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Unformatted text preview: STATISTICS
HIGHER SECONDARY  SECOND YEAR Untouchability is a sin
Untouchability is a crime
Untouchability is inhuman TAMILNADU
TEXTBOOK CORPORATION
College Road, Chennai 600 006 i ©Government of Tamilnadu
First Edition – 2005
Reprint – 2006 Chairperson
Dr. J. Jothikumar
Reader in Statistics
Presidency College
Chennai  600 005.
Reviewers and Authors
Thiru K.Nagabushanam
Dr. R.Ravanan
S.G.Lecturer in Statistics
Presidency College
Chennai  600 005. Reader in Statistics
Presidency College
Chennai  600 005. Authors
Thiru G.Gnana Sundaram Tmt.N.Suseela P.G.Teacher
S.S.V. Hr. Sec. School
Parktown, Chennai  600 003. P.G.Teacher
Anna Adarsh Matric HSS
Annanagar, Chennai 600 040 Tmt.S.Ezhilarasi
P.G.Teacher
P.K.G.G. Hr. Sec. School
Ambattur, Chennai 600 053. Thiru A. S. Sekar
P.G.Teacher
O.R.G.N. Govt Boys HSS
Redhills, Chennai  600 052 Price: Rs.
This book has been prepared by the Directorate of School
Education on behalf of the Government of Tamilnadu.
This book has been printed on 60 G.S.M paper
Printed by offset at:
ii PREFACE
We take great pleasure in presenting the book on
Statistics to the students of the Second year Higher Secondary
classes.
This book has been written in conformity with the
revised syllabus. The book is designed to be selfcontained and
comprises of ten chapters and includes two new chapters
Association of attributes and Decision Theory. Besides the
additional (new) topics covered in this book, all the chapters
have been completely rewritten and simplified in many ways.
The book covers the theoretical, practical and applied
aspects of statistics as far as possible in a clear and exhaustive
manner. Every chapter in this book explains the principles
through appropriate examples in a graded manner. A set of
exercise concludes each chapter to provide an opportunity for
the students to reinforce what they learn, to test their progress
and increase their confidence.
The book is very helpful to the students who take their
higher studies and the professional courses like Charted
Accountants and ICWA
At the end of this textbook, necessary statistical tables
are included for the convenience of the students.
We welcome suggestions from students, teachers and
academicians so that this book may further be improved upon.
We thank everyone who has a helping hand in the lent
preparation of this book. Dr. J. Jothikumar
Chairperson
Writing team
iii CONTENTS
Probability 1. Page
1 1.0 Introduction 1 1.1 Definition and basic concepts 1 1.2 Definitions of Probability 3 1.3 Addition theorem on probability 5 1.4 Conditional Probability 8 1.5 Multiplication theorem on Probability 9 1.6 Bayes’ theorem 10 1.7 11 2.0 Basic principles of permutation and
combination
Random Variable and Mathematical
Expectation
Introduction 2.1 Random Variable 37 2.2 Probability mass function 39 2.3 Properties of distribution function 41 2.4 45 2.5 An introduction to elementary
calculus
Mathematical Expectation 2.6 Moment generating function 61 2.7 Characteristic function 62 3.1 Some important Theoretical
Distributions
Binomial Distribution 3.2 Poisson Distribution 2. 3. iv 37
37 53 68
68
78 3.3
4. Normal Distribution 87 Test of significance (Basic Concepts) 110 4.0 Introduction 110 4.1 Parameter and statistic 110 4.2 Sampling Distribution 110 4.3 Standard Error 111 4.4
4.5 Null hypothesis and Alternative
hypothesis
Level of significance and critical value 4.6 One Tailed and Two Tailed Tests 115 4.7 Type I and Type II errors 117 4.8 Test procedure 118 Test of Significance (Large Samples) 122 5.0 Introduction 122 5.1 Large Samples (n >30) 122 5.2 Test of significance for proportion 123 5.3 Test of significance for difference
between two proportions 126 5.4 Test of significance for Mean 131 5.5 Test of significance for difference
between two means
Test of significance (Small Samples) 135 6.0 Introduction 143 6.1 tstatistic definition 143 6.2 Test of significance for Mean 146 5. 6. v 112
113 143 6.3 Test of significance for difference
between two means 150 6.4 Chisquare distribution 159 6.5 161 6.6 Testing the Goodness of fit (Binomial
and Poisson distribution)
Test of Independence 6.7 Test for Population variance 172 6.8 Fstatistic definition 174 Analysis of Variance 186 7.0 Introduction 186 7.1 Definition 186 7.2 Assumptions 187 7.3 Oneway classification 187 7.4 Test Procedure 188 7.5 Twoway Classification 194 7.6 Test Procedure for twoway
classification 195 Time Series 207 8.0 Introduction 207 8.1 Definition 207 8.2 Components of Time Series 208 8.3 Method of Least Square 217 8.4 Seasonal Variation 222 8.5 Forecasting 227 7. 8. vi 165 9. 235 9.0 Introduction 235 9.1 Notations 235 9.2 Classes and class frequencies 235 9.3 Consistency of the data 237 9.4 Independence of Attributes 238 9.5
10. Theory of Attributes Yules’ coefficient of Association 241 Decision Theory 249 10.0 Introduction 249 10.1 Payoff 252 10.2 Decision making under certainty
(without probability)
10.3 Decision making under risk (with
probability)
10.4 Decision Tree Analysis 256 vii 262
266 1. PROBABILITY
1.0 Introduction:
The theory of probability has its origin in the games of
chance related to gambling such as tossing of a coin, throwing of a
die, drawing cards from a pack of cards etc. Jerame Cardon, an
Italian mathematician wrote ‘ A book on games of chance’ which
was published on 1663. Starting with games of chance, probability
has become one of the basic tools of statistics. The knowledge of
probability theory makes it possible to interpret statistical results,
since many statistical procedures involve conclusions based on
samples.
Probability theory is being applied in the solution of social,
economic, business problems. Today the concept of probability
has assumed greater importance and the mathematical theory of
probability has become the basis for statistical applications in both
social and decisionmaking research. Probability theory, in fact, is
the foundation of statistical inferences.
1.1 Definitions and basic concepts:
The following definitions and terms are used in studying
the theory of probability.
Random experiment:
Random experiment is one whose results depend on
chance, that is the result cannot be predicted. Tossing of coins,
throwing of dice are some examples of random experiments.
Trial:
Performing a random experiment is called a trial.
Outcomes:
The results of a random experiment are called its outcomes.
When two coins are tossed the possible outcomes are HH, HT,
TH, TT.
1 Event:
An outcome or a combination of outcomes of a random
experiment is called an event. For example tossing of a coin is a
random experiment and getting a head or tail is an event.
Sample space:
Each conceivable outcome of an experiment is called a
sample point. The totality of all sample points is called a sample
space and is denoted by S. For example, when a coin is tossed, the
sample space is S = { H, T } . H and T are the sample points of the
sample space S.
Equally likely events:
Two or more events are said to be equally likely if each one
of them has an equal chance of occurring. For example in tossing
of a coin, the event of getting a head and the event of getting a tail
are equally likely events.
Mutually exclusive events:
Two or more events are said to be mutually exclusive,
when the occurrence of any one event excludes the occurrence of
the other event. Mutually exclusive events cannot occur
simultaneously.
For example when a coin is tossed, either the head or the
tail will come up. Therefore the occurrence of the head completely
excludes the occurrence of the tail. Thus getting head or tail in
tossing of a coin is a mutually exclusive event.
Exhaustive events:
Events are said to be exhaustive when their totality includes
all the possible outcomes of a random experiment. For example,
while throwing a die, the possible outcomes are {1, 2, 3, 4, 5, 6}
and hence the number of cases is 6.
Complementary events:
The event ‘ A occurs’ and the event ‘ A does not occur’ are
called complementary events to each other. The event ‘ A does not
occur’ is denoted by A′ o rA or Ac. The event and its complements
are mutually exclusive. For example in throwing a die, the event
of getting odd numbers is { 1, 3, 5 } and getting even numbers is
2 { 2, 4, 6} .These two events are mutually exclusive and complement
to each other.
Independent events:
Events are said to be independent if the occurrence of one
does not affect the others. In the experiment of tossing a fair coin,
the occurrence of the event ‘ head’ in the first toss is independent of
the occurrence of the event ‘ head’ in the second toss, third toss and
subsequent tosses.
1.2 Definitions of Probability:
There are two types of probability. They are Mathematical
probability and Statistical probability.
1.2.1 Mathematical Probability (or a priori probability):
If the probability of an event can be calculated even before
the actual happening of the event, that is, even before conducting
the experiment, it is called Mathematical probability.
If the random experiments results in
exhaustive,
mutually exclusive and equally likely cases, out of which
are
favourable to the occurrence of an event A, then the ratio m/n is
called the probability of occurrence of event A, denoted by P(A), is
given by
m
Number of cases favourable to the event A
P(A) =
=
n
Total number of exhaustive cases
Mathematical probability is often called classical
probability or a priori probability because if we keep using the
examples of tossing of fair coin, dice etc., we can state the answer
in advance (prior), without tossing of coins or without rolling the
dice etc.,
The above definition of probability is widely used, but it
cannot be applied under the following situations:
(1) If it is not possible to enumerate all the possible outcomes for
an experiment.
(2) If the sample points(outcomes) are not mutually independent.
(3) If the total number of outcomes is infinite.
(4) If each and every outcome is not equally likely.
3 Some of the drawbacks of classical probability are removed
in another definition given below:
1.2.2 Statistical Probability (or a posteriori probability):
If the probability of an event can be determined only after
the actual happening of the event, it is called Statistical
probability.
If an event occurs m times out of n, its relative frequency is
m/n.
In the limiting case, when n becomes sufficiently large it
corresponds to a number which is called the probability of that
event.
In symbol, P(A) = Limit (m/n)
n→∞
The above definition of probability involves a concept
which has a long term consequence. This approach was initiated by
the mathematician Von Mises .
If a coin is tossed 10 times we may get 6 heads and 4 tails
or 4 heads and 6 tails or any other result. In these cases the
probability of getting a head is not 0.5 as we consider in
Mathematical probability.
However, if the experiment is carried out a large number of
times we should expect approximately equal number of heads and
tails and we can see that the probability of getting head approaches
0.5. The Statistical probability calculated by conducting an actual
experiment is also called a posteriori probability or empirical
probability.
1.2.3 Axiomatic approach to probability:
The modern approach to probability is purely axiomatic
and it is based on the set theory. The axiomatic approach to
probability was introduced by the Russian mathematician A.N.
Kolmogorov in the year 1933.
Axioms of probability:
Let S be a sample space and A be an event in S and P(A) is
the probability satisfying the following axioms:
4 (1)
(2)
(3) The probability of any event ranges from zero to one.
i.e
0 ≤ P(A) ≤ 1
The probability of the entire space is 1.
i.e
P(S) = 1
If A1, A2,…is a sequence of mutually exclusive events in
S, then
P (A1 ∪ A2 ∪ … = P(A1) + P(A2) +...
) Interpretation of statistical statements in terms of set theory:
S ⇒ Sample space
A ⇒ A does not occur
A ∪ A = S
A ∩ B = φ ⇒ A and B are mutually exclusive.
A ∪ B ⇒ Event A occurs or B occurs or both A and B occur.
(at least one of the events A or B occurs)
A ∩ B ⇒ Both the events A and B occur.
A ∩ B ⇒ Neither A nor B occurs
A ∩ B ⇒ Event A occurs and B does not occur
A ∩ B ⇒ Event A does not occur and B occur.
1.3 Addition theorem on probabilities:
We shall discuss the addition theorem on probabilities for
mutually exclusive events and not mutually exclusive events.
1.3.1 Addition theorem on probabilities for mutually exclusive
events:
If two events A and B are mutually exclusive, the
probability of the occurrence of either A or B is the sum of
individual probabilities of A and B. ie P(AUB) = P(A) + P(B)
This is clearly stated in axioms of probability. A B 5 1.3.2 Addition theorem on probabilities for notmutually
exclusive events:
If two events A and B are notmutually exclusive, the
probability of the event that either A or B or both occur is given as
P(AUB) = P(A) + P(B) – P(A I B)
Proof:
Let us take a random experiment with a sample space S of
N sample points.
Then by the definition of probability ,
P(AUB) = n (AUB)
n (AUB)
=
n (S)
N
S B
A A I B AI B AI B From the diagram, using the axiom for the mutually
exclusive events, we write
n( A) + n( A I B)
P(AUB) =
N
Adding and subtracting n( A I B ) in the numerator,
n( A) + n( A I B) + n( A I B) − n( A I B)
=
N
n( A) + n( B) − n( A I B)
=
N
n( A) n( B) n( A I B)
=
+
−
N
N
N
P(AUB) = P(A) + P(B) – P(A I B)
6 Note:
In the case of three events A,B,C, P(AUBUC) = P(A) +
P(B) + P(C) – P( A I B ) – P( A I B ) – P( B I C ) + P ( A I B I C )
Compound events:
The joint occurrence of two or more events is called
compound events. Thus compound events imply the simultaneous
occurrence of two or more simple events.
For example, in tossing of two fair coins simultaneously,
the event of getting ‘ atleast one head’ is a compound event as it
consists of joint occurrence of two simple events.
Namely,
Event A = one head appears ie A = { HT, TH} and
Event B = two heads appears ie B = {HH}
Similarly, if a bag contains 6 white and 6 red balls and we
make a draw of 2 balls at random, then the events that ‘ both are
white’ or one is white and one is red’ are compound events.
The compound events may be further classified as
(1) Independent event
(2) Dependent event
Independent events:
If two or more events occur in such a way that the
occurrence of one does not affect the occurrence of another, they
are said to be independent events.
For example, if a coin is tossed twice, the results of the
second throw would in no way be affected by the results of the first
throw.
Similarly, if a bag contains 5 white and 7 red balls and then
two balls are drawn one by one in such a way that the first ball is
replaced before the second one is drawn. In this situation, the two
events, ‘ the first ball is white’ and ‘ second ball is red’ , will be
independent, since the composition of the balls in the bag remains
unchanged before a second draw is made.
Dependent events:
If the occurrence of one event influences the occurrence of
the other, then the second event is said to be dependent on the first.
7 In the above example, if we do not replace the first ball
drawn, this will change the composition of balls in the bag while
making the second draw and therefore the event of ‘ drawing a red
ball’ in the second will depend on event (first ball is red or white)
occurring in first draw.
Similarly, if a person draw a card from a full pack and does
not replace it, the result of the draw made afterwards will be
dependent on the first draw.
1.4 Conditional probability:
Let A be any event with p(A) >0. The probability that an
event B occurs subject to the condition that A has already occurred
is known as the conditional probability of occurrence of the event
B on the assumption that the event A has already occurred and is
denoted by the symbol P(B/A) or P(BA) and is read as the
probability of B given A.
The same definition can be given as follows also:
Two events A and B are said to be dependent when A can
occur only when B is known to have occurred (or vice versa). The
probability attached to such an event is called the conditional
probability and is denoted by P(B/A) or, in other words,
probability of B given that A has occurred.
If two events A and B are dependent, then the conditional
probability of B given A is
P(B/A) = P(A I B)
P ( A) Similarly the conditional probability of A given B is given as P(A/B) = P(A I B)
P(B) Note:
If the events A and B are independent, that is the
probability of occurrence of any one of them P(A/B) = P(A) and
P(B/A) = P(B)
8 1.5 Multiplication theorem on probabilities:
We shall discuss multiplication theorem on probabilities for
both independent and dependent events.
1.5.1 Multiplication theorem on probabilities for independent
events:
If two events A and B are independent, the probability that
both of them occur is equal to the product of their individual
probabilities. i.e P(A I B) = P(A) .P(B)
Proof:
Out of n1 possible cases let m1 cases be favourable for the
occurrence of the event A.
m1
∴P(A) =
n1
Out of n2 possible cases, let m2 cases be favourable for the
occurrence of the event B
m
∴ P(B) = 2
n2
Each of n1 possible cases can be associated with each of the
n2 possible cases.
Therefore the total number of possible cases for the
occurrence of the event ‘ A’ and ‘ B’ is n1 × n2 . Similarly each of
the m1 favourable cases can be associated with each of the m2
favourable cases. So the total number of favourable cases for the
event ‘ A’ and ‘ B’ is m1 × m2
mm
∴ P(A I B) = 1 2
n1 n 2
m
m2
= 1.
n1
n2
= P(A).P(B)
Note:
The theorem can be extended to three or more independent
events.
If
A,B,C….
… be
independent
events,
then
P(A I B I C….) = P(A).P(B).P(C)….
…
…
9 Note:
If A and B are independent then the complements of A and
B are also independent. i.e P( A I B ) = P( A ) . P( B )
1.5.2 Multiplication theorem for dependent events:
If A and B be two dependent events, i.e the occurrence of
one event is affected by the occurrence of the other event, then the
probability that both A and B will occur is
P(A I B) = P(A) P(B/A)
Proof:
Suppose an experiment results in n exhaustive, mutually
exclusive and equally likely outcomes, m of them being favourable
to the occurrence of the event A.
Out of these n outcomes let m1 be favourable to the
occurrence of another event B.
Then the outcomes favourable to the happening of the
events ‘ A and B’ are m1.
m
∴ P(A I B) = 1
n
m1
m m1
m
=
=
×
n
m
nm
m1
m
=
×
n
m
∴ P(A I B) = P(A) . P(B/A)
Note:
In the case of three events A, B, C, P(A I B I C) = P(A).
P(B/A). P(C/A I B). ie., the probability of occurrence of A, B and
C is equal to the probability of A times the probability of B given
that A has occurred, times the probability of C given that both A
and B have occurred.
1.6 BAYES’ Theorem:
The concept of conditional probability discussed earlier
takes into account information about the occurrence of one event to
10 predict the probability of another event. This concept can be
extended to revise probabilities based on new information and to
determine the probability that a particular effect was due to
specific cause. The procedure for revising these probabilities is
known as Bayes theorem.
The Principle was given by Thomas Bayes in 1763. By this
principle, assuming certain prior probabilities, the posteriori
probabilities are obtained. That is why Bayes’ probabilities are
also called posteriori probabilities.
Bayes’ Theorem or Rule (Statement only):
Let A1, A2, A3, ….Ai, …An be a set of n mutually
…
…
exclusive and collectively exhaustive events and P(A1), P(A2)…
,
P(An) are their corresponding probabilities. If B is another event
such that P(B) is not zero and the priori probabilities P(BAi)
i =1,2…n are also known. Then
,
P(B  A i ) P(A i )
P(Ai  B) = k
∑ P(B  A i ) P(A i )
i =1 1.7 Basic principles of Permutation and Combination:
Factorial:
The consecutive product of first n natural numbers is
known as factorial n and is denoted as n! o r ∠n
That is n! = 1 × 2 × 3 × 4 × 5 ×...× n
3! = 3 × 2 × 1
4! = 4 × 3 × 2 × 1
5! = 5 × 4 × 3 × 2 × 1
Also 5! = 5 × ( 4 × 3 × 2 × 1 ) = 5 × ( 4! )
Therefore this can be algebraically written as n! = n × (n – 1)!
Note that 1! = 1 and 0! = 1.
Permutations:
Permutation means arrangement of things in different ways.
Out of three things A, B, C taking two at a time, we can arrange
them in the following manner.
AB
BA
11 AC
CA
BC
CB
Here we find 6 arrangements. In these arrangements order
of arrangement is considered. The arrangement AB and the other
arrangement BA are different.
The number of arrangements of the above is given as the
number of permutations of 3 things taken 2 at a time which gives
the value 6. This is written symbolically, 3P2 = 6
Thus the number of arrangements that can be made out of
n things taken r at a time is known as the number of permutation of
n things taken r at a time and is denoted as nPr.
The expansion of nPr is given below:
nPr = n(n1)(n2)……… – ( r – 1)]
……[n
The same can be written in factorial notation as follows:
n!
nPr =
(n − r )!
For example, to find 10P3 we write this as follows:
10P3 = 10(101)(102)
= 10 × 9 × 8
= 720
[To find 10P3, Start with 10, write the product of 3 consecutive
natural numbers in the descending order]
Simplifying 10P3 using factorial notation:
10P3 = 10!
10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
=
(10 − 3)!
7 × 6 × 5 × 4 × 3 × 2 ×1 = 10 × 9 × 8
= 720
Note that nPo = 1, nP1 = n, nPn = n!
Combinations:
A combination is a selection of objects without considering
the order of arrangements.
12 For example, out of three things A,B,C we have to select
two things at a time.
This can be selected in three different ways as follows:
AB
AC
BC
Here the selection of the object A B and B A are
one and the same. Hence the order of arrangement is not
considered in combination. Here the number of combinations
from 3 different things taken 2 at a time is 3.
This is written symbolically 3C2 = 3
Thus the number of combination of n different things, taken
n Pr
r at a time is given by nCr =
r!
n!
Or nCr =
(n − r )!r!
Note that nC0 =1,
Find 10C3. Find 8C 4. nC1 = n,
nCn = 1
10 × 9 × 8
10 P3
=
= 120
10C3 =
3!
1× 2 × 3
8C 4 = 8×7×6×5
= 70
1× 2 × 3 × 4 [ To find 8 C 4 : In the numerator, first write the product of 4
natural numbers starting with 8 in descending order and in the
denominator write the factorial 4 and then simplify.]
Compare 10C 8 and 10 C 2
10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 10 × 9
= 45
=
10 C 8 =
1× 2 × 3 × 4 × 5 × 6 × 7 × 8
1× 2
10 × 9
= 45
10 C 2 =
1× 2
From the above, we find 10C 8 =
13 10 C2 This can be got by the following method also:
10C 8 = 10 C (10 – 8) = 10 C 2
This method is very useful, when the difference between n and r is
very high in nCr.
This property of the combination is written as nCr = nC (nr).
To find 200C198 we can use the above formula as follows:
200 × 199
= 19900.
200C198 = 200C(200 – 198) = 200C2 =
1× 2
Example:
Out of 13 players, 11 players are to be selected for a cricket
team. In how many ways can this be done?
Out of 13 players, 11 players are selected in 13 C 11 ways
13 x 12
i.e. 13 C 11 = 13 C 2 =  = 78.
1x2
Example 1:
Three coins are tossed simultaneously Find the probability that
(i) no head
(ii) one head
(iii)
two
heads
(iv) atleast two heads. (v) atmost two heads appear.
Solution:
The sample space for the 3 coins is
S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} ; n(S) = 8
(i)
No head appear A = {TTT}; n(A) = 1
1
∴P(A) =
8
(ii)
One head appear B = {HTT, THT, TTH}; n (B) = 3
3
∴ P(B) =
8
(iii)
Two heads appear C = {HHT, HTH, THH}; n(c)=3
3
∴ P(C) =
8
(iv)
Atleast two heads appear
D = { HHT, HTH, THH, HHH}; n(D) = 4
14 4
= 1/2
8
(v)
Atmost two heads appear E = { TTT, HTT, THT,
TTH,HHT, HTH,THH}
n(E)= 7
7
∴P(E) =
8
Example 2:
When two dice are thrown, find the probability of getting
doublets (Same number on both dice)
∴ P(D) = Solution:
When two dice are thrown, the number of points in the
sample space is n(S) = 36
Getting doublets: A = {(1,1) , (2,2) , (3,3) , (4,4) , (5,5) , (6,6)}
61
∴P(A) =
=
36 6
Example 3:
A card is drawn at random from a well shuffled pack of 52
cards. What is the probability that it is (i) an ace (ii) a diamond
card
Solution:
We know that the Pack contains 52 cards ∴n(S)= 52
(i) There are 4 aces in a pack. n(A) = 4
4
1
∴P(A) =
=
52 13
(ii) There are 13 diamonds in a pack ∴n(B) = 13
13 1
∴ P(B) =
=
52 4
Example 4:
A ball is drawn at random from a box containing 5 green, 6
red, and 4 yellow balls. Determine the probability that the ball
drawn is (i) green (ii) Red (iii) yellow (iv) Green or Red (v) not
yellow.
15 Solution:
Total number of balls in the box = 5+6+4 = 15 balls
51
(i)
Probability of drawing a green ball =
=
15 3
62
(ii)
Probability of drawing a red ball =
=
15 5
4
(iii)
Probability of drawing a yellow ball =
15
(iv)
Probability of drawing a Green or a Red ball
5 6 11
=
+
=
15 15 15
(v)
Probability of getting not yellow = 1 − P (yellow)
4
=1 −
15
11
=
15
Example 5:
Two dice are thrown, what is the probability of getting the sum
being 8 or the sum being 10?
Solution:
Number of sample points in throwing two dice at a time is
n(S)=36
Let A= {the sum being 8}
5
∴ A= {(6,2), (5,3) , (4,4), (3,5) , (2,6)}; P(A) =
36
B = { the sum being 10}
3
P(B) =
∴ B = {(6,4), (5,5) (4,6)} ;
36
A I B = { 0 } ; n(A I B) = 0
∴ The two events are mutually exclusive
∴P(AUB) = P(A) + P(B)
5
3
+
=
36
36
16 = 8
2
=
36
9 Example 6 :
Two dice are thrown simultaneously. Find the probability
that the sum being 6 or same number on both dice.
Solution:
n(S) = 36
The total is 6:
∴ A = {(5,1) , (4,2), (3,3) , (2,4) , (1,5)}; P(A) = 5
36 Same number on both dice:
∴B = {(1,1) (2,2), (3,3), (4,4), (5,5), (6,6)};
A I B = {(3,3)} ; 6
36
1
P(A B) =
36
P(B) = Here the events are not mutually exclusive.
∴ P(AUB) = P(A) + P(B) – P(A I B)
5
6
1
+
=
−
36
36
36
5 + 6 −1
=
36
11 − 1
=
36
10
=
36
5
=
18
Example 7:
Two persons A and B appeared for an interview for a job. The
probability of selection of A is 1/3 and that of B is 1/2. Find the
probability that
(i)
both of them will be selected
(ii)
only one of them will be selected
(iii)
none of them will be selected
17 Solution:
1
1
P(A)=
, P(B) =
3
2
2
1
P( A ) =
and P( B ) =
3
2
Selection or nonselection of any one of the candidate is not
affecting the selection of the other. Therefore A and B are
independent events.
(i) Probability of selecting both A and B
P(A I B) = P(A).P(B)
1
1
=×
3
2
1
=
6
(ii) Probability of selecting any one of them
= P (selecting A and not selecting B) + P(not selecting A and
selecting B)
_
_
_
_
i.e P (A I B) + P (A I B) = P(A). P(B) + P(A). P(B)
1
1
2
1
=
+
×
×
3
2
3
2
1
2
+
=
6
6
1
3
=
=
6
2
(iii) Probability of not selecting both A and B
i.e P( A I B )
= P( A ). P( B )
21
.
=
32
1
=
3
18 Example 8:
There are three T.V programmes A , B and C which can be
received in a city of 2000 families. The following information is
available on the basis of survey.
1200 families listen to Programme A
1100 families listen to Programme B
800 families listen to Programme C
765 families listen to Programme A and B
450 families listen to Programme A and C
400 families listen to Programme B and C
100 families listen to Programme A, B and C
Find the probability that a family selected at random listens atleast
one or more T.V Programmes.
Solution:
Total number of families n(S)= 2000
Let n(A) = 1200
n(B) = 1100
n(C) = 800
n(A I B) = 765
n(A I C) = 450
n(B I C) = 400
n(A I B I C) = 100
Let us first find n(AUBUC).
n(AUBUC) = n(A) + n(B)+ n(C) – n(A I B)–n(A I C)− n (B I C) + n(A I B I C) = 1200 + 1100 + 800 − 765 − 450 − 400 + 100
n(AUBUC) = 1585
n (AUBUC)
now P(AUBUC) =
n (S)
1585
=
= 0.792
2000
Therefore about 79% chance that a family selected at random
listens to one or more T.V. Programmes.
Example 9:
A stockist has 20 items in a lot. Out of which 12 are nondefective and 8 are defective. A customer selects 3 items from the
19 lot. What is the probability that out of these three items (i) three
items are nondefective (ii) two are non defective and one is
defective
Solution:
(i) Let the event, that all the three items are nondefective, be
denoted by E1. There are 12 nondefective items and out of them 3
can be selected in 12C3 ways ie n(E1)=12C3
Total number of ways in which 3 items can be selected are 20C3
i.e n(S) = 20C3
12C 3
n (E 1 )
=
∴ P(E1) =
n (S)
20C 3
12 × 11 × 10
=
20 × 19 × 18
= 0.193
ii) Let the event, that two items are nondefective and one is
defective be denoted by E2.
Two nondefective items out of 12 can be selected in12C2
ways. One item out of 8 defective can be selected in 8C1 ways.
Thus n(E2) =12C2 . 8C1
n (E 2 )
12C 2 . 8C1
=
Then the probability P(E2) =
n (S)
20C 3
12 × 11 × 8 × 3
=
20 × 19 × 18
= 0.463
Example 10:
A test paper containing 10 problems is given to three
students A,B,C. It is considered that student A can solve 60%
problems, student B can solve 40% problems and student C can
solve 30% problems. Find the probability that the problem chosen
from the test paper will be solved by all the three students.
Solution:
Probability of solving the problem by A = 60%
Probability of solving the problem by B = 40%
Probability of solving the problem by C = 30%
20 Solving the problem by a student is independent of solving the
problem by the other students.
Hence, P(A I B I C) = P(A). P(B). P(C)
60 40 30
=
×
×
100 100 100
= 0.6 × 0.4 × 0 .3
= 0.072
Example 11:
From a pack of 52 cards, 2cards are drawn at random. Find
the probability that one is king and the other is queen.
Solution:
From a pack of 52 cards 2 cards are drawn n(S)=52C2
Selection of one king is in 4C1 ways
Selection of one queen is in 4C1 ways
Selection of one king and one queen is in 4C1.4C1 ways
ie n(E) = 4C1.4C1
n (E) 4C1 .4C1
∴ P(E) =
=
n (S)
52C 2
52 × 51
= 4×4÷
1× 2
4× 4× 2
=
52 × 51
8
=
663
Example 12:
An urn contains 4 black balls and 6 white balls. If 3 balls
are drawn at random, find the probability that (i) all are black
(ii) all are white
Solution:
Total number of balls = 10
Total number ways of selecting 3 balls = 10C3
(i) Number of ways of drawing 3 black balls = 4C3
4C 3
Probability of drawing 3 black balls =
10C 3
21 4 × 3 × 2 10 × 9 × 8
÷
1× 2 × 3 1 × 2 × 3
4× 3× 2
=
10 × 9 × 8
1
=
30
(ii) Number of ways of drawing 3 white balls = 6C3
6C 3
Probability of drawing 3 white balls
=
10C 3
6× 5× 4
=
10 × 9 × 8
1
=
6
Example 13:
A box containing 5 green balls and 3 red colour balls. Find
the probability of selecting 3 green colour balls one by one
(i) without replacement (ii) with replacement
= Solution:
(i) Selection without replacement
Selecting 3 balls out of 8 balls = 8C3 ways
i.e n(S) = 8C3
Selecting 3 green balls in 5C3 ways
5C 3
5× 4× 3
5
=
=
∴ P(3 green balls) =
8×7×6
28
8C 3
(ii) Selection with replacement
When a ball is drawn and replaced before the next draw,
the number of balls in the box remains the same. Also the 3 events
of drawing a green ball in each case is independent. ∴Probability
5
of drawing a green ball in each case is
8
The event of selecting a green ball in the first, second and third
event are same,
22 ∴ Probability of drawing
5
5
5 125
3 green balls =
×
×=
8
8
8
512
Example 14:
A box contains 5 red and 4 white marbles. Two marbles are
drawn successively from the box without replacement and it is
noted that the second one is white. What is the probability that the
first is also white?
Solution:
If w1, w2 are the events ‘ white on the first draw’ , ‘ white on
the second draw’ respectively.
Now we are looking for P(w1/w2)
P(w 1 I w 2 )
P(w 1 ).P( w 2 )
P(w1/w2) =
=
P( w 2 )
P( w 2 )
(4 / 9)(3 / 8)
=
(3 / 8)
4
=
9
Example 15:
A bag contains 6 red and 8 black balls. Another bag
contains 7 red and 10 black balls. A bag is selected and a ball is
drawn. Find the probability that it is a red ball.
Solution:
There are two bags
1
2
Let A denote the first bag and B denote the second bag.
1
Then P(A) = P(B)=
2
Bag ‘ A’ contains 6 red and 8 black balls.
6
∴ Probability of drawing a red ball is
14
∴ probability of selecting a bag = 23 Probability of selecting bag A and drawing a red ball from that bag
1
6
3
is P(A). P(R/A) =
=
×
2 14 14
Similarly probability of selecting bag B and drawing a red ball
1
7
7
from that bag is P(B). P(R/B) =
=
×
2 17
34
All these are mutually exclusive events
∴ Probability of drawing a red ball either from the bag A or B is
P(R) = P(A) P(R/A) + P(B) P(R/B)
3
7
+
=
14
34
17 × 3 + 7 × 7
=
238
51 + 49
=
238
100 50
=
=
238 119
Example 16:
If P(A I B) = 0.3, P(A) = 0.6, P(B) = 0.7 Find the value of P(B/A)
and P(A/B)
Solution:
P(A I B)
P(B/A) =
P (A )
0.3
=
0.6
1
=
2
P(A I B)
P(A/B) =
P(B)
0.3
=
0.7
3
=
7
24 Example 17:
In a certain town, males and females form 50 percent of the
population. It is known that 20 percent of the males and 5 percent
of the females are unemployed. A research student studying the
employment situation selects unemployed persons at random.
What is the probability that the person selected is (i) a male
(ii) a female?
Solution:
Out of 50% of the population 20% of the males are
50 20
10
unemployed. i.e
×
=
= 0.10
100 100 100
Out of 50% the population 5% of the females are unemployed.
50
5
25
= 0.025
i.e
×
=
100 100 1000
Based on the above data we can form the table as follows:
Employed
Unemployed
Total
Males
0.400
0.100
0.50
Females
0.475
0.025
0.50
Total
0.875
0.125
1.00
Let a male chosen be denoted by M and a female chosen be
denoted by F
Let U denotes the number of unemployed persons then
P(M ∩ U)
0.10
(i) P(M/U) =
=
= 0.80
P(U)
0.125
P(F ∩ U)
0.025
(ii) P(F/U) =
=
= 0.20
P(U)
0.125
Example 18:
Two sets of candidates are competing for the positions on
the Board of directors of a company. The probabilities that the first
and second sets will win are 0.6 and 0.4 respectively. If the first set
wins, the probability of introducing a new product is 0.8, and the
corresponding probability if the second set wins is 0.3. What is the
probability that the new product will be introduced?
25 Solution:
Let the probabilities of the possible events be:
P(A1) = Probability that the first set wins
= 0.6
P(A2) = Probability that the second set wins = 0.4
P(B)
= Probability that a new product is introduced
P(B/A1) = Probability that a new product is introduced given
that the first set wins = 0.8
P(B/A2) = Probability that a new product is introduced given
that the second set wins = 0.3
Then the rule of addition gives:
P(new product) = P(first set and new product) + P(second set and
new product)
i.e P(B) = P(A1 B) + P(A2 B)
= P(A1) P(B/A1) + P(A2).P(B/A2)
= 0.6 × 0.8 + 0.4 × 0.3
= 0.60
Example 19:
Three persons A, B and C are being considered for the
appointment as the chairman for a company whose chance of being
selected for the post are in the proportion 4:2:3 respectively. The
probability that A, if selected will introduce democratization in the
company structure is 0.3 the corresponding probabilities for B and
C doing the same are respectively 0.5 and 0.8. What is the
probability that democratization would be introduced in the
company?
Solution:
Let A1 and A2 and A3 denote the events that the persons A, B and
C respectively are selected as chairman and let E be the event of
introducing democratization in the company structure.
Then we are given
4
2
3
P(A1) =
P(A2) =
P(A3) =
9
9
9
P(E/A1) = 0.3 P(E/A2) = 0.5 26 P(E/A3) = 0.8 The event E can materialize in the following mutually exclusive
ways:
(i)
Person A is selected and democratization is introduced
ie A1 I E happens
(ii)
Person B is selected and democratization is introduced
ieA2 I E happens
(iii)
Person C is selected and democratization is introduced
ie A3 I E happens
Thus E = (A1 I E) U (A2 I E) U(A3 I E) , where these sets are
disjoint
Hence by addition rule of probability we have
P(E) = P(A1 I E) + P(A2 I E) + P(A3 I E)
= P(A1) P(E/A1) + P(A2) P(E/A2) + P(A3) P(E/A3)
4
2
3
=
× 0.3 +
× 0.5 + × 0.8
9
9
9
46
23
=
=
90
45
Example 20:
In a bolt factory machines A1, A2, A3 manufacture
respectively 25%, 35% and 40% of the total output. Of these 5, 4,
and 2 percent are defective bolts. A bolt is drawn at random from
the product and is found to be defective. What is the probability
that it was manufactured by machine A2 ?
Solution:
P(A1 ) = P( that the machine A1 manufacture the bolts) = 25%
= 0.25
Similarly P(A2) = 35% = 0.35
and
P(A3) = 40% = 0.40
Let B be the event that the drawn bolt is defective.
P(B/ A1 ) = P (that the defective bolt from the machine A1 )
= 5 % = 0.05
Similarly, P(B/ A2 ) = 4% = 0.04
And
P(B/ A3) = 2% = 0.02
27 We have to find P(A2/ B).
Hence by Bayes’ theorem, we get
P ( A 2 ) P( B / A 2 )
P(A2/ B).=
P ( A 1 ) P ( B / A 1 ) + P ( A 2 ) P ( B / A 2 ) + P ( A 3 ) P( B / A 3 )
(0.35)(0.04)
=
(0.25)(0.05) + (0.35)(0.04) + (0.4)(0.02)
28
=
69
= 0.4058
Example 21:
A company has two plants to manufacture motorbikes.
Plant I manufactures 80 percent of motor bikes, and plant II
manufactures 20 percent. At Plant I 85 out of 100 motorbikes are
rated standard quality or better.
At plant II only 65 out of 100 motorbikes are rated standard quality
or better.
(i) What is the probability that the motorbike, selected at
random came from plant I. if it is known that the motorbike is
of standard quality?
(ii) What is the probability that the motorbike came from plant II
if it is known that the motor bike is of standard quality?
Solution:
Let A1 be the event of drawing a motorbike produced by plant I.
A2 be the event of drawing a motorbike produced by plant II.
B be the event of drawing a standard quality motorbike produced
by plant I or plant II.
Then from the first information, P(A1) = 0.80, P(A2) = 0.20
From the additional information
P(B/A1) = 0.85
P(B/A2) = 0.65
The required values are computed in the following table.
The final answer is shown in last column of the table.
