comparing two proportionsvirtual.yosemite.cc.ca.us/jcurl/math 134 3 s... · 2004. 9. 2. ·...

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CHAPTER

19

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Comparing TwoProportions

In this chapter we cover. . .Two-sample problems:

proportions

The sampling distribution ofa difference betweenproportions

Large-sample confidenceintervals for comparingproportions

Using technology

Accurate confidenceintervals for comparingproportions*

Significance tests forcomparing proportions

In a two-sample problem, we want to compare two populations or the responsesto two treatments based on two independent samples. When the comparisoninvolves the mean of a quantitative variable, we use the two-sample t methodsof Chapter 17. To compare the standard deviations of a variable in two groups, weuse (under restrictive conditions) the F statistic also described in Chapter 17.Now we turn to methods to compare the proportions of successes in two groups.

Two-sample problems: proportionsWe will use notation similar to that used in our study of two-sample t statistics.The groups we want to compare are Population 1 and Population 2. We have aseparate SRS from each population or responses from two treatments in a ran-domized comparative experiment. A subscript shows which group a parameteror statistic describes. Here is our notation:

Population Sample SamplePopulation proportion size proportion

1 p1 n1 p̂12 p2 n2 p̂2

492



493The sampling distribution of a difference between proportions

We compare the populations by doing inference about the difference p1 −p2 between the population proportions. The statistic that estimates this differ-ence is the difference between the two sample proportions, p̂1 − p̂2.

EXAMPLE 19.1 Does preschool help?

To study the long-term effects of preschool programs for poor children, theHigh/Scope Educational Research Foundation has followed two groups of Michiganchildren since early childhood.1 One group of 62 attended preschool as 3- and 4-year-olds. This is a sample from Population 2, poor children who attend preschool.A control group of 61 children from the same area and similar backgrounds rep-resents Population 1, poor children with no preschool. Thus the sample sizes aren1 = 61 and n2 = 62.

One response variable of interest is the need for social services as adults. In thepast ten years, 38 of the preschool sample and 49 of the control sample have neededsocial services (mainly welfare). The sample proportions are

p̂1 = 4961 = 0.803

p̂2 = 3862 = 0.613

That is, about 80% of the control group uses social services, as opposed to about 61%of the preschool group.

To see if the study provides significant evidence that preschool reduces the laterneed for social services, we test the hypotheses

H0: p1 − p2 = 0 (the same as H0: p1 = p2)Ha: p1 − p2 > 0 (the same as Ha: p1 > p2)

To estimate how large the reduction is, we give a confidence interval for the differ-ence, p1 − p2. Both the test and the confidence interval start from the difference ofsample proportions:

p̂1 − p̂2 = 0.803 − 0.613 = 0.190

The sampling distribution of a differencebetween proportionsTo use p̂1 − p̂2 for inference, we must know its sampling distribution. Here arethe facts we need:� When the samples are large, the distribution of p̂1 − p̂2 is approximately

Normal.� The mean of p̂1 − p̂2 is p1 − p2. That is, the difference between sample

proportions is an unbiased estimator of the difference between populationproportions.



494 CHAPTER 19 � Comparing Two Proportions

Values of p1 − p2

Mean p1 − p2

Standard deviationSampling distribution

of p1 − p2 p1(1 − p1) p2(1 − p2)n1 n2

+

Figure 19.1 Select independent SRSs from two populations having proportions ofsuccesses p1 and p2. The proportions of successes in the two samples are p̂1 and p̂2.When the samples are large, the sampling distribution of the difference p̂1 − p̂2 isapproximately Normal.

� The standard deviation of the difference is√p1(1 − p1)

n1+ p2(1 − p2)

n2

Figure 19.1 displays the distribution of p̂1 − p̂2. The standard deviation ofp̂1 − p̂2 involves the unknown parameters p1 and p2. Just as in the previouschapter, we must replace these by estimates in order to do inference. And justas in the previous chapter, we do this a bit differently for confidence intervalsand for tests.

