statistical sampling - icediced.cag.gov.in/wp-content/uploads/2016-17/ntp 01/iced_statistical... ·...

Statistical Sampling

Amit Bhola

2

Source : Internet

There are better ways of Sampling!

GENERALIZATION

3

Representative SampleFaithful Generalization

POPULATION

Population is ‘observations’ or ‘measurements’ of property

under consideration

DO not confuse this with No. Of Objects.(Objects may or may not be property under study)

Population

(a) ‘Height’ of Plants in a garden ‘Height’

(b) ‘No. of Plants’ in a garden ‘No. of plants’4

SAMPLE

A small part of a population

SAMPLING

Process of obtaining samples

STATISTICAL INFERENCE

Process of inferring facts about population from the results

found in samples

Aim of Sampling

5

POPULATION, SAMPLE & GENERALIZATION

SamplePopulation

6

Sampling Frame : Population accessible for sampling

SAMPLING METHODS & GENERALIZATION

7

PROBABILITY SAMPLING – Aim of Generalization

- Best effort is made to draw sample representative of

population

NON PROBABILITY SAMPLING – When?

- Generalization is not the aim

- Qualitative study, Pilot study, Demonstration of population trait

- Probability sampling is infeasible

- Inaccessible sampling frame, constraints of time, money, etc.

- Initial study to be followed by probability sampling

Focus

TYPES OF PROBABILITY SAMPLING

8

Probability Sampling

Simple Random

Systematic Random

Stratified Cluster

POPULATIONS & SAMPLE SIZE

EXAMPLE Population Sample

Estimate Avg. weight of college students by studying only 100

FiniteN

Finiten

Estimate Head or Tail of a coin tossInfinite Finite

n8

9

SAMPLING WITH REPLACEMENT

EXAMPLE Replacement

Estimate Avg. weight of college students by studying only 100 different students

No

Estimate how many bolts in a bin of 500 are defective by : picking one bolt checking it returning it …. (repeat say 20 times)

Yes

10

SAMPLING THEORY

N

n

88-

n

No Replacement Replacement

Practically

N small N large

11

RANDOMNESS

Being Random means being equally probable

A sample is random if there is no bias in selecting its n objects

Each object has equal chance of getting selected

To effectively represent a population, a sample should be

random

For getting a random sample of size n,

n random nos. should be obtained first

12

RANDOM NUMBER – MS EXCEL

14

BLINDING

15

Blinding is done to eliminate psychological bias

• Single Blinding : The participants (i.e. sample) are completely unaware of which group they are in and what intervention they are receiving until conclusion of the study.

• Double Blinding : Neither the participants nor the researcher knows to which group the participant belongs and what intervention the participant is receiving until the conclusion of study

SIMPLE RANDOM SAMPLING

STEP 1 : Obtain the approx size of population N

STEP 2 : Label the population items 1,2,… N

STEP 3 : Find n random nos.

STEP 4 : Select items labeled as nos. got in [3]

16

SIMPLE RANDOM SAMPLING

SYSTEMATIC RANDOM SAMPLING

N

Random Derived = pick every (N/n)th element

N

n sections

17

STEP 2 : Label the population items 1,2,… N

18

PROBLEM WITH SIMPLE / SYSTEMATIC

1. Need availability of complete list of population.

For a large population, this may not be

available!

2. Although highly unlikely, Systematic sampling

carries risk of collecting a poor sample if (A)

there exist some periodic traits in the

population, and at the same time (B) The

period of the trait is a multiple of common

difference!

Sampling Frame Errors…

19

PROBLEM WITH SYSTEMATIC SAMPLING

20


random

Trait period =CD

21


random

Trait period = 2*CD

22

random

Trait period ≠ n*CD

DEALING PERIODICITY PROBLEM – 1

23

DEALING PERIODICITY PROBLEM – 2

random

Repeated sampling and combining two samples into one single sample

random

LINEAR SYSTEMATIC RANDOM SAMPLING

N = ?

