Syllabus: Introduction, Null hypothesis, Alternative hypothesis, Type –I,

II errors, Level of significance, Critical Region. Confidence interval, one sided

test, two sided test, Small Sample Tests. : Students t - distribution , its
properties;

Test of significance difference between sample mean and population

mean; difference between means of two small samples , F-Distribution, Test
of equality of two population variances, Chi-square test of goodness of fit .

Large sample Tests : Test of Significance of Large Samples – Tests of

significance difference between sample proportion and population
proportion & difference between two sample proportions , Tests of
significance difference between sample mean and population mean &
difference between two sample means.

Population (Universe)
A population consists of the totality of the observations with which we are
concerned.

(or) A group of individuals under study is called Population. It may be finite

or infinite

Population means aggregate of all possible units. It need not be

human population. It may be population of plants, population of insects,
population of fruits, etc.

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Finite population:
When the number of observation can be counted and is definite, it is
known as finite population

Infinite population
When the number of units in a population is innumerably large, that
we cannot count all of them, it is known as infinite population.

Size of the population:

It is the number of observations in population and is denoted by 𝑁.

Eg: 1.The number of students in a college 4000.Here 𝑁= 4000,which is

finite.

2. The number of stars in the sky .Here 𝑁 is infinite.

Sample:
It is a subset of the population and size of the sample is denoted by 𝑛.

A portion or small number of unit of the total population is known as

sample

A selected group of some elements from the totality of the

population is known as the sample. It is from the study of this sample that
something is known and said about the whole population

Example:

1.All the farmers in a village(population) and a few farmers(sample)

2.All plants in a plot is a population of plants. A small number of plants

selected out of that population is a sample of plants

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 Parameter:
It is a statistical measure based on all units of a population

Population are described in terms of certain measures like mean, standard

deviation etc. These measures of the population are called parameter and
are usually denoted by Greek letters

1.Population Mean: 𝜇

2. Population variance: 𝜎 2

3. Population Proportion: 𝑃

Statistic:
It is a statistical measure based on all units of selected sample

Sample measures are

1.Sample Mean: 𝑥̅

2. Sample variance: 𝑠 2

3. Sample Proportion: 𝑝

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For example, say you want to know the mean income of the subscribers to
a particular magazine—a parameter of a population. You draw a random
sample of 100 subscribers and determine that their mean income is Rs
30,500 (a statistic). You conclude that the population mean income μ is
likely to be close to Rs30,500 as well. This example is one of statistical
inference.

Symbols for Population and Samples

Characteristic Population Sample

Parameter Statistic

Symbols 1. Population Mean= 𝜇 1. Sample mean =𝑥̅

2. Population SD = 𝜎 2.Sample SD = 𝑠

3.Population Proportion=𝑃 3.Sample proportion = 𝑝

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Classification of sample:
(1)Large sample : If 𝑛 ≥ 30 ,then sample is called lagre sample.
i.e., if the size of the sample is greater than or equal to 30 , then
the sample is considered as large sample.

(2) Small sample: : If 𝑛 < 30 ,then sample is called sample sample.

i.e., if the size of the sample is less than 30 , then the sample is
considered as small sample.

Sampling:
The process of selection of a sample is called sampling.

Eg. To assess the quality of a bag of rice ,we examine only a portion of it by
taking a handful of it from the bag and then decide to purchase it or not.

Thus in enumerating the characteristics of the population ,instead

of enumerating the entire population ,only the individuals in the sample are
examined .Then sample characteristics are utilised to estimate the
population.

To eliminate any possibility of bias in the sampling procedure choose a

random sample where observations are made independently and at
random

The primary purpose of a statistical study is to draw conclusions about the

population (parameters) based on sample (information) drawn from the
population. The sampling theory can be employed to obtain information
about samples drawn from a known population.
The theory of statistical inference can be divided into two major areas:
(i) Estimation of parameters (ii) Test of Hypothesis

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A study of either type of inferences about a population may lead to correct
conjectures about the population. Procedure of estimating a population
(parameter) by using sample information is referred as Estimation.

Procedures which enables one to decide whether to accept or reject

hypothesis (the conjectures about the population) are called tests of
hypothesis.
Simply Hypothesis means Quantitative statement about the population

Tests of hypothesis
The main object of the sampling theory is the study of the Tests of
hypothesis/Tests of significance

Statistical hypothesis is a statement about the population which we want

to verify on the basis of information available.
A statistical hypothesis is stated in such a way that they may be evaluated
by appropriate statistical techniques.
There are two types of statistical hypothesis.
1) Null Hypothesis
2) Alternative Hypothesis

Null hypothesis: For applying tests of significance, we first set up a

hypothesis a definite statement about the population parameter. Such a
hypothesis is usually a hypothesis of no-difference, is called Null
Hypothesis.

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Definition: A null hypothesis is the hypothesis which states that there is no
significant difference between the statistic and the population parameter.
Null hypothesis is usually denoted by the symbol 𝐻0 .

We can also say that

Null hypothesis is a claim or statement about the population

parameter that it is assumed to be true until it is declared to be false.

If we want to test any statement about the population ,we set up the
null hypothesis that it is true.

For example ,if we want to find if the population mean has specified
value 𝜇0 ,then we set up the null hypothesis

𝐻0 : 𝜇 = 𝜇0
For example ,if we want to test the null hypothesis that the average
salary of a ANITS students in the placements is 5 Lakhs per year.

In this case we assume the null hypothesis as

𝐻0 : 𝜇 = 5 𝑙𝑎𝑘ℎ𝑠

Alternate Hypothesis: Any hypothesis which contradicts (or)

alternative to the Null hypothesis is called an Alternate Hypothesis, usually
denoted by H1. The two hypothesis H0 and H1are such that if one is true, the
other is false and vice versa.

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Example: if we want to test the null hypothesis that the population has a
specified mean 0 say i.e., H0 :  = 0 then the alternate hypothesis would
be
(i) H1 :   0  either  > 0 or  < 0 or
(ii) H1 :  > 0 ( Right tailed test)
(iii) H1 :  < 0 ( Left tailed test)
The alternate hypothesis is very important to decide whether we
have to use a single – tailed (right or left ) or two-tailed test.

The alternative statement to the null hypothesis that

𝐻0 : 𝜇 = 5 𝑙𝑎ℎℎ𝑠
Could be :
i)H1 :   5 𝑙𝑎ℎℎ𝑠  either  > 5 𝑙𝑎ℎℎ𝑠 or  < 5 𝑙𝑎ℎℎ𝑠 or
ii)H1 :  > 5 𝑙𝑎ℎℎ𝑠 ( Right tailed test)
iii) H1 :  < 5 𝑙𝑎ℎℎ𝑠 ( Left tailed test)
The alternative hypothesis (i) is known as a two tailed alternative,
(ii) is known as right tailed and (iii) is called left tailed.

Mathematical Symbols Used in H0 and H1 :

𝑯𝟎 H1

equal (=) not equal (≠) or greater than (>) or less than (<)
greater than or equal to (≥) less than (<)
less than or equal to (≤) more than (>)

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Identification of the test statisitic

There are several tests of significance namely 𝑧, 𝑡 , 𝐹 etc.

