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. What is one way to accomplish that? e. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ±3%. In one to three complete sentences, explain what the ±3% represents. 124. Refer to Exercise 8.123. Another question in the poll was “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools. a. Define the random variables X and P′ in words. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 95% confidence interval for the population proportion of adult Americans who are worried a lot about the quality of education in our schools. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. d. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ±3%. In one to three complete sentences, explain what the ±3% represents. Use the following information to answer the next three exercises: According to a Field Poll, 79% of California adults (actual results are 400 out of 506 surveyed) feel that “education and our schools” is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education and the schools is one of the top issues facing California. 125. A point estimate for the true population proportion is: a. 0.90 b. 1.27 c. 0.79 d. 400 126. A 90% confidence interval for the population proportion is _______. a. b. c. d. (0.761, 0.820) (0.125, 0.188) (0.755, 0.826) (0.130, 0.183) 127. The error bound is approximately _____. a. 1.581 b. 0.791 c. 0.059 d. 0.030 Use the following information to answer the next two exercises: Five hundred and eleven (511) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed met the minimum recommendations for earthquake preparedness, and 338 did not. 128. |
Find the confidence interval at the 90% Confidence Level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness. a. b. c. d. (0.2975, 0.3796) (0.6270, 0.6959) (0.3041, 0.3730) (0.6204, 0.7025) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 129. The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______. CHAPTER 8 | CONFIDENCE INTERVALS 457 a. 0.6614 b. 0.3386 c. 173 d. 338 130. On May 23, 2013, Gallup reported that of the 1,005 people surveyed, 76% of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at 95% with a ±3% margin of error. a. Determine the estimated proportion from the sample. b. Determine the sample size. c. d. Calculate the error bound based on the information provided. e. Compare the error bound in part d to the margin of error reported by Gallup. Explain any differences between the Identify CL and α. values. f. Create a confidence interval for the results of this study. g. A reporter is covering the release of this study for a local news station. How should she explain the confidence interval to her audience? 131. A national survey of 1,000 adults was conducted on May 13, 2013 by Rasmussen Reports. It concluded with 95% confidence that 49% to 55% of Americans believe that big-time college sports programs corrupt the process of higher education. a. Find the point estimate and the error bound for this confidence interval. b. Can we (with 95% confidence) conclude that more than half of all American adults believe this? c. Use the point estimate from part a and n = 1,000 to calculate a 75% confidence interval for the proportion of American adults that believe that major college sports programs corrupt higher education. d. Can we (with 75% confidence) conclude that at least half of all American adults believe this? 132. Public Policy Polling recently conducted a survey asking adults across the U.S. about music preferences |
. When asked, 80 of the 571 participants admitted that they have illegally downloaded music. a. Create a 99% confidence interval for the true proportion of American adults who have illegally downloaded music. b. This survey was conducted through automated telephone interviews on May 6 and 7, 2013. The error bound of the survey compensates for sampling error, or natural variability among samples. List some factors that could affect the survey’s outcome that are not covered by the margin of error. c. Without performing any calculations, describe how the confidence interval would change if the confidence level changed from 99% to 90%. 133. You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2012 presidential election with 95% confidence and a margin of error no greater than five percent. How many students must you interview? 134. In a recent Zogby International Poll, nine of 48 respondents rated the likelihood of a terrorist attack in their community as “likely” or “very likely.” Use the “plus four” method to create a 97% confidence interval for the proportion of American adults who believe that a terrorist attack in their community is likely or very likely. Explain what this confidence interval means in the context of the problem. REFERENCES 8.1 A Single Population Mean using the Normal Distribution “American Fact Finder.” U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/nav/jsf/pages/ searchresults.xhtml?refresh=t (accessed July 2, 2013). “Disclosure Data Catalog: Candidate Summary Report 2012.” U.S. Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2, 2013). “Headcount Enrollment Trends by Student Demographics Ten-Year Fall Trends to Most Recently Completed Fall.” Foothill De Anza Community College District. Available online at http://research.fhda.edu/factbook/FH_Demo_Trends/ FoothillDemographicTrends.htm (accessed September 30,2013). Kuczmarski, Robert J., Cynthia L. Ogden, Shumei S. Guo, Laurence M. Grummer-Strawn, Katherine M. Flegal, Zuguo |
Mei, Rong Wei, Lester R. Curtin, Alex F. Roche, Clifford L. Johnson. “2000 CDC Growth Charts for the United States: Methods and Development.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/growthcharts/ 2000growthchart-us.pdf (accessed July 2, 2013). 458 CHAPTER 8 | CONFIDENCE INTERVALS La, Lynn, Kent German. "Cell Phone Radiation Levels." c|net part of CBX Interactive Inc. Available online at http://reviews.cnet.com/cell-phone-radiation-levels/ (accessed July 2, 2013). “Mean Income in the Past 12 Months (in 2011 Inflaction-Adjusted Dollars): 2011 American Community Survey 1-Year Estimates.” American Fact Finder, U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/ tableservices/jsf/pages/productview.xhtml?pid=ACS_11_1YR_S1902&prodType=table (accessed July 2, 2013). “Metadata Description of Candidate Summary File.” U.S. Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml (accessed July 2, 2013). “National Health and Nutrition Examination Survey.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed July 2, 2013). 8.2 A Single Population Mean using the Student t Distribution “America’s Best Small Companies.” Forbes, 2013. Available online at http://www.forbes.com/best-small-companies/list/ (accessed July 2, 2013). Data from Microsoft Bookshelf. Data from http://www.businessweek.com/. Data from http://www.forbes.com/. “Disclosure Data Catalog: Leadership PAC and Sponsors Report, 2012.” Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2,2013). “Human Toxome Project: Mapping the Pollution in People.” Environmental Working Group. Available online at http://www. |
ewg.org/sites/humantoxome/participants/participant-group.php?group=in+utero%2Fnewborn (accessed July 2, 2013). “Metadata Description of Leadership PAC List.” Federal Election Commission. Available online at http://www.fec.gov/ finance/disclosure/metadata/metadataLeadershipPacList.shtml (accessed July 2, 2013). 8.3 A Population Proportion Jensen, Tom. “Democrats, Republicans Divided on Opinion of Music Icons.” Public Policy Polling. Available online at http://www.publicpolicypolling.com/Day2MusicPoll.pdf (accessed July 2, 2013). Madden, Mary, Amanda Lenhart, Sandra Coresi, Urs Gasser, Maeve Duggan, Aaron Smith, and Meredith Beaton. “Teens, Social Media, and Privacy.” PewInternet, 2013. Available online at http://www.pewinternet.org/Reports/2013/Teens-SocialMedia-And-Privacy.aspx (accessed July 2, 2013). Prince Survey Research Associates International. “2013 Teen and Privacy Management Survey.” Pew Research Center: Internet and American Life Project. Available online at http://www.pewinternet.org/~/media//Files/Questionnaire/2013/ Methods%20and%20Questions_Teens%20and%20Social%20Media.pdf (accessed July 2, 2013). Saad, Lydia. “Three in Four U.S. Workers Plan to Work Pas Retirement Age: Slightly more say they will do this by choice rather than necessity.” Gallup® Economy, 2013. Available online at http://www.gallup.com/poll/162758/three-fourworkers-plan-work-past-retirement-age.aspx (accessed July 2, 2013). The Field Poll. Available online at http://field.com/fieldpollonline/subscribers/ (accessed July 2, 2013). Zogby. “New SUNYIT/Zogby Analytics Poll: Few Americans Worry about Emergency Situations Occurring in Their Community; Only one in three have an Emergency Plan; 70% Support Infrastructure ‘Investment’ for National Security.” Zogby Analytics, 2013. Available online at http://www.zogbyanalytics.com/ |
news/299-americans-neither-worried-norprepared-in-case-of-a-disaster-sunyit-zogby-analytics-poll (accessed July 2, 2013). “52% Say Big-Time College Athletics Corrupt Education Process.” Rasmussen Reports, 2013. Available online at http://www.rasmussenreports.com/public_content/lifestyle/sports/may_2013/ 52_say_big_time_college_athletics_corrupt_education_process (accessed July 2, 2013). SOLUTIONS 1 a. 244 b. 15 c. 50 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 459 3 N⎛ ⎝244, 15 50 ⎞ ⎠ 5 As the sample size increases, there will be less variability in the mean, so the interval size decreases. ¯ 7 X is the time in minutes it takes to complete the U.S. Census short form. X people to complete the U.S. Census short form. is the mean time it took a sample of 200 9 CI: (7.9441, 8.4559) Figure 8.11 EBM = 0.26 11 The level of confidence would decrease because decreasing n makes the confidence interval wider, so at the same error bound, the confidence level decreases. 13 a. x¯ = 2.2 b. σ = 0.2 c. n = 20 ¯ 15 X is the mean weight of a sample of 20 heads of lettuce. 17 EBM = 0.07 CI: (2.1264, 2.2736) Figure 8.12 19 The interval is greater because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals. 21 The confidence level would increase. 460 CHAPTER 8 | CONFIDENCE INTERVALS 23 30.4 25 σ 27 μ 29 normal 31 0.025 33 (24.52,36.28) 35 We are 95% confident that the true mean age for Winger Foothill College |
students is between 24.52 and 36.28. 37 The error bound for the mean would decrease because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean. ¯ 39 X is the number of hours a patient waits in the emergency room before being called back to be examined. X mean wait time of 70 patients in the emergency room. is the 41 CI: (1.3808, 1.6192) Figure 8.13 EBM = 0.12 43 a. b. x¯ = 151 s x = 32 c. n = 108 d. n – 1 = 107 ¯ 45 X is the mean number of hours spent watching television per month from a sample of 108 Americans. 47 CI: (142.92, 159.08) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 461 Figure 8.14 EBM = 8.08 49 a. 3.26 b. 1.02 c. 39 51 μ 53 t38 55 0.025 57 (2.93, 3.59) 59 We are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors. 60 The error bound would become EBM = 0.245. This error bound decreases because as sample sizes increase, variability decreases and we need less interval length to capture the true mean. 63 It would decrease, because the z-score would decrease, which reducing the numerator and lowering the number. 65 X is the number of “successes” where the woman makes the majority of the purchasing decisions for the household. P′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household. 67 CI: (0.5321, 0.6679) Figure 8.15 EBM: 0.0679 69 X is the number of “successes” where an executive prefers a truck. P′ is the percentage of executives sampled who prefer a truck. 462 CHAPTER 8 | CONFIDENCE INTERVALS 71 CI: (0.19432, 0.33068) Figure 8.16 EBM: 0.0707 73 The sampling error means that the true mean can be 2% above or below |
the sample mean. 75 P′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election. 77 CI: (0.62735, 0.67265) EBM: 0.02265 79 The number of girls, ages 8 to 12, in the 5 P.M. Monday night beginning ice-skating class. 81 a. x = 64 b. n = 80 c. p′ = 0.8 83 p 85 P′ ~ N⎛ ⎝0.8, (0.8)(0.2) 80 ⎞ ⎠. (0.72171, 0.87829). 87 0.04 89 (0.72; 0.88) 91 With 92% confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72% and 88%. 93 The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error. 95 a. i. 71 ii. 3 iii. 48 b. X is the height of a Swiss male, and is the mean height from a sample of 48 Swiss males. c. Normal. We know the standard deviation for the population, and the sample size is greater than 30. d. i. CI: (70.151, 71.49) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 463 ii. Figure 8.17 iii. EBM = 0.849 e. The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean. 97 a. i. ii. x¯ = 23.6 σ = 7 iii. n = 100 ¯ b. X is the time needed to complete an individual tax form. X is the mean time to complete tax forms from a sample of 100 customers. c. N ⎛ ⎝23.6, ⎞ ⎠ 7 100 because we know sigma. d. |
i. (22.228, 24.972) ii. Figure 8.18 iii. EBM = 1.372 e. It will need to change the sample size. The firm needs to determine what the confidence level should be, then apply the error bound formula to determine the necessary sample size. f. The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval. 464 CHAPTER 8 | CONFIDENCE INTERVALS g. According to the error bound formula, the firm needs to survey 206 people. Since we increase the confidence level, we need to increase either our error bound or the sample size. 99 a. i. 7.9 ii. 2.5 iii. 20 ¯ b. X is the number of letters a single camper will send home. X is the mean number of letters sent home from a sample of 20 campers. c. N 7.9 ⎛ ⎝ ⎞ ⎠ 2.5 20 d. i. CI: (6.98, 8.82) ii. Figure 8.19 iii. EBM: 0.92 e. The error bound and confidence interval will decrease. 101 a. x¯ = $568,873 b. CL = 0.95 α = 1 – 0.95 = 0.05 z α 2 = 1.96 909200 EBM = z0.025 σ n 40 = 1.96 = $281,764 c. x¯ − EBM = 568,873 − 281,764 = 287,109 x¯ + EBM = 568,873 + 281,764 = 850,637 Alternate solution: 1. Press STAT and arrow over to TESTS. 2. Arrow down to 7:ZInterval. 3. Press ENTER. 4. Arrow to Stats and press ENTER. 5. Arrow down and enter the following values: σ : 909,200 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 465 x¯ : 568,873 n: 40 CL: 0.95 6. Arrow down to Calculate and press ENTER. 7. The confidence interval is ($287, |
114, $850,632). 8. Notice the small difference between the two solutions–these differences are simply due to rounding error in the hand calculations. d. We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637. 103 Use the formula for EBM, solved for n: n = z2 σ 2 EBM 2 From the statement of the problem, you know that σ = 2.5, and you need EBM = 1. z = z0.035 = 1.812 (This is the value of z for which the area under the density curve to the right of z is 0.035.) ≈ 20.52 You need to measure at least 21 male students to achieve your goal. z2 σ 2 EBM 2 = 1.8122 2.52 12 n = 105 a. i. 8629 ii. 6944 iii. 35 iv. 34 t34 i. CI: (6244, 11,014) b. c. ii. Figure 8.20 iii. EB = 2385 d. It will become smaller 107 a. i. ii. x¯ = 2.51 s x = 0.318 466 CHAPTER 8 | CONFIDENCE INTERVALS iii. n = 9 iv. n - 1 = 8 b. c. the effective length of time for a tranquilizer the mean effective length of time of tranquilizers from a sample of nine patients d. We need to use a Student’s-t distribution, because we do not know the population standard deviation. e. i. CI: (2.27, 2.76) ii. Check student's solution. iii. EBM: 0.25 f. If we were to sample many groups of nine patients, 95% of the samples would contain the true population mean length of time. 109 x¯ = $251, 854.23 s = $521, 130.41 Note that we are not given the population standard deviation, only the standard deviation of the sample. There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29 CL = 0.96, so α = 1 - CL = 1 - 0.96 = 0.04 α 2 ⎞ ⎠ ~ $204, 561.66 x¯ ⎛ ⎞ ⎠ = 2.150 ⎝ |
= 2.150 EBM = t α 2 521, 130.41 30 = 0.02t α 2 = t0.02 s n ⎛ ⎝ - EBM = $251,854.23 - $204,561.66 = $47,292.57 x¯ + EBM = $251,854.23+ $204,561.66 = $456,415.89 We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89. Alternate Solution Enter the data as a list. Press STAT and arrow over to TESTS. Arrow down to 8:TInterval. Press ENTER. Arrow to Data and press ENTER. Arrow down and enter the name of the list where the data is stored. Enter Freq: 1 Enter C-Level: 0.96 Arrow down to Calculate and press Enter. The 96% confidence interval is ($47,262, $456,447). The difference between solutions arises from rounding differences. 111 a. i. ii. x¯ = s x = iii. n = iv. n - 1 = ¯ b. X is the number of unoccupied seats on a single flight. X is the mean number of unoccupied seats from a sample of 225 flights. c. We will use a Student’s-t distribution, because we do not know the population standard deviation. d. i. CI: (11.12, 12.08) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 467 ii. Check student's solution. iii. EBM: 0.48 113 a. i. CI: (7.64, 9.36) ii. Figure 8.21 iii. EBM: 1.13 b. The sample should have been increased. c. Answers will vary. d. Answers will vary. e. Answers will vary. 115 b 117 a. 1,068 b. The sample size would need to be increased since the critical value increases as the confidence level increases. 119 a. X = the number of people who feel that the president is doing an acceptable job; P′ = the proportion of people in a sample who feel that |