28 A1 0.80 0.85 0.68 Posterior
(revised)
probability
P(Ai/B) =
P(A ∩ B)
P(B)
0.68 68
=
0.81 81 A2 0.20 0.65 0.13 0.13 13
=
0.81 81 Total 1.00 Event Prior
probability
P(Ai) Conditional
probability
of event B
given Ai Joint
probability
P(Ai I B) =
P(Ai)P(B/Ai) P(B/Ai) P(B) = 0.81 1 Without the additional information, we may be inclined to
say that the standard motor bike is drawn from plant I output, since
P(A1) = 80% is larger than P(A2) =20%
Remark:
The above answer may be verified by actual number of
motorbikes as follows:
Suppose 10,000 motorbikes were produced by the two
plants in a given period, the number of motorbikes produced by
plant I is
10,000 × 80% = 8000
and number of motorbikes produced by plant II is
10000 × 20% = 2000 =2000
The number of standard quality motorbikes produced by plant I is
85
8000 ×
= 6800
100
And by plant II is
2000 × 65
= 1300
100
29 The probability that a standard quality motor bike was produced by
plant I is
6800
6800 68
=
=
6800 + 1300 8100 81
And that by plant II is
1300
1300 13
=
=
6800 + 1300 8100 81
The process of revising a set of prior probabilities may be
repeated if more information can be obtained. Thus Bayes’
theorem provides a powerful method in improving quality of
probability for aiding the management in decision making under
uncertainty. Exercise  1
I. Choose the best answer :
1. Probability is expressed as
(a) ratio
(b) percentage
(c) Proportion
(d) all the above
2. Probability can take values from
(a)  ∞ to +∞
(b)  ∞ to 1
(c) 0 to +1
(d) –1 to +1
3. Two events are said to be independent if
(a) each out come has equal chance of occurrence
(b) there is the common point in between them
(c) one does not affect the occurrence of the other.
(d) both events have only one point
4. Classical probability is also known as
(a) Statistical probability
(b) A priori probability
(c) Empirical probability
(d) None of the above
5. When a coin and a die are thrown, the number of all possible
cases is
(a) 7
(b) 8
(c)12
(d) 0
30 6. Probability of drawing a spade queen from a well shuffled pack
of cards is
1
1
4
(a)
(b)
(c)
(d) 1
13
52
13
7. Three dice are thrown simultaneously the probability that sum
being 3 is
(a) 0
(b) 1 / 216
(c) 2 / 216
(d) 3 / 216
8. An integer is chosen from 1 to 20. The probability that the
number is divisible by 4 is
1
1
1
1
(a)
(b)
(c)
(d)
4
3
2
10
9. The conditional probability of B given A is
P(A ∩ B)
P(A ∩ B)
(b)
(a)
P(B)
P(B)
P(A)
P(AUB)
P(AUB)
(c)
(d)
P(B)
P ( A)
10. P(X) = 0.15, P(Y) = 0.25, P(X I Y) = 0.10 then P(XUY) is
(a) 0.10
(b) 0.20
(c) 0.30
(d) 0.40
11. If P(A) = 0.5, P(B) = 0.3 and the events A and B are
independent then P(A I B) is
(a) 0.8
(b) 0.15
(c) 0.08
(d) 0.015
12. If P(A) = 0.4 P(B) = 0.5 and P(A I B) = 0.2 then P(B/A) is
1
1
4
2
(b)
(c)
(d)
(a)
2
3
5
5
13. A coin is tossed 6 times. Find the number of points in the
sample space.
(a) 12
(b)16
(c) 32
(d) 64
14. When a single die is thrown the event of getting odd number or
even number are
(a) Mutually exclusive events
(b) Notmutually exclusive events
(c) Independent event
(d) Dependent event
31 15. The probability of not getting 2, when a die is thrown is
1
2
1
5
(a)
(b)
(c)
(d)
3
3
6
6
II. Fill in the blanks:
16. The probability of a sure event is __________
17. The probability of an impossible event is _________
18. Mathematical probability is also called a ______________
probability.
19. The joint occurrence of two or more events is called
___________
20. If A and B are mutually exclusive events, then P(AUB) = ____
21. If A and B are independent events then P(A I B) = _________
22. If A and B are dependent events then P(A/B) = ___________
23. If A and B are mutually exclusive events P(A I B) =
_________
24. When three coins are tossed the probability of getting 3 heads
is _________
25. When three dice are thrown the probability of sum being 17 is
__________
26. The probability getting the total is 11 when two dice are throws
____________
III Answer the following:
27. Define the following terms:
Event, equally likely events, mutually exclusive events,
exhaustive events, sample space.
28. Define dependent and independent events.
29. Define mathematical probability.
30. Define statistical probability.
31. State the axioms of probability.
32. Explain addition theorem on probability for any two events.
33. State the multiplication theorem on probability.
34. Define conditional probability.
35. State Bayes’ Rule.
32 36. There are 5 items defective in a sample of 30 items. Find the
probability that an item chosen at random from the sample is
(i) defective, (ii) nondefective
37. Four coins are tossed simultaneously what is the probability of
getting (i) 2 heads
(ii) 3 heads
(iii) atleast 3 heads
38. Two dice are thrown. What is probability of getting (i) the sum
is 10 (ii) atleast 10
39. Three dice are rolled once. What is the chance that the sum of
the face numbers on the dice is (i) exactly 18 (ii) exactly 17
(iii) atmost 17.
40. An integer is chosen from 20 to 30. Find the probability that it
is a prime number.
41. An integer is chosen from 1 to 50. Find the probability that it is
multiple of 5 or multiply of 7
42. From a pack of cards find the probability of drawing a spade
card or a diamond card.
43. Find the probability that a leap year selected at random will
contain 53 Sundays.
44. Find the probability that a nonleap year selected at random
will contain either 53 Sundays or 53 Mondays.
45. If two events A and B are not mutually exclusive and are not
connected with one random experiment P(A)= 1/4, P(B) =2/5
and P(AUB) = 1/2 then find the value of P(B/A)
46. For two independent events A and B for which P(A) = 1/2 and
P(B) = 1/3. Find the probability that one of them occur.
47. For two events A and B, P(A) =1/3 = P( B ), P(B/A) =1/4
find P(A/B)
48. A box contains 4 red pens and 5 black pens. Find the
probability of drawing 3 black pens one by one (i) with
replacement (ii) without replacement
49. An urn contains 5 red and 7 green balls. Another urn contains 6
red and 9 green balls. If a ball is drawn from any one of the two
urns, find the probability that the ball drawn is green.
50. Two cards are drawn at random from a pack of 52 cards. Find
the probability that the cards drawn are (i) a diamond and a
spade (ii) a king and a queen (iii) 2 aces
33 51. A problem in statistics is given to two students A and B. The
probability that A solves the problem is 1/2 and that of B’ s to
solve it is 2/3. Find the probability that the problem is solved.
52. A bag contains 6 white, 4 green and 10 yellow balls. Two balls
are drawn at random. Find the probability that both will be
yellow.
53. In a certain class there are 21 students in subject A, 17 in
subject B and 10 in subject C. Of these 12 attend subjects A
and B, 5 attend subjects B and C, 6 attend subjects A and C.
These include 2 students who attend all the three subjects.
Find the probability that a student studies one subject alone.
54. If P(A) = 0.3, P(B) = 0.2 and P(C) = 0.1 and A,B,C are
independent events, find the probability of occurrence of at
least one of the three events A , B and C
55. The odds that A speaks the truth are 3:2 and the odds that B
speaks the truth 5:3. In what percentage of cases are they
likely to contradict each other on an identical point?
56. The chances of X, Y and Z becoming managers of a certain
company are 4:2:3. The probabilities that bonus scheme will
be introduced if X, Y and Z become managers are 0.3, 0.5 and
0.4 respectively. If the bonus scheme has been introduced
what is the probability that Z is appointed as the manager?
57. A manufacturing firm produces steel pipes in three plants with
daily production volumes of 500, 1000 and 2000 units
respectively. According to past experience, it is known that the
fractions of defective outputs produced by the three plants are
respectively, 0.005, 0.008 and 0.010. If a pipe is selected from
days total production and found to be defective, what is the
probability that it came from the (i) first plant (ii) the second
plant (iii) the third plant? 34 Answers:
I.
1. (d)
6. (b)
11. (b) 2. (c)
7. (b)
12.(a) II.
16. 1
17. zero
20. P(A) +P(B)
P(A ∩ B)
22.
P(B)
3
25.
216 3. (c)
8. (a)
13.(d) 4. (b)
9. (b)
14.(a) 5. (c)
10. (c)
15. (d) 18. priori
19. Compound events
21. P(A). P(B)
1
23. 0
24.
8
1
26.
18 III
15
,
66
2
40.
11 36. 31 5
;;
8 4 16
8
41.
25 37. 11
;
12 6
1
42.
2
38. 1
3 215
,
,
216 216 216
2
2
43.
44.
7
7
39. 45. P(A I B) = 3/20 ; P(B/A) = 3/5
1
2
125 5
,
48.
729 42
5
51.
6
46. 54. 0.496
56. 6/17 47. P(A I B) = 1/12
71
120
9
52.
38
3 3 2 5 19
55. × + × =
5 8 5 8 40
124
57. (a) ; ;
777
49. 35 P(A/B)=1/8
13 8
1
;
;
102 663 221
8
53.
27 50. = 47.5%
(b) 5 16 40
;;
61 61 61 Activity:
We know that, when a coin is tossed, the probability of getting
a head is 0.5 mathematically.
Now do the following experiment.
1. Toss a fair coin 10 times. Record the event that the number
of heads occur in this experiment
2. Toss a fair coin 100 time with the help of your friends
group and record the same event that the number of heads
appearing.
3. Now compare all the three above mentioned and write your
inference. 36 2. RANDOM VARIABLE AND
MATHEMATICAL EXPECTATION
2.0 Introduction:
It has been a general notion that if an experiment is
conducted under identical conditions, values so obtained would be
similar. Observations are always taken about a factor or character
under study, which can take different values and the factor or
character is termed as variable.
These observations vary even though the experiment is
conducted under identical conditions. Hence, we have a set of
outcomes (sample points) of a random experiment. A rule that
assigns a real number to each outcome (sample point) is called
random variable.
From the above discussion, it is clear that there is a value
for each outcome, which it takes with certain probability. Hence a
list of values of a random variable together with their
corresponding probabilities of occurrence, is termed as Probability
distribution.
As a tradition, probability distribution is used to denote the
probability mass or probability density, of either a discrete or a
continuous variable.
The formal definition of random variable and certain
operations on random variable are given in this chapter prior to the
details of probability distributions.
2.1 Random variable:
A variable whose value is a number determined by the
outcome of a random experiment is called a random variable.
We can also say that a random variable is a function defined
over the sample space of an experiment and generally assumes
different values with a definite probability associated with each
value. Generally, a random variable is denoted by capital letters
like X, Y, Z… where as the values of the random variable are
..,
denoted by the corresponding small letters like x, y, z ….
…
37 Suppose that two coins are tossed so that the sample space
is
S = {HH, HT, TH, TT}
Suppose X represent the number of heads which can come up, with
each sample point we can associate a number for X as shown in the
table below:
Sample
point
X HH HT TH TT 2 1 1 0 Thus the random variable X takes the values 0, 1,2 for this
random experiment.
The above example takes only a finite number of values and
for each random value we can associate a probability as shown in
the table.
Usually, for each random variable xi, the probability of
respective random variable is denoted by p(xi) or simply pi .
X
x1 = 0
x2 = 1
x3 = 2
p(xi) p(xi) = 1
4 p(xi) = 2
4 p(xi) = 1
4 Observe that the sum of the probabilities of all the random variable
121
is equal to one. ie p(x1) + p(x2) + p(x3) = + + = 1
444
Thus the probability distribution for a random variable
provides a probability for each possible value and that these
probabilities must sum to 1.
Similarly if 3 coins are tossed, the random variable for
getting head will be X=0, X=1, X=2, X=3 and sum of their
respective probabilities i.e Σp(xi) =1
If two dice are rolled then the sample space S consists of 36
sample points. Let X denote the sum of the numbers on the two
dice. Then X is a function defined on S by the rule X(i,j) = i+j .
Then X is a random variable which can takes the values
2,3,4…12. That is the range of X is {2,3,4…12}
…
…
38 2.1.1 Discrete random variable:
If a random variable takes only a finite or a countable
number of values, it is called a discrete random variable.
For example, when 3 coins are tossed, the number of heads
obtained is the random variable X assumes the values 0,1,2,3 which
form a countable set. Such a variable is a discrete random variable.
2.1.2 Continuous random variable:
A random variable X which can take any value between
certain interval is called a continuous random variable.
Note that the probability of any single value at x, value of X
is zero. i.e P(X = x) = 0 Thus continuous random variable takes
value only between two given limits.
For example the height of students in a particular class lies
between 4 feet to 6 feet.
We write this as X = {x4 ≤ x ≤ 6}
The maximum life of electric bulbs is 2000 hours. For this
the continuous random variable will be X = {x  0 ≤ x ≤ 2000}
2.2 Probability mass function:
Let X be a discrete random variable which assumes the
values x1, x2, ...xn with each of these values, we associate a number
called the probability Pi= P(X=xi), i = 1,2,3… This is called
n
probability of xi satisfying the following conditions.
(i)
Pi ≥ 0 for all i, ie Pi’ s are all nonnegative
(ii)
p
Σpi = p1 + p2 + …n =1
ie the total probability is one.
This function pi or p(xi) is called the probability mass
function of the discrete random variable X.
The set of all possible ordered pairs (x, p(x)) is called the
probability distribution of the random variable X.
Note:
The concept of probability distribution is similar to that of
frequency distribution. Just as frequency distribution tells us how
the total frequency is distributed among different values (or classes)
of the variable, a probability distribution tells us how total
39 probability 1 is distributed among the various values which the
random variable can take. It is usually represented in a tabular form
given below:
X
x1
x2
x3
…
.
xn
P(X = x) P(x1) P(x2) P(x3) …
. P(xn) 2.2.1 Discrete probability distribution:
If a random variable is discrete in general, its distribution
will also be discrete. For a discrete random variable X, the
distribution function or cumulative distribution is given by F(x) and
is written as F(x) = P(X ≤ x) ;  ∞ < x < ∞
Thus in a discrete distribution function, there are a
countable number of points x1, x2,… and their probabilities pi such
..
that
F(xi) = ∑ pi , i = 1, 2, ….n
…
xi < x Note:
For a discrete distribution function, F(xj) – F(xj1) = p(xj)
2.2.2 Probability density function (pdf):
A function f is said to be the probability density function of
a continuous random variable X if it satisfies the following
properties.
(i) f(x) ≥ 0 −∞ < x < ∞
∞ (ii) ∫ f ( x) dx = 1
−∞ Remark:
In case of a discrete random variable, the probability at a
point ie P(x = a) is not zero for some fixed ‘ a’ However in case of
continuous random variables the probability at a point is always
zero
a ie P(X = a) = ∫ f ( x) dx =0
a
40 Hence P( a ≤ X ≤ b) = P(a < X < b) = P(a ≤ X < b) =
P(a < X ≤ b)
The probability that x lies in the interval (a,b) is given by
b P( a < X < b) = ∫ f ( x) dx
a Distribution function for continuous random variable.
If X is a continuous random variable with p.d.f f(x), then the
distribution function is given by
x ∫ (i) F(x) = f ( x ) dx = P(X ≤ x) ; ∞ < x < ∞ −∞
b (ii) F(b) – F(a) = ∫ f ( x) dx = P(a ≤ X ≤ b)
a 2.3 Properties of distribution function:
Suppose that X be a discrete or continuous random variable, then
(i)
F(x) is a non  decreasing function of x
(ii)
0 ≤ F(x) ≤ 1 , −∞ < x < ∞
(iii)
F( ∞) = limit F(x) =0
xà−∞
(iv)
F(∞) = limit F(x) =1
xà∞
(v)
If F(x) is the cumulative distribution function of a
continuous random variable X with p.d.f f(x) then
F′(x) = f(x)
Example 1:
A random variable has the following probability distribution
Values
of X
P(x) 0 1 2 3 4 5 6 7 8 a 3a 5a 7a 9a 11 a 13 a 15 a 17 a (1) Determine the value of a
(2) Find (i) P( x < 3)
(ii) P(x ≤ 3)
(iii)
(iv)P( 2 ≤ x ≤ 5),
(v) P(2 < x <5)
(3) Find the cumulative distribution function of x.
41 P(x >7) Solution:
(1) Since pi is the probability mass function of discrete
random variable X,
We have Σpi = 1
∴ a + 3 a + 5a + 7a + 9a +11a + 13a + 15a + 17a = 1
81a = 1
a = 1/81
(2)
(i) P(x <3) = P(x=0) + P(x=1) + P(x=2)
= a + 3 a + 5a
= 9a
1
=9( )
81
1
=
9
(ii) P(x ≤ 3) = P (x=0) + P(x=1) + P(x=2) +P(x=3)
=a+3a+5a+7a
= 16 a
16
=
81
iii) P(x >7) = P(x = 8)
= 17 a
17
=
81
iv) P ( 2 ≤ x ≤ 5) = P(x=2) +P(x=3) + P( x = 4) +P(x=5)
= 5 a + 7a +9a +11a
= 32a
32
=
81
v) P(2 < x < 5 ) = P(x = 3) + P(x = 4)
= 7a + 9a
= 16a
16
=
81
42 3) The distribution function is as follows:
X=x
0
F(x)=
a
P (X≤ x)
(or)
1
F(x)
81 1
4a 2
9a 3
16a 4
25a 5
36a 6
49a 7
64a 8
81a 4
81 9
81 16
81 25
81 36
81 49
81 64
81 81
=1
81 Example 2:
Find the probability distribution of the number of sixes in
throwing two dice once.
Solution:
When two dice are thrown the total number of sample
points are 36.
Let X denote the number of sixes obtained in throwing two
dice once. Then X is the random variable, which can take the
values 0,1,2.
Let A denote the success of getting a six in throwing a die
and A denote not getting a six.
Then probability getting a six
P(A) = 1
6 Probability not getting a six
P( A ) = 5
6 No sixes:
∴ P(x = 0) = P( A and A )
= P( A ) . P( A )
5
5
=
.
6
6
25
=
36
43 P(x = 1) = P(A and A ) or P( A and A)
= P(A) . P( A ) + P( A ) .P(A)
15
51
=.
+
66
66
5
5
=
+
36
36
10
=
36
5
=
18
P(x = 2) = P( A and A)
= P(A) .P(A)
11
=.
66
1
=
36
Hence the probability distribution of X is given by
X= x
P(X = x) 0 1 2 25
36 10
36 1
36 Example 3:
An urn contains 6 red and 4 white balls. Three balls are
drawn at random. Obtain the probability distribution of the number
of white balls drawn.
Solution:
The total number of balls in the urn is 10
Let X denote the number of white balls drawn
If three balls are drawn, the random variable takes the value
X= 0, 1, 2, 3
Probability of getting white balls from the urn containing 10
balls (red and white) with the following combination are
44 4C 0 6C 3
1 × 120
5
=
=
720
30
10C 3
4C1 .6C 2
15
=
P (1 white, 2 red) =
30
10C 3
4C 2 6C1
9
=
P (2 white, 1 red) =
30
10C 3
4C 3 6C 0
1
P (3 white, no red) =
=
30
10C 3
Hence the probability distribution of X is given by
C
0
1
2
3
P(X=x)
5
15
9
1
30
30
30
30
2.4 An introduction to elementary calculus:
Before going to see the problems on continuous random
variables, we need to know some fundamental ideas about
differentiation and integration, which are part of calculus in higherlevel mathematics.
Hence we introduce some simple techniques and formulae
to calculate the problems in statistics, which involve calculus
methods.
2.4.1 Differentiation:
1. Functional value is an exact value. For some function f(x),
when x = a, we obtain the functional value as f(a) = k.
2. Limiting value is an approximate value. This value
approaches the nearest to the exact value k.
Suppose the exact value is 4. Then the limiting value will be
4.000000001 or 3.999999994. Here the functional value and
limiting value are more or less same.
Hence in many occasions we use the limiting values for
critical problems.
The limiting value of f(x) when x approaches a number 2 is
given as
Limit f(x) = f(2) = l
xà 2
45
P (no white, 3 red balls) = f (x + h ) − f ( x )
is
h →0
h
called the derivative of the function f with respect to x and
is denoted by f ′(x). If y is the function x then we say the
differential coefficient of y with respect to x and is denoted
dy
as
dx
4. Some rules on differentiation:
(i) Derivative of a constant function is zero. f ′(c)=0 where
c is some constant.
(ii) If u is a function of x and k is some constant and dash
denotes the differentiation, [ku]′ = k[u]′
(iii) (u ± v)′ = u′ ± v′
(iv) (uv)′ = u′v +uv′
3. The special type of existing limit, limit ′ u ' v − uv'
u
(v) =
v2
v
5. Important formulae:
(i) (xn)′ = nxn1
(ii) (ex)′ = ex
1
(iii) (logx)′ =
x
Example 4:
Evaluate the following limits:
x 2 + 5x
x 2 −1
(i) Limit
(ii) Limit
x →1
x+2
x −1
x→2
Solution:
(i) Limit
x→2 x 2 + 5x
x+2 = (2) 2 + 5(2)
4 + 10 14
7
=
=
=
2+2
4
4
2 x 2 −1
12 − 1
0
=
= . This is an indeterminate form.
x →1
x −1
1−1
0
Therefore first factorise and simplify and then apply the same limit
to get the limiting value
46
(ii) Limit x2 −1
( x − 1)( x + 1)
=
= x +1
x −1
( x − 1) ∴
∴ Limit
x →1 x 2 −1
= Limit (x+1) = 1+1 = 2
x →1
x −1 Example 5:
Find the derivative of the following with respect to x.
x2 +1
12
4
2
3
x
(i) x + 7 (ii) (x + 4x –5)
(iii) (x ) (e ) (iv)
x −5
Solution:
(i) Let y = x12 + 7
dy
= 12x121 + 0 = 12x11
∴
dx
(ii) Let y = x3 + 4x2 –5
y′ = 4x3 +4(2x) – 0
= 4x3 +8
(iii) Let y = x3 ex
(uv)′= u′v +uv′
= [x3]′ (ex) + (x3) [ex]′
= 3x2 ex +x3 ex
= ex (3x2 + x3)
′ (iv) y = u ' v − uv'
x2 +1
u
. This is of the type =
x −5
v2
v
′
′
dy x 2 + 1 (x − 5) − (x 2 + 1)[x − 5 ]
∴
=
dx
( x − 5) 2 [ [2x (]x − 5) − ( x 2 + 1)[1]
=
( x − 5) 2 2x 2 − 10 x − x 2 − 1
=
(x − 5) 2
x 2 − 10 x − 1
=
( x − 5) 2
47 2.4.2 Integration:
Integration is known as the reverse process of
differentiation. Suppose the derivative of x3 is 3x2. Then the
integration of 3x2 with respect to x is x3 . We write this in symbol
as follows:
d4
3
( x ) = 4x3
⇒ 4 ∫ x dx = x4
dx
Similarly
d8
( x ) = 8x7
dx
dx
(e ) = ex
dx ⇒8
⇒ ∫x ∫e x 7 dx = x8 dx = ex Note:
While differentiating the constant term we get zero. But in
the reverse process, that is on integration, unless you know the
value of the constant we cannot include. That is why we include an
arbitrary constant C to each integral value.
Therefore the above examples, we usually write
x
7
x
8
∫ e dx = e +c and ∫ 8x dx = x + c
These integrals are also called improper integrals or
indefinite integrals
Rules and formulae on integration:
(i) ∫ k dx = kx
+ xn 1
(ii) ∫ x dx =
n +1
x
(iii) ∫ e dx = ex
n 1 ∫ x dx = log x
(v) ∫ (u ± v)dx = ∫ u dx ± ∫ v dx
(iv) 48 Example 6:
Integrate the following with respect to x:
+
x6 1
x7
6
=
(i) ∫ x dx =
+c
6 +1
7
−+
−
11
1
x51
x4
5
=−
+c
=
=(ii) ∫ x dx =
4
4x
− 5 +1
−4
4x 4
1
(iii) ∫ dx = log x +c
x
+
2 3/ 2
x1/ 2 1
x3/ 2
1/ 2
(iv) ∫ x dx = ∫ x dx =
x +c
=
=
1
3
3
+1
2
2
4
2
(v) ∫ (x +2x + 4x + 8) dx
x5
x3
x2
+2
+4
+ 8x + c
5
3
2
(ex + x4 + 1/x3 +10) dx
= (vi) = ex + x5 /5 − 1/2x2 + 10x + c
The above discussed integrals are known as improper
integrals or indefinite integrals. For the proper or definite integrals
we have the limiting point at both sides. ie on the lower limit and
the upper limit.
This integral ∫ f ( x ) dx is an indefinite integral
Integrating the same function within the given limits a and
b is known as the definite integral.
b ie ∫ f (x )dx = k (a constant value) is a definite integral where a is a known as lower limit and b is known as the upper limit of the
definite integral.
To find the value of definite integral we use the formulae:
b Suppose ∫ f ( x ) dx = F(x) then ∫ f ( x ) dx = F(b) – F(a)
a 49 An important note to the Teachers and students
As per as statistics problems concerned, the differentiation
and integration methods restricted to simple algebraic functions
only.
Example 7:
Evaluate the following definite integrals.
4 3 (i) ∫ 3x 2 dx (ii) 0 3
∫ x dx
1 Solution:
4 3x 3 34
(i) ∫ 3x dx = = [x ]0 3 0
0
3
= 4 –03
= 64
4 2 3 3 (ii) ∫ x dx
3 1 x4 = 4 1 1 43
[ x ]1
4
1
= [3 4 − 14 ]
4
1
= [81 − 1]
4
1
=
[80]
4
= 20
= 5 x2 (iii) ∫ xdx = 2 2
2
1
= [5 2 − 2 2 ]
2
1
21
= [25 − 4] =
2
2
5 50 5 (iii) ∫ xdx
2 Example 8:
Examine whether f(x) = 5x4 , 0 < x < 1 can be a p.d.f of a
continuous random variable x.
Solution:
∞ For a probability density function, to show that ∫ f ( x ) dx =1 −∞
1 That is to show that ∫ 5( x) 4 dx = 1 0 1 x5 5( x) dx
=5 ∫
0 5 0
1
5
= x5 0
5
= [15 –0]
=1
∴ f(x) is a p.d.f
1 4 Example 9:
A continuous random variable x follows the rule
f(x) = Ax2, 0 < x < 1. Determine A
Solution:
Since f(x) is a p.d.f, ∫ ∞ −∞ f ( x ) dx =1 1 Therefore ∫ Ax2 dx = 1 0 1 x3 A 3 0
A 31
x0
3
A
[1]
3
A =1 [ ]= 1
=1
=3
51 Example 10:
Let f(x) = c(1x) x2 , 0 < x < 1 be a probability density
function of a random variable x. Find the constant c
Solution:
f(x) = c(1x)x2 , 0 < x < 1
since f(x) is a p.d.f ∫ ∞ −∞ f ( x ) dx = 1 1 ∴ ∫ c(x2 –x3)dx =1 0 1 x3 x4 =1
c −
3
4 0 13 14 c − − (0 − 0) = 1 3 4 1 1
c − =1
3 4 4 −1
c =1 12 1
c =1 12 c = 12
Example 11:
A random variable x has the density function
1
2< x<2
,
f(x) = 4
0,
else where obtain (i) P (−1 < x < 2) (ii) P (x >1) Solution:
2 (i) P(1 < x < 2) = ∫ f(x) dx −1 52 2 +2
1
1
dx =
[ x]−1
∫4
4
−1
1
=
[2 – (1)]
4
1
=
[3]
4
3
=
4
(ii) Here the upper limit of the p.d.f is 2∴the probability for the
given random variable.
2
1
P( x > 1)
=∫
dx
4
1
1
2
= [x ]
1
4
1
[2 –1]
=
4
1
=
[1]
4
1
=
4
2.5 Mathematical Expectation:
Expectation is a very basic concept and is employed widely
in decision theory, management science, system analysis, theory of
games and many other fields. Some of these applications will be
discussed in the chapter on Decision Theory.
The expected value or mathematical expectation of a
random variable X is the weighted average of the values that X can
assume with probabilities of its various values as weights.
Thus the expected value of a random variable is obtained by
considering the various values that the variable can take
multiplying these by their corresponding probabilities and summing
these products. Expectation of X is denoted by E(X) 53 2.5.1 Expectation of a discrete random variable:
Let X be a discrete random variable which can assume any
of the values of x1, x2, x3…..xn with respective probabilities p1,
…
p2, p3…pn. Then the mathematical expectation of X is given by
…
E(x) = x1p1 + x2p2 + x3p3 +……npn
…x
n n = ∑
i =1 xipi , where ∑ pi = 1 i =1 Note:
Mathematical expectation of a random variable is also
known as its arithmetic mean. We shall give some useful theorems
on expectation without proof.
2.5.2 Theorems on Expectation:
1. For two random variable X and Y if E(X) and E(Y)
exist, E(X + Y) = E(X) + E(Y) . This is known as
addition theorem on expectation.
2. For two independent random variable X and Y,
E(XY) = E(X).E(Y) provided all expectation exist. This
is known as multiplication theorem on expectation.
3. The expectation of a constant is the constant it self.
ie E(C) = C
4. E(cX) = cE(X)
5. E(aX+b) = aE(X) +b
6. Variance of constant is zero. ie Var(c) = 0
7. Var(X+c) = Var X
Note: This theorem gives that variance is independent
of change of origin.
8. Var (aX) = a2 var(X)
Note: This theorem gives that change of scale affects
the variance.
9. Var (aX+b) = a2Var(X)
10. Var (bax) = a2 Var(x)
Definition:
Let f(x) be a function of random variable X. Then
expectation of f(x) is given by E(f(x)) = Σ f(x) P(X=x) , where
P(X=x) is the probability function of x.
54 Particular cases:
1. If we take f(x) = Xr, then E(Xr) = Σxrp(x) is defined as the
rth moment about origin or rth raw moment of the
probability distribution. It is denoted by µ′r
Thus µ′r = E(Xr)
µ′1 = E(X)
µ′2 = E(X2)
Hence mean = X = µ′1 = E(X)
2 Σ X 2 ΣX 
N
N 2
= E ( X ) – [E (X)]2
= µ′2 – (µ′1)2
Variance is denoted by µ2
2. If we take f(x) = (X – X )r then E(X – X )r = Σ(X – X )r p(x)
which is µr, the rth moment about mean or rth central moment.
In particular if r = 2, we get
µ2 = E (X – X )2
= Σ (X – X )2 p(X)
= E [X – E(X)]2
These two formulae give the variance of probability distribution in
terms of expectations.
Variance = Example 12:
Find the expected value of x, where x represents the
outcome when a die is thrown.
Solution:
Here each of the outcome (ie., number) 1, 2, 3, 4, 5 and 6
1
occurs with probability . Thus the probability distribution of
6
X will be
x
1
2
3
4
5
6
P(x) 1
6 1
6 1
6
55 1
6 1
6 1
6 Thus the expected value of X is
E(X) = Σxipi
= x1p1 + x2p2 + x3p3 + x4p4+ x5p5 + x6p6 1 1 1 1
E(X) = 1× + 2 × + 3 × + 4 × 6 6 6 6 1 1
+ 5 × + 6 × 6 6
7
=
2
E(X) = 3.5
Remark:
In the games of chance, the expected value of the game is
defined as the value of the game to the player.
The game is said to be favourable to the player if the
expected value of the game is positive, and unfavourable, if value
of the game is negative. The game is called a fair game if the
expected value of the game is zero.
Example 13:
A player throws a fair die. If a prime number occurs he wins
that number of rupees but if a nonprime number occurs he loses
that number of rupees. Find the expected gain of the player and
conclude.
Solution:
Here each of the six outcomes in throwing a die have been
assigned certain amount of loss or gain. So to find the expected
gain of the player, these assigned gains (loss is considered as
negative gain) will be denoted as X.
These can be written as follows:
Outcome on a die
1
2
3
4
5
6
Associated gain to
1
2
3
4
5
6
the outcome (xi)
1
1
1
1
1
1
P(xi)
6
6
6
6
6
6
56 Note that 2,3 and 5 are prime numbers now the expected gain is
66 E (x)∑ EΣ1 xi pi
==
i =1 1
1
1
1
1
1
= (–1) + (2) + (3) + (– 4) + (5) + (– 6) 6
6
6
6
6
6
1
=– 6
Since the expected value of the game is negative, the game
is unfavourable to the player.
Example 14:
An urn contains 7 white and 3 red balls. Two balls are
drawn together at random from the urn. Find the expected number
of white balls drawn.
Solution:
From the urn containing 7 white and 3 red balls, two balls
can be drawn in 10C2 ways. Let X denote the number of white balls
drawn, X can take the values 0, 1 and 2.
The probability distribution of X is obtained as follows:
P(0) = Probability that neither of two balls is white.
= Probability that both balls drawn are red.
3× 2
1
3C 2
=
=
=
10 × 9
15
10C 2
P(1) = Probability of getting 1 white and 1 red ball.
7 × 3× 2
7
= 7C1 × 3C1 =
=
10 × 9
15
10C 2
P(2) = Probability of getting two white balls
7× 6
7
7C 2
=
=
=
10C 2 10 × 9 15
Hence expected number of white balls drawn is
1 7 7 E(x) = Σ xi p(xi) = 0 × + 1 × + 2 × 15 15 15 7
=
= 1.4
5
57 Example 15:
A dealer in television sets estimates from his past
experience the probabilities of his selling television sets in a day is
given below. Find the expected number of sales in a day.
Number of
TV Sold in
a day
Probability 0 1 2 3 4 5 6 0.02 0.10 0.21 0.32 0.20 0.09 0.06 Solution:
We observe that the number of television sets sold in a day
is a random variable which can assume the values 0, 1,2, 3,4,5,6
with the respective probabilities given in the table.
Now the expectation of x = E(X) = Σxipi
= x1p1 + x2p2 + x3p3 + x4p4+ x5p5 + x6p6
= (0) (0.02) + (1) (0.010) + 2(0.21) + (3) (0.32) + 4(0.20)
+(5) (0.09) + (6) (0.06)
E(X) = 3.09
The expected number of sales per day is 3
Example 16:
Let x be a discrete random variable with the following
probability distribution
X 3 6
1/2 P(X= x)
1/6
Find the mean and variance
Solution:
E (x) 9
1/3 = Σ xi pi
1
1
1 = (3) + (6) + (9) 6 2
3
11
= 2
58 = Σ xi2 pi
1
1
1 93 = (3)2 + (6)2 + (9)2 = 6 2
3
2
2
2
Var (X) = E (X )  [E(X)]
E(x2) 2 93 11 =  2 2 93 121
= 
2 4 186 − 121
=
4
65
=
4
2.5.3 Expectation of a continuous random variable:
Let X be a continuous random variable with probability
density function f(x), then the mathematical expectation of x is
defined as
∞ E(x) = ∫ xf ( x)dx , provided the integral exists. −∞ Remark:
If g(x) is function of a random variable and E[g(x)] exists,
∞ then E[(g(x)] = ∫ g(x) f(x)dx −∞ Example 17:
Let X be a continuous random variable with p.d.f given by
f(x) = 4x3, 0 < x < 1. Find the expected value of X.
Solution:
∞ We know that E(X) = ∫ xf ( x)dx −∞
1 In this problem E(X) = ∫ x(4 x ) dx
3 0 59 1 = 4 ∫ x ( x 3 )dx
0 1 x5 =4 5 0
1
4 x5 = 0
5
4
= [ 15  05]
5
4
= [1]
5
4
=
5
Example 18:
Let x be a continuous random variable with pdf. given by
f(x) = 3x2 , 0 < x < 1 Find mean and variance
Solution:
∞ E(x) ∫ xf (x ) dx = −∞
1 E(x) = ∫ x[3x 2 dx 0 1 =3 ∫ x dx
3 0 1 =
=
=
= x4 3 4 0
3 41
x0
4
34
1 −0
4
3
4 [ 60 ∞ ∫x 2 E(x ) = 2 f ( x ) dx −∞
1 = ∫x 2 [3x 2 ] dx 0 1 =3 ∫x 4 dx 0 1 x5 =3 5 0
1
3
= x5 0
5
3
= 15 − 0
5
3
=
5
Variance = E(x2) – [E(x)]2 [ 2 3 3
− 5 4
3
9
=
−
5 16
48 − 45
3
=
=
80
80
2.6 Moment generating function (M.G.F) (concepts only):
To find out the moments, the moment generating function is
a good device. The moment generating function is a special form of
mathematical expectation and is very useful in deriving the
moments of a probability distribution.
Var (x)= Definition:
If X is a random variable, then the expected value of etx is
known as the moment generating functions, provided the expected
value exists for every value of t in an interval,  h < t < h , where h
is some positive real value.
61 The moment generating function is denoted as Mx(t)
For discrete random variable
Mx(t) = E(etx)
= Σ etx p(x) (tx ) 2 (tx ) 3
= Σ 1 + tx +
+
+ ....... px(x) 2!
3! 2
3 ∞ tr
t
t
1 + tµ1 '+ µ 2 '+ µ 3 '+....... = ∑ µ r '
Mx(t) = = r!
2!
3! r0
In the above expression, the rth raw moment is the
tr
coefficient of
in the above expanded sum. To find out the
r!
moments differentiate the moment generating function with respect
to t once, twice, thrice… and put t = 0 in the first, second, third,
…
… derivatives to obtain the first, second, third,….. moments.
..
…
From the resulting expression, we get the raw moments
about the origin. The central moments are obtained by using the
relationship between raw moments and central moments.
2.7 Characteristic function:
The moment generating function does not exist for every
distribution. Hence another function, which always exists for all the
distributions is known as characteristic function.
It is the expected value of eitx, where i = − 1 and t has a
real value and the characteristic function of a random variable X is
denoted by φx(t)
For a discrete variable X having the probability function
p(x), the characteristic function is φx(t) = Σ eitx p(x)
For a continuous variable X having density function f(x),
b such that a < x < b , the characteristic function φx(t) = ∫
a 62 eitx f(x)dx Exercise  2
I. Choose the best answer:
n 1. ∑ p( x ) is equal to
i =1 i (a) 0
(b) 1
(c) –1
(d) ∞
2. If F(x) is distribution function, then F(∞) is
(a) –1
(b) 0
(c) 1
(d) ∞
3. From the given random variable table, the value of a is (a) 1 X=x
pi
1
(b)
2 0
a 1
2a
(c) 4 2
a
(d) 1
4 4. E(2x+3) is
(a) E(2x)
(b) 2E(x) +3 (c) E(3)
(d) 2x+3
5. Var(x+8) is
(a) var (8)
(b) var(x)
(c) 8 var(x) (d) 0
6. Var(5x+2) is
(a) 25 var (x) (b) 5 var (x) (c) 2 var (x) (d) 25
7. Variance of the random variable X is
(a) E(x2)  [E(x)]2
(b) [E(x)]2  E(x2)
2
(c) E(x )
(d) [E(x)]2
1
8. Variance of the random variable x is
; its standard
16
deviation is
1
1
(a)
(b)
256
32
1
1
(c)
(d)
64
4
9. A random variable X has E(x) = 2 and E(x2) = 8 its variance
is
(a) 4
(b) 6
(c) 8
(d) 2 63 10. If f(x) is the p.d.f of the continuous random variable x, then
E(x2) is
∞ (a) ∞ ∫ f ( x) dx (b) ∫x (d) −∞
∞ (c) 2 ∫ xf (x ) dx −∞
∞ f ( x ) dx ∫ f (x 2 ) dx −∞ −∞ II. Fill in the blanks:
11. If f(x) is a distribution function, then F(+∞) is equal to
________
12. If F(x) is a cumulative distribution function of a continuous
random variable x with p.d.f f(x) then F′(x) = __________
13. f(x) is the probability density function of a continuous
∞ random variable X. Then ∫ f ( x) dx is equal to ________ −∞ 14. Mathematical expectation of a random variable X is also
known as _____________
15. Variance of a constant is _____________
16. Var (12x) is _____________
17. Var (4x+7) is _________
18. If x is a discrete random variable with the probabilities pi ,
then the expected value of x2 is ________
19. If f(x) is the p.d.f of the continuous random variable X,
then the expectation of X is given by __________
20. The moment generating function for the discrete random
variable is given by ____________
III. Answer the following:
21. Define random variable.
22. Define discrete random variable
23. Define continuous random variable
24. What is probability mass function?
25. What is discrete probability distribution?
26. Define probability density function.
64 27. Write the properties of distribution function.
