all units for managerial statistics (mgmt 222)

UNIT 1: INTRODUCTIONUNIT 1: INTRODUCTION

ContentsContents

1.01.0 Aims and ObjectivesAims and Objectives

1.11.1 IntroductionIntroduction

1.21.2 Statistics Defined Statistics Defined

1.31.3 Importance of StatisticsImportance of Statistics

1.41.4 Types of Statistics Types of Statistics

1.4.11.4.1 Descriptive StatisticsDescriptive Statistics

1.4.21.4.2 Inferential Statistics Inferential Statistics

1.51.5 Model Examination Questions Model Examination Questions

1.01.0 AIMS AND OBJECTIVESAIMS AND OBJECTIVES

This unit will introduce you to statistics and its uses and importance. After completing theThis unit will introduce you to statistics and its uses and importance. After completing the

unit you will be able to:unit you will be able to:

•• define statisticsdefine statistics

•• identify the types of statisticsidentify the types of statistics

•• know the benefits of managerial statistics.know the benefits of managerial statistics.

1.1 INTRODUCTION1.1 INTRODUCTION

Governments, businesses, researchers and scientists in the Natural or Social science needGovernments, businesses, researchers and scientists in the Natural or Social science need

information for their activities. Most of these information requirements are quantitative andinformation for their activities. Most of these information requirements are quantitative and

need a scientific approach or technique to gather and use.need a scientific approach or technique to gather and use.

1.2 STATISTICS DEFINED1.2 STATISTICS DEFINED

The world statistics is an Italian word composed of two words, stato, which means the stateThe world statistics is an Italian word composed of two words, stato, which means the state

and statista-refers to a person involved with the affairs of the state. Therefore statistics wasand statista-refers to a person involved with the affairs of the state. Therefore statistics was

meant the collection of facts useful to the state. meant the collection of facts useful to the state.

Nowadays statistics in not restricted to information about the state. It extends to almost everyNowadays statistics in not restricted to information about the state. It extends to almost every

realm of human endeavor. realm of human endeavor.

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Statistics is defined as a science or process of collecting, organizing, presenting, analyzingStatistics is defined as a science or process of collecting, organizing, presenting, analyzing

and interpreting data to assist in making effective decision. and interpreting data to assist in making effective decision.

1.3 IMPORTANCE OF STATISTICS1.3 IMPORTANCE OF STATISTICS

Statistics is useful for:Statistics is useful for:

-- Government officials for making policy decisions in unemployment,Government officials for making policy decisions in unemployment,

inflation, health, education, infrastructure etc…inflation, health, education, infrastructure etc…

-- Financial planners for trend analysis, stock market, future investment etc..Financial planners for trend analysis, stock market, future investment etc..

-- Businesses, for product development, customer satisfaction, Risk Businesses, for product development, customer satisfaction, Risk

-- Production supervisors for quality control, improve product quality etc. Production supervisors for quality control, improve product quality etc.

-- Politicians for legislation campaign strategy Politicians for legislation campaign strategy

-- Physicians and Hospitals on effectiveness of drugs and disease surveillancePhysicians and Hospitals on effectiveness of drugs and disease surveillance

etc. etc.

Managerial statistical analysis of data used to help in improving business processes to. Managerial statistical analysis of data used to help in improving business processes to.

1-1- Demonstrate the need for improvements Demonstrate the need for improvements

2-2- Identity ways to make improvements Identity ways to make improvements

3-3- Asses weather or not improvement activities have been successful and Asses weather or not improvement activities have been successful and

4-4- Estimate the benefits of improvement strategies Estimate the benefits of improvement strategies

Statistical methods are used for learning about population, which is a set of existingStatistical methods are used for learning about population, which is a set of existing

units (people, objects or events) units (people, objects or events)

Often the population that we want to study is very large, time consuming or costly to conductOften the population that we want to study is very large, time consuming or costly to conduct

a census. In such a situation we select and analyze a subset (or portion) of the populationa census. In such a situation we select and analyze a subset (or portion) of the population

units. This subset of the units in a population is called sample.units. This subset of the units in a population is called sample.

1.4 TYPE OF STATISTICS 1.4 TYPE OF STATISTICS

There are tow types of StatisticsThere are tow types of Statistics

1.4.11.4.1 Descriptive Statistics Descriptive Statistics

It is the science of describing the important aspects of a set of measurements eg. If we areIt is the science of describing the important aspects of a set of measurements eg. If we are

studying a set of starting salaries we might wish to describe. studying a set of starting salaries we might wish to describe.

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-- How large or small they tend to be How large or small they tend to be

-- What a typical Salary should be What a typical Salary should be

-- How much the salaries differ from each other How much the salaries differ from each other

When the population of interest is small and we can conduct a census of the population weWhen the population of interest is small and we can conduct a census of the population we

will be able to directly describe the important aspects of the population measurement. Thewill be able to directly describe the important aspects of the population measurement. The

subject area of descriptive statistics includes procedures used to summarize masses of datasubject area of descriptive statistics includes procedures used to summarize masses of data

and present them in an understandable manner. However it has nothing to do with the future. and present them in an understandable manner. However it has nothing to do with the future.

1.4.21.4.2 Inferential Statistics Inferential Statistics

A Conclusion drawn about a population based on information in a sample drawn from theA Conclusion drawn about a population based on information in a sample drawn from the

population is called statistical inference. Statistics is usually concerned with inference. Thepopulation is called statistical inference. Statistics is usually concerned with inference. The

population we want to study is usually large or infinite. So we need to select a sample since itpopulation we want to study is usually large or infinite. So we need to select a sample since it

is impossible to study the population. is impossible to study the population.

1.51.5 MODEL EXAMINATION QUESTIONSMODEL EXAMINATION QUESTIONS

Answer the following questions. Do not look into the text while writing the answers. HoweverAnswer the following questions. Do not look into the text while writing the answers. However

at the end refer to the text and see how you answered the questions.at the end refer to the text and see how you answered the questions.

a)a) Why governments, businesses, researchers need information?Why governments, businesses, researchers need information?

b)b) Define statistics.Define statistics.

c)c) What are the types of statistics?What are the types of statistics?

d)d) What are the particular benefits or importance of managerial statistics in improvingWhat are the particular benefits or importance of managerial statistics in improving

business processes?business processes?

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UNIT 2:UNIT 2:PROBABILITY AND PROBABILITY DISTRIBUTION PROBABILITY AND PROBABILITY DISTRIBUTION

ContentsContents



2.22.2 Probability Defined Probability Defined

2.32.3 Approaches in ProbabilityApproaches in Probability

2.3.12.3.1 Objective ProbabilityObjective Probability

2.3.1.12.3.1.1 Classic probabilityClassic probability

2.3.1.22.3.1.2 Long-term Relative Frequency ProbabilityLong-term Relative Frequency Probability

2.3.22.3.2 Subjective ProbabilitySubjective Probability

2.42.4 Sample Space and Sample Space OutcomeSample Space and Sample Space Outcome

2.52.5 Probability RuleProbability Rule

2.5.12.5.1 Addition Rule for Independent EventsAddition Rule for Independent Events

2.5.22.5.2 Addition Rule for Mutually Exclusive EventsAddition Rule for Mutually Exclusive Events

2.62.6 Complement of an EventComplement of an Event

2.72.7 Conditional Probability and Statistical IndependenceConditional Probability and Statistical Independence

2.7.12.7.1 Conditional ProbabilityConditional Probability

2.7.22.7.2 Statistical IndependenceStatistical Independence

2.7.32.7.3 Independent and Mutually Exclusive EventsIndependent and Mutually Exclusive Events

2.7.3.12.7.3.1 Multiplication Rule for independent EventsMultiplication Rule for independent Events

2.7.3.22.7.3.2 Union Rule for Independent EventsUnion Rule for Independent Events

2.82.8 The Total Probability and Baye’s TheoremThe Total Probability and Baye’s Theorem

2.8.12.8.1 Total ProbabilityTotal Probability

2.8.22.8.2 Baye’s TheoremBaye’s Theorem

2.9 Answers to Check Your progress2.9 Answers to Check Your progress

2.10 Model Examination Questions2.10 Model Examination Questions

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2.0 AIMS AND OBJECTIVES2.0 AIMS AND OBJECTIVES

Probability theory forms the basis for inferential statistics as well as other fields that requireProbability theory forms the basis for inferential statistics as well as other fields that require

quantitative assessment of chance occurrences; such as quality control, management decisionquantitative assessment of chance occurrences; such as quality control, management decision

analysis; and in areas of the natural sciences, engineering, economics etc. analysis; and in areas of the natural sciences, engineering, economics etc.

After completing this unit, you will be able toAfter completing this unit, you will be able to

•• define probabilitydefine probability

•• define important terms in probabilitydefine important terms in probability

•• identify the approaches in probabilityidentify the approaches in probability

•• list sample space of an experimentlist sample space of an experiment

•• identify the types of eventsidentify the types of events

•• calculate probabilities using deferent rules.calculate probabilities using deferent rules.


Since life is full of uncertainties, people have always been interest in evaluating probabilities.Since life is full of uncertainties, people have always been interest in evaluating probabilities.

The theory of probability is an in indispensable tool in the analysis of situations involvingThe theory of probability is an in indispensable tool in the analysis of situations involving

uncertainty.uncertainty.

2.2 PROBABILITY DEFINED2.2 PROBABILITY DEFINED

Probability can be defined asProbability can be defined as

-- A mathematical means of studying uncertainty and variability.A mathematical means of studying uncertainty and variability.

-- A number that conveys the strength of our belief in the occurrence of anA number that conveys the strength of our belief in the occurrence of an

uncertain eventuncertain event

From the above definitions you can differentiate probability to chances or possibilities. As theFrom the above definitions you can differentiate probability to chances or possibilities. As the

latter cannot be quantified. latter cannot be quantified.

Probability is a number between zero and one inclusive. The probability of zero representsProbability is a number between zero and one inclusive. The probability of zero represents

something that cannot happen and the probability of one represents something that is certainsomething that cannot happen and the probability of one represents something that is certain

to happen. The closer a probability is to zero, the more improbable it is that something willto happen. The closer a probability is to zero, the more improbable it is that something will

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happen the closer the probability is to one the more sure we are it will happen. Whenhappen the closer the probability is to one the more sure we are it will happen. When

probability is 0.5 uncertainty will reach its maximum.probability is 0.5 uncertainty will reach its maximum.

Important Terms Important Terms

1.1. Experiment Experiment

A process that leads to the occurrence of one and only one of several possible observations or A process that leads to the occurrence of one and only one of several possible observations or

A process of observation that has an uncertain outcome. eg Tossing a coin; answering aA process of observation that has an uncertain outcome. eg Tossing a coin; answering a

question where the answer can be correct or incorrect; drawing a card from a deck of playingquestion where the answer can be correct or incorrect; drawing a card from a deck of playing

card. card.

2.2. Event Event

A collection of one or more outcomes of an experiment or A collection of one or more outcomes of an experiment or

An experimental outcome that may or may not occur. If the experiment is tossing a coin theAn experimental outcome that may or may not occur. If the experiment is tossing a coin the

events are Head, or Tail.events are Head, or Tail.

3.3. Outcome Outcome

A particular result of an experiment. In case of tossing a coin, If head face up we will considerA particular result of an experiment. In case of tossing a coin, If head face up we will consider

head as the out come of the experiment. head as the out come of the experiment.

2.2.3 3 APPROACHES IN PROBABILITYAPPROACHES IN PROBABILITY

2.3.1Objective Probability 2.3.1Objective Probability

2.3.3.12.3.3.1 Classic Probability Classic Probability

It is probability based on the symmetry of games of chance or similar situations. ThisIt is probability based on the symmetry of games of chance or similar situations. This

probability is based on the idea that certain occurrences are equally likely. eg. The numbersprobability is based on the idea that certain occurrences are equally likely. eg. The numbers

1,2,3,4,5,and 6 on fair die are equally likely to occur i.e they do have equal chance of1,2,3,4,5,and 6 on fair die are equally likely to occur i.e they do have equal chance of

occurrence.occurrence.

2.3.1.22.3.1.2 Long-term Relative Frequency ProbabilityLong-term Relative Frequency Probability

The probability of an event happening in the long-term is determined by observing whatThe probability of an event happening in the long-term is determined by observing what

fraction of the time similar events happened in the past. We often think of a probability infraction of the time similar events happened in the past. We often think of a probability in

terms of the percentage of the time the event would occur in many repetition of theterms of the percentage of the time the event would occur in many repetition of the

experiment. Suppose that A is an event that might occur when a particular experiment isexperiment. Suppose that A is an event that might occur when a particular experiment is

performed then the probability that the event A will occur, P (A), can be interpreted to be theperformed then the probability that the event A will occur, P (A), can be interpreted to be the

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number that would be approached by the relative frequency of the event A If we perform thenumber that would be approached by the relative frequency of the event A If we perform the

experiment an indefinitely large number of times. experiment an indefinitely large number of times.

eg. When we say that the probability of obtaining a head when we toss a coin is 0.5 we areeg. When we say that the probability of obtaining a head when we toss a coin is 0.5 we are

saying that, when we repeatedly toss the coin an indefinitely large number of times, we willsaying that, when we repeatedly toss the coin an indefinitely large number of times, we will

obtain a head 50% of the repetition.obtain a head 50% of the repetition.

In terms of formula In terms of formula

Probability of an event happening =Probability of an event happening = Number of times occurred in pastNumber of times occurred in past Total number of observation Total number of observation

If a truck operator experienced 5 accidents out of 50 truck last year, then the probability that If a truck operator experienced 5 accidents out of 50 truck last year, then the probability that

a truck will have an accident next year can be 5/50 = 0.10a truck will have an accident next year can be 5/50 = 0.10

2.3.2 Subjective Probability 2.3.2 Subjective Probability

When there is no past experience or little on which to base a probability, personal judgment,When there is no past experience or little on which to base a probability, personal judgment,

experience, intuition or expertise or any other subjective evaluation criteria will be applied toexperience, intuition or expertise or any other subjective evaluation criteria will be applied to

estimating or assigning probability. This probability is subjective probability. estimating or assigning probability. This probability is subjective probability.

It is also called personal probability. Unlike objective probability one person’s subjectiveIt is also called personal probability. Unlike objective probability one person’s subjective

probability may very well different from another person’s subjective probability of the sameprobability may very well different from another person’s subjective probability of the same

event.event.

eg. A physician assessing the probability of a patient’s recovery and an expert in the nationaleg. A physician assessing the probability of a patient’s recovery and an expert in the national

bank assessing probability of currency devaluation are both making a personal judgmentbank assessing probability of currency devaluation are both making a personal judgment

based on what they know and feel about the situation and other group of physicians or expertsbased on what they know and feel about the situation and other group of physicians or experts

will arrive with different probability, though both can employee identical techniques orwill arrive with different probability, though both can employee identical techniques or

approaches and information. approaches and information.

Both classic and long-term relative frequency probabilities are objective in the sense that noBoth classic and long-term relative frequency probabilities are objective in the sense that no

personal judgment is involved.personal judgment is involved.

Whatever the kind of probability involved /subjective or objective/ the same set ofWhatever the kind of probability involved /subjective or objective/ the same set of

mathematical rules holds for manipulating and analyzing probability. mathematical rules holds for manipulating and analyzing probability.

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2.4 SAMPLE SPACE AND SAMPLE SPACE OUTCOME2.4 SAMPLE SPACE AND SAMPLE SPACE OUTCOME

In order to calculate and interpreter probabilities it is important to understand and use the ideaIn order to calculate and interpreter probabilities it is important to understand and use the idea

of sample space. of sample space.

The sample space of an experimentThe sample space of an experiment is the set of all of the distinct possible outcomes of the is the set of all of the distinct possible outcomes of the

experiment. Each distinct out come is called sample space out come or sample point orexperiment. Each distinct out come is called sample space out come or sample point or

elementary event. elementary event.

Example 1Example 1

A newly married couple plans to have two children. Naturally, they are curious about whetherA newly married couple plans to have two children. Naturally, they are curious about whether

their children will be boys or girls. Therefore, we consider the experiment of having twotheir children will be boys or girls. Therefore, we consider the experiment of having two

children. children.

In order to find the sample spaces of this experiment, of having two children, we let ‘B’In order to find the sample spaces of this experiment, of having two children, we let ‘B’

denote that child is a boy and ‘G’ denotes that child is a girl. denote that child is a boy and ‘G’ denotes that child is a girl.

This experiment is a two-step process i.e having the first child, which could be a boy or a girlThis experiment is a two-step process i.e having the first child, which could be a boy or a girl

and having the second child, which could also be either a boy or a girl. and having the second child, which could also be either a boy or a girl.

This can be constructed by a tree diagram. Each branch of the tree leads us to a distinctThis can be constructed by a tree diagram. Each branch of the tree leads us to a distinct

sample space outcome. sample space outcome.

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We see that there are four sample space outcomes. Therefore the sample space (i.e the set ofWe see that there are four sample space outcomes. Therefore the sample space (i.e the set of

all of the distinct samples space outcomes is all of the distinct samples space outcomes is BBBB BGBG GBGB GGGG. .

In order to consider the probabilities of these outcomes, suppose that boys and girls areIn order to consider the probabilities of these outcomes, suppose that boys and girls are

equality likely each time a child is born. This says that each of the sample space out comes isequality likely each time a child is born. This says that each of the sample space out comes is

equally likely. i.e. equally likely. i.e.

P(BB) = p(BG)=p(GB)=p(GG)= P(BB) = p(BG)=p(GB)=p(GG)= 11//44 This says that there is a 25%, chance that each of these This says that there is a 25%, chance that each of these

outcomes will occur. Since we are certain that there is no other option or combinationoutcomes will occur. Since we are certain that there is no other option or combination

remaining, the probability that the couple will have any one of the sample space outcomes isremaining, the probability that the couple will have any one of the sample space outcomes is

one. i.e. P(BB) + P(BG) + P(EB) + P(EG) = 1one. i.e. P(BB) + P(BG) + P(EB) + P(EG) = 1

Notice that these probabilities sum one i.e the sum of the probabilities of all sample spaceNotice that these probabilities sum one i.e the sum of the probabilities of all sample space

outcomes is one. outcomes is one.

Therefore the sampleTherefore the sample space (space (that is, the set of all of the distinct sample space out comes) is that is, the set of all of the distinct sample space out comes) is

BB, BG, GB, GG BB, BG, GB, GG

Example 2Example 2

A student takes a quiz that consist of three true or false questions. If we consider ourA student takes a quiz that consist of three true or false questions. If we consider our

experiment to be answering the three questions, each question can be answered correctly orexperiment to be answering the three questions, each question can be answered correctly or

incorrectly. incorrectly.

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Boy(B)

Girl (G) Boy (B)

Girl (G)

Boy(B)

Girl (G)

2nd Child 1st Child

BB Sample outcome

BG - samples space outcomes

GB sample space outcome

GG sample space outcome sample space

Correct (c)

Incorrect I

Correct (c)

Incorrect I

Correct (c)

Correct (c)Incorrect (I)

Incorrect (I)

Cor

rect

(c)

Correct (c)

Incorrect (I)

Incorrect (I)Correct (c)

Incorrect (I)

CCC

CCI

CII

CII ICC

ICI IIC

III Sample space

Step IAnswering the 1st

questionStep IIAnswering the 2nd

questionStep IIIAnswering the 3rd

question

Let c denote answering a question correctly and I denote answering a question incorrectly.Let c denote answering a question correctly and I denote answering a question incorrectly.

Then we can depict a tree diagram of the sample space out come for the experiment. Then we can depict a tree diagram of the sample space out come for the experiment.

This diagram portrays the experiment as a three-step process This diagram portrays the experiment as a three-step process

Step I – answering the 1Step I – answering the 1stst question (Correctly or incorrectly) (C or I) question (Correctly or incorrectly) (C or I)

Step II – answering the 2Step II – answering the 2ndnd question (Correctly or incorrectly). question (Correctly or incorrectly).

Step III – answering the 3Step III – answering the 3rdrd question (Correctly or incorrectly). question (Correctly or incorrectly).

The tree diagram has eight different branches and the eight distinct sample space outcomesThe tree diagram has eight different branches and the eight distinct sample space outcomes

are listed at the end of the branches. We see the sample space is are listed at the end of the branches. We see the sample space is

CCC CCI CIC CIICCC CCI CIC CII

ICC ICI IIC III ICC ICI IIC III

Now suppose that the student was totally unprepared for the test, and has to blindly guess theNow suppose that the student was totally unprepared for the test, and has to blindly guess the

answer to each question that is the student has a 50-50 chance or 0.5 probability of correctlyanswer to each question that is the student has a 50-50 chance or 0.5 probability of correctly

answering each question. This means that each of the eight sample space outcomes is equallyanswering each question. This means that each of the eight sample space outcomes is equally

likely to occur. likely to occur.

i.e i.e

P(ccc) = P(ccI) ------P(III) =1/8 P(ccc) = P(ccI) ------P(III) =1/8

Here also the sum of the probabilities of the sample space out comes is one.Here also the sum of the probabilities of the sample space out comes is one.

In General the sum of the probabilities of all the sample space is equal to 1. In General the sum of the probabilities of all the sample space is equal to 1.

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Finding Probabilities by using Sample SpaceFinding Probabilities by using Sample Space

If all of the sample space out comes are equally likely, then the probability that an event willIf all of the sample space out comes are equally likely, then the probability that an event will

occur is equal to the ratio:occur is equal to the ratio:

The number of sample space outcomes that correspond to the eventThe number of sample space outcomes that correspond to the event

The total number of sample space outcomes. The total number of sample space outcomes.

Consider the couple planning to have two children to find the probability of two boys first weConsider the couple planning to have two children to find the probability of two boys first we

have to find the sample space outcome corresponding to the event of having the first child ahave to find the sample space outcome corresponding to the event of having the first child a

boy and the second child also a boy.boy and the second child also a boy.

There is only one sample space outcome corresponding to this event i.e. BB so the probabilityThere is only one sample space outcome corresponding to this event i.e. BB so the probability

will be: will be: 4

1= 0.25 the probability that the couple will have a boy and a girls is similarly= 0.25 the probability that the couple will have a boy and a girls is similarly

calculated by first identifying the sample space outcomes corresponding to the event ofcalculated by first identifying the sample space outcomes corresponding to the event of

having a boy and a girls. The sample space outcomes are BG and GB. So the probability willhaving a boy and a girls. The sample space outcomes are BG and GB. So the probability will

be be 4

2= 0.5= 0.5

Check Your Progress – 1 Check Your Progress – 1

1.1. Suppose that a couple will have three children. Letting B denote a boy and G denote a Suppose that a couple will have three children. Letting B denote a boy and G denote a

girl. girl.

a)a) Draw a tree diagram depicting the sample space out come for this Draw a tree diagram depicting the sample space out come for this

experiment. experiment.

b)b) list the sample space outcomes that correspond to each of the following events.list the sample space outcomes that correspond to each of the following events.

1)1) All three children will have the same gender All three children will have the same gender

2)2) Exactly two of the three children will be girls. Exactly two of the three children will be girls.

3)3) Exactly one of the three children will be a girl. Exactly one of the three children will be a girl.

4)4) None of the tree children will be a girl. None of the tree children will be a girl.

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2.2. Four people will enter an automobile show Room and each will either purchase a car Four people will enter an automobile show Room and each will either purchase a car

(P) or will not purchase a car (N) (P) or will not purchase a car (N)

a)a) Draw a tree diagram depicting the sample space of all possible purchase Draw a tree diagram depicting the sample space of all possible purchase

decision that could potentially be made by the four people.decision that could potentially be made by the four people.

b)b) List the sample space out comes that correspond to each of the following List the sample space out comes that correspond to each of the following

events. events.

1)1) Exactly three people will purchase a car Exactly three people will purchase a car

2)2) Two or fewer will purchase a car Two or fewer will purchase a car

3)3) One or more people will purchase a carOne or more people will purchase a car

4)4) All four people will make the same purchase decisionAll four people will make the same purchase decision

Often time it may be practically impossible to list all possible sample space outcomes of an Often time it may be practically impossible to list all possible sample space outcomes of an

experiment. Under such circumstances we can find the probability of an event by identifying experiment. Under such circumstances we can find the probability of an event by identifying

the number of sample space outcomes /without listing/ corresponding to the event.the number of sample space outcomes /without listing/ corresponding to the event.

ExampleExample - Suppose that 650.000 of 1,000,000 households in Addis subscribe to a newspaper - Suppose that 650.000 of 1,000,000 households in Addis subscribe to a newspaper

called Addis Zemen, and consider randomly selecting one of the Households in this city. Thatcalled Addis Zemen, and consider randomly selecting one of the Households in this city. That

is consider selecting one household & giving each and every household in the city the sameis consider selecting one household & giving each and every household in the city the same

chance of being selected. Let A be the event that the randomly selected household subscribeschance of being selected. Let A be the event that the randomly selected household subscribes

to the Addis Zemen. Then since the sample space of this experiment consists of 1,000,000to the Addis Zemen. Then since the sample space of this experiment consists of 1,000,000

equally likely sample space outcomes (households). It follows that equally likely sample space outcomes (households). It follows that

P(A) = P(A) = the number of Households that subscribe to the Addis Zementhe number of Households that subscribe to the Addis Zemen The total number of households in the city The total number of households in the city

= = 650,000650,000 = 0.65 = 0.65 1000,000 1000,000

Now also suppose that 500,000 households in the city subscribe to the Ethiopian Herald (H)Now also suppose that 500,000 households in the city subscribe to the Ethiopian Herald (H)

and further suppose that 250,000 households subscribe to both the newspapers.and further suppose that 250,000 households subscribe to both the newspapers.

We consider randomly selecting one household in the city, and we define the following eventsWe consider randomly selecting one household in the city, and we define the following events

A = The random of selected house hold subscribes to the Addis Zemen. A = The random of selected house hold subscribes to the Addis Zemen.

Ā = The randomly selected, hose hold does not subscribe to the Addis Zemen.Ā = The randomly selected, hose hold does not subscribe to the Addis Zemen.

H = The randomly selected household subscribes to the Ethiopian Herald. H = The randomly selected household subscribes to the Ethiopian Herald.

H = The randomly selected household does not subscribe the Herald.= The randomly selected household does not subscribe the Herald.

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Using the notation AnH to denote both A& H we also define.Using the notation AnH to denote both A& H we also define.

AnH = The randomly selected household subscribes both to Addis Zemen & Herald. AnH = The randomly selected household subscribes both to Addis Zemen & Herald.

Since 650,000 of the 1,000,0000 households subscribe to the Addis Zemen (that is correspondSince 650,000 of the 1,000,0000 households subscribe to the Addis Zemen (that is correspond

to the event Occurring). Then 350,000 households do not subscribe to Zemen (Ā) i.e.to the event Occurring). Then 350,000 households do not subscribe to Zemen (Ā) i.e.

1,000,000 – 650,000.1,000,000 – 650,000.

Similarly since 500,000 households subscribe to Herald (H) 500,000 households do notSimilarly since 500,000 households subscribe to Herald (H) 500,000 households do not

subscribe to herald (subscribe to herald ( H ).).

Next consider the events Next consider the events

AnAn H = the randomly selected household subscribes to Zemen and does not subscribe to = the randomly selected household subscribes to Zemen and does not subscribe to

Herald;Herald;

ĀnH = the randomly selected household does not subscribe to Zemen and does subscribe to ĀnH = the randomly selected household does not subscribe to Zemen and does subscribe to

Herald. Herald.

A summary of the number of house holds corresponding to the events A, Ā, H, A summary of the number of house holds corresponding to the events A, Ā, H, H and and

AnH AnH

EventsEvents Subscribe to Subscribe to

HeraldHerald

Does not subscribe Does not subscribe

to Herald to Herald

Total Total

Subscribe & Addis ZemenSubscribe & Addis Zemen 250,000250,000 650,000650,000Does not subscribe to Addis Does not subscribe to Addis

Zemen Zemen

350,000350,000

Total Total 500,000500,000 500,000500,000 1,000,0001,000,000

Define the event ĀDefine the event Ā n n H ,,

ĀĀ n n H = the randomly selected household does not subscribe to both newspaper. = the randomly selected household does not subscribe to both newspaper.

Since 650,000 households subscribe to the Addis Zemen (A) and 250,000 householdsSince 650,000 households subscribe to the Addis Zemen (A) and 250,000 households

subscribe to both Zemen and Herald (AnH) it follows that 650,000 – 250,000 = 40,000 housesubscribe to both Zemen and Herald (AnH) it follows that 650,000 – 250,000 = 40,000 house

holds subscribe to Addis Zemen but do not subscribe to Herald, (Anholds subscribe to Addis Zemen but do not subscribe to Herald, (An H ). This subtraction is). This subtraction is

illustrated in the table below. illustrated in the table below.

By similar logicBy similar logic

a.a. 500,000 – 250,000 = 25,000 households do not subscribe to Addis Zemen but do 500,000 – 250,000 = 25,000 households do not subscribe to Addis Zemen but do

subscrige to Herald (subscrige to Herald (ĀĀ n nHH) )

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b.b. 350,000 – 250,000 = 100,000 households do not subscribe the Addis Zemen and also 350,000 – 250,000 = 100,000 households do not subscribe the Addis Zemen and also

do not subscribe the Herald (do not subscribe the Herald (ĀĀ n n H ))

c.c. Subtracting to find the number of households corresponding to the events. Subtracting to find the number of households corresponding to the events.

d.d. AnH, AnAnH, An H ,,

EventEvent HH H

AA 250,000250,000 650,000-250,000650,000-250,000 650,000650,000ĀĀ 350,000350,000Total Total 500,000500,000 500,000500,000 1,000,0001,000,000

e. (e. (ĀĀ n H) = 5000,000-250,000 n H) = 5000,000-250,000

= 250,000= 250,000

f(f(ĀĀ n n H ) ) = 350,000 – 250,000= 350,000 – 250,000

= 100,000 = 100,000

A contingency table summarizing subscription data for Addis Zemen and HeraldA contingency table summarizing subscription data for Addis Zemen and Herald

EventEvent Subscribe to HeraldSubscribe to Herald

(H)(H)

Does not Subscribe to HeraldDoes not Subscribe to Herald

(( H ))

Total Total

Subscribe to Addis Zemen Subscribe to Addis Zemen

(A)(A)

250,000250,000 400,000400,000 650,000650,000

Does not subscribe to Addis Does not subscribe to Addis

Zemen (Zemen (Ā)Ā)

250,000250,000 100,0000100,0000 350,000350,000

Total Total 500,000500,000 500,000500,000 1,000,0001,000,000

Now since we will randomly select one household (making all the households equally likelyNow since we will randomly select one household (making all the households equally likely

to be chosen), the probability of any of the previously defined events is the ration of theto be chosen), the probability of any of the previously defined events is the ration of the

number of households corresponding to the event’s occurrence to the total number ofnumber of households corresponding to the event’s occurrence to the total number of

households in the city. households in the city.

Therefore Therefore

P(A) = P(A) = 650,000650,000 = 0.65 = 0.65 1,000,000 1,000,000

P(H) = P(H) = 500,000500,000 = 0.5 = 0.5 1,000,000 1,000,000

P(AnH) = P(AnH) = 250,000250,000 = 0.25 = 0.25 1,000,000 1,000,000

Next letting AUH denote either A or H, we consider finding the probability of the event Next letting AUH denote either A or H, we consider finding the probability of the event

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AUH = the randomly selected household subscribes to either the Addis Zemen or Herald. (i.e AUH = the randomly selected household subscribes to either the Addis Zemen or Herald. (i.e

subscribe to at least one of the two newspapers).subscribe to at least one of the two newspapers).

We see that the households subscribing to either Addis Zemen or Herald:We see that the households subscribing to either Addis Zemen or Herald:

a)a) The 400,000 households that subscribe to only Addis Zemen, AnThe 400,000 households that subscribe to only Addis Zemen, An H

b)b) The 250,000 house holds that subscribe to only the Herald, The 250,000 house holds that subscribe to only the Herald, ĀĀnH andnH and

c)c) The 250,000 households that subscribes to both Addis Zemen and Herald, AnH. The 250,000 households that subscribes to both Addis Zemen and Herald, AnH.

Therefore since a total of 900,000 households subscribe to either the Addis Zemen Therefore since a total of 900,000 households subscribe to either the Addis Zemen

or Herald it follows: -or Herald it follows: -

P(AUH) = P(AUH) = 900,000900,000 = 0.9 = 0.9 1,000,000 1,000,000

i.e 90% of the house holds in the city subscribe to either Addis Zemen or Herald. i.e 90% of the house holds in the city subscribe to either Addis Zemen or Herald.

Notice that P(AUH) = 0.9 does not equal Notice that P(AUH) = 0.9 does not equal

P(A) +P(H) = 0.65 +0.5 = 1.15 P(A) +P(H) = 0.65 +0.5 = 1.15

Logically the reason for this is that both P(A) = 0.65 and P(H) = 0.5 count the 25% of theLogically the reason for this is that both P(A) = 0.65 and P(H) = 0.5 count the 25% of the

households that subscribe to both newspapers. Therefore;households that subscribe to both newspapers. Therefore;

the sum of P(A) and P(H) counts this 25% of the households once to oftenthe sum of P(A) and P(H) counts this 25% of the households once to often

It follows that if we subtract P(AnH) = 0.25 from the sum of P(A) and P(H) then we willIt follows that if we subtract P(AnH) = 0.25 from the sum of P(A) and P(H) then we will

obtain P(AuH) i.eobtain P(AuH) i.e

P(AuH) = P(A)+P(H) – P(AnH) P(AuH) = P(A)+P(H) – P(AnH)

= 0.65+0.5-0.25 = 0.90= 0.65+0.5-0.25 = 0.90

The intersection and union of Two events. The intersection and union of Two events.

Given two events A&BGiven two events A&B

1)1) The Intersection of A&B is the event consisting of the sample space outcomesThe Intersection of A&B is the event consisting of the sample space outcomes

belonging to both A&B, denoted AnB. Further more P(AnB) denotes the probabilitybelonging to both A&B, denoted AnB. Further more P(AnB) denotes the probability

that Both A&B will simultaneously Occur. that Both A&B will simultaneously Occur.

2)2) The union of A&B is the event consisting of sample space outcomes belonging toThe union of A&B is the event consisting of sample space outcomes belonging to

either A or B. The union is denoted AUB Further more P(AUB) denotes theeither A or B. The union is denoted AUB Further more P(AUB) denotes the

probability that either A or B will occur.probability that either A or B will occur.

2.5 PROBABILITY RULES2.5 PROBABILITY RULES

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2.5.1 The Addition Rule2.5.1 The Addition Rule

2.5.1.1 Addition Rule for two Dependent Events2.5.1.1 Addition Rule for two Dependent Events

Let A and B be events then the probability that either A or B will occur is Let A and B be events then the probability that either A or B will occur is

P(AUB) = P(A)+P(B)-P(AnB)P(AUB) = P(A)+P(B)-P(AnB)

2.5.1.2 Addition Rule for Two Mutually Exclusive Events2.5.1.2 Addition Rule for Two Mutually Exclusive Events

Two events are said mutually exclusive if they have no sample space outcomes inTwo events are said mutually exclusive if they have no sample space outcomes in

common. In this case the event A&B cannot occur simultaneously and thus. common. In this case the event A&B cannot occur simultaneously and thus.

P(AnB) = 0 P(AnB) = 0

Let A&B Mutually exclusive events then, the probability that either A or B will occur is Let A&B Mutually exclusive events then, the probability that either A or B will occur is

P(AUB) = P(A) + P(B) P(AUB) = P(A) + P(B)

Example Example - consider randomly selecting a card from a standard deck of 52 playing cards and - consider randomly selecting a card from a standard deck of 52 playing cards and

define the events.define the events.

J, a randomly drawn card is Jack; Q, a randomly drawn card is Queen; and K, a randomly J, a randomly drawn card is Jack; Q, a randomly drawn card is Queen; and K, a randomly

drawn card is a king.drawn card is a king.

Since there are 4 Jacks, 4 Queens and 4 Kings in the deck.Since there are 4 Jacks, 4 Queens and 4 Kings in the deck.

P(Q) = P(Q) = 524 P(K) = P(K) = 52

4 P(J) = P(J) = 524

Since there is no card that is both a J & Q the event J and Q are mutually exclusive and thusSince there is no card that is both a J & Q the event J and Q are mutually exclusive and thus

P(JnQ) = 0 it follows that the probability that the randomly selected card is either J or Q is P(JnQ) = 0 it follows that the probability that the randomly selected card is either J or Q is

P(JUQ) = P(T) + PQ P(JUQ) = P(T) + PQ

= 4/52 + 4/52 = 2/13= 4/52 + 4/52 = 2/13

2.5.1.3 The Addition Rule for N mutually exclusive events. 2.5.1.3 The Addition Rule for N mutually exclusive events.

The event AThe event A11, A, A22, ------A, ------Ann are mutually exclusive if no two of the events have any sample are mutually exclusive if no two of the events have any sample

space out come in common. In this case no two of the events can occur simultaneously and space out come in common. In this case no two of the events can occur simultaneously and

P(AP(A11UAUA22U-----UAU-----UAnn) = P(A) = P(A11)+P(A)+P(A22)+-----+P(A)+-----+P(Ann) )

ExampleExample P(JuQUKU nine) = P(JuQUKU nine) =

P(J)+P(Q) +P(K) +P(nine) P(J)+P(Q) +P(K) +P(nine)

= = 44//5252 + + 44//5252 + + 44//5252 + + 44//5252 = = 52

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2.6 THE COMPLEMENT OF AN EVENT 2.6 THE COMPLEMENT OF AN EVENT

Given an event A, the complement of A is the event consisting of all sample space outcomes Given an event A, the complement of A is the event consisting of all sample space outcomes

that do not correspond to the occurrence of A. that do not correspond to the occurrence of A.

The complement of A is denoted The complement of A is denoted ĀĀ

Furthermore P(Furthermore P(ĀĀ) denotes the probability that A will not occur. ) denotes the probability that A will not occur.

In any probability situation, either an event A or its compliment A must occur. In any probability situation, either an event A or its compliment A must occur.

Therefore we have Therefore we have

P(A) + P(P(A) + P(ĀĀ) = 1 ) = 1

This implies This implies

P(P(ĀĀ) = 1-P(A) ) = 1-P(A)

ExampleExample – If team A and B are playing for a final cup we can say that the events that team. A – If team A and B are playing for a final cup we can say that the events that team. A

will win is complement to the event that B will win. i.e., if A wins B will lose. Under no will win is complement to the event that B will win. i.e., if A wins B will lose. Under no

circumstance that A will win and looses at the same time winning and losing are mutually circumstance that A will win and looses at the same time winning and losing are mutually

exclusive. exclusive.

2.7 CONDITIONAL PROBABILITY AND INDEPENDENCE 2.7 CONDITIONAL PROBABILITY AND INDEPENDENCE

2.7.1 Conditional Probability2.7.1 Conditional Probability

Probability is conditional upon information. We may define the probability of event AProbability is conditional upon information. We may define the probability of event A

conditional upon the occurrence of event B. conditional upon the occurrence of event B.

If we think about two adjacent rooms, RIf we think about two adjacent rooms, R11 and R and R22, the probability that R1 will be caught by fire, the probability that R1 will be caught by fire

is highly conditional on the probability of the other room.is highly conditional on the probability of the other room.

Example 1Example 1. Suppose that we randomly select a household, and that the chosen house hold. Suppose that we randomly select a household, and that the chosen house hold

reports it subscribes to Herald. Given this new information we wish to find the probabilityreports it subscribes to Herald. Given this new information we wish to find the probability

that this household subscribes to Addis Zemen. The new probability is called a conditionalthat this household subscribes to Addis Zemen. The new probability is called a conditional

probability. probability.

