Problems with solutions -MS Exam: Option I. Probability and Statistics- Fall 2007

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1 Problems with solutions -MS Exam: Option I Probability and Statistics- Fall 2007 [1] Math 310 -MS Exam, Fall 2007 If A is a real symmetric matrix then show that its range (column space) R(A) and null space N(A) have only the null vector in common, i.e., R(A) N(A) = {0}. Solution to Math 310 -MS Exam, Fall 2007-Majumdar Let y R(A) N(A). Then y = Ax, and Ay = 0. i.e AAx = 0 i.e A T Ax = 0(sinceA = A T ) i.e x T A T Ax = 0 i.e Ax = 0 i.e y = 0 [2] Math MS Exam, Fall 2007 Find x sin x lim. x 0 x 3 Solution to Math MS Exam, Fall 2007-Miescke We use L Hospital s Rule three times: x sin x lim x 0 x 3 = lim x 0 1 cos x 3x 2 [3] Stat MS Exam, Fall 2007 = lim x 0 sin x 6x = lim cos x x 0 6 = 1 6. If X and Y are independent random variables each uniform on the interval [0, 1] then find (i). p.d.f. of X + Y. (ii). E(X + Y ) 2 1

2 (i.) Solution to Math MS Exam, Fall 2007-Majumdar { 1 if 0 < x < 1, f X (x) = 0 Otherwise { 1 if f Y (y) = 0 Otherwise 0 < y < 1 { 1 if f X,Y (x, y) = 0 Otherwise 0 < x, y < 1 Let Z = X + Y. Let F Z (z) be the cumulative distribution function of Z. case 1. 0 < z < 1. In this case 0 < x, y < 1. Thus F Z (z) = P (Z z) = P (X + Y z) = z dx (z x) dy = z case 2. 1 z < 2. In This case F Z (z) = P (Z z) = P (X + Y z) = 1 P (X + Y > z) Now P (X + Y > z) for z > 1 is simply the area of the triangle formed by the intersection of the sets A = {(x, y) : x + y > z} and B = {(x, y) :, 0 x, y 1} which is (2 z)2. Thus 2 F Z (z) = 1 (2 z)2. Now the density of Z is given by 2 (ii.) f X (x) = z if 0 z 1, 2 z if 1 < z 2, 0 Otherwise By symmetry and by the independence of X, Y we have E(X) = E(Y ), E(X 2 ) = E(Y 2 ). Now E(X+Y ) 2 = E(X 2 )+E(Y 2 )+2E(XY ) = 2E(X 2 )+2[E(X)] 2 = 2 [ 1 ] x2 dx + 2 xdx) 0 = 7 6 2

3 [4] Stat (Chapter 5,6,7):-MS Exam, Fall 2007 Let X 1, X 2,..., X n, n > 2, be a random sample from the discrete distribution b(1, θ) that has the pdf where 0 < θ < 1. f(x; θ) = { θ x (1 θ) 1 x, x = 0, 1; 0, elsewhere. (a) Show that Y 1 = n i=1 X i is a complete sufficient statistic for θ. (b) Find the function ϕ(y 1 ) which is the MVUE of θ. Solution to Stat MS Exam, Fall 2007-Yang (a) Proof: For 0 < θ < 1, rewrite f(x; θ) = exp {x log θ + (1 x) log(1 θ)} { } θ = exp log x + log(1 θ), x {0, 1} 1 θ So b(1, θ) belongs to the regular exponential class with Therefore p(θ) = log θ, q(θ) = log(1 θ), K(x) = x, S(x) = 0 1 θ Y 1 = n K(X i ) = i=1 is a complete sufficient statistic for θ. (b) Note that n i=1 X i E( Y 1 n ) = 1 n ne(x 1) = θ So ϕ(y 1 ) = Y 1 /n is the MVUE of θ. 3

