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CEE 202 Homework 11 solution

# 2806 04042 or 16848 08764 from the table above

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Unformatted text preview: iginal Data 60 Linear Regression w/o outlier 50 40 30 20 10 0 0 c) 10 20 30 40 Deformation (mm) The original data and linear regression are plotted. Our assumption about the outlier is valid. Give 90% confidence intervals for the two parameters you found in (b). From the table above, we can get i 9/19 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) n 2 H Homework 11 Diff. between Means; Least ­Squares 2 SST = ( n − 1) S = ∑ ( H i − H ) = 1046 i =1 n ˆ SSE = ∑ H i − H i i =1 ( n − 1) S 2 D ( = 53.56 2 = ∑ ( Di − D ) = 605.20 i =1 ∑(D − D)(H i − H ) = −775 i i =1 n ∑(H i =1 s= 2 n n ) n 2 i − H ) − b ∑ ( Di − D ) ( H i − H ) i =1 n−2 1046 − ( −1.2806 ) × ( −775 ) = 4.2253 5−2 = Or you can use SSE 53.56 s= = = 4.2253 n−2 3 α = 1 − 0.9 = 0.1 , in each tail α 2 = 0.05 ; d.f.=n ­2=5 ­2=3 Using Table A4. or function tinv in Excel, we find that t0.05 = 2.353 for 3 degrees of freedom. Therefore, a 90% confidence interval for b is s 4.2253 β = b ± tα /2 = −1.2806 ± 2.353 × = n 605.20 2 ∑ ( Di − D ) i =1 −1.2806 ± 0.4042 or ( −1.6848, −0.8764 ) From the table above, we can also get n 2 i ∑D = 2335 , i =1 A 90% confidence interval for a is n s α = a ± tα /2 2 i ∑D i =1 n n∑ ( Di − D ) = 74.8186 ± 2.353 × 2 4.2253 2335 = 5 × 605.2 i =1 74.8186 ± 8.7349or(66.0837,83.5534) 10/19 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) d) e) Homework 11 Diff. between Means; Least ­Squares What is the coefficient of determination for the relationship you determined in (b)? SSE 53.56 The coefficient of determination: R 2 = 1 − = 1− = 0.9488 SST 1046 What is the correlation coefficient between H and D? The correlation coefficient between H and...
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