1. Analysis of Variance (ANOVA) ANOVA can be used to test for the equality of three or more population means We want to use the sample results to test the following hypotheses. H 0 :  1  =  2  =  3  = ... =  k  H a : Not all population means are equal Rejecting H 0 means that at least two population means have different values.

2. Assumptions of ANOVA The values from each population are normally distributed. The variance,  2, is the same for all of the populations. The observations must be independent.

3. Impact of Assumptions Given these assumptions, what is the source of variation among the observed variables? Two values could come from different populations, or distributions, which could have different means. Two values could come from the same distribution.

4. ANOVA:Testing for the Equality of k Population Means The ANOVA table Between-treatments estimate of  2 Within-treatments estimate of  2 Comparing these two estimates: The F Test

5. Preliminaries Compute the jth sample mean and variance, j = 1,…,k Compute the overall sample mean Note: the book calls this n T

6. Between-Treatment Estimate of  2 Sum of squares due to treatments A between-treatments estimate of  2, MSTR, can be computed by

7. Within-Treatment Estimate of  2 The within-treatments estimate of  2

8. The ANOVA Table Source of Sum of Degrees of Mean Variation Squares Freedom Squares F Between SSTR k - 1 MSTR MSTR/MSE Within SSE n T - k MSE Total SSTn T - 1

9. Example: Kelli Home Products, Inc. Kelli Home Products, Inc. is considering marketing a long lasting car wax. Three different waxes (Type 1, Type 2, and Type 3) have been developed. In order to test the durability of these waxes, 5 new cars were waxed with Type 1, 5 with Type 2, and 5 with Type 3. Each car was then repeatedly run through an automatic carwash until the wax coating showed signs of deterioration. The number of times each car went through the carwash is shown on the next slide. Kelli Home Products, Inc. must decide which wax to market. Are the three waxes equally effective?

10. Example continued ObservationType 1 Type 2 Type 3 1 48 73 51 2 54 63 63 3 57 66 61 4 54 64 54 5 62 74 56 Sample Mean 55 68 57 Sample Variance 26.0 26.5 24.5

11. The F-Test If the null hypothesis is true and the ANOVA assumptions are valid, the sampling distribution of MSTR/MSE is an F distribution with k – 1 numerator and n T – k denominator df If the means of the k populations are not equal, the value of MSTR/MSE will be inflated because MSTR overestimates  2.

12. Sampling Distribution of MSTR/MSE The figure below shows the rejection region associated with a level of significance equal to  where F  denotes the critical value. Do Not Reject H 0 Reject H 0 MSTR/MSE FF FF Critical Value

13. The F-Test We will reject H 0 if the resulting value of MSTR/MSE appears to be too large to have been selected at random from the appropriate F distribution. Our decision rule becomes

14. Assuming  =.05, F.05 = 3.89 (2 d.f. numerator, 12 d.f. denominator). Therefore, reject H 0 if F > 3.89 Test Statistic: F = MSTR/MSE = 245/25.667 = 9.55 Decision:Reject H 0 There appears to be a difference in the types of wax at a 5% level of significance. Example continued

15. Multiple Comparison Procedures If ANOVA provides statistical evidence to reject the null hypothesis of equal population means, then Fisher’s least significance difference (LSD) procedure can be used to determine where the differences occur.

16. Fisher’s LSD Procedure Pairwise hypothesis tests of the form Test Statistic

17. Fisher’s LSD Procedure Rejection rule:

18. Alternative Fisher’s LSD Procedure Hypotheses Rejection Rule where

19. Using LSD in our Example Assuming  =.05, t.025,12 = 2.179. – Test Statistic – Conclusion The mean number of hours worked at Plant 1 is not equal to the mean number worked at Plant 2.

20. Using LSD in our Example Note that Since 13 > 6.98, and 11 > 6.98, we conclude that types 1 and 2 have different means, and types 2 and 3 have different means, but there is no significant difference between the means of types 1 and 3.

21. Getting it Done in EXCEL Use Tools/Data Analysis/ANOVA: Single Factor Data should be in columns which represent each treatment

22. Experimental Design Statistical studies can be classified as being either experimental or observational. In an experimental study, one or more factors are controlled so that data can be obtained about how the factors influence the variables of interest. In an observational study, no attempt is made to control the factors. Cause-and-effect relationships are easier to establish in experimental studies than in observational studies.