The probability of the event A, given the condition that the event H has occurred, is written The probability of the event A, given the condition that the event H has occurred, is written

P(A/H) = the probability of A given H. We often refer to such a probability as the P(A/H) = the probability of A given H. We often refer to such a probability as the

conditional probability of A given H .conditional probability of A given H .

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In order to find the conditional probability that a household subscribes to Addis Zemen givenIn order to find the conditional probability that a household subscribes to Addis Zemen given

that it subscribes to Herald we know that we are considering one of 500,000 households.that it subscribes to Herald we know that we are considering one of 500,000 households.

Since 250,000 of these 500,000 Herald subscribers also subscribe to Addis Zemen we have Since 250,000 of these 500,000 Herald subscribers also subscribe to Addis Zemen we have

P(A/H/ = P(A/H/ = 250,000250,000 =0.5 =0.5 500,000 500,000

i.e 50% of the Herald subscribers also subscribe to Addis Zemen:i.e 50% of the Herald subscribers also subscribe to Addis Zemen:

Example 2.Example 2. Next suppose that we randomly select another household from the 1,000,000 Next suppose that we randomly select another household from the 1,000,000

house holds and suppose that this newly chosen household reports that it subscribes to Addis house holds and suppose that this newly chosen household reports that it subscribes to Addis

ZemenZemen

Now find the probability that this house hold subscribes to Herald Now find the probability that this house hold subscribes to Herald

P(H/A)P(H/A)

= = 250,000250,000 = 0.3846 = 0.3846 650,000 650,000

This says that the probability that the randomly selected household subscribes to herald givenThis says that the probability that the randomly selected household subscribes to herald given

that the household subscribes to Addis Zemen is 0.3846. ie., 38.46% of Addis Zementhat the household subscribes to Addis Zemen is 0.3846. ie., 38.46% of Addis Zemen

subscribers also subscribe to Herald. subscribers also subscribe to Herald.

We have We have

P(A) = P(A) = 650,000650,000 =0.65 =0.65 1,000,000 1,000,000

P(AnH) = P(AnH) = 250,000 250,000 = 0.25 = 0.25 1,000,000 1,000,000

P(H/A) = P(H/A) = 250,000250,000 = 0.3846 = 0.3846 650,000 650,000

P(H) = P(H) = 500,000500,000 = 0.5 = 0.5 1,000,000 1,000,000

P(A/H) = P(A/H) = 250,000250,000 = 0.5 = 0.5 500,000 500,000

If we divide both the numerator and denominator of each conditional probability by If we divide both the numerator and denominator of each conditional probability by

1,000,0001,000,000

P(A/H) = P(A/H) = 250,000250,000 = = 250,000/1,000,000250,000/1,000,000 = = P(AnH) P(AnH) 500,000 500,000/1,000,000 P(H) 500,000 500,000/1,000,000 P(H)

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P(H/A) = P(H/A) = 250,000250,000 = = 250,000/1,000,000250,000/1,000,000 = = P(AnHP(AnH) ) 650,000 650,000/1,000,000 P(A) 650,000 650,000/1,000,000 P(A)

We express these conditional probabilities in terms of P(A), P(H) and P(AnH) We express these conditional probabilities in terms of P(A), P(H) and P(AnH)

Given the sample space outcomes are equally likely.Given the sample space outcomes are equally likely.

P(A/H) = P(A/H) = P(AnH)P(AnH) , then P(AnH) = P(H) P(A/H), by simple cross multiplication , then P(AnH) = P(H) P(A/H), by simple cross multiplication P(H) P(H)

P(H/A) = P(H/A) = P(AnH)P(AnH) = then P(AnH) = P(A) P(H/A) = then P(AnH) = P(A) P(H/A) P(A) P(A)

The General Multiplication Rule The General Multiplication Rule

(two ways to calculate P(AnH) (two ways to calculate P(AnH)

P(AnH) = P(A) P(H/A)=P(H) P(A/H )P(AnH) = P(A) P(H/A)=P(H) P(A/H )

Example 1. In a firm 20% of the employees have an accounting background, while 5% of theExample 1. In a firm 20% of the employees have an accounting background, while 5% of the

employees are executives and have an accounting backgrounds. If an employee hasemployees are executives and have an accounting backgrounds. If an employee has

accounting background, what is the probability that the employee is an executive.accounting background, what is the probability that the employee is an executive.

Let us define the eventsLet us define the events

E, an employee is an executive andE, an employee is an executive and

A, an employee has an accounting backgroundA, an employee has an accounting background

P(A) = 0.2P(A) = 0.2

P(AnE) = 0.05P(AnE) = 0.05

then then

P(E/A) = P(E/A) = P(AnE)P(AnE) = = 0.050.05 = = 0.250.25 P(A) 0.2 P(A) 0.2

Example 2Example 2. A contractor is bidding for two projects with Co. A and Co. B. The contractor. A contractor is bidding for two projects with Co. A and Co. B. The contractor

estimates that the probability of obtaining the project with Co. A is 0.45. He also fells that ifestimates that the probability of obtaining the project with Co. A is 0.45. He also fells that if

he should get the project with Co. A then there is a 0.90 probability that Co. B will also givehe should get the project with Co. A then there is a 0.90 probability that Co. B will also give

him the project. What are the contractor’s chances of getting both projects:him the project. What are the contractor’s chances of getting both projects:

Solution:Solution: We are given We are given

P(A) = 0.45P(A) = 0.45

P(B/A) = 0.90 and we are looking for P(AnB), which is the probability that both A andP(B/A) = 0.90 and we are looking for P(AnB), which is the probability that both A and

B will occur. From the equation we haveB will occur. From the equation we have

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P(AnB) = P(B/A) P(A) = 0.9 x 0.45 = 0.405P(AnB) = P(B/A) P(A) = 0.9 x 0.45 = 0.405

Check Your Progress –2Check Your Progress –2

21% of the executive in a large firm are at the top salary level. It is further known that 40% of21% of the executive in a large firm are at the top salary level. It is further known that 40% of

all the executives at the firm are women. Also 6.4% of all executives are women and are atall the executives at the firm are women. Also 6.4% of all executives are women and are at

the top salary level. Recently among executives at the firm arose a question as to whetherthe top salary level. Recently among executives at the firm arose a question as to whether

there is any evidence of salary inequality. Check. there is any evidence of salary inequality. Check.

ClueClue. To solve this problem, pose a question in terms of probabilities. I.e., ask whether the. To solve this problem, pose a question in terms of probabilities. I.e., ask whether the

probability that the executive will be at the top salary level given the executive is a woman. Ifprobability that the executive will be at the top salary level given the executive is a woman. If

this probability is less than 16% (the average) you can conclude that salary inequity does existthis probability is less than 16% (the average) you can conclude that salary inequity does exist

because of gender.because of gender.

2.7.2 Statistical Independence2.7.2 Statistical Independence

If the occurrence of events A and B have nothing to do with each other, then we know that AIf the occurrence of events A and B have nothing to do with each other, then we know that A

and B are independent events. and B are independent events.

i.e the probability of occurrence of A well not influence the probability of occurrence of B. i.e the probability of occurrence of A well not influence the probability of occurrence of B.

This implies that This implies that

P(A/B)= p(A) and that P(A/B)= p(A) and that

P(B/A) = p(B)P(B/A) = p(B)

Further more the general multiplication rule tells us that, for any two events A and B we canFurther more the general multiplication rule tells us that, for any two events A and B we can

say thatsay that

P(A n B) = p(A) p(B/A) there fore if p(B/A)= p(B) if follows that P(A n B) = p(A) p(B/A) there fore if p(B/A)= p(B) if follows that

P(AnB) = p(A) p(B) P(AnB) = p(A) p(B)

This is called the multiplication rule for two independent events.This is called the multiplication rule for two independent events.

However, if the probability of an event is influenced by whether or not another event occurs,However, if the probability of an event is influenced by whether or not another event occurs,

we say the two events are we say the two events are dependentdependent. .

eg. Define the events C and P as follows eg. Define the events C and P as follows

C= your favorite college football team C= your favorite college football team will winwill win its first match next season. its first match next season.

P= Your favorite professional football team will win its first match next season. P= Your favorite professional football team will win its first match next season.

20

Suppose that you believe that for next season p(c) = 0.6 and p(p) =0.6 then since the outcomesSuppose that you believe that for next season p(c) = 0.6 and p(p) =0.6 then since the outcomes

of a college football games and a professional football game would probably have noting toof a college football games and a professional football game would probably have noting to

do with each other, it is reasonable to assume that C and P are independent events. do with each other, it is reasonable to assume that C and P are independent events.

It follows that : Both your favorite teams will win their first match next season, It follows that : Both your favorite teams will win their first match next season,

P(CnP)= p(c) p(p)=0.6(0.6)=0.36 P(CnP)= p(c) p(p)=0.6(0.6)=0.36

When two events are independent, neither are their complements. When two events are independent, neither are their complements.

2.7.3 Independent and Mutually Exclusive Events2.7.3 Independent and Mutually Exclusive Events

When two events are mutually exclusive they are not independent. In fact they are veryWhen two events are mutually exclusive they are not independent. In fact they are very

dependent events in the sense that if one happens the other cannot happen. The intersection ofdependent events in the sense that if one happens the other cannot happen. The intersection of

two mutually exclusive events is zero but the probability of the intersection of twotwo mutually exclusive events is zero but the probability of the intersection of two

independent events is not zero. It is equal to the product of the probabilities of the separateindependent events is not zero. It is equal to the product of the probabilities of the separate

events. events.

2.7.3.1 The multiplication rule for N independent events2.7.3.1 The multiplication rule for N independent events

The events AThe events A11, A, A22 …. An are independent events if the occurrence of these events have …. An are independent events if the occurrence of these events have

nothing to do with each other. if events Anothing to do with each other. if events A11, A, A22, …, A, …, Ann are independent events, then are independent events, then

P(AP(A11 nA nA22 n. . . nA n. . . nAnn)= P(A)= P(A11) P(A) P(A22). . . P(A). . . P(Ann) )

Example 1Example 1. An electronic devise has four independent components C. An electronic devise has four independent components C11, C, C22, C, C33, C4, with a, C4, with a

reliability of 0.85 each. The device works only if all four components are functional. reliability of 0.85 each. The device works only if all four components are functional.

What is the probability that the device will work when needed?What is the probability that the device will work when needed?

P(the device will work) = P(all components will work ) = P(cP(the device will work) = P(all components will work ) = P(c11, nc, nc22,nc,nc33,nc,nc44) )

= p(C= p(C11) p(C) p(C22) p(C) p(C33) p(C) p(C44) )

= 0.85 x 0.85 x 0.85 x 0.85= 0.85 x 0.85 x 0.85 x 0.85

=0.85 x 0.85 x 0.85 x 0.85 = =0.85 x 0.85 x 0.85 x 0.85 = 0.5220.522

Example 2.Example 2. The rate of defects in corks of wine is 0.75. Assuming independence, if four The rate of defects in corks of wine is 0.75. Assuming independence, if four

bottles are opened (Bbottles are opened (B11, B, B22, B, B33, B, B44), what is probability that four corks are defective. ), what is probability that four corks are defective.

P(all 4 are defective)= P(BP(all 4 are defective)= P(B11 n B n B22 n B n B33 n B n B44) = P(B) = P(B11) P(B) P(B22) P(B) P(B33) P(B) P(B44))

= 0.75 x 0.75 x 0.75 x 0.75=0.316 = 0.75 x 0.75 x 0.75 x 0.75=0.316

2.7.3.2 Union rule 2.7.3.2 Union rule

The union of several independent events is the event that at least one of the events happens. The union of several independent events is the event that at least one of the events happens.

21

The probability of the union of several independent events AThe probability of the union of several independent events All, A, A22, … A, … Ann is is

P(A, uAP(A, uA22 u. . . uA u. . . uAnn) = 1- P(Ā) = 1- P(Ā11) P(Ā) P(Ā 2 2). . . p(Ā). . . p(Ā n n) )

Example 1:Example 1: A device similar to the above one has three components, but the device works as A device similar to the above one has three components, but the device works as

long as at least one of the components is functional. The reliability of the components arelong as at least one of the components is functional. The reliability of the components are

0.96, 0.91 and 0.80what is the probability that the device will work when needed?0.96, 0.91 and 0.80what is the probability that the device will work when needed?

P(The device will work) = p(at least one will work) = 1 – p(all will fail) =1–p(P(The device will work) = p(at least one will work) = 1 – p(all will fail) =1–p( 1c )P()P( 2c )P()P(

3c ) = 1-(0.04) (0.09)0.02) = 1-(0.04) (0.09)0.02

Example 3:Example 3: In the developing world a woman’s adds of dying from problems related to In the developing world a woman’s adds of dying from problems related to

pregnancy is 1 in 51. If three women are pregnant what is the probability that at least one willpregnancy is 1 in 51. If three women are pregnant what is the probability that at least one will

diedie

p(at least one will die)= 1- p(all will survive) p(at least one will die)= 1- p(all will survive)

1-(50/51)1-(50/51)33 = 0.0577 = 0.0577

2.8 THE TOTAL PROBABILITY AND BYE’S THEOREM2.8 THE TOTAL PROBABILITY AND BYE’S THEOREM

2.8.1 Total Probability 2.8.1 Total Probability

What ever may be the relationship between two events we can always say that the probabilityWhat ever may be the relationship between two events we can always say that the probability

of A is equal to the probability of the intersection of A and B plus the probability of theof A is equal to the probability of the intersection of A and B plus the probability of the

intersection of A and the complement of B (eventintersection of A and the complement of B (eventB) B)

P(A) = P(AnB) + P(A nP(A) = P(AnB) + P(A nB) B)

Total probabilityTotal probability

Consider the households subscribing to the two newspapers. Consider the households subscribing to the two newspapers.

P(A) = 0.65 P(A) = 0.65

This probability includes the households subscribing to both the newspapers P(AnB) or theThis probability includes the households subscribing to both the newspapers P(AnB) or the

households subscribing to Addis Zemen and not for Herald. I.e.,households subscribing to Addis Zemen and not for Herald. I.e.,

P(A) = P(AnN) + P(AnP(A) = P(AnN) + P(AnH)H)

= 0.25 + 0.40 = 0.25 + 0.40

= = 0.650.65

22

The law of total probability may be extended to more complex situations, where the sampleThe law of total probability may be extended to more complex situations, where the sample

space X is portioned into more then two events. Say we partition the sample space in to aspace X is portioned into more then two events. Say we partition the sample space in to a

collection of n sets Bcollection of n sets B11, B, B22…B…Bnn The law of total probability in this situation is The law of total probability in this situation is

P(A) = P(A) = ∑=

n

i

p1

(AnB(AnBii))

Example 1:Example 1: Suppose A is the event that a picture card is drawn out of a standard deck of 52 Suppose A is the event that a picture card is drawn out of a standard deck of 52

cards Letting H.C.D and S denote the events that the card drawn is a Heart, Club, Diamond orcards Letting H.C.D and S denote the events that the card drawn is a Heart, Club, Diamond or

Spade respectively. Spade respectively.

In a standard deck there are 12 picture cards. The probability will then be In a standard deck there are 12 picture cards. The probability will then be 1212//52. 52. Following theFollowing the

law of total probability. This probability can be obtained as the sample of the intersections oflaw of total probability. This probability can be obtained as the sample of the intersections of

the four events with A. In the deck there are three pictured cards and Heart (Jack heart, queenthe four events with A. In the deck there are three pictured cards and Heart (Jack heart, queen

hearts and king heart), three pictured and club; there pictured and diamond and three picturedhearts and king heart), three pictured and club; there pictured and diamond and three pictured

and at the same time spade. and at the same time spade.

We find the probability of a picture card, P(A)We find the probability of a picture card, P(A)

P(A) = P(AnH) + P(AnC) + P(AnD) +P(AnS) P(A) = P(AnH) + P(AnC) + P(AnD) +P(AnS)

= = 33//5252 + + 33//5252 + + 33//5252= = 1212//5252

The law of total probability can be extended using the definition of conditional probability. The law of total probability can be extended using the definition of conditional probability.

P(AnB) = P(A/B) p(B) similarly P(AnB) = P(A/B) p(B) similarly

P(AnP(AnB) = P(A/B) = P(A/B) p(B) p(B) B)

Substituting this formula to the addition Substituting this formula to the addition

i.e P(A)= p(AnB) + p(Ani.e P(A)= p(AnB) + p(AnB) B)

P(A) = P(A/B) p(B) + p(A/P(A) = P(A/B) p(B) + p(A/B)(B)(B) B)

For more than two sets For more than two sets

P(A) = P(A) = ∑=

n

i

p1

(A/B(A/Bii) p(B) p(Bii) )

Where there are n sets in the partition Where there are n sets in the partition

Example 1:Example 1: An analyst believes that the market has a 0.75 probability of going up in the next An analyst believes that the market has a 0.75 probability of going up in the next

year if the economy should do well, and a 0.30 probability of going up if the economy shouldyear if the economy should do well, and a 0.30 probability of going up if the economy should

23

not do well during the year. The analyst further believes there is a 0.80 probability that thenot do well during the year. The analyst further believes there is a 0.80 probability that the

economy will do well in the coming year.economy will do well in the coming year.

What is the probability that the market will go up next year? What is the probability that the market will go up next year?

Define the events Define the events

U= The Market will go up U= The Market will go up

W= The economy will do well W= The economy will do well

Find p(U) Find p(U)

P(u) = P(u/W)p(w) + p(u/P(u) = P(u/W)p(w) + p(u/w ) p(w ) p(w)w)

=0.75(080) + 03(0.2) =0.75(080) + 03(0.2)

=0.66 =0.66

This means the market can go up in two ways i.e if the economy will do well and the marketThis means the market can go up in two ways i.e if the economy will do well and the market

will go up and if the economy will not do well and the market will go up. will go up and if the economy will not do well and the market will go up.

2.8.2 Baye’s Theorem2.8.2 Baye’s Theorem

Baye’s Theorem is a very important theorem to revise probabilities using some additionalBaye’s Theorem is a very important theorem to revise probabilities using some additional

information. First let us define to important terms.information. First let us define to important terms.

2.8.2.1 Prior Probability /Initial Probability)2.8.2.1 Prior Probability /Initial Probability)

It is a given probability before any empirical data is observedIt is a given probability before any empirical data is observed

2.8.2.2 Posterior Probability 2.8.2.2 Posterior Probability

Is revised probability based on new information. Prior probabilities can be reviized as weIs revised probability based on new information. Prior probabilities can be reviized as we

have additional or new information about the events. have additional or new information about the events.

DerivationDerivation

P(B/A) = PP(B/A) = P(AnB)(AnB) P(A) P(A)

By another definition By another definition

i.e., P(AnB) = Pi.e., P(AnB) = P(A/B) P(B) (A/B) P(B) P(A) P(A)

From the law of total probability From the law of total probability

P(A) = P(A/B) P(B) + p(A/P(A) = P(A/B) P(B) + p(A/ B) P(B) P( B)B)

24

P(U/W) = 0.75 If P(W) = 0.8 thenP(W) = 0.80 P(W = 1 – 0.8 = 0.2P(U/W ) = 0.3

Substituting this expression for P(A) in the denominator Substituting this expression for P(A) in the denominator

P(B/A) = P(B/A) = P(A/B) p(B) P(A/B) p(B) P(A/B) p(B)+p(A/ P(A/B) p(B)+p(A/B)p(B)p(B)B)

The probabilities p(B) and p(The probabilities p(B) and p(B) are called prior probabilities of the events B andB) are called prior probabilities of the events B andB. TheB. The

probability P(B/A) is called the posterior probability of B. probability P(B/A) is called the posterior probability of B.

The theorem allows us to reverse the conditional it of events. The theorem allows us to reverse the conditional it of events.

We can obtain the probability of B given A from the probability of A given B. We can obtain the probability of B given A from the probability of A given B.

Baye’s theorem may be viewed as a means of transforming one prior probability of an eventBaye’s theorem may be viewed as a means of transforming one prior probability of an event

B into a posterior probability of the event B posterior to the known occurrence of event A. B into a posterior probability of the event B posterior to the known occurrence of event A.

Example1Example1. Let A be the event that a randomly selected American has the deadly disease. Let A be the event that a randomly selected American has the deadly disease

AIDS. And letAIDS. And letA be the event that the randomly selected American does not have AIDS.A be the event that the randomly selected American does not have AIDS.

Since it is estimated that 0.6 percent of the American population have AIDS. Since it is estimated that 0.6 percent of the American population have AIDS.

P(A) = 0.006 and P(P(A) = 0.006 and P(A)=0.994 A)=0.994

There is a test that attempts to detect whether a person has AIDS. According to historical dataThere is a test that attempts to detect whether a person has AIDS. According to historical data

99.9% of people with AIDS react positively (RP) to the test. 99.9% of people with AIDS react positively (RP) to the test.

i.e P(RA/A)=0.999 i.e P(RA/A)=0.999

Further more 1% of people with out AIDS react positively. Further more 1% of people with out AIDS react positively.

i.e., P(RP/i.e., P(RP/A) = 0.01A) = 0.01

If we give a randomly selected American the test and the person reacts positively, what is theIf we give a randomly selected American the test and the person reacts positively, what is the

probability that the person actually has Aids?probability that the person actually has Aids?

The idea of Bay’es theorem is that we can find P(A/RP) by thinking as follows. A person willThe idea of Bay’es theorem is that we can find P(A/RP) by thinking as follows. A person will

react positively (RP) if the person react positively and actually has AIDS (AnRP) or if thereact positively (RP) if the person react positively and actually has AIDS (AnRP) or if the

person react positively and does not actually have AIDS.person react positively and does not actually have AIDS.

((A nRP) A nRP)

Therefore, Therefore,

P(RP) = P(AnRP) + P(P(RP) = P(AnRP) + P(A nRP) A nRP)

This implies that This implies that

25

Baye’s TheoremBaye’s Theorem

P(A/RP)= P(A/RP)= P(AnRP)P(AnRP) P(RP) P(RP)

= = P(AnRP)P(AnRP)P(AnRP) + p(P(AnRP) + p(AnRP)AnRP)

= = P(A) P(RP/A)P(A) P(RP/A)P(A) p(RP/A) + P(P(A) p(RP/A) + P(A) P(RP/A) P(RP/A)A)

= = (0.006) (0.999)(0.006) (0.999)(0.006)(0.999)+(0.994)(0.01)(0.006)(0.999)+(0.994)(0.01)

= = 0.380.38

This probability says that, if all Americans were given an AIDS test only 38%of the peopleThis probability says that, if all Americans were given an AIDS test only 38%of the people

who would react positively to the test would actually have AIDS. who would react positively to the test would actually have AIDS.

Bay’es theorem may be extended to a partition of more than two sets. This is done using theBay’es theorem may be extended to a partition of more than two sets. This is done using the

law of total probability involving a partition of sets Blaw of total probability involving a partition of sets B11, B, B22, . . . . B, . . . . Bnn. .

The theorem gives the probability of one of the sets in the partition B, Given the occurrenceThe theorem gives the probability of one of the sets in the partition B, Given the occurrence

of event A. of event A.

Extended Bayes’ theorem. Extended Bayes’ theorem.

P(BP(B11/A) = /A) = ∑=

n

iii BPBAP

BPBAP

1

11

)()/(

)()/(

Example 1.Example 1. An Economist believes that during periods of high economic growth the U.S An Economist believes that during periods of high economic growth the U.S

dollar appreciates with probability 0.70; in periods of moderate economic growth the dollardollar appreciates with probability 0.70; in periods of moderate economic growth the dollar

appreciates with probability 0.40; and during periods of low economic growth the dollarappreciates with probability 0.40; and during periods of low economic growth the dollar

appreciates with probability 0.20. During any period of time the probability of high economicappreciates with probability 0.20. During any period of time the probability of high economic

growth is 0.30, the probability of moderate growth is 0.50 and the probability of lowgrowth is 0.30, the probability of moderate growth is 0.50 and the probability of low

economic growth is 0.2. Suppose the dollar has been appreciating during the present period.economic growth is 0.2. Suppose the dollar has been appreciating during the present period.

What is the probability that the economy is experiencing a period of high growth. Define theWhat is the probability that the economy is experiencing a period of high growth. Define the

three events, three events,

High economic growth (H) High economic growth (H)

Moderate economic growth(M) Moderate economic growth(M)

26

Prior probabilities

Low economic growth (L) Low economic growth (L)

The prior probabilities of the three states of the economy are P(H) =0.3 P(M)= 0.5 P(L)=0.2The prior probabilities of the three states of the economy are P(H) =0.3 P(M)= 0.5 P(L)=0.2

Let A denote the event that the dollar appreciate. We have the following conditionalLet A denote the event that the dollar appreciate. We have the following conditional

probabilities. probabilities.

P(A/H)= 0.70P(A/H)= 0.70 P(A/M) = 0.40P(A/M) = 0.40 P(A/L)= 0.20 P(A/L)= 0.20

Find P(H/A)Find P(H/A)

= = P(H/A) = P(A/H) P(H) P(H/A) = P(A/H) P(H) = = P(A/H) P(H) P(A/H) P(H)

P(A/H) P(H) + P(A/M)+P(M)+P(M)+P(A/L)P(L) P(AnH) + P(AnM) + P(AnL) P(A/H) P(H) + P(A/M)+P(M)+P(M)+P(A/L)P(L) P(AnH) + P(AnM) + P(AnL)

= = 0.70(0.30) 0.70(0.30)

0.70(0.30) + 0.4(0.5) + 0.2(0.2) 0.70(0.30) + 0.4(0.5) + 0.2(0.2)

= = 0.4670.467

We can obtain this answer along with posterior probabilities of the other two states of theWe can obtain this answer along with posterior probabilities of the other two states of the

economy M and L. i.e P(M/A) and P(L/A) economy M and L. i.e P(M/A) and P(L/A)

EventEvent

______________________

PriorPrior

probabilityprobability

ConditionalConditional


JointJoint


PosteriorPosterior

probabilityprobabilityH H P(H) 0.30P(H) 0.30 P(A/H)=70P(A/H)=70 P((AnH)=0.21P((AnH)=0.21

++

P(H/A)=P(H/A)=0.210.21=0.467=0.467

0.45 0.45MM P(M)=0.50P(M)=0.50 P(A/M)=0.40P(A/M)=0.40 P(AnM)= 0.20P(AnM)= 0.20

++

P(M/A)=P(M/A)=0.200.20= 0.444= 0.444 0.450.45

L L P(L)=0.20P(L)=0.20 P(A/L) =0.2 P(A/L) =0.2 P(AnL)=0.04P(AnL)=0.04

==

P(L/A)=P(L/A)=0.040.04=0.089=0.089 0.45 0.45

Sum Sum 11 P(A)=P(A)=0.450.45 Sum =Sum =11

Note that both the prior probabilities and the posterior probabilities of the three states add toNote that both the prior probabilities and the posterior probabilities of the three states add to

one.one.

Tree Diagram for the above exampleTree Diagram for the above example

Posterior probabilitiesPosterior probabilities

Joint ProbabilitiesJoint Probabilities

Conditional prob. Conditional prob.

27

P(HnA) = (0.3)(0.7)= 0.21

P(A/H)=0.70

P(AA /H)=0.30

P(H/A) = 45.021.0 = 0.467

BBGG

GG

BB

GG

BB

BB

GG

BB

GG BB

GG

BB

GG

PP

PP

NN

PP

NN

PP

NN

PP

NNPP

2.9 ANSWERS TO CHECK YOUR PROGRESS2.9 ANSWERS TO CHECK YOUR PROGRESS

1)1)

b) 1) Bb) 1) B BB, BB, GGGGGG

2) GBG, 2) GBG, BGG, BGG, GGBGGB

3) BBG 3) BBG BGBBGB GBBGBB

4) BBB 4) BBB

2) a)2) a)

28

P(M)=0.50

P(MnA) (0.5)(0.4) = 0.20

P(LnA) (0.2)(0.2)= 0.04P(L)=0.2

P(A/L)=0.2

P(A/M)=0.4

P(AA /M)=0.6

P(AA /L)=0.8

P(H)=0.30P(M/A) = 45.0

20.0 = 0.444

P(L/A) = 45.004.0 = 0.089

P(A) = 0.45Sum 1 Sum =1

NN

PP

NN

NN PP

NN

PP

NN

b) 1) PNPP, PPNP, PPPN, NPPPb) 1) PNPP, PPNP, PPPN, NPPP

2) NNNN, NNNP, NNPN, NPNN, PNNN, NNPP, NPNP, PNNP, PPNN, NPPN, PPNN 2) NNNN, NNNP, NNPN, NPNN, PNNN, NNPP, NPNP, PNNP, PPNN, NPPN, PPNN

3) NNNP, NNPN, NPNN, PNNN, NNPP, NPNP, PNNP, PPNN, NPPN, PPNN, PNPP,3) NNNP, NNPN, NPNN, PNNN, NNPP, NPNP, PNNP, PPNN, NPPN, PPNN, PNPP,

PPNP, PPPN, NPPP, PPPP.PPNP, PPPN, NPPP, PPPP.

4) PPPP, NNNN4) PPPP, NNNN

2.10 MODEL EXAMINATION QUESTION2.10 MODEL EXAMINATION QUESTION

Part I. Define the following terms of wordsPart I. Define the following terms of words

1.1. ProbabilityProbability

2.2. an experimentan experiment

3.3. an evenan even

4.4. an outcomean outcome

5.5. objective probabilityobjective probability

6.6. is subjective probabilityis subjective probability

7.7. sample space outcomesample space outcome

8.8. sample spacesample space

9.9. mutually exclusive eventsmutually exclusive events

10.10. Independent eventsIndependent events

11.11. Dependent eventsDependent events

12.12. Complement of an eventComplement of an event

13.13. Prior probabilitiesPrior probabilities

14.14. Posterior probabilities.Posterior probabilities.

Part II. Workout the following questionsPart II. Workout the following questions

29

Clearly show the stepsClearly show the steps

1.1. A newly established company is planning to recruit trainees for four jobs in theA newly established company is planning to recruit trainees for four jobs in the

marketing department. The marketing manager contacted an employment agency. Themarketing department. The marketing manager contacted an employment agency. The

agency has selected four candidates and send them to the Company. agency has selected four candidates and send them to the Company.

The company will hire those who fulfill the requirement of the job. Assuming that aThe company will hire those who fulfill the requirement of the job. Assuming that a

candidate’s chance to pass the final evaluation is 0.5.candidate’s chance to pass the final evaluation is 0.5.

a.a. List all the sample space outcomes of the experimentList all the sample space outcomes of the experiment

b.b. Identity the sample space outcomes corresponding to the following eventsIdentity the sample space outcomes corresponding to the following events

i.i. All of them will qualifyAll of them will qualify

ii.ii. Only two of them will qualifyOnly two of them will qualify

iii.iii. None of them will qualifyNone of them will qualify

iv.iv. There of them will qualityThere of them will quality

c.c. Assuming the probability that a candidate will be qualified for job is 0.5, findAssuming the probability that a candidate will be qualified for job is 0.5, find

the probability for the events listed in part D. the probability for the events listed in part D.

2.2. The personnel manager of a company construct the following summary table about theThe personnel manager of a company construct the following summary table about the

efficiency of Company employees.efficiency of Company employees.

EventEvent EfficiencyEfficiencyHigh, HHigh, H Average, AAverage, A Low, LLow, L TotalTotal

MenMen 120120 100100 8080 300300WomenWomen 4545 3535 2020 100100Total Total 165165 135135 100100 400400

Find the probability that a randomly selected employeeFind the probability that a randomly selected employee

a)a) has high efficiencyhas high efficiency

b)b) has average efficiencyhas average efficiency

c)c) has low efficiencyhas low efficiency

d)d) has high efficiency given that this employee ishas high efficiency given that this employee is

i.i. a mana man

ii.ii. a womana woman

e)e) is a woman and has high efficiencyis a woman and has high efficiency

f)f) is a man and has low efficiency is a man and has low efficiency

g)g) has high or low efficiencyhas high or low efficiency

30

3.3. A firm is planning to introduce a new product. The probability that the product will beA firm is planning to introduce a new product. The probability that the product will be

successful if a competitor does not come up with a similar product is 0.67. Thesuccessful if a competitor does not come up with a similar product is 0.67. The

probability that the new product will be successful in the presence of a competitor newprobability that the new product will be successful in the presence of a competitor new

product is 0.42. The probability that the competing firm will come out with a newproduct is 0.42. The probability that the competing firm will come out with a new

product during the period is question is 0.35.product during the period is question is 0.35.

What is the probability that the product will be a success?What is the probability that the product will be a success?

4.4. 25% of college class graduated with honors, while 20% of the class were honors25% of college class graduated with honors, while 20% of the class were honors

graduates and obtained good jobs. What is the probability that a person got a good job ifgraduates and obtained good jobs. What is the probability that a person got a good job if

he graduated with honors?he graduated with honors?

5.5. A contractor is bidding for four-construction project. He assesses his chances of winningA contractor is bidding for four-construction project. He assesses his chances of winning

the projects at 0.6, 0.75, 0.9 and 0.5. Assuming independence.the projects at 0.6, 0.75, 0.9 and 0.5. Assuming independence.

a)a) What is the probability that the contractor will win all projects?What is the probability that the contractor will win all projects?

b)b) What is the probability that the contractor will win at least one project?What is the probability that the contractor will win at least one project?

c)c) What is the probability that he will win none of the projects?What is the probability that he will win none of the projects?

6.6. A package of documents needs to be sent to a given destination and it is important that itA package of documents needs to be sent to a given destination and it is important that it

arrive with in one day. To maximize the chance of on time delivery, three copies of thearrive with in one day. To maximize the chance of on time delivery, three copies of the

document are sent via three different delivery services. Service A is known to have adocument are sent via three different delivery services. Service A is known to have a

90% on time delivery record, service B has an 88% on time delivery record, and service90% on time delivery record, service B has an 88% on time delivery record, and service

C has 91% on time delivery record. What is the probability that at least one copy of theC has 91% on time delivery record. What is the probability that at least one copy of the

documents will arrive at its destination on time?documents will arrive at its destination on time?

7.7. Three secretaries, SThree secretaries, S11, S, S22 and S and S33 do office work for a company, mainly filling papers, of do office work for a company, mainly filling papers, of

all the papers that come into the office, Sall the papers that come into the office, S11 files 50% S files 50% S22 files 30% and S files 30% and S33 files the rest. files the rest.

Each secretary occasionally misfiles a paper SEach secretary occasionally misfiles a paper S11 misfiles 5% of the papers she files, S misfiles 5% of the papers she files, S22

misfiles 7% of the papers she files and Smisfiles 7% of the papers she files and S3 3 misfiles 10% of the papers she files. Themisfiles 10% of the papers she files. The

manager has been looking for a particular paper and has found that it has been misfiled.manager has been looking for a particular paper and has found that it has been misfiled.

He decides to give warning to the one who most likely filed it. Who most likely filed it?He decides to give warning to the one who most likely filed it. Who most likely filed it?

Draw a tree diagram.Draw a tree diagram.

8.8. A manufacturing Co. purchases a component form three different suppliers. WhenA manufacturing Co. purchases a component form three different suppliers. When

components arrive at the warehouse of the co. they are placed in a bin withoutcomponents arrive at the warehouse of the co. they are placed in a bin without

inspection or otherwise identified by supplier. The materials manager does know thatinspection or otherwise identified by supplier. The materials manager does know that

45% of the components are purchased from S45% of the components are purchased from S11, 35% purchased from S, 35% purchased from S22 and the and the

31

remaining from Sremaining from S33. From past records it is also known that 6% of components purchased. From past records it is also known that 6% of components purchased

form Sform S11 are below standard, 8% of the components purchased from S are below standard, 8% of the components purchased from S22 are below are below

standard and 11% of the components purchased from Sstandard and 11% of the components purchased from S33 are below standard. The are below standard. The

materials manager randomly selects a component and found it below standard. Frommaterials manager randomly selects a component and found it below standard. From

which supplier the component is most likely purchased? Draw a tree diagram. which supplier the component is most likely purchased? Draw a tree diagram.

UNIT 3: PROBABILITY DISTRIBUTION UNIT 3: PROBABILITY DISTRIBUTION

ContentsContents



3.23.2 Random variablesRandom variables

3.2.13.2.1 Discrete Random VariableDiscrete Random Variable

3.2.23.2.2 Continuous Random VariableContinuous Random Variable

3.33.3 Discrete Probability DistributionDiscrete Probability Distribution

3.3.13.3.1 Constructing Probability DistributionConstructing Probability Distribution

3.3.23.3.2 Mean and Advance of a Discrete Probability DistributionMean and Advance of a Discrete Probability Distribution

32

3.3.33.3.3 Binomial Probability DistributionBinomial Probability Distribution

3.3.43.3.4 Hypergeometric Probability DistributionHypergeometric Probability Distribution

3.3.53.3.5 Poisson Probability DistributionPoisson Probability Distribution

3.43.4 Continuous /Normal/ Probability DistributionContinuous /Normal/ Probability Distribution

3.4.13.4.1 Normal Approximation to the BinomialNormal Approximation to the Binomial

3.4.23.4.2 Normal Approximation to the PoissonNormal Approximation to the Poisson

3.5 Answers to Check Your Progress3.5 Answers to Check Your Progress

3.6 Model Examination Question3.6 Model Examination Question


In this unit, you will be introduced to repeated experiments where the result of the experimentIn this unit, you will be introduced to repeated experiments where the result of the experiment

produces two different and many possible outcomes. You will learn how to computeproduces two different and many possible outcomes. You will learn how to compute

probabilities involving two-outcome situation using special probability formulas.probabilities involving two-outcome situation using special probability formulas.

After completing this unit you will be able: After completing this unit you will be able:

to understand the types of random variablesto understand the types of random variables

to calculate the expected value and variance of a discrete random variableto calculate the expected value and variance of a discrete random variable

to identity the characteristics of the binomial, hyper geometric and poison probabilityto identity the characteristics of the binomial, hyper geometric and poison probability

distributionsdistributions

to calculate probabilities for random variables following the binomial, hyper geometricto calculate probabilities for random variables following the binomial, hyper geometric

and poison distributionsand poison distributions

to calculate the mean and variance of the binomial, hyper geometric and positionto calculate the mean and variance of the binomial, hyper geometric and position

distribution distribution

to identity the characteristics of the continuous probability distribution and itsto identity the characteristics of the continuous probability distribution and its

accompanying normal curveaccompanying normal curve

to calculate probabilities of a continuous random variableto calculate probabilities of a continuous random variable

to approximate the normal distribution to the binomial and the poison distributions. to approximate the normal distribution to the binomial and the poison distributions.

3.1 INTRODUCTION 3.1 INTRODUCTION

Probability distribution is listing all possible values of the random variable withProbability distribution is listing all possible values of the random variable with

corresponding probabilities. The outcome of the experiment is either a success or failure. Thecorresponding probabilities. The outcome of the experiment is either a success or failure. The

33

number of ways to get certain number of successes will determine the value that the randomnumber of ways to get certain number of successes will determine the value that the random

variable will assume.variable will assume.

3.2 RANDOM VARIABLE3.2 RANDOM VARIABLE

Random variable is a variable whose value is determined by the out come of an experiment. Random variable is a variable whose value is determined by the out come of an experiment.

That is random variable represents an uncertain outcome or it can be defined as a quantityThat is random variable represents an uncertain outcome or it can be defined as a quantity

resulting from a random experiment that by chance, can assume different values. resulting from a random experiment that by chance, can assume different values.

A random variable may be either discrete or continuousA random variable may be either discrete or continuous

3.2.1 Discrete Random Variable 3.2.1 Discrete Random Variable

Is a variable that can assume only certain clearly separated values resulting from account ofIs a variable that can assume only certain clearly separated values resulting from account of

some item of interest?some item of interest?

Example:Example:

-- The No. of employees absent in a given day The No. of employees absent in a given day

-- Toss two coins and count the number of heads Toss two coins and count the number of heads

-- Number of defective products produced in a factory at a given shift or day orNumber of defective products produced in a factory at a given shift or day or

month. month.

