CHAPTER. Inference about a Population Proportion. In this chapter we cover

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CHAPTER

Blaine Harrington III/CORBIS

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Inference about a Population Proportion Our discussion of statistical inference to this point has concerned making inferences about population means. Now we turn to questions about the proportion of some outcome in a population. Here are some examples that call for inference about population proportions.

EXAMPLE 20.1

Risky behavior in the age of AIDS

20

In this chapter we cover... The sample proportion pˆ The sampling distribution of pˆ Large-sample confidence intervals for a proportion Accurate confidence intervals for a proportion Choosing the sample size Significance tests for a proportion

How common is behavior that puts people at risk of AIDS? The National AIDS Behavioral Surveys interviewed a random sample of 2673 adult heterosexuals. Of these, 170 had more than one sexual partner in the past year. That’s 6.36% of the sample.1 Based on these data, what can we say about the percent of all adult heterosexuals who have multiple partners? We want to estimate a single population proportion. This chapter concerns inference about one proportion.

EXAMPLE 20.2

Young adults living at home

A surprising number of young adults (ages 19 to 25) still live at home with their parents. A random sample of 2253 men and 2629 women in this age group found that 44% of the men but only 35% of the women lived at home. Is this significant evidence that the proportions living at home differ in the populations of all young men and all young women? We want to compare two population proportions. This is the topic of Chapter 21. 491

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C H A P T E R 20 • Inference about a Population Proportion

To do inference about a population mean μ, we use the mean x of a random sample from the population. The reasoning of inference starts with the sampling distribution of x. Now we follow the same pattern, replacing means by proportions.

ˆ The sample proportion p

sample proportion

We are interested in the unknown proportion p of a population that has some outcome. For convenience, call the outcome we are looking for a “success.” In Example 20.1, the population is adult heterosexuals, and the parameter p is the proportion who have had more than one sexual partner in the past year. To estimate p, the National AIDS Behavioral Surveys used random dialing of telephone numbers to contact a sample of 2673 people. Of these, 170 said they had multiple sexual partners. The statistic that estimates the parameter p is the sample proportion number of successes in the sample total number of individuals in the sample 170 = 0.0636 = 2673

pˆ =

Read the sample proportion pˆ as “p-hat.”

APPLY YOUR KNOWLEDGE In each of the following settings: (a) Describe the population and explain in words what the parameter p is. (b) Give the numerical value of the statistic pˆ that estimates p.

20.1 Do college students pray? A study of religious practices among college students interviewed a sample of 127 students; 107 of the students said that they prayed at least once in a while. 20.2 Playing games online. A random sample of 1100 teenagers (ages 12 to 17) were asked whether they played games online; 775 said that they did.

ˆ The sampling distribution of p How good is the statistic pˆ as an estimate of the parameter p? To find out, we ask, “What would happen if we took many samples?” The sampling distribution of pˆ answers this question. Here are the facts.

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The sampling distribution of pˆ

SAMPLING DISTRIBUTION OF A SAMPLE PROPORTION Draw an SRS of size n from a large population that contains proportion p of successes. Let pˆ be the sample proportion of successes, pˆ =

number of successes in the sample n

Then: • As the sample size increases, the sampling distribution of pˆ becomes approximately Normal. • The mean of the sampling distribution is p. • The standard deviation of the sampling distribution is  p(1 − p) n

Figure 20.1 summarizes these facts in a form that helps you recall the big idea of a sampling distribution. The behavior of sample proportions pˆ is similar to the behavior of sample means x. When the sample size n is large, the sampling distribution is approximately Normal. The larger the sample, the more nearly Normal the distribution is. Don’t use the Normal approximation to the distribution of pˆ when the sample size n is small. The mean of the sampling distribution of pˆ is the true value of the population proportion p. That is, pˆ is an unbiased estimator of p. The standard deviation of pˆ gets smaller as the sample size n gets larger, so that estimation is likely to be more

SRS size n SRS size n SRS size n

Standard deviation p(1 − p) n

p p p

• • • Population proportion p of successes

CAUTION UTION

Mean p

Values of p

F I G U R E 2 0 . 1 Select a large SRS from a population of which the proportion p are successes. The sampling distribution of the proportion pˆ of successes  in the sample is approximately Normal. The mean is p and the standard deviation is p(1 − p)/n.

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C H A P T E R 20 • Inference about a Population Proportion

accurate when the sample √is larger. As is the case for x, the standard deviation gets smaller only at the rate n. We need four times as many observations to cut the standard deviation in half.

