Download Solutions to PSTAT 120C Assignment #1: Hypothesis Testing and Power Calculations and more Assignments Asian literature in PDF only on Docsity! PSTAT 120C: Solutions to Assignment # 1 April 14, 2009 1. We have 9 independent observations from a normal distribution with standard deviation 10, and we want ttest H0 : µ = 150 vs. HA : µ 6= 150 The GLRT with level α = 0.05 has the critical set C1 = {x̄ < 143.5 or x̄ > 156.5}. In all of these calculations x̄ ∼ N ( µ, 1009 ) . (a) The power of this test against the alternative µ = 151 is P {x̄ < 143.5}+ P {x̄ > 156.5} = P { x̄− 151 10/3 < 143.5− 151 10/3 } + P { x̄− 151 10/3 > 156.5− 151 10/3 } = P{Z < −2.25}+ P{Z > 1.65} = .0122 + 0.0495 = 0.0617 (b) The power of this test against the alternative µ = 123 is P {x̄ < 143.5}+ P {x̄ > 156.5} = P { x̄− 123 10/3 < 143.5− 123 10/3 } + P { x̄− 123 10/3 > 156.5− 123 10/3 } = P{Z < 6.15}+ P{Z > 10.05} ≈ 1 (c) Alternatively, consider the test with C2 = {x̄ < 144.5 or x̄ > 160.9}. i. The level of this second test is the probability when the mean is 150: P {x̄ < 144.5}+ P {x̄ > 160.9} = P { x̄− 150 10/3 < 144.5− 150 10/3 } + P { x̄− 150 10/3 > 160.9− 150 10/3 } = P{Z < −1.65}+ P{Z > 3.27} = 0.0495 + 0.0005 = 0.05 ii. The power against the alternatives µ = 151 is P {x̄ < 144.5}+ P {x̄ > 160.9} = P { x̄− 151 10/3 < 144.5− 151 10/3 } + P { x̄− 151 10/3 > 160.9− 151 10/3 } = P{Z < −1.95}+ P{Z > 2.97} = 0.0256 + 0.0015 = 0.0271 and against µ = 123 is P {x̄ < 144.5}+ P {x̄ > 160.9} = P { x̄− 123 10/3 < 144.5− 123 10/3 } + P { x̄− 123 10/3 > 160.9− 123 10/3 } = P{Z < 6.45}+ P{Z > 11.37} ≈ 1 1 (d) Both of these test have level α = 0.05. Just comparing at the two points µ = 151 and 123, the first test has greater power at 151, and the second test might have negligibly more power at 123. Thus, it seems that the first test is the better of the two. However, if we compared the powers at µ = 149 the second test would look a little better, but the power of test 1 at 149 is higher than the power of test 2 at 151. It still would be the case that we should prefer test 1. 2. Next week 3. Suppose that Y is a random sample of size 1 from a population with density f(y | θ) = { θyθ−1, 0 ≤ y ≤ 1 0, elsewhere. where θ > 0. (a) The power is P{Y > 0.5 | θ} = ∫ 1 .5 θyθ−1 dy = θ ( 1 θ yθ ) ∣∣∣1 y=.5 = 1− .5θ The picture is 2
