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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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You didn't specify how you choose which hypothesis to accept -- I assume by higher likelihood? – joriki Apr 20 '11 at 13:22
You also didn't specify a prior. The probability of errors is undetermined unless you make an assumption about the prior. Are you assuming a flat (50/50) prior? – joriki Apr 20 '11 at 13:25
Type 1 error is rejecting a true $H_0$ hypothesis and Type 2 error is falsely accepting $H_0$ hypothesis. I'm quite not sure what do you mean by specifying prior. – Paul S Apr 20 '11 at 13:52
The prior is the probability distribution that you assign to the hypotheses prior to taking the sample. For instance, if you know that someone determined the mean $\mu$ by rolling a $6$-sided die and choosing $\mu=1$ if it came up $6$ and $\mu=2$ otherwise, then your prior would be $p(H_0)=1/6$ and $p(H_1)=5/6$. Often one does not have this sort of information about the hypotheses to be tested, and a usual "solution" to this problem is to arbitrarily assume a "flat" prior, in this case $p(H_0)=p(H_1)=1/2$. Without some assumption about the prior, it's meaningless to state error probabilities. – joriki Apr 20 '11 at 13:59
Also you didn't respond to my first question -- do you choose the hypothesis with the higher likelihood? If not, how do you choose? – joriki Apr 20 '11 at 14:01

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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Am I right in concluding that you're interpreting "the probability of Type 1 and Type 2 errors" to mean "the conditional probability of a Type 1 error given $H_0$ and the conditional probability of a Type 2 error given $H_1$"? – joriki Apr 20 '11 at 14:12
@joriki: I believe that conditional probabilities are part of the standard definitions, though I may be wrong. – Henry Apr 20 '11 at 14:17
Part of the standard definitions of which terms? There must be some way of talking about the probability of a Type 1 error occurring (i.e. of $H_0$ holding but being rejected) without assuming that $H_0$ holds. (At least in a Bayesian framework there must be.) So the expression "the probability of a Type 1 error" can't simply be defined to mean "the probability of a Type 1 error given $H_0$". The expression that is defined to mean that is "false positive rate":… (and analogously for "false negative rate"). – joriki Apr 20 '11 at 14:24
@joriki: I don't think the original question was posed in Bayesian terms. – Henry Apr 20 '11 at 14:42

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 problem – iostream007 May 18 '13 at 10:19

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