# A simple question from ANOVA

As per definition of $$F$$ statistic, $$F= \frac{MST}{MSE}$$ where MST and MSE denote mean square due to treatment and error respectively. From this definition of $$F$$ am I right in saying that $$F$$ increases with decrease in variability due to chance (error) and decrease with decrease in variability between groups (treatment).

My question stems from the confusion that if a random variable $$Z$$ is a ratio of two other random variables, then, can I say a similar relation as mentioned above.

## 1 Answer

Under the assumptions of a simple one-way ANOVA model, when the Mean Square due to Treatment increases, this leads to a decrease in the Mean Square due to Error and an increase in corresponding $$F$$-statistic (since $$MST+MSE = MS Total$$).

However, a ratio of any two random variables does not necessarily follow an $$F$$ distribution. The $$F$$ distribution is formed by taking the ratio of two independent $$\chi^2$$ random-variables, divided by their degrees of freedom. It just so happens from the Normality assumption of the ANOVA (and the centering and squaring of terms) that the ratio of $$MST$$ and $$MSE$$ follows an $$F$$ distribution.

That being said, in general, if you have any ratio, $$Z=X/Y$$, then $$Z$$ increases if $$X$$ increases while $$Y$$ remains unchanged and $$Z$$ decreases if $$Y$$ increases while $$X$$ remains unchanged.

See here https://en.m.wikipedia.org/wiki/F-distribution for more details on $$F$$ distribution.

• So, is it wrong to say that F stat increases with decrease in variability due to chance if nothing about the error due to treatments is mentioned? – user46697 Jun 18 at 6:34
• In single-factor ANOVA, if there is a "decrease in variability due to chance" this implies that the $MSE$ decreases, which in turn implies an increase in $MST$, and this implies an increase in the $F$-statistic. In SINGLE factor when $MSE$ increases, $MST$ decreases, since the two values are linked via the formula $MSTotal=MSE+MST$ as I stated above. You cannot increase one without decreasing the other and vice versa. – StatsStudent Jun 18 at 15:24
• So the number of factors is to be taken into consideration. In a case where number of factors is unknown, or more than one, nothing can be said. Right? – user46697 Jun 18 at 15:30
• I'm not sure what you mean by the number of factors is to be taken into consideration. For what purpose? If you have another factor, say Treatment 2, then in a two-way anova, the $MSE = MSTotal-MSTreatment_1-MSTreament_2-InteractionEffects$ – StatsStudent Jun 18 at 15:39
• I meant, is this relation between mse and F obeyed only in a single factor case? – user46697 Jun 18 at 15:40