2. Statistics
Q1 I want to invest Rs 1000 , in which company
should I invest
Growth
Returns
Risk
3. Statistics
• Q2 How do I know which Company will
give me
• High Return
• Or
• Growth
• But Risk should be Low
4. Statistics
• Collect Information from the Past
• Qualitative
• Quantitative ( Data)
• Analyze Information ( Data) to provide
patterns of past performance ( Descriptive
Statistics)
• Project these patterns to answer your
questions ( Inference)
5. Types of data
Primary: You do a survey to find out
the percentage of people living below
the poverty line in Allahabad
Secondary: You are interested in
studying the performance of banks
and for that you study the RBI
published documents
7. Non frequency data
Time series representation of data
BSE(30) Close
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
3-Jan-94
5-Jan-94
7-Jan-94
11-Jan-94
13-Jan-94
17-Jan-94
19-Jan-94
24-Jan-94
27-Jan-94
31-Jan-94
2-Feb-94
4-Feb-94
8-Feb-94
10-Feb-94
14-Feb-94
16-Feb-94
18-Feb-94
22-Feb-94
24-Feb-94
28-Feb-94
1-Mar-94
3-Mar-94
7-Mar-94
9-Mar-94
15-Mar-94
17-Mar-94
21-Mar-94
23-Mar-94
25-Mar-94
29-Mar-94
31-Mar-94
Date
BSE(30)Close
8. Non frequency data
Spatial series representation of
data
Fertiliser Consumption for few Indian states for 1999-2000 (in tonnes)
0
500000
1000000
1500000
2000000
2500000
3000000
3500000
A
nd
hra
P
radeshK
arnata
ka
K
eralaT
am
ilN
adu
G
ujarat
M
adh
ya
P
rad
esh
M
aha
rashtraR
ajasthan
H
aryan
a
P
un
jab
U
ttar
P
radesh
B
ihar
O
rissa
W
e
stB
e
ngal
A
ssam
States
Fertiliserconsumption(intonnes)
Fertiliser Consumption
9. Frequency data:
Tabular representation
India at a glance Year
(% of GDP) 1983 1993 2002 2003
Agriculture 36.6 31.0 22.7 22.2
Industry 25.8 26.3 26.6 26.6
Mfg 16.3 16.1 15.6 15.8
Services 37.6 42.8 50.7 51.2
Pvt Consump 71.8 37.4 65.0 64.9
GOI consump 10.6 11.4 12.5 12.8
Import 8.1 10.0 15.6 16.0
Domes save 17.6 22.5 24.2 22.2
Interests paid 0.4 1.3 0.7 18.3
Note: 2003 refers to 2003-2004; data are preliminary. Gross domestic
savings figures are taken directly from India′s central statistical
organization.
10. Frequency data
Line diagram representation
Fertiliser Consumption for few Indian states for 1999-2000 (in tonnes)
0
500000
1000000
1500000
2000000
2500000
3000000
3500000
Andhra
PradeshKarnataka
KeralaTam
ilN
adu
G
ujarat
M
adhya
P
radesh
M
aharashtraR
ajasthan
H
aryana
Punjab
U
ttarPradesh
Bihar
O
rissa
W
estBengal
Assam
States
Fertiliserconsumption(intonnes)
Fertiliser Consumption
11. Frequency data
Bar diagram (histogram)
representation
World Population (projected mid 2004)
0
1000000000
2000000000
3000000000
4000000000
5000000000
6000000000
7000000000
1950 1960 1970 1980 1990 2000
Year
Population
World Population
12. Frequency data
Bar diagram (histogram)
representation
Height and Weight of individuals
0
20
40
60
80
100
120
140
160
180
200
Ram Shyam Rahim Praveen Saikat Govind Alan
Individual
Height/Weight
Height (in cms) Weight (in kgs)
13. Frequency data
Pie diagram/chart representation
Median marks in JMET (2003)
Verbal Quantitative Analytical Data Interpresentation
14. Frequency data
Box plot representation
The box plot is also called the box whisker
plot. A box plot is a set of five summary
measures of distribution of the data which
are
median
lower quartile
upper quartile
smallest observation
largest observation.
16. Frequency data
Box plot representation
Here:
UQ – LQ = Inter quartile range
(IQR)
X = Smallest observation within
certain percentage of LQ
Y = Largest observation within
certain percentage of UQ
17. Important note
A cumulative frequency diagram
is called the ogive. The abscissa
of the point of intersection
represents the median of the
data.
20. Measurements of uncertainty
Concept of uncertainty and different measures
of uncertainty, Probability as a measure of
uncertainty, Description of qualitative as well
as quantitative probability, Assessment of
probability, Concepts of Decision trees,
Random Variables, Distributions, Expectations,
Probability plots, etc.
21. Definitions
Quantitative variable: It can be described by a number
for which arithmetic operations such as averaging
make sense.
Qualitative (or categorical) variable: It simply records a
qualitative, e.g., good, bad, right, wrong, etc.
We know statistics deals with measurements, some
being qualitative others being quantitative. The
measurements are the actual numerical values of a
variable. Qualitative variables could be described by
numbers, although such a description might be
arbitrary, e.g., good = 1, bad = 0, right = 1, wrong = 0,
etc.
22. Scales of measurement
Nominal scale: In this scale numbers are used
simply as labels for groups or classes. If we
are dealing with a data set which consists of
colours blue, red, green and yellow, then we
can designate blue = 3, red = 4, green = 5 and
yellow = 6. We can state that the numbers
stand for the category to which a data point
belongs. It must be remembered that nothing
is sacrosanct regarding the numbering against
each category. This scale is used for
qualitative data rather than quantitative data.
