DATA ANALYSIS
Module Code :CA660
(Application Areas: Bio

, Business,
Social, Environment etc.)
2
STRUCTURE of Investigation/DA
Level of
Measurement
Distributional Assumptions, Probability , Estimation properties
Basis: Size/Type of Data Set/Tools
Parametric
Non

Parametric
Study techniques
Lab. techniques
Estimation/H.T.
H.T.
1,2, many samples
E.D., Reg
n.
, C.T.
Replication,
Assays, Counts
Probability & Statistics Primer

overview
Note:
Short overview. Other statistical distributions in lectures
3
Summary Statistics

Descriptive
In analysis of practical sets of data, useful to define a small number of values that
summarise main features present. We derive (
i
) representative values, (ii)
measures of spread and (iii) measures of
skewness
and other characteristics.
Representative Values
Sometimes called measures of
location
or
measures of central tendency
.
1. Random Value
Given a set of data S = { x
1
, x
2
, … ,
x
n
}, we select a random number, say k, in the
range 1 to n and return the value
x
k
. 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 a
rithmetic
mean (or just mean)
is
x = {x
1
+ x
2
+ … +
x
n
}/ n.
If x
1
occurs f
1
times, x
2
occurs f
2
times and so on, we get the formula
x = { f
1
x
1
+ f
2
x
2
+ … +
f
n
x
n
} / { f
1
+ f
2
+ … +
f
n
} ,
written
4
Example 1.
D
ata
are student marks in an examination. Find the average mark for the class.
N
ote 1
:
M
arks
are given
as ranges, so care
Mark
Mid

Point
Number
needed
in range interpretation
of Range of Students
All intervals must be of equal rank and there
x
i
f
i
f
i
x
i
must be no gaps in the classification
0

19
10
2
20
We 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,
mid

point
is 10. The other intervals
are
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
2
:
Pivot. If
weights of size f
i
are suspended
x
1
x
2
x
x
n
from a metre stick at the points x
i
, then the
average is the
centre of gravity
of
the
distribution. Consequently, it is very sensitive
f
1
f
2
f
n
to outlying values.
Note 3:
P
opulation
should be
homogenous
for
average
to be meaningful. For example, if
assume
that
typical
height of girls in a class is less than that of boys,
then
average
height
of all students
is neither indicative of the girls nor of the boys.
5
3. The Mode
This is
the value that
occurs
most
frequently
. By common agreement,
it is calculated from the histogram
using
linear
interpolation on the modal class.
The various similar triangles in the diagram
generate the common ratios. In our case,
the mode is
60 + 13 / 33 (20) = 67.8 marks.
4. The Median
T
he
middle point
of the distribution.
If
{ x
1
, x
2
, … ,
x
n
} are marks of students
in
a class,
arranged
in
nondecreasing
order
, then the median is the mark of
the
(n + 1)/2 student.
O
ften use the
ogive
or
cumulative
frequency
Diagram to calculate.
In our case,
the median is
60 + 5.5 / 25 (20) = 64.4 marks.
50
Frequency
20
20
40
60
80
100
6
12
25
5
2
13
13
20
Cumulative
Frequency
100
80
60
40
20
50
25.5
6
Measures of Dispersion or Scattering
Example 2. The
distribution shown has
the same
Marks
Frequency
arithmetic mean as
Example
1, but
values
are
more
x
j
f
j
f
j
x
j
dispersed
.
Illustrates that an average value alone
may not adequately
d
e
scribe statistical
10 6 60
distributions
.
30 8 240
50 6 300
To devise a formula that
captures degree
to which a
70 15 1050
distribution is concentrated about the average, we
90 15 1350
consider the deviations of the values from the average.
Sums
50
3000
If distribution
is concentrated around the mean,
then
deviations
will be small, while
if it is very scattered,
then 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
the
standard
deviation
, has same
units of measurement as
the original values and is the preferred measure of
dispersion in many applications.
x
1
x
2
x
3
x
4
x
5
x
6
x
7
Variance & Standard Deviation
s
2
VAR[X] = Average of the Squared Deviations
=
S
f { Squared Deviations } /
S
f
=
S
f
i
{ x
i

x }
2
/
S
f
i
=
S
f x
i
2
/
S
f

x
2
, called the
product moment
formula.
s
Standard Deviation =
Variance
Example 1
Example
2
f
x
f x
f x
2
f
x
f x
f x
2
2
10
20
200
6
10
60
600
6
30
180
5400
8
30
240
7200
12
50
600
30000
6
50
300
15000
25
70
1750
122500
15
70
1050
73500
5
90
450
40500
15
90
1350
121500
50
3000
198600
50
3000
217800
VAR [X] = 198600 / 50

