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Unformatted text preview: PSTAT 120C: Solutions to Assignment # 7 June 5, 2006 1. Most of this problem is just working out the various sums. (a) Starting from the righthand side of the first equation, 1 n k X ` =1 n ` (¯ y i ¯ y ` ) = 1 n k X ` =1 n ` ¯ y i 1 n k X ` =1 n ` ¯ y ` = ¯ y i n k X ` =1 n ` ¯ y = ¯ y i n n ¯ y = ¯ y i ¯ y where we have used the identity ∑ n ` = n . For the second half of this problem we go back to the sum ∑ k ` =1 n ` (¯ y i ¯ y ` ) and note that for ` = i the term in the sum is 0 and can be dropped. Thus, 1 n k X ` =1 n ` (¯ y i ¯ y ` ) = 1 n X ` 6 = i n ` (¯ y i ¯ y ` ) = 1 n X ` 6 = i n ` ¯ y i 1 n X ` 6 = i n ` ¯ y ` = ¯ y i n X ` 6 = i n ` 1 n X ` 6 = i n ` ¯ y ` = n n i n ¯ y i 1 n X ` 6 = i n ` ¯ y ` . The last equality follow because X ` 6 = i n ` = k X ` =1 n ` n i = n n i . Equating the two displayed results above ¯ y i ¯ y = n n i n ¯ y i 1 n X ` 6 = i n ` ¯ y ` . (b) The motivation for part 1a) is to write ¯ y i ¯ y as a sum of independent components. The 1 variance of this sum is the sum of the variances. Var (¯ y i ¯ y ) = Var n n i n ¯ y i 1 n X ` 6 = i n ` ¯ y ` = n n i n 2 Var (¯ y i ) + 1 n 2 X ` 6 = i...
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 Spring '07
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 Degrees Of Freedom, Null hypothesis, Statistical hypothesis testing

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