Practice Exam #1 CAS Exam 3L

Practice Exam #1 CAS Exam 3L These practice exams should be used during the last month prior to your exam. This practice exam contains 25 questions, of equal value, corresponding to an exam of about 2.5 hours. Prior to the start of this exam, give yourself a 1 minute reading period in which you can silently read the questions. Writing is not permitted during this time and you are not permitted to hold pens or pencils. You are also not allowed to use calculators during the reading period. The mix of questions by topic is similar to what you can expect on your exam. Solutions to problems are at the end of each practice exam. prepared by Robert W. Batten, FSA and Howard C. Mahler, FCAS Copyright 212 by Howard C. Mahler. Robert Batten Rwbatten@aol.com Howard Mahler hmahler@mac.com www.howardmahler.com/teaching

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 2 CAS Exam 3L, Practice Exam #1 1. Given: (i) s(x) = Find ω. ω - x ω (ii) For a particular value of y, µ(y) =.3 and e o y = 25. (iii) 5 p 1 =.125. k, x ω, for some k >. (A) 8 (B) 9 (C) 1 (D) 11 (E) 12 2. Given: µ(x) = Evaluate e o - e o 1. 2 1 + x, x >. (A) -2 (B) -1 (C) (D) 1 (E) 2 3. Given: s(x) = 1 - x 2, x 5. 5 The probability that a 4-year old will die in the n years following age 41 is 14 75. Find n. (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 4. If the force of mortality is constant for all ages, find the probability that exactly one of two newborns will survive for at least their expected lifetime. (A) e -1 - e -2 (B) 2(e -1 - e -2 ) (C) 2e -1 - e -2 (D) 2e -1 (E) The answer depends on the value of the constant force of mortality. 5. A husband aged x and his wife aged y purchase a deferred temporary life annuity. Continuous annual payments of 1 will begin at the first death and will continue for ten years, or until death of the survivor, if sooner. Find the single benefit premium, if µ =.2 and δ =.8. (A) 1.5 (B) 1.25 (C) 1.75 (D) 2.1 (E) 2.5

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 3 6. The single benefit premium for a $1 whole life policy to (2), payable at the instant of death, is $.15. Given the following values, find n. µ(2 + t) = δ =.4., t n.6, t > n (A) 34. (B) 34.33 (C) 34.5 (D) 34.66 (E) 35. 7. A three-year temporary life annuity-due is issued to a woman aged 6. The scheduled payment amounts are 1, 3, and 5, respectively. If Y is the present value random variable for the annuity, find the variance of Y, if d =.1 and klq 6 =.5(k + 1), k =, 1, 2, 3. (A) 3.4 (B) 3.6 (C) 3.8 (D) 4. (E) 4.2 8. A $1 fully discrete whole life policy to (6) requires benefit premiums only for the first four policy years, of amounts 4k, 3k, 2k, and k. 1 If µ(x) =, x < 7, and i =.5, find the value of k. 7 - x (A) 88. (B) 88.5 (C) 89. (D) 89.5 (E) 9. 9. Find the 1 th terminal benefit reserve for a $1 fully discrete whole life policy issued to (3). Assume demoivre's Law with ω = 1 and i =.4. (A).6 (B).62 (C).64 (D).66 (E).68 1. A $1 1-year term insurance policy is issued to (5). An additional $2 will be paid if death is caused by an accident. Find the expected amount to be paid by the insurer if (i) the force of mortality for accidental death is µ (1) 5 (t) =.1t, t. (ii) the force of mortality for death due to all other causes is µ (2) 5 (t) =.4t, t. (A) 1285 (B) 1295 (C) 13 (D) 135 (E) 1315

