Master s Written Examination

Transcription

1 Master s Written Examination Option: Statistics and Probability Fall 2013 Full points may be obtained for correct answers to eight questions. Each numbered question (which may have several parts) is worth the same number of points. All answers will be graded, but the score for the examination will be the sum of the scores of your best eight solutions. Use separate answer sheets for each question. DO NOT PUT YOUR NAME ON YOUR ANSWER SHEETS. When you have finished, insert all your answer sheets into the envelope provided, then seal it. 1

2 Problem 1 Stat 401. Let X be a negative binomial random variable with pmf where 0 < p < 1 is the success rate. f(x) = p(1 p) x, x = 0, 1, 2..., (a) Compute the moment generating function of X, M X (t) = E(e tx ). (b) Let Y = 2pX be a new random variable. Show that, as p 0, lim M Y (t) = (1 2t) 1, when t < 1 p 0 2. Solution to Problem 1. (a) The mgf of X is M X (t) = E(e tx ) = e tx p(1 p) x = x=1 p ( e t (1 p) ) x x=1 = p(1 e t (1 p)) 1 (1 e t (1 p)) ( e t (1 p) ) x = p(1 e t (1 p)) 1, x=1 where the last equality is due to the fact that (1 e t (1 p)) ( e t (1 p) ) x is the pmf of a negative binomial distribution with success rate 1 e t (1 p). (b) The mgf of Y is When p 0, M Y (t) = E(e ty ) = E(e 2ptX ) = M X (2pt) = p(1 e 2pt (1 p)) 1. lim M p Y (t) = lim p 0 p 0 1 e 2pt (1 p) 1 = lim p 0 2te 2pt + e 2pt + 2pte = 1, when t < 1/2. 2pt 1 2t [Note: This shows that Y converges in distribution to χ 2 (2) as p 0.] 2

3 Problem 2 Stat 401. For two continuous variables x 1 and x 2 and a response variable Y, a general polynomial regression model of degree 2 can be written as E(Y ) = β 0 + β 1 x 1 + β 2 x 2 + β 3 x 1 x 2 + β 4 x β 5 x 2 2. There is one parameter, β 0, of order 0; two parameters, β 1, β 2, of order 1; and three parameters, β 3, β 4, and β 5, of order 2. In a scientific study, suppose there are k continuous variables x 1,..., x k and a response variable Y. A statistician plans to fit a general polynomial regression model of degree d. (a) How many parameters are of order d 1 (d 1 d)? (b) What is the total number of parameters? Solution to Problem 2. (a) We only need to find out the number of different k xl i i with k l i = d 1. This is equivalent to all possible nonnegative integer solutions of l 1 + l l k = d 1. Each nonnegative integer solution corresponds to a permutation of d 1 balls and k 1 sticks in a line. Each permutation of d 1 balls and k 1 sticks in a line also corresponds to a nonnegative integer solution. There are ( d 1 ) +k 1 d 1 possible different permeations. So there are ( d 1 ) +k 1 d 1 nonnegative integer solutions, which mean there are ( d 1 ) +k 1 d 1 parameters are of order d1. (b) We only need to find out the number of different k xl i i with k l i = d 1, d 1 = 0, 1,..., d. The all possible nonnegative integer solutions of l 1 + l l k = d 1, d 1 = 0, 1,..., d is equivalent to all possible nonnegative integer solutions of l 1 + l l k + l k+1 = d. By the same argument as that of (a), there are ( ) d+k d parameters. 3

