Characteristics of F -Distribution

1. 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. • F cannot be negative, and it is a continuous distribution. • The F distribution is positively skewed. • Its values range from 0 to . As F   the curve approaches the X-axis.

2. Test for Equal Variances • For the two tail test, the test statistic is given by: and are the sample variances for the two samples. • The null hypothesis is rejected if the computed value of the test statistic is greater than the critical value.

3. 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?

4. EXAMPLE 1 continued • Step 1: The hypotheses are: • Step 2: The significance level is .05. • Step 3: The test statistic is the F distribution.

5. Example 1 continued • Step 4:H0 is rejected if F>3.68. The degrees of freedom are 9 in the numerator and 7 in the denominator. • Step 5: The value of F is computed as follows. H0 is not rejected. There is insufficient evidence to show more variation in the internet stocks.

6. 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. • This technique is called analysis of variance or ANOVA.

7. 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. • The Decision rule is to reject the null hypothesis if F (computed) is greater than F (table) with numerator and denominator degrees of freedom.

8. 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:

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

10. 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.

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

12. 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.

13. 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

14. Example 2 continued • The SS total is:

15. Example 2 continued • The SST is:

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

17. EXAMPLE 2 continued • Step 1:H0: 1 = 2 = 3 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.

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

19. 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.

20. 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.

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

22. 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?

23. Example 3continued • 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.