatest power?
A) = 0.03, n = 250
B) = 0.01, n = 400
C) The power will be the same as long as the true proportion p remains 26%
D) = 0.01, n = 250
E) = 0.03, n = 400
Answer: E
Explanation: A)
B)
C)
D)
E)
12) We will test the hypothesis that p = 60% versus p > 60%. We don’t know it, but actually p is 70%.
With which sample size and significance level will our test have the greatest power?
A) = 0.05, n = 200
B) = 0.01, n = 500
C) The power will be the same so long as the true proportion p remains 70%.
D) = 0.05, n = 500
E) = 0.01, n = 200
Answer: D
Explanation: A)
B)
C)
D)
E)
6
10)
11)
12)13) Suppose that a device advertised to increase a car’s gas mileage really does not work. We test it on a
small fleet of cars (with H0: not effective), and our data results in a P–value of 0.004. What
probably happens as a result of our experiment?
A) We reject H0, making a Type II error.
B) We correctly reject H0.
C) We reject H0, making a Type I error.
D) We fail to reject H0, committing aType II error.
E) We correctly fail to reject H0.
13)
Answer: C
Explanation: A)
B)
C)
D)
E)
14) A P–value indicates 14)
A) the probability of the observed statistic given that the null hypothesis is true.
B) the probability the null is true given the observed statistic.
C) the probability of the observed statistic given that the alternative hypothesis is true.
D) the probability that the null hypothesis is true.
E) the probability that the alternative hypothesis is true.
Answer: A
Explanation: A)
B)
C)
D)
E)
715) A truck company wants on–time delivery for 98% of the parts they order from a metal
manufacturing plant. They have been ordering from Hudson Manufacturing but will switch to a
new, cheaper manufacturer (Steel–R–Us) unless there is evidence that this new manufacturer
cannot meet the 98% on–time goal. As a test the truck company purchases a random sample of
metal parts from Steel–R–Us, and then determines if these parts were delivered on–time. Which
hypothesis should they test?
A) H0: p = 0.98
HA: p > 0.98
B) H0: p > 0.98
HA: p = 0.98
C) H0: p = 0.98
HA: p < 0.98
D) H0: p < 0.98
HA: p > 0.98
E) H0: p = 0.98
HA: p 0.98
Answer: C
Explanation: A)
B)
C)
D)
E)
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
16) Approval rating A newspaper article reported that a poll based on a sample of 1150
residents of a state showed that the state’s Governor’s job approval rating stood at 58%.
They claimed a margin of error of ±3%. What level of confidence were the pollsters using?
^^
Answer:
Since ME = z* pq
, we have 0.03 = z* (0.58)(0.42)
n
1150 or z* 2.06. Confidence level is
16)
96%.
15)
Explanation:
The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. A
marketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latest
jeans…). He loans her several different “people” to scatter around the store for a two–week trial period. The owner carefully
counts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’ll
purchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy something
increases.
17) Over the trial month the rate of in–store sales rose to 30% of shoppers. The store’s owner
decided this increase was statistically significant. Now that she’s convinced the
mannequins work, why might she still choose not to purchase them?
Answer: Although statistically significant, the small increase in sales from 28% to 30% of
shoppers might not be enough to justify the expense of purchasing the mannequins.
Explanation:
17)
8A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concerned
that employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defective
chips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentive
program is instituted for one month. If successful, the company will continue with the incentive program.
18) Based on the data they collected during the trial program, management found that a 95%
18)
confidence interval for the percentage of defective chips was (5.0%, 7.0%). What conclusion
should management reach about the new incentive program? Explain.
Answer: The confidence interval contains values that are all below the hypothesized value of
8%, so the data provide convincing evidence that the incentive program lowers the
defect rate of the computer chips.
Explanation:
19) Births A city has two hospitals, with many more births recorded at the larger hospital than
at the smaller one. Records indicate that in general babies are about equally likely to be
boys or girls, but the actual gender ratio varies from week to week. Which hospital is more
likely to report a week when over two–thirds of the babies born were girls? Explain.
Answer: The smaller hospital should experience greater variability in the weekly percentage
of male babies. We’d expect the larger hospital to stay closer to the expected 50–50
ratio.
19)
Explanation:
A state’s Department of Education reports that 12% of the high school students in that state attend private high schools. The
State University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a random
sample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attend
private schools.
20) Interpret the confidence interval in this context. 20)
Answer: We are 90% confident that between 7.9% and 12.6% of the applicants attend private
high schools.
Explanation:
921)
P = P(T > 126) = P(z > 1.5)
21) Pumpkin pie A can of pumpkin pie mix contains a mean of 30 ounces and a standard
deviation of 2 ounces. The contents of the cans are normally distributed. What is the
probability that four randomly selected cans of pumpkin pie mix contain a total of more
than 126 ounces?
Answer: Two methods are shown below to solve this problem:
Method 1:
Let P = one can of pumpkin pie mix and T = four cans of pumpkin pie mix.
We are told that the contents of the cans are normally distributed, and can assume
that the content amounts are independent from can to can.
E(T) = E(P1 + P2 + P3 + P4) = E(P1) + E(P2) + E(P3) + E(P4) = 120 ounces
Since the content amounts are independent,
Var(T) = Var(P1 + P2 + P3 + P4) = Var(P1) + Var(P2) + Var(P3) + Var(P4) = 16
SD(T) = Var(T) = 16 = 4 ounces
We model T with N(120, 4) z = 126 – 120
4 = 1.5 = 0.067
There is a 6.7% chance that four randomly selected cans of pumpkin pie mix contain
more than 126 ounces.
Method 2:
Using the Central Limit Theorem approach, let y = average content of cans in
sample
Since the contents are Normally distributed, y is modeled by N 30, 2
P y > 126
4 = P(y > 31.5) = P z > 31.5 – 30
1 = P(z > 1.5) = 0.067
There is about a 6.7% chance that 4 randomly selected cans will contain a total of
over 126 ounces.
Explanation:
.
4
1022) Employment program A city council must decide whether to fund a new
“welfare–to–work” program to assist long–time unemployed people in finding jobs. This
program would help clients fill out job applications and give them advice about dealing
with job interviews. A six–month trial has just ended. At the start of this trial a number of
unemployed residents were randomly divided into two groups; one group went through
the help program and the other group did not. Data about employment at the end of this
trial are shown in the table. Should the city council fund this program? Test an appropriate
hypothesis and state your conclusion.
22)
Answer:
H0 : p1 – p2 = 0 HA : p1 – p2 > 0
People were randomly assigned to groups, we assume the groups are independent,
and 20, 34, 13, 33 are all 10. OK to do a 2–proportion z–test.
^ ^ ^
p1 = 0.370, p2 = 0.283, p = 20 + 13
54 + 46 = 0.33
z = (0.370 – 0.283) – 0
(0.33)(0.67)
54 – (0.33)(0.67)
46
= 0.93
^ ^
P = P(p1 – p2 > 0.087) = P(z > 0.93) = 0.176
We fail to reject the null hypothesis because P is very large. We do not have
evidence that the help program is beneficial, so it should not be funded.
Explanation:
A state’s Department of Education reports that 12% of the high school students in that state attend private high schools. The
State University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a random
sample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attend
private schools.
23) The admissions officers want to estimate the true percentage of private school applicants to
within ±4%, with 90% confidence. How many applications should they sample?
23)
Answer:
^^
ME = z
* pq
n
0.04 = 1.645 (0.12)(0.88)
n
n = 1.645 (0.12)(0.88)
0.04
n = 178.60 179
^
They should sample at least 179 applicants. (423 if p = 0.5 is used)
Explanation:
1124) Depression A recent psychiatric study from the University of Southampton observed a
higher incidence of depression among women whose birth weight was less than 6.6
pounds than in women whose birth weight was over 6.6 pounds. Based on a P–value of
0.0248 the researchers concluded there was evidence that low birth weights may be a risk
factor for susceptibility to depression. Explain in context what the reported P–value means.
Answer: If birth weight was not a risk factor for susceptibility to depression, an observed
difference in incidence of depression this large (or larger) would occur in only 2.48%
of such samples.
Explanation:
24)
The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. A
marketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latest
jeans…). He loans her several different “people” to scatter around the store for a two–week trial period. The owner carefully
counts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’ll
purchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy something
increases.
25) In this context describe a Type I error and the impact such an error would have on the
25)
store.
Answer: A Type I error would be deciding the percentage of customers who’ll make
purchases will go up when in fact it won’t. The store’s owner would waste money
buying useless mannequins.
Explanation:
The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of females
in the labor force is the same in the United States. Representatives from the United States Department of Labor plan to check
a random sample of over 10,000 employment records on file to estimate a percentage of females in the United States labor
force.
26) Interpret the confidence interval in this context. Answer: We are 90% confident that between 40.0% and 47.2% of the employment records
from the United States labor force are for females.
26)
Explanation:
12Great Britain has a great literary tradition that spans centuries. One might assume, then, that Britons read more than citizens
of other countries. Some Canadians, however, feel that a higher percentage of Canadians than Britons read. A recent Gallup
Poll reported that 86% of 1004 randomly sampled Canadians read at least one book in the past year, compared to 81% of 1009
randomly sampled Britons. Do these results confirm a higher reading rate in Canada?
27) Find a 99% confidence interval for the difference in the proportion of Britons and
27)
Canadians who read at least one book in the last year. Interpret your interval.
Answer:
With the conditions satisfied (from Problem 1), the sampling distribution of the
difference in
proportions is approximately Normal with a mean of pB – pC, the true difference
between the
population proportions. We can find a two–proportion z–interval.
^ ^
We know: nB = 1009, pB = 0.81, nC = 1004, pC = 0.86
^ ^
We estimate SD(pB – pC) as
^ ^ ^ ^
^ ^
SE(pB – pC) = pB qB
nB + pC qC
nC = (0.81)(0.19)
1009 + (0.86)(0.14)
1004 = 0.0165
^ ^
ME = z* × SE(pB – pC) = 2.576(0.0165) = 0.0425
^ ^
The observed difference in sample proportions = pB – pC = 0.81 – 0.186 = –0.05, so
the 99%
confidence interval is –0.05 ± 0.0425, or –9.3% to –0.8%.
We are 99% confident that the proportion of Britons who read at least one book in
the past year is
between 0.8–percentage points and 9.3–percentage points lower than the proportion
of Canadians who read at least one book in the past year.
Explanation:
The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Crest
toothpaste. A marketing director suggests that the company invest in a new marketing campaign which will include
advertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for one
month to determine if the market share increases with new labels.
28) Write the company’s null and alternative hypotheses. 28)
Answer: The null hypothesis is that 19.5% of all people who use toothpaste buy Crest. The
alternative hypothesis is that the percentage of all people who use toothpaste who
use Crest is greater than 19.5%. In symbols: H0: p = 0.195 and HA: p > 0.195
Explanation:
13A report on health care in the US said that 28% of Americans have experienced times when they haven’t been able to afford
medical care. A news organization randomly sampled 801 black Americans, of whom 38% reported that there had been
times in the last year when they had not been able to afford medical care. Does this indicate that this problem is more severe
among black Americans?
29) Test an appropriate hypothesis and state your conclusion. (Make sure to check any
29)
necessary conditions and to state a conclusion in the context of the problem.)
Answer:
Hypotheses: H0 : p = 0.28. The proportion of all black Americans that were unable
to afford medical care in the last year is 28%.
HA : p > 0.28. The proportion of all black Americans that were unable
to afford medical care in the last year is greater than 28%.
Model: Okay to use the Normal model because the sample is random, these 801
black Americans are less than 10% of all black Americans, and
np = (801)(0.28) = 224.28 10 and nq = (801)(0.72)= 576.72 10.
We will do a one–proportion z–test.