4 Problem 3 Stat 401. A random variable X has a log-normal distribution, if Y = log X has a normal distribution with density function f Y (y) = 1 (y µ)2 2πσ exp{ }. 2 2σ 2 (a) Find the density function f X (x) of X. (b) Find E(X) and Var(X). (c) If the X i s are independent lognormal random variables, then show that the product X = n X i is also a lognormal random variable. Solution to Problem 3. (a) log(x) = Y implies that X = exp(y ) and J = Y X f X (x) = (b) For first two moments of X we get and 1 1 { 2πσ 2 x exp (log(x) } µ)2, x > 0. 2σ 2 = 1/X. Thus, the pdf of X is E(X) = E(exp{Y }) + = (2πσ 2 ) 1/2 (y } µ)2 exp { + y dy 2σ 2 + = (2πσ 2 ) 1/2 exp { (y µ σ2 ) 2 + µ 2 (µ + σ 2 ) 2 } dy 2σ 2 1 } = exp{ 2 (σ2 + 2µ) E(X 2 ) = E(exp{2Y }) + = (2πσ 2 ) 1/2 (y } µ)2 exp { + 2y dy 2σ 2 + = (2πσ 2 ) 1/2 exp { (y µ 2σ2 ) 2 + µ 2 (µ + 2σ 2 ) 2 } dy 2σ 2 = exp(2σ 2 + 2µ) These two moments could be obtained directly from the moment-generating function of Y, i.e., E(X) = M Y (1) and E(X 2 ) = M Y (2), where M Y (t) = E(exp{tY }) = exp(µt+σ 2 t 2 /2). Next, for the variance, we get Var(X) = exp(2µ+σ 2 ) (exp(σ 2 ) 1). (c) Let Y i = log(x i ), i = 1,..., n. Then Y i, i = 1,..., n are independent normal distributed. Thus n Y i is still normally distributed. On the other hand, log(x) = n log(x i) = n Y i, so the conclusion follows. 4

5 Problem 4 Stat 411. Let X 1,..., X n be iid from a continuous uniform distribution on the interval ( θ, θ), where θ > 0 is unknown. (a) Show that the maximum likelihood estimator for θ is ˆθ = max 1 i n X i. (b) Suppose cˆθ is an unbiased estimator of θ. Determine the constant c. Solution to Problem 4. (a) Note that a pdf of X i is Consider the likelihood function and get L(θ) = f θ (x) = 1 2θ I [ θ,θ](x) = 1 2θ I [0,θ]( x ). n f θ (X i ) = ( 1 ) n ( ) I[0,θ] max 2θ X i, 1 i n l(θ) = n log(2θ) I [maxi X i, )(θ). Since l (θ) = n θ < 0, for θ max 1 i n X i > 0, then the MLE of θ is ˆθ = max 1 i n X i. (b) Denote Y = max 1 i n X i. Then the support of Y is [0, θ). For 0 y < θ, P θ (Y y) = Therefore, a pdf of Y (or ˆθ) is n P θ ( X i y) = n y θ = yn θ n. h θ (y) = dp θ(y y) dy = nyn 1 θ n, 0 y < θ. Since E θ (ˆθ) = E θ (Y ) = θ 0 y nyn 1 dy = n θ y n dy = θ n θ n 0 n n + 1 θ θ, ˆθ is not an unbiased estimator of θ. Nevertheless, if c = n+1, then cˆθ is unbiased. n 5

6 Problem 5 Stat 411. Let X 1,..., X n be iid from a continuous uniform distribution on the interval (0, θ). For a given θ 0 > 0, the goal is to test H 0 : θ = θ 0 versus H 1 : θ > θ 0. Consider a test that rejects H 0 if and only if X (n) > c, where X (n) = max{x 1,..., X n } is the sample maximum and c > 0 is a constant to be determined. (a) Find c such that the size of the test is α. (b) Find the power function pow(θ) for the size-α test. n for all θ > θ 0. Show that pow(θ) 1 as (c) Suppose θ 0 = 1 and α = 0.1. Find n such that pow(2) 0.9 Solution to Problem 5. (a) The distribution function F n,θ of X (n) satisfies ( x ) n. F n,θ (x) = P θ (X (n) x) = P θ (X 1 x) n = θ In this case, α set = P θ0 (X (n) > c) = 1 F n,θ0 (c) = 1 ( c θ 0 ) n. Solving this equation for c gives c = θ 0 (1 α) 1/n. (b) Using our calculations from above, the power function looks like ( θ0 (1 α) 1/n ) n ( θ0 ) n. pow(θ) = P θ (X (n) > c) = 1 = 1 (1 α) θ θ Since θ > θ 0 the ratio (θ 0 /θ) n 0, so pow(θ) 1 as n. (c) If θ 0 = 1 and α = 0.1, then pow(2) = 1 0.9(0.5) n. Then Therefore, take n = 4. pow(2) 0.9 (0.5) n 1/9 n log 1 9 log

