DATA ANALYSIS
Module Code: CA660
STRUCTURE of Investigation/DA
1,2, many samples
Replication,
E.D., Regn., C.T.
Assays, Counts
Estimation/H.T.
H.T.
Study techniques
Lab. techniques
Non-Parametric
Parametric
Distributional Assumptions, Probability
Basis: Size/Type of Data Set
DESCRIPTIVE
ORDERED
2
‘BIO’ CONTEXT here
• GENETICS : 5 branches; aim = ‘Laws’ of Chemistry,
Physics, Maths. for Biology
• GENOMICS : Study of Genomes (complete set of
DNA carried by Gamete) by integration of 5 branches
of Genetics with ‘Informatics and Automated systems’
• PURPOSE of GENOME RESEARCH : Info. on
Structure, Function, Evolution of all Genomes – past
and present
• Techniques of Genomics from molecular,
quantitative, population genetics: Concepts and
Terminology from Mendelian genetics and
cytogenetics
3
CONTEXT: GENETICS - BRANCHES
• Classical Mendelian – Gene and Locus, Allele,
Segregation, Gamete, Dominance, Mutation
• Cytogenetics – Cell, Chromasome, Meiosis and
Mitosis, Crossover and Linkage
• Molecular – DNA sequencing, Gene Regulation
and Transcription, Translation and Genetic Code
Mutations
• Population – Allelic/Genotypic Frequencies,
Equilibrium, Selection, Drift, Migration, Mutation
• Quantitative – Heritability/Additive, Non-additive
Genetic Effects, Genetic by Environment
Interaction, Plant and Animal Breeding
4
CONTEXT+ : GENOMICS -LINKAGES
Mendelian
Cytogenetics
Molecular
GENOMICS
Genetic markers
DNA Sequences
Linkage/Physical Maps
Gene Location
Population
QTL Mapping
Quantitative
5
GENOMICS - FOCUS
CLASSICAL
Genetic Markers,
Linkage Analysis, Gene
Ordering, Multipoint
Analysis, Genetic and
QTL mapping
INFORMATICS
Databases, Sequence
Comparison,Data
Communications,
Automation
DNA SEQUENCE
ANALYSIS
Sequence Assembly,
Placement, Comparison
6
GENOMICS – some KEY QUESTIONS
• HOW do Genes determine total phenotype?
• HOW MANY functional genes necessary and
sufficient in a given system?
• WHAT are necessary Physical/Chemical aspects of
gene structure?
• IS gene location in Genome specific?
• WHAT DNA sequences/structures are needed for
gene-specific functions?
• HOW MANY different functional genes in whole
biosphere?
• WHAT MEASURES of essential DNA sameness in
different species?
7
‘DATA’ : STATISTICAL GENOMICS
Some UNUSUAL/SPECIAL FEATURES
• Size – databases very large e.g. molecular marker
and DNA / protein sequence data
• Mixtures of variables - discrete/continuous e.g.
combination of genotypes of genetic markers (D) and
values quantitative traits (C)
• Empirical Distributions needed for some Test
Statistics e.g. QTL analysis, H.T. of locus order
• Intensive Computation e.g. Linkage Analysis, QTL
and computationally greedy algorithms in locus
ordering, derivation of empirical distributions etc.
• Likelihood Analysis - Linear Models typically
insufficient alone
8
EXAMPLE – Mendelian Genetics Cytogenetics
GENE = unit of heredity. Single gene passed between
generations by Mendelian Inheritance
DIPLOID Individual - two copies (alleles) of a gene (A)
HOMOZYGOSITY
AA aa (genotypes)
HETEROZYGOSITY
Aa
(genotype)
(multiple alleles possible for a gene)
PHENOTYPE -appearance /measurement gene
characteristic
- AA,Aa,aa
(co-dominant)
- AA,Aa same (A the dominant allele)
9
Example – contd.
Common Mating schemes
Single gene haploid gene cell is a gamete
AA
A(gamete) while aa a
F1 hybrid gives diploid genotype Aa
• If F1 x 1 parent (AA or aa)
Backcross
• If F1 x F1 = (self-pollination - or sibs if 2 sexes)
continuing this example
F2
10
Mendelian Laws
1. Segregation Single gene trait, simple heredity
Genotypic segregation ratio 1:1 co-dominant(Backcross)
G.S.R.
1:2:1 (F2)
Phenotypic (P.S.R.)
