GAGE UNIVERSITY COLLEGE

FACULTY OF BUSINESS AND ECONOMICS

STATISTICS FOR MANAGEMENT II

(MGMT 2022)

Address

https://gagecollege.net/

Tel: +251 11 465 2067 /09 05818181 / 09 88415600

P.O. Box: 23115

Email: - adtsm@gagecollege.net

ADDIS ABABA, ETHIOPIA

JANUARY, 2024
Contents
CHAPTER ONE ......................................................................................................................... 1
SAMPLING AND SAMPLING DISTRIBUTION ...................................................................... 1
1.1 Sampling Theory ............................................................................................................... 1
1.1.1 Basic Definitions ............................................................................................................ 1
1.1.2. The need for samples ..................................................................................................... 2
1.1.3. Designing and conducting a sampling study ................................................................... 2
1.1.4. Bias and errors in sampling, non-sampling errors ........................................................... 3
1.1.5. Types of sampling random and non-random sampling.................................................... 7
1.2 Sampling Distribution ..................................................................................................... 10
1.2.1 Definition ......................................................................... Error! Bookmark not defined.
1.2.2 Sampling Distribution of the mean and proportion ................................................. 11
1.2.3 Sampling Distribution of the difference between two means and two proportions ... 19
CHAPTER TWO ...................................................................................................................... 23
STATISTICAL ESTIMATIONS .............................................................................................. 23
2.1. Basic concepts ................................................................................................................ 23
2.2. Point estimators of the mean and proportion ................................................................... 23
2.3. Interval estimators of the mean and proportion ............................................................... 23
2.4 Interval estimation of the difference between two independent means .............................. 24
2.5 Student’s t-distribution .................................................................................................... 26
2.6. Determining the sample size ........................................................................................... 28
CHAPTER THREE .................................................................................................................. 31
HYPOTHESIS TESTING ......................................................................................................... 31
3.1 Basic Concepts ................................................................................................................ 31
3.2 Steps to Hypothesis Testing ............................................................................................ 32
3.3 Type I and Type II errors ................................................................................................. 35
3.4 One tailed/Is two tailed hypothesis tests ........................................................................... 36
3.5. Hypothesis testing .......................................................................................................... 39
3.5.1 Hypothesis testing of the population mean .................................................................... 43

CHAPTER FOUR..................................................................................................................... 52

CHI-SQUARE DISTRIBUTIONS ............................................................................................ 52

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 4.1 Areas of application......................................................................................................... 52

4.1.1 Test for independence between two variables ................................................................ 53

4.1.2 Testing for the equality of several proportions .............................................................. 60
4.1.3 Goodness of fit tests (binomial, Normal, poison)........................................................... 65
CHAPTER FIVE ...................................................................................................................... 84
ANALYSIS OF VARIANCE .................................................................................................... 84
5.1. Areas of applications ...................................................................................................... 84
5.1.1. Comparison of the mean of more than two populations ................................................ 84
5.1.2 Variance test .......................................................................................................... 92
CHAPTER SIX......................................................................................................................... 96
REGRESSION AND CORRELATION .................................................................................... 96
6.1 Linear correlations........................................................................................................... 96
6.2.1 The coefficient of correlation ........................................................................................ 96
6.1.2 Rank correlation coefficient .......................................................................................... 99
6.2 Simple linear regression ................................................................................................ 100
6.2.1 Curve fitting ............................................................................................................... 100
6.2.2 The method of least square ......................................................................................... 103
Answer key ............................................................................................................................. 108

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 CHAPTER ONE

SAMPLING AND SAMPLING DISTRIBUTION

1.1 Sampling Theory

1.1.1 Basic Definitions

1. Statistics: we can define it in two senses
a) In the plural sense : statistics are the raw data themselves , like statistics of
births, statistics of deaths, statistics of students, statistics of imports and exports,
etc.
b) In the singular sense statistics is the subject that deals with the collection,
organization, presentation, analysis and interpretation of numerical data.
2. A (statistical) population: is the complete set of possible measurements for which
inferences are to be made. The population represents the target of an investigation,
and the objective of the investigation is to draw conclusions about the population
hence we sometimes call it target population.

Examples

 Population of trees under specified climatic conditions

 Population of animals fed a certain type of diet
 Population of farms having a certain type of natural fertility
 Population of households, etc. ƒ
 The population could be finite or infinite (an imaginary collection of units) ƒ
 There are two ways of investigation: Census and sample survey.
3. Census: a complete enumeration of the population. But in most real problems it
cannot be realized, hence we take sample.
4. Sample: A sample from a population is the set of measurements that are actually
collected in the course of an investigation. It should be selected using some pre-
defined sampling technique in such a way that they represent the population very
well.

Examples:

 Monthly production data of a certain factory in the past 10 years.

 Small portion of a finite population.

In practice, we don’t conduct census, instead we conduct sample survey

5. Parameter: Characteristic or measure obtained from a population.

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 6. Statistic: Characteristic or measure obtained from a sample.
7. Sampling: The process or method of sample selection from the population.
8. Sampling unit: the ultimate unit to be sampled or elements of the population to be
sampled.
Examples:
 If somebody studies Scio-economic status of the households, households is
the sampling unit.
 If one studies performance of freshman students in some college, the student
is the sampling unit

1.1.2. The need for samples

 Reduced cost
 Greater speed
 Greater accuracy
 Greater scope
 Avoids destructive test
 The only option when the population is infinite

Because of the above consideration, in practice we take sample and make conclusion
about the population values such as population mean and population variance, known as
parameters of the population.

1.1.3. Designing and conducting a sampling study

Sometimes taking a census makes more sense than using a sample. Some of the reasons
include:

 Universality
 Qualitativeness
 Detailedness
 Non-representativeness
Exercises:

1. Describe briefly the difference between ƒ

 Population and sample ƒ
 Parameter and statistic ƒ
 Census and sample survey ƒ
 Sampling and non-sampling error ƒ
 Sampling frame and sampling units
2. Why do researchers usually select sample elements from a given population?
3. Mention some of the disadvantage of sampling.

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 4. An insurance company has insured 300,000 cars over the last six years. The company
would like to know the number of cars involved in one or more accidents over this
period. The manger selected 1000 cars from the files and made a record of cars that
were involved in one or more accidents.
a) What is the population?
b) What is the sample?
c) What is the variable of interest to the insurance company?

1.1.4. Bias and errors in sampling, non-sampling errors

Sampling error is the proportion of overall error in research attributable to the sampling
procedure. Several reasons contribute to a sample not representing its population, including
sampling bias. Differentiation between a sample and a population is solely due to the
method of selecting certain units.

There are two fundamental reasons for sampling error:

 First, there is a chance that some unusual or variable units exist in a population and
are picked randomly, as this has a probability of occurring. Researchers can prevent
this mistake by expanding the sample size, which reduces the likelihood of selecting
rare or variant units.
 The second cause of sampling mistakes is sampling bias, which is the inclination
to pick units with specific features. The most obvious type of sampling bias is
known as selection bias, resulting from an improper sampling design. When a
subset of possible units is excluded from the sample, this is known as selection bias.

What Are Sampling Biases?

When data is obtained unfairly, some individuals in the target population will have a lower
or greater sampling probability than others. The data collected from the whole population
may not accurately reflect the information collected in the systematic study. It often occurs
unintentionally and is often missed by the researcher.

As was emphasized, sampling error occurs when an unrepresentative sample is selected.

The limited sample size is the only way sampling error can occur, leading to poor accuracy.
Nevertheless, sampling bias can be assessed after a study is completed.

Sampling Bias Types

Some of the key sampling bias types are:

 Self-Selection bias: People with particular qualities are more likely than others to
consent to participate in research.

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  Coverage bias: People who decline to participate or drop out of a study differ
systemically from those who participate.
 Under-coverage bias: Some population members are insufficiently represented in
the sample.
 Advertising or pre-screening bias: A sample may be biased based on how
volunteers are pre-screened or where research is publicized.

How to Prevent Sampling Bias:

 Using a well-thought-out study design and method of sampling can assist you in
preventing sample bias.
 Define a population of interest and a sampling frame
 Match the sample frame as closely as feasible to the target population to minimize
the possibility of sampling bias.
 Make user-friendly questionnaires.
 Follow up with non-respondents.

In contrast, the discrepancy between a survey’s outcome and the population value is known
as sampling error. The assessment of sampling error can be anticipated, evaluated, and
accounted for in contrast to bias. Sampling error is quantified using P-values, standard
errors, confidence intervals, and coefficients of variation. These metrics are employed to
compute sample size before sampling and assess our confidence level in the results
following analysis.

Choosing the appropriate sampling procedure and sample size helps reduce sampling error.
Although sample bias cannot be precisely quantified, sampling error can be precisely
estimated. Preventing and reducing sample bias must be a top priority. Sampling error
happens when the confidence intervals are too broad because of the limited sample size.

What are the most prevalent types of sampling errors?

If researchers are not certain who to interview, they make a mistake in defining the
population.

 Sampling frame error: Sampling frame mistakes occur when researchers incorrectly
target a subpopulation when choosing samples.
 Selection error: A selection error happens when respondents choose to engage in
research on their own. Only those who are interested react. By requesting replies
from the complete sample, it is possible to minimize selection mistakes. The
response rate will be increased by pre-survey planning, follow-ups, and a well-
organized survey design. Also, attempt CATI surveys and in-person interviews to
increase response rates.

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  Sampling mistakes: Sampling errors are caused by a mismatch in the respondents’
representativeness. It typically occurs when the researcher does not thoroughly
arrange his sample. To minimize these errors, it is necessary to use a rigorous
sample design. This sample size represents the total population, an online sample,
or survey respondents.

How to Minimize Sampling Errors?

 Increase sample size: A bigger sample yields more accurate results since the
research becomes more representative of the total population.
 Instead of using a random sample, test groups are proportional to their size in the
population.
 Know your audience: Examine your population and determine its demographic
composition.

Non-sampling error refers to all sources of error that are unrelated to sampling. Non-
sampling errors are present in all types of survey, including censuses and administrative
data. They arise for a number of reasons: the frame may be incomplete, some respondents
may not accurately report data, data may be missing for some respondents, etc.

Non-sampling errors can be classified into two groups: random errors and systematic
errors.

 Random errors are errors whose effects approximately cancel out if a large enough
sample is used, leading to increased variability.
 Systematic errors are errors that tend to go in the same direction, and thus
accumulate over the entire sample leading to a bias in the final results. Unlike
random errors, this bias is not reduced by increasing the sample size. Systematic
errors are the principal cause of concern in terms of a survey’s data quality.
Unfortunately, non-sampling errors are often extremely difficult, if not impossible,
to measure.

Types of non-sampling error

Non-sampling error can occur in all aspects of the survey process, and can be classified
into the following categories: coverage error, measurement error, nonresponse error and
processing error.

Coverage error

Coverage error consists of omissions (under coverage), erroneous inclusions, duplications

and misclassifications (over coverage) of units in the survey frame. Since it affects every
estimate produced by the survey, they are one of the most important types of error. In the

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case of a census, it may be the main source of error. Coverage error can have both spatial
and temporal dimensions, and may cause bias in the estimates. The effect can vary for
different subgroups of the population. This error tends to be systematic and is usually due
to under coverage, which is why it’s important to reduce it as much as possible.

Measurement error

Measurement error, also called response error, is the difference between measured values
and true values. It consists of bias and variance, and it results when data are incorrectly
requested, provided, received or recorded. These errors may occur because of inefficiencies
with the questionnaire, the interviewer, the respondent or the survey process.

 Poor questionnaire design it is essential that sample survey or census questions

are worded carefully in order to avoid introducing bias. If questions are misleading
or confusing, then the responses may end up being distorted.
 Interviewer bias an interviewer can influence how a respondent answers the
survey questions. This may occur when the interviewer is too friendly or aloof or
prompts the respondent. To prevent this, interviewers must be trained to remain
neutral throughout the interview. They must also pay close attention to the way they
ask each question. If an interviewer changes the way a question is worded, it may
impact the respondent’s answer.
 Respondent error respondents can also provide incorrect answers. Faulty
recollections, tendencies to exaggerate or underplay events, and inclinations to give
answers that appear more socially acceptable are several reasons why a respondent
may provide a false answer.
 Problems with the survey process errors can also occur because of a problem with
the actual survey process. Using proxy responses, meaning taking answers from
someone other than the respondent, or lacking control over the survey procedures
are just a few ways of increasing the risk of response errors.

Non-response error
Estimates obtained after nonresponse has been observed and imputation has been used to
deal with this nonresponse are usually not equivalent to the estimates that would have been
obtained had all the desired values been observed without error. The difference between
these two types of estimates is called the nonresponse error. There are two types of non-
response errors: total and partial.

 Total nonresponse error occurs when all or almost all data for a sampling unit are
missing. This can happen if the respondent is unavailable or temporarily absent, the
respondent is unable to participate or refuses to participate in the survey, or if the
dwelling is vacant. If a significant number of sampled units do not respond to a

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 survey, then the results may be biased since the characteristics of the non-
respondents may differ from those who have participated.
 Partial nonresponse error occurs when respondents provide incomplete
information. For certain people, some questions may be difficult to understand; they
may refuse or forget to answer a question. Poorly designed questionnaire or poor
interviewing techniques can also be reasons which result partial nonresponse error.
To reduce this form of error, care should be taken in designing and testing
questionnaires. Adequate interviewer training and appropriate edits and imputation
strategies will also help minimize this error.

Processing error

Processing error occurs during data processing. It includes all data processing activities
after collection and prior to estimation, such as errors in data capture, coding, editing and
tabulation of the data as well as in the assignment of survey weights.

 Coding errors occur when different coders code the same answer differently,
which can be caused by poor training, incomplete instructions, variance in coder
performance (i.e. tiredness, illness), data entry errors, or machine malfunction
(some processing errors are caused by errors in the computer programs).
 Data capture errors result when data are not entered into the computer exactly as
they appear on the questionnaire. This can be caused by the complexity of
alphanumeric data and by the lack of clarity in the answer provided. The physical
layout of the questionnaire itself or the coding documents can cause data capture
errors. The method of data capture, manual or automated (for example, using an
optical scanner), can also result in errors.
 Editing and imputation errors can be caused by the poor quality of the original
data or by its complex structure. When the editing and imputation processes are
automated, errors can also be the result of faulty programs that were insufficiently
tested. The choice of an inappropriate imputation method can introduce bias. Errors
can also result from incorrectly changing data that were found to be in error, or by
erroneously changing correct data.

1.1.5. Types of sampling random and non-random sampling

- There are two types of sampling techniques.

A. Random Sampling or probability sampling.

- Is a method of sampling in which all elements in the population have a pre-assigned non
zero probability to be included in to the sample.

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 Examples:

 Simple random sampling

 Stratified random sampling
 Cluster sampling
 Systematic sampling

1. Simple Random Sampling:

- Is a method of selecting items from a population such that every possible sample of
specific size has an equal chance of being selected, In this case, sampling may be with or
without replacement. Or

- All elements in the population have the same pre-assigned non zero probability to be
included in to the sample.

- Simple random sampling can be done either using the lottery method or table of random
numbers.

Table of Random Numbers

Table of random numbers are tables of the digits 0, 1, 2,…,, 9, each digit having an equal
chance of selection at any draw. For convenience, the numbers are put in blocks of five. In
using these tables to select a simple random sample, the steps are:

i. Number the units in the population from 1 to N (prepare frame of the population).
ii. Then proceed in the following way If the first digit of N is a number between 5 and
9 inclusively, the following method of selection is adequate. Suppose N=528 and we
want n=10. Select three columns from the table of random numbers, say columns 25 to
27. Go down the three columns selecting the first 10 distinct numbers between 001 &
528. These are 36, 509, 364, 417, 348, 127, 149, 186, 439, and 329. Then the units with
these roll numbers are our samples.
Note: If sampling is without replacement, reject all the numbers that comes more
than once.

2. Stratified Random Sampling:

- The population will be divided in to non-overlapping but exhaustive groups

called strata.
- Simple random samples will be chosen from each stratum.
- Elements in the same strata should be more or less homogeneous while different
in different strata.
- It is applied if the population is heterogeneous.

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 - Some of the criteria for dividing a population into strata are: Sex (male, female);
Age (under 18, 18 to 28, and 29 to 39); Occupation (blue-collar, professional, and
other).

3. Cluster Sampling:

- The population is divided in to non-overlapping groups called clusters.

- A simple random sample of groups or cluster of elements is chosen and all the
sampling units in the selected clusters will be surveyed.

- Clusters are formed in a way that elements within a cluster are heterogeneous, i.e.
observations in each cluster should be more or less dissimilar.

- Cluster sampling is useful when it is difficult or costly to generate a simple random

sample. For example, to estimate the average annual household income in a large city we
use cluster sampling, because to use simple random sampling we need a complete list of
households in the city from which to sample. To use stratified random sampling, we would
again need the list of households. A less expensive way is to let each block within the city
represent a cluster. A sample of clusters could then be randomly selected, and every
household within these clusters could be interviewed to find the average annual household
income.

4. Systematic Sampling:

- A complete list of all elements within the population (sampling frame) is required.

- The procedure starts in determining the first element to be included in the sample.

- Then the technique is to take the kth item from the sampling frame.
𝑁
- Let 𝑁 = 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑖𝑧𝑒 , 𝑛 = 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒, 𝑘= = 𝑠𝑎𝑚𝑝𝑙𝑖𝑛𝑔 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙.
𝑛

- Chose any number between 1 and k .Suppose it is j ( 1≤ j ≤ k).

- The unit is selected at first and then (𝑗 + 𝑘)𝑡ℎ , (𝑗 + 2𝑘)𝑡ℎ , . . . . . 𝑒𝑡𝑐 until the required
sample size is reached.

B. Non Random Sampling or non-probability sampling.

- It is a sampling technique in which the choice of individuals for a sample depends on the
basis of convenience, personal choice or interest.

Examples:

 Judgment sampling.

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  Convenience sampling
 Quota Sampling.

1. Judgment Sampling

- In this case, the person taking the sample has direct or indirect control over which
items are selected for the sample.

2. Convenience Sampling

- In this method, the decision maker selects a sample from the population in a manner
that is relatively easy and convenient.

3. Quota Sampling

- In this method, the decision maker requires the sample to contain a certain number of
items with a given characteristic. Many political polls are, in part, quota sampling.

Note: let N = population , n= sample size

1. Suppose simple random sampling is used

 We have Nn possible samples if sampling is with replacement.
𝑁
 We have ( ) possible samples if sampling is without replacement.
𝑛

1.2 Sampling Distribution

1.2.1 Definition
The normal probability distribution is used to determine probabilities for the normally
distributed individual measurements, given the mean and the standard deviation.
Symbolically, the variable is the measurement X, with the population mean µ and
population standard deviation δ. In contrast to such distributions of individual
measurements, a sampling distribution is a probability distribution for the possible values
of a sample statistic.

Population distribution: Is the distribution of measured values of its members and have
mean denoted by𝜇 and variance 𝛿 2 and standard deviation 𝜎. The population standard
deviation describes the variation among values of members of the population; where as the
standard deviation of sampling distribution measures the variability among values of the
statistics (sample) such as mean values, proportion values due to sampling errors.

Sample distribution: Is the distribution of measured values of sample in random samples

drawn from a given population. Each sample mean would vary from sample to sample.
This variability serves as the basis for random sampling distribution. A sampling

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distribution is a probability distribution for the possible values of a sample statistic, such
as a sample mean.

1.2.2 Sampling Distribution of the mean and proportion

- sampling distribution of the mean
Sampling distribution of the mean: Is the probability distribution of all possible values of
a given statistic (sample) from all distinct possible samples of equal size drawn from a
population or a process. The sampling distribution of the mean values has its own arithmetic
mean denoted by 𝜇𝑥̅ (read as mu sub x bar) and standard deviation 𝛿𝑥̅ (sigma sub x bar). The
sampling distribution of the mean is the probability distribution of the means, X of all simple
random samples of a given sample size n that can be drawn from the population.

NB: The sampling distribution of the mean is not the sample distribution, which is the
distribution of the measured values of X in one random sample. Rather, the sampling
distribution of the mean is the probability distribution for X , the sample mean.

For any given sample size n taken from a population with mean µ and standard deviation
δ, the value of the sample mean would vary from sample to sample if several random
samples were obtained from the population. This variability serves as the basis for
sampling distribution.

The sampling distribution of the mean is described by two parameters: the expected value
( X ) = X , or mean of the sampling distribution of the mean, and the standard deviation of
the mean  x , the standard error of the mean.

Properties of the Sampling Distribution of Means

1. The arithmetic mean 𝜇𝑥̅ of the sampling distribution of mean values is equal to the
population mean 𝜇 regardless of the form of population distribution .i.e. 𝜇𝑥̅ =𝜇
2. The sampling distribution has a standard deviation (also called standard error) equal
to the population standard deviation divided by the square root of the sample size i.e.,
σ
δx̅ = n . This holds true if and only of n<0.05N and N is very large. If N is finite and
√
𝑆 𝑆 𝑁−𝑛
𝛿𝑥̅ = Or𝑆𝛿𝑥̅ = √𝑁−1 , 𝑛 ≥ 0.05𝑁
√𝑛 √𝑛

𝑁−𝑛
3. The expression√ 𝑁−1 is called finite population correction factor/finite population
multiplier. In the calculation of the standard error of the mean, if the population

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 
standard deviation δ is unknown, the standard error of the mean x , can be estimated
S
by using the sample standard error of the mean X which is calculated as follows:
𝑆 𝑆 𝑁−𝑛
𝑆𝑥̅ = or 𝑆𝑥̅ = √𝑁−1
√𝑛 √𝑛

4. A sample size n≥30 is generally considered to be a large sample for statistical

analysis where as a sample of size n< 30 is considered to be a small sample. The
sampling distribution of means is approximately normal for sufficiently large sample
sizes (n≥ 30).
5. When standard deviation of population σ is not known, the standard deviation of the
sample s which closely approximates σ value is used to compute standard error,
s
i.e. δx̅ = .
√n

A population consists of the following ages: 10, 20, 30, 40, and 50. A random sample of
three is to be selected from this population and mean computed. Develop the sampling
distribution of the mean.

