Question

Let \(X_{1}, X_{2}, \ldots, X_{10}\) denote a random sample of size 10 from a Poisson distribution with mean \(\theta .\) Show that the critical region \(C\) defined by \(\sum_{1}^{10} x_{i} \geq 3\) is a best critical region for testing \(H_{0}: \theta=0.1\) against \(H_{1}: \theta=0.5 .\) Determine, for this test, the significance level \(\alpha\) and the power at \(\theta=0.5 .\) Use the \(\mathrm{R}\). function Ppois.

Short answer

To sum up, calculate the significance level with R function Ppois and check the critical region's appropriateness. The power of the test when \(\theta = 0.5\) is also based on the Poisson probabilities which are computed with Ppois in R.

Step-by-step solution

Define The Null and Alternative Hypotheses

The null hypothesis \(H_0\) is that the mean of the Poisson distribution, \(\theta\), is 0.1. The alternative hypothesis \(H_1\) is that \(\theta\) is 0.5.

Define the Critical Region

The critical region \(C\) is defined by the condition where the sum of all the observations in the sample, denoted by \(\sum_{1}^{10} x_{i}\), is greater than or equal to 3, i.e. \(\sum_{1}^{10} x_{i} \geq 3\).

Check the Critical Region

Now check that the critical region \(C\) is best for the given test. With \(H_0\), and hence \(\theta = 0.1\), the probability of being in \(C\) should be as small as possible. With \(H_1\), and hence \(\theta = 0.5\), it should be as large as possible. In other words, any sample point with a larger likelihood under \(H_1\) than under \(H_0\) should be in \(C\).

Determine the Significance Level (\(\alpha\))

The significance level \(\alpha\) of a statistical hypothesis test is the probability of rejecting the null hypothesis, given that it is true. In other words, \(\alpha\) is the probability of making a Type I error. You can calculate \(\alpha\) by summing up the probability of each sample point in \(C\) under \(H_0\) using the Ppois function in R. Since \(\theta = 0.1\) under \(H_0\), the number of occurrences \(k\) in a Poisson distribution has the probability: \(P(k) = e^{-0.1} * 0.1^k / k!\). So, \(\alpha = 1 - Ppois(2, 1)\), because sample values are 0, 1, or 2 with \(\sum_{1}^{10} x_{i} < 3\).

Calculate the Power of the Test

The power of a statistical test at a specific alternative value of the parameter is the probability of rejecting the null hypothesis when that value of the parameter is true. For this case, calculate the power of the test at \(\theta = 0.5\) by adding up the probabilities of each sample point in \(C\) using the Ppois function in R. The number of occurrences \(k\) in a Poisson distribution when \(\theta = 0.5\) has probability: \(P(k) = e^{-0.5} * 0.5^k / k!\). The power of the test at \(\theta = 0.5\) is \(1 - Ppois(2, 5)\), because sample values are 0, 1, or 2 with \(\sum_{1}^{10} x_{i} < 3\).