homework problems stat 479 - purdue universityjbeckley/q/wd/stat479… · · 2012-03-29homework...

Homework Problems

Stat 479

March 28, 2012

Chapter 2

1. Model 1 is a uniform distribution from 0 to 100. Determine the table entries for a

generalized uniform distribution covering the range from a to b where a < b.

2. Let X be a discrete random variable with probability function p(x) = 2(1/3)x for x

= 1, 2, 3, … What is the probability that X is odd?

3. * For a distribution where x > 2, you are given:

The hazard rate function: h(x) = z2/2x , for x > 2

F(5) = 0.84.

Calculate z.

4. FX(t) = (t2-1)/9999 for 1<t<100. Calculate fX(50).

5. You are given that the random variable X is distributed as a Weibull distribution

with parameters θ = 3 and τ = 0.5. Calculate:

a. Pr[X < 5]

b. Pr[3 < X < 5]

6. (Spreadsheet Problem) You are given that the random variable X is distributed as

a Geometric distribution with parameters β = 3. Calculate:

a. Pr[X < 5]

b. Pr[3 < X < 5]

7. A Weibull Distribution with parameter τ = 1 becomes what distribution?

8. A random variable X has a density function f(x) = 4x(1+x2)-3

, for x > 0.

Determine the mode of X.

Chapter 3

9. Determine the following for a generalized uniform distribution covering the range

from a to b where a < d < b:

a. E[Xk]

b. E[X]

c. Var(X)

d. e(d)

e. VaRp

f. TVaRp

10. For the Pareto distribution, determine E[X], Var(X), and the coefficient of

variation in terms of α and θ.

Homework Problems

Stat 479

March 28, 2012

11. For the Gamma distribution, determine E[X], Var(X), the coefficient of variation,

and skewness in terms of α and θ.

12. For the Exponential distribution, determine E[X], Var(X), and e(d) in terms of θ.

13. You are given:

x3/27, for 0 < x < 3

F(x) =

1, for x >3

Calculate:

a. E[X]

b. Var(X)

c. e(1)

d. E[(X-1)+]

e. E[X Λ 2]

f. The Median

g. The standard deviation principle with k = 1

h. VaR.80

i. TVaR.80

14. (Spreadsheet) If you roll two fair die, X is the sum of the dice. Calculate:

a. E[X]

b. Var(X)

c. e(4)

d. E[(X-4)+]

e. E[X Λ 10]

f. The Mode

g. π20

15. You are given a sample of 2, 2, 3, 5, 8. For this empirical distribution, determine:

a. The mean

b. The variance

c. The standard deviation

d. The coefficient of variation

e. The skewness

f. The kurtosis

g. The probability generating function

h. VaR.80

i. TVaR.80

Homework Problems

Stat 479

March 28, 2012

16. The random variable X is distributed exponentially with θ = 0.2. Calculate

E[e2X

].

17. * Losses follow a Pareto distribution has parameters of α = 7 and θ = 10,000.

Calculate e(5,000)

18. The amount of an individual claim has a Pareto distribution with θ = 8000 and α =

9. Use the central limit theorem to approximate the probability that the sum of

500 independent claims will exceed 550,000.

Chapter 4

19. The distribution function for losses from your renter’s insurance is the following:

F(x) = 1 – 0.8[1000/(1000+x)]5 – 0.2[12000/(12000+x)]

3

Calculate:

a. E[X]

b. Var(X)

c. Use the normal approximation to determine the probability that the sum of

100 independent claims will not exceed 200,000.

20. * X has a Burr distribution with parameters α = 1, γ = 2, and θ = 10000.5

. Y has a

Pareto distribution with parameters α = 1 and θ = 1000. Z is a mixture of X and Y

with equal weights on each component. Determine the median of Z.

21. The random variable X is distributed as a Pareto distribution with parameters α

and θ. E(X) = 1 and Var(X) = 3. The random variable Y is equal to 2X.

Calculate the Var(Y).

22. * Claim severities are modeled using a continuous distribution and inflation

impacts claims uniformly at an annual rate of s. Which of the following are true

statements regarding the distribution of claim severities after the effect of

inflation?

i. An exponential distribution will have a scale parameter of (1+s)θ.

ii. A Pareto distribution will have scale parameters (1+s)α and (1+s)θ.

iii. An Inverse Gaussian distribution will have a scale parameter of

(1+s)θ.

23. * The aggregate losses of Eiffel Auto Insurance are denoted in euro currency and

follow a Lognormal distribution with μ = 8 and σ = 2. Given that 1 euro = 1.3

dollars, determine the lognormal parameters for the distribution of Eiffel’s losses

in dollars.

Homework Problems

Stat 479

March 28, 2012

Chapter 5

24. (Spreadsheet) Calculate Γ(1.2).

25. The random variable X is the number of dental claims in a year and is distributed

as a gamma distribution given parameter θ and with parameter α = 1. θ is

distributed uniformly between 1 and 3. Calculate E(X) and Var(X).

26. A dental insurance company has 1000 insureds. Assume the number of claims

from each insured is independent. Using the information in Problem 25 and the

normal approximation, calculate the probability that the Company will incur more

than 2100 claims.

27. * Let N have a Poisson distribution with mean Λ. Let Λ have a uniform

distribution on the interval (0,5). Determine the unconditional probability that

N > 2.

Chapter 6

28. The number of hospitalization claims in a year follows a Poisson distribution with

a mean of λ. The probability of exactly three claims during a year is 60% of the

probability that there will be 2 claims. Determine the probability that there will

be 5 claims.

