ST102 Outline solutions to Exercise 13 (201415)
as the estimator of the population mean, which is the
1. (a) We use the sample mean, X,
method of moments estimator (MME), the least sqaures estimator (LSE) and also
the maximum likelihood estimator (MLE).
ST102 Outline solutions to Exercise 12 (201415)
1. (a) We have E
k
P
k
P
=
ai Xi
E(ai Xi ) =
i=1
i=1
k
P
ai E(Xi ).
i=1
(b) We have:
Var
k
X
!
ai Xi
= E
i=1
=
k
X
k
X
i=1
a2i E (Xi E(Xi )2 +
i=1
=
k
X
i=1
ai Xi
k
X
!2
ai E(Xi )
i=1
X
= E
k
X
!2
ai (Xi
ST102 Outline solutions to Exercise 3 (201415)
1. (a) The size of the sample space is the total number of ordered sequences with
replacement, which is 65 = 7776. To count the sequences which satisfy the
requirement, it is easiest to simply list them:
1
1
ST102 Outline solutions to Exercise 5 (201415)
1. We have:
p(x) = P (X = x) = P (X x) P (X x 1).
However, the event cfw_X x occurs when all three dice are x and that can occur in x3
ways. Therefore:
P (X x) = x3 /216.
Similarly:
P (X x 1) = (x 1)3 /216.
T
ST102 Outline solutions to Exercise 2 (201415)
1. There is more than one way to answer this question, because the sets can be expressed
in different, but logically equivalent, forms. One way to do so is the following:
(a) A B c C c .
(b) Ac B c C c .
(c)
ST102 Outline solutions to Exercise 8 (201415)
1. These should be simple exercises in using statistical tables of the standard normal distribution. For (a), you can
use the table directly. For (b), you use it for the the standardised
variable Z = (X 2)/
ST102 Outline solutions to Exercise 17 (201314)
1. (a) c = F0.01, 7, 8 = 6.18.
(b) = 1 P (F5, 3 > 28.2) = 1 0.01 = 0.99.
(c) Since P (F6, 10 c) = P (F10, 6 1/c), 1/c = F0.05, 10, 6 = 4.06, c = 0.2463.
2 = 2 against H : 2 6= 2 . Under H , T = S 2 /S 2 F
2.
Summer 2010 examination
ST102
Elementary Statistical Theory
Suitable for all candidates
Instructions to candidates
Time allowed: 3 hours
This paper contains ten questions. Answer not more than THREE questions from
Section A, and not more than THREE questi
2004 examination
ST102
Elementary Statistical Theory
(1 Unit)
Instructions to candidates
Time allowed: 3 hours.
Full marks may be obtained for fully correct answers to FIVE
questions.
Answer not more than THREE questions from Section A and not
more than T
SOLUTIONS to ST102
1.
a)
2005
pairwise independence for all pairs e.g P( A B ) = P( A) P ( B)
plus P( A B C ) = P( A) P ( B) P( C )
Not independent.
Each of the three pairs can be shown to be pairwise
independent but A B C
P( C A B) = 1 P( A B C ) = P( A
1.
a) P( B A) =
P( A B )
provided P( A) > 0.
P( A)
P ( B c Ac ) =
P( Ac B c ) P( A B ) c ) 1 P( A) P( B ) + P( A B )
=
=
P ( Ac )
P ( Ac )
P ( Ac )
Since P ( B A) > P ( B ) P ( A B > P ( A). P ( B )
P ( B c Ac ) >
1 P( A) P( B ) + P( A). P( B ) (1 P( A)(1
[SE
2005 examination
ST102
Elementary Statistical Theory
(1 Unit)
jarred? FOE. Au, ngwﬁ-g)
Instructions to candidates
Time allowed: 3 hours.
Full marks may be obtained for good answers to FIVE questions.
Answer not more than THREE questions from Section A
[SE
2007 examination
ST102
Elementary Statistical Theory
(1 Unit)
Suitable for all Candidates
Instructions to candidates
Time allowed: 3 hours.
Full marks may be obtained for good answers to FIVE questions.
