Solutions to Statistics 131C Homework 3: Probability Distributions and Hypothesis Testing, Assignments of Mathematical Statistics

Download Solutions to Statistics 131C Homework 3: Probability Distributions and Hypothesis Testing and more Assignments Mathematical Statistics in PDF only on Docsity! Spring 2009 Statistics 131C Solution : Homework 3 1. Problem 8.4.4 : If we choose c1 and c2 to be symmetric about the value 0.15 = (0.1 + 0.2)/2, then π(0.1|δ) = π(0.2|δ). Accordingly, let c1 = 0.15 − k and c2 = 0.15 + k for some k > 0. When µ = 0.1, the random variable Z = 5(Xn − 0.1) has N(0, 1) distribution. Hence, π(0.1|δ) = P (Xn ≤ c1|δ) + P (Xn ≥ c2|δ) = P (Z ≤ 0.25− 5k) + P (Z ≥ 0.25 + 5k) = Φ(0.25− 5k) + Φ(−0.25− 5k). We must choose k such that π(0.1|δ) = 0.07. By trial and error, using the table of standard normal distribution, or by using a statistical software package like R, we obtain that when 5k = 1.867, π(0.1|δ) = Φ(−1.617) + Φ(−2.117) = 0.0529 + 0.0171 = 0.07. Thus, k = 0.3734 and so c1 = −0.2234 and c2 = 0.5234. 2. Problem 8.4.5 : As in Problem 8.4.4, π(0.1|δ) = P (Z ≤ 5(c1 − 0.1)) + P (Z ≥ 5(c2 − 0.1)) = Φ(5c1 − 0.5) + Φ(0.5− 5c2). Similarly, π(0.2|δ) = P (Z ≤ 5(c1 − 0.2)) + P (Z ≥ 5(c2 − 0.2)) = Φ(5c1 − 1) + Φ(1− 5c2). Thus, c1 and c2 are solutions of the following equations: Φ(5c1 − 0.5) + Φ(0.5− 5c2) = 0.02 Φ(5c1 − 1) + Φ(1− 5c2) = 0.07 By trial and error, using the table for standard normal probability distribution, or using a statistical software package, it is found that if 5c1 = −2.12 and 5c2 = 2.655 (so that c1 = −0.424 and c2 = 0.531), then Φ(5c1 − 0.5) + Φ(0.5− 5c2) = Φ(−2.62) + Φ(−2.155) = 0.0044 + 0.0155 ≈ 0.02 and Φ(5c1 − 1) + Φ(1− 5c2) = Φ(−3.12) + Φ(−1.655) = 0.0009 + 0.0490 ≈ 0.07. 3. Problem 8.4.10 : Let α0 = 0.05 and let c1 = 3α 1/n 0 . Also let c2 = 3. Then, with T = max{X1, . . . , Xn}, π(θ|δ) = P (T ≤ 3α1/n0 |θ) + P (T ≥ 3|θ). Since for θ ≤ 3, P (T ≥ 3|θ) = 0 and P (T ≤ 3α1/n0 |θ) = (3α1/n0 /θ)n, it follows that π(3|δ) = α0 = 0.05. If θ > 3, then π(θ|δ) = ( 3α1/n0 θ )n + 1− ( 3 θ )n = 1− (1− α0) ( 3 θ )n > α0. 1