2. Introduction Analysis of variance compares two or more populations of interval data. Specifically, we are interested in determining whether differences exist between the population means. The procedure works by analyzing the sample variance.
3. One Way Analysis of Variance The analysis of variance is a procedure that tests to determine whether differences exits between two or more population means. To do this, the technique analyzes the sample variances
4. One Way Analysis of Variance Example An apple juice manufacturer is planning to develop a new product -a liquid concentrate. The marketing manager has to decide how to market the new product. Three strategies are considered Emphasize convenience of using the product. Emphasize the quality of the product. Emphasize the product’s low price.
5. One Way Analysis of Variance Example continued An experiment was conducted as follows: In three cities an advertisement campaign was launched . In each city only one of the three characteristics (convenience, quality, and price) was emphasized. The weekly sales were recorded for twenty weeks following the beginning of the campaigns.
6. One Way Analysis of Variance Weekly sales Weekly sales Weekly sales
7. One Way Analysis of Variance Solution The data are interval The problem objective is to compare sales in three cities. We hypothesize that the three population means are equal
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9. 1 2 k First observation, first sample Second observation, second sample Independent samples are drawn from k populations (treatments). X11 x21 . . . Xn1,1 X12 x22 . . . Xn2,2 X1k x2k . . . Xnk,k Sample size Sample mean X is the “response variable”. The variables’ value are called “responses”. Notation
10. Terminology In the context of this problem… Response variable – weekly salesResponses – actual sale valuesExperimental unit – weeks in the three cities when we record sales figures.Factor – the criterion by which we classify the populations (the treatments). In this problems the factor is the marketing strategy. Factor levels – the population (treatment) names. In this problem factor levels are the marketing strategies.
11. Two types of variability are employed when testing for the equality of the population means The rationale of the test statistic
13. 30 25 20 19 12 10 9 7 1 Treatment 3 Treatment 1 Treatment 2 20 16 15 14 11 10 9 The sample means are the same as before, but the larger within-sample variability makes it harder to draw a conclusion about the population means. A small variability within the samples makes it easier to draw a conclusion about the population means. Treatment 1 Treatment 2 Treatment 3
14. The rationale behind the test statistic – I If the null hypothesis is true, we would expect all the sample means to be close to one another (and as a result, close to the grand mean). If the alternative hypothesis is true, at least some of the sample means would differ. Thus, we measure variability between sample means.
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16. There are k treatments The mean of sample j The size of sample j Sum of squares for treatments (SST) Note: When the sample means are close toone another, their distance from the grand mean is small, leading to a small SST. Thus, large SST indicates large variation between sample means, which supports H1.
17. Solution – continuedCalculate SST = 20(577.55 - 613.07)2 + + 20(653.00 - 613.07)2 + + 20(608.65 - 613.07)2 = = 57,512.23 The grand mean is calculated by Sum of squares for treatments (SST)
18. Large variability within the samples weakens the “ability” of the sample means to represent their corresponding population means. Therefore, even though sample means may markedly differ from one another, SST must be judged relative to the “within samples variability”. The rationale behind test statistic – II
19. The variability within samples is measured by adding all the squared distances between observations and their sample means. This sum is called the Sum of Squares for Error SSE In our example this is the sum of all squared differences between sales in city j and the sample mean of city j (over all the three cities). Within samples variability
20. Solution – continuedCalculate SSE Sum of squares for errors (SSE) = (n1 - 1)s12 + (n2 -1)s22 + (n3 -1)s32 = (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24 = 506,983.50
21. To perform the test we need to calculate the mean squaresas follows: The mean sum of squares Calculation of MST - Mean Square for Treatments Calculation of MSE Mean Square for Error
22. Calculation of the test statistic Required Conditions: 1. The populations tested are normally distributed. 2. The variances of all the populations tested are equal. with the following degrees of freedom: v1=k -1 and v2=n-k
23. H0: m1 = m2 = …=mk H1: At least two means differ Test statistic: R.R: F>Fa,k-1,n-k the hypothesis test: And finally The F test rejection region
24. The F test Ho: m1 = m2= m3 H1: At least two means differ Test statistic F= MST/ MSE= 3.23 Since 3.23 > 3.15, there is sufficient evidence to reject Ho in favor of H1,and argue that at least one of the mean sales is different than the others.
