3: Large-Sample Inferences Concerning the Difference Between Two Population or Treatment Proportions
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550
Chapter 11
Comparing Two Populations or Treatments
When comparing two populations or treatments on the basis of “success” proportions, it is common to focus on the quantity p1 Ϫ p2, the difference between the two
proportions. Because p^ 1 provides an estimate of p1 and p^ 2 provides an estimate of p2,
the obvious choice for an estimate of p1 Ϫ p2 is p^ 1 Ϫ p^ 2.
Because p^ 1 and p^ 2 each vary in value from sample to sample, so will the difference
p^ 1 2 p^ 2. For example, a first sample from each of two populations might yield
p^ 1 ϭ .69
p^ 2 ϭ .70
p^ 1 Ϫ p^ 2 ϭ .01
A second sample from each might result in
p^ 1 ϭ .79
p^ 2 ϭ .67
p^ 1 Ϫ p^ 2 ϭ .12
and so on. Because the statistic p^ 1 Ϫ p^ 2 is the basis for drawing inferences about
p1 Ϫ p2, we need to know something about its behavior.
Properties of the Sampling Distribution of
p^ 1
2
p^ 2
If two random samples are selected independently of one another, the following
properties hold:
1. m p^ 12p^ 2 5 p1 2 p2
This says that the sampling distribution of p^ 1 2 p^ 2 is centered at p1 2 p2,
so p^ 1 2 p^ 2 is an unbiased statistic for estimating p1 2 p2.
2. s2p^ 12p^ 2 5 s2p^ 1 1 s2p^ 2 5
p1 1 1 2 p 1 2
p2 11 2 p22
1
n1
n2
and
s p^ 12p^ 2 5
p1 1 1 2 p 1 2
p2 1 1 2 p 2 2
1
n1
n2
Å
3. If both n1 and n2 are large (that is, if n1 p1 $ 10, n1(1 2 p1) $ 10, n2 p2 $
10, and n2(1 2 p2) $ 10), then p^ 1 and p^ 2 each have a sampling distribution that is approximately normal, and their difference p^ 1 2 p^ 2 also has a
sampling distribution that is approximately normal.
The properties in the box imply that when the samples are independently selected and when both sample sizes are large, the distribution of the standardized
variable
z5
p^ 1 2 p^ 2 2 1p1 2 p22
p1 1 1 2 p 1 2
p2 11 2 p22
1
n1
n2
Å
is described approximately by the standard normal (z) curve.
A Large-Sample Test Procedure
Comparisons of p1 and p2 are often based on large, independently selected samples,
and we restrict ourselves to this case. The most general null hypothesis of interest has
the form
H0: p1 Ϫ p2 ϭ hypothesized value
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11.3 Large Sample Inferences Concerning a Difference Between Two Population or Treatment Proportions
551
However, when the hypothesized value is something other than 0, the appropriate test
statistic differs somewhat from the test statistic used for H0: p1 2 p2 5 0. Because this
H0 is almost always the relevant one in applied problems, we focus exclusively on it.
Our basic testing principle has been to use a procedure that controls the probability
of a Type I error at the desired level ␣. This requires using a test statistic with a sampling
distribution that is known when H0 is true. That is, the test statistic should be developed under the assumption that p1 5 p2 (as specified by the null hypothesis
p1 2 p2 5 0). In this case, p is used to denote the common value of the two population
proportions. The z variable obtained by standardizing p^ 1 2 p^ 2 then simplifies to
z5
p^ 1 2 p^ 2
p 11 2 p2
p 11 2 p2
1
n2
Å n1
Unfortunately, this cannot serve as a test statistic, because the denominator cannot be
computed. H0 says that there is a common value p, but it does not specify what that
value is. A test statistic can be obtained, though, by first estimating p from the sample
data and then using this estimate in the denominator of z.
When p1 ϭ p2, both p^ 1 and p^ 2 are estimates of the common proportion p. However, a better estimate than either p^ 1 or p^ 2 is a weighted average of the two, in which
more weight is given to the sample proportion based on the larger sample.