Random Derived = pick every (N/n)th element

n sections

24

First Random is

Chosen from

1st section

( 1 ~ N/n )

CIRCULAR SYSTEMATIC RANDOM SAMPLING

26

Random

from

1 ~ N

Derived

till n samples

are obtained

Selection is

done with

continuing

counting at the

end of the list

TYPES OF PROBABILITY SAMPLING

27

Probability Sampling

Simple Random

Systematic Random

Linear

Repeated

Circular

Stratified Cluster

PROBLEM WITH SIMPLE / SYSTEMATIC

N

Random Derived

N

0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%

E.g. :Mon Tue Wed Thu/Fri/Sat

28

STRATIFIED SAMPLING

N

0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%

1. Divide N by category of Strata

Strata:

1. Divide N by category of Strata2. Select random samples from each Strata – in same ratio as that Strata

Mon Tue Wed Thu / Fri / Sat

29

STRATIFIED SAMPLING

30

Proportionate

=>

Representative of Population

STRATIFIED SAMPLING – ADVANTAGES

1. Ensures the presence of each subgroup within

the sample – better representation of population.

31

Especially useful for population with highly skewed strata eg. A:B::70:30

2. Permits analyses of within-stratum patterns and

separate reporting of the results for each stratum.

STRATIFIED SAMPLING – DIFFICULTIES

1. Requires information on the proportion of the

total population that belongs to each stratum.

32

2. More expensive, time-consuming, and

complicated than simple random sampling.

3. In order to calculate sampling estimates, at least

two elements must be taken in each stratum.

DISPROPORTIONATE STRATIFIED SAMPLING

33

Proportionate

=>

Representative of Population – Fine

But considering the cost of the sampling, studies other than the Generalization may separately

be performed at same time


34

Dis-Proportionate

to

Represent all strata sufficiently

Example 1 :-Analysis of variation within

strata


35

Dis-Proportionate

to

Represent all strata equally

Example 2 :-Analysis of variation among

strata


36

Example 3 :-Optimization of Cost and/or

Precision

j 1/j 1200

18 0.055556 299.1371

10 0.1 538.4468

39 0.025641 138.0633

24 0.041667 224.3528

s 1200

4.3 189.7059

6.4 282.3529

9.4 414.7059

7.1 313.2353

TYPES OF STRATIFIED SAMPLING

37

Stratified Sampling

Proportionate Disproportionate

Within Strata analysis

Among Strata analysis

Optimization

Useful, but not as good

representative of

population

as Proportionate

CLUSTER SAMPLING

N

STEP 1 : Divide N into homogenous clusters

(Clusters : different from each other but same within)

Eg. Subjects , Districts , Offices-shops-showrooms38

CLUSTER SAMPLING

N

STEP 2 : Choose some clusters randomly

N

39

CLUSTER SAMPLING

N

STEP 3 : Recommendation but not must, select

whole of the selected clusters.N

40

(STEP 4) : Sampling within clusters may be done

Multi-Stage Sampling…

CLUSTER SAMPLING

N

N

Usual motive is to avoid the high cost of a geographical survey

Accuracy is not as good as Stratified, but in comparison to No Survey at all (due to high cost), it is better to do Cluster Sampling

E.g. TRPs , IQS . . .

41

Section 4

Section 5

Section 3

Section 2Section 1

CLUSTER SAMPLING

Source : Internet

CLUSTER SAMPLING vs. STRATIFIED

1. In stratified random sampling, all the strata of

the population are sampled while in cluster

sampling, only a part of clusters are sampled.

43

2 With stratified sampling, the best survey results

occur when elements within strata are internally

homogeneous. However, with cluster sampling,

the best results occur when elements within

clusters are internally heterogeneous.

CLUSTER SAMPLING – DISADVANTAGES

44

There is tendency for the clusters to display similar characteristics within themselves – especially so in case the clusters are regional

Statistically, it is the least precise compared to the

Simple, Systematic and Stratified sampling.

COMPARISON OF SAMPLING TECHNIQUES

EASYEASY

45


>> ACCURATE

> ACCURATE

STRATA has to be known

46


ECONOMICAL

QUICK

47

SELECTION OF SAMPLING TECHNIQUES

48


49

Merit based Sections

Gender based Sections


50

Merit based Sections

No particular criteria

CASE STUDY

THERE ARE AROUND 8,000 FIRMS ACROSS INDIA

PROVIDING A CAB SERVICE.

AN AUDIT IS TO BE PLANNED TO CHECK THE

CONFORMANCE OF TAX PAYMENTS.