28. Define mathematical expectation for discrete random
variable.
29. Define the expectation of a continuous random variable.
30. State the moment generating function.
31. State the characteristic function for a discrete random
variable.
32. State the characteristic function for the continuous random
variable.
33. Write short note on moment generating function.
34. Write a short note on characteristic function.
35. Find the probability distribution of X when 3 coins are
tossed, where x is defined as getting head.
36. Two dice are thrown simultaneously and getting three is
termed as success. Obtain the probability distribution of the
number of threes.
37. Three cards are drawn at random successively, with
replacement, from a well shuffled pack of 52 cards. Getting
a card of diamond is termed as success. Obtain the
probability distribution of the number of success.
38. A random variable X has the following probability
distribution
Value of x
0
1
2
3
4
P(X=x)
3a
4a
6a 7a 8a
(a) determine the value of a
(b) Find p( 1 < x < 4 )
(c) P(1 ≤ x ≤ 4)
(d) Find P(x >2)
(e) Find the distribution function of x
39. A random variable X has the following probability
function.
Values
0
1
2
3
4
5
6
7
of X, x
P(x)
0k
(i) Find k
(iii) Find p(x ≤ 6) 2k
2k
3k
k2
(ii) Find p(0 < x < 5)
65 2k2 7k2+k 40. Verify whether the following are probability density
function
(i) f(x) = 6x5 ,
0<x<1
2x
(ii) f(x) =
,
0<x<3
9
41. A continuous random variable x follows the probability
law. f(x) = Ax3, 0 < x < 1 determine A
42. A random variable X has the density function f(x) = 3x2 ,
0 < x < 1 Find the probability between 0.2 and 0.5
43. A random variable X has the following probability
distribution
X=x 5 2 1 1
1
1
4
2
4
Find the expected value of x
44. A random variable X has the following distribution
x
1
0
1
2
P(x) 1
1
3
6
2
Find E(x) , E(x ) and Var (x) P(x) 45. A random variable X has E(x) = 1
6 1
3 1
1
and E(x2) =
find its
2
2 variance and standard deviation.
46. In a continuous distribution, whose probability density
3
function is given by f(x) = x(2x) , 0 < x < 2. Find the
4
expected value of x.
47. The probability density function of a continuous random
x
variable X is given by f(x) =
for 0 < x < 2. Find its mean
2
and variance 66 Answers
I.
1. (b)
6. (a)
II.
11. 1
15. zero
18. Σxi2 pi 2. (b)
7. (a) 3. (d)
8. (d) 12. f(x)
16. 144 var(x) 4. (b)
9. (a)
13. 1 ∞ 19. ∫ x f (x ) dx −∞ 5. (b)
10. (c) 14. Mean
17. 16 var(x)
∞
tr
20. ∑ µ!r
r = 0 r! III.
35.
X=x 0 1 2 3 P(xi) 1/8 3/8 3/8 1/8 36.
X=x 0 37.
X=x
P(xi) 0 2 25
36 P(x=x) 1
10
36 1
36 1 2 3 27
27
9
1
64
64
64
64
38.
(i) a = 1/28
(ii) 13/28
(iii) 25/28 (iv) 15/28
(v)
x
0
1
2
3
4
F(x)
28
3
7
13
20
=1
28
28
28
28
28
39. (i) k = 1/10 (ii) 4/5 (iii) 83/100
40. (i) p.d.f (ii) p.d.f
41. A = 4 42. P(0.2 < x , 0.5) = 0.117 43. 2.5
44. E(x) = 1/2 , var (x) = 19/12
45. 1/4 , 1/2 46. E(x) = 1
47. E(x) = 4/3 , var (x) = 2/9
67 3. SOME IMPORTANT
THEORETICAL DISTRIBUTIONS
3.1 BINOMIAL DISTRIBUTION
3.1.0 Introduction:
In this chapter we will discuss the theoretical discrete
distributions in which variables are distributed according to some
definite probability law, which can be expressed mathematically.
The Binomial distribution is a discrete distribution expressing the
probability of a set of dichotomous alternative i.e., success or
failure. This distribution has been used to describe a wide variety of
process in business and social sciences as well as other areas.
3.1.1 Bernoulli Distribution:
A random variable X which takes two values 0 and 1 with
probabilities q and p i.e., P(x=1) = p and P(x=0) = q, q = 1−p, is
called a Bernoulli variate and is said to be a Bernoulli Distribution,
where p and q takes the probabilities for success and failure
respectively. It is discovered by Swiss Mathematician James
Bernoulli (16541705).
Examples of Bernoulli’ s Trails are:
1) Toss of a coin (head or tail)
2) Throw of a die (even or odd number)
3) Performance of a student in an examination (pass or fail)
3.1.2 Binomial Distribution:
A random variable X is said to follow binomial distribution,
if its probability mass function is given by
nCx px qnx ; x = 0, 1,2, …
,n
0
; otherwise
Here, the two independent constants n and p are known as the
‘ parameters’ of the distribution. The distribution is completely
determined if n and p are known. x refers the number of successes.
68
P (X = x) = P(x) = If we consider N sets of n independent trials, then the number of
times we get x success is N(nCx px qnx). It follows that the terms in
the expansion of N (q + p)n gives the frequencies of the occurrences
of 0,1,2,...,x,...,n success in the N sets of independent trials.
3.1.3 Condition for Binomial Distribution:
We get the Binomial distribution under the following
experimental conditions.
1) The number of trials ‘ n’ is finite.
2) The trials are independent of each other.
3) The probability of success ‘ p’ is constant for each trial.
4) Each trial must result in a success or a failure.
The problems relating to tossing of coins or throwing of
dice or drawing cards from a pack of cards with replacement lead to
binomial probability distribution.
3.1.4 Characteristics of Binomial Distribution:
1. Binomial distribution is a discrete distribution in which the
random variable X (the number of success) assumes the
values 0,1, 2, … where n is finite.
.n,
2. Mean = np, variance = npq and
standard deviation σ = npq ,
q− p
Coefficient of skewness =
,
npq
1  6pq
, clearly each of the
npq
probabilities is nonnegative and sum of all probabilities is
1 ( p < 1 , q < 1 and p + q =1, q = 1− p ).
3. The mode of the binomial distribution is that value of the
variable which occurs with the largest probability. It may
have either one or two modes.
4. If two independent random variables X and Y follow
binomial distribution with parameter (n1, p) and (n2, p)
respectively, then their sum (X+Y) also follows Binomial
distribution with parameter (n1 + n2, p)
69
Coefficient of kurtosis = 5. If n independent trials are repeated N times, N sets of n
trials are obtained and the expected frequency of x success
is N(nCx px qnx). The expected frequencies of 0,1,2… n
success are the successive terms of the binomial distribution
of N(q + p)n.
Example 1:
Comment on the following: “ The mean of a binomial
distribution is 5 and its variance is 9”
Solution:
The parameters of the binomial distribution are n and p
We have mean ⇒ np = 5
Variance ⇒ npq = 9
npq
9
=
∴q=
np
5
9
q = >1
5
Which is not admissible since q cannot exceed unity. Hence
the given statement is wrong.
Example 2:
Eight coins are tossed simultaneously. Find the probability
of getting atleast six heads.
Solution:
Here number of trials, n = 8, p denotes the probability of
getting a head.
1
1
and q =
∴p=
2
2
If the random variable X denotes the number of heads, then
the probability of a success in n trials is given by
P(X = x) = ncx px qnx , x = 0 , 1, 2, ..., n
x 1 1
= 8Cx 2 2
1
= 8 8Cx
2 8− x 1
= 8Cx 2 70 8 Probability of getting atleast six heads is given by
P(x ≥ 6) = P(x = 6) + P(x = 7) + P(x = 8)
1
1
1
= 8 8C6 + 8 8C7 + 8 8C8
2
2
2
1
= 8 [ 8C6 + 8C7 + 8C8]
2
1
37
= 8 [ 28 +8 +1] =
256
2
Example 3:
Ten coins are tossed simultaneously. Find the probability
of getting (i) atleast seven heads (ii) exactly seven heads
(iii) atmost seven heads
Solution:
1
2
1
q = Probability of not getting a head =
2
The probability of getting x heads throwing 10 coins
simultaneously is given by
P(X = x) = nCx px qnx.
, x = 0 , 1, 2 , ..., n
p = Probability of getting a head x 10 − x 1 1
= 10Cx 2 2 = = 1
10Cx
210 i) Probability of getting atleast seven heads
P(x ≥ 7) = P (x = 7) + P(x = 8) + P (x = 9) + P (x =10)
1
= 10 [ 10C7 + 10C8 + 10C9+ 10C10]
2
1
176
=
[ 120 + 45 + 10 +1] =
1024
1024
ii) Probability of getting exactly 7 heads
1
1
P ( x = 7) = 10 10C7 = 10 (120)
2
2
120
=
1024
71 iii) Probability of getting atmost 7 heads
P( x ≤ 7) = 1 – P(x > 7)
= 1 − { P(x = 8) + P (x = 9) + P(x = 10)}
1
= 1− 10 {10C8 + 10C9 + 10C10}
2
1
= 1 − 10 [45 +10+1]
2
56
=1−
1024
968
=
1024
Example 4:
20 wrist watches in a box of 100 are defective. If 10
watches are selected at random, find the probability that (i) 10 are
defective (ii) 10 are good (iii) at least one watch is defective
(iv) at most 3 are defective.
Solution:
20 out of 100 wrist watches are defective
20
1
Probability of defective wrist watch , p =
=
100
5
∴ q = 1− p =
Since 10 watches are selected at random, n =10
P(X = x) = nCx px qnx , x = 0 , 1, 2, ...,10
10 − x x 1 4
= 10Cx 5 5
i) Probability of selecting 10 defective watches
10 P( x =10) = 10C10
= 1. 1 5 1
.1
510 0 4 5
1
= 10
5 72 4
5 ii) Probability of selecting 10 good watches (i.e. no defective)
0 10 1 4
P(x = 0) = 10C0 5 5
10 10 4 4
= 1.1. = 5 5
iii) Probability of selecting at least one defective watch
P(x ≥ 1) = 1 – P(x < 1)
= 1 – P(x = 0)
0 10 1 4
= 1 − 10C0 5 5
10 4
=1− 5
iv) Probability of selecting at most 3 defective watches
P (x ≤ 3) = P (x = 0) + P(x =1) + P(x = 2) + P(x = 3)
0 10 1 9 2 1 4
1 4
1 4
= 10C0 +10C1 + 10C2 5 5
5 5 5 5
3 1 4
+10C3 5 5
10 7 1 9 2 10.9 1 4 4
1 4
= 1.1. + 10 + 1.2 5 5 5
5 5
3 8 8 7 10.9.8
1 4
+ 1.2.3
5 5
= 1. (0.107) + 10 (0.026) + 45 (0.0062) + 120 (0.0016)
= 0.859 (approx)
Example 5:
With the usual notation find p for binomial random variable
X if n = 6 and 9P(X = 4) = P(X = 2)
Solution:
The probability mass function of binomial random variable X is
given by
P(X = x) = nCx px qnx.
, x = 0 , 1, 2, ...,n
73 ∴ P(X = x) = 6Cx px q6x
P (x = 4) = 6C4 p4 q2
P (x = 2) = 6C2 p2 q4
Given that,
9. P(x = 4) = P(x = 2)
9. 6C4 p4q2 = 6C2 p2q4
⇒ 9 × 15p2 = 15q2
9p2 = q2
Taking positive square root on both sides we get,
3p = q
= 1− p
4p = 1
1
= 0.25
∴p=
4
Here n = 6 3.1.5 Fitting of Binomial Distribution:
When a binomial distribution is to be fitted to an observed
data, the following procedure is adopted.
Σfx
1. Find Mean = x =
= np
Σf
x
where n is number of trials
⇒p=
n
2. Determine the value, q = 1− p.
3. The probability function is P(x) = nCx px qnx put x = 0, we
set P(0) = qn and f(0) = N × P(0)
4. The other expected frequencies are obtained by using the
recurrence formula is given by
nx p
f(x+1) =
f(x)
x+1 q
Example 6:
A set of three similar coins are tossed 100 times with the
following results
Number of heads :
0
1
2
3
Frequency : 36 40 22
74 2 Solution:
X
0
1
2
3 f
fx
36
0
40
40
22
44
2
6
Σf =100 Σfx = 90
90
Σfx
=
Mean = x =
= 0.9
100
Σf
x
p=
n
0.9
= 0.3
=
3
q = 1 –0.3
= 0.7
The probability function is P(x) = nCx px qnx
Here n = 3, p = 0.3 q = 0.7
∴P(x) = 3Cx (0.3)x (0.7)3x
P(0) = 3C0 (0.3)0 (0.7)3
= (0.7)3 = 0.343
∴ f(0) = N × P(0) = 0.343 × 100 = 34.3
The other frequencies are obtained by using the recurrence formula
nx p f(x+1) =
f(x). By putting x = 0, 1, 2 the expected
x+1 q frequencies are calculated as follows.
30 p f(1) =
× 34.3
0 +1 q = 3 × (0.43) × 34.3 = 44.247
3 1 p f(2) =
f(1)
1+1 q 2
=
(0.43) × 44.247
2
= 19.03
75 3− 2 p f(2)
2 +1 q 1
= (0.43) × 19.03
3
= 2.727
The observed and theoretical (expected) frequencies are tabulated
below:
Total
Observed
36
40
22
2
100
frequencies
Expected
34
44
19
3
100
frequencies
f(3) = Example 7:
4 coins are tossed and number of heads noted. The
experiment is repeated 200 times and the following distribution is
obtained .
x: Number of heads
0
1
2
3
4
f: frequencies
62
85
40
11
2
Solution:
X
f
fx 0
62
0 1
85
85 2
40
80 3
11
33 206
Σfx
=
= 1.03
Σf
200
x
1.03
p=
=
= 0.2575
n
4
∴ q = 1− 0.2575 = 0.7425
Here n = 4 , p = 0.2575 ; q = 0.7425
The probability function of binomial distribution is
P(x) = nCx px qnx
Mean = x = 76 4
2
8 Total
200
206 The binomial probability function is
P(x) = 4Cx (0.2575)x (0.7425)4x
P(0) = (0.7425)4
= 0.3039
∴ f(0) = NP(0)
= 200 × 0.3039
= 60.78
The other frequencies are calculated using the recurrence formula
nx p f(x+1) =
f(x). By putting x = 0,1, 2, 3 then the expected
x+1 q frequencies are calculated as follows:
Put x = 0, we get
40
f(1) =
(0.3468) (60.78)
0 +1
= 84.3140
4 1
f(2) =
(0.3468) (84.3140)
1+1
= 43.8601
42
f(3) =
(0.3468) (43.8601)
2 +1
= 10.1394
43
f(4) =
(0.3468) (10.1394)
3 +1
= 0.8791
The theoretical and expected frequencies are tabulated below:
Total
Observed
frequencies
Expected
frequencies 62 85 40 11 2 200 61 84 44 10 1 200 77 3.2 POISSON DISTRIBUTION:
3.2.0 Introduction:
Poisson distribution was discovered by a French
MathematiciancumPhysicist Simeon Denis Poisson in 1837.
Poisson distribution is also a discrete distribution. He derived it as a
limiting case of Binomial distribution. For ntrials the binomial
distribution is (q + p)n ; the probability of x successes is given by
P(X=x) = nCx px qnx . If the number of trials n is very large and the
probability of success ‘ p’ is very small so that the product np = m is
non – negative and finite.
The probability of x success is given by
e−m m x
P( X = x ) =
for x = 0,1,2, …
x!
0
; otherwise
Here m is known as parameter of the distribution so that m >0
Since number of trials is very large and the probability of
success p is very small, it is clear that the event is a rare event.
Therefore Poisson distribution relates to rare events.
Note:
1) e is given by e = 1 + 1
11
+
+ +… = 2.71828
..
1! 2! 3! e− m m 0
, 0! = 1 and 1! = 1
0!
e− m m1
3) P(X=1) =
1!
Some examples of Poisson variates are :
1. The number of blinds born in a town in a particular year.
2. Number of mistakes committed in a typed page.
3. The number of students scoring very high marks in all
subjects
4. The number of plane accidents in a particular week.
5. The number of defective screws in a box of 100,
manufactured by a reputed company.
6. Number of suicides reported in a particular day.
2) P(X=0) = 78 3.2.1 Conditions:
Poisson distribution is the limiting case of binomial
distribution under the following conditions:
1. The number of trials n is indefinitely large i.e., n à ∞
2. The probability of success ‘ p’ for each trial is very small;
i.e., p à 0
3. np = m (say) is finite , m > 0
3.2.2 Characteristics of Poisson Distribution:
The following are the characteristics of Poisson distribution
1. Discrete distribution: Poisson distribution is a discrete
distribution like Binomial distribution, where the random
variable assume as a countably infinite number of values
0,1,2 …
.
2. The values of p and q: It is applied in situation where the
probability of success p of an event is very small and that of
failure q is very high almost equal to 1 and n is very large.
3. The parameter: The parameter of the Poisson distribution is
m. If the value of m is known, all the probabilities of the
Poisson distribution can be ascertained.
4. Values of Constant: Mean = m = variance; so that standard
deviation = m
Poisson distribution may have either one or two modes.
5. Additive Property: If X and Y are two independent Poisson
distribution with parameter m1 and m2 respectively. Then
(X+Y) also follows the Poisson distribution with parameter
(m1 + m2)
6. As an approximation to binomial distribution: Poisson
distribution can be taken as a limiting form of Binomial
distribution when n is large and p is very small in such a
way that product np = m remains constant.
7. Assumptions: The Poisson distribution is based on the
following assumptions.
i)
The occurrence or non occurrence of an event does
not influence the occurrence or nonoccurrence of
any other event.
79 ii) iii) The probability of success for a short time interval
or a small region of space is proportional to the
length of the time interval or space as the case may
be.
The probability of the happening of more than one
event is a very small interval is negligible. Example 8:
Suppose on an average 1 house in 1000 in a certain district
has a fire during a year. If there are 2000 houses in that district,
what is the probability that exactly 5 houses will have a fire during
the year? [given that e2 = 0.13534]
1
Mean, x = np , n = 2000 and p =
1000
1
= 2000 ×
1000
m=2
The Poisson distribution is
e−m m x
P(X=x) =
x!
−2 5
e2
∴P(X =5) =
5!
(0.13534) × 32
=
120
= 0.036
(Note: The values of em are given in Appendix )
Example 9:
In a Poisson distribution 3P(X=2) = P(X=4)
parameter ‘ m’ .
Solution:
e−m m x
Poisson distribution is given by P(X=x) =
x!
Given that 3P(x=2) = P(x= 4) 80 Find the e−m m2
e−m m4
=
2!
4!
3× 4!
m2 =
2!
∴ m= ±6
Since mean is always positive ∴ m = 6
3. Example 10:
If 2% of electric bulbs manufactured by a certain company
are defective. Find the probability that in a sample of 200 bulbs
i) less than 2 bulbs ii) more than 3 bulbs are defective.[e4 = 0.0183]
Solution:
2
= 0.02
100
Given that n = 200 since p is small and n is large
We use the Poisson distribution
mean, m = np = 200 × 0.02 = 4
e−m m x
Now, Poisson Probability function, P(X = x) =
x!
i)
Probability of less than 2 bulbs are defective
= P(X<2)
= P(x = 0) + P(x = 1)
e−4 40 e−4 41
=
+
0!
1!
4
4
= e + e (4)
= e 4 (1 + 4) = 0.0183 × 5
= 0.0915
ii)
Probability of getting more than 3 defective bulbs
P(x > 3) = 1− P(x ≤ 3)
= 1− {P(x = 0) + P(x =1) + P(x=2) + P(x=3)}
4 2 43
4
= 1− e {1+ 4+
+}
2! 3!
= 1− {0.0183 × (1 + 4 + 8 + 10.67)}
= 0.567
81 The probability of a defective bulb = p = 3.2.3 Fitting of Poisson Distribution:
The process of fitting of Poisson distribution for the
probabilities of x = 0, 1,2,... success are given below :
∑ fx
i) First we have to calculate the mean = x =
=m
∑f
ii) The value of em is obtained from the table (see Appendix )
e − m .m x
iii) By using the formula P(X=x) =
x!
Substituting x = 0, P(0) = em
Then f(0) = N×P(0)
The other expected frequencies will be obtained by using the
recurrence formula
m
f(x+1) =
f(x) ; x = 0,1,2, …
x +1
Example 11:
The following mistakes per page were observed in a book.
Number of mistakes ( per page)
0
1
2
3
4
Number of pages
211 90 19
5
0
Fit a Poisson distribution to the above data.
Solution:
xi
0
1
2
3
4 fi
211
90
19
5
0
N = 325 fx
N
143
=
= 0 .44 = m
325
Then em ⇒ e 0.44 = 0.6440
82
Mean = x = fixi
0
90
38
15
0
fx = 143 Probability mass function of Poisson distribution is
mx
P(x) = em
x!
0
0.44 44
Put x = 0,
P(0) = e
0!
= e0.44
= 0.6440
∴ f(0) = N P(0)
= 325 × 0.6440
= 209.43
The other expected frequencies will be obtained by using the
recurrence formula
m
f(x+1) =
f(x). By putting x = 0,1,2,3 we get the
x +1
expected frequencies and are calculated as follows.
f(1) = 0.44 × 209.43 = 92.15
0.44
f(2) =
× 92.15 = 20.27
2
0.44
f(3) =
× 20.27 = 2.97
3
0.44
= 0.33
f(4) =
× 2.97
4
Total
Observed
211
90
19
5
0
325
frequencies
Expected
210
92
20
3
0
325
frequencies
Example 12:
Find mean and variance to the following data which gives the
frequency of the number of deaths due to horse kick in 10 corps per
army per annum over twenty years.
X
0
1
2
3
4
Total
F
109
65
22
3
1
200
83 Solution:
Let us calculate the mean and variance of the given data
xi
fi
fixi
fixi2
0
109
0
0
1
65
65
65
2
22
44
88
3
3
9
27
4
1
4
16
Total
N = 200
fx = 122
fx2 = 196
i Mean = x = fix N
122
=
200
= 0.61 Variance = σ2 = Hence, i f i2x () −x 2 N
196
=
− (0.61)2
200
= 0.61
mean = variance = 0.61 Example 13:
100 car radios are inspected as they come off the production
line and number of defects per set is recorded below
No. of
defects
No. of
sets 0 1 2 3 4 79 18 2 1 0 Fit a Poisson distribution and find expected frequencies
84 Solution:
x
0
1
2
3
4 f
fx
79
0
18
18
2
4
1
3
0
0
N = 100
f x = 25
fx
Mean = x =
N
25
=
100
∴m = 0.25
Then em = e 0.25 = 0.7788 = 0.779
Poisson probability function is given by
e− m m x
P(x) =
x!
−0.25
e (0.25)0
P(0) =
= (0.779)
0!
∴ f(0) = N.P(0) = 100 × (0.779) = 77.9
Other frequencies are calculated using the recurrence formula
m
f(x+1) =
f(x).
x +1
By putting x = 0,1,2,3, we get the expected frequencies and are
calculated as follows.
m
f(1) = f(0+1) =
f(0)
0+1
0.25
(77.9)
f(1) =
1
= 19.46
0.25
f(2) =
(19.46)
2
= 2.43
85 0.25
(2.43)
3
= 0.203
0.25
f(4) =
(0.203)
4
= 0.013
f(3) = Observed
frequencies
Expected
frequencies 79 18 2 1 0 100 78 20 2 0 0 100 Example 14:
Assuming that one in 80 births in a case of twins, calculate
the probability of 2 or more sets of twins on a day when 30 births
occurs. Compare the results obtained by using (i) the binomial and
(ii) Poisson distribution.
Solution:
(i) Using Binomial distribution
1
= 0.0125
80
∴ q = 1− p = 1 – 0.0125
= 0.9875
n = 30
Binomial distribution is given by
P(x) = nCx px qnx
P(x ≥ 2) = 1 – P(x < 2)
= 1 – {P(x = 0) + P(x =1)}
= 1 – {30C0(0.0125)0 (0.9875)30
+ 30C1 (0.0125)1(0.9875)29}
= 1– {1.1(0.9875)30 + 3 (0.125) (0.9875)29}
= 1 – { 0.6839 + 0.2597}
= 1 – 0.9436
P( x ≥ 2) = 0.0564
Probability of twins birth = p = 86 (ii) By using Poisson distribution:
The probability mass function of Poisson distribution is given by
e− m m x
P(x) =
x!
Mean = m = np
= 30 (0.0125) = 0.375
P(x ≥2) = 1− P(x <2)
= 1 − { P( x = 0) + P( x = 1)}
e −0.375 (0.375)0
e −0.375 (0.375)1
=1–{
+
}
0!
1!
= 1 − e 0.375 ( 1 + 0.375)
= 1 – (0.6873) (1.375) = 1 – 0.945 = 0.055 3.3 NORMAL DISTRIBUTION:
3.3.0 Introduction:
In the preceding sections we have discussed the discrete
distributions, the Binomial and Poisson distribution.
In this section we deal with the most important continuous
distribution, known as normal probability distribution or simply
normal distribution. It is important for the reason that it plays a
vital role in the theoretical and applied statistics.
The normal distribution was first discovered by DeMoivre
(English Mathematician) in 1733 as limiting case of binomial
distribution. Later it was applied in natural and social science by
Laplace (French Mathematician) in 1777. The normal distribution
is also known as Gaussian distribution in honour of Karl Friedrich
Gauss(1809).
3.3.1 Definition:
A continuous random variable X is said to follow normal
distribution with mean µ and standard deviation σ, if its
probability density function
1 x −µ 2
− 1
2 σ ;−∞ < x < ∞ , − ∞ < µ < ∞, σ > 0.
f(x) =
e
σ 2π
87 Note:
The mean µ and standard deviation σ are called the
parameters of Normal distribution. The normal distribution is
expressed by X ∼ N(µ, σ2)
3.3.2 Condition of Normal Distribution:
i) Normal distribution is a limiting form of the binomial
distribution under the following conditions.
a) n, the number of trials is indefinitely large ie., nà ∞ and
b) Neither p nor q is very small.
ii) Normal distribution can also be obtained as a limiting form of
Poisson distribution with parameter m à ∞
iii) Constants of normal distribution are mean = µ, variation =σ2,
Standard deviation = σ.
3.3.3 Normal probability curve:
The curve representing the normal distribution is called the
normal probability curve. The curve is symmetrical about the mean
(µ), bellshaped and the two tails on the right and left sides of the
mean extends to the infinity. The shape of the curve is shown in the
following figure. ∞ x=µ 88 ∞ 3.3.4 Properties of normal distribution:
1. The normal curve is bell shaped and is symmetric at x = µ.
2. Mean, median, and mode of the distribution are coincide
i.e., Mean = Median = Mode = µ
3. It has only one mode at x = µ (i.e., unimodal)
4. Since the curve is symmetrical, Skewness = β1 = 0 and
Kurtosis = β2 = 3.
5. The points of inflection are at x = µ ± σ
6. The maximum ordinate occurs at x = µ and
1
its value is =
σ 2π
7. The x axis is an asymptote to the curve (i.e. the curve
continues to approach but never touches the x axis)
8. The first and third quartiles are equidistant from median.
9. The mean deviation about mean is 0.8 σ
10. Quartile deviation = 0.6745 σ
11. If X and Y are independent normal variates with mean µ1
and µ2, and variance σ12 and σ22 respectively then their sum
(X + Y) is also a normal variate with mean (µ1 + µ2) and
variance (σ12 + σ22)
12. Area Property
P(µ  σ < × < µ + σ) = 0.6826
P(µ  2σ < × < µ + 2σ) = 0.9544
P(µ  3σ < × < µ + 3σ) = 0.9973
3.3.5 Standard Normal distribution:
Let X be random variable which follows normal distribution
with mean µ and variance σ2 .The standard normal variate is
X−µ
defined as Z =
which follows standard normal distribution
σ
with mean 0 and standard deviation 1 i.e., Z ∼ N(0,1). The standard
1 −1 2
Z
2 ;  ∞ < z< ∞
e
2π
The advantage of the above function is that it doesn’ t contain any
parameter. This enable us to compute the area under the normal
probability curve.
89
normal distribution is given by φ(z) = 3.3.6 Area properties of Normal curve:
The total area under the normal probability curve is 1. The
curve is also called standard probability curve. The area under the
curve between the ordinates at x = a and x = b where a < b,
represents the probabilities that x lies between x = a and x = b i.e.,
P(a ≤ x ≤ b) ∞ x = µ x=a x=b +∞ To find any probability value of x, we first standardize it by
X−µ
using Z =
, and use the area probability normal table. (given
σ
in the Appendix).
For Example: The probability that the normal random variable x to
lie in the interval (µ−σ , µ+σ) is given by ∞ x=µ−σ
z = 1 x=µ x=µ+σ
z=0 z=+1 90 +∞ P( µ − σ < x < µ+σ) = P(−1 ≤ z ≤ 1 )
= 2P(0 < z < 1)
= 2 (0.3413) (from the area table)
= 0.6826
P( µ  2σ < x < µ+2σ) = P(2 < z < 2 )
= 2P(0 < z < 2)
= 2 (0.4772) = 0.9544 ∞ x=µ−2σ
z = 2 x=µ
z=0 x=µ+2σ
z = +2 +∞ P(µ − 3σ < x < µ + 3σ) = P(−3 < z < 3 )
= 2P(0 < z < 3)
= 2 (0.49865) = 0.9973 ∞ x=µ−3σ
z = 3 x=µ
z =0
91 x=µ+3σ
z=+3 +∞ The probability that a normal variate x lies outside the range µ ± 3σ
is given by
P(x −µ  > 3σ) = P(z >3)
= 1 – P(−3 ≤ z ≤ 3)
= 1 − 0.9773 = 0.0027
Thus we expect that the values in a normal probability curve will
lie between the range µ ± 3σ, though theoretically it range
from − ∞ to ∞.
Example 15:
Find the probability that the standard normal variate lies
between 0 and 1.56
Solution: 0.4406 ∞ z =0 z = 1.56 +∞ P(0<z<1.56) = Area between z = 0 and z = 1.56
= 0.4406 (from table)
Example 16:
Find the area of the standard normal variate from –1.96 to 0.
Solution: 0.4750 ∞ z = 1.96 z=0
92 +∞ Area between z = 0 & z =1.96 is same as the area z = −1.96 to z = 0
P(1.96 < z < 0) = P(0 < z < 1.96) (by symmetry)
= 0.4750
(from the table)
Example 17:
Find the area to the right of z = 0.25
Solution: 0.4013 ∞ +∞ z = 0 z = 0.25 P(z >0.25) = P(0<z < ∞) – P(0<z<0.25)
= 0.5000  0.0987 (from the table) = 0.4013
Example 18:
Find the area to the left of z = 1.5
Solution: 0.9332 ∞ z=0 z = 1.5 +∞ P(z < 1.5) = P( − ∞ < z < 0 ) + P( 0 < z < 1.5 )
= 0.5 + 0.4332 (from the table)
= 0.9332
93 Example 19:
Find the area of the standard normal variate between –1.96 and 1.5
Solution: 0.9082 ∞ z= 1.96 z=0 z=1.5 +∞ P(1.96 < z < 1.5) = P(1.96 < z < 0) + P(0 < z < 1.5)
= P(0 < z < 1.96) + P(0 < z < 1.5)
= 0.4750 + 0.4332
(from the table)
= 0.9082
Example 20:
Given a normal distribution with µ = 50 and σ = 8, find the
probability that x assumes a value between 42 and 64
Solution: 0.8012 ∞ z= 1 z=0 Given that µ = 50 and σ = 8
The standard normal variate z =
94 z=1.75 x−µ
σ +∞ 42 − 50 − 8
= −1
=
8
8
64 − 50 14
If X = 64, Z2 =
= = 1.75
8
8
∴ P(42 < x < 64) = P(−1 < z <1.75)
= P(−1< z < 0) + P(0 < z <1.95)
= P(0<z<1) + P (0 < z <1.75) (by symmetry)
= 0.3413 +0 .4599 (from the table)
= 0 .8012
Example 21:
Students of a class were given an aptitude test. Their marks
were found to be normally distributed with mean 60 and standard
deviation 5. What percentage of students scored.
i) More than 60 marks
(ii) Less than 56 marks
(iii) Between 45 and 65 marks
If X = 42 , Z1 = Solution:
Given that mean = µ = 60 and standard deviation = σ = 5
x−µ
i) The standard normal varaiate Z =
σ 0.5 ∞ z=0 z>0 +∞ x−µ
60 − 60
=
=0
5
σ
∴P(x > 60) = P(z > 0)
= P(0 < z < ∞ ) = 0.5000
Hence the percentage of students scored more than 60
marks is 0.5000(100) = 50 %
If X = 60, Z = 95 ii) If X = 56, Z = 56 − 60 − 4
=
= − 0.8
5
5 0.2119 ∞ z= 0.8 +∞ z=0 P(x < 56) = P(z < −0.8)
= P( ∞ < z < 0) – P(−0.8 < z < 0) (by symmetry)
= P(0 < 2 < ∞) – P(0 < z < 0.8)
= 0.5 − 0.2881
(from the table)
= 0.2119
Hence the percentage of students score less than 56 marks is
0.2119(100) = 21.19 %
45 − 60 − 15
iii) If X = 45, then z =
=
= −3
5
5 0.83995 ∞ z= 3 z=0 z=1 65 − 60 5
= =1
5
5
P(45 < x < 65) = P(−3 < z < 1)
= P(−3 < z < 0 ) + P ( 0 < z < 1)
X = 65 then z = 96 +∞ = P(0 < z < 3) + P(0 < z < 1)
( by symmetry)
= 0.4986 + 0.3413
(from the table)
= 0.8399
Hence the percentage of students scored between 45 and 65
marks is 0.8399(100) = 83.99 %
Example 22:
X is normal distribution with mean 2 and standard deviation
3. Find the value of the variable x such that the probability of the
interval from mean to that value is 0.4115
Solution:
Given µ = 2, σ = 3
Suppose z1 is required standard value,
Thus P (0 < z < z1) = 0.4115
From the table the value corresponding to the area 0.4115 is 1.35
that is z1 = 1.35
x−µ
Here z1 =
σ
x−2
1.35 =
3
x = 3(1.35) + 2
= 4.05 + 2 = 6.05
Example 23:
In a normal distribution 31 % of the items are under 45 and
8 % are over 64. Find the mean and variance of the distribution.
Solution:
Let x denotes the items are given and it follows the normal
distribution with mean µ and standard deviation σ
The points x = 45 and x = 64 are located as shown in the figure.
i)
Since 31 % of items are under x = 45, position of x into
the left of the ordinate x = µ
ii)
Since 8 % of items are above x = 64 , position of this x
is to the right of ordinate x = µ 97 ∞ z = z1 z=0
x = 45 x = µ z = z2
x = 64 +∞ x−µ
45 − µ
=
= − z1 (say)
σ
σ
Since x is left of x = µ , z1 is taken as negative
64 − µ
= z2 (say)
When x = 64, z =
σ
From the diagram P(x < 45) = 0.31
P(z <  z1) = 0.31
P( z1 < z < 0) = P( ∞ < z < 0) – p( ∞ < z < z1)
s
= 0.5  0.31 = 0.19
P(0 < z < z1) = 0.19
(by symmetry)
z1 = 0.50
(from the table)
Also from the diagram p(x > 64) = 0.08
P(0 < z < z2) = P(0 < z < ∞) – P(z2 < z < ∞)
= 0.5  0.08 = 0.42
z2 = 1.40
(from the table)
Substituting the values of z1 and z2 we get
45 − µ
64 − µ
= − 0.50 and
= 1.40
σ
σ
Solving µ  0.50 σ = 45  (1)
µ + 1.40 σ = 64  (2)
(2) – (1) ⇒ 1.90 σ = 19 ⇒ σ = 10
Substituting σ = 10 in (1)
µ = 45 + 0.50 (10)
= 45 + 5.0 = 50.0
Hence mean = 50 and variance = σ2 = 100
When x = 45, z = 98 Exercise – 3
I. Choose the best answer:
1. Binomial distribution applies to
(a) rare events
(b) repeated alternatives
(c) three events
(d) impossible events
2. For Bernoulli distribution with probability p of a success
and q of a failure, the relation between mean and variance
that hold is
(a) mean < variance
(b) mean > variance
(c) mean = variance
(d) mean < variance
3. The variance of a binomial distribution is
(a) npq (b) np (c) npq (d) 0
x 4. 5. 6. 7. 15 − x 2 1
The mean of the binomial distribution 15Cx 3 3
2
in which p =
is
3
(a) 5
(b) 10
(c) 15
(d) 3
The mean and variance of a binomial distribution are 8 and
4 respectively. Then P(x = 1) is equal to
1
1
1
1
(a) 12
(b) 4
(c) 6
(d) 8
2
2
2
2
If for a binomial distribution , n = 4 and also
P(x = 2) = 3P(x=3) then the value of p is
9
1
(a)
(b) 1
(c)
(d) None of the above
11
3
The mean of a binomial distribution is 10 and the number of
trials is 30 then probability of failure of an event is
(a) 0.25
(b) 0.333
(c) 0.666
(d) 0.9 99 8. The variance of a binomial distribution is 2. Its standard
deviation is
(a) 2
(b) 4
(c) 1/2
(d) 2
9. In a binomial distribution if the numbers of independent
trials is n, then the probability of n success is
(a) nCxpxqnx
(b) 1
(c) pn
(d)qn
10. The binomial distribution is completely determined if it is
known
(a) p only
(b) q only
(c) p and q (d) p and n
11. The trials in a binomial distribution are
(a) mutually exclusive
(b) nonmutually exclusive
(c) independent
(d) nonindependent
12. If two independent variables x and y follow binomial
distribution with parameters,(n1, p) and (n2, p) respectively,
their sum(x+y) follows binomial distribution with
parameters
(a) (n1 + n2, 2p)
(b) (n, p)
(c) (n1 + n2, p)
(d) (n1 + n2, p + q)
13. For a Poisson distribution
(a) mean > variance
(b) mean = variance
(c) mean < variance
(d) mean < variance
14. Poisson distribution correspondents to
(a) rare events
(b) certain event
(c) impossible event
(d) almost sure event
15. If the Poisson variables X and Y have parameters m1 and m2
then X+Y is a Poisson variable with parameter.
