Stat HW5

# Stat HW5 - Statistics 285 Solutions to HW#5 Problems 6.12...

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Unformatted text preview: Statistics 285 Solutions to HW #5 Problems 6.12 Let p = proportion of U.S. companies that have formal, written travel and entertainment policies for their employees. The null hypothesis would be: H : p = .80 6.13 Let μ = mean caloric content of Virginia school lunches. To test the claim that after the testing period ended, the average caloric content dropped, we test: H : μ = 863 H a : μ < 863 6.18 a. The null hypothesis is: H o : There is no intrusion. b. The alternative hypothesis is: H a : There is an intrusion. c. α = P (warning | no intrusion) = 1 .001 1000 = . β = P (no warning | intrusion) = 500 .5 1000 = . 6.24 a. To determine whether the mean July, 2006 dealer price of the Toyota Prius differs from \$25,000, we test: H : μ = 25,000 H a : μ ≠ 25,000 b. The sample mean is 69 . 476 , 25 160 271 , 076 , 4 = = = ∑ n x x i The sample variance is: ( 29 862 . 057 , 904 , 5 1 160 160 271 , 076 , 4 115 , 653 , 788 , 104 1 2 2 2 2 =-- =-- = ∑ ∑ n n x x s i i The sample standard deviation is: 2 5,904, 057.862 2, 429.8267 s s = = = c. The test statistic is 25, 476.69 25, 000 2.48 2, 429.8267 160 o x x z μ σ-- = ≈ = d. The rejection region requires α /2 = .05/2 = .025 in each tail of the z-distribution. From Table IV, Appendix B, z .025 = 1.96. The rejection region is z < - 1.96 or z > 1.96. e. Since the observed value of the test statistic falls in the rejection region ( z = 2.48 > 1.96), H o is rejected. There is sufficient evidence to indicate the mean July, 2006 dealer price of the Toyota Prius differs from \$25,000 at α = .05. 6.26 a. A Type I error is rejecting H when H is true. In this case, we would conclude that the mean number of carats per diamond is different from .6 when, in fact, it is equal to .6. A Type II error is accepting H when H is false. In this case, we would conclude that the mean number of carats per diamond is equal to .6 when, in fact, it is different from .6. b. From Exercise 5.18, the random sample of 30 diamonds yielded x = .691 and s = .262. Let μ = mean number of carats per diamond. To determine if the mean number of carats per diamond is different from .6, we test: H : μ = .6 H a : μ ≠ .6 The test statistic is .691 .6 1.90 .262 30 x x z μ σ-- = ≈ = The rejection region requires α /2 = .05/2 = .025 in each tail of the z- distribution. From Table IV, Appendix B, z .025 = 1.96. The rejection region is z > 1.96 or z < - 1.96. Since the observed value of the test statistic does not fall in the rejection region ( z = 1.90 / 1.96), H is not rejected. There is insufficient evidence to indicate the mean number of carats per diamond is different from .6 carats at α = .05. c. When α is changed, H , H a , and the test statistic remain the same....
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Stat HW5 - Statistics 285 Solutions to HW#5 Problems 6.12...

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