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3: Large-Sample Inferences Concerning the Difference Between Two Population or Treatment Proportions

# 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

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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556

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

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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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

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.

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

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.

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

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

Video Solution available

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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

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.

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

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.

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