Solution: The number of simple random samples of size n that can be drawn without
𝑁!
replacement from a population of size is 𝑁𝐶𝑛 = 𝑛!(𝑁−𝑛)! With N= 5 and n = 3, 5C3 = 10
samples can be drawn from the population as:

Sampled items Sample means ( X )

10, 20, 30 20.00
10, 20, 40 23.33
10, 20, 50 26.67
10, 30, 40 26.67
10, 30, 50 30.00
10, 40, 50 33.33
20, 30, 40 30.00
20, 30, 50 33.33
20, 40, 50 36.67
30, 40, 50 40.00
300.00
A systematic organization of the above figures gives the following:

Sample mean ( X ) Frequency Prob. (relative freq.) of X

20.00 1 0.1

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 23.33 1 0.1
26.67 2 0.2
30.00 2 0.2
33.33 2 0.2
36.67 1 0.1
40.00 1 0.1
TOTAL 10.00 1.00
Columns 1 and 2 show frequency distribution of sample means.
Columns 1 and 3 show sampling distribution of the mean.


 X   x  30, Regardless of the sample size   X
N n
𝑥 (Observation) 𝑥−𝜇 (𝑥 − 𝜇)2

10 -20 400

20 -10 100

30 0 0

40 10 100

50 20 400

∑(𝑥 − 𝜇)2 1,000

 X 
2
X 1000
   14.142
i

N 5
 N  n 14.142 53
X  *  *  5.774
n N 1 3 5 1
2

  X i  X  333.4

   5.774
N 10
Since averaging reduces variability  x < δ except the cases where δ = 0 and n = 1.
Central Limit Theorem and the Sampling Distribution of the Mean

The Central Limit Theorem (CLT) states that:

1. If the population is normally distributed, the distribution of sample means is normal

regardless of the sample size.
2. If the population from which samples are taken is not normal, the distribution of
sample means will be approximately normal if the sample size (n) is sufficiently
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 large (n ≥ 30). The larger the sample size is used, the closer the sampling
distribution is to the normal curve.

The relationship between the shape of the population distribution and the shape of the
sampling distribution of the mean is called the Central Limit Theorem.

The significance of the Central Limit Theorem is that it permits us to use sample statistics
to make inference about population parameters without knowing anything about the shape
of the frequency distribution of that population other than what we can get from the sample.
It also permits us to use the normal distribution curve for analyzing distributions whose
shape is unknown. It creates the potential for applying the normal distribution to many
problems when the sample is sufficiently large.

As mentioned earlier the above properties must exist, given this value of sample mean 𝑋̅
is first converted in to a value Z on the standard normal distribution to know how any single
value deviates from 𝑋̅ of sample mean values (𝜇𝑥̅ ), by using the formula;

𝑋̅−𝜇𝑥̅ 𝑋̅−𝜇
𝑍= = 𝛿 because 𝜇𝑥̅ = 𝜇
𝛿𝑥̅
√𝑛

If the population is finite and samples of fixed size n are drawn without replacement, then
the standard error of sampling distribution of mean can be modified to adjust the continued
change in the size of population 𝜇 due to the several draws of samples of size n is as follows:
Example 1. The mean length of a certain tool is 41.5 hours with a standard
deviation of 2.5 hours. What is the probability that a simple random sample
of size 50 drawn from this population will have a mean between 40.5 hours
and 42 hours?

𝜇=41.5 𝛿=2.5 n=50

P (40.5≤ 𝑋̅ ≤42.0) =?
𝛿 2.5 2.5
𝜇𝑥̅ = 𝜇 𝛿𝑥̅ = = =7.0711 = 0.3536
√𝑛 √50
The population distribution is unknown, but sample size n=50 is large enough to apply the
central limit theorem. Hence the normal distribution can be used to find the required
probability.
𝑋̅1 −𝜇 𝑋̅2 −𝜇
P (40.5≤ 𝑋̅ ≤420) = P ( ≤𝑍≤ )
𝛿𝑥̅ 𝛿𝑥̅
40.5−41.5 42−41.5
=P( ≤𝑍≤ )
0.3536 0.3536
= P (−2.8281 ≤ 𝑍 ≤ 1.4140)
=P (𝑍 ≥ −2.8281) + P (𝑍 ≤ 1.4140)
=0.4977+0.4207=0.9184

14 | P a g e
 Thus 0.9184 is the probability of the tool having mean life between the required hours.
𝛿 = 2.5
0.4977
0.4207

𝑥̅ = 40.5 𝜇 = 41.5 𝑥̅ = 40.5

Example 2. A continuous manufacturing process produces items whose weights

are normally distributed with a mean weight of 800gms and a standard
deviation of 300gms. A random sample of 16 items is to be selected from the
process.
A. What is the probability that the arithmetic mean of the sample exceeds 900gms? Interpret the
result.
B. Find the values of the sample arithmetic mean within which the middle 95% of all sample
means will fall.

Solution:

A. P (𝑥̅ ≥ 900) =?
𝜇𝑋̅ = 𝜇=800gms 𝛿=300gms
n=16
P (𝑥̅ ≥ 900) =?
𝛿 300 300
𝛿𝑥̅ = = = = 75
√𝑛 √16 4

0.0918

𝜇𝑋̅ = 800 ̅ = 900

𝑋

𝑋̅−𝜇𝑥̅ 900−800
P (𝑥̅ ≥ 900) =P (Z≥ = )
𝛿𝑥̅ 75
=P (Z≥ 1.33)
=0.5000-0.4082
=0.0918

15 | P a g e
B. Since Z=1.96 for the middle 95% area under the normal curve, therefore using the formula
for z to solve for the values of x̅ in terms of the known values are as follows.
𝑥̅1 =𝜇𝑋̅ -Z𝛿𝑥̅ 𝑥̅ 2 =𝜇𝑋̅ +Z𝛿𝑥̅
=800-1.96(75) =800+1.96(75)
=653gms 0.95
=947gms
𝛿=300

𝑋̅ = 653 𝜇𝑋̅ = 800 ̅ = 947

𝑋

- Sampling Distribution of Sample Proportions

The sample proportion 𝑃̅ having the characteristic of interest (success or failure, accept or
reject, head or tail) is the best use for statistical inferences about the population parameter P.
the sample proportion can be defined as:
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠,𝑋
𝑃̅= 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒,𝑛

With same logic of sampling distribution of mean, the sampling distribution of sample
proportions with mean 𝜇𝑃̅ and standard deviation also called standard error) 𝛿𝑃̅ is given by:

𝑝𝑞 𝑝(1−𝑃)
𝜇𝑃̅ = P and 𝛿𝑃̅ =√ 𝑛 =√ 𝑛

If a large ample size (n≥30) satisfying following two conditions

A. np≥5
B. nq≥5
Then the sampling distribution of proportions is very closely normally distributed. It may be
noted that the sampling distribution of the proportion would actually follow binomial
distribution because population is binomially distributed.
For finite population in which sampling is done without replacement we have;
𝑝𝑞 𝑁−𝑛
𝜇𝑃̅ = P and 𝛿𝑃̅ =√ 𝑛 *√𝑁−1
Under the same guidelines as mentioned in the previous sections, for a large sample size
n≥30, the sampling distribution of proportion is closely approximated by a normal

16 | P a g e
distribution with a mean and standard deviation as stated above. Hence, to standardize
sample proportion𝑃̅, the standard normal variable
𝑃̅ −𝜇𝑃
̅ 𝑃̅ −𝑃
Z= =
𝛿𝑃
̅ 𝑝𝑞
√
𝑛

Example 3. Few years back, a policy was introduced to give loans to unemployed
engineers to start their own business. Out of 1,000,000 engineers, 600,000
accepted the policy and got the loan. A sample of 100 unemployed engineers
is taken at the same time of allotment of loans. What is the probability that
sample portion would have exceeded 50% acceptance?
Solution:

𝜇𝑃̅ = P=0.60 N=1,000,000

n=100 P (𝑃̅ ≥ 0.5) =?
𝑝𝑞 𝑁−𝑛) (0.6)(0.4) 1,000,000−100)
𝛿𝑃̅ =√ 𝑛 √ 𝑁−1 = (√ )(√ )
100 1,000,000−1

𝛿𝑃̅ =0.0489
𝑃̅ −𝜇 0.50−0.60
P (𝑃̅ ≥ 0.5) =P (Z≥ 𝛿 𝑃̅) =P (Z≥ 0.0489 ) =0.4793+0.5000=0.9793
̅
𝑃

0.4793
0.5000

𝑃̅ = 0.5 P=0.60

Example 4. A population proportion is 0.40. A simple random sample of size 200

will be taken and the sample proportion will be used to estimate the
population proportion, what is the probability that the sample proportion will
be with in ±0.03 of the population proportion.
Given:

𝜇𝑃̅ = P=0.40 n=200

(0.4)(0.6) 𝑃̅ −𝑃
𝛿𝑃̅ =√ =0.0346 P (-0.03≤ 𝑃̅ ≤ 0.03) = 2P (Z≥ 𝛿 )
200 ̅
𝑃
= 2P (Z ≤ 0.87)
=2x0.3078
=0.6156

17 | P a g e
 0.3078 0.3078

𝑃̅ = −0.03 P=0.40 𝑃̅ = 0.03

Example 5. A manufacturer of watches has determined from past experience that

3% of the watches he produces are defective. If a random sample of 300
watches is examined, what is the probability that the proportion of defective
is between 0.02 and 0.035?

𝜇𝑃̅ = P=0.03 𝑃̅2 =0.035

𝑃̅1 =0.02 n=300

(0.03)(0.97)
𝛿𝑃̅ =√ =0.0098
300

𝑃̅ −𝑃 𝑃̅ −𝑃
P (-0.03≤ 𝑃̅ ≤ 0.03) = P ( 𝛿 ≤𝑍≤ )
̅
𝑃 𝛿𝑃
̅
0.02−0.03 0.035−0.03
=P( ≤𝑍≤ )
0.0098 0.0098
= P (-1.02≤ 𝑍 ≤ 0.51)
=P (Z≥ −1.02) + P (Z≤ 0.51)
=0.3461+0.1950
= 0.5411
Hence the probability that the proportion of defective will lie between 0.02 and 0.035 is
0.5411

0.3461 0.1950

𝑃̅1 =0.02 P=0.03 𝑃̅2 =0.035

18 | P a g e
 1.2.3 Sampling Distribution of the difference between two means and two
proportions
- Sampling Distribution Of The Difference Between Two Means

The concept of sampling distribution of sample mean introduced earlier can also be used to
compare a population of size 𝑁1 having mean 𝜇1 and standard deviation 𝛿1 with another
similar type of population of size 𝑁2 having mean 𝜇2 and standard deviation𝛿2 .

Let 𝑋̅1 𝑎𝑛𝑑 𝑋̅2 be the mean of sampling distribution of the mean of two populations,
respectively. Then the difference between their mean values 𝜇1 and 𝜇2 can be estimated by
generalizing the formula of standard normal variable as follows;
(𝑋̅1 −𝑋̅2 ) − (𝜇𝑋
̅ 1 − 𝜇𝑋
̅2 ) (𝑋̅1 −𝑋̅2 ) − (𝜇1− 𝜇2 )
Z= =
𝛿 (𝑋
̅ −𝑋̅ ) 𝛿(𝑋
̅ 1−𝑋
̅2 )
1 2

Where: 𝜇𝑋̅1 − 𝜇𝑋̅2 = 𝜇1 − 𝜇2 (mean of sampling distribution of sample mean)

𝛿 2 𝛿2 2
𝛿(𝑋̅1−𝑋̅2 ) = √𝛿𝑋̅1 2 + 𝛿𝑋̅2 2 = √ 𝑛1 + (standard error of sampling distribution of difference of two means)
1 𝑛2

𝑛1 And 𝑛2 are independent random samples drawn from first and second population
, respectively.

Example 6. Car stereos of manufacturer A have a mean lifetime of 1,400 hours

with a standard deviation of 200 hours, while those of manufacturer B have a
mean life time of 1,200 hours with a standard deviation of 100 hours. If a
random sample of 125 stereos of each manufacturer are tested, what is the
probability that manufacturer A’s stereos will have a mean life time which is
at least;
A. 160 hours more than manufacturer B’s stereos?
B. 250 hours more than manufacturer B’s stereos?
Solution:

Manufacturer A 𝜇1 =1,400 hours

𝛿1 = 200 hours 𝑛1 =125
Manufacturer B 𝜇1 =1,200 hours
𝛿1 = 100 hours 𝑛2 =125
a)
𝛿 2 𝛿2 2 (200)2 (100)2
𝛿(𝑋̅1 − 𝑋̅2 ) =√ 𝑛1 + =√ + = √80 + 320 = √400 =20
1 𝑛2 125 125
(𝑋1− 𝑋̅2 ) (𝜇1_ 𝜇2 )
̅
P (𝑋̅1 − 𝑋̅2 ≥160) = P (𝑍 ≥ )
𝛿(𝑋
̅ ̅2 )
1− 𝑋
160−200
=P (𝑍 ≥ )
20

19 | P a g e
 =P (𝑍 ≥ −2)
=0.5000+0.4772
=0.9772 (area under normal curve)

0.9772

𝑋̅1 − 𝑋̅2 = 160 𝜇𝑋̅ = 200

1 − 𝑋̅2

Hence, the probability is very high that the life time of the stereos of A is 160
hours more than that of b.

b) Proceeding in the same manner as in part a) as follows:

(𝑋̅ − 𝑋̅ ) (𝜇 − 𝜇 )
P (𝑋̅1 − 𝑋̅2 ≥250) = P (𝑍 ≥ 1 2 1 2 𝛿(𝑋
̅ 1 −𝑋
̅2 )
250−200
=P (𝑍 ≥ )
20
=P (𝑍 ≥ −2.5)
=0.5000 - 0.4938
=0.0062 (area under normal curve)

Example 7. The strength of a wire produced by company has a mean of 4,500kg

and a 𝛿1 of 200 kg. Company B has a mean of 4,000 kg and a 𝛿2 of 300 kg. if
50 wires of company A and 100 wires of company B are selected at random
and tested for strength, what is the probability that the sample mean strength
of A will be at least 600gk more than that of B?

Given:
𝜇1 = 4,500 𝜇2 = 4,000
𝛿1 =200 𝛿2 =300
𝑛1 =50 𝑛2 =100

𝛿 2 𝛿2 2 (200)2 (300)2
𝛿(𝑋̅1 − 𝑋̅2 ) =√ 𝑛1 + =√ + = =41.23
1 𝑛2 50 100
(𝑋̅1− 𝑋̅2 ) (𝜇1_ 𝜇2 )
P (𝑋̅1 − 𝑋̅2 ≥600) = P (𝑍 ≥ )
𝛿(𝑋
̅ ̅2 )
1− 𝑋

20 | P a g e
 600−500
=P (𝑍 ≥ )
41.23
=P (𝑍 ≥ 2.43)
=0.4925
=0.5000 - 0.4925=0.0075 (area under normal curve)

- Sampling Distribution of the Difference of Two Proportions

Suppose two populations of size 𝑁1 and 𝑁2 are given. For each sample of size 𝑛1 from the first
population, compute sample proportion 𝑃̅1 and standard deviation𝛿𝑃̅1 . Similarly for each
sample size of 𝑛2 from the second population, compute sample proportion𝑃̅2 and standard
deviation𝛿𝑃̅2 .

For all combinations of these samples from these populations, we can obtain a sampling
distribution of the difference 𝑃̅1 − 𝑃̅2 of sample proportion. Such a distribution is called
sampling distribution of the difference of two proportions. The mean and standard deviation
of this distribution are given by;

𝜇𝑃̅1 − 𝜇𝑃̅2 = 𝑃1 − 𝑃2

𝑃1𝑞1 𝑃2 𝑞2
𝛿(𝑃̅1−𝑃̅2 ) = √𝛿𝑃̅1 2 + 𝛿𝑃̅2 2 = √ +
𝑛1 𝑛2

If sample size 𝑛1 𝑎𝑛𝑑 𝑛1 are large i.e. 𝑛1 ≥30, then the sampling distribution of difference
of proportions is closely approximated by a normal distribution.

Example 1. 10% of the machines produced by company A are defective and 5%

of those produced by company B are defective. A random sample of 250 machines is
taken from company A and a random sample of 300 machines is taken from company
B. What is the probability that the difference in sample proportion is less than or
equal to0.02?

𝜇𝑃̅1 − 𝜇𝑃̅2 = 𝑃1 − 𝑃2 = 0.10 − 0.05=0.05

𝑛1 =250 𝑛2 =300
The standard error of the difference in a sample proportions is given by

𝑃1𝑞1 𝑃2 𝑞2
𝛿(𝑃̅1−𝑃̅2 ) = √𝛿𝑃̅1 2 + 𝛿𝑃̅2 2 = √ +
𝑛1 𝑛2

𝛿(𝑃̅1−𝑃̅2 ) = √0.0052 = 0.0228

The desired probability of the difference in sample proportion is given by

(𝑃̅ − 𝑃̅ )−(𝑃1− 𝑃2 )
P(𝑃̅1 − 𝑃̅2 ≤0.02) =P (𝑍 ≥ 1 𝛿 2
̅ 1−𝑃
(𝑃 ̅2 )
0.02−0.05
=P (𝑍 ≥ )
0.0228
21 | P a g e
 =P (𝑍 ≥ −1.32)
=0.5000 - 0.4066=0.0934 (area under normal curve)
Hence the desired probability for the difference in sample proportions is 0.0934

Exercise:

1. A population has mean 75 and standard deviation 12

a) Random samples of size 8181 are taken. Find the mean and standard deviation of
the sample mean
b) How would the answers to part (a) change if the size of the samples
were 400 instead of 121?
2. Suppose that 8% of all males suffer some form of color blindness. Find the
probability that in a random sample of 250 men at least 10% will suffer some form
of color blindness. First verify that the sample is sufficiently large to use the normal
distribution

3. Suppose that 2% of all cell phone connections by a certain provider are dropped.
Find the probability that in a random sample of 1,500 calls at most 40 will be
dropped. First verify that the sample is sufficiently large to use the normal
distribution

4. An airline claims that 72% of all its flights to a certain region arrive on time. In a
random sample of 30 recent arrivals, 19 were on time. You may assume that
the normal distribution applies.
a. Compute the sample proportion.
b. Assuming the airline’s claim is true, find the probability of a sample of
size 30 producing a sample proportion so low as was observed in this sample
5. In one study it was found that 86% of all homes have a functional smoke detector.
Suppose this proportion is valid for all homes. Find the probability that in a
random sample of 600 homes, between 80% and 90% will have a functional smoke
detector. You may assume that the normal distribution applies.
6. An outside financial auditor has observed that about 4% of all documents he
examines contain an error of some sort. Assuming this proportion to be accurate, find
the probability that a random sample of 700 documents will contain at least 30 with
some sort of error. You may assume that the normal distribution applies.

22 | P a g e
 CHAPTER TWO

STATISTICAL ESTIMATIONS

2.1. Basic concepts

This is one way of making inference about the population parameter where the investigator does
not have any prior notion about values or characteristics of the population parameter.

2.2. Point estimators of the mean and proportion

Another term for statistic is point estimate, since we are estimating the parameter value. A point
estimator is the mathematical way we compute the point estimate. For instance, sum of 𝑋𝑖 over n
∑𝑋
is the point estimator used to compute the estimate of the population means, 𝜇 .That is ̅
X = 𝑖 is
𝑛
a point estimator of the population mean.

2.3. Interval estimators of the mean and proportion

It is the procedure that results in the interval of values as an estimate for a parameter, which is
interval that contains the likely values of a parameter. It deals with identifying the upper and lower
limits of a parameter. The limits by themselves are random variable.

Estimator is the rule or random variable that helps us to approximate a population parameter. But
estimate is the different possible values which an estimator can assume.
∑𝑋
Example: The sample mean ̅ X = 𝑛 𝑖 is an estimator for the population mean and X =10 is an
estimate, which is one of the possible values of ̅
X

Properties of best estimator

The following are some qualities of an estimator

 It should be unbiased.
 It should be consistent.
 It should be relatively efficient.
To explain these properties let 𝜃̂ be an estimator of θ

1. Unbiased Estimator: An estimator whose expected value is the value of the parameter being
estimated. i.e. 𝐸(𝜃̂) = 𝜃
2. Consistent Estimator: An estimator which gets closer to the value of the parameter as the
sample size increases. I.e. 𝜃̂ gets closer to θ as the sample size increases.
3. Relatively Efficient Estimator: The estimator for a parameter with the smallest variance.
This actually compares two or more estimators for one parameter.

23 | P a g e
2.4 Interval estimation of the difference between two independent means
Although X ̅ possesses nearly all the qualities of a good estimator, because of sampling error, we
know that it's not likely that our sample statistic will be equal to the population parameter, but
instead will fall into an interval of values. We will have to be satisfied knowing that the statistic is
"close to" the parameter. That leads to the obvious question, what is "close"?

We can phrase the latter question differently: How confident can we be that the value of the statistic
falls within a certain "distance" of the parameter? Or, what is the probability that the parameter's
value is within a certain range of the statistic's value? This range is the confidence interval.

The confidence level is the probability that the value of the parameter falls within the range
specified by the confidence interval surrounding the statistic. There are different cases to be
considered to construct confidence intervals.

Case 1:

If sample size is large or if the population is normal with known variance

Recall the Central Limit Theorem, which applies to the sampling distribution of the mean of a
sample. Consider a sample of size n drawn from a population, whose mean, is μ and standard
deviation is σ with replacement and order important. The population can have any frequency
̅ will have a mean
distribution. The sampling distribution of X
𝜎
𝜇X̅ = 𝜇 and a standard deviation 𝜎x̅ = , and approaches a normal distribution as n gets large.
√𝑛

This allows us to use the normal distribution curve for computing confidence intervals.
𝑋̅−𝜇
⟹ Z = 𝜎/ has a normal distribution with mean=0 and variance=1
√𝑛

⟹μ=̅
X ± 𝑍𝜎/√𝑛
̅ ± 𝜀 , 𝑤ℎ𝑒𝑟𝑒 𝜀 𝑖𝑠 𝑎 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟
=X

⟹ ε = 𝑍𝜎/√𝑛

- For the interval estimator to be good the error should be small. How it is small?

