Business-Statistics-munotes

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INTRODUCTION TO STATISTICS
Unit Structure
1.1Meaning
1.2Statistical Methods
1.3Importance ofStatistics
1.4Functions ofStatistics
1.5Limitations ofStatistics
1.6Branches inStatistics
1.7Characteristics ofStatistics
1.8Basic Definitions inStatistics
1.9Exercise
1.1 MEANING :
The word Statistics describes several concepts of importance to decision -
maker. It is important for a beginner to have an understanding of these
different concepts.
STATISTICAL METHODS V/S EXPERIMENTAL METHODS
We try to get the knowledge of any phenom enon through direct
experiment. There may be many factors affecting a certain phenomenon
simultaneously. If we want to study the effect of a particular factor, we
keep other factors fixed and study the effect of only one factor. This is
possible in all exa ct sciences like Physics, Chemistry etc. This method
cannot be used in many sciences, where all the factors cannot be isolated
and controlled. This difficulty is particularly encountered in social
sciences, where we deal with human beings. No two persons a re exactly
alike. Besides the environment also changes and it has its effect on every
human being and therefore it is not possible to study one factor keeping
other conditions fixed. Here we use statistical methods. The results
obtained by the use of this science will not be as accurate as those
obtained by experimental methods. Even then they are of much use and
they have a very important role to play in the modern World. Even in
exact sciences some of the statistical methods are made use of.
The word Sta tistics is derived from the Latin word "statis' which means a
political state. The word Statistics was originally applied to only such facts
and figures that were required by the state for official purposes. The
earliest form of statistical data is related to census of population and
property, through the collection of data for other purposes was not
completely ruled out. The word has now acquired a wider meaning.munotes.in

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2STATISTICS IN PLURAL
Statistics in plural refer to any set of data or information. The preside nt of
a company may call for 'statistics on the sales of northern region' or an MP
may quote the statistics on price -rise in agricultural products. More
familiar examples for the students will be the marks of students in a class,
the ages of children in pr imary school.
Prof. Secrist defines the word 'Statistics' in the first sense as follows"
"ByStatistics we mean aggregate of facts affected to a marked extend by
multiplicity of causes, numerically expressed, enumerated or estimated
according to reasonab le standards of accuracy, collected in a systematic
manner for a predetermined purpose and placed in relation to each other."
This definition gives all the characteristics of Statistics:
i. Aggregate of Facts: A single isolated figure is not 'Statistics. ' Marks of
one student in one subject will not be called Statistics. But, if we consider
the marks of all the students in the class in a particular subject, they will
be called 'Statistics.'
ii. Affected by Multiplicity of causes: There are various causes for the
changes in the data, the marks of the students depend upon, the
intelligence of students, their capacity and desire to work etc.
iii. Numerically expressed: Unless the characteristics have some
numerical measurement they will not be called Statis tics. The statement 'A
student writes very good English' is not Statistics. But if marks of the
whole class in 'English' are given they will be called 'Statistics.'
iv. Enumerated or Estimated according to reasonable standards of
accuracy: However much a person tries, it is not possible to attain perfect
accuracy whether we actually measure or estimate the characteristic. But a
certain standard of accuracy should be set up according to the problem
under consideration. The estimate for the cost of big proje ct may be
correct up to Rs. 1, 000 but for household expenses it should be correct up
to a rupee.
v. Collected in a systematic manner: There should be a method in the
manner of collection, and then only the figures will be reliable and useful.
vi. Collec ted for a predetermined purpose: Unless we know the
purpose, the data collected may not be sufficient. Besides some
unnecessary information may be collected which will be a waste of time
and money.
vii. Placed in relation to each other: Only when we want t o compare
characteristics, which have some relation with each other, we collect
Statistics. The wages of fathers and ages of sons should not be collected
together. But we can have ages and heights of a group of persons, so that
we can find the relation bet ween the two.munotes.in

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31.2 STATISTICAL METHODS
The word Statistics used in the second sense means the set of techniques
and principles for dealing with data.
1.Suppose you have the data about production profits and sales for a
number of years of a company. Sta tistics in this sense is concerned
with questions such as
(i)What is the best way to present these data for review?
(ii)What processing is required to reveal more details about the data?
(iii)What ratios should be obtained and reported?
2.A public ag ency wants to estimate the number of fish in a lake. Five
hundred fish are captured in a net tagged and returned to the lake. One
week later 1, 000 fish are captured from the same lake in nets and 40
are found to be with tags. Here Statistics in this secon d sense deals
with questions such as:
(i)What is a good estimate of the number of fish in the lake?
(ii)What is our confidence in it and how much error can be expected?
and
(iii)Can we have a method, which will make a better estimate?
Statisticians ha ve defined this in various ways. Bowley says, "Statistics
may rightly be called the science of averages." But this definition is not
correct. Statistics has many refined techniques and it does much more than
just averaging the data.
Kendall defines it as, "The branch of scientific methods that deals with the
data obtained by counting or measuring the properties of population of
natural phenomena." This definition does not give the idea about the
functions of Statistics. It is rather vague.
Seligman define s it as, "The science which deals with the methods of
collecting, classifying, presenting, comparing and interpreting numerical
data collected to throw some light on any sphere of inquiry." Croxton,
Cowden and Klein define it as, "The last two definitions can be considered
to be proper which explain the utility of 'statistics'. We will examine the
four procedures mentioned in the definition in brief.
Collection: The day may be collected from various published and
unpublished sources, or the investigator ca n collect his own information.
Collecting first hand information is a very difficult task. The usefulness of
the data collected depends to a very great extent upon the manner in which
they are collected. Though theoretical knowledge is necessary for the
proper collection of data, much can be learnt through experience and
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4Presentation: The data collected, to be understood, should be presented in
a suitable form. Just a given mass of figures signifies nothing to a person
and they can lead only t o confusion. They are usually presented in a
tabular form and represented by diagrams.
Analysis: Presentation of data in a tabular form is one elementary step in
the analysis of the collected data. If we want to compare two series, a
typical value for eac h series is to be calculated. If we want to study some
characteristic of a big group, exhaustive study is not possible. We take a
sample, study it and inferences are drawn on the basis of sample studies.
Sometimes forecasting is necessary. The management o f a firm may be
interested in future sales. For that it has to analyse the past data. We are
going to study some of these methods of analysing the data in this book.
Interpretation: This is the final step in an investigation. Based upon the
analysis of th e data, we draw certain conclusions. While drawing these
conclusions, we must consider that nature of the original data. Experts in
the particular field of activity must make the final interpretation. The
statistical methods are not like experimental metho ds, which are exact. For
interpreting the analysis of the data dealing with some psychological
problems, a psychologist is right person. (An economist, though well
versed in statistical methods will not be of any use there).
STATISTICAL MEASURES
Statistic s also has a precise technical meaning. Measures derived from the
sample data are referred to as Statistics. If only one measure is obtained it
is called a Statistic.
A magazine takes a sample of 100 readers. 15 of them are over 30 years of
age. The sampl e proportion of readers over 30 years of age is 0.15. This
sample proportion is referred to as a statistic obtained by this survey.
The weekly sales for 5 weeks for a salesman are Rs. 2, 000, Rs. 2, 500, Rs.
15, 000, Rs. 3000 and Rs. 1, 800. As a measure of the spread of the values
the difference between the smallest and the largest value (called the range)
is calculated. This range is a statistic.
1.3 IMPORTANCE OF STATISTICS
Statistics is not studied for its own sake. It is employed as a tool to study
the problems in various natural and social sciences. The analysis of data is
used ultimately for forecasting, controlling and exploring.
Statistics is important because it makes the data comprehensible. Without
its use the information collected will hardl y be useful. To understand the
economic condition of any country we must have different economic
aspects quantitatively expressed and properly presented. If we want to
compare any two countries, statistics is to be used. For studyingmunotes.in

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5relationship between t wo phenomena, we have to take the help of
statistics, which explains the correlation between the two.
People in business can study past data and forecast the condition of their
business, so that they can be ready to handle the situations in future.
Nowada ys a businessman has to deal with thousands of employees under
him and cannot have direct control over them. Therefore, he can judge
them all and control their performance using statistical methods e.g., he
can set up certain standards and see whether the final product conforms to
them. He can find out the average production per worker and see whether
any one is giving less, i.e., he is not working properly.
Business must be planned properly and the planning to be fruitful must be
based on the right analys is of complex statistical data. A broker has to
study the pattern in the demand for money by his clients, so that he will
have correct amount of reserves ready.
Scientific research also uses statistical methods. While exploring new
theories, the validity of the theory is to be tested only by using statistical
methods. Even in business many new methods are introduced. Whether
they are really an improvement over the previous ones, can be tested using
statistical techniques.
We can see many more examples fro m almost all sciences, like biology,
physics, economics, psychology and show that statistical methods are used
in all sciences. The point here is that 'Statistics' is not an abstract subject.
It is a practical science and it is very important in the modern World.
1.4FUNCTIONS OF STATISTICS
1. Statistics presents the data in numerical form: Numbers give the
exact idea about any phenomenon. We know that India is overpopulated.
But only when we see the census figure, 548 millions, we have the real
idea abou t the population problem. If we want to compare the speed of
two workmen working in the same factory, with the same type of machine,
we have to see the number of units they turn out every day. Only when we
express the facts with the help of numbers, they a re convincing.
2. It simplifies the complex data: The data collected are complex in
nature. Just by looking at the figures no person can know the real nature of
the problem under consideration. Statistical methods make the data easy to
understand. When we have data about the students making use of the
college library, we can divide the students according to the number of
hours spent in the library. We can also see how many are studying and
how many are sitting there for general reading.
3. It facilitates comparison: We can compare the wage conditions in two
factories by comparing the average wages in the two factories. We can
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6during that period. Such comparisons are very useful in many social
sciences.
4. It studies relationship between two factors: The relationship between
two factors, like, height and weight, food habits and health, smoking and
occurrence of cancer can be studied using statistical techniques. We can
estimate one fact or given the other when there is some relationship
established between two factors.
5. It is useful for forecasting: We are interested in forecasting using the
past data. A shopkeeper may forecast the demand for the goods and store
them when they are easi ly available at a reasonable price. He can store
only the required amount and there will not be any problem of goods
being wasted. A baker estimates the daily demand for bread, and bakes
only that amount so that there will be no problem of leftovers.
6. It helps the formulation of policies: By studying the effect of policies
employed so far by analysing them, using statistical methods, the future
policies can be formulated. The requirements can be studied and policies
can be determined accordingly. The imp ort policy for food can be
determined by studying the population figures, their food habits etc.
1.5LIMITATIONS OF STATISTICS
Though Statistics is a very useful tool for the study of almost all types of
data it has certain limitations.
1. It studies on ly quantitative data: A very serious drawback is that
statistics cannot study qualitative data. Only when we have data expressed
in numerical form we can apply statistical methods for analysing them.
Characteristics like beauty, cruelty, honesty or intelli gence cannot be
studied with the help of statistics. But in some cases we can relate the
characteristics to number and try to study them. Intelligence of students
can be studied by the marks obtained by them in various tests; we can
compare the intelligenc e of students or arrange them in order if we take
marks as an indicator of intelligence. Culture of a society or the lack of it
can be studied considering the number of charitable institutions, their sizes
and number of crimes.
2. It cannot be used for an individual: The conclusions drawn from
statistical data are true for a group of persons. They do not give us any
knowledge about an individual. Though Statistics can estimate the number
of machines in a certain factory that will fail after say, 5 years, i t cannot
tell exactly which machines will fail. One in 2, 000 patients may die in a
particular operation. Statistically this proportion is very small and
insignificant. But for the person who dies and his family, the loss is total.
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73. It gives results only on an average: Statistical methods are not exact.
The results obtained are true only on an average in the long run. When we
say that the average student studies for 2 hours daily there may not be a
singly st udent studying for 2 hours, not only that, every day the average
will not be 2 hours. In the long run, if we consider a number of students,
the daily average will be 2 hours.
4. The results can be biased: The data collected may sometimes be
biased which w ill make the whole investigation useless. Even while
applying statistical methods the investigator has to be objective. His
personal bias may unconsciously make him draw conclusions favourable
in one way or the other.
5. Statistics can be misused: It is s aid that statistics can prove or disprove
anything. It depends upon how the data are presented. The workers in a
factory may accuse the management of not providing proper working
conditions, by quoting the number of accidents. But the fact may be that
most of the staff is inexperienced and therefore meet with an accident.
Besides only the number of accidents does not tell us anything. Many of
them may be minor accidents. With the help of the same data the
management can prove that the working conditions are very good. It can
compare the conditions with working conditions in other factories, which
may be worse. People using statistics have to be very careful to see that it
is not misused.
Thus, it can be seen that Statistics is a very important tool. But its
usefulness depends to a great extent upon the user. If used properly, by an
efficient and unbiased statistician, it will prove to be a wonderful tool.
1.6BRANCHES IN STATISTICS
Statistics may be divided into two main branches:
1. Descriptive Statistic s:In descriptive statistics, it deals with collection
of data, its presentation in various forms, such as tables, graphs and
diagrams and findings, averages and other measures which would describe
the data.
For example, Industrial Statistics, population statistics, trade statistics
etc....Such as businessmen make to use descriptive statistics in presenting
their annual reports, final accounts and bank statements.
2. Inferential Statistics: In inferential statistics deals with techniques
used for analysis of data, making the estimates and drawing conclusions
from limited information taken on sample basis and testing the reliability
of the estimates.
For example, suppose we want to have an idea about the percentage of
illiterates in our country. We take a sample from the population and findmunotes.in

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8the proportion of illiterates in the sample. This sample proportion with the
help of probability enables us to make some inferences about the
population proportion. This study belongs to inferential statistics.
1.7CHAR ACTERISTICS OF STATISTICS
1.Statistics are aggregates of facts.
2.Statistics are numerically expressed.
3. Statistics are affected to a marked extent by multiplicity of causes.
4.Statistics are enumerated or estimated according to a reasonable
standa rd of accuracy.
5.Statistics are collected for a predetermined purpose.
6.Statistics are collected in a systematic manner.
7.Statistics must be comparable to each other.
1.8SOME BASIC DEFINITIONS IN STATISTICS
Constant: A quantity which can be ass uming only one value is called
a constant. It is usually denoted by the first letters of alphabets a, b, c.
For example value of π= 22/7 = 3.14159.... and value of e =
2.71828....
Variable: A quantity which can vary from one individual or object to
andother is called a variable. It is usually denoted by the last letters of
alphabets x, y, z.
For example, heights and weights of students, income, temperature,
number of children in a family etc.
Continuous variable: A variable which can assume each and e very
value within a given range is called a continuous variable. It can occur
in decimals.
For example, heights and weights of students, speed of a bus, the age
of a shopkeeper, the life time of a T.V. etc.
Continuous Data: Data which can be described by a continuous
variable is called continuous data.
For example: Weights of 50 students in a class.
Discrete Variable: A variable which can assume only some specific
values within a given range is called discrete variable. It cannot occur
in decimals. It ca n occur in whole numbers.
For example: Number of students in a class, number of flowers on the
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9Discrete Data: Data which can be described by a discrete variable is
called discrete data .
For example, Number of students in a College.
Quantitative Variable: A characteristic which varies only in
magnitude from an individual to another is called quantitative variable.
It can be measurable.
For example, wages, prices, heights, weights etc.
Qualitative Variable: A characteristic which varies only in quality
from one individual to another is called qualitative variable. It cannot
be measured.
For example, beauty, marital status, rich, poor, smell etc.
1.9EXERCISE
1. Explain the meaning o f statistics.
2. Give a definition of statistics and discuss it.
3. Explain the functions of statistics.
4. What are the limitations of statistics?
5. Define the term Statistics and discuss its characteristics.
6. Enumerate with example some terms of Stati stics.
7. Discuss on the different branches of Statistics.
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DATA COLLECTION AND
PRESENTATION
Unit Structure :
2.1Data
2.1.1Statistical Data
2.1.2 Collection ofData
2.1.3 Types of Data
2.1.4 Methods ofCollecting Data
2.2Classification ofData
2.2.1 Bases ofClassification
2.2.2 Types ofClassification
2.3Tabulation of Data
2.3.1 Types ofTabulation
2.4Frequency Distribution
2.4.1 Construction of Frequency Distribution
2.4.2 Cumulative Frequency Distribution
2.5Types of Graphs
2.6Exercise
2.1 DATA
2.1.1 STATISTICAL DATA :
A sequence of observati on made on a set of objects included in the sample
drawn from population is known as statistical data.
1. Ungrouped Data: Data which have been arranged in a systematic order
are called raw data or ungrouped data.
2. Grouped Data: Data presented in the fo rm of frequency distribution is
called grouped data.
2.1.2 COLLECTION OF DATA
The first step in any enquiry (investigation) is collection of data. The data
may be collected for the whole population or for a sample only. It is
mostly collected on sample b asis. Collection of data is very difficult job.
The enumerator or investigator is the well trained person who collects the
statistical data. The respondents (information) are the persons whom the
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112.1.3 TYPES OF DATA
There are t wo types (sources) for the collection of data:
1. Primary Data: The primary data are the first hand information
collected, compiled and published by organisation for some purpose. They
are most original data in character and have not undergone any sort of
statistical treatment.
For example, Population census reports are primary data because these are
collected, compiled and published by the population census organisation.
2. Secondary Data: The secondary data are second hand information
which are already collected by someone (organisation) for some purpose
and are available for the present study. The secondary data are not pure in
character and have undergone some treatment at least once.
For example, Economics survey of England is secondary data because
these are collected by more than one organisation like Bureau of Statistics,
Board of Revenue, the Banks etc.
2.1.4 METHODS OF DATA COLLECTION
A)METHODS OF COLLECTING PRIMARY DATA
Primary data are collected by the following methods:
1.Personal Inves tigation: The researcher conducts the survey
him/herself and collects data from it. The data collected in this way is
usually accurate and reliable. This method of collecting data is only
applicable in case of small research projects.
2.Through Investiga tion: Trained investigators are employed to collect
the data. These investigators contact the individuals and fill in
questionnaire after asking the required information. Most of the
organisations implied this method.
3.Collection through questionnaire: The researchers get the data from
local representation or agents that are based upon their own
experience. This method is quick but gives only rough estimate.
4.Through Telephone: The researchers get information through
telephone. This method is quick.
B)METHODS OF COLLECTING SECONDARY DATA
The secondary data are collected by the following sources:
Official: The publications of Statistical Division, Ministry of Finance,
the Federal Bureaus of Statistics, Ministries of Food, Agriculture,
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12Semi -Official: State Bank, Railway Board, Central Cotton
Committee, Boards of Economic Enquiry etc....
Publication of Trade Associations, Chambers of Commerce etc....
Technical and Trade Journals and Newspapers.
Research Organisations such as Un iversities and other Institutions.
C)DIFFERENCE BETWEEN PRIMARY AND SECONDARY
DATA
The difference between primary and secondary data is only a change of
hand. The primary data are the first hand information which is directly
collected from one source. T hey are most original data in character and
have not undergone any sort of statistical treatment while the secondary
data are obtained from some other sources or agencies. They are not pure
in character and have undergone some treatment at least once.
Forexample, suppose we are interested to find the average age of MS
students. We collect the age's data by two methods; either by directly
collecting from each student himself personally or getting their ages from
the University record. The data collected by the direct personal
investigator is called primary data and the data obtained from the
University record is called Secondary data.
D)EDITING OF DATA
After collecting the data either from primary or secondary source, the next
step is its editing. Editin g means the examination of collected data to
discover any error before presenting it. It has to be decided before hand
what degree of accuracy is wanted and what extent of errors can be
tolerated in the inquiry. The editing of secondary data is simpler tha n that
of primary data.
2.2CLASSIFICATION OF DATA
The process of arranging data into homogenous group or classes according
to some common characteristics present in the data is called classification.
For example, the process of sorting letters in a pos t office, the letters are
classified according to the cities and further arranged according to streets.
2.2.1 BASES OF CLASSIFICATION
There are four important bases of classification:
1. Qualitative Base
2. Quantitative Base
3. Geographical Base
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131. Qualitative Base: When the data are classified according to some
quality or attributes such as sex, religion, literacy, intelligence etc..
.
2. Quantitative Base: When the data are classified by quantitative
characteristics li ke heights, weights, ages, income etc..
3. Geographical Base: When the data are classified by geographical
regions or location, like states, provinces, cities, countries etc.
4. Chronological or Temporal Base: When the data are classified or
arranged by their time of occurrence, such as years, months, weeks, days
etc.... For example, Time Series Data.
2.2.2 TYPES OF CLASSIFICATION :
1. One -way classification: If we classify observed data keeping in view
single characteristic, this type of classification is known as one -way
classification.
For example, the population of world may be classified by religion as
Muslim, Christian etc.
2. Two -way classification: If we consider two characteristics at a time in
order to classify the observed data then we are d oing two -way
classification.
For example, the population of world may be classified by Religion and
Sex.
3. Multi -way classification: We may consider more than two
characteristics at a time to classify given data or observed data. In this way
we deal in multi -way classification.
For example, the population of world may be classified by Religion, Sex
and Literacy.
2.3TABULATION OF DATA
The process of placing classified data into tabular form is known as
tabulation. A table is a symmetric arrangement of statistical data in rows
and columns. Rows are horizontal arrangements whereas columns are
vertical arrangements. It may be simple, double or complex depending
upon the type of classification.
2.3.1 TYPES OF TABULATION
1. Simple Tabulation or One -way ta bulation: When the data are
tabulated to one characteristic, it is said to be simple tabulation or one -way
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14For example, tabulation of data on population of world classified by one
characteristic like Religion is example of simple tabulation.
2. Double Tabulation or Two -way tabulation: When the data are
tabulated according to two characteristics at a time. It is said to be double
tabulation or two -way tabulation.
For example, tabulation of data on population of world classified by two
character istics like religion and sex is example of double tabulation.
3. Complex Tabulation: When the data are tabulated according to many
characteristics, it is said to be complex tabulation.
For example, tabulation of data on population of world classified by two
characteristics like Religion, Sex and Literacy etc... is example of complex
tabulation.
DIFFERENCES BETWEEN CLASSIFICATION AND
TABULATION
1.First the data are classified and then they are presented in tables, the
classification and tabulation in fac t goes together. So classification is
the basis for tabulation.
2.Tabulation is a mechanical function of classification because in
tabulation classified data are placed in row and columns.
3.Classification is a process of statistical analysis where as tabulation is
a process of presenting the data in suitable form.
2.4FREQUENCY DISTRIBUTION
A frequency distribution is a tabular arrangement of data into classes
according to the size or magnitude along with corresponding class
frequencies (the number o f values fall in each class).
Ungrouped data or Raw Data
Data which have not been arranged in a systemic order is called
ungrouped or raw data.
Grouped Data
Data presented in the form of frequency distribution is called grouped
data.
Array
The numerical raw data is arranged in ascending or descending order is
called an array.
Example
Array the following data in ascending or descending order 6, 4, 13, 7, 10,
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15Solution
Array in ascending order is 4, 6, 7, 10, 13, 16 and 19.
Array in descending ord er is 19, 16, 13, 10, 7, 6, and 4.
CLASS LIMITS
The variant values of the classes or groups are called the class limits. The
smaller value of the class is called lower class limit and larger value of the
class is called upper class limit. Class limits are also called inclusive
classes.
For example, let us take class 10 -19, the smaller value 10 is lower class
limit and larger value 19 is called upper class limit.
CLASS BOUNDARIES
The true values, which describe the actual class limits of a class, are call ed
class boundaries. The smaller true value is called the lower class boundary
and the larger true value is called the upper class boundary of the class. It
is important to note that the upper class boundary of a class coincides with
the lower class bounda ry of the next class. Class boundaries are also
known as exclusive classes.
For example,
Weights in Kg Number of Students
60-65
65-70
70-758
12
5
25
A student whose weights are between 60 kg and 64.5 kg would be
included in the 60 -65 class. A stude nt whose weight is 65 kg would be
included in next class 65 -70.
A class has either no lower class limit or no upper class limit in a
frequency table is called an open -end class. We do not like to use open -
end classes in practice, because they create probl ems in calculation.
For example,
Weights (Pounds) Number of Persons
Below -110
110-120
120-130
130-140
140-above6
12
20
10
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16Class Mark or Mid Point
The class marks or mid point is the mean of lower and upper class limits
or boundaries. So it divide s the class into two equal parts. It is obtained by
dividing the sum of lower and upper -class limit or class boundaries of a
class by 2.
For example, the class mark or mid -point of the class 60 -69 is 60+69/2 =
64.5
Size of Class Interval
The difference b etween the upper and lower class boundaries (not between
class limits) of a class or the difference between two successive mid points
is called size of class interval.
2.4.1 CONSTRUCTION OF FREQUENCY DISTRIBUTION
Following steps are involved in the const ruction of a frequency
distribution.
1.Find the range of the data: The range is the difference between the
largest and the smallest values.
2.Decide the approximate number of classes: Which the data are to be
grouped. There are no hard and first rule s for number of classes. Most
of the cases we have 5 to 20 classes. H. A. Sturges has given a formula
for determining the approximation number of classes.
K = 1 + 3.322 log N
where K = Number of classes
where log N = Logarithm of the total number of obser vations
For example, if the total number of observations is 50, the number of
classes would be:
K = 1 + 3.322 log N
K = 1 + 3.322 log 50
K = 1 + 3.322 (1.69897)
K = 1 + 5.644
K = 6.644 or 7 classes approximately.
3. Determine the approximate class inter val size: The size of class
interval is obtained by dividing the range of data by number of classes and
denoted by h class interval size
(h) = Range/Number of Classes
In case of fractional results, the next higher whole number is taken as the
size of the class interval.
4. Decide the starting Point: The lower class limits or class boundary
should cover the smallest value in the raw data. It is a multiple of class
interval.munotes.in

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17For example, 0, 5, 10, 15, 20 etc... are commonly used.
5. Determine the remaining class limits (boundary): When the lowest
class boundary of the lowest class has been decided, then by adding the
class interval size to the lower class boundary, compute the upper class
boundary. The remaining lower and upper class limits may be determine d
by adding the class interval size repeatedly till the largest value of the data
is observed in the class.
6. Distribute the data into respective classes: All the observations are
marked into respective classes by using Tally Bars (Tally Marks) methods
which is suitable for tabulating the observations into respective classes.
The number of tally bars is counted to get the frequency against each
class. The frequency of all the classes is noted to get grouped data or
frequency distribution of the data. The total of the frequency columns
must be equal to the number of observations.
Example, Construction of Frequency Distribution
Construct a frequency distribution with suitable class interval size of
marks obtained by 50 students of a class are given below:
23, 50, 38, 42, 63, 75, 12, 33, 26, 39, 35, 47, 43, 52, 56, 59, 64, 77, 15, 21,
51, 54, 72, 68, 36, 65, 52, 60, 27, 34, 47, 48, 55, 58, 59, 62, 51, 48, 50, 41,
57, 65, 54, 43, 56, 44, 30, 46, 67, 53.
Solution
Arrange the marks in ascending order as:
12,15, 21, 23, 26, 27, 30, 33, 34, 35, 36, 38, 39, 41, 42, 43, 43, 44, 46, 47,
47, 48, 48, 50, 50, 51, 51, 52, 52, 53, 54, 54, 55, 56, 56, 57, 58, 59, 59, 60,
62, 63, 64, 65, 65, 67, 68, 72, 75, 77.
Minimum value = 12; Maximum value = 77
Range = Maximum val ue-Minimum value = 77 -12 = 65
Number of classes = 1 + 3.322 log N
= 1 + 3.322 log 50
= 1 + 3.322 (1.69897)
= 1 + 5.64 = 6.64 or 7 approximate
class interval size (h) = Range/No. of classes = 65/7 = 9.3 or 10.munotes.in

