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Statistics Practical B.Sc. II

 B.Sc. II Statistics Practical Paper -II     

In this article we see all the practical problem Of  B.Sc. II  in Practical Paper -II

Practical Number 1.


Padmbhushan Vasantraodada Patil Mahavidyalaya, Kavathe Mahankal

Department of Statistics

Title: Model sampling from Discrete Uniform distribution

                                                                                                               

Questions: 

1. Draw a model sample of size 10 from the following Discrete Uniform Distribution

P(X) = 1/8; x=1,2,3,...,8

            = 0 otherwise.

Calculate A.M. and H. M. of your sample.


2. Draw a model sample of size 15 from the following Discrete Uniform Distribution Taking values 10,15,20,25,30,35,40,45,50,55. Find mean deviation from mode of your sample.


3. Draw a model sample of size 8 from the following Discrete Uniform Distribution

P(X) = 1/13; x=1,2,3,...,13

            = 0 otherwise.

Obtain the quartiles of your sample.


4. Draw a model sample of size 10 from the Discrete Uniform Distribution over the set of first 25 natural numbers. Find C.V. of your sample.                                                                            

5. Draw a random sample of size 11 from discrete uniform distribution taking values 1, 2,     3, 4, 5. find coefficient of M.D. about mean of your sample.

6. Draw a model sample of size 10 from the Discrete Uniform Distribution over values -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. find mean and variance of your sample.

                                                                                                               

ANSWERS Practical NO. 1.

Q. 1. Draw a model sample of size 10 from the following Discrete Uniform Distribution

P(X) = 1/8; x=1,2,3,...,8

            = 0 otherwise.

Calculate A.M. and H. M. of your sample.

Answer : 

first we see the discrete uniform distribution.

If  X ~ DU(1,…., n ), Then the p.m.f. of discrete uniform distribution is given  as,     

P(X=x) = 1/n ; x=1,2,3,...,n

            = 0 otherwise.

and it's mean = ( n+1)/2

& variance = (n²-1) /12.

then  we, write the Aim  of the question.

Aim: To Draw a model sample of size 10 from the following Discrete Uniform Distribution  and Calculate A.M. and H. M. of your sample.

 Statistical Formula:

A.M. = ∑Xi / n

H. M.  = n /  ∑(1/ Xi)

Given,

P(X) = 1/8; x=1,2,3,...,8

            = 0 otherwise.

here n = 8

and P(X) = 0.125

 now we draw random sample so we create observation table.

Observation Table: 

Sr. No.

xi

p(xi)

F(xi)

Rand. No.

xi(Random Sample)

1/xi

1

1

0.125

0.125

0.180575

2

0.5

2

2

0.125

0.250

0.005194

1

1

3

3

0.125

0.375

0.091513

1

1

4

4

0.125

0.500

0.228582

2

0.5

5

5

0.125

0.625

0.437781

4

0.25

6

6

0.125

0.75

0.900141

8

0.125

7

7

0.125

0.875

0.001238

1

1

8

8

0.125

1

0.931718

8

0.125

9

 

0.079296

1

1

10

0.892168

8

0.125

Total

36

5.625

note: here P(X) is given for X= 1,2,3,4,5,6,7,8. and it is P(x) = 1/8  = 0.125 is same for all 8  values.

then F(X) is obtained as sum of probability, we known that F(X) = P(X≤ x )

if x =1 then F( 1) =  P(X≤ 1 )  =  P(X =1 )= 0.125

if x = 2  then F( 2 ) =  P(X≤ 2 )  =  P(X =1 )+ P(X = 2) = 0.125+0.125 = 0.250 

if x = 3  then F( 3 ) =  P(X≤ 3. )  =  P(X =1 )+ P(X = 2)+ P(X =3)  = 0.1225+0.125+0.125 = 0.375

and so on

and the Rand .No. means random number that are obtained as i. in MS- Excel write  =RAND()   and enter we get random number.

ii. in scientific Calculator we first press shift button  then Rand# ( . button on  scientific calculator) then = button.

the random number obtained as a value is slightly greater than random number should be seen in the column of F(X) and corresponding value of x in front of F(X) is selected as random sample. 

i.e. first random number is 0.180 then we see the value slightly grater than 0.180 in column of F(X) that is 0.250  and corresponding value of x in front of F(X) is 2 and it is selected as random sample.

that are in column of X. and the next column is created for calculating H.M 

the observation table is contains mainly five columns that are 

xi

p(xi)

F(xi)

Rand. No.

xi

and the  next column is depended on what you are told to find. 