the president is doing an acceptable job. b. N⎛ ⎝0.61, (0.61)(0.39) 1200 ⎞ ⎠ c. i. CI: (0.59, 0.63) ii. Check student’s solution iii. EBM: 0.02 121 a. i. ii. (0.72, 0.82) (0.65, 0.76) iii. (0.60, 0.72) b. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap. c. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families. 468 CHAPTER 8 | CONFIDENCE INTERVALS d. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a black person into their families. 123 a. X = the number of adult Americans who feel that crime is the main problem; P′ = the proportion of adult Americans who feel that crime is the main problem b. Since we are estimating a proportion, given P′ = 0.2 and n = 1000, the distribution we should use is N⎛ ⎝0.2, (0.2)(0.8) 1000 ⎞ ⎠. c. i. CI: (0.18, 0.22) ii. Check student’s solution. iii. EBM: 0.02 d. One way to lower the sampling error is to increase the sample size. e. The stated “± 3%” represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3%. Thus, they estimate the percentage of adult Americans who feel that crime is the main problem to be between 18% and 22%. 125 c 127 d 129 a 131 a. p′ = (0.55 + 0.49) 2 = 0.52; EBP = 0.55 - 0.52 = 0.03 b. No |
, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this. c. CL = 0.75, so α = 1 – 0.75 = 0.25 and α 2 = 0.125 z α 2 = 1.150. (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.) EBP = (1.150) 0.52(0.48) 1, 000 ≈ 0.018 (p′ - EBP, p′ + EBP) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538) Alternate Solution STAT TESTS A: 1-PropZinterval with x = (0.52)(1,000), n = 1,000, CL = 0.75. Answer is (0.502, 0.538) d. Yes – this interval does not fall less than 0.50 so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75% confidence. 133 CL = 0.95 α = 1 – 0.95 = 0.05 α 2 = 0.025 z α 2 = 1.96. Use p′ = q′ = 0.5. n = 2 p′ q′ z α 2 EBP2 = 1.962(0.5)(0.5) 0.052 within 5% at 95% confidence. = 384.16 You need to interview at least 385 students to estimate the proportion to This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 469 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Figure 9.1 You can use a hypothesis test to decide if a dog breeder’s claim that every Dalmatian has 35 spots is statistically sound. (Credit: Robert Neff) Introduction By the end of this chapter, the student should be able to: Chapter Objectives • Differentiate between Type I and Type II Errors • Describe hypothesis testing in general and in practice • Conduct and interpret hypothesis tests for a single population mean, population standard |
deviation known. • Conduct and interpret hypothesis tests for a single population mean, population standard deviation unknown. • Conduct and interpret hypothesis tests for a single population proportion. One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to 470 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE make a decision about a parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of $60,000 per year. A statistician will make a decision about these claims. This process is called " hypothesis testing." A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analyses of the data, to reject the null hypothesis. In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also learn about the errors associated with these tests. Hypothesis testing consists of two contradictory hypotheses or statements, a decision based on the data, and a conclusion. To perform a hypothesis test, a statistician will: 1. Set up two contradictory hypotheses. 2. Collect sample data (in homework problems, the data or summary statistics will be given to you). 3. Determine the correct distribution to perform the hypothesis test. 4. Analyze sample data by performing the calculations that ultimately will allow you to reject or decline to reject the null hypothesis. 5. Make a decision and write a meaningful conclusion. NOTE To do the hypothesis test homework problems for this chapter and later chapters, make copies of the appropriate special solution sheets. See the 47882 (http://cnx.org/content/47882/latest/). 9.1 | Null and Alternative Hypotheses The actual test begins by considering two hypotheses. They are called the null hypothesis and the alternative hypothesis. These hypotheses contain opposing viewpoints. H0: The null hypothesis: It is a statement about the population that either is believed to be true or is used to put forth an argument unless it can be shown to be incorrect beyond a reasonable doubt. Ha: The alternative hypothesis: It is a claim about the population that is contradictory to H0 and what |
we conclude when we reject H0. Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data. After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are "reject H0" if the sample information favors the alternative hypothesis or "do not reject H0" or "decline to reject H0" if the sample information is insufficient to reject the null hypothesis. Mathematical Symbols Used in H0 and Ha: H0 equal (=) Ha not equal (≠) or greater than (>) or less than (<) greater than or equal to (≥) less than (<) less than or equal to (≤) more than (>) Table 9.1 NOTE H0 always has a symbol with an equal in it. Ha never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers (including one of the co-authors in research work) use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 471 Example 9.1 H0: No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ 30 Ha: More than 30% of the registered voters in Santa Clara County voted in the primary election. p > 30 9.1 A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses. Example 9.2 We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are: H0: μ = 2.0 Ha: μ ≠ 2.0 9.2 We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative |
hypotheses. a. H0: μ __ 66 b. Ha: μ __ 66 Example 9.3 We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are: H0: μ ≥ 5 Ha: μ < 5 9.3 We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. a. H0: μ __ 45 b. Ha: μ __ 45 Example 9.4 In an issue of U. S. News and World Report, an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses. 472 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE H0: p ≤ 0.066 Ha: p > 0.066 9.4 On a state driver’s test, about 40% pass the test on the first try. We want to test if more than 40% pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. a. H0: p __ 0.40 b. Ha: p __ 0.40 Bring to class a newspaper, some news magazines, and some Internet articles. In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class. 9.2 | Outcomes and the Type I and Type II Errors When you perform a hypothesis test, there are four possible outcomes depending on the actual truth (or falseness) of the null hypothesis H0 and the decision to reject or not. The outcomes are summarized in the following table: ACTION H0 IS ACTUALLY... True False Do not reject H0 Correct Outcome Type II error Reject H0 Type I Error Correct Outcome Table 9.2 The four possible outcomes in the table are: 1. The decision is not to reject H0 when H0 is true (correct decision). 2. The decision |
is to reject H0 when H0 is true (incorrect decision known as a Type I error). 3. The decision is not to reject H0 when, in fact, H0 is false (incorrect decision known as a Type II error). 4. The decision is to reject H0 when H0 is false (correct decision whose probability is called the Power of the Test). Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities. α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true. β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false. α and β should be as small as possible because they are probabilities of errors. They are rarely zero. The Power of the Test is 1 – β. Ideally, we want a high power that is as close to one as possible. Increasing the sample size can increase the Power of the Test. The following are examples of Type I and Type II errors. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 473 Example 9.5 Suppose the null hypothesis, H0, is: Frank's rock climbing equipment is safe. Type I error: Frank thinks that his rock climbing equipment may not be safe when, in fact, it really is safe. Type II error: Frank thinks that his rock climbing equipment may be safe when, in fact, it is not safe. α = probability that Frank thinks his rock climbing equipment may not be safe when, in fact, it really is safe. β = probability that Frank thinks his rock climbing equipment may be safe when, in fact, it is not safe. Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.) 9.5 Suppose the null hypothesis, H0, is: the blood cultures contain no traces of pathogen X. State the Type I and Type II errors. Example 9.6 Suppose the null hypothesis, H0, is: The victim of an automobile accident is alive when he arrives at the emergency room of a hospital. Type I error: |
The emergency crew thinks that the victim is dead when, in fact, the victim is alive. Type II error: The emergency crew does not know if the victim is alive when, in fact, the victim is dead. α = probability that the emergency crew thinks the victim is dead when, in fact, he is really alive = P(Type I error). β = probability that the emergency crew does not know if the victim is alive when, in fact, the victim is dead = P(Type II error). The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim is dead, they will not treat him.) 9.6 Suppose the null hypothesis, H0, is: a patient is not sick. Which type of error has the greater consequence, Type I or Type II? Example 9.7 It’s a Boy Genetic Labs claim to be able to increase the likelihood that a pregnancy will result in a boy being born. Statisticians want to test the claim. Suppose that the null hypothesis, H0, is: It’s a Boy Genetic Labs has no effect on gender outcome. Type I error: This results when a true null hypothesis is rejected. In the context of this scenario, we would state that we believe that It’s a Boy Genetic Labs influences the gender outcome, when in fact it has no effect. The probability of this error occurring is denoted by the Greek letter alpha, α. Type II error: This results when we fail to reject a false null hypothesis. In context, we would state that It’s a Boy Genetic Labs does not influence the gender outcome of a pregnancy when, in fact, it does. The probability of this error occurring is denoted by the Greek letter beta, β. The error of greater consequence would be the Type I error since couples would use the It’s a Boy Genetic Labs product in hopes of increasing the chances of having a boy. 474 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 9.7 “Red tide” is a bloom of poison-producing algae–a few different species of a class of plankton called dinoflagellates. When the weather and water conditions cause these blooms, shellfish such as clams living in the area develop dangerous levels of a paralysis-inducing toxin. In Massachusetts, the Division of Marine Fisheries (DMF) monitors levels of the toxin in shellfish by regular sampling of shellfish along the |
coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kg of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequence. Example 9.8 A certain experimental drug claims a cure rate of at least 75% for males with prostate cancer. Describe both the Type I and Type II errors in context. Which error is the more serious? Type I: A cancer patient believes the cure rate for the drug is less than 75% when it actually is at least 75%. Type II: A cancer patient believes the experimental drug has at least a 75% cure rate when it has a cure rate that is less than 75%. In this scenario, the Type II error contains the more severe consequence. If a patient believes the drug works at least 75% of the time, this most likely will influence the patient’s (and doctor’s) choice about whether to use the drug as a treatment option. 9.8 Determine both Type I and Type II errors for the following scenario: Assume a null hypothesis, H0, that states the percentage of adults with jobs is at least 88%. Identify the Type I and Type II errors from these four statements. a. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88% b. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%. c. Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%. d. Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88%. 9.3 | Distribution Needed for Hypothesis Testing Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student's t-distribution. (Remember, use a Student's t-distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually n is large or the sample size is large). If you are testing a single population |
mean, the distribution for the test is for means: ¯ X ~ N⎛ ⎝µ X, σ X n ⎞ ⎠ or td f The population parameter is μ. The estimated value (point estimate) for μ is x¯, the sample mean. If you are testing a single population proportion, the distribution for the test is for proportions or percentages: This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 P′ ~ N⎛ ⎝p, p ⋅ q n ⎞ ⎠ The population parameter is p. The estimated value (point estimate) for p is p′. p′ = x n where x is the number of successes CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 475 and n is the sample size. Assumptions When you perform a hypothesis test of a single population mean μ using a Student's t-distribution (often called a t-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. (Note that if the sample size is sufficiently large, a t-test will work even if the population is not approximately normally distributed). When you perform a hypothesis test of a single population mean μ using a normal distribution (often called a z-test), you take a simple random sample from the population. The population you are testing is normally distributed or your sample size is sufficiently large. You know the value of the population standard deviation which, in reality, is rarely known. When you perform a hypothesis test of a single population proportion p, you take a simple random sample from the population. You must meet the conditions for a binomial distribution which are: there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success p. The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five (np > 5 and nq > 5). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ |
= p and σ = pq n. Remember that q = 1 – p. 9.4 | Rare Events, the Sample, Decision and Conclusion Establishing the type of distribution, sample size, and known or unknown standard deviation can help you figure out how to go about a hypothesis test. However, there are several other factors you should consider when working out a hypothesis test. Rare Events Suppose you make an assumption about a property of the population (this assumption is the null hypothesis). Then you gather sample data randomly. If the sample has properties that would be very unlikely to occur if the assumption is true, then you would conclude that your assumption about the population is probably incorrect. (Remember that your assumption is just an assumption—it is not a fact and it may or may not be true. But your sample data are real and the data are showing you a fact that seems to contradict your assumption.) For example, Didi and Ali are at a birthday party of a very wealthy friend. They hurry to be first in line to grab a prize from a tall basket that they cannot see inside because they will be blindfolded. There are 200 plastic bubbles in the basket and Didi and Ali have been told that there is only one with a $100 bill. Didi is the first person to reach into the basket and pull out a bubble. Her bubble contains a $100 bill. The probability of this happening is 1 200 = 0.005. Because this is so unlikely, Ali is hoping that what the two of them were told is wrong and there are more $100 bills in the basket. A "rare event" has occurred (Didi getting the $100 bill) so Ali doubts the assumption about only one $100 bill being in the basket. Using the Sample to Test the Null Hypothesis Use the sample data to calculate the actual probability of getting the test result, called the p-value. The p-value is the probability that, if the null hypothesis is true, the results from another randomly selected sample will be as extreme or more extreme as the results obtained from the given sample. A large p-value calculated from the data indicates that we should not reject the null hypothesis. The smaller the p-value, the more unlikely the outcome, and the stronger the evidence is against the null hypothesis. We would reject the null hypothesis if the evidence is strongly against it. Draw a graph that shows the p-value. The hypothesis test is easier to perform if you use a graph because you see the problem |