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18Marks
Class
Limits
C.L.Tally Ma rks Number
of
Students
ƒClass
Boundary
C.B.Class Marks x
10-19
20-29
30-39
40-49
50-59
60-69
70-79II
IIII
IIIIII
IIII IIII
IIII IIII IIII I
IIIIIII
III2
4
7
10
16
8
39.5-19.5
19.5-29.5
29.5-39.5
39.5-49.5
49.5-59.5
59.5-69.5
69.5-79.510 + 19 /2 = 14.5
20 + 29/2 = 24.5
30 + 39/2 = 34.5
40 + 49/2 = 44.5
50 + 59/2 = 54.5
60 + 69/2 = 64.5
70 + 79/2 = 74.5
50
Note: For finding the class boundaries, we take half of the difference
between lower class limit of the 2nd class and upper class limi t of the 1st
class 20 -19/2 = 1/2 = 0.5 This value is subtracted from lower class limit
and added in upper class limit to get the required class boundaries.
Frequency Distribution by Exclusive Method
Class Boundary C.B. Tally Marks Frequency ƒ
10-19
20-29
30-39
40-49
50-59
60-69
70-79II
IIII
IIIIII
IIII IIII
IIII IIII IIII I
IIIIIII
III2
4
7
10
16
8
3
50
2.4.2 CUMULATIVE FREQUENCY DISTRIBUTION
The total frequency of all classes less than the upper class boundary of a
given cl ass is called the cumulative frequency of the class. "A table
showing the cumulative frequencies is called a cumulative frequency
distribution". There are two types of cumulative frequency distribution.
Less than cumulative frequency distribution
It is ob tained by adding successively the frequencies of all the previous
classes including the class against which it is written. The cumulate is
started from the lowest to the highest size.
More than cumulative frequency distribution
It is obtained by finding t he cumulative total of frequencies starting from
the highest to the lowest class. The less than cumulative frequency
distribution and more than cumulative frequency distribution for the
frequency distribution given below are:munotes.in

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19Less than C.F. More than C .F.
Class
LimitƒC.B. Marks C.F Marks C.F.
10-19 2 9.5 -
19.5Less than 19.5 2 9.5 or more 48 + 2 = 50
20-29 4 19.5 -
29.5Less than 29.5 2+4=6 19.5 or more 44 + 4 = 48
30-39 7 29.5 -
39.5Less than 39.5 6+7=1 3 29.5 or more 37 + 7 = 44
40-49 10 39.5 -
49.5Less than 49.5 13 + 10 = 23 39.5 or more 27 + 10 = 37
50-59 16 49.5 -
59.5Less than 59.5 23 + 16 = 39 49.5 or more 11 + 16 = 27
60-69 8 59.5 -
69.5Less than 69.5 39 + 8 = 47 59.5 or more 3+8=1 1
70-79 3 69.5 -
79.5Less than 79.5 47 + 3 = 50 69.5 or more 3
DIAGRAMS AND GRAPHS OF STATISTICAL DATA
We have discussed the techniques of classification and tabulation that help
us in organising the collected data in a meaningful fashion. However, this
way of presentation of st atistical data dos not always prove to be
interesting to a layman. Too many figures are often confusing and fail to
convey the message effectively.
One of the most effective and interesting alternative way in which a
statistical data may be presented is t hrough diagrams and graphs. There
are several ways in which statistical data may be displayed pictorially such
as different types of graphs and diagrams. The commonly used diagrams
and graphs to be discussed in subsequent paragraphs are given as under:
2.5TYPES OF DIAGRAMS/CHARTS
1. Simple Bar Chart
2. Multiple Bar Chart or Cluster Chart
3. Staked Bar Chart or Sub -Divided Bar Chart or Component Bar Chart
a. Simple Component Bar Chart
b. Percentage Component Bar Chart
c. Sub -Divided Rectangular Bar Chart
d. Pie Chart
4. Histogram
5. Frequency Curve and Polygon
6. Lorens Curve
1. SIMPLE BAR CHART
A simple bar chart is used to represent data involving only one variable
classified on spatial, quantitative or temporal basis. In simple bar chart, wemunotes.in

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20make bars of equal width but variable length, i.e. the magnitude of a
quantity is represented by the height or length of the bars. Following steps
are undertaken in drawing a simple bar diagram:
Draw two perpendicular lines one horizontally and the other vertically
at an appropriate place of the paper.
Take the basis of classification along horizontal line (X -axis) and the
observed variable along vertical line (Y -axis) or vice versa.
Mark signs of equal breadth for each class and leave equal or not less
than half br eadth in between two classes.
Finally mark the values of the given variable to prepare required bars.
Sample problem: Make a bar graph that represents exotic pet ownership
in the United States. There are 8,000,000 fish, 1,500,000 rabbits,
1,300,000 turtle s, 1,000,000 poultry and 900,000 hamsters.
Step 1:Number the Y -axiswith the dependent variable. The dependent
variable is the one being tested in an experiment. In this sample question,
the study wanted to know how many pets were in U.S. households. So the
number of pets is the dependent variable. The highest number in the study
is 8,000,000 and the lowest is 1,000,000 so it makes sense to label the Y -
axis from 0 to 8.
Step 2:Draw your bars. The height of the bar should be even with the
correct number on the Y -axis. Don’t forget to label each bar under the x -
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21
Step 3:Label the X -axiswith what the bars represent. For this sample
problem, label the x -axis “Pet Types” and then label the Y -axis with what
the Y -axis represents: “Number of pets (per 1,000 households).” Finally,
give your graph a name. For this sample problem, call the graph “Pet
ownership (per 1,000 households).
Optional: In the above graph, I chose to write the actual numbers on the
bars themselves. You don’t have to do this, but if you have numbers than
don’t fall on a line (i.e. 900,000), then it can help make the graph clearer
for a viewer.
Tips:
1.Line the numbers up on the lines of the graph paper, not the spaces.
2.Make all your bars the same width.
2. MULTIPLE BAR CHART
By multiple bars diagram two or more sets of inter related data are
represented (multiple bar diagram facilities comparison between more
than one phenomena). The technique of simple bar chart is used to draw
this diagram but the difference is that we use different sha des, colours or
dots to distinguish between different phenomena. We use to draw multiple
bar charts if the total of different phenomena is meaningless.munotes.in

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22Sample Example
Draw a multiple bar chart to represent the import and export of Pakistan
for the years 1982 -1988.
YearsImports Exports
Rs.(billion) Rs.(billion)
1982 -83 68.15 34.44
1983 -84 76.71 37.33
1984 -85 89.78 37.98
1985 -86 90.95 49.59
1986 -87 92.43 63.35
1987 -88 111.38 78.44
3. a. COMPONENT BAR CHART
Sub-divided or component bar chart is used to represent data in which the
total magnitude is divided into different components.
In this diagram, first we make simple bars for each class taking total
magnitude in that class and then divide these simple bars into parts in the
ratio of variou s components. This type of diagram shows the variation in
different components without each class as well as between different
classes. Sub -divided bar diagram is also known as component bar chart or
staked chart.munotes.in

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23Current andDevelopment Expenditure –Pakistan (Allfigures inRs.
Billion)
YearsCurrent
ExpenditureDevelopment
ExpenditureTotal
Expenditure
1988 -89 153 48 201
1989 -90 166 56 222
1990 -91 196 65 261
1991 -92 230 91 321
1992 -93 272 76 348
1993 -94 294 71 365
1994 -95 346 82 428
3.b. PERCENTAGE COMPONENT BAR CHART
Sub-divided bar chart may be drawn on percentage basis. to draw sub -
divided bar chart on percentage basis, we express each component as the
percentage of its respective total. In drawing percentage bar chart, bars of
lengt h equal to 100 for each class are drawn at first step and sub -divided in
the proportion of the percentage of their component in the second step.
The diagram so obtained is called percentage component bar chart or
percentage staked bar chart. This type of c hart is useful to make
comparison in components holding the difference of total constant.munotes.in

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24Areas Under Crop Production (1985 -90)
(‘000hectors)
Year Wheat Rice Others Total
1985 -86 7403 1863 1926 11192
1986 -87 7706 2066 1906 11678
1987 -88 7308 1963 1612 10883
1988 -89 7730 2042 1966 11738
1989 -90 7759 2107 1970 11836
Percentage Areas Under Production
Year Wheat Rice Others Total
1985 -86 66.2% 16.6% 17.2% 100%
1986 -87 66.0 17.7 16.3 100
1987 -88 67.2 18.0 14.8 100
1988 -89 65.9 17.4 16.7 100
1989 -90 65.6 17.8 16.6 100
3. d. PIE -CHART
Pie chart can be used to compare the relation between the whole and its
components. Pie chart is a circular diagram and the area of the sector of a
circle is used in pie chart. Circles are drawn with rad ii proportional to the
square root of the quantities because the area of a circle is
munotes.in

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25To construct a pie chart (sector diagram), we draw a circle with radius
(square root of the total). The total angle of the circle is 360°. The angles
of eac h component are calculated by the formula:
Angle of Sector =
These angles are made in the circle by means of a protractor to show
different components. The arrangement of the sectors is usually anti -clock
wise.
Example
2.6EXERCISES
1.Draw a histogram of the following data:
Weekly Wages 1-10 11-20 21-30 31-40 41-50
No. of Workers 14 28 36 12 10munotes.in

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262. The following table shows the temperature for the consecutive five days
in a particular week. Draw range graph.
Day M T W Th F
High° C
Low° C40
2535
2050
4060
5525
15
3. The following is the distribution of total house hold expenditure (in Rs.)
of 202 workers in a city.
Expenditure in Rs. 100-150 150-200 200-250 250-300
No. of Workers 25 40 33 28
Expenditure i n Rs. 300-350 350-400 400-450 450-500
No. of Workers 30 22 16 8

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273
MEASURES OF DISPERSION
Unit Structure
3.1Introduction tomeasure ofdispersion
3.1.1 Dispersion
3.2Absolute measure ofdispersion
3.3Relative measure ofdispersion
3.4Range andcoefficient ofrange
3.5Quartile deviation anditscoefficient
3.6Themean deviation
3.7Standard deviation
3.7.1coefficient ofstandard deviation
3.8Coefficient ofvariation
3.8.1 Uses ofcoefficient ofvariation
3.9The variance
3.10 Skewness andkurtosis
3.11 Exercise
3.1 INTRODUCTION TO MEASURE OF DISPERSION
Am o dern student of statistics is mainly interested in the study of
variability and uncertainty. We live in a changing world. Changes are
taking place in every sphere of life. A man of Statistics does not show
much interest in those things which are constant. The total area of the
earth may not be very important to a research minded person but the area
under different crops, areas covered by forests, area covered by residential
and commercial buildings are figures of great importance because these
figures keep on changing from time to time and from place to place. Very
large number of experts is engaged in the study of changing phenomenon.
Experts working in different countries of the world keep a watch on forces
which are responsible for bringing changes in the fields of human interest.
The agricultural, industrial and mineral production and their transportation
from one part to the other parts of the world are the matters of great
interest to the economists, statisticians and other experts. The changes in
human population, the changes in standard of living, and changes in
literacy rate and the changes in price attract the experts to make detailed
studies about them and then correlate these changes with the human life.
Thus variability or variation is something c onnected with human life and
study is very important for mankind.
3.1.1 DISPERSION
The word dispersion has a technical meaning in Statistics. The average
measures the centre of the data. It is one aspect observations. Another
feature of the observations is as to how the observations are spread about
the centre. The observation may be close to the centre or they may bemunotes.in

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28spread away from the centre. If the observation is close to the centre
(usually the arithmetic mean or median), we say that dispersion, sca tter or
variation is small. If the observations are spread away from the centre, we
say dispersion is large. Suppose we have three groups of students who
have obtained the following marks in a test. The arithmetic means of the
three groups are also given b elow:
Group A: 46, 48, 50, 52, 54
A= 50
Group B: 30, 40, 50, 60, 70
B= 50
Group C; 40, 50, 60, 70, 80
C= 60
In a group A and B arithmetic means are equal i.e.
A=
B= 50. But in
group A the observations are concentrated on the centre. All students of
group A have almost the same level of performance. We say that there is
consistence in the observations in group A. In group B the mean is 50 but
the observations are not close to the cen tre. One observation is as small as
30 and one observation is as large as 70. Thus, there is greater dispersion
in group B. In group C the mean is 60 but the spread of the observations
with respect to the centre 60 is the same as the spread of the observat ions
in group B with respect to their own centre which is 50. Thus in group B
and C the means are different but their dispersion is the same. In group A
and C the means are different and their dispersions are also different.
Dispersion is an important feat ure of the observations and it is measured
with the help of the measures of dispersion, scatter or variation. The word
variability is also used for this idea of dispersion.
The study of dispersion is very important in statistical data. If in a certain
factory there is consistence in the wages of workers, the workers will be
satisfied. But if workers have high wages and some have low wages, there
will be unrest among the low paid workers and they might go on strikes
and arrange demonstrations. If in a certa in country some people are very
poor and some are very rich, we say there is economic disparity. It means
that dispersion is large. The idea of dispersion is important in the study of
wages of workers, prices of commodities, standard of living of different
people, distribution of wealth, distribution of land among framers and
various other fields of life. Some brief definitions of dispersion are:
1.The degree to which numerical data tend to spread about an average
value is called the dispersion or variat ion of the data.
2.Dispersion or variation may be defined as a statistics signifying the
extent of the scattered items around a measure of central tendency.
3.Dispersion or variation is the measurement of the scattered size of the
items of a series abo ut the average.
For the study of dispersion, we need some measures which show whether
the dispersion is small or large. There are two types of measures of
dispersion, which are:
a. Absolute Measure of Dispersion
b. Relative Measure of Dispersion.munotes.in

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293.2 ABS OLUTE MEASURE OF DISPERSION
These measures give us an idea about the amount of dispersion in a set of
observations. They give the answers in the same units as the units of the
original observations. When the observations are in kilograms, the
absolute mea sure is also in kilograms. If we have two sets of observations,
we cannot always use the absolute measures to compare their dispersion.
We shall explain later as to when the absolute measures can be used for
comparison of dispersions in two or more than tw o sets of data. The
absolute measures which are commonly used are:
1. The Range
2. The Quartile Deviation
3. The Mean Deviation
4. The Standard Deviation and Variance
3.3 RELATIVE MEASURE OF DISPERSION
These measures are calculated for the comparison of dispersion in two or
more than two sets of observations. These measures are free of the units in
which the original data is measured. If the original data is in dollar or
kilometres, we do not use these units with relative measure of dispersion.
These mea sures are a sort of ratio and are called coefficients. Each
absolute measure of dispersion can be converted into its relative measure.
Thus, the relative measures of dispersion are:
1.Coefficient of Range or Coefficient of Dispersion.
2.Coefficient o f Quartile Deviation or Quartile Coefficient of
Dispersion.
3.Coefficient of Mean Deviation or Mean Deviation of Dispersion.
4.Coefficient of Standard Deviation or Standard Coefficient of
Dispersion.
5.Coefficient of Variation (a special case of Stan dard Coefficient of
Dispersion).
3.4RANGE AND COEFFICIENT OF RANGE
The Range
Range is defined as the difference between the maximum and the
minimum observation of the given data. If X mdenotes the maximum
observation X odenotes the minimum observation t hen the range is defined
as Range = X m-Xo.
In case of grouped data, the range is the difference between the upper
boundary of the highest class and the lower boundary of the lowest class.
It is also calculated by using the difference between the mid poi nts of themunotes.in

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30highest class and the lowest class. it is the simplest measure of dispersion.
It gives a general idea about the total spread of the observations. It does
not enjoy any prominent place in statistical theory. But it has its
application and utility in quality control methods which are used to
maintain the quality of the products produced in factories. The quality of
products is to be kept within certain range of values.
The range is based on the two extreme observations. It gives no weight to
the c entral values of the data. It is a poor measure of dispersion and does
not give a good picture of the overall spread of the observations with
respect to the centre of the observations. Let us consider three groups of
the data which have the same range:
Group A: 30, 40, 40, 40, 40, 40, 50
Group B: 30, 30, 30, 40, 50, 50, 50
Group C: 30, 35, 40, 40, 40, 45, 50
In all the three groups the range is 50 -30 = 20. In group A there is
concentration of observations in the centre. In group B the observations
are friendly with the extreme corner and in group C the observations are
almost equally distributed in the interval from 30 to 50. The range fails to
explain these differences in the three groups of data. This defect in range
cannot be removed even if we calcul ate the coefficient of range which is a
relative measure of dispersion. If we calculate the range of a sample, we
cannot draw any inferences about the range of the population.
Coefficient of Range
It is relative measure of dispersion and is based on the v alue of
range. It is also called range coefficient of dispersion. It is defined as:
Coefficient of Range = .
The range X m-Xois standardised by the total X m+X o.
Let us take two sets of observations. Set A contains marks of five
students in Mathematics out of 25 marks and group B contains marks of
the same student in English out of 100 marks.
Set A: 10, 15, 18, 20, 20
Set B: 30, 35, 40, 45, 50munotes.in

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Measure ofDispersion
31The values of range and coefficient of range are calculated as
Range Coefficient of Range
Set A: (Mathematics)
Set B: (English)20-10 = 10
50-30 = 20
= 0.33
= 0.25
In set A the range is 10 and in set B the range is 20. Apparently it seems as
if there is greater dispersion in set B. But th is is not true. The range of 20
in set B is for large observations and the range of 10 in set A is for small
observations. Thus 20 and 10 cannot be compared directly. Their base is
not the same. Marks in Mathematics are out of 25 and marks of English
are o ut of 100. Thus, it makes no sense to compare 10 with 20. When we
convert these two values into coefficient of range, we see that coefficient
of range for set A is greater than that of set B. Thus, there is greater
dispersion or variation in set A. The mar ks of students in English are more
stable than their marks in Mathematics.
Example
Following are the wages of 8 workers of a factory. Find the range and
coefficient of range. Wages in ($) 1400, 1450, 1520, 1380, 1485, 1495,
1575, 1440.
Solution:
Here Lar gest Value = X m= 1575 and Smallest Value = X o= 1380
Range = X m-Xo= 1575 -1380 = 195.
Coefficient of Range =
=
=
= 0.66
Example
The following distribution gives the numbers of houses and the
number of perso ns per house.
Number of
Persons1 2 3 4 5 6 7 8 9 10
Number of
Houses26 113 120 95 60 42 21 14 5 4
Calculate the range and coefficient of range.
Solution:
Here Largest Value = X m= 10 and Smallest Value = X o=1
Range = X m-Xo= 10 -1 = 9.
Coeffici ent of Range =
=
=
= 0.818munotes.in

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32Example
Find the range of the weight of the students of a University.
Weights (Kg) 60-62 63-65 66-68 69-71 72-74
Number of Students 5 18 42 27 8
Calculate the range and coefficient of range.
Solution:
Weights (Kg) Class Boundaries Mid Value No. of Students
60-62
63-65
66-68
69-71
72-7459.5 -62.5
62.5 -65.5
65.5 -68.5
68.5 -71.5
71.5 -74.561
64
67
70
735
18
42
27
8
Method 1
Here X m= Upper class boundary of the highest class = 74.5; X o= Lower
Class Boundary of the lowest class = 59.5
Range = X m-Xo= 74.5 -59.5 = 15 Kilogram.
Coefficient of Range =
=
=
= 0.1119.
Method 2
Here X m= Mid value of the highest class = 73; X o= Mid Va lue of
the lowest class = 61
Range = X m-Xo= 73 -61 = 12 Kilogram.
Coefficient of Range =
=
=
= 0.0895.
3.5QUARTILE DEVIATION AND ITS COEFFICIENT
Quartile Deviation
It is based on the lower Quartile Q1 and the upper quartile Q3. The
difference Q3 -Q1 is called the inter quartile range. The difference Q3 -
Q1 divided by 2 is called semi -inter-quartile range or the quartile
deviation. Thus
Quartile Deviation (Q.D) =
. The quartile deviation is a slightly
better measure of absolute dispersion than the range. But it ignores the
observation on the tails. If we take different samples from a population
and calculate their quartile deviations, their values are quite likely to be
sufficiently different . This is called sampling fluctuation. It is not a
popular measure of dispersion. The quartile deviation calculated from themunotes.in

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Measure ofDispersion
33sample data does not help us to draw any conclusion (inference) about the
quartile deviation in the population.
Coefficient of Qua rtile Deviation
A relative measure of dispersion based on the quartile deviation is
called the coefficient of quartile deviation. It is defined as
Coefficient of Quartile Deviation ==
=
.
It is pure number free of any units of measurement. It can be sued
for comparing the dispersion in two or more than two sets of data.
Example
The Wheat production (in Kg) of 20 acres is given as: 1120, 1240,
1320, 1040, 1080, 1200, 1440, 1360, 1680, 1730, 1785, 1342, 1960, 1880,
1755, 1600, 1470, 1750 and 1885. Find the quartile deviation and
coefficient of quartile deviation.
Solution
After arranging the observation in ascending order, we get, 1040,
1080, 1120, 1200, 1240, 1320, 1342, 1360, 1440, 1470, 1600, 1680, 1720,
1730, 1750, 1755, 17 85, 1880, 1885, 1960.
Q1 = Value of th item
= Value of
th item
= Value of (5.25)th item
= 5th item + 0.25 (6th item -5th item) = 1240 + 0.25 (1320 -1240)
Q1 = 1240 + 20 = 1260
Q3 = Value of
th item
= Value of
th item
= Value od (15.75) th item
15th item + 0.75 (16th item -15th item) = 1750 + 0.75 (1755 -1750)
Q3 = 1750 + 3.75 = 1753.75munotes.in

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34Quartile Deviation (QD) =
=
=
=
246.875
Coefficient of Quartile Deviation =
=
= 0.164.
Example
Calculate the quartile deviation and coefficient of quartile
deviation from the data given below:
Maximum Load (Short tons) Number of Cables
9.3-9.7
9.8-10.2
10.3-10.7
10.8 -11.2
11.3 -11.7
11.8 -12.2
12.3 -12.7
12.8 -13.22
5
12
17
14
6
3
1
Solution
The necessary calculations are given below:
Maximum Load
(Short Tons)Number of Cables
FClass Boundaries Cumulative
Frequencies
9.3-9.7
9.8-10.2
10.3 -10.7
10.8 -11.2
11.3 -11.7
11.8 -12.2
12.3 -12.7
12.8 -13.22
5
12
17
14
6
3
19.25 -9.75
9.75 -10.25
10.25 -10.75
10.75 -11.25
11.25 -11.75
11.75 -12. 25
12.25 -12. 75
12. 75 -13.252
2+5=7
7 + 12 = 19
19 + 17 = 36
36 + 14 = 50
50+ 6 = 56
56 + 3 = 59
59 + 1 = 60
Q1= Value of [
]t hi t e m
= Value of [
]t hi t e m
= 15th item
Q1lies in the class 10.25 -10.75
⸫Q1=1+
[
]
Where 1 = 10.25, h = 0.5, f = 12, n/4 = 15 and c = 7
Q1= 10.25 +
(15-7)munotes.in

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Measure ofDispersion
35= 10.25 + 0.33
= 10.58
Q1= Value of [
]t hi t e m
= value of [
]t hi t e m
= 45th item
Q3lies in th e class 11.25 -11.75
⸫Q3=1+
[
]
where 1 = 11.25, h = 0.5, f = 14, 3n/4 = 45 and c = 36
⸫Q1= 11.25 +
(45-36)
= 11.25 + 0.32
= 11.57
Quartile Deviation (Q.D) =
=
=
=0.495
Coefficient of Quartile Deviation =
=
=
= 0.045
3.6THE MEAN DEVIATION
The mean deviation or the average deviation is defined as the mean of the
absolute deviations of observations from some suitable average which
may be arithmetic mean, the median or the mode. The difference (X -
average) is called deviation and when we ignore the negative sign, this
deviation is written as
and is read as mod deviations. The
mean of these more or absolute devia tions is called the mean deviation or
the mean absolute deviation. Thus for sample data in which the suitable
average is the
, the mean deviation (M.D) is given by the relation
M.D =
For frequency distribution, the mean deviati on is given by
M.D =
When the mean deviation is calculated about the median, the
formula becomes
M.D. (about median) =
The mean deviation about the mode ismunotes.in

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Business Statistics
36M.D (about mode) =
For a population data the mean devi ation about the population mean
µi s
M.D =
The mean deviation is a better measure of absolute dispersion than the
range and the quartile deviation.
A drawback in the mean deviation is that we use the absolute deviations
which does not seem logical. The reason for this is that
(X-
is always equal to zero. Even if we use median or more in place
of
, even then the summation (X -median) or (X -mode) will be zero or
approximately zero with the resul t that the mean deviation would always
be better either zero or close to zero. Thus, the very definition of the mean
deviation is possible only on the absolute deviations.
The mean deviation is based on all the observations, a property which is
not posses sed by the range and the quartile deviation. The formula of the
mean deviation gives a mathematical impression that it is a better way of
measuring the variation in the data. Any suitable average among the mean,
median or more can be used in its calculatio n but the value of the mean
deviation is minimum if the deviations are taken from the median. A
drawback of the mean deviation is that it cannot be used in statistical
inference.
Coefficient of the Mean Deviation
A relative measure of dispersion based on the mean deviation is called the
coefficient of the mean deviation or the coefficient of dispersion. It is
defined as the ratio of the mean deviation to the average used in the
calculation of the mean deviation.
Thus,
Coefficient of M.D (about mean) = Mea n Deviation from Mean/Mean
Coefficient of M.D (about median) = Mean Deviation from
Median/Median
Coefficient of M.D (about mode) = Mean Deviation from Mode/Mode
Example
Calculate the mean deviation from (1) Arithmetic Mean (2) Median (3)
Mode in respect of the marks obtained by nine students given below and
show that the mean deviation from median is minimum.
Marks out of 25: 7, 4, 10, 9, 15, 12, 7, 9, 7
Solution
After arranging the observations in ascending order, we get
Marks 4, 7, 7, 7, 9, 9, 10, 12, 15munotes.in

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Measure ofDispersion
37Mean =
=
= 8.89
Median = Value of (
)t hi t e m
= Value of (
)t hi t e m
= Value of (5) the item = 9
Mode = 7 (Since 7 is repeated maximum number of times)
Marks X
4
7
7
7
9
9
10
12
154.89
1.89
1.89
1.89
0.11
0.11
1.11
3.11
6.115
2
2
2
0
0
1
3
63
0
0
0
2
2
3
5
8
Total 21.11 21 23
M.D from mean =
=
= 2.35
M.D from Median =
=
= 2.33
M.D from Mode =
=
= 2.56
From the above calculations, it is clear that the mean deviation from the
median has the least value.
Example
Calculate the mean deviation from mean and its coefficients from the
following data:
Size of items 3-44-55-66-77-88-99-10
Frequency 3 7 22 60 85 32 8munotes.in

