Solutions to the Spring 2017 CAS Exam S

Transcription

1 Solutions to the Spring 2017 CAS Exam S (Incorporating the Final CAS Answer Key) There were 45 questions in total, of equal value, on this 4 hour exam. There was a 15 minute reading period in addition to the 4 hours. The Exam S is copyright 2017 by the Casualty Actuarial Society. The exam is available from the CAS. The solutions and comments are solely the responsibility of the author. CAS Exam S prepared by Howard C. Mahler, FCAS Copyright 2017 by Howard C. Mahler. Howard Mahler hmahler@mac.com

2 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 1 1. You are given the following information about a random variable, X. For all r, s 0, Pr(X > r + s) = Pr(X > r) Pr(X > s). E[X I X > 10] = 30. Calculate the result for the expression: E[X I X > 20]. A. Less than 30 B. At least 30, but less than 40 C. At least 40, but less than 50 D. At least 50, but less than 60 E. At least C. Pr(X > r + s) = Pr(X > r) Pr(X > s) is one way to write the memoryless property, which means we have an Exponential Distribution, with mean θ. 30 = E[X I X > 10] = E[X - 10 I X > 10] + 10 = e(10) + 10 = θ θ = 20. E[X I X > 20] = e(20) + 20 = θ + 20 = = 40. Comment: See Equation 5.3 in Introduction to Probability Models, by Ross. For the Exponential: Pr(X > r + s) = e -λ(r+s) = e -λr e -λs = Pr(X > r) Pr(X > s). For the Exponential, due the memoryless property: Prob[X > r + s X > r] = Prob[X > s]. Pr(X > r + s) / Pr(X > r) = Prob[X > s]. Pr(X > r + s) = Pr(X > r) Prob[X > s].

3 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 2 2. You are given the following information about a senior-year college student from an actuarial science program: Job offers arrive according to a Poisson process with a rate of three per month. A job offer is acceptable if the salary offered is at least 50,000 per year. The salaries offered are mutually independent and follow a lognormal distribution with µ = 10.5 and σ = 0.2. Calculate the probability that it will take this senior-year college student more than four months to receive an acceptable job offer. A. Less than B. At least 0.505, but less than C. At least 0.515, but less than D. At least 0.525, but less than E. At least C. The number of job offer over 4 months is Poisson with mean: (4)(3) = 12. ln[50,000] For the LogNormal, F[50,000] = Φ[ ] = Φ[1.60] = Thus the number of acceptable job offers during 4 months is Poisson with mean: ( )(12) = The chance of no acceptable job offers during 4 months is: e = Comment: Using a computer with no rounding, the answer is As shown in the tables attached to your exam, for the LogNormal Distribution: F(x) = Φ ln(x) - µ σ.

4 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 3 3. You are given the following information about emergency room visits at a local hospital: The amount of time between emergency room visits for broken bones is exponentially distributed with a mean of two hours. The amount of time between emergency room visits for the flu is exponentially distributed with a mean of five hours. Emergency room visits for broken bones and emergency room visits for the flu are independent. Calculate the probability that two patients will come into the hospital with broken bones before three patients come into the hospital with the flu. A. Less than B. At least 0.875, but less than C. At least 0.900, but less than D. At least 0.925, but less than E. At least D. λ bones = 1/2. λ flu = 1/5. Prob[bones visit] = λ bones λ bones + λ flu = 1/ 2 1/ 2 + 1/ 5 = 5/7. Prob[at least 2 of first 4 visits are due to broken bones] = Prob[exactly 2 due to broken bones] + Prob[exactly 3 due to broken bones] + Prob[exactly 4 due to broken bones] = 4 2 (5/7) 2 (2/7) (5/7) 3 (2/7) (5/7) 4 = (6)(100/2401) + (4)(250/2401) + (1)(625/2401) = 2225/2401 = Alternately, define flu as a failure and broken bones as a success. Then the chance that we observe 2 broken bones before seeing 3 flus, is the chance of fewer than 3 failures prior to our 2nd success, in a series of independent Bernoulli trials. The number of failures before 2 successes is given by a Negative Binomial with r = 2 and chance of failure β = = (2/7) / (5/7) = 0.4. chance of success We want the probability of fewer than three failures : f(0) + f(1) + f(2) = (2)(3) 2! = Comment: Since the mean time between visits due to broken bones is shorter than that due to flu, more than half of the visits are due to broken bones. For this type of problem one can either add up Binomial or Negative Binomial densities. The chance of x failures prior to success number r is given by a Negative Binomial Distribution.

5 4. Auto physical damage claims from 100 policyholders of an insurance company arise according to a Poisson process with rate 20 per year. You are given the following: Claim sizes are independent and follow an exponential distribution with mean 3,000. Claim sizes and claim counts are independent. For claims less than 10,000, the insurance company pays the policyholder the claim amount. For claims greater than or equal to 10,000, the insurance company pays the policyholder 10,000. Calculate the expected cost to the insurance company per policyholder per year. A. Less than 560 B. At least 560, but less than 570 C. At least 570, but less than 580 D. At least 580, but less than 590 E. At least C. Frequency per policyholder is: 20 / 100 = 0.2. In the tables attached to the exam, the Iimited expected value of the Exponential Distribution is given as: E[X x] = θ (1- e -x/θ ). Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 4 E[X 10,000] = (3000) (1- e /3000 ) = Expected cost per policyholder is: (0.2) (2893.0) = Alternately, the average payment per loss is: 10,000 x e - x / 3000 / 3000 dx + 10,000 e -10,000/3000 = 0 {-x e - x / e - x / 3000 x = }] x = ,000 e -10/3 = {-10,000 e -10/ e -10/3-3000} + 10,000 e -10/3 = e -10/3 = Expected cost per policyholder is: (0.2) (2893.0) = Alternately, the average payment per loss is: 10,000 10,000 S(x) dx = x e - x / 3000 / 3000 dx = e -10/3 = Proceed as before. 0 0