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EXAMPLE 20.3

Asking about risky behavior

STATE: Suppose that in fact 6% of all adult heterosexuals had more than one sexual partner in the past year (and would admit it when asked). The National AIDS Behavioral Surveys interviewed a random sample of 2673 people from this population. What is the probability that at least 5% of such a sample admit to having more than one partner? FORMULATE: Take pˆ to be the proportion of individuals among the 2673 in the sample who had more than one partner. We want to find P ( pˆ ≥ 0.05). SOLVE: The sample size is n = 2673 and the population proportion is p = 0.06. So the sample proportion pˆ is approximately Normal with mean 0.06 and standard deviation   p(1 − p) (0.06)(0.94) = n 2673  = 0.0000211 = 0.00459 Standardize pˆ by subtracting the mean 0.06 and dividing by the standard deviation 0.00459. The standardized statistic has approximately the standard Normal distribution. In terms of a standard Normal variable Z , 

0.05 − 0.06 pˆ − 0.06 ≥ 0.00459 0.00459 = P ( Z ≥ −2.18) = 1 − 0.0146 = 0.9854



P ( pˆ ≥ 0.05) = P

Figure 20.2 shows this probability as an area under the standard Normal curve.

Standard Normal curve Probability = 0.9854

Probability = 0.0146

z = −2.18

z

F I G U R E 2 0 . 2 Probabilities in Example 20.3 as areas under the standard Normal curve.

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The sampling distribution of pˆ

CONCLUDE: If we repeat the National AIDS Behavioral Surveys many times, more than 98% of all the samples will contain at least 5% of respondents who admit to more than one sexual partner.

The Normal approximation for the sampling distribution of pˆ is least accurate when p is close to 0 or 1. If p = 0, successes are impossible. Every sample has pˆ = 0 and there is no Normal distribution in sight. In the same way, the approximation works poorly when p is close to 1. In practice, this means that we need larger n for values of p near 0 or 1. In Example 20.3, p is small but n is large. The exact probability is P ( pˆ ≥ 0.05) = 0.9843, so the Normal approximation is quite accurate. Inference about a population proportion p starts by using the sample proportion pˆ to estimate p. Confidence levels and P-values are probabilities calculated from the sampling distribution of pˆ . We will consider only situations that allow us to use the Normal approximation to this sampling distribution. Here is a summary of the conditions we need.

CAUTION UTION

CONDITIONS FOR INFERENCE ABOUT A PROPORTION • •

We can regard our data as a simple random sample (SRS) from the population. This is, as usual, the most important condition. The sample size n is large enough to ensure that the distribution of pˆ is close to Normal. We will see that different inference procedures require different answers to the question “how large is large enough?”

Remember also that all of our inference procedures require that the population be much larger than the sample.2 This condition is usually satisfied in practice and is satisfied in all of our examples and exercises.

APPLY YOUR KNOWLEDGE 20.3 Student drinking. The College Alcohol Study interviewed an SRS of 14,941 college students about their drinking habits. Suppose that half of all college students “drink to get drunk” at least once in a while. That is, p = 0.5. (a) What are the mean and standard deviation of the proportion pˆ of the sample who drink to get drunk? (b) Use the Normal approximation to find the probability that pˆ is between 0.49 and 0.51. 20.4 Students on diets. A sample survey interviews an SRS of 267 college women. Suppose (as is roughly true) that 70% of all college women have been on a diet within the past 12 months. What is the probability that 75% or more of the women in the sample have been on a diet? 20.5 Student drinking, continued. Suppose that half of all college students drink to get drunk at least once in a while. Exercise 20.3 asks for the probability that the sample proportion pˆ estimates p = 0.5 within ±1 percentage point. Find this probability for SRSs of sizes 1000, 4000, and 16,000. What general fact do your results illustrate?

CAUTION UTION

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20.6 No inference. A local television station conducts a call-in poll about a proposed city tax increase to buy natural areas and protect them from development. Of the 2372 calls, 1921 support the proposal. We can’t use these data as the basis for inference about the proportion of all citizens who support the tax increase. Why not?

Large-sample confidence intervals for a proportion

standard error of pˆ

To estimate a population proportion p, use the sample proportion pˆ . If our conditions for inference apply, the sampling distribution of pˆ is close to Normal with  mean p and standard deviation p(1 − p)/n. To obtain a level C confidence interval for p, we would like to use  p(1 − p) pˆ ± z ∗ n ∗ with the critical value z chosen to cover the central area C under the standard Normal curve. Figure 20.3 shows why. Because we don’t know the value of p, we replace the standard deviation by the standard error of pˆ  pˆ (1 − pˆ ) SE = n

Sampling distribution of p

Probability C

p − z* p(1 − p) n

p (unknown)

p + z*

p(1 − p) n

 F I G U R E 2 0 . 3 With probability C, pˆ lies within ±z ∗ p(1 − p)/n of the unknown population proportion p. That is to say that in these samples p lies within  ±z ∗ p(1 − p)/n of pˆ .

