1. Copyright © Cengage Learning. All rights reserved. 12 Analysis of Variance

2. Copyright © Cengage Learning. All rights reserved. 12.3 Applications of Single-Factor ANOVA

3. 3 Applications of Single-Factor ANOVA A mathematical model (equation) is often used to express a particular situation. The equation served as the model when we believed that a straight-line relationship existed. The probability functions studied in Chapter 5 are also examples of mathematical models. For the single-factor ANOVA, the mathematical model, formula (12.13), is an expression of the composition of each data value, x c,k, entered in our data table.

4. 4 Applications of Single-Factor ANOVA We interpret each term of this model as follows: x c,k is the value of the variable at the kth replicate of level c.  is the mean value for all the data without respect to the test factor. F c is the effect that the factor being tested has on the response variable at each different level c.

5. 5 Applications of Single-Factor ANOVA  k(c) (  is the lowercase Greek letter epsilon) is the experimental error that occurs among the k replicates in each of the c columns.

6. 6 Example 6 – Hypothesis Test for the Equality of Several Means A rifle club performed an experiment with a randomly selected group of first-time shooters. The purpose of the experiment was to determine whether shooting accuracy is affected by the method of sighting used: only the right eye open, only the left eye open, or both eyes open. Fifteen first-time shooters were selected and divided into three groups. Each group experienced the same training and practicing procedures, with one exception—the method of sighting used.

7. 7 Example 6 – Hypothesis Test for the Equality of Several Means After completing training, each shooter was given the same number of rounds and asked to shoot at a target. Their scores are listed in Table 12.8. At the 0.05 level of significance, is there sufficient evidence to reject the claim that the three methods of sighting are equally effective? Table 12.8 Sample Results on Target Shooting [TA12-8] cont’d

8. 8 Example 6 – Solution In this experiment the factor is method of sighting and the levels are the three different methods of sighting (right eye, left eye, and both eyes open). The replicates are the scores received by the shooters in each group. The null hypothesis to be tested is “The three methods of sighting are equally effective, or the mean scores attained using each of the three methods are the same.”

9. 9 Example 6 – Solution Step 1 a. Parameter of interest: The “mean” at each level of the test factor is of interest: the mean score using the right eye  R ; the mean score using the left eye  L ; and the mean score using both eyes,  B. The factor being tested, “method of sighting,” has three levels: right, left, and both. b. Statement of hypotheses: H o :  R =  L =  B H a : The means are not all equal (i.e., at least one mean is different). cont’d

10. 10 Example 6 – Solution Step 2 a. Assumptions: The shooters were randomly assigned to the method, and their scores are independent of each other. The effects due to chance and untested factors are assumed to be normally distributed. b. Test statistic: The F-distribution and formula (12.12) will be used with df (numerator) = df (method) = 2 and df (denominator) = df (error) = 12. c. Level of significance:  = 0.05 cont’d

11. 11 Example 6 – Solution Step 3 a. Sample information: See Table 12.8. cont’d Table 12.8 Sample Results on Target Shooting [TA12-8]

12. 12 Example 6 – Solution b. Calculated test statistic: The test statistic is F : Table 12.9 is used to find the column totals. cont’d Table 12.9 Sample Results for Target Shooting

13. 13 Example 6 – Solution First, the summations  x and  x 2 need to be calculated:  x = 12 + 10 + 18 + 12 + 14 + 10 + 17 +... + 21 = 220 (Or 66 + 56 + 98 = 220)  x 2 = 12 2 + 10 2 + 18 2 + 12 2 + 14 2 + 10 2 +... + 21 2 = 3392 Using formula (12.2), we find = 3392 – 3226.67 cont’d = 165.33

14. 14 Example 6 – Solution Using formula (12.3), we find = (871.2 + 784 + 1600.67) – 3226.67 = 3255.87 – 3226.67 = 29.20 cont’d

15. 15 Example 6 – Solution To find SS (error) we first need:  (x 2 ) = 3392 (found previously) 3255.87 (found previously) Then using formula (12.4), we have cont’d

16. 16 Example 6 – Solution SS (error) = 3392 – 3255.87 = 136.13 We use formula (12.8) to check the sum of squares: SS (method) + SS (error) = SS (total): 29.20 + 136.13 = 165.33 cont’d

17. 17 Example 6 – Solution The degrees of freedom are found using formulas (12.5), (12.6), and (12.7): df (method) = c – 1 = 3 – 1 = 2 df (total) = n – 1 = 15 – 1 = 14 df (error) = n – c = 15 – 3 = 12 cont’d

18. 18 Example 6 – Solution Using formulas (12.10) and (12.11), we find The results of these computations are recorded in the ANOVA table in Table 12.10. cont’d = 14.60 = 11.34 Table 12.10 ANOVA Table for Example 12.6

19. 19 The calculated value of the test statistic is then found using formula (12.12): Step 4 Probability Distribution: p-Value: a. Use the right-hand tail : P = P(F > 1.287, df n = 2 and df d = 12), as shown on the figure. Example 6 – Solution cont’d = 1.287

20. 20 Example 6 – Solution To find the p-value, you have two options: 1. Use Table 9A (Appendix B) to place bounds on the p-value: P > 0.05. 2. Use a computer or calculator to find the p-value: P = 0.312. b. The p-value is not smaller than the level of significance,  (0.05). cont’d

21. 21 Example 6 – Solution Classical: a. The critical region is the right-hand tail; the critical value is obtained from Table 9A. b. F is not in the critical region, as shown in red in the figure. cont’d

22. 22 Example 6 – Solution Step 5 a. Decision: Fail to reject H o. b. Conclusion: The data show no evidence to reject the null hypothesis that the three methods are equally effective. cont’d