1. © The McGraw-Hill Companies, Inc., 2000 12-1 Chapter 12 Analysis of Variance (ANOVA)

2. © The McGraw-Hill Companies, Inc., 2000 12-1 Objectives Use ANOVA technique to determine differences among three or more means when data are ratio or interval.

3. © The McGraw-Hill Companies, Inc., 2000 12-3 Analysis of Variance (ANOVA) analysis of variance (ANOVA) When an F test is used to test a hypothesis concerning the means of three or more populations, the technique is called analysis of variance (ANOVA).

4. © The McGraw-Hill Companies, Inc., 2000 12-4 Assumptions for the F Test for Comparing Three or More Means The populations from which the samples were obtained must be normally or approximately normally distributed. The samples must be independent of each other and randomly sampled. The variances of the populations must be equal.

5. © The McGraw-Hill Companies, Inc., 2000 12-5 Although means are being compared in this F test, variances are used in the test instead of the means. Two different estimates of the population variance are made. Analysis of Variance

6. © The McGraw-Hill Companies, Inc., 2000 12-6 Between-group variance Between-group variance - this involves computing the variance by using the means of the groups or between the groups. Within-group variance Within-group variance - this involves computing the variance by using all the data and is not affected by differences in the means. Analysis of Variance

7. © The McGraw-Hill Companies, Inc., 2000 12-7 The following hypotheses should be used when testing for the difference between three or more means. H 0 :    =   =   = … =  k H 1 : At least one mean is different from the others. Analysis of Variance

8. © The McGraw-Hill Companies, Inc., 2000 12-8 d.f.N. = k – 1, where k is the number of groups. d.f.D. = N – k, where N is the sum of the sample sizes of the groups. Note: The formulas for this test are tedious to work through, so examples will be done in MINITAB. See text for formulas. Analysis of Variance

9. © The McGraw-Hill Companies, Inc., 2000 12-9 A marketing specialist wishes to see whether there is a difference in the average time a customer has to wait in a checkout line in three large self-service department stores. The times (in minutes) are shown on the next slide. Is there a significant difference in the mean waiting times of customers for each store using  = 0.05? Analysis of Variance- Analysis of Variance - Example

10. © The McGraw-Hill Companies, Inc., 2000 12-10 Analysis of Variance- Analysis of Variance - Example

11. © The McGraw-Hill Companies, Inc., 2000 12-11 Step 1: Step 1: State the hypotheses and identify the claim. H 0 :    =   =   H 1 : At least one mean is different from the others (claim). Analysis of Variance- Analysis of Variance - Example

12. © The McGraw-Hill Companies, Inc., 2000 12-12 Step 2: Step 2: Find the critical value. Since k = 3, N = 18, and  = 0.05, d.f.N. = k – 1 = 3 – 1= 2, d.f.D. = N – k = 18 – 3 = 15. The critical value is 3.68. Step 3: Step 3: Compute the test value. From the SPSS output, F = 2.70. (See your text for computations). Analysis of Variance- Analysis of Variance - Example

13. © The McGraw-Hill Companies, Inc., 2000 12-13 Step 4: Step 4: Make a decision. Since 2.70 < 3.68, the decision is not to reject the null hypothesis. Step 5: Step 5: Summarize the results. There is not enough evidence to support the claim that there is a difference among the means. The ANOVA summary table is given on the following slides. Analysis of Variance- Analysis of Variance - Example

14. © The McGraw-Hill Companies, Inc., 2000 12-14 Analysis of Variance- Analysis of Variance - Example Before analyzing the ANOVA table check that there is “homogeneity of variance”. Below are the results. Since Sig. (P) is greater than 0.05 the variances are equal (i.e., homogeneous). Test of Homogeneity of VariancesTime Levenedf1df2Sig. Statistic 0.1872150.832

15. © The McGraw-Hill Companies, Inc., 2000 12-14 Analysis of Variance- Analysis of Variance - Example Below is the SPSS output from Compare Means | One-way ANOVA… Since P = 0.100 (Sig.) is > 0.05 the null hypothesis is not rejected, thus all group means are equal. ANOVATimeSum of dfMeanFSig. SquaresSquare Between Groups25.000212.5002.6980.100 Within Groups69.500154.633 Total94.50017