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Large-sample confidence intervals for a proportion

to get the confidence interval  pˆ ± z



pˆ (1 − pˆ ) n

This interval has the form estimate ± z ∗ SEestimate Notice that we don’t change z ∗ to t ∗ when we replace the standard deviation by the standard error. When the sample mean x estimates the population mean μ, a separate parameter σ describes the spread of the distribution of x. We separately estimate σ , and this leads to a t distribution. When the sample proportion pˆ estimates the population proportion p, the spread depends on p, not on a separate parameter. There is no t distribution—we just make the Normal approximation slightly less accurate when we replace p in the standard deviation by pˆ . We now have a confidence interval for a proportion. This interval can be trusted only for quite large samples. A rule of thumb for “how large” must take into account the fact that n must be larger if the sample proportion pˆ suggests that p may be close to 0 or 1. Because npˆ is just the number of successes in the sample and n(1 − pˆ ) is the number of failures, we can count successes and failures to get a simple guideline for “n is large and pˆ is not too close to 0 or 1.”

CAUTION UTION

LARGE-SAMPLE CONFIDENCE INTERVAL FOR A POPULATION PROPORTION Draw an SRS of size n from a large population that contains an unknown proportion p of successes. An approximate level C confidence interval for p is  pˆ (1 − pˆ ) pˆ ± z ∗ n ∗ where z is the critical value for the standard Normal density curve with area C between −z ∗ and z ∗ . Use this interval only when the numbers of successes and failures in the sample are both at least 15.3

EXAMPLE 20.4

Estimating risky behavior

The four-step process for any confidence interval is outlined on page 350.

STATE: The National AIDS Behavioral Surveys found that 170 of a sample of 2673 adult heterosexuals had multiple partners. That is, pˆ = 0.0636. What can we say about the population of all adult heterosexuals? FORMULATE: We will give a 99% confidence interval to estimate the proportion p of all adult heterosexuals who have multiple partners.

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SOLVE: First verify the conditions for inference: •



The sampling design was a complex stratified sample, and the survey used inference procedures for that design. The overall effect is close to an SRS, however. The sample is large enough: the numbers of successes (170) and failures (2503) in the sample are both much larger than 15.

The sample size condition is easily satisfied. The condition that the sample be an SRS is only approximately met. A 99% confidence interval for the proportion p of all adult heterosexuals with multiple partners uses the standard Normal critical value z ∗ = 2.576. The confidence interval is   ˆ ˆ p (1 − p ) (0.0636)(0.9364) pˆ ± z ∗ = 0.0636 ± 2.576 n 2673 = 0.0636 ± 0.0122 = 0.0514 to 0.0758

CONCLUDE: We are 99% confident that the percent of adult heterosexuals who have had more than one sexual partner in the past year lies between about 5% and 7.6%.

CAUTION UTION

As usual, the practical problems of a large sample survey weaken our confidence in the AIDS survey’s conclusions. Only people in households with telephones could be reached. This is acceptable for surveys of the general population, because about 95% of American households have telephones. However, some groups at high risk for AIDS, like intravenous drug users, often don’t live in settled households and therefore are underrepresented in the sample. About 30% of the people reached refused to cooperate. A nonresponse rate of 30% is not unusual in large sample surveys, but it may cause some bias if those who refuse differ systematically from those who cooperate. The survey used statistical methods that adjust for unequal response rates in different groups. Finally, some respondents may not have told the truth when asked about their sexual behavior. The survey team tried hard to make respondents feel comfortable. For example, Hispanic women were interviewed only by Hispanic women, and Spanish speakers were interviewed by Spanish speakers with the same regional accent (Cuban, Mexican, or Puerto Rican). Nonetheless, the survey report says that some bias is probably present: It is more likely that the present figures are underestimates; some respondents may underreport their numbers of sexual partners and intravenous drug use because of embarrassment and fear of reprisal, or they may forget or not know details of their own or of their partner’s HIV risk and their antibody testing history.4

Reading the report of a large study like the National AIDS Behavioral Surveys reminds us that statistics in practice involves much more than formulas for inference.