First we have to select the right test depending on the nature of the
information given in the problem. Then we construct the test criterion and
select the appropriate probability distribution

Error sampling: The main objective in sampling theory is to draw valid

inferences about the population parameters on the basis of the sample
results. In practice we decide to accept or reject the lot after examining a
sample from it. As such we have two types of errors.

(i) Type I error: reject H0 when it is true. If the null hypothesis H0 is

true but it rejected by test procedure, then the error made is called
type I error or  error.
(ii) Type II Error: Accept H0 when it is wrong i.e., accept H0 when H1 is
true. If the null hypothesis is false but it is accepted by test, then
error committed is called type II error or  error.

if we write P(Reject H0 when it is true) = P (type I error) =

and P(Reject H0 when it is false) = P (type II error) =

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Level of significance
The maximum probability of making type-I error specified in a test of
hypothesis is called the level of significance.

Level of significance is the size of test. Level of significance denoted by a 

is the confidence with which we rejects or accepts the null hypothesis H 0
i.e., it is the maximum possible probability with which we are willing to risk
an error in rejecting H0 when it is true

The commonly used levels of significance are 5%

(0.05) and 1% (0.01).
 If we adopt 5% level significance ,it implies that in 5 out of 100 are
likely to reject a correct 𝐻0. In other words this implies that we are
95% confident that our decision to reject 𝐻0 is correct.
 If we select 0.01(99%) as the significance level ,it means that we are
99% confident in our decision but still there is 1% chance for our
decision being false.

Critical Region: A region corresponding to a statistic ‘t’ in the

sample space S which leads to the rejection of H0 is called critical region and
rejection region. Those region which lead to the acceptance of H0 give us a
region called acceptance region.

Critical values or Significant values: It is the value of the test

statistic z, which divides the area under the probability curve into critical
region, for given level of significance . It is also known as significance
value.

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 Critical Value (CV) – separates the critical region from the non-critical
region, when we should accept or reject 𝐻0 .

 The location of the critical value depends on the inequality sign of the
alternative hypothesis.

 Depending on the distribution of the test value, you will use different
tables to find the critical value.

Rejection region
Acceptance region

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One-tailed or Two-tailed Tests
In Statistics hypothesis testing, we need to judge whether it is a one-tailed
or a two-tailed test so that we can find the critical values in tables such as
Standard Normal z Distribution Table and t Distribution Table. And then, by
comparing test statistic value with the critical value or whether statistic
value falls in the critical region, we make a conclusion either to reject the
null hypothesis or to fail to reject the null hypothesis.

How can we tell whether it is a one-tailed or a two-tailed test? It depends

on the original claim in the question

The critical region may be represented by a portion of the area under the
normal curve in two ways: (i) Two ‘tails’ or sides under the normal curve (ii)
one ‘tail’ or side under the normal curve which is either the right tail or the
left tail.

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In a one-tailed test, the critical region has just one part (the green area
below). It can be a left tailed test or a right-tailed test. Left-tailed test: The
critical region is in the extreme left region (tail) under the curve .Right-
tailed test: The critical region is in the extreme right region (tail) under the
curve

In two-tailed test, the critical region has two parts (the red areas below)
which are in the two extreme regions (tails) under the curve.

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Critical values (𝒁𝜶 ) of 𝒁

Critical Values
(𝒁𝜶 ) Level of Significance (𝜶)

Two –Tailed |𝒁𝜶 | = 𝟐. 𝟓𝟖 |𝒁𝜶 | = 𝟏. 𝟗𝟔

test

Right- tailed test 𝒁𝜶 = 𝟐. 𝟑𝟑 𝒁𝜶 = 𝟏. 𝟔𝟒𝟓

Left –tailed test 𝒁𝜶 = −𝟐. 𝟑𝟑 𝒁𝜶 = −𝟏. 𝟔𝟒𝟓

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 Large sample Tests
If the size of the sample 𝑛 ≥ 30, then the sample is called
Large sample.

In large samples we have the following four important tests.

1. Test of significance for single mean

2. Test of significance for difference of means

3. Test of significance for single proportion

4. Test of significance for difference of proportion

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 Write given Data in symbolic
form.

Set up Null Hypothesis 𝐻0

Set up Alternative Hypothesis 𝐻1

Choose level of significance 1% or 5%

Select test statistic and calculate

Accept Statistical Reject

𝐻0 Decision 𝐻0

Conclude 𝐻0 is true Conclude 𝐻1 is true

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Procedure for Testing of Hypothesis
1.NullHypothesis: Set up the Null Hypothesis 𝐻0
2. Alternative Hypothesis: Set up the Alternative Hypothesis 𝐻1 .
This will enable us to decide whether we have to use a single-tailed (right
or left) test or two-tailed test.

3. Level of Significance. Choose the appropriate level of Significance (𝛼)

Generally we choose either 5% or 1% level of significance.

4. Test Statistic : Compute the test statistic

𝑡−𝐸(𝑡)
𝑍 = , under the null hypothesis 𝐻0
𝑆.𝐸 (𝑡)

Where 𝑡 is a statistic, like 𝑥̅ , 𝑝, 𝑒𝑡𝑐.

𝐸 (𝑡) = Expectation of a statistic 𝑡.

𝐸 (𝑥̅ ) = 𝜇, 𝐸 (𝑝) = 𝑃,…

𝑆. 𝐸 (𝑡) = standard error of a statistic 𝑡

𝜎 𝑃𝑄
𝑆. 𝐸 (𝑥̅ ) = , 𝑆. 𝐸 (𝑝) = ,… …
√𝑛 √𝑛

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5. Conclusion.
We compare 𝑧 the computed value of Z in step 4 with the significant value
(tabulated value) 𝑧𝛼 at the given level of significance, '𝛼'

If |𝑧|< 𝑧𝛼 , we accept Null Hypothesis 𝐻0 at level of significance 𝛼.

If |𝑧|> 𝑧𝛼 , we reject Null Hypothesis 𝐻0 at level of significance 𝛼 with a

confidence coefficient (1 − 𝛼).

Let 𝜇 and 𝜎 2 be the mean and variance of a population from which a

random sample of size n is drawn .

Let 𝑥̅ be the mean of the sample.

