All non parametric test

all non parametric methods

1. Mann – Whitney U test

2. It is a non-parametric statistical method that compares two groups that are independent of sample data. It is used to test the null hypothesis that the two samples have similar median or whether observations in one sample are likely to have larger values than those in other sample The parametric equivalent of Mann-Whitney U test is t- test of unrelated sample

3. Assumption The two samples are random Two samples are independent of each other Measurement is of ordinal type thus observations are arranged in ranks

4. Steps to perform The null hypothesis and alternative hypothesis are identified. The significance level [alpha] related with null hypothesis is stated. Usually alpha is set at 5% and therefore, the confidence level is 95 % All of the observations are arranged in terms of magnitude.

5. The Ra denotes the sum of the ranks in group a The Rb denotes the sum of ranks in group b U statistics is determined by Verify Ua + Ub = nanb

6. Evaluate U = min [ Ua,Ub ] The obtained value is smaller of the two statistics Using table of critics evaluate the possibility of obtaining value of U or lower The critical value is compared with the obtained value. The results are then interpreted to draw conclusion.

7. Perform the Mann-Whitney U test Treatment A Treatment B 3 9 4 7 2 5 6 10 2 6 5 8

8. Why Mann-Whitney U test Student t test I s preferred for this data but Data are not normal Sample size is small

9. Obsevations Arranged in order 1 2 2 2 3 3 4 4 5 5 6 5 7 6 8 6 9 7 10 8 11 9 12 10

10. There is no difference between the rank of each treatment Rank observation 1.5 2 1.5 2 3 3 4 4 5.5 5 5.5 5 7.5 6 7.5 6 9 7 10 8 11 9 12 10

11. TA Rank a Tb Rank b 3 3 9 11 4 4 7 9 2 1.5 5 5.5 6 7.5 10 12 2 1.5 6 7.5 5 5.5 8 10 Sum of Ra 23 Sum of Rb 55

12. Cross check

13. Ua = 23- 6[6+1]/2 = 23- 42/2 = 23-21 = 2 Ub = 55- 6[ 6+1]/2 = 55-21 = 34

14. We have to choose lowest value hence U= 2 Use u table= critical value N1=6 n2=6 U critics from table = 5 We should get the calculated value as equal to or greater than table value. Here we got lesser value than table value hence null hypothesis is rejected.

15. Wilcoxon Rank sum test It is non-parametric dependent samples t test that can be performed on ranked or ordinal data. Mann-Whitney Wilcoxon test It is used to test null hypothesis It is used to assess whether the distribution of observations obtained between two separate groups on a dependent variable are systematically different from one another.

16. It is used to evaluate the populations that are equally distributed or not A population is set of similar items or data obtained from experiment Rank basically two types of rank given Ra large and Rb small.

17. It can be used in the place of One sample t test Paired t test For ordered categorical data where a numerical scale is in appropriate but where it is possible to rank the observations

18. General way to perform test State the null hypothesis Ho and the alternative hypothesis H1 Define alpha level Define decision rule Calculate Z statistics Calculate results Make conclusion

19. For paired data State the null hypothesis Calculate each paired difference Rank di ignoring signs [ assign rank 1 to the smallest , rank 2 to the next etc ] Designate each rank along with its sign. Based on the sign of di

20. Calculate W+ the sum of the ranks of positive di and W- the sum of the ranks of the negative di. [W+] + [W-] = n [n+1] 2

21. Problem Group A p1 Group B p2 41 66 56 43 64 72 42 62 50 55 70 80 44 74 57 75 63 77 78 N1=9 N2=10

22. Group s = P1 + P2 Group Rank 41 A 1 42 A 2 43 B 3 44 A 4 50 A 5 55 B 6 56 A 7 57 A 8 62 B 9 63 64 A A 10 11 66 B 12 70 A 13 72 B 14

23. Group s = P1 + P2 Group Rank 74 B 15 75 B 16 77 B 17 78 B 18 80 B 19

24. Group A Rank sum Group s = P1 + P2 Group Rank 41 A 1 42 A 2 44 A 4 50 A 5 56 A 7 57 A 8 63 64 A A 10 11 70 A 13 SUM OF Rank a 61

25. Group B Rank sum Group s = P1 + P2 Group Rank 43 B 3 55 B 6 62 B 9 66 B 12 72 B 14 74 B 15 75 B 16 77 B 17 78 B 18 80 B 19 Sum of Group B 129

26. Small Rank sum is chosen: 61 μ r=n1 [n1+ n2 + 1] /2 μ r= 9 [ 9+ 10+1 ] /2 μ r= 9 [20]/2 = 180/2 = 90 σ r =

27. Krushal –Wallis H-test H test Non parametric statistical procedure used for comparing more than two independent sample Parametric equivalent to this test is one way ANOVA H test is for non-normally distributed data.

