409Quiz2Cans - α-α-χ ⋅ χ ⋅ 2 2 1 2 2 2 2 s s 1 1 n n 90 confidence level α = 0.10 α 2 = 0.05 n-1 = 15-1 = 14 degrees of freedom 2 05 2 2

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STAT 409 Fall 2011 Version C Name ANSWERS . Quiz 5 (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. A sample of 15 adult white rhinoceroses had an average weight of 5,800 pounds, with a sample standard deviation of 410 pounds. Assume that the population of adult white rhino weights is normally distributed. a) (3) Construct a 95% confidence interval for the overall average weight of adult white rhinos. σ is unknown The confidence interval: X t s ± α 2 n . n - 1 = 15 - 1 = 14 degrees of freedom α = 0.05 t 0.025 ( 14 ) = 2.145 15 410 145 . 2 800 , 5 ± 5,800 ± 227 ( 5,573 ; 6,027 ) b) (4) Construct a 90% confidence interval for the overall standard deviation of the weights of adult white rhinos. The confidence interval : ( ) ( )
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Unformatted text preview: α-α--χ ⋅ χ ⋅ 2 2 1 2 2 2 2 s s 1 , 1 n n . 90% confidence level α = 0.10 α 2 = 0.05 . n-1 = 15 -1 = 14 degrees of freedom. 2 05 . 2 2 χ χ = α = 23.68. 2 95 . 2 2 1 χ χ =-= 6.571. ( ) ( ) --⋅ ⋅ 571 . 6 410 1 15 , 68 . 23 410 1 15 2 2 ( 315.2514 , 598.4559 ) 1. (continued) c) (3) What is the minimum sample size required if we wish to estimate the overall average weight of adult white rhinos to within 70 pounds with 95% confidence, if the overall standard deviation of the weights is 400 pounds? ε = 70 σ = 400 α = 0.05 z 0.025 = 1.96 2 2 2 70 400 96 . 1 α ε σ z = = ⋅ ⋅ n = 125.44. Round up. n = 126 ....
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This note was uploaded on 09/23/2011 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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409Quiz2Cans - α-α-χ ⋅ χ ⋅ 2 2 1 2 2 2 2 s s 1 1 n n 90 confidence level α = 0.10 α 2 = 0.05 n-1 = 15-1 = 14 degrees of freedom 2 05 2 2

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