29. If the number of claims is distributed as a Poison distribution with λ = 3,

calculate:

a. Pr(N = 0)

b. Pr(N = 1)

c. Pr(N = 2)

d. E(N)

e. Var(N)

30. If the number of claims is distributed as a zero truncated Poison distribution with

λ = 3, calculate:

a. Pr(N = 0)

b. Pr(N = 1)

c. Pr(N = 2)

d. E(N)

e. Var(N)

31. If the number of claims is distributed as a zero modified Poison distribution with

λ = 3 and p0M

= 0.5, calculate:

a. Pr(N = 0)

b. Pr(N = 1)

c. Pr(N = 2)

d. E(N)

Homework Problems

Stat 479

March 28, 2012

e. Var(N)

32. If the number of claims is distributed as a Geometric distribution with β = 3,

calculate:

a. Pr(N = 0)

b. Pr(N = 1)

c. Pr(N = 2)

d. E(N)

e. Var(N)

33. (Spreadsheet) If the number of claims is distributed as a Binomial distribution

with m = 6 and q = 0.4, calculate:

a. Pr(N = 0)

b. Pr(N = 1)

c. Pr(N = 2)

d. E(N)

e. Var(N)

34. (Spreadsheet) If the number of claims is distributed as a Negative Binomial

distribution with γ = 3 and β = 2, calculate:

a. Pr(N = 0)

b. Pr(N = 1)

c. Pr(N = 2)

d. E(N)

e. Var(N)

35. (Spreadsheet) If the number of claims is distributed as a Logarithmic distribution

with β = 3, calculate:

a. Pr(N = 0)

b. Pr(N = 1)

c. Pr(N = 2)

d. E(N)

e. Var(N)

36. * The Independent Insurance Company insures 25 risks, each with a 4%

probability of loss. The probabilities of loss are independent. What is the

probability of 4 or more losses in the same year? (Hint: Use the binomial

distribution.)

37. * You are given a negative binomial distribution with γ = 2.5 and β = 5. For what

value of k does pk take on its largest value?

Homework Problems

Stat 479

March 28, 2012

38. * N is a discrete random variable form the (a, b, 0) class of distributions. The

following information is known about the distribution:

P(N = 0) = 0.327680

P(N = 1) = 0.327680

P(N = 2) = 0.196608

E[N] = 1.25

Based on this information, which of the following are true statements?

I. P(N = 3) = 0.107965

II. N is from a binomial distribution.

III. N is from a Negative Binomial Distribution.

39. * You are given:

Claims are reported at a Poisson rate of 5 per year.

The probability that a claim will settle for less than 100,000 is 0.9.

What is the probability that no claim of 100,000 or more will be reported in the

next three years?

40. N is distributed as a zero modified geometric distribution with β = 3 and p0M

=

0.5. Calculate:

a. e(2)

b. [ 3]Var N

c. E[(N-2)+]

Homework Problems

Stat 479

March 28, 2012

Chapter 8

41. Losses are distributed Exponentially with θ = 1000. Losses are subject to an

ordinary deductible of 500. Calculate:

a. fYL(y)

b. FYL(y)

c. SYL(y)

d. E(YL)

e. The loss elimination ratio

42. Losses are distributed Exponentially with parameter θ = 1000. Losses are subject

to an ordinary deductible of 500. Calculate:

a. fYP(y)

b. FYP(y)

c. E(YP)

43. Losses are distributed Exponentially with θ = 1000. Losses are subject to a

franchise deductible of 500. Calculate:

a. fYL(y)

b. FYL(y)

c. SYL(y)

d. E(YL)

e. The loss elimination ratio

44. Losses are distributed Exponentially with parameter θ = 1000. Losses are subject

to a franchise deductible of 500. Calculate:

a. fYP(y)

b. FYP(y)

c. E(YP)

45. Last year, losses were distributed Exponentially with θ = 1000. This year losses

are subject to 10% inflation. Losses in both years are subject to an ordinary

deductible of 500. Calculate the following for this year:

a. E(YL)

b. E(YP)

46. * Losses follow an exponential distribution with parameter θ. For a deductible of

100, the expected payment per loss is 2000. What is the expected payment per

loss in terms of θ for a deductible of 500? (Note: Assume it is an ordinary

deductible.)

Homework Problems

Stat 479

March 28, 2012

47. * In year 2007, claim amounts have the following Pareto distribution

F(x) = 1 – (800/[x+800])3

The annual inflation rate is 8%. A franchise deductible of 300 will be

implemented in 2008. Calculate the loss elimination ratio of the franchise

deductible.

48. (Spreadsheet) If you roll two fair die, X is the sum of the dice. Let X represent

the distribution of losses from a particular event. If an ordinary deductible of 3 is

applied, calculate:

a. E(YL)

b. E(YP)

c. Var(YL)

d. Var(YP)

49. Losses are distributed uniformly between 0 and 100,000. An insurance policy

which covers the losses has a 10,000 deductible and an 80,000 upper limit. The

upper limit is applied prior to applying the deductible. Calculate:

a. E(YL)

b. E(YP)

c. Var(YL)

d. Var(YP)

50. Last year, losses were distributed Exponentially with θ = 1000. This year losses

are subject to 25% inflation. Losses in both years are subject to an ordinary

deductible of 500, an upper limit of 4000, and coinsurance where the company

pays 80% of the claim. Calculate the following for this year:

a. E(YL)

b. E(YP)

51. * An insurance company offers two types of policies: Type Q and Type R. Type

Q has no deductible, but has a policy limit of 3000. Type R has no limit, but has

an ordinary deductible of d. Losses follow a Pareto distribution with θ = 2000

and α = 3.

Calculate the deductible d such that both policies have the same expected cost per

loss.

Homework Problems

Stat 479

March 28, 2012

52. * Well Traveled Insurance Company sells a travel insurance policy that

reimburses travelers for any expense incurred for a planned vacation that is

cancelled because of airline bankruptcies. Individual claims follow a Pareto

distribution with α = 2 and θ = 500. Well Traveled imposes a limit of 1000 on

each claim. If a planned policyholder’s planned vacation is cancelled due to

airline bankruptcy and he or she has incurred more than 1000 of expenses, what is

the expected non-reimbursable amount of the claim?

53. If X is uniformly distributed on from 0 to b. An ordinary deductible of d is

applied. Calculate:

a. E[YL]

b. Var[YL]

c. E[YP]

d. Var[YP]

54. Losses follow a Pareto distribution with α = 3 and θ = 200. A policy covers the

loss except for a franchise deductible of 50. X is the random variable

representing the amount paid by the insurance company per payment. Calculate

the expected value of the X.

55. Losses in 2007 are uniformly distributed between 0 and 100,000. An insurance

policy pays for all claims in excess of an ordinary deductible of $20,000. For

2008, claims will be subject to uniform inflation of 20%. The Company

implements an upper limit u (without changing the deductible) so that the

expected cost per loss in 2008 is the same as the expected cost per loss in 2007.

Calculate u.

56. Losses follow an exponential distribution with a standard deviation of 100. An

insurance company applies an ordinary policy deductible d which results in a Loss

Elimination Ratio of 0.1813. Calculate d.

57. Losses are distributed following the single parameter Pareto (Not the same as the

Pareto distribution in Question 54) with α = 4 and θ = 90. An insurance policy is

issued with an ordindary deductible of 100. Calculate the Var(YL).