51

ZONE OPERATORS

NORTH 4627

SOUTH 3423

CENTRAL 1488

EAST 891

WEST 2396

REVENUE NO. OF FIRMS

< 1 CRORE 4221

1 CR – 5 CR 3217

5 CR – 50 CR 770

50 CR – 250 CR 145

≥ 250 CR 13

CASE STUDY

52

ZONE OPERATORS

NORTH 4627

SOUTH 3423

CENTRAL 1488

EAST 891

WEST 2396

REVENUE NO. OF FIRMS

< 1 CRORE 4221

1 CR – 5 CR 3217

5 CR – 50 CR 770

50 CR – 250 CR 145

≥ 250 CR 13

HETROGENEOUS WITHIN

GEOGRAPHICAL

MUTUALLY EXCLUSIVE ?MUTUALLY EXCLUSIVE

HETROGENEOUS WITHIN

SKEWED PROPORTIONS

MUTUALLY EXCLUSIVE

PERIODIC TRAITS – NO

SKEWED PROPORTIONS

BUDGET ? / MULTI-STAGE ?

CLUSTERCLUSTER

SYSTEMATI

STRATIFIED STRATIFIED

SIMPLE SIMPLE

WHAT NEXT ?

53

PROBABILITY

Trial 1 Trial 2 Trial 3 Trial 4 Trial 5

H T

54

PROBABILITY

Trial 1 Trial 2 Trial 3 Trial 4 Trial 5

H

T

0% 20% 40% 60% 80% 100%

p q = (1-p)

55

PROBABILITY

0

0.5

1

1.5

2

2.5

3

3.5

1 2 3 4 5 656

PROBABILITY

0

0.5

1

1.5

2

2.5

3

3.5

1 2 3 4 5 6

0% 20% 40% 60% 80% 100%

p1 p2 p3 p6

p4 p5

p1 + p2 + p3 + p4 + p5 + p6 = 1

57

PROBABILITY

Lucky No?

Pollution Level?

Height?

Person 1 2 3 4 5 6 7 8 9 10 11 12 … …

Response 5 73 46 20 4 97 5 38 13 51 … … …

The response / outcome possible is

NOT fixed / restricted.

Theoretically EVERY real no. has a

Non Zero Probability

58

PROBABILITY

Person 1 2 3 4 5 6 7 8 9 10 11 12 … …

Response 5 73 46 20 4 97 5 38 13 51 … … …

0 20 40 60 80 100 120

59

PROBABILITY

Person 1 2 3 4 5 6 7 8 9 10 11 12 … …

Response 5 73 46 20 4 97 5 38 13 51 … … …

0

1

2

3

4

5

6

0 20 40 60 80 100 120

HISTOGRAMS

TELL

PROBABILITY

60

WHAT NEXT ?

61

PROBABILITY

BINOMIAL MULTINOMIAL CONTINUOUS

p p1 p (R1) = f(R1)

q=1-p p2 ... p (R2) = f(R2) …

p+q=1 p1+p2+p3+...+pn= 1 ∑ p (all R) = 162

PROBABILITY

BINOMIAL MULTINOMIAL CONTINUOUS

p p1 p (R1)

q=1-p p2 ... p (R2) …

p+q=1 p1+p2+p3+...+pn= 1 ∑ p (all R) = 1

RARE

AND

COMPLEX

OUT

OF

SCOPE

Formulae 1 Formulae 2

63

HISTOGRAMS

0

2

4

6

8

10

12

14

16

-4 6 15 24 33 42 51 60 69 78 87 96

-8 - 1 1 - 10 10 - 19 19 - 28 28 - 37 37 - 46 46 - 55 55 - 64 64 - 73 73 - 82 82 - 91 91 - 100

64

PROBABILITY (DENSITY) FUNCTION

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0

2

4

6

8

10

12

14

16

-4 6 15 24 33 42 51 60 69 78 87 96

-8 - 1 1 - 10 10 - 19 19 - 28 28 - 37 37 - 46 46 - 55 55 - 64 64 - 73 73 - 82 82 - 91 91 - 100

65

PROBABILITY FUNCTION

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

-8 1 10 19 28 37 46 55 64 73 82 91 100

66

CUMULATIVE DENSITY FUNCTION

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

-4-8 - 1

61 - 10

1510 - 19

2419 - 28

3328 - 37

4237 - 46

5146 - 55

6055 - 64

6964 - 73

7873 - 82

8782 - 91

9691 - 100

67

P.D.F. AS PROBABILITY INDICATOR

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

-4-8 - 1

61 - 10

1510 - 19

2419 - 28

3328 - 37

4237 - 46

5146 - 55

6055 - 64

6964 - 73

7873 - 82

8782 - 91

9691 - 100

More Area = More Probability

68

RANGE PROBABILITY FROM C.D.F.

