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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 10.16 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 551 KEY TERMS Degrees of Freedom (df) the number of objects in a sample that are free to vary. Pooled Proportion estimate of the common value of p1 and p2. 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. Variable (Random Variable) a characteristic of interest in a population being studied. Common notation for variables are upper-case Latin letters X, Y, Z,... Common notation for a specific value from the domain (set of all possible values of a variable) are lower-case Latin letters x, y, z,.... For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3,.... Variables in statistics differ from variables in intermediate algebra in two following ways. • The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color, then the domain is {black, blond, gray, green, orange}. • We can tell what specific value x of the random variable X takes only after performing the experiment. CHAPTER REVIEW 10.1 Two Population Means with Unknown Standard Deviations Two population means from independent samples where the population standard deviations are not known ¯ • Random Variable: X ¯ 1 − X 2 = the difference of the sampling means • Distribution: Student's t-distribution with degrees of freedom (variances not pooled) 10.2 Two Population Means with Known Standard Deviations A hypothesis test of two population means from independent samples where the population standard deviations are known (typically approximated with |
the sample standard deviations), will have these characteristics: ¯ • Random variable: X ¯ 1 − X 2 = the difference of the means • Distribution: normal distribution 10.3 Comparing Two Independent Population Proportions Test of two population proportions from independent samples. • Random variable: p^ A – p^ B = difference between the two estimated proportions • Distribution: normal distribution 10.4 Matched or Paired Samples A hypothesis test for matched or paired samples (t-test) has these characteristics: • Test the differences by subtracting one measurement from the other measurement • Random Variable: x¯ d = mean of the differences • Distribution: Student’s-t distribution with n – 1 degrees of freedom • If the number of differences is small (less than 30), the differences must follow a normal distribution. • Two samples are drawn from the same set of objects. 552 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES • Samples are dependent. FORMULA REVIEW 10.1 Two Population Means with Unknown Standard Deviations Standard error: SE = (s1)2 n1 (s2)2 n2 + Test statistic (t-score): t = ( x¯ Degrees of freedom: 1 − x¯ 2) − (µ 1 − µ 2) (s2)2 n2 + (s1)2 n1 (s1)2 n1 ⎛ ⎜ ⎝ (s1)2 n1 ⎛ ⎞ ⎜ ⎠ ⎝ 2 (s2)2 n2 ⎞ ⎟ ⎠ + 2 ⎞ ⎟ ⎠ + ⎛ ⎝ 1 n2 − 1 ⎛ ⎞ ⎜ ⎠ ⎝ (s2)2 n2 ⎞ ⎟ ⎠ 2 d f = where: ⎛ ⎝ 1 n1 − 1 s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes. x¯ 1 and x¯ 2 are the sample means. Cohen’s d is the measure of effect size: d = x¯ 1 − x¯ s pooled 2 where s pooled = (n1 − 1)s1 2 + (n2 − 1)s2 2 n1 + n2 − 2 Generally µ1 - µ2 = 0. where: σ1 |
and σ2 are the known population standard deviations. n1 and x¯ and n2 are the sample sizes. x¯ 2 means. μ1 and μ2 are the population means. are the sample 1 10.3 Comparing Two Independent Population Proportions Pooled Proportion: pc = xF + x M nF + n M Distribution for the differences: ⎡ ⎛ ⎣0, pc(1 − pc) ⎝ p′ A − p′B ∼ N 1 n A + 1 nB ⎤ ⎞ ⎦ ⎠ where the null hypothesis is H0: pA = pB or H0: pA – pB = 0. Test Statistic (z-score): z = (p′ A − p′ B) ⎛ pc(1 − pc) ⎝ n A + 1 1 nB ⎞ ⎠ where the null hypothesis is H0: pA = pB or H0: pA − pB = 0. where p′A and p′B are the sample proportions, pA and pB are the population proportions, Pc is the pooled proportion, and nA and nB are the sample sizes. 10.2 Two Population Means with Known Standard Deviations 10.4 Matched or Paired Samples Normal Distributionµ 1 − µ 2, ⎣ (σ1)2 n1 + (σ2)2 n2 ⎤ ⎥. ⎦ Generally µ1 – µ2 = 0. Test Statistic (z-score): z = ( x¯ 1 − x¯ 2) − (µ 1 − µ 2) (σ2)2 n2 + (σ1)2 n1 Test Statistic (t-score): t = x¯ d − µ d sd ⎛ ⎞ ⎝ ⎠ n where: x¯ d is the mean of the sample differences. μd is the mean of the population differences. sd is the sample standard deviation of the differences. n is the sample size. PRACTICE 10.1 Two Population Means with Unknown Standard Deviations Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for a. b. independent group means, population standard deviations, and/or variances known independent group means, population standard deviations, and/or variances unknown c. matched or paired samples 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 553 d. single mean e. two proportions f. single proportion 1. It is believed that 70% of males pass their drivers test in the first attempt, while 65% of females pass the test in the first attempt. Of interest is whether the proportions are in fact equal. 2. A new laundry detergent is tested on consumers. Of interest is the proportion of consumers who prefer the new brand over the leading competitor. A study is done to test this. 3. A new windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted. 4. The known standard deviation in salary for all mid-level professionals in the financial industry is $11,000. Company A and Company B are in the financial industry. Suppose samples are taken of mid-level professionals from Company A and from Company B. The sample mean salary for mid-level professionals in Company A is $80,000. The sample mean salary for mid-level professionals in Company B is $96,000. Company A and Company B management want to know if their midlevel professionals are paid differently, on average. 5. The average worker in Germany gets eight weeks of paid vacation. 6. According to a television commercial, 80% of dentists agree that Ultrafresh toothpaste is the best on the market. 7. It is believed that the average grade on an English essay in a particular school system for females is higher than for males. A random sample of 31 females had a mean score of 82 with a standard deviation of three, and a random sample of 25 males had a mean score of 76 with a standard deviation of four. 8. The league mean batting average is 0.280 with a known standard deviation of 0.06. The Rattlers and the Vikings belong to the league. The mean batting average for a sample of eight Rattlers is 0.210, and the mean batting average for a sample of eight Vikings is 0.260. There are 24 players on the Rattlers and 19 players on the Vikings. Are the batting averages of the Rattlers and Vikings statistically different? 9. In a random sample of 100 |
forests in the United States, 56 were coniferous or contained conifers. In a random sample of 80 forests in Mexico, 40 were coniferous or contained conifers. Is the proportion of conifers in the United States statistically more than the proportion of conifers in Mexico? 10. A new medicine is said to help improve sleep. Eight subjects are picked at random and given the medicine. The means hours slept for each person were recorded before starting the medication and after. 11. It is thought that teenagers sleep more than adults on average. A study is done to verify this. A sample of 16 teenagers has a mean of 8.9 hours slept and a standard deviation of 1.2. A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of 0.6. 12. Varsity athletes practice five times a week, on average. 13. A sample of 12 in-state graduate school programs at school A has a mean tuition of $64,000 with a standard deviation of $8,000. At school B, a sample of 16 in-state graduate programs has a mean of $80,000 with a standard deviation of $6,000. On average, are the mean tuitions different? 14. A new WiFi range booster is being offered to consumers. A researcher tests the native range of 12 different routers under the same conditions. The ranges are recorded. Then the researcher uses the new WiFi range booster and records the new ranges. Does the new WiFi range booster do a better job? 15. A high school principal claims that 30% of student athletes drive themselves to school, while 4% of non-athletes drive themselves to school. In a sample of 20 student athletes, 45% drive themselves to school. In a sample of 35 non-athlete students, 6% drive themselves to school. Is the percent of student athletes who drive themselves to school more than the percent of nonathletes? Use the following information to answer the next three exercises: A study is done to determine which of two soft drinks has more sugar. There are 13 cans of Beverage A in a sample and six cans of Beverage B. The mean amount of sugar in Beverage A is 36 grams with a standard deviation of 0.6 grams. The mean amount of sugar in Beverage B is 38 grams with a standard deviation of 0.8 grams. The researchers believe that Beverage B has more sugar than Beverage A, on average. Both populations have normal |
distributions. 16. Are standard deviations known or unknown? 17. What is the random variable? 18. Is this a one-tailed or two-tailed test? 554 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. 19. Is this a test of means or proportions? 20. State the null and alternative hypotheses. a. H0: __________ b. Ha: __________ 21. Is this a right-tailed, left-tailed, or two-tailed test? 22. In symbols, what is the random variable of interest for this test? 23. In words, define the random variable of interest for this test. 24. Which distribution (normal or Student's t) would you use for this hypothesis test? 25. Explain why you chose the distribution you did for Exercise 10.24. 26. Calculate the test statistic and p-value. 27. Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value. 28. Find the p-value. 29. At a pre-conceived α = 0.05, what is your: a. Decision: b. Reason for the decision: c. Conclusion (write out in a complete sentence): 30. Does it appear that the means are the same? Why or why not? 10.2 Two Population Means with Known Standard Deviations Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. Table 10.18 shows the result. Scouters believe that Rodriguez pitches a speedier fastball. Pitcher Sample Mean Speed of Pitches (mph) Population Standard Deviation |
Wesley 86 Rodriguez 91 Table 10.18 3 7 31. What is the random variable? 32. State the null and alternative hypotheses. 33. What is the test statistic? 34. What is the p-value? 35. At the 1% significance level, what is your conclusion? Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller. Plant Group Sample Mean Height of Plants (inches) Population Standard Deviation Food No food Table 10.19 16 14 2.5 1.5 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 555 36. Is the population standard deviation known or unknown? 37. State the null and alternative hypotheses. 38. What is the p-value? 39. Draw the graph of the p-value. 40. At the 1% significance level, what is your conclusion? Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. Sample Mean Melting Temperatures (°F) Population Standard Deviation Alloy Gamma 800 Alloy Zeta 900 Table 10.20 95 105 41. State the null and alternative hypotheses. 42. Is this a right-, left-, or two-tailed test? 43. What is the p-value? 44. Draw the graph of the p-value. 45. At the 1% significance level, what is your conclusion? 10.3 Comparing Two Independent Population Proportions Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random |
sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1. 46. Is this a test of means or proportions? 47. What is the random variable? 48. State the null and alternative hypotheses. 49. What is the p-value? 50. What can you conclude about the two operating systems? Use the following information to answer the next twelve exercises. In the recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota. 51. Is this a test of means or proportions? 52. State the null and alternative hypotheses. a. H0: _________ b. Ha: _________ 53. Is this a right-tailed, left-tailed, or two-tailed test? How do you know? 54. What is the random variable of interest for this test? 55. In words, define the random variable for this test. 56. Which distribution (normal or Student's t) would you use for this hypothesis test? 57. Explain why you chose the distribution you did for the Exercise 10.56. 556 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 58. Calculate the test statistic. 59. Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value. Figure 10.17 60. Find the p-value. 61. At a pre-conceived α = 0.05, what is your: a. Decision: b. Reason for the decision: c. Conclusion (write out in a complete sentence): 62. Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not? 10.4 Matched or Paired Samples Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a |
software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table 10.21. The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level. Installation A B C D E F G H Before After Table 10.21 63. What is the random variable? 64. State the null and alternative hypotheses. 65. What is the p-value? 66. Draw the graph of the p-value. 67. What conclusion can you draw about the software patch? Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level. Subject A B C D E F Before After Table 10.22 68. State the null and alternative hypotheses. 69. What is the p-value? 70. What is the sample mean difference? 71. Draw the graph of the p-value. 72. What conclusion can you draw about the juggling class? 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 557 Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level. Patient A B C D E F Before 161 162 165 162 166 171 After 158 159 166 160 167 169 Table 10.23 73. State the null and alternative hypotheses. 74. What is the test statistic? 75. What is the p-value? 76. What is the sample mean difference? 77. What is the conclusion? HOMEWORK 10.1 Two Population Means with Unknown Standard Deviations DIRECTIONS: 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 a homework problem in what follows, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) 78. The mean number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of three English courses with a standard deviation of 0.8. The females took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same? 79. A student at a four-year college claims that mean enrollment at four–year colleges is higher than at two–year colleges in the United States. Two surveys are conducted. Of the 35 two–year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191. 80. At Rachel’s 11th birthday party, eight girls were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The girls thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis. Relaxed time (seconds) Jumping time (seconds) 26 47 30 22 Table 10.24 21 40 28 21 558 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES Relaxed time (seconds) Jumping time (seconds) 23 45 37 29 Table 10.24 25 43 35 32 81. Mean entry-level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. A recruiting office thinks that the mean mechanical engineering salary is actually lower than the mean electrical engineering salary. The recruiting office randomly surveys 50 entry level mechanical engineers and 60 entry level electrical engineers. Their mean salaries were $46,100 and $46,700, respectively. Their standard deviations were $3,450 and $4,210, respectively. Conduct a hypothesis test to determine if you agree that the mean |
entry-level mechanical engineering salary is lower than the mean entry-level electrical engineering salary. 82. Marketing companies have collected data implying that teenage girls use more ring tones on their cellular phones than teenage boys do. In one particular study of 40 randomly chosen teenage girls and boys (20 of each) with cellular phones, the mean number of ring tones for the girls was 3.2 with a standard deviation of 1.5. The mean for the boys was 1.7 with a standard deviation of 0.8. Conduct a hypothesis test to determine if the means are approximately the same or if the girls’ mean is higher than the boys’ mean. Use the information from Appendix C to answer the next four exercises. 83. Using the data from Lap 1 only, conduct a hypothesis test to determine if the mean time for completing a lap in races is the same as it is in practices. 84. Repeat the test in Exercise 10.83, but use Lap 5 data this time. 85. Repeat the test in Exercise 10.83, but this time combine the data from Laps 1 and 5. 86. In two to three complete sentences, explain in detail how you might use Terri Vogel’s data to answer the following question. “Does Terri Vogel drive faster in races than she does in practices?” Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals. Western Eastern Los Angeles 9 D.C. United 9 FC Dallas 3 Chicago 8 Chivas USA 4 Columbus 7 Real Salt Lake 3 New England 6 Colorado 4 MetroStars 5 San Jose 4 Kansas City 3 Table 10.25 Conduct a hypothesis test to answer the next two exercises. 87. The exact distribution for the hypothesis test is: a. b. c. d. the normal distribution the Student's t-distribution the uniform distribution the exponential distribution 88. If the level of significance is 0.05, the conclusion is: a. There is sufficient evidence to conclude that the W Division teams score fewer goals, on average, than the E teams b. There is insufficient evidence to conclude that the W Division teams score more goals, on average, than the E teams. 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 559 c. There is insufficient evidence to conclude that the W teams score fewer goals, on average, than the E teams score. d. Unable to determine 89. Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. A concluding statement is: a. There is sufficient evidence to conclude that statistics night students' mean on Exam 2 is better than the statistics day students' mean on Exam 2. b. There is insufficient evidence to conclude that the statistics day students' mean on Exam 2 is better than the statistics night students' mean on Exam 2. c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. d. There is sufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. 90. Researchers interviewed street prostitutes in Canada and the United States. The mean age of the 100 Canadian prostitutes upon entering prostitution was 18 with a standard deviation of six. The mean age of the 130 United States prostitutes upon entering prostitution was 20 with a standard deviation of eight. Is the mean age of entering prostitution in Canada lower than the mean age in the United States? Test at a 1% significance level. 91. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet group had a mean weight loss of 45 pounds with a standard deviation of 14 pounds. 92. Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean and standard |
deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. An appropriate alternative hypothesis for the hypothesis test is: a. μday > μnight b. μday < μnight c. μday = μnight d. μday ≠ μnight 10.2 Two Population Means with Known Standard Deviations DIRECTIONS: 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, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) 93. A study is done to determine if students in the California state university system take longer to graduate, on average, than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3. 94. Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls, it was $559. From past years, it is known that the population standard deviation for each group is $180. Determine whether or not you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls. 560 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 95. A group of transfer bound students wondered if they will spend the same mean amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four- |
year university. The sample means were $947 and $1,011, respectively. The population standard deviations are known to be $254 and $87, respectively. Conduct a hypothesis test to determine if the means are statistically the same. 96. Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of seven mpg. Thirtyone non-hybrid sedans get a mean of 22 mpg with a standard deviation of four mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test to evaluate the manufacturers claim. 97. A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test. 98. One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from one (strongly agree) to five (strongly disagree). Table 10.26 contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife). Wife’s Score Husband’s Score Table 10.26 10.3 Comparing Two Independent Population Proportions DIRECTIONS: 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, including for paired data, you may assume |
that the underlying population is normally distributed. (In general, you must first prove that assumption, however.) 99. A recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them. 100. We are interested in whether the proportions of female suicide victims for ages 15 to 24 are the same for the whites and the blacks races in the United States. We randomly pick one year, 1992, to compare the races. The number of suicides estimated in the United States in 1992 for white females is 4,930. Five hundred eighty were aged 15 to 24. The estimate for black females is 330. Forty were aged 15 to 24. We will let female suicide victims be our population. larger + smaller dimension larger dimension 101. Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ⎛ ⎝ to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation? was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 ⎞ ⎠ 102. A recent year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic students out of a total 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 561 of 2,441 students. In general, do you think that the percent of Hispanic students at the two colleges is basically the same or different? Use the following information to answer the next two exercises. Neuroinvasive West Nile virus is a severe disease that affects a person’s nervous system. It is spread by the Culex species of mosquito |
