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Unformatted text preview:  2 2 1 2 2 2 2 s s 1 , 1 n n . 95% confidence level = 0.05 2 = 0.025 . n1 = 6 1 = 5 degrees of freedom. 2 025 2 2 . = = 12.83. 2 975 2 2 1 . = = 0.831. ( ) ( )  831 . 420 1 6 , 83 . 12 420 1 6 2 2 ( 262.193 , 1030.229 ) 1. (continued) c) (3) What is the minimum sample size required if we wish to estimate the overall average weight of adult elephants to within 100 pounds with 90% confidence, if the overall standard deviation of the weights is 400 pounds? = 100 = 400 = 0.10 z 0.05 = 1.645 2 2 2 100 400 645 . 1 z = = n = 43.2964. Round up. n = 44 ....
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 Spring '08
 Kim

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