To test the significant difference between the sample mean and the
population mean μ, the statistic is given by

𝑥̅ −𝜇
𝑧= 𝜎
√𝑛

If the population S.D. is not known, then we can use sample standard
deviation ‘s’ in place of 𝜎 ,then statistic is given by

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 𝑥̅ −𝜇
𝑧= 𝑠
√𝑛

IMP

Test statistic for single Mean is

𝑥̅ − 𝜇
𝑧= 𝜎
√𝑛
𝑥̅ −𝜇
(or) 𝑧= 𝑠
√𝑛

Population
Sample mean
mean

𝑥̅ −𝜇
𝑧= 𝜎
Population √𝑛
Standard
deviation Sample
size

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95%confidence limits for population mean 𝜇 are given by
𝜎
𝑥̅ ± 1.96
√𝑛
𝜎 𝜎 𝑥̅ − 𝜇
i.e 𝑥̅ − 1.96 < 𝜇 < 𝑥̅ + 1.96 𝑧= 𝜎
√𝑛 √𝑛
√𝑛

99%confidence limits for population mean 𝜇 are given by

𝜎
𝑥̅ ± 2.58
√𝑛
𝜎 𝜎
i.e 𝑥̅ − 2.58 < 𝜇 < 𝑥̅ + 2.58
√𝑛 √𝑛

I. 𝐻0: 𝜇 = 𝜇0 II. 𝐻0: 𝜇 = 𝜇0 III. 𝐻0: 𝜇 = 𝜇0

𝐻1: 𝜇 > 𝜇0 𝐻1: 𝜇 < 𝜇0 𝐻1: 𝜇 ≠ 𝜇0

Right- tailed test Left-tailed test Two- tailed test

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Example-1
A sample of 400 items is taken from a population whose standard
deviation is 10.The mean of sample is 40.Test whether the
sample has come from a population with mean 38 also calculate
95% confidence interval for the population.

Solution:

Step-1: Write given data in terms symbols

Given 𝑛=400, 𝑥̅  40 and  = 38 𝑎𝑛𝑑  = 10

Step:2 Set up Null hypothesis(𝐻0 )

𝐻0 :  =38 ,

i.e assume that sample has come from a population with mean 38.

Step:3 . Set up Alternative hypothesis(𝐻1 ).

𝐻1 :  38 ( Two tailed )

Sample is not from that population

Step:4. Choose level of significance .

Level of significance:  =0.05 and 𝑍𝛼 =1.96

Step:5 Apply the test statistic
𝑥̅ −𝜇
Test statistic: 𝑧= 𝜎
√𝑛

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 40 − 38 2 20
𝑧= = =
10 0.5 5
√400
𝑧=4
|𝑧 | = 4
Step-6: Conclusion . Compare the calculated value of z with

tabulated value 𝑍𝛼 =1.96

Acceptance region

|𝑧| = 4 > 𝑍𝛼 =1.96

|𝑧| > 𝑍𝛼=0.05

∴ we reject the Null Hypothesis 𝐻0 at 5% level of

significance .

(i)95% confidence interval for the population 𝜇 are

𝜎 𝜎
( 𝑥̅ − 𝑍0.05 , 𝑥̅ + 𝑍0.05 )
√𝑛 √𝑛

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 𝜎 𝜎
i.e. 𝑥̅ − 𝑍0.05 < 𝜇 < 𝑥̅ + 𝑍0.05
√𝑛 √𝑛

10 10
40 − 1.96 < 𝜇 < 40 + 1.96
√400 √400

39 .02< 𝜇 <40 .98

(ii)99% confidence interval for the population 𝜇 are
𝜎 𝜎
( 𝑥̅ − 𝑍0.01 , 𝑥̅ + 𝑍0.01 )
√𝑛 √𝑛
𝜎 𝜎
i.e. 𝑥̅ − 𝑍0.01 < 𝜇 < 𝑥̅ + 𝑍0.01
√𝑛 √𝑛

10 10
40 − 2.57 < 𝜇 < 40 + 2.
√400 √400

1 1
40 − 2.57 ( ) < 𝜇 < 40 + 2.57 ( )
2 2

38.715 .02< 𝜇 <41.285

Example-2:

The length of life of certain computer is approximately normally

distributed with mean 800 hours and standard deviation 40
hours. If a random sample of 30 computers has an average life of
788 hours, test the null hypothesis that 𝜇 = 800 hours against
the alternative hypothesis 𝜇 ≠ 800hours at (a) 5% (b) 1% level
of significance.

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Solution:

Step-1: Write given data in terms symbols

Given 𝑛=30 , 𝑥̅  788 hours and  = 800 𝑎𝑛𝑑  = 40

Step:2 Set up Null hypothesis(𝐻0 ) .

𝐻0 :  =800

i.e assume that average life of a computer is 800 hrs..

Step:3 . Set up Alternative hypothesis 𝐻1 .

𝐻1 :   800 ( Two tailed )

average life of a computer is not 800 hrs.

Step:4. Choose level of significance 𝛼.

(a)  =0.05 and 𝑍𝛼 =1.96

(b)  =0.01 and 𝑍𝛼 =2.58

Step:5 Apply the test statistic

𝑥̅ −𝜇
𝑧= 𝜎
√𝑛

788 − 800
𝑧= = −1.643
40
√30
𝑧 = −1.643

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 |𝑧| = 1.643
Step-6: Conclusion . Compare the calculated value of z with

tabulated value 𝑍𝛼 .

|𝑧| = 1.643 < 𝑍𝛼 =1.96

|𝑧| < 𝑍𝛼=0.05

∴ we accept the Null Hypothesis 𝐻0 at 5% level of
significance .

(b)𝛼 = level of significance 1%

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 |𝑧| = 1.643 < 𝑍0.01 =2.58

|𝑧| < 𝑍𝛼=0.01

∴ we accept the Null Hypothesis 𝐻0 at 1% level of significance .

Example-3

A random sample of 100 students is taken from a

large population. The mean height of the students in this
sample is 160 cm. Can it be reasonably regarded that ,in the
population ,the mean height is 165 cm, and the SD is 10cm?

Solution:

Step-1: Write given data in terms symbols

Given 𝑛=100 , 𝑥̅  160 hours and  = 165 𝑎𝑛𝑑  = 10

Step:2 Set up Null hypothesis 𝐻0 .

𝐻0 :  =165.

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There is no significance difference between sample mean and population
mean. I.e., sample is taken from a population with mean 165 and SD is 10

Step:3 . Set up Alternative hypothesis 𝐻1 .

𝐻1 :  165 ( Two tailed )

sample is not taken from a population with mean 165 and SD is 10.

Step:4. Choose level of significance 𝛼.

 =0.05 and 𝑍𝛼 =1.96

Step:5 Apply the test statistic

𝑥̅ −𝜇
𝑧= 𝜎
√𝑛

160 − 165
𝑧= = −5
10
√100
𝑧 = −5 , |𝑧| = 5
Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 .

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 |𝑧| = 5 > 𝑍𝛼 =1.96

|𝑧| > 𝑍𝛼=0.05

∴ we reject the Null Hypothesis 𝐻0 at 5% level of
significance .

Example:4

Solution: Step-1: Write given data in terms symbols

Given 𝑛=50 , 𝑥̅  1850 and  = 1800 𝑎𝑛𝑑  = 100

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Step:2 Set up Null hypothesis 𝐻0 .

𝐻0 :  =1800.

There is no significance difference between sample mean and population

mean. I.e., mean breaking strength=1800

Step:3 . Set up Alternative hypothesis 𝐻1

𝐻1 :  > 1800 ( Right tailed test )

i.e. sample is not taken from that population.

