1. Analysis of Variance Chapter 15 - continued

2. 15.5 Two-Factor Analysis of Variance - Example 15.3 –Suppose in Example 15.1, two factors are to be examined: The effects of the marketing strategy on sales. –Emphasis on convenience –Emphasis on quality –Emphasis on price The effects of the selected media on sales. –Advertise on TV –Advertise in newspapers

3. Solution –We may attempt to analyze combinations of levels, one from each factor using one-way ANOVA. –The treatments will be: Treatment 1: Emphasize convenience and advertise in TV Treatment 2: Emphasize convenience and advertise in newspapers ……………………………………………………………………. Treatment 6: Emphasize price and advertise in newspapers Attempting one-way ANOVA

4. Solution –The hypotheses tested are: H 0 :  1 =  2 =  3 =  4 =  5 =  6 H 1 : At least two means differ. Attempting one-way ANOVA

5. City1 City2 City3 City4City5City6 Convnce Convnce Quality Quality Price Price TVPaper TV Paper TV Paper – In each one of six cities sales are recorded for ten weeks. – In each city a different combination of marketing emphasis and media usage is employed. Solution Attempting one-way ANOVA

6. The p-value =.0452. We conclude that there is evidence that differences exist in the mean weekly sales among the six cities. City1 City2 City3 City4City5City6 Convnce Convnce Quality Quality Price Price TVPaper TV Paper TV Paper Solution Xm15-03 Attempting one-way ANOVA

7. These result raises some questions: –Are the differences in sales caused by the different marketing strategies? –Are the differences in sales caused by the different media used for advertising? –Are there combinations of marketing strategy and media that interact to affect the weekly sales? Interesting questions – no answers

8. The current experimental design cannot provide answers to these questions. A new experimental design is needed. Two-way ANOVA (two factors)

9. City 1 sales City3 sales City 5 sales City 2 sales City 4 sales City 6 sales TV Newspapers ConvenienceQualityPrice Are there differences in the mean sales caused by different marketing strategies? Factor A: Marketing strategy Factor B: Advertising media

10. Test whether mean sales of “Convenience”, “Quality”, and “Price” significantly differ from one another. H 0 :  Conv. =  Quality =  Price H 1 : At least two means differ Calculations are based on the sum of square for factor A SS(A) Two-way ANOVA (two factors)

11. City 1 sales City 3 sales City 5 sales City 2 sales City 4 sales City 6 sales Factor A: Marketing strategy Factor B: Advertising media Are there differences in the mean sales caused by different advertising media? TV Newspapers ConvenienceQualityPrice

12. Test whether mean sales of the “TV”, and “Newspapers” significantly differ from one another. H 0 :  TV =  Newspapers H 1 : The means differ Calculations are based on the sum of square for factor B SS(B) Two-way ANOVA (two factors)

13. City 1 sales City 5 sales City 2 sales City 4 sales City 6 sales TV Newspapers ConvenienceQualityPrice Factor A: Marketing strategy Factor B: Advertising media Are there differences in the mean sales caused by interaction between marketing strategy and advertising medium? City 3 sales TV Quality

14. Test whether mean sales of certain cells are different than the level expected. Calculation are based on the sum of square for interaction SS(AB) Two-way ANOVA (two factors)

15. Graphical description of the possible relationships between factors A and B.

16. Levels of factor A 123 Level 1 of factor B Level 2 of factor B 123 123123 Level 1and 2 of factor B Difference between the levels of factor A No difference between the levels of factor B Difference between the levels of factor A, and difference between the levels of factor B; no interaction Levels of factor A No difference between the levels of factor A. Difference between the levels of factor B Interaction M R e s a p n o n s e M R e s a p n o n s e M R e s a p n o n s e M R e s a p n o n s e

17. Sums of squares

18. F tests for the Two-way ANOVA Test for the difference between the levels of the main factors A and B F= MS(A) MSE F= MS(B) MSE Rejection region: F > F ,a-1,n-ab F > F , b-1, n-ab Test for interaction between factors A and B F= MS(AB) MSE Rejection region: F > F  a-1)(b-1),n-ab SS(A)/(a-1) SS(B)/(b-1) SS(AB)/(a-1)(b-1) SSE/(n-ab)

19. Required conditions: 1.The response distributions is normal 2.The treatment variances are equal. 3.The samples are independent.

