# 61 a h µ 900 h a µ 900 b 025 x z n σ 180 935 196

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61. a. H 0 : µ = 900 H a : µ 900 b. .025 x z n σ ± 180 935 1.96 200 ± 935 ± 25 (910 to 960) c. Reject H 0 because µ = 900 is not in the interval. d. 0 935 900 2.75 / 180/ 200 x z n µ σ = = = p -value = 2(.5000 - .4970) = .0060 62. a. H 0 : µ 45,250 H a : µ > 45,250 b. 0 47,000 45,250 2.71 / 6300/ 95 x z n µ σ = = = p -value = .5000 - .4966 = .0034 p -value .01; reject H 0 . Conclude New York City has higher mean salary. 63. a. H 0 : µ 37,000 H a : µ > 37,000 b. 0 38,000 37,000 1.47 / 5200/ 48 x t s n µ = = = Degrees of freedom = n – 1 = 47 Using t table, p -value is between .05 and .10 Actual p -value = .0747 c. p -value > .05, do not reject H 0 . Cannot conclude mean greater than \$37,000. A larger sample is desirable.

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Chapter 9 9 - 26 64. H 0 : µ = 6000 H a : µ 6000 0 5812 6000 .93 / 1140/ 32 x t s n µ = = = − Degrees of freedom = n – 1 = 31 Using t table, area in tail is between .10 and .20 p -value is between .20 and .40 Actual p -value = .3581 Do not reject H 0 . There is no evidence to conclude that the mean number of freshman applications has changed. 65. a. H 0 : µ 30 H a : µ < 30 b. 0 29.6 30 1.57 / 1.8/ 50 x t s n µ = = = − Degrees of freedom = 50 – 1 = 49 Using t table, p -value is between .05 and .10 Actual p -value = .0613 c. p -value > .01; do not reject H 0 . Cannot conclude miles per gallon is less than 30. 66. H 0 : µ 125,000 H a : µ > 125,000 0 130,000 125,000 2.26 / 12,500/ 32 x t s n µ = = = Degrees of freedom = 32 – 1 = 31 Using t table, p -value is between .01 and .025 Actual p -value = .0154 p -value .05; reject H 0 . Conclude that the mean cost is greater than \$125,000 per lot. 67. a. H 0 : µ = 3 H a : µ 3
Hypothesis Testing 9 - 27 b. 2.8 i x x n Σ = = c. ( ) 2 .70 1 i x x s n Σ = = d. 0 2.8 3 .90 / .70/ 10 x t s n µ = = = − Degrees of freedom = 10 - 1 = 9 Using t table, area in tail is between .10 and .20 p -value is between .20 and .40 Actual p -value = .3902 e. p -value > .05; do not reject H 0 . There is no evidence to conclude a difference compared to prior year. 68. a. H 0 : p .50 H a : p > .50 b. 64 .64 100 p = = c. 0 0 0 .64 .50 2.80 (1 ) .50(1 .50) 100 p p z p p n = = = Area in tail = .4974 p -value = .5000 - .4974 = .0026 p -value .01; reject H 0 . College graduates have a greater stop-smoking success rate. 69. a. H 0 : p = .6667 H a : p .6667 b. 355 .6502 546 p = = c. 0 0 0 .6502 .6667 .82 (1 ) .6667(1 .6667) 546 p p z p p n = = = − p -value = 2(.5000 - .2939) = .4122

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Chapter 9 9 - 28 p -value > .05; do not reject H 0 ; Cannot conclude that the population proportion differs from 2/3. 70. a. H 0 : p .50 H a : p > .50 b. 67 .6381 105 p = = (64%) c. 0 0 0 .6381 .50 2.83 (1 ) .50(1 .50) 105 p p z p p n = = = p -value = .5000 - .4977 = .0023 p -value .01; reject H 0 . Conclude that the four 10-hour day schedules is preferred by more than 50% of the office workers. 71. a. 330 .825 400 p = = b. H 0 : p = .78 H a : p .78 0 0 0 .825 .78 2.17 (1 ) .78(1 .78) 400 p p z p p n = = = p -value = 2(.5000 - .4850) = .03 c. p -value .05; reject H 0 . The on-time arrival record has changed. It is improving. 72. H 0 : p .90 H a : p < .90 49 .8448 58 p = = 0 0 0 .8448 .90 1.40 (1 ) .90(1 .90) 58 p p z p p n = = = − p -value = .5000 - .4192 = .0808 p -value > .05; do not reject H 0 . Claim of at least 90% cannot be rejected. 73. a. H 0 : p .47 H a : p < .47
Hypothesis Testing 9 - 29 b. 44 .352 125 p = = c. 0 0 0 .352 .47 2.64 (1 ) .47(1 .47) 125 p p z p p n = = = − p -value = .5000 - .4959 = .0041 d. p -value .01; reject H 0 . The proportion of foods containing pesticides has declined.

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• Spring '08
• Staff
• Statistical hypothesis testing, β

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