Homework Problems

Stat 479

March 28, 2012

58. Losses follow a Pareto distribution with α = 5 and θ = 2000. A policy pays 100%

of losses from 500 to 1000. In other words, if a loss occurs for less than 500, no

payment is made. If a loss occurs which exceeds 1000, no payment is made.

However, if a loss occurs which is between 500 and 1000, then the entire amount

of the loss is paid. Calculate E(YP).

59. * You are given the following:

Losses follow a lognormal distribution with parameters μ =10 and σ = 1.

For each loss less than or equal to 50,000, the insurer makes no payment.

For each loss greater than 50,000, the insurer pays the entire amount of the

loss up to the maximum covered loss of 100,000.

Determine the expected amount of the loss.

60. You are given the following:

Losses follow a distribution prior to the application of any deductible with

a mean of 2000.

The loss elimination ratio at a deductible of 1000 is 0.3.

60% of the losses in number are less than the deductible of 1000.

Determine the average size of the loss that is less than the deductible of 1000.

61. Losses follow a Pareto distribution with α = 5 and θ = 2000. An insurance policy

covering these losses has a deductible of 100 and makes payments directly to the

physician. Additionally, the physician is entitled to a bonus if the loss is less than

500. The bonus is 10% of the difference between 500 and the amount of the loss.

The following table should help clarify the arrangement:

Amount of Loss Loss Payment Bonus

50 0 45

100 0 40

250 150 25

400 300 10

500 400 0

1000 900 0

Calculate the expected total payment (loss payment plus bonus) from the

insurance policy to the physician per loss.

Homework Problems

Stat 479

March 28, 2012

Chapter 6

62. * For an insured, Y is the random variable representing the total amount of time

spent in a hospital each year. The distribution of the number of hospital

admissions in a year is:

Number of Admissions Probability

0 0.60

1 0.30

2 0.10

The distribution of the length of each stay for an admission follows a Gamma

distribution with α = 1 and θ = 5.

Calculate E[Y] and Var[Y].

63. For an insurance company, each loss has a mean of 100 and a variance of 100.

The number of losses follows a Poisson distribution with a mean of 500. Each

loss and the number of losses are mutually independent.

The loss ratio for the insurance company is defined as the ratio of aggregate losses

to the total premium collected.

The premium collected is 110% of the expected aggregate losses.

Using the normal approximation, calculate the probability that the loss ratio will

exceed 95%.

64. An automobile insurer has 1000 cars covered during 2007. The number of

automobile claims for each car follows a negative binomial distribution with β = 1

and γ = 1.5. Each claim is distributed exponentially with a mean of 5000.

Assume that the number of claims and the amount of the loss are independent and

identically distributed.

Using the normal distribution as an approximating distribution of aggregate

losses, calculate the probability that losses will exceed 8 million.

Homework Problems

Stat 479

March 28, 2012

65. X1, X2, and X3 are mutually independent random variables. These random

variables have the following probability functions:

x f1(x) f2(x) f3(x)

0 0.6 0.4 0.1

1 0.4 0.3 0.4

2 0.0 0.2 0.5

3 0.0 0.1 0.0

Calculate fS(x).

66. The number of claims follows a Poisson distribution with a mean of 3. The

distribution of claims is fX(1) = 1/3 and fX(2) = 2/3. Calculate fS(4).

67. (Spreadsheet) The number of claims and the amount of each claim is mutually

independent. The frequency distribution is:

n pn

1 0.05

2 0.20

3 0.30

4 0.25

5 0.15

6 0.05

The severity distribution is:

x fX(x)

100 0.25

200 0.20

300 0.15

400 0.12

500 0.10

600 0.08

700 0.06

800 0.04

Calculate fS(x). (See Example 6.6 in the book for an example.)

Homework Problems

Stat 479

March 28, 2012

68. You are given the following table for aggregate claims:

S FS(s) E[(X-d)+]

0

100

200 0.50

300 0.65 60

400 0.75

500

600

Losses can only occur in multiples of 100. Calculate the net stop loss premium

for stop loss insurance covering losses in excess of 325.

69. * Losses follow a Poisson frequency distribution with a mean of 2 per year. The

amount of a loss is 1, 2, or 3 with each having a probability of 1/3. Loss amounts

are independent of the number of losses and from each other.

An insurance policy covers all losses in a year subject to an annual aggregate

deductible of 2. Calculate the expected claim payments for this insurance policy.

70. Purdue University has decided to provide a new benefit to each class at the

university. Each class will be provided with group life insurance. Students in

each class will have 10,000 of coverage while the professor for the class will have

20,000. STAT 490C has 27 students all age 22 split 12 males and 15 females.

The class also has an aging professor. The probability of death for each is listed

below:

Age Gender Probability of Death

22 Male .005

22 Female .003

52 Male .020

Purdue purchases a Stop Loss Policy such with an aggregate deductible of 20,000.

Calculate the net premium that Purdue will pay for the stop loss coverage.

Homework Problems

Stat 479

March 28, 2012

71. For an automobile policy, the severity distribution is Gamma with α = 3 and θ =

1000. The number of claims in a year is distributed as follows:

Number of Claims Probability

1 .65

2 .20

3 .10

4 .05

Calculate:

a. E(S)

b. Var(S)

c. MS(t)

72. (Spreadsheet) You are given that the number of losses follow a Poisson with λ =

6. Losses are distributed as follows: fX(1) = fX(2) = fX(4) = 1/3. Find fS(x) for

total losses of 100 or less.

73. On a given day, a doctor provides medical care for NA adults and NC children.

Assume that NA and NC have Poisson distributions with parameters 3 and 2

respectively. The distributions for the length of care per patient are as follows:

Let NA, NC, and the lengths of stay for all patients be independent. The doctor

charges 200 per hour. Determine the probability that the office income on a given

day will be less than or equal to 800.

74. (Spreadsheet) Losses follow a Pareto distribution with α = 3 and θ = 1000.

a. Using the Method of Rounding with a span of 100, calculate fj for 0 < j

<1000. Calculate the mean of the discretized distribution. Compare this

mean to the actual mean.

b. Using the Method of Local Moment Matching and matching the mean

(with k = 1) and a span of 100, calculate fj for 0 < j <1000. Calculate the

mean of the discretized distribution. Why does this mean not equal the

actual mean?

75. Using the example that was handed out in class, find fS(6) and fS(7).


a. S has a compound Poisson distribution with λ = 2; and

b. Individual claim amounts x are distributed as follows:

fX(1) = 0.4 and fX(2) = 0.6

Determine fS(4).