more clearly. 476 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Example 9.9 Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The baker knows from baking hundreds of loaves of bread that the standard deviation for the height is 0.5 cm. and the distribution of heights is normal. The null hypothesis could be H0: μ ≤ 15 The alternate hypothesis is Ha: μ > 15 The words "is more than" translates as a ">" so "μ > 15" goes into the alternate hypothesis. The null hypothesis must contradict the alternate hypothesis. Since σ is known (σ = 0.5 cm.), the distribution for the population is known to be normal with mean μ = 15 and standard deviation σ n = 0.5 10 = 0.16. Suppose the null hypothesis is true (the mean height of the loaves is no more than 15 cm). Then is the mean height (17 cm) calculated from the sample unexpectedly large? The hypothesis test works by asking the question how unlikely the sample mean would be if the null hypothesis were true. The graph shows how far out the sample mean is on the normal curve. The p-value is the probability that, if we were to take other samples, any other sample mean would fall at least as far out as 17 cm. The p-value, then, is the probability that a sample mean is the same or greater than 17 cm. when the population mean is, in fact, 15 cm. We can calculate this probability using the normal distribution for means. Figure 9.2 p-value= P( x¯ > 17) which is approximately zero. A p-value of approximately zero tells us that it is highly unlikely that a loaf of bread rises no more than 15 cm, on average. That is, almost 0% of all loaves of bread would be at least as high as 17 cm. purely by CHANCE had the population mean height really been 15 cm. Because the outcome of 17 cm. is so unlikely (meaning it is happening NOT by chance alone), we conclude that the evidence is strongly against the null hypothesis (the mean height is at most 15 cm.). There is sufficient evidence that the true mean height for the population |
of the baker's loaves of bread is greater than 15 cm. 9.9 A normal distribution has a standard deviation of 1. We want to verify a claim that the mean is greater than 12. A sample of 36 is taken with a sample mean of 12.5. H0: μ ≤ 12 Ha: μ > 12 The p-value is 0.0013 Draw a graph that shows the p-value. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 477 Decision and Conclusion A systematic way to make a decision of whether to reject or not reject the null hypothesis is to compare the p-value and a preset or preconceived α (also called a "significance level"). A preset α is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem. When you make a decision to reject or not reject H0, do as follows: • • If α > p-value, reject H0. The results of the sample data are significant. There is sufficient evidence to conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct. If α ≤ p-value, do not reject H0. The results of the sample data are not significant.There is not sufficient evidence to conclude that the alternative hypothesis,Ha, may be correct. • When you "do not reject H0", it does not mean that you should believe that H0 is true. It simply means that the sample data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of Ho. Conclusion: After you make your decision, write a thoughtful conclusion about the hypotheses in terms of the given problem. Example 9.10 When using the p-value to evaluate a hypothesis test, it is sometimes useful to use the following memory device If the p-value is low, the null must go. If the p-value is high, the null must fly. This memory aid relates a p-value less than the established alpha (the p is low) as rejecting the null hypothesis and, likewise, relates a p-value higher than the established alpha (the p is high) as not rejecting the null hypothesis. Fill in the blanks |
. Reject the null hypothesis when ______________________________________. The results of the sample data _____________________________________. Do not reject the null when hypothesis when __________________________________________. The results of the sample data ____________________________________________. Solution 9.10 Reject the null hypothesis when the p-value is less than the established alpha value. The results of the sample data support the alternative hypothesis. Do not reject the null hypothesis when the p-value is greater than the established alpha value. The results of the sample data do not support the alternative hypothesis. 9.10 It’s a Boy Genetics Labs claim their procedures improve the chances of a boy being born. The results for a test of a single population proportion are as follows: H0: p = 0.50, Ha: p > 0.50 α = 0.01 p-value = 0.025 Interpret the results and state a conclusion in simple, non-technical terms. 478 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 9.5 | Additional Information and Full Hypothesis Test Examples • In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is the preconceived or preset α. • The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data. • If no level of significance is given, a common standard to use is α = 0.05. • When you calculate the p-value and draw the picture, the p-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed. • The alternative hypothesis, Ha, tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test. • Ha never has a symbol that contains an equal sign. • Thinking about the meaning of the p-value: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller p-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p-value such as 0.4, as opposed to a p-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct |
decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules. The following examples illustrate a left-, right-, and two-tailed test. Example 9.11 Ho: μ = 5, Ha: μ < 5 Test of a single population mean. Ha tells you the test is left-tailed. The picture of the p-value is as follows: Figure 9.3 9.11 H0: μ = 10, Ha: μ < 10 Assume the p-value is 0.0935. What type of test is this? Draw the picture of the p-value. Example 9.12 H0: p ≤ 0.2 Ha: p > 0.2 This is a test of a single population proportion. Ha tells you the test is right-tailed. The picture of the p-value is as follows: This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 479 Figure 9.4 9.12 H0: μ ≤ 1, Ha: μ > 1 Assume the p-value is 0.1243. What type of test is this? Draw the picture of the p-value. Example 9.13 H0: p = 50 Ha: p ≠ 50 This is a test of a single population mean. Ha tells you the test is two-tailed. The picture of the p-value is as follows. Figure 9.5 9.13 H0: p = 0.5, Ha: p ≠ 0.5 Assume the p-value is 0.2564. What type of test is this? Draw the picture of the p-value. 480 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Full Hypothesis Test Examples Example 9.14 Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the |
16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal. Solution 9.14 Set up the Hypothesis Test: Since the problem is about a mean, this is a test of a single population mean. H0: μ = 16.43 Ha: μ < 16.43 For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed. Determine the distribution needed: ¯ Random variable: X = the mean time to swim the 25-yard freestyle. ¯ Distribution for the test: X is normal (population standard deviation is known: σ = 0.8) ¯ X ~ N⎛ ⎝µ, σ X n ¯ ⎞ ⎠ Therefore, X ~ N⎛ ⎝16.43, 0.8 15 ⎞ ⎠ μ = 16.43 comes from H0 and not the data. σ = 0.8, and n = 15. Calculate the p-value using the normal distribution for a mean: p-value = P( x¯ < 16) = 0.0187 where the sample mean in the problem is given as 16. p-value = 0.0187 (This is called the actual level of significance.) The p-value is the area to the left of the sample mean is given as 16. Graph: Figure 9.6 μ = 16.43 comes from H0. Our assumption is μ = 16.43. Interpretation of the p-value: If H0 is true, there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event. Compare α and the p-value: α = 0.05 p-value = 0.0187 α > p-value Make a decision: Since α > p-value, reject H0. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 481 This means that |
you reject μ = 16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles. Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. The p-value can easily be calculated. Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 16.43 for μ0 (null hypothesis),.8 for σ, 16 for the sample mean, and 15 for n. Arrow down to μ : (alternate hypothesis) and arrow over to < μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.0187) but it also calculates the test statistic (zscore) for the sample mean. μ < 16.43 is the alternative hypothesis. Do this set of instructions again except arrow to Draw(instead of Calculate). Press ENTER. A shaded graph appears with z = -2.08 (test statistic) and p = 0.0187 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. When the calculator does a Z-Test, the Z-Test function finds the p-value by doing a normal probability calculation using the central limit theorem: P( x¯ < 16) = 2nd DISTR normcdf ⎛ ⎝ − 10 ^ 99, 16, 16.43, 0.8 / 15⎞ ⎠. The Type I and Type II errors for this problem are as follows: The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.) The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. ( |
Do not reject the null hypothesis when the null hypothesis is false.) 9.14 The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal. First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion. Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for μ0 (null hypothesis), 2 for σ, 45 for the sample mean, and 20 for n. Arrow down to μ: (alternative hypothesis) and set it either as <, ≠, or >. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value but it also calculates the test statistic (z-score) for the sample mean. Select <, ≠, or > for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of 482 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Calculate). Press ENTER. A shaded graph appears with test statistic and p-value. Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. HISTORICAL NOTE (EXAMPLE 9.11) The traditional way to compare the two probabilities, α and the p-value, is to compare the critical value (z-score from α) to the test statistic (z-score from data). The calculated test statistic for the p-value is –2.08. (From the Central Limit Theorem, the test statistic formula is z = x¯ − µ X σ X n ) (. For this problem, x¯ = 16, μX = 16.43 from the null hypothes is, σX = 0.8, and n = 15.) You can find the critical value for α = 0.05 in the normal table (see 15.Tables in the Table of Contents). The z- |
score for an area to the left equal to 0.05 is midway between –1.65 and –1.64 (0.05 is midway between 0.0505 and 0.0495). The z-score is –1.645. Since –1.645 > –2.08 (which demonstrates that α > p-value), reject H0. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities α and the p-value is very common. For this problem, the p-value, 0.0187 is considerably smaller than α, 0.05. You can be confident about your decision to reject. The graph shows α, the p-value, and the test statistics and the critical value. Figure 9.7 Example 9.15 A college football coach thought that his players could bench press a mean weight of 275 pounds. It is known that the standard deviation is 55 pounds. Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2); 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1). Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds. Solution 9.15 Set up the Hypothesis Test: Since the problem is about a mean weight, this is a test of a single population mean. H0: μ = 275 Ha: μ > 275 This is a right-tailed test. Calculating the distribution needed: Random variable: X ¯ = the mean weight, in pounds, lifted by the football players. Distribution for the test: It is normal because σ is known. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 483 ¯ X ~ N ⎛ ⎝275, 55 30 ⎞ ⎠ x¯ = 286.2 pounds (from the data). |
σ = 55 pounds (Always use σ if you know it.) We assume μ = 275 pounds unless our data shows us otherwise. Calculate the p-value using the normal distribution for a mean and using the sample mean as input (see Appendix G for using the data as input): p-value = P( x¯ > 286.2) = 0.1323. Interpretation of the p-value: If H0 is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event. Figure 9.8 Compare α and the p-value: α = 0.025 p-value = 0.1323 Make a decision: Since α <p-value, do not reject H0. Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds. The p-value can easily be calculated. Put the data and frequencies into lists. Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Data and press ENTER. Arrow down and enter 275 for μ0, 55 for σ, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to μ: and arrow over to > μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.1331, a little different from the previous calculation - in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic (z-score) for the sample mean, the sample mean, and the sample standard deviation. μ > 275 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z = 1.112 (test statistic) and p = 0.1331 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. 484 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Example 9.16 Statistics students believe that the mean score on the first statistics test |
is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65; 65; 70; 67; 66; 63; 63; 68; 72; 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution. Solution 9.16 Set up the hypothesis test: A 5% level of significance means that α = 0.05. This is a test of a single population mean. H0: μ = 65 Ha: μ > 65 Since the instructor thinks the average score is higher, use a ">". The ">" means the test is right-tailed. Determine the distribution needed: Random variable: X ¯ = average score on the first statistics test. Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given. You are only given n = 10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's t. Use tdf. Therefore, the distribution for the test is t9 where n = 10 and df = 10 - 1 = 9. Calculate the p-value using the Student's t-distribution: p-value = P( x¯ > 67) = 0.0396 where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data. Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more. Figure 9.9 Compare α and the p-value: Since α = 0.05 and p-value = 0.0396. α > p-value. Make a decision: Since α > p-value, reject H0. This means you reject μ = 65. In other words, you believe the average test score is more than 65. Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks. The p-value can easily be calculated. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 485 Put |
the data into a list. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 65 for μ0, the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to > μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.0396) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standard deviation. μ > 65 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with t = 1.9781 (test statistic) and p = 0.0396 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. 9.16 It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors. Example 9.17 Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50%. Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance. Solution 9.17 Set up the hypothesis test: The 1% level of significance means that α = 0.01. This is a test of a single population proportion. H0: p = 0.50 Ha: p ≠ 0.50 The words "is the same or different from" tell you this is a two-tailed test. Calculate the distribution needed: Random variable: P′ = the percent of of first-time brides who are younger than their grooms |
. Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′, the estimated proportion. P′ ~ N⎛ ⎝p, p ⋅ q n ⎞ ⎠ Therefore, P′ ~ N⎛ ⎝0.5, 0.5 ⋅ 0.5 100 ⎞ ⎠ where p = 0.50, q = 1−p = 0.50, and n = 100 Calculate the p-value using the normal distribution for proportions: p-value = P (p′ < 0.47 or p′ > 0.53) = 0.5485 where x = 53, p′ = x n = 53 100 = 0.53. Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion p'is 0.53 or more OR 0.47 or less (see the graph in Figure 9.9). 486 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Figure 9.10 μ = p = 0.50 comes from H0, the null hypothesis. p′ = 0.53. Since the curve is symmetrical and the test is two-tailed, the p′ for the left tail is equal to 0.50 – 0.03 = 0.47 where μ = p = 0.50. (0.03 is the difference between 0.53 and 0.50.) Compare α and the p-value: Since α = 0.01 and p-value = 0.5485. α < p-value. Make a decision: Since α < p-value, you cannot reject H0. Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%. The p-value can easily be calculated. Press STAT and arrow over to TESTS. Press 5:1-PropZTest. Enter.5 for p0, 53 for x and 100 for n. Arrow down to Prop and arrow to not equals p0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator calculates the p-value (p = 0.5485) and the test statistic (z-score). |
Prop not equals.5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z = 0.6 (test statistic) and p = 0.5485 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.) 9.17 A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance. First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 487 Example 9.18 Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones. Solution 9.18 Set up the Hypothesis Test: H0: p = 0.30 Ha: p ≠ 0.30 Determine the distribution needed: The random variable is P′ = proportion of households that have three cell phones. The distribution for the hypothesis test is P'~ N⎛ (0.30) ⋅ (0.70) 150 |
