How to find a confidence interval for a contrast? I have a latin square with a particular treatment; I have done a contrast so I have a value for the mean of the contrast $$\hat \theta_1 = \mu_1-\frac{1} {3}(u_2+u_3+u_4) \>.$$ 
I also have a response error from the whole sample $\sigma^2$.
I have the variance of the contrast mean $$\mathrm{Var}(\hat \theta_1) \>,$$
the total N = 48, each treatment group size is 12, hence 4 groups.
to take a confidence interval I am trying things like
$$\bar x \pm t_{n-1}^{(1-\alpha/2)}\frac{s} {\sqrt{n} } \>,  $$
but it's not giving me anything like I am expecting. 
Is this the correct formula for a confidence interval of a contrast mean?
What is the sample variance $s^2$ in the context of the contrast?
 A: So basically, your model is given by:
$$\begin{array}{c c}Y_{ij}=\mu_{i}+\sigma\epsilon_{ij} & \epsilon_{ij}\sim N(0,1) & j=1,\dots,n_i & i=1,\dots,I\end{array}$$
Where $\epsilon_{ij}$ are pairwise independent.  And you have $n_i=12,I=4$.  Now, to create a confidence interval, we note that, because the $\epsilon_{ij}$ are independent, the within group means, $\overline{Y}_{i}=n_i^{-1}\sum_{j=1}^{n_i}Y_{ij}$ are independent, and also independent of the sum of squared errors.  This also means that any linear combination can be dealt with easily.  So we have:
$$\overline{Y}_{i}\sim N(\mu_i,n_i^{-1}\sigma^2)$$
Which means that, in your case, after a bit of calculation, we have
$$M=\overline{Y}_{1}-\frac{1}{3}(\overline{Y}_{2}+\overline{Y}_{3}+\overline{Y}_{4})
\sim N\left(\theta_1,\frac{1}{9}\sigma^2\right)$$
Where $\theta_1$ is as you have defined it in your question.  This implies that $\frac{M-\theta_1}{\frac{1}{3}\sigma}\sim N(0,1)$.  We also have:
$$\frac{(n-I)s^2}{\sigma^2}=\frac{\sum_{i=1}^{I}\sum_{j=1}^{n_i}(Y_{ij}-\overline{Y}_i)^2}{\sigma^2}\sim\chi^2_{n-I}$$
We also have the standard result that $Z\sqrt{\frac{v}{\chi^2_v}}\sim T_v$ (i.e. a T-distribution with v degrees of freedom),  so we get:
$$\frac{M-\theta_1}{\frac{1}{3}\sigma}\sqrt{\frac{n-I}{\frac{(n-I)s^2}{\sigma^2}}}=\frac{M-\theta_1}{\frac{1}{3}s}\sim T_{n-I}$$
So a confidence interval would be (noting that you have $48-4=44$ degrees of freedom)
$$\overline{Y}_{1}-\frac{1}{3}(\overline{Y}_{2}+\overline{Y}_{3}+\overline{Y}_{4})\pm 
t_{1-\alpha/2}(44)\frac{s}{3}$$