(60)
2
VAR [X] = 217800 / 50

(60)
2
= 372 marks
2
= 756 marks
2
8
Other
Summary Statistics
Skewness
An important attribute of a statistical distribution
is its
degree of symmetry
. The
“
skew
”
means a tail, so
distributions with a
large tail of outlying values on the right

hand

side
are
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
for Example
1 is:
Skewness
= (60

67.8) / 19.287 =

0.4044.
Coefficient of Variation
This formula
was devised to ‘standardise’
the arithmetic mean so
comparisons
can be
drawn between different
distributions. Not universally
used.
Coefficient of Variation = Mean / Standard Deviation.
Semi

Interquartile Range
T
he
M
edian is the mid or 0.5 point
in a
distribution.
T
he
quartiles Q
1
, Q
2
, Q
3
correspond
to the 0.25, 0.50 and 0.75 points. An alternative measure of dispersion
is thus
Semi

Interquartile Range = ( Q
3

Q
1
) / 2.
Geometric Mean
For data that
grow
geometrically,
e.g.
economic
data
with high
inflation effect
,
another
mean is sometimes used.
The G.M.
is defined for a product of frequencies, where N
=
S
f
G. M.
=
N
x
1
f1
x
2
f2
…
x
k
fk
9
Regression
[Example 3.]
As a motivating example, suppose
we model
sales data over time.
SALES
3
5
4
5
6
7
TIME
1990
1991
1992
1993
1994
1995
Want the
straight line “Y = m X + c” that best
approximates the data.
“Best
” in this
case
is
the
line which minimizes
the sum
of squares
of vertical deviations of points from
the line:
SSQ
= SS =
S
( Y
i

[
mX
i
+ c ] )
2
Setting
partial
derivatives of SS
w.r.t.
m
and c to zero leads to the “Normal Equations”
S
Y
= m
S
X + n
c
S
X
Y
= m
S
X
2
+ c
S
X ,
where n = # points
Let
1990 correspond to Year 0
.
X.X X X.Y
Y
Y.Y
0 0 0 3 9
1 1 5 5 25
4 2 8 4 16
9 3 15 5 25
16 4 24 6 36
25 5 35 7 49
55 15 87 30 160
X
Y
Y
i
= m X
i
+ c
m X
i
+ c
Y
i
0
X
i
Time
Sales
10
5
0
5
10
Example 3

Working
The normal equations are:
30 = 15 m + 6 c
=>
150 = 75 m + 30 c
87 = 55 m + 15 c
174 = 110 m + 30 c
=>
24 = 35 m
=>
30 = 15 (24 / 35) + 6
c => c
= 23/7
Thus the regression line of Y on X is
Y = (24/35) X + (23/7)
and to plot the line
just need
two points, so
X = 0
=> Y = 23/7
and X = 5 => Y = (24/35) 5 + 23/7 = 47/7.
E
asy
to see that ( X, Y ) satisfies the normal equations, so that the regression line
of Y on X passes
through “
Centre of Gravity” of the data. By expanding terms,
get
S
( Y
i

Y )
2 =
S
(
Y
i

[ m X
i
+ c ] )
2
+
S
( [ m X
i
+ c ]

Y )
2
Total Sum
Error Sum
Regression Sum
of Squares
of Squares
of Squares
SST
= SSE
+
SSR
Distinguish
the
independent and dependent variables
(
X and
Y
respectively)
X
Y
Y
i
mX
i
+C
Y
X
Y
11
Correlation
The
coefficient of determination
r
2
( 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:
r
2
=
SSR/ SST =
1

SSE / SST.
The
coefficient of correlation
‘r’
(values
in the range

1 to +1 ) is
a more common
measure
of the degree to which a
mathematical relationship
exists between X
and Y.
It can
be
calculated as:
r
=
( X

X ) ( Y

Y )
( X

X )
2
( Y

Y )
2
=
n
X Y

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 = + 1
r = 0
12
Col
l
inearity
For correlation
coefficient
value >
0.9
or
<