Calculation:
for calculating A. M. i.e. arithmetic mean we first find  ∑Xi = total of Xi 
and for H.. Harmonic Mean we first find    ∑(1/ Xi) = total of (1/ Xi) both totals are calculating in observation table

A.M. = ∑Xi / n ;  HERE Xi = total of Xi = 36 and  n = total number of observations = 10

A.M. =  36/10

A.M. = 3.6

H. M.  = n /  ∑(1/ Xi)

H. M. = 10 / 5.625

H. m. = 1.7777

Result : Random sample of size 10 is 

Sample

2

1

1

2

4

8

1

8

1

8

A.M. of sample is 3.6

H. M. Of sample is 1.7777

                                                                                                               

Q. 2. Draw a model sample of size 15 from the following Discrete Uniform Distribution Taking values 10,15,20,25,30,35,40,45,50,55. Find mean deviation from mode of your sample.

Answer:

first we see the discrete uniform distribution.

If  X ~ DU(1,…., n ), Then the p.m.f. of discrete uniform distribution is given  as,     

and it's mean = ( n+1)/2

& variance = (n²-1) /12.

here x takes values 10, 15, 20, 25, 30, 35, 40, 45, 50, 55.

then total number of observation is 10, i.e. n=10

therefore the p.m.f. of discrete uniform distribution is 

P(X=x) = 1/10 = 0.1; x=10, 15, 20, 25, 30, 35, 40, 45, 50, 55.

             = 0 otherwise.

then  we, write the Aim  of the question.

Aim: Draw a model sample of size 15 from the following Discrete Uniform Distribution. and Find mean deviation from mode of your sample. 

Statistical Formula:

Mean Deviation from mode  = M.D. = (1/n) ∑| Xi  - Mode|

therefore the p.m.f. of discrete uniform distribution is 

P(X=x) = 1/10 = 0.1; x=10, 15, 20, 25, 30, 35, 40, 45, 50, 55.

             = 0 otherwise.

now we draw random sample so we create observation table.

Observation Table: 

Sr. No.

xi

p(xi)

F(xi)

Rand. No.

Random Sample

| Xi  - Mode|

1

10

0.1

0.1

0.235467

20

15

2

15

0.1

0.2

0.230688

20

15

3

20

0.1

0.3

0.585224

35

0

4

25

0.1

0.4

0.144258

15

20

5

30

0.1

0.5

0.574728

35

0

6

35

0.1

0.6

0.094317

10

25

7

40

0.1

0.7

0.445045

30

5

8

45

0.1

0.8

0.125616

15

20

9

50

0.1

0.9

0.872183

50

15

10

55

0.1

1

0.776099

45

10

11

 

0.024739

10

25

12

0.975556

55

20

13

0.569214

35

0

14

0.621989

40

5

15

0.460018

30

5

Total

180

mode is 35 because it repeated 3 times. 35 has maximum frequency.

Calculation: 

Mean Deviation from mode  = M.D. = (1/n) ∑| Xi  - Mode|

M.D. = (1/15) x 180

M. D. = 12

Result: Random sample of size 15 is 

Random Sample

20

20

35

15

35

10

30

15

50

45

10

55

35

40

30

Mean Deviation from mode  =12.

                                                                                                               

Q. 3. Draw a model sample of size 8 from the following Discrete Uniform Distribution

P(X) = 1/13; x=1,2,3,...,13

            = 0 otherwise.

Obtain the quartiles of your sample.

Answer:

here given that x has discrete uniform distribution

P(X) = 1/13; x=1,2,3,...,13

            = 0 otherwise.

Aim: To Draw a model sample of size 8 and find the quartiles of your sample.

Statistical Formula:

quartiles = Qi = Size of (i(N+1)/4}th observation, for i = 1, 2, 3.

here given that x has discrete uniform distribution

P(X) = 1/13 = 0.0769; x=1,2,3,...,13

            = 0 otherwise.

now we draw random sample so we create observation table.

