Download Scientific Research Methods Lecture-13

RESEARCH
METHODOLOGY
LECTURE 13
SAMPLING & PROBABILITY
DISTRIBUTIONS
1
Mazhar Hussain
Dept of Computer Science
ISP,Multan
Mazhar.hussain@isp.edu.pk
ROAD MAP
Introduction
Chosing your research problem
Chosing your research advisor
Literature Review
Plagiarism
Variables in Research
Construction of Hypothesis
Research Design
Writing Research Proposal
Writing your Thesis
Data Collection
Data Representation
Sampling and Distributions
Paper Writing
Ethics of Research
2
SAMPLING

How to find average student age in the
university?
Ask each student and compute the average
 Randomly select 3 to 4 students from each discipline
and find their average age – Estimation of the
average age of student in the university

3
SAMPLING

Why sampling?


Efforts and resources required to carry out the study
on the population
Examples
Average income of families living in a city
 Results of an election
 Opinion about the a problem

4
SAMPLING
Sampling is the process of selcetion a few (a
sample) from a bigger group (the sampling
population) to become the basis for estimating
or predicting the prevalence of an unknown
piece of information, situation or outcome
regarding the bigger group
5
RECAP – MEAN & STANDARD DEVIATION

Mean/Average

Standard Deviation

On the average, how far the data values are from the
mean
6
POPULATION VS SAMPLE
7
Gaussian
Distribution
Karl Friedrich Gauss 1777-1855
8
GAUSSIAN/NORMAL PROBABILITY
DISTRIBUTION

Most of the naturally occurring processes can be
modeled by a bell shaped curve
9
GAUSSIAN/NORMAL PROBABILITY
DISTRIBUTION
The Gaussian probability distribution is perhaps
the most used distribution in all of science.
 Sometimes it is called the “bell shaped curve” or
normal distribution.

N ( , 2 )

1
p( x) 
e
 2
( x   )2
2 2
= mean of distribution
 = standard deviation of distribution
x is a continuous variable (-∞x ∞
10
GAUSSIAN/NORMAL PROBABILITY
DISTRIBUTION
The area within +/- σ is ≈ 68%
The area within +/- 2σ is ≈ 95%
The area within +/- 2σ is ≈ 99.7%
11
GAUSSIAN/NORMAL PROBABILITY
DISTRIBUTION

Probability (P) of x being in the range [a, b] is
given by an integral:
b
1
P(a  x  b)   p ( x)dx 
 2
a
b
e

( x   )2
2 2
dx
a
Gaussian pdf with =0 and =1
12
95% of area within 2
Only 5% of area outside 2
GAUSSIAN/NORMAL PROBABILITY
DISTRIBUTION
Standard Normal Distribution
13
STANDARD NORMAL DISTRIBUTION

Normal distribution with mean of zero and
standard deviation of one

Since mean and standard deviation define any
normal distribution…

Standard normal distribution can be used for any
normally distributed variable by converting mean
to zero and standard deviation to one—z scores
14
Z SCORES
By itself, a raw score or X value provides very little
information about how that particular score
compares with other values in the distribution.
 A score of X = 53, for example, may be a relatively
low score, or an average score, or an extremely
high score depending on the mean and standard
deviation for the distribution from which the score
was obtained.
 If the raw score is transformed into a z-score,
however, the value of the z-score tells exactly
where the score is located relative to all the other
scores in the distribution.

15
Z SCORES

The process of changing an X value into a z-score
involves creating a signed number, called a zscore, such that
The sign of the z-score (+ or –) identifies whether the
X value is located above the mean (positive) or below
the mean (negative).
 The numerical value of the z-score corresponds to the
number of standard deviations between X and the
mean of the distribution.
 Thus, a score that is located two standard deviations
above the mean will have a z-score of +2.00

16
Z SCORES
In addition to knowing the basic definition of a zscore and the formula for a z-score, it is useful to
be able to visualize z-scores as locations in a
distribution.
 Remember, z = 0 is in the center (at the mean),
and the extreme tails correspond to z-scores of
approximately –2.00 on the left and +2.00 on the
right.
 Although more extreme z-score values are
possible, most of the distribution is contained
between z = –2.00 and z = +2.00.