. In the United States in 2010 there were 629 reported cases of neuroinvasive West Nile virus out of a total of 1,021 reported cases and there were 486 neuroinvasive reported cases out of a total of 712 cases reported in 2011. Is the 2011 proportion of neuroinvasive West Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus cases? Using a 1% level of significance, conduct an appropriate hypothesis test. • “2011” subscript: 2011 group. • “2010” subscript: 2010 group 103. This is: a. a test of two proportions b. a test of two independent means c. a test of a single mean d. a test of matched pairs. 104. An appropriate null hypothesis is: a. p2011 ≤ p2010 b. p2011 ≥ p2010 c. μ2011 ≤ μ2010 d. p2011 > p2010 105. The p-value is 0.0022. At a 1% level of significance, the appropriate conclusion is a. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease. b. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease. c. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease. d. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease. 106. Researchers conducted a study to find out if there is a difference in the use of eReaders by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7% of the 628 surveyed use eReaders, while 11% of the 2,309 participants 30 years old and older use eReaders. 107. Adults aged 18 years old and older were randomly selected for a survey on obesity. Adults are considered obese if their body mass |
index (BMI) is at least 30. The researchers wanted to determine if the proportion of women who are obese in the south is less than the proportion of southern men who are obese. The results are shown in Table 10.27. Test at the 1% level of significance. Number who are obese Sample size Men 42,769 Women 67,169 Table 10.27 155,525 248,775 108. Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than the proportion of people age 30 and older. Table 10.28 details the number of tablet owners for each age group. Test at the 1% level of significance. 16–29 year olds 30 years old and older Own a Tablet 69 231 Table 10.28 562 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 16–29 year olds 30 years old and older Sample Size 628 2,309 Table 10.28 109. A group of friends debated whether more men use smartphones than women. They consulted a research study of smartphone use among adults. The results of the survey indicate that of the 973 men randomly sampled, 379 use smartphones. For women, 404 of the 1,304 who were randomly sampled use smartphones. Test at the 5% level of significance. 110. While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey. 111. We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of $4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard deviation of $10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software. 112. Joan Nguyen recently claimed that the proportion of college-age males with at least one pierced ear is as high as the proportion of college-age females. She conducted a survey in her classes. Out of 107 males, 20 had at least |
one pierced ear. Out of 92 females, 47 had at least one pierced ear. Do you believe that the proportion of males has reached the proportion of females? 113. Use the data sets found in Appendix C to answer this exercise. Is the proportion of race laps Terri completes slower than 130 seconds less than the proportion of practice laps she completes slower than 135 seconds? 114. "To Breakfast or Not to Breakfast?" by Richard Ayore In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, …, birthdays. During this time, one looks back to see what he or she has achieved for the past year and also focuses ahead for more to come. If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had with my brothers and sister while we did our daily routine. Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!” And so, to prove him wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we worked again before getting tired and stopping. Of interest was our mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in Table 10.29, solve our problem. Work hours with breakfast Work hours without breakfast 8 7 9 5 9 8 10 7 Table 10.29 6 5 5 4 7 7 7 5 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 563 Work hours with breakfast Work hours without breakfast 6 9 Table 10.29 6 |
5 10.4 Matched or Paired Samples DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in m47889 (http://cnx.org/content/m47889/latest/). 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 the homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) 115. Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. The data are recorded in Table 10.30. Do you think that their cholesterol levels were significantly lowered? Starting cholesterol level Ending cholesterol level 140 220 110 240 200 180 190 360 280 260 Table 10.30 140 230 120 220 190 150 200 300 300 240 Use the following information to answer the next two exercises. A new AIDS prevention drug was tried on a group of 224 HIV positive patients. Forty-five patients developed AIDS after four years. In a control group of 224 HIV positive patients, 68 developed AIDS after four years. We want to test whether the method of treatment reduces the proportion of patients that develop AIDS after four years or if the proportions of the treated group and the untreated group stay the same. Let the subscript t = treated patient and ut = untreated patient. 116. The appropriate hypotheses are: a. H0: pt < put and Ha: pt ≥ put b. H0: pt ≤ put and Ha: pt > put c. H0: pt = put and Ha: pt ≠ put d. H0: pt = put and Ha: pt < put 117. If the p-value is 0.0062 what is the conclusion (use α = 0.05)? a. The method has no effect. b. There is sufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years. 564 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES c. There is sufficient evidence to conclude that the method increases the proportion of HIV positive patients who develop AIDS after four years. d. There is insufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years. Use the following information to |
answer the next two exercises. An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a “biofeedback exercise program.” Six subjects were randomly selected and blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated (after - before) producing the following results: x¯ d = −10.2 sd = 8.4. Using the data, test the hypothesis that the blood pressure has decreased after the training. 118. The distribution for the test is: t5 t6 a. b. c. N(−10.2, 8.4) d. N(−10.2, 8.4 6 ) 119. If α = 0.05, the p-value and the conclusion are a. 0.0014; There is sufficient evidence to conclude that the blood pressure decreased after the training. b. 0.0014; There is sufficient evidence to conclude that the blood pressure increased after the training. c. 0.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training. d. 0.0155; There is sufficient evidence to conclude that the blood pressure increased after the training. 120. A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as follows. Player 1 Player 2 Player 3 Player 4 Mean score before class 83 Mean score after class 80 78 80 93 86 87 86 Table 10.31 The correct decision is: a. Reject H0. b. Do not reject the H0. 121. A local cancer support group believes that the estimate for new female breast cancer cases in the south is higher in 2013 than in 2012. The group compared the estimates of new female breast cancer cases by southern state in 2012 and in 2013. The results are in Table 10.32. Southern States 2012 2013 Alabama Arkansas Florida Georgia Kentucky Louisiana 3,450 3,720 2,150 2,280 15,540 15,710 6,970 7,310 3,160 3,300 3,320 3,630 Mississippi 1,990 2,080 North Carolina 7,090 7,430 Oklahoma 2,630 2,690 Table 10.32 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 565 Southern States 2012 2013 South Carolina 3,570 3,580 Tennessee 4,680 5,070 Texas Virginia Table 10.32 15,050 14,980 6,190 6,280 122. A traveler wanted to know if the prices of hotels are different in the ten cities that he visits the most often. The list of the cities with the corresponding hotel prices for his two favorite hotel chains is in Table 10.33. Test at the 1% level of significance. Hyatt Regency prices in dollars Hilton prices in dollars Cities Atlanta Boston Chicago Dallas Denver Indianapolis Los Angeles New York City Philadelphia 107 358 209 209 167 179 179 625 179 Washington, DC 245 Table 10.33 169 289 299 198 169 214 169 459 159 239 123. A politician asked his staff to determine whether the underemployment rate in the northeast decreased from 2011 to 2012. The results are in Table 10.34. Northeastern States 2011 2012 Connecticut Delaware Maine Maryland Massachusetts New Hampshire New Jersey New York Ohio Pennsylvania Rhode Island Vermont West Virginia Table 10.34 17.3 17.4 19.3 16.0 17.6 15.4 19.2 18.5 18.2 16.5 20.7 14.7 15.5 16.4 13.7 16.1 15.5 18.2 13.5 18.7 18.7 18.8 16.9 22.4 12.3 17.3 566 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES BRINGING IT TOGETHER: HOMEWORK Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. b. independent group means, population standard deviations and/or variances known independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion 124. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. 125. A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it |
. 126. The mean number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from nine males and 16 females. 127. A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased. 128. A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and one year, respectively. 129. According to a YWCA Rape Crisis Center newsletter, 75% of rape victims know their attackers. A study is done to verify this. 130. According to a recent study, U.S. companies have a mean maternity-leave of six weeks. 131. A recent drug survey showed an increase in use of drugs and alcohol among local high school students as compared to the national percent. Suppose that a survey of 100 local youths and 100 national youths is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. 132. A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected: Pre-course score Post-course score 1 960 1010 840 1100 1250 860 1330 790 990 1110 Table 10.35 300 920 1100 880 1070 1320 860 1370 770 1040 1200 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 567 Pre-course score Post-course score 740 850 Table 10.35 133. University of Michigan researchers reported in the Journal of the National Cancer Institute that quitting smoking is especially beneficial for those under age 49. In this American Cancer Society study, the risk (probability) of dying of lung cancer was about the same as for those who had never smoked. 134. Lesley E. Tan investigated the relationship between left-handedness vs. right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and |
41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown Table 10.36. Determine the appropriate test and best distribution to use for that test. Left-handed Right-handed Sample size Sample mean 41 97.5 Sample standard deviation 17.5 41 98.1 19.2 Table 10.36 a. Two independent means, normal distribution b. Two independent means, Student’s-t distribution c. Matched or paired samples, Student’s-t distribution d. Two population proportions, normal distribution 135. A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four (4) new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as Table 10.37. Player 1 Player 2 Player 3 Player 4 Mean score before class 83 Mean score after class 80 78 80 93 86 87 86 Table 10.37 This is: a. a test of two independent means. b. a test of two proportions. c. a test of a single mean. d. a test of a single proportion. REFERENCES 10.1 Two Population Means with Unknown Standard Deviations Data from Graduating Engineer + Computer Careers. Available online at http://www.graduatingengineer.com Data from Microsoft Bookshelf. Data from the United States Senate website, available online at www.Senate.gov (accessed June 17, 2013). “List of current United States Senators by Age.” Wikipedia. Available online at http://en.wikipedia.org/wiki/ List_of_current_United_States_Senators_by_age (accessed June 17, 2013). 568 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES “Sectoring sectors.aspx?page=sectors&base=industry (accessed June 17, 2013). Industry Groups.” Nasdaq. Available by online at http://www.nasdaq.com/markets/barchart- “Strip Clubs: Where Prostitution and Trafficking Happen.” Prostitution Research and Education, 2013. Available online at www.prostitutionresearch.com/ProsViolPosttrauStress.html (accessed June |
17, 2013). “World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseball-almanac.com/ws/wsmenu.shtml (accessed June 17, 2013). 10.2 Two Population Means with Known Standard Deviations Data from the United States Census Bureau. Available online at http://www.census.gov/prod/cen2010/briefs/c2010br-02.pdf Hinduja, Sameer. “Sexting Research and Gender Differences.” Cyberbulling Research Center, 2013. Available online at http://cyberbullying.us/blog/sexting-research-and-gender-differences/ (accessed June 17, 2013). “Smart Phone Users, By the Numbers.” Visually, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed June 17, 2013). Smith, Aaron. “35% of American adults own a Smartphone.” Pew Internet, 2013. Available online http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf (accessed June 17, 2013). at “State-Specific Prevalence of Obesity AmongAduls—Unites States, 2007.” MMWR, CDC. Available online at http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm (accessed June 17, 2013). “Texas Crime Rates 1960–1012.” FBI, Uniform Crime Reports, 2013. Available online at: http://www.disastercenter.com/ crime/txcrime.htm (accessed June 17, 2013). 10.3 Comparing Two Independent Population Proportions Data from Educational Resources, December catalog. Data from Hilton Hotels. Available online at http://www.hilton.com (accessed June 17, 2013). Data from Hyatt Hotels. Available online at http://hyatt.com (accessed June 17, 2013). Data from Statistics, United States Department of Health and Human Services. Data from Whitney Exhibit on loan to San Jose Museum of Art. Data from the American Cancer Society. Available online at http://www.cancer.org/index (accessed June 17, 2013). Data from the Chancellor’s Office, California Community Colleges, November 1994 |
. “State States.aspx?ref=interactive (accessed June 17, 2013). States.” Gallup, the of 2013. Available online at http://www.gallup.com/poll/125066/State- “West Nile Virus.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/ncidod/dvbid/ westnile/index.htm (accessed June 17, 2013). SOLUTIONS 1 two proportions 3 matched or paired samples 5 single mean 7 independent group means, population standard deviations and/or variances unknown 9 two proportions 11 independent group means, population standard deviations and/or variances unknown 13 independent group means, population standard deviations and/or variances unknown 15 two proportions 17 The random variable is the difference between the mean amounts of sugar in the two soft drinks. 19 means 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 569 21 two-tailed 23 the difference between the mean life spans of whites and nonwhites 25 This is a comparison of two population means with unknown population standard deviations. 27 Check student’s solution. 29 a. Reject the null hypothesis b. p-value < 0.05 c. There is not enough evidence at the 5% level of significance to support the claim that life expectancy in the 1900s is different between whites and nonwhites. 31 The difference in mean speeds of the fastball pitches of the two pitchers 33 –2.46 35 At the 1% significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s. 37 Subscripts: 1 = Food, 2 = No Food H0: μ1 ≤ μ2 Ha: μ1 > μ2 39 Figure 10.18 41 Subscripts: 1 = Gamma, 2 = Zeta H0: μ1 = μ2 Ha: μ1 ≠ μ2 43 0.0062 45 There is sufficient evidence to reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma. 47 P′OS1 – P′OS2 = difference in the proportions of phones that had system failures within the |
first eight hours of operation with OS1 and OS2. 49 0.1018 51 proportions 53 right-tailed 55 The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota. 570 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 57 Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test. 59 Check student’s solution. 61 a. Reject the null hypothesis. b. p-value < alpha c. At the 5% significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota. 63 the mean difference of the system failures 65 0.0067 67 With a p-value 0.0067, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures. 69 0.0021 71 Figure 10.19 73 H0: μd ≥ 0 Ha: μd < 0 75 0.0699 77 We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective. 79 Subscripts: 1: two-year colleges; 2: four-year colleges a. H0: μ1 ≥ μ2 b. Ha: μ1 < μ2 ¯ c. X ¯ 1 – X 2 d. Student’s-t is the difference between the mean enrollments of the two-year colleges and the four-year colleges. e. test statistic: -0.2480 f. p-value: 0.4019 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject iii. Reason for Decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges. 81 Subscripts: 1: mechanical engineering; 2: electrical engineering 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 571 a. H0: µ1 ≥ µ2 b. |
Ha: µ1 < µ2 ¯ c. X ¯ 1 − X 2 is the difference between the mean entry level salaries of mechanical engineers and electrical engineers. d. e. t108 test statistic: t = –0.82 f. p-value: 0.2061 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for Decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean entry-level salaries of mechanical engineers is lower than that of electrical engineers. 83 a. H0: µ1 = µ2 b. Ha: µ1 ≠ µ2 ¯ c. X ¯ 1 − X 2 is the difference between the mean times for completing a lap in races and in practices. d. e. t20.32 test statistic: –4.70 f. p-value: 0.0001 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 mean time for completing a lap in races is different from that in practices. 85 a. H0: µ1 = µ2 b. Ha: µ1 ≠ µ2 c. d. e. is the difference between the mean times for completing a lap in races and in practices. t40.94 test statistic: –5.08 f. p-value: zero 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 mean time for completing a lap in races is different from that in practices. 88 c 572 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES ¯ 90 Test: two independent sample means, population standard deviations unknown. Random variable: X 2 Distribution: H0: μ1 = μ2 Ha: μ1 < μ2 The mean age of entering prostitution in Canada is lower than the mean age in the United States. ¯ 1 − X Figure 10.20 Graph: left-tailed p-value : 0.0151 Decision: Do not |
reject H0. Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of entering prostitution in Canada is lower than the mean age in the United States. 92 d 94 Subscripts: 1 = boys, 2 = girls a. H0: µ1 ≤ µ2 b. Ha: µ1 > µ2 c. The random variable is the difference in the mean auto insurance costs for boys and girls. d. normal e. test statistic: z = 2.50 f. p-value: 0.0062 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 mean cost of auto insurance for teenage boys is greater than that for girls. 96 Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans a. H0: µ1 ≥ µ2 b. Ha: µ1 < µ2 c. The random variable is the difference in the mean miles per gallon of non-hybrid sedans and hybrid sedans. d. normal e. test statistic: 6.36 f. p-value: 0 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 mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans. 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 573 98 a. H0: µd = 0 b. Ha: µd < 0 c. The random variable Xd is the average difference between husband’s and wife’s satisfaction level. d. e. t9 test statistic: t = –1.86 f. p-value: 0.0479 g. Check student’s solution h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis, but run another test. iii |
. Reason for Decision: p-value < alpha iv. Conclusion: This is a weak test because alpha and the p-value are close. However, there is insufficient evidence to conclude that the mean difference is negative. 100 a. H0: PW = PB b. Ha: PW ≠ PB c. The random variable is the difference in the proportions of white and black suicide victims, aged 15 to 24. d. normal for two proportions e. test statistic: –0.1944 f. p-value: 0.8458 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 insufficient evidence to conclude that the proportions of white and black female suicide victims, aged 15 to 24, are different. 102 Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College a. H0: p1 = p2 b. Ha: p1 ≠ p2 c. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College. d. normal for two proportions e. test statistic: 4.29 f. p-value: 0.00002 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: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different. 104 a 106 Test: two independent sample proportions. Random variable: p′1 - p′2 Distribution: H0: p1 = p2 574 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES Ha: p1 ≠ p2 The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users. Graph: two-tailed Figure 10.21 p-value : 0.0033 Decision: Reject the null hypothesis. Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of eReader users 16 to 29 years old is different from the proportion of eReader users 30 and older. 108 Test: two independent sample proportions Random variable: p′1 − p′2 |