-- Number of customers entering to a bank in an hour time.Number of customers entering to a bank in an hour time.

Is should be noted that a discrete random variable can in some cases assume fractional orIs should be noted that a discrete random variable can in some cases assume fractional or

decimal values. These values must be separated i.e have distance between them eg. The scoredecimal values. These values must be separated i.e have distance between them eg. The score

of a student in a given test can be 8.5 or 7.5 such values are discrete b/se there is a distanceof a student in a given test can be 8.5 or 7.5 such values are discrete b/se there is a distance

b/n scores. There is a fixed gap between scores. You can easily list all possible values clearlyb/n scores. There is a fixed gap between scores. You can easily list all possible values clearly

and separately. If the number of students in a classroom is 35, you know the next succeedingand separately. If the number of students in a classroom is 35, you know the next succeeding

value will be 36 there is no another value in between.value will be 36 there is no another value in between.

3.2.2 Continuous Random Variable3.2.2 Continuous Random Variable

A variable that can assume any value in an interval. It can assume one of an infinitely largeA variable that can assume any value in an interval. It can assume one of an infinitely large

number of values. Mostly results of measurement number of values. Mostly results of measurement

Example Example - The distance b/n two cities - The distance b/n two cities

- The weight of a person. - The weight of a person.

- The rate of return on investment - The rate of return on investment

- The time that a customer must wait to receive his changes. - The time that a customer must wait to receive his changes.

34

The values are not clearly separated. It is not possible to exhaustively list possible values ofThe values are not clearly separated. It is not possible to exhaustively list possible values of

the random variable. If the distance between two cities is 300 km. You cannot estimate orthe random variable. If the distance between two cities is 300 km. You cannot estimate or

identify the next higher distance. There are infinitely very large number of values.identify the next higher distance. There are infinitely very large number of values.

3.3 DISCRETE PROBABILITY DISTRIBUTIONS 3.3 DISCRETE PROBABILITY DISTRIBUTIONS

The values assumed by a discrete random variable depends upon the out come of anThe values assumed by a discrete random variable depends upon the out come of an

experiment. Since the out come of the experiment will be uncertain the value assumed by theexperiment. Since the out come of the experiment will be uncertain the value assumed by the

random variable will also be uncertain. random variable will also be uncertain.

The probability distribution of a discrete random variable is listing of all the outcomes of anThe probability distribution of a discrete random variable is listing of all the outcomes of an

experiment and the probabilities associated with each out come The probability distribution ofexperiment and the probabilities associated with each out come The probability distribution of

a discrete random variable is a table, graph or formula that gives the probability associateda discrete random variable is a table, graph or formula that gives the probability associated

with each possible value that a random variable can assume or if we organize the value of awith each possible value that a random variable can assume or if we organize the value of a

discrete random variable in a probability distribution the distribution is called a Discretediscrete random variable in a probability distribution the distribution is called a Discrete

Probability distribution. In this unit we will discuss three types of discrete probabilityProbability distribution. In this unit we will discuss three types of discrete probability

distribution.distribution.

Binomial, Hyper geometric and PoissonBinomial, Hyper geometric and Poisson

We denote probability distribution of a random variable x as p(x). We can sometimes use theWe denote probability distribution of a random variable x as p(x). We can sometimes use the

sample space of an experiment and probability rules. sample space of an experiment and probability rules.

Example: Consider a test consisting of three true or false questions Example: Consider a test consisting of three true or false questions

The sample space consists The sample space consists

CCC CCC CC| C|C |CC CC| C|C |CC

C| | |C| | |C | | | C| | |C| | |C | | |

We assume; We assume;

-- The student blindly guesses the answer to each question. Then each outThe student blindly guesses the answer to each question. Then each out

come will be equally tickly i.e each having a probability 1/8. come will be equally tickly i.e each having a probability 1/8.

-- Since the student guesses blindly then the probability of answering eachSince the student guesses blindly then the probability of answering each

question correctly is ½ and the probability of answering incorrectly is also ½question correctly is ½ and the probability of answering incorrectly is also ½

35

-- Since each question will be answered independently it follows that we canSince each question will be answered independently it follows that we can

obtain the probability of each sample space out come by multiplying together theobtain the probability of each sample space out come by multiplying together the

probabilities of correctly ( or incorrectly) answering individual questions. probabilities of correctly ( or incorrectly) answering individual questions.

-- There fore, by independence, the probability of the samples space out come. There fore, by independence, the probability of the samples space out come.

CCC, answering all the three questions correctly, CCC, answering all the three questions correctly,

P(CCC) = p(c) p(c) p(c) = (½) (½) (½) = P(CCC) = p(c) p(c) p(c) = (½) (½) (½) = 11//88

Similarly the probability of the sample space outcome CCI is Similarly the probability of the sample space outcome CCI is

P(CCI) = (1/2 ) (1/2 ) (1/2 )= 1/8P(CCI) = (1/2 ) (1/2 ) (1/2 )= 1/8

We define the random variable X to be the number of questions that the student answersWe define the random variable X to be the number of questions that the student answers

correctly. X can assume the values 0 , 1 , 2 , or 3 . Then if x = 1 one question will becorrectly. X can assume the values 0 , 1 , 2 , or 3 . Then if x = 1 one question will be

answeredanswered

Correctly if and only if we obtain one of the sample space outcomes C| | Correctly if and only if we obtain one of the sample space outcomes C| | |C||C| | |C then | |C then

P(X=1) = P(C| |) + P(|C|) + P(| |C) P(X=1) = P(C| |) + P(|C|) + P(| |C)

=1/8 + 1/8 + 1/8 = 3/8 =1/8 + 1/8 + 1/8 = 3/8

Finding the probability distribution Finding the probability distribution

Value of XValue of X

= = The No. of correctThe No. of correct

AnswersAnswers

Sample spaceSample space

out comesout comes

correspondingcorresponding

to Xto X

probability of sampleprobability of sample

space out comespace out come

P(X) P(X) = = probability of theprobability of the

value of Xvalue of X

X=0(no correct answer) X=0(no correct answer) | | || | | ½ x ½ x ½=1/8½ x ½ x ½=1/8 P(0) = 1/8 P(0) = 1/8 X=1(one correct answer) X=1(one correct answer) C| |C| | ½ x ½ x ½=1/8½ x ½ x ½=1/8

|C||C| ½ x ½ x ½=1/8½ x ½ x ½=1/8 P(1)= 1/8 + 1/8 +1/8 =3/8 P(1)= 1/8 + 1/8 +1/8 =3/8 | |C| |C ½ x ½ x ½=1/8½ x ½ x ½=1/8

X=2(two correct answers)X=2(two correct answers) CC1CC1 ½ x ½ x ½=1/8½ x ½ x ½=1/8C1CC1C ½ x ½ x ½=1/8½ x ½ x ½=1/8 P(2)= 1/8 + 1/8 +1/8 =3/8P(2)= 1/8 + 1/8 +1/8 =3/8

36

1CC1CC ½ x ½ x ½=1/8½ x ½ x ½=1/8X=3(three correct answers)X=3(three correct answers) CCCCCC ½ x ½ x ½=1/8½ x ½ x ½=1/8 P(3) = 1/8P(3) = 1/8

Summary: probability distribution of xSummary: probability distribution of x

X, number of question X, number of question P(X) , probability of XP(X) , probability of X

Answered correctly Answered correctly

00 P(0) = P( X=0) 1/8 P(0) = P( X=0) 1/8

11 P(1) = P(X=1) 1/8 P(1) = P(X=1) 1/8

22 P(2) = P(X=2) 1/8P(2) = P(X=2) 1/8

33 P(3)= P(X=3) 1/8 P(3)= P(X=3) 1/8

Sum ISum I

Example: 2Example: 2 Suppose that the student taking the test has studied hard and does not have to Suppose that the student taking the test has studied hard and does not have to

guess at the answer, suppose that there is now a 90% chance that the student will answer eachguess at the answer, suppose that there is now a 90% chance that the student will answer each

of the questions correctly. The probability distribution will be:of the questions correctly. The probability distribution will be:

XX Sample spaceSample space Probability of sampleProbability of sample

spacespace

P(X)P(X)

X=0X=0 111111 0.1 X 0.1 X 0.1 = 0.0010.1 X 0.1 X 0.1 = 0.001 P(0) =0.001P(0) =0.001X=1X=1 C11C11 0.9 X0.1 X 0.1 =0.0090.9 X0.1 X 0.1 =0.009

1C11C1 0.1 X 0.9 X 0.1 = 0.0090.1 X 0.9 X 0.1 = 0.009 P(1)=0.009+0.009+0.009 = 0.027P(1)=0.009+0.009+0.009 = 0.02711C11C 0.1 X 0.1 X0.9 =0.0090.1 X 0.1 X0.9 =0.009

X=2X=2 CC1CC1 0.9 X 0.9 X0.1 =0.0810.9 X 0.9 X0.1 =0.081 P(2)=0.081 +0.081 +0.081= 0.243P(2)=0.081 +0.081 +0.081= 0.243C1CC1C 0.9 x0.1 x0.9=0.0810.9 x0.1 x0.9=0.0811CC1CC 0.1 x0.9 x0.9=0.0810.1 x0.9 x0.9=0.081

X=3X=3 CCCCCC 0.9 x 0.9x 0.9 = 0.7290.9 x 0.9x 0.9 = 0.729 P(3)=0.729P(3)=0.729Similarly the distribution can be summarized Similarly the distribution can be summarized Sum 1Sum 1

X X P(X)P(X)

00 P (0) = P (X=0) 0.001P (0) = P (X=0) 0.001

11 P (1) = P (X=1) 0.027P (1) = P (X=1) 0.027

22 P (2) = P (X=2) 0.243P (2) = P (X=2) 0.243

37

33 P (3) = P (X=3) 0.729P (3) = P (X=3) 0.729

Sum 1 Sum 1

Properties of discrete probability distribution Properties of discrete probability distribution

1.1. P (X) P (X) ≥≥ 0 for each value of X 0 for each value of X

2.2. ∑∑ P (X)=1 P (X)=1

Check Your Progress -1Check Your Progress -1

Suppose a newly married couple plans to have four children. Naturally they are curious aboutSuppose a newly married couple plans to have four children. Naturally they are curious about

the sex of their children and want to estimate the outcome. Defining the event ‘G’ that thethe sex of their children and want to estimate the outcome. Defining the event ‘G’ that the

child will be a girl and ‘B’ that the child is a boy, construct the probability distribution for thechild will be a girl and ‘B’ that the child is a boy, construct the probability distribution for the

number of Boys and Girls.number of Boys and Girls.

3.3.2 The Mean, Variance, and Standard Deviation of a Discrete Probability3.3.2 The Mean, Variance, and Standard Deviation of a Discrete Probability

Distribution Distribution

3.3.2.1 Mean 3.3.2.1 Mean

If the values of the random variable X are observed on the repetition and recorded, we wouldIf the values of the random variable X are observed on the repetition and recorded, we would

obtain the population of all possible observed values of the random variable X. Thisobtain the population of all possible observed values of the random variable X. This

population has a mean or expected value of X. population has a mean or expected value of X.

µµxx denotes the mean of the random variable X. It is also called the expected Value of X as denotes the mean of the random variable X. It is also called the expected Value of X as

denoted by E(x)denoted by E(x)

µµxx = Multiply each value of X by its probability P(X) and then sum the resulting products over = Multiply each value of X by its probability P(X) and then sum the resulting products over

all possible value of X. all possible value of X.

That is That is

µµx x = = ∑xAll

xpx )(

Example.Example. A car dealer has established the following probability distribution for the number A car dealer has established the following probability distribution for the number

of cars he expects to sell on a particular Saturday. of cars he expects to sell on a particular Saturday.

38

Number of cars sold (X)Number of cars sold (X) Probability P(x)Probability P(x)

00 0.100.10

11 0.200.20

22 0.300.30

33 0.300.30

44 0.100.10

Sum .1 Sum .1

On a typical Saturday ,how many cars should the dealer expect to sell? On a typical Saturday ,how many cars should the dealer expect to sell?

µµ = E(x) = = E(x) =∑∑ [xp(x)] [xp(x)]

= 0.(0.1) + 1(0.2) + 2(0.3) +3(0.3) + 4(0.1) = = 0.(0.1) + 1(0.2) + 2(0.3) +3(0.3) + 4(0.1) = 2.12.1 cars. cars.

In the long run the dealer expects to sell 2.1 cars. On a large number of Saturdays.In the long run the dealer expects to sell 2.1 cars. On a large number of Saturdays.

Example 2:Example 2:

Monthly sales of a certain product are believed to follow the following probabilityMonthly sales of a certain product are believed to follow the following probability

distribution. Suppose that the company has fixed monthly production cost $8,000 and thatdistribution. Suppose that the company has fixed monthly production cost $8,000 and that

each item brings $2. Find the expected monthly profit from product saleseach item brings $2. Find the expected monthly profit from product sales

No. of items xNo. of items x p(x)p(x)

50005000 0.20.2

60006000 0.30.3

70007000 0.20.2

80008000 0.20.2

90009000 0.10.1

1 1

E/h(x) = E/h(x) = ∑xall

xpxh )()(

Solution:Solution:

h(x) = 2x – 8000h(x) = 2x – 8000

xx h(x)h(x) p(x)p(x) h(x)p(x)h(x)p(x)

50005000 2000 2000 0.20.2 400 400

39

60006000 4000 4000 0.30.3 1200 1200

70007000 6000 6000 0.20.2 1200 1200

80008000 8000 8000 0.20.2 1600 1600

90009000 10000 10000 0.10.1 1000 1000

1E[h(x)] = 1E[h(x)] = 54005400

The expected value of a linear function of a random variableThe expected value of a linear function of a random variable

E(ax + b) = aE(x) + bE(ax + b) = aE(x) + b

Where a and b are fixed numbers once we know the expected value of x, the expected valueWhere a and b are fixed numbers once we know the expected value of x, the expected value

of ax + b is just aE(x) + b. of ax + b is just aE(x) + b.

In the above example we could have obtained the expected profit by finding the mean of xIn the above example we could have obtained the expected profit by finding the mean of x

first and then multiplying the mean of x by 2 and subtracting from this the fixed cost of 8000.first and then multiplying the mean of x by 2 and subtracting from this the fixed cost of 8000.

The mean The mean χχ is 6, 700 and the expected profit is therefore E[h(x)] = is 6, 700 and the expected profit is therefore E[h(x)] =

E(2x – 8000) = 2E(x) – 8000 = 2(67,000) – 8000 = E(2x – 8000) = 2E(x) – 8000 = 2(67,000) – 8000 = 54005400

3.3.2.2 Variance and Standard Deviation of the Discrete Probability Distribution3.3.2.2 Variance and Standard Deviation of the Discrete Probability Distribution

The mean does not describe the amount of spread or variation of a distribution. The varianceThe mean does not describe the amount of spread or variation of a distribution. The variance

and standard deviation allows us to compare the variation in two distributions having theand standard deviation allows us to compare the variation in two distributions having the

same mean but different spread. same mean but different spread.

The formula for the variance of a discrete probability distribution isThe formula for the variance of a discrete probability distribution is

δδ22 = = ∑∑[(x - [(x - µµ))22 p(x)] p(x)] δδ = = ∑ − )()( 2 xpx µ

oror

E(xE(x22) – [E(x)]) – [E(x)]22 where where

ExEx22 = the expected value of x = the expected value of x22 i.e., i.e., ∑∑xx22 p(x) p(x)

E(x) = the expected value of xE(x) = the expected value of x

Example.Example. For the car dealer find the variance and standard deviation For the car dealer find the variance and standard deviation

XX p(x)p(x) (x - (x - µµ )) (x - (x - µµ )) 22 (x - (x - µµ )) 22 p(x) p(x)

00 0.10.1 0 – 2.100 – 2.10 4.41 4.41 0.441 0.441

11 0.20.2 1 – 2-101 – 2-10 1.21 1.21 0.224 0.224

22 0.30.3 2 – 2.102 – 2.10 0.01 0.01 0.003 0.003

33 0.30.3 3 – 2.103 – 2.10 0.81 0.81 0.243 0.243

40

44 0.10.1 4 – 2.14 – 2.1 3.61 3.61 0.3610.361

1 1 δδ22 = 1.29 = 1.29

δδ22 = 1.29 = 1.29

δδ = = 29.1

= = 1.1361.136 cars cars

Using the other formula we will have the same variance and standard deviationUsing the other formula we will have the same variance and standard deviation

XX p(x)p(x) xx22 x p(x)x p(x) xx22 p(x) p(x)

00 0.100.10 00 0 0 0 0

11 0.200.20 11 0.20.2 0.2 0.2

22 0.300.30 44 0.60.6 1.201.20

33 0.300.30 99 0.90.9 1.701.70

44 0.100.10 16 16 0.40.4 1.601.60

µµ = 2.1 = 2.1 Ex Ex22 = 5.7 = 5.7

δδ22 = E(x = E(x22) – [E(x)]) – [E(x)]22

= 5.7 – (2.1) = 5.7 – (2.1)22

= 5.7 – 4.41 = 5.7 – 4.41

= 1.29 = 1.29

δδ = = 29.1 = = 1.1361.136


Find the variance and standard deviation of the distribution of correct answer answered by theFind the variance and standard deviation of the distribution of correct answer answered by the

student with 0.90 probability of answering each of the three questions correctly. student with 0.90 probability of answering each of the three questions correctly.

3.3.3 The Binomial Distribution3.3.3 The Binomial Distribution

The binomial distribution is a discrete probability distribution The binomial distribution is a discrete probability distribution

The binomial distribution has the following characteristics.The binomial distribution has the following characteristics.

1.1. The experiment consists of N identical trials and the data collected are theThe experiment consists of N identical trials and the data collected are the

results of counts. results of counts.

2.2. An out come of an experiment is classified into one of two mutuallyAn out come of an experiment is classified into one of two mutually

exclucle categories a success or failure. i.e each trial results in a success orexclucle categories a success or failure. i.e each trial results in a success or

failure. failure.

41

3.3. The probability of success remains the same for each trial. So does theThe probability of success remains the same for each trial. So does the

probability of a failure. This implies that the probability of failure of anyprobability of a failure. This implies that the probability of failure of any

trial is 1- (probability of success). Probability of success is denoted by p andtrial is 1- (probability of success). Probability of success is denoted by p and

probability of failure by q of then q = 1 - pprobability of failure by q of then q = 1 - p

4.4. The trials are independent i.e the outcome of one trial does not affect theThe trials are independent i.e the outcome of one trial does not affect the

outcome of any other trial. outcome of any other trial.

Example 1Example 1. Suppose that 40% of all customers who enter a department store make a. Suppose that 40% of all customers who enter a department store make a

purchase. purchase.

What is the probability that 2 of the next 3 customers will make a purchase? What is the probability that 2 of the next 3 customers will make a purchase?

Note that this problem qualifies all the characteristics of the binomial distribution Note that this problem qualifies all the characteristics of the binomial distribution

-- The trials are three and each of the three customers will either purchase or not purchaseThe trials are three and each of the three customers will either purchase or not purchase

so the three trials are identical so the three trials are identical

-- The outcome of each trial will result in either a purchase (success) or not purchaseThe outcome of each trial will result in either a purchase (success) or not purchase

(failure)(failure)

-- The probability of purchase is the same 0.4 for each of the three customers. AndThe probability of purchase is the same 0.4 for each of the three customers. And

probability of failure (not purchase) will be 1 – 0.4 = 0.6 for each.probability of failure (not purchase) will be 1 – 0.4 = 0.6 for each.

-- The decision of one customer will not affect the decision of others. I.e., decision toThe decision of one customer will not affect the decision of others. I.e., decision to

purchase or not to purchase by each customer is independent. purchase or not to purchase by each customer is independent.

The sample space of this trial consist of eight-sample space out comes. The sample space of this trial consist of eight-sample space out comes.

SSS SSS SSF SSF SFSSFS FSS FSS

FFS FFS FSF FSF SFF SFF FFF FFF

S is a success (purchase)S is a success (purchase)

F is a failure (not purchase)F is a failure (not purchase)

Two out of three customers make a purchase if one of the sample space out come SSF, SFS,Two out of three customers make a purchase if one of the sample space out come SSF, SFS,

FSS occurs. By independentFSS occurs. By independent

P(SSF)= P(S) P (S) P(F) = (4) (4) (.6) = (0.4)P(SSF)= P(S) P (S) P(F) = (4) (4) (.6) = (0.4)22 (0.6) (0.6)

P(SFS)= P(S) P(F) P(S) = (0.4) (0.6) (0.4) = (0.4)P(SFS)= P(S) P(F) P(S) = (0.4) (0.6) (0.4) = (0.4)22 (0.6) (0.6)

P(SSF)= P(F) P (S) P(S) = (0.6) (0.4) (0.4) = (0.4)P(SSF)= P(F) P (S) P(S) = (0.6) (0.4) (0.4) = (0.4)22 (0.6) (0.6)

Then the probability that two out of the three customers make a purchase is Then the probability that two out of the three customers make a purchase is

42

P(SSF) + P (SFS) + P (FSS) P(SSF) + P (SFS) + P (FSS)

= (0.4)= (0.4)22 (0.6) + (0.4) (0.6) + (0.4)22(0.6) +(0.4)(0.6) +(0.4)22(0.6) (0.6)

= 3(0.4) = 3(0.4)22(0.6)(0.6)

Note that:Note that:

1.1. The 3 is the number of sample space out come (SSF, SFS and FSS) thatThe 3 is the number of sample space out come (SSF, SFS and FSS) that

correspond to the event i.e., two out of the three customers make a purchase. Thiscorrespond to the event i.e., two out of the three customers make a purchase. This

equals the number of ways we can arrange two successes among three trials. equals the number of ways we can arrange two successes among three trials.

2.2. 0.4 is P, the probability that a customer makes a purchase 0.4 is P, the probability that a customer makes a purchase

3.3. 0.6 is q = 1 – P , the probability that a customer does not make a purchase. 0.6 is q = 1 – P , the probability that a customer does not make a purchase.

Therefore, the probability that two of the next three customers make a purchase isTherefore, the probability that two of the next three customers make a purchase is

= (the number of ways to arrange 2 success among 3 trials) P= (the number of ways to arrange 2 success among 3 trials) P22qq11

Notice that SSF, SFS, FSS each of these sample space out comes consists of two successesNotice that SSF, SFS, FSS each of these sample space out comes consists of two successes

and one failure. The probability of each of these sample space out comes equals (0.4 ) and one failure. The probability of each of these sample space out comes equals (0.4 ) 22(0.6)(0.6)11==

pp22qq1 1

P is raised to a power that equals the number of successes (2) in the three trials and q is raisedP is raised to a power that equals the number of successes (2) in the three trials and q is raised

to a power of failures (1) in the three trials. to a power of failures (1) in the three trials.

In general, each of the sample space out comes describing the occurrence of X successesIn general, each of the sample space out comes describing the occurrence of X successes

(purchase) in n trials represents a different arrangements of X success in n trials. However(purchase) in n trials represents a different arrangements of X success in n trials. However

each of these sample space outcomes consist of X successes and n – X failures. There fore,each of these sample space outcomes consist of X successes and n – X failures. There fore,

the probability of each sample space outcome is the probability of each sample space outcome is

PPxxqqn-xn-x it follows by analogy that the probability that X of the next n trials are successes it follows by analogy that the probability that X of the next n trials are successes

(purchase) is (purchase) is

(The number of ways to arrange X successes among n trials) (P(The number of ways to arrange X successes among n trials) (Pxxq q n-xn-x))

The number of ways to arrange X successes among n trials equals. The number of ways to arrange X successes among n trials equals.

)!(!

!

xnX

n

− n! is read n factorial n! is read n factorial

n! = n(n – 1) (n – 2) … (n – n)!n! = n(n – 1) (n – 2) … (n – n)!

(n – n) = 0; 0! = 1 by definition (n – n) = 0; 0! = 1 by definition

43

Then we call x a binomial random variable and the probability of obtaining X success in nThen we call x a binomial random variable and the probability of obtaining X success in n

trials is trials is

P (X) = P (X) = ( )xnxqpxnX

n −

− )!(!

! ⇒⇒ The Binomial formula The Binomial formula

For the above example we can solve for p(x = 2) as followsFor the above example we can solve for p(x = 2) as follows

n = 3n = 3

p = 0.4p = 0.4

q = 0.6q = 0.6

p(x = 2) = p(x = 2) = )!13(!2

!3

− 0.40.422 0.6 0.611

= = 0.2880.288

Example 2:Example 2: An examination consists of four true or false question and student has no An examination consists of four true or false question and student has no

knowledge of the subject matter. The chance that the student will guess the correct answer toknowledge of the subject matter. The chance that the student will guess the correct answer to

the first question is 0.5. a) What is the probability of getting exactly none out of four correct? the first question is 0.5. a) What is the probability of getting exactly none out of four correct?

N = 4N = 4 p = 0.5p = 0.5 q = 0.5q = 0.5 x = 0x = 0

P(X)= n! P(X)= n! Px qPx qn-x n-x

x! x! ( n –x ) ! ( n –x ) !

4! 4!P(X =0) = 0!(4-0)! 0.5P(X =0) = 0!(4-0)! 0.500 0.5 0.544 = 0.0625 = 0.0625

b) What is the probability of getting exactly one out of four correct b) What is the probability of getting exactly one out of four correct

P(1) = 4! (0.5P(1) = 4! (0.511) (1-0.5)) (1-0.5)4-14-1 = 0.2500 = 0.2500 1!(4-1)! 1!(4-1)!

The probability of getting exactly 0, 1 , 2 , 3 or 4 correct out of a total of four questions isThe probability of getting exactly 0, 1 , 2 , 3 or 4 correct out of a total of four questions is

shown in the table for the Binomial probability distribution. shown in the table for the Binomial probability distribution.

Number of correct guess (x) Number of correct guess (x) Probability (x) Probability (x) 00 1/6 = 0.6251/6 = 0.62511 4/16=0.25004/16=0.250022 6/16=0.37506/16=0.375033 4/16 = 0.25004/16 = 0.250044 1/161/16= = 0.06250.0625

44

Total Total 16/16 = 116/16 = 1


A truck operator has determined that a car repair shop derivers maintained trucks on scheduleA truck operator has determined that a car repair shop derivers maintained trucks on schedule

60% of the time. If the operator has 6 trucks under maintenance (a) construct the probability60% of the time. If the operator has 6 trucks under maintenance (a) construct the probability

distribution for the number of truck to be delivered on time. (b) Find the expected value anddistribution for the number of truck to be delivered on time. (b) Find the expected value and

the standard deviation of the distribution.the standard deviation of the distribution.

Using the Binomial Probability Table: Using the Binomial Probability Table:

A binomial probability distribution is a theoretical distribution, can be generatedA binomial probability distribution is a theoretical distribution, can be generated

mathematically. However except for problems involving small n the calculations for themathematically. However except for problems involving small n the calculations for the

probabilities of 0, 1, 2 success can be rather tedious. As an aid in finding the neededprobabilities of 0, 1, 2 success can be rather tedious. As an aid in finding the needed

probabilities of 0,1,2,3 . . . successes for various values of n and P an extensive table has beenprobabilities of 0,1,2,3 . . . successes for various values of n and P an extensive table has been

developed. developed.

The table has up to n =25 or 30 The table has up to n =25 or 30

P from 0.05 .0.1,0.2. . . 0.90 , 0.99 P from 0.05 .0.1,0.2. . . 0.90 , 0.99

X= from 0-25 or 30X= from 0-25 or 30

Example.Example. 25% of college students in a classroom join the HIV AIDS prevention club. If 20 25% of college students in a classroom join the HIV AIDS prevention club. If 20

students are enrolled in the class, what is the probability that two or fewer will join the club?students are enrolled in the class, what is the probability that two or fewer will join the club?

Solution:Solution:

P = 0.25P = 0.25

n = 20 thenn = 20 then

p(x p(x ≤≤ 2) = p(0) + p(1) + p(2) from the table 2) = p(0) + p(1) + p(2) from the table

p(0) = 0.0032p(0) = 0.0032

p(1) = 0.0211p(1) = 0.0211

p(2) = p(2) = 0.06600.0660

Sum = p(x Sum = p(x ≤≤ 2) = 2) = 0.09090.0909

In similar fashion you can find the probability for any value of x using the table.In similar fashion you can find the probability for any value of x using the table.

45

The mean, variance and standard deviation of a Binomial Random VariableThe mean, variance and standard deviation of a Binomial Random Variable

If X is a binomial Random variable then If X is a binomial Random variable then µµxx = np the mean = np the mean µµ, of the distribution is equal to np, of the distribution is equal to np

The mean is equal to the number of trials n, times the probability of success in a single trial, p.The mean is equal to the number of trials n, times the probability of success in a single trial, p.

Example 1.Example 1. The number of heads appearing in five tosses of a fair coin. The number of heads appearing in five tosses of a fair coin.

E(x) = n p = 5(0.5) =2.5 E(x) = n p = 5(0.5) =2.5

As a long run average, we expect that 2.5 out of 5 tosses of a fair coin will result in heads. As a long run average, we expect that 2.5 out of 5 tosses of a fair coin will result in heads.

The variance of a binomial X is, The variance of a binomial X is, σσ22 and and σσ22 = npq = npq

The standard deviation is The standard deviation is σσ = = 2σ = = npq

Example 2Example 2. 35% of the students registered in the 1. 35% of the students registered in the 1stst semester join the marketing department. semester join the marketing department.

If 1000 students are registered, If 1000 students are registered,

(a) How many of them are expected to join the marketing department (a) How many of them are expected to join the marketing department

µµ = np = np

µµ - 1000 (0.35) = 560 - 1000 (0.35) = 560

(b) What is the standard deviation?(b) What is the standard deviation?

δδ = = npq

= = )65.0)(35.0(1600

= = 364

3.3.4 Hyper Geometric Distribution3.3.4 Hyper Geometric Distribution

The binomial distribution is appropriate when we are sampling from a population that is muchThe binomial distribution is appropriate when we are sampling from a population that is much

larger than the sample. The Binomial assumes sampling with replacement.larger than the sample. The Binomial assumes sampling with replacement.

We sample an item, whether it is a success or failure, returne or put it back to the populationWe sample an item, whether it is a success or failure, returne or put it back to the population

before the next item is selected for the sample, then we are sampling with replacement. before the next item is selected for the sample, then we are sampling with replacement.

46

Sampling with replacement is not a frequently used procedure and most sampling is doneSampling with replacement is not a frequently used procedure and most sampling is done

without replacement. Thus the outcomes are not independent and the probability for eachwithout replacement. Thus the outcomes are not independent and the probability for each

successive observation or trial will change. successive observation or trial will change.

Since the probability of success, does not remain the same from trial to trial the binomialSince the probability of success, does not remain the same from trial to trial the binomial

distribution should not be used. distribution should not be used.

Example.Example. If you draw a card (without replacement) from a standard deck of 52 playing cards If you draw a card (without replacement) from a standard deck of 52 playing cards

what is the probability of getting the first card a king and the second a queen? P(1what is the probability of getting the first card a king and the second a queen? P(1ststk n 2k n 2ndndQ) Q)

= p(k) p(Q/ 1= p(k) p(Q/ 1ststk) = k) =

=

2652

16

51

4

52

4

Note that probability of success for the 1Note that probability of success for the 1stst card was card was 52

4 while for the 2 while for the 2ndnd card card

51

4 i.e., i.e.,

probability of success changes.probability of success changes.

If a sample is selected from a small population with out replacement the hyper geometricIf a sample is selected from a small population with out replacement the hyper geometric

distribution should be applied.distribution should be applied.

Since we sample from a large population the hyper geometric distribution is less use full thanSince we sample from a large population the hyper geometric distribution is less use full than

the binomial. the binomial.

Derivation of the hyper geometric distributionDerivation of the hyper geometric distribution

Consider a collection of N objects which S of these objects have a certain attribute and theConsider a collection of N objects which S of these objects have a certain attribute and the

remaining N – S objects do not have this attribute. If a sample of n objects is chosen atremaining N – S objects do not have this attribute. If a sample of n objects is chosen at

random and with out replacement from this collection of objects, then the number of objectsrandom and with out replacement from this collection of objects, then the number of objects

in the sample having the attribute is a random variable having a hyper geometric distribution. in the sample having the attribute is a random variable having a hyper geometric distribution.

To find the probability distribution for X we follow the following arguments. To find the probability distribution for X we follow the following arguments.

Since the n objects are chosen randomly from the N objects available, there are Since the n objects are chosen randomly from the N objects available, there are

NNCCnn different possible subset of n objects that could be chosen. To find p(x) we need to know different possible subset of n objects that could be chosen. To find p(x) we need to know

the number of these subsets that have X objects having the attribute ( and n – x objects notthe number of these subsets that have X objects having the attribute ( and n – x objects not

having the attribute) . There are having the attribute) . There are SSCCXX ways of choosing X objects from the S having the ways of choosing X objects from the S having the

attribute in the population. attribute in the population.

47

N-SN-S C C n-xn-x ways of choosing n – x objects from the N-S not having the attribute. The quantities n, ways of choosing n – x objects from the N-S not having the attribute. The quantities n,

N and S and parameters of this distribution as indicated by the following notationN and S and parameters of this distribution as indicated by the following notation

P ( X) = P ( X) = SS CC X ( N-s C n – X ) X ( N-s C n – X ) N NCCn n

Where: Where:

N: the size of the population N: the size of the population

S- the number of success (objects with certain attributes) is the populationS- the number of success (objects with certain attributes) is the population

X- the number of success (of interest) objects in the sample having the attribute n is theX- the number of success (of interest) objects in the sample having the attribute n is the

size of the sample (objects chosen randomly from the population) size of the sample (objects chosen randomly from the population)

Example 1Example 1. An inspector is to examine a population of 20 shipping orders to check for. An inspector is to examine a population of 20 shipping orders to check for

authorized credit approval. If 15 of these have authorized credit approval and if a sample of 4authorized credit approval. If 15 of these have authorized credit approval and if a sample of 4

orders is to be randomly chosen, what is the probability that exactly 3 will have authorizedorders is to be randomly chosen, what is the probability that exactly 3 will have authorized

credit approval? credit approval?

Since the orders are chosen, at random, we know that all subsets of 4 orders from the 20 areSince the orders are chosen, at random, we know that all subsets of 4 orders from the 20 are

equally likely to be chosen. By using the equally likely outcomes approach, we see that thereequally likely to be chosen. By using the equally likely outcomes approach, we see that there

are are

20 C 4 = 20 C 4 = 20! 20! = = 20! 20! = 4845 = 4845 4!(20-4)! 4! (16!) 4!(20-4)! 4! (16!)

Ways that a sample of four can be chosen out of 20.Ways that a sample of four can be chosen out of 20.

There are 15CThere are 15C3 3 = 455 ways that three credit approved orders can be selected from 15 credit = 455 ways that three credit approved orders can be selected from 15 credit

approved orders and approved orders and

SSCC11 = 5 ways that one non –approved order can be selected from five non-approved order = 5 ways that one non –approved order can be selected from five non-approved order

consequently. consequently.

P(x=3) = P(x=3) = (15C3) ( 5C(15C3) ( 5C11)) = = 455 (5) 455 (5) = = 0.46960.4696 20c4 4845 20c4 4845

Example 2.Example 2. Suppose that automobiles arrive at a dealer's shop in lots of 10 and that for time Suppose that automobiles arrive at a dealer's shop in lots of 10 and that for time

and resource considerations only 5 out of each 10 are inspected for safety. The 5 cars areand resource considerations only 5 out of each 10 are inspected for safety. The 5 cars are

randomly chosen from the 10 in the lot. randomly chosen from the 10 in the lot.

If 2 out of the 10 cars in the lot are bellow standards for safety, what is the probability that atIf 2 out of the 10 cars in the lot are bellow standards for safety, what is the probability that at

least 1 out of the 5 cars to be inspected will be found not meeting the safety standard? least 1 out of the 5 cars to be inspected will be found not meeting the safety standard?

48

N = 10N = 10

S = 2S = 2

N = 5N = 5

X = at least one i.e., one or two X = at least one i.e., one or two

p(at least one) = p(1) + p(2)p(at least one) = p(1) + p(2)

= 0.556 + 0.222 = = 0.556 + 0.222 = 0.7780.778


Suppose 50 TV sets were manufactured during the week. 40 operated perfectly and 10 had at Suppose 50 TV sets were manufactured during the week. 40 operated perfectly and 10 had at

least one defect. least one defect.

A sample of 5 is selected at random. What is the probability that 4 of the 5 will operateA sample of 5 is selected at random. What is the probability that 4 of the 5 will operate

perfectly?perfectly?

Mean and Variance of the Hyper Geometric DistributionMean and Variance of the Hyper Geometric Distribution

If X is a random variable having a hyper geometric distribution with parameters n, N and SIf X is a random variable having a hyper geometric distribution with parameters n, N and S

then. E(x) = nthen. E(x) = n

N

Sand and

−−

−

=

112

N

nN

N

S

N

Snxδ

Example.Example. If 180 out of 200 shipping orders that the inspector will examine have authorized If 180 out of 200 shipping orders that the inspector will examine have authorized

credit approval what are the mean and variance of the number in a sample of 40 randomlycredit approval what are the mean and variance of the number in a sample of 40 randomly

chosen orders that will have credit approvals? chosen orders that will have credit approvals?

E(x) = 40 (180/200) =36 E(x) = 40 (180/200) =36

δδ22x = 4(180/200) (20/200) (160/199)=2.8945x = 4(180/200) (20/200) (160/199)=2.8945

3.3.5 The Poison Probability Distribution3.3.5 The Poison Probability Distribution

The third important discrete probability distribution is the Poisson. The Poisson distributionThe third important discrete probability distribution is the Poisson. The Poisson distribution

counts the number of successes in a fixed interval of time or with in specified regions. counts the number of successes in a fixed interval of time or with in specified regions.

Eg. The number of machine failure in a weekEg. The number of machine failure in a week

-- the number of traffic accidents per month in townthe number of traffic accidents per month in town

49

P(x=1) = P(x=1) = 2C2C11(10(10-2-2CC5-15-1)) = 0.556 = 0.556 10C 10C55

p(x = 2) = p(x = 2) = 2c2c22((10-210-2CC5.25.2)) = 0.222 = 0.222

10C 10C55

-- the number of emergency patients arriving at a hospital in an hourthe number of emergency patients arriving at a hospital in an hour

-- the number of orders received per day the number of orders received per day

-- the number of defects in a square metere metal sheet. the number of defects in a square metere metal sheet.

To apply the Poisson distribution the following condition are required To apply the Poisson distribution the following condition are required

1.1. The probability of success in a short interval of time (or space) isThe probability of success in a short interval of time (or space) is

proportional to the size of the interval. If we count 6 patients arriving in an hourproportional to the size of the interval. If we count 6 patients arriving in an hour

then we expect 3 in half an hour and 2 in 20 minutes. then we expect 3 in half an hour and 2 in 20 minutes.

2.2. In a very small interval, the probability of successes is close to zero. If 6In a very small interval, the probability of successes is close to zero. If 6

patients arrive in an hour we expect none in 10 seconds. patients arrive in an hour we expect none in 10 seconds.

3.3. The probability of success in a given interval is independent of where theThe probability of success in a given interval is independent of where the

interval begins. interval begins.

4.4. The probability of success over a given interval is independent of the number ofThe probability of success over a given interval is independent of the number of

the events that occurred prior to the interval. the events that occurred prior to the interval.

The Poisson distribution is described mathematically by the formula. The Poisson distribution is described mathematically by the formula.

P ( x) = P ( x) = µµ xx ee -- µµ X! X!