17
Z SCORES


z-score for a sample value in a data set is obtained by
subtracting the mean of the data set from the value
and dividing the result by the standard deviation of
the data set.
NOTE: When computing the value of the z-score,
the data values can be population values or sample
values. Hence we can compute either a population zscore or a sample z-score.
18
Z SCORES

The Sample z-score for a value x is given by the
following formula:
xx
z
s

Where x is the sample mean and s is the sample
standard deviation.
19
Z SCORES

The Population z-score for a value x is given by
the following formula:
z

x

Where  is the population mean and  is the
population standard deviation.
20
EXAMPLE

Example: What is the z-score for the value of 14
in the following sample values?
3 8 6 14 4 12 7 10
Thus, the data value of 14 is 1.57 standard deviations
above the mean of 8, since the z-score is positive.
21
EXAMPLE

Dot Plot of the data points with the location of
the mean and the data value of 14.
22
Z SCORE & PROBABILITY

What is the probability of finding a value
between 100 and 110?
How to calculate
this area using z
scores?
23
Reading area under curve for z=1.55
Z SCORE CHART
0.9394
24
Z SCORE & PROBABILITY
0.45
0.40
P=1-0.9394
0.35
P=0.0606
P=.0606
0.30
0.9394
0.25
0.20
0.15
0.10
0.05
0.00
-3.0
-2.0
-1.0
0.0
1.0
2.0
1.55
Probability of z>1.55 (Area in tail)
3.0
25
Z SCORE & PROBABILITY
0.45
0.40
0.35
P=.0606+.0606
P=.1212
0.30
0.25
0.20
0.15
0.10
0.05
0.00
-3.0
-2.0
-1.55
-1.0
0.0
1.0
2.0
3.0
1.55
Probability of z>1.55 + z<-1.55 (Area in both the tails)
26
Z SCORE & PROBABILITY
0.45
0.40
P=.5-.0606=.4394
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
-3.0
-2.0
-1.0
0.0
1.0
2.0
1.55
Probability of z>0 and z<-1.55 )
3.0
27
EXAMPLE: 50 MEASURES OF POLLUTION
Histogram
14
12
10
8
6
4
2
  40.68
or
e
M
65
60
55
50
45
40
35
30
25
0
20
Frequency
V a lu e F r e q u e n c y
20
0
25
3
30
5
35
6
40
8
45
13
50
5
55
6
60
3
65
1
M o re
0
Value
  9.88
28
EXAMPLE: 50 MEASURES OF POLLUTION
•

Probability value > 45
45  40.68
z
 .4372
9.88
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
-3
-2
-1
0
1
2
3
29
P=.3300
.4372
EXAMPLE: 50 MEASURES OF POLLUTION

Probability from 35 to 45
•
35  40.68
z
 .5749
9.88
0.45
45  40.68
z
 .4372
9.88
0.30
0.40
0.35
0.25
0.20
0.15
0.10
0.05
0.00
-3
P=.2157+.1700=.3857
-2
-1
-.5749
P=.5-.2843=.2157
0
1
2
3
.4372
P=.5-.3300=.1700
30
Sampling
31
SAMPLING

Pros
Saves time
 Resources – financial, human


Cons
Not exact value for the population
 An estimate or prediction
 Compromise on accuracy of findings

32
SAMPLING – TERMINOLOGY

Examples
Average student age in the university
 Average income of families living in a city
 Results of an election


Population or study population (N)


The university students, families living in the city,
electors
Sample

The small group of students, families or electors you
chose to collect the required information
33
SAMPLING – TERMINOLOGY

Sample size (n)


Sampling design or strategy


The way you select the students, families or electors
Sampling unit or sampling element


