STAT 2006 Assignment 1
Due Time and Date: 5 p.m., 4th Feb, 2016
1. A company starts a fund of M dollars from which it pays $1000 to each employee who achieves
high performance during the year. The probability of each employee achieving this goal is 0.10 a
STAT 2006 Assignment 1
Due Time and Date: 5 p.m., 28th Jan, 2015
1. Let X and Y are independent continuous real-valued random variables with pdf fX , fY respectively.
0
(a) Show that E[X] =
(1 FX (x)dx
FX (x)dx.
0
(b) Let Z = X + Y . Show that fZ (z
STAT 2006 Assignment 4
Due Date: 17:00, 21th April, 2017
1. Let X and Y denotethe tarsus lengths of male and female grackles, respectively. Assume that X is
N (X , 2 X ) and Y is N (Y , 2 Y ). Given that n = 25, x = 33.80, s2x = 4.88, m = 29, y = 31.66, a
STAT 2006 Assignment 2 Suggested Solution
i.i.d.
1. Let X1 , X2 , . . . , Xn Bin(1, p) where 0 < p < 1.
n
i=1
(a) E X = E
Xi
=
n
n
i=1
(b) Var X = Var
Xi
n
n
i=1
E [Xi ]
np
=
= p. Therefore, X is an unbiased estimator of p.
n
n
=
n
i=1
n2
Var (
Xi )
n
i=1
STAT2006 Tutorial 4
3rd/4th/5th, Feb 2015
1. t-distribution
1. Let T = Z , where Z N (0, 1), U 2 (r), Z and U are independent. Then T
U/r
has a t-distribution with degree of freedom r and the p.d.f.
(r + 1)/2)
1
f (t) =
.
2 /r)(r+1)/2
r(r/2) (1 + t
2. If
STAT 2006 Tutorial 4 Solution
3rd/4th/5th Feb 2015
Q1.
2
2
(2).
(a) Since Z1 N (0, 1), Z2 + Z3 2 (2), so V T
2
2
(b) Let U = (Z1 + Z2 )/2, note U > 0, W ( 2, 2). Therefore, Z1 = W U , Z2 = U (2 W 2 ),
i.e. the transformation is not one-to-one. However,
STAT 2006 Tutorial 1
19/20/21 Jan 2016
1. Sample space, outcome, event and probability
Recall some denitions, notations and results:
Sample space S: set of possible outcomes s. Write s S.
Event A (measurable) subset of S.
Probability P: a function meas
STAT2006 Tutorial 2
26th/27th/28th Jan 2016
1. Change-of-variables technique: Suppose the joint pdf fX1 ,X2 ,.,Xn (x1 , x2 , ., xn ) and the
transformation Yi = ui (X1 , X2 , ., Xn ), i = 1, 2, ., n are given. We want to nd out the joint pdf
fY1 ,Y2 ,.,Yn
Review, Topics on Multivariate Random
Variables, and Beyond
Dr Phillip YAM
2015/2016 Spring Semester
Reference: Chapters 1 to 5 and Chapter 10 of Probability and
Statistical Inference by Hogg and Tanis.
Preface
After taking STAT2001, you are supposed to b
STAT 2006A/2006B
Basic Concepts in Statistics and Probability II
2015/16 Spring Semester
1. Description:
This course covers basic theories in statistical estimation of distribution parameters
and hypothesis testing. Topics include such as: point estimatio
STAT 2006 Assignment 1 Suggested Solution
1. Let X be the number of employees getting paid for high performance. Therefore X Bin(100,0.1)
and 1000X is the amount that company needs to pay. We want to find any M 0 such that
P(1000X M ) 0.90.
P(1000X M ) 0.
Reading Guide for I. Bernard Cohen, The Birth of a New Physics and Isaac Newton, Principia / A New
Translation by I. Bernard Cohen and Anne Whitman.
Reading Guide for
Text 3a: I. Bernard Cohen, The Birth of a New Physics. W. W. Norton & Company,
1985.
Tex
STAT2006 Tutorial 2
20th/21th/22th Jan 2015
1. Covariance and correlation, etc.
Tower Property - E(X) = E(E(X|Y ).
Nested Variance formula (EVE) - V ar(X) = E(V ar(X|Y ) + V ar(E(X|Y ).
Covariance - a measure of how much two random variables change tog
STAT 2006 Assignment 2
Due Time and Date: 3 p.m., 17st Feb, 2015
1. Let X1 , X2 , . . . , Xn be random samples from bin(1, p) where 0 < p < 1.
(a) Show that X =
n
i=1
Xi
is an unbiased estimator of p.
n
p(1 p)
(b) Show that Var X =
.