Where; Where;

µµ is the mean number of success /average rate/ is the mean number of success /average rate/

e is the base of natural logarithm or mathematical constant with value 2.7183 e is the base of natural logarithm or mathematical constant with value 2.7183

X is the number of success in the interval X is the number of success in the interval

P (X) is the probability of X successes in an intervalP (X) is the probability of X successes in an interval

The Poisson distribution can be used to approximate the binomial distribution when theThe Poisson distribution can be used to approximate the binomial distribution when the

probability of a success is small and the number of trial is very large. probability of a success is small and the number of trial is very large.

Usually the probability of success become quite small after few occurrences as the randomUsually the probability of success become quite small after few occurrences as the random

variable X for a Poisson distribution assume an infinite number of values. variable X for a Poisson distribution assume an infinite number of values.

Example1.Example1. Assume that billing clerks rarely make errors in data entry on the billing Assume that billing clerks rarely make errors in data entry on the billing

statements of a co. Many statements have no mistakes; some have one, a very few have towstatements of a co. Many statements have no mistakes; some have one, a very few have tow

mistakes; rarely will a statement have three mistakes; and soon. A random sample of 1000mistakes; rarely will a statement have three mistakes; and soon. A random sample of 1000

50

statements revealed 300 errors. What is the probability of no mistakes appearing in astatements revealed 300 errors. What is the probability of no mistakes appearing in a

statement statement µµ = 300/1000=0.3 = 300/1000=0.3

P(0) = P(0) = 0.30.3 00 (2.7183)(2.7183) -0.3-0.3 = 0.7408 = 0.7408 0! 0!

Example 2.Example 2. A bank manger wants to provided prompt service for customers at the banks A bank manger wants to provided prompt service for customers at the banks

drive up window. The bank currently can serve up to10 customers per 15-minute period withdrive up window. The bank currently can serve up to10 customers per 15-minute period with

out significant delay. The average arrival rate is 7 customers per 15minute period. A assumingout significant delay. The average arrival rate is 7 customers per 15minute period. A assuming

X has a Poisson distribution find the probability that 10 customers. Will arrive in a particularX has a Poisson distribution find the probability that 10 customers. Will arrive in a particular

15-minute period. 15-minute period.

µµ = 7 = 7

X= 10 X= 10

P(10) = P(10) = 77 1010 2.7183 2.7183 -7-7 = 0.710 = 0.710 10! 10!

Check Your Progress –5 Check Your Progress –5

A telephone company’s goal is not to have more than there line failures in a particular 1kmA telephone company’s goal is not to have more than there line failures in a particular 1km

line. Currently the company is experiencing four line failures in 1km line.line. Currently the company is experiencing four line failures in 1km line.

a)a) what is the probability that the company will meet its goal?what is the probability that the company will meet its goal?

b)b) what is the probability that the company will not meet its goal?what is the probability that the company will not meet its goal?

Variance and standard deviation of the passion probability distributionVariance and standard deviation of the passion probability distribution

The variance of the poison distribution is equal to the mean of the distribution.The variance of the poison distribution is equal to the mean of the distribution.

δδ22 = = µµ then then

δδ = = µ

3.4 THE NORMAL / CONTINIOUS / PROBABILITY DISTRIBUTION 3.4 THE NORMAL / CONTINIOUS / PROBABILITY DISTRIBUTION

As noted earlier in this unit a continuous random variable is one that can assume an infiniteAs noted earlier in this unit a continuous random variable is one that can assume an infinite

number of possible values with in a specified range. It usually results from measuring somenumber of possible values with in a specified range. It usually results from measuring some

thing. thing.

51

It is not possible to list every possible value of the continuous random variable along with aIt is not possible to list every possible value of the continuous random variable along with a

corresponding probability. corresponding probability.

The most convenient approach is to construct a probability curve. The proportion of areaThe most convenient approach is to construct a probability curve. The proportion of area

included between any two point under the probability curve identified the probability that aincluded between any two point under the probability curve identified the probability that a

randomly selected continuous variable has a value between those points. randomly selected continuous variable has a value between those points.

Characteristics of a normal probability distribution and its accompanying normal curveCharacteristics of a normal probability distribution and its accompanying normal curve

1.1. The normal curve is bell – shaped and has a single peak at the exact centerThe normal curve is bell – shaped and has a single peak at the exact center

of the distribution. The arithmetic mean median and the mode are equal and areof the distribution. The arithmetic mean median and the mode are equal and are

located at peak. Thus half the area under the curve is above this center point, and thelocated at peak. Thus half the area under the curve is above this center point, and the

other half is below it. other half is below it.

2.2. The normal probability distribution is symmetrical about its mean. If we cutThe normal probability distribution is symmetrical about its mean. If we cut

the normal curve vertically at this central value, the two halves will be mirror images. the normal curve vertically at this central value, the two halves will be mirror images.

3.3. The normal curve falls of smoothly in either direction from the central value.The normal curve falls of smoothly in either direction from the central value.

It is asymptotic, meaning that the curve gets closer and closer to the X – axis but neverIt is asymptotic, meaning that the curve gets closer and closer to the X – axis but never

actually touches it. In real world problems, however, this is somewhat unrealistic. The actually touches it. In real world problems, however, this is somewhat unrealistic. The

f(x)

XX

The Normal Curve The Normal Curve

The normal probability distribution is important in statistical inference for three distinctThe normal probability distribution is important in statistical inference for three distinct

reasons:reasons:

1.1. The measurements produced in many random processes are known to follow thisThe measurements produced in many random processes are known to follow this

distribution. distribution.

2.2. Normal probability can often be used to approximate other probability distribution,Normal probability can often be used to approximate other probability distribution,

such as the binomial and Poisson distributions.such as the binomial and Poisson distributions.

52

3.3. Distribution of such statistics as the sample mean and sample proportion often followDistribution of such statistics as the sample mean and sample proportion often follow

the normal distribution regardless of the distribution of the population. the normal distribution regardless of the distribution of the population.

Constructing the Probability CurveConstructing the Probability Curve

There is not just one normal probability distribution. There is a family of them we night haveThere is not just one normal probability distribution. There is a family of them we night have

one of the following: one of the following:

a.a. Equal means and different standard deviations eg. AverageEqual means and different standard deviations eg. Average

age of students in three sections S1, S2, S3 is equal 24 years. But the standardage of students in three sections S1, S2, S3 is equal 24 years. But the standard

deviation for S1 =2.5, S2 = 3.1 and S3 = 4.deviation for S1 =2.5, S2 = 3.1 and S3 = 4.

1.3=σ

4=σ

5.2=σ

The shape of the curves is determined by the standard deviation. The smaller theThe shape of the curves is determined by the standard deviation. The smaller the

standard deviation the more packed the curve will be and the larger the standardstandard deviation the more packed the curve will be and the larger the standard

deviation the more flat and wider the curve will bedeviation the more flat and wider the curve will be

b. different means but equal standard deviation. Both sections have equal standard b. different means but equal standard deviation. Both sections have equal standard

deviation 3.1 but different means S1=23 S2=26 S3=28deviation 3.1 but different means S1=23 S2=26 S3=28

1.3=σ 1.3=σ 1.3=σ

23=µ 26=µ 28=µ

c. Different means and different standard deviationsc. Different means and different standard deviations

53

years24=µ

For S1 For S1 µµ = 22 and = 22 and δδ =2.8 =2.8

S2 S2 µµ =24 and =24 and δδ =2.1 =2.1

S3 S3 µµ =27 and =27 and δδ =3.1 =3.1

1.2=σ

8.2=σ

1.3=σ

years22=µ years24=µ years27=µ

The number of normal distributions is unlimited. It would be practically impossible to provideThe number of normal distributions is unlimited. It would be practically impossible to provide

a table of probabilities (as the binomial and Poisson) for each combination of a table of probabilities (as the binomial and Poisson) for each combination of µµ and and δδ or using or using

a formula. a formula.

One member of the families of normal distributions can be used for all problems where theOne member of the families of normal distributions can be used for all problems where the

normal distribution is applicable. normal distribution is applicable.

It has a mean of 0 and a standard deviation of 1 and is called Standard Normal Distribution. It has a mean of 0 and a standard deviation of 1 and is called Standard Normal Distribution.

First it is necessary to convert or standardize the actual distribution to a standard normalFirst it is necessary to convert or standardize the actual distribution to a standard normal

distribution using Z value. Z is called the normal deviate. distribution using Z value. Z is called the normal deviate.

Z value is the distance between a selected value and the population mean in units of theZ value is the distance between a selected value and the population mean in units of the

standard deciation. standard deciation.

Transformation of the Normal Random Variable Transformation of the Normal Random Variable

Since there are infitely many possible normal random variables one of them is selected toSince there are infitely many possible normal random variables one of them is selected to

serve as our standard. serve as our standard.

We want to transform X in to the standard normal random variable Z. We want to transform X in to the standard normal random variable Z.

Example.Example. We have a normal random variable X with We have a normal random variable X with µµ =50 and =50 and σσ =10 we want to convert =10 we want to convert

this random variable with this random variable with µµ =0 and =0 and σσ =1. =1.

54

We move the distribution from its center of 50 to a center of 0. this is done by subtracting 50We move the distribution from its center of 50 to a center of 0. this is done by subtracting 50

from all the values of X. Thus we shift the distribution 50 units back so that its new center isfrom all the values of X. Thus we shift the distribution 50 units back so that its new center is

0. If we subtract the mean from all values of X, the new distribution (X-0. If we subtract the mean from all values of X, the new distribution (X-µµ) will have a mean) will have a mean

of zero.of zero.

The second thing we need to do is to make the width of the distribution, standard deviationThe second thing we need to do is to make the width of the distribution, standard deviation

equal to 1. This is done by squeezing the width of the distribution down from 10 to 1. Becauseequal to 1. This is done by squeezing the width of the distribution down from 10 to 1. Because

the total probability under the curve must remain 1. the distribution must grow up ward tothe total probability under the curve must remain 1. the distribution must grow up ward to

maintain the same area. maintain the same area.

Mathematically, squeezing the curve to make the width 1 is equivalent to dividing the randomMathematically, squeezing the curve to make the width 1 is equivalent to dividing the random

variable by its standard deviation. The area under the curve adjusted so that the total remainsvariable by its standard deviation. The area under the curve adjusted so that the total remains

the same. the same.

The mathematical transformation from X to Z is thus achieved by first subtracting The mathematical transformation from X to Z is thus achieved by first subtracting µµ from X from X

and then dividing the result by and then dividing the result by σσ. .

Z = Z = X – X – µµ σσ

1=µ

10=σ

50=µ 0=µ

ExampleExample – The weekly incomes of a large group of middle managers are normally distributed – The weekly incomes of a large group of middle managers are normally distributed

with a mean of 1000 Br. and standard deviation of Br. 100. What is the Z value for an incomewith a mean of 1000 Br. and standard deviation of Br. 100. What is the Z value for an income

ofof

a) Br. 1100?a) Br. 1100? Z = Z = X - X - µµ µµ = 1000 = 1000 σσ σσ = 100 = 100

Z = Z = 1100 – 10001100 – 1000 = 1 = 1100100

55

This means an income of 1100 is one standard deviation above the mean.This means an income of 1100 is one standard deviation above the mean.

b) Br 900?b) Br 900?

Z = Z = 900 – 1000 900 – 1000 = -1 = -1 100 100

This implies that an income of Br. 900 is one standard deviation (Br. 100) below the mean.This implies that an income of Br. 900 is one standard deviation (Br. 100) below the mean.

c) Br. 1250?c) Br. 1250?

Z = Z = 1250 – 10001250 – 1000 = 2.5 = 2.5100100

This implies that an income of Br. 1250 is 2.5 standard deviations above the mean This implies that an income of Br. 1250 is 2.5 standard deviations above the mean

d) Br. 850?d) Br. 850?Z = Z = 850 – 1000850 – 1000 = -1.5 = -1.5 100 100

This means an income of Br. 850 is 1.5 standard deviations below the meanThis means an income of Br. 850 is 1.5 standard deviations below the mean

Finding probabilities using the normal probability tableFinding probabilities using the normal probability table

For any value of Z calculated the corresponding probability can be easily found from the ZFor any value of Z calculated the corresponding probability can be easily found from the Z

table.table.

Example 1:Example 1: The lifetime of an electrical component is known to follow normal distribution The lifetime of an electrical component is known to follow normal distribution

with mean 2000 hr and standard deviation 200 hr with mean 2000 hr and standard deviation 200 hr

(a) What is the probability that a randomly selected component will last between 2000 and(a) What is the probability that a randomly selected component will last between 2000 and

2400 hr?2400 hr?

XX hrs hrs

1400 1600 1800 1400 1600 1800 2000=µ 2200 2400 2600 2200 2400 2600

-3 -2 -1 -3 -2 -1 0=µ +1 +2 +3 +1 +2 +3

The lower boundary of the interval is at the mean of the distribution and therefore at Z = 0.The lower boundary of the interval is at the mean of the distribution and therefore at Z = 0.

The upper boundary of the interval in terms of Z isThe upper boundary of the interval in terms of Z is

56

Z (Standard Normal Unit)

Z = Z = 2200

20002400 =−=−σ

µχ

By reference to the probability tableBy reference to the probability table

p(0 p(0 ≤≤ Z Z ≤≤ + 2) = 0.4772 + 2) = 0.4772

p(2000 p(2000 ≤≤ x x ≤≤ 2400) = 0.4772 2400) = 0.4772

This means a randomly selected component will have a probability of 0.4772 to last betweenThis means a randomly selected component will have a probability of 0.4772 to last between

2000 to 2400 hr. Or we can say 47.72% of all components will last between 2000 to 2400 hr. 2000 to 2400 hr. Or we can say 47.72% of all components will last between 2000 to 2400 hr.

(b) What is the probability that a randomly selected component will last more than 2200 hrs?(b) What is the probability that a randomly selected component will last more than 2200 hrs?

Note that the total area to the right of the mean 2000 is 0.5. Therefore if we determine theNote that the total area to the right of the mean 2000 is 0.5. Therefore if we determine the

proportion between the mean and 2200, we can subtract this value from 0.50 to obtain theproportion between the mean and 2200, we can subtract this value from 0.50 to obtain the

probability of the hrs x being greater than 2200.probability of the hrs x being greater than 2200.

Z = Z = 2200 – 20002200 – 2000 = 1 = 1 200 200

p(0 p(0 ≤≤ Z Z ≤≤ +1.0) = 0.3413 +1.0) = 0.3413

p(Z > +1) = 0.5000 – 0.3413p(Z > +1) = 0.5000 – 0.3413

= 0.1587 = 0.1587

This means 15.87% of the components will last more than 2200 hrs.This means 15.87% of the components will last more than 2200 hrs.

P= 0.90 P= 0.90

45 X 45 X X minX min

Example 2:Example 2: The amount of time required for a certain type of car repair at a service guarage The amount of time required for a certain type of car repair at a service guarage

is normally distributed with the is normally distributed with the µµ = 45 min. And the standard deviation = 45 min. And the standard deviation σσ = 8 min. The = 8 min. The

service manage plans to have work begin on a customers car 10 min after the car is droppedservice manage plans to have work begin on a customers car 10 min after the car is dropped

off and he tells the customer that the car will be ready with in 1 hrs total time. off and he tells the customer that the car will be ready with in 1 hrs total time.

a) What is the probability that he will be wrong? a) What is the probability that he will be wrong?

57

P(error) = p ( x > 50 min) , since actually work is to begin in 10 min, the actual repair must P(error) = p ( x > 50 min) , since actually work is to begin in 10 min, the actual repair must

be completed in the remaining 50 min. And the manager will be wrong if the repair takesbe completed in the remaining 50 min. And the manager will be wrong if the repair takes

more than 50 minutes.more than 50 minutes.

Z = Z = X – X – µµ = = 50-4050-40= + 0.62 p(Z = 0.62) = 0.2324 then,= + 0.62 p(Z = 0.62) = 0.2324 then, σσ 8 8

P( x > 50) = P (Z > + 0.62 )= 0.5000 –0.2324 = 0.2676P( x > 50) = P (Z > + 0.62 )= 0.5000 –0.2324 = 0.2676

b) What is the required working time allotment such that there is a 90%chance that the repainb) What is the required working time allotment such that there is a 90%chance that the repain

will be completed with in that time?will be completed with in that time?

If the proportion of the area is 0.90, then because a proportion of 0.5 is to the left of theIf the proportion of the area is 0.90, then because a proportion of 0.5 is to the left of the

mean, it follows that a proportion of 0.4 is between the mean and the unknown value of X. mean, it follows that a proportion of 0.4 is between the mean and the unknown value of X.

By looking the table the closest we can come to a proportion of 0.40 is 0.3997 and the Z valueBy looking the table the closest we can come to a proportion of 0.40 is 0.3997 and the Z value

associated with this proportion is Z = + 1.28 associated with this proportion is Z = + 1.28

Now convert Z value to a value of X Now convert Z value to a value of X

Z = Z = X – X – µµ , , Z (Z (σσ) = x - ) = x - µµ, x = , x = µµ + Z + Zσσ σσ

X = 45 + (+1.28) (8.00)= 45 +10.24=X = 45 + (+1.28) (8.00)= 45 +10.24=55.2455.24 min min

2000 2200 2000 2200 XX

0 1 0 1 ZZ

This means if the service manager allots 55.24 minutes for the repair he will have a 90%This means if the service manager allots 55.24 minutes for the repair he will have a 90%

chance to complete the repair with in 55.24 minutes.chance to complete the repair with in 55.24 minutes.

C) What is the working time allotment such that there is a probability of just 30% that theC) What is the working time allotment such that there is a probability of just 30% that the

repair can be completed with in that time? repair can be completed with in that time?

58

Since a proportion of area of 0.3 is to the left of the unknown value of X it follows that aSince a proportion of area of 0.3 is to the left of the unknown value of X it follows that a

portion of 0.20 is between the unknown value and the mean. By reference to the table theportion of 0.20 is between the unknown value and the mean. By reference to the table the

proportion of area closest to this is 0.1985 and the Z value corresponding to this probability isproportion of area closest to this is 0.1985 and the Z value corresponding to this probability is

0.52. The Z value is negative because the unknown value is to the left of the mean. 0.52. The Z value is negative because the unknown value is to the left of the mean.

X = X = µµ + 2 + 2σσ

X = 45 + (-0.52)(8) = 40.84 min. The service manager will have a 30% chance to completeX = 45 + (-0.52)(8) = 40.84 min. The service manager will have a 30% chance to complete

the repair with in 40.84 min. the repair with in 40.84 min.

Example 3.Example 3. Returning again to the weekly incomes illustration, Returning again to the weekly incomes illustration, µµ = 1000 and = 1000 and σσ =100 =100

(a)(a) What percent of the executive earn weekly incomes of 1245 or more?What percent of the executive earn weekly incomes of 1245 or more?

X X ≥≥ 1245 1245

Z= Z= 1245 – 10001245 – 1000 = 2.45 = 2.45 100 100

The area associated with Z = 2.4 is 0.4929. This is the probability between 1000 and 1245.The area associated with Z = 2.4 is 0.4929. This is the probability between 1000 and 1245.

The probability for 1245 and beyond is found by subtracting 0.4929 from 0.5. This is equal toThe probability for 1245 and beyond is found by subtracting 0.4929 from 0.5. This is equal to

= 0.0075. That only 0.71% of the executives earn weekly incomes of 1245 or more. = 0.0075. That only 0.71% of the executives earn weekly incomes of 1245 or more.

(b)(b) What is the probability of selecting an income between 840 and 1200What is the probability of selecting an income between 840 and 1200

This problem is divided in to two partsThis problem is divided in to two parts

1) for the probability between 840 and the mean1) for the probability between 840 and the mean

Z = Z = 840 – 1000840 – 1000 = -1.60 = -1.60 100 100

2) for the probability between the mean 1000 and 1200 2) for the probability between the mean 1000 and 1200

Z = Z = 1200 – 10001200 – 1000 = 2 = 2 100 100

The probability of Z = -1.60 is 0.4452 The probability of Z = -1.60 is 0.4452

The probability of Z = 2 is 0.4772 The probability of Z = 2 is 0.4772

0.4452 + 0.4772 = 0.9224 or 92.24% i.e.,0.4452 + 0.4772 = 0.9224 or 92.24% i.e.,

92.24% of the managers have weekly incomes between 840 and 1200. 92.24% of the managers have weekly incomes between 840 and 1200.

59

0.4452 0.4772 0.4452 0.4772

840 1000 1200 840 1000 1200 X BirrX Birr

0.9224 0.9224

c)c) What is the probability that a randomly selected middle manager will have an incomeWhat is the probability that a randomly selected middle manager will have an income

between 1150 and 1250between 1150 and 1250

This problem is separated in two parts. First find the Z value associated with 1250 This problem is separated in two parts. First find the Z value associated with 1250

Z = Z = 1250 –100 1250 –100 = 2.5 = 2.5 100 100

Next find the Z value for 1150Next find the Z value for 1150

Z = Z = 1150 – 10001150 – 1000 = 1.5 = 1.5100100

p(Z = 2.5) = 0.4935p(Z = 2.5) = 0.4935

Similarly p(Z = 1.5) = 0.4332Similarly p(Z = 1.5) = 0.4332

So the probability between 1150 and 1250 equals So the probability between 1150 and 1250 equals

0.4938 – 0.4332 = 0.06060.4938 – 0.4332 = 0.0606

0.06060.0606

1000 1500 1250 1000 1500 1250


60

Service life of truck tires for heavy-duty trucks follows the normal distribution with meanService life of truck tires for heavy-duty trucks follows the normal distribution with mean

50000 km and standard deviation 5000 km.50000 km and standard deviation 5000 km.

a)a) What is the probability that a tyre will last between 47,000 km and 60000 km?What is the probability that a tyre will last between 47,000 km and 60000 km?

b)b) What percentage of the tyres will last below 48500 km?What percentage of the tyres will last below 48500 km?

c)c) If the supplier of the tyres is planning to replace only 1% of those tyres with theIf the supplier of the tyres is planning to replace only 1% of those tyres with the

minimum performance what should be the service life for warranty?minimum performance what should be the service life for warranty?

Computing unknown Mean and unknown Standard deviationComputing unknown Mean and unknown Standard deviation

Some times the mean and the standard deviation of normal probability distribution may not beSome times the mean and the standard deviation of normal probability distribution may not be

given or known. In such situations the probability of two unknown variables (xgiven or known. In such situations the probability of two unknown variables (x 11 and x and x22) is) is

used to compute the mean and standard deviation. used to compute the mean and standard deviation.

Example 1:Example 1: The construction time for a certain building is normally distributed with an The construction time for a certain building is normally distributed with an

unknown mean and unknown variance. We do know, however, that 75% of the timeunknown mean and unknown variance. We do know, however, that 75% of the time

construction takes less than 12 months and 45% of the time construction takes less than 12construction takes less than 12 months and 45% of the time construction takes less than 12

months and 45% of the time construction takes less than deviation of the construction time.months and 45% of the time construction takes less than deviation of the construction time.

We have p(x < 12) = 0.75 andWe have p(x < 12) = 0.75 and

p(x < 10) = 0.45, this follows that p(x < 10) = 0.45, this follows that

pp

−<

δµ12

1Z = 0.75 and= 0.75 and

pp

−<

δµ10

2Z = 0.45= 0.45

0.750.75

0.450.45

61

10 10 µ 1212 XX

Z Z1 1 Z Z22 ZZ

From the table we find that ZFrom the table we find that Z11 = -0.12 and Z = -0.12 and Z22 = 0.67 substituting these two values for = 0.67 substituting these two values for µµ and and σσ

we get: we get: σµ−10

= -0.12 and= -0.12 and

σµ−10

= 0.67 = 0.67

by cross multiplication,by cross multiplication,

-0.12-0.12σσ = 10 - = 10 - µµ

0.67 0.67σσ = 12 - = 12 - µµ

µµ = 10 + 0.1 = 10 + 0.1σσ

µµ = 12 – 0.67 = 12 – 0.67σσ

We have two equation with two unknown and it follows thatWe have two equation with two unknown and it follows that

10 + 0.1210 + 0.12σσ = 12 – 0.67 = 12 – 0.67σσ

0.790.79σσ = 2 = 2

σσ = 279 = 2.53 = 279 = 2.53

µµ = 10 + 0.12 (2.53) = 10 + 0.12 (2.53)

= = 10.3010.30


A machine is to be designed so that only 2.5% of the length of bolts made are more than 0.01A machine is to be designed so that only 2.5% of the length of bolts made are more than 0.01

mm above the mean and only 2.5% are more than 0.01 below the mean. What standardmm above the mean and only 2.5% are more than 0.01 below the mean. What standard

deviation must the machine have to meet these objectives? deviation must the machine have to meet these objectives?

Normal ApproximationNormal Approximation

One of the reasons why we apply the normal probability distribution is that it is more efficientOne of the reasons why we apply the normal probability distribution is that it is more efficient

than the binomial or poisson when these distributions involve larger n or than the binomial or poisson when these distributions involve larger n or µµ values values

respectively.respectively.

62

3.4.1 The Normal approximation to the Binomial 3.4.1 The Normal approximation to the Binomial

The table of the binomial probabilities goes successively from an n of 1 to n of 25 or 30.The table of the binomial probabilities goes successively from an n of 1 to n of 25 or 30.

Suppose a problem involved taking a sample of 60. Generating a binomial distribution for thatSuppose a problem involved taking a sample of 60. Generating a binomial distribution for that

large a number using the formula would be very time consuming. A more efficient approachlarge a number using the formula would be very time consuming. A more efficient approach

is to apply the normal approximation. This seems reasonable because as n increases, ais to apply the normal approximation. This seems reasonable because as n increases, a

binomial distribution gets closer and closer to a normal distribution. binomial distribution gets closer and closer to a normal distribution.

The normal probability distribution is generally deemed a good approximation to the binomialThe normal probability distribution is generally deemed a good approximation to the binomial

probability distribution when np and nq are both greater than 5.probability distribution when np and nq are both greater than 5.

Since there is no area under the normal curve at a single point, we assign interval on the realSince there is no area under the normal curve at a single point, we assign interval on the real

line to the discrete value of X by making what we call a line to the discrete value of X by making what we call a continuity correction factor.continuity correction factor.

Continuity correction factor is subtracting or adding, depending on the problem, the value 0.5Continuity correction factor is subtracting or adding, depending on the problem, the value 0.5

to a selected value when a binomial probability distribution is being approximated by ato a selected value when a binomial probability distribution is being approximated by a

normal distribution. We add 0.5 to x when x normal distribution. We add 0.5 to x when x ≤≤ and x > a certain value we subtract 0.5 from x and x > a certain value we subtract 0.5 from x

when x < and when x < and ≥≥ a certain value. a certain value.

Example1:Example1: supposes that the management of a restaurant found that 70% of their new supposes that the management of a restaurant found that 70% of their new

customers return for another meal. For a week in which 80 new (first time) customers dined atcustomers return for another meal. For a week in which 80 new (first time) customers dined at

the restaurant, what is the probability that 60 or more will return for another meal?the restaurant, what is the probability that 60 or more will return for another meal?

Notice that the binomial conditions are met.Notice that the binomial conditions are met.

To calculate this probability using the binomial formula means computing the probabilities ofTo calculate this probability using the binomial formula means computing the probabilities of

60 , 61 , 62 ….. 80 and adding them to arrive at probability of 60 or more. This is quick ward60 , 61 , 62 ….. 80 and adding them to arrive at probability of 60 or more. This is quick ward

the practically impossible. So the most appropriate solution is the normal approximation. the practically impossible. So the most appropriate solution is the normal approximation.

Step 1: compute the arithmetic mean and the standard deviation of the binomial distribution Step 1: compute the arithmetic mean and the standard deviation of the binomial distribution

µµ = np = 80 (0.70) = 56 = np = 80 (0.70) = 56

σσ = = )3.0)(7.0(80=npq = 4.0988= 4.0988

Step 2. Apply continuity correction factor for x. x = 60 for the discrete random variableStep 2. Apply continuity correction factor for x. x = 60 for the discrete random variable

60 or more means 60 inclusive. Since the lower limit for 60 is 59.5, Sixty starts from 59.5.60 or more means 60 inclusive. Since the lower limit for 60 is 59.5, Sixty starts from 59.5.

This is similar to rounding number between 59.5 and 60.5 to 60. 60 is a value b/n 59.5 andThis is similar to rounding number between 59.5 and 60.5 to 60. 60 is a value b/n 59.5 and

60.5 60.5

63

Step 3: Determine the standard normal value, Z, Step 3: Determine the standard normal value, Z,

Z = Z = σNX −

= = 0998.4

565.59 −= 0.85 = 0.85

Step 4: calculate the probability of a Z value greater then 0.85 Step 4: calculate the probability of a Z value greater then 0.85

The probability of Z value between O and 0.85 is 0.3023The probability of Z value between O and 0.85 is 0.3023

To determine the probability of a Z value greater than 0.85 subtract 0.3023 from 0.50To determine the probability of a Z value greater than 0.85 subtract 0.3023 from 0.50

0.5000 – 0.3023 = 0.1977. So the probability that 60 or more customers will come again is0.5000 – 0.3023 = 0.1977. So the probability that 60 or more customers will come again is

19.77%19.77%

Example 2:Example 2: For a large group of sales prospects it is known that 20% of those contacted For a large group of sales prospects it is known that 20% of those contacted

personally by a sales representative will make a purchase. If a sales representative contacts 30personally by a sales representative will make a purchase. If a sales representative contacts 30

prospects, what is the probability that 10 or more will make a purchase?prospects, what is the probability that 10 or more will make a purchase?

µµ = np = (30) (0.2) = 6.00 = np = (30) (0.2) = 6.00

σσ = = )8.0)(2.0)(30(=npq = 2.19 = 2.19

10 or more is assumed to begin at 9.5. i.e., x = 9.510 or more is assumed to begin at 9.5. i.e., x = 9.5

Z = Z = 9.5 – 6.009.5 – 6.00 = = 3.53.5 = + 1.60 = + 1.60 2.19 2.19 2.19 2.19

The probability for Z = 1.60 = 0.4452 The probability for Z = 1.60 = 0.4452

p(p(≥≥1.6) = 0.5000 – 0.4452 =1.6) = 0.5000 – 0.4452 =0.0548 0.0548

3.4.2 Normal Approximation of Poisson Distribution3.4.2 Normal Approximation of Poisson Distribution

When the mean of a Poisson distribution is relatively large, the normal probability distributionWhen the mean of a Poisson distribution is relatively large, the normal probability distribution

can be used to approximate the Poisson distribution. For a good normal approximation to thecan be used to approximate the Poisson distribution. For a good normal approximation to the

poisson poisson µµ must be greater than or equal to 10. must be greater than or equal to 10.

Example:Example: The average number of calls for a service received by a machine repair shop per 8 The average number of calls for a service received by a machine repair shop per 8

hr shift is 10.00. What is the probability that more than 15 calls will be received during ahr shift is 10.00. What is the probability that more than 15 calls will be received during a

randomly selected 8 hr shiftrandomly selected 8 hr shift

µµ = 10 = 10

σσ = = 10 = 3.16 = 3.16

Z = Z = 15.5 – 1015.5 – 10 = = 5.55.5 = 1.71 = 1.71

64

3.16 6 3.16 3.16 6 3.16

The probability for Z = 1.74 = 0.4591 The probability for Z = 1.74 = 0.4591

p(Zp(Z>> 1.74) = 0.5000 – 0.4591 = 0.0409 1.74) = 0.5000 – 0.4591 = 0.0409


Patients arrive at a hospital at an average rate of 25 per a day. What is the probability thatPatients arrive at a hospital at an average rate of 25 per a day. What is the probability that

more than 22 patients will arrive in a day? Assuming arrival of patients follow the poissonmore than 22 patients will arrive in a day? Assuming arrival of patients follow the poisson



1.1. P(E) = 0.5P(E) = 0.5

P(B) = 0.5P(B) = 0.5

XX P(x)P(x)

00 0.06250.0625

11 0.2500.250

22 0.3750.375

33 0.2500.250

44 0.06250.0625

Sum 1Sum 1

2.2. σσ22 = 0.27 = 0.27

σσ = 0.5196 = 0.5196

3.3. p = 0.60p = 0.60 q = 0.4q = 0.4 n = 6n = 6

a) a) XX p(x)p(x)

00 0.00410.0041

11 0.036840.03684

22 0.138240.13824

33 0.276480.27648

44 0.311040.31104

55 0.18660.1866

66 0.046660.04666 1 1

65

b) E(x) = np = 3.6b) E(x) = np = 3.6

Standard deviation = Standard deviation = npq

= = 44.1)4.0(6.3 = = = 1.21.2

4.4. 1) N = 501) N = 50

S = 40 S = 40

n = 5 n = 5

x = 4 x = 4 P(x = 4) = 0.4313P(x = 4) = 0.4313

5.5. a) The company will meet its goal if line failures do not exceed three so thea) The company will meet its goal if line failures do not exceed three so the

probability that the company will meet its goal isprobability that the company will meet its goal is

P 6 + 1 + 2 + 3 line failuresP 6 + 1 + 2 + 3 line failures

= 0.4335= 0.4335

b) p(will not meet its goal)b) p(will not meet its goal)

= 1 – p (will meet its goal)= 1 – p (will meet its goal)

= 1 – 0.4335= 1 – 0.4335

= = 0.56650.5665

6.6. a) 0.7029a) 0.7029

b) 0.3821b) 0.3821

c) 38350 lmc) 38350 lm

7.7. σσ = 0.005 mm = 0.005 mm

8.8. P(x > 22) = 0.6915P(x > 22) = 0.6915

3.6 MODEL EXAMINATION QUESTIONS3.6 MODEL EXAMINATION QUESTIONS

Answer the following questions (clearly show your steps)Answer the following questions (clearly show your steps)

1.1. List the characteristics of the normal or continuous probability distribution and itsList the characteristics of the normal or continuous probability distribution and its

accompanying normal curve. accompanying normal curve.

2.2. Why we apply the normal probability distribution.Why we apply the normal probability distribution.

3.3. What determines the shape of the normal curve why?What determines the shape of the normal curve why?

4.4. Service life of truck tyres for heavy-duty trucks follows the normal distribution withService life of truck tyres for heavy-duty trucks follows the normal distribution with

mean 50,000km and standard deviation 5000km. mean 50,000km and standard deviation 5000km.

66

a)a) Calculate Z value for 60,000km, 48,000km, 63,000km, 58,000km, 39,000km,Calculate Z value for 60,000km, 48,000km, 63,000km, 58,000km, 39,000km,

62,750km.62,750km.

b)b) What is the probability that a tyre will last What is the probability that a tyre will last

i) between 47,000km and 50,000km?i) between 47,000km and 50,000km?

ii) between 50,000 and 60,000kmii) between 50,000 and 60,000km

iii) between 45,000 and 57,500km?iii) between 45,000 and 57,500km?

iv) less than 48,000km?iv) less than 48,000km?

v) greater than 45,000km?v) greater than 45,000km?

vi) less than 63000km?vi) less than 63000km?

vii) between 53,000 and 62,000km?vii) between 53,000 and 62,000km?

viii) between 55,000 and 63,000km?viii) between 55,000 and 63,000km?

c)c) The supplier of the tyres is planning to replace only 1% of those tyres with theThe supplier of the tyres is planning to replace only 1% of those tyres with the

least performance. What should be the service life for warranty? least performance. What should be the service life for warranty?

d)d) Tyres with less than 38500km performance are considered below standards orTyres with less than 38500km performance are considered below standards or

defective. How many tyres will be below standards, if 2500 tyres are made?defective. How many tyres will be below standards, if 2500 tyres are made?

5.5. Sales at a department store follow the normal distribution with an unknown mean andSales at a department store follow the normal distribution with an unknown mean and

unknown standard deviation. The retailing manager does know, however that, 16% ofunknown standard deviation. The retailing manager does know, however that, 16% of

the time he sells more than 2200 assortments and 34% of the time he sells less thanthe time he sells more than 2200 assortments and 34% of the time he sells less than

1800 assortments. Find the mean and standard deviation for the number of items sold.1800 assortments. Find the mean and standard deviation for the number of items sold.

6.6. For an Airline 80% of the time seats in all flights are occupied. If a particular AirFor an Airline 80% of the time seats in all flights are occupied. If a particular Air

plane has 180 sealsplane has 180 seals

a)a) What is the expected number of occupied seats?What is the expected number of occupied seats?

b)b) What is the probability (applying normal approximation to the binomial) thatWhat is the probability (applying normal approximation to the binomial) that

i.i. More than 150 seats will be occupied More than 150 seats will be occupied

ii.ii. Less than 175 seats will be occupiedLess than 175 seats will be occupied

iii.iii. 190 or more seats will be occupied190 or more seats will be occupied

7.7. Customers arrivals at a bank follow the poisson distribution with an average rate of 45Customers arrivals at a bank follow the poisson distribution with an average rate of 45

in an hour. What is the probability that in a particular one hour timein an hour. What is the probability that in a particular one hour time

a)a) more than 50 will arrivemore than 50 will arrive

b)b) 55 or more will arrive55 or more will arrive

c)c) 35 to 55 will arrive35 to 55 will arrive

67

UNIT 4: UNIT 4: SAMPLING AND SAMPLING DISTRIBUTIONSAMPLING AND SAMPLING DISTRIBUTION

ContentsContents



4.24.2 Why SamplingWhy Sampling

4.34.3 ErrorsErrors

4.44.4 Probability SamplingProbability Sampling

4.54.5 Method of Probability SamplingMethod of Probability Sampling

4.64.6 Sampling DistributionSampling Distribution

4.74.7 Central Limits TheoremCentral Limits Theorem

4.84.8 Distribution of the Standardized StatisticsDistribution of the Standardized Statistics

4.94.9 EstimatesEstimates

4.9.14.9.1 Point Estimates and their PropertiesPoint Estimates and their Properties

4.9.24.9.2 Interval EstimatesInterval Estimates

4.9.2.14.9.2.1 Constructing Confidence IntervalConstructing Confidence Interval

4.9.2.24.9.2.2 Finite Population Correction FactorFinite Population Correction Factor

4.104.10 Selecting A Sample SizeSelecting A Sample Size

4.10.14.10.1 Sample Size for the MeanSample Size for the Mean

4.10.24.10.2 Sample Size for ProportionSample Size for Proportion


4.12 Model Examination Question4.12 Model Examination Question


Usually the population under study is very large or infinite which makes studding it veryUsually the population under study is very large or infinite which makes studding it very

difficult or impossible. Under such circumstances we take a sample or a subset of thedifficult or impossible. Under such circumstances we take a sample or a subset of the

population to study the population. After completing this unit, you will be able topopulation to study the population. After completing this unit, you will be able to

understand why we sampleunderstand why we sample

identify types of probability sampling techniquesidentify types of probability sampling techniques

define sampling distribution and the central limit theoremdefine sampling distribution and the central limit theorem

estimate the population mean and population proportion estimate the population mean and population proportion

69

identify the types of estimates and construct confidence interval for the mean andidentify the types of estimates and construct confidence interval for the mean and

proportionproportion

determine the sample size for the mean and the proportiondetermine the sample size for the mean and the proportion

4.1 INTRODUCTION 4.1 INTRODUCTION

Statistics is a science of inference. It is the science of making general conclusion about theStatistics is a science of inference. It is the science of making general conclusion about the

entire group (the population) based on information obtained from a small group or sample. entire group (the population) based on information obtained from a small group or sample.

4.2 WHY SAMPLING4.2 WHY SAMPLING

It is often not feasible to study the entire population. The following are some of the majorIt is often not feasible to study the entire population. The following are some of the major

reasons why sampling is necessary. reasons why sampling is necessary.

4.2.1 The Destructive Nature of Certain Testes4.2.1 The Destructive Nature of Certain Testes

Many experiments especially in quality control demand destructing outputs. Consider theMany experiments especially in quality control demand destructing outputs. Consider the

following tests: following tests:

-- Testing wine or coffee Testing wine or coffee

-- Blood test for a patient Blood test for a patient

-- Testing strength of light bulbs Testing strength of light bulbs

-- Seed test for germination etc.Seed test for germination etc.