Step:4. Choose level of significance 𝛼.

. Level of significance  =0.01 and 𝑍𝛼 =2.33( Right tailed )

Step:5 Apply the test statistic

𝑥̅ −𝜇
Test statistic: 𝑧= 𝜎
√𝑛

1850 − 1800
𝑧= = 3.54
100
√50
Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

𝑧 = 3.54 > 𝑍0.01 = 2.33

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 |𝑧| > 𝑍𝛼=0.01
∴ we reject the Null Hypothesis 𝐻0 at 1% level of
significance and 𝐻1 is accepted. We conclude that there is increase in
breaking strength.

Example-5

A normal population has a mean of 0.1 and SD of 2.1. Find the

probability that the mean of a sample of size 900 drawn from
this population will be negative

Solution. Given 𝑛 = 900, 𝜇 = 0.1 , 𝜎 = 2.1

The standard normal variate corresponding to sample mean 𝑥̅

Is given by
𝑥̅ −𝜇
𝑧= 𝜎
√𝑛

Substituting values in the above formula ,

𝑥̅ − 0.1 𝑥̅ − 0.1
𝑧= =
2.1 0.07
√900
⇒ 𝑥̅ = 0.1 + 0.07𝑧 where𝑧~𝑁(0,1)

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Practice Problems
1. A sample of 900 items is found to have a mean of 3.47cm.Can it be
reasonably regarded as a simple sample from a population with mean
3.23 cm and SD 2.31 cm?
2. A manufacturer claims that,the mean breaking strength of belts for air
passengers produced in his factory is 1275 kgs.A sample of 100 belts
was tested and mean breaking strength and SD were found to be 1258
and 90kg respectively.Test the manufacturer’s claim at 5% level.
3. A random sample of 100 students gave a mean weight of 58kg with SD
of 4kg.Find 95% and 99% confidence limits of the mean of the
population.

4.It is claimed that a random sample of 49 tyres has a mean life of 15200
kms.This sample was drawn from a population whose mean is
15,150 km and aS.D.of 1200 km.Test the significance at 0.05 level

Population Mean=𝜇1 Population Mean=𝜇2

Population variance =𝜎1 2 Population variance =𝜎2 2
Sample mean = 𝑥̅1 Sample mean = 𝑥̅2
Sample size= 𝑛1 Sample size= 𝑛2

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 Let 𝑥̅1 be the mean of a random sample of size 𝑛1 from a
population with mean 𝜇1 and variance 𝜎1 2 and let 𝑥̅2 be the mean of an
independent random sample of size 𝑛1 from another population with mean
𝜇2 and variance 𝜎2 2 .

Then, under the assumption that

𝐻0 : 𝜇1 = 𝜇2
i.e by assuming that there is no significance
difference between two means,

the test statistic for difference means is given by

(𝑥̅1 − 𝑥̅2 ) − (𝜇1 − 𝜇2 )

𝑧=
𝜎1 2 𝜎2 2
√( + )
𝑛1 𝑛2

But ,when 𝜇1 = 𝜇2 , the above formula becomes

(𝑥̅1 − 𝑥̅2 )
𝑧=
𝜎1 2 𝜎2 2
√(
𝑛1 + 𝑛2 )
Note-1: If population variances 𝜎1 2 and 𝜎2 2 are not known ,then we can use
sample variances 𝑠1 2 and 𝑠2 2 in place of population variances.

In this case test statistic is given by

MVS Page 32
 𝑥̅1 −𝑥̅2
𝑧=
𝑠1 2 𝑠2 2
√( + )
𝑛1 𝑛2

Note-2: If the two samples are taken from same population variance 𝜎 2
then the test statistic is given by
𝑥̅1 −𝑥̅2
𝑧= =
𝜎2 𝜎2 IMP
√( + )
𝑛1 𝑛2

Test statistic for difference of means , when

𝐻0 : 𝜇1 = 𝜇2 is
(𝑥̅1 − 𝑥̅2 )
𝑧=
𝜎1 2 𝜎2 2
√( + )
𝑛1 𝑛2
When population variances 𝜎1 2 and 𝜎2 2 are not known, we
can sample variances.
𝑥̅ 1−𝑥̅ 2
Then, test statistic 𝑧=
𝑠1 2 𝑠2 2
ඩቌ
𝑛1 + 𝑛2 ቍ

MVS Page 33
 I. 𝐻0: 𝜇1 = 𝜇2 II. 𝐻0: 𝜇1 = 𝜇2 III. 𝐻0: 𝜇1 = 𝜇2

𝐻1: 𝜇1 > 𝜇2 𝐻1: 𝜇1 < 𝜇2 𝐻1: 𝜇1 ≠ 𝜇2

Right- tailed test Left-tailed test Two- tailed test

Example-1.

In a random sample of 100 tube lights produced by a company A,

the mean lifetime of tube light is 1190 hours with standard
deviation 90 hours. Also in a random sample of 75 tube lights from
company B, the mean lifetime is 1230 hours with standard
deviation of 120 hours. Is there a difference between the mean life
times of two brands of tube lights at a significant levels of (a) 0.05
(b) 0.01?

Solution:
Step-1: Write given data in terms symbols

𝑥̅1 = mean lifetime of tube lights of company A =1190

𝑛1 = Size of the sample taken from population A=100

𝑠1 = standard deviation of the tube lights of A= 90

MVS Page 34
 𝑥̅2 = mean lifetime of tube lights of company B =1230

𝑛2 = Size of the sample taken from population A=75

𝑠2 = standard deviation of the tube lights of A= 120

Step:2 Set up Null hypothesis 𝐻0 .

𝐻0 : 𝜇1 = 𝜇2
i.e by assuming that there is no significance
difference between two means,

Step:3 Set up Alternative hypothesis 𝐻1

𝐻1 : 𝜇1 ≠ 𝜇2 (two tailed test)

Step:4. Choose level of significance 𝛼.

Level of significance  =0.05 and 𝑍𝛼 =1.96

Step:5 Apply the test statistic

𝑥̅ 1 −𝑥̅2
Test statistic: 𝑧=
𝑠 2 𝑠 2
√( 1 + 2 )
𝑛1 𝑛2

1190−1230
𝑧= = -2.421
902 752
√( + )
100 120

Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

MVS Page 35
 (a) For 𝛼 = 0.05
|𝑧| = 2.421 > 𝑍0.05 = 1.96
We reject Null Hypothesis 𝐻0 . we conclude that
there is a difference in lifetimes of the tube lights produced by A and B .

(a) For 𝛼 = 0.01

|𝑧| = 2.421 < 𝑍0.01 = 2.58

We accept Null Hypothesis 𝐻0 . we conclude that there is no significant

difference in lifetimes of the tube lights produced by A and B .

MVS Page 36
Example-2

In a random sample of size 500,the mean is found to be 20.In

another independent sample of size 400,the mean is 15.Could
the sample have been drawn from the same population with
SD 4?