^
Mechanics: n = 801, p = 0.38
z = 0.38 – 0.28
(0.28)(0.72)
801
^
P(p > 0.38) = P z > 6.29 0
Conclusion: With a P–value so small (just about zero), I reject the null hypothesis.
There is enough evidence to suggest that the proportion of black Americans who
were not able to afford medical care in the past year is more than 28%.
Explanation:
1430) It is generally believed that nearsightedness affects about 12% of children. A school district
gives vision tests to 133 incoming kindergarten children.
a. Describe the sampling distribution model for the sample proportion by naming the
model and telling its mean and standard deviation. Justify your answer.
b. Sketch and clearly label the model.
c. What is the probability that in this group over 15% of the children will be found to be
nearsighted?
Answer:
a. We can assume these kids are a random sample of all children, and
certainly less than 10% of them. We expect np = (133)(0.12) = 15.96
successes and 117.04 failures so the sample size is large enough to use
the sampling model N(0.12, 0.028).
b.
30)
c. z = 0.15 – 0.12
0.028 = 1.07
^
P(p > 0.15) = P(z > 1.07) = 0.142
Explanation:
The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of females
in the labor force is the same in the United States. Representatives from the United States Department of Labor plan to check
a random sample of over 10,000 employment records on file to estimate a percentage of females in the United States labor
force.
31) Explain what 90% confidence means in this context. Answer: If many random samples were taken, 90% of the confidence intervals produced
would contain the actual percentage of all female employment records in the United
States labor force.
31)
Explanation:
The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. A
marketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latest
jeans…). He loans her several different “people” to scatter around the store for a two–week trial period. The owner carefully
counts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’ll
purchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy something
increases.
32) What alpha level did the store’s owner use? Answer: = 1 – 0.98
2 = 0.01 = 1%
Explanation:
32)
1533) In this context describe a Type II error and the impact such an error would have on the
33)
store.
Answer: A Type II error would be deciding the percentage of customers who’ll make
purchases won’t go up when in fact it would have. The store’s owner would miss an
opportunity to increase sales.
Explanation:
34) Sleep Do more than 50% of U.S. adults feel they get enough sleep? According to Gallup’s
December 2004 Lifestyle poll, 55% of U.S. adults said that that they get enough sleep. The
poll was based on a random sample of 1003 U.S. adults. Test an appropriate hypothesis
and state your conclusion in the context of the problem.
Answer:
Hypothesis: H0 : p = 0.50 HA : p > 0.50
Plan: Okay to use the Normal model because the trials are independent (random
sample of U.S. adults), these 1003 U.S. adults are less than 10% of all U.S. adults, and
np0 = (1003)(0.50) = 501.5 10 and nq0 = (1003)(0.50) = 501.5 10.
We will do a one–proportion z–test.
34)
Mechanics: SD(p0) = p0q0
n = (0.50)(0.50)
1003 ^
= 0.0158; sample proportion: p = 0.55
^
P(p > 0.55) = P(z > 0.55 – 0.50
0.0158 ) = P(z > 3.16) = 0.0008
With a P–value of 0.0008, I reject the null hypothesis. There is strong evidence that
the proportion of U.S. adults who feel they get enough sleep is more than 50%.
Explanation:
The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Crest
toothpaste. A marketing director suggests that the company invest in a new marketing campaign which will include
advertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for one
month to determine if the market share increases with new labels.
35) The board of directors asked the research department to extend the trial period so that the
35)
decision can be made on two months worth of data. Will the power increase, decrease, or
remain the same?
Answer: The power would increase because of the larger sample size.
Explanation:
A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concerned
that employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defective
chips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentive
program is instituted for one month. If successful, the company will continue with the incentive program.
36) In this context describe a Type I error and the impact such an error would have on the
36)
company.
Answer: A Type I error would be deciding the percentage of defective chips has decreased,
when in fact it has not. The company would waste money on a new incentive
program that does not decrease the defect rate of the chips.
Explanation:
16A report on health care in the US said that 28% of Americans have experienced times when they haven’t been able to afford
medical care. A news organization randomly sampled 801 black Americans, of whom 38% reported that there had been
times in the last year when they had not been able to afford medical care. Does this indicate that this problem is more severe
among black Americans?
37) Was your test one–tail upper tail, one–tail lower tail, or two–tail? Explain why you chose
37)
that kind of test in this situation.
Answer: One–tail, upper tail test. We are concerned that the proportion of people who are not
able to afford medical care is higher among black Americans.
Explanation:
38) Cereal A box of Raspberry Crunch cereal contains a mean of 13 ounces with a standard
deviation of 0.5 ounce. The distribution of the contents of cereal boxes is approximately
Normal. What is the probability that a case of 12 cereal boxes contains a total of more than
160 ounces?
Answer: Two methods can be used to solve this problem:
38)
Method 1:
Let B = weight of one box of cereal and T = weight of 12 boxes of cereal. We are told
that the contents of the boxes are approximately Normal, and we can assume that
the content amounts are independent from box to box.
E(T) = E(B1 + B2 +…+ B12 ) = E(B1) + E(B2) + … + E(B12 ) = 156 ounces
Since the content amounts are independent,
Var(T) = Var(B1 + B2 +…+ B12 ) = Var(B1) + Var(B2) + … + Var(B12 ) = 3
SD(T) = Var(T) = 3 = 1.73 oz.
We model T with N(156, 1.73).
z = 160 – 156
1.73 = 2.31 and P(T > 160) = P(z > 2.31) = 0.0104
There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160
ounces.
Method 2:
Using the Central Limit Theorem approach, let y = average content of boxes in the
case. Since the contents are Normally distributed, y is modeled by N 13, 0.5
12
P y > 160
12 = P(y > 13.33) = P z > 13.33 – 13
0.5
= P(z > 2.31) = 0.0104.
.
12
There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160
ounces.
Explanation:
17The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Crest
toothpaste. A marketing director suggests that the company invest in a new marketing campaign which will include
advertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for one
month to determine if the market share increases with new labels.
39) Based on the data they collected during the trial the research department found that a 98%
39)
confidence interval for the proportion of all consumers who might buy Crest toothpaste
was (16%, 28%). What conclusion should the company reach about the new marketing
campaign? Explain.
Answer: The confidence interval contains the current value of 19.5%, so the product trials do
not provide convincing evidence that the new marketing campaign would increase
sales.
Explanation:
40) Dolphin births A state has two aquariums that have dolphins, with more births recorded
at the larger aquarium than at the smaller one. Records indicate that in general babies are
equally likely to be male or female, but the gender ratio varies from season to season.
Which aquarium is more likely to report a season when over two–thirds of the dolphins
born were males? Explain.
Answer: The smaller aquarium would experience more variability in the season percentage
of male births. We would expect the larger aquarium to stay more consistent and
closer to the 50–50 ratio for gender births.
Explanation:
40)
A state’s Department of Education reports that 12% of the high school students in that state attend private high schools. The
State University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a random
sample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attend
private schools.
41) They actually select a random sample of 450 applications, and find that 46 of those
41)
students attend private schools. Create the confidence interval.
Answer: We have a random sample of less than 10% of the applicants, with 46 successes and
404 failures, so a Normal model applies. The confidence interval is (0.079, 0.126).
Explanation:
The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. A
marketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latest
jeans…). He loans her several different “people” to scatter around the store for a two–week trial period. The owner carefully
counts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’ll
purchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy something
increases.
42) Describe to the owner an advantage and a disadvantage of using an alpha level of 5%
instead.
Answer: Advantage: the test would have greater power to detect a positive effect of the
mannequins. Disadvantage: she’d be more likely to think the mannequins were
effective even if they were not.
Explanation:
42)
18A statistics professor asked her students whether or not they were registered to vote. In a sample of 50 of her students
(randomly sampled from her 700 students), 35 said they were registered to vote.
43) What is the probability that the true proportion of the professor’s students who were
43)
registered to vote is in your confidence interval?
Answer: There is no probability involved–once the interval is constructed, the true
proportion of the professor’s students who were registered to vote is in the interval
or it is not.
Explanation:
44) Exercise A random sample of 150 men found that 88 of the men exercise regularly, while a
random sample of 200 women found that 130 of the women exercise regularly.
44)
a. Based on the results, construct and interpret a 95% confidence interval for the difference
in the proportions of women and men who exercise regularly.
b. A friend says that she believes that a higher proportion of women than men exercise
regularly. Does your confidence interval support this conclusion? Explain.
Answer:
a. Conditions:
* Randomization Condition: We are told that we have random samples.
* 10% Condition: We have less than 10% of all men and less than 10% of all women.
* Independent samples condition: The two groups are clearly independent of each
other.
* Success/Failure Condition: Of the men, 88 exercise regularly and 62 do not; of the
women, 130 exercise regularly and 70 do not. The observed number of both
successes and failures in both groups is at least 10.
With the conditions satisfied, the sampling distribution of the difference in
proportions is approximately Normal with a mean of pM – pW , the true difference
between the population proportions. We can find a two–proportion z–interval.
We know: nM = 150, pM = 88
^ ^
150 = 0.587, nW = 200, pW = 130
200 = 0.650
^ ^
We estimate SD(pM – pW) as
^ ^ ^ ^
^ ^
SE(pM – pW) = pM qM
nM + pW qW
nW = (0.587)(0.413)
150 + (0.65)(0.35)
200 = 0.0525
^ ^
ME = z* × SE(pM – pW) = 1.96(0.0525) = 0.1029
^ ^
The observed difference in sample proportions = pM – pW = 0.587 – 0.650 = –0.063,
so the 95%
confidence interval is –0.063 ± 0.1029, or –16.6% to 4.0%.
We are 95% confident that the proportion of women who exercise regularly is
between 4.0% lower and 16.6% higher than the proportion of men who exercise
regularly.
b. Since zero is contained in my confidence interval, I cannot say that a higher
proportion of women than men exercise regularly. My confidence interval does not
support my friend’s claim.
Explanation:
19In 2000, the United Nations claimed that there was a higher rate of illiteracy in men than in women from the country of
Qatar. A humanitarian organization went to Qatar to conduct a random sample. The results revealed that 45 out of 234 men
and 42 out of 251 women were classified as illiterate on the same measurement test. Do these results indicate that the United
Nations findings were correct?
45) Test an appropriate hypothesis and state your conclusion. 45)
Answer:
H0 : pm – pw = 0, HA: pm – pw > 0, where m = men and w = women
* Randomization Condition: The men and women in the sample were randomly
selected by the
humanitarian organization.
* 10% Condition: The number of men and women in Qatar is greater than 2340 (10 ×
234) and
2510 (10 × 251), respectively.
* Independent samples condition: The two groups are independent of each other
because the
samples were selected at random.
* Success/Failure Condition: In men, 45 were illiterate and 189 were not. In women,
42 were
illiterate and 209 were not. The observed number of both successes and failures in
both groups is larger than 10.
Because the conditions are satisfied, we can model the sampling distribution of the
difference in proportions with a Normal model. We can perform a two–proportion
z–test.
0.179.
^ ^
SEpooled (pm – pw) = ^ ^ ^ ^
ppooled qpooled
+ ppooled qpooled
nm
nw
=
(0.179)(0.821)
234 + (0.179)(0.821)
251 = 0.0349
^ ^ ^
We know: nm = 234, pm = 0.192, nw = 251, pw = 0.167, and ppooled = 45 + 42
234 + 251 =
^ ^
The observed difference in sample proportions = pm – pw = 0.192 – 0.167 = 0.025
^ ^
z = (pm – pw) – 0
^ ^
= 0.025 – 0
0.0349 = 0.716
SEpooled (pm – pw)
P = P(z > 0.716) = 0.24
The P–value of 0.24 is high, so we fail to reject the null hypothesis. There is
insufficient evidence to conclude that illiteracy rate in men is higher than for women
in Qatar.