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

-4-8 - 1

61 - 10

1510 - 19

2419 - 28

3328 - 37

4237 - 46

5146 - 55

6055 - 64

6964 - 73

7873 - 82

8782 - 91

9691 - 100

16%

84%

More Area = More Probability

69


0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

-4-8 - 1

61 - 10

1510 - 19

2419 - 28

3328 - 37

4237 - 46

5146 - 55

6055 - 64

6964 - 73

7873 - 82

8782 - 91

9691 - 100

60% 40%

70

It means 60% of values are likely to be < 51


0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

-4-8 - 1

61 - 10

1510 - 19

2419 - 28

3328 - 37

4237 - 46

5146 - 55

6055 - 64

6964 - 73

7873 - 82

8782 - 91

9691 - 100

87%

13%

71

It means 13% of values are likely to be > 78


0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

-4-8 -1

61 -10

1510 -19

2419 -28

3328 -37

4237 -46

5146 -55

6055 -64

6964 -73

7873 -82

8782 -91

9691 -100

87% - 16%= 71%

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

-4-8 -1

61 -10

1510 -19

2419 -28

3328 -37

4237 -46

5146 -55

6055 -64

6964 -73

7873 -82

8782 -91

9691 -100

87% - 60%

27%

72

It means 27% of values are likely to be between 51 and 78

NORMAL DISTRIBUTION

73

It means:-

Knowing μ and σ of a normally distributed variable, one can determine how much probable is it to lie between a range.

Most common

MEAN AND STANDARD DEVIATION

-0.005

0

0.005

0.01

0.015

0.02

0.025

-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100 110 120

μ = 40σ = 20

74

MEAN AND STANDARD DEVIATION

-0.005

0

0.005

0.01

0.015

0.02

0.025

-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100 110 120

-0.005

0

0.005

0.01

0.015

0.02

0.025

-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100 110 120

μ = 40σ = 20 vs. 30

More σ =>Wider range of values are more probable

σ = 20

σ = 30

75

AREA COVERED BETWEEN STD DEV

-0.005

0

0.005

0.01

0.015

0.02

0.025

-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100 110 120

68.27%

± σ

76

σ σ

-0.005

0

0.005

0.01

0.015

0.02

0.025

-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100 110 120

68.27%


-0.005

0

0.005

0.01

0.015

0.02

0.025

-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100

68.27%

± σ

77

σ σ


± σ

± 2σ

± 3σ± 1.96σ

78

1.96σ

1.96σ


± σ

± 2σ

± 3σ± 2.58σ

79

2.58σ

2.58σ

SAMPLING THEORY

N

n

88-

n


Practically

N small N large

80

SAMPLING INFERENCE - MEAN

N

n

88-

n

μ


μ

μ* μ*

μ* = μ

Sample mean = Population mean

With some error81


82

I sampled

the ABC

tax

payments. I

conclude

that Avg.

tax as 1.2

MRs

How

much

sure are

you

about

your

inference

?

NORMAL DISTRIBUTION

83

It means:-


SAMPLING INFERENCE – MEAN’s DISTRIBU.

N

n

88-

n

μ


μ

μ* μ*

84

Sampling Sampling 1 Sampling 2 Sampling 3 Sampling 4 …

Sample size n (say 100) n n n …

Sample mean μ*

μ1 μ2 μ3 μ4 …

μ* = sample mean is normally distributed N(μ*, σ*)

Even if

population is

not normal

!

NORMAL DISTRIBUTION

85

It means:-


MARGIN OF ERROR

μ* ± 1.96σ*

This value is Margin of error at 95% confidence

σ* has special calculation86

MARGIN OF ERROR - CALCULATION

N

n

88-

n

σ


σ

σσ * = ----

√n

σ √(N-n)

σ * = ---------√n (N-1)

87

MARGIN OF ERROR - CALCULATION

N

n

88-

n

σ


σ

1.96 σE = ---------

√n

1.96σ √(N-n)

E = ---------√n (N-1)

88

μ* ± 1.96σ*

POPULATION STD DEV

N

n

88-

n

σ


σ

Sample σs ≈ Population σ

If n ≥ 100

σs σs

89


90

I sampled

240 firms

so n=240.

