Stat 411 Homework 10: Statistical Inference and Hypothesis Testing, Assignments of Statistics

Download Stat 411 Homework 10: Statistical Inference and Hypothesis Testing and more Assignments Statistics in PDF only on Docsity! Stat 411 – Homework 10 Due: Wednesday 04/11 Undergraduates may solve the “Graduate only” problem(s) for possible extra credit. 1. Problem 6.3.5 on page 339. Take µ0 = 0 for simplicity (and no loss of generality). 2. Let X1, . . . , Xn iid∼ fθ(x) = θxθ−1, where x ∈ (0, 1) and θ > 0. Use Wilk’s theorem (i.e., Theorem 6.3.1 in the text) to derive an approximate size α test of H0 : θ = θ0 versus H1 : θ 6= θ0. 3. Problem 6.3.18 on page 342. Also, in part (a), give an intuitive explanation for setting Λ = 0 when Yn > θ0. Hints for part (b), where θ = θ0: (i) Yn is definitely less than θ0; (ii) Z = Yn/θ0 has the same distribution as that of the maximum of n iid Unif(0, 1) random variables; (iii) we know the CDF for Z; (iv) the chi-square distribution with 2 degrees of freedom is the same as an exponential distribution with mean 0.5. 4. Let X1, . . . , Xn iid∼ N(µ, σ2) where both µ and σ2 are unknown. Derive the exact size α likelihood ratio test of H0 : µ = µ0 versus H1 : µ 6= µ0—Wilk’s theorem isn’t allowed here! 5. (Graduate only) According to Mendel’s law, the genotypes aa, Aa, and AA in a population in equilibrium (with respect to a single gene with two alleles) have proportions θ2, 2θ(1 − θ), and (1 − θ)2, respectively, where θ ∈ (0, 1). From this population, a sample of n individuals is taken, of which Naa, NAa, and NAA in- dividuals have genotypes aa, Aa, and AA, respectively. Note that, by definition, Naa +NAa +NAA = n. (a) Find the MLE θ̂. (Hint: This is a special kind of multinomial distribution; see page 138.) (b) Use Wilk’s theorem to derive an approximate size α likelihood ratio test of H0 : θ = θ0 versus H1 : θ 6= θ0. An optional problem involving software. Plot the power function for the likelihood ratio test derived Problem 1 above; take θ0 = 1, µ0 = 0, and n = 10. (Hint: In R, you’ll need the pchisq function, which returns the CDF for a chi-square distribution.) 1 Stat 411 – Homework 10 Solutions 1. Problem 6.3.5 from the text. Let X1, . . . , Xn iid∼ N(0, θ) where θ is the variance.1 We’ve shown several times that the MLE of θ is θ̂ = 1 n ∑n i=1X 2 i . Therefore, the likelihood ratio statistic is Λ = L(θ0) L(θ̂) = θ−n0 e −(nθ̂/2θ0) θ̂−ne−n/2 = ( θ̂ θ0 )n e− n 2 (θ̂/θ0−1). If W = nθ̂/θ0 = θ −1 0 ∑n i=1X 2 i , then Λ = g(W ) = n−nen/2W ne−W/2. Since Λ is a function of W , the likelihood ratio test can be based on W . That is, a rule which rejects H0 iff Λ is too small is equivalent to a test which rejects H0 iff W falls in some set to be determined. The null distribution of W (i.e., its distribution when the null hypothesis H0 : θ = θ0 is true) can be found by writing W as W = 1 θ0 n∑ i=1 X2i = n∑ i=1 (Xi/ √ θ0) 2. The random variables Xi/ √ θ0, i = 1, . . . , n are independent standard normal. So their squares are independent chi-square with one degree of freedom. From Cochran’s theorem,2 the sum of n independent ChiSq(1) random variables is a ChiSq(n) random variable. Therefore, W ∼ ChiSq(n) is the null distribution.3 To find the rejection region for W , we need to understand the shape of the function g(w) given above. Ignoring the parts not involving w, g(w) resembles the PDF formula for a gamma random variable. These PDFs start at 0, increase for a while, and then decrease to 0 as w → ∞. Therefore, Λ ≤ k iff W ≤ k′1 or W ≥ k′2. To find k′1, k ′ 2 we need to solve the equation: α = Pθ0(W ≤ k′1 or W ≥ k′2) = Pθ0(W ≤ k′1) + Pθ0(W ≥ k′2), where α is the desired size. Let’s do this by making each term in the sum equal α/2. Then we should take k′1 = χ 2 n,1−α/2 and k ′ 2 = χ 2 n,α/2, the upper percentile points from the ChiSq(n) distribution. For example, if n = 10 and α = 0.05, then k′1 = 3.247 and k′2 = 20.483, from Table II on page 673. Finally, we have the following testing rule for the exact size α likelihood ratio test in terms of W : Reject H0 iff W ≤ χ2n,1−α/2 or W ≥ χ2n,α/2. 1I’ve taken the mean µ0 to be zero without loss of generality. 2That is, if Ui ∼ ChiSq(di), i = 1, . . . , n, are independent, then ∑n i=1 Ui ∼ ChiSq(d) with d = ∑n i=1 di; see Theorem 3.5.4 in the text. 3The distribution of W under a general θ value is a multiple of ChiSq(n). 1