= Solution
Group means are <independent random variables> with <normal distributions>
$$
\overline Y_i\sim N(\mu+\alpha_i,\sigma^2/J).
$$
The resulting univariate <sampling distributions> are
$$
\boxed{\widehat\mu\sim N(\mu,\sigma^2/J),\qquad
\widehat\alpha_2\sim N(\alpha_2,2\sigma^2/J).}
$$
The same <variance> calculation holds for every $i\geq2$, since the difference involves two independent group means. Therefore the <standard errors> satisfy
$$
\frac{\operatorname{se}(\widehat\alpha_i)}{\operatorname{se}(\widehat\mu)}
=\frac{\sqrt{2}\sigma/\sqrt J}{\sigma/\sqrt J}=\boxed{\sqrt2}.
$$
Replacing $\sigma$ by a common estimated error <standard deviation> preserves this ratio. Although the individual group means are independent, the contrasts $\widehat\alpha_i$ share the baseline mean and have <covariance> $\sigma^2/J$ for distinct $i\geq2$.
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