4 [5] Stat (Chapter 8,9):-MS Exam, Fall 2007 Let X = (X 1,..., X n ) denote a random sample from the distribution N(0, θ) that has the pdf f(x; θ) = 1 ) exp ( x2, < x <. 2πθ 2θ The purpose is to test the hypothesis H 0 : θ = 1 against the alternative hypothesis H 1 : θ > 1. (a) Show that the likelihood ratio L(θ = 1; X)/L(θ = θ 1 ; X) is based upon the statistic Y = n i=1 X2 i, where θ 1 > 1 is fixed. (b) If n = 15, find a uniformly most powerful critical region of size α = 0.05 for the hypothesis test. (Hint: Use the chi-square table attached.) Solution to Stat (chapters 8,9):- MS Exam, Fall 2007-Yang (a) Proof: The likelihood function of θ given x = (x 1,..., x n ) is L(θ; x) = Therefore the likelihood ratio ( ) { n/2 1 exp 1 2πθ 2θ n i=1 { ( ) n L(θ = 1; X) L(θ = θ 1 ; X) = θ1 1 θn/2 1 exp 2θ 1 So it is based upon the statistic Y = n i=1 X2 i. (b) For any fixed θ 1 > 1 and for any positive constant k, x 2 i i=1 } X 2 i } L(θ = 1; x) L(θ = θ 1 ; x) k n i=1 x 2 i 2θ [ 1 n θ log θ 1 log k] = c By the Neyman-Pearson Theorem, the best critical region C for testing H 0 : θ = 1 against H 1 : θ = θ 1 4

5 takes the form of {(x 1,..., x n ) : n i=1 x2 i c}, where the constant c is determined by P H0 (X C) = α If n = 15, Y χ 2 (15) under the null hypothesis. By the chi-square table attached, c = if α = Note that the best critical region C does not depend on θ 1. So { } n C = (x 1,..., x n ) : x 2 i is a uniformly most powerful critical region of size α = 0.05 for testing H 0 : θ = 1 against H 1 : θ > 1. [6] Stat MS Exam, Fall 2007 i=1 In a study of effectiveness of hypnosis the emotions of fear, happiness, depression, and calmness were requested in random order for each of five subjects during hypnosis. The following table gives the resulting measurements of skin potential (adjusted for initial level) in millivolts. Subject A B C D E Fear Happiness Depression Calmness Assuming that the skin potential of a person has different levels in the four types of emotions suppose we want to test whether hypnosis can cause these emotions. (a) Describe the basic model of the Friedman test, give its test statistic S, and compute S. (b) Explain why the Friedman test is here more appropriate than the Kruskal- Wallis test. 5

6 Solution to Stat MS Exam, Fall 2007-Miescke The basic underlying model is X i,j = θ + β i + τ j + e i,j, i = 1,..., n, j = 1,..., k, where the e i,j are i.i.d. with a common continuous c.d.f. F. The β i are the block effects, and the τ j are the treatment effects. For testing H 0 : τ 1 = = τ k versus H A : τ [1] < τ [k] we reject H 0 for large values of the test statistic S = 12n k(k + 1) k ( R j k j=1 where the cut-off value is taken from a table with quantiles of the distribution of S under H 0. For the given data, n = 5, k = 4, and the ranks are Subject A B C D E Fear Happiness Depression Calmness This gives S = 3 [( )2 + ( )2 + ( )2 + ( ] )2 = 5.88 [7] Stat MS Exam, Fall 2007 This refers to sampling from a finite population of size N using auxiliary size measures given in terms of values of an auxiliary character X while the study variable is denoted by Y. The sample size is n > 2. First one unit is randomly selected by adopting PPS method. Denote by i the selected unit. Then unit i is removed and the remaining (n 1) units are selected by adopting SRSWOR method from the remaining (N 1) population units. Denote by s = (i,.) the sample so drawn. ) 2 6