7 Problem 6 Stat 411. Let X 1,..., X n be iid Poisson random variables with probability mass function p θ (x) = e θ θ x /x!, x = 0, 1, 2,.... Write η = e θ and set T = n X i. (a) Show that ˆη n = (1 1 n )T is the minimum variance unbiased estimator of η. (b) Show that E(ˆη 2 n) = e θ/n 2θ and, then find the variance of ˆη n. Use Chebyshev s inequality to prove that ˆη n is a consistent estimator of η. (c) Compare the variance of ˆη n with the approximate variance of the maximum likelihood estimator η n = e T/n obtained from the Delta Theorem. Solution to Problem 6. (a) Since ˆη n is a function of the complete sufficient statistic T, if it s unbiased then it must be minimum variance unbiased by Lehmann Scheffe. To compute the expectation of ˆη n, recall that T Pois(nθ), a fact that can be verified with momentgenerating functions. Then we have E(ˆη n ) = ( 1 1 ) t e nθ (nθ)t n t! t=0 = e nθ+(nθ θ) t=0 (b) We get E(ˆη 2 n) in a similar way: = t=0 (nθ θ) (nθ θ)t e t! nθ (nθ θ)t e t! = e θ η = unbiased! ( E(ˆη n) 2 = 1 1 ) 2te nθ (nθ)t = e nθ [(1 1 n )2 nθ] t n t! t! t=0 t=0 = e nθ+(1 1 n )2 nθ e [(1 1 n )2 nθ] [(1 1 n )2 nθ] t = e nθ+(1 1 n )2nθ = e θ/n 2θ. t! t=0 Then the variance is V(ˆη n ) = e θ/n 2θ e 2θ = e 2θ (e θ/n 1). Using Chebyshev s inequality, we get P{ ˆη n η > ε} ε 2 e 2θ (e θ/n 1). As n, the upper bound vanishes, so ˆη n is a consistent estimator of η. (c) By the Delta Theorem, the MLE η n = e T/n is asymptotically unbiased with variance e 2θ θ n. Since ex 1 x for all x, it follows that the MLE has smaller asymptotic variance than the MVUE. 7

8 Problem 7 Stat 416. A group of researchers want to investigate the relationship between math and computer anxiety. The test scored are shown in the table below, with larger scores for indicating greater amount of the trait. Student A B C D E F Math Anxiety Computer Anxiety (a) Suppose both scores are symmetrically distributed. State your hypothesis and computer p-value for the signed rank test, what is your conclusion? (b) Use Spearman s Rho test to test if there is significant association between computer anxiety as math anxiety. Solution to Problem 7. (a) Let D = Y X, both hypotheses: H 0 : M D = 0 vs. H 1 : M D 0. Signed-rank test: Student A B C D E F X i Y i D i = Y i X i r( D i ) T + = N r( D i )I {Di >0} = = 12 p-value = 2P{T + 13} = = The large p-value shows that there is no significant difference between the medians of the two anxiety scores. (b) Use Spearman s test statistic for association measure, first rank the two scores respectively Spearman s Rho test Its p-value = P(R 0.829) < Student A B C D E F S i = rank(x i ) R i = rank(y i ) D i = S i R i R = 1 6 n D2 i n(n 2 1) = =