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APPLY YOUR KNOWLEDGE 20.7 No confidence interval. In the National AIDS Behavioral Surveys sample of 2673 adult heterosexuals, 0.2% (that’s 0.002 as a decimal fraction) had both received a blood transfusion and had a sexual partner from a group at high risk of AIDS. Explain why we can’t use the large-sample confidence interval to estimate the proportion p in the population who share these two risk factors. 20.8 How common is SAT coaching? A random sample of students who took the SAT college entrance examination twice found that 427 of the respondents had paid for coaching courses and that the remaining 2733 had not.5 Give a 99% confidence interval for the proportion of coaching among students who retake the SAT. Follow the four-step process as illustrated in Example 20.4. 20.9 Deaths from guns. The Harris Poll asked a random sample of 1009 adults which causes of death they thought would become more common in the future. Topping the list was gun violence: 70% of the sample thought deaths from guns would increase. (a) How many of the 1009 people interviewed thought deaths from gun violence would increase? (b) Harris says that the margin of error for this poll is plus or minus 3 percentage points. Explain to someone who knows no statistics what “margin of error plus or minus 3 percentage points” means. (c) Give a 95% confidence interval for this survey. Does your margin of error agree with the 3 percentage points announced by Harris?

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Accurate confidence intervals for a proportion

 The confidence interval pˆ ± z ∗ pˆ (1 − pˆ )/n for a sample proportion p is easy to calculate. It is also easy to understand because it rests directly on the approximately Normal distribution of pˆ . Unfortunately, confidence levels from this interval are often quite inaccurate unless the sample is very large. The actual confidence level is usually less than the confidence level you asked for in choosing the critical value z ∗ . That’s bad. What is worse, accuracy does not consistently get better as the sample size n increases. There are “lucky” and “unlucky” combinations of the sample size n and the true population proportion p. Fortunately, there is a simple modification that is almost magically effective in improving the accuracy of the confidence interval. We call it the “plus four” method because all you need to do is add four imaginary observations, two successes and two failures. With the added observations, the plus four estimate of p is number of successes in the sample + 2 p˜ = n+4 The formula for the confidence interval is exactly as before, with the new sample size and number of successes.6 You do not need software that offers the plus four interval—just enter the new sample size (actual size + 4) and number of successes (actual number +2) into the large-sample procedure.

plus four estimate

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PLUS FOUR CONFIDENCE INTERVAL FOR A PROPORTION Draw an SRS of size n from a large population that contains an unknown proportion p of successes. To get the plus four confidence interval for p, add four imaginary observations, two successes and two failures. Then use the large-sample confidence interval with the new sample size (n + 4) and count of successes (actual count + 2). Use this interval when the confidence level is at least 90% and the sample size n is at least 10.

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EXAMPLE 20.5

Blinding in medical trials

STATE: Many medical trials randomly assign patients to either an active treatment or a placebo. These trials are always double-blind. Sometimes the patients can tell whether or not they are getting the active treatment. This defeats the purpose of blinding. Reports of medical research usually ignore this problem. Investigators looked at a random sample of 97 articles reporting on placebo-controlled randomized trials in the top five general medical journals. Only 7 of the 97 discussed the success of blinding—and in 5 of these the blinding was imperfect.7 What proportion of all such studies discuss the success of blinding? FORMULATE: Take p to be the proportion of articles that discuss the success of blinding. Give a 95% confidence interval for p.

Who is a smoker? When estimating a proportion p, be sure you know what counts as a “success.” The news says that 20% of adolescents smoke. Shocking. It turns out that this is the percent who smoked at least once in the past month. If we say that a smoker is someone who smoked on at least 20 of the past 30 days and smoked at least half a pack on those days, fewer than 4% of adolescents qualify.

SOLVE: The conditions for use of the large-sample interval are not met because there are fewer than 15 successes in the sample. Add two successes and two failures to the original data. The plus four estimate of p is 7+2 9 p˜ = = = 0.0891 97 + 4 101 The plus four confidence interval is the same as the large-sample interval based on 9 successes in 101 observations. Here it is:   ˜ ˜ p(1 − p) (0.0891)(0.9109) p˜ ± z ∗ = 0.0891 ± 1.960 n+4 101 = 0.0891 ± 0.0556 = 0.0335 to 0.1447 CONCLUDE: We estimate with 95% confidence that between about 3.4% and 14.5% of all such articles discuss whether the blinding succeeded.

For comparison, the ordinary sample proportion is pˆ =

7 = 0.0722 97

The plus four estimate p˜ = 0.0891 in Example 20.5 is farther away from zero than pˆ = 0.0722. The plus four estimate gains its added accuracy by always moving toward 0.5 and away from 0 or 1, whichever is closer. This is particularly helpful

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when the sample contains only a few successes or a few failures. The numerical difference between a large-sample interval and the corresponding plus four interval is often small. Remember that the confidence level is the probability that the interval will catch the true population proportion in very many uses. Small differences every time add up to accurate confidence levels from plus four versus inaccurate levels from the large-sample interval. How much more accurate is the plus four interval? Computer studies have asked how large n must be to guarantee that the actual probability that a 95% confidence interval covers the true parameter value is at least 0.94 for all samples of size n or larger. If p = 0.1, for example, the answer is n = 646 for the large-sample interval and n = 11 for the plus four interval.8 The consensus of computational and theoretical studies is that plus four is very much better than the large-sample interval for many combinations of n and p. We recommend that you always use the plus four interval.

