Clarification on ANOVA mechanism This website explains ANOVA and F ratio as follows:
"ANOVA partitions the variability among all the values into one component that is due to variability among group means (due to the treatment) and another component that is due to variability within the groups (also called residual variation)... Each sum-of-squares is associated with a certain number of degrees of freedom... and the mean square (MS) is computed by dividing the sum-of-squares by the appropriate number of degrees of freedom... The F ratio is the ratio of two mean square values..."
My questions are:
How can ANOVA know to partition the variability into 2 distinct components (due to the treatment and due to inherent variation)?
What two mean squares does it refer to (The F ratio is the ratio of two mean square values). Are they mean squares due to treatment and due to inherent variation?
Thanks in advance.
 A: Consider the following data simulated in R according to the model
for a one-factor ANOVA with three levels of the factor and ten
replications at each level. Each level has variance $\sigma^2 = 3^2 = 9.$
set.seed(2020)
x1 = rnorm(10, 20, 3)
x2 = rnorm(10, 21, 3)
x3 = rnorm(10, 22, 4)
x = c(x1,x2,x3)
gp = as.factor(rep(1:3, each=10))

Here is a stripchart in R showing the ten observations in each group.
stripchart(x ~ gp, pch="|", ylim=c(.5,3.5))


The ANOVA table is given below:
anova(lm(x~gp))
Analysis of Variance Table

Response: x
          Df Sum Sq Mean Sq F value  Pr(>F)  
gp         2 140.48  70.240   4.463 0.02115 *
Residuals 27 424.93  15.738                  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

MSA = $15.7382$ is the average of the variances within each of the three groups. This is
one way to estimate $\sigma^2.$ [Never mind that it is not a very good estimate;
with only 30 observations altogether, we can't expect a really close estimate.]
mean(c(var(x1),var(x2),var(x3)))
[1] 15.7382

If all three groups had the same mean $\mu$ (the assumption of the null hypothesis), then the three group means $(\bar X_1,\bar X_2, \bar X_3)$ would each would have a normal distribution with mean
$\mu$ and variance $\sigma^2/10.$ So, if $H_0$ were true, we could also estimate
$\sigma^2$ as the $10$ times the variance of the 'sample' of three $\bar X_i$s:
10*var(c(mean(x1),mean(x2),mean(x3)))
[1] 70.23971

Thus MS(Group) = $70.2397.$ [Because $H_0$ is false, this estimate is much too
large; the three means also express the differences among groups.]
So the way ANOVA "knows" how to get the two variances is because of the two
procedures we have just seen.
If $H_0$ is true the two variance estimates tend to be about the same so that
the F-ratio would tend to be about $1.$ The larger the F-ratio is above $1,$ the stronger the evidence against $H_0.$ In our case $F = 4.463 > 1.$ Taking numerator and denominator degrees
of freedom into account, $4.463$ is judged to be "significantly" larger than $1.$
The variance estimate in the numerator of $F$ involves both $\sigma^2$ and the
difference in group population means $\mu_.$ The variance estimate in the denominator
involves only $\sigma^2.$
Here is a plot of the density function of the distribution $\mathsf{F}(2, 27).$
The (tiny) area under the density curve to the right of the vertical dotted line
is the P-value $0.02115.$

curve(df(x, 2, 27), 0, 10, lwd=2, ylab="PDF", xlab="F", 
         main="Density of F(2,27)")
  abline(v = 4.463, col="red", lwd=2, lty="dotted")
  abline(h=0, col="green2"); abline(v=0, col="green2")

