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I have a question about something I found in the book Frailty models in survival analysis (Wieneke, 2011). He talks about testing signficance of the frailty term. According to him, Claeskens et al. (2008) argued that one should use a mixture of chi-squared dsitributions, as

$$\tfrac{1}{2}\chi^2_0+\tfrac{1}{2}\chi^2_1,$$

to find the p-value corresponding to the Likelihood ratio test. He found a test statistic of 1.324. This should correspond to a p-value of 0.10 in a one-sided test (We test whether $\sigma^2 >0$). How did he find this p-value? Hope somebody can help me out. Kind regards.

kjetil b halvorsen
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Cardinal
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    Unless I misunderstand this notation, the correct p-value appears to be $0.125$ rather than $0.10.$ Do you have a reference we can access? – whuber Apr 24 '20 at 20:27
  • Page 82 of Frailty models in survival analysis (Wieneke, 2011). "Consequently, the models can be compared by the likelihood ratio test, which yields a test statistic χ2 = −2(−220.294+ 219.622) = 1.324. Because of the nonstandard test situation, a mixture of a (1/2)X^2_0 and (1/2)X^2_1 distribution should be used (Claeskens et al. 2008), which results in a p-value of p = 0.10. Hence, the additional parameter σ2 in the frailty model does not cause a significant improvement in the fit of the data." – Cardinal Apr 24 '20 at 20:37
  • But I agree with the 0.125 @whuber. I had something similar. Probably a mistake on the author's part(?). Maybe he meant to say it is larger than a p-value of p=0.10. – Cardinal Apr 24 '20 at 21:13

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