6 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 5 5. You are given: R and S are two independent components in a series system. X and Y are the time-to-failure random variables of R and S, respectively. X and Y both follow the exponential distribution but with different hazard rates. The probability that R fails before S is 0.2. The mean lifetime of X is 1. R and S start operation at the same time. Calculate the expected time until the system fails. A. Less than 0.25 B. At least 0.25, but less than 0.50 C. At least 0.50, but less than 0.75 D. At least 0.75, but less than 1.00 E. At least A. 0.2 = Prob[R fails before S] = λ R λ R + λ S. λ S = 4 λ R. The mean lifetime of X is 1. 1/λ R = 1. λ R = 1. λ S = 4. The series system fails when the first one fails: S(t) = exp[-λ R t] exp[-λ S t] = exp[-(λ R +λ S ) t] = e -5t. This is an Exponential Distribution with hazard rate 5. Mean time until system fails = 1/5 = 0.2. Alternately, we want the minimum of two independent Exponentials. This is another Exponential with the sum of the hazard rates: = 5. Mean time until system fails = 1/5 = 0.2.

7 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 6 6. X and Y are two independent exponential random variables with hazard rates λ X = 2 and λ Y = 8, respectively. Calculate the expected value of X, conditional on 1 < X < Y. A. Less than 1.20 B. At least 1.20, but less than 1.40 C. At least 1.40, but less than 1.60 D. At least 1.60, but less than 1.80 E. At least 1.80

8 6. A. Truncate and shift each of X and Y at 1; subtract 1 from each value and only keep the positive results. Call these new variables X* and Y*. Then due to memoryless property of the Exponential, X* and Y* are two independent exponential random variables with hazard rates 2 and 8, respectively. E[X 1 < X < Y] = 1 + E[X* 0 < X* < Y*] = 1 + E[X* X* < Y*]. It can be shown that for two independent Exponentials U and V, E[U U < V] = E[V V < U] = E[ Min[U, V] ]. In other words, the minimum is independent of the rank order of the two Exponentials; this is another special property of the Exponential Distribution. In this case, Min[X*, Y*] is Exponential with hazard rate = 10, and thus mean 1/10. Thus, E[X 1 < X < Y] = 1 + 1/10 = 1.1. Alternately, the joint density of X and Y is: f(x, y) = 2 e -2x 8 e -8y. The numerator of E[X 1 < X < Y] is: Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 7 y x 2 e - 2x 8 e - 8y dx dy = 1 1 x = y 8 e - 8y (-x e - 2x - e - 2x / 2) ] 1 x = 1 dy = 8 e - 8y (-y e - 2y - e - 2y / e -2 ) dy = 1-8 y e - 10y - 4 e - 10y + e e - 8y dy = 1 y = (0.8 y e - 10y e -10y )] - (4/10)e e -2 (12/8) e -8 = y = e e e e -10 = 0.22 e -10. The denominator of E[X 1 < X < Y] is: y 2 e - 2x 8 e - 8y dx dy = 8 e - 8y (e -2 - e - 2y ) dy = e -2 8 e - 8y - 8 e - 10y dy = e -2 e -8 - (8/10) e -10 = 0.2 e Thus E[X 1 < X < Y] = (0.22 e -10 ) / (0.2 e -10 ) = 1.1. Comment: Difficult! Assume that we have two independent Exponentials X and Y, with hazard rates λ X and λ Y. The conditional density of X, given X < Y is proportional to: Prob[X=x] Prob[Y > x] = f x (X) S Y (x) = λ X exp[- λ X x] λ Y exp[- λ Y x] = λ X λ Y exp[-(λ X + λ Y )x]. Since this is proportional to the density of an Exponential Distribution with hazard rate λ X + λ Y, this must be the conditional density of X, given X < Y. Thus, X given X < Y follows the same distribution as the minimum of X and Y.

9 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 8 Therefore, the minimum is independent of the rank order of the two Exponentials. x For an Exponential with mean θ, t f(t) dt = θ {1 - e -x/θ - (x/θ)e -x/θ }. 0 Thus, y y 1 x 2 e -2x dx = x 2 e -2x dx - x 2 e -2x dx = (1/2)(1 - e -2y - 2y e -2y ) - (1/2)(1 - e -2-2e -2 ) = -ye -2y - e -2y / e -2.

10 7. You are given: A factory machine is processing parts sequentially in an assembly line. After a part is processed, the machine moves to the next part in line. There is a chance that while waiting to be processed, a part will fall to the floor and drop out of the processing queue. The probability a part drops on the floor in a given time is distributed exponentially with rate θ for each part. Once a part falls on the floor, it drops out of the processing line permanently. The time to complete processing, once the machine has begun work on a part, is exponentially distributed with mean 5 and is independent for each part. A part is now 5 th in line for processing and has a probability of being processed of P 7 is the probability that the part that is seventh in line for processing will eventually be processed. Calculate P 7. A. Less than B. At least 0.060, but less than C. At least 0.065, but less than D. At least 0.070, but less than E. At least Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 9

11 7. D. I will use λ for the hazard rate of the Exponential Distribution of falling to the floor. (θ is usually the mean.) For the first part in line, there are two independent Exponential Distributions competing to operate on it first; will it be processed or will it fall to the floor. Prob[first part is processed] = (1/5) / (λ + 1/5) = 0.2 / (λ + 0.2). The second part has to wait for something to happen to the first part; something happens to the first part via an Exponential with rate: λ Thus the probability that the 2nd part makes it to the head of the line prior to falling to the floor is: λ (λ + 0.2) + λ = λ λ By the memoryless property, if the 2nd part makes it to the head of the line, then its probability of being processed is equal to that calculated previously for the first part: 0.2 / (λ + 0.2). Prob[2nd part is processed] = Prob[2nd part makes it to head of line] Prob[2nd part is processed makes it to head of line] = λ λ λ = 0.2 2λ In order for the third part to get into the 2nd spot in line, something has to happen to either the first or second part before the third part falls on the floor. The combined things happening to the first part (λ+ 0.2) and second part λ has a rate of: 2λ The rate at which the third part falls on the floor is λ. Thus the chance that the 3rd part makes it into the second spot in line is: 2λ λ Based on the memoryless property, if the third part makes it into the second spot in line, then its 0.2 chance of being processed is the previously calculated: 2λ Thus the chance of the third part being processed is: Prob[3rd part makes it to 2nd spot in line] Prob[processed in 2nd spot in line] = 2λ λ λ = 0.2 3λ The pattern is: Prob[part n is processed] = 0.2 nλ We are given: 0.1 = Prob[5th part is processed] = Prob[7th part is processed] = Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page λ λ = λ = 0.2 (7)(0.36) = Comment: Difficult! This is an example of continuous time Markov Chain.

