  Given $n=mk$ $iid$ samples from $N(\theta,\sigma^2)$ with $\theta$ and
$\sigma^2$ unknown, we have two ways to construct $t$-based confidence
intervals for $\theta$. The traditional method is to treat these $n$ samples as
$n$ groups and calculate the intervals. The second, and less frequently used,
method is to divide them into $m$ groups with each group containing $k$
elements. For this method, we calculate the mean of each group, and these $k$
mean values can be treated as $iid$ samples from $N(\theta,\sigma^2/k)$. We can
use these $k$ values to construct $t$-based confidence intervals. Intuition
tells us that, at the same confidence level $1-\alpha$, the first method should
be better than the second one. Yet if we define "better" in terms of the
expected length of the confidence interval, then the second method is better
because the expected length of the confidence interval obtained from the first
method is shorter than the one obtained from the second method. Our work proves
this intuition theoretically. We also specify that when the elements in each
group are correlated, the first method becomes an invalid method, while the
second method can give us correct results. We illustrate this with analytical
expressions.