 By making n large
 Small variability
 Taking Z small

- To obtain the value of Z, we have to attach this to a theory of chance. That is, there is an area of
size 1−α such

𝑃(−𝑍𝛼/2 < 𝑍 < 𝑍𝛼/2 ) = 1 − 𝛼

Where 𝛼 = 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑙𝑖𝑒𝑠 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙

24 | P a g e
 𝑍𝛼 = 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓
2

𝑤ℎ𝑖𝑐ℎ 𝛼/2 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑙𝑖𝑒𝑠, 𝑖. 𝑒, 𝑃(𝑍 > 𝑍𝛼/2 ) = 𝛼/2

𝑋̅−𝜇
⟹ 𝑃(−𝑍𝛼/2 < 𝜎/ < 𝑍𝛼/2 ) = 1 − 𝛼
√𝑛

⟹ 𝑃(𝑋̅ − 𝑍𝛼/2 𝜎/√𝑛 < μ < 𝑋̅ + 𝑍𝛼/2 𝜎/√𝑛) = 1 − 𝛼

⟹ (𝑋̅ − 𝑍𝛼/2 𝜎/√𝑛, 𝑋̅ + 𝑍𝛼/2 𝜎/√𝑛) Is a 100(1 − 𝛼 )% confidence interval for𝜇, But usually σ2
is not known, in that case we estimate by its point estimator S 2.

⟹ (𝑋̅ − 𝑍𝛼/2 𝑆/√𝑛, 𝑋̅ + 𝑍𝛼/2 𝑆/√𝑛) Is a 100(1 − 𝛼 )% confidence interval for𝜇,

Here are the z values corresponding to the most commonly used confidence levels.

100(1 − 𝛼 )% 𝛼 𝛼/2 𝑍𝛼/2

90 0.10 0.05 1.645

95 0.05 0.025 1.96

99 0.01 0.005 2.58

Case 2:

If sample size is small and the population variance, 𝜎 2 is not known

𝑋̅−𝜇
t = 𝑆/ Has t distribution with n-1 degrees of freedom
√𝑛

⟹ (𝑋̅ − 𝑡𝛼/2 𝑆/√𝑛, 𝑋̅ + 𝑡𝛼/2 𝑆/√𝑛) Is a 100(1 − 𝛼 )% confidence interval for𝜇. The unit of
measurement of the confidence interval is the standard error. This is just the standard deviation of
the sampling distribution of the statistic.

Example: From a normal sample of size 25 a mean of 32 was found .Given that the population
standard deviation is 4.2. Find

a) A 95% confidence interval for the population mean.

b) A 99% confidence interval for the population mean.

Solution:

a)

𝑋̅ = 32, 𝜎 = 4.2 1 − 𝛼 = 0.95 ⟹ 𝛼 = 0.05, 𝛼/2 = 0.025

⟹ 𝑍𝛼/2 = 1.96 from table

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⟹ 𝑇ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑋̅ + 𝑡𝛼/2 𝜎/√𝑛

= 32 ± 1.96 × 4.2/√25

= 32 ± 1.65

= (30.35,33.65)

b)

𝑋̅ = 32, 𝜎 = 4.2 1 − 𝛼 = 0.99 ⟹ 𝛼 = 0.01, 𝛼/2 = 0.005

⟹ 𝑍𝛼/2 = 2.58 from table

⟹ 𝑇ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑋̅ + 𝑍𝛼/2 𝜎/√𝑛

= 32 ± 2.58 × 4.2/√25

= 32 ± 2.17

= (29.83,34.17)

Example2.A Drug Company is testing a new drug which is supposed to reduce blood pressure.
From the six people who are used as subjects, it is found that the average drop in blood pressure
is 2.28 points, with a standard deviation of 0.95 points. What is the 95% confidence interval for
the mean change in pressure?

𝑋̅ = 32, 𝑆 = 0.95 1 − 𝛼 = 0.95 ⟹ 𝛼 = 0.05, 𝛼/2 = 0.025

⟹ 𝑡𝛼/2 = 2.571 With df =5

⟹ 𝑇ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑋̅ + 𝑡𝛼/2 𝑆/√𝑛

= 28 ± 2.571 × 0.95/√6

= 2.28 ± 1.008

= (1.28,3.28)

That is, we can be 95% confident that the mean decrease in blood pressure is between 1.28 and
3.28 points.

2.5 Student’s t-distribution

When sampling is from a normal distribution with 𝜎 2 unknown and small sample size

- The relevant test statistic is

𝑋̅ − 𝜇
t= ~𝑡 𝑤𝑖𝑡ℎ 𝑛 − 1 𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚
𝑆/√𝑛
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- After specifying α we have the following regions on the student t-distribution corresponding
H0 Reject H0 if Accept H0 if Inconclusive if

𝜇 ≠ 𝜇0 |𝑡𝑐𝑎𝑙 | > 𝑡𝛼/2 |𝑡𝑐𝑎𝑙 | < 𝑡𝛼/2 𝑡𝑐𝑎𝑙 = 𝑡𝛼/2 = −𝑡𝛼/2

𝜇 < 𝜇0 𝑡𝑐𝑎𝑙 < −𝑡𝛼 𝑡𝑐𝑎𝑙 > −𝑡𝛼 𝑡𝑐𝑎𝑙 = −𝑡𝛼

𝜇 > 𝜇0 𝑡𝑐𝑎𝑙 > 𝑡𝛼 𝑡𝑐𝑎𝑙 < 𝑡𝛼 𝑡𝑐𝑎𝑙 = 𝑡𝛼

𝑋̅−𝜇0
Where: 𝑡𝑐𝑎𝑙 = 𝑆/√𝑛
Example1: Test the hypotheses that the average height content of containers of certain lubricant
is 10 liters if the contents of a random sample of 10 containers are 10.2, 9.7, 10.1, 10.3, 10.1,
9.8, 9.9, 10.4, 10.3, and 9.8 liters. Use the 0.01 level of significance and assume that the
distribution of contents is normal.
Solution:
Let μ =Population mean, 𝜇0 = 10
Step 1: Identify the appropriate hypothesis
H0 ∶ 𝜇 = 10 𝑉𝑆 H1 : 𝜇 ≠ 10
Step 2: select the level of significance, α =0.01 (given)
Step 3: Select an appropriate test statistics
t- Statistic is appropriate because population variance is not known and the sample size is
also small.
Step 4: identify the critical region.
Here we have two critical regions since we have two tailed hypothesis
The critical region is
|t cal | > t 0.005 (9) = 3.2498
⟹ (−3.2498,3.2498) is acceptance region
Step 5: Computations:
̅ = 10.06, S = 0.25
X

𝑋̅ − 𝜇0 10.06 − 10
⟹ 𝑡𝑐𝑎𝑙 = = = 0.76
𝑆 0.25
√𝑛 √10
Step 6: Decision

Accept H0 , since tcal is in the acceptance region.

Step 7: Conclusion: At 1% level of significance, we have no evidence to say that the average
height content of containers of the given lubricant is different from 10 litters, based on the given
sample data.

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2.6. Determining the sample size
Sample size is a research term used for defining the number of individuals included in a research
study to represent a population. The sample size references the total number of respondents
included in a study, and the number is often broken down into sub-groups by demographics such
as age, gender, and location so that the total sample achieves represents the entire population.
Determining the appropriate sample size is one of the most important factors in statistical analysis.
If the sample size is too small, it will not yield valid results or adequately represent the realities of
the population being studied. On the other hand, while larger sample sizes yield smaller margins
of error and are more representative, a sample size that is too large may significantly increase the
cost and time taken to conduct the research.

This article will discuss considerations to put in place when determining your sample size and how
to calculate the sample size.

Confidence Interval and Confidence Level

As we have noted before, when selecting a sample there are multiple factors that can impact the
reliability and validity of results, including sampling and non-sampling errors. When thinking
about sample size, the two measures of error that are almost always synonymous with sample sizes
are the confidence interval and the confidence level.

Confidence Interval (Margin of Error)

Confidence intervals measure the degree of uncertainty or certainty in a sampling method and how
much uncertainty there is with any particular statistic. In simple terms, the confidence interval tells
you how confident you can be that the results from a study reflect what you would expect to find
if it were possible to survey the entire population being studied. The confidence interval is usually
a plus or minus (±) figure. For example, if your confidence interval is 6 and 60% percent of your
sample picks an answer, you can be confident that if you had asked the entire population, between
54% (60-6) and 66% (60+6) would have picked that answer.

Confidence Level
The confidence level refers to the percentage of probability, or certainty that the confidence
interval would contain the true population parameter when you draw a random sample many times.
It is expressed as a percentage and represents how often the percentage of the population who
would pick an answer lies within the confidence interval. For example, a 99% confidence level
means that should you repeat an experiment or survey over and over again, 99 percent of the time,
your results will match the results you get from a population.

The larger your sample size, the more confident you can be that their answers truly reflect the
population. In other words, the larger your sample for a given confidence level, the smaller your
confidence interval.

28 | P a g e
Standard Deviation
Another critical measure when determining the sample size is the standard deviation, which
measures a data set’s distribution from its mean. In calculating the sample size, the standard
deviation is useful in estimating how much the responses you receive will vary from each other
and from the mean number, and the standard deviation of a sample can be used to approximate the
standard deviation of a population.

The higher the distribution or variability is the greater the standard deviation and the greater the
magnitude of the deviation. For example, once you have already sent out your survey, how much
variance do you expect in your responses? That variation in responses is the standard deviation.

Population Size
The other important consideration to make when determining your sample size is the size of the
entire population you want to study. A population is the entire group that you want to draw
conclusions about. It is from the population that a sample is selected, using probability or non-
probability samples. The population size may be known (such as the total number of employees in
a company), or unknown (such as the number of pet keepers in a country), but there’s a need for a
close estimate, especially when dealing with a relatively small or easy to measure groups of people.

As demonstrated through the calculation below, a sample size of about 385 will give you a
sufficient sample size to draw assumptions of nearly any population size at the 95% confidence
level with a 5% margin of error, which is why samples of 400 and 500 are often used in research.
However, if you are looking to draw comparisons between different sub-groups, for example,
provinces within a country, a larger sample size is required. GeoPoll typically recommends a
sample size of 400 per country as the minimum viable sample for a research project, 800 per
country for conducting a study with analysis by a second-level breakdown such as females versus
males, and 1200+ per country for doing third-level breakdowns such as males aged 18-24 in
Nairobi.

How to Calculate Sample Size

As we have defined all the necessary terms, let us briefly learn how to determine the sample size
using a sample calculation formula known as Andrew Fisher’s Formula.

1. Determine the population size (if known).

2. Determine the confidence interval.
3. Determine the confidence level.
4. Determine the standard deviation (a standard deviation of 0.5 is a safe choice where the
figure is unknown)
5. Convert the confidence level into a Z-Score. This table shows the z-scores for the most
common confidence levels:

Confidence level z-score

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 80% 1.28

85% 1.44

90% 1.65

95% 1.96

99% 2.58

6. Put these figures into the sample size formula to get your sample size.

Sample Size Calculation

RANGE 2
( )
2
Sample size = ACCURACY LEVEL 2
( )
CONFIDENCE LEVEL

Example: Say you choose to work with a 95% confidence level, a standard deviation of 0.5, and
a confidence interval (margin of error) of ± 5%, you just need to substitute the values in the
formula:

= ((1.96)2 x .5(.5)) / (.05)2

= (3.8416 x .25) / .0025

= 0.9604 / .0025

= 384.16

Your sample size should be 385.

Exercises:

1. An electrical firm manufactures light bulbs that have a length of life that is approximately
normally distributed with a standard deviation of 40 hours. If a random sample of 30 bulbs
has an average life of 780 hours, find a 99% confidence interval for the population mean
of all bulbs produced by this firm.
2. A random sample of 400 households was drawn from a town and a survey generated data
on weekly earning. The mean in the sample was Birr 250 with a standard deviation Birr
80. Construct a 95% confidence interval for the population mean earning.
3. A major truck has kept extensive records on various transactions with its customers. If a
random sample of 16 of these records shows average sales of 290 liters of diesel fuel with
a standard deviation of 12 liters, construct a 95% confidence interval for the mean of the
population sampled.

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 CHAPTER THREE

HYPOTHESIS TESTING

3.1 Basic Concepts

Hypothesis testing or significance testing is a method for testing a claim or hypothesis about a
parameter in a population, using data measured in a sample. In this method, we test some
hypothesis by determining the likelihood that a sample statistic could have been selected, if the
hypothesis regarding the population parameter were true.

Hypothesis is a statement about the value of a population parameter developed for testing.

Hypothesis is an assertion or tentative claim which requires justification.

The method of hypothesis testing can be summarized in four steps. We will describe each of
these four steps in greater detail in the next section.

1. To begin, we identify a hypothesis or claim that we feel should be tested. For example, we
might want to test the claim that the mean number of hours that children in Ethiopia watch
TV is 3 hours.

2. We select a criterion upon which we decide that the claim being tested is true or not. For
example, the claim is that children watch 3 hours of TV per week. Most samples we select
should have a mean close to or equal to 3 hours if the claim we are testing is true. So at
what point do we decide that the discrepancy between the sample mean and 3 is so big that
the claim we are testing is likely not true? We answer this question in this step of hypothesis
testing.

3. Select a random sample from the population and measure the sample mean. For example,
we could select 20 children and measure the mean time (in hours) that they watch TV per
week.

4. Compare what we observe in the sample to what we expect to observe if the claim we are
testing is true. We expect the sample mean to be around 3 hours. If the discrepancy between
the sample mean and population mean is small, then we will likely decide that the claim

31 | P a g e
 we are testing is indeed true. If the discrepancy is too large, then we will likely decide to
reject the claim as being not true.

3.2 Steps to Hypothesis Testing

The goal of hypothesis testing is to determine the likelihood that a population parameter, such as
the mean, is likely to be true. In this section, we describe the four steps of hypothesis testing that
were briefly introduced in Section 8.1:

Step 1: State the hypotheses.

Step 2: Set the criteria for a decision.

Step 3: Compute the test statistic.

Step 4: Make a decision.

Step 1: State the hypotheses. We begin by stating the value of a population mean in a null
hypothesis, which we presume is true. For the children watching TV example, we state the null
hypothesis that children in the United States watch an average of 3 hours of TV per week. This is
a starting point so that we can decide whether this is likely to be true, similar to the presumption
of innocence in a courtroom. When a defendant is on trial, the jury starts by assuming that the
defendant is innocent. The basis of the decision is to determine whether this assumption is true.
Likewise, in hypothesis testing, we start by assuming that the hypothesis or claim we are testing
is true. This is stated in the null hypothesis. The basis of the decision is to determine whether this
assumption is likely to be true.

The null hypothesis (H0), stated as the null, is a statement about a population parameter,
such as the population mean, that is assumed to be true.

The null hypothesis is a starting point. We will test whether the value stated in the null
hypothesis is likely to be true.

Keep in mind that the only reason we are testing the null hypothesis is because we think it is wrong.
We state what we think is wrong about the null hypothesis in an alternative hypothesis. For the
children watching TV example, we may have reason to believe that children watch more than (>)
or less than (<) 3 hours of TV per week. When we are uncertain of the direction, we can state that
the value in the null hypothesis is not equal to (≠) 3 hours.

32 | P a g e
In a courtroom, since the defendant is assumed to be innocent (this is the null hypothesis so to
speak), the burden is on a prosecutor to conduct a trial to show evidence that the defendant is not
innocent. In a similar way, we assume the null hypothesis is true, placing the burden on the
researcher to conduct a study to show evidence that the null hypothesis is unlikely to be true.
Regardless, we always make a decision about the null hypothesis (that it is likely or unlikely to be
true). The alternative hypothesis is needed for Step 2.

An alternative hypothesis (H1) is a statement that directly contradicts a null hypothesis by stating
that that the actual value of a population parameter is less than, greater than, or not equal to the
value stated in the null hypothesis.
The alternative hypothesis states what we think is wrong about the null hypothesis, which is needed
for Step 2.

Step 2: Set the criteria for a decision. To set the criteria for a decision, we state the level of
significance for a test. This is similar to the criterion that jurors use in a criminal trial. Jurors
decide whether the evidence presented shows guilt beyond a reasonable doubt (this is the
criterion). Likewise, in hypothesis testing, we collect data to show that the null hypothesis is not
true, based on the likelihood of selecting a sample mean from a population (the likelihood is the
criterion). The likelihood or level of significance is typically set at 5% in behavioral research
studies. When the probability of obtaining a sample mean is less than 5% if the null hypothesis
were true, then we conclude that the sample we selected is too unlikely and so we reject the null
hypothesis.

Level of significance, or significance level, refers to a criterion of judgment upon which a

decision is made regarding the value stated in a null hypothesis. The criterion is based on the
probability of obtaining a statistic measured in a sample if the value stated in the null
hypothesis were true.

In behavioral science, the criterion or level of significance is typically set at 5%. When the
probability of obtaining a sample mean is less than 5% if the null hypothesis were true, then
we reject the value stated in the null hypothesis.

The alternative hypothesis establishes where to place the level of significance. Remember that
we know that the sample mean will equal the population mean on average if the null hypothesis is
true. All other possible values of the sample mean are normally distributed (central limit theorem).

33 | P a g e
The empirical rule tells us that at least 95% of all sample means fall within about 2 standard
deviations (SD) of the population mean, meaning that there is less than a 5% probability of
obtaining a sample mean that is beyond 2 SD from the population mean. For the children watching
TV example, we can look for the probability of obtaining a sample mean beyond 2 SD in the upper
tail (greater than 3), the lower tail (less than 3), or both tails (not equal to 3). Figure 8.2 shows that
the alternative hypothesis is used to determine which tail or tails to place the level of significance
for a hypothesis test.

Step 3: Compute the test statistic. Suppose we measure samples mean equal to 4 hours per week
that children watch TV. To make a decision, we need to evaluate how likely this sample outcome
is, if the population mean stated by the null hypothesis (3 hours per week) is true. We use a test
statistic to determine this likelihood. Specifically, a test statistic tells us how far, or how many
standard deviations, a sample mean is from the population mean. The larger the value of the test
statistic, the further the distance, or number of standard deviations, a sample mean is from the
population mean stated in the null hypothesis. The value of the test statistic is used to make a
decision in Step 4.

The test statistic is a mathematical formula that allows researchers to determine the likelihood of
obtaining sample outcomes if the null hypothesis were true. The value of the test statistic is used
to make a decision regarding the null hypothesis.

Step 4: Make a decision. We use the value of the test statistic to make a decision about the null
hypothesis. The decision is based on the probability of obtaining a sample mean, given that the
value stated in the null hypothesis is true. If the probability of obtaining a sample mean is less than
5% when the null hypothesis is true, then the decision is to reject the null hypothesis. If the
probability of obtaining a sample mean is greater than 5% when the null hypothesis is true, then
the decision is to retain the null hypothesis. In sum, there are two decisions a researcher can make:

1. Reject the null hypothesis. The sample mean is associated with a low probability of
occurrence when the null hypothesis is true.
2. Retain the null hypothesis. The sample mean is associated with a high probability of
occurrence when the null hypothesis is true.
The probability of obtaining a sample mean, given that the value stated in the null hypothesis is
true, is stated by the p value. The p value is a probability: It varies between 0 and 1 and can never
34 | P a g e
be negative. In Step 2, we stated the criterion or probability of obtaining a sample mean at which
point we will decide to reject the value stated in the null hypothesis, which is typically set at 5%
in behavioral research. To make a decision, we compare the p value to the criterion we set in Step
2.

A p value is the probability of obtaining a sample outcome, given that the value stated in
the null hypothesis is true. The p value for obtaining a sample outcome is compared to the
level of significance.

Significance, or statistical significance, describes a decision made concerning a value

stated in the null hypothesis. When the null hypothesis is rejected, we reach significance.
When the null hypothesis is retained, we fail to reach significance.

When the p value is less than 5% (p < .05), we reject the null hypothesis. We will refer to p < .05
as the criterion for deciding to reject the null hypothesis, although note that when p = .05, the
decision is also to reject the null hypothesis. When the p value is greater than 5% (p > .05), we
retain the null hypothesis. The decision to reject or retain the null hypothesis is called significance.
When the p value is less than .05, we reach significance; the decision is to reject the null
hypothesis. When the p value is greater than .05, we fail to reach significance; the decision is to
retain the null hypothesis. Figure 8.3 shows the four steps of hypothesis testing.

3.3 Type I and Type II errors

In Step 4, we decide whether to retain or reject the null hypothesis. Because we are observing a
sample and not an entire population, it is possible that a conclusion may be wrong. Table 8.3 shows
that there are four decision alternatives regarding the truth and falsity of the decision we make
about a null hypothesis:

1. The decision to retain the null hypothesis could be correct.

2. The decision to retain the null hypothesis could be incorrect.
3. The decision to reject the null hypothesis could be correct.
4. The decision to reject the null hypothesis could be incorrect.
Four outcomes for making a decision. The decision can be either correct (correctly reject or
retain null) or wrong (incorrectly reject or retain null).

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DECISION: RETAIN THE NULL HYPOTHESIS

When we decide to retain the null hypothesis, we can be correct or incorrect. The correct decision
is to retain a true null hypothesis. This decision is called a null result or null finding. This is usually
an uninteresting decision because the decision is to retain what we already assumed: that the value
stated in the null hypothesis is correct. For this reason, null results alone are rarely published in
behavioral research.

The incorrect decision is to retain a false null hypothesis. This decision is an example of a Type
II error, or b error. With each test we make, there is always some probability that the decision
could be a Type II error. In this decision, we decide to retain previous notions of truth that are in
fact false. While it’s an error, we still did nothing; we retained the null hypothesis. We can always
go back and conduct more studies.

Type II error, or beta (b) error, is the probability of retaining a null hypothesis that is actually
false.

3.4 One tailed/Is two tailed hypothesis tests

DECISION: REJECT THE NULL HYPOTHESIS

When we decide to reject the null hypothesis, we can be correct or incorrect. The incorrect decision
is to reject a true null hypothesis. This decision is an example of a Type I error. With each test
we make, there is always some probability that our decision is a Type I error. A researcher who
makes this error decides to reject previous notions of truth that are in fact true. Making this type
of error is analogous to finding an innocent person guilty. To minimize this error, we assume a

36 | P a g e
defendant is innocent when beginning a trial. Similarly, to minimize making a Type I error, we
assume the null hypothesis is true when beginning a hypothesis test.

Type I error is the probability of rejecting a null hypothesis that is actually true.
Researchers directly control for the probability of committing this type of error.

