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IEOR 165 Lecture 7 June 9 2009

IEOR 165 Lecture 7 June 9 2009 - Lec t ur e No t es IEOR...

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Lecture Notes IEOR 165 | George Shanthikumar Tu W Th 2-4:30 pm | 3113 Etcheverry Quiz 1 June 16 (2 – 3 pm) 2 Questions on Estimation HW 1 and 2 2 Sided Notes Quiz 2 June 23 (2 – 3:30 pm) 2 Questions on Hypothesis Testing HW 3 2 Sided Notes Regression Analysis Recap: Y = a + bx + ε , where x is the input and Y is the output = , = 0 Varε σ2 Data , , = , …, xk Yk k 1 n Objectives: : Find Estimates for a and b AND : Find Estimates that minimizes the sum of squared deviations SSD = = - - SSD k 1nYk A Bxk2 = = Ybar 1nk 1nYk = = xbar 1nk 1nxk = = - Sxx k 1 nxk xbar2 = = - ( - ) SxY k 1 nxk xbar yk ybar = = - SYY k 1 nyk ybar2 = - Ahat Ybar Bhatxbar | forecast for the mean response = = + * μx EYxis Yhat Ahat Bhat x = Bhat SxYSxx = + μx a bx = , = = We need to see if EYhat μx EAhat a?EBhat b?are they unbiased?
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Lecture Notes IEOR 165 | George Shanthikumar Tu W Th 2-4:30 pm | 3113 Etcheverry (1) : = Let's calculate EAhat first to show it a = - * EAhat EYbar EBhat x We we first calculate E[Ybar] = = = = + = + * EYbar E1nk 1nYk 1nk 1na bxk a b xbar (2) Substitute: (3) = - * EAhat EYbar EBhat x = + * - * a b xbar EBhat xbar = + - * a b EBhat xbar , We conclude that if Bhat is an unbiased estimator of b then Ahat is unbiased estimator of a (4) = * EBhat 1Sxx ESxY (5) Now what is SxY? = = - - SxY k 1nxk xbarYk Ybar : Substitute = = - + - - * ESxY k 1nxk xbara bxk a b xbar = = - - = k 1nxk xbarxk xbarb bSxx : Therefore = * = EBhat 1Sxx bSxx b (6) So we are able to make the original conclusion. (7) (8) (9) So: Ahat is an unbiased estimator of a AND Bhat is an unbiased estimator of b (10) (11) Next Objective: So now we need to construct CI for a and b. To do so, we need
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