You may use your calculator and a single page of notes.

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1 LAST NAME (Please Print): KEY FIRST NAME (Please Print): HONOR PLEDGE (Please Sign): Statistics 111 Midterm 2 This is a closed book exam. You may use your calculator and a single page of notes. The room is crowded. Please be careful to look only at your own exam. Try to sit one seat apart; the proctors may ask you to randomize your seating a bit. Report all numerical answers to at least two correct decimal places or (when appropriate) write them as a fraction. All question parts count for 1 point. 1

2 1. Let f(x,y) = c(y x) 8 for 0 < x < y < 1, and it is 0 otherwise. 90 What is c? 1 = c 1 y 0 0 (y x)8 dxdy = c 1 0 [ 1 9 (y x)9 ] y 0 dy = c y9 dy = so c = Let f(x,y) = 6(y x) on 1 < x < y < 2, and it is 0 otherwise. What is the marginal density for X? (Remember to indicate where it is non-zero.) f X (x) = 3x 2 12x + 12 for 1 x 2 and 0 else. Integrate wrt y, so f X (x) = 2 x 6(y x)dy = 3x2 12x + 12 for 1 x What is the expected value of X? IE[X] = 2 1 x f X(x)dx = 2 1 3x3 12x xdx = 1.25 What is the conditional density of Y given that X = 1.5? (Indicate the support.) 4(2y 3) on 3/2 y 2 and 0 else. f Y X (y x) = f(x,y)/f X (x). For X = 1.5 this is 6(y 1.5)/[3(1.5) 2 12(1.5) + 12] = 4(2y 3) on 3/2 y 2 and 0 else. 3. Let f(x) = 2x/θ 2 for 0 x θ with θ > 0. The density is 0 elsewhere. You observe X 1,...,X n. IE[X] = θ 0 xf(x)dx = θ 0 2x2 /θ 2 dx = (2/3)θ. So an unbiased estimator is (3/2) X. θ 2 /(8n) What is the variance of your estimator? 2

3 Var(X) = IE[X 2 ] (IE[X]) 2. IE[X 2 ] = θ 2 /2. So Var(X) = θ 2 /2 (2/3) 2 θ 2 = θ 2 /18. The variance of X the variance of X over n. And then multiply this by (3/2) 2 to get θ 2 /(8n). 4. A Farlie-Gumbel-Morganstern distribution has mean 17 and variance 20. You draw a random sample of size What is the expected value of X? 0.63 What is the standard error of X? The standard error is σ/ n = What is IP[ X > 17.2] (approximately)? Use the CLT. This is the probability that Z > ( )/(0.6324) = From the table, this is You will inherit a fortune if you spend the night in a haunted house. Ghosts will not attack you as long as there is a candle burning. The lifetime of a candle is exponentially distributed with mean 30 minutes and standard deviation 30 minutes. You have 15 candles, and the night is 8 hours long. What is the (approximate) chance you survive? 0.40 This uses the CLT for sums. The total lifetime of 15 candles is approximately normal with mean 15 * (1/2) = 7.5 hours and standard error 15 (1/2) = Thus the chance that you survive is the probability that the sum exceeds 8, or the approximate chance Z > (8 7.5)/1.936 = From the table, this is You toss a fair coin 10 times. Let X be the number of heads in the first 4 tosses, and let Y be the total number of heads. What is the correlation between X and Y. (Hint: Write out the definition of Cov(A,B + C).) 3

4 0.63 Cov(A,B + C) = IE[A (B + C)] IE[A] IE[B + C] = IE[AB + BC] IE[A] (IE[B]+ IE[C]) = IE[AB] IE[A] IE[B] + IE[AC] IE[A] IE[C] = Cov(A,B) + Cov(A,C). So now let Z = Y X be the number of heads in the last 6 tosses. Then Cov(X,Y ) = Cov(X,X + Z) = Cov(X,X) + Cov(X,Z) = Var(X) since X and Z are independent. For a binomial, we know that Var(X) = n (1/2) (1/2) so the correlation is Cov(X,Y )/ Var(X) Var(Y ) = (4 (1/2) (1/2))/ (4 (1/2) (1/2)) (10 (1/2) (1/2)) = Suppose that Y i has Poisson distribution with mean λx i where the x i are fixed, known positive values (e.g., x i = i). You observe a random sample Y 1,...,Y n. Write the likelihood function. f(y 1,...,y n ;λ) = n i=1 (λx i ) y i (y i!) 1 exp( λx i ) What is the maximum likelihood estimator ˆλ? Take the logarithm of the likelihood to get n n n lnf = y i lnλ + y i lnx i λ x i c. i=1 i=1 i=1 Take the derivative, set it to zero, and solve: so ˆλ = y i / x i. ˆλ = y i / x i 0 = dln f dλ = λ 1 n n y i x i i=1 i=1 What is the bias of this estimator? Bias is 0. IE[ˆλ] = 1 xi IE[yi ] = 1 xi λxi = λ. 4

