If there is a significant difference between treatment groups there may be a interest in what that difference is. The difference between two treatment groups is a contrast and often called pairwise comparison. The difference between the means of two treatment groups is defined as:
\begin{equation*}
d=\mu_i-\mu_j
\end{equation*}
and is estimated as
\begin{equation*}
\hat{d}=\hat{\mu}_i-\hat{\mu}_j=\bar{y}_i.-\bar{y}_j.
\end{equation*}
The expected value of $\hat{d}$ is $E[\hat{d}]=\mu_i-\mu_j$ and the variance of $\hat{d}$ is $V[\hat{d}]=\sigma^2(\frac{1}{n_i}+\frac{1}{n_j})$ which is estimated by $MSE(\frac{1}{n_i}+\frac{1}{n_j})$. The difference $d$ is normally distributed and therefore a confidence interval of the difference is:
\[\hat{d} \pm t_{1-\alpha/2,N-I}\times \sqrt{MSE \bigg(\frac{1}{n_i}+\frac{1}{n_j} \bigg)}\]

If the interest is in testing if there is a difference between more than two treatment group means methods of multiple comparisons should be applied to control the type I error rate.

In the rat diet experiment a difference between the protein and carbohydrate diet group is:
\[\hat{d}=\hat{y}_2.-\hat{y}_3.=62.6-61.3=1.3\]
$MSE$ was found to be 14.9 and the table value of the t-distribution with 13 degrees of freedom is 2.16 and hence the confidence interval of the difference is:
\[1.3 \pm 2.16 \times \sqrt{14.9\bigg(\frac{1}{5}+\frac{1}{6}\bigg)}\]
\[-3.7 \le \mu_2-\mu_3 \le 6.3\]
A significance test can also be carried out to test if the treatment group means are the same. $\frac{\hat{d}-d}{ \sqrt{MSE (\frac{1}{n_i}+\frac{1}{n_j})}}$ follows a t-distribution with $N-I$ degrees of freedom. The null and alternative hypothesis are \[H_0: \mu_i-\mu_j=0\]
\[H_1: \mu_i-\mu_j\ne 0\]
The t statistic is then
\[t=\frac{\hat{d}-0}{ \sqrt{MSE \bigg(\frac{1}{n_i}+\frac{1}{n_j} \bigg)}}\]
In the rat diet experiment the test statistic for testing if the protein and carbohydrate group means are the same is:
\[t=\frac{1.3-0}{\sqrt{14.9\bigg(\frac{1}{5}+\frac{1}{6}\bigg)}}=0.56\]
the table value is $t_{0.975,13}=2.16$ and because $0.56=t<t_{0.975,13}=2.16$ the null hypothesis can not be rejected.\\