DEFINITION
The combined estimate of the common population proportion is
p^ c 5
n1 p^ 1 1 n2 p^ 2
total number of S’s in the two samples
5
n1 1 n 2
total of the two sample sizes
The test statistic for testing H0: p1 Ϫ p2 ϭ 0 results from using p^ c, the combined
estimate, in place of p in the standardized variable z given previously. This z statistic
has approximately a standard normal distribution when H0 is true, so a test that has
the desired significance level ␣ can be obtained by calculating a P-value using the
z table.
Summary of Large-Sample z Tests for
Null hypothesis:
Test statistic:
p1
2
p2
50
H0: p1 2 p2 5 0
z5
p^ 1 2 p^ 2
p^ c 11 2 p^ c2
p^ c 11 2 p^ c2
1
n1
n2
Å
Alternative hypothesis:
Ha: p1 2 p2 . 0
P-value:
Area under the z curve to the right of the computed z
Ha: p1 2 p2 , 0
Area under the z curve to the left of the computed z
Ha: p1 2 p2 2 0
(1) 2(area to the right of z) if z is positive
or
(2) 2(area to the left of z) if z is negative
(continued)
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552
Chapter 11
Comparing Two Populations or Treatments
Assumptions:
1. The samples are independently chosen random samples, or
treatments were assigned at random to individuals or objects
(or subjects were assigned at random to treatments).
2. Both sample sizes are large:
n1 p^ 1 $ 10 n1 11 2 p^ 12 $ 10 n2 p^ 2 $ 10 n2 11 2 p^ 22 $ 10
E X A M P L E 1 1 . 9 Duct Tape to Remove Warts?
Some people seem to believe that you can fix anything with duct tape. Even so, many
were skeptical when researchers announced that duct tape may be a more effective
and less painful alternative to liquid nitrogen, which doctors routinely use to freeze
warts. The article “What a Fix-It: Duct Tape Can Remove Warts” (San Luis Obispo
Tribune, October 15, 2002) described a study conducted at Madigan Army Medical
Center. Patients with warts were randomly assigned to either the duct tape treatment
or the more traditional freezing treatment. Those in the duct tape group wore duct
tape over the wart for 6 days, then removed the tape, soaked the area in water, and
used an emery board to scrape the area. This process was repeated for a maximum of
2 months or until the wart was gone. Data consistent with values in the article are
summarized in the following table:
Treatment
Liquid nitrogen freezing
Duct tape
n
Number with Wart
Successfully Removed
100
104
60
88
Do these data suggest that freezing is less successful than duct tape in removing
warts? Let p1 represent the true proportion of warts that would be successfully removed by freezing, and let p2 represent the true proportion of warts that would be
successfully removed with the duct tape treatment. We test the relevant hypotheses
H0: p1 Ϫ p2 ϭ 0
versus
Ha: p1 2 p2 Ͻ 0
using a ϭ .01. For these data,
p^ 1 5
60
5 .60
100
p^ 2 5
88
5 .85
104
Suppose that p1 ϭ p2 and let p denote the common value. Then p^ c, the combined
estimate of p, is
p^ c 5
n1 p^ 1 1 n2 p^ 2
100 1.602 1 104 1.852
5
5 .73
n1 1 n2
100 1 104
The nine-step procedure can now be used to perform the hypothesis test:
1. p1 Ϫ p2 is the difference between the true proportions of warts removed by
freezing and by the duct tape treatment.
(p1 ϭ p2)
2. H0: p1 Ϫ p2 ϭ 0
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11.3 Large Sample Inferences Concerning a Difference Between Two Population or Treatment Proportions
553
3. Ha: p1 2 p2 Ͻ 0
(p1 Ͻ p2, in which case the proportion of warts removed
by freezing is lower than the proportion by duct tape.)
4. Significance level: a ϭ .01
p^ 1 2 p^ 2
5. Test statistic: z 5
p^ c 11 2 p^ c2
p^ c 11 2 p^ c2
1
n1
n2
Å
6. Assumptions: The subjects were assigned randomly to the two treatments.
Checking to make sure that the sample sizes are large enough, we have
n1 p^ 1
n1 11 2 p^ 12
n2 p^ 2
n2 11 2 p^ 22
5 100 1.602
5 100 1.402
5 104 1.852
5 104 1.152
5 60 $ 10
5 40 $ 10
5 88.4 $ 10
5 15.6 $ 10
7. Calculations:
n1 5 100
n2 5 104
p^ 1 ϭ .60
p^ 2 ϭ .85
p^ c ϭ.73
and so
z5
.60 2 .85
1.732 1.272
1.732 1.272
1
Å 100
104
5
2.25
5 24.03
.062
8. P-value: This is a lower-tailed test, so the P-value is the area under the z curve
and to the left of the computed z ϭ Ϫ4.03. From Appendix Table 2,
P-value Ϸ 0.