0.9, we would take this to mean that
there is a
mathematical relationship
between the variables. D
oes
not imply that
a
cause

and

effect relationship
exists.
E.g.
c
onsider
a country with a slowly changing population size, where a certain political
party retains a relatively stable
percentage
of the poll in elections. Let
X = Number of people that vote for the party in an election
Y = Number of people that die
of a
given disease in a year
Z = Population size.
Then,
correlation
coefficient between X and Y is
~1
, indicating
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 incidence of
disease
is caused by the number of people that vote for the
given political party. This is known as the problem of
col
l
inearity
.
Spotting
hidden
dependencies is
non

trivial.
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
of 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, though may be useful to guide research.
13
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.
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
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 E
1
E
2
is the set of all outcomes that belong to E
1
or
to E
2
or
to both.
The
intersection
of two events E
1
E
2
is the set of all events that belong to both E
1
and
E
2.
Two events are
mutually exclusive
if occurrence
of either precludes
occurrence
of the
other (
i.e
) their intersection is the empty
set .
Two events are
independent
if
occurrence
of either is un
a
ffected
by
occurrence
or non

occur
r
ence
of the other event.
Theorem of Total Probability.
P {E
1
E
2
} = P{E
1
} + P{E
2
}

P{E
1
E
2
}
Proof.
P{E
1
E
2
} = (n
1, 0
+ n
1, 2
+ n
0, 2
) / n
= (n
1, 0
+ n
1, 2
) / n + (n
1, 2
+ n
0, 2
) / n

n
1, 2
/ n
= P{E
1
} + P{E
2
}

P{E
1
E
2
}
Corollary.
If E
1
and E
2
are mutually exclusive, P{E
1
E
2
} = P{E
1
} + P{E
2
}

see
Axioms
and
Addition Rule
E
S
n = n
0, 0
+ n
1, 0
+ n
0, 2
+ n
1, 2
E
1
E
2
S
n
1, 0
n
1, 2
n
0, 2
n
0, 0
14
The probability P{E
1
 E
2
} that
E
1
occurs,
given that E
2
has occurred (or must occur)
is called the
conditional probability
of E
1
. Note
: only
possible outcomes of the experiment are confined to
E
2
and not to S.
Theorem of Compound Probability
Multiplication Rule.
P{E
1
E
2
} = P{E
1
 E
2
}
P{E
2
}.
Proof.
P{E
1
E
2
} = n
1, 2
/ n
= {n
1, 2
/ (n
1, 2
+ n
0, 2
) }
{ n
1, 2
+ n
0, 2
) / n}
Corollary
If E
1
and E
2
are independent, P{E
1
E
2
} = P{E
1
}
P{E
2
}.
Special case
of Multiplication Rule
Note:
If
E
itself compound, expands further =
Chain Rule: P{E
7
E
8
E
9
} =P{E
7
(E
8
E
9
)}
C
ounting
possib
le
outcomes
of
an event is crucial to calculating pro
babilities
.
A
permutation
of
size r of n different items,
is
an
arrangement
of r of the items,
where
order
of
arrangement is
important. If
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.
n
1, 0
n
1, 2
n
0, 2
n
0, 0
E
1
E
2
S
15
Bayes’ Rule (Theorem):
For a series of
mutually exclusive
and
exhaustive events
B
r
,
where union of the B
r
= B
1
B
2
B
3
…….
B
r
= all possibilities for B,
Then:
Where
the 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[E  man is father}
P{man is father}
P{man not father  E} P{E  man not father} P{man not father}
Written in terms of ratio of
posterior probs
. (= LHS), paternity index (L say) and ratio
of
prior probs.
(RHS). Rearrange and substitute in above to give prob. of an alleged
man with
particular genotype ‘
C’ being
the true father
NB
: L is a way of ‘weighting’ the genetic evidence; the
issue
is setting a
prior
.
=
16
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
X
1
, X
2
, … ,
X
n
with probabilities
p
1
, p
2
, … ,
p
n
then the
expected
(
average
)
value of X is defined to be
E[X] =
p
j
X
j
and its variance is
VAR[X] = E[X
2
]

E[X]
2
=
p
j
X
j
2

E[X]
2
.
Example. Let X be a random variable measuring
Prob.
Distance
the distance in Kilometres travelled by children
p
j
X
j
p
j
X
j
p
j
X
j
2
to a school and suppose that the following data
applies. Then the mean and variance are
0.15
2.0 0.30 0.60
E[X] = 5.30 Kilometres
0.40
4.0 1.60 6.40
VAR[X] =
33.80