9 Problem 8 Stat 431. It is stated that there is an even proportion of left-handed males and females in the neighboring society. To verify this statistically, a statistician confines her study to a neighboring township which happens to have 8 elementary schools. Clearly, a complete frame of the population of left-handed students in all the schools combined is not readily available. However, the school district provided information on the number of registered students in each school. The data are given below. School ID Registered Students The survey statistician decided to use the following fixed-size (3) sampling plan: s {1, 2, 4} {2, 3, 5} {3, 4, 6} {4, 5, 7} {1, 5, 6} {2, 6, 7} P (s) s {1, 3, 7} {1, 2, 8} {3, 4, 8} {5, 6, 8} {2, 7, 8} P (s) (a) The statistician reported that she used the school size as an informative auxiliary size measure in designing her survey. (i) Verify that, indeed, this survey uses school size as a size measure. (ii) What is the reason behind using this size measure for the survey? (b) Upon implementation of the sampling plan, the sample {1, 3, 7} was selected for the interviews. The survey team visited these three schools and found the following information: School ID Left-handed boys Left-handed girls (i) Use the Horvitz Thompson method and construct unbiased estimates for the total number of left-handed boys and left-handed girls in these 8 schools. (ii) Use the survey plan and the selected sample and exhibit the variances of your estimates in part (i). Represent your variances in the format of Sen Yates Grundy. (iii) Use the collected data based on this survey and construct unbiased estimates of the variances in part (ii) via the Sen Yates Grundy procedure. (c) What is the name of this survey in the survey book/literature? 9

10 Solution to Problem 8. (a) (i) The survey design proposed by the statistician is a π ps design. To see this, the first-order inclusion probabilities are π 1 = 0.45, π 2 = 0.37, π 3 = 0.40, π 4 = 0.28 π 5 = 0.37, π 6 = 0.43, π 7 = 0.50, π 8 = 0.20, and the number of registered students in the 8 schools are X 1 = 450, X 2 = 370, X 3 = 400, X 4 = 280, X 5 = 370, X 6 = 430, X 7 = 500, X 8 = 200, with T X = X X 8 = Then it s easy to see that π l = n Xl X l = 3 T X 3000 = X l, l = 1, 2,..., (ii) The statistician was guided that the number of left-handed students is roughly proportional to the school size and, thus, using this fixed size sampling plan together with Sen Yates Grundy variance estimation will produce a more precise estimate of the number of left-handed students. (b) (i) Since {1, 3, 7} was selected, we get ˆT B HT E = l S T B l π l = T B 1 π 1 + T B 3 π 3 + T B 7 π 7, where Then and, similarly, π 1 = = 0.45 π 3 = = 0.40 π 7 = = ˆT B HT E = , ˆT G HT E = T G 1 π 1 + T G 3 π 3 + T G 7 π 7 = (ii) We use the survey plan and the selected sample to exhibit the variances of the estimates provided above. The estimator used above is HTE, and this is a fixed-size design, so we can use the Sen Yates Grundy formula, i.e., V( ˆT B HT E) = i V( ˆT G HT E) = i ( T B (π i π j π ij ) j>i i π i ( T G (π i π j π ij ) j>i i π i T B j π j ) 2 T G j π j ) 2. 10

11 (iii) Based on the formulae above, we can check that all T i T j are included, so we need all π ij > 0 and we can easily see that this is true. So, in order to construct unbiased estimators of the variances, we can simply use the HTE (of course, there are other unbiased estimators). Indeed, we get V( ˆT B HT E) = and V( ˆT G HT E) = i,j S,j>i (π i π j π ij ) π ij ( T B i π i T B j π j ) 2 = (π 1π 3 π 13 ) π 13 ( T B 1 π 1 T B 3 π 3 ) 2 + (π 1 π 7 π 17 ) π 17 ( T B 1 π 1 T B 7 π 7 ) 2 + (π 3π 7 π 37 ) ( T B 3 T B ) 2 7 π 37 π 3 π ( 20 = ) ( ) ( ) , i,j S,j>i (π i π j π ij ) π ij ( T G i π i T G j π j ) 2 = (π 1π 3 π 13 ) π 13 ( T G 1 π 1 T G 3 π 3 ) 2 + (π 1 π 7 π 17 ) π 17 ( T G 1 π 1 T G 7 π 7 ) 2 + (π 3π 7 π 37 ) ( T G 3 T G ) 2 7 π 37 π 3 π ( 17 = ) ( ) ( ) (c) Fixed-size first-stage cluster sampling. 11

