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Dr. Hackney STA Solutions pg 143

# Dr. Hackney STA Solutions pg 143 - Chapter 9 Interval...

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Chapter 9 Interval Estimation 9.1 Denote A = { x : L ( x ) θ } and B = { x : U ( x ) θ } . Then A B = { x : L ( x ) θ U ( x ) } and 1 P { A B } = P { L ( X ) θ or θ U ( X ) } ≥ P { L ( X ) θ or θ L ( X ) } = 1, since L ( x ) U ( x ). Therefore, P ( A B ) = P ( A )+ P ( B ) - P ( A B ) = 1 - α 1 +1 - α 2 - 1 = 1 - α 1 - α 2 . 9.3 a. The MLE of β is X ( n ) = max X i . Since β is a scale parameter, X ( n ) is a pivot, and . 05 = P β ( X ( n ) c ) = P β (all X i ) = ± β ² α 0 n = c α 0 n implies c = . 05 1 0 n . Thus, . 95 = P β ( X ( n ) /β > c ) = P β ( X ( n ) /c > β ), and { β : β < X ( n ) / ( . 05 1 0 n ) } is a 95% upper conﬁdence limit for β . b. From 7.10, ˆ α = 12 . 59 and X ( n ) = 25. So the conﬁdence interval is (0 , 25 / [ . 05 1 / (12 . 59 · 14) ]) = (0 , 25 . 43). 9.4 a. λ ( x,y ) = sup λ = λ 0 L ( σ 2 X 2 Y ³ ³ x,y ) sup λ (0 , + ) L ( σ 2 X 2 Y | x,y ) The unrestricted MLEs of σ 2 X and σ 2 Y are ˆ σ 2 X = Σ X 2 i n and ˆ σ 2 Y = Σ Y 2 i m , as usual. Under the restriction, λ = λ 0 , σ 2 Y = λ 0 σ 2 X , and L ( σ 2 X 0 σ 2 X ³ ³ x,y ) = ( 2 πσ 2 X ) - n/ 2 ( 2 πλ 0 σ 2 X ) - m/ 2 e - Σ x 2 i / (2 σ 2 X ) · e - Σ y 2 i / (2 λ 0 σ 2 X ) = ( 2 πσ 2 X ) - ( m + n ) / 2 λ - m/ 2 0 e - ( λ 0 Σ x 2 i y 2 i ) / (2 λ 0 σ 2 X ) Diﬀerentiating the log likelihood gives
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