APPLY YOUR KNOWLEDGE 20.10 Drug-detecting rats? Dogs are big and expensive. Rats are small and cheap. Might rats be trained to replace dogs in sniffing out illegal drugs? A first study of this idea trained rats to rear up on their hind legs when they smelled simulated cocaine. To see how well rats performed after training, they were let loose on a surface with many cups sunk in it, one of which contained simulated cocaine. Four out of six trained rats succeeded in 80 out of 80 trials.9 How should we estimate the long-term success rate p of a rat that succeeds in every one of 80 trials? (a) What is the rat’s sample proportion pˆ ? What is the large-sample 95% confidence interval for p? It’s not plausible that the rat will always be successful, as this interval says. (b) Find the plus four estimate p˜ and the plus four 95% confidence interval for p. These results are more reasonable. This example illustrates how p˜ improves on pˆ when a sample has almost all successes or almost all failures. 20.11 Whelks and mussels. Sample surveys usually contact large samples, so we can use the large-sample confidence interval if the sample design is close to an SRS. Scientific studies often use small samples that require the plus four method. For example, the small round holes you often see in sea shells were drilled by other sea creatures, who ate the former owners of the shells. Whelks often drill into mussels, but this behavior appears to be more or less common in different locations. Investigators collected whelk eggs from the coast of Oregon, raised the whelks in the laboratory, then put each whelk in a container with some delicious mussels. Only 9 of 98 whelks drilled into a mussel.10 (a) Why can’t we use the large-sample confidence interval? (b) Give the plus four 90% confidence interval for the proportion of Oregon whelks that will spontaneously drill mussels. 20.12 High-risk behavior. In the National AIDS Behavioral Surveys sample of 2673 adult heterosexuals, 5 respondents had both received a blood transfusion and had a sexual partner from a group at high risk of AIDS. (a) You should not use the large-sample confidence interval for the proportion p in the population who share these two risk factors. Why not?

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(b) The plus four method adds four observations, two successes and two failures. What are the sample size and the count of successes after you do this? What is the plus four estimate p˜ of p? (c) Give the plus four 95% confidence interval for p.

Choosing the sample size In planning a study, we may want to choose a sample size that will allow us to estimate the parameter within a given margin of error. We saw earlier (page 355) how to do this for a population mean. The method is similar for estimating a population proportion. The margin of error in the large-sample confidence interval for p is  pˆ (1 − pˆ ) m = z∗ n

New York, New York New York City, they say, is bigger, richer, faster, ruder. Maybe there’s something to that. The sample survey firm Zogby International says that as a national average it takes 5 telephone calls to reach a live person. When calling to New York, it takes 12 calls. Survey firms assign their best interviewers to make calls to New York and often pay them bonuses to cope with the stress.

Here z ∗ is the standard Normal critical value for the level of confidence we want. Because the margin of error involves the sample proportion of successes pˆ , we need to guess this value when choosing n. Call our guess p ∗ . Here are two ways to get p ∗: 1. Use a guess p ∗ based on a pilot study or on past experience with similar studies. You can do several calculations to cover the range of values of pˆ you might get. 2. Use p ∗ = 0.5 as the guess. The margin of error m is largest when pˆ = 0.5, so this guess is conservative in the sense that if we get any other pˆ when we do our study, we will get a margin of error smaller than planned. Once you have a guess p ∗ , the recipe for the margin of error can be solved to give the sample size n needed. Here is the result for the large-sample confidence interval. For simplicity, use this result even if you plan to use the plus four interval.

