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- 1. Chapter 8Hypothesis TestingSection 8-2Steps in HypothesisTesting – TraditionalMethod
- 2. Chapter 8Hypothesis TestingSection 8-2Exercise #12b
- 3. Using the z table find the critical value (or values). = 0.01 Left - tailed test. 0.49 – 2.33 0.01 – 2.33 0
- 4. Chapter 8Hypothesis TestingSection 8-2Exercise # 12g
- 5. Using the z table find the critical value (or values). = 0.05 Right - tailed test. 0.45 1.65 0.05 0 1.65
- 6. Chapter 8Hypothesis TestingSection 8-2Exercise #12h
- 7. Using the z table find the critical value (or values). = 0.01 Two - tailed test. 0.495 0.005 – 2.58 0 2.58 2.58, – 2.58
- 8. Chapter 8Hypothesis TestingSection 8-3z Test for a Mean
- 9. Chapter 8Hypothesis TestingSection 8-3Exercise #5
- 10. A report in USA TODAY stated that the average age ofcommercial jets in the United States is 14 years. Anexecutive of a large airline company selectsa sample of 36 planes and finds the averageage of the planes is 11.8 years. Thestandard deviation of the sample is2.7 years. At = 0.01, can it beconcluded that the average ageof the planes in his Company isless than the national average?
- 11. H0 : 14 H1 : < 14 (claim)C.V. = – 2.33 X – 11.8 – 14 z= = = – 4.89 s 2.7 n 36 – 4.89 – 2.33 0
- 12. H0 : 14 H1 : < 14 (claim)C.V. = – 2.33 X – 11.8 – 14 z= = = – 4.89 s 2.7 n 36 Reject the null hypothesis. There is enough evidence to support the claim that the average age of the planes in the executive’s airline is less than national average.
- 13. Chapter 8Hypothesis TestingSection 8-3Exercise #7
- 14. 25 32 35 25 30 26.5 26 25.5 29.5 3230 28.5 30 32 28 31.5 29 29.5 30 3429 32 27 28 33 28 27 32 29 29.5The average one-year-old (both sexes) is 29 inches tall.A random sample of 30 one-year-olds in a large daycare franchise resulted in the followingheights. At = 0.05, can it be concludedthat the average height differsfrom 29 inches?
- 15. H0 : = 29 H1 : 29 (claim)C.V. = ± 1.96 X = 29.45 s = 2.61 X – 29.45 – 29z= = = 0.944 2.61 n 30 0.944 – 1.96 0 1.96
- 16. H0 : = 29 H1 : 29 (claim)C.V. = ± 1.96 X = 29.45 s = 2.61 X – 29.45 – 29z= = = 0.944 2.61 n 30Do not reject the null hypothesis.There is not enough evidence tosupport the claim that the averageheight differs from 29 inches.
- 17. Chapter 8Hypothesis TestingSection 8-3Exercise #13
- 18. To see if young men ages 8 through 17 years spendmore or less than the national average of $24.44 pershopping trip to a local mall, the managersurveyed 33 young men and found theaverage amount spent per visit was $22.97.The standard deviation of the sample was$3.70. At = 0.02, can it be concludedthat the average amount spent at a localmall is not equal to the nationalaverage of $24.44.
- 19. H0 : = $24.44 H1 : $24.44 (claim)n = 33 X = $22.97s = $3.70 = 0.02C.V. = ± 2.33 – 2.33 0 2.33 X – 22.97 – 24.44z= = = – 2.28 3.70 n 33
- 20. H0 : = $24.44 H1 : $24.44 (claim)n = 33 X = $22.97s = $3.70 = 0.02C.V. = ± 2.33 – 2.28 – 2.33 0 2.33Do not reject the null hypothesis.
- 21. H0 : = $24.44 H1 : $24.44 (claim)n = 33 X = $22.97s = $3.70 = 0.02C.V. = ± 2.33 – 2.28 There is not enough evidence to support the claim that the amount spent at a local mall is not equal to the national average.