Observation Table: 

Sr. No.

xi

p(xi)

F(xi)

Rand. No.

Random Sample

ascending

1

1

0.0769

0.0769

0.5956

8

2

2

2

0.0769

0.1538

0.9960

13

5

3

3

0.0769

0.2308

0.8944

12

7

4

4

0.0769

0.3077

0.3768

5

8

5

5

0.0769

0.3846

0.7382

10

10

6

6

0.0769

0.4615

0.7611

10

10

7

7

0.0769

0.5385

0.0930

2

12

8

8

0.0769

0.6154

0.4907

7

13

9

9

0.0769

0.6923

 

 

 

10

10

0.0769

0.7692

 

 

 

11

11

0.0769

0.8462

 

 

 

12

12

0.0769

0.9231

 

 

 

13

13

0.0769

1

 

 

 


Calculation:

Q1 = Size of (N+1)/4 th observation

    =Size of (8+1)/4 th observation

    =Size of (2.25) th observation

    =5+0.25*(7-5)

Q1  = 5.5

Q2=Size of 2* (N+1)/4 th observation    

    =Size of 2*(8+1)/4 th observation

    =Size of (4.5) th observation

    =8+0.5*(10-8)

    Q2 =9

Q3 =Size of 3*(N+1)/4 th observation

    =Size of 3*(8+1)/4 th observation

    =Size of (6.75) th observation

    =10+0.75*(12-10)

    Q3 =11.5


Result:  The Quartiles of  sample is Q1  = 5.5, Q2 =9, Q3 =11.5.

                                                                                                               

Q. 4. Draw a model sample of size 10 from the Discrete Uniform Distribution over the set of first 25 natural numbers. Find C.V. of your sample.     

Answer:

here given that x has discrete uniform distribution

and the Discrete Uniform Distribution over the set of first 25 natural numbers.

i.e. x take values 1, 2, 3, 4, ......., 25

P(X) = 1/25; x=1,2,3,...,25

            = 0 otherwise.

Aim: To Draw a model sample of size 10 and Find C.V. of your sample.

Statistical Formula:

C. V. = {(S.D.) / (Mean)} x 100

here Mean = ∑Xi / n

S.D. = √(Variance)

Variance = {∑X²i / n} - (x̄)²

here given that x has discrete uniform distribution

P(X) = 1/25 = 0.04 ; x=1,2,3,...,25

            = 0 otherwise.

now we draw random sample so we create observation table.

Observation Table: 

Sr.No.

xi

p(xi)

F(xi)

Rand.No.

xi(Random sample)

Xi^2

1

1

0.04

0.04

0.594799

15

225

2

2

0.04

0.08

0.25552

7

49

3

3

0.04

0.12

0.472988

12

144

4

4

0.04

0.16

0.62064

16

256

5

5

0.04

0.2

0.899819

23

529

6

6

0.04

0.24

0.104979

3

9

7

7

0.04

0.28

0.902166

23

529

8

8

0.04

0.32

0.178694

5

25

9

9

0.04

0.36

0.379054

10

100

10

10

0.04

0.4

0.102018

3

9

11

11

0.04

0.44

Total

117

1875

12

12

0.04

0.48

13

13

0.04

0.52

14

14

0.04

0.56

15

15

0.04

0.6

16

16

0.04

0.64

17

17

0.04

0.68

18

18

0.04

0.72

19

19

0.04

0.76

20

20

0.04

0.8

21

21

0.04

0.84

22

22

0.04

0.88

23

23

0.04

0.92

24

24

0.04

0.96

25

25

0.04

1

Calculation:

Mean = ∑Xi / n

    = 117/10

 Mean    =11.7


Variance = {∑X²i / n} - (x̄)²

        = (1875/10) -(11.7)²

      Variance   = 50.61

S.D. = √(Variance) 

S.D. = √(50.61)

        S.D.= 7.1141

C. V. = {(S.D.) / (Mean)} x 100

C. V. = {(7.1141) / (11.7)} x 100

C. V. = 60.8040%

Result: 

C. V. of sample is = 60.8040%

                                                                                                               

Q. 5. Draw a random sample of size 11 from discrete uniform distribution taking values 1, 2, 3, 4, 5. find coefficient of M.D. about mean of your sample.