20. Example 15.3 – continued( Xm15-03) Xm15-03 F tests for the Two-way ANOVA

21. Example 15.3 – continued –Test of the difference in mean sales between the three marketing strategies H 0 :  conv. =  quality =  price H 1 : At least two mean sales are different F tests for the Two-way ANOVA Factor A Marketing strategies

22. Example 15.3 – continued –Test of the difference in mean sales between the three marketing strategies H 0 :  conv. =  quality =  price H 1 : At least two mean sales are different F = MS(Marketing strategy)/MSE = 5.33 F critical = F ,a-1,n-ab = F.05,3-1,60-(3)(2) = 3.17; (p-value =.0077) –At 5% significance level there is evidence to infer that differences in weekly sales exist among the marketing strategies. F tests for the Two-way ANOVA MS(A)  MSE

23. Example 15.3 - continued –Test of the difference in mean sales between the two advertising media H 0 :  TV. =  Nespaper H 1 : The two mean sales differ F tests for the Two-way ANOVA Factor B = Advertising media

24. Example 15.3 - continued –Test of the difference in mean sales between the two advertising media H 0 :  TV. =  Nespaper H 1 : The two mean sales differ F = MS(Media)/MSE = 1.42 F critical = F  a-1,n-ab = F.05,2-1,60-(3)(2) = 4.02 (p-value =.2387) –At 5% significance level there is insufficient evidence to infer that differences in weekly sales exist between the two advertising media. F tests for the Two-way ANOVA MS(B)  MSE

25. Example 15.3 - continued –Test for interaction between factors A and B H 0 :  TV*conv. =  TV*quality =…=  newsp.*price H 1 : At least two means differ F tests for the Two-way ANOVA Interaction AB = Marketing*Media

26. Example 15.3 - continued –Test for interaction between factor A and B H 0 :  TV*conv. =  TV*quality =…=  newsp.*price H 1 : At least two means differ F = MS(Marketing*Media)/MSE =.09 F critical = F  a-1)(b-1),n-ab = F.05,(3-1)(2-1),60-(3)(2) = 3.17 (p-value=.9171) –At 5% significance level there is insufficient evidence to infer that the two factors interact to affect the mean weekly sales. MS(AB)  MSE F tests for the Two-way ANOVA

27. 15.7 Multiple Comparisons When the null hypothesis is rejected, it may be desirable to find which mean(s) is (are) different, and at what ranking order. Three statistical inference procedures, geared at doing this, are presented: –Fisher’s least significant difference (LSD) method –Bonferroni adjustment –Tukey’s multiple comparison method

28. Two means are considered different if the difference between the corresponding sample means is larger than a critical number. Then, the larger sample mean is believed to be associated with a larger population mean. Conditions common to all the methods here: –The ANOVA model is the one way analysis of variance –The conditions required to perform the ANOVA are satisfied. –The experiment is fixed-effect 15.7 Multiple Comparisons

29. Fisher Least Significant Different (LSD) Method This method builds on the equal variances t-test of the difference between two means. The test statistic is improved by using MSE rather than s p 2. We can conclude that  i and  j differ (at  % significance level if |  i -  j | > LSD, where

30. Experimentwise Type I error rate (  E ) (the effective Type I error) The Fisher’s method may result in an increased probability of committing a type I error. The experimentwise Type I error rate is the probability of committing at least one Type I error at significance level of  It is  calculated by  E = 1-(1 –  ) C where C is the number of pairwise comparisons (I.e. C = k(k-1)/2 The Bonferroni adjustment determines the required T ype I error probability per pairwise comparison (  ), to secure a pre- determined overall  E 

31. The procedure: –Compute the number of pairwise comparisons (C) [C=k(k-1)/2], where k is the number of populations. –Set  =  E /C, where  E is the true probability of making at least one Type I error (called experimentwise Type I error). –We can conclude that  i and  j differ (at  /C% significance level if Bonferroni Adjustment

32. Example 15.1 - continued –Rank the effectiveness of the marketing strategies (based on mean weekly sales). –Use the Fisher’s method, and the Bonferroni adjustment method Solution (the Fisher’s method) –The sample mean sales were 577.55, 653.0, 608.65. –Then, Fisher and Bonferroni Methods

33. Solution (the Bonferroni adjustment) –We calculate C=k(k-1)/2 to be 3(2)/2 = 3. –We set  =.05/3 =.0167, thus t.0167  2, 60-3 = 2.467 (Excel). Again, the significant difference is between  1 and  2. Fisher and Bonferroni Methods

34. The test procedure: –Find a critical number  as follows: k = the number of samples =degrees of freedom = n - k n g = number of observations per sample (recall, all the sample sizes are the same)  = significance level q  (k, ) = a critical value obtained from the studentized range table Tukey Multiple Comparisons

35. If the sample sizes are not extremely different, we can use the above procedure with n g calculated as the harmonic mean of the sample sizes. Repeat this procedure for each pair of samples. Rank the means if possible. Select a pair of means. Calculate the difference between the larger and the smaller mean. If there is sufficient evidence to conclude that  max >  min. Tukey Multiple Comparisons

36. City 1 vs. City 2: 653 - 577.55 = 75.45 City 1 vs. City 3: 608.65 - 577.55 = 31.1 City 2 vs. City 3: 653 - 608.65 = 44.35 Example 15.1 - continued We had three populations (three marketing strategies). K = 3, Sample sizes were equal. n 1 = n 2 = n 3 = 20,  = n-k = 60-3 = 57, MSE = 8894. Take q.05 (3,60) from the table. Population Sales - City 1 Sales - City 2 Sales - City 3 Mean 577.55 653 698.65 Tukey Multiple Comparisons

37. Excel – Tukey and Fisher LSD method Xm15-01 Fisher’s LDS Bonferroni adjustments  =.05  =.05/3 =.0167