- 22. Chapter 8Hypothesis TestingSection 8-3Exercise #17
- 23. A study found that the average stopping distance of a schoolbus traveling 50 miles per hour was 264 feet (Snapshot, USATODAY, March12, 1992). A group of automotiveengineers decided to conduct a study of itsschool buses and found that for 20 buses,the average stopping distance of busestraveling 50 miles per hour was 262.3 feet.The standard deviation of the populationwas 3 feet. Test the claim that the averagestopping distance of the company’sbuses is actually less than 264 feet.Find the P-value. On the basis of theP-value, should the null hypothesis berejected at = 0.01? Assume that thevariable isnormally distributed.
- 24. H0 : 264 H1 : < 264 (claim) X– 262.3 – 264z= = = – 2.53 3 n 20 The area corresponding to z = 2.53 is 0.4943. The P-value is 0.5 – 0.4943 = 0.0057. The decision is to reject the null hypothesis since 0.0057 < 0.01. There is enough evidence to support the claim that the average stopping distance is less than 264 feet.
- 25. Chapter 8Hypothesis TestingSection 8-4t Test for a Mean
- 26. Chapter 8Hypothesis TestingSection 8-4Exercise #3a
- 27. Find the critical value (or values) for the t test for each. n = 10 = 0.05 Right - tailed d.f . = 9 C.V. = + 1.833
- 28. Chapter 8Hypothesis TestingSection 8-4Exercise #3b
- 29. Find the critical value (or values) for the t test for each. n = 18 = 0.10 Two - tailed d.f . = 17 C.V = ± 1.740
- 30. Chapter 8Hypothesis TestingSection 8-4Exercise #3c
- 31. Find the critical value (or values) for the t test for each. n=6 = 0.01 Left - tailed d.f .= 5 C.V.= –3.365
- 32. Chapter 8Hypothesis TestingSection 8-4Exercise #7
- 33. The average salary of graduates entering the actuarialfield is reported to be $40,000. To test this,a statistics professor surveys 20 graduatesand finds their average salary to be $43,228with a standard deviation of $4,000. Using = 0.05, has he shown the reportedsalary incorrect?
- 34. H0 : = $40,000 H1 : $40,000 (claim) = $40,000 n = 20 = 0.05 X = $43,228 s = $4,000C.V. = ± 2.093 d.f . = 19 X– 43,228 – 40,000 t= = = 3.61 4000 n 20 3.61 2.093 0 2.093
- 35. H0 : = $40,000 H1 : $40,000 (claim) = $40,000 n = 20 = 0.05 X = $43,228 s = $4,000C.V. = ± 2.093 d.f . = 19 Reject the null hypothesis. There is enough evidence to support the claim that the average salary is not $40,000.
- 36. Chapter 8Hypothesis TestingSection 8-4Exercise #9
- 37. A researcher estimates that the average height of thebuildings of 30 or more stories in a large city is at least700 feet. A random sample of 10 buildingsis selected, and the heights in feetare shown:485 511 841 725 615520 535 635 616 582At = 0.025, is there enoughevidence to reject the claim?
- 38. H0 : 700 (claim) H1 : < 700 = 700 ft. n = 10 buildings X = 606.5 s = 109.1 = 0.025C.V. = – 2.262 d.f. = 9 X – 606.5 – 700t= = = – 2.71 S 109.1 n 10– 2.71 – 2.262 0
- 39. H0 : 700 (claim) H1 : < 700 = 700 ft. n = 10 buildings X = 606.5 s = 109.1 = 0.025C.V. = – 2.262 d.f. = 9 Reject the null hypothesis. There is enough evidence to reject the claim that the average height of the building is at least 700 feet
- 40. Chapter 8Hypothesis TestingSection 8-4Exercise #13
- 41. Last year the average cost of making a movie was $54.8million. This year, a random sample of 15 recent actionmovies had an average production cost of$62.3 million with a variance of $90.25million.At the 0.05 level of significance,can it be concluded that it costs morethan average to produce an action movie?
