The document discusses hypothesis testing and provides examples. It can be summarized as follows: 1. Hypothesis testing involves making a tentative null hypothesis (H0) about a population parameter and defining an alternative hypothesis (HA). Sample data is collected and used to either reject or fail to reject the null hypothesis. 2. There are two types of possible errors in hypothesis testing - Type I errors where the null hypothesis is incorrectly rejected, and Type II errors where it is incorrectly failed to be rejected. 3. Examples are provided to demonstrate hypothesis testing for a single mean where the population standard deviation is both known and unknown. The examples show how to formulate hypotheses, calculate test statistics, determine critical values, and

1. Lecture 6,7 & 8
Prepared By:
Mohammad Kamrul Arefin
Lecturer, School of Business, North South University

2. Hypothesis Testing
In hypothesis testing we begin by making a tentative assumption
about a population parameter (mean, proportions, or variance). This
i i i ll d N ll H h i d i d d b
tentative assumption is called Null Hypothesis and is denoted by
H0. We then define the alternative hypothesis HA, which is the
opposite of what is stated in the null hypothesis
opposite of what is stated in the null hypothesis.
Once we have specified null and alternative hypothesis and
collected sample data, a decision concerning the null hypothesis
collected sample data, a decision concerning the null hypothesis
must be made.
2
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

3. We can either reject the null hypothesis and accept the
alternative or fail to reject the null hypothesis.
Many statisticians prefer not to say “accept the null hypothesis”;
instead they say “fail to reject the null hypothesis.”
When we fail to reject the null hypothesis then either the null
hypothesis is true or our test procedure was not strong enough to
j t d h itt d
reject and we have committed an error.
Depending on the situation, hypothesis tests about a population
parameter may take one of three forms: two use inequalities in
parameter may take one of three forms: two use inequalities in
the null hypothesis; the third uses an equality in the null
hypothesis. Null hypothesis should always accompany an equal
yp yp y p y q
sign
0 0 0 0 0 0
: : :
H H H
H H H
        
0 0 0 0
: : :
A A
H H H
        
3
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

4. Type I and Type II errors
The null and alternative hypotheses are competing statements
about the population. Either the null hypothesis H0 is true or the
alternative hypothesis HA is true, but not both.
Ideally the hypothesis testing procedure should lead to the
acceptance of H0 when H0 is true and the rejection of H0 when HA
is true. Unfortunately, the correct conclusions are not always
possible Because hypothesis tests are based on sample
possible. Because hypothesis tests are based on sample
information, we must allow for the possibility of errors.
P l i di i
Population condition
Decision H0 is True H0 is False
Reject H0 Type I error Correct Decision
Reject H0 Type I error
Probability = α
(α = level of significance)
Correct Decision
Probability = 1‐β
Accept H or fail to reject H Correct Decision Type II error
Accept H0 or fail to reject H0 Correct Decision
Probability = 1‐α
Type II error
Probability = β
4
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

5. Hypothesis test of the single population mean(σ
known) 0 0
( ) =
x
x x
Z statistic
 



n
0 0
:
H    0 0
:
H    0 0
:
H   
0
:
A
H    0
:
A
H    0
:
A
H   
l l l
0
Reject H
If ,
stat
z z
 0
RejectH
If ,
stat
z z

P‐value
=P[z<(z‐stat value)]
P‐value
=P[z>(z‐stat value)]
P‐value
=P[z>(z‐stat value)]
+P[z<(‐z‐stat value)]
If P‐value ≤ α, Reject H0
5
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

6. Example:
Hilltop Coffee states that their coffee can contains 3 pounds of
coffee. The Federal Trade Commission (FTC) collected a sample
of 36 cans to investigate the claim where previous FTC study
show that the population standard deviation can be assumed with
a al e of 0 18 The sample of cans pro ides a sample mean of
a value of σ=0.18. The sample of cans provides a sample mean of
2.92 pounds. Formulate hypothesis and perform a test.
: 3
H     h 2 92 d 0 18 36
0 0
0
: 3
: 3
A
H
H
   