Explanation:
2046) According to Gallup, about 33% of Americans polled said they frequently experience stress
in their daily lives. Suppose you are in a class of 45 students.
46)
a. What is the probability that no more than 12 students in the class will say that they
frequently experience stress in their daily lives? (Make sure to identify the sampling
distribution you use and check all necessary conditions.)
b. If 20 students in the class said they frequently experience stress in their daily lives,
would you be surprised? Explain, and use statistics to support your answer.
Answer:
a. We want to find the probability that no more than 12 students in the class will say
that they frequently experience stress. This is the same as asking the probability of
finding less than 26.7% of “stressed” students in a class of 45 students.
Check the conditions:
1. 10% condition: 45 students is less than 10% of all students who could take the
class
2. Success/failure cond.: np = 45(0.33) = 14.85, nq = 45(0.67) = 30.15, which both
exceed 10
We can use the N(0.33 (0.33)(0.67)
45 = 0.070) to model the sampling distribution.
We need to standardize the 26.7% and then find the probability of getting a z–score
less than or equal to the one we find: z = 0.267 – 0.33
0.070 = –0.90
^
P(p < 0.267) = P(z < –0.90) = 0.1841, so the probability is about 18.4% that no more
than 12 students will say that they frequently experience stress in their daily lives.
b. From part a, we can use N(0.33, 0.070) to model the sampling distribution. Twenty
students is about 44.4% of the class. This is about 1.63 standard deviations above
what we would expect, which is not a surprising result.
Explanation:
A statistics professor asked her students whether or not they were registered to vote. In a sample of 50 of her students
(randomly sampled from her 700 students), 35 said they were registered to vote.
47) Explain what 95% confidence means in this context. 47)
Answer: If many random samples were taken, 95% of the confidence intervals produced
would contain the actual percentage of the professor’s students who are registered to
vote.
Explanation:
48) Wildlife scientists studying a certain species of frogs know that past records indicate the
adults should weigh an average of 118 grams with a standard deviation of 14 grams. The
researchers collect a random sample of 50 adult frogs and weigh them. In their sample the
mean weight was only 110 grams. One of the scientists is alarmed, fearing that
environmental changes may be adversely affecting the frogs. Do you think this sample
result is unusually low? Explain.
Answer: We have a random sample of frogs drawn from a much larger population. With a
sample of size 50 the CLT says that the approximate sampling model for sample
means will be N(118, 1.98). A sample mean of only 110 grams is about 4 standard
deviations below what we expect, a very unusual result.
Explanation:
48)
21A company claims to have invented a hand–held sensor that can detect the presence of explosives inside a closed container.
Law enforcement and security agencies are very interested in purchasing several of the devices if they are shown to perform
effectively. An independent laboratory arranged a preliminary test. If the device can detect explosives at a rate greater than
chance would predict, a more rigorous test will be performed. They placed four empty boxes in the corners of an otherwise
empty room. For each trial they put a small quantity of an explosive in one of the boxes selected at random. The company’s
technician then entered the room and used the sensor to try to determine which of the four boxes contained the explosive.
The experiment consisted of 50 trials, and the technician was successful in finding the explosive 16 times. Does this indicate
that the device is effective in sensing the presence of explosives, and should undergo more rigorous testing?
49) Explain what your P–value means in this context. 49)
Answer: Even if the device actually performs no better than guessing, we could expect to find
the explosives 16 or more times out of 50 about 13% of the time.
Explanation:
A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concerned
that employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defective
chips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentive
program is instituted for one month. If successful, the company will continue with the incentive program.
50) Describe to management an advantage and disadvantage of using a 1% alpha level of
50)
significance instead.
Answer: Advantage: management would be less likely to conclude the incentive program
was effective if it really were not. Disadvantage: the test would have less power to
detect a positive effect of the new incentive program.
Explanation:
51) Tax advice Each year people who have income file income tax reports with the
government. In some instances people seek advice from accountants and financial advisors
regarding their income tax situations. This advice is meant to lower the percentage of taxes
paid to the government each year. A random sample of people who filed tax reports
resulted in the data in the table below. Does this data indicate that people should seek tax
advice from an accountant or financial advisor? Test an appropriate hypothesis and state
your conclusion.
51)
Answer: We want to know whether the percentage of people who paid lower taxes was
different based on whether they sought tax advice or not.
H0: padvice – pno advice = 0, HA: padvice – pno advice > 0
Conditions:
* Independence: The people who filed tax reports were randomly selected and do
not influence each other.
* Random Condition: The people who filed tax reports were randomly selected.
* 10% Condition: 72 is less than 10% of people who didn’t get tax advice, and 105 is
less than 10% of people who did get tax advice.
* Success/Failure: All observed counts (48, 19, 24, and 86) are at least 10.
Because all conditions have been satisfied, we can model the sampling distribution
of the difference in proportions with a Normal model. We can perform a
two–proportion z–test.
Let ‘Advice’ group be the people who sought tax advice and the ‘No advice’ group
22= 0.49 – 0
0.074 = 6.62
Answer:
be the people who did not seek tax advice.
^ ^
We know: nadvice = 105, nno advice = 72, padvice = 0.819, pno advice = 0.333.
^
ppooled = 24 + 86
72 + 105 = 0.621
^ ^
SEpooled (padvice – pno advice) = (0.62)(0.38)
72 + (0.62)(0.38)
105 = 0.074
^ ^
The observed difference in sample proportions is padvice – pno advice = 0.819 – 0.333
= 0.486
z = ^ ^
(padvice – pno advice) – 0
^ ^
SEpooled (padvice – pno advice)
P = P(z > 6.62) < 0.0001
The P–value is small, so we reject the null hypothesis. There is strong evidence of a
difference in the tax percentages paid between the group who has tax advice and the
group who did not have tax advice. It appears that if you have tax advice you are
more likely to pay a lower percentage of taxes to the government.
Explanation:
The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of females
in the labor force is the same in the United States. Representatives from the United States Department of Labor plan to check
a random sample of over 10,000 employment records on file to estimate a percentage of females in the United States labor
force.
52) The representatives from the Department of Labor want to estimate a percentage of
females in the United States labor force to within ±5%, with 90% confidence. How many
employment records should they sample?
^^
Answer:
ME = z* pq
n
52)
0.05 = 1.645 (0.46)(0.54)
n
n = 1.645 (0.46)(0.54)
0.05
n = 268.87 269
They should sample at least 269 employment records.
Explanation:
23Great Britain has a great literary tradition that spans centuries. One might assume, then, that Britons read more than citizens
of other countries. Some Canadians, however, feel that a higher percentage of Canadians than Britons read. A recent Gallup
Poll reported that 86% of 1004 randomly sampled Canadians read at least one book in the past year, compared to 81% of 1009
randomly sampled Britons. Do these results confirm a higher reading rate in Canada?
53) Test an appropriate hypothesis and state your conclusions. 53)
Answer:
H0 : pB – pC = 0, HA: pB – pC < 0, where B = Britons and C = Canadians
Conditions:
* Randomization Condition: The Britons and Canadians were randomly sampled by
Gallup.
* 10% Condition: The number of Britons and Canadians is greater than 10,090 (10 ×
1009) and
10,040 (10 × 1004), respectively.
* Independent samples condition: The two groups are clearly independent of each
other.
* Success/Failure Condition: Of the Britons, approximately 817 read at least one
book and 192 did not; of the Canadians, approximately 863 read at least one book
and 141 did not. The observed number of both successes and failures in both groups
is larger than 10.
Because the conditions are satisfied, we can model the sampling distribution of the
difference
in proportions with a Normal model. We can perform a two–proportion z–test.
^ ^
We know: nB = 1009, pB = 0.81, nC = 1004, pC = 0.86, and
^
ppooled = 817 + 863
1009 + 1004 = 1680
2013 = 0.835
^ ^ ^ ^
^ ^
SEpooled (pB – pC) = ppooled qpooled
nB + ppooled qpooled
nC =
(0.835)(0.165)
1009 + (0.835)(0.165)
1004 = 0.0165
^ ^
The observed difference in sample proportions = pB – pC = 0.81 – 0.86 = –0.05
^ ^
z = (pB – pC) – 0
^ ^
= –0.05 – 0
0.0165 = –3.03, so the P–value = P(z < –3.03) = 0.0012
SEpooled (pB – pC)
The P–value of 0.0012 is low, so we reject the null hypothesis.There is strong
evidence that the percentage of Britons who read at least one book in the past year is
less than the percentage of Canadians who read at least one book in the past year.
Explanation:
24A company claims to have invented a hand–held sensor that can detect the presence of explosives inside a closed container.
Law enforcement and security agencies are very interested in purchasing several of the devices if they are shown to perform
effectively. An independent laboratory arranged a preliminary test. If the device can detect explosives at a rate greater than
chance would predict, a more rigorous test will be performed. They placed four empty boxes in the corners of an otherwise
empty room. For each trial they put a small quantity of an explosive in one of the boxes selected at random. The company’s
technician then entered the room and used the sensor to try to determine which of the four boxes contained the explosive.
The experiment consisted of 50 trials, and the technician was successful in finding the explosive 16 times. Does this indicate
that the device is effective in sensing the presence of explosives, and should undergo more rigorous testing?
54) Was your test one–tail upper tail, lower tail, or two–tail? Explain why you chose that kind
54)
of test in this situation.
Answer: One–tail, upper tail. The device is effective only if it can detect explosives at a rate
higher than chance (25%).
Explanation:
The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. A
marketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latest
jeans…). He loans her several different “people” to scatter around the store for a two–week trial period. The owner carefully
counts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’ll
purchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy something
increases.
55) The owner talked the salesman into extending the trial period so that she can base her
decision on data for a full month. Will the power of the test increase, decrease, or remain
the same?
Answer: The power of the test will increase.
Explanation:
55)
The International Olympic Committee states that the female participation in the 2004 Summer Olympic Games was 42%,
even with new sports such as weight lifting, hammer throw, and modern pentathlon being added to the Games. Broadcasting
and clothing companies want to change their advertising and marketing strategies if the female participation increases at the
next games. An independent sports expert arranged for a random sample of pre–Olympic exhibitions. The sports expert
reported that 202 of 454 athletes in the random sample were women. Is this strong evidence that the participation rate may
increase?
56) Was your test one–tail upper tail, lower tail, or two–tail? Explain why you choose that kind
of test in this situation.
Answer: One–tail, upper test. The companies will change strategies only if there is strong
evidence of an increase in female participation rate from the current rate of 42%.
Explanation:
56)
The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Crest
toothpaste. A marketing director suggests that the company invest in a new marketing campaign which will include
advertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for one
month to determine if the market share increases with new labels.
57) What level of significance did the research department use? 57)
Answer: = 1 – 0.98
2 = 0.01 = 1%
Explanation:
2558) Describe to the board of directors an advantage and a disadvantage of using a 5% alpha
level of significance instead.
Answer: Advantage: the test would have greater power to detect a positive effect of the new
marketing campaign. Disadvantage: they would be more likely to think the
marketing campaign was effective even if it were not.
Explanation:
59) Truckers On many highways state police officers conduct inspections of driving logbooks
from large trucks to see if the trucker has driven too many hours in a day. At one truck
inspection station they issued citations to 49 of 348 truckers that they reviewed.
a. Based on the results of this inspection station, construct and interpret a 95% confidence
interval for the proportion of truck drivers that have driven too many hours in a day.
b. Explain the meaning of “95% confidence” in part A.
Answer:
a. Conditions:
* Independence: We assume that one trucker’s driving times do not influence other
trucker’s driving times.
* Random Condition: We assume that trucks are stopped at random.
* 10% Condition: This sample of 348 truckers is less than 10% of all truckers.
