HW5Solution - ESI 6321 APPLIED PROBABILITY METHODS FOR...

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Prof. Edwin Romeijn ESI 6321 Applied Probability Methods For Engineers Homework 5 - Solutions Spring 2008 1/8 ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 5 – Spring 2008 Solutions Exercise 12.8.5 To obtain the linear model, the logarithmic transformation is used in the following way: ( ) ( ) ( ) ( ) ) ( ln ln ln ln ) ln( 1 0 ) ( 0 ) ( 0 1 1 temprature e e yield temprature temprature γ + = + = = Letting ) ln( yield y = , temprature x = , ( ) 0 0 ln β = and 1 1 = gives us: x y 1 0 + = . This allows us to use the given values for 0 ˆ and 1 ˆ to find 0 ˆ and 1 ˆ : 85 . 13 ˆ 628 . 2 ˆ 0 0 = = = e e , 341 . 0 ˆ ˆ 1 1 = = A 95% confidence interval for 1 using 069 . 2 23 , 025 . 0 = t would be: ( ) ( ) 393 . 0 , 289 . 0 025 . 0 069 . 2 341 . 0 , 025 . 0 069 . 2 341 . 0 1 1 = × + × - = Exercise 13.1.1 a) Coefficient of determination: 886 . 0 9 . 108 4 . 12 9 . 108 2 = - = = SST SSR R b) Source D.O.F. Sum of Squares Mean Squares F -Stat p -value Regression 3 96.5 32.17 67.4 0.000 Error 26 12.4 0.477 Total 29 108.9 Note that in the above table, ) ( F X P value p > = - in which X has an F -distribution with degrees of freedom 3 and 26. c) The estimate of error variance: 477 . 0 26 4 . 12 1 3 30 ˆ 2 = = - - = = SSE MSE σ d) The p -value for this hypothesis is equal to: ) ( F X P value p > = - in which the random variable X has an F -distribution with degrees of freedom 3 and 26. This value is very small and can be estimated by zero. e) A 95% confidence interval for 1 using 056 . 2 26 , 025 . 0 = t would be: ( ) ( ) 8 . 21 , 2 . 11 6 . 2 056 . 2 5 . 16 , 6 . 2 056 . 2 5 . 16 1 = × + × - Exercise 13.2.3 a) To fit the regression model, we need to solve the normal equations. Solving these equations gives us the following values:
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Prof. Edwin Romeijn ESI 6321 Applied Probability Methods For Engineers Homework 5 - Solutions Spring 2008 2/8 x1 x2 x3 x4 y Depth Geology Downtime Rig-index Cost ($1000) y*x1 y*x2 y*x3 y*x4 5000 319 0 0.62 2596.80 12984000.24 828379.2156 0.00 1620.40 5200 241 160 1.01 3328.00 17305600 802048 532480.00 3361.28 6000 268 121 0.63 3181.10 19086600.59 852534.8262 384913.11 2016.82 6538 345 48 0.90 3198.40 20911138.56 1103447.966 153523.20 2888.16 7109 549 120 0.57 4779.90 33980308.41 2624165.046 573587.99 2705.42 7556 552 608 1.03 5905.60 44622714.34 3259891.254 3590604.86 6106.39 8005 417 391 1.21 5769.20 46182447.56 2405756.481 2255757.28 6951.89 8207 568 471 0.91 8089.50 66390526.5 4594836 3810154.50 7329.09 8210 618 80 1.04 4813.10 39515551.8 2974495.86 385048.01 5010.44 8600 767 140 1.04 5618.70 48320821.68 4309543.05 786618.03 5843.45 9026 675 51 2.54 7736.00 69825136 5221800 394536.00 19610.76 9197 504 137 1.21 6788.30 62431993.3 3421303.102 929997.07 8179.90 9926 639 0 2.76 7840.80 77827778.86 5010271.075 0.00 21664.13 10813 641 158 0.66 8882.50 96046472.5 5693682.5 1403435.00 5871.33 13800 953 0 1.20 10489.50 144755100 9996493.5 0.00 12534.95 14311 1165 56 0.77 12506.60 178981947 14570188.54 700369.58 9642.59 x1^2 x2*x1 x3*x1 x4*x1 x2^2 x3*x2 x4*x2 x3^2 x4*x3 x4^2 25000000 1595000 0 3120.00 101761.00 0 199.0560041 0 0.00 0.39 27040000 1253200 832000 5252.00 58081.00 38560 243.4099977 25600 161.60 1.02 36000000 1608000 726000 3804.00 71824.00 32428 169.9120009 14641 76.71 0.40 42745444 2255610 313824 5903.81 119025.00 16560 311.534999 2304 43.34 0.82 50537881 3902841 853080 4023.69 301401.00 65880 310.7339916 14400 67.92 0.32 57093136 4170912 4594048 7812.90 304704.00 335616 570.7680216 369664 628.67 1.07 64080025
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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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HW5Solution - ESI 6321 APPLIED PROBABILITY METHODS FOR...

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