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38Solution
The necessary calculation is given below:
Size of Items X F fX
f
3-4
4-5
5-6
6-7
7-8
8-9
9-103.5
4.5
5.5
6.5
7.5
8.5
9.53
7
22
60
85
32
810.5
31.5
121.0
390.0
637.5
272.0
76.03.59
2.59
1.59
0.59
0.41
1.41
2.4110.77
18.13
34.98
35.40
34.85
45.12
19.28
Total 217 1538.5 198.53
Mean =
=
= 7.09
M.D from Mean =
= 0.915
Coefficient of M.D (Mean) =
=
= 0.129
3.7STANDARD DEVIATION
The standard deviation is defined as the positive square root of the mean
of the square deviations taken from arithmetic mean of the data.
For the sample data the standard deviation is denoted by S and is defined
as
S=
For a population data the standard deviation is denoted by σ
(sigma) and is defined as:
σ=
For frequency distribution the formulas become
S=
orσ=
The standard deviation is in the same units as the units of the original
observations. If the original observations are in grams, the value of the
standard deviation will also be in grams.
The standard deviation plays a dominating role for the study of variation
in the data. It is a very widely used measure of dispersion. It stands like a
tower among measure of dispersion. As far as the important statisticalmunotes.in

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Measure ofDispersion
39tools are concerned, the first important tool is the mean and the second
important tool is the standard deviation S. It is based on the observa tions
and is subject to mathematical treatment. It is of great importance for the
analysis of data and for the various statistical inferences.
However, some alternative methods are also available to compute standard
deviation. The alternative methods simp lify the computation. Moreover in
discussing these methods we will confirm ourselves only to sample data
because sample data rather than whole population confront mostly a
statistician.
Actual Mean Method
In applying this method first of all we compute ar ithmetic mean of the
given data either ungroup or grouped data. Then take the deviation from
the actual mean. This method is already being defined above. The
following formulas are applied:
For Ungrouped Data For Grouped Data
S=
2 S=
This method is also known as direct method.
Assumed Mean Method
a. We use the following formulas to calculate standard deviation:
For Ungrouped Data For Grouped Data
S=
S=
where D = X -A and A is any assumed me an other than zero. This method
is also known as short -cut method.
b. If A is considered to be zero then the above formulas are reduced to the
following formulas:
For Ungrouped Data For Grouped Data
S=
-(
S=
c. If we are in a position to simplify the calculation by taking some
common factor or divisor from the given data the formulas for computing
standard deviation are:munotes.in

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Business Statistics
40For Ungrouped Data For Grouped Data
S=
-(
XcS=
X c or h
Where u =
=
; h = Class Interval and c = Common Divisor.
This method is also called method of step -deviation.
Examples of Standard Deviation
This tutorial is about some examples of standard deviation using a ll
methods which are discussed in the previous tutorial.
Example
Calculate the standard deviation for the following sample data using all
methods: 2, 4, 8, 6, 10 and 12.
Solution:
Method -1 Actual mean Method
X (X-
)2
2
4
8
6
10
12
ƩX = 42(2-7)2= 25
(4-7)2=9
(8-7)2=1
(6-7)2=1
(10-7)2=9
(12-7)2= 25
Ʃ(X-
)2= 70
=
=
=7
S=
2
S=
=
= 3.42
Method 2: Taking assumed mean as 6.
X D=( X -6) D22
4
8
6
10
12-4
-2
2
0
4
616
4
4
0
16
36
Total ƩD=6 ƩD2= 76munotes.in

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Measure ofDispersion
41S=
2
S=
2=
=
= 3.42
Method 3: Taking Assumed Mean as Zero
X X22
4
8
6
10
124
16
64
36
100
144
ƩX = 42 ƩX2= 364
S=
-(
2
S=
-(
2
S=
=
= 3.42
Method 4: Taking 2 as common divisor or factor
X u=( X -4)/2 u22
4
8
6
10
12-1
0
2
1
3
41
0
4
1
9
16
Total Ʃu=9 Ʃu2= 31
S=
-(
Xc
S=
-(
2X 2
S=
X 2 = 3.42.munotes.in

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42Example
Calculate standard deviation from the following distribution of marks by
using all the methods:
Marks No. of Student s
1-3
3-5
5-7
7-940
30
20
10
Solution
Method 1: Actual mean method
Marks f X fX (X-
)2f( X-
)2
1-3
3-5
5-7
7-940
30
20
102
4
6
880
120
120
804
0
4
16160
0
80
160
Total 100 400 400
=
=
=4
S=
S=
=
= 2 Marks
Method 2: Taking assumed mean as 2
Marks f X D=( X -
2)fD fD21-3
3-5
5-7
7-940
30
20
102
4
6
80
2
4
60
60
80
600
120
320
160
Total 100 200 800
S=
2
S=
2
S=
=
= 2 Marksmunotes.in

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43Method 3: Using Assumed Mean as Zero
Marks f X fX fX21-3
3-5
5-7
7-940
30
20
102
4
6
880
120
120
80160
480
720
640
Total 100 400 2000
S=
2
S=
2
S=
=
= 2 marks.
Method 4: By taking 2 as the Common Divisor
Marks f X u=( X -
2)/2Fu fu21-3
3-5
5-7
7-940
30
20
102
4
6
8-2
-1
0
1-80
-30
0
10160
30
0
10
Total 100 -100 200
S=
2Xh
S=
2X2
S=
X2=
X 2 = 2 marks.
3.7.1 COEFFICIENT OF STANDARD DEVIATION
The standard deviation is the absolute measure of dispersion. Its relative
measure is called stan dard coefficient of dispersion or coefficient of
standard deviation, It is defined as
Coefficient of Standard Deviation =
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443.8 COEFFICIENT OF VARIATION
The most important of all the relative measure of dispersion is the
coefficient of varia tion. This word is variation and not variance. There is
no such thing as coefficient of variance. The coefficient of variation (CV)
is defined as
Coefficient of Variation (C.V) =
x 100
Thus C.V is the value of S when
is assumed equal to 100. It is a pure
number and the unit of observations is not mentioned with its value. It is
written in percentage form like 20% or 25%. When its value is 20%, it
means that when the mean of the observation is assumed equal to 100,
their standard deviation will be 20. The C.V is used to compare the
dispersion in different sets of data particularly the data which differ in
their means or differ in the units of measurement. The wages of workers
may be in dollars and the consumption of meat in their families may be in
kilograms. The standard deviation of wages in dollars cannot be compared
with the standard deviation of amount of meat in kilograms. Both the
standard deviations need to be converted into coefficient of variation for
comparison. Suppose the value of C.V for wages is 10% and the values of
C.V for kilograms of meat is 25%. This means that the wages of workers
are consistent. Their wages are close to the overall average of their wages.
But the families consume meat in quite different quantit ies. Some families
use very small quantities of meat and some others use large quantities of
meat. We say that there is greater variation in their consumption of meat.
The observations about the quantity of meat are more dispersed or more
variant.
Example
Calculate the coefficient of standard deviation and coefficient of variation
for the following sample data: 2, 4, 8, 6, 10 and 12.
Solution
X (X-
)2
2
4
8
6
10
12(2-7)2= 25
(4-7)2=9
(8-7)2=1
(6-7)2=1
(10-7)2=9
(12-7)2= 25
ƩX = 42 Ʃ(X-
)2= 70
=
=7
S=
(X-
)2/nmunotes.in

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45S=
=
= 3.42
Coefficient of Standard Deviation =
=
= 0.49
Coefficient of Variation (C. V) =
X 100 =
X 100 = 48 .86%
3.8.1 USES OF COEFFICIENT OF VARIATION
Coefficient of variation is used to know the consistency of the data. By
consistency we mean the uniformity in the values of the
data/distribution from arithm etic mean of the data/distribution. A
distribution with smaller C.V than the other is taken as more consistent
than the other.
C.V is also very useful when comparing two or more sets of data that
are measured in different units of measurement.
3.9THE VAR IANCE
Variance is another absolute measure of dispersion. It is defined as the
average of the squared difference between each of the observation in a set
of data and the mean. For a sample data the variance is denoted by S2and
the population variance is denoted by σ2(sigma square).
The sample variance S2has the formula
S2=
2
where
is sample mean and n is the number of observations in the
sample.
The population variance σ2is defined as
σ2=
where µ is the mea n of the population and N is the number of observations
in the data. It may be remembered that the population variance σ2is
usually not calculated. The sample variance S2is calculated and if need be,
this S2is used to make inference about the population variance.
The term =(X-
)2is positive, therefore S2is always positive. If
the original observations are in centimetre, the value of the variance will
be (Centimetre) 2. Thus the unit of S2is the square of the units of the
original measur ement.
For a frequency distribution the sample variance S2is defined as
S2=
For a frequency distribution the population variance σ2is defined asmunotes.in

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Business Statistics
46σ2=
In simple words we can say that variance is the square root of standard
deviation.
Variance = (Standard Deviation)2
Example
Calculate variance from the following dis tribution of marks:
Marks No. of Students
1-3
3-5
5-7
7-940
30
20
10
Solution
Marks F X fX (X -
2f( X -
2
1-3
3-5
5-7
7-940
30
20
102
4
6
880
120
120
804
0
4
16160
0
80
160
Total 100 400 400
=
=
=4
S2=
=
=4
Variance S2= 4.
3.10SKEWNESS AND KURTOSIS
Skewness is the absence of symmetry in a distribution. Though averages
and measures of dispersion are useful in studying the d ata, the shape of the
frequency curve may also be equally important to the statistician. If we are
studying a certain phenomenon over a period of time, the average may
remain the same, but the structure of the distribution may change. Two
distributions may have identical averages, yet one my tail off towards the
higher values and the other towards the lower values.
To study the distribution we need a measure of this tendency which will
give us the direction and degree of this tendency which is called skewn ess.munotes.in

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47A fundamental task in many statistical analyses is to characterizethelocation andvariabilityof a data set. A further characterizationof the data includes skewness and kurtosis.Skewness is a measure of symmetry, or more precisely, the lack ofsymmetry. A distribution, or data set, is symmetric if it looks thesame to the left and right of the centre point.
Kurtosis is a measure of whether the data are heavy -tailed or light -tailed relative to a normal distribution. That is, data sets with highkurtosis tend to have heavy tails, or outliers. Data sets with lowkurtosis tend to have light tails,or lack of outliers. A uniformdistribution would be the extreme case.
Thehistogramis an effective graphical technique for showing boththe skewness and kurtosis of data se t.
For univariate data Y1,Y2, ...,YN, the formula for skewness is:
g1=∑Ni=1(Yi−Y¯)3/Ns3
where Y¯is the mean, sis the standard deviation, and Nis thenumber of data points. Note that in computing the skewness, the siscomputed with Nin the denominat or rather than N-1.
The above formula for skewness is referred to as the Fisher -Pearsoncoefficient of skewness. Many software programs actually computethe adjusted Fisher -Pearson coefficient of skewness
G1=N(N −1)−−−−−−−−√N−2∑Ni=1(Yi−Y¯)3/Ns3
This isan adjustment for sample size. The adjustment approaches 1asNgets large. For reference, the adjustment factor is 1.49 for N=5, 1.19 for N= 10, 1.08 for N= 20, 1.05 for N= 30, and 1.02forN= 100.
The skewness for a normal distributionis zero, and any symmetricdata should have a skewness near zero. Negative values for theskewness indicate data that are skewed left and positive values forthe skewness indicate data that are skewed right. By skewed left, wemean that the left tail is long relative to the right tail. Similarly,skewed right means that the right tail is long relative to the left tail.If the data are multi -modal, then this may affect the sign of theskewne ss.Some measurements have a lower bound and are skewed right. Forexample, in reliability studies, failure times cannot be negative.It should be noted that there are alternative definitions of skewnessin the literature. For example, the Galton skewness (also known asBowley's skewness) is defined asmunotes.in

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Business Statistics
48Galton skewness=Q1+Q3 −2Q2Q3 −Q1
where Q1is the lower quartile, Q3is the upper quartile, and Q2is themedian.
The Pearson 2 skewness coefficient is defined as
Sk2=3(Y¯ −Y~)s
where Y~is the sample median.There are many other definitions for skewness that will not bediscussed here.
KURTOSIS
For univariate data Y1,Y2, ..., YN, the formula for kurtosis is:
kurtosis= ∑Ni=1(Yi−Y¯)4/Ns4
where Y¯is the mean, sis the standard deviation, and Nis thenumberof data points. Note that in computing the kurtosis, thestandard deviation is computed using Nin the denominator ratherthan N-1.
The kurtosis for a standard normal distrib utionis three. For thisreason, some sources use the following definition of kurtosis (oftenreferred to as "excess kurtosis"):
kurtosis= ∑Ni=1(Yi−Y¯)4/Ns4 −3This definition is used so that the standard normal distribution has akurtosis of zero. In addition, with the second definition positivekurtosis indicates a "heavy -tailed" distribution and negative kurtosisindicates a "ligh t tailed" distribution.Which definition of kurtosis is used is a matter of convention (thishandbook uses the original definition). When using software tocompute the sample kurtosis, you need to be aware of whichconvention is being followed. Many sources use the term kurtosiswhen they are actually computing "excess kurtosis", so it may notalways be clear.The following example shows histograms for 10,000 randomnumbers generated from a normal, a double exponential, a Cauchy,and a Weibull distributio n.munotes.in

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Measure ofDispersion
49
The first histogram is a sample from a normal distribution. Thenormal distribution is a symmetric distribution with well -behavedtails. This is indicated by the skewnessof 0.03. The kurtosis of 2.96is near the expected value of 3. The histogram verifies thesymmetry.
The second histogram is a sample from adouble exponentialdistribution . Thedouble exponential is a symmetric distribution.Compared to the normal, it has a stronger peak, more rapid decay,and heavier tails. That is, we would expect a skewness near zero anda kurtosis higher than 3. The skewness is 0.06 and the kurtosisis 5.9.
The fourth histogram is a sample from a Weibull distribution withshape parameter 1.5. The Weibull distribution is a skeweddistribution with the amount of skewness depending on the value ofthe shape parameter. The degree of decay as we move away fromthe center also depends on the value of the shape parameter. For thisdata set, the skewness is 1.08 and the kurtosis is 4.46, whichindicates moderate skewness and kurtosis.Many classical statistical tests and intervals depend on normalityassumptions. Significant skewness and kurtosis clearly indicate thatdata are not normal. If a data set exhibits significant skewness orkurtosis (as indicated by a histogramor the numerical measures),what can we do about it?munotes.in

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50One approach is to apply some type of transformation to try to makethe data normal, or more nearly normal. The Box-Coxtransformationis a useful technique for trying to normalize a dataset. In particular, taking the log or square root of a data set is oftenuseful for data that exhibit moderate right skewness.
Another approach is to use techniques based on distributions otherthan the normal. For example, in reliability studies, the exponential,Weibull, and lognormal distributions are typically used as a basisfor modelling rather than using the normal distribution.Theprobability plot correlation coefficient plot and theprobabilityplotare useful tools for determining a good distributional model forthe data.
3.11EXERCISES
Q1. Calculate the range and quartile deviation for wages: Also calculate
coefficient of quartile deviation:
Wages 30 -
3232 -
3434 -
3636 -
3838 -
4040 -
4242 -
44
Labourers 12 18 16 14 12 8 6
Hint: Coefficient of Q.D. =
= 0.081
Q2. Calculate the standard deviation from the following:
Marks 10 20 30 40 50 60
No. of
Students8 12 20 10 7 3
Hint: σ=
2XC
=
2X 10 = 13.5
Q3. Find the mean and standard deviation of the following obser vations:
X: 1 2 4 6 8 9munotes.in

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51Transform the above observation such that the mean of transformed
observations becomes double the mean of X, standard deviation remains
unchanged.
Hint: Mean =
= 30/6 = 5 Let d = X -5. Then
Ʃd2= 52. σ=
)2=
= 2.94.
Q4. Explain positive and negative skewness with the help of sketches.
Q5. Write short notes on skewness and kurtosis.

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524
PERMUTATIONS AND COMBINATIONS
Unit Structure :
4.1Counting Techniques
(A) Permutation
(B) Combination
4.2Exercise
4.1 COUNTING TECHNIQUES
Fundamental Principle of Counting
If the first operation can be performed in any one of the m ways and then a
second operation can be performed in any one of the n ways, then both can
be performed together in any one of the m × n ways. This rule can be
generalised. If first operation can be performed in any one of the n1 ways,
second operation in any one of the n2 ways, ...... kth operation in any one
of the nk ways, then together these can be performed in any one of the n1
× n2 × ...... × nk ways.
Factorial Notation:
The product of the first natural number is called factorial n,
denoted by n!.
Thus n! =1x2x3… …..x(n -1) x n Note: 0!=1, n!=n (n -1)!
e.g. 4! =1x2x3x4=24; n! = n(n -1)(n-2)……..(n -r+1)(n -r)!;r≤n
(A)Permutations:
Each of the different arrangements, which can be made out of a given
number of things by taking some or all of them at a time is called a
permutation.
Note: In Permutations, th e order in which the things are arranged is
important.
(a)Permutations of n objects: The total number of permutations of n
distinct objects is n! Using symbols, we can writenPn=n ! ,( w h e r en
denotes the permutations of n objects, all taken together).
Let us assume there are n persons to be seated on n chairs. The first chair
can be occupied by any one of the n persons and hence, there are n ways
in which it can be occupied. Similarly, the second chair can be occupied in
n-1 ways and so on. Using the fundamental principle of counting, the
total number of ways in which n chairs can be occupied by n persons or
the permutations of n objects taking all at a time is given by:munotes.in

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Permutations A nd
Combinations
53nPn=n ( n –1)(n–2) ...... 3.2.1 = n!
(b)Permutations of n objects taking r at at i m e :
e.g. If there are three things a, b, c then the number of permutations
i.e. arrangements of these three things taken two at a time is denoted by
3P2and is given by3P2=3(3 -1)=3.2=6 as follows (a, b); (a ,c); (b ,c); (b, a);
(c, a); (c, b)
In terms of the example, considered above, now we have n persons to be
seated on r chairs, where r ≤n.
The number of permutations of n persons to be seated on r chairs, where r
≤n.
,nPr== n(n –1)(n –2) ...... [n –(r–1)] = n(n –1)(n –2) ...... (n –r+1 ) .
On multiplication and division of the R.H.S. by (n -r)!, we get
nPr=
=
,r≤
n
e.g. If there are three things a, b, c then the number of permutations i.e.
arrangements of these three things taken two at a time is denoted by3P2
and is given by3P2=3(3 -1)=3.2=6 as follows (a, b); (a ,c); (b ,c); (b, a); (c,
a); (c, b)
(c)Permutations of n objects taking r at a time when any object may
be repeated any number of times:
Here, each of the r places can be filled in n ways. The refore, total number
of permutations is nr.
Examples:
[1] What are the total number of ways of simultaneous throwing of (i)
3 coins, (ii) 2 dice and (iii) 2 coins and a die?
Solution:
(i)Each coin can be thrown in any one of the two ways, i.e., a head or a
tail,
Therefore, the number of ways of simultaneous throwing of 3 coins
=23= 8.
(ii)Each die can be thrown in any one of the 6 ways,
Therefore the total number of ways of simultaneous throwing of two dice
=62= 36.
(iii)The to tal number of ways of simultaneous throwing of 2 coins and a
die
=22× 6 = 4x6=24.munotes.in

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54[2] In how many ways 5 passengers can sit in a compartment having
15 vacant seats?
Solution:
No. of ways of 5 passengers can sit in a compartment having 15
vacant seats=15P5
=
=
=
=
[3] In how many ways can the letters of the word MATHEMATICS
be arranged?
Solution:
The word MATHEMATICS has 11 letters in which there are 2M 's, 2T's,
2A's,1H,1E,1I,1Cand 1S. Thus, the required number of permutations
=4989600
[4] (a) In how many ways 4 men and 3 women can be seated in a row
such that women occupy the even places?
(b) In how many ways 4 men and 4 women ca n be seated such that
men and women occupy alternative places?
Solution:
(a) 4 men can be seated in 4! ways and 3 women can be seated in 3! ways.
Since each arrangement of men is associated with each arrangement of
women, therefore, the required number of permutations = 4! 3! = 144.
(b) There are two ways in which 4 men and 3 women can be seated
The required number of permutations = 2 .4! 3! = 288
[5] How many five digits numbers can be made by using the digits 1,
2, 3,4,5,6 and 7? When repetition i s not allowed? How many of these
will be greater than 50,000.
Solution:
There are 7 given digits 1, 2, 3,4,5,6 and 7.
No. of ways to form five digit number out of given digits=7P5
A five digit number out of given digit is greater than 50,000 if it begin
with 5, 6 and 7.
Therefore digit at ten thousand’s place can be chosen in 3 ways. The other
four digits can be chosen out of the remaining 6 digits in6P4ways.munotes.in

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Combinations
55Therefore, required no. of five digit numbers
=3x6P4=3x
=
=3x6x5x4x3= 1080
[6] How many four digits numbers can be made by using the digits 0,
1, 2 and 4? When (i) repetition is allowed and (ii) repetition is not
allowed.
Solution:
(i)Repetition is allowed:
To make four digit number we cannot take 0 at thousand’ place.
Therefore digit at thousand’s place can be chosen in 3 ways. The digit at
hundred’s place can be chosen in 4 ways. The digit at ten’s place can be
chosen in 4 ways. The digit at unit’s place can be chosen in 4 ways.
Therefore, four digit number can be made by 3x4x4x4=192 ways.
(ii) Repetition is not allowed.
To make four digit number we cannot take 0 at thousand’ place.
Therefore digit at thousand’s place can be chosen in 3 ways. The
digit at hundred’s place can be chosen in 3 ways. The digit at ten’s pla ce
can be chosen in 2 ways. The digit at unit’s place can be chosen in 1way.
Therefore, four -digit number can be made by 3x3x2x1=18 ways.
[7] In how many ways can 5 men, 4 women and 3 children be
arranged for photographs so that (i) all men are together and so are
all women and children. (ii)all men are together and so are all
women. (iii) only all men are together and so are all women and
children.
Solution:
(i)all men are together and so are all women and children.
Now, 5 men forms one group, 4 women forms one group and 3
children also forms one group. These three groups with all men, women
and children are together can be arranged in 3! Ways. Further, 5 men
among themselves in 5! ways, 4 women among themselves in 4! ways, 3
children among the mselves in 3! ways.
Therefore, 5 men, 4 women and 3 children be arranged for photographs so
that all men are together and so are all women and children in 3!5!4!3!
ways.munotes.in

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56(ii)all men are together and so are all women.
Now, 5 men forms one group, 4 wom en forms one group and 3 children
can be arranged in 5! Ways. . Further, 5 men among themselves in 5!
Ways, 4 women among themselves in 4! ways.
Therefore, 5 men, 4 women and 3 children be arranged for photographs so
that all men are together and so are a ll women in 5!5!4! ways.
(iii) only all men are together.
Now, 5 men forms one group, 4 women and 3 children can be arranged in
8! ways.
Further, 5 men among themselves in 5! Ways.
Therefore, 5 men, 4 women and 3 children be arranged for photographs so
that only all men are together in 8!5! ways.
[8] How many different 7 -place license plates are possible if the first 3
places are occupied by letters and the final 4 by numbers under the
assumption that no letter or number can be repeated in a single
license plate?
Solution:
7-places license plates with the first 3 places are occupied by letters
selecting from the A to Z (26 letters) and the final 4 by numbers selecting
from the numbers 0, 1, 2……9 (10 numbers)
If no letter or number can be repeated in a single license plate, the first
three places are occupied out of 26 letters in26p3ways and the final 4
numbers are occupied from 10 numbers in10p4ways.
Therefore, different 7 -place license plates are possible if the first 3 places
are occupied by lett ers and the final 4 by numbers in
26p3x10p4=26x25x24x10x9x8x7= 78624000 ways.
[9] When Dr. Shah arrives in his dispensary, he finds 12 patients
waiting to see him. If he can see only one patient at a time, find the
number of ways, he can schedule his pati ents (a) if they all want their
turn, and (b) if 3 leave in disgust before Dr. Shah get around to seeing
them.
Solution:
(a)if they all want their turn, there are 12 patients and all wait to see the
doctor.
Therefore, number of ways Dr. Shahcan schedule h is patients
=12P12=12!=479,001,600
(b) if 3 leave in disgust before Dr. Shah get around to seeing them.
There are 12 -3=9 patients. They can be seen in12P9ways = 79,833,600
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Combinations
57(B)Combi nation:
When no attention is given to the order of arrangement of the selected
objects, we get a combination. We know that the number of permutations
of n objects taking r at a time is,nPr. Since r objects can be arranged in r!
ways, therefore, there are r! Permutations corresponding to one
combination. Thus, the number of combinations of n objects taking r at a
time, denoted by,nCr, can be obtained by dividing,nPrby r!,
i.e.nCr=
=nPrr! ,nPr
Note:
(a)Since,nCr=nCn-r, therefore,nCris also equal to the combinations
of n objects taking (n -r) at a time.
(b)(b) The total number of combinations of n distinct objects taking
1, 2, ...... n respectively, at a time isnC1+nC2+nC3+nC4………. +nCn=
2n-1.
Examples:
[1]In how many ways two balls can be selected from 8 balls?
Solution:
2 balls can be selected from 8 balls in8C2ways.
8C2=
=
=
=28
[2]In how many ways can 5 students be selected for a student's
committee o ut of 7 students?
Solution:
No. of ways 5 students can be selected for a student's committee
out of 7 students=7C5
=
=
=
=
=21
[3]In how many ways can 2 boys and 2 girls be selected fr om a group
of 6 boys and 5 girls?
Solution:
No. of ways 2 boys can be selected from a group of 6 boys=6C2and No.
of ways 2 girls can be selected from a group of 5 girls=5C2munotes.in

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58Therefore,
No. of ways 2 boys and 2 girls can be selected from a group of 6 boys and
5g i r l s
=6C2x5C2
=
x
=
x
=
x
=15x10=150
[4] The staff of a department consists of a manager, an officer and 10
clerks. A committee of 4 i s to be selected from the department. Find
the number of ways in which this can be done so as to always include
(1) the manager, (2) the manager but not the officer, (3) neither the
manager nor the officer.
Solution:
(1)The number of ways in which this can be done so as to always
include the manager:
The manager can be selected in only one way and the remaining three
members of the committee can be selected from amongst the 11 members
of the department in11C3ways.
Therefore, number of ways in which the co mmittee can be selected so as
to include the manager=1 x11C3=
=1 6 5
(2)The number of ways in which this can be done so as to always
include the manager but not the officer:
The manager can be selected in only one way. Since officer is not
to be included in the committee, the remaining three members can be
selected amongst 10 clerks in10C3ways.
Therefore, number of ways in which committee can be selected so as to
include the manager but not the officer= 1x10C3=
=1 2 0
(3)The number o f ways in which this can be done so as to always
include neither the manager nor the officer:
As all the four members are to be selected from 10 clerks and neither the
manager nor the officer is to be included in the committee,
Number of possible ways =10C4=
=2 1 0
[5] In a party, every person shakes hands with every other person
present. If the total number of handshakes was 105, find the total
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Combinations
59Solution:
Assume there are n persons in the party, t hen the number of handshakes
isnC2.
AsnC2= 105 i.e.
=1 0 5
Therefore, n (n -1) = 210
Thus, factors of 210 such that they are successive integers are 15 x 14 =
210
So, the number of persons present at the party are 15.
[6] A person has 12 friends of whom 8 are relatives. In how many way
can he invite7 guests such that 5 of them are relatives?
Solution:
Of the 12 friends, 8 are relatives and the remaining 4 are not relatives. He
has to invite 5 relatives and 2 friends as his guests. 5 relative can be
chosen out of 8 in8C5ways; 2 friends can be chosen out of 4 in4C2ways.
Hence, by the fundamental principle, the number of ways in which he can
invite 7 guests such that 5 of them are relatives and two of them are
friends =8C5x4C2
=
x
=
x
= 8x7x6
=336
[7] A box contains 7 red, 6 white and 4 blue balls. How many
selections of three balls can be made so that (a) all three are red, (b)
none is red, (c) one is each colour?
Solution:
(a)all three are red:
All three balls will be red colour if taken out from 7 red balls and this can
be done in7C3ways i.e.
=
= 35
(b)none is red :
None of three balls is red if these are chosen from 6 white and 4 blue balls
and this can be done in10C3ways i.e.
=
=
120
(c) one is each colour:
The number of groups of three balls such that one is of each colour
=7C1x6C1x4C1= 168munotes.in

