I'm reading a paper on Truncated distributions, where some asymptotic tests are considered to determine if a data sample is truncated or not. I'm confused about the following line:

An asymptotic test based on $L(\hat{\alpha})$ (this is a known statistic) rejects $H_0$ on asymptotic level $q$ when \begin{equation} T_B = \sqrt{12}L(\hat{\alpha}) < -z_{q} \end{equation} with $P(N(0,1)>z_q) = q.$ The p-value is then given by $\phi(\sqrt{12}L(\hat{\alpha}))$.

$L(\hat{\alpha})$ is a known statistic, but that's all.

The problem is, I don't know how to compute this p-value. I thought that this is just a lower tailed Z-test, so that $\phi$ is just the normal cumulative distribution function. I'm working with a dataset in R in order to calculate the p-value and I tried it like this:

p = pnorm(sqrt{12}L(\hat{\alpha}),0,1, lower.tail=TRUE),

this however, doesn't give any right results. I'm sure it's not a programming fault in previous code, hence the question on this site, so it has to be something I don't understand or don't see.

What am I missing here?


The statement of the test is a bit confusing because the authors change the sign on the critical value. Your doing a left tailed test and the critical value is given as -z_q. Then the authors switch to giving the formula for q in terms of z_q instead of -z_q. Then they give the formula for p-value in terms of phi, which conventionally refers to the cumulative distribution, i.e. phi(x)=P(N(0,1) < x).

To answer your question, you don't want lower.tail=FALSE in the R code since you want to do a left tailed test.

  • $\begingroup$ thanks for the answer, I've changed it in my code. Unfortunately, it still doesn't add up. $\endgroup$ – Riley Dec 5 '15 at 18:04
  • $\begingroup$ @Riley, how is $L(\hat{\alpha})$ calculated? Could you post your code and the results from calculations? $\endgroup$ – KMcC Dec 6 '15 at 22:54

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