⎞ ⎠ ⎝0.30, a. The value that helps determine the p-value is p′. Calculate p′. Solution 9.18 a. p′ = x n where x is the number of successes and n is the total number in the sample. x = 43, n = 150 p′ = 43 150 b. What is a success for this problem? Solution 9.18 b. A success is having three cell phones in a household. c. What is the level of significance? Solution 9.18 c. The level of significance is the preset α. Since α is not given, assume that α = 0.05. d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately. Calculate the p-value. Solution 9.18 d. p-value = 0.7216 e. Make a decision. _____________(Reject/Do not reject) H0 because____________. Solution 9.18 e. Assuming that α = 0.05, α < p-value. The decision is do not reject H0 because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%. 9.18 Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% 488 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE level of significance. State the null and alternative hypothesis, find the p-value, state your conclusion, and identify the Type I and Type II errors. The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter p. The distribution for the test is normal. The estimated proportion p′ is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived α = 0.01, for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it! Example 9.19 My dog has so many fleas, They do not come off with ease. As for shampoo, I have tried many types Even one called Bubble Hype, Which only killed 25% of the |
fleas, Unfortunately I was not pleased. I've used all kinds of soap, Until I had given up hope Until one day I saw An ad that put me in awe. A shampoo used for dogs Called GOOD ENOUGH to Clean a Hog Guaranteed to kill more fleas. I gave Fido a bath And after doing the math His number of fleas Started dropping by 3's! Before his shampoo I counted 42. At the end of his bath, I redid the math And the new shampoo had killed 17 fleas. So now I was pleased. Now it is time for you to have some fun With the level of significance being.01, You must help me figure out Use the new shampoo or go without? Solution 9.19 Set up the hypothesis test: H0: p ≤ 0.25 Ha: p > 0.25 Determine the distribution needed: In words, CLEARLY state what your random variable X ¯ or P′ represents. P′ = The proportion of fleas that are killed by the new shampoo State the distribution to use for the test. Normal: N⎛ (0.25)(1 − 0.25) 42 ⎝0.25, ⎞ ⎠ This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 489 Test Statistic: z = 2.3163 Calculate the p-value using the normal distribution for proportions: p-value = 0.0103 In one to two complete sentences, explain what the p-value means for this problem. If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 ⎛ ⎝ or more. ⎞ ⎠ 17 42 Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p-value. Figure 9.11 Compare α and the p-value: Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences. alpha decision reason for decision 0.01 Do not reject H0 α |
< p-value Table 9.3 Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%. Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. 490 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Figure 9.12 Confidence Interval: (0.26,0.55) We are 95% confident that the true population proportion p of fleas that are killed by the new shampoo is between 26% and 55%. NOTE This test result is not very definitive since the p-value is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return. Example 9.20 The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass. 1.11; 1.07; 1.11; 1.07; 1.12; 1.08;.98;.98 1.02;.95;.95 Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal. Solution 9.20 Let’s follow a four-step process to answer this statistical question. 1. State the Question: We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be a. H0: μ ≤ 1 b. Ha: μ > 1 2. Plan: We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal. 3. Do the calculations: We will input the sample data into the TI-83 as follows. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 491 Figure 9.13 Figure 9.14 Figure 9.15 Figure 9.16 492 CHAPTER 9 | HYP |
OTHESIS TESTING WITH ONE SAMPLE 4. State the Conclusions: Since the p-value* (p = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one. Example 9.21 In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error. Solution 9.21 We will follow the four-step process. 1. We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be a. H0: p ≤ 0.00034 b. Ha: p > 0.00034 If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancercausing environments, we want to minimize the chances of incorrectly identifying causes of cancer. 2. We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np = 420,019(0.00034) = 142.8, nq = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p = 0.00034. Thus we will be able to generalize our results to the population. 3. The associated TI results are Figure 9.17 Figure 9.18 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 493 4. Since the p-value = 0.0073 is greater than our alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users. Example 9.22 According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that |
, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01. Solution 9.22 We will follow the four-step plan. 1. We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average. 2. Since we are presented with proportions, we will use a one-proportion z-test. The hypotheses for the test will be a. H0: p = 0.00078 b. Ha: p ≠ 0.00078 3. The following screen shots display the summary statistics from the hypothesis test. Figure 9.19 Figure 9.20 4. Since the p-value, p = 0.00063, is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion. 494 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 9.1 Hypothesis Testing of a Single Mean and Single Proportion Class Time: Names: Student Learning Outcomes • The student will select the appropriate distributions to use in each case. • The student will conduct hypothesis tests and interpret the results. Television Survey In a recent survey, it was stated that Americans watch television on average four hours per day. Assume that σ = 2. Using your class as the sample, conduct a hypothesis test to determine if the average for students at your school is lower. 1. H0: _____________ 2. Ha: _____________ 3. In words, define the random variable. __________ = ______________________ 4. The distribution to use for the test is _______________________. 5. Determine the test statistic using your data. 6. Draw a graph and label it appropriately.Shade the actual level of significance. a. Graph: Figure 9.21 b. Determine the p-value. 7. Do you or do you not reject the null hypothesis? Why? 8. Write a clear conclusion using a |
complete sentence. Language Survey About 42.3% of Californians and 19.6% of all Americans over age five speak a language other than English at home. Using your class as the sample, conduct a hypothesis test to determine if the percent of the students at your school who speak a language other than English at home is different from 42.3%. 1. H0: ___________ 2. Ha: ___________ This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 495 3. In words, define the random variable. __________ = _______________ 4. The distribution to use for the test is ________________ 5. Determine the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 9.22 b. Determine the p-value. 7. Do you or do you not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Jeans Survey Suppose that young adults own an average of three pairs of jeans. Survey eight people from your class to determine if the average is higher than three. Assume the population is normal. 1. H0: _____________ 2. Ha: _____________ 3. In words, define the random variable. __________ = ______________________ 4. The distribution to use for the test is _______________________. 5. Determine the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 9.23 b. Determine the p-value. 7. Do you or do you not reject the null hypothesis? Why? 496 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 8. Write a clear conclusion using a complete sentence. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 497 KEY TERMS p-value the probability that an event will happen purely by chance assuming the null hypothesis is true. The smaller the p-value, the stronger the evidence is against the null |
hypothesis. Binomial Distribution a discrete random variable (RV) that arises from Bernoulli trials. There are a fixed number, n, of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV Χ is defined as the number of successes in n trials. The notation is: X ~ B(n, p) μ = np and the standard deviation is σ = npq. The probability of exactly x successes in n trials is P(X = x) = ⎠p x qn − x. ⎞ n ⎛ x ⎝ Central Limit Theorem sampling with size n and we are interested in two new RVs - the sample mean, X Given a random variable (RV) with known mean µ and known standard deviation σ. We are ¯, and the sample sum, ΣX. If the ¯ size n of the sample is sufficiently large, then X ⎞ ⎠ and ΣX ~ N(nµ, nσ). If the size n of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample means will equal the population mean and the mean of the sample sums will equal n times the population mean. The standard deviation of the distribution of the sample means, σ n, is called the standard error of the mean. ~ N⎛ σ n ⎝µ, Confidence Interval (CI) an interval estimate for an unknown population parameter. This depends on: • The desired confidence level. • Information that is known about the distribution (for example, known standard deviation). • The sample and its size. Hypothesis Testing Based on sample evidence, a procedure for determining whether the hypothesis stated is a reasonable statement and should not be rejected, or is unreasonable and should be rejected. Hypothesis a statement about the value of a population parameter, in case of two hypotheses, the statement assumed to be true is called the null hypothesis (notation H0) and the contradictory statement is called the alternative hypothesis (notation Ha). Level of Significance of the Test probability of a Type I error (reject the null hypothesis when it is true). Notation: α. In |
hypothesis testing, the Level of Significance is called the preconceived α or the preset α. Normal Distribution a continuous random variable (RV) with pdf f (x) = −(x − µ)2 2σ 2 1 σ 2π e, where μ is the mean of the distribution, and σ is the standard deviation, notation: X ~ N(μ, σ). If μ = 0 and σ = 1, the RV is called the standard normal distribution. p-value the probability that an event will happen purely by chance assuming the null hypothesis is true. The smaller the p-value, the stronger the evidence is against the null hypothesis. Standard Deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation. Student's t-Distribution investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student. The major characteristics of the random variable (RV) are: • It is continuous and assumes any real values. • The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution. • It approaches the standard normal distribution as n gets larger. • There is a "family" of t distributions: every representative of the family is completely defined by the number of degrees of freedom which is one less than the number of data items. Type 1 Error The decision is to reject the null hypothesis when, in fact, the null hypothesis is true. 498 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Type 2 Error The decision is not to reject the null hypothesis when, in fact, the null hypothesis is false. CHAPTER REVIEW 9.1 Null and Alternative Hypotheses In a hypothesis test, sample data is evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we: 1. Evaluate the null hypothesis, typically denoted with H0. The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality (=, ≤ or ≥) 2. Always write the alternative hypothesis, typically denoted with Ha or H1, using less than, greater than, or not equals symbols, i.e., (≠, > |
, or <). 3. If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. 4. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties. 9.2 Outcomes and the Type I and Type II Errors In every hypothesis test, the outcomes are dependent on a correct interpretation of the data. Incorrect calculations or misunderstood summary statistics can yield errors that affect the results. A Type I error occurs when a true null hypothesis is rejected. A Type II error occurs when a false null hypothesis is not rejected. The probabilities of these errors are denoted by the Greek letters α and β, for a Type I and a Type II error respectively. The power of the test, 1 – β, quantifies the likelihood that a test will yield the correct result of a true alternative hypothesis being accepted. A high power is desirable. 9.3 Distribution Needed for Hypothesis Testing In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied. When testing for a single population mean: 1. A Student's t-test should be used if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with an unknown standard deviation. 2. The normal test will work if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with a known standard deviation. When testing a single population proportion use a normal test for a single population proportion if the data comes from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of success and the mean number of failures satisfy the conditions: np > 5 and nq > n where n is the sample size, p is the probability of a success, and q is the probability of a failure. 9.4 Rare Events, the Sample, Decision and Conclusion When the probability of an event occurring is low, and it happens, it is called a rare event. Rare events are important to consider in hypothesis testing because they can inform your willingness not to reject or to reject a null hypothesis. To test a null hypothesis, find the p-value for the sample data and graph the results. When deciding whether or not to reject the null the hypothesis, keep these two parameters in mind: 1. α > |
p-value, reject the null hypothesis 2. α ≤ p-value, do not reject the null hypothesis 9.5 Additional Information and Full Hypothesis Test Examples The hypothesis test itself has an established process. This can be summarized as follows: 1. Determine H0 and Ha. Remember, they are contradictory. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 499 2. Determine the random variable. 3. Determine the distribution for the test. 4. Draw a graph, calculate the test statistic, and use the test statistic to calculate the p-value. (A z-score and a t-score are examples of test statistics.) 5. Compare the preconceived α with the p-value, make a decision (reject or do not reject H0), and write a clear conclusion using English sentences. Notice that in performing the hypothesis test, you use α and not β. β is needed to help determine the sample size of the data that is used in calculating the p-value. Remember that the quantity 1 – β is called the Power of the Test. A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be. less than (<) greater than (>) • Single population mean, known population variance (or standard deviation): Normal test. Types of Hypothesis Tests β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false. 9.3 Distribution Needed for Hypothesis Testing If there is no given preconceived α, then use α = 0.05. • Single population mean, unknown population variance (or standard deviation): Student's t-test. • Single population proportion: Normal test. • For a single population mean, we may use a normal distribution with the following mean and standard deviation. Means: µ = µ x¯ and σ x¯ = σ x n • A single population proportion, we may use a normal distribution with the following mean and standard deviation. Proportions: µ = p and σ = pq n. FORMULA REVIEW 9.1 Null and Alternative Hypotheses |
H0 and Ha are contradictory. greater than or equal to (≥) less than or equal to (≤) If Ho has: then Ha has: equal (=) not equal (≠) or greater than (>) or less than (<) Table 9.4 If α ≤ p-value, then do not reject H0. If α > p-value, then reject H0. α is preconceived. Its value is set before the hypothesis test starts. The p-value is calculated from the data. 9.2 Outcomes and the Type I and Type II Errors α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true. PRACTICE 9.1 Null and Alternative Hypotheses 1. You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. What is the random variable? Describe in words. 2. You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. State the null and alternative hypotheses. 3. The American family has an average of two children. What is the random variable? Describe in words. 4. The mean entry level salary of an employee at a company is $58,000. You believe it is higher for IT professionals in the company. State the null and alternative hypotheses. 5. A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the proportion is actually less. What is the random variable? Describe in words. 6. A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the claim is correct. State the null and alternative hypotheses. 500 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 7. In a population of fish, approximately 42% are female. A test is conducted to see if, in fact, the proportion is less. State the null and alternative hypotheses. 8. Suppose that a recent article stated that the mean time spent in jail by a first–time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean |
