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CEE 202 Homework 11 solution

# 0430 3707 2 n vi v 03122 1983904 8 37647950 i

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Unformatted text preview: 4 689.06 0.43 17.29 4.18 0.16 501 4.89 52.75 0.45 251001 2782.56 0.20 23.80 4.97 0.01 605 6.5 156.75 2.06 366025 24570.56 4.25 323.10 6.01 0.24 688 6.88 239.75 2.44 4733440 57480.06 5.96 585.29 6.84 0.00 755 7.24 306.75 2.80 570025 94095.56 7.85 859.28 7.51 0.07 1983904 376479.50 38.22 3764.13 Average 448.25 4.44 Sum 0.58 16/19 i CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 11 Diff. between Means; Least ­Squares n ∑ (V − V ) ( pH i b= − pH i i =1 n ∑ (V − V ) 2 ) = 3764.13 = 0.0100 376479.50 i i =1 a = pH − bV = 4.44 − ( 0.0100 ) × 448.25 = −0.0430 The linear relationship between output voltage and pH is: pH= ­0.0430+0.0100V 8 7 Original Data Linear regression 6 pH 5 4 3 2 1 0 0 200 400 600 Output Voltage (mv) 800 b) What is a 99% confidence interval for those coefficients? From the table above, we can get n ( n − 1) SV2 = ∑ (Vi − V ) 2 = 376479.50 , i =1 n ( 2 SST = ( n − 1) S pH = ∑ pH i − pH i =1 n ∑ (V − V ) ( pH i i ˆ SSE = ∑ pH i − pH i i =1 ( 2 = 38.22 , ) − pH = 3764.13 , i =1 n ) ) 2 = 0.58 . 17/19 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) n ∑( pH i − pH i =1 s= ) 2 n ( Homework 11 Diff. between Means; Least ­Squares − b∑ (Vi − V ) pH i − pH i =1 ) n−2 38.22 − ( 0.0100 ) × ( 3764.13) 8−2 = 0.3122 Or you can use SSE 0.58 s= = = 0.3122 n−2 6 α = 1 − 0.99 = 0.01 , in each tail α 2 = 0.005; d.f.=n ­2=8 ­2=6 Using Table A4. or function tinv in Excel, we find that t0.005 = 3.707 for 6 degrees of freedom. Therefore, a 99% confidence interval for b is s 0.3122 b ± tα /2 = 0.0100 ± 3.707 × = n 376479.50 2 ∑ (Vi − V ) = i =1 0.0100 ± 0.0019or ( 0.0081,0.0119) From the table above, we can also get n ∑V i 2 = 1983904 , i =1 A 99% confidence inte...
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