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

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Unformatted text preview: 36-220 Engineering Stats, Fall 2005Homework 8 SolutionsDue: November 9, 2005Exercise 8.4We consider these two hypothesis tests:Test 1Test 2H:= 5= 5H1: >5 <5A desirable hypothesis test keeps the occurrences of type I error (false positives) and type II error(false negatives) low. However, decreasingPr(type I error) increasesPr(type II error). So we willchoose the test that keeps the more serious error (between type I and II) low. In the context of thisproblem, a type I error in test 2 is the event concluding that the water is safe when thats not thecase. The type II error in test 2 is judging the water is unsafe when that is not true. It should beclear that type I error in test 2 is more serious than type II error. A false positive here, thinkingthe water is safe when its not, is more likely to result in deaths. So if we pick test 2, we need totry to control (i.e. reduce) the type I error instead of the type II. Its the other way around if wepick test 1. In that case, we need to control type II error instead of type I.Exercise 8.8Parameters involved1= the true average amount of warpage for the regular laminate2= the true average amount of warpage for the special laminateHypotheses being testedH:1=2vs.H1:1> 2The ErrorsType I: Concluding that the special laminate produces less warpage than the regular laminate whenit does not. Type II: Concluding that there is no difference in the two laminates when in reality,the special laminate produces less warpage.1Exercise 8.10The rejection region is defined:{X1331.26}.a)H:= 1300 vs.H1: >1300b) By CLT,XN(,2/n= 13.4162)XN(1300,13.4162) underH. Then, it follows,Pr(type I error)=Pr(rejectHwhenHis true) =Pr(X1331.26 when= 1300)=Pr(Z2.33) = 0.01c)XN(1350,13.4162) when= 1350. The following results:Pr(type II error) =Pr(X <1331.26 when= 1350) =Pr(Z -1.40) = 0.0808d) Recall that the probability of type I error is the same as the significance level of a hypothesis test.Thus, the test in part b) is a hypothesis test with 1% significance level. In order to derive a test with5% significance level, we just need to change the rejection region so thatPr(type I error) = 0.05....
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hw8 solutions - 36-220 Engineering Stats, Fall 2005Homework...

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