n
p(1 p)(n 1)
(c) Sho
STAT 2006 Assignment 2 Suggested Solution
1. (a)
E[
n
X
2
(Xi X) ] =E[
i=1
=E[
n
X
i=1
n
X
(Xi (X )2 ]
(Xi )2 + (X )2 2(Xi )(X )]
i=1
=E[
n
X
2
(Xi ) 2
n
X
i=1
=E[
n
X
2
2
(Xi ) n(X ) ], Since
n
X
Xi = n
i=1
n
X
(Xi ) = 0
i=1
Pn
V ar(Xi ), E[(X )2 ] = V
STAT2006 Assignment 3 Solution
h
i
s
s
1. Note that x z0.05 n , x + z0.05 n is an approximate 90% confidence interval for . Therefore,
[
x , x + ] is an approximate 90% confidence for if and only if = z0.05 sn . Since z0.05 1.64,
s = 36, = 8, we have the
STAT 2006 Tutorial 9 Solution
5th/7th April 2017
1. For a sample size of n = 27, we will have df = n 1 = 26 degrees of
freedom. For a 95% confidence interval, we have = 0.05, which gives
2.5% of the area at each end of the chi-square distribution. We find
STAT2006 Tutorial 10
11th/12th April 2017
1. Terminologies of Hypothesis Testing
Statistical Hypothesis: A statement about population(s) parameter(s) . There are two
complementary (mutually exclusive) hypotheses in a hypothesis testing problem, namely Nu
Theory of Statistical Inference
Dr. Phillip YAM
2016/2017 Spring Semester
Reference: Chapter 10: Section 7 of Some Theory by Hogg and
Tanis.
Section 10.7 ASYMPTOTIC DISTRIBUTIONS OF
MAXIMUM LIKELIHOOD ESTIMATORS
I
I
Recall that the maximum likelihood esti
STAT 2006 Tutorial 9
5th/7th April 2017
1. Confidence interval for variances
2
We can construct a 100(1 )% confidence interval for X
based on the chi-square distribution
with n 1 degrees of freedom:
"
#
2
2
(n 1)SX
(n 1)SX
,
2 (n 1) 21 (n 1)
2
2
If U 2
STAT2006 Tutorial 8 Solution
28th/29th/31th 2017
1. [5.845, 0.845]
2. [179.148, 607.480]
3. [0, 1.905]
4. (a) The mgf of X is (1 t) , so the mgf of Y =
Pn
i=1 Xi
is
n
X
Pn
1
MY (t) = E exp t
Xi = (1 t) i=1 i , t < .
i=1
Since the mgf uniquely
determines
Statistical Hypothesis Testing
Dr. Phillip YAM
2016/2017 Spring Semester
Reference: Chapter 7 of Tests of Statistical Hypotheses by Hogg
and Tanis.
Section 7.1 Tests about Proportions
I
A statistical hypothesis test is a formal method of making
decisions,
STAT 2006 Assignment 4
Due Date: 24th April, 2013
1. Let X equal the number of male children in a four-child family. Among sudents who were taking
statistics, 79 came from families with four children. For these families, x = 0, 1, 2, 3, and 4 for
13, 22,
STAT 2006 Assignment 1 Suggested Solution
1. (a)
0
(1 FX (x)dx
= [x(1
FX (x)]
0
FX (x)dx
0
0
xd(1 FX (x)
[xFX (x)]0
xfX (x)dx =
0
(b)
0
xfX (x)dx +
=
xdFX (x)
+
0
xfX (x)dx = E[X].
The Jacobian approach: Note Z = X + Y, W = X X = W, Y = Z W .
Hence th
STAT 2006 Assignment 2
Due Time and Date: 3 p.m., 17st Feb, 2015
1. Let X1 , X2 , . . . , Xn be random samples from bin(1, p) where 0 < p < 1.
(a) Show that X =
n
i=1
Xi
is an unbiased estimator of p.
n
p(1 p)
(b) Show that Var X =
.
n
p(1 p)(n 1)
(c) Sho
STAT2006 Assignment 3
Due Date: 17:00, 27th March, 2017
1. A manufacturer sells a light bulb that has a mean life of 1551 hours with a standard deviation of 36
hours. A new manufacturing process is being tested, and there is an interest in knowing the mea
Review, Topics on Multivariate Random
Variables, and Beyond
Dr Phillip YAM
2016/2017 Spring Semester
Reference: Chapters 1 to 5 and Chapter 10 of Probability and
Statistical Inference by Hogg and Tanis.
Preface
After taking STAT2001, you are supposed to b
STAT 2006 Assignment 2
Due Time and Date: 5 p.m., 6th March, 2017
1. Let X1 ,P
X2 , . . . , Xn be randomP
samples from Normal distribution with mean and variance
X = n1 ni=1 Xi and S 2 = n 1 1 ni=1 (Xi X).
P
P
(a) Show that E[ ni=1 (Xi X)2 ] = ni=1 V ar(X
STAT2006 Assignment 3
Due Date: 17:00, 27th March, 2017
1. A manufacturer sells a light bulb that has a mean life of 1551 hours with a standard deviation of 36
hours. A new manufacturing process is being tested, and there is an interest in knowing the mea
STAT 2006 Assignment 2 Suggested Solution
i.i.d.
1. Let X1 , X2 , . . . , Xn Bin(1, p) where 0 < p < 1.
Pn
Pn
E [Xi ]
np
i=1 Xi
is an unbiased estimator of p.
(a) E X = E
= i=1
=
= p. Therefore, X
n
n
n
Pn
Pn
Pn
X
Var
(
X
)
Var (Xi )
np(1 p)
p(1 p)
i