Distribution: H0: p1 = p2 Ha: p1 > p2 A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. Graph: righttailed Figure 10.22 p-value: 0.2354 Decision: Do not reject the H0. Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. 110 Subscripts: 1: men; 2: women a. H0: p1 ≤ p2 b. Ha: p1 > p2 c. P′1 − P′2 is the difference between the proportions of men and women who enjoy shopping for electronic equipment. d. normal for two proportions e. test statistic: 0.22 f. p-value: 0.4133 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for Decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women. 112 a. H0: p1 = p2 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 575 b. Ha: p1 ≠ p2 c. P′1 − P′2 is the difference between the proportions of men and women that have at least one pierced ear. d. normal for two proportions e. test statistic: –4.82 f. p-value: zero 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 proportions of males and females with at least one pierced ear is different. 114 a. H0: µd = 0 b. Ha: µd > 0 c. The random variable Xd is the mean difference in work times on days when eating breakfast and on days when not eating breakfast. t9 test |
statistic: 4.8963 d. e. f. p-value: 0.0004 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% level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased. 115 p-value = 0.1494 At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks. 117 b 119 c ¯ 121 Test: two matched pairs or paired samples (t-test) Random variable: X d Distribution: t12 H0: μd = 0 Ha: μd > 0 The mean of the differences of new female breast cancer cases in the south between 2013 and 2012 is greater than zero. The estimate for new female breast cancer cases in the south is higher in 2013 than in 2012. Graph: right-tailed p-value: 0.0004 Figure 10.23 576 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES Decision: Reject H0 Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in 2013 than in 2012. 123 Test: matched or paired samples (t-test) Difference data: {–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, ¯ 1.8} Random Variable: X d Distribution: H0: μd = 0 Ha: μd < 0 The mean of the differences of the rate of underemployment in the northeastern states between 2012 and 2011 is less than zero. The underemployment rate went down from 2011 to 2012. Graph: left-tailed. Figure 10.24 p-value: 0.1207 Decision: Do not reject H0. Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from 2011 to 2012. 125 e 127 d 129 f 131 e 133 f |
135 a This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 577 11 | THE CHI-SQUARE DISTRIBUTION Figure 11.1 The chi-square distribution can be used to find relationships between two things, like grocery prices at different stores. (credit: Pete/flickr) Introduction Chapter Objectives By the end of this chapter, the student should be able to: Interpret the chi-square probability distribution as the sample size changes. • • Conduct and interpret chi-square goodness-of-fit hypothesis tests. • Conduct and interpret chi-square test of independence hypothesis tests. • Conduct and interpret chi-square homogeneity hypothesis tests. • Conduct and interpret chi-square single variance hypothesis tests. Have you ever wondered if lottery numbers were evenly distributed or if some numbers occurred with a greater frequency? How about if the types of movies people preferred were different across different age groups? What about if a coffee machine was dispensing approximately the same amount of coffee each time? You could answer these questions by conducting a hypothesis test. 578 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION You will now study a new distribution, one that is used to determine the answers to such questions. This distribution is called the chi-square distribution. In this chapter, you will learn the three major applications of the chi-square distribution: 1. 2. 3. the goodness-of-fit test, which determines if data fit a particular distribution, such as in the lottery example the test of independence, which determines if events are independent, such as in the movie example the test of a single variance, which tests variability, such as in the coffee example NOTE Though the chi-square distribution depends on calculators or computers for most of the calculations, there is a table available (see Appendix G). TI-83+ and TI-84 calculator instructions are included in the text. Look in the sports section of a newspaper or on the Internet for some sports data (baseball averages, basketball scores, golf tournament scores, football odds, swimming times, and the like). Plot a histogram and a boxplot using your data. See if you can determine a probability distribution that your data fits. Have a discussion with the class about your choice. 11.1 | Facts About the Chi-Square Distribution The |
notation for the chi-square distribution is: 2 χ ∼ χd f where df = degrees of freedom which depends on how chi-square is being used. (If you want to practice calculating chisquare probabilities then use df = n - 1. The degrees of freedom for the three major uses are each calculated differently.) For the χ2 distribution, the population mean is μ = df and the population standard deviation is σ = 2(d f ). The random variable is shown as χ2, but may be any upper case letter. The random variable for a chi-square distribution with k degrees of freedom is the sum of k independent, squared standard normal variables. χ2 = (Z1)2 + (Z2)2 +... + (Zk)2 1. The curve is nonsymmetrical and skewed to the right. 2. There is a different chi-square curve for each df. Figure 11.2 3. The test statistic for any test is always greater than or equal to zero. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 4. When df > 90, the chi-square curve approximates the normal distribution. For X ~ χ1,000 2 the mean, μ = df = 1,000 and the standard deviation, σ = 2(1,000) = 44.7. Therefore, X ~ N(1,000, 44.7), approximately. 5. The mean, μ, is located just to the right of the peak. CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 579 Figure 11.3 11.2 | Goodness-of-Fit Test In this type of hypothesis test, you determine whether the data "fit" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities. The test statistic for a goodness-of-fit test is: (O − E)2 E Σ k where: • O = observed values (data) • E = expected values (from theory) • k = the number |
of different data cells or categories The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true. There are n terms of the form (O − E)2. E The number of degrees of freedom is df = (number of categories – 1). The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve. NOTE The expected value for each cell needs to be at least five in order for you to use this test. Example 11.1 Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to Table 11.1. 580 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION Number of absences per term Expected number of students 0–2 3–5 6–8 9–11 12+ Table 11.1 50 30 12 6 2 A random survey across all mathematics courses was then done to determine the actual number (observed) of absences in a course. The chart in Table 11.2 displays the results of that survey. Number of absences per term Actual number of students 0–2 3–5 6–8 9–11 12+ Table 11.2 35 40 20 1 4 Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test. H0: Student absenteeism fits faculty perception. The alternative hypothesis is the opposite of the null hypothesis. Ha: Student absenteeism does not fit faculty perception. a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test? Solution 11.1 a. No. Notice that the expected number of absences for the "12+" entry is less than five (it is two). Combine that group with the "9–11" group to create new tables where the number of students for each entry are at least five. The new results are in Table 11.2 and Table 11.3. Number of absences per term Expected number of students 0–2 3–5 6–8 9+ Table 11.3 50 30 12 8 Number of absences |
per term Actual number of students 0–2 35 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 581 Number of absences per term Actual number of students 3–5 6–8 9+ Table 11.4 40 20 5 b. What is the number of degrees of freedom (df)? Solution 11.1 b. There are four "cells" or categories in each of the new tables. df = number of cells – 1 = 4 – 1 = 3 11.1 A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in Table 11.5. Number produced Number defective 0–100 101–200 201–300 301–400 401–500 Table 11.5 5 6 7 8 10 A random sample was taken to determine the actual number of defects. Table 11.6 shows the results of the survey. Number produced Number defective 0–100 101–200 201–300 301–400 401–500 Table 11.6 5 7 8 9 11 State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom. 582 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION Example 11.2 Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in Table 11.6. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level. Monday Tuesday Wednesday Thursday Friday Number of Absences 15 12 9 9 15 Table 11.7 Day of the Week Employees were Most Absent Solution 11.2 The null and alternative hypotheses are: • H0: The absent days occur with equal frequencies, that is, they fit a uniform distribution. • Ha: The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution. If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 |
+ 9 + 15 = 60), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected (E) values. The values in the table are the observed (O) values or data. This time, calculate the χ2 test statistic by hand. Make a chart with the following headings and fill in the columns: • Expected (E) values (12, 12, 12, 12, 12) • Observed (O) values (15, 12, 9, 9, 15) • • • (O – E) (O – E)2 (O – E)2 E Now add (sum) the last column. The sum is three. This is the χ2 test statistic. To find the p-value, calculate P(χ2 > 3). This test is right-tailed. (Use a computer or calculator to find the p-value. You should get p-value = 0.5578.) The dfs are the number of cells – 1 = 5 – 1 = 4 Press 2nd DISTR. Arrow down to χ2cdf. Press ENTER. Enter (3,10^99,4). Rounded to four decimal places, you should see 0.5578, which is the p-value. Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 583 Figure 11.4 The decision is not to reject the null hypothesis. Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies. TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-offit test. The next example Example 11.3 has the calculator instructions. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi |
2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start. To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of the particular list. Press CLEAR and then arrow down. The list will be cleared. Alternatively, you can press STAT and press 4 (for ClrList). Enter the list name and press ENTER. 11.2 Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 49 students were asked on which night of the week they did the most homework. The results were distributed as in Table 11.8. Sunday Monday Tuesday Wednesday Thursday Friday Saturday 11 8 10 7 10 5 5 Number of Students Table 11.8 From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use? 584 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION Example 11.3 One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as in Table 11.9. Number of Televisions Percent 0 1 2 3 4+ Table 11.9 10 16 55 11 8 The table contains expected (E) percents. A random sample of 600 families in the far western United States resulted in the data in Table 11.10. Number of Televisions Frequency 0 1 2 3 4+ Table 11.10 66 119 340 60 15 Total = 600 The table contains observed (O) frequency values. At the 1% significance level, does it appear that the distribution "number of televisions" of far western United States families is different from the distribution for the American population as a whole? Solution 11.3 This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed. The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are shown in Table 11.10. Number of Televisions Percent Expected Frequency 0 1 2 3 over 3 Table 11.11 10 16 55 11 8 (0.10)(600) = 60 (0.16)(600) = |
96 (0.55)(600) = 330 (0.11)(600) = 66 (0.08)(600) = 48 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 585 Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10*600. H0: The "number of televisions" distribution of far western United States families is the same as the "number of televisions" distribution of the American population. Ha: The "number of televisions" distribution of far western United States families is different from the "number of televisions" distribution of the American population. Distribution for the test: χ4 2 where df = (the number of cells) – 1 = 5 – 1 = 4. NOTE df ≠ 600 – 1 Calculate the test statistic: χ2 = 29.65 Graph: Figure 11.5 Probability statement: p-value = P(χ2 > 29.65) = 0.000006 Compare α and the p-value: • α = 0.01 • p-value = 0.000006 So, α > p-value. Make a decision: Since α > p-value, reject Ho. This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole. Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the "number of televisions" distribution for the far western United States is different from the "number of televisions" distribution for the American population as a whole. 586 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION Press STAT and ENTER. Make sure to clear lists L1, L2, and L3 if they have data in them (see the note at the end of Example 11.2). Into L1, put the observed frequencies 66, 119, 349, 60, 15. Into L2, put the expected frequencies.10*600,.16*600,.55*600,.11*600,.08*600. Arrow over to list L3 and up |
to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum" (Enter L3). Rounded to 2 decimal places, you should see 29.65. Press 2nd DISTR. Press 7 or Arrow down to 7:χ2cdf and press ENTER. Enter (29.65,1E99,4). Rounded to four places, you should see 5.77E-6 =.000006 (rounded to six decimal places), which is the p-value. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start. 11.3 The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in Table 11.12. Number of Pets Percent 0 1 2 3 4+ Table 11.12 18 25 30 18 9 A random sample of 1,000 students from the Eastern United States resulted in the data in Table 11.13. Number of Pets Frequency 0 1 2 3 4+ Table 11.13 210 240 320 140 90 At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is 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 11 | THE CHI-SQUARE DISTRIBUTION 587 Example 11.4 Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the coins fair? Test at a 5% significance level. Solution 11.4 This problem can be set up as a goodness-of-fit problem. The sample space for |
flipping two fair coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25 HH, 25 HT, 25 TH, and 25 TT. This is the expected distribution. The question, "Are the coins fair?" is the same as saying, "Does the distribution of the coins (20 HH, 27 HT, 30 TH, 23 TT) fit the expected distribution?" Random Variable: Let X = the number of heads in one flip of the two coins. X takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the number of cells is three. Since X = the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed. H0: The coins are fair. Ha: The coins are not fair. Distribution for the test: χ2 Calculate the test statistic: χ2 = 2.14 2 where df = 3 – 1 = 2. Graph: Figure 11.6 Probability statement: p-value = P(χ2 > 2.14) = 0.3430 Compare α and the p-value: • α = 0.05 • p-value = 0.3430 α < p-value. Make a decision: Since α < p-value, do not reject H0. Conclusion: There is insufficient evidence to conclude that the coins are not fair. Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed frequencies 20, 57, 23. Into L2, put the expected frequencies 25, 50, 25. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum".Enter L3. Rounded to two decimal places, you should see 2.14. Press 2nd DISTR. Arrow down to 7:χ2cdf (or press 7). Press ENTER. Enter 2.14,1E99,2). Rounded to four places, you |
should see.3430, which is the p-value. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) 588 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start. 11.4 Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. Table 11.14 shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom? MDG Region Adult Literacy Rate (%) Developed Regions 99.0 Commonwealth of Independent States 99.5 Northern Africa Sub-Saharan Africa Latin America and the Caribbean Eastern Asia Southern Asia South-Eastern Asia Western Asia Oceania Table 11.14 67.3 62.5 91.0 93.8 61.9 91.9 84.5 66.4 11.3 | Test of Independence Tests of independence involve using a contingency table of observed (data) values. The test statistic for a test of independence is similar to that of a goodness-of-fit test: (O – E)2 E Σ (i ⋅ j) where: • O = observed values • E = expected values • i = the number of rows in the table • j = the number of columns in the table There are i ⋅ j terms of the form (O – E)2 E. A test of independence determines whether two factors are independent or not. You first encountered the term independence in Section 3.. As a review, consider the following example. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 NOTE The expected value for each cell needs to be at least five in order for you to use this test. CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 589 Example 11.5 Suppose A = a speeding violation in the last year and B = a cell phone user while driving. If A and B are independent then P |
(A AND B) = P(A)P(B). A AND B is the event that a driver received a speeding violation last year and also used a cell phone while driving. Suppose, in a study of drivers who received speeding violations in the last year, and who used cell phone while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305 used cell phones while driving and 450 did not. Let y = expected number of drivers who used a cell phone while driving and received speeding violations. If A and B are independent, then P(A AND B) = P(A)P(B). By substitution, y 755 = ⎛ ⎝ 70 755 ⎛ ⎞ ⎝ ⎠ 305 755 ⎞ ⎠ Solve for y: y = (70)(305) 755 = 28.3 About 28 people from the sample are expected to use cell phones while driving and to receive speeding violations. In a test of independence, we state the null and alternative hypotheses in words. Since the contingency table consists of two factors, the null hypothesis states that the factors are independent and the alternative hypothesis states that they are not independent (dependent). If we do a test of independence using the example, then the null hypothesis is: H0: Being a cell phone user while driving and receiving a speeding violation are independent events. If the null hypothesis were true, we would expect about 28 people to use cell phones while driving and to receive a speeding violation. The test of independence is always right-tailed because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chi-square curve, as it is in a goodness-of-fit. The number of degrees of freedom for the test of independence is: df = (number of columns - 1)(number of rows - 1) The following formula calculates the expected number (E): E = (row total)(column total) total number surveyed 11.5 A sample of 300 students is taken. Of the students surveyed, 50 were music students, while 250 were not. Ninetyseven were on the honor roll, while 203 were not. If we assume being a music student and being on the honor roll are independent events, what is the expected number of music students who are also on the honor roll? Example 11.6 |
In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend time with a disabled senior citizen. The program recruits among community college students, four-year college students, and nonstudents. In Table 11.15 is a sample of the adult volunteers and the number of hours they volunteer per week. 590 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION Type of Volunteer 1–3 Hours 4–6 Hours 7–9 Hours Row Total Community College Students 111 Four-Year College Students Nonstudents Column Total 96 91 298 96 133 150 379 48 61 53 162 255 290 294 839 Table 11.15 Number of Hours Worked Per Week by Volunteer Type (Observed) The table contains observed (O) values (data). Is the number of hours volunteered independent of the type of volunteer? Solution 11.6 The observed table and the question at the end of the problem, "Is the number of hours volunteered independent of the type of volunteer?" tell you this is a test of independence. The two factors are number of hours volunteered and type of volunteer. This test is always right-tailed. H0: The number of hours volunteered is independent of the type of volunteer. Ha: The number of hours volunteered is dependent on the type of volunteer. The expected result are in Table 11.15. Type of Volunteer 1-3 Hours 4-6 Hours 7-9 Hours Community College Students 90.57 Four-Year College Students 103.00 Nonstudents 104.42 115.19 131.00 132.81 49.24 56.00 56.77 Table 11.16 Number of Hours Worked Per Week by Volunteer Type (Expected) The table contains expected (E) values (data). For example, the calculation for the expected frequency for the top left cell is E = (row total)(column total) total number surveyed = (255)(298) 839 = 90.57 Calculate the test statistic: χ2 = 12.99 (calculator or computer) 2 Distribution for the test: χ4 df = (3 columns – 1)(3 rows – 1) = (2)(2) = 4 Graph: Figure 11.7 Probability statement: p-value=P(χ2 > 12.99) = 0.0113 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562 |
