Consider the following hypothetical samples:

Given data:

A number of samples $k=3$.

Each sample has a set of values $n_1=3,$

$n_2=5,$

$n_3=4$.

The total of each sample's values $T_1=9,$

$T_2=15,$

$T_3=24$

To calculate the mean of each sample:

$=\frac{9}{3}$

$=3$

$=\frac{15}{5}$

$=3$

$=\frac{24}{4}$

$=6$

To find the sample variance:

$=2$

$=2.449$

$=4.243$

Given data:

The overall mean is

$\overline{x}=\frac{\sum x_i}{n}$

$=\frac{9+15+24}{12}$

$=4$

The total sum of squares is

$SST=\sum {\left({\overline{x}}_i-\overline{x}\right)}^2$

The treatment sum of squares is

$SSTR=\sum n_i{\left({\overline{x}}_i-\overline{x}\right)}^2$

$=3(3-4{)}^2+5(3-4{)}^2+4(6-4{)}^2$

$=24$

The error sum of squares is

$SSE=\sum \left(n_i-1\right){\left(s_i\right)}^2$

$=(3-1)(2{)}^2+(5-1\left)\right(2.4{)}^2+(4-1)(4.2{)}^2$

$=86$

$SST=SSTR+SSE$

$=24+86$

$=110$

Given data:

The overall mean is

$\overline{x}=\frac{\sum x_i}{n}$

$=\frac{9+15+24}{12}$

$=4$

The total sum of squares is

$SST=\sum x_i^2-\frac{{\left(\sum x_i\right)}^2}{n}$

$=302-\frac{{48}^2}{12}$

$=110$

The treatment sum of squares is

$SSTR=\sum \frac{T_i^2}{n_i}-\frac{{\left(\sum x_i\right)}^2}{n}$

$=\frac{81}{3}+\frac{225}{5}+\frac{24\left(24\right)}{4}+\frac{48\left(48\right)}{12}$

$=24$

The error sum of squares is

$SSE=SST-SSTR$

$=110-24$

Given data:

The mean treatment of the sum of squares is

$MSTR=\frac{SSTR}{k-1}$

$=\frac{24}{3-1}$

$=12$

The mean error of the sum of squares is

$MSE=\frac{SSE}{n-k}$

$=\frac{86}{12-3}$

$=9.556$

The F- static is

$F-\text{static}=\frac{MSTR}{MSE}$

$=\frac{12}{9.556}$

$=1.26$

The ANOVA table is

$\begin{array}{|lllll|}\hline \text{Source}& df& SS& MS=SS/df& \text{F-statistic}\\ \text{Treatment}& 2& 24& 12& 1.256\\ \text{Error}& 9& 86& 9.56& \\ \text{total}& 11& 110& & \\ \hline\end{array}$