length of time in jail from the survey was 3 years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. If you were conducting a hypothesis test to determine if the mean length of jail time has increased, what would the null and alternative hypotheses be? The distribution of the population is normal. a. H0: ________ b. Ha: ________ 9. A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis test to determine if the population mean time on death row could likely be 15 years, what would the null and alternative hypotheses be? a. H0: __________ b. Ha: __________ 10. The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. If you were conducting a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population, what would the null and alternative hypotheses be? a. H0: ________ b. Ha: ________ 9.2 Outcomes and the Type I and Type II Errors 11. The mean price of mid-sized cars in a region is $32,000. A test is conducted to see if the claim is true. State the Type I and Type II errors in complete sentences. 12. A sleeping bag is tested to withstand temperatures of –15 °F. You think the bag cannot stand temperatures that low. State the Type I and Type II errors in complete sentences. 13. For Exercise 9.12, what are α and β in words? 14. In words, describe 1 – β For Exercise 9.12. 15. A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H0, is: the surgical procedure will go well. State the Type I and Type II errors in complete sentences. 16. A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H0, is: the surgical procedure will go well. Which is the error with the greater consequence? 17. The power of a test is 0.981 |
. What is the probability of a Type II error? 18. A group of divers is exploring an old sunken ship. Suppose the null hypothesis, H0, is: the sunken ship does not contain buried treasure. State the Type I and Type II errors in complete sentences. 19. A microbiologist is testing a water sample for E-coli. Suppose the null hypothesis, H0, is: the sample does not contain E-coli. The probability that the sample does not contain E-coli, but the microbiologist thinks it does is 0.012. The probability that the sample does contain E-coli, but the microbiologist thinks it does not is 0.002. What is the power of this test? 20. A microbiologist is testing a water sample for E-coli. Suppose the null hypothesis, H0, is: the sample contains E-coli. Which is the error with the greater consequence? 9.3 Distribution Needed for Hypothesis Testing 21. Which two distributions can you use for hypothesis testing for this chapter? 22. Which distribution do you use when you are testing a population mean and the standard deviation is known? Assume sample size is large. 23. Which distribution do you use when the standard deviation is not known and you are testing one population mean? Assume sample size is large. 24. A population mean is 13. The sample mean is 12.8, and the sample standard deviation is two. The sample size is 20. What distribution should you use to perform a hypothesis test? Assume the underlying population is normal. 25. A population has a mean is 25 and a standard deviation of five. The sample mean is 24, and the sample size is 108. What distribution should you use to perform a hypothesis test? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 501 26. It is thought that 42% of respondents in a taste test would prefer Brand A. In a particular test of 100 people, 39% preferred Brand A. What distribution should you use to perform a hypothesis test? 27. You are performing a hypothesis test of a single population mean using a Student’s t-distribution. What must you assume about the distribution of the data? 28. You are performing a hypothesis test |
of a single population mean using a Student’s t-distribution. The data are not from a simple random sample. Can you accurately perform the hypothesis test? 29. You are performing a hypothesis test of a single population proportion. What must be true about the quantities of np and nq? 30. You are performing a hypothesis test of a single population proportion. You find out that np is less than five. What must you do to be able to perform a valid hypothesis test? 31. You are performing a hypothesis test of a single population proportion. The data come from which distribution? 9.4 Rare Events, the Sample, Decision and Conclusion 32. When do you reject the null hypothesis? 33. The probability of winning the grand prize at a particular carnival game is 0.005. Is the outcome of winning very likely or very unlikely? 34. The probability of winning the grand prize at a particular carnival game is 0.005. Michele wins the grand prize. Is this considered a rare or common event? Why? 35. It is believed that the mean height of high school students who play basketball on the school team is 73 inches with a standard deviation of 1.8 inches. A random sample of 40 players is chosen. The sample mean was 71 inches, and the sample standard deviation was 1.5 years. Do the data support the claim that the mean height is less than 73 inches? The p-value is almost zero. State the null and alternative hypotheses and interpret the p-value. 36. The mean age of graduate students at a University is at most 31 y ears with a standard deviation of two years. A random sample of 15 graduate students is taken. The sample mean is 32 years and the sample standard deviation is three years. Are the data significant at the 1% level? The p-value is 0.0264. State the null and alternative hypotheses and interpret the p-value. 37. Does the shaded region represent a low or a high p-value compared to a level of significance of 1%? Figure 9.24 38. What should you do when α > p-value? 39. What should you do if α = p-value? 40. If you do not reject the null hypothesis, then it must be true. Is this statement correct? State why or why not in complete sentences. Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2 |
.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume the distribution of the jail times is approximately normal. 41. Is this a test of means or proportions? 42. What symbol represents the random variable for this test? 43. In words, define the random variable for this test. 502 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 44. Is the population standard deviation known and, if so, what is it? 45. Calculate the following: x¯ _______ a. b. σ _______ c. sx _______ d. n _______ 46. Since both σ and s x are given, which should be used? In one to two complete sentences, explain why. 47. State the distribution to use for the hypothesis test. 48. A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.4 years with a standard deviation of 6.3 years. Conduct a hypothesis test to determine if the population mean time on death row could likely be 15 years. a. Is this a test of one mean or proportion? b. State the null and alternative hypotheses. H0: ____________________ Ha : ____________________ Is this a right-tailed, left-tailed, or two-tailed test? c. d. What symbol represents the random variable for this test? e. f. g. Calculate the following: In words, define the random variable for this test. Is the population standard deviation known and, if so, what is it? x¯ = _____________ i. ii. s = ____________ iii. n = ____________ h. Which test should be used? i. State the distribution to use for the hypothesis test. j. Find the p-value. k. At a pre-conceived α = 0.05, what is your: i. Decision: ii. Reason for the decision: iii. Conclusion (write out in a complete sentence): 9.5 Additional Information and Full Hypothesis Test Examples 49. Assume H0: μ = |
9 and Ha: μ < 9. Is this a left-tailed, right-tailed, or two-tailed test? 50. Assume H0: μ ≤ 6 and Ha: μ > 6. Is this a left-tailed, right-tailed, or two-tailed test? 51. Assume H0: p = 0.25 and Ha: p ≠ 0.25. Is this a left-tailed, right-tailed, or two-tailed test? 52. Draw the general graph of a left-tailed test. 53. Draw the graph of a two-tailed test. 54. A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use? 55. Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use? 56. A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use? 57. You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use? 58. If the alternative hypothesis has a not equals ( ≠ ) symbol, you know to use which type of test? 59. Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test? 60. Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test? 61. Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 503 HOMEWORK 9.1 Null and Alternative Hypotheses 62. Some of the following statements refer to the null hypothesis, some to the alternate hypothesis. State the null hypothesis, H0, and the alternative hypothesis. Ha, in terms of the appropriate |
parameter (μ or p). a. The mean number of years Americans work before retiring is 34. b. At most 60% of Americans vote in presidential elections. c. The mean starting salary for San Jose State University graduates is at least $100,000 per year. d. Twenty-nine percent of high school seniors get drunk each month. e. Fewer than 5% of adults ride the bus to work in Los Angeles. f. The mean number of cars a person owns in her lifetime is not more than ten. g. About half of Americans prefer to live away from cities, given the choice. h. Europeans have a mean paid vacation each year of six weeks. i. The chance of developing breast cancer is under 11% for women. j. Private universities' mean tuition cost is more than $20,000 per year. 63. Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is: a. p < 0.30 b. p ≤ 0.30 c. p ≥ 0.30 d. p > 0.30 64. A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is: a. p = 0.20 b. p > 0.20 c. p < 0.20 d. p ≤ 0.20 65. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are: a. Ho: x¯ = 4.5, Ha : x¯ > 4.5 b. Ho: μ ≥ 4.5, Ha: μ < 4.5 c. Ho: μ = 4.75, Ha: μ > 4.75 d. Ho |
: μ = 4.5, Ha: μ > 4.5 9.2 Outcomes and the Type I and Type II Errors 66. State the Type I and Type II errors in complete sentences given the following statements. a. The mean number of years Americans work before retiring is 34. b. At most 60% of Americans vote in presidential elections. c. The mean starting salary for San Jose State University graduates is at least $100,000 per year. d. Twenty-nine percent of high school seniors get drunk each month. e. Fewer than 5% of adults ride the bus to work in Los Angeles. f. The mean number of cars a person owns in his or her lifetime is not more than ten. g. About half of Americans prefer to live away from cities, given the choice. h. Europeans have a mean paid vacation each year of six weeks. i. The chance of developing breast cancer is under 11% for women. j. Private universities mean tuition cost is more than $20,000 per year. 67. For statements a-j in Exercise 9.109, answer the following in complete sentences. a. State a consequence of committing a Type I error. b. State a consequence of committing a Type II error. 504 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 68. When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error? a. To conclude the drug is safe when in, fact, it is unsafe. b. Not to conclude the drug is safe when, in fact, it is safe. c. To conclude the drug is safe when, in fact, it is safe. d. Not to conclude the drug is unsafe when, in fact, it is unsafe. 69. A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________. a. at least 20%, when in fact, it is less than 20%. b. 20%, when in fact, it is 20%. c. d. less than 20%, when in fact, it is |
at least 20%. less than 20%, when in fact, it is less than 20%. 70. It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours a. b. c. d. is more than seven hours. is at most seven hours. is at least seven hours. is less than seven hours. 71. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is: a. b. c. d. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same to conclude that the mean hours per week currently is 4.5, when in fact, it is higher to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher 9.3 Distribution Needed for Hypothesis Testing 72. It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is X ¯ ~ ________________ ) a. N(7.24, 1.93 22 b. N(7.24, 1.93) c. d. t22 t21 9.4 Rare Events, the |
Sample, Decision and Conclusion 73. The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population. a. Is this a test of one mean or proportion? b. State the null and alternative hypotheses. H0: ____________________ Ha: ____________________ Is this a right-tailed, left-tailed, or two-tailed test? c. d. What symbol represents the random variable for this test? e. f. Calculate the following: In words, define the random variable for this test. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 505 i. x = ________________ ii. n = ________________ iii. p′ = _____________ g. Calculate σx = __________. Show the formula set-up. h. State the distribution to use for the hypothesis test. i. Find the p-value. j. At a pre-conceived α = 0.05, what is your: i. Decision: ii. Reason for the decision: iii. Conclusion (write out in a complete sentence): 9.5 Additional Information and Full Hypothesis Test Examples For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the.doc or the.pdf files. NOTE If you are using a Student's-t distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.) 74. A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 |
tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly inconsistent with the claim? 75. From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level? 76. The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level? 77. An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level? 78. The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let x = the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten? 79. In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level? 80. Your statistics instructor claims that 60 |
percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think? 81. A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief. 82. Refer to Exercise 9.119. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four. 506 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 83. According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7? 84. A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results. 85. The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups |
for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours? Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55. 86. Use the “Lap time” data for Lap 4 (see Appendix C) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given. 87. Use the “Initial Public Offering” data (see Appendix C) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices. NOTE The following questions were written by past students. They are excellent problems! 88. "Asian Family Reunion," by Chau Nguyen Every two years it comes around. We all get together from different towns. In my honest opinion, It's not a typical family reunion. Not forty, or fifty, or sixty, But how about seventy companions! The kids would play, scream, and shout One minute they're happy, another they'll pout. The teenagers would look, stare, and compare From how they look to what they wear. The men would chat about their business That they make more, but never less. Money is always their subject And there's always talk of more new projects. The women get tired from all of the chats They head to the kitchen to set out the mats. Some would sit and some would stand Eating and talking with plates in their hands. Then come the games and the songs And suddenly, everyone gets along! With all that laughter, it's sad to say That it always ends in the same old way. They hug and kiss and say "good-bye" And then they all begin to cry! I say that 60 percent shed their tears This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 507 But my mom counted 35 people this year. She said that boys and men will always have their pride, So we won't ever see them cry. I myself don't think she's correct, So could you please try this problem to see if you object? 89. "The Problem with Angels," by Cy |
hundred one pups: Dalmatians that fetched the most shekels. They asked Mr. Spreckles to name An average spot count he'd claim To bring in big bucks. Said Spreckles, “Well, shucks, It's for one hundred one that I aim.” Said an amateur statistician Who wanted to help with this mission. “Twenty-one for the sample Standard deviation's ample: They examined one hundred and one Dalmatians that fetched a good sum. They counted each spot, Mark, freckle and dot And tallied up every one. Instead of one hundred one spots They averaged ninety six dots Can they muzzle Spreckles’ Obsession with freckles Based on all the dog data they've got? 92. "Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 509 As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value. One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02!!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook: Price per box of Mac and Cheese: • 5 stores @ $2.02 • 15 stores @ $0.25 • 3 stores @ $1.29 • 6 stores @ $0.35 • 4 stores @ $2.27 • 7 stores @ $1.50 • 5 stores @ $1. |