/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 591 Compare α and the p-value: Since no α is given, assume α = 0.05. p-value = 0.0113. α > p-value. Make a decision: Since α > p-value, reject H0. This means that the factors are not independent. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer are dependent on one another. For the example in Table 11.15, if there had been another type of volunteer, teenagers, what would the degrees of freedom be? Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 3 ENTER 3 ENTER. Enter the table values by row from Table 11.15. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 12.9909 and the p-value = 0.0113. Do the procedure a second time, but arrow down to Draw instead of calculate. 11.6 The Bureau of Labor Statistics gathers data about employment in the United States. A sample is taken to calculate the number of U.S. citizens working in one of several industry sectors over time. Table 11.17 shows the results: Industry Sector Nonagriculture wage and salary Goods-producing, excluding agriculture Services-providing Agriculture, forestry, fishing, and hunting Nonagriculture self-employed and unpaid family worker 2000 2010 2020 Total 13,243 13,044 15,018 41,305 2,457 1,771 1,950 6,178 10,786 11,273 13,068 35,127 240 931 214 894 201 972 655 2,797 Secondary wage and salary jobs in agriculture and private household industries 14 11 11 36 Secondary jobs as a self-employed or unpaid family worker 196 144 152 492 Total Table 11.17 27,867 27,351 31,372 86,590 We want to know if the change in the number of jobs is independent of the change in years. State the null and alternative hypotheses and the degrees of freedom. Example 11.7 De Anza College is interested in the relationship between anxiety |
level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. Table 11.18 shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events. 592 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION Need to Succeed in School High Anxiety Medhigh Anxiety Medium Anxiety Medlow Anxiety Low Anxiety Row Total High Need Medium Need Low Need Column Total 35 18 4 57 42 48 5 95 53 63 11 127 15 33 15 63 10 31 17 58 155 193 52 400 Table 11.18 Need to Succeed in School vs. Anxiety Level a. How many high anxiety level students are expected to have a high need to succeed in school? Solution 11.7 a. The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400. E = (row total)(column total) total surveyed = 155 ⋅ 57 400 = 22.09 The expected number of students who have a high anxiety level and a high need to succeed in school is about 22. b. If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety? Solution 11.7 b. The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400. c. E = (row total)(column total) total surveyed = ________ Solution 11.7 c. E = (row total)(column total) total surveyed = 8.19 d. The expected number of students who have a med-low anxiety level and a low need to succeed in school is about ________. Solution 11.7 d. 8 11.7 Refer back to the information in Try It. How many service providing jobs are there expected to be in 2020? How many nonagriculture wage and salary jobs are there expected to be in 2020? 11.4 | Test for Homogeneity The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a |
test for homogeneity, follow the same procedure as with the test of independence. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 NOTE The expected value for each cell needs to be at least five in order for you to use this test. CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 593 Hypotheses H0: The distributions of the two populations are the same. Ha: The distributions of the two populations are not the same. Test Statistic Use a χ 2 test statistic. It is computed in the same way as the test for independence. Degrees of Freedom (df) df = number of columns - 1 Requirements All values in the table must be greater than or equal to five. Common Uses Comparing two populations. For example: men vs. women, before vs. after, east vs. west. The variable is categorical with more than two possible response values. Example 11.8 Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are shown in Table 11.18. Do male and female college students have the same distribution of living arrangements? Dormitory Apartment With Parents Other Males 72 Females 91 84 86 49 88 45 35 Table 11.19 Distribution of Living Arragements for College Males and College Females Solution 11.8 H0: The distribution of living arrangements for male college students is the same as the distribution of living arrangements for female college students. Ha: The distribution of living arrangements for male college students is not the same as the distribution of living arrangements for female college students. Degrees of Freedom (df): df = number of columns – 1 = 4 – 1 = 3 2 Distribution for the test: χ3 Calculate the test statistic: χ2 = 10.1287 (calculator or computer) Probability statement: p-value = P(χ2 >10.1287) = 0.0175 594 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 2 ENTER 4 ENTER. Enter the table values by row. Press ENTER after |
each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 10.1287 and the p-value = 0.0175. Do the procedure a second time but arrow down to Draw instead of calculate. Compare α and the p-value: Since no α is given, assume α = 0.05. p-value = 0.0175. α > p-value. Make a decision: Since α > p-value, reject H0. This means that the distributions are not the same. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the distributions of living arrangements for male and female college students are not the same. Notice that the conclusion is only that the distributions are not the same. We cannot use the test for homogeneity to draw any conclusions about how they differ. 11.8 Do families and singles have the same distribution of cars? Use a level of significance of 0.05. Suppose that 100 randomly selected families and 200 randomly selected singles were asked what type of car they drove: sport, sedan, hatchback, truck, van/SUV. The results are shown in Table 11.20. Do families and singles have the same distribution of cars? Test at a level of significance of 0.05. Sport Sedan Hatchback Truck Van/SUV Family 5 Single 45 Table 11.20 15 65 35 37 17 46 28 7 Example 11.9 Both before and after a recent earthquake, surveys were conducted asking voters which of the three candidates they planned on voting for in the upcoming city council election. Has there been a change since the earthquake? Use a level of significance of 0.05. Table 11.20 shows the results of the survey. Has there been a change in the distribution of voter preferences since the earthquake? Perez Chung Stevens Before 167 After 214 128 197 135 225 Table 11.21 Solution 11.9 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 595 H0: The distribution of voter preferences was the same before and after the earthquake. Ha: The distribution of |
voter preferences was not the same before and after the earthquake. Degrees of Freedom (df): df = number of columns – 1 = 3 – 1 = 2 2 Distribution for the test: χ2 Calculate the test statistic: χ2 = 3.2603 (calculator or computer) Probability statement: p-value=P(χ2 > 3.2603) = 0.1959 Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 2 ENTER 3 ENTER. Enter the table values by row. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 3.2603 and the p-value = 0.1959. Do the procedure a second time but arrow down to Draw instead of calculate. Compare α and the p-value: α = 0.05 and the p-value = 0.1959. α < p-value. Make a decision: Since α < p-value, do not reject Ho. Conclusion: At a 5% level of significance, from the data, there is insufficient evidence to conclude that the distribution of voter preferences was not the same before and after the earthquake. 11.9 Ivy League schools receive many applications, but only some can be accepted. At the schools listed in Table 11.22, two types of applications are accepted: regular and early decision. Application Type Accepted Brown Columbia Cornell Dartmouth Penn Yale Regular Early Decision Table 11.22 2,115 1,792 577 627 5,306 1,228 1,734 444 2,685 1,245 1,195 761 We want to know if the number of regular applications accepted follows the same distribution as the number of early applications accepted. State the null and alternative hypotheses, the degrees of freedom and the test statistic, sketch the graph of the p-value, and draw a conclusion about the test of homogeneity. 11.5 | Comparison of the Chi-Square Tests You have seen the χ2 test statistic used in three different circumstances. The following bulleted list is a summary that will help you decide which χ2 test is the appropriate one to use. • Goodness-of-Fit: Use the goodness-of-fit test to decide whether a population with an unknown distribution |
"fits" a known distribution. In this case there will be a single qualitative survey question or a single outcome of an experiment 596 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION from a single population. Goodness-of-Fit is typically used to see if the population is uniform (all outcomes occur with equal frequency), the population is normal, or the population is the same as another population with a known distribution. The null and alternative hypotheses are: H0: The population fits the given distribution. Ha: The population does not fit the given distribution. • Independence: Use the test for independence to decide whether two variables (factors) are independent or dependent. In this case there will be two qualitative survey questions or experiments and a contingency table will be constructed. The goal is to see if the two variables are unrelated (independent) or related (dependent). The null and alternative hypotheses are: H0: The two variables (factors) are independent. Ha: The two variables (factors) are dependent. • Homogeneity: Use the test for homogeneity to decide if two populations with unknown distributions have the same distribution as each other. In this case there will be a single qualitative survey question or experiment given to two different populations. The null and alternative hypotheses are: H0: The two populations follow the same distribution. Ha: The two populations have different distributions. 11.6 | Test of a Single Variance A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance (or population standard deviation). The test statistic is: ⎞ ⎠s2 ⎛ ⎝n - 1 σ 2 where: • n = the total number of data • s2 = sample variance • σ2 = population variance You may think of s as the random variable in this test. The number of degrees of freedom is df = n - 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. Example 11.10 will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance. Example 11.10 Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance (or standard deviation) may be more important than the average. Suppose a math instructor believes that the standard deviation for his final exam is five |
points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be? Solution 11.10 Even though we are given the population standard deviation, we can set up the test using the population variance as follows. • H0: σ2 = 52 • Ha: σ2 > 52 11.10 A SCUBA instructor wants to record the collective depths each of his students dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be? CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 597 Example 11.11 With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes. With a significance level of 5%, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers. Solution 11.11 Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ2, or the population standard deviation, σ. Random Variable: The sample standard deviation, s, is the random variable. Let s = standard deviation for the waiting times. • H0: σ2 = 7.22 • Ha: σ2 < 7.22 The word "less" tells you this is a left-tailed test. Distribution for the test: χ24 2, where: • n = the number of customers sampled • df = n – 1 = 25 – 1 = 24 Calculate the test statistic: χ 2 = (n − 1)s2 σ 2 = (25 − 1)(3.5)2 7.22 = 5.67 where n = 25, s = 3. |
5, and σ = 7.2. Graph: Figure 11.8 Probability statement: p-value = P ( χ2 < 5.67) = 0.000042 Compare α and the p-value: α = 0.05; p-value = 0.000042; α > p-value Make a decision: Since α > p-value, reject H0. This means that you reject σ2 = 7.22. In other words, you do not think the variation in waiting times is 7.2 minutes; you think the variation in waiting times is less. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes. 598 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION In 2nd DISTR, use 7:χ2cdf. The syntax is (lower, upper, df) for the parameter list. For Example 11.11, χ2cdf(-1E99,5.67,24). The p-value = 0.000042. 11.11 The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. As of August of 2012, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the p-value, and draw a conclusion. Test at the 1% significance level. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 599 11.1 Lab 1: Chi-Square Goodness-of-Fit Class Time: Names: Student Learning Outcome • The student will evaluate data collected to determine if they fit either the uniform or exponential distributions. Collect the Data Go to your local supermarket. Ask 30 people as they leave for the total amount on their grocery receipts. (Or, ask three cashiers for the |
last ten amounts. Be sure to include the express lane, if it is open.) NOTE You may need to combine two categories so that each cell has an expected value of at least five. 1. Record the values. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 11.23 2. Construct a histogram of the data. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Figure 11.9 3. Calculate the following: a. b. x¯ = ________ s = ________ 600 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION c. s2 = ________ Uniform Distribution Test to see if grocery receipts follow the uniform distribution. 1. Using your lowest and highest values, X ~ U (_______, _______) 2. Divide the distribution into fifths. 3. Calculate the following: a. lowest value = _________ b. 20th percentile = _________ c. 40th percentile = _________ d. 60th percentile = _________ e. 80th percentile = _________ f. highest value = _________ 4. For each fifth, count the observed number of receipts and record it. Then determine the expected number of receipts and record that. Fifth Observed Expected 1st 2nd 3rd 4th 5th Table 11.24 5. H0: ________ 6. Ha: ________ 7. What distribution should you use for a hypothesis test? 8. Why did you choose this distribution? 9. Calculate the test statistic. 10. Find the p-value. 11. Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to 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 11 | THE CHI-SQUARE DISTRIBUTION 601 Figure 11.10 12. State your decision. 13. State your conclusion in a complete sentence |
. Exponential Distribution Test to see if grocery receipts follow the exponential distribution with decay parameter 1 x¯. 1. Using 1 x¯ as the decay parameter, X ~ Exp(_________). 2. Calculate the following: a. b. lowest value = ________ first quartile = ________ c. 37th percentile = ________ d. median = ________ e. 63rd percentile = ________ f. 3rd quartile = ________ g. highest value = ________ 3. For each cell, count the observed number of receipts and record it. Then determine the expected number of receipts and record that. Cell Observed Expected 1st 2nd 3rd 4th 5th 6th Table 11.25 4. H0: ________ 5. Ha: ________ 602 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 6. What distribution should you use for a hypothesis test? 7. Why did you choose this distribution? 8. Calculate the test statistic. 9. Find the p-value. 10. Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the p-value. Figure 11.11 11. State your decision. 12. State your conclusion in a complete sentence. Discussion Questions 1. Did your data fit either distribution? If so, which? 2. In general, do you think it’s likely that data could fit more than one distribution? In complete sentences, explain why or why not. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 603 11.2 Lab 2: Chi-Square Test of Independence Class Time: Names: Student Learning Outcome • The student will evaluate if there is a significant relationship between favorite type of snack and gender. Collect the Data 1. Using your class as a sample, complete the following chart. Ask each other what your favorite snack is, then total the results. NOTE You may need to combine two food categories so that each cell has an expected value of at least five. sweets (candy & baked goods) ice cream chips & pretzels fruits & vegetables Total male female Total Table 11.26 Favorite type of snack 2. Looking at Table 11.26, does it appear to you that there is a dependence between gender and favorite type |
of snack food? Why or why not? Hypothesis Test Conduct a hypothesis test to determine if the factors are independent: 1. H0: ________ 2. Ha: ________ 3. What distribution should you use for a hypothesis test? 4. Why did you choose this distribution? 5. Calculate the test statistic. 6. Find the p-value. 7. Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the p-value. 604 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION Figure 11.12 8. State your decision. 9. State your conclusion in a complete sentence. Discussion Questions 1. Is the conclusion of your study the same as or different from your answer to answer to question two under Collect the Data? 2. Why do you think that occurred? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 605 KEY TERMS Contingency Table a table that displays sample values for two different factors that may be dependent or contingent on one another; it facilitates determining conditional probabilities. CHAPTER REVIEW 11.1 Facts About the Chi-Square Distribution The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population. An important parameter in a chi-square distribution is the degrees of freedom df in a given problem. The random variable in the chi-square distribution is the sum of squares of df standard normal variables, which must be independent. The key characteristics of the chi-square distribution also depend directly on the degrees of freedom. The chi-square distribution curve is skewed to the right, and its shape depends on the degrees of freedom df. For df > 90, the curve approximates the normal distribution. Test statistics based on the chi-square distribution are always greater than or equal to zero. Such application tests are almost always right-tailed tests. 11.2 Goodness-of-Fit Test To assess whether a data set fits a specific distribution, you can apply the goodness-of-fit hypothesis |
test that uses the chi-square distribution. The null hypothesis for this test states that the data come from the assumed distribution. The test compares observed values against the values you would expect to have if your data followed the assumed distribution. The test is almost always right-tailed. Each observation or cell category must have an expected value of at least five. 11.3 Test of Independence To assess whether two factors are independent or not, you can apply the test of independence that uses the chi-square distribution. The null hypothesis for this test states that the two factors are independent. The test compares observed values to expected values. The test is right-tailed. Each observation or cell category must have an expected value of at least 5. 11.4 Test for Homogeneity To assess whether two data sets are derived from the same distribution—which need not be known, you can apply the test for homogeneity that uses the chi-square distribution. The null hypothesis for this test states that the populations of the two data sets come from the same distribution. The test compares the observed values against the expected values if the two populations followed the same distribution. The test is right-tailed. Each observation or cell category must have an expected value of at least five. 11.5 Comparison of the Chi-Square Tests The goodness-of-fit test is typically used to determine if data fits a particular distribution. The test of independence makes use of a contingency table to determine the independence of two factors. The test for homogeneity determines whether two populations come from the same distribution, even if this distribution is unknown. 11.6 Test of a Single Variance To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance (or standard deviation). FORMULA REVIEW 11.1 Facts About the Chi-Square Distribution χ2 = (Z1)2 + (Z2)2 + … (Zdf)2 chi-square distribution random variable μχ2 = df chi-square distribution population mean ⎝d f ⎞ ⎠ Chi-Square distribution population standard σ χ 2 = 2⎛ deviation 606 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 11.2 Goodness-of-Fit Test (O − E)2 E ∑ k goodness-of-fit test statistic where: O: observed values E: expected values k: number |
