# 409Quiz3Bans - z 0.10 = – 1.282 Decision The value of the...

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STAT 409 Fall 2011 Version B Name ANSWERS . Quiz 3 (10 points) Be sure to show all your work; your partial credit might depend on it. No credit will be given without supporting work. 1. In a random sample of 80 female customers at Burger Queen , 48 ordered fries with their burgers. a) (5) Test H 0 : p = 0.70 vs. H 1 : p < 0.70 at a 10% level of significance, where p is the proportions of female customers who order fries with their burgers. Y = 48. n = 80. 80 48 Y ˆ = = n p = 0.60. Test Statistic: Z = ( ) 80 30 . 0 70 . 0 70 . 0 60 . 0 1 0 0 0 ˆ - = - - n p p p p = 1.95 . P-value: Left – tailed. P-value = P( Z 1.95 ) = 0.0256 . Decision: P-value < α = 0.10. Reject H 0 at α = 0.10. OR Rejection Region: Reject H 0 if Z < – z α

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Unformatted text preview: z 0.10 = – 1.282. Decision: The value of the test statistic is in the Rejection Region. Reject H at α = 0.10. b) (5) Suppose also that in a random sample of 120 male customers, 84 ordered fries with their burgers. Find the p-value of the test H : p M = p F vs. H 1 : p M > p F . 120 84 Y 1 1 1 ˆ = = n p = 0.70. 80 48 Y 2 2 2 ˆ = = n p = 0.60. 66 . ˆ 200 132 80 120 48 84 Y Y 2 1 2 1 = = + + = + + = n n p . Test Statistic: Z = +-⋅ ⋅ 80 1 120 1 34 . 66 . 60 . 70 . = 1.46 . P-value: Right – tailed. P-value = ( area of the right tail ) = P( Z ≥ 1.46 ) = 0.0721 ....
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## This note was uploaded on 10/13/2011 for the course STAT 409 taught by Professor Stephanov during the Fall '11 term at University of Illinois at Urbana–Champaign.

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409Quiz3Bans - z 0.10 = – 1.282 Decision The value of the...

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