(a) m1m2
(b) m1+m2
(c) m1−m2
(d)m1/m2
16. Poisson distribution is a
(a) Continuous distribution
(b) discrete distribution
(c) either continuous or discrete
(d) neither continue nor discrete
100 17. Poisson distribution is a limiting case of Binomial
distribution when
(a) n à ∞ ; pà 0 and np = m
(b) n à 0 ; pà ∞ and p=1/m
(c) n à ∞ ; pà ∞ and np=m
(d) n à ∞ ; pà 0 ,np=m
18. If the expectation of a Poisson variable (mean) is 1 then
P(x < 1) is
(a) e1
(b) 12e1
(c) 1 5/2e1
(d) none of these
19. The normal distribution is a limiting form of Binomial
distribution if
(a) nà ∞ pà0
(b) nà0 , pàq
(c) nà∞ , pà n
(d) nà ∞ and neither p nor q is small.
20. In normal distribution, skewness is
(a) one
(b) zero
(c) greater than one
(d) less than one
21. Mode of the normal distribution is
1
(a) σ
(b)
(c) µ
(d) 0
2π
22. The standard normal distribution is represented by
(a) N(0,0)
(b) N(1,1)
(c) N(1,0)
(d) N(0,1)
23. Total area under the normal probability curve is
(a) less than one
(b) unity (c) greater than one (d) zero
24. The probability that a random variable x lies in the interval
(µ  2σ , µ + 2σ) is
(a) 0.9544
(b) 0.6826
(c) 0.9973
(d) 0.0027
25. The area P( ∞ < z < 0) is equal to
(a) 1
(b) 0.1
(c) 0.5
(d) 0
26. The standard normal distribution has
(a) µ =1, σ = 0
(b) µ = 0, σ = 1
(c) µ = 0 ,σ = 0
(d) µ =1, σ = 1
101 27. The random variable x follows the normal distribution
− 1 ( x −100 ) 2
2
25 then the value of C is
1
1
(a) 5 2π
(c)
(d) 5
(b)
5 2π
2π
28. Normal distribution has
(a) no mode
(b) only one mode
(c) two modes
(d) many mode
29. For the normal distribution
(a) mean = median =mode
(b) mean < median < mode
(c) mean > median > mode (d) mean > median < mode
30. Probability density function of normal variable
f(x) = C. e P(X = x) = 1
5 2π − e 1 ( x −30 )
2
25 2 ; α < x < α then mean and variance are
(a) mean = 30 variance = 5 (b) mean = 0, variance = 25
(c) mean = 30 variance = 25 (d) mean = 30, variance = 10
31. The mean of a Normal distribution is 60, its mode will be
(a) 60
(b) 40
(c) 50
(d) 30
2
32. If x is a normal variable with µ =100 and σ = 25 then
P(90 < x < 120) is same as
(a) P(1 < z < 1)
(b) P(2 < z < 4)
(c) P(4 < z < 4.1)
(d) P(2 < z < 3)
33. If x is N(6, 1.2) and P(0 ≤ z ≤1) = 0.3413 then
P(4.8 ≤ x ≤ 7.2) is
(a) 0.3413
(b) 0.6587
(c) 0.6826
(d) 0.3174
II. Fill in the blanks:
34. The probability of getting a head in successive throws of a
coin is _________
35. If the mean of a binomial distribution is 4 and the variance
is 2 then the parameter is __________
102 9 2 1
36. + refers the binomial distribution and its standard 3 3
deviation is _________
37. In a binomial distribution if number of trials to be large and
probability of success be zero, then the distribution becomes
________.
38. The mean and variance are _______ in Poisson distribution
39. The mean of Poisson distribution is 0.49 and its standard
deviation is ________
40. In Poisson distribution, the recurrence formula to calculate
expected frequencies is ______.
2
2
∑ fx
41. The formula
− x is used to find ________
N
42. In a normal distribution, mean takes the values from
__________to ________
43. When µ = 0 and σ = 1 the normal distribution is called
________
44. P( − ∞ < z < 0) covers the area ______
45. If µ = 1200 and σ = 400 then the standard normal variate z
for x = 800 is _________
46. At x = µ ± σ are called as __________ in a normal
distribution.
47. P(−3 < z < 3) takes the value __________
48. X axis be the ________to the normal curve. () III. Answer the following
49. Comment the following
“ For a binomial distribution mean = 7 and variance = 16
50. Find the binomial distribution whose mean is 3 and
variance 2
51. In a binomial distribution the mean and standard deviation
are 12 and 2 respectively. Find n and p
52. A pair of dice is thrown 4 times. If getting a doublet is
considered a success, find the probability of 2 success.
53. Explain a binomial distribution.
103 54. State the characteristics of a binomial distribution.
55. State the conditions for a binomial variate.
56. Explain the fitting of a binomial distribution.
57. For the binomial distribution (0.68+0.32)10 find the
probability of 2 success.
58. Find the mean of binomial distribution of the probability of
occurrence of an event is 1/5 and the total number of trials
is 100
59. If on an average 8 ships out of 10 arrive safely at a port,
find the mean and standard deviation of the number of ships
arriving safely out of total of 1600 ships.
60. The probability of the evening college student will be a
graduate is 0.4. Determine the probability that out of 5
students (i) none (ii) one (iii) atleast one will be a graduate
61. Four coins are tossed simultaneously. What is the
probability of getting i) 2 heads and 2 tails ii) atleast 2 heads
iii) atleast one head.
62. 10% of the screws manufactured by an automatic machine
are found to be defective. 20 screws are selected at random.
Find the probability that i) exactly 2 are defective ii) atmost
3 are defective iii) atleast 2 are defective.
63. 5 dice are thrown together 96 times. The numbers of getting
4, 5 or 6 in the experiment is given below. Calculate the
expected frequencies and compare the standard deviation of
the expected frequencies and observed frequencies.
Getting 4 ,5 or 6 : 0
1
2
3
4
5
Frequency : 1 10 24 35 18 8 64. Fit a binomial distribution for the following data and find
the expected frequencies.
X:
0
1
2
3
4
f 18 35 30 13 4 65. Eight coins are tossed together 256 times. Number of heads
observed at each toss is recorded and the results are given
104 below. Find the expected frequencies. What are the
theoretical value of mean and standard deviation? Calculate
also mean and standard deviation of the observed
frequencies.
Number of heads: 0 1 2 3 4
5
6
7
8
32 10
1
Frequencies
: 2 6 39 52 67 56
66. Explain Poisson distribution.
67. Give any two examples of Poisson distribution.
68. State the characteristics of Poisson distribution.
69. Explain the fitting of a Poisson distribution
70. A variable x follows a Poisson distribution with mean 6
calculate i) P(x = 0) ii) P(x = 2)
71. The variance of a Poisson Distribution is 0.5. Find P(x = 3).
[e 0.5 = 0.6065]
72. If a random variable X follows Poisson distribution such
that P(x =1) = P(x = 2) find (a) the mean of the distribution
and P(x = 0). [e2 = 0.1353]
73. If 3% of bulbs manufactured by a company are defective
then find the probability in a sample of 100 bulbs exactly
five bulbs are defective.
74. It is known from the past experience that in a certain plant
there are on the average 4 industrial accidents per month.
Find the probability that in a given year there will be less
than 3 accidents. Assume Poisson distribution.[e4 = 0.0183]
75. A manufacturer of television sets known that of an average
5% of this product is defective. He sells television sets in
consignment of 100 and guarantees that not more than 4
sets will be defective. What is the probability that a
television set will fail to meet the guaranteed quality?
[e5 = 0.0067]
76. One fifth percent of the blades produced by a blade
manufacturing factory turns out to be a defective. The
blades are supplied in pockets of 10. Use Poisson
distribution to calculate the approximate number of pockets
containing i) no defective (ii) all defective (iii) two
defective blades respectively in a consignment of 1,00,000
pockets.
105 77. A factory employing a huge number of workers find that
over a period of time, average absentee rate is three workers
per shift. Calculate the probability that in a given shift
i) exactly 2 workers (ii) more than 4 workers will be absent.
78. A manufacturer who produces medicine bottles finds that
0.1 % of the bottles are defective. They are packed in boxes
containing 500 bottles. A drag manufactures buy 100 boxes
from the producer of bottles. Using Poisson distribution find
how many boxes will contain (i) no defective ii) exactly 2
(iii) atleast 2 defective.
79. The distribution of typing mistakes committed by a typist is
given below:
Mistakes per page: 0
1
2
3
4
5
No of pages
: 142 156 69
57
5
1
Fit a Poisson distribution.
80. Fit a Poisson distribution to the following data:
01
2
3
45
6 7 8 Total
x:
229 325 257 119 50 17
2 1 0 1000
f:
81. The following tables given that number of days in a 50,
days period during which automatically accidents occurred
in city. Fit a Poisson distribution to the data
No of accidents :
0
1
2
3
4
No of days
:
21
18
7
3
1
82. Find the probability that standard normal variate lies
between 0.78 and 2.75
83. Find the area under the normal curve between z = 0 and
z = 1.75
84. Find the area under the normal curve between z = 1.5 and
z = 2.6
85. Find the area to the left side of z = 1.96
86. Find the area under the normal curve which lies to the right
of z = 2.70
87. A normal distribution has mean = 50 and standard deviation
is 8. Find the probability that x assumes a value between 34
and 62
88. A normal distribution has mean = 20 and S.D = 10. Find
area between x =15 and x = 40
106 89. Given a normal curve with mean 30 and standard deviation
5. Find the area under the curve between 26 and 40
90. The customer accounts of a certain departmental store have
an average balance of Rs.1200 and a standard deviation of
Rs.400. Assuming that the account balances are normally
distributed. (i) what percentage of the accounts is over
Rs.1500? (ii) What percentage of the accounts is between
Rs.1000 and Rs.1500? iii) What percentage of the accounts
is below Rs.1500?
91. The weekly remuneration paid to 100 lecturers coaching for
professional entrance examinations are normally distributed
with mean Rs.700 and standard deviation Rs.50. Estimate
the number of lecturers whose remuneration will be i)
between Rs.700 and Rs.720 ii) more than Rs.750 iii) less
than Rs.630
92. x is normally distributed with mean 12 and standard
deviation 4. Find the probability of the following i) x ≥
20 ii) x ≤ 20
iii) 0 < x < 12
93. A sample of 100 dry cells tested to find the length of life
produced the following results µ =12 hrs, σ = 3 hrs.
Assuming the data, to be normally distributed. What
percentage of battery cells are expressed to have a life.
i) more than 15 hrs ii) between 10 and 14 hrs as iii) less
than 6 hrs?.
94. Find the mean and standard deviation of marks in an
examination where 44 % of the candidates obtained marks
below 55 and 6 % got above 80 marks.
95. In a normal distribution 7 % of the items are under 35 and
89 % of the items are under 63. Find its mean as standard
deviation.
Note: For fitting a binomial distribution in the problem itself, if it is
given that the coin is unbiased, male and female births are equally
probable, then we consider p = q = ½ . All other cases we have to
find the value of p from the mean value of the given data.
107 Answers
I.
1. b 2. b 3. a 4. b 5. a 6. c 7. c 8. d 9. c 10. d 11.c 12.c 13. b 14. a 15. b 16. b 17. d 18. a 19. d 20. b 21. c 22. d 23. b 24. a 25. c 26. b 27. b 28. b 31. a 32. b 33. c 29. a
30. c
1
1
34.
35. (8, )
2
2
38. equal
39. 0.7 36. 37. Poisson distribution
m
40. f(x+1) =
f(x) 41. variance
42.  ∞, + ∞
x +1
43. Standard normal distribution
44. 0.5
45. –1
46. Point of inflections
47. 0.9973
48. Asymptote
16
49. This is not admissible . Since q =
>1
7
9
2
1 2 1
50. + , p = , q = and n = 9
3
3 3 3
2
25
51. n = 18, p = .
52.
3
216
57. 10 C2 (0.32)2+(0.68)8
58. 20
59. 1280
3
11
15
60. i) 0.08 ii) 0.259 iii) 0.92
61. i) ii)
iii)
8
16
16
18
9
1
62. (i) 190 × 20 (ii) 20 [920 + 20 × 919 +190× 918 +1140 × 917]
10
10
1
(iii) 1 20 [920 + 20 × 919 +190× 918]
10
2 108 63. Observed S.D. = 1.13 and expected S.D.= 1.12
65. Observed mean = 4.0625 and S.D. = 1.462
70. i).0.00279 ii) 0.938 71. 0.0126 72. a) Mean = 2 b) P(x=0) = 0.1353
73. P(x = 5) = 0.1008
76. i) 98,020 ii)1960
78. i) 61 ii) 76
80. P(x) = e 74. 0.2379
iii) 20 77. i) 0.2241 ii) 0.1846
−
e 1 1x
79. P(x) =
x!
−
e 0.9 (0.9) x
81. P(x) =
x!
83. 0.4599 iii) 9 −1. 5 (1.5) x
x! 82. 0.2147
84. 0.9285 85. 0.9750 86. 0.0035 87. 0.9104 88. 0.6687 89. 0.7653 90. i) 22.66 % ii) 46.49 % iii) 77.34 % 91. i) 16 iii) 8 ii)16 75. 0.9598 92. i) 0.0228 ii) 0.9772 iii) 0.4987 93. i) 15.87 % ii) 49.72 % iii) 2.28 % 94. Mean = 57.21 and SD = 14.71
95. Mean = 50.27 and SD = 10.35 109 4. TEST OF SIGNIFICANCE (Basic Concepts)
4.0 Introduction:
It is not easy to collect all the information about population
and also it is not possible to study the characteristics of the entire
population (finite or infinite) due to time factor, cost factor and
other constraints. Thus we need sample. Sample is a finite subset of
statistical individuals in a population and the number of individuals
in a sample is called the sample size.
Sampling is quite often used in our daytoday practical life.
For example in a shop we assess the quality of rice, wheat or any
other commodity by taking a handful of it from the bag and then to
decide to purchase it or not.
4.1 Parameter and Statistic:
The statistical constants of the population such as mean, (µ),
variance (σ2), correlation coefficient (ρ) and proportion (P) are
called ‘ Parameters’ .
Statistical constants computed from the samples
corresponding to the parameters namely mean ( x ), variance (S2),
sample correlation coefficient (r) and proportion (p) etc, are called
statistic.
Parameters are functions of the population values while
statistic are functions of the sample observations. In general,
population parameters are unknown and sample statistics are used
as their estimates.
4.2 Sampling Distribution:
The distribution of all possible values which can be
assumed by some statistic measured from samples of same size ‘ n’
randomly drawn from the same population of size N, is called as
sampling distribution of the statistic (DANIEL and FERREL).
Consider a population with N values .Let us take a random
sample of size n from this population, then there are
110 N!
= k (say), possible samples. From each of
n!(N  n)!
these k samples if we compute a statistic (e.g mean, variance,
correlation coefficient, skewness etc) and then we form a frequency
distribution for these k values of a statistic. Such a distribution is
called sampling distribution of that statistic.
For example, we can compute some statistic
t = t(x1, x2,… n) for each of these k samples. Then t1, t2 … tk
..x
..,
determine the sampling distribution of the statistic t. In other words
statistic t may be regarded as a random variable which can take the
values t1, t2 … tk and we can compute various statistical constants
..,
like mean, variance, skewness, kurtosis etc., for this sampling
distribution.
The mean of the sampling distribution t is
1k
1
t = [t1 + t2 + ..... + tk ] =
∑ ti
K i =1
K
1
(t1  t ) 2 + (t2  t) 2 +......... + (tk  t )2 and var (t) = K
1
=
( ti  t ) 2
K
4.3 Standard Error:
The standard deviation of the sampling distribution of a
statistic is known as its standard error. It is abbreviated as S.E. For
example, the standard deviation of the sampling distribution of the
mean x known as the standard error of the mean, x + x + ...........xn Where v( x ) = v 1 2 n v( x1 ) v( x2 )
v( x )
=
+ 2 + ....... + 2n
2
n
n
n
2
2
2
nσ 2
σ
σ
σ
= 2 + 2 + ...... + 2 = 2
n
n
n
n
σ
∴ The S.E. of the mean is
n
NCn = 111 The standard errors of the some of the well known statistic
for large samples are given below, where n is the sample size, σ2 is
the population variance and P is the population proportion and
Q = 1−P. n1 and n2 represent the sizes of two independent random
samples respectively.
Sl.No Statistic
Standard Error
1.
σ
Sample mean x
n
2. Observed sample proportion p PQ
n 3. Difference between of two
samples means ( x 1 − x 2) σ112 σ2 2
+
n1
n2 4. Difference of two
proportions p1 – p2 P1Q1 P 2Q 2
+
n1
n2 sample Uses of standard error
i) Standard error plays a very important role in the large
sample theory and forms the basis of the testing of
hypothesis.
ii) The magnitude of the S.E gives an index of the precision of
the estimate of the parameter.
iii) The reciprocal of the S.E is taken as the measure of
reliability or precision of the sample.
iv) S.E enables us to determine the probable limits within
which the population parameter may be expected to lie.
Remark:
S.E of a statistic may be reduced by increasing the sample
size but this results in corresponding increase in cost, labour and
time etc.,
4.4 Null Hypothesis and Alternative Hypothesis
Hypothesis testing begins with an assumption called a
Hypothesis, that we make about a population parameter. A
hypothesis is a supposition made as a basis for reasoning. The
conventional approach to hypothesis testing is not to construct a
112 single hypothesis about the population parameter but rather to set
up two different hypothesis. So that of one hypothesis is accepted,
the other is rejected and vice versa.
Null Hypothesis:
A hypothesis of no difference is called null hypothesis and
is usually denoted by H0 “ Null hypothesis is the hypothesis” which
is tested for possible rejection under the assumption that it is true “
by Prof. R.A. Fisher. It is very useful tool in test of significance.
For example: If we want to find out whether the special classes (for
Hr. Sec. Students) after school hours has benefited the students or
not. We shall set up a null hypothesis that “H0: special classes after
school hours has not benefited the students”.
Alternative Hypothesis:
Any hypothesis, which is complementary to the null
hypothesis, is called an alternative hypothesis, usually denoted by
H1, For example, if we want to test the null hypothesis that the
population has a specified mean µ0 (say),
i.e., : Step 1: null hypothesis H0: µ = µ0
then
2. Alternative hypothesis may be
i)
H1 : µ ≠ µ0 (ie µ > µ0 o r µ < µ0)
ii)
H1 : µ > µ0
iii)
H1 : µ < µ0
the alternative hypothesis in (i) is known as a two – tailed
alternative and the alternative in (ii) is known as righttailed (iii) is
known as left –tailed alternative respectively. The settings of
alternative hypothesis is very important since it enables us to decide
whether we have to use a single – tailed (right or left) or two tailed
test.
4.5 Level of significance and Critical value:
Level of significance:
In testing a given hypothesis, the maximum probability with
which we would be willing to take risk is called level of
significance of the test. This probability often denoted by “ α” is
generally specified before samples are drawn.
113 The level of significance usually employed in testing of
significance are 0.05( or 5 %) and 0.01 (or 1 %). If for example a
0.05 or 5 % level of significance is chosen in deriving a test of
hypothesis, then there are about 5 chances in 100 that we would
reject the hypothesis when it should be accepted. (i.e.,) we are
about 95 % confident that we made the right decision. In such a
case we say that the hypothesis has been rejected at 5 % level of
significance which means that we could be wrong with probability
0.05.
The following diagram illustrates the region in which we
could accept or reject the null hypothesis when it is being tested at
5 % level of significance and a twotailed test is employed.
Accept the null hypothesis if the
sample statistics falls in this region Reject the null hypothesis if the sample
Statistics falls in these two region
Note: Critical Region: A region in the sample space S which
amounts to rejection of H0 is termed as critical region or region of
rejection.
Critical Value:
The value of test statistic which separates the critical (or
rejection) region and the acceptance region is called the critical
value or significant value. It depends upon i) the level of
114 significance used and ii) the alternative hypothesis, whether it is
twotailed or singletailed
For large samples the standard normal variate corresponding to the
t  E (t)
statistic t, Z =
~ N (0,1)
S.E. (t)
asymptotically as n à ∞
The value of z under the null hypothesis is known as test
statistic. The critical value of the test statistic at the level of
significance α for a two  tailed test is given by Zα/2 and for a one
tailed test by Zα. where Zα is determined by equation P(Z >Zα)= α
Zα is the value so that the total area of the critical region
α
α
on both tails is α . ∴ P(Z > Zα) = . Area of each tail is .
2
2
Zα is the value such that area to the right of Zα and to the
α
left of − Zα is
as shown in the following diagram.
2 /2 /2 4.6 One tailed and Two Tailed tests:
In any test, the critical region is represented by a portion of
the area under the probability curve of the sampling distribution of
the test statistic.
One tailed test: A test of any statistical hypothesis where the
alternative hypothesis is one tailed (right tailed or left tailed) is
called a one tailed test.
115 For example, for testing the mean of a population H0: µ = µ0,
against the alternative hypothesis H1: µ > µ0 (right – tailed) or
H1 : µ < µ0 (left –tailed)is a single tailed test. In the right – tailed
test H1: µ > µ0 the critical region lies entirely in right tail of the
sampling distribution of x , while for the left tailed test H1: µ < µ0
the critical region is entirely in the left of the distribution of x .
Right tailed test: Left tailed test: Two tailed test:
A test of statistical hypothesis where the alternative hypothesis
is two tailed such as, H0 : µ = µ0 against the alternative hypothesis
H1: µ ≠µ0 (µ > µ0 and µ < µ0) is known as two tailed test and in
such a case the critical region is given by the portion of the area
lying in both the tails of the probability curve of test of statistic.
116 For example, suppose that there are two population brands of
washing machines, are manufactured by standard process(with
mean warranty period µ1) and the other manufactured by some new
technique (with mean warranty period µ2): If we want to test if the
washing machines differ significantly then our null hypothesis is
H0 : µ1 = µ2 and alternative will be H1: µ1 ≠ µ2 thus giving us a two
tailed test. However if we want to test whether the average
warranty period produced by some new technique is more than
those produced by standard process, then we have H0 : µ1 = µ2 and
H1 : µ1 < µ2 thus giving us a lefttailed test.
Similarly, for testing if the product of new process is
inferior to that of standard process then we have, H0 : µ1 = µ2 and
H1 : µ1>µ2 thus giving us a righttailed test. Thus the decision about
applying a two – tailed test or a single –tailed (right or left) test will
depend on the problem under study.
Critical values (Zα) of Z
0.05 or 5%
0.01 or 1%
Level of
significance
Left
Right
Left
Right
α
Critical
2.33
values of
1.645
−2.33
−1.645
Zα for one
tailed Tests
Critical
2.58
values of
1.96
−2.58
−1.96
Zα/2 for two
tailed tests
4.7 Type I and Type II Errors:
When a statistical hypothesis is tested there are four
possibilities.
1. The hypothesis is true but our test rejects it ( Type I error)
2. The hypothesis is false but our test accepts it (Type II error)
3. The hypothesis is true and our test accepts it (correct
decision)
4. The hypothesis is false and our test rejects it (correct
decision)
117 Obviously, the first two possibilities lead to errors.
In a statistical hypothesis testing experiment, a Type I error is
committed by rejecting the null hypothesis when it is true. On the
other hand, a Type II error is committed by not rejecting
(accepting) the null hypothesis when it is false.
If we write ,
α = P (Type I error) = P (rejecting H0  H0 is true)
β = P (Type II error) = P (Not rejecting H0  H0 is false)
In practice, type I error amounts to rejecting a lot when it is good
and type II error may be regarded as accepting the lot when it is
bad. Thus we find ourselves in the situation which is described in
the following table.
Accept H0
Reject H0
H0 is true
Correct decision
Type I Error
H0 is false
Type II error
Correct decision
4.8 Test Procedure :
Steps for testing hypothesis is given below. (for both large
sample and small sample tests)
1. Null hypothesis : set up null hypothesis H0.
2. Alternative Hypothesis: Set up alternative hypothesis H1,
which is complementry to H0 which will indicate whether
one tailed (right or left tailed) or two tailed test is to be
applied.
3. Level of significance : Choose an appropriate level of
significance (α), α is fixed in advance.
4. Test statistic (or test of criterian):
t  E (t)
Calculate the value of the test statistic, Z =
under
S.E. (t)
the null hypothesis, where t is the sample statistic
5. Inference: We compare the computed value of Z (in
absolute value) with the significant value (critical value)
Zα/2 (or Zα). If Z > Zα, we reject the null hypothesis
H0 at α % level of significance and if Z ≤ Zα, we accept
H0 at α % level of significance.
118 Note:
1. Large Sample: A sample is large when it consists of more than
30 items.
2. Small Sample: A sample is small when it consists of 30 or less
than 30 items. Exercise 4
I. Choose the best answers:
1. A measure characterizing a sample such as x or s is called
(a). Population
(b). Statistic (c).Universe (d).Mean
2. The standard error of the mean is
n
σ
σ
(a). σ2
(b).
(d).
(c).
n
σ
n
3. The standard error of observed sample proportion “P” is
P(1 − Q)
PQ
(1 − P)Q
PQ
(a).
(b).
(d).
(c).
n
n
n
n
4. Alternative hypothesis is
(a). Always Left Tailed
(b). Always Right tailed
(c). Always One Tailed
(d). One Tailed or Two Tailed
5. Critical region is
(a). Rejection Area
(b). Acceptance Area
(c). Probability
(d). Test Statistic Value
6. The critical value of the test statistic at level of significance
α for a two tailed test is denoted by
(b).Zα
(c). Z2α
(a). Zα/2
(d). Zα/4
7. In the right tailed test, the critical region is
(a). 0
(b). 1
(c). Lies entirely in right tail (d). Lies in the left tail
8. Critical value of Zα at 5% level of significance for two
tailed test is
(a). 1.645
(b). 2.33
(c). 2.58
(d). 1.96 119 9. Under null hypothesis the value of the test statistic Z is
t  S.E. (t)
t + E(t)
t  E (t)
PQ
(a).
(b).
(c).
(d).
E (t)
S.E. (t)
S.E. (t)
n
10. The alternative hypothesis H1: µ ≠ µ0 (µ >µ0 or µ < µ0)
takes the critical region as
(a). Right tail only
(b). Both right and left tail
(c). Left tail only
(d). Acceptance region
11. A hypothesis may be classified as
(a). Simple
(b). Composite
(c). Null
(d). All the above
12. Whether a test is one sided or two sided depends on
(a). Alternative hypothesis
(b). Composite hypothesis
(c). Null hypothesis
(d). Simple hypothesis
13. A wrong decision about H0 leads to:
(a). One kind of error
(b). Two kinds of error
(c). Three kinds of error
(d). Four kinds of error
14. Area of the critical region depends on
(a). Size of type I error
(b). Size of type II error
(c). Value of the statistics
(d). Number of observations
15. Test of hypothesis H0 : µ = 70 vs H1 = µ > 70 leads to
(a). One sided left tailed test (b). One sided right tailed test
(c). Two tailed test
(d). None of the above
16. Testing H0 : µ = 1500 against µ < 1500 leads to
(a). One sided left tailed test (b). One sided right tailed test
(c). Two tailed test
(d). All the above
17. Testing H0 : µ = 100 vs H1: µ ≠ 100 lead to
(a). One sided right tailed test (b). One sided left tailed test
(c). Two tailed test
(d). None of the above
II. Fill in the Blanks
18. n1 and n2 represent the _________ of the two independent
random samples respectively.
19. Standard error of the observed sample proportion p is
_______
20. When the hypothesis is true and the test rejects it, this is
called _______
120 21. When the hypothesis is false and the test accepts it this is
called _______
22. Formula to calculate the value of the statistic is __________
III. Answer the following
23. Define sampling distribution.
24. Define Parameter and Statistic.
25. Define standard error.
26. Give the standard error of the difference of two sample
proportions.
27. Define Null hypothesis and alternative hypothesis.
28. Explain: Critical Value.
29. What do you mean by level of significance?
30. Explain clearly type I and type II errors.
31. What are the procedure generally followed in testing of a
hypothesis ?
32. What do you mean by testing of hypothesis?
33. Write a detailed note on one tailed and twotailed tests.
Answers:
I.
1. (b) 2. (c) 3. (b) 4.(d) 5. (a) 6. (a) 7. (c) 8. (d) 9. (c) 10 (b) 11.(d) 12.(a) 13.(b) 14.(a) 15.(b) 16.(a)
II. 17.(c) 18. size 19. PQ
n 20. Type I error 21. Type II error
t  E (t)
22. Z =
S.E. (t) 121 5. TEST OF SIGNIFICANCE
(Large Sample)
5.0 Introduction:
In practical problems, statisticians are supposed to make
tentative calculations based on sample observations. For example
(i)
The average weight of school student is 35kg
(ii)
The coin is unbiased
Now to reach such decisions it is essential to make certain
assumptions (or guesses) about a population parameter. Such an
assumption is known as statistical hypothesis, the validity of which
is to be tested by analysing the sample. The procedure, which
decides a certain hypothesis is true or false, is called the test of
hypothesis (or test of significance).
Let us assume a certain value for a population mean. To test
the validity of our assumption, we collect sample data and
determine the difference between the hypothesized value and the
actual value of the sample mean. Then, we judge whether the
difference is significant or not. The smaller the difference, the
greater the likelihood that our hypothesized value for the mean is
correct. The larger the difference the smaller the likelihood, which
our hypothesized value for the mean, is not correct.
5.1 Large samples (n > 30):
The tests of significance used for problems of large samples
are different from those used in case of small samples as the
assumptions used in both cases are different. The following
assumptions are made for problems dealing with large samples:
(i) Almost all the sampling distributions follow normal
asymptotically.
(ii) The sample values are approximately close to the
population values.
The following tests are discussed in large sample tests.
(i) Test of significance for proportion
(ii) Test of significance for difference between two
proportions
122 (iii) Test of significance for mean
(iv) Test of significance for difference between two means.
5.2 Test of Significance for Proportion:
Test Procedure
Set up the null and alternative hypotheses
H0 : P =P0
H1 = P ≠ P0 (P>P0 or P <P0)
Level of significance:
Let α = 0 .05 or 0.01
Calculation of statistic:
Under H0 the test statistic is
Z0 = p−P
PQ
n Expected value:
p−P ∼ N (0, 1)
PQ
n
= 1.96 for α = 0.05 (1.645)
= 2.58 for α = 0.01 (2.33) Ze = Inference:
(i) If the computed value of Z0 ≤ Ze we accept the null
hypothesis and conclude that the sample is drawn from the
population with proportion of success P0
(ii) If Z0 > Ze we reject the null hypothesis and conclude that
the sample has not been taken from the population whose
population proportion of success is P0.
Example 1:
In a random sample of 400 persons from a large population
120 are females.Can it be said that males and females are in the
ratio 5:3 in the population? Use 1% level of significance
123 Solution:
We are given
n = 400 and
x = No. of female in the sample = 120
p = observed proportion of females in the sample = 120
= 0.30
400 Null hypothesis:
The males and females in the population are in the ratio 5:3
3
i.e., H0: P = Proportion of females in the population =
= 0.375
8
Alternative Hypothesis:
H1 : P ≠ 0.375 (twotailed)
Level of significance:
α = 1 % o r 0.01
Calculation of statistic: Under H0, the test statistic is Z0 = = = p−P
PQ
n 0.300 − 0.375
0.375 × 0.625
400
0.075
0.075
= 3.125
=
0.024
0.000586 Expected value: Ze = p−P
PQ
n
124 ∼ N(0,1) = 2.58 Inference :
Since the calculated Z0 > Ze we reject our null hypothesis at
1% level of significance and conclude that the males and females in
the population are not in the ratio 5:3
Example 2:
In a sample of 400 parts manufactured by a factory, the
number of defective parts was found to be 30. The company,
however, claimed that only 5% of their product is defective. Is the
claim tenable?
Solution:
We are given n = 400
x = No. of defectives in the sample = 30
p= proportion of defectives in the sample
x 30
=
= 0.075
=
n 400
Null hypothesis:
The claim of the company is tenable H0: P= 0.05
Alternative Hypothesis:
H1 : P > 0.05 (Right tailed Alternative)
Level of significance: 5%
Calculation of statistic:
Under H0, the test statistic is
Z0 = = = p−P
PQ
n
0.075 − 0.050
0.05 × 0.95
400
0.025
0.0001187 125 = 2.27 Expected value:
p−P ∼ N(0, 1)
PQ
n
= 1.645 (Single tailed) Ze = Inference :
Since the calculated Z0 > Ze we reject our null hypothesis at
5% level of significance and we conclude that the company’ s claim
is not tenable.
5.3 Test of significance for difference between two proportion:
Test Procedure
Set up the null and alternative hypotheses:
H0 : P1 =P2 = P (say)
H1 : P1 ≠ P2 (P1>P2 or P1 <P2)
Level of significance:
Let α = 0.05 or 0.01
Calculation of statistic:
Under H0, the test statistic is Z0 = p1 − p 2
P1Q1 P2 Q 2
+
n1
n2 = p1 − p 2 (P1 and P2 are known) (P1 and P2 are not known)
1
1
PQ + n n 2
1
n p + n 2 p 2 x1 + x 2
=
where P = 1 1
n1 + n 2
n1 + n 2
Q =1− P
126 Expected value:
Ze = p1 − p 2
S.E(p1 − p 2 ) ∼ N(0,1) Inference:
(i) If Z0 ≤ Ze we accept the null hypothesis and conclude that
the difference between proportions are due to sampling
fluctuations.
(ii) If Z0 > Ze we reject the null hypothesis and conclude that
the difference between proportions cannot be due to sampling
fluctuations
Example 3:
In a referendum submitted to the ‘ student body’ at a
university, 850 men and 550 women voted. 530 of the men and 310
of the women voted ‘ yes’ . Does this indicate a significant
difference of the opinion on the matter between men and women
students?
Solution:
We are given
n1= 850
n2 = 550
530
310
p1 =
= 0.62
p2 =
= 0.56
850
550
x + x2
530 + 310
=
P= 1
= 0.60
n1 + n 2
1400 x1= 530 x2=310 Q = 0.40
Null hypothesis:
H0: P1= P2 ie the data does not indicate a significant difference of
the opinion on the matter between men and women students.
Alternative Hypothesis:
H1 : P1≠ P2 (Two tailed Alternative)
Level of significance:
Let α = 0.05 127 Calculation of statistic:
Under H0, the test statistic is Z0 = = p1 − p 2
1
1 PQ +
n
n2 1 0.62 − 0.56 1
1
0.6 × 0.4
+ 850 550 0.06
=
= 2.22
0.027 Expected value: Ze = p1 − p 2
1
1 PQ +
n 1 n2 ∼ N(0,1) = 1.96 Inference :
Since Z0 > Ze we reject our null hypothesis at 5% level of
significance and say that the data indicate a significant difference of
the opinion on the matter between men and women students.
Example 4:
In a certain city 125 men in a sample of 500 are found to be
self employed. In another city, the number of self employed are 375
in a random sample of 1000. Does this indicate that there is a
greater population of self employed in the second city than in the
first?
Solution:
We are given
n1= 500
n2 = 1000
x1 = 125
x2 = 375
128 125
375
= 0.25
p2 =
= 0.375
500
1000
x + x2
125 + 375
P= 1
=
n 1 + n 2 500 + 1000
500 1
=
=
1500 3
12
Q = 1− =
33
p1 = Null hypothesis:
H0: P1= P2 There is no significant difference between the two
population proportions.
Alternative Hypothesis:
H1 : P1< P2 (left tailed Alternative)
Level of significance: Let α = 0.05
Calculation of statistic:
Under H0, the test statistic is Z0 = = p1 − p 2
1
1 PQ +
n 1 n2 0.25 − 0.375
1 2 1
1
×
+ 3 3 500 1000 = 0.125
= 4.8
0.026 Expected value: Ze = p1 − p 2
1
1 PQ +
n
n2 1 129 ∼ N(0,1) = 1.645 Inference :
Since Z0 > Ze we reject the null hypothesis at 5% level of
significance and say that there is a significant difference between
the two population proportions.
Example 5:
A civil service examination was given to 200 people. On the
basis of their total scores, they were divided into the upper 30% and
the remaining 70%. On a certain question 40 of the upper group
and 80 of the lower group answered correctly. On the basis of this
question, is this question likely to be useful for discriminating the
ability of the type being tested?
Solution:
We are given
30 × 200
70 × 200
n1 =
= 60
n2 =
= 140
100
100
x1 = 40
x2 = 80
40 2
80 4
p1=
p2 =
=
=
60 3
140 7
x + x2
40 + 80
P= 1
=
n 1 + n 2 60 + 140
120 6
=
=
200 10
14
Q = 1− P = 1− =
6 10
Null hypothesis:
H0: P1= P2 (say) The particular question does not discriminate the
abilities of two groups.
Alternative Hypothesis:
H1 : P1≠ P2 (two tailed Alternative)
Level of significance:
Let α = 0.05
Calculation of statistics
Under H0, the test statistic is
130 Z0 = = = p1 − p 2
1
1 PQ +
n
n2 1 24
−
37
6 41
1
×+ 10 10 60 140 10
21 3 = 1.3 Expected value: Ze = p1 − p 2 1
1 PQ +
n
n2 1 = 1.96 for α=0.05 ∼ N(0,1) Inference :
Since Z0 < Ze we accept our null hypothesis at 5% level of
significance and say that the particular question does not
discriminate the abilities of two groups.
5.4 Test of significance for mean:
Let xi (i = 1,2… be a random sample of size n from a
..n)
population with variance σ2, then the sample mean x is given by
1
(x1 + x2 +… n)
..x
x=
n
E( x ) = µ
1
V( x ) = V [ (x1 + x2 +… n)]
..x
n
131 1
[(V(x1) + V(x2) +…
..V(xn)]
n2
2
1
σ
= 2 nσ2 =
n
n
σ
∴ S.E ( x ) =
n
= Test Procedure:
Null and Alternative Hypotheses:
H0:µ = µ0.
H1:µ ≠ µ0 (µ > µ0 o r µ < µ0)
Level of significance:
Let α = 0.05 or 0.01
Calculation of statistic:
Under H0, the test statistic is
x −µ
x − E( x )
Z0 =
=
S.E( x )
σ/ n
Expected value:
x −µ
Ze =
∼ N(0,1)
σ/ n
= 1.96 for α = 0.05 (1.645)
or
= 2.58 for α = 0.01 (2.33)
Inference :
If Z0 < Ze, we accept our null hypothesis and conclude that
the sample is drawn from a population with mean µ = µ0
If Z0 > Ze we reject our H0 and conclude that the sample is
not drawn from a population with mean µ = µ0
Example 6:
The mean lifetime of 100 fluorescent light bulbs produced
by a company is computed to be 1570 hours with a standard
deviation of 120 hours. If µ is the mean lifetime of all the bulbs
produced by the company, test the hypothesis µ=1600 hours against
132 the alternative hypothesis µ ≠ 1600 hours using a 5% level of
significance.
Solution:
We are given
x = 1570 hrs
µ = 1600hrs s =120 hrs n=100 Null hypothesis:
H0: µ= 1600.ie There is no significant difference between the
sample mean and population mean.