of different data cells or categories df = k − 1 degrees of freedom 11.3 Test of Independence Test of Independence • The number of degrees of freedom is equal to (number of columns - 1)(number of rows - 1). • The test statistic is Σ (i ⋅ j) (O – E)2 E where O = observed values, E = expected values, i = the number of rows in the table, and j = the number of columns in the table. • If the null hypothesis is true, the expected number E = (row total)(column total) total surveyed. 11.4 Test for Homogeneity PRACTICE Homogeneity test statistic where: O = (O − E)2 E ∑ i ⋅ j observed values E = expected values i = number of rows in data contingency table j = number of columns in data contingency table df = (i −1)(j −1) Degrees of freedom 11.6 Test of a Single Variance χ 2 = (n − 1) ⋅ s2 σ 2 Test of a single variance statistic where: n: sample size s: sample standard deviation σ: population standard deviation df = n – 1 Degrees of freedom Test of a Single Variance • Use the test to determine variation. • The degrees of freedom is the number of samples – 1. • The test statistic is (n – 1) ⋅ s2, where n = the total σ 2 number of data, s2 = sample variance, and σ2 = population variance. • The test may be left-, right-, or two-tailed. 11.1 Facts About the Chi-Square Distribution 1. If the number of degrees of freedom for a chi-square distribution is 25, what is the population mean and standard deviation? 2. If df > 90, the distribution is _____________. If df = 15, the distribution is ________________. 3. When does the chi-square curve approximate a normal distribution? 4. Where is μ located on a chi-square curve? 5. Is it more likely the df is 90, 20, or two in the graph? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 607 Figure 11.13 11.2 Goodness-of-Fit |
Test Determine the appropriate test to be used in the next three exercises. 6. An archeologist is calculating the distribution of the frequency of the number of artifacts she finds in a dig site. Based on previous digs, the archeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, she compares the actual number of artifacts found in each grid section to see if her expectation was accurate. 7. An economist is deriving a model to predict outcomes on the stock market. He creates a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, he records the actual points on the index. He wants to see how well his model matched what actually happened. 8. A personal trainer is putting together a weight-lifting program for her clients. For a 90-day program, she expects each client to lift a specific maximum weight each week. As she goes along, she records the actual maximum weights her clients lifted. She wants to know how well her expectations met with what was observed. Use the following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in Table 11.27. Grade Proportion A B C D 0.25 0.30 0.35 0.10 Table 11.27 The actual distribution for a class of 20 is in Table 11.28. Grade Frequency A B C D 7 7 5 1 Table 11.28 608 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 9. d f = ______ 10. State the null and alternative hypotheses. 11. χ2 test statistic = ______ 12. p-value = ______ 13. At the 5% significance level, what can you conclude? Use the following information to answer the next nine exercises: The following data are real. The cumulative number of AIDS cases reported for Santa Clara County is broken down by ethnicity as in Table 11.29. Ethnicity White Hispanic Number of Cases 2,229 1,157 Black/African-American 457 Asian, Pacific Islander 232 Total = 4,075 Table 11.29 The percentage of each ethnic group in Santa Clara County is as in Table 11.30. Ethnicity White Hispanic Black/AfricanAmerican Asian, Pacific Islander Table 11.30 Percentage of total county population Number expected (round to two decimal places) 1748.18 42.9% 26.7% 2.6% 27 |
.8% Total = 100% 14. If the ethnicities of AIDS victims followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group. Perform a goodness-of-fit test to determine whether the occurrence of AIDS cases follows the ethnicities of the general population of Santa Clara County. 15. H0: _______ 16. Ha: _______ 17. Is this a right-tailed, left-tailed, or two-tailed test? 18. degrees of freedom = _______ 19. χ2 test statistic = _______ 20. p-value = _______ 21. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to 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 11 | THE CHI-SQUARE DISTRIBUTION 609 Figure 11.14 Let α = 0.05 Decision: ________________ Reason for the Decision: ________________ Conclusion (write out in complete sentences): ________________ 22. Does it appear that the pattern of AIDS cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not? 11.3 Test of Independence Determine the appropriate test to be used in the next three exercises. 23. A pharmaceutical company is interested in the relationship between age and presentation of symptoms for a common viral infection. A random sample is taken of 500 people with the infection across different age groups. 24. The owner of a baseball team is interested in the relationship between player salaries and team winning percentage. He takes a random sample of 100 players from different organizations. 25. A marathon runner is interested in the relationship between the brand of shoes runners wear and their run times. She takes a random sample of 50 runners and records their run times as well as the brand of shoes they were wearing. Use the following information to answer the next seven exercises: Transit Railroads is interested in the relationship between travel distance and the ticket class purchased. A random sample of 200 passengers is taken. Table 11.31 shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance they must travel. Traveling Distance Third class Second class First class Total 1–100 miles 101–200 miles 201–300 miles 301–400 miles 401–500 miles Total Table |
11.31 21 18 16 12 6 73 14 16 17 14 6 67 6 8 15 21 10 60 41 42 48 47 22 200 26. State the hypotheses. H0: _______ Ha: _______ 27. df = _______ 28. How many passengers are expected to travel between 201 and 300 miles and purchase second-class tickets? 29. How many passengers are expected to travel between 401 and 500 miles and purchase first-class tickets? 610 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 30. What is the test statistic? 31. What is the p-value? 32. What can you conclude at the 5% level of significance? Use the following information to answer the next eight exercises: An article in the New England Journal of Medicine, discussed a study on smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans and 7,650 whites. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 whites. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 whites. Of the people smoking at least 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 whites. 33. Complete the table. Smoking Level Per Day African American Native Hawaiian Latino Japanese Americans White TOTALS 1-10 11-20 21-30 31+ TOTALS Table 11.32 Smoking Levels by Ethnicity (Observed) 34. State the hypotheses. H0: _______ Ha: _______ 35. Enter expected values in Table 11.32. Round to two decimal places. Calculate the following values: 36. df = _______ 37. χ 2 test statistic = ______ 38. p-value = ______ 39. Is this a right-tailed, left-tailed, or two-tailed test? Explain why. 40. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in |
the region corresponding to the p-value. Figure 11.15 State the decision and conclusion (in a complete sentence) for the following preconceived levels of α. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 611 41. α = 0.05 a. Decision: ___________________ b. Reason for the decision: ___________________ c. Conclusion (write out in a complete sentence): ___________________ 42. α = 0.01 a. Decision: ___________________ b. Reason for the decision: ___________________ c. Conclusion (write out in a complete sentence): ___________________ 11.4 Test for Homogeneity 43. A math teacher wants to see if two of her classes have the same distribution of test scores. What test should she use? 44. What are the null and alternative hypotheses for Exercise 11.43? 45. A market researcher wants to see if two different stores have the same distribution of sales throughout the year. What type of test should he use? 46. A meteorologist wants to know if East and West Australia have the same distribution of storms. What type of test should she use? 47. What condition must be met to use the test for homogeneity? Use the following information to answer the next five exercises: Do private practice doctors and hospital doctors have the same distribution of working hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are selected at random and asked about the number of hours a week they work. The results are shown in Table 11.33. 20–30 30–40 40–50 50–60 Private Practice 16 Hospital 8 40 44 38 59 6 39 Table 11.33 48. State the null and alternative hypotheses. 49. df = _______ 50. What is the test statistic? 51. What is the p-value? 52. What can you conclude at the 5% significance level? 11.5 Comparison of the Chi-Square Tests 53. Which test do you use to decide whether an observed distribution is the same as an expected distribution? 54. What is the null hypothesis for the type of test from Exercise 11.53? 55. Which test would you use to decide whether two factors have a relationship? 56. Which test would you use to decide if two populations have the same distribution? 57. How are tests |
of independence similar to tests for homogeneity? 58. How are tests of independence different from tests for homogeneity? 11.6 Test of a Single Variance Use the following information to answer the next three exercises: An archer’s standard deviation for his hits is six (data is measured in distance from the center of the target). An observer claims the standard deviation is less. 59. What type of test should be used? 60. State the null and alternative hypotheses. 61. Is this a right-tailed, left-tailed, or two-tailed test? Use the following information to answer the next three exercises: The standard deviation of heights for students in a school 612 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION is 0.81. A random sample of 50 students is taken, and the standard deviation of heights of the sample is 0.96. A researcher in charge of the study believes the standard deviation of heights for the school is greater than 0.81. 62. What type of test should be used? 63. State the null and alternative hypotheses. 64. df = ________ Use the following information to answer the next four exercises: The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought. 65. What type of test should be used? 66. What is the test statistic? 67. What is the p-value? 68. What can you conclude at the 5% significance level? HOMEWORK 11.1 Facts About the Chi-Square Distribution Decide whether the following statements are true or false. 69. As the number of degrees of freedom increases, the graph of the chi-square distribution looks more and more symmetrical. 70. The standard deviation of the chi-square distribution is twice the mean. 71. The mean and the median of the chi-square distribution are the same if df = 24. 11.2 Goodness-of-Fit Test For each problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. 72. A six-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a hypothesis test |
to determine if the die is fair. The data in Table 11.34 are the result of the 120 rolls. Face Value Frequency Expected Frequency 1 2 3 4 5 6 Table 11.34 15 29 16 15 30 15 73. The marital status distribution of the U.S. male population, ages 15 and older, is as shown in Table 11.35. Marital Status Percent Expected Frequency never married married widowed Table 11.35 31.3 56.1 2.5 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 613 Marital Status Percent Expected Frequency divorced/separated 10.1 Table 11.35 Suppose that a random sample of 400 U.S. young adult males, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in Table 11.36, rounding to two decimal places. Marital Status Frequency never married married widowed 140 238 2 divorced/separated 20 Table 11.36 Use the following information to answer the next two exercises: The columns in Table 11.37 contain the Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for the Advanced Placement Examinee Population for that class, and the Overall Student Population. Suppose the right column contains the result of a survey of 1,000 local students from that year who took an AP Exam. Race/Ethnicity AP Examinee Population Overall Student Population Survey Frequency Asian, Asian American, or Pacific Islander Black or African-American Hispanic or Latino 10.2% 8.2% 15.5% American Indian or Alaska Native 0.6% White Not reported/other Table 11.37 59.4% 6.1% 5.4% 14.5% 15.9% 1.2% 61.6% 1.4% 113 94 136 10 604 43 74. Perform a goodness-of-fit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity. 75. Perform a goodness-of-fit test to determine whether the local results follow the distribution of U.S. AP examine |
e population, based on ethnicity. 76. The City of South Lake Tahoe, CA, has an Asian population of 1,419 people, out of a total population of 23,609. Suppose that a survey of 1,419 self-reported Asians in the Manhattan, NY, area yielded the data in Table 11.38. Conduct a goodness-of-fit test to determine if the self-reported sub-groups of Asians in the Manhattan area fit that of the Lake Tahoe area. Race Lake Tahoe Frequency Manhattan Frequency Asian Indian 131 Chinese 118 Filipino 1,045 Table 11.38 174 557 518 614 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION Race Lake Tahoe Frequency Manhattan Frequency Japanese Korean 80 12 Vietnamese 9 Other 24 Table 11.38 54 29 21 66 Use the following information to answer the next two exercises: UCLA conducted a survey of more than 263,000 college freshmen from 385 colleges in fall 2005. The results of students' expected majors by gender were reported in The Chronicle of Higher Education (2/2/2006). Suppose a survey of 5,000 graduating females and 5,000 graduating males was done as a follow-up last year to determine what their actual majors were. The results are shown in the tables for Exercise 11.77 and Exercise 11.78. The second column in each table does not add to 100% because of rounding. 77. Conduct a goodness-of-fit test to determine if the actual college majors of graduating females fit the distribution of their expected majors. Major Women - Expected Major Women - Actual Major Arts & Humanities 14.0% Biological Sciences 8.4% Business Education Engineering 13.1% 13.0% 2.6% Physical Sciences 2.6% Professional 18.9% Social Sciences 13.0% Technical Other Undecided Table 11.39 0.4% 5.8% 8.0% 670 410 685 650 145 125 975 605 15 300 420 78. Conduct a goodness-of-fit test to determine if the actual college majors of graduating males fit the distribution of their expected majors. Major Men - Expected Major Men - Actual Major Arts & Humanities 11.0% Biological Sciences 6.7% Business Education Engineering 22.7% 5.8% 15.6% Physical Sciences 3.6% Professional Social Sciences Technical Table 11.40 9.3% 7.6% 1.8% 600 330 1130 305 800 175 460 370 90 This content is available for free |
at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 615 Major Other Undecided Table 11.40 Men - Expected Major Men - Actual Major 8.2% 6.6% 400 340 Read the statement and decide whether it is true or false. 79. In a goodness-of-fit test, the expected values are the values we would expect if the null hypothesis were true. 80. In general, if the observed values and expected values of a goodness-of-fit test are not close together, then the test statistic can get very large and on a graph will be way out in the right tail. 81. Use a goodness-of-fit test to determine if high school principals believe that students are absent equally during the week or not. 82. The test to use to determine if a six-sided die is fair is a goodness-of-fit test. 83. In a goodness-of fit test, if the p-value is 0.0113, in general, do not reject the null hypothesis. 84. A sample of 212 commercial businesses was surveyed for recycling one commodity; a commodity here means any one type of recyclable material such as plastic or aluminum. Table 11.41 shows the business categories in the survey, the sample size of each category, and the number of businesses in each category that recycle one commodity. Based on the study, on average half of the businesses were expected to be recycling one commodity. As a result, the last column shows the expected number of businesses in each category that recycle one commodity. At the 5% significance level, perform a hypothesis test to determine if the observed number of businesses that recycle one commodity follows the uniform distribution of the expected values. Business Type Office Retail/ Wholesale Food/ Restaurants Manufacturing/ Medical Hotel/Mixed Table 11.41 Number in class Observed Number that recycle one commodity Expected number that recycle one commodity 35 48 53 52 24 19 27 35 21 9 17.5 24 26.5 26 12 85. Table 11.42 contains information from a survey among 499 participants classified according to their age groups. The second column shows the percentage of obese people per age class among the study participants. The last column comes from a different study at the national level that shows the corresponding percentages of obese people in the same age classes in the USA. Perform a |
hypothesis test at the 5% significance level to determine whether the survey participants are a representative sample of the USA obese population. Age Class (Years) Obese (Percentage) Expected USA average (Percentage) 20–30 31–40 41–50 51–60 61–70 Table 11.42 75.0 26.5 13.6 21.9 21.0 32.6 32.6 36.6 36.6 39.7 616 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 11.3 Test of Independence For each problem, use a solution sheet to solve the hypothesis test problem. Go to the Appendix (http://cnx.org/ content/m47157/latest/) for the chi-square solution sheet. Round expected frequency to two decimal places. 86. A recent debate about where in the United States skiers believe the skiing is best prompted the following survey. Test to see if the best ski area is independent of the level of the skier. U.S. Ski Area Beginner Intermediate Advanced Tahoe Utah Colorado Table 11.43 20 10 10 30 30 40 40 60 50 87. Car manufacturers are interested in whether there is a relationship between the size of car an individual drives and the number of people in the driver’s family (that is, whether car size and family size are independent). To test this, suppose that 800 car owners were randomly surveyed with the results in Table 11.44. Conduct a test of independence. Family Size Sub & Compact Mid-size Full-size Van & Truck 1 2 3–4 5+ Table 11.44 20 20 20 20 35 50 50 30 40 70 100 70 35 80 90 70 88. College students may be interested in whether or not their majors have any effect on starting salaries after graduation. Suppose that 300 recent graduates were surveyed as to their majors in college and their starting salaries after graduation. Table 11.45 shows the data. Conduct a test of independence. Major < $50,000 $50,000 – $68,999 $69,000 + English 5 Engineering 10 Nursing Business 10 10 Psychology 20 Table 11.45 20 30 15 20 30 5 60 15 30 20 89. Some travel agents claim that honeymoon hot spots vary according to age of the bride. Suppose that 280 recent brides were interviewed as to where they spent their honeymoons. The information is given in Table 11.46. Conduct a test of independence. Location 20–29 30–39 40–49 50 and |
over Niagara Falls 15 Poconos Europe 15 10 Table 11.46 25 25 25 25 25 15 20 10 5 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 617 Location 20–29 30–39 40–49 50 and over Virgin Islands 20 25 15 5 Table 11.46 90. A manager of a sports club keeps information concerning the main sport in which members participate and their ages. To test whether there is a relationship between the age of a member and his or her choice of sport, 643 members of the sports club are randomly selected. Conduct a test of independence. Sport 18 - 25 26 - 30 31 - 40 41 and over racquetball 42 tennis 58 swimming 72 Table 11.47 58 76 60 30 38 65 46 65 33 91. A major food manufacturer is concerned that the sales for its skinny french fries have been decreasing. As a part of a feasibility study, the company conducts research into the types of fries sold across the country to determine if the type of fries sold is independent of the area of the country. The results of the study are shown in Table 11.48. Conduct a test of independence. Type of Fries Northeast South Central West skinny fries curly fries steak fries Table 11.48 70 100 20 50 60 40 20 15 10 25 30 10 92. According to Dan Lenard, an independent insurance agent in the Buffalo, N.Y. area, the following is a breakdown of the amount of life insurance purchased by males in the following age groups. He is interested in whether the age of the male and the amount of life insurance purchased are independent events. Conduct a test for independence. Age of Males None < $200,000 $200,000–$400,000 $401,001–$1,000,000 $1,000,001+ 20–29 30–39 40–49 50+ Table 11.49 40 35 20 40 15 5 0 30 40 20 30 15 0 20 0 15 5 10 30 10 93. Suppose that 600 thirty-year-olds were surveyed to determine whether or not there is a relationship between the level of education an individual has and salary. Conduct a test of independence. Annual Salary Not a high school graduate High school graduate College graduate Masters or doctorate < $30,000 15 $30,000–$40, |
