EXAMPLE 12.13
The ANOVA F test for the Facebook friends study. In the Facebook friends study, we found F = 4.14. (Note that it is standard practice to round F statistics to two places after the decimal point.) There were five populations, so the degrees of freedom in the numerator are DFG = I − 1 = 4. For this example, the degrees of freedom in the denominator are DFE = N − I = 134 − 5 = 129. Software provided a P-value of 0.003, so at the 0.05 significance level, we reject H0 and conclude that the population means are not all the same.
662
| df = (4, 100) | |
|---|---|
| p | Critical value |
| 0.100 | 2.00 |
| 0.050 | 2.46 |
| 0.025 | 2.92 |
| 0.010 | 3.51 |
| 0.001 | 5.02 |
Suppose that P = 0.003 was not provided. We’ll now run through the process of using the table of F critical values to approximate the P-value. Although you will rarely need to do this in practice, the process will help you to understand the P-value calculation.
In Table E, we first find the column corresponding to 4 degrees of freedom in the numerator. For the degrees of freedom in the denominator, we see that there are entries for 100 and 200. The values for these entries are very close. To be conservative, we use critical values corresponding to 100 degrees of freedom in the denominator because these are slightly larger.
We have F = 4.14. This is in between the critical value for P = 0.010 and P = 0.001. Using the table, we can conclude only that 0.001 < P < 0.010.