* Success/Failure: 49 tickets and 299 tickets are both at least 10, so our sample is large
enough.
Under these conditions the sampling distribution of the proportion can be modeled
by a Normal model. We will find a one–proportion z–interval.
^^
^ ^
We know n = 348 and p = 0.14 , so SE(p) = pq
n = (0.14)(0.86)
348 = 0.0186
The sampling model is Normal, for a 95% confidence interval the critical value is z*
= 1.96.
^
The margin of error is ME = z* × SE(p) = 1.96(0.0186) = 0.0365.
The 95% confidence interval is 0.14 ± 0.0365 or (0.1035, 0.1765).
We are 95% confident that between 10.4% and 17.7% of truck drivers have driven
too many hours in a day.
b. If we repeated the sampling and created new confidence intervals many times we
would expect about 95% of those intervals to contain the actual proportion of truck
drivers that have driven too many hours in a day.
Explanation:
58)
59)
2660) The average composite ACT score for Ohio students who took the test in 2003 was 21.4.
Assume that the standard deviation is 1.05. In a random sample of 25 students who took
the exam in 2003, what is the probability that the average composite ACT score is 22 or
more? (Make sure to identify the sampling distribution you use and check all necessary
conditions.)
Answer: Check the conditions:
1. Random sampling condition: We have been told that this is a random sample.
2. Independence assumption: It’s reasonable to think that the scores of the 25
students are mutually independent.
3. 10% condition: 25 students is certainly less than 10% of all students who took the
exam.
We’re assuming that the model for composite ACT scores has mean µ = 21.4 and
standard deviation = 1.05. Since the sample size is large enough and the
distribution of ACT scores is most likely unimodal and symmetric, CLT allows us to
describe the sampling distribution of y with a Normal model with mean 21.4 and SD
= (y) = 1.05
25 = 0.21.
An average score of 22 is z = 22 – 21.4
0.21 = 2.86 SDs above the mean.
P(x > 22) = P(Z > 2.86) = 0.0021, so the probability that the average composite ACT
score for a sample of 25 randomly selected students is 22 or more is 0.0021.
60)
Explanation:
61) Approval rating The President’s job approval rating is always a hot topic. Your local
paper conducts a poll of 100 randomly selected adults to determine the President’s job
approval rating. A CNN/USA Today/Gallup poll conducts a poll of 1010 randomly selected
adults. Which poll is more likely to report that the President’s approval rating is below
50%, assuming that his actual approval rating is 54%? Explain.
Answer: The smaller poll would have more variability and would thus be more likely to vary
from the actual approval rating of 54%. We would expect the larger poll to be more
consistent with the 54% rating. So, it is more likely that the smaller poll would
report that the President’s approval rating is below 50%.
Explanation:
61)
A statistics professor asked her students whether or not they were registered to vote. In a sample of 50 of her students
(randomly sampled from her 700 students), 35 said they were registered to vote.
62) If the professor only knew the information from the September 2004 Gallup poll and
62)
wanted to estimate the percentage of her students who were registered to vote to within
±4% with 95% confidence, how many students should she sample?
Answer: ME = z
* pq
n
0.04 = 1.96 (0.73)(0.27)
n
n = (1.96)2(0.73)(0.27)
(0.04)2 = 473.24 n = 474
Note: Since there are only 700 students in the professor’s class, she cannot sample
this many students without violating the 10% condition!
Explanation:
27The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Crest
toothpaste. A marketing director suggests that the company invest in a new marketing campaign which will include
advertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for one
month to determine if the market share increases with new labels.
63) In this context describe a Type I error and the impact such an error would have on the
63)
company.
Answer: A Type I error would be concluding the proportion of people who will buy Crest
toothpaste will go up when in fact it won’t. The company would waste money on a
new marketing campaign that will not increase sales.
Explanation:
The International Olympic Committee states that the female participation in the 2004 Summer Olympic Games was 42%,
even with new sports such as weight lifting, hammer throw, and modern pentathlon being added to the Games. Broadcasting
and clothing companies want to change their advertising and marketing strategies if the female participation increases at the
next games. An independent sports expert arranged for a random sample of pre–Olympic exhibitions. The sports expert
reported that 202 of 454 athletes in the random sample were women. Is this strong evidence that the participation rate may
increase?
64) Explain what your P–value means in this context. Answer: If the proportion of female athletes has not increased, we could expect to find at least
202 females of 454 pre–Olympic athletes about 13.8% of the time.
64)
Explanation:
The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. A
marketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latest
jeans…). He loans her several different “people” to scatter around the store for a two–week trial period. The owner carefully
counts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’ll
purchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy something
increases.
65) Write the owner’s null and alternative hypotheses. Answer: The null hypothesis is that the percentage of all customers who buy something is
28%. The alternative hypothesis is that the percentage of all customers who buy
something is greater than 28%. In symbols: H0: p = 0.28 HA: p > 0.28
65)
Explanation:
2866) Roadblocks From time to time police set up roadblocks to check cars to see if the safety
inspection is up to date. At one such roadblock they issued tickets for expired inspection
stickers to 22 of 628 cars they stopped.
66)
a. Based on the results at this roadblock, construct and interpret a 95% confidence interval
for the proportion of autos in that region whose safety inspections have expired.
b. Explain the meaning of “95% confidence” in part a.
Answer:
a. We assume the cars stopped are representative of all cars in the area, and that 628
< 10% of the cars. The 22 successes (expired stickers) and 608 failures are both
greater than 10. It’s okay to use a Normal model and find a one–proportion
z–interval.
^
p = 0.035; 0.035 ± 1.96 (0.035)(0.965)
628 = (0.021, 0.049)
The interval (0.021,0.049) means that we are 95% confident that between 2% and 5%
of local cars have expired safety inspection stickers.
b. If we repeated the sampling and created new confidence intervals many times
we’d expect about 95% of those intervals to contain the actual proportion of cars
with expired stickers.
Explanation:
A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concerned
that employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defective
chips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentive
program is instituted for one month. If successful, the company will continue with the incentive program.
67) What level of significance did management use? 67)
Answer: = 1 – 0.95
2 = 0.025 2.5%
Explanation:
29Researchers conduct a study to test a potential side effect of a new allergy medication. A random sample of 160 subjects with
allergies was selected for the study. The new “improved” Brand I medication was randomly assigned to 80 subjects, and the
current Brand C medication was randomly assigned to the other 80 subjects. 14 of the 80 patients with Brand I reported
drowsiness, and 22 of the 80 patients with Brand C reported drowsiness.
68) Compute a 95% confidence interval for the difference in proportions of subjects reporting
68)
drowsiness. Show all steps.
Answer:
Conditions:
* Randomization Condition: The treatments were randomly assigned to subjects.
* 10% Condition: The subjects were randomly selected. We assume it is from a large
population of
allergy sufferers.
* Independent samples condition: The two groups are independent of each other
because the treatments were assigned at random.
* Success/Failure Condition: For Brand C, 22 were drowsy and 58 were not. For
Brand I, 14 were
drowsy and 66 were not. The observed number of both successes and failures in
both groups is larger than 10.
Because the conditions are satisfied, we can model the sampling distribution of the
difference in
proportions with a Normal model.
^ ^
We know: nI = 80, pI = 0.175, nC = 80, pC = 0.275
^ ^
We estimate SD(pC – pI) as
^ ^ ^ ^
^ ^
SE(pC – pI) = pC qC
nC + pI qI
nI = (0.275)(0.725)
80 + (0.175)(0.825)
80 = 0.066
^ ^
ME = z* × SE(pC – pI) = 1.96(0.066) = 0.128
^ ^
The observed difference in sample proportions = pC – pI = 0.275 – 0.175 = 0.10, so
the 95% confidence interval is 0.10 ± 0.128, or (–0.028, 0.228)
We are 95% confident that the difference between the population proportions of
patients that reported drowsiness for Brand C and Brand I is between –2.8% and
22.8%.
Explanation:
The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Crest
toothpaste. A marketing director suggests that the company invest in a new marketing campaign which will include
advertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for one
month to determine if the market share increases with new labels.
69) In this context describe a Type II error and the impact such an error would have on the
69)
company.
Answer: A Type II error would be deciding the proportion of people who will buy Crest
toothpaste won’t go up when in fact it would have. The company would miss an
opportunity to increase sales.
Explanation:
3070) Herpetologists (snake specialists) found that a certain species of reticulated python has an
average length of 20.5 feet with a standard deviation of 2.3 feet. The scientists collect a
random sample of 30 adult pythons and measure their lengths. In their sample the mean
length was 19.5 feet long. One of the herpetologists fears that pollution might be affecting
the natural growth of the pythons. Do you think this sample result is unusually small?
Explain.
70)
Answer: We have a random sample of adult pythons drawn from a much larger population.
With a sample size of 30, the CLT says that the approximate sampling model for
sample means will be N(20.5, 0.42). A sample mean of only 19.5 feet is about 2.38
standard deviations below what we expect. The sample mean of 19.5 feet is
unusually small.
Explanation:
A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concerned
that employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defective
chips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentive
program is instituted for one month. If successful, the company will continue with the incentive program.
71) In this context describe a Type II error and the impact such an error would have on the
71)
company.
Answer: A Type II error would be deciding the percentage of defective chips has not
decreased, when in fact it has. The company would miss an opportunity to decrease
the defect rate of the chips.
Explanation:
Researchers conduct a study to test a potential side effect of a new allergy medication. A random sample of 160 subjects with
allergies was selected for the study. The new “improved” Brand I medication was randomly assigned to 80 subjects, and the
current Brand C medication was randomly assigned to the other 80 subjects. 14 of the 80 patients with Brand I reported
drowsiness, and 22 of the 80 patients with Brand C reported drowsiness.
72) Would you make the same conclusion as question 2 if you conducted a hypothesis test?
72)
Explain.
Answer: A hypothesis test at a 0.05 level should reach the same conclusion if a two–sided test
is used. If the
alternate hypothesis were “less drowsy”, the significance level would need to be
changed, or a
different conclusion could occur.
Explanation:
31A company claims to have invented a hand–held sensor that can detect the presence of explosives inside a closed container.
Law enforcement and security agencies are very interested in purchasing several of the devices if they are shown to perform
effectively. An independent laboratory arranged a preliminary test. If the device can detect explosives at a rate greater than
chance would predict, a more rigorous test will be performed. They placed four empty boxes in the corners of an otherwise
empty room. For each trial they put a small quantity of an explosive in one of the boxes selected at random. The company’s
technician then entered the room and used the sensor to try to determine which of the four boxes contained the explosive.
The experiment consisted of 50 trials, and the technician was successful in finding the explosive 16 times. Does this indicate
that the device is effective in sensing the presence of explosives, and should undergo more rigorous testing?
73) Test an appropriate hypothesis and state your conclusion. 73)
Answer:
Hypotheses: H0: p = 0.25. The device can detect explosives at the same level as
guessing. HA: p > 0.25. The device can detect explosives at a level greater
than chance.
Model: OK to use a Normal model because trials are independent (box is randomly
chosen each time), and np = 12.5, nq = 37.5. Do a 1–proportion z–test
^
Mechanics: p = 0.32, z = 1.14, P = 0.13
Conclusion: With a P–value so high I fail to reject the null hypothesis. This test does
not provide convincing evidence that the sensor can detect the presence of
explosives inside a box.
Explanation:
The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of females
in the labor force is the same in the United States. Representatives from the United States Department of Labor plan to check
a random sample of over 10,000 employment records on file to estimate a percentage of females in the United States labor
force.
74) They actually select a random sample of 525 employment records, and find that 229 of the
people are females. Create the confidence interval.
Answer:
We have a random sample of less than 10% of the employment records, with 229
successes (females) and 296 failures (males), so a Normal model applies.