Large-sample confidence intervals forcomparing proportionsTo obtain a confidence interval, replace the population proportions p1 and p2in the standard deviation by the sample proportions. The result is the standardstandard errorerror of the statistic p̂1 − p̂2:

SE =√

p̂1(1 − p̂1)n1

+ p̂2(1 − p̂2)n2

The confidence interval again has the form

estimate ± z∗SEestimate



495Large-sample confidence intervals for comparing proportions

LARGE-SAMPLE CONFIDENCE INTERVAL FOR COMPARINGTWO PROPORTIONS

Draw an SRS of size n1 from a population having proportion p1 ofsuccesses and draw an independent SRS of size n2 from anotherpopulation having proportion p2 of successes. When n1 and n2 are large,an approximate level C confidence interval for p1 − p2 is

(p̂1 − p̂2) ± z∗SEIn this formula the standard error SE of p̂1 − p̂2 is

SE =√

p̂1(1 − p̂1)n1

+ p̂2(1 − p̂2)n2

and z∗ is the critical value for the standard Normal density curve witharea C between −z∗ and z∗.Use this interval only when the populations are at least 10 times as largeas the samples and the counts of successes and failures are each 10 ormore in both samples.

EXAMPLE 19.2 How much does preschool help?

Example 19.1 describes a study of the effect of preschool on later use of social ser-vices. Here are the facts:

Population Sample Count of SamplePopulation description size successes proportion

1 control n1 = 61 49 p̂1 = 49/61 = 0.8032 preschool n2 = 62 38 p̂2 = 38/62 = 0.613

To check that our approximate confidence interval is safe, look at the counts ofsuccesses and failures in the two samples. The smallest of these four quantities is thecount of failures in the control group, 61 − 49 = 12. This is larger than 10, so theinterval will be reasonably accurate.

The difference p1 − p2 measures the effect of preschool in reducing the propor-tion of people who later need social services. To compute a 95% confidence intervalfor p1 − p2, first find the standard error

SE =√

p̂1(1 − p̂1)n1

+ p̂2(1 − p̂2)n2

=√

(0.803)(0.197)61

+ (0.613)(0.387)62

=√

0.00642 = 0.0801




The 95% confidence interval is

( p̂1 − p̂2) ± z∗SE = (0.803 − 0.613) ± (1.960)(0.0801)= 0.190 ± 0.157= 0.033 to 0.347

We are 95% confident that the percent needing social services is somewhere between3.3 and 34.7 percentage points lower among people who attended preschool. Theconfidence interval is wide because the samples are quite small.

The researchers in the study of Example 19.1 selected two separate samplesfrom the two populations they wanted to compare. Many comparative studiesstart with just one sample, then divide it into two groups based on data gath-ered from the subjects. For example, an opinion poll may contact a randomsample of adults, then compare the opinions of the women and the men in thesample. The two-sample z procedures for comparing proportions are valid insuch situations. This is an important fact about these methods.

Using technologyFigure 19.2 displays software output for Example 19.2 from Minitab and the TI-83 calculator. As usual, you can understand the output even without knowledgeof the program that produced it. The Excel spreadsheet lacks menu items forinference about proportions. You must use the spreadsheet’s formula capabilityto program the confidence interval or test statistic and then to find the P-valueof a test.

APPLY YOUR KNOWLEDGE

19.1 Fear of crime among older blacks. The elderly fear crime more thanyounger people, even though they are less likely to be victims of crime.One of the few studies that looked at older blacks recruited randomsamples of 56 black women and 63 black men over the age of 65 fromAtlantic City, New Jersey. Of the women, 27 said they “felt vulnerable”to crime; 46 of the men said this.2

(a) What proportion of women in the sample feel vulnerable? Of men?Men are victims of crime more often than women, so we expect ahigher proportion of men to feel vulnerable.

(b) Give a 95% confidence interval for the difference (men minuswomen).