9. Conclusion: Since P-value Յ a, the null hypothesis is rejected at significance
level .01. There is convincing evidence that the proportion of warts successfully
removed is lower for freezing than for the duct tape treatment.
Minitab can also be used to carry out a two-sample z test to compare two population proportions, as illustrated in the following example.
EXAMPLE 11.10
Not Enough Sleep?
Do people who work long hours have more trouble sleeping? This question was examined in the paper “Long Working Hours and Sleep Disturbances: The Whitehall
II Prospective Cohort Study” (Sleep [2009]: 737–745). The data in the accompanying table are from two independently selected samples of British civil service workers,
all of whom were employed full-time and worked at least 35 hours per week. The
authors of the paper believed that these samples were representative of full-time
British civil service workers who work 35 to 40 hours per week and of British civil
service workers who work more than 40 hours per week.
Work over 40 hours per week
Work 35–40 hours per week
n
Number who usually
get less than 7 hours
of sleep a night
1501
958
750
407
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554
Chapter 11
Comparing Two Populations or Treatments
Do these data support the theory that the proportion that usually get less than 7
hours of sleep a night is higher for those who work more than 40 hours per week than
for those who work between 35 and 40 hours per week? Let’s carry out a hypothesis
test with a 5 .01. For these samples
Over 40 hours per week
n1 5 1501
p^ 1 5
750
5 .500
1501
Between 35 and 40 hours per week
n2 5 958
p^ 2 5
407
5 .425
958
1. p1 5 proportion of those who work more than 40 hours per week who get less
than 7 hours of sleep
p2 5 proportion of those who work between 35 and 40 hours per week who get
less than 7 hours of sleep
2. H0: p1 Ϫ p2 ϭ 0
3. Ha: p1 2 p2 Ͼ 0
4. Significance level: a ϭ .01
p^ 1 2 p^ 2
p^ c 11 2 p^ c2
p^ c 11 2 p^ c2
1
n
n2
Å
1
6. Assumptions: The two samples were independently selected. It is reasonable
to regard the samples as representative of the two populations of interest. Checking to make sure that the sample sizes are large enough by using n1 5 1501,
p^ 1 5 .500, n2 5 958, and p^ 2 5 .425, we have
5. Test statistic: z 5
n1 p^ 1
n1 11 2 p^ 12
n2 p^ 2
n2 11 2 p^ 22
5 750.50 $ 10
5 750.50 $ 10
5 407.15 $ 10
5 550.85 $ 10
7. Calculations: Minitab output is shown below. From the output, z ϭ 3.64.
Test for Two Proportions
Sample
1
2
X
750
407
N
1501
958
Sample p
0.499667
0.424843
Difference ϭ p (1) Ϫ p (2)
Estimate for difference: 0.0748235
Test for difference ϭ 0 (vs . 0): Z ϭ 3.64 P-Value ϭ 0.000
8. P-value: From the computer output, P-value ϭ 0.000
9. Conclusion: Because P-value Յ a, the null hypothesis is rejected at significance
level .01.
There is strong evidence that the proportion that gets less than 7 hours of sleep a
night is higher for British civil service workers who work more than 40 hours per
week than it is for those who work between 35 and 40 hours per week. Note that
because the data were from an observational study, we are not able to conclude that
there is a cause and effect relationship between work hours and sleep. Although we
can conclude that a higher proportion of those who work long hours get less than
7 hours of sleep a night, we can’t conclude that working long hours is the cause of
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11.3 Large Sample Inferences Concerning a Difference Between Two Population or Treatment Proportions
555
shorter sleep. We should also note that the sample was selected from British civil
service workers, so it would not be a good idea to generalize this conclusion to all
workers.