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 4 11. In a double-decrement model, If l (T) 73 µ (1) x = 2 75 - x, x < 75. µ (2) x = 1 75 - x, x < 75. = 12, find d(2) 74. (A) 5 (B) 75 (C) 1 (D) 125 (E) 15 12. A tyrannosaur eats only scientists. It eats scientists at a Poisson rate per hour that varies by the time of day: From 12 midnight to 6 a.m. λ(t) =.2. From 6 a.m. to 9 a.m. λ(t) =.15. From 9 a.m. to 6 p.m. λ(t) =.1. From 6 p.m. to 12 midnight λ(t) =.2. What is the probability that tomorrow the tyrannosaur eats more scientists before noon than after noon? (A) 37% (B) 39% (C) 41% (D) 43% (E) 45% 13. The credit rating scores of private passenger automobile insureds are in one of four classes: A, B, C or D. Transitions between classes occur at the end of each six month period, with transition matrix:.9.1 A.1.8.1 B.1.8.1 C.1.9 D Currently an insurer writes 2 insureds rated A, 4 insureds rated B, 3 insureds rated C, and 1 insureds rated D. Assuming that the insurer continues to write all of these insureds, of the original 1, insureds what is the expected number of insureds who are rated C in two years? (A) Less than 265 (B) At least 265, but less than 27 (C) At least 27, but less than 275 (D) At least 275, but less than 28 (E) At least 28

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 5 14. For a large group of insured vehicles, collisions claims occur at a Poisson rate of 4 per year. Collisions claims fall into three categories: 6% of the claims are minor and cost $3. 3% of the claims are serious and cost $2. 1% of the claims are total and cost $8. Using the normal approximation, calculate the probability that the total claims cost in one year is more than $1,. (A) 1.4% (B) 1.5% (C) 1.6% (D) 1.7% (E) 1.8% 15. An insurance company offers two different three-year insurance policies to insureds who have not made a claim in the previous year: Policy #1 charges premium X at the start of each year, but also pays to the insured a dividend of $7 at the end of each year in which no claim is made. Policy #2 charges premium Y at the start of each year. You are also given: The probability that no claim is made in a year is 75% if no claim was made in the previous year. The probability that no claim is made in a year is 6% if at least one claim was made in the i = 6%. previous year. Premiums are calculated using the equivalence principle. Calculate the difference X - Y. (A) Less than $35 (B) At least $35, but less than $4 (C) At least $4, but less than $45 (D) At least $45, but less than $5 (E) At least $5

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 6 16. Use the following information: Stefan and Damon arrive at the bus terminal and each wait for their bus to leave. Stefan is going to Lakeside and Damon is going to Bellwood. Lakeside buses leave the terminal via a Poisson Process at a rate of 7 per hour. Bellwood buses leave the terminal via a Poisson Process at a rate of 1 per hour. Lakeside buses and Bellwood buses are independent. Determine the average time in minutes that Damon waits conditional on Stefanʼs bus leaving first. (A) Less than 8 (B) At least 8, but less than 9 (C) At least 9, but less than 1 (D) At least 1, but less than 11 (E) At least 11 17. You are given a random sample of size 1 from a Normal Distribution with σ = 8. You test H : µ = 75 versus H 1 : µ > 75. The critical region has the form X c. If the test has a significance level of 5%, what is the power of the text at µ = 85? (A) 95% (B) 96% (C) 97% (D) 98% (E) 99% 18. A random sample of five losses: 6, 8, 11, 2, 35. Losses are assumed to follow a Pareto distribution with parameters θ = 1 and α. Determine the maximum likelihood estimate of α. (A) Less than 1. (B) At least 1., but less than 1.1 (C) At least 1.1, but less than 1.2 (D) At least 1.2, but less than 1.3 (E) At least 1.3

