Thus n 675 25 2 196 1 0 675 implies n

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Unformatted text preview: = −2.33. n(.51)(.49) Solving these equations gives n = 13,567 and c = 6,783.5. 8.15 From the Neyman-Pearson lemma the UMP test rejects H0 if ￿ ￿ ￿n ￿ ￿￿ 2 2 −n/2 −Σi x2 /(2σ1 ) i f (x | σ1 ) (2πσ1 ) e σ0 1￿ 2 1 1 = = exp x >k 2 2 − σ2 2 −n/2 e−Σi x2 /(2σ0 ) f (x | σ0 ) σ1 2 i i σ0 i (2πσ0 ) 1 for some k ≥ 0. After some algebra, this is equivalent to rejecting if ￿ ￿ n ￿ 1 2log (k (σ1 /σ0 ) ) 1 ￿ ￿ x2 > =c because 2 − 2 > 0 . i 1 1 σ0 σ1 − σ2 i σ2 2 0 1 ￿2 This is the UMP test￿ size α, where α = Pσ0 ( i Xi > c). To determine c to obtain a specified of 2 2 α, use the fact that i Xi /σ0 ∼ χ2 . Thus n ￿ ￿ ￿ ￿ ￿ √ ￿ ￿ Φ(−.675) ≈ .25￿ 2 1.96 − (σ1 /σ0 ).675 implies n = 6.943 ≈ 7. implies 2log (k n = −n ) 1 1 ￿ ￿ =c because 2 2 > 0 . xi > 8.19 The pdf of Y is 1 1 σ0 σ1 2 − σ2 i σ0 1 1 (1/θ )−1 −y 1/θ f (y |θ) = y e , y > 0. ￿2 θ This is the UMP test￿ size α, where α = Pσ0 ( i Xi > c). To determine c to obtain a specified of Byuse the fact that the UMP 2 2 α, the Neyman-Pearson Lemma, χ2 . Thus test will reject if i X i /σ 0 ∼ n ￿1 ￿ 1/2 ￿ 1/2 ey−y = f (y |2) > k. ￿ ￿ y− 2 2 2 2 f |1) α = Pσ0 2 Xi /σ0 > c/σ0 (y= P χ2 > c/σ0 , n Problem 4[C&B] 8.16 i To see the form of this rejection region, we compute 2 2 so we must have c/σ0 = χ2 α , which means c = σ0 χ2 α . ￿ n, ￿ n, ￿ ￿ d 1 −1/2 y−y1/2 1 −3/2 y−y1/2 y 1/2...
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This note was uploaded on 08/09/2013 for the course STATS 10-705 taught by Professor Wasserman during the Fall '12 term at Carnegie Mellon.

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