This preview shows page 1. Sign up to view the full content.
Unformatted text preview: MindExpanding Exercises 1298. Because R2 = SS R S yy of regression. Then, SS E SS R / k , F0 = and this is the usual Ftest 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. 1273 1299. 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. 12100. Because ^ ^ = ( X ' X ) 1 X 'Y , e = Y  X = Y  X ( X ' X ) 1 X 'Y = ( I  H )Y 12101. From Exercise 1276, ei is ith element of (IH)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 (IH)(IH') = IH 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 (IH)(IH') = IH. Therefore, Cov(ei , e j ) = hij 2 . 12102. ^ = ( X ' X ) 1 X 'Y = (X' X) 1 X ' ( X + ) = + (X' X) 1 X ' = + R 12103. 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 ) 1274 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. 12104. a) For the piecewise linear function to be continuous at x = show that x* , the pointslope 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 1280. 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 1280. One could estimate x* as a parameter in the model. A simple approach is to refit the model in Exercise 1280 with different choices for x* and to select the value for x* that minimizes the residual sum of squares. 1275 ...
View
Full
Document
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.
 Spring '09
 KAILASHKAPUR

Click to edit the document details