P c0 y c1 2 y n y x l xi c0 c0 2 i j 820

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Unformatted text preview: 1 8.16 a. y e =y e y− − dy 2 2 2 2 Size = P (reject H0 | H0 is true) = 1 ⇒ Type I error = 1. which is negative for y < 1 and positive for y > 1. Thus f (y |2)/f (y |1) is decreasing for y ≤ 1 Power = P (reject H0 | HA is true) = 1 ⇒ Type II error = 0. and increasing for y ≥ 1. Hence, rejecting for f (y |2)/f (y |1) > k is equivalent to rejecting for y b. ≤ c0 or y ≥ c1 . To obtain a size α test, the constants c0 and c1 must satisfy Size = + P (Y H 1 = is = 1 − − + ⇒ Type f (c0 |2) = f (c |2) α = P (Y ≤ c0 |θ = 1)P (reject≥ c0 |θH0 1) true) e=c0 0 e−c1 and I error = 0. 1 . f (c0 |1) f (c |1) Power = P (reject H0 | HA is true) = 0 ⇒ Type II error = 1. 1 Solving these two equations numerically, for α = .10, yields c0 = .076546 and c1 = 3.637798. 8.17 a. The likelihood function is is The Type II error probability θ−1 ￿µ−1 ￿ c1 ￿ ￿ 1 −1/2 −y1/2 ￿ ￿ −y1/2 ￿c1 Problem 5[C&B] 8.20 n e dy = −e yj ￿ .= .609824. P (c0 < Y < c1(θ = |2) y) = µn y | µ, θ x, = L xi θ c0 c0 2 i j 8.20 By the Neyman-Pearson Lemma, the UMP test rejects for large values of f (x|H1 )/f (x|H0 ). Computing this ratio we obtain x 1 2 3 4 5 6 f (x|H1 ) f (x|H0 ) 7 6 5 4 3 2 1 .84 The ratio is decreasing in x. So rejecting for large values of f (x|H1 )/f (x|H0 ) corresponds to reject...
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