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 7 19. You are given the following information: X 1,..., X 11, is a random sample, where X is normally distributed, the sample mean is 5, and the sample variance is 22. The mean and variance for the random sample is unknown. H : µ = 9. H 1 : µ < 9. Let T be the critical value for a test with significance level α =.5. Let t be the test statistic calculated from the sample. Calculate the test statistic, t, and the critical value, T, then find the range below that contains the absolute value of the difference of the test statistic, t, and the critical value, T, t - T. (A) Less than.6 (B) At least.6, but less than 1.2 (C) At least 1.2, but less than 1.8 (D) At least 1.8, but less than 2.4 (E) At least 2.4 2. You are modeling Y as a function of X, using a linear regression. You are given the following data points and summary statistics: X Y 13 27 18 23 29 17 36 11 4 4 X i = 96. X 2 i = 263. i=1 i=1 Calculate the residual at (36, 11). (A) Less than -1. (B) At least -1., but less than -.3 (C) At least -.3, but less than.3 (D) At least.3, but less than 1. (E) At least 1. 4 4 Y i = 78. Y 2 i = 1668. i=1 i=1 4 X i Y i = 1654. i=1

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 8 21. Two independent populations X and Y have density functions f(x) = λx λ 1 for < x < 1, and g(y) = µy µ 1 for < y < 1, respectively. Let X 1, X 2,..., X n and Y 1, Y 2,..., Y m be random samples from X and Y. What is the form of the critical region for the best (Neyman-Pearson) test of H : λ = 2, µ = 4 against the alternative H 1 : λ = 3, µ = 6? (A) (B) n m lnx i + 2 lny i c i=1 i =1 n m lnx i + 2 lny i c i=1 i =1 n m (C) 2 lnx i - lny i c i=1 i = 1 n m (D) 2 lnx i - lny i c i=1 i = 1 (E) None of A, B, C, or D 22. Two thousand insureds are assigned to one of four classes based on prior period claim experience, as follows: Class Number of Claims in Prior Period Number of Insureds in Class A No Claims 125 B One Claim 75 C Two Claims 215 D Three or More Claims 55 Let H be the hypothesis that claim frequency follows a Poisson distribution with mean.7. Using the Chi-Square test, which of the following are true? (A) Reject H at.5%. (B) Do not reject H at.5%. Reject H at 1.%. (C) Do not reject H at 1.%. Reject H at 2.5%. (D) Do not reject H at 2.5%. Reject H at 5.%. (E) Do not reject H at 5.%.

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 9 Use the following information for the next two questions: Claim sizes are uniformly distributed over the interval [, θ]. Let Y = X (7) be the 7 th order statistic from a sample of 8 claims. 23. Which of the following is the probability density function for Y? (A) 7y 6 /θ 7, < y < θ. (B) 8y 7 /θ 8, < y < θ. (C) 42y 5 /θ 6-42y 6 /θ 7, < y < θ. (D) 56y 6 /θ 7-56y 7 /θ 8, < y < θ. (E) None of the above. 24. An estimate of θ was obtained as θ^ = 2Y. Calculate the mean squared error of θ^. (A) Less than.3θ 2 (B) At least.3θ 2, but less than.35θ 2 (C) At least.35θ 2, but less than.4θ 2 (D) At least.4θ 2, but less than.45θ 2 (E) At least.45θ 2 25. One has observed the following distribution of insureds by number of claims: Number of Claims 1 2 3 4 5&+ All Number of Insureds 4 7 5 2 3 183 A Negative Binomial Distribution with r = 3 is fit via the Method of Moments. Which of the following is an approximate 9% confidence interval for β? (A) [.415,.467] (B) [.45,.477] (C) [.395,.487] (D) [.385,.497] (E) [.375,.57] END OF PRACTICE EXAM

