Answer Key for STAT 200B HW No. 7

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1 Answer Key for STAT 200B HW No. 7 May 5, 2007 Problem 2.2 p. 649 Assuming binomial 2-sample model ˆπ =.75, ˆπ 2 =.6. a ˆτ = ˆπ 2 ˆπ =.5. From Ex. 2.5a on page 644: ˆπ ˆπ + ˆπ 2 ˆπ = + =.087; n n ˆτ ± z α/2.5 ± =.3,.0 ; Wald [ test of the ] hypothesis [ H 0 :] π = π 2, equiv. H 0 : τ = 0 vs. H : π π 2 : p = 2 Φ ˆτ = 2 Φ b τ = h π, π 2 = π 2 π, thus ˆτ = ˆπ 2 ˆπ =.8, and h π = π 2, h π 2 π 2 = π. Hence, from Ex. 2.5b on page 644: ˆπ 2 2 ˆπ ˆπ ˆπ 2 + ˆπ 2 2 ˆπ 2 ˆπ 2 n ˆπ 2 n 2 = ˆτ ˆπ n ˆπ + ˆπ 2 n 2ˆπ 2 = ; ˆτ ± z α/2.8 ± =.598,.002 ; Wald test for H 0 : τ = vs. H : τ : p = 2 Φ ˆτ = 2 Φ c τ = h π, π 2 = π 2/ π 2 π / π, thus ˆτ = ˆπ 2/ ˆπ 2 ˆπ / ˆπ =.5. From Ex. 2.5b on page 644: ˆτ n ˆπ ˆπ + n 2ˆπ 2 ˆπ 2 = ; ˆτ ± z α/2.5 ± =.38,.862 ; Wald test for H 0 : τ = vs. H : τ : p = 2 Φ ˆτ = 2 Φ d Accept H a 0 : π = π 2, accept H b 0 : π 2 π =, but reject H c 0 : π 2/ π 2 π / π that the 2 cancer treatments are equivalent. Why is there such a large dierence between the p-values in a and b

2 vs. c: the odds-ratio distribution is not as well approximated by the normal distribution as the risk dierence and the relative risk distributions are for this sample size. Problem 2.6 p. 650 Assuming Poisson 2-sample model ˆλ = ˆµ n = y n = 00, ˆλ = 60 a ˆτ = ˆλ 2 ˆλ = τ λ 2 =, τ λ =, thus the standard error of the estimate ˆτ is: ˆλ n + ˆλ 2 n 2 = ; ˆτ ± z α/ ± = , ; Wald [ test of the ] hypothesis [ H 0 : λ ] = λ 2, equiv. H 0 : τ = 0 vs. H : λ λ 2 : p = 2 Φ ˆτ = 2 Φ b τ = h π, π 2 = λ 2 λ, thus ˆτ = ˆλ 2 =.8, and ˆλ h π = λ 2, h π 2 = λ. Hence: = ˆτ ˆλ 2 ˆλ 2 λ ˆλ + ˆλ ˆλ2 n n ˆλ + n 2ˆλ2 =.8 n 2 + ˆµ.02; ˆµ 2 ˆτ ± z α/2.8 ± =.6,.9998 ; Wald test for H 0 : τ = vs. vs. H : τ : p = 2 Φ ˆτ = 2 Φ c Reject H a 0 : λ = λ 2, but accept H b 0 : λ 2 λ = that the country rates are equal. Similar to the rst problem the dierence in the p-values can be explained with the fact that the normal distribution does not approximate the relative risk distribution very well for such small sample size. Problem 2.6 p. 659 Assuming binomial -sample model with n = 24 observations we want to test H 0 : π = π 0 = 6 vs. H : π π 0. a Exact p-value using density c = 0, and c 2 = 8, because f z; n, π 0 f 8; n, π 0 for z {0, 8, 9,..., 24}: p f = f 0; n, π z=8 f z; n, π 0 =.04802; b Exact p-value using c.d.f: p F = F 0; n, π 0 + F 8 ; n, π 0 =.04802; Problem 2.25 a p. 66 2