23. Scales of measurement
Ordinal Scale: In this scale of
measurement, data elements may be
ordered according to relative size or
quality. For example a customer or a
buyer can rank a particular
characteristics of a car as good,
average, bad and while doing so he/she
can assign some numeric value which
may be as follows, characteristic good =
10, average = 5 and bad = 0.
24. Scales of measurement
Interval Scale: For the interval scale we specify intervals
in a way so as to note a particular characteristic, which
we are measuring and assign that item or data point
under a particular interval depending on the data point.
Consider we are measuring the age of school going
students between classes 5 to 12 in the city of Kanpur.
We may form intervals 10-12 years, 12-14 years,....., 18-
20 years. Now when we have one data point, i.e., the
age of a student we put that data under any one
particular interval, e.g. if the student's age is 11 years,
we immediately put that under the interval 10-12 years.
25. Scales of measurement
Ratio Scale: If two measurements are in
ratio scale, then we can take ratios of
measurements. The ratio scale
represents the reading for each recorded
data in a way which enables us to take a
ratio of the readings in order to depict it
either pictorially or in figures. Examples
of ratio scale are measurements of
weight, height, area, length etc.
26. Definitions: different measures
1) Measure of central tendency
Mean (Arithmetic mean (AM), Geometric
mean (GM), Harmonic mean (HM))
Median
Mode
1) Measure of dispersion
Variance or Standard deviation
Skewness
Kurtosis
27. Definition: Mean
Given N number of observations, x1, x2,
….., xN we define the following
].....[
1
21 NXXX
N
AM +++=
N
NXXXGM
1
21 ]*.....**[=
1
21
)]
1
.....
11
(
1
[ −
+++=
NXXXN
HM
28. Definition: Median and Mode
Median(µe) : The median of a data set is
the value below which lies half of the data
points. To find the median we use F (µe) =
0.5.
Mode (µo): The mode of a data set is the
value that occurs most frequently. Hence
f(µo) ≥ f(x); ∀ x.
30. Example
Consider we have the following data
points:
5, 7, 10, 7, 10, 11, 3, 5, 5
For these data points we have
µ = 7; µe = 10; µo = 5; σ2
= 6.89
31. Descriptive statistics
Suppose the data are available in the form of a
frequency distribution. Assume there are k
classes and the mid-points of the corresponding
class intervals being x1, x2,…., xk. While the
corresponding frequencies are f1, f2,….., fk, such
that n = f1+f2+…..+fk
Then: ∑
=
=
k
i
ii fx
n 1
1
µ ∑
=
−=
k
i
ii fxx
n 1
2
1
2
})(
1
{σ
32. Consider m groups of observations with
respective means µ1, µ2,….., µm and standard
deviations σ1, σ2,….., σm. Let the group sizes be
n1, n2,….., nm such that n = n1, n2,….., nm.
Then:
Descriptive statistics
∑
=
=
m
i
iiOVERALL f
n 1
1
µµ
2
1
1
2
1
2
}])({
1
[ ∑∑
==
−+=
m
i
OVERALLii
m
i
iiOVERALL nn
n
µµσσ
34. Random event
Random experiment: Is an experiment whose
outcome cannot be predicted with certainty.
Sample space (Ω) : The set of all possible
outcomes of a random experiment
Sample point (ωi): The elements of the
sample space
Event (A): Is a subset of the sample space
such that it is a collection of sample point(s).
35. Probability
Probability (P(A)): Of an event is defined as a quantitative
measure of uncertainty of the occurrence of the event
Objective probability: Based on game of chance and which
can be mathematically proved or verified . If the experiment
is the same for two different persons, then the value of
objective probability would remain the same. It is the
limiting definition of relative frequency. Example: be
probability of getting the number 5 when we roll a fair die.
Subjective probability: Based on personal judgment,
intuition and subjective criteria. Its value will change from
person to person. Example one person sees the chance of
India winning the test series with Australia high while the
other person sees it to be low.
36. Random event
For a random experiment, we denote
P(ωi) = pi
Where:
P(ωi) = pi = Probability of occurrence of the sample
point ωi
P(A) = Probability of occurrence of the event
∑
∈
=
A
i
i
pAP
ϖ
)(
1)( ==Ω ∑
Ω∈i
ipP
ϖ
37. Example 1
Suppose there are two dice each with faces 1, 2,....., 6
and they are rolled simultaneously. This rolling of the
two dice would constitute our random experiment
Then we have:
Ω = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1),.…..,
(5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}.
ωi = (1,1), (1,2),…., (6,5), (6,6)
We define the event is such that the outcomes for
each die are equal in one simultaneous throw, then A
= {(1, 1), (2, 2),….., (6, 6)}
P(ωi): p1 = p2 = ….. = p36 = 1/36
P(A) = p1 + p8 + p15 + p22 + p29 + p36 = 6/36
38. Example 2
Suppose a coin is tossed repeatedly till the first
head is obtained.
Then we have:
Ω = {(H), (T,H), (T,T,H),………}
ωi = (H), (T,H), (T,T,H),…..
