Testing of Hypothesis combined with tests.pdf

HYPOTHESIS AND ITS TESTING PROCEDURES
B. S. Parajuli

Hypothesis
➢An assumption that we make about the population
parameter
➢A statistical hypothesis is a statement about population
usually concerned with the value of one or more
parameters of the population.
i. logically conjectured relationship
ii. Relationship between two or more variable
iii. expressed in the form of testable statement(s)
iv. Formal statement(s) / conjectured statement
v. Variables (independent and dependent)

 The average monthly income of the people of this city is
Rs. 17,000. i.e. 𝝁 = Rs. 17,000
 The proportion of people with cell phones in this rural
municipality is 0.62
i.e. P = 0.62
 More than 20% of the adults watch children’s program
in the television.
 Is the average waiting time for the customers of Smart
Supermarket at the checkouts greater than 15
minutes?
 Is there any difference between the percentage of
internet users between Kathmandu and Pokhara ?

Purpose
will identify
independent
variables
influence upon
dependent
variable
investigate a
research
problem
enhances the
objectivity of
the study
understand
what data is
to be
collected
increase the
objectivity of
research.
concludes
what is true
or what is
false

Null Hypothesis: A statistical hypothesis or assumption made about the
population parameter to testing its validity for the possible acceptance is
called null hypothesis. It is also called hypothesis of no difference. It is
denoted by H0.
• Always contains “=” , “≤” or “” sign
Example : The mean age of MBM students of SMC is equal to 23 years
(H0 ∶ 𝝁 = 23 years)
Alternative Hypothesis: Any hypothesis which is complementary to the
null hypothesis is called an alternative hypothesis and is denoted by H1.
 Always contains “≠” , “>” or “<” sign
H1 ∶ 𝝁 ≠ 23 years (TwoTailed)
H1 ∶ 𝝁 > 23 years (RightTailed)
H1 ∶ 𝝁 < 23 years(LeftTailed)

To determine whether the test is one-tailed or two tailed depends
upon the alternative hypothesis. If the alternative hypothesis is two
tailed we apply two-tailed test and if the alternative hypothesis is
one-tailed, we apply one-tailed test.
Two Tailed Test : A test of statistical hypothesis in which the alternati
ve hypothesis looks for the values are significantly different or not
but the direction of the difference(more or less) is not specified.
Key words like:
 differs from, is not equal to, differs significantly, same,
whether any difference, unbiased, whether equal to, not
equals to

One Tailed Test: A test of statistical hypothesis in which the alternative
hypothesis H1 looks for a definite direction of the difference (more or less)
with parameter is specified.
Thus for the comparative study of the population parameter identifying the
words such as:
less than, more than, greater than, no more than, taller than,
smaller than, better than , at least, at most , only, improved,
superior, inferior, below, above, high, low, increase, decrease,
better than, up to, majority, minority etc.
After identifying one-tailed test, the left tailed test and right
tailed tests can be identified in following way:
i. If sample statistic < Population parameter- left tailed test
ii. If sample statistic > Population parameter- Right tailed test
iii. If first sample statistic < Second sample statistic- Left tailed test
iv. If first sample statistic > Second sample statistic- Right tailed test

Level of Significance: The maximum size of type I error that we are
prepared to take risk is called the level of significance and is denoted by 𝛼.
Symbolically it is defined as
𝛼 = P(type I error)
= P (Rejecting a true null hypothesis)
If we allow α = 5%, we are likely to reject 5 cases in making decision 100
times under the same conditions.
Critical Region and Rejection Region:
The region where the null hypothesis is rejected is called critical region or
rejection region and the region for accepting true null hypothesis is called
acceptance region. If the statistic belongs to the critical region, H0 is rejected
and if the statistic belongs to the acceptance region H0 is accepted.
The critical region is established on the basis of the choice of level of
significance. If 𝛼 = 0.05(5%) then the total rejection area under the
normality curve will be 0.05(5%) and the acceptance area under the normal
probability curve will be 0.95(95%).

Critical / Significant Value:
The value which separates the acceptance region and the
rejection region is called critical value or significant value. It
depends upon
(i) The level of significance used and
(ii) The alternative hypothesis whether it is two tailed or one
tailed (left or right ) test.
Critical values
Rejection Region
/2
0
a
/2
a
Acceptance Region
Two Tailed Test
0.025 0.025
0.95

For large sample, the critical value of ‘Z’ obtained from the statistical table
similarly for small sample the critical value of ‘t’ obtained from the
statistical table.
When the computed test statistic lies in the acceptance region, it is
reasonable to accept the null hypothesis as it is believed to be probable true.
Acceptance
Region
0
a
Critical values
Rejection Region
0
a
Acceptance
Region
RightTailedTest LeftTailedTest
0.95
0.95
0.05

Blood
Pressure
Heridity
Regular
Exercise
Food
Habit
Weight
Age

Contd…
➢ Example
➢ H0: Food habit does not affect blood pressure.
➢ H1: Food habits affects blood pressure.
➢ Food habit: independent variable
➢ Blood pressure: dependent variable.
➢ H0: Age does not affect blood pressure.
➢ H1: As age increases blood pressure increases.
➢ age: independent variable
➢ Blood pressure: dependent variable.
➢ H0: Regular exercise does not affect blood pressure.
➢ H1: Regular exercise may decrease blood pressure.
➢ regular exercise: independent variable
➢ blood pressure: dependent variable.

Parametric and Non-parametric test
 ParametricTest:
i. Model specifies certain conditions about the parameters of the
populations from which the samples are drawn
ii. Implies that there is normal distribution
iii. The data used in this test are continuous
 Non- ParametricTest:
i. does not depend on the particular form of the basic frequency
function from which the samples are drawn
ii. do not implies that there is a normal distribution
iii. The data used in this test are frequency or ranks
iv. measuremental data: data obtained by actual measurement.

Parametric test Non-parametric test
1. Parametric tests specify certain
conditions about the parameters of
the population from which the
sample has been drawn
1. Non-parametric test do not specify
any condition about the population
parameter of the population from which
the sample has been drawn.
2. It is used in testing of hypothesis
and estimation of parameters.
2. It is used in testing of hypothesis and
but not estimation of parameters.
3. Parametric test are mostly applied
only to the data which are measured
in interval and ratio scale.
3. Non-parametric test are applied only
to the data which are measured in
nominal and ordinal data.
4. Parametric test are most powerful
test.
4. Non-parametric test are less powerful
test than parametric test.
5. It requires complicated sampling
technique.
5. It does not require complicated
sampling technique.

