IEOR 165 Notes by Andrew Wang

IEOR 165 Notes by Andrew Wang - √ 20 =(332 9892 337...

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IEOR 165: Engineering Statistics, Quality Control and Forecasting, Summer 2009 Homework 1 Solution Chapter 7 Question 8 Noting ¯ X = 3 . 1502 (a) 95 percent CI: ¯ X ± 1 . 96 σ n = 3 . 1502 ± 1 . 96( . 1) / 5 = (3 . 0625 , 3 . 2379) (b) 99 percent CI: ¯ X ± z . 005 σ n = 3 . 1502 ± 12 . 58( . 1) / 5 = (3 . 0348 , 3 . 2656) Question 13 99 percent CI: ¯ X ± z . 005 σ n = 1 . 2 ± z . 005 0 . 2 / 20 = (1 . 0848 , 1 . 3152) Question 14 99 percent CI: ¯ X ± t . 005 ,n - 1 S n = 1 . 2 ± t . 005 , 19 0 . 2 / 20 = (1 . 0720 , 1 . 3280) Question 17 Using the two-sided CI on mean with unknown variance ¯ X ± t α/ 2 ,n - 1 S n (a) 95 percent CI: (331.0572,336.9345) (b) 99 percent CI: (330.0082,337.9836) Question 18 Noting ¯ X = 133 . 22, S = 10 . 2127 (a) 95 percent CI: ¯ X ± t . 025 ,n - 1 S n = (128 . 14 , 138 . 30) (b) 95 percent lower CI: ( -∞ , ¯ X + t . 05 ,n - 1 S n ) = ( -∞ , 137 . 41) (c) 95 percent upper CI: ( ¯ X - t . 05 ,n - 1 S n , ) = (129 . 03 , ) Question 22 (a) 95 percent CI: ¯ X ± t . 025 ,n - 1 S n = 330 . 2 ± 2 . 094(15 . 4) /
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Unformatted text preview: √ 20 = (332 . 9892 , 337 . 4108) (b) 99 percent CI: ¯ X ± t . 05 ,n-1 S √ n = 330 . 2 ± 2 . 861(15 . 4) / √ 20 = (320 . 3480 , 340 . 0520) Question 23 95 percent CI: ¯ X ± t . 025 ,n-1 S √ n = 1220 ± 1 . 968(840) / √ 300 = (1124 . 5571 , 1315 . 4429) Question 26 Using the same CI calculation method as question 17 (a) (2013 . 9 , 2111 . 6) (b) (1996 . , 2129 . 6) (c) This is just the left hand side of a 95 percent upper CI: 2022.4 Question 30 Noting ¯ X = 11 . 5667 ,S = 3 . 98, 95 percent CI: ¯ X ± t . 025 ,n-1 S √ n = (10 . 08 , 13 . 05) Question 31 95 percent CI: ¯ X ± t . 025 ,n-1 S √ n = (85442 . 15 , 95457 . 85) 1...
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This note was uploaded on 07/22/2009 for the course IEOR 165 taught by Professor Shanthikumar during the Summer '08 term at Berkeley.

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