1. Chapter 19 Analysis of Variance (ANOVA)

2. ANOVA How to test a null hypothesis that the means of more than two populations are equal. H 0 :  1 =  2 =  3 H 1 : Not all three populations are equal Test hypothesis with ANOVA procedure (Analysis of variance) ANOVA tests use the F distribution

3. F Distribution F distribution has 2 numbers of degree of freedom (DF) -- numerator and denominator. –EXAMPLE: df = (8,14) Change in numerator df has a greater effect on the shape of the distribution. Properties: –Continuous and skewed to the right –Has 2 df numbers –Nonnegative unites.

4. Finding the F value Example 19.1 SITUATION: Find the F value for 8 degrees of freedom for the numerator, 14 degrees of freedom for the denominator and.05 area in the right tail of the F curve. Consult Table V of Appendix A –corresponding to.05 area. –Locate the numerator on the top row, and the denominator along the left. –Find where they intersect. This will give the critical value of F. Excel: FDIST (x, df1, df2), FINV(prob., df1, df2)

5. Assumptions in ANOVA To test H 0 :  1 =  2 =  3 H 1 : Not all three populations are equal –The following must be true: Population from which samples are drawn are normally distributed Population from which the samples are drawn have the same variance (or standard deviation) The samples are drawn from different populations that are random and independent.

6. How does ANOVA work? The purpose of ANOVA is to test differences in means (for groups or variables) for statistical significance. By partitioning the total variance into the component that is due to true random error (i.e., within-group SSE) and the components that are due to differences between groups (SSG). SSG is then tested for statistical significance, and, if significant, the null hypothesis of no differences between means is rejected. Always right-tailed with the rejection region in the right tail

7. Types of ANOVA One-way ANOVA: Only one factor is considered Two-way ANOVA –Answer the question if the two categorical variables act together to impact the averages for the various groups? –If the two factors do not act together to impact the averages, does at least one of the factors have an impact on the averages for the various groups? N-way ANOVA –Looking for interaction of multiple factors. –Requires more data –Always right-tailed with the rejection region in the right tail

8. ANOVA Notation and Formulas x i = sample mean for group (or treatment) i k = the number of groups (or treatments) n i = sample size of group i x = the average (the grand mean) of all of the observations in all groups n = sum of the k sample sizes = n 1 + n 2 + n 3 …. + n k s i 2 = the sample variance for group (or treatment) i

9. MSG and MSE Sum of squares for groups (SSG) Mean squares for groups (MSG) Sum of squared error (SSE) Mean squared error (MSE)

10. SST and relationship among the SSs Total sum of squares (SST) –SST is the numerator when calculating sample variance –Does not include a group distinction –Dividing SST by its df  sample variance Relationships SSG + SSE = SST GroupsErrorTotal dfk-1n-kn-1

11. ANOVA Tables It is common practice to report results using an ANOVA table: Source Sum of Squares dfMean SquareFP GroupsSSGk-1P-value ErrorSSEn-k TOTALSSTn-1

12. ANOVA process by hand Example 19.2 SITUATION: Soap manufacturer wants to test 3 new machines that should fill a jug. They tested for 5 hours and recorded the number of jugs filled by each per hour: –At the 10% significance level can we reject the null hypothesis that the mean number of jugs filled per hour by each machine is the same? k = 3 n 1 = n 2 =n 3 = 5 continued…. Machine 1Machine 2Machine 3 545349 5653 525747 555150 485954

13. ANOVA process by hand Example 19.2 continued We now need to calculate the ANOVA table For machine 1: Now do the same for machine 2 & 3 Then for 1-3 combined

14. ANOVA process by hand Example 19.2 continued Then we can calculate SSG/E/T: SST =SSG + SSE = 58.5335 + 111.2 = 169.7335 Degrees of freedom –Group df = k-1 = 3-1 = 2 –Error df = n-k = 15-3 = 12 –Total df = n-1 = 15-1 = 14 continued….

15. ANOVA process by hand Example 19.2 continued Now calculate MSG, MSE, and F Determine if the assumption that the three populations have the same population variance are valid. The assumption is reasonable if: Now, look in Table V of Appendix A. Use numerator df=2, denominator df=12 …. continued….