Criteria of Good Hypothesis Statement
• Should be stated in declarative form
• Should articulate (describe) a relationship between two or
more variables
• Should be testable empirically
• Useless if it cannot be tested empirically
• Should be limited in scope
• Should be clearly and precisely stated
• Should no ambiguity (vegness) in the variables
• Should state the conditions to apply.
• Context and study units must be clear
• Should reflect a guess at a solution
• Should outcome to a problem based upon some knowledge,
previous research, or identified needs
• Should be consistent with most known facts

Actual
situation
Decision from sample
Accept H₀ Reject H₀
H₀ is true Correct decision
No error
Probability = 1 – α
Wrong decision
Type I error
Probability = α
(Not so harmful and can
be corrected)
H₀ is false
(H₁ true)
Wrong decision
Type II error
Probability = β
(Degrees of impact of
Type II error is very
harmful and can’t be
corrected)
Correct decision
No error
Probability = 1 – β
1 – β = power of test
Types of Errors inTesting of Hypothesis

➢ If a medicine is administered to a few patients of a particular
disease to cure them and the medicine is curing the disease, but it
is claimed that it has no effect and hence it is discontinued. This
Type I error.
➢ If a medicine is administered to a few patients of a particular
disease to cure them and the medicine is not curing the disease,
but it is claimed that it have good effect and the treatment is
continued. This type II error.
Type I error : Rejecting the good lot of products
(Producer’s risk)
Type II error: Accepting the bad lot of products
(Consumer’s risk)

Steps in Hypothesis Testing
a. Critical Value approach:
Depends upon the size of
sample and nature of data
Use 𝛼 = 0.05 unless we are
given
Formulate the Null and Alternative
Hypothesis
Select the Right test statistic and
compute test statistic
Decide the level of significance
Determine Critical value
Decision: compare computed test
statistic with critical value
If Calculated Value ≤ Tabulated Value
Yes
No
It is not Significant Accept HO
It is significant Reject HO

b. p-value approach:
The p-value is the probability of getting the observed value of the
computed test statistic or a value with even greater evidence against
null hypothesis H0 if the null hypothesis is true.
In other words, the observed value of level of significance for the
testing of hypothesis assuming null hypothesis is true is called p-
value.
The p- value is the maximum rejection area under the normal
probability curve under the null hypothesis H0, when the some test
(or procedure) is applied.
➢ If p-value ≥ 𝛼 (level of significance) then accept null hypothesis
(H0).
➢ If p-value < 𝛼 (level of significance) then accept alternative
hypothesis (H1).
Mostly p-value can be obtained from different statistical software
based on the general output for the given problem.

Finding p-value:
(i) For Left Tailed Test:
p = P( Z < Zcal)
= 0.5 – P(0 < Z < Z cal)
(ii) For Right Tailed Test
p = P( Z >Zcal)
= 0.5 – P(0 < Z < Z cal)
(iii) For Two Tailed Test:
p = P( Z < Zcal) + P( Z >Zcal)
= 2 P(Z >Zcal)
= 2[0.5 – P(0 < Z < Z cal)] Z = 0
Zcal
Zcal
Zcal
Zcal
Z = 0
Z = 0

Two tailed test and Confidence interval
• If the reference value specified in H0 lies outside the
interval (that is, is less than the lower bound or
greater than the upper bound), you can reject H0.
• If the reference value specified in H0 lies within the
interval (that is, is not less than the lower bound or
greater than the upper bound), you fail to reject H0.
If hypothesis value 𝐻0: 𝜇 = 𝜇0lies within calculated
confidence interval estimate
[ ത
𝑋 − 𝑍 Τ
𝛼
2
𝑆. 𝐸. ( ത
𝑋) < 𝜇0 < ത
𝑋 − 𝑍 Τ
𝛼
2
𝑆. 𝐸. ( ത
𝑋)]
then null hypothesis is accepted otherwise rejected.

27
• Z-test is important parametric test and based on normal
probability distribution. When the samples are selected from
population of known parameter with sample size more than
30 , Z test is used. We consider that if sample size is more
than 30 then sample selected non-normal population is also
approximately normally distributed.
• Z- test is defined as: 𝑍 =
𝑡 − 𝐸(𝑡)
𝑆.𝐸.(𝑡)
• 𝑇𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
Statistic – Parameter
Standard error
Application of Z test:
(i) Test of significance of single mean
(ii) Test of significance of difference between two means
(iii) Test of significance of single proportion
(iv) Test of significance of difference between two proportions
Introduction

Critical value
Critical
values
Level of significance (α)
Zα 1% 2% 5% 10%
Two tailed
test
ǀ𝑍𝛼ǀ = 2.58 ǀ𝑍𝛼ǀ = 2.33 ǀ𝑍𝛼ǀ = 1.96 ǀ𝑍𝛼ǀ = 1.645
Right tailed
test
ǀ𝑍𝛼ǀ = 2.33 ǀ𝑍𝛼ǀ = 2.05 ǀ𝑍𝛼ǀ = 1.645 ǀ𝑍𝛼ǀ = 1.28
Left tailed
test
ǀ𝑍𝛼ǀ
= −2.33
ǀ𝑍𝛼ǀ
= −2.05
ǀ𝑍𝛼ǀ =
−1.645
ǀ𝑍𝛼ǀ
= −1.28
28

Procedure
Step 1: Formulate the Null (H0) and Alternative (H1) Hypothesis.
H0: 𝝁 = 𝝁𝟎 i. e. there is no significant difference between the sample mean (𝑋) and the
population mean
H1: 𝝁 ≠ 𝝁𝟎 (two tailed test), or H1 : 𝝁 > 𝝁𝟎 (right tailed test),or H1 : 𝝁 < 𝝁𝟎 (left tailed
test)
If sample mean( ത
𝑋) > Population mean (𝜇 ) : Right Tailed Test
If sample mean( ത
𝑋) < Population mean (𝜇 ) : Left Tailed Test
Step 2: Select test statistic: 𝒁 =
ഥ
𝑿 − 𝝁
ൗ
𝝈
𝒏
If 𝜎 is unknown for large samples ො
𝜎 = s
Step 3: Fix the level of significance (𝛼)
Step 4: Write the critical value (Z ) for 𝛼 % level of significance from the Z table.
Step 5: Make decision:
Critical value approach:
If ǀZǀ(calculated) ≤ Zα(table or critical ) we say it is not significant (insignificant). So H₀ is accepted.
If ǀZǀ(calculated) > Zα(table or critical) we say it is significant. So H₀ is rejected, then H₁ is accepted.
P-value approach:
If p-value ≥ 𝛼 (level of significance) then accept null hypothesis (H0).
If p-value < 𝛼 (level of significance) then accept alternative hypothesis (H1). 29
Case I: Test of Significance of Single Mean

Example : A samples of 50 pieces of certain type of string was tested. The mean
breaking strength turned out to be 14.5 kgs. Test whether the sample is from a
batch of strings having a mean breaking strength of 15.6 kgs and standard
deviation of 2.2 kgs.
Here, Number of sample(n) = 50 Sample mean ( ത
𝑋) = 14.5 kgs.
Population mean () = 15.6 kg Population standard deviation (𝜎) = 2.2 kgs
Formulation of Hypothesis
• Null Hypothesis: Ho :  = 15.6 kgs i.e. the Population mean breaking strength of
the strings is 15.6 kg.
• Alternative Hypothesis: H1 :  ≠15.6 kgs i.e. the mean breaking strength is not
equal to 15.6 kg. (two tailed test)
Test Statistics: Z =
ഥ
𝑿 − 𝝁
ൗ
𝝈
𝒏
=
𝟏𝟒.𝟓− 𝟏𝟓.𝟔
ൗ
𝟐.𝟐
𝟓𝟎
= = - 3.55
| zcal | = | - 3.55 | = 3.55
Level of significance (a) = 5%
Critical value: Tabulated value of z at 5% level of significance for two tailed test Ztab
= 1.96
Decision: Since Zcal  Ztab , Ho is rejected i.e. H1 is accepted.
Conclusion: Hence we conclude that the sample has not been drawn from the
batch with mean breaking strength of 15.6 kgs and standard deviation 2.2 kgs.
30

