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Section 7–4 Conﬁdence Intervals and Sample Size for Proportions 369 3. Enter 563.2 for the mean of the data, and then select t. Enter 87.9 for the standard deviation and 10 for n, the sample size. 4. Either type in or scroll to 95% for the conﬁdence level, and then click [OK]. The result of the procedure will show the output in a new chart as indicated. Conﬁdence interval - mean 95% 563.2 87.9 10 2.262 62.880 626.080 500.320 7–4 Objective 4 Find the conﬁdence interval for a proportion. confidence level mean std. dev. n t (df = 9) half-width upper confidence limit lower confidence limit Conﬁdence Intervals and Sample Size for Proportions A USA TODAY Snapshots feature stated that 12% of the pleasure boats in the United States were named Serenity. The parameter 12% is called a proportion. It means that of all the pleasure boats in the United States, 12 out of every 100 are named Serenity. A proportion represents a part of a whole. It can be expressed as a fraction, decimal, or percentage. In 12 this case, 12% 0.12 100 or 253 . Proportions can also represent probabilities. In this case, if a pleasure boat is selected at random, the probability that it is called Serenity is 0.12. Proportions can be obtained from samples or populations. The following symbols will be used. Symbols Used in Proportion Notation p population proportion p̂ (read “p hat”) sample proportion For a sample proportion, X nX and q̂ or 1 p̂ n n where X number of sample units that possess the characteristics of interest and n sample size. p̂ For example, in a study, 200 people were asked if they were satisﬁed with their job or profession; 162 said that they were. In this case, n 200, X 162, and p̂ X兾n 162兾200 0.81. It can be said that for this sample, 0.81, or 81%, of those surveyed were satisﬁed with their job or profession. The sample proportion is p̂ 0.81. The proportion of people who did not respond favorably when asked if they were satisﬁed with their job or profession constituted q̂, where q̂ (n X)兾n. For this survey, q̂ (200 162)兾200 38兾200, or 0.19, or 19%. When p̂ and q̂ are given in decimals or fractions, p̂ q̂ 1. When p̂ and q̂ are given in percentages, p̂ q̂ 100%. It follows, then, that q̂ 1 p̂, or p̂ 1 q̂, when p̂ and q̂ are in decimal or fraction form. For the sample survey on job satisfaction, q̂ can also be found by using q̂ 1 p̂, or 1 0.81 0.19. 7–23 370 Chapter 7 Conﬁdence Intervals and Sample Size Similar reasoning applies to population proportions; that is, p 1 q, q 1 p, and p q 1, when p and q are expressed in decimal or fraction form. When p and q are expressed as percentages, p q 100%, p 100% q, and q 100% p. Example 7–8 In a recent survey of 150 households, 54 had central air conditioning. Find p̂ and q̂ , where p̂ is the proportion of households that have central air conditioning. Solution Since X 54 and n 150, X 54 p̂ 0.36 36% n 150 n X 150 54 96 q̂ 0.64 64% n 150 150 One can also ﬁnd q̂ by using the formula q̂ 1 p̂. In this case, q̂ 1 0.36 0.64. As with means, the statistician, given the sample proportion, tries to estimate the population proportion. Point and interval estimates for a population proportion can be made by using the sample proportion. For a point estimate of p (the population proportion), p̂ (the sample proportion) is used. On the basis of the three properties of a good estimator, p̂ is unbiased, consistent, and relatively efﬁcient. But as with means, one is not able to decide how good the point estimate of p is. Therefore, statisticians also use an interval estimate for a proportion, and they can assign a probability that the interval will contain the population proportion. The conﬁdence interval for a particular p is based on the sampling distribution of p̂. When the sample size n is no more than 5% of the population size, the sampling distribution of p̂ is approximately normally distributed with a mean of p and a standard deviation of 2pqn, where q 1 p. Conﬁdence Intervals To construct a conﬁdence interval about a proportion, one must use the maximum error of estimate, which is 兹 E z Ⲑ2 p̂q̂ n Conﬁdence intervals about proportions must