12 8. You are given the following information: Engineers are comparing two systems. System U is a 4-out-of-5 system consisting of 5 identical components. For each component of System U, its lifetime follows the exponential distribution with a hazard fate of 0.5. System V is a series system consisting of 10 identical components. For each component of System V, its lifetime follows the exponential distribution with a hazard rate of 1.0. T U and T V are the lifetimes of Systems U and V, respectively. Calculate the results for the expression: Min[E[T U ], E[T V ]]. A. Less than 0.05 B. At least 0.05, but less than 0.15 C. At least 0.15, but less than 0.25 D. At least 0.25, but less than 0.35 E. At least 0.35 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page B. V is the minimum of 10 Exponentials each with mean of 1. E[T V ] = 1/10 = 0.1. U fails when the second component fails. Thus T U is the second order statistic of a sample of size 5 from an Exponential Distribution with mean 2. E[T U ] = (2)(1/5 + 1/4) = 0.9. Min[E[T U ], E[T V ]] = Min[0.9, 0.1] = 0.1. Comment: One could get this question correct even if one could only roughly estimate E[T U ]. A 4-out-of-5 system functions when at least 4 of the components operate; thus it fails when the second component fails.

13 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page You are given a system whose structure is represented in the diagram below: The structure has identical components, with probabilities of working all equal to 0.6. u is the upper bound reliability of the system by using the first two inclusion-exclusion bounds, defining the events in terms of minimal path sets. Calculate u. A. Less than B. At least 0.150, but less than C. At least 0.250, but less than D. At least 0.350, but less than E. At least 0.450

14 9. C. (See Comment) The minimal path sets are: {1, 2, 3, 6, 7}, {1, 2, 3, 6, 8}, {1, 4, 5, 6, 7}, {1, 4, 5, 6, 8}. Let A i be the minimal path sets for a system, and E i = {all components of A i functioning}. The probabilities that each minimal path set works is: = n Thus using r(p) P(E i ), the upper bound is: (4)( ) = , choice C. i=1 A 1 A 2 = {1, 2, 3, 6, 7, 8}. Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 13 P(E 1 E 2 ) = Prob[all components work in the union of minimal path set 1 and 2] = = A 1 A 3 = {1, 2, 3, 4, 5, 6, 7}. P(E 1 E 3 ) = = A 1 A 4 = {1, 2, 3, 4, 5, 6, 7, 8}. P(E 1 E 4 ) = = A 2 A 3 = {1, 2, 3, 4, 5, 6, 7, 8}. P(E 2 E 3 ) = = A 2 A 4 = {1, 2, 3, 4, 5, 6, 8}. P(E 2 E 4 ) = = A 3 A 4 = {1, 4, 5, 6, 7, 8}. P(E 3 E 4 ) = = A 1 A 2 A 3 = {1, 2, 3, 4, 5, 6, 7, 8}. P(E 1 E 2 E 3 ) = = A 1 A 2 A 4 = {1, 2, 3, 4, 5, 6, 7, 8}. P(E 1 E 2 E 4 ) = = A 1 A 3 A 4 = {1, 2, 3, 4, 5, 6, 7, 8}. P(E 1 E 3 E 4 ) = = A 2 A 3 A 4 = {1, 2, 3, 4, 5, 6, 7, 8}. P(E 2 E 3 E 4 ) = = n n n Thus using r(p) P(E i ) - P(E i Ej) + P(E i EjE k ), the upper bound on reliability is: i=1 i=1 j>i i=1 j>i k>j (4)( ) - {(2)( ) + (2)( ) + (2)( )} + (4)( ) = This is answer choice B. Comment: The CAS only allowed answer choice C, although I think B should also be acceptable. The problem was poorly worded; I think of these two upper bounds as the first or first three rather than the first two inclusion-exclusion bounds. These can also be thought as the first and second upper bounds gotten via inclusion-exclusion. (Apparently they were trying to test whether you knew that the first set of terms gives an upper bound while using the first two sets of terms would give a lower bound. However, their wording confused many people who knew the material.) We have a series of four subsystems. The second subsystem consists of components 2 through 5. The chance of 2 and 3 both working is: = The chance of 4 and 5 both working is: = 0.36.

15 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 14 Thus the chance of second subsystem failing is: (1-0.36)(1-0.36) = Thus the chance of second subsystem working is: The fourth subsystem consists of components 7 and 8 in parallel. Thus the chance of the fourth subsystem working is: 1 - (1-0.6)(1-0.6) = Thus the exact answer is: (0.6) ( ) (0.6) ( ) = Note that since there are only four minimal path sets, the exact answer is: n n n n r(p) = P(E i ) - P(E i Ej) + P(E i EjE k ) - P(E i EjE k E l ) = i=1 i=1 j>i i=1 j>i k>j i=1j>ik>j l>k (4)( ) - {(2)( ) + (2)( ) + (2)( )} + (4)( ) = n n r(p) P(E i ) - P(E i Ej). Thus a lower bound on reliability is: i=1 i=1 j>i (4)( ) - {(2)( ) + (2)( ) + (2)( )} =

16 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page You are given: A basic series system has 3 components. The series system can be simplified to 2 components. All components are independent and function for an amount of time that is uniformly distributed over (0, 1). Calculate the increase in expected system life by simplifying from 3 components to 2 components. A. Less than 0.03 B. At least 0.03, but less than 0.05 C. At least 0.05, but less than 0.07 D. At least 0.07, but less than 0.09 E. At least D. The expected value of a minimum of three uniform distributions on (0, 1) is: 1/(3+1) = 1/4. The expected value of a minimum of two uniform distributions on (0, 1) is: 1/(2+1) = 1/3. 1/3-1/4 = 1/12 = Comment: See Order Statistics in Mahlerʼs Guide to Statistics.