Unless sample is taken from the entire population the wine tester should drink all the wine, allUnless sample is taken from the entire population the wine tester should drink all the wine, all

the blood from the patient should be poured-out, all the light bulbs produced should bethe blood from the patient should be poured-out, all the light bulbs produced should be

destroyed and nothing would remain for sale. Here sample is a must.destroyed and nothing would remain for sale. Here sample is a must.

4.2.2 The Physical Impossibility of Checking all Items in the Population4.2.2 The Physical Impossibility of Checking all Items in the Population

The populations of fish, birds and other wild lives are large and are constantly moving beingThe populations of fish, birds and other wild lives are large and are constantly moving being

born and dying. There is no mechanism to contact all items or individual members of theborn and dying. There is no mechanism to contact all items or individual members of the

population. population.

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4.2.3 The Cost of Studying all the Items in a Population is Often Prohibitive4.2.3 The Cost of Studying all the Items in a Population is Often Prohibitive

Public opinion polls and consumer testing organizations usually contact fewer families out ofPublic opinion polls and consumer testing organizations usually contact fewer families out of

millions. Consider a multi national corporation with 50 million customers world wide. If thismillions. Consider a multi national corporation with 50 million customers world wide. If this

company plans to undertake market survey out of the 50 million it will take 2000 samples, ifcompany plans to undertake market survey out of the 50 million it will take 2000 samples, if

it takes 20 br. to mail samples and tabulate the responses of 2000 samples, total survey willit takes 20 br. to mail samples and tabulate the responses of 2000 samples, total survey will

cost Br. 40000. While the same survey involving 50 million population would cost about onecost Br. 40000. While the same survey involving 50 million population would cost about one

billion br. billion br.

4.2.4 The Adequacy of Sample Results4.2.4 The Adequacy of Sample Results

Even if funds were available, it is doubtful whether the additional accuracy of 100% sampleEven if funds were available, it is doubtful whether the additional accuracy of 100% sample

i.e., studying the entire population is essential in most problems. To determine monthly indexi.e., studying the entire population is essential in most problems. To determine monthly index

of food prices, bread, beans, milk etc, it is unlikly that the inclusion of all grocery stores andof food prices, bread, beans, milk etc, it is unlikly that the inclusion of all grocery stores and

shops would significantly affect the index, Since, the prices of such commodities usually doshops would significantly affect the index, Since, the prices of such commodities usually do

not vary by more than a few cents form one store to another. 100% accuracy cannot be allnot vary by more than a few cents form one store to another. 100% accuracy cannot be all

ways guaranteed by studying the entire population. The chance of error in collecting andways guaranteed by studying the entire population. The chance of error in collecting and

analyzing bulk data has its own disadvantage. analyzing bulk data has its own disadvantage.

4.2.5 To Contact the Whole Population Would Often be Time Consuming4.2.5 To Contact the Whole Population Would Often be Time Consuming

A market survey may take two or three days for field interviews by taking a sample of 2000A market survey may take two or three days for field interviews by taking a sample of 2000

customers. By using the same staff and interviewers and working seven days a week it wouldcustomers. By using the same staff and interviewers and working seven days a week it would

take nearly 200 years to contact 50 million customers.take nearly 200 years to contact 50 million customers.

4.3 ERRORS4.3 ERRORS

Avery important consideration in sampling is to select the sample in such a way that it is veryAvery important consideration in sampling is to select the sample in such a way that it is very

likely to have characteristics similar to the population as a whole. Other wise, the samplelikely to have characteristics similar to the population as a whole. Other wise, the sample

could have characteristics quite different form the population. In that case you could drawcould have characteristics quite different form the population. In that case you could draw

erroneous conclusions about the population on the basis of improperly chosen sample. Errorerroneous conclusions about the population on the basis of improperly chosen sample. Error

can be sampling or non-sampling error.can be sampling or non-sampling error.

Sampling error is related with the sampling technique and approaches while non-samplingSampling error is related with the sampling technique and approaches while non-sampling

error is related with administering the survey. Sampling errors can be identified and rectifiederror is related with administering the survey. Sampling errors can be identified and rectified

71

using some mathematical techniques. While the non-sampling errors are very difficult tousing some mathematical techniques. While the non-sampling errors are very difficult to

identify and rectify before making conclusions. identify and rectify before making conclusions.

4.4 PROBABILITY SAMPLE 4.4 PROBABILITY SAMPLE

Probability sample is a sample selected in such away that each item or person in theProbability sample is a sample selected in such away that each item or person in the

population being studied has a known (nonzero) likelihood of being included in the sample.population being studied has a known (nonzero) likelihood of being included in the sample.

Non-probability sample is a sample selected based on contingency and judgment.Non-probability sample is a sample selected based on contingency and judgment.

If non-probability methods are used, not all items or people have a chance of being includedIf non-probability methods are used, not all items or people have a chance of being included

in the sample. In such instances the result may be biased, the sample result may not bein the sample. In such instances the result may be biased, the sample result may not be

representative of the population.representative of the population.

Panel sampling and convenience sampling are non-probability sampling. They are based onPanel sampling and convenience sampling are non-probability sampling. They are based on

convenience to the statistician. Statistical procedures used to evaluate sample results based onconvenience to the statistician. Statistical procedures used to evaluate sample results based on

probability sampling. probability sampling.

4.5 METHODS OF PROBABILITY SAMPLING4.5 METHODS OF PROBABILITY SAMPLING

All probability sampling methods have one goal, to allow chance to determine the items orAll probability sampling methods have one goal, to allow chance to determine the items or

persons to be included in the sample. There are different types of sampling techniques.persons to be included in the sample. There are different types of sampling techniques.

However tHowever there is no one best method of selecting a probability sample. A technique best for ahere is no one best method of selecting a probability sample. A technique best for a

given circumstance or situation may fail in another situations. given circumstance or situation may fail in another situations.

Commonly used probability sampling techniques are the following:Commonly used probability sampling techniques are the following:

4.5.1 Simple Random Sampling4.5.1 Simple Random Sampling

A sample formulated in such a manner that each item or person in the population has the sameA sample formulated in such a manner that each item or person in the population has the same

chance of being included in the sample. We can easily list the name or identification of allchance of being included in the sample. We can easily list the name or identification of all

items i.e. the population on a piece of paper and properly fold and mixing and ruing the lotitems i.e. the population on a piece of paper and properly fold and mixing and ruing the lot

until we have the required sample size. This method is time consuming and awkward. until we have the required sample size. This method is time consuming and awkward.

More convenient method of selecting a random sample is to use a table of random numbers. ItMore convenient method of selecting a random sample is to use a table of random numbers. It

is necessary first to give identification for all elements in the population. We will select the is necessary first to give identification for all elements in the population. We will select the

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starting point arribitrarily and continue to take the sample until we have the required sample starting point arribitrarily and continue to take the sample until we have the required sample

size.size.

This method may be to use in certain research situations. Mostly difficult when the populationThis method may be to use in certain research situations. Mostly difficult when the population

is very larger. is very larger.

4.5.2 Systematic Random Sampling4.5.2 Systematic Random Sampling

The items or individuals of the population are arranged in some way (alphabetical) or someThe items or individuals of the population are arranged in some way (alphabetical) or some

other method. A random starting point is selected and then every Kother method. A random starting point is selected and then every K thth member of the member of the

population is selected for the sample. population is selected for the sample.

A systematic random sample should not be used, if there is a predetermined pattern to theA systematic random sample should not be used, if there is a predetermined pattern to the

population. Like inventory control, or if values are listed in ascending or descending orders.population. Like inventory control, or if values are listed in ascending or descending orders.

4.5.3 Stratified Random Sample 4.5.3 Stratified Random Sample

A population is first divided into subgroups called strata, and a sample is selected form each A population is first divided into subgroups called strata, and a sample is selected form each

stratum. Stratum can be stratum. Stratum can be

- Proportional sample / to the population or- Proportional sample / to the population or

- Non-proportional - Non-proportional

Example.Example. Studying advertising expenditure of 352 large companies. Profitability percentage Studying advertising expenditure of 352 large companies. Profitability percentage

is used to stratify this population. We need to select 50 samples. is used to stratify this population. We need to select 50 samples.

StratumStratum(0)(0)

ProfitabilityProfitability(1)(1)

Number ofNumber of (2)(2)

% of total% of total(3)(3)

Number (4)Number (4) (50x(3))(50x(3))

11 30 % and over 30 % and over 88 22 1122 20-30%20-30% 3535 1010 5533 10-20%10-20% 189189 5454 272744 0 up to 10%0 up to 10% 115115 3333 161655 deficit deficit 55 11 11

352352 150150 5050

Stratified sampling has the advantage, in some cases, of more accuracy reflecting theStratified sampling has the advantage, in some cases, of more accuracy reflecting the

characteristics of the population than dose simple random or systematic random sampling. characteristics of the population than dose simple random or systematic random sampling.

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4.5.4 Cluster Sampling 4.5.4 Cluster Sampling

It is dividing the population in to small units. These units are called primary units. There It is dividing the population in to small units. These units are called primary units. There

select at random certain groups or clusters. This technique isselect at random certain groups or clusters. This technique is

Often employed to reduce cost of sampling a population scattered over a large geographic Often employed to reduce cost of sampling a population scattered over a large geographic

area. area.

4.6 SAMPLING DISTRIBUTION 4.6 SAMPLING DISTRIBUTION

Two important terms in sampling distributive:Two important terms in sampling distributive:

a)a) Population parameter – A numerical measure of a population, population mean, Population parameter – A numerical measure of a population, population mean, µµ

population variance, population variance, σσ22, population standard deviation, , population standard deviation, σσ, population proportion, p etc., population proportion, p etc.

b)b) Sample statistics / Statistic/ - A numerical measure of the sample Sample statistics / Statistic/ - A numerical measure of the sample

Sample mean, Sample mean, x , sample variance S, sample variance S22 sample standard deviation S, sample proportion sample standard deviation S, sample proportion p ,,

etc.etc.

Sampling Distribution of the means (Sampling Distribution of the means ( x ))

Sampling distribution of the sample means, Sampling distribution of the sample means, x , is the probability distribution consisting of a, is the probability distribution consisting of a

list of all possible sample means of a given sample size selected from a population, and thelist of all possible sample means of a given sample size selected from a population, and the

probability of occurrence associated with each sample mean. probability of occurrence associated with each sample mean.

Example.Example. The following distribution is the hourly wage of seven employees The following distribution is the hourly wage of seven employees

EmployeeEmployee Hourly wageHourly wageAA 77BB 77CC 88DD 88EE 77FF 88GG 99

This population has a mean of 7.71 hoary wage i.e. This population has a mean of 7.71 hoary wage i.e. 5454//77

If we are planning to take sample of two employees, we will have 21 (If we are planning to take sample of two employees, we will have 21 (77CC22) possible samples) possible samples

and corresponding sample means. The 21 possible samples with their mean are theand corresponding sample means. The 21 possible samples with their mean are the

following:-following:-

74

PossiblePossibleSample Sample Sample mean (Sample mean ( x ))ABAB 7.0 7.0

ACAC 7.5 7.5

ADAD 7.5 7.5

AEAE 7.0 7.0

AFAF 7.5 7.5

AGAG 8.0 8.0

BCBC 7.5 7.5

BDBD 7.5 7.5

BEBE 7.0 7.0

BFBF 7.5 7.5

BGBG 8.0 8.0

CDCD 8.0 8.0

CECE 7.5 7.5

CFCF 8.0 8.0

CGCG 8.5 8.5

DEDE 7.5 7.5

DFDF 8.0 8.0

DGDG 8.5 8.5

EFEF 7.5 7.5

EGEG 8.0 8.0

FGFG 8.5 8.5

∑∑ x = 162= 162

Summary of sampling distribution of the means for n=2 will beSummary of sampling distribution of the means for n=2 will be

SampleSample

meanmean

No of meansNo of means ProbabilityProbability

75

77 33 0.14290.14297.57.5 99 0.42850.42858.008.00 66 0.28570.28578.508.50 33 0.14290.1429TotalTotal 2121 11

The mean of the distribution of sample means is obtained by summing the various sampleThe mean of the distribution of sample means is obtained by summing the various sample

means and dividing the sum by the number of samples. The mean of all the sample means ismeans and dividing the sum by the number of samples. The mean of all the sample means is

usually written usually written µµx reminds us that it is a population value because we have considered all reminds us that it is a population value because we have considered all

possible samples. The subscript possible samples. The subscript x indicates that it is a sampling distribution of means. indicates that it is a sampling distribution of means.

xµ = = 71.7

21

162

21

5.8...5.77 ==+++

The following graphs represent the population distribution and the distribution of the sampleThe following graphs represent the population distribution and the distribution of the sample

means. means.

Population Distribution Probability Sampling Distribution Population Distribution Probability Sampling Distribution

0.4 0.4 0.4 0.4

0.3 0.3 0.3 0.3

0.1 0.1 0.1 0.1

7 8 9 7 8 9 Hourly Wage Hourly Wage 7 7.5 8 8.5 7 7.5 8 8.5 X Hourly rate X Hourly rate

From the above graphs / distributions we can understand that:From the above graphs / distributions we can understand that:

a)a) The mean of the sample means (7.71) is equal to the mean of the population. This isThe mean of the sample means (7.71) is equal to the mean of the population. This is

always true if all possible samples of a given size are selected from the population ofalways true if all possible samples of a given size are selected from the population of

interest interest

b)b) The range of sample means is less than the range in the population. The sample meansThe range of sample means is less than the range in the population. The sample means

range form 7 to 8.5 where as the population vary form 7 to 9.00.range form 7 to 8.5 where as the population vary form 7 to 9.00.

c)c) The graph representing the distribution of the population and that of the sample meansThe graph representing the distribution of the population and that of the sample means

shows the change in shape from the population to the sample. The graph representingshows the change in shape from the population to the sample. The graph representing

the distribution of the sample means looks like a normal curve.the distribution of the sample means looks like a normal curve.

76

0.2 0.2

4.7 THE CENTRAL LIMIT THEOREM 4.7 THE CENTRAL LIMIT THEOREM

For a population with mean For a population with mean µµ and Variance and Variance σσ22, the sampling distribution of the means of all, the sampling distribution of the means of all

possible samples of size n generated from the population will be approximately normallypossible samples of size n generated from the population will be approximately normally

distributed with the mean of the sampling distribution equal to distributed with the mean of the sampling distribution equal to µµ and the variance equal to and the variance equal to

n2σ , assuming that the sample size is sufficiently large. , assuming that the sample size is sufficiently large.

The important facets of the central limit theorem bear repeating. The important facets of the central limit theorem bear repeating.

1.1. if the sample size n is sufficiently large, the sampling distribution of theif the sample size n is sufficiently large, the sampling distribution of the

means will be approximately normal regardless of the distribution of the populationmeans will be approximately normal regardless of the distribution of the population

form which the random sample is drawn form which the random sample is drawn

2.2. if a population is large and a large number of samples are selected from theif a population is large and a large number of samples are selected from the

population then the means of the sample means will be close to the population mean.population then the means of the sample means will be close to the population mean.

3.3. the variance of the distribution of sample means is determined by the variance of the distribution of sample means is determined by σσ22/n. This/n. This

implies that as the sample size increases the variation of implies that as the sample size increases the variation of x about its mean decrease. about its mean decrease.

Note that a sample of 30 or more elements is considered sufficiently large for theNote that a sample of 30 or more elements is considered sufficiently large for the

central limit theorem to take effect. central limit theorem to take effect.

A larger minimum sample size may be required for a good normal approximation when theA larger minimum sample size may be required for a good normal approximation when the

population distribution is very different from a normal distribution. While a smaller minimumpopulation distribution is very different from a normal distribution. While a smaller minimum

sample size may suffice for a good normal approximation when the population distribution issample size may suffice for a good normal approximation when the population distribution is

close to a normal distribution. close to a normal distribution.

4.8 DISTRIBUTION OF THE STANDARDIZED STATISTICS FOR THE SAMPLE MEAN4.8 DISTRIBUTION OF THE STANDARDIZED STATISTICS FOR THE SAMPLE MEAN

In order to use the central limit theorem, we need to know the population standard deviationIn order to use the central limit theorem, we need to know the population standard deviation

when it is not know the standard deviation of the sample, designated by S is used towhen it is not know the standard deviation of the sample, designated by S is used to

approximate it. The standardized distribution of the sample means is Z andapproximate it. The standardized distribution of the sample means is Z and

77

Z = Z = n

xσ

µ−, if the population standard deviation is known or , if the population standard deviation is known or

ns

x µ−, if the population, if the population

standard deviation is unknown.standard deviation is unknown.

Example 1: Example 1:

The annual wages of all employees of a company has a mean of 20,400 per year with standardThe annual wages of all employees of a company has a mean of 20,400 per year with standard

deviation of 3200. The personnel manager is going to take a random sample of 36 employeesdeviation of 3200. The personnel manager is going to take a random sample of 36 employees

and calculate the sample mean wage. What is the probability that the sample mian will exceedand calculate the sample mean wage. What is the probability that the sample mian will exceed

21.000? 21.000?

n= 36 n= 36 µµ = 20,400 and = 20,400 and σσ =3200 =3200

P[P[ x > 21,000] = > 21,000] = n

xσ

µ−= =

363200

2040021000 − = 1.125 = 1.125

P(Z > 1.13) = 0.1292P(Z > 1.13) = 0.1292

Example. 2 Example. 2

A company makes engine used in speedboats. The company’s engineers believe that theA company makes engine used in speedboats. The company’s engineers believe that the

engine delivers an average power of 220 horse power / HP/ and that the standard deviation ofengine delivers an average power of 220 horse power / HP/ and that the standard deviation of

power delivered is 15 HP. A potential buyer intends to sample 100 engines (each engine to bepower delivered is 15 HP. A potential buyer intends to sample 100 engines (each engine to be

run a single time ) . What is the probability that the sample mean, run a single time ) . What is the probability that the sample mean, x , will be less than 217, will be less than 217

HP. HP.

P(P( x <2/7)= P <2/7)= P 100

15

220217217 −=

−<n

Zσ

µ = -2 = -2

P(Z < -2) = 0.0228P(Z < -2) = 0.0228

Thus if the population mean is indeed Thus if the population mean is indeed µµ = 220 HP and the standard deviation is = 220 HP and the standard deviation is σσ = 15 HP, = 15 HP,

there is a rather small probability that the potential buyer’s tests will result in a sample meanthere is a rather small probability that the potential buyer’s tests will result in a sample mean

lower than 217HP. lower than 217HP.


78

The average GPA of all graduating students in a college is 2.85 with a standard deviation ofThe average GPA of all graduating students in a college is 2.85 with a standard deviation of

0.96. The placement unit randomly selects 64 graduating students. What is the probability that0.96. The placement unit randomly selects 64 graduating students. What is the probability that

the sample mean will be greater than 3.00?the sample mean will be greater than 3.00?

One important application of the central limit theorem is in the area of quality control. TheOne important application of the central limit theorem is in the area of quality control. The

manufacturing process is variable and be monitored to be sure that the variability does not getmanufacturing process is variable and be monitored to be sure that the variability does not get

beyond acceptable levels. beyond acceptable levels.

A control chart is used to assist in monitoring the variability A control chart is used to assist in monitoring the variability x chart is used to control chart is used to control

variation in the sample means.variation in the sample means.

The Chart has two limits about the mean The Chart has two limits about the mean µµ

c)c) Upper control limit (UCL) Upper control limit (UCL)

d)d) Lower control limit (LCL) Lower control limit (LCL)

The centerline is the desired mean, The centerline is the desired mean, µµ. .

UCL(Upper Control Limit)UCL(Upper Control Limit)

µ

LCL (Lower Control Limit)LCL (Lower Control Limit)

1 2 3 4 5 6……………………….. 50 …………………… Sample number

If a point is observed above UCL or below LCL the process is stopped and find the problem.If a point is observed above UCL or below LCL the process is stopped and find the problem.

The upper and lower control limits are generally located one, two, or three times The upper and lower control limits are generally located one, two, or three times xσ above above

and below and below µµ depending on the nature of the product and the process. depending on the nature of the product and the process.

4.9 ESTIMATES 4.9 ESTIMATES

Inferential statistics is concerned with estimation.Inferential statistics is concerned with estimation.

79

Sam

plin

g M

ean

In many cases values for a population parameter are unknown. If parameters are unknown it isIn many cases values for a population parameter are unknown. If parameters are unknown it is

generally not sufficient to make some convenient assumption about their values, rather thosegenerally not sufficient to make some convenient assumption about their values, rather those

unknown parameters should be estimated. unknown parameters should be estimated.

In business many decision are made with out complete information.In business many decision are made with out complete information.

A firm does not know exactly what will be its sales volume next year or next month. AA firm does not know exactly what will be its sales volume next year or next month. A

college does not know exactly how many students will enroll next year. Both must estimate tocollege does not know exactly how many students will enroll next year. Both must estimate to

make decision about the future.make decision about the future.

Types of Estimates Types of Estimates

4.9.1 Point estimate 4.9.1 Point estimate

A number or a simple number is used to estimate a population parameter. A number or a simple number is used to estimate a population parameter.

A random sample of observations is taken from the population of interest and the observedA random sample of observations is taken from the population of interest and the observed

values are used to obtain a point estimate of the relevant parameter. values are used to obtain a point estimate of the relevant parameter.

a. The ample mean, a. The ample mean, x , is the best estimator of the population mean , is the best estimator of the population mean µµ. .

Different samples from a population yield different point estimates of Different samples from a population yield different point estimates of µµ, ,

b. Sample proportion b. Sample proportion p is a good estimator of population proportion, p. is a good estimator of population proportion, p.

- Population proportion P is equal to the number of elements in the population belonging to- Population proportion P is equal to the number of elements in the population belonging to

the category of interest divided by the total number of elements in the population p = the category of interest divided by the total number of elements in the population p = N

X

Where: X is the number of success in the population and Where: X is the number of success in the population and

N population size N population size

Sample proportion, Sample proportion, p = = n

x where; where;

x is the number of elements in the sample found to belong to the category of interest and n isx is the number of elements in the sample found to belong to the category of interest and n is

the sample size. the sample size.

or or p = = Number of success in a sampleNumber of success in a sample number sampled number sampled

80

ExampleExample of 2000 persons sampled 1600 favored more strict environmental protection of 2000 persons sampled 1600 favored more strict environmental protection

measures, what is the estimated population proportion. measures, what is the estimated population proportion.

p = = 1600016000 = 0.80 = 0.80 2000 2000

80% is an estimate of the proportion in the population that favor more strict measures80% is an estimate of the proportion in the population that favor more strict measures

In general: In general:

The statistic The statistic x estimates estimates µµ

S estimates S estimates σσ

S S22 estimates estimates σσ22

p estimates pestimates p

Estimators and their properties / Goodness of an estimator/ Estimators and their properties / Goodness of an estimator/

The properties of good estimators areThe properties of good estimators are

a)a) Un biasedness Un biasedness

b)b) Efficiency Efficiency

c)c) Consistency and Consistency and

d)d) Sufficiency Sufficiency

a) An estimator is said to be unbiased if its expected value is equal to the populationa) An estimator is said to be unbiased if its expected value is equal to the population

parameter it estimates. parameter it estimates.

E(E( x ) = ) = µµ The sample mean , The sample mean , x , is therefore, an unbiased estimator of the population mean., is therefore, an unbiased estimator of the population mean.

Any systematic deviation of the estimator away from the parameter of interest is called Bias. Any systematic deviation of the estimator away from the parameter of interest is called Bias.

b) An estimator is efficient if it has a relatively small variance (as standard deviation). Theb) An estimator is efficient if it has a relatively small variance (as standard deviation). The

sample means have a variance of sample means have a variance of σσ//nn value is less than value is less than σσ. So the sample mean is an. So the sample mean is an

efficient estimator of the population mean. efficient estimator of the population mean.

c) An estimator is said to be consistent if its probability of being close to the parameter itc) An estimator is said to be consistent if its probability of being close to the parameter it

estimates increases as the sample size increases. estimates increases as the sample size increases.

81

The sample mean is a consistent estimator of The sample mean is a consistent estimator of µµ. This is so because the standard deviation of. This is so because the standard deviation of

x is is nx

σσ = . As the sample size n increases, the standard deviation of . As the sample size n increases, the standard deviation of x decreases and decreases and

hence the probability that hence the probability that x will be closes to its expected value, will be closes to its expected value, µµ, increases. , increases.

d) An estimator is said to be sufficient if it contains all the information in the data about thed) An estimator is said to be sufficient if it contains all the information in the data about the

parameter it estimates. The sample mean is sufficient estimator of parameter it estimates. The sample mean is sufficient estimator of µµ. Other estimators like. Other estimators like

the median and mode do not consider all values. But the mean considers all values (addedthe median and mode do not consider all values. But the mean considers all values (added

and divided by the sample size).and divided by the sample size).

4.9.2 Interval Estimates4.9.2 Interval Estimates

Interval estimate states the range within which a population parameter probably lies. TheInterval estimate states the range within which a population parameter probably lies. The

interval with in which a population parameter is expected to lie is usually referred to as theinterval with in which a population parameter is expected to lie is usually referred to as the

confidence intervalconfidence interval. .

The confidence interval for the population mean is the interval that has a high probability ofThe confidence interval for the population mean is the interval that has a high probability of

containing the population mean, containing the population mean, µµ

Two confidence intervals are used extensively. Two confidence intervals are used extensively.

1.1. 95% confidence interval and 95% confidence interval and

2.2. 99% confidence interval 99% confidence interval

A 95% confidence interval means that about 95% of the similarly constructed intervals willA 95% confidence interval means that about 95% of the similarly constructed intervals will

contain the parameter being estimated. If we use the 99% confidence interval we expect aboutcontain the parameter being estimated. If we use the 99% confidence interval we expect about

99% of the intervals to contain the parameter being estimated. 99% of the intervals to contain the parameter being estimated.

Another interpretation of the 95 % confidence interval is that 95 % of the sample means for aAnother interpretation of the 95 % confidence interval is that 95 % of the sample means for a

specified sample size will lie with in 1.96standred deviations of the hypothesized populationspecified sample size will lie with in 1.96standred deviations of the hypothesized population

mean. For 99% the sample means will lie, with in 2.58 standard deviations of themean. For 99% the sample means will lie, with in 2.58 standard deviations of the

hypothesized population mean.hypothesized population mean.

Where do the values 1.96 and 2.58 come form? Where do the values 1.96 and 2.58 come form?

82

The middle 95% of the sample mean lie equally on either side of the mean. And logicallyThe middle 95% of the sample mean lie equally on either side of the mean. And logically

0.95/2=0.4750 or 47.5% of the area is to the right of the mean and the area to the left of the0.95/2=0.4750 or 47.5% of the area is to the right of the mean and the area to the left of the

mean is 0.4750. mean is 0.4750.

The Z value for this probability is 1.96. The Z value for this probability is 1.96.

The Z to the right of the mean is + 1.96 and Z to the left is – 1.96. The Z to the right of the mean is + 1.96 and Z to the left is – 1.96.

4.9.2.1 Constructing Confidence Interval 4.9.2.1 Constructing Confidence Interval

a) Compute the standard error of the meana) Compute the standard error of the mean

Standard error of the mean is the standard deviation of the sample means. Standard error of the mean is the standard deviation of the sample means.

nx

σσ =

If the population standard deviation is not know, the standard deviation of the sample s, isIf the population standard deviation is not know, the standard deviation of the sample s, is

used to approximate the population standard deviation. used to approximate the population standard deviation. n

SS

x=

This indicates that the error in estimating the population mean decreases as the sample sizeThis indicates that the error in estimating the population mean decreases as the sample size

increases. increases.

b) The 95% and 99% confidence intervals are constructed as follows when n b) The 95% and 99% confidence intervals are constructed as follows when n >> 30. 30.

95% confidence interval 95% confidence interval x ±± 1.96 1.96 n

S

99% confidence interval 99% confidence interval x ±± 2.58 2.58 n

S

1.96 and 2.58 indicate the Z values corresponding to the middle 95% or 99% of the1.96 and 2.58 indicate the Z values corresponding to the middle 95% or 99% of the

observation respectively.observation respectively.

In general a confidence interval for the mean is computed byIn general a confidence interval for the mean is computed byn

SZx ± , Z reflects the selected, Z reflects the selected

level of confidence. level of confidence.

Example.Example. An experiment involves selecting a random sample of 256 middle managers for An experiment involves selecting a random sample of 256 middle managers for

studying their annual income. The sample mean is computed to be Br. 35,420 and the samplestudying their annual income. The sample mean is computed to be Br. 35,420 and the sample

standard deviation is Br. 2,050. standard deviation is Br. 2,050.

a.a. What is the estimated mean income of all middle managers ( the population ) ? What is the estimated mean income of all middle managers ( the population ) ?

b.b. What is the 95% confidence interval c(rounded to the nearest 10)What is the 95% confidence interval c(rounded to the nearest 10)

83

σσ = population standard = population standard deviation deviation

n = sample sizen = sample size

c.c. What are the 95% confidence limits? What are the 95% confidence limits?

d.d. Interpret the finding. Interpret the finding.

Solution Solution

a.a. Sample mean is 35 420 so this will approximate the population mean so Sample mean is 35 420 so this will approximate the population mean so µµ = 35420. It = 35420. It

is estimated from the sample mean.is estimated from the sample mean.

b.b. The confidence interval is between 35170 and 35670 found by The confidence interval is between 35170 and 35670 found by

n

SX 96.1± = 35420 = 35420 ±± 1.96 1.96

256

2050= 35168.87 and 35671.13= 35168.87 and 35671.13

c.c. The end points of the confidence interval are called the confidence limits. In this caseThe end points of the confidence interval are called the confidence limits. In this case

they are rounded to 35170 and 35670. 35170 is the lower limit and 35070 is the upperthey are rounded to 35170 and 35670. 35170 is the lower limit and 35070 is the upper

limit. limit.

d.d. Interpretation Interpretation

If we select 100 samples of size 256 form the population of all middle managers and computeIf we select 100 samples of size 256 form the population of all middle managers and compute

the sample means and confidence intervals, the population mean annual income would bethe sample means and confidence intervals, the population mean annual income would be

found in about 95 out of the 100 confidence intervals. About 5 out of the 100 confidencefound in about 95 out of the 100 confidence intervals. About 5 out of the 100 confidence

intervals would not contain the population mean annual income. intervals would not contain the population mean annual income.


A research firm conducted a survey to determine the mean amount smokers spend on cigaretteA research firm conducted a survey to determine the mean amount smokers spend on cigarette

during a week. A sample of 49 smokers revealed that the sample mean is Br. 20 with standardduring a week. A sample of 49 smokers revealed that the sample mean is Br. 20 with standard

deviation of Br. 5. Construct 95% confidence interval for the mean amount spent.deviation of Br. 5. Construct 95% confidence interval for the mean amount spent.

Confidence interval for a population proportion Confidence interval for a population proportion

The confidence interval for a population proportion is estimated The confidence interval for a population proportion is estimated

p ±± Z Zσσpp

Where Where σσpp is the standard error of the proportion and is the standard error of the proportion and

n

ppp

)1( −=σ

Therefore the confidence interval for population proportion is constructed by Therefore the confidence interval for population proportion is constructed by

p ±± Z Z n

pp )1( −

84

Example.Example. Suppose 1600 of 2000 union members sampled said they plan to vote for the Suppose 1600 of 2000 union members sampled said they plan to vote for the

proposal to merge with a national union. Union by laws state that at least 75% of all membersproposal to merge with a national union. Union by laws state that at least 75% of all members

must approve for the merger to be enacted. Using the 0.95 degree of confidence, what is themust approve for the merger to be enacted. Using the 0.95 degree of confidence, what is the

interval estimate for the population proportion? Based on the confidence interval, whatinterval estimate for the population proportion? Based on the confidence interval, what

conclusion can be drawn? conclusion can be drawn? p = = 2000

1600= 0.8. The sample proportion is 80%= 0.8. The sample proportion is 80%

The interval is computed as follows. The interval is computed as follows. p ±± Z Z n

pp )1( − = 0.80 = 0.80 ±± 1.96 1.96 2000

)8.01(80.0 −==

0.08 0.08 ±± 1.96 1.96 00008.0

= 0.78247 and 0 – 81753 rounded to 0.782 and 0.818.= 0.78247 and 0 – 81753 rounded to 0.782 and 0.818.

Based on the sample results when all union members vote, the proposal will probably passBased on the sample results when all union members vote, the proposal will probably pass

because 0.75 lie below the interval between 0.782 and 0.818. because 0.75 lie below the interval between 0.782 and 0.818.


A sample of 200 people were assumed to identify their major source of news information; 110A sample of 200 people were assumed to identify their major source of news information; 110

stated that their major source was television news coverage. Construct a 90% confidencestated that their major source was television news coverage. Construct a 90% confidence

interval for the proportion of people in the population who consider television their majorinterval for the proportion of people in the population who consider television their major

source of news information. source of news information.

4.9.2.2 Finite Population Correction Factor4.9.2.2 Finite Population Correction Factor

The population we have sampled so far has been very large, or assumed to be infinite. The population we have sampled so far has been very large, or assumed to be infinite.

If the sampled population is not infinite or not larger we need to make some adjustments inIf the sampled population is not infinite or not larger we need to make some adjustments in

the standard error of the mean and the standard error of the proportion. This is done to reducethe standard error of the mean and the standard error of the proportion. This is done to reduce

the error we committee in estimating a parameter.the error we committee in estimating a parameter.

A population that has a fixed upper bond is said to be finite. A finite population can be smallA population that has a fixed upper bond is said to be finite. A finite population can be small

or can be very large. or can be very large.

For a finite population, where the total number of objects is N, and the size of the sample is nFor a finite population, where the total number of objects is N, and the size of the sample is n

the following adjustment is made to the standard errors of the mean and the proportion.the following adjustment is made to the standard errors of the mean and the proportion.

Standard error of the mean Standard error of the mean

85

1−−=

N

nN

nx

σσ

Standard error of the proportion Standard error of the proportion

1

)1(

−−−=

N

nN

n

ppp

σ

This adjustment is called finite population correction factor. This adjustment is called finite population correction factor.

Why is it necessary to apply a factor and what is its effect?Why is it necessary to apply a factor and what is its effect?

Logically, if a sample is a substantial percentage of the population, then we would expect anyLogically, if a sample is a substantial percentage of the population, then we would expect any

estimate to be more precise than those for a smaller sample. estimate to be more precise than those for a smaller sample.

Suppose the population is 1000 and the sample is 100. Then this ratio is Suppose the population is 1000 and the sample is 100. Then this ratio is 11000

1001000

−−

or or 999

900

. Taking the square root gives the correction factor 0.9492. Multiplying the standard error. Taking the square root gives the correction factor 0.9492. Multiplying the standard error

reduces the error by about 5% or (1-0.9492)= 0.5. This reduction of the size of the standardreduces the error by about 5% or (1-0.9492)= 0.5. This reduction of the size of the standard

error yields a smaller range of values in estimating the population mean. If the sample size iserror yields a smaller range of values in estimating the population mean. If the sample size is

200 the correction factor is 0.8949. Meaning that the standard error has been reduced by more200 the correction factor is 0.8949. Meaning that the standard error has been reduced by more

than 10%. than 10%.

The usual rule is that If the ratio of the sample to the population, The usual rule is that If the ratio of the sample to the population, nn//N,N, is less than 0.05, the is less than 0.05, the

finite population correction factor is ignored. finite population correction factor is ignored.

ExampleExample. There are 250 families in a small town A poll of 40 families revealed that the mean. There are 250 families in a small town A poll of 40 families revealed that the mean

annual community contribution is 450 with a standard deviation of 75. Construct a 95%annual community contribution is 450 with a standard deviation of 75. Construct a 95%

confidence interval for the mean annual contribution. confidence interval for the mean annual contribution.

Solution: - Solution: -

First note that the population is finite.First note that the population is finite.

Second the sample constitute more than 5% of the population n/N = 40/250 =0.16 Hence theSecond the sample constitute more than 5% of the population n/N = 40/250 =0.16 Hence the

finite population correction factor is applied. finite population correction factor is applied.

−−±

1N

nN

n

SZx = 450 = 450 ±± 1.96 1.96

1250

40250

40

75

−−

= 450 = 450 ±± 23.24 23.24 8433.0

= 450 = 450 ±± 21.34 21.34

86

= 428.66 and 471.34= 428.66 and 471.34

Confidence interval for small sample (Student’s Distribution)Confidence interval for small sample (Student’s Distribution)

When the population is large and normal and the standard deviation is known the standardWhen the population is large and normal and the standard deviation is known the standard

normal distribution is employed to construct the confidence interval for the mean andnormal distribution is employed to construct the confidence interval for the mean and

proportion. If the sample size is at least 30, the sample standard deviation can substitute theproportion. If the sample size is at least 30, the sample standard deviation can substitute the

population standard deviation and the results are deemed satisfactory.population standard deviation and the results are deemed satisfactory.

If the sample size is less than 30 and population standard deviation is unknown, the standardIf the sample size is less than 30 and population standard deviation is unknown, the standard

normal distribution, Z, is not appropriate. The student’s t or the t distribution is used.normal distribution, Z, is not appropriate. The student’s t or the t distribution is used.

Characteristics of the Student’s t DistributionCharacteristics of the Student’s t Distribution

Assuming that the population of interest is normal or approximately normal, the following Assuming that the population of interest is normal or approximately normal, the following

are the characteristics of the t distributionare the characteristics of the t distribution

1.1. It is a continuous distributionIt is a continuous distribution

2.2. It is bell-shaped and symmetricalIt is bell-shaped and symmetrical

3.3. There is not one t distribution, but rather a ‘family’ of t distribution. All have the sameThere is not one t distribution, but rather a ‘family’ of t distribution. All have the same

mean of zero but their standard deviation differ according to the sample size, n. The tmean of zero but their standard deviation differ according to the sample size, n. The t

distribution differs for different sample size.distribution differs for different sample size.

4.4. It is more spread out and flatter at the center than is the Z. However as the sample sizeIt is more spread out and flatter at the center than is the Z. However as the sample size

increases the curve representing t distribution approaches the Z distribution.increases the curve representing t distribution approaches the Z distribution.

t distribution for sample size of 28 t distribution for sample size of 28

t distribution for sample size of 20t distribution for sample size of 20

87

t distribution for sample size of 10 t distribution for sample size of 10

tt

As the sample size decreases the curve representing the t distribution will have wider tails andAs the sample size decreases the curve representing the t distribution will have wider tails and

will be more flat at the center.will be more flat at the center.

Z DistributionZ Distribution

t Distribution t Distribution

For a given confidence level, say 95%, the t value is greater than the Z value. This is soFor a given confidence level, say 95%, the t value is greater than the Z value. This is so

because there is more variability in sample means computed from smaller samples. Thus ourbecause there is more variability in sample means computed from smaller samples. Thus our

confidence in the resulting estimate is not strong. t values are found referring to theconfidence in the resulting estimate is not strong. t values are found referring to the

appropriate degrees of freedom in the t table. Degrees of freedom means the freedom to freelyappropriate degrees of freedom in the t table. Degrees of freedom means the freedom to freely

move data points or the freedom to freely assign values arbitrarily.move data points or the freedom to freely assign values arbitrarily.

Degrees of freedom (df) = n – 1 where n is the sample size. Degrees of freedom (df) = n – 1 where n is the sample size.

This implies that we can freely move or assign values for all data points except the last nThis implies that we can freely move or assign values for all data points except the last n thth

value. If the mean of the distribution is specified there is a freedom to assign any value for allvalue. If the mean of the distribution is specified there is a freedom to assign any value for all

data points except the lost point. data points except the lost point.