Answer not more than THREE questions from Sectio
ST102 Exercise 1, Outline solutions (201415)
1. (a) Insert your guess here. Chances are that you got the average correct, without
even thinking that it could possibly be anything else. The standard deviation was
probably less obvious.
(b)
i. Using the rul
ST102 Outline solutions to Exercise 11 (201415)
1. Each possible sample has an equal probability of occurrence of 1/6.
is:
distribution of X
(1, 2)
(2, 1)
(1, 3)
(3, 1)
(2, 3)
(3, 2)
=x
Sample mean, X
1.5
2
2.5
=x
Relative frequency, P (X
)
1/3
1/3
1/3
Examiners commentaries 2014
Examiners commentary 2014
ST102 Elementary Statistical Theory
General remarks
Learning outcomes
By the end of this module you should:
be able to summarise the ideas of randomness and variability, and the way in which these lin
ST102
Elementary Statistical Theory
Example workshop:
Multivariate random variables
Dr James Abdey
Department of Statistics
London School of Economics and Political Science
ST102 Elementary Statistical Theory
Dr James Abdey
MT 2014
Example workshop: Mult
ST102 Outline solutions to Exercise 14 (201314)
1. (a) x
= 2.115, and the top 0.5% point of N (0, 1) is z0.005 = 2.576. Hence a 99%
confidence interval for is
x
z0.005 / n = 2.115 2.576 1.1/ 11 = (1.260, 2.969).
(b) The sample median is 1.79, and a plot
ST102 Outline solutions to Exercise 18 (201314)
1. (a) The total number of observations is n = 4 + 8 + 17 + 37 + 62 + 45 + 17 + 10 = 200.
The expected frequencies are calculated according to the fitted Normal distribution
N (54.9, 223.11). Hence the expec
ST102 Outline solutions to Exercise 13 (201314)
1. (a) We use the sample mean as the estimator for the population mean, which is the
MME and also the LSE. For the given sample, it gives a point estimate of
x
=
94 + 100 + 85 + 94 + 92
= 93.
5
(b) The sampl
ST102
Elementary Statistical Theory
Example workshop:
Some common distributions of random variables
Dr James Abdey
Department of Statistics
London School of Economics and Political Science
ST102 Elementary Statistical Theory
Dr James Abdey
MT 2014
Exampl
ST102 Outline solutions to Exercise 11 (201314)
1. The sampling distribution is:
(1, 2)
(2, 1)
(1, 3)
(3, 1)
(2, 3)
(3, 2)
=x
Sample mean, X
1.5
2
2.5
=x
Relative frequency, P (X
)
1/3
1/3
1/3
Possible samples
2. Each combination has equal probability 1
Week 9 MC
1.
High interest rates in the Euro Area periphery in the 1990s were mostly due to
a)
Risk of bank insolvencies
b)
The value of the currency
c)
Tight monetary policy
d) Risk of sovereign default
SOLUTION: During the 1990s the high interest rate i
ST102 Outline solutions to Exercise 16 (201314)
1. (a) We test H0 : X = Y against H1 : X 6= Y . Under H0 , the test statistic
Y
X
N (0, 1).
T =q
2 /n + 2 /m
X
Y
At the 5% significance level we reject H0 if |t| > z0.025 = 1.96. With the given data,
t = 2
ST102
Elementary Statistical Theory
Example workshop:
Introduction to probability theory
Dr James Abdey
Department
of Statistics
London School of Economics and Political Science
ST102 Elementary Statistical Theory
Dr James Abdey
MT 2014
Example workshop:
ST102 Outline solutions to Exercise 12 (201314)
1. (a) E
(b)
k
P
ai Xi
i=1
Var
k
X
i=1
=
k
X
=
k
P
E(ai Xi ) =
i=1
a i Xi
!
a2i E (Xi
k
P
0
= E@
k
X
=
i=1
k
X
ai Xi
i=1
E(Xi )2 +
i=1
k
X
ai E(Xi ).
i=1
ai E(Xi )
i=1
X
!2 1
ai aj E (Xi
0
A = E@
k
X
ai (Xi
Examiners commentaries 2012
Examiners commentary 2012
ST102 Elementary Statistical Theory
General remarks
Learning outcomes
By the end of this module you should:
be able to summarise the ideas of randomness and variability, and the way in which these lin