Alternative Hypothesis:
H1: µ ≠ 1600 (two tailed Alternative)
Level of significance:
Let α = 0.05
Calculation of statistics
Under H0, the test statistic is
x −µ
Z0 =
s/ n
1570 − 1600
120
100
30 × 10
=
120
= 2.5
= Expected value:
x −µ
∼ N(0,1)
s/ n
= 1.96 for α = 0.05 Z0 = Inference :
Since Z0 > Ze we reject our null hypothesis at 5% level of
significance and say that there is significant difference between the
sample mean and the population mean.
133 Example 7:
A car company decided to introduce a new car whose mean
petrol consumption is claimed to be lower than that of the existing
car. A sample of 50 new cars were taken and tested for petrol
consumption. It was found that mean petrol consumption for the 50
cars was 30 km per litre with a standard deviation of 3.5 km per
litre. Test at 5% level of significance whether the company’ s claim
that the new car petrol consumption is 28 km per litre on the
average is acceptable.
Solution:
We are given x = 30 ; µ =28 ; n=50 ; s=3.5
Null hypothesis:
H0: µ = 28. i.e The company’ s claim that the petrol consumption of
new car is 28km per litre on the average is acceptable.
Alternative Hypothesis:
H1: µ < 28 (Left tailed Alternative)
Level of significance:
Let α = 0.05
Calculation of statistic:
Under H0 the test statistics is
x −µ
Z0 =
s/ n = 30 − 28
3.5
50 2 × 50
3.5
= 4.04
= Expected value:
x −µ
∼ N(0,1) at α = 0.05
s/ n
= 1.645 Ze = 134 Inference :
Since the calculated Z0 > Ze we reject the null hypothesis at
5% level of significance and conclude that the company’ s claim is
not acceptable.
5.5 Test of significance for difference between two means:
Test procedure
Set up the null and alternative hypothesis
H0: µ1 = µ2 ; H1: µ1 ≠ µ2 (µ1 > µ2 or µ1 < µ2)
Level of significance:
Let α%
Calculation of statistic:
Under H0 the test statistic is Z0 = x1 − x 2
σ1
σ
+2
n1
n2
2 2 If σ12 = σ22 = σ2 (ie) If the samples have been drawn from
the population with common S.D σ then under H0 : µ1 = µ2 Z0 = x1 − x 2
1
1
σ
+
n1 n 2 Expected value:
Ze = x1 − x 2
∼N(0,1)
S.E( x 1 − x 2 ) Inference:
(i) If Z0 ≤ Ze we accept the H0 (ii) If Z0 > Ze we reject the H0
Example 8:
A test of the breaking strengths of two different types of cables
was conducted using samples of n1 = n2 = 100 pieces of each type
of cable.
135 Cable I
Cable II
x 1 =1925
x 2 = 1905
σ1= 40
σ2 = 30
Do the data provide sufficient evidence to indicate a
difference between the mean breaking strengths of the two cables?
Use 0.01 level of significance.
Solution:
We are given
x1
=1925
x 2 = 1905
σ1= 40
σ2 = 30
Null hypothesis
H0:µ1 = µ2 .ie There is no significant difference between the mean
breaking strengths of the two cables.
H1 : µ1 ≠ µ2 (Two tailed alternative)
Level of significance:
Let α = 0.01 or 1%
Calculation of statistic:
Under H0 the test statistic is Z0 = = x1 − x 2
σ1
σ
+2
n1
n2
2 2 1925 − 1905
40 2 30 2
+
100 100 = 20
=4
5 Expected value: Ze = x1 − x 2
σ1
σ
+2
n1
n2
2 2 136 ∼ N (0,1) = 2.58 Inference:
Since Z0 > Ze, we reject the H0. Hence the formulated null
hypothesis is wrong ie there is a significant difference in the
breaking strengths of two cables.
Example 9:
The means of two large samples of 1000 and 2000 items are
67.5 cms and 68.0cms respectively. Can the samples be regarded as
drawn from the population with standard deviation 2.5 cms. Test at
5% level of significance.
Solution:
We are given
n1 = 1000 ; n2 = 2000 x 1 = 67.5 cms ; x 2 = 68.0 cms σ = 2.5 cms
Null hypothesis
H0: µ1 = µ2 (i.e.,) the sample have been drawn from the same
population.
Alternative Hypothesis:
H1: µ1 ≠ µ2 (Two tailed alternative)
Level of significance:
α = 5%
Calculation of statistic:
Under Ho the test statistic is Z0 = = x1 − x 2
1
1
σ
+
n1 n 2
67.5 − 68.0
1
1
+
1000 2000
0.5 × 20 2.5
= 2.5 3 / 5
= 5.1
137 Expected value: Ze = x1 − x 2
1
1
σ
+
n1 n 2 ∼ N(0,1) = 1.96 Inference :
Since Z0 > Ze we reject the H0 at 5% level of significance
and conclude that the samples have not come from the same
population. Exercise – 5
I. Choose the best answer:
1. Standard error of number of success is given by
pq
np
(a)
(c) npq
(d)
(b) npq
n
q
2. Large sample theory is applicable when
(a) n > 30
(b) n < 30
(c) n < 100
(d) n < 1000
3. Test statistic for difference between two means is
x −µ
p−P
(a)
(b)
PQ
σ/ n
n
x1 − x 2
p1 − p 2
(d)
(c)
2
2
1
1
σ1
σ2 PQ +
+
n
n2 n1
n2 1
4. Standard error of the difference of proportions (p1p2) in two
classes under the hypothesis H0:p1 = p2 with usual notation is
1
1
1
1
(b) p( + )
(a) pq ( + )
n1 n 2
n1 n 2
(c) pq ( 1
1
+)
n1 n 2 (d) p 1q 1 p 2 q 2
+
n1
n2 138 5. Statistic z = x−y
1
1
σ
+
n1 n 2 is used to test the null hypothesis (a) H0: µ1 + µ 2 = 0
(c) H0: µ = µ 0 ( a constant) (b) H0: µ1 − µ 2 = 0
(c) none of the above. II. Fill in the blanks:
2
6. If P = , then Q = _________
3
7. If z0 < ze then the null hypothesis is ____________
8. When the difference is __________, the null hypothesis is
rejected.
9. Test statistic for difference between two proportions is
________
10. The variance of sample mean is ________
III. Answer the following
11. In a test if z0 ≤ ze, what is your conclusion about the null
hypothesis?
12. Give the test statistic for
(a) Proportion
(b) Mean
(c) Difference between two means
(d) Difference between two proportions
13. Write the variance of difference between two proportions
14. Write the standard error of proportion.
15. Write the test procedure for testing the test of significance for
(a) Proportion
(b) mean
(c) difference between two proportions
(d) difference between two mean
16. A coin was tossed 400 times and the head turned up 216 times.
Test the hypothesis that the coin is unbiased.
17. A person throws 10 dice 500 times and obtains 2560 times 4, 5
or 6. Can this be attributed to fluctuations of sampling?
139 18. In a hospital 480 female and 520 male babies were born in a
week. Do these figure confirm the hypothesis that males and
females are born in equal number?
19. In a big city 325 men out of 600 men were found to be selfemployed. Does this information support the conclusion that the
majority of men in this city are selfemployed?
20. A machine puts out 16 imperfect articles in a sample of 500.
After machine is overhauled, it puts out 3 imperfect articles in a
batch of 100. Has the machine improved?
21. In a random sample of 1000 persons from town A , 400 are
found to be consumers of wheat. In a sample of 800 from town
B, 400 are found to be consumers of wheat. Do these data
reveal a significant difference between town A and town B, so
far as the proportion of wheat consumers is concerned?
22. 1000 articles from a factory A are examined and found to have
3% defectives. 1500 similar articles from a second factory B are
found to have only 2% defectives. Can it be reasonably
concluded that the product of the first factory is inferior to the
second?
23. In a sample of 600 students of a certain college, 400 are found
to use blue ink. In another college from a sample of 900
students 450 are found to use blue ink. Test whether the two
colleges are significantly different with respect to the habit of
using blue ink.
24. It is claimed that a random sample of 100 tyres with a mean life
of 15269kms is drawn from a population of tyres which has a
mean life of 15200 kms and a standard deviation of 1248 kms.
Test the validity of the claim.
25. A sample of size 400 was drawn and the sample mean B was
found to be 99. Test whether this sample could have come from
a normal population with mean 100 and variance 64 at 5% level
of significance.
26. The arithmetic mean of a sample 100 items drawn from a large
population is 52. If the standard deviation of the population is 7,
test the hypothesis that the mean of the population is 55 against
the alternative that the mean is not 55. Use 5% level of
significance.
140 27. A company producing light bulbs finds that mean life span of
the population of bulbs is 1200 hrs with a standard deviation of
125hrs. A sample of 100 bulbs produced in a lot is found to
have a mean life span of 1150hrs. Test whether the difference
between the population and sample means is statistically
significant at 5% level of significance.
28. Test the significance of the difference between the means of the
samples from the following data
Size of sample Mean 100
150 50
51 Sample A
Sample B Standard
deviation
4
5 29. An examination was given to two classes consisting of 40 and
50 students respectively. In the first class the mean mark was 74
with a standard deviation of 8, while in the second class the
mean mark was 78 with a standard deviation of 7. Is there a
significant difference between the performance of the two
classes at a level of significance of 0.05?
30. If 60 M.A. Economics students are found to have a mean height
of 63.60 inches and 50 M.Com students a mean height of 69.51
inches. Would you conclude that the commerce students are
taller than Economics students? Assume the standard deviation
of height of postgraduate students to be 2.48 inches.
Answers:
I.
1. (b) 2.(a) 3.(c) 7.accepted 8.significant II.
6. 1
3 141 4.(a) 5.(b) 9. p1 − p 2
1
1 PQ +
n
n2 1 σ
n 2 10. III.
16. z = 1.6 Accept H0 17. z = 1.7 Accept H0 18. z = 1.265Accept H0 19. z = 2.04 Accept H0 20. z = 0.106 Accept H0 21. z = 4.247 Reject H0 22. z = 1.63 Accept H0 23. z = 6.38 Reject H0 24. z = 0.5529 Accept H0 25. z = 2.5 Reject H0 26. z = 4.29 Reject H0 27. z = 4 Reject H0 28. z = 1.75 Accept H0 29. z = 2.49 Reject H0 30. z = 12.49 Reject H0 142 6. TESTS OF SIGNIFICANCE
(Small Samples)
6.0 Introduction:
In the previous chapter we have discussed problems relating
to large samples. The large sampling theory is based upon two
important assumptions such as
(a) The random sampling distribution of a statistic is
approximately normal and
(b) The values given by the sample data are sufficiently close
to the population values and can be used in their place for
the calculation of the standard error of the estimate.
The above assumptions do not hold good in the theory of small
samples. Thus, a new technique is needed to deal with the theory of
small samples. A sample is small when it consists of less than 30
items. ( n< 30)
Since in many of the problems it becomes necessary to take
a small size sample, considerable attention has been paid in
developing suitable tests for dealing with problems of small
samples. The greatest contribution to the theory of small samples is
that of Sir William Gosset and Prof. R.A. Fisher. Sir William
Gosset published his discovery in 1905 under the pen name
‘ Student’ and later on developed and extended by Prof. R.A.Fisher.
He gave a test popularly known as ‘ ttest’ .
6.1 t  statistic definition:
If x1, x2, …xn is a random sample of size n from a normal
…
population with mean µ and variance σ2, then Student’ s tstatistic is
x−µ
defined as t =
S
n
where x = ∑x
is the sample mean
n
143 2
1
∑( x − x )
n −1
is an unbiased estimate of the population variance σ2 It follows
student’ s tdistribution with ν = n −1 d.f and S2 = 6.1.1 Assumptions for students ttest:
1. The parent population from which the sample drawn is
normal.
2. The sample observations are random and independent.
3. The population standard deviation σ is not known.
6.1.2 Properties of t distribution:
1. tdistribution ranges from −∞ to ∞ just as does a normal
distribution.
2. Like the normal distribution, tdistribution also symmetrical
and has a mean zero.
3. tdistribution has a greater dispersion than the standard
normal distribution.
4. As the sample size approaches 30, the tdistribution,
approaches the Normal distribution.
Comparison between Normal curve and corresponding t curve: 144 6.1.3 Degrees of freedom (d.f):
Suppose it is asked to write any four number then one will
have all the numbers of his choice. If a restriction is applied or
imposed to the choice that the sum of these number should be 50.
Here, we have a choice to select any three numbers, say 10, 15, 20
and the fourth number is 5: [50 − (10 +15+20)]. Thus our choice of
freedom is reduced by one, on the condition that the total be 50.
therefore the restriction placed on the freedom is one and degree of
freedom is three. As the restrictions increase, the freedom is
reduced.
The number of independent variates which make up the
statistic is known as the degrees of freedom and is usually denoted
by ν (Nu)
The number of degrees of freedom for n observations is
n − k where k is the number of independent linear constraint
imposed upon them.
For the student’ s tdistribution. The number of degrees of
freedom is the sample size minus one. It is denoted by ν = n −1
The degrees of freedom plays a very important role in χ2
test of a hypothesis.
When we fit a distribution the number of degrees of
freedom is (n– k−1) where n is number of observations and k is
number of parameters estimated from the data.
For e.g., when we fit a Poisson distribution the degrees of
freedom is ν = n – 1 −1
In a contingency table the degrees of freedom is (r−1) (c −1)
where r refers to number rows and c refers to number of columns.
Thus in a 3 × 4 table the d.f are (3−1) (4−1) = 6 d.f In a 2 ×
2 contingency table the d.f are (2−1) (2−1) = 1
In case of data that are given in the form of series of
variables in a row or column the d.f will be the number of
observations in a series less one ie ν = n−1
Critical value of t:
The column figures in the main body of the table come
under the headings t0.100, t0.50, t0.025, t0.010 and t0.005. The subscripts
145 give the proportion of the distribution in ‘ tail’ area. Thus for twotailed test at 5% level of significance there will be two rejection
areas each containing 2.5% of the total area and the required
column is headed t0.025
For example,
tν (.05) for single tailed test = tν (0.025) for two tailed test
tν (.01) for single tailed test = tν (0.005) for two tailed test
Thus for one tailed test at 5% level the rejection area lies in
one end of the tail of the distribution and the required column is
headed t0.05.
Critical value of t – distribution tα
t=0
tα
+∞
∞
6.1.4 Applications of tdistribution:
The tdistribution has a number of applications in statistics,
of which we shall discuss the following in the coming sections:
(i) ttest for significance of single mean, population variance being
unknown.
(ii) ttest for significance of the difference between two sample
means, the population variances being equal but unknown.
(a) Independent samples
(b) Related samples: paired ttest
6.2 Test of significance for Mean:
We set up the corresponding null and alternative hypotheses
as follows: 146 H0: µ = µ0; There is no significant difference between the sample
mean and population Mean.
H1: µ ≠ µ0 ( µ < µ0 (or) µ > µ0)
Level of significance:
5% or 1%
Calculation of statistic:
Under H0 the test statistic is
x −µ
x−µ
or
S
s / n −1
n
∑x
where x =
is the sample mean
n
2
2
1
1
and S2 =
∑( x − x ) (or) s2 = ∑( x − x )
n −1
n
Expected value :
t0 = te = x−µ
∼ student’ s tdistribution with (n1) d.f
S
n Inference :
If t0 ≤ te it falls in the acceptance region and the null
hypothesis is accepted and if to > te the null hypothesis H0 may be
rejected at the given level of significance.
Example 1:
Certain pesticide is packed into bags by a machine. A
random sample of 10 bags is drawn and their contents are found to
weigh (in kg) as follows:
50
49
52
44
45
48
46
45
49 45
Test if the average packing can be taken to be 50 kg.
Solution:
Null hypothesis:
H0 : µ = 50 kgs in the average packing is 50 kgs.
147 Alternative Hypothesis:
H1 : µ ≠ 50kgs (Two tailed )
Level of Significance:
Let α = 0.05
Calculation of sample mean and S.D
X
d = x –48
50
2
49
1
52
4
44
–4
45
–3
48
0
46
–2
45
–3
49
+1
45
–3
Total
–7
∑d
x =A+
n
−7
= 48 +
10
= 48–0.7 =47.3
1
(∑ d ) 2
[∑d 2 −
S2 =
n −1
n
2
1
(7 )
=
[69 −
9
10
64.1
=
= 7.12
9
Calculation of Statistic:
Under H0 the test statistic is :
t0 = x −µ
S2 / n 148 d2
4
1
16
16
9
0
4
9
1
9
69 =
= 47.3 − 50.0
7.12 / 10
2.7
0.712 = 3.2 Expected value:
te = x −µ follows t distribution with (10–1) d.f
S2 / n
= 2.262
Inference:
Since t0 > te , H0 is rejected at 5% level of significance and
we conclude that the average packing cannot be taken to be 50 kgs.
Example 2:
A soap manufacturing company was distributing a
particular brand of soap through a large number of retail shops.
Before a heavy advertisement campaign, the mean sales per week
per shop was 140 dozens. After the campaign, a sample of 26 shops
was taken and the mean sales was found to be 147 dozens with
standard deviation 16. Can you consider the advertisement
effective?
Solution:
We are given
n = 26;
x = 147dozens;
s = 16
Null hypothesis:
H0: µ = 140 dozens i.e. Advertisement is not effective.
Alternative Hypothesis:
H1: µ > 140kgs (Right tailed)
Calculation of statistic:
Under the null hypothesis H0, the test statistic is
x−µ
t0 =
s / n −1
= 147 − 140
16 / 25
149 = 7×5
16 = 2.19 Expected value:
x−µ
te =
follows tdistribution with (261) = 25d.f
s / n −1
= 1.708
Inference:
Since t0 > te, H0 is rejected at 5% level of significance.
Hence we conclude that advertisement is certainly effective in
increasing the sales.
6.3 Test of significance for difference between two means:
6.3.1 Independent samples:
Suppose we want to test if two independent samples have
been drawn from two normal populations having the same means,
the population variances being equal. Let x1, x2,…x n1 and y1, y2,
.
… y n 2 be two independent random samples from the given
…
normal populations.
Null hypothesis:
H0 : µ1 = µ2 i.e. the samples have been drawn from the normal
populations with same means.
Alternative Hypothesis:
H1 : µ1 ≠ µ2 (µ1 < µ2 o r µ1 > µ2)
Test statistic:
Under the H0, the test statistic is t0 = where x = x−y
1
1
S
+
n1 n 2
∑x
∑y
; y=
n1
n2 2
n s + n 2s 2
1
[ ∑ ( x − x ) + ∑( y − y ) 2 ] = 1 1
n1 + n 2 − 2
n1 + n 2 − 2
2 and S2 = 150 2 Expected value: te = x−y
1
1
S
+
n1 n 2 follows tdistribution with n1+ n2 –2 d.f Inference:
If the t0 < te we accept the null hypothesis. If t0 > te we reject
the null hypothesis.
Example 3:
A group of 5 patients treated with medicine ‘ A’ weigh 42,
39, 48, 60 and 41 kgs: Second group of 7 patients from the same
hospital treated with medicine ‘ B’ weigh 38, 42 , 56, 64, 68, 69 and
62 kgs. Do you agree with the claim that medicine ‘ B’ increases the
weight significantly?
Solution:
Let the weights (in kgs) of the patients treated with
medicines A and B be denoted by variables X and Y respectively.
Null hypothesis:
H0 : µ1 = µ2
i.e. There is no significant difference between the medicines A and
B as regards their effect on increase in weight.
Alternative Hypothesis:
H1 : µ1 < µ2 (lefttail) i.e. medicine B increases the weight
significantly.
Level of significance : Let α = 0.05
Computation of sample means and S.Ds
Medicine A
X
x – x ( x = 46)
(x – x )2
42
–4
16
39
–7
49
48
2
4
60
14
196
41
–5
25
230
0
290
151 x= y= ∑ x 230
=
= 46
n1
5
Medicine B
Y
y – y ( y = 57)
38
–19
42
–15
56
–1
64
7
68
11
69
12
62
5
399
0
∑y
n2 = (y – y )2
361
225
1
49
121
144
25
926 399
= 57
7 2
1
[ ∑ ( x − x ) + ∑( y − y ) 2 ]
n1 + n 2 − 2
1
=
[ 290 + 926] = 121.6
10
Calculation of statistic:
Under H0 the test statistic is S2 = t0 = = x−y
1
1 S2 + n 1 n2 46 − 57
1 1
121.6 + 5 7
152 = 11
121.6 × = 12
35 11
= 1.7
6.57 Expected value: te = x−y
1
1
S2 +
n 1 n2 follows tdistribution with (5+7–2) =10 d.f = 1.812
Inference:
Since t0 < te it is not significant. Hence H0 is accepted and
we conclude that the medicines A and B do not differ significantly
as regards their effect on increase in weight.
Example 4:
Two types of batteries are tested for their length of life and
the following data are obtained:
No of samples
Mean life
Variance
(in hrs)
Type A
9
600
121
Type B
8
640
Is there a significant difference in the two means? 144 Solution:
We are given
s22=144
n1=9; x 1=600hrs; s12 =121; n2 =8; x 2 =640hrs;
Null hypothesis:
H0 : µ1 = µ2 i.e. Two types of batteries A and B are identical i.e.
there is no significant difference between two types of batteries.
153 Alternative Hypothesis:
H1 : µ1 ≠ µ2 (Two tailed)
Level of Significance:
Let α = 5%
Calculation of statistics:
Under H0, the test statistic is t0 = x−y
1
1 S2 +
n 1 n2 n s + n 2s 2
where S = 1 1
n1 + n 2 − 2
9 × 121 + 8 × 144
=
9+8− 2
2241
=
= 149.4
15
2 2 2 ∴ t0 = = 600 − 640
1 1
149.4 + 9 8
40
40
= 6.735
=
5.9391 17 149.4 72 Expected value:
x−y te = 1
1 S +
n 1 n2 = 2.131 follows tdistribution with 9+8–2 =15 d.f 2 154 Inference:
Since t0 > te it is highly significant. Hence H0 is rejected and
we conclude that the two types of batteries differ significantly as
regards their length of life.
6.3.2 Related samples –Paired ttest:
In the ttest for difference of means, the two samples were
independent of each other. Let us now take a particular situations
where
(i)
The sample sizes are equal; i.e., n1 = n2 = n(say), and
(ii)
The sample observations (x1, x2, …..xn) and (y1, y2,
…
….yn) are not completely independent but they are
…
dependent in pairs.
That is we are making two observations one before treatment
and another after the treatment on the same individual. For example
a business concern wants to find if a particular media of promoting
sales of a product, say door to door canvassing or advertisement in
papers or through T.V. is really effective. Similarly a
pharmaceutical company wants to test the efficiency of a particular
drug, say for inducing sleep after the drug is given. For testing of
such claims gives rise to situations in (i) and (ii) above, we apply
paired ttest.
Paired – t –test:
Let di = Xi – Yi (i = 1, 2, …n) denote the difference in
…
the observations for the ith unit.
Null hypothesis:
H0 : µ1 = µ2 ie the increments are just by chance
Alternative Hypothesis:
H1 : µ1 ≠ µ2 ( µ1 > µ2 (or) µ1 < µ2)
Calculation of test statistic:
d
t0 =
S/ n
1
(∑ d )
1
∑d
2
where d =
[∑ d 2 −
and S2 =
∑ (d − d ) =
n −1
n −1
n
n
2 155 Expected value:
d
follows tdistribution with n –1 d.f
te =
S/ n
Inference:
By comparing t0 and te at the desired level of significance,
usually 5% or 1%, we reject or accept the null hypothesis.
Example 5:
To test the desirability of a certain modification in typists
desks, 9 typists were given two tests of as nearly as possible the
same nature, one on the desk in use and the other on the new type.
The following difference in the number of words typed per minute
were recorded:
Typists
A
B
C
D
E
F
G
H
I
Increase in
5
number of words
2
4
0
3 −1 4 −3 2
Do the data indicate the modification in desk promotes speed in
typing?
Solution:
Null hypothesis:
H0 : µ1 = µ2 i.e. the modification in desk does not promote speed in
typing.
Alternative Hypothesis:
H1 : µ1 < µ2 (Left tailed test)
Level of significance: Let α = 0.05
Typist
A
B
C
D
E
F
G
H
I d
2
4
0
3
−1
4
−3
2
5
Σd = 16 156 d2
4
16
0
9
1
16
9
4
25
Σd2 = 84 d=
S=
= 16
∑d
=
= 1.778
n
9
1
( ∑ d) 2
[∑ d 2 −
n −1
n
1
(16) 2
[84 −=
8
9 6.9 = 2.635 Calculation of statistic:
Under H0 the test statistic is
t0 = d. n
1.778 × 3
=
= 2.024
S
2.635 Expected value:
te = d. n
S follows t distribution with 9 –1 = 8 d.f = 1.860
Inference:
When t0 < te t he null hypothesis is accepted. The data does
not indicate that the modification in desk promotes speed in typing.
Example 6:
An IQ test was administered to 5 persons before and after
they were trained. The results are given below:
Candidates
I
II
III
IV
V
IQ before
110
120
123
132
125
training
IQ
after
120
118
125
136
121
training
Test whether there is any change in IQ after the training
programme (test at 1% level of significance)
Solution:
Null hypothesis:
H0 : µ1 = µ2 i.e. there is no significant change in IQ after the
training programme.
157 Alternative Hypothesis:
H1 : µ1 ≠ µ2 (two tailed test)
Level of significance :
α = 0.01
x
110
120
y
120
118
2
d = x−y
−10
2
d
100
4 123
125
−2
4 132
136
−4
16 125
121
4
16 Total
−10
140 ∑d
− 10
=
= −2
n
5
1
(∑ d ) 2
2
2
S=
[∑ d −
n −1
n
1
100
= [140 − = 30
4
5
Calculation of Statistic:
Under H0 the test statistic is
d
t0 =
S/ n
d= = −2 30 / 5
2
=
2.45
= 0.816
Expected value:
te = d follows tdistribution with 5 –1 = 4 d.f
S2 / n
= 4.604
Inference:
Since t0 < te at 1% level of significance we accept the null
hypothesis. We therefore, conclude that there is no change in IQ
after the training programme.
158 6.4 Chi square statistic:
Various tests of significance described previously have
mostly applicable to only quantitative data and usually to the data
which are approximately normally distributed. It may also happens
that we may have data which are not normally distributed.
Therefore there arises a need for other methods which are more
appropriate for studying the differences between the expected and
observed frequencies. The other method is called Nonparametric
or distribution free test. A non parametric test may be defined as a
statistical test in which no hypothesis is made about specific values
of parameters. Such nonparametric test has assumed great
importance in statistical analysis because it is easy to compute.
6.4.1 Definition:
The Chi square (χ2) test (Chipronounced as ki) is one of
the simplest and most widely used nonparametric tests in statistical
work. The χ2 test was first used by Karl Pearson in the year 1900.
The quantity
χ2 describes the magnitude of the discrepancy
between theory and observation. It is defined as
n
(Oi − Ei) 2 χ2 = ∑ Ei
i =1 Where O refers to the observed frequencies and E refers to
the expected frequencies.
Note:
If χ2 is zero, it means that the observed and expected
frequencies coincide with each other. The greater the discrepancy
between the observed and expected frequencies the greater is the
value of χ2.
Chi square  Distribution:
The square of a standard normal variate is a Chisquare
variate with 1 degree of freedom i.e., If X is normally distributed
x −µ
with mean µ and standard deviation σ, then is a Chiσ
square variate (χ2) with 1 d.f. The distribution of Chisquare
depends on the degrees of freedom. There is a different distribution
for each number of degrees of freedom.
159
2 6.4.2 properties of Chisquare distribution:
1. The Mean of χ2 distribution is equal to the number of
degrees of freedom (n)
2. The variance of χ2 distribution is equal to 2n
3. The median of χ2 distribution divides, the area of the curve
into two equal parts, each part being 0.5.
4. The mode of χ2 distribution is equal to (n−2)
5. Since Chisquare values always positive, the Chi square
curve is always positively skewed.
6. Since Chisquare values increase with the increase in the
degrees of freedom, there is a new Chisquare distribution
with every increase in the number of degrees of freedom.
7. The lowest value of Chisquare is zero and the highest value
is infinity ie χ2 ≥ 0.
8. When Two Chi squares χ12 and χ22 are independent χ2
distribution with n1 and n2 degrees of freedom and their sum
χ12 + χ22 will follow χ2 distribution with (n1 + n2) degrees of
freedom.
9. When n (d.f) > 30, the distribution of 2χ 2 approximately
follows normal distribution. The mean of the distribution
2χ 2 is 2n  1 and the standard deviation is equal to 1.
160 6.4.3 Conditions for applying χ 2 test:
The following conditions should be satisfied before
applying χ2 test.
1. N, the total frequency should be reasonably large, say
greater than 50.
2. No theoretical cellfrequency should be less than 5. If it is
less than 5, the frequencies should be pooled together in
order to make it 5 or more than 5.
3. Each of the observations which makes up the sample for
this test must be independent of each other.
4. χ2 test is wholly dependent on degrees of freedom.
6.5 Testing the Goodness of fit (Binomial and Poisson
Distribution):
Karl Pearson in 1900, developed a test for testing the
significance of the discrepancy between experimental values and
the theoretical values obtained under some theory or hypothesis.
This test is known as χ2test of goodness of fit and is used to test if
the deviation between observation (experiment) and theory may be
attributed to chance or if it is really due to the inadequacy of the
theory to fit the observed data.
Under the null hypothesis that there is no significant
difference between the observed and the theoretical values. Karl
Pearson proved that the statistic
n
(Oi − Ei) 2 χ2 = ∑ Ei
i =1 = (O 1 − E 1 ) 2 (O 2 − E 2 ) 2
(O − E n ) 2
+
+ ........ n
E1
E2
En Follows χ2distribution with ν = n – k – 1 d.f. where 01, 02,
...0n are the observed frequencies, E1 , E2… n, corresponding to the
E
expected frequencies and k is the number of parameters to be
estimated from the given data. A test is done by comparing the
computed value with the table value of χ2 for the desired degrees of
freedom.
161 Example 7:
Four coins are tossed simultaneously and the number of
heads occurring at each throw was noted. This was repeated 240
times with the following results.
No. of heads 0 1 2 3 4 No. of throws 13 64 85 58 20
Fit a Binomial distribution assuming under the hypothesis
that the coins are unbiased.
Solution:
Null Hypothesis:
H0: The given data fits the Binomial distribution. i.e the coins are
unbiased.
p = q = 1/2
n=4
N = 240
Computation of expected frequencies:
No. of P(X= x ) = 4 Cx px q nx
heads
0
1
2
3
4 1 1 1 4Co 0 4 = 2 2 16 1 1 4 4C1 1 3 = 2 2 16 1 1 6 4C2 2 2 = 2 2 16 1 1 4 4C3 3 1 = 2 2 16 1 1 1 4C4 4 0 = 2 2 16 162 Expected
Frequency
N. P(X = x)
1 x 240 = 15 16 4 x 240 = 60 16 6 x 240 = 90 16 4 x 240 = 60 16 1 x 240 = 15 16 240 Computation of chi square values
( O – E)2
Expected
Observed
Frequency
Frequency
E
O
13
15
4
64
60
16
85
90
25
58
60
4
20
15
25 ( O  E) 2
χ0 = Σ E 2 ( O  E) 2 E 0.27
0.27
0.28
0.07
1.67
2.56 =2.56 Expected Value: ( O  E) 2 2 follows χ2distribution with ( n−k−1) d.f.
χe = Σ E (Here k = 0, since no parameter is estimated from the data)
= 9.488 for ν = 5−1= 4 d.f.
Inference:
Since χ02 < χe2 we accept our null hypothesis at 5% level of
significance and say that the given data fits Binomial distribution.
Example 8:
The following table shows the distribution of goals in a foot ball
match.
No. of goals
0
1
2
3
4
5
6
7
No. of matches 95 158 108 63 40 9
5
2
Fit a Poisson distribution and test the goodness of fit.
Solution:
Null Hypothesis :
The given data fits the Poisson distribution.
Level of significance :
Let α = 0.05
163 Computation of expected frequencies:
812
m=
= 1.7
480
0
– 1.7 (1 .7 )
P(0) = e
= 0.183.
0!
f (o) = N.P(o) = 480 x 0.183 = 87.84
The other expected frequencies will be obtained by using the
recurrence formula
m
f (x+1) =
x f(x)
x +1
Putting x = 0, 1, 2, ... we obtain the following frequencies.
f (1) = 1.7 x 87.84
= 149.328
1.7
f (2) =
x 149.328
2
= 126.93
1.7
f (3) =
x 126.93
3
= 71.927
1.7
f (4) =
x 71.927
4
= 30.569
1.7
f (5) =
x 30.569
5
= 10.393
1.7
f (6) =
x 10.393
6
= 2.94
1.7
f (7) =
x 2.94
7
= 0.719
No. of goals
0
1
2
3
4
5
6
7 Total
Expected
1
480
88 149 127 72 30 10 3
frequency
164 Computation of statistic:
Observed
Frequency
O
95
158
108
63
40
9
5
16
2 ( O – E)2 Expected
Frequency
E
88
150
126
72
30
10
3
14
1 49
64
324
81
100
4 ( O  E) 2 E 0.56
0.43
2.57
1.13
3.33 0.29
8.31 ( O  E)
χo2 = Σ E 2 = 8.31 Expected Value: ( O  E) 2
χe2 = Σ E 2 χ distribution with ( n – k – 1) d.f for ν = 6 −1−1 = 4 d.f. = 9.488
Inference:
Since χ02 < χe2 , we accept our null hypothesis at 5% level
of significance and say that the given data fits the Poisson
distribution.
6.6 Test of independence
Let us suppose that the given population consisting of N
items is divided into r mutually disjoint (exclusive) and exhaustive
classes A1, A2, … Ar with respect to the attribute A so that
,
randomly selected item belongs to one and only one of the
attributes A1, A2, … Ar .Similarly let us suppose that the same
,
population is divided into c mutually disjoint and exhaustive
classes B1, B2, … Bc w.r.t another attribute B so that an item
,
selected at random possess one and only one of the attributes
B1, B2, … Bc. The frequency distribution of the items belonging to
,
165 the classes A1, A2, … Ar and B1, B2, … Bc can be represented in
,
,
the following r × c manifold contingency table.
r × c manifold contingency table
B B1 B2 A1
A2
.
.
.
Ai
.
.
.
Ar
Total (A1B1)
(A2B1)
.
.
.
(AiB1)
.
.
.
(ArB1)
(B1) (A1B2)
(A2B2)
.
.
.
(AiB2)
.
.
.
(ArB2)
(B2) … Bj … Bc Total …
…
.
.
.
…
.
.
.
…
… (A1Bc)
(A2Bc)
.
.
.
(AiBc)
.
.
.
(ArBc)
(Bc) (A1)
(A2)
.
.
.
(Ai)
.
.
.
(Ar) A
… (A1Bj)
… (A2Bj)
.
.
.
.
.
.
… (AiBj)
.
.
.
.
.
.
… (ArBj)
… (Bj) ΣAi =
ΣBj = N (Ai) is the number of persons possessing the attribute Ai,
(i=1,2,… (Bj) is the number of persons possing the attribute
r),
Bj,(j=1,2,3,… and (Ai Bj) is the number of persons possessing
.c)
both the attributes Ai and Bj (i=1,2,… ,j=1,2,…
r
c).
Also ΣAi = ΣBj = N
Under the null hypothesis that the two attributes A and B
are independent, the expected frequency for (AiBj) is given by
(Ai)( Bj)
=
N
Calculation of statistic:
Thus the under null hypothesis of the independence of attributes,the
expected frequencies for each of the cell frequencies of the above
table can be obtained on using the formula ( O  Ei )2 χ02 = Σ i Ei 166 Expected value: ( O  Ei )2 follows χ2distribution with (r−1) (c−1) d.f
χe2 = Σ i Ei Inference:
Now comparing χ02 with χe2 at certain level of significance
,we reject or accept the null hypothesis accordingly at that level of
significance.
6.6.1 2× 2 contingency table :
Under the null hypothesis of independence of attributes, the
value of χ2 for the 2×2 contingency table Total a
c
a+c b
d
b+d Total
a+b
c+d
N is given by
χo2 = N (ad − bc) 2
(a + c)(b + d )(a + b)(c + d ) 6.6.2 Yate’ s correction
In a 2×2 contingency table, the number of d.f. is (2−1)(2−1) =1. If
any one of the theoretical cell frequency is less than 5,the use of
pooling method will result in d.f = 0 which is meaningless. In this
case we apply a correction given by F.Yate (1934) which is usually
known as “Yates correction for continuity”. This consisting adding
0.5 to cell frequency which is less than 5 and then adjusting for the
remaining cell frequencies accordingly. Thus corrected values of χ2
is given as
1
1
1
1 N (a m )(d m ) − (b ± )(c ± )
2
2
2
2 χ2 =
(a + c)(b + d )(a + b)(c + d)
167 2 Example 9:
1000 students at college level were graded according to
their I.Q. and the economic conditions of their homes. Use χ2 test
to find out whether there is any association between economic
condition at home and I.Q.
Economic
Conditions
Rich
Poor
Total IQ
High
460
240
700 Total
Low
140
160
300 600
400
1000 Solution:
Null Hypothesis:
There is no association between economic condition at home and
I.Q. i.e. they are independent.
(A )(B)
600 × 700
E11 =
=
= 420
N
1000
The table of expected frequencies shall be as follows.
Total
420
180
600
280
120
400
Total
700
300
1000
Observed
Frequency
O
460
240
140
160 ( O – E)2 Expected
Frequency
E
420
280
180
120 χo2 1600
1600
1600
1600 ( O  E) 2
=Σ E 168 ( O  E) 2 E 3.81
5.714
8.889
13.333
31.746 = 31.746 Expected Value: ( O  E) 2 follow χ2 distribution with (2−1) (2−1) = 1 d.f
χe2 = Σ E = 3.84
Inference :
χo2 > χe2, hence the hypothesis is rejected at 5 % level of
significance. ∴ t here is association between economic condition at
home and I.Q.
Example 10:
Out of a sample of 120 persons in a village, 76 persons were
administered a new drug for preventing influenza and out of them,
24 persons were attacked by influenza. Out of those who were not
administered the new drug ,12 persons were not affected by
influenza.. Prepare
(i)
2x2 table showing actual frequencies.
(ii)
Use chisquare test for finding out whether the new drug
is effective or not.
Solution:
The above data can be arranged in the following 2 x 2 contingency
table.
Table of observed frequencies
Effect of Influenza
New drug
Total
Attacked
Not attacked
Administered
24
76 – 24 = 52
76
Not
44 –12 = 32
12
120 – 76 = 44
administered
Total
120 – 64 = 56 52 + 12 = 64
120
24 + 32 = 56
Null hypothesis:
‘ Attack of influenza’ and the administration of the new drug are
independent. 169 Computation of statistic:
χo2 = N (ad − bc) 2
(a + c)(b + d )(a + b)(c + d ) 120(24 × 12 − 52 × 32) 2
56 × 64 × 76 × 44
120(−1376) 2
120(1376) 2
=
=
56 × 64 × 76 × 44
56 × 64 × 76 × 44
= Anti log [log 120 + 2log1376 –(log56 +log64+log76+log44)]
= Antilog (1.2777) = 18.95
= Expected value: ( O  E) 2
2
χe = Σ E = 3.84 follows χ2distribution with ( 2−1) × (2−1) d.f Inference:
Since χo2 > χe2, H0 is rejected at 5 % level of significance.