17 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page You are given: Loan borrowers move among three states at the end of each month: State 1: Current State 2: Delinquent State 3: Default 75% of Current borrowers remain Current; 25% move to Delinquent. 60% of Delinquent borrowers remain Delinquent; 20% move to Current; 20% move to Default. Once in Default, the loan is written off so that the borrower remains in Default forever. Calculate the probability that a Current borrower at the beginning of a calendar quarter will not be in Default by the end of the same quarter. A. Less than B. At least 0.200, but less than C. At least 0.400, but less than D. At least 0.600, but less than E. At least E. The transition matrix is: P = The chain starts in State 1, Current. (1, 0, 0) P = (0.75, 0.25, 0). (0.75, 0.25, 0) P = (0.6125, , 0.05). (0.6125, , 0.05) P = ( , , ). Prob[not in default state 3] = = =

18 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page You are given the following information about a homogeneous Markov chain: There are three daily wildfire risk states: Green (state 0), Yellow (state 1) and Red (state 2). Transition between states occurs at the end of each day m n The daily transition matrix, P = The wildfire risk was Yellow on Wednesday. Today is Thursday and the wildfire risk is Green. The probability that the wildfire risk will be Red on Saturday is Calculate the absolute difference between m and n. A. Less than B. At least 0.065, but less than C. At least 0.075, but less than D. At least 0.085, but less than E. At least C. Since each row of P adds to one: m + n = 1. n = m. Green 0.82 m m P = Yellow Red We start in Green, state 0. (1, 0, 0) P = (0.82, m, m). (0.82, m, m) P = ( m, m, m). Set 0.07 = m. m = n = = m - n = = Comment: Since a Markov Chain is memoryless, the information given about the state on Wednesday does not affect the answer.

19 13. You are given the following information about a Markov chain with a transition probability matrix: P = The three states are 0, 1, and 2. Calculate the long run proportion of time in state 2. A. Less than B. At least 0.680, but less than C. At least 0.700, but less than D. At least 0.720, but less than E. At least A. Going down the columns of P, the balance equations are: π 0 = π π π π 1 = π π π 2 0. π 0 = 3 π 1. π 2 = π π π Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 18 From the first equation: π 2 = 4 π 0-4 π 1 = 8 π 1. Also, π 0 + π 1 + π 2 = 1. 3 π 1 + π π 1 = 1. π 1 = 1/12. π 2 = 8/12 = π 0 = 3/12.

20 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page An insurance company is considering either charging a cancellation fee for customers who cancel their policies in the first three years, or charging an additional fixed premium of 10 in each of the first three years for customers that renew to cover the cost of writing a new policy. Customers only cancel their policy at the end of a year. The cancellation fee is paid to the company upon cancellation. The remaining customers pay the additional fixed premium of 10 at the end of each year for the first three years their policy is in effect. The survival distribution of a policy is uniform between 0 and 5 years. Annual interest rate: i = 5%. Calculate the cancellation fee required so that the expected present value is equivalent to the expected present value of the additional fixed premium of 10 for new customers that renew for at least three years. A. Less than 20 B. At least 20, but less than 25 C. At least 25, but less than 30 D. At least 30, but less than 35 E. At least D. (See Comment) Let the unknown cancelation fee be x. Since the distribution is uniform from 0 to 5, the chance that there is a fee paid at the end each of the first three years is: 1/5 = 0.2. Then the actuarial present value of the cancellation fees is: x 0.2/ x 0.2/ x 0.2/ = x. The probability that the policy will continue in effect at the end of year one is 8/10, at the end of year 2 is 6/10, and at the end of year three is 4/10. Thus the actuarial present value of the fixed payments is: 10 {0.8/ / / } = Setting the two actuarial present values equal: x = x = Comment: The CAS declared this a defective question; the last sentence was confusing. The final sentence should have read Calculate the cancellation fee required so that the actuarial present value of cancellation fees is equal to the actuarial present value of the additional fixed premiums for new customers. For property/casualty insurance It would be impractical to try to collect a cancellation fee at the end of a year of coverage.

21 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page You are given the following information: There are two independent lives (40) and (60). Mortality is assumed to follow the Illustrative Life Table. Calculate the probability that life (40) lives to age (55) and life (60) dies before age (70). A. Less than B. At least 0.170, but less than C. At least 0.190, but less than D. At least 0.210, but less than E. At least B. 15 p 40 = l 55 / l 40 = 8,640,861 / 9,313,166 = q 60 = 1 - l 70 / l 60 = 1-6,616,155 / 8,188,074 = p q 60 = (0.9288)(0.1920) =

22 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page Buses arrive at a bus stop according to a Poisson process with parameter λ. Starting from time t = 0, you observe bus inter arrival times of t = 7, 3, 5, 3, 2. Calculate the maximum likelihood estimate of λ. A. Less than 0.1 B. At least 0.1, but less than 0.2 C. At least 0.2, but less than 0.3 D. At least 0.3, but less than 0.4 E. At least C. The waiting times are Exponential. Maximum likelihood is equal to the method of moments. Thus, θ^ = X = 20/5 = 4. λ^ = 1/θ^ = 0.25.

23 17. Let y 1, y 2,..., y 10 be a random sample from a population with mean, µ, and variance σ 2. Consider the following two unbiased estimators for µ: µ^ 1 = (y 1 + y 2 )/2. µ^ 2 = 1 3 y (y y 9 ) y 10. Calculate the efficiency of µ^ 1 relative to µ^ 2. A. Less than 0.15 B. At least 0.15, but less than 0.35 C. At least 0.35, but less than 0.55 D. At least 0.55, but less than 0.75 E. At least 0.75 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page C. Var[µ^ 1] = (1/2) 2 σ 2 +(1/2) 2 σ 2 = σ 2 /2. Var[µ^ 2] = (1/3) 2 σ 2 + (8)(1/24 2 )σ 2 + (1/3) 2 σ 2 = σ 2. The efficiency of µ^ 1 relative to µ^ 2 is: Var[µ^ 2] / Var[µ^ 1] = / 0.5 = Comment: µ^ 2 with the smaller variance is more efficient than µ^ 1. For two unbiased estimators of the same quantity, define the relative efficiency of estimator B compared to estimator A as: Variance[A] / Variance[B]. Any estimator that is a weighted average of the y i, with the weights adding to one, would be an unbiased estimator of µ. X would be the most efficient of this class of estimators.