89 • 8 stores @ 0.75. I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!) 93. "William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi THE CHARACTERS (in order of appearance): • HAMLET, Prince of Denmark and student of Statistics • POLONIUS, Hamlet’s tutor • HOROTIO, friend to Hamlet and fellow student Scene: The great library of the castle, in which Hamlet does his lessons Act I (The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.) POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination! HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable. POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true. HORATIO (to Hamlet): What should we do, my Lord? HAMLET: Go to thy purpose, Horatio. HORATIO: To what |
end, my Lord? HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no. (Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.) Act II (The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.) POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations? HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes. POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.) 510 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.) (Curtain falls) 94. "Untitled," by Stephen Chen I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%. So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out |
how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right? 95. "Japanese Girls’ Names" by Kumi Furuichi It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on. However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children. I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation. Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko |
, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko. 96. "Phillip’s Wish," by Suzanne Osorio My nephew likes to play Chasing the girls makes his day. He asked his mother If it is okay To get his ear pierced. She said, “No way!” To poke a hole through your ear, Is not what I want for you, dear. He argued his point quite well, Says even my macho pal, Mel, Has gotten this done. It’s all just for fun. C’mon please, mom, please, what the hell. Again Phillip complained to his mother, Saying half his friends (including their brothers) Are piercing their ears This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 511 And they have no fears He wants to be like the others. She said, “I think it’s much less. We must do a hypothesis test. And if you are right, I won’t put up a fight. But, if not, then my case will rest.” We proceeded to call fifty guys To see whose prediction would fly. Nineteen of the fifty Said piercing was nifty And earrings they’d occasionally buy. Then there’s the other thirty-one, Who said they’d never have this done. So now this poem’s finished. Will his hopes be diminished, Or will my nephew have his fun? 97. "The Craven," by Mark Salangsang Once upon a morning dreary In stats class I was weak and weary. Pondering over last night’s homework Whose answers were now on the board This I did and nothing more. While I nodded nearly napping Suddenly, there came a tapping. As someone gently rapping, Rapping my head as I snore. Quoth the |
teacher, “Sleep no more.” “In every class you fall asleep,” The teacher said, his voice was deep. “So a tally I’ve begun to keep Of every class you nap and snore. The percentage being forty-four.” “My dear teacher I must confess, While sleeping is what I do best. The percentage, I think, must be less, A percentage less than forty-four.” This I said and nothing more. “We’ll see,” he said and walked away, And fifty classes from that day He counted till the month of May The classes in which I napped and snored. The number he found was twenty-four. At a significance level of 0.05, 512 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Please tell me am I still alive? Or did my grade just take a dive Plunging down beneath the floor? Upon thee I hereby implore. 98. Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%. 99. Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data. 100. According to an article in Bloomberg Businessweek, New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased. 101. The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly |
surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test. 102. Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test. 103. La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old. 104. Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? After conducting the test, your decision and conclusion are a. Reject H0: There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin. b. Do not reject H0: There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin. c. Do not reject H0: There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin. d. Reject H0: There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin. 105. A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. At a 1% level of significance, an appropriate conclusion is: a. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%. b. There is sufficient evidence to conclude that the percent of EV |
C students who attended the midnight showing of Harry Potter is more than 20%. c. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%. d. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%. 106. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. At a significance level of a = 0.05, what is the correct conclusion? a. There is enough evidence to conclude that the mean number of hours is more than 4.75 b. There is enough evidence to conclude that the mean number of hours is more than 4.5 c. There is not enough evidence to conclude that the mean number of hours is more than 4.5 d. There is not enough evidence to conclude that the mean number of hours is more than 4.75 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 513 Instructions: For the following ten exercises, Hypothesis testing: For the following ten exercises, answer each question. a. State the null and alternate hypothesis. b. State the p-value. c. State alpha. d. What is your decision? e. Write a conclusion. f. Answer any other questions asked in the problem. 107. According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions. 108. A recent survey in the N.Y. Times Almanac indicated that 48.8 |
% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate? 109. Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using α = 0.05, is the AAA proportion accurate? 110. The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the α = 0.05 level in Kentucky? Are the results applicable across the country? Why? 111. For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use α = 0.01 level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library? 112. The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the α = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if α = 0.01? Assume the amount of summer rainfall follows a normal distribution. 113. A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the α = 0.10 level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities? 114. A report by the Gallup Poll found that a woman visits her doctor, |
on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals 3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1 At the α = 0.05 level can it be concluded that the sample mean is higher than 5.8 visits per year? 115. According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes: 5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2 At α = 0.05 level, is the class’ mean family size greater than the national average? Does the Almanac result remain valid? Why? 116. The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct? REFERENCES 9.1 Null and Alternative Hypotheses Data from the National Institute of Mental Health. Available online at http://www.nimh.nih.gov/publicat/depression.cfm. 514 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 9.5 Additional Information and Full Hypothesis Test Examples Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC. Data from Bloomberg Businessweek. Available online at http://www.businessweek.com/news/2011- 09-15/nyc-smokingrate-falls-to-record-low-of-14-bloomberg-says.html. Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013). Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013). Data from Growing by Degrees by Allen and Seaman. Data from La Leche League International. Available online at http://www.l |
alecheleague.org/Law/BAFeb01.html. Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013). Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013). Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm. Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013) Data from the U.S. Census Bureau, available online at http://quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013). Data from the United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/. Data detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1. International. Toastmasters Available from online at http://toastmasters.org/artisan/ Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013). Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/ 3868.htm (accessed June 27, 2013). “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online http://research.fhda.edu/factbook/DAdemofs/Fact_sheet_da_2006w.pdf. at Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013). Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at http://www.rainn.org/get-information/statistics |
/frequency-of-sexual-assault (accessed June 27, 2013). SOLUTIONS 1 The random variable is the mean Internet speed in Megabits per second. 3 The random variable is the mean number of children an American family has. 5 The random variable is the proportion of people picked at random in Times Square visiting the city. 7 a. H0: p = 0.42 b. Ha: p < 0.42 9 a. H0: μ = 15 b. Ha: μ ≠ 15 11 Type I: The mean price of mid-sized cars is $32,000, but we conclude that it is not $32,000. Type II: The mean price of mid-sized cars is not $32,000, but we conclude that it is $32,000. 13 α = the probability that you think the bag cannot withstand -15 degrees F, when in fact it can β = the probability that you think the bag can withstand -15 degrees F, when in fact it cannot This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 515 15 Type I: The procedure will go well, but the doctors think it will not. Type II: The procedure will not go well, but the doctors think it will. 17 0.019 19 0.998 21 A normal distribution or a Student’s t-distribution 23 Use a Student’s t-distribution 25 a normal distribution for a single population mean 27 It must be approximately normally distributed. 29 They must both be greater than five. 31 binomial distribution 33 The outcome of winning is very unlikely. 35 H0: μ > = 73 Ha: μ < 73 The p-value is almost zero, which means there is sufficient data to conclude that the mean height of high school students who play basketball on the school team is less than 73 inches at the 5% level. The data do support the claim. 37 The shaded region shows a low p-value. 39 Do not reject H0. 41 means 43 the mean time spent in jail for 26 first time convicted burglars 45 a. 3 b. 1.5 c. 1.8 d. 26 ¯ 47 X ~ N⎛ ⎝2.5, 1.5 26 ⎞ ⎠ |
49 This is a left-tailed test. 51 This is a two-tailed test. 53 516 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Figure 9.25 55 a right-tailed test 57 a left-tailed test 59 This is a left-tailed test. 61 This is a two-tailed test. 62 a. H0: μ = 34; Ha: μ ≠ 34 b. H0: p ≤ 0.60; Ha: p > 0.60 c. H0: μ ≥ 100,000; Ha: μ < 100,000 d. H0: p = 0.29; Ha: p ≠ 0.29 e. H0: p = 0.05; Ha: p < 0.05 f. H0: μ ≤ 10; Ha: μ > 10 g. H0: p = 0.50; Ha: p ≠ 0.50 h. H0: μ = 6; Ha: μ ≠ 6 i. H0: p ≥ 0.11; Ha: p < 0.11 j. H0: μ ≤ 20,000; Ha: μ > 20,000 64 c 66 a. Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years. b. Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do. c. Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000. d. Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%. e. Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. |
Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do. f. Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 517 g. Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half. h. Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not. i. Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%. j. Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000. 68 b 70 d 72 d 74 a. H0: μ ≥ 50,000 b. Ha: μ < 50,000 c. Let X ¯ = the average lifespan of a brand of tires. d. normal distribution e. z = -2.315 f. p-value = 0.0103 g. Check student’s solution. h. i. alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than |
0.05. iv. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles. i. (43,537, 49,463) 76 a. H0: μ = $1.00 b. Ha: μ ≠ $1.00 c. Let X ¯ = the average cost of a daily newspaper. d. normal distribution e. z = –0.866 f. p-value = 0.3865 g. Check student’s solution. h. i. Alpha: 0.01 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.01. iv. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1. i. ($0.84, $1.06) 78 518 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE a. H0: μ = 10 b. Ha: μ ≠ 10 ¯ c. Let X the mean number of sick days an employee takes per year. d. Student’s t-distribution e. t = –1.12 f. p-value = 0.300 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten. i. (4.9443, 11.806) 80 a. H0: p ≥ 0.6 b. Ha: p < 0.6 c. Let P′ = the proportion of students who feel more enriched as a result of taking Elementary Statistics. d. normal for a single proportion e. 1.12 f. p-value = 0.1308 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched. i. Confidence Interval: (0.409, 0.654) The “plus-4s” confidence interval |
is (0.411, 0.648) 82 a. H0: μ = 4 b. Ha: μ ≠ 4 ¯ c. Let X the average I.Q. of a set of brown trout. d. e. two-tailed Student's t-test t = 1.95 f. p-value = 0.076 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05 iv. Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four. i. (3.8865,5.9468) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 519 84 a. H0: p ≥ 0.13 b. Ha: p < 0.13 c. Let P′ = the proportion of Americans who have seen or sensed angels d. normal for a single proportion e. –2.688 f. p-value = 0.0036 g. Check student’s solution. h. i. alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%. i. (0, 0.0623). The“plus-4s” confidence interval is (0.0022, 0.0978) 86 a. H0: μ ≥ 129 b. Ha: μ < 129 c. Let X ¯ = the average time in seconds that Terri finishes Lap 4. d. Student's t-distribution e. t = 1.209 f. 0.8792 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds. i. (128.63, 130.37) 88 a. |
H0: p = 0.60 b. Ha: p < 0.60 c. Let P′ = the proportion of family members who shed tears at a reunion. d. normal for a single proportion e. –1.71 f. 0.0438 g. Check student’s solution. h. i. alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the p-value and alpha are quite close, so other tests should be done. 520 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE i. We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. (0.3829, 0.6171). The“plus-4s” confidence interval (see chapter 8) is (0.3861, 0.6139) Note that here the “large-sample” 1 – PropZTest provides the approximate p-value of 0.0438. Whenever a p-value based on a normal approximation is close to the level of significance, the exact p-value based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course. 90 a. H0: μ ≥ 22 b. Ha: μ < 22 c. Let X ¯ = the mean number of bubbles per blow. d. Student's t-distribution e. –2.667 f. p-value = 0.00486 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22. i. (18.501, 21.499) 92 a. H0: μ ≤ 1 b. Ha: μ > 1 c. Let X ¯ = the mean cost in dollars of macaroni and cheese in a certain town. d. Student's t-distribution e. t = 0.340 f. p-value = 0.36756 g. Check student’s |
solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05 iv. Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1. i. (0.8291, 1.241) 94 a. H0: p = 0.01 b. Ha: p > 0.01 c. Let P′ = the proportion of errors generated d. Normal for a single proportion e. 2.13 f. 0.0165 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 521 iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01. i. Confidence interval: (0, 0.094). The“plus-4s” confidence interval is (0.004, 0.144). 96 a. H0: p = 0.50 b. Ha: p < 0.50 c. Let P′ = the proportion of friends that has a pierced ear. d. normal for a single proportion e. –1.70 f. p-value = 0.0448 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis iii. Reason for decision: The p-value is less than 0.05. (However, they are very close.) iv. Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears. i. Confidence Interval: (0.245, 0.515): The “plus-4s” confidence interval is (0.259, 0.519). 98 a. H0: p = 0.40 b. Ha: p < 0.40 c. Let P′ = |
the proportion of schoolmates who fear public speaking. d. normal for a single proportion e. –1.01 f. p-value = 0.1563 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking. i. Confidence Interval: (0.3241, 0.4240): The “plus-4s” confidence interval is (0.3257, 0.4250). 100 a. H0: p = 0.14 b. Ha: p < 0.14 c. Let P′ = the proportion of NYC residents that smoke. d. normal for a single proportion e. –0.2756 f. p-value = 0.3914 g. Check student’s solution. h. i. alpha: 0.05 ii. Decision: Do not reject the null hypothesis. 522 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE iii. Reason for decision: The p-value is greater than 0.05. iv. At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14. i. Confidence Interval: (0.0502, 0.2070): The “plus-4s” confidence interval (see chapter 8) is (0.0676, 0.2297). 102 a. H0: μ = 69,110 b. Ha: μ > 69,110 c. Let X ¯ = the mean salary in dollars for California registered nurses. d. Student's t-distribution e. t = 1.719 f. p-value: 0.0466 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110. i. ($68,757, $73,485) 104 c 106 c 108 a. H0: p = 0.488 Ha: p |
