Module 13: F-Distribution and One-Way ANOVA
Facts about the F Distribution
Barbara Illowsky & OpenStax et al.
Here are some 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.
try it
MRSA, or Staphylococcus aureus, can cause a serious bacterial infections in hospital patients. This table shows various colony counts from different patients who may or may not have MRSA.
| 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 |
Plot of the data for the different concentrations:
Test whether the mean number 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% significance level.
While there are differences in the spreads between the groups, the differences do not appear to be big enough to cause concern.
We test for the equality of mean number of colonies:
H0 : μ1 = μ2 = μ3 = μ4 = μ5Ha: μi ≠ μj some i ≠ j
The one-way ANOVA table results are shown in below.
| Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F |
|---|---|---|---|---|
| Factor (Between) | 10,233 | 5 – 1 = 4 | [latex]displaystylefrac{{{10},{233}}}{{4}}={2},{558.25}[/latex] | [latex]displaystylefrac{{{2},{558.25}}}{{{4},{194.9}}}={0.6099}[/latex] |
| Error (Within) | 41,949 | 15 – 5 = 10 | ||
| Total | 52,182 | 15 – 1 = 14 | [latex]displaystylefrac{{{41},{949}}}{{10}}={4},{194.9}[/latex] |
Distribution for the test: F4,10Probability Statement: p-value = P(F > 0.6099) = 0.6649.
Compare α and the p-value: α = 0.05, p-value = 0.669, α > p-value
Make a decision: Since α > p-value, we do not reject H0.
Conclusion: At the 5% significance level, there is insufficient evidence from these data that different levels of tryptone will cause a significant difference in the mean number of bacterial colonies formed.
Example
Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in the table.
Mean Grades 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 mean grades among the sororities?
Solution:
Let μ1, μ2, μ3, μ4 be the population means of the sororities. Remember that the null hypothesis claims that the sorority gr