3:1 (dominant alleles in F2)
2. Independent Assortment (Inheritance of unlinked multiple
genes). Each pair of alleles of a gene segregate independently of the
segregation of alleles of another gene)
e.g. A, B ; 2 alleles for both
9 genotypes F2 progeny
AABB, AABb, AAbb, AaBB, AaBb, Aabb,aaBB,aaBb & aabb
Expect G.S.R. = [1:2:1]2 = 1:2:1:2:4:2:1:2:1
For Dominant genes there are 4 phenotypes, thus
P.S.R. = 9:3:3:1
(i.e. A_B_, A_bb, aaB_, aabb)
where _ either dominant or recessive
11
EXPERIMENTAL OUTCOMES
Estimation of Expected “frequency” of specific
genotype/ phenotype in population
e.g. Frequency A_b_ in F2 = 9/16 in previous example
e.g. 4 independent loci
AaBBccDd in F2
(with parental groups of cross AAbbCCdd & aaBBccDD)
P{AaBBccDd} = (1/2)(1/4)(1/4)(1/2) = 1/64
Note: Basis for estimation/testing how closely observed
fits expected segregation = chi-squared (2)
12
Genotypes/phenotypes for genes
in F2 progeny under Hardy-Weinberg m
Genotype
Expected
frequency

pi

p1= 0.25
p2= 0.25
p3= 0.25
p4= 0.25
p1= 0.3
p2= 0.3
p3= 0.2
p4= 0.2
p1= 0.4
p2= 0.4
p3= 0.1
p4= 0.1
p1= 0.4
p2= 0.3
p3= 0.2
p4= 0.1
p1= 0.7
p2= 0.1
p3= 0.1
p4= 0.1
A1A1
p1p1
0.0625
0.09
0.16
0.16
0.49
A1A2
2p1p2
0.125
0.18
0.32
0.24
0.14
A1A3
2p1p3
0.125
0.12
0.08
0.16
0.14
A1A4
2p1p4
0.125
0.12
0.08
0.08
0.14
A2A2
p2p2
0.0625
0.09
0.16
0.09
0.01
A2A3
A2A4
A3A3
A3A4
A4A4
2p2p3
2p2p4
p3p3
2p3p4
p4p4
0.125
0.125
0.0625
0.125
0.0625
0.12
0.12
0.04
0.08
0.04
0.08
0.08
0.01
0.02
0.01
0.12
0.06
0.04
0.04
0.01
0.02
0.02
0.01
0.02
0.01
pH
0.75
0.74
0.66
0.70
0.48
13
Mechanisms of MENDELIAN HEREDITY:
• Cell division - mitosis, meiosis
• Genetic Linkage = association of genes on same
chromasome
Effects: Seg. Ratios no longer Mendelian. Result of
recombination (meiosis) is the existence of non-parental
chromasomes in cellular meiotic products).
Each crossover (exchange of chromasomal segments
between homologs) gives 2 reciprocal recombinant (nonparental) gametes. Measurement : recombinant fraction
• -Recombination generally random on chromasomes:
recombination between loci associated with distance apart.
(Basic premise - genetic/genomic mapping).
• Models: Linkage phase (co-dominance, experimental
data), factors affecting recombination
14
Example
• For 2 loci, A and B, same chromasome:
A a
B b
segregation - two alleles each locus
Ab, Ab, aB, ab gametes by meiosis
if AB, ab
possible Parents
Ab, aB
non-parental recombinants
• Sampling from population, observe nr recombinant
gametes ( Ab and aB) out of total of n samples
Recombinant Fraction
r = nr/n
Notes: Usually observe phenotypic rather than gamete frequencies.
Estimation of R.F. using phenotypic data involves constructing the
likelihood and estimation using maximum likelihood
More than 2 or 3 loci - several R.F. & crossover interference;
complex so use a Mapping Function, e.g. based on Poisson.
15
Population Genetics
Focus - frequencies, distributions, origins of genes in
populations & changes -due to mutation, migration, selection
Allelic frequency (Prob) of crossover between two parents
Cross
Possible
Ai (Allelic frequencies (Probs.))
abxcd
abxcc
abxab
abxaa
Aaxaa
4
3
2 (F2)
Backcross
Fixed
A1(a)
0.25
0.25
0.5
0.75
1.0
A2(b)
0.25
0.25
0.5
0.25
0
A3(c)
0.25
0.5
0
0
0
A4 (d)
0.25
0
0
0
0
16
Quantitative Genetics
• Focus - inheritance of quantitative traits. As number of genes
controlling a trait increases, as effects on phenotype increase, ability
to model through Mendelian inheritance diminishes.
• Single Gene Model
Single-locus A two alleles, A and a, so 3 possible
genotypes AA, Aa & aa.
Three values a,d, and -a assigned arbitrarily to the 3
genotypes. Population assumed in Hardy-Weinberg equilibrium
(gene and genotypic frequencies constant, generation to generation).
The two alleles have frequencies of p and (1-p) = q. Set up :
aa
AA
Aa
-a
d
a
where population mean - in terms of allelic frequencies and genotypic
values is
(probability primer)
  p 2 a  2 pqd  q 2 a
17
Quantitative Genetics measures - examples
Deviation genotypic value from mean of population (a - )
Average effect of gene substitution
Breeding Value = average genotypic value of progeny
Dominance deviation (D.D.) = part of genotypic value not
explained by breeding value.
• Total Genetic Variance in a Population = Variance of
genotypic values = Sum of variances for B.V. and D.D.
•
•
•
•
• Heritability = ratio of genotypic/phenotypic variances
• Trait Models (e.g. Linear Model of what influences a trait)
yij    Gi  ij
where yij is trait type for genotype i in replication j,  the population
mean, Gi the genetic effect for i and ij the error term associated with
genotype i in replication j
18
Probability & Statistics Primer
-appendix, if needed
Note: Short overview. Other statistical distributions in lectures
Summary Statistics- Descriptive
When analysing practical sets of data, it is useful to be able to define a small
number of values that summarise the main features present. We will derive (i)
representative values, (ii) measures of spread and (iii) measures of skewness and
other characteristics.