19.2 How to quit smoking. Nicotine patches are often used to help smokersquit. Does giving medicine to fight depression help? A randomizeddouble-blind experiment assigned 244 smokers who wanted to stop to


7314F-19 PB286-Moore-V5.cls May 6, 2003 20:4

497Accurate confidence intervals for comparing proportions

Microsoft Excel

Minitab

Texas Instruments TI-83 Plus

Figure 19.2 The 95% confidence interval of Example 19.2. Output from Minitaband the TI-83 graphing calculator.

receive nicotine patches and another 245 to receive both a patch andthe antidepression drug bupropion. Results: After a year, 40 subjects inthe nicotine patch group had abstained from smoking, as had 87 in thepatch-plus-drug group.3 Give a 99% confidence interval for thedifference (treatment minus control) in the proportion of smokers whoquit.

Accurate confidence intervals forcomparing proportions*Like the large-sample confidence interval for a single proportion p , the large-sample interval for p1 − p2 generally has true confidence level less than thelevel you asked for. The inaccuracy is not as serious as in the one-sample case,at least if our guidelines for use are followed. Once again, a simple modificationimproves the accuracy, so much so that the resulting interval can be used for

*Although this material is optional, it is essential if you wish to calculate confidence intervals for a differencebetween proportions for small samples.




sample sizes as small as 5 in each group, with no restrictions on the counts ofsuccesses and failures.

We call this interval the plus four interval because you add four imaginaryplus four intervalobservations, one success and one failure in each of the two samples. Then just usethe large-sample formula for the new data.4 Let’s apply the plus four method tothe data on the effect of preschool.

Computer-assistedinterviewing

The days of the interviewerwith a clipboard are past.Contemporary interviewersread questions from a computerscreen and use the keyboard toenter responses. The computerskips irrelevant items—once awoman says that she has nochildren, further questionsabout her children neverappear. The computer caneven present questions inrandom order to avoid bias dueto always following the sameorder. Computer softwarekeeps records of who hasresponded and prepares a file ofdata from the responses. Thetedious process of transferringresponses from paper tocomputer, once a source oferrors, has disappeared.

EXAMPLE 19.3 How much does preschool help?

In Example 19.2, we applied the large-sample interval to a study of the effect ofpreschool on later use of social services. Here are the facts again:

Population Sample Count of SamplePopulation description size successes proportion

1 control n1 = 61 49 p̂1 = 49/61 = 0.8032 preschool n2 = 62 38 p̂2 = 38/62 = 0.613

The plus four method adds four imaginary observations. The new facts are:

Population Sample Count of Plus four samplePopulation description size successes proportion

1 control n1 + 2 = 63 49 + 1 = 50 p̃1 = 50/63 = 0.7942 preschool n2 + 2 = 64 38 + 1 = 39 p̃2 = 39/64 = 0.609

The standard error based on the new facts is

SE =√

p̃1(1 − p̃1)n1 + 2 +

p̃2(1 − p̂2)n2 + 2

=√

(0.794)(0.206)63

+ (0.609)(0.391)64

=√

0.006317 = 0.0795

The plus four 95% confidence interval is

( p̃1 − p̃2) ± z∗SE = (0.794 − 0.609) ± (1.960)(0.0795)= 0.185 ± 0.156= 0.029 to 0.341



499Accurate confidence intervals for comparing proportions

The plus four interval does not differ greatly from the large-sample interval in Ex-ample 19.2 because the sample sizes are large enough to satisfy our guidelines for useof the large-sample method.

The plus four interval may be conservative (that is, the true confidence levelmay be higher than you asked for) for very small samples and population p ’s closeto 0 or 1. It is generally much more accurate than the large-sample intervalwhen the samples are small.


19.3 Drug testing in schools. In 2002 the Supreme Court ruled that schoolscould require random drug tests of students participating in competitiveafter-school activities such as athletics. Does drug testing reduce use ofillegal drugs? A study compared two similar high schools in Oregon.Wahtonka High School tested athletes at random and Warrenton HighSchool did not. In a confidential survey, 7 of 135 athletes at Wahtonkaand 27 of 141 athletes at Warrenton said they were using drugs.5 Regardthese athletes as SRSs from the populations of athletes at similarschools with and without drug testing.(a) You should not use the large-sample confidence interval. Why not?(b) The plus four method adds two observations, a success and a failure,

to each sample. What are the sample sizes and the counts of drugusers after you do this?