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604.2EXERCISES
(A)Multiple ChoicesChoose the correct alternative from the following:a)Permutation is used when………
(A)Order is not important (B)order is important
(C)either (A) OR (B) (D) neither (A) OR (B)
b)Combination is used when………
(A)order is not important (B)order is important
(C)either (A) OR (B) (D) neither (A) OR (B)
c)Ifnp3=6nc4, then value of n is
(A)10 (B)8 (C)7 (D)5
d)Number of way four different books can be arranged
(A)4 (B) 24 (C) 1 (D) 0
e)In how many different ways can 6 people be photographed, if only 2
can be seated at a time?
(A)6 (B) 2 (C) 15 (D) 30
f)The value of r if 54Pr=65Pr-1:
(A)5 (B) 3 (C) 2 (D) 4
g)In how many four digit numbers can be formed from the digits 0, 1,
2… to 9, if no digit is repeated?
(A) 2688 (B) 4536 (C) 5040 (D) 3024
h)How many words can be formed of the letter ARTICLE, so that the
vowels occupy only the even position?
(A) 576 (B) 840 (C) 600 (D) None
j)There are 8 questions in a question paper. In how many different ways
can a student attempt 7 questions from the paper?
(A) 1 (B) 8 (C) 7 (D) 15
k)There are 10 people in a room. Each person shakes hand with every
other person in the room. The total number of hand shake is:
(A) 10 (B) 100 (C) 20 (D) 45
l)A Supreme court bench consist of 5 judges. The number of ways in
which the bench can give majority decision is
(A) 10 (B) 18 (C) 17 (D) 16
m) Relationship betwee nnprandncris
(A)npr=ncr+r! (B)npr=ncrxr ! (C)npr=ncr-r! (D)npr=ncr/r!munotes.in

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61(B)Problems
[1]In how many ways can the letters of the word COMMITTEE be
arranged?
[2]A group of 6 students comprised of 3 boys and 3 girls. In how many
ways could they be arranged in a straight line such that (1) the girls and
boys occupy alternate position? (2) All three boys were sitting together?
[3]There are 3 different books of economics, 4 different books of
commerce and 5 different books of statistics. In how many ways these can
be arranged on a shelf when
(a)all the books are arranged at random ,
(b)books of each subject are arranged together,
(c)books of only statistics are arranged together, and
(d)books of statistics and books of other subjects are arranged together?
[4]How many 4 lettered different words can be formed by using the
letters a, b, c and d? When (i) repetition is allowed and (ii) repetition is
not allowed
[5]In how many different ways can 6 people be photographed, if only 4
can be seated at a time?
[6]A cricket team of 11 members is to be selected from 18 players of
whom 10 are batsmen, 6 are bowlers and 2 are wicket -keepers. In how
many ways can the tea m be formed if at least one wicket keeper and at
least 5 bowlers are to be included?
[7] In how many ways can a pack of 52 cards be divided equally among 4
players in order?
[8]There are 8 professors and 12 students out of whom a committee of 2
professo rs and 3 students is to be formed. Find the number of ways in
which the committee can be formed such that (i) a particular professor is
included. (ii) a particular student is to be excluded.
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PROBABILITY
Unit Structure :
5.1Introduction to Probability
5.2Concept of Probability
5.3Basic rules of Probability
5.4Solved Examples
5.5Exercises
5.1INTRODUCTION TO PROBABILITY
UNCERTAINTY AND MYSTERY ARE ENERGIES OF LIFE.DON’T
LET THEM SCARE YOU UNDULY, FOR THEY KEEP BOREDOM
AT BAY AND SPARK CREATIVITY.
R.I.FITZHENRY
Uncertainty is an important part of everyday life. Some events are certain
like the Sun rises every day, but many events like winning or losing the
game, stock market etc. involved uncertainty or chance. When there is no
guarantee of an event it belongs to the domain of uncertainty. Though
such uncertainty is involved in most of the situations, such happenings
have some “chance” or “probabi lity” of their occurrence. To find a
measure for probability, it is necessary to perform certain experiments,
visualize the possible outcomes of the experiment. The following terms
are used to describe the result of an experiment:
Statistical Experiment :A statistical experiment is a random i.e.
nondeterministic experiment. An experiment is a process of measurement
or observation which has number of possible results (outcomes). The
experiments like tossing of an unbiased coin, throwing a fair dice,
Selecti ng a committee of three persons from a group of ten persons, ,
Measuring copper content of brass, Number of customers enters a mall are
examples statistical experiment. Measuring no. of hands of undergraduate
students, measuring density of pure gold are ex amples non -statistical
experiment.
Sample Space of an Experiment : A set of all possible outcomes of an
experiment is called sample space of the experiment. It is denoted by S.
e.g. sample space for the random experiment of tossing of an unbiased
coin is S = {H, T}, where H denotes “head” and T denotes “tail”.
Similarly, sample space for the random experiment of throwing a fair dice
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63Sample points : The possible outcomes in a sample space of an
experiment are called Sample points. The number of sample points in
sample space is denoted by n(S).e.g. in case of tossing of an unbiased
coin, n(S)=2 and in case of throwing a fair dice, n(S)=6
An Event: Any subset (say A) of sample space S of an experiment is
called an Event A and the number of sa mple points in event A is denoted
by n(A).e.g. If a fair dice is rolled and event A is that even number occurs
on the uppermost face of the dice, then A={2,4,6}and n(A) =3.
Impossible Event: An event corresponding to null set is called an
impossible event . e.g. If a fair dice is rolled and event A is that number 7
occurs on the uppermost face of the dice , then event A is impossible
event.
Certain Event: An event that is certain to occur is called certain event.
Certain event contains all the elemen ts of the sample space.
Complementary Event: If A is any event from sample space of an
experiment S, then non -occurrence of event A is an event called
complementary event of event A and is denoted by Acor A’.e.g. If
S={1,2,3,4,5,6} and A={2,4,6} then Ac={1,3,5}
Algebra of Events:
(i)A U B: Either event A or event B or both A and B occur.
(ii)A
: Occurrence of both event A and B simultaneously.
Equally Likely Events: The events of a sample space which have same
chance of occurring are cal led Equally Likely Events. Eg. Consider an
experiment of throwing a die. Any one of the numbers cannot be expected
to turn in preference to any other of them. These sample points are equally
likely. The event corresponding to these sample points.
i.e.{1}, {2},{3},{4},{5},{6} are equally likely events.
Mutually Exclusive Events:
If two events A and B cannot occur together, then A and B are said to be
mutually exclusive events. In such a case A
=φ. e.g. AnAc=φ, hence
A and Acare mutually ex clusive Events. Thus complementary events are
mutually exclusive Events. Eg. In one throw of a fair die, A be the event
obtaining even numbers and B be the event obtaining odd numbers on the
uppermost face. Then S={1,2,3,4,5,6}; A={2,4,6} ; B={1,3,5}.
Here A
=φ. Hence events A and B are called mutually exclusive
events.
Exhaustive events: If two events A and B are such that AUB =S, then A
and B are called Exhaustive Events. e.g. AUAc=S, hence A and Acare
exhaustive Events. Thus complementary events are ex haustive Events. . In
one throw of a fair die, A be the event obtaining even numbers and B bemunotes.in

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64the event obtaining odd numbers on the uppermost face. Then
S={1,2,3,4,5,6}; A={2,4,6} ; B={1,3,5}.
Here AUB =S. Hence events A and B are called exhaustive events.
5.2CONCEPT OF PRBABILITY
Mathematical or Classical definition of Probability of an event :
If sample space S of an experiment has n equally likely ,Mutually
exclusive and Exhaustive outcomes of which mare favourable to the
occurrence of event A ,then probability of event A ,denoted by P(A) is
given by
P( A )=
=
=
It is obvious from the definition that, 0 ≤P(A)≤1
Probability Assigning Techniques:
The assignment of probabilities to various elementary events of a sample
space can be done in any one of the following three ways:
[1]Subjective probability Assignment: This is the technique of
assigning probabilities on the basis of personal judgement. Under the
subjective assignment, th e probabilities to various elementary events are
assigned on the basis of the expectations or the degree of belief of the
statistician. Such assignment may differ from individual to individual and
much depends upon the expertise of the statistician an assi gning the
probability. These probabilities, also known as personal probabilities, are
very useful in the analysis of various business and economic problems.
[2]Classical/Mathematical Probability Assignment:
This is the technique under which the proba bility is assigned by
calculating the ratio of the number of ways in which a given outcome can
occur to the total number of possible outcomes. We know that various
elementary events of a random experiment, under the classical definition,
are equally likely and, therefore, can be assigned equal probabilities.
[3]Empirical/Statistical probability assignment: This is an objective
method of assigning probabilities and is popular with the decision maker.
Under this technique the probability is assigned by ca lculating the relative
frequency of occurrence of of a given event over an infinite no. of
occurrences. The probability assignment through this technique may well
be unrealistic if future conditions do not happen to be a reflection of the
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655.3SOME BASIC RULES OF PROBABILITY
(i)Addition Theorem (General): If A and B are any two events
associated with an experiment, then the probability of occurrence of
events A or B or both A and B is given by
P (AUB) =P (A) + P (B) –P(AB)
If A, B and C are three events associated with an experiment, then
the probability of occurrence of atleast one of the three events A, B and C
is given by
P(AUBUC)=P(A) + P(B)+P(C) –P(AB)-P(BC)-P(AC)+P(A BC)
(ii)Addition Theorem (Mutuall y exclusive event):
If A and B are mutually exclusive events then A B=φ.
In this case Addition theorem becomes P (AUB) =P (A) + P (B)
If A, B and C are mutually exclusive events, then
P( A U B U C )=P( A )+P( B )+ P(C)
(iii)If A and Acare complementary events, then P (Ac)= 1-P( A )
Conditional Probability : The probability of event A given that Event B
has already occurred is called conditional probability of event A given that
the event B has already occurred and is denoted by P(A/B).Similarly ,
conditional probability of event B given that A has already occurred is
denoted by P(B/A).
Independent Event : If probability of event A is not affected by
occurrence or non -occurrence of event B, then the events A and B a re
independent events. In such a case,
P (A/B) = P (A) and P (B/A) =P (B).
Multiplication Theorem: If A and B are any two events associated with
an experiment, then the probability of simultaneous occurrence of events
A and B is given by
P( AB) =P (A) .P (B/A) = P (B) .P (A/B)
Multiplication theorem (Independent Events): If events A and B are
independent events then, P (A/B) = P (A) and P (B/A) =P (B). Hence
Multiplication theorem is
P( AB) =P (A) .P (B)
If A, B and C are three events associated with an experiment, then
the probability of occurrence of atleast one of the three events A, B and C
is given by
P( ABC)=P( A ). P( B ) . P ( C )munotes.in

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66Addition theorem (Independent Events):
P( A U B )=P ( A )+P ( B ) –P(A).P(B)
Baye’s Theorem On Inverse Probability:
Let A 1,A 2,A 3, ……..A nbe n mutually exclusive and exhaustive
events defined on a sample space S. Let H be an event defined on S, such
that P(H) 0.
Then
P(A i/H) =
Bayes' Theorem is based on the concept of conditional probability. The
revised probabilities, thus obtained, are known as posterior or inverse
probabilities. Using this theorem it is possible to revise various business
decisions in the light of additional information.
5.4SOLVED EXAMPLES:
[1] If P( A U B )=
,P( ͞A) =
,P (B )=
,Find P (͞B), P( A B ) and
P(A/B)
Solution:
P(B) =1 -P(B) =
P(A) =1 -P(A) = 1 -
=
P( AUB)=P ( A )+P ( B ) –P(AB)
Therefo re, P (A B)= P ( A ) +P ( B ) -P( AUB)
=
+
-
=
[2] Given : P(A) =
,P ( B )=
,P ( C )=
,P(A B) =
,P(A C)
=
,P ( B C) =
,
P(AB
)=
Find P (A U B U C)
Solution:
P(AUBUC)=P(A) + P(B)+P(C) –P(AB)-P(BC)-
P(AC)+P(A BC)
=
+
+
-
-
-
+
=
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67[3] Determine the probability of the following events in drawing a
card from a standard desk of 52 cards. (a) a seven (b) A black card
(c) An ace or a king (d) a black two or a black three (e) a red face
card
Solution:
Let S be the sample space of the experiment. Therefore, n(S) = 5
(a)A: a card with number 7 is drawn
There can be 4 ways to get a seven from a d eck of 52 cards. n(A) = 4
by definition, P(A) =
=
(b)B: a black card is drawn
There are 26 black cards in a deck. n(B) = 26
by definition, P(B) =
=
=
(c)C:an ace or a king is drawn
There are 4 ace cards and 4 kings. So number of ways in which ace or
king is drawn
= n(C) = 4+4= 8
by definition, P(C) =
=
=
(d)D : a black 2 or black 3 is drawn.
There are 2 black cards having number 2 (i.e. spade and club)
There are 2 black cards having number 3 (i.e. spade and club)
n(C)=2+2=4
by definition, P(D) =
=
=
(e)E: a red face card is drawn
There are 12 fa ce cards (4x Jack, Queen, King)
Therefore, there are 6 red face cards. n(E)=6
by definition, P(E) =
=
=
[4] If 2 fair dice are rolled, find the probability that sum of the
numbers appearing on the u ppermost faces of the dice is sum is (i)
greater than 8 (ii) an odd number (iii) a perfect square
Solution:
Let S be the sample space of the experiment
S= {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
(2,1),(2,2),2,3),(2,4),(2,5),(2,6),
(3,1),(3,2 ),3,3),(3,4),(3,5),(3,6),
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}munotes.in

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68(i)A: sum greater than 8
A = {(3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4),(5,6),(6,5);(6,6 )}
Therefore, n(A)= 10
by definition, P(A) =
=
=
(ii) B: sum is an odd number (i.e. sum is 3,5,7 or 9)
Sum 3: {(1,2),(2,1)}
Sum 5: {(1,4),(2,3),(3,2),(4,1)}
Sum 7: {(1,6),(2,5),(3,4),(4 ,3),(5,2),(6,1)}
Sum 9: {(3,6),(4,5),(5,4),(6,3)}
Sum 11:{(5,6),(6,5)}
B = {(1,2),(2,1),(1,4),(2,3),(3,2),(4,1), (1,6),(2,5),(3,4),(4,3),(5,2),(6,1),
(3,6),(4,5),(5,4),(6,3), (5,6)(6,5)}
Therefore, n(B)= 18
by definition, P(B) =
=
=
(iii) C: sum is a perfect square (i.e. sum is 4 or 9)
C={(1,3),(2,2),(3,1),(3,6),(4,5),(5,4),(6,3)}
by definition, P(C) =
=
[5] A coin is tossed thrice. What is the probability of getting 2 or more
heads?
Solution:
The sample space S ={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
2 or more heads imply 2 or 3 heads as only 3 coins are tossed
A : Occurrence of 2 heads
B : Occurrence of 3 heads
A ={HHT,HTH,THH}
B = {HHH}
P(A) =
=
and P(B) =
=
As A and B are mutually exclusive, probability of getting 2 or
more heads is
P(A
P(A) + P(B)
=
=
[6] From the past experience it is known that A can solve 3 examples
out of given 5 and B can solve 4 examples out of given 7. An example
is given to both of them to solve independently. Find t he probability
that a) the example remains unsolved b) the example is solved c) only
one of them solves the example. [Ans. 6/35, 29/35, 17/35]munotes.in

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69Solution:
X : A solves an example
Y: B solves an example
P(X)=
Therefore, P( X)=1 -P(X)=1 -
And P(Y)=
P(Y)=1 -P(Y)=1 -
(a)P(Example remains unsolved):
=P(A cannot solve the problem and B cannot solve the problem)
=P(XY)
=P(X).P( Y)……[As X and Y are independent, X and Ya r e
independent
=
x
=
(b)P(Example is solved):
=P(A solves the problem or B solves the problem or both of them
solve)
=P(XUY)
=P(X)+P(Y) -P(XY)
=P(X)+P(Y) -P(X).P(Y)
=
+
-
x
=
(c)P(Only one of them solves):
=P(A solves the problem and B cannot solve it)+
P(B solves the problem and A cannot solve it)
=P(X Y) + P( XY)
=
+
+
x
=
[7] Shankar is known to hit a target in 5 out of 9 shots whereas Ha ri is
known to hit the same target in 6 out of 11 shots. What is the
probability that the target would be hit once they both try?
Solution:
Let A : Shankar hits the target
B : Hari hits the target
P(A)=
P(A)=
And P (A B)=P ( A )xP ( B )
=
x
=
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70The probability that the target would be hit is given by
P(A U B) = P(A) + P(B) –P(AB)
=
+
-
=
[8] the following data pertains to analysis of 200 engineers
Age(years) Bachelor’s
degree onlyMaster’s
degreeTotal
Under 30 90 10 100
30 to 40 20 30 50
Over 40 40 10 50
Total 150 50 200
If one engineer is selected at random from the company. Find
the probability that:
(a) he has only a bachelor’s degree.
(b) he has a master’s degree given that he is over 40.
(c) he is under 30 g iven that he has only a bachelor’s degree .
Solution:
Let A: Engineer has a bachelor’s degree only
B: Engineer has a master’s degree
C: Engineer is under 30 years of age
D: Engineer is over 40 years of age
(a)P(A)=
=
(b)P(B/D)=
=
(c)P(C/A)=
=
[9] A suitcase consists of 7 marbles, of which 2 are red and 5 are
green. Two marbles are removed at random and their colour noted.
The first marble is not replaced before the second is selected. Find the
probability that the marbles will be (a) both red, (b) of different
colours, (c) the same colour
Solution:
Let R 1: The first marble removed is red
R2: The second marble removed is red
G1: The first marble removed is green
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71Therefore,
R2 P(R 1R2)=P ( R 1). P( R 2/R1)
=
x
=
R1
G2 P(R 1G2)=P ( R 1). P(
G2/R1)=
x
=
R2P(G 1R2)=P ( G 1). P( R 2/G1)
=
x
=
G1
G2P(G 1G2)=P ( G 1). P( G 2/G1)=
x
=
(a)P( Both marbles are red) = P(R 1R2)=P ( R 1). P( R 2/R1)=
x
=
(b)P ( Both marbles are of different colours) = P(R 1G2)+P ( G 1R2)
P(G 1R2)=P ( G 1). P( R 2/G1)=
x
=
P(R 1G2)=P ( R 1). P( G 2/R1)=
x
=
P ( Both marbles are of different colours) = P(R 1G2)+P ( G 1R2)
=
+
=
(c)P(Both marbles are of same colour) = P(R 1R2)+P ( G 1G2)
=
+
=
[10] Two sets of candidates are competing for the positions of board of
directors in a company. The probabilities That the first and second setmunotes.in

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72will win are 0.6 and 0.4 resp ectively. If the first set wins, the
probability of introducing a new product is 0.8 and corresponding
probability if the second set wins is 0.3. The new product is
introduced .What is the probability that the first set won?
Solution:
Let A: First set win s P(A) = 0.6
B: Second set wins P(B) = 0.4
X: Product is introduced
Y: Product is not introduced.
X
A
Y
X
B
Y
By Baye’s theorem,
P(First set wins given that the product is introduced) = P( A/X) =
=
=
5.5EXERCISES
Do as Directed:
(1) Choose the correct alternative from the following
a)The value of probability is:
(A)Less than 0 (B) between 1and 10
(C) more then 0 (D) between 0 and1
b)All possible outcomes of a statistical experiment are called:
(A)Sample space (B) Hyper space
(C) Cyber space (D) Virtual spacemunotes.in

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73c)Complimentary events ar e:
(A)Exhaustive (B) exclusive
(C) both A and B (D) none
d)IfABSthen A and B are said to be
(A)Exhaustive (B) exclusive
(C) both A and B (D) none
e)IfABthen A and B are said to be
(A)Exhaustive (B) exclu sive
(C) both A and B (D) none
(2)State whether the following are statistical experiment or not.
(a)Tossing of a coin.
(b)Throwing a fair die.
(c)Measuring no. of hands of undergraduate students.
(d)Selecting a committee of three persons from a group of eight
persons.
(e)Measuring density of pure gold.
(f)Number of customers enters a mall.
(3)Write sample space for the following random experiment.
(a)Tossing of two coins.
(b)Throwing a fair dice.
(c)Selecting a card from pack of 52 cards.
(d)Number of customers enters a mall.
(4)In experiment of throwing a fair die:
A = {3,4} B = {4,5,6}
Give (a)cA(b)ccAB(c)cAB(d)cAB(5)An insurance company insured 2,000 scooter drivers, 4,000 car drivers
and 6,000 truck drivers. The probability of an accident is 0.01, 0.03 and
0.15 in the respective category. One of the insured driver meets an
accident. What i s the probability that he is a scooter driver?
(6)Consider a population of consumers consisting of two types. The upper
class of consumers comprise 35% of the population and each member has
a probability 0.8 of purchasing brand A of a product. Each member of the
rest of the population has a probability 0.3 of purchasing brand A of the
product. A consumer, chosen at random, is found to be buyer of brand A.
What is the probability that the buyer belongs to the middle and lower
class of consumers?
Problems :
[1]A bag contains 7 white balls, 5 black balls and 4 red balls. If two balls
are drawn at random from the bag, find the probability that a) both the
balls are white b) One is black and other is red.
[Ans: 7/40, 1/6]munotes.in

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74[2]A room has 3 lamps. From a collection of 10 light bulbs of which 6 are
not good, a person select 3 at random and puts them in the sockets. What
is the probability that he will have light from all the three lamps?
[Ans.: 1/30]
[3]P(A) = 2/3, P(B) = 1 /4, P(AUB) = 5/12. Find P(A B), P(A/B) and
P(B/A).
[Ans.: 1/6, 2/3, 1/2]
[4]A box contains 5 red and 4 blue balls and other box contains 4 red and
7 blue balls. A ball is selected at random from the first box and without
noting the color put in the ot her. A ball is then drawn from the second box.
What is the probability that it is blue?
[Ans.: 67/108]
[5]The probability that A will be alive 30 year hence is 0.3 and that B will
be alive 30 years hence is 0.4 what is the probability that a) both w ill be
alive b) only A will be alive c) both will not be alive d) at least one of
them will be alive?
[Ans.: 0.12, 0.18, 0.42, 0.58]
[6]Students of a class were given a test in Economics & Accountancy. Of
these students 20% failed in Economics, 15% fai led in Accountancy and
5% failed in both. Find the chance that a student selected at random a)
failed in at least one of the two subjects. b) failed in only Accountancy c)
passed in both the subjects.
[Ans.: 0.3, 0.1, 0.7]
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RANDOM VARIABLE AND ITS
PROBABILITY DISTRIBUTION
Unit Structure
6.1 Introduction to Probability Distribution
6.2 Standard Distributions
6.3Exercises
Aim and Objective :
1)Probability distribution are used both in theoretical aspects as well as
practical aspects.
2)They are the fundamental concepts of statistics.
6.1INTRODUCTION :
Probability distribution can either be discrete or continuous. A discrete
probability distribution is sometimes called a probability density function.
The stu dy of a population can be done either by constructing an observed
(or empirical) frequency distribution, often based on a sample from it, or
by using a theoretical distribution. If a random variable satisfies the
conditions of a theoretical probability dis tribution, then this distribution
can be fitted to the observed data.
It finds applications in the understanding and analysis of a large number of
business and economic situations. E.g. It is possible to test a hypothesis
about a population, to take deci sion in face of uncertainty, to make
forecasts, etc.
Probability distribution of a random variable X :
All possible values of a random variable ‘X’ along with their
corresponding probabilities such that the sum of all these probabilities is
unity, is called a probability distribution of the random variables ‘X’.munotes.in

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76In general,
X1X2X3X….. ….. …..nXPX1P2P3P….. ….. …..nPWhereiPOfor all''iand
11n
i
iP

Probability Mass Function :
ProbabilityPX, such that the discrete random variable ‘X’ takes
correspondence to every sample point in sample space S is called
probability Mass Function (p, m, f,) of a random variable ‘X’.
If123, , ,....nxxx xare different values of a discrete rand om variable X and123,,, . . . . .nPx Px Px Pxare their respective probabilities such that
i)0iPx1, 2, 3....inii)1iPx1, 2, 3....inThen,Pxis known as the Probability Mass function of variable X.
Probability Density Function : The probabilities associated with a
continuous random variable X are determined by the probability density
functionfxof a random variable X.
Properties of Probability density function ‘fx’:
1)0fxfor all values ofx2)The probability thatxwill lie between two numbers ‘a’ and ‘b’ is
equal to the area under t he curveyf xbetweenxaandxb.
3)The total area under the entire curveyf xis equal to one.
Expectation and Variance of a random variable ‘X’ :
If a random variable X takes values12, ,....nxx xwith corresponding
probabilities12, ,.....nPP Prespectively, expectation of the random variable
X is denoted byEXand is given by11 2 2
1
22....nnn
ii
iXx p x p x pxp p x
xp x


EXis also known as expected value of X. If population mean is, then the expected value ofxis given by,EX=Variance of ‘x’munotes.in

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77

222
22
22
2xVE x E xpx px
Ex
Ex





Standard Deviation 22..xSD V p x p x 
I)Whenxis discrete random variable with probability mass functionPx, then its expected value is given by :xP xand its variance is22 2Ex where22Ex x Px.
II)For a continuous random variablexdefined in,, its expected
value (i.e. mean) and variance is given by,
Ex x f xd x
and22 2Ex where22Ex xf xd x
 
Properties of Expected values
1)Expectation of a constant K is K i.e.Ex K, for any constant K.
2)Expectation of sum of t wo random variables is sum of their
expectation i.e.Ex y Ex Eyfor any two random variable ‘x’
&‘y’.
3)Expectation of the product of a constant and a random variable is the
product of co nstant and the expectation of random variable. i.e.,.EKx KExfor any constant K.
4)Expectation of product of two random variables is the product of
expectations of two random variables, provided the two variables are
independent. i.e.Ex y Ex Eywherex&yare independent.
E.g.
1)Two fair dice are rolled. If X denotes the sum of the numbers
appearing on the uppermost faces on the dice, find
1)4PX2)10PX 
3)37PX4)3PXmunotes.in

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78If S is a sample space of experiment.
Solution :



1,1 1, 2 1, 3 1, 4 1, 5 1, 62,1 2, 2 2,3 2, 4 2,5 2, 63,1 3, 2 3, 3 3, 4 3, 5 3, 64,1 4, 2 4,3 4, 4 4,5 4, 65,1 5, 2 5, 3 5, 4 5, 5 5, 66,1 6, 2 6,3 6, 4 6,5 6, 6S  

