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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.

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  • $\begingroup$ It is not possible to answer this until you define your terms! $\endgroup$ Oct 28, 2015 at 10:03
  • $\begingroup$ @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. $\endgroup$
    – SmallChess
    Oct 28, 2015 at 10:26

1 Answer 1

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$$ \begin{align} Var(\bar{X}) = \frac{1}{n^2} Var( \sum_{i=1}^n X_i ) & = \frac{1}{n^2}\left( \sum_{i=1}^n var(X_i) + \sum _{i \neq j} cov\left(X_i , X_j \right) \right) \\ &= \frac{1}{n^2} \left( n \sigma^2 + n (n-1) \rho \sigma^2 \right) \\ & =\frac{\sigma^2 }{n} \left( 1 + \left(n-1 \right) \rho \right) \end{align}$$

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!

On a side note, Walsh is a genius and his papers are always worth reading no matter how frustrating it can be at times.

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  • $\begingroup$ or rather, standardize by dividing by the square root of that. $\endgroup$
    – Glen_b
    Oct 28, 2015 at 10:31
  • $\begingroup$ @Glen_b Yes, I thought that this was implied :) $\endgroup$
    – JohnK
    Oct 28, 2015 at 10:32

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