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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 Spring '08
 Clark
 α, Pallavolo Modena, the00

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