24 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page Let y 1, y 2,..., y n be a random sample from an exponential distribution with density: f(y) = λ e -λy, for y > 0 Determine the Rao-Cramer lower bound for the variance of any unbiased estimator of λ. A. λ/n B. λ/n 2 C. λ 2 /n D. (λ/n) 2 E. nλ C. ln[f(y)] = lnλ - yλ. ln[f(y)] λ = 1/λ - y. 2 ln[f(y)] λ 2 = -1/λ 2. E[ 2 ln[f(y)] λ 2 ] = -1/λ 2. Thus, the Rao-Cramer lower bound is: -1 (-1/ λ 2 ) n = λ2 /n. Comment: The Rao-Cramer lower bound goes down as 1/n, eliminating choices B, D, and E.

25 19. Let 150, 204, 310, 480, 500 be a random sample from the density function given by: y α - 1 exp[-y / θ] f(y α,θ) = Γ(α) θ α, for y > 0 0, elsewhere α = 20 The coefficient of variation (CV) is the standard deviation divided by the mean. The mean of θ is estimated using maximum likelihood. Calculate the CV of the maximum likelihood estimate. A. Less than 0.05 B. At least 0.05, but less than 0.15 C. At least 0.15, but less than 0.25 D. At least 0.25, but less than 0.35 E. At least B. (See Comment) The given density is that of a Gamma Distribution. For the Gamma Distribution with α fixed, maximum likelihood is equal to the method of moments. θ^ = X /α = 328.8/20 = Var[θ^ ] = Var[X ] / 20 2 = (Var[X]/n) / 400 = {(αθ 2 )/5} / 400 = (20)( ) / 2000 = The CV of the maximum likelihood estimate of θ is: 2.70 / = Alternately, ln[f(y)] = (α-1)ln[y] - y/θ - ln[γ(α)] - α ln[θ]. ln[f(y)] θ = y/θ 2 - α/θ. Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 24 2 ln[f(y)] θ 2 = -2y/θ 3 + α/θ 2. E[ 2 ln[f(y)] θ 2 ] = -2E[Y]/θ 3 + α/θ 2 = -2(αθ) / θ 3 + α/θ 2 = -α/θ 2. Thus, Var[θ^ ] = Rao-Cramer lower bound = -1 (-α / θ 2 ) n = θ2 / (αn) = / {(20)(5)} = Proceed as before. Comment: The CAS also allowed choice C. The mean of θ is estimated should have read θ is estimated. Thus some people computed the CV of the fitted Gamma Distribution. E[X] = αθ = 20θ. Var[X] = αθ 2 = 20θ 2. StdDev[X] = 20 θ. Cv of Gamma Distribution = StdDev[X] /E[X] = 20 θ / (20θ) = 1/ 20 = 0.224, choice C.

26 20. You are given: The current manufacturing process for bullets has a mean exit velocity of 2900 MPH and a standard deviation of 200 MPH. A new process for manufacturing bullets was developed. 100 bullets using the new process were tested and had a mean exit velocity of 2950 MPH. H 0 : Mean exit velocity = 2900 MPH H 1 : Mean exit velocity > 2900 MPH The standard deviation is the same as the existing manufacturing process. Exit velocities are assumed to follow the normal distribution. Calculate the minimum significance level at which you would reject H 0. A. Less than B. At least 0.010, but less than C. At least 0.025, but less than D. At least 0.050, but less than E. At least Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page A. For this one-sided test with a known variance, the p-value is: Prob[X 2950 H 0 ] = 1 - Φ[ ] = 1 - Φ[2.5] = = 0.62%. 200 / 100 Comment: The variance of X is: 200 / 100 = 20.

27 21. You are given: x 1, x 2,..., x 10 is a random sample from a Bernoulli distribution with parameter p = Pr(X= 1) H 0 : p = 0.5 H 1 : p > The critical region, C = { x j 7}. j=1 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 26 Calculate the probability of a Type I error. A. Less than 0.05 B. At least 0.05, but less than 0.10 C. At least 0.10, but less than 0.15 D. At least 0.15, but less than 0.20 E. At least D. The sum of 10 independent Bernoullis follows a Binomial Distribution with m = Prob[Type I error] = Prob[reject when H 0 is true] = Prob[ x j 7 p = 1/2] = j= = 10 ( ) ( ) = 17.19%.

28 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page Two independent random samples of size 12 and 8 are drawn from two normal populations, X and Y. Let S X 2 and SY 2 be the unbiased sample variances. Using the test statistic W = S X 2 / SY 2 you wish to test: H 0 : the variance of population Y is one-third the variance of population X. H 1 : the variance of population Y is less than one-third the variance of population X. Calculate the critical value for a one-sided test of significance A. Less than 1.20 B. At least 1.20, but less than 1.40 C. At least 1.40, but less than 1.60 D. At least 1.60, but less than 1.80 E. At least E. H 0 : σ Y 2 = σ X 2 /3. H 1 : σ Y 2 < σ X 2 /3. The F-statistic is: (S X 2 /3) / S Y 2, with 12-1 = 11 and 8-1 = 7 degrees of freedom. Looking in the F-Table, the 2% critical value is We reject when (S X 2 /3) / S Y 2 > W = S X 2 / SY 2 > Comment: The answer is way outside the range of the answer choices! Assume X has sample size n and Y has sample size m. Then, in order to test H 0 : σ X 2 = kσ Y 2 versus H 1 : σ X 2 > kσ Y 2. The test statistic is: S X 2 / (ks Y 2 ), which has an F-Distribution with n -1 and m - 1 degrees of freedom. As always we put in the numerator the larger quantity assuming H 1 is true.