   
0 0 2.92 3
( ) = 2 67
x x
Z statistic
  
  
here, 2.92pounds, =0.18, n=36
x  
( ) = 2.67
0.18
36
x
Z statistic
n
   


α= level of significance = 5%=0.05
Decision Rule: If ,
here, 2.67>1.645
stat
z z

1.645 1.645
z z
 
   
Decision: Reject Null
6
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

7. If α= level of significance = 1%=0 01
If α= level of significance = 1%=0.01
2.33 2.33
z z
 
   
Decision Rule: If ,
here 2 67>2 33
stat
z z

P-value approach: Decision: Reject Null
here, 2.67>2.33
P-value = P(z<z value)=P(z< 2.67)=0.0038
stat 
Decision rule: If P-value ≤ α Reject Null Hypothesis H0
Decision rule: If P value ≤ α, Reject Null Hypothesis, H0
When α= level of significance = 5%=0.05, P-value<0.05;
th f j t th ll h th i t 5% i ifi l l
therefore reject the null hypothesis at 5% significance level.
When α= level of significance = 1%=0.01, P-value<0.01;
therefore reject the null hypothesis at 1% significance level
therefore reject the null hypothesis at 1% significance level.
Hence we can see that, null hypothesis is rejected at both 5%
and 1% significance level and can conclude that Hilltop Coffee is
and 1% significance level and can conclude that Hilltop Coffee is
under filling cans.
7
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

8. Example:
0 0
0
: 1030
: 1030
A
H
H
   
   
here, 1061.6, =90, n=40
x  
0
A  
0 0 1061.6 1030
( ) = 2.22
90
x x
Z statistic
  
  
 90
40
x
n


l l f i ifi 5% 0 05
α= level of significance = 5%=0.05
1.645 1.645
z z
 
   
 
Decision Rule: If ,
stat
z z

D i i R j N ll
here, 2.22>1.645
Decision: Reject Null
8
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

9. If α= level of significance = 1%=0 01
If α= level of significance = 1%=0.01
2.33 2.33
z z
 
   
Decision Rule: If ,
here 2 22<2 33
stat
z z

P-value approach: Decision: Fail to Reject Null
here, 2.22<2.33
P-value = P(z>z value)=P(z>2.22)
1 P( 2 22) 1 0 9868 0 0132 1 32%
stat
=1-P(z<2.22)=1-0.9868=0.0132=1.32%
Decision rule: If P-value ≤ α, Reject Null Hypothesis, H0
j yp 0
When α= level of significance = 5%=0.05, P-value<0.05;
therefore reject the null hypothesis at 5% significance level.
therefore reject the null hypothesis at 5% significance level.
When α= level of significance = 1%=0.01, P-value>1%;
therefore fail to reject the null hypothesis at 1% significance
j yp g
level.
9
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

10. Example:
0 0
0
: 450
: 450
A
H
H
   
   
here, 458, =20.5, n=35
x  
0
A  
0 0 458 450
( ) = 2.31
20 5
x x
Z statistic
  
  
 20.5
35
x
n


l l f i ifi 5% 0 05
α= level of significance = 5%=0.05
1.96 1.96
z z z
  
    
  
Decision Rule: If ,
stat
z z

D i i R j N ll
here, 2.31>1.96
Decision: Reject Null
10
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

11. If α= level of significance = 1%=0 01
If α= level of significance = 1%=0.01
2.58 2.58
z z z
  
    
Decision Rule: If ,
here 2 31<2 58
stat
z z

P-value approach: Decision: Fail to Reject Null
here, 2.31<2.58
P-value P(z 2.31) P(z 2.31)=
0 0104 0 0104 0 0208 2 08%
    
=0.0104+0.0104=0.0208=2.08%
Decision rule: If P-value ≤ α, Reject Null Hypothesis, H0
j yp 0
When α= level of significance = 5%=0.05, P-value<5%;
therefore reject the null hypothesis at 5% significance level.
therefore reject the null hypothesis at 5% significance level.
When α= level of significance = 1%=0.01, P-value>1%;
therefore fail to reject the null hypothesis at 1% significance
j yp g
level.
11
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

12. Hypothesis test of the single population mean(σ
not known)
0 0
( ) =
statistic
x x
t
s
 

x
s
n

:
H  
0 0
0
:
:
A
H
H
  
  
0 0
0
:
:
A
H
H
  
  
0 0
0
:
:
A
H
H
  
  
0
n-1)df RejectH
If ,
stat
t t

(n-1)df
Reject H
If stat
t t

0
(n-1)df
Reject H
If stat
t t

P‐value
=P[t<(t‐stat value)]
P‐value
=P[t>(t‐stat value)]
P‐value
=P[t>(t‐stat value)]
[ ( l )]
0
Reject H
0
j
+P[t<(‐t‐stat value)]
If P‐value ≤ α Reject H
If P value ≤ α, Reject H0
12
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