7 (a). Work out first and second order inclusion probabilities of the units in the population, to be denoted as Π j s and Π jk s. (b). Show that Π j Π k > Π jk for all pairs of units (j, k), j k. (c). Suggest an unbiased estimate of the population mean Ȳ, i. based only on the first unit selected from the population; ii. based on all the n units selected from the population. Solution to Stat MS Exam, Fall 2007-Hedayat We denote by p j the normed size of j-th population unit, so that j p j = 1. j = 1, 2,..., N (a). In standard notations : (b). Easy (c). y i p i (i). Π j = p j + (1 p j ). (n 1) (N 1) (ii). Π jk = p j. (n 1) (N 1) + p k. (n 1) (N 1) + (1 p j p k ). (n 1)(n 2) (N 1)(N 2) y i Np i serves as an unbiased estimate of the population total. Hence, serves as an unbiased estimate of Ȳ. We may suggest the Horvitz- Thompson Estimate [HTE] of the population mean Ȳ which is given by 1 y j N j Π j sum being over all sample units in s. [8] Stat MS Exam, Fall 2007 Consider a three dimensional Poisson process of particles in the space with intensity parameter ν. Fix a particle and let D be the distance from this particle to its nearest neighbor. Find (i). P (D > d). (ii). E(D). Solution to Stat MS Exam, Fall 2007-El-Neweihi 7

8 P (D > d) = P (no particles in a sphere with center at chosen particle and radius d). =e 4 3 πd3ν, d > 0. E(D) = = e 4 3 πνx3 dx e 4 3 πνy y 2 3 dy = 1 3 Γ(1 3 ) πν 3 [9] Stat 481 Problem 1 - MS Exam, Fall 2007 In studying the thrust force developed by a drill press, it is thought that drilling speed and feed rate of the material are the most important factors. An experiment is conducted with 4 feed rates and two drill speeds. The experiment is conducted by using a completely randomized design with 2 replicates for each level combination. The measurements(thrust force) are presented in the table below: Drill speed Feed Ratio (i.) Complete the df and MS column for the following ANOVA table. 8

9 Source df SS MS Drill speed (A).148 Feed Rate (B).092 Interaction (A*B).042 Error Total(corrected).303 (ii.) The two factor interaction turned out to have a p-value of Since feed rate is a factor with quantitative levels the interaction was partitioned into a component A B L and a reminder where B L refers to the linear component of B. Compute the sums of squares for both of these components. (iii.) By studying the main factors and two factor interaction (all of which are significant), it became clear that a succinct description of the relationship between the response and the two factors was difficult. It is therefore decided to make a two way table of average responses. From this table an estimate for the effect of A was computed separately at every level of B. Compute the common estimated standard error for these four effect estimators. [10] Stat 481 Problem 2 - MS Exam, Fall 2007 This problem with 3 parts deals with 2 3 factorial experiment that uses a completely randomized design in which each of the 8 treatments is replicated twice. (i.) In this situation, would you make a normal probability plot of the factorial effect estimates? Why or why not? (ii.) The observed treatment means from this experiment, each of which is 9

10 based on two replicates, are shown in the following table. Treatment mean (1) 22 a 25 b 12 ab 35 c 23 ac 30 bc 15 abc 38 For the contrast corresponding to the main effect of factor A, compute an estimate and the corresponding standard error. (iii.) It turned out that A and AB were the only significant factorial effects. With this information and the treatment means given in part (ii.) discuss in as much detail as you can at how the three factors A, B, and C affect the response. Also decide which treatment(s) you would recommend if it was desirable to maximize the response. (Hint: An appropriate two- way table might be helpful). Solution to Stat 481 Problem [9] - MS Exam, Fall 2007 Solution: Solution to Stat 481 Problem [10] - MS Exam, Fall

11 Appendix A: Table I Chi-Square Distribution The following table presents selected quantiles of chi-square distribution; i.e., the value x such that P (X x) = for selected degrees of freedom r. x 0 1 Γ(r/2)2 r/2 wr/2 1 e w/2 dw, 11

12 P (X x) r

13 Appendix B: Table II Normal Distribution The following table presents the standard normal distribution. The probabilities tabled are P (X x) = Φ(x) = x 1 2π e w2 /2 dw Note that only the probabilities for x 0 are tabled. To obtain the probabilities for x < 0, use the identity Φ( x) = 1 Φ(x). 13

14 x