Consider $X$ is a Poisson random variable with distribution $X$~$Pois(\theta)$. I define the mean in my hypothesis as $\lambda$ and nominal significance level $\alpha$.
Null hypothesis $H_0 : \lambda=\theta $
Alternative hypothesis $H_1 : \lambda>\theta$
Reject $H_0$ if $P(X\geq \theta_u )\leq \alpha$
I'm trying to find $\theta_u$ using Chebyshev's inequality.
Here is my approach: According to Chebyshev's inequality $P(|X-\theta|\geq k\sqrt \theta)\leq \frac{1}{k^2}$
Since $X\geq \theta_u>\theta$ , $|X-\theta|=(X-\theta)$
So $P((X-\theta)\geq k\sqrt \theta)\leq \frac{1}{k^2}$
$P(X\geq \theta + k\sqrt \theta)\leq \frac{1}{k^2}$
Let's assume $\theta_u=\theta + k\sqrt \theta$
So, $k=\frac{\theta_u-\theta}{\sqrt \theta}$,
If $\alpha=\frac{1}{k^2}$ then $\theta_u=\theta + \sqrt\frac{\theta}{\alpha}$; Is it correct?
Direct approach: $P(X\geq \theta_u)=e^\theta\sum_{k=\theta_u}^{\infty}\frac{\theta^k}{k!}\leq \alpha$
I believe it's not possible to find a solution for $\theta_u$ through direct approach.
What is the best way to find a simplified (which can be computed easily) expression for $\theta_u$ ?
Chebyshev's Inequality is useful in proofs, but it applies to any distribution with a mean and variance, so it does not give a very sharp bound for such purposes as getting a CI. If this is a drill problem on use of Chebyshev's Inequality, then OK. But if you want a CI for the Poisson mean that has anywhere near it's advertised coverage probability, this is not a good procedure.
First, let's make it clear we have a sample of "size 1." This not a severe restriction because if we had 10 iid observations, then their sum would again be Poisson. So the $\lambda$ we wish to estimate is for a single (perhaps combined) observation.
Suppose our observation is $X \sim Pois(\lambda)$ with $\lambda$ unknown. If $X$ is reasonably large, then a good 95% CI is $X + 2 \pm 1.96\sqrt{X+1}.$
Another method is to make a frequentist CI using a Bayesian argument with an "improper" gamma prior. It requires software to compute. A statement in R that gives a 95% CI is
qgamma(c(.025,.975), x, 1). This method works well for confidence levels other than 95%.For example, if $X = 24$, then the first method gives $26 \pm 1.96\sqrt{25}$ or $(16.2, 35.8).$ and the second gives $(16.18, 35.71).$ A simulation with a million values of $X$ from $Pois(24)$ showed that 95% of CIs generated by each method actually contained $\lambda=24$ very nearly 95% of the time.
If $X = 8$, then the first gives $( 4.12, 15.88)$, and the second $(3.45, 14.42).$ Here I would prefer the second interval because of the relatively small value of $X$.
If $X$ is the sum of $n = 10\,$ observations, then divide the endpoints of the above CIs by 10, to get a CI for the per-observation rate $\lambda/10$.
Notes: (1) Wikipedia on 'Poisson' has CI based on the chi-squared distribution. (2) 'Exact' CIs based on the likelihood ratio are a little difficult to compute. One problem is that the Poisson distribution is discrete, so that it is difficult to apportion exactly 2.5% in each tail of the distribution. (3) One sometimes sees the CI $X \pm 1.96\sqrt{X}$ recommended. This method has been shown to give relatively poor results. For $\lambda = 24$, its coverage probability was only about 93%.