Representative Values
These are sometimes called measures of location or measures of central tendency.
1. Random Value
Given a set of data S = { x1, x2, … , xn }, we select a random number, say k, in the
range 1 to n and return the value xk. This method of generating a representative
value is straightforward, but it suffers from the fact that extreme values can occur
and successive values could vary considerably from one another.
2. Arithmetic Mean
For the set S above, the arithmetic mean (or just mean) is
x = {x1 + x2 + … + xn }/ n.
If x1 occurs f1 times, x2 occurs f2 times and so on, we get the formula
x = { f1 x1 + f2 x2 + … + fn xn } / { f1 + f2 + … + fn } ,
written
x =  fx / f
Example 1.
The data refers to the marks that students in a class obtained in an examination. Find the
average mark for the class. The first
point to note is that the marks are presented as Mark
Mid-Point Number
ranges, so we must be careful in our
of Range of Students
interpretation of the ranges. All the intervals
xi
fi
fi xi
must be of equal rank and their must be no
gaps in the classification. In our case, we
0 - 19
10
2
20
interpret the range 0 - 19 to contain marks
21 - 39
30
6
180
greater than 0 and less than or equal to 20.
40 - 59
50
12
600
Thus, its mid-point is 10. The other intervals
60 - 79
70
25
1750
are interpreted accordingly.
80 - 99
90
5
450
Sum
50
3000
The arithmetic mean is x = 3000 / 50 = 60 marks.
Note that if weights of size fi are suspended
from a metre stick at the points xi, then the
average is the centre of gravity of the
distribution. Consequently, it is very sensitive
to outlying values.
x1
x2
f1
x
xn
fn
f2
Equally, the population should be homogenous for the average to be meaningful. For
example, if we assume that the typical height of girls in a class is less than that of boys,
then the average height of all students is neither representative of the girls or the boys.
3. The Mode
Frequency
50
This is the value in the distribution that occurs
most frequently. By common agreement,
it is calculated from the histogram using linear
interpolation on the modal class.
13 20
25
13
The various similar triangles in the diagram
generate the common ratios. In our case,
the mode is
60 + 13 / 33 (20) = 67.8 marks.
20
12
6
2
20
5
40
60
80
40
60
80
100
4. The Median
Cumulative
50
This is the middle point of the distribution. It
is used heavily in educational applications. If
{ x1, x2, … , xn } are the marks of students in a
class, arranged in nondecreasing order, then 25.5
the median is the mark of the (n + 1)/2 student.
It is often calculated from the ogive or
cumulative frequency diagram. In our case,
the median is
60 + 5.5 / 25 (20) = 64.4 marks.
Frequency
20
100
Measures of Dispersion or Scattering
Example 2. The following distribution has the same
arithmetic mean as example 1, but the values are more
dispersed. This illustrates the point that an average
value on its own may not be adequately describe
statistical distributions.
To devise a formula that traps the degree to which a
distribution is concentrated about the average, we
consider the deviations of the values from the average.
If the distribution is concentrated around the mean,
then the deviations will be small, while if the distribution
is very scattered, then the deviations will be large.
The average of the squares of the deviations is called
the variance and this is used as a measure of dispersion.
The square root of the variance is called the standard
deviation and has the same units of measurement as
the original values and is the preferred measure of
dispersion in many applications.
Marks
x
Frequency
f
10
30
50
70
90
Sums
6
8
6
15
15
50
fx
60
240
300
1050
1350
3000
x6
x5
x4
x3
x2
x1
x
Variance & Standard Deviation
s2 
VAR[X] = Average of the Squared Deviations
= S f { Squared Deviations } / S f
= S f { xi - x } 2 / S f
= S f xi 2 / S f - x 2
, called the product moment formula.
s  Standard Deviation =  Variance
Example 1
f
x
2
10
6
30
12
50
25
70
5
90
50
fx
20
180
600
1750
450
3000
f x2
200
5400
30000
122500
40500
198600
VAR [X] = 198600 / 50 - (60) 2
= 372 marks2
Example 2
f
x
6
10
8
30
6
50
15
70
15
90
50
fx
60
240
300
1050
1350
3000
VAR [X] = 217800 / 50 - (60)2
= 756 marks2
f x2
600
7200
15000
73500
121500
217800
Other Summary Statistics
Skewness
An important attribute of a statistical distribution relates to its degree of symmetry. The word
“skew” means a tail, so that distributions that have a large tail of outlying values on the righthand-side are called positively skewed or skewed to the right. The notion of negative
skewness is defined similarly. A simple formula for skewness is
Skewness = ( Mean - Mode ) / Standard Deviation
which in the case of example 1 is:
Skewness = (60 - 67.8) / 19.287 = - 0.4044.
Coefficient of Variation
This formula was devised to standardise the arithmetic mean so that comparisons can be
drawn between different distributions.. However, it has not won universal acceptance.
Coefficient of Variation = Mean / standard Deviation.
Semi-Interquartile Range
Just as the median corresponds to the 0.50 point in a distribution, the quartiles Q1, Q2, Q3
correspond to the 0.25, 0.50 and 0.75 points. An alternative measure of dispersion is
Semi-Interquartile Range = ( Q3 - Q1 ) / 2.