^^
^ ^ ^
n = 525, p = 0.436 and q = 0.564, so SE(p) = pq
n = (0.436)(0.564)
525 = 0.022
74)
^
margin of error: ME = z* × SE(p) =(1.645)(0.022) = 0.0362.
^
Confidence interval: p ± ME = 0.436 ± 0.0362 or (0.3998, 0.4722)
Explanation:
32In 2000, the United Nations claimed that there was a higher rate of illiteracy in men than in women from the country of
Qatar. A humanitarian organization went to Qatar to conduct a random sample. The results revealed that 45 out of 234 men
and 42 out of 251 women were classified as illiterate on the same measurement test. Do these results indicate that the United
Nations findings were correct?
75) Find a 95% confidence interval for the difference in the proportions of illiteracy in men and
75)
women from Qatar. Interpret your interval.
Answer:
With the conditions satisfied (from Problem 1), the sampling distribution of the
difference in
proportions is approximately Normal with a mean of pm – pw, the true difference
between the
population proportions.
We can find a two–proportion z–interval.
^ ^
We know: nm = 234, pm = 0.192, nw = 251, pw = 0.167
^ ^
We estimate SD(pm – pw) as
^ ^ ^ ^
^ ^
SE(pm – pw) = pm qm
+ pw qw
nm
nw
= (0.192)(0.808)
234 + (0.167)(0.833)
251 = 0.0349
^ ^
ME = z* × SE(pm – pw) = 1.96(0.0349) = 0.0684
^ ^
The observed difference in sample proportions = pm – pw = 0.192 – 0.167 = 0.025, so
the 95%
confidence interval is 0.025 ± 0.0684, or –4% to 9%.
We 95% confident that the proportion of illiterate men in Qatar is between
4–percentage points
lower and 9–percentage points higher than the proportion of illiterate women.
Explanation:
A state’s Department of Education reports that 12% of the high school students in that state attend private high schools. The
State University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a random
sample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attend
private schools.
76) Explain what 90% confidence means in this context. Answer: If many random samples were taken, 90% of the confidence intervals produced
would contain the actual percentage of all applicants who attend private schools.
76)
Explanation:
33The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. A
marketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latest
jeans…). He loans her several different “people” to scatter around the store for a two–week trial period. The owner carefully
counts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’ll
purchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy something
increases.
77) Based on data that she collected during the trial period the store’s owner found that a 98%
confidence interval for the proportion of all shoppers who might buy something was (27%,
35%). What conclusion should she reach about the mannequins? Explain.
Answer: Because the hypothesized value of 28% is in the confidence interval the trial results
do not provide convincing evidence that the mannequins would help increase sales.
Explanation:
77)
A report on health care in the US said that 28% of Americans have experienced times when they haven’t been able to afford
medical care. A news organization randomly sampled 801 black Americans, of whom 38% reported that there had been
times in the last year when they had not been able to afford medical care. Does this indicate that this problem is more severe
among black Americans?
78) Explain what your P–value means in this context. 78)
Answer: If the proportion of black Americans was 28%, we would almost never expect to find
at least 38% of 801 randomly selected black Americans responding “yes”.
Explanation:
A state’s Department of Education reports that 12% of the high school students in that state attend private high schools. The
State University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a random
sample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attend
private schools.
79) Should the admissions officers conclude that the percentage of private school students in
79)
their applicant pool is lower than the statewide enrollment rate of 12%? Explain.
Answer: No. Since 12% lies in the confidence interval it’s possible that the percentage of
private school students in the applicant pool matches the statewide enrollment rate.
Explanation:
80) Internet access A recent Gallup poll found that 28% of U.S. teens aged 13–17 have a
computer with Internet access in their rooms. The poll was based on a random sample of
1028 teens and reported a margin of error of ±3%. What level of confidence did Gallup use
for this poll?
80)
Answer:
Since ME = z * ^^
pq
, we have 0.03 = z* n
(0.28)(0.72)
1028 or z* = 0.03
(0.28)(0.72)
1028
2.14.
Our confidence level is approximately P(–2.14 < z < 2.14) = 0.9676, or 97%.
Explanation:
81) Baldness and heart attacks A recent medical study observed a higher frequency of heart
attacks among a group of bald men than among another group of men who were not bald.
Based on a P–value of 0.062 the researchers concluded there was some evidence that male
baldness may be a risk factor for predicting heart attacks. Explain in context what their
P-value means.
Answer: If baldness is not a risk factor an observed level of heart attacks this much higher (or
more) would occur in only 6% of such samples.
Explanation:
81)
34A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concerned
that employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defective
chips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentive
program is instituted for one month. If successful, the company will continue with the incentive program.
82) Management decided to extend the incentive program so that the decision can be made on
82)
three months of data instead. Will the power increase, decrease, or remain the same?
Answer: The power would increase because of the larger sample size.
Explanation:
83) Write the company’s null and alternative hypotheses. Answer: The null hypothesis is that the defect rate is 8%. The alternative hypothesis is that
the defect rate is lower than 8%. In symbols: H0: p = 0.08 and HA: p < 0.08
83)
Explanation:
The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of females
in the labor force is the same in the United States. Representatives from the United States Department of Labor plan to check
a random sample of over 10,000 employment records on file to estimate a percentage of females in the United States labor
force.
84) Should the representatives from the Department of Labor conclude that the percentage of
females in their labor force is lower than Europe’s rate of 46%? Explain.
Answer: No. Since 46% lies in the confidence interval, (0.3998, 0.4722), it is possible that the
percentage of females in the labor force matches Europe’s rate of 46% females in the
labor force.
84)
Explanation:
85) Egg weights The weights of hens’ eggs are normally distributed with a mean of 56 grams
and a standard deviation of 4.8 grams. What is the probability that a dozen randomly
selected eggs weighs over 690 grams?
Answer: P y > 690
12 = P z > 57.5 – 56
48
85)
= P(z > 1.08) = 14%
12
Explanation:
86) Approval rating A newspaper article reported that a poll based on a sample of 800 voters
showed the President’s job approval rating stood at 62%. They claimed a margin of error of
±3%. What level of confidence were the pollsters using?
Answer: 0.03 = z* (0.62)(0.38)
800 ; z* = 1.75; P(–1.75 < z < 1.75) = 92% confidence
86)
Explanation:
Researchers conduct a study to test a potential side effect of a new allergy medication. A random sample of 160 subjects with
allergies was selected for the study. The new “improved” Brand I medication was randomly assigned to 80 subjects, and the
current Brand C medication was randomly assigned to the other 80 subjects. 14 of the 80 patients with Brand I reported
drowsiness, and 22 of the 80 patients with Brand C reported drowsiness.
87) Does the interval in question 1 provide evidence that the side effect of drowsiness is
87)
different with the new medication?
Answer: There is not sufficient evidence because 0 is contained in the interval. There may be
no difference in the proportion of drowsiness reported.
Explanation:
35A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concerned
that employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defective
chips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentive
program is instituted for one month. If successful, the company will continue with the incentive program.
88) Over the trial month, 6% of the computer chips manufactured were defective. Management
88)
decided that this decrease was significant. Why might management might choose not to
permanently institute the employee incentive program?
Answer: Although statistically significant, the practical significance (cost of the incentive
program compared to the savings due to the decrease in defect rate of the chips)
might not be great enough to warrant instituting the program permanently.
Explanation:
The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Crest
toothpaste. A marketing director suggests that the company invest in a new marketing campaign which will include
advertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for one
month to determine if the market share increases with new labels.
89) Over the trial month the market share in the sample rose to 22% of shoppers. The
89)
company’s board of directors decided this increase was significant. Now that they have
concluded the new marketing campaign works, why might they still choose not to invest in
the campaign?
Answer: Although statistically significant, the small increase in sales from 19.5% to 22%
might not be enough to justify the expense of the new marketing campaign
nationwide.
Explanation:
A statistics professor asked her students whether or not they were registered to vote. In a sample of 50 of her students
(randomly sampled from her 700 students), 35 said they were registered to vote.
90) According to a September 2004 Gallup poll, about 73% of 18– to 29–year–olds said that
90)
they were registered to vote. Does the 73% figure from Gallup seem reasonable for the
professor’s class? Explain.
Answer: The 73% figure from Gallup seems reasonable since 73% lies in our confidence
interval.
Explanation:
36The International Olympic Committee states that the female participation in the 2004 Summer Olympic Games was 42%,
even with new sports such as weight lifting, hammer throw, and modern pentathlon being added to the Games. Broadcasting
and clothing companies want to change their advertising and marketing strategies if the female participation increases at the
next games. An independent sports expert arranged for a random sample of pre–Olympic exhibitions. The sports expert
reported that 202 of 454 athletes in the random sample were women. Is this strong evidence that the participation rate may
increase?
91) Test an appropriate hypothesis and state your conclusion. Answer:
Hypotheses: H0: p = 0.42 . The female participation rate in the Olympics is 42%.
HA: p > 0.42 . The female participation rate in the Olympics is greater
91)
than 42%.
Model: Okay to use the Normal model because the sample is random, these 454
athletes are less than 10% of all athletes at exhibitions, and np=(454)(0.42) = 190.68
10 and nq = (454)(0.58) = 263.32 10. Use a N(0.42, 0.023) model, do a 1–proportion
z–test.
^
Mechanics: n = 454, x = 202, p = 202
454 = 0.445
z = 0.445 – 0.42
= 1.09
(0.42)(0.58)
454
^
P = P(p > 0.445) = P z > 1.09 = 0.138
Conclusion: With a p–value (0.138) so large, I fail to reject the null hypothesis that
the proportion of female athletes is 0.42. There is not enough evidence to suggest
that the proportion of female athletes will increase.
Explanation:
37A statistics professor asked her students whether or not they were registered to vote. In a sample of 50 of her students
(randomly sampled from her 700 students), 35 said they were registered to vote.
92) Find a 95% confidence interval for the true proportion of the professor’s students who were
92)
registered to vote. (Make sure to check any necessary conditions and to state a conclusion
in the context of the problem.)
Answer:
We have a random sample of less than 10% of the professor’s students, with 35
expected successes (registered) and 15 expected failures (not registered), so a
Normal model applies.
n = 50, p = 35
^ ^ ^ ^
50 = 0.70, q = 1 – p = 0.30, so SE(p) = ^^
pq
n = (0.70)(0.30)
50 = 0.065
Our 95% confidence interval is:
^ ^
p ± z*SE(p) = 0.70 ± 1.96(0.065) = 0.70 ± 0.127 = 0.573 to 0.827
We are 95% confident that between 57.3% and 82.7% of the professor’s students are
registered to vote.
Explanation:
93) It is generally believed that electrical problems affect about 14% of new cars. An
automobile mechanic conducts diagnostic tests on 128 new cars on the lot.
a. Describe the sampling distribution for the sample proportion by naming the model and
telling its mean and standard deviation. Justify your answer.
b. Sketch and clearly label the model.
c. What is the probability that in this group over 18% of the new cars will be found to have
electrical problems?
Answer:
a. We can assume these cars are a representative sample of all new cars, and
certainly less than 10% of them. We expect np = (128)(0.14)= 17.92 successes
(electrical problems) and nq = (128)(0.86) = 110.08 failures (no problems) so the
sample is large enough to use the sampling model N(0.14, 0.031).
^
SD(p) = pq
n = (0.14)(0.86)
128 = 0.031
b.
93)
^
c. z = p – p
^
SD(p)
^
P(p > 0.18) = P(z > 0.18 – 0.14
0.031 ) =
P(z > 1.30) = 0.096, about 10%
Explanation:
38MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
94) We are about to test a hypothesis using data from a well–designed study. Which is true?