A Confidence Interval
A large-sample confidence interval for p1 Ϫ p2 is a special case of the general z interval
formula
point estimate Ϯ (z critical value)(estimated standard deviation)
The statistic p^ 1 Ϫ p^ 2 gives a point estimate of p1 Ϫ p2, and the standard deviation of
this statistic is
s p^ 12p^ 2 5
p1 1 1 2 p 1 2
p2 1 1 2 p 2 2
1
n1
n2
Å
An estimated standard deviation is obtained by using the sample proportions p^ 1 and
p^ 2 in place of p1 and p2, respectively, under the square-root symbol. Notice that this
estimated standard deviation differs from the one used previously in the test statistic.
When constructing a confidence interval, there isn’t a null hypothesis that claims
p1 ϭ p2, so there is no assumed common value of p to estimate.
A Large-Sample Confidence Interval for p1 2 p2
When
1. the samples are independently selected random samples or treatments were
assigned at random to individuals or objects (or vice versa), and
2. both sample sizes are large:
n1 p^ 1 $ 10 n1 11 2 p^ 12 $ 10 n2 p^ 2 $ 10 n2 11 2 p^ 22 $ 10
a large-sample conﬁdence interval for p1 2 p2 is
1 p^ 1 2 p^ 22 6 1z critical value2
EXAMPLE 11.11
p^ 2 11 2 p^ 22
p^ 1 11 2 p^ 12
1
n1
n2
Å
Opinions on Freedom of Speech
The article “Freedom of What?” (Associated Press, February 1, 2005) described a
study in which high school students and high school teachers were asked whether they
agreed with the following statement: “Students should be allowed to report controversial issues in their student newspapers without the approval of school authorities.” It
was reported that 58% of the students surveyed and 39% of the teachers surveyed
agreed with the statement. The two samples—10,000 high school students and 8000
high school teachers—were selected from 544 different schools across the country.
We will use the given information to estimate the difference between the proportion of high school students who agree that students should be allowed to report
controversial issues in their student newspapers without the approval of school
authorities, p1, and the proportion of high school teachers who agree with the statement, p2.
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Chapter 11 Comparing Two Populations or Treatments
The sample sizes are large enough for the large-sample interval to be valid
(n1 p^ 1 5 10,000(.58) $ 10, n1(1 2 p^ 1) ϭ 10,000(.42) Ն 10, etc.). A 90% confidence
interval for p1 Ϫ p2 is
1p^ 1 2 p^ 22 6 1z critical value2
5 1.58 2 .392 6 11.6452
p^ 2 11 2 p^ 22
p^ 1 11 2 p^ 12
1
n1
n2
Å
1.582 1.422
1.392 1.612
1
Å 10,000
8000
5 .19 6 11.6452 1.00742
5 .19 6 .012
1.178, .2022
Statistical software or a graphing calculator could also have been used to compute the
endpoints of the confidence interval. Minitab output is shown here.
Test and CI for Two Proportions
Sample
1
2
X
5800
3120
N
10000
8000
Sample p
0.580000
0.390000
Difference = p (1) – p (2)
Estimate for difference: 0.19
90% CI for difference: (0.177902, 0.202098)
Test for difference = 0 (vs not = 0): Z = 25.33 P-Value = 0.000
Assuming that it is reasonable to regard these two samples as being independently
selected and also that they are representative of the two populations of interest, we
can say that we believe that the proportion of high school students who agree that
students should be allowed to report controversial issues in their student newspapers
without the approval of school authorities exceeds that for teachers by somewhere
between .178 and .202. We used a method to construct this estimate that captures
the true difference in proportions 90% of the time in repeated sampling.
EX E RC I S E S 1 1 . 3 7 - 1 1 . 5 7
11.37 A hotel chain is interested in evaluating reserva-
11.38 The authors of the paper “Adolescents and MP3
tion processes. Guests can reserve a room by using either
a telephone system or an online system that is accessed
through the hotel’s web site. Independent random samples of 80 guests who reserved a room by phone and 60
guests who reserved a room online were selected. Of
those who reserved by phone, 57 reported that they were
satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. Based
on these data, is it reasonable to conclude that the proportion who are satisfied is higher for those who reserve
a room online? Test the appropriate hypotheses using
a ϭ .05.