Hence we conclude that the new drug is definitely effective in
controlling (preventing) the disease (influenza).
Example 11:
Two researchers adopted different sampling techniques
while investigating the same group of students to find the number
of students falling in different intelligence levels. The results are as
follows
Researchers X
Y
Total No. of Students
Below
Average
average
86
60
40
33
126
93 Total
Above Genius
average
44
10
25
2
69
12 200
100
300 Would you say that the sampling techniques adopted by the two
researchers are independent?
170 Solution:
Null Hypothesis:
The sampling techniques adopted by the two researchers are
independent.
126 × 200
E(86) =
= 84
300
93 × 200
E(60) =
= 62
300
69 × 200
E(44) =
= 46
300
The table of expected frequencies is given below.
Below
average
84 Average
62 Above
average
46 Y 126 – 84
= 42 Total 126 X Genius Total 200−192
=8 200 93 – 62
= 31 69 – 46 12 – 8 = 4
= 23 100 93 69 300 12 Computation of chisquare statistic:
Observed
Frequency
O
86
60
44
10
40
33 25 2 27 300 Expected
Frequency
E
84
62
46
8
42
31 23 4 27 300 ( O − E) ( O – E)2 2
−2
−2
2
−2
2 4
4
4
4
4
4 0 0 0
171 ( O  E) 2 E 0.048
0.064
0.087
0.500
0.095
0.129
0
0.923 χo2 ( O  E) 2
= Σ E = 0.923 Expected value: ( O  E) 2
χe2 =Σ E follows χ2distribution with (4−1)( 2−1) = 3 –1 = 2 df = 5.991
Inference:
Since χo2 < χe2, we accept the null hypothesis at 5 % level of
significance. Hence we conclude that the sampling techniques by
the two investigators, do not differ significantly.
6.7 Test for population variance:
Suppose we want to test if the given normal population has a
specified variance σ2 = σo2
Null Hypothesis:
Ho : σ2 = σo2 if x1, x2 … n
x
Level of significance:
Let α = 5% or 1%
Calculation of statistic:
χo2 = n
∑ (x i
i =1 Where s2 = − x) 2 σ02 = ns 2
σ0 2 1n
2
∑ (x i − x)
i =1
n Expected Value:
χe2 = n
∑ (x i
i =1 − x) 2 σ0 2 follows χ2 distribution with ( n –1) degrees of freedom.
Inference:
If χo2 ≤ χe2 we accept the null hypothesis otherwise if
χo2 > χe2 we reject the null hypothesis.
172 Example 12:
A random sample of size 20 from a population gives the
sample standard deviation of 6. Test the hypothesis that the
population Standard deviation is 9.
Solution:
We are given n = 20 and s = 6
Null hypothesis:
H0 The population standard deviation is σ= 9.
Level of significance:
Let α = 5 %
Calculation of statistic:
Under null hypothesis H 0 :
ns 2
20 × 36
2
= 8.89
χo = 2 =
9×9
σ
Expected value:
ns 2
2
χe = 2 follows χ2 distribution 20 –1 = 19 d.f.
σ
= 30.144
Inference:
Since χo2 < χe2 , we accept the null hypothesis at 5 % level
of significance and conclude that the population standard deviation
is 9.
Example 13:
Weights in kgs of 10 students are given below:
38, 40, 45, 53, 47, 43, 55, 48, 52 and 49.
Can we say that the variance of distribution of weights of all the
students from which the above sample of 10 students was drawn is
equal to 20 sq kg?
Solution:
Null hypothesis :
H0 : σ2 = 20
173 Computation of sample variance
Weight
in Kg x
38
40
45
53
47
43
55
48
52
49 x − x = x −47
−9
−7
−2
6
0
−4
8
1
5
2 Sample mean is (x − x )2
81
49
4
36
0
16
64
1
25
4
280 470
∑x
=
= 47
n
10
Calculation of statistic:
Test statistic is
2
ns 2
280
∑( x − x )
2
= 14
=
χo = 2 =
2
20
σ
σ
Expected value:
ns 2
2
χe = 2 follows χ2 distribution with 10 – 1 = 9 d.f.
σ
= 16.919
Inference:
Sinceχ02 < χe2 we accept Ho and we conclude that the
variance of the distribution of weights of all the students in the
population is 20 sq. kgs.
6.8 F – Statistic Definition:
If X is a χ2 variate with n1d.f. and Y is an independent χ2X / n1
variate with n2 d.f., then F  statistic is defined as F =
Y / n2
x= 174 i.e. F  statistic is the ratio of two independent chisquare variates
divided by their respective degrees of freedom. This statistic
follows G.W. Snedocor’ s Fdistribution with ( n1, n2) d.f.
6.8.1 Testing the ratio of variances:
Suppose we are interested to test whether the two normal
population have same variance or not. Let x1, x2, x3 … x n 1 , be a
..
random sample of size n1, from the first population with variance
σ12 and y1, y2, y3 … y n 2 , be random sample of size n2 form the
second population with a variance σ22. Obviously the two samples
are independent.
Null hypothesis:
H0 = σ12 = σ22 = σ2
i.e. population variances are same. In other words H0 is that the two
independent estimates of the common population variance do not
differ significantly.
Calculation of statistics:
Under H0, the test statistic is
2
S
F0= 1 2
S2
2 ns
1
2
Where S1 =
∑( x − x ) = 1 1
n1 − 1
n1 − 1
2 2 ns
1
2
∑ ( y − y) = 2 2
n 2 −1
n 2 −1
It should be noted that numerator is always greater than the
denominator in Fratio
L arg er Variance
F=
Samller Variance
ν1 = d.f for sample having larger variance
ν2 = d.f for sample having smaller variance
Expected value :
2
S1
Fe = 2 follows F distribution with ν1 = n1 – 1 , ν2 = n2−1 d.f
S2
S2 2 = 175 The calculated value of F is compared with the table value
for ν1 and ν2 at 5% or 1% level of significance If F0 > Fe then we
reject H0. On the other hand if F0 < Fe we accept the null hypothesis
and it is a inferred that both the samples have come from the
population having same variance.
Since F test is based on the ratio of variances it is also
known as the variance Ratio test. The ratio of two variances follows
a distribution called the F distribution named after the famous
statisticians R.A. Fisher.
Example 14:
Two random samples drawn from two normal populations
are :
Sample I:
20 16 26 27 22 23 18 24 19 25
Sample II:
27 33 42 35 32 34 38 28 41 43 30 37
Obtain the estimates of the variance of the population and
test 5% level of significance whether the two populations have the
same variance.
Solution:
Null Hypothesis:
H0: σ12 = σ22 i.e. The two samples are drawn from two populations
having the same variance.
Alternative Hypothesis:
H1: σ12 ≠ σ22 (two tailed test)
∑ x1
x1 =
n1
220
=
10
= 22
∑ x2
x2 =
n2
420
=
12
= 35
176 x1 x1 − x 1
−2
−6
4
5
0
1
−4
2
−3
3
0 20
16
26
27
22
23
18
24
19
25
220 (x1 − x 1 )2
4
36
16
25
0
1
16
4
9
9
120 x2
27
33
42
35
32
34
38
28
41
43
30
37
420 x2 − x 2
−8
−2
7
0
−3
−1
3
−7
6
8
−5
2
0 (x2− x 2 )2
64
4
49
0
9
1
9
49
36
64
25
4
314 Level of significance :
0.05
The statistic F is defined by the ratio
2
S1
F0 = 2
S2
Where S12 = ∑(x 1 − x 1 )
n1 − 1 2 = 120
= 13.33
9 ∑( x 2 − x 2 )
314
=
= 28.54
S2 =
11
n2 −1
Since S22 > S12 larger variance must be put in the numerator and
smaller in the denominator
28.54
= 2.14
∴F0 =
13.33
Expected value:
2
S2
follows F distribution with
Fe =
2
S1
ν1 = 12−1 = 11 ; ν2 = 10−1= 9 d.f = 3.10
2 2 177 Inference :
Since F0 < Fe we accept null hypothesis at 5% level of
significance and conclude that the two samples may be regarded as
drawn from the populations having same variance.
Example 15:
The following data refer to yield of wheat in quintals on
plots of equal area in two agricultural blocks A and B Block A was
a controlled block treated in the same way as Block B expect the
amount of fertilizers used.
No of plots
Mean yield
Variance
Block A
8
60
50
Block B
6
51
40
Use F test to determine whether variance of the two blocks differ
significantly?
Solution:
We are given that
n1 = 8 n2 =6
x 1 = 60 x 2 = 51 s12 =50
s22 = 40
Null hypothesis:
H0: σ12 = σ22 ie there is no difference in the variances of yield of
wheat.
Alternative Hypothesis:
H1: σ12 ≠ σ22 (two tailed test)
Level of significance:
Let α = 0.05
Calculation of statistic:
2
ns
8 × 50
S1 2 = 1 1 =
7
n1 − 1
= 57.14
2
n 2 s2
6 × 40
2
=
S2 =
5
n 2 −1
= 48
178 Since S12 > S22
2
S
57.14
F0 = 1 2 =
= 1.19
48
S2
Expected value:
2
S1
Fe = 2 follows F distribution with ν1 = 8−1 =7 ν2 = 6−1 = 5 d.f
S2
= 4.88
Inference:
Since F0 < Fe, we accept the null hypothesis and hence infer
that there is no difference in the variances of yield of wheat.
Exercise 6
I. Choose the best answer:
1. Student’ s ‘ t’ distribution was pioneered by
(a) Karl Pearson
(b) Laplace
(c)R.A. Fisher
(d) William S.Gosset
2. t  distribution ranges from
(a) − ∞ to 0
(b) 0 to ∞
(c) − ∞ to ∞
(d) 0 to 1
3. The difference of two means in case of a small samples is tested
by the formula
x1 − x 2
x1 − x 2 n1 + n2
(a) t =
(b)
s
s
n1 + n2
x1 − x 2
n1n2
n1n2
(d) t =
s
n1 + n2
n1 + n2
4. While testing the significance of the difference between two
sample means in case of small samples, the degree of freedom
is
(a) n1+n2
(b) n1+n2 –1
(c) n1+n2 –2
(d) n1+n2 +2
5. Paired ttest is applicable when the observations in the two
samples are
(a) Paired
(b) Correlated
(c) equal in number
(d) all the above
179
(c) t = 6. The mean difference between a paired observations is 15.0 and
the standard deviation of differences is 5.0 if n = 9, the value of
statistic t is
(a) 27
(b) 9
(c) 3
(d) zero
7. When observed and expected frequencies completely coincide
χ2 will be
(a) –1
(b) +1
(c) greater than 1 (d) 0
2
8. For ν =2, χ 0.05 equals
(a) 5.9
(b) 5.99
(c) 5.55
(d) 5.95
2
9. The calculated value of χ is
(a) always positive
(b) always negative
(c ) can be either positive or negative
(d) none of these
10. The Yate’ s corrections are generally made when the cell
frequency is
(a) 5
(b) < 5
(c) 1
(d) 4
2
11. The χ test was derived by
(a) Fisher
(b) Gauss
(c) Karl Pearson
(d) Laplace
12. Degrees of freedom for Chisquare in case of contingency table
of order (4 ×3) are
(a) 12
(b) 9
(c) 8
(d) 6
13. Customarily the larger variance in the variance ratio for Fstatistic is taken
(a) in the denominator
(b) in the numerator
(c) either way
(d) none of the above
2
S
14. The test statistic F = 1 2 is used for testing
S2
(a) H0: µ1 = µ2
(b) H0: σ12 = σ22
(c) H0: σ1 = σ2
(d) H0: σ2 = σ02
15. Standard error of the sample mean in testing the difference
between population mean and sample mean under t statistic
s
σ2
(b)
(a)
n
n
σ
s
(d)
(c)
n
n
180 II. Fill in the blanks:
16. The assumption in t test is that the population standard
deviation is ____________
17. t values lies in between ____________
18. Paired t test is applicable only when the observations are
_____________
19. Student t test is applicable in case of ___________ samples
20. The value of χ2 statistic depends on the difference between
__________ and _________ frequencies
21. The value of χ2 varies from ___________to ___________
22. Equality of two population variances can be tested by
___________
23. The χ2 test is one of the simplest and most widely used
_________test.
24. The greater the discrepancy between the observed and expected
frequency _________the value of χ2
25. In a contingency table ν _________
26. The distribution of the χ2 depends on the __________
27. The variance of the χ2 distribution is equal to _______ the d.f
28. One condition for application of χ2 t est is that no cell
frequency should be _________
29. In a 3 × 2 contingency table, there are ________ cells
30. F test is also known as __________ ratio test.
III. Answer the following
31. Define students ‘ t’ – statistic
32. State the assumption of students ‘ t’ test
33. State the properties of t distribution
34. What are the applications of t distribution
35. Explain the test procedure to test the significance of mean in
case of small samples.
36. What do you understand by paired ‘ t’ test > What are its
assumption.
37. Explain the test procedure of paired – t test
38. Define Chi square test
39. Define Chi square distribution
181 40. What is χ2 test of goodness of fit.
41. What are the precautions are necessary while applying χ2 test?
42. Write short note on Yate’ s correction.
43. Explain the term ‘ Degrees of freedom’
44. Define nonparametric test
45. Define χ2 test for population variance
46. Ten flower stems are chosen at random from a population and
their heights are found to be (in cms) 63 , 63, 66, 67, 68, 69, 70,
70, 71 and 71. Discuss whether the mean height of the
population is 66 cms.
47. A machine is designed to produce insulating washers for
electrical devices of average thickness of 0.025cm. A random
sample of 10 washers was found to have an average thickness
of 0.024cm with a standard deviation of 0.002cm. Test the
significance of the deviation.
48. Two types of drugs were used on 5 and 7 patients for reducing
their weight.
Drug A was imported and drug B indigenous. The decrease
in the weight after using the drugs for six months was as
follows:
Drug A :
10 12 13 11 14
Drug B :
8
9 12 14 15 10 9
Is there a significant difference in the efficiency of the two
drugs? If not, which drug should you buy?
49. The average number of articles produced by two machines per
day are 200 and 250 with standard deviations 20 and 25
respectively on the basis of records 25 days production. Can
you conclude that both the machines are equally efficient at 1%
level of significance.
50. A drug is given to 10 patients, and the increments in their blood
pressure were recorded to be 3, 6, 2 , +4, −3, 4, 6, 0, 0, 2. Is it
reasonable to believe that the drug has no effect on change of
blood pressure?
51. The sales data of an item in six shops before and after a special
promotional campaign are as under:
182 Shops:
A
B
C
D
E
F
Before
Campaign: 53
28
31
48
50
42
After
Campaign: 58
29
30
55
56
45
Can the campaign be judges to be a success? Test at 5% level of
significance.
52.
A survey of 320 families with 5 children each revealed the
following distribution.
No of
5
4
3
2
1
0
boys
No of
0
1
2
3
4
5
Girls
No of
14
56
110
88
40
12
Families
Is the result consistent with the hypothesis that the male and female
births are equally probable?
53. The following mistakes per page were observed in a book. No of
mistakes
per page
No of
pages 0 1 2 3 4 Total 211 90 19 5 0 325 Fit a Poisson distribution and test the goodness of fit.
54. Out of 800 persons, 25% were literates and 300 had travelled
beyond the limits of their district 40% of the literates were
among those who had not travelled. Test of 5% level whether
there is any relation between travelling and literacy 183 55.
Fathers You are given the following
Intelligent
Boys
24
32 Not intelligent Total
boys
12
36
32
64 Skilled father
Unskilled
Father
Total
56
44
100
Do these figures support the hypothesis that skilled father have
intelligent boys?
56. A random sample of size 10 from a normal population gave the
following values
65 , 72, 68, 74, 77, 61,63, 69 , 73, 71
Test the hypothesis that population variance is 32.
57. A sample of size 15 values shows the s.d to be 6.4. Does this
agree with hypothesis that the population s.d is 5, the
population being normal.
58. In a sample of 8 observations, the sum of squared deviations of
items from the mean was 94.5. In another sample of 10
observations, the value was found to be 101.7 test whether the
difference in the variances is significant at 5% level.
59. The standard deviations calculated from two samples of sizes 9
and 13 are 2.1 and 1.8 respectively. May the samples should be
regarded as drawn from normal populations with the same
standard deviation?
60. Two random samples were drawn from two normal populations
and their values are
A
66 67 75 76 82 84 88 90 92 −
B
64 66 74 78 82 85 87 92 93 95 97
Test whether the two populations have the same variance at 5%
level of significance.
61. An automobile manufacturing firm is bringing out a new
model. In order to map out its advertising campaign, it wants to
determine whether the model will appeal most to a particular
184 age – group or equal to all age groups. The firm takes a random
sample from persons attending a preview of the new model and
obtained the results summarized below:
Person
who
Liked
the car
Disliked
the car
Total Age groups
4050
60 and
over
48
28 Under
20
146 2039 54 52 32 62 200 200 130 80 90 500 78 Total
300 What conclusions would you draw from the above data?
Answers:
I.
1. (d)
2.(c)
3. (c)
4. (c)
5. (d)
6. (b)
7. (d)
8. (b)
9. (a)
10. (c)
11. (c)
12. (d)
13. (b)
14. (b)
15. (b)
II.
16. not known
17. − ∞ t o ∞
18. paired
19 small
20.observed, expected
21. 0, ∞
22. F test
23.non parametric
24. greater
25. ( r−1 ) ((−1))
26. degrees of freedom
27. d.f. twice
28. less than 5
29. 6
30. variance
III.
46. t = 1.891 H0 is accepted
47. t = 1.5 H0 is accepted
48. t = 0.735 H0 is accepted
49. t = 7.65 H0 is rejected
50. t = 2, H0 is accepted
51. t= 2.58 H0 is rejected
52. χ2 = 7.16 H0 is accepted
53. χ2 = 0.068 H0 is accepted
2
54. χ = 0.0016 H0 is accepted
55. χ2 = 2.6 H0 is accepted
56. χ2 = 7.3156 H0 is accepted
57. χ2 = 24.58 H0 is rejected
58. χ2 = 24.576 H0 is rejected
59. F = 1.41 H0 is accepted
60. F = 1.415 H0 is accepted
61. χ2 = 7.82 , H0 is rejected
185 7. ANALYSIS OF VARIANCE
7.0 Introduction:
The analysis of variance is a powerful statistical tool for
tests of significance. The term Analysis of Variance was introduced
by Prof. R.A. Fisher to deal with problems in agricultural research.
The test of significance based on tdistribution is an adequate
procedure only for testing the significance of the difference
between two sample means. In a situation where we have three or
more samples to consider at a time, an alternative procedure is
needed for testing the hypothesis that all the samples are drawn
from the same population, i.e., they have the same mean. For
example, five fertilizers are applied to four plots each of wheat and
yield of wheat on each of the plot is given. We may be interested in
finding out whether the effect of these fertilizers on the yields is
significantly different or in other words whether the samples have
come from the same normal population. The answer to this problem
is provided by the technique of analysis of variance. Thus basic
purpose of the analysis of variance is to test the homogeneity of
several means.
Variation is inherent in nature. The total variation in any set
of numerical data is due to a number of causes which may be
classified as:
(i) Assignable causes and (ii) Chance causes
The variation due to assignable causes can be detected and
measured whereas the variation due to chance causes is beyond the
control of human hand and cannot be traced separately.
7.1 Definition:
According to R.A. Fisher , Analysis of Variance (ANOVA)
is the “ Separation of Variance ascribable to one group of causes
from the variance ascribable to other group”. By this technique the
total variation in the sample data is expressed as the sum of its nonnegative components where each of these components is a measure
of the variation due to some specific independent source or factor
or cause.
186 7.2 Assumptions:
For the validity of the Ftest in ANOVA the following
assumptions are made.
(i)
The observations are independent
(ii)
Parent population from which observations are taken is
normal and
(iii)
Various treatment and environmental effects are
additive in nature.
7.3 One way Classification:
Let us suppose that N observations xij , i = 1, 2, … k ;
…
j = 1,2… i) of a random variable X are grouped on some basis,
.n
k into k classes of sizes n1, n2 , … k respectively ( N = ∑ n i ) as
..n
i =1 exhibited below
x11 x12
x22 … x2n2 .
.
.
xi1 .
.
.
xi2 .
.
.
xk1 .
.
.
xk2 … .
.
.
xini Total
T1. x2
.
.
. … x1n1 x21 Mean
x1 . T2. . xi .
.
.
.
xk . .
.
.
… xknk
… .
.
.
Ti.
.
.
.
Tk.
G The total variation in the observation xij can be spilit into
the following two components :
(i)
The variation between the classes or the variation due to
different bases of classification, commonly known as
treatments.
187 (ii) The variation within the classes i.e., the inherent
variation of the random variable within the observations
of a class.
The first type of variation is due to assignable causes which
can be detected and controlled by human endeavour and the second
type of variation due to chance causes which are beyond the control
of human hand.
In particular, let us consider the effect of k different rations on
the yield in milk of N cows (of the same breed and stock) divided
k into k classes of sizes n1, n2 , …nk respectively. N = ∑ n i . Hence
..
i =1 the sources of variation are
(i)
Effect of the rations
(ii)
Error due to chance causes produced by numerous
causes that they are not detected and identified.
7.4 Test Procedure:
The steps involved in carrying out the analysis are:
1) Null Hypothesis:
The first step is to set up of a null hypothesis
H0: µ1 = µ2 = … µk
=
Alternative hypothesis H1: all µi ‘ s are not equal (i = 1,2,…
,k)
2) Level of significance : Let α : 0.05
3) Test statistic:
Various sum of squares are obtained as follows.
a) Find the sum of values of all the (N) items of the
given data. Let this grand total represented by ‘ G’ .
G2
Then correction factor (C.F) =
N
b) Find the sum of squares of all the individual items (xij)
and then the Total sum of squares (TSS) is
TSS = Σ xi2j – C.F
c) Find the sum of squares of all the class totals (or each
treatment total) Ti (i:1,2,… and then the sum of
.k)
squares between the classes or between the treatments
(SST) is
188 Ti 2
∑ n  C.F
i =1
i
Where ni (i: 1,2,… is the number of observations in
..k)
the ith class or number of observations received by ith
treatment
d) Find the sum of squares within the class or sum of
squares due to error (SSE) by subtraction.
SSE = TSS  SST
4) Degrees of freedom (d.f):
The degrees of freedom for total sum of squares (TSS) is
(N−1). The degrees of freedom for SST is (k−1) and the degrees of
freedom for SSE is (N−k)
5) Mean sum of squares:
SST
The mean sum of squares for treatments is
and mean
k −1
SSE
sum of squares for error is
N−k
6) ANOVA Table
The above sum of squares together with their respective
degrees of freedom and mean sum of squares will be summarized
in the following table.
k SST = ANOVA Table for oneway classification
Sources of
variation
Between
treatments
Error
Total N−1 d.f S.S M.S.S F ratio K−1 SST MST
= FT
MSE N−k SSE SST
= MST
k −1
SSE
=MSE
N−k Calculation of variance ratio:
Variance ratio of F is the ratio between greater variance
and smaller variance, thus
189 Variance between the treatments
Variance within the treatment
MST
=
MSE
If variance within the treatment is more than the variance
between the treatments, then numerator and denominator should be
interchanged and degrees of freedom adjusted accordingly.
7) Critical value of F or Table value of F:
The Critical value of F or table value of F is obtained from
F table for (k1, Nk) d.f at 5% level of significance.
8) Inference:
If calculated F value is less than table value of F, we may
accept our null hypothesis H0 and say that there is no significant
difference between treatments.
If calculated F value is greater than table value of F, we
reject our H0 and say that the difference between treatments is
significant.
F= Example 1:
Three processes
outputs are equivalent.
made:
A
10
12
B
9
11
C
11
10 A, B and C are tested to see whether their
The following observations of outputs are
13
10
15 11
12
14 10
13
12 14 15 13 13 Carry out the analysis of variance and state your conclusion.
Solution:
To carry out the analysis of variance, we form the following
tables
Total
Squares
A
10 12 13 11 10 14 15 13
98
9604
B
9 11 10 12 13
55
3025
C
11 10 15 14 12 13
75
5625
G = 228
190 Squares:
A
B
C 100 144 169 121 100 196 225 169
81 121 100 144 169
121 100 225 196 144 169
Total = 2794 Test Procedure:
Null Hypothesis: H0: µ1 = µ2 = µ3
i.e., There is no significant difference between the three processes.
Alternative Hypothesis H1: µ1 ≠ µ2 ≠ µ3
Level of significance : Let α : 0.05
Test statistic
G2
Correct factor (c.f) =
N
2282
=
19
51984
=
19
= 2736
Total sum of squares (TSS) = xij2 − C. F
= 2794 – 2736
= 58
Sum of squares due to processes = (SST)
3 ∑T . 2 i − C.F
ni
9604 3025 5625
=
+
+
− 2736
8
5
6
= (1200.5 + 605 + 937.5) − 2736
= 2743 − 2736
=7
Sum of squares due to error (SSE) = TSS – SST
= 58 − 7 = 51
= 191 i =1 ANOVA Table
Sources of
variation
Between
Processes
Error
Total d.f S.S M.S.S F ratio 3 −1 = 2 7 3.5
= 1.097
3.19 16 51 7
= 3.50
2
51
= 3.19
16 19 −1 = 18 Table Value:
Table value of Fe for (2,16) d.f at 5% level of significance is
3.63
Inference:
Since calculated F0 is less than table value of Fe, we may
accept our H0 and say that there is no significant difference
between the three processes.
Example 2:
A test was given to five students taken at random from the
fifth class of three schools of a town. The individual scores are
School I
9
7
6
5
8
School II
7
4
5
4
5
School III
6
5
6
7
6
Carry out the analysis of variance
Solution:
To carry out the analysis of variance, we form the following
tables. School I
School II
School III 9
7
6 7
4
5 6
5
6 5
4
7
192 Total Squares
8
35
1225
5
25
625
6
30
900
Total G=90 2750 Squares:
School I
School II
School III 81
49
36 49
16
25 36
25
36 25
64
16
25
49
36
Total = 568 Test Procedure :
Null Hypothesis: H0: µ1 = µ2 = µ3 i.e., There is no significant
difference between the performance of schools.
Alternative Hypothesis: H1: µ1 ≠ µ2 ≠ µ3
Level of significance: Let α :0.05
Test Statistic:
G2
Correct factor (c.f) =
N
902
=
15
8100
=
= 540
15
Total sum of squares (TSS) = xij2 − C. F
= 568 – 540 = 28
ΣTi 2
Sum of squares between schools =
− C.F
ni
2750
=
− 540
5
= 550 – 540 = 10
Sum of squares due to error (SSE) = TSS – SST
= 2810 = 18
ANOVA TABLE:
Source of
d.f
S.S
M.S.S
F ratio
variation
Between
31 = 2
10
10
5
= 5.0
= 3.33
Schools
2
1.5
Error
12
18
18
= 1.5
12
Total
15 1 = 14
193 Table Value:
Table value of Fe for (2,12) d.f at 5% level of significance is
3.8853
Inference:
Since calculated F0 is less than table value of Fe, we may
accept our H0 and say that there is no significant difference
between the performance of schools
7.5 Two way classification:
Let us consider the case when there are two factors which
may affect the variate values xij, e.g the yield of milk may be
affected by difference in treatments i.e., rations as well as the
difference in variety i.e., breed and stock of the cows. Let us now
suppose that the N cows are divided into h different groups or
classes according to their breed and stock, each group containing k
cows and then let us consider the effect of k treatments (i.e., rations
given at random to cows in each group) on the yield of milk.
Let the suffix i refer to the treatments (rations) and j refer
to the varieties (breed of the cow), then the yields of milk xij (i:1,2,
… j:1,2… of N = h × k cows furnish the data for the
..k;
.h)
comparison of the treatments (rations). The yields may be
expressed as variate values in the following k × h two way table.
Mean
Total
T1.
x11 x12 x1j … 1h
x
x1 .
T2.
x21 x22 x2j … 2h
x
x.
2 .
. .
. xi1 .
.
xi2 xk1 .
.
.
xk2 .
.
.
.
.
.
xkj … kh
x .
.
. .
. .
.
xi .
.
.
.
xk . xij … ih
x Mean x .1 . x .2 …x .j …x .h
. Total T.1 T.2. … T.j… h
..
.T. .
.
Ti.
.
.
.
Tk. x
G
194 The total variation in the observation xij can be split into
the following three components:
(i)
The variation between the treatments (rations)
(ii)
The variation between the varieties (breed and stock)
(iii)
The inherent variation within the observations of
treatments and within the observations of varieties.
The first two types of variations are due to assignable
causes which can be detected and controlled by human endevour
and the third type of variation due to chance causes which are
beyond the control of human hand.
7.6 Test procedure for Two  way analysis:
The steps involved in carrying out the analysis are:
1. Null hypothesis:
The first step is to setting up a null hypothesis H0
Ho : µ1. = µ2. = … µk. = µ
…
Ho : µ . 1 = µ . 2 = … . h = µ
µ
i.e., there is no significant difference between rations
(treatments) and there is no significant difference between
varieties ( breed and stock)
2.Level of significance: Let α : 0.05
3.Test Statistic:
Various sums of squares are obtained as follows:
a) Find the sum of values of all the N (k×h) items of
the given data. Let this grand total represented by
G2
‘ G’ Then correction factor (C.F) =
N
b) Find the sum of squares of all the individual items
(xij) and then the total sum of squares (TSS)
k k i −1 j−1 ∑∑ x 2ij − C.F c) Find the sum of squares of all the treatment (rations)
totals, i.e., sum of squares of row totals in the h × k
twoway table. Then the sum of squares between
treatments or sum of squares between rows is
195 Ti . 2
SST = SSR = ∑
− C.F
h
i −1
where h is the number of observations in each row
d) Find the sum of squares of all the varieties (breed
and stock) totals, in the h × k two  way table. Then
the sum of squares between varieties or sum of
squares between columns is
k k ∑T 2 .j j−1 − C.F where k is the number
k
of observations in each column.
e) Find the sum of squares due to error by subtraction:
i.e., SSE = TSS – SSR − SSC
4. Degrees of freedom:
(i)
The degrees of freedom for total sum of squares is
N−1 = hk−1
(ii)
The degrees of freedom for sum of squares between
treatments is k−1
(iii)
The degree of freedom for sum of squares between
varieties is h – 1
(iv)
The degrees of freedom for error sum of squares is
(k−1) (h−1)
5. Mean sum of squares (MSS)
SST
(i) Mean sum of squares for treatments (MST) is
k −1
SSV
(ii) Mean sum of squares for varieties (MSV) is
h −1
SSE
(iii) Mean sum of squares for error (MSE) is
(h − 1)(k − 1)
6. ANOVA TABLE
The above sum of squares together with their respective
degrees of freedom and mean sum of squares will be summarized
in the following table.
SSV = SSC = 196 ANOVA Table for Twoway classification
Sources of
d.f
SS
variation
Between
k1
SST
Treatments
Between
h1
SSV
Varieties
Error
(h1) (k1)
SSE
Total
N1 MSS F0  ratio MST MST
= FR
MSE
MSV
= Fc
MSE MSV
MSE 7. Critical values Fe or Table values of F:
(i) The critical value or table value of ‘ F’ for between treatments is
obtained from F table for [(k−1, (k−1) (h−1)] d.f at 5% level of
significance.
(ii) The critical value or table value of Fe for between varieties is
obtained from F table for [ (h−1), (k−1) (h−1)] d.f at 5% level
of significance.
8. Inference:
(i) If calculated F0 value is less than or greater than the table value
of Fe for between treatments (rows) H0 may be accepted or
rejected accordingly.
(ii) If calculated F0 value is less than or greater than the table value
of Fe for between varieties (column), H0 may be accepted or
rejected accordingly.
Example 3:
Three varieties of coal were analysed by four chemists and
the ashcontent in the varieties was found to be as under.
Chemists
Varieties
1
2
A
8
5
B
7
6
C
3
6
Carry out the analysis of variance.
197 3
5
4
5 4
7
4
4 Solution:
To carry out the analysis of variance we form the following
tables Varieties 1 2 A 8 5 B 7 C Chemists
3 4 Total 5 7 25 625 6 4 4 21 441 3 6 5 4 18 324 Total 18 17 14 15 G = 64 1390 Squares 324 289 196 225 1034 Varieties
A
B
C 1
64
49
9 Individual squares
Chemists
2
25
36
36 3
25
16
25 Squares 4
49
16
16
Total = 366 Test Procedure:
Null hypothesis:
H0 : µ1. = µ2. = µ3. = µ
H0 : µ.1 = µ.2 = µ.3 = µ.4 = µ
(i)
i.e., there is no significant difference between varieties
(rows)
(ii)
i.e., there is no significant difference between chemists
(columns)
Alternative hypothesis H1:
(i)
not all µi. ’ s equal
(ii)
not all µ.j’ s equal
Level of significance :
Let α : 0.05
198 Test statistic:
G2
G2
=
N
h×k
2
(64)
(64)2
=
=
3× 4
12
4096
=
= 341.33
12 Correction factor (c.f) = k k i −1 Total sum of squares (TSS) = j−1 ∑∑ x 2ij − C.F = 366 – 341.33
= 24.67
Sum of squares between varieties (Rows)
2
∑ Ti.
=
− C.F
4
1390
=
− 341.33
4
= 347.5 – 341.33
= 6.17
Sum of squares between chemists (columns)
2
∑ T. j
=
− C.F
3
1034
=
341.33
3
= 344.67 − 341.33
= 3.34
Sum of square due to error (SSE)
= TSS – SSR – SSC
= 24.67 – 6.17 – 3.34
= 24.67 – 9.51
= 15.16
199 ANOVA TABLE
Sources of
d.f
SS
MSS
F ratio
variation
6.17
3.085
Between
3.085
3 −1 = 2
= 1.22
Varieties
2.527
3.34
1.113
Between
2.527
4 −1 = 3
= 2.27
Chemists
1.113
Error
6
15.16
2.527
Total
12−1 = 11
Table value :
(i)
Table value of Fe for (2,6) d.f at 5% level of significance
is 5.14
(ii)
Table value of Fe for (6,3) d.f at 5% level of significance
is 8.94
Inference:
(i)
Since calculated F0 is less than table value of Fe, we
may accept our H0 for between varieties and say that
there is no significant difference between varieties.
(ii)
Since calculated F0 is less than the table value of Fe for
chemists, we may accept our Ho and say that there is no
significant difference between chemists. Exercise – 7
I. Choose the best answers:
1. Equality of several normal population means can be tested by
(a). Bartlet’ s test (b) F  test
(c) χ2 test (d) t test
2. Analysis of variance technique was developed by
(a) S. D. Poisson (b) Karl  Pearson
(c) R.A. Fisher
(d) W. S. Gosset
3. Analysis of variance technique originated in the field of
(a) Agriculture
(b) Industry (c) Biology
(d) Genetics
4. One of the assumption of analysis of variance is that the
population from which the samples are drawn is
(a) Binomial
(b) Poisson
(c) Chisquare (d) Normal
200 5. In the case of oneway classification the total variation can be
split into
(a) Two components
(b) Three components
(c) Four components
(d) Only one component
6. In the case of oneway classification with N observations and t
treatments, the error degrees of freedom is
(a) N−1
(b) t −1
(c) N − t
(d) Nt
7. In the case of oneway classification with t treatments, the mean
sum of squares for treatment is
(a) SST/N−1
(b) SST/ t−1 (c) SST/N−t (d) SST/t
8. In the case of twoway classification with r rows and c
columns, the degrees of freedom for error is
(a) (rc) – 1
(b) (r1).c
(c) (r1) (c1) (d) (c1).r
9. In the case of twoway classification, the total variation (TSS)
equals.
(a) SSR + SSC + SSE
(b) SSR − SSC + SSE
(c) SSR + SSC – SSE
(d) SSR + SSC.
10. With 90, 35, 25 as TSS, SSR and SSC respectively in case of
two way classification, SSE is
(a) 50
(b) 40
(c) 30
(d) 20
I. Fill in the blanks
11. The technique of analysis of variance was developed by
____________
12. One of the assumptions of Analysis of variance is:
observations are ______________
13. Total variation in two – way classification can be split into
____________ components.
14. In the case of one way classification with 30 observations and
5 treatment, the degrees freedom for SSE is _____________
15. In the case of twoway classification with 120, 54, 45
respectively as TSS, SSC, SSE, the SSR is _____________
III. Answer the following:
16. What is analysis of variance?
17. Distinguish between ttest for difference between means and
ANOVA.
201 18.
19.
20.
21.
22.
23.
24.
25.
26. State all the assumptions involved in analysis of variance
technique.
Explain the structure for oneway classification.
Write down the ANOVA table for oneway classification.
Distinguish between one  way classification and twoway
classification.
Explain the structure of twoway classification data.
Explain the procedure of obtaining various sums of squares in
oneway classification.
Write down ANOVA table for twoway classification.
Explain the procedure of obtaining various sums of squares in
twoway classification.
A test was given to a number of students taken at random
from the eighth class from each of the 5 schools. The Individual Scores are:
Schools
I
II
III
IV
V
8
9
12
10
12
9
7
14
11
11
10
11
15
9
10
7
12
12
12
9
8
13
11
10
13
Carry out the analysis of variance and give your conclusions.
27. The following figures relate to production in kg of three
varieties A, B and C of wheat shown in 12 plots.
A: 20 18 19 B: 17 16 19 18 C: 20
21
20
19
18
Is there any significant difference in the production of the three
varieties
202 28. 29. A special type of fertilizer was used in four agricultural fields
A,B,C and D each field was divided into four beds and the
fertilizer was applied over them. The respective yields of the
beds of four fields are given below. Find whether the
difference in mean yields of fields is significant or not?
Plot yield
A
B
C
D
8
9
3
3
12
4
8
7
1
7
2
8
9
1
5
2
The following table gives the retail prices of a commodity in
(Rs. Per Kg) in some shops selected at random in four cities.
A
22
24
20
21
CITY
B
20
19
21
22
C
19
17
21
18
D
20
22
21
22 Analysis the data to test the significance of the differences between
the price of commodity in four cities.
30. For experiments determine the moisture content of sample of a
powder, each man taking a sample from each of six
consignments Their assessments are: Observer
1
2
3
4 1
9
12
11
12 Consignment
2
3
10
9
11
9
10
10
13
11 4
10
11
12
14 5
11
10
11
12 6
11
10
10
10 Perform an analysis of variance of these data and discuss if
there is any significant difference between consignments or
between observers. 203 31. The following are the defective pieces produced by four
operators working in turn, on four different machines:
Operator
Machine
I
II
III
A
3
2
3
B
3
2
3
C
2
3
4
D
3
4
3 IV
2
4
3
2 Perform analysis of variance at 5% level of significance to ascertain
whether variability in production is due to variability in operator’ s
performance or variability in machine’ s performance.
32. Apply the technique of Analysis of variance to the following
data relating to yields of 4 varieties of wheat in 3 blocks.
Blocks
Varieties
1
2
3
I
10
9
8
II
7
7
6
III
8
5
4
IV
5
4
4
33. Four Varieties of potato are planted, each on five plots of
ground of the same size and type and each variety is treated with
five different fertilizers. The yields in tons are as follows. Varieties
V1
V2
V3
V4 F1
1.9
2.5
1.7
2.1 Fertilizers
F2
F3
2.2
2.6
1.9
2.2
1.9
2.2
1.8
2.5 F4
1.8
2.6
2.0
2.2 Perform an analysis of variance and test whether there is any
significant difference between yields of different varieties and
fertilizers.