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Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 11 Solutions: 1. D. Basic facts: Under Modified demoivre's Law (MDML), s(x) = ω - x ω k ; µ(x) = k ω - x ; eo x = ω - x k + 1 ; np x = ω - x - n k. ω - x We are given that for a particular value of y, µ(y) =.3 and e o y = 25. (.3) (25) = µ(y) e o y = 3/4 = k k + 1. k = 3. k ω - y ω - y k + 1 = ω - 5-1 Now,.125 = 5 p 1 = ω - 1 k k + 1. 3. ω = 11. 2. B. Key Fact: µ(x) = 1 x + 1. s(x) = 1 x + 1. If one multiplies the force of mortality by a constant, then the survival function is taken to that power. Therefore, since the force of mortality we are given is twice the above force of mortality, the survival 1 function is the above survival function squared: s(x) = (1 + x) 2. Let l x = 1 (1 + x) 2. e o x = l x + t dt = l x (x +1) 2 t = (x +1)2 (x+ t + 1) 2 dt = - x + t +1 t = = x + 1. Therefore, e o - e o 1 = 1-2 = -1. Comment: e o x is generally expected to be greater than e o x+1. Here, however, mortality improves with advancing age, and thus e o x increases as x does. The given survival function is for a Pareto Distribution with α = 2 and θ = 1. For a Pareto Distribution, µ(x) = α θ + x, and eo x = θ + x α - 1.

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 12 3. B. Assume for convenience that: l = 5 2 = 25. Then, l x = s(x) l = 25 - x 2 ; construct a crutch (a numerical representation of the survival model). x 4 41 42 43 l x 9 819 736 651 For example, l 4 = 25-4 2 = 9. We have 1 n q 4 = l 41 - l 41+ n l 4 = 819 - l 41+ n 9 = 14 75. l 41+n = 651. 41 + n = 43. n = 2. 4. B. Key Fact: Under constant force, n p x = e -nµ for all n, and e x = 1/µ. Accordingly, 1/µ p x = e -1. Thus the required probability is: 2 1/µ p x (1-1/µ p x ) = 2 e -1 (1 - e -1 ). 5. D. Key Formula: Under constant force of mortality, since A x = µ µ + δ, we have A xy = 2µ 2µ + δ. Key Concept: At the first death, the actuarial present value of the benefit would be: a if (x) is the survivor and : x:1 a y if (y) is the survivor. :1 But, under constant force, these two annuities are equal. Accordingly, the single benefit premium is: A x y a = x:1 2µ 2µ + δ 1 µ + δ {1 - e-1(µ+δ) } = 1 3 1.1 (1 - e-1 ) = 2.1.

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 13 6. D. Key Concept: Whole Life = Term Plus Deferred. A x = A 1 _ x:n + ne x A x+n. Letting x = 2,.15 = + ( n E 2 ) (.6) =.6v n (since n p 2 = 1)..25 = v n = e -.4n. Solving, n = 34.66. 7. A. Construct a table showing the probabilities for each of the three possible values of Y. Number of Payments Made y Pr[Y = y] 1 1 q 6 =.5 2 1 + 3v = 3.7 1lq 6 =.1 3 1 + 3v + 5v 2 = 7.75 2p 6 =.85 Appealing to the definitions of first and second moments: E[Y] = (1)(.5) + (3.7)(.1) + (7.75)(.85) = 7.75. E[Y 2 ] = (1) 2 (.5) + (3.7) 2 (.1) + (7.75) 2 (.85) = 52.472125. Thus VAR[Y] = E[Y 2 ] - E[Y] 2 = 52.472125-7.75 2 = 3.367. 8. D. Key Formula: A x = a ω - x ω - x. (DML) A 6 = a 1 1 =.77217. Since "benefit premiums", by definition, are equivalence principle premiums, we have: k {4 + 3v p 6 + 2v 2 2 p 6 + v 3 3 p 6 } = 1A 6 = 772.17. k {4 + (3)(.9) / 1.5 + (2)(.8) / 1.5 2 +.7 / 1.5 3 } = 772.17 Solving, k = 772.17 8.62736 = 89.5. Comment: It is assumed that students can calculate, in a matter of seconds, the present or accumulated value of any level annuity certain, immediate or due, with a financial calculator such as the BA II Plus. This is especially useful with most questions which involve DeMoivre's Law and a rate of interest.