Time Adult Child

1 hour 0.4 0.9

2 hours 0.6 0.1

Homework Problems

Stat 479

March 28, 2012

77. * Aggregate claims S has a compound Poisson distribution with discrete

individual claim amount distributions of fX(1) = 1/3 and fX(3) = 2/3.

Also, fS(4) = fS(3) + 6fS(1).

Calculate the Var(S).

78. * For aggregate claims S, you are given:

a. fS(X) =

0

50*

!

)50()(

n

nn

X nx ef

b. Losses are distributed as follows:

x fX(x)

1 0.4

2 0.5

3 0.1

Determine Var(S)

79. With no deductible, the number of payments for losses under warranty coverage

for an Iphone follows a negative binomial distribution with a mean of 0.25 and a

variance of 0.375.

The company imposes a deductible of d such that the number of expected

payments for losses is reduced to 50% of the prior expected payments.

Calculate the Var (NP) after the imposition of the deductible.

80. * Prior to the application of any deductible, aggregate claim counts during 2005

followed a Poisson with λ = 14. Similarly, individual claim sizes followed a

Pareto with α = 3 and θ = 1000.

Annual inflation for the claim sizes is 10%.

All policies in 2005 and 2006 are subject to a 250 ordinary deductible.

Calculate the increase in the number of claims that exceed the deductible in 2006

when compared to 2005.

Homework Problems

Stat 479

March 28, 2012

81. * An insurance portfolio produces N claims with the following distribution:

n P(N=n)

0 0.1

1 0.5

2 0.4

Individual claim amounts have the following distribution:

x fX(x)

0 0.7

10 0.2

20 0.1

Individual claim amounts and counts are independent.

Stop Loss insurance is purchased with an aggregate deductible of 400% of

expected claims.

Calculate the net stop loss premium.

82. Losses are modeled assuming that the amount of all losses is 40 and that the

number of losses follows a geometric distribution with a mean of 4. Calculate the

net stop loss premium for coverage with an aggregate deductible of 100.

83. * An aggregate claim distribution has the following characteristics:

P[S = i] = 1/6 for i = 1, 2, 3, 4, 5, or 6. A stop loss insurance with a deductible

amount of d has an expected insurance payment of 1.5

Determine d.

84. Purdue University has decided to provide a new benefit to each class at the

university. Each class will be provided with group life insurance. Students in

each class will have 10,000 of coverage while the professor for the class will have

20,000. STAT 490C has 27 students all age 22 split 12 males and 15 females.

The class also has an aging professor. The probability of death for each is listed

below:

Age Gender Probability of Death

22 Male .005

22 Female .003

52 Male .020

The insurance company providing the coverage charges a premium equal to the

expected claims plus one standard deviation. Calculate the premium.

Homework Problems

Stat 479

March 28, 2012

85. * An insurer provides life insurance for the following group of independent lives:

Number of Lives Death Benefit Probability of Death

100 1 0.01

200 2 0.02

300 3 0.03

The insurer purchases reinsurance with a retention of 2 on each life.

The reinsurer charges a premium of H equal to the its expected claims plus the

standard deviation of its claims.

The insurer charges a premium of G which is equal to its expected retained claims

plus the standard deviation of the retained claims plus H

Calculate G.

86. * Two portfolios of independent insurance policies have the following

characteristics:

Portfolio A

Class Number in

Class

Probability of

Claim

Claim Amount

Per Policy

1 2000 0.05 1

2 500 0.10 2

Portfolio B

Class Number in

Class

Probability of

Claim

Claim Amount

Distribution

Mean Variance

1 2000 0.05 1 1

2 500 0.10 2 4

The aggregate claims in the portfolios are denoted by SA and SB, respectively.

Calculate Var[SB]/Var[SA].

Homework Problems

Stat 479

March 28, 2012

87. * An insurance company is selling policies to individuals with independent future

lifetimes and identical mortality profiles. For each individual, the probability of

death by all causes is 0.10 and the probability of death due to an accident is 0.01.

Each insurance policy pays a benefit of 10 for an accidental death and 1 for non-

accidental death.

The company wishes to have at least 95% confidence that premiums with a

relative security loading of 0.20 are adequate to cover claims. (In other words,

the premium is 1.20E(S).)

Using the normal approximation, determine the minimum number of policies that

must be sold.

Chapter 12

88. * A random sample, X1, X2, …, Xn, is drawn from a distribution with a mean of

2/3 and a variance of 1/18.

= (X1 + X2 + … + Xn)/(n-1) is the estimator of the distribution mean θ.

Find MSE( ).

89. * Claim sizes are uniformly distributed over the interval [0, θ]. A sample of 10

claims, denoted by X1, X2, …,X10 was observed and an estimate of θ was obtained

using:

= Y = max(X1, X2, …,X10)

Recall that the probability density function for Y is:

fY(y) = 10y9/θ

10

Calculate the mean square error for for θ =100.

90. * You are given two independent estimates of an unknown quantity θ:

a. Estimator A: E( A) = 1000 and σ( A) = 400

b. Estimator B: E( B) = 1200 and σ( B) = 200

Estimator C is a weighted average of Estimator A and Estimator B such that:

C = (w) A + (1-w) B

Determine the value of w that minimizes σ( C).

Homework Problems

Stat 479

March 28, 2012


x Pr(X = x)

0 0.5

1 0.3

2 0.1

3 0.1

Using a sample size of n, the population mean is estimated by the sample

mean X . The variance is estimated by:

S n

2 =

n

XX i 2

)(

Calculate the bias of S2n when n = 4.

92. * For the random variable X, you are given:

a. E(X) = θ, θ > 0

b. Var(θ) = θ2/25

c. = kX/(k+1)

d. MSEθ(θ) = 2[bias θ(θ)]2

Determine k.

93. You are given the following sample of claims:

X: 12, 13, 16, 16, 22, 24, 26, 26, 28, 30

The sum of X is 213 and the sum of X2 is 4921.

H0 is that μx = 17 and H1 is that μx > 17.

Calculate the z statistic, the critical value(s) assuming a significance level of 1%,

and the p value. State your conclusion with regard to the Hypothesis Testing.

