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Disclaimer: This is a question I couldn't solve. I feel it is ethical to state this here so you can evaluate how much of your solution will share through either statements or hints.

Let $X_{1},…,X_n $ be an $i.i.d.$ $Poiss(\lambda)$ distributed random variable for some unknown $\lambda . 0$. Now consider the following hypothesis test:

$H_0:λ=λ_0$ v.s. $H_1:λ≠λ_0$ where $λ_0>0$.

Question: What is the asymptotic p-value?

Here's what I know:

Let $\overline{X}_n$ be the average of the random variable $X$. Thanks to the Central Limit Theorem, I can write:

$\dfrac{|\overline{X}_n - \lambda_0|}{\sigma/\sqrt{n}} \xrightarrow{(d)} N(0,1)$ as $n$ goes to infinity.

Given it's a two-sided test, according to [1] I can write

$P_{\lambda_0}\left(\dfrac{|\overline{X}_n - \lambda_0|}{\sigma/\sqrt{n}} > z_{\alpha/2}\right)$ $\rightarrow \alpha$

$P_{\lambda_0}\left(|W| > z_{\alpha/2}\right)$ $\rightarrow \alpha$

where $W$ is the Wald Test and $z_{\alpha/2} = \Phi^{-1}(1 - \alpha/2)$

Also from [1], I can write

$p-value = P_{\lambda_0}(|W| > |z_{\alpha/2}|) = 2.\Phi(-|z_{\alpha/2}|)$

This is where I'm stuck. I know all of these statements are true, however, I cannot work with the knowledge I have to obtain the answer I need. I'm still re-reading about hypothesis testing and p-values to have a better grip on problems like this.


References

[1] Wasserman, Larry. All of statistics: a concise course in statistical inference. Springer Science & Business Media, 2004, second edition.

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    $\begingroup$ The p-value is the biggest $\alpha$ such that the hypothesis test doesn‘t reject, or the smallest $\alpha$ such that the test rejects. $\endgroup$ – AlexR Jun 17 at 21:02
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Significance level. Let $n = 100$ (sufficiently large that an asymptotic test has a chance) and $\lambda_0 = 10.$ Then it seems safe to use $Z = \frac{\bar X - \mu_0}{\sqrt{\mu_0/n}}$ as a test statistic, rejecting $H_0: \mu = 10$ against $H_a: \mu \ne 10,$ at the 5% level if $|Z| \ge 1.96.$ [Notice that if $X \sim \mathsf{Pois}(\lambda),$ then $E(X) = Var(X) = \lambda.]$

set.seed(617)
n = 100;  lam.0 = 10;  se = sqrt(lam.0/n)
a = replicate(10^6, mean(rpois(n,lam.0)))
z = (a-lam.0)/se
mean(abs(z)>=1.96)
[1] 0.051348

This test has nearly the 5% level, as anticipated.

P-value. The P-value is the probability under $H_0$ that the test statistic would take a value farther from $0$ than observed. Using the R function pnorm (for which the default, without specifying parameters other than 0 or 1, is $\Phi)$ we get the following:

pv = pnorm(-abs(z)) + 1 - pnorm(abs(z))
summary(pv)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.0000  0.2549  0.5066  0.5000  0.7518  1.0000

In general, under $H_0$ for a continuous test statistic, the P-value has a standard uniform distribution. Our test statistic, based on Poisson data, is actually discrete with small increments. Among the $100\,000$ simulated tests, there were only 286 uniquely different values of the test statistic $Z.$

length(unique(z))
[1] 286

Thus, the histogram of our P-value is only roughly standard uniform. The important thing for testing at the 5% level is that the probability of the left-most bar is nearly 5%.

hist(pv, prob=T, col="skyblue2", main="Histogram of P-values")
  curve(dunif(x), 0, 1, add=T, col="red")

enter image description here

Power against specific alternatives. The power of this test against the particular alternative $\lambda = 12$ is above 99%.

set.seed(2020)
n = 100;  lam.0 = 10;  se = sqrt(lam.0/n)
a = replicate(10^6, mean(rpois(n,12)))   # alternative 12
z = (a-lam.0)/se
mean(abs(z)>=1.96)
[1] 0.99998

And its power against the particular alternative $\lambda = 9$ is about 90%.

set.seed(2021)
n = 100;  lam.0 = 10;  se = sqrt(lam.0/n)
a = replicate(10^6, mean(rpois(n,9)))     # alternative 9
z = (a-lam.0)/se
mean(abs(z)>=1.96)
[1] 0.899814

So for the parameters used in the simulation, the test seems to work as intended.

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