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Unformatted text preview: Solutions to Homework #9 (W4150 ‘Intro to Probability and Statistics’, S04) Sec. 9.11. (1) (a) n = 200, 05 . 2 , 43 . ˆ , 57 . 200 / 114 ˆ 02 . 2 / = = = = = z z q p α Hence using the formula on p.258 we obtain the interval of the form: 200 / ) 43 )(. 57 (. 05 . 2 57 . 200 / ) 43 )(. 57 (. 05 . 2 57 . + < < p or .498 < p < .642 (b) By Theorem 9.3 we can be 96% sure that the error will not exceed 072 . 200 / ) 43 )(. 57 (. 05 . 2 = Sec.9.11. (3) n = 1000, 575 . 2 , 772 . ˆ , 228 . 1000 / 228 ˆ 005 . 2 / = = = = = z z q p α Hence by the same formula we obtain the following interval: or .194 < p < .262 Sec. 9.11. (9) By Theorem 9.4 we have the size of sample equal to n = (2.05) 2 (.57)(.43)/(.02) 2 = 2576 when rounded up Sec. 9.11 (10) Applying the same theorem we have n = (2.575) 2 (.228)(.772)/(.05) 2 = 467 after rounding up Sec. 9.11. (17) Here we have: 645 . 1 , 804 . ˆ , 196 . 500 / 98 ˆ 76 ....
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This note was uploaded on 06/02/2010 for the course IEOR SIEO W4150 taught by Professor Gallego during the Spring '04 term at Columbia.
 Spring '04
 GALLEGO

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