Homework Problems

Stat 479

March 28, 2012

Wang Warranty Corporation is testing iPods. Wang starts with 100 iPods and tests

them by dropping them on the ground. Wang records the number of drops before each

iPod will no longer play. The following data is collected from this test:

Drops to Failure Number Drops to Failure Number

1 6 9 7

2 2 10 6

4 3 11 7

6 4 12 7

7 4 13 48

8 5 22 1

Ledbetter Life Insurance Company is completing a mortality study on a 3 year term

insurance policy. The following data is available:

Life Date of Entry Date of Exit Reason for Exit

1 0 0.2 Lapse

2 0 0.3 Death

3 0 0.4 Lapse

4 0 0.5 Death

5 0 0.5 Death

6 0 0.5 Lapse

7 0 1.0 Death

8 0 3.0 Expiry of Policy








16 0.5 2.0 Lapse

17 0.5 1.0 Death

18 1.0 3.0 Expiry of Policy

19 1.0 3.0 Expiry of Policy

20 2.0 2.5 Death

Schneider Trucking Company had the following losses during 2006:

Amount

of Claim

Number

of Payments

Total Amount

of Losses

0 – 10 8 60

10-20 5 70

20-30 4 110

30 + 3 200

Total 20 440

Homework Problems

Stat 479

March 28, 2012

Chapter 13

94. Using the data from Wang Warranty Corporation, calculate:

a. 100 ( )p x

b. 100 ( )F x

c. The empirical mean

d. The empirical variance

e. Ĥ(x) where Ĥ(x) is the cumulative hazard function from the Nelson Åalen

estimate

f. Ŝ(x) where Ŝ(x) is the survival function from the Nelson Åalen estimate

95. Using the data from Schneider Trucking Company, calculate:

a. The ogive, F20(x)

b. The histogram, f20(x)

c. E(XΛ30) minus E[(X-20)+]


Claim Size (X) Number of Claims

(0,25] 30

(25,50] 32

(50,100] 20

(100,200] 8

Assume a uniform distribution of claim sizes within each interval.

Estimate the mean of the claim size distribution.

Estimate the second raw moment of the claim size distribution.

Homework Problems

Stat 479

March 28, 2012

Chapter 14

97. A mortality study is completed on 30 people. The following deaths occur during

the five years:

3 deaths at time 1.0





There were no other terminations and no lives entered the study after the start of

the study.

The data was smoothed using a uniform kernel with a bandwidth of 1. Calculate

f (x) and F (x) for all x > 0.

98. * From a population having a distribution function F, you are given the following

sample:

2.0, 3.3, 3.3, 4.0, 4.0, 4.7, 4.7, 4.7

Calculate the kernel density estimate of F(4), using a uniform kernel with

bandwidth of 1.4.

99. * You study five lives to estimate the time from the onset of a disease until death.

The times to death are:

2 3 3 3 7

Using a triangular kernel with a bandwidth of 2, estimate the density function at

2.5.

100. You are given the following random sample:

12 15 27 42

The data is smoothed using a uniform kernel with a bandwidth of 6.

Calculate the mean and variance of the smoothed distribution.

Homework Problems

Stat 479

March 28, 2012


12 15 27 42

The data is smoothed using a triangular kernel with a bandwidth of 12.



12 15 27 42

The data is smoothed using a gamma kernel with a bandwidth of 3.


Homework Problems

Stat 479

March 28, 2012

Chapter 15

103. You are given the following sample of claims obtained from an inverse gamma

distribution:

X: 12, 13, 16, 16, 22, 24, 26, 26, 28, 30

The sum of X is 213 and the sum of X2 is 4921.

Calculate α and θ using the method of moments.

104. * You are given the following sample of five claims:

4 5 21 99 421

Find the parameters of a Pareto distribution using the method of moments.

105. * A random sample of death records yields the follow exact ages at death:

30 50 60 60 70 90

The age at death from which the sample is drawn follows a gamma distribution.

The parameters are estimated using the method of moments.

Determine the estimate of α.


The random variable X has the density function

f(x) = αx-α-1

, 1 < x < ∞, α >1

A random sample is taken of the random variable X.

Calculate the estimate of α in terms of the sample mean using the method of

moments.


The random variable X has the density function f(x) = {2(θ – x)}/θ2, 0 < x < θ

A random sample of two observations of X yields values of 0.50 and 0.70.

Determine θ using the method of moments.

Homework Problems

Stat 479

March 28, 2012

108. You are given the following sample of claims:

X: 12, 13, 16, 16, 22, 24, 26, 26, 28, 30

Calculate the smoothed empirical estimate of the 40th

percentile of this

distribution.

109. * For a complete study of five lives, you are given:

a. Deaths occur at times t = 2, 3, 3, 5, 7.

b. The underlying survival distribution S(t) = 4-λt

, t > 0

Using percentile matching at the median, calculate the estimate of λ.

110. You are given the following 9 claims:

X: 10, 60, 80, 120, 150, 170, 190, 230, 250

The sum of X = 1260 and the sum of X2 = 227,400.

The data is modeled using an exponential distribution with parameters

estimated using the percentile matching method.

Calculate θ based on the empirical value of 120.

111. * For a sample of 10 claims, x1 < x2 < … < x10 you are given:

a. The smoothed empirical estimate of the 55th

percentile is 380.

b. The smoothed empirical estimate of the 60th

percentile is 402.

Determine x6.

112. You are given the following:

a. Losses follow a Pareto distribution with parameters α and θ.

b. The 10th

percentile of the distribution is θ – k, where k is a constant.

c. The 90th

percentile of the distribution is 5θ – 3k.

Determine α.

Homework Problems

Stat 479

March 28, 2012

113. You are given the following random sample of 3 data points from a population

with a Pareto distribution with θ = 70:

X: 15 27 43

Calculate the maximum likelihood estimate for α.


a. Losses follow an exponential distribution with mean θ.

b. A random sample of 20 losses is distributed as follows:

Range Frequency

[0,1000] 7

(1000, 2000) 6

(2000,∞) 7

Calculate the maximum likelihood estimate of θ.


The random variable X has the density function f(x) = {2(θ – x)}/θ2, 0 < x < θ

A random sample of two observations of X yields values of 0.50 and 0.90.

Determine the maximum likelihood estimate for θ.


a. Ten lives are subject to the survival function S(t) = (1-t/k)0.5

, 0 < t < k

b. The first two deaths in the sample occured at time t = 10.

c. The study ends at time t = 10.

Calculate the maximum likelihood estimate of k.


a. The random variable X follows the exponential distribution with

parameter θ.

b. A random sample of three observations of X yields values of 0.30, 0.55,

and 0.80

Determine the maximum likelihood estimate of θ.