28. Krushal –Wallis H-test It is a generalization of the M ann- W hitney test which is a test for determining whether the two samples selected are taken from the same population. The p values in both the K rushal –Wallis and the Mann-Whitney tests are equal It is used for samples to evaluate their degree of association.

29. Description of sample 3 independently drawn sample. Data in each sample should be more than 5 Both distribution and population have same shape Data must be ranked Samples must be independent K independent sample k> 3 or K=3

30. Characteristics Test statistics is applied when data is not normally distributed Test uses k samples of data. Test can be used for one nominal and one ranked variable Significance level is denoted with α Data is ranked and d f is n-1 The rank of each sample is calculated Average rank is applied in case if there is tie

31. Problem Null hypothesis K independent sample drawn from population which are identically distributed. Alternative hypothesis K independent sample drawn from population which are not identically distributed.

32. Notation Sampl1 obseravtion 1 Xxx Xxx Xxx Xxx Xxx 2 Xxx Xxx Xxx Xxx Xxx Xxx xxx 3 Xxx Xxx Xxx xxx Xxx Xxx Xxx Xxx K Xxx Xxx Xxx Xxx Xxx Xxx xxx Observation more than five K =3 or K>3

33. Procedure Define null H0 and alternative H1 hypothesis. Rank the sample observations in the combined series. Compute Ti sum of ranks

34. Apply chi square variate with K-1 degree of freedom K = number of sample Conclusion Take the table value from Chi 2 [k-1][ α ] If calculated H value > Chi 2 [k-1][ α ] We reject H0

35. Use krushal wallis H test at 5 % level of significance if three methods are equally effective Method 1 99 64 101 85 79 88 97 95 90 100 Method 2 83 102 125 61 91 96 94 89 93 75 Method 3 89 98 56 105 87 90 87 101 76 89

36. Step I Null hypotheis H0 : μ a = μ b = μ c Three methods are equally effective Alternative hypothesis H1= at least two of the μ are different Three methods are not equally effective n 1+n2+n3=30

37. Step II Method 1 99 64 101 85 79 88 97 95 90 100 T1 Rank 24 3 26.5 8 6 11 22 20 15.5 25 161 Method 2 83 102 125 61 91 96 94 89 93 75 T2 Rank 7 28 30 2 17 21 19 13 18 4 159 Method 3 89 98 56 105 87 90 87 101 76 89 T3 Rank 13 23 1 29 9.5 15.5 9.5 26.5 5 13 145 n1 10 n2 10 n3 10

38. Compute statistics H

39. Compute statistics H

40. Df = K-1 =3-1=2 Table value = 5.99 Calculated value is 0.196 Since the calculated H value 0.196 < 5.99 We fail to reject H0 All the teaching methods are equal

41. Friedman test It is a non parametric test developed and implemented by M ilton F riedman. It is used for finding differences in treatments across multiple attempts by comparing three or more dependent samples. It is an alternative to ANOVA when the assumption of normality is not met

42. Friedman test The test is calculated using ranks of data instead of unprocessed data It is used to test for differences between groups when the dependent variable being measured is ordinal. It can also be used for continuous data that has marked as deviations from normality with repeated measures.

43. It is a repeated measures of ANOVA that can be performed on the ordinal data.

44. Descriptions and requirements Dependent variable should be measured at the ordinal or continuous level Data comes from a single group measured on at least three different occasions. Random sampling method must be used All of the pairs are independent. Observations are ranked within blocks with no ties Samples need not be normally distributed.

45. Problem ordinal data is given .is there a difference between weeks 1,2,3 using alpha as 0.05 week1 Week 2 Week 3 27 20 34 2 8 31 4 14 3 18 36 23 7 21 30 9 22 6

46. Steps Define null and alternative hypothesis. State alpha Calculate degree of freedom State decision rule Calculate the statistic State result conclusion

47. Step 1 Null hypothesis H0 there is no difference between three conditions. Alternative hypotheis H1 there is a difference between three conditions

48. Step 2 State alpha 0.05

49. Step3 Degree of freedom’ df =K-1 K = no of groups =3-1=2 df =2

50. Step 4 Decision rule – chi square table Chi square is greater than 5.991 than you can reject the null hypothesis

51. Step 5Rank the value Week 1 Week 2 Week 3 2 1 3 1 2 3 2 3 1 1 3 2 1 2 3 2 3 1 R=9 R=14 R=13

52. Step 6 Calculation of statistics

53. Step 7.state result If chi square is greater than 5.991 than reject the null hypothesis. Calculated chi square value is 2.33 Calculated value is lesser than table value hence fail to reject null hypotheis . Hence there is no difference among the three group.

54. Thank you