mont4e_sm_ch12_mind

# mont4e_sm_ch12_mind - Mind-Expanding Exercises 12-98....

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Unformatted text preview: Mind-Expanding Exercises 12-98. Because R2 = SS R S yy of regression. Then, SS E SS R / k , F0 = and this is the usual F-test for significance S yy SS E /(n - k - 1) 0.90 / 4 F0 = = 33.75 and the critical value is f.05,4,15 = 3.06 . Because (1 - 0.9) /( 20 - 4 - 1) and 1- R2 = 33.75 > 3.06, regression is significant. 12-73 12-99. Using n = 20, k = 4, f. 05, 4 ,15 = 3. 06. Reject H0 if R2 / 4 3.06 (1 - R 2 ) / 15 R2 0.816 (1 - R 2 ) Then, R 2 0.449 results in a significant regression. 12-100. Because ^ ^ = ( X ' X ) -1 X 'Y , e = Y - X = Y - X ( X ' X ) -1 X 'Y = ( I - H )Y 12-101. From Exercise 12-76, ei is ith element of (I-H)Y. That is, ei = -hi ,1Y1 - hi , 2 Y2 - ... - hi ,i -1Yi -1 + (1 - hi ,i )Yi - hi ,i +1Yi +1 - ... - hi , n Yn and V (ei ) = ( hi2,1 + hi2, 2 + ... + hi2,i -1 + (1 - hi2,i ) + hi2,i +1 + ... + hi2,n ) 2 The expression in parentheses is recognized to be the ith diagonal element of (I-H)(I-H') = I-H by matrix multiplication. Consequently, V (ei ) = (1 - hi ,i ) 2 . Assume that i < j. Now, ei = -hi ,1Y1 - hi , 2 Y2 - ... - hi ,i -1Yi -1 + (1 - hi ,i )Yi - hi ,i +1Yi +1 - ... - hi , n Yn e j = -h j ,1Y1 - h j , 2Y2 - ... - h j , j -1Y j -1 + (1 - h j , j )Y j - h j , j +1Y j +1 - ... - h j , n Yn Because the yi`s are independent, Cov(ei , e j ) = (hi ,1 h j ,1 + hi , 2 h j , 2 + ... + hi ,i -1 h j ,i -1 + (1 - hi ,i )h j ,i + hi ,i +1 h j ,i +1 + ... + hi , j (1 - h j , j ) + hi , j +1 h j , j +1 + ... + hi , n h j , n ) 2 The expression in parentheses is recognized as the ijth element of (I-H)(I-H') = I-H. Therefore, Cov(ei , e j ) = -hij 2 . 12-102. ^ = ( X ' X ) -1 X 'Y = (X' X) -1 X ' ( X + ) = + (X' X) -1 X ' = + R 12-103. a) Min L = ( y - X )' ( y - X ) subject to :T = c This is equivalent to Min Z = ( y - X )'( y - X ) + 2 '( T - c ) where ' = [ , ,..., ] is a vector of La Grange multipliers. 1 2 p Z = -2 X ' y + 2 ( X ' X ) + 2 T ' Z Z = 0, = 2(T - c ) . Set ^ ( X ' X ) c + T ' = X ' y ^ T = c c Z = 0. Then we get where c is the constrained estimator. From the first of these equations, c = ( X ' X ) -1 ( X ' y - T ' ) = - ( X ' X ) -1 T ' From the second, T - T ( X ' X ) -1 T ' = c or = [ T ( X ' X ) -1 T ' ]-1(T - c) Then c = - ( X ' X ) -1 T '[ T ( X ' X ) -1 T ' ] -1 ( T - c ) = + ( X ' X ) -1 T '[ T ( X ' X ) -1 T ' ] -1 ( c - T ) 12-74 b) This solution would be appropriate in situations where you have tested the hypothesis that T = c and concluded that this hypothesis cannot be rejected. 12-104. a) For the piecewise linear function to be continuous at x = show that x* , the point-slope formula for a line can be used to where 0 , 1 , 2 are arbitrary constants. + ( x - x* ) 1 0 y= 0 + 2 ( x - x* ) 0, z= 1, x x* x > x x x x > x . Let Then, y can be written as y = 0 + 1 ( x - x ) + ( 2 - 1 )( x - x ) z . Let x1 = x - x x2 = (x - x )z 0 = 0 1 = 1 2 = 2 - 1 Then, y = 0 * + 1 * x1 + 2 * x2 . b) Should refer to exercise 12-80. If there is a discontinuity at x = x , then a model that can be used is + x 1 0 y= 0 + 1 x 0, z= Let 1, 0 = 0 1 = 1 2 = 0 - 0 3 = 1 - 1 x1 = x x 2 = xz x x* x > x x x x > x Then, y can be written as y = 0 + 1x + [( 0 - 0 ) + (1- 1 ) x ] z = 0 + 1x1 + 2z + 3x2 where c) Should refer to exercise 12-80. One could estimate x* as a parameter in the model. A simple approach is to refit the model in Exercise 12-80 with different choices for x* and to select the value for x* that minimizes the residual sum of squares. 12-75 ...
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## This note was uploaded on 10/12/2009 for the course IND E 315 taught by Professor Kailashkapur during the Spring '09 term at University of Washington.

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