The number of entities in your sample
Each student, family or elector in your study
Sample statistics

Your findings based on infomration obtained from
your sample
34
SAMPLING – TERMINOLOGY

Population Parameters
Aim of research – find answers to research question
for study population not the sample
 Use sample statistics to estimate answers to research
questions in study population
 Estimates arrived at from sample statistics –
population parameters


Saturation Point

When no new information is coming from your
respondents
35
SAMPLING – TERMINOLOGY

Sampling Frame

A list identifying each student, family or elector in
the study population
36
PRINCIPLES OF SAMPLING

Example – Four individuals A,B,C, D
A = 18 years
 B = 20 years
 C = 23 years
 D = 25 years


Average age


(18+20+23+25) / 4 = 21.5 years
Use a sample of two indivudals to estimate the
average age of your study population (4
individuals)
37
PRINCIPLES OF SAMPLING

How many possible combinations of two
individuals?
A and B
 A and C
 A and D
 B and C
 B and D
 C and D

38
PRINCIPLES OF SAMPLING
A+B = 18+20 = 38/2 = 19.0 years
 A+C = 18+23 = 41/2 = 20.5 years
 A+D = 18+25 = 43/2 = 21.5 years
 B+C = 20+23 = 43/2 = 21.5 years
 B+D = 20+25 = 45/2 = 22.5 years
 C+D = 23+25 = 48/2 = 24.0 years



In two cases – no difference between sample
statistics and population parameters
Difference – Sampling error
39
PRINCIPLES OF SAMPLING
Sample
Sample
Statistics
Population
Parameters
Difference
1
19.0
21.5
-2.5
2
20.5
21.5
-1.5
3
21.5
21.5
0.0
4
21.5
21.5
0.0
5
22.5
21.5
+1.0
6
24
21.5
+2.5
40
PRINCIPLES OF SAMPLING

Principle I
In majority of cases of sampling, there will
be a difference between sample statistics
and the true population parameters which
is attribuatable to the selection of the units
in the sample
41
PRINCIPLES OF SAMPLING
Instead of samples of two – take a sample of
three
 Four possible combinations

A+B+C = 18+20+23 = 61/3 = 20.33 years
 A+B+D = 18+20+25 = 63/3 = 21.00 years
 A+C+D = 18+23+25 = 66/3 = 22.00 years
 B+C+D = 20+23+25 = 68/3 = 22.67 years

42
PRINCIPLES OF SAMPLING
Sample
Sample
Statistics
Population
Parameters
Difference
1
20.33
21.5
-1.17
2
21.00
21.5
-0.5
3
22.00
21.5
+0.5
4
22.67
21.5
+1.17
43
PRINCIPLES OF SAMPLING
Sample
Sample
Statistics
Population
Parameters
Difference
1
19.0
21.5
-2.5
2
20.5
21.5
-1.5
3
21.5
21.5
0.0
4
21.5
21.5
0.0
5
22.5
21.5
+1.0
6
24
21.5
+2.5
Sample
Sample
Statistics
Population
Parameters
Difference
1
20.33
21.5
-1.17
2
21.00
21.5
-0.5
3
22.00
21.5
+0.5
4
22.67
21.5
+1.17
-2.5 to +2.5
-1.17 to +1.17
44
PRINCIPLES OF SAMPLING


The gap between sample statistics and population
parameters is reduced
Principle II
The greater the sample size, the more
accurate will be the estimate of the true
population statistics
45
PRINCIPLES OF SAMPLING

Same Example – Different Data
A =18 years
 B = 26 years
 C = 32 years
 D = 40 years


Variable (age) – markedly different
46
PRINCIPLES OF SAMPLING

Estimate average using
Samples of two
 Samples of three


Difference in the average age:
Sample size of 2: -7.00 to +7.00 years
 Sample size of 3: -3.67 to +3.67 years


