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- 1. The Kruskal-Wallis H Test • The Kruskal-Wallis H Test is a nonparametric procedure that can be used to compare more than two populations in a completely randomized design. • All n = n1+n2+…+nk measurements are jointly ranked (i.e.treat as one large sample). • We use the sums of the ranks of the k samples to compare the distributions.
- 2. The Kruskal-Wallis H Test Here are the steps in doing the Kruskal- Wallis test: Here are the steps in doing the Kruskal- Wallis test: a. State the null hypothesis. a. State the null hypothesis. Ho: The kk population distributions are Ho: The population distributions are identical. identical. b. State the alternative. b. State the alternative. Ha: at least two of the kk population Ha: at least two of the population distributions differ. distributions differ.
- 3. Example Four groups of students were randomly assigned to be taught with four different techniques, and their achievement test scores were recorded. Are the distributions of test scores the same, or do they differ in location? 1 2 3 4 65 75 59 94 87 69 78 89 73 83 67 80 79 81 62 88
- 4. Teaching Methods 1 65 2 (3) 75 87 (13) 69 3 4 (7) 59(1) 94 (16) (5) 78 (8) 89 (15) 73 79 Ti (6) 83 (12) 67 (4) 80 (10) (9) 81 (11) 62 (2) 88 (14) 31 35 15 55 Rank the 16 H00:the distributions of scores are the same Rank the 16 H : the distributions of scores are the same measurements Ha::the distributions differ in location measurements Ha the distributions differ in location from 1 to 16, from 1 to 16, 12 Ti 2 and calculate and calculate Test statistic: H = ∑ − 3(n + 1) n(n + 1) ni the four rank the four rank sums. sums. 12 312 + 352 + 152 + 552 − 3(17) = 8.96 = 16(17) 4
- 5. Teaching Methods H00:the distributions of scores are the same H : the distributions of scores are the same Ha::the distributions differ in location H the distributions differ in location a 12 Ti 2 Test statistic: H = ∑ − 3(n + 1) n(n + 1) ni 12 312 + 352 + 152 + 552 − 3(17) = 8.96 = 16(17) 4 Rejection region: For aarightRejection region: For righttailed chi-square test with α = .. tailed chi-square test with α = 05 and df = 4-1 =3, reject H00if H 05 and df = 4-1 =3, reject H if H ≥ 7.81. ≥ 7.81. Reject H00.There is sufficient Reject H . There is sufficient evidence to indicate that there evidence to indicate that there is aadifference in test scores for is difference in test scores for the four teaching techniques. the four teaching techniques.
- 6. Activity Four different teaching techniques in Physics were compared with one another. Four classes were randomly assigned to undergo one of the four teaching techniques. After being taught for one month, the students were given in achievement test. The ff. are their achievement scores. Teaching Technique 1 2 3 4 65 75 59 94 87 69 78 89 73 83 67 80 79 81 62 88 81 72 83 95 69 79 76 90 Test the null hypothesis that four teaching techniques in Physics do not differ in effectiveness.

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