Hw6Sol - 1 4 = f Y 1 y 1 f Y 2 y 2 = 1 2 1 2 By...

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Homework Solutions Assignment Number 6 Due February 26, 2007 Section 6.4. Problem 10. (a) P (Type I error) = P (reject H 0 | H 0 is true) = P ( Y 3 . 20 | λ = 1) = Z 3 . 20 e - y dy = 0 . 04 . (b) P (Type II error) = P (accept H 0 | H 1 is true) = P Y 3 . 20 λ = 4 3 = Z 3 . 20 0 3 4 e - 3 y/ 4 dy = Z 2 . 4 0 = 0 . 91 . Problem 16. If H 0 is true, X = X 1 + X 2 has a binomial distribution with n = 6 and p = 1 2 . Therefore α = P (reject H 0 | H 0 is true) = P X 5 p = 1 2 = 6 X k =5 6 k 1 2 k 1 - 1 2 6 - k = 7 2 6 = 0 . 11 . Problem 21. α = P (reject H 0 | H 0 is true) = P ( Y 1 + Y 2 k | θ = 2) . When H 0 is true, Y 1 and Y 2 are uniformly distributed over the square defined by 0 Y 1 2 and 0 Y 2 2, so the joint pdf of Y 1 and Y 2 is a plane parallel to the Y 1 Y 2 -axis at height
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Unformatted text preview: 1 4 ( = f Y 1 ( y 1 ) · f Y 2 ( y 2 ) = 1 2 · 1 2 ) . By geometry, α is the volume of the triangular wedge in the lower left-hand corner of the square over which Y 1 and Y 2 are defined. The hypotenuse of the triangle in the Y 1 Y 2-plane has the equation y 1 + y 2 = k . Therefore α =area of triangle × height of wedge= 1 2 · k · k · 1 4 = k 2 8 . For α to be 0.05, k = √ . 04 = 0 . 63 . 1...
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