Players: Too Many Risks, Too Few Precautions” (Pediatrics [2009]: e953–e958) concluded that more boys
Bold exercises answered in back
Data set available online
than girls listen to music at high volumes. This conclusion
was based on data from independent random samples of
764 Dutch boys and 748 Dutch girls age 12 to 19. Of the
boys, 397 reported that they almost always listen to music
at a high volume setting. Of the girls, 331 reported listening to music at a high volume setting. Do the sample data
support the authors’ conclusion that the proportion of
Dutch boys who listen to music at high volume is greater
than this proportion for Dutch girls? Test the relevant
hypotheses using a .01 significance level.
Video Solution available
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.3 Large Sample Inferences Concerning a Difference Between Two Population or Treatment Proportions
11.39 After the 2010 earthquake in Haiti, many charitable organizations conducted fundraising campaigns to
raise money for emergency relief. Some of these campaigns allowed people to donate by sending a text message using a cell phone to have the donated amount
added to their cell-phone bill. The report “Early Signals
on Mobile Philanthropy: Is Haiti the Tipping Point?”
(Edge Research, 2010) describes the results of a national
survey of 1526 people that investigated the ways in
which people made donations to the Haiti relief effort.
The report states that 17% of Gen Y respondents (those
born between 1980 and 1988) and 14% of Gen X respondents (those born between 1968 and 1979) said that
they had made a donation to the Haiti relief effort via
text message. The percentage making a donation via text
message was much lower for older respondents. The report did not say how many respondents were in the Gen
Y and Gen X samples, but for purposes of this exercise,
suppose that both sample sizes were 400 and that it is
reasonable to regard the samples as representative of the
Gen Y and Gen X populations.
a. Is there convincing evidence that the proportion of
those in Gen Y who donated to Haiti relief via text
message is greater than the proportion for Gen X?
Use ␣ ϭ .01.
b. Estimate the difference between the proportion of
Gen Y and the proportion of Gen X that made a donation via text message using a 99% confidence interval. Provide an interpretation of both the interval and
the associated confidence level.
11.40 Common Sense Media surveyed 1000 teens and
1000 parents of teens to learn about how teens are using
social networking sites such as Facebook and MySpace
(“Teens Show, Tell Too Much Online,” San Francisco
Chronicle, August 10, 2009). The two samples were
independently selected and were chosen in a way that
makes it reasonable to regard them as representative of
American teens and parents of American teens.
a. When asked if they check their online social networking sites more than 10 times a day, 220 of the
teens surveyed said yes. When parents of teens were
asked if their teen checked his or her site more than
10 times a day, 40 said yes. Use a significance level
of .01 to carry out a hypothesis test to determine if
there is convincing evidence that the proportion of
all parents who think their teen checks a social networking site more than 10 times a day is less than
the proportion of all teens who report that they
check more than 10 times a day.
Bold exercises answered in back
Data set available online
557
b. The article also reported that 390 of the teens surveyed said they had posted something on their networking site that they later regretted. Would you use
the two-sample z test of this section to test the hypothesis that more than one-third of all teens have
posted something on a social networking site that
they later regretted? Explain why or why not.
c. Using an appropriate test procedure, carry out a test
of the hypothesis given in Part (b). Use ␣ ϭ .05 for
this test.
11.41 The report “Audience Insights: Communicating
to Teens (Aged 12–17)” (www.cdc.gov, 2009) described teens’ attitudes about traditional media, such as
TV, movies, and newspapers. In a representative sample
of American teenage girls, 41% said newspapers were
boring. In a representative sample of American teenage
boys, 44% said newspapers were boring. Sample sizes
were not given in the report.
a. Suppose that the percentages reported had been
based on a sample of 58 girls and 41 boys. Is there
convincing evidence that the proportion of those
who think that newspapers are boring is different for
teenage girls and boys? Carry out a hypothesis test
using ␣ ϭ .05.
b. Suppose that the percentages reported had been
based on a sample of 2000 girls and 2500 boys. Is
there convincing evidence that the proportion of
those who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using ␣ ϭ .05.
c. Explain why the hypothesis tests in Parts (a) and (b)
resulted in different conclusions.