36
121 , 136231 , 2 , 136nS
PX P
PX P
 
 
The Probability distribution of X is givenX2 3 4 5 6 7 8 9 10 11 12Px136236336436536636536436336236136Here, 12
21xPX

1)42 3PX PX o r X2312
36 36
112PX PX

2)10 10 11 12PX PX o r X o r X 10 11 12321
36 36 36
16PX PX PX

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793)37 4 5 6PX P X o r X o r X456345
36 36 36
13PX PX PX

4)345 . . . 1 2PX PX PX PX 
But 12
21xPx
23 . . . . 1 2 145 . . . . 1 2PX PX PX
PX PX PX12 312136 36
1112PX PX

Q.2 A random variable x has following probability distribution.X0 1 2 3 4 5 6 7Px0 2K 3K K 2K2K27K22KKFind :
1)Value of K
2)3PX3)4PX4)25PxSolution :
1)1Px222
202 3 2 7 2 110 9 1 0
11 0 1 0
110KK KK K K KK
KK
KK
K


2)30 1 2PX PX PX PX02 3
50.5 0.1KK
K
K
munotes.in

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803)3407PX PX PX PX222
2
227 2
10 3
10 0.1 0.1KK K K KKK
 

4)25 3 4 5Px P X P X P X222
3KK KKK


Q.3If X is a random variable having probability mass function.
;0 , 18,24;316xPX x x
Kx
KxxFind the value of K &EXSolution :PX xis p.m.f.
3
01xPX x
 0123 113018 4 16
7 2 162PX PX PX PX
KK
KK
;0 , 18
1;22;38xPX x x
x
xx


Now,  3
0xEx xPX x01 12 23 3OP X P X P X P Xmunotes.in

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8111301 2 382819188
94

Q.4If X is a number appearing on the uppermost face of a fair dice. FindEXandVX.
Solution :
Since X is a no. appearing on uppermost face of a fair dice, the
probability distribution of X is given byX1 2 3 4 5 6PX x161616161616 11111112345666666611234566
21
6
72EX x PX x 


Now,22Ex x PX x  222222
22222211111112345666666611234566
916

22VX EX EX 291 76291 49643512
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82Q.5In a cer tain lottery, one price of `1000, 2 prizes of `500 each and 5
prizes of `100 each are to be awarded to 8 tickets drawn from the total
number of 10,000 tickets sold at a price of `1 per ticket. Find the expected
net gain to a person buying a particular tick et.
Solution :
Let the person get a prize of `X with probabilityPXPx1000 500 100 01100002100005100008110000   Ex x PX x125 81000 500 100 0 110000 10000 10000 100001
40.25   the person has to spend `1 for buying the ticket, his net expected gain =
0.25 -1= `-0.75
i.e. the person will incur an expected loss of 0.75`.
6.2SOME STANDARD DISTRIBUTIO NS :
1)Binomial Distribution :
i)It is named after Swiss Mathematician Jacob James Bernoulli and is
also known as “Bernoulli Trial”.
ii)Trial or Process, the literal meaning of word ‘Binomial is two groups’.
Hence, in this distribution, frequencies ar e divided on basis of two
aspects or two possible outcomes, called as “Success” and “Failure”.
iii)Each trial is associated with two mutually exclusive and exhaustive
events i.e.1pq, one of them is called success (p) and other is
called failure (q) E.g. When we toss a coin, there are only two possible
outcomes -Head or tail and one of them must happen.
iv)Binomial distribution is symmetric when0.5pq.
v)The general form of binomial distribution is the expan sion of npq,
in which the no. of success is written in a descending order. If the no.
of success is written in ascending order, then nqpwill be
expanded.
vi)The trials are independent.
vii)The random exper iment is performed repeatedly for a fixed number of
times. For example, a coin is tossed 8 times. In other words, the
number of trialsnare finite and fixed and it is a positive integer.munotes.in

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83viii) If,xB n p0,1, 2,...,nx n x
xPx s u c c e s s f x C P q x n 
p : probability of success
q : probability of failure (q = 1 –p)
n : number of trials
x : no. of successes in ‘n’ trials
ix)It is a probability distribution in which each0pr11Op x
px 
x)Mean ofnPbinomial distribution.
xi)Variance of Binomial Distribution
 2xx rV npq V
 xxVn p q
Variance is always i.e.npq npless than mean
Eg (i) For a Binomial distri bution, mean = S and standard deviation = 2.
Find n and p.
Solution :
Mean = S;
S.d. = 2np…..(1)
Variance =2244npq…….(2)
Dividing (2) by (1), we get45q115pqMean152 5 &5np n pii)If X has a Binomial distributed with2FXand 43Var X ,
find the probability distribution of X.munotes.in

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84Solution :2434263113
26E X np
Var X npq
npqqnp
pq
np n


Hence, the distribution is 66 12
33xxnx n x
xx PX x C p q C Putting0,1, 2...6x, we get the probability distribution of X.X0 1 2 3 4 5 6PX x6472919272924072916072960729127291729iii)Twelve cards, 4 Kings 4 Queens and 4 Jacks, are removed from a pack
of cards and put in a box A card is picked up at r andom from this box,
noted down and is replaced back in the box. Find the probability for each
of the following when 6 cards are taken out, one by one.
a)P (at least one is a King)
b)P (S are Kings)
c)P (at most S are Kings)
d)P (none is a king)
Solution : Success is drawing a King out of 12 cards in a box.

4112 3121336pP s u c c e s s
qP f a i l u r en  a)P (at least one is a King)60 6
00610
112133PCpq
641729665729
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85b)P(5 are Kings) =5P
65 1
5
5
6
512334243Cpq
C
c)P(at most 5 are Kings)56PP
56
66
5612 1233 3313
729CC 
d)P(none is a King)0P566
5123364
729C
2)Poisson Distribution :
In events such as the accident in a factory, or the number of deat hs
in a city in one year by a rare disease, binomial distribution is not
applicable because we don’t know the value of ‘n’ in expression nqp.
In order to deal such situations. Poisson distribution comes to our rescue.
Here, one canno t count the number of times an event occurs or does not
occur.
i)Poisson Distribution was originated by French Mathematician Simon
Denis Poisson.
ii)The Poisson distribution is a discrete probability distribution in which
the number of successes are g iven in whole numbers such as
0, 1,..,..,..,etc.
iii)Poisson distribution is used in those conditions where value of P is
very smallpoand value of q is at most equal to 11qand
value of n is large. E.g. behavior of rare events.
iv)A random variable x is defined to follow Poisson distribution with
parameter ‘m’ and is given byxP mthen !mxemPxx
.
v)‘m’ is called uniparameter.
vi)ei stranscendental value. It is an irrational number lying between 2
and 32.718280emunotes.in

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86vii)Poisson distribution applies for discrete probability distribution i.e.11OP r P xviii) In practice, Poisson distribution may be used i n place of binomial
where100nand0.1pand01m.
ix)Mean of Poisson Distribution is given by,xmx)Variance of Poisson Distribution

p xpxVm nmn
m is always positive.
Mean = Expected valuexE x
Coefficient of =100 100mmmVariation.
xi)Applications are
a)In Biology, to count the number of bacteria.
b)In counting the number of defects per item in sta tistical quality
control.
c)In determining the number of deaths due to a rare disease.
d)In insurance problems, to count the number of casualties.
E.g.
i)For a Poisson distribution with0.7m, find2, 2PX PX ,
given0.70.497e.
Solution :
!mxemPX x Pxx

i.e. 20.7
20.722!ePX P

0120.7 0.7 0.720 1 2
0120.7 0.7 0.70! 1! 2!
0.497 0.348 0.1720.967PX PX PX PX
PP P
eee  

  
ii)If a random variable x follows Poisson distribution such that12PP , find its mean and variance .munotes.in

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87Solution :
!mxemPX x Pxx
12PP ………………. [given]121! 2!2mmem emm 


Mean & variance are both 2.
iii)The average number of incoming telephone calls per minute in a stock
firm is 2. Find the probability tha t during a given minute, 2 or more calls
are received.20.135e
Solution :
This is a Poisson distribution with mean m = 2.
!mxemPxx


 20 212 2 3 ......2 3 ....
101
2210! 1!
1 0.135 0.270
1 0.4050.595PX PX PX
PP
PP
ee 

  
   

3)Normal Distribution :
i)It was originated by Carl Gauss and is also known as ‘Gaussian
distribution’.
ii)The normal distribution is symmetrical about x=0.
Properties of Normal Distribution :
i)It is a bell shaped, symmetrical curve
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88iii)Quart iles are situated at equal distances from median31312QMM QQQM
iv)
The curve is asymptotic (i.e. Curve comes closer & closer but never
touches the base)
v)Semi Inter Quartile range (QD) is always 0.6745 times SD
310.6745 . .0.67452QD S DQQ 
i.e.23QD
3
10.6750.675Q
Q
vi)Mean deviation M.D. = 0.7979 S.D.4..5.. 0 . 8MDMDvii)QD:MD:SD = 10:12:15
viii) MD about mean, median and mode are equal.munotes.in

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89ix)Normal Distribution is biparam etric i.e.and S.D.are the
parametersxNwhere x is a continuous variable E.g. Height,
profit, rainfall, etc.
x)Area under standard 0.5 normal curve is 1
Area to left of the o rdinate = Area to the right of the ordinate
= 0.5
xi)Probability density function of x is given by PDF.
The equation of normal
curve 2221
2x
yf x e



0x

     Where= Standard deviationthe arithmetic mean
N= the number of observations
We can transform the variable x toxZ, then the equation takes the
form 2
21;
2Z
fZ e Z
 It is called P.D.F. for the standard normal variab le Z. It is a normal variate
with mean0and S.D.1.
xii)The standard normal distribution is also known as Unit Normal
Distribution or Z -distribution.
Note :
1)x is any real number.
2)is always positive
3)Coefficient100 100SD mMean mof variation Mean.
4)1mean area below Z=1 i.e. area betweento 1.
5)Probability = Area under the curve
6)%A r e a=A r e a1007)Standard Deviation2212munotes.in

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90Methods of fitting a Normal Curve
1)Area method
2)Ordinate Method
Relation between Binomial & Poisson Distribution
Conditions under which Binomial Distribution tends to becom e
Poisson distribution are as follows :
i)When number of trials is very largenii)When the occurrence of every trial is very small0, 1pq
iii)When average number of success (m) is equal to positive finite
quantitynpmn pRelationship between Binomial & Normal Distribution
Conditions under which Binomial distribution tends to become
Normal distribution.
i)n approaches to infinitynii)Values of p and q are moderate
Difference between Binomial, Poisson and Normal Distribution
1)Number of Distributions
Binomial, PoissonDiscrete Probability Distribution
NormalContinuous Prob ability Distribution
2)Definitions of Probability functions
Binomial Distributions =nx n xxxPC p q
Poisson Distributions =!mxxemPx

Normal Distribution = 2221
2x
fx e
r


3)Value of no. of trials ‘n’
Binomia l : finite
Poisson, Normal : Infinite / Very largePx4)Parameters
2 in Binomial :,np2 in Normal :,m1 in Poisson : m
Basic Comparison :Binomial ,Poisson,Normal Distribu tion
1)Types of Discrete Discrete Continuous
Distribution
2)Parameters : Two One Two,npm,munotes.in

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913)Restriction1OP0m01on parameter
4)Mean:npXmn p15)Variance :2npq2mn p226)Probability Function:xn xx xPC P q!0,1,2....mxxemPx
xm Mean

221
2xxxPe
where SD
mean





E.g.
1)If X follows normal distribution with mean 120 and variance 1600,
find,
i)140PX 
ii)110PX 
iii)100 130PX 
Area between Z = 0 and Z = 0.25 = 0.0987
Area between Z = 0 and Z = 0.5 = 0.1916
Solution :X follows normal distribution with mean 120 and variance 1600.120 & 1600 40 
i)140PX 
120 140 12040 40
0.5XP
PZ     
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92= (Area to left of Z = 0) + (Area between Z = 0 & Z = 0.5)
= 0.5 + 0.1961
= 0.6916
ii)110PX 

120 110 12040 40
0.25
0.25XP
PZ
PZ     

= (Area to left of Z = 0) + (Area between Z = 0 & Z = 0.25)
= 0.5 + 0.0987
= 0.598
iii)100 130PX 
 100 120 120 130 12040 40 40
0.5 0.25XP
PZ       
= Area between Z = -0.5 and Z = 0.25
= (Area between Z = -0.5 and Z = 0) + (Area between Z = 0 and Z =
0.25)
= 0.1916 + 0.0987
= 0.2903
2)If the weights of 10,000 soldiers in a regiment are normally distributed
with a mean of 72kgs and a standard deviation of 5kgs. How many
soldiers have weights above 80 kgs? Also find percentage of soldiers with
weights between 70 and 77kgs. Given that
Area (Z = 0 to Z = 1.6) = 0.4452
Area (Z = 0 to Z = 0.4) = 0.1554
Area (Z = 0 to Z = 1) = 0.3413
Solution :
Let X be the weights of soldiers in the regiment. Let N be no. of soldiers.
Let&be the mean weight and SD of weights of the solders.munotes.in

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9310,000, 72, 5Ni)To find no of soldiers with weights above 80kgs, we find the
probability that a soldier has weight above 80kgs.
80PX 
72 80 7255
1.6XP
PZ    
= (Area of right to Z = 0) -(Area between Z = 0 and Z = 1.6)
= 0.5 -0.4452
= 0.0548No. of soldiers with weight above 80kgs8010,000 0.0548548NP X


ii)To find the percenta ge of soldiers with weights between 70 and 77kgs.
i.e.70 77Px 
 70 72 72 77 72555
0.4 1XP
PZ    
= (Area between Z = 0.4 and Z = 0) + (Area between Z = 0 and Z = 1)
= 0.1554 + 0.3413
= 0.4967
% of soldiers with weight between 70 & 77kgs100 70 77100 0.4967
49.67PX 

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943)The incomes of a group of 10,000 persons were found to be normally
distributed with mean Rs. 520 and standard deviation Rs. 60. Find the
number of persons having incomes between Rs. 400 and 55 0 given that
Area (Z = 0 to Z = 2) = 0.4772 & Area (Z = 0 to Z = 0.5) = 0.1915
Solution :
Standard Normal Variate52060Xm XZ
When400 520400 , 260xZ 550 520550 , 0.560xZ 400 550PX
=A r e a2Zt o o+ Area (Z = 0 to Z = 0.5)
= 0.4772 + 0.1913
= 0.6687
The number of persons whose incomes are between Rs. 400 and
Rs. 550.10000 0.668766.87Np

6.3EXERCISES
(A) Multiple Choice Questions
[1]For a Binomial distribution, mean is 10 and n=30 then the probability
of failure q is
(a)1/3 (b)3 (c)2/3 (d)-1/3
[2]For a poission distribution with P (2) =P (3), then its Mean and
Variance are
(a)2 and 3 (b)3 and 3 (c)3 and 2 (d)2 and 2
[3]The average number of incoming telephone calls at a switch board is 2
per minute. Find the probability that, during a given minute, less than
2 calls are received. (e-2=0.135)
(a)0.5 (b)0.135 (c)0.27 (d)0.405munotes.in

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95[4]When X is continuous function, then f(x) is called
(a)Probability mass function (b)probability density function
(c)both (d)None of these
[5]If x and y are independent, then
(a)E(xy)=E(x) x E(y) (b)E(xy)=E(x)+E(y)
(c)E(x +y)=E(x)+E(y) (d)E(x-y)=E(x) -E(y)
[6]Number of misprints per page of a thick book follows
(a)Binomial distribution (b)Normal distribution
(c)Poission distribution (d)Standard Normal distributio n
[7] The total area of normal curve is
(a)one (b)50 per cent
(c)0.50 (d)any value between 0 and1
[8]The mean and mode of a normal distribution
(a)may be equal (b)may be different
(c)are always equal (d)(a) or (b)
[9]For a poission distribution,
(a)mean and standard deviation are equal
(b)mean and variance are equal
(c)standard deviation and variance are equal
(d)both (a) and (b)
[10] Which one is uniparametric dist ribution?
(a)Binomial distribution (b)Normal distribution
(c)Poission distribution (d)Hyper geometric distribution
[11] Probability Distribution may be
(a)discrete (b)continuous (c)infinite (d)(a) or (b)
[12] An unbiased coin is tossed 6 times. Find the probability of getting at
least 4 tails.
(a)
(b)
(c)
(d)
(B) Problems
[1]Write down probability distribution of random variables Xdefined as
number of heads appeare d when four coins are tossed simultaneously.
[2]Two fair dice are rolled. X denotes the sum of the numbers appearing
on the uppermost faces of the dices. From the probability distribution of X
find a) P(X is a multiple of 3) b) P(X<5) c) P (54).
[3]A distributer makes a profit of Rs. 30 on each item that is received in
perfect condition and suffers a loss of Rs. 6 on each item that is received
in less -than-perfect condition. If each item received is in perfect condition
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96[4]In a game of throwing a fair dice, A wins Rs.60/ -if a 6 is thrown. He
gains Rs.30/ -if the dice show 2 or 4 and he loses Rs.30/ -if odd numbers
occurs on the uppermost face of the dice. Find the expected gain of A?
[5] If the mean and variance of a Binomial distribution are 4 and 2.4
respectively, find the probability of (1) 5 successes (2) at least nine
successes.
[6]An unbiased cubical dice is thrown 5times and the number appearing
on its u ppermost face is noted. Find the probability that the number of
times an even appear is (1) 3 times (2) all 5 times.
[7] A student attempts an online test of 20 multiple -choice independent
questions. Each question has 4 possible answers of which only one is
correct. Find probability that (i) he has exactly 2 answers correct (ii) 3 or 4
answers correct(iii) none of the answer is correct.
[8] A Poisson distribution has standard deviation 3.Find P(o) and P
(1).(Given: e-9=0.000123).
[9] It is observed that 1% of mangoes in a box are bad. Find the
probability that in a box of 100 mangoes, number of bad mangoes is (i) nil
(ii) only 1 (iii) less than 2 (iv) more than 2. (Given: e-1=0.3679)
[10] In an intelligence test administered to 1000 persons, the average I.Q
was 100 with a standard deviation of 15. (a) How many people had their
I.Q. between 70 and 110? (b) What is the percentage of person with I.Q.
above 110?
Given : Area between z=0 and z=2 is 0.4772
: Area between z=0 and z=0.67 is 0.2486
[11] 1500 candidates appeared for a certain examination. The mean marks
were 58 with a standard deviation of 5 marks. Assuming that the
distribution of marks to be Normal. Find (i) the proportion of students
securing more than 63 marks. (ii) the number of students securing marks
between 60 and 68. (iii) the percentage of students with marks below 53.Given : Area between z=0 and z=1 is 0.3413: Area between z=0 and z=2 is 0.4772
: Area between z=0 and z=0.4 is 0.1554
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INTRODUCTION TO SAMPLING AND
REASONS FOR SAMPLING
Unit Structure :
7.1 Objectives
7.2 Introduction to Sampling
7.3 Reasons for Sampling
7.4 Sampling Design Process
7.5 Random Sampling
7.6 Non-Random Sampling
7.7 Exercise
Sampling & Sampling Distri bution
Introduction to Sampling, Reason for Sampling, Sampling Design
Process, Random Sampling vs Non Random Sampling, Sampling Types
& Methods, Sampling Distribution, Central Limit Theorem
7.1OBJECTIVES
At the end of this unit the learners will be abl et o
Understand the meaning of Sampling
Understand the Reasons for Sampling.
7.2INTRODUCTION TO SAMPLING
In statistics population refers to the total universe of objects being studied.
Examples include: all votes in India or all possible outcomes when a dice
is thrown. In reality it is not always possible to study the whole population
and practically such study is not feasible. Sampling is widely used by
researchers in business as a means of gathering useful information about a
population. Data are gat hered from samples and conclusions are drawn
about the population as a part of the inferential statistics process.
A sample can be cheaper to obtain than a census for a given magnitude of
questions. For example, if an eight -minute telephone interview is b eing
undertaken, conducting the interviews with a sample of 100 customers
rather than with a population of 100,000 customers obviously is less
expensive. In addition to the cost savings, the significantly smaller number
of interviews usually requires less total time. Thus, if obtaining the results
is a matter of urgency, sampling can provide them more quickly. With the
volatility of some markets and the constant barrage of new competition
and new ideas, sampling has a strong advantage over a census in terms of
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98
7.3REASONS FOR SAMPLING
Researchers usually cannot make direct observations of every individual
in the population they are studying. Instead, they collect data from a subset
of individuals –asample –and use those obse rvations to make inferences
about the entire population.
Our knowledge, our attitudes, and our actions are based to a very large
extension samples. This is equally true in everyday life and in scientific
research. A person's opinion of an institution that conducts thousands of
transactions everyday is often determined by the one or two encounters he
has had with the institution in the course of several years; Travelers who
spend 10 days in a foreign country and then proceed to write a book telling
the inha bitants how to revive their industries, reform their political system,
balance their budget, and improve the food in their hotels are a familiar
figure of fun. But in a real sense they differ from the political scientist
who devotes 20 years to living and studying in the country only in that
they base their conclusions on a much smaller sample of experience and
are less likely to be aware of the extent of their ignorance. In science and
human affairs alike we lack the resources to study more than a fragment of
the phenomena that might advance our knowledge.
Ideally, the sample corresponds to the larger population on the
characteristic(s) of interest. In that case, the researcher's conclusions from
the sample are probably applicable to the entire population. This type of
correspondence between the sample and the larger population is most
important when a researcher wants to know what proportion of the
population has a certain characteristic –like a particular opinion or a
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99percentage of the population that plans to vote for a particular candidate,
for example, require a sample that is highly representative of the
population.
Taking a sample instead of conducting a census offers several advant ages.
1.The sample can save money.
2.The sample can save time.
3.For given resources, the sample can broaden the scope of the study.
4.Because the research process is sometimes destructive, the sample can
save product.
5.If accessing the population is impossible, th e sample is the only
option.
A sample can be cheaper to obtain than a census for a given magnitude of
questions.
For example, if an eight -minute telephone interview is being undertaken,
conducting the interviews with a sample of 100 customers rather than with
a population of 100,000 customers obviously is less expensive. In addition
to the cost savings, the significantly smaller number of interviews usually
requires less total time. Thus, if obtaining the results is a matter of
urgency, sampling can provi de them more quickly. With the volatility of
some markets and the constant barrage of new competition and new ideas,
sampling has a strong advantage over a census in terms of research
turnaround time.
If the resources allocated to a research project are fixed, more detailed
information can be gathered by taking a sample than by conducting a
census. With resources concentrated on fewer individuals or items, the
study can be broadened in scope to allow for more specialized questions.
One organization budget ed Rs.100, 000 for a study and opted to take a
census instead of a sample by using a mail survey. The researchers mass -
mailed thousands of copies of a computer card that looked like a Major
League Baseball all -star ballot. The card contained 20questions to which
the respondent could answer Yes or No by punching out a perforated hole.
The information retrieved amounted to the percentages of respondents
who answered Yes and No on the 20 questions. For the same amount of
money, the company could have taken a random sample from the
population, held interactive one -on-one sessions with highly trained
interviewers, and gathered detailed information about the process being
studied. By using the money for a sample, the researchers could have
spent significantly mor e time with each respondent and thus increased the
potential for gathering useful information.
Some research processes are destructive to the product or item being
studied. For example, if light bulbs are being tested to determine how long
they burn or if candy bars are being taste tested to determine whether the
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100If a census were conducted for this type of research, no product would be
left to sell. Hence, taking a sample is the only realistic option for test ing
such products.
Sometimes a population is virtually impossible to access for research. For
example, some people refuse to answer sensitive questions, and some
telephone numbers are unlisted. Some items of interest are so scattered
that locating all of them would be extremely difficult. When the
population is inaccessible for these or other reasons, sampling is the only
option.
7.4SAMPLING DESIGN PROCESS
Distinguishing Between a Sample and a Population
Before describing sampling procedures, we need t o define a few key
terms. The term population means all members that meet a set of
specifications or a specified criterion. For example, the population of India
is defined as all people residing in India. The population of Tamil Nadu
means all people livin g within the Tamil Nadu state boundary. A
population of inanimate objects can also exist, such as all automobiles
manufactured in India in the year 2017. A single member of any given
population is referred to as an element. When only some elements are
selected from a population, we refer to that as a sample; when all
elements are included, we call it a census .
Let’s look at a few examples that will clarify these terms.
Two research psychologists were concerned about the different kinds of
training that graduate students in clinical psychology were receiving. They
knew that different programs emphasized different things, but they did not
know which clinical orientations were most popular. Therefore, they
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101and sent each of them a questionnaire regarding aspects of their program.
The response to the survey was excellent; nearly 95% of the directors of
these programs returned the completed questionnaire. The researchers then
began anal yzing their data and also classifying schools into different
clinical orientations: psychoanalytic, behaviorist, humanistic, rogueries,
and so on. When the task was complete, they reported the percentage of
schools having these different orientations and d escribed the orientations
that were most popular, which were next, and so on. They also described
other aspects of their data. The study was written up and submitted for
publication to one of the professional journals dealing with matters of
clinical psych ology. The editor of the journal read the report and then
returned it with a letter rejecting the manuscript for publication. In part,
the letter noted that the manuscript was not publishable at this time
because the proper statistical analyses had not bee n performed. The editor
wanted to know whether the differences in orientation found among the
different schools were significant or if they were due to chance
The researchers were unhappy, and rightly so. They wrote back to the
editor, pointing out that t heir findings were not estimates based on a
sample. They had surveyed all training programs (that is, the population).
In other words, they had obtained a census rather than a sample.
Therefore, their data were exhaustive; they included all programs and
described what existed in the real world. The editor would be correct only
if they had sampled some schools and then wanted to generalize to all
schools. There searchers were not asking whether a sample represented the
population; they were dealing with the population.
A comparable example would be to count all students (the population)
enrolled in a particular university and then report the number of male and
female students. If we found that 60% of the students were female, and
40% male, it would be improp er and irrelevant to ask whether this
difference in percentage is significantly different from chance. The fact is
that the percentages that exist in the school population are parameters.
They are not estimates derived from a sample. Had we taken a small
sample of students and found this 60/40 split, it would then be appropriate
to ask whether differences this large could have occurred by chance alone.
Data derived from a sample are treated statistically. Using sample data, we
calculate various statistics, such as the mean and standard deviation. These
sample statistics summarize (describe) aspects of the sample data. These
data, when treated with other statistical procedures, allow us to make
certain inferences. From the sample statistics, we make correspo nding
estimates of the population. Thus, from the sample mean, we estimate the
population mean; from the sample standard deviation, we estimate the
population standard deviation.
The above examples illustrate a problem that can occur when the terms
popula tion and sample are confused. The accuracy of our estimates
depends on the extent to which the sample is representative of the
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102Definitions
Population: It is the totality of the objects or individuals regarding
inferences are made in a sampling study.
Sample : It is the smaller representation of a large whole.
Sampling: It is a process of selecting a subset of randomized number of
the members of the population of a study.
Sampling Frame / Source List: It is a com plete list of all the members/
units of the population from which each sampling unit is selected.
Sample Design / Sample Plan: It is a definite plan for obtaining a sample
from a given population.
Sampling Unit: It is a geographical one (state, district) .
Sample Size: It is the number of items selected for the study.
Sampling Error: It is the difference between population value and sample
value.
Sampling Distribution: It is the relative frequency distribution of
samples.
Census: It is the collection o f data from whole population.
Sampling: It is taking any portion of a population or universe as
representative of that population.
Sampling method has been using in social science research since1754 by
A.L. Bowley.
Thesampling design process includes f ive steps which are closely related
and are important to all aspect of any research project. The five steps are:
defining the target population ; determining the sample frame; selecting
asampling technique; determining the sample size; and executing
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103
Examples of Sampling Designing Process
Example 1
Example 2
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104Example 3
Example 5
From the definition of aims of the study and from secondary research the
researcher would already have a group of people, organizations or mayb e
companies (so -called ‘elements’) in mind from whom the data can be
collected for statistical investigation. This group is called as the
‘population of interest ’ (in some textbooks it is called the ‘ target
population ’).If the group population of interest is rather small, it might be
possible to collect data from all elements in this group –this is called a
census . However, in most cases the researcher will collect data from only
a few elements –these then make up the sample . During sampling, the
research er will also have to decide about the sampling unit : For example,
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105want to know about household spending on organic products. In this case,
the sampling unit would be households. Sam pling is easiest when the
researcher already has an existing sampling frame , which is a list of all
the elements included in the population of interest (i.e. a list of all organic
farmers in the country). If such a list does not yet exist, the phrase
‘samp ling frame’ refers to the procedure of creating this list.
7.5RANDOM SAMPLING
Random sampling is a part of the sampling technique in which each
sample has an equal probability of being chosen. A sample chosen
randomly is meant to be an unbiased represen tation of the total population.
If for some reasons, the sample does not represent the population, the
variation is called a sampling error.
Random sampling is one of the simplest forms of collecting data from the
total population. Under random sampli ng, each member of the subset
carries an equal opportunity of being chosen as a part of the sampling
process. For example, the total workforce in organizations is 300 and to
conduct a survey, a sample group of 30 employees is selected to do the
survey. In this case, the population is the total number of employees in the
company and the sample group of 30 employees is the sample. Each
member of the workforce has an equal opportunity of being chosen
because all the employees which were chosen to be part of th e survey
were selected randomly. But, there is always a possibility that the group or
the sample does not represent the population as a whole, in that case, any
random variation is termed as a sampling error.
An unbiased random sample is important for dra wing conclusions. For
example when we took out the sample of 30 employees from the total
population of 300 employees, there is always a possibility that a researcher
might end up picking over 25 men even if the population consists of 200
men and 100 women. Hence, some variations when drawing results can
come up, which is known as a sampling error. One of the disadvantages of
random sampling is the fact that it requires a complete list of population.
For example, if a company wants to carry out a survey and intends to
deploy random sampling, in that case, there should be total number of
employees and there is a possibility that all the employees are spread
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106
7.6NON-RANDOM SAMPLI NG
In random sampling, every item in a population has a known chance of
being included in a sample.
In non -random sampling this is not the case. Indeed, one of the main
criticisms of non -random sampling is: because it’s non -random, biasis
almost certain ly introduced.
A sample is said to be biased if:
Not all outcomes have a known chance of occurring or if some outcomes
have a zero chance of occurring.
Non-random sampling is useful when descriptive comments about
thesample itself are desired.
Howev er, it can be difficult to draw conclusions about the population
based on information derived from a sample, as samples are
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107
7.7EXERCISE
Q.1. Explain the concept and the need of sampling.
Q.2. Explain with ex amples sampling design process.
Q.3. What do you understand by random sampling?
Q.4. What do you understand by non -random sampling?
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SAMPLING TYPES AND METHODS –
RANDOM SAMPLING & NON -RANDOM
SAMPLING
Unit Structure :
8.1 Sampling types and Methods -Random Sampling
a.Simple Random Sampling
b. Stratified Random Sampling
c.Systematic Sampling
d.Cluster (or Area) Sampling
e.Multi -Stage Sampling
8.2 Sampling Types and Methods –Non-Random Sampling
a. Convenience sampling
b. Judgment sampling
c. Quota Sampling
d. Snowball Sampling
e. Purposive Sampling
f. Haphazard Sampling
8.3 a.Sampling Distribution
b. Central Limit Theore m
8.4 Exercise
8.1SAMPLING TYPES AND METHODS –RANDOM
SAMPLING
The two main types of sampling are random and non -random. In random
sampling, every unit of the population has the same probability of being
selected into the sample. Random sampling impli es that chance enters into
the process of selection. For example, most Indians would like to believe
that winners of nationwide magazine sweepstakes or numbers selected as
state lottery winners are selected by some random draw of numbers.
Sometimes random sampling is called probability sampling and non -
random sampling is called non probability sampling. Because every unit
of the population is not equally likely to be selected, assigning a
probability of occurrence in non -random sampling is impossible.
The statistical methods presented and discussed are based on the
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109
The four basic random sampling techniques are simple random sampling,
stratified random sampling, systematic random sampling, and cluster (or
area) random sampling. Each technique offers advantages and
disadvantages. Some techniques are simpler to use, some are less costly,
and others show greater potential for reducing sampling error.
A.Simple Random Sampling
The most elementary random s ampling technique is simple random
sampling . Simple random sampling can be viewed as the basis for the
other random sampling techniques.
With simple random sampling, each unit of the frame is numbered from 1
toN(where Nis the size of the population). N ext, a table of random
numbers or a random number generator is used to select nitems into the
sample. A random number generator is usually a computer program that
allows computer -calculated output to yield random numbers.
A brief table of random numbers is shown in the following lines in this
section. The spaces in the table are there only for ease of reading the
values. For each number, any of the 10 digits(0 –9) is equally likely, so
getting the same digit twice or more in a row is possible.
A Brief Tab le of Random Numbers
91567 42595 27958 30134 04024 86385 29880 99730
46503 18584 18845 49618 02304 51038 20655 58727
34914 63974 88720 82765 34476 17032 87589 40836
57491 16703 23167 49323 45021 33132 12544 41035
30405 83946 23792 14422 15059 45799 22716 19792
09983 74353 68668 30429 70735 25499 16631 35006
85900 07119 97336 71048 08178 77233 13916 47564
As an example, from the population frame of companies listed in
following table, we will use simple random sampling to select a sample of
six companies. First, we number every member of the population. We
select as many digits for each unit sampled as there are in the largest
number in the population. For example, if a population has 2,000
members, we select four -digit numbers. Because the population in th is
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110number. The population is numbered from 01 to 30, as shown in the
Numbered Population of 30 Companies table.
A Population Frame of 30 Companies
Alaska Airlines DuPont Lubrizol
Alcoa ExxonMobil Mattel
Ashland General Dynamics Merck
Bank of America General Electric Microsoft
BellSouth General Mills Occidental Petroleum
Chevron Halliburton JCPenney
Citigroup IBM Procter & Gamble
Clorox Kellogg Ryder
Delta Air Lines Kmart Sears
Disney Lowe’s Time Warner
The population is numbered from 01 to 30, as shown in the
Numbered Population of 30 Companies table.
Numbered Population of 30 Companies
01 Alaska Airlines 11 DuPont 21 Lubrizo l
02 Alcoa 12 ExxonMobil 22 Mattel
03 Ashland 13 General Dynamics 23 Merck
04 Bank of America 14 General Electric 24 Microsoft
05 BellSouth 15 General Mills 25Occidental Petroleum
06 Chevron 16 Halliburton 26 JCPenney
07 Citigroup 17 IBM 27Proct er & Gamble
08 Clorox 18 Kellogg 28 Ryder
09 Delta Air Lines 19 Kmart 29 Sears
10 Disney 20 Lowe’s 30 Time Warner
The object is to sample six companies, so six different two -digit numbers
must be selected from the table of random numbers. Because th is
population contains only 30 companies, all numbers greater than 30 (31 –
99) must be ignored. If, for example, the number 67 is selected, the
process is continued until a value between 1 and 30 is obtained. If the
same number occurs more than once, we pro ceed to another number. For
ease of understanding, we start with the first pair of digits in the above
table and proceed across the first row until n= 6different values between
01 and 30 are selected. If additional numbers are needed, we proceed
across th e second row, and so on. Often a researcher will start at some
randomly selected location in the table and proceed in a predetermined
direction to select numbers. In the first row of digits in the above table of
random numbers, the first number is 91. This number is out of range so it
is cast out. The next two digits are 56. Next is 74, followed by 25, which is
the first usable number. From the table of numbered population of 30
companies, we see that 25 is the number associated with Occidental
Petroleum, s o Occidental Petroleum is the first company selected into the
sample. The next number is 95, unusable, followed by 27, which is usable.
Twenty -seven is the number for Procter & Gamble, so this company is
selected. Continuing the process, we pass over the n umbers 95 and 83.munotes.in

