# Least Square Regression Homework - Version 035 EXAM03...

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Unformatted text preview: Version 035 – EXAM03 – gilbert – (57245) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points If v 1 , v 2 , . . ., v p span a subspace W of R n and if x is a vector in R n such that x · v j = 0 , j = 1 , . . ., p, then x is in W ⊥ . True or False? 1. TRUE correct 2. FALSE Explanation: By definition W ⊥ consists of all vectors in R n that are orthogonal to any set that spans W . So if Span { v 1 , v 2 , . . ., v p } = W and if x is a vector in R n such that x · v j = 0 for all 1 ≤ j ≤ p , then x is in W ⊥ . Consequently, the statement is TRUE . 002 10.0 points Determine the singular value σ 1 for the ma-trix A = − 2 2 2 3 . 1. σ 1 = 2 √ 5 2. σ 1 = 3 √ 2 3. σ 1 = √ 21 4. σ 1 = √ 19 5. σ 1 = √ 17 correct Explanation: By definition, σ is a singular value of A when σ = √ λ and λ is an eigenvalue of A T A ; σ 1 is the largest of these singular values. Now A T A = bracketleftbigg − 2 2 2 3 bracketrightbigg − 2 2 2 3 = bracketleftbigg 8 6 6 13 bracketrightbigg . But then det( A T A − λI ) = bracketleftbigg 8 − λ 6 6 13 − λ bracketrightbigg = λ 2 − 21 λ + 68 = ( λ − 17)( λ − 4) . Consequently, σ 1 = √ 17 . 003 10.0 points Find the x-intercept of the Least Squares Regression line y = mx + b that best fits the data points ( − 2 , − 1) , ( − 1 , 1) , (0 , 3) , (1 , 4) . 1. x-intercept = − 24 17 2. x-intercept = − 26 17 correct 3. x-intercept = − 22 17 4. x-intercept = − 23 17 5. x-intercept = − 25 17 Explanation: The design matrix and list of observed val-ues for the data ( − 2 , − 1) , ( − 1 , 1) , (0 , 3) , (1 , 4) . Version 035 – EXAM03 – gilbert – (57245) 2 are given by A = 1 − 2 1 − 1 1 1 1 , b = − 1 1 3 4 . The least squares regression line for this data is y = mx + b where ˆ x is the solution of the normal equation A T A ˆ x = A T b , ˆ x = bracketleftbigg b m bracketrightbigg . Now A T b = bracketleftbigg 1 1 1 1 − 2 − 1 1 bracketrightbigg − 1 1 3 4 = bracketleftbigg 7 5 bracketrightbigg , while A T A = bracketleftbigg 1 1 1 1 − 2 − 1 1 bracketrightbigg 1 − 2 1 − 1 1 1 1 = bracketleftbigg 4 − 2 − 2 6 bracketrightbigg . Thus the normal equation is bracketleftbigg 4 − 2 − 2 6 bracketrightbigg ˆ x = bracketleftbigg 4 − 2 − 2 6 bracketrightbigg bracketleftbigg b m bracketrightbigg = bracketleftbigg 7 5 bracketrightbigg . So bracketleftbigg b m bracketrightbigg = bracketleftbigg 4 − 2 − 2 6 bracketrightbigg-1 bracketleftbigg 7 5 bracketrightbigg = 1 20 bracketleftbigg 6 2 2 4 bracketrightbiggbracketleftbigg 7 5 bracketrightbigg = bracketleftBigg 13 5 17 10 bracketrightBigg ....
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