(c) Give the plus four 95% confidence interval for the differencebetween the proportion of athletes using drugs at schools with andwithout testing.

(Ron Chapple/Thinkstock/PictureQuest)

19.4 In-line skaters. A study of injuries to in-line skaters used data from theNational Electronic Injury Surveillance System, which collects datafrom a random sample of hospital emergency rooms. The researchersinterviewed 161 people who came to emergency rooms with injuriesfrom in-line skating. Wrist injuries (mostly fractures) were the mostcommon.6

(a) The interviews found that 53 people were wearing wrist guards and6 of these had wrist injuries. Of the 108 who did not wear wristguards, 45 had wrist injuries. Why should we not use thelarge-sample confidence interval for these data?

(b) Give the plus four 95% confidence interval for the differencebetween the two population proportions of wrist injuries. Statecarefully what populations your inference compares. We would liketo draw conclusions about all in-line skaters, but we have data onlyfor injured skaters.




Significance tests for comparing proportionsAn observed difference between two sample proportions can reflect a differ-ence in the populations, or it may just be due to chance variation in randomsampling. Significance tests help us decide if the effect we see in the samplesis really there in the populations. The null hypothesis says that there is nodifference between the two populations:

H0: p1 = p2The alternative hypothesis says what kind of difference we expect.

Statisticians honestand dishonest

Developed nations rely ongovernment statisticians toproduce honest data. We trustthe monthly unemploymentrate and Consumer Price Indexto guide both public andprivate decisions. Honestycan’t be taken for grantedeverywhere, however. In 1998,the Russian governmentarrested the top statisticians inthe State Committee forStatistics. They were accusedof taking bribes to fudge datato help companies avoid taxes.“It means that we knownothing about the performanceof Russian companies,” saidone newspaper editor.

EXAMPLE 19.4 Choosing a mate

“Would you marry a person from a lower social class than your own?” Researchersasked this question of a sample of 385 black, never-married students at two his-torically black colleges in the South. We will consider this to be an SRS of blackstudents at historically black colleges. Of the 149 men in the sample, 91 said “Yes.”Among the 236 women, 117 said “Yes.”7 (The men and women in a single SRS canbe treated as if they were separate SRSs of men and women students.)

The sample proportions who would marry someone from a lower social classare

p̂1 = 91149 = 0.611 (men)

p̂2 = 117236 = 0.496 (women)

That is, about 61% of the men but only about 50% of the women would marrybeneath their class. Is this apparent difference statistically significant? We had nodirection for the difference in mind before looking at the data, so we have a two-sided alternative:

H0: p1 = p2Ha: p1 �= p2

To do a test, standardize p̂1 − p̂2 to get a z statistic. If H0 is true, all the ob-servations in both samples come from a single population of students of whoma single unknown proportion p would marry someone from a lower social class.So instead of estimating p1 and p2 separately, we pool the two samples and usethe overall sample proportion to estimate the single population parameter p .Call this the pooled sample proportion. It ispooled sample proportion

p̂ = count of successes in both samples combinedcount of observations in both samples combined

Use p̂ in place of both p̂1 and p̂2 in the expression for the standard error SEof p̂1 − p̂2 to get a z statistic that has the standard Normal distribution whenH0 is true. Here is the test.



501Significance tests for comparing proportions

SIGNIFICANCE TEST FOR COMPARING TWO PROPORTIONS

To test the hypothesis

H0: p1 = p2first find the pooled proportion p̂ of successes in both samples combined.Then compute the z statistic

z = p̂1 − p̂2√p̂(1 − p̂)

(1n1

+ 1n2

)

In terms of a variable Z having the standard Normal distribution, theP-value for a test of H0 against

Ha: p1 > p2 is P(Z ≥ z)z

Ha: p1 < p2 is P(Z ≤ z)z

Ha: p1 �= p2 is 2P(Z ≥ |z|)|z|

Use this test when the populations are at least 10 times as large as thesamples and the counts of successes and failures are each 5 or more inboth samples.

