All non parametric test

All non parametric test

• 1.
  Mann – WhitneyU test
• 2.
  • It isa 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 twosamples 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 Radenotes 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-WhitneyU test Treatment A Treatment B 3 9 4 7 2 5 6 10 2 6 5 8
• 8.
  Why Mann-Whitney Utest • Student t test I s preferred for this data but • Data are not normal • Sample size is small
• 9.
  Obsevations Arranged inorder 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 nodifference 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 aTb 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 tochoose 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 sumtest • 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 isused 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 beused 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 toperform 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 p1Group 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 Ranksum 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 Ranksum 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 sumis chosen: 61 • μr=n1 [n1+ n2 + 1] /2 • μr= 9 [ 9+ 10+1 ] /2 • μr= 9 [20]/2 = 180/2 = 90 • σr =
• 29.
  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.
• 30.
  Krushal –Wallis H-test •It is a generalization of the Mann- Whitney test which is a test for determining whether the two samples selected are taken from the same population. • The p values in both the Krushal –Wallis and the Mann-Whitney tests are equal • It is used for samples to evaluate their degree of association.
• 31.
  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
• 32.
  Characteristics • Test statisticsis 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 df is n-1 • The rank of each sample is calculated • Average rank is applied in case if there is tie
• 33.
  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.
• 34.
  Notation Sampl1 obseravtion 1 XxxXxx 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
• 35.
  Procedure • Define nullH0 and alternative H1 hypothesis. • Rank the sample observations in the combined series. • Compute Ti sum of ranks
• 36.
  • Apply chisquare 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
• 37.
  Use krushal wallisH 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
• 38.
  Step I • Nullhypotheis • H0 : μ a = μ b = μc • Three methods are equally effective • Alternative hypothesis H1= at least two of the μ are different • Three methods are not equally effective • n1+n2+n3=30
• 39.
  Step II Met hod 1 99 64101 85 79 88 97 95 90 100 T1 Rank 24 3 26.5 8 6 11 22 20 15.5 25 161 Met hod 2 83 102 125 61 91 96 94 89 93 75 T2 Rank 7 28 30 2 17 21 19 13 18 4 159 Met hod 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
• 40.
  Compute statistics H
• 41.
  Compute statistics H
• 42.
  • 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
• 43.
  Friedman test • Itis a non parametric test developed and implemented by Milton Friedman. • 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
• 44.
  Friedman test • Thetest 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.
• 45.
  • It isa repeated measures of ANOVA that can be performed on the ordinal data.
• 46.
  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.
• 47.
  Problem ordinal datais 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
• 48.
  Steps • Define nulland alternative hypothesis. • State alpha • Calculate degree of freedom • State decision rule • Calculate the statistic • State result • conclusion
• 49.
  Step 1 • Nullhypothesis • H0 there is no difference between three conditions. • Alternative hypotheis • H1 there is a difference between three conditions
• 50.
  Step 2 • Statealpha • 0.05
• 51.
  Step3 • Degree offreedom’ • df=K-1 • K = no of groups • =3-1=2 • df=2
• 52.
  Step 4 • Decisionrule – chi square table Chi square is greater than 5.991 than you can reject the null hypothesis
• 53.
  Step 5Rank thevalue 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
• 54.
  Step 6 Calculationof statistics
• 55.
  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.