# PS8s - SOLUTIONS TO ISMT111 PROBLEM SET#8(1 p1 = 24/80 p2 =...

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Unformatted text preview: SOLUTIONS TO ISMT111 PROBLEM SET #8 (1) p1 = 24/80; p2 = 15/150; 24 80 15 150 24 80 56 80 1 80 15 150 135 150 1 150 − ± (1.645) + (2) Answer: 385 5 2 = (1.96) 1.96 2.5 202 n 2 + 152 n n= (625) = 384.16 (3) 2 Small sample size, use Sp and t (df = 5) : 2 Sp = 6 1 3 (3 − 1)(3)2 + (4 − 1)(2)2 3+4−2 + 1 4 = 1.87083 =6 (36.9 − 25.2) ± (2.571)(1.8708) (4) Answer: (B) Large sample size, right-tailed: zobs = (X 1 − X 2 ) − 0 2 S1 n1 + 2 S2 n2 = (37723 − 36916) − 0 33002 80 + 39602 80 = 1.40026 p-value is (0.5 − 0.4192) = 0.0808, greater than α of 0.05. (5) Answer: 0.0768 p1 = 0.85; p2 = (160 − 85)/100 = 0.75; Sp 1 −p2 p = 160/200 = 0.8; = 0.05657 = (0.8)(0.2) 1 100 + 1 100 (0.85 − 0.75) − 0 = 1.7677 . 0.05657 p − value = 2 × (0.5 − 0.4616) = 0.0768. zobs = ∗ ∗ 1 ∗ ATChan 2008 ...
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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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