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Suppose the number of customers \(X\) that enter a store between the hours \(9: 00\) a.m. and \(10: 00\) a.m. follows a Poisson distribution with parameter \(\theta\). Suppose a random sample of the number of customers that enter the store between \(9: 00\) a.m. and \(10: 00\) a.m. for 10 days results in the values $$ \begin{array}{llllllllll} 9 & 7 & 9 & 15 & 10 & 13 & 11 & 7 & 2 & 12 \end{array} $$ (a) Determine the maximum likelihood estimate of \(\theta\). Show that it is an unbiased estimator. (b) Based on these data, obtain the realization of your estimator in part (a). Explain the meaning of this estimate in terms of the number of customers.

Short Answer

Expert verified
The maximum likelihood estimate for the Poisson distribution parameter θ is 9.3. This estimate is unbiased. The interpretation of θ^ is that, on average, we expect about 9.3 customers to enter the store between 9:00 a.m. and 10:00 a.m.

Step by step solution

01

Understand the Poisson distribution and its MLE

A random variable X follows a Poisson distribution if its probability mass function is given by \(P(x; θ) = e^{-θ} * θ^{x} / x!\), where x>=0 and θ>0. Here, θ is the mean and variance of the distribution. The maximum likelihood estimate (MLE) θ^ for a Poisson distribution is given by \(\θ^ = ∑x / n\), where ∑x is the sum of the data points and n is the number of data points.
02

Show θ^ is an unbiased estimator

An estimator is unbiased if its expected value is equal to the parameter it is estimating. For a Poisson distribution, θ is the population mean. So, we want to show that E[θ^] = θ. Since θ^ is the sample mean, E[θ^] = θ by the law of large numbers, which states that the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed.
03

Calculate θ^ using the data

We can now use the MLE formula to calculate the MLE of θ using the given data: θ^ = (9 + 7 + 9 + 15 + 10 + 13 + 11 + 7 + 2 + 12)/10 = 93/10 = 9.3
04

Interpret the estimate θ^

The estimate θ^ = 9.3 suggests that on average, we expect about 9.3 customers to enter the store between 9:00 a.m. and 10:00 a.m.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{9}\) be a random sample of size 9 from a distribution that is \(N\left(\mu, \sigma^{2}\right)\) (a) If \(\sigma\) is known, find the length of a \(95 \%\) confidence interval for \(\mu\) if this interval is based on the random variable \(\sqrt{9}(\bar{X}-\mu) / \sigma\) (b) If \(\sigma\) is unknown, find the expected value of the length of a \(95 \%\) confidence interval for \(\mu\) if this interval is based on the random variable \(\sqrt{9}(\bar{X}-\mu) / S\). Hint: \(\quad\) Write \(E(S)=(\sigma / \sqrt{n-1}) E\left[\left((n-1) S^{2} / \sigma^{2}\right)^{1 / 2}\right]\). (c) Compare these two answers.

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