/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Each of 150 newly manufactured i... [FREE SOLUTION] | 91Ó°ÊÓ

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Each of 150 newly manufactured items is examined and the number of scratches per item is recorded (the items are supposed to be free of scratches), yielding the following data: \begin{tabular}{llllllllll} Number of scratches per item & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline Observed frequency & 18 & 37 & 42 & 30 & 13 & 7 & 2 & 1 \\ \hline \end{tabular} Let \(X=\) the number of scratches on a randomly chosen item, and assume that \(X\) has a Poisson distribution with parameter \(\lambda\). a. Find an unbiased estimator of \(\lambda\) and compute the estimate for the data. [Hint: \(E(X)=\lambda\) for \(X\) Poisson, so \(E(\bar{X}=\) ?)] b. What is the standard deviation (standard error) of your estimator? Compute the estimated standard error. [Hint: \(\sigma_{X}^{2}=\lambda\) for \(X\) Poisson.]

Short Answer

Expert verified
\(\lambda = 2.113\); standard error \(\approx 1.453\).

Step by step solution

01

Understanding the Poisson Distribution

For a random variable \(X\) following a Poisson distribution with parameter \(\lambda\), the expected value \(E(X)\) is equal to \(\lambda\). The task is to estimate \(\lambda\) based on the sample data provided.
02

Setting Up the Equation for \(\lambda\)

The unbiased estimator for \(\lambda\) in a Poisson distribution when given sample data is the sample mean. Therefore, \(\hat{\lambda} = \bar{X}\), where \(\bar{X}\) is the average number of scratches per item.
03

Calculating the Sample Mean \(\bar{X}\)

To compute \(\bar{X}\), use the formula: \[ \bar{X} = \frac{1}{n} \sum_{i=0}^{7} (x_i \cdot f_i) \] where \(n = 150\) is the total number of items, \(x_i\) is the number of scratches corresponding to each observed frequency \(f_i\). Compute as follows:\[\bar{X} = \frac{1}{150} (0 \times 18 + 1 \times 37 + 2 \times 42 + 3 \times 30 + 4 \times 13 + 5 \times 7 + 6 \times 2 + 7 \times 1)\] Calculating the result gives:\[ \bar{X} = \frac{1}{150} (0 + 37 + 84 + 90 + 52 + 35 + 12 + 7) = \frac{317}{150} \approx 2.113 \]
04

Computing the Standard Deviation of the Estimator

For a Poisson distribution, the variance \(\sigma_X^2 = \lambda\). Therefore, the standard deviation (standard error) of \(\lambda\) is equal to \(\sqrt{\lambda}\). Using the estimated value \(\hat{\lambda} = 2.113\), the standard error is calculated as:\[ \text{Standard Error} = \sqrt{2.113} \approx 1.453 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unbiased Estimator
When dealing with statistical data, it's crucial to have estimates that accurately reflect the true population parameters without any systematic error. An unbiased estimator is a statistic used to estimate a parameter, where the expected value of the estimator is equal to the true parameter. For the Poisson distribution with parameter \( \lambda \), the sample mean \( \bar{X} \) is an unbiased estimator. This means that, given a large number of samples, the average of these sample means will converge to the actual parameter \( \lambda \) of the population. This property is highly desirable because it indicates that our estimation method does not systematically overestimate or underestimate \( \lambda \). In this case, the unbiased estimate for the average number of scratches on a newly manufactured item is obtained by calculating the sample mean from the observed data.
Sample Mean
The sample mean, denoted as \( \bar{X} \), is a measure of the central tendency of a set of data. It is calculated by summing all observed values and dividing by the number of observations. In the context of the Poisson distribution, the sample mean serves as an unbiased estimator for the parameter \( \lambda \).
  • The formula for the sample mean is \( \bar{X} = \frac{1}{n} \sum_{i=0}^{k} (x_i \cdot f_i) \), where \( n \) is the total number of observations, \( x_i \) is each distinct value observed (number of scratches), and \( f_i \) is the frequency of each value.
  • This formula allows you to compute the average number of scratches per item by considering both the value and frequency of occurrences.
  • In our example, the calculated sample mean \( \bar{X} \approx 2.113 \), which serves as the estimate for \( \lambda \).
The sample mean provides a snapshot of the data's central location, helping you understand the average behavior of the observed phenomenon.
Standard Deviation
Standard deviation is a statistical measure that quantifies the amount of variation or dispersion in a set of values. In the case of a Poisson distribution, the variance \( \sigma_X^2 \) is equal to the parameter \( \lambda \). Consequently, the standard deviation is precisely \( \sqrt{\lambda} \).
  • The standard error of the estimate, \( \sqrt{\hat{\lambda}} \), represents how much the estimated \( \lambda \) might differ from the actual population \( \lambda \) purely due to random sampling variability.
  • In practice, the standard error is used to assess the reliability of an estimate; a smaller standard error indicates a more precise estimate.
  • For our example, the calculated standard error of the sample mean is \( \sqrt{2.113} \approx 1.453 \).
Understanding standard deviation helps in grasping how tightly clustered around the mean your data values are. A smaller standard deviation suggests that the data points tend to be closer to the mean, while a larger one indicates more spread out data.
Parameter Estimation
Parameter estimation involves using sample data to infer the values of the population parameters. In statistical modeling, estimating parameters accurately is critical as it allows for the characterization of the entire population based on a sample. In a Poisson distribution, the primary parameter of interest is \( \lambda \), which represents the average rate of occurrence of an event.
  • For a Poisson process, \( \lambda \) can be estimated using the sample mean, \( \bar{X} \).
  • Accurate parameter estimation ensures that predictions and inferences made based on the distribution are as valid and reliable as possible.
  • In this process, analysts often aim to minimize biases and errors in their estimates to produce more reliable results.
Parameter estimation is fundamental in statistical analysis, as it directly impacts the accuracy and validity of any conclusions drawn from data. By understanding and applying these concepts carefully, one can gain deeper insights into the nature and patterns within larger populations based on sample observations.