We define the event such that at most 3 tosses
are needed to obtain the first head, then A =
{(H), (T,H), (T,T,H)}
P(ωi): p1 = ½, p2 = (½)2
, p3 = (½)3
, p4 = (½)4
,..…
P(A) = p1 + p2 + p3 = 7/8
39. Example 3
In a club there 10 members of whom 5 are
Asians and the rest are Americans. A committee
of 3 members has to be formed and these
members are to be chosen randomly. Find the
probability that there will be at least 1 Asian and
at least 1 American in the committee
Total number of cases = 10
C2 and the number of
cases favouring the formation of the committee
is 5
C2*5
C1 + 5
C1*5
C2
Hence P(A) = 100/120
40. Example 4
Suppose we continue with example 2
which we have just discussed and we
define the event B , that al least 5 tosses
are needed to produce the first head
Ω = {(H), (T,H), (T,T,H),………}
ωi = (H), (T,H), (T,T,H),…..
P(ωi): p1 = ½, p2 = (½)2
, p3 = (½)3
, p4 = (½)4
,..…
P(B) = p5+p6+p7+ ….. = 1 – (p1+p2+p3+p4)
41. Theorem in probability
For any event A, B ∈ Ω
0 ≤ P(A) ≤ 1
If A ⊂ B, then P(A) ≤ P(B)
P(A U B) = P(A) + P(B) – P(A ∩ B)
P(AC
) = 1 – P(A)
P(Ω) = 1
P(φ) = 0
42. Definitions
Mutually exclusive: Consider n events A1,
A2,….., An. They are mutually exclusive if no
two of them can occur together, i.e., P(Ai ∩
Aj) = 0. ∀ i, j (i ≠ j) ∈ n
Mutually exhaustive: Consider n events A1,
A2,….., An. They are mutually exhaustive if at
least one of them must occur and
P(A1UA2U…..UAn) = 1
43. Example 5
Suppose a fair die with faces 1, 2,….., 6 is rolled.
Then Ω = {1, 2, 3, 4, 5, 6}. Let us define the events
A1 = {1, 2}, A2 = {3, 4, 5, 6} and A3 = {3, 5}
The events A2 and A3 are neither mutually exclusive
nor exhaustive
A1 and A3 are mutually exclusive but not exhaustive
A1, A2 and A3 are not mutually exclusive but are
exhaustive
A1 and A2 are mutually exclusive and exhaustive
44. Conditional probability
Let A and B be two events such that P(B) > 0.
Then the conditional probability of A given B is
Assume Ω = {1, 2, 3, 4, 5, 6}, A = {2}, B = {2, 4, 6}.
Then A ∩ B = {2} and
)(
)(
)|(
BP
BAP
BAP
∩
=
3
1
63
61
)|( ==BAP 1
61
61
)|( ==ABP
45. Baye′s Theorem
Let B1, B2,….., Bn be mutually exclusive and
exhaustive events such that P(Bi) > 0, for
every i =1, 2,…., n and A be any event
such that then we have∑
=
=
n
i
ii BPBAPAP
1
)()|()(
∑
=
=
n
i
ii
jj
j
BPBAP
BPBAP
ABP
1
)()|(
)()|(
)|(
48. Distribution
Depending what are the outcomes of an
experiment a random variable (r.v) is used
to denote the outcome of the experiment
and we usually denote the r.v using X, Y
or Z and the corresponding probability
distribution is denoted by f(x), f(y) or f(z)
Discrete: probability mass function (pmf)
Continuous: probability density function
(pdf)
49. Discrete distribution
1) Uniform discrete distribution
2) Binomial distribution
3) Negative binomial distribution
4) Geometric distribution
5) Hypergeometric distribution
6) Poisson distribution
7) Log distribution
50. Bernoulli Trails
1) Each trial has two possible outcomes,
say a success and a failure.
2) The trials are independent
3) The probability of success remains the
same and so does the probability of
failure from one trial to another
51. Uniform discrete distribution
[X ~ UD (a , b) ]
f(x) = 1/n x = a, a+k, a+2k,….., b
a and b are the parameters where a, b ∈
R
E[X] =
V[X] =
Example: Generating the random
numbers 1, 2, 3,…, 10. Hence
X~UD(1,10) where a=1, k=1, b=10. Hence
n=10.
)1(
2
−+ n
k
a
)
12
1
(
2
2 −n
k
53. Binomial distribution
[X ~ B (p , n)]
f(x) = n
Cxpx
qn-x
x = 0, 1, 2,…..,n
n and p are the parameters where p ∈ [0, 1] and n ∈ Z+
E[X] = np
V[X] = npq
Example: Consider you are checking the quality of the
product coming out of the shop floor. A product can
either pass (with probability p = 0.8) or fail (with
probability q = 0.2) and for checking you take such 50
products (n = 50). Then if X is the random variable
denoting the number of success in these 50 inspections,
we have
X~50
Cx(0.8)x
(0.2)50-x
55. Negative binomial distribution
[X ~ NB (p , r)]
f(x) = r+x-1
Cr-1pr
qx
x = r, r+1,.…..
p and r are the parameters where p ∈ [0, 1] and r ∈ Z+
E[X] = rq/p
V[X] = rq/p2
Example: Consider the example above where you are
still inspecting items from the production line. But now
you are interested in finding the probability distribution
of the number of failures preceding the 5th
success of
getting the right product. Then, we have, considering
p=0.8, q=0.2
X ~ 5+x-1
C5-1(0.8)5
(0.2)x
57. Geometric distribution
[X ~ G (p)]
f(x) = pqx
x = 0,1,2,…..
p is the parameter where p ∈ [0, 1]
E[X] = q/p (r = 1 in the Negative Binomial distribution
case)
V[X] = q/p2
(r = 1 in the Negative Binomial distribution
case)
Example: Consider the example above. But now you
are interested in finding the probability distribution of
the number of failures preceding the 1st
success of
getting the right product. Then, we have considering
p=0.8, q=0.2
X~ (0.8)(0.2)x
59. Hypergeometric distribution
[X ~ HG (N, n, p)]
f(x) = Np
Cx
Nq
Cn-x/N
Cn 0 ≤ x ≤ Np and 0 ≤ (n – x) ≤ Nq
N, n and p are the parameters
E[X] = np
V[X] = npq{(N – n)/(N – 1)}
Example: Consider the example above. But now you are interested in
finding the probability distribution of the number of failures(success) of
getting the wrong(right) product when we choose n number of products
for inspection out of the total population N. If the population is 100 and
we choose 10 out of those, then the probability distribution of getting the
right product, denoted by X is given by
X~85
Cx
15
C10-x/100
C10
Remember
p (0.85) and q (0.15) are the proportions of getting a good item and
bad item respectively.