000 20 Table 11.50 25 40 10 70 5 30 618 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION Annual Salary Not a high school graduate High school graduate College graduate Masters or doctorate $40,000–$50,000 10 $50,000–$60,000 5 $60,000+ 0 Table 11.50 20 10 5 40 20 10 55 60 150 Read the statement and decide whether it is true or false. 94. The number of degrees of freedom for a test of independence is equal to the sample size minus one. 95. The test for independence uses tables of observed and expected data values. 96. The test to use when determining if the college or university a student chooses to attend is related to his or her socioeconomic status is a test for independence. 97. In a test of independence, the expected number is equal to the row total multiplied by the column total divided by the total surveyed. 98. An ice cream maker performs a nationwide survey about favorite flavors of ice cream in different geographic areas of the U.S. Based on Table 11.51, do the numbers suggest that geographic location is independent of favorite ice cream flavors? Test at the 5% significance level. Strawberry Chocolate Vanilla Rocky Road Mint Chocolate Chip Pistachio 12 10 8 15 45 21 32 31 28 22 22 27 30 112 101 19 11 8 8 46 15 15 15 15 60 8 6 7 6 27 Row total 97 96 96 102 391 U.S. region/ Flavor West Midwest East South Column Total Table 11.51 99. Table 11.52 provides a recent survey of the youngest online entrepreneurs whose net worth is estimated at one million dollars or more. Their ages range from 17 to 30. Each cell in the table illustrates the number of entrepreneurs who correspond to the specific age group and their net worth. Are the ages and net worth independent? Perform a test of independence at the 5% significance level. Age Group\ Net Worth Value (in millions of US dollars) 1–5 6–24 ≥25 Row Total 17–25 26–30 Column Total Table 11.52 8 6 7 5 14 12 5 9 14 20 20 40 100. A 2013 poll in California surveyed people about taxing sugar-sweetened beverages. The results are presented in Table 11.53, and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a test of independence at the 5% significance level. This content is available for free at http://textbo |
okequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 619 AsianAmerican White/NonHispanic AfricanAmerican 433 234 43 710 41 24 16 71 Opinion/ Ethnicity Against tax In Favor of tax No opinion 48 54 16 Column Total 118 Table 11.53 11.4 Test for Homogeneity Latino 160 147 19 272 Row Total 628 459 84 1171 For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. 101. A psychologist is interested in testing whether there is a difference in the distribution of personality types for business majors and social science majors. The results of the study are shown in Table 11.54. Conduct a test of homogeneity. Test at a 5% level of significance. Open Conscientious Extrovert Agreeable Neurotic Business 41 Social Science 72 52 75 46 63 61 80 58 65 Table 11.54 102. Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place is shown in Table 11.55. Conduct a test for homogeneity at a 5% level of significance. French Toast Pancakes Waffles Omelettes Men 47 Women 65 Table 11.55 35 59 28 55 53 60 103. A fisherman is interested in whether the distribution of fish caught in Green Valley Lake is the same as the distribution of fish caught in Echo Lake. Of the 191 randomly selected fish caught in Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 were bass, and 24 were catfish. Of the 293 randomly selected fish caught in Echo Lake, 115 were rainbow trout, 58 were other trout, 67 were bass, and 53 were catfish. Perform a test for homogeneity at a 5% level of significance. 104. In 2007, the United States had 1.5 million homeschooled students, according to the U.S. National Center for Education Statistics. In Table 11.56 you can see that parents decide to homeschool their children for different reasons, and some reasons are ranked by parents as more important than others. According to the survey results shown in the table, is the distribution of applicable reasons the same as the distribution of the most important reason? Provide your assessment at the 5% significance level. Did you expect the |
result you obtained? Reasons for Homeschooling Applicable Reason (in thousands of respondents) Most Important Reason (in thousands of respondents) Concern about the environment of other schools 1,321 309 Row Total 1,630 Table 11.56 620 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION Reasons for Homeschooling Dissatisfaction with academic instruction at other schools To provide religious or moral instruction Child has special needs, other than physical or mental Nontraditional approach to child’s education Other reasons (e.g., finances, travel, family time, etc.) Column Total Table 11.56 Applicable Reason (in thousands of respondents) Most Important Reason (in thousands of respondents) 1,096 1,257 315 984 485 5,458 258 540 55 99 216 1,477 Row Total 1,354 1,797 370 1,083 701 6,935 105. When looking at energy consumption, we are often interested in detecting trends over time and how they correlate among different countries. The information in Table 11.57 shows the average energy use (in units of kg of oil equivalent per capita) in the USA and the joint European Union countries (EU) for the six-year period 2005 to 2010. Do the energy use values in these two areas come from the same distribution? Perform the analysis at the 5% significance level. European Union United States Row Total Year 2010 2009 2008 2007 2006 2005 3,413 3,302 3,505 3,537 3,595 3,613 7,164 7,057 7,488 7,758 7,697 7,847 Column Total 45,011 20,965 Table 11.57 10,557 10,359 10,993 11,295 11,292 11,460 65,976 106. The Insurance Institute for Highway Safety collects safety information about all types of cars every year, and publishes a report of Top Safety Picks among all cars, makes, and models. Table 11.58 presents the number of Top Safety Picks in six car categories for the two years 2009 and 2013. Analyze the table data to conclude whether the distribution of cars that earned the Top Safety Picks safety award has remained the same between 2009 and 2013. Derive your results at the 5% significance level. Year \ Car Type 2009 2013 Column Total Table 11.58 Small 12 31 43 MidSize 22 30 52 Large Small SUV Mid-Size SUV Large SUV 10 19 29 10 11 21 27 29 56 6 4 10 Row Total 87 124 211 11. |
5 Comparison of the Chi-Square Tests This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 621 For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. 107. Is there a difference between the distribution of community college statistics students and the distribution of university statistics students in what technology they use on their homework? Of some randomly selected community college students, 43 used a computer, 102 used a calculator with built in statistics functions, and 65 used a table from the textbook. Of some randomly selected university students, 28 used a computer, 33 used a calculator with built in statistics functions, and 40 used a table from the textbook. Conduct an appropriate hypothesis test using a 0.05 level of significance. Read the statement and decide whether it is true or false. 108. If df = 2, the chi-square distribution has a shape that reminds us of the exponential. 11.6 Test of a Single Variance Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes. 109. Is the traveler disputing the claim about the average or about the variance? 110. A sample standard deviation of 15 minutes is the same as a sample variance of __________ minutes. 111. Is this a right-tailed, left-tailed, or two-tailed test? 112. H0: __________ 113. df = ________ 114. chi-square test statistic = ________ 115. p-value = ________ 116. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the p-value. 117. Let α = 0.05 Decision: ________ Conclusion (write out in a complete sentence.): ________ 118. How did you know to test the variance instead of the mean? 119. If an additional test were done on the claim of the average delay |
, which distribution would you use? 120. If an additional test were done on the claim of the average delay, but 45 flights were surveyed, which distribution would you use? For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. 121. A plant manager is concerned her equipment may need recalibrating. It seems that the actual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard deviation should be at most 0.5 oz. In order to determine if the machine needs to be recalibrated, 84 randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be recalibrated? 122. Consumers may be interested in whether the cost of a particular calculator varies from store to store. Based on surveying 43 stores, which yielded a sample mean of $84 and a sample standard deviation of $12, test the claim that the standard deviation is greater than $15. 123. Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time to run the 7.5 mile race is at most three minutes. To test her claim, Rupinder looks up five of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57 minutes. 124. Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. They are also interested in the variation of the number of babies. Suppose that an airline executive believes the average number of babies on flights is six with a variance of nine at most. The airline conducts a survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief. 125. The number of births per woman in China is 1.6 down from 5.91 in 1966. This fertility rate has been attributed to the law passed in 1979 restricting births to one per woman. Suppose that a group of students studied whether or not the standard deviation of births per woman was greater than 0.75. They asked 50 women across China the number of births they had had. The results are shown in Table 11.59. Does the students’ survey indicate that the standard deviation is greater than 0.75? |
622 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION # of births Frequency 0 1 2 3 Table 11.59 5 30 10 5 126. According to an avid aquarist, the average number of fish in a 20-gallon tank is 10, with a standard deviation of two. His friend, also an aquarist, does not believe that the standard deviation is two. She counts the number of fish in 15 other 20-gallon tanks. Based on the results that follow, do you think that the standard deviation is different from two? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11 127. The manager of "Frenchies" is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a ten-ounce order of fries is at most 1.5 oz., but the manager thinks that it may be higher. He randomly weighs 49 orders of fries, which yields a mean of 11 oz. and a standard deviation of two oz. 128. You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of $1,249 with a very narrow standard deviation of $25. You find a website that has a price comparison for the same computer at a series of stores as follows: $1,299; $1,229.99; $1,193.08; $1,279; $1,224.95; $1,229.99; $1,269.95; $1,249. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the 5% significance level. As a potential buyer, what would be the practical conclusion from your analysis? 129. A company packages apples by weight. One of the weight grades is Class A apples. Class A apples have a mean weight of 150 g, and there is a maximum allowed weight tolerance of 5% above or below the mean for apples in the same consumer package. A batch of apples is selected to be included in a Class A apple package. Given the following apple weights of the batch, does the fruit comply with the Class A grade weight tolerance requirements. Conduct an appropriate hypothesis test. (a) at the 5% significance level (b) at the 1% significance level Weights in selected apple batch (in grams): 158; 167; |
149; 169; 164; 139; 154; 150; 157; 171; 152; 161; 141; 166; 172; BRINGING IT TOGETHER: HOMEWORK 130. a. Explain why a goodness-of-fit test and a test of independence are generally right-tailed tests. b. If you did a left-tailed test, what would you be testing? REFERENCES 11.1 Facts About the Chi-Square Distribution Data from Parade Magazine. “HIV/AIDS Epidemiology Santa Clara County.”Santa Clara County Public Health Department, May 2011. 11.2 Goodness-of-Fit Test Data from the U.S. Census Bureau Data from the College Board. Available online at http://www.collegeboard.com. Data from the U.S. Census Bureau, Current Population Reports. Ma, Y., E.R. Bertone, E.J. Stanek III, G.W. Reed, J.R. Hebert, N.L. Cohen, P.A. Merriam, I.S. Ockene, “Association between Eating Patterns and Obesity in a Free-living US Adult Population.” American Journal of Epidemiology volume 158, no. 1, pages 85-92. Ogden, Cynthia L., Margaret D. Carroll, Brian K. Kit, Katherine M. Flegal, “Prevalence of Obesity in the United States, 2009–2010.” NCHS Data Brief no. 82, January 2012. Available online at http://www.cdc.gov/nchs/data/databriefs/db82.pdf (accessed May 24, 2013). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 623 Stevens, Barbara J., “Multi-family and Commercial Solid Waste and Recycling Survey.” Arlington Count, VA. Available online at http://www.arlingtonva.us/departments/EnvironmentalServices/SW/file84429.pdf (accessed May 24,2013). 11.3 Test of Independence DiCamilo, Mark, Mervin Field, “Most Californians See a Direct Linkage between Obesity and Sugary Sodas. Two in Three Voters Support Tax |
ing Sugar-Sweetened Beverages If Proceeds are Tied to Improving School Nutrition and Physical Activity Programs.” The Field Poll, released Feb. 14, 2013. Available online at http://field.com/fieldpollonline/subscribers/ Rls2436.pdf (accessed May 24, 2013). Harris Interactive, “Favorite Flavor of Ice Cream.” Available online at http://www.statisticbrain.com/favorite-flavor-of-icecream (accessed May 24, 2013) “Youngest Online Entrepreneurs List.” Available online at http://www.statisticbrain.com/youngest-online-entrepreneur-list (accessed May 24, 2013). 11.4 Test for Homogeneity Data from the Insurance Institute for Highway Safety, 2013. Available online at www.iihs.org/iihs/ratings (accessed May 24, 2013). “Energy use (kg of oil equivalent per capita).” The World Bank, 2013. Available online at http://data.worldbank.org/ indicator/EG.USE.PCAP.KG.OE/countries (accessed May 24, 2013). “Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubsearch/ pubsinfo.asp?pubid=2009030 (accessed May 24, 2013). “Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubs2009/ 2009030_sup.pdf (accessed May 24, 2013). 11.6 Test of a Single Variance “AppleInsider Price Guides.” Apple Insider, 2013. Available online at http://appleinsider.com/mac_price_guide (accessed May 14, 2013). Data from the World Bank, June 5, 2012. SOLUTIONS 1 mean = 25 and standard deviation = 7.0711 3 when the number of degrees of freedom is greater than 90 5 df = 2 7 a goodness-of-fit test 9 3 11 2.04 13 We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different |
from the expected test scores. 15 H0: the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County. 17 right-tailed 19 88,621 21 Graph: Check student’s solution. Decision: Reject the null hypothesis. Reason for the Decision: p-value < alpha Conclusion (write out in complete sentences): The make-up of AIDS cases does not fit the ethnicities of the general population of Santa Clara County. 624 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 23 a test of independence 25 a test of independence 27 8 29 6.6 31 0.0435 33 Smoking Level Per Day African American Native Hawaiian Latino Japanese Americans 12,831 8,378 4,932 1,406 800 10,680 4,715 2,305 White Totals 7,650 9,877 6,062 3,970 41,490 35,065 15,273 8,622 19,969 26,078 27,559 10,0450 9,886 6,514 1,671 759 18,830 2,745 3,062 1,419 788 8,014 1-10 11-20 21-30 31+ Totals Table 11.60 35 White 11383.01 9620.27 4190.23 2365.49 Smoking Level Per Day African American Native Hawaiian Latino Japanese Americans 7777.57 6573.16 2863.02 1616.25 3310.11 2797.52 1218.49 687.87 8248.02 10771.29 6970.76 9103.29 3036.20 3965.05 1714.01 2238.37 1-10 11-20 21-30 31+ Table 11.61 37 10,301.8 39 right 41 a. Reject the null hypothesis. b. p-value < alpha c. There is sufficient evidence to conclude that smoking level is dependent on ethnic group. 43 test for homogeneity 45 test for homogeneity 47 All values in the table must be greater than or equal to five. 49 3 51 0.00005 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 625 53 a goodness-of-fit test 55 a test for independence 57 Answers will |
vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way ∑ (i j) (O - E)2 E. In addition, all values must be greater than or equal to five. 59 a test of a single variance 61 a left-tailed test 63 H0: σ2 = 0.812; Ha: σ2 > 0.812 65 a test of a single variance 67 0.0542 69 true 71 false 73 Marital Status Percent Expected Frequency never married married widowed 31.3 56.1 2.5 divorced/separated 10.1 Table 11.62 125.2 224.4 10 40.4 a. The data fits the distribution. b. The data does not fit the distribution. c. 3 d. chi-square distribution with df = 3 e. 19.27 f. 0.0002 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Reject null iii. Reason for decision: p-value < alpha iv. Conclusion: Data does not fit the distribution. 75 a. H0: The local results follow the distribution of the U.S. AP examinee population b. Ha: The local results do not follow the distribution of the U.S. AP examinee population c. df = 5 d. chi-square distribution with df = 5 e. chi-square test statistic = 13.4 626 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION f. p-value = 0.0199 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Reject null when a = 0.05 iii. Reason for Decision: p-value < alpha iv. Conclusion: Local data do not fit the AP Examinee Distribution. v. Decision: Do not reject null when a = 0.01 vi. Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution. 77 a. H0: The actual college majors of graduating females fit the distribution of their expected majors b. Ha: The actual college majors of graduating females do not fit the distribution of their expected majors c. df = 10 d. chi-square distribution with df = 10 e. test statistic = 11.48 f. p-value = 0.3211 g. Check student’s |
solution. h. i. Alpha = 0.05 ii. Decision: Do not reject null when a = 0.05 and a = 0.01 iii. Reason for decision: p-value > alpha iv. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females fits the distribution of their expected majors. 79 true 81 true 83 false 85 a. H0: Surveyed obese fit the distribution of expected obese b. Ha: Surveyed obese do not fit the distribution of expected obese c. df = 4 d. chi-square distribution with df = 4 e. test statistic = 54.01 f. p-value = 0 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% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese. 87 a. H0: Car size is independent of family size. b. Ha: Car size is dependent on family size. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 627 c. df = 9 d. chi-square distribution with df = 9 e. test statistic = 15.8284 f. p-value = 0.0706 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent. 89 a. H0: Honeymoon locations are independent of bride’s age. b. Ha: Honeymoon locations are dependent on bride’s age. c. df = 9 d. chi-square distribution with df = 9 e. test statistic = 15.7027 f. p-value = 0.0734 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5% significance level, |
there is insufficient evidence to conclude that honeymoon location and bride age are dependent. 91 a. H0: The types of fries sold are independent of the location. b. Ha: The types of fries sold are dependent on the location. c. df = 6 d. chi-square distribution with df = 6 e. test statistic =18.8369 f. p-value = 0.0044 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 that types of fries and location are dependent. 93 a. H0: Salary is independent of level of education. b. Ha: Salary is dependent on level of education. c. df = 12 d. chi-square distribution with df = 12 e. test statistic = 255.7704 628 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION f. p-value = 0 g. Check student’s solution. h. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p-value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent. 95 true 97 true 99 a. H0: Age is independent of the youngest online entrepreneurs’ net worth. b. Ha: Age is dependent on the net worth of the youngest online entrepreneurs. c. df = 2 d. chi-square distribution with df = 2 e. test statistic = 1.76 f. p-value 0.4144 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that age and net worth for the youngest online entrepreneurs are dependent. 101 a. H0: The distribution for personality types is the same for both majors b. Ha: The distribution for personality types is not the same for both majors c. df = 4 d. chi-square with df = 4 e. test statistic = 3.01 f. p-value = 0.5568 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. |