Example : In the past a blending process has produced an average of 5 Kg.
of waste material for every batch with standard deviation 5 Kg. From a
sample of 100 batches an average of 7 Kg. of waste per batch is obtained. At
5% level of significance is it reasonable to believe that the average has
increased?
Solution: with usual notations, n = 100, ത
𝑋= 7 kg,  = 5 kg,  = 5 kg
Formulation of Hypothesis
• H0:  = 5 kg, i.e. the average of waste material has not been increased
• H1:   5 kg, i.e. the average has increased (Right tailed test)
Test Statistics: Z =
ഥ
𝑿 − 𝝁
ൗ
𝝈
𝒏
=
𝟕− 𝟓
ൗ
𝟓
𝟏𝟎𝟎
= 4 zcal = 4
Critical value: Tabulated value of z at 5% level of significance for right tailed
test Ztab = 1.645
Decision: Since Zcal  Ztab , Ho is rejected i.e. H1 is accepted. Hence we conclude
that the average waste material has been increased.
31

Example : An insurance agent has claimed that the average age of the policy holders
who have insured through him is less than the average age for all the agents, which
is 30.55 years. A random sample of 100 policy holders who had insured through him
gave the following age distribution.
Calculate the arithmetic mean and standard deviation of this distribution and use
these values to test his claim at 5% level of significance.
Solution:
32
Class limits 16-20 21-25 26-30 31-35 36-40
Frequency 12 22 20 30 16
Age f x 𝒅′
=
𝒙−𝟐𝟖
𝟓
f𝒅′
f𝒅′𝟐
16-20 12 18 -2 -24 48
21-25 22 23 -1 -22 22
26-30 20 28 0 0 0
31-35 30 33 1 30 30
36-40 16 38 2 32 64
Σfd’= 16 Σfd’2=164

x̄ = A +
Σfd’
n
× h = 28+
16
100
×5 = 28 + 0.08 = 28.8 years
𝑠 =
σ 𝑓d’2
𝑛
− (
Σfd’
n
)2 × h =
164
100
− (
16
100
)2 × 5 = 6.352 years
Setting of Hypothesis
H0: 𝝁 = 30.5 years i.e. the average age of the policy holders for all agents is 30.5
years.
H1: 𝝁 < 30.5 years (left tailed test) i.e. the average age of the policy holders
who have insured through him is less than the average age for all the agents.
Test Statistics: Under H0, the test statistic is Z =
ഥ
𝑿 − 𝝁
ൗ
𝝈
𝒏
=
ഥ
𝑿 − 𝝁
ൗ
𝒔
𝒏
=
𝟐𝟖.𝟖 − 𝟑𝟎.𝟓
ൗ
𝟔.𝟑𝟓𝟑
𝟏𝟎𝟎
= -2.68
| zcal | = | - 2.68 | = 2.68
Critical value: Tabulated value of z at 5% level of significance for left tailed test
Ztab = -1.645 i.e. | ztab | = 1.645
Decision: Since | zcal |> | ztab | , So Ho is rejected i.e. H1 is accepted. Hence we
conclude that the average age of the policy holders who have insured through
him is less than the average age of the policy holders of all the agents.
33

Case II: Test of Significance of Difference between Two Means
34
Procedure
H0:𝛍₁ = 𝛍₂ i. e. Two samples have been drawn from the sample parent population
H1:𝛍₁ ≠ 𝛍₂(two tailed test), H1:μ₁ > μ₂(right tailed test), H1:μ₁ < μ₂(left tailed test)
If First sample mean(ഥ
X1) > Second sample mean(ഥ
X2) : Right Tailed Test
X1) < Second sample mean(ഥ
X2) : Left Tailed Test
Step 2: Select test statistic: Z =
ഥ
X1− ഥ
X2
σ1
2
n1
+
σ2
2
n2
𝑜𝑟 Z =
(ഥ
X1−ഥ
X2)
s1
2
n1
+
s2
2
n2
𝑜𝑟Z =
(ഥ
X1− ഥ
X2)
𝜎2 1
n1
+
1
n2
Step 3: Fix the level of significance (𝜶)
If ǀZǀ(calculated) ≤ Zα(table or critical ) we say it is not significant (insignificant). So H₀ is
accepted.
If ǀZǀ(calculated) > Zα(table or critical) we say it is significant. So H₀ is rejected, then H₁ is
accepted.
Step 6: Write Conclusion:

Example: An examination was given to 50 students at college A and 60
students at college B. At college A, the average marks of the students was
found to be 75 with a standard deviation of 9. At college B, the average
marks of the students was 79 with a standard deviation of 7. Is there a
significant difference between the average marks of the students at
these two colleges?
Solution: with usual notations,
College A College B
n1 = 50 n2 = 60
ഥ
X1= 75 ഥ
X2 =79
s1 = 9 s2 = 7
Setting of Hypothesis:
H0 : 𝛍₁ = 𝛍₂ i. e. there is no significant difference between the average
marks of the students at these two colleges
H1: 𝛍₁ ≠ 𝛍₂ (Two tailed test) There is significant difference between the
average marks of the students at these two colleges

Test Statistic: Under H0 the test statistic is
Z =
(ഥ
X1−ഥ
X2)
s1
2
n1
+
s2
2
n2
=
(75−79)
92
50
+
72
60
= -2.56
ǀZǀcal= 2.56
Level of significance(𝜶) = 5%
Critical value: Ztab= 1.96
Decision: Since Zcal > Ztab so H0 is rejected and H1 is accepted.
Conclusion: Hence we conclude that There is significant
difference between the average marks of the students at these
two colleges
36

Example: In a random sample of 500 the mean is found to be 20. In another
independent sample of 400 the mean is 15. Could the samples have been drawn from
the same population with standard deviation 4 ?
Solution: With usual notations,
Common Population S. D. (𝜎) = 4
H0:𝛍₁ = 𝛍₂ (Both samples have been
Drawn from the same population)
H1:𝛍₁ ≠ 𝛍₂ (Two tailed test)
Test statistic: Z =
ഥ
X1− ഥ
X2
𝜎2 1
n1
+
1
n2
contd…
Example : If 60 MA students are found to have a mean height of 63.60 inches and 50
MBS students have mean height of 69.51 inches, would you conclude that the
MBS students are taller than MA students? Assume the standard deviation of height
of post-graduate students to be 2.48 inches.
For MA: n1 = 60 , ഥ
X1= 63.60 For MBS n2 = 50 , ഥ
X2 = 69.51 common population s.d.
(𝜎) = 2.48
H0:𝛍₁ = 𝛍₂ (There is no difference between height of MA and MBS students)
H1:𝛍₁ < 𝛍₂ (MBS students are taller than MA students) Left Tailed Test
Test statistic: Z =
ഥ
X1− ഥ
X2
𝜎2 1
n1
+
1
n2
contd…
37
First Sample Second Sample
n1 = 500 n2 = 400
ഥ
X1= 20 ഥ
X2 = 15

