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Unformatted text preview: PSTAT 120C: Practice Midterm Solutions May 3, 2006 1. One of the main facts that we have learned is that the residuals are uncorrelated with the parameters in the model and thus the fitted values. Cov( e i , ˆ y i ) = 0 = ⇒ Cov( y i ˆ y i , ˆ y i ) = 0 = ⇒ Cov( y i , ˆ y i ) Cov(ˆ y i , ˆ y i ) = 0 . This implies that Cov( y i , ˆ y i ) = Var(ˆ y i ) = σ 2 1 n + ( x i ¯ x ) 2 ∑ n i =1 ( x i ¯ x ) 2 from the variance we use in calculating the confidence interval. 2. (a) ˆ β 1 = ∑ n i =1 ( x i ¯ x )( y i ¯ y ) ∑ n i =1 ( x i ¯ x ) 2 = 365(5 . 67) 364(100) = 0 . 0569 ˆ β = ¯ y ˆ β 1 ¯ x = 2 15(0 . 0569) = 1 . 147 Thus the linear model is wave height = 1 . 147 + 0 . 0569(wind speed) (b) The variance of the residuals is s 2 e = n 1 n 2 ( s 2 wave ˆ β 2 1 s 2 wind ) = 364 363 ( . 81 (0 . 0569) 2 (100) ) = 0 . 488 (c) The appropriate t statistic is t = ˆ β 1 s/ p ( n 1) s 2 wind = . 0569 p . 488 / 36400 = 15 . 546 which means there is a significant relationship. (15which means there is a significant relationship....
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 Spring '07
 idk
 Statistics, Normal Distribution, 3 degrees, 11.5Year

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