Comparing k Populations

1. Comparing k Populations Means – One way Analysis of Variance (ANOVA)

2. The F test – for comparing k means Situation • We have k normal populations • Let miand s denote the mean and standard deviation of population i. • i = 1, 2, 3, … k. • Note: we assume that the standard deviation for each population is the same. s1 = s2 = … = sk = s

3. We want to test against

4. The data • Assume we have collected data from each of th k populations • Let xi1, xi2 , xi3 , … denote the ni observations from population i. • i = 1, 2, 3, … k. Let

5. One possible solution (incorrect) • Choose the populations two at a time • then perform a two sample t test of • Repeat this for every possible pair of populations

6. The flaw with this procedure is that you are performing a collection of tests rather than a single test • If each test is performed with a= 0.05, then the probability that each test makes a type I error is 5% but the probability the group of tests makes a type I error could be considerably higher than 5%. • i.e. Suppose there is no different in the means of the populations. The chance that this procedure could declare a significant difference could be considerably higher than 5%

7. The Bonferoni inequality If N tests are preformed with significance level a. then P[group of N tests makes a type I error] ≤ 1 – (1 – a)N Example: Suppose a. = 0.05, N = 10 then P[group of N tests makes a type I error] ≤ 1 – (0.95)10= 0.41

8. For this reason we are going to consider a single test for testing: against Note: If k = 10, the number of pairs of means (and hence the number of tests that would have to be performed ) is:

9. The F test

10. To test against use the test statistic

11. the statistic is called the Between Sum of Squares and is denoted by SSBetween It measures the variability between samples k – 1 is known as the Between degrees of freedom and is called the Between Mean Square and is denoted by MSBetween

12. the statistic is called the Within Sum of Squares and is denoted by SSWithin is known as the Within degrees of freedom and is called the Within Mean Square and is denoted by MSWithin

13. then

14. The Computing formula for F: Compute 1) 2) 3) 4) 5)

15. Then 1) 2) 3)

16. The critical region for the F test We reject if Fais the critical point under theF distribution with n1 = k - 1degrees of freedom in the numerator and n2 = N – k degrees of freedom in the denominator

17. Example In the following example we are comparing weight gains resulting from the following six diets • Diet 1 - High Protein , Beef • Diet 2 - High Protein , Cereal • Diet 3 - High Protein , Pork • Diet 4 - Low protein , Beef • Diet 5 - Low protein , Cereal • Diet 6 - Low protein , Pork

19. Hence

20. Thus Thus since F > 2.386 we reject H0

21. The ANOVA Table A convenient method for displaying the calculations for the F-test

22. Anova Table

23. The Diet Example

24. Equivalence of the F-test and the t-test when k = 2 the t-test

25. the F-test

27. Hence

28. Using SPSS Note: The use of another statistical package such as Minitab is similar to using SPSS

29. Assume the data is contained in an Excel file

30. Each variable is in a column • Weight gain (wtgn) • diet • Source of protein (Source) • Level of Protein (Level)

31. After starting the SSPS program the following dialogue box appears:

32. If you select Opening an existing file and press OK the following dialogue box appears

33. The following dialogue box appears:

34. If the variable names are in the file ask it to read the names. If you do not specify the Range the program will identify the Range: Once you “click OK”, two windows will appear

35. One that will contain the output:

36. The other containing the data:

37. To perform ANOVA select Analyze->General Linear Model-> Univariate

38. The following dialog box appears

39. Select the dependent variable and the fixed factors Press OK to perform the Analysis

40. The Output

41. Comments • The F-test H0: m1 = m2 = m3 = … = mk against HA: at least one pair of means are different • If H0 is accepted we know that all means are equal (not significantly different) • If H0 is rejected we conclude that at least one pair of means is significantly different. • The F – test gives no information to which pairs of means are different. • One now can use two sample t tests to determine which pairs means are significantly different

42. Fishers LSD (least significant difference) procedure: • Test H0: m1 = m2 = m3 = … = mk against HA: at least one pair of means are different, using the ANOVA F-test • If H0 is accepted we know that all means are equal (not significantly different). Then stop in this case • If H0 is rejected we conclude that at least one pair of means is significantly different, then follow this by • using two sample t tests to determine which pairs means are significantly different

43. Example In the following example we are comparing weight gains resulting from the following six diets • Diet 1 - High Protein , Beef • Diet 2 - High Protein , Cereal • Diet 3 - High Protein , Pork • Diet 4 - Low protein , Beef • Diet 5 - Low protein , Cereal • Diet 6 - Low protein , Pork

45. Hence

46. Thus

47. The ANOVA Table Thus since F > 2.386 we reject H0 Conclusion: There are significant differences amongst the k = 6 means

48. Now we want to perform t tests to compare the k = 6 means with t0.025= 2.005 for 54 d.f.

49. Table of means t test results Critical value t0.025= 2.005 for 54 d.f. t values that are significant are indicated in bold.

50. Conclusions: • There is no significant difference between diet 1 (high protein, pork)and diet 3 (high protein, pork). • There are no significant differences amongst diets 2, 4, 5 and 6. (i. e. high protein, cereal (diet 2) and the low protein diets (diets 4, 5 and 6)). • There are significant differences between diets 1and 3 (high protein, meat) and the other diets (2, 4, 5, and 6). • Major conclusion: High protein diets result in a higher weight gain but only if the source of protein is a meat source.