Homework Problems

Stat 479

March 28, 2012

118. * Ten laboratory mice are observed for a period of five days. Seven mice die

during the observation period, with the following distribution of deaths:

Time of Death in Days Number of Deaths

2 1

3 2

4 1

5 3

The lives in the study are subject to an exponential survival function with mean of

θ.

Calculate the maximum likelihood estimate of θ.

119. * A policy has an ordinary deductible of 100 and a policy limit of 1000. You

observe the following 10 payments:

15 50 170 216 400 620 750 900 900 900

An exponential distribution is fitted to the ground up distribution function, using

the maximum likelihood estimate.

Determine the estimated parameter θ.

120. * Four lives are observed from time t = 0 until death. Deaths occur at t = 1, 2, 3,

and 4. The lives are assumed to follow a Weibull distribution with τ = 2.

Determine the maximum likelihood estimator for θ.

121. * The random variable X has a uniform distribution on the interval [0,θ]. A

random sample of three observations of X are recorded and grouped as follows:

Interval

Number of

Observations

[0,k) 1

[k,5) 1

[5,θ] 1

Calculate the maximum likelihood estimate of θ

Homework Problems

Stat 479

March 28, 2012

122. * A random sample of three claims from a dental insurance plan is given below:

225 525 950

Claims are assumed to follow a Pareto distribution with parameters θ = 150 and α.

Determine the maximum likelihood estimate of α.

123. * The following claim sizes are experienced on an insurance coverages:

100 500 1,000 5,000 10,000

You fit a lognormal distribution to this experience using maximum likelihood.

Determine the resulting estimate of σ.

You are given the following data from a sample:

k nk

0 20

1 25

2 30

3 15

4 8

5 2

Use this data for the next four problems.

124. Assuming a Binomial Distribution, estimate m and q using the Method of

Moments.

125. Assuming a Binomial Distribution, find the MLE of q given that m = 6.

126. (Spreadsheet) Assuming a Binomial Distribution, find the MLE of m and q.

127. Assuming a Poisson Distribution, approximate the 90% confidence interval for

the true value of λ.

Homework Problems

Stat 479

March 28, 2012

Chapter 16


X: 10, 40, 60, 65, 75, 80, 120, 150, 170, 190, 230, 340, 430, 440, 980, 600,

675, 950, 1250, 1700

The data is being modeled using an exponential distribution with θ = 427.5.

Calculate D(200).


X: 10, 40, 60, 65, 75, 80, 120, 150, 170, 190, 230, 340, 430, 440, 980, 600,

675, 950, 1250, 1700

The data is being modeled using an exponential distribution with θ = 427.5.

You are developing a p-p plot for this data. What are the coordinates for x7 =

120.

130. Mark the following statements True or False with regard to the Kolmogorov-

Smirnov test:

The Kolmogorov-Smirnov test may be used on grouped data as well as

individual data.

If the parameters of the distribution being tested are estimated, the critical

values do not need to be adjusted.

If the upper limit is less than ∞, the critical values need to be larger.

Homework Problems

Stat 479

March 28, 2012

131. Balog’s Bakery has workers’ compensation claims during a month of:

100, 350, 550, 1000

Balog’s owner, a retired actuary, believes that the claims are distributed

exponentially with θ = 500.

He decides to test his hypothesis at a 10% significance level.

Calculate the Kolmogorov-Smirnov test statistic.

State the critical value for his test and state his conclusion.

He also tests his hypothesis using the Anderson-Darling test statistic. State the

values of this test statistic under which Mr. Schmidt would reject his hypothesis.

132. * The observations of 1.7, 1.6, 1.6, and 1.9 are taken from a random sample. You

wish to test the goodness of fit of a distribution with probability density function

given by f(x) = 0.5x for 0 < x < 2.

Using the Kolmogorov-Smirnov statistic, which of the following should you do?

a. Accept at both levels

b. Accept at the 0.01 level but reject at the 0.10 level

c. Accept at the 0.10 level but reject at the 0.01 level

d. Reject at both levels

e. Cannot be determined.

133. * Two lives are observed beginning at time t=0. One dies at time 5 and the other

dies at time 9. The survival function S(t) = 1 – (t/10) is hypothesized.

Calculate the Kolmogorov-Smirnov statistic.

134. * From a laboratory study of nine lives, you are given:

a. The times of death are 1, 2, 4, 5, 5, 7, 8, 9, 9

b. It has been hypothesized that the underlying distribution is uniform

with ω = 11.

Calculate the Kolmogorov-Smirnov statistic for the hypothesis.

Homework Problems

Stat 479

March 28, 2012

135. You are given the following data:

Claim Range Count

0-100 30

100-200 25

200-500 20

500-1000 15

1000+ 10

H0: The data is from a Pareto distribution.

H1: The data is not from a Pareto distribution.

Your boss has used the data to estimate the parameters as α = 4 and θ = 1200.

Calculate the chi-square test statistic.

Calculate the critical value at a 10% significance level.

State whether you would reject the Pareto at a 10% significance level.

136. During a one-year period, the number of accidents per day in the parking lot of

the Steenman Steel Factory is distributed:

Number of Accidents Days

0 220

1 100

2 30

3 10

4+ 5

H0: The distribution of the number of accidents is distributed as Poison with a

mean of 0.625.

H1: The distribution of the number of accidents is not distributed as Poison with a

mean of 0.625.

Calculate the chi-square statistic.

Calculate the critical value at a 10% significance level.

State whether you would reject the H0 at a 10% significance level.

Homework Problems

Stat 479

March 28, 2012

137. * You are given the following random sample of automobile claims:

54 140 230 560 600 1,100 1,500 1,800 1,920 2,000

2,450 2,500 2,580 2,910 3,800 3,800 3,810 3,870 4,000 4,800

7,200 7,390 11,750 12,000 15,000 25,000 30,000 32,200 35,000 55,000

You test the hypothesis that automobile claims follow a continuous distribution

F(x) with the following percentiles:

x 310 500 2,498 4,876 7,498 12,930

F(x) 0.16 0.27 0.55 0.81 0.90 0.95

You group the data using the largest number of groups such that the expected

number of claims in each group is at least 5.

Calculate the Chi-Square goodness-of-fit statistic.

138. Based on a random sample, you are testing the following hypothesis:

H0: The data is from a population distributed binomial with m = 6 and q = 0.3.

H1: The data is from a population distributed binomial.

