1. Inferential statistic –Non Parametric test BY- DR HARSHAL P. BHUMBAR

2. DEFINITION A statistical method wherein the data is not required to fit a normal distribution. Nonparametric statistics uses data that is often ordinal, meaning it does not rely on numbers, but rather a ranking or order of sorts.

3. S. No. Parametric Test Non- parametric Test 1Study of two independent samples Student t testWilcoxon- Mann-Whitney test 2Study of two matched samples Paired t testWilcoxon signed rank test 3Study of two or more independent samples One way ANOVA Kruskal-Wallis test 4Study of two or more matched samples Two way ANOVA Friedman test

4. Difference between Parametric & Non-parametric test Parametric testNon parametric test 1.Used for ratio or interval dataFor ordinal or nominal data 2.Used for Normal distributionAny distribution 3.Mean is usual central measureMedian is usual central measure 4.Information about population is completely known No information available 5.Specific assumptions made regarding population Assumption free test

5. 6.Null hypothesis based on parameters of population Null hypothesis free of parameters 7.Applicable only for variableFor both variable & attribute 8.More efficientLess efficient 9.More powerful if existsLess powerful

6. Wilcoxon-Mann-Whitney test Example - The effectiveness of advertising for two rival products (Brand X and Brand Y) was compared. Market research at a local shopping centre was carried out, with the participants being shown adverts for two rival brands of coffee, which they then rated on the overall likelihood of them buying the product (out of 10, with 10 being "definitely going to buy the product"). Half of the participants gave ratings for one of the products, the other half gave ratings for the other product.

7. Brand XBrand Y participantratingparticipantrating 1319 2427 3235 46410 5256 6568

8. We have two conditions, with each participant taking part in only one of the conditions. The data are ratings (ordinal data), and hence a nonparametric test is appropriate - the Mann-Whitney U test (the non- parametric counterpart of an independent measures t-test).

9. STEP ONE Rank all scores together, ignoring which group they belong to.

10. Brand XBrand Y Participa nt RatingRankParticipa nt RatingRank 1331911 244279 321.5355.5 467.541012 521.5567.5 655.56810

11. STEP TWO: Add up the ranks for Brand X, to get T1 Therefore, T1 = 3 + 4 + 1.5 + 7.5 + 1.5 + 5.5 = 23 STEP THREE: Add up the ranks for Brand Y, to get T2 Therefore, T2 = 11 + 9 + 5.5 + 12 + 7.5 + 10 = 55

12. STEP FOUR: Select the larger rank. In this case it’s T2 STEP FIVE: Calculate n1, n2 and nx These are the number of participants in each group, and the number of people in the group that gave the larger rank total. Therefore n1 = 6 n2 = 6 nx = 6

13. STEP SIX: Find U (Note: Tx is the larger rank total) U = n1*n2 + nx *(nx+1)/2 – Tx U = 6*6+6*(6+1)/2- 55 U = 2 STEP SEVEN : Use a table of critical U values for the Mann-Whitney U Test

14. For n1 = 6 and n2=6, the critical value of U is 5 at the 0.05 significance level. For n1 = 6 and n2=6, the critical value of U is 2 at the 0.01 significance level. STEP EIGHT : To be significant, our obtained U has to be equal to or LESS than this critical value. Our obtained U = 2

15. Our obtained U = 2 The critical value for a two tailed test at.05 significance level = 5 The critical value for a two tailed test at.01 significance level = 2 So, our obtained U is less than the critical value of U for a 0.05 significance level. It is also equal to the critical value of U for a 0.01 significance level.

16. But what does this mean? We can say that there is a highly significant difference (p<.01) between the ratings given to each brand in terms of the likelihood of buying the product.

17. Wilcoxon sign rank test Example - To know effectiveness of new drug designed to reduce repetitive behaviors in children affected with autism. A total of 8 children with autism enroll in study and amount of time that each is engaged in repetitive behaviour during three hour observation periods are measured both before treatment and then again after taking new medication for a period of 1 week. The data shown below -.

18. childBefore treatmentAfter 1 week treatment 18575 27050 34050 46540 58020 67565 75540 82025

19. First we compute difference score for each child childBefore treatment After 1 week treatment Difference (before- after) 1857510 2705020 34050- 10 4654025 5802060 6756510 7554015 82025- 5

20. Next step to rank difference scores. First order absolute values of difference scores and assigned rank from 1 to lowest and n to highest for difference scores and assigned mean rank when there are ties in absolute values of different scores.