I. A small P–value would be strong evidence against the null hypothesis.
II. We can set a higher standard of proof by choosing = 10% instead of 5%.
III. If we reduce the alpha level, we reduce the power of the test.
A) None
B) I and III only
C) II only
D) III only
E) I only
Answer: B
Explanation: A)
B)
C)
D)
E)
95) We have calculated a confidence interval based on a sample of size n = 100. Now we want to get a
better estimate with a margin of error that is only one–fourth as large. How large does our new
sample need to be?
A) 50 B) 1600 C) 25 D) 400 E) 200
Answer: B
Explanation: A)
B)
C)
D)
E)
96) We are about to test a hypothesis using data from a well–designed study. Which is true?
I. A large P–value would be strong evidence against the null hypothesis.
II. We can set a higher standard of proof by choosing = 10% instead of 5%.
III. If we reduce the risk of committing a Type I error, then the risk of a Type II error will also
decrease.
A) I and II only
B) II only
C) III only
D) none
E) I only
Answer: D
Explanation: A)
B)
C)
D)
E)
94)
95)
96)
3997) We have calculated a 95% confidence interval and would prefer for our next confidence interval to
have a smaller margin of error without losing any confidence. In order to do this, we can
I. change the z* value to a smaller number.
II. take a larger sample.
III. take a smaller sample.
A) I only B) I and II C) II only D) I and III E) III only
Answer: C
Explanation: A)
97)
B)
C)
D)
E)
98) Which is true about a 99% confidence interval based on a given sample?
I. The interval contains 99% of the population.
II. Results from 99% of all samples will lie in this interval.
III. The interval is wider than a 95% confidence interval would be.
A) III only
B) I only
C) II only
D) none
E) II and III only
Answer: A
Explanation: A)
B)
C)
D)
E)
99) A certain population is strongly skewed to the right. We want to estimate its mean, so we will
collect a sample. Which should be true if we use a large sample rather than a small one?
I. The distribution of our sample data will be closer to normal.
II. The sampling model of the sample means will be closer to normal.
III. The variability of the sample means will be greater.
A) II only
B) II and III only
C) I and III only
D) I only
E) III only
Answer: A
Explanation: A)
B)
C)
D)
E)
98)
99)
40100) Which of the following is true about Type I and Type II errors?
I. Type I errors are always worse than Type II errors.
II. The severity of Type I and Type II errors depends on the situation being tested.
III. In any given situation, the higher the risk of Type I error, the lower the risk of Type II error.
A) I only B) II and III C) II only D) I and III E) III only
Answer: B
Explanation: A)
B)
C)
D)
E)
101) Which is true about a 98% confidence interval for a population proportion based on a given
sample?
I. We are 98% confident that other sample proportions will be in our interval.
II. There is a 98% chance that our interval contains the population proportion.
III. The interval is wider than a 95% confidence interval would be.
A) III only B) I and II C) None D) II only E) I only
Answer: A
Explanation: A)
B)
C)
D)
E)
102) Not wanting to risk poor sales for a new soda flavor, a company decides to run one more taste test
on potential customers, this time requiring a higher approval rating than they had for earlier tests.
This higher standard of proof will increase
I. the risk of Type I error
II. the risk of Type II error
III. power
A) I and II B) I and III C) I only D) II only E) III only
Answer: D
Explanation: A)
B)
C)
D)
E)
103) The manager of an orchard expects about 70% of his apples to exceed the weight requirement for
“Grade A” designation. At least how many apples must he sample to be 90% confident of
estimating the true proportion within ± 4%?
A) 19 B) 356 C) 505 D) 89 E) 23
Answer: B
Explanation: A)
B)
C)
D)
E)
100)
101)
102)
103)
41104) A certain population is bimodal. We want to estimate its mean, so we will collect a sample. Which
should be true if we use a large sample rather than a small one?
I. The distribution of our sample data will be clearly bimodal.
II. The sampling distribution of the sample means will be approximately normal.
III. The variability of the sample means will be smaller.
A) II only
B) I only
C) II and III
D) I, II, and III
E) III only
Answer: D
Explanation: A)
B)
C)
D)
E)
105) We have calculated a confidence interval based upon a sample of n = 200. Now we want to get a
better estimate with a margin of error only one fifth as large. We need a new sample with n at least
…
A) 450 B) 40 C) 5000 D) 240 E) 1000
Answer: C
Explanation: A)
B)
C)
D)
E)
106) A certain population is strongly skewed to the left. We want to estimate its mean, so we collect a
sample. Which should be true if we use a large sample rather than a small one?
I. The distribution of our sample data will be more clearly skewed to the left.
II. The sampling model of the sample means will be more skewed to the left.
III. The variability of the sample means will greater.
A) III only
B) I only
C) I and III only
D) II and III only
E) II only
Answer: B
Explanation: A)
B)
C)
D)
E)
42
104)
105)
106)107) We have calculated a confidence interval based on a sample of n = 180. Now we want to get a better
estimate with a margin of error only one third as large. We need a new sample with n at least…
A) 60 B) 1620 C) 20 D) 540 E) 312
Answer: B
Explanation: A)
B)
C)
D)
E)
108) Which is true about a 95% confidence interval based on a given sample?
I. The interval contains 95% of the population.
II. Results from 95% of all samples will lie in the interval.
III. The interval is narrower than a 98% confidence interval would be.
A) II only
B) III only
C) II and III only
D) None
E) I only
Answer: B
Explanation: A)
B)
C)
D)
E)
107)
108)
43Answer Key
Testname: PART5
1) B
2) E
3) C
4) D
5) A
6) E
7) E
8) E
9) A
10) D
11) E
12) D
13) C
14) A
15) C
^^
16)
Since ME = z* pq
, we have 0.03 = z* (0.58)(0.42)
n
1150 or z* 2.06. Confidence level is 96%.
17) Although statistically significant, the small increase in sales from 28% to 30% of shoppers might not be enough to
justify the expense of purchasing the mannequins.
18) The confidence interval contains values that are all below the hypothesized value of 8%, so the data provide
convincing evidence that the incentive program lowers the defect rate of the computer chips.
19) The smaller hospital should experience greater variability in the weekly percentage of male babies. We’d expect the
larger hospital to stay closer to the expected 50–50 ratio.
20) We are 90% confident that between 7.9% and 12.6% of the applicants attend private high schools.
21) Two methods are shown below to solve this problem:
Method 1:
Let P = one can of pumpkin pie mix and T = four cans of pumpkin pie mix.
We are told that the contents of the cans are normally distributed, and can assume that the content amounts are
independent from can to can.
E(T) = E(P1 + P2 + P3 + P4) = E(P1) + E(P2) + E(P3) + E(P4) = 120 ounces
Since the content amounts are independent,
Var(T) = Var(P1 + P2 + P3 + P4) = Var(P1) + Var(P2) + Var(P3) + Var(P4) = 16
SD(T) = Var(T) = 16 = 4 ounces
We model T with N(120, 4) z = 126 – 120
4 = 1.5 P = P(T > 126) = P(z > 1.5) = 0.067
There is a 6.7% chance that four randomly selected cans of pumpkin pie mix contain more than 126 ounces.
Method 2:
Using the Central Limit Theorem approach, let y = average content of cans in sample
Since the contents are Normally distributed, y is modeled by N 30, 2
4
P y > 126
4 = P(y > 31.5) = P z > 31.5 – 30
1 = P(z > 1.5) = 0.067
There is about a 6.7% chance that 4 randomly selected cans will contain a total of over 126 ounces.
.
44Answer Key
Testname: PART5
22)
H0 : p1 – p2 = 0 HA : p1 – p2 > 0
People were randomly assigned to groups, we assume the groups are independent, and 20, 34, 13, 33 are all 10. OK to
do a 2–proportion z–test.
^ ^ ^
p1 = 0.370, p2 = 0.283, p = 20 + 13
54 + 46 = 0.33
z = (0.370 – 0.283) – 0
(0.33)(0.67)
54 – (0.33)(0.67)
46
= 0.93
23)
^ ^
P = P(p1 – p2 > 0.087) = P(z > 0.93) = 0.176
We fail to reject the null hypothesis because P is very large. We do not have evidence that the help program is
beneficial, so it should not be funded.
^^
ME = z
* pq
n
0.04 = 1.645 (0.12)(0.88)
n
n = 1.645 (0.12)(0.88)
0.04
n = 178.60 179
^
They should sample at least 179 applicants. (423 if p = 0.5 is used)
24) If birth weight was not a risk factor for susceptibility to depression, an observed difference in incidence of depression
this large (or larger) would occur in only 2.48% of such samples.
25) A Type I error would be deciding the percentage of customers who’ll make purchases will go up when in fact it won’t.
The store’s owner would waste money buying useless mannequins.
26) We are 90% confident that between 40.0% and 47.2% of the employment records from the United States labor force are
for females.
27)
With the conditions satisfied (from Problem 1), the sampling distribution of the difference in
proportions is approximately Normal with a mean of pB – pC, the true difference between the
population proportions. We can find a two–proportion z–interval.
^ ^
We know: nB = 1009, pB = 0.81, nC = 1004, pC = 0.86
^ ^
We estimate SD(pB – pC) as
^ ^ ^ ^
^ ^
SE(pB – pC) = pB qB
nB + pC qC
nC = (0.81)(0.19)
1009 + (0.86)(0.14)
1004 = 0.0165
^ ^
ME = z* × SE(pB – pC) = 2.576(0.0165) = 0.0425
^ ^
The observed difference in sample proportions = pB – pC = 0.81 – 0.186 = –0.05, so the 99%
confidence interval is –0.05 ± 0.0425, or –9.3% to –0.8%.
We are 99% confident that the proportion of Britons who read at least one book in the past year is
between 0.8–percentage points and 9.3–percentage points lower than the proportion of Canadians who read at least
one book in the past year.
28) The null hypothesis is that 19.5% of all people who use toothpaste buy Crest. The alternative hypothesis is that the
percentage of all people who use toothpaste who use Crest is greater than 19.5%. In symbols: H0: p = 0.195 and HA: p
> 0.195
45Answer Key
Testname: PART5
29)
30)
Hypotheses: H0 : p = 0.28. The proportion of all black Americans that were unable to afford medical care in the last
year is 28%.
HA : p > 0.28. The proportion of all black Americans that were unable to afford medical care in the last
year is greater than 28%.
Model: Okay to use the Normal model because the sample is random, these 801 black Americans are less than 10% of
all black Americans, and
np = (801)(0.28) = 224.28 10 and nq = (801)(0.72)= 576.72 10.
We will do a one–proportion z–test.
^
Mechanics: n = 801, p = 0.38
z = 0.38 – 0.28
(0.28)(0.72)
801
^
P(p > 0.38) = P z > 6.29 0
Conclusion: With a P–value so small (just about zero), I reject the null hypothesis. There is enough evidence to suggest
that the proportion of black Americans who were not able to afford medical care in the past year is more than 28%.
a. We can assume these kids are a random sample of all children, and certainly less than 10% of them. We
expect np = (133)(0.12) = 15.96 successes and 117.04 failures so the sample size is large enough to use the
sampling model N(0.12, 0.028).
b.
c. z = 0.15 – 0.12
0.028 = 1.07
^
P(p > 0.15) = P(z > 1.07) = 0.142
31) If many random samples were taken, 90% of the confidence intervals produced would contain the actual percentage of
all female employment records in the United States labor force.
32) = 1 – 0.98
2 = 0.01 = 1%
33) A Type II error would be deciding the percentage of customers who’ll make purchases won’t go up when in fact it
would have. The store’s owner would miss an opportunity to increase sales.