SAMPLE SIZE FOR DESIRED MARGIN OF ERROR The level C confidence interval for a population proportion p will have margin of error approximately equal to a specified value m when the sample size is  ∗ 2 z p ∗ (1 − p ∗ ) n= m where p ∗ is a guessed value for the sample proportion. The margin of error will be less than or equal to m if you take the guess p ∗ to be 0.5. Which method for finding the guess p ∗ should you use? The n you get doesn’t change much when you change p ∗ as long as p ∗ is not too far from 0.5. You can

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Choosing the sample size

use the conservative guess p ∗ = 0.5 if you expect the true pˆ to be roughly between 0.3 and 0.7. If the true pˆ is close to 0 or 1, using p ∗ = 0.5 as your guess will give a sample much larger than you need. Try to use a better guess from a pilot study when you suspect that pˆ will be less than 0.3 or greater than 0.7. EXAMPLE 20.6

Planning a poll

STATE: Gloria Chavez and Ronald Flynn are the candidates for mayor in a large city. You are planning a sample survey to determine what percent of the voters plan to vote for Chavez. You will contact an SRS of registered voters in the city. You want to estimate the proportion p of Chavez voters with 95% confidence and a margin of error no greater than 3%, or 0.03. How large a sample do you need? FORMULATE: Find the sample size n needed for margin of error m = 0.03 and 95% confidence. The winner’s share in all but the most lopsided elections is between 30% and 70% of the vote. You can use the guess p ∗ = 0.5.

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VOTE CHAVEZ

SOLVE: The sample size you need is   1.96 2 n= (0.5)(1 − 0.5) = 1067.1 0.03 Round the result up to n = 1068. (Rounding down would give a margin of error slightly greater than 0.03.) CONCLUDE: An SRS of 1068 registered voters is adequate for margin of error ±3%.

If you want a 2.5% margin of error rather than 3%, then (after rounding up)   1.96 2 (0.5)(1 − 0.5) = 1537 n= 0.025 For a 2% margin of error the sample size you need is   1.96 2 n= (0.5)(1 − 0.5) = 2401 0.02 As usual, smaller margins of error call for larger samples.

APPLY YOUR KNOWLEDGE 20.13 Canadians and doctor-assisted suicide. A Gallup Poll asked a sample of Canadian adults if they thought the law should allow doctors to end the life of a patient who is in great pain and near death if the patient makes a request in writing. The poll included 270 people in Que´ bec, 221 of whom agreed that doctor-assisted suicide should be allowed.11 (a) What is the margin of error of the large-sample 95% confidence interval for the proportion of all Que´ bec adults who would allow doctor-assisted suicide? (b) How large a sample is needed to get the common ±3 percentage point margin of error? Use the previous sample as a pilot study to get p ∗ . 20.14 Can you taste PTC? PTC is a substance that has a strong bitter taste for some people and is tasteless for others. The ability to taste PTC is inherited. About

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C H A P T E R 20 • Inference about a Population Proportion

75% of Italians can taste PTC, for example. You want to estimate the proportion of Americans with at least one Italian grandparent who can taste PTC. Starting with the 75% estimate for Italians, how large a sample must you collect in order to estimate the proportion of PTC tasters within ±0.04 with 90% confidence?

Significance tests for a proportion The test statistic for the null hypothesis H 0: p = p 0 is the sample proportion pˆ standardized using the value p 0 specified by H 0 , z=

pˆ − p 0 p 0 (1 − p 0 ) n

This z statistic has approximately the standard Normal distribution when H 0 is true. P-values therefore come from the standard Normal distribution. Because H 0 fixes a value of p, the inaccuracy that plagues the large-sample confidence interval does not affect tests. Here is the procedure for tests. SIGNIFICANCE TESTS FOR A PROPORTION Draw an SRS of size n from a large population that contains an unknown proportion p of successes. To test the hypothesis H 0: p = p0 , compute the z statistic pˆ − p 0 z= p 0 (1 − p 0 ) n In terms of a variable Z having the standard Normal distribution, the approximate P-value for a test of H 0 against Ha : p > p 0

is

P ( Z ≥ z) z

Ha : p < p 0

is

H a : p = p 0

is

P ( Z ≤ z)

z

2P ( Z ≥ |z|) |z|

Use this test when the sample size n is so large that both np 0 and n(1 − p 0 ) are 10 or more.12

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Significance tests for a proportion

EXAMPLE 20.7

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Are boys more likely?

STEP

The four-step process for any significance test is outlined on page 372.

STATE: We hear that newborn babies are more likely to be boys than girls, presumably to compensate for higher mortality among boys in early life. Is this true? A random sample found 13,173 boys among 25,468 firstborn children.13 The sample proportion of boys was pˆ =

13,173 = 0.5172 25,468

Boys do make up more than half of the sample, but of course we don’t expect a perfect 50-50 split in a random sample. Is this sample evidence that boys are more common than girls in the entire population?

FORMULATE: Take p to be the proportion of boys among all firstborn children of American mothers. (Biology says that this should be the same as the proportion among all children, but the survey data concern first births.) We want to test the hypotheses H0: p = 0.5 Ha : p > 0.5

SOLVE: The conditions for inference are met, so we can go on to the z test statistic: z= 

pˆ − p 0 p 0 (1 − p 0 ) n

0.5172 − 0.5 =  = 5.49 (0.5)(0.5) 25,468 The P-value is the area under the standard Normal curve to the right of z = 5.49. We know that this is very small; Table C shows that P < 0.0005. Software (see Figure 20.4) tells us that in fact P < 0.0001.