5.30
2
=5.71 km
2
0.20
6.0 1.20 7.20
0.15 8.0 1.
20 9.60
Similar concepts apply to continuous distributions.
0.10 10.0 1.00 1.00
The
distribution function
is defined by
1.00

5.30 33.80
F(t) = P{ X t}
and its
derivative
is the
frequency function
f(t) = d F(t) /
dt
so that
F(t) =
f(x
) dx.
17
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] = k
2
.E[X] for a constant k.
Example. A company records the journey time X
X=
1
2
3
4
Totals
of a lorry from a depot to customers and
Y =1
7
5 4 4
20
the unloading times Y, as shown.
2
2
6 8 3
19
E[X] = {
1
(
10
)+
2
(
13
)+
3
(
17
)+
4
(
10
)}/
50
= 2.54
3
1
2 5 3
11
E[X
2
] = {1
2
(10+2
2
(13)+3
2
(17)+4
2
(10)}/50 = 7.5
VAR[X] = 7.5

(2.54)
2
=
1.0484
Totals
10 13 17 10 50
E[Y] = {1(20)+2(19)+3(11)}/50 = 1.82 E[Y
2
] = {1
2
(20)+2
2
(19)+3
2
(11)}/50 = 3.9
VAR[Y] = 3.9

(1.82)
2
= 0.5876
E[X+Y] = { 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
E[(X + Y)
2
] = {2
2
(7)+3
2
(5)+4
2
(4)+5
2
(4)+3
2
(2)+4
2
(6)+5
2
(8)+6
2
(3)+4
2
(1)+5
2
(2)+6
2
(5)+7
2
(3)}/50 = 21.04
VAR[(X+Y)] = 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
18
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
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 formally
expectation
of the distribution) and
s
(SD) or variance (
s
2
).
For
small samples, or when
s
2
not known but must be estimated from
a sample
,
a
slightly more conservative distribution

the Student’s T or just ‘t’
distribution, applies.
Introduces the
degrees of freedom
concept.
19
Student’s t Distribution
A random
var
i
able X has a t distribution with n
degrees of freedom
(
t
n
)
.
The t distribution is symmetrical about the origin, with
E[X]
= 0
VAR [X] = n / (n

2
).
For
small values of n, the
t
n
distribution is very flat. As n is increased the density
assumes a bell shape. For values of n
25, the
t
n
distribution is practically
indistinguishable from the
S
tandard Normal
curve
.
O If X and Y are independent random variables
If X has a standard normal distribution and Y has a
c
n
2
distribution
then
X
has a
t
n
distribution
(
Y / n
)
O If x
1
, x
2
, … ,
x
n
is a random sample from a normal distribution, with
mean
and variance
s
2
and if we define s
2
= 1 / ( n

1)
( x
i

x )
2
then
( x

) / ( s /
n) has a
t
n

1
distribution
Estimated Sample variance

see calculators
,tables etc.
+
Many other standard distributions
20
Sampling Theory
To draw a
random sample
from a distribution, assign numbers
1, 2, … to the
elements of the
distribution, use random number
tabes
or generated set 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
.
U
sual
convention in sampling is that lower case letters
designate
the sample
characteristics, with capital letters
used
for the
(finite) parent population and
greek
letters for the infinite.
Thus if
sample
size =
n, its elements are designated, x
1
, x
2
, …,
x
n
, its mean is x and its modified variance is
s
2
=
(x
i

x )
2
/ (n

1
).
C
orresponding
parent population characteristics =
N, X
and
S
2
or (
,
and
s
2
)
Suppose
we
repeatedly draw random samples of size n (with replacement) from a
distribution with mean
and
variance
s
2
. Let x
1
, x
2
, …
be the collection of sample
means
and let
x
i
’ =
x
i

