Dr. Hackney STA Solutions pg 186

Dr. Hackney STA Solutions pg 186 - 11-14 Solutions Manual...

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Unformatted text preview: 11-14 Solutions Manual for Statistical Inference 11.34 a. ANOVA table for height data Source Regression Residual Total df 1 6 7 SS 60.36 7.14 67.50 MS 60.36 1.19 F 50.7 The least squares line is y = 35.18 + .93x. ^ b. Since yi - y = (yi - yi ) + (^i - y ), we just need to show that the cross term is zero. ^ y n n (yi - yi )(^i - y ) ^ y i=1 = i=1 n yi - (^ + xi ) ^ (^ + xi ) - y ^ ^ (xi - x) n = i=1 n ^ (^i - y ) - (xi - x) y ^ (xi - x)(yi - y ) - 2 i=1 (^ = y - x) ^ ^ = ^ from the definition of . (xi - x)2 = 0, i=1 c. ^ (^i - y )2 = 2 y 11.35 a. For the least squares estimate: d d which implies ^ = b. The log likelihood is log L = - i (xi - x)2 = 2 Sxy . Sxx (yi - x2 )2 = 2 i i i (yi - x2 )x2 = 0 i i yi x2 i . x4 i i n 1 log(2 2 ) - 2 2 2 (yi - x2 )2 , i i and maximizing this is the same as the minimization in part (a). c. The derivatives of the log likelihood are d log L = d d2 log L = d2 so the CRLB is 2 / i 1 2 -1 2 (yi - x2 )x2 i i i x4 , i i ^ x4 . The variance of is i i ^ Var = Var yi x2 i 4 i xi = i x2 i 4 j xj 2 = 2 / i x4 , i ^ so is the best unbiased estimator. ...
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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