In this distribution we are considering the choosing is done without
replacement
61. Poisson distribution
[X ~ P (λ)]
f(x) = e-λ
λx
/x! x = 0,1,2,…..
λ is the parameter where λ > 0
E[X] = λ
V[X] = λ
Example: Consider the arrival of the number of
customers at the bank teller counter. If we are
interested in finding the probability distribution of the
number of customers arriving at the counter in specific
intervals of time and we know that the average
number of customers arriving is 5, then, we have
X~ e-5
5x
/x!
63. Log distribution
[X ~ L (p)]
f(x) = -(loge p)-1
x-1
(1 – p)x
x = 1,2,3,…..
p is the parameter where p ∈ (0, 1)
E[X] = -(1-p)/(plogep)
V[X] = -(1-p)[1 + (1 - p)/logep]/(p2
logep)
Example
1) Emission of gases from engines against fuel type
2) Used to represent the distribution of the number of
items of a product purchased by a buyer in a
specified period of time
65. Continuous distribution
1) Uniform distribution
2) Normal distribution
3) Exponential distribution
4) Chi-Square distribution
5) Gamma distribution
6) Beta distribution
7) Cauchy distribution
66. Continuous distribution
8) t-distribution
9) F-distribution
10)Log-normal distribution
11)Weibull distribution
12)Double exponential distribution
13)Pareto distribution
14)Logistic distribution
67. Uniform distribution
[X ~ U (a, b)]
f(x) = 1/(b – a) a ≤ x ≤ b
a and b are the parameters where a, b ∈
R and a < b
E[X] = (a+b)/2
V[X] = (b-a)2
/12
Example: Choosing any number between
1 and 10, both inclusive, from the real line
69. Normal distribution
[X ~ N ( µ, σ2
)]
-∞ < x < ∞
µX, σ2
X are the parameters where µX ∈ R and σ2
X >
0
E[X] = µX
V[X] = σ2
X
Example: Consider the average age of a student
between class VII and VIII selected at random
from all the schools in the city of Kanpur
2
2
2
)(
2
1
)( X
Xx
X
exf σ
µ
σπ
−
−
=
71. Log-normal distribution
[X ~ LN ( µ, σ2
)]
0 < x < ∞
µX, σ2
X are the parameters where µX ∈ R
and σ2
X > 0
E[X] = exp(µX+σ2
X/2)
V[X] = exp(2µX+σ2
X){exp(σ2
X)-1}
Example: Stock prices return distribution
2
2
2
)(log
1
2
1
)( X
Xe x
X
e
x
xf σ
µ
σπ
−
−
=
73. Relationship between Poisson and
Exponential distribution
If a process has the intervals
between successive events as
independent and identical and it is
exponentially distributed then the
number of events in a specified
time interval will be a Poisson
distribution
74. Normal distribution results
Normal distribution
0
0.05
0.1
0.15
0.2
0.25
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
x
f(x)
a b
76. Standard Normal distribution
Z ~ N(0,1) given by the equation
The area within an interval (a,b) is given by
which is not integrable
algebraically. The Taylor’s expansion of the above assists in
speeding up the calculation, which is
2
2
2
1
)(
z
ezf
−
=
π
dzebZaF
b
a
z
∫
−
=≤≤ 2
2
2
1
)(
π
∑
∞
=
+
+
−
+=≤
0
12
!2)12(
)1(
2
1
2
1
)(
k
k
kk
kk
z
zZF
π
77. Cumulative distribution function
(cdf) or the distribution function
We denote the distribution function by F(x)
∑
≤
=≤=
xx
i
i
xfxXPxF )()()(
∫∫
∞−∞−
==≤=
xx
xdFdxxfxXPxF )()()()(
78. Properties of distribution function
1) F(x) is non-decreasing in x, i.e., if x1 ≤
x2, then F(x1) ≤ F(x2)
2) Lt F(x) = 0 as x → - ∞
3) Lt F(x) = 1 as x → + ∞
4) F(x) is right continuous
79. Standard normal distribution
Putting Z=(X-µX)/σX in the normal distribution we
have the standard normal distribution
Where: µZ = 0 and σZ = 1
Remember
• F(x) = P(X ≤ x) = F(z) = F(Z ≤ z)
• f(z) = φ(z)
•
2
2
2
1
)(
z
ezf
−
=
π
)()()()()( zxdFdxzfzZPzF
zz
Φ===≤= ∫∫
∞−∞−
80. Standard normal distribution
Standard Normal distribution
-0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
z
f(z)
81. Exponential distribution
[ X ~ E (a, θ)]
a < x < ∞
a and θ are the parameters where a ∈ R
and θ > 0
E[X] = a + θ
V[X] = θ2
`
Example: The life distribution of the
number of hours a electric bulb survives.
