1 / 10

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

khuyen
Download Presentation

Kruskal Wallis

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related