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-Random Sampling andNon-Random Sampling
111The next usable number is 01, which is the value for Alaska Airlines.
Thirty -four is next, followed by 04 and 02, both of which are usable.
These numbers are associated with Bank of America and Alcoa,
respectively. Continuing along the fi rst row, the next usable number is 29,
which is associated with Sears. Because this selection is the sixth, the
sample is complete. The following companies constitute the final sample:
Simple random sampling is easier to perform on small than on large
populations. The process of numbering all the members of the population
and selecting items is cumbersome for large populations.
B.Stratified Random Sampling
A second type of random sampling is stratified random sampling ,i n
which the population is divi ded into non overlapping subpopulations
called strata. The investigator then extracts a random sample from each of
the subpopulations. The main reason for using stratified random sampling
is that it has the potential for reducing sampling error. Sampling e rror
occurs when, by chance, the sample does not represent the population.
With stratified random sampling, the potential to match the sample closely
to the population is greater than it is with simple random sampling because
portions of the total sample a re taken from different population subgroups.
However, stratified random sampling is generally more costlier than
simple random sampling because each unit of the population must be
assigned to a stratum before the random selection process begins.
Strata s election is usually based on available information. Such
information may have been gleaned from previous censuses or surveys.
Stratification benefits increase as the strata differ more. Internally, a
stratum should be relatively homogeneous; externally, st rata should
contrast with each other. Stratification is often done by using demographic
variables, such as sex, socioeconomic class, geographic region, religion,
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112Stratified Random Sampling of FM Radio Listeners
In FM radio market s, age of listener is an important determinant of the
type of programming used by a station. Figure contains stratification by
age with three strata, based on the assumption that age makes a difference
in preference of programming. This stratification impl ies that listeners 20
to 30 years of age tend to prefer the same type of programming, which is
different from that preferred by listeners 30 to 40 and 40 to 50 years of
age. Within each age subgroup (stratum), homogeneity or alikeness is
present; between e ach pair of subgroups a difference, or heterogeneity, is
present.
Stratified random sampling can be either proportionate or
disproportionate. Proportionate stratified random sampling occurs
when the percentage of the sample taken from each stratum is
proportionate to the percentage that each stratum is within the whole
population. For example, suppose voters are being surveyed in Boston and
the sample is being stratified by religion as Catholic, Protestant, Jewish,
and others. If Boston’s population is 90% Catholic and if a sample ofmunotes.in

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-Random Sampling andNon-Random Sampling
1131,000 voters is taken, the sample would require inclusion of 900 Catholics
to achieve proportionate stratification. Any other number of Catholics
would be disproportionate stratification. The sample proportion of other
religions would also have to follow population percentages. Or consider
the city of El Paso, Texas where the population consists of approximately
77% Hispanic people. If a researcher is conducting a citywide poll in El
Paso and if stratification is by ethnicity , a proportionate stratified random
sample should contain 77% Hispanics. Hence, an ethnically proportionate
stratified sample of 160 residents from El Paso’s 600,000 residents should
contain approximately 123Hispanics. Whenever the proportions of the
strat a in the sample are different from the proportions of the strata in the
population, disproportionate stratified random sampling occurs.
C.Systematic Sampling
Systematic sampling is a third random sampling technique. Unlike
stratified random sampling, systematic sampling is not done in an attempt
to reduce sampling error. Rather, systematic sampling is used because of
its convenience and relative ease of administration.
With systematic sampling ,every k th item is selected to produce a sample
of size n from a population of size N. The value of k, sometimes called the
sampling cycle, can be determined by the following formula. If kis not an
integer value, the whole -number value should be used.
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114As an example of systematic sampling, a management infor mation
systems researcher wanted to sample the manufacturers in Texas. He had
enough financial support to sample 1,000 companies ( n). The Directory of
Texas Manufacturers listed approximately 17,000 total manufacturers in
Texas ( N) in alphabetical order. T he value of kwas 17 (17,000/1,000) and
the researcher selected every 17th company in the directory for his sample.
Did the researcher begin with the first company listed or the 17th or one
somewhere between? In selecting every kth value, a simple random
number table should be used to select a value between 1 and kinclusive as
a starting point. The second element for the sample is the starting point
plus k. In the example, k=17, so the researcher would have gone to a
table of random numbers to determin e a starting point between 1 and 17.
Suppose he selected the number 5. He would have started with the 5th
company, then selected the 22nd(5+17), and then the 39th, and so on.
Besides convenience, systematic sampling has other advantages. Because
systema tic sampling is evenly distributed across the frame, a
knowledgeable person can easily determine whether a sampling plan has
been followed in a study. However, a problem with systematic sampling
can occur if the data are subject to any periodicity, and the sampling
interval is in syncopation with it. In such a case, the sampling would be
non-random. For example, if a list of 150 college students is actually a
merged list of five classes with 30 students in each class and if each of the
lists of the five cla sses has been ordered with the names of top students
first and bottom students last, then systematic sampling of every 30th
student could cause selection of all top students, all bottom students, or all
mediocre students; that is, the original list is subj ect to a cyclical or
periodic organization. Systematic sampling methodology is based on the
assumption that the source of population elements is random.
D.Cluster (or Area) Sampling
Cluster (or area) sampling is a fourth type of random sampling. Cluster
(or area) sampling involves dividing the population into non -overlapping
areas, or clusters. However, in contrast to stratified random sampling
where strata are homogeneous within, cluster sampling identifies clusters
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115wide variety of elements, and the cluster is a miniature, or microcosm, of
the population.
Examples of clusters are towns, companies, homes, colleges, areas of a
city, and geographic regions. Often clusters are n aturally occurring groups
of the population and are already identified, such as states or Standard
Metropolitan Statistical Areas. Although area sampling usually refers to
clusters that are areas of the population, such as geographic regions and
cities, th et e r m s cluster sampling and area sampling are used
interchangeably in this chapter.
After randomly selecting clusters from the population, the business
researcher either selects all elements of the chosen clusters or randomly
selects individual elements into the sample from the clusters. One example
of business research that makes use of clustering is test marketing of new
products. Often in test marketing, the United States is divided into clusters
of test market cities, and individual consumers within t he test market cities
are surveyed.
Sometimes the clusters are too large, and a second set of clusters is taken
from each original cluster. This technique is called two-stage sampling .
For example, a researcher could divide the United States into clusters of
cities. She could then divide the cities into clusters of blocks and randomly
select individual houses from the block clusters. The first stage is selecting
the test cities and the second stage is selecting the blocks.
Cluster or area sampling offers several advantages. Two of the foremost
advantages are convenience and cost. Clusters are usually convenient to
obtain, and the cost of sampling from the entire population is reduced
because the scope of the study is reduced to the clusters. The cost per
element is usually lower in cluster or area sampling than in stratified
sampling because of lower element listing or locating costs. The time and
cost of contacting elements of the population can be reduced, especially if
travel is involved, because cluster ing reduces the distance to the sampled
elements. In addition, administration of the sample survey can be
simplified. Sometimes cluster or area sampling is the only feasible
approach because the sampling frames of the individual elements of the
population are unavailable and therefore other random sampling
techniques cannot be used.
Cluster or area sampling also has several disadvantages. If the elements of
a cluster are similar, cluster sampling may be statistically less efficient
than simple random sampl ing. In an extreme case —when the elements of
a cluster are the same —sampling from the cluster maybe no better than
sampling a single unit from the cluster. Moreover, the costs and problems
of statistical analysis are greater with cluster or area sampling t han with
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116
E.Multi -Stage Sampling
TheMultistage Sampling is the probability sampling technique wherein
the sampling is carried out in several stages such that the sample size gets
reduced at each stage.
The multistage sampli ng is a complex form of cluster sampling . The
cluster sampling is yet another random sampling technique wherein the
population is divided into subgroups called as clusters; then few clusters
are chosen randomly for the survey.
While in the multistage samp ling technique, the first level is similar to that
of the cluster sampling, where the clusters are formed out of the
population, but further, these clusters are sub -divided into smaller
targeting groups, i.e. sub -clusters and then the subject from each sub -
clusters are chosen randomly. Further, the stages can be added depending
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117For example, if the government wants to take a sample of 10,000
households residing in Gujarat state. At the fi rst stage, the state can be
divided into the number districts, and then few districts can be selected
randomly. At the second -stage, the chosen districts can be further sub -
divided into the number of villages and then the sample of few villages
can be take n at random. Now at the third -stage, the desired number of
households can be selected from the villages chosen at the second stage.
Thus, at each stage the size of the sample has become smaller and the
research study has become more precise.
8.2 SAMPLI NG TYPES AND METHODS –NON-
RANDOM SAMPLING
Innonrandom sampling not every unit of the population has the same
probability of being selected into the sample. Members of nonrandom
samples are not selected by chance. For example, they might be selected
because they are at the right place at the right time or because they know
the people conducting the research.
Non-random sampling methods are not appropriate techniques for
gathering data to be analyzed by most of the statistical methods presented
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118Sampling techniques used to select elements from the population by any
mechanism that does not involve a random selection process are called
nonrandom sampling techniques. Because chance is not used to select
items from the samples, these techniques are no n-probability techniques
and are not desirable for use in gathering data to be analyzed by the
methods of inferential statistics presented. Sampling error cannot be
determined objectively for these sampling techniques.
Four nonrandom sampling techniques are presented here:
Convenience sampling
Judgment sampling
Quota sampling, and
Snowball sampling
Purposive sampling
Haphazard sampling.
A.Convenience sampling
Inconvenience sampling ,elements for the sample are selected for the
convenience of the s earcher. The researcher typically chooses elements
that are readily available, nearby, or willing to participate. The sample
tends to be less variable than the population because in many
environments the extreme elements of the population are not readily
available. There searcher will select more elements from the middle of the
population. For example, a convenience sample of homes for door -to-door
interviews might include houses where people are at home, houses with no
dogs, houses near the street, first -floor apartments, and houses with
friendly people. In contrast, a random sample would require the researcher
to gather data only from houses and apartments that have been selected
randomly, no matter how inconvenient or unfriendly the location. If a
researc h firm is located in a mall, a convenience sample might be selected
by interviewing only shoppers who pass the shop and look friendly.
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119
B.Judgment sampling
Judgment sampling occurs when elements selected for the sample ar e
chosen by the judgment of the researcher. Researchers often believe they
can obtain a representative sample by using sound judgment, which will
result in saving time and money. Sometimes ethical, professional
researchers might believe they can select a more representative sample
than the random process will provide. They might be right! However,
some studies show that random sampling methods outperform judgment
sampling in estimating the population mean even when the researcher who
is administering the judgment sampling is trying to put together a
representative sample. When sampling is done by judgment, calculating
the probability that an element is going to be selected into the sample is
not possible. The sampling error cannot be determined objectively because
probabilities are bas ed on nonrandom selection. Other problems are
associated with judgment sampling. The researcher tends to make errors of
judgment in one direction. These systematic errors lead to what are called
biases . The researcher also is unlikely to include extreme el ements.
Judgment sampling provides no objective method for determining whether
one person’s judgment is better than another’s.
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120
C.Quota Sampling
A third nonrandom sampling technique is quota sampling , which appears
to be similar to stratified rando m sampling. Certain population subclasses,
such as age group, gender, or geographic region, are used as strata.
However, instead of randomly sampling from each stratum, the researcher
uses a nonrandom sampling method to gather data from one stratum until
the desired quota of samples is filled. Quotas are described by quota
controls, which set the sizes of the samples to be obtained from the
subgroups. Generally, a quota is based on the proportions of the subclasses
in the population. In this case, the quota concept is similar to that of
proportional stratified sampling.
Quotas often are filled by using available, recent, or applicable elements.
For example, instead of randomly interviewing people to obtain a quota of
Italian Americans, there searcher would go to the Italian area of the city
and interview there until enough responses are obtained to fill the quota.
In quota sampling, an interviewer would begin by asking a few filter
questions; if the respondent represents a subclass whose quota has been
filled, the interviewer would terminate the interview.
Quota sampling can be useful if no frame is available for the population.
For example, suppose a researcher wants to stratify the population into
owners of different types of cars but fails to find any lis ts of Toyota van
owners. Through quota sampling, the researcher would proceed by
interviewing all car owners and casting out non –Toyota van owners until
the quota of Toyota van owners is filled.
Quota sampling is less expensive than most random sampling t echniques
because it essentially is a technique of convenience. However, cost may
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121Another advantage of quota sampling is the speed of data gathering. The
researcher does not have to call back or send out a second questionnaire if
he does not receive a response; he just moves on to the next element. Also,
preparatory work for quota sampling is minimal.
The main problem with quota sampling is that, when all is said and done,
it still is only a nonrandom sampling technique. Some researchers believe
that if the quota is filled by randomly selecting elements and discarding
those not from a stratum, quota sampling is essentially a version of
stratified random sa mpling. However, most quota sampling is carried out
by the researcher going where the quota can be filled quickly. The object
is to gain the benefits of stratification without the high field costs of
stratification. Ultimately, it remains a non probability sampling method.
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122D.Snowball Sampling
Another nonrandom sampling technique is snowball sampling , in which
survey subjects are selected based on referral from other survey
respondents. The researcher identifies a person who fits the profile of
subjects wanted for the study. The researcher then asks this person for the
names and locations of others who would also fit the profile of subjects
wanted for the study. Through these referrals, survey subjects can be
identified cheaply and efficiently, which is particularly useful when survey
subjects are difficult to locate. It is the main advantage of snowball
sampling; its main disadvantage is that it is nonrandom.
E.Purposive Sampling
A purposive sample is a non -probability sample that is selected based on
characteristics of a population and the objective of the study. Purposive
sampling is also known as judgmental, selective, or subjective sampling.
This type of sampling can be very useful in situations when you need to
reach a targeted sample qui ckly, and where sampling for proportionality is
not the main concern. There are seven types of purposive samples, each
appropriate to a different research objective.
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123F.Haphazard Sampling
It is a non -probability sample selection method in which the in terviewer
arbitrarily selects respondents for the survey without using systematic or
random selection methods. There is no way to ensure that the estimates
derived from a haphazard sample will be unbiased.
An auditor may choose from several methodologies for determining what
to look at when auditing a company. One of the frequently employed
techniques is called haphazard sampling used by auditors to simulate a
variety of random sampling techniques when testing for potential errors in
various accounting pop ulations such as inventory and accounts receivable.
8.3 A. SAMPLING DISTRIBUTION
Sampling Error
Sampling error occurs when the sample is not representative of the
population. When random sampling techniques are used to select elements
for the sample , sampling error occurs by chance. Many times the statistic
computed on the sample is not an accurate estimate of the population
parameter because the sample was not representative of the population.
This result is caused by sampling error. With random sam ples, sampling
error can be computed and analyzed.
Non-sampling Errors
All errors other than sampling errors arenon-sampling errors . The many
possible non -sampling errors include missing data, recording errors, input
processing errors, and analysis error s. Other non -sampling errors result
from the measurement instrument, such as errors of unclear definitions,
defective questionnaires, and poorly conceived concepts. Improper
definition of the frame is a non -sampling error. In many cases, finding a
frame th at perfectly fits the population is impossible. Insofar as it does not
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124Sampling Distribution of sample -mean (µ)
In the inferential statistics process, a researcher selects a random sample
from population, comp utes a statistic on the sample, and reaches
conclusions about the population parameter from the statistic. In
attempting to analyze the sample statistic, it is essential to know the
distribution of the statistic.
The sample -mean is one of the more commo n statistics used in the
inferential process. To compute and assign the probability of occurrence
of a particular value of a sample mean, the researcher must know the
distribution of the sample means. One way to examine the distribution
possibilities is to take a population with a particular distribution, randomly
select samples of a given size, compute the sample means, and attempt to
determine how the means are distributed.
Suppose a small finite population consists of only N = 8 numbers:
54 55 59 63 64 68 69 70
Using an Excel -produced histogram, we can see the shape of the
distribution of this population of data.
Suppose we take all possible samples of size n=2 from this population
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125The result is the following pairs of data :
(54,54) (55,54) (59,54) (63,54)
(54,55) (55,55) (59,55) (63,55)
(54,59) (55,59) (59,59) (63,59)
(54,63) (55,63) (59,63) (63,63)
(54,64) (55,64) (59,64) (63,64)
(54,68) (55,68) (59,68) (63,68)
(54,69) (55,69) (59,69) (63,69)
(54,70) (55,70) (59,70) (63,70)
(64,54) (68,54) (69,54) (70,54)
(64,55) (68,55) (69,55) (70,55)
(64,59) (68,59) (69,59) (70,59)
(64,63) (68,63) (69,63) (70,63)
(64,64) (68,64) (69,64) (70,64)
(64,68) (68,68) (69,68) (70,68)
(64,69) (68,69) (69,69) (70,69)
(64,70) (68,70) (69,70) (70,70).
The means of each of these samples are:
5454.5 56.5 58.5 59 61 61.5 62
54.5 55 57 59 59.5 61.5 62 62.5
56.5 57 59 61 61.5 63.5 64 64.5
58.5 59 61 63 63.5 65.5 66 66.5
5959.5 61.5 63.5 64 66 66.5 67
6061.5 63.5 65.5 66 68 68.5 69
61.5 62 64 66 66.5 68.5 69 69.5
6262.5 64.5 66.5 67 69 69.5 70
Again using an Excel -produced histogram, we can see the shape of
the distribution of these sample means.
8.3 B. CENTRAL LIMIT THEOREM
Notice that the shape of the histogram for sample means is quite unlike the
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126“pile up” toward the middle of the distribution and “tail off” toward the
extremes.
The sample means form a distribution that approaches a symmetrical,
nearly normal -curve -type distribution.
Observe the shape of the distributions. Notice that even for small sample
sizes, the distributions of sample means for samples taken from t he
uniformly distributed population begin to “pile up” in the middle. As
sample sizes become much larger, the samples mean distributions begin to
approach a normal distribution and the variation among the means
decreases.
However, the sample means for sam ples taken from these
populations appear to be approximately normally distributed, especially as
the sample sizes become larger. What would happen to the distribution of
sample means if we studied populations with differently shaped
distributions?
The ans wer to that question is given in the central limit theorem .
Central Limit Theorem
The central limit theorem creates the potential for applying the normal
distribution to many problems when sample size is sufficiently large.
Sample means that have been computed for random samples drawn from
normally distributed populations are normally distributed.
However, the real advantage of the central limit theorem comes when
sample data drawn from populations not normally distributed or from
populations of unkno wn shape also can be analyzed by using the normal
distribution because the sample means are normally distributed for
sufficiently large sample sizes.
The central limit theorem states that sample means are normally
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127for any sample size with normally distributed populations. Thus, sample
means can be analyzed by using z scores.
The formula to determine z scores for individual values from a normal
distribution:
If sample means are normally distributed, the z score formula applied to
sample means would be
z formula for sample mean
Suppose, for example, that the mean expenditure per customer at a tire
store is 85.00, with a standard deviation of 9.00. If a random sample of 4 0
customers is taken, what is the probability that the sample average
expenditure per customer for this sample will be 87.00 or more? Because
the sample size is greater than 30, the central limit theorem can be used,
and the sample means are normally distr ibuted. With mean = 85.00,
standard deviation = 9.00, and the z formula for sample means, z is
computed as
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128From the zdistribution table z=1.41 produces a probability of 0.4207.
This number is the probability of getting a sample mean between 87.00
and 85.00 (the population mean). Solving for the tail of the distribution
yields
0.5000 -0.4207 =0.0793
8.4EXERCISES
Q.1 Develop a frame for the population of each of the following research
projects:
a.Measuring the job satisfaction of all union e mployees in a company
b.Conducting a telephone survey in a city, to determine the level of
interest in opening a new hunting and fishing specialty store in the
mall
c.Interviewing passengers of a major airline about its food service
d.Studying the quality con trol programs of boat manufacturers
e.Attempting to measure the corporate culture of cable television
companies.
Q.2Make a list of 20 people you know. Include men and women, various
ages, various educational levels, and so on. Number the list and then
usethe random number table to select six people randomly from your
list. How representative of the population is the sample? Find the
proportion of men in your population and in your sample. How do the
proportions compare? Find the proportion of20 -year-olds in your
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129Q.3 For each of the following research projects, list three variables for
stratification of the sample:
a.A nationwide study of motels and hotels is being conducted. An
attempt will be made to determine the extent of the availability of
online links for customers. A sample of motels and hotels will be
taken.
b.A consumer panel is to be formed by sampling people in a town.
Members of the panel will be interviewed periodically in an effo rt to
understand current consumer attitudes and behaviors.
c.A large soft drink company wants to study the characteristics of the
bottlers of its products, but the company does not want to conduct a
census.
d.The business research bureau of a large universit y is conducting a
project in which the bureau will sample paper -manufacturing
companies.
Q.4 In each of the following cases, the variable represents one way that a
sample can be stratified in a study. For each variable, list some strata
into which the var iable can be divided.
a.Age of respondent (person)
b.Size of company (sales volume)
c.Size of retail outlet (square feet)
d.Geographic location
e.Occupation of respondent (person)
f.Type of business (company).
Q.5 A city’s telephone book lists 100,0 00 people. If the telephone book
is the frame for a study, how large would the sample size be if
systematic sampling were done on every200th person?
Q.6 Give an example of how judgment sampling could be used in a
study to determine how district attorneys feel about attorneys
advertising on television.
Q.7 Give an example of how convenience sampling could be used in a
study of Fortune 500 executives to measure corporate attitude
toward paternity leave for employees.
Q.8 Give an example of how quota sampli ng could be used to conduct
sampling by a company test marketing a new personal computer.
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130Q.10 A population has a mean of 50 and a standard deviation of 10. If a
random sample of 64 is taken, what is the proba bility that the sample
mean is each of the following?
a.Greater than 52
b.Less than 51
c.Less than 47
d.Between 48.5 and 52.4
e.Between 50.6 and 51.3
Q.10 --Answer:
a.0.0548
b.0.7881
c.0.0082
d.0.8575
e.0.1664
Q.11 Suppose a random samp le of size 36 is drawn from a population
with a mean of 278. If 86% of the time the sample mean is less
than 280, what is the population standard deviation?
Q.11 --Answer:
11.11
Q.12 Suppose a subdivision on the southwest side of Denver, Colorado,
contains 1,500houses. The subdivision was built in 1983. A sample
of 100 houses is selected randomly and evaluated by an appraiser.
If the mean appraised value of a house in this subdivision for all
houses is 177,000, with a standard deviation of 8,500, wha t is the
probability that the sample average is greater than 185,000?
Q.12 --Answer:
0.0000
Q.13 According to Nielsen Media Research, the average number of
hours of TV viewing per household per week in the United States
is 50.4 hours. Suppose the standard deviation is 11.8 hours and a
random sample of 42 U.S. households is taken.
a.What is the probability that the sample average is more than 52
hours?
b.What is the probability that the sample average is less than 47.5
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131c.What is the probab ility that the sample average is less than 40
hours? If the sample average actually is less than 40 hours, what
would it mean in terms of the Nielsen Media Research figures?
d.Suppose the population standard deviation is unknown. If 71% of
all sample mean s are greater than 49 hours and the population
mean is still 50.4 hours, what is the value of the population
standard deviation?
Q.13 --Answers:
a.0.1894
b.0.0559
c.0.0000
d.16.4964