iii. Reason for decision: p-value > alpha iv. Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors. 103 a. H0: The distribution for fish caught is the same in Green Valley Lake and in Echo Lake. b. Ha: The distribution for fish caught is not the same in Green Valley Lake and in Echo Lake. c. 3 d. chi-square with df = 3 e. 11.75 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION 629 f. p-value = 0.0083 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: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake 105 a. H0: The distribution of average energy use in the USA is the same as in Europe between 2005 and 2010. b. Ha: The distribution of average energy use in the USA is not the same as in Europe between 2005 and 2010. c. df = 4 d. chi-square with df = 4 e. test statistic = 2.7434 f. p-value = 0.7395 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the average energy use values in the US and EU are not derived from different distributions for the period from 2005 to 2010. 107 a. H0: The distribution for technology use is the same for community college students and university students. b. Ha: The distribution for technology use is not the same for community college students and university students. c. 2 d. chi-square with df = 2 e. 7.05 f. p-value = 0.0294 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: There is sufficient evidence |
to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities. 110 225 112 H0: σ2 ≤ 150 114 36 116 Check student’s solution. 118 The claim is that the variance is no more than 150 minutes. 120 a Student's t- or normal distribution 122 630 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION a. H0: σ = 15 b. Ha: σ > 15 c. df = 42 d. chi-square with df = 42 e. test statistic = 26.88 f. p-value = 0.9663 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Do not reject null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15. 124 a. H0: σ ≤ 3 b. Ha: σ > 3 c. df = 17 d. chi-square distribution with df = 17 e. test statistic = 28.73 f. p-value = 0.0371 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: There is sufficient evidence to conclude that the standard deviation is greater than three. 126 a. H0: σ = 2 b. Ha: σ ≠ 2 c. df = 14 d. chi-square distiribution with df = 14 e. chi-square test statistic = 5.2094 f. p-value = 0.0346 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: There is sufficient evidence to conclude that the standard deviation is different than 2. 128 The sample standard deviation is $34.29. H0 : σ2 = 252 Ha : σ2 > 252 df = n – 1 = 7. test statistic: x2 = x7 2 = (n – 1)s2 252 ⎠ = 1 – P⎛ ⎞ 2 > 13.169 ⎝x7 (8 – 1)(34.29)2 252 ⎞ 2 ≤ 13.169 |
⎠ = 0.0681 = p-value: P⎛ ⎝x7 = 13.169 ; 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 11 | THE CHI-SQUARE DISTRIBUTION 631 Decision: Do not reject the null hypothesis. Reason for decision: p-value > alpha Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625. 130 a. The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected. b. Testing to see if the data fits the distribution “too well” or is too perfect. 632 CHAPTER 11 | THE CHI-SQUARE DISTRIBUTION This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION 633 12 | LINEAR REGRESSION AND CORRELATION Figure 12.1 Linear regression and correlation can help you determine if an auto mechanic’s salary is related to his work experience. (credit: Joshua Rothhaas) Introduction Chapter Objectives By the end of this chapter, the student should be able to: • Discuss basic ideas of linear regression and correlation. • Create and interpret a line of best fit. • Calculate and interpret the correlation coefficient. • Calculate and interpret outliers. Professionals often want to know how two or more numeric variables are related. For example, is there a relationship between the grade on the second math exam a student takes and the grade on the final exam? If there is a relationship, what is the relationship and how strong is it? In another example, your income may be determined by your education, your profession, your years of experience, and your ability. The amount you pay a repair person for labor is often determined by an initial amount plus an hourly fee. 634 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION The type of data described in the examples is bivariate data — "bi" for two variables. In reality, statisticians use multivariate data, meaning many variables. In |
this chapter, you will be studying the simplest form of regression, "linear regression" with one independent variable (x). This involves data that fits a line in two dimensions. You will also study correlation which measures how strong the relationship is. 12.1 | Linear Equations Linear regression for two variables is based on a linear equation with one independent variable. The equation has the form: y = a + bx where a and b are constant numbers. The variable x is the independent variable, and y is the dependent variable. Typically, you choose a value to substitute for the independent variable and then solve for the dependent variable. Example 12.1 The following examples are linear equations. y = 3 + 2x y = –0.01 + 1.2x 12.1 Is the following an example of a linear equation? y = –0.125 – 3.5x The graph of a linear equation of the form y = a + bx is a straight line. Any line that is not vertical can be described by this equation. Example 12.2 Graph the equation y = –1 + 2x. Figure 12.2 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION 635 12.2 Is the following an example of a linear equation? Why or why not? Figure 12.3 Example 12.3 Aaron's Word Processing Service (AWPS) does word processing. The rate for services is $32 per hour plus a $31.50 one-time charge. The total cost to a customer depends on the number of hours it takes to complete the job. Find the equation that expresses the total cost in terms of the number of hours required to complete the job. Solution 12.3 Let x = the number of hours it takes to get the job done. Let y = the total cost to the customer. The $31.50 is a fixed cost. If it takes x hours to complete the job, then (32)(x) is the cost of the word processing only. The total cost is: y = 31.50 + 32x 12.3 Emma’s Extreme Sports hires hang-gliding instructors and pays them a fee of $50 per class as well as $20 per student in the class. The total cost Emma pays depends on the number of students in a class |
. Find the equation that expresses the total cost in terms of the number of students in a class. Slope and Y-Intercept of a Linear Equation For the linear equation y = a + bx, b = slope and a = y-intercept. From algebra recall that the slope is a number that describes the steepness of a line, and the y-intercept is the y coordinate of the point (0, a) where the line crosses the y-axis. Figure 12.4 Three possible graphs of y = a + bx. (a) If b > 0, the line slopes upward to the right. (b) If b = 0, the line is horizontal. (c) If b < 0, the line slopes downward to the right. 636 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION Example 12.4 Svetlana tutors to make extra money for college. For each tutoring session, she charges a one-time fee of $25 plus $15 per hour of tutoring. A linear equation that expresses the total amount of money Svetlana earns for each session she tutors is y = 25 + 15x. What are the independent and dependent variables? What is the y-intercept and what is the slope? Interpret them using complete sentences. Solution 12.4 The independent variable (x) is the number of hours Svetlana tutors each session. The dependent variable (y) is the amount, in dollars, Svetlana earns for each session. The y-intercept is 25 (a = 25). At the start of the tutoring session, Svetlana charges a one-time fee of $25 (this is when x = 0). The slope is 15 (b = 15). For each session, Svetlana earns $15 for each hour she tutors. 12.4 Ethan repairs household appliances like dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work. A linear equation that expresses the total amount of money Ethan earns per visit is y = 25 + 20x. What are the independent and dependent variables? What is the y-intercept and what is the slope? Interpret them using complete sentences. 12.2 | Scatter Plots Before we take up the discussion of linear regression and correlation, we need to examine a way to display the relation between two variables x and y. The most common and easiest way is |
a scatter plot. The following example illustrates a scatter plot. Example 12.5 In Europe and Asia, m-commerce is popular. M-commerce users have special mobile phones that work like electronic wallets as well as provide phone and Internet services. Users can do everything from paying for parking to buying a TV set or soda from a machine to banking to checking sports scores on the Internet. For the years 2000 through 2004, was there a relationship between the year and the number of m-commerce users? Construct a scatter plot. Let x = the year and let y = the number of m-commerce users, in millions. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION 637 x (year) y (# of users) 0.5 20.0 33.0 47.0 2000 2002 2003 2004 Table 12.1 (a) Table showing the number of mcommerce users (in millions) by year. Figure 12.5 (b) Scatter plot showing the number of m-commerce users (in millions) by year. To create a scatter plot: 1. Enter your X data into list L1 and your Y data into list L2. 2. Press 2nd STATPLOT ENTER to use Plot 1. On the input screen for PLOT 1, highlight On and press ENTER. (Make sure the other plots are OFF.) 3. For TYPE: highlight the very first icon, which is the scatter plot, and press ENTER. 4. For Xlist:, enter L1 ENTER and for Ylist: L2 ENTER. 5. For Mark: it does not matter which symbol you highlight, but the square is the easiest to see. Press ENTER. 6. Make sure there are no other equations that could be plotted. Press Y = and clear any equations out. 7. Press the ZOOM key and then the number 9 (for menu item "ZoomStat") ; the calculator will fit the window to the data. You can press WINDOW to see the scaling of the axes. 638 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION 12.5 Amelia plays basketball for her high school. She wants to improve to play at the college level. She notices that the number of points she scores in a game goes up in response to the number of hours she practices |
her jump shot each week. She records the following data: X (hours practicing jump shot) Y (points scored in a game) 5 7 9 10 11 12 Table 12.2 15 22 28 31 33 36 Construct a scatter plot and state if what Amelia thinks appears to be true. A scatter plot shows the direction of a relationship between the variables. A clear direction happens when there is either: • High values of one variable occurring with high values of the other variable or low values of one variable occurring with low values of the other variable. • High values of one variable occurring with low values of the other variable. You can determine the strength of the relationship by looking at the scatter plot and seeing how close the points are to a line, a power function, an exponential function, or to some other type of function. For a linear relationship there is an exception. Consider a scatter plot where all the points fall on a horizontal line providing a "perfect fit." The horizontal line would in fact show no relationship. When you look at a scatterplot, you want to notice the overall pattern and any deviations from the pattern. The following scatterplot examples illustrate these concepts. Figure 12.6 Figure 12.7 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION 639 Figure 12.8 In this chapter, we are interested in scatter plots that show a linear pattern. Linear patterns are quite common. The linear relationship is strong if the points are close to a straight line, except in the case of a horizontal line where there is no relationship. If we think that the points show a linear relationship, we would like to draw a line on the scatter plot. This line can be calculated through a process called linear regression. However, we only calculate a regression line if one of the variables helps to explain or predict the other variable. If x is the independent variable and y the dependent variable, then we can use a regression line to predict y for a given value of x 12.3 | The Regression Equation Data rarely fit a straight line exactly. Usually, you must be satisfied with rough predictions. Typically, you have a set of data whose scatter plot appears to "fit" a straight line. This is called a Line of Best Fit or Least-Squares Line. If you know a person's pinky (smallest |
) finger length, do you think you could predict that person's height? Collect data from your class (pinky finger length, in inches). The independent variable, x, is pinky finger length and the dependent variable, y, is height. For each set of data, plot the points on graph paper. Make your graph big enough and use a ruler. Then "by eye" draw a line that appears to "fit" the data. For your line, pick two convenient points and use them to find the slope of the line. Find the y-intercept of the line by extending your line so it crosses the y-axis. Using the slopes and the y-intercepts, write your equation of "best fit." Do you think everyone will have the same equation? Why or why not? According to your equation, what is the predicted height for a pinky length of 2.5 inches? Example 12.6 A random sample of 11 statistics students produced the following data, where x is the third exam score out of 80, and y is the final exam score out of 200. Can you predict the final exam score of a random student if you know the third exam score? 640 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION x (third exam score) y (final exam score) 65 67 71 71 66 75 67 70 71 69 69 175 133 185 163 126 198 153 163 159 151 159 Table 12.3 (a) Table showing the scores on the final exam based on scores from the third exam. Figure 12.9 (b) Scatter plot showing the scores on the final exam based on scores from the third exam. 12.6 SCUBA divers have maximum dive times they cannot exceed when going to different depths. The data in Table 12.4 show different depths with the maximum dive times in minutes. Use your calculator to find the least squares regression line and predict the maximum dive time for 110 feet. X (depth in feet) Y (maximum dive time) 50 60 70 80 90 100 Table 12.4 80 55 45 35 25 22 The third exam score, x, is the independent variable and the final exam score, y, is the dependent variable. We will plot a regression line that best "fits" the data. If each of you were to fit a line "by eye," you would draw different lines. We can use what is called a least-squares regression line to obtain the best fit line. Consider the following diagram. Each point of data is |
of the the form (x, y) and each point ofthe line of best fit using leastsquares linear regression has the form (x, ŷ). The ŷ is read "y hat" and is the estimated value of y. It is the value of y obtained using the regression line. It is not generally equal to y from data. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION 641 Figure 12.10 The term y0 – ŷ0 = ε0 is called the "error" or residual. It is not an error in the sense of a mistake. The absolute value of a residual measures the vertical distance between the actual value of y and the estimated value of y. In other words, it measures the vertical distance between the actual data point and the predicted point on the line. If the observed data point lies above the line, the residual is positive, and the line underestimates the actual data value for y. If the observed data point lies below the line, the residual is negative, and the line overestimates that actual data value for y. In the diagram in Figure 12.10, y0 – ŷ0 = ε0 is the residual for the point shown. Here the point lies above the line and the residual is positive. ε = the Greek letter epsilon For each data point, you can calculate the residuals or errors, yi - ŷi = εi for i = 1, 2, 3,..., 11. Each |ε| is a vertical distance. For the example about the third exam scores and the final exam scores for the 11 statistics students, there are 11 data points. Therefore, there are 11 ε values. If yousquare each ε and add, you get 11 (ε1)2 + (ε2)2 +... + (ε11)2 = Σ i = 1 ε2 This is called the Sum of Squared Errors (SSE). Using calculus, you can determine the values of a and b that make the SSE a minimum. When you make the SSE a minimum, you have determined the points that are on the line of best fit. It turns out that the line of best fit has the equation: where a = y¯ − b x¯ and |
b = Σ(x − x¯ )(y − y¯ ) Σ(x − x¯ ) 2. y^ = a + bx The sample means of the x values and the y values are x¯ and y¯, respectively. The best fit line always passes through the point ( x¯, y¯ ). The slope b can be written as b = r⎛ ⎝ s y s x ⎞ ⎠ where sy = the standard deviation of the y values and sx = the standard deviation of the x values. r is the correlation coefficient, which is discussed in the next section. Least Squares Criteria for Best Fit The process of fitting the best-fit line is called linear regression. The idea behind finding the best-fit line is based on the assumption that the data are scattered about a straight line. The criteria for the best fit line is that the sum of the squared errors (SSE) is minimized, that is, made as small as possible. Any other line you might choose would have a higher SSE than the best fit line. This best fit line is called the least-squares regression line. 642 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION NOTE Computer spreadsheets, statistical software, and many calculators can quickly calculate the best-fit line and create the graphs. The calculations tend to be tedious if done by hand. Instructions to use the TI-83, TI-83+, and TI-84+ calculators to find the best-fit line and create a scatterplot are shown at the end of this section. THIRD EXAM vs FINAL EXAM EXAMPLE: The graph of the line of best fit for the third-exam/final-exam example is as follows: Figure 12.11 The least squares regression line (best-fit line) for the third-exam/final-exam example has the equation: y^ = − 173.51 + 4.83x REMINDER Remember, it is always important to plot a scatter diagram first. If the scatter plot indicates that there is a linear relationship between the variables, then it is reasonable to use a best fit line to make predictions for y given x within the domain of x-values in the sample data, but not necessarily for x-values outside that domain. You could use the line to predict the final exam score for a student who earned a grade of 73 on the third exam. You should NOT |
use the line to predict the final exam score for a student who earned a grade of 50 on the third exam, because 50 is not within the domain of the x-values in the sample data, which are between 65 and 75. UNDERSTANDING SLOPE The slope of the line, b, describes how changes in the variables are related. It is important to interpret the slope of the line in the context of the situation represented by the data. You should be able to write a sentence interpreting the slope in plain English. INTERPRETATION OF THE SLOPE: The slope of the best-fit line tells us how the dependent variable (y) changes for every one unit increase in the independent (x) variable, on average. THIRD EXAM vs FINAL EXAM EXAMPLE Slope: The slope of the line is b = 4.83. Interpretation: For a one-point increase in the score on the third exam, the final exam score increases by 4.83 points, on average. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION 643 Using the Linear Regression T Test: LinRegTTest 1. In the STAT list editor, enter the X data in list L1 and the Y data in list L2, paired so that the corresponding (x,y) values are next to each other in the lists. (If a particular pair of values is repeated, enter it as many times as it appears in the data.) 2. On the STAT TESTS menu, scroll down with the cursor to select the LinRegTTest. (Be careful to select LinRegTTest, as some calculators may also have a different item called LinRegTInt.) 3. On the LinRegTTest input screen enter: Xlist: L1 ; Ylist: L2 ; Freq: 1 4. On the next line, at the prompt β or ρ, highlight "≠ 0" and press ENTER 5. Leave the line for "RegEq:" blank 6. Highlight Calculate and press ENTER. Figure 12.12 The output screen contains a lot of information. For now we will focus on a few items from the output, and will return later to the other items. The second line says y = a + bx. Scroll down |
to find the values a = –173.513, and b = 4.8273; the equation of the best fit line is ŷ = –173.51 + 4.83x The two items at the bottom are r2 = 0.43969 and r = 0.663. For now, just note where to find these values; we will discuss them in the next two sections. Graphing the Scatterplot and Regression Line 1. We are assuming your X data is already entered in list L1 and your Y data is in list L2 2. Press 2nd STATPLOT ENTER to use Plot 1 3. On the input screen for PLOT 1, highlight On, and press ENTER 4. For TYPE: highlight the very first icon which is the scatterplot and press ENTER 5. Indicate Xlist: L1 and Ylist: L2 6. For Mark: it does not matter which symbol you highlight. 7. Press the ZOOM key and then the number 9 (for menu item "ZoomStat") ; the calculator will fit the window to the data 8. To graph the best-fit line, press the "Y=" key and type the equation –173.5 + 4.83X into equation Y1. (The X key is immediately left of the STAT key). Press ZOOM 9 again to graph it. 9. Optional: If you want to change the viewing window, press the WINDOW key. Enter your desired window using Xmin, Xmax, Ymin, Ymax 644 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION NOTE Another way to graph the line after you create a scatter plot is to use LinRegTTest. 1. Make sure you have done the scatter plot. Check it on your screen. 2. Go to LinRegTTest and enter the lists. 3. At RegEq: press VARS and arrow over to Y-VARS. Press 1 for 1:Function. Press 1 for 1:Y1. Then arrow down to Calculate and do the calculation for the line of best fit. 4. Press Y = (you will see the regression equation). 5. Press GRAPH. The line will be drawn." The Correlation Coefficient r Besides looking at the scatter plot and seeing that a line seems reasonable, how can you tell if the line is a good predictor? Use the correlation coefficient as another indicator (besides the scatterplot) of the strength of the relationship |
