The following are facts about the F distribution:

  • 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. Two-way analysis is beyond the scope of this chapter.
The curve one the left is a nonsymmetrical F distribution curve skewed to the right, more values in the right tail and the peak is closer to the left. This curve is different from the graph on the right because of the different dfs. The curve on the right shows a nonsymmetrical F distribution curve skewed to the right. This curve is different from the graph on the left because of the different dfs. Because its dfs are larger, it more closely resembles a normal distribution curve.
Figure 13.3

Example 13.2

Let’s return to the slicing tomato exercise in Try It. The means of the tomato yields under the five mulching conditions are represented by μ1, μ2, μ3, μ4, μ5. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5 percent, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest.

Solution 13.2

The null and alternative hypotheses are as follows:

H0: μ1 = μ2 = μ3 = μ4 = μ5

Ha: μi ≠ μj for some i ≠ j

The one-way ANOVA results are shown in Table 13.5

Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F
Factor (Between) 36,648,561 5 – 1 = 4 36,648,5614 = 9,162,14036,648,5614 = 9,162,140 9,162,1402,044,672.6 = 4.48109,162,1402,044,672.6 = 4.4810
Error (Within) 20,446,726 15 – 5 = 10 20,446,72610 = 2,044,672.620,446,72610 = 2,044,672.6  
Total 57,095,287 15 – 1 = 14    
Table 13.5

Distribution for the test: F4,10

df(num) = 5 – 1 = 4

df(denom) = 15 – 5 = 10

Test statistic: F = 4.4810

This graph shows a nonsymmetrical F distribution curve. The horizontal axis extends from 0 - 5, and the vertical axis ranges from 0 - 0.7. The curve is strongly skewed to the right.
Figure 13.4

Probability statement: p-value = P(F > 4.481) = 0.0248

Compare α and the p-value: α = 0.05, p-value = 0.0248

Make a decision: Since α > p-value, we reject H0.

Conclusion: At the 5 percent significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of the mulches led to different mean yields.

Using the TI-83, 83+, 84, 84+ Calculator

To find these results on the calculator:

Press STAT. Press 1:EDIT. Put the data into the lists L1, L2, L3, L4, L5.

Press STAT, arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter (L1, L2, L3, L4, L5). Press ENTER. You will see that the values in the foregoing ANOVA table are easily produced by the calculator, including the test statistic and the p-value of the test.

The calculator displays:

F = 4.4810

p = 0.0248 (p-value)


df = 4

SS = 36648560.9

MS = 9162140.23


df = 10

SS = 20446726

MS = 2044672.6
Try It 13.2

MRSA, or Staphylococcus aureus, can cause serious bacterial infections in hospital patients. Table 13.6 shows various colony counts from different patients who may or may not have MRSA. The data from the table is plotted in Figure 13.5.

Conc = 0.6 Conc = 0.8 Conc = 1.0 Conc = 1.2 Conc = 1.4
9 16 22 30 27
66 93 147 199 168
98 82 120 148 132
Table 13.6

Plot of the data for the different concentrations:

This graph is a scatterplot for the data provided. The horizontal axis is labeled 'Colony counts' and extends from 0 - 200. The vertical axis is labeled 'Tryptone concentrations' and extends from 0.6 - 1.4.
Figure 13.5

Test whether the mean numbers of colonies are the same or are different. Construct the ANOVA table by hand or by using a TI-83, 83+, or 84+ calculator, find the p-value, and state your conclusion. Use a 5 percent significance level.

Example 13.3

Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in Table 13.7.

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
Table 13.7 Mean Grades for Four Sororities

Using a significance level of 1 percent, is there a difference in mean grades among the sororities?

Solution 13.3

Let μ1, μ2, μ3, μ4 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 five.


This is an example of a balanced design, because each factor (i.e., sorority) has the same number of observations.

H0: μ1 = μ2 = μ3 = μ4

Ha: Not all of the means μ1, μ2, μ3, μ4 are equal.

Distribution for the test: F3,16

where k = 4 groups and n = 20 samples in total.

df(num)= k – 1 = 4 – 1 = 3

df(denom) = nk = 20 – 4 = 16

Calculate the test statistic: F = 2.23


This graph shows a nonsymmetrical F distribution curve with values of 0 and 2.23 on the x-axis representing the test statistic of sorority grade averages. The curve is slightly skewed to the right, but is approximately normal. A vertical upward line extends from 2.23 to the curve and the area to the right of this is shaded to represent the p-value.
Figure 13.6

Probability statement: p-value = P(F > 2.23) = 0.1241

Compare α and the p-value: α = 0.01

p-value = 0.1241

α p-value

Make a decision: Since α p-value, you cannot reject H0.

Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities.

Using the TI-83, 83+, 84, 84+ Calculator

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 calculator displays the F statistic, the p-value, and the values for the one-way ANOVA table:

F = 2.2303

p = 0.1241 (p-value)


df = 3

SS = 2.88732

MS = 0.96244


df = 16

SS = 6.9044

MS = 0.431525
Try It 13.3

Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in Table 13.8.

Basketball Baseball Hockey Lacrosse
3.6 2.1 4.0 2.0
2.9 2.6 2.0 3.6
2.5 3.9 2.6 3.9
3.3 3.1 3.2 2.7
3.8 3.4 3.2 2.5
Table 13.8 GPAs for four sports teams

Use a significance level of 5 percent and determine if there is a difference in GPA among the teams.

Example 13.4

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 five plants. At the end of the growing period, each plant was measured, producing the data in inches in Table 13.9.

Tommy's Plants Tara's Plants Nick's Plants
24 25 23
21 31 27
23 23 22
30 20 30
23 28 20
Table 13.9

Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3 percent level of significance.

Solution 13.4

This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants, so we will use the formula F' = nsx¯2s2poolednsx¯2s2pooled .

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
Table 13.10

Next, calculate the variance of the three group means by calculating the variance of 24.2, 25.4, and 24.4. Variance of the group means = 0.413 = sx¯2sx¯2,

then MSbetween = nsx¯2nsx¯2 = (5)(0.413) where n = 5 is the sample size (number of plants each child grew).

Calculate the mean of the three sample variances (calculate the mean of 11.7, 18.3, and 16.3). Mean of the sample variances = 15.433 = s2pooled,

then MSwithin = s2pooled = 15.433.

The F statistic (or F ratio) is F=MSbetweenMSwithin=nsx¯2s2pooled=(5)(0.413)15.433=0.134.F=MSbetweenMSwithin=nsx¯2s2pooled=(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 F2,12 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 H0. Why?

Conclusion: With a 3 percent level of significance from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.

Using the TI-83, 83+, 84, 84+ Calculator

To calculate the p-value:

•Press 2nd DISTR,

•Arrow down to Fcdf and press ENTER,

•Enter 0.134, E99, 2, 12, and

•Press ENTER.

The p-value is 0.8759.

Try It 13.4

Another fourth grader also grew bean plants, but in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in Example 13.4.

Collaborative Exercise

From the class, create four groups of the same size as follows: men under 22, men at least 22, women under 22, women at least 22. 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 percent level of significance. Use one of the solution sheets in Appendix E.