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 14 9. C. Key Approach: When demoivre's Law is assumed, annuity values, benefit premiums, and terminal reserves should generally be approached using single benefit premiums for life insurances, as their determination is routine, i.e., a A x = ω - x. ω - x Therefore, in a question such as this, it is efficient to use the formula: kv x = A x + k - A x 1 - A x. Accordingly, since A 4 = 1V 3 = A 4 - A 3 1 - A 3 = a 6 6 =.3776 and A 3 =.3776 -.33421 1 -.33421 =.644. a 7 7 =.33421, we have: 1. A. Key Observation: Since the force for decrement 2 is four times as large as that for decrement 1: 1 q (1) 1 = 5 5 1q (T) 5, and 1 q( 2) 5 = 4 5 1q (T) 5. The required value is therefore: 1 1 q (2) 5 + 3 1 q( 1) 5 = 14 1 q (T) 5. 1 (T) But 1 q 5 = 1 - (T) 1 p 5 = 1 - exp[-.5t dt ] = 1 - e -2.5 =.9179, producing a final answer of: (14)(.9179) = 1285.

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 15 11. A. Key Observations: The total force of decrement is a modified demoivre (MDML) with a modification factor of 3, i.e., µ (T) (1) x = µ x + µ (2) x = 3 75 - x, x < 75. (T) 1 3 Making p 73 = 2 = 1 8, and 2 p (T) 73 =. Constructing a crutch, and realizing that decrement 1 is twice as strong as decrement 2, and therefore d (1) x = 2 d (2) x : x l x (T) d (1) x d (2) x 73 12 7 35 74 15 1 5 75 d (2) 74 = 5. 12. A. The number of scientists eaten before noon is Poisson with mean: (6)(.2) + (3)(.15) + (3)(.1) =.87. The number of scientists eaten after noon is Poisson with mean: (6)(.1) + (6)(.2) =.72. Prob[more before noon than after noon] = Prob[ after noon] Prob[at least 1 before noon] + Prob[1 after noon] Prob[at least 2 before noon] + Prob[2 after noon] Prob[at least 3 before noon] + Prob[3 after noon] Prob[at least 4 before noon] +... = e -.72 (1 - e -.87 ) +.72 e -.72 (1 - e -.87 -.87e -.87 ) + (.72 2 e -.72 / 2!) (1 - e -.87 -.87e -.87 -.87 2 e -.87 / 2!) + (.72 3 e -.72 / 3!) (1 - e -.87 -.87e -.87 -.87 2 e -.87 / 2! -.87 3 e -.87 / 3!) +... = (.4868)(.581) + (.355)(.2166) + (.1262)(.58) + (.33)(.12) = 36.6%. Comment: See Section 1 of Mahlerʼs Guide to Stochastic Models. With a small number of expected events, do not use the Normal Approximation unless told to. In this case, the result from the Normal Approximation would be:.5 -.15 1 - Φ[ ] = 1 - Φ[.28] = 39%..72 +.87