(
i
= 1, 2, … )
s
n
The collection x
1
’, x
2
’, … is called the
sampling distribution of means, (usual U or Z)
Central Limit Theorem.
In the limit, as
sample size
n tends to infinity,
the sampling distribution of means
has
a
S
tandard Normal
distribution
.
Basis for
Statistical
I
nference
.
21
Attribute and Proportionate Sampling
If
sample
elements
are a
measurement of some characteristic
,
then have
attribute sampling
.
However, if
all
sample
elements are 1 or 0 (success/failure,
agree/
do

not agree
), we have
proportionate sampling
.
For
proportionate sampling, the sample average x and the sample proportion p
are
synon
y
mous
,
(just
as
for mean
and proportion P for the parent
population).
From our results on the
B
inomial
distribution, the sample variance is p (1

p) and
the variance of the parent distribution is P (1

P
) in the proportionate case.
T
he ‘sampling distribution’
of means
concept generalizes 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 w
e
have sampling
with replacement
the
fpc
= 1.
Lemma.
E [s] = S
fpc
for estimated sample S.D. with
fpc
22
Confidence Intervals
From the statistical tables for a Standard Normal (Gaussian)
distribution, we note that
Area Under
From
To
Density Function
0.90

1.64
1.64
0.95

1.96
1.96
0.99

2.58
2.58
From the
central limit theorem
, if x and s
2
are the mean and variance of a random sample
of size n (with n greater than 25) drawn from a large parent population,
size N
,
then
the
following statement
,about
the unknown parent mean
,
applies
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.
N
(0,1)
0

1.96
+1.96
0.95
23
Small Sampling Theory
For reference purposes, it is useful to regard the expression
x
1.96 s /
n
as
“
default formula” for a confidence interval and
to modify
it
for 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
fol
l
ows
:
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
For sample
size
n 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 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
: 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%”.
S
implifies
calculations in the case of
pu
b
lic
opinion polls for large political parties.
24
Tests of Hypothesis
[Motivational Example]. It is claimed that
average
grade of all 12 year
olds in
a country in
a particular aptitude test is 60%. A random sample of n= 49
students gives a mean x =
55% with
standard
deviation s = 2%. Is the sample
finding consistent with the claim
?
T
he
original claim
regarded as a
null
hypothesis
(H
0
)
which is tentatively accepted as TRUE
:
H
0
:
If the null hypothesis is true, the
test statistic
T
S
=
x

s
n
is a Random Variable with a
Normal
(0, 1)
i.e.
Standardised Normal
Z(0,1) (or U(0,1
))
distribution.
Thus
55

60 =

35 / 2 =

17.5
2/
49
rejection regions
is a random value from
Z
(0, 1).
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,
reject
the null hypothesis, knowing
a probability of 0.05
exists
that we are
in error.
Technically
: reject the null hypothesis at the 0.05
level of significance.
The alternative to rejecting H
0
, is to declare the
test to be inconclusive.
This
mean
s
that
there is some tentative evidence to support the view that H
0
is
approximately correct
.
Z
(0,1)
0.95
1.96

1.96
25
Modifications
Based on the properties of the
N
ormal
,
Student ‘t’
and other distributions, we can
generalise these ideas. If the sample size n < 25,
a t
n

1
distribution should be used
;
the
level of significance of the test
may also be varied or the test applied to a
proportionate
sampling
environment.
Example. 40% of a random sample of 1000 people in a country indicate
satisfaction
with
government policy. Test at the 0.01 level of significance if this consistent with the claim that 45% of
the people support government policy?
Here, H
0
: 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 H
0
is rejected at the
0.01
level of significance.
One

Tailed Tests
If the null hypothesis is of the form H
0
: P
> 0.45
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 0.05 level of significance if this consistent with the claim that
at least 45% of the people support government policy?
Here the critical value is

1.64, so
the null hypothesis H
0
: P
is rejected at the
0.05
level of
significance
N
(0,1
)

1.64
0.95
Rejection region
26
Suppose
that
x
1
x
2
…
x
m
is a random
sample,
mean x and standard deviation
s
1
drawn
from a distribution with mean
1
and
y
1
y
2
…
y
n
is
a random
sample, mean
y and standard deviation
s
2
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.)
H
0
:
1
=
2.
The pooled estimate of the parent variance is
s
*
2
=
s
p
2
= {
(m

1) s
1
2
+ (n

1) s
2
2
} / ( m + n

2)
and the variance of
(
x
–
y), is the variance
of the difference of
two
independent random
variables,
i.e.
s
diff
2
= s
p
2
/
m + s
p
2
/ n.
This allows us to construct the test statistic, which
under H
0
has a t
m+n

