/*! 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: Number of scratches \begin{tabular}{lllllllll} per item & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline Observed frequency & 18 & 37 & 42 & 30 & 13 & 7 & 2 & 1 \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
The estimate for \(\lambda\) is 2.07 and the standard error is approximately 0.118.

Step by step solution

01

Identify the Observed Data

We have a data set showing the number of scratches per item and their corresponding observed frequencies. The observed frequencies for the number of scratches (0 to 7) are: - 0 scratches: 18 - 1 scratch: 37 - 2 scratches: 42 - 3 scratches: 30 - 4 scratches: 13 - 5 scratches: 7 - 6 scratches: 2 - 7 scratches: 1.
02

Calculate the Mean Number of Scratches

To find the mean number of scratches, calculate the weighted average using the observed frequencies. The formula is:\[ \bar{X} = \frac{1}{n} \sum_{i=0}^{7} x_i \cdot f_i \]where \(n\) is the total number of observations (150 items), \(x_i\) is the number of scratches, and \(f_i\) is the frequency. Plugging in the values:\[ \bar{X} = \frac{1}{150} (0\cdot18 + 1\cdot37 + 2\cdot42 + 3\cdot30 + 4\cdot13 + 5\cdot7 + 6\cdot2 + 7\cdot1) = 2.07 \]
03

Identify the Unbiased Estimator for \(\lambda\)

For a Poisson distribution, the mean \(E(X)\) is equal to \(\lambda\), so an unbiased estimator for \(\lambda\) is the sample mean, \(\bar{X}\). Thus, \(\lambda = \bar{X} = 2.07\).
04

Calculate the Standard Error of the Estimator

The standard deviation (or standard error) of a Poisson distribution's estimator is the square root of the parameter \(\lambda\). Thus, the standard error is given by:\[ \sigma_{\bar{X}} = \sqrt{\frac{\lambda}{n}} = \sqrt{\frac{2.07}{150}} \approx 0.118 \].

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Key Concepts

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

Unbiased Estimator
An unbiased estimator is a statistical tool used to estimate a parameter of a population. In simple terms, it is a way to make an educated guess about a measurement if you studied every single example in your population based on a smaller sample. The beauty of an unbiased estimator lies in its property that, on average, it hits the true value of the population parameter.The Poisson distribution, which is often used to model the number of times an event happens in a fixed interval of time or space, has a parameter denoted by \( \lambda \). This parameter corresponds to the mean of the distribution. For a Poisson distribution, the expected value \( E(X) \) is equal to \( \lambda \). Consequently, the sample mean \( \bar{X} \) serves as an unbiased estimator for \( \lambda \).In the case of our exercise, we calculated the mean number of scratches, \( \bar{X} = 2.07 \). Therefore, this sample mean is our unbiased estimator for \( \lambda \), letting us estimate that the true average number of scratches per item for the entire population of items is around 2.07. This is helpful for understanding the overall quality control at the manufacturing site.
Standard Error
The standard error is a crucial concept in statistics that provides insight into how much the estimate of the mean (our estimator) will vary from one sample to another. In essence, it quantifies the variability of our sample mean estimator.For a Poisson distribution, the variance is equal to \( \lambda \). The standard error of the mean is therefore given by the formula \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation of our population, and \( n \) is the sample size.In our example, since \( \sigma^2 = \lambda = 2.07 \), the standard deviation \( \sigma \) is \( \sqrt{2.07} \). Hence, the standard error of the estimator is calculated as \( \sigma_{\bar{X}} = \sqrt{\frac{2.07}{150}} \approx 0.118 \).Having a small standard error, such as 0.118, suggests that if we took multiple samples from the population, the calculated means of those samples would be fairly consistent and close to the actual mean, \( \lambda \). This gives us more confidence in the reliability of our estimation process.
Parameter Estimation
Parameter estimation is a branch of statistics that involves using sample data to estimate the parameters of a chosen statistical model. Essentially, it seeks to provide the "best guess" for the unknown parameters based on the observed data.In the case of a Poisson distribution, the primary parameter we aim to estimate is \( \lambda \), which represents both the mean and variance of the distribution. By using methods such as Maximum Likelihood Estimation (MLE), we derive estimators like the sample mean \( \bar{X} \) to infer \( \lambda \).Our task was to estimate \( \lambda \) through an unbiased estimator. We found \( \bar{X} = 2.07 \), giving us a reasonable parameter estimate. This estimation tells us about the general behavior or characteristics of the model or phenomenon we're observing—in this case, the number of scratches in a batch of items. The estimate helps professionals in quality control decide whether the production process needs adjustment.Overall, the estimation of parameters such as \( \lambda \) for a Poisson distribution can guide decision-making in various fields, ensuring processes are effective and efficient.