between x and y. The correlation coefficient, r, developed by Karl Pearson in the early 1900s, is numerical and provides a measure of strength and direction of the linear association between the independent variable x and the dependent variable y. The correlation coefficient is calculated as r = nΣ(xy) − (Σx)(Σy) ⎡ ⎡ ⎣nΣx2 − (Σx)2⎤ ⎣nΣy2 − (Σy)2⎤ ⎦ ⎦ where n = the number of data points. If you suspect a linear relationship between x and y, then r can measure how strong the linear relationship is. What the VALUE of r tells us: • The value of r is always between –1 and +1: –1 ≤ r ≤ 1. • The size of the correlation r indicates the strength of the linear relationship between x and y. Values of r close to –1 or to +1 indicate a stronger linear relationship between x and y. • • If r = 0 there is absolutely no linear relationship between x and y (no linear correlation). If r = 1, there is perfect positive correlation. If r = –1, there is perfect negativecorrelation. In both these cases, all of the original data points lie on a straight line. Of course,in the real world, this will not generally happen. What the SIGN of r tells us • A positive value of r means that when x increases, y tends to increase and when x decreases, y tends to decrease (positive correlation). • A negative value of r means that when x increases, y tends to decrease and when x decreases, y tends to increase (negative correlation). • The sign of r is the same as the sign of the slope, b, of the best-fit line. NOTE Strong correlation does not suggest that x causes y or y causes x. We say "correlation does not imply causation." Figure 12.13 (a) A scatter plot showing data with a positive correlation. 0 < r < 1 (b) A scatter plot showing data with a negative correlation. –1 < r < 0 (c) A scatter plot showing data with zero correlation. r = 0 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 | LINE |
AR REGRESSION AND CORRELATION 645 The formula for r looks formidable. However, computer spreadsheets, statistical software, and many calculators can quickly calculate r. The correlation coefficient r is the bottom item in the output screens for the LinRegTTest on the TI-83, TI-83+, or TI-84+ calculator (see previous section for instructions). The Coefficient of Determination The variable r2 is called the coefficient of determination and is the square of the correlation coefficient, but is usually stated as a percent, rather than in decimal form. It has an interpretation in the context of the data: • r 2, when expressed as a percent, represents the percent of variation in the dependent (predicted) variable y that can be explained by variation in the independent (explanatory) variable x using the regression (best-fit) line. • 1 – r 2, when expressed as a percentage, represents the percent of variation in y that is NOT explained by variation in x using the regression line. This can be seen as the scattering of the observed data points about the regression line. Consider the third exam/final exam example introduced in the previous section • The line of best fit is: ŷ = –173.51 + 4.83x • The correlation coefficient is r = 0.6631 • The coefficient of determination is r2 = 0.66312 = 0.4397 • Interpretation of r2 in the context of this example: • Approximately 44% of the variation (0.4397 is approximately 0.44) in the final-exam grades can be explained by the variation in the grades on the third exam, using the best-fit regression line. • Therefore, approximately 56% of the variation (1 – 0.44 = 0.56) in the final exam grades can NOT be explained by the variation in the grades on the third exam, using the best-fit regression line. (This is seen as the scattering of the points about the line.) 12.4 | Testing the Significance of the Correlation Coefficient The correlation coefficient, r, tells us about the strength and direction of the linear relationship between x and y. However, the reliability of the linear model also depends on how many observed data points are in the sample. We need to look at both the value of the correlation coefficient r and the sample size n, together. We perform a hypothesis test of the "significance of the correlation coefficient" to decide whether the linear relationship in |
the sample data is strong enough to use to model the relationship in the population. The sample data are used to compute r, the correlation coefficient for the sample. If we had data for the entire population, we could find the population correlation coefficient. But because we have only have sample data, we cannot calculate the population correlation coefficient. The sample correlation coefficient, r, is our estimate of the unknown population correlation coefficient. The symbol for the population correlation coefficient is ρ, the Greek letter "rho." ρ = population correlation coefficient (unknown) r = sample correlation coefficient (known; calculated from sample data) The hypothesis test lets us decide whether the value of the population correlation coefficient ρ is "close to zero" or "significantly different from zero". We decide this based on the sample correlation coefficient r and the sample size n. If the test concludes that the correlation coefficient is significantly different from zero, we say that the correlation coefficient is "significant." • Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero. • What the conclusion means: There is a significant linear relationship between x and y. We can use the regression line to model the linear relationship between x and y in the population. If the test concludes that the correlation coefficient is not significantly different from zero (it is close to zero), we say that correlation coefficient is "not significant". • Conclusion: "There is insufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is not significantly different from zero." • What the conclusion means: There is not a significant linear relationship between x and y. Therefore, we CANNOT use the regression line to model a linear relationship between x and y in the population. 646 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION NOTE • • • If r is significant and the scatter plot shows a linear trend, the line can be used to predict the value of y for values of x that are within the domain of observed x values. If r is not significant OR if the scatter plot does not show a linear trend, the line should not be used for prediction. If r is significant and if the scatter plot shows a linear trend, the line may NOT be appropriate or reliable for prediction OUTSIDE the domain of observed x values in the data. PERFORMING THE HYPOTHESIS TEST • Null Hypothesis: H0: ρ = 0 • Alternate Hypothesis: |
Ha: ρ ≠ 0 WHAT THE HYPOTHESES MEAN IN WORDS: • Null Hypothesis H0: The population correlation coefficient IS NOT significantly different from zero. There IS NOT a significant linear relationship(correlation) between x and y in the population. • Alternate Hypothesis Ha: The population correlation coefficient IS significantly DIFFERENT FROM zero. There IS A SIGNIFICANT LINEAR RELATIONSHIP (correlation) between x and y in the population. DRAWING A CONCLUSION: There are two methods of making the decision. The two methods are equivalent and give the same result. • Method 1: Using the p-value • Method 2: Using a table of critical values In this chapter of this textbook, we will always use a significance level of 5%, α = 0.05 NOTE Using the p-value method, you could choose any appropriate significance level you want; you are not limited to using α = 0.05. But the table of critical values provided in this textbook assumes that we are using a significance level of 5%, α = 0.05. (If we wanted to use a different significance level than 5% with the critical value method, we would need different tables of critical values that are not provided in this textbook.) METHOD 1: Using a p-value to make a decision To calculate the p-value using LinRegTTEST: On the LinRegTTEST input screen, on the line prompt for β or ρ, highlight "≠ 0" The output screen shows the p-value on the line that reads "p =". (Most computer statistical software can calculate the p-value.) If the p-value is less than the significance level (α = 0.05): • Decision: Reject the null hypothesis. • Conclusion: "There is sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero." If the p-value is NOT less than the significance level (α = 0.05) • Decision: DO NOT REJECT the null hypothesis. • Conclusion: "There is insufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is NOT significantly different from zero." You will use technology to calculate the p-value. The following describes the calculations to compute the test statistics and the p-value: The p-value is calculated using a t-distribution with n - 2 degrees of freedom. This content is |
available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION 647 The formula for the test statistic is. The value of the test statistic, t, is shown in the computer or calculator output along with the p-value. The test statistic t has the same sign as the correlation coefficient r. The p-value is the combined area in both tails. An alternative way to calculate the p-value (p) given by LinRegTTest is the command 2*tcdf(abs(t),10^99, n-2) in 2nd DISTR. THIRD-EXAM vs FINAL-EXAM EXAMPLE: p-value method • Consider the third exam/final exam example. • The line of best fit is: ŷ = -173.51 + 4.83x with r = 0.6631 and there are n = 11 data points. • Can the regression line be used for prediction? Given a third exam score (x value), can we use the line to predict the final exam score (predicted y value)? H0: ρ = 0 Ha: ρ ≠ 0 α = 0.05 • The p-value is 0.026 (from LinRegTTest on your calculator or from computer software). • The p-value, 0.026, is less than the significance level of α = 0.05. • Decision: Reject the Null Hypothesis H0 • Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between the third exam score (x) and the final exam score (y) because the correlation coefficient is significantly different from zero. Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores. METHOD 2: Using a table of Critical Values to make a decision The 95% Critical Values of the Sample Correlation Coefficient Table can be used to give you a good idea of whether the computed value of r is significant or not. Compare r to the appropriate critical value in the table. If r is not between the positive and negative critical values, then the correlation coefficient is significant. If r is significant, then you may want to use the line for prediction. Example 12.7 Suppose you computed r = 0.801 using n = 10 data points.df |
= n - 2 = 10 - 2 = 8. The critical values associated with df = 8 are -0.632 and + 0.632. If r < negative critical value or r > positive critical value, then r issignificant. Since r = 0.801 and 0.801 > 0.632, r is significant and the line may be usedfor prediction. If you view this example on a number line, it will help you. Figure 12.14 r is not significant between -0.632 and +0.632. r = 0.801 > +0.632. Therefore, r is significant. 12.7 For a given line of best fit, you computed that r = 0.6501 using n = 12 data points and the critical value is 0.576. Can the line be used for prediction? Why or why not? 648 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION Example 12.8 Suppose you computed r = –0.624 with 14 data points. df = 14 – 2 = 12. The critical values are –0.532 and 0.532. Since –0.624 < –0.532, r is significant and the line can be used for prediction Figure 12.15 r = –0.624-0.532. Therefore, r is significant. 12.8 For a given line of best fit, you compute that r = 0.5204 using n = 9 data points, and the critical value is 0.666. Can the line be used for prediction? Why or why not? Example 12.9 Suppose you computed r = 0.776 and n = 6. df = 6 – 2 = 4. The critical values are –0.811 and 0.811. Since –0.811 < 0.776 < 0.811, r is not significant, and the line should not be used for prediction. Figure 12.16 -0.811 < r = 0.776 < 0.811. Therefore, r is not significant. 12.9 For a given line of best fit, you compute that r = –0.7204 using n = 8 data points, and the critical value is = 0.707. Can the line be used for prediction? Why or why not? THIRD-EXAM vs FINAL-EXAM EXAMPLE: critical value method Consider the third exam/final exam example. The |
line of best fit is: ŷ = –173.51+4.83x with r = 0.6631 and there are n = 11 data points. Can the regression line be used for prediction? Given a third-exam score (x value), can we use the line to predict the final exam score (predicted y value)? H0: ρ = 0 Ha: ρ ≠ 0 α = 0.05 • Use the "95% Critical Value" table for r with df = n – 2 = 11 – 2 = 9. • The critical values are –0.602 and +0.602 • Since 0.6631 > 0.602, r is significant. • Decision: Reject the null hypothesis. • Conclusion:There is sufficient evidence to conclude that there is a significant linear relationship between the third exam score (x) and the final exam score (y) because the correlation coefficient is significantly different from zero. Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION 649 Example 12.10 Suppose you computed the following correlation coefficients. Using the table at the end of the chapter, determine if r is significant and the line of best fit associated with each r can be used to predict a y value. If it helps, draw a number line. a. b. c. d. r = –0.567 and the sample size, n, is 19. The df = n – 2 = 17. The critical value is –0.456. –0.567 < –0.456 so r is significant. r = 0.708 and the sample size, n, is nine. The df = n – 2 = 7. The critical value is 0.666. 0.708 > 0.666 so r is significant. r = 0.134 and the sample size, n, is 14. The df = 14 – 2 = 12. The critical value is 0.532. 0.134 is between –0.532 and 0.532 so r is not significant. r = 0 and the sample size, n, is five. No matter what the dfs are, r = 0 is |
between the two critical values so r is not significant. 12.10 For a given line of best fit, you compute that r = 0 using n = 100 data points. Can the line be used for prediction? Why or why not? Assumptions in Testing the Significance of the Correlation Coefficient Testing the significance of the correlation coefficient requires that certain assumptions about the data are satisfied. The premise of this test is that the data are a sample of observed points taken from a larger population. We have not examined the entire population because it is not possible or feasible to do so. We are examining the sample to draw a conclusion about whether the linear relationship that we see between x and y in the sample data provides strong enough evidence so that we can conclude that there is a linear relationship between x and y in the population. The regression line equation that we calculate from the sample data gives the best-fit line for our particular sample. We want to use this best-fit line for the sample as an estimate of the best-fit line for the population. Examining the scatterplot and testing the significance of the correlation coefficient helps us determine if it is appropriate to do this. The assumptions underlying the test of significance are: • There is a linear relationship in the population that models the average value of y for varying values of x. In other words, the expected value of y for each particular value lies on a straight line in the population. (We do not know the equation for the line for the population. Our regression line from the sample is our best estimate of this line in the population.) • The y values for any particular x value are normally distributed about the line. This implies that there are more y values scattered closer to the line than are scattered farther away. Assumption (1) implies that these normal distributions are centered on the line: the means of these normal distributions of y values lie on the line. • The standard deviations of the population y values about the line are equal for each value of x. In other words, each of these normal distributions of y values has the same shape and spread about the line. • The residual errors are mutually independent (no pattern). • The data are produced from a well-designed, random sample or randomized experiment. 650 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION Figure 12.17 The y values for each x value are normally distributed about the line with the same standard deviation. For each x value, the mean of the y values lies on the regression line. |
More y values lie near the line than are scattered further away from the line. 12.5 | Prediction Recall the third exam/final exam example. We examined the scatterplot and showed that the correlation coefficient is significant. We found the equation of the best-fit line for the final exam grade as a function of the grade on the third-exam. We can now use the least-squares regression line for prediction. Suppose you want to estimate, or predict, the mean final exam score of statistics students who received 73 on the third exam. The exam scores (x-values) range from 65 to 75. Since 73 is between the x-values 65 and 75, substitute x = 73 into the equation. Then: y^ = − 173.51 + 4.83(73) = 179.08 We predict that statistics students who earn a grade of 73 on the third exam will earn a grade of 179.08 on the final exam, on average. Example 12.11 Recall the third exam/final exam example. a. What would you predict the final exam score to be for a student who scored a 66 on the third exam? Solution 12.11 a. 145.27 b. What would you predict the final exam score to be for a student who scored a 90 on the third exam? Solution 12.11 b. The x values in the data are between 65 and 75. Ninety is outside of the domain of the observed x values in the data (independent variable), so you cannot reliably predict the final exam score for this student. (Even though it is possible to enter 90 into the equation for x and calculate a corresponding y value, the y value that you get will not be reliable.) To understand really how unreliable the prediction can be outside of the observed x values observed in the data, make the substitution x = 90 into the equation. y^ = –173.51 + 4.83 ⎛ ⎝90 ⎞ ⎠ = 261.19 The final-exam score is predicted to be 261.19. The largest the final-exam score can be is 200. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 | LINEAR REGRESSION AND CORRELATION 651 NOTE The process of predicting inside of the observed x values observed in the data is called interpolation. The process of |
predicting outside of the observed x values observed in the data is called extrapolation. 12.11 Data are collected on the relationship between the number of hours per week practicing a musical instrument and scores on a math test. The line of best fit is as follows: ŷ = 72.5 + 2.8x What would you predict the score on a math test would be for a student who practices a musical instrument for five hours a week? 12.6 | Outliers In some data sets, there are values (observed data points) called outliers. Outliers are observed data points that are far from the least squares line. They have large "errors", where the "error" or residual is the vertical distance from the line to the point. Outliers need to be examined closely. Sometimes, for some reason or another, they should not be included in the analysis of the data. It is possible that an outlier is a result of erroneous data. Other times, an outlier may hold valuable information about the population under study and should remain included in the data. The key is to examine carefully what causes a data point to be an outlier. Besides outliers, a sample may contain one or a few points that are called influential points. Influential points are observed data points that are far from the other observed data points in the horizontal direction. These points may have a big effect on the slope of the regression line. To begin to identify an influential point, you can remove it from the data set and see if the slope of the regression line is changed significantly. Computers and many calculators can be used to identify outliers from the data. Computer output for regression analysis will often identify both outliers and influential points so that you can examine them. Identifying Outliers We could guess at outliers by looking at a graph of the scatterplot and best fit-line. However, we would like some guideline as to how far away a point needs to be in order to be considered an outlier. As a rough rule of thumb, we can flag any point that is located further than two standard deviations above or below the best-fit line as an outlier. The standard deviation used is the standard deviation of the residuals or errors. We can do this visually in the scatter plot by drawing an extra pair of lines that are two standard deviations above and below the best-fit line. Any data points that are outside this extra pair of lines are flagged as potential outliers. Or we can |
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