100BHWS5

100BHWS5 - STAT 100B HW5 Solution March 18, 2008 Problem 1...

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STAT 100B HW5 Solution March 18, 2008 Problem 1 Hardy-Weinberg model 1. L ( θ ) = θ 2 n AA (2 θ (1 - θ )) n Aa (1 - θ ) 2 n aa l ( θ ) = 2 n AA log θ + n Aa log(2 θ (1 - θ )) + 2 n aa log(1 - θ ) ∂l ( θ ) ∂θ = 2 n AA θ + n Aa θ - n Aa 1 - θ - 2 n aa 1 - θ = 0 ˆ θ MLE = 2 n AA + n Aa 2 n = 0 . 5753 2. Likelihood ratio test L ( p AA ,p Aa ) = p n AA AA p n Aa Aa (1 - p AA - p aa ) n aa l ( p AA ,p Aa ) = n AA log p AA + n Aa log p Aa + n aa log(1 - p AA - p Aa ) ∂l ∂p AA = n AA p AA - n aa 1 - p AA - p Aa = 0 ∂l ∂p Aa = n Aa p Aa - n aa 1 - p AA - p Aa = 0 ˆ p AA = n AA /n, ˆ p Aa = n Aa /n, ˆ p aa = 1 - ˆ p AA - ˆ p Aa = n aa /n. Λ = ˆ p n AA AA ˆ p n Aa Aa ˆ p naa aa θ 2 n AA (2 θ (1 - θ )) n Aa (1 - θ ) 2 n aa = ( ˆ p AA ˆ θ 2 ) n AA ( ˆ p Aa 2 ˆ θ (1 - ˆ θ ) ) n Aa ( ˆ p aa (1 - ˆ θ ) 2 ) n aa = 1 . 0164 2log Λ = 0 . 0325 p - value = p ( χ 2 1 > 0 . 0325) = 2 Pr ( Z > 0 . 1803) = 0 . 8569 No reason to doubt the Hardy-Weinberg model! Problem 2
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100BHWS5 - STAT 100B HW5 Solution March 18, 2008 Problem 1...

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