2. SAMPLING
Definition :
Sampling is the process by which inference is made to the
whole by examining a part. Or the process of drawing the
sample from the population is called sampling.
For examples
(1)With a single grain of rice, an Asian housewife tests if all
the rice in the pot has boiled
(2) from a cup of tea, a tea-taster determines the quality of the
brand of tea; and
(3) a sample of moon rocks provides scientists with
information on the origin of the moon.
This process of testing some data based on a small sample is
called sampling.
Types of sampling:
Probability sampling (random sampling):
In probability sampling all the items in the population have a
chance of being chosen in the sample.
Non Probability sampling (non-random or judgement
sampling):
In non- probability sampling personal knowledge and opinion
are used to identify the items from the population that are to
be included in the sample.
3. Simple Random Sampling:
(i) The population should be homogenious
(ii) Each element of population has an equal chance being
include in the sample.
(iii) Different sample of same size have equal chance.
On the basis of above assumption sample being
selected is called simple random sampling.
Sampling With Replacement or Without Replacement:
Suppose we have a bowl of 100 unique numbers from 0 to 99.
We want to select a random sample of numbers from the
bowl. After we pick a number from the blow, we can put the
number aside or we can put it back into the bowl. If we put
the number back in the bowl, it may be selected more than
once; if we put it aside, it can selected only one time.
When a population element can be selected more than one
time, we are sampling with replacement. When a population
element can be selected only one time, we are sampling
without replacement.
Example-1: Draw all possible samples of size 2 without
replacement from the population 2, 5 , 6 , 8 , 9 .
Solution:
2,5 2,6 2,8 2,9 5,6 5,8 5,9 6,8 6,9 8,9
4. Example-2: Draw all possible samples of size 3 without
replacement from the population 2, 5 , 6 , 8 , 9 .
Solution:
2,5,6 2,5,8 2,5,9 2,6,8 2,6,9
2,8,9 5,6,8 5,6,9 5,8,9 6,8,9
Example-3: Draw all possible samples of size 2 with
replacement from the population 2, 5 , 6 , 8 .
Solution:
2,2 2,5 2,6 2,8
5,2 5,5 5,6 5,8
6,2 6,5 6,6 6,8
8,2 8,5 8,6 8,8
Example-4: Draw all possible samples of size 3 with
replacement from the population 4 , 6 .
Solution:
4,4,4 6,6,6
4,4,6 6,6,4
4,6,4 6,4,6
6,4,4 4,6,6
5. Sampling distribution:
The probability distribution of statistic is called sampling
distribution.
Standard Error:
The standard deviation of sampling distribution of statistic is
called the standard error of statistic.
Sampling distribution of sample mean:
The probability distribution of sample mean is called a
sampling distribution of mean.A sampling distribution of
sample mean have the following properties.
Following are the properties of Sampling With-out
Replacement:
(1) 𝑬( 𝑿̅) = 𝝁
(2) 𝑽( 𝑿̅) =
𝝈 𝟐
𝒏
×
𝑵−𝒏
𝑵−𝟏
.
(3) The sampling distribution of sample mean will be normal
or approximately normal for reasonably large sample.
Following are the properties of Sampling With
Replacement:
(1) 𝑬( 𝑿̅) = 𝝁
(2)
𝑽( 𝑿̅) =
𝝈 𝟐
𝒏
(3) The sampling distribution of sample mean will be normal
or approximately normal for reasonably large sample.
6. Example-5: Draw all possible samples of size 2 without
replacement from the population 2, 5 , 6 , 8 , 9 .
Show that 𝐸( 𝑋̅) = 𝜇 and 𝑉( 𝑋̅) =
𝜎2
𝑛
×
𝑁−𝑛
𝑁−1
.
Solution:
Population : 2,5,6,8,9
Size of population =N= 5
Size of sample = n = 2
2 4
5 25
6 36
8 64
9 81
∑x=30 ∑x2
=
210
→ (1)
x 2
x
30
6
5
x
N
= = =
2
2
2
N
x
N
x
−=
2
2 210 30
5 5
= −
2
2
42 36
6
= −
=
7. Samples
Samples
2, 5 3.5 12.25
2, 6 4 16
2, 8 5 25
2, 9 5.5 30.25
5, 6 5.5 30.25
5, 8 6.5 42.25
5, 9 7 49
6, 8 7 49
6, 9 7.5 56.25
8, 9 8.5 72.25
60 382.5
By comparing (1) and (3)
2
2
6 5 2
1 2 5 1
2.25 (2
1
N n
n N
N n
n N
− −
=
− −
−
= →
−
10CCm 2
5
n
N
===
x 2
x
x = 2
x =
( ) ( )
60
6 3
10
x
E x
m
= = = →
( ) =xE
8. By comparing (2) and (4)
Example-6: Draw all possible samples of size 3 without
replacement from the population 1, 5 , 6 , 8 , 9 .