meet the criteria that np 5 and nq 5. Formula for a Speciﬁc Conﬁdence Interval for a Proportion 兹 兹 p̂q̂ p̂q̂ p p̂ zⲐ2 n n when np and nq are each greater than or equal to 5. p̂ zⲐ2 Rounding Rule for a Conﬁdence Interval for a Proportion Round off to three decimal places. Example 7–9 7–24 A sample of 500 nursing applications included 60 from men. Find the 90% conﬁdence interval of the true proportion of men who applied to the nursing program. Section 7–4 Conﬁdence Intervals and Sample Size for Proportions 371 Solution Since a 1 0.90 0.10 and za兾2 1.65, substituting in the formula 兹 p̂ z Ⲑ2 兹 p̂q̂ p p̂ z Ⲑ2 n p̂q̂ n when p̂ 60兾500 0.12 and q̂ 1 0.12 0.88, one gets 兹 0.12 1.65 冸 0.12 冹冸 0.88 冹 兹 p 0.12 1.65 500 0.12 0.024 p 0.12 0.024 冸 0.12 冹冸 0.88 冹 500 0.096 p 0.144 9.6% p 14.4% or Hence, one can be 90% conﬁdent that the percentage of applicants who are men is between 9.6 and 14.4%. When a speciﬁc percentage is given, the percentage becomes p̂ when it is changed to a decimal. For example, if the problem states that 12% of the applicants were men, then p̂ 0.12. Example 7–10 A survey of 200,000 boat owners found that 12% of the pleasure boats were named Serenity. Find the 95% conﬁdence interval of the true proportion of boats named Serenity. Source: USA TODAY Snapshot. Solution From the Snapshot, p̂ 0.12 (i.e., 12%), and n 200,000. Since za兾2 1.96, substituting in the formula 兹 p̂ z Ⲑ2 yields 兹 0.12 1.96 兹 p̂q̂ p p̂ z Ⲑ2 n 冸 0.12 冹冸 0.88 冹 p̂q̂ n 兹 p 0.12 1.96 200,000 0.119 p 0.121 冸 0.12 冹冸 0.88 冹 200,000 Hence, one can say with 95% conﬁdence that the true percentage of boats named Serenity is between 11.9 and 12.1%. Sample Size for Proportions To ﬁnd the sample size needed to determine a conﬁdence interval about a proportion, use this formula: Objective 5 Determine the minimum sample size for ﬁnding a conﬁdence interval for a proportion. Formula for Minimum Sample Size Needed for Interval Estimate of a Population Proportion n p̂q̂ 冸 zEⲐ 冹 2 2 If necessary, round up to obtain a whole number. 7–25 372 Chapter 7 Conﬁdence Intervals and Sample Size This formula can be found by solving the maximum error of estimate value for n: 兹 p̂q̂ n There are two situations to consider. First, if some approximation of p̂ is known (e.g., from a previous study), that value can be used in the formula. Second, if no approximation of p̂ is known, one should use p̂ 0.5. This value will give a sample size sufﬁciently large to guarantee an accurate prediction, given the conﬁdence interval and the error of estimate. The reason is that when p̂ and q̂ are each 0.5, the product p̂q̂ is at maximum, as shown here. E z Ⲑ2 Example 7–11 p̂ q̂ p̂q̂ 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.09 0.16 0.21 0.24 0.25 0.24 0.21 0.16 0.09 A researcher wishes to estimate, with 95% conﬁdence, the proportion of people who own a home computer. A previous study shows that 40% of those interviewed had a computer at home. The researcher wishes to be accurate within 2% of the true proportion. Find the minimum sample size necessary. Solution Since za兾2 1.96, E 0.02, p̂ 0.40, and q̂ 0.60, then n p̂q̂ 冸 冹 冸 冹 z Ⲑ2 2 1.96 2 冸0.40冹冸0.60 冹 2304.96 E 0.02 which, when rounded up, is 2305 people to interview. Example 7–12 The same researcher wishes to estimate the proportion of executives who own a car phone. She wants to be 90% conﬁdent and be accurate within 5% of the true proportion. Find the minimum sample size necessary. Solution Since there is no prior knowledge of p̂, statisticians assign the values p̂ 0.5 and q̂ 0.5. The sample size obtained by using these values will be large enough to ensure the speciﬁed degree of conﬁdence. Hence, n p̂q̂ 冸 zEⲐ 冹 2 2 冸0.5冹冸0.5 冹 冸 1.65 0.05 冹 2 272.25 which, when rounded up, is 273 executives to ask. In determining the sample size, the size of the population is irrelevant. Only the degree of conﬁdence and the maximum error are necessary to make the determination. 