46Answer Key
Testname: PART5
^
= 0.0158; sample proportion: p = 0.55
34)
Hypothesis: H0 : p = 0.50 HA : p > 0.50
Plan: Okay to use the Normal model because the trials are independent (random sample of U.S. adults), these 1003 U.S.
adults are less than 10% of all U.S. adults, and np0 = (1003)(0.50) = 501.5 10 and nq0 = (1003)(0.50) = 501.5 10.
We will do a one–proportion z–test.
Mechanics: SD(p0) = p0q0
n = (0.50)(0.50)
1003 ^
P(p > 0.55) = P(z > 0.55 – 0.50
0.0158 ) = P(z > 3.16) = 0.0008
With a P–value of 0.0008, I reject the null hypothesis. There is strong evidence that the proportion of U.S. adults who
feel they get enough sleep is more than 50%.
35) The power would increase because of the larger sample size.
36) A Type I error would be deciding the percentage of defective chips has decreased, when in fact it has not. The
company would waste money on a new incentive program that does not decrease the defect rate of the chips.
37) One–tail, upper tail test. We are concerned that the proportion of people who are not able to afford medical care is
higher among black Americans.
38) Two methods can be used to solve this problem:
Method 1:
Let B = weight of one box of cereal and T = weight of 12 boxes of cereal. We are told that the contents of the boxes are
approximately Normal, and we can assume that the content amounts are independent from box to box.
E(T) = E(B1 + B2 +…+ B12 ) = E(B1) + E(B2) + … + E(B12 ) = 156 ounces
Since the content amounts are independent,
Var(T) = Var(B1 + B2 +…+ B12 ) = Var(B1) + Var(B2) + … + Var(B12 ) = 3
SD(T) = Var(T) = 3 = 1.73 oz.
We model T with N(156, 1.73).
z = 160 – 156
1.73 = 2.31 and P(T > 160) = P(z > 2.31) = 0.0104
There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160 ounces.
Method 2:
Using the Central Limit Theorem approach, let y = average content of boxes in the case. Since the contents are
Normally distributed, y is modeled by N 13, 0.5
12
P y > 160
12 = P(y > 13.33) = P z > 13.33 – 13
0.5
= P(z > 2.31) = 0.0104.
.
12
There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160 ounces.
39) The confidence interval contains the current value of 19.5%, so the product trials do not provide convincing evidence
that the new marketing campaign would increase sales.
40) The smaller aquarium would experience more variability in the season percentage of male births. We would expect the
larger aquarium to stay more consistent and closer to the 50–50 ratio for gender births.
41) We have a random sample of less than 10% of the applicants, with 46 successes and 404 failures, so a Normal model
applies. The confidence interval is (0.079, 0.126).
42) Advantage: the test would have greater power to detect a positive effect of the mannequins. Disadvantage: she’d be
more likely to think the mannequins were effective even if they were not.
47Answer Key
Testname: PART5
43) There is no probability involved–once the interval is constructed, the true proportion of the professor’s students who
were registered to vote is in the interval or it is not.
44)
a. Conditions:
* Randomization Condition: We are told that we have random samples.
* 10% Condition: We have less than 10% of all men and less than 10% of all women.
* Independent samples condition: The two groups are clearly independent of each other.
* Success/Failure Condition: Of the men, 88 exercise regularly and 62 do not; of the women, 130 exercise regularly and
70 do not. The observed number of both successes and failures in both groups is at least 10.
With the conditions satisfied, the sampling distribution of the difference in proportions is approximately Normal with
a mean of pM – pW , the true difference between the population proportions. We can find a two–proportion
z–interval.
We know: nM = 150, pM = 88
^ ^
150 = 0.587, nW = 200, pW = 130
200 = 0.650
^ ^
We estimate SD(pM – pW) as
^ ^ ^ ^
^ ^
SE(pM – pW) = pM qM
nM + pW qW
nW = (0.587)(0.413)
150 + (0.65)(0.35)
200 = 0.0525
^ ^
ME = z* × SE(pM – pW) = 1.96(0.0525) = 0.1029
^ ^
The observed difference in sample proportions = pM – pW = 0.587 – 0.650 = –0.063, so the 95%
confidence interval is –0.063 ± 0.1029, or –16.6% to 4.0%.
We are 95% confident that the proportion of women who exercise regularly is between 4.0% lower and 16.6% higher
than the proportion of men who exercise regularly.
b. Since zero is contained in my confidence interval, I cannot say that a higher proportion of women than men exercise
regularly. My confidence interval does not support my friend’s claim.
48Answer Key
Testname: PART5
45)
H0 : pm – pw = 0, HA: pm – pw > 0, where m = men and w = women
* Randomization Condition: The men and women in the sample were randomly selected by the
humanitarian organization.
* 10% Condition: The number of men and women in Qatar is greater than 2340 (10 × 234) and
2510 (10 × 251), respectively.
* Independent samples condition: The two groups are independent of each other because the
samples were selected at random.
* Success/Failure Condition: In men, 45 were illiterate and 189 were not. In women, 42 were
illiterate and 209 were not. The observed number of both successes and failures in both groups is larger than 10.
Because the conditions are satisfied, we can model the sampling distribution of the difference in proportions with a
Normal model. We can perform a two–proportion z–test.
^ ^ ^
We know: nm = 234, pm = 0.192, nw = 251, pw = 0.167, and ppooled = 45 + 42
234 + 251 = 0.179.
^ ^
SEpooled (pm – pw) = ^ ^ ^ ^
ppooled qpooled
+ ppooled qpooled
nm
nw
=
(0.179)(0.821)
234 + (0.179)(0.821)
251 = 0.0349
46)
^ ^
The observed difference in sample proportions = pm – pw = 0.192 – 0.167 = 0.025
^ ^
z = (pm – pw) – 0
^ ^
= 0.025 – 0
0.0349 = 0.716
SEpooled (pm – pw)
P = P(z > 0.716) = 0.24
The P–value of 0.24 is high, so we fail to reject the null hypothesis. There is insufficient evidence to conclude that
illiteracy rate in men is higher than for women in Qatar.
a. We want to find the probability that no more than 12 students in the class will say that they frequently experience
stress. This is the same as asking the probability of finding less than 26.7% of “stressed” students in a class of 45
students.
Check the conditions:
1. 10% condition: 45 students is less than 10% of all students who could take the class
2. Success/failure cond.: np = 45(0.33) = 14.85, nq = 45(0.67) = 30.15, which both exceed 10
We can use the N(0.33 (0.33)(0.67)
45 = 0.070) to model the sampling distribution.
We need to standardize the 26.7% and then find the probability of getting a z–score less than or equal to the one we
find: z = 0.267 – 0.33
0.070 = –0.90
^
P(p < 0.267) = P(z < –0.90) = 0.1841, so the probability is about 18.4% that no more than 12 students will say that they
frequently experience stress in their daily lives.
b. From part a, we can use N(0.33, 0.070) to model the sampling distribution. Twenty students is about 44.4% of the
class. This is about 1.63 standard deviations above what we would expect, which is not a surprising result.
49Answer Key
Testname: PART5
47) If many random samples were taken, 95% of the confidence intervals produced would contain the actual percentage of
the professor’s students who are registered to vote.
48) We have a random sample of frogs drawn from a much larger population. With a sample of size 50 the CLT says that
the approximate sampling model for sample means will be N(118, 1.98). A sample mean of only 110 grams is about 4
standard deviations below what we expect, a very unusual result.
49) Even if the device actually performs no better than guessing, we could expect to find the explosives 16 or more times
out of 50 about 13% of the time.
50) Advantage: management would be less likely to conclude the incentive program was effective if it really were not.
Disadvantage: the test would have less power to detect a positive effect of the new incentive program.
51)
We want to know whether the percentage of people who paid lower taxes was different based on whether they sought
tax advice or not.
H0: padvice – pno advice = 0, HA: padvice – pno advice > 0
Conditions:
* Independence: The people who filed tax reports were randomly selected and do not influence each other.
* Random Condition: The people who filed tax reports were randomly selected.
* 10% Condition: 72 is less than 10% of people who didn’t get tax advice, and 105 is less than 10% of people who did
get tax advice.
* Success/Failure: All observed counts (48, 19, 24, and 86) are at least 10.
Because all conditions have been satisfied, we can model the sampling distribution of the difference in proportions
with a Normal model. We can perform a two–proportion z–test.
Let ‘Advice’ group be the people who sought tax advice and the ‘No advice’ group be the people who did not seek tax
advice.
^ ^
We know: nadvice = 105, nno advice = 72, padvice = 0.819, pno advice = 0.333.
^
ppooled = 24 + 86
72 + 105 = 0.621
^ ^
SEpooled (padvice – pno advice) = (0.62)(0.38)
72 + (0.62)(0.38)
105 = 0.074
^ ^
The observed difference in sample proportions is padvice – pno advice = 0.819 – 0.333 = 0.486
^ ^
z = (padvice – pno advice) – 0
^ ^
= 0.49 – 0
0.074 = 6.62
SEpooled (padvice – pno advice)
P = P(z > 6.62) < 0.0001
52)
The P–value is small, so we reject the null hypothesis. There is strong evidence of a difference in the tax percentages
paid between the group who has tax advice and the group who did not have tax advice. It appears that if you have tax
advice you are more likely to pay a lower percentage of taxes to the government.
^^
ME = z* pq
n
0.05 = 1.645 (0.46)(0.54)
n
n = 1.645 (0.46)(0.54)
0.05
n = 268.87 269
They should sample at least 269 employment records.
50Answer Key
Testname: PART5
53)
H0 : pB – pC = 0, HA: pB – pC < 0, where B = Britons and C = Canadians
Conditions:
* Randomization Condition: The Britons and Canadians were randomly sampled by Gallup.
* 10% Condition: The number of Britons and Canadians is greater than 10,090 (10 × 1009) and
10,040 (10 × 1004), respectively.
* Independent samples condition: The two groups are clearly independent of each other.
* Success/Failure Condition: Of the Britons, approximately 817 read at least one book and 192 did not; of the
Canadians, approximately 863 read at least one book and 141 did not. The observed number of both successes and
failures in both groups is larger than 10.
Because the conditions are satisfied, we can model the sampling distribution of the difference
in proportions with a Normal model. We can perform a two–proportion z–test.
^ ^
We know: nB = 1009, pB = 0.81, nC = 1004, pC = 0.86, and
^
ppooled = 817 + 863
1009 + 1004 = 1680
2013 = 0.835
^ ^ ^ ^
^ ^
SEpooled (pB – pC) = ppooled qpooled
nB + ppooled qpooled
nC = (0.835)(0.165)
1009 + (0.835)(0.165)
1004 = 0.0165
^ ^
The observed difference in sample proportions = pB – pC = 0.81 – 0.86 = –0.05
^ ^
z = (pB – pC) – 0
^ ^
= –0.05 – 0
0.0165 = –3.03, so the P–value = P(z < –3.03) = 0.0012
SEpooled (pB – pC)
The P–value of 0.0012 is low, so we reject the null hypothesis.There is strong evidence that the percentage of Britons
who read at least one book in the past year is less than the percentage of Canadians who read at least one book in the
past year.
54) One–tail, upper tail. The device is effective only if it can detect explosives at a rate higher than chance (25%).
55) The power of the test will increase.
56) One–tail, upper test. The companies will change strategies only if there is strong evidence of an increase in female
participation rate from the current rate of 42%.
57) = 1 – 0.98
2 = 0.01 = 1%
58) Advantage: the test would have greater power to detect a positive effect of the new marketing campaign.
Disadvantage: they would be more likely to think the marketing campaign was effective even if it were not.
51Answer Key
Testname: PART5
59)
a. Conditions:
* Independence: We assume that one trucker’s driving times do not influence other trucker’s driving times.
* Random Condition: We assume that trucks are stopped at random.
* 10% Condition: This sample of 348 truckers is less than 10% of all truckers.