Geometric Mean
For data that is growing geometrically, such as economic data with a high inflation effect, an
alternative to the the arithmetic mean is preferred. It involves getting the root to the power
N = S f of a product of terms
Geometric Mean = N x1f1 x2 f2 … xk fk
Regression
[Example 3.] As a motivating example, suppose we are modelling sales data over time.
SALES
3
5
4
5
6
7
TIME
1990
1991
1992
1993
1994
1995
Y=mX+c
We seek the straight line “Y = m X + c” that best
Y
Yi
approximates the data. By “best” in this case, we
mean the line which minimizes the sum of squares
of vertical deviations of points from the line:
m Xi + c
SS = S ( Yi - [ mXi + c ] ) 2.
Setting the partial derivatives of SS with respect to m
and c to zero leads to the “normal equations”
0
X
S Y = m S X + n .c
, where n = # points
S X .Y= m S X2 + c S X .
Xi
Let 1990 correspond to Year 0.
X.X
0
1
4
9
16
25
X
0
1
2
3
4
5
X.Y
0
5
8
15
24
35
Y
3
5
4
5
6
7
Y.Y
9
25
16
25
36
49
55 15
87
30
160
Sales
10
5
Time
0
5
Example 3 - Workings.
The normal equations are:
30 = 15 m + 6 c
87 = 55 m + 15 c
=>
24 = 35 m=>
=>
150 = 75 m + 30 c
174 = 110 m + 30 c
30 = 15 (24 / 35) + 6 c
=>
c = 23/7
Thus the regression line of Y an X is
Y = (24/35) X + (23/7)
and to plot the line we need two points, so
X=0
=> Y = 23/7
and X = 5 => Y = (24/35) 5 + 23/7 = 47/7.
It is easy to see that ( X, Y ) satisfies the normal equations, so that the regression
line of Y on X passes through the “Center of Gravity” of the data. By expanding
terms, we also get
S ( Yi - Y ) 2 = S ( Yi - [ m Xi + c ] ) 2
+
S ( [ m Xi + c ] - Y ) 2
Y
Yi
Total SumErrorSum
Regression Sum
of Squares
of Squares
SST
=
SSE
+
of Squares
SSR
Y
In regression, we refer to the X variable as the independent
variable and Y as the dependent variable.
mXi +C
Y
X
X
Correlation
The coefficient of determination r2 ( which takes values in the range 0 to 1) is a
measure of the proportion of the total variation that is associated with the regression
process:
r2
=
SSR/ SST
=
1 - SSE / SST.
The coefficient of correlation r ( which takes values in the range -1 to +1 ) is more
commonly used as a measure of the degree to which a mathematical relationship exists
between X and Y. (Dashed lines show association).It can be calculated from the formula:
r
=
(X-X)(Y-Y)

=
( X - X )2 ( Y - Y ) 2
nXY-XY
 [{ n X 2 - ( X ) 2 } { n  Y 2 - ( Y ) 2 }]
Example. In our case r = {6(87) - (15)(30)}/  { 6(55) - (15)2 } { 6 (160) - (30)2 } = 0.907.
r=-1
r=0
r=+1
Collinearity
If the value of the correlation coefficient is greater than 0.9 or less than - 0.9, we
would take this to mean that there is a mathematical relationship between the
variables. This does not imply that a cause-and-effect relationship exists.
Consider a country with a slowly changing population size, where a certain political
party retains a relatively stable per centage of the poll in elections. Let
X = Number of people that vote for the party in an election
Y = Number of people that die due to a given disease in a year
Z = Population size.
Then, the correlation coefficient between X and Y is likely to be close to 1, indicating
that there is a mathematical relationship between them (i.e.) X is a function of Z and
Y is a function of Z also. It would clearly be silly to suggest that the indicence of the
disease is caused by the number of people that vote for the given political party.
This is known as the problem of collinearity.
Spotting hidden dependencies between distributions can be difficult. Statistical
experimentation can only be used to disprove hypotheses, or to lend evidence to
support the view that reputed relationships between variables may be valid. Thus,
the fact that we observe a high correlation coefficient between deaths due to heart
failure in a given year with the number of cigarettes consumed twenty years earlier
does not establish a cause-and-effect relationship. However, this result may be of
value in directing biological research.
Overview of Probability Theory
In statistical theory, an experiment is any operation that can be replicated infinitely often and
gives rise to a set of elementary outcomes, which are deemed to be equally likely. The
sample space S of the experiment is the set of all possible outcomes of the experiment. Any
subset E of the sample space is called an event. We say that an event E occurs whenever
any of its elements is an outcome of the experiment. The probability of occurrence of E is
P {E} =
Number of elementary outcomes in E
Number of elementary outcomes in S
S
E
The complement E of an event E is the set of all elements that belong to S but not to E. The
union of two events E1  E2 is the set of all outcomes that belong to E1 or to E2 or to both. The
intersection of two events E1  E2 is the set of all events that belong to both E1 and E2.
Two events are mutually exclusive if the occurrence of either precludes the occurrence of the
 events are independent if the occurrence
other (i.e) their intersection is the empty set . Two
of either is unaffected by the occurrence or non-occurrence of the other event.