7–26 Section 7–4 Conﬁdence Intervals and Sample Size for Proportions 373 Speaking of Statistics Does Success Bring Happiness? W. C. Fields said, “Start every day off with a smile and get it over with.” Do you think people are happy because they are successful, or are they successful because they are just happy people? A recent survey conducted by Money magazine showed that 34% of the people surveyed said that they were happy because they were successful; however, 63% said that they were successful because they were happy individuals. The people surveyed had an average household income of $75,000 or more. The margin of error was 2.5%. Based on the information in this article, what would be the conﬁdence interval for each percent? Applying the Concepts 7–4 Contracting Inﬂuenza To answer the questions, use the following table describing the percentage of people who reported contracting inﬂuenza by gender and race/ethnicity. Inﬂuenza Characteristic Gender Men Women Race/ethnicity Caucasian African American Hispanic Other Total Percent (95% CI) 48.8 51.5 (47.1–50.5%) (50.2–52.8%) 52.2 33.1 47.6 39.7 50.4 (51.1–53.3%) (29.5–36.7%) (40.9–54.3%) (30.8–48.5%) (49.3–51.5%) Forty-nine states and the District of Columbia participated in the study. Weighted means were used. The sample size was 19,774. There were 12,774 women and 7000 men. 1. 2. 3. 4. 5. 6. Explain what (95% CI) means. How large is the error for men reporting inﬂuenza? What is the sample size? How does sample size affect the size of the conﬁdence interval? Would the conﬁdence intervals be larger or smaller for a 90% CI, using the same data? Where does the 51.5% under inﬂuenza for women ﬁt into its associated 95% CI? See page 389 for the answers. 7–27 374 Chapter 7 Conﬁdence Intervals and Sample Size Exercises 7–4 1. In each case, ﬁnd p̂ and q̂. a. b. c. d. e. n 80 and X 40 n 200 and X 90 n 130 and X 60 n 60 and X 35 n 95 and X 43 2. (ans) Find p̂ and q̂ for each percentage. (Use each percentage for p̂.) a. b. c. d. e. 15% 37% 71% 51% 79% 3. A U.S. Travel Data Center survey conducted for Better Homes and Gardens of 1500 adults found that 39% said that they would take more vacations this year than last year. Find the 95% conﬁdence interval for the true proportion of adults who said that they will travel more this year. Source: USA TODAY. 4. A recent study of 100 people in Miami found 27 were obese. Find the 90% conﬁdence interval of the population proportion of individuals living in Miami who are obese. Source: Based on information from the Center for Disease Control and Prevention, USA TODAY. 5. The proportion of students in private schools is around 11%. A random sample of 450 students from a wide geographic area indicated that 55 attended private schools. Estimate the true proportion of students attending private schools with 95% conﬁdence. How does your estimate compare to 11%? Source: National Center for Education Statistics (www.nces.ed.gov). 6. The Gallup Poll found that 27% of adults surveyed nationwide said they had personally been in a tornado. How many adults should be surveyed to estimate the true proportion of adults who have been in a tornado with a 95% conﬁdence interval 5% wide? Source: www.pollingreport.com. 7. A survey found that out of 200 workers, 168 said they were interrupted three or more times an hour by phone messages, faxes, etc. Find the 90% conﬁdence interval of the population proportion of workers who are interrupted three or more times an hour. Source: Based on information from USA TODAY Snapshot. 8. A CBS News/New York Times poll found that 329 out of 763 adults said they would travel to outer space in their 7–28 lifetime, given the chance. Estimate the true proportion of adults who would like to travel to outer space with 92% conﬁdence. Source: www.pollingreport.com. 9. A study by the University of Michigan found that one in ﬁve 13- and 14-year-olds is a sometime smoker. To see how the smoking rate of the students at a large school district compared to the national rate, the superintendent surveyed two hundred 13- and 14-year-old students and found that 23% said they were sometime smokers. Find the 99% conﬁdence interval of the true proportion, and compare this with the University of Michigan study. Source: USA TODAY. 