3A03 test 1a - STAT 3A03 Applied Regression With SAS Fall 2014 Term Test 1-A Solution Set Q 1 a The sum of squared errors is n(yi 2 1 xi)2 S(1 = i=1[1

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STAT 3A03 Applied Regression With SAS Fall 2014 Term Test 1-A Solution Set Q. 1 a) The sum of squared errors is S ( β 1 ) = n X i =1 ( y i - 2 - β 1 x i ) 2 [1 marks] To minimize this we take the derivative with respect to β 1 dS ( β 1 ) 1 = - 2 n X i =1 x i ( y i - 2 - β 1 x i ) Now we set that equal to 0 when β 1 = ˆ β 1 and solve. dS ( β 1 ) 1 β 1 = ˆ β 1 = 0 - 2 n X i =1 x i ( y i - 2 - ˆ β 1 x i ) = 0 n X i =1 x i y i - 2 n X i =1 x i - ˆ β 1 n X i =1 x 2 i = 0 ˆ β 1 = x i y i - 2 x i x 2 i [1 marks] b) E( ˜ β 1 | x 1 , . . . , x n ) = E x i Y i - 2 x i x 2 i x 1 , . . . , x n = x i E[ Y i | x i ] - 2 x i x 2 i = x i (2 + β 1 x i ) - 2 x i x 2 i = 2 x i + β 1 x 2 i - 2 x i x 2 i = β 1 [2 marks] 1
c) Var( ˜ β 1 | x 1 , . . . , x n ) = Var x i Y i - 2 x i x 2 i | x 1 , . . . , x n = Var n X i =1 x i x 2 j ! Y i - 2 x i x 2 i | x 1 , . . . , x n ! = n X i =1 x i x 2 j ! 2 Var( Y i | x i ) = n X i =1 x i x 2 j ! 2 Var(2 + x i β 1 + ε i | x i ) = n X i =1 x i x 2 j ! 2 Var( ε i ) = x 2 i ( x 2 j ) 2 ! σ 2 = σ 2 x 2 i [2 marks] Q. 2 interpretations a) β 0 the long run average petal length when petal width is zero. [2 marks] β 1