Hypothesis testing with Poisson Data

Question

The random variable $X$ is $Po(\theta)$ distributed, with an observed value of $x=5$.

Im asked to test the hypothesis:

$H_0: \quad \theta \leq 3$

$H_1: \quad \theta >3$

I haven't really understood the whole concept behind hypothesis testing. But since $x=5$ is an observed value, $\hat{\theta}=5 $ can be used as a point estimator of $\theta$. How do I proceed now? Do I reject the null hypothesis since $\hat{\theta} >3$ ?

Answer 1

Several approaches are possible. One is to get an exact P-value and reject $H_0$ at the 5% level if it is smaller than 0.05.

Intuitively, you have observed $X = 5$ which might be taken as evidence that $\theta > 3.$ The question is whether 5 is enough bigger than 3 to be considered 'significantly' bigger and thus reject $H_0.$

Formaly, the $=$-sign in the null hypothesis determines the 'null distribution' used in testing. Here that's $\mathsf{Pois}(\theta = 3.)$ The P-value is the probability of a result 'as extreme or more extreme' than 5 (in the direction of $H_1.)$ That means we want $P(X \ge 5\,|\,\theta = 3).$ You can evaluate that using the Poisson PDF function, using a printed table of Poisson probabilities (if available), or using software. (I don't think this is a good situation for a normal approximation.) In R statistical software (where ppois is a Poisson CDF) we use $P(X \ge 5) = 1 - P(X \le 4) = 0.1845.$ Thus the P-value exceeds 5% and we do not reject $H_0.$

1 - ppois(4, 3)
## 0.1847368
x = 0:4; 1- sum(dpois(x,3))
## 0.1847368

The second computation in R sums five probabilities: $P(X = 0), \dots, P(X=4),$ where $X \sim \mathsf{POIS}(3),$ which may be mildly tedious but certainly possible to do on a calculator.

In the figure below, the P-value is the sum of the heights of the black bars to the right of the vertical red dashed line.

Note: You might be wondering just how large $X$ would have to be in order to reject $H_0.$ The computation in R below shows that $X = 7$ would lead to rejection at the $3.34\%$ level.

 qpois(.95, 3)     # Inverse CDF or quantile function
 ## 6              
 1 - ppois(6, 3)
 ## 0.03350854     # P(X >= 7) = 0.034