Answer:

here given that x has discrete uniform distribution

and the Discrete Uniform Distribution over the set of values 1, 2, 3, 4, 5.

i.e. x take values 1, 2, 3, 4, 5.

P(X) = 1/ 5 = 0.2; x=1,2,3,4, 5.

            = 0 otherwise.

Aim: To Draw a model sample of size 11 and find coefficient of M.D. about mean. 

Statistical Formula:

Mean = ∑Xi / n

Mean Deviation about Mean = M.D.(x̄)= 𝟏 /𝒏 ∑ |𝑿𝒊 − x̄|

Coeff. of  Mean Deviation about Mean = M.D.(x̄) / x̄ 

now we draw random sample so we create observation table.

Observation Table: 

Sr.No.

xi

p(xi)

F(xi)

Rand.No.

xi(Random sample)

| Xi  - Mean|

1

1

0.2

0.2

0.357684

2

0.73

2

2

0.2

0.4

0.199077

1

1.73

3

3

0.2

0.6

0.802135

5

2.27

4

4

0.2

0.8

0.331298

2

0.73

5

5

0.2

1

0.485472

3

0.27

0.372287

2

0.73

0.643617

4

1.27

0.322358

2

0.73

0.570065

3

0.27

0.253577

2

0.73

0.790086

4

1.27

Total

30

10.73
Calculation: 

Mean = ∑Xi / n
            =30/11
           =2.7273
Mean Deviation about Mean = M.D.(x̄)= (𝟏 /𝒏) ∑ |𝑿𝒊 − x̄|
        M.D.(x̄) = (1/11) x (10.73)
        M.D.(x̄) =  0.9755
Coeff. of  Mean Deviation about Mean = M.D.(x̄) / x̄ 
 Coeff. of  Mean Deviation about Mean    = 0.9755 / 2.7273
Coeff. of  Mean Deviation about Mean    = 0.3577

Result: 
Sample of size 11is

Random Sample

2

1

5

2

3

2

4

2

3

2

4
Coeff. of  Mean Deviation about Mean    = 0.3577
                                                                                                               

Q.6. Draw a model sample of size 10 from the Discrete Uniform Distribution over values -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. find mean and variance of your sample.

Answer:

here given that x has discrete uniform distribution

and the Discrete Uniform Distribution over the set of values -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

i.e. x take values -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

P(X) = 1/ 21 = 0.0476; x= -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

            = 0 otherwise.

Aim: To Draw a model sample of size 10 and find mean and variance of your sample. 

Statistical Formula:

Mean = ∑Xi / n

Variance = {∑X²i / n} - (x̄)²

now we draw random sample so we create observation table.

Observation Table: 

Sr. No.

xi

p(xi)

F(xi)

Rand.       No.

xi (Random sample)

Xi^2

1

-10

0.0476

0.0476

0.497454

0

0

2

-9

0.0476

0.0952

0.334415

-3

9

3

-8

0.0476

0.1429

0.978027

10

100

4

-7

0.0476

0.1905

0.919356

9

81

5

-6

0.0476

0.2381

0.63215

3

9

6

-5

0.0476

0.2857

0.066648

-9

81

7

-4

0.0476

0.3333

0.373673

-3

9

8

-3

0.0476

0.381

0.235007

-6

36

9

-2

0.0476

0.4286

0.9047

8

64

10

-1

0.0476

0.4762

0.757458

5

25

11

0

0.0476

0.5238

Total

14

414

12

1

0.0476

0.5714

13

2

0.0476

0.619

14

3

0.0476

0.6667

15

4

0.0476

0.7143

16

5

0.0476

0.7619

17

6

0.0476

0.8095

18

7

0.0476

0.8571

19

8

0.0476

0.9048

20

9

0.0476

0.9524

21

10

0.0476

1

Calculation: 

Mean = ∑Xi / n

        =14/10

Mean =1.4

Variance = {∑X²i / n} - (x̄)²

                ={414/10} - (1.4)²

 Variance =39.44

Result: 

Random Sample is 

Random Sample

0

-3

10

9

3

-9

-3

-6

8

5

 The mean of sample= 1.4 and variance of sample= 39.44 

                                                                                                               


Padmbhushan Vasantraodada Patil Mahavidyalaya, Kavathe Mahankal

Department of Statistics

Title: Model sampling from Binomial distribution

                                                                                                               

1. Draw a model sample of size 10 from the binomial Distribution with parameters n = 8 and p = 0.6 also find A.M., G.M. and H. M. of your sample.

