Math 466 Sample Problems for Midterm I
1. Three random variables W N (0, 2), Y1 N (1, 1), Y2 N (2, 1), and they are mutually independent. Find the distribution of the following rv (justication is needed).
(2Y1 Y2 )2
, T3 = W/
T1 = 2Y1 Y2 , T2 =
5
W2
.
2T2
MATH 466: Sample Problems for Midterm 2
1. Let X1 , , Xn be a random sample from a Bernoulli distribution with probability of
success p. Let p = X be the sample proportion of successes.
(a) Show that p is an unbiased estimator of p.
(b) Show that p is a c
8
Parameter Estimation
8
8.1
Parameter Estimation
Introduction
Parameter estimation is one important type of statistical
inference: use the sample (observations) to estimate the value of
unknown parameter which characterizes the population.
Examples of pr
MATH 466: Sample Problems for Midterm 2
1. Let X1 , , Xn be a random sample from a Bernoulli distribution with probability of
success p. Let p = X be the sample proportion of successes.
(a) E() = E(X) = p.
p
P
(b) V ar(X) = p(1 p) < . By WLLN, p p.
(c) p
Math 466/566 - Final Do six of the seven problems. 1. A manufacturer of a brand of light bulbs claims that the mean life-time of their bulbs is more than one year. (a) Find an appropriate test with significance level = 0.05 of the null hypothesis H0 : = 1
10
Hypothesis Testing
10
10.1
Hypothesis Testing
Introduction
Def 1: A hypothesis is a statement about a population parameter .
Def 2: The two complementary hypotheses in a hypothesis testing
problem a are called the null hypothesis and the alternative
hy
9
Properties of Point Estimators & Methods of Estimation
9
9.1
Properties of Point Estimators &
Methods of Estimation
Consistency
Example: Toss a fair coin n times. Denote Yi = 1 if the ith outcome
is head; 0 otherwise. The sample mean
= Y1 + + Yn .
Y
n
1
MATH 466: Theory of Statistics
http:/d2l.arizona.edu
Yue Selena Niu
[email protected]
Oce: 520 Math Building
2
Course Outline:
1. Sampling Distributions and the Central Limit Theorem
2. Estimation and Condence Intervals
3. Properties of Point Esti
1
Review of Chapter 7
Three methods to nd the sampling distributions of statistics:
Method of Distribution Functions
Obtain the cdf FY (y) = P (Y y), and then dierentiate F
with respect to y to obtain the pdf of Y , fY (y) = F (y).
Method of Transformat
ANSWERS
Chapter 1
1.5 a
b
c
1.9 a
b
c
d
1.13 a
b
2.45 2.65, 2.65 2.85
7/30
16/30
Approx. .68
Approx. .95
Approx. .815
Approx. 0
y = 9.79; s = 4.14
k = 1: (5.65, 13.93); k = 2: (1.51,
18.07); k = 3: (2.63, 22.21)
1.15 a y = 4.39; s = 1.87
b k = 1: (2.52, 6
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Math 466 Sample Problems for Final Exam
1. (a) X Y N (0, 7/3).
(b)
2
SX
2
2
SX
2
=
+
2
2SX
2
4 2
2
3SY
2
4 5 .
and
2
3SY
4
2
2
2 . Since Xs and Y s are independent, so are SX and SY . So
3
(c) Note that X1 , X2 , Y1 are mutually independent. So
2
[ (X1
MATH466 Practice Final Exam
1. Let X1 , X2 , X3 , Y1 , Y2 , Y3 , Y4 be independent random variables from N ( =
3, 2 = 4). Dene
1
X=
3
3
i=1
1
Xi , Y =
4
4
Yi ,
2
SX
i=1
1
=
2
3
1
2
(Xi X)2 , SY =
3
i=1
4
(Yi Y )2
i=1
Find the distribution of the following
Math 466/566 - Quiz 5
1. The population has a uniform distribution on [0, ] with unknown. So
1
f (x|) = 1(0 x )
We take a Bayesian point of view, and assume the prior distribution of is
() =
3 2
0
if 0 1
if [0, 1]
/
We consider a random sample of size n
Math 466 - Homework 1 1. Consider the experiment of throwing a fair die n times. Let X1 , X2 , Xn be the results. (So Xi can be 1, 2, 3, 4, 5 or 6.) Let
n
Y =
i=1
Xi
(1)
Find the following: (a) the mean and variance of Xi . 1 7 E[Xi ] = (1 + 2 + 3 + 4 + 5
Math 466/566 - Homework 2 Solutions 1. Problem 1 from chapter 2 in the book. Solution: E[1Ai ] = 2/3. So E[12 i ] = 2/3. So var(1Ai ) = 2/3 - (2/3)2 = 2/9. A 2. Problem 2 from chapter 2 in the book. Solution: f = 2/3. f = var(1Ai ) 1 = 15 n
3. Problem 4 f
Math 466/566 - Homework 3 1. We defined the chi-squared distribution as follows. Let Z1 , Z2 , , Zn be independent standard normal RV's. Let
n
X=
i=1
Zi2
(1)
Then X has the chi-squared distribution with n degrees of freedom. Recall that you computed the d
Math 466/566 - Homework 4 1. We want to test a hypothesis involving a population proportion. The unknown population proportion is p. The null hypothesis is p = 1/2 and the alternative hypothesis is p > 1/2. We want the level of the test to be 0.01, and we
Math 466/566 - Homework 5 Solutions 1. Book, chapter 7, problem 4. Solution: The expected value of the sample mean is always the population mean, so the sample mean is always an unbiased estimator. The variance of a Poisson RV is equal to its mean, . So t
Math 466/566 - Homework 6 - Solutions 1. Book, chapter 10, number 3. Solution: f (x|) = -1 1(0 x ) and () = 2 e- So the joint density of x and is f (x, ) = 1(0 x ) 2 e- We compute f (x) by integrating out . So f (x) = =
x
1(0 x ) 2 e- d
2 e- d
= e-x (1)
Math 466/566 - Homework 7 1. (a) The population has a gamma distribution with = 1 and unknown. We want to test the null hypothesis = 0 against the alternative > 0 with a sample size of 100 and a significance level of 5%. Find the uniformly most powerful t
Math 466/566 - Midterm Solutions NOTE: These solutions are for both the 466 and 566 exam. The problems are the same until questions 4 and 5. 1. The moment generating function of a random variable X is M (t) = 1 1-t
4
(a) Find the mean and variance of X. S
Math 466/566 - Quiz 1 Solutions 1. A random variable X is uniformly distributed between 0 and 1 if the pdf is fX (x) = 1 for 0 x 1. The moment generating function of such a random variable is et - 1 MX (t) = (1) t You can use this fact without deriving it
Math 466/566 - Quiz 4 - Solutions 1. Consider the following density f (x|) = 2 x e-x , x 0; f (x|) = 0, x<0
It is easy to show the integral of this is 1, the mean is = 2/ and the variance is 2 = 2/2 . ^ (a) Find the maximum likelihood estimator of .
n n
f
Math 466/566 - Quiz 5 1. The population has a uniform distribution on [0, ] with unknown. So 1 f (x|) = 1(0 x ) We take a Bayesian point of view, and assume the prior distribution of is () = 32 0 if 0 1 if [0, 1] /
We consider a random sample of size n =