ecn 102 hw 4 #2 - 1 ) Y(bar) = (8+12+11+5)/4 = 9 X(bar) =...

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Unformatted text preview: 1 ) Y(bar) = (8+12+11+5)/4 = 9 X(bar) = (1+3+2+2)/4 = 2 β = ¡¢ £ ¤¢¥¦§¨©ª¡« £ ¤«¥¦§¨©ª ¬ £­® ¥ ¯ £ ¬ £­® ¤«¥¦§¨©©°± = ¥²¤³©¥´¤±©µ¥´±¤³©¥¶¤±©µ¥´´¤³©¥±¤±©µ¥·¤³©¥±¤±© ¥´¤±© ¸ µ¥¶¤±© ¸ µ¥±¤±© ¸ µ¥±¤±©°± = ¥¤´©¥¤´©µ¥¶©¥´©µ¹µ¹ ¥´©µ´µ¹µ¹ = º ± =2 » = ¼¥½¾¿© À ÁÂ¥½¾¿© = (9) – (2)(2) = 5 Therefore Ã Ä (hat)= 5+2 Å Ä ¡ (hat)= 5+2 ¢ ¡ Plugging in the values of ¢ ¡ [1, 3, 2, 2] into the above equation, I got the following values for ¡ (hat) 5+2(1) = 7 5+2(3) = 11 5+2(2) = 9 5+2(2) = 9 TSS = £ ¤ ¡ ¥ ¡¦§ ¨ ¤©ª«¬¬­®¯ =(8-9)^2+(12-9)^2+(11-9)^2+(5-9)^2 =(-1)^2+(3)^2+(2)^2+(-4)^2 =1+9+4+16 =30 RSS = £ ¤ ¡ ¥ ¡¦§ ¤°ª±¬ ¨ ¤©ª«¬¬­® = (7-9)^2+(11-9)^2+(9-9)^2+(9-9)^2 = (-2)^2+(2)^2+(0)+(0) = 4+4 = 8 ² ³ = ´µµ ¶µµ = · ¸¹ = 0.266 An R^2 of 0.266 means that almost 27% of the variation in X can be explained by the An R^2 of 0....
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This note was uploaded on 01/25/2012 for the course ECN 102 taught by Professor Saler during the Fall '11 term at University of Colorado Denver.

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ecn 102 hw 4 #2 - 1 ) Y(bar) = (8+12+11+5)/4 = 9 X(bar) =...

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