1
$\begingroup$

This question is based on this question: How to calculate a confidence level for a Poisson distribution? and its answers.

In that post, the question is "How do I calculate the confidence interval of a poisson distribution with $n = 88$ and $\lambda =47.18$?"

The answer came as

$$ \lambda \pm 1.96\sqrt{\dfrac{\lambda}{n}}, $$

for the upper and lower bounds. It might be noted that this is an approximation which is okay when $n\lambda$ is big enough -- whatever big enough might be, apparently $4152$ is big enough.

Now, as far as I know -- this might be completely wrong since I haven't really properly studied statistics yet -- the confidence interval gives you an interval such that the probability that the mean is in this interval is $95\%$. So, I'd think the probability of being outside this interval would be $2.5\%$? But it's not. So I'm confused.

Improve this question
$\endgroup$
  • 1
    $\begingroup$ "the confidence interval gives you an interval such that the probability that the mean is in this interval is 95%" is not quite true. Indeed, in this framework, the unknown mean is fixed (either it lies inside the interval or it lies outside, there is no probability attached to it). Instead, you should say that the probability that the interval (which is random) covers the true mean equals 95%. Or that there is a 5% chance that the interval does not cover the true mean. $\endgroup$ – ocram Apr 19 '18 at 11:49
  • $\begingroup$ Ok, but the confidence interval with using the approximate technique above gives (45.7467, 48.617). Yet for $\lambda = 47.18$ I get $P(X \geq 49) = 0.41469$. That I don't get. $\endgroup$ – PaleBlueDot Apr 19 '18 at 11:55
  • $\begingroup$ see my answer below $\endgroup$ – ocram Apr 19 '18 at 11:59
  • $\begingroup$ You appear to be confusing the distribution of any single observation with the distribution of the sample mean. This confusion is furthered by conflating an apparent estimate of the Poisson parameter $\lambda$ with the value of the parameter itself (which is unknown). These considerations suggest you might find enlightenment by reviewing the meanings of confidence intervals and sampling distributions. $\endgroup$ – whuber♦ Apr 19 '18 at 13:47
1
$\begingroup$
  • "the confidence interval gives you an interval such that the probability that the mean is in this interval is 95%"

This is not quite true. Indeed, in this framework, the unknown mean is fixed (either it lies inside the interval or it lies outside, there is no probability attached to it). Instead, you should say that the probability that the interval (which is random) covers the true mean equals 95%. Or that there is a 5% chance that the interval does not cover the true mean.

  • "Ok, but the confidence interval with using the approximate technique above gives (45.7467, 48.617). Yet for λ=47.18 I get P(X≥49)=0.41469. That I don't get." (cf. comment below question)

$(45.7467, 48.617)$ is meant to provide information concerning the mean of the Poisson distribution, i.e. $\lambda$. Not about an observation from the Poisson distribution.

Perhaps you are interested in a prediction interval...

  • my answer to your comment below

$\Pr(X > 330)$ is the probability that the first Poisson variable (with mean 326) returns a count larger than the mean of the second Poisson variable. It does not answer your question (which I can rephrase: are the two means equivalent?).

To convince yourself, you can compute Pr(Pois(326) > 326), i.e. just as if the second Poisson has the same mean as the first ("equivalence").

What you might want to do instead is to construct a (90%) confidence interval for the mean difference and check that it lies within some equivalence margins. This is called an equivalence test (for means) and this approach is known as TOST.

Improve this answer
$\endgroup$
  • $\begingroup$ Thank you for your answer, but I'm still quite confused. I'll tell you what I'm looking for: I've been doing measurments with radioactive sources and am trying to use decent statistics to determine that I've only got gamma radiation. So, without an absorbing material I've got a mean of 326 measurments in 10 seconds. This is poisson distributed because it's just counts (on a GM-tube) in 10 second intervals. The mean comes from 20 measurments. Then I've put a piece of paper in between the source and the GM-tube, and now I've got a mean of 330 counts in 10 seconds (also in 20 readings). $\endgroup$ – PaleBlueDot Apr 19 '18 at 12:12
  • $\begingroup$ -- Continued here: I'm trying to show that the new mean, that is $ m =330$, is nothing new really. I want to show that the average counts per 10 seconds hasn't really changed. What I've done is: $P(X \geq 330) = 0.42 \geq 0.05$, and from that I've concluded that with 95% certainty, the mean hasn't changed. Can I also just calculate an approximate confidence level, which would be (317.89, 333.712) and say that since 330 is in there, the mean hasn't changed with 95% certainty? $\endgroup$ – PaleBlueDot Apr 19 '18 at 12:17
  • $\begingroup$ I have edited my answer to address your comment. $\endgroup$ – ocram Apr 19 '18 at 12:25
  • $\begingroup$ I'm sorry, but I'm not convinced that the proposition "$Pr(X>330)$ is the probability that a Poisson variable with mean 330 returns a count larger than its mean" is correct. I've calculated $Pr(X \geq 330)$ for a poisson distribution with a mean of 326. So isn't the probability I'm calculating the probability that in a poisson distribution with a mean of 326 the chance that I get counts more than or equal to 330? $\endgroup$ – PaleBlueDot Apr 19 '18 at 12:39
  • $\begingroup$ OK, I misunderstood. It returns the probability that the first Poisson gives a count larger than the mean of the second Poisson. Still, it does not answer your question. $\endgroup$ – ocram Apr 19 '18 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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