* Success/Failure: 49 tickets and 299 tickets are both at least 10, so our sample is large enough.
Under these conditions the sampling distribution of the proportion can be modeled by a Normal model. We will find a
one–proportion z–interval.
^^
^ ^
We know n = 348 and p = 0.14 , so SE(p) = pq
n = (0.14)(0.86)
348 = 0.0186
The sampling model is Normal, for a 95% confidence interval the critical value is z* = 1.96.
^
The margin of error is ME = z* × SE(p) = 1.96(0.0186) = 0.0365.
The 95% confidence interval is 0.14 ± 0.0365 or (0.1035, 0.1765).
We are 95% confident that between 10.4% and 17.7% of truck drivers have driven too many hours in a day.
b. If we repeated the sampling and created new confidence intervals many times we would expect about 95% of those
intervals to contain the actual proportion of truck drivers that have driven too many hours in a day.
60) Check the conditions:
1. Random sampling condition: We have been told that this is a random sample.
2. Independence assumption: It’s reasonable to think that the scores of the 25 students are mutually independent.
3. 10% condition: 25 students is certainly less than 10% of all students who took the exam.
We’re assuming that the model for composite ACT scores has mean µ = 21.4 and standard deviation = 1.05. Since the
sample size is large enough and the distribution of ACT scores is most likely unimodal and symmetric, CLT allows us
to describe the sampling distribution of y with a Normal model with mean 21.4 and SD = (y) = 1.05
25 = 0.21.
An average score of 22 is z = 22 – 21.4
0.21 = 2.86 SDs above the mean.
P(x > 22) = P(Z > 2.86) = 0.0021, so the probability that the average composite ACT score for a sample of 25 randomly
selected students is 22 or more is 0.0021.
61) The smaller poll would have more variability and would thus be more likely to vary from the actual approval rating of
54%. We would expect the larger poll to be more consistent with the 54% rating. So, it is more likely that the smaller
poll would report that the President’s approval rating is below 50%.
62) ME = z
* pq
n
0.04 = 1.96 (0.73)(0.27)
n
n = (1.96)2(0.73)(0.27)
(0.04)2 = 473.24 n = 474
Note: Since there are only 700 students in the professor’s class, she cannot sample this many students without violating
the 10% condition!
63) A Type I error would be concluding the proportion of people who will buy Crest toothpaste will go up when in fact it
won’t. The company would waste money on a new marketing campaign that will not increase sales.
64) If the proportion of female athletes has not increased, we could expect to find at least 202 females of 454 pre–Olympic
athletes about 13.8% of the time.
65) The null hypothesis is that the percentage of all customers who buy something is 28%. The alternative hypothesis is
that the percentage of all customers who buy something is greater than 28%. In symbols: H0: p = 0.28 HA: p > 0.28
52Answer Key
Testname: PART5
66)
a. We assume the cars stopped are representative of all cars in the area, and that 628 < 10% of the cars. The 22 successes
(expired stickers) and 608 failures are both greater than 10. It’s okay to use a Normal model and find a one–proportion
z–interval.
^
p = 0.035; 0.035 ± 1.96 (0.035)(0.965)
628 = (0.021, 0.049)
The interval (0.021,0.049) means that we are 95% confident that between 2% and 5% of local cars have expired safety
inspection stickers.
b. If we repeated the sampling and created new confidence intervals many times we’d expect about 95% of those
intervals to contain the actual proportion of cars with expired stickers.
67) = 1 – 0.95
2 = 0.025 2.5%
68)
Conditions:
* Randomization Condition: The treatments were randomly assigned to subjects.
* 10% Condition: The subjects were randomly selected. We assume it is from a large population of
allergy sufferers.
* Independent samples condition: The two groups are independent of each other because the treatments were assigned
at random.
* Success/Failure Condition: For Brand C, 22 were drowsy and 58 were not. For Brand I, 14 were
drowsy and 66 were not. The observed number of both successes and failures in both groups is larger than 10.
Because the conditions are satisfied, we can model the sampling distribution of the difference in
proportions with a Normal model.
^ ^
We know: nI = 80, pI = 0.175, nC = 80, pC = 0.275
^ ^
We estimate SD(pC – pI) as
^ ^ ^ ^
^ ^
SE(pC – pI) = pC qC
nC + pI qI
nI = (0.275)(0.725)
80 + (0.175)(0.825)
80 = 0.066
^ ^
ME = z* × SE(pC – pI) = 1.96(0.066) = 0.128
^ ^
The observed difference in sample proportions = pC – pI = 0.275 – 0.175 = 0.10, so the 95% confidence interval is 0.10 ±
0.128, or (–0.028, 0.228)
We are 95% confident that the difference between the population proportions of patients that reported drowsiness for
Brand C and Brand I is between –2.8% and 22.8%.
69) A Type II error would be deciding the proportion of people who will buy Crest toothpaste won’t go up when in fact it
would have. The company would miss an opportunity to increase sales.
70) We have a random sample of adult pythons drawn from a much larger population. With a sample size of 30, the CLT
says that the approximate sampling model for sample means will be N(20.5, 0.42). A sample mean of only 19.5 feet is
about 2.38 standard deviations below what we expect. The sample mean of 19.5 feet is unusually small.
71) A Type II error would be deciding the percentage of defective chips has not decreased, when in fact it has. The
company would miss an opportunity to decrease the defect rate of the chips.
72) A hypothesis test at a 0.05 level should reach the same conclusion if a two–sided test is used. If the
alternate hypothesis were “less drowsy”, the significance level would need to be changed, or a
different conclusion could occur.
53Answer Key
Testname: PART5
73)
Hypotheses: H0: p = 0.25. The device can detect explosives at the same level as guessing. device can detect explosives at a level greater than chance.
HA: p > 0.25. The
Model: OK to use a Normal model because trials are independent (box is randomly chosen each time), and np = 12.5, nq
= 37.5. Do a 1–proportion z–test
^
Mechanics: p = 0.32, z = 1.14, P = 0.13
74)
Conclusion: With a P–value so high I fail to reject the null hypothesis. This test does not provide convincing evidence
that the sensor can detect the presence of explosives inside a box.
We have a random sample of less than 10% of the employment records, with 229 successes (females) and 296 failures
(males), so a Normal model applies.
^ ^ ^
n = 525, p = 0.436 and q = 0.564, so SE(p) = ^^
pq
n = (0.436)(0.564)
525 = 0.022
^
margin of error: ME = z* × SE(p) =(1.645)(0.022) = 0.0362.
75)
^
Confidence interval: p ± ME = 0.436 ± 0.0362 or (0.3998, 0.4722)
With the conditions satisfied (from Problem 1), the sampling distribution of the difference in
proportions is approximately Normal with a mean of pm – pw, the true difference between the
population proportions.
We can find a two–proportion z–interval.
^ ^
We know: nm = 234, pm = 0.192, nw = 251, pw = 0.167
^ ^
We estimate SD(pm – pw) as
^ ^ ^ ^
^ ^
SE(pm – pw) = pm qm
+ pw qw
nm
nw
= (0.192)(0.808)
234 + (0.167)(0.833)
251 = 0.0349
^ ^
ME = z* × SE(pm – pw) = 1.96(0.0349) = 0.0684
^ ^
The observed difference in sample proportions = pm – pw = 0.192 – 0.167 = 0.025, so the 95%
confidence interval is 0.025 ± 0.0684, or –4% to 9%.
We 95% confident that the proportion of illiterate men in Qatar is between 4–percentage points
lower and 9–percentage points higher than the proportion of illiterate women.
76) If many random samples were taken, 90% of the confidence intervals produced would contain the actual percentage of
all applicants who attend private schools.
54Answer Key
Testname: PART5
77) Because the hypothesized value of 28% is in the confidence interval the trial results do not provide convincing
evidence that the mannequins would help increase sales.
78) If the proportion of black Americans was 28%, we would almost never expect to find at least 38% of 801 randomly
selected black Americans responding “yes”.
79) No. Since 12% lies in the confidence interval it’s possible that the percentage of private school students in the applicant
pool matches the statewide enrollment rate.
^^
80)
Since ME = z * pq
, we have 0.03 = z* n
(0.28)(0.72)
1028 or z* = 0.03
(0.28)(0.72)
1028
2.14.
Our confidence level is approximately P(–2.14 < z < 2.14) = 0.9676, or 97%.
81) If baldness is not a risk factor an observed level of heart attacks this much higher (or more) would occur in only 6% of
such samples.
82) The power would increase because of the larger sample size.
83) The null hypothesis is that the defect rate is 8%. The alternative hypothesis is that the defect rate is lower than 8%. In
symbols: H0: p = 0.08 and HA: p < 0.08
84) No. Since 46% lies in the confidence interval, (0.3998, 0.4722), it is possible that the percentage of females in the labor
force matches Europe’s rate of 46% females in the labor force.
85) P y > 690
12 = P z > 57.5 – 56
= P(z > 1.08) = 14%
48
12
86) 0.03 = z* (0.62)(0.38)
800 ; z* = 1.75; P(–1.75 < z < 1.75) = 92% confidence
87) There is not sufficient evidence because 0 is contained in the interval. There may be no difference in the proportion of
drowsiness reported.
88) Although statistically significant, the practical significance (cost of the incentive program compared to the savings due
to the decrease in defect rate of the chips) might not be great enough to warrant instituting the program permanently.
89) Although statistically significant, the small increase in sales from 19.5% to 22% might not be enough to justify the
expense of the new marketing campaign nationwide.
90) The 73% figure from Gallup seems reasonable since 73% lies in our confidence interval.
55Answer Key
Testname: PART5
91)
Hypotheses: H0: p = 0.42 . The female participation rate in the Olympics is 42%.
HA: p > 0.42 . The female participation rate in the Olympics is greater than 42%.
Model: Okay to use the Normal model because the sample is random, these 454 athletes are less than 10% of all athletes
at exhibitions, and np=(454)(0.42) = 190.68 10 and nq = (454)(0.58) = 263.32 10. Use a N(0.42, 0.023) model, do a
1–proportion z–test.
^
Mechanics: n = 454, x = 202, p = 202
454 = 0.445
z = 0.445 – 0.42
= 1.09
(0.42)(0.58)
454
^
P = P(p > 0.445) = P z > 1.09 = 0.138
92)
Conclusion: With a p–value (0.138) so large, I fail to reject the null hypothesis that the proportion of female athletes is
0.42. There is not enough evidence to suggest that the proportion of female athletes will increase.
We have a random sample of less than 10% of the professor’s students, with 35 expected successes (registered) and 15
expected failures (not registered), so a Normal model applies.
^^
n = 50, p = 35
^ ^ ^ ^
50 = 0.70, q = 1 – p = 0.30, so SE(p) = pq
n = (0.70)(0.30)
50 = 0.065
Our 95% confidence interval is:
^ ^
p ± z*SE(p) = 0.70 ± 1.96(0.065) = 0.70 ± 0.127 = 0.573 to 0.827
We are 95% confident that between 57.3% and 82.7% of the professor’s students are registered to vote.
56Answer Key
Testname: PART5
93)
a. We can assume these cars are a representative sample of all new cars, and certainly less than 10% of them. We
expect np = (128)(0.14)= 17.92 successes (electrical problems) and nq = (128)(0.86) = 110.08 failures (no problems) so the
sample is large enough to use the sampling model N(0.14, 0.031).
^
SD(p) = pq
n = (0.14)(0.86)
128 = 0.031
b.
^
c. z = p – p
^
SD(p)
^
P(p > 0.18) = P(z > 0.18 – 0.14
0.031 ) =
P(z > 1.30) = 0.096, about 10%
94) B
95) B
96) D
97) C
98) A
99) A
100) B
101) A
102) D
103) B
104) D
105) C
106) B
107) B
108) B
57
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