Example: Two random samples of Nepalese people taken from rural and urban region
gave the following data of income.
Test whether the average
monthly income of urban
region people is significantly
more than rural region
people.
Solution:
With usual notations,
Test Statistic: Under H0 the test statistic is
𝐙 =
(ഥ
𝑿𝟏−ഥ
𝑿𝟐)
𝒔𝟏
𝟐
𝒏𝟏
+
𝒔𝟐
𝟐
𝒏𝟐
contd…..
38
Sample size Average
monthly
income
Standard
deviation
From Urban Region 100 1250 30
From Rural Region 150 800 50
Urban Region Rural Region
n1 = 100 n2 = 150
ഥ
X1= 1250 ഥ
X2 = 800
s1 = 30 s2 = 50
H0:𝛍₁ = 𝛍₂ (There is no significant difference
between the average income of the people of
urban and rural region)
H1:𝛍₁ > 𝛍₂ (The average monthly income of urban
region people is more than rural region people)
Right Tailed Test

Case III: Test of significance of Single Proportion
39
Procedure
H0:P = 𝑃0i. e. the sample has been drawn from a population with proportion H1:P ≠
𝑃0(two tailed test), H1:P > 𝑃0(right tailed test), H1:𝑃 < 𝑃0(left tailed test)
If sample proportion(p) > Population Proportion(P) : Right Tailed Test
If Sample Proportion(p) < Population proportion(P) : Left tailed Test
Step 2: Select test statistic:𝑍 =
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝑆.𝐸.
=
𝑝 − 𝑃
𝑃𝑄
𝑛
[Note: if we have sampling from a
finite population of size N, then 𝑆. 𝐸. 𝑝 =
𝑁−𝑛
𝑁−1
𝑃(1−𝑃)
𝑛
=
𝑁−𝑛
𝑁−1
𝑃𝑄
𝑛
]
Step 3: Fix the level of significance (𝛼)
If ǀZǀ(calculated) ≤ Zα(table or critical ) we say it is not significant (insignificant).So H₀ is
accepted.
If ǀZǀ(calculated) > Zα(table or critical) we say it is significant. So H₀ is rejected, then H₁ is
accepted.

Example 1 : An airline must allocate available seating space between first
class passengers and economy class passengers. The null hypothesis is that
20% of the passengers fly first class but management recognizes the
possibility that the percentage could be more or less. A random sample of
400 passengers includes 70 passengers holding first class tickets. Can the
null hypothesis be rejected at the 10% level of significance?
Solution:
With usual notations, P = 0.20 , n = 400 , Q = 1- P = 1 – 0.20 = 0.80
sample proportion(p) =
𝑥
𝑛
=
70
400
= 0.175
H0: P = 0.20 (20% of the passengers fly first class)
H1: P ≠ 0.20 ( more or less than 20% passengers fly first class)
Test Statistics: Under H0 ,the test statistic is Z =
𝑝 − 𝑃
𝑃𝑄
𝑛
=
0.175−0.20
0.20 ×0.80
400
= - 1.25
Zcal =1.25
Level of significance (𝜶) = 10%
Critical value: Ztab= 1.645
Decision : Since Zcal < Ztab so H0 is accepted
Conclusion: Hence we conclude that 20% of the passengers fly first class.
40

Example 2: A manufacturer claimed that at least 90% of the machine parts that is
supplied by that factory conformed to specifications. An examination of 200 such
parts revealed that 160 parts were not faulty. Determine if the manufactuer’s
claim is legitimate at 1% level of significance.
Solution: With usual notations P = 0.90 , Q = 1- P = 1- 0.90 = 0.10 n = 200 ,
x = 160, p =
𝑥
𝑛
=
160
200
= 0.80
H0: P ≥0.90 (At least 90% of the machine parts is conformed to specifications)
H1:P<0.90 [Left tailed Test](Less than 90% of the machine parts is conformed to
specifications)
Test Statistics: Zcal =
𝑝 − 𝑃
𝑃𝑄
𝑛
=
0.80−0.90
0.90 ×0.10
200
= -4.71 ǀZcalǀ = 4.71
Level of Significance(𝜶): 1%
Critical Value: Ztab = -2.33 ǀZtabǀ = 2.33
Decision: Since | zcal |> | ztab | So H0 is rejected and H1 is accepted.
Conclusion:
41

Example 3: A cellphone provider has the business objective of wanting to
determine the proportion of subscribers who would upgrade to a new
cellphone with improved features if it were made available at a substantially
reduced cost. Data are collected from a random sample of 500 subscribers.
The results indicate that 135 of the subscribers would upgrade to a new
cellphone at a reduced cost. At the 0.05 level of significance, is there evidence
that more than 20% of the customers would upgrade to a new cellphone at a
reduced cost?
Solution:
With usual notations, n = 500 , x = 135 , p =
𝑥
𝑛
=
135
500
= 0.27, P = 0.20 , Q = 0.80
H0: P =0.20 (20% of the customers would upgrade to a new cellphone at a
reduced cost
H1: P>0.20 [Right tailed Test](more than 20% of the customers would upgrade
to a new cellphone at a reduced cost)
𝑝 − 𝑃
𝑃𝑄
𝑛
=
0.27−0.20
0.20 ×0.80
500
= 3.91
Level of Significance(𝜶): 0.05
Critical value: Ztab = 1.645 Decision: Since Zcal > Ztab so H0 is rejected
42

43
Procedure
H0:𝑷𝟏 = 𝑷𝟐i. e. the sample has been drawn from a population with proportion P.
H1: 𝑷𝟏 ≠ 𝑷𝟐(two tailed test), H1 :𝑷𝟏 > 𝑷𝟐(right tailed test), H1 :𝑷𝟏 < 𝑷𝟐(left tailed test)
If First sample proportion (p1) > Second sample proportion (p2) : Right Tailed
If First sample proportion (p1) < Second sample proportion (p2): Left Tailed
Step 2: Select test statistic: 𝑍 =
𝑝1 − 𝑝2
𝑃1𝑄1
𝑛1
+
𝑃2𝑄2
𝑛2
𝑜𝑟 𝑍 =
(𝑝1− 𝑝2)
෠
𝑃 ෠
𝑄
1
𝑛1
+
1
𝑛2
Note: If P1 and P2 are not known, each of them is estimated by ෡
P =
𝑥1+𝑥2
𝑛1+𝑛2
=
n1p1+n2p2
n1+ n2
෡
Q = 1 - ෡
P
If ǀZǀ(calculated) ≤ Zα(table or critical ) we say it is not significant (insignificant). So H₀ is accepted.
If ǀZǀ(calculated) > Zα(table or critical) we say it is significant. So H₀ is rejected, then H₁ is accepted.
Case IV: Test of Significance for Difference of Two Proportions

Example: An online survey asked 1,000 adults “What do you buy from your
mobile device?”. Both sample sizes were 500 and that 195 out of 500 males and
305 out of 500 females reported they buy clothing from their mobile device. Is
there evidence of a difference between males and females in the proportion who
said they buy clothing from their mobile device at the 0.01 level of significance?
Solution: With usual notations, For Male: 𝑛1= 500 , 𝑝1=
195
500
=0.39
For Female: 𝑛2= 500 , 𝑝2=
305
500
=0.61
H0: 𝑃1 = 𝑃2 i. e. there is no significant difference between males and females in the
proportion of clothes buying from mobile device.
H1: 𝑃1 ≠ 𝑃2 (Two tailed test).
p1− p2
෡
P෡
Q
1
n1
+
1
n2
෡
P =
n1p1+n2p2
n1+n2
=
500×0.39+500×0.61
500+500
= 0.5
෡
Q = 1 − 0.5 = 0.5
Hence Zcal = -6.957
Level of significance(𝜶) = 0.01 (1%)
Critical value: Ztab= 2.58 Decision:
44