12 Problem 9 Stat 451. Suppose that x 1,..., x m+n are iid N 2 (µ, Σ) with ( ) ( ) µ1 1 1/2 µ = and Σ = 1/2 1 µ 2 Let x i = (x i1, x i2 ), i = 1,..., n. For some reason, only x 11,..., x m1 and x m+1,2,..., x m+n,2 are observed, while all the other values are missing. Devise the EM algorithm for finding the maximum likelihood estimate of µ. Solution to Problem 9. The log-likelihood for complete data is l(x comp ; µ) = (m + n) log(2π) m + n 2 where Σ 1 = = const 2 3 m+n log Σ 1 m+n (x i µ) T Σ 1 (x i µ), 2 ( (x i1 µ 1 ) 2 + (x i2 µ 2 ) 2 (x i1 µ 1 )(x i2 µ 2 ) ( ) 4/3 2/3 2/3 4/3. Given the current estimate µ, the E step yields that Q(µ µ) = E{l(x comp ; µ) x obs, µ} = const 2 m ( ) (x i1 µ 1 ) 2 + E((x i2 µ 2 ) 2 x i1, µ) (x i1 µ 1 )E(x i2 µ 2 x i1, µ) m+n i=m+1 ( E((x i1 µ 1 ) 2 x i2, µ) + (x i2 µ 2 ) 2 E(x i1 µ 1 x i2, µ)(x i2 µ 2 ) where E(x i2 x i1, µ) = µ 2 + (x i1 µ 1 )/2 and E(x 2 i2 x i1, µ) = (E(x i2 x i1, µ)) 2 + 3/4 for i = 1,..., m, and E(x i1 x i2, µ) = µ 1 +(x i2 µ 2 )/2 and E(x 2 i1 x i2, µ) = (E(x i1 x i2, µ)) 2 +3/4 for i = m + 1,..., m + n. Next, the M step maximizes Q(µ µ) and yields that µ 1 = 1 m + n µ 2 = 1 m + n ( m x i1 + m+n i=m+1 ( m E(x i1 x i2, µ) + E(x i2 x i1, µ) m+n i=m+1 ) x i2 ) Finally, the EM algorithm iterates the E step and the M step until convergence.,. ), ), 12

13 Problem 10 Stat 461. matrix Is there a limiting distribution for the Markov chain? distribution. A Markov chain X 0, X 1,... has the transition probability P = If yes, determine the limiting Solution to Problem 10. It is easy to check that P 2 has all of its entries strictly positive. Hence P is a regular matrix (or the Markov chain X 0, X 1,... is regular). So there is a limiting distribution π = (π 0, π 1, π 2 ). By solving the line system of equations satisfied by π π j = 2 π k P kj, j = 0, 1, 2; and π 0 + π 1 + π 2 = 1, k=0 one obtains π 0 = 1 3, π 1 = 1 3, π 2 = 1 3. Note: Since in this problem, the Markov chain is doubly stochastic, one can tell immediately that π 0 = π 1 = π 2 = 1 without solving the above equations. 3 13

