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"I suppose that the model Wald test and the coefficient Wald test may even be the same test" You are right on it. In fact you can check that $z^2=$ Chi square, $(-1.196)^2 = 1.430$.

But for LRT, its validity also depends on the large sample if response variable is not normal. See Is the likelihood ratio test a large sample inference method?.

You want to select LRT because LRT produceproduces the p value that you wanted. In fact, when sample size = 6, the reliability of statistical results is low.

"I suppose that the model Wald test and the coefficient Wald test may even be the same test" You are right on it. In fact you can check that $z^2=$ Chi square, $(-1.196)^2 = 1.430$.

But for LRT, its validity also depends on the large sample if response variable is not normal. See Is the likelihood ratio test a large sample inference method?.

You want to select LRT because LRT produce the p value that you wanted. In fact, when sample size = 6, the reliability of statistical results is low.

"I suppose that the model Wald test and the coefficient Wald test may even be the same test" You are right on it. In fact you can check that $z^2=$ Chi square, $(-1.196)^2 = 1.430$.

But for LRT, its validity also depends on the large sample if response variable is not normal. See Is the likelihood ratio test a large sample inference method?.

You want to select LRT because LRT produces the p value that you wanted. In fact, when sample size = 6, the reliability of statistical results is low.

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source | link

"I suppose that the model Wald test and the coefficient Wald test may even be the same test" You are right on it. In fact you can check that $z^2=$ Chi square, $(-1.196)^2 = 1.430$.

But for LRT, its validity also depends on the large sample if response variable is not normal. See Is the likelihood ratio test a large sample inference method?.

You want to select LRT because LRT produce the p value that you wanted. In fact, when sample size = 6, the reliability of statistical results is low.