10. A survey of 50 ﬁrst-time white-water canoers showed that 23 did not want to repeat the experience. Find the 90% conﬁdence interval of the true proportion of canoers who did not wish to canoe the rapids a second time. If a rafting company wants to distribute brochures for repeat trips, what is the minimum number it should print? 11. A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% conﬁdence interval of the true proportion of families who own at least one DVD player. If another survey in a different location found that the proportion of families who owned at least one DVD player was 0.52, would you consider that the proportion of families in this area was larger than in the area where the original survey was done? 12. In a certain countrywide school district, a survey of 350 students showed that 28% carried their lunches to school. Find the 95% conﬁdence interval of the true proportion of students who carried their lunches to school. If the cafeteria manager wanted to be reasonably sure that all the children who didn’t bring their lunches could purchase a lunch, how many lunches should she plan to make each day? Explain your answer. 13. In a Gallup Poll of 1005 individuals, 452 thought they were worse off ﬁnancially than a year ago. Find the 95% conﬁdence interval for the true proportion of individuals that feel they are worse off ﬁnancially. Source: Gallup Poll. 14. In a poll of 1000 likely voters, 560 say that the United States spends too little on ﬁghting hunger at home. Find a 95% conﬁdence interval for the true proportion of voters who feel this way. Source: Alliance to End Hunger. Section 7–4 Conﬁdence Intervals and Sample Size for Proportions 15. A medical researcher wishes to determine the percentage of females who take vitamins. He wishes to be 99% conﬁdent that the estimate is within 2 percentage points of the true proportion. A recent study of 180 females showed that 25% took vitamins. a. How large should the sample size be? b. If no estimate of the sample proportion is available, how large should the sample be? 16. A recent study indicated that 29% of the 100 women over age 55 in the study were widows. a. How large a sample must one take to be 90% conﬁdent that the estimate is within 0.05 of the true proportion of women over age 55 who are widows? b. If no estimate of the sample proportion is available, how large should the sample be? 17. A researcher wishes to estimate the proportion of adult males who are under 5 feet 5 inches tall. She wants to be 90% conﬁdent that her estimate is within 5% of the true proportion. 375 a. How large a sample should be taken if in a sample of 300 males, 30 were under 5 feet 5 inches tall? b. If no estimate of the sample proportion is available, how large should the sample be? 18. Obesity is deﬁned as a body mass index (BMI) of 3 kg/m2 or more. A 95% conﬁdence interval for the percentage of U.S. adults aged 20 years and over who were obese was found to be 22.4% to 23.5%. What was the sample size? Source: National Center for Health Statistics (www.cdc.gov/nchs). 19. How large a sample should be surveyed to estimate the true proportion of college students who do laundry once a week within 3% with 95% conﬁdence? 20. A federal report indicated that 27% of children ages 2 to 5 years had a good diet—an increase over previous years. How large a sample is needed to estimate the true proportion of children with good diets within 2% with 95% conﬁdence? Source: Federal Interagency Forum on Child and Family Statistics, Washington Observer-Reporter. Extending the Concepts 21. If a sample of 600 people is selected and the researcher decides to have a maximum error of estimate of 4% on the speciﬁc proportion who favor gun control, ﬁnd the degree of conﬁdence. A recent study showed that 50% were in favor of some form of gun control. 22. In a study, 68% of 1015 adults said that they believe the Republicans favor the rich. If the margin of error was 3 percentage points, what was the conﬁdence interval used for the proportion? Source: USA TODAY. Technology Step by Step MINITAB Step by Step Find a Conﬁdence Interval for a Proportion MINITAB will calculate a conﬁdence interval, given the statistics from a sample or given the raw data. In Example 7–9, in a sample of 500 nursing applications 60 were from men. Find the 90% conﬁdence interval estimate for the true proportion of male applicants. 