EXAMPLE 19.5 Choosing a mate, continued

The pooled proportion of students who would marry beneath their own social classis

p̂ = count of “Yes” responses among men and women combinedcount of subjects among men and women combined

= 91 + 117149 + 236

= 208385

= 0.5403




Standard Normalcurve

z = 2.21− 2.21 z

P-value is double thearea under the curveto the right of z = 2.21.

Figure 19.3 The P-value for the two-sided test of Example 19.5.

The z test statistic is

z = p̂1 − p̂2√p̂(1 − p̂)

(1n1

+ 1n2

)

= 0.611 − 0.496√(0.5403)(0.4597)

(1

149+ 1

236

)

= 0.1150.05215

= 2.21

The two-sided P-value is the area under the standard Normal curve more than2.21 distant from 0. Figure 19.3 shows this area. Table A gives P = 0.0272. Alter-natively, z = 2.21 lies between the critical values for tail areas 0.01 and 0.02 in thelast row of Table C. The two-sided P-value is therefore between 0.02 and 0.04.

Because P < 0.05, the results are statistically significant at the α = 0.05 level.There is good evidence that men are more likely than women to say they will marrysomeone from a lower social class.


19.5 Drug testing in schools, continued. Exercise 19.3 describes a studythat compared the proportions of athletes who use illegal drugs in twosimilar high schools, one that tests for drugs and one that does not.



503Chapter 19 Summary

Drug testing is intended to reduce use of drugs. Do the data give goodreason to think that drug use among athletes is lower in schools thattest for drugs? State hypotheses, find the test statistic, and use Table Cto approximate the P-value. Be sure to state your conclusion. (Becausethe study is not an experiment, the conclusion depends on theassumption that athletes in these two schools can be considered SRSsfrom all similar schools.)

19.6 How to quit smoking, continued. Exercise 19.2 describes arandomized comparative experiment to test whether adding medicineto fight depression increases the effectiveness of nicotine patches inhelping smokers to quit. How significant is the evidence that themedicine increases the success rate? State hypotheses, calculate a teststatistic, use Table A to give its P-value, and state your conclusion.

19.7 The Gold Coast. A historian examining British colonial records forthe Gold Coast in Africa suspects that the death rate was higher amongAfrican miners than among European miners. In the year 1936, therewere 223 deaths among 33,809 African miners and 7 deaths among1541 European miners on the Gold Coast.8

Consider this year as a sample from the pre-war era in Africa. Is theregood evidence that the proportion of African miners who died washigher than the proportion of European miners who died? Statehypotheses, calculate a test statistic, use Table C to give a P-value, andstate your conclusion.

Chapter 19 SUMMARY

We want to compare the proportions p1 and p2 of successes in twopopulations. The comparison is based on the difference p̂1 − p̂2 between thesample proportions of successes. When the sample sizes n1 and n2 are largeenough, we can use z procedures because the sampling distribution ofp̂1 − p̂2 is close to Normal.The level C large-sample confidence interval for p1 − p2 is

(p̂1 − p̂2) ± z∗SEwhere the standard error of p̂1 − p̂2 is

SE =√

p̂1(1 − p̂1)n1

+ p̂2(1 − p̂2)n2

and z∗ is a standard Normal critical value.The true confidence level of the large-sample interval can be substantially lessthan the planned level C. Use this interval only if the counts of successes andfailures in both samples are 10 or greater.




To get a more accurate confidence interval, add four imaginary observations,one success and one failure in each sample. Then use the same formula for theconfidence interval. This is the plus four confidence interval. You can use itwhenever both samples have 5 or more observations.Significance tests of H0: p1 = p2 use the pooled sample proportion

p̂ = count of successes in both samples combinedcount of observations in both samples combined

and the z statistic

z = p̂1 − p̂2√p̂(1 − p̂)

(1n1

+ 1n2

)

P-values come from the standard Normal distribution. Use this test whenthere are 5 or more successes and 5 or more failures in both samples.

