Chapter Eleven

1. Chapter Eleven Analysis of Variance GOALS When you have completed this chapter, you will be able to: ONEDiscuss the general idea of analysis of variance. TWOList the characteristics of the F distribution. THREEConduct a test of hypothesis to determine whether the variances of two populations are equal.

2. Chapter Eleven continued Analysis of Variance GOALS When you have completed this chapter, you will be able to: FOUR Organize data into a one-way ANOVA table. FIVEConduct a test of hypothesis among three or more treatment means. SIXDevelop confidence intervals for the difference between treatment means.

3. Characteristics of F-Distribution • There is a “family” of F Distributions. Each member of the family is determined by two parameters: the numerator degrees of freedom and the denominator degrees of freedom.

4. Characteristics of F-Distribution continued • The F distribution is continuous. • 3. The F distribution cannot be negative. • 4. It is positively skewed. • 5. It is asymptotic.

5. Test for Equal Variances • For the two tail test, the test statistic is given by: • and are the sample variances for the two samples.

6. Test for Equal Variances • The null hypothesis is rejected if the computed value of the test statistic is greater than the critical value.

7. EXAMPLE 1 • Colin, a stockbroker at Critical Securities, reported that the mean rate of return on a sample of 10 internet stocks was 12.6 percent with a standard deviation of 3.9 percent. The mean rate of return on a sample of 8 utility stocks was 10.9 percent with a standard deviation of 3.5 percent. At the .05 significance level, can Colin conclude that there is more variation in the software stocks?

8. EXAMPLE 1 continued • Step 1: The hypotheses are:

9. EXAMPLE 1 continued The significance level is .05. The test statistic is the F distribution. H0 is rejected if F>3.68. The degrees of freedom are 9 in the numerator and 7 in the denominator.

10. Example 1 continued The value of F is computed as follows. H0 is not rejected. There is insufficient evidence to show more variation in the internet stocks.

11. Underlying Assumptions for ANOVA • The F distribution is also used for testing whether two or more sample means came from the same or equal populations.

12. Underlying Assumptions for ANOVA ANOVA requires the following conditions: • The samples are randomly selected and are independent. • The populations have equal standard deviations. • The sampled populations follow the normal distribution.

13. Analysis of Variance Procedure • TheNull Hypothesis is that the population means are the same. The Alternative Hypothesis is that at least one of the means is different. TheTest Statistic is the F distribution. TheDecision rule is to reject the null hypothesis if F (computed) is greater than F (table) with numerator and denominator degrees of freedom.

14. Analysis of Variance Procedure • If there are k populations being sampled, the numerator degrees of freedom is k – 1. • If there are a total of n observations the denominator degrees of freedom is n – k. • The test statistic is computed by:

15. Analysis of Variance Procedure • SS Total is the total sum of squares.

16. Analysis of Variance Procedure • SST is the treatment sum of squares. • TC is the column total, nc is the number of observations in each column, X the sum of all the observations, and n the total number of observations.

17. Analysis of Variance Procedure • SSE is the sum of squares error.

18. EXAMPLE 2 • Rosenbaum Restaurants specialize in meals for senior citizens. Katy Polsby, President, recently developed a new meat loaf dinner. Before making it a part of the regular menu she decides to test it in several of her restaurants. She would like to know if there is a difference in the mean number of dinners sold per day at the Anyor, Loris, and Lander restaurants. Use the .05 significance level.

19. Example 2 continued • Aynor Loris Lander • 13 10 18 • 12 12 16 • 14 13 17 • 12 11 17 • 17 • Tc 51 46 85 • nc 4 4 5

20. Example 2 continued • The SS total is:

21. Example 2 continued • The SST is:

22. Example 2 continued • The SSE is: • SSE = SS total – SST • = 86 – 76.25 = 9.75

23. EXAMPLE 2 continued • Step 1:H0: μ1 = μ2 = μ3 • H1: Treatment means are not the same Step 2:H0 is rejected if F>4.10. There are 2 df in the numerator and 10 df in the denominator.

24. Example 2 continued • To find the value of F:

25. Example 2 continued • The decision is to reject the null hypothesis. The treatment means are not the same. The mean number of meals sold at the three locations is not the same. The ANOVA table on the next slide is from the Minitab system.

26. Example 2 continued • Analysis of Variance • Source DF SS MS F P • Factor 2 76.250 38.125 39.10 0.000 • Error 10 9.750 0.975 • Total 12 86.000 • Individual 95% CIs For Mean • Based on Pooled StDev • Level N Mean StDev ---------+---------+---------+------- • Aynor 4 12.750 0.957 (---*---) • Loris 4 11.500 1.291 (---*---) • Lander 5 17.000 0.707 (---*---) • ---------+---------+---------+------- • Pooled StDev = 0.987 12.5 15.0 17.5

27. Inferences About Treatment Means • When we reject the null hypothesis that the means are equal, we may want to know which treatment means differ. One of the simplest procedures is through the use of confidence intervals.

28. Confidence Interval for the Difference Between Two Means • where t is obtained from the t table with degrees of freedom (n - k) and MSE = [SSE/(n - k)]

29. EXAMPLE 3 • From EXAMPLE 2 develop a 95% confidence interval for the difference in the mean number of meat loaf dinners sold in Lander and Aynor. Can Katy conclude that there is a difference between the two restaurants?

30. EXAMPLE 3

31. Example 3 continued • Because zero is not in the interval, we conclude that this pair of means differs. The mean number of meals sold in Aynor is different from Lander.