# Computing the F-ratio under null hypothesis

Statistical model for a Completely Randomized Design:

$$y_{ij}=\mu+\tau_{i}+\epsilon_{ij}\quad i=1,2,\ldots,a; j=1,2,\ldots,n$$

where,

$y_{ij}$ is the $j$th observation of $i$th treatment effect

$\mu$ is the overall mean

$\tau_{i}$ is the $i$th treatment effect

$\epsilon_{ij}$ random error term and $\epsilon_{ij}\sim NID(0,\sigma^2)$

For testing the hypothesis of no difference in treatment means

$$H_0:\tau_1=\tau_2=\ldots=\tau_a = 0 \quad vs.\quad H_1:\tau_i\ne 0\quad \text{for some } i.$$

Under null hypothesis , the ratio $$F=\frac{\frac{SS_{treatments}}{a-1}}{\frac{SS_{error}}{N-a}}$$ is distributed as F with $a-1$ and $N-a$ degrees of freedom .

• My question is: Why do i have to compute the F-ratio under null hypothesis, why do not under alternative hypothesis ?
• Which alternative will you compute it under? – Glen_b Aug 29 '14 at 15:06
• Exactly what @Glen_b said. There are literally, not figuratively, an infinite number of alternative hypotheses to consider. – Alexis Aug 29 '14 at 15:13
• In a linear models course you would find the distribution of $F$ under the alternative. (It's a noncentral $F$ distribution, if you care.) Now, as to why not: the goal of statistical hypothesis testing is determine if the null hypothesis is consistent with the observed data. The null occupies a privileged position in this schema and so is used as a standard of comparison. – Dennis Aug 29 '14 at 23:31
• Leaf - please check you're happy with Dennis' edit. If not you can edit again, but keep his comments on the reasons for the edit in mind. – Glen_b Aug 30 '14 at 0:00
• My teacher said something like that(I can not recall it exactly):"we need the F-ratio under null hypothesis so that both numerator and denominator have variance $\sigma^2$, which cancels out. If we don't compute the F-ratio under null hypothesis, then the numerator has variance more than $\sigma^2$.Something about $\frac{SS_{treatments}}{\sigma^2}\sim\chi^2_{a-1}$ and Cochran theorem." I couldn't able to visualize. – user 31466 Aug 30 '14 at 1:35