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Most popular questions from this chapter

The accompanying data on IQ for first-graders at a university lab school was introduced in Example 1.2. \(\begin{array}{rrrrrrrrrrr}82 & 96 & 99 & 102 & 103 & 103 & 106 & 107 & 108 & 108 & 108 \\ 108 & 109 & 110 & 110 & 111 & 113 & 113 & 113 & 113 & 115 & 115 \\\ 118 & 118 & 119 & 121 & 122 & 122 & 127 & 132 & 136 & 140 & 146\end{array}\) a. Calculate a point estimate of the mean value of IQ for the conceptual population of all first graders in this school, and state which estimator you used. [Hint: \(\mathbf{\Sigma} x_{i}=3753\) ] b. Calculate a point estimate of the IQ value that separates the lowest \(50 \%\) of all such students from the highest \(50 \%\), and state which estimator you used. c. Calculate and interpret a point estimate of the population standard deviation \(\sigma\). Which estimator did you use? [Hint: \(\Sigma x_{i}^{2}=432,015\) ] d. Calculate a point estimate of the proportion of all such students whose IQ exceeds 100 . [Hint: Think of an observation as a "success" if it exceeds 100.] e. Calculate a point estimate of the population coefficient of variation \(\sigma / \mu\), and state which estimator you used.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a continuous distribution with pdf \(f(x ; \theta)\). For large \(n\), the variance of the sample median is approximately \(1 /\left\\{4 n[f(\tilde{\mu} ; \theta)]^{2}\right\\}\). If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from the normal distribution with known standard deviation \(\sigma\) and unknown \(\mu\), determine the efficiency of the sample median.

Each of \(n\) specimens is to be weighed twice on the same scale. Let \(X_{i}\) and \(Y_{i}\) denote the two observed weights for the \(i\) th specimen. Suppose \(X_{i}\) and \(Y_{i}\) are independent of each other, each normally distributed with mean value \(\mu_{i}\) (the true weight of specimen \(i\) ) and variance \(\sigma^{2}\). a. Show that the maximum likelihood estimator of \(\sigma^{2}\) is \(\hat{\sigma}^{2}=\sum\left(X_{i}-Y_{i}\right)^{2} /(4 n)\) [Hint: If \(\bar{z}=\left(z_{1}+z_{2}\right) / 2\), then \(\sum\left(z_{i}-\bar{z}\right)^{2}=\) \(\left.\left(z_{1}-z_{2}\right)^{2} / 2 .\right]\) b. Is the mle \(\hat{\sigma}^{2}\) an unbiased estimator of \(\sigma^{2}\) ? Find an unbiased estimator of \(\sigma^{2}\). [Hint: For any rv \(Z, E\left(Z^{2}\right)=V(Z)+[E(Z)]^{2}\). Apply this to \(Z=X_{i}-Y_{i}\).]

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a gamma distribution with parameters \(\alpha\) and \(\beta\). a. Derive the equations whose solution yields the maximum likelihood estimators of \(\alpha\) and \(\beta\). Do you think they can be solved explicitly? b. Show that the mle of \(\mu=\alpha \beta\) is \(\hat{\mu}=\bar{X}\).

Here is a result that allows for easy identification of a minimal sufficient statistic: Suppose there is a function \(t\left(x_{1}, \ldots, x_{n}\right)\) such that for any two sets of observations \(x_{1}, \ldots, x_{n}\) and \(y_{1}, \ldots, y_{n}\), the likelihood ratio \(f\left(x_{1}, \ldots, x_{n} ; \theta\right) / f\left(y_{1}, \ldots, y_{n} ; \theta\right)\) doesn't depend on \(\theta\) if and only if \(t\left(x_{1}, \ldots, x_{n}\right)\) \(=t\left(y_{1}, \ldots, y_{n}\right)\). Then \(T=t\left(X_{1}, \ldots, X_{n}\right)\) is a minimal sufficient statistic. The result is also valid if \(\theta\) is replaced by \(\theta_{1}, \ldots, \theta_{k}\), in which case there will typically be several jointly minimal sufficient statistics. For example, if the underlying pdf is exponential with parameter \(\lambda\), then the likelihood ratio is \(\lambda^{\sum x_{i}-\Sigma y_{i}}\), which will not depend on \(\lambda\) if and only if \(\sum x_{i}=\sum y_{i}\), so \(T=\sum x_{i}\) is a minimal sufficient statistic for \(\lambda\) (and so is the sample mean). a. Identify a minimal sufficient statistic when the \(X_{i}\) 's are a random sample from a Poisson distribution. b. Identify a minimal sufficient statistic or jointly minimal sufficient statistics when the \(X_{i}\) 's are a random sample from a normal distribution with mean \(\theta\) and variance \(\theta\). c. Identify a minimal sufficient statistic or jointly minimal sufficient statistics when the \(X_{i}\) 's are a random sample from a normal distribution with mean \(\theta\) and standard deviation \(\theta\).

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