45
Example: At a certain date in a large city 400 out of a random sample of 500
men were found to be smokers. After the tax on tobacco had been heavily
increased another random sample of 600 men in the same city included 400
smokers. Was the observed decrease in the proportion of the smokers
significant?
Solution:
With usual notations,
• Null Hypothesis (H0): 𝑃1 = 𝑃2 i. e. there is no significant difference
between the proportion of the smokers before and after the increase in
the tax.
• Alternative Hypothesis (H1): 𝑃1 > 𝑃2 (right tailed test). There is a
significant decrease in the proportion of the smokers after the increase in
the tax.
Before Tax increased After Tax increased
𝑛1= 500 𝑛2= 600
𝑝1=
𝑥1
𝑛1
=
400
500
= 0.8 𝑝2=
𝑥2
𝑛2
=
400
600
= 0.667

Test statistic: Zcal =
p1− p2
෡
P෡
Q
1
n1
+
1
n2
෡
P =
n1p1 + n2p2
n1 + n2
=
500 x 0.8 + 600 x 0.667
500 + 600
=
400 + 400
1100
=
800
1100
=
8
11
& ෡
Q = 1 −
8
11
=
3
11
Zcal =
p1− p2
෡
P෡
Q
1
n1
+
1
n2
=
0.8 − 0.667
8
11
𝑥
3
11
1
500
+
1
600
=
0.133
0.027
= 4.926
Level of significance (α) = 0.05,
Critical value: then tabulated value of Z(α = 0.05) for right tailed Ztab = 1.645
Decision: Since Zcal = 4.926 > Ztab = 1.645 then, H0 is rejected i. e. Calculated
value of Z is greater than tabulated value of Z at 0.05 level of significance.
Then H0 is rejected.
Conclusion: Then there is a significant decrease in the proportion of the
smokers after the increase in the tax.
46

Example: A firm found with the help of a sample survey of a city (size of
sample 900) that
𝟑
𝟒
𝒕𝒉
of them consumes things produced by them. The firm
then advertised the goods in paper and on radio. After one year a sample of
size 1000 reveals that proportion of consumers of the goods produced by
the firm is
𝟒
𝟓
𝒕𝒉
. Is this rise significant to indicate that the advertisement was
effective?
Solution:
With usual notations
H0: 𝑃1 = 𝑃2 The advertisement was not effective
H1: 𝑃1 < 𝑃2 (Left tailed test). The advertisement was effective.
p1− p2
෡
P෡
Q
1
n1
+
1
n2
෡
P =
𝑛1𝑝1+𝑛2𝑝2
𝑛1+𝑛2
=
900×
3
4
+(1000×
4
5
)
900+1000
= 0.77
෡
Q = 1 - ෡
P = 1 – 0.77 = 0.23
Hence Zcal = -2.59 |𝑍𝑡𝑎𝑏| = 2.59
Level of Significance (𝜶) = 5%
Critical Value: Ztab = -1.645, |𝑍𝑡𝑎𝑏| = 1.645
Decision: H0 is rejected and H1 is accepted. 47
Before Ad After Ad
𝑛1= 900 𝑛2= 1000
𝑝1=
3
4
𝑝2=
4
5

Student’s ‘t’: Definition
• If x1, x2,….,xn is a random sample of size ‘n’ from a
normal population with mean μ and population
variance σ2, then Student’s ‘t’ statistic is defined as:
• 𝐭 =
ഥ
𝐗 − 𝛍
ൗ
𝐒
𝐧
=
(ഥ
𝐗 − 𝛍)
ൗ
𝐒𝟐
𝐧
=
𝑛(ഥ
X − μ)
S
~𝑡(𝑛−1) … … (𝟏)
• Where: ത
x =
σ X
n
, is the sample mean and S2 =
1
𝑛−1
σ(𝑥 − 𝑥)2 = unbiased estimate of the population
variance σ2
.
• Note: ‘t’ defined in the (1) follows Student’s ‘t’-
distribution with (n – 1) degrees of freedom
• Degrees of freedom is defined as number of
independent observations.

Assumptions
1. The parent population from which the sample is
drawn is normal.
2. Sample size is generally less than or equals to 30.
(which is considered as general traditional
convention)
3. All the observations in the samples are independent.
4. Samples are drawn randomly from normal
populations.
5. Population standard deviation (σ) is unknown.
6. The sample values are correctly measured and
recorded.
7. The hypothetical value of population mean (𝜇0) of 𝜇
is correct values of the population mean.

‘t’ distribution
t = 0
-1
-3 -2
f(t)
Normal curve
t- distribution for n = 25
+2 +3
+1
Fig. 1
t- distribution for n = 5

Applications of t-test
(i) Test of significance of a single mean
(ii) Test of significance of the difference
between two independent means
(iii)Paired t test for difference of two means
(iv)Test of significance of an observed sample
correlation coefficient and regression
coefficients

Case I: Test of significance of a single mean
Procedure
H0: 𝜇 = 𝜇0
H1: 𝜇 ≠ 𝜇0 (Two tailed test), : 𝜇 > 𝜇0 (Right tailed test), : 𝜇 < 𝜇0 (Left tailed test)
If sample mean ( ത
𝑋) > Population mean (𝜇) : Right Tailed Test
If sample mean ( ത
𝑋) < Population mean (𝜇) : Left Tailed Test
Step 2: Select test statistic:
t =
ഥ
X − μ
ൗ
S
n
=
n (ഥ
X − μ)
S
[𝑓𝑜𝑟 𝑢𝑛𝑏𝑖𝑎𝑠𝑒𝑑( 𝐴𝑐𝑡𝑢𝑎𝑙 𝑑𝑎𝑡𝑎 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛)]
t=
ഥ
X − μ
ൗ
s
n−1
=
(n−1) (ഥ
X − μ)
s
[𝑓𝑜𝑟 𝑏𝑖𝑎𝑠𝑒𝑑 (𝑠𝑎𝑚𝑝𝑙𝑒 𝑠. 𝑑. 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛)]
Step 4: degrees of freedom(d.f.) = (n – 1)
Step 5: Write the critical value (t ) for 𝛼 % level of significance from the t- table.
Step 6: Make decision: Critical value approach:
If ǀtǀ(calculated) ≤ tα(table or critical ) : it is not significant (insignificant). So H₀ is accepted.
If ǀtǀ(calculated) > tα(table or critical) : it is significant. So H₀ is rejected, then H₁ is accepted.