11.42 The director of the Kaiser Family Foundation’s
Program for the Study of Entertainment Media and
Health said, “It’s not just teenagers who are wired up and
tuned in, its babies in diapers as well.” A study by Kaiser
Foundation provided one of the first looks at media use
among the very youngest children—those from 6 months
to 6 years of age (Kaiser Family Foundation, 2003, www
.kff.org). Because previous research indicated that children who have a TV in their bedroom spend less time
reading than other children, the authors of the Foundation study were interested in learning about the proportion of kids who have a TV in their bedroom. They collected data from two independent random samples of
parents. One sample consisted of parents of children age 6
months to 3 years old. The second sample consisted of
parents of children age 3 to 6 years old. They found that
the proportion of children who had a TV in their bedVideo Solution available
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558
Chapter 11
Comparing Two Populations or Treatments
room was .30 for the sample of children age 6 months to
3 years and .43 for the sample of children age 3 to 6 years
old. Suppose that the two sample sizes were each 100.
a. Construct and interpret a 95% confidence interval
for the proportion of children age 6 months to 3
years who have a TV in their bedroom. Hint: This is
a one-sample confidence interval.
b. Construct and interpret a 95% confidence interval
for the proportion of children age 3 to 6 years who
have a TV in their bedroom.
c. Do the confidence intervals from Parts (a) and (b)
overlap? What does this suggest about the two population proportions?
d. Construct and interpret a 95% confidence interval
for the difference in the proportion that have TVs in
the bedroom for children age 6 months to 3 years
and for children age 3 to 6 years.
e. Is the interval in Part (d) consistent with your answer in Part (c)? Explain.
11.43 The Insurance Institute for Highway Safety issued a press release titled “Teen Drivers Often Ignoring
Bans on Using Cell Phones” (June 9, 2008). The following quote is from the press release:
Just 1–2 months prior to the ban’s Dec. 1, 2006
start, 11 percent of teen drivers were observed using
cell phones as they left school in the afternoon.
About 5 months after the ban took effect, 12% of
teen drivers were observed using cell phones.
Suppose that the two samples of teen drivers (before the
ban, after the ban) can be regarded as representative of
these populations of teen drivers. Suppose also that 200
teen drivers were observed before the ban (so n1 5 200
and p^ 1 5 .11) and 150 teen drivers were observed after
the ban.
a. Construct and interpret a 95% confidence interval
for the difference in the proportion using a cell
phone while driving before the ban and the proportion after the ban.
b. Is zero included in the confidence interval of Part
(c)? What does this imply about the difference in the
population proportions?
11.44 The press release referenced in the previous exercise also included data from independent surveys of teenage drivers and parents of teenage drivers. In response to
a question asking if they approved of laws banning the
use of cell phones and texting while driving, 74% of the
teens surveyed and 95% of the parents surveyed said they
approved. The sample sizes were not given in the press
Bold exercises answered in back
Data set available online
release, but for purposes of this exercise, suppose that
600 teens and 400 parents of teens responded to the
surveys and that it is reasonable to regard these samples
as representative of the two populations. Do the data
provide convincing evidence that the proportion of teens
that approve of cell-phone and texting bans while driving
is less than the proportion of parents of teens who approve? Test the relevant hypotheses using a significance
level of .05.
11.45 The article “Fish Oil Staves Off Schizophrenia”
(USA Today, February 2, 2010) describes a study in
which 81 patients age 13 to 25 who were considered atrisk for mental illness were randomly assigned to one of
two groups. Those in one group took four fish oil capsules daily. The other group took a placebo. After 1 year,
5% of those in the fish oil group and 28% of those in the
placebo group had become psychotic. Is it appropriate to
use the two-sample z test of this section to test hypotheses about the difference in the proportions of patients
receiving the fish oil and the placebo treatments who
became psychotic? Explain why or why not.
11.46 The report “Young People Living on the Edge”
(Greenberg Quinlan Rosner Research, 2008) summarizes a survey of people in two independent random
samples. One sample consisted of 600 young adults (age
19 to 35) and the other sample consisted of 300 parents
of children age 19 to 35. The young adults were presented with a variety of situations (such as getting married or buying a house) and were asked if they thought
that their parents were likely to provide financial support
in that situation. The parents of young adults were presented with the same situations and asked if they would
be likely to provide financial support to their child in
that situation.
a. When asked about getting married, 41% of the
young adults said they thought parents would provide financial support and 43% of the parents said
they would provide support. Carry out a hypothesis
test to determine if there is convincing evidence that
the proportion of young adults who think parents
would provide financial support and the proportion
of parents who say they would provide support are
different.
b. The report stated that the proportion of young
adults who thought parents would help with buying
a house or apartment was .37. For the sample of
parents, the proportion who said they would help
with buying a house or an apartment was .27. Based
on these data, can you conclude that the proportion
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11.3 Large Sample Inferences Concerning a Difference Between Two Population or Treatment Proportions
of parents who say they would help with buying a
house or an apartment is significantly less than the
proportion of young adults who think that their
parents would help?