29 23. You are given the mid-term and final exam scores of 10 students, and you would like to test whether the mid-term and final scores have the same distribution using the Wilcoxon Signed-Rank Test for matched pairs. Student Mid-Term Score Final Score Calculate the test statistic. A. Less than 10 B. At least 10, but less than 15 C. At least 15, but less than 20 D. At least 20, but less than 25 E. At least 25 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page D. We take the absolute differences, noting the sign of the differences. Student Mid-Term Score Final Score Sign Absolute Difference Rank We rank these absolute differences. The sum of the ranks for the positive differences is: = The sum of the ranks for the negative differences is: = Check: = 55 = (10)(11)/2. For this two-sided test, the test statistic is: Min[31.5, 23.5] = Comment: I prefer questions that ask one to perform the test rather than calculate the test statistic. Looking in the table attached to the exam, for n = 10 the two-sided 10% critical value is 10. Since 23.5 > 10, we do not reject the null hypothesis at 10%.

30 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page You are given the following information: Wilcoxon's rank test and Mann-Whitney U test are both used for comparing two independent random samples from two populations: Population I and Population II. Both random samples have a sample size of 10. For the sample from Population I, the Wilcoxon's rank-sum statistic equals Calculate the Mann-Whitney U statistic for the sample from Population I. A. Less than 60 B. At least 60, but less than 90 C. At least 90, but less than 120 D. At least 120, but less than 150 E. At least B. (See Comment) Mann-Whitney U Statistic: U = n 1 n 2 + n 1 (n 1 +1)/2 - (sum of the ranks of the first sample of size n 1 ) = (10)(10) + (10)(10+1)/ = Comment: The CAS also allowed answer A. The sum of the ranks for Population II is: (10+10)( )/ = If one uses the sum of the ranks for Population II, then: U = n 2 n 1 + n 2 (n 2 +1)/2 - (sum of the ranks of the second sample of size n 2 ) = (10)(10) + (10)(10+1)/ = 30.5, which corresponds to choice A.

31 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page You are given the following information about two samples and the test parameters: Sample X is of size 5 Sample Y is of size 9 The rank sum of Sample Y is 84 H 0 : The two samples are from the same distribution You are to do a two-sided test of H 0 The Mann-Whitney-U procedure without a normal approximation is used Calculate the p-value for this test. A. Less than B. At least 0.005, but less than C. At least 0.010, but less than D. At least 0.015, but less than E. At least E. Treating Y as the first sample, Mann-Whitney U Statistic: U = n 1 n 2 + n 1 (n 1 +1)/2 - (sum of the ranks of the first sample of size n 1 ) = (9)(5) + (9)(9+1)/2-84 = 6. For samples of size 9 and 5, Prob[U 6] = For a two-sided test, the p-value is: (2)(0.014) = 2.8%. Alternately, we can treat X as the first sample. The sum of the ranks for X is: (5+9)(5+9+1)/2-84 = 21. Mann-Whitney U Statistic: U = n 1 n 2 + n 1 (n 1 +1)/2 - (sum of the ranks of the first sample of size n 1 ) = (5)(9) + (5)(5+1)/2-21 = 39. For samples of size 9 and 5, Prob[U 39] = Prob[U (9)(5) - 39 = 6] = For a two-sided test, the p-value is: (2)(0.014) = 2.8%.

32 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page You are given the following information: Y has a binomial distribution with parameters m and q. The prior distribution of λ is beta with parameters a = 5, b = 5, and θ = 1. A sample of size 26 is drawn. The number of successes in the sample was 19. Calculate the posterior mean of q. A. Less than 0.50 B. At least 0.50, but less than 0.57 C. At least 0.57, but less than 0.64 D. At least 0.64, but less than 0.71 E. At least D. Assume that when they said λ they meant q. The term successes is used with a Bernoulli. Therefore, I will also assume that they meant Y was Bernoulli, m = 1; since they do not give us m, one can not do the question unless you assume a value of m. Then the posterior distribution is also Beta. Add the number of successes to a: aʼ = = 24. Add the number of failures to b: bʼ = = 12. Posterior mean is: aʼ / (aʼ + bʼ) = 24/( ) = 2/3. Comment: The CAS did not declare this a defective question. For a Beta-Binomial, the posterior beta has parameters: aʼ = a + number of claims, bʼ = b + m (number of years) - number of claims.

33 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page For a portfolio of homeowner's insurance contracts, q represents the proportion of policies for which there were claims made within one year where q is a random variable with a beta prior distribution with parameters a, b and θ = 1. Prior to reviewing the results from a random sample of policies described below, it has been estimated that the mean and variance of q are 2 3 and 1 72, respectively. In a given year, for 150 randomly selected policies from the portfolio, there were 70 claims. Calculate the posterior mean of q. A. Less than 0.45 B. At least 0.45, but less than 0.50 C. At least 0.50, but less than 0.55 D. At least 0.55, but less than 0.60 E. At least 0.60

34 27. B. We have a Beta-Binomial situation with m = 1. (See Comment.) For the Beta Distribution with θ = 1: mean = a/(a+b), and second moment = a(a +1) (a+ b)(a + b+ 1). Setting these equal to the given values for the prior Beta Distribution: 2 = a/(a+b). 2a + 2b = 3a. a = 2b ( 2 3 )2 = 33/72 = Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 33 a(a +1). 33 (a+b)(a+b+1) = 72 a (a+1). (a+ b)(a + b+ 1) 33 (3b)(3b+1) = 72 (2b) (2b+1). 297b b = 288b b. b = 45/9 = 5. a = (2)(5) = 10. The posterior Beta has: aʼ = = 80, and bʼ = 5 + (150-70) = 85. The posterior mean of q is: aʼ / (aʼ + bʼ) = 80 / ( ) = Alternately, K = (a + b)/m = (10 + 5) / 1 = 15. Z = N / (N+K) = 150/(150+15) = 10/11. The prior mean of q is given as 2/3. Thus the posterior mean of q is: (10/11)(70/150) + (1-10/11)(2/3) = Comment: We have solved for the parameters of the prior Beta Distribution, a and b, using the Method of Moments. The next to last sentence in this exam question should have read: In a given year, for 150 randomly selected policies from the portfolio, there were 70 which had claims. The derivation of the conjugate prior situation assumes that we observe either one policy or several policies of the same type, in other words with the same q. If we observe several policies picked at random, then the mathematical result does not hold; the posterior distribution is not another Beta. However, one can still use Buhlmann Credibility; for the conjugate prior setup Bayes Analysis and Buhlmann Credibility give the same answer for the estimate of the posterior mean of q. The CAS did not declare this a defective question.