An alpha (𝛼) level is the level of significance or criterion for a hypothesis test. It is the
largest probability of committing a Type I error that we will allow and still decide to
reject the null hypothesis.

Since we assume the null hypothesis is true, we control for Type I error by stating a level of
significance. The level we set, called the alpha level (symbolized as a), is the largest probability
of committing a Type I error that we will allow and still decide to reject the null hypothesis. This
criterion is usually set at .05 (𝛼 = 0.05), and we compare the alpha level to the p value. When the
probability of a Type I error is less than 5% (p < .05), we decide to reject the null hypothesis;
otherwise, we retain the null hypothesis.

The correct decision is to reject a false null hypothesis. There is always some probability that
we decide that the null hypothesis is false when it is indeed false. This decision is called the power
of the decision-making process. It is called power because it is the decision we aim for. Remember
that we are only testing the null hypothesis because we think it is wrong. Deciding to reject a false
null hypothesis, then, is the power, inasmuch as we learn the most about populations when we
accurately reject false notions of truth. This decision is the most published result in behavioral
research.

The power in hypothesis testing is the probability of rejecting a false null hypothesis. Specifically,
it is the probability that a randomly selected sample will show that the null hypothesis is false
when the null hypothesis is indeed false.

Example 1. Templer and Tomeo (2002) reported that the population mean score on the
quantitative portion of the Graduate Record Examination (GRE) General Test for
students taking the exam between 2018 and 2020 was 558 ± 139 (m ± s). Suppose
we select a sample of 100 participants (n = 100). We record a sample mean equal
to 585 (M = 585). Compute the one–independent sample z test for whether or not
we will retain the null hypothesis (m = 558) at a .05 level of significance (a = .05).

37 | P a g e
Step 1: State the hypotheses. The population mean is 558, and we are testing whether the null
hypothesis is (=) or is not (≠) correct:

H0: m = 558 Mean test scores are equal to 558 in the population.

H1: m ≠ 558 Mean test scores are not equal to 558 in the population.

Step 2: Set the criteria for a decision. The level of significance is .05, which makes the alpha level
a = .05. To locate the probability of obtaining a sample mean from a given population, we use the
standard normal distribution. We will locate the z scores in a standard normal distribution that are
the cutoffs, or critical values, for sample mean values with less than a 5% probability of
occurrence if the value stated in the null (m = 558) is true.

A critical value is a cutoff value that defines the boundaries beyond which less than 5%
of sample means can be obtained if the null hypothesis is true. Sample means obtained
beyond a critical value will result in a decision to reject the null hypothesis.

In a non-directional two-tailed test, we divide the alpha value in half so that an equal proportion
of area is placed in the upper and lower tail. Table 8.4 gives the critical values for one-and two-
tailed tests at a .05, .01, and .001 level of significance. Figure 8.4 displays a graph with the
critical values for Example 8.1 shown. In this example a = .05, so we split this probability in
half:

𝛼 0.05
Splitting 𝛼 in half = = 0.0250 in each tail
2 2

TABLE Critical values for one- and two-tailed tests at three commonly used levels of
significance

38 | P a g e
Example 2. A soft drink bottling company’s advertisement states that a bottle of its product

contains 330 milliliters (ml). But customers are complaining that the company is

under filling its products. To check whether the complaint is true or not, an

inspector may test the following hypothesis:

HO: The average content of a bottle of this product equals 330 ml against

H1: The average content of a bottle of this product is less than 330 ml.

Or symbolically,

HO:  ≤ 330 ml

H1:  < 330 ml

If the inspector takes a random sample of bottles of this product and finds that the mean content

per bottle is much less than 330 ml, then he may conclude that the complaint of the customers is

correct.

3.5. Hypothesis testing

Hypothesis testing is a procedure based on sample evidence and probability theory used to

determine whether the hypothesis is reasonable statement and should not be rejected, or is

unreasonable and should be rejected. Or hypothesis testing is a procedure for checking the validity

of a statistical hypothesis. It is the process by which we decide whether the null hypothesis should

be rejected or not. The value, computed from sample information, used to determine whether or
not to reject the null hypothesis is called test statistic.

ACCEPTANCE AND REJECTION (CRITICAL) REGIONS

As mentioned earlier, in order to determine whether HO is accepted or not, one computes a test

statistic from the sample data. Our decision is then based on where this figure (the test statistic)

falls.

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Depending on the sampling distribution of the statistic under consideration, one can identify the
acceptance region and the rejection region. If the test statistic falls in the acceptance region, then
HO is accepted and if the test statistic falls in the rejection region also called (the critical region),
then we reject HO and accept H1. The value that is a borderline between acceptance and rejection
is called the critical value. The critical value is obtained from appropriate statistical table such as
standard normal distribution table, the student t distribution table and others.

Some of these tables are indicated in the appendix.

The following three charts indicate the acceptance and rejection regions for a test of significance

that will be conducted later in the unit.

Figure 1Sampling Distribution for statistic Z for a one-Tailed Test [right], 0.05 level of
significance.

𝑧 = 1.64
Note in the chart that:

1. The area where the null hypothesis is not rejected includes the area to the left of 1.645.
Where the value 1.645 is Z.05 read from the table of standard normal distribution.

2. The area of rejection is to the right of 1.645

3. A one-tailed test (right) is being applied (this will be explained soon in the unit)

4. The .05 level of significance is chosen.

5. The sampling distribution of the statistic Z is normally distributed.

6. The value 1.645 separates the regions where the null hypothesis is rejected and where it is

not rejected.

7. The vale 1.645 is called the critical value.

Figure 2 Sampling Distribution for the statistic Z, one-tailed (Left) test, 0.05 level of

significance.

40 | P a g e
 𝑧 = −1.64

Figure 3 Regions of non-rejection and Rejection for a Two-Tailed Test, Z-statistics, 0.05 level of
significance.

𝑧 = −1.64 𝑧 = 1.64

We can construct a similar acceptance and rejection region while using the t-statistics.

Types of tests

Based on the form of the null and alternative hypothesis, there are two types of tests: a one-sided

test and a two-sided test.

a) One-tailed test a type of hypothesis test that involves a directional research

hypothesis. It specifies that the values of one group are either larger or smaller than
some specified population value.
A one-sided test: a test is said to be one sided (one tailed) when the alternative hypothesis,
H1, states a direction, such as:

HO ; The mean income of females is less-than or equal to the mean income of

males.

41 | P a g e
 H1: The mean income of males is greater than the mean income of females.

There are two kinds of one-sided test

i) A left-tailed test: This is a type of test in which the less than sign is involved in the
alternative hypothesis. It has one rejection region at the left tail of the appropriate

distribution.

Left-tailed test a one-tailed test in which the sample outcome is hypothesized to

be at the left tail of the sampling distribution.

Example 3. Suppose O is the hypothesized (assumed) mean.

A left-tailed test for  O is

HO:   O

H1:  < O

Figure 2 indicates the acceptance and rejection region of left-tailed test using the Z-distribution.

ii) A right-tailed test: This is a type of test in which the greater than sign is involved in the

alternative hypothesis. It has one rejection region at the right tail of the appropriate

distribution.

Right-tailed test a one-tailed test in which the sample outcome is hypothesized to be at the
right tail of the sampling distribution.
Example 4. Suppose  O is the assumed mean. A right-tailed test for  O is

HO:   O

H1:  >  O

b) Two-tailed test A type of hypothesis test that involves a non-directional research

hypothesis. We are equally interested in whether the values are less than or greater
than one another. The sample outcome may be located at both the lower and the higher
ends of the sampling distribution.

42 | P a g e
3.5.1 Hypothesis testing of the population mean
Mean of the population can be tested presuming different situations such as the population may be

normal or other than normal, it may be finite or infinite, sample size may be large or small, variance

of the population may be known or unknown and the alternative hypothesis many be two-sided or

one-sided. Our testing technique will differ in different situations. We may consider some of the

important situations.

Case I Population normal, large sample and population standard deviation is known. The
alternative hypothesis may be one-sided or two-sided.

In such a situation, Z-test (Z statistics) is used for testing hypothesis of the mean and the test
statistic Z is worked out as under

𝑥̅ − 𝜇0
𝑍=
𝜎/√𝑛

Where Z is the standard normal distribution

𝑥̅ is the sample mean

𝜇0 is the hypothesized mean

 is the population standard deviation

n is the sample size.

Example 5. Selam Hotel has been having average sales of 500 teacups per day. Because of the

development of bus stand nearby, he expects to increase its sales. For the first 50

days after the start of the bus stand, he recorded an average daily sale of 550 teacups

per day. From the past records, it is known that the sales standard deviation is 50.

On the bases of this sample information, can one conclude that Selam’s Hotel sales have increased?
Use 5 percent level of significance

Solution:
43 | P a g e
Taking the null hypothesis that sales average 500 teacups per day and they have not increased
unless proved, we can write:

HO: ≤500 cups per day

H1:  > 500 (as we want to conclude that sales have increased)

As the sample size is 50 > 30 (n > 30), and the population standard deviation  = 50 is known, we

shall use Z-test assuming normal population and shall work out the test statistic Z as:

𝑥̅ − 𝜇0
𝑍=
𝜎/√𝑛

550−500 50 1
= = = = ඥ50 = 7.03
50/ඥ50 50/ඥ50 1/ඥ50

As H1 is one-sided, we shall determine the rejection region applying one-tailed test (in the right

tail because H1 is of more than type) at 5 percent level of significance and it comes to as under,
using table of Z-distribution, Z0.05 = 1.96

Z0.05 = 1.96

The rejection region is Z > 1.96, and as the observed (calculated) test statistic is 7.03 which is
greater than 1.96, i.e. which is not in the rejection region, thus, there is an evidence to reject H O.

I.e. HO is rejected and we can conclude that the sample data indicate that Selam’s Hotel sales have
shown a considerable increase.

Example 6. The mean of a certain production process is known to be 50 with a standard

deviation of 3. The production manager may welcome any change is mean value

towards higher side but would like to safeguard against decreasing values of mean.

44 | P a g e
He takes a sample of 36 items that gives a mean value of 48.5. What inference should the manager

take for the production process on the basis of sample results? Use 1 percent level of significance

for the purpose.

Solution: -

Taking the mean value of the population to be 50, we may write:

HO: O ≤50

H1 :  < 50 (since the manager wants to safeguard against decreasing values of

mean)

And the given information as 𝑋̅ = 48.5,  = 3,

n = 36, assuming the population to be normal, we can work out the test statistic Z as under:

𝑥̅ − 𝜇0
𝑍=
𝜎/√𝑛

48.5−50 −1.5
= = = −3.00
3/ඥ36 3/6

As H1 is one-sided in the given question, we shall determine the rejection region applying one-
tailed test (in the left tail as H1 is of less than type) at 1 percent level of significance and it comes

to as under, using normal curve area table:

Rejection region is Z < Z   R: Z < -Z 0.01

 R: Z < -2.33

45 | P a g e
 Z = -2.33

The observed value of Z which we call Z calculated is –3 which is < - 2.33 i.e. it is in the rejection

region, and thus, HO is rejected at 1 percent level of significance. We can conclude that the

production process is showing mean which is significantly less than the population mean and this
calls for some corrective action concerning the said process.

Example 7. A sample of 400 male students is found to have a mean height of 67.47 inches. Can

it be reasonably regarded as a sample from a large population with mean height

67.39 inches and standard deviation 1.30 inches? Test at 5% level of significance.
Solution: Taking the null hypothesis that the mean height of the population is equal to 67.39

inches, we can write:

HO: O = 67.39

H1:   67.39

And the given information as 𝑋̅ = 67.47” = 1.30” n = 400. Assuming the population to be normal,

we can work out the test statistic Z as under:

𝑥̅ − 𝜇0
𝑍=
𝜎/√𝑛

67.47−67.39 0.08
= = 0.065 = 1.231
1.30/ඥ400

As H1 is two-sided in the given question, we shall be applying a two-tailed test. In the two-tailed

test, the rejection region at 5% level is

R: Z > 1.96

46 | P a g e
 Z = 1.96

As the observed value of Z is 1.231 which is less than 1.96, 1.231, is in the acceptance region for

the rejection region is R: Z > 1.96. Therefore, Ho is accepted. i.e. we may conclude that the given

sample (with mean height = 67.47”) can be regarded to have been taken from a population with
mean height 67.39” and standard deviation 1.30” at 5% level of significance.

* In case if the population is finite but population standard deviation is known, one can use a test-

statistic

𝑥̅ − 𝜇0
𝑍=
𝜎 𝑁−𝑛
ቆ√ ቇ
√ 𝑛 𝑁−1

N n
Where is what we call the finite population correction
N 1

And the procedure of testing is the same as in the above three examples.

Example 8. A workers’ union is on strike for higher wages. The union claims that the mean
salary for workers is at most Birr 8,400 per year. The legislator does not want to
reject the union’s claim, however, unless the evidence is very strong against it.
Assume that salaries follow a normal distribution and the population standard
deviation is known to be Birr 3000. A random sample of 64 workers is obtained,
and the sample mean is Br, 9,400. Test if the state legislator accepts the unions’
claim or not at 1% significance level.
Solution:

HO :   8,400

H1:  > 8,400

47 | P a g e
𝛿 = 3,000

𝑥̅ = 9,400

𝑛 = 64

𝛼 = 0.01

𝑧0.01 = 2.33

𝑥̅ −𝜇
Z= 𝛿
⁄
√𝑛

9,400−8,400
Z= 3,000
⁄
√64

Z= 2.67

𝑧0.01 = 2.33

Case II testing for the population mean; large sample, population standard deviation is
unknown

In such a case, the population standard deviation  will be approximated by the sample standard

deviation S provided the sample size is considerably greater than 30. And the test statistic will be:

And the testing procedure will be the same as in case I for different level of significance and

different types of tests (one-tailed or two-tailed) as mentioned in the three examples in case I.

Example 9. The Thompson’s discount store chain issues its own credit card. The credit manager
wants to find out if the mean monthly-unpaid balance is more than $ 400. The level

of significance is set at .05. A random check of 172 unpaid balances revealed the

sample mean to be $407 and the standard deviation of the sample to be $ 38. Should

48 | P a g e
 the credit manager conclude that the population mean is greater than $ 400, or is it

reasonable to assume that the difference of $ 7, ($ 407 - $ 400) is due to chance?

Solution:

The null and alternative hypotheses are stated as:

HO:   $ 400

H1:  > $ 400

Because the alternative hypothesis states a direction, a one-tailed test is applied. The critical value

of Z, Z = Z0.05 = 1.645. The computed value of Z, using the statistic

X  $ 407  $ 400
Z=   2.42
S n 38 172

As the computed value of the test statistic (2.42) is larger than the critical value (1.645) or as 2.42

is in the rejection region R: Z > 1.645, the null hypothesis is rejected and the credit manager can
conclude that the mean unpaid balance is greater than $ 400.

Case III Population Normal, Small sample and Standard deviation of the population is
Unknown

1) If the population is infinite, we use the test statistic

𝑥̅ −𝜇0
𝑡= Which follows a student t-distribution with (n – 1) degree of freedom.
𝑆/√𝑛

And S is the sample standard deviation given by the formula,

 Xi  X 
2

S=
n 1

2) If the population is finite, using the finite population correction, the test statistic used is modified

as:

𝑥̅ −𝜇0
𝑡 = 𝑆 𝑁−𝑛
With (n – 1) df.
ቆ√ ቇ
√𝑛 𝑁−1

49 | P a g e
While testing, the value of the test statistic calculated from the sample result will be compared
with the tabulated value of the t-distribution at the given level of significance.

Example 10. The specimen of copper wires drawn from a large lot has the following breaking

strength (in kg. Weight): 578, 572, 570, 568, 572, 578, 570, 572, 596, 544.

Test whether the mean breaking strength of the lot may be taken to be 578 kg. Weight at 5% level

of significance

Solution: -

Taking the null hypothesis that the population mean is equal to hypothesized mean of 578 kg., we

can write:

HO:  = O = 578 kg.

H1:   578 kg.

As the sample size is small (n = 10) and the population standard deviation is not known, we shall

use t-test assuming normal population and shall work out the test statistic t as under:

𝑥̅ − 𝜇0
𝑡=
𝑆/√𝑛

Calculating the sample mean, and sample standard deviation, one can obtain

𝑥̅ = 572 kg. And S = 12.72 kg.

572−578
Hence 𝑡=
12.72/ඥ10

Degree of freedom = (n – 1) = (10 – 1) = 9.

As H1 is two-sided, we shall determine the rejection regions applying two-tailed test at 5% level
of significance, and it comes to as under, using table of t-distribution for 9.d.f.

R:  t  > 2.262 t > t  / 2 or t < -t / 2

or  t  > t / 2
50 | P a g e
 Acceptance region – 2.262 < t < 2.262

As the observed value of t (i.e. –1.488) is in the acceptance region, we accept HO at 5% level and
conclude that the mean breaking strength of copper wires lot may be taken as 578 kg. Weight

* For two-tailed test in t-distribution, if the level is , the area to the right tail and the area to the

left tail must add up to . Thus we consider them each  / 2

EXERCISE:

1. A sample of 36 observations is selected from a normal population. The sample mean is 21,

and the sample standard deviation is 5. Conduct the following test of hypothesis using the
.05 level of significance.

HO :   20

H1 :  > 20

a) Is this a one-tailed or a two-tailed test?

b) State the decision rule?

c) What is your decision regarding HO?

2. A machine is set to fill a small bottle with 9.0 gms of medicine. It is claimed that the mean

weight is less than 9 grams. The hypothesis is to be tested at the .01 significance level.

A sample revealed these weights (in gms): 9.2, 8.7, 8.9, 8.6 8.8, 8.5, 8.7, and 9.0.

a) State the null and alternative hypotheses

b) How many degrees of freedom are there?

c) Compute the test statistic and arrive at a decision.

d) An electronic company wants to check if advertisement has a significant effect on the

number of TV sets that are sold within six months of production. A random sample of 600

TV sets reveals the following results.

51 | P a g e
 CHAPTER FOUR

CHI-SQUARE DISTRIBUTIONS

4.1 Areas of application

A Chi-square (𝑥 2 ) distribution is a continuous distribution ordinarily derived as the sampling
distribution of s sum of squares of independent standard normal variables.

Characteristics of the Chi-square distribution

1. It is a continuous distribution.
2. The 𝑥 2 distribution has a single parameter; the degree of freedom, 𝑣
3. The mean of the Chi-square distribution is 𝑣
4. The variance of the Chi-square distribution is2𝑣. Thus the mean and the variance depend
on the degree of freedom.
52 | P a g e
 5. It is based on a comparison of the sample of observed data (results) with the expected
results under the assumption that the null hypothesis is true.
6. It is skewed distribution and only non-negative values of the variable 𝑥 2 are possible. The
skewness decreases as 𝑣 increases; and when 𝑣 increases without limit it approaches a
normal distribution. It extends indefinitely in the positive direction.
7. The area under the curve is 1.0
Having the above characteristics, 𝑥 2 distribution has the following areas of application:
1. Test for independence between two variables
2. Testing for equality of several proportions
3. Goodness of fit tests (Binomial, Normal and Poisson)

4.1.1 Test for independence between two variables

A 𝑥 2 test of independence is used to analyze the frequencies of two variables with multiple
categories to determine whether the two variables are independent. That is, the Chi-square
distribution involves using sample data to test for the independence of two variables. The sample
data is given in two way table called a contingency table. Because the 𝑥 2 test of independence uses
a contingency table, the test is sometimes referred to as CONTINGENCY ANALYSIS (contingency
table test). The 𝑥 2 test is used to analyze, for example, the following cases:

 Whether employee absenteeism is independent of job classification

 Whether beer preference is independent of sex (gender)
 Whether favorite sport is independent of nationality
 Whether type of financial investment is independent of Geographic region.
The steps and procedures are similar with hypothesis testing.

Example:
1. A company planning a TV advertising campaign wants to determine which TV shows its
target audience watches and thereby to know whether the choice of TV program an
individual watches is independent of the individuals income. The table supporting this is
shown below. Use a 5% level of significance and the null hypothesis.

Income Type of show

Sport Entertainment News Total

53 | P a g e
 Low 143 70 37 250

Medium 90 67 43 200

High 17 13 20 50

Total 250 150 100 500

Solution :

I. 𝐻𝑜 :Choice of TV program an individual watches is independent of the individuals income

𝐻1 : income and choice of TV program are not independent

II. Decision rule:

𝛼 = 0.05
𝑣 = (𝑅 − 1)(𝐶 − 1)1
𝑣 = (3 − 1)(3 − 1) = 4
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,4 = 9.49
Reject 𝐻𝑜 if sample 𝑥 2 is greater than 9.49
III. Compute the test statistic
In computing the test statistic our first task is to estimate the expected frequencies (𝑒𝑖𝑗 =
𝑟𝑖 𝑐𝑗
⁄𝑛) where
𝑟𝑖 = 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠 𝑡𝑜𝑡𝑎𝑙 𝑓𝑜𝑟 𝑟𝑜𝑤 𝑖
𝑐𝑗 = 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠 𝑡𝑜𝑡𝑎𝑙 𝑓𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛 𝑗
𝑛 = 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒

𝐞𝟏𝟏 = 𝟐𝟓𝟎 × 𝟐𝟓𝟎⁄𝟓𝟎𝟎 = 𝟏𝟐𝟓 𝒆𝟏𝟐 = 𝟐𝟓𝟎 × 𝟏𝟓𝟎⁄𝟓𝟎𝟎 = 𝟕𝟓 𝒆𝟏𝟑 = 𝟐𝟓𝟎 × 𝟏𝟎𝟎⁄𝟓𝟎𝟎 = 𝟓𝟎

e21 = 200 × 250⁄500 = 100 𝒆𝟐𝟐 = 𝟐𝟎𝟎 × 𝟏𝟓𝟎⁄𝟓𝟎𝟎 = 60 𝑒23 = 200 × 100⁄500 = 40

e31 = 50 × 250⁄500 = 25 𝑒32 = 50 × 150⁄500 = 15 𝑒33 = 50 × 100⁄500 = 10

A test of the null hypothesis that variables are independent of one another is based on the
magnitude of the differences between the observed frequencies and the expected frequencies.
Large differences between 𝑂𝑖𝑗 𝑎𝑛𝑑 𝑒𝑖𝑗 provide evidence that the null hypothesis is false. The test is
based on the following chi-square test statistic.