5 What is the standard deviation of this estimator? (Hint: Recall that if X Pois(λ), then IE[X] = Var [X] = λ.) Var [ˆλ] = Var [ yi xi ] = ( x i ) 2 Var [y i ] = ( x i ) 2 λ x i = λ/ x i so the standard deviation is the square root of this. ˆ sd(λ) = λ/ x i 8. Let X 1,...,X n be the lifespans of n lightbulbs, each independent, and each of which has the exponential distribution with parameter λ. Find the cumulative distribution function for the lifespan of the first bulb to fail, i.e., the distribution of min{x 1,...,X n }. IP[min{X 1,...X n } < x] = 1 IP[X 1 > x and X 2 > x and... and X n > x] = 1 i = 1 n IP[X i > x] = 1 n i=1 exp λx = 1 exp nλx. So the distribution of the minimum of a set of n exponentials has the exponential distribution with parameter nλ. 1/(nλ) What is the expected lifetime of the first bulb to fail? Recall that the mean of an exponential distribution with parameter θ is 1/θ. 9. Let f(x;θ) = (θ)x θ 1 for 0 x 1. You draw two random values: 1/4 and 1/ What is the maximum likelihood estimate of θ? The likelihood is so the log-likelihood is f(x 1,x 2 ;θ) = θ 2 x θ 1 1 x θ 1 2 l(θ) = 2ln θ + (θ 1)(ln x 1 + ln x 2 ). Take the derivative, set it to zero and solve, so dl(θ) dθ = 2 θ + (ln x 1 + ln x 2 ) = 0 which implies ˆθ = 2/(ln x 1 + ln x 2 ). So, for these numbers, ˆθ =

6 0.90 What is the maximum likelihood estimate of θ The MLE of the function is the function of the MLE, or = Let p be the probability that a female child learns to talk before a male child does. Before you collect data, you believe that p is likely to be greater than 0.5, and your prior on p is a beta distribution with parameters α = 3 and β = 1. You now observe 3 pairs of opposite-gender twins, and see that in each case, the girl spoke first. This is the beta-binomial example, so p has the Beta distribution with parameters α = = 6 and β = What is your posterior probability that p > 0.75? IP[p > 0.75] = p5 (1 p) 0 dp = (α+β)/(α+β + n) For general parameters α and β, suppose you observe x successes in n tries. Write the posterior mean as a linear combination of the following form: c * prior mean + (1 c) (x/n). What is c? The posterior mean is α /(α +β ) = (α+x)/(n+α+β). Writing this as c implies, after some algebra, that c = (α + β)/(α + β + n). α α+β+(1 c) x n From the preceding calculation, how does your inference change as one moves from a samll sample size to a large one? For small n, most of the weight is on the prior mean. But as n increases, the weight on the prior goes to 0 and the weight on the sample mean goes to 1. So the data dominates whatever opinions one originally had. 11. You have 80 cards, numbered 1 through 80. You shuffle them and deal fifty to yourself and thirty to a friend. What is the appoximate distribution of the average value of the cards you hold? (You get three points for this: specifying the family of distributions, the mean, and the standard deviation.) (Hint: n i=1 i = n(n + 1)/2 and n i=1 i 2 = n(n + 1)(2n + 1)/6.) 6

7 This uses the CLT and the Finite Population Correction Factor. The mean of the tickets 40.5, and the sd of the tickets is (1/80) ( )/6 (40.5) 2 = By the CLT, the family is normal and the mean is The sd is ( / 50) FPCF = ( / 50) (80 50)/(80 1) = family: Normal mean: 40.5 standard deviation: Write the letters for all (and only) the true statements. A, C, D, F A. The finite population correction factor adjusts for drawing a large fraction of the population without replacement. B. Maximum likelihood estimators are unbiased. C. Maximum likelihood estimators have approximate normal distributions as the sample size increases. D. As the sample size increases, maximum likelihood estimators have the minimum possible variance. E. If ˆθ is unbiased for θ, then cos ˆθ is unbiased for cosθ. F. A 10% trimmed mean throws out the largest 5% of the values and the smallest 5% of the values. G. Nonresponse bias occurs if the interviewee refuses to answer your questions. H. Respondent bias occurs if the wording of the question encourages certain kinds of answers. I. Household bias implies that residents in large families are overrepresented. 13. The map on the following page was drawn by Charles-Joseph Minard. The thickness of the curved bands conveys the same kind of information as did his graph describing Napoleon s attack upon Moscow. This graph shows immigration flows around the world in Describe the information about the sources and approximate relative magnitudes of immigration into the U.S. (Note: Don t overlook the Pacific coast.) Germany had the largest number of immigrants, followed by the U.K., then France, then China. The total from the U.K. and France was about equal to the German influx. 7

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