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 16 13. B. (.2,.4,.3,.1) Q = (.22,.37,.29,.12). After 6 months. (.22,.37,.29,.12) Q = (.235,.347,.281,.137). After 12 months. (.235,.347,.281,.137) Q = (.2462,.3292,.2732,.1514). After 18 months. (.2462,.3292,.2732,.1514) Q = (.2545,.3153,.26662,.16358). After 2 years. (1,)(.26662) = 2666.2. Comment: Similar to CAS3L, 5/1, Q.11 in Section 14 of Mahlerʼs Guide to Stochastic Models. 14. E. The first moment of severity is: (.6)(3) + (.3)(2) + (.1)(8) = 158. The second moment of severity is: (.6)(3 2 ) + (.3)(2 2 ) + (.1)(8 2 ) = 7,654,. Mean Aggregate = (4)(158) = 63,2. Variance of Aggregate = (4)(7,654,) = 36.16 million. Prob[Aggregate > 1,] 1 - Φ[(1, - 63,2) / 36.16 million ] = 1 - Φ[2.13] = 1.78%. Comment: Similar to CAS3L, 5/11, Q.11 in Section 9 of Mahlerʼs Guide to Stochastic Models. 15. D. Q =.75.25.6.4 No Claim At Least 1 Claim (1, )Q = (.75,.25). (.75,.25)Q = (.7125,.2875). (.7125,.2875)Q = (.76875,.293125). For policy #1, the actuarial present value of premiums is: X(1 + 1/1.6 + 1/1.6 2 ) = 2.833X. There is 75% chance of a dividend paid at the end of year one. The chance of a dividend at the end of year two is.7125. The chance of a dividend at the end of year three is.76875. Thus the APV of dividends is: ($7)(.75/1.6 +.7125/1.6 2 +.76875/1.6 3 ) = 135.46. For policy #2, the actuarial present value of premiums is: Y(1 + 1/1.6 + 1/1.6 2 ) = 2.833Y. The two policies provide the same coverage, in other words they pay the same benefits, so the actuarial present value of their premiums minus dividends must be the same. 2.833X - 135.46 = 2.833Y. X - Y = 135.46/2.833 = $47.82. Comment: Similar to CAS3L, 5/9, Q.16 in Section 15 of Mahlerʼs Guide to Stochastic Models.

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 17 16. C. Assume Damonʼs bus leaves at time t. S(t) = e -1t. f(t) = 1e -1t. The probability Stefanʼs bus has left by time t is: 1 - e -7t. Prob[Stefanʼs bus left first] = (1 - e - 7t ) 1e - 1t dt = 1 (e - 1t - e - 17t ) dt = (1)(1/1-1/17) = 7/17. The average time Damon waits if Stefanʼs bus leaves first is: (1 - e - 7t ) 1e - 1t t dt / (7/17) = (17/7) (t e - 1t - t e - 17t ) dt = (17/7) (1/1 2-1/17 2 ) =.1588 hours = 9.53 minutes. Comment: See Section 3 of Mahlerʼs Guide to Stochastic Models. The unconditional average time that Damon waits for a Bellwood bus is: 6/1 = 6 minutes. The probability that Poisson Process 1 occurs before Poisson Process 2 is: λ 1 /(λ 1 + λ 2 ) = 7/(7 + 1) = 7/17. t e - λt dt = t λe - λt dt / λ = (mean of Exponential with hazard rate λ) / λ = 1/λ 2. 17. E. 5% = Prob[X c µ = 75] = 1 - Φ[ c - 75 8 1 ]. 1.645 = c - 75 8 1. c = 79.162. Power of test at µ = 85 is: Prob[reject µ = 85] = Prob[X > 79.162 µ = 85] 79.162-85 = 1 - Φ[ ] = 1 - Φ[-2.31] = 98.96%. 8 1 Comment: See Section 2 of Mahlerʼs Guide to Statistics.

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 18 18. C. f(x) = α θ α (θ+x) (α+1). Σ ln (f(xi)) = Σ {ln(α) + α ln(θ) (α+1)ln(θ + xi)}. The derivative with respect to α is: Σ {(1/α) + ln(θ) - ln(θ + xi)} = N/α - Σ ln {(θ +xi)/θ}. Setting this derivative equal to zero: = (N/α) Σ ln[(θ +xi)/θ]. Solving for alpha: α = N / Σln[(θ + x i )/θ] = 5 / {ln((1 + 6)/1) + ln((1 + 8)/1) + ln((1 + 11)/1) + ln((1 + 2)/1) + ln((1 + 35)/1)} = 5 / {ln(1.6) + ln(1.8) + ln(2.1) + ln(3) + ln(4.5)} = 1.136. Comment: Similar to 4, 11/2, Q.1 in Section 4 of Mahlerʼs Guide to Statistics. On the exam, unless they say Single Parameter Pareto, they mean a two-parameter Pareto with support starting at zero. 19. B. t = 5-9 22 / 11 = -4 2 = -2.828. For 11-1 = 1 degrees of freedom, Prob[t < -1.812] = 5%. This is the critical region for this test, and -1.812 is the critical value. -2.828 - (-1.812) = 1.16. Comment: Similar to question 2, 5/88, Q.19 in Section 16 of Mahlerʼs Guide to Statistics. Since -3.169 < -2.828 < -2.764, we reject the null hypothesis at 1% and do not reject at 1/2% (one-sided test.)

