One of the most common statistical data analyzes is one that studies the effect that a factor (for example, a treatment) has on a continuous variable (for example, the weight of the piglets). To do this, we can perform a Student’s t-test or an Analysis of Variance (ANOVA).
While the first is only useful when the factor studied has two levels (for example, the treatment factor with the groups “with probiotics” in the feed and “without probiotics”), the ANOVA can be done when there are more than two levels (for example, to compare three types of probiotics).
The ANOVA makes it possible to relate an independent categorical variable with a dependent continuous quantitative variable through the comparison of “c” means.
The ANOVA explains the total variation and how it is distributed within the groups. Thus, it tells us if the variability of the groups is significantly different from each other.
In this case, we could answer the question: is part of the variability in weight among piglets explained by the fact that they are subjected to different probiotics? OR… does the treatment have any effect on the weight of the piglets? We can answer all this with an ANOVA, where we obtain a probability value (p-value) whose interpretation is very simple:
- p-value <0.005: There are significant differences among groups.
- p-value> 0.05: There are no significant differences among groups.
Therefore, when we compare our p.value (suppose it is 2 · 10-6) with a significance level that is 0.05 (5% error) we can accept the starting hypothesis. Following the example: a treatment affects the average profit of the piglets. On the other hand, if the value had been equal to or greater than 0.05 (5% error), we must accept the hypothesis that the treatment does not affect weight.
In this example that we present below, we are going to explain all the numbers that appear in an ANOVA, not just the p-value. We will perform the analysis with the Síagro program.
Suppose we have three vaccines for any given disease and that we record the difference in temperature between the time of application and six hours after application. The question we are going to answer is …