Unformatted text preview: y ) 2 n = 9231089 11 = 824 . 00 SS x = s ( x 2 )( ∑ x ) 2 n = 13349 11 = 128 . 55 r = SP xy r SS x SS y = 7 . 00 r (128 . 55)(824 . 00) = 0.02 Regression: b = SP xy SS x = 7 . 00 128 . 55 = 0 . 05 ¯ Y = ∑ y n = 33 . 00 / 11 = 3 . 00 , ¯ X = ∑ x n =7 . 00 / 11 =. 64 a = ¯ Y( b )( ¯ X ) = 3 . 00(0 . 05)(. 64) = 3 . 03 Y p = bX + a = Y p =(0.05)(X)+(3.03) Prediction: at x = 3, the prediction for y = (0 . 05)(3)+ 3 . 03 = 3.20 Standard Error of Prediction: s y  x = R SS y (1r 2 ) n2 = R 824 . 00(1(0 . 02) 2 ) 112 = 9.57 Note: On the exam you should round all intermediate results to 2 decimal places. However the fnal answers on this key were computed using higher precision, thereFore there may be rounding discrepancies with solutions computed rounding at every step, especially For prediciton....
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 Fall '08
 Ard
 Statistics, Normal Distribution, Regression Analysis, Standard Deviation, major mode

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