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Chapter 6: Problem 5

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.)

Short Answer

Expert verified
Propose three formulas: 1. \( N \times \bar{X} \), 2. \( T - N \times \bar{D} \), 3. \( T \times \frac{\bar{X}}{\bar{Y}} \). Insert given values to complete calculations.

Step by step solution

01

Define the Statistics

To find the total audited value, we propose the following statistics: 1. \( N \times \bar{X} \) for using the sample mean audited value directly.2. \( T - N \times \bar{D} \) for using the sample mean error to adjust the total book value.3. \( T \times \frac{\bar{X}}{\bar{Y}} \) for using the ratio of sample means.
02

Calculate Estimate Using Sample Mean Audited Value

The first statistic is \( N \times \bar{X} \). Here, we assume \( \bar{X} \) is known from the audited sample. Calculate it by multiplying the size of the population \( N \) by the sample mean audited value \( \bar{X} \).
03

Calculate Estimate Using Sample Mean Error

The second statistic is \( T - N \times \bar{D} \). Calculate it by subtracting the total error estimated as \( N \times \bar{D} \) from the total book value \( T \). This statistic uses the average error to adjust the estimated total value.
04

Calculate Estimate Using Ratio of Means

The third statistic is \( T \times \frac{\bar{X}}{\bar{Y}} \). Here, we adjust the total book value \( T \) by the ratio of the audited value to the book value from the sample, reflecting the proportion of error correction.
05

Insert Given Values

Given that \( N = 5000 \) and \( T = 1,761,300 \), replace these values into the formulas from Steps 2, 3, and 4. The sample mean values \( \bar{X} \), \( \bar{Y} \), and \( \bar{D} \) need to be provided to calculate specific numerical estimates.

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Key Concepts

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

Point Estimate
In the world of Statistics for Auditing, a point estimate is a critical term you will come across. A point estimate gives us a single value that serves as an educated guess or estimate of a population parameter.Let's break it down further. Consider that you've received numerous invoices and want to determine the total audited value of all invoices. To do this, calculations known as point estimates are used.
  • The first statistic involves multiplying the size of the population by the sample mean audited value, denoted as \( N \times \bar{X} \).

  • The second one uses the total book value adjusted by the sample mean error, \( T - N \times \bar{D} \).

  • The third applies the ratio of sample means, adjusting our book value to \( T \times \frac{\bar{X}}{\bar{Y}} \).
These different calculations help in understanding how varied methods can lead to the estimation of the same parameter. It represents a fundamental approach to auditing, adding value by increasing accuracy and reliability in financial evaluations.
Sample Mean
The concept of the sample mean is quite simple yet incredibly powerful. It's a measure of the average value of the sample data collected. For instance, in an auditing scenario:
  • \( \bar{X} \) is the sample mean audited value describing average values after the audit.

  • \( \bar{Y} \) is the sample mean book value, representing average values from the records prior to auditing.

  • \( \bar{D} \) is the sample mean error, calculated as the average discrepancy between audited and book values.
By using these sample means, we can derive insights about the population without examining every single record. The sample mean helps generalize findings from a smaller set to a larger population. It simplifies calculations and is crucial for various statistical estimations during audits.
Error Correction
When handling financial data, having a method to correct errors is essential. Error correction helps align the recorded values with the true values. Auditors need to determine the overall correctness of financial documentation.By using the formula \( T - N \times \bar{D} \), we correct errors by subtracting the total estimated error from the total book value. This method provides an adjusted estimate of the true total value.Ultimately, the point of error correction is to ensure that records reflect accurate figures. It aids auditors in adjusting reported values, ideally closing the gap between what is reported and the financial truth.This step is crucial in building trust in financial statements, ensuring stakeholders can make informed decisions based on the revised and corrected data.
Statistical Estimators
Statistical estimators incorporate tools and methods for estimating population parameters. They are frameworks used to derive insights from sample data during audits.Different types of statistical estimators include:
  • Estimators using sample means (like \( \bar{X} \) and \( \bar{Y} \)).

  • Estimators involving correction mechanisms (employing \( \bar{D} \)).

  • Ratio estimators adjusting values through comparisons (such as \( \frac{\bar{X}}{\bar{Y}} \)).
Statistical estimators are indispensable in audits, as they allow professionals to evaluate extensive data and propound reasonably accurate conclusions without examining the whole data set.These methods provide a vital link in comprehending complex data, offering strategies to enhance decision-making and support accountable fiscal management.

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Most popular questions from this chapter

In a random sample of 80 components of a certain type, 12 are found to be defective. a. Give a point estimate of the proportion of all such components that are not defective. b. A system is to be constructed by randomly selecting two of these components and connecting them in series, as shown here. The series connection implies that the system will function if and only if neither component is defective (i.e., both components work properly). Estimate the proportion of all such systems that work properly. [Hint: If \(p\) denotes the probability that a component works properly, how can \(P\) (system works) be expressed in terms of \(p\) ?]

Two different computer systems are monitored for a total of \(n\) weeks. Let \(X_{i}\) denote the number of breakdowns of the first system during the \(i\) th week, and suppose the \(X_{i}\) s are independent and drawn from a Poisson distribution with parameter \(\lambda_{1}\). Similarly, let \(Y_{i}\) denote the number of breakdowns of the second system during the \(i\) th week, and assume independence with each \(Y_{i}\) Poisson with parameter \(\lambda_{2}\). Derive the mle's of \(\lambda_{1}, \lambda_{2}\), and \(\lambda_{1} \quad \lambda_{2}\). [IIint: Using independence, write the joint pmf (likelihood) of the \(X_{i} \mathrm{~s}\) and \(Y_{i}\) s together.]

Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from the shifted exponential pdf $$ f(x ; \lambda, \theta)=\left\\{\begin{array}{cl} \lambda e^{-\lambda(x-\theta)} & x \geq \theta \\ 0 & \text { otherwise } \end{array}\right. $$ Taking \(\theta=0\) gives the pdf of the exponential distribution considered previously (with positive density to the right of zero). An example of the shifted exponential distribution appeared in Example \(4.5\), in which the variable of interest was time headway in traffic flow and \(\theta=.5\) was the minimum possible time headway. a. Obtain the maximum likelihood estimators of \(\theta\) and \(\lambda\). b. If \(n=10\) time headway observations are made, resulting in the values \(3.11, .64,2.55,2.20,5.44,3.42,10.39,8.93\), \(17.82\), and \(1.30\), calculate the estimates of \(\theta\) and \(\lambda\).

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: \(\begin{array}{lllll}16.88 & 10.23 & 4.59 & 6.66 & 13.68 \\ 14.23 & 19.87 & 9.40 & 6.51 & 10.95\end{array}\)

Of \(n_{1}\) randomly selected male smokers, \(X_{1}\) smoked filter cigarettes, whereas of \(n_{2}\) randomly selected female smokers, \(X_{2}\) smoked filter cigarettes. Let \(p_{1}\) and \(p_{2}\) denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes. a. Show that \(\left(X_{1} / n_{1}\right)-\left(X_{2} / n_{2}\right)\) is an unbiased estimator for \(p_{1}-p_{2}\). [Hint: \(E\left(X_{i}\right)=n_{i} p_{i}\) for \(i=1,2\).] b. What is the standard error of the estimator in part (a)? c. How would you use the observed values \(x_{1}\) and \(x_{2}\) to estimate the standard error of your estimator? d. If \(n_{1}=n_{2}=200, x_{1}=127\), and \(x_{2}=176\), use the estimator of part (a) to obtain an estimate of \(p_{1}-p_{2}\). e. Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator.
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