Chapter 19 EXERCISES

We recommend using the plus four method for all confidence intervals forproportions. However, the large-sample method is acceptable when theguidelines for its use are met.

19.8 Information online. A random-digit dialing sample of 2092 adultsfound that 1318 used the Internet.9 Of the users, 1041 said that theyexpect businesses to have Web sites that give product information;294 of the 774 nonusers said this.(a) Give a 95% confidence interval for the proportion of all adults

who use the Internet.(b) Give a 95% confidence interval to compare the proportions of

users and nonusers who expect businesses to have Web sites.Call a statistician. Does involving a statistician to help with statisticalmethods improve the chance that a medical research paper will bepublished? A study of papers submitted to two medical journals found that135 of 190 papers that lacked statistical assistance were rejected withouteven being reviewed in detail. In contrast, 293 of the 514 papers withstatistical help were sent back without review.10 Exercises 19.9 to 19.11are based on this study.

19.9 Does statistical help make a difference? Is there a significantdifference in the proportions of papers with and without statisticalhelp that are rejected without review? Use Table C to get anapproximate P-value. (This observational study does not establishcausation: the studies that include statistical help may also be betterthan those that do not in other ways.)



505Chapter 19 Exercises

19.10 How often are statisticians involved? Give a 95% confidenceinterval for the proportion of papers submitted to these journals thatinclude help from a statistician.

19.11 How big a difference? Give a 95% confidence interval for thedifference between the proportions of papers rejected without reviewwhen a statistician is and is not involved in the research.

19.12 Satisfaction with high schools. A sample survey asked 202 blackparents and 201 white parents of high school children, “Are the publichigh schools in your state doing an excellent, good, fair or poor job, ordon’t you know enough to say?” The investigators suspected that blackparents are generally less satisfied with their public schools than arewhites. Among the black parents, 81 thought high schools were doinga “good” or “excellent” job; 103 of the white parents felt this way.11 Isthere good evidence that the proportion of all black parents whothink their state’s high schools are good or excellent is lower than theproportion of white parents with this opinion?

19.13 College is important. The sample survey described in the previousexercise also asked respondents if they agreed with the statement “Acollege education has become as important as a high school diplomaused to be.” In the sample, 125 of 201 white parents and 154 of 202black parents said that they “strongly agreed.” Is there good reason tothink that different percents of all black and white parents wouldstrongly agree with the statement?

(Bob Llewellyn/ImageState-Pictor/PictureQuest)

19.14 Seat belt use. The proportion of drivers who use seat belts dependson things like age (young people are more likely to go unbelted) andgender (women are more likely to use belts). It also depends on locallaw. In New York City, police can stop a driver who is not belted. InBoston (as of late 2000), police can cite a driver for not wearing a seatbelt only if the driver has been stopped for some other violation. Hereare data from observing random samples of female Hispanic drivers inthese two cities:12

City Drivers Belted

New York 220 183Boston 117 68

(a) Is this an experiment or an observational study? Why?(b) Comparing local law suggests the hypothesis that a smaller

proportion of drivers wear seat belts in Boston than in New York.Do the data give good evidence that this is true for femaleHispanic drivers?

19.15 Ethnicity and seat belt use. Here are data from the study described inthe previous exercise for Hispanic and white male drivers in Chicago:




Group Drivers Belted

Hispanic 539 286White 292 164

Is there a significant difference between Hispanic and white drivers?How large is the difference? Do inference to answer both questions. Besure to explain exactly what inference you choose to do.

