Hw6Sol - 1 4 ( = f Y 1 ( y 1 ) f Y 2 ( y 2 ) = 1 2 1 2 ) ....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ² ² ² ² λ = 4 3 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 ² ² ² ² p = 1 2 ³ = 6 X k =5 ± 6 k ³± 1 2 ³ k ± 1 - 1 2 ³ 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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 4 ( = f Y 1 ( y 1 ) f Y 2 ( y 2 ) = 1 2 1 2 ) . By geometry, is the volume of the triangular wedge in the lower left-hand corner of the square over which Y 1 and Y 2 are dened. The hypotenuse of the triangle in the Y 1 Y 2-plane has the equation y 1 + y 2 = k . Therefore =area of triangle height of wedge= 1 2 k k 1 4 = k 2 8 . For to be 0.05, k = . 04 = 0 . 63 . 1...
View Full Document

This note was uploaded on 03/23/2008 for the course MATH 181A taught by Professor Xu,lily during the Spring '07 term at UCSD.

Ask a homework question - tutors are online