11.47 Some commercial airplanes recirculate approximately 50% of the cabin air in order to increase fuel efficiency. The authors of the paper “Aircraft Cabin Air
Recirculation and Symptoms of the Common Cold”
(Journal of the American Medical Association [2002]:
483–486) studied 1100 airline passengers who flew from
San Francisco to Denver between January and April
1999. Some passengers traveled on airplanes that recirculated air and others traveled on planes that did not recirculate air. Of the 517 passengers who flew on planes that
did not recirculate air, 108 reported post-flight respiratory symptoms, while 111 of the 583 passengers on
planes that did recirculate air reported such symptoms. Is
there sufficient evidence to conclude that the proportion
of passengers with post-flight respiratory symptoms differs for planes that do and do not recirculate air? Test the
appropriate hypotheses using a 5 .05. You may assume
that it is reasonable to regard these two samples as being
independently selected and as representative of the two
populations of interest.
11.48 “Doctors Praise Device That Aids Ailing Hearts”
(Associated Press, November 9, 2004) is the headline
of an article that describes the results of a study of the
effectiveness of a fabric device that acts like a support
stocking for a weak or damaged heart. In the study, 107
people who consented to treatment were assigned at random to either a standard treatment consisting of drugs or
the experimental treatment that consisted of drugs plus
surgery to install the stocking. After two years, 38% of
the 57 patients receiving the stocking had improved and
27% of the patients receiving the standard treatment had
improved. Do these data provide convincing evidence
that the proportion of patients who improve is higher for
the experimental treatment than for the standard treatment? Test the relevant hypotheses using a significance
level of .05.
11.49 The article “Portable MP3 Player Ownership
Reaches New High” (Ipsos Insight, June 29, 2006) reported that in 2006, 20% of those in a random sample
of 1112 Americans age 12 and older indicated that they
owned an MP3 player. In a similar survey conducted in
2005, only 15% reported owning an MP3 player. Suppose that the 2005 figure was also based on a random
sample of size 1112. Estimate the difference in the proBold exercises answered in back
Data set available online
559
portion of Americans age 12 and older who owned an
MP3 player in 2006 and the corresponding proportion
for 2005 using a 95% confidence interval. Is zero included in the interval? What does this tell you about the
change in this proportion from 2005 to 2006?
11.50 The article referenced in the previous exercise
also reported that 24% of the males and 16% of the females in the 2006 sample reported owning an MP3
player. Suppose that there were the same number of
males and females in the sample of 1112. Do these data
provide convincing evidence that the proportion of females that owned an MP3 player in 2006 is smaller than
the corresponding proportion of males? Carry out a test
using a significance level of .01.
11.51 Public Agenda conducted a survey of 1379 parents and 1342 students in grades 6–12 regarding the importance of science and mathematics in the school curriculum (Associated Press, February 15, 2006). It was
reported that 50% of students thought that understanding science and having strong math skills are essential for
them to succeed in life after school, whereas 62% of the
parents thought it was crucial for today’s students to learn
science and higher-level math. The two samples—parents
and students—were selected independently of one another. Is there sufficient evidence to conclude that the
proportion of parents who regard science and mathematics as crucial is different than the corresponding proportion for students in grades 6–12? Test the relevant hypotheses using a significance level of .05.