Solution.
Step-1: Write given data in terms symbols

𝑥̅1 = 20 𝑥̅2 = 15

𝑛1 =500 𝑛1 = 400

𝜎=4
Step:2 Set up Null hypothesis 𝐻0 :

𝐻0 : 𝜇1 = 𝜇2
i.e., Assume that two samples are drawn from same population.

Step:3 Set up Alternative hypothesis 𝐻1

𝐻1 : 𝜇1 ≠ 𝜇2 (two tailed test)

two samples are not drawn from same population.

Step:4. Choose level of significance 𝛼.

Level of significance  =0.05 and 𝑍𝛼 =1.96

Step:5 Apply the test statistic

𝑥̅1 −𝑥̅ 2
. Test statistic: 𝑧 = 1 1
𝜎 √( + )
𝑛1 𝑛2

MVS Page 37
 20−15
𝑧= = 18.6
1 1
4√(500 +400 )

Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

(a) For 𝛼 = 0.05

|𝑧| = 18.6 > 𝑍0.05 = 1.96

We reject Null Hypothesis 𝐻0 . we conclude that two

samples are not drawn from same population.

Example-3

MVS Page 38
 Solution.
Step-1: Write given data in terms symbols

𝑥̅1 = 170 𝑥̅2 = 172

𝑛1 =6400 𝑛2= 1600 Americans data

English men data

𝑠1 = 6.4 𝑠2 = 6.3

Step:2 Set up Null hypothesis 𝐻0 :

𝐻0 : 𝜇1 = 𝜇2

i.e., Assume that both English men and American men have same mean
height.

Step:3 Set up Alternative hypothesis 𝐻1

𝐻1 : 𝜇1 < 𝜇2 (Left tailed test)

Americans are taller than Englishmen.

Step:4. Choose level of significance 𝛼.

Level of significance  =0.01 and 𝑍𝛼 = −2.33

Step:5 Apply the test statistic

𝑥̅1 −𝑥̅2
. Test statistic: 𝑧=
𝑠 2 𝑠 2
√( 1 + 2 )
𝑛1 𝑛2

MVS Page 39
 170−172
𝑧= 2 2
= −11.32
(6.4) (6.3)
6400 + 1600 )
√(

Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

For 𝛼 = 0.01
|𝑧| = 11.32 > 𝑍0.01 = 2.33
We reject Null Hypothesis 𝐻0 and we accept
𝐻1 and conclude that Americans are taller than Englishmen.

Example-4

Test the significance difference between the means of the

samples,drawn from two normal populations using the following data

MVS Page 40
Solution.
𝑥̅1 = 61 𝑥̅2 = 63

𝑛1 =100 𝑛2= 200

𝑠1 = 4 𝑠2 = 6

1.Null hypothesis(𝐻0 ): 𝐻0 : 𝜇1 = 𝜇2

i.e., there is no significance difference between two means.

2.Alternative hypothesis(𝐻1 ): 𝐻1 : 𝜇1 ≠ 𝜇2 (Two tailed test)

3. Level of significance  =0.05 and 𝑍𝛼 = 1.96

𝑥̅1 −𝑥̅2
4. Test statistic: 𝑧=
𝑠1 2 𝑠2 2
√( + )
𝑛1 𝑛2

61−63
𝑧= 2 2
= −3.02
(4) (6)
√( +
100 200
)

5. Conclusion:

MVS Page 41
 For 𝛼 = 0.05
|𝑧| = 3.02 > 𝑍0.01 = 1.96
We reject Null Hypothesis 𝐻0 and we accept 𝐻1 and conclude that
there is a significance difference between two means.

Example-5

The means of simple samples of sizes 1000 and 2000 are 67.5
and 68.0cm respectively. Can the sample be regarded as
drawn from the same population of S.D 2.5cm

Solution.
𝑥̅1 = 67.5 𝑥̅2 = 68

𝑛1 =1000 𝑛1 = 2000 𝜎 = 2.5

1.Null hypothesis(𝐻0 ): 𝐻0 : 𝜇1 = 𝜇2

i.e., Assume that two samples are drawn from same population.

2.Alternative hypothesis(𝐻1 ): 𝐻1 : 𝜇1 ≠ 𝜇2 (two tailed test)

two samples are not drawn from same population.

3. Level of significance  =0.05 and 𝑍𝛼 =1.96

MVS Page 42
 𝑥̅1 −𝑥̅2
4. Test statistic: 𝑧= 1 1
𝜎 √( + )
𝑛1 𝑛2

67.5−68
𝑧= = 5.1
1 1
2.5√(1000 +2000 )

5. Conclusion:

For 𝛼 = 0.05
|𝑧| = 5.1 > 𝑍0.05 = 1.96

We reject Null Hypothesis 𝐻0 . we conclude that two

samples are not drawn from same population.

Practice problems
1. The average marks scored by 32 boys is 72 with SD 0f 8 ,while that for
36 girls is 70 with SD of 6.Test at 1% LOS where the boys perform
better than girls.

2.

MVS Page 43
3.

4.

5.

6.

MVS Page 44
 .
If 𝑋 is the number of successes in 𝑛 independent trials with constant
probability P of success for each trial.

Then we have
𝐸(𝑋) = 𝑛𝑃 𝑎𝑛𝑑 𝑉(𝑋) = 𝑛𝑃𝑄,
where 𝑄 = 1 − 𝑃, is the probability of failure.

It has been proved that for large n, the binomial distribution tends to
normal distribution. Hence for large n ,

We have test statistic ,

MVS Page 45
 𝑥−𝐸(𝑋) 𝑋−𝑛𝑃
𝑧= =
√𝑉(𝑋) √𝑛𝑃𝑄

𝑋 − 𝑛𝑃
𝑧=
√𝑛𝑃𝑄

In a sample of size n, let 𝑋 be the number of observations

possessing the given attribute. Then
𝑋
Observed proportion of successes = = 𝑝, (say)
𝑛

𝑋 𝐸(𝑋) 𝑛𝑃
∴ 𝐸(𝑝) = 𝐸 ( ) = = =𝑃
𝑛 𝑛 𝑛

𝐸 (𝑝) = 𝑃
𝑋 1 𝑛𝑃𝑄
𝑉(𝑝) = 𝑉 ( ) = ( )
2𝑉 𝑋 =
𝑛 𝑛 𝑛2

𝑃𝑄
𝑉(𝑝) =
𝑛
MVS Page 46
 We have test statistic for proportion 𝑝, is
𝑝−𝐸(𝑝) 𝑝−𝑃
𝑧= = where 𝑄 = 1 − 𝑃
√𝑉(𝑝) √
𝑃𝑄
𝑛
IMP

Test statistic for proportion 𝑝, is

𝑝−𝑃
𝑧=
√𝑃𝑄
𝑛

Population
Sample proportion
proportion

𝑝−𝑃
Test statistic
𝑧=
for single √𝑃𝑄⁄𝑛
Proportion
Sample
size

𝑄 = 1−𝑃

MVS Page 47
1.If 𝑃 is not known then taking 𝑝 (the sample proportion) as an estimate of
𝑃, the probable limits for the proportion in the population are :