35 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page You are given: A random sample is drawn from a Poisson distribution with parameter λ, where the prior density of λ is proportional to the following function: g(λ) = λ 2 e 2λ, for λ > 0 The sample size is n = 98. The sum of the sampled values equals 100. Calculate the mean of the posterior distribution of λ. A. Less than 0.98 B. At least 0.98, but less than 1.00 C. At least 1.00, but less than 1.02 D. At least 1.02, but less than 1.04 E. At least D. This is a Gamma-Poisson with: α - 1 = 2, and 1/θ = 2. αʼ = α + C = = /θʼ = 1/θ + E = = 100. Posterior mean is: αʼ θʼ = 103/100 = 1.03.

36 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page You are given the following output from a GLM to estimate an insured's pure premium: Response variable pure premium Response distribution gamma Link log Dispersion parameter (φ) 1 Parameter Df β_^ Intercept Risk Group 2 Group Group Group Vehicle Symbol Group 1 Group Group Calculate the variance of the pure premium for an insured in Risk Group 3 with Vehicle Symbol Group 2. A. Less than 3,000 B. At least 3,000, but less than 4,000 C. At least 4,000, but less than 5,000 D. At least 5,000, but less than 6,000 E. At least 6, B. The fitted mean pure premium for an insured in Risk Group 3 with Vehicle Symbol Group 2 is: exp[ ] = For the Gamma Distribution, the variance is proportional to the square of the mean. Variance is: φ µ 2 = (1) ( ) = Comment: For the Gamma Distribution, the dispersion parameter is: φ = 1/α. For the Gamma Distribution, the variance is: αθ 2 = (αθ) 2 /α = µ 2 /α = φ µ 2. Neither the syllabus reading by Dobson and Barnett nor the CAS study note by Larsen mention dispersion parameters. For a (linear) exponential family: Var[Y] = φ µ p.

37 30. You are given the following information for a fitted GLM: Response variable Y Response distribution Poisson Link log AIC Parameter Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 36 β_^ s.e.(β_^ ) Intercept Gender Male Female Age Calculate the predicted value of Y for a Female with an Age value of 22. A. Less than 1,000 B. At least 1,000, but less than 1,500 C. At least 1,500, but less than 2,000 D. At least 2,000, but less than 2,500 E. At least 2, B. exp[ (22)(0.107)] = 1364.

38 31. If Y has a distribution from the exponential family, and f(y) is in canonical form f(y) = exp[a(y)b(θ) + c(θ) + d(y)], then: I. E[a(y)] = - c'(θ) b'(θ) Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 37 II. Var[a(y)] = b''(θ)c'(θ) - c''(θ)b'(θ) [b'(θ)] 3 III. E[(y 2 )] = - c''(θ) b''(θ) Determine which of the above equations are correct. A. None are correct B. I and II C. I and Ill D. II and III E. The answer is not given by (A), (B), (C), or (D) 31. B. I is true. See equation 3.9 in Dobson and Barnett. II is true. See equation 3.11 in Dobson and Barnett. III is not true. Comment: Since f(y) is in canonical form, a(y) = y. One could derive from the first two equations: E[a(y) 2 ] = Var[a(y)] + E[a(y)] 2 = b''(θ)c'(θ) + b'(θ) c'(θ) 2 - c''(θ)b'(θ) [b'(θ)] 3. b''(θ)c'(θ) - c''(θ)b'(θ) [b'(θ)] 3 + c'(θ)2 b'(θ) 2 =

39 32. You are given: A generalized linear model includes three explanatory variables: X 1, X 2, and X 3 R 2 (j) is the coefficient of determination obtained from regressing the jth explanatory variable against all other explanatory variables. R 2 (1) = 0.05, R2 (2) = 0.83, and R2 (3) = 0.47 The threshold of the Variance Inflation Factor for variable j (VIF j ) for determining excessive collinearity is VIF j > 5 Determine which of the 3 variables under consideration will exceed the threshold established above. A. X 1 only B. X 2 only C. X 3 only D. X 1, X 2 and X 3 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 38 E. The answer is not given by (A), (B), (C), or (D) 32. B. VIF i = 1/(1 - R i 2 ). VIF 1 = 1/(1-0.05) = VIF 2 = 1/(1-0.83) = 5.88 > 5. VIF 3 = 1/(1-0.47) = Comment: If we regress the i th independent variable against all of the other independent variables, then the Variance Inflation Factor is: VIF i = 1/(1 - R i 2 ) 1. If one or more of the VIFs is large, that is an indication of multicollinearity. Note that the closer R i 2 is to one, the larger is VIF i. VIF i > 5. 1/(1 - R i 2 ) > R i 2 < 0.2. R i 2 > 0.8.