3 The data is generated according to the following model Y P oissonnλ. We test H 0 : λ = λ 0 vs. H : λ = λ using likelihood ratio test function: Λ y = f y; n, λ y f y; n, λ 0 = λ e nλ 0 λ, λ 0 which for λ λ 0 > is strictly increasing function of y. Thus, the rejection region C where Λ y takes values larger than in the acceptance region must be of the form {c, c +,...} for some critical value c and vice versa. Problem 2.34 p. 67 Assuming binomial 2-sample model the estimates of the probabilities of catching cold are ˆπ = 3 40, ˆπ 2 = a ˆτ = ˆθ 2 ˆθ = logitˆπ 2 logitˆπ =.738. From Ex. 2.6b on pp : n ˆπ ˆπ + n 2ˆπ 2 ˆπ 2 = =.329; Nominal 95% upper condence bound: ˆτ ± z α.738 ± =.725; b Similar to Ex. 2.8 a,b and using the bound found in a the nominal 95% upper conf. bound for the odds ratio is exp ˆθ2 ˆθ = exp.725 =.842. c Wald test for H 0 : θ 2 θ vs. H : θ 2 θ : p = Φ ˆθ2 ˆθ = Φ reject H 0 at α =.05 level. d The exact p-values are more conservative due to the discreteness of the underlying binomial distributions: in our case we get p exact =.02. On the other hand the nominal p-value for testing the null hypothesis that τ = π 2/ π 2 π / π is Φ =.002 is quite optimistic. Problem 2.37 p. 672 In canonical form the density of the univariate normal is f y = exp µy n µ2 r y; n, where σ 2 2σ 2 r y; n = exp y 2 and 2πσ 2 2σ 2 a θ = µ θ,, and C θ = µ2 ; σ 2 2σ 2 b µ = σ 2 θ; c C θ = µ2 d C θ = σ 2 θ Range C θ, and C θ is strictly increasing; e C θ = σ 2 > 0; = θ2 σ 2 2σ 2 2 ; f t = C θ = σ 2 θ θ = t σ 2 = C t, t R; g nc θ = nσ 2 θ = nµ and nc θ = nσ 2. Problem 3.2 p. 685 a Solve the following linear system of equations: ED50 = logit.5 β 0 β = 0 ED90 = logit.9 β 0 β = 20, 3

4 to get the estimates ˆβ = log 9 /0.22 and ˆβ 0 = log ; b c ED99 = logit.99 ˆβ 0 ˆβ 30.9; ˆπ x = 25 = d Figure shows the plot of ˆπ x = exp ˆβ0 + ˆβ 25 + exp ˆβ0 + ˆβ.964; 25 [. + exp ˆβ 0 ˆβ x] pi_hatx x Figure : Plot of ˆπ x for x [0, 25]. Problem 3.5 p. 685 Let π = Pr Y = denote the probability that a random individual from society Y has the disease. This probability varies between the 4 subpopulation stratied by the behavioural factor X, hence: Pr Y = X = x = π = 3 x=0 exp θx +exp θx = +exp5 x, thus: Pr Y = X = x Pr X = x. ˆπ =.6 + exp exp Similarly, ˆπ 2.048, and, nally, the relative risk estimate is ˆπ2 ˆπ Problem 3.0 p

5 Plug-in X = D X t into formula 2 on page 68 to show the desired result. Problem 3.2 p. 698 a L Y,..., Y n ; µ, σ 2 = { [ 2πσ 2 ] n exp i b l Y,..., Y n ; µ, σ 2 = 2 c [Y i µx i ] 2 2σ 2 }. { n log 2π + n log σ 2 + i [Y i µx i ] 2 2σ 2 }. arg max l Y,..., Y n ; µ, σ 2 [Y i µ x i ] 2 = arg max µ,σ 2 µ,σ 2 2σ i 2. µ minimizes the denominator alone so it follows that the maximum likelihood estimate ˆµ equals the least squares one. Because l Y,..., Y n ; µ, σ 2 is convex function of µ considered as a n dimensional vector of values..., µ x i,... n it follows that this maximum is unique as well. The likelihood function is monotonic transformation of the log-likelihood, thus it follows that it is also uniquely maximized and the point of maximum is the same. d Dierentiating the log-likelihood with respect to σ 2 and setting it to 0 leads to ˆσ 2 = n i [Y i ˆµ x i ] 2 = RSS/n, assuming we are maximizing with respect to ˆµ x i found in c. Problem 3.27 p. 699 a It's the same as in Ex. 3.4 a; b Using theorem 3. on page 90, the MLE for θ is ˆθ = log Ȳ : ˆθ = log Ȳ =.28, ˆθ 2 =.26, ˆθ 3 =.5, ˆθ 4 =.772; c ˆβ = X ˆθ =.95,.44,.74,.275 t. 5

Answer Key for STAT 200B HW No. 8

Answer Key for STAT 200B HW No. 8 May 8, 2007 Problem 3.42 p. 708 The values of Ȳ for x 00, 0, 20, 30 are 5/40, 0, 20/50, and, respectively. From Corollary 3.5 it follows that MLE exists i G is identiable