Also

240>100,

so ‘σ’ can be taken =

σs

1.96 σE = ---------

√n


91

In my

sample

σs came

out to be

0.8 MRs

σs ≈ σ ≈

0.8

1.96 σE = ---------

√n


92

So I can

say that

Margin of

Error = ± E

= ± 1.96*(0.8) /

(240^0.5)

≈ 0.1

1.96 σE = ---------

√n


93

Thus tax

Avg.

=1.2±E

=1.2±0.1

It lies b/w

1.1 to 1.3

MRs

μ* ± 1.96σ*

Always

?


94

No,

Not always,

but (almost)

95% times

you sample

n=240, it

would lie b/w

1.1 to 1.3

MRs

μ* ± 1.96σ*

So, tax

would lie

b/w 1.1

~ 1.3

95% of

times..


95

Strictly

speaking -

No, Not the

tax.

We are

talking about

a particular

sample

statistic

here…

μ* ± 1.96σ*


96

We are

discussing

the statistic

“AVG.”

So there’s

95%

chance that

sample

AVG. is

1.1~1.3

μ* ± 1.96σ*


97

If you ask me

99% confidence

level, my range

would be

E= ± 2.58*(0.8)

/ (240^0.5)

Avg.

=1.2 ± 0.133

b/w

1.067 to 1.333

μ* ± 2.58σ*

How to

further

reduce

the

margin

of Error

E ?

SAMPLE SIZE - CALCULATION

N

n

88-

n

σ


σ

98

21.96 σ

n = ---------

E

1.96 σE = ---------

√n

1.96σ √(N-n)

E = ---------√n (N-1)

Calculate n using the desired E

SAMPLE SIZE - CALCULATION

N

n

88-

n

σ


σ

21.96 σ

n = ---------E

1.96σ √(N-n)

E = ---------√n (N-1)

N large

99

MARGIN OF ERROR - BINOMIAL

N

n

88-

n

p


p

√pqσ* = -----

√n

√pq (N-n)σ* = ------------

√n (N-1)

100

MARGIN OF ERROR - BINOMIAL

N

n

88-

n

p


p

1.96 √pqE = ----------

√n

1.96 √pq (N-n)

E= ----------------√n (N-1)

101

SAMPLE SIZE - BINOMIAL

N

n

88-

n

p


p

1.96 √pqE = ----------

√n

1.96 √pq (N-n)

E= ----------------√n (N-1)

102

21.96 √pq

n = ---------

E

Calculate n using the desired E


88-

n

p

103

21.96 √pq

n = ---------

E

p & q are expected

ideal population

proportions here

When not known

beforehand, assumed

p = q = 0.5


88-

n

p

104

21.96 √pq

n = ---------

E

Note that in case of

proportions, the E is

also in terms of

proportions, e.g. 5%

So, in the formula it is

entered in %age

e.g. 0.05 for ±5% error


88-

n

p

105

n = 384

So, example

calculation for E=5%

and unknown p & q,

& 95% confidence

21.96 √0.5*0.5

n = --------------

0.05

21.96 √0.5*0.5

n = --------------

E

CAUTION ! APPLICABILITY

106

μ* ± 1.96σ*

106

Sampling Sampling 1 Sampling 2 Sampling 3 Sampling 4 …

Sample size n (say 100) n n n …

Sample mean μ*

μ1 μ2 μ3 μ4 …

μ* = sample mean is normally distributed N(μ*, σ*)

The Formulas of Margin of Error E studied are valid only for the sampling statistic = AVERAGE μ* And not for other sampling statistics like STD DEV. etc.

REPORTING STATISTICS

PARAMETER MENTIONED IN SAMPLING REPORT

CHECK

Basic Assumptions about Population

Sampling Technique used

Sample Size

Sampling Inference (outcome, eg. μ)

Margin of Error (± E)

Confidence Level (eg. 95%)

WHAT NEXT ?