19.16 Lyme disease. Lyme disease is spread in the northeastern UnitedStates by infected ticks. The ticks are infected mainly by feeding onmice, so more mice result in more infected ticks. The mousepopulation in turn rises and falls with the abundance of acorns, theirfavored food. Experimenters studied two similar forest areas in a yearwhen the acorn crop failed. They added hundreds of thousands ofacorns to one area to imitate an abundant acorn crop, while leavingthe other area untouched. The next spring, 54 of the 72 mice trappedin the first area were in breeding condition, versus 10 of the 17 micetrapped in the second area.13

(a) The large-sample confidence interval may not be accurate forthese data. Why not?

(b) Give the plus four 90% confidence interval for the differencebetween the proportions of mice ready to breed in good acornyears and bad acorn years.

19.17 Steroids in high school. A study by the National Athletic TrainersAssociation surveyed 1679 high school freshmen and 1366 highschool seniors in Illinois. Results showed that 34 of the freshmen and24 of the seniors had used anabolic steroids. Steroids, which aredangerous, are sometimes used to improve athletic performance.14

(a) In order to draw conclusions about all Illinois freshmen andseniors, how should the study samples be chosen?

(b) Give a 95% confidence interval for the proportion of all highschool freshmen in Illinois who have used steroids.

(c) Is there a significant difference between the proportions offreshmen and seniors who have used steroids?

(Richard Hamilton Smith/CORBIS)

19.18 Detecting genetically modified soybeans. Exercise 18.27 (page 488)describes a study in which batches of soybeans containing somegenetically modified (GM) beans were submitted to 23 grain-handlingfacilities. When batches contained 1% of GM beans, 18 of thefacilities detected the presence of GM beans. Only 7 of the facilitiesdetected GM beans when they made up one-tenth of 1% of the beansin the batches. Explain why we cannot use the methods of this chapterto compare the proportions of facilities that will detect the two levelsof GM soybeans.



507Chapter 19 Exercises

19.19 Are genetically modified foods risky? Europe and the United Statesdiffer considerably in their attitudes toward food made from crops thathave been genetically modified (GM) to, for example, resist pests orcontain more protein. A random sample of 12,178 European adultsfound that 63% thought such foods were risky. In the United States, arandom sample of 863 adults who were asked the same questionsfound that 46% considered GM foods risky.15

(a) What are the counts of people in each sample who thought GMfoods were risky?

(b) Give a 95% confidence interval to compare Europe and theUnited States.

19.20 Are urban students more successful? North Carolina StateUniversity looked at the factors that affect the success of students in arequired chemical engineering course. Students must get a C or betterin the course in order to continue as chemical engineering majors.There were 65 students from urban or suburban backgrounds, and 52of these students succeeded. Another 55 students were from rural orsmall-town backgrounds; 30 of these students succeeded in thecourse.16

(a) Is there good evidence that the proportion of students whosucceed is different for urban/suburban versus rural/small-townbackgrounds? State hypotheses, give the P-value of a test, andstate your conclusion.

(b) Give a 90% confidence interval for the size of the difference.

19.21 Small-business failures. The study of small-business failuresdescribed in Exercise 18.34 (page 489) looked at 148 food-and-drinkbusinesses in central Indiana. Of these, 106 were headed by men and42 were headed by women. During a three-year period, 15 of the men’sbusinesses and 7 of the women’s businesses failed. Is there a significantdifference between the rate at which businesses headed by men andthose headed by women fail?

19.22 Female and male students. The North Carolina State Universitystudy (Exercise 19.20) also looked at possible differences in theproportions of female and male students who succeeded in the course.They found that 23 of the 34 women and 60 of the 89 men succeeded.Is there evidence of a difference between the proportions of womenand men who succeed?

(Dallas and John Heaton/CORBIS)

19.23 Nobody is home in July. Nonresponse to sample surveys may differwith the season of the year. In Italy, for example, many people leavetown during the summer. The Italian National Statistical Institutecalled random samples of telephone numbers between 7 p.m. and10 p.m. at several seasons of the year. Here are the results for twoseasons:17




Dates Number of calls No answer Total nonresponse

Jan. 1 to Apr. 13 1558 333 491July 1 to Aug. 31 2075 861 1174

(a) How much higher is the proportion of “no answers” in July andAugust compared with the early part of the year? Give a 99%confidence interval.