1
For the 𝑅 × 𝐶 contingency table, the degree of freedom are calculated as (𝑅 − 1)(𝐶 − 1). The degrees of freedom
refers to the number of expected frequencies that can be chosen freely provided the row and column totals of expected
frequencies are identical to the row and column totals of the observed frequency table.
54 | P a g e
 2
(𝑂𝑖𝑗 − 𝑒𝑖𝑗 )2 2
(𝑓𝑜 − 𝑓𝑒 )2
𝑥 =∑ 𝑜𝑟 𝑥 = ∑
𝑒𝑖𝑗 𝑓𝑒

Where 𝑂𝑖𝑗 (𝑓𝑜 ) =

𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑓𝑜𝑟 𝑐𝑜𝑛𝑡𝑖𝑛𝑔𝑒𝑛𝑐𝑦 𝑡𝑎𝑏𝑙𝑒 𝑐𝑎𝑡𝑒𝑔𝑜𝑟𝑦 𝑖𝑛 𝑟𝑜𝑤 𝑖 𝑎𝑛𝑑 𝑐𝑜𝑙𝑢𝑚𝑛 𝑗.

𝐸𝑖𝑗 (𝑓𝑒 ) = 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑓𝑜𝑟 𝑐𝑜𝑛𝑡𝑖𝑛𝑔𝑒𝑛𝑐𝑦 𝑡𝑎𝑏𝑙𝑒 𝑖𝑛 𝑟𝑜𝑤 𝑖 𝑎𝑛𝑑 𝑐𝑜𝑙𝑢𝑚𝑛 𝑗.

(143−125)2 (70−75)2 (37−50)2 (90−100)2 (67−60)2 (17−25)2 (13−15)2 (20−10)2

𝑥2 = + + + + + + + +
125 75 50 100 60 25 15 10
(43−40)2
= 21.174
40

IV. Reject the null hypothesis that choice of TV program is independent from income level

2. A human resource manager at EAGEL Inc. was interested in knowing whether the
voluntary absence behavior of the firm’s employees was independent of the marital status.
The employee files contained data on material status and on voluntary absenteeism
behavior for a sample of 500 employees is shown below.
Marital status

Absence of behavior Married Divorced Widowed Single Total

Often absent 36 16 14 34 100

Seldom absent 64 34 20 82 200

Never absent 50 50 16 84 200

Total 150 100 50 200 500

Test the hypothesis that the absence behavior is independent of marital status at a significance
level of 1%.

Solution:

I. 𝐻𝑜 : Voluntary absence behavior is independent of the marital status

𝐻1 : Voluntary absence behavior and marital status are dependent

II. Decision rule:

55 | P a g e
 𝛼 = 0.01
𝑣 = (𝑅 − 1)(𝐶 − 1)
𝑣 = (3 − 1)(4 − 1) = 6
𝑥 2 𝛼,𝑣 = 𝑥 2 0.01,6 = 16.81
Reject 𝐻𝑜 if sample 𝑥 2 > 16.81
III. Compute the test statistic

𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝑓𝑜 ) 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝐸𝑖𝑗 ) (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

𝑓𝑒

36 30 36 1.200

64 60 16 0.267

50 60 100 1.667

16 20 16 0.800

34 40 36 0.900

50 40 100 2.500

14 10 16 1.600

20 20 0 0.000

16 20 16 0.800

34 40 36 0.900

82 80 4 0.500

84 80 16 0.200

(𝑓𝑜 − 𝑓𝑒 )2 10.883
∑
𝑓𝑒

IV. Do not reject 𝐻𝑜 because 10.883 < 16.81

Voluntary absence behavior and marital status are independent

3. The personnel administrator of XYZ Company provided the following data as an example
of selection among 40 male and 40 female applicants for 12 open positions.
Applicant Status

Selected Not Selected Total

56 | P a g e
 Male 7 33 40

Female 5 35 40

Total 12 68 80

A. The 𝑥 2 test of independence was suggested as a way of determining if the decision to hire
7 males and females should be interpreted as having a selection bias in favor of males.
Conduct the test of independence using𝛼 = 0.10. What is your conclusion?
B. Using the same test, would the decision to hire 8 males and 4 females suggest concern for
a selection bias?
C. How many males could be hired for the 12 open positions before the procedure would
concern for a selection bias?
Solution:

A.
I.
Ho : There is no selection bias in favor of males. (Selection status and gender of the applicant are
independent).

𝐻1 : There is selection bias in favor of males. (Selection status and gender of the applicant are not
independent).

II. Decision rule:

𝛼 = 0.10
𝑣 = (𝑅 − 1)(𝐶 − 1)
𝑣 = (2 − 1)(2 − 1) = 1
𝑥 2 𝛼,𝑣 = 𝑥 2 0.10,1 > 2.71
Reject 𝐻𝑜 if sample 𝑥 2 > 2.71
III. Sample 𝑥 2
𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝐸𝑖𝑗 ) 𝑓𝑒

7 6 1 0.1667

33 34 1 0.0294

5 6 1 0.1667

35 34 1 0.0294

57 | P a g e
 (𝑓𝑜 − 𝑓𝑒 )2 0.3922
∑
𝑓𝑒

IV. Do not reject 𝐻𝑜 because 0.392 < 2.71

There is no selection bias in favor of males applicants
B.
I. 𝐻𝑜 There is no selection bias in favor of males. (Selection status and gender of the applicant
are independent).
𝐻1 : There is selection bias in favor of males. (Selection status and gender of the applicant are not
independent).

II. Decision rule:

𝛼 = 0.10
𝑣 = 𝑅 − 1)(𝐶 − 1)
(
𝑣 = (2 − 1)(2 − 1) = 1
𝑥 2 𝛼,𝑣 = 𝑥 2 0.10,1 = 2.71
Reject 𝐻𝑜 if sample 𝑥 2 > 2.71
III. Sample 𝑥 2
𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝐸𝑖𝑗 ) 𝑓𝑒

8 6 4 0.6667

32 34 4 0.1176

4 6 4 0.6667

36 34 4 0.1176

(𝑓𝑜 − 𝑓𝑒 )2 1.5686
∑
𝑓𝑒
IV. Do not reject 𝐻𝑜 because 1.569< 2.71
There is no selection bias in favor of males applicants
C. There is no shortcut method to answer this question. Therefore, let’s try by increasing the
number of male applicants who are accepted and decreasing the number of female
applicants who are females.
I.
There is no selection bias in favor of males. (Selection status and gender of the applicant are
independent).

58 | P a g e
𝐻1 : There is selection bias in favor of males. (Selection status and gender of the applicant are not
independent).

II. Decision rule:

𝛼 = 0.10
𝑣 = (𝑅 − 1)(𝐶 − 1)
𝑣 = (2 − 1)(2 − 1) = 1
𝑥 2 𝛼,𝑣 = 𝑥 2 0.10,1 = 2.71
Reject 𝐻𝑜 if sample 𝑥 2 > 2.71
III. Sample 𝑥 2
𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝐸𝑖𝑗 ) 𝑓𝑒

9 6 9 1.5000

31 34 9 0.2647

3 6 9 1.5000

37 34 9 0.2647

(𝑓𝑜 − 𝑓𝑒 )2 3.5294
∑
𝑓𝑒
IV. Reject 𝐻𝑜 because 3.5294 > 2.71
There is no selection bias in favor of males applicants
Therefore, 8 male and 4 female applicants must be hired for the 12 open positions so as to
avoid selection bias in favor of males.

The chi-square test for independence is useful in helping to determine whether a relationship exists
between two variables, but it does not enable us to estimate or predict the values of one variable
based on the value of the other. If it is determine that dependence does exist between two
quantitative variables, then the techniques of regression analysis are useful in helping to find a
mathematical formula that expresses the nature of mathematical relationship.

Small expected frequencies can lead to inordinately large chi-square values with the chi-square
test of independence. Hence contingency tables should not be used with expected cell values of
less than 5 one way to avoid small expected values is to combine columns or rows whenever
possible and whenever doing so makes sense.

59 | P a g e
 4.1.2 Testing for the equality of several proportions
Testing for the equality of several proportions emphasizes on whether several proportions are equal
or not; and hence the null hypothesis takes the following form:

𝐻𝑜 : 𝑃1 = 𝑃2 = 𝑃3 = 𝑃20 ; 𝑃3 = 𝑃30 ; … 𝑃𝑘
= 𝑃𝑘𝑜 ; 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑣𝑒 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠 𝑡𝑎𝑘𝑒𝑠 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑓𝑜𝑟𝑚

𝐻1 : 𝑡ℎ𝑒 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑧𝑒𝑑 𝑣𝑎𝑙𝑢𝑒𝑠.

The degree of freedom is determined as 𝑣 = 𝑘 − 1; where 𝑘 referes to the number of proportions

and all expected cell values must be greater than or equal to 5

Example:

1. In the business credit institution industry the accounts receivable for companies are
classified as being “current”. “Moderately late”, “very late” and “uncollectible”. Industry
figure shows that the ratio of these four classes is 9: 3: 3: 1
I. ENDURANCE firm has 800 accounts receivable, with 439, 168, 133, and 60 failing in each
class. Are these proportions in agreement with the industry ratio? 𝐿𝑒𝑡 𝛼 = 0.05
Solution:

I. 𝐻𝑜 : : 𝑃1 = 9⁄16 𝑃2 = 3⁄16 ; 𝑃3 = 3⁄16 ; 𝑃4 = 1⁄16

𝐻1 : at least one account is different from the other.

II. 𝛼 = 0.05
𝑣 = (𝐾 − 1) = (4 − 1) = 3
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,3 = 7.81
Reject 𝐻𝑜 if sample 𝑥 2 > 7.81
III. Test statistic (sample 𝑥 2 )

Class 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑒 = 𝑛𝑝𝑖 ) 𝑓𝑒

Current 439 450 121 0.269

Moderately late 168 150 324 2.160

Very late 133 150 289 1.927

Uncollectible 60 50 100 2.000

(𝑓𝑜 − 𝑓𝑒 )2 6.356
∑
𝑓𝑒

IV. Do not reject

60 | P a g e
 2. ETHIO PLASTIC factory sells its products in three primary colors: Red, Blue, and yellow.
The marketing manager of feels that customers have no color preference for the product. To
set this hypothesis the manger set up a test in which 120 purchases were given equal
opportunity to buy the product in each of the three colors. The results were that 60 bought
red, 20 bought blue, and 40 bought yellow. Test the marketing manger’s null hypothesis,
using𝛼 = 0.05.
Solution:

I. 𝐻𝑜 : : 𝑃𝑒𝑜𝑝𝑙𝑒 ℎ𝑎𝑣𝑒 no color preference for with this product 𝑃1 = 𝑃2 = 𝑃3 = 1⁄3

𝐻1 : 𝑃𝑒𝑜𝑝𝑙𝑒 ℎ𝑎𝑣𝑒 color preference for with this product

II. 𝛼 = 0.05
𝑣 = (𝐾 − 1) = (3 − 1) = 2
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,2 = 5.99
Reject 𝐻𝑜 if sample 𝑥 2 > 5.99
III. Test statistic (sample 𝑥 2 )

Class 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑒 𝑓𝑒
1
= 𝑛𝑝𝑖 ); 𝑝𝑖 = 2
3
Red 60 40 400 10

Blue 20 40 400 10

Yellow 40 40 0 0.00

(𝑓𝑜 − 𝑓𝑒 )2
∑ = 20
𝑓𝑒

IV. Reject 𝐻𝑜 ; because 20 > 5.99. This means that customers do have color preference. It
appears that red is the most popular color and blue is the least popular.

3. Rating sciences, Inc., a TV program-rating service, surveyed 600 families where the
television was turned on during the prime time on week nights. They found the following
numbers of people turned to the various networks.

2
Since the null hypothesis states that there is no color preference, each of the three colors is preferred by one third
of the purchases.
61 | P a g e
 Name of the network Type Number of viewers

EBS Commercial 210

Arts 170

Balageru 165

EBC Non commercial 55

600

A. Test the hypothesis that all four networks have equal proportions of viewers during this
prime time period. using𝛼 = 0.05.
B. Eliminate the results for EBC and repeat the test of hypothesis for the three commercial
networks, using𝛼 = 0.05.
C. Test the hypothesis that each of the three major networks has 30% of the weeknight prime
time market and EBC has 10% using𝛼 = 0.005.
Solution:

A.
I. 𝐻𝑜 : : 𝐴𝑙𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑢𝑟 𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠 𝑑𝑜 ℎ𝑎𝑣𝑒 𝑒𝑞𝑢𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑖𝑒𝑤𝑒𝑟𝑠; 𝑃1 = 𝑃2 = 𝑃3 =
𝑃4 = 1⁄4
𝐻1 : 𝐴𝑙𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑢𝑟 𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠 𝑑𝑜 𝑛𝑜𝑡 ℎ𝑎𝑣𝑒 𝑒𝑞𝑢𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑖𝑒𝑤𝑒𝑟𝑠

II. 𝛼 = 0.05
𝑣 = (𝐾 − 1) = (4 − 1) = 3
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,3 = 7.81
Reject 𝐻𝑜 if sample 𝑥 2 > 7.81

III. Test statistic (sample 𝑥 2 )

Class 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑒 𝑓𝑒
1
= 𝑛𝑝𝑖 ); 𝑝𝑖 = 3
4
EBS 210 150 3,600 24.0000

3
Since equal numbers of viewers are expected to watch each network, each of the four networks is watched by one
fourth of the viewers.
62 | P a g e
 Arts 170 150 400 2.6667

Balageru 165 150 225 1.5000

EBC 55 150 9,025 60.1667

(𝑓𝑜 − 𝑓𝑒 )2 88.3334
∑
𝑓𝑒

IV. Reject𝐻𝑜 ; because88.3334 > 7.81.

B.
I. 𝐻𝑜 : : 𝐴𝑙𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑢𝑟 𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠 𝑑𝑜 ℎ𝑎𝑣𝑒 𝑒𝑞𝑢𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑖𝑒𝑤𝑒𝑟𝑠; 𝑃1 = 𝑃2 = 𝑃3 =
1⁄
3
𝐻1 : 𝐴𝑙𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑢𝑟 𝑛𝑒𝑡𝑤𝑜𝑟𝑘𝑠 𝑑𝑜 𝑛𝑜𝑡 ℎ𝑎𝑣𝑒 𝑒𝑞𝑢𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑖𝑒𝑤𝑒𝑟𝑠

II. 𝛼 = 0.05
𝑣 = (𝐾 − 1) = (3 − 1) = 2
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,2 = 5.99
Reject 𝐻𝑜 if sample 𝑥 2 > 5.99
III. Test statistic (sample 𝑥 2 )

Class 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑒 𝑓𝑒
1
= 𝑛𝑝𝑖 ); 𝑝𝑖 =
3
EBS 210 181.67 802.62 4.4179

Arts 170 181.67 136.20 0.7497

Balageru 165 181.67 277.90 1.5270

(𝑓𝑜 − 𝑓𝑒 )2 6.6946
∑
𝑓𝑒

IV. Reject 𝐻𝑜 ; because 6.70 > 5.99

C.
Solution:

I. 𝐻𝑜 : : 𝑃1 = 𝑃2 = 𝑃3 = 0.30 ; 𝑃4 = 0.10

𝐻1 : 𝑂𝑛𝑒 𝑜𝑟 𝑚𝑜𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑠 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠.

63 | P a g e
 II. 𝛼 = 0.005
𝑣 = (𝐾 − 1) = (4 − 1) = 3
𝑥 2 𝛼,𝑣 = 𝑥 2 0.005,3 = 12.838
Reject 𝐻𝑜 if sample 𝑥 2 > 12.838
III. Test statistic (sample 𝑥 2 )

Class 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑒 = 𝑛𝑝𝑖 ); 𝑝𝑖 𝑓𝑒
1
=
3
EBS 210 180 900 5.00
Arts 170 180 100 0.55
Balageru 165 180 225 1.25
EBC 55 60 25 0.42
2 7.22
(𝑓𝑜 − 𝑓𝑒 )
∑
𝑓𝑒

IV. Do not Reject 𝐻𝑜 ; because 7.22 < 12.838

4. Suppose that three companies, A, B, C, have recently conducted aggressive advertising

campaigns in order to maintain and possibly increase their respective shares of the market
for a particular product. The market shares prior to the campaigns were 𝑃1 = 0.45 for
company A, 𝑃2 = 0.40 for company B, 𝑃3 = 0.13 for company C, and 𝑃4 = 0.02 for other
competitors. To determine if these market shares changed after the advertising campaigns,
a marketing analyst solicited the preferences of a random sample of 200 customers of this
product. Of these 200 customers 95 indicated a preference for company A’s product, 85
preferred company B’s product , 18 preferred company C’s product and the remainder
preferred one or another of the products distributed by the competitors conduct a test, at the
5% level of significance. If the market shares have changed from the levels what they were
at before the advertising campaigns.
Solution:

I. 𝐻𝑜 : : 𝑃1 = 0.45 , 𝑃2 = 0.40 , 𝑃3 = 0.13 ; 𝑃4 = 0.02

𝐻1 : 𝐴𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑃𝑖 𝑖𝑠 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑖𝑡𝑠 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝑣𝑎𝑙𝑢𝑒

II. 𝛼 = 0.05

𝑣 = (𝐾 − 1) = (3 − 1) = 2

64 | P a g e
 𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,2 = 5.99
Reject 𝐻𝑜 if sample 𝑥 2 > 5.99
Because of the above change 𝐻𝑜 is translated as:

𝐻𝑜 : : 𝑃1 = 0.45 , 𝑃2 = 0.40 , 𝑃3 = 0.15

𝐻1 : 𝐴𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑃𝑖 𝑖𝑠 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑖𝑡𝑠 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝑣𝑎𝑙𝑢𝑒

III. Test statistic (sample 𝑥 2 )

Class 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝑓𝑒 ) (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝑓𝑜 ) 𝑓𝑒

A 95 90 25 0.2778

B 85 80 25 0.1250

Others 20 304 100 3.3333

(𝑓𝑜 − 𝑓𝑒 )2 3.9236
∑
𝑓𝑒

IV. Do not reject 𝐻𝑜

There is no sufficient evidence at the 5% level of significance to conclude that the market
shares have changed from the levels they were at before the advertising campaign.

4.1.3 Goodness of fit tests (binomial, Normal, poison)

The chi-square test is widely used for a variety of analysis. One of the more important uses of chi-
square is the goodness-of-fit-test. That is, it can be used to decide whether a particular probability
distribution, such as the binomial, Poisson, or normal distribution. This is an important ability,
because as decision makers using statistics, we will need to choose a certain probability
distribution to represent the distribution of the data we happen to be considering.

In tests of hypothesis (Previous chapter), we assumed that the population was normal and tested
the hypothesis𝜇 = 𝜇𝑜 , 𝜌 = 𝜌𝑜 , , etc. but what if we want to check on the assumption of normality
itself? The multinomial 𝑥 2 goodness-of-fit-test can be applied.

4
Expected frequency value for 𝑃4 is less than 5 (200*0.02=4). So, we have to combine 𝑃4 with one of other expected
frequencies, say 𝑃3 , to obtain a combined expected frequency of 30 (200*0.15). it can also be combined with other
expected frequency values.
65 | P a g e
The null hypothesis for a goodness-of-fit-test is test in that the distribution of the population from
which a sample it taken is the one specified. The alternative hypothesis is that the actual
distribution is not the specified distribution. Generally, a researcher specifies only the name of
distribution and uses the sample data to estimate the particular parameters of the distribution. In
this situation one degree of freedom is test for each parameter that has to be estimated. However,
if the research completely specifies the distribution including parameter values, then no additional
degrees of freedom is lost.