13. A b i t l i t ff d l f 60
Example:
A business travel magazine staff surveyed a sample of 60
business travelers at each airport to obtain the ratings data. The
sample for London’s Heathrow Airport provided a sample mean
sample for London s Heathrow Airport provided a sample mean
rating of 7.25 with a sample std. deviation of s =1.052. Airports
with a population mean rating greater than 7 will be designated as
superior service airports. Do the data indicate that Heathrow
should be designated as a superior service airport?
h 7 25 1 052 60
0 0
0
: 7
: 7
A
H
H
   
   
0 0 7 25 7
x x
  
here, 7.25, s=1.052, n=60
x 
0 0 7.25 7
( ) = 1.84
1.052
60
x
x x
t statistic
s
n
 
  

60
n
α= level of significance = 5%=0.05
Decision Rule: If ,
here, 1.84>1.671
stat
t t

n-1)df df 1.671
t t
 
  Decision: Reject Null
13
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

14. If α= level of significance = 1%=0 01
If α= level of significance = 1%=0.01
n-1)df df 2.391
t t
 
 
Decision Rule: If ,
here 1 84<2 391
stat
t t

P-value approach:
Decision: Fail to Reject Null
here, 1.84<2.391
P-value = P(t>1.84)=0.025<p<0.05=2.5%<p<5%
( ) p p
Decision rule: If P-value ≤ α, Reject Null Hypothesis, H0
j yp 0
When α= level of significance = 5%=0.05, P-value<0.05;
therefore reject the null hypothesis at 5% significance level.
therefore reject the null hypothesis at 5% significance level.
When α= level of significance = 1%=0.01, P-value>1%;
therefore fail to reject the null hypothesis at 1% significance
j yp g
level.
14
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

15. Example:
0 0
: 410
: 410
H
H
   
   
here, 511.33, s=183.75, n=18
x 
0
: 410
A
H    
0 0 511.33 410
( ) = 2.34
183 75
x x
t statistic
s
  
  
183.75
18
x
s
n

α= level of significance = 5%=0.05
1)df df df 2.11
t t t
        
  
n-1)df df df 2.11
t t t
  
Decision Rule: If ,
stat
t t
 ,
here, 2.34>2.11
stat 
Decision: Reject Null
15
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

16. Hypothesis test of the population proportion
0 0
ˆ 0 0
ˆ ˆ
( ) =
(1 )
p p p p
Z statistic
p p
 

 
0 0
(1 )
p p p
n

0 0
0
:
:
A
H p p
H p p


0 0
0
:
:
A
H p p
H p p


0 0
0
:
:
A
H p p
H p p


0
Reject H
If ,
stat
z z
 0
RejectH
If ,
stat
z z

P‐value
=P[z<(z‐stat value)]
P‐value
=P[z>(z‐stat value)]
P‐value
=P[z>(z‐stat value)]
P[ ( t t l )]
+P[z<(‐z‐stat value)]
If P‐value ≤ α, Reject H
If P value ≤ α, Reject H0
16
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

17. 20% f th l t Pi k lf I
Example:
20% of the players at Pine creek golf course were women. In an
effort to increase the proportion of women players, Pine Creek
implemented a special promotion designed to attract women
implemented a special promotion designed to attract women
golfers. One month after the promotion was implemented, the
course manager requested a statistical study to determine whether
the proportion of women players at Pine Creek had increased.
Suppose a random sample of 400 players was selected, and that
100 f th l Th ti f lf
100 of the players were women. The proportion of women golfers
in the sample is 100
ˆ
here, p =0.25, n=400
400

0 20
H 400
0 0
0
: 0.20
: 0.20
A
H p p
H p p
 
 
0
ˆ
ˆ 0.25 0.20
( ) = 2.50
0.20(1 0.20)
p
p p
Z statistic
 
 
 
400
17
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

18. l l f i ifi 5% 0 05 Decision Rule: If z z

α= level of significance = 5%=0.05
1.645 1.645
z z
 
   
Decision Rule: If ,
here, 2.50>1.645
stat
z z

P-value approach:
P-value = P(z>z value)=P(z>2 50)=1-P(z<2 50)
Decision: Reject Null
P value P(z>z value) P(z>2.50) 1 P(z<2.50)
=1-0.9938=0.0062
stat
Decision rule: If P-value ≤ α, Reject Null Hypothesis, H0
When α= level of significance = 5%=0.05, P-value<0.05;
th f j t th ll h th i t 5% i ifi l l
therefore reject the null hypothesis at 5% significance level.
When α= level of significance = 1%=0.01, P-value<0.01;
therefore reject the null hypothesis at 1% significance level
therefore reject the null hypothesis at 1% significance level.
Hence we can see that, null hypothesis is rejected at both 5%
and 1% significance level and can conclude that promotion
and 1% significance level and can conclude that promotion
activity increased the women players at Pine Creek Golf Course.
18
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