(b) The difference between the proportions of “no answers” is so largethat it is clearly statistically significant. How can you tell from yourwork in (a) that the difference is significant at the α = 0.01 level?

(c) Use the information given to find the counts of calls that hadnonresponse for some reason other than “no answer.” Do the ratesof nonresponse due to other causes also differ significantly for thetwo seasons?

19.24 Who responds? Example 17.4 (page 446) describes a study comparingcollege students who have volunteered for community service workwith those who have not. A weakness in the study is that the responserates for the two groups appear to differ. Students were placed in thetwo groups on the basis of a questionnaire: 39 in the “no service”group and 71 in the “service” group. The data were gathered from afollow-up survey two years later; 17 of the 39 “no service” studentsresponded (44%), compared with 80% response (57 of 71) in the“service” group. How significant is this difference in response rates?

19.25 Significant does not mean important. Never forget that even smalleffects can be statistically significant if the samples are large. Toillustrate this fact, return to the study of 148 small businesses inExercise 19.21.(a) Find the proportions of failures for businesses headed by women

and businesses headed by men. These sample proportions are quiteclose to each other. Give the P-value for the z test of thehypothesis that the same proportion of women’s and men’sbusinesses fail. (Use the two-sided alternative.) The test is very farfrom being significant.

(b) Now suppose that the same sample proportions came from asample 30 times as large. That is, 210 out of 1260 businessesheaded by women and 450 out of 3180 businesses headed by menfail. Verify that the proportions of failures are exactly the same asin (a). Repeat the z test for the new data, and show that it is nowsignificant at the α = 0.05 level.

(c) It is wise to use a confidence interval to estimate the size of aneffect, rather than just giving a P-value. Give 95% confidenceintervals for the difference between the proportions of women’s



509Chapter 19 Media Exercises

and men’s businesses that fail for the settings of both (a) and (b).What is the effect of larger samples on the confidence interval?

EESEE

Chapter 19 MEDIA EXERCISES

19.26 Police radar and speeding. Do drivers reduce excessive speed whenthey encounter police radar? The EESEE story “Radar Detectors andSpeeding” describes a study of the behavior of drivers on a ruralinterstate highway in Maryland where the speed limit was 55 miles perhour. The researchers measured speed with an electronic devicehidden in the pavement and, to eliminate large trucks, considered onlyvehicles less than 20 feet long. During some time periods, police radarwas set up at the measurement location. Here are some of the data:

Number Numberof vehicles over 65 mph

No radar 12,931 5,690Radar 3,285 1,051 (Joseph Sohm/ChromoSohm

Inc./CORBIS)

(a) Give a 95% confidence interval for the proportion of vehiclesgoing faster than 65 miles per hour when no radar is present.

(b) Give a 95% confidence interval for the effect of radar, as measuredby the difference in proportions of vehicles going faster than65 miles per hour with and without radar.

(c) The researchers chose a rural highway so that cars would beseparated rather than in clusters, where some cars might slowbecause they see other cars slowing. Explain why such clustersmight make inference invalid.

EESEE

19.27 Reducing nonresponse? Telephone surveys often have high rates ofnonresponse. When the call is handled by an answering machine,perhaps leaving a message on the machine will encourage people torespond when they are called again. The data below are from theEESEE story “Leave Survey after the Beep.” In this study, a messagewas or was not left (at random) when an answering machine picked upthe first call from a survey.

Total Eventual Completedhouseholds contact survey

No message 100 58 33Message 291 200 134




(a) Is there good evidence that leaving a message increases theproportion of households that are eventually contacted?

(b) Is there good evidence that leaving a message increases theproportion who complete the survey?

(c) If you find significant effects, look at their size. Do you think theseeffects are large enough to be important to survey takers?

comparing two proportionsvirtual.yosemite.cc.ca.us/jcurl/math 134 3 s... · 2004. 9. 2. ·...

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