Null hypothesis Parameters to be Degrees of

estimated freedom lost

𝐻𝑜 : population is normal 𝜇, 𝜎 2

𝐻𝑜 : population is normal with 𝜇 = 𝑥 𝜎 1

𝐻𝑜 : population is normal with 𝜎 = 𝑦 𝜇 1

𝐻𝑜 : population is normal with 𝜇 = None 0

𝑥, 𝜎 = 𝑦

𝐻𝑜 : population is Poisson 𝜆 1

𝐻𝑜 : population is Poisson with 𝜆 = Ζ None 0

𝐻𝑜 : population is Binomial 𝑝, 𝑞 = w None 0

Example (Binomial)

1. Miss Tsion, saleswoman for Moon Paper Company, has five accounts to visit per day. It is
suggested that sales by Miss Tsion May be described by the binomial distribution, with the
probability of selling each account being 0.40. Given the following frequency distribution
of Miss Tsion’s number of sales per day, can we conclude that the data do in fact follow
the binomial distribution? Uses 0.05 significance level
No of sales per day 0 1 2 3 4 5

66 | P a g e
 Frequency 10 41 60 20 6 3

Solution:

I. 𝐻𝑜 : : 𝑇ℎ𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝑏𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑤𝑖𝑡ℎ 𝑛 = 5 𝑎𝑛𝑑 𝑝 = 0.40

𝐻1 : 𝑇ℎ𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑜𝑡 𝑏𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑤𝑖𝑡ℎ 𝑛 = 5 𝑎𝑛𝑑 𝑝 = 0.40

II. 𝛼 = 0.05
𝑣 =𝐾−1−𝑚 =5−1−0 =4

𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,4 = 9.49

Reject 𝐻𝑜 if sample 𝑥 2 > 9.49

Because of the above change 𝐻𝑜 is translated as :

III. Test statistic (sample 𝑥 2 )

No Prob. With 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

of 𝑛 = 5 ,𝑝 ( )
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑓𝑜 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑛𝑝𝑖 ) 𝑓𝑒
sales = 0.40
per
day
0 0.0778 10 10.892 0.7957 0.0731

1 0.2592 41 36.288 22.2029 0.6119

2 0.3456 60 48.384 134.9315 2.7888

3 0.2304 20 32.256 150.2095 4.6567

4 & 0.0870 9 12.18 10.1124 0.8302

5
(𝑓𝑜 − 𝑓𝑒 )2 8.9607
∑
𝑓𝑒

IV. Do not reject 𝐻𝑜 the data are well described by the binomial distribution with
𝑛 = 5 , 𝑝 = 0.40

2. A professional baseball player, Philippos, was at bat five times in each of 100 games.
Philippos claims that he has a probability of 0.40 of getting a hit each time he goes to bat.
Test his claim at the 0.05 level by seeing if the following data are distributed binomially.
No of hits/game 0 1 2 3 4 5

67 | P a g e
 No of games with that number of hits 12 38 27 17 5 1

Solution:

I. 𝐻𝑜 : 𝑇ℎ𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑐𝑎𝑛 𝑏𝑒 𝑏𝑒𝑠𝑡 𝑑𝑒𝑠𝑐𝑟𝑖𝑏𝑒𝑑 𝑏𝑦 𝑏𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑛 =

5 𝑎𝑛𝑑 𝑝 = 0.40

𝐻1 : 𝑇ℎ𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑐𝑎𝑛′𝑡 𝑏𝑒 𝑏𝑒𝑠𝑡 𝑑𝑒𝑠𝑐𝑟𝑖𝑏𝑒𝑑 𝑏𝑦 𝑏𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑛 =

5 𝑎𝑛𝑑 𝑝 = 0.40

II. 𝛼 = 0.05
𝑣 =𝐾−1−𝑚 =5−1−0 =4
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,4 = 9.49
Reject 𝐻𝑜 if sample 𝑥 2 > 9.49
Because of the above change 𝐻𝑜 is translated as:

III. Test statistic (sample 𝑥 2 )

No of Prob. With No of games with 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

hits/game 𝑛 = 5 , 𝑝 that number of hits 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑒 𝑓𝑒
= 0.40 (𝑓𝑜 ) = 𝑛𝑝𝑖 )
0 0.0778 12 7.78 17.8084 2.2890

1 0.2592 38 25.92 145.9264 5.6249

2 0.3456 27 34.56 57.1536 1.6538

3 0.2304 17 23.04 36.4816 1.5834

4&5 0.0870 6 8.70 4.2900 0.8379

(𝑓𝑜 − 𝑓𝑒 )2 11.9940
∑
𝑓𝑒

IV. Reject 𝐻𝑜 the # of hit over the same is not normally distributed

3. The Ethiopian postal service is interested in modeling the “mangled letter” problem. It has
been suggested that any letter sent to a certain area has a 0.15 chance of being mangled.
Since the post office is so big, it can be assumed that two letters chances of being mangled
are independent. A sample of 310 people was selected, and two test letters were mailed to
each of them. The number of people receiving zero, one, or two mangled letters was 260,40
68 | P a g e
 and 10, respectively. At 0.10 level of significance, is it reasonable to conclude that the
number of mangled letters received by people follows a binomial distribution with𝑃 =
0.15?
Solution:
𝐻𝑜: The number of mangled letters received by people follows a binomial distribution with
𝑛=2 𝑎𝑛𝑑 𝑝=0.15
𝐻1: The number of mangled letters received by people doesn’t′ follow a binomial
distribution with 𝑛=2 𝑎𝑛𝑑 𝑝=0.15

Solution:

I. 𝐻𝑜 : The number of mangled letters received by people follows a binomial distribution with 𝑛 =
2 𝑎𝑛𝑑 𝑝 = 0.15
𝐻1 : The number of mangled letters received by people doesn′ t follow a binomial distribution with 𝑛
= 2 𝑎𝑛𝑑 𝑝 = 0.15

II. 𝛼 = 0.10
𝑣 =𝐾−1−𝑚 =3−1−0 =2
𝑥 2 𝛼,𝑣 = 𝑥 2 0.10,2 = 4.61
Reject 𝐻𝑜 if sample 𝑥 2 > 4.61
III. Test statistic (sample 𝑥 2 )

No of Prob. With Observed frequency 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

mangled 𝑛 = 2 ,𝑝 (𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑒 𝑓𝑒
letters = 0.15 = 𝑛𝑝𝑖 )
0 0.7225 260 223.9750 1297.8006 5.7944

1 0.2550 40 79.0500 1524.9025 19.2904

2 0.0225 10 6.9750 9.1506 1.3119

(𝑓𝑜 − 𝑓𝑒 )2 26.3967
∑
𝑓𝑒
IV. Reject 𝐻𝑜 .
The number of mangled letters received by people doesn′ t follow a binomial distribution with 𝑛 =
2 𝑎𝑛𝑑 𝑝 = 0.15
Example (Poisson)

1. It is hypothesized that the number of breakdowns per month of a computer system at a

major university follows a Poisson distribution with 𝜇 = 2. The data below show the

69 | P a g e
 observed number of breakdowns per month during a sample of 100 months. Use a 5% level
of significance and test the null hypothesis.
Breakdowns 0 1 2 3 4 5 and above

Observed frequency 14 20 34 22 5 3

Solution:

I. 𝐻𝑜 : The population distribution is Poisson with 𝜆 = 2.

𝐻1 : The population distribution is not Poisson with 𝜆 = 2.

II. 𝛼 = 0.05
𝑣 =𝐾−1−𝑚 =6−1−0 =5
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,5 = 11.07
Reject 𝐻𝑜 if sample 𝑥 2 > 11.07
III. Test statistic (sample 𝑥 2 )

Breakdowns Observed Prob. With 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

frequency (𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑒 𝑓𝑒
𝜆 = 2.
= 𝑛𝑝𝑖 )

0 14 0.1353 13.53 0.2209 0.0163

1 20 0.2707 27.07 49.9849 1.8465

2 34 0.2707 27.07 48.0249 1.7741

3 22 0.1804 18.04 15.6816 0.8693

4 5 0.0902 9.02 16.1604 1.7916

5 or more 3 0.0527 5.27 0.0729 0.0138

(𝑓𝑜 − 𝑓𝑒 )2 6.3117
∑
𝑓𝑒

IV. Do not reject𝐻𝑜 . The number of breakdowns per month of a computer system at the
university follows a Poisson distributionwith 𝜇 = 2. .
2. Suppose that a teller supervisor believes that the distribution of random arrivals at local
bank is Poisson and sets out to test the hypothesis by gathering information. The following
data represent a distribution of frequency of arrivals during one minute intervals at a bank.
Use 𝛼 = 0.05 to test these data in an effort to determine whether they are Poisson
Distributed.

70 | P a g e
 No of arrivals 0 1 2 3 4 5 and above

Observed frequency 7 18 25 17 12 5

Solution:
Before we solve the question, first we have to compute the arrival rate per minute, and
hence one degree of freedom is lost.
∑(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑟𝑟𝑖𝑣𝑎𝑙𝑠 ∗ 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦)
𝜆=
∑(𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦)
(0 ∗ 7) + (18 ∗ 1) + (25 ∗ 2) + (17 ∗ 3)(12 ∗ 4) + (5 ∗ 5) 192
= = = 2.3𝑐𝑢𝑠𝑡/𝑚𝑖𝑛
84 84
I. 𝐻𝑜 : The arrival of customers at a bank is Poisson distributed with 𝜆 = 2.3.
𝐻1 : The arrival of customers at a bank is not Poisson distributed with 𝜆 = 2.3.

II. 𝛼 = 0.05
𝑣 =𝐾−1−𝑚 =6−1−1 =4
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,4 = 9.488
Reject 𝐻𝑜 if sample 𝑥 2 > 9.488
III. Test statistic (sample 𝑥 2 )

Breakdowns Observed Prob. With 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

frequency (𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑒 𝑓𝑒
𝜆 = 2.3
= 𝑛𝑝𝑖 )

0 7 0.1003 8.4252 2.0312 0.2411

1 18 0.2306 19.3704 1.8778 0.0969

2 25 0.2652 22.2768 7.4158 0.3329

3 17 0.2033 17.0772 0.0060 0.0003

4 12 0.1169 9.8196 4.7541 0.4841

5 or more 5 0.0837 7.0308 4.1241 0.5866

(𝑓𝑜 − 𝑓𝑒 )2 1.795
∑
𝑓𝑒

IV. Do not reject𝐻𝑜 . The arrival of customers at a bank follows a Poisson distribution with 𝜆 =
2.3

71 | P a g e
 3. The number of automobile accidents occurring per day in a particular city is believed to
have a Poisson distribution. A sample of 80 days during the past year gives the data shown
below. Do the data support the belief that the number of accidents per day has a poison
distribution? Use 𝛼 = 0.05

No of accidents 0 1 2 3 4

Observed frequency(days) 34 25 11 7 3

Solution:
Before we solve the question, first we have to compute the occurrence rate per day, and
hence one degree of freedom is lost.
∑(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑟𝑟𝑖𝑣𝑎𝑙𝑠 ∗ 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦)
𝜆=
∑(𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦)
(0 ∗ 34) + (25 ∗ 1) + (11 ∗ 2) + (7 ∗ 3)(3 ∗ 4) 80
= = 1𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡/𝑑𝑎𝑦
80 80
I. 𝐻𝑜 : The occurence of acciddents per day follows a Poisson distribution with 𝜆 = 1.0
𝐻1 : : The occurence of acciddents per day does not follow a Poisson distribution with 𝜆
= 1.0

II. 𝛼 = 0.05
𝑣 =𝐾−1−𝑚 =4−1−1 =2
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,2 = 5.99
Reject 𝐻𝑜 if sample 𝑥 2 > 5.99
III. Test statistic (sample 𝑥 2 )

Breakdowns Observed Prob. With 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 (𝑓𝑜 − 𝑓𝑒 )2 (𝑓𝑜 − 𝑓𝑒 )2

frequency (𝑓𝑜 ) 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑒 𝑓𝑒
𝜆 = 1.0
= 𝑛𝑝𝑖 )

0 34 0.3679 29.4320 20.8666 0.7090

1 25 0.3679 29.4320 19.6426 0.6674

2 11 0.1839 14.7120 13.7789 0.9366

3 or more 10 0.0803 6.4240 12.7878 1.9906

(𝑓𝑜 − 𝑓𝑒 )2 4.3036
∑
𝑓𝑒

72 | P a g e
 IV. Do not reject𝐻𝑜 .
The occurence of acciddents follows a Poisson distribution with 𝜆 = 1.0
Example (Normal)

1. Suppose that Ato Paulos developed an overall attitude scale to determine how his
company’s employees feel toward their company. In theory the scores can vary from 0 to
50. Ato Paulos retests his measurement instrument on a randomly selected group of 100
employees. He tallies the scores and summarizes them in to six categories as shown below.
Are these retest scores approximately normally distributed with 𝜇 = 24.9 𝑎𝑛𝑑 𝜎 = 7.194?
Use 𝛼 = 0.05

Score category 10-15 15-20 20-25 25-30 30-35 35-40

Frequency 11 14 24 28 13 10

I. 𝐻𝑜 : The attitude scoress are normally distributed with 𝜇 = 24.9 𝑎𝑛𝑑 𝜎 = 7.194
𝐻1 : The attitude scoress are not normally distributed with 𝜇 = 24.9 𝑎𝑛𝑑 𝜎 = 7.194

II. 𝛼 = 0.05
𝑣 =𝐾−1−𝑚 =6−1−0 =5
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,5 = 11.07
Reject 𝐻𝑜 if sample 𝑥 2 > 11.07
III. Test statistic (sample 𝑥 2 )
𝑋−𝜇
With𝑧 = , the expected probability of each category can be obtained as follows:
𝜎

For category 10-15 Probability

10−24.9
𝑧10 = = −2.07 0.48077
7.194
15−24.9 −0.41621
𝑧15 = = −1.38
7.194

Expected probability 0.06456

For category 15-20 Probability

15−24.9 0.41621
𝑧15 = = −1.38
7.194
20−24.9
𝑧20 = = -0.68 −0.25175
7.194

Expected probability 0.16446

For category 20-25 Probability

20−24.9
𝑧20 = = −0.68 0.25175
7.194

73 | P a g e
 25−24.9 +0.00399
𝑧25 = = +0.01
7.194

Expected probability 0.25574

For category 25-30 Probability

25−24.9 0.00399
𝑧25 = = +0.01
7.194
30−24.9
𝑧30 = = +0.71 +0.26115
7.194

Expected probability 0.25716

For category 30-35 Probability

30−24.9
𝑧30 = = +0.71 0.26115
7.194
35−24.9 0.41924
𝑧35 = = +1.40
7.194

Expected probability 0.25716

For category 35-40 Probability

35−24.9 0.41924
𝑧35 = = +1.40
7.194
40−24.9
𝑧40 = = +2.10 0.48214
7.194

Expected probability 0.06290

The six probabilities do not sum to 1.00. even though observed frequencies were obtained only for
these six categories, getting a score less than 10 or greater than 40 was also possible. Because 0.50
of the probabilities liee in each half of a normal distribution utilizing the sum of expected
probabilities on each side of the mean, 24.9, we can obtain a probability of the <
10 𝑐𝑎𝑡𝑒𝑔𝑜𝑟𝑦: 0.5 − (0.06456 + 0.16446 + 0.25175) =
0.01923. 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦, 𝑤𝑤𝑒 𝑐𝑎𝑛 𝑜𝑏𝑡𝑎𝑖𝑛 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 > 40 𝑐𝑎𝑡𝑒𝑒𝑔𝑜𝑟𝑦: 0.5 −
(0.00399 + 0.25716 + 0.15809 + 0.06290) = 0.01786. expected frequencies can then be
obtained by multiplying each expected probability by thee total frequency (100), as shown below.

Score category Probability Expected frequency

(𝑓𝑒 = 𝑛𝑝𝑖 )
< 10 0.01923 1.923

10 − 15 0.06456 6.456 8.379

15 − 20 0.16446 16.446 16.446

74 | P a g e
 20 − 25 0.16845 25.574 25.574

25 − 30 0.25716 25.716 25.716

30 − 35 0.15809 15.809 15.809

35 − 40 0.06290 6.290 8.076

> 40 0.01786 1.786

As the < 10 𝑎𝑛𝑑 > 40 categories have values of less than 5, each must be combined
with the adjacent category. As a result, the < 10 category becomes part of the 10-15
categories and the > 40 category becomes part of the 35-40 categories.

Score category Probability Expected frequency (𝑓𝑒 =

𝑛𝑝𝑖 )

10 − 15 0.08379 8.379

15 − 20 0.16446 16.446

20 − 25 0.25574 25.574

25 − 30 0.25716 25.716

30 − 35 0.15809 15.809

35 − 40 0.08076 8.076

The value of the chi-square can then be computed.

Score Observed Probability Expected (𝑓𝑜 − 𝑓𝑒 )2 𝑓𝑜 − 𝑓𝑒 )2

category frequency frequency 𝑓𝑒
(𝑓𝑜 ) (𝑓𝑒 = 𝑛𝑝𝑖 )
10 − 15 11 0.08379 8.379 6.8696 0.8199

15 − 20 14 0.16446 16.446 5.9829 0.3638

20 − 25 24 0.25574 25.574 2.4775 0.0964

25 − 30 28 0.25716 25.716 5.2167 0.2029

30 − 35 13 0.15809 15.809 7.8905 0.4991

75 | P a g e
 35 − 40 10 0.08076 8.076 3.7018 0.4584

𝑓𝑜 − 𝑓𝑒 )2 2.4409
∑
𝑓𝑒

IV. Do not reject 𝐻𝑜 . The attitude score are normally distributed with mean 24.9 and standard
deviation 7.194.
2. The director of a major soccer team believes that the ages f purchasers of game tickets are
normally distributed. If the following data represent the distribution of ages for a sample
of observed purchasers of major soccer game tickets, use the chi-square goodness-of-fit-
test to determine whether this distribution is significantly different from the normal
distribution. Assume that 𝛼 = 0.05.

Age of purchaser 10-20 20-30 30-40 40-50 50-60 60-70

Frequency 16 44 61 56 35 19

I. 𝐻𝑜 : The ages of purchasers of soccer game tickets are normally distributed

𝐻1 : The ages of purchasers of soccer game tickets are not normally distributed

II. 𝛼 = 0.05
𝑣 =𝐾−1−𝑚 =6−1−2 =3
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,3 = 7.81
Reject 𝐻𝑜 if sample 𝑥 2 > 7.81

III. Test statistic (sample 𝑥 2 )

Age Observed Mid-point (m) 𝑓𝑚 𝑓𝑚2

category frequency (𝑓)
10-20 16 15 240 3600

20-30 44 25 1100 27500

30-40 61 35 2135 74725

40-50 56 45 2520 113400

50-60 35 55 1925 105875

60-70 19 65 1235 80275

231 ∑ 𝑓𝑚 =9155 ∑ 𝑓𝑚2 =405375

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 ∑ 𝑓𝑚 9155
𝑥̅ = = = 39.63
𝑛 231
2
(∑ 𝑓𝑚)
2 (9155)2
√ ∑ 𝑓𝑚 − √405375 −
𝑠= 𝑛 = 231 = 13.60
𝑛−1 231 − 1
𝑋−𝜇
With 𝑧 = , the expected probability of each category can be obtained as follows:
𝜎

For category 10-20 Probability

10−39.63
𝑧10 = = −2.18 0.4854
13.6
20−39.63
𝑧20 = = −1.44 −0.4251
13.6

Expected probability 0.06456

For category 20-30 Probability

20−39.63
𝑧20 = = −1.44 0.4251
13.6
30−39.63
𝑧30 = = −0.71 −0.26115
13.6

Expected probability 0.16446

For category 30-40 Probability

30−39.63
𝑧30 = = −0.71 0.26115
13.6
40−39.63
𝑧40 = = +0.03 +0.01197
13.6

Expected probability 0.27312

For category 40-50 Probability

40−39.63
𝑧40 = = +0.03 −0.01197
13.6
50−39.63 0.27637
𝑧50 = =+0.76
13.6

Expected probability 0.26440

For category 50-60 Probability

50−39.63 −0.27637
𝑧50 = = +0.76
13.6
60−39.63
𝑧60 = = +1.50 0.43319
13.6

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 Expected probability 0.15682

For category 60-70 Probability

20−39.63
𝑧60 = = +1.50 −0.43319
13.6
20−39.63
𝑧70 = = +2.33 0.48713
13.6

Expected probability 0.05394

The six probabilities do not sum to 1.00. Even though the observed frequencies were obtained only
for these six categories, getting a score less than 10 or greater than 70 was also possible.

For < 10

Probability between 10 and the mean =0.06030+0.16392+0.26115=0.48537

Probability < 10 = 0.5 − 0.48537 = 0.01463

For > 70

Probability between 70 and the mean =0.05394+0.15682+0.2640+0.01197=0. 48713

Probability > 70 = 0.5 − 0.48713 = 0.01287

Then, the expected frequencies can be obtained by multiplying each expected probability by thee
total frequency (231), as shown below.

Age Probability Expected frequency

category (𝑓𝑒 = 𝑛𝑝𝑖 )
< 10 0.01463 3.380

10-20 0.06030 13.929 17.309

20-30 0.16392 37.866 37.866

30-40 0.27312 63.091 63.091

40-50 0.26440 61.076 61.076

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 50-60 0.15682 36.225 36.225

60-70 0.05394 12.460

> 70 0.01287 2.973 15.433

As the < 10 𝑎𝑛𝑑 > 70 categories have values of less than 5, each must be combined
with the adjacent category. As a result, the < 10 category becomes part of the 10-20
categories and the > 70 category becomes part of the 60-70 categories.

Age Probability Expected frequency

category (𝑓𝑒 = 𝑛𝑝𝑖 )

10-20 0.07493 0.07493

20-30 0.16392 0.16392

30-40 0.27312 0.27312

40-50 0.26440 0.26440

50-60 0.15682 0.15682

60-70 0.06681 0.06681

The value of the chi-square can then be computed.

Score Observed Probability Expected (𝑓𝑜 − 𝑓𝑒 )2 𝑓𝑜 − 𝑓𝑒 )2

category frequency frequency 𝑓𝑒
(𝑓𝑜 ) (𝑓𝑒 = 𝑛𝑝𝑖 )
10-20 16 0.07493 0.07493 17.30883 1.7135 0.0990

20-30 44 0.16392 0.16392 37.6260 0.9937

30-40 61 0.27312 0.27312 4.3723 0.0693

40-50 56 0.26440 0.26440 25.7658 0.4219

50-60 35 0.15682 0.15682 1.5006 0.0414

60-70 19 0.06681 0.06681 12.72.35 0.8244

𝑓𝑜 − 𝑓𝑒 )2 2.4497
∑
𝑓𝑒
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IV. Do not reject 𝐻𝑜 . The age of purchasers of soccer game tickets are normally distributed.

3. The instructor for introductory statistics course attempts to construct the final examination
so that the grades are normally distributed with a mean of 65. From the sample of grades
appearing in the accompanying frequency distribution table, can you conclude that they
have achieved his objective? Use 𝛼 = 0.05.

Grade 30-40 40-50 50-60 60-70 70-80 80-90

Frequency 4 17 29 49 33 18

I. 𝐻𝑜 : The grades of students are normally distributed with a mean 65.

𝐻1 : The grades of students are not normally distributed with a mean 65.

II. α = 0.05
𝑣 =𝐾−1−𝑚 =5−1−1 =3
𝑥 2 𝛼,𝑣 = 𝑥 2 0.05,3 = 7.81
Reject 𝐻𝑜 if sample 𝑥 2 > 7.81
III. Test statistic (sample 𝑥 2 )

Grade Observed Mid-point 𝑓𝑚 𝑓𝑚2

frequency (𝑓) (m)
30-40 4 35 140 4,900

40-50 17 45 765 34,425

50-60 29 55 1595 87,725

60-70 49 65 3185 207,025

70-80 33 75 2475 185,625

80-90 18 85 1530 130,050

150 ∑ 𝑓𝑚 =9690 ∑ 𝑓𝑚2 =649,750

∑ 𝑓𝑚 9690
𝑥̅ = = = 64.60~65
𝑛 150
2
2 (∑ 𝑓𝑚) (9690)2
√∑ 𝑓𝑚 − 𝑛 √649,750 − 150 = 12.63
𝑠= =
𝑛−1 150 − 1
𝑋−𝜇
With𝑧 = , the expected probability of each category can be obtained as follows:
𝜎
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 For category 30-40 Probability
30−65 0.49720
𝑧30 = = −2.77
12.63
40−65 0.47615
𝑧40 = = −1.98
12.63

Expected probability 0.02105

For category 40-50 Probability

40−65 0.47615
𝑧40 = = −1.98
12.63
50−65 −0.38298
𝑧50 = = −1.19
12.63

Expected probability 0.09317

For category 50-60 Probability

50−65 0.38298
𝑧50 = = −1.19
12.63
60−65 −0.15542
𝑧60 = = −0.40
12.63

Expected probability 0.22756

For category 60-70 Probability

60−65 0.15542
𝑧60 = =−0.40
12.63
70−65 +0.15542
𝑧70 = = +0.40
12.63

Expected probability 0.31084

For category 70-80 Probability

70−65 −0.15542
𝑧70 = = +0.40
12.63
80−65 0.38298
𝑧80 = = +1.19
12.63

Expected probability 0.22756

For category 80-90 Probability

80−65 −0.38298
𝑧80 = = +1.19
12.63
90−65 0.47615
𝑧90 = = +1.98
12.63

Expected probability 0.09317

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The six probabilities do not sum to 1.00. Even though the observed frequencies were obtained only
for these six categories, getting a score less than 30 or greater than 90 was also possible.