θ
θ
)(
1
)(
ax
exf
−
−
=
85. Arrival time problem # 1
In a factory shop floor for a certain CNC machine (machine
marked # 1) the number of jobs arriving per unit time are given
below
# of Arrivals Frequency
0 2
1 4
2 4
3 1
4 2
5 1
6 4
7 6
8 4
9 1
86. Arrival time problem # 1
Frequency distribution of arrivals
0
1
2
3
4
5
6
7
0 1 2 3 4 5 6 7 8 9
# of Arrivals
Frequency
87. Arrival time problem # 1
Relative Frequency of Number of Arrivals
0
0.05
0.1
0.15
0.2
0.25
0
1
2
3
4
5
6
7
8
9
# of Arrivals
RelativeFrequency
88. Arrival time problem # 1
Cumulative Relative Frequency of Number of Arrivals
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
1
2
3
4
5
6
7
8
9
10
# of Arrivals
CumulativeRelativeFrequency
89. Arrival time problem # 1
1) The probability of number of arrivals of
jobs being equal to or more than 7 is
about 0.18.
2) The average number of arrival of jobs is
5.
3) The probability of number of arrivals of
jobs being less than or equal to 4 is
about 0.45.
91. SAT problem
This data set [ ] includes eight variables:
1) STATE: Name of state
2) COST: Current expenditure per pupil (measured in thousands of
dollars per averagedaily attendance in public elementary and
secondary schools)
3) RATIO: Average pupil/teacher ratio in public elementary and
secondary schools during Fall 1994
4) SALARY: Estimated average annual salary of teachers in public
elementary and secondary schools during 1994-95 (in thousands of
dollars)
5) PERCENT: percentage of all eligible students taking the SAT in
1994-95
6) VERBAL: Average verbal SAT score in 1994-95
7) MATH: Average math SAT score in 1994-95
8) TOTAL: Average total score on the SAT in 1994-95
93. SAT problem
Histograms for Cost and Ratio
0
5
10
15
20
25
30
Alabama
Arkansas
Connecticut
Georgia
Illinois
Kansas
Maine
Michigan
Missouri
Nevada
NewMexico
NorthDakota
Oregon
SouthCarolina
Texas
Virginia
Wisconsin
State
CostandRatio
COST RATIO
94. SAT problem
Histogramof Cost, Ratio and Salary
0
10
20
30
40
50
60
Alabama
Arizona
California
Connecticut
Florida
Hawaii
Illinois
Iowa
Kentucky
Maine
Massachusetts
Minnesota
Missouri
Nebraska
New
NewMexico
NorthCarolina
Ohio
Oregon
RhodeIslan
SouthDakota
Texas
Vermont
Washington
Wisconsin
State
Cost,RatioandSalary
COST RATIO SALARY
95. SAT problem
Average value is given by
Variance is given by
Covariance is given by
( ) ∑
=
=
n
i
iX
n
XE
1
1
( ) ( ){ }∑
=
−=
n
i
i XEX
n
XV
1
21
( ) ( ){ } ( ){ }[ ] ( ) ( )YVXVYEYXEXEYXCov YX ,, ρ=−−=
96. SAT problem
COST RATIO SALARY PERCENT VERBAL MATH TOTAL
Mean 5.90526 16.858 34.82892 35.24 457.14 508.78 965.92
Median 5.7675 16.6 33.2875 28 448 497.5 945.5
Maximum 9.774 24.3 50.045 81 516 592 1107
Minimum 3.656 13.8 25.994 4 401 443 844
Standard Deviation(1) 1.362807 2.266355 5.941265 26.76242 35.17595 40.20473 74.82056
Standard Deviation(2) 1.34911 2.243577 5.881552 26.49344 34.82241 39.80065 74.06857
97. SAT problem
COST RATIO SALARY PERCENT VERBAL MATH TOTAL
COST 1.820097 -1.12303 6.901753 21.18202 -19.2638 -18.7619 -38.0258
RATIO -1.12303 5.033636 -0.01512 -12.6639 4.98188 8.52076 13.50264
SALARY 6.901753 -0.01512 34.59266 96.10822 -97.6868 -93.9432 -191.63
PERCENT 21.18202 -12.6639 96.10822 701.9024 -824.094 -916.727 -1740.82
VERBAL -19.2638 4.98188 -97.6868 -824.094 1212.6 1344.731 2557.331
MATHS -18.7619 8.52076 -93.9432 -916.727 1344.731 1584.092 2928.822
TOTAL -38.0258 13.50264 -191.63 -1740.82 2557.331 2928.822 5486.154
98. SAT problem
COST RATIO SALARY PERCENT VERBAL MATH TOTAL
COST 1 -0.37103 0.869802 0.592627 -0.41005 -0.34941 -0.38054
RATIO -0.37103 1 -0.00115 -0.21305 0.063767 0.095422 0.081254
SALARY 0.869802 -0.00115 1 0.61678 -0.47696 -0.40131 -0.43988
PERCENT 0.592627 -0.21305 0.61678 1 -0.89326 -0.86938 -0.88712
VERBAL -0.41005 0.063767 -0.47696 -0.89326 1 0.970256 0.991503
MATHS -0.34941 0.095422 -0.40131 -0.86938 0.970256 1 0.993502
TOTAL -0.38054 0.081254 -0.43988 -0.88712 0.991503 0.993502 1
101. Types of sampling
• Probability Sampling
– Simple Random Sampling
– Stratified Random Sampling
– Cluster Sampling
– Multistage Sampling
– Systematic Sampling
• Judgement Sampling
– Quota Sampling
– Purposive Sampling
102. Simple Random Sampling
A simple random sampling of size (n)
from a finite population (N) is a
sample selected such that each
possible sample of size n has he same
probability of being selected. This
would be akin to SRSWOR
103. Simple Random Sampling
A simple random sampling of size (n) from
an infinite population (N) is a sample
selected such that the following conditions
hold
Each element selected comes from the
population.
