Example 54.
Consider the statistical inferences given in the regression output for the peak
particle velocity. Conduct two hypothesis tests: one for the constant term of the regression
line, and the other for the slope of the regression line.
For the cons
8.3 Multiple Comparisons
Anova will not be able to tell you which groups are dierent. To do this, we use multiple
comparisons - pairwire t tests. Using the Bonferonni adjusted t test we get df=n-k
xs xt
T =
The level of signcance is given by
1
M SE( ns +
MATH2801/2901 Solutions to Assignment 1 Random variables
1. Moment generating functions (mgfs) contain more information about probability distributions than just the moments. For example, we can nd the (so-called) negative
moments of a non-negative random
Solutions to Assignment 2 Transformations, Convergence, Inference
1. Benfords Law is named after an American engineer Frank Benford, who in 1938 set
out to nd the relative frequency of the leading digit in naturally occurring numbers.
He was curious about
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2010
Assignment 1 Random variables
Important skills to demonstrate:
Ability to use key theoretical tools to derive and explore the
properties of random variables.
Q1a,b
/6
Q1c*
/2*
Demonstrated und
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2010
Assignment 2 Transformations, Convergence, Inference
Please include this cover sheet with your submission.
Submission date: Monday lecture, 10am, week 10.
Assignment length: No more than 8 pages
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2009
Chapter 2 exercises solutions
1. end line
Distribution
Discrete?
Parameters
fX (x)
px (1
Bernoulli
Bin(n, p)
Geometric
yes
yes
yes
p
n, p
p
Hypergeometric
yes
n, m, N
Poisson
Exponential
yes
no
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2009
Chapter 1 exercises solutions
1.
fX (1) = P (X = 1) = P (winner found in rst round)
= P (dierent items in rst round)
= 1 P (same items in rst round)
= 1 P (both rock or both paper or both scisso
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2009
Chapter 3 exercises solutions
1. (a) The probability function of X is
1
fX (x) =
fX,Y (x, y).
y=1
This leads to:
x
fX (x)
0
1
0.1 0.5
2
0.4
y
1 0
fY (y) 0.1 0.8
1
0.1
Similarly,
(b) Hence:
2
E(X
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2009
Chapter 5 exercises solutions
1. X N (68, 162 ).
(a) P (X > 80) = P (Z > 0.75) = 0.2266
(b)
i. 18 68 = 1224
ii. 18 162 = 4608
67.88
iii. Y18 N (1224, 4608)
iv. P (Y18 > 1224) = P (Z > 0) = 0.5
(
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2009
Chapter 6 exercises solutions
1. Each of the required distributions can be found using results such as that for
the sum of linear combinations and sum of squares of normal random variables.
For
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2009
Chapter 4 exercises solutions
1. The transformation is dened by the relationship
y = x1/3 .
The inverse transformation is then given by
x = y 3 =
dx
= 3y 2 .
dy
Hence,
fY (y) = fX (x)
dx
= 2y 3
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2009
Chapter 9 exercises solutions
1. (Step 1 in notes: state the hypotheses to be tested.)
The hypotheses to be tested are:
H0 : = 78.1 versus H1 : = 78.1
=mean % nitrogen in the Cretaceous atmosph
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2009
Chapter 8 exercises solutions
1. (a) Let X be a Poisson() random variable. From results for Poisson random
variables, E(X) = so the method of moments estimator for is
= X.
(b) The log-likelihoo
THE UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
June 2008 Examination
MATH2901
HIGHER THEORY OF STATISTICS
(1) TIME ALLOWED - 2 hours
(2) TOTAL NUMBER OF QUESTIONS - 4
(3) ANSWER ALL QUESTIONS
(4) THE QUESTIONS ARE OF EQUAL VALUE
(5
9 Correlation & Regression
9.1 Correlation
A multivariate data set consists of measurements or observations on each of two or more
variables. One important special case, bivariate data, involves only two variables,
x
and
y.
We can represent bivariate data
9.4 Inuential Observations
An outlier is a point in the scatterplot which falls outside the overall pattern suggested by the
other variables. An
observation is inuential for statistical calculation if removing it would
markedly changed the result of the c
8 ANOVA
ANOVA seeks to test the signcance of dierences between three of more sampling means, or
equivalently, to test the null hypothesis that the sample means are all equal. The assumptions
with this test include the normality assumption and constant var
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2009
Chapter 6 exercises solutions
1. Each of the required distributions can be found using results such as that for
the sum of linear combinations and sum of squares of normal random variables.
For
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2009
Chapter 0 exercises solutions
1. (a) A B C
(b) A B C
(c) (A B) (A C) (B C)
(d) (A B C) (A B C) (A B C)
(e) (A B C) (A B C) (A B C)
2. P (B|A) is larger.
3. (a)
19
45
(b)
11
25
(c)
6
11
4. (a) 0.
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2011
Chapter 9 exercises solutions
1.
H0 : = 78.1 versus H1 : 6= 78.1
=mean % nitrogen in the Cretaceous atmosphere.
The test statistic we will use is
T =
X 78.1
S/ 9
which has a t8 distribution if
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2009
Chapter 5 exercises solutions
1. X N (68, 162 ).
(a) P (X > 80) = P (Z > 0.75) = 0.2266
(b)
i. 18 68 = 1224
ii. 18 162 = 4608 ' 67.88
iii. Y18 N (1224, 4608)
iv. P (Y18 > 1224) = P (Z > 0) = 0.5
MATH2801/2901 (Higher) Theory of Statistics
Semester 1, 2011
Chapter 8 exercises solutions
1. (a) E(X) = so the method of moments estimator for is
= X.
(b)
`() = ln
n
Y
e Xi
Xi !
i=1
= n + ln()
n
X
Xi
i=1
n
X
ln(Xi !).
i=1
2
Differentiating and solving
Consumer behaviour refers to the buying behaviour consumers. Through its survey Nike
needs to analyze buying behaviour for:
Buyers reactions to its marketing strategy will reflect on the firms success
ould help Nike to determine:
Nature and scope of targe