Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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36-220 Engineering Stats, Fall 2005 Homework 8 Solutions Due: November 9, 2005 Exercise 8.4 We consider these two hypothesis tests: Test 1 Test 2 H 0 : μ = 5 μ = 5 H 1 : μ > 5 μ < 5 A desirable hypothesis test keeps the occurrences of type I error (false positives) and type II error (false negatives) low. However, decreasing Pr (type I error) increases Pr (type II error). So we will choose the test that keeps the more serious error (between type I and II) low. In the context of this problem, a type I error in test 2 is the event concluding that the water is safe when that’s not the case. The type II error in test 2 is judging the water is unsafe when that is not true. It should be clear that type I error in test 2 is more serious than type II error. A false positive here, thinking the water is safe when it’s not, is more likely to result in deaths. So if we pick test 2, we need to try to control (i.e. reduce) the type I error instead of the type II. It’s the other way around if we pick test 1. In that case, we need to control type II error instead of type I. Exercise 8.8 Parameters involved μ 1 = the true average amount of warpage for the regular laminate μ 2 = the true average amount of warpage for the special laminate Hypotheses being tested H 0 : μ 1 = μ 2 vs. H 1 : μ 1 > μ 2 The Errors Type I: Concluding that the special laminate produces less warpage than the regular laminate when it does not. Type II: Concluding that there is no difference in the two laminates when in reality, the special laminate produces less warpage. 1
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Exercise 8.10 The rejection region is defined: { ¯ X 1331 . 26 } . a) H 0 : μ = 1300 vs. H 1 : μ > 1300 b) By CLT, ¯ X N ( μ, σ 2 /n = 13 . 416 2 ) ¯ X N (1300 , 13 . 416 2 ) under H 0 . Then, it follows, Pr (type I error)= Pr (reject H 0 when H 0 is true) = Pr ( ¯ X 1331 . 26 when μ = 1300) = Pr ( Z 2 . 33) = 0 . 01 c) ¯ X N (1350 , 13 . 416 2 ) when μ = 1350. The following results: Pr (type II error) = Pr ( ¯ X < 1331 . 26 when μ = 1350) = Pr ( Z ≤ - 1 . 40) = 0 . 0808 d) Recall that the probability of type I error is the same as the significance level of a hypothesis test.
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