𝑝𝑞
𝑝±3√
𝑛

95%confidence limits for population proportion 𝑃 are given

𝑝𝑞
by 𝑝 ± 1.96 √
𝑛

𝑝𝑞 𝑝𝑞
i.e 𝑝 − 1.96 √ < 𝑃 < 𝑝 + 1.96 √
𝑛 𝑛

99%confidence limits for population proportion 𝑃 are given

𝑝𝑞
by 𝑝 ± 2.58 √
𝑛

𝑝𝑞 𝑝𝑞
i.e 𝑝 − 2.58 √ < 𝑃 < 𝑝 + 2.58 √
𝑛 𝑛

Example-1

A dice is thrown 9,000 times and a throw of 3 or 4 is

observed 3240 times . Show that the dice cannot be
regarded as an unbiased one and find the limits between
MVS
which the probability of a throw of 3 or 4 lies. Page 48
Solution.
Step-1: Write given data in terms symbols

𝑛= number of times experiment conducted=9000

Getting 3 or 4 is considered as success.
𝑋 = Observed Number of success=3240
𝑋 3240
Proportion of success= 𝑝 = =
𝑛 9000
2 1
𝑃=Probability of success = =
6 3

Step:2 Set up Null hypothesis 𝐻0 :

1
Null hypothesis(𝐻0 ): 𝐻0 : 𝑃 =
3

i.e., Assume that the coin is unbiased .

Step:3 Set up Alternative hypothesis 𝐻1

1
𝐻1 : 𝑃 ≠ (two tailed test)
3

coin is biased one.

Step:4. Choose level of significance 𝛼.

Level of significance  =0.05 and 𝑍𝛼 =1.96

Step:5 Apply the test statistic

MVS Page 49
 𝑝−𝑃
Test statistic: 𝑧 =
𝑃𝑄
√
𝑛

3240 1
−
𝑧= 9000 3
1 2
√3 (3)
9000
𝑧 = = 5.36
Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

For 𝛼 = 0.05
|𝑧| = 5.36 > 𝑍0.05 = 1.96

We reject Null Hypothesis 𝐻0 . we conclude that coin is biased.

The probable limits for the population proportion of success are

𝑝𝑞
𝑝±3√
𝑛
3240
Here 𝑝 =
9000
= 0.36, 𝑞 = 1 − 𝑝 = 0.64

0.36(0.64)
0.36 ± 3√ = 0·360 ± 0·015 =
9000

0·345 and 0·375.

MVS Page 50
∴Limits of population 𝑃 are (0.345, 0.375)

Example-2
New-born babies are more likely to be boys than girls. A random
sample found 13,173 boys were born among 25,468 new-born children.
The sample proportion of boys was 0.5172. Is this sample evidence
that the birth of boys is more common than the birth of girls in the entire
population?

Solution.
Step-1: Write given data in terms symbols

𝑛= number of new born babies = 25468

𝑋 = Number of new born babies are boys =13,173
𝑋 13173
Observed Proportion of boys= 𝑝 = = = 0.5
𝑛 25468

𝑃= Population proportion=
1
Probability that a new born baby to be a boy =
2

Step:2 Set up Null hypothesis 𝐻0 :

1
Null hypothesis(𝐻0 ): 𝐻0 : 𝑃 =
2

i.e. boys and girls are equal in the entire population.

Step:3 Set up Alternative hypothesis 𝐻1

1
𝐻1 : 𝑃 > ( right tailed test)
2

MVS Page 51
 boys are more common than girls in the entire population .

Step:4. Choose level of significance 𝛼.

Level of significance  =0.05 and 𝑍𝛼 =1.645

Step:5 Apply the test statistic

𝑝−𝑃
Test statistic: 𝑧 =
𝑃𝑄
√
𝑛

Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

For 𝛼 = 0.05
|𝑧| = 5.49 > 𝑍0.05 = 1.645
We reject Null Hypothesis 𝐻0 .

We accept alternative hypothesis. we conclude boys are more common than

girls in the entire population

Example-3

Test the claim of a manufacturer that 95% of his stabilizers

confirm to ISI specifications. If out of a random sample of 200
stabilizers produced by this manufacturer 18 were faulty.

Use (a) 0.01(b) 0.05 L.O.S

MVS Page 52
Solution.

Step-1: Write given data in terms symbols

𝑛= size of the sample=200

𝑋 = Number of good stabilizers =200-18=182
𝑋 182
Observed Proportion of good stabilizers = 𝑝 = =
𝑛 200

𝑃= Population proportion=95%=0.95
Step:2 Set up Null hypothesis 𝐻0 :

Null hypothesis(𝐻0 ): 𝐻0 : 𝑃 = 0.95

Let us accept manufacturer claim.

Step:3 Set up Alternative hypothesis 𝐻1

𝐻1 : 𝑃 < 0.95 ( left tailed test)

Assume that manufacturer claim is false.

Step:4. Choose level of significance 𝛼.

Level of significance 

(a) = 0.01 (b)0.05

Critical region (a) 𝑧 < −2.33 (b) 𝑧 < −1.645

Step:5 Apply the test statistic

𝑝−𝑃
Test statistic: 𝑧 =
𝑃𝑄
√
𝑛

MVS Page 53
 182
−0.95 −0.04
200
𝑧= =
0.95(0.05) 0.0154
√
200

= −2.595
Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

For 𝛼 = 0.05
|𝑧| = 2.595 > 𝑍0.05 = 1.645
We reject Null Hypothesis 𝐻0 and accept alternative Hypothesis and
conclude that manufacturer claim is wrong.

Example-4

Solution:

Step-1: Write given data in terms symbols

𝑛= size of the sample=400

𝑋 = Number of top quality production =50
𝑋 50 1
Observed Proportion of top quality production = 𝑝 = = =
𝑛 400 8
20
𝑃= Population proportion= 20% = = 0.20
100

MVS Page 54
Step:2 Set up Null hypothesis 𝐻0 :

Null hypothesis(𝐻0 ): 𝐻0 : 𝑃 = 20% = 0.20

Let us accept 20% manufactured product is of top quality .

Step:3 Set up Alternative hypothesis 𝐻1

𝐻1 : 𝑃 ≠ 20% = 0.20 ( two- tailed test)

Assume that 20% manufactured product is not of top quality.

Step:4. Choose level of significance 𝛼.

Level of significance =0.05 ,𝑧𝛼 = 1.96

Critical region (a) |𝑧| > 1.96

Step:5 Apply the test statistic

𝑝−𝑃
Test statistic: 𝑧 =
𝑃𝑄
√
𝑛

1
− 0.20
8
𝑧= = −3.75
0.20(0.80)
√
400

Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

For 𝛼 = 0.05
|𝑧| = 3. .75 > 𝑍0.05 = 1.96
We reject Null Hypothesis 𝐻0 and accept alternative Hypothesis and
conclude that20% manufactured product is not of top quality.

MVS Page 55
And production of the particular day chosen is not a representative sample.

Example-5

The fatality rate of typhoid patients is believed to be 17.26%.

