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> 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
The Null Hypothesis is that the population means are the same.
The Alternative Hypothesis is that at least one of the means is different.
The Test 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 17 Tc nc

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 – = 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> There are 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.