4494 2132 n 61483 2 di d i 1 14494 14610 or

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Unformatted text preview: ) = −891.17 2 n i − H ) − b ∑ ( Di − D ) ( H i − H ) i =1 n−2 2446.83 − ( −1.4494 ) × ( −891.17 ) = 16.9937 6−2 5/19 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 11 Diff. between Means; Least ­Squares Or you can use SSE 1155.14 s= = = 16.9937 n−2 6−2 α = 1 − 0.9 = 0.1 , in each tail α 2 = 0.05 ; d.f.=n ­2=6 ­2=4 Using Table A4. or function tinv in Excel, we find that t0.05 = 2.132 for 4 degrees of freedom. Therefore, a 90% confidence interval for b is s 16.9937 β = b ± tα /2 = −1.4494 ± 2.132 × = n 614.83 2 ∑ ( Di − D ) i =1 −1.4494 ± 1.4610 or ( −2.9105,0.0116 ) From the table above, we can also get n 2 i ∑D = 2819 , i =1 A 90% confidence interval for a is n s 2 i ∑D i =1 α = a ± tα /2 = 71.9477 ± 2.132 × n n∑ ( Di − D ) 2 16.9937 2819 = 6 × 614.83 i =1 d) e) 71.9477 ± 31.6691or(40.2786,103.6168) What is the coefficient of determination for the relationship you determined in (b)? SSE 1155.14 The coefficient of determination: R 2 = 1 − = 1− = 0.5279 SST 2446.83 What is the correlation coefficient between H and D? The correlation coefficient between H and D: n ∑( D − D)(H i i =1 i −H) −891.17 n −1 = = −0.7266 n 614.83 × 2446.83 2 Ⱥ Ⱥ 2 Ⱥ Ⱥ n Ⱥ ∑ ( Di − D ) Ⱥ Ⱥ ∑ ( H i − H ) Ⱥ Ⱥ i =1 Ⱥ Ⱥ i =1 Ⱥ n −1 n −1 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ ˆ Or ρ = r = − 0.5279 = −0.0.7266 ˆ ρ= 6/19 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 11 Diff. between Means; Least ­Squares Comments: 2 a) i) No comments =  ­1; ii) no mention of linear relationship =  ­1 b) Wrong use of independent variable =  ­2 c) Wrong formula of s or wrong d.f. or t value from table =  ­2 (if more wrong formula and wrong values were used =  ­4) d) Completely wrong formula =  ­4 e) No negative sign but right value =  ­...
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