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# 8.2 - i = 9 30 60 and 125 and with t 0.95,10 = 1.812 from...

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8.2 (a) (b) Vehicle No. Speed Stopping Distance (kph) (m) Xi Yi Xi 2 Yi 2 XiYi Yi'=a+bXi (Yi-Yi') 2

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Total: On the basis of calculations in the above table we obtain the respective sample means of X and Y as, X = 679/12 = 56.6 kph, Y = 238/12 = 19.8 m and corresponding sample variances, 84 . 1289 ) 6 . 56 12 52631 ( 11 1 2 2 = × = x S , 50 . 200 ) 8 . 19 12 6910 ( 11 1 2 2 = × = y S From Eq. 8.4 & 8.3, we also obtain, 388 . 0 6 . 56 12 52631 8 . 19 6 . 56 12 18953 2 = × × × = β ± , α ± = 19.8 – 0.388 × 56.6 = -2.161 From Eq. 8.6a, the conditional variance is, = = n i i i X Y y y n S 1 2 ' 2 | ) ( 2 1 = 71.716 / (12 – 2) = 7.172 and the corresponding conditional standard deviation is = 2.678 m x Y S | From Eq. 8.9, the correlation coefficient is, 98 . 0 50 . 200 84 . 1289 8 . 19 6 . 56 12 18953 11 1 1 1 ˆ 1 = × × = = = y x n i i i s s y x n y x n ρ (c) To determine the 90% confidence interval, let us use the following selected values of X
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Unformatted text preview: i = 9, 30, 60 and 125, and with t 0.95,10 = 1.812 from Table A.3, we obtain, At X i = 9; m X Y ) 723 . 3 063 . 1 ( ) 6 . 56 12 52631 ( ) 6 . 56 9 ( 12 1 679 . 2 812 . 1 33 . 1 2 2 90 . → − = × − − + × ± = > < µ At X i = 30; m X Y ) 252 . 11 708 . 7 ( ) 6 . 56 12 52631 ( ) 6 . 56 30 ( 12 1 679 . 2 812 . 1 48 . 9 2 2 90 . → = × − − + × ± = > < At X i = 60; m X Y ) 528 . 22 712 . 19 ( ) 6 . 56 12 52631 ( ) 6 . 56 60 ( 12 1 679 . 2 812 . 1 12 . 21 2 2 90 . → = × − − + × ± = > < At X i = 125; m X Y ) 460 . 49 220 . 43 ( ) 6 . 56 12 52631 ( ) 6 . 56 125 ( 12 1 679 . 2 812 . 1 34 . 46 2 2 90 . → = × − − + × ± = > <...
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8.2 - i = 9 30 60 and 125 and with t 0.95,10 = 1.812 from...

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