1. 1 HYPOTHESIS TESTING: ABOUT MORE THAN TWO (K) INDEPENDENT POPULATIONS

2. 2 ONE-WAY ANALYSIS OF VARIANCE (ANOVA) Analysis of variance is used for two different purposes: 1.To estimate and test hypotheses about population variances 2.To estimate and test hypotheses about population means We are concerned here with the latter use.

3. 3 H 0 :  1 =  2 =  3 =...=  k H a : Not all the  i are equal. Assumptions: We have K independent samples, one from each of K populations. Each population has a normal distribution with unknown mean  i All of the populations have the same standard deviation  (unknown)

4. 4 Mean T.. T.k T.3 T.2 T.1 Total x 3k x 33 x 32 x 31 x 2k x 23 x 22 x 21 x 1k x 13 x 12 x 11 k321 Treatment

5. 5 The Total Sum of Squares The Within Groups Sum of Squares The Among Groups Sum of Squares SST=SSA+SSW

6. 6 Within groups mean square Among groups mean square Variance Ratio (F)

7. 7 SourceSSdfMSF (VR) Among samplesSSAk-1MSAMSA/MSW Within samplesSSWN-kMSW TotalSSTN-1 ANOVA TABLE

8. 8 Testing for Significant Differences Between Individual Pairs of Means Whenever the analysis of variance leads to a rejection of the null hypothesis of no difference among population means, the question naturally arise regarding just which pairs of means are different. Over the years several procedures for making individual comparisons have been suggested. The oldest procedure, and perhaps the one most widely used in the past, is the Least Significant Difference (LSD) procedure. LSD () LSD (Least Significant Difference ) Tukey Tukey Bonferroni Bonferroni Sidak Sidak Dunnett’s C Dunnett’s C Dunnett’s T3 Dunnett’s T3

9. 9 When sample sizes are equal (n 1 =n 2 =n 3 =...=n k =n) Least Significant Difference (LSD) p<0.05 When sample sizes are not equal (n 1  n 2  n 3 ...  n k ) p<0.05

10. 10 Example In a study of the effect of glucose on insulin release, specimens of pancreatictissue from experimental animals were randomly assigned to be treated with one of five different stimulants. Later, a determination was made on the amount of insulin released. The experimenters wished to know if they could conclude that there is a difference among the five treatments with respect to the mean amount of insulin released. The resulting measurements of amount of insulin released following treatment are displayed in the table. The five sets of observed data constitute five independent samples from the respective populations. Each of the populations from which he samples come is normally distributed with mean,  i, and variances  i 2. Each population has the same variance.

11. 11 Stimulant 12345 1.533.153.898.185.86 1.613.963.685.645.46 3.753.595.707.365.69 2.891.895.625.336.49 3.261.455.798.827.81 1.565.335.269.03 7.107.49 8.98 Total13.0415.6030.0147.6956.81163.15 Mean2.612.605.006.817.105.10

12. 12 H 0 :  1 =  2 =  3 =  4 =  5 H a : Not all the  i are equal. SSA=SST-SSW=162.54282-41.35739=121.18543

13. 13 ANOVA TABLE 121,185430,29619,779,000 41,357271,532 162,54331 Between Groups Within Groups Total Sum of SquaresdfMean SquareFSig. MSW=SSW/27=41.357/27=1.532 MSA=SSA/(5-1)=121.185/4=30.296 F=MSA/MSW=30.296/1.532=19.779 We conclude that not all population means are equal.

14. 14 Since n 1  n 2  n 3  n 4  n 5 ), reject H 0 if HypothesisLSDStatistical Decision H 0 :  1 =  2 0.01<1.538, accept H 0. H 0 :  1 =  3 2.39  1.538, reject H 0. H 0 :  4 =  5 0.29<1.314, accept H 0.

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17. 17 KRUSKAL- WALLIS ONE-WAY ANOVA When the assumptions underlying One-way ANOVA are not met, that is, when the populations from which the samples are drawn are not normally distributed with equal variances, or when the data for analysis consist only of ranks, a nonparametric alternative to the one- way analysis of variance may be used to test the hypothesis of equal location parameters.

