prob4sol

# prob4sol - The Hong Kong University of Science and...

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The Hong Kong University of Science and Technology ISMT 111 - Business statistic Problem Sheet 4 Solutions 1. A 95% C. I. for μ 1 - μ 2 is ( x 1 - x 2 ) ± t α /2, n 1 + n 2 – 2 s p 2 1 1 1 n n + where t 0.025, 12 = 2.179 s p = ) 2 ( ) 1 ( ) 1 ( 2 1 2 2 2 2 1 1 - + - + - n n s n s n = {[(7 - 1)(4) 2 + (7 - 1) (4) 2 ] / (7 + 7 - 2)} ½ = 2 95% C. I. is (7 - 3) ± (2.179)(2)(1/7 + 1/7) ½ = 4 ± 2.3294 = (1.6706, 6.3294) 2. (25.42, 36.58) 3. (69.9141, 93.0859) 4. = (620.13, 8879.87) 5. A 98% C. I. for μ 1 - μ 2 is ( x 1 - x 2 ) ± z α /2 2 2 2 1 2 1 n s n s + where α = 0.02, z 0.01 = 2.326 98% C. I. is (82700 - 78100) ± 2.326 [(7100)2 / 66 + (6300)2 / 57] ½ = 4600 ± 2810.62 = (1789.38, 7410.62)

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6. Let x be the number of consumers out of 200 who identify Pepsi. x ~ Bin (200, ½ ) p ˆ = x / n ~ N ( p , pq / n ) p ˆ ~ N (1 / 2, 1 / 800) (a) P(0.45 < p ˆ < 0.6) = 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 p ˆ = 120 / 200 = 0.6 The point estimate is good because i. it is unbiased ii. it has small standard error Actually standard error of p ˆ = ( pq / n ) 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 p ˆ ± z α /2 n p p ) ˆ 1 ( ˆ - where α = 0.02 z 0.01 = 2.326 z α / 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. A 90% C. I. for p 1 - p 2 is 2 1 ˆ ˆ p p - ± z α /2 2 2 2 1 1 1 ) ˆ 1 ( ˆ ) ˆ 1 ( ˆ n p p n p p - + - = (3 / 50 – 8 / 50) ± z 0.05 [(0.06)(0.94) / 50 + (0.16)(0.84) / 50] ½ = -0.1 ± 1.645 (0.062) = (-0.2016, 0.0016)

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