课程名称︰统计学与计量经济学暨实习上
课程性质︰必修
课程教师︰张胜凯
开课学院：社会科学院
开课系所︰经济系
考试日期（年月日）︰2014/12/03
考试时限（分钟）：180分钟
试题 :
Econ 2014 Statistics and Econometrics I
Second Midterm Exam (12/3/2014)
Problem 1. (20 points(10,10))
(a) We want to know the quality of bulbs produced by a factory, so we randomly
select 50 bulbs produced by the factory. The average lifetime of the 50
bulbs is 80 hours, and summation of squares of lifetime is 320784. Find an
unbiased estimator of variance of bulb's lifetime.
(b) The pair of random variables X and Y is bivariate-normally distributed with
parameters μ_X = 3, μ_Y = 2, (σ_X)^2 = 9, (σ_Y)^2 = 16, and σ_XY = 2.
What is E[Y|x = 4] and E[Y^2|x = 4].
Sol: (a) We use sample variance to be the unbiased estimator for variance.
n _ n _
(S_X)^2 = [1/(n-1)]Σ[(x_i - X)^2] = [1/(50-1)][Σ (x_i)^2 - nX^2]
i=1 i=1
= (1/49)(320784-50˙80˙80) = 16
So we use 16 to be the estimated value for variance of bulb's lifetime.
(5 points if you write the sample variance formula.)
(b) (X,Y) ~ BVN, Y|X~N(α + βX,σ^2),
β = σ_XY/(σ_X)^2 = 2/9, α = μ_Y - βμ_X = 2 - (2/9)˙3 = 4/3
σ^2 = (σ_Y)^2 - β^2(σ_X)^2 = 16 - (2/9)^2˙9 = 140/9,
Y|X ~ N(4/3 + [2/9]X,140/9), Y|X = 4 ~ N(20/9,140/9), E[Y|x=4] = 20/9
E[Y^2|x=4] = Var(Y|x=4) + (E[Y|x=4])^2 = 140/9 +(20/9)^2 = 1660/81
(Get each point for α,β,σ^2, 1 point for linear function, 3 point
for E[Y|x=4] and 3 for E[Y^2|x=4] )
Problem 2. (20 points (5,5,5,5))
A random sample of 121 checking accounts at a bank showed an average daily
balance of 280. The standard deviation of the population is known to be 66.
(a) Is it necessary to know anything about the shape of the distribution of the
account balances in order to make an interval estimate of the mean of all
the account balances? Explain.
(b) Find the standard error of the mean.
(c) Give a point estimate of the population mean.
(d) Construct a 80% confidence interval estimates for the mean.
Sol: (a) (1) Is it necessary to know anything about the shape of the
distribution?
ex: Unimodal; thick tail; ...etc.
(2) Is it necessary to know everything about the shape of the
distribution?
No, according to CLT, and population standard deviation is known,
we can do estimation.
(Explanation for 2 points.)
(b) Standard error for the mean = √(σ^2/n) = 66/11 = 6
(c) We can use sample mean to be the unbiased estimator of population
mean. Sample mean = 280.
_
(Get 3 points if you only write X or "sample mean")
_
(d) X~N(280,6^2),
_ _ _ _
80% confident interval = (E[X] - Z_0.1˙σ_X,E[X] + Z_0.1˙σ_X)
= [280 - 1.28˙6,280 + 1.28˙6]
= [273.31,287.69] (Z_0.1 ≒ 1.28)
(Get 2 points if you write down the formula of conditional interval)
Problem 3. (15 points (5,5,5))
The daily dinner bills in a local restaurant are normally distributed with a
mean of 28 dollars and a standard deviation of 6 dollars.
(a) What is the probability that a randomly selected bill will at least 39.10
dollars?
(b) What are the minimum and maximum of the middle 95% of the bills?
(c) If twelve of one day's bills had a value of at least 43.06 dollars, how
many bills did the restaurant collect on that day?
Sol: (a) (x*-μ)/σ = (39.1-28)/6 = 1.85 = z*,from standard normal distribution
table, we can find when z* = 1.85, Pr(0≦Z≦z*) = 0.4678,
Pr(X≧39.1) = Pr(Z≧1.85) = 0.5 -0.4678 = 0.0322
(b) (1) Pr(0≦Z≦z*) = 0.95/2 = 0.475, z* = 1.96,
the minimum = 28 - 1.96˙6 = 16.24
the maximum = 28 + 1.96˙6 = 39.76
(2) min = 28 - 2˙6 = 12, max = 28 + 2˙6 = 40
(Get 2 points if you write down the formula)
(c) Let total number of the daily dinner bill is N.
Pr(X≧43.06)˙N = 12, Pr(X≧43.06)˙N = Pr[(X-μ)/σ ≧ (43.06-28)/6]˙N
= P(Z≧2.51)˙N = [0.5 - Pr(0≦Z≦2.51)]˙N = (0.5 - 0.494)˙N = 0.006N
N = 12/0.006 = 2000
(Get 2 points if you write down the formula)
Problem 4. (15 points (5,5,5))
Students of a larg university spend an average of 5 a day on lunch. The
standard deviation of the expenditure is 3. A simple random sample of 36
students is taken.
(a) What are the expected value, standard deviation, and shape of the sampling
distribution of the sample mean?
(b) What is the probability that the sample mean will be at least 4?
(c) What is the probability that the sample mean will be at most 5.90?
_
Sol: (a) E[X] = μ = 5 (2 points)
_
Std.Dev(X) = 3/(√36) = 0.5 (2 points)
Sample distribution is a normal distribution, Bell-shape.
(2pts for expected value, 2pts for standard deviation, 1pt for the shape)
_ _ _ _
(b) Pr(X≧4) = Pr[(X-E[X])/σ_X ≧ (4-5)/0.5] = Pr(Z≧-2)
= 0.5 + Pr(0≦Z≦2) = 0.5 + 0.4772 = 0.9772
(Get 2 points if you write down the formula)
(Get 3 points if you fail in the last step)
_ _ _ _
(c) Pr(X≦5.9) = Pr[(X-E[X])/σ_X ≦ (5.9-5)/0.5] = Pr(Z≦1.8)
= 0.5 + Pr(0≦Z≦1.8) = 0.5 + 0.4641 = 0.9641
(Get 2 points if you write down the formula)
(Get 3 points if you fail in the last step)
Problem 5.(30 points (10,10,10))
Let X_i ~ Bernoulli(p), i = 1,2,3,..., n and X_i s are i.i.d. random variables.
_
(a) Show that the sample average X_n is an unbiased estimator and consistent
estimator of p.
(b) Find the maximum likelihood estimator of p.
_
(c) Show that variance of X_n attains the Cramer-Rao Lower Bound, therefore,
_
X_n is Minimum Variance Unbiased Estimator (MVUE) of p.
(Hint: if X ~ Bernoulli(p), f_X(x) = p^x˙(1-p)^(1-x), x = 0, 1)