Given the following hypothesis test: $H_0 : \lambda =65, H_1 : \lambda >65$ , where $\lambda$ is the parameter of an $X$ distributed as a Poisson $\alpha=0.05$ . We have n=10 samples. Using as statistics the mean $\bar X$ Find the rejection region.

I did this: Let $S = \bar Xn$ , $S $ is distributed as a $Poiss(n\lambda=650)$

Let the rejection region be $R= \{\underline X: T(\underline X)=\bar X \ge c \}$ Since the test has level $\alpha$

$\mathbb{P}( \bar X\ge \gamma|\lambda=65)\le \alpha$

$\mathbb{P}(S \ge n\gamma|\lambda=65)\le \alpha$

$\mathbb{P}(S \ge \gamma^*|\lambda=65)\le \alpha$

$nc=c^*=\min\{\gamma^* : \mathbb{P}(S \ge \gamma^*|\lambda=65)\le \alpha \}$

Can I claim $c$ is the quantile of level $1-\alpha=0.95$ ? With R:

> qpois(0.95,650)
[1] 692
> 1-ppois(692,650)
[1] 0.0488844
> 1-ppois(691,650)
[1] 0.05290045

Since 1-ppois(692,650) is the probability from 692 onwards, not including 692. I claim $c^*=693 $ as oposed to the result of the qpois command qpois(0.95,650). So the answer is $R= \{\underline X: T(\underline X)=\bar X \ge 69.3 \}$

My lecturer did this: Let $S = \bar Xn$ , $S $ is distributed as a $Poiss(n\lambda=650)$

Let the rejection region be $R= \{\underline X: T(\underline X)=\bar X \ge c \}$ Since the test has level $\alpha$

$\mathbb{P}( \bar X\ge \gamma|\lambda=65)\le \alpha$

$\mathbb{P}(S \ge n\gamma|\lambda=65)\le \alpha$

$\mathbb{P}(S \ge \gamma^*|\lambda=65)\le \alpha$

$nc=c^*=\min\{\gamma^* : \mathbb{P}(S \ge \gamma^*|\lambda=65)\le \alpha \}$

$nc=c^*=\min\{\gamma^* : 1- \mathbb{P}(S \le \gamma^*|\lambda=65)\le \alpha \}$

$nc=c^*=\min\{\gamma^* : \mathbb{P}(S \le \gamma^*|\lambda=65)\ge 1-\alpha \}$...(1)

And said that (1) was the definition of the $0.95$-quantile and that therefore $c^*$=qpois(0.95,650)$=692$ and therefore that the answer was $R= \{\underline X: T(\underline X)=\bar X \ge 69.2 \}$?

But I disagree, I think (1) is wrong and it should be

$nc=c^*=\min\{\gamma^* : \mathbb{P}(S \le \gamma^*-1 |\lambda=65)\ge 1-\alpha \}$...(1') in addition to my different answer

So I claim:

1 The R output proves (1) is not the definition of the 0.95-quantile? It's just somefing else, but I can't called it the 0.95-quantile, can I?. Of course in the continuous case it would certainly be definition of the 0.95-quantile

2 (1) is incorrect, it should be (1')

3 The answer to the problem is $R= \{\underline X: T(\underline X)=\bar X \ge 69.3 \}$?

Are my claims correct? If not, why?

  • 1
    $\begingroup$ Your version of $(1^\prime)$ is not equivalent to anything that preceded it, so how did you derive it? $\endgroup$
    – whuber
    Commented May 3, 2022 at 20:03
  • $\begingroup$ @whuber Since $\gamma^*$ is a value of a Poisson, it's an integer: $ \mathbb{P}(S \ge \gamma^*|\lambda=65)= 1- \mathbb{P}(S < \gamma^*|\lambda=65) = 1-\mathbb{P}(S \le \gamma^*-1|\lambda=65)$ and then solving $1-\mathbb{P}(S \le \gamma^*-1|\lambda=65) \le \alpha$ for the probability yields (1') $\endgroup$ Commented May 3, 2022 at 21:49
  • $\begingroup$ So what you are really complaining about--and it looks legitimate--is not $(1),$ but rather the line that immediately precedes it. $\endgroup$
    – whuber
    Commented May 3, 2022 at 22:40
  • $\begingroup$ @whuber Yes, I am saying the final result is wrong, but I wasn't sure if arguing that $\gamma^*$ is an integer was ok, after all if S is a Pois(650) ,$ P(S \le 5.9)=P(S \le 5)$ for instance $\endgroup$ Commented May 3, 2022 at 22:51

