/*! 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 14 A sample of \(n\) captured Pande... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 14

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.

Short Answer

Expert verified
Estimate: 324. The estimate is exact if both smallest \(\alpha\) and largest \(\beta\) numbers are in the sample. The estimator is biased and usually underestimates.

Step by step solution

01

Understand the Estimator Formula

The estimator used to estimate the total number of planes is \(\max(X_j) - \min(X_j) + 1\). This formula represents the range of captured serial numbers, adjusted by adding 1 to account for inclusive counting.
02

Identify Given Sample Data

In this problem, the sampled data is given as follows: \(n = 5\), \(x_1 = 237\), \(x_2 = 375\), \(x_3 = 202\), \(x_4 = 525\), and \(x_5 = 418\). These are the serial numbers of the captured planes.
03

Calculate Maximum and Minimum of Sample Data

To find the range for estimation, first determine the maximum and minimum serial numbers in the sample. Here, \(\max(X_j) = 525\) and \(\min(X_j) = 202\).
04

Compute the Estimate

Using the formula from Step 1, calculate the estimate as follows: \[ \mathrm{Estimate} = \max(X_j) - \min(X_j) + 1 = 525 - 202 + 1 = 324. \]
05

Analysis of the Estimator Conditions

To determine if the estimate is exact:- The estimate will equal the true number of planes if the sample includes both the smallest numbered (\(\alpha\)) and the largest numbered (\(\beta\)) planes.The estimate could be larger if:- The sample's minimum and maximum are bigger and smaller, respectively, than \(\alpha\) and \(\beta\).The estimator is biased because it tends not to capture the entire range of manufactured plane serials unless specifically inclusive of \(\alpha\) and \(\beta\). Therefore, it typically underestimates the total number of planes.

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Key Concepts

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

Estimator Bias
Bias in an estimator occurs when the estimator does not perfectly measure what it is intended to estimate. Here, when estimating the total number of planes, an estimator is used that takes the maximum and minimum observed serial numbers and adds one to facilitate inclusive counting.
This estimator is biased because it assumes that capturing just the smallest and largest numbers seen in the sample (with no gaps) equals the actual total number of planes. However, the reality is often different because:
  • The sample might not include all numbers, possibly missing some serials both lower and higher than those seen.
  • The gap between lowest and highest numbers may not reflect missing numbers.
Therefore, it often leads to an underestimation of the total number of planes.
Range Estimator
A range estimator is used to gauge the spread or size of a set of data points.
In the context of this exercise, the range estimator applied is \(\max(X_j) - \min(X_j) + 1\). This represents the range of observed serial numbers from the minimum to the maximum, adding 1 to make sure both endpoints are counted.
It's important to note:
  • This estimator presumes that serial numbers are consecutive without missing numbers between the maximum and minimum values.
  • While sometimes accurate, other times it can underestimate, especially if there are unobserved digits at both ends (beyond the minimum and maximum found in the sample).
Understanding these characteristics allows a clearer interpretation when utilizing the range estimator in practice.
Sampling Techniques
Sampling techniques determine how data points are selected from a population and are pivotal in statistical studies. For our case with fighter jet serial numbers, how these numbers are sampled affects the estimator's accuracy.
The sampling in this exercise was random, capturing a set of five serial numbers. The challenges linked with sampling arise because:
  • Random sampling can miss boundary values (i.e., the smallest, \(\alpha\), or the largest, \(\beta\), numbers), making the range estimator biased.
  • Larger sample sizes might increase accuracy but are not always feasible.
The choice of sampling directly impacts the data set's ability to mirror the population and can influence how reliable the estimator is in practice.
Serial Number Estimation
Serial number estimation involves predicting the total number manufactured items using observed samples. In the exercise, the estimate was based on captured fighter jet serial numbers. The formula was \(\max(X_j) - \min(X_j) + 1\).
This estimation does work under the condition:
  • The sample perfectly includes the lowest (\(\alpha\)) and highest (\(\beta\)) numbers.
However, this ideal scenario rarely occurs. Thus, the estimation may fail if the sample does not reflect the whole range.
This method is simple but can often lead to inaccuracies due to gaps in the capture of serial numbers, and so additional strategies or more extensive sampling might be necessary for precise estimation.