Show that 𝐸( 𝑋̅) = 𝜇 and 𝑉( 𝑋̅) =
𝜎2
𝑛
×
𝑁−𝑛
𝑁−1
.
Solution:
Population : 1,5,6,8,9
Size of population =N= 5
Size of sample = n = 3
1 1
5 25
6 36
8 64
9 81
∑x=29 ∑x2
=
207
→ (1)
( )
( )
( )
( ) ( )
22
2
382.5 60
10 10
38.25 36
2.25 4
x x
V x
m m
V x
V x
V x
= −
= −
= −
= →
( )
1N
nN
n
xV
2
−
−
=
x 2
x
30
6
5
x
N
= = =
9. 5
3 10 samplesN
nm c c= = =
Samples
1, 5, 6 4 16
1, 5, 8 4.67 21.81
1, 5, 9 5 25
1, 6, 8 5 25
1, 6, 9 5.33 28.44
1, 8, 9 6 36
5, 6, 8 6.33 40.07
5, 6, 9 6.67 44.49
5, 8, 9 7.33 53.73
6, 8, 9 7.67 58.83
58 349.37
By comparing (1) and (3)
2
2
2
N
x
N
x
−=
2
2
2
2
207 29
5 5
41.4 33.64
7.76
= −
= −
=
2
2
7.76 5 3
1 3 5 1
1.29 (2
1
N n
n N
N n
n N
− −
=
− −
−
= →
−
x 2
x
x = 2
x =
( ) ( )
58
5.8 3
10
x
E x
m
= = = →
( ) =xE
10. By comparing (2) and (4)
Example-7: Draw all possible samples of size 2 with
replacement from the population 2, 5 , 6 , 8 .
Show that 𝐸( 𝑋̅) = 𝜇 and 𝑉( 𝑋̅) =
𝜎2
𝑛
.
Solution:
Population : 2,5,6,8
Size of population =N= 4
Size of sample = n = 2
2 4
5 25
6 36
8 64
∑x=21 ∑x2
=
129
( )
( )
( )
( ) ( )
22
2
349.37 58
10 10
34.93 33.64
1.29 4
x x
V x
m m
V x
V x
V x
= −
= −
= −
= →
( )
1N
nN
n
xV
2
−
−
=
x 2
x
( )
21
5.25 1
4
x
N
= = = →
2
2
2
N
x
N
x
−=
11. 2
4 16 samplesn
m N= = =
Samples
2,2 2 4
2,5 3.5 12.25
2,6 4 16
2,8 5 25
5,2 3.5 12.25
5,5 5 25
5,6 5.5 30.25
5,8 6.5 42.25
6,2 4 16
6,5 5.5 30.25
6,6 6 36
6,8 7 49
8,2 5 25
8,5 6.5 42.25
8,6 7 49
8,8 8 64
84 478.5
By comparing (1) and (3)
2
2
2
2
129 21
4 4
32.25 27.56
4.69
= −
= −
=
2
2
4.69
2
2.35 (2
n
n
=
= →
x 2
x
x = 2
x =
( ) ( )
84
5.25 3
16
x
E x
m
= = = →
12. By comparing (2) and (4)
Example-8: Draw all possible samples of size 3 with
replacement from the population 4, 6 .
Show that 𝐸( 𝑋̅) = 𝜇 and 𝑉( 𝑋̅) =
𝜎2
𝑛
.
Solution:
Population : 4,6
Size of population =N= 2
Size of sample = n = 3
4 16
6 36
∑x=10 ∑x2
=
52
( ) =xE
( )
( )
( )
( ) ( )
22
2
478.5 84
16 16
29.91 27.56
2.35 4
x x
V x
m m
V x
V x
V x
= −
= −
= −
= →
( )
2
V x
n
=
x 2
x
( )
10
5 1
2
x
N
= = = →
13. Samples
Samples
4,4,4 4 16
4,4,6 4.67 21.81
4,6,4 4.67 21.81
6,4,4 4.67 21.81
6,6,6 6 36
6,6,4 5.33 28.41
6,4,6 5.33 28.41
4,6,6 5.33 28.41
40 202.66
By comparing (1) and (3)
2
2
2
N
x
N
x
−=
2
2
2
2
52 10
2 2
26 27.56
1
= −
= −
=
2
2
1
3
0.33 (2
n
n
=
= →
3
2 8n
m N= = =
x 2
x
x = 2
x =
( ) ( )
40
5 3
8
x
E x
m
= = = →
( ) =xE
14. ( )
( )
( )
( ) ( )
22
2
202.66 40
8 8
25.33 25
0.33 4
x x
v x
m m
v x
v x
v x
= −
= −
= −
= − − − − −
By comparing (2) and (4)
( )
2
V x
n
=