≠ 0.488 b. p-value = 0.0114 c. alpha = 0.05 d. Reject the null hypothesis. e. At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks. f. The survey does not appear to be accurate. 110 a. H0: p = 0.517 Ha: p ≠ 0.517 b. p-value = 0.9203. c. alpha = 0.05. d. Do not reject the null hypothesis. e. At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517. f. However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation. 112 a. H0: µ ≥ 11.52 Ha: µ < 11.52 b. p-value = 0.000002 which is almost 0. c. alpha = 0.05. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE 523 d. Reject the null hypothesis. e. At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average. f. We would make the same conclusion if alpha was 1% because the p-value is almost 0. 114 a. H0: µ ≤ 5.8 Ha: µ > 5.8 b. p-value = 0.9987 c. alpha = 0.05 d. Do not reject the null hypothesis. e. At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year. 116 a. H0: µ ≥ 150 Ha: µ < 150 b. p-value = 0.0622 c. alpha = 0.01 d. Do not |
reject the null hypothesis. e. At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average. f. The student academic group’s claim appears to be correct. 524 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 525 10 | HYPOTHESIS TESTING WITH TWO SAMPLES Figure 10.1 If you want to test a claim that involves two groups (the types of breakfasts eaten east and west of the Mississippi River) you can use a slightly different technique when conducting a hypothesis test. (credit: Chloe Lim) Introduction By the end of this chapter, the student should be able to: Chapter Objectives • Classify hypothesis tests by type. • Conduct and interpret hypothesis tests for two population means, population standard deviations known. • Conduct and interpret hypothesis tests for two population means, population standard deviations unknown. • Conduct and interpret hypothesis tests for two population proportions. • Conduct and interpret hypothesis tests for matched or paired samples. Studies often compare two groups. For example, researchers are interested in the effect aspirin has in preventing heart attacks. Over the last few years, newspapers and magazines have reported various aspirin studies involving two groups. 526 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES Typically, one group is given aspirin and the other group is given a placebo. Then, the heart attack rate is studied over several years. There are other situations that deal with the comparison of two groups. For example, studies compare various diet and exercise programs. Politicians compare the proportion of individuals from different income brackets who might vote for them. Students are interested in whether SAT or GRE preparatory courses really help raise their scores. You have learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two means or two proportions to each other. The general procedure is still the same, just expanded. To compare two means or two proportions, you work with two groups. The groups are classified either as independent or matched pairs. Independent groups consist of two samples that are independent, that is, sample values selected from one population are not related in any way |
to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions. NOTE This chapter relies on either a calculator or a computer to calculate the degrees of freedom, the test statistics, and p-values. TI-83+ and TI-84 instructions are included as well as the test statistic formulas. When using a TI-83+ or TI-84 calculator, we do not need to separate two population means, independent groups, or population variances unknown into large and small sample sizes. However, most statistical computer software has the ability to differentiate these tests. This chapter deals with the following hypothesis tests: Independent groups (samples are independent) • Test of two population means. • Test of two population proportions. Matched or paired samples (samples are dependent) • Test of the two population proportions by testing one population mean of differences. 10.1 | Two Population Means with Unknown Standard Deviations 1. The two independent samples are simple random samples from two distinct populations. 2. For the two distinct populations: if the sample sizes are small, the distributions are important (should be normal) if the sample sizes are large, the distributions are not important (need not be normal) ◦ ◦ NOTE The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch. The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual ¯ samples. In order to account for the variation, we take the difference of the sample means, X ¯ – X 2 1, and divide by the standard error in order to standardize the difference. The result is a t-score test statistic. Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, ¯ of the difference in sample means, X ¯ – X. 2 1 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTH |
ESIS TESTING WITH TWO SAMPLES 527 The standard error is: (s1 )2 n1 (s2 )2 n2 + The test statistic (t-score) is calculated as follows: ( x¯ 1 – x¯ 2 ) – (µ 1 – µ 2 ) (s2 )2 n2 + (s1 )2 n1 where: • s1 and s2, the sample standard deviations, are estimates of σ1 and σ2, respectively. • σ1 and σ1 are the unknown population standard deviations. • x¯ 1 and x¯ 2 are the sample means. μ1 and μ2 are the population means. The number of degrees of freedom (df) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df are not always a whole number. The test statistic calculated previously is approximated by the Student's t-distribution with df as follows: Degrees of freedom 2 + (s2)2 n2 ⎞ ⎟ ⎠ (s1)2 n1 ⎛ ⎜ ⎝ (s1)2 n1 2 ⎞ ⎟ ⎠ + ⎛ ⎝ 1 n2 – 1 ⎛ ⎞ ⎜ ⎠ ⎝ (s2)2 n2 ⎞ ⎟ ⎠ 2 d f = ⎛ ⎝ 1 n1 – 1 ⎛ ⎞ ⎜ ⎠ ⎝ When both sample sizes n1 and n2 are five or larger, the Student's t approximation is very good. Notice that the sample variances (s1)2 and (s2)2 are not pooled. (If the question comes up, do not pool the variances.) NOTE It is not necessary to compute this by hand. A calculator or computer easily computes it. Example 10.1 Independent groups The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in Table 10.1. Each populations has a normal distribution. Sample Size Average Number of Hours Playing Sports Per Day Sample Standard Deviation Girls 9 Boys 16 Table 10.1 2 3.2 0.866 1.00 Is there a difference in the mean amount of time boys and girls aged seven to 11 play |
sports each day? Test at the 5% level of significance. Solution 10.1 The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μg is the population mean for girls and μb is the population mean for boys. This is a test of two independent groups, two population means. 528 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES ¯ Random variable: X ¯ g − X b = difference in the sample mean amount of time girls and boys play sports each day. H0: μg = μb Ha: μg ≠ μb The words "the same" tell you H0 has an "=". Since there are no other words to indicate Ha, assume it says "is different." This is a two-tailed test. H0: μg – μb = 0 Ha: μg – μb ≠ 0 Distribution for the test: Use tdf where df is calculated using the df formula for independent groups, two population means. Using a calculator, df is approximately 18.8462. Do not pool the variances. Calculate the p-value using a Student's t-distribution: p-value = 0.0054 Graph: Figure 10.2 sg = 0.866 sb = 1 So, x¯ Half the p-value is below –1.2 and half is above 1.2. b = 2 – 3.2 = –1.2 g – x¯ Make a decision: Since α > p-value, reject H0. This means you reject μg = μb. The means are different. Press STAT. Arrow over to TESTS and press 4:2-SampTTest. Arrow over to Stats and press ENTER. Arrow down and enter 2 for the first sample mean, 0.866 for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER. Arrow down to Pooled: and No. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw. Conclusion: At the 5% level of significance, the sample data show there is |
sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 529 10.1 Two samples are shown in Table 10.2. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance. Sample Size Sample Mean Sample Standard Deviation Population A 25 Population B 16 Table 10.2 5 4.7 1 1.2 NOTE When the sum of the sample sizes is larger than 30 (n1 + n2 > 30) you can use the normal distribution to approximate the Student's t. Example 10.2 A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions. a. Is this a test of two means or two proportions? Solution 10.2 a. two means b. Are the populations standard deviations known or unknown? Solution 10.2 b. unknown c. Which distribution do you use to perform the test? Solution 10.2 c. Student's t d. What is the random variable? Solution 10.2 ¯ d. X ¯ A - X B e. What are the null and alternate hypotheses? Write the null and alternate hypotheses in words and in symbols. Solution 10.2 e. 530 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES • Ho : µ A ≤ µ B • Ha : µ A > µ B f. Is this test right-, left-, or two-tailed |
? Solution 10.2 f. Figure 10.3 right g. What is the p-value? Solution 10.2 g. 0.1928 h. Do you reject or not reject the null hypothesis? Solution 10.2 h. Do not reject. i. Conclusion: Solution 10.2 i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B. 10.2 A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed. a. Are the population standard deviations known? b. Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 531 Example 10.3 A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in Table 10.3 and Table 10.4. 67.6 41.2 85.3 55.9 82.4 91.2 73.5 94.1 64.7 64.7 70.6 38.2 61.8 88.2 70.6 58.8 91.2 73.5 82.4 35.5 94.1 88.2 64.7 55.9 88.2 97.1 85.3 61.8 79.4 79.4 Table 10.3 Online Class 77.9 95.3 81.2 74.1 98.8 88.2 85.9 92.9 87.1 88.2 69.4 57.6 |
69.4 67.1 97.6 85.9 88.2 91.8 78.8 71.8 98.8 61.2 92.9 90.6 97.6 100 95.3 83.5 92.9 89.4 Table 10.4 Face-to-face Class Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions: a. Is this a test of two means or two proportions? b. Are the population standard deviations known or unknown? c. Which distribution do you use to perform the test? d. What is the random variable? e. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. f. Is this test right, left, or two tailed? g. What is the p-value? h. Do you reject or not reject the null hypothesis? i. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. (See the conclusion in Example 10.2, and write yours in a similar fashion) First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to μ1: and arrow to ≠ μ2 (does not equal). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER. NOTE Be careful not to mix up the information for Group 1 and Group 2! Solution 10.3 532 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES a. two means b. unknown c. Student's t ¯ d. X ¯ 1 – X 2 e. 1. H0: μ1 = μ2 Null hypothesis: the means of the final exam scores are equal for the online and face-to- face statistics classes. 2. Ha: μ1 < μ2 Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class. f. left-tailed g. p-value = 0.0011 |
Figure 10.4 h. Reject the null hypothesis i. The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class. At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the faceto-face class. Cohen's Standards for Small, Medium, and Large Effect Sizes Cohen's d is a measure of effect size based on the differences between two means. Cohen’s d, named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes. Size of effect d Small medium Large 0.2 0.5 0.8 Table 10.5 Cohen's Standard Effect Sizes Cohen's d is the measure of the difference between two means divided by the pooled standard deviation: d = 1 – x¯ x¯ s pooled 2 where s pooled = (n1 – 1)s1 2 2 + (n2 – 1)s2 n1 + n2 – 2 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 533 Example 10.4 Calculate Cohen’s d for Example 10.2. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem. Solution 10.4 μ1 = 4 s1 = 1.5 n1 = 11 μ2 = 3.5 s2 = 1 n2 = 9 d = 0.384 The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small indicating that there is not a significant difference between them. Example 10.5 Calculate Cohen’s d for Example 10.3. Is the size of the effect small, medium or large? Explain what the size of the effect means for this problem. |
Solution 10.5 d = 0.834; Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference. 10.5 Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq on May 24, 2013 are listed in Table 10.6 and Table 10.7, respectively. 94.2 75.2 69.6 52.0 48.0 41.9 36.4 33.4 31.5 27.6 77.3 71.9 67.5 50.6 46.2 38.4 35.2 33.0 28.7 26.5 76.3 71.7 56.3 48.7 43.2 37.6 33.7 31.8 28.5 26.0 Table 10.6 Northeast 126.0 70.6 65.2 51.4 45.5 37.0 33.0 29.6 23.7 22.6 116.1 70.6 58.2 51.2 43.2 36.0 31.4 28.7 23.5 21.6 78.2 68.2 55.6 50.3 39.0 34.1 31.0 25.3 23.4 21.5 Table 10.7 West Is there a difference in the weighted alpha of the top 30 stocks of banks in the northeast and in the west? Test at a 5% significance level. Answer the following questions: a. Is this a test of two means or two proportions? b. Are the population standard deviations known or unknown? c. Which distribution do you use to perform the test? d. What is the random variable? e. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. 534 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES f. Is this test right, left, or two tailed? g. What is the p-value? h. Do |
you reject or not reject the null hypothesis? i. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. j. Calculate Cohen’s d and interpret it. 10.2 | Two Population Means with Known Standard Deviations Even though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal and both populations must be normal. The random variable is X1 – X2 normal distribution has the following format:. The Normal distribution isµ 1 – µ 2, ⎣ (σ1)2 n1 + (σ2)2 n2 ⎤ ⎥ ⎦ The standard deviation is: (σ1)2 n1 (σ2)2 n2 + The test statistic (z-score) is: z = Example 10.6 ( x¯ 1 – x¯ 2) – (µ 1 – µ 2) (σ2)2 n2 + (σ1)2 n1 Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have a normal distributions. The data are recorded in Table 10.8. Wax Sample Mean Number of Months Floor Wax Lasts Population Standard Deviation 1 2 3 2.9 Table 10.8 0.33 0.36 Does the data indicate that wax 1 is more effective than wax 2? Test at a 5% level of significance. Solution 10.6 This is a test of two independent groups, two population means, population standard deviations known. ¯ Random Variable: X ¯ 1 – X 2 = difference in the mean number of months the competing floor waxes last. H0: μ1 ≤ μ2 Ha: μ1 > μ2 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 535 The words "is more effective" says that wax 1 lasts longer than wax 2, on average. "Longer" is a “>” symbol and goes into Ha. Therefore, this is a right-tailed test. Distribution for the |
test: The population standard deviations are known so the distribution is normal. Using the formula, the distribution is⎛ ⎝0, 0.332 20 + 0.362 20 ⎞ ⎠ Since μ1 ≤ μ2 then μ1 – μ2 ≤ 0 and the mean for the normal distribution is zero. Calculate the p-value using the normal distribution: p-value = 0.1799 Graph: Figure 10..9 = 0.1 Compare α and the p-value: α = 0.05 and p-value = 0.1799. Therefore, α < p-value. Make a decision: Since α < p-value, do not reject H0. Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts. Press STAT. Arrow over to TESTS and press 3:2-SampZTest. Arrow over to Stats and press ENTER. Arrow down and enter.33 for sigma1,.36 for sigma2, 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to μ1: and arrow to > μ2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1799 and the test statistic is 0.9157. Do the procedure again, but instead of Calculate do Draw. 10.6 The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines are randomly assigned to be tested. Both populations have normal distributions. Table 10.9 shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance. 536 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES Engine Sample Mean Number of RPM Population Standard Deviation 1 2 1,500 1,600 Table 10.9 50 60 Example 10.7 An interested citizen wanted to know if Democratic U. S. senators are older than Republican U.S. senators, on average. On May 26 2013, the mean age of 30 randomly selected Republican Senators was 61 years 247 days old (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years |