In a certain year 640 patients suffering from typhoid were treated in
a metropolitan hospital and only 63 patients died.Can you consider
the hospital efficient?

MVS Page 56
Practce problems:
1. A manufacturer claims that only 4% of his products are defective.
A random sample of 500 were taken among which 100 were
defective. Test the hypothesis at 0.05level.
2. A random sample of 500 computers were taken from a large
consignment and 60 wee found to be defective.Find 99% confidence
limits for the percentage number of defective computes in the
consignment.

3 . Test the claim of a manufacturer that 95%of his stabilizers conform to

its specifications if out of a random sample 200 stabilizers produced

by the manufacturer 18 were faulty. Use (a) 0.01 (b) 0.05 L.O.S.

4.A manufacturer of light bulbs claims on the average 2% of the bulbs

MVS Page 57
 manufactured by his firm are defective.A random sample of 400 bulbs
contained 13 defective bulbs.On the basis of this sample ,can you
support the manufacture’s claim at 5% LOS?.

5. A coin is thrown 400 times and is found to result in Head 245

times.Test whether the coin is a fair one.

6.A random sample of 400 mangoes was taken from a big consignment
and 40 were found to be bad. Prove that the percentage of bad
mangoes in the consignment will,lie between 5.5 and 14.5

4.Test of Significance for Difference of

Proportions.

Population Proportion=𝑃1 Population Proportion=𝑃2

Sample proportion = 𝑝1 Sample proportion = 𝑝2
Sample size= 𝑛1 Sample size= 𝑛2

Suppose we want to compare two distinct populations with respect to the

prevalence of a certain attribute, say A, among their members.

Let 𝑋1 , 𝑋2 be the number of persons possessing the given attribute A in

random samples of sizes 𝑛1 and 𝑛2 from the two populations respectively.

MVS Page 58
 Then sample proportions are given by
𝑋1 𝑋2
𝑝1 = and 𝑝2 =
𝑛1 𝑛2

If 𝑃1 𝑎𝑛𝑑 𝑃2 are the population proportions.

then 𝐸(𝑝1 ) = 𝑃1 𝐸(𝑝2 ) = 𝑃2

Now the test statistic for difference of proportions is given by

(𝑝1 − 𝑝2 ) − (𝑃1 − 𝑃2 )
𝑧=
1 1
√𝑃𝑄 ( + )
𝑛1 𝑛2

Under 𝐻0 : 𝑃1 = 𝑃2 , the test statistic for the'difference of proportions

becomes
(𝑝1 − 𝑝2 ) − 0
𝑧=
1 1
√𝑃𝑄 ( + )
𝑛1 𝑛2
(𝑝1 − 𝑝2 )
∴𝑍=
1 1
√𝑃𝑄 ( + )
𝑛1 𝑛2
𝑛1 𝑝1 +𝑛2 𝑝2 𝑋1 +𝑋2
Where 𝑃 = = ,𝑄 =1−𝑃
𝑛1 +𝑛2 𝑛1 +𝑛2

MVS Page 59
 Test statistic for difference of proportions, when 𝐻0 : 𝑃1 = 𝑃2 is
(𝑝1 −𝑝2 )
𝑍= 1 1
√𝑃𝑄(𝑛 +𝑛 )
1 2

where 𝑃 = 𝑛1𝑛𝑝1+𝑛
+𝑛2𝑝2 𝑋 +𝑋
= 1 2
𝑛 +𝑛
1 2 1 2

𝑄 = 1−𝑃

I. 𝐻0: 𝑃1 = 𝑃2 II. 𝐻0: 𝑃1 = 𝑃2 III. 𝐻0: 𝑃1 = 𝑃2

𝐻1: 𝑃1 > 𝑃2 𝐻1: 𝑃1 < 𝑃2 𝐻1: 𝑃1 ≠ 𝑃2

Right- tailed test Left-tailed test Two- tailed test

MVS Page 60
Example-1

In a large city A,20% of a random sample of 900 school boys had a

slight physical defect. In another large city B, 18.5 % of a random
sample of 1600 school boys had the same defect. Is the
difference between the proportions significant.?

Solution:

Step-1: Write given data in terms symbols

𝑛1 = size of the first sample taken from city A =900

𝑋 20
Observed Proportion defect boys in city A = 𝑝1 = = = 0.2
𝑛 100

𝑛2 = size of the first sample taken from city A =1600

𝑋 18.5
Percentage of defective boys in city B = 𝑝2 = = = 0.185
𝑛 100

Step:2 Set up Null hypothesis 𝐻0 :

Null hypothesis(𝐻0 ): 𝐻0 : 𝑃1 = 𝑃2

Proportion of physical defective boys in both cities are equal

Step:3 Set up Alternative hypothesis 𝐻1

𝐻1 : 𝑃1 ≠ 𝑃2 ( two- tailed test)

Proportion of physical defective boys in both cities are not equal

Step:4. Choose level of significance 𝛼.

MVS Page 61
Level of significance =0.05 ,𝑧𝛼 = 1.96

Critical region (a) |𝑧| > 1.96

Step:5 Apply the test statistic

𝑝1 −𝑝2
Test statistic: 𝑍= 1 1
-----(1)
√𝑃𝑄(𝑛 +𝑛 )
1 2

𝑛1 𝑝1 +𝑛2 𝑝2
𝑃=
𝑛1 +𝑛2
,𝑄 =1−𝑃

900(0.2) + 1600(0.185)
𝑃=
900 + 1600
0.2 − 0.185
𝑧=
√0.1904 × 0.8096 ( 1 + 1 )
900 1600
𝑧 = 0.92

Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

For 𝛼 = 0.05
|𝑧| = 0.92 < 𝑍0.05 = 1.96
We accept Null Hypothesis 𝐻0 and conclude that percentage of physical
defect boys in both cities are equal

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Example-2

In a survey of mobiles produced by company A it was found that

19 mobiles were defective in a random sample of 200 while for
company B,5 were defective out of 100. At 0.05 level of
significant is there reason to believe that
(a) there is significant difference in performance of mobiles
between the
two companies A and B.
(b) Products of B are superior to products of A.

Solution (a)

Step-1: Write given data in terms symbols

𝑛1 = size of the first sample taken from company A =200

𝑋1 = observed no. of good mobiles from company A=200-19=181

𝑛2 = size of the second sample taken from company A =100

𝑋2 = observed no. of good mobiles from company B= 95

𝑋1 181
Observed Proportion defectives from A = 𝑝1 = =
𝑛1 200

𝑋2 95
Observed Proportion defectives from B = 𝑝2 = =
𝑛2 100

Step:2 Set up Null hypothesis 𝐻0 :

Null hypothesis(𝐻0 ): 𝐻0 : 𝑃1 = 𝑃2

Proportion of good mobiles from both companies are equal

i.e performance of both are same

Step:3 Set up Alternative hypothesis 𝐻1

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𝐻1 : 𝑃1 ≠ 𝑃2 ( two- tailed test)

performance of both are same not equal

Step:4. Choose level of significance 𝛼.

