Non parametric test

Non parametric tests with examples

1. Non – Parametric Test By Dr. Dinesh kumar Meena, Pharm.D Ph.D Research Scholar Department of Pharmacology, Jawaharlal Institute of Postgraduate Medical Education & Research (JIPMER), Puducherry, India

2. Parameter ( S omething that decides or limits the way in which something can be done ) Introduction Fixed parameter No fixed parameter Parametric statistics Non-Parametric statistics Parametric test Non – Parametric test

3. Parametric test Non-Parametric test Parameter Information about population and parameter is known Information about population and parameter is known Assumptions Assumptions are made No assumptions are made Value for central tendency Mean Median Probability distribution Normal distribution Arbitrary distribution Power More powerful Less powerful Applicable for Variables Variables & Attributes Null hypothesis Made on parameters of population distribution Free from parameters Difference between parametric and non-parametric test

4. Parametric test Non-Parametric test Used for Ratio or Interval data Nominal or Ordinal data Independent measures (2 groups) Independent measure t – test Mann Whitney U test Independent measures (>2 groups) One way independent measure ANOVA Kruskal Wallis test Repeated measures (2 conditions) Matched pair t-test Wilcoxon signed rank test

5. Non-parametric tests Non-parametric test also known as distribution free test. They do not assume that outcome is approximately normally distributed. Situations in which outcome does not follow normal distribution. When the outcome is an ordinal variable or Rank When there are definite outlier When the outcome has clear limit of detection

6. Mann Whitney U test Kruskal Wallis test Wilcoxon signed rank test Commonly used non-parametric test

7. Mann Whitney U test

8. Example 1: With the help of M ann Whitney U test, find if significant difference exist between the scores of treatment A and Treatment B. Treatment A 3, 4, 2, 6, 2, 5 Treatment B 9, 7, 5, 10, 6, 8

9. Steps to be followed in Mann Whitney U Test Generate the hypothesis (Null & Alternative) Arrange data Assign ranks Determine the sum of rank for each sample Calculate U (test) for each sample Take the smaller U(test) value Compare smaller U (test) value with U (Critical) to accept/reject null hypothesis.

10. Step 1. Generate the hypothesis (Null & Alternative ) Null hypothesis = There is no significant difference between Treatment A and Treatment B. Alternative hypothesis = There is significant difference between Treatment A and Treatment B Step 2. Arrange data

11. Step 3. Assign ranks : combine rank for data 1 and data 2 to be assigned in Mann Whitney U test. Data Rank 2 1 2 2 3 3 4 4 5 5 5 6 6 7 6 8 7 9 8 10 9 11 10 12 Data Rank 2 1.5 2 1.5 3 3 4 4 5 5.5 5 5.5 6 7.5 6 7.5 7 9 8 10 9 11 10 12 Data 1 R 1 3 3 4 4 2 1.5 6 7.5 2 1.5 5 5.5 Data 2 R 2 9 11 7 9 5 5.5 10 12 6 7.5 8 10 1 st Arrange all data in ascending order and give ranks 2 nd Wherever any number is repeating more than one time, calculate the average of their ranking 3 rd Arrange rankings as per their data in different data groups

12. Step 4. Calculate the sum of ranks of each sample Σ R1= 23 Σ R1= 55 Step 5. Calculate U (test) for each sample F or treatment A = 23 – 6 (6+1) /2 = 2.0 For treatment B = 55 – 6 (6+1) / 2 = 34.0 Data 1 R 1 3 3 4 4 2 1.5 6 7.5 2 1.5 5 5.5 Data 2 R 2 9 11 7 9 5 5.5 10 12 6 7.5 8 10 U (test) = Rank sum – n (n+1) 2

13. Step 4. Take the smaller U(test) value Smaller U (Test ) value is 2.0 for treatment A Step 5. Compare smaller U (test) value with U (Critical) to accept/reject null hypothesis. U (test ) = 2 U (Critical ) = 6 U (test) < U (critical) Step 6. Interpretation if Null hypothesis rejected and alternative hypothesis is accepted “ There is significant difference between Treatment A and Treatment B” H (test) < H (Critical) H (test) > H (Critical) Reject null hypothesis Accept null hypothesis

14. Example 2. Researchers recruited 19 participants to study the effect of two different analgesics i.e. A & B. he divided patients in two groups . One group consists of 10 patients and was given drug A while another group consists of 9 patients and was given drug B. With the help of Mann Whitney U test, find if significant difference exist between the scores of group 1 and group 2. Group 1: 25, 44, 23, 50, 48, 29, 60, 75, 49, 66 Group 2: 17, 23, 13, 24, 23, 21, 18, 16, 32

15. Kruskal Wallis test

16. Example 1 : 15 patients having complaint of tooth decay were randomly assigned to 3 groups of 5 each. One group ( group A) was given no treatment. Second groups (group B) was given drug A and third group (group C) was given drug B. at the end of six weeks, the extent of tooth decay was evaluated as % of tooth decay. We wish to know if there is a difference among three groups. Control group 87 76 65 81 75 Drug A 63 70 87 92 70 Drug B 45 60 43 56 60

17. Steps to be followed in Kruskal Wallis test Generate the hypothesis (Null & Alternative) Arrange data Assign ranks Determine the sum of rank for each sample Calculate H (test) Calculate degree of freedom Compare H (test) value with H (Critical) to accept/reject null hypothesis.