19. Example:
0 70
H
ˆ
Suppose, p 0.66, n=200
 0 0
0
: 0.70
: 0.70
A
H p p
H p p
 
 
0
ˆ 0.66 0.70
( ) = 1.23
0 70(1 0 70)
p p
Z statistic
 
  
ˆ 0.70(1 0.70)
200
p
 
l l f i ifi 5% 0 05
α= level of significance = 5%=0.05
1.96 1.96
z z z
  
    
  
Decision Rule: If ,
stat
z z

Decision: Fail to Reject N ll
here, 1.23<1.96
Decision: Fail to Reject Null
19
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

20. Hypothesis test for the difference between two
( )
population mean(σ known)
0 0
( ) ( )
( ) =
x y D x y D
Z statistic
   

2
2
( )
( )
x y y
x
x y
Z statistic
n n

 


x y
0 0
0
:
:
x y
A
H D
H D
  
  
0 0
:
:
x y
H D
H D
  
  
0 0
:
:
x y
H D
H D
  
  
0
:
A x y
H D
  
Reject H
If z z
 RejectH
If ,
t t
z z 

0
:
A x y
H D
   0
:
A x y
H D
  
P‐value P‐value P‐value
0
Reject H
If ,
stat
z z
 0
RejectH
If ,
stat
z z

=P[z<(z‐stat value)] =P[z>(z‐stat value)] =P[z>(z‐stat value)]
+P[z<(‐z‐stat value)]
If P‐value ≤ α, Reject H0
20
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

21. A t f t d t l t diff i d ti lit
Example:
As part of a study to evaluate differences in education quality
between two training centre, a standardized examination is given
to individuals who are trained at the centre The difference
to individuals who are trained at the centre. The difference
between the mean examination scores is used to assess quality
differences between the centre. The population means for the two
centre are as follows.
= the mean examination score for the population
of individuals trained at centre A
A

of individuals trained at centre A.
= the mean examination score for the population
of individuals trained at centre B.
B

We begin with the tentative assumption that no difference exists
between the training quality provided at the two centre. Hence, in
g q y p ,
terms of the mean examination scores, the null hypothesis is that
0 0
: 0
A B
H D
   
0
: 0
A A B
H D
   
21
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

22. S 6 4 2 7
0 0
0
: 0
: 0
A B
A A B
H D
H D
   
   
Suppose, =6.4, 2.7
and sample data shows:
x y
  
0
: 0
A A B
H D
  
n 31, 33.5; n 30, 27.6
x y
x y
   
( ) (33 5 27 6) 0
D0
2 2
( )
( ) (33.5 27.6) 0
( ) = 4.72
6.4 2.7
x y
x y D
Z statistic

   
 


31 30
α= level of significance = 1%=0.01
2.58 2.58
z z z
  
    
Decision Rule: If ,
stat
z z

ec s o u e: ,
here, 4.72>2.58
stat 
Decision: Reject Null
22
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

23. Hypothesis test for the difference between two
( )
population mean(σ unknown)
2 2
When  
0 0
When = ,
( ) ( )
x y
x y D x y D
 
   
0 0
( 2) . 2 2
( )
( ) ( )
=
x y
n n d f
x y p p
x y D x y D
t
s s
 




2 2
( 1) ( 1)
x y
n n

2 2
2
( 1) ( 1)
Where,
2
x x y y
p
x y
n s n s
s
n n
  

 
x y
23
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

24. Hypothesis test for the difference between two
( )
population mean(σ unknown)
2 2
When   
0 0
When ,
( ) ( )
x y
x y D x y D
t
  
   
0 0
. 2
2
( )
( ) ( )
=
v d f
x y y
x
y y
t
s
s




2
2
x y
n n
s
2
2
[ ]
Where
y
x
x y
s
s
n n
v

2
2
2 2
Where,
[ ] / ( 1) [ ] / ( 1)
y
y
x
x y
v
s
s
n n

  
y
x y
n n
24
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

25. Hypothesis test for the difference between two
( )
population mean(σ unknown)
0 0
0
:
:
x y
A x y
H D
H D
  
  
0 0
0
:
:
x y
A x y
H D
H D
  
  
0 0
0
:
:
x y
A x y
H D
H D
  
  
0
df RejectH
If ,
stat
t t

(n-1)df
Reject H
If stat
t t

0
(n-1)df
Reject H
If stat
t t

P‐value
=P[t<(t‐stat value)]
P‐value
=P[t>(t‐stat value)]
P‐value
=P[t>(t‐stat value)]
0
Reject H 0
j
+P[t<(‐t‐stat value)]
If P‐value ≤ α, Reject H0
25
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