You are also given:

L(θ0) = .1 and L(θ1) = .3

Calculate the test statistic for the Likelihood Ratio Test

State the critical value at the 10% significance level

139. State whether the following are true or false

The principle of parsimony states that a more complex model is better

because it will always match the data better.

In judgment-based approaches to determining a model, a modeler’s

experience is critical.

In most cases, judgment is required in using a score-based approach to

selecting a model.

Homework Problems

Stat 479

March 28, 2012

Chapter 21

140. A random number generated from a uniform distribution on (0, 1) is 0.6. Using

the inverse transformation method, calculate the simulated value of X assuming:

X is distributed Pareto with α = 3 and θ = 2000

0.5x 0 < x < 1.2

F(x) = 0.6 1.2 < x < 2.4

0.5x -0.6 2.4 < x < 3.2

F(x) = 0.1x -1 10 < x < 15

0.05x 15 < x < 20

141. * You are given that f(x) = (1/9)x2 for 0 < x < 3.

You are to simulate three observations from the distribution using the inversion

method. The follow three random numbers were generated from the uniform

distribution on [0,1]:

0.008 0.729 0.125

Using the three simulated observations, estimate the mean of the distribution.

142. * You are to simulate four observations from a binomial distribution with two

trials and probability of success of 0.30. The following random numbers are

generated from the uniform distribution on [0,1]:

0.91 0.21 0.72 0.48

Determine the number of simulated observations for which the number of

successes equals zero.

Homework Problems

Stat 479

March 28, 2012

143. Kyle has an automobile insurance policy. The policy has a deductible of 500 for

each claim. Kyle is responsible for payment of the deductible.

The number of claims follows a Poison distribution with a mean of 2.

Automobile claims are distributed exponentially with a mean of 1000.

Kyle uses simulation to estimate the claims. A random number is first used to

calculate the number of claims. Then each claim is estimated using random

numbers using the inverse transformation method.

The random numbers generated from a uniform distribution on (0, 1) are 0.7, 0.1,

0.5, 0.8, 0.3, 0.7, 0.2.

Calculate the simulated amount that Kyle would have to pay in the first year.

144. * Insurance for a city’s snow removal costs covers four winter months.

You are given:

There is a deductible of 10,000 per month.

The insurer assumes that the city’s monthly costs are independent and

normally distributed with mean of 15,000 and standard deviation of 2000.

To simulate four months of claim costs, the insurer uses the inversion

method (where small random numbers correspond to low costs).

The four numbers drawn from the uniform distribution on [0,1] are:

0.5398 0.1151 0.0013 0.7881

Calculate the insurer’s simulated claim cost.

145. * Annual dental claims are modeled as a compound Poisson process where the

number of claims has mean of 2 and the loss amounts have a two-parameter

Pareto distribution with θ = 500 and α = 2.

An insurance pays 80% of the first 750 and 100% of annual losses in excess of

750.

You simulate the number of claims and loss amounts using the inversion method.

The random number to simulate the number of claims is 0.80. The random

numbers to simulate the amount of claims are 0.60, 0.25, 0.70, 0.10, and 0.80.

Calculate the simulated insurance claims for one year.

Homework Problems

Stat 479

March 28, 2012

146. A sample of two selected from a uniform distribution over (1,U) produces the

following values:

3 7

You estimate U as the Max(X1, X2).

Estimate the Mean Square Error of your estimate of U using the bootstrap

method.

147. * Three observed values from the random variable X are:

1 1 4

You estimate the third central moment of X using the estimator:

g(X1, X2, X3) = 1/3 Σ(Xj - X )3

Determine the bootstrap estimate of the mean-squared error of g.

Chapter 3

148. * Using the criterion of existence of moments, determine which of the following

distributions have heavy tails.

a. Normal distribution with mean μ and variance of σ2.

b. Lognormal distribution with parameters μ and σ2.

c. Single Parameter Pareto.

Homework Problems

Stat 479

March 28, 2012

Answers

1. Not Provided

2. ¾

3. 2

4. 100/9999

5.

a. 0.725

b. 0.093

6.

a. 0.822

b. 0.244

7. No Answer Given

8. 1/√5

9.

a. (bk+1

– ak+1

)/(b-a)(k+1)

b. (b+a)/2

c. (b-a)2/12

d. (b-d)/2

e. (1-p)a+pb

f. [b+(1-p)a+pb]/2

10.

a. θ/(α-1)

b. αθ2/[(α-1)

2(α-2)]

c. [α/( α-2)]0.5

11.

a. θα

b. θ2α

c. 1/√α

d. 2/√α

12.

a. θ

b. θ2

c. θ

13.

a. 2.25

b. 27/80

c. 17/13

d. 34/27

e. 50/27

f. 2.3811

g. 2.83

h. 2.785

i. 2.895

Homework Problems

Stat 479

March 28, 2012

Answers

14.

a. 7

b. 5.8333

c. 3.7333

d. 3.1111

e. 6.8888

f. 7

g. 5

15.

a. 4

b. 5.2

c. 2.28

d. 0.57

e. 0.8096

f. 2.145

g. 0.4z2 + 0.2z

3 + 0.2z

5 + 0.2z

8

h. 5

i. 8

16. 5/3

17. 2500

18. 0.0244

19.

a. 1400

b. 26,973,333

c. 87.7%

20. 100

21. 12

22. i only. Be sure to indicate how to make the others true.

23. μ = 8.26 and σ = 2.00

24. 0.918169

25. 2 and 14/3

26. 7.2%

27. 0.6094

28. 2.6%

29.

a. 0.0498

b. 0.1494

c. 0.2240

d. 3

e. 3

30.

a. 0

b. 0.1572

c. 0.2358

Homework Problems

Stat 479

March 28, 2012

d. 3.1572

e. 2.6609

Answers

31.

a. 0.5000

b. 0.0786

c. 0.1179

d. 1.5786

e. 3.8224

32.

a. 0.2500

b. 0.1875

c. 0.1406

d. 3

e. 12

33.

a. 0.0467

b. 0.1866

c. 0.3110

d. 2.4

e. 1.44

34.

a. 0.0370

b. 0.0741

c. 0.0988

d. 6

e. 18

35.

a. 0

b. 0.5410

c. 0.2029

d. 2.1640

e. 3.9731

36. 0.0165

37. 7

38. Statement III is only true statement

39. 0.2231

40.

a. 4

b. 1.6943

c. 1.125

Homework Problems

Stat 479

March 28, 2012

Answers

41.