26. Fixed effects If all possible levels of a factor are included in our analysis we have a fixed effect ANOVA. The conclusion of a fixed effect ANOVA applies only to the levels studied. Random effects If the levels included in our analysis represent a random sample of all the possible levels, we have a random-effect ANOVA. The conclusion of the random-effect ANOVA applies to all the levels (not only those studied). Models of Fixed and Random Effects
27. In some ANOVA models the test statistic of the fixed effects case may differ from the test statistic of the random effect case. Fixed and random effects - examples Fixed effects - The advertisement Example .All the levels of the marketing strategies were included Random effects - To determine if there is a difference in the production rate of 50 machines, four machines are randomly selected and there production recorded. Models of Fixed and Random Effects.
29. One - way ANOVA Single factor Two - way ANOVA Two factors Response Response Treatment 3 (level 1) Treatment 2 (level 2) Treatment 1 (level 3) Level 3 Level2 Factor A Level 1 Level 1 Level2 Factor B
30. Two-Factor Analysis of Variance - Example Suppose in the Example, 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
31. Attempting one-way ANOVA 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
32. Attempting one-way ANOVA Solution The hypotheses tested are: H0: m1= m2= m3= m4= m5= m6 H1: At least two means differ.
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34. In each one of six cities sales are recorded for ten weeks.
35. In each city a different combination of marketing emphasis and media usage is employed. City1City2City3City4City5City6Convnce Convnce Quality Quality Price Price TV Paper TV Paper TV Paper
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38. The current experimental design cannot provide answers to these questions. A new experimental design is needed. Two-way ANOVA (two factors)
39. Two-way ANOVA (two factors) Factor A: Marketing strategy Factor B: Advertising media Convenience Quality Price City 1 sales City3 sales City 5 sales TV City 2 sales City 4 sales City 6 sales Newspapers Are there differences in the mean sales caused by different marketing strategies?
40. Calculations are based on the sum of square for factor ASS(A) Test whether mean sales of “Convenience”, “Quality”, and “Price” significantly differ from one another. H0: mConv.= mQuality = mPrice H1: At least two means differ Two-way ANOVA (two factors)
41. Two-way ANOVA (two factors) Factor A: Marketing strategy Convenience Quality Price City 1 sales City 3 sales City 5 sales TV Factor B: Advertising media City 2 sales City 4 sales City 6 sales Newspapers Are there differences in the mean sales caused by different advertising media?
42. Calculations are based onthe sum of square for factor BSS(B) Test whether mean sales of the “TV”, and “Newspapers” significantly differ from one another. H0: mTV = mNewspapers H1: The means differ Two-way ANOVA (two factors)
43. Two-way ANOVA (two factors) Quality TV Factor A: Marketing strategy Convenience Quality Price City 1 sales City 5 sales City 3 sales TV Factor B: Advertising media City 2 sales City 4 sales City 6 sales Newspapers Are there differences in the mean sales caused by interaction between marketing strategy and advertising medium?
44. 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)
49. F tests for the Two-way ANOVA Example – continued Test of the difference in mean sales between the three marketing strategies H0: mconv. = mquality = mprice H1: At least two mean sales are different Factor A Marketing strategies
50. F tests for the Two-way ANOVA Example – continued Test of the difference in mean sales between the three marketing strategies H0: mconv. = mquality = mprice H1: At least two mean sales are different F = MS(Marketing strategy)/MSE = 5.33 Fcritical = Fa,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. MS(A)/MSE
51. F tests for the Two-way ANOVA Example - continued Test of the difference in mean sales between the two advertising media H0: mTV. = mNespaper H1: The two mean sales differ Factor B = Advertising media
52. F tests for the Two-way ANOVA Example - continued Test of the difference in mean sales between the two advertising media H0: mTV. = mNespaper H1: The two mean sales differ F = MS(Media)/MSE = 1.42 Fcritical = Fa,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. MS(B)/MSE
53. F tests for the Two-way ANOVA Example - continued Test for interaction between factors A and B H0: mTV*conv. = mTV*quality =…=mnewsp.*price H1: At least two means differ Interaction AB = Marketing*Media
54. F tests for the Two-way ANOVA Example - continued Test for interaction between factor A and B H0: mTV*conv. = mTV*quality =…=mnewsp.*price H1: At least two means differ F = MS(Marketing*Media)/MSE = .09 Fcritical = Fa,(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
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