2. Draw a model sample of size 13 from the binomial Distribution with parameter n =10 and p = 0.438 also find C.V. of your sample and compare it with theoretical values.

3. Draw a model sample of size 10 from the binomial Distribution with parameter n = 12 and p = 0.25 also find A.M., Median of your sample.

4. Draw a model sample of size 13 from the binomial Distribution with parameter n = 7 and p = 0.336. Find variance of your sample and compare it with theoretical value.

5. Draw a model sample of size 14 from the binomial Distribution with parameter n = 10 and p = 0.62. Find mean and variance of your sample and compare it with theoretical value.

6. If X has binomial distribution such that P{X = 0} = P{X = 5} and n = 5 then obtain a random sample of size 11 from this distribution. Obtain sample median and sample mode.

7. If x has binomial distribution such that E(X) =2 and V(X) =1.5 then obtain a random sample of size 25 from this distribution. Obtain sample mean and sample variance. Compare these with theoretical mean and variance of the distribution. 

                                                                                                               

ANSWERS Practical NO. 1.

Q1. Draw a model sample of size 10 from the binomial Distribution with parameters n = 8 and p = 0.6 also find A.M., G.M. and H. M. of your sample.

Answer : 








Binomial Distribution PMF Calculator Binomial Distribution PMF

Binomial Distribution PMF

Enter the parameters of the binomial distribution:



Enter the value of k to calculate P(X = k):


PMF for X =

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Basic Concepts of Probability and Binomial Distribution , Poisson Distribution.

 Probability:  Basic concepts of Probability:  Probability is a way to measure hoe likely something is to happen. Probability is number between 0 and 1, where probability is 0 means is not happen at all and probability is 1 means it will be definitely happen, e.g. if we tossed coin there is a 50% chance to get head and 50% chance to get tail, it can be represented in probability as 0.5 for each outcome to get head and tail. Probability is used to help us taking decision and predicting the likelihood of the event in many areas, that are science, finance and Statistics.  Now we learn the some basic concepts that used in Probability:  i) Random Experiment OR Trail: A Random Experiment is an process that get one or more possible outcomes. examples of random experiment include tossing a coin, rolling a die, drawing  a card from pack of card etc. using this we specify the possible outcomes known as sample pace.  ii)Outcome: An outcome is a result of experi...
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Statistical Inference: Basic Terms and Definitions.

  📚📖 Statistical Inference: Basic Terms. The theory of estimation is of paramount importance in statistics for several reasons. Firstly, it allows researchers to make informed inferences about population characteristics based on limited sample data. Since it is often impractical or impossible to measure an entire population, estimation provides a framework to generalize findings from a sample to the larger population. By employing various estimation methods, statisticians can estimate population parameters such as means, proportions, and variances, providing valuable insights into the population's characteristics. Second, the theory of estimating aids in quantifying the estimates' inherent uncertainty. Measures like standard errors, confidence intervals, and p-values are included with estimators to provide  an idea of how accurate and reliable the estimates are. The range of possible values for the population characteristics and the degree of confidence attached to those est...
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Index Number

 Index Number      Introduction  We seen in measures of central tendency the data can be reduced to a single figure by calculating an average and two series can be compared by their averages. But the data are homogeneous then the average is meaningful. (Data is homogeneous means data in same type). If the two series of the price of commodity for two years. It is clear that we cannot compare the cost of living for two years by using simple average of the price of the commodities. For that type of problem we need type of average is called Index number. Index number firstly defined or developed to study the effect of price change on the cost of living. But now days the theory of index number is extended to the field of wholesale price, industrial production, agricultural production etc. Index number is like barometers to measure the change in change in economics activities.   An index may be defined as a " specialized  average designed to measure the...
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B. Com. -I Statistics Practical No. 1 Classification, tabulation and frequency distribution –I: Qualitative data.

