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Kruskal Wallis

Kruskal Wallis. First rank all of the data Sum the ranks of each column. Kruskal Wallis. Find the square of each R Ra 2 = 210.3 Rb 2 =576 Rc 2 =1089 Rd 2 =42.25 H = 12 (Sum [R 2 /(number in column ]) – 3(N +1) N(N +1) H= {12/156(70.1 +192+363+14.08)} -39 H=10.2

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Kruskal Wallis

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  1. Kruskal Wallis • First rank all of the data • Sum the ranks of each column

  2. Kruskal Wallis • Find the square of each R • Ra2=210.3 Rb2=576 Rc2=1089 Rd2=42.25 • H = 12 (Sum [R 2/(number in column]) – 3(N +1) N(N +1) • H= {12/156(70.1+192+363+14.08)} -39 • H=10.2 • Df=(#of columns)-1 • Use the critical value from the chi square table df3=7.82 • If value is greater than critical than there are differences between groups

  3. Kruskal Wallis multiple comparisons • |Ra-Rb| >= Z alpha/k(k-1)sqr.root[N(N+1)/12{1/Na+1/Nb}] • |Ra-Rb| = 9.5 • Z alpha/k(k-1) Is found in the Z distribution • Use table (table Aii, pg320) • C=k(k-1)/2=6 Z=2.638 • 9.5>7.76 • Repeat procedure for all six combinations

  4. Chicken Lab Statistics

  5. Basic stats • Event behaviors- Frequency per unit time • State behaviors – • average duration (use any of the compairisons we already taught you) • Duration percentage (useful for when the duration of observation is not always the same, no good statistic) v

  6. Non Ratio data • Frequency data is not ratio data • Therefore you cannot use any of the tests we have discussed that are based on a normal distribution • Frequency data utilzes a poisson distribution and hypothesis are tested using a chi square statistic.

  7. Goodness of fit test • Different levels of a single variable • Comparing the frequency of behavior of one chick across days X2 = Sum (observed frequency – Expected frequency)2 Expected frequency Expected frequencies = assume equal distribution of frequencies Degrees of freedom - # categories - 1

  8. Goodness of fit • X2 = (39-25)2/25 + (15-25)2/25 + (28-25)2/25 + (18-25)2/25 • = 196/25 + 100/25 + 9/25 +49/25 • =14.96 • Critical values are found in the table on pg 153 of Brown and Downhower • If your value exceeds the table value (7.8) than we can conclude that the frequencies are not evenly distributed across categories

  9. Test of independence • Test based on two categorical variables • Tests that the two categories are not related in any way • Comparing the frequencies of your two different chicks across days X2 = Sum (observed frequency – Expected frequency)2 Expected frequency Degrees of freedom – (# rows – 1)(#columns –1)

  10. Test of independence • Expected frequency = row total x column total N X2 = 9 + 1 + 1 + 9 + 9 + 1 + 1 + 9 = 40 Df = 3 crit value = 7.8 Therefore there is a significant difference in the development of the two groups of chickens

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