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52 2 b0 y 2 b1 x 252 255 2 x2 2 y b1 x

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Unformatted text preview: 1 = ¯ (X 1 X 2.51. n(1) 0 } =i E {nb0 − 1 X } Xi = 0 E {b Y − Y − b b1¯ 2.51. E1(2)} =Yi XY− b0 1 XXi − b1 Xi2 = 0 {b0 E { i¯ − b ¯ } ¯1 (β01+ β1E {) − X β¯ {b } Xi Y − XE = i 1 n =1Yi − b0 − b}Xi )2 (n 1 (3) σ2 ˆ ¯ =b1 } { = ¯ E¯ Yi } − XE { n = β0 + βnX − X β1 = β0 11 ¯ Equations (1)(β0 + (2)Xi ) − X β1 and β1 are the same as the least squares normal equations (1.9), = 2 2 the maximum likelihood estimators b and b are the same as those in (1.27). 1n b X } ¯ hence ¯ ¯ 0 1 2.52. σ {b0 } = σ={Y − (1 0 + β1 Xi ) − X β1 β nβ0 + β1 X − X β1 = β0 ¯ ¯ = ¯2 2 2.54. Yes,= σ 2 {Y } + X σ ¯ b1 }¯ 2X σ {Y , b1 } no ¯ { −¯ ¯ 2 = β0 + β...
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