MATH264: Exercise Sheet 9 Solutions
1. (a) Told that X N (21.37, 0.16), so
20.857 21.37
P (X < 20.857) = P Z <
where Z N (0, 1)
0.4
= P (Z < 1.2825) = P (Z 1.2825) 1 (1.28) = 1 0.8997 = 0.1003
N (21.37, 0.16/100) = N 21.37, 0.042 , so
(b) With n = 100 th

MATH264: Exercise Sheet 2 Solutions
1. For the given random variable X,
E[X] = 0 0.5 + 1 0.25 + 2 0.25 = 0.75
E X 2 = 02 0.5 + 12 0.25 + 22 0.25 = 1.25
2
2
Var[X] = E X 2 (E[X]) = 1.25 (0.75) = 0.6875
p
SD(X) =
Var[X] 0.8292
2. If X = number of heads

MATH264: Exercise Sheet 6 Solutions
1. Marginal probability mass function of X: Pr(X = 1) =
7
16 ,
Pr(X = 2) =
9
16 .
Conditional mass function of Y given that X = 2:
Pr(Y = y | X = 2)
For y = 1, 2, 3, 4,
Pr(X = 2, Y = y)
Pr(X = 2)
=
so that
Pr(Y = 1 | X

MATH264: Exercise Sheet 10 Solutions
1. We know that density of V is
fV (v) =
1 n/2
2
v n/21 ev/2 .
n2
Consider A = V /n. Using transformation method
1 n/2
2
(na)n/21 ena/2 n.
n2
fA (a) = fV (na)n =
Similarly, setting B = U/m, we obtain
fB (b) =
1 m/2

MATH264: Exercise Sheet 4 Solutions
1. (a) Pdf must integrate to 1, and in this case
Z
1
Z
cx3 dx =
f (x) dx =
0
cx4
4
1
=
0
c
4
so must have c = 4.
(b)
f (x)
=
0
4x3
0
x0
0<x<1
x1
Figure 1: The graph of f (x)
(c) For 0 x 1,
x
Z
x
Z
F (x) =
f (u) du =
0

MATH264: Exercise Sheet 1 Solutions
1.
(i) True: Pr(A B) = Pr(A) Pr(B) if A, B are independent.
(ii) False. Correct version: Pr(A B) = Pr(A) + Pr(B) if A, B are mutually exclusive.
(iii) False. Correct version: In a sequence of n independent identical tri

MATH264: Exercise Sheet 5 Solutions
1. Pdf must integrate to 1, and
Z 1
Z
kx(1 x) dx =
0
1
k xx
2
0
2
1
x
x3
k
dx = k
=
2
3
6
0
so k = 6.
Figure 1: The graph of f (x)
(a) Y = 3X + 3.
Since 0 < X < 1, have 0 < Y < 3.
X=
y3
3
and
fY (y)
=
dy
= 3
dx
1
y 3

MATH264: Statistical Theory and Methods II
Lecturer:
Dr. Kai Liu
Room 219, Mathematical Sciences Building
Tel: 0151-7944759
Email: k.liu@liv.ac.uk
Lectures:
Henry Cohen Lecturer Theatre, Duncan Building
Monday 15:00-15:50,
Wednesday 11:00-11:50,
Thursday

MATH264: Exercise Sheet 3 Solutions
1. For a valid probability mass function, probabilities must add up to 1.
k
X
X
p
Pr(X = k) = A
k
k=1
k=1
1
= A ln
= A ln(1 p)
1p
So A = 1/ ln (1 p).
Expectation:
E[X]
X
k
k
p
1
ln(1 p)
k
k=1
!,
X
pk
ln (1 p)
=
=
p
=

MATH264: Exercise Sheet 8 Solutions
1. PGF:
GX (s)
k1
X
2s X s k
1
2
=
= E sX =
sk
3
3
3
3
k=0
k=1
2s
1
=
for 3 < s < 3
3 1 (s/3)
2s
=
3s
(Note range of validity 3 < s < 3.)
To find mean and variance:
G0X (s)
=
G00X (s)
=
E [X]
Var [X]
6
(3 s) 2 2s (1)

MATH264: Exercise Sheet 7
Set: Monday 16th November; Hand in: not later than 4pm, Monday 23th November
1. Discrete random variables X and Y have joint probability mass function given by
Y =1 Y =2 Y =3 Y =4
0.25
0.25
0
0
0
0
0.25
0.25
X=2
X=4
Find the corr

