hw8 solutions

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

• Homework Help
• PresidentHackerCaribou10582
• 5

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

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.
This is the end of the preview. Sign up to access the rest of the document.
• Fall '05
• Shalizi
• Statistics, Null hypothesis, Statistical hypothesis testing, Type I and type II errors, significance level

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern