HW9Solutions

HW9Solutions - 3. a. b. H o : = 1300 v. H a : > 1300 x is...

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3. a. 1300 : = μ o H v. 1300 : > a H b. x is normally distributed with mean ( ) = x E and standard deviation . 416 . 13 20 60 = = n σ When H 0 is true, ( ) 1300 = x E . Thus 26 . 1331 ( = x P α when H 0 is true) = () 01 . 33 . 2 416 . 13 1300 26 . 1331 = = z P z P c. When 1350 = , x has a normal distribution with mean 1350 and standard deviation 13.416, so 26 . 1331 ( 1350 < = x P β when ) 1350 = = 0808 . 40 . 1 416 . 13 1350 26 . 1331 = = z P z P d. Replace 1331.26 by c, where c satisfies 645 . 1 416 . 13 1300 = c (since . Thus c = 1322.07. Increasing ) 05 . ) 645 . 1 ( = z P gives a decrease in ; now ( ) 0188 . 08 . 2 1350 = = z P . e. 26 . 1331 x iff 416 . 13 1300 26 . 1331 z i.e. iff . 33 . 2 z 4. a. H 0 : 5 . 5 = v.. H a : 5 . 5 ; for a level .01 test, (not specified in the problem description), reject H 0 if either or 58 . 2 z 58 . 2 z . Since 58 . 2 33 . 3 075 . 5 . 5 25 . 5 = = z , reject H 0 . b. ( ) ( ) + Φ + + Φ = 075 . 1 . 58 . 2 075 . 1 . 58 . 2 1 6 . 5 1 ( ) ( ) 105 . 91 . 3 25 . 1 1 = Φ + Φ = c. 97 . 216 1 . 33 . 2 58 . 2 3 . 2 = + = n , so use n = 217. 5. a. Let X = the number of couples who lean more to the right when they kiss. If n = 124 and p = 2/3, then E[X] = 124(2/3) = 82.667. The researchers observed x = 80, for a difference of 2.667. The probability in question is P(|X – 82.667| 2.667) = P(X 80 or X 85.33) = P(X 80) + [1 – P(X 85)] = Bin(80;124,2/3) + [1 – Bin(85;124,2/3)] = 0.634. (Using a large- sample z -based calculation gives a probability of 0.610.) b. We wish to test H 0 : p = 2/3 v. H a : p 2/3. From the data, 645 . 124 80 ˆ = = p , so our test statistic is 124 / ) 333 (. 667 .
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This note was uploaded on 09/06/2009 for the course ENGRD 2700 taught by Professor Staff during the Spring '05 term at Cornell.

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HW9Solutions - 3. a. b. H o : = 1300 v. H a : > 1300 x is...

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