Look at the example from the lecture about the blood pressure medicine.
The data are the following:
\begin{center}
\begin{tabular}{ccc}
Medicine 1 & Medicine 2 & Medicine 3 \\
\hline
4.29 & 10.32 & 12.89 \\
11.28 & 3.23 & 15.68 \\
5.37 & 4.51 & 16.03 \\
7.89 & 4.57 & 9.43 \\
8.10 & 8.85 & 12.86 \\
11.93 & 6.23 & 11.15 \\
\end{tabular}
\end{center}

\begin{enumerate}
\item We would like to compare three mean values, the samples are independent, the variance is similar so we use analysis of variance.
\item %Ákveða hvaða $\alpha$ stigs við krefjumst.
%\\
$\alpha = 0.05$ að venju.
\item
The hypotheses are
\[
H_0: \mu_1 = \mu_2 = \mu_3
\]
og
\[
H_1: \text{at least one mean different from the others.}
\]
\item
We need to calculate the sums of squares.

We have three groups so $a = 3$. We have six measurements per group so $n_1 = n_2 = n_3 = 6$ and $N = 6 + 6 + 6 = 18$.
The grand mean is:
\[
 \bar{y}_{..} = \frac{\sum_{i=1}^{a}\sum_{j=1}^{n_i}y_{ij}}{N} = \frac{4.29+10.32+12.98+11.28+...+11.15}{18} = 9.15
\]
and the averages within the groups:
\[
\bar{y}_{1.} = \frac{\sum_{j=1}^{n_1} y_{1j}}{n_1} = \frac{4.29+11.28+...+11.93}{6} = 8.14,
 \]
\[
\bar{y}_{2.} = \frac{\sum_{j=1}^{n_2} y_{2j}}{n_2} = \frac{10.32+3.23+...+6.23}{6} = 6.29,
 \]
\[
\bar{y}_{3.} = \frac{\sum_{j=1}^{n_3} y_{3j}}{n_3} = \frac{12.89+15.68+...+11.15}{6} = 13.01.
 \]

\begin{align*}
SS_T = & \sum_{i = 1}^{a} \sum_{j = 1}^{n_i} (y_{ij} - \bar{y}_{..})^2 \\
     = & (4.29-9.15)^2 + (11.28-9.15)^2 + ... + (11.15-9.15)^2 = 262.16.
\end{align*}

\begin{align*}
 SS_{Tr} = & \sum_{i = 1}^{a} n_i (\bar{y}_{i.} - \bar{y}_{..})^2 \\
         = & 6\cdot(8.14 - 9.15)^2 + 6\cdot(6.29 - 9.15)^2 + 6\cdot(13.01 - 9.15)^2 = 144.53.
\end{align*}

\[
SS_E = SS_{T} - SS_{Tr} = 117.63.
\]

Lets make a SS table:
\begin{center}
\begin{tabular}{|ccc|}
\hline
 SS & DF & MS \\
\hline
 $SS_{Tr}$ = 144.53 & $a - 1$ = 2 & $MS_{Tr} = 72.27$\\
\hline
 $SS_E$ = 117.63& $N - a$ = 15& $MS_E = 7.84$\\
\hline
 $SS_T$ = 262.16 & $N - 1$ = 17& \\
\hline
\end{tabular}
\end{center}
The value of the test statistic:
\[
F = \frac{MS_{Tr}}{MS_E} = \frac{72.27}{7.84} = 9.21.
\]
\item
We look up for $a - 1$ = 2 og $N - a$ = 15 degrees of freedom.
$F_{1-\alpha,((a-1),(N-a))}$ = $F_{0.95,(2,15)}$ = 3.68. We see that $F > 3.68$.
\item
We reject the null hypothesis and conclude that at leas one of the mean values is different from the others.
\end{enumerate}