One sample Proportion with summary Hypothesis test results: p : proportion of successes for population H0 : p = 0.5 HA : p > 0.5 Proportion

Count

Total

Sample Prop.

p

13173

25468

0.5172373

Std. Err.

Z-Stat

P-value

0.003133088

5.5017023

0.5. (b) H0: p > 0.5, Ha : p = 0.5. (c) H0: p = 0.5, Ha : p = 0.5. 20.26 The value of the z statistic for the test of the previous exercise is about (a) z = 12. (b) z = 6. (c) z = 0.6. 20.27 A Harris Poll found that 54% of American adults do not think that human beings developed from earlier species. The poll’s margin of error was 3%. This means that (a) the poll used a method that gets an answer within 3% of the truth about the population 95% of the time. (b) we can be sure that the percent of all adults who feel this way is between 51% and 57%. (c) if Harris takes another poll using the same method, the results of the second poll will lie between 51% and 57%.

C H A P T E R 20 EXERCISES We recommend using the plus four method for all confidence intervals for a proportion. However, the large-sample method is acceptable when the guidelines for its use are met. 20.28 Reporting cheating. Students are reluctant to report cheating by other students. A student project put this question to an SRS of 172 undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only 19 answered “Yes.”15 Give a 95% confidence interval for the proportion of all undergraduates at this university who would report cheating.

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Chapter 20 Exercises

20.29 Do college students pray? Social scientists asked 127 undergraduate students “from courses in psychology and communications” about prayer and found that 107 prayed at least a few times a year.16 (a) Give the plus four 99% confidence interval for the proportion p of all students who pray. (b) To use any inference procedure, we must be willing to regard these 127 students, as far as their religious behavior goes, as an SRS from the population of all undergraduate students. Do you think it is reasonable to do this? Why or why not? 20.30 Which font? Plain type fonts such as Times New Roman are easier to read than fancy fonts such as Gigi. A group of 25 volunteer subjects read the same text in both fonts. (This is a matched pairs design. One-sample procedures for proportions, like those for means, are used to analyze data from matched pairs designs.) Of the 25 subjects, 17 said that they preferred Times New Roman for Web use. But 20 said that Gigi was more attractive.17 (a) Because the subjects were volunteers, conclusions from this sample can be challenged. Show that the sample size condition for the large-sample confidence interval is not met, but that the condition for the plus four interval is met. (b) Give a 95% confidence interval for the proportion of all adults who prefer Times New Roman for Web use. Give a 90% confidence interval for the proportion of all adults who think Gigi is more attractive. 20.31 Seat belt use. The proportion of drivers who use seat belts depends on things like age, gender, ethnicity, and local law. As part of a broader study, investigators observed a random sample of 117 female Hispanic drivers in Boston; 68 of these drivers were wearing seat belts.18 Give a 95% confidence interval for the proportion of all female Hispanic drivers in Boston who wear seat belts. Follow the four-step process as illustrated in Example 20.5. 20.32 Running red lights. A random digit dialing telephone survey of 880 drivers asked, “Recalling the last ten traffic lights you drove through, how many of them were red when you entered the intersections?” Of the 880 respondents, 171 admitted that at least one light had been red.19 (a) Give a 95% confidence interval for the proportion of all drivers who ran one or more of the last ten red lights they met. (b) Nonresponse is a practical problem for this survey—only 21.6% of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: do you think more or fewer than 171 of the 880 respondents really ran a red light? Why? 20.33 Seat belt use, continued. Do the data in Exercise 20.31 give good reason to conclude that more than half of Hispanic female drivers in Boston wear seat belts? Follow the four-step process as illustrated in Example 20.7. 20.34 Seat belt use: planning a study. How large a sample would be needed to obtain margin of error ±0.05 in the study of seat belt use among Hispanic females? Use the pˆ from Exercise 20.31 as your guess for the unknown p. 20.35 Detecting genetically modified soybeans. Most soybeans grown in the United States are genetically modified to, for example, resist pests and so reduce use of pesticides. Because some nations do not accept genetically modified (GM) foods,

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C H A P T E R 20 • Inference about a Population Proportion

grain-handling facilities routinely test soybean shipments for the presence of GM beans. In a study of the accuracy of these tests, researchers submitted lots of soybeans containing 1% of GM beans to 23 randomly selected facilities. Eighteen detected the GM beans.20 (a) Show that the conditions for the large-sample confidence interval are not met. Show that the conditions for the plus four interval are met. Wesley Hitt/Alamy

(b) Use the plus four method to give a 90% confidence interval for the percent of all grain-handling facilities that will correctly detect 1% of GM beans in a shipment.