Computation of S2 for Numerical Problems:
• 𝑆2 =
1
(𝑛−1)
σ(𝑥 − 𝑥 )2 where σ(𝑥 − 𝑥 )2= sum of the squares of
deviation taken from mean
• S2 =
1
(n− 1)
σ 𝑥2 −
σ 𝑥 2
𝑛
• 𝑆2
=
1
(𝑛−1)
σ 𝑑2
−
σ 𝑑 2
𝑛
where 𝑥 = 𝐴 +
σ 𝑑
𝑛
How can we see critical value:
d.f.
Level of significance for one tailed test
0.10 0.05 0.025 0.01 0.005 0.0005
Level of significance for two tailed test
0.20 0.10 0.05 0.02 0.01 0.001
1 3.078 6.314 12.706 31.821 63.657 636.619
24 2.064

Confidence intervals using ‘t’ or Confidence interval for
Small Samples
• Confidence intervals = estimator ± (reliability
constant) x (standard error of the estimate).
• Reliability coefficient is obtained from the table of t-
distribution.
• C. I. (μ) = ത
𝐱 ± 𝐭𝜶,𝒏−𝟏.
𝐒
𝐧
. [ For unbiased estimator,
When actual data is given]
• C. I. (μ) = ത
𝐱 ± 𝐭𝜶,𝒏−𝟏.
𝐬
𝐧−𝟏
. [ For biased estimator
when sample s.d. or variance is given]

Example: The average cost of a hotel room in New York is said to be $168 per
night. To determine if this is true, a random sample of 25 hotels is taken and
resulted in average of $172.50 with standard deviation of $15.40. Test the
appropriate hypothesis at 0.05 level of significance.
Solution: With usual notations: 𝜇 =$ 168 , n = 25 , ത
𝑋= $ 172.50 , s = $ 15.40
Formulation of Hypothesis:
H0: μ = $ 168 . The average cost of a hotel room in New York is $168 per night.
H1: μ  168 (Two tailed test). The average cost of a hotel room in New York is not
equals to $168 per night.
Test statistic: Under H0 ,the test statistics is 𝑡 =
ത
𝑋 − 𝜇
ൗ
𝑆
𝑛−1
=
172.50 − 168
ൗ
15.40
25−1
= 1.43 tcal =1.43
Level of significance(𝜶) = 0.05
Degree of freedom(d.f.) = n-1 = 25-1 = 24
Critical value: The tabulated value of t for two tailed test at 0.05 level of significance
with 24 d.f. is 2.064 i.e. ttab = 2.064
Decision: Since tcal < ttab so H0 is accepted.
Conclusion: Hence we conclude that the average cost of a hotel room in New York is
$168 per night

Contd..
95% confidence interval for population mean:
n= 25, ത
𝑋 = $ 172.50 , s = $ 15.40
1- 𝛼 = 0.95 , 𝛼 = 0.05 d.f. = n-1 = 25-1 = 24
𝑡𝛼,𝑛−1= 𝑡0.05,24 = 2.064
95% CI for Population mean( 𝜇) is
= ത
𝑋 ± 𝑡𝛼,𝑛−1 .
𝑠
𝑛−1
= 172.50 ± 2.064 .
15.40
25−1
= 172.50 ± 6.48
= (172.50 – 6.48 , 172.50 + 6.48)
= (166.01 , 178.98)

Example : The following are weight (gm.) values for
sample of 21 chocolates. Can we conclude from these
data that the mean of the population from which the
sample was drawn is greater than 14.0? Use α = 0.05.
14.5 12.9 14.0 16.1 12.0 17.5 14.1
12.9 17.9 12.0 16.4 24.2 12.2 14.4
17.0 10.0 18.5 20.8 16.2 14.9 19.6

Solution:
Calculation of S2:
X X2
14.5 210.25
10.9 166.41
14.0 196.00
16.1 259.21
12.0 144.00
17.5 306.25
14.1 198.81
12.9 166.41
17.9 320.41
12.0 144.00
16.4 268.96
24.2 585.64
12.2 148.84
14.4 207.36
17.0 289.00
10.0 100.00
18.5 342.25
20.8 432.64
16.2 262.44
14.9 222.01
19.6 384.16
ΣX = 328.1 ΣX2 = 5355.05

• Now 𝑆2 =
1
(𝑛−1)
σ 𝑋2 −
(σ 𝑋)
2
𝑛
=
1
21−1
5355.05 −
328.1 2
21
= 11.44
• ∴ 𝑠 = 11.44 = 3.38
• Sample size (n) = 21, sample mean (𝑥) =
σ 𝑋
𝑛
=328.1/21 = 15.62, sample
standard deviation (s) = 3.38,, population mean (μ) = 14,
• Hypothesis setting:
• Null Hypothesis: H0:μ = 14 gm i. e.
• Alternative hypothesis: H1: μ > 14 gm(Right tailed test) i. e.
• Test statistic: 𝑡 =
ഥ
X − μ
ൗ
S
n
=
15.62−14
ൗ
3.38
21
=
1.62
Τ
3.38
4.58
=
1.62
0.73
= 2.21
• Level of significance(𝜶 )= 0.05
• Degree of freedom(d.f.) = n -1 = 21 – 1 = 20
• Critical value: 𝑡𝛼=0.05 𝑓𝑜𝑟 𝑜𝑛𝑒 𝑡𝑎𝑖𝑙𝑒𝑑 ,20 𝑑𝑓 = 1.725
• Decision: since 𝑡𝑐𝑎𝑙 = 2.21 > 𝑡𝛼 = 1.725 then the test is significant so
we reject hull hypothesis and then accept alternative hypothesis (H1).
• Conclusion:

Case II: Test of significance of difference between two independent
means (Pooled t- test) independent t -test
Procedure
H0: 𝝁𝟏 = 𝝁𝟐 i. e. there is no significant difference between the two population means
H1: 𝝁𝟏 ≠ 𝝁𝟐 (two tailed test), H1: 𝝁𝟏 > 𝝁𝟐 (right tailed test), H1:𝝁𝟏 < 𝝁𝟐(left tailed test)
X1) > Second sample mean(ഥ
X2) : Right Tailed Test
X1) < Second sample mean(ഥ
X2) : Left Tailed Test
Step 2: Test Statistics , t =
(𝑋1 − 𝑋2 )
𝑆2 1
𝑛1
+
1
𝑛2
=
(𝑋1 − 𝑋2 )
𝑆
1
𝑛1
+
1
𝑛2
~𝑡(𝑛1+𝑛2−2)
Step 4: Degrees of freedom(d.f.) = (𝑛1 + 𝑛2 − 2)
Step 5: Write the critical value (t ) for 𝛼 % level of significance at (n1 + n2 – 2) d. f. from the t- table.
If ǀtǀ(calculated) ≤ tα(table or critical ) we say it is not significant (insignificant). So H₀ is accepted.
If ǀtǀ(calculated) > tα(table or critical) we say it is significant. So H₀ is rejected, then H₁ is
accepted.
Step 7: Conclusion:

Calculation of S2
• Actual mean method: 𝑆2
=
1
(𝑛1+ 𝑛2−2)
[σ(𝑋1 − 𝑋1)2
+ σ(𝑋2 − 𝑋2)2
]
• Direct method: 𝑆2 =
1
(𝑛1+ 𝑛2−2)
ቂ
ቃ
σ 𝑋1
2
−
σ 𝑋1
2
𝑛1
+ σ 𝑋2
2
−
σ 𝑋2
2
𝑛2
• Short cut method: 𝑆2
=
1
(𝑛1+ 𝑛2−2)
ቂ
ቃ
σ 𝑑1
2
−
σ 𝑑1
2
𝑛1
+
σ 𝑑2
2
−
σ 𝑑2
2
𝑛2
• Where d1 = X – A, d2 = X – B, A = assumed mean of series X,
B = assumed mean of series Y.
• 𝑺𝟐
=
𝒏𝟏.𝒔𝟏
𝟐+ 𝒏𝟐.𝒔𝟐
𝟐
(𝒏𝟏+ 𝒏𝟐−𝟐)
[ when the sample variance (or sample
standard deviation) i. e. biased estimates are given]

Example: Two different types of drugs D1 and D2 were
administrated on certain patients for increasing weight at
interval of one week time period. From the following
observation, can you conclude that Test at 5% level of
significance.
i. Whether two drugs differ significantly?
ii. The second drug is more effective in increasing
weight.
D1 8 12 13 9 3 8 10 9
D2 10 8 12 15 6 11 12 12

Solution
Calculation of S2
X1 X2 X1
2 X2
2
8 10 64 100
12 8 144 64
13 12 169 144
9 15 81 225
3 6 9 36
8 11 64 121
10 12 100 144
9 12 81 144
ΣX1 = 72 ΣX2 = 86 ΣX1
2 = 712 ΣX2
2 = 978
• Here n1 = n2 = 8 then 𝑥1= 72/8 = 9, 𝑥2= 86/8= 10.75
• 𝑆2 =
1
𝑛1+ 𝑛2−2
σ 𝑋1
2
−
(σ 𝑋1)2
𝑛1
+ σ 𝑋2
2
−
(σ 𝑋2)2
𝑛2
• =
1
8+ 8−2
712 −
(712)2
8
+ 978 −
(86)2
8
= 8.4

Contd…
(i) Setting of Hypothesis:
H0: μ1 = μ2 (i. e. both drugs are equally effective)
H1: μ1 ≠ μ2 (i. e. both drugs are not equally effective)
Test statistic: 𝑡 =
ത
𝑋1− ത
𝑋2
𝑆2 1
𝑛1
+
1
𝑛2
=
9 − 10.75
8.4
1
8
+
1
8
= −1.208
then 𝑡𝑐𝑎𝑙 = −1.208 = 1.208
Degree of freedom(d.f.) = (𝑛1 + 𝑛2 − 2) = 8+8- 2 = 14
Critical value:
Tabulated value of t for two tailed test at 0.05 level of
significance with 14 d.f. is 2.145 i.e. 𝑡𝑡𝑎𝑏= 2.145
Decision: 𝑡𝑐𝑎𝑙 = 1.208 < 𝑡𝑡𝑎𝑏 = 2.145 so we accept H0 .
Conclusion: we can conclude that both drugs are equally
effective.

Contd…
(ii) Setting of Hypothesis:
H0: μ1 = μ2 (i. e. both drugs are equally effective)
H1: μ1 < μ2 (i. e. second drug is effective than first drug)
Test statistic: 𝑡 =
ത
𝑋1− ത
𝑋2
𝑆𝑝
2 1
𝑛1
+
1
𝑛2
=
9 − 10.75
8.4
1
8
+
1
8
= −1.208
then 𝑡𝑐𝑎𝑙 = −1.208 = 1.208
Critical value: Tabulated value of t for two tailed test at 0.05
level of significance with 14 d.f. is 2.145 i.e. 𝑡𝑡𝑎𝑏= 1.761
Decision: 𝑡𝑐𝑎𝑙 = 1.208 < 𝑡𝑡𝑎𝑏 =1.761 so we accept H0 .
Conclusion: we can conclude that both drugs are equally
effective

Example: Two salesmen A and B are working in a certain district. From a sample
survey conducted by the head office, the following results were obtained. State
whether there is any significant difference in the average sales between the two
salesmen.
Solution: With the usual notations, we have,
For A For B
𝑛1 = 20 𝑛2 = 18
𝑥1 = 170 𝑥2= 205
𝑠1 = 20 𝑠2 = 25
H0: μ1 = μ2 , i.e. there is no significant difference between the average sales of
two salesmen
H1: μ1 ≠ μ2 i.e. there is a significant difference between the average sales of two
salesmen.
Salesmen A Salesmen B
No of sales: 20 18
Average sales: 170 205
Standard deviation 20 25

Test statistic: under H0the test statistic is 𝑡 =
ത
𝑋1− ത
𝑋2
𝑆2 1
𝑛1
+
1
𝑛2
𝑆2 =
𝑛1.𝑠1
2+ 𝑛2.𝑠2
2
(𝑛1+ 𝑛2−2)
=
𝑛1.𝑠1
2+ 𝑛2.𝑠2
2
(𝑛1+ 𝑛2−2)
= 534.75
t=
170−205
534.72
1
20
+
1
18
= -4.56 Hence tcal = 4.56
Level of significance( ) = 0.05
Degree of freedom(d.f.) = n1 + n2 – 2 = 20 + 18 – 2 = 36
Critical Value: Since d.f. =36 > 30, it follows normal distribution. Hence the
tabulated value of t at 5% level of significance for two tailed test is 1.96.
Decision: Since the calculated value of t is greater than the tabulated value of
t, it is significant, hence the null hypothesis is rejected i.e. alternative
hypothesis is accepted, which means that there is a significant difference
between the average sales of two salesmen.

When Z-test and When t-test is
suitable?
• In case of test of significance of single mean
and double(two independent means)
• You can use t test if population standard
deviation is not given(unknown) in case of
sample size greater than 30
• Use tabulated value from table (also in t-table)
• See next example

• Two independent samples z-test:
• Large sample size (typically > 30)
• Known population standard deviation
• Normally distributed population
• Two independent samples (unpaired)
t-test:
• Small sample size (typically < 30)
• Unknown population standard deviation
• Population may not be normally distributed

Question: Read the following information perform the following:
(i) Test whether two means are significantly difference at 0.05
level of significance using independent t-test.
(ii) Compute 95% confidence interval estimation for the
difference of means
(iii) Show the linkage between testing of hypothesis and interval
estimation in this problem
Group I Group II
Sample means 10 15
Sample standard deviation 3 5
Sample size 49 64

Solution:
• (i) as same process solving above
• (ii) C. I.
• (𝑿𝟏 − 𝑿𝟐) ± t𝛂, 𝒏𝟏 + 𝒏𝟐 − 𝟐 ∗ 𝒔𝟐(
𝟏
𝒏𝟏
+
𝟏
𝒏𝟐
)
• 𝑆2
=
𝑛1.𝑠1
2+ 𝑛2.𝑠2
2
(𝑛1+ 𝑛2−2)
see ttab two tailed test
• (iii) if hypothetical value of mean difference
zero (0) lies between confidence interval we
accept null hypothesis otherwise reject

Case III : Paired t-test for difference of two means
Procedure
H0: 𝝁𝒙 = 𝝁𝒚 or 𝝁𝒅 = 𝟎 i. e. there is no significant difference between the two population means
H1: 𝝁𝒙 ≠ 𝝁𝒚 or 𝝁𝒅 ≠ 𝟎 (two tailed test), : 𝝁𝒙 > 𝝁𝒚 (right tailed test), :𝝁𝒙 < 𝝁𝒚(left tailed test)
Step 2: Select test statistic:𝒕 =
ഥ
𝒅
ൗ
𝒔𝒅
𝟐
𝒏
~𝒕(𝒏−𝟏) Where d = X – Y
X = value before treatment Y = value after treatment
Treatment means performing advertisement for promoting sales of goods or performing, training for
increasing memory capacity of person etc…..
𝑑 =
σ 𝑑
𝑛
𝑎𝑛𝑑 𝑠𝑑
2
=
σ(𝑑− ത
𝑑)2
(𝑛−1)
=
1
(𝑛−1)
σ 𝑑2
−
σ 𝑑 2
𝑛
Step 4: degrees of freedom = (n – 1)
Step 5: Write the critical value (t ) for 𝛼 % level of significance and (n – 1)d. f. from the t- table.
If ǀtǀ(calculated) ≤ tα(table or critical ) we say it is not significant (insignificant). So H₀ is
accepted.
accepted.

