HW4Solutions - ESI 6321 APPLIED PROBABILITY METHODS FOR...

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Prof. Edwin Romeijn ESI 6321 Applied Probability Methods For Engineers Homework 4 - Solutions Spring 2008 1/8 ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 4 – Spring 2008 Solutions Exercise 12.2.4 Cost is the dependent variable, y , and Depth is the explanatory variable, x . We need ∑ ∑ ∑ ∑ ∑ 2 2 , , , , , , i i i i i i i i i i i y x x y x y x y X Y Depth (feet) Cost ($1000) X * Y X^2 Y^2 5000 2596.800049 12984000.24 25000000 6743370.494 5200 3328 17305600 27040000 11075584 6000 3181.100098 19086600.59 36000000 10119397.83 6538 3198.399902 20911138.56 42745444 10229761.94 7109 4779.899902 33980308.41 50537881 22847443.08 7556 5905.600098 44622714.34 57093136 34876112.51 8005 5769.200195 46182447.56 64080025 33283670.89 8207 8089.5 66390526.5 67354849 65440010.25 8210 4813.100098 39515551.8 67404100 23165932.55 8600 5618.700195 48320821.68 73960000 31569791.88 9026 7736 69825136 81468676 59845696 9197 6788.299805 62431993.3 84584809 46081014.24 9926 7840.799805 77827778.86 98525476 61478141.58 10813 8882.5 96046472.5 116920969 78898806.25 13800 10489.5 144755100 190440000 110029610.3 14311 12506.59961 178981947 204804721 156415033.8 Average 8593.625 6345.249985 n 16 Sum x*y 979168137.4 SumX * SumY 13959346918 Sum(X^2) 1287960086 Sum(Y^2) 762099377.5 (SumX)^2 18905700004 Sxy 106708955 Sxx 106353835.8 B1^Hat 1.003339035 B0^Hat -2277.069432 SSE 10838959.72 SigmaHat^2 774211.4086
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ESI 6321 Applied Probability Methods For Engineers Homework 4 - Solutions Spring 2008 2/8 a) So, our estimated model for the cost as a function of depth is: Cost = -2277.0694 + 1.0033*Depth b) The model predicts a cost increase of $1.0033 * 1000 = $1003.3 for an additional depth of 1000 feet. c) Using the model, = - + × = 10000 ˆ 2277.0694 1.0033 10000 $7756.32 y Since 10,000 is within the range of observed values for the Depth, this cost estimate is reasonable. d) The estimate of the error variance is 2 ˆ σ = 774211.41 e) We could predict the cost for a depth of 20,000 feet as = - + × = 20000 ˆ 2277.0694 1.0033 20000 $17789.71 y However, since 20,000 feet is outside of the range of the depth data we collected, this result may not be accurate since we’re extrapolating. Exercise 12.3.3 a) The standard error for the point estimate ( ) β = = = 1 ˆ 774211.41 ˆ s.e. 0.08532 106353835.8 xx S b) The 2-sided 95% confidence interval for 1 is ( ) α - - - × + × 1 1 / 2, 2 1 1 / 2, 2 1 ˆ ˆ ˆ ˆ s.e.( ), s.e.( ) n n t t with = 0.05 Noting that n =16, a table lookup yields that - = = / 2, 2 0.025,14 2.145 n t t . Therefore, the desired confidence interval for 1 is ( ) ( ) ( ) - + = - + = 1 1.0033 0 1 1 1 : 0 vs. : 0 H H The corresponding t -statistic is
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This note was uploaded on 05/12/2010 for the course ESI 6321 taught by Professor Josephgeunes during the Spring '07 term at University of Florida.

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HW4Solutions - ESI 6321 APPLIED PROBABILITY METHODS FOR...

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