Theorem of Total Probability.
P {E1  E2} = P{E1} + P{E2} - P{E1 E2}
S
n = n0, 0 + n1, 0 + n0, 2 + n1, 2
E1
n1, 0
E2
n1, 2
P{E1  E2} = (n1, 0 + n1, 2 + n0, 2) / n
n0, 2
= (n1, 0 + n1, 2) / n + (n1, 2 + n0, 2) / n - n1, 2 / n
n0, 0
= P{E1} + P{E2} - P{E1 E2}
Corollary.
If E1 and E2 are mutually exclusive, P{E1  E2} = P{E1} + P{E2} - see Axioms and Addition Rule
Proof.
The probability P{E1 | E2} that E1 occurs, given that E2 has occurred (or must occur) is called the
conditional probability of E1. Note that in this case, the only possible
S outcomes of the
experiment are confined to E2 and not to S.
E2
E1
Theorem of Compound Probability Multiplication Rule.
n1, 0
n1, 2
P{E1  E2} = P{E1 | E2} * P{E2}.
n0, 2
Proof.
P{E1  E2} = n1, 2 / n
n0, 0
= {n1, 2 / (n1, 2 + n0, 2) } * { n1, 2 + n0, 2) / n}
Corollary
If E1 and E2 are independent, P{E1  E2} = P{E1} * P{E2}. Special case of Multiplication Rule
Note: If an E itself compound, expands further = Chain Rule: P{E7  E8  E9} =P{E7  (E8  E9)}
Ability to count possible outcomes in an event is crucial to calculating probs. By a permutation of
size r of n different items, we mean an arrangement of r of the items, where the order of the
arrangement is important. If the order is not important, the arrangement is called a combination.
Example. There are 5*4 permutations and 5*4 / (2*1) combinations of size 2 of A, B, C, D, E
Permutations:
AB, BA, AC, CA, AD, DA, AE, EA
CD, DC, CE, EC
BC, CB, BD, DB, BE, EB
DE, ED
Combinations:
AB, AC, AD, AE, BC, BD, BE, CD, CE, DE
Standard reference books on probability theory give a comprehensive treatment of how these
ideas are used to calculate the probability of occurrence of the outcomes of games of chance.
Bayes’ Rule (Theorem): For a series of mutually exclusive and exhaustive
events Br, where union of the Br = B1  B2  B3 …….Br = all possibilities for B,
Then:
PA | Bs  PBs 
PBs | A 
 PA | Br  PBr 
r
Where denominator is the Total probability of A occurring.
Ex. Paternity indices: based on actual genotypes of mother, child, and
alleged father. Before collection of any evidence have a prior probability of
paternity P{C}. So, what is the situation after the genetic evidence E is in?
From Bayes’: P {man is father | E}
P{man not father | E}
P[E | man is father}
=
P{E | man not father}

P{man is father}
P{man not father}
Written in terms of ratio of posterior probs. (= LHS), paternity index (L say) and
prior probs. (RHS). Rearrange and substitute in above to give prob. of an
alleged man with
particular genotype
being the true father
PC | E 
PL PC 
L  P  (1  PC )
NB: L is a way of ‘weighting’ the genetic evidence; the issue is setting a prior.
Statistical Distributions- Characterisation
If a statistical experiment only gives rise to real numbers, the outcome of the experiment is
called a random variable. If a random variable X
takes values
X1, X2, … , Xn
with probabilities
p1, p2, … , pn
then the expected (average) value of X is defined to be
n
E[X] =
 p j Xj
j 1
and its variance is
n
VAR[X] = E[X2] - E[X]2 =
 pj Xj2 - E[X]2.
j 1
Example. Let X be a random variable measuring
the distance in Kilometres travelled by children
to a school and suppose that the following data
applies. Then the mean and variance are
E[X]
= 5.30 Kilometres
VAR[X] = 33.80 - 5.302
= 5.71 Kilometres2
Prob. Distance
pj
Xj
pj Xj
pj Xj2
0.15
2.0
0.30 0.60
0.40
4.0
1.60 6.40
0.20
6.0
1.20 7.20
0.15
8.0
1.20 9.60
0.10
10.0
1.00 1.00
1.00
5.30 33.80
Similar concepts apply to continuous distributions. The distribution function is defined by
F(t) = P{ X  t}
and its derivative is the frequency function
f(t) = d F(t) / dt
t
so that
F(t) =


f(x) dx.
Sums and Differences of Random Variables
Define the covariance of two random variables to be
COVAR [ X, Y]
= E [(X - E[X]) (Y - E[Y]) ] = E[X Y] - E[X] E[Y].
If X and Y are independent, COVAR [X, Y] = 0.
Lemma
E[ X + Y] = E[X] + E[Y]
VAR [ X + Y]
= VAR [X] + VAR [Y] + 2 COVAR [X, Y]
E[ k. X] = k .E[X] VAR[ k. X] = k2 .E[X] for a constant k.
Example. A company records the journey time X
of a lorry from a depot to customers and
the unloading times Y, as shown.
