Download Stat 411 Homework 09: Power Functions and Hypothesis Testing and more Assignments Statistics in PDF only on Docsity! Stat 411 – Homework 09 Due: Wednesday 04/04 Undergraduates may solve the “Graduate only” problem(s) for possible extra credit. Some of the problems ask for plots; you’re welcome (actually encouraged) to use a com- puter to produce these plots. 1. Problem 5.5.1 on page 269.1 2. Let X1, . . . , Xn iid∼ Unif(0, θ). The goal is to test H0 : θ = θ0 versus H1 : θ > θ0, where θ0 > 0 is some fixed value. The rule is to reject H0 if and only if X(n) > c, where c is a constant to be determined. (a) Find c such that the size is 0.05. (b) Find the power function pow(θ) of the test. (c) Sketch your power function when θ0 = 1 and n = 10. 3. Problem 5.6.6 on page 277. (Hint: The differences W1, . . . ,W51 are a sample from a distribution with unknown mean µW and unknown variance σ 2 W .) 4. Problem 5.6.8 on page 277. (I’d recommend writing “θ” for the parameter instead of “p” to be consistent with our general notation.) 5. Let X1, . . . , Xn iid∼ Ber(θ). The goal is to test H0 : θ = 0.49 versus H1 : θ = 0.51. Using the Central Limit Theorem-based test, find n such that the Type I and Type II error probabilities are both approximately 0.01. 6. (Graduate only) Reconsider the setup in Problem 2 above. (a) Find a test with size 0.05 based on the Central Limit Theorem approximation of the sampling distribution of X (b) Find the power function pow0(θ) for this new test. (c) Compare the power function pow0(θ) to pow(θ) from Problem 2(b) by sketching a graph of the two, overlaid on the same plot, with θ0 = 1 and n = 10. 1This problem explains why it is typical to consider tests like H0 : θ = θ0 versus H1 : θ > θ0 rather than H0 : θ ≤ θ0 versus H1 : θ > θ0. The point is that the power function is monotone and, therefore, achieves its maximum over the null parameter space at the boundary point θ0. 1 Stat 411 – Homework 09 Solutions 1. Problem 5.5.1 from the text. For testing H0 : µ = µ0 vs. H1 : µ > µ0, the power function for under consideration in equation (5.5.12) is pow(µ) = Φ ( −zα − √ n(µ0 − µ) σ ) , where zα satisfies Φ(−zα) = α, and both µ0 and σ are fixed and known constants. To show that the same test has size α for the different null hypothesis H ′0 : µ ≤ µ0, we need to show that maxµ≤µ0 pow(µ) = α. By the property of zα, it is clear that pow(µ0) = α; moreover, pow(µ) → 0 as µ → −∞. So if we can show that pow(µ) is increasing for µ ∈ (−∞, µ0], then we’re done. To accomplish this, show that the derivative of pow(µ) is positive: dpow(µ) dµ = ϕ ( −zα − √ n(µ0 − µ) σ ) · √ n σ , where ϕ is the standard normal PDF, the derivative of Φ. Since the PDF ϕ is positive and so are n and σ, the derivative must be positive. Therefore, the power function is strictly increasing, hence, the size of the test for H ′0 : µ ≤ µ0 is size α. 2. (a) The size for the test (which rejects H0 when X(n) > c) is size = Pθ0(X(n) > c) = 1− Pθ0(X(n) ≤ c) = 1− (c/θ0)n, where we have used the fact that the CDF of X(n), the maximum of the Xi’s, is the CDF of X1 to power n. Therefore, if we want the size to be equal to 0.05, take c = 0.951/nθ0. (b) The power function is given by pow(θ) = Pθ(X(n) > 0.95 1/nθ0) = 1− 0.95(θ0/θ)n. (c) Picture below plots pow(θ) when n = 10 and θ0 = 1. 1.0 1.5 2.0 2.5 0 .2 0 .4 0 .6 0 .8 1 .0 θ p o w (θ ) 1