1. Select Stat >Basic Statistics>1 Proportion. 2. Click on the button for Summarized data. No data will be entered in the worksheet. 3. Click in the box for Number of trials and enter 500. 4. In the Number of events box, enter 60. 5. Click on [Options]. 6. Type 90 for the conﬁdence level. 7–29 376 Chapter 7 Conﬁdence Intervals and Sample Size 7. Check the box for Use test and interval based on normal distribution. 8. Click [OK] twice. The results for the conﬁdence interval will be displayed in the session window. Test and CI for One Proportion Test of p = 0.5 vs p not = 0. Sample X N Sample p 90% CI 1 60 500 0.120000 (0.096096, 0.143904) TI-83 Plus or TI-84 Plus Step by Step Z-Value -16.99 P-Value 0.000 Finding a Conﬁdence Interval for a Proportion 1. Press STAT and move the cursor to TESTS. Input 2. Press A (ALPHA, MATH) for 1-PropZlnt. 3. Type in the appropriate values. 4. Move the cursor to Calculate and press ENTER. Example TI7–3 Find the 95% conﬁdence interval of p when X 60 and n 500, as in Example 7–9. The 95% conﬁdence level for p is 0.09152 p 0.14848. Also p̂ is given. Excel Step by Step Output Finding a Conﬁdence Interval for a Proportion Excel does not produce conﬁdence intervals for a proportion. However, you may determine conﬁdence intervals for a proportion by using the MegaStat Add-in available on your CD and Online Learning Center. If you have not installed this add-in, do so by following the instructions on page 24. There were 500 nursing applications in a sample, including 60 from men. Find the 90% conﬁdence interval for the true proportion of male applicants. 1. From the toolbar, select MegaStat>Conﬁdence Intervals/Sample Size. 2. In the dialog box, select the Conﬁdence interval—p. 3. Enter 60 in the ﬁrst box; p will automatically switch to x. 4. Enter 500 in the second box for n. 5. Either type in or scroll to 90% for the conﬁdence level, and then click [OK]. 7–30 Section 7–5 Conﬁdence Intervals for Variances and Standard Deviations Speaking of Statistics 377 OTHER PEOPLE’S MONEY Here is a survey about college students’ credit card usage. Suggest several ways that the study could have been more meaningful if conﬁdence intervals had been used. Undergrads love their plastic. That means—you guessed it—students are learning to become debtors. According to the Public Interest Research Groups, only half of all students pay off card balances in full each month, 36% sometimes do and 14% never do. Meanwhile, 48% have paid a late fee. Here's how undergrads stack up, according to Nellie Mae, a provider of college loans: Undergrads with a credit card 78% Average number of cards owned 3 Average student card debt $1236 Students with 4 or more cards 32% Balances of $3000 to $7000 13% Balances over $7000 9% Reprinted with permission from the January 2002 Reader’s Digest. Copyright © 2002 by The Reader’s Digest Assn. Inc. The result of the procedure will show the output in a new chart, as indicated here. Conﬁdence interval - proportion 90% 0.12 500 1.645 0.024 0.144 0.096 7–5 Objective 6 Find a conﬁdence interval for a variance and a standard deviation. confidence level proportion n z half-width upper confidence limit lower confidence limit Conﬁdence Intervals for Variances and Standard Deviations In Sections 7–2 through 7–4, conﬁdence intervals were calculated for means and proportions. This section will explain how to ﬁnd conﬁdence intervals for variances and standard deviations. In statistics, the variance and standard deviation of a variable are as important as the mean. For example, when products that ﬁt together (such as pipes) are manufactured, it is important to keep the variations of the diameters of the products as small as possible; otherwise, they will not ﬁt together properly and will have to be scrapped. In the manufacture of medicines, the variance and standard deviation of the medication in the pills play an important role in making sure patients receive the proper dosage. For these reasons, conﬁdence intervals for variances and standard deviations are necessary. 7–31