- 42. H0 : $54.8 H1: > $54.8 (claim) = $ 54.8 million n = 15 movies X = $ 62.3 million s 2 = $ 90.5 million = 0.05 C.V. = 1.761 d.f . = 14 X– $62.3 – $54.8t= = = 3.06 S 9.5 n 15 3.06 0 1.761
- 43. H0 : $54.8 H1: > $54.8 (claim) = $ 54.8 million n = 15 movies X = $ 62.3 million s 2 = $ 90.5 million = 0.05 C.V. = 1.761 d.f . = 14 Reject the null hypothesis. There is enough evidence to support the claim that the cost to produce and action movie is more than $54.8 million.
- 44. Chapter 8Hypothesis TestingSection 8-4Exercise #17
- 45. A report by the Gallup Poll stated that on average a woman visits her physician 5.8 times a year. A researcher randomly selected 20 women and the following data was obtained. 3 2 1 3 7 2 9 4 6 6 8 0 5 6 4 2 1 3 4 1 At = 0.05 can it be concluded that the average is still 5.8? Use the P - value method.H0 : = 5.8 (claim) H1 : 5.8
- 46. H0 : = 5.8 (claim) H1 : 5.8 P - value < 0.05 = 5.8 X = 3.85 n = 20 women s = 2.52 = 0.05 d.f. = 19 X– 3.85 – 5.8 t= = = – 3.46 S 2.52 n 20
- 47. H0 : = 5.8 (claim) H1 : 5.8 P - value < 0.05 = 5.8 X = 3.85 n = 20 women s = 2.52 = 0.05 d.f. = 19 Since the P - value is less than 0.05, reject the null hypothesis. There is not enough evidence to support the claim that the average number of visits is the same as in the Gallup Poll.
- 48. Chapter 8Hypothesis TestingSection 8-5z Test for a Proportion
- 49. Chapter 8Hypothesis TestingSection 8-5Exercise #7
- 50. It has been reported that 40% of the adult populationparticipates in computer hobbies during their leisuretime. A random sample of 180 adults foundthat 65 engaged in computer hobbies.At = 0.01, is there sufficient evidenceto conclude that the proportion differsfrom 40%?H0 : p = 0.40 H1 : p 0.40 (claim)
- 51. H0 : p = 0.40 H1 : p 0.40 (claim) p = 65 = 0.361 p = 0.40 ˆ q = 0.60 180 C.V . = ± 2.58 ˆ p–p = 0.361 0.40 = – 1.07 – z= pq (0.40)(0.60) n 180 – 1.07 – 2.58 0 2.58
- 52. H0 : p = 0.40 H1 : p 0.40 (claim) p = 65 = 0.361 ˆ p = 0.40 q = 0.60 180 C.V . = ± 2.58 Do not reject the null hypothesis. There is not enough evidence to support the claim that the proportion is different from 0.40 (40%).
- 53. Chapter 8Hypothesis TestingSection 8-5Exercise #9
- 54. An item in USA TODAY reported that 63% of Americansowned an answering machine. A survey of 143employees at a large school showed that85 owned an answering machine. At = 0.05, test the claim that thepercentage is the same asstated in USA TODAY .H0 : p = 0.63 (claim) H1 : p 0.63
- 55. H0 : p = 0.63 (claim) H1 : p 0.63 p = 85 = 0.5944 ˆ p = 0.63 q = 0.37 143 C.V . = ± 1.96 ˆ p–p 0.5944 – 0.63 z= = = – 0.88 pq (0.63)(0.37) n 143 – 0.88 – 1.96 0 1.96
- 56. H0 : p = 0.63 (claim) H1 : p 0.63 p = 85 = 0.5944 ˆ p = 0.63 q = 0.37 143 C.V . = ± 1.96 ˆ p–p 0.5944 – 0.63 z= = = – 0.88 pq (0.63)(0.37) n 143 Do not reject the null hypothesis. There is not enough evidence to reject the claim that the percentage is the same.
- 57. Chapter 8Hypothesis TestingSection 8-5Exercise #15
- 58. Researchers suspect that 18% of all high school studentssmoke at least one pack of cigarettes a day. At WilsonHigh School, with an enrollment of 300students, a study found that 50 studentssmoked at least one pack of cigarettesa day. At = 0.05, test the claim that18% of all high school students smokeat least one pack of cigarettes a day.Use the P - value method.