MATH264: Exercise Sheet 8
Set: Monday 23th November; Hand in: not later than 4pm, Monday 30st November
1. Find the PGF of the discrete random variable X with probability mass function
P (X = k)
=
(2/3) (1/3)k1 for k = 1, 2, . . .
Hence find the mean and v

MATH264: Exercise Sheet 1
Set: Monday 5th October; Hand in: not later than 4pm, Monday 12th October
1. For each of the following statements, say whether true or false. For false statements, give the correct
version of the statement.
(i) Pr(A B) = Pr(A) Pr

MATH264: Exercise Sheet 3
Set: Monday 19th October; Hand in: not later than 4pm, Monday 26th October
1. Random variable X is said to follow the logarithmic distribution with parameter p, where 0 < p < 1,
if X has probability mass function
k
p
Pr(X = k) =

MATH264: Exercise Sheet 4
Set: Monday 26rd October; Hand in: not later than 4pm, Monday 2rd November
1. For the probability density function
f (x)
= cx3
(0 < x < 1)
(a) find the value of the constant c;
(b) sketch f (x);
(c) obtain the distribution functi

MATH264: Exercise Sheet 9
Set: Monday 30st November; Hand in: not later than 4pm, Monday 7th December
1. A sweet maker produces mints whose weights are normally distributed with mean 21.37 and variance 0.16.
(a) Let X denote the weight of a single mint se

MATH264: Exercise Sheet 6
Set: Monday 9rd November; Hand in: not later than 4pm, Monday 16th November
1. Discrete random variables X and Y have joint probability mass function given by
X=1
X=2
Y =1 Y =2 Y =3 Y =4
2/32
3/32
4/32
5/32
3/32
4/32
5/32
6/32
Ob

MATH264: Exercise Sheet 2
Set: Monday 12th October; Hand in: not later than 4pm, Monday 19th October
1. For a discrete random variable X with Pr(X = 0) = 0.5, Pr(X = 1) = 0.25 and Pr(X = 2) = 0.25,
compute the expectation E[X] and the standard deviation o

MATH264: Exercise Sheet 10
Set: Monday 7th December; Hand in: not later than 4pm, Monday 14th December
1. Suppose U and V are independent chi-square random variables with m and n degrees of freedom,
respectively. Show that the random variable W defined by

MATH264: Exercise Sheet 5
Set: Monday 2rd November; Hand in: not later than 4pm, Monday 9th November
1. Suppose X has pdf
fX (x)
=
kx(1 x)
0
0<x<1
otherwise
Evaluate the constant k.
Sketch the pdf.
Find the range of possible Y values and the pdf fY (y) of

Estimating the error variance
The distributions of the estimators b0 , b1 each have variance
proportional to the unknown parameter 2
To compute confidence intervals or carry out hypothesis tests for
0 , 1 , we need to estimate 2
In a similar way to ANOVA,

Comparison of k population means: ANOVA
Example:
Activity levels (measured in non-standard units) of rats on various
doses of THC (tetrahydrocannabinol, the major active ingredient
of marijuana).
ANOVA Example
ni
x
i
Placebo
30
27
52
38
20
26
8
41
49
49
1

Hypothesis testing
Formulate a null hypothesis H0 and an alternative H1 , e.g.
H0 :
=0
H1 :
6= 0
Hypothesis testing
Formulate a null hypothesis H0 and an alternative H1 , e.g.
H0 :
=0
H1 :
6= 0
Collect the data and calculate an appropriate test stati

ANOVA hypothesis test
Want to test whether group means i are all the same.
ANOVA hypothesis test
Want to test whether group means i are all the same.
That is, see if differences between group sample means yi are big
enough to be statistically significant.

Statistical inference
Two kinds of question:
1. Estimation
Statistical inference
Two kinds of question:
1. Estimation
Point estimation
Confidence interval
Statistical inference
Two kinds of question:
1. Estimation
Point estimation
Confidence interval

Simple Linear Regression
Estimating the parameters 0 , 1 is useful, but not enough.
Simple Linear Regression
Estimating the parameters 0 , 1 is useful, but not enough.
Also need confidence intervals for 0 , 1 , and need to be able to
test hypotheses such

ANOVA assumptions
Samples are independent
Responses are Normally distributed
Each group has same variance
ANOVA assumptions
Samples are independent
Responses are Normally distributed
Each group has same variance
How to check assumptions?
Robustness?

Two sample tests: Independent samples
Now suppose want to compare two independent samples, from
different populations.
Do the population means differ?
Sample 1: X11 , . . . , X1n1
Sample 2: X21 , . . . , X2n2
1, X
2
Sample sizes n1 , n2 with sample means