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TESTING OF HYPOTHESIS -ONE
SAMPLE
Unit Structure :
9.1Introduction
9.2Hypothesis Testing
9.3Test of Hypothesis Examples: Error Type I and Type 2
9.4Procedure of Testing a Hypothesis
9.5Standard Error
9.6Testing of Hypothesis Using Various Di stribution Tests
9.7Exercises
9.1INTRODUCTION
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. According to Prof. Morris Hamburg, "A
Hypothesis in statistics is simply a quantitative statement about a
population." Palmer O. Johnson has beautifully described hypothesis as
"islands in the uncharted seas of thought to be used as bases for
consolidation and recuperation as we advance in to the unknown."
In order to test a hypothesis, we collect sample data, produce sample
statistics, and use this information to decide how likely it is that our
hypothesized population parameter is correct. Say that we assume a
certain value for a populati on mean. To test the validity of our assumption,
we gather 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. 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.
Unfortunately, the difference between the hypothesized population
parameter and the actual sample statistic is more often n either so large that
we automatically reject our hypothesis nor so small that we just as quickly
accept it. So in hypothesis testing as in most significant real -life decisions,
clear -cut solutions are the exception, not the rule.
There can be several type s of hypotheses. For example, a coin may be
tossed 200 times and we may get heads 80 times and tails 120 times. We
may now be interested in testing the hypothesis that the coin is unbiased.
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133students of a particular college and may get the result as 110 lb. We may
now be interested in testing the hypothesis that the sample has been drawn
from a population with average weight 115 lb. Similarly, we may be
interested in testing the hypothesis th at the variables in the population are
uncorrelated.
Suppose a manager of a large shopping mall tells us that the average work
efficiency of her employees is 90%. How can we test the validity of her
hypothesis? using the sampling methods we learnt earlier , we could
calculate the efficiency of a sample of her employees. If we did this and
the sample statistic came out to be 93%, we would readily accept the
manager's statement. However, if the sample statistic were 46 percent, we
would reject her assumption as untrue. We can interpret both these
outcomes, 93 percent and 46 percent, using our common sense.
Now suppose that our sample statistic reveals an efficiency of 81 percent.
This value is relatively close to 90%. But is it close enough for us to
accept t he manager's hypothesis? Whether we accept or reject the
manager's hypothesis, we cannot be absolutely certain that our decision is
correct; therefore, we will have to learn to deal with uncertainty in our
decision making. We cannot accept or reject a hypo thesis about a
population parameter simply by intuition. Instead, we need to learn how to
decide objectively, on the basis of sample information, whether to accept
or reject a hunch.
9.2 HYPOTHESIS TESTING
Use a statistic calculated from the sample to te st an assertion about the
value of a population parameter.
STEP 1: Determine the sample statistic to be calculated and formulate the
hypothesis.
1. The decision about which sample statistic to calculate depends upon the
scale used to measure the variabl e.
a proportion ( π) is calculated for nominal scaled variables.
a median (med) is calculated for ordinal scaled variables.
a mean (µ) is calculated for interval or ratio scaled variables.
2. The hypotheses are:
Null Hypothesis (H 0):H0specifies a value for the population parameter
against which the sample statistic is tested. H 0always includes an equality.
Alternative Hypothesis (H a):Haspecifies a competing value for the
population parameter. H a
is formulated to reflect the proposition the researcher wan ts to
verify.
includes a non -equality that is mutually exclusive of H 0.
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134The decision about using a one tailed vs. two tailed test depends upon the
proposition the researcher wants to verify. For example, if the mean age of
the students in this class is tested against the value 21, the hypotheses
could be:
ONE TAILED TEST TWO TAILED TEST
H0: µ = 21 or H 0: µ = 21
Ha: µ > 21 or Ha: µ > 21H0: µ = 21
Ha:µ
21
STEP 2: Conduct the tes t.
1.All hypothesis tests take action on H 0.H0is either rejected or not
rejected. When H 0is rejected (not rejected), the proposition in H ais
verified (not verified).
2.Conducting the test involves deciding if H 0should be rejected or not to
be reje cted.
3.There is always a chance a mistake will be made when H 0is rejected
or not rejected. This is because the decision is based on information
obtained from a sample rather than the entire target population, i.e.,
sampling error. Hypothesis tests are designed to control for Type I
error: rejecting a true null hypothesis.
4.One approach to deciding if H 0should be rejected or not rejected is the
critical value approach. The researcher controls the chance of Type I
error by setting the test's level of significance ( α). Traditionally, αis
set at either .01, .05, or .10.
With the critical value approach:
Rejecting H 0when the researcher sets α= .01 means the researcher is
willing to accept no more than a 1% chance that a true null hypothesis
is being r ejected. The results of a test at the 1% level of significance
are highly significant.
Rejecting H 0when the researcher sets α= .05 means the researcher is
willing to accept no more than a 5% chance that a true null hypothesis
is being rejected. The resul ts of a test at the 5% level of significance
are significant.
Rejecting H 0when the researcher sets α= .10 means the researcher is
willing to accept no more than a 10% chance that a true null
hypothesis is being rejected. The results of a test at the 10% level of
significance are marginally significant.
5.An alternative approach to deciding if H 0should be rejected or not
reject is the p -value approach. The researcher knows precisely the
chance of Type I error because the statistical package calculates the
exact probability that a true null hypothesis is being rejected. This
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135With the p -value approach:
The researcher sets the test's αlevel based on how much risk of Type I
error the researcher is willing to tolerate. The αlevel can be set at any
value as long as it is less than or equal to 0.10.
The researcher rejects H 0if the p -value < α.
The Methods section of a research report that uses the p -value
approach should include a statement about the level that has been set
forα.
Most Statistical packages calculate the p -value for a 2 -tailed test. If
you're conducting a 1 -tailed test you must divide p -value by 2 before
deciding if it is acceptable.
In SPSS output, the p -value is labelled "Sig(2 -tailed)".
An Interesting Note
Because the p -value precisely measures the test's chances of Type I
error, it measures the exact α. level the test obtains. Consequent ly:
The p -value is also called the "obtained α. level".
The smaller (larger) the obtained α. level, the more (less) statistically
significant the results.
STEP 3: State the results of the test as they relate to the problem under
study. When H 0is rejecte d, there is sufficient "evidence" in the data to
support the assertion made in H a. When H 0is not rejected, the data do not
contain sufficient "evidence" to support the assertion made in H a.
EXAMPLE RESEARCH PROBLEM
An ongoing concern of University of Wis consin System
administrators is one frequently expressed by students and their parents:
earning a degree from a System University takes longer than the
advertised four years. As aspiring UW -L Bus 230 team decides to look
into the problem. Their research is guided by the hypothesis that the
problem, at least in part, is due UW -L students' lack of commitment. The
team reasons that for students to be committed to graduating "on time"
they must average 15 credit hours a semester (the minimum number
needed to gr aduate in four years), and study hard enough so they won't
have to repeat classes. The team hypothesises that UW -L students are
averaging fewer than 15 credit hours per semester, and are studying less
than most faculty recommend: two hours per week for eac h credit hour
attempted. The team interviews 200 randomly selected CBA
undergraduates. Their questionnaire asks:
1. How many credits are you taking this semester?
2. In a typical week, how many hours do you study?
The results of the analysis of these dat a appear below. Do these data
confirm the research team's hypothesis?munotes.in

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136Step 1: Determine the sample statistic to calculate and formulate the
hypotheses.
The sample statistic is a mean (µ) because the variable is measured
with a ratio scale.
The test is set up as a one -tail test to evaluate the researchers' position
that students are averaging fewer than 15 credits per semester.
Null Hypothesis H 0: µ credits = 15
Alternative Hypothesis H a:µ credits < 15 1 tailed test divide
Sig (2 -tailed) by 2.
Step 2: Conduct the test.
One-Sample Test
Test Value = 15
t DfSig.
(2-tailed)Mean
Differenc
e95% Confidence
Interval of the
Difference
Lower Upper
Credits -4.096 199 .000 -.8850 -1.3111 -.4589
One-Sample Statistics
N Mean Std.
DeviationStd. Error
Mean
Credits 200 14.1150 3.0559 .2161
SPSS OUTPUT: Analyse>Compare Means>One Sample t -test:
p-value/2
.0005/2 = .000 the chance a true null hypothesis is
being rejected is less than .025%.
.005 < .05 reject H 0because the p -value is less than the αlevel.
Note: These results are highly significant because the test's obtained α
level is almost zero.
Step 3: State the Results
The data contain sufficient evidence to conclude that UW -L
students are averaging fewer than 15 credit hours per semester.
Step 1: Determine the sample statistic to calculate and formulate the
hypotheses.
The sample statistic is a mean (µ) because the variable is measured
with a ratio scale.
The test is set up as a one -tail test to evaluate the researchers' position
that students are averaging fewer than 28 hours of studying per week.
Null Hypothesis H 0:µ study = 28
Alternative Hypothesis H a: µ study < 28 1 tailed test divide Sig
(2-tailed) by 2.munotes.in

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137Step 2: Conduct the test
Setα= .05
One Sample Statistics
N Mean Std.
DeviationStd.
Error Mean
STUDY 200 20.7000 11.8619 .8388
SPSS OUTPUT: Analyse>Compare Means>One Sample t -test:
One-Sample Test
Test Value = 15
t DfSig.
(2-
tailed)Mean
Difference95% Confidence
Interval of the
Difference
Lower Upper
Credits -8.703 199 .000 -7.3000 -8.9540 -5.6460
p-value/2
.0005/2 = .000 the chance a true null hypothesis is
being rejected is less than .025%.
.005 < .05 reject H 0because the p -value is less than the αlevel.
Note: These results are highly significant because the test's obtained α
level is almost zero.
Step 3: State the results.
The data contain sufficient evidence to conclude that on average UW -L
students study fewer than 28 ho urs per week.
PRESENTING STATISTICAL RESULTS
The sample estimate for the average number of credit hours UW -L
students take per semester is 14.2 (Figure 1). This value is statistically less
than 15 (p -value/2
.00025, Appendix 2, p.1), the mi nimum number of
credit hours needed per semester to graduate in four years. Students study
an average of 20.7 hours per week (Figure 1). This value is statistically
less than 28 (p -value/2
.00025, Appendix 2, p.2), the number of study
hours pe r week faculty would recommend for a 14 hour credit load.
DISCUSSING RESEARCH RESULTS
The results indicate that UW -L student behaviour contributes to
terms to graduation that exceed four years. Students average only 14.2
credit hours per semester. This va lue is statistically less than 15
(p-value/2
.00025), the minimum number of credit hours per semester
needed to graduate on time. Also, students study less than the amount
most faculty recommend. Given a 14 credit hou r load, faculty recommend
that students study 28 hours per week. The 20.7 hours UW -L students
study is statistically less than 28 (p -value/2
.00025). While UW -L,
students may be brighter than most thereby needing to study less, it is
more like ly that the lack of study effort leads to poor classroom
performance and a need to retake some classes. This would extend the
number of semester needed to graduate.munotes.in

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138EXAMPLE RESEARCH PROBLEM
One objective of the authors of "Alcohol Consumption and College Life"
was to evaluate the UW -L Spring Core Alcohol and Drug Survey finding
that "Most UW -L students have 0 -5 drink a week." To do so their
questionnaire asked:
During a typical week, how many days per week do you consume
alcoholic beverages?
On average, how many drinks do you consume each time you drink?
To do the analysis, the authors multiplied the responses to Q2 and Q3, and
used SPSS to generate a frequency table of the product, which they
labelled Weekly Consumption:
SPSS Output: Analyse
Descriptive Statistics > Frequencies:
Weekly Consumption
Valid Frequency Percent Valid
PercentCumulative
Percent
0
1
2
3
4
5
6
7
8
9
10
12
14
15
16
18
20
21
24
27
30
33
36
39
40
45
54
60
72
75
12024
2
7
11
10
7
7
1
10
1
10
8
4
4
7
6
3
1
3
2
6
1
2
2
3
1
1
1
1
1
116.2
1.4
4.7
7.4
6.8
4.7
4.7
0.7
6.8
0.7
6.8
5.4
2.7
2.7
4.7
4.1
2.0
.07
.20
1.4
4.1
0.7
1.4
1.4
2.0
0.7
0.7
0.7
0.7
0.7
0.716.2
1.4
4.7
7.4
6.8
4.7
4.7
0.7
6.8
0.7
68
5.4
2.7
2.7
4.7
4.1
2.0
0.7
2.0
1.4
4.1
0.7
1.4
1.4
2.0
0.7
0.7
0.7
0.7
0.7
0.716.2
17.6
22.3
29.7
36.5
41.2
45.9
46.6
60.8
54.1
60.8
66.2
68.9
71.6
76.4
80.4
82.4
83.1
85.1
86.5
90.5
91.2
92.6
93.9
95.9
96.6
97.3
98.0
98.6
99.3
100.0
Total 148 100.0 100.0munotes.in

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139Using the same approach as the Core Study, the authors concluded that
most UW -L students have 0 -8 drinks per week.
EXAMPLE RESEARCH PROBLEM CONTINUED
The authors of "Alcohol Consumption and College Life" wanted to test
the hypothesis that the average number of drinks UW -L students consume
was greater than 8.6, the value tha t was found in the Core Study.
Step 1: Determine the sample statistic to calculate and formulate the
hypotheses.
The sample statistic is a mean (µ) because the variable is measured
with a ratio scale.
The test is set up as a one -tail test to evaluate the researchers' position
that students drink more than 8.6 drinks per week.
Null Hypothesis H 0: µ Weekly Consumption = 8.6
Alternative Hypothesis H a:µ Weekly Consumption> 8.6 1
tailed test divide Sig (2 -tailed) by 2.
Step 2: Conduct the te st.
Setα= .05
SPSS OUTPUT: Analyse
Compare Means > One Sample t -test:
One-Sample Test
Test Value = 8.6
T DfSig.
(2-
tailed)Mean
Difference95% Confidence
Interval of the
Difference
Lower Upper
Weekly
Consumption3.179 147 .002 4.31 1.63 6.98
One-Sample Statistics
N Mean Std.
DeviationStd.
Error Mean
Weekly
Consumption148 12.91 16.48 1.35
p-value/2 = .002/2 = .001 the chance a true null hypothesis is
being rejected is less than -1%.
.001 < .05 reject H 0because the p -value is less than the αlevel.
Note: These results are highly significant because the test's obtained α
level is almost .001.
Step 3: State the results.
The data contain sufficient evidence to conclude that on average UW -L
students are co nsuming on average more than 8.6 drinks per week.munotes.in

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140PRESENTING STATISTICAL RESULTS
Frequency Percent Valid Percent Cumulative
Percent
Valid 0
1
2
3
4
5
6
7
8
924
2
7
11
10
7
7
1
10
116.2
1.4
4.7
7.4
6.8
4.7
4.7
.7
6.8
.716.2
1.4
4.7
7.4
6.8
4.7
4.7
.7
6.8
.716.2
17.6
22.3
29.7
36.5
41.2
45.9
46.6
53.4
54.1
Weekly Consumption
Figure 6
Another hypothesis tested was that most UW -L students consume five or
less drinks per week. According to the cumulative frequency observed,
most (53.4%) UW -L students drink zero to eight alcoholic beverages per
week (Figure 6). Furthermore, the sample estimate for the average number
of drinks consumed per week is 12.91. A one sample t -test found this
figure to be statistically larger than 8.6, the mean figure reported i n the
Core Study (p -value/2 = .001, Appendix B, page 30).
DISCUSSING RESEARCH RESULTS
There are some stark differences in the findings of this study and those of
the Core Study. In contrast to the Core Study, which concluded that most
UW-L students have 0 -5 drinks a week, this study found that most
students have 0 -8 drinks a week. Using the same methodology as the Core
Study, the cumulative frequency for drinks per week exceeded 50%
(53.4%) at eight drinks. Furthermore, statistical evidence exists to estim ate
of 12.91 is clearly statistically larger than the value reported in the Core
Study. These differences may be the consequence of how the samples
were chosen. This study's sample was randomly chosen from a list of all
UW-L students. The Core's sample was a "modified stratified random
sampling of classes" in which General Education, junior and senior level
classes were randomly selected in an effort to reflect the proportion of
freshmen, sophomore etc., in the population. All students in attendance at
24 of the 60 classes selected were surveyed. While this procedure may
result in a sample that is a fair representation of the academic classes, the
time of day the surveyed class met may have influenced the results. For
example, 7:45 a.m. classes may be those that students skip most,
especially if the preceding night involved drinking. A sampling procedure
that might miss drinkers would bias the consumption numbers downward,
and lead to the differences in the findings of the two studies.
EXAMPLE RESEARCH PROBL EM
Ted Skemp, a La Crosse area attorney, was startled when he read the
April, 1996 edition of the ABA Journal. It reported that lawyers' "[clients
complained most often about being ignored.....more than 20% believedmunotes.in

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141their lawyers failed to return phone cal ls promptly, [and] more than 20%
believed their lawyers did not pay adequate attention to their cases." To
make sure he was treating his clients right, Mr. Skemp commissioned a
Bus 230 team to survey his clients. The research team prepared a
questionnaire that included the questions:
1. When you call Mr. Skemp and have to leave a message, does he return
your calls promptly?
0. No 1. Yes
2. Does Mr. Skemp pay adequate attention to your case?
0. No 1. Yes
The team named the variable measured by Q1 "PHO NE," and the variable
measured by Q2 "ATTENTION."
Output from statistical analysis of these variables appears below. Present
the statistical results and discuss them in terms of whether or not they are
favourable for Mr. Skemp.
Statistician with Mr. Skem p
1.00
0.90 .91
0.80
0.70 7.8
0.60
0.50
0.40
0.30
0.20
0.10
0.00
Phone Attention
9.3 TEST OF HYPOTHESIS EXAMPLES: ERROR TYPE
I & TYPE II
Example 1
The Alpha -Fetoprotein (AFP) Test has both Type I and Type II error
possibilities. This test screens the mother's blood during pregnancy
for AFP and determine risk. A bnormally high or low levels may
indicate Down Syndrome.
H0:patient is healthy
Ha:patient is unhealthymunotes.in

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142Error Type I (False Positive) is: Test wrongly indicates that patient has a
Down Syndrome, which means that pregnancy must be aborted for no
reason.
Error Type II (False Negative) is: Test is negative and the child will be
born with multiple anomalies.
Example 2
The Head of the Cartel is trying to uncover the mole from within his crew.
H0:The henchman was not an undercover Miami Dade Police Officer
Ha:The henchman was an undercover Miami Dade Police Officer
Error Type 1: (False Positive)
The head of the Cartel ended up murdering the henchman that was not an
undercover Miami Dade Police Officer. Although the henchman was
innocent, he was killed pre venting him from ever flipping and giving the
government information.
Error Type 2: (False Negative)
The head of the Cartel interviews a henchman that wan an undercover
Miami Dade Police Officer, but fails to unveil his true identity.
Consequently, he con tinues to allow exposure of his operation to the
undercover Miami Dade Police officer, and further reveals the ins and outs
of his operation, that will eventually bring him to his demise.
Example 3
Airplane mechanic inspects plane for any irregularities o r malfunction.
H0:Plane seems to meet all standards of FAA and is ok -ed to fly.
Ha:Plane seems to NOT meet all standards of FAA and is AOG (airplane
on the ground).
Error Type 1: (False Positive): Airplane Reverse Thruster is visually
fine and operable but while check testing light indicator states it is not, it
is replaced even though thruster was fine and operable, thus avoiding any
accident or problem.
Error Type 2: (False Negative): Airplane Reverse Thruster seems
visually to be malfunctioning but c heck testing light indicator states it is
Fine & Operable, it is NOT replaced. At landing a pilot reports a
malfunction with the thruster and cannot reduce speed at landing, plane is
involved in accident and many innocent lives are lost.
Example 4
The mec hanic inspects the brake pads for the minimum allowable
thickness.munotes.in

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143H0:Vehicles breaks meet the standard for the minimum allowable
thickness.
Ha:Vehicles brakes do not meet the standard for the minimum allowable
thickness.
Error Type 1: (False Positive)
The brakes are fine, but the check indicates you need to replace the brake
pads; therefore any possible problems with brakes are avoided even
though the brakes were not worn.
Error Type 2: (False Negative)
The brake pads are worn to beyond the minimum all owable thickness, but
the mechanic does not find anything wrong with them and does not
replace them. Consequently, the driver of the vehicle gets into an accident
because she was unable to break effectively and gets into a fatal accident.
Example 5
During a boxing match, two contenders bump heads. The referee
checks the concussion on one of the boxers.
H0:The boxer is find and able to continue boxing.
Ha:The boxer is injured and must call the bout.
Error Type 1
The boxer is fine and not seriously injure d but the referee finds the
concussion too severe and stops the fight.
Error Type 2
The boxer is seriously injured and the concussion is detrimental to
his health, but the referee does not find the concussion severe, and allows
the fight to continue. Due to the severity of the cut, the boxer faints in mid
fight and goes into a coma.
9.4 PROCEDURE OF TESTING A HYPOTHESIS
Following are the steps required for testing a hypothesis:
1. Setting up of the hypothesis.
2. Test Statistic
3. Type I & Type II Error
4. Level of Significance
5. Critical Region and Rejection Region
6. Tailed Test Observation
7. Taking a Decision
1. Setting up of the hypothesis : A statistical hypothesis or simply a
hypothesis is a tentative solution logically drawn concerning any
param eter or the population.munotes.in

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144Generally, two hypotheses are set up. They are referred to as,
a) Null Hypothesis (H 0):A statistical hypothesis which is stated for the
purpose of possible acceptance is called null hypothesis. It is usually
referred to by the sym bol (H 0). In the words of FISHER, "Null
Hypothesis is the hypothesis which is tested for possible rejection
under the assumption that it is true."
b) Alternative Hypothesis (H a):Any hypothesis which is set up as a
complementary to the null hypothesis is called as alternate hypothesis and
is denoted by (Ha).
For example, Null Hypothesis and Alternative Hypothesis in the
above examples would be as follows:
i) H 0:µ = µ 0and H a:µ>µ 0or µ < µ 0.
ii) H 0: There is no difference between the two Drugs A a nd B.
Or H a: Drug A is better than Drug B.
Or H a: Drug A is inferior to Drug B.
Then from the above, it is clear that the null hypothesis indicates no
preferential attitude. Hence a null hypothesis is a hypothesis of no
difference. The main problem of t he testing of hypothesis is to accept or
reject the null hypothesis. As against the null hypothesis, the alternative
hypothesis specifies a range of the other values that the statistician
believes to be true. Only one alternative hypothesis is tested again st the
null hypothesis.
2. Test Static: The next step is to compute an appropriate test static which
is based upon an appropriate probability distribution. It is used to test
whether the null hypothesis set up should be accepted or rejected.
3. Type I an d Type II Errors: Acceptance or rejection of a hypothesis is
based on the result of the sample information which may not always be
consistent with the population. The decision may be correct in two ways:
Accepting the null hypothesis when it is true.
Rejec ting the null hypothesis when it is false.
The decision may be wrong in two ways:
1. Rejecting the null hypothesis when it is true.
2. Accepting the null hypothesis when it is false.
Actual Decision
Accept Reject
H0is true Correct Decision (No
error)Wrong (Type I Error)
H0is false Wrong Decision (Type
II Error)Correct Decision (No
Error)munotes.in