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Most popular questions from this chapter

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.

As an example of a situation in which several different statistics could reasonably be used to calculate a point estimate, consider a population of \(N\) invoices. Associated with each invoice is its "book value," the recorded amount of that invoice. Let \(T\) denote the total book value, a known amount. Some of these book values are erroneous. An audit will be carried out by randomly selecting \(n\) invoices and determining the audited (correct) value for each one. Suppose that the sample gives the following results (in dollars). Let $$ \begin{aligned} &\bar{Y}=\text { sample mean book value } \\ &\bar{X}=\text { sample mean audited value } \\ &\bar{D}=\text { sample mean error } \end{aligned} $$ Propose three different statistics for estimating the total audited (i.e., correct) value-one involving just \(N\) and \(\bar{X}\), \(\frac{\text { another involving } T, N \text {, and } \bar{D} \text {, and the last involving } T \text { and }}{}\) \(\bar{X} / \bar{Y}\). If \(N=5000\) and \(T=1,761,300\), calculate the three corresponding point estimates. (The article "Statistical Models and Analysis in Auditing," Statistical Science, 1989: 2-33). discusses properties of these estimators.)

Suppose the true average growth \(\mu\) of one type of plant during a l-year period is identical to that of a second type, but the variance of growth for the first type is \(\sigma^{2}\), whereas for the second type, the variance is \(4 \sigma^{2}\). Let \(X_{1}, \ldots, X_{m}\) be \(m\) independent growth observations on the first type [so \(E\left(X_{i}\right)=\mu\), \(V\left(X_{i}\right)=\sigma^{2}\), and let \(Y_{1}, \ldots, Y_{n}\) be \(n\) independent growth observations on the second type \(\left[E\left(Y_{i}\right)=\mu, V\left(Y_{i}\right)=4 \sigma^{2}\right]\). a. Show that for any \(\delta\) between 0 and 1 , the estimator \(\hat{\mu}=\) \(\delta \bar{X}+(1-\delta) \bar{Y}\) is unbiased for \(\mu\). b. For fixed \(m\) and \(n\), compute \(V(\hat{\mu})\), and then find the value of \(\delta\) that minimizes \(V(\hat{\mu})\). [Hint: Differentiate \(V(\hat{\mu})\) with respect to \(\delta\).]

Suppose a certain type of fertilizer has an expected yield per acre of \(\mu_{1}\) with variance \(\sigma^{2}\), whereas the expected yield for a second type of fertilizer is \(\mu_{2}\) with the same variance \(\sigma^{2}\). Let \(S_{1}^{2}\) and \(S_{2}^{2}\) denote the sample variances of yields based on sample sizes \(n_{1}\) and \(n_{2}\), respectively, of the two fertilizers. Show that the pooled (combined) estimator $$ \hat{\sigma}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2} $$ is an unbiased estimator of \(\sigma^{2}\).

Consider the following sample of observations on coating thickness for low- viscosity paint ("Achieving a Target Value for a Manufacturing Process: A Case Study,".J. of Quality Technology, 1992: 22-26): \(\begin{array}{rrrrrrrr}.83 & .88 & .88 & 1.04 & 1.09 & 1.12 & 1.29 & 1.31 \\\ 1.48 & 1.49 & 1.59 & 1.62 & 1.65 & 1.71 & 1.76 & 1.83\end{array}\) Assume that the distribution of coating thickness is normal (a normal probability plot strongly supports this assumption). a. Calculate a point estimate of the mean value of coating thickness, and state which estimator you used. b. Calculate a point estimate of the median of the coating thickness distribution, and state which estimator you used. c. Calculate a point estimate of the value that separates the largest \(10 \%\) of all values in the thickness distribution from the remaining \(90 \%\), and state which estimator you used. [Hint: Express what you are trying to estimate in terms of \(\mu\) and \(\sigma .]\) d. Estimate \(P(X<1.5)\), i.e., the proportion of all thickness values less than 1.5. [Hint: If you knew the values of \(\mu\) and \(\sigma\), you could calculate this probability. These values are not available, but they can be estimated.] e. What is the estimated standard error of the estimator that you used in part (b)?
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