# Standardization for correlated observations

I am interested in a paper by John Walsh (1947), which demonstrates the effect of intraclass correlation on significance testing. The full text is available here.

In outlining a proof, Walsh states "It is easily seen from elementary considerations that

$$\frac{(\bar{x} - \mu)\sqrt{n}} {\sigma\sqrt{1 + (n - 1)\rho}}$$

has the distribution N(0 , 1)" Where $\rho$ is the intraclass correlation.

Unfortunately, this is not so elementary for me. I would be grateful if someone can explain why this random variable comes from the standard normal distribution.

• It is not possible to answer this until you define your terms! Oct 28, 2015 at 10:03
• @kjetilbhalvorsen It's not necessary to define the terms. It's obviously a standardised version of the average. JohnK has a simple but nice proof. Oct 28, 2015 at 10:26

\begin{align} Var(\bar{X}) = \frac{1}{n^2} Var( \sum_{i=1}^n X_i ) & = \frac{1}{n^2}\left( \sum_{i=1}^n var(X_i) + \sum _{i \neq j} cov\left(X_i , X_j \right) \right) \\ &= \frac{1}{n^2} \left( n \sigma^2 + n (n-1) \rho \sigma^2 \right) \\ & =\frac{\sigma^2 }{n} \left( 1 + \left(n-1 \right) \rho \right) \end{align}
where we have used the fact that $cov\left(X_i, X_j \right) = \rho \sigma_i \sigma_j$ and the equal variances of your observations. So standardize by that in order to get the standard normal distribution!