204 F5
2.1
2.2
2.1
2.5 34. In an experiment on the effects of temperature conditions in
human performance, 8 persons were given a test on 4 temperature
conditions. The scores in the test are shown in the following table. Temperature
1
2
3
4 1
70
70
75
65 2
80
80
85
75 Persons
3
4
70
90
80
90
80
95
70
85 5
80
80
75
80 6
100
100
85
90 7
90
90
95
80 8
80
80
75
75 Perform the analysis of variance and state whether there is any
significant difference between persons and temperature conditions.
35. The following table gives the number of refrigerators sold by 4
salesmen in three months may, June and July
Sales Man
Months
A
B
C
D
May
50
40
48
39
June
46
48
50
45
July
39
44
40
39
Carry out the analysis
Answers
I.
1. b 2. c 3. a 4. d 5. a 6. c 7. b 8. c 9. a 10.c II.
11. R.A. Fisher 12. Independent 14. 25 15. 21
205 13. Three III.
26. Calculated F = 4.56, Table value of F (4,20) = 2.87 27. Calculated F = 9.11, Table value of F (9,2) = 19.39 28. Calculated F = 1.76, Table value of F (12,3) = 8.74 29. Calculated F = 3.29, Table value of F (3,12) = 3.49 30. Calculated FR = 5.03, Table value of F (3,15) = 3.29 Calculated FC = 2.23, Table value of F (5,15) = 2.90 31. Calculated FR = 2.76, FC Table value of F (9,3) = 8.81
32. Calculated FR = 18.23, Table value of F (3,6) = 4.77 Calculated FC = 6.4, Table value of F (2,6) = 5.15 33. Calculated FR = 1.59, Table value of F (3,12) = 3.49 Calculated FC = 3.53, Table value of F (4,12) = 3.25 34. Calculated FR = 3.56, Table value of F (3,21) = 3.07 Calculated FC = 14.79, Table value of F (7,21) = 2.49 35. Calculated FR = 3.33, Table value of F (2,6) = 5.15 Calculated FC = 1.02, Table value of F (3,6) = 4.77 206 8. TIME SERIES
8.0 Introduction:
Arrangement of statistical data in chronological order ie., in
accordance with occurrence of time, is known as “Time Series”.
Such series have a unique important place in the field of Economic
and Business statistics. An economist is interested in estimating the
likely population in the coming year so that proper planning can be
carried out with regard to food supply, job for the people etc.
Similarly, a business man is interested in finding out his likely sales
in the near future, so that the businessman could adjust his
production accordingly and avoid the possibility of inadequate
production to meet the demand. In this connection one usually deal
with statistical data, which are collected, observed or recorded at
successive intervals of time. Such data are generally referred to as
‘ time series’ .
8.1 Definition:
According to Mooris Hamburg “A time series is a set of
statistical observations arranged in chronological order”
YaLun chou defining the time series as “A time series
may be defined as a collection of readings belonging to different
time periods, of some economic variable or composite of variables.
A time series is a set of observations of a variable usually at equal
intervals of time. Here time may be yearly, monthly, weekly, daily
or even hourly usually at equal intervals of time.
Hourly temperature reading, daily sales, monthly production
are examples of time series. Number of factors affect the
observations of time series continuously, some with equal intervals
of time and others are erratic studying, interpreting analyzing the
factors is called Analysis of Time Series.
The Primary purpose of the analysis of time series is to
discover and measure all types of variations which characterise a
time series. The central objective is to decompose the various
elements present in a time series and to use them in business
decision making.
207 8.2 Components of Time series:
The components of a time series are the various elements
which can be segregated from the observed data. The following are
the broad classification of these components.
Components Long Term Secular Trend Short Term Cyclical Seasonal Irregular
(or)
Erratic Regular
In time series analysis, it is assumed that there is a
multiplicative relationship between these four components.
Symbolically,
Y=T×S×C×I
Where Y denotes the result of the four elements; T = Trend ;
S = Seasonal component; C = Cyclical components; I = Irregular
component
In the multiplicative model it is assumed that the four
components are due to different causes but they are not necessarily
independent and they can affect one another.
Another approach is to treat each observation of a time
series as the sum of these four components. Symbolically
Y = T + S+ C+ I
The additive model assumes that all the components of the
time series are independent of one another.
1) Secular Trend or Long  Term movement or simply Trend
2) Seasonal Variation
3) Cyclical Variations
4) Irregular or erratic or random movements(fluctuations)
8.2.1 Secular Trend:
It is a long term movement in Time series. The general
tendency of the time series is to increase or decrease or stagnate
208 during a long period of time is called the secular trend or simply
trend. Population growth, improved technological progress,
changes in consumers taste are the various factors of upward trend.
We may notice downward trend relating to deaths, epidemics, due
to improved medical facilities and sanitations. Thus a time series
shows fluctuations in the upward or downward direction in the long
run.
8.2.2 Methods of Measuring Trend:
Trend is measured by the following mathematical methods.
1. Graphical method
2. Method of Semiaverages
3. Method of moving averages
4. Method of Least Squares
Graphical Method:
This is the easiest and simplest method of measuring trend.
In this method, given data must be plotted on the graph, taking time
on the horizontal axis and values on the vertical axis. Draw a
smooth curve which will show the direction of the trend. While
fitting a trend line the following important points should be noted
to get a perfect trend line.
(i)
The curve should be smooth.
(ii)
As far as possible there must be equal number of points
above and below the trend line.
(iii)
The sum of the squares of the vertical deviations from
the trend should be as small as possible.
(iv)
If there are cycles, equal number of cycles should be
above or below the trend line.
(v)
In case of cyclical data, the area of the cycles above and
below should be nearly equal.
Example 1:
Fit a trend line to the following data by graphical method.
Year
Sales
(in Rs
‘ 000) 1996 1997 1998 1999 2000 2001 2002 60 72 75 65 80 85 95 209 Solution:
100 Sales 80
60
40
20
0 1996 1997 1998 1999 2000 2001 2002 2003 Year The dotted lines refers trend line
Merits:
1. It is the simplest and easiest method. It saves time and labour.
2. It can be used to describe all kinds of trends.
3. This can be used widely in application.
4. It helps to understand the character of time series and to select
appropriate trend.
Demerits:
1. It is highly subjective. Different trend curves will be obtained by
different persons for the same set of data.
2. It is dangerous to use freehand trend for forecasting purposes.
3. It does not enable us to measure trend in precise quantitative
terms.
Method of semi averages:
In this method, the given data is divided into two parts,
preferably with the same number of years. For example, if we are
given data from 1981 to 1998 i.e., over a period of 18 years, the
two equal parts will be first nine years, i.e., 1981 to 1989 and from
1990 to 1998. In case of odd number of years like 5,7,9,11 etc, two
equal parts can be made simply by omitting the middle year. For
example, if the data are given for 7 years from 1991 to 1997, the
two equal parts would be from 1991 to 1993 and from 1995 to
1997, the middle year 1994 will be omitted.
210 After the data have been divided into two parts, an average
of each parts is obtained. Thus we get two points. Each point is
plotted at the midpoint of the class interval covered by respective
part and then the two points are joined by a straight line which
gives us the required trend line. The line can be extended
downwards and upwards to get intermediate values or to predict
future values.
Example 2:
Draw a trend line by the method of semiaverages.
Year
1991
1992
1993
1994
1995
Sales
Rs in
60
75
81
110
106
(1000) 1996
117 Solution:
Divide the two parts by taking 3 values in each part.
Year
Sales (Rs)
Semi total
Semi
Trend
average
values
59
60
1991
72
216
72
75
1992
85
81
1993
98
110
1994
111
333
111
106
1995
124
117
1996
Difference in middle periods = 1995 –1992 = 3 years
Difference in semi averages = 111 –72 = 39
∴ Annual increase in trend = 39/3 = 13
Trend of 1991 = Trend of 1992 13
= 7213 = 59
Trend of 1993 = Trend of 1992 +13
= 72 + 13 = 85
Similarly, we can find all the values
The following graph will show clearly the trend line.
211 140
120
Sales 100
80
60
40
20
0
1991 1992 1993 1994 1995 1996
Year Example 3 :
Calculate the trend value to the following data by the
method of semi averages.
Year
1995 1996 1997 1998 1999 2000 2001
Expenditure
(Rs in
1.5
1.8
2.0
2.3
2.4
2.6
3.0
Lakhs)
Solution:
Year
1995
1996
1997
1998
1999
2000
2001 Expenditure
(Rs)
1.5
1.8
2.0
2.3
2.4
2.6
3.0 Semi total Semi
average 5.3 1.77 8.0 2.67 Difference between middle periods = 2000 – 1996
= 4 years
Difference between semiaverages = 2.67  1.77
= 0.9
212 Trend
values
1.545
1.770
1.995
2.220
2.445
2.670
2.895 0.9
4
= 0.225
Trend of 1995 = Trend of 1996 – 0.225
= 1.77 – 0.225
= 1.545
Trend of 1996 = 1.77
Trend of 1997 = 1.77 + 0.225
= 1.995
Similarly we can find all the trend values
∴ Annual trend values = 3.5 Expenditure 3
2.5
2
1.5
1
0.5
0 1995 1996 1997 1998 1999 2000 2001 Year Merits:
1. It is simple and easy to calculate
2. By this method every one getting same trend line.
3. Since the line can be extended in both ways, we can find the
later and earlier estimates.
Demerits:
1. This method assumes the presence of linear trend to the values
of time series which may not exist.
2. The trend values and the predicted values obtained by this
method are not very reliable.
Method of Moving Averages:
This method is very simple. It is based on Arithmetic mean.
Theses means are calculated from overlapping groups of successive
213 time series data. Each moving average is based on values covering
a fixed time interval, called “period of moving average” and is
shown against the center of the interval.
The method of ‘ odd period of moving average is as follows.
a+b+c
( 3 or 5) . The moving averages for three years is
,
3
b+c +d c+d+e
,
etc
3
3
a +b+c+d+e
The formula for five yearly moving average is
,
5
b+c+d+e+f c+d+e+f +g
,
etc.
5
5
Steps for calculating odd number of years.
1. Find the value of three years total, place the value against the
second year.
2. Leave the first value and add the next three years value (ie 2nd,
3rd and 4th years value) and put it against 3rd year.
3. Continue this process until the last year’ s value taken.
4. Each total is divided by three and placed in the next column.
These are the trend values by the method of moving averages
Example 4 :
Calculate the three yearly average of the following data.
Year
1975 1976 1977 1978 1979 1980
Production
in (tones) 50 36 43 45 Year
1981 1982 1983 1984
Production 33
42
41
34
in (tones) 214 39 38 Solution:
Year Production
(in tones) 3 years
moving total 3 years moving
average as
Trend values
43.0
41.3
42.3
40.7
36.7
37.7
38.7
39.0
 50
1975
129
36
1976
124
43
1977
127
45
1978
122
39
1979
110
38
1980
113
33
1981
116
42
1982
117
41
1983
34
1984
Even Period of Moving Averages:
When the moving period is even, the middle period of each
set of values lies between the two time points. So we must center
the moving averages.
The steps are
1. Find the total for first 4 years and place it against the middle of
the 2nd and 3rd year in the third column.
2. Leave the first year value, and find the total of next fouryear
and place it between the 3rd and 4th year.
3. Continue this process until the last value is taken.
4. Next, compute the total of the first two four year totals and
place it against the 3rd year in the fourth column.
5. Leave the first four years total and find the total of the next two
four years’ totals and place it against the fourth year.
6. This process is continued till the last two four years’ total is
taken into account.
7. Divide this total by 8 (Since it is the total of 8 years) and put it
in the fifth column.
These are the trend values.
Example 5 :
The production of Tea in India is given as follows.
Calculate the Fouryearly moving averages
215 Year 1993 1994 1995 1996 1997 1998 Production
(tones) 464 518 502 540 Year 1999 2000 2001 2002 Production
(tones) 557 586 515 571 467 612 Solution:
Year Production
(in tones) 1993
1994 Trend
Values  3966 495.8 4029 503.6 4093 511.6 4236 529.5 4424 553.0 4580 464 Total of
Two four
years
 4 years
Moving
total 572.5 515 1964
1995 518
2002 1996 467
2027 1997 502
2066 1998 540 1999 557 2000 571 2170
2254
2326
2001 586
 2002 612
216 Merits:
1. The method is simple to understand and easy to adopt as
compared to other methods.
2. It is very flexible in the sense that the addition of a few more
figures to the data, the entire calculations are not changed. We
only get some more trend values.
3. Regular cyclical variations can be completely eliminated by a
period of moving average equal to the period of cycles.
4. It is particularly effective if the trend of a series is very irregular.
Demerits:
1. It cannot be used for forecasting or predicting future trend,
which is the main objective of trend analysis.
2. The choice of the period of moving average is sometimes
subjective.
3. Moving averages are generally affected by extreme values of
items.
4. It cannot eliminate irregular variations completely.
8.3 Method of Least Square:
This method is widely used. It plays an important role in
finding the trend values of economic and business time series. It
helps for forecasting and predicting the future values. The trend
line by this method is called the line of best fit.
The equation of the trend line is y = a + bx, where the
constants a and b are to be estimated so as to minimize the sum of
the squares of the difference between the given values of y and the
estimate values of y by using the equation. The constants can be
obtained by solving two normal equations.
y = na + b x …… (1)
….
xy = a x + b x2 ……
… (2)
Here x represent time point and y are observed values. ‘ n’ is the
number of pair values.
When odd number of years are given
Step 1: Writing given years in column 1 and the corresponding
sales or production etc in column 2.
217 Step 2: Write in column 3 start with 0, 1, 2 .. against column 1 and
denote it as X
Step 3: Take the middle value of X as A
Step 4: Find the deviations u = X A and write in column 4
Step 5: Find u2 values and write in column 5.
Step 6: Column 6 gives the product uy
Now the normal equations become
y = na + b u
(1)
where u = XA
uy = a u + b u2
(2)
Since u = 0 ,
From equation (1)
Σy
a=
n
From equation (2)
uy = b u2
uy
∴b=
Σu 2
∴ The fitted straight line is
y = a + bu = a + b ( X  A)
Example 6:
For the following data, find the trend values by using the
method of Least squares
Year
1990
1991
1992
1993
1994
Production
50
55
45
(in tones)
Estimate the production for the year 1996
Solution: 52 54 u2 uy Trend
values 2 4 100 50.2 1 1 1 55 50.7 45 2A 0 0 0 51.2 1993 52 3 1 1 52 51.7 1994 54 4 2 4 108 52.2 Total 256 10 5 X= x 1990 u = XA Year
(x) Production
(y) 1990 50 0 1991 55 1992 = X2 218 Where A is an assumed value
The equation of straight line is
Y = a + bX
= a + bu , where u = X  2
the normal equations are
….(1)
Σy = na+ bΣu……
..(2)
Σuy = aΣu +bΣu2 …
since Σu = 0 from(1) Σy = na
Σy 256
=
a=
= 51.2
n
5
From equation (2)
Σuy = bΣu2
5 = 10b
5
= 0.5
b=
10
The fitted straight line is
y = a+ bu
y = 51.2 + 0.5 (X2)
y = 51.2 + 0.5X –1.0
y = 50.2 + 0.5X
Trend values are, 50.2, 50.7, 51.2, 51.7, 52.2
The estimate production in 1996 is put X = x – 1990
X = 1996 –1990 = 6
Y = 50.2 + 0.5X = 50.2 +0.5(6)
= 50.2 +3.0 = 53.2 tonnes.
60 production 50
40
30
20
10
0
1990 1991 1992 Year 219 1993 1994 When even number of years are given
Here we take the mean of middle two values of X as A
X−A
Then u =
= 2 (XA). The other steps are as given in the
1/ 2
odd number of years.
Example 7:
Fit a straight line trend by the method of least squares for the
following data.
Year
Sales
(Rs. in lakhs) 1983 1984 1985 1986 1987 1988 3 8 7 9 11 14 Also estimate the sales for the year 1991
Solution:
Sales X =
Year
u
(x)
(y)
x1983 =2X5
1983
3
0
5
1984
8
1
3
1985
7
2
1
1986
9
3
1
1987
11
4
3
1988
14
5
5
Total
52
0
X−A
u=
1/ 2
= 2 (X − 2.5) = 2X − 5
The straight line equation is
y = a + bX = a + bu
The normal equations are
…
Σy = na ….(1)
2
…
Σuy = bΣu …(2)
From (1) 52 = 6a
220 u2
25
9
1
1
9
25
70 uy
15
24
7
9
33
70
66 Trend
values
3.97
5.85
7.73
9.61
11.49
13.37 52
6
= 8.67
From (2) 66 = 70 b
66
b=
70
= 0.94
The fitted straight line equation is
y = a+bu
y = 8.67+0.94(2X5)
y = 8.67 + 1.88X  4.7
y = 3.97 + 1.88X (3)
The trend values are
Put X = 0, y = 3.97
X = 1, y = 5.85
X = 2, y = 7.73
X = 3, y = 9.61
X = 4, y = 11.49
X = 5, y = 13.37
The estimated sale for the year 1991 is; put X = x –1983
= 1991 – 1983 = 8
y = 3.97 + 1.88 × 8
= 19.01 lakhs
The following graph will show clearly the trend line. Sales a= 16
14
12
10
8
6
4
2
0
1983 1984 1985 1986 Year 221 1987 1988 Merits:
1. Since it is a mathematical method, it is not subjective so it
eliminates personal bias of the investigator.
2. By this method we can estimate the future values. As well as
intermediate values of the time series.
3. By this method we can find all the trend values.
Demerits:
1. It is a difficult method. Addition of new observations makes recalculations.
2. Assumption of straight line may sometimes be misleading since
economics and business time series are not linear.
3. It ignores cyclical, seasonal and irregular fluctuations.
4. The trend can estimate only for immediate future and not for
distant future.
8.4 Seasonal Variations:
Seasonal Variations are fluctuations within a year during the
season. The factors that cause seasonal variation are
i) Climate and weather condition.
ii) Customs and traditional habits.
For example the sale of icecreams increase in summer, the
umbrella sales increase in rainy season, sales of woolen clothes
increase in winter season and agricultural production depends upon
the monsoon etc.,
Secondly in marriage season the price of gold will increase,
sale of crackers and new clothes increase in festival times.
So seasonal variations are of great importance to
businessmen, producers and sellers for planning the future. The
main objective of the measurement of seasonal variations is to
study their effect and isolate them from the trend.
Measurement of seasonal variation:
The following are some of the methods more popularly used
for measuring the seasonal variations.
1. Method of simple averages.
2. Ratio to trend method.
3. Ratio to moving average method.
4. Link relative method
222 Among the above four methods the method of simple averages
is easy to compute seasonal variations.
8.4.1 Method of simple averages
The steps for calculations:
i) Arrange the data season wise
ii) Compute the average for each season.
iii) Calculate the grand average, which is the average of
seasonal averages.
iv) Obtain the seasonal indices by expressing each season
as percentage of Grand average
The total of these indices would be 100n where ‘ n’ is the
number of seasons in the year.
Example 8:
Find the seasonal variations by simple average method for
the data given below.
Quarter
Year
I
II
III
IV
34
36
40
30
1989
44
50
52
34
1990
48
54
58
40
1991
62
68
76
54
1992
82
86
92
80
1993
Solution:
Year
1989
1990
1991
1992
1993
Total
Average
Seasonal
Indices I
30
34
40
54
80
238
47.6
85 Quarter
II
40
52
58
76
92
318
63.6
113.6
223 III
36
50
54
68
86
294
58.8 IV
34
44
48
62
82
270
54 105 96.4 47.6 + 63.6 + 58.8 + 54
4
224
=
= 56
4
Seasonal Index for
First quarterly Average
I quarter =
× 100
Grand Average
47.6
=
× 100 = 85
56
Seasonal Index for
Second quarterly Average
II quarter =
× 100
Grand Average
63.6
=
× 100 = 113.6
56
Seasonal Index for
Third quarterly Average
III quarter =
× 100
Grand Average
58.8
=
× 100 = 105
56
Seasonal Index for
Fourth quarterly Average
IV quarter =
× 100
Grand Average
54
=
×100 = 96.4
56
Example 9:
Calculate the seasonal indices from the following data using
simple average method.
Year
Quarter
1974
1975
1976
1977
1978
I
72
76
74
76
74
II
68
70
66
74
74
III
80
82
84
84
86
IV
70
74
80
78
82
Grand average = 224 Solution:
Year
1974
1975
1976
1977
1978
Total
Average
Seasonal
Indices I
72
76
74
76
74
372
74.4
97.6 Quarter
II
68
70
66
74
74
352
70.4
92.4 III
80
82
84
84
86
416
83.2
109.2 74.4 + 70.4 + 83.2 + 76.8
4
304.8
= 76.2
=
4
Seasonal Index for
First quarterly Average
I quarter =
× 100
Grand Average
74.4
=
× 100
76.2
= 97.6
Seasonal Index for
Second quarterly Average
II quarter =
× 100
Grand Average
70.4
=
× 100
76.2
= 92.4
Seasonal Index for
Third quarterly Average
III quarter =
× 100
Grand Average
83.2
=
× 100
76.2
= 109.2
Grand Average = 225 IV
70
74
80
78
82
384
76.8
100.8 Seasonal Index for
Fourth quarterly Average
IV quarter =
× 100
Grand Average
76.8
=
×100
76.2
= 100.8
The total of seasonal indices calculated must be equal to
400 here we have = 97.6 + 92.4 + 109.2 + 100.8
= 400 hence verified.
Cyclical variations:
The term cycle refers to the recurrent variations in time
series, that extend over longer period of time, usually two or more
years. Most of the time series relating to economic and business
show some kind of cyclic variation. A business cycle consists of
the recurrence of the up and down movement of business activity. It
is a fourphase cycle namely.
1. Prosperity 2. Decline
3. Depression
4. Recovery
Each phase changes gradually into the following phase. The
following diagram illustrates a business cycle. The study of cyclical variation is extremely useful in
framing suitable policies for stabilising the level of business
activities. Businessmen can take timely steps in maintaining
business during booms and depression.
Irregular variation:
Irregular variations are also called erratic. These variations
are not regular and which do not repeat in a definite pattern.
226 These variations are caused by war, earthquakes, strikes
flood, revolution etc. This variation is shortterm one, but it affect
all the components of series. There is no statistical techniques for
measuring or isolating erratic fluctuation. Therefore the residual
that remains after eliminating systematic components is taken as
representing irregular variations.
FORECASTING
8.5 Introduction:
A very important use of time series data is towards
forecasting the likely value of variable in future. In most cases it is
the projection of trend fitted into the values regarding a variable
over a sufficiently long period by any of the methods discussed
latter. Adjustments for seasonal and cyclical character introduce
further improvement in the forecasts based on the simple projection
of the trend. The importance of forecasting in business and
economic fields lies on account of its role in planning and
evaluation. If suitably interpreted, after consideration of other
forces, say political, social governmental policies etc., this
statistical technique can be of immense help in decision making.
The success of any business depends on its future estimates.
On the basis of these estimates a business man plans his production
stocks, selling market, arrangement of additional funds etc.
Forecasting is different from predictions and projections.
Regression analysis, time series analysis, Index numbers are some
of the techniques through which the predictions and projections are
made. Where as forecasting is a method of foretelling the course of
business activity based on the analysis of past and present data
mixed with the consideration of ensuring economic policies and
circumstances. In particularly forecasting means forewarning.
Forecasts based on statistical analysis are much reliable than a
guess work.
According to T.S.Levis and and R.A. Fox, “ Forecasting is
using the knowledge we have at one time to estimate what will
happen at some future movement of time”.
227 8.5.1 Methods of Business forecasting:
There are three methods of forecasting
1. Naive method
2. Barometric methods
3. Analytical Methods
1. Naive method :
It contains only the economic rhythm theory.
2. Barometric methods:
It covers
i)
Specific historical analogy
ii)
Lead Lag relationship
iii)
Diffusion method
iv)
Action –reaction theory
3. Analytical Methods:
It contains
i)
The factor listing method
ii)
Crosscut analysis theory
iii)
Exponential smoothing
iv)
Econometric methods
The economic rhythm theory:
In this method the manufactures analysis the timeseries data
of his own firm and forecasts on the basis of projections so
obtained. This method is applicable only for the individual firm for
which the data are analysed, The forecasts under this method are
not very reliable as no subjective matters are being considered.
Diffusion method of Business forecasting
The diffusion index method is based on the principle that
different factors, affecting business, do not attain their peaks and
troughs simultaneously. There is always timelog between them.
This method has the convenience that one has not to identify which
series has a lead and which has a lag. The diffusion index depicts
the movement of broad group of series as a whole without
bothering about the individual series. The diffusion index shows
the percentage of a given set of series as expanding in a time
period. It should be carefully noted that the peaks and troughs of
diffusion index are not the peaks troughs of the business cycles. All
228 series do not expand or contract concurrently. Hence if more than
50% are expanding at a given time, it is taken that the business is in
the process of booming and vice  versa.
The graphic method is usually employed to work out the
diffusion index. The diffusion index can be constructed for a group
of business variables like prices, investments, profits etc.
Cross cut analysis theory of Business forecasting:
In this method a thorough analysis of all the factors under
present situations has to be done and an estimate of the composite
effect of all the factors is being made. This method takes into
account the views of managerial staff, economists, consumers etc.
prior to the forecasting. The forecasts about the future state of the
business is made on the basis of over all assessment of the effect of
all the factors. Exercise – 8
I. Choose the best answer:
1. A time series consists of
a) Two components
b) Three Components
c) Four components
d) Five Components
2. Salient features responsible for the seasonal variation are
a) Weather
b) Social customers
c) Festivals
d) All the above
3. Simple average method is used to calculate
a) Trend Values
b) Cyclic Variations
c) Seasonal indices
d) None of these
4. Irregular variations are
a) Regular
b) Cyclic
c) Episodic
d) None of the above
5. If the slope of the trend line is positive it shows
a) Rising Trend
b) Declining trend
c) Stagnation
d) None of the above
6. The sales of a departmental store on Diwali are associated with
the component of timeseries
a) Secular trend
b) Seasonal variation
c) Irregular variation
d) All the above
229 7. The component of timeseries attached to long term variation is
termed as
a) Secular Trend
b) Seasonal Variation
c) Irregular variation
d) Cyclic variation
8. Business forecasts are made on the basis of
a) Present Data
b) Past data
c) Polices and circumstances
d) All the above
9. Econometric methods involve
a) Economics and mathematics b) Economics and Statistics
c) Economics, Statistics and Mathematics
d) None of the above
10. The economic rhythm theory comes under the category of
a) Analytical methods
b) Naive method
c) Barometric methods
d) None of the above
II. Fill in the blanks:
11. A time series in a set of values arranged in ________ order
12. Quarterly fluctuations observed in a time series represent
_______ variation
13. Periodic changes in a business time series are called _________
14. A complete cycle passes through _____________stages of
phenomenon.
15. An overall tendency of rise and fall in a time series represents
___________
16. The trend line obtained by the method of least square is known
as the ___________
17. Forecasting is different from ________ and _________.
18. No statistical techniques measuring or isolating _________
III. Answer the following questions
19. What is a time series?
20. What are the components of time series.
21. Write briefly about seasonal variation.
22. What is cyclic variation.
23. Give the names of different methods of measuring trend.
24. What are the merits and demerits of the semiaverage method.
25. Discuss the mathematical models for a time series analysis.
230 26. Discuss irregular variation in the context of time series.
27. What do you understand by business forecasting
28. Give the names of different methods of fore casting.
29. Write briefly about any one method of forecasting?
30. In what sense forecasting differ from prediction and projection.
IV. Problems
31. With the help of graph paper obtain the trend values.
Year
1996 1997
1998
1999
2000
2001
Value
65
85
95
75
100
80 2002
130 32. Using graphical method, fit a trendline to the following data.
Year 1982
1983
1984
1985
1986
1987
Value
24
22
25
26
27
26
33. Draw a trend line by the method of semiaverages.
98 99
Year 1993 94 95 96 97
Sales 210 200 215 205 220 235
210
34. The following figures are given relating to
factory. Draw a trendline with the help of
averages.
Year
1996 1997 1998 1999 2000
Output 600 800 1000 800 1200
35. 2000
235 the output in a
method of semi2001 2002
1000 1400 Calculate three yearly moving average of the following data Year
No of
students 91 92 93 94 95 96 97 98 99 00 15 18 17 20 23 25 29 33 36 40 36. The following figures relating to the profits of a commercial
concern for 8 years. Find the 3yearly moving averages.
Years
Profits
Years
Profits
1995
1996
1997
1998 15,420
14,470
15,520
21,020 1999
2000
2001
2002 231 26,120
31,950
35,370
35,670 37. Construct a four yearly centered moving average from the
following data.
Year
1940 1950 1960 1970 1980 1990 2000
Imported
129 131 106
91
95
84
93
cotton
consumption
(‘ 000)
38. From the following data calculate the 4yearly moving average
and determine the trend values.
Find the shortterm
fluctuations. Plot the original data and the trend on a graph.
Year
Value 93
50 94
36.5 95
43 96
44.5 97
38.9 98
38.1 99
32.6 00
41.7 01
41.1 02
33.8 39. Calculate trend value by taking 5 yearly period of moving
average from the data given below
Year
1987
Production
4
in tones
Year
95 96
Production 8 7
in tones 88 89
5 90 91 92 93 94 7 9 6 5 7 6
97 98 99 2000 01 02 6 8 9 10 7 9 40. Fit a straight line trend by the method of least squares to the
following data and calculate trend values.
Year
1996
1997
Sales of
TV Sets
4
6
(Rs ‘ 000 )
Estimate the sales for the year 2005
232 1998
7 1999
8 2000
10 41. Below are given the figures of production in ‘ 000 quintals of a
sugar factory.
Year
1994
95
96
97
98
99 2000
Production
80
90
92
83
94
99
92
in tones
42. Fit a straight line trend by the method of least square to the
following data.
Year
1996 97
98
99
2000 2001
Profit
300
700 600 800 900 700
43. Fit a straight line trend by the method of least squares to the
following data. Estimate the earnings for the year 2002.
Year
1993 94 95
96
97
98
99 2000
Earnings 38
40 65
72
69
60
87 95
44. Compute the average seasonal movement for the following
series.
Year Ist quarter
1999
3.5
2000
3.5
2001
3.5
2002
4.0
2003
4.1 IInd quarter
3.9
4.1
3.9
4.6
4.4 IIIrd quarter
3.4
3.7
3.7
3.8
4.2 IVth quarter
3.6
4.0
4.2
4.5
4.5 45. Obtain seasonal fluctuations from the following timeseries
Quarterly output of coal for four years.
Year
I
II
III
IV 2000
65
58
56
61 2001
58
63
63
67 233 2002
70
59
56
52 2003
60
55
51
58 Answers
I.
1. (c)
7. (a) 2.(d)
8.(d) 3. (c)
9. (c) 4. (c)
10.(b) 5. (a) 6. (b) II.
11. Chronological
14. four
16. line of best fit
18. Erratic Fluctuation. 12. Seasonal
13. Cycles
15. secular trend
17. Prediction, projection IV.
33. Trend values are 200.94, 205.31, 209.69, 214.06, 218.43,
222.80, 227.19, 231.56
34. 700, 800, 900, 1000, 1100, 1200, 1300
35. 16.7, 18.3, 20, 22.7, 25.7, 29, 32.7, 36.3
36. 15137, 17003, 20363, 26363, 31.147, 34330
37. 110.0, 99.88, 92.38
38. 42.1, 40.9, 39.8, 38.2, 38.1, 37.8,
39. 6.2, 6.6, 6.6, 6.8, 7.0, 6.6, 6.6, 7.2, 7.6, 8.0, 8.0
40. Trend values are 4.2, 5.6, 7; 8.4, 9.8
41. 84, 86, 88, 90, 92, 94, 96
42. 446.67, 546.67, 626.67, 706.67, 786.67, 866.67
43. 40.06, 47.40, 54.74, 62.08, 69.42, 76.76, 84.10, 91.44
44. 94.18, 105.82, 95.19, 105.32
45. 106.4, 98.7, 94.9,100 234 9. THEORY OF ATTRIBUTES
9.0 Introduction:
Generally statistics deal with quantitative data only. But in
behavioural sciences, one often deals with the variable which are
not quantitatively measurable. Literally an attribute means a
quality on characteristic which are not related to quantitative
measurements. Examples of attributes are health, honesty,
blindness etc. They cannot be measured directly. The observer may
find the presence or absence of these attributes. Statistics of
attributes based on descriptive character.
9.1 Notations:
Association of attribute is studied by the presence or
absence of a particular attribute. If only one attribute is studied, the
population is divided into two classes according to its presence or
absence and such classification is termed as division by dichotomy.
If a class is divided into more than two scaleclasses, such
classification is called manifold classification.
Positive class which denotes the presence of attribute is
generally denoted by Roman letters generally A,B,… and the
.etc
negative class denoting the absence of the attribute and it is
denoted by the Greek letters α, β… For example, A represents
.etc
the attribute ‘ Literacy’ and B represents ‘ Criminal’ . α and β
represents the ‘ Illiteracy’ and ‘ Not Criminal’ respectively.
9.2 Classes and Class frequencies:
Different attributes, their subgroups and combinations are
called different classes and the number of observations assigned to
them are called their class frequencies.
If two attributes are studied the number of classes will be 9.
(i.e.,) (A) , (α), (B), (β), (A β) (α β), (α B) and N. 235 The chart given below illustrate it clearly.
N
(α) (A) (AB) (Aβ) (αB) (αβ) The number of observations or units belonging to class is
known as its frequency are denoted within bracket. Thus (A) stands
for the frequency of A and (AB) stands for the number objects
possessing the attribute both A and B. The contingency table of
order (2×2) for two attributes A and B can be displayed as given
below
A
B
β
Total α Total (AB)
(Aβ)
(A) (αB)
(αβ)
(α) (B)
(β )
N Relationship between the class frequencies:
The frequency of a lower order class can always be
expressed in terms of the higher order class frequencies.
i.e.,
N = ( A ) + (α ) = (B) + (β)
(A) = (AB) + (Aβ)
(α) = (αB) + (αβ)
(B) = (AB) + (α B)
(β) = (Aβ) + (α β)
If the number of attributes is n, then there will be 3n classes
and we have 2n cell frequencies. 236 9.3 Consistency of the data:
In order to find out whether the given data are consistent or
not we have to apply a very simple test. The test is to find out
whether any or more of the ultimate classfrequencies is negative or
not. If none of the class frequencies is negative we can safely
calculate that the given data are consistent (i.e the frequencies do
not conflict in any way each other). On the other hand, if any of the
ultimate class frequencies comes to be negative the given data are
inconsistent.
Example 1:
Given N = 2500, (A) = 420, (AB) = 85 and
Find the missing values.
Solution:
We know N = (A) +(α) = (B) + (β)
(A) = (AB) + (Aβ)
(α) = (αB) + (αβ)
(B) = (AB) + (α B)
(β) = (Aβ) + (α β)
From (2) 420 = 85 + (Aβ)
∴ (Aβ) = 420 –85
(A β) = 335
From (4) 670 = 85 + (αB)
∴ (αB) = 670 − 85
(αB) = 585
From (1) 2500 = 420 + (α)
∴ (α) = 2500 − 420
(α) = 2080
From (1) (β) = 2500 −670
(β) = 1830
From (3) = 2080 = 585 + (αβ)
∴(αβ) = 1495
237 (B) = 670. Example 2:
Test the consistency of the following data with the symbols
having their usual meaning.
N = 1000 (A) = 600 (B) = 500 (AB) = 50
Solution:
A α Total B 50 450 500 β 550 50 500 Total 600 400 1000 Since (αβ)) = −50, the given data is inconsistent.
Example 3:
Examine the consistency of the given data. N = 60 (A) = 51
(B) = 32 (AB) = 25
Solution:
A α Total B 25 7 32 β 26 2 28 Total 51 9 60 Since all the frequencies are positive, it can be concluded
that the given data are consistent.
9.4 Independence of Attributes:
If the attributes are said to be independent the presence or
absence of one attribute does not affect the presence or absence of
the other. For example, the attributes skin colour and intelligence of
persons are independent.
238 If two attributes A and B are independent then the actual
frequency is equal to the expected frequency
(A).(B)
(AB) =
N
(α).(β)
Similarly
(α β ) =
N
9.4.1 Association of attributes:
Two attributes A and B are said to be associated if they are
not independent but they are related with each other in some way or
other.
The attributes A and B are said to be positively associated if
(A).(B)
(AB) >
N
(A).(B)
If (AB) <
,. then they are said to be negatively associated.
N
Example 4:
Show that whether A and B are independent, positively
associated or negatively associated.
(AB) = 128, (αB) = 384, (Aβ) = 24 and (αβ) = 72
Solution:
(A) = (AB) + (Aβ)
= 128 + 24
(A) = 152
(B) = (AB) + (αB)
= 128 +384
(B) = 512
(α) = (αB) + (αβ)
= 384 + 72
∴(α) = 456
(N) = (A) + (α)
= 152 + 456
= 608
239 (A) × (B)
152 × 512
=
N
608
= 128
(AB) = 128
(A) × (B)
∴ (AB) =
N
Hence A and B are independent
Example 5:
From the following data, find out the types of association of
A and B.
1) N = 200
(A) = 30
(B) = 100
(AB) = 15
2) N = 400
(A) = 50
(B) = 160
(AB) = 20
3) N = 800
(A) = 160
(B) = 300
(AB) = 50
Solution:
(A).(B)
N
(30)(100)
=
= 15
200
Since the actual frequency is equal to the expected frequency,
ie 15 = 15, therefore A and B are independent.
1. Expected frequency of (AB) = (A).(B)
N
(50)(160)
=
= 20
400
Since the actual frequency is greater than expected frequency. i.e.,
25 > 20, therefore A and B are positively associated. 2. Expected frequency of (AB) = 3. Expected frequency of (AB) = (A).(B)
(160)(300)
=
= 60
N
800 Since Actual frequency is less than expected frequency i.e., 50 < 60
therefore A and B are negatively associated.
240 9.5 Yules’ coefficient of association:
The above example gives a rough idea about association but
not the degree of association. For this Prof. G. Undy Yule has
suggested a formula to measure the degree of association. It is a
relative measure of association between two attributes A and B.
If (AB), (αB), (Aβ) and (αβ) are the four distinct
combination of A, B, α and β then Yules’ coefficient of
association is
(AB)(αβ) − (Aβ).(αB)
Q=
(AB)(αβ) + (Aβ).(αB)
Note:
I. If Q = +1 there is perfect positive association
If Q = 1 there is perfect negative association
If Q = 0 there is no association (ie) A and B are independent
1. For rememberance of the above formula , we use the table
below
A
α
B AB αB β Aβ αβ Example 6:
Investigate the association between darkness of eye colour
in father and son from the following data.
Fathers’ with dark eyes and sons’ with dark eyes = 50
Fathers’ with dark eyes an sons’ with no dark eyes = 79
Fathers’ with no dark eyes and sons with dark eyes = 89
Neither son nor father having dark eyes
= 782
Solution:
Let A denote the dark eye colour of father and B denote
dark eye colour of son.
