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# ISE361_HW6_Answers_01 - Homework Six Chapter 7 4 a...

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Unformatted text preview: Homework Six Chapter 7 4) a. 58.3 +/- (1.96)(3)/ (25) 1/2 = 58.3 +/- 1.18 = (57.1, 59.5) b. 58.3 +/- (1.96)(3)/ (100) 1/2 = 58.3 +/- .59 = (57.7, 58) c. 58.3 +/- (2.58)(3)/ (100) 1/2 = 58.3 +/- 0.77 = (57.5, 59.10) e. n = [(2)(2.58)(3) / (1)] 2 = 239.62 so next highest whole number n = 240 6) a. 8439 +/- (1.645)(100)/ (25) 1/2 = 8439 +/- 32.9 = (8406.1, 8471.9) b. 1 - o = .92 >>> o = 0.08 >>> o /2 = 0.04 so z o/2 = z 0.04 = 1.75 12) x +/- 2.58 (s)/ (n) 1/2 = 0.81 +/- (2.58)(0.34)/ (110) 1/2 = 0.81 +/- 0.08 = (0.73, 0.89) 14) a. 89.10 +/- (1.96)(3.73)/ (169) 1/2 = 89.10 +/- 0.56 = (88.54, 89.66) Yes, this is a very narrow interval. It appears quite precise. b. n = [(1.96)(0.16)/ (0.5)] 2 = 245.86 >>> n = 246 16) n = 46, x = 382.1, s = 31.5; The 95% Upper Confidence Bound = x + z o (s)/ (n) 1/2 = = 382.1 + (1.645)(31.5)/ (46) 1/2 = 382.1 + 7.64 = 389.74 18) 90% Lower Bound: x – z 0.10 (s)/ (n) 1/2 = 4.25 – (1.28)(1.30)/ (0.75) 1/2 = 4.06 24) n = 56, x = 8.17, s = 1.42; For a 95% C.I., For a 95% C....
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ISE361_HW6_Answers_01 - Homework Six Chapter 7 4 a...

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