Each element is selected independently
This would be akin to SRSWR
106. Chi-square distribution
0 ≤ x <∞
n is the parameter (degree of freedom) where n
∈ Z+
E[X] = n
V[X] = 2n
)
2
(1)
2
(
22)
2
(
1
)(
xn
n
ex
n
xf
−−
Γ
=
]~[ 2
nX χ
116. Some results
If X1, X2,….., Xn are ′mX′ observations from X ~
N(µX, σ2
X) and Y1, Y2,….., Ym are ′mY′ observations
from Y ~ N(µY, σ2
Y) and more over these samples
are independent then
1,12
2
2
2
2
2
2
2
~))((
)1(
)1(
)1(
)1(
−−=
−
−
−
−
YX mm
Y
X
X
Y
Y
Y
YY
X
X
XX
F
S
S
m
Sm
m
Sm
σ
σ
σ
σ
118. Estimators and their properties
Estimator: Any statistic (a random
function) which is used to estimate the
population parameter
Unbiasedness
Eθ (tn) = θ
Consistency
P[|tn - θ| < ε] = 1 as n → ∞
119. Estimators (Discrete distribution)
1) X ~ UD (a , b) then and
2) X ~ P (λ), then
3) X ~ B (n, p), then
),...,min(ˆ 1 nXXa =
),...,max(ˆ
1 nXXb =
nX=λˆ
n
favouring
p
#
ˆ =
120. Estimators (Continuous distribution)
1) X ~ N(µ, σ2
), then
2) X ~ N(µ, σ2
) and if µ is known then
3) X ~ N(µ, σ2
) and if µ is unknown then
4) X ~ E(θ), then
∑
=
−=
n
i
iX
n 1
22
}{
1
ˆ µσ
nX=µˆ
nX=θˆ
∑
=
−
−
=
n
i
ni XX
n 1
22
}{
)1(
1
ˆσ
121. Examples (Estimation)
Number of jobs arriving in a unit time for a CNC
machine Consider we choose from a discrete
distribution whose population distribution [X ~
UD (20, 35)] is not know. We select the values
from the distribution and the numbers sampled
are 22, 34, 33, 21, 29, 29, 30. Then the best
estimate for a, i.e., and the best estimate
for b, i.e.,
21ˆ =a
34ˆ =b
122. Examples (Estimation)
You are testing the components coming
out of the shop floor and find that 9 out of
30 components fail. Then the estimated
value of p (proportion of bad items in the
population) = 9/30.
123. Examples (Estimation)
At a particular teller machine in one of the
bank branches it was found that the
number of customers arriving in an unit
time span were 4, 6, 7, 4, 3, 5, 6, and 5.
Then for this Poisson process the
estimated value of λ is 5.
124. Examples (Estimation)
Suppose it is known that the survival time
of a particular type of bulb has the
exponential distribution. You test 10 such
bulbs and find their respective survival
times as 150, 225, 275, 300, 95, 155, 325,
75, 20 and 400 hours respectively. Then
the estimated value of θ = 202.
126. Multiple Linear Regression
Given ′k′ independent variables X1, X2,….., Xk
and one dependent variable Y we predict the
value of Y given by or y using the values of
Xi ′s. We need ′n′ ( n ≥ k+1) data points and the
multiple linear regression (MLR) equation is as
follows:
Yj = β1x1,j + β2x2,j +…..+ βkxk,j + εj
∀ j = 1, 2,….., n
Yˆ
127. Multiple Linear Regression
Note
There is no randomness in measuring Xi
The relationship is linear and not non-
linear. By non-linear we mean that at least
one derivative of Y wrt βi′s is a function of
at least one of the parameters. By
parameters we mean the βi′s.
128. Multiple Linear Regression
Assumptions for the MLR
Xi, Y are normally distributed
Xi are all non-stochastic
εj ~ N(0,σ2
I)
rank(X) = k
n ≥ k
No dependence between the Xj′s, i.e., the rank
of the matrix X is
E(εjεl) = 0 ∀ i, j = 1, 2,….., n
Cov(Xi,εj) = 0 ∀ i ≠ j, i, j = 1, 2,….., n
129. Multiple Linear Regression
Find β1, β2,….., βk using the concept of
minimizing the sum of square of errors. This
is also known as least square method or
method of ordinary least square. The
estimates found are the
estimates of β1, β2,….,βk respectively.
Utilize these estimates to find the forecasted
value of Y (i.e., or y) and compare those
with actual values of Y obtained in future.
kβββ ˆ,.....,ˆ,ˆ
21
Yˆ
130. To check for normality of data
We need to check for the normality of Xi′s and Y
1) List the observation number in the column # 1,call it i.
2) List the data in column # 2.
3) Sort the data from the smallest to the largest and place in
column # 3.
4) For each ith
of the n observations, calculate the
corresponding tail area of the standard normal distribution
(Z) as follows, A = (i – 0.375)/(n + 0.25). Put the values in
column # 4.
5) Use NORMSINV(A) function in MS-EXCEL to produce a
column of normal scores. Put these values in column # 5.
6) Make a copy of the sorted data (be sure to use paste
special and paste only the values) in column # 6.
7) Make a scatter plot of the data in columns # 5 and # 6.