For < 30

Probability between 30 and the mean =0.02105+0.09317+0.22756+0.15542=0.49720

Probability < 30 = 0.5 − 0.49720 = 0.00280

For > 90

Probability between 90 and the mean =0.15542+0.022756+0.09317=0...47615

Probability > 90 = 0.5 − 0.0.47615 = 0.02385

Then, the expected frequencies can be obtained by multiplying each expected probability by thee
total frequency (150), as shown below.

Grade Probability Expected frequency

(𝑓𝑒 = 𝑛𝑝𝑖 )
< 30 0.00280 0.42

30-40 0.02105 3.1575 17.553

40-50 0.09317 13.9755

50-60 0.22756 34.134 34.134

60-70 0.31084 46.626 46.626

70-80 0.22756 34.134 34.134

80-90 0.09317 13.9755

> 90 0.02385 3.5775 17.553

Since the < 30, 30 − 40 𝑎𝑛𝑑 > 90 categories have values of less than 5, each must be combined
with the adjacent category. As a result, the < 30 𝑎𝑛𝑑 30 − 40 category becomes part of the 40-
50 categories and the > 90 category becomes part of the 80-90 category.

Grade Probability Expected frequency

(𝑓𝑒 = 𝑛𝑝𝑖 )

40-50 0.11702 17.553

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 50-60 0.22756 52.5664

60-70 0.31084 71.8040

70-80 0.22756 52.5664

80-90 0.11702 17.553

The value of the chi-square can then be computed.

Score Observed Probability Expected (𝑓𝑜 − 𝑓𝑒 )2 𝑓𝑜 − 𝑓𝑒 )2

category frequency frequency 𝑓𝑒
(𝑓𝑜 ) (𝑓𝑒 = 𝑛𝑝𝑖 )
40-50 21 0.11702 17.553 11.8818 0.6769

50-60 29 0.22756 52.5664 26.3580 0.7722

60-70 49 0.31084 71.8040 5.6359 0.1209

70-80 33 0.22756 52.5664 1.2860 0.0377

80-90 18 0.11702 17.553 0.1998 0.0114

𝑓𝑜 − 𝑓𝑒 )2 1.6190
∑
𝑓𝑒

IV. Do not reject𝐻𝑜 . The grades of students are normally distributed with a mean 65.

Exercise

1. The theory predicts the proportion of beans in the 4 groups A, B, C, D should be 9:3:3:1. In
an experiment among 1600 beans, the numbers in the four groups were 882, 313, 287 and 118.

2. A study was conducted, among 100 professors from 3 different divisions for the preference
on beverages of 3 categories test if there is any relationship between the field of teaching and
preference of beverage.

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3. A random sample of 600 students from Delhi University are selected and asked their opinion
ab out autonomous Status of Colleges. The results were given below. Test the hypothesis at 5%
level that opinions are independent of class groupings.

CHAPTER FIVE

ANALYSIS OF VARIANCE

5.1. Areas of applications

5.1.1. Comparison of the mean of more than two populations

There are situations where more than two populations are involved and we need to test the
significance of differences between three or more sample means. We also need to test the null
hypothesis that three or more populations from which independent samples are drawn have equal
[or homogeneous] means against the alternative hypothesis that population means are not equal.

An F test is used to test a hypothesis concerning the means of three or more populations, the
technique is called analysis of variance (commonly abbreviated as ANOVA). At first glance, you
might think that to compare the means of three or more samples, you can use the t test, comparing
two means at a time. But there are several reasons why the t test should not be done.

First, when you are comparing two means at a time, the rest of the means under study are ignored.
With the F test, all the means are compared simultaneously. Second, when you are comparing two
means at a time and making all pairwise comparisons, the probability of rejecting the null
hypothesis when it is true is increased, since the more t tests that are conducted, the greater is the
likelihood of getting significant differences by chance alone. Third, the more means there are to
compare, the more t tests are needed. For example, for the comparison of 3 means two at a time, 3
t tests are required. For the comparison of 5 means two at a time, 10 tests are required. And for the
comparison of 10 means two at a time, 45 tests are required.
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With the F test, two different estimates of the population variance are made. The first estimate is
called the between-group variance, and it involves finding the variance of the means. The second
estimate, the within-group variance, is made by computing the variance using all the data and is
not affected by differences in the means. If there is no difference in the means, the between-group
variance estimate will be approximately equal to the within-group variance estimate, and the F test
value will be approximately equal to 1. The null hypothesis will not be rejected. However, when
the means differ significantly, the between-group variance will be much larger than the within-
group variance; the F test value will be significantly greater than 1; and the null hypothesis will be
rejected. Since variances are compared, this procedure is called analysis of variance (ANOVA).
Let 𝜇1 , 𝜇2 , 𝜇3 ,…..,= 𝜇𝑘 be the mean value for population 1,2,3,….K respectively. Then from
sample data we intend to test the following hypothesis.

𝐻0 =𝜇1 = 𝜇2 = 𝜇3 =…..,= 𝜇𝑘
𝐻1 = Not all 𝜇𝑗 are equal j=1, 2, 3….K
I.e. the null hypothesis should be rejected if any of the n sample means is different from others. A
significant test value means that there is a high probability that this difference in means is not due
to chance, but it does not indicate where the difference lies.

The following are few examples involving more than two populations where it is necessary to
conduct a comparison to arrive at a statistical inference:

 Effectiveness of different promotional devices in terms of sales.

 Quality of a product produced by different manufacturers in terms of an attribute.
 Production volume in different shifts in a factory.
 Yield from plots of land due to varieties of seeds, fertilizers and cultivation methods.

Assumptions for analysis of variance

1. Each population has a normal distribution

2. The populations from which the samples are drawn have equal variance, i.e.
𝛿 21 = 𝛿 2 2 = 𝛿 2 3 =…... = 𝛿 2 𝑘
3. Each sample is drawn randomly and is independent of other samples.

Analysis of variance approach

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The first step in the analysis of variance is to partition the total variation in the sample data in to
the following two component variations in such a way that it is possible to estimate the contribution
of factors that may cause variation.

1. The amount of variation among the sample mean or the variation attributed to the
difference among sample means. This variation is due to assignable causes.
2. The amount of variation within the sample observations. This difference is considered due
to chance causes or experimental [random] errors.

One-way classification to test equality of population means

The classification is based on one factor or attribute of interest such as

i. Range of marks scored by students in a course.

ii. Favor preference of ice-cream by customers.
iii. Yield of crop due to variety of seeds and so on.

Suppose our main aim is to make inferences about 𝑘 population means based on sample data,
where, 𝜇𝑗 is the mean of the population of measurements associated with the treatment [𝑗 =
1,2,3, … 𝑘]. The null and alternative hypothesis to be tested is stated as:

𝐻0 =𝜇1 = 𝜇2 = 𝜇3 =…..,= 𝜇𝑘
𝐻1 = Not all 𝜇𝑗 are equal j=1,2,3….K
Let:
𝑛𝑗 = size of the 𝑗𝑡ℎ sample [𝑗 = 1,2,3, … 𝑘]
𝑛𝑗 = total number of observations in all samples combined i.e. [𝑛 = 𝑛1 + 𝑛2 + 𝑛3 … 𝑛𝑘 ]or [𝑛 =
𝑟𝑘] if 𝑖 = 𝑗
𝑥𝑖𝑗 = 𝑡ℎ𝑒 𝑖 𝑡ℎ Observation in 𝑗𝑡ℎ sample
Observation Populations [number of samples]

[measurements] 1 2 3

1 𝑥11 𝑥12 𝑥13

2 𝑥21 𝑥22 𝑥23

3 𝑥31 𝑥32 𝑥33

. . . .

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 . . . .

. . . .

r 𝑥𝑟1 𝑥𝑟2 𝑥𝑟𝑘

Sum 𝑇1 𝑇2 𝑇𝑘 = 𝑇

Arithmetic mean x̅1 x̅2 x̅k =𝑥̿

Where:
𝑟 𝑘 𝑟
1
𝑇𝑖 = ∑ 𝑥𝑖𝑗 𝑇 = ∑ 𝑇𝑖 x̅i = ∑ 𝑥𝑖𝑗
𝑟
𝑖=1 𝑗=1 𝑖=1
𝑘 𝑟 𝑘
1 1
𝑥̿ = ∑ 𝑥̅𝑗 = ∑ ∑ 𝑥𝑖𝑗
𝑟𝑘 𝑛
𝑗=1 𝑖=1 𝑗=1
The values of x̅ are called sample means and 𝑥̿ is the grand mean of all observations (or
measurements) in all the samples. Since there are r rows and k columns in the table above, then
total number of observations is = 𝑛 , provided each row has equal number of observations. But if
the number of observations in each row varies, then the total number of observations is 𝑛 = 𝑛1 +
𝑛2 + 𝑛3 … 𝑛𝑘 = 𝑛
Example 1. Three brands of tires, A, B, C were tested for durability. A sample of four types of
each brand is subjected to the same test and the number of kilometers until wear out was noted
from each brand of tires. The data in thousand kilometers is given below
Observations Samples [number of brands]

A B C

1 26 18 23

2 25 16 19

3 28 17 26

4 12 18 30

Mean 22.75 17.25 24.50

𝑠𝑖 2 𝑠𝐴 2 = 52.917 𝑠𝐵 2 = 0.917 𝑠𝐶 2 = 21.67

Since the same number of observations is obtained from each brand of tires [population], therefore
the number of observations in the table is 𝑛 = 𝑟𝑘
𝑛 = 4 × 3 = 12
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 The sample [population] mean of the three samples are given by
𝑟
1
x̅j = ∑ 𝑥𝑖𝑗
𝑟
𝑖=1
4 4 4
1 1 1 1 1 1
x̅A = ∑ 𝑥𝑖1 = (91) x̅B = ∑ 𝑥𝑖2 = (69) x̅C = ∑ 𝑥𝑖3 = (98)
4 4 4 4 4 4
𝑖=1 𝑖=1 𝑖=1

= 𝟐𝟐. 𝟕𝟓 = 𝟏𝟕. 𝟐𝟓 = 𝟐𝟒. 𝟓𝟎

The grand mean for all samples is
𝑘 𝑟 𝑘
1 1
𝑥̿ = ∑ 𝑥̅𝑗 = ∑ ∑ 𝑥𝑖𝑗
𝑟𝑘 𝑛
𝑗=1 𝑖=1 𝑗=1

4 3
1 1
𝑥̿ = ∑ ∑ 𝑥𝑖𝑗 = (26 + 25 + 28 + ⋯ 19 + 30) = 𝟐𝟏. 𝟓𝟎
12 12
𝑖=1 𝑗=1

Steps for testing the null hypothesis

STEP 1. State the hypothesis to test equality of population means as

𝐻0 =𝜇1 = 𝜇2 = 𝜇3 =…..,= 𝜇𝑘
𝐻1 = Not all 𝜇𝑗 are equal j=1, 2, 3….K
STEP 2. Find the critical values. Since 𝑘 = 3 𝑎𝑛𝑑 𝑛 = 12 and 𝛼 = 0.05

. The degrees of freedom for this application are the denominators in the mean squares, that is,
𝑉1 = 𝑘 − 1 and 𝑉2 = 𝑛 − 𝑘
𝑉1 𝑑𝑓 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 = 𝑘 − 1 ,𝑉1 = 3 − 1 = 𝟐
𝑉2 𝑑𝑓 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 = 𝑛 − 𝑘, 𝑉2 = 12 − 3 = 𝟗
𝐹𝛼,(𝑘−1,𝑛−𝑘) = 𝐹0.05,(3−1,12−3) = 𝟒. 𝟐𝟔

𝑟𝑒𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑔𝑖𝑜𝑛

4.26

STEP 3. Compute the test statistics, using the procedure outlined here

88 | P a g e
 a) Find the mean and variance of each sample
b) Find the grand mean, denoted by 𝑥̿ , it is the mean of values in the samples

∑ 𝑥 26 + 25 + 28 + ⋯ … … … … + 30
𝑥̿ = = = 21.5
𝑛 12
1 22.75+17.25+24.50
If the samples are equal in size then, 𝑥̿ = 𝑘 (𝑥̅1 , 𝑥̅ 2 , 𝑥̅ 3 … … . 𝑥̅ 𝑘 ) = = 21.5
3
c) Find the between group variance, denoted by MSB

If the null hypothesis is true, the population means would all be equal. We would then expect
that the sample means would be close to one another. If the alternative hypothesis is true,
however, there would be large differences between some of the sample means.

Sum of squares between samples [SSB] measures the proximity of the sample means to each other.

Calculate the difference between the mean of each sample and the grand mean as x̅1 − 𝑥̿ , x̅2 −
𝑥̿ , x̅3 − 𝑥̿ … x̅k − 𝑥̿ . Multiply each of these by the number of observations in the corresponding
sample and sum. The total gives the sum of the squared differences between the sample means in
each group and is denoted by𝑆𝑆𝐵.

The variation between sample means:

𝑘

𝑆𝑆𝐵 = ∑ 𝑛𝑗 (𝑥̅ − 𝑥̿ )2
𝑗=1
= 𝑛𝐴 (𝑥̅𝐴 − 𝑥̿ )2 + 𝑛𝐵 (𝑥̅ 𝐵 − 𝑥̿ )2 + 𝑛𝐶 (𝑥̅𝐶 − 𝑥̿ )2
= 4(22.75 − 21.50)2 + 4(17.25 − 21.50)2 + 4(24.50 − 21.50)2
= 6.25 + 72.25 + 36
=𝟏𝟏𝟒. 𝟓𝟎
𝑆𝑆𝐵 ∑𝑘𝑗=1 𝑛𝑗 (𝑥̅ − 𝑥̿ )2 114.50
𝑀𝑆𝐵 = = 𝑀𝑆𝐵 = = = 𝟓𝟕. 𝟐𝟓
𝐾−1 𝐾−1 3−1
4(22.75 − 21.5)2 + 4(17.25 − 21.5)2 + 4(24.5 − 21.5)2 114.5
= = = 57.25
3−1 2
d) Find within group variance -within sample variability

If large differences exist between sample means, at least some sample means differ considerably
from the grand mean, producing a large value of SSB. It is then rationale to reject the null
hypothesis in favor of the alternative hypothesis. Sum of square within sample [SSW] provides a
measure of the amount of variation in the response variable that is not caused by the samples.

89 | P a g e
 𝑟 𝑘

𝑆𝑆𝑊 = ∑ ∑(𝑥𝑖𝑗 − 𝑥̅𝑗 )2

𝑖=1 𝑗=1

Where:
∑𝑘𝑗=1(𝑥1𝑗 − 𝑥̅𝐴 )2 + ∑𝑘𝑗=1(𝑥2𝑗 − 𝑥̅ 𝐵 )2 + ∑𝑘𝑗=1(𝑥3𝑗 − 𝑥̅ 𝐶 )2 +… … … … . ∑𝑘𝑗=1(𝑥𝑖𝑗 − 𝑥̅𝑗 )2
= [(26 − 22.75)2 +(25 − 22.75)2 (28 − 22.75)2 (12 − 22.75)2 ]+[(18 − 17.25)2 + (16 −
17.25)2 + (17 − 17.25)2 + (18 − 17.25)2 ] + [(23 − 24.50)2 + (19 − 24.50)2 + (26 −
24.50)2 + (30 − 24.50)2 ]

= 10.5625 + 5.0625 + 27.5625 + 115.5625 + 0.5625 + +1.5625 + 0.0625 + 0.5625

+ 2.25 + 30.25 + 2.25 + 30.25

=𝟐𝟐𝟔. 𝟓𝟎

STEP 4. Calculate average between and within variation

𝑆𝑆𝑊 226.50
𝑀𝑆𝑊 = 𝑀𝑆𝑆𝐸 = 𝑛−𝑘 = =𝟐𝟓. 𝟏𝟔𝟕
12−3

Or in the easiest way

𝑀𝑆𝑊 = 𝑀𝑆𝑆𝐸

∑(𝑛𝑖 − 1)𝑠𝑖 2 = ∑𝑟𝑖=1 ∑𝑘𝑗=1(𝑥𝑖𝑗 − 𝑥̅𝑗 )2

∑(𝑛𝑖 − 1) = 𝑛 − 𝑘
𝑆𝑆𝐸 ∑𝑟𝑖=1 ∑𝑘𝑗=1(𝑥𝑖𝑗 − 𝑥̅𝑗 )2
𝑀𝑆𝑆𝐸 = =
𝑛−𝑘 𝑛−𝑘
So
∑(𝑛𝑖 − 1)𝑠𝑖 2
𝑀𝑆𝑆𝐸 =
∑(𝑛𝑖 − 1)

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 ∑(𝑛𝑖 − 1)𝑠𝑖 2 ∑(𝑛𝑖 − 1)𝑠𝑖 2
𝑀𝑆𝑊 = = 𝑀𝑆𝑆𝐸 =
∑(𝑛𝑖 − 1) ∑(𝑛𝑖 − 1)
∑(𝑛𝑖 − 1)𝑠𝑖 2
𝑀𝑆𝑊 =
∑(𝑛𝑖 − 1)
(4 − 1)(52.917) + (4 − 1)(0.9167) + (4 − 1)(21.67) 226.50
𝑀𝑆𝑊 = =
(4 − 1) + (4 − 1) + (4 − 1) 9
= 𝟐𝟓. 𝟏𝟔𝟕
Note:

This formula finds an overall variance by calculating a weighted average of the variances. It does
not involve using differences of the means.

e) Find the F test statistics value

𝑀𝑆𝐵 57.25
𝐹= = = 𝟐. 𝟐𝟕
𝑀𝑆𝑊 25.167
STEP 5. Make the decision. The decision is to accept 𝐻0 Since F test statistic value is less
than F table i.e. 2.27 <4.26
STEP 6. Summarize the results. There is enough evidence to accept the claim and conclude
that the means from the three populations are approximately equal.

Example 2: Lowering Blood Pressure

A researcher wishes to try three different techniques to lower the blood pressure of individuals
diagnosed with high blood pressure. The subjects are randomly assigned to three groups; the first
group takes medication, the second group exercises, and the third group follows a special diet.
After four weeks, the reduction in each person’s blood pressure is recorded. At 𝛼 = 0.05, test the
claim that there is no difference among the means. The data are shown.

Medication Exercise Diet

10 6 5

12 8 9

9 3 12

15 0 8

13 2 4

𝒙
̅ 𝟏 = 𝟏𝟏. 𝟖 𝒙
̅ 𝟐 = 𝟑. 𝟖 𝒙
̅ 𝟑 = 𝟕. 𝟔

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 𝒔𝟏 𝟐 = 𝟓. 𝟕 𝑠2 2 = 10.2 𝑠3 2 = 10.3

5.1.2 Variance test

STEP 1. State the hypothesis to test equality of population means as
𝐻0 =𝜇1 = 𝜇2 = 𝜇3 (claim)
𝐻1 = At least one mean is different from the others
STEP 2. Find the critical values. Since 𝑘 = 3 𝑎𝑛𝑑 𝑛 = 15 and 𝛼 = 0.05

𝑉1 𝑑𝑓 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 = 𝑘 − 1 ,𝑉1 = 3 − 1 = 𝟐
𝑉2 𝑑𝑓 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 = 𝑛 − 𝑘, 𝑉2 = 15 − 3 = 𝟏𝟐
𝐹𝛼,(𝑘−1,𝑛−𝑘) = 𝐹0.05,(3−1,15−3) = 𝟑. 𝟖𝟗

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 𝑟𝑒𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑔𝑖𝑜𝑛

3.89

STEP 3. Compute the test statistics, using the procedure outlined here

a) Find the mean and variance of each sample

b) Find the grand mean, denoted by 𝑥̿ , it is the mean of values in the samples

∑ 𝑥 10 + 12 + 9 + ⋯ … … … … + 4 116
𝑥̿ = = = = 𝟕. 𝟕𝟑
𝑛 12 15
c) Find the between group variance, denoted by MSB

∑𝑘𝑗=1 𝑛𝑗 (𝑥̅ − 𝑥̿ )2
𝑀𝑆𝐵 =
𝐾−1
5(11.8 − 7.73)2 + 5(3.8 − 7.73)2 + 5(7.6 − 7.73)2 160.13
= = = 𝟖𝟎. 𝟎𝟕
3−1 2
d) Find within group variance -within sample variability, denoted by 𝑀𝑆𝑊
∑(𝑛𝑖 − 1)𝑠𝑖 2
𝑀𝑆𝑊 =
∑(𝑛𝑖 − 1)
(5 − 1)(5.7) + (5 − 1)(10.2) + (5 − 1)(10.3) 104.80
𝑀𝑆𝑊 = = = 𝟖. 𝟕𝟑
(5 − 1) + (5 − 1) + (5 − 1) 12
Note:
This formula finds an overall variance by calculating a weighted average of the variances. It does
not involve using differences of the means.

e) Find the F test statistics value

𝑀𝑆𝐵 80.07
𝐹= = = 𝟗. 𝟏𝟕
𝑀𝑆𝑊 8.73
STEP 4. Make the decision. The decision is to reject the null hypothesis, since F table i.e.
9.17 > 3.89
STEP 5. Summarize the results. There is enough evidence to reject the claim and conclude
that at least one mean is different from the others.

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Exercise

1. The correlation coefficient is used to determine:

a. A specific value of the y-variable given a specific value of the x-variable

b. A specific value of the x-variable given a specific value of the y-variable
c. The strength of the relationship between the x and y variables
d. None of these

2. If there is a very strong correlation between two variables then the correlation coefficient
must be

a. any value larger than 1

b. much smaller than 0, if the correlation is negative

c. much larger than 0, regardless of whether the correlation is negative or positive

d. None of these alternatives is correct.

