2. Analysis of Variance Chapter 15

3. 15.1 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.

4. 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 15.2 One Way Analysis of Variance

5. Example 15.1 –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. One Way Analysis of Variance

6. Example 15.1 - 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. One Way Analysis of Variance

7. See file Xm15 -01 Weekly sales

8. Solution –The data are interval –The problem objective is to compare sales in three cities. –We hypothesize that the three population means are equal One Way Analysis of Variance

9. H 0 :  1 =  2 =  3 H 1 : At least two means differ To build the statistic needed to test the hypotheses use the following notation: Solution Defining the Hypotheses

10. Independent samples are drawn from k populations (treatments). 12k X 11 x 21. X n1,1 X 12 x 22. X n2,2 X 1k x 2k. X nk,k Sample size Sample mean First observation, first sample Second observation, second sample X is the “response variable”. The variables’ value are called “responses”. Notation

11. Terminology In the context of this problem… Response variable – weekly sales Responses – actual sale values Experimental 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 trategies.

12. Two types of variability are employed when testing for the equality of the population means The rationale of the test statistic

13. Graphical demonstration: Employing two types of variability

14. 20 25 30 1 7 Treatment 1Treatment 2 Treatment 3 10 12 19 9 Treatment 1Treatment 2Treatment 3 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.

15. 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.

16. The variability between the sample means is measured as the sum of squared distances between each mean and the grand mean. This sum is called the S um of S quares for T reatments SST In our example treatments are represented by the different advertising strategies. Variability between sample means

17. There are k treatments The size of sample j The mean of sample j Sum of squares for treatments (SST) Note: When the sample means are close to one 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 H 1.

18. Solution – continued Calculate 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)

19. Is SST = 57,512.23 large enough to reject H 0 in favor of H 1 ? See next. Sum of squares for treatments (SST)

20. 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

21. The variability within samples is measured by adding all the squared distances between observations and their sample means. This sum is called the S um of S quares for E rror 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

22. Solution – continued Calculate SSE Sum of squares for errors (SSE)  (n 1 - 1)s 1 2 + (n 2 -1)s 2 2 + (n 3 -1)s 3 2 = (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24 = 506,983.50

23. Is SST = 57,512.23 large enough relative to SSE = 506,983.50 to reject the null hypothesis that specifies that all the means are equal? Sum of squares for errors (SSE)

24. mean squares To perform the test we need to calculate the mean squares as follows: The mean sum of squares Calculation of MST - M ean S quare for T reatments Calculation of MSE M ean S quare for E rror

25. Calculation of the test statistic with the following degrees of freedom: v 1 =k -1 and v 2 =n-k Required Conditions: 1. The populations tested are normally distributed. 2. The variances of all the populations tested are equal.

26. And finally the hypothesis test: H 0 :  1 =  2 = …=  k H 1 : At least two means differ Test statistic: R.R: F>F ,k-1,n-k The F test rejection region

27. The F test H o :  1 =  2 =  3 H 1 : At least two means differ Test statistic F= MST  MSE= 3.23 Since 3.23 > 3.15, there is sufficient evidence to reject H o in favor of H 1, and argue that at least one of the mean sales is different than the others.

28. Use Excel to find the p-value – f x Statistical FDIST(3.23,2,57) =.0467 The F test p- value p Value = P(F>3.23) =.0467

29. Excel single factor ANOVA SS(Total) = SST + SSE Xm15-01.xls

30. 15.3 Analysis of Variance Experimental Designs Several elements may distinguish between one experimental design and others. –The number of factors. Each characteristic investigated is called a factor. Each factor has several levels.

31. Factor A Level 1Level2 Level 1 Factor B Level 3 Two - way ANOVA Two factors Level2 One - way ANOVA Single factor Treatment 3 (level 1) Response Treatment 1 (level 3) Treatment 2 (level 2)

32. Groups of matched observations are formed into blocks, in order to remove the effects of “unwanted” variability. By doing so we improve the chances of detecting the variability of interest. Independent samples or blocks

33. 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

34. 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 (15.1): 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.

35. 15.4 Randomized Blocks (Two-way) Analysis of Variance The purpose of designing a randomized block experiment is to reduce the within-treatments variation thus increasing the relative amount of between treatment variation. This helps in detecting differences between the treatment means more easily.

36. Treatment 4 Treatment 3 Treatment 2 Treatment 1 Block 1Block3Block2 Block all the observations with some commonality across treatments Randomized Blocks

37. Block all the observations with some commonality across treatments Randomized Blocks

38. The sum of square total is partitioned into three sources of variation –Treatments –Blocks –Within samples (Error) SS(Total) = SST + SSB + SSE Sum of square for treatmentsSum of square for blocksSum of square for error Recall. For the independent samples design we have: SS(Total) = SST + SSE Partitioning the total variability

39. Calculating the sums of squares Formulai for the calculation of the sums of squares SST = SSB=

40. Calculating the sums of squares Formulai for the calculation of the sums of squares SST = SSB=

41. To perform hypothesis tests for treatments and blocks we need Mean square for treatments Mean square for blocks Mean square for error Mean Squares

42. Test statistics for the randomized block design ANOVA Test statistic for treatments Test statistic for blocks

43. Testing the mean responses for treatments F > F ,k-1,n-k-b+1 Testing the mean response for blocks F> F ,b-1,n-k-b+1 The F test rejection regions

44. Example 15.2 –Are there differences in the effectiveness of cholesterol reduction drugs? –To answer this question the following experiment was organized: 25 groups of men with high cholesterol were matched by age and weight. Each group consisted of 4 men. Each person in a group received a different drug. The cholesterol level reduction in two months was recorded. –Can we infer from the data in Xm15-02 that there are differences in mean cholesterol reduction among the four drugs?Xm15-02 Randomized Blocks ANOVA - Example

45. Solution –Each drug can be considered a treatment. –Each 4 records (per group) can be blocked, because they are matched by age and weight. –This procedure eliminates the variability in cholesterol reduction related to different combinations of age and weight. –This helps detect differences in the mean cholesterol reduction attributed to the different drugs. Randomized Blocks ANOVA - Example

46. BlocksTreatmentsb-1MST / MSEMSB / MSE Conclusion: At 5% significance level there is sufficient evidence to infer that the mean “cholesterol reduction” gained by at least two drugs are different. K-1 Randomized Blocks ANOVA - Example