Level of significance =0.05 ,𝑧𝛼 = 1.96

Critical region (a) |𝑧| > 1.96

Step:5 Apply the test statistic

𝑝1 −𝑝2
Test statistic: 𝑍= 1 1
-----(1)
√𝑃𝑄(𝑛 +𝑛 )
1 2

𝑛1 𝑝1 +𝑛2 𝑝2 𝑋1 +𝑋2
𝑃=
𝑛1 +𝑛2
=
𝑛1 +𝑛2
,𝑄 =1−𝑃
181+95
𝑃= = 0.92, 𝑄 = 1 − 𝑃 = 0.08
200+100

0.905 − 0.95
𝑧=
√0.92 × 0.08 ( 1 + 1 )
200 100
𝑧 = −1.3636
Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

For 𝛼 = 0.05
|𝑧| = 1.3636 < 𝑍0.05 = 1.96
We accept Null Hypothesis 𝐻0 and conclude that performance of both
are same.

(b) 𝐻1 : 𝑃1 < 𝑃2 left- tailed test

i.e Products of company B is superior

. Choose level of significance 𝛼.

MVS Page 64
 Level of significance =0.05 ,𝑧𝛼 = −1.645

Critical region (a) |𝑧| > 1.645

|𝑧| = 1.3636 < 1.645

We accept Null Hypothesis.

We conclude that Products of company B is not superior.

Example-3

!5.5% of a random sample of 1600 undergraduates were smokers,

whereas 20% of a random sample of 900 postgraduates were smokers in a
state. Can we conclude that less number of undergraduates are smokers
than the postgraduates?

Solution

Step-1: Write given data in terms symbols

𝑛1 = size of the Sample taken from undergraduates =1600

15.5
Observed Proportion smokers in under graduates= 𝑝1 = 15.5% = =
100
= 0.155

𝑛2 = size of the Sample taken from postgraduates =900

20
Observed Proportion smokers in post graduates= 𝑝2 = 20% = =
100
= 0.2
Step:2 Set up Null hypothesis 𝐻0 :

Null hypothesis(𝐻0 ): 𝐻0 : 𝑃1 = 𝑃2

Proportion of smokers in both undergraduates and postgraduates are equal

Step:3 Set up Alternative hypothesis 𝐻1

MVS Page 65
𝐻1 : 𝑃1 < 𝑃2 ( left- tailed test) ( 𝑃1 < 𝑃2why?)

Assume that less number of undergraduates are smokers than the

postgraduates.

Step:4. Choose level of significance 𝛼.

Level of significance =0.05 ,𝑧𝛼 = −1.645

Critical region (a) |𝑧| > 1.645

Step:5 Apply the test statistic

𝑝1 −𝑝2
Test statistic: 𝑍= 1 1
-----(1)
√𝑃𝑄(𝑛 +𝑛 )
1 2

𝑛1 𝑝1+𝑛2 𝑝2 1600(0.155) +900(0.2)

𝑃= = = 0.1712
𝑛1 +𝑛2 1600+900

𝑄 = 1 − 𝑃 = 0.8288
0.155 − 0.2
𝑧=
√0.1712 × 0.8288 ( 1 + 1 )
1600 900
𝑧 = −2.87

Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

For 𝛼 = 0.05
|𝑧| = 2.87 > 𝑍0.05 = 1.645
We rejct Null Hypothesis 𝐻0 .

Therefore, we accept 𝐻1 and conclude that less number of undergraduates

are smokers than the postgraduates.

MVS Page 66
Example-4

Before an increase in excise duty on tea,800 people out of a

sample 1000 were consumers of tea. After the increase in duty
,800people were consumers of tea in a sample of 1200 persons.
Find whether there is significant decrease in the consumption of
tea after the increase in duty.

Solution.

Solution (a)

Step-1: Write given data in terms symbols

𝑛1 = size of the first sample taken =1000

𝑋1 = observed no. tea drinkers before increase in excise

duty=800

𝑛2 = size of the second sample taken after increase in excise

=1200

𝑋2 = observed no. tea drinkers after increase in excise

duty=800
𝑋1
Observed Proportion tea drinkers in the beginning = 𝑝1 =
𝑛1

800
= = 0.8
1000
Observed Proportion tea drinkers after increase in excise
𝑋2 800 2
duty= 𝑝2 = = =
𝑛2 1200 3

MVS Page 67
Step:2 Set up Null hypothesis 𝐻0 :

Null hypothesis(𝐻0 ): 𝐻0 : 𝑃1 = 𝑃2

Proportion of tea drinkers before and after the increase excise duty are
equal.

Step:3 Set up Alternative hypothesis 𝐻1

𝐻1 : 𝑃1 > 𝑃2 ( right - tailed test)( 𝑃1 > 𝑃2 why?)

performance of both are same not equal

Step:4. Choose level of significance 𝛼.

Level of significance =0.05 ,𝑧𝛼 = 1.645

Critical region (a) |𝑧| > 1.645

Step:5 Apply the test statistic

𝑝1 −𝑝2
Test statistic: 𝑍= 1 1
-----(1)
√𝑃𝑄(𝑛 +𝑛 )
1 2

𝑛1 𝑝1 +𝑛2 𝑝2 800+800
𝑃= =
𝑛1 +𝑛2 1000+1200

1600 8
𝑃= = = 0.7273
2200 11

𝑄 = 1 − 𝑃 = 0.2727
0.8 − 2⁄3
𝑧=
√0.7273 1 1
× 0.2727 ( + )
1000 1200
𝑧 = 6.82

MVS Page 68
Step-6: Conclusion .

Compare the calculated value of z with tabulated value 𝑍𝛼 :

For 𝛼 = 0.05
|𝑧| = 6.82 > 𝑍0.05 = 1.645
We reject Null Hypothesis 𝐻0 and accept𝐻1

We conclude that there is significant decrease in the consumption of tea

after the increase in duty.

Practice problems.

1. Random samples of 200 bolts manufactured by machine A and 100

bolts manufactured by machine B shows 19 and 5 defective bolts
respectively. Test hypothesis that machine B is performing better than
A. Use 0.05 Level of significance.
2. A manufacturer of electronic equipment subject’s samples of two
competing brands of transistors to an accelerated performance test.
If 45 of 180 transistors of the first kind and 34 of 120 transistors of
the second kind fail the test. What can be conclude at the level of
significance 𝛼 = 0.05 about the difference between the
corresponding sample proportions.?

3. If 57 out of 150 patients suffering with certain disease are cured by

allopathy and 33out of 100 patients with same disease are cured by
homeopathy, is there reason to believe that allopathy is better than
homeopathy at 0.05 L.O.S.
4. A machine produces16 defective bolts in a batch of 500 bolts. After
the machine is overhauled , it produces 3 defective bolts in a batch of
100 bolts. Has the machine improved?

MVS Page 69
 𝐻0 : 𝑃1 = 𝑃2 , 𝐻1 : 𝑃1 > 𝑃2 (i.e machine is improved)

5.

6.

MVS Page 70