ps4sol_F08

ps4sol_F08 - The Hong Kong University of Science and...

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The Hong Kong University of Science and Technology IS O M 111 - Business statistics Problem Sheet 4 Suggested Solutions 1. (1.6706, 63294) 2. (25.42, 36.58) 3. (69.9141, 93.0859) 4. (620.13, 8879.87). Since 0 does not included in the interval, the employee can say that there is a real difference between them. 5. (1789.38, 7410.62) 6. Let x be the number of consumers out of 200 who identify Pepsi. x ~ Bin (200, ½ ) p = x / n ~ N ( p , pq / n ) ˆ p ~ N (1 / 2, 1 / 800) ˆ (a) P(0.45 < < 0.6) p ˆ = P({(0.45 - 0.5) / (0.00125) 1/2 } < z < { (0.6 - 0.5) / (0.00125) 1/2 }) = P( -1.414 < z < 2.828 ) = 0.4207 + 0.4977 = 0.9184 (b) Point estimate = 120 / 200 = 0.6 p ˆ The point estimate is good because i. it is unbiased ii. it has small standard error Actually standard error of = ( pq / n ) p ˆ 1/2 [(0.5)(0.5) / 200] 1/2 = 0.0354 (c) Maximum error of estimation =1.645 [(0.5)(0.5) / 200] 1/2 = 0.05816 (OR) =1.645 [(0.6)(0.4) / 200] 1/2 = 0.05698 (d) A 98% C. I. for p is ± z p ˆ α /2 n p p ) ˆ 1 ( ˆ where = 0.02 z 0.01 = 2.326 z / 2 n p p ) ˆ 1 ( ˆ = 0.02 2.33 [(0.6)(0.4) / n ] 1/2 = 0.02 n = 3257.34 Therefore, number of additional consumers is at least = 3258 - 200 = 3058 7. (-0.2016, 0.0016) Page 1 of 6
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8. P(| x - μ | 0.8) = 0.95 P(| z | 0.8 / ( s / n )) = 0.95 0.8 / ( s / n ) = 1.96 n = 150.0625 min. n = 151 9. (a) = 101 / 165 = 0.6121 1 ˆ p 95% C. I. is ± z 1 ˆ p 0.025 1 1 1 ) ˆ 1 ( ˆ n p p = 0.6121 ± 1.96 [(0.6121)(0.3879) / 165] ½ = 0.6121 ± 0.0744 = (0.5377, 0.6865) (b) The interval has 95% c onfidence that it covers the true proportion p 1 . (c) P(| - p 1 ˆ p 1 | 0.05) = 0.95 P(| z | (0.05 / 1 1 1 ) 1 ( n p p )) = 0.95 (0.05 / 1 1 1 ) 1 ( n p p ) z 0.025 = 1.96 n 1 (1.96/0.05) 2 ( p 1 (1- p 1 )) Choose
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This note was uploaded on 09/18/2010 for the course BBA ISOM111 taught by Professor Hu during the Fall '08 term at HKUST.

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ps4sol_F08 - The Hong Kong University of Science and...

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