2
distribution.
Example. A random sample of size m = 25 has mean x = 2.5 and standard deviation s
1
= 2, while a
second sample of size n = 41 has mean y = 2.8 and standard deviation s
2
= 1. Test at the 0.05 level of
significance if the means of the parent populations are identical.
Here
H
0
:
1
=
2
x

y =

0.3 and
s
p
2
=
{24(4) + 40(1)} / 64 = 2.125
so the test statistic is

0.3 /
22 2 22
8
The 0.05 critical value for
Z
(0, 1) is
,
so the test is
inconclusive
27
Testing Differences between Means
Paired Tests
If the sample values
(
x
i
,
y
i
) are paired, such as the marks of students in two examinations,
then let
d
i
= x
i

y
i
be their differences and treat these values as the elements of a sample to
generate a test statistic for the hypothesis
H
0
:
1
=
2.
The test statistic
d /
s
d
/
n
has a t
n

1
distribution if H
0
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
0.01
level of significance if the true mean mark for both components is the
same.
Here
n = 100, d = 5,
s
d
/
n = 2/10 = 0.2
so the test
statistic
is
then
5 / 0.2 = 10.
the 0.01 critical value for a
Z
(0, 1) distribution is 2.58, so H
0
is rejected at the
0.01
level of significance.
Tests for the Variance.
For normally distributed random variables, given
H
0
:
s
2
= k, a constant, then (n

1) s
2
/ k
has a
c
2
n

1
distribution.
Example. A random sample of size 30 drawn from a normal distribution has variance s
2
= 5.
Test at the
0.05
level of significance if this is consistent with H
0
:
s
2
= 2 .
Test statistic = (29) 5 /2 = 72.5, while the 0.05 critical value for
c
2
29
is 45.72,
so H
0
is rejected at the 0.05 level of significance.
28
Chi

Square Test of Goodness of Fit
C
an
be used to test the hypothesis H
0
that a set of observations is consistent with a given
probability distribution. G
iven
a set of categories
with observed (
O
j
) and
expected
(
E
j
)
number
of observations
(frequency) in
each category.
Under
H
0
T
est Statistic
S
(
O
j
E
j
)
2
/
E
j
has a
c
2
n

1
distribution,
with
n
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 0.05 level of
significance,
if the results are consistent with the hypothesis
that the outcomes are randomly distributed.
Range
1

25 26

50 51

75 76

100 Total
Observed Number
6 12 14 8 40
Expected Number 10 10 10 10 40
Test statistic = (6

10)
2
/10 + (12

10)
2
/10 + (14

10)
2
/10 + (8

10)
2
/10 = 4.
The 0.05 critical value of
c
2
3
= 7.81, so the test is inconclusive.
Chi

Square Contingency Test
To test that two random variables are
statistically independent
, a set of
obs
e
rvations
can
be
tabled,
with m rows corresponding to categories for one random variable and
n columns for the other.
Under H
0
, the expected number of observations for the cell in row
i
and column j =
appropriate (row
total
column total)
(G
rand total).
Under
H
0
T
est
S
tatistic
S
S
(
O
ij
E
ij
)
2
/
E
ij
has a
c
2
(m

1)(n

1)
distribution.
29
Chi

Square Contingency Test

Example
In the following table, the
Results Maths
History
Geography Totals
figures in brackets are the
Honours
100 (50)
70 (67)
30 (83) 200
expected values.
Pass 130 (225
)
320 (300) 450 (375) 900
Fail
70 (25)
10 (33)
20 (42) 100
The test statistic
is
Totals
300
400
500
1200
S
[
(
O
ij
E
ij
)
2
/
E
ij
]
= (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 0.05 critical value for
c
2
2 * 2
is 9.49 so H
0
is rejected at the 0.05 level of significance.
Note:
In general, chi

squared
tests tend to be very conservative
vis

a

vis
other tests of
hypothesis
, (
i.e.) they tend to give inconclusive results.
T
he 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
cons
t
raint
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
).
30
31
Analysis of Variance/Experimental Design