Range of difference is greater than previously
calculated
47
PRINCIPLES OF SAMPLING

Principle III
The greater the difference in the variable
under study in a population for a given
sample size, the greater will be the
difference between the sample statistics
and the true population parameters
48
FACTORS AFFECTING THE INFERENCE

Principles suggest that two factors may influence
the degree of certainity about the inferences
drawn from a sample

Size of sample


Larger the sample size, the more accurate will be the
findings
The extent of variation in the sampling population

Greater the variation in the study population w.r.t. the
chracteristics under study, the greater will be the
uncertainity for a given sample size
49
AIMS IN SELECTING A SAMPLE
Achieve maximum precision in your estimate
 Avoid bias in selection


Bias can occur if:
Non-random sampling – consciously or unconsciously affected
by human choice
 Sampling frame does not cover the sampling population
accurately or completely
 A section of sampling population is impossible to find or
refuses to cooperate

50
SAMPLING METHODS
 Probability

Sampling
Used to generate random/non-biased samples
required for conducting inferential analyses
 Non-probability

Sampling
Used mostly in qualitative analysis
 Mixed
Sampling
51
SAMPLING METHODS
 Probability
Sampling
 Non-probability Sampling
 Mixed Sampling
52
PROBABILITY SAMPLING


Each element in the population has an equal and
independent chance of selection in the sample
Equal:
Probability of selection of each element is the same
 Choice is not affected by other considerations –
human preferences


Independent:
Choice of one element is not dependent upon the
choice of another element
 Selection or rejection of one element does not affect
the inclusion or exclusion of another

53
PROBABILITY SAMPLING

Example – Equal Chance
80 students in the class
 20 refuse to participate in your study
 Each of 80 students (population) does not have an
equal chance of selection
 Sample is not representative of your class

54
PROBABILITY SAMPLING

Example – Independence





Three close friends in the class
One is selected – Two are not
Refuses to participate without friends
Forced to chose all three or none
Not independent sampling
Inferences drawn from random samples
can be generalized to the total sampling
population
55
PROBABILITY SAMPLING

Simple Random Sampling
Fishbowl draw
 Computer program
 Table of random numbers


Stratified Sampling
Proportional
 Disproportional


Cluster Sampling
56
SIMPLE RANDOM SAMPLING

The Fishbowl Draw






Small population
Number each element on separate slips of paper for
each element
Put them in a box
Pick out one by one until you get desired sample size
Similar to lotteries
Computer Program

Write a program to select a random sample
57
SIMPLE RANDOM SAMPLING

Table of random numbers
Random
Number Table
58
How to Use Random Number Tables
________________________________________________
1. Assign a unique number to each population element in the
sampling frame. Start with serial number 1, or 01, or 001,
etc. upwards depending on the number of digits required.
2. Choose a random starting position.
3. Select serial numbers systematically across rows or down
columns.
4. Discard numbers that are not assigned to any population
element and ignore numbers that have already been
selected.
5. Repeat the selection process until the required number of
sample elements is selected.
59
How to Use a Table of Random Numbers to Select a Sample
Your supervisor wants to randomly select 20 students from your class of 100
students. Here is how he can do it using a random number table.
Step 1: Assign all the 100 members of the population a unique number.You may
identify each element by assigning a two-digit number. Assign 01 to the first name
on the list, and 00 to the last name. If this is done, then the task of selecting the
sample will be easier as you would be able to use a 2-digit random number table.
NAME
Adam, Tan
………………
Carrol, Chan
……………….
Jerry Lewis
……………….
Lim Chin Nam
……………….
Singh, Arun
……………….
NUMBER
01
08
…
18
…
26
…
30
NAME
Tan Teck Wah
……………………
…………..
Tay Thiam Soon
………………..
Teo Tai Meng
………………….
……………………
Yeo Teck Lan
Zailani bt Samat
NUMBER
42
…
61
…
87
…
…
99
00
60
Step 2: Select any starting point in the Random Number Table and find the first number that
corresponds to a number on the list of your population. In the example below, # 08 has been
chosen as the starting point and the first student chosen is Carol Chan.
Starting point:
move right to the end
of the row, then down
to the next row row;
move left to the end,
then down to the next
row, and so on.
10 09 73 25 33 76
37 54 20 48 05 64
08 42 26 89 53 19
90 01 90 25 29 09
12 80 79 99 70 80
66 06 57 47 17 34
31 06 01 08 05 45
Step 3: Move to the next number, 42 and select the person corresponding to that number into
the sample. #42 – Tan Teck Wah
Step 4: Continue to the next number that qualifies and select that person into the sample.
# 26 -- Jerry Lewis, followed by #89, #53 and #19
Step 5: After you have selected the student # 19, go to the next line and choose #90. Continue
in the same manner until the full sample is selected. If you encounter a number selected
earlier (e.g., 90, 06 in this example) simply skip over it and choose the next number.
61
TABLE OF RANDOM NUMBERS