18. 18 The application of the test involves the following steps: 1.The n1, n2,..., nk observations from the k groups are combined into a single series of size n and arranged in order of magnitude from smallest to largest. The observations are then replaced by ranks. 2.The ranks assigned to observations in each of the k groups are added separately to give k rank sums. 3.The test statistic is computed. # of groups # of obs. in jth group Sum of ranks in jth group

19. 19 4.When there are three groups and five and fewer observations in each group, the significance of the computed KW is determined by using special tables. When there are more than five observations in one or more of the groups, KW is compared with the tabulated values of  2 with k-1 df.

20. 20 Determing which groups are significantly different Like the one-way ANOVA, the Kruskal-Wallis test is an overall test of significant result, the test does not indicate where the differences are among the groups. To determine which groups are significantly different from one another, it is necessary to undertake multiple comparisons.  p<0.05

21. 21 Example The effect of two drugs on reaction time to a certain stimulus were studied in three groups of experimental animals. Group III served as a control while the animals in group I treated with drug A and those in group II were treated with drug B prior to the application of the stimulus. Table shows the reaction times in seconds of 13 animals. Can we conclude that the three populations represented by the three samples differ with respect to reaction time? H 0 : The population distributions are all identical. H a : At least one of the populations tends to exhibit larger values than at least one of the other populations.

22. 22 Group IIIIII 1782 2075 4094 3183 35 Rank 96.51 1054 1383 116.52 12 R i 552610 KW (5,4,4;0.05) =5.617<KW cal p<0.05, reject H 0.

23. 23 GroupsStatistical Decision 1-24.52.115p<0.05 1-38.52.115p<0.05 2-342.229p<0.05 Multiple Comparisons Table

24. 24 We can use the chi-square test to compare frequencies or proportions in two or more groups. The classification according to two criteria, of a set of entities, can be shown by a table in which the r rows represents the various levels of one criterion of classification and c columns represent the various levels of the second criterion. Such a table is generally called a contingency table. We will be interested in testing the null hypothesis that in the population the two criteria of classification are independent or associated. rxc Chi Square Test

25. 25 Second Criteria First Criteria 12cTotal 1O 11 O 12 O 1c O 1. 2O 21 O 22 O 2c O 2. rO r1 O r2 O rc O r. TotalO.1 O.2 O.c N

26. 26 No more than 20% of the cells should have expected frequencies of less than 5. df = (r-1)(c-1)

27. 27 Example A research team studying the relationship between blood type and severity of a certain condition in a population collected data on 1500 subjects as displayed in the below contingency table. The researchers wished to know if these data were compatible with the hypothesis that severity of condition and blood type are independent. Severity of Condition Blood Type ABAB0Total Absent543211904761320 Mild4422831105 Severe28973175 Total6152421055381500

28. 28 543211904761320 41,1%16,0%6,8%36,1%100,0% 4422831105 41,9%21,0%7,6%29,5%100,0% 28973175 75,0 37,3%12,0%9,3%41,3%100,0% 6152421055381500 41,0%16,1%7,0%35,9%100,0% Severity of condition Count % within severity Count % within severity Count % within severity Count % within severity Absent Mild Severe Total ABABO Blood Type Total 541,2213,092,4473,41320,0 43,116,97,437,7105,0 30,812,15,326,9 615,0242,0105,0538,01500,0 Expected Count 0 cells (,0%) have expected count less than 5. The minimum expected count is 5,25.

29. 29  2 (6,0.05) =12.592>  2 (calculated), accept H 0, p>0.05 We conclude that these data are compatible with the hypothesis that severity of the condition and blood type are independent. H 0 : severity of condition and blood type are independent. H a : severity of condition and blood type are not independent.

30. 30 Assumption is violated We decide to merge two conditions When the sample size is small and assumption about expected frequencies is not met;

31. 31 After combining mild and severe groups in one group, no more than 20% of the cells have expected frequencies less than 5.

32. 32 12,375 a 3,006 200 Chi-Square N of Valid Cases Valuedf Asymp. Sig. (2-sided) 1 cells (12,5%) have expected count less than 5. The minimum expected count is 3,30. a.  2 = 5,118261  2 = 0,204807  2 = 0,067016  2 = 7,038861 Reject H 0. Which type of blood group(s) is/are different from the others ? Exclude Type O from the analysis If null hypothesis is rejected, how can we find the group which is different?

33. 33 p>0.05 Except for blood type O, distribution of tromboembolism is similar within the others.