Associate Professor S. Fan Sample Midterm II STAT 6205 Winter 2013 Instructions: You have 110 minutes to complete the exam. Please write all solutions in the space provided, including all pertinent work for full credit. Partial credit will only be given where work is shown. Please box all answers. You are allowed one sheet of notes, handwritten, two-sided, for the exam along with a calculator. You are also allowed to bring a text book for use of the tables only. 1. Suppose that the time (in minutes) until the n Statistics professors show up for their respective classes (starting from the time the class begins) independently follow an exponential distribution. Concisely, X1 , X2 , . . . , Xn are independently exponential with p.d.f. fXi (xi ) = 1 − (axiθ) e i , (ai θ) xi > 0, θ > 0, where a1 , a2 , . . . , an are known constants. ˆ (a) Find the maximum likelihood estimator of θ, say θ. Solution: L(θ) = n Y i=1 1 − (axiθ) e i (ai θ) giving `(θ) = − X ln(ai ) − n ln(θ) − i 1X θ i xi ai . Taking the derivative yields d`(θ) n 1 X xi =− + 2 . dθ θ θ ai i Setting the above to zero yields 1X θˆ = n i xi ai . Checking the second derivative, we get 2 X xi 2n n n d2 `(θ) = − = − <0 2 ˆ dθ2 θ=θˆ θˆ2 θˆ3 ai θ θˆ2 i Xi 1 P Therefore the m.l.e. is θˆ = n provided that at least one of the professors i ai shows up late (i.e. at least one of the xi s is greater than 0). ˆ (b) Find the expected value and variance of θ. Solution: Here, X ˆ = 1 E(θ) n i Page 1 of 4 E(Xi ) ai 1X = n i (ai θ) ai =θ Continue on next page . . . Associate Professor S. Fan Sample Midterm II and ˆ = V ar(θ) STAT 6205 Winter 2013 2 1 X V ar(Xi ) 1 X (a2i θ2 ) = = θn . 2 2 2 2 n n ai ai i i (c) Now suppose for n = 4 professors, a1 = 1, a2 = 2, a3 = 3, a4 = 4 with observed values 1, 3.5, 4, and 2.5 respectively. Find the point estimate of θ. Solution: Using the result from part (a), 1 X xi 1 1 3.5 4 2.5 113 ˆ θ= = = + + + = 1.177083. n ai 4 1 2 3 4 96 i (d) Derive the exact distribution of θˆ using the moment generating function. Solution: Mθˆ(t) = n Y MXi /(nai ) (t) i=1 where MXi /(nai ) (t) = E exp tXi nai so = MXi t nai = ai θt 1− nai −1 θt −1 = 1− n θt −n Mθˆ(t) = 1 − n which is the m.g.f. for a gamma random variable with parameters θ0 = respectively. θ n and α = n (e) Use part (d) to show that in the limit, as n → ∞, the distribution of θˆ is degenerate at θ with probability one. Solution: θt −n = eθt . lim Mθˆ(t) = lim 1 − n→∞ n→∞ n This then shows that in the limit, θˆ will be equal to θ with probability one. 2. Suppose we have a random sample X1 , X2 , . . . , Xn from a distribution with p.d.f. f (xi ; θ) = θxiθ−1 , (a) Show that Y = − you use. Page 2 of 4 Pn i=1 ln(Xi ) 0 < x < 1, 0 < θ < ∞ is a sufficient statistic for 1/θ and name any theorem that Continue on next page . . . Associate Professor S. Fan Sample Midterm II STAT 6205 Winter 2013 Solution: Using exponential family and writing the p.d.f. as f (x; θ) = exp {ln(θ) + (θ − 1) ln(x)} , we have that U = also be sufficient. Pn i=1 ln(Xi ) is sufficient for 1/θ and so Y = − Pn i=1 ln(Xi ) must (b) Find the distribution of Y [Hint: find the distribution of V = − ln(X) first]. Solution: −v fV (v) = fX (e −v de = θe−v(θ−1) e−v = θe−θv , ) dv v > 0. This means V has an exponential distribution with mean 1/θ. G(n, 1/θ). Therefore, Y ∼ ˆ that is a linear function of the sufficient statistic such that it is (c) Find an estimator, say θ, unbiased for 1/θ. You must show that it is unbiased. Solution: Since E(Y ) = nθ , θˆ = Y /n is an unbiased estimator of 1/θ. Concisely, ˆ = E(θ) ˆ − 1/θ = 0. Bias1/θ (θ) (d) Calculate the mean square error for θˆ as an estimator of 1/θ Solution: Since θˆ is unbiased, the MSE is equal to the variance. In this case, the variance is ˆ = V ar(Y /n) = n = 1 . V ar(θ) n2 θ 2 nθ2 3. A team of industrial designers want to determine the average time it takes an adult to assemble an “easy to assemble” toy. In a random sample of 28 adults, the mean is 19.9 minutes with standard deviation 5.73 minutes. (a) Define, with notation, the parameter of interest and create a 95% confidence interval for it. Give the interpretation for your interval in terms of the problem. Solution: Here, the parameter of interest is µ, the average time of all adults to assemble this class of toys. We have t.025 (27) = 2.052 and the resulting 95% CI for µ is 5.73 19.9 ± (2.052) √ 28 or (17.68, 22.12). We are 95% confident that the average time it takes for adults to assemble “easy to assemble” toys is between 17.68 and 22.12 minutes. Page 3 of 4 Continue on next page . . . Associate Professor S. Fan Sample Midterm II STAT 6205 Winter 2013 (b) How would increasing the sample size affect the size of the interval in (a)? Explain without calculations. Solution: Increasing the sample size would have two effects on our original CI, both making the CI shorter. First, the t–multiplier would have been smaller since the ¯ would degrees of freedom are larger and secondly, the estimated standard error of X also be smaller (assuming s is about the same). (c) If companies claim that the average time is 15 minutes, does the confidence interval you created above support this? Why or why not? Solution: Since 15 is not in the CI we constructed, the companies’ claim is not supported. Page 4 of 4 End of Exam