F Distribution and ANOVA: Facts About the F Distribution 1.8 2008/06/23 17:13:55 GMT-5 2008/10/27 14:44:49.133 GMT-5 Susan Dean deansusan@deanza.edu Barbara Illowsky illowskybarbara@deanza.edu Susan Dean deansusan@deanza.edu Barbara Illowsky illowskybarbara@deanza.edu Connexions cnx@cnx.org ANOVA curve degrees of freedom F Distribution skew statistics Two-Way Analysis of Variance This module states the factors associated with F Distributions and provides students with some examples to help further understand the concept. Students will be given the opportunity to see F Distributions in action through participation in an optional classroom exercise. The curve is not symmetrical but skewed to the right. There is a different curve for each set of dfs . The F statistic is greater than or equal to zero. As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal. Other uses for the F distribution include comparing two variances and Two-Way Analysis of Variance. Comparing two variances is discussed at the end of the chapter. Two-Way Analysis is mentioned for your information only.

One-Way ANOVA: Four sororities took a random sample of sisters regarding their grade averages for the past term. The results are shown below: GRADE AVERAGES FOR FOUR SORORITIES Sorority 1 Sorority 2 Sorority 3 Sorority 4 2.17 2.63 2.63 3.79 1.85 1.77 3.78 3.45 2.83 3.25 4.00 3.08 1.69 1.86 2.55 2.26 3.33 2.21 2.45 3.18 Using a significance level of 1%, is there a difference in grade averages among the sororities? Let μ 1 , μ 2 , μ 3 , μ 4 be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each size 5. H o : μ 1 = μ 2 = μ 3 = μ 4 H a : Not all of the means μ 1 , μ 2 , μ 3 , μ 4 are equal. Distribution for the test: F 3 , 16 where k = 4 groups and N = 20 samples in total df(num) = k - 1 = 4 - 1 = 3 df(denom) = N - k = 20 - 4 = 16 Calculate the test statistic: F = 2.23 Graph:

Probability statement: p-value = P ( F > 2.23 ) = 0.1241 Compare α and the p-value : α = 0.01 p-value = 0.1242 α < p-value . Make a decision: Since α < p-value , you cannot reject H o . This means that the population averages appear to be the same. Conclusion: There is not sufficient evidence to conclude that there is a difference among the grade averages for the sororities. TI-83+ or TI 84: Put the data into lists L1, L2, L3, and L4. Press STAT and arrow over to TESTS. Arrow down to F:ANOVA. Press ENTER and Enter (L1,L2,L3,L4). The F statistic is 2.2303 and the p-value is 0.1241. df(numerator) = 3 (under "Factor") and df(denominator) = 16 (under Error). A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew 5 plants. At the end of the growing period, each plant was measured, producing the following data (in inches): Tommy's Plants Tara's Plants Nick's Plants 24 25 23 21 31 27 23 23 22 30 20 30 23 28 20 Does it appear that the three media in which the bean plants were grown produce the same average height? Test at a 3% level of significance. This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants. First, calculate the sample mean and sample variance of each group. Tommy's Plants Tara's Plants Nick's Plants Sample Mean 24.2 25.4 24.4 Sample Variance 11.7 18.3 16.3 Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). Variance of the group means = 0.413 Then MS between = ( 5 ) ( 0.413 ) where the 5 is the sample size (number of plants each child grew). Calculate the average of the three sample variances (Calculate the average of 11.7, 11.3, and 16.3). Average of the sample variances = 15.433 Then MS within = 15.433 . The F statistic (or F ratio) is F = MS between MS within = ( 5 ) ⋅ ( 0.413 ) 15.433 = 0.134 The dfs for the numerator = the number of groups - 1 = 3 - 1 = 2 The dfs for the denominator = the total number of samples - the number of groups = 15 - 3 = 12 The distribution for the test is F 2 , 12 and the F statistic is F = 0.134 The p-value is P ( F > 0.134 ) = 0.8759 . Decision: Since α = 0.03 and the p-value = 0.8759 , do not reject H o . (Why?) Conclusion: With a 3% the level of significance, from the sample data, the evidence is not sufficient to conclude that the average heights of the bean plants are not different. Of the three media tested, it appears that it does not matter which one the bean plants are grown in. (This experiment was actually done by three classmates of the son of one of the authors.) Another fourth grader also grew bean plants but this time in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do an ANOVA test on the 4 groups. You may use your calculator or computer to perform the test. Are the heights of the bean plants different? Use a solution sheet. F = 0.9496 p-value = 0.4401 The heights of the bean plants are the same.

Optional Classroom ActivityRandomly divide the class into four groups of the same size. Have each member of each group record the number of states in the United States he or she has visited. Run an ANOVA test to determine if the average number of states visited in the four groups are the same. Test at a 1% level of significance. Use one of the solution sheets at the end of the chapter (after the homework).