a. fYL(y) = 1 – e-0.5

, y = 0

fYL(y) = 0.001e-(0.001y + 0.5)

, y > 0

b. FYL(y) = 1 – e-0.5

, y = 0

FYL(y) = 1 – e-(.001 y + 0.5)

, y > 0

c. SYL(y) = e-0.5

, y = 0

SYL(y) = e-(.001 y + 0.5)

, y > 0

d. 1000e-0.5

e. 1 – e-0.5

42.

a. fYP(y) = 0.001e-0.001y

b. FYP(y) = 1 – e-.001 y

c. 1000

43.

a. fYL(y) = 1 – e-0.5

, y = 0

fYL(y) = 0.001e-0.001y

, y > 500

b. FYL(y) = 1 – e-0.5

, y = 0

FYL(y) = 1 – e-.001 y

, y > 500

c. SYL(y) = e-0.5

, y = 0

SYL(y) = e-.001 y

, y > 500

d. 1500e-0.5

e. 1 – 1.5e-0.5

44.

a. fYP(y) = 0.001e-0.001y+0.5

b. FYP(y) = 1 – e-.001 y+0.5

c. 1500

Homework Problems

Stat 479

March 28, 2012

Answers

45.

a. 1100e-(0.5/1.1)

b. 1100

46. 2000e-400/θ

47. 0.165

48.

a. 4.0278

b. 4.39

c. 5.58

d. 4.48

49.

a. 38,500

b. 42,777.78

c. 641,083,333

d. 529,320,798

50.

a. 629.56

b. 939.19

51. 182.18

52. 1500

53.

a. (b-d)2/2b

b. [(b-d)3/3b] – [(b-d)

2/2b]

2

c. (b-d)/2

d. (b-d)2/12

54. 175

55. 71,833.62

56. 20

57. 1708.70

58. 705.06

59. 16,224

60. 333.33

61. 431.83

62. 2.5 and 23.75

63. 0.1587

64. 0.068

Homework Problems

Stat 479

March 28, 2012

65.

x fS(x)

0 0.024

1 0.130

2 0.280

3 0.280

4 0.180

5 0.086

6 0.020

66. 0.151436

67. None provided

68. 51.25

69. 2.360894

70. 22.60

71.

a. 4,650

b. 11,377,500

c. 0.65(1-1000t)-3

+ 0.2(1-1000t)-6

+ 0.1(1-1000t)-9

+ 0.05(1-1000t)-12

72. No Answer Provided

73. 0.2384

74.

a. Mean = 498.61

b. Mean = 499.85

75. 0.106978 and 0.060129

76. 0.1517

77. 76

78. 165

79. 0.15625

80. 0.406

81. 0.224

82. 92.16

83. 2.25

84. 5,727

85. 46.13

86. 2.1

87. 1975

88. (n+8)/[18(n-1)2]

89. 151.52

90. 1/5

91. -0.24

92. 5

93. z = 2.0814; critical value = 2.33; Since 2.0814 is less than 2.33, we cannot reject

the null hypothesis; p = 0.0188

94.

x p100(x) x F100(x) Ĥ(x) Ŝ(x)

Homework Problems

Stat 479

March 28, 2012

x<1 0 0 1

1 0.06 1<x<2 0.06 0.060000 0.941765

2 0.02 2<x<4 0.08 0.081277 0.921939

4 0.03 4<x<6 0.11 0.113885 0.892360

6 0.04 6<x<7 0.15 0.158829 0.853142

7 0.04 7<x<8 0.19 0.205888 0.813924

8 0.05 8<x<9 0.24 0.267616 0.765201

9 0.07 9<x<10 0.31 0.359722 0.697871

10 0.06 10<x<11 0.37 0.446678 0.639750

11 0.07 11<x<12 0.44 0.557789 0.572473

12 0.07 12<x<13 0.51 0.682789 0.505206

13 0.48 13<x<22 0.99 1.662381 0.189687

22 0.01 x>22 1.00 2.662381 0.069782

Empirical Mean = 10.44 and Empirical Variance = 14.4064

95.

a. 0.04x for 0 < x < 10

0.15 + 0.025x for 10 < x < 20

0.25 + 0.02x for 20 < x < 30

Undefined for x > 30

b. 0.04 for 0 < x < 10

0.025 for 10 < x < 20

0.02 for 20 < x < 30

Undefined for x > 30

c. 8

96. 47.50 and 3958.333333333

97.

x f(x) F(x)

x < 0 0 0

0 < x < 1 3/60 3x/60

1 < x < 2 7/60 (7x-4)/60

2 < x < 2.8 9/60 (9x-8)/60

2.8 < x < 3.0 17/60 (17x-30.4)/60

3.0 < x < 3.5 13/60 (13x-18.4)/60

3.5 < x < 4 23/60 (23x-53.4)/60

4 < x < 4.8 18/60 (18x-33.4)/60

4.8 < x < 5.5 10/60 (10x+5)/60

5.5 < x 0 1

98. 0.53125

99. 0.3

100. Mean = 24 and Variance = 151.50



103. α = 13.812 and θ = 272.89

Homework Problems

Stat 479

March 28, 2012

104. α = 3.81889 and θ = 310.0782

105. 10.8

106. )1/( XX

107. 1.8

108. 18.4

109. 1/6

110. 234.91

111. 378

112. 2

113. 3.002

114. 1996.90

115. 1.5

116. 25

117. 0.55

118. 6

119. 703

120. 2.7386

121. 7.5

122. 0.6798

123. 1.6436

124. m = 30 and q = 0.0573333333

125. 0.28667

126. m = 26 and q = 0.066154

127. (1.50426, 1.93574)

128. 0.1264

129. (1/3,0.2447)

130. All Statements are false.

131. 0.2534, 0.61, Cannot reject,

10% => Reject at A2 > 1.933

5% => Reject at A2 > 2.492

1% => Reject at A2 > 3.857

132. B

133. 0.5

134. 0.1818

135. χ2 = 5.5100; critical value = 4.605; Reject H0

136. χ2 = 18.5; critical value = 7.779; Reject H0

137. 6.6586

138. T = 2.197; critical value = 4.605

139. False; True; True

Homework Problems

Stat 479

March 28, 2012

140. 714.42, 2.4, 15

141. 1.6

142. 2

143. 1105.36

144. 14,400

145. 630.79

146. 4

147. 4 8/9

148. C only

homework problems stat 479 - purdue universityjbeckley/q/wd/stat479… · · 2012-03-29homework...

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