3044hw4s09sols - E i = 10, the following table is...

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2 ISyE 3044/Spring 2009/Solutions to Exercises 3-4 from Homework #4 2. (a) We have the following table interval i [0 , 0 . 25) [0 . 25 , 0 . 50) [0 . 50 , 0 . 75) [0 . 75 , 1) O i 2 7 7 3 E i 5 5 5 5 The test statistic is χ 2 0 = (2 - 5) 2 5 + (7 - 5) 2 5 + (7 - 5) 2 5 + (3 - 5) 2 5 = 21 5 = 4 . 2 . Since χ 2 0 χ 2 3 , 0 . 10 = 6 . 25, we fail to reject H 0 . (b) The multipliers are a = 2 , 6 , 7 , 8. For each of these multipliers, verify that the smallest integer k such that a k - 1 is divisible by m = 11 is k = m - 1 = 10. (See page 255 of the text.) 7.4 X 0 = 27 , a = 8 , c = 47 , m = 100. X 1 = (8 × 27 + 47)mod 100 = 63 , R 1 = 63 / 100 = . 63. X 2 = (8 × 63 + 47)mod 100 = 51 , R 2 = 51 / 100 = . 51. X 3 = (8 × 51 + 47)mod 100 = 55 , R 3 = 55 / 100 = . 55. 7.8 Let ten intervals be de±ned each from (10 i - 9) to (10 i ) where i = 1 , 2 ,..., 10. By counting the numbers that fall within each interval and comparing this to the expected value for each interval,
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Unformatted text preview: E i = 10, the following table is generated: Interval O i ( O i-E i ) 2 /E i (01-10) 9 0.1 (11-20) 9 0.1 (21-30) 9 0.1 (31-40) 6 1.6 (41-50) 17 4.9 (51-60) 5 2.5 (61-70) 10 0.0 (71-80) 12 0.4 (81-90) 7 0.9 (91-00) 16 3.6 100 14 . 2 = χ 2 3 From Table A.6, χ 2 . 05 , 9 = 16 . 9. Since χ 2 < χ . 05 , 9 , then the null hypothesis of no di±erence between the sample distribution and the uniform distribution is not rejected. 7.10 X 1 = [7 × 37 + 29] mod 100 = 88 and R 1 = . 88. X 2 = [7 × 88 + 29] mod 100 = 45 and R 2 = . 45. X 3 = [7 × 45 + 29] mod 100 = 44 and R 3 = . 44....
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This note was uploaded on 09/17/2009 for the course ISYE 3044 taught by Professor Alexopoulos during the Spring '08 term at Georgia Institute of Technology.

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3044hw4s09sols - E i = 10, the following table is...

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