Assume the linear relationship between hardness h and

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Unformatted text preview: 3/19 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 11 Diff. between Means; Least ­Squares 80 Hardness (kgmm2) 70 60 50 40 30 20 10 0 0 b) 10 20 30 40 Deformation (mm) Hardness decreases with deformation. They look like linearly related. Assuming that the Brinell hardness varies linearly with the deformation (independent variable), determine a best estimate of the relationship between H and D. Assume the linear relationship between hardness (H) and deformation (D) is: H = a + bD 1n H = ∑ H i = 44.17 n i =1 1n D = ∑ Di = 19.17 n i =1 2 ˆ ( D − D ) ( H i − H ) ( Di − D) ( Hi − H ) Hi = a + bDi 2 ˆ (H − H ) 2 Di Di − D Hi − H 6 68 -13.17 23.83 36 173.36 568.03 -313.81 63.25 22.55 11 65 -8.17 20.83 121 66.69 434.03 -170.14 56.00 80.93 13 53 -6.17 8.83 169 38.03 78.03 -54.47 53.10 0.01 22 10 2.83 -34.17 484 8.03 1167.36 -96.81 40.06 903.60 28 37 8.83 -7.17 784 78.03 51.36 -63.31 31.36 31.77 35 Average Hi 32 15.83 -12.17 1225 250.69 148.03 -192.64 21.22 116.27 19.17 44.17 2819 614.83 2446.83 -891.17 Sum Di2 i i 1155.14 i 4/19 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) n ∑(D − D)(H i b= −H) i i =1 n ∑( D − D) = 2 Homework 11 Diff. between Means; Least ­Squares −891.17 = −1.4494 614.83 i i =1 a = H − bD = 44.17 − ( −1.2806 ) ×19.17 = 71.9477 The best estimate of the relationship is H=71.9477 ­1.4494D 80 70 Hardness (kgmm2) 60 50 40 30 20 Original Data 10 Linear Regression 0 0 c) 10 20 Deformation (mm) 30 40 Give 90% confidence intervals for the two parameters you found in (b). From the table above, we can get n 2 2 SST = ( n − 1) S H = ∑ ( H i − H ) = 2446.83 i =1 n ˆ SSE = ∑ H i − H i i =1 ( n ) 2 = 1155.14 2 ( n − 1) S D = ∑ ( Di − D ) 2 = 614.83 i =1 n ∑(D − D)(H i i =1 n s= = ∑(H i =1 i − H...
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