E[X]
= {1(10)+2(13)+3(17)+4(10)}/50 = 2.54
E[X2]
= {12(10+22(13)+32(17)+42(10)}/50 = 7.5
VAR[X] = 7.5 - (2.54)2 = 1.0484
E[Y]
= {1(20)+2(19)+3(11)}/50 = 1.82
VAR[Y] = 3.9 - (1.82)2 = 0.5876
E[X+Y]
E[(X + Y)2]
VAR[(X+Y)]
X=1
Y =1
7
2
2
3
1
Totals
10
2
5
6
2
13
3
4
8
5
17
4
4
3
3
10
Totals
20
19
11
50
E[Y2] = {12(20)+22(19)+32(11)}/50 = 3.9
= { 2(7)+3(5)+4(4)+5(4)+3(2)+4(6)+5(8)+6(3)+4(1)+5(2)+6(5)+7(3)}/50 = 4.36
= {22(7)+32(5)+42(4)+52(4)+32(2)+42(6)+52(8)+62(3)+42(1)+52(2)+62(5)+72(3)}/50 = 21.04
= 21.04 - (4.36)2 = 2.0304
E[X Y]
= {1(7)+2(5)+3(4)+4(4)+2(2)+4(6)+6(8)+8(3)+3(1)+6(2)+9(5)+12(3)}/50 = 4.82
COVAR (X, Y) = 4.82 - (2.54)(1.82) = 0.1972
VAR[X] + VAR[Y] + 2 COVAR[ X, Y] = 1.0484 + 0.5876 + 2 ( 0.1972) = 2.0304
Standard Statistical Distributions
Most elementary statistical books provide a survey of commonly used statistical
distributions.
Importantly, we can characterise them by their expectation and variance (as for random
variables) and by the parameters on which these are based; (see lecture notes for those
we will refer to).
So, e.g. for a Binomial distribution, the parameters are p the probability of success in an
individual trial and n the No. of trials.
The probability of success remains constant – otherwise, another distribution applies.
Use of the correct distribution is core to statistical inference – I.e. estimating what is
happening in the population on the basis of a (correctly drawn, probabilistic) sample.
The sample is then representative of the population.
Fundamental to statistical inference is the Normal (or Gaussian), with parameters,  the
mean (or more formally expectation of the distribution) and s (or
s 2), the Standard deviation ( Variance). For small samples, or when s 2 not known but
must be estimated from the sample, a slightly more conservative distribution applies = the
Student’s T or just ‘t’ distribution. Introduces the degrees of freedom concept.
Student’s t Distribution
A random variable X has a t distribution with n degrees of freedom ( tn ) .
The t distribution is symmetrical about the origin, with
E[X]
=0
VAR [X] = n / (n -2).
For small values of n, the tn distribution is very flat. As n is increased the density
assumes a bell shape. For values of n  25, the tn distribution is practically
indistinguishable from the standard normal curve.
O If X and Y are independent random variables
If X has a standard normal distribution and Y has a n2 distribution
then
X
has a tn distribution
(Y / n)
O If x1, x2, … , xn is a random sample from a normal distribution, with
mean  and variance s2 and if we define s2 = 1 / ( n - 1)  ( xi - x ) 2
then
( x -  ) / ( s /  n) has a tn- 1 distribution
Estimated Sample variance
see W,Y and S tables
+ Many other standard distributions
Sampling Theory
The procedure for drawing a random sample a distribution is that numbers 1, 2, …
are assigned to the elements of the distribution and tables of random numbers are
then used to decide which elements are included in the sample. If the same element
can not be selected more than once, we say that the sample is drawn without
replacement; otherwise, the sample is said to be drawn with replacement.
The usual convention in sampling is that lower case letters are used to designate
the sample characteristics, with capital letters being used for the parent population.
Thus if the sample size is n, its elements are designated, x1, x2, …, xn, its mean is x
and its modified variance is
s2 =
 (xi - x )2 / (n - 1).
The corresponding parent population characteristics are N (or infinity), X and S2.
Suppose that we repeatedly draw random samples of size n (with replacement) from
a distribution with mean  and variance s2. Let x1, x2, … be the collection of
sample means and let
xi’ =
xi - 
(i = 1, 2, … )
sn
The collection x1’, x2’, … is called the sampling distribution of means.
Central Limit Theorem.
In the limit, as sample size n tends to infinity, the sampling distribution of
means has a standard normal distribution. Basis for statistical inference.
Attribute and Proportionate Sampling
If the sample elements area measurement of some characteristic, we are said to
have attribute sampling. On the other hand if all the sample elements are 1 or 0
(success/failure, agree/ no-not-agree), we have proportionate sampling. For
proportionate sampling, the sample average x and the sample proportion p are
synonymous, just as are the mean  and proportion P for the parent population.
From our results on the Binomial distribution, the sample variance is p (1 - p) and
the variance of the parent distribution is P (1 - P).
We can generalise the concept of the sampling distribution of means to get the
sampling distribution of any statistic. We say that a sample characteristic is an
unbiased estimator of the parent population characteristic, i.e. the expectation of the
corresponding sampling distribution is equal to the parent characteristic.
Lemma. The sample average (proportion ) is an unbiased estimator of the parent
average (proportion):
E [ x] = ; so E [p] = P.