20.36 The IRS plans an SRS. The Internal Revenue Service plans to examine an SRS of individual federal income tax returns from each state. One variable of interest is the proportion of returns claiming itemized deductions. The total number of tax returns in a state varies from more than 15 million in California to about 240,000 in Wyoming. (a) Will the margin of error for estimating the population proportion change from state to state if an SRS of 2000 tax returns is selected in each state? Explain your answer. (b) Will the margin of error change from state to state if an SRS of 1% of all tax returns is selected in each state? Explain your answer. 20.37 Small-business failures. A study of the survival of small businesses chose an SRS from the telephone directory’s Yellow Pages listings of food-and-drink businesses in 12 counties in central Indiana. For various reasons, the study got no response from 45% of the businesses chosen. Interviews were completed with 148 businesses. Three years later, 22 of these businesses had failed.21 (a) Give a 95% confidence interval for the percent of all small businesses in this class that fail within three years. (b) Based on the results of this study, how large a sample would you need to reduce the margin of error to 0.04? (c) The authors hope that their findings describe the population of all small businesses. What about the study makes this unlikely? What population do you think the study findings describe? 20.38 Customer satisfaction. An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by the local dealer. The customer relations department will survey a random sample of customers and compute a 99% confidence interval for the proportion who are not satisfied. (a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of error of the confidence interval is to be about 0.015. (b) When the sample is actually contacted, 10% of the sample say they are not satisfied. What is the margin of error of the 99% confidence interval? 20.39 Surveying students. You are planning a survey of students at a large university to determine what proportion favor an increase in student fees to support an expansion of the student newspaper. Using records provided by the registrar, you can select a random sample of students. You will ask each student in the sample whether he or she is in favor of the proposed increase. Your budget will allow a sample of 100 students.

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Chapter 20 Exercises

(a) For a sample of size 100, construct a table of the margins of error for 95% confidence intervals when pˆ takes the values 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, and 0.9. (b) A former editor of the student newspaper offers to provide funds for a sample of size 500. Repeat the margin of error calculations in (a) for the larger sample size. Then write a short thank-you note to the former editor describing how the larger sample size will improve the results of the survey. In responding to Exercises 20.40 to 20.43, follow the Formulate, Solve, and Conclude steps of the four-step process. It may be helpful to restate in your own words the State information given in the exercise.

20.40 Student drinking. The College Alcohol Study interviewed a sample of 14,941 college students about their drinking habits. The sample was stratified using 140 colleges as strata, but the overall effect is close to an SRS of students. The response rate was between 60% and 70% at most colleges. This is quite good for a national sample, though nonresponse is as usual the biggest weakness of this survey. Of the students in the sample, 10,010 supported cracking down on underage drinking.22 Estimate with 99% confidence the proportion of all college students who feel this way. 20.41 Condom usage. The National AIDS Behavioral Surveys (Example 20.1) also interviewed a sample of adults in the cities where AIDS is most common. This sample included 803 heterosexuals who reported having more than one sexual partner in the past year. We can consider this an SRS of size 803 from the population of all heterosexuals in high-risk cities who have multiple partners. These people risk infection with the AIDS virus. Yet 304 of the respondents said they never use condoms. Is this strong evidence that more than one-third of this population never use condoms? 20.42 Online publishing. Publishing scientific papers online is fast, and the papers can be long. Publishing in a paper journal means that the paper will live forever in libraries. The British Medical Journal combines the two: it prints short and readable versions, with longer versions available online. Is this OK with authors? The journal asked a random sample of 104 of its recent authors several questions.23 One question was “Should the journal continue using this system?” In the sample, 72 said “Yes.” What proportion of all authors would say “Yes” if asked? (Estimate with 95% confidence.) Do the data give good evidence that more than two-thirds (67%) of authors support continuing this system? Answer both questions with appropriate inference methods. 20.43 More online publishing. The previous exercise describes a survey of authors of papers in a medical journal. Another question in the survey asked whether authors would accept a stronger move toward online publishing: “As an author, how acceptable would it be for us to publish only the abstract of papers in the paper journal and continue to put the full long version on our website?” Of the 104 authors in the sample, 65 said “Not at all acceptable.” What proportion of all authors feel that abstract-only publishing is not acceptable? (Estimate with 95% confidence.) Do the data provide good evidence that more than half of all authors feel that abstract-only publishing is not acceptable? Answer both questions with appropriate inference methods.

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