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Hey, I find this problem quite difficult to handle,I would be grateful to anyone who could at least lead on the path of solving it.

The random variable X has a Poisson distribution with mean µ. The value of µ is known to be either 1 or 2 so the following hypotheses are set up.

$H_0$ : µ = 1; $H_1$ : µ = 2

A random sample $x_1$ , $x_2$ , ..., $x_{10}$ of 10 observations is taken from the distribution of X and the following critical region is defined.

\begin{eqnarray*} \sum_{i\in 1}^{10}x_i \geqslant 15 \end{eqnarray*}

Determine the probability of a Type 1 and Type 2 errors.

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    @joriki, yes, I did change my name, sorry for the inconvience.2011-04-20

3 Answers 3

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What you are being asked it to find is

  1. The probability of being in the critical region if $H_0$ is true
  2. The probability of not being in the critical region if $H_1$ is true

The following property of Poisson distributions will also be useful:

If $X_i \sim \mathrm{Pois}(\lambda_i)$ follow a Poisson distribution with parameter $\lambda_i$, and $X_i$ are independent, then $Y = \sum_{i=1}^N X_i \sim \mathrm{Pois}\left(\sum_{i=1}^N \lambda_i\right)$ also follows a Poisson distribution whose parameter is the sum of the component parameters.

Perhaps you could try to take this forward and then we could comment.

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    @joriki: I don't think the original question was posed in Bayesian terms.2011-04-20
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1) you want to get $P$(rejecting $H_0$ | $H_0$ is true), so when will we reject $H_0$, we will do so if the sum of the 10 observations is bigger than or equal 15. and when does $H_0$ is true, it's true when $\mu = 1$. let the summation of the observations be $Y$. $\mu$ of $Y = E(10X) = 10E(X)$, since $\mu$ of $X=E(X)=1$, $\mu$ of $Y= 10 \times 1 =10$. since we want to reject it, therefore we want $P(Y \ge 15)=1-P(Y<14)$ where $Y$ is Poisson distributed with $\mu=10$ by using GDC --> 1-poissoncdf(10, 14)= 0.08346

2) follow the same method :)

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1-poissoncdf(10, 14)= 0.08346 is the correct answer

But only one issue, Why is it P(Y≥15)=1−P(Y<14), isn't this a discrete distribution? Why are we using cdf?

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    try to explain more so that OP can clearly get solution of their problem2013-05-18