131. To check for normality of data
Checking normality of data
0
50
100
150
200
250
300
350
400
450
-2.5 -1.6 -1.2 -1.0 -0.8 -0.6 -0.5 -0.3 -0.2 -0.1 0.1 0.2 0.3 0.5 0.6 0.8 1.0 1.2 1.6 3.0
Normal Score
Data
132. Basic transformation for MLR
Some transformation to convert to MLR
X → X(p)
= {Xp
– 1}/p, as p → 0 then X(p)
=
logeX
If the variability in Y increases with
increasing values of Y, then we use
logeY = β1logeX1 + β2logeX2 +….. + βklogeXk +
ε
133. Simple linear regression
In the simple linear regression we have
Yj = α + βXj + εj ∀ j = 1,2,…..,n
The question is how do we find α and β, provided we have ′n′
number of observations which constitutes the sample.
We minimize the sum of square of the error wrt to α and β
Finally: )(ˆˆ)( XEYE βα+=
∑
=
+−=∆
n
j
jj XY
1
2
)}ˆˆ({ βα
})()({ˆ)(ˆ)()(),cov( 2
XEXVXEYEXEYX ++=+ βα
134. Simple linear regression
After we have found out the estimators of α
and β, we use these values to predict/forecast
the subsequent future values of Y, i.e., we find
out y and compare those y′s with the
corresponding values of Y. Thus we find
and compare them with
corresponding values of Yk , for k = n+1, n+2,
…….
kk Xy βα ˆˆ +=
136. Non-linear regression
• y = (β + γX)/(1 + αX)
• y = α(X - β)γ
• y = α - βloge(X + γ)
• y = α - βloge(X + γ)
• y = α[1 – exp(-βX)]γ
NOTE: For all these and other models we
minimize the sum of squares and find the
parameters α, β and γ.
137. Simple Moving Average
3MA = 3 month Moving Average
5MA = 5 month Moving Average
Month Actual 3MA 5MA
Jan 266.00
Feb 145.90 198.33
Mar 183.10 149.43 178.92
Apr 119.30 160.90 159.42
May 180.30 156.03 176.60
Jun 168.50 193.53 184.88
Jul 231.80 208.27 199.58
Aug 224.50 216.37 188.10
Sep 192.80 180.07 221.70
Oct 122.90 217.40 212.52
Nov 336.50 215.10 206.48
Dec 185.90 238.90 197.82
Jan 194.30 176.57 215.26
Feb 149.50 184.63 202.62
Mar 210.10 210.97 203.72
Apr 273.30 224.93 222.26
May 191.40 250.57 237.56
Jun 287.00 234.80 256.26
Month Actual 3MA 5MA
Jul 226.00 272.20 259.58
Aug 303.60 273.17 305.62
Sep 289.90 338.37 301.12
Oct 421.60 325.33 324.38
Nov 264.50 342.80 331.60
Dec 342.30 315.50 361.70
Jan 339.7 374.13 340.56
Feb 440.4 365.33 375.52
Mar 315.9 398.53 387.32
Apr 439.3 385.50 406.86
May 401.3 426.00 433.88
Jun 437.4 471.40 452.22
Jul 575.5 473.50 500.76
Aug 407.6 555.03 515.56
Sep 682 521.63 544.34
Oct 475.3 579.53 558.62
Nov 581.3 567.83
Dec 646.9
138. Simple Moving Averages
Actual, 3MA, 5MA
Simple Moving Averages (Yellow is actual; Blue is 3MA; Red is 5MA)
0
100
200
300
400
500
600
700
800
Jan
M
ar
M
ay
Jul
Sep
N
ov
Jan
M
ar
M
ay
Jul
Sep
N
ov
Jan
M
ar
M
ay
Jul
Sep
N
ov
Month
Value
139. Centred Moving Average
4MA(1) = 4 month Moving Average, 4MA(2) = 4 month Moving Average
2X4MA=Averages of 2MA(1) and 4MA(2)
Mth Actual 4MA(1) 4MA(2) 2X4MA
Jan 266.00
Feb 145.90
Mar 183.10 178.58 157.15 167.86
Apr 119.30 157.15 162.80 159.98
May 180.30 162.80 174.98 168.89
Jun 168.50 174.98 201.28 188.13
Jul 231.80 201.28 204.40 202.84
Aug 224.50 204.40 193.00 198.70
Sep 192.80 193.00 219.18 206.09
Oct 122.90 219.18 209.53 214.35
Nov 336.50 209.53 209.90 209.71
Dec 185.90 209.90 216.55 213.23
Jan 194.30 216.55 184.95 200.75
Feb 149.50 184.95 206.80 195.88
Mar 210.10 206.80 206.08 206.44
Apr 273.30 206.08 240.45 223.26
May 191.40 240.45 244.43 242.44
Jun 287.00 244.43 252.00 248.21
Mth Actual 4MA(1) 4MA(2) 2X4MA
Jul 226.00 252.00 276.63 264.31
Aug 303.60 276.63 310.28 293.45
Sep 289.90 310.28 319.90 315.09
Oct 421.60 319.90 329.58 324.74
Nov 264.50 329.58 342.03 335.80
Dec 342.30 342.03 346.73 344.38
Jan 339.7 346.73 359.58 353.15
Feb 440.4 359.58 383.83 371.70
Mar 315.9 383.83 399.23 391.53
Apr 439.3 399.23 398.48 398.85
May 401.3 398.48 463.38 430.93
Jun 437.4 463.38 455.45 459.41
Jul 575.5 455.45 525.63 490.54
Aug 407.6 525.63 535.10 530.36
Sep 682 535.10 536.55 535.83
Oct 475.3 536.55 596.38 566.46
Nov 581.3 596.38
Dec 646.9
140. Centred Moving Average
Actual, 4MA(1), 4MA(2), 2X4MA
Centred Moving Average (Yellow is Actual; Red is 4MA(1); Blue is 4MA(2); Green is
2X4MA)
0
100
200
300
400
500
600
700
800
Jan
M
ar
M
ay
Jul
Sep
Nov
Jan
M
ar
M
ay
Jul
Sep
Nov
Jan
M
ar
M
ay
Jul
Sep
Nov
Month
Values
141. Centred Moving Average
Note:
4MA(1)=(Y1+Y2+Y3+Y4)/4
4MA(2)=(Y2+Y3+Y4+Y5)/4
2X4MA=(Y1+2*Y2+2*Y3+2*Y4+Y5)/8
Similarly we can have 2X6MA, 2X8MA,
2X12MA etc.
