Equicorrelation of deviations from a grand mean In the following article:
Nelson, L. S. "EXACT CRITICAL-VALUES FOR USE WITH THE ANALYSIS OF MEANS." Journal of Quality Technology 15.1 (1983): 40-44.
Nelson states that "When ANOM [Analysis of Means] is applied to k means based on equal sample sizes, their deviations from the grand mean are all equicorrelated with correlation -1/(k-1)."
What exactly does this mean? How can I calculate the correlation stated above? If I generate k vectors and each vector's mean, then calculate each mean's deviation from the grand mean, I end up with k deviations. What do I compare those to to get a correlation of -1/(k-1)?
 A: Lets do the algebra, let $X_{ji}, j=1,\dotsc,k ; i=1,\dotsc,n$ be independent observations from $k$ groups, each with expectation $\mu_j, j=1,\dotsc,k$ and variance $\sigma^2$.  
The group means is $\bar{X}_j= \frac1{n}\sum_i X_{ji}$ and the total mean is $\bar{\bar{X}} = \frac1{nk} \sum_j \sum_i X_{ji} = \frac1{k} \sum_j \bar{X}_j$  with variances $\sigma^2/n$ and $\sigma^2/nk$, respectively. We calculate (for $j\not = l$):
$$  \DeclareMathOperator{\var}{\mathbb{V}ar}
    \DeclareMathOperator{\cov}{\mathbb{C}ov}
    \DeclareMathOperator{\corr}{\mathbb{c}or}
   \cov(\bar{X}_j-\bar{\bar{X}},\bar{X}_l-\bar{\bar{X}}) = 
 \cov(\bar{X}_j,\bar{X}_l) - \cov(\bar{X}_j,\bar{\bar{X}})-\cov(\bar{X}_l,\bar{\bar{X}}) + \cov(\bar{\bar{X}},\bar{\bar{X}})
$$
Each of the terms:
$$
   \cov(\bar{X}_j,\bar{X}_l) = 0 \quad (j\not= l) \\
   \cov(\bar{X}_j,\bar{\bar{X}}) = \cov(\bar{X}_l,\bar{\bar{X}}) = \\\cov(\bar{X}_j, \frac1k(\bar{X}_1+\dotso+\bar{X}_j+\dotso+\bar{X}_k)) = 
  \cov(\bar{X}_j,\frac1k \bar{X}_j)= \frac1k \cov(\bar{X}_j,\bar{X}_j) =
\frac1k \var(\bar{X}_j )= \sigma^2/nk \\
\cov( \bar{\bar{X}}, \bar{\bar{X}})=\var(\bar{\bar{X}})=\sigma^2/nk
$$
Putting everything together, we have 
$$
\cov(\bar{X}_j-\bar{\bar{X}},\bar{X}_l-\bar{\bar{X}}) = 0 - \sigma^2/nk - \sigma^2/nk + \sigma^2/nk = -\sigma^2/nk
$$
and finally the correlation is 
$$
   \corr(\bar{X}_j-\bar{\bar{X}},\bar{X}_l-\bar{\bar{X}}) = 
\frac{-\sigma^2/nk}{\sigma^2/n}= -\frac1k
$$
and since this correlation is the same for each pair $(j,l) \quad j\not= l$ we say it is equicorrelated. Your source above gave $k-1$ not $k$ in the denominator, you should check if you cited correctly, because what we have here is correct.  You can check that intuitively by looking on the case $k=2$, your formula gives then a correlation $-1$ which cannot be correct. 