26. : 0
H D
   
Example:
0 0
0
: 0
: 0
A B
A A B
H D
H D
   
   
S k d l d h
Suppose, not known and sample data shows:
n 29, 3.9, s 1.2; n 25, 3.5, s 1.4
x y
x x y y
x y
   
     
x x y y
2 2
2 2
When = ,
x y
 
2 2 2 2
2
( 1) ( 1) (29 1)1.2 (25 1)1.4
1.68
2 29 25 2
x x y y
p
x y
n s n s
s
n n
     
  
   
0 0
( 2) . 2 2
( ) ( ) (3.9 3.5) 0
= 1.13
1 68 1 68
x y
n n d f
x y D x y D
t  
     
  
( 2) . 2 2
( ) 1.68 1.68
29 25
x y
n n d f
x y p p
x y
s s
n n





52 degrees of freedom
x y
26
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

27. α= level of significance = 5%=0.05
2 007
t t t
  
52df df df 2.007
t t t
  
  
Decision Rule: If ,
here, 1.13<2.007
stat
t t

Decision: Fail to Reject Null
,
27
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

28. : 0
H D
   
Example:
0 0
0
: 0
: 0
A B
A A B
H D
H D
   
   
S k d l d h
Suppose, not known and sample data shows:
n 29, 3.9, s 1.2; n 25, 3.5, s 1.4
x y
x x y y
x y
   
     
x x y y
2 2
When ,
x y
  
0 0
( ) . 2
2
( ) ( ) (3.9 3.5) 0
= 1.11
1 44 1 96
v d f
x y D x y D
t
     
  
 2
2
( ) 1.44 1.96
29 25
x y y
x
x y
s
s
n n




28
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

29. 2
2
2
[ ]
y
x
s
s

2
2
2 2
[ ]
[ ] / ( 1) [ ] / ( 1)
x y
y
x
n n
v
s
s
n n

  
2 2
2
[ ] / ( 1) [ ] / ( 1)
1.2 1.4
[ ]
x y
x y
n n
n n


2 2
2 2
[ ]
29 25 47.64 47 df
1.2 1.4
[ ] / (29 1) [ ] / (25 1)
29 25

  
  
α= level of significance = 5%=0.05
2 012
t t t
[ ] ( ) [ ] ( )
29 25
df df df 2.012
t t t
  
  
Decision Rule: If ,
stat
t t

ec s o u e: ,
here, 1.11<2.012
stat
t t
Decision: Fail to Reject Null
29
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

30. Hypothesis test for the difference between two
population proportions
0 0
ˆ ˆ ˆ ˆ
( ) ( )
( ) =
x y x y
p p D p p D
Z statistic
   

ˆ ˆ
( )
( )
ˆ
ˆ
x y
p p y y
x x
x y
Z statistic
p q
p q
n n



y
0 0
0
:
:
x y
A
H D
H D
  
  
0 0
:
:
x y
H D
H D
  
  
0 0
:
:
x y
H D
H D
  
  
0
:
A x y
H D
  
Reject H
If z z
 RejectH
If ,
t t
z z 

0
:
A x y
H D
   0
:
A x y
H D
  
P‐value P‐value P‐value
0
Reject H
If ,
stat
z z
 0
RejectH
If ,
stat
z z

=P[z<(z‐stat value)] =P[z>(z‐stat value)] =P[z>(z‐stat value)]
+P[z<(‐z‐stat value)]
If P‐value ≤ α, Reject H0
30
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University

31. : 0
H p p D
  
Example:
0 0
0
: 0
: 0
x y
A x y
H p p D
H p p D
  
  
ˆ ˆ ˆ ˆ
n 655, p 0.42, q 0.58; n 455, p 0.34, q 0.66
x x x y y y
Suppose      
0 0
ˆ ˆ ˆ ˆ
( ) ( )
( ) =
x y x y
p p D p p D
Z statistic
   

ˆ ˆ
( )
( ) =
ˆ
ˆ
x y
p p y y
x x
x y
Z statistic
p q
p q
n n




(0.42 0.34) 0
2.72
0 42 0 58 0 34 0 66
x y
x x
 
 
0.42 0.58 0.34 0.66
655 455
x x
 α= level of significance = 1%=0.01
Decision Rule: If z z

2.58 2.58
z z z
  
    
Decision Rule: If ,
here, 2.72>2.58
stat
z z

Decision: Reject Null
31
Mohammad Kamrul Arefin, Lecturer-School of Business, North South University