14 Problem 11 Stat 461. Customers enters a store according to a Poisson process of rate λ = per year. Suppose it is known that only 9 customers entered during the first ten days. What is the conditional probability that among the 9 customers, only 3 entered during the first five days? Solution to Problem 11. Let N(0, n] be the number of customers arriving from the starting date to day n. Let W i be the time that the i-th customer arrives. We know that conditioned on N(0, n] = k, the location of W 1,..., W k is like throwing darts (independent and have uniform distribution). Hence ( ) 9 (1 3 ( 1 ) 6 9! 1 ) 9 21 P{N(0, 5] = 3 N(0, 10] = 9} = = = 3 2) 2 6!3!(

15 Problem 12 Stat 471. Consider the linear programming problem max(2x 1 + 4x 2 + x 3 + x 4 ) subject to 2x 1 + 3x 2 + 4x 3 + x 4 4 2x 1 + 3x 2 + 4x 3 + x 4 3 2x 1 + 3x 2 + 4x 3 + x 4 3 x 1, x 2, x 3, x 4 0. (a) Check whether x 1 = 3 2, x 2 = 0, x 3 = 1 8, x 4 = 5 2 is feasible for the problem. (b) Is it basic feasible? (c) Is it optimal? Solution to Problem 12. See handwritten attachment. 15

16 Problem 13 Stat 473. An amount of money accumulates. In period t for t = 1, 2,... T its size is 2t. In each period two people decide simultaneously whether to claim the money or wait. If exactly one person claims the money they receive the entire amount. If two people claim the money each receives half. If either claims the money the game ends. If t = T and no one claims the money each receives half. If t < T and no one claims the money then the go on to the next period. Find all sub-game perfect equilibrium strategies for this game. Solution to Problem 13. Write C for CLAIM and W for WAIT. We consider periods t = 1, 2,..., T. We first consider the case T = 1. We are essentially playing the game C W C 1,1 2,0 W 0,2 1,1 We see that C strongly dominates W for each player, so the only equilibrium is (C, C). We next consider the game with T = 2. If either player claims at t = 1 the game is over. If not, at t = 2 the amount of money is $4 and the play the game C W C 2,2 4,0 W 0,4 2,2 Again, (C,C) is the only Nash equilibrium and they each get payoff 2. Now consider the game at t = 1. We know that if at this stage they both choose to W, at the next stage the payoff will be 2,2. Thus we are essentially playing the game C W C 1,1 2,0 W 0,2 2,2 This game has two pure strategy Nash equilibria (C, C) or (W, W). Thus there are two subgame perfect equilibria in the game. At stage 1, either both players C or both W and at the second stage both W. We will describe these strategies as ((C, C),(C, C)) and ((W, C), (W, C)) where the first sequence describes the actions of Player 1 and the second sequence describes the actions of Player 2. Let s do one more concrete case when T = 3. At t = 3 we are playing and the only equilibrium is (C, C) At t = 2 we are playing C W C 3,3 6,0 W 0,6 3,3 C W C 2,2 4,0 W 0,4 3,3 16

17 and there is a unique equilibrium where both C Thus at t = 1 we are playing C W C 1,1 2,0 W 0,2 2,2 and there are two equilibria one where both C and one where both W. Thus there are two subgame perfect equilibria ((W, C,C), (W, C,C)) This is the general picture for T > 2. There will be two subgame perfect equilibria ((W,C,...,C), (W, C,...,C)) and ((C,C,...,C), (C, C,...,C)) First note that in period t = T we play C W C T, T 2T, 0 W 0, 2T T, T Thus in period T (C, C) is the only equilibrium and the payoff is T, T. We will show by backward induction that both Player should claim in stages t = T, T 1,..., 2 and that the payoff will be (t, t). We have already established this for T. Suppose we know t 2 and we know that at stage t + 1 we should both claim and that if we do the expected payoff is t + 1, t + 1. Then at stage t we are playing the game. C W C t, t 2t, 0 W 0, 2t t + 1, t + 1 Since t 2, 2t > t + 1. Thus (C, C) is the only equilibrium and the payoff is t, t. Finally we consider the case t = 1. We have shown by induction that in a subgame perfect equilibrium both players will claim at t = 2 and the payoff will be 2,2. Thus at stage 1 we are playing the game. C W C 1, 1 2, 0 W 0, 2 2, 2 and (W, W) and (C, C) are both perfect equilibria. Thus there are two subgame perfect equilibria ((W,C,...,C), (W, C,...,C)) and ((C,C,...,C), (C, C,...,C)), i.e., both players take the same action at t = 1 and C in all later rounds. 17