21. Observed difference Ordered absolute value of difference rank 10- 51 20103 - 10 3 25103 60155 10206 15257 - 5608

22. Final step is to attach signs ( +, - ) of observed difference to each rank shown below. rankSigned rank 1 33 3- 3 33 55 66 77 88

23. Test statistics for Wilcoxon sign rank test is given by W. W+ ( sum of positive ranks ) W- ( sum of negative ranks ) If Ho – true then W+ = W- If research hypothesis true then W+ > W- In our example, W+ = 32 and W- = 4 Recall sum of ranks always equal to n(n+1)/2, In our assignment, ( 8*9)/2 = 36, Test statistics is W = 4,

24. If the absolute value of W less than or equal to critical value we reject null hypothesis and if observed value of W exceeds critical value we don’t reject null hypothesis.

25. Friedman test Example- Hall et all compared three methods of determining serum amylase values in patients with pancreatitis. The results are shown in following table.we wish to know whether these data indicates a difference among three methods. ( given @=0.05 )

26. Specim en Methods of determination ABC 1400032106120 2160010402410 316006472210 412005702060 58404451400 6352156249 7224155224 820099208 918470227 Following table shows serum amylase values ( enzyme units per 100 ml of serum ) in patients with pancreatitis.

27. Hypothesis –  Ho – MA = MB= MC  H1 - at least one equality is violated. Test statistics, b= 9 & k = 3 After converting original observations to ranks, we have

28. SpecimenMethods of determination ABC 1213 2213 3213 4213 5213 6312 72.51 8213 9213

29. So R A = 19.5, R B = 9, R c=25.5 So by equation we have, k= 3 & b=9 Friedman test statistics X r² = 12/bk(k+1) ∑R j²- 3b (k+1) X r²= 15.5 From table X ²(1-ά, k-1), where ά=0.05, k=3 X²(0.95, 2) = 5.991 Since 15.5> 5.991, we reject null hypothesis Conclusion- Enough evidence to support the claim that three methods do not yield identical results.

30. Kruskal Wallis test EXAMPLE- Does it make any difference to students comprehension of statistics whether the lectures are given in English, Serbo - croat or Cantonese?  Group A – lectures in English  Group B – lectures in Serbo-croat  Group C– Lectures in Cantonese DV : Students rating of lectures intelligibility on 100 point scale

31. English (Raw score) English (Rank) Serbo- croat (Raw score) Serbo- croat (Rank) Cantones e (Raw score) Cantones e (Rank) 203.5257.5191.5 2793310203.5 191.53511257.5 2363612225

32. Step 1 - Rank the scores ignoring which group they belong to. Lowest scores get lowest rank. Tied scores get average rank Step 2 – Tc - Total of rank for each group Tc1 – 20 Tc2 – 40.5 Tc3 – 17.5

33. Step 3 – Find H Where N- Total number of subjects Tc – Rank total for each group nc – Number of subjects in each group

34.  Hypothesis – H o – M A = M B= M C H 1 - At least one equality is violated. Test statistics, b= 9 & k = 3 After converting original observations to ranks, we have

35. ∑ T c ² / n c = 20²/4 + 40.5²/4 +17.5²/4 = 586.62 H = 6.12 Step 4 – Df are number of groups minus one Step 5 – For 2 Df a chi square of 5.99 has a p = 0.05 occurring by chance But our H is > 5.99 even so less likely occur by chance H Is 6.12, p< 0.05  Conclusion – Three groups differ significantly. Language in which statistics is taught does make a difference to students intelligibility.

36. Advantages Simple & easy to understand. Not involve complicated sampling theory. No assumption made regarding parent population. Disadvantages Applied for only nominal or ordinal scale. They uses less information than parametric test. They are not so efficient as of parametric test.

37. References Rao VK. Biostatistics: A manual of statistical method for use in health nutrition and anthropometry. 2 nd ed. New Delhi: Jaypee Brothers; 2007. Armitage P, Berry G. Statistical Method in Medical Research. 3 rd ed. London: Oxford Blackwell scientific publication; 1994 Swinskow TV, Campbell MJ. Statistics at Square One. 10 th ed. London: BMJ Books; 2002.

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