# Poisson distribution confidence interval

Edit : this is the data I used for the first part of the problem : \begin{matrix} Rocks & 0& 1& 2& 3& 4& 5& 6 & 7\\ samples & 12& 27& 28& 19& 8& 3& 1 & 1\\ \end{matrix}

I did hypothesis to test if it follows poisson or binomial and concluded it could follow both.

## Part 2

Let p be the probability that a sample taken on a land contains strictly more than one rock.

A) What is the p value if we suppose the following : finding rock in samples follows a Poisson P(2) with $\lambda = 2$ ?

For this part I just do Poisson average is 2 and random is 1 : $0.594$

B) What is the empirical proportion $\hat p$ of the samples that contains $> 1$ rock ?

For this part, I have, using the continuity correction ( +0.5) :

$P(X > 1 - 0.5 - 2) / \sqrt{2} = p(z > 0.5 - 2) / 1.41 = -1.0638$

$= p(z < 1.0638)$

my answer : everything below 1.06 standard deviation left from the mean. Am I right with my interpretation of empirical proportion ?

C) Give a 98 % confidence interval for $p$

I found an interval using this calculator but I'm not sure if it's right to use 1 as my observed events value.

$[0.01, 6.64]$

also found this is the formula to use $\lambda \pm 1.96\sqrt{\lambda / n}$

but in my case it's going to be $\lambda \pm 1.96 * \sqrt{1}$ ?

D) Test at $\alpha = 5%$ the hypothesis $p = 0.5$ versus $p > 0.5$

I don't know where to start this part.

• Because this is a homework question, you should add the self-study tag. See stats.stackexchange.com/tags/self-study/info – Patrick Coulombe Apr 17 '14 at 1:05
• Something seems to be missing here. Confidence intervals, hypothesis tests and empirical proportions (or estimators) are usually computed based on observed data. There does not seem to be any data here. Have you left out some part of the problem text? – MånsT Apr 17 '14 at 13:16
• @MånsT You're right. I added it now. I though it was used for that part of the problem. – Dave Apr 17 '14 at 14:51
• For (A) you haven't actually supplied a p-value. In (B), all you have to do is count and add the data. For (C), it's dangerous to use statistical tools you don't understand! (D) looks just like (A) but with $2$ replaced by $0.5$ and it's a one-sided instead of two-sided test. – whuber Apr 17 '14 at 15:44