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.


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.

  • $\begingroup$ 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? $\endgroup$ – Harry Jun 18 '19 at 6:34
  • $\begingroup$ 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. $\endgroup$ – StatsStudent Jun 18 '19 at 15:24
  • $\begingroup$ 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? $\endgroup$ – Harry Jun 18 '19 at 15:30
  • $\begingroup$ 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 $MSE = MSTotal-MSTreatment_1-MSTreament_2-InteractionEffects$ $\endgroup$ – StatsStudent Jun 18 '19 at 15:39
  • $\begingroup$ I meant, is this relation between mse and F obeyed only in a single factor case? $\endgroup$ – Harry Jun 18 '19 at 15:40

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