18 Problem 14 Stat 481. Consider a study on the tail lengths (in inches, x) and the weights (in pounds, Y ) of 10 wolves. A numerical summary is given below: n x i = 200, n y i = 1000, n (x i x) (y i ȳ) = 750, n (x i x) 2 = 250, n (y i ȳ) 2 = 6250 (a) If use simple linear regression to analyze the data, please specify the model and necessary assumptions. (b) Construct the ANOVA table. level 0.10? What conclusion will you draw given significance (c) What is the standard error of the least square estimator for the slope coefficient β 1? Use the standard error to find a 90% confidence interval for β 1. Solution to Problem 14. (a) Simple linear regression model Y i = β 0 + β 1 x i + ε i, i = 1,..., n. It is assumed that the errors {ε i, i = 1,..n} are iid, and ε i N(0, σ 2 ). (b) Estimator for slope ˆβ 1 = S xy /S xx = 750/250 = 3 and SSR = ˆβ 2 1 S xx = = ANOVA table: Source DF SS MS F Regression Error Total Hypothesis H 0 : β 1 = 0 vs H 1 : β 1 0. F-test statistic in the ANOVA table is F = 4.5 > F 0.10 (1, 8) = (c) We can show that Var( ˆβ 1 ) = n c 2 i Var(Y i ) = = se 2 ( ˆβ 1 ) = Var( ˆβ 1 ) = n [ n ] 1 c 2 i σ 2 = σ 2 (x i x) 2 It is known that the sampling distribution of ˆβ 1 MSE n (x i x) = MSE = S xx 250 = 2. ˆβ 1 β 1 se( ˆβ 1 ) = ( ˆβ 1 β 1 )/ σ2 S 1 xx SSE σ 2 /(n 2) t(n 2), the 90% confidence interval for β 1 is ˆβ 1 ± t (n 2) se( ˆβ 1 ) = 3 ± = 3 ±

19 Problem 15 Stat 481. A company wants to evaluate the effectiveness of three different training methods (factor B) designed to improve the job performance of its entering sales workforce. New workers are classified on the basis of their educational background (factor A): high school and college graduates. A random sample of 9 high school graduates is randomly and evenly divided into three groups to receive different kinds of training. An analogous procedure is followed for a random sample of 9 college graduates. The sales (in $10,000) for the first quarter after the training of the new employees are as follows: Training Education High School 8, 4, 3 10, 8, 6 8, 6, 7 College 14, 10, 6 8, 9, 10 15, 9, 12 (a) Construct the ANOVA table. For your convenience, it is known that SS Total = and SS Error = 76. (b) Report any significant effects at 5% level. For your reference, the corresponding critical values are F (0.05; 1, 12) = 4.75 and F (0.05; 2, 12) = (c) Write a short paragraph to the company based on your findings. For example, do you have any suggestion/comment on the methods of training? Shall the initial salary of new employees depend on their educational background? Solution to Problem 15. (a) The corresponding ANOVA table is listed as follows: Source SS d.f. MS F A (Education) B (Training) AB Error Total (b) Since 9.55 > 4.75 (factor A), 0.95 < 3.89 (factor B), and 1.26 < 3.89 (interaction AB), only factor A (Education) is significant. (c) Based on the current data, there is no significant difference among the three different methods of training. Overall, the college employees have better sales performance than the high school employees and thus should be paid with a higher initial salary. 19