16. Example 19.2 ANOVA tables Replace the calculations results in the table below: Do we reject the null hypothesis? H 0 :  1 =  2 =  3 H 1 : Not all three populations are equal Source Sum of Squares dfMean SquareFP GroupsSSGk-1P-value ErrorSSEn-k TOTALSSTn-1

17. Example 19.2 by Excel Anova: Single Factor SUMMARY GroupsCountSumAverageVariance M1525851.69.3 M2527655.210.2 M3525350.68.3 ANOVA Source of VariationSSdfMSFP-valueF crit Between Groups58.53333229.266673.1582730.0790732.806796 Within Groups111.2129.266667 Total169.733314 M1M2M3 545349 5653 525747 555150 485954

18. Example 19.2 by Minitab One-way ANOVA: P versus M Source DF SS MS F P M 2 58.53 29.27 3.16 0.079 Error 12 111.20 9.27 Total 14 169.73 S = 3.044 R-Sq = 34.49% R-Sq(adj) = 23.57% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -+---------+---------+---------+-------- M1 5 51.600 3.050 (---------*---------) M2 5 55.200 3.194 (---------*---------) M3 5 50.600 2.881 (---------*---------) -+---------+---------+---------+-------- 48.0 51.0 54.0 57.0 Pooled StDev = 3.044

19. Example 19.3 by Excel Anova: Single Factor SUMMARY GroupsCountSumAverageVariance A510821.611.3 B68714.57.5 C69315.513.1 D5110228.5 ANOVA Source of VariationSSdfMSFP-valueF crit Between Groups255.6182385.206068.4177230.0010432.416005 Within Groups182.21810.12222 Total437.818221

20. Example 19.3 by Minitab One-way ANOVA: Cus. versus Teller Source DF SS MS F P Teller 3 255.6 85.2 8.42 0.001 Error 18 182.2 10.1 Total 21 437.8 S = 3.182 R-Sq = 58.38% R-Sq(adj) = 51.45% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ------+---------+---------+---------+--- A 5 21.600 3.362 (--------*-------) B 6 14.500 2.739 (------*-------) C 6 15.500 3.619 (-------*-------) D 5 22.000 2.915 (--------*-------) ------+---------+---------+---------+--- 14.0 17.5 21.0 24.5 Pooled StDev = 3.182

21. Pairwise Comparisons If the result of ANOVA is to reject the null hypothesis, it does not identify which group means are significantly different. Most software packages include this comparison. –Calculate a confidence interval for the differences of each unique pair of means. –Check to see if ZERO falls in the interval, if not then they are significantly different.

22. Example 19.3 by Minitab Fisher 95% Individual Confidence Intervals All Pairwise Comparisons Simultaneous confidence level = 80.96% A subtracted from: Lower Center Upper ---------+---------+---------+---------+ B -11.147 -7.100 -3.053 (------*------) C -10.147 -6.100 -2.053 (------*------) D -3.827 0.400 4.627 (------*------) ---------+---------+---------+---------+ -6.0 0.0 6.0 12.0 B subtracted from: Lower Center Upper ---------+---------+---------+---------+ C -2.859 1.000 4.859 (------*-----) D 3.453 7.500 11.547 (------*-----) ---------+---------+---------+---------+ -6.0 0.0 6.0 12.0 C subtracted from: Lower Center Upper ---------+---------+---------+---------+ D 2.453 6.500 10.547 (------*------) ---------+---------+---------+---------+ -6.0 0.0 6.0 12.0

23. Pairwise Comparisons Fisher’s Least Significant Difference (LSD) Method Null Hypothesis: H 0 :  i =  j Least Significant Difference (LSD) : The pair of means  i and  j is declared significantly different if

24. Example 19.3 with LSD n Teller A521.6 Teller B614.5 Teller C615.5 Teller D522.0

25. Example 19.3 with LSD nini njnj LSD Teller A-B5621.67.14.71 Teller A-C566.14.71 Teller A-D550.44.91 Teller B-C6614.51.04.49 Teller B-D6514.57.54.71 Teller C-D6515.56.54.71

26. Welch’s Approach to Heterogeneity of Variance If Max(s j 2 )/Min(s j 2 )>2, the assumption of equal variance can not be used. Welch’s approach modifies the F-test with the following steps: –For each sample j, calculate w j –Calculate the summation of w from k samples –Calculate the weighted avg. of sample means –Calculate the test statistic F 0 and df