3. In regression, the equation that describes how the response variable (y) is related to the
explanatory variable (x) is:

a. the correlation model

b. the regression model

c. used to compute the correlation coefficient

d. None of these alternatives is correct.

4. The relationship between number of beers consumed (x) and blood alcohol content (y) was
studied in 16 male college students by using least squares regression. The following regression
equation was obtained from this study:

Y’= -0.0127 + 0.0180x the above equation implies that:

a. each beer consumed increases blood alcohol by 1.27%

b. on average it takes 1.8 beers to increase blood alcohol content by 1%

c. each beer consumed increases blood alcohol by an average of amount of 1.8%

d. each beer consumed increases blood alcohol by exactly 0.018

5. SSE can never be

a. larger than SST

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 b. smaller than SST

c. equal to 1

d. equal to zero

6. Regression modeling is a statistical framework for developing a mathematical equation that

describes how

a. One explanatory and one or more response variables are related

b. several explanatory and several response variables response are related

c. One response and one or more explanatory variables are related

d. All of these are correct.

7. In regression analysis, the variable that is being predicted is the

a. response, or dependent, variable

b. independent variable

c. intervening variable

d. is usually x

8. Regression analysis was applied to return rates of sparrow hawk colonies. Regression analysis
was used to study the relationship between return rate (x: % of birds that return to the colony in a
given year) and immigration rate (y: % of new adults that join the colony per year). The
following regression equation was obtained.

Y’ = 31.9 – 0.34x

Based on the above estimated regression equation, if the return rate were to decrease by 10% the
rate of immigration to the colony would:

a. increase by 34%

b. increase by 3.4%

c. decrease by 0.34%

d. decrease by 3.4%

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 CHAPTER SIX

REGRESSION AND CORRELATION

What is a regression model?

Regression analysis is almost certainly the most important tool at the econometrician’s disposal.
But what is regression analysis? In very general terms, regression is concerned with describing
and evaluating the relationship between a given variable and one or more other variables. More
specifically, regression is an attempt to explain movements in a variable by reference to
movements in one or more other variables.
To make this more concrete, denote the variable whose movements the regression seeks to explain
by y and the variables which are used to explain those variations by x1, x2, . . . , xk . Hence, in this
relatively simple setup, it would be said that variations in k variables (the xs) cause changes in
some other variable, y. This chapter will be limited to the case where the model seeks to explain
changes in only one variable y (although this restriction will be removed in chapter 6).

Regression versus correlation

All readers will be aware of the notion and definition of correlation. The correlation between two
variables measures the degree of linear association between them. If it is stated that y and x are
correlated, it means that y and x are being treated in a completely symmetrical way. Thus, it is not
implied that changes in x cause changes in y, or indeed that changes in y cause changes in x. Rather,
it is simply stated that there is evidence for a linear relationship between the two variables, and
that movements in the two are on average related to an extent given by the correlation coefficient.
Regression as a tool is more flexible and more powerful than correlation.

6.1 Linear correlations

6.2.1 The coefficient of correlation

Correlation coefficients (denoted r) are statistics that quantify the relation between X and Y in
unit-free terms. When all points of a scatter plot fall directly on a line with an upward incline, r =
+1; When all points fall directly on a downward incline, r = +1.

Such perfect correlation is seldom encountered. We still need to measure correlational strength, –
defined as the degree to which data point adhere to an imaginary trend line passing through the

96 | P a g e
“scatter cloud.” Strong correlations are associated with scatter clouds that adhere closely to the
imaginary trend line. Weak correlations are associated with scatter clouds that adhere marginally
to the trend line.

The closer r is to +1, the stronger the positive correlation.

The closer r is to +1, the stronger the negative correlation.

Examples of strong and weak correlations are shown below. Note: Correlational strength can not
be quantified visually. It is too subjective and is easily influenced by axis-scaling. The eye is not
a good judge of correlational strength.

Pearson’s Correlation Coefficient

To calculate a correlation coefficient, you normally need three different sums of squares (SS). The
sum of squares for variable X, the sum of square for variable Y, and the sum of the cross-product
of XY. The sum of squares for variable X is:

𝑆𝑆𝑋𝑋 = ∑(𝑥𝑖 − 𝑋̅ )2

This statistic keeps track of the spread of variable X. For the illustrative data, 𝑋̅ = 30.83 and SSXX
= (50-30.83)2 + (11-30.83)2 + . . . + (25-30.83)2 = 7855.67. Since this statistic is the numerator of
the variance of X (s ), it can also be calculated as SSXX = (𝑆𝑥2 )(𝑛 − 1). Thus, SSXX = (714.152)(12-
1) = 7855.67. 2

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The sum of squares for variable Y is:

𝑆𝑆𝑌𝑌 = ∑(𝑦𝑖 − 𝑦̅)2

This statistic keeps track of the spread of variable Y and is the numerator of the variance of 𝑌(𝑆𝑥2 ) .
For the 2 illustrative data 𝑦̅ = 30.883 and SSYY = (22.1-30.883)2 + (35.9-30.883)2 + . . . + (38.4-
30.883)2 = 3159.68. An alterative way to calculate the sum of squares for variable Y is SS YY =
(𝑆𝑥2 )(𝑛 − 1). Thus, SSYY = (287.243)(12-1) = 2 3159.68.

Finally, the sum of the cross-products (SSXY) is:

𝑆𝑆𝑋𝑌 = ∑(𝑥𝑖 − 𝑥̅ )(𝑦𝑖 − 𝑦̅)

For the illustrative data, SSXY = (50-30.83)(22.1-30.883) + (11-30.83)(35.9-30.883) + . . . + (25-

30.83)(38.4-30.883) = !4231.1333. This statistic is analogous to the other sums of squares except
that it is used to quantify the extent to which the two variables “go together”.

The correlation coefficient (r) is

𝑆𝑆𝑋𝑌
𝑟=
ඥ(𝑆𝑆𝑋𝑋 )(𝑆𝑆𝑌𝑌 )

For the illustrative data,

−4231.1333
𝑟= = −0.849
ඥ(7855.67)(3159.68)

Interpretation of Pearson’s Correlation Coefficient The sign of the correlation coefficient

determines whether the correlation is positive or negative. The magnitude of the correlation
coefficient determines the strength of the correlation. Although there are no hard and fast rules for
describing correlational strength, I [hesitatingly] offer these guidelines:

Coefficient of Determination

The coefficient of determination is the square of the correlation coefficient (r2). For illustrative
data, r2 = -0.8492 = 0.72. This statistic quantifies the proportion of the variance of one variable
“explained” (in a 2 2 statistical sense, not a causal sense) by the other. The illustrative coefficient
98 | P a g e
of determination of 0.72 suggests 72% of the variability in helmet use is explained by
socioeconomic status.

6.1.2 Rank correlation coefficient

When we need to estimate the correlation between the qualitative variables which can not be
quantified in numerical terms but can be ranked or ordered, the Karl Person’s method fails. In such
cases, Spearman Rank correlation method is applied.

▪ Formula

6 ∑ 𝐷2
𝑅𝑘 = 1 − 𝑁(𝑁2 −1)

 𝑅𝑘 = Rank correlation coefficient

 D = Difference of rank between paired item in two series.
 N = Total number of observation

Interpretation of Rank Correlation Coefficient (𝑅𝑘)

 The value of rank correlation coefficient, R ranges from -1 to +1

 If 𝑅𝑘 = +1, then there is complete agreement in the order of the ranks and the ranks are in
the same direction
 If 𝑅𝑘 = -1, then there is complete agreement in the order of the ranks and the ranks are in
the opposite direction
 If 𝑅𝑘 = 0, then there is no correlation

Rank Correlation Coefficient (𝑅𝑘)

a) Steps to Calculate 𝑅𝑘 where actual rank are given.

1. Calculate the difference ‘D’ of corresponding ranks of two series i.e. (𝑅1 – 𝑅2).
2. Square the difference & calculate the sum of the difference i.e. ∑D 2
3. Substitute the values obtained in the formula
b) Problems where Ranks are not given: If the ranks are not given, then we need to assign
ranks to the data series in ascending or descending order. For example, the lowest value in
the series can be assigned rank 1 or the highest value in the series can be assigned rank 1.
We need to follow the same pattern of ranking for both the series.
Then we follow the all the steps as we do when the ranks are given.
Equal Ranks or tie in Ranks: In such cases average ranks should be assigned to each
individual.
1 1 1
6 ∑ 𝐷2 [12 (𝑚13 − 𝑚1 ) + 12 (𝑚23 − 𝑚2 ) + ⋯ + 12 (𝑚𝑛3 − 𝑚𝑛 )]
Rk = 1 −
𝑁 (𝑁 2 − 1)
Where,
 m is number of times a particular item in the series is repeated
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 1
 The component (𝑚𝑖3 − 𝑚𝑖 ) is repeated as many times are the numbers of items
12
are repeated in both the series.

Limitations of Spearman’s Rank Correlation

 Cannot be used for finding out correlation in a grouped frequency distribution.

 This method should be applied where N exceeds 30

6.2 Simple linear regression

6.2.1 Curve fitting

For simplicity, suppose for now that it is believed that y depends on only one x variable. Again,
this is of course a severely restricted case, but the case of more explanatory variables will be
considered in the next chapter. Three examples of the kind of relationship that may be of interest
include:
 How asset returns vary with their level of market risk
 Measuring the long-term relationship between stock prices and dividends
 Constructing an optimal hedge ratio.
Suppose that a researcher has some idea that there should be a relationship between two variables
y and x, and that financial theory suggests that an increase in x will lead to an increase in y. A
sensible first stage to testing whether there is indeed an association between the variables would
be to form a scatter plot of them. Suppose that the outcome of this plot is figure 2.1.
In this case, it appears that there is an approximate positive linear relationship between x and y
which means that increases in x are usually accompanied by increases in y, and that the relationship
between them can be described approximately by a straight line. It would be possible to draw by
hand onto the graph a line that appears to fit the data. The intercept and slope of the line fitted by
eye could then be measured from the graph. However, in practice such a method is likely to be
laborious and inaccurate.
It would therefore be of interest to determine to what extent this relationship can be described by
an equation that can be estimated using a defined procedure. It is possible to use the general
equation for a straight line to get the line that best ‘fits’ the data. The researcher would then be
seeking to find the values of the parameters or coefficients, α and β, which would place the line as
close as possible to all of the data points taken together.

100 | P a g e
 y = α + βx
However, this equation (y = α + βx) is an exact one. Assuming that this equation is appropriate, if
the values of α and β had been calculated, then given a value of x, it would be possible to determine
with certainty what the value of y would be. Imagine -- a model which says with complete certainty
what the value of one variable will be given any value of the other!
Charles Spearman’s coefficient of correlation (or rank correlation) is the technique of determining
the degree of correlation between two variables in case of ordinal data where ranks are given to
the different values of the variables. The main objective of this coefficient is to determine the
extent to which the two sets of ranking are similar or dissimilar. This coefficient is determined as
under:
6∑𝑑 2
𝑖
Spearman's coefficient of correlation (or rs) = 1 − 𝑛(𝑛2−1)

Where:𝑑𝑖 = difference between ranks of 𝑖th pair of the two variables

𝑛 = number of pairs of observations.
As rank correlation is a non-parametric technique for measuring relationship between paired
observations of two variables when data are in the ranked form, we have dealt with this technique
in greater details later on in the book in chapter entitled ‘Hypotheses Testing II (Non-parametric
tests)’.
Karl Pearson’s coefficient of correlation (or simple correlation) is the most widely used method
of measuring the degree of relationship between two variables. This coefficient assumes the
following:
i. There is linear relationship between the two variables;
ii. The two variables are casually related which means that one of the variables is independent
and the other one is dependent; and
iii. A large number of independent causes are operating in both variables so as to produce a
normal distribution.
Karl Pearson’s coefficient of correlation can be worked out thus.
∑(𝑋𝑖 −𝑋̅)(𝑌𝑖 −𝑌̅ )
Karl Pearson’s coefficient of correlation (or r)* = 𝑛.𝜎𝑋 .𝜎𝑌

Alternatively, the formula can be written in a different way:

(This applies when we take zero as the assumed mean for both variables, X and Y.)
Where 𝑋𝑖 = ith value of X variable

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 𝑋̅ = mean of X
𝑌𝑖 = ith value of Y variable
𝑌̅ = Mean of Y
𝑛 = number of pairs of observations of X and Y
𝜎𝑋 = Standard deviation of X
𝜎𝑌 = Standard deviation of Y
In case we use assumed means (Ax and Ay for variables X and Y respectively) in place of true
means, then Karl Person’s formula is reduced to:

Example 1. Two points on the straight line that represents Mr. Jones’s salary are(x1 ;
y1) = (5 ; 140 000) and (x2 ; y2) = (9 ; 216 000).
The slope of the line is given by
𝑦2 − 𝑦1
𝑏=
𝑥2− 𝑥1
216,000−140,000
=
9−5
76,000
= 4

= 𝟏𝟗, 𝟎𝟎𝟎

Two points on the straight line that represents Mr. Brown’s salary are (x1 ; y1) = (6 ; 160 000) and
(x2 ; y2) = (9 ; 244 000). The slope of the line is

244,000−160,000
= 9−6

= 𝟐𝟖, 𝟎𝟎𝟎

Two points on the straight line that represents Mr. Smith’s salary are (x1 ; y1) = (7 ; 150 000) and
(x2 ; y2) = (9 ; 210 000).The slope of the line is
b = 30 000.
Mr. Smith’s salary increased at the highest rate.
A. The equation is y = a + bx with b = 28000, thus y = a + 28000x.
Substitute point (6 ; 160 000) into the equation:

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 160 000 = a + 28 000 × 6
a + 168 000 = 160 000
a = −8 000.
Thus y = −8000 + 28000x.
B.
The year 2008 corresponds to x = 10.
Substitute x = 10 into y = −8000 + 28000x which gives

y = −8 000 + 28 000 × 10
= 272 000.
In 2008 he will make R272 000.

Example 2. It looks as if there exists a positive linear correlation between average

interest rate and yearly investment. This means that if the average interest rate increases,
then yearly investment will also increase.

6.2.2 The method of least square

A. You must do these calculations directly on your calculator. We are only showing
mathematical how your calculator came to the answer.

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 Year (𝑖) Average Yearly 𝑥𝑖 2 𝑥𝑖 𝑦𝑖 𝑦𝑖 2
interest (𝑥𝑖 ) investment
(𝑦𝑖 )
1 13.8 1,060 190.44 14,628 1,123,600

2 14.5 940 210.25 13,630 883,600

3 13.7 920 187.69 12,604 846,400

4 14.7 1,110 216.09 16,317 1,232,100

5 14.8 1,550 219.04 22,940 2,402,500

6 15.5 1,850 240.25 28,675 3,422,500

7 16.2 2,070 262.44 33,534 4,284,900

8 15.9 2,030 252.81 32,277 4,120,900

9 14.9 1,780 222.01 26,522 3,168,400

10 15.1 1,420 228.01 21,442 2,016,400

𝑛 = 10 149.1 14,730 2,229.03 222,569 23,501,300

𝑛 ∑𝑛 𝑛 𝑛
𝑖=1 𝑥𝑖 𝑦𝑖 −∑𝑖=1 𝑥𝑖 ∑𝑖=1 𝑦𝑖
𝑟=
2 2
√𝑛 ∑𝑛 2 𝑛
𝑖=1 𝑥𝑖 −(∑𝑖=1 𝑥𝑖 )
√𝑛 ∑𝑛 2 𝑛
𝑖=1 𝑦𝑖 −(∑𝑖=1 𝑦𝑖 )

10(22,569)−(149.1)(14,730)
=
ඥ10(2,229.03)−(149.1)2ඥ10(23,501,300)(147,730)2

24,447
= 32,759.8161

= 𝟎. 𝟖𝟗𝟖𝟗

B. The coefficient of determination is𝑟 2 = 0.89892 = 0.8080. This means that

almost 81% of the variation in yearly investments can be declared by the average
interest rate.

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 C. The equation of the straight line is 𝑦 = 𝑎 + 𝑏𝑥 where

∑10 10 10
𝑖=1 𝑥𝑖 𝑦𝑖 −∑𝑖=1 𝑥𝑖 ∑𝑖=1 𝑦𝑖
𝑏=
√𝑛 ∑10 𝑥𝑖 2 −(∑10 𝑥𝑖 )2
𝑖=1 𝑖=1

10(22,569)−(149.1)(14,730)
= 10(2,229.03)−(149.1)2

24,447
= 59.49

= 𝟒𝟗𝟒. 𝟗𝟗
∑10
𝑖=1 𝑦𝑖 𝑏 ∑10
𝑖=1 𝑥𝑖
And 𝑎 = −
𝑛 𝑛

14,730 (494.99)(149.1)
= −
10 10

= −𝟓𝟗𝟎𝟕. 𝟑𝟎

Thus,y = −5907.30 + 494,99x

Example 3. The following table shows the number of loans approved for different
amounts during the second half of 2008.
Amount of loan in Birr Number of loans (y)
100,000 (x)
2 45

3 250

4 250

5 175

6 125

Example 4. A study was undertaken at eight garages to determine how the resale value
of a car is affected by its age. The following data was obtained:
Garage Age of car (in years) Resale value (in Birr)

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 1 1 41,250

2 6 10,250

3 4 24,310

4 2 38,720

5 5 8,740

6 4 26,110

7 1 38,650

8 2 36,200

The garage manager suspects a linear relationship between the two variables. Fit a curve of the
form y = a + bx to the data.
The equation for the regression line is

y = 48 644, 17− 6 596, 93, the correlation coefficient is r = −0, 9601

Exercise

1. The annual car sales of a small car manufacturer, c, and the annual advertising
expenditure, £ a, has product moment correlation coefficient r ac. The data is coded as
𝑎
x =c − 7000 and 𝑦 = 1000 and the summary is shown in the table below.
Year 2010 2011 2012 2013 2014 2015 2016 2017

X 52 340 511 621 444 700 805 921

y 120 126 134 138 132 146 153 160

a) Find, by a statistical calculator, the value of the product moment correlation coefficient
between x and y , denoted by rxy
b) State with full justification the value of r ac
c) Interpret the value of rac

2. The table below shows the number of Math’s teachers x, working in 8 different towns and
the number of burglaries y, committed in a given month in the same 8 towns.

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 town A B C D E F G H

X 35 42 21 55 33 29 39 40

Y 30 28 21 38 35 27 30 K

0.0934
0.0062̅0.0075
𝑃 1− a) Use a statistical calculator to find the product moment correlation coefficient between
̅(2𝑋̅ =
𝜇𝑃 ̅ 2) = 𝜇(𝑋1 − 𝑋̅2 ) the
̅ = number of math’s teachers and the number of burglaries, for the towns A to G.
1− 𝑋
500 ̅
200 𝜇𝑋(1𝑋
0.02 ̅𝑃̅ −
1 ̅ 2 ) = b) Interpret the value of the product moment correlation coefficient in the context of this
1−−
𝑃
̅2 = 600
𝑋
̅
𝑋2 = 250
0.05 question.

c) Test, at the 5% level of significance, whether there is evidence of positive correlation

between the number of math’s teachers and the number of burglaries, for the towns A to G.

d) Use linear regression to estimate the value of k , for town H.

3. The table below shows the maximum daytime temperature, in °C, at a certain city Centre, and
the amount of a certain pollutant in mg per liter.

Maximum Temperature 10 12 14 16 18 20 22 24

Amount of Pollutant 513 475 525 530 516 520 507 521

a) Find, using a statistical calculator, the value of the product moment correlation coefficient
for the above data.

b) Test, at the 10% level of significance, whether there is evidence of positive correlation in
these bivariate data.

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Answer key
Chapter one
1. a) 𝜇𝑋̅ = 75 , 𝜎𝑋̅ = 1.09

b) 𝜇𝑋̅ 𝑠𝑡𝑎𝑦𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑏𝑢𝑡 𝜎𝑋̅ 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝑡𝑜 0.6

𝑝𝑞
2. p ±3√ = 0.08±0.05 and [0.03,0.13]⊂[0,1],0.1210
𝑛
𝑝𝑞
3. P ±3√ = 0.02±0.05 and [0.01, 0.03] ⊂ [0, 1], 0.9671
𝑛
4. A) 0.63 b) 0.1446
5. 0.9977
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 6. 0.3483

Chapter Two
1. (761.19, 798.81)
2. (242.16, 257.84)
3. (283.61, 296.39)

Chapter Three
1. a) ‘One-tailed’’ the alternate hypothesis is greater than direction

b) Reject HO when z > 1.645

c) This is evidence to conclude that the population mean is greater than 20.
2. A machine is set to fill a small bottle with 9.0 grams of medicine. A sample of eight
bottles revealed the following amounts (grams) in each bottle.
9.2 8.7 8.9 8.6 8.8 8.5 8.7 9.0 At the .01 significance level, can we conclude that the mean
weight is less than 9.0 grams?
a) State the null hypothesis and the alternate hypothesis.
Ho: mu=9 grams
Ha: mu<9
b. How many degrees of freedom are there? n-1 = 7
c. Compute the value of t. What is your decision regarding the null
hypothesis? X-bar = 8.8 grams; df = 7; Alpha = 1%;
Test Statistic = (8.8-9)/[0.2268/sqrt(8))] = -.2/0.0802 = -2.4942
d. Estimate the p-value. P-value = 0.0207
Conclusion: Since p-value is greater than 1% there is insufficient evidence to reject Ho.

Chapter Four
1. 0.98743
2. 0.8634
3. 0.6543

Chapter Five

1. C 2.B 3.B 4.C 5.A 6.C 7.A 8.B

Chapter Six

1. a) rxy=0.969
b) rac=0.969

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 c) Strong positive correlation, i.e. The More spend on advertising the more the car sales

2. a) r= 0.792

b) As the number of Math increases, it is positive correction

c) The critical view for h=7, at 5% significance is 0.6694 as 0.792>0.6694, i.e sufficient to
reject H0

d) Y=0.408x+15.1

3. a) r=0.320

b) As 0.320<0.5067 it appears there is no positive correction between the Maximum

Temperature of the Amount of Pollutant insufficient

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