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problemset6-solution

# problemset6-solution - The Hong Kong University of Science...

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1 The Hong Kong University of Science and Technology ISMT 111 - Business statistics, Fall 2002 Problem Sheet 6 Solutions 1. (a) 1 ˆ β = S xy / S xx = 1.7483 0 ˆ β = y - 1 ˆ β x = - 8.0068 y = - 8.0068 + 1.7483 x When there is a 1% increase in the market return, the average return on Dalton Company’s stock tends to increase by 1.7483 %. (b) SSE = y i 2 – ( 0 ˆ β y i + 1 ˆ β x i y i ) = 29.3061 s 2 = SSE/(n - 2) s = 1.913966 Sxx = [ x i 2 – ( x i ) 2 / n ] = ( n - 1) s x 2 = 44.1 H 0 : β 1 = 1 H a : β 1 > 1 Test statistic t = Sxx s 1 ˆ 1 β = 2.5963 And t α , n – 2 = t 0.05, 8 = 1.86 Since t > t 0.05, 8 , reject H 0 at 0.05 level of significance. We have sufficient evidence that Dalton’s Company’s Stock is sensitive to the market. (c) A 90% P. I. for y given x = 10.5 is ( ) xx p n S x x n s t y 2 2 , 2 1 1 ˆ + + ± α where x p = 10.5, x = 10.7, n = 10, t α /2, n – 2 = t 0.05, 8 = 1.86 Hence 95% P. I. is 10.350340 ± 3.735274 = (6.6151, 14.0856) (d) r = yy xx xy S S S = 0.9063 R 2 = 0.8214 or 82.14%

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2 (e) A 98% C. I. for E( y | x = 14) is ( ) xx p n S x x n s t y 2 2 , 2 1 ˆ + ± α where x p = 14, x = 10.7, n = 10, t α /2, n – 2 = t 0.01, 8 = 2.896 Hence 98% C. I. is 16.469388 ± 3.264820 = (13.2046, 19.7342) 2. (a) 1 ˆ β = - 364.6 On average, the Calorie content decreases by 364.6 when there is an unit increase in the Price per ounce. (b) H 0 : β 1 = 0 H a : β 1 0 Test statistic t = ) ˆ ( ˆ 1 1 1 β β β sd = - 364.6 / 101.2 = - 3.603 And t α /2, n – 2 = t 0.025, 7 = 2.365 Since | t | > t 0.025, 7 , reject H 0 at 0.05 level of significance. The price is a good predictor of the calorie content. (c) R 2 = r 2 = ( yy xx xy S S S ) 2 = 0.6498 = 64.98% That means 64.98% of the variability of the calorie content can be explained by the price. 3. (a) Let X be the hours of training. Let Y be the average improvement. 1 ˆ β = S xy / S xx = 1622.5 / 2875 = 0.564347826 0 ˆ β = y - 1 ˆ β x = 7.086956522 So , y = 7.087 + 0.564x (b) y ˆ = 7.087 + 0.564 (50) = 35.30434783 (c) SSE = S yy S xy 2 / S xx = 956.75 – 1622.5 2 / 2875 = 41.096 s 2 = SSE / ( n - 2) s = 4.532971
3 A 95% P. I. for the score improvement after 30 (= x p ) hours of training is 2 ˆ 2 2 , 2 ˆ y n s s t y + ± α = ( ) xx p n S x x n s t y 2 2 , 2 1 1 ˆ + + ± α where x p = 30, x = 37.5, n = 4, t α /2, n – 2 = t 0.025, 2 = 4.303 Hence 95% P. I. is 24.0173913 ± 21.97767741 = (2.039713895, 45.99506871) 4. S xx = x i 2 – ( x i ) 2 / n = 793.5 – 137.5 2 / 25 = 37.25 S yy = y i 2 – ( y i ) 2 / n = 213.4 – 70.3 2 / 25 = 15.7164 S xy = x i y i – ( x i ) ( y i ) / n = 404.2 – (137.5)(70.3) / 25 = 17.55 (a) 1 ˆ β = S xy / S xx = 0.47114 0 ˆ β = y - 1 ˆ β x = 0.22072 y = 0.22072 + 0.47114 x (b) No. It is not meaningful to interpret. It is because x ranges from 3.5 to 7.8, and x = 0 to estimate y (estimated y-intercept)

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problemset6-solution - The Hong Kong University of Science...

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