143. Weighted Moving Averages
In general a weighted k-point moving
average can be written as
Note:
The total of the weights is equal to 1
Weights are symmetric, i.e., aj = a-j
∑
−=
+=
m
mj
jtjt YaT
145. Exponential Smoothing Methods
1) Single Exponential Smoothing (one
parameter, adaptive parameter)
2) Holt′s linear method (suitable for trends)
3) Holt-Winter′s method (suitable for trends
and seasonality)
146. Single Exponential Smoothing
The general equation is:
Ft+1 = Ft + α(Yt – Ft) = αYt + (1 - α)Ft,
Note:
Error term: Et = Yt - Ft
Forecast value: Ft
Actual value: Yt
Weight: α ∈ (0,1)
α is such that sum of square of errors is minimized
147. Single Exponential Smoothing
Month Y(t) F(t, 0.1) F(t.0.5) F(t,0.9)
Jan 200.0 200.0 200.0 200.0
Feb 135.0 200.0 200.0 200.0
Mar 195.0 193.5 167.5 141.5
Apr 197.5 193.7 181.3 189.7
May 310.0 194.0 189.4 196.7
Jun 175.0 205.6 249.7 298.7
Jul 155.0 202.6 212.3 187.4
Aug 130.0 197.8 183.7 158.2
Sep 220.0 191.0 156.8 132.8
Oct 277.5 193.9 188.4 211.3
Nov 235.0 202.3 233.0 270.9
Dec ------- 205.6 234.0 238.6
148. Single Exponential Smoothing
Single Exponential Smoothing
0
50
100
150
200
250
300
350
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Month
Values
Y(t) F(t,0.1) F(t,0.5) F(t,0.9)
149. Extension of Exponential Smoothing
The general equation is:
Ft+1 = α1Yt + α2Ft + α3Ft-1
Note:
Error term: Et = Yt – Ft
Forecast value: Ft
Actual value: Yt
Weights: αi ∈ (0,1) ∀ i = 1, 2 and 3
α1 + α2 + α3 = 1
αi′s are such that the sum of square of errors is
minimized
150. Extension of Exponential Smoothing
Extension of Exponential Smoothing
0
50
100
150
200
250
300
350
400
450
Jan Jul Jan Jul Jan Jul Jan Jul Jan Jul Jan Jul Jan Jul Jan Jul
Month
Values
Y(t) F(t)
151. Adaptive Exponential Smoothing
The general equation is:
Ft+1 = αtYt + (1 - αt)Ft
Note:
Error term: Et = Yt – Ft
Forecast value: Ft
Actual value: Yt
Smoothed Error: At = βEt + (1 - β)At-1
Absolute Smoothed Error: Mt = β|Et| + (1 - β)Mt-1
Weight: αt+1 = |At/Mt|
α and β are such that sum of square of errors is
minimized
153. Adaptive Exponential Smoothing
Month Y(t) F(t) E(t) A(t) M(t) α β
Jan 200.0 0.0 0.0 0.2
Feb 135.0 200.0 -65.0 -13.0 13.0 0.2
Mar 195.0 187.0 8.0 -8.8 12.0 1.0
Apr 197.5 188.6 8.9 -5.3 11.4 0.7
May 310.0 190.4 119.6 19.7 33.0 0.5
Jun 175.0 214.3 -39.3 7.9 34.3 0.6
Jul 155.0 206.4 -51.4 -4.0 37.7 0.2
Aug 130.0 196.2 -66.2 -16.4 43.4 0.1
Sep 220.0 182.9 37.1 -5.7 42.1 0.4
Oct 277.5 190.3 87.2 12.9 51.1 0.1
Nov 235.0 207.8 27.2 15.7 46.4 0.3
Dec 213.2 0.3
154. Adaptive Exponential Smoothing
Adaptive Exponential Smoothing
0
50
100
150
200
250
300
350
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Month
Values
Y(t) F(t)
155. Adaptive Exponential Smoothing
Adaptive Exponential Smoothing
-100.00
-50.00
0.00
50.00
100.00
150.00
200.00
250.00
300.00
350.00
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Time
Values
A(t) M(t) E(t) F(t) Y(t)
156. Holt-Winter′s Method
The general equations are:
1) Lt = αYt/St-s + (1-α)(Lt-1 + bt-1)
2) bt = β(Lt – Lt-1) + (1 - β)bt-1
3) St = γYt/Lt + (1 - γ )St-s for t > s
4) Ft+m = (Lt + btm)St-s+m
5) Si = Yi/Ls where Ls = (Y1+…+Ys)/s for i ≤ s
Note:
Forecast value: Ft
Actual value: Yt
Trend: bt
Seasonal component: St
Length of seasonality: s
α, β and γ are chosen such that the sum of square of errors is minimized