11.52 The article “Spray Flu Vaccine May Work Better
Than Injections for Tots” (San Luis Obispo Tribune,
May 2, 2006) described a study that compared flu vaccine administered by injection and flu vaccine administered as a nasal spray. Each of the 8000 children under
the age of 5 who participated in the study received both
a nasal spray and an injection, but only one was the real
vaccine and the other was salt water. At the end of the flu
season, it was determined that 3.9% of the 4000 children
receiving the real vaccine by nasal spray got sick with the
flu and 8.6% of the 4000 receiving the real vaccine by
injection got sick with the flu.
a. Why would the researchers give every child both a
nasal spray and an injection?
b. Use the given data to estimate the difference in the
proportion of children who get sick with the flu after
being vaccinated with an injection and the proportion of children who get sick with the flu after being
vaccinated with the nasal spray using a 99% confiVideo Solution available
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
560
Chapter 11 Comparing Two Populations or Treatments
dence interval. Based on the confidence interval,
would you conclude that the proportion of children
who get the flu is different for the two vaccination
methods?
11.53 “Smartest People Often Dumbest About Sunburns” is the headline of an article that appeared in the
San Luis Obispo Tribune (July 19, 2006). The article
states that “those with a college degree reported a higher
incidence of sunburn that those without a high school
degree—43% versus 25%.” For purposes of this exercise,
suppose that these percentages were based on random
samples of size 200 from each of the two groups of interest (college graduates and those without a high school
degree). Is there convincing evidence that the proportion
experiencing a sunburn is higher for college graduates
than it is for those without a high school degree? Answer
based on a test with a .05 significance level.
11.54 The following quote is from the article “Canadians Are Healthier Than We Are” (Associated Press,
May 31, 2006): “The Americans also reported more
heart disease and major depression, but those differences
were too small to be statistically significant.” This statement was based on the responses of a sample of 5183
Americans and a sample of 3505 Canadians. The proportion of Canadians who reported major depression
was given as .082.
a. Assuming that the researchers used a one-sided test
with a significance level of .05, could the sample
proportion of Americans reporting major depression
have been as large as .09? Explain why or why not.
b. Assuming that the researchers used a significance
level of .05, could the sample proportion of Americans reporting major depression have been as large as
.10? Explain why or why not.
11.55 “Mountain Biking May Reduce Fertility in Men,
Study Says” was the headline of an article appearing in
the San Luis Obispo Tribune (December 3, 2002). This
conclusion was based on an Austrian study that compared sperm counts of avid mountain bikers (those who
ride at least 12 hours per week) and nonbikers. Ninety
percent of the avid mountain bikers studied had low
sperm counts, as compared to 26% of the nonbikers.
Suppose that these percentages were based on independent samples of 100 avid mountain bikers and 100 nonBold exercises answered in back
Data set available online
bikers and that it is reasonable to view these samples as
representative of Austrian avid mountain bikers and
nonbikers.
a. Do these data provide convincing evidence that the
proportion of Austrian avid mountain bikers with
low sperm count is higher than the proportion of
Austrian nonbikers?
b. Based on the outcome of the test in Part (a), is it
reasonable to conclude that mountain biking 12
hours per week or more causes low sperm count?
Explain.
11.56 Women diagnosed with breast cancer whose tumors have not spread may be faced with a decision between two surgical treatments—mastectomy (removal of
the breast) or lumpectomy (only the tumor is removed).
In a long-term study of the effectiveness of these two
treatments, 701 women with breast cancer were randomly assigned to one of two treatment groups. One
group received mastectomies and the other group received lumpectomies and radiation. Both groups were
followed for 20 years after surgery. It was reported that
there was no statistically significant difference in the
proportion surviving for 20 years for the two treatments
(Associated Press, October 17, 2002). What hypotheses
do you think the researchers tested in order to reach the
given conclusion? Did the researchers reject or fail to
reject the null hypothesis?
11.57 In December 2001, the Department of Veterans
Affairs announced that it would begin paying benefits to
soldiers suffering from Lou Gehrig’s disease who had
served in the Gulf War (The New York Times, December 11, 2001). This decision was based on an analysis in
which the Lou Gehrig’s disease incidence rate (the proportion developing the disease) for the approximately
700,000 soldiers sent to the Gulf between August 1990
and July 1991 was compared to the incidence rate for the
approximately 1.8 million other soldiers who were not in
the Gulf during this time period. Based on these data,
explain why it is not appropriate to perform a formal
inference procedure (such as the two-sample z test) and
yet it is still reasonable to conclude that the incidence
rate is higher for Gulf War veterans than for those who
did not serve in the Gulf War.
Video Solution available
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.