Example - Example - the mean of five data points is 12. Then it follows that the sum of all the fivethe mean of five data points is 12. Then it follows that the sum of all the five

points is 60 = (5 x 12). Thus if five points are constrained to have a sum of 60 or a mean ofpoints is 60 = (5 x 12). Thus if five points are constrained to have a sum of 60 or a mean of

12, we have 5 – 1 = 4 degrees of freedom.12, we have 5 – 1 = 4 degrees of freedom.

If all the five data points are missing we are free to assign any value as long as their sum is 60If all the five data points are missing we are free to assign any value as long as their sum is 60

say 14, 12, 10, 9, 15.say 14, 12, 10, 9, 15.

88

If 4 are missing we are free to assign any value since 60 minus the known value of a dataIf 4 are missing we are free to assign any value since 60 minus the known value of a data

point is known.point is known.

If two are un know, 14, 16, 10, xIf two are un know, 14, 16, 10, x33, x, x44 since 14 + 16 + 10 + x since 14 + 16 + 10 + x33 + x + x44 = 60 = 60

Then xThen x33 + x + x44 = 60 – 40 = 20 = 60 – 40 = 20

xx33 + x + x44 = 20. We can assign any value as long as their sum is 20. 10, 10 or 9.11 or 15.5 etc… = 20. We can assign any value as long as their sum is 20. 10, 10 or 9.11 or 15.5 etc…

But if the four data points are known, (10, 14, 16, 12), the 5But if the four data points are known, (10, 14, 16, 12), the 5 thth data point will have a data point will have a

predetermined value i.e. 60 – 52 = 8. Now we are not free to assign arbitrary value for thispredetermined value i.e. 60 – 52 = 8. Now we are not free to assign arbitrary value for this

data point. Degrees of freedom can be obtained from the deviation based on the assumptiondata point. Degrees of freedom can be obtained from the deviation based on the assumption

that sum of the differences (d) between the mean and all values of the random variable (x) isthat sum of the differences (d) between the mean and all values of the random variable (x) is

zero. I.e., if we subtract the mean from all values of x the sum of the difference will be zerozero. I.e., if we subtract the mean from all values of x the sum of the difference will be zero

consider the above five data points. Their mean is 12 and their sum 60. Thus (xconsider the above five data points. Their mean is 12 and their sum 60. Thus (x 11 – 12) + (x – 12) + (x22 + +

12) + (x12) + (x33 – 12) + (x – 12) + (x44 – 12) + (x – 12) + (x55 – 12) = 0 = d – 12) = 0 = d11 + d + d22 + d + d33 + d + d44 + d + d55 = 0 = 0

Now we are free to assign any value for only four missing differences as long as this sum isNow we are free to assign any value for only four missing differences as long as this sum is

zero. So we have still n – 1 degrees of freedom.zero. So we have still n – 1 degrees of freedom.

Computing t valueComputing t value

The t variable representing the student’s t distribution is defined asThe t variable representing the student’s t distribution is defined as

t = t = ns

x

/

µ−where: where: x is the sample mean of n measurements, is the sample mean of n measurements, µµ is the population mean is the population mean

and s is the sample standard deviationand s is the sample standard deviation

Note that t is just like Z = Note that t is just like Z = n

x

/σµ−

except that we replace except that we replace σσ with s. unlike our methods of large with s. unlike our methods of large

samples, samples, σσ cannot be approximated by s when the sample size is less than 30 and we can not cannot be approximated by s when the sample size is less than 30 and we can not

use the normal distribution. The table for the t distribution is constructed for selected levels ofuse the normal distribution. The table for the t distribution is constructed for selected levels of

confidence for degree of freedom up to 30. To use the table we need to know two numbers,confidence for degree of freedom up to 30. To use the table we need to know two numbers,

the tail area, (1 minus confidence level selected), and the degree of freedom.the tail area, (1 minus confidence level selected), and the degree of freedom.

(1 – confidence level selected) is (1 – confidence level selected) is αα, the Greek letter alpha. This is the error we committee in, the Greek letter alpha. This is the error we committee in

estimating.estimating.

The confidence interval for the sample mean is The confidence interval for the sample mean is x ±± 2α (n – 1)(n – 1) n

S

89

ExampleExample. A traffic department in town is planning to determine mean number of accidents at. A traffic department in town is planning to determine mean number of accidents at

a high-risk intersection. Only a random sample of 10 days measurements were obtained.a high-risk intersection. Only a random sample of 10 days measurements were obtained.

Number of accidents per day wereNumber of accidents per day were

8, 8, 77 1010 1515 1111 66 88 55 1313 1212

Construct a 95% confidence interval for the mean number of accident per day. Construct a 95% confidence interval for the mean number of accident per day.

a) Compute a) Compute x and s and s

x = = 10

95= 9.5 per day= 9.5 per day

9

5.94

1

)( 2

=−−

= ∑n

xxS

x= 3.24 per day= 3.24 per day

The confidence level is 95% soThe confidence level is 95% so

αα = 1 – 0.95 = 0.05 = 1 – 0.95 = 0.05

2

05.0

2=α

= 0.025 = 0.025

The degree of freedom, df = n – 1 = 10 – 1 = 9 from the t table tThe degree of freedom, df = n – 1 = 10 – 1 = 9 from the t table t0.0250.025, df 9 = 2.76, df 9 = 2.76

The confidence interval isThe confidence interval is

x ±± t t.0025.0025 df(9) df(9) n

s

9.5 9.5 ±± (2.26) (2.26) 10

24.3

9.5 9.5 ±± 2.3 2.3

7.2 to 11.807.2 to 11.80

With 95% confidence the mean number of accident at this particular intersection is betweenWith 95% confidence the mean number of accident at this particular intersection is between

7.2 and 11.8.7.2 and 11.8.


A quality controller of a company plans to inspect the average diameter of small bolts made.A quality controller of a company plans to inspect the average diameter of small bolts made.

A random sample of 6 bolts was selected. The sample is computed to be 2.0016mm and theA random sample of 6 bolts was selected. The sample is computed to be 2.0016mm and the

sample standard deviation 0.0012mm. Construct the 99% confidence interval for all boltssample standard deviation 0.0012mm. Construct the 99% confidence interval for all bolts

made.made.

90

4.10 SELECTING A SAMPLE SIZE 4.10 SELECTING A SAMPLE SIZE

Size of a sample must be determined scientifically. Care must be taken not to select a sampleSize of a sample must be determined scientifically. Care must be taken not to select a sample

too large or too small. There are two misconceptions about how many to sampletoo large or too small. There are two misconceptions about how many to sample

a)a) Sample consisting 5% (or similar constant percentage) is adequate for all problems. Sample consisting 5% (or similar constant percentage) is adequate for all problems.

5% can be too much for a particular population say 10 million or can be too small for5% can be too much for a particular population say 10 million or can be too small for

another say 200.another say 200.

b)b) A sample, for example, must be selected form a heavily populated area.A sample, for example, must be selected form a heavily populated area.

The avoid such problems the sample size should be mathematically determined. The avoid such problems the sample size should be mathematically determined.

4.10.1 Sample Size for the Mean4.10.1 Sample Size for the Mean

There are three factors that determine the size of the sample. None of which has any directThere are three factors that determine the size of the sample. None of which has any direct

relation ship to the size of the population. relation ship to the size of the population.

a.a. The degree of confidence selected. The degree of confidence selected.

b.b. The maximum allowable error The maximum allowable error

c.c. The variation in the population The variation in the population

a.a. The degree of confidence,The degree of confidence, This is usually 95% or 99%. But it may be any level. It is This is usually 95% or 99%. But it may be any level. It is

specified by the statistician. The higher the degree of confidence, the larger the samplespecified by the statistician. The higher the degree of confidence, the larger the sample

required. If we want to be sure the true mean will lie between an interval, we would hve torequired. If we want to be sure the true mean will lie between an interval, we would hve to

survey the entire population. Example. Suppose the parameter to be estimated is thesurvey the entire population. Example. Suppose the parameter to be estimated is the

arithmetic mean, and the degree of confidence selected is 90%. Based on a sample, it wasarithmetic mean, and the degree of confidence selected is 90%. Based on a sample, it was

estimated that the population mean is in the interval between 850 and 1050. Logically, ifestimated that the population mean is in the interval between 850 and 1050. Logically, if

the degree of confidence were increased to 95% or 99% the sample size would have tothe degree of confidence were increased to 95% or 99% the sample size would have to

increase. increase.

b.b. Maximum error allowedMaximum error allowed. It is the maximum error that will be tolerable at a specified level. It is the maximum error that will be tolerable at a specified level

of confidence. Suppose a statistician is interested to estimate the mean income of residentsof confidence. Suppose a statistician is interested to estimate the mean income of residents

of an area. There are indications that the family incomes range from a probable low ofof an area. There are indications that the family incomes range from a probable low of

19000 to a high of about 39000. On the assumption that these are reasonable estimates ,19000 to a high of about 39000. On the assumption that these are reasonable estimates ,

does it seem likely that the statistician would be satisfied with this statement resultingdoes it seem likely that the statistician would be satisfied with this statement resulting

from a sample of area residents. “ The population mean is between 23,000 and 35,000”from a sample of area residents. “ The population mean is between 23,000 and 35,000”

Probability not. Because confidence limits that wide indicate little or nothing about theProbability not. Because confidence limits that wide indicate little or nothing about the

91

population mean. Instead, the statistician stated “using the 0.95 confidence level, the totalpopulation mean. Instead, the statistician stated “using the 0.95 confidence level, the total

error is predicting the population mean should not exceed by 200”. The maximumerror is predicting the population mean should not exceed by 200”. The maximum

allowable error is denoted ‘E’ = E = |allowable error is denoted ‘E’ = E = | x - - µµ|. This means based on a sample size n, if the|. This means based on a sample size n, if the

estimate of population mean is computed to be 35,000, then we will assure that theestimate of population mean is computed to be 35,000, then we will assure that the

population mean is in the interval between 34800and 35200. Found by 35,000 + 200 andpopulation mean is in the interval between 34800and 35200. Found by 35,000 + 200 and

35000-200. For the 0.95 degree of confidence selected the maximum error of 35000-200. For the 0.95 degree of confidence selected the maximum error of ++ 200 200

interms of Z is 1.96. To determine the value of one standard error of the mean interms of Z is 1.96. To determine the value of one standard error of the mean xσ simply simply

divide the total error of 200 by 1.96 = 102.04 divide the total error of 200 by 1.96 = 102.04

xσ = =

96.1

200 = 102.04 = 102.04

Error cannot exceed Error cannot exceed

200 200 200 200

97.96 102.04 102.04 97.96 97.96 102.04 102.04 97.96

-1.96 -1 0 +1 1.96 -1.96 -1 0 +1 1.96

Population mean must be in the Population mean must be in the Z Z

interval interval ++ 200 from the sample 200 from the sample

mean mean

The size of the sample is computed by solving for n in the formulaThe size of the sample is computed by solving for n in the formula

n

SS

x= , note that since we are using a sample standard deviation., note that since we are using a sample standard deviation.

i.e., i.e., xS is substituted for is substituted for x

σ and S for and S for σσ

sizesample

deviationSample

deviationZ

errorallowableTotal standard

standard=

92

Total allowable error let be represented by ‘E’ then it follows that,Total allowable error let be represented by ‘E’ then it follows that,

n

S

n

S

Z

E ===96.1

200= 102.04= 102.04

Since there are two unknowns for one equation we cannot solve for both.Since there are two unknowns for one equation we cannot solve for both.

c.c. Variation in the population.Variation in the population. There are still two unknowns. To solve for the number to be There are still two unknowns. To solve for the number to be

sampled we need to estimate the variation in the population. The standard deviation is asampled we need to estimate the variation in the population. The standard deviation is a

measure of variation. Thus the standard deviation of the population must be estimated.measure of variation. Thus the standard deviation of the population must be estimated.

This can be done either: This can be done either:

a-a- By taking a small pilot survey and using the standard deviation of the pilot sample as anBy taking a small pilot survey and using the standard deviation of the pilot sample as an

estimate of the population standard deviation or estimate of the population standard deviation or

b-b- By estimating the standard deviation based on knowledge of the population. By estimating the standard deviation based on knowledge of the population.

Suppose a pilot survey is conducted and sample standard deviation is computed to be 3000.Suppose a pilot survey is conducted and sample standard deviation is computed to be 3000.

The number to be sampled can now be estimated. The number to be sampled can now be estimated.

Z

E

n

SS

x==

n

3000

96.1

200 = n = 864.36n = 864.36

xS is standard error of the mean, the error we commit in estimating . From the above is standard error of the mean, the error we commit in estimating . From the above

computation we can learn that as the variation in the population increase the sample size willcomputation we can learn that as the variation in the population increase the sample size will

increase.increase.

A more convenient computational formula for determining n is. A more convenient computational formula for determining n is.

n = n = 2

.

E

SZ

where E = allowable errorwhere E = allowable error

Z = Z value for the degree of confidence selected Z = Z value for the degree of confidence selected

S = Sample deviation S = Sample deviation

For this example For this example n = n = 2

200

300096.1

×

= 864.36= 864.36

Example 1Example 1. A marketing research firm wants to conduct a survey to estimate the average. A marketing research firm wants to conduct a survey to estimate the average

amount spent on entertainment by each person visiting a popular pub. The people who planamount spent on entertainment by each person visiting a popular pub. The people who plan

the survey would like to be able to determine the average amount spent by all people visitingthe survey would like to be able to determine the average amount spent by all people visiting

93

the pub to within br. 120, with 95% confidence. From past operations of the pub, an estimatethe pub to within br. 120, with 95% confidence. From past operations of the pub, an estimate

of the population standard deviation is of the population standard deviation is σσ = br. 400 what is the minimum required sample = br. 400 what is the minimum required sample

sizes?sizes?

Z = 1.96Z = 1.96

E = 120E = 120

σσ = 400 = 400

Required, n?Required, n?

n = n = 2

120

40096.1

×

= 42.68 = 42.68 ≈≈ 43 43


A processor of carrots cuts the green top of each carrot washes the carrots, and inserts six to aA processor of carrots cuts the green top of each carrot washes the carrots, and inserts six to a

package. Twenty packages are inserted in a box for shipment. To test the Wight of the boxes,package. Twenty packages are inserted in a box for shipment. To test the Wight of the boxes,

a few were checked. The mean weight was 10kg and the standard deviation 0.25kg. Howa few were checked. The mean weight was 10kg and the standard deviation 0.25kg. How

many boxes must the processor sample to be 95% confident that the sample mean does notmany boxes must the processor sample to be 95% confident that the sample mean does not

differ from the population mean by more than 0.1 kg?differ from the population mean by more than 0.1 kg?

4.10.2 Sample size for proportion 4.10.2 Sample size for proportion

The procedure used to determine the sample size for the mean is applicable to determine whenThe procedure used to determine the sample size for the mean is applicable to determine when

proportions are involved. proportions are involved.

Three things must be specified. Three things must be specified.

- Decide on the level of confidence - Decide on the level of confidence

- Indicate how precise the estimate of the population proportion must be - Indicate how precise the estimate of the population proportion must be

- Approximate the population proportion, P, either from past experience or from a small - Approximate the population proportion, P, either from past experience or from a small

pilot survey pilot survey p

The formula for determining the sample size n for a proportionThe formula for determining the sample size n for a proportion

n = n = p (1 - (1 - p ) ) ( ) 2

EZ

where: where: p - estimated proportion- estimated proportion

Z = Z value for the selected confidence levelZ = Z value for the selected confidence level

E = the maximum tolerable errorE = the maximum tolerable error

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Example 1.Example 1. A member of parliament wants to determine her popularity in her a region. She A member of parliament wants to determine her popularity in her a region. She

indicates that the proportion of voters who will vote for her must be estimated with in indicates that the proportion of voters who will vote for her must be estimated with in ++ 2 2

percent of the population proportion. Further, the 95% degree of confidence is to be used. Inpercent of the population proportion. Further, the 95% degree of confidence is to be used. In

past elections she received 40% of the popular vote in that area. She doubts whether it haspast elections she received 40% of the popular vote in that area. She doubts whether it has

changed much. How many registered voters should be sampled? changed much. How many registered voters should be sampled?

Z = 1.96Z = 1.96

p = 0.40 = 0.40

E = 0.02E = 0.02

n = n = p (1 - (1 - p ) ) ( ) 2

EZ

= 0.40 (1 – 0.4) = 0.40 (1 – 0.4) 2

02.0

96.1

= 2,304.96 = 2,304.96 ≈≈ 2305 2305

This sample size might be too large, orThis sample size might be too large, or

too small or exactly correct depending on the accuracy of too small or exactly correct depending on the accuracy of p . .

Note: if there is no logical estimate of Note: if there is no logical estimate of p , the sample size can be estimated by letting , the sample size can be estimated by letting p =0.5 =0.5

Example 2.Example 2. Suppose the president wants an estimate of the proportion of the population that Suppose the president wants an estimate of the proportion of the population that

support this current policy on unemployment. The president wants the estimate to be with insupport this current policy on unemployment. The president wants the estimate to be with in

0.04 of the true proportion. Assume a 95% level of confidence and the proportion supporting0.04 of the true proportion. Assume a 95% level of confidence and the proportion supporting

current policy to be 0.60. current policy to be 0.60.

a) How large a sample is required a) How large a sample is required

b)b) How large would the sample have to be if the estimate were not available? How large would the sample have to be if the estimate were not available?

Solution:Solution:

a)a) E = 0.04E = 0.04

Z = 1.96Z = 1.96

p = 0.60= 0.60

b)b) E = 0.4E = 0.4

Z = 1.96Z = 1.96

p = 0.50 (since there is no estimate)= 0.50 (since there is no estimate)

n = 0.5 (1 – 0.5) n = 0.5 (1 – 0.5) 2

04.0

96.1

= 600 = 600

95

n = 0.6(1 – 0.6) n = 0.6(1 – 0.6) 2

04.0

96.1

= = 577577


The marketing department of a company wishes to study the loyalty pattern of consumers.The marketing department of a company wishes to study the loyalty pattern of consumers.

Loyalty patterns range from extremely loyal to brand snitcher. If the department wishes toLoyalty patterns range from extremely loyal to brand snitcher. If the department wishes to

estimate the proportion of consumers who are extremely loyal to this brand, what sample sizeestimate the proportion of consumers who are extremely loyal to this brand, what sample size

would be necessary to estimate this proportion with 0.05 with 95% confidence?would be necessary to estimate this proportion with 0.05 with 95% confidence?


1.1. 0.10560.1056

2.2. 18.60 and 21.4018.60 and 21.40

3.3. 49.21% and 60.79%49.21% and 60.79%

4.4. 1.9996 and 2.00361.9996 and 2.0036

5.5. 2424

6.6. 384384


Answer the following questionsAnswer the following questions

1.1. Explain the central limit theorem and its important facets.Explain the central limit theorem and its important facets.

2.2. An investment consultant reports that the average 12-month return on a randomAn investment consultant reports that the average 12-month return on a random

sample of 50 projects was 20.74%. If the standard deviation was 5% for the entiresample of 50 projects was 20.74%. If the standard deviation was 5% for the entire

large group of stocks from which the sample of projects was chosen, construct a 95%large group of stocks from which the sample of projects was chosen, construct a 95%

confidence interval for the average 12-month return for all projects in this group. confidence interval for the average 12-month return for all projects in this group.

3.3. An advertising executive thinks that the proportion of consumer’s who have seen hisAn advertising executive thinks that the proportion of consumer’s who have seen his

company’s advertisement in newspapers is around 0.65. The executive wants tocompany’s advertisement in newspapers is around 0.65. The executive wants to

estimate the customer population proportion to with in estimate the customer population proportion to with in ±± 0.05 and have a 98% 0.05 and have a 98%

confidence in the estimate. How large a sample should be taken.confidence in the estimate. How large a sample should be taken.

4.4. A company wants to estimate the proportion of its employees, who are satisfied with aA company wants to estimate the proportion of its employees, who are satisfied with a

new incentive scheme. Out of a total of 1,242 employees, 160 were randomly selectednew incentive scheme. Out of a total of 1,242 employees, 160 were randomly selected

and interviewed. Of the one interviewed, 85 indicated that they were satisfied with theand interviewed. Of the one interviewed, 85 indicated that they were satisfied with the

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new scheme. Construct a 90% confidence interval for the proportion of all employeesnew scheme. Construct a 90% confidence interval for the proportion of all employees

who are satisfied with the new decision.who are satisfied with the new decision.

5.5. A survey is being planned to determine the mean amount of time senior executivesA survey is being planned to determine the mean amount of time senior executives

watch TV. A pilot survey indicated that the mean time per week is 12 hrs with awatch TV. A pilot survey indicated that the mean time per week is 12 hrs with a

standard deviation of 3 hrs. It is desired to estimate the mean viewing time within 0.25standard deviation of 3 hrs. It is desired to estimate the mean viewing time within 0.25

hrs. The 95% degree of confidence is to be used. How many executives should behrs. The 95% degree of confidence is to be used. How many executives should be

surveyed?surveyed?

6.6. Why sampling?Why sampling?

7.7. What are the properties of good estimators? ExplainWhat are the properties of good estimators? Explain

8.8. A sample of 200 people were asked to identify their major source of news information.A sample of 200 people were asked to identify their major source of news information.

110 said their major source was radio.110 said their major source was radio.

a)a) Construct a 95% confidence interval for the proportion of people in theConstruct a 95% confidence interval for the proportion of people in the

population that consider radio their major source of news informationpopulation that consider radio their major source of news information

b)b) How large a sample would be necessary to estimate the population proportionHow large a sample would be necessary to estimate the population proportion

with a sampling error of 0.05 at 95% confidence. with a sampling error of 0.05 at 95% confidence.

9.9. What are the factors that determine the size of the sample?What are the factors that determine the size of the sample?

10.10. Under what circumstances the finite population correction factor should be applied?Under what circumstances the finite population correction factor should be applied?

11.11. The registrar of a college wants to estimate the arithmetic mean final GPA of allThe registrar of a college wants to estimate the arithmetic mean final GPA of all

graduating senior students. GPAs range between 2.0 and 4.0. The mean GPA is to begraduating senior students. GPAs range between 2.0 and 4.0. The mean GPA is to be

estimated with plus and minus 0.05 of the population mean. The 99% confidence is toestimated with plus and minus 0.05 of the population mean. The 99% confidence is to

be used. The standard deviation of a small pilot survey is 0.279.be used. The standard deviation of a small pilot survey is 0.279.

How many grade reports (transcripts) should be sampled?How many grade reports (transcripts) should be sampled?

12.12. In a small town there are 250 families. From 50 families sample 15 regularly attendIn a small town there are 250 families. From 50 families sample 15 regularly attend

community meetings. Construct a 95% confidence interval for the proportion ofcommunity meetings. Construct a 95% confidence interval for the proportion of

families attending the meeting regularly.families attending the meeting regularly.

13.13. A wine importer needs to report the average percentage of alcohol in bottles of newA wine importer needs to report the average percentage of alcohol in bottles of new

wine.wine.

From experience with various kinds of wines, the importer believes the populationFrom experience with various kinds of wines, the importer believes the population

standard deviation is 1.2%. The importer randomly sampled 60 bottles of the newstandard deviation is 1.2%. The importer randomly sampled 60 bottles of the new

wine and obtain a sample mean of 9.3%. Give a 90% confidence interval for thewine and obtain a sample mean of 9.3%. Give a 90% confidence interval for the

average percentage of alcohol in all bottles of the new wine.average percentage of alcohol in all bottles of the new wine.

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14.14. The manufacturers of a sports car want to estimate the proportion of people in a givenThe manufacturers of a sports car want to estimate the proportion of people in a given

income bracket, who are interested in a model. The company wants to know theincome bracket, who are interested in a model. The company wants to know the

population proportion to within 0.10 with 99% confidence. Current company recordspopulation proportion to within 0.10 with 99% confidence. Current company records

indicate that the proportion may be around 0.25. what is the minimum required sampleindicate that the proportion may be around 0.25. what is the minimum required sample

size for this survey.size for this survey.

15.15. A survey of a random sample of 1000 managers found that 81% of them had a highA survey of a random sample of 1000 managers found that 81% of them had a high

need for power. This led to a conclusion that power is a motivator for managers.need for power. This led to a conclusion that power is a motivator for managers.

Construct a 90% confidence interval for the proportion of all managers in theConstruct a 90% confidence interval for the proportion of all managers in the

population under study who are motivated by power. population under study who are motivated by power.

16.16. The average score of trainees who participated in a special training program is 120The average score of trainees who participated in a special training program is 120

with a standard deviation of 15. A company who sent its employees sampled 36with a standard deviation of 15. A company who sent its employees sampled 36

employees and calculates their mean scores. What is the probability that the sampleemployees and calculates their mean scores. What is the probability that the sample

mean will be less than 115? mean will be less than 115?

17.17. A business faculty in a university is planning to introduce a new performanceA business faculty in a university is planning to introduce a new performance

evaluation technique. Instructors are required to evaluate their respective departmentevaluation technique. Instructors are required to evaluate their respective department

heads. A random sample of 7 instructors from the marketing department was selectedheads. A random sample of 7 instructors from the marketing department was selected

and their evaluation recorded. The results wereand their evaluation recorded. The results were

72, 81, 69, 78, 80, 75, 7972, 81, 69, 78, 80, 75, 79

Construct a 90% confidence interval for the average performance evaluation of all theConstruct a 90% confidence interval for the average performance evaluation of all the

instructors in the department. instructors in the department.

UNIT 5: TESTS OF HYPOTHESES UNIT 5: TESTS OF HYPOTHESES

ContentsContents

5.0 Aims and Objectives5.0 Aims and Objectives

5.1 Introduction5.1 Introduction

5.2 Hypothesis and Hypothesis Testing Defined5.2 Hypothesis and Hypothesis Testing Defined

5.2.1 Hypothesis 5.2.1 Hypothesis

5.2.2 Hypothesis Testing5.2.2 Hypothesis Testing

5.3 Steps for Testing a Hypothesis5.3 Steps for Testing a Hypothesis

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5.4 Hypothesis Testing Involving Large samples5.4 Hypothesis Testing Involving Large samples

5.4.1 Testing for the Population Mean /Large Sample/5.4.1 Testing for the Population Mean /Large Sample/

5.4.1.1 Population Standard Deviation Known5.4.1.1 Population Standard Deviation Known

5.4.1.2 Population Standard Deviation Unknown5.4.1.2 Population Standard Deviation Unknown

5.4.2 Testing for Two Population means 5.4.2 Testing for Two Population means

5.4.3 Testing for a Population Proportion5.4.3 Testing for a Population Proportion

5.4.4 Testing for the Difference between Two Population Proportions5.4.4 Testing for the Difference between Two Population Proportions

5.5 Hypothesis Testing Involving Small Samples5.5 Hypothesis Testing Involving Small Samples

5.5.1 Characteristics of the student’s t Distribution5.5.1 Characteristics of the student’s t Distribution

5.5.2 Test for the Population Mean5.5.2 Test for the Population Mean

5.5.3 Test for Comparison of Two Population Means5.5.3 Test for Comparison of Two Population Means

5.5.4 Hypothesis Testing Involving Paired Observations5.5.4 Hypothesis Testing Involving Paired Observations

5.6 Testing for Difference of Variance Comparing Two Population Variances5.6 Testing for Difference of Variance Comparing Two Population Variances


5.8 Model Examination Questions5.8 Model Examination Questions


When we estimate the value of a parameter we are using methods of estimation. The unknownWhen we estimate the value of a parameter we are using methods of estimation. The unknown

value of a population parameter is estimated from sample information by constructingvalue of a population parameter is estimated from sample information by constructing

confidence interval estimate.confidence interval estimate.

Decision concerning the value of a population parameter are obtained by hypothesis testing,Decision concerning the value of a population parameter are obtained by hypothesis testing,

which is the topic of this chapter.which is the topic of this chapter.

After completing this unit, you will be able to:After completing this unit, you will be able to:

•• define hypothesis and testing hypothesisdefine hypothesis and testing hypothesis

•• test hypothesis involving large sampletest hypothesis involving large sample

•• test hypothesis involving small sampletest hypothesis involving small sample

•• understand the p-value in hypothesis testingunderstand the p-value in hypothesis testing

•• testing for differences of variancetesting for differences of variance


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Most statistical inference centers around the parameters of a population. In hypothesis testingMost statistical inference centers around the parameters of a population. In hypothesis testing

we start with an assumed value of a population parameter. Then a sample evidence is used towe start with an assumed value of a population parameter. Then a sample evidence is used to

decide whether the assumed value is unreasonable and should be rejected, or whether itdecide whether the assumed value is unreasonable and should be rejected, or whether it

should be accepted; Hence the statistical inferences made are referred to as hypothesis testing.should be accepted; Hence the statistical inferences made are referred to as hypothesis testing.

5.2 HYPOTHESIS AND HYPOTHESIS TESTING DEFINED5.2 HYPOTHESIS AND HYPOTHESIS TESTING DEFINED

5.2.1 Hypothesis is a statement or an assumption about the value of a population5.2.1 Hypothesis is a statement or an assumption about the value of a population

parameter or parameters.parameter or parameters.

ExamplesExamples

-- The mean monthly income of all employees of a company is br. 2000.The mean monthly income of all employees of a company is br. 2000.

-- The average age of students in a college is 22 yearsThe average age of students in a college is 22 years

-- 5% of the products of a firm are defective5% of the products of a firm are defective

All these hypothesis have one thing in common:All these hypothesis have one thing in common:

The population of interest are so large that for various reasons it would not be feasible toThe population of interest are so large that for various reasons it would not be feasible to

study all the items, or persons, in the populationstudy all the items, or persons, in the population

5.2.2 Hypothesis Testing Defined5.2.2 Hypothesis Testing Defined

Hypothesis testing is a procedure based on sample evidence and probability distribution usedHypothesis testing is a procedure based on sample evidence and probability distribution used

to determine whether the hypothesis is a reasonable statement and should not be rejected, or isto determine whether the hypothesis is a reasonable statement and should not be rejected, or is

unreasonable and should be rejected. unreasonable and should be rejected.

It is simply selecting a sample from the populations, calculate sample statistic and based onIt is simply selecting a sample from the populations, calculate sample statistic and based on

certain decision rules accept or reject the hypothesis. certain decision rules accept or reject the hypothesis.

Test statistic is a sample statistic computed from the sample data. The value of the testTest statistic is a sample statistic computed from the sample data. The value of the test

statistic is used in determining whether or not we may reject the hypothesis.statistic is used in determining whether or not we may reject the hypothesis.

Decision rule of a statistical hypothesis is rule that specifies the conditions under which theDecision rule of a statistical hypothesis is rule that specifies the conditions under which the

hypothesis may be rejected. We decide whether or not to reject the hypothesis by followinghypothesis may be rejected. We decide whether or not to reject the hypothesis by following

the decision rule. the decision rule.

5.3 STEPS FOR TESTING HYPOTHESIS5.3 STEPS FOR TESTING HYPOTHESIS

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There is a five-step procedure that systematize hypothesis testing.There is a five-step procedure that systematize hypothesis testing.

Hypothesis testing as used by the statisticians does not provide proof that something is true, inHypothesis testing as used by the statisticians does not provide proof that something is true, in

the manner in which a mathematician “proves” a statement. It does provide a kind of “proofthe manner in which a mathematician “proves” a statement. It does provide a kind of “proof

beyond a reasonable doubt” in the manner of an attorney.beyond a reasonable doubt” in the manner of an attorney.

Step I. Identity the null hypothesis and the alternate hypothesisStep I. Identity the null hypothesis and the alternate hypothesis

The first step is to state the hypothesis to be tested. It is called the The first step is to state the hypothesis to be tested. It is called the Null HypothesisNull Hypothesis, designated, designated

by Hby Hoo and read “H sub-zero”. The capital letter H stands for hypothesis and the subscript zero and read “H sub-zero”. The capital letter H stands for hypothesis and the subscript zero

implies “no difference or no change. There is usually a ‘not’ or a ‘no’ term in the nullimplies “no difference or no change. There is usually a ‘not’ or a ‘no’ term in the null

hypothesis meaning no change”. The null hypothesis is set up for the purpose of either tohypothesis meaning no change”. The null hypothesis is set up for the purpose of either to

rejecting or not to rejecting it. The null hypothesis is a statement that will be rejected it ourrejecting or not to rejecting it. The null hypothesis is a statement that will be rejected it our

sample information provide us with convincing evidence that it false. And it will not besample information provide us with convincing evidence that it false. And it will not be

rejected if our sample data fail to provide ample evidence that it is false.rejected if our sample data fail to provide ample evidence that it is false.

If the null hypothesis is not rejected based on sample data, in effect we are saying that theIf the null hypothesis is not rejected based on sample data, in effect we are saying that the

evidence does not allow us to reject it. We cannot state, however, that the null hypothesis isevidence does not allow us to reject it. We cannot state, however, that the null hypothesis is

true. This is the same as the situation in the courts. true. This is the same as the situation in the courts.

In courts we heard judges saying, “Found not guilty” when they release a suspect free. TheyIn courts we heard judges saying, “Found not guilty” when they release a suspect free. They

never say “he is innocent”. The suspect is released may be because the prosecutor or thenever say “he is innocent”. The suspect is released may be because the prosecutor or the

police fail to provide the court with convincing evidence beyond reasonable doubt that thepolice fail to provide the court with convincing evidence beyond reasonable doubt that the

suspect has committed the crime. The null hypothesis is a tentative assumption made aboutsuspect has committed the crime. The null hypothesis is a tentative assumption made about

the value of a population parameter. Usually it is a statement that the population parameterthe value of a population parameter. Usually it is a statement that the population parameter

has a specific value.has a specific value.

Failure to reject the null hypothesis does not prove that Ho is true. To prove with out anyFailure to reject the null hypothesis does not prove that Ho is true. To prove with out any

doubt that the null hypothesis is true, the population parameter would have to be known. Thisdoubt that the null hypothesis is true, the population parameter would have to be known. This

is usually not feasible. is usually not feasible.

The sample statistic is usually different from the hypothesized population parameter. For thisThe sample statistic is usually different from the hypothesized population parameter. For this

reason we have to make a judgment about the difference. reason we have to make a judgment about the difference.

If a hypothesized mean is 70 and the sample mean is 69.5 we musts make a judgment aboutIf a hypothesized mean is 70 and the sample mean is 69.5 we musts make a judgment about

the difference 0.5. Is it a true difference, i.e a significant difference, or is it due to chance /the difference 0.5. Is it a true difference, i.e a significant difference, or is it due to chance /

101

sampling. To answer this question we conduct a test of significance, commonly referred to assampling. To answer this question we conduct a test of significance, commonly referred to as

a test of hypothesis. a test of hypothesis.

Identify the Alternative hypothesis (HIdentify the Alternative hypothesis (H11)): Alliterate hypothesis is a statement describes what: Alliterate hypothesis is a statement describes what

we will believe if we reject the null hypothesis. It is designated Hwe will believe if we reject the null hypothesis. It is designated H 11 (H sub – one) the alternate (H sub – one) the alternate

hypothesis will be accepted if the sample data provide us with evidence that the nullhypothesis will be accepted if the sample data provide us with evidence that the null

hypothesis is false.hypothesis is false.

It is a statement that will be accepted if our sample data provide us with ample evidence thatIt is a statement that will be accepted if our sample data provide us with ample evidence that

the null hypothesis is false. the null hypothesis is false.

Step II: Determine the level of significance Step II: Determine the level of significance

After setting up the null hypothesis and alternate hypothesis, the next step is to state the levelAfter setting up the null hypothesis and alternate hypothesis, the next step is to state the level

of significance. It is the probability of rejecting the null hypothesis when it is actually true. of significance. It is the probability of rejecting the null hypothesis when it is actually true.

Level of significance is the risk we assume of rejecting the null hypothesis when it is aLevel of significance is the risk we assume of rejecting the null hypothesis when it is a

actually true. actually true.

The level of significance is designated by the Greek letter alpha, The level of significance is designated by the Greek letter alpha, αα, it is also referred to as the, it is also referred to as the

level of risk. level of risk.

Traditionally three levels of significance are knownTraditionally three levels of significance are known

0.05. level is selected for consumer research 0.05. level is selected for consumer research

0.01. for quality assurance 0.01. for quality assurance

0.10. for political polling 0.10. for political polling

The level of significance reflects the risk we want to assume A0.01 level of significance willThe level of significance reflects the risk we want to assume A0.01 level of significance will

yield smaller risk than 0.05 or 0.1.yield smaller risk than 0.05 or 0.1.

The researcher must decide on the level of significance before formulating a decision rule andThe researcher must decide on the level of significance before formulating a decision rule and

collecting sample data. This is very important to reduce bias. The level of significance can becollecting sample data. This is very important to reduce bias. The level of significance can be

any level between 0 and 1.any level between 0 and 1.

To illustrate how it is possible to reject a true hypothesis, suppose that a computeTo illustrate how it is possible to reject a true hypothesis, suppose that a compute

manufacturer purchase a component form a supplier. Suppose the contract specifies that themanufacturer purchase a component form a supplier. Suppose the contract specifies that the

102

manufacture’s quality assurance department will sample all incoming shipment of component.manufacture’s quality assurance department will sample all incoming shipment of component.

If more than 6% of the components sampled are substandard the shipment will be rejected.If more than 6% of the components sampled are substandard the shipment will be rejected.

The null hypothesis is:The null hypothesis is:

HHoo= the incoming shipment of components contains 6% or less substandard components. = the incoming shipment of components contains 6% or less substandard components.

The alternative hypothesis is:The alternative hypothesis is:

HH1:1: More than 6% of the components are defective. More than 6% of the components are defective.

A sample of 50 components just received revealed that 4 components or 8% wereA sample of 50 components just received revealed that 4 components or 8% were

substandard. substandard.

The shipment was rejected because it exceeded maximum of 6%. If the shipment was actuallyThe shipment was rejected because it exceeded maximum of 6%. If the shipment was actually

substandard then the decision to return the component to the supplier was correct. substandard then the decision to return the component to the supplier was correct.

However suppose the 4 components selected in the sample were the only substandardHowever suppose the 4 components selected in the sample were the only substandard

components in the shipment of 4000 components. Only 1% were defective. In that case lesscomponents in the shipment of 4000 components. Only 1% were defective. In that case less

than 6% of the entire shipment was substandard and rejecting the shipment was an error.than 6% of the entire shipment was substandard and rejecting the shipment was an error.

In terms of hypothesis testing we rejected the null hypothesis that the shipment was notIn terms of hypothesis testing we rejected the null hypothesis that the shipment was not

subitandard when we should not have rejected it. subitandard when we should not have rejected it.

By rejecting a true hypothesis we committed a type I error. By rejecting a true hypothesis we committed a type I error.

A type I error is designated by A type I error is designated by αα (alpha). (alpha).

Type I error is rejecting the null hypothesis, Ho, when it is actually true. Type I error is rejecting the null hypothesis, Ho, when it is actually true.

The probability of committing another type of error, Type II error, is designated The probability of committing another type of error, Type II error, is designated ββ, beta,, beta,

failure to reject Ho when it is actually false. failure to reject Ho when it is actually false.

The above firm would commit a type II error if, unknown to it, an incoming shipmentThe above firm would commit a type II error if, unknown to it, an incoming shipment

contained 600 substandard components yet the shipment was accepted. Suppose 2 of the 50contained 600 substandard components yet the shipment was accepted. Suppose 2 of the 50

component in the sample (4%) tested were substandard and 48 were good. Because thecomponent in the sample (4%) tested were substandard and 48 were good. Because the

sample contains less than 6% substandard components, the shipment was accepted. But of allsample contains less than 6% substandard components, the shipment was accepted. But of all

task the entire shipment 15% of the components we defective. task the entire shipment 15% of the components we defective.

