# PS6s - SOLUTIONS TO ISMT111 PROBLEM SET #6 (1) Answer:...

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SOLUTIONS TO ISMT111 PROBLEM SET #6 (1) Answer: 98.0% n = 9 , (small); x = 83 . 7 , s = 12 . 9 , σ = unknown; X approx . normal . B = 96 . 153 - 83 . 7 = 12 . 453 = t 12 . 9 9 ; t = 2 . 896 ( df = 8); Answer = 2 × (0 . 5 - 0 . 010) = 0 . 98 . (2) Answer: (8 . 1984 , 12 . 9416) n = 100 , ( > 30); x = 10 . 57; s = 12 . 10; σ = unknown; Distribution ? Use z = 1 . 96 10 . 57 ± (1 . 96) 12 . 10 100 = 10 . 57 ± 2 . 3716 (3) Answer: - 24 . 3/ c to 64 . 3/ c . From the given data: n = 9 , (small); x = 0 . 2 , s = 0 . 714 , σ = unknown; X assumed normal. Use t with df=8 x ± t s n ; 0 . 2 ± (1 . 86) 0 . 714 9 . (4) Answer: 71 n = ± z σ B ² 2 = ± (2 . 33)(18) 5 ² 2 = 70 . 358 (rounded up to 71) . Note: If you read z = 2 . 32 from the z-table for 0.4898, you don’t meet the required minimum of 98% conﬁdence level. (5) Answer: (2 , 120 . 72 , 2 , 679 . 28) x = 2400 , n = 17 , ( < 30) Given σ = 700 , and normality . Use z = 1 . 645 2400 ± (1 . 645) 700 17 = 2400 ± 279 . 28 (6) Answer: 0.3 ± 0.0635 95% Conﬁdence Interval for p = z = 1 . 96; b p = 60 / 200 = 0 . 3 , b p ± z p b p (1 - b p ) /n = 0 . 3 ± (1 . 96) p (0 . 3)(0 . 7) / 200 (7) Answer: .950 Given the interval (14.775, 17.225), n = 64 , ( > 30) , x = 16 ,s = 5 , what is z so that 17

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## This note was uploaded on 12/22/2009 for the course ISOM ISOM111 taught by Professor Anthonychan during the Fall '09 term at HKUST.

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PS6s - SOLUTIONS TO ISMT111 PROBLEM SET #6 (1) Answer:...

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