Suppose you are using a table like this:
62
TABLE OF RANDOM NUMBERS
Sampling population – 256 individuals
 Numbered from 1 to 256
 You chose to select 10% - 25 individuals
 Randomly select any starting point
 Pick last three digits of the number
 Select the valid ones (001-256) and skip the
invalid numbers (257-999)

63
DRAWING A RANDOM SAMPLE

Two ways of selecting a random sample
Sampling without replacement
 Sampling with replacement


Example
20 students to be selected out of 80
 First student is selected – Probability 1/80
 For second student – 79 left, Probability 1/79
 By the time you select the 20th – Probability 1/61

64
DRAWING A RANDOM SAMPLE

Sampling without replacement


Contrary to randomization – Each element should
have equal probability of selection
Sampling with replacement
Selected element is replaced in the population
 If it is selected again – it is discarded

65
PROBABILITY SAMPLING

Simple Random Sampling
Fishbowl draw
 Computer program
 Table of random numbers


Stratified Sampling
Proportional
 Disproportional


Cluster Sampling
66
STRATIFIED SAMPLING

Step 1- Divide the population into homogeneous,
mutually exclusive and collectively exhaustive
subgroups or strata using some stratification variable.

Step 2- Select an independent simple random sample
from each stratum.

Step 3- Form the final sample by consolidating all
sample elements chosen in step 2.
67
STRATIFIED SAMPLING

Example
Stratify on the basis of gender
 Two groups – male and female
 Select random samples from each group

68
STRATIFIED SAMPLING
Stratified samples can be:

Proportionate: involving the selection of sample
elements from each stratum, such that the ratio of sample
elements from each stratum to the sample size equals that
of the population elements within each stratum to the
total number of population elements.