The quantity  ( N - n) / ( N - 1) is called the finite population correction (fpc). If
the parent population is infinite or we have sampling with replacement the fpc = 1.
Lemma.
E [s] = S * fpc for estimated sample S.D. with fpc
Confidence Intervals
From the statistical tables for a standard normal
distribution, we note that
Area Under
Density Function
0.90
0.95
0.99
From
To
-1.64
-1.96
-2.58
1.64
1.96
2.58
n (0,1)
0.95
-1.96
0
+1.96
From the central limit theorem, if x and s2 are the mean and variance of a random sample of
size n (with n greater than 25) drawn from a large parent population, then we can make the
following statement about the unknown parent mean 
Prob { -1.64  x -     
s/n
i.e.
Prob { x - 1.64 s /  n    x   s /  n }  
The range x + 1.64 s /  n
is called a 90% confidence interval for the parent mean .
Example [ Attribute Sampling]
A random sample of size 25 has x = 15 and s = 2. Then a 95% confidence interval for  is
15 + 1.96 (2 / 5)
(i.e.) 14.22 to 15.78
Example [ Proportionate Sampling]
A random sample of size n = 1000 has p = 0.40  1.96  p (1 - p) / (n - 1) = 0.03.
A 95% confidence interval for P is 0.40 + 0.03 (i.e.) 0.37 to 0.43.
Small Sampling Theory
For reference purposes, it is useful to regard the expression
x + 1.96 s /  n
as the “default formula” for a confidence interval and to modify it to suit particular
circumstances.
O If we are dealing with proportionate sampling, the sample proportion is the
sample mean and the standard error (s.e.) term s /  n simplifies as follows:
x -> p
and
s /  n ->  p(1 - p) / (n -1). (Also n-1 -> n)
O A 90% confidence interval will bring about the swap
1.96 -> 1.64.
O If the sample size n is less than 25, the normal distribution must be replaced by
Student’s t n - 1 distribution.
O For sampling without replacement from a finite population, a fpc term must be
used.
The width of the confidence interval band increases with the confidence level.
Example. A random sample of size n = 10, drawn from a large parent population, has a mean
x = 12 and a standard deviation s = 2. Then a 99% confidence interval for the parent mean is
x + 3.25 s /  n
(i.e.)
12 + 3.25 (2)/3
(i.e.)
9.83 to 14.17
and a 95% confidence interval for the parent mean is
x + 2.262 s /  n
(i.e.)
12 + 2.262 (2)/3
(i.e.)
10.492 to 13.508.
Note that for n = 1000, 1.96  p (1 - p) / n   for values of p between 0.3 and 0.7. This
gives rise to the statement that public opinion polls have an “inherent error of 3%”. This
simplifies calculations in the case of public opinion polls for large political parties.
Tests of Hypothesis
[Motivational Example]. It is claimed that the average grade of all 12 year old children in a
country in a particular aptitude test is 60%. A random sample of n= 49 students gives a
mean x = 55% with a standard deviation s = 2%. Is the sample finding consistent with the
claim?
We regard the original claim as a null hypothises (H0) which is tentatively accepted as
Z(0,1)
TRUE:
H0 :   
Z(0,1)
0.95
If the null hypothesis is true, the test statistic
t= x-
-1.96
1.96
sn
is a random variable with a Normal (0, 1) =
Standardised Normal Z(0,1) (or U(0,1)distribution.
Thus
55 - 60 = - 35 / 2 = - 17.5
2/  49
is a random value from Z(0, 1).
rejection regions
But this lies outside the 95% confidence interval (falls in the rejection region), so either
(i) The null hypothesis is incorrect
or
(ii) An event with a probability of at most 0.05 has occurred.
Consequently, we reject the null hypothesis, knowing a probability of 0.05 exists that we are
in error. Technically, we say we reject the null hypothesis at the 0.05 level of significance.
The alternative to rejecting H0, is to declare the test to be inconclusive. This means that
there is some tentative evidence to support the view that H0 is approximately correct.
Modifications
Based on the properties of the Normal , student t and other distributions, we can generalise
these ideas. If the sample size n < 25, we should use a tn-1 distribution; we can also vary the
level of significance of the test and we can apply the tests to proportionate sampling
environments.
Example. 40% of a random sample of 1000 people in a country indicate satisfaction with
government policy. Test at the .01 level of significance if this consistent with the claim that
45% of the people support government policy?
Here,
H0: P = 0.45
p = 0.40,
n = 1000
so
 p (1-p) / n = 0.015
test statistic = (0.40 - 0.45) / 0.015 = - 3.33
99% critical value = 2.58
so H0 is rejected at the .01 level of significance.
One-Tailed Tests
If the null hypothesis is of the form H0 : P   then arbitrary large values of p are
acceptable, so that the rejection region for the test statistic lies in the left hand tail only.
Example. 40% of a random sample of 1000 people in a country indicate satisfaction with
government policy. Test at the .05 level of significance if this consistent with the claim that
at least 45% of the people support government policy?
n(0,1)
Here the critical value is -1.64, so the
0.95
the null hypothesis H0: P  
is rejected at the .05 level of
-1.64
significance
Rejection region
Testing Differences between Means
Suppose that
s1
x1 x2
…
xm
is a random sample with mean x and standard deviation
drawn from a distribution with mean 1 and
y1 y2 … yn
with mean y and standard deviation s1
y1 y2
…
yn is a random sample
drawn from a distribution with mean 2. Suppose that we wish to test the null hypothesis that
both samples are drawn from the same parent population (i.e.)