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Most popular questions from this chapter

Let \(X\) denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of \(X\) is $$ f(x ; \theta)=\left\\{\begin{array}{cl} (\theta+1) x^{\theta} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ where \(-1<\theta\). A random sample of ten students yields data \(x_{1}=.92, x_{2}=.79, x_{3}=.90, x_{4}=.65, x_{5}=.86, x_{6}=.47\), \(x_{7}=.73, x_{8}=.97, x_{9}=.94, x_{10}=.77\). a. Use the method of moments to obtain an estimator of \(\theta\), and then compute the estimate for this data. b. Obtain the maximum likelihood estimator of \(\theta\), and then compute the estimate for the given data.

A sample of \(n\) captured Pandemonium jet fighters results in serial numbers \(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\). The CIA knows that the aircraft were numbered consecutively at the factory starting with \(\alpha\) and ending with \(\beta\), so that the total number of planes manufactured is \(\beta-\alpha+1\) (e.g., if \(\alpha=17\) and \(\beta=29\), then \(29-17+1=13\) planes having serial numbers 17 , \(18,19, \ldots, 28,29\) were manufactured). However, the CIA does not know the values of \(\alpha\) or \(\beta\). A CIA statistician suggests using the estimator \(\max \left(X_{j}\right)-\min \left(X_{j}\right)+1\) to estimate the total number of planes manufactured. a. If \(n=5, x_{1}=237, x_{2}=375, x_{3}=202, x_{4}=525\), and \(x_{5}=418\). what is the corresponding estimate? b. Under what conditions on the sample will the value of the estimate be exactly equal to the true total number of planes? Will the estimate ever be larger than the true total? Do you think the estimator is unbiased for estimating \(\beta-\alpha+1\) ? Explain in one or two sentences.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) represent a random sample from a Rayleigh distribution with pdf $$ f(x ; \theta)=\frac{x}{\theta} e^{-x^{2} /(2 \theta)} \quad x>0 $$ a. It can be shown that \(E\left(X^{2}\right)=2 \theta\). Use this fact to construct an unbiased estimator of \(\theta\) based on \(\sum X_{i}^{2}\) (and use rules of expected value to show that it is unbiased). b. Estimate \(\theta\) from the following \(n=10\) observations on vibratory stress of a turbine blade under specified conditions: \(\begin{array}{lllll}16.88 & 10.23 & 4.59 & 6.66 & 13.68 \\ 14.23 & 19.87 & 9.40 & 6.51 & 10.95\end{array}\)

a. A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are \(103,156,118,89,125,147,122,109,138,99\). Let \(\mu\) denote the average gas usage during January by all houses in this area. Compute a point estimate of \(\mu\). b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let \(\tau\) denote the total amount of gas used by all of these houses during January. Estimate \(\tau\) using the data of part (a). What estimator did you use in computing your estimate? c. Use the data in part (a) to estimate \(p\), the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a). What estimator did you use?

A sample of 20 students who had recently taken elementary statistics yielded the following information on brand of calculator owned \((\mathrm{T}=\) Texas Instruments, \(\mathrm{H}=\) Hewlett Packard, \(\mathrm{C}=\) Casio, \(\mathrm{S}=\) Sharp): \(\begin{array}{llllllllll}\text { T } & \text { T } & H & \text { T } & \text { C } & \text { T } & \text { T } & \text { S } & \text { C } & \text { H }\end{array}\) \(\begin{array}{lllllllllll}S & S & T & H & C & T & T & T & H & T\end{array}\) a. Estimate the true proportion of all such students who own a Texas Instruments calculator. b. Of the 10 students who owned a TI calculator, 4 had graphing calculators. Estimate the proportion of students who do not own a TI graphing calculator.

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