Disproportionate: the sample is disproportionate when
the above mentioned ratio is unequal.
69
To select a stratified sample of 20 members of the Island Video Club which has 100 members
belonging to three language based groups of viewers i.e., English (E), Mandarin (M) and Others
(X).
Step 1: Identify each member from the membership list by his or her respective language groups
00 (E )
01 (E )
02 ( X )
03 (E )
04 (E )
05 (E )
06 (M)
07 (M)
08 (E )
09 (E )
10 (M)
11 (E )
12 ( X )
13 (M)
14 (E )
15 (M)
16 (E )
17 ( X )
18 ( X )
19 (M)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
(M)
(X)
(E )
(X)
(E )
(M)
(E )
(M)
(X)
(E )
(E )
(E )
(E )
(M)
(E )
(M)
(E )
(E )
(X)
(X)
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
(E )
(X)
(X)
(E )
(M)
(E )
(X)
(M)
(E )
(E )
(E )
(M)
(X)
(M)
(E )
(E )
(M)
(E )
(M)
(M)
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
(X)
(M)
(M)
(E )
(E )
(X)
(M)
(E )
(M)
(E )
(E )
(E )
(M)
(E )
(X)
(E )
(E )
(M)
(M)
(E )
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
(M)
(E )
(E )
(M)
(X)
(E )
(E )
(M)
(X)
(E )
(X)
(E )
(M)
(E )
(E )
(X)
(E )
(E )
(M)
(E )
70
Step 2: Sub-divide the club members into three homogeneous sub-groups or strata by the
language groups: English, Mandarin and others.
EnglishLanguage
Stratum
00 22 40 64 82
01 24 43 67 85
03 26 45 69 86
04 29 48 70 89
05 30 49 71 91
08 31 50 73 93
09 32 54 75 94
11 34 55 76 96
14 36 57 79 97
16 37 63 81 99
Mandarin Language
Stratum
06 35 66
07 44 68
10 47 72
13 51 77
15 53 78
19 56 80
20 58 83
25 59 87
27 61 92
33 62 98
Other Language
Stratum
.
02 42
12 46
17 52
18 60
21 65
23 74
28 84
38 88
39 90
41 95
1. Calculate the overall sampling fraction, f, in the following manner:
f = n = 20 = 1 =
N 100
5
0.2
where n = sample size and N = population size
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 Determine the number of sample elements (n1) to be selected from the English
language stratum. In this example, n1 = 50 x f = 50 x 0.2 =10. By using a simple
random sampling method [using a random number table] members whose numbers
are 01, 03, 16, 30, 43, 48, 50, 54, 55, 75, are selected.
 Next, determine the number of sample elements (n2) from the Mandarin language
stratum. In this example, n2 = 30 x f = 30 X 0.2 = 6. By using a simple random
sampling method as before, members having numbers 10,15, 27, 51, 59, 87 are
selected from the Mandarin language stratum.
 In the same manner, the number of sample elements (n3) from the ‘Other language’
stratum is calculated. In this example, n3 = 20 x f = 20 X 0.2 = 4. For this stratum,
members whose numbers are 17, 18, 28, 38 are selected’
 These three different sets of numbers are now aggregated to obtain the ultimate
stratified sample as shown below.
S = (01, 03, 10, 15, 16, 17, 18, 27, 28, 30, 38, 43, 48, 50, 51, 54, 55, 59, 75, 87)
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PROBABILITY SAMPLING

Simple Random Sampling
Fishbowl draw
 Computer program
 Table of random numbers


Stratified Sampling
Proportional
 Disproportional


Cluster Sampling
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CLUSTER SAMPLING

Simple Random, Symmetric and Stratified
sampling – based on researcher’s ability to
identify each element in population

Small population size – easy

Large population – country

Cluster sampling
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CLUSTER SAMPLING
Divide population into clusters
 Select elements wihtin each cluster
 Cluster formation

Geographical proximity
 Common characteristic – similar to stratified
sampling

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STRATIFIED VS CLUSTER SAMPLING


In startified sampling the target population is
sub-divided into a few subgroups or strata, each
containing a large number of elements.
In cluster sampling, the target population is subdivided into a large number of sub-population or
clusters, each containing a few elements.
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AREA SAMPLING
 A common form of cluster sampling where clusters consist of geographic areas, such as
districts, housing blocks or townships. Area sampling could be one-stage, two-stage, or
multi-stage.
How to Take an Area Sample Using Subdivisions
Your company wants to conduct a survey on the expected patronage of its new outlet in a new
housing estate. The company wants to use area sampling to select the sample households to be
interviewed. The sample may be drawn in the manner outlined below.
___________________________________________________________________________________
Step 1: Determine the geographic area to be surveyed, and identify its subdivisions. Each
subdivision cluster should be highly similar to all others. For example, choose ten housing
blocks within 2 kilometers of the proposed site [say, Model Town ] for your new retail outlet;
assign each a number.
Step 2: Decide on the use of one-step or two-step cluster sampling. Assume that you decide to
use a two-stage cluster sampling.
Step 3: Using random numbers, select the housing blocks to be sampled. Here, you select 4
blocks randomly, say numbers #102, #104, #106, and #108.
Step 4: Using some probability method of sample selection, select the households in each of the
chosen housing block to be included in the sample. Identify a random starting point (say,
apartment no. 103), instruct field workers to drop off the survey at every fifth house
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(systematic sampling).
SAMPLING METHODS
 Probability
Sampling
 Non-probability Sampling
 Mixed Sampling
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NON-PROBABILITY SAMPLING
Do not follow probability theory
 Useful when the number of elements in
population is unknown or cannot be individually
identified
 Four common designs

