1. Session 9: k Samples (Zar, Chapter 10)

2. (1) General Setup: Group 1Group 2…Group k x 11 x 21 x k1 x 12 x 22 x k2 x 13 x 23 x k3 x 1n 1 x 2n 2 x kn k H 0 : Group 1 = Group 2 = …=Group k H A : Not all groups = Problem: If we reject H 0, then we have to figure out which are similar & which are different! (Zar, Chapter 11)

3. Comparison of Means: Assumptions: Group 1Group 2…Group k Reformulated:

4. Why not multiple t-tests? Need to t-test: # i.e. 1 vs 2; 1 vs 3; …; 1 vs k k-1 2 vs 3; …; 2 vs k k-2 (k-1) vs k 1 Total=

5. What is the  of the test? Table 10.1 Probability of Committing a Type I Error By Using Multiple t Tests to Seek Diffeences Between All Pairs of k Means. Note: The particular values were derived from a table by Pearson (1942) by assuming equal population variances and large samples.

6. Another Way: 1-Way (One factor) Analysis of Variance Notation:

7. Under H 0, they should be about the same! A Derivation: Start with the overall variance (numerator): Remember

8. Last term = 0

9. So that Remember

10. So under H 0 : Cochran’s theorem: If a sum of squares is the sum of two or more sums of squares and they are independent, then the degrees of Freedom of the sums of squares add: DF=

11. In Summary: Under H 0 : among variance = within variance Test:

12. Problem: Show

13. Another Derivation of the Sums of Squares: Where Sum of Squares Due to the overall Mean!

14. Analysis of Variance with the “Correction Term”

15. Example 10.1:

18. To Model or Not To Model: Model I: Fixed Effects Groups are (1) feeds (2) treatments (3) race (4) concentrations of drug (5) diagnostic groups Model II: Random Effects Groups are randomly selected from the population

19. AmongWithin Under H 0, each individual in the population has the same variance as the population! Among Within Each individual has a smaller variance, the sum of which makes up the population variability!

20. For both models, the analysis is the same: If H 0 is rejected, then (1)Model I (fixed effects): Compare means (Chapter 10) (2)Model II (Random effects): Calculate where

21. Ex. 10.2: H 0 : Phosphorus content equal among technicians. H A : Phosphorus content differs among technicians.  = 0.05

23. Assumptions in the analysis of variance: (1)What if variances are not all equal? Usual Cause: Means and variances/s.d.’s are not independent (More in Chapter 13). Best Test: Plot means and variances/s.d.’s. The Analysis of Variance (AOV) is robust to variance differences Tests for homogeneity of variances: Bartlett’s test -- too sensitive to normality assumption Levene’s test -- not sensitive enough ROT: If no more than a 10 fold difference in variances, then AOV is correct.

24. 0 1 Geometric RelationshipBinomial Relationship y = log(x) y = arcsin

25. Graphical display:

29. Comparison of ranks: The Kruskal-Wallis Test (a)Rank the data without regard to group (smallest to largest). If ties, handle as with the Mann-Whitney. (b)Calculate Summed ranks: R 1, R 2,…,R k (c) Calculate (d)If k ≤ 5 and n i ‘s are small use Table B.13: If H ≥ Table B.13  (n 1, n 2, …), reject (e) If k > 5 or n i ‘s too large, So...

30. (f)If ties, calculate Example 10.10 H 0 : The fly abundance is equal in all three vegetation layers. H A : The fly abundance is not equal in all three vegetation layers.  Test at  = 0.05

31. N = 5 +5+5 = 15 = n 

32. Example 10.11 H 0 : pH same in four ponds H A : Not all the same

34. Test of Medians: (1) Calculate overall median (2) Calculate how many in each group are (a) above median (b) not above median (3) Calculate 2 x k Chi-Square

35. Test of Coded Data or Histograms: (1) If data is coded into “r” codes or (2) If continuous data has been “binned” into “r’ bins (3) Calculate

36. Summary -- Chapter 10 One-way Analysis of Variance (1-way ANOVA) –Means approximately normally distributed –Variances approximately equal -- transformation, if not One-way Rank Test (Kruskal-Wallis) –Rank all data –Calculate ranks of each group and then H~chi-square(k-1) Median Test –Calculate median of all data –2  k contingency table Test of histograms or coded data. –r(bins/codes)  k contingency table