108

HISTOGRAMS – BIN SIZE

39.185 39.147 39.229 39.205 39.246

39.257 39.304 39.278 39 39.17

39.243 39.309 39.264 39.315 39.203

39.287 39.185 39.276 39.232 39.387

39.253 39.251 39.345 39.353 39.255

39.292 39.251 39.177 39.28 39.391

39.245 39.197 39.148 39.293 39.255

39.18 39.072 39.317 39.177 39.119

39.269 39.071 39.236 39.351 39.294

39.271 39.27 39.39 39.12 39.401

38.58 0

38.75 0

38.92 0

39.08 3

39.25 19

39.42 28

0

10

20

30

38.58 38.75 38.92 39.08 39.25 39.42

109


38.58 0

38.75 0

38.92 0

39.08 3

39.25 19

39.42 28

0

5

10

15

20

25

30

38.

58

38.

75

38.

92

39.

08

39.

25

39.

42

38.58 0

38.72 0

38.86 0

39.00 0

39.14 5

39.28 28

39.42 17

38.58 0

38.69 0

38.79 0

38.90 0

39.00 0

39.10 3

39.21 13

39.31 25

39.42 9

38.58 0

38.67 0

38.75 0

38.83 0

38.92 0

39.00 0

39.08 3

39.17 4

39.25 15

39.33 21

39.42 7

110


0

10

20

303

8.5

8

38.

75

38.

92

39.

08

39.

25

39.

42

0

10

20

30

38.

58

38.

72

38.

86

39.

00

39.

14

39.

28

39.

42

05

1015202530

0

5

10

15

20

25

Thumb Rule

Bin Size = √n

111

HISTOGRAMS – FIT NORMAL DISTRIBUTION

-1

0

1

2

3

4

5

0

5

10

15

20

25

30

38.58 38.75 38.92 39.08 39.25 39.42

Find

Mean & Std Dev.

112

WHAT NEXT ?

113

MISUSE OF STATISTICS – CASE STUDY 1

114

I II III IV

x y x y x y x y

10 8.04 10 9.14 10 7.46 8 6.58

8 6.95 8 8.14 8 6.77 8 5.76

13 7.58 13 8.74 13 12.74 8 7.71

9 8.81 9 8.77 9 7.11 8 8.84

11 8.33 11 9.26 11 7.81 8 8.47

14 9.96 14 8.1 14 8.84 8 7.04

6 7.24 6 6.13 6 6.08 8 5.25

4 4.26 4 3.1 4 5.39 19 12.5

12 10.84 12 9.13 12 8.15 8 5.56

7 4.82 7 7.26 7 6.42 8 7.91

5 5.68 5 4.74 5 5.73 8 6.89

Mean 9 7.500909 9 7.500909 9 7.5 9 7.500909

Std Dev 3.316625 2.031568 3.316625 2.031657 3.316625 2.030424 3.316625 2.030579

Covar(x,y) 5.000909091 5 4.997272727 4.999090909


115

y = 0.5001x + 3.0001

0

5

10

15

0 5 10 15

y = 0.5x + 3.0009

0

5

10

15

0 5 10 15

y = 0.4997x + 3.0025

0

5

10

15

0 5 10 15

y = 0.4999x + 3.0017

0

5

10

15

0 5 10 15 20

WHAT NEXT ?

116


117

A B C D E F G H

Jan 8 2 7 9 8 2 8 5

Feb 2 3 7 4 9 1 9 9

Mar 8 3 8 8 1 8 2 3

Apr 9 3 9 3 7 2 9 6

May 3 4 8 7 2 2 9 8

Jun 9 2 8 3 9 2 9 7

Jul 2 3 9 2 3 1 8 2

Aug 8 3 7 2 8 9 3 6

Sep 3 4 8 4 6 2 9 7

Oct 2 2 8 2 7 1 9 7

Nov 9 2 9 5 8 1 8 6

Dec 2 3 7 3 8 2 9 4

A B C D E F G H

Average 5.42 2.83 7.92 4.33 6.33 2.75 7.67 5.83

0

5

10

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

F

B


118

EXCELLENTVERY

GOODGOOD AVERAGE POOR

+ + + 0 - - -

Thanks

119

statistical sampling - icediced.cag.gov.in/wp-content/uploads/2016-17/ntp 01/iced_statistical... ·...

Documents