18. Steps 1: Generate the hypothesis Null hypothesis : There is no difference in extent of tooth decay among three groups. Alternative hypothesis : There is a difference in extent of tooth decay among three groups. Step 2: Arrange data Data 1 Data 2 Data 3 87 63 45 76 70 60 65 87 43 81 92 56 75 70 60

19. Step 3. Assign ranks Step 4.Calculate the sum Σ R1= 53.5 Σ R2 = 51.5 Σ R3 = 15 of rank for each sample Data 1 R1 Data 2 R2 Data 3 R3 87 13.5 63 6 45 2 76 11 70 8.5 60 4.5 65 7 87 13.5 43 1 81 12 92 15 56 3 75 10 70 8.5 60 4.5

20. Step.5: Calculate H (test ) H (test) = 9.39 Step 6. Calculate degree of freedom = K (No. of groups) -1 = 3-1 = 2 H = [ 12 ] [ Σ(ΣR)2 ] – 3(N + 1) -------- ------------- N(N+1) n N = Total No. of participants = 15 n = no. of participant in one group = 5 H = [ 12 ] (53.5)2 (51.5)2 (15)2 - 3(15 + 1) -------- ----------- + ---------- + -------- 15(15+1) 5 5 5

21. Step 7. Compare H (test) value with H (Critical) by using chi-square table to accept/reject null hypothesis.

22. H (Critical ) H (Test) α = 0.05 5.99 9.39 H (test) > H (Critical ) There is a difference in extent of tooth decay among three groups. Step 7. Interpretation H (test) > H (Critical ) H (test) < H (Critical ) Reject the null hypothesis Accept the null hypothesis

23. Wilcoxon signed rank test

24. Wilcoxon signed rank test Also called the  Wilcoxon matched pairs test  or the  Wilcoxon signed rank test . Appropriate for a repeated measure design where the same subjects are evaluated under two different conditions For example, measurements taken every 15 min after dosing for 2 h, each animal will be sampled once for baseline, and 8 times for treatment resulting in 8 comparisons. However, these 9 groups are not independent as they are all obtained from the same set of animals after the same treatment 

25. Example 1 The table shows score of pain relief provided by physiotherapy in 13 patients at 4 weeks and 8 weeks suffering from arthritis. Find if significant difference exist in median of these scores at 4 weeks and 8 weeks ? Participants 1 2 3 4 5 6 7 8 9 10 11 12 13 4 weeks 18.3 13.3 16.5 12.6 9.5 13.6 8.1 8.9 10 8.3 7.9 8.1 13.4 8 weeks 12.7 11.1 15.3 12.7 10.5 15.6 11.2 14.2 16.2 15.5 19.9 20.4 36.8

26. Step 1. Arrange the data Step 2. Calculate each paired difference Step 3. Calculate the absolute difference Participants 1 2 3 4 5 6 7 8 9 10 11 12 13 Difference -5.6 -2.2 -1.2 0.1 1 2 3.1 5.3 6.2 7.2 12 12.3 23.4 Participants 1 2 3 4 5 6 7 8 9 10 11 12 13 Absolute Difference 5.6 2.2 1.2 0.1 1 2 3.1 5.3 6.2 7.2 12 12.3 23.4

27. Step 4. Rank the absolute difference Step 5. Calculate Rank rum of positive value (T+) and negative value (T-) T(-) = 8+5+3 = 16 T (+) = 1+2+4+6+7+9+10+11+12+13 = 75 Step 6. Smaller T value will be W (test) W (test) = 16 Step 7. Find out the W (critical ) value by using wilcoxan signed rank test table W (critical) = 17 Participants 1 2 3 4 5 6 7 8 9 10 11 12 13 Absolute Difference 5.6 2.2 1.2 0.1 1 2 3.1 5.3 6.2 7.2 12 12.3 23.4 Ranking 8 5 3 1 2 4 6 7 9 10 11 12 13

28. Step 8. Interpretation W (test) < W (critical) “There is a significant difference between the median score of pain relief at 4 hour and 8 hour duration” W (test) < W (critical ) W (test) > W (critical) Reject the null hypothesis Accept the null hypothesis If

29. Z test statistics Z test = T – T SE Z test = 75 – 45.5 / 14.30 = 2.06 T = T (max) T = T(max) + T(min)/2 SE = n (n + 1) ( 2n+1) 24 T = 75 T = 75 + 16 / 2 = 45.5 SE = 13 (13 + 1) ( 2x 13+1) = 14.30 24

30. If Z (test ) value is less than 1.96 than accept the null hypothesis at 5 % level of significance is greater than 1.96 at 5 % level of significance than reject the null hypothesis In our study Z (test) is 2.06 which is greater than 1.96 thus null hypothesis will be rejected

31. When parametric tests are not satisfied. When testing the hypothesis, it does not have any distribution. For quick data analysis. When unscaled data is available. Applications of Non-Parametric Test

32. Easily understandable Short calculations Assumption of distribution is not required Applicable to all types of data A dvantages of the non-parametric test

33. Less efficient as compared to parametric test. The results may or may not provide an accurate answer because they are distribution free. Disadvantages of the non-parametric test

34. T hank Y ou