Example: A company has reorganized its sale department. The following data
show the weekly sales figures (in Rs. lakhs) before and after the reorganization.
Can we say that sales have significantly changed due to the reorganization?
In this problem, the sales before reorganization (x) and after reorganization (y) are
not independent, but are paired together. Hence we shall apply paired t-test.
H0: 𝜇𝑥 = 𝜇𝑦 i.e. there is no significant change in sale after the reorganization.
H1: 𝜇𝑥 ≠ 𝜇𝑦 (Two tailed test), , i.e. there is a significant change in sale after the
reorganization.
Test Statistics: Under H0 , the test statistics is
𝑡 =
ത
𝑑
ൗ
𝑆2
𝑛
Week 1 2 3 4 5 6 7 8 9 10
Sales before: 12 15 13 11 17 15 10 11 18 19
Sales after: 16 17 14 13 15 14 12 11 17 22

𝑑 =
σ 𝑑
𝑛
=
−10
10
= -1
s2 =
1
(𝑛−1)
σ 𝑑2 −
σ 𝑑 2
𝑛
=
1
(10−1)
44 −
−10 2
10
= 3.778
Wee
k
Sales (before
reorganization)
(x)
Sales (after
reorganization)
(y)
d= X - Y d2
1
2
3
4
5
6
7
8
9
10
12
15
13
11
17
15
10
11
18
19
16
17
14
13
15
14
12
11
17
22
-4
-2
-1
-2
2
1
-2
0
1
- 3
16
4
1
4
4
1
4
0
1
9
Total σ 𝑑 = -10 σ 𝑑2 = 44

Hence , 𝑡 =
ത
𝑑
ൗ
𝑆2
𝑛
=
−1
Τ
3.778
10
= -1.629 , ǀtǀ(calculated) = 1.629
Level of significance = 0.05
Degree of freedom (d.f.) = n-1 = 10-1 = 9
The critical (tabulated) value of t for 9 d.f. at 5% level of
significance for two tailed test 2.262.
Decision: Since calculated value of t is less than critical value of t,
it is not significant at 5% level of significance. Hence, the data do
not provide any evidence against the null hypothesis which may
be accepted. It may therefore be concluded that reorganization
has no effect on sales.

Calculations:
S. N. Change in score (d) (d)2
1 8 64
2 10 100
3 -2 4
4 0 0
5 -5 25
6 -1 1
7 9 81
8 12 144
9 6 36
10 5 25
Total Σd = 42 Σd2 = 480
Example 2: A drug is given to 10 patients, on the increment in their blood pressure
were recorded to be 8, 10, -2, 0, - 5, - 1, 9, 12, 6, 5. Test at 5% level of significance
whether the drug is effective for either increasing or decreasing the blood pressure.

Contd…
• Number of observation (n) = 10,
• Now 𝑑 =
σ 𝑑
𝑛
=
42
10
= 4.2 and
• 𝑆2
=
1
(𝑛−1)
σ 𝑑2
−
σ 𝑑 2
𝑛
=
1
9
480 −
(42)2
10
= 33.73
• Null Hypothesis: H0: 𝜇𝑥 = 𝜇𝑦 i. e. there was no increase or decrease in blood
pressure before and after the treatment. The treatment was not effective
• Alternative Hypothesis: H1: 𝜇𝑥 ≠ 𝜇𝑦 i. e. there was increase or decrease in blood
pressure before and after the treatment. The treatment was effective. (two tailed
test)
• Test statistic: 𝑡 =
ത
𝑑
ൗ
𝑆2
𝑛
=
4.2
33.735
10
= 2.29
• level of significance(𝜶)= 0.05
• Degree of freedom(d.f.) = n - 1 = 10 – 1 = 9
• Critical value of t for 9 d. f. and two tailed test 0.05 level of significance, 𝑡𝛼=0.05 =
2.262
• Decision: 𝑡𝑐𝑎𝑙 = 2.29 > 𝑡𝛼=0.05 = 2.262 then it is significant so we reject H0 and
accept H1.
• Conclusion: The treatment was effective.

Case IV: Test of significance of an observed sample correlation coefficient
Let r be the observed sample correlation coefficient in a sample of n pairs of observations
from bivariate normal population. In order to test whether any correlation coefficient
between variables in population, t-test is applied.
Procedure:
H0: 𝝆 = 0 i.e. Population correlation coefficient is zero or Variables are uncorrelated in the
populations
H1: 𝝆 ≠ 𝟎 (two tailed test), variables are correlated in the population
H1 : 𝝆 > 𝟎 (right tailed test) , H1 :𝝆 < 𝟎(left tailed test)
Step 2: Select test statistic:𝑡 =
𝒓
𝟏−𝒓𝟐
. 𝒏 − 𝟐
Step 4: degrees of freedom = (n – 2)
Step 5: critical value (t ) for 𝛼 % level of significance and (n – 2)d. f. from the t- table.
If ǀtǀ(calculated) ≤ tα(table or critical ) we say it is not significant (insignificant). So H₀ is
accepted.
accepted.
step 7: Conclusion:

Example: A random sample of 27 pairs of observations from a normal population
gives a correlation coefficient of 0.42. Is it likely that the variables in the
population are uncorrelated? Also compute 95% confidence limit for the
population correlation coefficient.
Solution: with usual notations, n = 27, r = 0.42
H0: 𝜌 = 0 i.e. Variables in the populations are not correlated .
H1: 𝜌 ≠ 0 (Two tailed test) i.e. Variables in the populations are correlated .
Test Statistics: 𝑡 =
𝒓
𝟏−𝒓𝟐
. 𝒏 − 𝟐 =
𝟎.𝟒𝟐
𝟏−𝟎.𝟒𝟐𝟐
. 𝟐𝟕 − 𝟐 = 2.31
Degree of freedom(d.f.) = n-2 = 27-2 = 25
Critical value: ttab = 2.060 Decision:
For 95% confidence limits:
1- 𝛼= 95%, 𝛼= 5% d.f. = 27-2 =25 𝑡𝛼,𝑛−2 = 2.060
Hence 95% Confidence limits for population correlation coefficient are
= r ± 𝑡𝛼,𝑛−2 .
1−𝑟2
𝑛
= 0.42 ± 2.060 .
1−0.422
27
= 0.42 ±

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