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1454. Level of Significance: The next step is the fixation of the level of
significance. Level of significance is the maximum probability of making
Type I error. T hese types of risks should be kept low as far as possible say
at 5% or 1%.
5. Critical region or Rejection Region: Critical region is the region of
rejection of the null hypothesis. It is a region corresponding the value of
the sample observations in the sample space which leads to rejection of the
null hypothesis. A single function of the sample observations can be fixed
and we can determine a region or range of values which lead to rejection
of H 0whenever the value of the function fails in this region.
If the observed set of results has the probability of more than 5% then the
difference between the sample result and hypothetical parameter is not
significant at 5% level i.e. the difference is due to fluctuations of sampling
and H 0is accepted. It implie s that the sample result supports the
hypothesis. Similarly, if the observed set of results has the probability less
than 5% then the difference is significant at 5% level i.e. the difference is
not wholly due to fluctuations of sampling and H 0is rejected .
6. Tailed test observation: The critical region is represented by the
portion of the area under the normal curve. The test of hypothesis is
confirmed after looking into this table of hypothesis.
7. Taking the decision: Lastly the decision should be arr ived at as to the
accepting or rejecting the null hypothesis. If the computed value of the test
static is less than the critical value as per the table, the hypothesis should
be accepted or vice versa.
9.5 STANDARD ERROR
The standard deviation of the sam pling distribution of a statistic
such as mean, median etc. is known as standard error.
USES OF STANDARD ERROR
1. S.E. plays a vital role in the large sample theory and is used in testing of
hypothesis.
If the difference between the observed and theoreti cal value of a statistic is
greater than 1.96 times the S.E the hypothesis is rejected at 5% level of
significance and say that the difference is significant at 5% level.
2. The confidence or probable limits within which the population
parameter is expect ed to lie, can be determined with the help of S.E.
3. It serves as a measure of reliability: As the S.E. increases the deviation
of actual values from the expected one increase. This indicates that the
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1469.6 TESTING OF HYPOTHESIS USING VARIOUS
DISTRIBUTION TESTS
1. T-Distribution
W. S. Gosset under the nom de plume (pen name) of 'student' first found
the distribution t =
of R.A. Fisher later on defined t =
correctly by the equat ion and found its distribution in 1926.
Using the notation of the previous article, we define a new statistic t
by the equation
t=
=
or t =
where v = (n -1) denote the number of degrees of freedom of t.
Then i t may be shown that, for samples of size n from a normal
population, the distribution of t is given by
y=
If we choose y 0so that the total area under the curve is unity; we
shall get
y0=
We can easily study the form of th et-distribution. Since only even powers
of t appear in its equation it is symmetrical about t = 0 like the normal
distribution, but unlike the normal distribution, it has g 2> 0 so that it is
more peaked than the normal distribution with the same standard
deviation. Also y attain its maximum value at t = 0 so that the mode
coincides with the mean at t = 0. Again the limiting form of the
distribution when y ® ¥ is given by
y=y 0e-1/2t2
It follows that t is normally distributed for large samples.
USES OF T-DISTRIBUTION
We have seen that if the sample is large, the use is made of the tables of
the normal probability integral in interpreting the results of an experiment
and on the basis of that to reject or accept the null hypothesis.
If, however, the sampl e size n is small, the normal probability tables will
no longer be useful. Following are the uses of t -distribution:
a.To test the significance of the mean of a small random sample from a
normal population.
b.To test the significance of the difference between the means of two
samples taken from a normal population.munotes.in

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147c.To test the significance of an observed coefficient of correlation
including partial and rank correlations.
d.To test the significance of an observed regression coefficient.
2. z-TABL ES OF POINTS AND THE SIGNIFICANCE TEST
We take y 0so that the total area under the curve given by unity. The
probability that we get a given value z 0or greater on random sampling
will be given by the area to the right of the ordinate at z 0. Tables for thi s
probability for various values of z are not available, since this probability
is difficult to evaluate, since it depends upon two numbers v 1and v 2.
Fisher has prepared tables showing 5% and 1% points of significance for
z. Colcord and Deming have prepa red a table of 0.1 % points of
significance. Generally, these tables are sufficient to enable us to gause the
significance of an observed value of z.
It should be noted that the z -tables given only critical values
corresponding to right -tail areas. Thus 5 % points of z imply that the area
to the right of the ordinate at the variable z is 0.05. A similar remark
applies to 1% points of z. In other words, 5% and 1% points of z
correspond to 10% and 2% levels of significance respectively.
USES OF z -DISTRIBUTIO N
1. To test the significance of mean of various samples having two or more
than two values.
2. To test the significance of difference between two samples from given
population.
3. To test the significance of an observed coefficients based upon the table
prepared by "FISHER" since, the probability is difficult to evaluate
based upon two numbers.
4. To test the significance on any observed set of values deriving its
critical values corresponding to 5% and 1% of z (since it uses only
"Right Tailed Test" for v aluing the significance testing).
9.7 EXERCISES
Q1. Write Explanatory Notes on the following:
a. Type I Error
b. Type II Error
c. Procedure for hypothesis testing.
d. t-distribution test
e. z-distribution test
f. Uses of t -test and z -test
munotes.in

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14810
TESTING OF HYPOTHESIS -TWO
SAMPLES
(RELATED AND INDEPENDENT)
Unit Structure :
10.1Introduction
10.2Standard Error of the difference between The Means of TWO
SAMPLES
10.3 Standard Error of the difference between The Means of TWO
STANDARD DEVIATIO NS
10.4 Hypothesis Testing of the Difference between Two Population
Means
10.5Caution while using T -Test
10.6Limitations of the Tests of Significance
10.7Exercises
10.1INTRODUCTION
Having discussed the problems relating to sampling of attributes in the
previous section, we now come to the problems of sampling of variables
such as height, weight etc. which may take any value. It shall not,
therefore, be possible for us to classify each member of a sample under
one of two heads, success or failure. The values of the variables given by
different trials will spread over a range, which will be unlimited -limited
by practical considerations, as in the case of weight of people or limited by
theoretical considerations as in the case of correlation coeffi cient which
cannot lie outside the range +1 to -1.
There are three main objects in studying problems relating to sampling of
variables:
i.To compare observation with expectation and to see how far the
deviation of one from the other can be attributed to fluctuations of
sampling;
ii.To estimate from samples some characteristic of the parent population,
such as the mean of a variable; and
iii.To gauge the reliability of our estimates.
DIFFERENCES BETWEEN SMALL AND LARGE SAMPLES
In this section, we s hall be studying problems relating to large samples
only. Though it is difficult to draw a clear -cut line of demarcation between
large and small samples, it is normally agreed amongst statisticians that amunotes.in

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149sample is to be recorded as large only if its size exceeds 30. The tests of
significance used for dealing with problems relating to large samples are
different from the ones used for small samples for the reasons that the
assumptions that we make in case of large samples do not hold good for
small samples. The assumptions made while dealing with problems
relating to large samples are:
i.The random sampling distribution of a statistic is approximately normal;
and
ii. Values given by the samples are sufficiently close to the population
value and can be sue d in its place for calculating the standard error of the
estimate.
While testing the significance of a statistic in case of large samples, the
concept of standard error discussed earlier is used. The following is a list
of the formulae for obtaining stand ard error for different statistics :
1. Standard Error of Mean
i. When standard deviation of the population is known
S. E.
=
where S.E.
refers to the standard error of the mean
= Standard deviation o f the population
n = number of observations in the sample.
ii. When standard deviation of population is not known, we have to use
standard deviation of the sample in calculating standard error of mean.
Consequently, the formula for calculating standard er ror is
S. E.
=
where σdenotes standard deviation of the sample.
It should be noted that if standard deviation of both sample as well as
population are available then we should prefer standard deviation of the
population for c alculating standard error of mean.
Fiducial limits of population mean:
95% fiducial limits of population mean are
99% fiducial limits of population mean are
2. S.E. of Median or S.E. Med = 1.25331
3. S.E. of Quartiles or S.E. = 1.36263
4. S.E. of Quartile Deviation or S.E QD= 0.78672
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1505. S.E. of Mean Deviation or S.E. MD= 0.6028
6. S.E. of Standard Deviation or S.E. σ=
7. S.E of Regression Estimate of Y on X or S.E xy.=σx
2
8. S.E. of Regression Estimate of X on Y or S.E. yx=σy
The following examples will illustrate how standard error of some of the
statistics is calculated:
Examples
1. Calculate standard error of mean from the following data showing the
amount paid by 100 firms in Calcutta on the occasion of Durga Puja.
Mid Value (Rs.) 39 49 59 69 79 89 99
No. of firms 2 3 11 20 32 25 7
Solution:
S.E.
=
Calculation of Standard Deviation
Mid-value
mF (m-69)/10
d'fd' fd'239
49
59
69
79
89
992
3
11
20
32
25
7-3
-2
-1
0
+1
+2
+3-6
-6
-11
0
32
50
2118
12
11
0
32
100
63
N = 100Ʃfd' = 80 Ʃfd'2= 236
σ=
-(
2XC=
-(
2X 100 =
X 10 = 1.311 X 10 = 13.11
S.E.
=
=
= 1.311.
10.2 STANDARD ERROR OF THE DIFFERENCE
BETWEEN TH E MEANS OF TWO SAMPLES
i. If two independent random samples with n 1and n 2numbers respectively
are drawn from the same population of standard deviation σ, the standard
error of the difference between the sample means is given by the formula:
S.E. of the difference between sample means
=
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151Ifσis unknown, sample standard deviation for combined samples must be
substituted.
ii. If two random samples with
,σ1,n1and
,σ2,n2respectively are
drawn from different pop ulations, then S.E. of the different between the
means is given by the formula:
=
+
and where σ1andσ2are unknown.
S.E. of difference between means
=
+
where S 1and S 2represent standard deviation of the two samples.
EXAMPLES
1. Intelligence test on two groups of boys and girls gave the following
results:
Mean S.D. N
Girls 75 15 150
Boys 70 20 250
Is there a significant difference in the mean scores obtained by boys and
girls?
Solution:
Let us take the hypothesis that there is no significant difference in the
mean scored obtained by boys and girls.
S.E.(
-
)=
+
where σ1= 15, σ2= 20, n 1= 150 and n 2=
250
Substituting the values
S.E.(
-
)=
+
=
= 1.781
=
= 2.84
Since the difference is more than 2.58 (1% level of sig nificance) the
hypothesis is rejected. There seems to be a significant difference in the
mean score obtained by boys and girls.
10.3 STANDARD ERROR OF THE DIFFERENCE
BETWEEN TWO STANDARD DEVIATIONS
In case of two large random samples, each drawn from a n ormally
distributed population, the S.E. of the difference between the standard
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152S.E ( σ1-σ2)=
where population standard deviations are not known
S.E (S 1-S2)=
EXAMPLE
1. Intelligence test of two groups of boys and girls gave the following
results:
Girls: Mean = 84, S.D. = 10, n = 121
Boys: Mean = 8 1, S.D. = 12, n = 81
a. Is the difference is mean scores significant?
b. Is the difference between standard deviations significant?
SOLUTION:
a. Let us take the hypothesis that there is no difference in mean scores.
S.E.(
-
)=
+
where σ1= 10, σ2= 12, n 1= 121 and n 2= 81
Substituting the values
S.E.(
-
)=
+
=
=
= 1.61
Difference of means (84 -81) = 3
=
= 1.86
Since t he difference is less than 1.96 S.E. (5% level of significance) the
given factors support hypothesis. Hence the difference in mean scores of
boys and girls is not significant.
b. Let us take the hypothesis that there is no difference between the
standard deviation of the two samples.
S.E ( σ1-σ2)=
where
= 10,
= 12, n 1= 121, n 2= 81
S.E ( σ1-σ2)=
+
=
+
=
= 1.14
Difference between the two standard deviations -(12-10) = 2
=
= 1.75
Since the difference is less than 1.96 S.E. (5% level of significance) the
given factors support hypothesis. Hence the difference in mean scores of
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153A)TWO -SAMPLE Z -TEST FOR COMPARIN G TWO MEANS
Requirements : Two normally dis tributed but independent populations, σ
is known
Hypothesis test
Formula :
where
and
are the means of the two samples, Δ is the hypothesized
difference between the population means (0 if testing for equal means),
σ1and σ2are the standard deviatio ns of the two populations,
andn1andn2are the sizes of the two samples.
The amount of a certain trace element in blood is known to vary with a
standard deviation of 14.1 ppm (parts per million) for male blood donors
and 9.5 ppm for female donors. Rando m samples of 75 male and 50
female donors yield concentration means of 28 and 33 ppm, respectively.
What is the likelihood that the population means of concentrations of the
element are the same for men and women?
Null hypothesis :H0:μ1=μ2
orH0:μ1–μ2=0
alternative hypothesis :Ha:μ1≠ μ2
or:Ha:μ1–μ2≠ 0
The computed z-value is negative because the (larger) mean for females
was subtracted from the (smaller) mean for males. But because the
hypothesized differ ence between the populations is 0, the order of the
samples in this computation is arbitrary —
could just as well have been
the female sample mean and
the male sample mean, in which
casezwould be 2.37 instead of –2.37. An extreme z-score in either tai l
of the distribution (plus or minus) will lead to rejection of the null
hypothesis of no difference.
The area of the standard normal curve corresponding to a z-score of –
2.37 is 0.0089. Because this test is two -tailed, that figure is doubled to
yield a p robability of 0.0178 that the population means are the same. If the
test had been conducted at a pre -specified significance level of α< 0.05,
the null hypothesis of equal means could be rejected. If the specified
significance level had been the more conse rvative (more stringent) α<
0.01, however, the null hypothesis could not be rejected.
In practice, the two -sample z-test is not used often, because the two
population standard deviations σ1andσ2are usually unknown. Instead,
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154Inferences About the Difference Between Two Population Means for
Paired Data
Paired samples: The sample selected from the first population is related
to the corresponding sample from the second population.
It is importa nt to distinguish independent samples and paired samples.
Some examples are given as follows.
Compare the time that males and females spend watching TV.
Think about the following, then click on the icon to the left to compare
your answers.
A. We randomly select 20 males and 20 females and compare the average
time they spend watching TV. Is this an ind ependent sample or paired
sample?
B. We randomly select 20 couples and compare the time the husbands and
wives spend watching TV. Is this an independent sample or paired
sample?
The paired t-test will be used when handling hypothesis testing for paired
data.
The Paired t-Procedure
Assumptions :
1.Paired samples
2.The differences of the pairs follow a normal distribution or the number
of pairs is large (note here that if the number of pairs is < 30, we need to
check whether the differences are normal, but we do not need to check for
the normality of each population )
Hypothesis :
H0:μd=0
Ha:μd≠0
OR
H0:μd=0
Ha:μd<0
OR
H0:μd=0
Ha:μd>0
t-statistic :
Letd= differences between the pairs of data, then d¯= mean of these
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155The test statistics is: t∗=d¯−0sd/n √degrees of freedom = n-1
where ndenotes the number of pairs or the number of differences.
Paired t-interval :
d¯±tα/2⋅sdn−−√
Note: sd¯=sdn √where sd¯is the standard deviation of the sample
differences.
Example: Drinking Water
Trace metals in drinking water affect the flavour and an unusually high
concentration can pose a health hazard. Ten pai rs of data were taken
measuring zinc concentration in bottom water and surface water
(zinc_conc.txt ).
Does the da ta suggest that the true average concentration in the bottom
water exceeds that of surface water?
Location
1 2 3 4 5 6 7 8 9 10
Zinc
concentration in
bottom water.430 .266 .567 .531 .707 .716 .651 .589 .469 .723
Zinc
concentration in
surface water.415 .238 .390 .410 .605 .609 .632 .523 .411 .612
To perform a paired t-test for the previous trace metal example:
Assumptions :
1. Is this a paired sample? -Yes.
2. Is this a large sample? -No.
3. Since the sample size is not large enough (less than 30), we need to
check whether the differences follow a normal distribution.
In Minitab, we can use Calc > calculator to obtain diff = bottom -
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156
Thus, we conclude that the difference may com e from a normal
distribution.
Step 1 . Set up the hypotheses:
H0:μd=0
Ha:μd>0
where 'd' is defined as the difference of bottom -surface.
Step 2 . Write down the significance level α=0.05.
Step 3 . What is the critical value and the rejection region?
α=0.05,df=9
t0.05=1.833
rejection region: t>1.833
Step 4 . Compute the value of the test statistic:
t∗=d¯sdn √=0.08040.052310 √=4.86
Step 5 . Check whether the test statistic falls in the rejection region and
determine whether to reject H o.
t∗=4.86>1.833
rejectH0
Step 6 . State the conclusion in words.
Atα=0.05, we conclude that, on average, the bottom zinc concentration is
higher than the surface zinc concentration.
Using Minitab to Perform a Paired t-Test
You can use a paired t-test in Minitab to perform t he test. Alternatively,
you can perform a 1 -sample t-test on difference = bottom -surface.
1. Stat > Basic Statistics > Paired t
2. Click 'Options' to specify the confidence level for the interval and the
alternative hypothesis you want to test. The defa ult null hypothesis is 0.
The Minitab output for paired T for bottom -surface is as follows:
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157N Mean StDev SE Mean
Bottom 10 0.5649 0.1468 0.0464
Surface 10 0.4845 0.1312 0.0415
Difference 10 0.0804 0.0523 0.0165
95% l ower bound for mean difference: 0.0505
T-Test of mean difference = 0 (vs > 0): T -Value = 4.86 P -Value = 0.000
Note: In Minitab, if you choose a lower -tailed or an upper -tailed
hypothesis test, an upper or lower confidence bound will be constructed,
respec tively, rather than a confidence interval.
Click on the 'Minitab Movie' icon to display a walk
through of ' Condu cting a Paired t -Test'.
Using the p-value to draw a conclusion about our example:
p-value = 0.000 < 0.05
Reject H0and conclude that bottom zinc concentration is higher than
surface zinc concentration.
Note: For the zinc concentration problem, if you do not recognize the
paired structure, but mistakenly use the 2 -sample t-test treating them as
independent samples, you will not be able to reject the null hypothesis.
This demonstrates the importance of distinguishing the two types of
samples. Also, it is wi se to design an experiment efficiently whenever
possible.
What if the assumption of normality is not satisfied? In this case we would
use a nonparametric 1 -sample test on the difference.
10.4 HYPOTHESIS TESTING OF THE
DIFFERENCE BETWEEN TWO POPULATION ME ANS
B)Hypothesis testing of the difference between two population means
This is a two sample z test which is used to determine if two population
means are equal or unequal. There are three possibilities for formulating
hypotheses.
l.
:
=
:
2.
:
:
<
3.
:
:
>
Proceduremunotes.in

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158The same procedure is used in three different situations
Sampling is from normally distributed populations with known
variances
Sampling from normally distribu ted populations where population
variances are unknown
opopulation variances equal
This is with tdistributed as Student's tdistribution with (
+
-2)
degrees of freedom and a pooled variance.
opopulation variances unequal
When po pulation variances are unequal, a distribution of t'is used in a
manner similar to calculations of confidence intervals in similar
circumstances.
Sampling from populations that are not normally distributed
If both sample sizes are 30 or larger the centr al limit theorem is in effect.
The test statistic is
Ifthe population variances are unknown, the sample variances are used.
Sampling from normally distributed populations with population
variances known
Example 7.3.1
Serum uric acid level s
Isthere a difference between the means between individuals with Down's
syndrome and normal individuals?
(1) Data
=4.5
=12
=1
=3.4
=15
=1.5
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159(2) Assumptions
two independent random samples
each dr awn from a normally distributed population
(3) Hypotheses
:
=
:
(4) Test statistic
This is a two sample z test.
(a) Distribution of test statistic
Ifthe assumptions are correct and
is true, the test statistic is distributed
as the normal distribution.
(b)Decision rule
With
= .05, the critical values of z are -1.96 and +1.96. We reject
if
z<-1.96 or z > +1.96.
(5) Calculation of test statistic
(6) Statistical decision
Reject
because 2.57 > 1.96.
(7) Co nclusion
From these data, it can be concluded that the population means are not
equal. A 95% confidence interval would give the same conclusion.
p = .0102.
Sampling from normally distributed populations with unknown
variances
With equal populatio n variances, we can obtain a pooled value from the
sample variances.
Example 7.3.2
Lung destructive index
Wewish to know if we may conclude, at the 95% confidence level, that
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160(1) Data
Smokers:
= 17.5
= 16
= 4.4752
Non-Smokers:
= 12.4
=9
= 4.8492
= .05
Calculation of Pooled Variance:
(2) Assumptions
independent random samples
normal distribution of the population s
population variances are equal
(3) Hypotheses
:
:
>
(4) Test statistic
(a) Distribution of test statistic
If the assumptions are met and
is true, the test statistic is distributed as
Student's t distributio n with 23 degrees of freedom.
(b) Decision rule
With
= .05 and df= 23, the critical value of tis 1.7139. We
reject
if t > 1.7139.
(5) Calculation of test statistic
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161(6) Statistical decision
Reject
because 2.6563 > 1.7139.
(7) Conclus ion
On the basis of the data, we conclude that
>
.
Actual values
t = 2.6558
p = .014
Sampling from populations that are not normally distributed
Example 7.3.4
These data were obtained in a study comparing persons with disabilities
with persons without disabilities. A scale known as the Barriers to Health
Promotion Activities for Disabled Persons (BHADP) Scale gave the data.
We wish to know if we may conclude, at the 99% confidence level, that
persons with disabilities score higher than person s without disabilities.
(1) Data
Disabled:
= 31.83
= 132
= 7.93
Nondisabled:
= 25.07
= 137
= 4.80
= .01
(2) Assumptions
independent random samples
(3) Hypotheses
:
:
>
(4) Test statistic
Because of the large samples, the central limit theorem permits calculation
of the z score as opposed to using t.Thezscore is calculated using the
given sample standard deviations.
(a) Distribution of test statistic
If the assumptions are correct and
is true, the test statistic is
approximately normally distributed
(b) Decision rule
With
= .01 and a one tail test, the critical value of z is 2.33. We
reject
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162(5) Calculation of test statistic
(6) Statistical decision
Reject
because 8.42 > 2.33.
(7) Conclusion
On the basis of these data, the average persons with disabilities score
higher on the BHADP test than do the nondisabled persons.
Actual values
z = 8.42
p = 1.91 x 10 -17
Paired comparisons
Sometimes data comes from no independent samples. An example might
be testing "before and after" of cosmetics or consumer products. We
could use a single random sample and do "before and after" tests on each
person. A hypothesis test b ased on these data would be called a paired
comparisons test .Since the observations come in pairs, we can study the
difference, d, between the samples. The difference between each pair of
measurements is called di.
Test statistic
With a population of n pairs of measurements, forming a simple random
sample from a normally distributed population, the mean of the
difference,
, is tested using the following implementation of t.
Paired comparisons
Example 7.4.1
Very -low-calorie diet (VLCD) Trea tment
Table gives B (before) and A (after) treatment data for obese female
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163
We calculate di=A-B for each pair of data resulting in negative values
meaning that the participants lost weight.
We wish to know if we may conclude, at the 95% confidence level, that
the treatment is effective in causing weight reduction in these people.
(1)Data
Values of diare calculated by subtracting each A from each B to give a
negative number. On the TI -83 calculator place the A da ta in L1 and the
B data in L2. Then make L3 = L1 -L2 and the calculator does each
calculation automatically.
In Microsoft Excel put the A data in column A and the B data in column
B, without using column headings so that the first pair of data are on line
1.In cell C1, enter the following formula: =a1-b1. This calculates the
difference, di, for B -A.Then copy the formula down column C until the
rest of the differences are calculated.
n=9
= .05
(2) Assumptions
the observed differences are a simple random sample from a normally
distributed population of differences
(3) Hypotheses
:
0
:
< 0 (meaning that the patients lost weight)
(4) Test statistic
The test statistic is twhich is calculated as
(a) Distribution of test statistic
The test statistic is distributed as Student's t with 8 degrees of freedom
(b) Decision rule
With
= .05 and 8 dfthe critical value of t is -1.8595. We reject
if t
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164(5) Calculation of test statistic
(6) Statistical decision
Reject
because -12.7395 < -1.8595
p = 6.79 x 10 -7
(7) Conclusion
On the basis of these data, we conclude that the diet program is effective.
Other considerations
a confidence interval for
can be constructed
z can be used if the variance is known or if the sample is large.
10.5 CAUTION WHILE USING T -TEST
While drawing inferences on the basis of t -test it should be remembered
that the conclusions arrived at on the basis of the 't -test' are justif ied only
if the assumptions upon which the test is based are true. If the actual
distribution is not normally distributed then, strictly speaking, the t -test is
not justified for small samples. If it is not a random sample, then the
assumption that the obs ervations are statistically independent is not
justified and the conclusions based on the t -test may not be correct. The
effect of violating the normality assumption is slight when making
inference about means provided that the sampling is fairly large whe n
dealing with small samples. However, it is a good idea to check the
normality assumption, if possible. A review of similar samples or related
research may provide evidence as to whether or not the population is
normally distributed.
10.6 LIMITATIONS OF THE TESTS OF SIGNIFICANCE
In testing statistical significance the following points must be noted:
1. They should not be used mechanically: Tests of significance are simply
the raw materials from which to make decisions, not decisions in
themselves. There may be situations where real differences exist but do
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Testing of Hypothesis -TwoSamples
165round. In each case it is absolutely necessary to exercise great care before
taking a decision.
2. Conclusions are to be g iven in terms of probabilities and not certainties:
When a test shows that a difference was statistically significant, it suggests
that the observed difference is probably not due to chance. Thus
statements are not made with certainty but with a knowledge of
probability. "Unusual" events do happen once in a while.
3. They do not tell us "why" the difference exists: Though tests can
indicate that a difference has statistical significance, they do not tell us
why the difference exists. However, they do sugge st the need for further
investigation in order to reach definite answers.
4. If we have confidence in a hypothesis it must have support beyond the
statistical evidence. It must have a rational basis. This phrase suggests two
conditions: first, the hypothe sis must be 'reasonable' in the sense of
concordance with a prior expectation. Secondly, the hypothesis must fit
logically into the relevant body of established knowledge.
The above points clearly show that in problems of statistical
significance as in ot her statistical problems, technique must be combined
with good judgement and knowledge of the subject -matter.
10.7EXERCISES
1. Explain the concept of standard error and discuss its role in the large
sample theory.
2. Explain briefly the procedure follo wed in testing hypothesis.
3. Give some important applications of the t -test and explain how it helps
in making business decisions.
4. What is null hypothesis? How is it different from alternative
hypothesis?
5. The mean life of a sample of 10 electric light bulbs was found to be 1,
456 hours with standard deviation of 423 hours. A second sample of 17
bulbs chosen from a different batch showed a mean life of 1, 280 hours
with standard deviation of 398 hours. Is there a significant difference
between the means of the two batches?
6.Test the significance of the correlation r = 0.5 from a sample of size 18
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1667. A correlation coefficient of 0.2 is discovered in a sample of 28 pairs of
observations. Use z -test to find out if this is significantl y different from
zero.
8. How many pairs of observations must be included in a sample in order
that an observed correlation coefficient of value 0.42 shall have a
calculated value of t greater than 2.72?
9. State the cautions of using t -test.
10. State the limitations of tests of significance.


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