Quota Sampling
 Accidental Sampling
 Judgemental Sampling
 Snowball Sampling

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QUOTA SAMPLING
Main consideration – ease of access to sample
population
 Guided by some visible chracteristic of study
population – age, gender etc.
 Sample selection – location convenient to the
researcher
 Whenever a person with required characteristic
is seen – asked to participate in the study
 Process continues until the required number of
respondents (quota) is reached

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QUOTA SAMPLING - EXAMPLE
Average age of male students in the university
 Select a sample of 20 male students
 You decide to stand at the entrance of the
university - convenient
 Whenever a male student arrives – ask his age
 When you get 20 – Target is achieved

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QUOTA SAMPLING

Advantages
Convenient
 Less expensive


Disadvantages
Not probability based
 May not be generalized to the population

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ACCIDENTAL SAMPLING
Also based on convenience
 Quota sampling – include people with some
obvious characteristic
 Accidental sampling – no such attempt
 Common in market research and newspaper
reporters
 Since you just pick up the people – may not get
the required information

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JUDGEMENTAL SAMPLING


Judgement of the researcher as to who can
provide the best information to achieve the
objectives of the study
Sampling based on some judgment, gut-feelings
or experience of the researcher.
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SNOWBALL SAMPLING
Sample selection using network
 Start with few individuals or organizations and
collect the information
 They are then asked to identify other
participants – people selected by them become a
part of sample
 The process continues……

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SAMPLING METHODS
 Probability
Sampling
 Non-probability Sampling
 Mixed Sampling
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MIXED SAMPLING
Systematic sampling
 Divide the frame into segments or intervals
 Select one element from first interval using SRS
 Select elements from subsequent intervals
depending upon the element selected from the
first interval
 Example

5th element selected from first element
 Select the 5th from each interval

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SYSTEMATIC SAMPLING
 To use systematic sampling, a researcher needs:
[i] A sampling frame of the population; .
[ii] A skip interval calculated as follows:
Skip interval = population list size
Sample size
 Names are selected using the skip interval.
 If a researcher were to select a sample of 1000 people using the local telephone
directory containing 215,000 listings as the sampling frame, skip interval is
[215,000/1000], or 215. The researcher can select every 215th name of the entire
directory [sampling frame], and select his sample.
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Example: How to Take a Systematic Sample
Step 1: Select a listing of the population, say the City Telephone Directory, from which to
sample.
Step 2: Compute the skip interval by dividing the number of entries in the directory by the
desired sample size.
Example: 250,000 names in the phone book, desired a sample size of 2500,
So skip interval = every 100th name
Step 3: Using random number(s), determine a starting position for sampling the list.
Example: Select: Random number for page number. (page 01)
Select: Random number of column on that page. (col. 03)
Select: Random number for name position in that column (#38, say, A..Mahadeva)
Step 4: Apply the skip interval to determine which names on the list will be in the sample.
Example: A. Mahadeva (Skip 100 names), new name chosen is A Rahman b Ahmad.
Step 5: Consider the list as “circular”; that is, the first name on the list is now the initial name
you selected, and the last name is now the name just prior to the initially selected one.
Example: When you come to the end of the phone book names (Zs), just continue on 89
through the beginning (As).
CREDITS
Chapter 12, Research Methodology, Ranjit
Kumar
 Sampling in Market Research - APMF

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