Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with mean \(\theta>0\) (a) Show that the likelihood ratio test of \(H_{0}: \theta=\theta_{0}\) versus \(H_{1}: \theta \neq \theta_{0}\) is based upon the statistic \(Y=\sum_{i=1}^{n} X_{i} .\) Obtain the null distribution of \(Y\). (b) For \(\theta_{0}=2\) and \(n=5\), find the significance level of the test that rejects \(H_{0}\) if \(Y \leq 4\) or \(Y \geq 17\)

Short answer

The null distribution of \(Y\) is a Poisson distribution with mean \(n\theta_{0}\). For \(n=5\) and \(\theta_{0}=2\), the significance level of the test that rejects \(H_{0}\) if \(Y \leq 4\) or \(Y \geq 17\) can be calculated using the Poisson cumulative distribution function: \(F_{Y}(4) + [1 - F_{Y}(16)]\).

Step-by-step solution

Understand the likelihood ratio test

The likelihood ratio test is used to compare the fit of two models, one of which is a special case of the other. The test is based on the likelihood ratio, which expresses how many times more likely the data are under one model than the other. In this case, \(H_{0}: \theta = \theta_{0}\) and \(H_{1}: \theta \neq \theta_{0}\). We will be testing if the data are more likely under \(H_{0}\) or \(H_{1}\)

Determine the null distribution of \(Y=\sum_{i=1}^{n} X_{i}\)

Given that each \(X_{i}\) is drawn from a Poisson distribution with mean \(\theta\), the sum will also be a Poisson distribution. This holds true because Poisson distributions are 'well-behaved' under taking sums. Hence under the null hypothesis, \(Y\) has a Poisson distribution with mean \(n\theta_{0}\)

Calculate the significance level for \(Y \leq 4\) or \(Y \geq 17\)

The null hypothesis is rejected if \(Y \leq 4\) or \(Y \geq 17\). Thus, the significance level is the probability that \(Y \leq 4\) or \(Y \geq 17\) under the null hypothesis. That gives us \(P(Y \leq 4) + P(Y \geq 17)\). Here, \(n=5\) and \(\theta_{0}=2\), so under the null distribution, \(Y\) follows a Poisson distribution with mean \(5*2=10\). To solve this, we can make use of the cumulative distribution function (CDF) for a Poisson distribution. The CDF gives the cumulative probability \(P(X \leq x)\) for a Poisson distributed random variable \(X\), and denoted as \(F_{X}(x)\). Thus, we calculate the probability as \(F_{Y}(4) + [1 - F_{Y}(16)]\)

Key concepts

Poisson Distribution

Understanding the Poisson distribution is critical when dealing with certain types of data especially in contexts involving counts or events. It is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, provided that these events occur with a known constant rate and independently of the time since the last event. For example, the number of emails one receives in a day could be modeled by a Poisson distribution if the average rate at which emails arrive is consistent.

The key parameter of the Poisson distribution is its mean, denoted by \(\theta\), which is also its variance. If you're trying to predict the probability of a specific number of events, say \(k\), occurring within a particular period, the Poisson probability mass function can be used, given by:
\[ P(X=k) = \frac{e^{-\theta} \theta^k}{k!} \]
This function tells us the probability that a Poisson random variable \(X\), which represents the number of events, takes on the value \(k\).

Null Hypothesis

In statistical hypothesis testing, the null hypothesis \(H_{0}\) is a statement or default position that there is no effect or no difference, and it is what we assume to be true before we collect any data. In the context of the Poisson distribution, the null hypothesis could be that the mean number of events per time period is a certain value, denoted as \(\theta_{0}\).

The null hypothesis serves as a starting point for statistical testing. It's the benchmark that we compare our evidence against to see if what we observe significantly deviates from what we would expect under the null hypothesis. In the given exercise, the null hypothesis is \(H_{0}: \theta = \theta_{0}\). This means that we are initially assuming that the mean of our Poisson-distributed random variable is \(\theta_{0}\) and we will test whether the data suggest otherwise with enough confidence to reject \(H_{0}\).

If after conducting the test, the data seem highly unlikely under the assumption of \(H_{0}\), we may reject the null hypothesis in favor of an alternative hypothesis \(H_{1}\), which posits that the mean is different from \(\theta_{0}\).

Statistical Significance

Statistical significance is a determination about the non-randomness of the results of a statistical test. In simpler terms, it tells us whether the differences or relationships we observe in our data could have occurred by chance, or whether they are likely indicative of a true effect.

A significance level, often denoted by \(\alpha\), is the threshold for determining whether a test statistic is sufficiently extreme to reject the null hypothesis. It represents the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true. A common choice for \(\alpha\) is 0.05, suggesting a 5% risk of concluding that an effect exists when it does not. To say a result is statistically significant is to assert that it is unlikely to have occurred by chance alone.

In the case of our likelihood ratio test for the Poisson distribution with \(n=5\) and \(\theta_{0}=2\), we calculated the probability of observing data as extreme as, or more extreme than, what we observed, assuming that the null hypothesis is true. If this probability, the p-value, is less than \(\alpha\), we would have sufficient evidence to reject the null hypothesis and conclude that our findings are statistically significant.