Many samples, Means and Variances
•
Analysis of Variance (AOV or ANOVA) was
originally devised for agricultural statistics
on
crop yields etc.
Typically, row and column
format, = small plots of a fixed size. The yield
y
i
, j
within each plot was recorded.
One Way classification
Model:
y
i
, j
= +
i
+
i
, j
,
i
,j
~ N (0,
s
2
) in the limit
where
= overall
mean as sample size large
i
= effect of the
i
th
factor
i
, j
= error term.
Hypothesis:
H
0
:
1
=
2
= … =
m
y
1, 3
y
1
, 1
y
1, 2
y
2, 2
y
1, 4
y
2, 1
y
2, 3
y
3, 1
y
3, 2
1
2
3
y
1
, 5
y
3, 3
32
Totals
Means
Factor 1
y
1, 1
y
1, 2
y
1, 3
y
1, n1
T
1
= y
1, j
y
1
.
= T
1
/ n
1
2
y
2, 1
y
2,, 2
y
2, 3
y
2, n2
T
2
= y
2, j
y
2
.
= T
2
/ n
2
m
y
m
, 1
y
m
, 2
y
m
, 3
y
m
, nm
T
m
=
y
m
, j
y
m
.
= T
m
/ n
m
Overall mean
y =
y
i
, j
/ n,
where n =
n
i
Decomposition (Partition) of Sums of Squares:
(
y
i
, j

y )
2
=
n
i
(
y
i
.

y )
2
+ (
y
i
, j

y
i
.
)
2
Total Variation (Q) = Between Factors (Q
1
) + Residual Variation (Q
E
)
Under H
0
: Q / (n

1)

>
2
n

1
, Q
1
/ (m

1)

>
2
m

1
, Q
E
/ (n

m)

>
2
n

m
Q
1
/ ( m

1 )

>
F
m

1, n

m
Q
E
/ ( n

m )
AOV Table:
Variation D.F. Sums of Squares Mean Squares F
Between
m

1 Q
1
=
n
i
(
y
i
.

y )
2
MS
1
= Q
1
/(m

1) MS
1
/ MS
E
Residual
n

m Q
E
= (
y
i
, j

y
i
.
)
2
MS
E
= Q
E
/(n

m)
Total
n

1 Q
=
(
y
i
, j.

y )
2
Q
/( n

1)
33
Two

Way Classification
Factor I
Means
Factor II y
1, 1
y
1, 2
y
1, 3
y
1, n
y
1
.
: :
: :
y
m
, 1
y
m
, 2
y
m
, 3
y
m
, n
y
m
.
Means
y
.
1
y
.
2
y
.
3
y
. n
y
. .
So we write as y
Partition SSQ:
(
y
i
, j

y )
2
= n (
y
i
.

y )
2
+ m (y
.
j

y )
2
+ (
y
i
, j

y
i
.

y
.
j
+ y )
2
Total
Between
Between
Residual
Variation Rows Columns Variation
Model:
y
i
, j
= +
i
+
j
+
i
, j
with
i
, j
~ N ( 0,
s
2
)
H
0
:
All
i
are equal.
H
0
: all
j
are equal
AOV
Variation D.F. Sums of Squares Mean Squares F
Between
m

1 Q
1
= n
(
y
i
.

y )
2
MS
1
= Q
1
/(m

1) MS
1
/ MS
E
Rows
Between
n

1 Q
2
= m (y
.
j

y )
2
MS
2
= Q
2
/(n

1) MS
2
/ MS
E
Columns
Residual
(m

1)(n

1)
Q
E
= (
y
i
, j

y
i
.

y
.
j
+ y)
2
MS
E
= Q
E
/(m

1)(n

1
)
Total
mn

1 Q
=
(
y
i
, j.

y )
2
Q
/(
mn

1)
34
Two

Way Example
ANOVA outline
Factor I
1 2 3 4 5
Totals Means
Variation d.f. SSQ MSQ F
Fact II 1
20 18 21 23 20 102 20.4
Rows
3 76.95 25.65 18.86**
2
19 18 17 18 18 90 18.0
Columns
4 8.50 2.13 1.57
3
23 21 22 23 20 109 21.8
Residual
12 16.30
4
17 16 18 16 17 84 16.8
Totals
79 73 78 80 75 385
Total
19 101.75
Means
19.75 18.25 19.50 20.00 18.75 19.25
FYI
software such as R,SAS,SPSS, MATLAB is designed for analysing these
data, e.g. SPSS as spreadsheet recorded with variables in columns and
individual observations in the rows. Thus the ANOVA data above would be
written as a set of columns or rows, e.g.
Var. value 20 18 21 23 20 19 18 17 18 18 23 21 22 23 20 17 16 18 16 17
Factor 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4
Factor 2 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
35
DA APPLICATIONS
CONTEXT
e.g. BIO
•
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
36
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
37
CONTEXT+ : GENOMICS