A
Total
α
B
50
89
139
79
782
861
β
Total
129
871
1000
241 Yules’ coefficient of association is
(AB)(αβ) − (Aβ).(αB)
Q=
(AB)(αβ) + (Aβ).(αB)
50 × 782 − 79 × 89
=
50 × 782 + 79 × 89
32069
=
= 0.69
46131
∴ there is a positive association between the eye colour of fathers’
and sons’ .
Example 7 :
Can vaccination be regarded as a preventive measure of
small pox from the data given below.
Of 1482 persons in a locality, exposed to small pox, 368 in
all were attacked, among the 1482 persons 343 had been vaccinated
among these only 35 were attacked.
Solution:
Let A denote the attribute of vaccination and B denote that
of attacked.
A
Total
α
B
35
333
368
308
806
1114
β
Total
343
1139
1482
Yules’ coefficient of association is
(AB)(αβ) − (Aβ).(αB)
Q=
(AB)(αβ) + (Aβ).(αB)
35 × 806 − 308 × 333
=
35 × 806 + 308 × 333
−74354
=
= − 0.57
130774
i.e., there is a negative association between attacked and
vaccinated. In other words there is a positive association between
not attacked and vaccinated. Hence vaccination can be regarded as
a preventive measure for small pox.
242 Example 8:
In a coeducational institution, out of 200 students, 150
were boys. They took an examination and it was found that 120
passed, 10 girls failed. Is there any association between sex and
success in the examination.
Solution:
Let A denote boys and α denote girls. Let B denote those
who passed the examination and β denote those who failed.
We have given N = 200 (A) = 150 (AB) = 120
(αβ) = 10
Other frequencies can be obtained from the following table
α
40
10
50 A
120
30
150 B
β
Total Total
160
40
200 Yule’ s coefficient of association is
(AB)(αβ) − (Aβ).(αB)
Q=
(AB)(αβ) + (Aβ).(αB)
120 ×10 − 30 × 40
=
=0
120 × 10 + 30 × 40
Therefore, there is no association between sex and success
in the examination.
Recall
(A) (B) denote positive attributes
(α) (β) denote negative attributes
2 ×2 contingency table.
X
A
Total
α
B (AB) (αB) (B) β (Aβ) (αβ) (β ) Total (A) (α) N 243 Vertical Total
(AB) + (Aβ) = (A)
(αB) + (αβ) = (α)
(A) + (α) = N
Types of Association Horizontal Total
(AB) + (αB) = B
(Aβ) + (αβ) = β
(B) + (β) = N (A).(B)
N
(A).(B)
Negative Association if (AB) <
N
(A).(B)
Independent if (AB) =
N
Yule’ s coefficient of Association
(AB)(αβ) − (Aβ).(αB)
Q=
(AB)(αβ) + (Aβ).(αB)
Positive Association if (AB) > Exercise – 9
I. Choose the best answer:
1. Measures of association in usually deal with
(a) Attributes
(b) Quantitative factors
(c) Variables
(d) Numbers
2. The frequency of class can always be expressed as a sum of
frequencies of
(a) Lower order classes
(b) Higher order classes
(c) Zero order classes
(d) None of the above
3. With the two attributes the total number of class frequencies
is
(a) Two
(b) Four
(c) Eight
(d) Nine
(A).(B)
4. If for two the attributes are A and B, (AB) >
the
N
attributes are
(a) Independent
(b) Positively associated
(c) Negatively associated
(d) No conclusion
244 5. In case of two attributes A and B the class frequency
(AB) = 0 the value of Q is
(a) 1
(b) − 1
(c) 0
(d) −1 ≤ Q ≤ 1
II. Fill in the blanks:
6. If an attribute has two classes it is said to be ____________
7. In case of consistent data, no class frequency can be
_________
8. If A and B are independent Yule’ s coefficient is equal to
________
9. If A and B are negatively associated then __________
10. If N = 500, (A) = 300, (B) = 250 and (AB) = 40 the data are
________
III. Answer the following:
11. Give a brief idea of notations used in classification of
attributes
12. How can the frequencies for various attributes be displayed
in contingency table
13. What do you understand by consistency of data.
14. Write briefly about association of attributes.
15. Give Yule’ s coefficient of association
IV. Problems
16. For two attributes A and B, we have (AB) = 35, (A) = 55;
N=100 and (B) = 65. Calculate the missing values.
17. From the following ultimate class frequencies, find the
frequencies of positive and negative classes and the total
number of observations. (AB) = 9, (Aβ) = 14, (αB) = 4 and
(αβ) = 37
18. Verify whether the given data N = 100, (A) = 75, (B) = 60
and (AB) = 15 are consistent.
19. Find whether A and B are independent in the following data
(AB) = 256 (αB) = 768
(Aβ) = 48
(αβ) = 144
20. In a report on consumer’ s preference it was given that out of
500 persons surveyed 410 preferred variety A 380 preferred
245 variety B and 270 persons linked both. Are the data
consistent?
21. For two attributes A and B, we have (AB) = 35, (A) = 55,
N=100, (αβ) = 20. Calculate the Yule’ s coefficient of
association.
22. Given N = 1500, (A) = 383, (B) = 360 and (AB) = 35.
Prepare 2 × 2 contingency table and compute Yule’ s coefficient of association and interpret the result.
23. In an experiment on immunization of cattle from
tuberculosis the following results were obtained.
Affected
Unaffected
Inoculated
12
26
Not inoculated
16
6
By calculating Yule’ s coefficient of association, examine
the effect of vaccine is in controlling the disease.
24. Calculate the coefficient of association between the
intelligence of fathers and sons from the following data
Intelligent fathers with intelligent sons = 300
Intelligent fathers with dull sons
= 100
Dull fathers with intelligent sons
= 50
Dull fathers with dull sons
= 500
25. Out of 3000 unskilled workers of a factory, 2000 come
from rural area and out of 1200 skilled workers 300 come
from rural area. Determine the association between skill and
residence
26. In an antimalarial campaign in a certain area, quinine was
administrated to 812 persons out of a total population of
3428. The number of fever cases is shown below:
Treatment
Quinine
No quinine Fever
20
220 No Fever
792
2216 Examine the effect of quinine on controlling malaria.
27. 1500 candidates appeared for competitive examinations 425
were successful. 250 had attended a coaching class and of
246 these 150 came out successful. Estimate the utility of the
coaching class.
28. In an examination at which 600 candidates appeared of
them 348 were boys. Number of passed candidates
exceeded the number of failed candidates by 310. Boys
failing in the examination numbered 88. Find the coefficient of association between male sex and success in
examination.
29. Following data relate to literacy and unemployment in a
group of 500 persons. Calculate Yule’ s coefficient of
association between literacy and unemployment and
interpret it
Literate unemployed = 220
Literate employed
= 20
Illiterate Employed = 180
30. In a group of 400 students, the number of married is 160.
Out of 120 students who failed 48 belonged to the married
group. Find out whether the attributes of marriage and
failure are independent.
Answers
I.
1. (a) 2. (b) 3. (d) II.
6. Dichotomy
(A).(B)
9. AB <
N 4. (b) 7. Negative 8. 0 10. Inconsistent IV.
16
B
β
Total A
35
20
55 α
30
15
45
247 5. (b) Total
65
35
100 17.
A
α
B
9
4
14
37
β
Total
23
41
Total No of observations = 64 Total
13
51
64 18. Inconsistent
19. A and B are independent
20. Inconsistent
21. 0.167
22. – 0.606, Negative association
23. − 0.705, Vaccine is effective
24. + 0.935
25. Negative association between skill and residence.
26. – 0.59. Negative association ∴quinine is effective.
27. + 0.68. Coaching class are useful
28. – 0.07
29. 0.92 Positive association between literacy and unemployment
30. Q = 0, Marriage and failure are independent. 248 10. DECISION THEORY
10. 0 Introduction:
Decision theory is primarily concerned with helping people
and organizations in making decisions. It provides a meaningful
conceptual frame work for important decision making. The
decision making refers to the selection of an act from amongst
various alternatives, the one which is judged to be the best under
given circumstances.
The management has to consider phases like planning,
organization, direction, command and control. While performing so
many activities, the management has to face many situations from
which the best choice is to be taken. This choice making is
technically termed as “decision making” or decision taking. A
decision is simply a selection from two or more courses of action.
Decision making may be defined as  “ a process of best selection
from a set of alternative courses of action, that course of action
which is supposed to meet objectives upto satisfaction of the
decision maker.
The knowledge of statistical techniques helps to select the
best action. The statistical decision theory refers to an optimal
choice under condition of uncertainty. In this case probability
theory has a vital role, as such, this probability theory will be used
more frequently in the decision making theory under uncertainty
and risk.
The statistical decision theory tries to reveal the logical
structure of the problem into alternative action, states of nature,
possible outcomes and likely payoffs from each such outcome. Let
us explain the concepts associated with the decision theory
approach to problem solving.
The decision maker:
The decision maker refers to individual or a group of
individual responsible for making the choice of an appropriate
course of action amongst the available courses of action.
249 Acts (or courses of action):
Decision making problems deals with the selection of a
single act from a set of alternative acts. If two or more alternative
courses of action occur in a problem, then decision making is
necessary to select only one course of action.
Let the acts or action be a1, a2, a3,…then the totality of all
these actions is known as action space denoted by A. For three
actions a1, a2 a3; A = action space = (a1, a2, a3) or A = (A1, A2, A3).
Acts may be also represented in the following matrix form
i.e., either in row or column was
Acts
A1 Acts A1 A2 … An A2
.
.
An
In a tree diagram the acts or actions are shown as
A1
Start A2
A3 Events (or States of nature):
The events identify the occurrences, which are outside of
the decision maker’ s control and which determine the level of
success for a given act. These events are often called ‘ States of
nature’ or outcomes. An example of an event or states of nature is
the level of market demand for a particular item during a stipulated
time period.
A set of states of nature may be represented in any one of
the following ways:
250 S = {S1, S2, … n}
,S
or E = {E1, E2, … n}
,E
or = {θ1, θ2, θ3}
For example, if a washing powder is marketed, it may be
highly liked by outcomes (outcome θ1) or it may not appeal at all
(outcome θ2) or it may satisfy only a small fraction, say 25%
(outcome θ3)
∴ = { θ1 , θ2, θ3}
In a tree diagram the places are next to acts. We may also
get another act on the happening of events as follows:
Acts Events
E1 A1 E2
E1 A2 E2
E1
A3 E2
In matrix form, they may be represented as either of the two
ways: States of nature
S1
Acts
A1
A2
OR
251 S2 Acts
A1 A2,… n
,A
States of nature
S1
S2
10.1 Payoff:
The result of combinations of an act with each of the states
of nature is the outcome and momentary gain or loss of each such
outcome is the payoff. This means that the expression payoff
should be in quantitative form.
Pay off may be also in terms of cost saving or time saving.
In general, if there are k alternatives and n states of nature, there
will be k × n outcomes or payoffs. These k × n payoffs can be
very conveniently represented in the form of a k × n pay off table.
States of nature
E1
E2
.
.
.
En A1
a11
a21
.
.
.
an1 Decision alternative
A2 ……… Ak
……
a12 ……… a1k
……
a22 ……… a2k
……
. ……… .
……
. ……… .
……
. .……… .
……
an2 ……… ank
…… where aij = conditional outcome (payoff) of the ith event when jth
alternative is chosen. The above payoff table is called payoff
matrix.
For example,
A farmer can raise any one of three crops on his field. The
yields of each crop depend on weather conditions. We have to show
pay –off in each case, if prices of the three products are as indicated
in the last column of yield matrix.
252 Dry
(E1) Damp
(E3) Price
Rs.per .
kg Paddy
(A1) 500 1700 4500 1.25 Gound
nut (A2) 800 1200 1000 4.00 Tobacco
(A3) Yield in
kg per
hectare Weather
Moderate
(E2) 100 300 200 15.00 Solution:
Pay  off Table
E1 E2 E3 A1 500 ×1.25 = 625 1700 ×1.25= 2125 4500 ×1.25 = 5625 A2 800 × 4 = 3200 1200 ×4 = 4800 1000 × 4 A3 100 × 15 = 1500 300 × 15 = 4500 200 × 15 = 3000 = 4000 10.1.1 Regret (or Opportunity Loss):
The difference between the highest possible profit for a
state of nature and the actual profit obtained for the particular
action taken is known as opportunity loss. That is an opportunity
loss is the loss incurred due to failure of not adopting the best
possible course of action. Opportunity losses are calculated
separately for each state of nature. For a given state of nature the
opportunity loss of possible course of action is the difference
between the payoff value for that course of action and the payoff
for the best possible course of action that could have been selected.
Let the payoff of the outcomes in the 1st row be
P11,P12…… 1n and similarly for the other rows.
…P
253 Payoff table
States of nature
S1
S2 ……Sn
…
P11 P12 …… 1n
…P
P21 P22 ……P2n
…
.
.
. Acts
A1
A2
.
.
.
.
.
.
.
.
.
Am
Pm1 Pm2 …… mn
…P
Consider a fixed state of nature Si. The payoff
corresponding to the n strategies are given by Pi1, Pi2,… in. Suppose
,P
Mi is the maximum of these quantities. The Pi1 if A1 is used by the
decision maker there is loss of opportunity of M1 – Pi1, and so on
Then a table showing opportunity loss can be computed as
follows:
Regret (or opportunity loss table)
States of nature
Acts
S1
S2
… Sn
A1
M1 P11 M2  P12 …Mn  P1n
M1 P21 M2  P22 … Mn  P2n
A2
.
.
.
.
.
.
.
.
.
.
.
.
Am
M1Pm1 M2  Pm2 … Mn  Pmn
Types of decision making:
Decisions are made based upon the information data
available about the occurrence of events as well as the decision
situation. There are three types of decision making situations:
certainty , uncertainty and risk.
Decision making under certainty:
In this case the decision maker has the complete knowledge
of consequence of every decision choice with certainty. In this
254 decision model, assumed certainty means that only one possible
state of nature exists.
Example 1:
A canteen prepares a food at a total average cost of Rs. 4
per plate and sells it at a price of Rs 6. The food is prepared in the
morning and is sold during the same day. Unsold food during the
same day is spoiled and is to be thrown away. According to the past
sale, number of plates prepared is not less than 50 or greater than
53. You are to formulate the (i) action space (ii) states of nature
space (iii) payoff table (iv) loss table
Solution:
(i) The canteen will not prepare less than 50 plates or more than 53
plates. Thus the acts or courses of action open to him are
a1 = prepare 50 plates
a2 = prepare 51 plates
a3 = prepare 52 plates
a4 = prepare 53 plates
Thus the action space is
A = {a1, a2 , a3, a4}
(ii) The state of nature is daily demand for food plates. Then are
four possible state of nature ie
S1 = demand is 50 plates
S2 = demand is 51 plates
S3 = demand is 52 plates
S4 = demand is 53 plates
Hence the state of nature space, S= {S1, S2, S3, S4}
iii) The uncertainty element in the given problem is the daily
demand. The profit of the canteen is subject to the daily demand.
Let
n = quantity demanded
m = quantity produced
For n ≥ m, profit = (Cost price – Selling price) x m
= (6 – 4) x m = 2m
For m > n,
255 profit ={(Cost price – Selling price) x n} {Cost price x (mn) }
= 2n – 4 (mn) = 6n – 4m Supply (m)
( a1 )
( a2 )
( a3 )
( a4 ) 50
51
52
53 (S1)
50
100
96
92
88 Payoff table
Demand (n)
(S2)
(S3)
51
52
100
100
102
102
98
104
94
100 (S4)
53
100
102
104
106 (iv) To calculate the opportunity loss we first determine the
maximum payoff in each state of nature. In this state
First maximum payoff
= 100
Second maximum payoff = 102
Third maximum payoff = 104
Fourth maximum payoff = 106
Loss table corresponding to the above payoff table
Demand (n)
(S1)
(S2)
(S3)
(S4)
Supply
50
51
52
53
(m)
(a1) 50 100 100 = 0 102100 = 2 104100 = 4 106 100 = 6 (a2) 51 100  96 = 4 102102 = 0 104102 = 2 106 102 = 4 (a3) 52 100  92 = 8 102  98 = 4 104104 = 0 106 104 = 2 (a4) 53 100  88 =12 102  94 = 8 104100 = 4 106 106 = 0 10.2 Decision making under uncertainty (without probability):
Under conditions of uncertainty, only payoffs are known
and nothing is known about the lilkelihood of each state of nature.
Such situations arise when a new product is introduced in the
256 market or a new plant is set up. The number of different decision
criteria available under the condition of uncertainty is given below.
Certain of optimism (Maximax ):
The maximax criterion finds the course of action or
alternative strategy that maximizes the maximum payoff. Since
this decision criterion locates the alternative with the highest
possible gain, it has also been called an optimistic decision
criterion. The working method is
(i) Determine the best outcome for each alternative.
(ii) Select the alternative associated with the best of these.
Expected Monetary value (EMV):
The expected monetary value is widely used to evaluate the
alternative course of action (or act). The EMV for given course of
action is just sum of possible payoff of the alternative each
weighted by the probability of that payoff occurring.
The criteria of pessimism or Maximin:
This criterion is the decision to take the course of action
which maximizes the minimum possible payoff. Since this
decision criterion locates the alternative strategy that has the least
possible loss, it is also known as a pessimistic decision criterion.
The working method is:
1) Determine the lowest outcome for each alternative.
2) Choose the alternative associated with the best of these.
Minimax Regret Criterion (Savage Criterion):
This criterion is also known as opportunity loss decision
criterion because decision maker feels regret after adopting a wrong
course of action (or alternative) resulting in an opportunity loss of
payoff. Thus he always intends to minimize this regret. The
working method is
(a) Form the given payoff matrix, develop an opportunity loss
(or regret) matrix.
(i)
find the best payoff corresponding to each state
of nature and
(ii)
subtract all other entries (payoff values) in that
row from this value.
257 (b) Identify the maximum opportunity loss for each
alternatives.
(c) Select the alternative associated with the lowest of these.
Equally likely decision (Baye’ s or Laplace)Criterion:
Since the probabilities of states of nature are not known, it
is assumed that all states of nature will occur with equal
probability. ie., each state of nature is assigned an equal probability.
As states of nature are mutually exclusive and collectively
exhaustive, so the probability of each these must be 1 /(number of
states of nature). The working method is
(a) Assign equal probability value to each state of nature by
using the formula:
1/(number of states of nature)
(b) Compute the expected (or average) value for each
alternative by multiplying each outcome by its probability
and then summing.
(c) Select the best expected payoff value (maximum for profit
and minimum for loss)
This criterion is also known as the criterion of insufficient reason
because, expect in a few cases, some information of the likelihood
of occurrence of states of nature is available.
Criterion of Realism (Hurwicz Criterion):
This criterion is a compromise between an optimistic and
pessimistic decision criterion. To start with a coefficient of
optimism α (0 ≤ α ≤ 1) is selected.
When α is close to one, the decision maker is optimistic
about the future and when α is close to zero, the decision maker is
pessimistic about the future.
According to Hurwicz , select strategy which maximizes
H = α (maximum payoff in row) + (1  α) minimum payoff in
row. Example 2:
Consider the following payoff (profit) matrix
258 States Action (S1)
(S2)
(S3)
(S4)
A1
5
10
18
25
A2
8
7
8
23
A3
21
18
12
21
A4
30
22
19
15
No Probabilities are known for the occurrence of the nature states .
Compare the solutions obtained by each of the following criteria:
(i) Maximin (ii) Laplace (iii) Hurwicz (assume that α = 0.5)
Solution:
i) Maximin Criterion:
A1:
5
10
A2:
8
7
A3:
21
18
A4:
30
22
Best action is A4
ii) Laplace criterion 18
8
12
19 Minimum
5
7
12
15 maximum 25
23
21
15 E(A1) = 1/4 [5 +10+18+25] = 14.5
E(A2) = 1/4 [8 +7+8+23]
= 11.5
E(A3) = 1/4 [21 +18+12+21] = 18.0
E(A4) = 1/4 [30 +22+19+15] = 21.5 maximum
E(A4) is maximum. So the best action is A4
iv) Hurwicz Criterion (with α = 0.5)
Minimum Maximum α (max) + (1α) min
A1 5 25 0.5(25) + 0.5(5) A2 7 23 0.5(7) + 0.5 (23) = 15 A3 12 21 0.5(12) + 0.5 (21) = 16.5 A4 15 30 0.5(15) + 0.5 (30) = 22.5 maximum Best action is A4
259 = 15 Example 3:
Suppose that a decision maker faced with three decision
alternatives and two states of nature. Apply (i) Maximin and (ii)
Minimax regret approach to the following payoff table to
recommend the decisions.
States of Nature
Act
A1
A2
A3 S1 S2 10
20
30 15
12
11 Solution:
(i) Maximin
Act
Minimum
A1
10
12 maximum
A2
A3
11
Act A2 is recommended
ii) Minimax regret
States of Nature
S1
S2
Maximum
Regret
Act
A1
3010 = 20
1515 = 0
20
A2
3020 = 10
1512 = 3
10
A3
3030 = 0
1511 = 4
4
Minimum of the maximum regrets is 4 which corresponds
to the act A3. So the act A3 is recommended
Example 4:
A business man has to select three alternatives open to him
each of which can be followed by any of the four possible events.
The conditional payoff (in Rs) for each action event combination
are given below:
260 Alternative
X
Y
Z A
8
4
14 Payoffs conditional events
B
C
0
10
12
18
6
0 D
6
2
8 Determine which alternative should the businessman
choose, if he adopts the
a) Maximin criterion
b) Maximax criterion
c) Hurwicz criterion with degree of optimism is 0.7
d) Minimax regret Criterion
e) Laplace criterion
Solution:
For the given payoff martrix, the maximum assured and
minimum possible payoff for each alternative are as given below.
Minimum
(α =0.7)
Alternative Maximum
payoff
payoff
H = α (maximum payoff)
(Rs)
(Rs)
+ (1 α)(minimum payoff)
X
8
10
2.6
Y
18
4
11.4
Z
14
0
9.8
a) Since z yields the maximum of the minimum payoff, under
maximin criterion, alternative z would be chosen.
b) Under maximax criterion, the businessman would choose
the alternative Y.
c) It will be optimal to choose Y under Hurwicz Criterion.
d) For the given payoff matrix, we determine the regrets as
shown below, when the regret payoffs amounts when event
A occurs, are computed by the relation
Regret payoff = maximum payoffs from A – payoff.
Similarly for the other events. 261 Alternative
X
Y
Z
Maximum
payoff Pay off
amount
AB
C
8 0 10
 4 12 18
14 6
0
14 12 18 D Regret payoff
amount
A
BCD 66
 2 18
8
0
8 12 28
00
6 18 2
10
0 Maximum
Regret
28
18
18 Since alternative Y and Z both corresponding to the
minimal of the maximum possible regrets, the decision maker
would choose either of these two
(e) Laplace Criterion
In this method assigning equal probabilities to the payoff
of each strategy, results in the following expected payoff.
Payoff
Expected payoff value
Alternative
A
B
CD
P=1/4 P=1/4 P=1/4 P=1/4 X
¼[ 8 + 0 –10 + 6] = 1
8 0 −10 6
Y
 4 12 18  2
¼[ 4 + 12 +18 2] = 6
Z
14 6
0
8
¼[ 14 + 6 +0 + 8] = 7
Since the expected payoff value for z is the maximum the
businessman would choose alternative z
10.3 Decision making under risk (with probability):
Here the decision maker faces many states of nature. As
such, he is supposed to believe authentic information, knowledge,
past experience or happenings to enable him to assign probability
values to the likelihood of occurrence of each state of nature.
Sometimes with reference to past records, experience or
information, probabilities to future events could be allotted. On the
basis of probability distribution of the states of nature, one may
select the best course of action having the highest expected payoff
value.
262 Example 5:
The payoff table for three courses of action (A) with three
states of nature (E) (or events) with their respective probabilities (p)
is given. Find the best course of action.
Events
Probability
Acts
A1
A2
A3 E1 E2 E3 0.2 0.5 0.3 2
3
4 1
2
2 1
0
1 The expected value for each act is
A1 : 2(0.2) + 1(0.5)  1(0.3) = 0.6
A2 : 3(0.2) + 2(0.5) + 0(0.3) = 1.6
A3 : 4(0.2) + 2(0.5) + 1(0.3) = 2.1
The expected monetary value for the act 3 is maximum. Therefore
the best course of action is A3.
Example 6:
Given the following payoff of 3 acts: A1, A2, A3 and their
events E1, E2, E3.
Act
A1
A2
A3
States of Nature
E1
E2
E3 35
200
550 10
240
640 150
200
750 The probabilities of the states of nature are respectively 0.3,
0.4 and 0.3. Calculate and tabulate EMV and conclude which of the
acts can be chosen as the best.
263 Solution:
Events Prob.
A1
A2
A3
35 × 0.3 = 10.5 10 × 0.3 = 3 150 × 0.3 =  45
E1
0.3
200 × 0.4 = 80.0 240 × 0.4 = 96 200 × 0.4 = 80
E2
0.4
E3
0.3 550 ×0.3 = 165.0 640 × 0.3 = 192 750 × 0.3 = 225
285
260
255.5
EMV
The EMV of A2 is maximum, therefore to choose A2
Example 7:
A shop keeper has the facility to store a large number of
perishable items. He buys them at a rate of Rs.3 per item and sells
at the rate of Rs.5 per item. If an item is not sold at the end of the
day then there is a loss of Rs.3 per item. The daily demand has the
following probability distribution.
Number of
3
4
5
6
Items demanded
Probability
:
0.2
0.3
0.3
0.2
How many items should he stored so that his daily expected profit
is maximum?
Solution:
Let
m = number of items stocked daily
n = number of items demanded daily
Now, for n ≥ m, profit = 2m
And for m > n,
profit = 2n – 3(mn)
= 2n – 3m + 3n = 5n – 3m Stock (m)
3
4
5
6
Probability Pay  off table
Demand (n)
3
4
5
6
3
0
−3
0.2 6
8
5
2
0.3
264 6
8
10
7
0.3 6
6
8
10
12
0.2 Stock(m)
3
4
5
6
Thus
stocked. So,
daily. Expected gain
6 × 0.2 + 6 × 0.3 + 6 × 0.3 + 6 × 0.2 = Rs. 6.00
3 × 0.2 + 8 × 0.3 + 8 × 0.3 + 8 × 0.2 = Rs. 7.00
0 × 0.2 + 5 × 0.3 + 10 × 0.3 + 10 × 0.2 = Rs. 6.50
−3 × 0.2 + 2 × 0.3 + 7 × 0.3 + 12 × 0.2 = Rs. 4.50
the highest expected gain is Rs 7.00 when 4 units
he can store 4 items to get maximum expected profit Example 8:
A magazine distributor assigns probabilities to the demand
for a magazine as follows:
Copies demanded :
2
3
4
5
Probability
: 0.4
0.3
0.2
0.1
A copy of magazine which he sells at Rs.8 costs Rs6. How many
should he stock to get the maximum possible expected profit if the
distributor can return back unsold copies for Rs.5 each?
Solution:
Let m = no of magazines stocked daily
n = no of magazines demanded
Now,
For n ≥ m, profit = Rs 2m
and for m > n, profit = 8n –6m +5(mn)
= 8n –6m +5m – 5n
= 3n –m Stock (m) Payoff table
Demand (n)
3
4
4
4 2 2
4 5
4 3 3 6 6 6 4 2 5 8 8 5 1 4
265 7 10 Probability
Stock
2
3
4
5 0.4 0.3
0.2
0.1
Expected Profit (in Rs)
4 × 0.4 + 4 × 0.3 + 4 × 0.2 + 4 × 0.1 = 4.0
3 × 0.4 + 6 × 0.3 + 6 × 0.2 + 6 × 0.1 = 4.8
2 × 0.4 + 5 × 0.3 + 8 × 0.2 + 8 × 0.1 = 4.7
1 × 0.4 + 4 × 0.3 + 7 × 0.2 + 10 ×0.1= 4.0 Thus the highest expected profit is Rs. 4.8, when 3
magazines stocked. So, the distributor can stock 3 magazines to get
the maximum possible expected profit.
10.4 Decision Tree Analysis:
A decision problem may also be represented with the help
of a diagram. It shows all the possible courses of action, states of
nature, and the probabilities associated with the states of nature.
The ‘ decision diagram’ looks very much like a drawing of a tree,
therefore also called ‘ decision tree’ .
A decision tree consists of nodes, branches, probability
estimates and payoffs. Nodes are of two types, decision node
(designated as a square) and chance node (designated as a circle).
Alternative courses of action originate from decision node as the
main branches (decision branches). Now at the terminal point of
decision node, chance node exists from where chance nodes,
emanate as subbranches. The respective payoffs and the
probabilities associated with alternative courses, and the chance
events are shown alongside the chance branches. At the terminal of
the chance branches are shown the expected payoff values of the
outcome.
There are basically two types of decision treesdeterministic
and probabilistic. These can further be divided into single stage and
multistage trees. A single stage deterministic decision tree involves
making only one decision under conditions of certainty (no chance
events). In a multistage deterministic tree a sequence or chain of
decisions are to be made, The optimal path (strategy) is one that
corresponds to the maximum EMV.
266 In drawing a decision tree, one must follow certain basic rules and
conventions as stated below:
1. Identify all decisions (and their alternatives) to be made and
the order in which they must be made.
2. Identify the chance events or state of nature that might
occur as a result of each decision alternative.
3. Develop a tree diagram showing the sequence of decisions
and chance events. The tree is constructed starting from left
and moving towards right. The square box
denotes a
decision point at which the available courses of action are
considered. The circle O represents the chance node or
event, the various states of nature or outcomes emanate
from this chance event.
4. Estimate probabilities that possible events or states of nature
will occur as a result of the decision alternatives.
5. Obtain outcomes (usually expressed in economic terms) of
the possible interactions among decision alternatives and
events.
6. Calculate the expected value of all possible decision
alternatives.
7. Select the decision alternative (or course of action) offering
the most attractive expected value
Advantages of decision tree:
1. By drawing of decision tree, the decision maker will be in a
position to visualise the entire complex of the problem.
2. Enable the decision  maker to see the various elements of
his problem in content and in a systematic way.
3. Multidimensional decision sequences can be strung on a
decision tree without conceptual difficulties.
4. Decision tree model can be applied in various fields such as
introduction of a new product, marketing strategy etc…
Example 9:
A manufacturing company has to select one of the two
products A or B for manufacturing. Product A requires investment
267 of Rs.20,000 and product B Rs 40,000. Market research survey
shows high, medium and low demands with corresponding
probabilities and returns from sales in Rs. Thousand for the two
products in the following table.
Market demand
High
Medium
Low Probability
A
0.4
0.3
0.3 B
0.3
0.5
0.2 Return from sales
A
B
50
80
30
60
10
50 Construct an appropriate decision tree. What decision the company
should take? Market demand
High
Medium
Low
Total A
X(‘ 000) P
50
0.4
30
0.3
10
0.3
268 PX X(‘ 000)
20 80
9 60
3 50
32 B
P
0.3
0.5
0.2 PX
24
30
10
64 Product
Return (Rs)
Investment(Rs)
Profit (Rs)
A
32,000
20,000
12,000
B
64,000
40,000
24,000
Since the profit is high in case of product B, so the company’ s
decision in favour of B.
Example 10:
A farm owner is considering drilling a farm well. In the
past, only 70% of wells drilled were successful at 20 metres of
depth in that area. Moreover, on finding no water at 20 metres,
some person drilled it further up to 25 metres but only 20% struck
water at 25 metres. The prevailing cost of drilling is Rs.500 per
metres. The farm owner has estimated that in case he does not get
his own well, he will have to pay Rs.15000 over the next 10 years
to buy water from the neighbour.
Draw an appropriate decision tree and determine the farm owner’ s
strategy under EMV approach.
Solution:
The given data is represented by the following decision tree
diagram. Decision Event 1. Drill
upto 25
metres Water
struck
No water
struck Probability Cash out
flows Decision at point D2
0.2
Rs.12500 0.8 Expected
cash out
flow
Rs 2500 Rs 22000 EMV(out
flows)
269 Rs27500 Rs.24500 2. Do not
drill EMV (out flow) = Rs.25000 The decision at D2 is : Drill up to 25 metres
Decision at point D1
0.7
Rs.10000
Rs 7000
Water
1. Drill
upto 20 struck
metres
0.3
Rs.24500
Rs.7350
No water
struck
EMV(out
Rs.14,350
flows)
2. Do not EMV (out flow) = Rs.15000
drill
The decision at D1 is : Drill up to 20 metres.
Thus the optimal strategy for the farmowner is to drill the well up
to 20 metres. Exercise – 10
I. Choose the correct answers:
1. Decision theory is concerned with
(a) The amount of information that is available
(b) Criteria for measuring the ‘ goodness’ of a decision
(c) Selecting optimal decisions in sequential problems
(d) All of the above
2. Which of the following criteria does not apply to decision –
making under uncertainly
(a) Maximin return
(b) Maximax return
(c) Minimax return
(d) Maximize expected return
3. Maximin return, maximax return and minimax regret are
criteria that
(a) Lead to the same optimal decision.
(b) Cannot be used with probabilities
(c) Both a and b
(d) None of the above
270 4. Which of the following does not apply to a decision tree?
(a) A square node is a point at which a decision must be
made.
(b) A circular node represents an encounter with
uncertainty.
(c) One chooses a sequence of decisions which have the
greatest probability of success.
(d) One attempts to maximize expected return.
5. The criterion which selects the action for which maximum
payoff is lowest is known as
(a) Maxmin criterion
(b) Minmax criterion
(c) Max –max criterion
(d) None of these
II. Fill in the blanks:
6. Decision trees involve ____________ of decisions and
random outcomes.
7. One way to deal with decision making in the ‘ uncertainity’
context is to treat all states of nature as ___________ and
maximize expected return.
8. Maximizing expected net rupee return always yields the
same optimal policy as ___________ expected regret.
9. The different criteria for making decisions under risk
always yields the same _________ choice.
10. In decision under uncertainty, the Laplace criterion is the
least conservative while the ___________ criterion is the
most conservative.
III. Answer the following:
11. Explain the meaning of ‘ statistical decision theory’
12. What techniques are used to solve decision making
problems under uncertainty?
13. Write a note on decision tree.
14. What is a payoff matrix?
15. Describe how you would determine the best decision using
the EMV criterion with a decision tree.
271 IV. Problems:
16. The payoff table for three courses of action (A) with three
states of nature (E) (or events) with their respective
probabilities (P) are given. Find the best course of action.
Events
Acts
A1
A2
A3
Probability E1 E2 E3 2.5
4.0
3.0
0.2 2.0
2.6
1.8
0.6 −1
0
1
0.2 17. Calculate EMV and thus select the best act for the following
payoff table:
States of
nature
X
Y
Z Probability Payoff (Rs) by the player
A
B
C
0.3
20
−2
−5
0.4
20
−10
−5
0.3
40
60
30 18. Consider the payoff matrix
States of
nature Probability High
demand
Medium
demand
Low
demand 0.4 Act A1 do Act A2
Expand
not
200 units
expand
2500
3500 Act A3
Expand
400 units
5000 0.4 2500 3500 2500 0.2 2500 1500 1000 272 Using EMV criterion decide the best act.
19. Apply (i) maximin (ii) minimax regret to the following
payoff matrix to recommended the decisions without any
knowledge of probability.
States of nature
Act
S1
S2
S3
a1
14
8
10
a2
11
10
7
a3
9
12
13
20. A shop keeper of some highly perishable type of fruits sees
that the daily demand X of this fruit in his area the
following probability distribution.
Daily Demand
(in Dozen)
:
6
7
8
9
Probability
:
0.1
0.3
0.4
0.2
He sells for Rs10.00 a dozen while he buys each dozen at
Rs4.00. Unsold fruits in a day are traded on the next day at
Rs.2.00 per dozen, assuming that the stocks the fruits in dozen,
how many should he stock so that his expected profit will be
maximum?
[Hint: profit = 6m for n ≥ m
= 10n – 4m +2(mn)
= 8n – 2m for n< m]
21. A florist, in order to satisfy the needs of a number of
regular and sophisticated customers, stocks a highly
perishable flowers. A dozen flowers cost Rs 3 and sell at
Rs10.00 Any flower not sold on the day are worthless.
Demand distribution in dozen of flowers is as follows:
Demand
1
2
3
4
Probability 0.2 0.3
0.3 0.2
How many flowers should he stock daily in order to maximize
his expected net profit?
273 22. A florist stock highly perishable flower. A dozen of flower
costs Rs3.00 and sells for Rs.10.00 Any flower not sold the
day are worthless. Demand in dozen of flowers is as
follows:
Demand in 0
1
2
3
4
dozen
Probability 0.1 0.2
0.4 0.2
0.1
Assuming that failure to satisfy any one customer’ s request
will result in future lost profit amounting to Rs.5.00, in addition
to the lost profit on the immediate sale, how many flowers
should the florist stock to expect maximum profit?
23. A newspaper agent’ s experience shows that the daily
demand x of newspaper in his area has the following
probability distribution
Daily
Demand(x)
Probability 300 400 500 0.1 0.3 0.4 600
0.1 700
0.1 He sales the newspapers for Rs.2.00 each while he buys each at
Rs.1.00. Unsold copies are treated as scrap and each such copy
fetches 10 paisa. Assuming that he stocks the news papers
in multiple of 100 only. How many should he stock so that his
expected profit is maximum?
24. Suppose that a decision maker faced with three decision
alternatives and four states of nature. Given the following
profit payoff table.
States of nature
Acts
S1
S2
S3
S4
a1
16
10
12
7
a2
13
12
9
9
a3
11
14
15
14
Assuming that he has no knowledge of the probabilities of
occurrence of the states of nature, find the decisions to be
recommended under each of the following criteria.
274 (i) maximin (ii) maximax (iii) minimax Regret 25. Payoff of three acts A, B and C and states of nature X, Y
and Z are given below
Payoff (in Rs)
Acts
States of
A
B
C
nature
X
2000
− 20
−50
Y
200
−100
−50
Z
400
600
300
The probabilities of the states of nature are 0.3, 0.4 and 0.3.
Calculate the EMV for the above and select the best art.
Answers:
I.
1. (d)
II.
6. Sequence
9. Optimal 2. (d) 3. (b) 7. equally likely
10. Minimax 4. (c) 5. (a) 8. minimizing IV.
16. A2 is the best
17. Select A with the highest EMV Rs.194
18. EMV: 3200, decide, Act A3, expand 400 units
19. (i) maximin : Act a3
(ii) minimax regret Act a1
20. So the shop keeper should stock 8 dozen of fruits to get
maximum expected profit.
21. He should stock 3 dozen of flowers to get maximum expected
net profit.
22. He stocks 3 dozen of flowers to expect maximum profit Rs.9.50
23. To stock 405 copies so that his expected profit is maximum
24. (i) Act a3 is recommended
275 (ii) Act a1 is recommended
(iii) Act a3 is recommended
25. EMA for A is highest. So the best act is A is selected 276 277 278 279 280 281 282 283 284 285 286 287 288 289 ...
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This note was uploaded on 12/09/2011 for the course BBA ms 08 taught by Professor Sharma during the Spring '06 term at Amity University.
 Spring '06
 Sharma

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