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 19 2. B. β^ = N X i Y i - X i Y i N X 2 i - X i ( ) 2 X = 96/4 = 24. Y = 78/4 = 19.5. ^α = Y - β^ X = 19.5 - (-.6687)(24) = 35.549. ^ Y 4 = 35.549 + (-.6687)(36) = 11.476. = (4)(1654) - (96)(78) (4)(263) - 96 2 = -.6687. Residual = Observed - Fitted. ε^ 4 = 11-11.476 = -.476. Comment: Similar to CAS3L, 5/8, Q.9 in Section 28 of Mahlerʼs Guide to Statistics. ε^ 1 = 27 - {35.549 + (-.6687)(13)} =.144. ε^ 2 = 23 - {35.549 + (-.6687)(18)} = -.512. ε^ 3 = 17 - {35.549 + (-.6687)(29)} =.843. ε^ 1 + ε^ 2 + ε^ 3 + ε^ 4 =.144 -.512 +.843 -.476 =, subject to rounding. For the linear regression model with an intercept, the sum of the residuals is always zero. 21. B. f(x) = λx λ 1. ln f(x) = ln(λ) + (λ- 1)ln(x). g(y) = µy µ 1. ln g(y) = ln(µ) + (µ- 1)ln(y). Thus the loglikelihood is: n ln(λ) + (λ- 1)Σln(x i ) + m ln(µ) + (µ- 1)Σln(y i ). For H : λ = 2, µ = 4, the loglikelihood is: n ln(2) + Σln(x i ) + m ln(4) + 3Σln(y i ). For H 1 : λ = 3, µ = 6, the loglikelihood is: n ln(3) + 2Σln(x i ) + m ln(6) + 5Σln(y i ). The difference between the loglikelihoods for H 1 and H is: n ln(3/2) + Σln(x i ) + m ln(6/4) + 2Σln(y i ). Thus the critical region for the best (Neyman-Pearson) test of H against H 1 is: n ln(3/2) + Σln(x i ) + m ln(6/4) + 2Σln(y i ) b. Σln(x i ) + 2Σln(y i ) c. Comment: Similar to 2, 5/88, Q.3 in Section 25 of Mahlerʼs Guide to Statistics.