1 Answer 1


Preliminary. For samples and discrete distributions, there are various definitions of quantile. Here it seems useful to show the values returned by qpois, a Poisson quantile function. The 95th quantile of the random variable $X\sim \mathsf{Pois}(\lambda=5)$ is $9$ because that is the smallest value $q$ such that $P(X \le q) \ge 0.95.$ For a discrete distribution this is intended to be the closest useful inverse of the CDF.

qpois(.95, 5)
[1] 9
ppois(9, 5)
[1] 0.9681719  # just above 0.95, but exact unavailable
ppois(8, 5)
[1] 0.9319064  # just below 0.95.

Let's look at the graph of the CDF of $\mathsf{Pois}(5).$ (Open circles show the exact values of the CDF at relevant integer values.

k = 0:15;  CDF = ppois(k, 5)
plot(k, CDF, type="s");  points(k, CDF)
 abline(h = .95, col="red")
  abline(v = 9, col="red")
 abline(h = 0, col="green2")
  abline(v = 0, col="green2")

enter image description here

A graph for $\lambda = 65$ would be similar, but harder to read accurately.

For your question: You have the distributions of $S = \sum_{i=1}^{10} X_i \sim \mathsf{Pois}(65)$ and of the related discrete distribution $\bar X = S/n.$ You want to find $c$ such that the rejection region for $H_0: \lambda = 65$ against $H_1: \lambda > 65$ has level $\alpha = 0.05 = 5\%$ without exceeding $5\%.$

You have proposed a solution different from the one given in class; in terms of your work, @whuber has given you a fine clue as to the correct answer. Without going into details of your derivation, I will show relevant simulations. Then I hope you can finish the problem.

In the R code rej is a logical vector containing a million TRUEs and FALSEs, and mean(rej) gives the proportion of TRUEs, which is an approximation to the rejection probability when $H_0$ is true (accurate to two or three places).

rej = replicate(10^6, mean(rpois(10,65)) >= 69.1)
[1] 0.056708   # 69.1 gives answer that's too large

rej = replicate(10^6, mean(rpois(10,65)) >= 69.2)
[1] 0.052556  # 69.2: still too large

rej = replicate(10^6, mean(rpois(10,65)) >= 69.3)
[1] 0.04865    # 69.3: largest value below 5%

rej = replicate(10^6, mean(rpois(10,65)) >= 69.4)
[1] 0.045184   # below 5%, but too low
  • 1
    $\begingroup$ So are you saying my solution is wrong and the lecturer's is ok? Because I was pretty sure mine was ok, I just wanted confirmation $\endgroup$ Commented May 3, 2022 at 23:03
  • $\begingroup$ This comment/question appeared a few seconds before I finished. So not sure what you saw. Also, before you could have read it carefully. So I'm not sure what you concluded and why. $\endgroup$
    – BruceET
    Commented May 3, 2022 at 23:24
  • 1
    $\begingroup$ So for the quantile thing , you agree that, using your POISS(5) as example if I wanted a rejection region of the form $[c, +\infty]$, c = 10, and one of the points in my post was that it is NOT qpois(.95, 5)=9 directly so the set defining it as written by the lecturer did not implied to find the quantile, at least considering the definition you have given which was the same I was using. Note the lecturer had written using your example, c= qpois(.95, 5)=9 $\endgroup$ Commented May 3, 2022 at 23:35
  • 1
    $\begingroup$ Then you wrote that I should finish the problem, unless you mean I was wrong, the solution I posted was already complete, so I was looking for a correction/confirmation $\endgroup$ Commented May 3, 2022 at 23:37
  • 2
    $\begingroup$ Your answer seems OK to me. But for practical purposes, there is little difference between the two. You are working very nearly at the 5% level either way. // A more traditional approach may be to use a normal approximation to Poisson, which is not terrible for $\lambda$ as large as 65. // Also for simplicity of presentation it may be reasonable to ignore the slight inaccuracy using "95th quantile" as 692. // It is clever of you to notice the difference in order to get closer to 5% level, and you can show lecturer what you have done, but maybe not best to claim lecturer made a "mistake." $\endgroup$
    – BruceET
    Commented May 4, 2022 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.