40 33. You are given the following table of eight individual adults' heights in inches, y ij, categorized by their season of birth, and selected summary statistics: y = 65 Observation Summer/Fall Winter/Spring Season of birth is either Summer/Fall (i= 1) or Winter/Spring (i=2). Observation is indexed by j. 2 8 y 2 ij = 33,878 i=2 j=1 2 8 ( y ij ) 2 = 270,400 i=2 j=1 8 (y 1, j - y 1 ) 2 = j=1 8 (y 2, j - y 2 ) 2 = j=1 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 39 Calculate the mean square for error of the observations. A. Less than 7.5 B. At least 7.5, but less than 8.5 C. At least 8.5, but less than 9.5 D. At least 9.5, but less than 10.5 E. At least 10.5

41 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page C. (See Comment) With only two categories, the error (residual) sum of squares is: 8 (y 1, j - y 1 ) (y 2, j - y 2 ) 2 = = j=1 j=1 The total degrees of freedom (as per Wackerly, Mendenhall, and Scheaffer) is: 8-1 = 7. The treatment or model degrees of freedom is 1. Thus the error (residual) degrees of freedom is: 7-1 = 6. The mean square for error (residual) is: 53.5 / 6 = Comment: The CAS also allowed choice E, without any explanation. The textbooks do not use the phrase mean square for error of the observations. What Wackerly, Mendenhall, and Scheaffer call the error sum of squares is what Dobson and Barnett calls the residual sum of squares. The lack of context made this question more difficult for some people. y = 270,400 /8 = 520/8 = 65. In Wackerly, Mendenhall, and Scheaffer: total sum of squares = (y ij - y ) 2 = y 2 ij - n y 2 = 33,878 - (8)(65 2 ) = 78. i j i j Thus the between treatment sum of squares is: = An ANOVA table as per Wackerly, Mendenhall, and Scheaffer: Source Degrees of freedom Sum of Squares Mean Square F Treatment Error Total 8-1 = This is the type of ANOVA table that would be produced by software such as Mathematica. In Dobson and Barnett, total sum of squares = y 2 ij = 33,878. i j In Dobson and Barnett, the sum of squares due to the mean is: n y 2 = (8)(65 2 ) = 33,800. An ANOVA table as per Dobson and Barnett: Source Degrees of freedom Sum of Squares Mean Square F Mean 1 33,800 Treatment Residual Total 8 33,878 The F-Statistic of 2.75 has 1 and 6 degrees of freedom. In order to test the hypothesis that the null hypothesis that means by season are the same versus the alternative that they are different, using a computer the p-value is 14.8%.

42 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 41 With only two categories, this is equivalent to a t-test. The sample mean of {70, 67, 64, 66} is: The sample variance of {70, 67, 64, 66} is: The sample mean of {60, 62, 68, 63} is: The sample variance of {60, 62, 68, 63} is: The pooled sample variance is: {(4-1)(6.25) + (4-1)( )} / {(4-1) + (4-1)} = The t-statistic is: / / 4 = For a two-sided test with = 6 degrees of freedom, using a computer the p-value is 14.8%. (y 1, j - y 1 ) 2 = ( ) 2 + ( ) 2 + ( ) 2 + ( ) 2 = 18.75, as given. j (y 2,j - y 1 ) 2 = ( ) 2 + ( ) 2 + ( ) 2 + ( ) 2 = 34.75, as given. j I think these sums should have gone from j = 1 to 4, rather than j = 1 to 8 as in the exam question. Also rather than zeros in the table, there should have been blanks.

43 34. You are given the following information for a one-way ANOVA: The ANOVA F-statistic for a sample is The mean square for treatments is with 2 degrees of freedom. The total sum of squares is n is the number of observations in the sample. Calculate n. A. Less than 14 B. At least 14, but less than 17 C. At least 17, but less than 20 D. At least 20, but less than 23 E. At least A. Between treatment sum of squares = (2)(606.69) = = F = Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 42 mean between treatment sum of squares mean error (residual) sum of squares = mean error (residual) sum of squares. mean error (residual) sum of squares = /3.75 = The error (residual) sum of squares has n = n - 3 degrees of freedom. The error (residual) sum of squares = (n - 3) ( ) = Total sum of squares = (between treatment sum of squares) + (error or residual sum of squares) = (n - 3) ( ). n - 3 = ( ) / = n = 12. Comment: I do not like backward questions like this one. What is called the treatment sum of squares in Wackerly, Mendenhall, and Scheaffer is called the between treatment sum of squares in Dobson and Barnett; other books call it the between sum of squares. What is called the error sum of squares in Wackerly, Mendenhall, and Scheaffer is called the residual sum of squares in Dobson and Barnett. Note also that in Dobson and Barnett what is called the total sum of squares also includes the sum of squares due to the mean, while in Wackerly, Mendenhall, and Scheaffer it does not. In Wackerly, Mendenhall, and Scheaffer, total sum of squares = (y ij - y ) 2, i j while in Dobson and Barnett, total sum of squares = y 2 ij. i j A filled in ANOVA table for this situation as per Wackerly, Mendenhall, and Scheaffer: Source Degrees of freedom Sum of Squares Mean Square F Treatment Error Total 12-1 =

44 35. You have fit a generalized linear model using a Poisson distribution. One observation of the response variable is 59. The corresponding GLM fitted value is 71. The corresponding leverage of this point is 22%. Calculate the standardized deviance residual corresponding to this observation. A. Less than -1.8 B. At least -1.8, but less than -1.7 C. At least -1.7, but less than -1.6 D. At least -1.6, but less than -1.5 E. At least C. For the Poisson Distribution, the form of the deviance is: n D = 2 {y i ln[y i / λ^ i ] - (y i - λ^ i ) }. i=1 d i 2 = 2 {y i ln[y i / λ^ i ] - (y i - λ^ i )} = 2 {(59) ln[59/71] - (59-71)} = We take the sign of the deviance residual d i as the same as that of the (ordinary) residual y i - µ^ i = < 0. Thus in this case, d i = = The standardized deviance residual is the deviance residual divided by 1 - h ii : / = Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page 43 Comment: The i th leverage is h ii, the i th diagonal element of the Hat Matrix H.

45 Solutions Spring 2017 CAS Exam S, HCM 10/15/17, Page You are given the following information for a model fitted using ordinary least squares (OLS): Response variable Rating Response distribution normal Link identity MSE Parameter df β^ standard error (β^ ) Intercept Complaints Summary Statistic Rating Complaints N Min Max Sum Median Sample Std. Dev Calculate the upper bound of the 95% prediction interval for Rating, for an observation with a Complaints value of 50. A. Less than 50 B. At least 50, but less than 60 C. At least 60, but less than 70 D. At least 70, but less than 80 E. At least 80