  Shree GaneshA B. Com. Part – I: Semester – I OE–I    Semester – I (BASIC STATISTICS PRACTICAL-I) Practical: 60 Hrs. Marks: 50 (Credits: 02) Course Outcomes: After completion of this practical course, the student will be able to: i) apply sampling techniques in real life. ii) perform classification and tabulation of primary data. iii) represent the data by means of simple diagrams and graphs. iv) summarize data by computing measures of central tendency.   LIST OF PRACTICALS: 1. Classification, tabulation and frequency distribution –I: Qualitative data. 2. Classification, tabulation and frequency distribution –II : Quantitative data. 3. Diagrammatic representation of data by using Pie Diagram and Bar Diagrams. 4. Graphical representation of data by using Histogram, Frequency Polygon, Frequency Curve and     Locating Modal Value. 5. Graphical representation of data by using Ogive Curves and Locating Quartile Values....
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B. Com. I Practical No. 3 :Diagrammatic representation of data by using Pie Diagram and Bar Diagrams.

Practical No. 3 :Diagrammatic representation of data by using Pie Diagram and Bar Diagrams. Diagrammatic Presentation. We have observed the classification and tabulation method. We use this method to take a lot of information and make it fit into a small table. The reason we do this is to make the information more organized and easier to understand. Tabulation helps us arrange data neatly so that it's not messy and confusing. tabulation is a way to make big files of information look neat and tidy in a table.  but better and beautiful way to represent data using diagrams and graphs. the diagram and graph have some advantages because that used to visualise the data. that helps to understand and give information easily to any common man or any one, following are the some  advantages of diagram and graph.  I. Advantages i. Data Representation: Diagrams and graphs are excellent for presenting data visually, making trends, comparisons, and statistical information easier to...
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Method of Moment & Maximum Likelihood Estimator: Method, Properties and Examples.

 Statistical Inference I: Method Of Moment:   One of the oldest method of finding estimator is Method of Moment, it was discovered by Karl Pearson in 1884.  Method of Moment Estimator Let X1, X2, ........Xn be a random sample from a population with probability density function (pdf) f(x, θ) or probability mass function (pmf) p(x) with parameters θ1, θ2,……..θk. If μ r ' (r-th raw moment about the origin) then μ r ' = ∫ -∞ ∞ x r f(x,θ) dx for r=1,2,3,….k .........Equation i In general, μ 1 ' , μ 2 ' ,…..μ k ' will be functions of parameters θ 1 , θ 2 ,……..θ k . Let X 1 , X 2 ,……X n be the random sample of size n from the population. The method of moments consists of solving "k" equations (in Equation i) for θ 1 , θ 2 ,……..θ k to obtain estimators for the parameters by equating μ 1 ' , μ 2 ' ,…..μ k ' with the corresponding sample moments m 1 ' , m 2 ' ,…..m k ' . Where m r ' = sample m...
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Time Series

 Time series  Introduction:-         We see the many variables are changes over period of time that are population (I.e. population are changes over time means population increase day by day), monthly demand of commodity, food production, agriculture production increases and that can be observed over period of times known as time series. Time series is defined as a set of observation arranged according to time is called time series. Or a time Series is a set of statistical observation arnging chronological order. ( Chronological order means it is arrangements of variable according to time) and it gives information about variable.  Also we draw the graph of time series to see the behaviour of variable over time. It can be used of forecasting. The analysis of time series is helpful to economist, business men, also for scientist etc. Because it used to forecasting the future, observing the past behaviour of that variable or items. Also planning for future...
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Non- Parametric Test: Run Test

Non- Parametric Test  A Non-Parametric tests is a one of the part of Statistical tests that non-parametric test does not assume any particular distribution for analyzing the variable. unlike the parametric test are based on the assumption like normality or other specific distribution  of the variable. Non-parametric test is based on the rank, order, signs, or other non-numerical data. we know both test parametric and non-parametric, but when use particular test? answer is that if the assumption of parametric test are violated such as data is not normally distributed or sample size is small. then we use Non-parametric test they can used to analyse categorical data  or ordinal data and data are obtained form in field like psychology, sociology and biology. For the analysis use the  some non-parametric test that are Wilcoxon signed-ranked test, mann-whiteny U test, sign test, Run test, Kruskal-wallis test. but the non-parametric test have lower statistical power than ...
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