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I need help with the following exercise:

Assume the amount of apples falling to the ground from a single tree can be modeled by a poisson distributed random variable $X$ with expectation $m$.

The apples from 4 trees are collected and counted every day for 4 days with the results:

Day 1: 9

day 2: 9

day 3: 10

day 4: 8

The owner of the trees hopes that the average amount of apples falling to the ground per day per tree is less than or equal to 4. Help the owner test that assertion.

I have made my own attempt at a solution where i put the following hypothesis:

Design a test for the hypothesis: $$H_0:m=4$$ $$H_1: m> 4$$

While this is not exactly whats asked for it allows me to approximate the poisson distributions with normal distributions, which is what we will see next.

Let $X_i$ denote the fallen fruit from $4$ trees during day $i$, then $X_i\in Po(4m)$. Further let $\bar {X}= \frac{1}{4}\sum_{i=1}^4X_i$ be an estimator of $4m$.

If the null hypothesis is true we can approximate $\bar{X}$ with a normal distribution, $4m>15$.

$$X\in Po(16) \sim N(16,4)$$ $$Z=\frac{\bar X - 16}{2} \in N(0,1)$$

Next we define a critical region $C$ for which we reject the null hypothesis:

$$P(\bar{X}\in C|m=4)\sim P(Z>t')=\alpha \implies t'=\lambda_\alpha $$ $$P(Z>\lambda_\alpha)=P(\frac{\bar X - 16}{2} > \lambda_\alpha)=P(\frac{1}{4}\sum_{i=1}^4X_i>16 + 2\lambda_\alpha)$$

Hence our critical region $C$ can be set to $C=[16,\infty)$. Since our sample gives us $\bar x = 9$ our null hypothesis can not be rejected on any significance.

This gives us some notion of the amount of apples falling to the ground but I have this lingering feeling that I am missing the point and proceeding in the wrong way.

Any help is much appreciated.

Regards, Tobias

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Sum of independent poisson processes is still a poisson process. I don't know statistics and how to design a test. –  mez Jan 28 '13 at 12:04
    
Yeah I knew that, but thanks anyway –  user25470 Jan 28 '13 at 12:21

1 Answer 1

up vote 0 down vote accepted

As mezhang pointed out, the sum of independent Poisson processes is again a Poisson process. Thus, under the hypothesis that $4$ apples fall to the ground per day per tree, the total number of apples collected and counted would follow a Poisson distribution with mean $64$. A plot of the cumulative distribution function shows that the observed value of $36$ allows this hypothesis to be rejected with high significance; indeed a computation shows that the probability of at most $36$ apples being collected and counted is less than $0.0001$ under that hypothesis. Any hypothesis that even more apples fall could be rejected with even greater significance, so the owner can safely conclude that less than $4$ apples fall to the ground per day per tree.

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Thanks, I didn't realize I would have access to the distribution for Po(m) during the exam, but I will so this method is great. Just out of curiosity, is it possible to calculate the sum for the probability by hand? –  user25470 Jan 28 '13 at 13:34
    
@user25470: I doubt it. –  joriki Jan 28 '13 at 13:47

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