H0: 1 = 2.
The pooled estimate of the parent variance is
s* 2 = { (m - 1) s12 + (n - 1) s22 } / ( m + n - 2)
and the variance of x - y, being the variance of the difference of two independent random
variables, is
s ’ 2 = s *2 / m + s* 2 / n.
This allows us to construct the test statistic, which under H0 has a tm+n-2 distribution.
Example. A random sample of size m = 25 has mean x = 2.5 and standard deviation s1 = 2,
while a second sample of size n = 41 has mean y = 2.8 and standard deviation s 2 = 1. Test at
the .05 level of significance if the means of the parent populations are identical.
Here
H0 : 1 = 2
x - y = - 0.3 and
s*2 = {24(4) + 40(1)} / 64 = 2.125
so the test statistic is
- 0.3 /  22  2  22     8
The .05 critical value for Z(0, 1) is , so the test is inconclusive
Paired Tests
If the sample values
( xi , yi ) are paired, such as the marks of students in two
examinations, then let
di = xi - yi be their differences and treat these values as
the elements of a sample to generate a test statistic for the hypothesis
H0: 1 = 2.
The test statistic
d / sd / n
has a tn-1 distribution if H0 is true.
Example. In a random sample of 100 students in a national examination their examination mark
in English is subtracted from their continuous assessment mark, giving a mean of 5 and a
standard deviation of 2. Test at the .01 level of significance if the true mean mark for both
components is the same.
Here
n = 100, d = 5,
sd / n = 2/10 = 0.2
so the test statistic is
5 / 0.2 = 10.
the 0.1 critical value for a Z(0, 1) distribution is 2.58, so H0 is rejected at the .01 level of
significance.
Tests for the Variance.
For normally distributed random variables, given
H0: s 2 = k, a constant, then
(n-1) s2 / k
has a  2n - 1 distribution.
Example. A random sample of size 30 drawn from a normal distribution has variance s2 = 5.
Test at the .05 level of significance if this is consistent with H0 : s 2 = 2 .
Test statistic = (29) 5 /2 = 72.5, while the .05 critical value for  229 is 45.72,
so H0 is rejected at the .05 level of significance.
Chi-Square Test of Goodness of Fit
This can be used to test the hypothesis H0 that a set of observations is consistent with a
given probability distribution. We are given a set of categories and for each we record the
observed Oj nd expected Ej number of observations that fall in each category. Under H0,
the test statistic
S (Oj  Ej )2 / Ej
has a  2n - 1 distribution, where n is the number of
categories.
Example.A pseudo random number generator is used to used to generate 40 random
numbers in the range 1 - 100. Test at the .05 level of significance if the results are consistent
with the hypothesis that the outcomes are randomly distributed.
Range
1-25
Observed Number 6
Expected Number 10
26 - 50
12
10
51 - 75
14
10
76 - 100 Total
8
40
10
40
Test statistic = (6-10)2/10 + (12-10)2/10 + (14-10)2/10 + (8-10)2/10 = 4.
The .05 critical value of  23 = 7.81, so the test is inconclusive.
Chi-Square Contingency Test
To test that two random variables are statistically independent, a set of observations can be
recorded in a table with m rows corresponding to categories for one random variable and
n columns for the other. Under H0, the expected number of observations for the cell in row i
and column j is the appropriate row total by the column total divided by the grand total. Under
H0,
the test statistic
S (Oij  Eij )2 / Eij
has a  2(m -1)(n-1) distribution.
Chi-Square Contingency Test - Example
In the following table, the
figures in brackets are the
expected values.
The test statistic is
Results
Honours
Pass
Fail
Totals
Maths
100 (50)
130 (225)
70 (25)
300
History
Geography
70 (67)
30 (83)
320 (300) 450 (375)
10 (33)
20 (42)
400
500
Totals
200
900
100
1200
S (Oij  Eij )2 / Eij = (100-50)2/ 50 + (70 - 67)2/ 67 + (30-83)2/ 83 + (130-225)2/ 225
+ (320-300)2/ 300 + (450-375)2/375 + (70-25)2/ 25 + (10-33)2/ 33 + (20-42)2/ 42
= 248.976
The .05 critical value for  22 * 2 is 9.49 so H0 is rejected at the .05 level of significance.
In general the chi square tests tend to be very conservative vis-a-vis other tests of hypothesis,
(i.e.) they tend to give inconclusive results.
The meaning of the term “degrees of freedom” .
In simplified terms, as the chi-square distribution is the sum of, say k, squares of independent
random variables, it is defined in a k-dimensional space. When we impose a constraint of the
type that the sum of observed and expected observations in a column are equal or estimate a
parameter of the parent distribution, we reduce the dimensionality of the space by 1. In the case
of the chi-square contingency table, with m rows and n columns, the expected values in the final
row and column are predetermined, so the number of degrees of freedom of the test statistic is
(m-1) (n-1).