Email: support@essaywriterpros.com
Call Us: US - +1 845 478 5244 | UK - +44 20 7193 7850 | AUS - +61 2 8005 4826

Perform a 1-Way ANOVA Test

A 1-way ANOVA tests whether the means of all groups are equal for different levels of one factor, using some fairly lengthy calculations. You could do all the computations by hand as shown in the Appendix, but no one ever does. Here are some alternatives:

  • Excel’s Anova: Single Factor command is in the Tools » Data Analysis menu in Excel 2003 and below, or the Data » Analysis » Data Analysis menu in Excel 2007. If you don’t see it there, follow instructions in Excel help to load the Analysis Toolpak.
  • On a TI-83 or TI-84, enter each sample in a statistics list, then press [STAT] [] [] to select ANOVA, and enter the list names separated by commas.
  • There are even Web-based ANOVA calculators, such as Lowry 2001b.
  • There are many software packages for mathematics and statistics that include ANOVA calculations. One of them, R, is highly regarded and is open source.

When you use a calculator or computer program to do ANOVA, you get an ANOVA table that looks something like this:

SSdfMSFp
Between groups
(or “Factor”)
1636.53545.45.410.0069
Within groups
(or “Error”)
2018.020100.9
Total3654.523

Note that the mean square between treatments, 545.4, is much larger than the mean square within treatments, 100.9. That ratio, between-groups mean square over within-groups mean square, is called an F statistic (F = MSB/MSW = 5.41 in this example). It tells you how much more variability there is between treatment groups than within treatment groups. The larger that ratio, the more confident you feel in rejecting the null hypothesis, which was that all means are equal and there is no treatment effect.

But what you care about is the p-value of 0.0069, obtained from the F distribution. The p-value has the usual interpretation: the probability of the between-treatments MS being ≥5.41 times the within-treatments MS, if the null hypothesis is true, is p = 0.0069.

The p-value is below your significance level of 0.05: it would be quite unlikely to have MSB/MSW this large if there were no real difference among the means. Therefore you reject H0 and accept H1, concluding that the mean absorption of all the fats is not the same.