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Take a linear model with a block random effect ($b_i$) $$ y_{ij} = \mu + b_i + \beta x_{ij} + e_{ij} $$ with $b_i \sim N(0, \sigma_b^2)$, etc etc.

If $\sigma^2_b = 0$, then the model reduces to a very simple linear model. To test the null hypothesis $H_0: \sigma_b^2 = 0$ versus the alternative $H_1: \sigma_b^2 > 0$, one can use a likelihood ratio test. However, the limiting distribution under the null is not $\chi_1^2$, but rather $\tfrac{1}{2} \chi^2_0 + \tfrac{1}{2} \chi^2_1$, i.e., a 50:50 mixture of a point mass at zero and a chi square distribution with one degree of freedom.

How do we determine the critical value of (or the p-value using) that mixture?

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The area to the right of any point above 0 is half that of a $\chi_1^2$. So to get a level $\alpha$ test, look up the $2\alpha$ point of a $\chi_1^2$.

.... as long as $\alpha < 0.5$.

Of course, p-values work similarly. Look the value up as if it were a $\chi_1^2$ and halve the resulting p-value.`

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  • $\begingroup$ OUUUHHH, this was obvious!!! $\endgroup$ – ocram Mar 28 '13 at 10:30
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Some R code if interested:

Using TcGSA package:

ss <- 0.3
sample_mixt <- TcGSA:::rchisqmix(n=1e5, s=0, q=1)
TcGSA:::pval_simu(s=ss, sample_mixt)

Using base:

1/2*(1-pchisq(ss,df=1))
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