- 59. H0 :p = 0.18 (claim) H1 : p 0.18 50ˆp= = 0.1667 p = 0.18 q = 0.82 300 ˆ p–p 0.1667 – 0.18z= = = – 0.60 pq (0.18)(0.82) n 300Area = 0.2257P - value = 2 (0.5 – 0.2257) = 0.5486
- 60. H0 :p = 0.18 (claim) H1 : p 0.18 50ˆp= = 0.1667 p = 0.18 q = 0.82 300 ˆ p–p 0.1667 – 0.18z= = = – 0.60 pq (0.18)(0.82) n 300 Since P - value > 0.05, do not reject the null hypothesis. There is not enough evidence to reject the claim that 18% of all high school students smoke at least a pack of cigarettes a day.
- 61. Chapter 8Hypothesis TestingSection 8-5Exercise #19
- 62. A report by the NCAA states that 57.6% of footballinjuries occur during practices. A head trainer claimsthat this is too high for his conference, sohe randomly selects 36 injuries and findsthat 17 occurred during practices. Is hisclaim correct, using = 0.05 ?H0 : p 0.576 H1: p < 0.576 (claim)
- 63. H0 : p 0.576 H1: p < 0.576 (claim) = 0.05 C.V. = –1.65p = 17 = 0.472 , p = 0.576 q = 0.424 n = 36ˆ 36 ˆ p –p 0.472 – 0.576 = – 1.26z= = pq (0.576)(0.424) n 36 0.05 – 1.65 0
- 64. H0 : p 0.576 H1: p < 0.576 (claim) = 0.05 C.V. = –1.65 Do not reject the null hypothesis. There is not enough evidence to support the claim that the percentage of injuries during practice is below 57.6%. 0.05 – 1.26 – 1.65 0
- 65. Chapter 8Hypothesis TestingSection 8-6c2 Test for a Variance orStandard Deviation
- 66. Chapter 8Hypothesis TestingSection 8-6Exercise #5
- 67. Test the claim that the standard deviation of the numberof aircraft stolen each year in the United States is lessthan 15 if a sample of 12 years had astandard deviation of 13.6. Use = 0.05.H1 : < 15 (claim) H0 : 15
- 68. H1 : < 15 (claim) H0 : 15 = 0.05 C.V . = 4.575 d .f . 11 2c 2 = (n – 1)s = (12 – 1)(13.6)2 2 2 = 9.04 15 9.04 0.0250 4.575 Do not reject the null hypothesis. 9.04 There is not enough evidence to support the claim that the standard deviation is less than 15.
- 69. Chapter 8Hypothesis TestingSection 8-6Exercise #7
- 70. The manager of a large company claims that thestandard deviation of the time (in minutes) that it takesa telephone call to be transferred to the correct office inher company is 1.2 minutes or less. Asample of 15 calls is selected, and thecalls are timed. The standard deviationof the sample is 1.8 minutes. At = 0.01,test the claim that the standard deviationis less than or equal to 1.2 minutes. Usethe P-value method.H0 : 1.2 (claim) H1 : > 1.2
- 71. H0 : 1.2 (claim) H1 : > 1.2 = 0.01 s = 1.8 n = 15 d.f. = 14 (n – 1)s2 (15 – 1)(1.8)2c 2= = = 31.5 2 (1.2)2 31.5 0.0050 31.319P - value < 0.005 (0.0047)
- 72. Reject the null hypothesis. There is not enough evidence to support the claim that the standard deviation is less than or equal to 1.2 minutes. 31.5 0.0050 31.319P - value < 0.005 (0.0047)
- 73. Chapter 8Hypothesis TestingSection 8-6Exercise #9
- 74. 290 320 260 220 300 310 310 270 250 230270 260 310 200 250 250 270 210 260 300A random sample of 20 differentkinds of doughnuts had thefollowing calorie contents. At = 0.01, is there sufficientevidence to conclude that thestandard deviation is greaterthan 20 calories?