LINKAGES
Mendelian
Cytogenetics
Molecular
Population
Quantitative
GENOMICS
Genetic markers
DNA Sequences
Linkage/Physical Maps
Gene Location
QTL Mapping
38
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?
39
‘DATA’ : STATISTICAL
GENOMICS
Some UNUSUAL/SPECIAL FEATURES
•
Size
–
databases very large
e.g. molecular marker and DNA
/ protein sequence
data;
unreconciled
, Legacy
•
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, sequence match
etc.
•
Likelihood Analysis

Linear Models typically insufficient
alone
40
DA
APPLICATIONS
CONTEXT e.g. BUSINESS/ FINANCE
•
http://big.computing.dcu.ie
/
;
http://sci

sym.dcu.ie
•
Data

rich environments
–
under

utilisation of resources
•
RAW DATA
into useful information and knowledge
•
Similar underpinning:
(‘Laws’)
–
based on analysis
•
Purpose
–
Informed decision

making
•
Techniques
–
quantitative.
Concepts & Nature
–
Pervasive, Dynamic, ‘Health’ subject to Internal/External
environments.
Key elements

Systems and
people
•
Forecasting/Prediction/Trigger
41
41
CONTEXT+ : FACTORS
Supply
Chain
Capital
Knowledge
& Systems
Labour
Globalisation,
technology
HEALTH of ENTERPRISE
(governmental, corporate,
educational, non

profit)
Adaptability
FRAMEWORK
•
Status
: Huge array of information
systems
&product software.
C
hallenges:
include
development, delivery, adoption,
and
implementation
of IT solutions into usable and effective systems that
mimic/support organisational
processes.
‘KS alignment with work
practice.’
(
Toffler
&
Drucker
–
80’s
: organisations
of 20
th
Century

>
knowledge

based. Greater autonomy, revised management structures).
Opportunities
: KM popularity
grew
through 90’s, spawned ideas of 'KM
models',
‘KM
strategy
', concepts of 'organisational
learning',
'knowledge
/practice networks
', 'knowledge discovery',
‘intellectual capital‘).
•
Objectives:
To Plan
, develop,
implement,
operate, optimise,
cost information /
communication
systems and interpret use
.
•
S
tarting
point
:
understanding
ICT opportunities
requires
both
technological
and
organisational perspective
+
understanding
of benefits associated with data capture and analysis.
42
Data Mining & KM
•
The
Knowledge Discovery Process
•
Classification
e.g. clusters, trees
•
Exploratory Data Analysis
•
Models
(including Bayesian Networks
), Graphical or other.
•
Frequent Pattern Mining
and special groups/subgroups
Key Features:
•
‘Learning models’
from data can be an important part of
building
an
intelligent decision support system.
•
Sophistication of analyses
–
computationally expensive data
mining methods, complexity of algorithms, interpretation and
application of models.
43
Hot Topics in BI
•
Business
Process Management and
Modelling
•
Supply Chain Management and Logistics
•
Innovation and ICT
•
Analytical Information
Systems, Databases
and Data Warehousing
•
Knowledge Management and Discovery
•
Social Networks and Knowledge Communities
•
Performance Indicators &
Measurement systems/
Information Quality
•
Data Analytics, Integration and
Interpretation
•
Cost

benefit
and
Impact Analysis
•
Reference Models and Modelling
•
Process Simulation and Optimization
•
Security
and Privacy
•
IT and IS
Architectures/Management
•
Info. Sys. development, Tools
and Software Engineering
44
Example
Questions
•
What
are the
characteristics
of internet purchases for a given
age

group?
How
can this be used to develop further E

business?
•
What
are
key
risk factors
for
profit/loss
on
a product on the
basis of
historical data
and demographic
variables?
•
Can
we segment into/identify groups of similar on
the basis of their
characteristics and purchase
behaviour?
•
Which
products are typically
bought together
in one transaction by
customers
?
•
What
are financial projections, given
market volatility
and
knock

on
for recent shock?
•
What
data should an in

house
information
system collect?
What
design principles are involved for a large
database
?
•
What
is involved in modelling and IT

supported optimisation of key
business processes?
45
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