257 days old (61.704 years) with a standard deviation of 9.55 years. Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5% level of significance. Solution 10.7 This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. Subscripts: 1: Democratic senators 2: Republican senators ¯ Random variable: X ¯ 1 – X 2 H0: µ1 ≤ µ2 H0: µ1 – µ2 ≤ 0 Ha: µ1 > µ2 Ha: µ1 – µ2 > 0 = difference in the mean age of Democratic and Republican U.S. senators. The words "older than" translates as a “>” symbol and goes into Ha. Therefore, this is a right-tailed test. Distribution for the test: The distribution is the normal approximation to the Student’s t for means, independent ¯ groups. Using the formula, the distribution is: X ¯ 1 – X 2 ∼ N[0, (9.55)2 30 + (10.17)2 30 ] Since µ1 ≤ µ2, µ1 – µ2 ≤ 0 and the mean for the normal distribution is zero. (Calculating the p-value using the normal distribution gives p-value = 0.4040) Graph: Figure 10.6 Compare α and the p-value: α = 0.05 and p-value = 0.4040. Therefore, α < p-value. Make a decision: Since α < p-value, do not reject H0. Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 537 10.3 | Comparing Two Independent Population Proportions When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present: 1. The two independent samples are simple random samples that are independent. 2. |
The number of successes is at least five, and the number of failures is at least five, for each of the samples. 3. Growing literature states that the population must be at least ten or 20 times the size of the sample. This keeps each population from being over-sampled and causing incorrect results. Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions. The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, H0: pA = pB. To conduct the test, we use a pooled proportion, pc. The pooled proportion is calculated as follows: pc = x A + xB n A + nB The distribution for the differences is: P′ A − P′B ~ N[0, pc(1 − pc)( 1 n A + 1 nB )] The test statistic (z-score) is: z = (p′ A − p′B) − (p A − pB) n A + 1 nB) pc(1 − pc)( 1 Example 10.8 Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance. Solution 10.8 The problem asks for a difference in proportions, making it a test of two proportions. Let A and B be the subscripts for medication A and medication B, respectively. Then pA and pB are the desired population proportions. Random Variable: P′A – P′B = difference in the proportions of adult patients who did not react after 30 minutes to medication A and to medication B. H0: pA = pB pA – pB = 0 Ha: pA ≠ pB pA – pB ≠ 0 The words "is a difference" tell you the test is two-tailed. Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal: pc = x A + xB n |
A + nB = 20 + 12 200 + 200 = 0.08 1 – pc = 0.92 P′ A – P′B ~ N ⎡ ⎣0, (0.08)(0.92)( 1 200 + 1 200 ⎤ ) ⎦ 538 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES P′A – P′B follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.1404. Estimated proportion for group A: p′ A = Estimated proportion for group B: p′B = x A n A xB nB = 20 200 = 12 200 = 0.1 = 0.06 Graph: Figure 10.7 P′A – P′B = 0.1 – 0.06 = 0.04. Half the p-value is below –0.04, and half is above 0.04. Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α < p-value. Make a decision: Since α < p-value, do not reject H0. Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to medication A and medication B. Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 20 for x1, 200 for n1, 12 for x2, and 200 for n2. Arrow down to p1: and arrow to not equal p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1404 and the test statistic is 1.47. Do the procedure again, but instead of Calculate do Draw. 10.8 Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5% level of significance. Example 10.9 A research study was conducted about gender differences in “sexting.” The researcher believed that the proportion of girls involved in “sexting” is less than the proportion of boys involved |
. The data collected in the spring of 2010 among a random sample of middle and high school students in a large school district in the southern United States This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 539 is summarized in Table 10.9. Is the proportion of girls sending sexts less than the proportion of boys “sexting?” Test at a 1% level of significance. Males Females Sent “sexts” 183 Total number surveyed 2231 156 2169 Table 10.10 Solution 10.9 This is a test of two population proportions. Let M and F be the subscripts for males and females. Then pM and pF are the desired population proportions. Random variable: p′F − p′M = difference in the proportions of males and females who sent “sexts.” H0: pF = pM H0: pF – pM = 0 Ha: pF < pM Ha: pF – pM < 0 The words "less than" tell you the test is left-tailed. Distribution for the test: Since this is a test of two population proportions, the distribution is normal: pc = = 0.077 = 156 + 183 2169 + 2231 xF + x M nF + n M 1 − pc = 0.923 Therefore, p′F – p′ M ∼ N⎛ ⎝0, p′F – p′M follows an approximate normal distribution. ⎛ (0.077)(0.923) ⎝ 1 2169 + 1 2231 ⎞ ⎞ ⎠ ⎠ Calculate the p-value using the normal distribution: p-value = 0.1045 Estimated proportion for females: 0.0719 Estimated proportion for males: 0.082 Graph: Figure 10.8 Decision: Since α < p-value, Do not reject H0 Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of girls sending “sexts” is less than the proportion of boys sending “sexts.” 540 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAM |
PLES Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is P = 0.1045 and the test statistic is z = -1.256. Example 10.10 Researchers conducted a study of smartphone use among adults. A cell phone company claimed that iPhone smartphones are more popular with whites (non-Hispanic) than with African Americans. The results of the survey indicate that of the 232 African American cell phone owners randomly sampled, 5% have an iPhone. Of the 1,343 white cell phone owners randomly sampled, 10% own an iPhone. Test at the 5% level of significance. Is the proportion of white iPhone owners greater than the proportion of African American iPhone owners? Solution 10.10 This is a test of two population proportions. Let W and A be the subscripts for the whites and African Americans. Then pW and pA are the desired population proportions. Random variable: p′W – p′A = difference in the proportions of Android and iPhone users. H0: pW = pA H0: pW – pA = 0 Ha: pW > pA Ha: pW – pA > 0 The words "more popular" indicate that the test is right-tailed. Distribution for the test: The distribution is approximately normal: pc = xW + x A nW + n A = 134 + 12 1343 + 232 = 0.1077 1 − pc = 0.8923 Therefore, p′W – p′ A ∽ N ⎛ ⎝0, ⎛ (0.1077)(0.8923) ⎝ 1 1343 + 1 232 ⎞ ⎞ ⎠ ⎠ p′W – p′ A follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.0092 Estimated proportion for group A: 0.10 Estimated proportion for group B: 0.05 Graph: This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHES |
IS TESTING WITH TWO SAMPLES 541 Figure 10.9 Decision: Since α > p-value, reject the H0. Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of white cell phone owners use iPhones than African Americans. TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 135 for x1, 1343 for n1, 12 for x2, and 232 for n2. Arrow down to p1: and arrow to greater than p2. Press ENTER. Arrow down to Calculate and press ENTER. The P-value is P = 0.0092 and the test statistic is Z = 2.33. 10.10 A concerned group of citizens wanted to know if the proportion of forcible rapes in Texas was different in 2011 than in 2010. Their research showed that of the 113,231 violent crimes in Texas in 2010, 7,622 of them were forcible rapes. In 2011, 7,439 of the 104,873 violent crimes were in the forcible rape category. Test at a 5% significance level. Answer the following questions: a. Is this a test of two means or two proportions? b. Which distribution do you use to perform the test? c. What is the random variable? d. What are the null and alternative hypothesis? Write the null and alternative hypothesis in symbols. e. Is this test right-, left-, or two-tailed? f. What is the p-value? g. Do you reject or not reject the null hypothesis? h. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ____________. 10.4 | Matched or Paired Samples When using a hypothesis test for matched or paired samples, the following characteristics should be present: 1. Simple random sampling is used. 2. Sample sizes are often small. 3. Two measurements (samples) are drawn from the same pair of individuals or objects. 542 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 4. Differences are calculated from the matched or paired samples. 5. The differences form the sample that is used for the hypothesis test. 6. Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of |
the sample mean of differences is approximately normal. In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μd, is then tested using a Student's-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences. The test statistic (t-score) is: x¯ t = d − µ d sd ⎛ ⎞ ⎝ ⎠ n Example 10.11 A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in Table 10.10. A lower score indicates less pain. The "before" value is matched to an "after" value and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level. Subject: A B C D E F G H Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6 After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0 Table 10.11 Solution 10.11 Corresponding "before" and "after" values form matched pairs. (Calculate "after" – "before.") After Data Before Data Difference 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2 Table 10.12 6.6 6.5 9 10.3 11.3 8.1 6.3 11.6 0.2 -4.1 -1.6 -1.8 -3.2 -2 -2.9 -9.6 The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6} The sample mean and sample standard deviation of the differences are: xd = –3.13 and sd = 2.91 Verify these values. Let µ d be the population mean for the differences. We use the subscript d to denote "differences." ¯ Random variable: X d = the mean difference of the sensory measurements This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org |
/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 543 H0: μd ≥ 0 The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. μd is the population mean of the differences.) Ha: μd < 0 The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement. Distribution for the test: The distribution is a Student's t with df = n – 1 = 8 – 1 = 7. Use t7. (Notice that the test is for a single population mean.) Calculate the p-value using the Student's-t distribution: p-value = 0.0095 Graph: Figure 10.10 ¯ X d is the random variable for the differences. The sample mean and sample standard deviation of the differences are: x¯ s¯ d = –3.13 d = 2.91 Compare α and the p-value: α = 0.05 and p-value = 0.0095. α > p-value. Make a decision: Since α > p-value, reject H0. This means that μd < 0 and there is improvement. Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain. NOTE For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time (after before) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then go to a third list and arrow up to the name. Enter 1st list name - 2nd list name. The calculator will do the subtraction, and you will have the differences in the third list. Use your list of differences as the data. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 0 for µ 0, the name of the list where you put the data, 544 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES and 1 for Freq:. |
Arrow down to μ: and arrow over to < µ 0. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is 0.0094, and the test statistic is -3.04. Do these instructions again except, arrow to Draw (instead of Calculate). Press ENTER. 10.11 A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5% level. Subject A B C D E F G H I Before 209 210 205 198 216 217 238 240 222 After 199 207 189 209 217 202 211 223 201 Table 10.13 Example 10.12 A college football coach was interested in whether the college's strength development class increased his players' maximum lift (in pounds) on the bench press exercise. He asked four of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows: Weight (in pounds) Player 1 Player 2 Player 3 Player 4 Amount of weight lifted prior to the class 205 Amount of weight lifted after the class 295 241 252 338 330 368 360 Table 10.14 The coach wants to know if the strength development class makes his players stronger, on average. Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}. Assume the differences have a normal distribution. Using the differences data, calculate the sample mean and the sample standard deviation. x¯ d = 21.3, sd = 46.7 NOTE The data given here would indicate that the distribution is actually right-skewed. The difference 90 may be an extreme outlier? It is pulling the sample mean to be 21.3 (positive). The means of the other three data values are actually negative. Using the difference data, this becomes a test of a single __________ (fill in the blank). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHES |
IS TESTING WITH TWO SAMPLES 545 ¯ Define the random variable: X d mean difference in the maximum lift per player. The distribution for the hypothesis test is t3. H0: μd ≤ 0, Ha: μd > 0 Graph: Figure 10.11 Calculate the p-value: The p-value is 0.2150 Decision: If the level of significance is 5%, the decision is not to reject the null hypothesis, because α < p-value. What is the conclusion? At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average. 10.12 A new prep class was designed to improve SAT test scores. Five students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in Table 10.15. Are the scores, on average, higher after the class? Test at a 5% level. SAT Scores Student 1 Student 2 Student 3 Student 4 Score before class 1840 Score after class 1920 1960 2160 1920 2200 2150 2100 Table 10.15 Example 10.13 Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The data were collected and recorded in Table 10.16. Distance (in feet) using Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Student 7 Dominant Hand 30 26 34 17 19 26 20 Table 10.16 546 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES Distance (in feet) using Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Student 7 Weaker Hand 28 14 27 18 17 26 16 Table 10.16 Conduct a hypothesis test to determine whether the mean difference in distances between the children’s dominant versus weaker hands is significant. Record the differences data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: {2, 12, 7, –1, 2, 0, 4}. The differences have a normal distribution. Using the differences data, calculate the sample mean and the sample standard deviation. x¯ d = 3.71, sd = 4.5. ¯ Random variable: X d = mean difference in the distances between |
the hands. Distribution for the hypothesis test: t6 H0: μd = 0 Ha: μd ≠ 0 Graph: Figure 10.12 Calculate the p-value: The p-value is 0.0716 (using the data directly). (test statistic = 2.18. p-value = 0.0719 using ⎛ ⎝ x¯ d = 3.71, sd = 4.5. ⎞ ⎠ Decision: Assume α = 0.05. Since α < p-value, Do not reject H0. Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children’s weaker and dominant hands to push the shot-put. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 547 10.13 Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in Table 10.17. Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% level. Player 1 Player 2 Player 3 Player 4 Player 5 Dominant Hand 120 Off-hand 105 111 109 135 98 140 111 125 99 Table 10.17 548 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 10.1 Hypothesis Testing for Two Means and Two Proportions Class Time: Names: Student Learning Outcomes • The student will select the appropriate distributions to use in each case. • The student will conduct hypothesis tests and interpret the results. Supplies: • • • the business section from two consecutive days’ newspapers three small packages of M&Ms® five small packages of Reese's Pieces® Increasing Stocks Survey Look at yesterday’s newspaper business section. Conduct a hypothesis test to determine if the proportion of New York Stock Exchange (NYSE) stocks that increased is greater than the proportion of NASDAQ stocks that increased. As randomly as possible, choose 40 NYSE stocks, and 32 NASDAQ stocks and complete the following statements. 1. H0: _________ 2. Ha: _________ 3. In words, define the |
random variable. 4. The distribution to use for the test is _____________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 10.13 b. Calculate the p-value. 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 549 Decreasing Stocks Survey Randomly pick eight stocks from the newspaper. Using two consecutive days’ business sections, test whether the stocks went down, on average, for the second day. 1. H0: ________ 2. Ha: ________ 3. In words, define the random variable. 4. The distribution to use for the test is _____________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 10.14 b. Calculate the p-value: 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Candy Survey Buy three small packages of M&Ms and five small packages of Reese's Pieces (same net weight as the M&Ms). Test whether or not the mean number of candy pieces per package is the same for the two brands. 1. H0: ________ 2. Ha: ________ 3. In words, define the random variable. 4. What distribution should be used for this test? 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: 550 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES Figure 10.15 b. Calculate the p-value. 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Shoe Survey Test whether women have, on average, more pairs of shoes than men. Include all forms of sneakers, shoes, sandals, and boots. Use your class as the sample. 1. H0: ________ 2. Ha: ________ |
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