Assignment4SolutionsNEW

(c) First for Var( ˆ β W LS ): Var( ˆ β W LS ) = Var ( ( x 0 W x ) - 1 x 0 W Y ) = ( x 0 W x ) - 1 x 0 W Var( Y ) ( ( x 0 W x ) - 1 x 0 W ) 0 = ( x 0 W x ) - 1 x 0 W Var( Y ) W 0 x ( x 0 W x ) - 1 Note the following facts: 1) { ( x 0 W x ) - 1 } 0 = ( x 0 W x ) - 1 since it is a symmetric matrix 2) W 0 = W because W is diagonal 3) Var( Y ) = Var( ) = σ 2 V . Since V = W - 1 , σ 2 V = σ 2 W - 1 . So we have Var( ˆ β W LS ) = ( x 0 W x ) - 1 x 0 W ( σ 2 W - 1 ) W x ( x 0 W x ) - 1 = σ 2 ( x 0 W x ) - 1 x 0 W W - 1 W x ( x 0 W x ) - 1 = σ 2 ( x 0 W x ) - 1 x 0 W x ( x 0 W x ) - 1 = σ 2 ( x 0 W x ) - 1 And similarly, Var( ˆ β OLS ) = Var ( ( x 0 x ) - 1 x 0 Y ) = ( x 0 x ) - 1 x 0 Var( Y ) ( ( x 0 x ) - 1 x 0 ) 0 = ( x 0 x ) - 1 x 0 Var( Y ) x ( x 0 x ) - 1 Using the fact that Var( Y ) = σ 2 W - 1 : Var( ˆ β OLS ) = ( x 0 x ) - 1 x 0 σ 2 W - 1 x ( x 0 x ) - 1 3. (a) After entering the data in SAS and running proc reg , we obtain: Therefore the least-squares regression line is: ˆ y = 19 . 47269 + 3 . 26981 x. (b) Below is a plot of the response (computer time) versus the covariate (total number of responses): 3

(b) We fit the model using OLS and find the following diagnostic plot: which clearly shows violation of constant variance. This violation is also evident in the residual versus fitted values plot: Therefore, weighted least-squares is preferable since it will aim at correcting this. (c) The weights are unknown but we have replicated measures, so we can estimate the sample variance of the response at each dosage level. Using the reciprocal of these variances as weights, we fit the model and obtain the following output: which gives the fitted regression line as: ˆ y = 4 . 66419 - 0 . 00659 x. 6