Quiz_3_Solutions

Quiz_3_Solutions - again with a given degree of...

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IEOR 165: Engineering Statistics, Quality Control and Forecasting, Spring 2007 Quiz 3 Solution (a) Noting S xx = 227 . 5, S vv = 28601 . 7, S xy = 192 . 95, S vy = 1367 . 01 and S yy = 166 . 5893 B = S xy S xx = 192 . 95 227 . 5 = . 8481 A = ¯ Y - B ¯ x = 50 . 3929 - . 8481 · 7 . 5 = 44 . 0319 Thus, the estimated regression line is Y = 44 . 0319 + . 8481 x . The estimated response at x = 16 is 57 . 602. SS R 1 = S xx S yy - S 2 xy S xx = 227 . 5 · 166 . 5893 - 192 . 95 2 227 . 5 = 2 . 9422 Thus the variance is σ 2 1 = SS R 1 n - 2 = 2 . 9422 12 = . 2452 . (b) B = S vy S vv = 1367 . 01 28601 . 7 = . 0478 A = ¯ Y - B ¯ v = 50 . 3929 - . 0478 · 82 . 1429 = 46 . 4669 Thus, the estimated regression line is Y = 46 . 4669 + . 0478 v . The estimated response at v = 100 is 51 . 2463. SS 2 R = S vv S yy - S 2 vy S vv = 28601 . 7 · 166 . 5893 - 1367 . 01 2 28601 . 7 = 101 . 253 Thus the variance is σ 2 2 = SS R 2 n - 2 = 101 . 253 12 = 8 . 4378 . (c) A confidence interval is an interval that does contain, with a given degree of confidence, a fixed parameter of interest. A prediction interval, on the other hand, is an interval that will contain,
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Unformatted text preview: again with a given degree of confidence, a random variable of interest. (d) A + Bx ± t α/ 2 ,n-2 r ± n +1 n + ( x-¯ x ) 2 S xx ² SS R n-2 = 50 . 602 ± 2 . 179 r ± 15 14 + (16-7 . 5) 2 227 . 5 ² 2 . 9422 12 = (56 . 326 , 58 . 866) . (e) R 2 1 = 1-SS R 1 S yy = 1-2 . 9422 166 . 5893 = . 9823 r 1 = q R 2 1 = √ . 9823 = . 9911 R 2 2 = 1-SS R 2 S yy = 1-101 . 253 166 . 5893 = . 3922 r 2 = q R 2 2 = √ . 3922 = . 6263 . The value of R 2 is often used as an indicator of how well the regression model fits the data, while r provides a measure of the degree to which high values of x are paired with high values of Y and low values of x with low values of Y . 1...
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This note was uploaded on 07/22/2009 for the course IEOR 165 taught by Professor Shanthikumar during the Summer '08 term at Berkeley.

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