1.
Since ll = E[X] and a2 = Var[X]. are unchanged, the standard for ﬁill credibility will be
unchanged if no = (3,92 is unchanged. With k = .05, P = .9 we have yo = 1.645, and
if- = 34%?! = 32.9 . With 19 = .95 , 11.95 is the “2‘95 = .975 percentile of th

173B
Chapter 6 Review of Hypothesis Testing
Panayiotis Skordi
1. Formulate the hypothesis statement for the following claim: The average adult drinks 1.7
cups of coffee per day. A sample of 35 adults drank an average of 1.95

UCLA Chapter 6 Solutions
1. Formulate the hypothesis statement for the following claim: The average adult drinks 1.7
cups of coffee per day. A sample of 35 adults drank an average of 1.95 cups per day. Assume
the population standard deviation is 0.5

Problems 1 and 2 refer to the following situation. A11 insurance company has two group policies. The
aggregate claim amounts (in millions of dollars) for the ﬁrst three policy years are summarized in the
table below. Assume that the two groups have the sa

Hypothetical mean = p(/\) = E [X [A] = A , Process variance = v(/\) = Var[X|)\] = A .
EIX] = p = E[#(A)l = En] 1= .5. v = Elv(A)] = Eu] = .s ,
a. = Var{p(A)] = Var[A] = -1-2
(variance of the uniform distribution on interval (a, b) is
z: "u = "n =L. Answer

1. Assume that the number of claims each year for an individual insured has a Poisson distribution.
Assume also that the expected annual claim frequencies (the Poisson parameters A) of the members
of the population of insureds are uniformly distributed ov

We wish to ﬁnd Irmx(A|0) = = ——IXME,[:_?3r(M .
Since the model distribution is Poisson, we have fx-A(0|A) = e"‘ , and since the prior distribution
is uniform on (O, 1) we have 1r(A) = l for D < A < 1 . The marginal probability fx(0) is
fX(0) = fulﬁY.A(0:/

Problems 1 and 2 are based on the following situation. Assume that the number of claims each year for
an individual insured has a Poisson distribution. Assume also that the expected annual claim frequencies
(the Poisson parameters A) of the members of the

l.
2.
F(2=)=0ifa:so, F(a:)=1ifa:2 1, and Fm: {2
T=-716=F(3)=2$ - 3% -’ $=.603. Answer: B.
(a) The simulation of an exponential variable Y with mean 3 using the inversion method is
y = — 9111(1 — u) . In this case, 0 = 1, so the three simulated exponential

Chapter 14
Simulation
Simulation
The objective in performing a simulation is to reproduce the behavior of a random variable by
generating observations from another random variable which has the same distribution as the

Problems 1 to 3 refer the following random sample of 15 data points:
8.0, 5.1, 2.2, 8.6, 4.5, 5.6, 8.1, 6.4, 3.3 , 7.3, 8.0, 4.0, 6.5, 6.3 , 9.1
The following three bootstrap samples of the empirical distribution have been simulated. Each sample is
of siz

Chapter 15
Simulation
The Bootstrap Method And Statistical Analysis Using Simulation
In statistics, bootstrapping can refer to any test or metric that relies on random sampling with
replacement . Bootstrapping allows ass

1 if 0 S a:
The random variable X has probability density function f (9:) = { 2 if % S a:
0 otherwise
Using the inversion method of simulation, ﬁnd the random observation generated by the (uniform
decimal) random number 1' = .716 .
S
S
mu NIH
A) .358 B) .

1. Two bowls each contain 10 similarly shaped balls. Bo'wl 1 contains 5 red and 5 white balls (equally
likely to be chosen). Bowl 2 contains 2 red and 8 white balls (equally likely to be chosen). A bowl
is chosen at random with each bowl having the chance

2.
The pdf of X is f = % for 20 < cc < 30 , so the likelihood function is _
4 n . ‘
L(9) = Hf(a:i) — i . L09) will be maximized when 9 is minimized.
i=1
For each sci it must be true that < 6 < . 'The smallest possible value of 6 that is consistent
with

W
J
; r u , ., ,. an,“ m is“, g md'n‘m‘ ﬁrs V. i; 3‘ mg“ 1‘ .g' ‘57 ~ we; .Q ' 21; wu’:~*.1‘.$ a ‘ Liz-u 7. it‘suanmx‘Vgsértvt-4d‘p‘ainidﬁgAﬁhﬁwr
_ ~r 1‘ r '7 ' _ :2 r» A7 a ,Aw” ,' r r: r W, :‘« 7“ r r m) t r: n a me.“ A my - a: , , r ,m i m
4. 2. W~rm~n

4.
ii
. A 12
Themleofﬁts Hzi -———-.—
2;: - %+,’—2+-+
_ = 20.04. F(.e) = e‘B/I.
[1:1
P[X > 20] = 1 a 3"20/20 = .632. Answer: A
The mle of ,u is ﬂ = i (the sample mean) and the mle of 02 is
32 = % - 2(23 — m2 = — 332] (biased form of sample variance).
i=1

Questions 1 to 7 are based on the following random sample of 12 data points ﬁ-om a population
distribution X
7,12,15,19,26,27,29,29,30,33,38,53
1.
For the inverse exponential distribution, use the maximum likelihood estimate of the parameter to
estimMe

Chapter 2
Model Estimation
Maximum Likelihood Estimation
General Definition of Maximum Likelihood Estimation
Maximum likelihood estimation is a method that is applied to estimate the parameters in a parametric
distri