- 75. H0 : 20 H1 : > 20 (claim)s = 35.11 C.V. = 36.191 = 0.01 d.f . = 19c 2= (n – 1)s2 (20 – 1)(35.11)2 = 58.55 = 2 202 58.550 36.191
- 76. H0 : 20 H1 : > 20 (claim)s = 35.11 C.V. = 36.191 = 0.01 d.f . = 19c 2= (n – 1)s2 (20 – 1)(35.11)2 = 58.55 = 2 202 Reject the null hypothesis. There is enough evidence to support the claim that the standard deviation is greater than 20.
- 77. Chapter 8Hypothesis TestingSection 8-6Exercise #13
- 78. 34 47 43 23 36 50 4244 43 40 39 41 47 45A random sample of home runtotals for National League HomeRun Champions from 1938 to 2001is shown. At the 0.05 level ofsignificance, is there sufficientevidence to conclude that thevariance isgreater than 25?
- 79. H0 : 2 25 H1 : 2 > 25( claim) s 2 = 45.38 = 0.05 d.f . = 13 C.V. = 22.362 2 = (n – 1)s = 14 – 1 45.38 = 23.622 2c 2 25 23.6 0 22.362
- 80. H0 : 2 25 H1 : 2 > 25( claim) s 2 = 45.38 = 0.05 d.f . = 13 C.V. = 22.362 c 2 23.622 Reject the null hypothesis. There is evidence to support the claim that the variance is greater than 25. 23.6 0 22.362
- 81. Chapter 8Hypothesis TestingSection 8-7Additional TopicsRegarding HypothesisTesting
- 82. Chapter 8Hypothesis TestingSection 8-7Exercise #1
- 83. A ski shop manager claims that the average of the salesfor her shop is $1800 a day during the winter months.Ten winter days are selected at random,and the mean of the sales is $1830. Thestandard deviation of the population is$200. Can one reject the claim at = 0.05?Find the 95% confidence interval of themean. Does the confidence intervalInterpretation agree with thehypothesis test results? Explain.Assume that the variable isnormally distributed.H0 : = 1800 (claim) H1 : 1800
- 84. H0 : = 1800 (claim) H1 : 1800 C.V. = ± 1.96 X– 1830 – 1800 z= = = 0.47 200 n 10 0.47 – 1.96 0 1.96 Do not reject the null hypothesis. There is not evidence to reject the claim that the average of the sales in $1800.
- 85. The 95% confidence interval of the mean is: X – z < <X +z 2 n 2 n 200 2001830 – 1.96( ) < < 1830 + 1.96( ) 10 10 1706.04 < < 1953.96 The hypothesized mean is within the interval, thus we can be 95% confident that the average sales will be between $1,706.94 and $1,953.96.
- 86. Chapter 8Hypothesis TestingSection 8-7Exercise #5
- 87. From past studies the average time college freshmenspend studying is 22 hours per week. The standarddeviation is 4 hours. This year, 60 studentswere surveyed, and the average time thatthey spent studying was 20.8 hours. Testthe claim that the time students spendstudying has changed. Use = 0.01. It isbelieved that the standard deviationis unchanged. Find the 99%confidence interval of the mean.Do the results agree? Explain.H0 : = 22 H1 : 22 (claim)
- 88. H0 : = 22 H1 : 22 (claim) C.V. = ± 2.58 X– 20.8 – 22 = – 2.32 z= = 4 n 60 – 2.32 – 2.58 0 2.58
- 89. H0 : = 22 H1 : 22 (claim) C.V. = ± 2.58 Do not reject the null hypothesis. There is not enough evidence to support the claim that the average studying time has changed. – 2.32 – 2.58 0 2.58
- 90. The 99% confidence interval of the mean is: X – z < <X +z 2 n 2 n 4 420.8 – 2.58( ) < < 20.8 – 2.58( ) 60 60 19.47 < < 22.13The 99% confidence interval containsthe hypothesized mean of 22. Thereis not enough evidence to supportthe claim that the average studying time has changed.

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