We often refer to those two possible errors as the alpha error We often refer to those two possible errors as the alpha error αα, and the beta error , and the beta error ββ, ,

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αα error – the probability of making a type I error error – the probability of making a type I error

ββ error – the probability of making type II error error – the probability of making type II error

The following table shows the decision the researcher could make and the possibleThe following table shows the decision the researcher could make and the possible

consequences. consequences.

Null HypothesisNull Hypothesis The researcherThe researcher

does not reject Hdoes not reject Hoo

The ResearcherThe Researcher

rejects Hrejects Hoo

If Ho is true If Ho is true Correct decisionCorrect decision Type I errorType I errorIf Ho is falseIf Ho is false Type II errorType II error Correct decisionCorrect decision

Step III: Find the Test statistic Step III: Find the Test statistic

Test statistic – A value, determined from sample information, used to reject or not to rejectTest statistic – A value, determined from sample information, used to reject or not to reject

the null hypothesis. the null hypothesis.

There are many test statistics, Z (the normal distribution), the student t test, F, and XThere are many test statistics, Z (the normal distribution), the student t test, F, and X 22 or the or the

chi –square.chi –square.

The standard normal deviate, Z distribution is used as test statistic when the sample size isThe standard normal deviate, Z distribution is used as test statistic when the sample size is

large, n large, n ≥≥ 30. Based on the sample size and the parameter to be tested the statistician will 30. Based on the sample size and the parameter to be tested the statistician will

select the appropriate test statistic.select the appropriate test statistic.

Step IV: Determine the decision ruleStep IV: Determine the decision rule

A decision rule is a statement of the conditions under which the null hypothesis is rejectedA decision rule is a statement of the conditions under which the null hypothesis is rejected

and the conditions under which it is not rejected. and the conditions under which it is not rejected.

The region or area of rejection defines the location of all those values that are so large or soThe region or area of rejection defines the location of all those values that are so large or so

small that the probability of their occurrence under a true null hypothesis is rather remote. small that the probability of their occurrence under a true null hypothesis is rather remote.

Sampling distribution for the statistic Z, 0.05 level of significance. Sampling distribution for the statistic Z, 0.05 level of significance.

Non-rejection Non-rejection

Region or do not reject H Region or do not reject H00 Rejection regionRejection region

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Scale of Z Scale of Z

0 1.6 45 0 1.6 45

0.95 Probability 0.95 Probability 0.05 Probability 0.05 Probability

Initial Value Initial Value

The above chart portrays the rejection region for a test of significance. The level ofThe above chart portrays the rejection region for a test of significance. The level of

significance selected is 0.05.significance selected is 0.05.

1.1. The area where the null hypothesis is not rejected includes the area to the left of 1.645 The area where the null hypothesis is not rejected includes the area to the left of 1.645

2.2. The area of rejection is to the right of 1.645The area of rejection is to the right of 1.645

3.3. A one – tailed test is being applied /will be discussed latter on/A one – tailed test is being applied /will be discussed latter on/

4.4. The 0.05 level of significant was chosen The 0.05 level of significant was chosen

5.5. The sampling distribution is for the test statistic Z , the standard normal deviate. The sampling distribution is for the test statistic Z , the standard normal deviate.

6.6. The value 1.645 separates the regions where the null hypothesis is rejected and whereThe value 1.645 separates the regions where the null hypothesis is rejected and where

it is not rejected it is not rejected

7.7. The value 1.645 is called the critical value. It is the corresponding value of the testThe value 1.645 is called the critical value. It is the corresponding value of the test

statistic for the selected level of significance i.e. Z value at the 0.05 level ofstatistic for the selected level of significance i.e. Z value at the 0.05 level of

significance is 1.645.significance is 1.645.

Critical value:Critical value: The dividing point between the region where the null hypothesis is rejected The dividing point between the region where the null hypothesis is rejected

and the region where it is not rejected. and the region where it is not rejected.

Steps V: Take a sample and made a decision Steps V: Take a sample and made a decision

At this step a decision is made to reject or not to reject the null hypothesis. For the aboveAt this step a decision is made to reject or not to reject the null hypothesis. For the above

chart, if based on sample data or information, Z is computed be 2.34 the null hypothesis ischart, if based on sample data or information, Z is computed be 2.34 the null hypothesis is

rejected at the 0.05 level of significance. rejected at the 0.05 level of significance.

The decision to reject Ho is made because 2.34 lies in the region of rejection that is beyondThe decision to reject Ho is made because 2.34 lies in the region of rejection that is beyond

1.645. We would reject the null hypothesis reasoning that it is highly improbable that a1.645. We would reject the null hypothesis reasoning that it is highly improbable that a

computed Z value this large is due to sampling variation or chance. Had the computed valuecomputed Z value this large is due to sampling variation or chance. Had the computed value

105

been 1.645 or less say 0.71 then Ho would not be rejected. It would be reasoned that such abeen 1.645 or less say 0.71 then Ho would not be rejected. It would be reasoned that such a

small computed value could be attributed to chance that is sampling variation. small computed value could be attributed to chance that is sampling variation.

One – Tailed and Two – Tailed tests of significance One – Tailed and Two – Tailed tests of significance

One Tailed TestOne Tailed Test

The region of rejection is only in one tail of the curve. The above example indicates that theThe region of rejection is only in one tail of the curve. The above example indicates that the

region of rejection is in the right (upper) tail of the curve. region of rejection is in the right (upper) tail of the curve.


Rejection region Region or do not reject H Rejection region Region or do not reject H00

0.95 Probability0.95 Probability

Z Z

-1.6 45 0 -1.6 45 0


Initial Value Initial Value

Consider companies purchase larger quantities of tyre. Suppose they want the tires to anConsider companies purchase larger quantities of tyre. Suppose they want the tires to an

average mileage of 40,000 Km of wear under normal usage. They will therefore reject aaverage mileage of 40,000 Km of wear under normal usage. They will therefore reject a

shipment of tires if accelerated - life test reveal that the life of the tires is significantly belowshipment of tires if accelerated - life test reveal that the life of the tires is significantly below

40000 Km on the average. 40000 Km on the average.

The purchasers gladly accept a shipment if the mean life is greater than 40000 Kms, they areThe purchasers gladly accept a shipment if the mean life is greater than 40000 Kms, they are

not concerned with this possibility. not concerned with this possibility.

They are only concerned if they have sample evidence to conclude that the tires will averageThey are only concerned if they have sample evidence to conclude that the tires will average

less than 40000 Kms of useful life. less than 40000 Kms of useful life.

Thus the test is set up to satisfy the concern of the companies that the mean life of the tires isThus the test is set up to satisfy the concern of the companies that the mean life of the tires is

less than 40000Km. less than 40000Km.

106

The null and alternate hypotheses are written: -The null and alternate hypotheses are written: -

HHoo: : µµ = 40,000 km and = 40,000 km and

HH11: : µµ < 40000 km < 40000 km

One way to determine the location of the rejection region is to look at the direction in whichOne way to determine the location of the rejection region is to look at the direction in which

the inequality sign in the alternate hypothesis is pointing. the inequality sign in the alternate hypothesis is pointing.

Test is one – tailed, if HTest is one – tailed, if H11 states states µµ > or > or µµ < if < if µµ11 , states a direction, test is one - tailed. , states a direction, test is one - tailed.

Two-tailed testTwo-tailed test

A test is two - tailed if HA test is two - tailed if H11 does not state a direction. does not state a direction.

Consider the following example: Consider the following example:

Ho: there is no difference between the mean income of males and the mean income ofHo: there is no difference between the mean income of males and the mean income of

females. females.

HH11: there is a difference in the mean income of males and the mean income of females. : there is a difference in the mean income of males and the mean income of females.

If Ho is rejected and HIf Ho is rejected and H11 accepted the mean income of males could be greater than that of accepted the mean income of males could be greater than that of

females or vis versa. To accommodate these two possibilities, the 5 level of significancefemales or vis versa. To accommodate these two possibilities, the 5 level of significance

representing the area of rejection is divided equally in to two tails of the samplingrepresenting the area of rejection is divided equally in to two tails of the sampling

distribution. If the level of significant is 0.05 each rejection region will have 0.025distribution. If the level of significant is 0.05 each rejection region will have 0.025

probability.probability.

Note that the total area under the normal curve is one found by 0.95 + 0.025 + 0.025.Note that the total area under the normal curve is one found by 0.95 + 0.025 + 0.025.


Rejection region Region or do not reject H Rejection region Region or do not reject H00 Rejection regionRejection region

0.95 Probability0.95 Probability

Z Z

-1. 96 0 + 1. 96 -1. 96 0 + 1. 96


Initial Value Initial Value Initial Value Initial Value

107

5.4 HYPOTHESIS TESTING INVOLVING LARGE SAMPLE5.4 HYPOTHESIS TESTING INVOLVING LARGE SAMPLE

Note that a sample of 30 or more is considered large. Note that a sample of 30 or more is considered large.

5.4.1 Test for the Population Mean5.4.1 Test for the Population Mean

5.4.1.1 Population Standard Deviation Known5.4.1.1 Population Standard Deviation Known

Example.Example. The efficiency ratings of a company have been normally distributed over a period The efficiency ratings of a company have been normally distributed over a period

of many years. The arithmetic mean (of many years. The arithmetic mean (µµ) of the distribution is 200 and the standard deviation is) of the distribution is 200 and the standard deviation is

19. Recently, however, young employees have been hired and new training and production19. Recently, however, young employees have been hired and new training and production

methods introduced. Using the 0.01 level of significance, we want to test the hypothesis thatmethods introduced. Using the 0.01 level of significance, we want to test the hypothesis that

the mean is still 200.the mean is still 200.

Solution: Solution:

Step 1Step 1. The null hypothesis is " The population mean is still 200 " the alternative hypothesis. The null hypothesis is " The population mean is still 200 " the alternative hypothesis

is “The mean is different from 200 " or "The mean is not 200" is “The mean is different from 200 " or "The mean is not 200"

the two hypotheses are written as: the two hypotheses are written as:

Ho : Ho : µµ =200 =200

H1: H1: µµ ≠≠ 200 200

This is a two - tailed test because the alternate hypothesis does not state the direction of theThis is a two - tailed test because the alternate hypothesis does not state the direction of the

difference. difference.

That is, it does not state whether the mean is greater than or less than 200. That is, it does not state whether the mean is greater than or less than 200.

Step 2:Step 2: - As noted the 0.01 level of significance is to be used. This is - As noted the 0.01 level of significance is to be used. This is αα the probability of the probability of

committing a type I error. That is the probability of rejecting a true hypothesis. committing a type I error. That is the probability of rejecting a true hypothesis.

Step 3:Step 3: - The test statistic for this type of problem is Z, the standard normal deviate /you will - The test statistic for this type of problem is Z, the standard normal deviate /you will

see later on that the sample size is large/see later on that the sample size is large/

Z = Z = n

Xσ

µ−

108

Step 4Step 4: The decision rull is formulated by finding the critical values of Z from the table of: The decision rull is formulated by finding the critical values of Z from the table of

normal distribution. normal distribution.

Since this is a two - tailed test, half of 0.01 or 0.005 is in each tail. Each rejection region willSince this is a two - tailed test, half of 0.01 or 0.005 is in each tail. Each rejection region will

have a probability of 0.005. have a probability of 0.005.

The area where Ho is not rejected located between the two tails, is therefore, 0.99.The area where Ho is not rejected located between the two tails, is therefore, 0.99.

0.5000-0.005= 0.4950 so 0.4950 is the area between 0 and the critical value. The value0.5000-0.005= 0.4950 so 0.4950 is the area between 0 and the critical value. The value

nearest to 0.4950 is 0.495. The value for this probability is 2.58. nearest to 0.4950 is 0.495. The value for this probability is 2.58.


Rejection region with Region or do not reject HRejection region with Region or do not reject H00 Rejection regionRejection region

probability 0.99 Probability with probability 0.01÷2=0.005probability 0.99 Probability with probability 0.01÷2=0.005

0.01÷2=0.0050.01÷2=0.005 0.4950=0.5-0.0050.4950=0.5-0.005 0.4950=0.5-0.0050.4950=0.5-0.005

Z Z

It is not rejected It is not rejected

The decision rule is there fore: Reject the null hypothesis and accept the alternate hypothesisThe decision rule is there fore: Reject the null hypothesis and accept the alternate hypothesis

if the computed value of Z does not fall in the region between +2.58 and -2.58. Otherwise doif the computed value of Z does not fall in the region between +2.58 and -2.58. Otherwise do

not reject the null hypothesis. not reject the null hypothesis.

Step 5:Step 5: Take a sample and make a decision Take a sample and make a decision

Take a sample from the population (efficiently ratings) compute Z and based on the decisionTake a sample from the population (efficiently ratings) compute Z and based on the decision

rule, arrive at a decision to reject Ho or not reject Ho. rule, arrive at a decision to reject Ho or not reject Ho.

The efficenty ratings of 100 employees were analyzed. The mean of the sample was computedThe efficenty ratings of 100 employees were analyzed. The mean of the sample was computed

to be 203.5.to be 203.5.

Compute ZCompute Z

Z = Z = n

Xσ

µ− = = 6.1

2005.203

10016

2005.203 −=−203.5-200= 2.19 203.5-200= 2.19

109

Since 2.19 does not fall in the rejection region, Ho is not rejected. So we conclude that theSince 2.19 does not fall in the rejection region, Ho is not rejected. So we conclude that the

difference between 203.5, the sample mean, and 200 can be attributed to chance variation. difference between 203.5, the sample mean, and 200 can be attributed to chance variation.

Note: Selecting the level of significance before setting up the decision rule and sampling theNote: Selecting the level of significance before setting up the decision rule and sampling the

population is important not to be biased. population is important not to be biased.

Ho is not rejected at the 1% level. We would have biased the later decision by not initiallyHo is not rejected at the 1% level. We would have biased the later decision by not initially

selecting the 0.01 level. Instead we could have waited until after the sampling and selected aselecting the 0.01 level. Instead we could have waited until after the sampling and selected a

level of significance that would cause the null hypothesis to be rejected. We could havelevel of significance that would cause the null hypothesis to be rejected. We could have

chosen, for example , the 0.05 level. The critical value for that level are chosen, for example , the 0.05 level. The critical value for that level are ++ 1.96. 1.96.

Since the computed value of Z (2.19) lies beyond 1.96 the null hypothesis would be rejectedSince the computed value of Z (2.19) lies beyond 1.96 the null hypothesis would be rejected

and we could concluded that the mean efficiency rating is not 200.and we could concluded that the mean efficiency rating is not 200.

Example 2:Example 2: The mean annual turn over rate of a brand of chemical is 6.0 (this indicates that The mean annual turn over rate of a brand of chemical is 6.0 (this indicates that

the stock of the chemical turns over an average of six times a years) . The standard deviationthe stock of the chemical turns over an average of six times a years) . The standard deviation

is 0.5. It is suspected that the average turnover is not 6.0. The 0.05 level of significance is tois 0.5. It is suspected that the average turnover is not 6.0. The 0.05 level of significance is to

be used to test this hypothesis. be used to test this hypothesis.

1.1. State HState Hoo, ad H, ad H11

2.2. What is the value of What is the value of αα? ?

3.3. Give the formula for the test statistic Give the formula for the test statistic

4.4. State the decision rule State the decision rule

5.5. A random sample of 64 bottles of a brand was selected. The mean turn over rateA random sample of 64 bottles of a brand was selected. The mean turn over rate

computed to be 5.84. Shall we reject the null hypothesis at the 0.05 levels?computed to be 5.84. Shall we reject the null hypothesis at the 0.05 levels?

Interpret. Interpret.

Solution:Solution:

1.1. HHoo: : µµ = 6.00 = 6.00

HH11: : µµ ≠≠ 6.00 6.00

2.2. 0.05 0.05

3.3. Z = Z = n

Xσ

µ−

110

4.4. Do not reject the null hypothesis if the computed Z value fales between – 1.96 andDo not reject the null hypothesis if the computed Z value fales between – 1.96 and

+ 1.96 + 1.96

5.5. Z = Z = 64

5.000.684.5 −

= 2.56 = 2.56

6.6. reject Ho at the 0.05 level. Accept H1 the mean turnover is not equal to 6.00. reject Ho at the 0.05 level. Accept H1 the mean turnover is not equal to 6.00.

A one Tailed TestA one Tailed Test

If the alternate hypothesis states a direction (either greater than “ or “ less than”) the test isIf the alternate hypothesis states a direction (either greater than “ or “ less than”) the test is

one tailed. The hypothesis – testing procedure is generally the same as for a two – tailed test,one tailed. The hypothesis – testing procedure is generally the same as for a two – tailed test,

except that the critical value is different. except that the critical value is different.

Let us change the alternate hypothesis in the previous problem, involving efficing racting ofLet us change the alternate hypothesis in the previous problem, involving efficing racting of

worker worker

HH11: : µµ ≠≠ 200 (tow – tailed test) to 200 (tow – tailed test) to

HH11: : µµ > 200 ( a one – tailed test ) > 200 ( a one – tailed test )

The critical values for the two – tailed test were -2.58 and +2.58. The region of rejection for aThe critical values for the two – tailed test were -2.58 and +2.58. The region of rejection for a

one – tailed test is in the right tail of the curve one – tailed test is in the right tail of the curve

For a one-tailed test the critical value is found by For a one-tailed test the critical value is found by

a. 0.5000 – 0.01 = 0.4900 a. 0.5000 – 0.01 = 0.4900

b. The Z value for 0.4900 = probability is b. The Z value for 0.4900 = probability is ±± 2.33 2.33

111


The management of chain of restaurants claims that the mean waiting time of customers forThe management of chain of restaurants claims that the mean waiting time of customers for

service is normally distributed with a mean of 3 minutes and a standard deviation of oneservice is normally distributed with a mean of 3 minutes and a standard deviation of one

minute. The quality assurance department found a sample of 50 customers at a restaurant andminute. The quality assurance department found a sample of 50 customers at a restaurant and

that the mean waiting time was 2.75 minutes. At the 0.05 significance level is the meanthat the mean waiting time was 2.75 minutes. At the 0.05 significance level is the mean

waiting time less than 3 minutes? (Note that this test is one tailed)waiting time less than 3 minutes? (Note that this test is one tailed)

P – values is Hypothesis Testing P – values is Hypothesis Testing

Additional value is often reported on the strength of the rejection, or how confident we are inAdditional value is often reported on the strength of the rejection, or how confident we are in

rejecting the null hypothesis. This method reports the probability (assuming that the nullrejecting the null hypothesis. This method reports the probability (assuming that the null

hypothesis is true) of getting a value of the test statistic at least as exterm as that obtained. hypothesis is true) of getting a value of the test statistic at least as exterm as that obtained.

This procedure compares the probability, called P – Value, with the significance level. This procedure compares the probability, called P – Value, with the significance level.

If the P- value is smaller than the significance level, Ho is rejected. If it is larger than theIf the P- value is smaller than the significance level, Ho is rejected. If it is larger than the

significant level Ho is not rejected. This procedure not only results in decision regarding Hosignificant level Ho is not rejected. This procedure not only results in decision regarding Ho

but it gives us in sight into the strength of the decision. but it gives us in sight into the strength of the decision.

A very small P- values say 0.001, means that there is a very little likelihood that Ho is true.A very small P- values say 0.001, means that there is a very little likelihood that Ho is true.

On the other hand, a p- value of 0.4 means that Ho is not rejected, and we did not come veryOn the other hand, a p- value of 0.4 means that Ho is not rejected, and we did not come very

close to rejecting it.close to rejecting it.

Recall that for the efficiency ratings the computed value of Z was 2.19. The decision was notRecall that for the efficiency ratings the computed value of Z was 2.19. The decision was not

to reject Ho because the Z of 2.19 fall in the non-rejection area between 2.58 and + 2.58. Theto reject Ho because the Z of 2.19 fall in the non-rejection area between 2.58 and + 2.58. The

probability of obtaining a Z values of 2.19 or more is 0.0143 found by 0.5000 – 0.4857. Toprobability of obtaining a Z values of 2.19 or more is 0.0143 found by 0.5000 – 0.4857. To

compute the P – value, we need to be concerned with values less than -2.19 and values greatercompute the P – value, we need to be concerned with values less than -2.19 and values greater

than + 2.19. The p- value is 0.0286 found by 2(0.0143). The P – value of 0.0286 is greaterthan + 2.19. The p- value is 0.0286 found by 2(0.0143). The P – value of 0.0286 is greater

than the significance level (0.01) decided upon initially, so Ho is not rejected. than the significance level (0.01) decided upon initially, so Ho is not rejected.

5.4.1.2 Testing for the population mean: (standard deviation unknown) 5.4.1.2 Testing for the population mean: (standard deviation unknown)

112

In the preceding problems, we knew population standard deviation, In the preceding problems, we knew population standard deviation, σσ. In most cases,. In most cases,

however, it is unlikely that however, it is unlikely that σσ would be known. Thus it must be estimated using the sample would be known. Thus it must be estimated using the sample

standard deviation, S. Then the test statistic Z = standard deviation, S. Then the test statistic Z = n

SX µ−

Example: Example:

A department store issues it own credit card. The credit manger wants to find out if the meanA department store issues it own credit card. The credit manger wants to find out if the mean

monthly unpaid balance is more than Br. 400. The level of significance is set at 0.05. Amonthly unpaid balance is more than Br. 400. The level of significance is set at 0.05. A

random check of 172 unpaid balances revealed the sample mean to be 407 and the standardrandom check of 172 unpaid balances revealed the sample mean to be 407 and the standard

deviation of the sample 38. Should the credit manager conclude that the population mean isdeviation of the sample 38. Should the credit manager conclude that the population mean is

greater than 400, or is it reasonable to assume that the difference of 407- 400=7 is due togreater than 400, or is it reasonable to assume that the difference of 407- 400=7 is due to

chance: chance:

Solution Solution

Ho: Ho: µµ =400 =400

Hi: Hi: µµ > 400 > 400

Because HBecause Hll states a direction, a one tailed test is applied. The critical value of Z is 1.645 for states a direction, a one tailed test is applied. The critical value of Z is 1.645 for

0.05 level 0.05 level

Z = Z = n

SX µ−

= = 172

380400407 −

= 2.42 = 2.42

A value of this large (2.42) will occur less than 5% of the time. So the credit manager wouldA value of this large (2.42) will occur less than 5% of the time. So the credit manager would

reject the null hypothesis, Ho. that the mean unpaid balance is greater than 400, in favor ofreject the null hypothesis, Ho. that the mean unpaid balance is greater than 400, in favor of

HH11, which states that the mean is greater than 400. , which states that the mean is greater than 400.

The P – value, in this one – tailed test is the probability that Z is greater than 2.42. Found byThe P – value, in this one – tailed test is the probability that Z is greater than 2.42. Found by

0.5000-0.4922. 0.4922 is the probability that Z can assume a value of 2.420.0.5000-0.4922. 0.4922 is the probability that Z can assume a value of 2.420.


At the time a server was heired at a restaurant was told by the manager that she can averageAt the time a server was heired at a restaurant was told by the manager that she can average

more than 20 br a day in tips. Over the first 35 days she was employed at the restaurant, themore than 20 br a day in tips. Over the first 35 days she was employed at the restaurant, the

mean daily amount of her tips was 24.85 br with a standard deviation of 3.24 br. At the 0.01mean daily amount of her tips was 24.85 br with a standard deviation of 3.24 br. At the 0.01

113

significance level, can the manager conclude that she is earning more than 20 br. per day insignificance level, can the manager conclude that she is earning more than 20 br. per day in

tips?tips?

5.4.2 Hypothesis testing; Two-population means; Independent population 5.4.2 Hypothesis testing; Two-population means; Independent population

Assumption for two-sample testAssumption for two-sample test

1.1. The population should be normally distributedThe population should be normally distributed

2.2. The population standard deviations for both population should be known. If they areThe population standard deviations for both population should be known. If they are

not known, then both samples should contain at least 30 observations so that thenot known, then both samples should contain at least 30 observations so that the

sample standard deviation can be used to approximate the population standardsample standard deviation can be used to approximate the population standard

deviationdeviation

3.3. The samples should be drawn from independent population.The samples should be drawn from independent population.

If we select random samples from two normal population the distribution of the differencesIf we select random samples from two normal population the distribution of the differences

between the two means is also normal or if a large number of independent random samplesbetween the two means is also normal or if a large number of independent random samples

are selected from two population, the difference between the two means will be normallyare selected from two population, the difference between the two means will be normally

distributed. If these differences are divided by the standard error of the difference, the result isdistributed. If these differences are divided by the standard error of the difference, the result is

the standard normal distribution. the standard normal distribution.

The formula for the test statistic Z isThe formula for the test statistic Z is

Z = Z =

2

22

1

21

21

n

S

n

S

xx

+

−

Example:Example: Each patient at a hospital is asked to evaluate the service at the time of discharge. Each patient at a hospital is asked to evaluate the service at the time of discharge.

Recently there have been several complaints that resident physicians and nurses on theRecently there have been several complaints that resident physicians and nurses on the

surgical wing respond too slowly to the emergency calls of senior citizens. The administratorsurgical wing respond too slowly to the emergency calls of senior citizens. The administrator

of the hospital asked the quality assurance department to investigate. After studying theof the hospital asked the quality assurance department to investigate. After studying the

problem, the quality assurance department collected the following sample information. At theproblem, the quality assurance department collected the following sample information. At the

0.01 significance level, is the response time longer for the senior citizens, emergencies? 0.01 significance level, is the response time longer for the senior citizens, emergencies?

Patient type Patient type Smaple meaSmaple mean n Sample standardSample standard Sample Size Sample Size

deviation deviation

Senor Citizens Senor Citizens 5.5 Minutes 5.5 Minutes 0.40 minuets 0.40 minuets 50 50

Other Other 5.3 Minutes 5.3 Minutes 0.30 minutes 0.30 minutes 100 100

114

The difference between two The difference between two sample meanssample means

Standard error of the difference Standard error of the difference between two sample meansbetween two sample means

Solution:- Solution:-

The testing procedure is the same as for one sample test except the formula for the testThe testing procedure is the same as for one sample test except the formula for the test

statistic, Z:statistic, Z:

Step 1: Ho: there is no difference in the mean response time between the two groups ofStep 1: Ho: there is no difference in the mean response time between the two groups of

patients.patients.

i: e The difference of 0.2 minute, in the arithmetic mean response time is due to chances. i: e The difference of 0.2 minute, in the arithmetic mean response time is due to chances.

HH11: the mean response time is greater for the senior citizens: the mean response time is greater for the senior citizens

Because the quality assurance department is concerned that the response time is greater forBecause the quality assurance department is concerned that the response time is greater for

senior citizens, he wants to conduct a one – tailed test. There fore the null and alternatesenior citizens, he wants to conduct a one – tailed test. There fore the null and alternate

hypotheses are stated as follows. hypotheses are stated as follows.

HHoo: : µµ11 = = µµ22

HH11: : µµ11 > > µµ22

Step 2: The 0.01 significance level is selected. Step 2: The 0.01 significance level is selected.

Step 3: the test statistics is Z, the standard normal distribution, Z = Step 3: the test statistics is Z, the standard normal distribution, Z =

2

22

1

21

21

n

S

n

S

xx

+

−

Step 4: The decision rule is:Step 4: The decision rule is:

Reject the null hypothesis if the computed value of Z is greater then 2.33. Reject the null hypothesis if the computed value of Z is greater then 2.33.

The critical value for 0.01 cruel, one-tailed test is 2.33The critical value for 0.01 cruel, one-tailed test is 2.33

Step 5: Calculate the test statistic and make a decision.Step 5: Calculate the test statistic and make a decision.

The test statistic is Z = The test statistic is Z =

2

22

1

21

21

n

S

n

S

xx

+

−

Z = Z = 100

)30.0(

50

)40.0(

3.55.522

+

−

= 3.13 = 3.13

The computed value of 3.13 is beyond the critical value of 2:33. Therefore, the nullThe computed value of 3.13 is beyond the critical value of 2:33. Therefore, the null

hypothesis is rejected and the alternate hypothesis is accepted at the 0.01 significant level. hypothesis is rejected and the alternate hypothesis is accepted at the 0.01 significant level.

115

The quality assurance department will report to the administrator that the mean response timeThe quality assurance department will report to the administrator that the mean response time

of the nurses and resident physicians is longer for senior citizens than for other patients. of the nurses and resident physicians is longer for senior citizens than for other patients.

What is the P-value in this problem?What is the P-value in this problem?

P- Value is the probability of computing aZ value this large or larger when Ho is true. P- Value is the probability of computing aZ value this large or larger when Ho is true.

What is the likelihood of aZ value greater than 3.13 What is the likelihood of aZ value greater than 3.13

P(Z=3.13)= 0.4991P(Z=3.13)= 0.4991

So, P(Z)>31.13 ) =0.5000-0.44991=0.0009 So, P(Z)>31.13 ) =0.5000-0.44991=0.0009

Ho is very likely false and there is little likelihood of a type I error. Ho is very likely false and there is little likelihood of a type I error.


A peal Estate Association is preparing a pamphlet that they feel might be of interest toA peal Estate Association is preparing a pamphlet that they feel might be of interest to

prospective home buyers in the eastern and western areas of the city. One item of interest isprospective home buyers in the eastern and western areas of the city. One item of interest is

the length of time the seller occupied the home. A sample of 40 home sold recently in thethe length of time the seller occupied the home. A sample of 40 home sold recently in the

eastern areas revealed that the mean length of ownership was 7.6 years with standardeastern areas revealed that the mean length of ownership was 7.6 years with standard

deviation of 2.3 years. deviation of 2.3 years.

A sample of 55 homes in the western areas reaealled that the mean length of ownership wasA sample of 55 homes in the western areas reaealled that the mean length of ownership was

8.1 years with a standard deviation of 2.9 years. At the 0.05 significance level can we8.1 years with a standard deviation of 2.9 years. At the 0.05 significance level can we

conclude that the Eastern residents owned the homes for a shorter period of time? conclude that the Eastern residents owned the homes for a shorter period of time?

5.4.3 Testing for Population Proportion5.4.3 Testing for Population Proportion

In testing hypothesis for the population proportion the assumptions of the binomialIn testing hypothesis for the population proportion the assumptions of the binomial

distribution should be met. To test for the proportiondistribution should be met. To test for the proportion

a)a) np and n(1-p) both should be greater than 5. np and n(1-p) both should be greater than 5.

b)b) n should be at least 50n should be at least 50

Example:Example: suppose prior elections in a region indicated that it is necessary for a candidate for suppose prior elections in a region indicated that it is necessary for a candidate for

governor to receive at least 80% of the majority vote. The incumbent governor is interested ingovernor to receive at least 80% of the majority vote. The incumbent governor is interested in

assessing his chance of returning to office and plans to have a survey conducted consisting ofassessing his chance of returning to office and plans to have a survey conducted consisting of

2000 registered voters2000 registered voters

Using the five – step hypothesis testing procedure, asses the governor’s chances of reflectionUsing the five – step hypothesis testing procedure, asses the governor’s chances of reflection

116

np = 2000(0.8) = 1600 which is greater than 5np = 2000(0.8) = 1600 which is greater than 5

nq = n(1-p) = 2000(1-0.8) = 400 which is greater than 5nq = n(1-p) = 2000(1-0.8) = 400 which is greater than 5

both 1600 and 400 are greater than 5both 1600 and 400 are greater than 5

Step 1: The null hypothesis Ho is that the population proportions is 0.80 Step 1: The null hypothesis Ho is that the population proportions is 0.80

The alternate hypothesis, HThe alternate hypothesis, H11 is that the proportion is less than 0.80. is that the proportion is less than 0.80.

The incumbent governor is concerned only when the sample proportion is less than 0.8. If it isThe incumbent governor is concerned only when the sample proportion is less than 0.8. If it is

equal to or greater than 0.8 he will have no problem; that is the sample data would indicate heequal to or greater than 0.8 he will have no problem; that is the sample data would indicate he

will be probably be reelected. will be probably be reelected.

Ho: P = 0.80Ho: P = 0.80

HH11: P<0.80: P<0.80

Step 2: The level of significance is 0.05 Step 2: The level of significance is 0.05

Step 3: Z is the appropriate statisticStep 3: Z is the appropriate statistic

Z = Z = p

PP

σ−

where P – is the population proportion and where P – is the population proportion and

P is the sample proportion, is the sample proportion, σσpp is the standard error of the proportion is the standard error of the proportion

Pσ = =

n

pp )1( − so the formula for Z becomes : so the formula for Z becomes :

Z = Z = n

PP

pp

)1( −−

Step 4: Step 4:

The area between 0 and the critical value is, 1.645 obtained for the Z table 0.45000 = 0.5000 –The area between 0 and the critical value is, 1.645 obtained for the Z table 0.45000 = 0.5000 –

0.05 Z value for probability 0.450 is 1.645. 0.05 Z value for probability 0.450 is 1.645.

The decision rule is therefore reject the null hypothesis and accept the alternate hypothesis ifThe decision rule is therefore reject the null hypothesis and accept the alternate hypothesis if

the computed value of Z falls to the left of -1.645 otherwise do not reject Ho. the computed value of Z falls to the left of -1.645 otherwise do not reject Ho.

Step 5.Step 5. Take Sample and make a decision with respect to Ho. Take Sample and make a decision with respect to Ho.

The sample survey of 2000 potential voters revealed that 1550 planned to vote for theThe sample survey of 2000 potential voters revealed that 1550 planned to vote for the

incumbent governor. Is the proportion of 0.775 (found by 1550/2000) close enough to 0.80 toincumbent governor. Is the proportion of 0.775 (found by 1550/2000) close enough to 0.80 to

conclude that the difference if due to chance? conclude that the difference if due to chance?

117

n =2000 n =2000

2000

1550=P = 0.775 = 0.775

p = 0.80, the hypothesized population proportion p = 0.80, the hypothesized population proportion

Z = Z = 2000

801.01(8.0

80.0775.0

/)1( −−=

−−

nPP

PP

= -2.80= -2.80

The computed value of Z (-2.80) is in the rejection region. So the null hypothesis is rejected atThe computed value of Z (-2.80) is in the rejection region. So the null hypothesis is rejected at

the 0.05 level of significance. The difference of 2.5 percentage points between the samplethe 0.05 level of significance. The difference of 2.5 percentage points between the sample

(77.5) and the hypothesized population percentage (80.0) is statistically significance. It is(77.5) and the hypothesized population percentage (80.0) is statistically significance. It is

probably not due to sampling variation. probably not due to sampling variation.

To put it another way the evidence at this point does not support the claim that the incumbentTo put it another way the evidence at this point does not support the claim that the incumbent

governor will return to the office. governor will return to the office.

The p- Value is 0.0026 found by 0.5000-0.4974. 0.4974 is the probability of Z to assume The p- Value is 0.0026 found by 0.5000-0.4974. 0.4974 is the probability of Z to assume

–2.80 value. It is less than the significance level of 0.05. So Ho should be rejected. This–2.80 value. It is less than the significance level of 0.05. So Ho should be rejected. This

further indicates that the likelihood that Ho is ture is small. further indicates that the likelihood that Ho is ture is small.


This Claim is to be investigated at the 0.02 level “Forty percent of those persons who retiredThis Claim is to be investigated at the 0.02 level “Forty percent of those persons who retired

from an industrial job before the age of 60 would return to work if a suitable job werefrom an industrial job before the age of 60 would return to work if a suitable job were

available” 74 persons out of the 200 sampled said they would return to work. available” 74 persons out of the 200 sampled said they would return to work.

Can we conclude that the fraction returning to work is different from 0.40? Can we conclude that the fraction returning to work is different from 0.40?

1)1) Can the Z test be used? Why or why not? Can the Z test be used? Why or why not?

2)2) State the null hypothesis and the alternate hypothesis State the null hypothesis and the alternate hypothesis

3)3) Compute Z, and arrive at a decision Compute Z, and arrive at a decision

5.4.4 Testing for the Difference between two Population Proportions5.4.4 Testing for the Difference between two Population Proportions

Example:Example: - a company has developed a new perfume - a company has developed a new perfume

One of the questions is whether the perfume is preferred by a larger proportion of youngerOne of the questions is whether the perfume is preferred by a larger proportion of younger

women or a larger proportions of older women. A standard smell test is used. women or a larger proportions of older women. A standard smell test is used.

118

Women selected at random are asked to sniff several perfumes in succession, including theWomen selected at random are asked to sniff several perfumes in succession, including the

new. Each woman selects the perfume she likes best. new. Each woman selects the perfume she likes best.

Step 1 Step 1

Ho “ There is no difference between the proportion of younger women who prefer theHo “ There is no difference between the proportion of younger women who prefer the

perfume and the proportion of older women who prefer it” If the proportion of youngerperfume and the proportion of older women who prefer it” If the proportion of younger

women in the population is designated as Pwomen in the population is designated as P11 and the proportion of older women is P and the proportion of older women is P22 then; then;

Ho: PHo: P11= P= P22

The alternate hypothesis is that the two proportions are not equal or: The alternate hypothesis is that the two proportions are not equal or:

Hi: PHi: P11 ≠≠ P P22

Step 2:Step 2: It was decided to use the 0.05 level. It was decided to use the 0.05 level.

Step 3:Step 3: The test statistic is Z and the formula is: - The test statistic is Z and the formula is: -

where: nwhere: n11 , is the number of young women selected , is the number of young women selected

in the sample nin the sample n22 is the number of older women is the number of older women

selected in the sample, selected in the sample, cP = is the weighted mean = is the weighted mean

of the two sample proportion computed byof the two sample proportion computed by

cP == samples ofnumber Total

successe ofnumber Total = =

21

21

nn

xx

++

where xwhere x11 is the number of younger women is the number of younger women

(sample 1) who prefer the perfume, x(sample 1) who prefer the perfume, x22 is the is the

number of older women (sample 2) who prefer thenumber of older women (sample 2) who prefer the

perfume. perfume.

cP is generally referred to as the pooled estimate of the population proportion or it is a is generally referred to as the pooled estimate of the population proportion or it is a

combined estimate, combined proportion. combined estimate, combined proportion.

Step 4: The Formulate Decision Rule:Step 4: The Formulate Decision Rule:

The critical values for the 0.05 level two-tailed tests are -1.96 and +1.96. If the computed ZThe critical values for the 0.05 level two-tailed tests are -1.96 and +1.96. If the computed Z

value is in the region between +1.96 and -1.96, the null hypothesis will not be rejected. If itvalue is in the region between +1.96 and -1.96, the null hypothesis will not be rejected. If it

does occur it is assumed that any difference between the two proportions is due to chancedoes occur it is assumed that any difference between the two proportions is due to chance

variation. variation.

Two – tailed test, Areas of rejection and Non-rejection 0.05 level of significance. Two – tailed test, Areas of rejection and Non-rejection 0.05 level of significance.

119

Z = Z =

21

2

)1()1(

n

PP

n

PP

PP

cccc −+−

−

Step 5: The decision Step 5: The decision

A total of 100 young women selected at random, and each was given the standard smell test.A total of 100 young women selected at random, and each was given the standard smell test.