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 2 22. E. For the Poisson, f() = e -.7 = 49.6585%. f(1) =.7e -.7 = 34.761%. f(2) =.7 2 e -.7 /2 = 12.1663%. Prob[3 or more] = 1 - f() - f(1) - f(2) = 3.4142%. There are a total of 2 observed insureds, and therefore, the expected numbers by class are: (2)(49.6585%) = 993.17, 695.22, 243.33, and 68.28. Chi-square statistic is: Σ (Observed - Expected) 2 /Expected = 7.39. Class Observed Assumed Expected Chi-Square Number Poisson Number ((Observed - Expected)^2)/Expected A 125 49.6585% 993.17 1.2 B 75 34.761% 695.22.138 C 215 12.1663% 243.33 3.298 D 55 3.4142% 68.28 2.584 Sum 2 1.% 2 7.39 There are 4-1 = 3 degrees of freedom. Since 7.39 < 7.82, do not reject H at 5%. Comment: Similar to CAS3L, 5/8, Q.4 in Section 12 of Mahlerʼs Guide to Statistics. 23. D. Prob[X (7) < y] = Prob[at least 7 of 8 claims are < y] = Prob[exactly 7 of 8 claims are < y] + Prob[exactly 8 of 8 claims are < y] = 8 F(y) 7 S(y) + F(y) 8 = 8(y/θ) 7 (1 - y/θ) + (y/θ) 8 = 8y 7 /θ 7-7y 8 /θ 8. Taking the derivative, the probability density function of y is: 56y 6 /θ 7-56y 7 /θ 8, < y < θ. Alternately, for a sample of size N from a uniform distribution on [, θ], the distribution of X (r) is a Beta distribution as per Appendix A attached to the exam with a = r, b = N + 1 - r. Thus in this case, Y = X (7) has a Beta Distribution with parameters a = 7, b = 2, and θ. Y has density: {Γ[7 + 2]/(Γ[7] Γ[2])} (y/θ) 7 (1 - y/θ) 2-1 /y = {8! / (6! 1!)} (y 6 /θ 7 ) (1 - y/θ) = 56y 6 /θ 7-56y 7 /θ 8, < y < θ. Alternately, the sample must have 6 values less than y, one value equal to y, and one value greater than y. Thus the density of Y is proportional to: F(x) 6 f(x) S(x) = (y/θ) 6 (1/θ)(1 - y/θ). This can happen in 8! / (6! 1!) = 56 ways. Thus Y has density: 56 (y/θ) 6 (1/θ)(1 - y/θ) = 56y 6 /θ 7-56y 7 /θ 8, < y < θ. Comment: See Section 35 of Mahlerʼs Guide to Statistics.

Batten and Mahler CAS Exam 3L, Practice Exam #1 12/1/11, Page 21 24. C. The expected value of X (7) is: θ θ y f(y) dy = 56 (y 7 / θ 7 - y 8 / θ 8 ) dy = 56θ (1/8-1/9) = 7θ / 9. Therefore, E[θ^ ] = 2(7θ/9) = 14θ/9. The Bias is: 14θ/9 - θ = 5θ/9. The second moment of X (7) is: θ θ y 2 f(y) dy = 56 (y 8 / θ 7 - y 9 / θ 8 ) dy = 56θ 2 (1/9-1/1) = 28θ 2 / 45. Therefore, the variance of X (7) is: 28θ 2 /45 - (7θ/9) 2 = 7θ 2 /45. Var[θ^ ] = Var[2 X (7) ] = 4 Var[X (7) ] = 28θ 2 /45. Mean Square Error is: Variance + Bias 2 = 28θ 2 /45 + (5θ/9) 2 = 153θ 2 /45 = 17θ 2 /45 =.378θ 2. Alternately, for a sample of size 8 from the uniform distribution on (, θ), E[X (7) ] = 7θ/(8 + 1) = 7θ/9. X (7) has a Beta Distribution with parameters a = 7, b = 2, and θ. The second moment of this Beta Distribution is: θ 2 (a+ b -1)! (a +1)! (a+ b +1)! (a -1)! = θ 2 a (a +1) (a+ b) (a + b+ 1) = θ2 (7)(8) (9)(1) 28 = θ2. Proceed as before. 45 Comment: Similar to CAS3, 11/5, Q.6 in Section 37 of Mahlerʼs Guide to Statistics. 25. D. X = {(4)() + (7)(1) + (5)(2) + (2)(3) + (4)(3)}/183 = 1.3224. ^ β = X / r = 1.3224/3 =.448. Var[ ^β] = Var[X ]/3 2 = (Var[X]/183) / 9 = 3(β)(1+β) / 1647 = 3(.448)(1.448) / 1647 =.1157. Standard Deviation of ^β is:.1157 =.34. Thus an approximate 9% confidence interval for β is:.448 ± (1.645)(.34) =.441 ±.56. Comment: See Section 3 of Mahlerʼs Guide to Statistics.