The document discusses one-way analysis of variance (ANOVA), which compares the means of three or more populations. It provides an example where sales data from three marketing strategies are analyzed using ANOVA. The null hypothesis is that the population means are equal, and it is rejected since the F-statistic is greater than the critical value, indicating at least one mean is significantly different. Post-hoc comparisons using the Bonferroni method find that Strategy 2 (emphasizing quality) has significantly higher sales than Strategy 1 (emphasizing convenience).

1. Analysis of Variance Chapter 12

6. Idea Behind ANOVA Graphical demonstration: Employing two types of variability

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

11. One Way Analysis of Variance Weekly sales Weekly sales Weekly sales

14. Notation Independent samples are drawn from k populations (treatment groups). 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 X is the “response variable”. The variables’ value are called “responses”. 1 2 k First observation, first sample Second observation, second sample

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

19. Sum of squares for treatment groups (SSG) There are k treatments The size of sample j The mean of sample j Note: When the sample means are close to one another, their distance from the grand mean is small, leading to a small SSG. Thus, large SSG indicates large variation between sample means, which supports H 1 .

26. The mean sum of squares To perform the test we need to calculate the mean squares as follows: Calculation of MSG - M ean S quare for treatment Groups Calculation of MSE M ean S quare for E rror

27. 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.

28. The F test rejection region 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

29. The F test H o :  1 =  2 =  3 H 1 : At least two means differ Test statistic F= MSG  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.

31. Excel single factor ANOVA SS(Total) = SSG + SSE