Forty of the 100 young women chose the perfume, as they liked best Forty of the 100 young women chose the perfume, as they liked best

xx11 = 40 = 40

nn11= 100 and= 100 and

200 older women were selected at random and each was given the same standard smell test of200 older women were selected at random and each was given the same standard smell test of

the 200 women 100 preferred the perfume. the 200 women 100 preferred the perfume.

xx22 = 100 = 100

nn22=200 =200

The pooled or weighted proportion The pooled or weighted proportion cP is is

cP = = 21

21

nn

xx

++

= = 200100

10040

++

= 140 / 300 = 0.4667 = 140 / 300 = 0.4667

Z = Z = 64.1

200

5333.04667.0

100

)5333.0(4667.0

50.040

)1()1(

21

21 −=++

−=−+−

−

n

PP

n

PP

PP

cccc

The computed value of Z (-1.64) falls in the non-rejection region. Therefore we concludedThe computed value of Z (-1.64) falls in the non-rejection region. Therefore we concluded

that there is no difference in the proportion of younger and older women who prefer thethat there is no difference in the proportion of younger and older women who prefer the

perfume. In this case we expect the P- value to be greater than the significance level of 0.05,perfume. In this case we expect the P- value to be greater than the significance level of 0.05,

and it is. and it is.

for Z = -1.64 probability is 0.4495for Z = -1.64 probability is 0.4495

P value = 0.5000 – 0.4495 = P value = 0.5000 – 0.4495 = 0.05050.0505 for one tail only for one tail only

However the test was two tailed, so we must account for the area beyond 1.64 as well as theHowever the test was two tailed, so we must account for the area beyond 1.64 as well as the

area less than -1.64. Then area less than -1.64. Then

The P – value is 2(0.0505) = The P – value is 2(0.0505) = 0.10100.1010


120

40.0100

40

1

11 ===n

xP

50.0200

100

2

22 ===

n

xP

Of 150 girls who tried a new candy 87 rated it excellent of 200 boys sampled 123 rated itOf 150 girls who tried a new candy 87 rated it excellent of 200 boys sampled 123 rated it

excellent using the 0.10 level of significance, can we conclude that there is a difference in theexcellent using the 0.10 level of significance, can we conclude that there is a difference in the

proportion of girls versus boys who rate the candy excellent? proportion of girls versus boys who rate the candy excellent?

1.1. State the null and alternate hypotheses State the null and alternate hypotheses

2.2. What is the decision rule What is the decision rule

3.3. Compute the value of the test statistics Compute the value of the test statistics

4.4. State your decision granting Ho State your decision granting Ho

5.5. Compute the P – value Compute the P – value

5.5 STUDENT’S 5.5 STUDENT’S tt TEST/ SMALL SAMPLE/TEST/ SMALL SAMPLE/

When the population is normal and the standard deviation is known the Z distribution isWhen the population is normal and the standard deviation is known the Z distribution is

employed as a test statistic for a test. If the population standard deviation is not know theemployed as a test statistic for a test. If the population standard deviation is not know the

sample standard deviation is substituted for sample standard deviation is substituted for σσ. If the sample size is at least 30, the results are. If the sample size is at least 30, the results are

deemed satisfactory. deemed satisfactory.

If the sample size is less than 30 observations and If the sample size is less than 30 observations and σσ is unknown the Z distribution is not is unknown the Z distribution is not

appropriate. The student’s t or the t distribution is used as the test statistic. appropriate. The student’s t or the t distribution is used as the test statistic.

5.5.1 Characteristics of Student’s t Distribution5.5.1 Characteristics of Student’s t Distribution

Note:Note: The Characteristics of student’s distribution are discussed in unit 4. To mention some The Characteristics of student’s distribution are discussed in unit 4. To mention some

1.1. It is a continuous distribution.It is a continuous distribution.

2.2. It is bell- shaped and symmetrical,It is bell- shaped and symmetrical,

3.3. There is not one distribution, but rather a “family” of t distribution. All have the There is not one distribution, but rather a “family” of t distribution. All have the

small mean of zero but their standard deviations differ according to the sample size n.small mean of zero but their standard deviations differ according to the sample size n.

The t distribution for a sample size of 20,22, 25 are different. The t distribution for a sample size of 20,22, 25 are different.

4.4. It is more spread out and flat at the center than is the Z. However as the sample sizeIt is more spread out and flat at the center than is the Z. However as the sample size

increases, the curve representing the t distribution approaches the Z distribution. If theincreases, the curve representing the t distribution approaches the Z distribution. If the

sample size is 30 we will have approximately the same t distribution as the Z. sample size is 30 we will have approximately the same t distribution as the Z.

Since the t distribution has a greater spread or the tails are wide, the critical values of t for aSince the t distribution has a greater spread or the tails are wide, the critical values of t for a

given level of significance are larger in magnitude than the corresponding Z critical values. given level of significance are larger in magnitude than the corresponding Z critical values.

121

Region of rejection for the Z and t distribution 0.05 level, one tailed testRegion of rejection for the Z and t distribution 0.05 level, one tailed test

Why the critical value for a given level of significance is greater for small samples than forWhy the critical value for a given level of significance is greater for small samples than for

large samples? large samples?

a.a. The confidence interval will be wider than for large samples using the Z distribution The confidence interval will be wider than for large samples using the Z distribution

b.b. The region where Ho is not rejected is wider than for large samples using ZThe region where Ho is not rejected is wider than for large samples using Z


c.c. A larger t value will be needed to reject the null hypothesis than for large samplesA larger t value will be needed to reject the null hypothesis than for large samples

using Z. In other words because there is more variability in sample means computedusing Z. In other words because there is more variability in sample means computed

from smaller samples we are less apt to reject the null hypothesis. from smaller samples we are less apt to reject the null hypothesis.

122

5.5.2 A Test for the Population Mean5.5.2 A Test for the Population Mean

ExampleExample: Experience in investigating accident claims by an insurance company revealed that: Experience in investigating accident claims by an insurance company revealed that

it cost 60 on the average to handle the paper work, pay the investigator, and make a decision.it cost 60 on the average to handle the paper work, pay the investigator, and make a decision.

The cost compared with that of other insurance firms was deemed exorbitant, and cost cuttingThe cost compared with that of other insurance firms was deemed exorbitant, and cost cutting

measures were instituted. In order to evaluate the impact of these new measures, a sample ofmeasures were instituted. In order to evaluate the impact of these new measures, a sample of

26 recent claims was selected at random and cost studies were made. It was found that the26 recent claims was selected at random and cost studies were made. It was found that the

sample mean, sample mean, x , and the sample standard deviations, s, were 57 and 10 respectively. , and the sample standard deviations, s, were 57 and 10 respectively.

At the 0.01 level is there a reduction in the average cost, or can the difference of 3 = (60-57)At the 0.01 level is there a reduction in the average cost, or can the difference of 3 = (60-57)

be attributed to chance? be attributed to chance?

The usual five-step hypothesis testing procedure is used The usual five-step hypothesis testing procedure is used

Step 1:Step 1: - the null hypothesis, Ho: the population mean is 60 - the null hypothesis, Ho: the population mean is 60

The alternate hypothesis, HThe alternate hypothesis, H11 the population mean is less than 60. i.e. the population mean is less than 60. i.e.

Ho: Ho: µµ = 60 = 60

H1:- H1:- µµ < 60 < 60

Step 2:Step 2: The 0.01 level is to be used The 0.01 level is to be used

Step: 3Step: 3 the test statistic is student’s t distribution. Because the population standard deviation the test statistic is student’s t distribution. Because the population standard deviation

is unknown and the sample size is small (26 under 30) is unknown and the sample size is small (26 under 30)

t = t = nS

X

/

µ−

Step 4:Step 4: The critical value of t are given in table 4 The critical value of t are given in table 4

There are n -1 degrees of freedom for the test df (26-1= 25) There are n -1 degrees of freedom for the test df (26-1= 25)

The critical value for df = 25, a one tailed test and 0.01 level is 2.485 The critical value for df = 25, a one tailed test and 0.01 level is 2.485

The decision rule for this one tailed test is reject Ho if the computed value of t falls in any partThe decision rule for this one tailed test is reject Ho if the computed value of t falls in any part

of the tails to the left of –2.485 otherwise do not reject Ho. of the tails to the left of –2.485 otherwise do not reject Ho.

Ho: N= 60 Ho: N= 60

123

Ho; Ho; µµ = 60 = 60

H1: H1: µµ < 60 < 60

df = 26 – 1 = 25df = 26 – 1 = 25

Step 5:Step 5: Compute t, and arrive at a decision Compute t, and arrive at a decision

t = t = nS

X

/

µ−

X = 57 = 57

µµ = 67 = 67

S = 10S = 10

n = 26n = 26

Because -1.530 lies in the region to the right of the critical value –2.485 Ho is not rejected atBecause -1.530 lies in the region to the right of the critical value –2.485 Ho is not rejected at

the 0.01 level. the 0.01 level.

This indicates that the cost cutting measures have not reduced the mean cost per claim to lessThis indicates that the cost cutting measures have not reduced the mean cost per claim to less

than 60 based on sample results. than 60 based on sample results.


From past records it is known that the arithmetic mean life of a battery used in a digital clockFrom past records it is known that the arithmetic mean life of a battery used in a digital clock

is 305 days. The lives of the batteries is normally distributed. The battery was recentlyis 305 days. The lives of the batteries is normally distributed. The battery was recently

modified to last longer. A sample of 20 modified batteries were tested. It was discovered thatmodified to last longer. A sample of 20 modified batteries were tested. It was discovered that

the man life was 311 days and the sample standard deviation was 12 days. At the 0.05 level ofthe man life was 311 days and the sample standard deviation was 12 days. At the 0.05 level of

significance, did the modification increases the mean life of the battery? significance, did the modification increases the mean life of the battery?

1.1. State the null and alternate hypotheses State the null and alternate hypotheses

2.2. State the decisionState the decision

124

t = t = 26/10

6057 −= -1.530= -1.530

3.3. Compute t and make a decision Compute t and make a decision

5.5.3 Comparing two Population Mean5.5.3 Comparing two Population Mean

A test using the t distribution can also be applied to compare two sample means to determineA test using the t distribution can also be applied to compare two sample means to determine

if the samples were obtained from normal populations with the same mean. if the samples were obtained from normal populations with the same mean.

Three assumption are required to test for two population means.Three assumption are required to test for two population means.

1.1. The populations must be normally distributed (or approximately normally distributed) The populations must be normally distributed (or approximately normally distributed)

2.2. The populations must be independent The populations must be independent

3.3. The population variance must be equal The population variance must be equal

The statistic for the two sample is similar to that employed for the Z statistic except that anThe statistic for the two sample is similar to that employed for the Z statistic except that an

additional calculation is required. additional calculation is required.

The two-sample variance must be polled to form a single estimate of the unknown populationThe two-sample variance must be polled to form a single estimate of the unknown population

variance. Since the samples have fewer than 30 observations the population standardvariance. Since the samples have fewer than 30 observations the population standard

deviations, are not known. So, we substitute Sdeviations, are not known. So, we substitute S22 for for σσ22, because we assume that the two, because we assume that the two

populations have equal variances, the best estimate we can make of that value is to combinepopulations have equal variances, the best estimate we can make of that value is to combine

or pool all the information we have with respect to the population variance. or pool all the information we have with respect to the population variance.

The following formula is used to pool the sample variances. Notice that two factors make upThe following formula is used to pool the sample variances. Notice that two factors make up

the weights: - the number of observations in each sample and the sample variancesthe weights: - the number of observations in each sample and the sample variances

themselves. The pooled variance, themselves. The pooled variance, SpSp22 is is

SpSp22 = =

2

)1()1(

21

222

21

−+−+−

nn

SnSn

where Swhere S1122 – variance of sample one – variance of sample one

S S222 2 – variance of sample two and – variance of sample two and

n n11 + n + n22 – 2 is total df. – 2 is total df.

The value of t is then determined by the formulaThe value of t is then determined by the formula

t = t =

+

−

21

2

21

11

nnS

XX

p

125

where: where: 1X is sample mean one is sample mean one

2X is sample mean two is sample mean two

n n11 is sample size for first sample is sample size for first sample

n n22 is sample size for second sample is sample size for second sample

The number of degrees of freedom in the test is equal to the total number of items sampledThe number of degrees of freedom in the test is equal to the total number of items sampled

minus the number of sample. Since there are two samples, there are minus the number of sample. Since there are two samples, there are

nn11+ n+ n22 – 2 degrees of freedom. – 2 degrees of freedom.

Example:Example: Two different procedures are proposed for mounting engine on a frame. The Two different procedures are proposed for mounting engine on a frame. The

question is: ‘is there a difference in the mean time to mount the engine on the frame?’ Toquestion is: ‘is there a difference in the mean time to mount the engine on the frame?’ To

evaluate the two proposed methods, it was decided to conduct a time and motion study. Aevaluate the two proposed methods, it was decided to conduct a time and motion study. A

sample of five employees were timed using procedure 1 and 6 were timed using procedures 2.sample of five employees were timed using procedure 1 and 6 were timed using procedures 2.

The results in minutes, are:The results in minutes, are:

Procedure 1 Procedure 1

(Minutes)(Minutes)Procedure 2Procedure 2( Minutes )( Minutes )

3322 7744 5599 8833 4422 33

Is there a difference is the mean mounting times? Use the 0.10 significance level. Is there a difference is the mean mounting times? Use the 0.10 significance level.

Solution : Solution :

The null hypothesis states that there is no difference in mean mounting time between the twoThe null hypothesis states that there is no difference in mean mounting time between the two

procedures and the alternate hypothesis states that there in a difference is the mean mountingprocedures and the alternate hypothesis states that there in a difference is the mean mounting

time between the two procedures.time between the two procedures.

Step I.Step I. Ho: Ho: µµ11 = = µµ22 HH11: : µµ11 ≠≠ µµ22

The required assumptions are met. The required assumptions are met.

The degrees of freedom are determined by nThe degrees of freedom are determined by n11 + n + n22 – 2 there are 9 degrees of freedom – 2 there are 9 degrees of freedom

(5 + 6-2). (5 + 6-2).

Step II.Step II. The 0.01 level is to be used The 0.01 level is to be used

126

Step III.Step III. The test statistic is t = The test statistic is t = ( )21

112

21

nnps

xx

+

−

Step IVStep IV. The critical value of t for df = 9, a two tailed test, at the 0.10 level of significance,. The critical value of t for df = 9, a two tailed test, at the 0.10 level of significance,

are + 1.833 and -1.833 are + 1.833 and -1.833

We do not reject the null hypothesis if the computed t value falls between -1.853 and +1.833We do not reject the null hypothesis if the computed t value falls between -1.853 and +1.833

otherwise Ho is rejected. otherwise Ho is rejected.

Calculate t and make the decisionCalculate t and make the decision

(a)(a) Calculate the sample varianceCalculate the sample variance

Procedure 1 Procedure 1 Procedure 2Procedure 2

X X1 1 XX1122 XX22 XX22

22

2 4 2 4 3 9 3 9

4 16 4 16 7 49 7 49

9 81 9 81 5 25 5 25

3 9 3 9 8 64 8 64

2 2 4 4 4 16 4 16

∑∑xx11= 20 = 20 ∑∑xx1122 114 114 3 3 9 9

∑∑xx11 = 30 = 30 ∑∑xx2222 = 172 = 172

SS1122 = =

( )

11

2

121

−

−∑ ∑

n

n

xx

SS2211 = =

( )

12

2

2

222

−

−∑ ∑

n

n

xx

= = 155

)20(144

2

−

− = 8.5 = 8.5 = =

166

)30(172

2

−

−= 4.44= 4.44

(b)(b) Pool the variancesPool the variances

127

SSpp22 = = 2222.6

265

)4.4)(16()5.8)(15(

2

)1()1(

21

222

21 =

−+−+−=

−+−+−

nn

SnSn

(c)(c) Determine tDetermine t

5201 =X = 4 and = 4 and

6

302 =X = 5 = 5

t = t = ( ) ( )61

51112

21

2222.6

54

21+

−=+

−

nnSp

XX= -0.6626= -0.6626

The decision is not to reject Ho because -0.6620 falls in the region between -1.833 and +The decision is not to reject Ho because -0.6620 falls in the region between -1.833 and +

1.833.We conclude that there is no difference in the mean time to mount the engine on the1.833.We conclude that there is no difference in the mean time to mount the engine on the

frame. frame.


The net weight of sample of bottles filled by two different machines produced by twoThe net weight of sample of bottles filled by two different machines produced by two

different manufactures, are ( in grams ) different manufactures, are ( in grams )

Machine 1-5,8,7,6,9,7 Machine 1-5,8,7,6,9,7

Machnies 2-8,10,11,9,12,14,9 Machnies 2-8,10,11,9,12,14,9

At the 0.05 level is the mean might of the bottled filed by machine 2 are greater than the meanAt the 0.05 level is the mean might of the bottled filed by machine 2 are greater than the mean

weight of the bottles filled by machine 1? (Note that the test is one tailed)weight of the bottles filled by machine 1? (Note that the test is one tailed)

5.5.4 Hypothesis Testing Involving Paired Observations5.5.4 Hypothesis Testing Involving Paired Observations

There are situations where the samples are not independent. A particular group will beThere are situations where the samples are not independent. A particular group will be

exposed to two different experiments. In a sense the sample is one.exposed to two different experiments. In a sense the sample is one.

Example:Example: The production manager wants to find out whether a unique training program will The production manager wants to find out whether a unique training program will

increase employee efficiency.increase employee efficiency.

He plans to take a random sample of 10 employees and record their efficiency before theHe plans to take a random sample of 10 employees and record their efficiency before the

training starts. After completion of the program, the efficiency of the same sample oftraining starts. After completion of the program, the efficiency of the same sample of

employees will be recorded. employees will be recorded.

128

Thus there will be a pair of efficiency ratings for each member of the sample. A test ofThus there will be a pair of efficiency ratings for each member of the sample. A test of

hypothesis is conducted to find out if there is a difference between the ratings before and afterhypothesis is conducted to find out if there is a difference between the ratings before and after

the training program. It is called a paired difference testthe training program. It is called a paired difference test

The sample dates are The sample dates are

SampleSample member member

Efficiency RatingsEfficiency Ratings

Difference (d) Difference (d) difference /// difference /// squared dsquared d22

BeforeBefore AfterAfter

11 128128 135135 77 494922 105105 110110 55 252533 119119 131131 1212 14414444 140140 142142 22 4455 9898 105105 77 494966 123123 130130 77 494977 127127 131131 44 161688 115115 110110 -5-5 252599 122122 125125 33 991010 145145 149149 44 1616 ∑∑d = 46 d = 46 ∑∑dd22 = 386 = 386

For the test of hypothesis to be conducted, there is essentially only one sample, not two. WeFor the test of hypothesis to be conducted, there is essentially only one sample, not two. We

are testing the hypothesis that the distribution of the differences has a mean of 0. are testing the hypothesis that the distribution of the differences has a mean of 0.

The sample is made up of the differences b/n the efficiency ratings before the trainingThe sample is made up of the differences b/n the efficiency ratings before the training

program and the ratings after the program.program and the ratings after the program.

If production methods before and after the training program remain the same, one couldIf production methods before and after the training program remain the same, one could

logically expect some employees to benefit from the training program and to become morelogically expect some employees to benefit from the training program and to become more

efficient. Other employees would prefer the method used before the training program,. Andefficient. Other employees would prefer the method used before the training program,. And

their efficiency would remain the same or even decrease. Thus the mean of the difference intheir efficiency would remain the same or even decrease. Thus the mean of the difference in

efficiency ratings designated efficiency ratings designated µµdd would balance out and equal zero. would balance out and equal zero.

129

The production manager wants to know whether or not the new production technique affectThe production manager wants to know whether or not the new production technique affect

efficiency. If it does one would reasonably assume that most of the difference would beefficiency. If it does one would reasonably assume that most of the difference would be

positive i.e. increased efficiency.positive i.e. increased efficiency.

The null hypothesis to be tested is therefore; the mean difference is zero or there is noThe null hypothesis to be tested is therefore; the mean difference is zero or there is no

difference in the efficiency ratings before and after the training.difference in the efficiency ratings before and after the training.

Ho: Ho: µµd = 0. d = 0.

The alternate hypothesis is that the mean of the difference is greater than O The alternate hypothesis is that the mean of the difference is greater than O

HH11: : µµdd > 0, signifying that the differences are positive. > 0, signifying that the differences are positive.

The test statistic t is The test statistic t is

t = t = nS

d

d /

where where d = the mean difference i.e., = the mean difference i.e., n

d∑

S Sdd = standard deviation of the differences between the paired observations = standard deviation of the differences between the paired observations

The standard deviation of the differences is computed as The standard deviation of the differences is computed as

S Sdd = = ( )

1

2

2

−

−∑ ∑

nn

dd

The critical value of t for this one tailed test of paired difference for 9 degree of freedom atThe critical value of t for this one tailed test of paired difference for 9 degree of freedom at

the 0.05 level is 1.833the 0.05 level is 1.833

d = = n

d∑ = = 10

46 = 4.60 = 4.60

SSdd = = ( )

1

2

2

−

−∑ ∑

nn

dd = =

11010

)46(386

2

−

− = 4.40 = 4.40

130

t = t = nSd

d

/ = =

10/4.4

6.4 = 3.33 = 3.33

Because the value of t (3.30) lies in the rejection rejoin, that is beyond the critical value ofBecause the value of t (3.30) lies in the rejection rejoin, that is beyond the critical value of

1.833, the null hypothesis is rejected.1.833, the null hypothesis is rejected.

The production manger has convincing evidence that this special training program will beThe production manger has convincing evidence that this special training program will be

effective in increasing efficiency.effective in increasing efficiency.


AnAn Agricultural Experimental Station plans to test the effectiveness of two solutions for cornAgricultural Experimental Station plans to test the effectiveness of two solutions for corn

seeds to increases resistance for a particular type of pest and increase germination and growthseeds to increases resistance for a particular type of pest and increase germination and growth

times. The purpose of the experiment is to determine if there is a difference in effectivenesstimes. The purpose of the experiment is to determine if there is a difference in effectiveness

of two solutions, solution A and solution B.of two solutions, solution A and solution B.

Various corn seeds are to be used in the experiment. A pair of seeds is selected one is soakedVarious corn seeds are to be used in the experiment. A pair of seeds is selected one is soaked

in solution A, the other in solution B. Then they are planted and the germination and growthin solution A, the other in solution B. Then they are planted and the germination and growth

times in days are recorded.times in days are recorded.

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PairPairSolutionSolution 11 22 33 44 55 66 77 88 99

AA 11

66

99 22

11

11

44

22

66

22

77

11

88

11

44

3030

BB 11

88

77 22

66

11

11

22

66

22

77

11

99

22

00

2828

1.1. State the null and alternative hypothesis. State the null and alternative hypothesis.

2.2. Using the 0.05 level what is the critical value?Using the 0.05 level what is the critical value?

3.3. Using the above nine pairs of sample compute t and arrival at a decision.Using the above nine pairs of sample compute t and arrival at a decision.

5.6 TESTING FOR DIFFERENCES OF VARIANCES / THE ‘F’ DISTRIBUTION5.6 TESTING FOR DIFFERENCES OF VARIANCES / THE ‘F’ DISTRIBUTION

FOR COMPARING TWO POPULATION VARIANCESFOR COMPARING TWO POPULATION VARIANCES

Determining whether or not one normal population has more variation than the other isDetermining whether or not one normal population has more variation than the other is

important for many decision-making purposes in business. important for many decision-making purposes in business.

Suppose two machines are set to produce steel bars of the same length. The bars, therefore,Suppose two machines are set to produce steel bars of the same length. The bars, therefore,

should have the same mean length. We want to ensure that, in addition to having the sameshould have the same mean length. We want to ensure that, in addition to having the same

mean length, they have similar variation. ormean length, they have similar variation. or

The mean rate of return on investment of two types of projects may be the same. But thereThe mean rate of return on investment of two types of projects may be the same. But there

may be more variation in the return of one than the other. Decision, as to which project ismay be more variation in the return of one than the other. Decision, as to which project is

more feasible, is based on the level of variation. more feasible, is based on the level of variation.

The F distribution is used to test the hypothesis that the variation of one normally distributedThe F distribution is used to test the hypothesis that the variation of one normally distributed

population equals the variance of another normally distributed population.population equals the variance of another normally distributed population.

The major characteristics of the F distribution:The major characteristics of the F distribution:

a.a. There is A “Family” of F distribution. A particular member of the family isThere is A “Family” of F distribution. A particular member of the family is

determined by two parameters, the degree of freedom in the numerator, and the degreedetermined by two parameters, the degree of freedom in the numerator, and the degree

of freedom in the denominator. There is only one F distribution for the combination ofof freedom in the denominator. There is only one F distribution for the combination of

29 degree of freedom in the numerator and 28 degrees of freedom in the denominator.29 degree of freedom in the numerator and 28 degrees of freedom in the denominator.

The shape of the curves changes as the degrees of freedom change.The shape of the curves changes as the degrees of freedom change.

b.b. F cannot be negative; and it is a continuous distribution.F cannot be negative; and it is a continuous distribution.

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c.c. The curve representing an F distribution is positively skewed. F can not be negative. The curve representing an F distribution is positively skewed. F can not be negative.

d.d. Its value range from 0 to positive infinite (Its value range from 0 to positive infinite (∞∞. As the value of F increases the curve. As the value of F increases the curve

approaches the x-axis, but it never touches it.approaches the x-axis, but it never touches it.

For all investigations the null hypothesis is that ‘the variance of one normal population For all investigations the null hypothesis is that ‘the variance of one normal population σσ1122,,

equals the variance of the other normal population equals the variance of the other normal population σσ2222.’ To conduct the test, a random sample.’ To conduct the test, a random sample

of n, observations is obtained from one population and a sample of nof n, observations is obtained from one population and a sample of n22 observations is obtained observations is obtained

from the second population. The test statistics isfrom the second population. The test statistics is

22

21

SS

, where S , where S1122 and S and S22

22 are the respective sample variances. are the respective sample variances.

If the null hypothesis is true Ho: If the null hypothesis is true Ho: σσ1122 = = σσ22

22. .

The test statistic follows the F distribution with nThe test statistic follows the F distribution with n11 – 1 and n – 1 and n22 – n – n11 degrees of freedom degrees of freedom

The larger sample variance is placed in the numerator; hence, the F ratio is always positiveThe larger sample variance is placed in the numerator; hence, the F ratio is always positive

and greater than one. Thus, the upper-tail critical value is the only one required. The criticaland greater than one. Thus, the upper-tail critical value is the only one required. The critical

value of F is found by dividing the significance level in half value of F is found by dividing the significance level in half ( )2α and then referring to theand then referring to the

appropriate number of degrees of freedom in the F table. appropriate number of degrees of freedom in the F table.

Example:Example: A car rental offers limousine service from city center to the airport. The manager A car rental offers limousine service from city center to the airport. The manager

of the company is considering two routes. He wants to conduct a study of both routs and thenof the company is considering two routes. He wants to conduct a study of both routs and then

133

compare the results. He recorded the following data. Using the 0.10 significance level, iscompare the results. He recorded the following data. Using the 0.10 significance level, is

there a difference in the variation in the two routes?there a difference in the variation in the two routes?

RouteRoute Mean Mean Time Time StandardStandard SampleSample

(minutes) (minutes) deviation deviation sizesize

(Minutes)(Minutes)

11 56 56 12 12 7 7

22 59 59 5 5 8 8

The manager noted that the mean times seem very similar but there is more variation, asThe manager noted that the mean times seem very similar but there is more variation, as

measured by the standard deviation, in route 1,measured by the standard deviation, in route 1,

The reason can be route 1 contains more stoplights, while the distance is shorter but for rout 2The reason can be route 1 contains more stoplights, while the distance is shorter but for rout 2

the distance is longer but it is a limited access high way. So he decides to conduct a statisticalthe distance is longer but it is a limited access high way. So he decides to conduct a statistical

test to determine if there is really a difference in the variation of the two routes.test to determine if there is really a difference in the variation of the two routes.

The usual five-step hypothesis testing procedure will be employed.The usual five-step hypothesis testing procedure will be employed.

Step 1:Step 1: the test is two-tailed because we are looking for a difference in the variation of the the test is two-tailed because we are looking for a difference in the variation of the

two routes. We are not trying to show that one route has more variation than the other.two routes. We are not trying to show that one route has more variation than the other.

Ho: Ho: σσ1122 = = σσ22

22

HH11: : σσ1122 ≠σ≠σ22

22

Step 2:Step 2: A significant level of 0.10 is selected. A significant level of 0.10 is selected.

Step 3:Step 3: the appropriate test statistic is F distribution the appropriate test statistic is F distribution F = F = 22

21

S

S

Step 4:Step 4: The decision rule is obtained from the F table because we are using a two tailed test The decision rule is obtained from the F table because we are using a two tailed test

the significance level is 0.05 found by the significance level is 0.05 found by 2

10.02 =α there are n there are n11 – 1 = 7-1 = 6 degrees of – 1 = 7-1 = 6 degrees of

freedom in the numerator and nfreedom in the numerator and n22 – 1 = 8 – 1 =7 degrees of freedom in he denominator. The – 1 = 8 – 1 =7 degrees of freedom in he denominator. The

134

critical value for a 0.05 level and df(7,6) is 3.87. If the ratio of the sample variances critical value for a 0.05 level and df(7,6) is 3.87. If the ratio of the sample variances 22

21

SS

exceed 3.87 the null hypothesis is rejected.exceed 3.87 the null hypothesis is rejected.

Step 5:Step 5: the computed value of the tests statistic is 5.70 = the computed value of the tests statistic is 5.70 = 2

2

22

21

)5(

)12(=S

S

The null hypothesis is rejected and the alternate hypothesis accepted. The variation is not theThe null hypothesis is rejected and the alternate hypothesis accepted. The variation is not the

same in the two pouts. same in the two pouts.

The usual procedure is to determine the F ratio by putting the larger variance in theThe usual procedure is to determine the F ratio by putting the larger variance in the

numerator. This will force the F ratio to be larger than 1.0. Why is this necessary?numerator. This will force the F ratio to be larger than 1.0. Why is this necessary?

It allows us to always use the upper tail of the F statistic thus avoiding the need for moreIt allows us to always use the upper tail of the F statistic thus avoiding the need for more

extensive F tables.extensive F tables.

How a one-failed testis to be handled: Again we will arrange the F ratio so that it is alwaysHow a one-failed testis to be handled: Again we will arrange the F ratio so that it is always

greater that 1.00. Under these conditions it is not necessary to divide the level of significancegreater that 1.00. Under these conditions it is not necessary to divide the level of significance

in half. We are there fore restricted to the 0.05 of 0.1 level and significance for one-tailedin half. We are there fore restricted to the 0.05 of 0.1 level and significance for one-tailed

tests in the F table.tests in the F table.


A company assembles electrical components. For the last 10 days employee ‘A’ averaged 9A company assembles electrical components. For the last 10 days employee ‘A’ averaged 9

rejects per day with a standard deviation of 2 rejects. Employee ‘B’ averaged 8.5 rejects perrejects per day with a standard deviation of 2 rejects. Employee ‘B’ averaged 8.5 rejects per

day with a standard deviation of 1.5 rejects over that same period. At the 0.05 level, can weday with a standard deviation of 1.5 rejects over that same period. At the 0.05 level, can we

conclude that there is more variation in the number of rejects per day attributed to employedconclude that there is more variation in the number of rejects per day attributed to employed

A? (Note that the givens are standard deviations not variances. The test is one-tailed)A? (Note that the givens are standard deviations not variances. The test is one-tailed)


1.1. Reject HoReject Ho Z = 1.767 > 1.645Z = 1.767 > 1.645

2.2. Reject Ho;Reject Ho; Z = 8.86 > 2.33Z = 8.86 > 2.33

3.3. Do not reject Ho; Do not reject Ho; Z = 0.936 > 1.645Z = 0.936 > 1.645

4.4. 1) Yes, because Np and n(1 – p) or nq exceeds 1) Yes, because Np and n(1 – p) or nq exceeds

135

2) Ho: p = 0.402) Ho: p = 0.40

p p ≠≠ 0.40 0.40

3) Do not reject Ho 3) Do not reject Ho Z = -0.866 > -2.58Z = -0.866 > -2.58

5.5. 1) Ho: P1) Ho: P11 = P = P22 H: PH: P11 ≠≠ P P22

2) Reject Ho if Z < -1.645 or > 1.6452) Reject Ho if Z < -1.645 or > 1.645

3) 3) 20150

12387

++=cP = 0.6 = 0.6

Z = Z = 200

)4.0(6.0150

)4.0(6.0

615.055.0

+

−

4) Do not reject Ho4) Do not reject Ho

5) P-value = 2(0.5000 – 0.2454) = 5) P-value = 2(0.5000 – 0.2454) = 0.50920.5092

6.6. 1) Ho: 1) Ho: µµ = 305 = 305 HH11: : µµ > 305 > 305

2) Reject Ho if t > 1.7292) Reject Ho if t > 1.729

3) t = 2.2363) t = 2.236

reject Ho, the mean is greater than 305 agereject Ho, the mean is greater than 305 age

7.7. 1) Ho: 1) Ho: µµ11 = = µµ22 HH11 µµ11 < < µµ22

Reject Ho if t < -1.782Reject Ho if t < -1.782

t = -2.827t = -2.827

reject Horeject Ho

8. Ho: 8. Ho: µµdd = 0 = 0 HH11 µµdd ≠≠ 0 0

critical values are –2.306 and +2.306critical values are –2.306 and +2.306

t = t = 9

667.3

22.0=n

Sd

d= 0.180= 0.180

Do not reject Ho. There is no difference between the two solutionsDo not reject Ho. There is no difference between the two solutions

8.8. Critical value is 3.18Critical value is 3.18

If F > 3.18 reject HoIf F > 3.18 reject Ho

F = F = 2

2

)5.1(

)2(= 1.78= 1.78

Do not reject HoDo not reject Ho

136


1.1. Test for the population meanTest for the population mean

An educator claims that the average IQ of city collage students is no more than 110.An educator claims that the average IQ of city collage students is no more than 110.

To test this claim, a random sample of 150 was taken and given relevant tests. TheirTo test this claim, a random sample of 150 was taken and given relevant tests. Their

average IQ score came to 111.2 with a standard deviation of 7.2. At level ofaverage IQ score came to 111.2 with a standard deviation of 7.2. At level of

significance of 0.01, test if the claim of the educator is justified.significance of 0.01, test if the claim of the educator is justified.

2.2. Test for two population meansTest for two population means

A potential buyer of electric bulbs bought 100 bulbs each of two famous brands, AA potential buyer of electric bulbs bought 100 bulbs each of two famous brands, A

and B. Upon testing both these samples, he found that brand A had a mean life of 1500and B. Upon testing both these samples, he found that brand A had a mean life of 1500

hours with standard deviation of 50 hours whereas brand B had an average life of 1530hours with standard deviation of 50 hours whereas brand B had an average life of 1530

hours with a standard deviation of 60 hours. Can it be concluded at 5% level ofhours with a standard deviation of 60 hours. Can it be concluded at 5% level of

significance that the two brands differ significantly in quality?significance that the two brands differ significantly in quality?

3.3. Test for the population proportionTest for the population proportion

A sociologist taken a survey of the previous lottery winners of one million br. hasA sociologist taken a survey of the previous lottery winners of one million br. has

taken and found that 80% of these winners continue to work on their job. Ataken and found that 80% of these winners continue to work on their job. A

psychologist felt otherwise. To test the report of the state, he took a sample of 100psychologist felt otherwise. To test the report of the state, he took a sample of 100

such winners at random and found that only 25 winners of this sample had quit theirsuch winners at random and found that only 25 winners of this sample had quit their

jobs. At 95% confidence level, can we conclude that the state report is correct?jobs. At 95% confidence level, can we conclude that the state report is correct?

4.4. Test for the difference between two population proportionsTest for the difference between two population proportions

Random samples of 2000 people in town A and 3000 in town B were asked if theyRandom samples of 2000 people in town A and 3000 in town B were asked if they

thought there was too much violence on TV these days. 1400 people in town A andthought there was too much violence on TV these days. 1400 people in town A and

1800 people in town B replied in the affirmative. Can we conclude at 99% confidence1800 people in town B replied in the affirmative. Can we conclude at 99% confidence

level that proportions are significantly different?level that proportions are significantly different?

5.5. Test for the population meanTest for the population mean

A Home Owners’ Association has determined that the average number of days a houseA Home Owners’ Association has determined that the average number of days a house

was in the market for sale was 90 days, before it was sold. A real estate agencywas in the market for sale was 90 days, before it was sold. A real estate agency

believes that in certain section of Long Island, the average number of days the housesbelieves that in certain section of Long Island, the average number of days the houses

remained in the market before sales was less than 90. It selected a random sample ofremained in the market before sales was less than 90. It selected a random sample of

137

10 homes that were sold in this section in order to justify what it believes. The10 homes that were sold in this section in order to justify what it believes. The

following data represents the number of days that each of these 10 homes stayed in thefollowing data represents the number of days that each of these 10 homes stayed in the

market before sale?market before sale?

87, 95, 78, 83, 110, 75, 82, 92, 90, 8087, 95, 78, 83, 110, 75, 82, 92, 90, 80

At 0.01 level of confidence and assuming that population is approximately normal, isAt 0.01 level of confidence and assuming that population is approximately normal, is

the real estate agency justified in its belief?the real estate agency justified in its belief?

6.6. Test for the two population meansTest for the two population means

Two drug manufacturing companies produce headache remedies. Each companyTwo drug manufacturing companies produce headache remedies. Each company

claims that its drug bring faster-acting relief. A consumer protection agency wants toclaims that its drug bring faster-acting relief. A consumer protection agency wants to

test if one drug brings relief faster than the other. An experiment was performed totest if one drug brings relief faster than the other. An experiment was performed to

compare the mean lengths of time required for bodily absorption of both drugs. 12compare the mean lengths of time required for bodily absorption of both drugs. 12

people selected at random were given dosage of one drug and another 12 peoplepeople selected at random were given dosage of one drug and another 12 people

randomly selected wee given dosage of the second drug. The length of time in minutesrandomly selected wee given dosage of the second drug. The length of time in minutes

for the drugs to reach a specified level in the blood was recorded for both drugs. Thefor the drugs to reach a specified level in the blood was recorded for both drugs. The

means and standard deviations of the two samples are recorded as follows:means and standard deviations of the two samples are recorded as follows:

Drug (1)Drug (1) Drug (2)Drug (2)

X = 10.1X = 10.1 X = 8.9X = 8.9

s = 4.2s = 4.2 s = 3.8s = 3.8

Use a 5% level of significance to test the hypothesis that there is no difference in theUse a 5% level of significance to test the hypothesis that there is no difference in the

mean time required for bodily absorption of these two drugs.mean time required for bodily absorption of these two drugs.

7.7. Comparing two population meansComparing two population means

An industrial engineer consultant has conducted a time and motion study on aAn industrial engineer consultant has conducted a time and motion study on a

particular manufacturing assembly operation which he claims would save time. Theparticular manufacturing assembly operation which he claims would save time. The

production manager decides to test new procedure to see if it actually reduces theproduction manager decides to test new procedure to see if it actually reduces the

average assembly time. A random sample of ten assemblers is selected and eachaverage assembly time. A random sample of ten assemblers is selected and each

assembler is timed using the old procedure. Then the same assemblers are givenassembler is timed using the old procedure. Then the same assemblers are given

training in the new procedures and are timed again as they perform the sametraining in the new procedures and are timed again as they perform the same

operation. The following table shows the time in minutes taken for the operation underoperation. The following table shows the time in minutes taken for the operation under

previous procedure and the new procedure:previous procedure and the new procedure:

AssemblerAssembler Old ProcedureOld Procedure New ProcedureNew Procedure

1 1 10.6 10.6 10.0 10.0

138

2 2 6.7 6.7 7.4 7.4

3 3 8.0 8.0 6.1 6.1

4 4 9.5 9.5 6.0 6.0

5 5 7.1 7.1 7.1 7.1

6 6 7.0 7.0 6.0 6.0

7 7 6.4 6.4 5.5 5.5

8 8 6.9 6.9 7.0 7.0

9 9 9.0 9.0 8.6 8.6

10 10 10.0 10.0 9.6 9.6

Assuming that the times under the old procedure, times under the new procedures and henceAssuming that the times under the old procedure, times under the new procedures and hence

the data pf paired differences are all normally distributed, can we conclude at 99% confidencethe data pf paired differences are all normally distributed, can we conclude at 99% confidence

level that the new procedure reduces the average time required for the operation?level that the new procedure reduces the average time required for the operation?

139