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Solution
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Solution provided by AtoZmath.com
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Kruskal-wallis test calculator
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1. Non parametric test - Kruskal-wallis test for the following data 8,5,7,11,9,6 10,12,11,9,13,12 11,14,10,16,17,12 18,20,16,15,14,22, Significance Level `alpha=0.05` and One-tailed test
2. Non parametric test - Kruskal-wallis test for the following data 24,32,39,34,30,40,36,38 28,31,35,22,38,28,32,28 34,31,28,24,31,26,28,34, Significance Level `alpha=0.05` and One-tailed test
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Example1. Non parametric test - Kruskal-wallis test for the following data 8,5,7,11,9,6 10,12,11,9,13,12 11,14,10,16,17,12 18,20,16,15,14,22, Significance Level `alpha=0.05` and One-tailed testSolution:Step-1: Take the hypothesis Null Hypothesis `H_0` : All groups are equal Alternative Hypothesis `H_1` : Atleast one group is not equal Step-2: Ranking all group values Size in Ascending Order | Rank | Name of related sample A for sample-1 B for sample-2 C for sample-3 D for sample-4 | Rank for A | Rank for B | Rank for C | Rank for D | 5 | 1 | A | 1 | | | | 6 | 2 | A | 2 | | | | 7 | 3 | A | 3 | | | | 8 | 4 | A | 4 | | | | 9 | 5.5 | B | | 5.5 | | | 9 | 5.5 | A | 5.5 | | | | 10 | 7.5 | C | | | 7.5 | | 10 | 7.5 | B | | 7.5 | | | 11 | 10 | C | | | 10 | | 11 | 10 | B | | 10 | | | 11 | 10 | A | 10 | | | | 12 | 13 | C | | | 13 | | 12 | 13 | B | | 13 | | | 12 | 13 | B | | 13 | | | 13 | 15 | B | | 15 | | | 14 | 16.5 | D | | | | 16.5 | 14 | 16.5 | C | | | 16.5 | | 15 | 18 | D | | | | 18 | 16 | 19.5 | D | | | | 19.5 | 16 | 19.5 | C | | | 19.5 | | 17 | 21 | C | | | 21 | | 18 | 22 | D | | | | 22 | 20 | 23 | D | | | | 23 | 22 | 24 | D | | | | 24 | Total | | | 25.5 | 64 | 87.5 | 123 |
The rank total for A is `R_1=25.5` The rank total for B is `R_2=64` The rank total for C is `R_3=87.5` The rank total for D is `R_4=123` Step-3: Compute test statistic `sum R_j^2/n_j=(25.5)^2/6+(64)^2/6+(87.5)^2/6+(123)^2/6=4588.5833` n = total number of samples = 24 `H=12/(n(n+1)) sum R_j^2/n_j - 3(n+1)` `=12/(24(24+1)) (4588.5833) - 3(24+1)` `=12/600 * (4588.5833) - 75` `=16.7717` Step-4: `alpha=0.05` Step-5: Compute the degrees of freedom (df). `df=(4-1)=3` Step-6:The Critical value of chi-square is `chi^2(0.05,3)=7.8147` Since the computed `H`(16.7717) > critical `chi^2`(7.8147) So we reject the null hypothesis (`H_0`).
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