Let's say $P(X=0)$ for a Poisson distribution is $0.18,$ and the $P(X ≥ 5) = 0.04.$ [See OP clarification in Comment.]

Obviously, $X ≥ 5$ is a critical region. But because $X ≥ 0$ for a Poisson distribution, does that mean there is no lower bound critical region?

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    $\begingroup$ For $X \sim \mathsf{Pois}(\lambda),$ from $P(X = 0) = 0.18$ one deduces that $\lambda \approx 1.715.$ But then $P(X \ge 5) \approx 0.03$ because in R 1-ppois(4, 1.715) returns 0.03057812. So I wonder what null hypothesis you are trying to test and against what alternative. // Please clarify. $\endgroup$ – BruceET Jan 10 at 21:51
  • $\begingroup$ It was simply a hypothetical scenario, asking whether X ≥ 5 would be the only critical region even though the hypothesis test is two tailed $\endgroup$ – Sam Connell Jan 10 at 21:58
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    $\begingroup$ OK, but I'm saying that I can't reconcile numbers 0.18 and 0.04 stated in your question. How did those numbers arise? // One of us may be able to make up a similar example that has consistent values and illustrates difficulties finding critical region(s) with exact probabilities for tests on parameters of discrete distributions such as Poisson. But the question you ask is self-contradictory, so I can't see how anyone can answer it as it stands. $\endgroup$ – BruceET Jan 10 at 22:22
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    $\begingroup$ By contrast, it easier to get crit vals to test $H_0: \lambda = 30$ vs $H_a: \lambda \ne 30.$ Then $X \sim \mathsf{Pois}(\lambda = 30)$ has $P(X \le 19) \approx 0.022$ and $P(X \ge 42) \approx 0.022,$ so you could test at level $0.044 \approx 0.05$ using 19 and 42 as lwr & upr crit values. // As you suggest, the difficulty in achieving exactly 5% level test is the discreteness of Poisson dist'ns. Sometimes people use normal aprx to Poisson. With normal aprx you can fool yourself into thinking you have exactly 5% level, but then the discreteness difficulty is just "swept under the rug." $\endgroup$ – BruceET Jan 11 at 18:58
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    $\begingroup$ OP has clarified in a Comment that the inconsistent values are just made-up hypothetical values. With that clarification I have given 'answers' in a couple of Comments. Pretty clearly not a HW problem. Voting to 'reopen' so someone can write a complete Answer, perhaps somewhat in the spirit of my telegraphic comments. $\endgroup$ – BruceET Jan 11 at 19:26

The Question seems to ask for a discussion of the role the discreteness of the Poisson distribution plays in finding critical values of two-sided tests about the parameter $\lambda.$ This Answer slightly refines my Comments, mainly by providing R code to compute the key probabilities. [In R, ppois (with appropriate parameters) is a Poisson CDF and qpois is a Poisson quantile function (inverse CDF).]

Small $\lambda:$ Suppose we wish to test $H_0: \lambda = 3$ vs. $H_a: \lambda \ne 3$ at the 5% level. Quantiles .025 and .975 of the null distribution $\mathsf{Pois}(\lambda = 3)$ are $0$ and $7,$ respectively.

qpois(c(.025, .975), 3)
[1] 0 7

More precisely, if $X \sim \mathsf{Pois}(3),$ then $P(X \le 0) = P(X = 0) = 0.0498$ and $P(X \ge 7) = 1 - P(X \le 6) = 0.0335.$ So if we use lower critical value $c_L = 0$ and upper critical value $c_U = 7,$ the significance level of the two-sided test is $\alpha = 0.0498 + 0.0335 = 0.0833.$ This is about as close to a 5% level test, putting roughly equal probabilities in each tail, as we can get.

[Of course, we could use $c_U = 8$ with $P(X \ge 8) = 0.0119$ to get $\alpha = 0.067$ (nearer to 5%), but then the two tail-probabilities of about 5% and about 1% are far from being equal.]

ppois(0, 3);  1 - ppois(6, 3);  1 - ppois(7, 3)
[1] 0.04978707
[1] 0.03350854
[1] 0.0119045

Note: An approach often taken in elementary texts that ignore the use of software to get Poisson probabilities, is to use a normal approximation, pretending that $Z = \frac{X-3}{\sqrt{3}},$ under $H_0,$ is approximately normal and to reject when $|Z| \ge 1.96$ for a "test at level $\alpha = 5\%.$"

It is true that a standard normal random variable puts exactly 5% of its area to the left of -1.96 or to the right of +1.96, but our $Z$ is not normal and using the continuous normal distribution only disguises the difficulty of getting a test at exactly the 5% level.

In fact according to the normal approximation, one fails to reject when $-1.96 < \frac{X-3}{\sqrt{3}} < 1.96$ or when $-0.3948 < X < 6.3948.$ Thus with this criterion, we will reject only for $X \ge 7,$ so we are really doing a one sided test that rejects for large $X$ and has actual $\alpha = 0.0335.$ We might have avoided the awkwardness of discussing the discreteness of the Poisson distribution, but we are not doing the intended test.

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Moderate $\lambda:$ Now suppose we wish to test $H_0: \lambda = 30$ vs. $H_a: \lambda \ne 30$ at the 5% level. When $\lambda$ is larger, individual Poisson probabilities tend to be smaller, and it is often easier to find a sensible rejection region with probability near 5%.

Quantiles .025 and .975 of the null distribution $\mathsf{Pois}(\lambda = 30)$ are $20$ and $41,$ respectively. We can use lower critical value $c_L = 19$ and upper critical value $c_U = 41.$ Because $X \sim \mathsf{Pois}(\lambda=30)$ has $P(X \le 19) = 0.0219$ and $P(X \ge 42) = 0.0221$ we get $\alpha = 0.0440 \approx 0.05.$

[You can explore what happens if an additional value is included in the upper and/or lower part of the rejection region. There are several alternate possibilities that give significance levels near 5%; it is customary, but not mandatory, to make $\alpha \le 5\%.$ The normal approximation amounts to using $c_L = 19, c_U = 41.]$

qpois(c(.025, .975), 30)
[1] 20 41
ppois(19, 30);  1 - ppois(41, 30)
[1] 0.02187347
[1] 0.02210704

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